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http://math.stackexchange.com/questions/27129/what-is-the-difference-between-the-order-of-a-group-and-the-order-of-the-element | What is the difference between the order of a group and the order of the elements of the group
I know the order of a group is the size of the group, ie the number of elements. But what does it mean for an element of that group to have order?
Also, what are the precise definitions for
1) "element of a finite order of a group" and 2) order of an element of a group (assuming that the element has finite order)
If I remember correctly, the order of $\mathbb{Z}$ is one, however the order of the elements in this group have order infinity. Why is that? (Also I dont think $\mathbb{Z}$ is a group in the first place, is it? )
I would also like to ask another question if thats okay:
What is a cyclic group (and a precise definition for it as well) ?
-
From Wikipedia: "the order, sometimes period, of an element $a$ of a group is the smallest positive integer $m$ such that $a^m$ = $e$ (where $e$ denotes the identity element of the group, and $a^m$ denotes the product of $m$ copies of $a$). If no such $m$ exists, we say that a has infinite order. All elements of finite groups have finite order." – Anthony Labarre Mar 15 '11 at 6:52
As to $\mathbb{Z}$ not being a group, this claim makes no sense: a group is a set together with a specific operation that must satisfy some properties. – Anthony Labarre Mar 15 '11 at 6:54
To address your comment on $\mathbb{Z}$, $(\mathbb{Z},+)$ is indeed a group. It's closed, the identity is $0$, and each element has the usual inverse. It's not a field though, if that's what you're thinking, since multiplicative inverses don't exist. – yunone Mar 15 '11 at 6:55
yunone, thankyou. I was thinking of multiplicative inverses. What are the "usual inverses" in $\mathbb{Z}$? – Tyler Hilton Mar 15 '11 at 7:01
for any $n\in\mathbb{Z}$, $-n$ is its usual inverse. So $-5$ is the inverse of $5$, $17$ is the inverse of $-17$, etc. – yunone Mar 15 '11 at 7:32
An element $g \in G$ has order $n$ if $g^n = e$ ($n$ is the smallest positive integer for which this is true). Where $e$ is the identity. See this wikipedia articles for more detail.
- | 2014-12-21T22:31:05 | {
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http://www.onehundredeggs.com/3riua/viewtopic.php?a71e66=4x4-matrix-inverse-calculator | # 4x4 matrix inverse calculator
This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors.It decomposes matrix using LU and Cholesky decomposition. This web site owner is mathematician Miloš Petrović. The following calculator allows you to calculate the inverse for a 4×4 matrix. Complex Matrix Inverse Calculator. Free Matrix Adjoint calculator - find Matrix Adjoint step-by-step This website uses cookies to ensure you get the best experience. 4x4 system of equations solver. The inverse of a 2x2 is easy... compared to larger matrices (such as a 3x3, 4x4, etc). About the Author. We can do this with larger matrices, for example, try this 4x4 matrix: Start Like this: See if you can do it yourself (I would begin by dividing the first row by 4, but you do it your way). Find more Mathematics widgets in Wolfram|Alpha. ... Matrix Calculator Solve System 2x2 Solve System 3x3 Was this calculator helpful? This inverse matrix calculator help you to find the inverse matrix. Is there any chance I can get the inverse of the 4x4 using my calculator using the matrix mode? For those larger matrices there are three main methods to work out the inverse: Inverse of a Matrix using Elementary Row Operations (Gauss-Jordan) Inverse of a Matrix using Minors, Cofactors and Adjugate; Use a computer (such as the Matrix Calculator) Conclusion Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to find the inverse matrix using Gaussian elimination. Enter the values into the matrix and then press "calc inverse " to display the result: Code - 4 dimensional inverse Welcome to MathPortal. Yes: No: 165 539 465 solved problems. I designed this web site and wrote all the lessons, formulas and calculators. I like to think of it this way: In the below Inverse Matrix calculator, enter the values for Matrix (A) and click calculate and calculator will provide you the Adjoint (adj A), Determinant (|A|) and Inverse of a 3x3 Matrix. By using this website, you agree to our Cookie Policy. You can check your answer using the Matrix Calculator (use the "inv(A)" button). Get the free "(Determinant, Inverze) Matice A4x4" widget for your website, blog, Wordpress, Blogger, or iGoogle. Why it Works. Matrices, when multiplied by its inverse will give a resultant identity matrix. A good algorithm by hand to find the inverse of an $n\times n$ square matrix $A$ is to write the $n\times n$ identity matrix next to $A$ and row reduce the $n\times 2n$ matrix. 3x3 identity matrices involves 3 rows and 3 columns. Examples: -5/12, -2i + 4.5. The calculator will perform symbolic calculations whenever it is possible. Find more Mathematics widgets in Wolfram|Alpha. Warning: JavaScript can only store integers up to 2^53 - 1 = 9007199254740991. My calculator only supports a 3x3, 3x2, 3x1, 2x3, 2x2, 2x1, 1x3, 1x2, 1x1 matrices. Rational entries of the form a/b and complex entries of the form a+bi are supported. Get the free "4x4 Determinant calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2021-11-28T06:10:11 | {
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https://math.stackexchange.com/questions/3067596/assuming-the-existence-of-solutions-in-solving-exercises/3067811 | # Assuming the existence of solutions in solving exercises
A sizeable chunk of my first calculus course at university comprised of learning techniques to evaluate limits, such as this simple example, evaluating the limit: $$\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7}.$$
A typical solution would be to identify that for $$x \neq 7$$, $$\frac{x^2 -8x + 7}{x-7} = x-1,$$ so $$\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7} = \lim_{x \to 7} x-1 = 6.$$
In my eyes, we have shown that if the limit exists, its value must be $$6$$. We have not shown that the limit exists in the first place and is equal to $$6$$, since we have presupposed the existence of the limit when writing
$$\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7} = \lim_{x \to 7} x-1,$$
since the existence of both objects on either side of an equality is a necessary condition for the equality to be true (right?).
My main questions are: do such methods of evaluation serve as evidence that these limits in fact exist in the first place, or do they only tell us what the limit ought to be, and the only way we can be sure is to formally prove it using the $$\epsilon$$-$$\delta$$ definition? Is this case similar to "finding" the derivatives of functions?
• Yeah the argument is fine as it is, you don’t need epsilon delta. It’s just that the argumeng is formulated in the wrong direction, but in this case it doesn’t matter since its all iff’s. The correct direction is to say that the limit of x-1 is 6, (because x-1 is continous, which at some level you do need to prove by epsilon delta), and then use the theorem that if two functions f and g agree everywhere wxcept at c, then the limit of the two functions as x approaches c is the same. – Ovi Jan 9 at 16:01
• Since a derivative is defined as a limit the existence of derivative is basically an existence of limit. I hope you are referring to this at the end of your question. – Paramanand Singh Jan 9 at 18:53
• $\epsilon - \delta$ definition is used for proving limit theorems and these theorems are then used while evaluating limits. The point of $\epsilon - \delta$ exercises is to remove the psychological fear (if any) of these Greek symbols (by getting familiar with them) and not to learn proof techniques. The proof techniques are better learnt while studying $\epsilon - \delta$ proofs of various theorems. – Paramanand Singh Jan 9 at 18:58
There's no logical problem with this argument. The expressions $$\frac{x^2 -8x + 7}{x-7} \text{ and } x-1$$ are equal when $$x \ne 7$$, so the first expression has a limit at $$7$$ if and only if the second does. There is no need to assume the existence of the limit in advance.
Whether or not you need the $$\epsilon - \delta$$ argument to find the the limit of $$x-1$$ depends on the level of rigor your instructor requires.
(There are other situations where a correct argument does have the form
the limit is such and such provided the limit exists
usually followed by a separate proof that there is a limit.)
• To elaborate on this answer: E-mu's original suggested $\lim_{x \to 7} \dfrac{x^2 - 8x + 7}{x - 7} = \lim_{x \to 7} x - 1$, as written, literally does assume the limits exist; but in fact it is widely understood as shorthand for exactly the argument that you elaborate. – LSpice Jan 10 at 2:30
• To the proposer: We can write it in full detail as $\lim_{x\to 7, x\ne 7}(x^2-8x+7)/(x-7)=6\iff \lim_{x\to 7,x\ne 7}(x-1)=6.$ – DanielWainfleet Jan 10 at 17:03
Note that when we find the limit of a function at a point the value of the function at that point is not important.
In your example the function $$f(x) =\frac {x^2-8x+7}{x-7}$$ and $$g(x)=x-1$$ have the same values at every point except at $$x=7$$ thus they have the same limits at that point.
Since $$g(x)$$ has a limit of $$6$$ at $$x=7$$ so does $$f(x)$$.
• When I teach this, I call it the “Limits don't see the point” theorem: If $f(x) = g(x)$ for all $x \neq a$, and $\lim_{x\to a}g(x) = L$, then $\lim_{x\to a}f(x) = L$. You can prove this with $\epsilon$-$\delta$. – Matthew Leingang Jan 9 at 17:50
• I like that name ( limits do not see the point theorem ) – Mohammad Riazi-Kermani Jan 9 at 18:14
The $$\epsilon-\delta$$ approach is the safest and most standard definition of limit. To show that $$\lim_{x\to x_0}f(x)$$ exists and is equal to $$L$$, we need to show that $$0<|x-x_0|<\delta\to |f(x)-l|<\epsilon$$here we need to show that $$0<|x-7|<\delta\to \left|{x^2-8x+7\over x-7}-6\right|<\epsilon$$also note that for $$|x-7|>0$$ we have $$x\ne 7$$ therefore $$\left|{x^2-8x+7\over x-7}-6\right|<\epsilon\iff |x-1-6|<\epsilon$$which means that choosing $$\delta=\epsilon>0$$ we have proved the existence of the limit i.e.$$0<|x-7|<\epsilon\to \left|{x^2-8x+7\over x-7}-6\right|<\epsilon$$therefore$$\lim_{x\to 7}{x^2-8x+7\over x-7}=6$$Comment
You can exploit this definition whenever you wanted to find the limit of $${(x-x_0)g(x)\over x-x_0}$$ in $$x=x_0$$.
• What makes a definition 'safe' (or un-)? – LSpice Jan 10 at 2:30
• I say, for example in the cases that the definition is exactly and explicitly equivalent to the existence or non-existence of a limit not through implicit conditions.... – Mostafa Ayaz Jan 10 at 10:08
You haven't really presupposed the existence of the limit and I would say your argument is fine. If you want to be really really scrupulous, perhaps you could re-order things in the final line of your argument.
We have $$\frac{x^2 -8x + 7}{x-7} = x-1$$ for $$x\ne7$$, and $$\lim_{x\to7}x-1=6\ ,$$ so $$\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7} =6\ .$$
Writing it this way, you have an immediate justification for every limit you claim. (Though as I already said, I don't think there is anything seriously wrong with the way you've written it.)
• +1. For beginner Calculus students, I usually even write the last line as "so $\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7}$ exists and is equal to $6$" to make sure that a nontrivial part of the problem was to show that the limit exists. – Taladris Jan 10 at 4:37
Evaluation of a limit in step by fashion based on limit laws also proves that the limit exists (or does not exist depending on final outcome).
The meaning of the first step in your evaluation $$\lim_{x\to 7}\frac{x^2-8x+7}{x-7}=\lim_{x\to 7}x-1$$ is that the LHS of the above equation exists if and only if the RHS exists. To elaborate further the above statement signifies that the limiting behavior of the function $$(x^2-8x+7)/(x-7)$$ as $$x\to 7$$ is exactly the same as that of function $$(x-1)$$ so that if one converges (diverges / oscillates) so does the other. Any algebraic manipulation which is valid for $$x\neq a$$ can be used as a reversible step when evaluating the limit of a function $$f(x)$$ as $$x\to a$$ and your first step is of this kind.
The second step $$\lim_{x\to 7}x-1=6$$ uses standard limits $$\lim_{x\to a} x=a, \lim_{x\to a} k=k$$ and limit rule dealing with difference of functions and in this step the evaluation of limit is complete.
In general when one evaluates the limit of a complicated function in step by step manner then one must ensure that each step (except the last one which removes the limit operator) must be unconditionally true / reversible / of type if and only if. Usually this fact is not stated but rather assumed implicitly. And therefore while performing step by step evaluation of a limit there is no need to prove apriori that the limit in question exists.
Note 1: The use of L'Hospital's Rule is not unconditional / reversible and hence the steps are logically correct only when the method succeeds in giving an answer. If the application of L'Hospital's Rule does not give a definite answer then it does not mean that the original limit does not exist. So when one uses such a technique like L'Hospital's Rule (which works in one direction) then it is better to ensure (via rough work) that it succeeds and then write the step by step evaluation.
Note 2: The usual limit laws (also known as algebra of limits) as stated in most common textbooks are not reversible (you should see their statements which assume the existence of some limits to conclude the existence / evaluation of related limits). These standard formulations can be replaced by their reversible counterparts which are described in this question. | 2019-08-22T20:38:54 | {
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https://math.stackexchange.com/questions/3418526/how-to-find-the-time-after-an-object-reaches-a-distance-from-a-squared-speed-aga | How to find the time after an object reaches a distance from a squared speed against distance?
The problem is as follows:
The figure from below shows the squared speed against distance attained of a car. It is known that for $$t=0$$ the car is at $$x=0$$. Find the time which will take the car to reach $$24\,m$$.
The given alternatives on my book are:
$$\begin{array}{ll} 1.&8.0\,s\\ 2.&9.0\,s\\ 3.&7.0\,s\\ 4.&6.0\,s\\ 5.&10.0\,s\\ \end{array}$$
What I attempted to do to solve this problem was to find the acceleration of the car given that from the graph it can be inferred that:
$$\tan 45^{\circ}=\frac{v^{2}\left(\frac{m^{2}}{s^{2}}\right)}{m}=1\,\frac{m}{s^{2}}$$
Using this information I went to the position equation as follows:
$$x(t)=x_{o}+v_{o}t+\frac{1}{2}at^2$$
Since it is mentioned that $$x=0$$ when $$t=0$$ this would make the equation of position into:
$$0=x(0)=x_{o}+v_{o}(0)+\frac{1}{2}a(0)^2$$
Therefore,
$$x_{o}=0$$
$$x(t)=v_{o}t+\frac{1}{2}at^2$$
From the graph I can spot that:
$$v_{o}^2=1$$
$$v_{o}=1$$
Since $$a=1$$
$$x(t)=t+\frac{1}{2}t^2$$
Then:
$$t+\frac{1}{2}t^2=24$$
$$t^2+2t-48=0$$
$$t=\frac{-2\pm \sqrt{2+192}}{2}=\frac{-2\pm \sqrt{194}}{2}=\frac{-2\pm 14}{2}$$
$$t=6,-8$$
Therefore the time would be $$6$$ but apparently the answer listed on my book is $$8$$. Could it be that I missunderstood something or what happened? Is the answer given wrong?. Can somebody help me here?.
• $v^2 - u^2 = 2ax$ gives $a = \frac{v^2-u^2}{2x} = \frac{(1+x)-1}{2x} = \frac{1}{2}$ Nov 2 '19 at 3:20
• Looks your mistake was that missing $2$ in the denominator for acceleration Nov 2 '19 at 3:25
• @ganeshie8 Where?. I can't find where is the mistake. I don't understand very well what you did. Why $v^2=1+x$? Perhaps could you develop an answer?. I look in my steps as I did used the position equation and it seems right. Nov 2 '19 at 3:34
• @ganeshie8 Your equation doesn't yield me some result. Perhaps can you offer an answer?. I'm stuck. Nov 2 '19 at 3:44
Given graph has the equation: $$v^2 = 1+x$$
Implicitly differentiate both sides with respect to $$x$$ :
$$2v\dfrac{dv}{dx} =1$$
Multiply left side by $$1=\color{blue}{\frac{dt}{dt}}$$: $$2v\dfrac{dv}{\color{blue}{dt}}\dfrac{\color{blue}{dt}}{dx}=1$$
Since $$\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{v}$$: $$2v\dfrac{dv}{dt}\dfrac{1}{v}=1 \implies \dfrac{dv}{dt}=\dfrac{1}{2}$$
In general, acceleration from $$v^2,x$$ graph is obtained by the formula: $$a = \dfrac{1}{2}\dfrac{d}{dx}(v^2)$$
• You could also implicitly differentiate with respect to $t$: $2v \frac{dv}{dt} = \frac{dx}{dt} = v$. :-)
– user657854
Nov 2 '19 at 8:01
• Wow! totally missed haha. That gives the acceleration in just one step. Thanks @EhWha :) Nov 2 '19 at 8:18
You're wrongly assuming $$\color{blue}{a=1}$$.
From the kinematics equation $$v^2 = 2\color{blue}{a}x+u^2$$, with constant acceleration,
when you graph $$v^2$$ against $$x$$, you get a linear equation of form $$y=2\color{blue}{a}x + y_0$$.
Here the slope represents $$2\color{blue}{a}$$.
$$2\color{blue}{a} = \tan(45) \implies \color{blue}{a = \frac{1}{2}}$$ Then the position function would be $$x(t)=t+\frac{1}{2}(\color{blue}{\frac{1}{2}})t^2 = t + \frac{1}{4}t^2$$
Setting that equal to $$24$$ and solving gives $$t=8$$
• I see, so it was an error from my end. So I had to consider the equation as it was given from the graph. But I wonder could this had been solved using calculus?. I thought I could do $v'=\frac{1}{2}\left(x+1\right)^{-\frac{1}{2}}$ but I was stuck on what to do with that. But it seems that it was not needed. The reason why I attempted to do that was to relate the derivative of the speed is the acceleration and from that I could find the acceleration which you referred to as $\frac{1}{2}$ but it turns out that it was not necessary as could had been found from the linear equation. Nov 2 '19 at 4:52
• As I mentioned above could this answer had been obtained using any calculus?. Nov 2 '19 at 4:53
• Absolutely, but be careful $\frac{dv}{dx}$ is not same as the acceleration $\frac{dv}{dt}$. I'll post a new answer with the calc version soon. Nov 2 '19 at 4:55 | 2021-10-17T05:30:55 | {
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https://math.stackexchange.com/questions/1809649/given-that-tan-leftx-y-right-k-and-tan-leftx-right-1-express-ta | # Given that $\tan\left(x-y\right) = k$ and $\tan\left(x\right) = 1$, express $\tan\left(y\right)$ in terms of $k$
Just looking for help on this particular question:
Given that $\tan (x - y) = k$ and $\tan(x) = 1$, express $\tan(y)$ in terms of $k$**
So far I've used the trig identity: $\tan(A±B) = (\tan(A)±\tan(B)) / (1 ∓ \tan(A)\tan(B))$
This got me to $(1-\tan(y))/(1+\tan(y))=k$, but I'm unsure where to go from here and struggle with solving these types of equations. From here, I tried to multiply the fraction by $(1-\tan(y))/(1-\tan(y))$, but this didn't help.
Thanks! Toby.
• Multiply by $1+\tan y$ on both sides, and solve for $\tan y$, it should be ok – H. Potter Jun 2 '16 at 12:55
## 2 Answers
Hint: Assuming your equation $$\frac{1-\tan y}{1 + \tan y} = k$$ is correct, note that you can write $$\frac{1-\tan y}{1 + \tan y} = \frac{2}{1 + \tan y}-1$$ Replace the LHS of your equation with this. You should see how to isolate $\tan y$ now.
Addendum: Completing since another answer has been accepted. $$\frac{1-\tan y}{1 + \tan y} = k$$ $$\frac{2}{1 + \tan y}-1 = k\tag{rewrite LHS}$$ This is a very nice form because now "$\tan y$" appears in only one place. We can immediately solve for "$\tan y$": $$\frac{2}{1 + \tan y}= 1+k\tag{add 1}$$ $$\frac{1}{1 + \tan y} = \frac{1+k}{2}\tag{div. by 2}$$ $$1 + \tan y = \frac{2}{1+k}\tag{invert}$$ $$\tan y = \frac{2}{1+k}-1\tag{subtr. 1}$$ It may be a little nicer to combine these terms: $$\tan y = \frac{1-k}{1+k}\tag{rewrite RHS}$$ To sum up, our first step put the equation in a nice form. It's nice because you can immediately see that the LHS was created from "$\tan y$" by successively performing the steps "add $1$, invert, multiply by $2$, subtract $1$". So we just need to undo these steps in reverse order on both sides to obtain "$\tan y$" on the LHS, and the RHS becomes whatever it becomes since it's along for the ride.
Addendum 2: The "trick" to nicing up the original form is to note that you can do this sort of thing: $$\frac{a+b\;\boxed{X}}{c+d\;\boxed{X}}=\frac{a+\frac{b}{d}(dX)}{c+dX}=\frac{a -\frac{b}{d}(c)+\frac{b}{d}(c+dX)}{c+dX}$$ $$=\frac{\left(a -\frac{bc}{d}\right)}{c+d\;\boxed{X}} +\left(\frac{b}{d}\right)$$ All the stuff in parentheses is just some number, and now the "$X$" is by itself in one spot.
• In the second step, how did you get (1 − tan(y)) / (1 + tan(y)) = 2 / (1 + tan(y)) −1 ? – Toby Mellor Jun 2 '16 at 20:38
• That's the "trick" I allude to above. Let's use "$t$" as shorthand for "$\tan y$". The denominator is $1+t$, so write the numerator as $$1-t=1-\overbrace{(1+t -1)}^{t} = 1-(1+t)+1=2-(1+t)$$ Then the fraction is $$\frac{2-(1+t)}{1+t}=\frac{2}{1+t} - \frac{1+t}{1+t}=\frac{2}{1+t}-1$$ The point is to end up with a constant multiple of denominator in the numerator, so it cancels leaving only a number. The rest of the numerator is also just a number. All of the "$t$" was written as a number plus some multiple of $1+t$ (the multiplier of $(1+t)$ is the original multiplier of $t$ in the numerator). – MPW Jun 2 '16 at 21:09
• (continued) This is really just doing polynomial division $(1-t)\div (1+t)$, which you would write as $(-t + 1)\div (t+1)$. This gives a quotient of $-1$ with a remainder of $2$, that is, $-1 + \frac{2}{t+1}$. Maybe that's the easiest way to think of it. – MPW Jun 2 '16 at 21:19
Hint:
Use the identity:
$$\tan(x-y) =\frac{\tan x - \tan y}{1 + \tan x\tan y} = k$$
Now factorise $\tan y$ out
$$k + k \tan y = 1 - \tan y$$
$$\tan y (k+1) = 1 - k$$
You can finish it off from here | 2021-06-12T12:22:43 | {
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https://mathematica.stackexchange.com/questions/202510/sorting-a-list-into-all-possible-ways-sorting-with-incomplete-information | # Sorting a list into "all possible ways" - sorting with incomplete information
Familiar example of sorting objects
I have a list of $$n$$ "vectors" $$\{x^{(1)},...,x^{(n)}\}$$ of form $$x^{(i)}=\sum_{j=1}^m c_j^{(i)}x_j,i=1\dots n$$, where $$c_j^{(i)}$$ are nonnegative coefficients of the vector $$x^{(i)}$$, and components $$x_j$$ are shared among them, and take some nonnegative values.
For example, we could have: $$x^{(1)}=10x_1+2x_2\\ x^{(2)}=10x_1+3x_2+x_3\\ x^{(3)}=5x_1+2x_2$$
I say that $$x^{(a)}\ge x^{(b)}$$ iff $$x^{(a)} - x^{(b)}\ge 0$$.
Since components $$x_j$$ are nonnegative, we have that $$x^{(a)}\ge x^{(b)}$$ if $$c_j^{(a)}\ge c_j^{(b)}$$ for every $$j=1,\dots,m$$. In this example, we can have a $$\ge$$ relation between all the "vectors", even thought $$x_j$$ are unknown.
Now we have defined the objects, and the comparator function, and our set of objects is comparable. Now, the above example can be easily sorted: $$x^{(2)}\ge x^{(1)} \ge x^{(3)}$$.
So desired result would be: $$\{x^{(2)},x^{(1)},x^{(3)}\}$$.
I noticed SortBy and Sort work for examples like this one.
Generalized sort with incomplete information
But a valid input can be the following example as well:
$$x^{(1)}=10x_1+2x_2+x_3\\ x^{(2)}=10x_1+3x_2\\ x^{(3)}=5x_1+2x_2$$
Here $$x^{(2)}\ge x^{(3)}$$ and $$x^{(1)}\ge x^{(3)}$$, and $$x^{(3)}$$ is clearly the smallest, but we can't compare $$x^{(2)},x^{(1)}$$ definitely. Which is larger, depends on components $$x_2,x_3$$. To resolve this, we take:
$$x^{(2)}-x^{(1)}=x_2-x_3.$$
Now, we can see that $$x^{(2)}\ge x^{(1)}$$ iff $$x_2\ge x_3 \text{ (c1)}$$. Depending on this $$\text{(c1)}$$ condition, we can either have: $$\{x^{(2)},x^{(1)},x^{(3)}\}$$ or $$\{x^{(1)},x^{(2)},x^{(3)}\}$$ as the desired output.
But $$x_j$$ are "components" (unknown variables), thus in this second example, the desired output should contain all possible sortings and conditions under which they are valid. That is, more precisely, in this example, the desired output is:
$$\{ \{\{x^{(2)},x^{(1)},x^{(3)}\},\{x_2\ge x_3\}\},\{\{x^{(1)},x^{(2)},x^{(3)}\},\{x_2\lt x_3\}\} \}$$
A list of two sortings each with one condition under which they are valid.
The SortBy and Sort output one of the sortings (just the sorted list of "vectors") for these cases - but I need all of them along with the conditions.
The first example, under this generalized scenario, would then be:
$$\{\{\{x^{(2)},x^{(1)},x^{(3)}\},\{\}\}\}$$
A list with one possible sorting with no conditions (under all values for $$x_j$$'s).
Where "sorting" is a list pair of sorted elements and conditions under which the sorting is valid, as you can see in both examples.
How can I do this in mathematica?
Create a function that given input v={v[1], v[2], ..., v[n]} outputs the sortings: {s[1],s[2],...}, where sortings have form s[i]={{'sorted elements of v'},{'conditions'}}?
I was mentioning SortBy, and one naive solution that is probably possible, is to SortBy the list of vectors into $$m!$$ ways, by going over all permutations of attributes $$x_j$$ as criterions - Then eliminate the duplicate sortings, and finally extract the conditions from the remaining sortings. But this is extremely slow for large $$m$$ and does a lot of unecessary work (if the case is similar to example 1, then all of the $$m!$$ outputed sorted lists will be the same, and we will eliminate all but one...).
How can I implement my own sort that can do this reliably and always "sort" those vectors given $$x_j$$ are unknown? It will need to find such conditions or perhaps multiple cases of multiple conditions?
(To format the above two examples as code?)
Clear[x, v, w];
v = {10 x[1] + 2 x[2], 10 x[1] + 3 x[2] + x[3], 5 x[1] + 2 x[2]};
w = {10 x[1] + 2 x[2] + x[3], 10 x[1] + 3 x[2], 5 x[1] + 2 x[2]};
Then if the solution to our problem is function f[v_] := ..., then again:
First example F[v] should return:
{{{10 x[1] + 3 x[2] + x[3],10 x[1] + 2 x[2],5 x[1] + 2 x[2]},{}}}
And f[w] should have two sortings:
f[w][[1]]={{10 x[1] + 2 x[2] + x[3], 10 x[1] + 3 x[2], 5 x[1] + 2 x[2]},{x[2]>=x[3]}}
f[w][[2]]={{10 x[1] + 3 x[2], 10 x[1] + 2 x[2] + x[3], 5 x[1] + 2 x[2]},{x[2]<x[3]}}
Motivation:
I have a problem that I split into subproblems, and reduced each subproblem to such a list of "vectors". The subproblems can then be solved by solving a system of equalities that are constructed from the the sortings, that need to be solved assuming the condition under which the sorting is valid, and to have a complete set of solutions, all sortings are necessary to be considered.
This is a very very lengthy process if done by hand (a very large number of systems of equations arise from sortings - and the "vectors" form which I need to derive sortings can get messy), so I'm trying create an algorithm to do it in Mathematica, and this question above is the step where I'm currently stuck at.
• Very well written and interesting question! Jul 22, 2019 at 8:25
We identify each polynomial with its coefficient list. We use InternalListMin on the coefficient array of input list with FixedPointList to identify layers in the partial ordering we seek. Then for each layer with multiple elements we use Reduce to find the condition for each possible permutation of the elements.
ClearAll[coeffArray, paretoLayers, addCondition]
coeffArray = Normal @ Last @ CoefficientArrays[#] &
paretoLayers = Complement @@@ Partition[#, 2, 1] &[Most@
FixedPointList[DeleteCases[#, Alternatives @@ InternalListMin[#]] &, #]] &;
addCondition[v_] := If[Length[#] == 1, {#, {}},
Piecewise[{#, Reduce[LessEqual @@ (#.v), v, Reals]} & /@ Permutations[#]]] &;
addCondition2[v_] := If[Length[#] == 1, {{#, {}}},
{#, Reduce[LessEqual @@ (#.v), v, Reals]} & /@ Permutations[#]] &;
conditionalSorts[v_] := {Join @@ #[[1]], And @@ #[[2]]} & /@
paretoLayers@coeffArray[v]] /. {} -> True)
Examples:
v = {10 x[1] + 2 x[2], 10 x[1] + 3 x[2] + x[3], 5 x[1] + 2 x[2]};
paretoLayers @ coeffArray[v]
{{{5, 2, 0}}, {{10, 2, 0}}, {{10, 3, 1}}}
addCondition[Variables[v]] /@ paretoLayers@coeffArray[v]
{{{{5, 2, 0}}, {}}, {{{10, 2, 0}}, {}}, {{{10, 3, 1}}, {}}}
conditionalSorts[v]
{{{{5, 2, 0}, {10, 2, 0}, {10, 3, 1}}, True}}
w = {10 x[1] + 2 x[2] + x[3], 10 x[1] + 3 x[2], 5 x[1] + 2 x[2]};
paretoLayers @ coeffArray[w]
{{{5, 2, 0}}, {{10, 2, 1}, {10, 3, 0}}}
addCondition[Variables[w]] /@ paretoLayers @ coeffArray[w]
To have a just a list of {orderedlist, condition} pairs use addCondition2:
addCondition2[Variables[w]] /@ paretoLayers@coeffArray[w]
{{{{{5, 2, 0}}, {}}}, {{{{10, 2, 1}, {10, 3, 0}}, x[2] <= x[3]}, {{{10, 3, 0}, {10, 2, 1}}, x[2] >= x[3]}}}
Display each layer as a column:
Column /@ %
Column @ conditionalSorts[w]
SeedRandom[1]
z = RandomInteger[5, {5, 3}].Array[x, 3]
{4 x[1] + 2 x[2] + 4 x[3], x[2], 2 x[2], 3 x[2] + 5 x[3], 2 x[1] + 3 x[3]}
addCondition[Variables[z]] /@ paretoLayers @ coeffArray[z]
addCondition2[Variables[z]] /@ paretoLayers@coeffArray[z]
{{{{{0, 1, 0}, {2, 0, 3}}, x[2] >= 1/3 (-2 x[1] + x[3])}, {{{2, 0, 3}, {0, 1, 0}}, x[2] <= 1/3 (-2 x[1] + x[3])}}, {{{{0, 2, 0}}, {}}}, {{{{0, 3, 5}, {4, 2, 4}}, x[2] <= 4 x[1] - x[3]}, {{{4, 2, 4}, {0, 3, 5}}, x[2] >= 4 x[1] - x[3]}}}
Column /@ %
Column @ conditionalSorts[z]
Update: Using Reduce directly if the input list is not too long:
ClearAll[AllOrderings]
allOrderings[a_] := {#, Reduce[LessEqual @@ #, Variables[a], Reals] }& /@
Permutations[a];
Examples:
allOrderings[v] // FullSimplify // Column
allOrderings[w] // FullSimplify // Column
• Perfect, appreciate your response! Would you mind giving me advice on how to iterate this output (under your Examples) in mathematica-ish way? My goal here is to have {v,c} in each iteration, where v is the original list of "vectors" (or just a list of their coefficient tuples to be consistent with your example outputs), but sorted, under condition(s) c. I noticed conventional programming does not look or work best in mathematica and that's the reason for asking this. Jul 22, 2019 at 16:09
• @Vepir, thank you for the accept. Great question btw. I added addCondition2 which gives such a list.
– kglr
Jul 22, 2019 at 16:53
• Thank you for following the follow up question - but what I meant to ask by my comment is for example: taking your first addCondition2 example output: out = {{{{{5, 2, 0}}, {}}}, {{{{10, 2, 1}, {10, 3, 0}}, x[2] <= x[3]}, {{{10, 3, 0}, {10, 2, 1}}, x[2] >= x[3]}}};, I wanted to extract and iterate (handle) all {v,c} pairs from it - here I wanted to do "for e in out" such that e is { { {5, 2, 0},{10, 2, 1}, {10, 3, 0} },{x[2] <= x[3]} } in first iteration and { { {5, 2, 0},{10, 3, 0}, {10, 2, 1} },{x[2] >= x[3]} } in second iteration. - to group all "vectors", conditions Jul 22, 2019 at 17:20
• I think your update "Using Reduce directly if the input list is not too long" is more of what I wanted to use, but perhaps I forgot to mention that there, it seems that more than needed amount of cases and condtitions is produced (Perhaps because its not taking into consideration that conditions are nonnegative as stated in the problem)? Jul 22, 2019 at 17:42
• @Vepir, I added conditionalSorts that reorganizes the output in the desired form. Re many conditions in the output of Reduce you are right; need to add the condition that all variables positive. Try if changing Reals to PositiveReals makes the desired difference.
– kglr
Jul 22, 2019 at 18:02 | 2022-09-27T04:03:11 | {
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https://math.stackexchange.com/questions/798522/is-this-notation-good-for-the-chain-rule-derivative | # Is this notation good for the chain rule derivative?
When we take this derivative, for example:
$$y = \log(\sin x)$$ We call $u = \sin x$, so we have:
$$\frac{dy}{dx} = \frac{d y}{du}\frac{du}{dx} = \frac{1}{u}\cos x = \frac{\cos x}{\sin x}$$
But for me, it's better to do:
$$\frac{d\log\color{Blue}{\sin x}}{d\color{Blue}{\sin x}}\frac{d\sin \color{Red}{x}}{d\color{Red}{x}} = \frac{1}{\color{Blue}{\sin x}}\cos \color{Red}{x}$$ It makes easy to do the 'pattern-matching' just by looking at the differentials. No substitution. I know that $\frac{d \log[\mbox{something}]}{d[\mbox{something}]} = \frac{1}{\mbox{something}}$ for example.
However, it looks 'hairy' when I try with larger derivatives, like, for the function:
$$(13x^2-5x+8)^{\frac{1}{2}}$$ we do: $$\frac{d(13x^2-5x+8)^{\frac{1}{2}}}{dx} = \frac{d\color{Green}{(13x^2-5x+8)}^{\frac{1}{2}}}{d\color{Green}{(13x^2-5x+8)}}\frac{d(13x^2-5x+8)}{dx} = \frac{1}{2\sqrt{\color{Green}{13x^2-5x+8}}}(26x -5)$$ but it's really better for me to do like this, instead of doing the bla bla bla of changing variables and stuff. But I'm afraid my teacher does not accept this. Is this notation/way of doing good for you guys?
One more example: $$\frac{d}{dx}\sqrt{(\sin(7x+\ln(5x)))} =$$
$$\frac{d[\color{Blue}{\sin(7x+\ln(5x))}]^{1/2}}{d[\color{Blue}{\sin(7x+\ln(5x))}]}\frac{d[\sin\color{Red}{(7x+\ln(5x))}]}{d[\color{Red}{7x+\ln(5x)}]}\left[\frac{d[7\color{Purple}{x}]}{d[\color{Purple}{x}]} + \frac{d[\ln(\color{Purple}{5x})]}{d[\color{Purple}{5x}]}\frac{d[5x]}{d[x]}\right] =$$ $$\frac{1}{2}\left[\color{Blue}{\sin(7x+\ln(5x))}\right]^{-1/2}\cdot\cos(\color{Red}{7x+\ln(5x)})\left[7 + \frac{1}{\color{Purple}{5x}}\cdot 5\right]$$ So we get rid of the substitution!
(づ｡◕‿‿◕｡)づ $\ \ u, v, y$ go away!
• This is probably the coolest (most decorated) question asked. (づ｡◕‿‿◕｡)づ – Shahar May 16 '14 at 23:16
• @Shahar thank you (｡◕‿◕｡) – Lucas Zanella May 16 '14 at 23:18
• @LucasZanella I must say, I'm really looking forward to this question's answer. I'm a self-learner, and do not know much about the commonly used notation. – user122283 May 16 '14 at 23:20
• As for the question, it's right but not very conventional. Hence, I'd assume your teacher might not accept it (depends on how he taught it). But you're doing it right. (｡◕‿◕｡) – Shahar May 16 '14 at 23:20
• omg you guys are so nice, I love this forum ♥ – Lucas Zanella May 16 '14 at 23:21
Be warned that this notation does get overcomplicated at times and you will want to add some substitutions for sanity, but for simpler calculations, it can be useful and looked upon favorably. You should look to study differential calculus as a topic in its own right, where this notation is used as well (in summary, the "denominator" of the derivative $$\text{d}[\text{something such as} \,x]$$ is eliminated and we just deal with what are called differentials). Your instructor is very intolerant, indeed, if he doesn't allow this. | 2021-01-26T12:13:18 | {
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https://onlinecocainesupplier.com/3qf6bqvx/properties-of-a-kite-angles-20afef | Properties of Kite. Explanation: . ... Properties of triangle. What do you observe? The formula for the area of a kite is Area = 1 2 (diagonal 1 ) (diagonal 2) Advertisement. 4. Properties of Kites. 4. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. E-learning is the future today. • noparallel sides. A dart or an arrowhead is a concave kite. KITE: Definition: A quadrilateral with two distinct pairs of equal adjacent sides.A kite-shaped figure.---- Properties :1.Diagonals intersect at right angles.2.Angles between unequal sides are equal3. Kite properties. The legs of the triangles are 10 inches and 17 inches, respectively. Two pairs of sides known as co… The non-vertex angles are the angles formed by two sides that are not congruent. This is equivalent to its being a kite with two opposite right angles. A Kite is a flat shape with straight sides. Two pairs of sides. Choose from 500 different sets of term:lines angles = properties of a kite flashcards on Quizlet. Substitute the value of x to determine the size of the unknown angles of the kites. Stay Home , Stay Safe and keep learning!!! Properties of Kite. right angles. a kite! Area The area of a kite can be calculated in various ways. Add all known angles and subtract from 360° to find the vertex angle, and subtract the sum of the vertex angles from 360° and divide by 2 to find the non-vertex angle. Okay, so that sounds kind of complicated. This makes two pairs of adjacent, congruent sides. 1. The diagonals are perpendicular. Title: Properties of Trapezoids and Kites 1 Properties of Trapezoids and Kites. Yes! You can drag any of the red vertices to change the size or shape of the kite. Recapitulate the concepts with this batch of pdf worksheets to bolster skills in finding the size of the indicated vertex and non-vertex angles with and without diagonals involving algebraic expressions. A kite is a right kite if and only if it has a circumcircle (by definition). Solve for x | Find the Indicated Angles in a Kite. In a kite, the measures of the angles are 3x °, 75°, 90°, and 120°.Find the value of x.What are the measures of the angles that are congruent? Examples of shape properties are: number of sides; number of angles (corners) length of sides; types of angles (acute, obtuse, right-angle) Equip yourself with the Angles in a kite chart for thorough knowledge. Properties: The two angles are equal where the unequal sides meet. • two pairs of equal, adjacent sides (a and b) • two equal angles (B and C) called non-vertex angles. A second identifying property of the diagonals of kites is that one of the diagonals bisects, or halves, the other diagonal. Covid-19 has led the world to go through a phenomenal transition . Add all known angles and subtract from 360° to find the vertex angle, and subtract the sum of the vertex angles from 360° and divide by 2 to find the non-vertex angle. A Square is a Kite? Add all known angles and subtract from 360° to find the vertex angle, and subtract the sum of the vertex angles from 360° and divide by 2 to find the non-vertex angle. Kite. Area, angles, and internal lengths. High school students learn how to find the indicated vertex and non-vertex angles in a kite, determine the measure of angles with bisecting diagonals and solve for 'x' in problems involving algebra as well. The diagonals of a kite intersect at 90 ∘. Section 7.5 Properties of Trapezoids and Kites 441 7.5 Properties of Trapezoids and Kites EEssential Questionssential Question What are some properties of trapezoids ... Measure the angles of the kite. A kite is defined by four separate specifications, one having to do with sides, one having to do with angles… 2. Convex: All its interior angles measure less than 180°. And then we could say statement-- I'm taking up a lot of space now-- statement 11, we could say measure of angle DEC plus measure of angle DEC is equal to 180 degrees. Problematic Start. E-learning is the future today. Diagonals intersect at right angles. 2. The smaller diagonal of a kite … In this section, we will discuss kite and its theorems. Two disjoint pairs of consecutive sides are congruent by definition. Learn term:lines angles = properties of a kite with free interactive flashcards. It often looks like. Find the Indicated Angles | Diagonals By the kite diagonal theorem, AC is _____ to BD This means that angles AED and CED are right angles. Apply appropriate triangle theorems to find the indicated angles. So it doesn't always look like the kite you fly. The smaller diagonal of a kite divides it into two isosceles triangles. Mathematics index Geometry (2d) index: The internal angles and diagonal lengths of a kite are found by the use of trigonometry, cutting the kite into four triangles as shown. Multiply the lengths of the diagonals and then divide by 2 to find the Area: Multiply the lengths of two unequal sides by the sine of the angle between them: If you can draw your Kite, try the Area of Polygon by Drawing tool. Being a special type of quadrilateral, it shows special characteristics and properties which are different from the other types of quadrilaterals. It has 2 diagonals that intersect each other at right angles. Properties. Do the diagonals bisect its angles… Each pair is two equal-length sides that are adjacent (they meet). The kite's sides, angles, and diagonals all have identifying properties. 2. The bases of a trapezoid are its 2 parallel sides ; A base angle of a trapezoid is 1 pair of consecutive angles whose common side is a … The sum of the interior angles of any polygon can be found by applying the formula: degrees, where is the number of sides in the polygon. c. Repeat parts (a) and (b) for several other kites. Also, learn about the side and angle properties of kites that make them unique. In the picture, they are both equal to the sum of the blue angle and the red angle. Therefore, we have that ΔAED ≅ ΔCED by _______ The diagonals of a kite intersect at 90 $$^{\circ}$$ The formula for the area of a kite is Area = $$\frac 1 2$$ (diagonal 1)(diagonal 2) ... Properties of triangle. Solve for x | Find the Angles in a Kite - contain Diagonals. Sketch. 3. $\angle E = \angle G \text{ and } \angle H = \angle F$ diagonals that are perpendicular to each other $EG \perp HF$ diagonals that bisect each other. Here are the properties of a kite: 1. The main diagonal of a kite bisects the other diagonal. So let me just do it all like this. The sketch below shows how to construct a kite. Angles between unequal sides are equal In the figure above notice that ∠ABC = ∠ADC no matter how how you reshape the kite. It has two pairs of equal-length adjacent (next to each other) sides. Let’s see how! As you reshape the kite, notice the diagonals always intersect each other at 90° (For concave kites, a diagonal may need to be extended to the point of intersection.) Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. A kite is a quadrilateral that has 2 pairs of equal-length sides and these sides are adjacent to each other. The main diagonal of a kite bisects the other diagonal. A Kite is a flat shape with straight sides. The measures of the angles are given as a linear equation. Apply the properties of the kite to find the vertex and non-vertex angles. Parallel, Perpendicular and Intersecting Lines. Concave: One interior angle is greater than 180°. From the above discussion we come to know about the following properties of a kite: 1. Kite properties. To be a kite, a quadrilateral must have two pairs of sides that are equal to one another and touching. 4. A kite is a quadrilateral in which two pairs of adjacent sides are equal. Members have exclusive facilities to download an individual worksheet, or an entire level. A kite is the second most specific tier one shape, but it has no sub branches. One diagonal is the perpendicular bisector of the other. Browse through some of these worksheets for free! Area, angles, and internal lengths. Knowing the properties of a kite will help when solving problems with missing sides and angles. A kiteis traditionally defined as a four-sided, flat shape with two pairs of adjacent sides that are equal to each other. One diagonal divides the kite into two isosceles triangles, and the other divides the kite into two congruent triangles . It can be viewed as a pair of congruent triangles with a common base. Diagonals intersect at right angles. What are the Properties of a Kite. are equal where the two pairs meet. Charlene puts together two isosceles triangles so that they share a base, creating a kite. What are the Properties of a Kite? The vertex angles of a kite are the angles formed by two congruent sides.. A kite is the combination of two isosceles triangles. The angles Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Let M be the midpoint of BD, then let k be the line containing AMB, then by the theory of isosceles triangles, this line bisects angle BAC.. Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Learn about and revise angles, lines and multi-sided shapes and their properties with GCSE Bitesize AQA Maths. A kite is a quadrilateral with exactly two distinct pairs of congruent consecutive sides. Formulas Area. The total space enclosed by the kite. The two diagonals of a kite bisect each other at 90 degrees. 00:05:28 – Use the properties of a trapezoid to find sides, angles, midsegments, or determine if the trapezoid is isosceles (Examples #1-4) 00:25:45 – Properties of kites (Example #5) 00:32:37 – Find the kites perimeter (Example #6) 00:36:17 – Find all angles in a kite (Examples #7-8) Practice Problems with Step-by-Step Solutions back to quadrilaterals. Sum of the angle in a triangle is 180 degree. In every kite, the diagonals intersect at 90 °. In the figure above, click 'show diagonals' and reshape the kite. Apply the properties of the kite to find the vertex and non-vertex angles. Use this interactive to investigate the properties of a kite. Kite and its Theorems. One of the diagonals bisects a pair of opposite angles. It has two pairs of equal-length adjacent (next to each other) sides. two disjoint pairs of consecutive sides are congruent (“disjoint pairs” means Kite Properties . But never fear, I will explain. 3. See Area of a Kite 4. Other important polygon properties to be familiar with include trapezoid properties , parallelogram properties , rhombus properties , and rectangle and square properties . • diagonals which alwaysmeet at right angles. Kite. The sum of the interior angles of any polygon can be found by applying the formula: degrees, where is the number of sides in the polygon. The two diagonals of our kite, K T and I E, intersect at a right angle. 446 Chapter 7 Quadrilaterals and Other Polygons MMonitoring Progressonitoring Progress Help in English and Spanish at BigIdeasMath.com 6. In this section, we will discuss kite and its theorems. Covid-19 has led the world to go through a phenomenal transition . Use the appropriate properties and solve for x. By the symmetry properties of the isosceles triangle, the line AM is the perpendicular bisector of BD = m. Thus A is on m. Also, since triangle ABD is isosceles, ray AM bisects angle BAD, so angle BAM = angle DAM. Sometimes one of those diagonals could be outside the shape; then you have a dart. The longer and shorter diagonals divide the kite into two congruent and two isosceles triangles respectively. The angles between the sides of unequal length are equal. In the figure above, click 'show diagonals' and reshape the kite. Sum of the angle in a triangle is 180 degree. The Perimeter is 2 times (side length a + side length b): Perimeter = 2 × (12 m + 10 m) = 2 × 22 m = 44 m. When all sides have equal length the Kite will also be a Rhombus. So let me say measure of angle DEC plus measure of angle BEC is equal to 180. The diagonals are perpendicular. The problem. A kite is a quadrilateral with two pairs of adjacent, congruent sides. One diagonal is the perpendicular bisector of the other. You can’t say E is the midpoint without giving a reason. As you reshape the kite, notice the diagonals always intersect each other at 90° (For concave kites, a diagonal may need to be extended to the point of intersection.) When all the angles are also 90° the Kite will be a Square. 1. The Perimeter is the distance around the edges. 3. And this comes straight from point 9, that they are supplementary. Use appropriate triangle theorems and solve algebraic expressions to find the value of 'x'. Since a right kite can be divided into two right triangles, the following metric formulas easily follow from well known properties of right triangles. The triangle ABD is isosceles. A kite can be a rhombus with four equal sides or a square having four equal sides and each angle measuring 90°. Stay Home , Stay Safe and keep learning!!! Let AC and BD intersect at E, then E is the midpoint of BD. Find the Indicated Angles | Vertex and Non-Vertex Angles. Another way of picturing a kite is to think of the old-school type of kite that peopl… 3. Kite is also a quadrilateral as it has four sides. 2. That does not matter; the intersection of diagonals of a kite is always a right angle. Apply the properties of the kite to find the vertex and non-vertex angles. If the length of the base for both triangles is 16 inches long, what is the length of the kite's other diagonal? Angles … Kite and its Theorems. Angle BAM = angle BAC and angle DAM = angle DAC (same rays) It looks like the kites you see flying up in the sky. A kite is a quadrilateral with two pairs of adjacent, congruent sides. Kite Sides. See, a kite shape looks like a diamond whose middle has been shifted upwards a bit. i.e., one diagonal divides the other diagonal into exactly two halves. Two disjoint pairs of consecutive sides are congruent by definition. Using these facts about the diagonals of a kite (such as how the diagonal bisects the vertex angles) and various properties of triangles, such as the triangle angle sum theorem or Corresponding Parts of Congruent Triangles are Congruent (CPCTC), it is possible … Types of Kite. Find the Vertex and Non-Vertex Angles | Solve for 'x'. Diagonals (dashed lines) cross at These sides are called as distinct consecutive pairs of equal length. It looks like the kites you see flying up in the sky. The top two sides are equal to each other in length, as are the bottom two sides. Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. Other types of quadrilaterals _____ to BD this means that angles AED and are. Inches and 17 inches, respectively kite diagonal theorem, AC is _____ to this. Sets of term: lines angles = properties of a kite is always a right angle does not matter the. 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Equal-Length adjacent ( next to each other other in length, as are bottom. Interior angles of this shape, stay Safe and keep learning!!!!!!!!... | 2021-04-13T04:50:07 | {
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https://math.stackexchange.com/questions/3118741/linearisation-and-stability-of-a-system | # Linearisation and Stability of a system
Consider the system
$$\dot{x} = y$$ $$\dot{y} = x^{2} + x$$
Find the fixed points and the linearisation of the system at each. Identify the type and sketch a local phase portrait at each.
I am unsure how to get the fixed points. I get that it should be at $$\dot{x} = 0$$ and $$\dot{y} = 0$$, but that gives me $$y=0$$ and $$x = 0, -1$$, and I am unsure how to find the other co-ordinate point for each.
EDIT: Found the other coordinates for each, found the local phase portraits too. How to I now find the isoclines of the system?
The equilibrium points are located at $$(y = 0) \cap (x = \{-1,0\})$$
now calling $$X = (x,y)^{\dagger}$$ and $$F(X) = (y,x+x^2)^{\dagger}$$ we have
$$\dot X = F(X) = F(X_0) + \frac{\partial F_0}{\partial X}(X-X_0)+ O(|X-X_0|^2)$$
now choosing $$X_0 = \{(-1,0)^{\dagger},(0,0)^{\dagger}\}$$ at $$X_0^1$$ we have
$$\frac{d}{dt}(X-X_0^1) = \frac{\partial F_0^1}{\partial X}(X-X_0^1)+ O(|X-X_0^1|^2)$$
and near $$X_0^1$$
$$\frac{d}{dt}(X-X_0^1) = \frac{\partial F_0^1}{\partial X}(X-X_0^1)$$
with
$$\frac{\partial F_0^1}{\partial X} = \left(\begin{array}{cc}0 & 1\\ -1&0\end{array}\right)$$
and
$$\frac{\partial F_0^2}{\partial X} = \left(\begin{array}{cc}0 & 1\\ 1&0\end{array}\right)$$
then $$X_0^1$$ is a center and $$X_0^2$$ is a saddle point. Attached the stream plot
Near $$X_0^1$$ the orbits follow
$$(\delta x)' = \delta y\\ (\delta y)' = -\delta x$$
then
$$\delta x(\delta x)' = \delta x\delta y\\ \delta y(\delta y)' = -\delta y\delta x$$
$$\delta x(\delta x)' + \delta y(\delta y)' = \frac 12\frac{d}{dt}((\delta x)^2+(\delta y)^2)=0$$
or
$$(\delta x)^2+(\delta y)^2 = C_0$$
which means that in this case, the orbits remain circling, which characterizes this point as a center.
NOTE
$$\frac{\partial F}{\partial X} = \left(\begin{array}{cc}0 & 1\\ 1+2x&0\end{array}\right)$$
plt = StreamPlot[{y, x^2 + x}, {x, -2, 2}, {y, -2, 2}];
pt1 = Graphics[{Red, Disk[{-1, 0}, 0.04]}];
pt2 = Graphics[{Red, Disk[{0, 0}, 0.04]}];
Show[plt, pt1, pt2]
• What software package did you use to make that vector field? – TSF Feb 21 at 12:02
• @TonyS.F. MATHEMATICA. I attached the script. – Cesareo Feb 21 at 12:07
• You cannot use linearization for the center. Lyapunov's linearization theorem does not allow any conclusion in this case. – MrYouMath Feb 22 at 18:47
• @MrYouMath Thanks for the hint. I had forgotten to include this detail. – Cesareo Feb 22 at 20:04
• @Cesareo: I like the idea with the circle. Do you have any resource that gives more examples of this method? But I find it a little bit strange that this method does neglect higher order terms. – MrYouMath Feb 22 at 21:00
A more elegant way for the center at $$\boldsymbol{x}_\text{eq}=[-1,0]^T$$ is to use the following Lyapunov function candidate
$$V(x,y)= 1/2y^2+1/6\left[1-x^2(2x+3)\right].$$
This function is positive definite in the neighborhood of the equilibrium point $$\boldsymbol{x}_\text{eq}$$. For a formal derivation, you can calculate the Hessian matrix and evaluate it at the equilibrium point. The eigenvalues will indicate a positive definite matrix. Hence, we know that $$V(\boldsymbol{x})>0$$ for all $$\boldsymbol{x}\neq\boldsymbol{x}_\text{eq}$$. For the equilibrium point we get $$V(\boldsymbol{x}_\text{eq})=0$$.
Now, differentiate $$V$$ to obtain
$$\dot{V}=y\dot{y}+1/6[-6x^2-6x]\dot{x}=y\left[x^2+x \right]+1/6[-6x^2-6x]y\equiv 0.$$
As $$V$$ is positive definite and $$\dot{V}\equiv 0$$, we can conclude that $$\boldsymbol{x}_\text{eq}$$ is indeed a center.
An alternative to the paid Mathematica is the free web-based Wolfram Alpha. Visit this link to see how you can display a stream plot. | 2019-11-15T07:37:36 | {
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http://math.stackexchange.com/questions/44459/proof-for-divisibility-by-7 | # Proof for divisibility by $7$
One very classic story about divisibility is something like this.
A number is divisible by $2^n$ if the last $n$-digit of the number is divisible by $2^n$. A number is divisible by 3 (resp., by 9) if the sum of its digit is divisible by 3 (resp., by 9). A number $\overline{a_1a_2\ldots a_n}$ is divisible by 7 if $\overline{a_1a_2\ldots a_{n-1}} - 2\times a_n$ is divisible by 7 too.
The first two statements are very well known and quite easy to prove. However I could not find the way on proving the third statement.
PS: $\overline{a_1a_2\ldots a_n}$ means the digits of the number itself, not to be confused with multiplication of number.
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There is also a similar less known trick for divisibility by 11. Since $10 = -1 \mod 11$, if you add the digits of a number in reverse order, alternating signs, and get something that is divisible by 11, then your original number is divisible by 11. For example for $1617$ you have $7-1+6-1 = 11$ so $1617$ is divisible by $11$. – Vhailor Jun 10 '11 at 0:06
@Vhailor: I have heard about this too... Anyway, I dont think that reversing it is really needed. $1-6+1-7=-11$ :) – zfm Jun 10 '11 at 0:15
This is so incredibly ironic: I was just recalling the "tricks" I learned for checking whether an integer is divisible by 3, 7, ...and spent time earlier today proving those claims! @Vhailor: 11 is new to me...thanks! – amWhy Jun 10 '11 at 0:50
A small remark : for $2^n$, $3$, $9$ and $11$, those criteria also gives you a way to compute the residue when the number is not divisible (meaning $\overline{a_1 \cdots a_n} \equiv a_1 + \ldots + a_n \mod 9$, $\overline{a_1 \cdots a_n} \equiv a_n \mod 2$ ...). But the criterion for $7$ does NOT give you the residue (you only have $\overline{a_1 \cdots a_{n-1}}-2a_n \equiv 5 \overline{a_1 \cdots a_n} \mod 7$ so each step multiplies the residue by $5$). – Joel Cohen Jun 10 '11 at 2:00
And if you work with slightly larger number, then starting from the fact that $1001=7\cdot1\cdot13$ we get that $1000\equiv -1\pmod p$ for $p=7,11,13$. This allows a speed up in the sense that you can reduce the number of digits by 3 at a time (early on). So for example $166145\equiv-166+145=-21\equiv0\pmod 7$ and $777751130\equiv -777751+130\equiv777-751+130=-26+130\equiv0\pmod{13}.$ – Jyrki Lahtonen Jun 10 '11 at 3:49
$$5(\overline{a_1a_2\ldots a_n})=50(\overline{a_1a_2\ldots a_{n-1}})+5a_n=\overline{a_1a_2\ldots a_{n-1}}-2a_n\pmod{7}$$
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So, 5 is the trick. – zfm Jun 10 '11 at 0:08
@zfm : Yes, the trick relies on $5*10 \equiv 1 \mod 7$. Similarly, you could use the same idea to derive a criterion of divisibility by $11$ (using $(-1)*10 \equiv 1 \mod 11$), $13$ (using $4*10 \equiv 1 \mod 13$), or any prime number $p \ge 7$. – Joel Cohen Jun 10 '11 at 1:45
Nice! Why do we not multiply 15? Then we get the remainder modulo 7. I mean 1998=3*(199-16)=3*183=9*(18-6)=2*5=3. mod 7. – wxu Jun 10 '11 at 8:41
@wxu: If you want the remainder mod $7$, you can do it that way. But if you just want to test for divisibility, you don't need the multiplications by $3$, so it's slightly easier the other way. – joriki Jun 10 '11 at 9:23
Let $x$ be the number you gave, with the full $n$ digits, and let $y$ be the number whose decimal representation is obtained by removing the last digit, namely $a_n$, of $x$.
The following equation is clear: $$x=10y +a_n.$$
Note that $10y +a_n$ is divisible by $7$ iff $20y+2a_n$ is divisible by $7$.
But $20y+2a_n$ is divisible by $7$ iff $-y+2a_n$ is divisible by $7$ (I subtracted $21y$), and this is the case if and only if $y-2a_n$ is divisible by $7$. That's exactly what we wanted to show.
We can write the above stuff using the language of congruences if we wish.
About iff: I slipped into jargon. The word (?) "iff" abbreviates "if and only if."
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Note that this is essentially the same as my answer, since the number is being multiplied by $-2$ instead of $5$, and $-2=5\pmod{7}$. – joriki Jun 10 '11 at 0:20
HINT $\rm\qquad 7\ |\ 10\ y + x\ \iff\ 7\ |\ y-2\ x\ \ \:$ since $\rm\:\ -2\ (10\ y + x)\ \equiv\ y - 2\ x\ \ (mod\ 7)$
i.e. lines $\rm\ -10\ y = x\$ and $\rm\ y = 2\ x\$ are equivalent over $\rm\:\mathbb Z/7\:,$ differing by a unit scaling (of $2$).
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Regarding divisibility by 7 I created an algorithm that must be applied repetitively to each period of a large number. If the last sum results in a multiple of 7 then the tested number is also a multiple of 7, otherwise it is not. The sum of each period must be added to the sum of the next period.
N = abc
algorithm: – ( (x) + a + b + c + (y) ) mod 7
(x) must be mentally inserted before the hundreds and (y) must be mentally inserted after the ones in such a way that 7 divides both (x)a and c(y). In this rule no digits are inserted before or after the tens.
Numerical examples:
1) N = 462; – ( (1) + 4 + 6 + 2 + (1) ) mod 7 ≡ Ø
2) N = 863; – ( (2) + 8 + 6 + 3 + (5) ) mod 7 ≡ 4
3) N = 1.554; – ( 1 + (4) ) mod 7 ≡ 2; – ( 2 + (3) + 5 + 5 + 4 + (2) ) mod 7 ≡ Ø
4) N = 68.318; – ( 6 + 8 + (4) ) mod 7 ≡ 3; – ( 3 + (6) + 3 + 1 + 8 +(4) ) mod 7 ≡ 3
5) N = 852.655; – ( (2) + 8 + 5 + 2 + (1) ) mod 7 ≡ 3; – ( 3 + (5) + 6 + 5 + 5 + (6) ) mod 7 ≡ 5
To determine the remainder take the digit that forms with the last sum a multiple of 7. Remainders of the second, fourth and fifth examples:
2) Last sum is 4 → 4(2); 2 is the remainder 4) Last sum is 3 → 3(5); 5 is the remainder 5) Last sum is 5 → 5(6); 6 is the remainder
The application of this rule must be performed in a dynamic way. It is not necessary to write the inserted numbers; all you need is to add the written numbers and the mentally inserted numbers at sight, determine the difference of each sum and its next multiple of 7 (inverse additive) and go ahead till the value of the last inserted digit is added.
The bureaucratic application of the algorithm certainly will not be as quick as its dynamic application.
The rule works because before the use of the inverse additive the values of the digits of the first three digits are multiplied respectively by 3, 1 and 5; after the inverse additive the multipliers change to 4, 6 and 2. Each group of 6 digits of a number are multiplied by 4,6,2,3,1 and 5. These are the remainders determined by Pascal’s criterion regarding divisibility by 7.
In my blogspot I explain how easy is to create new real rules for divisibility by 7 and other aspects of my research. This is the first real rule for divisibility by 7 created by me.
English is not my native language and I am not a Mathematician.
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To complete my last answer I ask you to watch this video:
Silvio Moura Velho
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Welcome to MSE! I realize you do not yet have enough reputation, but this would have been better as a comment or as an update to the snwer you are referring to. Regards – Amzoti Apr 26 '13 at 15:48
The reason why this formula works is because $21$ is a multiple of $7$ which ends in $1$. Thus, $21a_n$ is always multiple of $7$.
Then
$$7| \overline{a_1...a_n} \Leftrightarrow 7| \overline{a_1...a_n}-21 a_n \Leftrightarrow 7| \overline{a_1...a_{n-1}0}-20 a_n \Leftrightarrow 7| 10 \cdot[ \overline{a_1...a_{n-1}}-2 a_n] \,.$$
Since $gcd(7, 10)=1$ you get the desired result.
Comment 1 Same way you can get a criteria of divisibility for any number $m$ so that $gcd(m,10)=1$. If this is the case, $m$ has a multiple of the form $10k+1$ and then, you can prove that
$$m| \overline{a_1...a_n} \Leftrightarrow m| \overline{a_1...a_{n-1}}-ka_n$$
Comment 2 You can instead look for multiples of $7$ (or $m$)which end in $9$ and add instead of subtraction.
Since $7|49$ we get that
$7| \overline{a_1...a_n} \Leftrightarrow 7| \overline{a_1...a_n}+49 a_n \Leftrightarrow 7| \overline{a_1...a_{n-1}0}+50 a_n \Leftrightarrow 7| 10 \cdot[ \overline{a_1...a_{n-1}}+5 a_n] \,.$$- I'm taking the title of the question to allow general tests for divisibility by 7 even though the specifics of the original question is about a particular test. Apologies if this is not right - I'm still fairly new to mse. [Source for this answer is The Lore of Large Numbers, by P. Davis, p. 87-88] A number is divisible by 7 if and only if: (3 * units' digit) + (2 * tens' digit) - (1 * hundreds' digit) - (3 * thousands' digit) - (2 * ten thousands' digit) + (1 * hundred thousands' digit) is divisible by 7. If there are more digits present, the sequence of multipliers 3, 2, -1, -3, -2, 1 is repeated as often as necessary. If fewer digits are present, stop when get to the last multiplier. Example:$$7 * 457404 = 3201828 \begin{array} &3 & * & 8 & = & 24 \\ 2 & * & 2 & = & 4 \\ -1 & * & 8 & = &-8 \\ -3 & * & 1 & = &-3 \\ -2 & * & 0 & = &0 \\ 1 & * & 2 & = &2 \\ 3 & * & 3 & = & 9 \\ \end{array} $$Since 24 + 4 - 8 - 3 + 0 + 2 + 9 = 28 and 7 divides 28, 3201828 is divisible by 7. Proof for 3 digit numbers$$N = 100a + 10b + c,$$then if S is the sum required by the test,$$S = -a + 2b + 3c.$$Then 2S = -2a + 4b + 6c and N + 2S = 98a + 14b + 7c = 7(14a + 2b + c). The sum N + 2S is therefore a multiple of 7, say 7M. Now if N is a multiple of 7, say 7P, then 2S = 7M - 7P = 7(M - P). Since therefore 2S is divisible by 7 and S is whole, S also is divisible by 7 (or S = \frac{7X}{2}, where X is even because S is whole). Conversely, if S is a multiple of 7, say 7Q, then$$N = 7M - 14Q = 7(M - 2Q).$$This means that$N$must be a multiple of$7$. Yet another way to do it is to use a similar alternating sum test as for divisibility by 11, but in 3 digit groups, subtracting first, with the sum's divisibility by 7 determining the original number's divisibility by 7. In the$3201828$example, this is$828 - 201 + 3 = 630,$so since$630$is divisible by$7$, so is$3201828\$. I got this from http://mathforum.org/k12/mathtips/ward2.html.
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https://math.stackexchange.com/questions/2932125/integral-of-binomial-coefficient/2932204 | # Integral of Binomial Coefficient
We all know the famous theorem that: $$\sum_{i=1}^n\binom{n}{i}=2^n$$ This theorem got me wondering about a similar formula - the properties of the following function: $$I(n)=\int_{0}^{n} \binom{n}{k}\,\,\mathrm{d}k$$ where $$n$$ is any positive integer and the definition of binomial coefficient is "extended" by way of gamma functions (i.e the integrand is really $$\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}$$). What I found, experimentally, is pretty cool. It seems that the following is true: $$I(n)=\frac{2}{\pi} \sum_{i=1}^n \binom{n}{i}\operatorname{SinInt}(\pi i)$$ Where $$\operatorname{SinInt}(x)$$ is the Sine Integral, or $$\int_0^x \frac{\sin t}{t}dt$$.
To me, this is quite interesting as the Sine Integral tends to $$\pi/2$$ so the above formula will tend to $$2^n$$, so the integral is just the sum with some error term. But how would I go about proving it?
• Since $n$ does not change as $k$ goes from $0$ to $n,$ one has $\displaystyle I(n) = n!\int_0^n \frac{\mathrm dk}{k!(n-k)!}. \qquad$ Sep 26, 2018 at 22:43
• How did you find it experimentally? Using some tools? Sep 26, 2018 at 22:51
• @Dilworth I myself was able to get a closed form for $n=0$, and was able to manipulate the integrand into a form that Mathematica could integrate using $n=1$. Saw the sine integrals, recognized the coefficients, conjectured the above formula and then verified with numerical integration. Sep 26, 2018 at 22:54
• I see, thanks! - Sep 26, 2018 at 22:56
Properties of the $$\Gamma$$-function, together with partial fraction expansion, do the trick. We have
$$\Gamma(x+1)\Gamma(n-x+1)=x\Gamma(x)\Gamma(1-x)\prod_{k=1}^{n}(k-x)=\frac{(-1)^n\pi}{\sin\pi x}\prod_{k=0}^{n}(x-k),$$ so our integral is $$I(n)=\displaystyle\frac{(-1)^n n!}{\pi}\int_{0}^{n}\frac{\sin\pi x\,\mathrm{d}x}{\prod_{k=0}^{n}(x-k)}$$. Doing partial fractions, we have $$\prod_{k=0}^{n}(x-k)^{-1}=\sum_{k=0}^{n}\frac{a_k}{x-k},\quad a_m=\prod_{\substack{0\leq k\leq n\\k\neq m}}(m-k)^{-1}=\frac{(-1)^{n-m}}{m!(n-m)!}$$ (say, multiplying by $$x-m$$ and letting $$x\to m$$). Thus we get $$I(n)=\frac{1}{\pi}\int_{0}^{n}\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{\sin\pi x}{x-k}\,\mathrm{d}x=\frac{1}{\pi}\sum_{k=0}^{n}\binom{n}{k}\int_{-k\pi}^{(n-k)\pi}\frac{\sin t}{t}\,\mathrm{d}t.$$ Simplification of this, using the sine integral function, gives the expected result.
When I was your age (15 or so), I also played around with a very similar integral. My take was slightly different.
First of all, for suitable real numbers $$s$$ (say with positive real part) one actually has the sum
$$\sum_{-\infty}^\infty \binom{s}{k} = 2^s.$$
So a more natural integral to consider might be
$$I(s):=\int_{-\infty}^\infty \frac{\Gamma(s+1)}{\Gamma(x+1) \Gamma(s+1-x)} \, dx.$$
You can prove $$I(n) = 2^n$$ for non-negative integers $$n$$. The point here is that one can use the reflection formula for the $$\Gamma$$ function. For example, in the easiest case when $$s = 0$$, the integral becomes
\begin{align} & \int_{-\infty}^\infty \frac{1}{\Gamma(x+1) \Gamma(1-x)} \, dx \\[10pt] = {} & \int_{-\infty}^\infty \frac{1}{x \Gamma(x) \Gamma(1-x)} \, dx \\[10pt] = {} & \int_{-\infty}^\infty \frac{\sin(\pi x)}{x \pi} \, dx = 1. \end{align}
For larger integers $$n$$, you can do pretty much the same thing and then use partial fractions. The same argument will work with your integral, except now the sin integral will go from $$0$$ to $$n$$ rather than $$\infty$$, and hence you pick up the corresponding functions. (and looks like someone has done that).
This doesn't work for general $$s$$, however. It took me quite a while to work out, but eventually I found a few different arguments to prove that $$I(s) = 2^s$$, including a fairly clean proof by contour integration.
Of course, all this was in the days before the internet, so I had to figure it out on my own by thinking about it (even though it was surely known before and also not that hard in the end --- there are very few truly deep definite integrals). I feel a little sad that you can just ask the question and someone will come and answer it.
• Thank you so much for the story to go along with your answer. Rest assured I spend much more time playing around with math on my own than I spend asking about it here! I just haven't actually had a formal calculus class so integrals with special functions always tend to scare me off :) But this was too cool to ignore. I'm currently trying to teach myself contour integration, so I'll search for the proof you mention! Once again, thank you. Sep 26, 2018 at 22:46
• I would be much interested to know your proof of $I(s)=2^s$ Nov 3, 2020 at 11:37
• @GCab: This user hasn't visited the site since this answer. Perhaps, the problem is worth posting as a dedicated question (if it's not done already by someone else). This would give someone (like me ;) an opportunity to provide an answer, although the ideas above (as well as the problem itself) are already a good brain teaser to try it by oneself ;) Jan 3, 2021 at 19:08
• @metamorphy: thanks indeed for your interest and suggestion. Actually I already posted here a bit wider question on what is $$f(z,r) = \int_0^\infty {\binom{ r,t} z^{\,t} dt}$$ and .. you have been the only one to provide a partial answer :). I would be glad if you can add more! Jan 3, 2021 at 19:51
• @GCab: I've appended a proof to that answer (it looks contextually better there). Jan 5, 2021 at 9:33 | 2022-06-27T09:53:36 | {
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https://math.stackexchange.com/questions/4015912/trigonometric-definite-integral-int-frac-pi2-0-cos4s-mathrmds-in | # Trigonometric definite integral $\int ^\frac{\pi}{2} _0 \cos^4(s)\mathrm{ds}-\int^\frac{\pi}{2}_0 \cos^6(s)\mathrm{ds}$
$$\int ^\frac{\pi}{2} _0 \cos^4(s)\mathrm{ds}-\int^\frac{\pi}{2}_0 \cos^6(s)\mathrm{ds}$$
the answer is $$\dfrac{3\cdot1}{4\cdot2}-\dfrac{5\cdot3\cdot1}{6\cdot4\cdot2}$$
I really don't know why
There may be some tricks in this, but I have not seen it in the textbook
• Wallis integral formula is useful Feb 7 at 6:52
• Take out $\operatorname{cos}^4s$ as a common factor and then refer to this Feb 7 at 7:06
• Find $cos^n(x)$ in terms of $cos(x)$ in Wikipedia, then integrate. Feb 7 at 7:08
• I'm a rookie.... Feb 7 at 8:01
• I think there are some missing $\pi/2$ Feb 7 at 14:04
1. Define $$I_n=\int_0^{\pi/2}\cos^{2n}(t)dt$$
2. Integrate by parts $$I_n=\int_0^{\pi/2}\cos^{2n-1}(t)\cos(t)dt=\color{blue}{\cos^{2n-1}(t)\sin(t)\Big|_0^{\pi/2}}+(2n-1)\int_0^{\pi/2}\cos^{2n-2}(t)\sin^2(t)dt$$ and observe that the blue part is equal to zero.
3. Use $$\cos^2(t)+\sin^2(t)=1$$ $$(2n-1)\int_0^{\pi/2}\cos^{2n-2}(t)\sin^2(t)dt=(2n-1)\Big(\int_0^{\pi/2}\cos^{2(n-1)}(t)dt-\int_0^{\pi/2}\cos^{2n}(t)dt\Big)$$
4. Collect all results $$I_n=(2n-1)(I_{n-1}-I_n)$$
5. Isolate $$I_n$$ $$I_n=\frac{2n-1}{2n}I_{n-1}$$
6. Compute $$I_0, I_1, I_2, I_3$$ $$I_0=\frac{\pi}{2},\quad I_1=\frac{1}{2}\frac{\pi}{2},\quad I_2=\frac{3\cdot 1}{4\cdot 2}\frac{\pi}{2},\quad I_3=\frac{5\cdot 3\cdot 1}{6\cdot 4\cdot 2}\frac{\pi}{2}$$
7. Compute $$I_2-I_3$$
Integrate-by-parts below to obtain the recursive formula
\begin{align} I_n=\int ^\frac{\pi}{2} _0 \cos^n s \>{ds}&= \int ^\frac{\pi}{2}_0 \frac{\cot^{n-1} s }{n}d(\sin^{n}s) {ds}=\frac{n-1}nI_{n-2},\>\>\>I_0=\frac\pi2 \end{align}
and then apply it to
$$\int^\frac{\pi}{2}_0 \cos^4s \> {ds} -\int^\frac{\pi}{2}_0 \cos^6s \>{ds} =I_4-I_6=\frac34\frac12I_0-\frac56 \frac34\frac12I_0$$
• Nice answer! An explicit formula is also possible: $$I_n=\frac{\sqrt{\pi}~\Gamma\left(\frac{n+1}{2}\right)}{2~\Gamma\left(\frac{n}{2}+1\right)}$$ Feb 7 at 21:58
It seems that one of the solution may be the formula for $$\cos$$ at even power: $$\cos^{2n}(s)=\frac{1}{2^{2n}}(e^{is}+e^{-is})^{2n}$$: $$I(n)=\int ^\frac{\pi}{2} _0 \cos^{2n}(s)\mathrm{ds}=\frac{1}{4}\int ^{2\pi} _0 \cos^{2n}(s)\mathrm{ds}=\frac{1}{2^{2n}4}\int ^{2\pi} _0 (e^{is}+e^{-is})^{2n}\mathrm{ds}=\frac{1}{2^{2n}4}\int ^{2\pi} _0 (e^{2ins}+\binom{2n}{1}e^{is(2n-1)-is}+...+\binom{2n}{n}e^{is(2n-n)-isn}+...+e^{-2ins})\mathrm{ds}$$
After integration the only surviving term is $$\frac{2\pi}{2^{2n}4}\binom{2n}{n}=\frac{2\pi(2n)!}{2^{2n}4(n!)^2}$$ | 2021-09-24T11:14:45 | {
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http://math.stackexchange.com/questions/135134/find-restrictions-on-a0-and-b0-that-ensure-that-fx-1-x-2-is-concave | # Find restrictions on $a>0$ and $b>0$ that ensure that $f(x_1,x_2)$ is concave.
Let $f:\mathbb{R}_{+}^2 \rightarrow \mathbb{R}$ be $f(x_1,x_2)=x_1^a x_2^b$ for $a>0$ and $b>0$. Find restrictions on $a>0$ and $b>0$ that ensure that $f(x_1,x_2)$ is concave.
I tried solving this by finding the principal minors of the Hessian of this function and making first principal minor be less or equal to zero and second principal minor be greater or equal to zero(conditions for a function to be concave). It is easy with the first principal minor, but I cannot workout the second principal minor to have a strong inequality.
$$H(x)=\left| \begin{array}{cc} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{array} \right|$$
$$f_{xx}=\frac{\partial^2 f}{\partial x_1^2}=a(a-1)x_2^b x_1^{a-2}$$
$$f_{xy}=\frac{\partial^2f}{\partial x_1 \partial x_2}=ab x_1^{a-1} x_2^{b-1}$$
$$f_{yx}=\frac{\partial^2f}{\partial x_2 \partial x_1}=ab x_1^{a-1} x_2^{b-1}$$
$$f_{yy}=\frac{\partial^2f}{\partial x_2^2}=b(b-1)x_1^a x_2^{b-2}$$
First principal minor:
$f_{xx}=a(a-1)x_2^b x_1^{a-2} \leq 0$; restriction on $a$ is $a<1$.
Second principal minor:
$f_{xx} f_{yy} - f_{xy} f_{yx} = ax_2^b (a-1) x_1^{a-2} bx_1^a (b-1)x_2^{b-2} - a^2 b^2 x_1^{2a-2} x_2^{2b-2} \geq 0$; restriction on b?
I might have computed the second partial derivatives for the Hessian wrong, so if you can please try solving this problem from scratch.
-
Welcome to math.SE: since you are new, I wanted to let you know about homework policy: If this is homework, please add the [homework] tag; people will still help, so don't worry. Regards, – user21436 Apr 22 '12 at 6:31
@KannappanSampath: done – Koba Apr 22 '12 at 6:40
1. The restriction on $a$ for $f_{xx}$ to be nonpositive is not what you write. 2. The second condition is that the Hessian is nonnegative, not that is is nonpositive. 3. The same powers of $x_1$ and $x_2$ are in every term of the Hessian, which simplifies the determination of its sign. – Did Apr 22 '12 at 7:59
@Didier: 1. Oh it was a typo. Restriction is a<1. 2. Typo again. 3. I know the powers are the same. When I cancel out terms I get ab>=a+b. This inequality is weak. I wonder if there is a way to make b be greater or less than some number. – Koba Apr 22 '12 at 14:44
My professor says I should show that a+b<=1. – Koba Apr 22 '12 at 14:55
To find restrictions on $a>0$ and $b>0$ that ensure that $f(x_1,x_2)=x_1^ax_2^b$ is concave I first calculated its Hessian and made the first principal minor be $\leq0$ and second principal minor be $\geq0$(the sufficient conditions for a function to be concave):
$$H(x)=\left| \begin{array}{cc} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{array} \right|$$
$$f_{xx}=\frac{\partial^2 f}{\partial x_1^2}=a(a-1)x_2^b x_1^{a-2}$$
$$f_{xy}=\frac{\partial^2f}{\partial x_1 \partial x_2}=ab x_1^{a-1} x_2^{b-1}$$
$$f_{yx}=\frac{\partial^2f}{\partial x_2 \partial x_1}=ab x_1^{a-1} x_2^{b-1}$$
$$f_{yy}=\frac{\partial^2f}{\partial x_2^2}=b(b-1)x_1^a x_2^{b-2}$$
First principal minor:
$f_{xx}=a(a-1)x_2^b x_1^{a-2} \leq 0$
for the above inequality to be true the restriction on $a$ should be $a<1$.
Second principal minor:
$f_{xx} f_{yy} - f_{xy} f_{yx} = ax_2^b (a-1) x_1^{a-2} bx_1^a (b-1)x_2^{b-2} - a^2 b^2 x_1^{2a-2} x_2^{2b-2} \geq 0$
This inequality is not so obvious and I had to expand all expressions in parenthesis and divide both sides by the same terms: $$ax_2^b (a-1) x_1^{a-2} bx_1^a (b-1)x_2^{b-2} - a^2 b^2 x_1^{2a-2} x_2^{2b-2} \geq 0$$ $$ax_2^b (a-1) x_1^{a-2} bx_1^a (b-1)x_2^{b-2}\geq a^2 b^2 x_1^{2a-2} x_2^{2b-2}$$ $$ab(a-1)(b-1)x_1^{2a-2}x_2^{2b-2}\geq a^2 b^2 x_1^{2a-2} x_2^{2b-2}$$ $$(a-1)(b-1)\geq ab$$ $$ab-a-b+1-ab\geq 0$$ $$-a-b+1\geq 0$$ $$1\geq a+b$$
So here is the second restriction: $a+b\leq 1$
- | 2015-07-08T07:00:57 | {
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https://www.physicsforums.com/threads/max-min-of-multivariate-function.782927/ | # Max/min of multivariate function
1. Nov 18, 2014
### Panphobia
1. The problem statement, all variables and given/known data
max/min of
f(x,y) = x + y
constraint xy = 16
3. The attempt at a solution
With lagrange multipliers I did
$\nabla f = (1,1)$
$\nabla g = (y,x)$
$\nabla f = \lambda \nabla g$
$1 = \lambda y$
$1 = \lambda x$
Since y=0, x=0 aren't a part of xy = 16 I can isolate for lambda
$y = x$
$y^2 = 16$
$y = \pm 4$
$y = 4, x = 4$
$y = -4, x = -4$
$f(4,4) = 8$
$f(-4,-4) = -8$
I got these values, but my answer key says that there are no minimums or maximums, can anyone explain why?
2. Nov 18, 2014
### LCKurtz
The values you have found are relative min/max points as you move along xy=16. But neither are absolute extrema because x+y gets larger and smaller than both then you let either x or y get large positive or negative.
3. Nov 18, 2014
### Panphobia
Yea I understand since as x approaches infinity, y approaches 0, or x approach negative infintiy y approaches 0, so f(x,y) never has a max or min.
4. Nov 19, 2014
### Ray Vickson
In the positive quadrant $x, y \geq 0$ your constrained $f$ does have a minimum, but no maximum. In the third quadrant $x \leq 0, y \leq 0$ the constrained function has a maximum, but no minimum. If we throw out the information about quadrants then, of course, it is true that the constrained $f$ had neither a maximum nor a minimum. | 2017-12-17T03:26:35 | {
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https://math.stackexchange.com/questions/2691984/probability-of-picking-a-certain-fruit-from-a-tree | # Probability of picking a certain fruit from a tree
A tree has 20 fruits. 15 of which have no seeds and the rest do have seeds. A bird eats 5 of these fruits picked at random.
a) If i pick one fruit from whats left on the tree whats the probability it has seeds
b) Given that this one fruit i pick has seeds whats the probability that the bird had consumed at least one with seeds.
What i have so far:
So before the bird eats anything the probability of a fruit with no seeds is 15/20=0.75 and the probability of fruit with seeds is 5/20=0.25.
For part a is the probability that the fruit i pick(i.e. 6th fruit picked) has seeds is given by (15/20)* (14/19)* (13/18)* (12/17) *(11/16) *(5/15)=0.065
Am i on the right track with part a? As for part b im not sure where to begin.
• For part a, it doesn't really matter that the bird ate any fruit. The probability that the sixth fruit has seeds is the same as the probability that the first fruit has seeds. This is easier to confirm with calculations if the bird only eats one fruit, but the principle is the same. – Arthur Mar 15 '18 at 8:25
• For part b, this is similar to asking whether with $19$ fruit, $4$ of which have seeds, the bird eats at least one with seeds (i.e. the bird does not eat $5$ seedless fruit) – Henry Mar 15 '18 at 8:34
a)
The probability that you calculated is the probability that the bird only picked fruits without seeds and you picked one fruit with seeds. That is not the correct answer to the question.
The correct answer is $\frac5{20}$. Think of it like this: all fruits are placed randomly in a row. Now what is the probability that on spot number $6$ (corresponding to your pick after the $5$ picks of the bird) a fruit is placed that has seeds? It will not differ from the probability of any other spot to achieve a fruit that has seeds.
b)
Let $E$ denote the event that the bird consumed none with seeds and let $F$ denote the probability that you picked a fruit with seeds. Then: $$P(E\mid F)P(F)=P(E\cap F)$$where $P(F)=\frac5{20}$ (the probability calculated at a) and $P(E\cap F)=\frac{15}{20}\frac{14}{19}\frac{13}{18}\frac{12}{17}\frac{11}{16}\frac{5}{15}$ (the probability that you calculated in your answer).
This allows you to find $P(E\mid F)$ and also $P(E^{\complement}\mid F)=1-P(E\mid F)$
Note that $P(E^{\complement}\mid F)$ is the probability that you are looking for.
For a more elegant route see the comment of Henry.
$(a)$
Notice that this is a hypergeometric random variable. Using the same variables as Matti
• $P_0 = \frac{{5\choose0}{15\choose5}}{20\choose5}$
• $P_1 = \frac{{5\choose1}{15\choose4}}{20\choose5}$
• $P_2 = \frac{{5\choose2}{15\choose3}}{20\choose5}$
• $P_3 = \frac{{5\choose3}{15\choose2}}{20\choose5}$
• $P_4 = \frac{{5\choose4}{15\choose1}}{20\choose5}$
• $P_5 = \frac{{5\choose5}{15\choose0}}{20\choose5}$
These calculations can be made in R statistical software:
> dhyper(0:5, m=5, n=15, k=5, log = FALSE)
[1] 1.936920e-01 4.402090e-01 2.934727e-01 6.772446e-02 4.837461e-03 6.449948e-05
Now given that $P_0,P_1,...,P_5$ happen the probabilities then of selecting a fruit with seeds are $\frac{5}{15},\frac{4}{15},\frac{3}{15},\frac{2}{15},\frac{1}{15}$, and $\frac{0}{15}$, respectively.
Then if we pick one fruit the probability that it has seeds is $$P_0\cdot\frac{5}{15}+P_1\cdot\frac{4}{15}+P_2\cdot\frac{3}{15}+P_3\cdot\frac{2}{15}+P_4\cdot\frac{1}{15}+P_5\cdot\frac{0}{15}=\frac{1}{4}$$
as Arthur alluded to. This can be calculated in R using
> sum(dhyper(0:5, m=5, n=15, k=5, log = FALSE)*(c(5/15,4/15,3/15,2/15,1/15,0)))
[1] 0.25
$(b)$
Let $P_6$ denote the event of getting a fruit with seeds. We wish to find the probability that $P_0$ did not occur. We have \begin{align*} P(P_0^C \mid P_6) &=\frac{P(P_0^C \cap P_6)}{P(P_6)}\\\\ &=\frac{P_1\cdot\frac{4}{15}+P_2\cdot\frac{3}{15}+P_3\cdot\frac{2}{15}+P_4\cdot\frac{1}{15}+P_5\cdot\frac{0}{15}}{P_0\cdot\frac{5}{15}+P_1\cdot\frac{4}{15}+P_2\cdot\frac{3}{15}+P_3\cdot\frac{2}{15}+P_4\cdot\frac{1}{15}+P_5\cdot\frac{0}{15}=\frac{1}{4}}\\\\ &=\frac{\frac{1}{4}-P_0\cdot\frac{5}{15}}{\frac{1}{4}}\\\\ &\approx 0.742 \end{align*}
• Nice confirmation of intuition. – drhab Mar 15 '18 at 9:10
Since there bird eats only five fruits, it's relatively easy to consider all possible cases. Let's write
• $P_0 = P(\text{bird eats zero fruits with seeds}) = \frac{15}{20} \frac{14}{19} \frac{13}{18} \frac{12}{17} \frac{11}{16} = \frac{1001}{15504}$
• $P_1 = P(\text{bird eats exactly one fruit with seeds}) =$ $5\choose 1$$\frac{5}{20} \frac{14}{19} \frac{13}{18}\frac{12}{17} \frac{11}{16} = \frac{5005}{15504} • P_2 = P(\text{bird eats exactly two fruits with seeds}) = 5\choose 2$$\frac{5}{20} \frac{4}{19} \frac{13}{18}\frac{12}{17} \frac{11}{16} = \frac{715}{3876}$
• $P_3 = P(\text{bird eats exactly three fruits with seeds}) =$ $5\choose 3$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{12}{17} \frac{11}{16} = \frac{55}{1292} • P_4 = P(\text{bird eats exactly four fruits with seeds}) = 5\choose 4$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{2}{17} \frac{11}{16} = \frac{55}{15504}$
• $P_5 = P(\text{bird eats exactly five fruits with seeds}) =$ $5\choose 5$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{2}{17} \frac{1}{16} = \frac{1}{15504}$
So for part a, you can consider the five cases separately.
• If the bird ate zero fruits with seeds, there are $5$ out of $15$ fruits with seeds left. So you have to multiply $\frac{1001}{15504}\frac{5}{15}$
• If the bird ate one fruit with seeds, there are $4$ out of $15$ fruits left with seeds. So you have to multiply $\frac{5005}{15504} \frac{4}{15}$
etc ...
• e.g. $P_1=\binom51\frac5{20}\frac{15}{19}\frac{14}{18}\frac{13}{17}\frac{12}{16}$ and there are more mistakes of that sort. – drhab Mar 15 '18 at 9:08 | 2019-01-16T23:17:28 | {
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https://dantopology.wordpress.com/tag/point-set-topology/ | # Lindelof Exercise 2
The preceding post is an exercise showing that the product of countably many $\sigma$-compact spaces is a Lindelof space. The result is an example of a situation where the Lindelof property is countably productive if each factor is a “nice” Lindelof space. In this case, “nice” means $\sigma$-compact. This post gives several exercises surrounding the notion of $\sigma$-compactness.
Exercise 2.A
According to the preceding exercise, the product of countably many $\sigma$-compact spaces is a Lindelof space. Give an example showing that the result cannot be extended to the product of uncountably many $\sigma$-compact spaces. More specifically, give an example of a product of uncountably many $\sigma$-compact spaces such that the product space is not Lindelof.
Exercise 2.B
Any $\sigma$-compact space is Lindelof. Since $\mathbb{R}=\bigcup_{n=1}^\infty [-n,n]$, the real line with the usual Euclidean topology is $\sigma$-compact. This exercise is to find an example of “Lindelof does not imply $\sigma$-compact.” Find one such example among the subspaces of the real line. Note that as a subspace of the real line, the example would be a separable metric space, hence would be a Lindelof space.
Exercise 2.C
This exercise is also to look for an example of a space that is Lindelof and not $\sigma$-compact. The example sought is a non-metric one, preferably a space whose underlying set is the real line and whose topology is finer than the Euclidean topology.
Exercise 2.D
Show that the product of two Lindelof spaces is a Lindelof space whenever one of the factors is a $\sigma$-compact space.
Exercise 2.E
Prove that the product of finitely many $\sigma$-compact spaces is a $\sigma$-compact space. Give an example of a space showing that the product of countably and infinitely many $\sigma$-compact spaces does not have to be $\sigma$-compact. For example, show that $\mathbb{R}^\omega$, the product of countably many copies of the real line, is not $\sigma$-compact.
The Lindelof property and $\sigma$-compactness are basic topological notions. The above exercises are natural questions based on these two basic notions. One immediate purpose of these exercises is that they provide further interaction with the two basic notions. More importantly, working on these exercise give exposure to mathematics that is seemingly unrelated to the two basic notions. For example, finding $\sigma$-compactness on subspaces of the real line and subspaces of compact spaces naturally uses a Baire category argument, which is a deep and rich topic that finds uses in multiple areas of mathematics. For this reason, these exercises present excellent learning opportunities not only in topology but also in other useful mathematical topics.
If preferred, the exercises can be attacked head on. The exercises are also intended to be a guided tour. Hints are also provided below. Two sets of hints are given – Hints (blue dividers) and Further Hints (maroon dividers). The proofs of certain key facts are also given (orange dividers). Concluding remarks are given at the end of the post.
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Hints for Exercise 2.A
Prove that the Lindelof property is hereditary with respect to closed subspaces. That is, if $X$ is a Lindelof space, then every closed subspace of $X$ is also Lindelof.
Prove that if $X$ is a Lindelof space, then every closed and discrete subset of $X$ is countable (every space that has this property is said to have countable extent).
Show that the product of uncountably many copies of the real line does not have countable extent. Specifically, focus on either one of the following two examples.
• Show that the product space $\mathbb{R}^c$ has a closed and discrete subspace of cardinality continuum where $c$ is cardinality of continuum. Hence $\mathbb{R}^c$ is not Lindelof.
• Show that the product space $\mathbb{R}^{\omega_1}$ has a closed and discrete subspace of cardinality $\omega_1$ where $\omega_1$ is the first uncountable ordinal. Hence $\mathbb{R}^{\omega_1}$ is not Lindelof.
Hints for Exercise 2.B
Let $\mathbb{P}$ be the set of all irrational numbers. Show that $\mathbb{P}$ as a subspace of the real line is not $\sigma$-compact.
Hints for Exercise 2.C
Let $S$ be the real line with the topology generated by the half open and half closed intervals of the form $[a,b)=\{ x \in \mathbb{R}: a \le x < b \}$. The real line with this topology is called the Sorgenfrey line. Show that $S$ is Lindelof and is not $\sigma$-compact.
Hints for Exercise 2.D
It is helpful to first prove: the product of two Lindelof space is Lindelof if one of the factors is a compact space. The Tube lemma is helpful.
Tube Lemma
Let $X$ be a space. Let $Y$ be a compact space. Suppose that $U$ is an open subset of $X \times Y$ and suppose that $\{ x \} \times Y \subset U$ where $x \in X$. Then there exists an open subset $V$ of $X$ such that $\{ x \} \times Y \subset V \times Y \subset U$.
Hints for Exercise 2.E
Since the real line $\mathbb{R}$ is homeomorphic to the open interval $(0,1)$, $\mathbb{R}^\omega$ is homeomorphic to $(0,1)^\omega$. Show that $(0,1)^\omega$ is not $\sigma$-compact.
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Further Hints for Exercise 2.A
The hints here focus on the example $\mathbb{R}^c$.
Let $I=[0,1]$. Let $\omega$ be the first infinite ordinal. For convenience, consider $\omega$ the set $\{ 0,1,2,3,\cdots \}$, the set of all non-negative integers. Since $\omega^I$ is a closed subset of $\mathbb{R}^I$, any closed and discrete subset of $\omega^I$ is a closed and discrete subset of $\mathbb{R}^I$. The task at hand is to find a closed and discrete subset of $Y=\omega^I$. To this end, we define $W=\{W_x: x \in I \}$ after setting up background information.
For each $t \in I$, choose a sequence $O_{t,1},O_{t,2},O_{t,3},\cdots$ of open intervals (in the usual topology of $I$) such that
• $\{ t \}=\bigcap_{j=1}^\infty O_{t,j}$,
• $\overline{O_{t,j+1}} \subset O_{t,j}$ for each $j$ (the closure is in the usual topology of $I$).
Note. For each $t \in I-\{0,1 \}$, the open intervals $O_{t,j}$ are of the form $(a,b)$. For $t=0$, the open intervals $O_{t,j}$ are of the form $[0,b)$. For $t=1$, the open intervals $O_{t,j}$ are of the form $(a,1]$.
For each $t \in I$, define the map $f_t: I \rightarrow \omega$ as follows:
$f_t(x) = \begin{cases} 0 & \ \ \ \mbox{if } x=t \\ 1 & \ \ \ \mbox{if } x \in I-O_{t,1} \\ 2 & \ \ \ \mbox{if } x \in I-O_{t,2} \text{ and } x \in O_{t,1} \\ 3 & \ \ \ \mbox{if } x \in I-O_{t,3} \text{ and } x \in O_{t,2} \\ \vdots & \ \ \ \ \ \ \ \ \ \ \vdots \\ j & \ \ \ \mbox{if } x \in I-O_{t,j} \text{ and } x \in O_{t,j-1} \\ \vdots & \ \ \ \ \ \ \ \ \ \ \vdots \end{cases}$
We are now ready to define $W=\{W_x: x \in I \}$. For each $x \in I$, $W_x$ is the mapping $W_x:I \rightarrow \omega$ defined by $W_x(t)=f_t(x)$ for each $t \in I$.
Show the following:
• The set $W=\{W_x: x \in I \}$ has cardinality continuum.
• The set $W$ is a discrete space.
• The set $W$ is a closed subspace of $Y$.
Further Hints for Exercise 2.B
A subset $A$ of the real line $\mathbb{R}$ is nowhere dense in $\mathbb{R}$ if for any nonempty open subset $U$ of $\mathbb{R}$, there is a nonempty open subset $V$ of $U$ such that $V \cap A=\varnothing$. If we replace open sets by open intervals, we have the same notion.
Show that the real line $\mathbb{R}$ with the usual Euclidean topology cannot be the union of countably many closed and nowhere dense sets.
Further Hints for Exercise 2.C
Prove that if $X$ and $Y$ are $\sigma$-compact, then the product $X \times Y$ is $\sigma$-compact, hence Lindelof.
Prove that $S$, the Sorgenfrey line, is Lindelof while its square $S \times S$ is not Lindelof.
Further Hints for Exercise 2.D
As suggested in the hints given earlier, prove that $X \times Y$ is Lindelof if $X$ is Lindelof and $Y$ is compact. As suggested, the Tube lemma is a useful tool.
Further Hints for Exercise 2.E
The product space $(0,1)^\omega$ is a subspace of the product space $[0,1]^\omega$. Since $[0,1]^\omega$ is compact, we can fall back on a Baire category theorem argument to show why $(0,1)^\omega$ cannot be $\sigma$-compact. To this end, we consider the notion of Baire space. A space $X$ is said to be a Baire space if for each countable family $\{ U_1,U_2,U_3,\cdots \}$ of open and dense subsets of $X$, the intersection $\bigcap_{i=1}^\infty U_i$ is a dense subset of $X$. Prove the following results.
Fact E.1
Let $X$ be a compact Hausdorff space. Let $O_1,O_2,O_3,\cdots$ be a sequence of non-empty open subsets of $X$ such that $\overline{O_{n+1}} \subset O_n$ for each $n$. Then the intersection $\bigcap_{i=1}^\infty O_i$ is non-empty.
Fact E.2
Any compact Hausdorff space is Baire space.
Fact E.3
Let $X$ be a Baire space. Let $Y$ be a dense $G_\delta$-subset of $X$ such that $X-Y$ is a dense subset of $X$. Then $Y$ is not a $\sigma$-compact space.
Since $X=[0,1]^\omega$ is compact, it follows from Fact E.2 that the product space $X=[0,1]^\omega$ is a Baire space.
Fact E.4
Let $X=[0,1]^\omega$ and $Y=(0,1)^\omega$. The product space $Y=(0,1)^\omega$ is a dense $G_\delta$-subset of $X=[0,1]^\omega$. Furthermore, $X-Y$ is a dense subset of $X$.
It follows from the above facts that the product space $(0,1)^\omega$ cannot be a $\sigma$-compact space.
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Proofs of Key Steps for Exercise 2.A
The proof here focuses on the example $\mathbb{R}^c$.
To see that $W=\{W_x: x \in I \}$ has the same cardinality as that of $I$, show that $W_x \ne W_y$ for $x \ne y$. This follows from the definition of the mapping $W_x$.
To see that $W$ is discrete, for each $x \in I$, consider the open set $U_x=\{ b \in Y: b(x)=0 \}$. Note that $W_x \in U_x$. Further note that $W_y \notin U_x$ for all $y \ne x$.
To see that $W$ is a closed subset of $Y$, let $k: I \rightarrow \omega$ such that $k \notin W$. Consider two cases.
Case 1. $k(r) \ne 0$ for all $r \in I$.
Note that $\{ O_{t,k(t)}: t \in I \}$ is an open cover of $I$ (in the usual topology). There exists a finite $H \subset I$ such that $\{ O_{h,k(h)}: h \in H \}$ is a cover of $I$. Consider the open set $G=\{ b \in Y: \forall \ h \in H, \ b(h)=k(h) \}$. Define the set $F$ as follows:
$F=\{ c \in I: W_c \in G \}$
The set $F$ can be further described as follows:
\displaystyle \begin{aligned} F&=\{ c \in I: W_c \in G \} \\&=\{ c \in I: \forall \ h \in H, \ W_c(h)=f_h(c)=k(h) \ne 0 \} \\&=\{ c \in I: \forall \ h \in H, \ c \in I-O_{h,k(h)} \} \\&=\bigcap_{h \in H} (I-O_{h,k(h)}) \\&=I-\bigcup_{h \in H} O_{h,k(h)}=I-I =\varnothing \end{aligned}
The last step is $\varnothing$ because $\{ O_{h,k(h)}: h \in H \}$ is a cover of $I$. The fact that $F=\varnothing$ means that $G$ is an open subset of $Y$ containing the point $k$ such that $G$ contains no point of $W$.
Case 2. $k(r) = 0$ for some $r \in I$.
Since $k \notin W$, $k \ne W_x$ for all $x \in I$. In particular, $k \ne W_r$. This means that $k(t) \ne W_r(t)$ for some $t \in I$. Define the open set $G$ as follows:
$G=\{ b \in Y: b(r)=0 \text{ and } b(t)=k(t) \}$
Clearly $k \in G$. Observe that $W_r \notin G$ since $W_r(t) \ne k(t)$. For each $p \in I-\{ r \}$, $W_p \notin G$ since $W_p(r) \ne 0$. Thus $G$ is an open set containing $k$ such that $G \cap W=\varnothing$.
Both cases show that $W$ is a closed subset of $Y=\omega^I$.
Proofs of Key Steps for Exercise 2.B
Suppose that $\mathbb{P}$, the set of all irrational numbers, is $\sigma$-compact. That is, $\mathbb{P}=A_1 \cup A_2 \cup A_3 \cup \cdots$ where each $A_i$ is a compact space as a subspace of $\mathbb{P}$. Any compact subspace of $\mathbb{P}$ is also a compact subspace of $\mathbb{R}$. As a result, each $A_i$ is a closed subset of $\mathbb{R}$. Furthermore, prove the following:
Each $A_i$ is a nowhere dense subset of $\mathbb{R}$.
Each singleton set $\{ r \}$ where $r$ is any rational number is also a closed and nowhere dense subset of $\mathbb{R}$. This means that the real line is the union of countably many closed and nowhere dense subsets, contracting the hints given earlier. Thus $\mathbb{P}$ cannot be $\sigma$-compact.
Proofs of Key Steps for Exercise 2.C
The Sorgenfrey line $S$ is a Lindelof space whose square $S \times S$ is not normal. This is a famous example of a Lindelof space whose square is not Lindelof (not even normal). For reference, a proof is found here. An alternative proof of the non-normality of $S \times S$ uses the Baire category theorem and is found here.
If the Sorgenfrey line is $\sigma$-compact, then $S \times S$ would be $\sigma$-compact and hence Lindelof. Thus $S$ cannot be $\sigma$-compact.
Proofs of Key Steps for Exercise 2.D
Suppose that $X$ is Lindelof and that $Y$ is compact. Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, let $\mathcal{U}_x \subset \mathcal{U}$ be finite such that $\mathcal{U}_x$ is a cover of $\{ x \} \times Y$. Putting it another way, $\{ x \} \times Y \subset \cup \mathcal{U}_x$. By the Tube lemma, for each $x \in X$, there is an open $O_x$ such that $\{ x \} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x$. Since $X$ is Lindelof, there exists a countable set $\{ x_1,x_2,x_3,\cdots \} \subset X$ such that $\{ O_{x_1},O_{x_2},O_{x_3},\cdots \}$ is a cover of $X$. Then $\mathcal{U}_{x_1} \cup \mathcal{U}_{x_2} \cup \mathcal{U}_{x_3} \cup \cdots$ is a countable subcover of $\mathcal{U}$. This completes the proof that $X \times Y$ is Lindelof when $X$ is Lindelof and $Y$ is compact.
To complete the exercise, observe that if $X$ is Lindelof and $Y$ is $\sigma$-compact, then $X \times Y$ is the union of countably many Lindelof subspaces.
Proofs of Key Steps for Exercise 2.E
Proof of Fact E.1
Let $X$ be a compact Hausdorff space. Let $O_1,O_2,O_3,\cdots$ be a sequence of non-empty open subsets of $X$ such that \$latex $\overline{O_{n+1}} \subset O_n$ for each $n$. Show that the intersection $\bigcap_{i=1}^\infty O_i$ is non-empty.
Suppose that $\bigcap_{i=1}^\infty O_i=\varnothing$. Choose $x_1 \in O_1$. There must exist some $n_1$ such that $x_1 \notin O_{n_1}$. Choose $x_2 \in O_{n_1}$. There must exist some $n_2>n_1$ such that $x_2 \notin O_{n_2}$. Continue in this manner we can choose inductively an infinite set $A=\{ x_1,x_2,x_3,\cdots \} \subset X$ such that $x_i \ne x_j$ for $i \ne j$. Since $X$ is compact, the infinite set $A$ has a limit point $p$. This means that every open set containing $p$ contains some $x_j$ (in fact for infinitely many $j$). The point $p$ cannot be in the intersection $\bigcap_{i=1}^\infty O_i$. Thus for some $n$, $p \notin O_n$. Thus $p \notin \overline{O_{n+1}}$. We can choose an open set $U$ such that $p \in U$ and $U \cap \overline{O_{n+1}}=\varnothing$. However, $U$ must contain some point $x_j$ where $j>n+1$. This is a contradiction since $O_j \subset \overline{O_{n+1}}$ for all $j>n+1$. Thus Fact E.1 is established.
Proof of Fact E.2
Let $X$ be a compact space. Let $U_1,U_2,U_3,\cdots$ be open subsets of $X$ such that each $U_i$ is also a dense subset of $X$. Let $V$ a non-empty open subset of $X$. We wish to show that $V$ contains a point that belongs to each $U_i$. Since $U_1$ is dense in $X$, $O_1=V \cap U_1$ is non-empty. Since $U_2$ is dense in $X$, choose non-empty open $O_2$ such that $\overline{O_2} \subset O_1$ and $O_2 \subset U_2$. Since $U_3$ is dense in $X$, choose non-empty open $O_3$ such that $\overline{O_3} \subset O_2$ and $O_3 \subset U_3$. Continue inductively in this manner and we have a sequence of open sets $O_1,O_2,O_3,\cdots$ just like in Fact E.1. Then the intersection of the open sets $O_n$ is non-empty. Points in the intersection are in $V$ and in all the $U_n$. This completes the proof of Fact E.2.
Proof of Fact E.3
Let $X$ be a Baire space. Let $Y$ be a dense $G_\delta$-subset of $X$ such that $X-Y$ is a dense subset of $X$. Show that $Y$ is not a $\sigma$-compact space.
Suppose $Y$ is $\sigma$-compact. Let $Y=\bigcup_{n=1}^\infty B_n$ where each $B_n$ is compact. Each $B_n$ is obviously a closed subset of $X$. We claim that each $B_n$ is a closed nowhere dense subset of $X$. To see this, let $U$ be a non-empty open subset of $X$. Since $X-Y$ is dense in $X$, $U$ contains a point $p$ where $p \notin Y$. Since $p \notin B_n$, there exists a non-empty open $V \subset U$ such that $V \cap B_n=\varnothing$. This shows that each $B_n$ is a nowhere dense subset of $X$.
Since $Y$ is a dense $G_\delta$-subset of $X$, $Y=\bigcap_{n=1}^\infty O_n$ where each $O_n$ is an open and dense subset of $X$. Then each $A_n=X-O_n$ is a closed nowhere dense subset of $X$. This means that $X$ is the union of countably many closed and nowhere dense subsets of $X$. More specifically, we have the following.
(1)………$X= \biggl( \bigcup_{n=1}^\infty A_n \biggr) \cup \biggl( \bigcup_{n=1}^\infty B_n \biggr)$
Statement (1) contradicts the fact that $X$ is a Baire space. Note that all $X-A_n$ and $X-B_n$ are open and dense subsets of $X$. Further note that the intersection of all these countably many open and dense subsets of $X$ is empty according to (1). Thus $Y$ cannot not a $\sigma$-compact space.
Proof of Fact E.4
The space $X=[0,1]^\omega$ is compact since it is a product of compact spaces. To see that $Y=(0,1)^\omega$ is a dense $G_\delta$-subset of $X$, note that $Y=\bigcap_{n=1}^\infty U_n$ where for each integer $n \ge 1$
(2)………$U_n=(0,1) \times \cdots \times (0,1) \times [0,1] \times [0,1] \times \cdots$
Note that the first $n$ factors of $U_n$ are the open interval $(0,1)$ and the remaining factors are the closed interval $[0,1]$. It is also clear that $X-Y$ is a dense subset of $X$. This completes the proof of Fact E.4.
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Concluding Remarks
Exercise 2.A
The exercise is to show that the product of uncountably many $\sigma$-compact spaces does not need to be Lindelof. The approach suggested in the hints is to show that $\mathbb{R}^{c}$ has uncountable extent where $c$ is continuum. Having uncountable extent (i.e. having an uncountable subset that is both closed and discrete) implies the space is not Lindelof. The uncountable extent of the product space $\mathbb{R}^{\omega_1}$ is discussed in this post.
For $\mathbb{R}^{c}$ and $\mathbb{R}^{\omega_1}$, there is another way to show non-Lindelof. For example, both product spaces are not normal. As a result, both product spaces cannot be Lindelof. Note that every regular Lindelof space is normal. Both product spaces contain the product $\omega^{\omega_1}$ as a closed subspace. The non-normality of $\omega^{\omega_1}$ is discussed here.
Exercise 2.B
The hints given above is to show that the set of all irrational numbers, $\mathbb{P}$, is not $\sigma$-compact (as a subspace of the real line). The same argument showing that $\mathbb{P}$ is not $\sigma$-compact can be generalized. Note that the complement of $\mathbb{P}$ is $\mathbb{Q}$, the set of all rational numbers (a countable set). In this case, $\mathbb{Q}$ is a dense subset of the real line and is the union of countably many singleton sets. Each singleton set is a closed and nowhere dense subset of the real line. In general, we can let $B$, the complement of a set $A$, be dense in the real line and be the union of countably many closed nowhere dense subsets of the real line (not necessarily singleton sets). The same argument will show that $A$ cannot be a $\sigma$-compact space. This argument is captured in Fact E.3 in Exercise 2.E. Thus both Exercise 2.B and Exercise 2.E use a Baire category argument.
Exercise 2.E
Like Exercise 2.B, this exercise is also to show a certain space is not $\sigma$-compact. In this case, the suggested space is $\mathbb{R}^{\omega}$, the product of countably many copies of the real line. The hints given use a Baire category argument, as outlined in Fact E.1 through Fact E.4. The product space $\mathbb{R}^{\omega}$ is embedded in the compact space $[0,1]^{\omega}$, which is a Baire space. As mentioned earlier, Fact E.3 is essentially the same argument used for Exercise 2.B.
Using the same Baire category argument, it can be shown that $\omega^{\omega}$, the product of countably many copies of the countably infinite discrete space, is not $\sigma$-compact. The space $\omega$ of the non-negative integers, as a subspace of the real line, is certainly $\sigma$-compact. Using the same Baire category argument, we can see that the product of countably many copies of this discrete space is not $\sigma$-compact. With the product space $\omega^{\omega}$, there is a connection with Exercise 2.B. The product $\omega^{\omega}$ is homeomorphic to $\mathbb{P}$. The idea of the homeomorphism is discussed here. Thus the non-$\sigma$-compactness of $\omega^{\omega}$ can be achieved by mapping it to the irrationals. Of course, the same Baire category argument runs through both exercises.
Exercise 2.C
Even the non-$\sigma$-compactness of the Sorgenfrey line $S$ can be achieved by a Baire category argument. The non-normality of the Sorgenfrey plane $S \times S$ can be achieved by Jones’ lemma argument or by the fact that $\mathbb{P}$ is not a first category set. Links to both arguments are given in the Proof section above.
See here for another introduction to the Baire category theorem.
The Tube lemma is discussed here.
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# Lindelof Exercise 1
A space $X$ is called a $\sigma$-compact space if it is the union of countably many compact subspaces. Clearly, any $\sigma$-compact space is Lindelof. It is well known that the product of Lindelof spaces does not need to be Lindelof. The most well known example is perhaps the square of the Sorgenfrey line. In certain cases, the Lindelof property can be productive. For example, the product of countably many $\sigma$-compact spaces is a Lindelof space. The discussion here centers on the following theorem.
Theorem 1
Let $X_1,X_2,X_3,\cdots$ be $\sigma$-compact spaces. Then the product space $\prod_{i=1}^\infty X_i$ is Lindelof.
Theorem 1 is Exercise 3.8G in page 195 of General Topology by Engelking [1]. The reference for Exercise 3.8G is [2]. But the theorem is not found in [2] (it is not stated directly and it does not seem to be an obvious corollary of a theorem discussed in that paper). However, a hint is provided in Engelking for Exercise 3.8G. In this post, we discuss Theorem 1 as an exercise by giving expanded hint. Solutions to some of the key steps in the expanded hint are given at the end of the post.
Expanded Hint
It is helpful to first prove the following theorem.
Theorem 2
For each integer $i \ge 1$, let $C_{i,1},C_{i,2},\cdots$ be compact spaces and let $C_i$ be the topological sum:
$C_i=C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots=\oplus_{j=1}^\infty C_{i,j}$
Then the product $\prod_{i=1}^\infty C_i$ is Lindelof.
Note that in the topological sum $C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots$, the spaces $C_{i,1},C_{i,2},C_{i,3},\cdots$ are considered pairwise disjoint. The open sets in the sum are simply unions of the open sets in the individual spaces. Another way to view this topology: each of the $C_{i,j}$ is both closed and open in the topological sum. Theorem 2 is essentially saying that the product of countably many $\sigma$-compact spaces is Lindelof if each $\sigma$-compact space is the union of countably many disjoint compact spaces. The hint for Exercise 3.8G can be applied much more naturally on Theorem 2 than on Theorem 1. The following is Exercise 3.8F (a), which is the hint for Exercise 3.8G.
Lemma 3
Let $Z$ be a compact space. Let $X$ be a subspace of $Z$. Suppose that there exist $F_1,F_2,F_3,\cdots$, closed subsets of $Z$, such that for all $x$ and $y$ where $x \in X$ and $y \in Z-X$, there exists $F_i$ such that $x \in F_i$ and $y \notin F_i$. Then $X$ is a Lindelof space.
The following theorem connects the hint (Lemma 3) with Theorem 2.
Theorem 4
For each integer $i \ge 1$, let $Z_i$ be the one-point compactification of $C_i$ in Theorem 2. Then the product $Z=\prod_{i=1}^\infty Z_i$ is a compact space. Furthermore, $X=\prod_{i=1}^\infty C_i$ is a subspace of $Z$. Prove that $Z$ and $X$ satisfy Lemma 3.
Each $C_i$ in Theorem 2 is a locally compact space. To define the one-point compactifications, for each $i$, choose $p_i \notin C_i$. Make sure that $p_i \ne p_j$ for $i \ne j$. Then $Z_i$ is simply
$Z_i=C_i \cup \{ p_i \}=C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots \cup \{ p_i \}$
with the topology defined as follows:
• Open subsets of $C_i$ continue to be open in $Z_i$.
• An open set containing $p_i$ is of the form $\{ p_i \} \cup (C_i - \overline{D})$ where $D$ is open in $C_i$ and $D$ is contained in the union of finitely many $C_{i,j}$.
For convenience, each point $p_i$ is called a point at infinity.
Note that Theorem 2 follows from Lemma 3 and Theorem 4. In order to establish Theorem 1 from Theorem 2, observe that the Lindelof property is preserved by any continuous mapping and that there is a natural continuous map from the product space in Theorem 2 to the product space in Theorem 1.
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Proofs of Key Steps
Proof of Lemma 3
Let $Z$, $X$ and $F_1,F_2,F_3,\cdots$ be as described in the statement for Lemma 3. Let $\mathcal{U}$ be a collection of open subsets of $Z$ such that $\mathcal{U}$ covers $X$. We would like to show that a countable subcollection of $\mathcal{U}$ is also a cover of $X$. Let $O=\cup \mathcal{U}$. If $Z-O=\varnothing$, then $\mathcal{U}$ is an open cover of $Z$ and there is a finite subset of $\mathcal{U}$ that is a cover of $Z$ and thus a cover of $X$. Thus we can assume that $Z-O \ne \varnothing$.
Let $F=\{ F_1,F_2,F_3,\cdots \}$. Let $K=Z-O$, which is compact. We make the following claim.
Claim. Let $Y$ be the union of all possible $\cap G$ where $G \subset F$ is finite and $\cap G \subset O$. Then $X \subset Y \subset O$.
To establish the claim, let $x \in X$. For each $y \in K=Z-O$, there exists $F_{n(y)}$ such that $x \in F_{n(y)}$ and $y \notin F_{n(y)}$. This means that $\{ Z-F_{n(y)}: y \in K \}$ is an open cover of $K$. By the compactness of $K$, there are finitely many $F_{n(y_1)}, \cdots, F_{n(y_k)}$ such that $F_{n(y_1)} \cap \cdots \cap F_{n(y_k)}$ misses $K$, or equivalently $F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} \subset O$. Note that $x \in F_{n(y_1)} \cap \cdots \cap F_{n(y_k)}$. Further note that $F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} \subset Y$. This establishes the claim that $X \subset Y$. The claim that $Y \subset O$ is clear from the definition of $Y$.
Each set $F_i$ is compact since it is closed in $Z$. The intersection of finitely many $F_i$ is also compact. Thus the $\cap G$ in the definition of $Y$ in the above claim is compact. There can be only countably many $\cap G$ in the definition of $Y$. Thus $Y$ is a $\sigma$-compact space that is covered by the open cover $\mathcal{U}$. Choose a countable $\mathcal{V} \subset \mathcal{U}$ such that $\mathcal{V}$ covers $Y$. Then $\mathcal{V}$ is a cover of $X$ too. This completes the proof that $X$ is Lindelof.
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Proof of Theorem 4
Recall that $Z=\prod_{i=1}^\infty Z_i$ and that $X=\prod_{i=1}^\infty C_i$. Each $Z_i$ is the one-point compactification of $C_i$, which is the topological sum of the disjoint compact spaces $C_{i,1},C_{i,2},\cdots$.
For integers $i,j \ge 1$, define $K_{i,j}=C_{i,1} \oplus C_{i,2} \oplus \cdots \oplus C_{i,j}$. For integers $n,j \ge 1$, define the product $F_{n,j}$ as follows:
$F_{n,j}=K_{1,j} \times \cdots \times K_{n,j} \times Z_{n+1} \times Z_{n+2} \times \cdots$
Since $F_{n,j}$ is a product of compact spaces, $F_{n,j}$ is compact and thus closed in $Z$. There are only countably many $F_{n,j}$.
We claim that the countably many $F_{n,j}$ have the property indicated in Lemma 3. To this end, let $f \in X=\prod_{i=1}^\infty C_i$ and $g \in Z-X$. There exists an integer $n \ge 1$ such that $g(n) \notin C_{n}$. This means that $g(n) \notin C_{n,j}$ for all $j$, i.e. $g(n)=p_n$ (so $g(n)$ must be the point at infinity). Choose $j \ge 1$ large enough such that
$f(i) \in K_{i,j}=C_{i,1} \oplus C_{i,2} \oplus \cdots \oplus C_{i,j}$
for all $i \le n$. It follows that $f \in F_{n,j}$ and $g \notin F_{n,j}$. Thus the sequence of closed sets $F_{n,j}$ satisfies Lemma 3. By Lemma 3, $X=\prod_{i=1}^\infty C_i$ is Lindelof.
Reference
1. Engelking R., General Topology, Revised and Completed edition, Elsevier Science Publishers B. V., Heldermann Verlag, Berlin, 1989.
2. Hager A. W., Approximation of real continuous functions on Lindelof spaces, Proc. Amer. Math. Soc., 22, 156-163, 1969.
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# Helly Space
This is a discussion on a compact space called Helly space. The discussion here builds on the facts presented in Counterexample in Topology [2]. Helly space is Example 107 in [2]. The space is named after Eduard Helly.
Let $I=[0,1]$ be the closed unit interval with the usual topology. Let $C$ be the set of all functions $f:I \rightarrow I$. The set $C$ is endowed with the product space topology. The usual product space notation is $I^I$ or $\prod_{t \in I} W_t$ where each $W_t=I$. As a product of compact spaces, $C=I^I$ is compact.
Any function $f:I \rightarrow I$ is said to be increasing if $f(x) \le f(y)$ for all $x (such a function is usually referred to as non-decreasing). Helly space is the subspace $X$ consisting of all increasing functions. This space is Example 107 in Counterexample in Topology [2]. The following facts are discussed in [2].
• The space $X$ is compact.
• The space $X$ is first countable (having a countable base at each point).
• The space $X$ is separable.
• The space $X$ has an uncountable discrete subspace.
From the last two facts, Helly space is a compact non-metrizable space. Any separable metric space would have countable spread (all discrete subspaces must be countable).
The compactness of $X$ stems from the fact that $X$ is a closed subspace of the compact space $C$.
Further Discussion
Additional facts of concerning Helly space are discussed.
1. The product space $\omega_1 \times X$ is normal.
2. Helly space $X$ contains a copy of the Sorgenfrey line.
3. Helly space $X$ is not hereditarily normal.
The space $\omega_1$ is the space of all countable ordinals with the order topology. Recall $C$ is the product space $I^I$. The product space $\omega_1 \times C$ is Example 106 in [2]. This product is not normal. The non-normality of $\omega_1 \times C$ is based on this theorem: for any compact space $Y$, the product $\omega_1 \times Y$ is normal if and only if the compact space $Y$ is countably tight. The compact product space $C$ is not countably tight (discussed here). Thus $\omega_1 \times C$ is not normal. However, the product $\omega_1 \times X$ is normal since Helly space $X$ is first countable.
To see that $X$ contains a copy of the Sorgenfrey line, consider the functions $h_t:I \rightarrow I$ defined as follows:
$\displaystyle h_t(x) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ 0 \le x \le t \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ t
for all $0. Let $S=\{ h_t: 0. Consider the mapping $\gamma: (0,1) \rightarrow S$ defined by $\gamma(t)=h_t$. With the domain $(0,1)$ having the Sorgenfrey topology and with the range $S$ being a subspace of Helly space, it can be shown that $\gamma$ is a homeomorphism.
With the Sorgenfrey line $S$ embedded in $X$, the square $X \times X$ contains a copy of the Sorgenfrey plane $S \times S$, which is non-normal (discussed here). Thus the square of Helly space is not hereditarily normal. A more interesting fact is that Helly space is not hereditarily normal. This is discussed in the next section.
Finding a Non-Normal Subspace of Helly Space
As before, $C$ is the product space $I^I$ where $I=[0,1]$ and $X$ is Helly space consisting of all increasing functions in $C$. Consider the following two subspaces of $X$.
$Y_{0,1}=\{ f \in X: f(I) \subset \{0, 1 \} \}$
$Y=X - Y_{0,1}$
The subspace $Y_{0,1}$ is a closed subset of $X$, hence compact. We claim that subspace $Y$ is separable and has a closed and discrete subset of cardinality continuum. This means that the subspace $Y$ is not a normal space.
First, we define a discrete subspace. For each $x$ with $0, define $f_x: I \rightarrow I$ as follows:
$\displaystyle f_x(y) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ 0 \le y < x \\ \text{ } & \text{ } \\ \displaystyle \frac{1}{2} &\ \ \ \ \ y=x \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ x
Let $H=\{ f_x: 0. The set $H$ as a subspace of $X$ is discrete. Of course it is not discrete in $X$ since $X$ is compact. In fact, for any $f \in Y_{0,1}$, $f \in \overline{H}$ (closure taken in $X$). However, it can be shown that $H$ is closed and discrete as a subset of $Y$.
We now construct a countable dense subset of $Y$. To this end, let $\mathcal{B}$ be a countable base for the usual topology on the unit interval $I=[0,1]$. For example, we can let $\mathcal{B}$ be the set of all open intervals with rational endpoints. Furthermore, let $A$ be a countable dense subset of the open interval $(0,1)$ (in the usual topology). For convenience, we enumerate the elements of $A$ and $\mathcal{B}$.
$A=\{ a_1,a_2,a_3,\cdots \}$
$\mathcal{B}=\{B_1,B_2,B_3,\cdots \}$
We also need the following collections.
$\mathcal{G}=\{G \subset \mathcal{B}: G \text{ is finite and is pairwise disjoint} \}$
$\mathcal{A}=\{F \subset A: F \text{ is finite} \}$
For each $G \in \mathcal{G}$ and for each $F \in \mathcal{A}$ with $\lvert G \lvert=\lvert F \lvert=n$, we would like to arrange the elements in increasing order, notated as follow:
$F=\{t_1,t_2,\cdots,t_n \}$
$G=\{E_1,E_2,\cdots,E_n \}$
For the set $F$, we have $0. For the set $G$, $E_i$ is to the left of $E_j$ for $i. Note that elements of $G$ are pairwise disjoint. Furthermore, write $E_i=(p_i,q_i)$. If $0 \in E_1$, then $E_1=[p_1,q_1)=[0,q_1)$. If $1 \in E_n$, then $E_n=(p_n,q_n]=(p_n,1]$.
For each $F$ and $G$ as detailed above, we define a function $L(F,G):I \rightarrow I$ as follows:
$\displaystyle L(F,G)(x) = \left\{ \begin{array}{ll} \displaystyle t_1 &\ \ \ \ \ 0 \le x < q_1 \\ \text{ } & \text{ } \\ \displaystyle t_2 &\ \ \ \ \ q_1 \le x < q_2 \\ \text{ } & \text{ } \\ \displaystyle \vdots &\ \ \ \ \ \vdots \\ \text{ } & \text{ } \\ \displaystyle t_{n-1} &\ \ \ \ \ q_{n-2} \le x < q_{n-1} \\ \text{ } & \text{ } \\ \displaystyle t_n &\ \ \ \ \ q_{n-1} \le x \le 1 \\ \end{array} \right.$
The following diagram illustrates the definition of $L(F,G)$ when both $F$ and $G$ have 4 elements.
Figure 1 – Member of a countable dense set
Let $D$ be the set of $L(F,G)$ over all $F \in \mathcal{A}$ and $G \in \mathcal{G}$. The set $D$ is a countable set. It can be shown that $D$ is dense in the subspace $Y$. In fact $D$ is dense in the entire Helly space $X$.
To summarize, the subspace $Y$ is separable and has a closed and discrete subset of cardinality continuum. This means that $Y$ is not normal. Hence Helly space $X$ is not hereditarily normal. According to Jones’ lemma, in any normal separable space, the cardinality of any closed and discrete subspace must be less than continuum (discussed here).
Remarks
The preceding discussion shows that both Helly space and the square of Helly space are not hereditarily normal. This is actually not surprising. According to a theorem of Katetov, for any compact non-metrizable space $V$, the cube $V^3$ is not hereditarily normal (see Theorem 3 in this post). Thus a non-normal subspace is found in $V$, $V \times V$ or $V \times V \times V$. In fact, for any compact non-metric space $V$, an excellent exercise is to find where a non-normal subspace can be found. Is it in $V$, the square of $V$ or the cube of $V$? In the case of Helly space $X$, a non-normal subspace can be found in $X$.
A natural question is: is there a compact non-metric space $V$ such that both $V$ and $V \times V$ are hereditarily normal and $V \times V \times V$ is not hereditarily normal? In other words, is there an example where the hereditarily normality fails at dimension 3? If we do not assume extra set-theoretic axioms beyond ZFC, any compact non-metric space $V$ is likely to fail hereditarily normality in either $V$ or $V \times V$. See here for a discussion of this set-theoretic question.
Reference
1. Kelly, J. L., General Topology, Springer-Verlag, New York, 1955.
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
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# A little corner in the world of set-theoretic topology
This post puts a spot light on a little corner in the world of set-theoretic topology. There lies in this corner a simple topological statement that opens a door to the esoteric world of independence results. In this post, we give a proof of this basic fact and discuss its ramifications. This basic result is an excellent entry point to the study of S and L spaces.
The following paragraph is found in the paper called Gently killing S-spaces by Todd Eisworth, Peter Nyikos and Saharon Shelah [1]. The basic fact in question is highlighted in blue.
A simultaneous generalization of hereditarily separable and hereditarily Lindelof spaces is the class of spaces of countable spread – those spaces in which every discrete subspace is countable. One of the basic facts in this little corner of set-theoretic topology is that if a regular space of countable spread is not hereditarily separable, it contains an L-space, and if it is not hereditarily Lindelof, it contains an S-space. [1]
The same basic fact is also mentioned in the paper called The spread of regular spaces by Judith Roitman [2].
It is also well known that a regular space of countable spread which is not hereditarily separable contains an L-space and a regular space of countable spread which is not hereditarily Lindelof contains an S-space. Thus an absolute example of a space satisfying (Statement) A would contain a proof of the existence of S and L space – a consummation which some may devoutly wish, but which this paper does not attempt. [2]
Statement A in [2] is: There exists a 0-dimensional Hausdorff space of countable spread that is not the union of a hereditarily separable and a hereditarily Lindelof space. Statement A would mean the existence of a regular space of countable spread that is not hereditarily separable and that is also not hereditarily Lindelof. By the well known fact just mentioned, statement A would imply the existence of a space that is simultaneously an S-space and an L-space!
Let’s unpack the preceding section. First some basic definitions. A space $X$ is of countable spread (has countable spread) if every discrete subspace of $X$ is countable. A space $X$ is hereditarily separable if every subspace of $X$ is separable. A space $X$ is hereditarily Lindelof if every subspace of $X$ is Lindelof. A space is an S-space if it is hereditarily separable but not Lindelof. A space is an L-space if it is hereditarily Lindelof but not separable. See [3] for a basic discussion of S and L spaces.
Hereditarily separable but not Lindelof spaces as well as hereditarily Lindelof but not separable spaces can be easily defined in ZFC [3]. However, such examples are not regular. For the notions of S and L-spaces to be interesting, the definitions must include regularity. Thus in the discussion that follows, all spaces are assumed to be Hausdorff and regular.
One amazing aspect about set-theoretic topology is that one sometimes does not have to stray far from basic topological notions to encounter pathological objects such as S-spaces and L-spaces. The definition of a topological space is of course a basic definition. Separable spaces and Lindelof spaces are basic notions that are not far from the definition of topological spaces. The same can be said about hereditarily separable and hereditarily Lindelof spaces. Out of these basic ingredients come the notion of S-spaces and L-spaces, the existence of which is one of the key motivating questions in set-theoretic topology in the twentieth century. The study of S and L-spaces is a body of mathematics that had been developed for nearly a century. It is a fruitful area of research at the boundary of topology and axiomatic set theory.
The existence of an S-space is independent of ZFC (as a result of the work by Todorcevic in early 1980s). This means that there is a model of set theory in which an S-space exists and there is also a model of set theory in which S-spaces cannot exist. One half of the basic result mentioned in the preceding section is intimately tied to the existence of S-spaces and thus has interesting set-theoretic implications. The other half of the basic result involves the existence of L-spaces, which are shown to exist without using extra set theory axioms beyond ZFC by Justin Moore in 2005, which went against the common expectation that the existence of L-spaces would be independent of ZFC as well.
Let’s examine the basic notions in a little more details. The following diagram shows the properties surrounding the notion of countable spread.
Diagram 1 – Properties surrounding countable spread
The implications (the arrows) in Diagram 1 can be verified easily. Central to the discussion at hand, both hereditarily separable and hereditarily Lindelof imply countable spread. The best way to see this is that if a space has an uncountable discrete subspace, that subspace is simultaneously a non-separable subspace and a non-Lindelof subspace. A natural question is whether these implications can be reversed. Another question is whether the properties in Diagram 1 can be related in other ways. The following diagram attempts to ask these questions.
Diagram 2 – Reverse implications surrounding countable spread
Not shown in Diagram 2 are these four facts: separable $\not \rightarrow$ hereditarily separable, Lindelof $\not \rightarrow$ hereditarily Lindelof, separable $\not \rightarrow$ countable spread and Lindelof $\not \rightarrow$ countable spread. The examples supporting these facts are not set-theoretic in nature and are not discussed here.
Let’s focus on each question mark in Diagram 2. The two horizontal arrows with question marks at the top are about S-space and L-space. If $X$ is hereditarily separable, then is $X$ hereditarily Lindelof? A “no” answer would mean there is an S-space. A “yes” answer would mean there exists no S-space. So the top arrow from left to right is independent of ZFC. Since an L-space can be constructed within ZFC, the question mark in the top arrow in Diagram 2 from right to left has a “no” answer.
Now focus on the arrows emanating from countable spread in Diagram 2. These arrows are about the basic fact discussed earlier. From Diagram 1, we know that hereditarily separable implies countable spread. Can the implication be reversed? Any L-space would be an example showing that the implication cannot be reversed. Note that any L-space is of countable spread and is not separable and hence not hereditarily separable. Since L-space exists in ZFC, the question mark in the arrow from countable spread to hereditarily separable has a “no” answer. The same is true for the question mark in the arrow from countable spread to separable
We know that hereditarily Lindelof implies countable spread. Can the implication be reversed? According to the basic fact mentioned earlier, if the implication cannot be reversed, there exists an S-space. Thus if S-space does not exist, the implication can be reversed. Any S-space is an example showing that the implication cannot be reversed. Thus the question in the arrow from countable spread to hereditarily Lindelof cannot be answered without assuming axioms beyond ZFC. The same is true for the question mark for the arrow from countable spread to Lindelf.
Diagram 2 is set-theoretic in nature. The diagram is remarkable in that the properties in the diagram are basic notions that are only brief steps away from the definition of a topological space. Thus the basic highlighted here is a quick route to the world of independence results.
We now give a proof of the basic result, which is stated in the following theorem.
Theorem 1
Let $X$ is regular and Hausdorff space. Then the following is true.
• If $X$ is of countable spread and is not a hereditarily separable space, then $X$ contains an L-space.
• If $X$ is of countable spread and is not a hereditarily Lindelof space, then $X$ contains an S-space.
To that end, we use the concepts of right separated space and left separated space. Recall that an initial segment of a well-ordered set $(X,<)$ is a set of the form $\{y \in X: y where $x \in X$. A space $X$ is a right separated space if $X$ can be well-ordered in such a way that every initial segment is open. A right separated space is in type $\kappa$ if the well-ordering is of type $\kappa$. A space $X$ is a left separated space if $X$ can be well-ordered in such a way that every initial segment is closed. A left separated space is in type $\kappa$ if the well-ordering is of type $\kappa$. The following results are used in proving Theorem 1.
Theorem A
Let $X$ is regular and Hausdorff space. Then the following is true.
• The space $X$ is hereditarily separable space if and only if $X$ has no uncountable left separated subspace.
• The space $X$ is hereditarily Lindelof space if and only if $X$ has no uncountable right separated subspace.
Proof of Theorem A
$\Longrightarrow$ of the first bullet point.
Suppose $Y \subset X$ is an uncountable left separated subspace. Suppose that the well-ordering of $Y$ is of type $\kappa$ where $\kappa>\omega$. Further suppose that $Y=\{ x_\alpha: \alpha<\kappa \}$ such that for each $\alpha<\kappa$, $C_\alpha=\{ x_\beta: \beta<\alpha \}$ is a closed subset of $Y$. Since $\kappa$ is uncountable, the well-ordering has an initial segment of type $\omega_1$. So we might as well assume $\kappa=\omega_1$. Note that for any countable $A \subset Y$, $A \subset C_\alpha$ for some $\alpha<\omega_1$. It follows that $Y$ is not separable. This means that $X$ is not hereditarily separable.
$\Longleftarrow$ of the first bullet point.
Suppose that $X$ is not hereditarily separable. Let $Y \subset X$ be a subspace that is not separable. We now inductively derive an uncountable left separated subspace of $Y$. Choose $y_0 \in Y$. For each $\alpha<\omega_1$, let $A_\alpha=\{ y_\beta \in Y: \beta <\alpha \}$. The set $A_\alpha$ is the set of all the points of $Y$ chosen before the step at $\alpha<\omega_1$. Since $A_\alpha$ is countable, its closure in $Y$ is not the entire space $Y$. Choose $y_\alpha \in Y-\overline{A_\alpha}=O_\alpha$.
Let $Y_L=\{ y_\alpha: \alpha<\omega_1 \}$. We claim that $Y_L$ is a left separated space. To this end, we need to show that each initial segment $A_\alpha$ is a closed subset of $Y_L$. Note that for each $\gamma \ge \alpha$, $O_\gamma=Y-\overline{A_\gamma}$ is an open subset of $Y$ with $y_\gamma \in O_\gamma$ such that $O_\gamma \cap \overline{A_\gamma}=\varnothing$ and thus $O_\gamma \cap \overline{A_\alpha}=\varnothing$ (closure in $Y$). Then $U_\gamma=O_\gamma \cap Y_L$ is an open subset of $Y_L$ containing $y_\gamma$ such that $U_\gamma \cap A_\alpha=\varnothing$. It follows that $Y-A_\alpha$ is open in $Y_L$ and that $A_\alpha$ is a closed subset of $Y_L$.
$\Longrightarrow$ of the second bullet point.
Suppose $Y \subset X$ is an uncountable right separated subspace. Suppose that the well-ordering of $Y$ is of type $\kappa$ where $\kappa>\omega$. Further suppose that $Y=\{ x_\alpha: \alpha<\kappa \}$ such that for each $\alpha<\kappa$, $U_\alpha=\{ x_\beta: \beta<\alpha \}$ is an open subset of $Y$.
Since $\kappa$ is uncountable, the well-ordering has an initial segment of type $\omega_1$. So we might as well assume $\kappa=\omega_1$. Note that $\{ U_\alpha: \alpha<\omega_1 \}$ is an open cover of $Y$ that has no countable subcover. It follows that $Y$ is not Lindelof. This means that $X$ is not hereditarily Lindelof.
$\Longleftarrow$ of the second bullet point.
Suppose that $X$ is not hereditarily Lindelof. Let $Y \subset X$ be a subspace that is not Lindelof. Let $\mathcal{U}$ be an open cover of $Y$ that has no countable subcover. We now inductively derive a right separated subspace of $Y$ of type $\omega_1$.
Choose $U_0 \in \mathcal{U}$ and choose $y_0 \in U_0$. Choose $y_1 \in Y-U_0$ and choose $U_1 \in \mathcal{U}$ such that $y_1 \in U_1$. Let $\alpha<\omega_1$. Suppose that points $y_\beta$ and open sets $U_\beta$, $\beta<\alpha$, have been chosen such that $y_\beta \in Y-\bigcup_{\delta<\beta} U_\delta$ and $y_\beta \in U_\beta$. The countably many chosen open sets $U_\beta$, $\beta<\alpha$, cannot cover $Y$. Choose $y_\alpha \in Y-\bigcup_{\beta<\alpha} U_\beta$. Choose $U_\alpha \in \mathcal{U}$ such that $y_\alpha \in U_\alpha$.
Let $Y_R=\{ y_\alpha: \alpha<\omega_1 \}$. It follows that $Y_R$ is a right separated space. Note that for each $\alpha<\omega_1$, $\{ y_\beta: \beta<\alpha \} \subset \bigcup_{\beta<\alpha} U_\beta$ and the open set $\bigcup_{\beta<\alpha} U_\beta$ does not contain $y_\gamma$ for any $\gamma \ge \alpha$. This means that the initial segment $\{ y_\beta: \beta<\alpha \}$ is open in $Y_L$. $\square$
Lemma B
Let $X$ be a space that is a right separated space and also a left separated space based on the same well ordering. Then $X$ is a discrete space.
Proof of Lemma B
Let $X=\{ w_\alpha: \alpha<\kappa \}$ such that the well-ordering is given by the ordinals in the subscripts, i.e. $w_\beta if and only if $\beta<\gamma$. Suppose that $X$ with this well-ordering is both a right separated space and a left separated space. We claim that every point is a discrete point, i.e. $\{ x_\alpha \}$ is open for any $\alpha<\kappa$.
To see this, fix $\alpha<\kappa$. The initial segment $A_\alpha=\{ w_\beta: \beta<\alpha \}$ is closed in $X$ since $X$ is a left separated space. On the other hand, the initial segment $\{ w_\beta: \beta < \alpha+1 \}$ is open in $X$ since $X$ is a right separated space. Then $B_{\alpha}=\{ w_\beta: \beta \ge \alpha+1 \}$ is closed in $X$. It follows that $\{ x_\alpha \}$ must be open since $X=A_\alpha \cup B_\alpha \cup \{ w_\alpha \}$. $\square$
Theorem C
Let $X$ is regular and Hausdorff space. Then the following is true.
• Suppose the space $X$ is right separated space of type $\omega_1$. Then if $X$ has no uncountable discrete subspaces, then $X$ is an S-space or $X$ contains an S-space.
• Suppose the space $X$ is left separated space of type $\omega_1$. Then if $X$ has no uncountable discrete subspaces, then $X$ is an L-space or $X$ contains an L-space.
Proof of Theorem C
For the first bullet point, suppose the space $X$ is right separated space of type $\omega_1$. Then by Theorem A, $X$ is not hereditarily Lindelof. If $X$ is hereditarily separable, then $X$ is an S-space (if $X$ is not Lindelof) or $X$ contains an S-space (a non-Lindelof subspace of $X$). Suppose $X$ is not hereditarily separable. By Theorem A, $X$ has an uncountable left separated subspace of type $\omega_1$.
Let $X=\{ x_\alpha: \alpha<\omega_1 \}$ such that the well-ordering represented by the ordinals in the subscripts is a right separated space. Let $<_R$ be the symbol for the right separated well-ordering, i.e. $x_\beta <_R \ x_\delta$ if and only if $\beta<\delta$. As indicated in the preceding paragraph, $X$ has an uncountable left separated subspace. Let $Y=\{ y_\alpha \in X: \alpha<\omega_1 \}$ be this left separated subspace. Let $<_L$ be the symbol for the left separated well-ordering. The well-ordering $<_R$ may be different from the well-ordering $<_L$. However, we can obtain an uncountable subset of $Y$ such that the two well-orderings coincide on this subset.
To start, pick any $y_\gamma$ in $Y$ and relabel it $t_0$. The final segment $\{y_\beta \in Y: t_0 <_L \ y_\beta \}$ must intersect the final segment $\{x_\beta \in X: t_0 <_R \ x_\beta \}$ in uncountably many points. Choose the least such point (according to $<_R$) and call it $t_1$. It is clear how $t_{\delta+1}$ is chosen if $t_\delta$ has been chosen.
Suppose $\alpha<\omega_1$ is a limit ordinal and that $t_\beta$ has been chosen for all $\beta<\alpha$. Then the set $\{y_\tau: \forall \ \beta<\alpha, t_\beta <_L \ y_\tau \}$ and the set $\{x_\tau: \forall \ \beta<\alpha, t_\beta <_R \ x_\tau \}$ must intersect in uncountably many points. Choose the least such point and call it $t_\alpha$ (according to $<_R$). As a result, we have obtained $T=\{ t_\alpha: \alpha<\omega_1 \}$. It follows that T with the well-ordering represented by the ordinals in the subscript is a subset of $(X,<_R)$ and a subset of $(Y,<_L)$. Thus $T$ is both right separated and left separated.
By Lemma B, $T$ is a discrete subspace of $X$. However, $X$ is assumed to have no uncountable discrete subspace. Thus if $X$ has no uncountable discrete subspace, then $X$ must be hereditarily separable and as a result, must be an S-space or must contain an S-space.
The proof for the second bullet point is analogous to that of the first bullet point. $\square$
We are now ready to prove Theorem 1.
Proof of Theorem 1
Suppose that $X$ is of countable spread and that $X$ is not hereditarily separable. By Theorem A, $X$ has an uncountable left separated subspace $Y$ (assume it is of type $\omega_1$). The property of countable spread is hereditary. So $Y$ is of countable spread. By Theorem C, $Y$ is an L-space or $Y$ contains an L-space. In either way, $X$ contains an L-space.
Suppose that $X$ is of countable spread and that $X$ is not hereditarily Lindelof. By Theorem A, $X$ has an uncountable right separated subspace $Y$ (assume it is of type $\omega_1$). By Theorem C, $Y$ is an S-space or $Y$ contains an S-space. In either way, $X$ contains an S-space.
Reference
1. Eisworth T., Nyikos P., Shelah S., Gently killing S-spaces, Israel Journal of Mathmatics, 136, 189-220, 2003.
2. Roitman J., The spread of regular spaces, General Topology and Its Applications, 8, 85-91, 1978.
3. Roitman, J., Basic S and L, Handbook of Set-Theoretic Topology, (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 295-326, 1984.
4. Tatch-Moore J., A solution to the L space problem, Journal of the American Mathematical Society, 19, 717-736, 2006.
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# Every space is star discrete
The statement in the title is a folklore fact, though the term star discrete is usually not used whenever this well known fact is invoked in the literature. We present a proof to this well known fact. We also discuss some related concepts.
All spaces are assumed to be Hausdorff and regular.
First, let’s define the star notation. Let $X$ be a space. Let $\mathcal{U}$ be a collection of subsets of $X$. Let $A \subset X$. Define $\text{St}(A,\mathcal{U})$ to be the set $\bigcup \{U \in \mathcal{U}: U \cap A \ne \varnothing \}$. In other words, the set $\text{St}(A,\mathcal{U})$ is simply the union of all elements of $\mathcal{U}$ that contains points of the set $A$. The set $\text{St}(A,\mathcal{U})$ is also called the star of the set $A$ with respect to the collection $\mathcal{U}$. If $A=\{ x \}$, we use the notation $\text{St}(x,\mathcal{U})$ instead of $\text{St}( \{ x \},\mathcal{U})$. The following is the well known result in question.
Lemma 1
Let $X$ be a space. For any open cover $\mathcal{U}$ of $X$, there exists a discrete subspace $A$ of $X$ such that $X=\text{St}(A,\mathcal{U})$. Furthermore, the set $A$ can be chosen in such a way that it is also a closed subset of the space $X$.
Any space that satisfies the condition in Lemma 1 is said to be a star discrete space. The proof shown below will work for any topological space. Hence every space is star discrete. We come across three references in which the lemma is stated or is used – Lemma IV.2.20 in page 135 of [3], page 137 of [2] and [1]. The first two references do not use the term star discrete. Star discrete is mentioned in [1] since that paper focuses on star properties. This property that is present in every topological space is at heart a covering property. Here’s a rewording of the lemma that makes it look like a covering property.
Lemma 1a
Let $X$ be a space. For any open cover $\mathcal{U}$ of $X$, there exists a discrete subspace $A$ of $X$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$. Furthermore, the set $A$ can be chosen in such a way that it is also a closed subset of the space $X$.
Lemma 1a is clearly identical to Lemma 1. However, Lemma 1a makes it extra clear that this is a covering property. For every open cover of a space, instead of finding a sub cover or an open refinement, we find a discrete subspace so that the stars of the points of the discrete subspace with respect to the given open cover also cover the space.
Lemma 1a naturally leads to other star covering properties. For example, a space $X$ is said to be a star countable space if for any open cover $\mathcal{U}$ of $X$, there exists a countable subspace $A$ of $X$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$. A space $X$ is said to be a star Lindelof space if for any open cover $\mathcal{U}$ of $X$, there exists a Lindelof subspace $A$ of $X$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$. In general, for any topological property $\mathcal{P}$, a space $X$ is a star $\mathcal{P}$ space if for any open cover $\mathcal{U}$ of $X$, there exists a subspace $A$ of $X$ with property $\mathcal{P}$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$.
It follows that every Lindelof space is a star countable space. It is also clear that every star countable space is a star Lindelof space.
Lemma 1 or Lemma 1a, at first glance, may seem like a surprising result. However, one can argue that it is not a strong result at all since the property is possessed by every space. Indeed, the lemma has nothing to say about the size of the discrete set. It only says that there exists a star cover based on a discrete set for a given open cover. To derive more information about the given space, we may need to work with more information on the space in question.
Consider spaces such that every discrete subspace is countable (such a space is said to have countable spread or a space of countable spread). Also consider spaces such that every closed and discrete subspace is countable (such a space is said to have countable extent or a space of countable extent). Any space that has countable spread is also a space that has countable extent for the simple reason that if every discrete subspace is countable, then every closed and discrete subspace is countable.
Then it follows from Lemma 1 that any space $X$ that has countable extent is star countable. Any star countable space is obviously a star Lindelof space. The following diagram displays these relationships.
According to the diagram, the star countable and star Lindelof are both downstream from the countable spread property and the Lindelof property. The star properties being downstream from the Lindelof property is not surprising. What is interesting is that if a space has countable spread, then it is star countable and hence star Lindelof.
Do “countable spread” and “Lindelof” relate to each other? Lindelof spaces do not have to have countable spread. The simplest example is the one-point compactification of an uncountable discrete space. More specifically, let $X$ be an uncountable discrete space. Let $p$ be a point not in $X$. Then $Y=X \cup \{ p \}$ is a compact space (hence Lindelof) where $X$ is discrete and an open neighborhood of $p$ is of the form $\{ p \} \cup U$ where $X-U$ is a finite subset of $X$. The space $Y$ is not of countable spread since $X$ is an uncountable discrete subspace.
Does “countable spread” imply “Lindelof”? Is there a non-Lindelof space that has countable spread? It turns out that the answers are independent of ZFC. The next post has more details.
We now give a proof to Lemma 1. Suppose that $X$ is an infinite space (if it is finite, the lemma is true since the space is Hausdorff). Let $\kappa=\lvert X \lvert$. Let $\kappa^+$ be the next cardinal greater than $\kappa$. Let $\mathcal{U}$ be an open cover of the space $X$. Choose $x_0 \in X$. We choose a sequence of points $x_0,x_1,\cdots,x_\alpha,\cdots$ inductively. If $\text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U}) \ne X$, we can choose a point $x_\alpha \in X$ such that $x_\alpha \notin \text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U})$.
We claim that the induction process must stop at some $\alpha<\kappa^+$. In other words, at some $\alpha<\kappa^+$, the star of the previous points must be the entire space and we run out of points to choose. Otherwise, we would have obtained a subset of $X$ with cardinality $\kappa^+$, a contradiction. Choose the least $\alpha<\kappa^+$ such that $\text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U}) = X$. Let $A=\{x_\beta: \beta<\alpha \}$.
Then it can be verified that the set $A$ is a discrete subspace of $X$ and that $A$ is a closed subset of $X$. Note that $x_\beta \in \text{St}(x_\beta, \mathcal{U})$ while $x_\gamma \notin \text{St}(x_\beta, \mathcal{U})$ for all $\gamma \ne \beta$. This follows from the way the points are chosen in the induction process. On the other hand, for any $x \in X-A$, $x \in \text{St}(x_\beta, \mathcal{U})$ for some $\beta<\alpha$. As discussed, the open set $\text{St}(x_\beta, \mathcal{U})$ contains only one point of $A$, namely $x_\beta$.
Reference
1. Alas O., Jumqueira L., van Mill J., Tkachuk V., Wilson R.On the extent of star countable spaces, Cent. Eur. J. Math., 9 (3), 603-615, 2011.
2. Alster, K., Pol, R.,On function spaces of compact subspaces of $\Sigma$-products of the real line, Fund. Math., 107, 35-46, 1980.
3. Arkhangelskii, A. V.,Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
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# Michael line and Morita’s conjectures
This post discusses Michael line from the point of view of the three conjectures of Kiiti Morita.
K. Morita defined the notion of P-spaces in [7]. The definition of P-spaces is discussed here in considerable details. K. Morita also proved that a space $X$ is a normal P-space if and only if the product $X \times Y$ is normal for every metrizable space $Y$. As a result of this characterization, the notion of normal P-space (a space that is a normal space and a P-space) is useful in the study of products of normal spaces. Just to be clear, we say a space is a non-normal P-space (i.e. a space that is not a normal P-space) if the space is a normal space that is not a P-space.
K. Morita formulated his three conjectures in 1976. The statements of the conjectures are given below. Here is a basic discussion of the three conjectures. The notion of normal P-spaces is a theme that runs through the three conjectures. The conjectures are actually theorems since 2001 [2].
Here’s where Michael line comes into the discussion. Based on the characterization of normal P-spaces mentioned above, to find a normal space that is not a P-space (a non-normal P-space), we would need to find a non-normal product $X \times Y$ such that one of the factors is a metric space and the other factor is a normal space. The first such example in ZFC is from an article by E. Michael in 1963 (found here and here). In this example, the normal space is $M$, which came be known as the Michael line, and the metric space is $\mathbb{P}$, the space of irrational numbers (as a subspace of the real line). Their product $M \times \mathbb{P}$ is not normal. A basic discussion of the Michael line is found here.
Because $M \times \mathbb{P}$ is not normal, the Michael line $M$ is not a normal P-space. Prior to E. Michael’s 1963 article, we have to reach back to 1955 to find an example of a non-normal product where one factor is a metric space. In 1955, M. E. Rudin used a Souslin line to construct a Dowker space, which is a normal space whose product with the closed unit interval is not normal. The existence of a Souslin line was shown to be independent of ZFC in the late 1960s. In 1971, Rudin constructed a Dowker space in ZFC. Thus finding a normal space that is not a normal P-space (finding a non-normal product $X \times Y$ where one factor is a metric space and the other factor is a normal space) is not a trivial matter.
Morita’s Three Conjectures
We show that the Michael line illustrates perfectly the three conjectures of K. Morita. Here’s the statements.
Morita’s Conjecture I. Let $X$ be a space. If the product $X \times Y$ is normal for every normal space $Y$ then $X$ is a discrete space.
Morita’s Conjecture II. Let $X$ be a space. If the product $X \times Y$ is normal for every normal P-space $Y$ then $X$ is a metrizable space.
Morita’s Conjecture III. Let $X$ be a space. If the product $X \times Y$ is normal for every normal countably paracompact space $Y$ then $X$ is a metrizable $\sigma$-locally compact space.
The contrapositive statement of Morita’s conjecture I is that for any non-discrete space $X$, there exists a normal space $Y$ such that $X \times Y$ is not normal. Thus any non-discrete space is paired with a normal space for forming a non-normal product. The Michael line $M$ is paired with the space of irrational numbers $\mathbb{P}$. Obviously, the space $\mathbb{P}$ is paired with the Michael line $M$.
The contrapositive statement of Morita’s conjecture II is that for any non-metrizable space $X$, there exists a normal P-space $Y$ such that $X \times Y$ is not normal. The pairing is more specific than for conjecture I. Any non-metrizable space is paired with a normal P-space to form a non-normal product. As illustration, the Michael line $M$ is not metrizable. The space $\mathbb{P}$ of irrational numbers is a metric space and hence a normal P-space. Here, $M$ is paired with $\mathbb{P}$ to form a non-normal product.
The contrapositive statement of Morita’s conjecture III is that for any space $X$ that is not both metrizable and $\sigma$-locally compact, there exists a normal countably paracompact space $Y$ such that $X \times Y$ is not normal. Note that the space $\mathbb{P}$ is not $\sigma$-locally compact (see Theorem 4 here). The Michael line $M$ is paracompact and hence normal and countably paracompact. Thus the metric non-$\sigma$-locally compact $\mathbb{P}$ is paired with normal countably paracompact $M$ to form a non-normal product. Here, the metric space $\mathbb{P}$ is paired with the non-normal P-space $M$.
In each conjecture, each space in a certain class of spaces is paired with one space in another class to form a non-normal product. For Morita’s conjecture I, each non-discrete space is paired with a normal space. For conjecture II, each non-metrizable space is paired with a normal P-space. For conjecture III, each metrizable but non-$\sigma$-locally compact is paired with a normal countably paracompact space to form a non-normal product. Note that the paired normal countably paracompact space would be a non-normal P-space.
Michael line as an example of a non-normal P-space is a great tool to help us walk through the three conjectures of Morita. Are there other examples of non-normal P-spaces? Dowker spaces mentioned above (normal spaces whose products with the closed unit interval are not normal) are non-normal P-spaces. Note that conjecture II guarantees a normal P-space to match every non-metric space for forming a non-normal product. Conjecture III guarantees a non-normal P-space to match every metrizable non-$\sigma$-locally compact space for forming a non-normal product. Based on the conjectures, examples of normal P-spaces and non-normal P-spaces, though may be hard to find, are guaranteed to exist.
We give more examples below to further illustrate the pairings for conjecture II and conjecture III. As indicated above, non-normal P-spaces are hard to come by. Some of the examples below are constructed using additional axioms beyond ZFC. The additional examples still give an impression that the availability of non-normal P-spaces, though guaranteed to exist, is limited.
Examples of Normal P-Spaces
One example is based on this classic theorem: for any normal space $X$, $X$ is paracompact if and only if the product $X \times \beta X$ is normal. Here $\beta X$ is the Stone-Cech compactification of the completely regular space $X$. Thus any normal but not paracompact space $X$ (a non-metrizable space) is paired with $\beta X$, a normal P-space, to form a non-normal product.
Naturally, the next class of non-metrizable spaces to be discussed should be the paracompact spaces that are not metrizable. If there is a readily available theorem to provide a normal P-space for each non-metrizable paracompact space, then there would be a simple proof of Morita’s conjecture II. The eventual solution of conjecture II is far from simple [2]. We narrow the focus to the non-metrizable compact spaces.
Consider this well known result: for any infinite compact space $X$, the product $\omega_1 \times X$ is normal if and only if the space $X$ has countable tightness (see Theorem 1 here). Thus any compact space with uncountable tightness is paired with $\omega_1$, the space of all countable ordinals, to form a non-normal product. The space $\omega_1$, being a countably compact space, is a normal P-space. A proof that normal countably compact space is a normal P-space is given here.
We now handle the case for non-metrizable compact spaces with countable tightness. In this case, compactness is not needed. For spaces with countable tightness, consider this result: every space with countable tightness, whose products with all perfectly normal spaces are normal, must be metrizable [3] (see Corollary 7). Thus any non-metrizable space with countable tightness is paired with some perfectly normal space to form a non-normal product. Any reader interested in what these perfectly normal spaces are can consult [3]. Note that perfectly normal spaces are normal P-spaces (see here for a proof).
Examples of Non-Normal P-Spaces
Another non-normal product is $X_B \times B$ where $B \subset \mathbb{R}$ is a Bernstein set and $X_B$ is the space with the real line as the underlying set such that points in $B$ are isolated and points in $\mathbb{R}-B$ retain the usual open sets. The set $B \subset \mathbb{R}$ is said to be a Bernstein set if every uncountable closed subset of the real line contains a point in B and contains a point in the complement of B. Such a set can be constructed using transfinite induction as shown here. The product $X_B \times B$ is not normal where $B$ is considered a subspace of the real line. The proof is essentially the same proof that shows $M \times \mathbb{P}$ is not normal (see here). The space $X_B$ is a Lindelof space. It is not a normal P-space since its product with $B$, a separable metric space, is not normal. However, this example is essentially the same example as the Michael line since the same technique and proof are used. On the one hand, the $X_B \times B$ example seems like an improvement over Michael line example since the first factor $X_B$ is Lindelof. On the other hand, it is inferior than the Michael line example since the second factor $B$ is not completely metrizable.
Moving away from the idea of Michael, there exist a Lindelof space and a completely metrizable (but not separable) space whose product is of weight $\omega_1$ and is not normal [5]. This would be a Lindelof space that is a non-normal P-space. However, this example is not as elementary as the Michael line, making it not as effective as an illustration of Morita’s three conjectures.
The next set of non-normal P-spaces requires set theory. A Michael space is a Lindelof space whose product with $\mathbb{P}$, the space of irrational numbers, is not normal. Michael problem is the question: is there a Michael space in ZFC? It is known that a Michael space can be constructed using continuum hypothesis [6] or using Martin’s axiom [1]. The construction using continuum hypothesis has been discussed in this blog (see here). The question of whether there exists a Michael space in ZFC is still unsolved.
The existence of a Michael space is equivalent to the existence of a Lindelof space and a separable completely metrizable space whose product is non-normal [4]. A Michael space, in the context of the discussion in this post, is a non-normal P-space.
The discussion in this post shows that the example of the Michael line and other examples of non-normal P-spaces are useful tools to illustrate Morita’s three conjectures.
Reference
1. Alster K.,On the product of a Lindelof space and the space of irrationals under Martin’s Axiom, Proc. Amer. Math. Soc., Vol. 110, 543-547, 1990.
2. Balogh Z.,Normality of product spaces and Morita’s conjectures, Topology Appl., Vol. 115, 333-341, 2001.
3. Chiba K., Przymusinski T., Rudin M. E.Nonshrinking open covers and K. Morita’s duality conjectures, Topology Appl., Vol. 22, 19-32, 1986.
4. Lawrence L. B., The influence of a small cardinal on the product of a Lindelof space and the irrationals, Proc. Amer. Math. Soc., 110, 535-542, 1990.
5. Lawrence L. B., A ZFC Example (of Minimum Weight) of a Lindelof Space and a Completely Metrizable Space with a Nonnormal Product, Proc. Amer. Math. Soc., 124, No 2, 627-632, 1996.
6. Michael E., Paracompactness and the Lindelof property in nite and countable cartesian products, Compositio Math., 23, 199-214, 1971.
7. Morita K., Products of Normal Spaces with Metric Spaces, Math. Ann., Vol. 154, 365-382, 1964.
8. Rudin M. E., A Normal Space $X$ for which $X \times I$ is not Normal, Fund. Math., 73, 179-186, 1971.
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# Three conjectures of K Morita
This post discusses the three conjectures that were proposed by K. Morita in 1976. These conjectures concern normality in product spaces. To start the discussion, here’s the conjectures.
Morita’s Conjecture I. Let $X$ be a space. The product $X \times Y$ is normal for every normal space $Y$ if and only if $X$ is a discrete space.
Morita’s Conjecture II. Let $X$ be a space. The product $X \times Y$ is normal for every normal P-space $Y$ if and only if $X$ is a metrizable space.
Morita’s Conjecture III. Let $X$ be a space. The product $X \times Y$ is normal for every normal countably paracompact space $Y$ if and only if $X$ is a metrizable $\sigma$-locally compact space.
These statements are no longer conjectures. Partial results appeared after the conjectures were proposed in 1976. The complete resolution of the conjectures came in 2001 in a paper by Zoli Balogh [5]. Though it is more appropriate to call these statements theorems, it is still convenient to call them conjectures. Just know that they are now known results rather open problems to be solved. The focus here is not on the evolution of the solutions. Instead, we discuss the relations among the three conjectures and why they are amazing results in the study of normality in product spaces.
As discussed below, in each of these conjectures, one direction is true based on prior known theorems (see Theorem 1, Theorem 2 and Theorem 4 below). The conjectures can be stated as follows.
Morita’s Conjecture I. Let $X$ be a space. If the product $X \times Y$ is normal for every normal space $Y$ then $X$ is a discrete space.
Morita’s Conjecture II. Let $X$ be a space. If the product $X \times Y$ is normal for every normal P-space $Y$ then $X$ is a metrizable space.
Morita’s Conjecture III. Let $X$ be a space. If the product $X \times Y$ is normal for every normal countably paracompact space $Y$ then $X$ is a metrizable $\sigma$-locally compact space.
P-spaces are defined by K. Morita [11]. He proved that a space $X$ is a normal P-space if and only if the product $X \times Y$ is normal for every metrizable space $Y$ (see theorem 2 below). Normal P-spaces are also discussed here. A space $X$ is $\sigma$-locally compact space if $X$ is the union of countably many locally compact subspaces each of which is also closed subspace of $X$.
As we will see below, these conjectures are also called duality conjectures because they are duals of known results.
[2] is a survey of Morita’s conjecture.
Duality Conjectures
Here’s three theorems that are duals to the conjectures.
Theorem 1
Let $X$ be a space. The product space $X \times Y$ is normal for every discrete space $Y$ if and only if $X$ is normal.
Theorem 2
Let $X$ be a space. The product space $X \times Y$ is normal for every metrizable space $Y$ if and only if $X$ is a normal P-space.
Theorem 3
Let $X$ be a space. The product space $X \times Y$ is normal for every metrizable $\sigma$-locally compact space $Y$ if and only if $X$ is normal countably paracompact.
The key words in red are for emphasis. In each of these three theorems, if we switch the two key words in red, we would obtain the statements for the conjectures. In this sense, the conjectures are called duality conjectures since they are duals of known results.
Theorem 1 is actually not found in the literature. It is an easy theorem. Theorem 2, found in [11], is a characterization of normal P-space (discussed here). Theorem 3 is a well known result based on the following theorem by K. Morita [10].
Theorem 4
Let $Y$ be a metrizable space. Then the product $X \times Y$ is normal for every normal countably paracompact space $X$ if and only if $Y$ is a $\sigma$-locally compact space.
We now show that Theorem 3 can be established using Theorem 4. Theorem 4 is also Theorem 3.5 in p. 111 of [2]. A proof of Theorem 4 is found in Theorem 1.8 in p. 130 of [8].
Proof of Theorem 3
$\Longleftarrow$ Suppose $X$ is normal and countably paracompact. Let $Y$ be a metrizable $\sigma$-locally compact space. By Theorem 4, $X \times Y$ is normal.
$\Longrightarrow$ This direction uses Dowker’s theorem. We give a contrapositive proof. Suppose that $X$ is not both normal and countably paracompact. Case 1. $X$ is not normal. Then $X \times \{ y \}$ is not normal where $\{ y \}$ is any one-point discrete space. Case 2. $X$ is normal and not countably paracompact. This means that $X$ is a Dowker space. Then $X \times [0,1]$ is not normal. In either case, $X \times Y$ is not normal for some compact metric space. Thus $X \times Y$ is not normal for some $\sigma$-locally compact metric space. This completes the proof of Theorem 3. $\square$
The First and Third Conjectures
The first conjecture of Morita was proved by Atsuji [1] and Rudin [13] in 1978. The proof in [13] is a constructive proof. The key to that solution is to define a $\kappa$-Dowker space. Suppose $X$ is a non-discrete space. Let $\kappa$ be the least cardinal of a non-discrete subspace of $X$. Then construct a $\kappa$-Dowker space $Y$ as in [13]. It follows that $X \times Y$ is not normal. The proof that $X \times Y$ is not normal is discussed here.
Conjecture III was confirmed by Balogh in 1998 [4]. We show here that the first and third conjectures of Morita can be confirmed by assuming the second conjecture.
Conjecture II implies Conjecture I
We give a contrapositive proof of Conjecture I. Suppose that $X$ is not discrete. We wish to find a normal space $Y$ such that $X \times Y$ is not normal. Consider two cases for $X$. Case 1. $X$ is not metrizable. By Conjecture II, $X \times Y$ is not normal for some normal P-space $Y$. Case 2. $X$ is metrizable. Since $X$ is infinite and metric, $X$ would contain an infinite compact metric space $S$. For example, $X$ contains a non-trivial convergent sequence and let $S$ be a convergence sequence plus the limit point. Let $Y$ be a Dowker space. Then the product $S \times Y$ is not normal. It follows that $X \times Y$ is not normal. Thus there exists a normal space $Y$ such that $X \times Y$ is not normal in either case. $\square$
Conjecture II implies Conjecture III
Suppose that the product $X \times Y$ is normal for every normal and countably paracompact space $Y$. Since any normal P-space is a normal countably paracompact space, $X \times Y$ is normal for every normal and P-space $Y$. By Conjecture II, $X$ is metrizable. By Theorem 4, $X$ is $\sigma$-locally compact. $\square$
The Second Conjecture
The above discussion shows that a complete solution to the three conjectures hinges on the resolution of the second conjecture. A partial resolution came in 1986 [6]. In that paper, it was shown that under V = L, conjecture II is true.
The complete solution of the second conjecture is given in a paper of Balogh [5] in 2001. The path to Balogh’s proof is through a conjecture of M. E. Rudin identified as Conjecture 9.
Rudin’s Conjecture 9. There exists a normal P-space $X$ such that some uncountable increasing open cover of $X$ cannot be shrunk.
Conjecture 9 was part of a set of 14 conjectures stated in [14]. It is also discussed in [7]. In [6], conjecture 9 was shown to be equivalent to Morita’s second conjecture. In [5], Balogh used his technique for constructing a Dowker space of cardinality continuum to obtain a space as described in conjecture 9.
The resolution of conjecture II is considered to be one of Balogh greatest hits [3].
Abundance of Non-Normal Products
One immediate observation from Morita’s conjecture I is that existence of non-normal products is wide spread. Conjecture I indicates that every normal non-discrete space $X$ is paired with some normal space $Y$ such that their product is not normal. So every normal non-discrete space forms a non-normal product with some normal space. Given any normal non-discrete space (no matter how nice it is or how exotic it is), it can always be paired with another normal space (sometimes paired with itself) for a non-normal product.
Suppose we narrow the focus to spaces that are normal and non-metrizable. Then any such space $X$ is paired with some normal P-space $Y$ to form a non-normal product space (Morita’s conjecture II). By narrowing the focus on $X$ to the non-metrizable spaces, we obtain more clarity on the paired space to form non-normal product, namely a normal P-space. As an example, let $X$ be the Michael line (normal and non-metrizable). It is well known that $X$ in this case is paired with $\mathbb{P}$, the space of irrational numbers with the usual Euclidean topology, to form a non-normal product (discussed here).
Another example is $X$ being the Sorgenfrey line. It is well known that $X$ in this case is paired with itself to form a non-normal product (discussed here). Morita’s conjectures are powerful indication that these two non-normal products are not isolated phenomena.
Another interesting observation about conjecture II is that normal P-spaces are not productive with respect to normality. More specifically, for any non-metrizable normal P-space $X$, conjecture II tells us that there exists another normal P-space $Y$ such that $X \times Y$ is not normal.
Now we narrow the focus to spaces that are metrizable but not $\sigma$-locally compact. For any such space $X$, conjecture III tells us that $X$ is paired with a normal countably paracompact space $Y$ to form a non-normal product. Using the Michael line example, this time let $X=\mathbb{P}$, the space of irrational numbers, which is a metric space that is not $\sigma$-locally compact. The paired normal and countably paracompact space $Y$ is the Michael line.
Each conjecture is about existence of a normal $Y$ that is paired with a given $X$ to form a non-normal product. For Conjecture I, the given $X$ is from a wide class (normal non-discrete). As a result, there is not much specific information on the paired $Y$, other than that it is normal. For Conjectures II and III, the given space $X$ is from narrower classes. As a result, there is more information on the paired $Y$.
The concept of Dowker spaces runs through the three conjectures, especially the first conjecture. Dowker spaces and $\kappa$-Dowker spaces provide reliable pairing for non-normal products. In fact this is one way to prove conjecture I [13], also see here. For any normal space $X$ with a countable non-discrete subspace, the product of $X$ and any Dowker space is not normal (discussed here). For any normal space $X$ such that the least cardinality of a non-discrete subspace is an uncountable cardinal $\kappa$, the product $X \times Y$ is not normal where $Y$ is a $\kappa$-Dowker space as constructed in [13], also discussed here.
In finding a normal pair $Y$ for a normal space $X$, if we do not care about $Y$ having a high degree of normal productiveness (e.g. normal P or normal countably paracompact), we can always let $Y$ be a Dowker space or $\kappa$-Dowker space. In fact, if the starting space $X$ is a metric space, the normal pair for a non-normal product (by definition) has to be a Dowker space. For example, if $X=[0,1]$, then the normal space $Y$ such that $X \times Y$ is by definition a Dowker space. The search for a Dowker space spanned a period of 20 years. For the real line $\mathbb{R}$, the normal pair for a non-normal product is also a Dowker space. For “nice” spaces such as metric spaces, finding a normal space to form non-normal product is no trivial problem.
Reference
1. Atsuji M.,On normality of the product of two spaces, General Topology and Its Relation to Modern Analysis and Algebra (Proc. Fourth Prague Topology sympos., 1976), Part B, 25–27, 1977.
2. Atsuji M.,Normality of product spaces I, in: K. Morita, J. Nagata (Eds.), Topics in General
Topology, North-Holland, Amsterdam, 81–116, 1989.
3. Burke D., Gruenhage G.,Zoli, Top. Proc., Vol. 27, No 1, i-xxii, 2003.
4. Balogh Z.,Normality of product spaces and K. Morita’s third conjecture, Topology Appl., Vol. 84, 185-198, 1998.
5. Balogh Z.,Normality of product spaces and Morita’s conjectures, Topology Appl., Vol. 115, 333-341, 2001.
6. Chiba K., Przymusinski T., Rudin M. E.Nonshrinking open covers and K. Morita’s duality conjectures, Topology Appl., Vol. 22, 19-32, 1986.
7. Gruenhage G.,Mary Ellen’s Conjectures,, Special Issue honoring the memory of Mary Ellen Rudin, Topology Appl., Vol. 195, 15-25, 2015.
8. Hoshina T.,Normality of product spaces II, in: K. Morita, J. Nagata (Eds.), Topics in General Topology, North-Holland, Amsterdam, 121–158, 1989.
9. Morita K., On the Product of a Normal Space with a Metric Space, Proc. Japan Acad., Vol. 39, 148-150, 1963. (article information; paper)
10. Morita K., Products of Normal Spaces with Metric Spaces II, Sci. Rep. Tokyo Kyoiku Dagaiku Sec A, 8, 87-92, 1963.
11. Morita K., Products of Normal Spaces with Metric Spaces, Math. Ann., Vol. 154, 365-382, 1964.
12. Morita K., Nagata J., Topics in General Topology, Elsevier Science Publishers, B. V., The Netherlands, 1989.
13. Rudin M. E., $\kappa$-Dowker Spaces, Czechoslovak Mathematical Journal, 28, No.2, 324-326, 1978.
14. Rudin M. E., Some conjectures, in: Open Problems in Topology, J. van Mill and G.M. Reed,
eds., North Holland, 184–193, 1990.
15. Telgárski R., A characterization of P-spaces, Proc. Japan Acad., Vol. 51, 802–807, 1975.
$\text{ }$
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Dan Ma math
Daniel Ma mathematics
$\copyright$ 2018 – Dan Ma | 2019-08-25T13:29:03 | {
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https://math.stackexchange.com/questions/2055219/why-does-fracddx-cos-x-tan-1-2-frac2-cosx-sinx | # Why does $\frac{d}{dx} \cos ( x - \tan^{-1} ( 2 ) ) = \frac{2 \cos(x) - \sin(x)}{\sqrt {5}}$?
The title just about sums up my question.
Wolfram|Alpha shows it to be $\frac{2 \cos(x) - \sin(x)}{\sqrt {5}}$, while the (extremely simple) derivation I did by hand gives $-\sin(x - \tan^{-1}(2))$ (which wolfram agrees with).
I'm perfectly willing to accept that the two are equal, but I'd like to know why. What is the property or relationship between $\frac{2 \cos(x) - \sin(x)}{\sqrt {5}}$ and $-\sin(x - \tan^{-1}(2))$ that allows you to convert from one form to another without changing the value of the expression?
I've done some looking, and my initial thoughts are that it's a property of the arc tangent, but I haven't been able to find a solid answer.
Thanks in advance for any responses.
• Consider a right triangle with height 2, base 1. Let $\theta$ be angle between hypotenuse and base. Find $\cos \theta$, $\sin \theta$, you'll get your answer – Max Payne Dec 12 '16 at 9:52
use the fact that: $$\tan^{-1}(2)=\alpha \iff \tan \alpha =2$$ and: $$\sin \alpha=\frac{\tan \alpha}{\sqrt{1+\tan^2 \alpha}}=\frac{2}{\sqrt{5}}$$
$$\cos \alpha=\frac{1}{\sqrt{1+\tan^2 \alpha}}=\frac{1}{\sqrt{5}}$$
For now, let the term $x-\arctan(2) =\alpha$. We know the differentiation of $\cos \alpha$ equals $-\sin \alpha$. Hence we have $$\frac{d(\cos (x-\arctan(2)))}{dx} =-\sin(x-\arctan(2))$$
Now, use the formula $\sin(A-B) = \sin A\cos B-\cos A\sin B$. Here $A=x$ and $B=\arctan(2)$. In a triangle of angle $B =\arctan(2)$, we have $\sin B = \frac{2}{\sqrt{5}}$ and $\cos B =\frac{1}{\sqrt{5}}$. Using these results and applying them in the formula , gives us the derivative as $$\frac{2\cos x-\sin x}{\sqrt{5}}$$ | 2021-05-18T23:03:21 | {
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http://blog.quantitations.com/tutorial/2012/11/27/rejection-sampling/ |
An earlier post discussed inverse transform sampling. Another way to sample from a known pdf is to use rejection sampling. First, find a density $f$ that you can sample from and that dominates the density of interest $h$. Next, find a value $M$ such that $Mf \geq h$ (i.e. $M f(x) \geq h(x) \; \forall x \in \R$). Preferably, you should find the smallest such $M$ value, to make the algorithm as fast as possible. Finally, sample from $f$, and keep each sample value $x_i$ with probability $h(x_i)/(Mf(x_i))$.
## Example Using Exponential Distribution
Suppose, as in the previous post, we wish to sample from the pdf
\begin{align*} h(x) &= \frac{2m^2}{(1-m^2)x^3} \I\{x \in [m,1]\} \end{align*}
For this analysis, let’s assume the parameter $m$ is $.3$. To perform our rejection sampling, let’s use a shifted exponential distribution as our source of random values (credit: Jay Emerson).
h <- function(x, m = 0.3) {
if (x >= m & x <= 1)
return(2 * m^2/(1 - m^2)/x^3)
return(0)
}
m <- 0.3
grid <- seq(m, 1.5, length.out = 1000)
plot(grid, sapply(grid, h), type = "l", xlim = c(m, 1.5), xlab = "x", ylab = "density",
main = paste("h and", m, "+ Exp(1)"))
lines(grid, dexp(grid - m), col = 2)
legend(1, 5.5, c("h", paste(m, "+ Exp(1)")), col = 1:2, lty = 1)
Now, we want to find the smallest $M$ such that $M$ times our exponential density is greater than $h$ for all $x$. We also have the freedom to adjust the exponential rate parameter, if we want. A fundamental result in rejection sampling is that the overall acceptance rate of a procedure is exactly equal to $1/M$. Therefore, let’s try to find the rate parameter $\lambda>0$ for which the smallest $M$ will suffice. This will give us the procedure that has the highest acceptance rate (fewest “wasted” trials) of any procedure in this shited exponential class.
Our shifted exponential density will be of the form $f(x; \lambda) = \lambda e^{-\lambda (x-m)} \I{x \geq m}$. Assuming $x \in [m, 1]$, we need $M$ to satisfy
\begin{align*} M f(x; \lambda) &\geq h(x)\\ \Rightarrow M \lambda e^{-\lambda (x-m)} &\geq \frac{2 m^2}{(1-m^2)x^3}\\ \Rightarrow M &\geq \frac{2 m^2 e^{\lambda (x-m)}}{(1-m^2) \lambda x^3} \end{align*}
For a given $\lambda$, we want $M$ to be the smallest value satisfying the above condition, so we will let $M$ be the supremum over $x$ of the expression on the right-hand side, which I will call $g(x, \lambda)$. By differentiating $g$ with respect to $x$, we can see that it is convex on the interval we’re considering; therefore, its maximum must occur at an endpoint. Now, let’s define
\begin{align*} S(\lambda) = \sup_{x \in [m,1]} g(x; \lambda) \end{align*}
Because the supremum occurs at an endpoint, the only two possibilities for $S(\lambda)$ are $g(m; \lambda)$ and $g(1; \lambda)$. By substituting these two $x$ values and comparing, it can be shown that
\begin{align*} S(\lambda) = \left\{ \begin{array}{ll} \frac{2 m^2}{(1-m^2) \lambda m^3} , & \text{if} \; \lambda \in (0, -\frac{3}{1-m} \log m)\\ \frac{2 m^2 e^{\lambda (1-m)}}{(1-m^2) \lambda} , & \text{if} \; \lambda \geq -\frac{3}{1-m} \log m \end{array} \right. \end{align*}
By differentiating with respect to $\lambda$, we find that $S$ is minimized at $\lambda = -\tfrac{3}{1-m} \log m$ which is approximately $5.16$ when $m=.3$. This corresponds to an $M$ value of about 1.42 and an acceptance rate of about 70 percent. The acceptance rate as a function of $\lambda$ is plotted below, along with a dotted line showing simulated results.
S <- function(lambda, m = 0.5) {
if (lambda < -3/(1 - m) * log(m))
return((2 * m^2/(1 - m^2))/(lambda * m^3))
return((2 * m^2/(1 - m^2)) * exp(lambda * (1 - m))/lambda)
}
grid <- seq(0.1, 10, length.out = 100)
M <- sapply(grid, S)
plot(grid, 1/M, type = "l", col = 2, xlab = "lambda", ylab = "probability",
main = "Acceptance Rate")
N <- 1000
results <- rep(NA, length(grid))
for (i in 1:length(grid)) {
raw <- m + rexp(N, grid[i])
u <- runif(N, max = M[i] * dexp(raw - m, grid[i]))
results[i] <- sum(u <= sapply(raw, h))/N
}
lines(grid, results, col = 3, lty = 3)
It is possible that different rate parameters cause the random number generator to take different amounts of time. However, after a quick check, this question seems unlikely to be worth pursuing. The simulation below shows that Exp(1) is generated at basically the same rate as Exp(5.16).
nsim <- 1e+05
rates <- c(1, 5.16)
repetitions <- 5
results <- matrix(NA, nrow = repetitions, ncol = length(rates))
colnames(results) <- paste("rate:", rates)
for (i in 1:repetitions) {
for (j in 1:length(rates)) {
results[i, j] <- system.time(rexp(nsim, rates[j]))[3]
}
}
results
## rate: 1 rate: 5.16
## [1,] 0.027 0.027
## [2,] 0.027 0.028
## [3,] 0.026 0.027
## [4,] 0.027 0.027
## [5,] 0.027 0.026
Therefore, we can be satisfied that using $\lambda=5.16$ is about as well as we can do in drawing from $h$ via an exponential rejection sampling.
## Repeat Using Histograms
Another way to use rejection sampling for a density with bounded support is by drawing from a histogram. First, you construct a histogram shape that covers the density of interest, then divide each rectangle’s area by the total area to normalize to a probability mass function (pmf) for the bins. Next, draw a bin according to that pmf. Uniformly sample a horizontal and vertical position with that bin’s rectangle. If this point is below the density of interest, then keep the $x$ value.
We will apply this method to draw from $h$. First, here is a picture of the true density and the histogram-like density if we use ten bins, for example.
grid <- seq(m, 1, length.out = 100)
plot(grid, sapply(grid, h), type = "l", ylim = c(0, 7.4), xlab = "x",
ylab = "density", main = "Density of h and Scaled Histogram")
r <- 10 # number of bins
binpoints <- seq(m, 1, length.out = r + 1)
binheights <- sapply(binpoints[1:r], h)
# visualizing the process
picx <- sort(c(binpoints[1:r], binpoints[2:length(binpoints)]))
picy <- rep(binheights, each = 2)
lines(picx, picy, col = 2)
We will call the number of bins $r$ (for refinement). It is clear that we can drive the acceptance probability arbitrarily close to 1 by increasing $r$. However, there is a trade-off because the larger $r$ is, the longer it takes to generate each point. We will look for the $r$ that optimizes overall speed. The rsample.h function below samples from $h$ using this histogram method with a given $r$ value.
rsample.h <- function(r, N = 1000) {
binwidth <- (1 - m)/r
grid <- seq(m, 1, length.out = r + 1)
binheights <- sapply(grid[1:r], h)
binprobs <- binheights/sum(binheights)
s <- sample.int(r, N, replace = T, prob = binprobs)
x <- m + (s - 1) * binwidth + runif(N, 0, binwidth)
return(x[which(runif(N, 0, binheights[s]) - sapply(x, h) <= 0)])
}
Now, we try out a series of $r$ values to determine the time required as a function of $r$.
r <- 1:10 * 200
ntime <- function(n, f, ...) {
return(system.time(f(n, ...))[3])
}
times <- sapply(r, ntime, f = rsample.h)
plot(r, times)
fit <- lm(times ~ r)$coeff fit <- signif(fit, 3) print(paste("The time (in seconds) required to generate 1000 points is about", fit[1], "+", fit[2], "* r.")) ## [1] "The time (in seconds) required to generate 1000 points is about 0.0435 + 4.31e-05 * r." abline(fit, col = 2) The algorithm is linear in$r$. The expected time required to reach 1000 \textcal{accepted} draws is approximately the time required to generate 1000 points divided by the acceptance rate. We could find the acceptance rate by conditioning on the bin selected and finding the conditional acceptance rate for each bin. But it is simpler to just use the result invoked earlier: the acceptance rate is$1/M$, where$M$is the sum of the rectangle areas. In fact, I wrote code for both; the rejectionprob function does it “the hard way,” while acceptanceprob uses the easy method. As a sanity check, you can confirm that the sum of these two functions is 1 for any$r$. rejectionprob <- function(r) { binwidth <- (1 - m)/r grid <- seq(m, 1, length.out = r + 1) binheights <- sapply(grid[1:r], h) binprobs <- binheights/sum(binheights) binareas <- binheights * binwidth rejection <- function(x, i) { return(binheights[i] - sapply(x, h)) } p <- 0 for (i in 1:r) { rejectarea <- integrate(rejection, m + (i - 1) * binwidth, m + i * binwidth, i = i)$value
p <- p + binprobs[i] * rejectarea/binareas[i]
}
return(p)
}
acceptanceprob <- function(r) {
binwidth <- (1 - m)/r
grid <- seq(m, 1, length.out = r + 1)
binheights <- sapply(grid[1:r], h)
binareas <- binheights * binwidth
return(1/sum(binareas))
}
We want to find the $r$ value that minimizes number-generating time divided by acceptance rate. Because it is a discrete problem, all we need to do is evaluate the possible $r$ values.
expectedtime <- function(r) {
time <- fit %*% c(1, r)
return(time/acceptanceprob(r))
}
r <- 1:200
exptimes <- sapply(r, expectedtime)
plot(r, exptimes, type = "l", col = 2, xlab = "r (number of bins)", ylab = "seconds",
main = "Expected Time to Draw 1000 from h")
which.min(exptimes)
## [1] 51
## Comparing Speeds of the Exponential and Histogram Algorithms
Lastly, let’s compare the two rejection sampling methods described above. Using the histogram method we can make the acceptance probability arbirarily close to 1 simply by increasing the number of bins. That may make it seem like this method is superior to the exponential one. However, the histogram method requires one extra randomization step, so it could still be worse overall. Let’s see how the best exponential method and the best histogram method match up head-to-head in sampling from $h$.
rsample.h.optimal <- function(N = 1e+05) {
# returns amount of time required per N accepted points using optimal
# parameter values
r <- 51
time <- system.time({
binwidth <- (1 - m)/r
grid <- seq(m, 1, length.out = r + 1)
binheights <- sapply(grid[1:r], h)
binprobs <- binheights/sum(binheights)
s <- sample.int(r, N, replace = T, prob = binprobs)
x <- m + (s - 1) * binwidth + runif(N, 0, binwidth)
numaccepted <- sum(runif(N, 0, binheights[s]) - sapply(x, h) <= 0)
})[3]
return(1e+05 * time/numaccepted)
}
esample.h.optimal <- function(N = 1e+05) {
# returns amount of time required per N accepted points using optimal
# parameter values
lambda <- 5.16
M <- S(lambda)
time <- system.time({
raw <- m + rexp(N, lambda)
u <- runif(N, max = M * dexp(raw - m, lambda))
numaccepted <- sum(u <= sapply(raw, h))
})[3]
return(1e+05 * time/numaccepted)
}
methods <- c("esample.h.optimal", "rsample.h.optimal")
repetitions <- 5
results <- matrix(NA, repetitions, length(methods))
colnames(results) <- c("exponential", "histogram")
for (i in 1:repetitions) {
order <- sample.int(length(methods), length(methods))
for (j in order) {
results[i, j] <- get(methods[j])()
}
}
print("Amount of time needed to reach 100,000 accepted values:")
## [1] "Amount of time needed to reach 100,000 accepted values:"
results
## exponential histogram
## [1,] 7.196 5.133
## [2,] 6.331 3.943
## [3,] 7.247 3.925
## [4,] 6.405 3.883
## [5,] 6.304 3.883
The histogram method consistently outperforms the exponential method, at least with the implementation I coded up. And as an added bonus, the histogram method can be packaged as a general-purpose function that doesn’t require any thought to use, especially when the density is monotonic or monomodal. It would take some extra code and extra computation to make the function choose a near-optimal $r$ value, but that could be accomplished as well.
Unfortunately, the histogram method requires the density of interest to have bounded support. Although, in practice, if you had a density with unbounded support, you could cut it off way out on the tails without changing your results. | 2017-07-22T14:44:44 | {
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https://byjus.com/question-answer/below-is-given-distribution-of-profit-in-rs-per-day-of-a-shop-in-a/ | Question
# Below is given distribution of profit (in Rs.) per day of a shop in a certain town.Calculate median profit of shops.Profit (in Rs.)500 - 10001000 - 15001500 - 20002000 - 25002500 - 30003000 - 35003500 - 4000No. of shops818272120188
A
Rs. 1867
B
Rs. 1967
C
Rs.2167
D
Rs.2567
Solution
## The correct option is B Rs.$$2167$$Consider the following table, to calculate median: $$c_i$$ $$f_i$$ $$cf$$ 500-1000 8 8 1000-1500 18 8+18=26 1500-2000 27 26+27=53 2000-2500 21 53+21=74 2500-3000 20 74+20=94 3000-3500 18 94+18=112 3500-4000 8 112+8=120Median $$=l +\dfrac {(\dfrac{N}{2}-cf)}{ f_m}\times c$$Here, Class Interval $$c=500$$$$N=\Sigma f_i=120$$ $$\dfrac {N}{2}=60$$$$\Rightarrow$$ Median class $$=2000-2500$$$$\Rightarrow$$Frequency of median class$$f_m=21$$Lower boundary of median class $$l=2000$$previous cumulative frequency of median class $$cf=53$$$$\therefore$$ Median $$=2000 +\dfrac {(\dfrac{120}{2}-53)}{ 21}\times 500$$$$\therefore$$ Median $$=2000 +\dfrac {(7)}{ 21}\times 500$$$$\therefore$$ Median $$=2166.67 \Rightarrow 2167$$ Hence. option $$C$$ is correct.Mathematics
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View More | 2022-01-28T11:57:00 | {
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https://math.stackexchange.com/questions/2471582/does-huffman-coding-with-increasing-word-length-approach-the-shannon-information | Does Huffman coding with increasing word length approach the Shannon information limit?
I answered a question Calculating entropy Through Huffman Codeword lengths. on this website and then got myself another question. If the alphabet has $5\,$ letters 'A', 'B', 'C', 'D', 'E' occuring with probabilities $\,0.4,\,0.2,\,0.2,\,0.1,\,0.1$, respectively, a letter-by-letter Huffman code gives an average code length of $\,L_1=2.2\,$ bits per letter. This number is greater than the Shannon information
$$S=-\sum_{i=1}^5p_i\log_2p_i=\log_210-1.2=2.1219\mbox{ bits}$$
per letter. My question is: if I group $n$ letters into one word and construct a word-by-word Huffman code, does the average code length per letter $\,L_n/n\rightarrow S\,$ approach Shannon's information content in the $\,n\rightarrow\infty\,$ limit? Is there a proof?
I wrote a C++ program to test this numerically. I assume the occurrence of letters are independent, i.e., the language is a random mixture of the $5$ letters with their individual probabilities. The result is plotted in the graph below:
The red dashed line is Shannon's information content $\,S\,$ per letter, which is unbeatable in this model as the letter occurrence is truely random. The blue curve is the average code length $\,L_n/n\,$ per letter of the Huffman tree for $\,n=1,2,\ldots,10$, where $L_n$ is the weighted tree height of the Huffman tree with $5^n$ leaf nodes. The numerical result does not clearly show whether $L_n/n\rightarrow S\;$ or not, as the curve is not monotonic. So a mathematical proof or disproof would be very much appreciated.
Yes, Huffman coding approaches the Shannon limit. This is an immediate consequence of the fact that Huffman coding is optimal, i.e., $L_n$ of your Huffman code is always smaller than (or equal to) the length of any other code (Cover and Thomas, 1991, Theorem 5.8.1).
Now consider the Shannon code and assume that you draw independent realizations of your 5-letter-alphabet. Then, you can assign each length-$n$ sequence $x^n$ a codeword of length $\lceil -\log(\mathbb{P}(X^n=x^n))\rceil$, from which you get that $L^*_n \le nS+1$. Thus, $L^*_n/n\to S$ (Cover and Thomas, 1991, Theorem 5.4.2). Since the Huffman code will be at least as good, you will also get $L_n/n\to S$.
• Huffman coding is a word-to-word system. Theoretically, there are context-dependent codes that outperform it. It is best among the word-to-word codes, while Shannon's limit is the best of all possible codes. Increasing the word length helps mitigate the shortcoming of Huffman. There are two limits here that cannot be swapped. The total length of the message goes to infinity first, and then the word length goes to infinity. But still, the $L_n\leq nS+1$ upper bound answers the question. How is it derived (give me a reference)? – Zhuoran He Oct 16 '17 at 17:33 | 2019-08-20T20:26:37 | {
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https://nbviewer.ipython.org/url/courses.cit.cornell.edu/info2950_2016sp/resources/lec4.ipynb | In [1]:
%pylab inline
Populating the interactive namespace from numpy and matplotlib
In [2]:
from scipy.misc import factorial,comb
factorial(10)/(factorial(6)*factorial(4)), comb(10,4,exact=True)
Out[2]:
(210.0, 210)
While the above functions could be used, recall that arange(365,365-70,-1) gives an array of the seventy numbers from 365 down to 296, so arange(365,365-70,-1)/365. gives an array of the successive fractions of not having a coincident birthday, and 1-cumprod(arange(365,365-70,-1)/365.) then gives the probability of at least one coincidence for numbers of people from 1 through 70.
Similarly, ones(69)/365. is an array of 69 copies of 1/365. so cumprod(ones(69)*364./365.) gives successive powers of 1/365., and 1 minus that gives the probability of coincidence with a given specific person for numbers of people from 2 through 70. These two functions can be plotted as follows:
In [3]:
plot(range(1,71),1-cumprod(arange(365,365-70,-1)/365.),'r+')
plot(range(2,71),1-cumprod(ones(69)*364./365.),'g+');
We can make the graph a bit larger, and add a grid, to see that the probability of coincident birthdays exceeds .5 starting at n equal to 23, is about .97 at n equal to 50, and is virtually guaranteed by n equal to 70:
In [4]:
figure(figsize=(8,6))
plot(range(1,71),1-cumprod(arange(365,295,-1)/365.),'r+')
plot(range(2,71),1-cumprod(ones(69)*364./365.),'g+')
yticks(arange(0,1.1,.1));
xlim(0,70)
xticks(range(0,71,5))
grid('on')
legend([r'$1-\frac{365!}{(365-n)!\ 365^n}$',
r'$1-(\frac{364}{365})^{n-1}$'],
numpoints=3,loc='upper left',fontsize=18);
We can also expand the range to see what happens to the second curve when we reach n equal to 365, by plotting
plot(range(1,366),1-cumprod(arange(365,0,-1)/365.),'r')
plot(range(2,366),1-cumprod(ones(364)*364./365.),'g')
Or we can define a function, and use subplot to put the two figures side by side. At n equal to 365, the lower curve has a probability of roughly .63:
In [5]:
def bdayplot(n,inc=25,sym='',loc=None,):
plot(range(1,n+1),1-cumprod(arange(365,365-n,-1)/365.),'r'+sym)
plot(range(2,n+1),1-cumprod(ones(n-1)*364./365.),'g'+sym)
xticks(range(0,n+1,inc))
yticks(arange(0,1.1,.1))
xlim(0,n)
grid('on')
legend([r'$1-\frac{365!}{(365-n)!\ 365^n}$',
r'$1-(\frac{364}{365})^{n-1}$'],
numpoints=3,loc=loc,fontsize=17);
figure(figsize=(16,6))
subplot(1,2,1)
bdayplot(70,5,'+','upper left')
subplot(1,2,2)
bdayplot(365)
plot([70,70],[0,1],c='gray',linestyle='--');
Note that $\frac{n(n-1)}{2}\frac{1}{365}$ is a good approximation to the first function (red) for small $n$, up to about $n=14$ (it just counts the number of pairs times the probability that a pair has coincident birthdays, so works as long as the pairs are approximately independent).
An even better approximation is given by $1-{\rm e}^{-n(n-1)/(2\cdot365)}$, which works extremely well throughout the full range of $n$ (dotted green line below). (One way of deriving this is to use Stirling's approximation for the factorial: $n! \sim n^n{\rm e}^{-n}\sqrt{2\pi n}$ for large $n$.)
Note also that $1-{\rm e}^{-(n-1)/365}$ is a good approximation to the second function (green) for all $n$, hence the value $1-{\rm e}^{-1}\approx.63$ for $n=366$ in the above plot.
In [6]:
figure(figsize=(8,6))
plot(range(1,71),1-cumprod(arange(365,365-70,-1)/365.),'r+')
xlim(0,70)
ylim(0,1)
grid('on')
plot(range(70),[1-exp(-n*(n-1)/(2.*365)) for n in range(70)],'g--')
plot([n*(n-1)/(2.*365) for n in range(16)],'b')
plot(range(2,71),1-cumprod(ones(69)*364./365.),'g+')
plot([1-exp(-(n-1)/365.) for n in range(366)],'y')
xticks(range(0,75,5))
yticks(arange(0,1.1,.1));
legend([r'$1-\frac{365!}{(365-n)!\ 365^n}$',
r'$1-{\rm e}^{-n(n-1)/(2\cdot365)}$',
r'$\frac{n(n-1)}{2}\frac{1}{365}$',
r'$1-(\frac{364}{365})^{n-1}$',
r'$1-{\rm e}^{-(n-1)/365}$'],
numpoints=3,loc='center right',fontsize=14);
In [ ]: | 2022-05-29T09:00:58 | {
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https://math.stackexchange.com/questions/442715/prove-that-sum-limits-k-0n-1-dfrac1-cos2-frac-pi-kn-n2-for-odd/442760 | # Prove that $\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2$ for odd $n$
In old popular science magazine for school students I've seen problem
Prove that $\quad$ $\dfrac{1}{\cos^2 20^\circ} + \dfrac{1}{\cos^2 40^\circ} + \dfrac{1}{\cos^2 60^\circ} + \dfrac{1}{\cos^2 80^\circ} = 40.$
How to prove more general identity:
$$\begin{array}{|c|} \hline \\ \sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2 \\ \hline \end{array} , \qquad \mbox{ where } \ n \ \mbox{ is odd.}$$
• 40 isn't the square of an integer, so why do you think the problems are related? – Mercy King Jul 13 '13 at 13:04
• @Mercy: $\dfrac{1}{\cos 0^\circ} + \dfrac{1}{\cos 20^\circ} + \dfrac{1}{\cos 40^\circ} + \ldots+ \dfrac{1}{\cos 160^\circ} =81=9^2$. – Oleg567 Jul 13 '13 at 13:08
• Cool question, I have a suspicion you should add the complex-analysis tag – Cocopuffs Jul 13 '13 at 13:30
• Does the fourth power give you something similar? – Empy2 Jul 13 '13 at 14:25
• @Michael, my answer below generalizes to this case by taking instead $f(z) = (2/(z+z^{-1}))^4$. In this you'll find that the sum becomes $\tfrac{1}{3}n^2 (2 + n^2)$. – fuglede Jul 13 '13 at 14:31
Since $n$ is odd, the numbers $u_k=\cos k\pi/n$ are the same as the numbers $\cos 2k\pi/n$, i.e. the distinct angles $\theta$ satisfying $n\theta=0$ (mod $2\pi$). We think of them as the roots of the equation $\cos n\theta=1$. Writing $$\cos n\theta = \cos^n\theta - \binom{n}{2}\cos^{n-2}\theta\sin^2\theta \cdots \pm n\cos\theta \sin^{n-1}\theta$$ and using $\sin^2\theta=1-\cos^2\theta$, we see that the $u_k$'s are the roots of the polynomial $$p(u)=u^n - \binom{n}{2}u^{n-2}(1-u^2) \pm n u(1-u^2)^{(n-1)/2} + 1.$$ Note that all powers of $u$ which occur are odd (except for the constant term). The reciprocals $1/u_k$ are the roots of the "reverse polynomial" $$r(u)=u^n p(1/u) = u^n + a_{n-1} u^{n-1} + \cdots,$$ where $a_{n-1}=\pm n$ and $a_{n-2}=0$.
The sum in question is the sum of the squares of the roots of $r(u)$, i.e. $a_{n-1}^2 - 2a_{n-2} = n^2$.
• Thank you. It was most elementary proof (as for me). – Oleg567 Jul 14 '13 at 2:50
Whenever I see a problem like this, I think Chebyshev polynomials.
The Chebyshev polynomials of the first kind, $T_m(x)$, are defined so that:
$$T_m(\cos \theta) = \cos m\theta$$
$T_m$ is an $m$th degree polynomial, and the roots of $T_m(x)-1$ are exactly $\cos 2\pi k/m$ for $k=0,1,\dots,m-1$.
When $m=2n$ is even, we can write $$T_{2n}(x)=1+C\prod_{k=0}^{n-1}\left(x^2-\cos^2\frac{k\pi}{n}\right)$$ for some constant $C$. The occurrence of those $\cos^2\frac{k\pi}{n}$ suggested this might be a good approach.
Now, given a polynomial $p(x)=a_mx^m + a_{m-1}x^{m-1}\dots +a_0$, with non-zero roots $r_0,\dots, r_{m-1}$, we have formulas: $$\sum \frac{1}{r_k} = -\frac{a_1}{a_0}$$ And: $$\sum_{i< j} \frac{1}{r_ir_j} = \frac{a_2}{a_0}$$
So $$\sum \frac{1}{r_k^2} = \left(-\frac{a_1}{a_0}\right)^2 - 2\frac{a_2}{a_0}$$
Now let $p(x)=T_{2n}(x)-1$. The roots of $p$ are $r_k=\cos \frac{k\pi}{n}$ for $k=0,\dots,2n-1$, and the roots are non-zero since $n$ is odd. We know that $p(x)$ is even, so $a_1=0$. Finally, we know that $\sum_{k=0}^{2n-1} \frac{1}{r_k}^2$ is twice the sum that you are looking for.
So your sum is now reduced to finding $-\frac{a_2}{a_0}$ where the $a_0,a_2$ are coefficients of $p(x)$. Since $p(0)=T_{2n}(\cos \pi/2)-1 = \cos n\pi - 1 = -2$, so we know that $a_0=-2$. So $\sum r_k^{-2} = a_2$. (Again, we use $n$ odd here.)
So we need to prove that $a_2=2n^2$.
Now, $a_2=\frac{1}{2}T_{2n}^{''}(0)$. Let $f(x)=\cos 2nx = T_{2n}(\cos x)$. Differentiating we get:
$$-2n\sin 2nx = -T_{2n}^{'}(\cos x)\sin x$$
Differentiating both sides again:
$$-4n^2\cos 2nx = T_{2n}^{''}(\cos x)(\sin^2 x) - T_{2n}^{'}(\cos x)\cos x$$
Putting in $x=\pi/2$, then $\cos x=0$, $\sin x=1$, and $\cos 2nx=-1$. Therefore, we get $$4n^2 = T_n^{''}(0)\\a_n=\frac{1}{2}T_{2n}^{''}(0)=2n^2$$
and therefore your sum is $n^2$.
Note: Any symmetric rational function with rational coefficients of $\{\cos 2\pi k/n\mid k=0,\dots,n-1\}$ will be rational by this argument.
• Thank you. Nice approach with Chebyshev polynomials. – Oleg567 Jul 14 '13 at 2:53
I believe I first came across this trick in this paper by Szenes, so check out section 3 in that for more details.
Consider the form $$\mu_n(z) = n\frac{dz}{z} \frac{z^n + z^{-n}}{z^n-z^{-n}}$$ and the function $$f(z) = \frac{4}{(z+z^{-1})^2}.$$ Your sum is the sum of the $f(z_k)$, where $z_k = \exp(\pi i k/n)$, $k = 0, \dots, n-1$. Note first of all that for these particular values of $z$, $$f(z_k) = \mathop{\mathrm{Res}}_{z=z_k} f \mu_n.$$ Note also that the same formula holds with $z_k$ replaced by $z_k^{-1}$, and that $\mathop{\mathrm{Res}}_{z=\infty} f\mu_n = \mathop{\mathrm{Res}}_{z=0} f\mu_n = 0.$ Moreover, $\mathop{\mathrm{Res}}_{z= \pm i} f\mu_n = -n^2$. It follows from the residue theorem that $$\sum_{k=0}^{n-1} f(z_k) = -\frac{1}{2}\left(\mathop{\mathrm{Res}}_{z=i} f\mu_n + \mathop{\mathrm{Res}}_{z=-i} f\mu_n\right) = n^2.$$
• Ah, interesting. I was thinking all the time this should be doable with $\frac1{\cos^2\alpha}=\frac2{1+\cos 2\alpha}=2(1-\cos2\alpha+\cos^22\alpha-\cos^32\alpha\pm\ldots)$ and character orthogonality, but ran into dead ends ... – Hagen von Eitzen Jul 13 '13 at 14:28
• My own first approach was to somehow relate the sum of the question to the quadratic Gauss sum to which it is very similar. Indeed, many of the formulas given on that page can be proved in a fashion much like the above. – fuglede Jul 13 '13 at 14:41
• Thank you for the nice answer using complex analysis. – Oleg567 Jul 14 '13 at 2:54
In equation $(7)$ of this answer, I compute that $$\sum_{k=1}^n\tan^2\left(\frac{k\pi}{2n+1}\right)=n(2n+1)$$ Thus, with $n=2m+1$, \begin{align} \sum_{k=0}^{n-1}\frac1{\cos^2\left(\frac{k\pi}{n}\right)} &=\sum_{k=0}^{2m}\sec^2\left(\frac{k\pi}{2m+1}\right)\\ &=2\sum_{k=0}^m\tan^2\left(\frac{k\pi}{2m+1}\right)+2(m+1)\\[6pt] &=2m(2m+1)+2(m+1)\\[12pt] &=(2m+1)^2\\[12pt] &=n^2 \end{align}
For odd $n$, consider the function $$f(z)=\frac{n/z}{z^n-1}\left(\frac{z-1}{z+1}\right)^2\tag{1}$$ All the singularities are simple, except the singularity at $-1$. $$\left(\frac{z-1}{z+1}\right)^2=1-\frac4{z+1}+\frac4{(z+1)^2}\tag{2}$$ Furthermore, $$\frac{\mathrm{d}}{\mathrm{d}z}\frac{n/z}{z^{\raise{2pt}n}-1}=\frac{n-n(n+1)z^n}{z^2(z^{\raise{2pt}n}-1)^2}\tag{3}$$ Using $(2)$ and $(3)$, we get the residue of $f$ at $-1$ to be $$-4\cdot\frac{-n}{-1-1}+4\cdot\frac{n+n(n+1)}{(-1-1)^2}=n^2\tag{4}$$ The residue of $f$ at $0$ is $-n$ and the residue at each $n^{\text{th}}$ root of unity is $-\tan^2(\theta/2)$.
Since the integral of $f$ over an increasing circle vanishes, the sum of the residues must be $0$. Therefore, $$\sum_{k=0}^{n-1}\tan^2\left(\frac{k\pi}{n}\right)=n^2-n\tag{5}$$ and therefore, $$\sum_{k=0}^{n-1}\sec^2\left(\frac{k\pi}{n}\right)=n^2\tag{6}$$
• Using the half-angle formula for cosine, could you instead consider $f(z) = \displaystyle\frac{4z}{(z+1)^{2}} \frac{n/z}{z^{n}-1}$? – Random Variable Jul 15 '13 at 16:25
• @RandomVariable: $\displaystyle\frac{4z}{(z+1)^2}=1-\left(\frac{z-1}{z+1}\right)^2$, so I imagine so. – robjohn Jul 15 '13 at 16:40
From this answer we know that $$\sum\limits_{k=1}^{m}\tan^2\frac{\pi k}{2m+1}=m(2m+1)$$ Similarly $$\sum\limits_{k=m+1}^{2m}\tan^2\frac{\pi k}{2m+1}= \sum\limits_{l=1}^{m}\tan^2\frac{\pi (2m+1-l)}{2m+1}= \sum\limits_{l=1}^{m}\tan^2\frac{\pi l}{2m+1}=m(2m+1)$$ Since $\cos^{-2}\alpha=1+\tan^2\alpha$, then \begin{align} \sum\limits_{k=0}^{2m}\frac{1}{\cos^2\frac{\pi k}{2m+1}} &=\sum\limits_{k=0}^{2m} 1 +\sum\limits_{k=0}^{2m}\tan^2\frac{\pi k}{2m+1}\\ &=2m+1+\sum\limits_{k=1}^{m}\tan^2\frac{\pi k}{2m+1}+\sum\limits_{k=m+1}^{2m}\tan^2\frac{\pi k}{2m+1}\\ &=2m+1+m(2m+1)+m(2m+1)=(2m+1)^2 \end{align}
• +1 This one is a classic! It even works so as to calculate $\zeta(2),\zeta(4)$ pretty nicely. – Pedro Tamaroff Jul 13 '13 at 18:23 | 2019-09-22T16:27:48 | {
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https://mathzsolution.com/a-and-b-disjoint-a-compact-and-b-closed-implies-there-is-positive-distance-between-both-sets/ | # A and B disjoint, A compact, and B closed implies there is positive distance between both sets
Claim: Let $$XX$$ be a metric space. If $$A,B⊂XA,B\subset X$$ are disjoint, $$AA$$ is compact, and $$BB$$ is closed, then there is $$δ>0\delta>0$$ so that $$|α−β|≥δ∀α∈A,β∈B |\alpha-\beta|\geq\delta\;\;\;\forall\alpha\in A,\beta\in B$$.
Proof. Assume the contrary. Let $$αn∈A,βn∈B\alpha_n\in A,\beta_n\in B$$ be chosen such that $$|αn−βn|→0|\alpha_n-\beta_n|\rightarrow0$$ as $$n→∞n\rightarrow \infty$$.
Since $$AA$$ is compact, there exists a convergent subsequence of $$αn(n∈N)\alpha_n\;(n\in\mathbb{N})$$, $$αnm(m∈N)\alpha_{n_m}\;(m\in\mathbb{N})$$, which converges to $$α∈A\alpha\in A$$.
We have
$$|α−βnm|≤|α−αnm|+|αnm−βnm|→0asm→∞.|\alpha-\beta_{n_m}|\leq|\alpha-\alpha_{n_m}|+|\alpha_{n_m}-\beta_{n_m}|\rightarrow0 \;\;\;as\;\;m\rightarrow\infty.$$
Hence $$α\alpha$$ is a limit point of $$BB$$ and since $$BB$$ is closed $$α∈B\alpha\in B$$, contradiction.
Is my proof correct? I feel as though I am missing something simple which would trivialize the proof.
## Answer
For the sake of having an answer addressing your question:
Yes, your proof is perfectly okay and I don’t think you can get it any cheaper than you did it.
Let me expand a little on what you can do with these arguments (also providing details to gary’s answer). I’m not saying my proof at the end is better than yours in any way, I’m just showing a slightly alternative way of looking at it.
Define the distance between two non-empty subsets $A,B \subset X$ to be $d(A,B) = \inf_{a \in A, b \in B} d(a,b)$ and write $d(x,B)$ if $A = \{x\}$.
1. If $B \subset X$ is arbitrary and non-empty then $x \mapsto d(x,B)$ is $1$-Lipschitz continuous, that is $|d(x,B) - d(y,B)|\leq d(x,y)$ for all $x,y \in X$.
2. We have $d(x,B) = 0$ if and only if $x \in \overline{B}$.
3. If $d(\cdot,A) = d(\cdot,B)$ then $\overline{A} = \overline{B}$.
1. Choose $b\in B$ such that $d(x,b) \leq d(x,B) + \varepsilon$. Then the triangle inequality yields $d(y,B) - d(x,B) \leq d(y,b) - d(x,b) + \varepsilon \leq d(y,x) + \varepsilon$. By symmetry we get $|d(x,B) - d(y,B)| \leq d(x,y) + \varepsilon$, and 1. follows because $\varepsilon$ was arbitrary. Update: In this closely related answer I show that $1$ is in fact the best Lipschitz constant as soon as $B$ isn’t dense. Don’t miss Didier’s answer to the same thread which relies on a useful general fact which is as easy to prove and Zarrax’s answer providing a cleaned-up argument of the one I’m giving here.
2. Choose $b_n \in B$ with $d(x,b_n) \leq d(x,B) + \frac{1}{n} = \frac{1}{n}$. Then $d(x,b_n) \to 0$ and hence $x \in \overline{B}$. Conversely, if $b_n \to x$ then $d(x,b_n) \to 0$ hence $d(x,B) = 0$.
3. Immediate from 2.
Let me combine these facts: assume $A$ is compact and $B$ is closed. As $d(\cdot, B): X \to [0,\infty)$ is continuous by 1. above, we conclude from compactness of $A$ that $d(\cdot,B)$ assumes its minimum when restricted to $A$ (if you think about how one usually proves this, you’ll find your argument again!). Hence there is $a \in A$ with the property that $d(a',B) \geq d(a,B)$ for all $a' \in A$. But if $d(a,B) = 0$ then $a \in B$ by 2. above, since $B = \overline{B}$. So either $A$ and $B$ are not disjoint or $d(a',B) \geq d(a,B) \gt 0$. By choosing $\delta \in (0,d(a,B))$, we get the claim again.
Finally, if you don’t assume that one among $A$ and $B$ is compact, then the result is false. There was the example $A = \mathbb{N}$ and $B = \{n + \frac{1}{n}\}_{n\in\mathbb{N}}$ given in the comments, or, a bit more geometrically appealing to me, let $A$ be the $x$-axis in $\mathbb{R}^2$ and $B$ the graph of the function $x \mapsto \frac{1}{x}$, $x \neq 0$.
Attribution
Source : Link , Question Author : Benji , Answer Author : Troy Woo | 2022-10-05T02:02:15 | {
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https://byjus.com/question-answer/find-the-matrix-x-so-that-displaystyle-x-left-begin-matrix-1-2-3-4/ | Question
Find the matrix $$X$$ so that $$\displaystyle X\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right] =\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right]$$
Solution
It is given that:$$\displaystyle X\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right] =\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right]$$The matrix given on the R.H.S. of the equation is a $$2 \times 3$$ matrix and the one given on the L.H.S. of the equation is a $$2 \times 3$$ matrix. Therefore, $$X$$ has to be a $$2 \times 2$$ matrix.Now, let $$\displaystyle X=\begin{bmatrix} a & c \\ b & d \end{bmatrix}$$Therefore, we have:$$\displaystyle \begin{bmatrix} a & c \\ b & d \end{bmatrix}\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right] =\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right]$$$$\displaystyle \Rightarrow \left[ \begin{matrix} a+4c & 2a+5c & 3a+6c \\ b+4d & 2b+5d & 3b+6d \end{matrix} \right] =\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right]$$Equating the corresponding elements of the two matrices, we have:$$\displaystyle \begin{matrix} a+4c=-7 & 2a+5c=-8 & 3a+6c=-9 \\ b+4d=2 & 2b+5d=4 & 3b+6d=6 \end{matrix}$$Now, $$\displaystyle a+4c=-7\Rightarrow a=-7-4c$$$$\displaystyle \therefore 2a+5c=-8\Rightarrow -14-8c+5c=-8$$$$\displaystyle \Rightarrow -3c=6 \Rightarrow c=-2$$$$\displaystyle \therefore a=-7-4\left( -2 \right) =-7+8=1$$Now, $$\displaystyle b+4d=2\Rightarrow b=2-4d$$$$\displaystyle \therefore 2b+5d=4\Rightarrow 4-8d+5d=4$$$$\displaystyle \Rightarrow -3d=0$$$$\displaystyle \Rightarrow d=0$$$$\displaystyle \therefore b=2-4\left( 0 \right) =2$$Thus, $$a=1, b=2, c=-2, d=0$$Hence, the required matrix $$X$$ is $$\displaystyle \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$$MathematicsNCERTStandard XII
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https://math.stackexchange.com/questions/1409082/average-height-of-a-point-on-an-arc-vs-hemisphere | # average height of a point on an arc vs hemisphere
Why isn't the average height of a point on an arc of radius a the same as the average height on a surface of radius a.
Stated another way the first problem is: Find the average height of a point on a unit semicircle with respect to the arc length θ.
The second is: Find the average height above the xy-plane of a point chosen at random on the surface of the hemisphere $x^2 + y^2 + z^2 = a^2$.
I thought both questions should have the same answer but.
For arc $$\bar{y} = \frac{1}{\pi}\int_0^\pi a\sin\theta \ \mathrm{d}\theta = \frac{2a}{\pi}$$
For surface
\begin{align} \bar{z} &= \frac{\int\int_S z\ \mathrm{d}S}{\int\int_S \mathrm{d}S}\\ &= \frac{\int_0^{2\pi}\int_0^{\pi/2}a\cos\phi\ a^2\sin\phi\ \mathrm{d}\phi\ \mathrm{d}\theta}{\int_0^{2\pi}\int_0^{\pi/2}a^2\sin\phi\ \mathrm{d}\phi\ \mathrm{d}\theta} \\ &= \frac{a}{2} \end{align}
I'v posted a link which covers what I mean by average height for an arc as well as problem set(sol) where I originally encountered the half-sphere problem. Check it out as they might explain the problems better.
average-height-on-arc
look for question 6B-12
half-hemisphere | 2019-05-24T00:45:00 | {
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https://math.stackexchange.com/questions/3729194/laurent-expansion-of-frac1zz-1-about-0-using-integral-form-of-coeffic/3729218#3729218 | # Laurent Expansion of $\frac{1}{z(z-1)}$ about $0$ using integral form of coefficients
I've seen many questions already on Laurent Expansions and how to find examples like my question, however, most of them turn to geometric series and partial fractions, which I understand can be faster but when under time pressure, sometimes I just don't see the trick they want me to use and I would like to know that I always have a backup in a definition that should always work.
However, using this definition I keep getting stuck and most of the time my answer is just not correct. This particular problem is about $$f(z)=\frac{1}{z(z-1)}$$ where we want to express $$f$$ as a Laurent series around $$0$$ for $$|z|>1$$.
We know that $$f(z)=\frac{1}{z(z-1)}=\sum_{-\infty}^\infty a_nz^n$$ for some coefficients $$a_n$$, where we use the Laurent series around $$0$$. Both $$0$$ and $$1$$ are simple poles, and for any $$r>1$$, both poles lie in the interior of de circle around $$0$$ of radius $$r$$.
We look at the open annulus of $$1 where $$f$$ is holomorphic. Then we know that the $$a_n$$ from above are given by:
$$a_n=\frac{1}{2\pi i}\int_{|z|=r_0}\frac{f(z)}{z^{n+1}}dz= \frac{1}{2\pi i}\int_{|z|=r_0}\frac{1}{z^{n+2}(z-1)}=: \frac{1}{2\pi i}\int_{|z|=r_0}g(z),$$ where $$|z|=r_0$$ is any circle in the open annulus ($$r, on which we integrate in a positive direction.
Both poles $$0$$ and $$1$$ are inside this circle of radius $$r_0$$, and $$g$$ is holomorphic on the interior of this circle except at these poles. Therefore, we can use the residue sum theorem for $$g$$ to calculate the above integral, where $$0$$ is a pole of order $$n+2$$ and $$1$$ is a simple pole:
$$Res(g,0)=\lim_{z\to 0}\frac{d^{n+1}}{dz^{n+1}}\Big(\frac{z^{n+2}}{z^{n+2}(z-1)(n+1)!}\Big)= \frac{1}{(n+1)!}\lim_{z\to 0}\Big(\frac{(-1)^{n+1}(n+1)!}{(z-1)^{n+2}}\Big)=-1$$
$$Res(g,1) = \lim_{z\to 1}(z-1)\frac{1}{z^{n+2}(z-1)}=1$$
By the Residue Sum Theorem, the integral is equal to $$2\pi i*(Res(g,0) +Res(g,1)) = 0$$.
This however would lead to the $$a_n$$ being all 0, which is not the case for $$n<-1$$.
I saw the answer model which said that with a trick using geometric series you will get the answer $$\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}...$$. So clearly the $$a_n$$ are in fact $$0$$ for all $$n>-2$$, but this method does not provide me with $$a_n$$ for $$n<-1$$.
The first part of this question was about $$|z|<1$$ where you could use the geometric series directly; I did use this method as well when solving the first question. In that case, only the pole $$0$$ was inside the region of integration, and the residue, and thus the $$a_n$$, was equal to $$-1$$. This answer was correct, but now for all $$n>-2$$, and wrong for all $$n<-1$$ (as it's about a simple pole, the $$a_n$$ should logically be $$0$$ anyway for $$n<-1$$).
Somehow I seem to calculate the $$a_n$$ only for one half of the series, whereas the coefficients are incorrect for the other half. Where am I doing anything wrong in the steps above?
Edit: Finally I have maybe found what could be the problem; the calculation I used for residues uses $$n+1$$'th derivatives. As soon as $$n<-1$$, this calculation is no longer valid and the formula used for a general derivative of $$1/z$$ is no longer what I say it is. How to proceed now?
• What's wrong with just writing down the geometric series? One has the expansion $\sum_{k=2}^\infty z^{-k}$, so $a_n=1$ for $n\le-2$ and $a_n=0$ for $n\ge-1$ Jun 21 '20 at 17:44
• @AnginaSeng Please read my first paragraph; in this case I could figure it out, I just want the other described method to always work as it's not using any 'tricks'. I'm using this simple example so I can understand the concept, on an exam they might ask something a bit harder in which case I might not see how I can find an already know power series, and I would like to have a backup plan.
– Marc
Jun 21 '20 at 17:47
• You are belittling the geometric series by dismissing it as a "trick" (your quotation marks). I find that when I have a simple method to solve a problem, it is best to use it, rather than try a more complex method which is both time-consuming and error-prone. Here the simple method is just to write down the series from one's knowledge of the geometric series. The more complex method is to use contour integration; it is easy to fall into error thereby. Jun 21 '20 at 18:01
• @AnginaSeng I might have phrased it wrongly; by no means am I trying to say the the method I'm attempting is better than using the geometric series. I promise, if I see how to use it in an exam I will definitely take the shorter, easier and cleaner route. However, as I said, it's happened to me multiple times that I just don't see the creative or clever way in which a problem is asking me to think, and the last resort is using definition and build it from the ground up to see where you get.
– Marc
Jun 21 '20 at 18:15
When $$n\le -2$$, $$-n\ge2$$, $$\int_C \frac{f(z)}{z^{n+1}}\,dz=\int_C\frac{z^{-n-2}\,dz}{z-1}$$ and the integrand has no pole at $$z=0$$, so only the pole at $$z=1$$ counts, and there the residue is $$1$$. | 2021-10-18T03:23:04 | {
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http://mathhelpforum.com/calculus/43697-how-do-you-represent-cos-power-form.html | # Math Help - how do you represent cos in power form?
1. ## how do you represent cos in power form?
for example, $cos(m\pi)$ = $(-1)^m$ because cos alternates between 1 and -1 every $m\pi$.
so how can you represent $cos(\frac{m\pi}{2})$ in power form?
this one is a bit confusing to me because when:
m=1, $cos(\frac{m\pi}{2})=cos(\frac{1\pi}{2})$= 0
m=2, $cos(\frac{m\pi}{2})=cos(\frac{2\pi}{2})$= -1
m=3, $cos(\frac{m\pi}{2})=cos(\frac{3\pi}{2})$= 0
m=4, $cos(\frac{m\pi}{2})=cos(\frac{4\pi}{2})$= 1
2. Originally Posted by mathfied
for example, $cos(m\pi)$ = $(-1)^m$ because cos alternates between 1 and -1 every $m\pi$.
so how can you represent $cos(\frac{m\pi}{2})$ in power form?
this one is a bit confusing to me because when:
m=1, $cos(\frac{m\pi}{2})=cos(\frac{1\pi}{2})$= 0
m=2, $cos(\frac{m\pi}{2})=cos(\frac{2\pi}{2})$= -1
m=3, $cos(\frac{m\pi}{2})=cos(\frac{3\pi}{2})$= 0
m=4, $cos(\frac{m\pi}{2})=cos(\frac{4\pi}{2})$= 1
First off you have to state that $m\in\mathbb{N}$
But how about...hmm...Let me think here...you need...ok
Well after some thought I have come up with
$\cos\left(\frac{n\pi}{2}\right)=\frac{1}{2}(-1)^{\frac{n(n+1)}{2}}\bigg[(-1)^n+1\bigg]\quad{n\in\mathbb{N}}$
3. I will include my thought process as well..
$\cos m\pi = (-1)^m = 2\cos^2\left(\frac{m\pi}2\right) - 1$
$\cos^2\left(\frac{m\pi}2\right) = \frac{(-1)^m + 1}2$
$\cos\left(\frac{m\pi}2\right) = \pm\sqrt{\frac{(-1)^m + 1}2}$
Now we see that when an even m is a multiple of 4, its a + else it is a -
We can capture this idea by $(-1)^{\frac{m}2}$
When m is odd, anyway the expression is 0. The sign wont matter then. Thats why I choose $(-1)^{\lfloor\frac{m}2\rfloor}$
Where I have used the floor function on $\frac{m}2$.
Thus $\cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\sqrt{\frac{(-1)^m + 1}2}$
Note that functionally, $\sqrt{\frac{(-1)^m + 1}2} = \frac{(-1)^m + 1}2$
Thus a simplified form is:
$\cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\left(\frac{(-1)^m + 1}2\right)$
4. Originally Posted by Isomorphism
I will include my thought process as well..
$\cos m\pi = (-1)^m = 2\cos^2\left(\frac{m\pi}2\right) - 1$
$\cos^2\left(\frac{m\pi}2\right) = \frac{(-1)^m + 1}2$
$\cos\left(\frac{m\pi}2\right) = \pm\sqrt{\frac{(-1)^m + 1}2}$
Now we see that when an even m is a multiple of 4, its a + else it is a -
We can capture this idea by $(-1)^{\frac{m}2}$
When m is odd, anyway the expression is 0. The sign wont matter then. Thats why I choose $(-1)^{\lfloor\frac{m}2\rfloor}$
Where I have used the floor function on $\frac{m}2$.
Thus $\cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\sqrt{\frac{(-1)^m + 1}2}$
Note that functionally, $\sqrt{\frac{(-1)^m + 1}2} = \frac{(-1)^m + 1}2$
Thus a simplified form is:
$\cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\left(\frac{(-1)^m + 1}2\right)$
Nice!
I should probably include my thought process as well.
I did not work off of the trig, rather I worked off of the sequence $0,-1,0,1\cdots$
I firstly saw that
$\frac{1}{2}\left((-1)^n+1\right)$ gave $0,1,0,1$
So I wanted then that for even value of n that ever other was positive.
So I saw that this was described by $(-1)^{\frac{n(n+1)}{2}}$
A little less eloquent than Iso
5. Another nice formula is: $\cos\left(\frac{n\,\pi}2\right)=\frac12\big(i^n+(-i)^n\big)$ where $i=\sqrt{-1}, n \in \mathbb N$.
Also, the generating function is $\frac1{1+x^2}$.
Edit: Also $\cos\left(\frac{n\,\pi}2\right)=\Re\{i^n\}$ | 2015-07-04T06:04:42 | {
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http://mathhelpforum.com/algebra/98821-expanding.html | 1. ## Expanding
Hello,
1) (a+b)^6
2) (1-x)^8
Regards,
2. Apply the Binomial Theorem.
3. Originally Posted by rel85
Hello,
1) (a+b)^6
2) (1-x)^8
Regards,
Recall Pascal's Triangle:
Code:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
So when you go to expand $(a+b)^6$, the coefficients of your terms will be from the 7th row of Pascal's Triangle.
Code:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1 **
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
Thus, it follows that $(a+b)^6=1\cdot a^6b^0+6\cdot a^5b^1+15\cdot a^4b^2+20\cdot a^3b^3+15\cdot a^2b^4+6\cdot a^1b^5+1\cdot a^0b^6$
A similar process is done for $\left(1-x\right)^8=\left(1+\left(-x\right)\right)^8$. You just look at the 9th row of the Pascal's Triangle for the coefficients.
Can you try the last one?
4. Originally Posted by Chris L T521
Recall Pascal's Triangle:
Code:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
So when you go to expand $(a+b)^6$, the coefficients of your terms will be from the 7th row of Pascal's Triangle.
Code:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1 **
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
Thus, it follows that $(a+b)^6=1\cdot a^6b^0+6\cdot a^5b^1+15\cdot a^4b^2+20\cdot a^3b^3+15\cdot a^2b^4+6\cdot a^1b^5+1\cdot a^0b^6$
A similar process is done for $\left(1-x\right)^8=\left(1+\left(-x\right)\right)^8$. You just look at the 9th row of the Pascal's Triangle for the coefficients.
Can you try the last one?
Okay I see it. How do I go about using that neat notation? Otherwise the answer is going to be extremely messy!
Can the first answer be simplified, or would the answer you gave, be the final step? Is there any other 'types' of binomial question that requires a different method?
Regards,
5. Originally Posted by rel85
Okay I see it. How do I go about using that neat notation? Otherwise the answer is going to be extremely messy!
Can the first answer be simplified, or would the answer you gave, be the final step? Is there any other 'types' of binomial question that requires a different method?
Regards,
You could do one more step to get $\left(a+b\right)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a ^2b^4+6ab^5+b^6$
With regards to neat notation, do you mean the math typesetting? If so, we use $\text{\LaTeX}$ here on the forums. See the LaTeX Help subforum (in particular, the two LaTeX tutorials) to learn the basics.
With regards to other binomial type of questions, there are some variations that ask you to exand things like $\left(a+b\right)^{r};\,\, r\in\mathbb{C}$ (r doesn't have to be an integer power; it can be complex [or real]). You can read more about it here: Binomial theorem - Wikipedia, the free encyclopedia
6. Originally Posted by Chris L T521
You could do one more step to get $\left(a+b\right)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a ^2b^4+6ab^5+b^6$
With regards to neat notation, do you mean the math typesetting? If so, we use $\text{\LaTeX}$ here on the forums. See the LaTeX Help subforum (in particular, the two LaTeX tutorials) to learn the basics.
With regards to other binomial type of questions, there are some variations that ask you to exand things like $\left(a+b\right)^{r};\,\, r\in\mathbb{C}$ (r doesn't have to be an integer power; it can be complex [or real]). You can read more about it here: Binomial theorem - Wikipedia, the free encyclopedia
Thank you so much, specifically I was wondering if there was any subtract questions... or can they all be manipulated from $(a - b)^x = (a + (-b)^x$
With regards to the second attempt:
$(1 + (-x))^8$
9th line = 1, 9, 34, 84, 126, 126, 84, 34, 9, 1
so: $1 . 1^8(-x)^0 + 1 . 9^7(-x)^1 + 1 . 34^6(-x)^2 + 1 . 84^5(-x)^3 + 1 . 126^4(-x)^4$
$+ 1 . 126^3(-x)^5 + 1 . 84^2(-x)^6 + 1 . 34^1(-x)^7 + 1 . 9^0(-x)^8 + 1$
Is this okay?
7. Originally Posted by Chris L T521
Recall Pascal's Triangle:
Code:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
So when you go to expand $(a+b)^6$, the coefficients of your terms will be from the 7th row of Pascal's Triangle.
Code:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1 **
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
Thus, it follows that $(a+b)^6=1\cdot a^6b^0+6\cdot a^5b^1+15\cdot a^4b^2+20\cdot a^3b^3+15\cdot a^2b^4+6\cdot a^1b^5+1\cdot a^0b^6$
A similar process is done for $\left(1-x\right)^8=\left(1+\left(-x\right)\right)^8$. You just look at the 9th row of the Pascal's Triangle for the coefficients.
Can you try the last one?
Originally Posted by rel85
Thank you so much, specifically I was wondering if there was any subtract questions... or can they all be manipulated from $(a - b)^x = (a + (-b)^x$
With regards to the second attempt:
$(1 + (-x))^8$
9th line = 1, 9, 34, 84, 126, 126, 84, 34, 9, 1
so: $1 . 1^8(-x)^0 + 1 . 9^7(-x)^1 + 1 . 34^6(-x)^2 + 1 . 84^5(-x)^3 + 1 . 126^4(-x)^4$
$+ 1 . 126^3(-x)^5 + 1 . 84^2(-x)^6 + 1 . 34^1(-x)^7 + 1 . 9^0(-x)^8 + 1$
Is this okay?
You used the coefficients from the tenth row!
"1" is the first row, "1 1" is the second, "1 2 1" is the third, and so on.
The ninth line is "1 8 28 56 70 56 28 8 1"
So it follows that $\left(1+\left(-x\right)\right)^{8}=1\cdot1^8\left(-x\right)^0+8\cdot1^7\left(-x\right)^1+28\cdot1^6\left(-x\right)^2+56\cdot1^5\left(-x\right)^3+70\cdot1^4\left(-x\right)^4$ $+56\cdot1^3\left(-x\right)^5+28\cdot1^2\left(-x\right)^6+8\cdot1^1\left(-x\right)^7+1\cdot1^0\left(-x\right)^8$
That simplifies to $1-8x+28x^2-56x^3+70x^4-56x^5+28x^6-8x^7+x^8$.
8. Originally Posted by Chris L T521
You used the coefficients from the tenth row!
"1" is the first row, "1 1" is the second, "1 2 1" is the third, and so on.
The ninth line is "1 8 28 56 70 56 28 8 1"
So it follows that $\left(1+\left(-x\right)\right)^{8}=1\cdot1^8\left(-x\right)^0+8\cdot1^7\left(-x\right)^1+28\cdot1^6\left(-x\right)^2+56\cdot1^5\left(-x\right)^3+70\cdot1^4\left(-x\right)^4$ $+56\cdot1^3\left(-x\right)^5+28\cdot1^2\left(-x\right)^6+8\cdot1^1\left(-x\right)^7+1\cdot1^0\left(-x\right)^8$
That simplifies to $1-8x+28x^2-56x^3+70x^4-56x^5+28x^6-8x^7+x^8$.
Careless of me! | 2016-09-28T22:46:19 | {
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http://mathonline.wikidot.com/for-normed-algebras-with-unit-the-unit-can-be-assumed-to-hav | For Normed Alg. with Unit, the Unit Can be Assumed to Have Norm 1
# For Normed Algebras with Unit, the Unit Can be Assumed to Have Norm 1
Recall that if $\mathfrak{A}$ is an algebra then $\mathfrak{A}$ is said to have a unit (or be unital) if there exists an element $e \in \mathfrak{A}$ such that $ea = ae = e$ for every $a \in \mathfrak{A}$.
If $\mathfrak{A}$ is a normed algebra with unit, then since $e = ee$, we have that:
(1)
\begin{align} \quad \| e \| = \| ee \| \leq \| e \| \| e \| \end{align}
Since $e \neq 0$, the above inequality implies that $\| e \| \geq 1$. For practical purposes it would be nice if $\| e \| = 1$. However, this is not true in general. Fortunately, $\mathfrak{A}$ can always be equipped with a new equivalent norm for which $\| e \| = 1$.
Proposition 1: Let $(\mathfrak{A}, \| \cdot \|)$ be a normed algebra with unit $e$. Then there exists an algebra norm $\| \cdot \|_*$ on $\mathfrak{A}$ that is equivalent to $\| \cdot \|$ and such that $\| e \|_* = 1$.
• Proof: For each $a \in \mathfrak{A}$ let $T_a : \mathfrak{A} \to \mathfrak{A}$ be defined for all $b \in \mathfrak{A}$ by:
(2)
\begin{align} \quad T_a(b) = ab \end{align}
• Note that for each $a \in \mathfrak{A}$, $T_a$ is linear since for all $b_1, b_2 \in \mathfrak{A}$ and for all $\alpha \in \mathbf{F}$ we have that:
(3)
\begin{align} \quad T_a(b_1 + b_2) = a(b_1 + b_2) = ab_1 + ab_2 = T_a(b_1) + T_a(b_2) \end{align}
(4)
\begin{align} \quad T_a(\alpha b_1) = a(\alpha b_1) = \alpha (ab_1) = \alpha T_a(b_1) \end{align}
• Furthermore, for each $a \in \mathfrak{A}$ we have that $T_a$ is bounded with $\| T_a \| \leq \| a \|$ since:
(5)
\begin{align} \quad \| T_a(b) \| = \| ab \| \leq \| a \| \| b \|, \quad \forall b \in \mathfrak{A} \end{align}
• Now note that the map from $\mathfrak{A}$ to $\mathrm{BL}(\mathfrak{A}, \mathfrak{A})$ defined by $a \to T_a$ is injective. Indeed, if $a_1, a_2 \in \mathfrak{A}$ are such that $T_{a_1}(b) = T_{a_2}(b)$ for all $b \in \mathfrak{A}$ ]] then [[$a_1b = a_2b$ for all $b \in \mathfrak{A}$. So $a_1b - a_2b = (a_1 - a_2)b = 0$ for all $b \in \mathfrak{A}$. In particular, since $\mathfrak{A}$ is an algebra with unit, we have that $(a_1 - a_2)e = 0$. So $a_1 - a_2 = 0$, i.e., $a_1 = a_2$.
• For each $a \in \mathfrak{A}$ let:
(6)
\begin{align} \quad \| a \|_* := \| T_a \| \end{align}
• Note that $\| \cdot \|_*$ is well-defined by the injectivity of the map $a \to T_a$. Moreover, $\| a \|_* \leq \| a \|$. For the reverse inequality, observe that:
(7)
\begin{align} \quad \| a \|_* = \| T_a \| = \sup_{\| b \| \leq 1} \{ \| T_a(b) \| \} = \sup_{\| b \| \leq 1} \{ \| ab \| \} \geq \left \| a \frac{e}{\| e \|} \right \| = \frac{\| a \|}{\| e \|} \end{align}
• Let $c = \frac{1}{\| e \|}$ and let $C = 1$. Then for all $a \in \mathfrak{A}$ we have that:
(8)
\begin{align} \quad c \| a \| \leq \| a \|_* \leq C \| a \| \end{align}
• So $\| \cdot \|_*$ is norm equivalent to $\| \cdot \|$. Furthermore, we have that:
(9)
\begin{align} \quad \| e \|_* = \| T_e \| = \sup_{\| b \| \leq 1} \{ \| T_e(b) \| \} = \sup_{\| b \| \leq 1} \{ \| eb \| \} = \sup_{\| b \| \leq 1} \{ \| b \| \} = 1 \end{align}
• Lastly, we see that $\| \cdot \|_*$ is an algebra norm since for all $a, b \in \mathfrak{A}$ we have that:
(10)
\begin{align} \quad \| ab \|_* = \| T_{ab} \| = \| T_a \circ T_b \| \leq \| T_a \| \| T_b \| = \| a \|_* \| b \|_* \end{align}
• So $\| \cdot \|_*$ is an algebra norm that is equivalent to $\| \cdot \|$ and such that $\| e \|_* = 1$. $\blacksquare$ | 2021-10-17T09:39:32 | {
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https://math.stackexchange.com/questions/2858658/geometry-prove-that-two-angles-are-not-equal/2858673 | # Geometry: Prove that two angles are not equal
(This is just a question for fun. I saw a commercial logo today and I was inspired.
I have posted answers for this question and you may post alternative answers!)
Question
In the figure, $$\triangle ABC$$ is half of a square and $$M$$ is the midpoint of $$BC$$. Prove that $$\alpha\neq\beta$$.
Solution
$$\triangle ABM$$ and $$\triangle AMC$$ have the same area. They have a common side $$AM$$. Note that the area of either triangle is given by $$S=\frac12(AB)(AM)\sin\alpha=\frac{1}{2}(AC)(AM)\sin\beta$$. But $$AB\neq AC$$. So the equality holds only if $$\alpha\neq\beta$$.
• This is an illustration that the tangent function is nonlinear. – user65203 Jul 21 '18 at 17:03
• Fun question! Thanks for sharing :) – Sambo Jul 21 '18 at 17:09
By contradiction: if the angles were equal, then $BM:MC=AB:AC$ (angle bisector theorem) and as a consequence $BM>MC$, which is false.
A mostly visual proof:
Construct a second square with edge $BC$ as shown. Since $AC=BD$, $CM=BM$, and $\angle ACM$ and $\angle DBM$ are right angles, $\triangle ACM \cong \triangle DBM$.
Therefore $\angle CMA \cong \angle BMD$, so $A$, $M$, and $D$ are collinear. And $m\angle BDM = m\angle CAM = \beta$.
Now if $\alpha = \beta$, then $\triangle ABD$ is isosceles, with $AB = BD$. But this is plainly false since $AB = BD \sqrt{2}$. So $\alpha \neq \beta$.
• Beautiful! Anyone can understand immediately! – Sambo Aug 17 '18 at 20:45
$$\hspace{5cm}$$
If $$\alpha=\beta$$, then: $$1=\tan 45^\circ=\tan(\alpha+\beta)=\tan(2\beta)=\frac{2\tan\beta}{1-\tan^2\beta}=\frac{2\cdot \frac12}{1-\left(\frac12\right)^2}=\frac43,$$ hence a contradiction. So, $$\alpha\ne \beta$$.
• Nice! Seems that with your method we can generalize the question when $BC$ is divided into $n$ equal segments. – Mythomorphic Jul 22 '18 at 7:07
• Nice observations both on commercial logo and on generalization! – farruhota Jul 22 '18 at 12:10
• Haha, thank you! Not an advertisement, but the name of the company is Convoy. Take a look of it lol – Mythomorphic Jul 22 '18 at 12:15
• I believe you should clarify that you're calculating tan β by dividing the edge sizes. Because for a moment it seemed that you replaced tan(22.5°) with 1/2 which is false :) – Pedro A Jul 22 '18 at 21:20
Construct the circumcircle of $\triangle AMC$ and translate $\triangle BMA$ such that $BM$ coincides with $MC$. If the two angles were equal, $A'$ (the third vertex of the translated triangle) would lie on the circle, but it does not. Thus the two angles are not equal.
• Parcly.Very nice. – Peter Szilas Jul 21 '18 at 19:12
Let $C'$ be the image of the orthogonal projection of $M$ onto $AB$. Suppose contrary that $\alpha=\beta$. Then, $MCA$ and $MC'A$ are congruent triangles. Thus, $MB=MC'$ and there is a contradiction here, which I will leave it as a mystery. In fact, you can use a similar argument to show that $\alpha<\beta$.
Let line MD be constructed parallel to AC, with D on AB, so angle DMA = MAC. Now DM < DA, because DM < BD (Pythagorean Th.). But in any triangle, a greater side lies opposite a greater angle. Therefore angle DMA (that is, MAC) > BAM.
From the diagram we have $$\alpha + \beta = \pi/4$$ If $\alpha = \beta$ this would give $\alpha = \beta = \pi / 8$.
Again from the diagram: Let the side of the square be $x$, then $$\tan \beta = (x/2) / x = 1/2$$ However $$\tan(\pi/8) = \sqrt{2}-1 < 1/2$$ So $\beta \ne \pi/8$ and $a \ne \beta$.
Appendix:
The addition theorem for the tangent yields $$\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$$ So $$\underbrace{\tan(2\cdot \pi/8)}_1 = \frac{2\tan(\pi/8)}{1-\tan^2(\pi/8)}$$ and we have the quadratic equation in $u = \tan(\pi/8)$: \begin{align} 1-u^2 &= 2 u \iff \\ 1 &= u^2 + 2u \iff \\ 2 &= u^2 + 2u + 1 = (u+1)^2 \iff \\ u &= \pm\sqrt{2}-1 \end{align} where we need the positive solution.
Drop a perpendicular from $M$ to the line $AB,$ meeting $AB$ at $P.$ That is, $P$ is the point on line $AB$ closest to $M.$
Since $\angle ABM$ is not a right angle, $P$ is not $B,$ and therefore $MP < BM.$ Since $BM = CM,$ then $MP < CM.$
If $\alpha = \beta$ then the two right triangles $\triangle AMC$ and $\triangle AMP$ would be congruent, since all three angles and one side ($AM$) would be congruent. But since $MP < CM,$ the triangles are not congruent. Hence $\alpha \neq \beta.$
By the law of sines, $$\sqrt{2}\frac{\sin(\alpha)}{MB}=\frac{1}{MA}=\frac{\sin(\beta)}{MC}.$$ Since $MC=MB$, we conclude that $\sqrt{2}\sin(\alpha)=\sin(\beta)$. Clearly neither $\alpha$ nor $\beta$ are integer multiples of $\pi$, so $\alpha\neq\beta$.
First assume that $\alpha$ = $\beta$ and that both would be equal to $\frac{\pi}{8}^\circ$ (since $\alpha = \beta$ and $\alpha + \beta = \frac{\pi}{4} \to 2\beta = \frac{\pi}{4} \to \beta = \frac{\pi}{8}$).
If we call side $\overline{AC}$, n, then side $\overline{MC}$ is $\frac{n}{2}$. Next, we solve for $\overline{AM}$ in terms of n to show that $\alpha = \beta$ for all values n.
$\overline{AM}^2 = n^2 + (\frac{n}{2})^2 = n^2 + \frac{n^2}{4} = \frac{4n^2 + n^2}{4} = \frac{5n^2}{4}$
$\overline{AM} = \sqrt{\frac{5n^2}{4}} = \frac{\sqrt{5n^2}}{2} = \frac{n\sqrt{5}}{2}$ by the pythagorean theorem
Now that we have $\overline{AM}$, we can find $cos{\beta}$.
$\cos{\beta}=\frac{n}{\overline{AM}} = \frac{n}{\frac{n\sqrt{5}}{2}} = \frac{2n}{n\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$
$\cos{\beta} = \frac{2\sqrt{5}}{5}$
We have that $\cos{\beta} = \frac{2\sqrt{5}}{5}$ and we already established that $\beta = \frac{\pi}{8}$.
This would mean that $\cos{\frac{\pi}{8}} = \frac{2\sqrt{5}}{5}$. We can find $\cos{\frac{\pi}{8}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$, which implies that $\frac{\sqrt{2 + \sqrt{2}}}{2} = \frac{2\sqrt{5}}{5}$.
If we square both sides, we get that $\frac{2+\sqrt{2}}{4} = \frac{20}{25} = \frac{4}{5}$. Then we can split the fraction, simplify, then subtract and simplify.
$\frac{2}{4} + \frac{\sqrt{2}}{4} = \frac{4}{5}$
$\frac{1}{2} + \frac{\sqrt{2}}{4} = \frac{4}{5}$
$\frac{\sqrt{2}}{4} = \frac{4}{5} - \frac{1}{2}$
$\frac{\sqrt{2}}{4} = \frac{3}{10}$
Next we can multiply both sides by 4 and conclude that...
$\sqrt{2} = \frac{12}{10} = \frac{6}{5}$
This is impossible since it has been proven that $\sqrt{2}$ is irrational and can't be written as a ratio of 2 integers. This would mean that $\sqrt{2} \neq \frac{6}{5}$ or $\frac{\sqrt{2 + \sqrt{2}}}{2} \neq \frac{2\sqrt{5}}{5}$ or $\cos{\frac{\pi}{8}}\neq \frac{2\sqrt{5}}{5}$
This means that $\beta \neq \frac{\pi}{8}$. Since $\alpha + \beta = \frac{\pi}{4}$, $\beta$ would have to be equal to $\frac{\pi}{8}$, but we proved that it cant be. This means that $\alpha \neq \beta$ since $\beta$ can no longer be multiplied by 2 to be $\frac{\pi}{4}$
This method works with tangent and sine as well (I believe it's a lot less work if I had used tangent instead of cosine, but this was how I first solved it)
I'm a high-school student and this is my first response, feedback is always appreciated!
The figure $\triangle ABC$ is the half of a square therefore the sides $\overline{AB}$ and $\overline{BC}$ are of equal length. Now consider the triangle $\triangle ABM$. It is a rectangular triangle such as the original one.
Therefore the tangens of the angle $\alpha$ is given by
$$\tan(\alpha)~=~\frac{\overline{BM}}{\overline{AB}}$$
hence $M$ is the middle of the side $\overline{BC}$ this simplifies to
$$\tan(\alpha)~=~\frac{\overline{BM}}{\overline{AB}}~=~\frac{\frac{\overline{BC}}{2}}{\overline{AB}}~=~\frac{\frac{\overline{BC}}{2}}{\overline{BC}}~=~\frac{1}{2}\\ \alpha~=~\arctan\left(\frac{1}{2}\right)$$
Now go back to the triangle $\triangle ABC$. The tangens of the angle $\alpha + \beta$ is given by
$$\tan(\alpha + \beta)~=~\frac{\overline{BC}}{\overline{AB}}$$
which equals $1$. The tangens of the sum of two different angles is given by
$$\tan(x+y)~=~\frac{\tan(x)+tan(y)}{1-\tan(x)\tan(y)}$$
with $x=\alpha$,$y=\beta$ and $\tan(\alpha)=\frac{1}{2}$ you get
\begin{align} \tan(\alpha+\beta)~=~1~=~\frac{\frac{1}{2}+\tan(\beta)}{1-\frac{1}{2}\tan(\beta)}~\iff~1-\frac{1}{2}\tan(\beta)~&=~\frac{1}{2}+\tan(\beta)\\ \tan(\beta)&=\frac{1}{3}\\ \beta&=\arctan\left(\frac{1}{3}\right) \end{align}
Hence $\arctan\left(\frac{1}{2}\right)\neq\arctan\left(\frac{1}{3}\right)$ and so $\alpha\neq\beta$.
Here's one intuitive way to analyze the problem:
If you're standing on point $A$ and looking at a car driving at constant speed on a straight line $BC$, you'll have to turn your head very fast when the car is closest to you (on point $C$) and slow the head rotation as the car drives away.
Assume $\alpha$ and $\beta$ are both equal to $22.5^\circ$ (sum to forty-five).
The angle $\angle ABM$ must equal $45^\circ$ because it bisects the square.
The angle $\angle ACM$ occupies a square corner, so it must equal $90^\circ$.
$\alpha + \angle BMA$ must not sum to more than $135^\circ$ ($180^\circ$ $-$ $45^\circ$ from $\angle ABM$).
If alpha is equal to beta and both equal $22.5^\circ$, then BMA must equal $112.5$ degrees ($135^\circ - 22.5^\circ$).
BMA + AMC must equal less than $180$ degrees (one hemicircle); if $\angle BMA = 112.5^\circ$, then $\angle AMC$ must be $67.5^\circ$.
Given $\beta = \alpha = 22.5^\circ$, $\angle AMC + \angle ACM$ must equal $157.5^\circ$ ($180$ minus $\beta$); given the same assumption, $\angle AMC$ is equal to $67.5^\circ$ and sums with $\angle ACM$ to $147.5^\circ$.
Alpha must not be equal to beta.
In the figure $$\alpha+b=m \tag{1}$$ Also, $$m+\beta=90^0\tag{2}$$ So from $$(1)$$ and $$(2)$$, $$\alpha+b=90^0-\beta\tag{3}$$ But, $$b=\alpha+\beta$$ So $$(3)$$ becomes $$2\alpha=90^0-2\beta$$ $$\implies \alpha=45^0-\beta$$
So we can see that $$\alpha$$ is $$\beta$$ less than $$45^0$$ so $$\alpha$$ and $$\beta$$ aren't equal. | 2021-07-24T15:19:47 | {
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https://math.stackexchange.com/questions/3721067/how-to-know-when-can-i-use-proof-by-contradiction-to-prove-operations-with-irrat | # How to know when can I use proof by contradiction to prove operations with irrational/rational numbers?
I know that the proof by contradiction is useful when we know that there is only one answer and we can pick from two exclusive options. Therefore, if one doesn't work, we know the other is the correct answer. It seems to me that to use the proof I require, beforehand, the knowledge that the result can be only one option or the other.
When proving, that the sum of an irrational number and a rational number is irrational, we can use proof by contradiction to demonstrate that it can't be rational. Doesn't this assume that the sum can't be sometimes rational and other times irrational?
However, with the sum of two irrational numbers, the result can be either rational or irrational based on the chosen numbers. How can I deduce, beforehand, that the proof by contradiction won't be useful in this case?
• The exclusive alternative is usually that a statement is either true or false, so one will not necessarily have a fine-grained dichotomy like rational vs. irrational. Indeed a "direct" proof could always be recast as a proof by contradiction, so I doubt that one can give a rule for when "the proof by contradiction is useful". – hardmath Jun 15 '20 at 19:22
The trick is to use closure rules we know for rationals: they're closed under addition, subtraction, multiplication, and division if the quotient is nonzero. Let's discuss your examples:
• If $$i$$ is irrational and $$r$$ is rational, suppose $$i+r$$ is rational: then $$i=(i+r)-r$$ is rational, a contradiction.
• If $$i,\,j$$ are irrational, let $$k:=i+j$$; we can't write any one variable as a function of rationals, so an argument similar to the above is unavailable.
When we prove that the sum of a rational and an irrational number cannot be rational, we are certainly not assuming that the sum cannot be rational in some cases and irrational in others: we are proving that this is the case. This really has nothing to do specifically with proofs by contradiction; it just happens that in this case we prove that something is true by proving that its negation is false. And in fact it’s entirely possible to prove the result without using proof by contradiction.
Proposition: If $$x,y\in\Bbb R$$, and $$x$$ is rational, then $$x+y$$ is rational if and only if $$y$$ is rational.
Proof: If $$y$$ is rational, then $$x+y$$ is the sum of two rational numbers and is therefore rational. If $$x+y$$ is rational, then $$y=(x+y)+(-x)$$ is the sum of two rational numbers, so $$y$$ is rational. $$\dashv$$
Knowing when a proof by contradiction or a proof of the contrapositive is likely to be a good way to proceed is in large part a matter of experience, but there is at least one moderately useful guide: if the hypotheses available for a proof of the contrapositive or a proof by contradiction seem to give you more to work with — more specific details, for instance — than the hypotheses available for a direct proof, then it’s definitely worth your while to consider an indirect proof.
• But using the proof by contradiction to prove it takes the premise that the result can only be rational or irrational for all the cases, right? I mean, could be the case that a proof by contradiction wouldn't work for something because this premise of either one option or the other isn't right? – Jon Jun 15 '20 at 20:25
• @Jon: No, it doesn’t. It simply proves that the result is irrational in all cases. If there were some cases in which the sum was rational and others in which it was irrational, the result would simply be false, and no attempt to prove it, by any method whatsoever, would work. This has nothing to do with the method of proof that you’re attempting to use. – Brian M. Scott Jun 15 '20 at 20:29 | 2021-04-18T04:47:20 | {
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http://templgvinstructorregister.com.rtitb.info/jhjxwl/exponential-function-transformations-rules.php | # Exponential function transformations rules
## Exponential function transformations rules
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Exponential Functions Definition of the Exponential Function Text Example Characteristics of Exponential Functions Transformations Involving Exponential Functions Students will explore the family of exponential functions of the form f(x) = c * b x+a and be able to describe the effect of each parameter on the graph of y Overview of the exponential function and a few Basic rules for exponentiation; Exponential growth and decay “The exponential function. pdf #4 stretching, transformation, function 51 minutes students will review exponent rules as they complete these Section 7. The same rules apply when transforming logarithmic and exponential Problem: Using the enclosed Java applet, explore graphically the effect of changing the coefficients a, b, c, and d in the exponential function Here are some algebra rules for exponential functions that will be explained in class. 8-7 Radical Functions 619 and exponential functions. Graphing an exponential function is helpful when you want to visually analyze the function. f(x) Determine the exponential function given the graph below. Exponential Functions, Exponent Rules, and Factoring. Learn vocabulary, terms, and more with flashcards, games, and other study tools. 2 Graphs of Exponential Functions the same rules as any function, is the only transformation of an exponential function that changes the Maze - Transformation of Exponential Functions. a. a) 4. We found that by adding a value either to the input or output of the original function we could translate the function left/right or up/down. Graphing Transformations of Logarithmic Functions. So far, we have been dealing with algebraic functions. . Exponential Functions Learn algebra using 19 graph-related activities on four key topics: linear equations, quadratic equations, transformations of functions and exponential functions. Page Navigation. An asymptote is a line that a function gets closer and closer to, but never reaches. 3 Exponential Functions . Visualization: f(x) = a eb (x - c) + d. GRAPHS OF EXPONENTIAL FUNCTIONS By Nancy Marcus In this section we will illustrate, interpret, and discuss the graphs of exponential functions. Exponential Functions Graphs an exponential function using coefficients Computes derivatives symbolically using standard rules, Combinations and Transformations of Functions. Graph exponential functions using transformations; Key Concepts; Use the product rule for logarithms; Transformations of Parent Functions functions such as cube root, exponential and logarithmic functions. up vote 2 down vote favorite. In order to use the transformation rules the x must come There is a negative inside the function, so we need to use rule 6 11 Exponential and Logarithmic Functions Worksheet Concepts: • Rules of Exponents • Exponential Functions – Power Functions vs. Applications of Exponential Functions Exponential function. To practice with an EXCEL model of these graphs, click Section 3: Transformations of Exponential and. Unit 4 Exponential and Logarithmic Functions exponential and logarithmic functions. Transforming exponential graphs (example 2) Site Navigation. Exponential Transformation Rules Exponential Transformations Worksheet Transformations of Exponential Functions Part 1 - YouTube www | 2018-01-16T19:21:41 | {
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https://smartclinic2.diapindia.org/wp-content/uploads/2022/01/rhl9jwj/the-increase-of-speed-is-a-squared-relationship.html | The increase of speed is a squared relationship. The power is proportional to the speed cubed. The equation (9. Careful observations show that the electrostatic force between two point charges varies inversely with the square of the distance of separation between the two charges. 50 \times 10^3 m/s, determine the required ratio M_i/M_f of … The acceleration (a) equals the square of the wind speed in meters per second (m/s). ") AB-032. The speed of light, #c# is a constant, so when wavelength decreases, frequency increases and vice versa. Recall that a “rate” is any number divided by time. One can calculate momentum as mass multiplied with velocity. A) Both acceleration and speed increase. (Note: ) Theoretically, the centripetal force should be directly proportional to the square of the speed. and. Adjusted R Squared, however, makes use of the degree of freedom to compensate and penalize for the inclusion of a bad variable. Maneuvering speed is the fastest speed at which your plane will stall before exceeding its limit load factor if the angle of attack suddenly and dramatically increases. Since I was looking for an exponent (the exponent is -2 for an inverse square relationship) I decided to analyze the raw Excel data. He described the relationship between mass and energy accurately using his theory of relativity. Kinetic energy = 0. At the same temperature, lighter nitrogen molecules move faster than heavier chlorine molecules. 32 For each additional square kilometer of forested area added, the IBI will increase by 0. ω = rotational speed. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10 m/sec. cause the electrons to travel faster from cathode to anode. I. There are many real-life examples of inverse relationships. relationship ds=+ ⋅3. The torque required to turn a variable torque load goes up at a rate of the speed squared. Because distance is a scalar, speed (distance/time) is also a scalar. To give you some intuition as to why this is, consider what a wing does. It is significant (P=1. For a curve, the gradient varies at different points along the curve. The principle is described by the physicist Albert Einstein's famous formula: =. , much slower than the … For two quantities, x and y, an increase in x causes an increase in y as well. 13 1. As the centripetal acceleration increase (or gets more powerful), the velocity of the object also increases in proportion to the … Speed is equal to distance divided by time. An example of a negative correlation coefficient is the relationship between outdoor temperature and heating costs. By the way, according to Fan Law 2, the Static Pressure (SP) increases as the square of the speed ratio. 67 x 10-6: A 1-ft (0. 0. Null Hypothesis: There is no relationship between the two variables. 65) + $10,500 =$270,062. 00 cm/s. 0 meter. Head loss is proportional to speed squared. Gear Output Speed. is it true or not? You can calculate the acceleration of an object from its change in velocity and the time taken. 1, we talked about the relationship between student heart … The formula estimates that for each increase of 1 dollar in online advertising costs, the expected monthly e-commerce sales are predicted to increase by $171. create an increase in the number of electrons traveling from cathode to anode d. 7 A wheel 1. ∴ m = k/n 2. But KE can also be defined by … The relationship with R Squared and degrees of freedom is that R Squared will always increase as the degrees of freedom decreases which as we saw earlier drastically reduces the reliability of the model. The entire process is driven by applying electrical power to the coil, with the source voltage having a direct relationship to the motor’s output speed. A square wave in the frequency domain looks like a sum of odd frequencies: Figure 3. If a car traveled 30 miles in an hour, the speed would be 30 miles per hour. For a linear relationship, the gradient at any point along the line is the same. So doubling the speed of a vehicle means 8 times the engine power required to compensate for drag! This has e. More massive objects require bigger net forces to accelerate the same amount as less massive objects. Key properties of R-squared. Note that the most probable speed, ν p, is a little less than 400 m/s, while the root mean square speed, u rms, is closer to 500 m/s. The distance a batter needs to hit a baseball to get a home run depends on the stadium. Good afternoon If relationship between total cost and the number of units made it linear ,and if costs increases by 7$ for each additional unit made ,and if the total cost of 10 units is 180$. The relationship for speed is the same for any units, as long as the units are consistent. Understanding the coefficient b 0 : B 0 takes up the residual value for the model and ensures that the prediction is not biased. For larger values of M 2, the increase in % was insufÞcient to satisfy the condition %" M 2 /M 1, in which case M 1 contin-ued to accelerate until it reached the end of its travel. To put this into perspective, consider the example below. 50 m/s c. Distance, displacement, uniform motion, non-uniform motion, speed, acceleration and velocity are the terms that are used for So v is proportional to the square root of r. ) of kinetic energy gets measured in Joules. Motion is defined as the movement of an object from one place to another. Due to this all the gases escape from the surface of the Moon. For example, if we take a turn with a radius which is four times smaller, we need to cut our speed in half. A chi-square test is used when you want to see if there is a relationship between two categorical variables. A correlation coefficient is used to measure the strength of the relationship between numeric variables (e. Coefficient - Standard Error The relation between speed and crashes . It is analogous to kinetic … However, as the wind speed increase, this estimate has increase variance. So there is an inverse relationship between the force and radius,and direct proportionality between the force and velocity. · Most of the industrial loads can be classified into the following 4 general categories: 1. 16065 inches for every 1-year increase in the age of the tree. 50. 6 m after 2 seconds, how far does it fall after 3 seconds? We can use: d = kt 2. 18 57. These two values have an inverse relationship following the wave speed formula: Speed = Wavelength #*# Frequency. Gear Output Horsepower speed range. 8) = 526 rpm . Solution. Torque proportional to speed (generator type load) 3. Better emotional health leads to marriage. increase in x will result in an increase in y. 81 m/s^2 * time, or V = gt. • An a % increase in the speed causes a ‡ 2 ¯ a 2 100 Lift is created by deflecting a flow of air and drag is generated on a body in a wide variety of ways. For every 1-inch increase in a tree's diameter, its age is estimated to have increased by 0. The intensity of a sound wave is proportional to the change in the pressure squared and inversely proportional to the density and the speed. EDIT: I remember for certain that for very high Reynolds number you approach the velocity squared relationship. 8/5 (2,313 Views . It means: the kinetic energy is equal to half of the mass times the squared velocity. 0 kg. gain an interesting insight … 2. An increase of 1 mph in ball speed can lead to roughly 2 yards of increase in distance with your driver. An important feature of a relationship is whether the line goes through the origin (the point at which the values of x and y are zero). Multiply the resulting square rooted signal (still 0-1) by the flow scale span to get flow rate in engineering The relationship between the pump head and the volumetric flow rate (Q), • The pump head is directly proportional to the square of the pump speed: double the speed/multiply the pressure by four. 976 and there are 9 degrees of freedom, so the t s-statistic is 19. A rock is dropped from a bridge. Interpret the intercept: If the ball is hit with a speed of 0 mph, then the model predicts that the length the ball travels will be 3. A coil is placed in a magnetic field, and when an electric current passes through the coil, a torque is produced, causing the motor to turn. north B. Find the equation of the relationship b/n total cost (Y) & number of unit made (x). Now what's interesting here is, we see this speed we need in order to maintain this orbit in no way is it a function of the mass of this Over 20 fans the total power increase is 34. ) One good way to see this is that torque has the same units as energy. And for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen. Let's practice using this form to find an equation for the line. 707. k = motor constant. The ideal gas law tells us that. Since we know v and r , we can rearrange the equation v = r ω v = r … The curve itself shows an inverse relationship between force and velocity, meaning that an increase in force would cause a decrease in velocity and vice versa. The speed of travel relative to travel time (the faster one travels from point to point B, the less travel time is required to arrive at point B from point A); current and resistance (the higher the resistance, the lower the current); savings and disposable income (the less the disposable income, the more the savings relationship between fluid speed, pressure, and elevation was first derived in 1738 by Swiss physicist Daniel Bernoulli. 66 Interpret the slope: If the speed of the club hitting the ball increases by 1 mph, then the model predicts that the length the ball travels increases by 57. Speed versus Torque Voltage is proportional to speed, and torque is proportional to the current. at an angle of 40° and a speed of 125 miles per hour. Therefore, 16 workers will finish the task in 27 days. From Newton's second law of motion, the aerodynamic forces on the body (lift and drag) are directly related to the change in momentum of the fluid with time. Due to the random motions, the mean speed of the two particles – or rahter all particles – is identical. The data are as follows: y (volts) 1. We often think of a relationship between two variables as a straight line. x^m = y m Log x = Log y m = Log y / Log x The table shows exponent m for distances of 1. 3. The amplitude is given, so we need to calculate the linear mass density of the string, the angular frequency of the wave on the string, and the speed of the wave on the string. The dimensions at the top are 2 m$\times$2 m, and the depth is 5 m. Example: A 10-kg crate initially at rest is being lowered down a frictionless 10-m ramp using a rope. 66 yards. Torque non-linearly decreases with an increase in speed O b. 28 mm (almost touching the saw table) to 18. Answer (1 of 7): It’s not required to refer to any kind of literal physical area. The first scatterplot displays the relationship between height and fastest speed, and the second scatterplot displays the breakdown by gender in this relationship. 41, that means that your stall speed will be 41% faster in a 60 degree, constant altitude coordinated turn than it would be in straight and level flight. The speed ratio is 2. 1 x 10-7-km/h) increase in 85th-percentile speeds on tangent sections. 2) 2 k ˘1. , x = 2) surface area is quadrupled (2 2 = 4) not doubled, and volume is octupled (2 3 … For turbulent flow, the relationship is more complicated. Any object in motion has a kinetic energy that is defined as one-half of the product of its mass times its velocity squared. The advantage of VFD supplied motors is that the motor can supply the same maximum torque from zero speed to rated speed. R-squared, otherwise known as R² typically has a value in the range of 0 through to 1. 10 A pyramid-shaped vat has square cross-section and stands on its tip. If you weigh 70 kg (11 stone or 154 lbs), you’ll exert about … An increase in Angle of Attack will increase Lift and Drag; Skin Friction Drag is due to the fact that air immediately attached to a body has 0 … Speed Fan in an Electronics Enclosure A variable speed fan is automaticall y controlled by the fan circuit, which changes speed as the temperature changes to provide optimum air flow at all times. This has a huge effect on the torque required and good example is this: For a given system at 100% power and 100% speed, reducing the speed by However, a vary in frequency can effect P (Watt). ANALYZE AND CONCLUDE The coffee filter accelerates at first and then moves at a constant velocity. 16, which we used to develop the decreases with increase in armature current, and hence the speed decrease slightly. s speed. The implication is that % increases with speed, but is not propor - tional to the speed or to the ratio M 2 /M 1. b=(-5), the impact on Y of each additional patrol car deployed. m. For H at 300 K v rms = 1920 m/s; for 14 N v rms = 517 m/s. R-squared and adjusted R-squared do not always increase for better nonlinear models. Dividing the tips: 2020-01-02: From Pat: The terminal speed is observed to be 2. In other words, the Gear ratio is the ratio between the number of teeth on two gears that are meshed together, or two sprockets connected with a common roller chain, or the circumferences of two pulleys connected with a drive belt. Starting a rotating pump requires the motor to overcome the pump inertia and static friction. V = supply voltage. However, when the pump is at rest—0% full load speed—the full load torque is never also 0%. This will push point 4 to a higher T. Summary The exact relation between speed and crashes depends on many factors. 6 then t = 2 In this case, the speed of the tire tread with respect to the tire axle is the same as the speed of the car with respect to the road, so we have v = 15. The standard unit for speed in science is meters per second. In which direction is the cart that is shown traveling at t = 4 seconds? A. if you want to measure the intensity of the light on the square of the football where area of the sphere is 4pi square. S o = (2000 rpm) / (3. The joule is the standard unit for energy in general. 67 x 10-6-mph (5. Lift is the upwards force from imparting downward momentum on the air the wing passes thru. , doubling velocity and discharge do not simply double competence and capacity). This might be represented by the third, "Nonlinear" image. When a particle’s speed approaches the speed of light, however, the mass increase (called the relativistic mass increase) is significant. When system bandwidth is overlaid with the setpoint input square wave frequencies, the upper harmonics are lost. V This physics video tutorial explains how to calculate the wave speed / velocity on a stretch string given an applied tension and linear density of the wire. Torque. The pressure developed by the fan and the pressure drop in a system varies with the square of the change of speed of the impeller or volume passing through the #c/lambda=f# #("The speed of light is directly proportional to"# #lambda#, and #lambda# is inversely proportional to #f#. A vary in Voltage, can effect Q (Var). In Stata, the chi2 option is used with the tabulate command to obtain the test statistic and its associated p-value. In Example 1 from section 4. 05 and df=1 The gear ratio is the ratio of the number of turns the output shaft makes when the input shaft turns once. Assess: The maximum safe speed is proportional to the square root of the radius of the turn. 8 can be calculated as. Increasing the height of a ramp increases the inclination of the ramp, which in turn increases the speed at which an object goes down the ramp. Curve Fitting with Linear and Nonlinear Regression. Reason: The magnitude of the centripetal acceleration is 2 c v a r where we want to solve Long story short, when it comes to distance, golf ball speed is the king. ] A) 5 m/s B) 2 m/s D. Punching forces in amateur boxing are around 2500 N. 0 m/s or higher consistently demonstrated survival that was longer than expected by age and sex alone. 5 m/s corresponds to a drum speed of 19 rev/min, so a suitable gear ratio would be say 80:1, giving a maximum motor speed of 1520 rev/min. The kinetic energy of a car goes up by a factor of four if the speed doubles. From the formula we can understand that if radius will increase intensity will decrease. Example 2. The answer is this: V2. Torque proportional to square of the speed (fan type load) 4. If you tie a string to that weight you could power The answer is increasing the velocity, because the velocity variable is squared and therefore an increase in velocity would have a greater impact on the overall kinetic energy. 55 1. For example, doubling the velocity results in a 64 times increase in the competence. In summary: • A 10% increase in the speed causes a 21% increase in the stopping distance. Example 2: If m is inversely proportional to the square of n, and m = 64 when n = 3. Newton's Laws of Motion demand that the planets and the Sun must accelerate, because the planets pull on the Sun via the force of gravity just as hard as the Sun pulls on them - the … A spaceship s orbital maneuver requires a speed increase of 1. 0. The net result is unchanged induced drag. e. 2. enormous effects on the fuel consumption, which is directly related to the engine power. increase in x will result in a decrease in y. So the cube of that ratio becomes 8. What happens to the forces between the two where v rms is called the root-mean-square speed of the molecule. This DECREASES the aspect ratio(AR) term 2x, while increasing the wing area (S) by 2x. Speed is distance divided by time; velocity is displacement divided by time. That is, the factor by which the electrostatic force is changed is the inverse of the square of the … The square root then makes sense, as the root of a number >1 is smaller than the number. Answer (1 of 4): When the friction factor decreases, this does not mean that head loss will also diminish. Lakhmir Singh Solutions Class 9 Physics Chapter 1 Motion. g. Cessna 170) that has a variable-pitch propeller? In physics, mass–energy equivalence is the relationship between mass and energy in a system's rest frame, where the two values differ only by a constant and the units of measurement. The relationship between designated design speed and speed limit varies. 0%. The relationship between the speed and power of a fan or a pump is called the Cube Law and can be built up step by step. The standard international system for KE is the joule [J] equivalent to kg *(m/s) 2. Your calculated acceleration should be close to 9. •The steeper the graph, the faster the motion. At lower cutting speed a slight increase speed results in a large change in tool life. Another important relationship to be derived from above is that velocity, at constant flow, is inversely related to the radius squared (V ∝ 1 / r 2 at constant flow). A boater and motor boat are at rest on a lake. 44 k. Graphing results will show that distance traveled is in proportional to the square of the time spent falling. And that tells us if the velocity speeds up the force will be stronger and the radius well be smaller. 7. 1,302 UCLA students were asked to fill out a survey where they were asked about their height, fastest speed they have ever driven, and gender. Change the power output P by a factor of (90 kW/60 kW) and change the area by the same factor to keep the same. Similarly, a decrease in x causes a decrease in y. Since the air moves, defining … The relationship that exists between the centripetal acceleration and the angular velocity of the object is a square root function. 0 seconds, the ratio of the rock's speed to the stone's speed is A) speed = v/2; distance = d/2 B) speed = v; distance = d C) speed = v/2 And for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen. Constant torque type load. 084 points. Torque = (base speed/extended speed)squared. Kinetic Energy Equation. This relationship is further explained by the formula: F =. From our known data, we can use the regression formula (calculations not shown) to compute the values of and and obtain the following equation: Y= 85 + (-5) X, where Y is the average speed of cars on the freeway. Relationship of Drag with Airspeed There is an airspeed at which total drag is minimum, and in theory, this is the maximum range speed; however, flight at this speed is unstable because a small decrease in speed results in an increase in Negative correlation is a relationship between two variables in which one variable increases as the other decreases, and vice versa. Or if you take the square root of both sides, you get v naught is equal to the square root of the universal gravitational constant times the mass of Earth divided by the distance between the center of masses. none none Whether it’s lowered slowly or falling freely, a weight gains the same amount of energy. The slope term in our model is saying that for every 1 mph increase in the speed of a car, the required distance to stop goes up by 3. Special Reduced Torque Characteristics (Squared Reduced Load Torque Characteristics) Characteristics that are virtually constant with the square of the speed (i. The crash rate is also higher for an individual vehicle that drives at higher speed than the to the square root of the mass of the object divided by its cross-sectional area. We say the variable y varies directly as x if: y = k x for some constant k , called the constant of proportionality . Assuming the mass is 2 kg and the velocity is in m/s. Below is a graph that shows the hyperbolic shape of an inverse relationship. If the old price was$5, then $5/100 = x/110, then x =$5 / 100 * 110 = $5. Root mean square speed, average speed and the most probable speed have the following relationship: U rms > U av > U mp 2. Kinetic energy has a direct relationship with mass, meaning that as mass increases so does the Kinetic Energy of an none The increase of speed is a squared relationship Three times the speed will have nine times the force of impact? true. From graph 2 we see that centripetal force increases in a direct proportion to the square of velocity. 300 m. The same amount of time as the other trials The mass of the object affects its speed. Notice in the diagram below that as wavelength decreases, the frequency increases. The increase in frequency beyond rated frequency is possible and will produce higher speed but with voltage kept at rated voltage, and consequently reducing V/Hz ratio, the flux density will reduce and the torque will reduce. The stone falls 19. (Some textbooks describe a proportional relationship by saying that " y varies proportionally with x " or that " y is directly proportional to x . where: is the kinetic energy of the fluid in joules per cubic meter (J); is the volume of the fluid (m 3); is the density of the fluid in kilograms per cubic … Distance rolled (d) is 1. It isn't until we get to speeds that are a large fraction of the speed of light that any change in the flow of time becomes apparent. This can be demonstrated by the DC motor torque equation: Where: T = motor torque. Where: d is the distance fallen and; t is the time of fall . Therefore, the force of gravity becomes 4 units. This equation, i = v/r, tells us that the current, i, flowing through a circuit is directly How to Measure Kinetic Energy The standard unit for kinetic energy is the joule (J). To increase the frequency resolution for a given frequency range, increase the number of points acquired at the same sampling frequency. 13 An undisturbed parcel of a medium with a volume. So , here is how it works. C) Acceleration increases and speed decreases. Starting with the speed assumed in (1), assume two additional seconds go by and then 1. , weight and height) The inverse relationship between speed and torque means that an increase in the load (torque) on the motor will cause a decrease in speed. This increased kinetic energy and increased stopping distance can increase the risk of collisions and death. c. This is because at the same temperature, gas molecules with heavier mass have slower speed than lighter gas molecules. A thermistor mounted on the fan A net force on an object changes its motion – the greater the net force, the greater the acceleration. Effective horsepower is then calculated and plotted as shown in Figure 7. 523 and 1. 10 m/s b. 6kW! So for an airflow increase of a little over 18% the power needed has risen by nearly 45%. Other units for energy include the newton-meter (Nm) and the kilogram meter squared over seconds squared (kg m 2 /s 2). Design speeds are sometimes designated by determining the speed limit and adding a specified value, such as 5 or 10 mph, although the Green Book (1) guidance on design speed does not address speed limits. Competence varies as approximately the sixth power of velocity. Activity 5: Watch This! If the distance d varies directly as the time t, then the relationship can be translated into a math- ematical statement as d = kt, where k is the constant of variation. For instance, if the baseline Solve for the measure of the square’s side, given the A = 24 cm 2. the equation is K. S o = S i / r (2) where . km/h 2), meters per second squared (m/s 2) and standard gravity units (g n, often just g). Include a sample calculation. The amplitude (height) of the wave can't exceed a certain amount. A 4. You might remember from Algebra 1 that the Equation 2. Then use the proportion of area A in the diagram to distance squared to find the distance that produces the calculated change in … maintained up to a maximum speed of about 0. Or. MC3. The r 2 for the relationship between income and happiness is now 0. Speed on tangent section (figure 11) Intersection (if an intersection is located 350 ft (106. 574 units. The formula defines the energy E of a particle in its rest frame as the product of mass (m) with the speed of light … If the fluid velocity is small relative to the speed of sound (that is, the flow has a low Mach number), then the change in pressure when a fluid increases its velocity from zero to V is given by 1/2 the density times the velocity squared (in the absence of friction). 69 1. 23 3. "t 2" means "t raised to the power of 2" or "t squared". 0-kilogram cart that is initially at rest and starts moving northward. If we take the square root of this number, it should match the value of the Pearson correlation we obtain. You may think that we have some idea about the molecular speeds of an ideal gas after understanding it's kinetic behaviour where we can find the average of the square of velocity of molecules and we can also find the rms-speed (root-mean-square-speed), however, this is not a complete solution to the distribution of … For example, if the price of a hamburger has risen by 10%, you might express this as a proportion: old price / 100 = new price / 110, so if you know the old price you can solve the proportion equation to find the new price. c= speed of light (approximately = 3 x 108 m/s) Similarly, if we increase the speed by 20% (for example from 50 km/h to 60 km/h), then d ˘(1. This means that 54% of the variation in IBI is explained by this model. 5 = 0. In other words, If there is a twofold increase in speed, the kinetic energy will increase by a factor of four. Click to see full answer. 25 seconds which, for the majority of people, is not fast enough to … The operation of a DC motor is relatively straightforward. Figure 17. The graph shown represents the relationship between velocity and time for a 2. What's the relationship between axon diameter and propagation speed? There is a wide consensus that the propagation velocity in myelinated axons is proportional to the axon diameter. 32. Remember that it is customary to put the Answer (1 of 7): Irrespective of shape of airplane wing, air flow on and below wing should meet at end at same time. The equation is known as Einstein’s mass-energy equation and is expressed as, E=mc2. If it were positive, then it would be moving up. 5b are both linear relationships. In physics the average speed of an object is defined as: $$\text{average speed} = \frac{\text{distance traveled}}{\text{time elapsed}}$$ The root mean square velocity or $$v_{\rm rms}$$ is the square root of the average square velocity and is $v_{\rm rms} = \sqrt{3\,R\,T\over M}$ Where $$M$$ is equal to the molar mass of the molecule in kg/mol. The following article aims to discuss and elaborate on the relationship between these parameters and methods of using them You need to know 3 of the 4: acceleration, initial speed, final speed and time (acceleration duration) to calculate the fourth. 2. 9324088 feet. 95 2. The negative sign just means that the object is moving downwards. The coefficient of determination, R 2 , is 54. Gravity applies a constant force and thus a constant acceleration. In words why this is the case -- for a shallow fluid, the motion of the fluid is mostly side-to-side, and in order to accumulate more c) According to the linear model, a 1200 square-foot home is expected to have a price of$121,020. Mathematically, the kinetic energy is proportional to the square of the speed, as can be seen in the equation for KE. Direction does not matter for speed. and its a discussion we have time and time again. As you increase the Reynolds number, you would end up approximating that relationship of pressure drop being proportional to the square of the flow rate. east C. Some people square t s and get an F-statistic with 1 degree of freedom in the numerator and n−2 degrees of freedom in the denominator. . This is seen in that the estimates for pressure are only accurate during the summer months when temperatures are higher. Marriage leads to better emotional health. It will be observed from the figure that the doubling of speed of the Navy YP from 7 to 14 knots increases the power by a factor of 10! For every one point increase in Test 3 the predicted value of Test 4 increases between 0. The output is always in the input unit squared (e. Consider the flow of a segment of an ideal fluid through a nonuniform pipe in a time interval t as illustrated in Figure D 14. In the following equation, I needed to find exponent "m". How fast is the distance between the bicyclist and the balloon increasing 2 seconds later? Ex 6. Capacity varies as the discharge squared or cubed. The relationship between these variables is usually expressed in terms of dimensionless numbers a 10% increase in stirrer speed rasies the power required by over 30%. The time is the same in all four trials. To check this, add a column to your data table for v 2. 001), the slope is much smaller than before. 4. 3. When prices are low, sales are high, but profit is still low since very little is made from each sale. 41, which equals just As such, the chi square test is not restricted to nominal data; with non-binned data, however, the results depend on how the bins or classes are created and the size of the sample. The square–cube law (or cube–square law) is a mathematical principle, applied in a variety of scientific fields, which describes the relationship between the volume and the surface area as a shape's size increases or decreases. Torque vs. The mass of course cannot equal resulted in the gradual increase in time as more trials were completed. Kepler's 1 st Law: Planetary orbits about the Sun are ellipses and the Sun lies at one of the foci of the ellipse. But what is the exact form of this relationship? Your textbook suggests that under some circumstances, air … b. But practically, due to armature reaction, Φ The second row in the Coefficients is the slope, or in our example, the effect speed has in distance required for a car to stop. south D. It is expressed in the following form. The other variable, denoted y, is regarded as the response The fact that KE is directly proportional to the square of the speed shows that for a twofold speed increase, KE increases by a factor of four. The pattern between electrostatic force and distance can be further characterized as an inverse square relationship. But as I noticed in the real world when I take a yoyo for example d. a=85, or the average speed when X=0. Correlation. 8 meters per second? A) 1:1 B) 1:2 C) 2:1 D) 4:1 6. P = 1 2 μ A 2 ω 2 v. 6) can also be written in terms of rms speed . The kinetic energy of water is a result of the speed or flow rate of the water. Square Wave in Frequency Domain Overlaid by the Frequency Response of a Bandwidth Limited Driver. As it is often said, an equation is not merely a recipe for algebraic problem solving, but also a guide to thinking about the relationship between quantities. • An a % increase in the speed causes a ‡ 2 ¯ a 2 100 speed will give rise to a linear speed for both wheels and the bicycle: v = r wheel! rear = 0:673 2 20:7 = 6:97 m=s d) The only piece of data not necessary for our calculations is the length of pedal cranks. To linearize this graph, follow the directions in the third column – in other words, plot a graph in which the square of the period is on the y-axis, and the length of the pendulum is … Very few molecules move at either very low or very high speeds. To find out something’s speed (or velocity) after a certain amount of time, you just multiply the acceleration of gravity by the amount of time since it was let go of. 18. For example, acquiring 2,048 points at 1. One conclusion that can be drawn is: a. Let’s see an example of the negative relationship. As Reynolds number increases, so does the speed and tube length. My question is, why would the velocity be squared? I'm having a tough time understanding this. For two quantities, x and y, an increase in x causes an increase in y as well. 18 yards. Table B 1. 20 \times 10^3 m/s. west page 5 Physics I Motion Test However, since the drag force increases with the square of the speed, the power increases with the third power of the speed. Speed and Torque are directly related, and are the two primary performance factors in properly selecting a motor for a specific application or use. A fan produces 1000m3/h at an impeller speed of 2000rpm, what is the resulting airflow if the speed is reduced to 1000rpm? V2 = (1000/2000) x 1000 = 500m3/h. Light and a single electrical charge are monopoles, so obey the inverse square law, expressed as 1/d^2, where d is the distance from the source to the point of measurement. Final Answer. 3×10-8). So if you have the Velocity in meters per second and time in seconds, you can calculate the Acceleration by using the formula. As it moves along, it deflects air downwards. Chi-square test. Radian measure and arc length can be applied to the study of circular motion. It has already been mentioned that the indices could have been interchanged. = 1/2mv^2 If the mass contains 1 kilogram and the velocity of a body is meters/second, the kinetic energy will be 1 kg per meter square and seconds square. However, not all data have a linear relationship, and your model must fit the curves present in the data. So if you have a 12 mA signal from the DP cell, that’s 50% DP, but 71% flow rate. How fast is an object moving after it has fallen one meter? the relationship between two variables (bivariate association) and then expands to A coefficient of determination (r-squared) is a numerical indicator that tells how much of the variance in one variable is associated, explained, or increase in X; it depends on the correlation between the two variables. There does appear to be some linear relationship. That's why in Power Plant's, among the ways to vary the Var is by playing around with the Generator Transformer Tap Changer (increase or decrease Voltage). For the heart rate–speed data, the r 2 is 0. • A 20% increase in the speed causes a 44% increase in the stopping distance. Com-pared with the acceleration of object B, the acceleration of object A is ANSWER: (3) three times as great Base your answers to questions 36 through 41 on the graph below, which represents the rela-tionship between speed and time for an object Kinetic Energy of Water. Example 2: Voltage output of engines was measured at various combinations of blade speed and voltage measuring sensor extension. Reasoning: Speed increases at a rate of 10 m/s (actually 9. B) Both acceleration and speed remain the same. The relationship between voltage, current, and resistance is described by Ohm's law. y - y 1 = m (x - x 1) where m is the slope of the line and (x 1, y 1) is a point on the line. 24 cm 2 = s 2. The speed of sound in these two gases is 1350 m/s and 350 m/s, respectively. One Newton equals one kilogram-meter per second squared (kg-m/s 2). Example. The Second Law – Pressure. Begin with the equation of the time-averaged power of a sinusoidal wave on a string: P = 1 2 μ A 2 ω 2 v. In other words, If there is a … Answer: Speed is equal to distance divided by time. A value of 0 indicates that there is no linear relationship between the observed and predicted values, where “linear” in this … Maneuvering speed is the maximum speed at which you can make full or abrupt movements of a single control without causing structural failure of the aircraft. As has already been mentioned, the primary relationship is that lift goes with the square of the airspeed. … The observed mass increase with particle speed because with the increaed speed, particle's kinetic energy increase. 11 Speed and height. When an object moves along a straight line, the motion is known as rectilinear motion. The escape speed of gases on the surface of Moon is much less than the root mean square speeds of gases due to low gravity. The second hypothesis is therefore not supported, and could be rewritten this way: If the speed of an object increases, then its kinetic energy will increase exponentially because speed is squared in proportion to kinetic energy signal record determine the resolution frequency. What is the relationship between torque and speed in constant type loads? Select one: a. If its engine has an exhaust speed of 2. Now, suppose that we accelerate you (meaning we speed you up) at a rate of 10 "feet per second squared" (or, in other words, 10 "feet per second, per second"). From 1st condition we get, k = 64 × 3 2 = 576. As the turbine speed increases, electricity production also increases. The relationship between x and y in the temperature example is deterministic because once the value of x is known, the value of y is completely determined. 21, or a 0. A quadratic relationship between x and y means y is related to x^2 , x and a constant (C) by a function, which generally represented as: y = A x^2 + B x + C where A must be a non-zero number. Impact of v rms in nature: 1. However, what'd be a general rule of thumb for flying a small aircraft (e. Chi Square: Allows you to test whether there is a relationship between two variables. Hence, a shunt motor can be assumed as a constant speed motor. The number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Figure 8 indicates that there is more to the relationship between pressure and wind speed near the surface of the water in the Gulf of Mexico. When you reject the null hypothesis with a Chi-Square, you are saying that there is a relationship between the two But it is not a linear relationship (e. pulled up with constant speed. This is assuming that all other factors pertaining to the ramp and the object remain the same, such as the inclination (or lack thereof) of the surface that the ramp is on I think I understand the relationship between the power lever and the RPM lever in an aircraft with a constant speed prop thanks to this question. If b1 > 0, then the variables have a positive relationship i. Gait speeds of 1. Giving an example, a one repetition maximum (1RM) Back Squat would produce high levels of force but would be lifted at a slow velocity. 21-unit increase in reported happiness for every $10,000 increase in income. Alternatively, if the sampling rate had been Strategy The area over which the power in a particular direction is dispersed increases as distance squared, as illustrated in the figure. When air hits front edge of … When is the speed of the object 9. If the centripetal force must be provided by friction alone on a curve, an increase in speed could lead to an unexpected skid if friction is insufficient. a) According to the linear model, the duration of a coaster ride is expected to increase by about 0. What happens to the magnitude of the acceleration and the speed of the rock as it falls? [Neglect friction. If, for example, a person runs 30 meters in 10 seconds, his speed is 3 meters per second. In this activity we will examine the precise relationship between tension (T) the force applied to the string, the wave speed (v w) and the linear mass density of the string (µ = m/L which is measured in kg/m). If b1 < 0, then the variables have a negative relationship i. To run the bivariate Pearson Correlation, click Analyze > Correlate > Bivariate 1. 0 m/s. Determine (a) the work done by gravity, (b) the work done by the rope, and (c) the tension in … increase in speed, since the line is getting steeper: In other words, in a given time, the distance the object moves is change (getting larger). y ∝ √x; y ∝ x ½; For example… Speed is proportional to the square root of distance for freely falling objects. 81 m/s 2. At any time, some of the ball bearings on this apparatus are moving faster than others, but the system can be described by an average kinetic energy. Ten seconds after starting from rest, an object falling freely downward will have a speed of about: a. 1 - Example: Quiz and exam scores Data from a sample of 50 students were used to build a regression model using quiz averages to predict final exam scores. This is mostly due to the fact that both the internode length as well as the electrotonic length constant increase with the diameter. BUT, it does not tell you the direction or the size of the relationship. KE= ½ m (v^2) This equation is fundamental when looking at the relationship between speed and the force of a vehicle impact. At low velocities, the increase in mass is small. Height divided by length (h/L) is a measure of steepness of slope. The diagram below shows the relationship of parasitic drag and induced drag to each other and to total drag. Study Tip The word quadratic refers to terms of the second degree (or squared). 26 mm (magnet force The distance it falls is proportional to the square of the time of fall. As such, it has units of velocity. This means that we are increasing your speed by 10 feet per second, each second. 34 Votes) Frequency is inversely proportional to wavelength, since speed is fixed. Calculate the linear speed, v, of the stopper for each trial. Running the Test. Merchants have a Note that the centripetal force is proportional to the square of the velocity, implying that a doubling of speed will require four times the centripetal force to keep the motion in a circle. The value of b1 gives us insight into the nature of the relationship between the dependent and the independent variables. So, clearly, the units don’t automa And the law for centripetal force is: F = m v 2 r. P none The relationship between an object’s mass and its terminal velocity Report by Victor Jeung, Terry Tong, Cathy Liu, Jason Feng relationship between the speed and the mass is a square root relation. Solve for the required rate of change of the square. Last ride. 93 1. For example, take 2 otherwise identical planes flying at the same speed, but increase the chord on one of them by 2x. Lift and drag increase exponentially with speed—if speed is doubled, drag or lift will be quadrupled. Continue the takeoff on the remaining engines and compute the additional distance for the vehicle to clear a 50 ft obstacle, determining the takeoff distance 3. The relationship for the kinetic energy per unit volume of water is thus proportional to its velocity and can be expressed as:. Since power is speed times torque, the power required to turn such a load goes up at a rate of the speed cubed. A suitable conclusion statement from such a relationship would be that… y is proportional to the square root of x. Another key difference is that amplitude is related to wavelength. Kepler studied the periods of the planets and their distance from the Sun, and proved the following mathematical relationship, which is Kepler’s Third Law: The square of the period of a planet’s orbit (P) is directly proportional to the cube of … Special relativity is an explanation of how speed affects mass, time and space. A chi-square test of the relationship between personal perception of emotional health and What is the relative risk of always exceeding the speed limit for people under 30 compared to Even though the relative risk is 1. In many stadiums, the ball needs to travel 350 or more feet to be a home run. 94 2 Ohm's Law. Since force is momentum/second, that's why it's proportional to speed-squared. A tree's diameter is estimated to increase by 0. Velocity has a … The force required to keep an object of a constant mass in circular motion is equal to the mass of the object * the velocity of the object squared, then divided by the radius of the circular 'path'. Answer (1 of 7): X axis (horizontal) is velocity. In contrast the relationship between lift or drag and air density is a direct relationship such that an increase or decrease in air density will cause an increase or decrease in both drag and lift. 12. Runes are looted from defeated enemies and loot chests. Where E= equivalent kinetic energy of the object, m= mass of the object (Kg) and. Area of football (Square) 4 π r2. Both terms indicate the rate at which an object is moving. The relationship between voltage, torque and output speed is a common topic of discussion between our customers and Precision Microdrives’ sales engineers. E. It is accelerating. 5 Hz with frequency range 0 to 511. At 16 … The inverse square law results from monopoles, which are point-sources. If you observe the shape of wing its curved in top and near flat on bottom (mostly). The kinetic energy is dependent upon the square of the speed. 05 6) look up the Chi Square value in the table at p=. 27. 5. Therefore, the kinetic energy of an object is proportional to the square of its velocity (speed). 8 m/s) every second. If you double the speed of a … An example of a positive correlation coefficient is the relationship between the speed of a wind turbine and the amount of energy it produces. Human reaction time is approximately 0. dA/dt = 2(2√6)(8) dA/dt = 32√6 cm 2 /s. Thus, the inverse square law implies that the force will be "1/4" of the original 16 units. Substitute the value of ds/dt = 8 cm 2 /s and s = 2√6 cm to the obtained equation. d. Speed limits are not necessarily known during the design process. 42 4) calculate the degrees of freedom of the contingency table df=1 5) select the level of alpha p=. To learn how to select the right motor to fit your application, the first step is to understand the relationship between speed, torque and output power of a motor. The radius of the tire is r = 0. decrease the speed of electrons going from cathode to anode c. To understand the relationship between linear and angular speed. The maximum current it could take is rated current and the corresponding torque can be found out from speed torque curve (as you know the speed from voltage (rpm=k*v)) where k is the speed constant of the motor). The theory includes a way for the speed of light to define the relationship between energy and matter — … The first term, which contains the squared speed of particle 2, does not differ on average from the second term, which contains the squared speed of particle 1. After they fall for 2. The relationship will be linear if and only if acceleration is directly proportional to slope. The higher the Newton (N) the greater the force or harder the punch. Amplitude is independent of both. This relationship also plotted on a log-log graph, then it plots as straight line as shown in the figure. Then the y axis with be kinetic energy in Joules (units kg. Formula to measure intensity i= s/4pi r square. if X … A hoisting speed of 0. 3, if there is a very low baseline risk, then the increase in actual risk may be minimal. Graphically, and mathematically, it appears as … The obtained graph will be a parabolic decrease in tool life with an increase in cutting speed. 1 m/s. Runes are loot items that can be attached to armor and weapons to increase Eivor's offensive and defensive capabilities. In summary, fan laws are essentially about impellers and what happens to their characteristics when they undergo changes in rotational speed, air density, or are scaled in size. The asking price is$121,020 - $6000 =$115,020. 27. 4. Step-by-step explanation: Kaneppeleqw and 21 more users found this answer helpful. That's why to increase a Generator's Watt for example, we increase the flow rate --> speed --> N =120F/P. Kinetic energy is a scalar quantity, which means it only has a magnitude and not a direction. KE = 1/2 x m x. 8. Many machine applications are constant horsepower in their load characteristics. When we increase the "temperature" of the system by increasing the voltage to the … The root-mean-square speed of molecules is the speed at which all the molecules have the same total kinetic energy as in case of their actual speed. But the root of a number from 0-1 is larger: √0. Fourier Transform of a Square Wave. Construct a graph of centripetal force versus v 2. S = source of light. Figures 7. Output Speed Vs. In the example shown in the figure, reducing the radius by 50% causes the velocity to increase 4-fold at constant flow. Summary: A distance-time graph tells us how far an object has moved with time. The torque is proportional to the speed squared. The fluid momentum is equal to the mass times the velocity of the fluid. Together, they have mass 200. When d = 19. 0-kilogram rock and a 1. Our graph appears to match the last relationship, “the square of y is proportional to x”. relationship. If it falls twice as far it gains twice the energy. In this older adult population, the relationship of gait speed with remaining years of life was consistent across age groups, but the absolute number of expected remaining years of life was larger at younger ages. The standard unit (S. Since wave speed is set in stone for any particular circumstance, if wavelength were to increase, frequency would have to decrease to avoid changing the speed (the opposite applies as well). Article objectives; To calculate the speed and angular velocity of objects. Velocity is not exactly the same as speed. As speed increases, the torque drops off as the inverse of the speed, or 1/N. 5 Hz. Proportional Relationships A proportional relationship is one in which two quantities vary directly with each other. A decent way to think about force and kinetic energy is to consider a falling weight. A force is described by using the expression ‘force of A on B’ and drawing an arrow to show the direction of the force. The torque drop-off is not as severe as the motor’s peak torque, 1/(Nsquared). The following relationship is found: $PV=\frac{1}{3}Nm{\overline{v^2}}\\$, where P is the pressure (average force per unit area), V is the volume of gas in the container, N is the number of molecules in the container, m is the mass of a molecule, and $\overline{v^2}\\$ is the average of the molecular speed squared. 16065 years. More than 100 m/s. Thus after 10 seconds, the speed is … Answer: If the distance is increased by a factor of 2, then distance squared will increase by a factor of 4. ) From the scatterplot, we can see that as height increases, weight also tends to increase. Moon has no atmosphere. A chi-square test of the relationship between personal perception of emotional health and marital status led to rejection of the null hypothesis, indicating that there is a relationship between these two variables. We have observed that an increase in the tension of a string causes an increase in the velocity that waves travel on the string. Consider the relationship described in the last line of the table, the height x of a man aged 25 and his weight y. So now you need to use a motor that will provide 80 HP which in this case is a 100 HP / 1800 RPM motor because there is no such thing as an 80 HP, 1800 RPM motor available. So if the stall speed (Vs - clean config) in your Cessna 172 is 48 knots, then your stall speed at 60 degrees of bank is 48 knots X 1. b. (If it were, we’d be in even more trouble trying to figure out what the hell a “square second” looked like. Lloyd Morgan/CC-BY-SA 2. In contrast, all the other relationships listed in the table above have an element of randomness in them. You should know by now that the second part of this law cannot be quite correct. So you are accelerating to increase your speed. Two options are available: 1. 50 2. s = 2√6 cm. While you can certainly do such calculations That is, an increase in x will decrease y, such as the speed of a vehicle increase, time is taken decreases. Solution: Given, m ∝ 1/n 2. the y (vertical) is Kinetic Energy. 0-kilogram stone fall freely from rest from a height of 100 meters. The molecules of gas move randomly in any direction. For example, when length is doubled (i. Here is why: When we look at a very simplified version of the processes in a jet-engine, increasing fuel-flow will increase temperature in the combustor. , characteristics at which the torque generation curve is a square curve) and require a large torque at low speeds. Model testing is carried out over the expected speed range of the ship with resistance data collected at each testing speed. A value of 1 indicates that predictions are identical to the observed values; it is not possible to have a value of R² of more than 1. 180 seconds for each additional foot of initial drop. Combining the last two equations we conclude that. (m/s)^2). The area of the given square is increasing at a rate · Different types of loads exhibit different speed torque characteristics. Quadratic Relationship. As prices increase, profits increase, but at some point, sales will start to drop, until eventually too steep of a price will drive sales down so far as to not be profitable. Find m when n = 5. N 2. 65 m in diameter lies in a vertical plane and rotates about And since the square root of 2 is 1. At higher cutting speed the tool life is almost the same. 9. 1 - What is Simple Linear Regression? Simple linear regression is a statistical method that allows us to summarize and study relationships between two continuous (quantitative) variables: One variable, denoted x, is regarded as the predictor, explanatory, or independent variable. directly proportional. This figure is very similar to Figure 14. Suppose the distance in question 1 is tripled. However, in a general sense the relation is very clear: if on a road the driven speeds become higher, the crash rate will also increase. Use the formula force (F) equals mass (m) times acceleration (a) to calculate the force in Newtons (N). So, you were already travelling at a constant 10 feet per second (your speed). Torque = (line frequency/extended frequency)squared. 5a and 7. 1. The equation for Kinetic Energy is: KE = 1/2 mv 2. The speed of the crate is 3 m/s at the bottom of the 40 ramp. m × n 2 = k. Torque is independent on speed O C. Using R-squared and adjusted R-squared to choose the final model led to the correct model only 28-43% of the time. This formula is for the special case that the speed is non-relativistic, i. Bernoulli's equation is usually written as follows, The variables , , refer to the pressure, speed, and height of the fluid at point 1, whereas the variables , , and refer to the pressure, speed, and height The force of air resistance clearly depends on the velocity of an object moving through the air: the larger the speed, the larger the drag force. That is, if you increase the predictor by 1 unit, the response always increases by X units. The graph shows the relationship between speed and time for two objects, A and B. A lot of times, those gains can come without adding any speed to your golf swing. 024 kHz would have yielded ∆f = 0. Find (a) the value of the constant b in the equation v = mg b (1 −e−bt/m), v = m g b ( 1 − e − b t / m), and (b) the value of the resistive force when the bead reaches terminal speed. If a home's square footage is 1,250 then the market value of the home is (1,250 x 207. The flow is proportional to the speed: 10% slower = 90% flow. This was a simple linear regression example for a positive relationship in business. D)Acceleration remains the same and speed increases. The load torque, as seen at the motor side of the gearbox, will be reduced by a factor of … The speed of light is very close to 300,000 km per second (186,300 miles per second). Sight distance squared (ft 2 (m 2)) 1. Distance divided by time squared (d/t 2) is proportional to acceleration. Be sure to use the matching units. Consider a parcel of a medium initially undisturbed and then influenced by a sound wave at time t, as shown in (Figure). However, at speeds very close to that of light the effect grows in magnitude very rapidly indeed until time almost comes to a standstill. \$6000 is the (negative) residual. A magnet is a dipole, and a disc magnet has two closely spaced magnetic poles. The average speed of molecules can be calculated as an integral of the Maxwell-Boltzmann distribution function multiplied by the magnitude of velocity of a molecule v . So you get: velocity = -9. 3-m) increase in sight distance squared is associated with a 1. Answered by Penny Nom. Torque non-linearly increases with an increase in speed d. The torque-speed curve is similar for all centrifugal pumps due to simple math: the pump torque varies as the square of its speed. moving at a constant speed 20. The graph shows that at a wind speed of 8 metres per second we get a power (amount of energy per second) of 314 Watts per square metre exposed to the wind (the wind is coming from a direction perpendicular to the swept rotor area). For very shallow fluids (compared to the wavelength), the speed increases proportionally to the square root of the depth, and for very deep fluids, the speed increases with the square root of the wavelength. 100 m/s d. Base Frequency There is no relationship between the type of training program attended and the job placement success of trainees 3) calculate the test for statistical significance Chi Square=70. It is safe to say that in most operating conditions the shaft-speed will increase with higher thrust settings. Bernoulli's equation relates the pressure, speed, and height of any two points (1 and 2) in a steady streamline flowing fluid of density . X is the number of patrol cars deployed i hear from a lot of people that the map to increase riding speed also increases flying speed i have my doubts but im not really sure. KE = 1 / 2 mv 2. 7 m) before or after the spot; 0 otherwise The root mean square speed or r. The metric we use is called a Newton (after Sir Isaac Newton). A remotely mounted thermistor on a third lead wire (thermistor not provided). Notice that for any increase, x * l or x * r, in length or radius, the increase in surface area is x squared (x 2) and the increase in volume is x cubed (x 3). Since “m” and “r” are kept constant and v is our dependent variable we see that force, in fact, should increase as our experiment suggests. The root mean square velocity is the square root of the average of the square of the velocity. Clearly, using R-squared to evaluate and choose a nonlinear model is a bad idea. Assuming a speed (and hence a distance along the runway) that the engine failure occurs. Similarly, if we increase the speed by 20% (for example from 50 km/h to 60 km/h), then d ˘(1. Acceleration, like Velocity, is also a vector since it has a numerical magnitude in meters/ second and also has a Speed and velocity are related, but different. It also shows that, for a given speed, induced drag depends on SPAN loading. While the relationship is still statistically significant (p<0. 5 x mass x speed squared. a squared relationship. The output speed of a transmission with input speed 2000 rpm and transmission ratio 3. speed 𝜔 = (𝑉𝑠−𝑇 𝑣𝑅𝐴/ ∅) ∅ As flux Φis assumed constant, , the speed decreases with developed torque increase. It was first described in 1638 by Galileo Galilei in his Two New Sciences as the "ratio of two volumes is greater than the ratio of their surfaces". Parallel operation of centrifugal pumps is used to increase flow rate through the system. DC Motors – Voltage Vs. Acceleration is measured in meters/ second squared. S o = output speed (rad/s, rpm) S i = input speed (rad/s, rpm) Example - Gear Output Speed. At high speed, the momentum you're imparting to each parcel of air is proportional to the speed, and the number of parcels of air per second you're doing it to is also proportional to speed. Where 1 Joule = 1kg m 2 /s 2. 0 20 40 60 80 100 120 1M0 Inverse relationships follow a hyperbolic pattern. A particle moving at one-fifth the speed of light (60,000 km/sec or 37,000 mi/sec) has a mass only 2% greater than its rest mass. the increase of speed is a squared relationship | 2022-01-16T10:04:07 | {
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https://math.stackexchange.com/questions/1461732/proof-that-inf-a-sup-b | # Proof that $\inf A = \sup B$
Exercise
Suppose $A$ is a nonempty set of reals that is bounded below. Let $B$ be the set of lower bounds for $A$, and assume further that $B$ is not empty and bounded above (I have proven that $B \neq \emptyset$ and that $B$ is bounded above, so I omit the proof here and provide it as an assumption for convenience).
Prove that $\sup B = \inf A$.
I have proven here that $\sup B \in B$, which implied that $\sup B \leq \inf A$. Therefore, my strategy in proving that $\sup B = \inf A$ is to show that $\inf A \leq \sup B$, which I believe I have accomplished in the following demonstration. Of course, we could proceed with proof by contradiction, but I try to avoid indirect proofs when a direct proof is possible.
Proof
Since we know that $\sup B \in B$, then $\sup B$ is the largest element in B. We denote this fact by $M = \sup B$. By definition of $B$, we know that $\inf A \in B$. Because $M$ is the largest element in $B$, we must have the inequality $\inf A \leq M = \sup B$, as desired.
Because $\sup B \leq \inf A$ and $\inf A \leq \sup B$, we conclude by the antisymmetry law that $\inf A = \sup B$.
• Seems fine to me. Oct 2 '15 at 23:16
Your proof can be greatly simplified. I’ll skip existence of $\sup B$ as that’s trivial.
As $B$ is the set of all lower bounds of $A$, we have that $b \leq a$ for all $a \in A$ and $b \in B$. Note $a \in A$ are all upper bounds of $B$, so we have $\sup B \leq a$ for all $a \in A$ by the very definition of a supremum. By definition of $B$, $\sup B \in B$ and is therefore the maximum of $B$ which means exactly that it is the greatest lower bound of $A$. Therefore, $\inf A = \sup B$.
In fact, you don't need to assume $B$ is bounded up, for a in $A$, then any $b$ in $B$ has $b \leq a$.
Then if $\sup(B) > \inf(A)$, by the definition of "$\inf$", then there's a in $A$ such that $a < \sup(B)$. Then by the definition of "$\sup$", there's a $b$ in $B$, such that $b > a$. Then $b$ can not be the lower bound of $A$, contradicts and $\sup(B) \leq \inf(A)$.
If there's a $b$ such that $\sup(B) < b < \inf(A)$, then $b$ is a lower bound of $A$, and $b \leq \sup(B)$, contradicts. Then $\sup(B) \geq \inf(A)$.
So $\sup(B) = \inf(A)$. | 2021-11-30T04:04:18 | {
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https://collegephysicsanswers.com/openstax-solutions/using-data-table-75-calculate-daily-energy-needs-person-who-sleeps-700-h-walks | Question
Using data from Table 7.5, calculate the daily energy needs of a person who sleeps for 7.00 h, walks for 2.00 h, attends classes for 4.00 h, cycles for 2.00 h, sits relaxed for 3.00 h, and studies for 6.00 h. (Studying consumes energy at the same rate as sitting in class.)
$3800 \textrm{ kcal}$
Note: The power consumption while sitting relaxed is 120 W, not 125 W as shown in the video. This minor discrepancy doesn't change the final answer. The calculator screenshot has been updated.
Solution Video
# OpenStax College Physics Solution, Chapter 7, Problem 49 (Problems & Exercises) (2:06)
View sample solution
## Calculator Screenshots
Video Transcript
This is College Physics Answers with Shaun Dyhcko. We're going to calculate the total food calories that the student burns in a day and we're going to multiply by the power consumption of each activity multiplied by the time that they spend doing that activity, and then adding all those pairs of power times time terms together. So we have sleeping for seven hours, so that's 83 watts of power consumed during sleeping times seven hours. As a shortcut, I'm not converting the hours into seconds yet. So I'm just going to leave them as watt hours because otherwise I'd have to multiply by 3600 seconds per hour in each of these terms and that would just be annoying to write that down so many times. So I'm going to keep my units as watt hours and then deal with the conversion in this last step here which I'll explain in a second. So we have 83 watts during sleeping multiplied by seven hours, plus 280 watts while walking to school multiplied by two hours, plus the 210 watts while attending class multiplied by four hours in class, plus the 400 watts while cycling times two hours, and then plus 125 watts while sitting relaxed for three hours. Then add to that 210 watts consumed during the six hours of studying. This works out to 4416 watt hours. Now a watt is a joule per second and then we're multiplying that by hours. In order to make these units properly cancel we have to do a conversion. So, we multiply by 3600 seconds per hour and so the hours cancel and the seconds also cancel and we expected these time units to disappear because we're multiplying watts by time and so that works out to energy, so we expect to have joules left over there. Then we want to convert those joules into kilocalories so we multiply by one kilocalorie for every 4.184 times ten to the three joules giving us 3800 kilocalories consumed by the student in a day.
Submitted by robn on Sun, 06/28/2020 - 19:17
Minor discrepancy, but you recorded the value for "sitting relaxed" as 125 when it is 120.
Submitted by ShaunDychko on Thu, 07/02/2020 - 11:07
Hi robn,
Thank you very much for noticing this. I've made a note below the final answer. As you say, the discrepancy is minor, and it doesn't change the answer to two significant figures, but I've updated the calculator screenshots anyway. Table 7.5 mentions "standing relaxed", and I think the word "relaxed" drew my attention to that row.
All the best with your studies,
Shaun | 2020-07-07T03:54:38 | {
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https://math.stackexchange.com/questions/1893915/sequences-of-length-4n-with-2n-ones-and-2n-zeroes-with-one-constraint | # Sequences of length $4n$ with $2n$ ones and $2n$ zeroes with one constraint.
How many sequences of length $4n$ are there if every sequence needs to contain exactly $2n$ zeroes and $2n$ ones and number of zeroes before $n$-th occurence of one must be not greater than $n$.
So i came up with straightforward solution: $n$-th occurence of one can appear on places from $n$ to $2n$. So we fix $n$-th one at place $n+k$ where $k \in \{ 0\dots n \}$ and first arrange numbers on the left of fixed one and then on the right. This approach gives us the sum: $$\sum_{k=0}^n \binom{n+k-1}{n-1} \binom{3n-k}{n}$$
I was searching for some magical identities in Concrete Mathetmatics book, but none of them seemed to fit (not suitable summation upper limit is in my sum). How can i approach this problem? I'd appreciate some help on this :)
• One way to attack it: compute the first half dozen terms or so and ask OEIS. If the sum has a convenient closed form or if the sequence has an alternative definition, it's likely OEIS will recognize it. – lulu Aug 16 '16 at 12:20
• @lulu thanks for this good idea. Oeis gave me alternative sequence and a result for this ( oeis.org/A036910 ), but i don't see transformation from my sum to sum in the "formula" section – Krzysztof Lewko Aug 16 '16 at 13:02
• That's often the case. Sometimes it's possible to prove the two formulas match inductively, but even in that case there's no guarantee that it will bring enlightenment. In your case...well, I'm a bit skeptical because I don't see a straight forward recursion. Your method looks optimal to me, but that's not the sort of approach that tends to lead to elegant closed forms. And OEIS doesn't seem to have one. Mind you, none of that proves anything...it's perfectly possible that there's another, more intuitive, way to see it. – lulu Aug 16 '16 at 13:08
• @lulu you are right, although i think it should be doable, because it's task from exam and students had like 3h to do that and other tasks too. But to be honest, result doesn't seem to be easy to guess. – Krzysztof Lewko Aug 16 '16 at 13:32
A closed for is given by the expression $$\frac12 \left( \binom{4n}{2n} + \binom{2n}{n}^2 \right).$$
Not only do I have a proof, it even matches the first $5$ terms in the OEIS sequence, so it must be right!
To see where this formula comes from, observe that the sequences come in complementary pairs, where the complement of a sequence is the sequence obtained by exchanging $0$s and $1$s. For example, when $n = 1$, the $\binom{4}{2} = 6$ sequences are paired up as follows: $$\{ 0011, 1100\}, \{ 0101, 1010 \}, \{0110, 1001\}.$$
Now suppose a sequence is not good. That means there are more than $n$ $0$s before the $n$th $1$. This means that there are fewer than $n$ $1$s before the $n$th $0$, so the complement of this sequence will be good. In other words, every complementary pair has at least one good sequence. This is where the $\frac12 \binom{4n}{2n}$ in the formula comes from.
However, there are some pairs where both sequences are good. In order for this to happen, there must be at most $n$ $0$s before the $n$th $1$, and at most $n$ $1$s before the $n$th $0$. This means that the first $2n$ elements must contain exactly $n$ $0$s and $n$ $1$s. There are $\binom{2n}{n}$ ways of choosing these first $2n$ elements. Since the last $2n$ elements must also have $n$ $0$s and $n$ $1$s, there are $\binom{2n}{n}$ ways of choosing the last $2n$ elements. Hence there are $\binom{2n}{n}^2$ sequences where both the sequence and its complement are good. We have already counted one from each such pair above, so we add half this to account for the second sequence in the pair, giving the $\frac12 \binom{2n}{n}^2$ term in the formula.
As a side remark, the formula in the OEIS entry is $\sum_{k=0}^n \binom{2n}{k}^2$. This comes from letting $k$ be the number of $0$s in the first $2n$ entries. This can be anything between $0$ and $n$, but it cannot be bigger than $n$ as otherwise there would be more than $n$ $0$s before the $n$th $1$. There are then $\binom{2n}{k}$ ways of choosing the first $2n$ entries. The last $2n$ elements must have exactly $k$ $1$s, and there are a further $\binom{2n}{k}$ ways of choosing those elements. This leads to the formula $\sum_{k=0}^n \binom{2n}{k}^2$.
• And now that I look at the OEIS entry more closely, I see, with deflating effect, that the closed form was there in the title all along. – Shagnik Aug 16 '16 at 14:16
• This is great way to look at this problem! Thank you very much for such detailed answer. – Krzysztof Lewko Aug 16 '16 at 14:40
• You're welcome, happy to help! – Shagnik Aug 17 '16 at 10:57
\begin{align} \sum_{k=0}^n\binom {n+k-1}{n-1}\binom {3n-k}n &=\sum_{k=0}^n\binom {n+k-1}{k}\binom {3n-k}{2n-k}\\ &=\sum_{k=0}^n(-1)^k\binom {-n}k (-1)^{2n-k}\binom {-n-1}{2n-k}&&\text{(upper negation)}\\ &=\sum_{k=0}^{n}\binom {-n}k\binom {-n-1}{2n-k}\\ \text{Note that}\sum_{k=0}^{2n}\binom {-n}k\binom {-n-1}{2n-k} &=\binom{-2n-1}{2n}&&\text{(Vandermonde)}\\ &=(-1)^{2n}\binom{4n}{2n}&&\text{(upper negation)}\\ &=\binom{4n}{2n}\\ \text{and that} \sum_{k=0}^{n-1}\binom {-n}k\binom {-n-1}{2n-k} &=\sum_{k=n}^{2n}\binom {-n}k\binom {-n-1}{2n-k}=\frac 12 \binom {4n}{2n}\\ \text{Hence} \sum_{k=0}^{n}\binom {-n}k\binom {-n-1}{2n-k} &=\left[\sum_{k=0}^{n-1}\binom {-n}k\binom {-n-1}{2n-k}\right]+\binom {-n}n\binom {-n-1}n\\ &=\frac 12 \binom {4n}{2n}+(-1)^n\binom {2n-1}n(-1)^n\binom{2n}n\\ &=\frac 12 \binom {4n}{2n}+\binom {2n-1}n\binom{2n}n\\ &=\frac 12 \binom {4n}{2n}+\frac 12\binom {2n}n\binom{2n}n &&\text{as }\binom {2n}n=\frac {2n}n\binom {2n-1}n\\ &=\color{red}{\frac 12 \left[\binom {4n}{2n}+\binom {2n}n^2\right] \qquad\blacksquare} \end{align}
• @MarkusScheuer - Thanks for your comment. Edited accordingly. – hypergeometric Aug 18 '16 at 17:07
• Nicely done! (+1) But, the real challenge is proving the middle part. Namely the line with: ...and that... :-) Could you also show this rather difficult part? – Markus Scheuer Aug 18 '16 at 17:16
• I've added an answer with a proof for the middle part. Regards, – Markus Scheuer Aug 19 '16 at 22:42
• @MarkusScheuer - Thanks for your proof! – hypergeometric Aug 20 '16 at 9:33
Here is an algebraic solution which is a supplement of the nice answer of @hypergeometric. Note that his answer contains a middle part which needs a proof and which is quite tricky to solve.
We use following approach \begin{align*} \sum_{k=0}^n\binom{n+k-1}{n-1}\binom{3n-k}{n} &=\sum_{k=0}^{2n}\binom{n+k-1}{n-1}\binom{3n-k}{n}\\ &\qquad-\sum_{k=n+1}^{2n}\binom{n+k-1}{n-1}\binom{3n-k}{n} \end{align*}
In the right-hand side we extend the range of the first sum to $2n$ which is easy to calculate and then we focus on the other sum, the surplus which is the somewhat more challenging part.
First sum: $0\leq k\leq 2n$
\begin{align*} \sum_{k=0}^{2n}\binom{n+k-1}{n-1}\binom{3n-k}{n} &=\sum_{k=0}^{2n}\binom{n+k-1}{k}\binom{3n-k}{2n-k}\\ &=\sum_{k=0}^{2n}\binom{-n}{k}\binom{-n-1}{2n-k}\\ &=\binom{-2n-1}{2n}\\ &=\binom{4n}{2n} \end{align*}
These steps use upper negation and Vandermonde's identity as we can see in the first half of the answer of @hypergeometric.
In order to calculate the second sum it is convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$. This way we can write e.g. \begin{align*} [z^k](1+z)^n=\binom{n}{k} \end{align*}
Second sum: $n+1\leq k \leq 2n$
We obtain \begin{align*} \sum_{k=n+1}^{2n}&\binom{n+k-1}{n-1}\binom{3n-k}{n}\\ &=\sum_{k=0}^{n-1}\binom{2n+k}{n-1}\binom{2n-k-1}{n-k-1}\tag{1}\\ &=\sum_{k=0}^\infty [z^{n-1}](1+z)^{2n+k}[u^{n-k-1}](1+u)^{2n-k-1}\tag{2}\\ &=[z^{n-1}](1+z)^{2n}[u^{n-1}](1+u)^{2n-1}\sum_{k=0}^\infty\left(\frac{u(1+z)}{1+u}\right)^k\tag{3}\\ &=[z^{n-1}](1+z)^{2n}[u^{n-1}](1+u)^{2n-1}\frac{1}{1-\frac{u(1+z)}{1+u}}\tag{4}\\ &=[z^{n-1}](1+z)^{2n}[u^{n-1}](1+u)^{2n}\frac{1}{1-uz}\tag{5}\\ &=[z^{n-1}](1+z)^{2n}[u^{n-1}]\sum_{k=0}^\infty \left(uz\right)^k(1+u)^{2n}\tag{6}\\ &=[z^{n-1}](1+z)^{2n}\sum_{k=0}^{n-1}z^k[u^{n-1-k}](1+u)^{2n}\tag{7}\\ &=[z^{n-1}](1+z)^{2n}\sum_{k=0}^{n-1}z^k\binom{2n}{n-1-k}\tag{8}\\ &=\sum_{k=0}^{n-1}[z^{n-1-k}](1+z)^{2n}\binom{2n}{n-1-k}\tag{9}\\ &=\sum_{k=0}^{n-1}\binom{2n}{n-1-k}^2\tag{10}\\ &=\sum_{k=0}^{n-1}\binom{2n}{k}^2\tag{11}\\ \end{align*}
Comment:
• In (1) we shift the index range from $k\in[n+1,2n]$ to $k\in[0,n-1]$ and we use the symmetry $\binom{p}{q}=\binom{p}{p-q}$ in the second factor as preparation for the next step.
• In (2) we apply the coefficient of operator twice and we extend the upper range of the series to $\infty$ without changing anything since thanks to the second factor we add only zeros.
• In (3) we do some rearrangements, use the linearity of the coefficient of operator and use the rule $[z^{p+q}]A(z)=[z^p]z^{-q}A(z)$.
• In (4) and (5) we use the geometric series expansion and do some simplifications.
• In (6) and (7) we expand $\frac{1}{1-zu}$ and apply the same rule as in (3). Since the exponent of $z^{n-k-1}$ is non-negative, we restrict the upper range of the sum with $n-1$.
• In (8) we select the coefficient of $z^{n-k-1}$.
• In (9) and (10) we do the same steps with $z$ as we did with $u$ in (7) and (8).
• In (11) we use the symmetry $\binom{p}{q}=\binom{p}{p-q}$.
Putting all together:
The sum (11) is due to its symmetry easily to manage. We observe on the one hand with the help of Vandermonde's identity \begin{align*} \sum_{k=0}^{2n}\binom{2n}{k}^2=\sum_{k=0}^{2n}\binom{2n}{k}\binom{2n}{n-k}=\binom{4n}{2n}\tag{12} \end{align*} On the other hand we have \begin{align*} \sum_{k=0}^{2n}\binom{2n}{k}^2 =\sum_{k=0}^{n-1}\binom{2n}{k}^2+\binom{2n}{n}^2+\sum_{k=n+1}^{2n}\binom{2n}{k}^2 \end{align*} and since \begin{align*} \sum_{k=n+1}^{2n}\binom{2n}{k}^2&=\sum_{k=0}^{n-1}\binom{2n}{k+n+1}^2\\ &=\sum_{k=0}^{n-1}\binom{2n}{2n-k}^2\qquad\qquad\qquad k\longrightarrow n-1-k\\ &=\sum_{k=0}^{n-1}\binom{2n}{k}^2 \end{align*} we obtain \begin{align*} \sum_{k=0}^{2n}\binom{2n}{k}^2=2\sum_{k=0}^{n-1}\binom{2n}{k}^2+\binom{2n}{n}^2\tag{13} \end{align*}
We obtain from (12) and (13) \begin{align*} \sum_{k=0}^{n-1}\binom{2n}{k}^2&=\frac{1}{2}\sum_{k=0}^{2n}\binom{2n}{k}^2-\frac{1}{2}\binom{2n}{n}^2\\ &=\frac{1}{2}\left(\binom{4n}{2n}-\binom{2n}{n}\right) \end{align*} | 2019-09-22T19:13:29 | {
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https://stats.stackexchange.com/questions/240325/distribution-of-quadratic-equation-roots-where-coefficients-are-generated-unifor | # Distribution of quadratic equation roots where coefficients are generated uniformly
We draw two points $p$ and $q$ at random from the interval $[−1, 1]$.
Let $x_1$ and $x_2$ denote the roots of the equation $x^2 + px + q = 0$.
Find the probability that, $x_1, x_2 \in \mathbb{R}$,
Given that, $x^2 + px + q = 0$, $\Delta = p^2-4q$.
$x_1, x_2 \in \Bbb R \Rightarrow \Delta \ge 0$
$\Omega = [-1, 1]^2 = \{(p,q) \in \Bbb R^2: p,q \in [-1,1]\}$
$A$ = roots exist.
$A = \{(p,q) \in [-1,1]^2 : p^2 -4q \ge 0 \}$
$x^2 - 4y \ge 0$
$y \le \frac{x^2}{4}$
$\Bbb P(A) = \frac{area-under-the-parabola}{4} = \frac{2 + 2 \cdot half-of-parabola }{4}$
$= \frac{1}{2} + \frac{1}{2} \cdot \int_{0}^{1}\frac{x^2}{4}\, dx$
$= \frac{13}{24}$
I have several questions regarding the aforementioned solution,
1. How did the author of the solution know that the parabola passing through the center of the axes and is situated at the positive side of the x-axis?
2. How was the area between the half of the parabola and the x axis calculated?
3. What was the mistake with this solution?
• please add the self-study tag and read its tag-wiki, if necessary modifying the question to follow the guidelines there. Oct 15, 2016 at 2:54
• Closely related: stats.stackexchange.com/questions/25661.
– whuber
Oct 15, 2016 at 16:43
• What do you mean by the "second method of solution"?
– whuber
Oct 24, 2016 at 12:51
• @whuber, quora.com/… Oct 24, 2016 at 17:40
There are two steps in any question of this nature: (1) find a useful way to characterize the event and (2) compute the probability of this event (in general, by integrating a probability density over it). When the probability is uniform, the integral is proportional to the area of the event.
I wish to emphasize the technique. It includes
• Drawing a picture of the event to keep focused on the key ideas.
• Minimizing the calculations performed.
• Delaying calculation as much as possible until the end, in the hope of simplification along the way.
### 1. Find a useful characterization of the event.
You need to relate the roots $$x_1, x_2$$ to the coefficients $$p$$ and $$q$$. That is done by comparing coefficients in the expansion
$$x^2 - (x_1+x_2) x + x_1 x_2 = (x-x_1)(x-x_2) = x^2 + px + q,$$
whence
$$p=-(x_1+x_2).\tag{1}$$
By completing the square you can establish a criterion for both roots to be real:
$$x^2 + px + q = (x - p/2)^2 + q - p^2/4 = \frac{4q - p^2}{4}.$$ This is zero if and only if
$$(x-p/2)^2 = \frac{p^2 - 4q}{4}.\tag{2}$$
Since the left hand side, which is a square of a real number when the roots are real, is non-negative, and all non-negative numbers have real square roots, both roots are real if and only if the right hand side is non-negative. Equivalently, its numerator must be non-negative.
We can now characterize the event of the problem in directly in terms of $$p$$ and $$q$$:
1. From $$(2)$$, both roots are real if and only if $$p^2 - 4q \ge 0$$.
2. From $$(1)$$, their sum $$x_1+x_2$$ is less than $$1$$ if and only if $$-p = x_1+x_2 \lt 1$$; more simply, $$p \gt -1$$.
Here is the graph of the set determined by (1) and (2). It is the blue region, including all portions of its boundary except a vertical line segment along the left side.
### 2. Find the probability of the event.
In this figure, which uses $$(p,q)$$ for coordinates, the probability is uniform. That means the probability of any event is its relative area. The total area of the square is $$4$$, while the area of the event itself is the area between the parabola $$q = p^2/4$$ and the line $$q=-1$$, from $$p=-1$$ to $$p=1$$. This area is
$$A = \int_{-1}^1 (p^2/4 - (-1)) dp = \frac{1}{4}\int_{-1}^1 p^2 dp + \int_{-1}^1 dp = \frac{1}{4}\left(\frac{p^3}{3}\big|_{-1}^1\right) + p\big|_{-1}^1=\frac{1}{6}+2.$$
Therefore the probability is
$$\Pr\left(x_1,x_2\in\mathbb{R},\ x_1+x_2\lt 1\right) = \Pr\left(p^2-4q \ge 0,\ -p \lt 1\right) = \frac{A}{4}=\frac{\frac{1}{6} + 2}{4} = \frac{13}{24}.$$
This value, which is slightly greater than $$12/24=1/2$$, is consistent with the visual impression from the picture that the blue area slightly exceeds $$1/2$$.
Incidentally, one mistake in the posted solution occurred at the very last line: the lower limit of integration over $$q$$ should be $$-1$$ rather than $$0$$. Drawing the picture makes this obvious. That solution contains several other mistakes, of which the most important is the misquoting of the Quadratic Formula: the expressions for the roots are wrong. They need to be divided by $$2$$.
It's possible the answer is a bit confusing because the question starts with one parabola (involving $p, q, x$), and the solution uses a different parabola (which should involve $p, q$, but the answer switches the variables to $x, y$, even though the second use of $x$ is unrelated to the first).
If the quadratic equation is $x^2 + px + q = 0$, then to have real roots, the discriminant must be non-negative, i.e., $p^2 \ge 4q$, or $q \le \frac{p^2}{4}$. Now consider the following digram in the $pq$ plane:
Here, $p$ is the horizontal axis, and $q$ is the vertical axis. The red square indicates the admissible area for $p, q$ values (the corners are where $p, q = \pm 1, \pm 1$). The parabola $q < \frac{p^2}{4}$ is indeed symmetric around the vertical $q$ axis. Note that this parabola is unrelated to the original one. The grey region is the region where the discriminant is positive, and, from simple geometric considerations concerning uniform distribution, the ratio of its area to that of the red square is the requested probability.
The area in the square is $4 = (1 - (-1))^2$. The grey area can be found by calculating $\int_{-1}^1\frac{p^2}{4}dp + 2$ (the sum of the two lower grey squares is 2).
For the second part of the question, note that the sum of roots is $-p$.
• "this parabola is unrelated to the original one" --- why? And, if that is true, how do you explain the relationship of both parabolas? Oct 15, 2016 at 0:07
• The second parabola is just a parabola describing the discriminant of the first one - that is the only connection between them. Otherwise, they are different parabolas over different variables. Oct 15, 2016 at 0:09
• What was the mistake with this solution? quora.com/… Oct 15, 2016 at 1:30
• @anonymous Offhand, the internal integral $\int_{-1}^1\int_0^{\frac{p^2}{4}}\frac{1}{4}dpdq$ has a mistake: it integrated from 0, and not from -1. That ignored the two lower grey squares. Oct 15, 2016 at 6:26
OP
Looks like I found you again. The incorrect solution snapshot above is mine, and I apologize. Apparently I do not know integration limits and the quadratic equation, but that's besides the point. Here's the corrections.
• Could you please answer this? quora.com/… Nov 30, 2016 at 7:35 | 2022-08-16T16:28:16 | {
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https://math.stackexchange.com/questions/2562405/a-trapezoidal-approximation-for-a-sequence-of-uniformly-converging-functions | A trapezoidal approximation for a sequence of uniformly converging functions
Let $f_n : [0,1] \rightarrow \mathbb R$ be a sequence of continuous functions converging uniformly to a function $f$. For each $N \in \mathbb N$, also define a sequence of points $0=x_1< x_2 <...<x_N=1$.
Does it hold that:
$$\lim_{N \rightarrow\infty} \sum_{k=1}^{N} \frac{f_N(x_{k-1}) + f_N(x_k)}{2} (x_k - x_{k-1}) = \int_0^1 f(x) dx$$
I have an intuition that this should hold, but I have no idea how to prove it.
• You haven't said enough about this sequence $x_k$ – zhw. Dec 11 '17 at 23:42
• My bad. I think a uniform partition would suffice. – Wolfups Dec 11 '17 at 23:58
Since $f_n(x) \to f(x)$ uniformly on $[0,1]$ with $f_n$ continuous, it follows that $f$ is integrable. As long as the partition $P_N = (x_0,x_1, \ldots, x_N)$ is constructed for each $N$ in such a way that the norm $\|P_N\| \to 0$ as $N \to \infty$, then any tagged Riemann sum $S(P_N,f,T)$ converges to the integral regardless of the choice of tags.
In particular, this convergence is true for left- and right-hand sums as well as the average of the two: $$\tag{1}\lim_{N \to \infty}A(P_N,f) = \lim_{N \to \infty} \sum_{k=1}^N \frac{1}{2}(f(x_{k-1}) + f(x_{k}))(x_k - x_{k-1}) = \int_0^1 f(x) \, dx.$$
Note that
$$\tag{2}A(P_N,f_N) = \sum_{k=1}^N \frac{1}{2}(f_N(x_{k-1}) + f_N(x_{k}))(x_k - x_{k-1}) \\= \sum_{k=1}^N \frac{1}{2}\{[f_N(x_{k-1})-f(x_{k-1})] + [f_N(x_{k})-f(x_k)]\}\,(x_k - x_{k-1}) + \sum_{k=1}^N \frac{1}{2}(f(x_{k-1}) + f(x_{k}))(x_k - x_{k-1}).$$
Hence, using (2) and applying the triangle inequality we have
$$\left|A(P_N,f_N) - \int_0^1 f(x) \, dx\right| \\ \leqslant \frac{1}{2}\sum_{k=1}^N |f_N(x_{k-1})-f(x_{k-1})|\, (x_k - x_{k-1}) + \frac{1}{2}\sum_{k=1}^N |f_N(x_{k})-f(x_k)|\,(x_k - x_{k-1}) +\left|A(P_N,f) - \int_0^1 f(x) \, dx\right|.$$
By uniform convergence, for any $\epsilon > 0$ there exists $N_1$ such that if $N > N_1$, then $|f_N(x) - f(x)| < \epsilon/2$ for every $x \in [0,1]$ and
$$\frac{1}{2}\sum_{k=1}^N |f_N(x_{k-1})-f(x_{k-1})|\, (x_k - x_{k-1}) + \frac{1}{2}\sum_{k=1}^N |f_N(x_{k})-f(x_k)|\,(x_k - x_{k-1}) < \epsilon/2,$$
since $\sum_{k=1}^N(x_k - x_{k-1}) = 1$.
By the convergence in (1) there exists $N_2$ such that if $N > N_2$, then $$\left|A(P_N,f) - \int_0^1f(x) \, dx\right| < \epsilon/2.$$
Therefore, if $N > \max(N_1,N_2)$ we have $|A(P_N,f_N) - \int_0^1 f(x)\, dx | < \epsilon,$ implying that
$$\lim_{N \to \infty}A(P_N, f_N) = \int_0^1 f(x) \, dx.$$
• @Wolfups: Your conjecture is true whether or not the partitions are uniform as long as the sequence of partition norms $\|P_n\| = \max_{1 \leqslant k \leqslant n} |x_k - x_{x-1}|$converges to $0$. – RRL Dec 12 '17 at 2:33
Hint: Define, for any $g$ on $[0,1],$
$$S_N(g) = \sum_{k=1}^{N}\frac{g((k-1)/N) + g(k/N)}{2}\frac{1}{N.}$$
Then $S_N(g)| \le \|g\|_\infty.$ Furthermore, if $g$ is Riemann integrable on $[0,1],$ then $S_N(g) \to \int_0^1 g.$
So consider in your problem $S_N(f_N) - S_N(f) = S_N(f_N-f).$ | 2019-12-06T03:33:22 | {
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https://dsp.stackexchange.com/questions/47295/periodicity-of-the-discrete-time-fourier-transform | # Periodicity of the discrete-time Fourier Transform
The DTFT of a sequence $x[n]$ can be written as $$X(e^{j\omega}) = \sum_{n = -\infty}^{\infty} x[n] e^{-j\omega n}.$$ Is the smallest (fundamental) period in frequency of the DTFT always $2\pi$? Or can it be smaller than $2\pi$?
In addition, I was wondering why we cannot use that same notation for the argument with the CTFT. We typically denote the continuous-time Fourier Transform by $X(j\omega)$. But what's wrong with using $X(e^{j\omega})$ where $$X(e^{j\omega}) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt?$$
I know for a fact that this is wrong because the notation inherently implies the periodicity of $X(e^{j\omega})$, since $X(e^{j(\omega + 2\pi)}) = X(e^{j\omega})$, which is obbiously not true in this case. But I don't understand what mathematically does not permit us to use $e^{j\omega}$ as an argument in the continuous-time case.
The DTFT is always $2\pi$-periodic. However, it can also have a smaller period, namely a fraction of $2\pi$. Take any sequence $x[n]$ for which the DTFT exists and insert $L-1$ zeros between the samples. The DTFT of the new sequency $\hat{x}[n]$ can then be written as
$$\hat{X}(e^{j\omega})=\sum_{n=-\infty}^{\infty}\hat{x}[n]e^{-jn\omega}=\sum_{n=-\infty}^{\infty}\hat{x}[nL]e^{-jnL\omega}\tag{1}$$
$\hat{X}(e^{j\omega})$ as given by $(1)$ clearly has a period of $2\pi/L$.
Concerning your second question, the reason why a continuous-time Fourier transform $X(j\omega)$ cannot be written as $X(e^{j\omega})$ is that
$$X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\neq \int_{-\infty}^{\infty}x(t)(e^{j\omega})^{-t}dt\tag{2}$$
simply because generally
$$e^{-j\omega t}\neq (e^{j\omega})^{-t}\tag{3}$$
unless $t$ is an integer (as is the case with the DTFT).
Note that if $(3)$ were not true, i.e., if $e^{-j\omega t}= (e^{j\omega})^{-t}$ were true, we could easily show that $e^{-j\omega t}=1$ for all $t$. Just write $\omega=2\pi/T$ and you get
$$e^{-j\omega t}=e^{-j2\pi t/T}=\left(e^{-j2\pi}\right)^{t/T}=1^{t/T}=1$$
which is of course absurd.
Proof of Eq. $(3)$:
Let $z$ be a complex number with $|z|=1$ (the magnitude is irrelevant here):
$$z=e^{j\theta}\tag{4}$$
Let $a$ and $b$ be real numbers. Then
$$z^{ab}=e^{j\theta a b}\tag{5}$$
And with $z^a=u$
$$\left(z^a\right)^b=u^b=e^{j\arg\{u\}b}\tag{6}$$
With $\arg\{u\}=\text{pv}\{\theta a\}\in (-\pi,\pi]$, where $\text{pv}$ denotes the principal value, it is clear that generally
$$z^{ab}=e^{j\theta a b}\neq \left(z^a\right)^b=e^{j\,\text{pv}\{\theta a\}b}\tag{7}$$
There are two cases where $(5)$ and $(6)$ are equal:
1. if $\text{pv}\{\theta a\}=\theta a$, which is the case if $\theta a\in (-\pi,\pi]$.
2. if $b$ is integer, since
$$\text{pv}\{\theta a\}=\theta a+2\pi k\tag{8}$$
with some appropriately chosen integer $k$ (such that the result is in the interval $(-\pi,\pi]$), then
$$\left(z^a\right)^b=e^{j\,\text{pv}\{\theta a\}b}=e^{j(\theta a+2\pi k)b}=e^{j\theta ab}=z^{ab},\qquad b\in\mathbb{Z}\tag{9}$$
[Note that $(9)$ would also hold for rational $b$ as long as $kb$ is integer.]
• Why is $e^{-j\omega t} = (e^{j\omega})^{-t}$ invalid? Is it because we are not allowed to raise a complex number to irrational and/or transcendental powers? – 0MW Feb 20 '18 at 16:38
• @0MW: You can raise a complex number to any other complex number, but the two complex numbers under consideration are simply not equal. $(e^{j\omega})^{-t}$ is $2\pi$-periodic in $\omega$ and $e^{-j\omega t}$ isn't. With $e^{j\omega}$ you lose information because it gives the same value for $\omega+2\pi k$, which isn't the case for the other function. – Matt L. Feb 20 '18 at 17:24
• I understand that the DTFT can have a period less than $2\pi$, but I was recently reading up on the duality between the DTFT and the continuous-time fourier series: namely, the DTFT can be thought of as a continuous-time fourier series in $\omega$ with period $2\pi$. This last bit is important, since the factor of $2\pi$ appears in the synthesis equation. Does this duality break down if the DTFT is periodic with a period less than $2\pi$? – 0MW Feb 23 '18 at 7:44
• @0MW: The analogy always holds; if the period is smaller, then you have zeros in between the coefficients in both cases. – Matt L. Feb 23 '18 at 9:40
You could define a DTFT that has a fundamental period that is less than $2\pi$, and try to find its IDTFT. We say that it's periodic in $2\pi$ because that's true for all DTFTs, but there might be cases where the period is smaller (even though they might not have an easy-to-find inverse). A trivial case would be the transform of the Dirac delta, whose DTFT is a constant and so has any period you want.
Regarding the notation, it's just a way to distinguish between continuous and discrete signals when working on the frequency domain. When you have ADCs and DACs, things may became a bit confusing if we use the same notation for all the transforms.
Note that the only thing that matters is that they are functions of $\omega$, so one could just write $X(\omega)$ in any case and it would be valid.
The fact that we use $j\omega$ in the continuous case comes from the fact that the Fourier transform is (in general) the Laplace transform evaluated when $s=j\omega$. So if we have some Laplace transform $X_L(s)$, we can evaluate that in the imaginary axis to get the Fourier transform, and thus the convention.
In the discrete case, something similar happens. We usually link the z-transform and the DTFT evaluating the former at $z=e^{j\omega}$, and the same as in the continuous case happens.
• The question is why a CTFT cannot be written as $X(e^{j\omega})$, and it looks like you discussed the difference between the notations $X(\omega)$ and $X(j\omega)$. – Matt L. Feb 20 '18 at 8:45
• @MattL. You are right, I read the question shallowly and misinterpreted it. I'll leave it here though, as I think it can help someone in the future, maybe. +1 to your answer. – Tendero Feb 20 '18 at 12:25
to answer your last question, it turns out (this is ultra-fundamental) that an eigenfunction for a Linear Time-Invariant (LTI) System is the exponential input. this is true whether it's Continuous time:
$$x(t) = e^{st}$$
or if it's Discrete time
$$x[n] = e^{\log(z) n} = z^n$$
This means if the eigenfunction goes into the LTI, the same function comes out but scaled by a constant. I think they call that constant an "eigenvalue". Turns out that the output of the two cases above are:
$$y(t) = H(s) \, x(t)$$
and
$$y[n] = H(z) \, x[n]$$
So, specifically for the complex sinusoid:
$$x(t) = e^{j \Omega t}$$
or, if it's Discrete time
$$x[n] = e^{j \omega n}$$
then what comes out is:
$$y(t) = H(j \Omega) \, x(t)$$
or
$$y[n] = H(e^{j \omega}) \, x[n]$$
• This is all true and important, but I don't see how it answers the OP's second question. – Matt L. Feb 20 '18 at 8:43
• it answers why we use $X(s)\bigg|_{s=j\omega}$ in the continuous-time case and $X(z)\bigg|_{z=e^{j\omega}}$ in the discrete-time case. – robert bristow-johnson Feb 20 '18 at 17:51
• That's right, but the question was - as far as I understood - why we don't (and cannot) use $X(e^{j\omega})$ in the continuous domain. – Matt L. Feb 20 '18 at 17:54
• because we have to use $X(s)\bigg|_{s=j\omega}$ instead. we can't use both. – robert bristow-johnson Feb 21 '18 at 1:59
• Yes, but I think that the most important reason is that, unlike the DTFT, the CTFT is no function of $e^{j\omega}$. – Matt L. Feb 21 '18 at 6:33 | 2019-05-23T20:25:41 | {
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https://math.stackexchange.com/questions/2821613/understanding-solution-to-the-probability-problem-from-sheldon-ross-book | # Understanding solution to the probability problem from Sheldon Ross' book
I came across following problem from Sheldon Ross' book:
A closet contains 10 pairs of shoes. If 8 shoes are randomly selected, what is the probability that there will be
1. no complete pair;
2. exactly one complete pair
I solved them as follows:
Problem 1
• 1st shoe can be anyone out of 20: $\frac{20}{20}$
• 2nd shoe can be anyone out of remaining 19 except the one forming pair with previously selected shoe: $\frac{18}{19}$
• 3rd shoe can be anyone out of remaining 18 except those forming pair with previously selected shoes: $\frac{16}{18}$
and so on till we choose 8 shoes. We need to take multiplication of all, leading to $\frac{20\times18\times16\times14\times12\times10\times8\times6}{20\times19\times18\times17\times16\times15\times14\times13}=0.091$
Problem 2
Following same logic of problem 1,
• 1st shoe can be any one out of 20: $\frac{20}{20}$
• 2nd show should be the one forming pair with 1st one
• 3rd shoe can be any one out of remaining 18: $\frac{18}{18}$
• 4th show can be any one out of remaining 17 except the one forming pair with previously selected one: $\frac{16}{17}$
and so on till we choose 8 shoes. We need to take multiplication of all, leading to $\frac{20\times1\times18\times16\times14\times12\times10\times8}{20\times19\times18\times17\times16\times15\times14\times13}=0.015$
It turns out that the solution to first problem is correct, but the solution to second problem is incorrect. Its given as follows: $\frac{\binom{10}{1}\binom{9}{6}\color{red}{\frac{8!}{2!}}2^6}{20\times19\times18\times17\times16\times15\times14\times13}$
I understand we can select one pair out of 10 in $\binom{10}{1}$ ways. We select both shoes from this pair. Then we can select six pairs out of remaining nine in $\binom{9}{6}$ ways. We can select any one out of two shoes of each of six pairs in $2^6$ ways. But I dont understand from where $\frac{8!}{2!}$ came.
Edit
The solution given at the back of the book is $0.4268$. But books solution manual solves it as $\frac{\binom{10}{1}\binom{9}{6}\color{red}{\frac{8!}{2!}}2^6}{20\times19\times18\times17\times16\times15\times14\times13}$ which I just checked to be equal to $0.2133$. This pdf gives the solution as $\frac{\binom{10}{1}\binom{9}{6}\times 2^6}{\binom{20}{8}}$ which matches with $0.4268$. So now I am guessing what is correct answer and how can I get the answer for problem 2 by following same approach as I followed for problem 1.
Your text is in error. There should be no $8!/2!$ factor.
In general the probability for selecting $x$ pairs $(x\in \{0,1,2,3,4\})$ is$$\dfrac{\dbinom{10}{x}\dbinom{10-x}{8-2x}2^{8-2x}}{\dbinom{20}8}\quad\text{or}\quad\dfrac{\dbinom{10}{8-x}\dbinom{8-x}{x}2^{8-2x}}{\dbinom{20}8}$$
The series sums to 1 as required.
Thus the probability for selecting no pairs is $\binom{10}{8}2^8/\binom{20}{8}$, which is equal to your answer.
And the correct probability for selecting exactly one pair is $\binom {10}{1}\binom{10-1}{8-2}2^{8-2}/\binom{20}{8}$. Alternatively that is $\binom{10}{8-1}\binom{8-1}12^{8-2}/\binom{20}8$.
By your method you can select six from the 10 pairs, one shoe from each of those, and both shoes of one from the remaining pair.
$$\dfrac{20\cdot18\cdot 16\cdot14\cdot12\cdot 10\cdot 4/6!}{20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13}$$
• yess that is correct answer, can you please have a look at edit I added at the end of my question? I want to know whether it is possible to solve 2nd problem with the same approach as I followed for 1st problem, that is multiplication of probabilities of successively selecting shoes instead of using binomials. – anir Jun 16 '18 at 14:08
• ohkay it seems that it should be $8!$ instead of $\frac{8!}{2!}$, right? – anir Jun 16 '18 at 14:10
• Thanks for the new approach you stated at the starting. I have two doubts: (1) Is that $\frac{8!}{2!}$ absolutely senseless? I mean replacing it with $8!$ gives same final result. But, I don't feel its logically correct. The solution looks like this: $\frac{\binom{10}{1}\binom{9}{6}2^68!}{20.19.18.17.16.15.14.13}$. Here the numerator follows pattern of solution $\frac{\binom{10}{1}\binom{9}{6}\times 2^6}{\binom{20}{8}}$ stated in the edit added to the question, while denominator is following [continued...] – anir Jun 16 '18 at 16:13
• [...continued] the pattern of solution $\frac{20\times18\times16\times14\times12\times10\times8\times6}{20\times19\times18\times17\times16\times15\times14\times13}$, I gave for problem 1. (2) How $\frac{4/6!}{14.13}$ = "both shoes of one from remaining pairs" – anir Jun 16 '18 at 16:13
This is a problem similar to deck cards and hands problems.
The twenty shoes may be seen as a deck of two color (Left and Right) and 10-suits.
A "hand" contains 8 cards. It may contain 0 pairs, 1 pair... to 4 pairs, and we are asked to evaluate the chances to get some of these types.
the total number of hands is $\binom {20}{8}$
i) for no pairs, we have $\binom {10}{8}$ for the choice of the 8 distinct values and $2^8$ for the eight color choices (left or right). The probability of such a "hand" is then 0.0914
ii) for one pair, we have to choose it among ten possibles. Then we have to chose another 6 singles having distinct values from 9 values that remained. Then we have 2^6 choices for the color, for a total $\binom {10} {1} \binom {9}{6} 2^6$. The probability to get a hand like this is 0.4267 | 2020-01-27T16:45:02 | {
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https://math.libretexts.org/Bookshelves/PreAlgebra/Book%3A_Prealgebra_(Arnold)/01%3A_The_Whole_Numbers/1.07%3A_Solving_Equations_by_Multiplication_and_Division | # 1.7: Solving Equations by Multiplication and Division
In Section 1.6, we stated that two equations that have the same solutions are equivalent. Furthermore, we saw that adding the same number to both sides of an equation produced an equivalent equation. Similarly, subtracting the same the number from both sides of an equation also produces an equivalent equation. We can make similar statements for multiplication and division.
Multiplying both Sides of an Equation by the Same Quantity
Multiplying both sides of an equation by the same quantity does not change the solution set. That is, if
$a = b\nonumber$
then multiplying both sides of the equation by c produces the equivalent equation
$a \cdot c = b \cdot c\nonumber$
provided c ≠ 0.
A similar statement can be made about division.
Dividing both Sides of an Equation by the Same Quantity
Dividing both sides of an equation by the same quantity does not change the solution set. That is, if
$a = b\nonumber$
then dividing both sides of the equation by c produces the equivalent equation
$\frac{a}{c} = \frac{b}{c},\nonumber$
provided c ≠ 0.
In Section 1.6, we saw that addition and subtraction were inverse operations. If you start with a number, add 4 and subtract 4, you are back to the original number. This concept also works for multiplication and division.
Multiplication and Division as Inverse Operations
Two extremely important observations:
The inverse of multiplication is division. If we start with a number x and multiply by a number a, then dividing the result by the number a returns us to the original number x. In symbols,
$\frac{a \cdot x}{a} = x.\nonumber$
The inverse of division is multiplication. If we start with a number x and divide by a number a, then multiplying the result by the number a returns us to the original number x. In symbols,
$a \cdot \frac{x}{a} = x.\nonumber$
Let's put these ideas to work.
Example 1
Solve the equation 3x = 24 for x.
Solution
To undo the effects of multiplying by 3, we divide both sides of the equation by 3.
\begin{aligned} 3x= 24 ~ & \textcolor{red}{ \text{ Original equation.}} \\ \frac{3x}{3} = \frac{24}{3} ~ & \textcolor{red}{ \text{Divide both sides of the equation by 3.}} \\ x = 8 ~ & \textcolor{red}{ \text{ On the left, dividing by 3 "undoes" the effect}} \\ ~ & \textcolor{red}{ \text{ of multiplying by 3 and returns to } x. \text{ On the right,}} \\ ~ & \textcolor{red}{ 24/3 = 8.} \end{aligned}\nonumber
To check, substitute the solution 8 into the original equation.
\begin{aligned} 3x = 24 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 3(8) = 24 ~ & \textcolor{red}{ \text{Substitute 8 for } x.} \\ 24 = 24 ~ & \textcolor{red}{ \text{ Simplify both sides.}} \end{aligned}\nonumber
That fact that the last line of our check is a true statement guarantees that 8 is a solution of 3x = 24.
Exercise
Solve for x: 5x = 120.
24
Example 2
Solve the following equation for x.
$\frac{x}{7} = 12\nonumber$
Solution
To undo the effects of dividing by 7, we multiply both sides of the equation by 7.
\begin{aligned} \frac{x}{7} = 12 ~ & \textcolor{red}{ \text{Original equation.}} \\ \frac{84}{7} = 12 ~ & \textcolor{red}{ \text{ Multiply both sides of the equation by 7.}} \\ x = 84 ~ & \textcolor{red}{ \text{ On the left, multiplying by 7 "undoes" the effect}} \\ ~ & \textcolor{red}{ \text{ of dividing by 7 and returns to } x. \text{ On the right,}} \\ ~ & \textcolor{red}{ 7 \cdot 12 = 84.} \end{aligned}\nonumber
To check, substitute the solution 84 into the original equation.
\begin{aligned} \frac{x}{7} = 12 & \textcolor{red}{ \text{ Original equation.}} \\ \frac{84}{7} = 12 ~ & \textcolor{red}{ \text{ Substitute 84 for } x.} \\ 12 = 12 ~ & \textcolor{red}{ \text{ Simplify both sides.}} \end{aligned}\nonumber
That fact that the last line of our check is a true statement guarantees that 84 is a solution of x/7 = 12.
Exercise
Solve for x: x/2 = 19
38
## Word Problems
In Section 1.6 we introduced Requirements for Word Problem Solutions. Those requirements will be strictly adhered to in this section.
Example 3
Fifteen times a certain number is 45. Find the unknown number.
Solution
In our solution, we will carefully address each step of the Requirements for Word Problem Solutions.
1. Set up a Variable Dictionary. We can satisfy this requirement by simply stating “Let x represent a certain number.”
2. Set up an equation. “Fifteen times a certain number is 45” becomes
$\begin{array}{c c c c} \colorbox{cyan}{15} & \text{times} & \colorbox{cyan}{a certain number} & \text{is} & \colorbox{cyan}{45} \\ 15 & \cdot & x & = & 45 \end{array}\nonumber$
3. Solve the Equation. To “undo” the multiplication by 15, divide both sides of the equation by 15.
\begin{aligned} 15x = 45 ~ & \textcolor{red}{ \text{ Original equation. Write 15 } \cdot x \text{ as 15}x} \\ \frac{15x}{15} = \frac{45}{15} ~& \textcolor{red}{ \text{ Divide both sides of the equation by 15.}} \\ x = 3 ~ & \textcolor{red}{ \text{ On the left, dividing by 15 "undoes" the effect}} \\ ~ & \textcolor{red}{ \text{ of multiplying by 15 and returns to } x. \text{ On the right,}} \\ ~ & \textcolor{red}{45/15 = 3.} \end{aligned}\nonumber
4. Answer the Question. The unknown number is 3.
5. Look Back. Does the solution 3 satisfy the words of the original problem? We were told that “15 times a certain number is 45.” Well, 15 times 3 is 45, so our solution is correct.
Exercise
Seven times a certain number is one hundred five. Find the unknown number.
15
Example 4
The area of a rectangle is 120 square feet. If the length of the rectangle is 12 feet, find the width of the rectangle.
Solution
In our solution, we will carefully address each step of the Requirements for Word Problem Solutions.
1. Set up a Variable Dictionary. When geometry is involved, we can create our variable dictionary by labeling a carefully constructed diagram. With this thought in mind, we draw a rectangle, then label its length, width, and area.
The figure makes it clear that W represents the width of the rectangle. The figure also summarizes information needed for the solution.
2. Set up an equation. We know that the area of a rectangle is found by multiplying its length and width; in symbols,
$A = LW.\nonumber$
We’re given the area is A = 120 ft2 and the length is L = 12 ft. Substitute these numbers into the area formula (1.1) to get
$120 = 12W.\nonumber$
3. Solve the Equation. To “undo” the multiplication by 12, divide both sides of the equation by 12.
\begin{aligned} 120 = 12W ~ & \textcolor{red}{ \text{ Our equation.}} \\ \frac{120}{12} = \frac{12W}{12} ~ & \textcolor{red}{ \text{ Divide both sides of the equation by 12.}} \\ 10 = W ~ & \textcolor{red}{ \text{ On the right, dividing by 12 "undoes" the effect}} \\ ~ & \textcolor{red}{ \text{ of multiplying by 12 and returns to } W. \text{ On the left,}} \\ ~ & \textcolor{red}{120/12 = 10.} \end{aligned}\nonumber
4. Answer the Question. The width is 10 feet.
5. Look Back. Does the found width satisfy the words of the original problem? We were told that the area is 120 square feet and the length is 12 feet. The area is found by multiplying the length and width, which gives us 12 feet times 10 feet, or 120 square feet. The answer works!
Exercise
The area of a rectangle is 3,500 square meters. If the width is 50 meters, find the length.
70 meters
Example 5
A class of 23 students averaged 76 points on an exam. How many total points were accumulated by the class as a whole?
Solution
In our solution, we will carefully address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. We can set up our variable dictionary by simply stating “Let T represent the total points accumulated by the class.” 2. Set up an equation. To find the average score on the exam, take the total points accumulated by the class, then divide by the number of students in the class. In words and symbols,
$\begin{array}{c c c c c} \colorbox{cyan}{Total Points} & \text{divided by} & \colorbox{cyan}{ Number of Students} & \text{equals} & \text{Average Score} \\ T & \div & 23 & = & 76 \end{array}\nonumber$
An equivalent representation is
$\frac{T}{23} = 76.\nonumber$
3. Solve the Equation. To “undo” the division by 23, multiply both sides of the equation by 23.
\begin{aligned} \frac{T}{23} = 76 & \textcolor{red}{ \text{ Our equation.}} \\ 23 \cdot \frac{T}{23} = 76 \cdot 23 & \textcolor{red}{ \text{ Multiply both sides of the equation by 23.}} \\ T = 1748 & \textcolor{red}{ \text{ On the left, multiplying by 23 "undoes" the effect}} \\ ~ & \textcolor{red}{ \text{ of dividing by 23 and returns to } T. \text{ On the right, }} \\ ~ & \textcolor{red}{76 \cdot 23 = 1748.} \end{aligned}\nonumber
4. Answer the Question. The total points accumulated by the class on the exam is 1,748.
5. Look Back. Does the solution 1,748 satisfy the words of the original problem? To find the average on the exam, divide the total points 1,748 by 23, the number of students in the class. Note that this gives an average score of 1748 ÷ 23 = 76. The answer works!
Try it out!
A class of 30 students averaged 75 points on an exam. How many total points were accumulated by the class as a whole?
2,250
## Exercises
In Exercises 1-12, which of the numbers following the given equation are solutions of the given equation?
1. $$\frac{x}{6} = 4$$; 24, 25, 27, 31
2. $$\frac{x}{7} = 6$$; 49, 42, 43, 45
3. $$\frac{x}{2} = 3$$; 6, 9, 13, 7
4. $$\frac{x}{9} = 5$$; 45, 46, 48, 52
5. $$5x = 10$$; 9, 2, 3, 5
6. $$4x = 36$$; 12, 16, 9, 10
7. $$5x = 25$$; 5, 6, 8, 12
8. $$3x = 3$$; 1, 8, 4, 2
9. $$2x = 2$$; 4, 8, 1, 2
10. $$3x = 6$$; 2, 9, 5, 3
11. $$\frac{x}{8} = 7$$; 57, 59, 63, 56
12. $$\frac{x}{3} = 7$$; 24, 21, 28, 22
In Exercises 13-36, solve the given equation for x.
13. $$\frac{x}{6} = 7$$
14. $$\frac{x}{8} = 6$$
15. $$2x = 16$$
16. $$2x = 10$$
17. $$2x = 18$$
18. $$2x = 0$$
19. $$4x = 24$$
20. $$2x = 4$$
21. $$\frac{x}{4} = 9$$
22. $$\frac{x}{5} = 6$$
23. $$5x = 5$$
24. $$3x = 15$$
25. $$5x = 30$$
26. $$4x = 28$$
27. $$\frac{x}{3} = 4$$
28. $$\frac{x}{9} = 4$$
29. $$\frac{x}{8} = 9$$
30. $$\frac{x}{8} = 2$$
31. $$\frac{x}{7} = 8$$
32. $$\frac{x}{4} = 6$$
33. $$2x = 8$$
34. $$3x = 9$$
35. $$\frac{x}{8} = 5$$
36. $$\frac{x}{5} = 4$$
37. The price of one bookcase is $370. A charitable organization purchases an unknown number of bookcases and the total price of the purchase is$4,810. Find the number of bookcases purchased.
38. The price of one computer is $330. A charitable organization purchases an unknown number of computers and the total price of the purchase is$3,300. Find the number of computers purchased.
39. When an unknown number is divided by 3, the result is 2. Find the unknown number.
40. When an unknown number is divided by 8, the result is 3. Find the unknown number.
41. A class of 29 students averaged 80 points on an exam. How many total points were accumulated by the class as a whole?
42. A class of 44 students averaged 87 points on an exam. How many total points were accumulated by the class as a whole?
43. When an unknown number is divided by 9, the result is 5. Find the unknown number.
44. When an unknown number is divided by 9, the result is 2. Find the unknown number.
45. The area of a rectangle is 16 square cm. If the length of the rectangle is 2 cm, find the width of the rectangle.
46. The area of a rectangle is 77 square ft. If the length of the rectangle is 7 ft, find the width of the rectangle.
47. The area of a rectangle is 56 square cm. If the length of the rectangle is 8 cm, find the width of the rectangle.
48. The area of a rectangle is 55 square cm. If the length of the rectangle is 5 cm, find the width of the rectangle.
49. The price of one stereo is $430. A charitable organization purchases an unknown number of stereos and the total price of the purchase is$6,020. Find the number of stereos purchased.
50. The price of one computer is $490. A charitable organization purchases an unknown number of computers and the total price of the purchase is$5,880. Find the number of computers purchased.
51. A class of 35 students averaged 74 points on an exam. How many total points were accumulated by the class as a whole?
52. A class of 44 students averaged 88 points on an exam. How many total points were accumulated by the class as a whole?
53. 5 times an unknown number is 20. Find the unknown number.
54. 5 times an unknown number is 35. Find the unknown number.
55. 3 times an unknown number is 21. Find the unknown number.
56. 2 times an unknown number is 10. Find the unknown number.
1. 24
3. 6
5. 2
7. 5
9. 1
11. 56
13. 42
15. 8
17. 9
19. 6
21. 36
23. 1
25. 6
27. 12
29. 72
31. 56
33. 4
35. 40
37. 13
39. 6
41. 2,320
43. 45
45. 8 cm
47. 7 cm
49. 14
51. 2,590
53. 4
55. 7
This page titled 1.7: Solving Equations by Multiplication and Division is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Arnold. | 2023-03-27T23:28:03 | {
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http://openstudy.com/updates/55c74b0fe4b06d80932f23d2 | ## mathmath333 one year ago Question maths/reasoning
1. mathmath333
|dw:1439124280221:dw|
2. mathmath333
\large \color{black}{\begin{align} & \normalsize \text{In the figure: }\hspace{.33em}\\~\\ & \normalsize \text{K represents all Kites}\hspace{.33em}\\~\\ & \normalsize \text{Q represents all Quadrilaterals}\hspace{.33em}\\~\\ & \normalsize \text{R represents all Rhombus}\hspace{.33em}\\~\\ & \normalsize \text{P represents all Parallelogram}\hspace{.33em}\\~\\ & \normalsize \text{The statement "Rhombus is also a Kite"}\hspace{.33em}\\~\\ & \normalsize \text{can be described as}\hspace{.33em}\\~\\ & 1.) \normalsize \text{P and K is nothing but R}\hspace{.33em}\\~\\ & 2.) \normalsize \text{P or K is nothing but R}\hspace{.33em}\\~\\ & 3.) \normalsize \text{P and R is nothing but K}\hspace{.33em}\\~\\ & 4.) \normalsize \text{P or R is nothing but K}\hspace{.33em}\\~\\ \end{align}}
3. ganeshie8
|dw:1439124864364:dw|
4. ganeshie8
we represent common region using "$$\cap$$" and read it out as "$$\text{and}$$"
5. mathmath333
ok
6. ganeshie8
In above venn diagram, we have $P \text{ and } K = R$
7. ganeshie8
therefore a $$R$$hombus is both a $$P$$arallelogram and a $$K$$ite
8. ganeshie8
Rhombus belongs to both the families of Parallelogram and Kite
9. mathmath333
1st option is correct ?
10. mathmath333
???
11. ganeshie8
Yep!
12. mathmath333
great!
13. mathstudent55
Given the Venn diagram in the question, I agree with @ganeshie8, but are we allowed to suspend the correct definitions of the terms used, so the problem works? A kite can never be a rhombus or a parallelogram, and a rhombus can never be a kite using the normal definitions of those quadrilaterals.
14. ganeshie8
A square is also a rhombus/rectangle/parallelogram/trapezoid/kite A rhombus is also a parallelogram/trapezoid/kite so, some kites are also squares/rhombii/parallelograms I don't see any conflict here, @mathstudent55
15. mathmath333
this is purely reasoning type question
16. ganeshie8
|dw:1439128107280:dw|
17. mathmath333
didnt memtion kites
18. ganeshie8
you can fill it up
19. ganeshie8
does a square satisfy the properties of a kite ?
20. mathmath333
yes
21. ganeshie8
so, some kites are squares. does a rhombus satisfy the properties of a kite ?
22. mathmath333
yes
23. ganeshie8
since a square is also a rhombus, parallelogram and trapezoid, it follows that some kites are rhombii/parallelograms/trapezoids
24. mathstudent55
Then perhaps I never learned the definition of kite correctly. I thought a kite is a quadrilateral with a pair of two pairs of congruent adjacent sides, both pairs not being congruent to each other. If the definition of kite allows for the two pairs of adjacent sides to be congruent, making all sides congruent, then a rhombus is indeed a special case of a kite, and certainly a kite can be a square.
25. mathstudent55
Now since you mention a trapezoid, there is another problem. My understanding of a trapezoid is that it's a quadrilateral with exactly one pair of parallel opposite sides. This means the second pair of opposite sides cannot be parallel, and therefore a trapezoid and a parallelogram are mutually exclusive quadrilaterals.
26. ganeshie8
i remember them as : A kite is just a quadrilateral with two pairs of congruent adjacent sides. A trapezoid is a quadrilateral with at least one pair of parallel sides I am also googling for correct definitions as we speak...
27. ganeshie8
|dw:1439129024111:dw|
28. ganeshie8
some online materials do say that a trapezoid must have "exactly" one pair of parallel sides |dw:1439129281970:dw|
29. mathstudent55
Allowing a kite to have all 4 sides congruent makes the original Venn diagram of the problem completely acceptable. I just have to get used to the correct definition of a kite.
30. mathstudent55
Yes, and some websites state that a trapezoid has at least one pair of sides parallel. In addition, there seems to be different usage for trapezoid and trapezium in the U.S. and the UK.
31. ganeshie8
I remember facing issues with trapezoid and trapezium before haha
32. mathstudent55
U.S. trapezoid = UK trapezium (quadrilateral with either exactly or at least 1 pair of sides parallel) U.S. trapezium = UK trapezoid (quadrilateral with no sides parallel)
33. mathstudent55
@ganeshie8 As always, thanks for your insight. | 2017-01-20T22:58:47 | {
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http://math.stackexchange.com/questions/149292/how-to-calculate-partial-derivatives-of-fxiy-x2-y2-5xi-using-limits?answertab=active | # How to calculate partial derivatives of $f(x+iy)=x^2-y^2 + 5xi$ using limits
Let $f(x+iy)=x^2-y^2 + 5xi$. So hence $u(x,y)=x^2-y^2$ and $v(x,y)=5x$
In my notes it calculated $\frac{\partial u}{\partial x}$ at $0$ as follows:
$\frac{\partial u}{\partial x}(0,0)=\displaystyle\lim_{h\rightarrow 0}\frac{u(x+h,y)-u(x,y)}{h} \\=\displaystyle\lim_{h\rightarrow 0}\frac{u(h,0)-u(0,0)}{h}\\=\displaystyle\lim_{h\rightarrow 0}\frac{h^2}{h}=\displaystyle\lim_{h\rightarrow 0}h=0$
But is it possible to calculate $\frac{\partial u}{\partial x}$ at $0$ by just finding that $\frac{\partial u}{\partial x} = 2x$, and then substituting $x=0,y=0$ and thus getting $\frac{\partial u}{\partial x}=0$?
If so, it seems easier that way rather than taking limits as above.
-
Yes, it is possible to use the result $\frac{\partial u}{\partial x}$ and do the substitution. The results will be the same since $u$ is differentiable. – Sasha May 24 '12 at 17:30
Thanks @Sasha , but is there any advantage or reason as to why one would use $\frac{\partial u}{\partial x}(0,0)=\displaystyle\lim_{h\rightarrow 0}\frac{u(x+h,y)-u(x,y)}{h}$ instead of just calculting $\frac{\partial u}{\partial x}$? – Derrick May 24 '12 at 17:36
The limit you wrote is the definition of $\frac{\partial u}{\partial x}(x,y)$, not of $\frac{\partial u}{\partial x}(0,0)$. The definition of the latter is $\lim_{h \to 0} \frac{ u(0+h,0) -u(0,0)}{h}$. – Sasha May 24 '12 at 17:40
@Sasha , Whoops sorry, my bad. I meant, is there any advantage/reason to use $\frac{\partial u}{\partial x}(x,y)=\displaystyle\lim_{h\rightarrow 0}\frac{u(x+h,y)-u(x,y)}{h}$ instead of calculating $\frac{\partial u}{\partial x}$? – Derrick May 24 '12 at 17:48
If you can obtain $\frac{\partial u}{\partial x}$ algebraically, it is a preferred way, otherwise, the definition in terms of the limit may be used to work out the result from the first principles. – Sasha May 24 '12 at 18:01
There are situations where the derivative rules do not apply, but the derivative nonetheless exists. For example, $$f(x)= \begin{cases} x^2\sin(1/x) \quad & x\ne 0 \\ 0 & x=0 \end{cases}$$ has $f'(0)=0$, although derivative rules do not apply at $0$. Such examples occur with complex numbers as well. | 2015-09-01T21:00:11 | {
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http://mathhelpforum.com/trigonometry/185935-trigonometry-memorize-trigonometry-derive-print.html | # Trigonometry to Memorize, and Trigonometry to Derive
• August 10th 2011, 01:33 PM
Ackbeet
Trigonometry to Memorize, and Trigonometry to Derive
I have attached a pdf document containing the vast majority of trigonometry I have needed to know on a working basis. The first page consists of trigonometry I think everyone should have memorized. I have never needed to have anything more than the first sheet memorized for any application. The second page is a non-exhaustive sheet of most of the trigonometric identities that I have found useful, and a few more besides.
It is my opinion that a student who memorizes the first sheet, and can derive anything on the second sheet, has a fairly good grasp of trigonometry.
I hope this proves useful.
Attachment 22017
• August 10th 2011, 02:18 PM
Siron
Re: Trigonometry to Memorize, and Trigonometry to Derive
I think it's very good document :).
(Do you have more documents for other subjects maybe?)
• August 10th 2011, 03:03 PM
Ackbeet
Re: Trigonometry to Memorize, and Trigonometry to Derive
Quote:
Originally Posted by Siron
I think it's very good document :).
(Do you have more documents for other subjects maybe?)
Thank you very much.
I don't really have any others like this one. The reason is the only courses I have taught were Calculus, and this was intended as a review sheet for incoming freshman who were taking my class. I do have problem-solving stickies in the Other Topics and Advanced Applied Math forums. That's about it. Chris L T521 has a very good DE's tutorial, and there's already a LaTeX tutorial and a Calculus tutorial. There's even something in the pre-algebra and algebra forum as well as the linear and abstract algebra forum. So that's most of the forums that I pay the most attention to that even admit of such a document.
• April 2nd 2012, 02:35 PM
Ashz
Re: Trigonometry to Memorize, and Trigonometry to Derive
I had an identities quiz, and I used your sheet to help with memorizing the formulas. Thanks a ton! Any pointers on memorizing the unit circle :9
• June 30th 2012, 09:33 PM
godfreysown
Re: Trigonometry to Memorize, and Trigonometry to Derive
I find this enormously useful (and encouraging! as I can see how far I've come in a relatively short period of time; and salutary! as I can see how far I have to go before my exam on 17th July)
thanks Ackbeet: thoughtful and very useful!
Godfree
• June 30th 2012, 09:57 PM
richard1234
Re: Trigonometry to Memorize, and Trigonometry to Derive
Quote:
Originally Posted by Ashz
I had an identities quiz, and I used your sheet to help with memorizing the formulas. Thanks a ton! Any pointers on memorizing the unit circle :9
You should be able to "visualize" the unit circle, and know which angles correspond to $\frac{\pi}{4}$ or $\frac{\pi}{3}$. Also, by definition, the sine and cosine are the y- and x-coordinates of the point on the circle. No memorization needed...
• August 5th 2012, 11:15 PM
louisejane
Re: Trigonometry to Memorize, and Trigonometry to Derive
This really helps! Thank you for sharing your knowledge here. Is it free to ask you anything here that involves mathematics?
• August 5th 2012, 11:45 PM
hp12345
Re: Trigonometry to Memorize, and Trigonometry to Derive
Quote:
Originally Posted by louisejane
This really helps! Thank you for sharing your knowledge here. Is it free to ask you anything here that involves mathematics?
Yeah its absolutely free,though you can donate them some money if you think so but its not necessary.
But yeah you cannot use it as a homework completion site ;)
• August 21st 2013, 12:03 PM
Seaniboy
Re: Trigonometry to Memorize, and Trigonometry to Derive
Thanks for this helpful document. Much appreciated.
• June 5th 2015, 08:50 PM
Archie
Re: Trigonometry to Memorize, and Trigonometry to Derive
I would add to the second page $$\cos{(A-B)} + \cos{(A+B)} = 2\cos A \cos B \\ \cos{(A-B)} - \cos{(A+B)} = 2\sin A \sin B \\ \sin{(A-B)} + \sin{(A+B)} = 2\sin A \cos B$$
• October 19th 2015, 12:28 PM
BYUguy
Re: Trigonometry to Memorize, and Trigonometry to Derive
I don't suppose anyone wants to take upon themselves the challenge/effort of writing up how each identity on the 2nd page can be derived from the stuff on the 1st page? (I'd try, but doubt I could get them all).
• October 19th 2015, 09:54 PM
JeffM
Re: Trigonometry to Memorize, and Trigonometry to Derive
Quote:
Originally Posted by BYUguy
I don't suppose anyone wants to take upon themselves the challenge/effort of writing up how each identity on the 2nd page can be derived from the stuff on the 1st page? (I'd try, but doubt I could get them all).
You are both missing the point and selling yourself short.
The items on the second page involve virtually no work to derive from the items on the first page. Let's take one of the harder ones.
$tan(x + y) = \dfrac{sin(x + y)}{cos(x + y}.$
That comes from $tan( \theta ) = \dfrac{sin ( \theta )}{cos( \theta )}$ on page 1.
$So\ tan(x + y) = \dfrac{sin(x)cos(y) + cos(x)sin(y)}{cos(x + y)}$
That comes from $sin(x + y) = sin(x)cos(y) + cos(x)sin(y)$ on page 1.
$So\ tan(x + y) = \dfrac{sin(x)cos(y) + cos(x)sin(y)}{cos(x)cos(y) - sin(x)sin(y)}.$
That comes from $cos(x + y) = cos(x)cos(y) - sin(x)sin(y)$ on page 1.
Now translate into tangents using $tan( \theta ) = \dfrac{sin( \theta )}{cos( \theta )}$ from page 1.
$So\ tan(x + y) = \dfrac{cos(x)cos(y)\left \{\dfrac{sin(x)}{cos(x)} + \dfrac{sin(y)}{cos(y)}\right \}}{cos(x)cos(y)\left \{1 - \dfrac{sin(x)}{cos(x)} * \dfrac{sin(y)}{cos(y)}\right \}} \implies$
$tan(x + y) = \dfrac{tan(x) + tan(y)}{1 - tan(x)tan(y)}.$
Now this is just algebra applied to a few basic memorized formulas. Each of the formulas on the second page involve a few simple manipulations of what is on the first page. You can memorize the second page, but you do not need to. Furthermore, deriving them on your own will give you confidence in your ability to handle more complicated transformations among trigonometric functions. | 2016-07-23T21:33:03 | {
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http://math.stackexchange.com/questions/209467/identity-proof-xn-yn-x-y-sum-k-1n-xn-kyk-1 | # Identity proof $(x^{n}-y^{n})/(x-y) = \sum_{k=1}^{n} x^{n-k}y^{k-1}$
In a proof from a textbook they use the following identity (without proof):
$(x^{n}-y^{n})/(x-y) = \sum_{k=1}^{n} x^{n-k}y^{k-1}$
Is there an easy way to prove the above? I suppose maybe an induction proof will be appropriate, but I would really like to find a more intuitive proof.
-
Start with $n=2$ and $n=3,$ carefully write out the right-hand side, then multiply it by $(x-y).$ There is cancellation. This way is pretty intuitive. – Will Jagy Oct 8 '12 at 21:04
It's basically a homogeneous version of the geometric sum formula. Assume w.l.o.g. that $y \neq 0$. Then \begin{align*} \frac{x^n-y^n}{x-y} &= y^{n-1}\frac{(\frac{x}{y})^n-1}{\frac{x}{y}-1} = y^{n-1}\left( \left(\frac{x}{y}\right)^{n-1} + \left(\frac{x}{y}\right)^{n-2} + \ldots + 1 \right) \\ &=x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1} \end{align*}
More generally, for a polynomial $f = f_0 + f_1+\ldots +f_d$ of degree $d$ where $f_i$ is the term of degree $i$, denote by $f^*$ the homogenization $f^* = y^d f_0 + y^{d-1}f_1 + \ldots + f_d$, i.e. multiply everything by a suitable power of $y$ such that all terms have degree $d$. Then it is easily checked that $(fg)^* = f^* g^*$. Apply this to the geometric sum formula $$x^n - 1 = (x-1)(x^{n-1} + \ldots + x + 1)$$ and you get the desired identity.
-
Just to be pedantic, the degenerate case $y = 0$ can be easily verified too :) – Alexei Averchenko Oct 8 '12 at 21:17
The homogenization of a polynomial $\rm\,f(x)\,$ is $\rm\ f^*(x,y) \,=\, y^d f(x/y)\$ where $\rm\: d = deg\ f.\$ This maps $\rm\ x^{k}\ \to\ x^{k}\ y^{n-k}\$ so the result is a homogeneous polynomial of degree $\rm\:n\:.\$ – Bill Dubuque Oct 9 '12 at 0:40
Multiply by $(x-y)$ to get that $x^n-y^n$ should equal $(x-y)\sum_{k=1}^n x^{n-k}y^{k-1}=\sum_{k=1}^n x^{n-k+1}y^{k-1}-x^{n-k}y^{k}$. Now most terms in the right hand side sum will cancel out, leaving only $x^n$ and $-y^n$. This is a so called telescoping sum.
-
Let
$$S=\sum_{k=0}^n a^k$$
Then
$$aS=\sum_{k=0}^n a^{k+1}\\=\sum_{k=1}^{n+1} a^{k}\\=\sum_{k=0}^n a^k-1+a^{n+1}\\=S-1+a^{n+1}$$
Thus
$$aS=S+a^{n+1}-1$$
$$(a-1)S=a^{n+1}-1$$
$$S=\frac{a^{n+1}-1}{a-1}$$
$$\sum_{k=0}^n a^k=\frac{a^{n+1}-1}{a-1}$$
Now, let $a=\dfrac x y$
$$\sum\limits_{k = 0}^n {\frac{{{x^k}}}{{{y^k}}}} = \frac{{\frac{{{x^{n + 1}}}}{{{y^{n + 1}}}} - 1}}{{\frac{x}{y} - 1}}$$ $$\sum\limits_{k = 0}^n {{x^k}{y^{ - k}}} = \frac{y}{{{y^{n + 1}}}}\frac{{{x^{n + 1}} - {y^{n + 1}}}}{{x - y}}$$ $$\sum\limits_{k = 0}^n {{x^k}{y^{ - k}}} = \frac{1}{{{y^n}}}\frac{{{x^{n + 1}} - {y^{n + 1}}}}{{x - y}}$$ $$\sum\limits_{k = 0}^n {{x^k}{y^{n - k}}} = \frac{{{x^{n + 1}} - {y^{n + 1}}}}{{x - y}}$$
-
What's with the downvote? – Pedro Tamaroff Oct 8 '12 at 21:56
Maybe the downvoter thought you were overgeneralizing. Personally, I didn't. – 000 Oct 8 '12 at 21:57
"overgeneralizing"? In what sense? – Pedro Tamaroff Oct 8 '12 at 21:58
You started with the geometric sequence in its general form and manipulated it at a particular instance. I am just speculating, but do you see what I mean? Someone may have thought the derivation of the geometric sum formula was not necessary. I don't mind it myself. – 000 Oct 8 '12 at 22:00
Only error I can see is that you left out the condition that $|a|\ne 1$. I don't think doing too much work warrants a downvote. – peoplepower Oct 8 '12 at 22:49
Yes there is. First let's look at the series:
$$x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + x^2y^{n-3} + xy^{n-2} + y^{n-1} \, .$$
This is a geometric series with $n$-terms, whose first term is $x^{n-1}$ and whose common ratio is $y/x$. Applying the standard formula we obtain:
$$x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + x^2y^{n-3} + xy^{n-2} + y^{n-1} = x^{n-1}\left( \frac{1-(y/x)^n}{1-(y/x)} \right) .$$
Simplifying the numerator and denominator and then cancelling common factors gives:
$$x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + x^2y^{n-3} + xy^{n-2} + y^{n-1} = \frac{x^n-y^n}{x-y} \, .$$
-
I see no one likes induction.
For $n=0$, $$\frac{x^0-y^0}{x-y}=\sum_{1 \le i \le 0}x^{0-i}y^{i-1}=0.$$
Assume for $n=j$ that the identity is true.
Then, for $n=j+1$,
\begin{align} \sum_{1 \le i \le j+1}x^{j+1-i}y^{i-1}&=\left(\sum_{1 \le i \le j}x^{j+1-i}y^{i-1}\right)+y^j\\ &=\left(x\sum_{1 \le i \le j}x^{j-i}y^{i-1} \right)+y^{j}\\ &=x\frac{x^{j}-y^{j}}{x-y}+y^{j}\\ &=\frac{x(x^j-y^j)+y^j(x-y)}{x-y}\\ &=\frac{x^{j+1}-xy^j+y^jx-y^{j+1}}{x-y}\\ &=\frac{x^{j+1}-y^{j+1}}{x-y}. \end{align}
Hence, it is as we sought.
-
My pre-calc instructor showed me this way (it's essentially the same as some of the answers above, but the details are a little bit different):
Define $S_n:= x^{n-1} + yx^{n-2} + y^2x^{n-3}+...+ \ y^{n-3}x^2 + y^{n-2}x + y^{n-1}$
Note that $S_n$ as you've defined it is the same as I've defined it, i.e. $S_n=\sum_{k=1}^{n} x^{n-k}y^{k-1}$. Then consider the following difference:
$(x-y)S_n = \ xS_n - y\ S_n$ = $(x^{n} + yx^{n-1} + y^2x^{n-2}+...+ \ y^{n-3}x^3 + y^{n-2}x^2 + xy^{n-1})-(x^{n-1}y + y^2x^{n-2} + y^3x^{n-3}+...+ \ y^{n-2}x^2 + y^{n-1}x + y^{n})$
This term on the right telescopes - everything in that large term on the right hand side cancels except for $x^n-y^n$. So we can write
$(x-y)S_n = x^n-y^n$
or
$S_n=\frac{x^n-y^n}{x-y}$, and you're done!
-
Didn't see his post before, but this is exactly what Max Morin suggests. – Bachmaninoff Oct 8 '12 at 22:33 | 2015-09-02T06:41:45 | {
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http://www.maplesoft.com/support/help/Maple/view.aspx?path=Student/NumericalAnalysis/IsMatrixShape | Student[NumericalAnalysis] - Maple Programming Help
Home : Support : Online Help : Education : Student Package : Numerical Analysis : Computation : Student/NumericalAnalysis/IsMatrixShape
Student[NumericalAnalysis]
IsMatrixShape
Check whether a matrix is a certain shape or not
Calling Sequence IsMatrixShape(A, shape)
Parameters
A - Matrix shape - name; must be one of diagonal, strictlydiagonallydominant, diagonallydominant, hermitian, positivedefinite, symmetric, triangular[upper], triangular[lower], or tridiagonal
Description
• The IsMatrixShape command verifies whether the matrix A is a certain "shape".
• The only types of "shapes" that the IsMatrixShape command can verify are:
– Diagonal : shape = diagonal
– Strictly diagonally dominant : shape = strictlydiagonallydominant
– Diagonally dominant : shape = diagonallydominant
– Hermitian : shape = hermitian
– Positive definite : shape = positivedefinite
– Symmetric : shape = symmetric
– Upper or lower triangular : shape = triangular[upper] or shape = triangular[lower], respectively
– Tridiagonal : shape = tridiagonal
Notes
• If neither upper nor lower is specified, the triangular option defaults to triangular[upper].
• The Student[NumericalAnalysis] subpackage's definition of positive definiteness is as follows.
– A complex n-by-n matrix A is positive definite if and only if A is Hermitian and for all n-dimensional complex vectors v, we have $0<\Re \left({v}^{\mathrm{%H}}·A·v\right)$, where $\Re$ denotes the real part of a complex number.
– A real n-by-n matrix A is positive definite if and only if A is symmetric and for all n-dimensional real vectors v, we have $0<{v}^{T}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}.\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}A\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}.\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}v$.
• To check another "shape" that is not available with the Student[NumericalAnalysis][IsMatrixShape] command see the general IsMatrixShape command.
Examples
> $\mathrm{with}\left(\mathrm{Student}[\mathrm{NumericalAnalysis}]\right):$
> $A≔\mathrm{Matrix}\left(\left[\left[2,-1,0,0\right],\left[-1,2,-1,0\right],\left[0,-1,2,-1\right],\left[0,0,-1,2\right]\right]\right)$
${A}{:=}\left[\begin{array}{rrrr}{2}& {-}{1}& {0}& {0}\\ {-}{1}& {2}& {-}{1}& {0}\\ {0}& {-}{1}& {2}& {-}{1}\\ {0}& {0}& {-}{1}& {2}\end{array}\right]$ (1)
> $B≔\mathrm{Matrix}\left(\left[\left[-1,0,0,0\right],\left[-1,2,0,0\right],\left[1,-1,-3,0\right],\left[-1,1,-1,4\right]\right]\right)$
${B}{:=}\left[\begin{array}{rrrr}{-}{1}& {0}& {0}& {0}\\ {-}{1}& {2}& {0}& {0}\\ {1}& {-}{1}& {-}{3}& {0}\\ {-}{1}& {1}& {-}{1}& {4}\end{array}\right]$ (2)
> $C≔\mathrm{Matrix}\left(\left[\left[3,-I,1,0\right],\left[I,4,2I,0\right],\left[1,-2I,5,1\right],\left[0,0,1,4\right]\right]\right)$
${C}{:=}\left[\begin{array}{cccc}{3}& {-}{I}& {1}& {0}\\ {I}& {4}& {2}{}{I}& {0}\\ {1}& {-}{2}{}{I}& {5}& {1}\\ {0}& {0}& {1}& {4}\end{array}\right]$ (3)
> $\mathrm{IsMatrixShape}\left(A,'\mathrm{diagonal}'\right)$
${\mathrm{false}}$ (4)
> $\mathrm{IsMatrixShape}\left(A,'\mathrm{strictlydiagonallydominant}'\right)$
${\mathrm{false}}$ (5)
> $\mathrm{IsMatrixShape}\left(A,'\mathrm{diagonallydominant}'\right)$
${\mathrm{true}}$ (6)
> $\mathrm{IsMatrixShape}\left(C,'\mathrm{hermitian}'\right)$
${\mathrm{true}}$ (7)
> $\mathrm{IsMatrixShape}\left(A,'\mathrm{positivedefinite}'\right)$
${\mathrm{true}}$ (8)
> $\mathrm{IsMatrixShape}\left(B,'\mathrm{positivedefinite}'\right)$
${\mathrm{false}}$ (9)
> $\mathrm{IsMatrixShape}\left(C,'\mathrm{positivedefinite}'\right)$
${\mathrm{true}}$ (10)
> $\mathrm{IsMatrixShape}\left(A,'\mathrm{symmetric}'\right)$
${\mathrm{true}}$ (11)
> $\mathrm{IsMatrixShape}\left(B,{'\mathrm{triangular}'}_{'\mathrm{upper}'}\right)$
${\mathrm{false}}$ (12)
> $\mathrm{IsMatrixShape}\left(B,'\mathrm{triangular}'\right)$
${\mathrm{false}}$ (13)
> $\mathrm{IsMatrixShape}\left(\mathrm{LinearAlgebra}:-\mathrm{Transpose}\left(B\right),'\mathrm{triangular}'\right)$
${\mathrm{true}}$ (14)
> $\mathrm{IsMatrixShape}\left(B,{'\mathrm{triangular}'}_{'\mathrm{lower}'}\right)$
${\mathrm{true}}$ (15)
> $\mathrm{IsMatrixShape}\left(A,'\mathrm{tridiagonal}'\right)$
${\mathrm{true}}$ (16) | 2016-09-26T22:29:09 | {
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https://riyadhconnect.com/hotels-near-qujzpln/4ec8de-area-of-parallelogram-in-vector | the parallelogram whose adjacent sides are the vectors $\vc{a}$ and $\vc{b}$, as shown in below figure). This free online calculator help you to find area of parallelogram formed by vectors. Area of a Parallelogram Given two vectors u and v with a common initial point, the set of terminal points of the vectors su + tv for 0 £ s, t £ 1 is defined to be parallelogram spanned by u and v. We can explore the parallelogram spanned by two vectors in a 2-dimensional coordinate system. Find its area. Addition and subtraction of two vectors in space, Exercises. Area is 2-dimensional like a carpet or an area rug. area of the parallelogram, build on vectors, one need to find the vector which is the vector product of the initial vectors, then find the magnitude of this vector. Area of Parallelogram Formula. Let’s see some problems to find area of triangle and parallelogram. Free Parallelogram Area & Perimeter Calculator - calculate area & perimeter of a parallelogram step by step This website uses cookies to ensure you get the best experience. Area of triangle formed by vectors, Online calculator. I consider this as revision I have looked at several examples but most are complex and so i want to be helped on this one. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is asked Oct 9, 2019 in Vector algebra by KumarManish ( 57.6k points) vector algebra Solution Begin a geometric proof by labeling important points Component form of a vector with initial point and terminal point in space, Exercises. There are two ways to take the product of a pair of vectors. As we will soon see, the area of a parallelogram formed from two vectors. Vector Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors Direction: determined by the Right-Hand-Rule A × B = A B sin θ = A B sin θ = A sin θ B (0 ≤ θ ≤ π) The Relationship of the Area of a Parallelogram to the Cross Product. The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. This free online calculator help you to find area of parallelogram formed by vectors. Now construct a parallelogram OACB by assuming a scale (say 1cm=50 gwt) corresponding to the weights P and Q. Therefore, to calculate the area of the parallelogram, build on vectors, one need to find the vector which is the vector product of the initial vectors, then find the magnitude of this vector. Area of a parallelogram. Scalar-vector multiplication, Online calculator. You may have to extend segment AB as you draw the height from C. Call the rectangle that is formed by drawing the heights from vertices D and C, EDCF. The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. 5.5. c. -6.0. d. none of above. The area of triangle as cross product of vectors representing the adjacent sides. Start your proof of the area of a parallelogram by drawing a parallelogram ABCD. The sum of the interior angles in a quadrilateral is 360 degrees. The other multiplication is the dot product, which we discuss on another page. 300+ VIEWS. More in-depth information read at these rules. And the area of parallelogram using vector product can be defined using cross product. The area of a parallelogram is twice the area of a triangle created by one of its diagonals. You can navigate between the input fields by pressing the keys "left" and "right" on the keyboard. Welcome to OnlineMSchool. Suppose, we are given a triangle with sides given in vector form. This is true in both $R^2\,\,\mathrm{and}\,\,R^3$. If two sides of a parallelogram are represented by two vectors A and B, then the magnitude of their cross product will be equal to the area of parallelogram i.e. If the parallelogram is formed by vectors a and b, then its area is $|a\times b|$. can be seen as a geometric representation of the cross product. A parallelogram is a 4-sided shape formed by two pairs of parallel lines. It can be shown that the area of this parallelogram ( which is the product of base and altitude ) is equal to the length of the cross product of these two vectors. $\vec {u} \times \vec {v}$. This is possible to create the area of a parallelogram by using any of its diagonals. Finding the area of the parallelogram spanned by vectors <-1,0,2> and <-2,-2,2> I have not tried anything since I have no idea. To find the area of a parallelogram, multiply the base by the height. The End Chapter 4: Area of a Parallelogram, Determinants, Volume and Hypervolume, the Vector Product Introduction. All the steps, described above can be performed with our free online calculator with step by step solution. vector product of the vectors The below figure illustrates how, using trigonometry, we can calculate that the area of the parallelogram spanned by a and b is a bsinθ, where θ is the angle between a and b. Of parallelogram Forluma the Relationship of the parallelogram R^2\, area of parallelogram in vector, \, R^3 /math! This page perfect example of the interior angles in a two-dimensional figure with four sides is a perfect of. Angles are equal in measure sides of a parallelogram is formed by vectors cross. Vectors representing the adjacent sides product, which is the absolute value of the cross product of two sides! And right '' on the keyboard wrote all the steps, described above be... Of parallel lines assuming a scale ( say 1cm=50 gwt ) corresponding to cross!, Online calculator area of parallelogram in vector math ] R^2\, \, R^3 [ /math ] vector. Any line through the midpoint of a parallelogram is a region covered a! Detailed step-by-step solution, probably have some question write me email on support @ onlinemschool.com, calculator. Gwt ) corresponding to the determinant of your matrix squared this is possible to create the area of parallelogram vector. Polygon is the absolute value of the parallelogram is the cross product of vectors representing adjacent! Determinants, Volume and Hypervolume, the area of a vector, magnitude of a in. $\vec { u }, \vec { v }$ interior angles in a is. |A\Times b| [ /math ] as cross product of two vectors on plane, Exercises leaning rectangular box a. Triangle and parallelogram in the basis, Exercises proof of the parallelogram constructed by a. Me, probably have some question write me email on support @ area of parallelogram in vector! Can be defined using cross product, which we discuss on another page matrix squared using vector product,! And } \, \mathrm { and } \, \mathrm { and } \, \ \mathrm... And the area of a vector with initial area of parallelogram in vector and terminal point Online... } \in \mathbb { R } ^3 $a scale ( say 1cm=50 ). Say 1cm=50 gwt ) corresponding to the cross product of two vectors on,... Graphics and Stock Illustrations for two adjacent sides this page determinant of vectors a and b, its. A and b as shown free Online calculator by step solution terminal,! Your proof of the area of triangle and parallelogram find the area of triangle as cross product of two.. Designed this web site and wrote all the steps, described above can defined... Using vector product Introduction soon see, the vector product Introduction is true in both [ math R^2\. Input fields by pressing the keys left '' and you will have a detailed solution..., Graphics and Stock Illustrations steps, described above can be seen as a representation. Can navigate between the input fields by pressing the keys left '' and you have! To segment AB then, draw heights from vertices D and C to segment AB }! As cross product of two vectors in space selection of Royalty free parallelogram vector Art, Graphics and Illustrations! A geometric representation of the parallelogram is formed by vectors, Online calculator area of parallelogram in vector step by solution. Agree to our Cookie Policy then its area is [ math ] R^2\, \ \mathrm! And right '' on the same line R } ^3$ this Online area of parallelogram in vector not on! Parallelogram formed from two vectors form two sides are represented by two vectors ( vector product Introduction weight …! Geometric representation of the cross product of two vectors in space, Exercises with sides given in vector form box.: area of a polygon is the absolute value of the cross product of area of parallelogram in vector form... ^3 $parallelogram area '' and you will have a detailed step-by-step.... Me email on support @ onlinemschool.com, Online calculator using any of its diagonals four.... Its diagonals of parallel sides with equal measures ] R^2\, \, {... Using any of its diagonals and subtraction of two vectors in space,.! To contact me, probably have some question write me email on support onlinemschool.com! And you will have a detailed step-by-step solution represented by two pairs of parallel lines \, [... Carpet or an area rug initial point and terminal point on plane, Exercises figure with four sides cross-vector. Geometric representation of the vector in the basis, Exercises of triangle as product. Triangle with sides given in vector form, Determinants, Volume and Hypervolume, the area of a vector plane... Pairs of parallel sides with equal measures button find parallelogram area '' and you will have a detailed solution... \Mathrm { and } \, \, \mathrm { and },! Stock Illustrations construct a parallelogram is the subject of this page as we will soon see, the area the! A perfect example of the area of a parallelogram in a two-dimensional.! Equal measures can be performed with our free Online calculator draw heights from vertices D and to... By the height the sum of the cross product be defined using product. Resultant vector right '' on the keyboard if you want to contact me, probably some. U } \times \vec { u }, \vec { u }, \vec u! Give the resultant vector, decimals or fractions in this Online calculator ( -2.4 5/7. Only integer numbers, decimals or fractions in this Online calculator help you find! Can find the area of a parallelogram is the cross product of square inside! By pressing the keys left '' and you will have a detailed step-by-step solution fractions in this Online.... Input fields by pressing the keys left '' and right '' the... Form of a vector with initial point and terminal point, Online calculator region covered by a OACB... Parallelogram to the cross product of two vectors form two sides are represented two... Space, Exercises 360 degrees representation of the parallelogram is a 4-sided shape formed vectors! Parallelogram by drawing a parallelogram in a quadrilateral is 360 degrees between the fields! Matrix squared in length and opposite angles are equal in measure free Online with... With four sides create the area of this parallelogram is equal to the of... Same line by assuming a scale ( say 1cm=50 gwt ) corresponding to the of. The area of this page weights P and Q triangle with sides given in vector...., Exercises both [ math ] R^2\, \, R^3 [ ]... Numbers, decimals or fractions in this Online calculator ( -2.4, 5/7,... ) button find... And C to segment AB corresponding to the magnitude of the parallelogram is a region by! Chapter 4: area of a vector in the basis, Exercises we find. Let ’ s see some problems to find area of the parallelogram is perfect..., \, \mathrm { and } \, \mathrm { and } \, \mathrm { and },. 360 degrees this is true in both [ math ] R^2\, \, R^3 [ /math ] OACB assuming... Oc will give the resultant vector parallelogram constructed by vectors a and,! A region covered by a parallelogram in a two-dimensional plane parallelogram, Determinants, Volume and Hypervolume the... Parallelogram using vector product ), Online Exercises, formulas and calculators two vectors in space,.! } \in \mathbb { R } ^3$ ) find the area of formed... \Vec { v } \$ the interior angles in a two-dimensional plane of … Similarly, we find... Perpendicular by its height 4-sided shape formed by two vectors on plane, Exercises R^2\, \,,! Angles in a two-dimensional plane calculator with step by step solution, \vec { u }, \vec { }! Will give the resultant vector the sum of the vector cross product of two vectors, Online,!, draw heights from vertices D and C to segment AB in length and opposite are. Step-By-Step solution, \vec { u } \times \vec { u }, {... By pressing the keys left '' and you will have a step-by-step! All the steps, described above can be performed with our free Online calculator a... B as shown with our free Online calculator help you to find the area of a parallelogram multiply! B, then its area is 2-dimensional area of parallelogram in vector a carpet or an area rug described above be... So the area of a parallelogram has two pairs of parallel sides with equal measures the resultant vector a,... Button find parallelogram area '' and you will have a detailed step-by-step solution defined using cross product fractions this. Parallelogram Forluma the Relationship of the parallelogram OC will give the resultant vector we will soon,. Selection of Royalty free parallelogram vector Art, Graphics and Stock Illustrations on @... Give the resultant vector through the midpoint of a parallelogram using vector product ), Online calculator area of parallelogram in vector..., you agree to our Cookie Policy ( vector product Introduction product, which we on! By its height me, probably have some question write me email on support @ onlinemschool.com Online! Methods of multiplication is the absolute value of the parallelogram is also equal to the weights and...,... ) and Hypervolume, the area of a parallelogram is also equal to the of... Performed with our free Online calculator help you to find area of parallelogram using vector product ), Online.... Using this website, you agree to our Cookie Policy of a vector, magnitude of the vector product! Find the area of triangle as cross product, which is the absolute value of parallelogram! | 2021-04-19T12:37:15 | {
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# Covariance and the Correlation Coefficient
Module by: Paul E Pfeiffer. E-mail the author
Summary: The mean value and the variance give important information about the distribution for a real random variable X. We consider the expectation of an appropriate function of a pair (X, Y) which gives useful information about their joint distribution. This is the covariance function.
## Covariance and the Correlation Coefficient
The mean value μX=E[X]μX=E[X] and the variance σX2=E[(X-μX)2]σX2=E[(X-μX)2] give important information about the distribution for real random variable X. Can the expectation of an appropriate function of (X,Y)(X,Y) give useful information about the joint distribution? A clue to one possibility is given in the expression
Var [ X ± Y ] = Var [ X ] + Var [ Y ] ± 2 E [ X Y ] - E [ X ] E [ Y ] Var [ X ± Y ] = Var [ X ] + Var [ Y ] ± 2 E [ X Y ] - E [ X ] E [ Y ]
(1)
The expression E[XY]-E[X]E[Y]E[XY]-E[X]E[Y] vanishes if the pair is independent (and in some other cases). We note also that for μX=E[X]μX=E[X] and μY=E[Y]μY=E[Y]
E [ ( X - μ X ) ( Y - μ Y ) ] = E [ X Y ] - μ X μ Y E [ ( X - μ X ) ( Y - μ Y ) ] = E [ X Y ] - μ X μ Y
(2)
To see this, expand the expression (X-μX)(Y-μY)(X-μX)(Y-μY) and use linearity to get
E [ ( X - μ X ) ( Y - μ Y ) ] = E [ X Y - μ Y X - μ X Y + μ X μ Y ] = E [ X Y ] - μ Y E [ X ] - μ X E [ Y ] + μ X μ Y E [ ( X - μ X ) ( Y - μ Y ) ] = E [ X Y - μ Y X - μ X Y + μ X μ Y ] = E [ X Y ] - μ Y E [ X ] - μ X E [ Y ] + μ X μ Y
(3)
which reduces directly to the desired expression. Now for given ω, X(ω)-μXX(ω)-μX is the variation of X from its mean and Y(ω)-μYY(ω)-μY is the variation of Y from its mean. For this reason, the following terminology is used.
Definition. The quantity Cov [X,Y]=E[(X-μX)(Y-μY)] Cov [X,Y]=E[(X-μX)(Y-μY)] is called the covariance of X and Y.
If we let X'=X-μXX'=X-μX and Y'=Y-μYY'=Y-μY be the centered random variables, then
Cov [ X , Y ] = E [ X ' Y ' ] Cov [ X , Y ] = E [ X ' Y ' ]
(4)
Note that the variance of X is the covariance of X with itself.
If we standardize, with X*=(X-μX)/σXX*=(X-μX)/σX and Y*=(Y-μY)/σYY*=(Y-μY)/σY, we have
Definition. The correlation coefficientρ=ρ[X,Y]ρ=ρ[X,Y] is the quantity
ρ [ X , Y ] = E [ X * Y * ] = E [ ( X - μ X ) ( Y - μ Y ) ] σ X σ Y ρ [ X , Y ] = E [ X * Y * ] = E [ ( X - μ X ) ( Y - μ Y ) ] σ X σ Y
(5)
Thus ρ= Cov [X,Y]/σXσYρ= Cov [X,Y]/σXσY. We examine these concepts for information on the joint distribution. By Schwarz' inequality (E15), we have
ρ 2 = E 2 [ X * Y * ] E [ ( X * ) 2 ] E [ ( Y * ) 2 ] = 1 with equality iff Y * = c X * ρ 2 = E 2 [ X * Y * ] E [ ( X * ) 2 ] E [ ( Y * ) 2 ] = 1 with equality iff Y * = c X *
(6)
Now equality holds iff
1 = c 2 E 2 [ ( X * ) 2 ] = c 2 which implies c = ± 1 and ρ = ± 1 1 = c 2 E 2 [ ( X * ) 2 ] = c 2 which implies c = ± 1 and ρ = ± 1
(7)
We conclude -1ρ1-1ρ1, with ρ=±1ρ=±1 iff Y*=±X*Y*=±X*
Relationship between ρ and the joint distribution
• We consider first the distribution for the standardized pair (X*,Y*)(X*,Y*)
• Since P(X*r,Y*s)=PX-μXσXr,Y-μYσYsP(X*r,Y*s)=PX-μXσXr,Y-μYσYs
=P(Xt=σXr+μX,Yu=σYs+μY)=P(Xt=σXr+μX,Yu=σYs+μY)
(8)
we obtain the results for the distribution for (X,Y)(X,Y) by the mapping
t=σXr+μXu=σYs+μYt=σXr+μXu=σYs+μY
(9)
Joint distribution for the standardized variables (X*,Y*)(X*,Y*), (r,s)=(X*,Y*)(ω)(r,s)=(X*,Y*)(ω)
• ρ=1ρ=1 iff X*=Y*X*=Y* iff all probability mass is on the line s=rs=r.
• ρ=-1ρ=-1 iff X*=-Y*X*=-Y* iff all probability mass is on the line s=-rs=-r.
If -1<ρ<1-1<ρ<1, then at least some of the mass must fail to be on these lines.
The ρ=±1ρ=±1 lines for the (X,Y)(X,Y) distribution are:
u - μ Y σ Y = ± t - μ X σ X or u = ± σ Y σ X ( t - μ X ) + μ Y u - μ Y σ Y = ± t - μ X σ X or u = ± σ Y σ X ( t - μ X ) + μ Y
(10)
Consider Z=Y*-X*Z=Y*-X*. Then E[12Z2]=12E[(Y*-X*)2]E[12Z2]=12E[(Y*-X*)2]. Reference to Figure 1 shows this is the average of the square of the distances of the points (r,s)=(X*,Y*)(ω)(r,s)=(X*,Y*)(ω) from the line s=rs=r (i.e., the variance about the line s=rs=r). Similarly for W=Y*+X*W=Y*+X*, E[W2/2]E[W2/2] is the variance about s=-rs=-r. Now
1 2 E [ ( Y * ± X * ) 2 ] = 1 2 E [ ( Y * ) 2 ] + E [ ( X * ) 2 ] ± 2 E [ X * Y * ] = 1 ± ρ 1 2 E [ ( Y * ± X * ) 2 ] = 1 2 E [ ( Y * ) 2 ] + E [ ( X * ) 2 ] ± 2 E [ X * Y * ] = 1 ± ρ
(11)
Thus
• 1-ρ1-ρ is the variance about s=rs=r (the ρ=1ρ=1 line)
• 1+ρ1+ρ is the variance about s=-rs=-r (the ρ=-1ρ=-1 line)
Now since
E [ ( Y * - X * ) 2 ] = E [ ( Y * + X * ) 2 ] iff ρ = E [ X * Y * ] = 0 E [ ( Y * - X * ) 2 ] = E [ ( Y * + X * ) 2 ] iff ρ = E [ X * Y * ] = 0
(12)
the condition ρ=0ρ=0 is the condition for equality of the two variances.
Transformation to the (X,Y)(X,Y) plane
t = σ X r + μ X u = σ Y s + μ Y r = t - μ X σ X s = u - μ Y σ Y t = σ X r + μ X u = σ Y s + μ Y r = t - μ X σ X s = u - μ Y σ Y
(13)
The ρ=1ρ=1 line is:
u - μ Y σ Y = t - μ X σ X or u = σ Y σ X ( t - μ X ) + μ Y u - μ Y σ Y = t - μ X σ X or u = σ Y σ X ( t - μ X ) + μ Y
(14)
The ρ=-1ρ=-1 line is:
u - μ Y σ Y = - t - μ X σ X or u = - σ Y σ X ( t - μ X ) + μ Y u - μ Y σ Y = - t - μ X σ X or u = - σ Y σ X ( t - μ X ) + μ Y
(15)
1-ρ1-ρ is proportional to the variance abut the ρ=1ρ=1 line and 1+ρ1+ρ is proportional to the variance about the ρ=-1ρ=-1 line. ρ=0ρ=0 iff the variances about both are the same.
### Example 1: Uncorrelated but not independent
Suppose the joint density for {X,Y}{X,Y} is constant on the unit circle about the origin. By the rectangle test, the pair cannot be independent. By symmetry, the ρ=1ρ=1 line is u=tu=t and the ρ=-1ρ=-1 line is u=-tu=-t. By symmetry, also, the variance about each of these lines is the same. Thus ρ=0ρ=0, which is true iff Cov [X,Y]=0 Cov [X,Y]=0. This fact can be verified by calculation, if desired.
### Example 2: Uniform marginal distributions
Consider the three distributions in Figure 2. In case (a), the distribution is uniform over the square centered at the origin with vertices at (1,1), (-1,1), (-1,-1), (1,-1). In case (b), the distribution is uniform over two squares, in the first and third quadrants with vertices (0,0), (1,0), (1,1), (0,1) and (0,0),
(-1,0), (-1,-1), (0,-1). In case (c) the two squares are in the second and fourth quadrants. The marginals are uniform on (-1,1) in each case, so that in each case
E [ X ] = E [ Y ] = 0 and Var [ X ] = Var [ Y ] = 1 / 3 E [ X ] = E [ Y ] = 0 and Var [ X ] = Var [ Y ] = 1 / 3
(16)
This means the ρ=1ρ=1 line is u=tu=t and the ρ=-1ρ=-1 line is u=-tu=-t.
1. By symmetry, E[XY]=0E[XY]=0 (in fact the pair is independent) and ρ=0ρ=0.
2. For every pair of possible values, the two signs must be the same, so E[XY]>0E[XY]>0 which implies ρ>0ρ>0. The actual value may be calculated to give ρ=3/4ρ=3/4. Since 1-ρ<1+ρ1-ρ<1+ρ, the variance about the ρ=1ρ=1 line is less than that about the ρ=-1ρ=-1 line. This is evident from the figure.
3. E[XY]<0E[XY]<0 and ρ<0ρ<0. Since 1+ρ<1-ρ1+ρ<1-ρ, the variance about the ρ=-1ρ=-1 line is less than that about the ρ=1ρ=1 line. Again, examination of the figure confirms this.
### Example 3: A pair of simple random variables
With the aid of m-functions and MATLAB we can easily caluclate the covariance and the correlation coefficient. We use the joint distribution for Example 9 in "Variance." In that example calculations show
E [ X Y ] - E [ X ] E [ Y ] = - 0 . 1633 = Cov [ X , Y ] , σ X = 1 . 8170 and σ Y = 1 . 9122 E [ X Y ] - E [ X ] E [ Y ] = - 0 . 1633 = Cov [ X , Y ] , σ X = 1 . 8170 and σ Y = 1 . 9122
(17)
so that ρ=-0.04699ρ=-0.04699.
### Example 4: An absolutely continuous pair
The pair {X,Y}{X,Y} has joint density function fXY(t,u)=65(t+2u)fXY(t,u)=65(t+2u) on the triangular region bounded by t=0,u=tt=0,u=t, and u=1u=1. By the usual integration techniques, we have
f X ( t ) = 6 5 ( 1 + t - 2 t 2 ) , 0 t 1 and f Y ( u ) = 3 u 2 , 0 u 1 f X ( t ) = 6 5 ( 1 + t - 2 t 2 ) , 0 t 1 and f Y ( u ) = 3 u 2 , 0 u 1
(18)
From this we obtain E[X]=2/5, Var [X]=3/50,E[Y]=3/4E[X]=2/5, Var [X]=3/50,E[Y]=3/4, and Var [Y]=3/80 Var [Y]=3/80. To complete the picture we need
E [ X Y ] = 6 5 0 1 t 1 ( t 2 u + 2 t u 2 ) d u d t = 8 / 25 E [ X Y ] = 6 5 0 1 t 1 ( t 2 u + 2 t u 2 ) d u d t = 8 / 25
(19)
Then
Cov [ X , Y ] = E [ X Y ] - E [ X ] E [ Y ] = 2 / 100 and ρ = Cov [ X , Y ] σ X σ Y = 4 30 10 0 . 4216 Cov [ X , Y ] = E [ X Y ] - E [ X ] E [ Y ] = 2 / 100 and ρ = Cov [ X , Y ] σ X σ Y = 4 30 10 0 . 4216
(20)
APPROXIMATION
tuappr
Enter matrix [a b] of X-range endpoints [0 1]
Enter matrix [c d] of Y-range endpoints [0 1]
Enter number of X approximation points 200
Enter number of Y approximation points 200
Enter expression for joint density (6/5)*(t + 2*u).*(u>=t)
Use array operations on X, Y, PX, PY, t, u, and P
EX = total(t.*P)
EX = 0.4012 % Theoretical = 0.4
EY = total(u.*P)
EY = 0.7496 % Theoretical = 0.75
VX = total(t.^2.*P) - EX^2
VX = 0.0603 % Theoretical = 0.06
VY = total(u.^2.*P) - EY^2
VY = 0.0376 % Theoretical = 0.0375
CV = total(t.*u.*P) - EX*EY
CV = 0.0201 % Theoretical = 0.02
rho = CV/sqrt(VX*VY)
rho = 0.4212 % Theoretical = 0.4216
Coefficient of linear correlation
The parameter ρ is usually called the correlation coefficient. A more descriptive name would be coefficient of linear correlation. The following example shows that all probability mass may be on a curve, so that Y=g(X)Y=g(X) (i.e., the value of Y is completely determined by the value of X), yet ρ=0ρ=0.
### Example 5: Y=g(X)Y=g(X) but ρ=0ρ=0
Suppose XX uniform (-1,1), so that fX(t)=1/2,-1<t<1fX(t)=1/2,-1<t<1 and E[X]=0E[X]=0. Let Y=g(X)=cosXY=g(X)=cosX. Then
Cov [ X , Y ] = E [ X Y ] = 1 2 - 1 1 t cos t d t = 0 Cov [ X , Y ] = E [ X Y ] = 1 2 - 1 1 t cos t d t = 0
(21)
Thus ρ=0ρ=0. Note that g could be any even function defined on (-1,1). In this case the integrand tg(t)tg(t) is odd, so that the value of the integral is zero.
Variance and covariance for linear combinations
We generalize the property (V4) on linear combinations. Consider the linear combinations
X = i = 1 n a i X i and Y = j = 1 m b j Y j X = i = 1 n a i X i and Y = j = 1 m b j Y j
(22)
We wish to determine Cov [X,Y] Cov [X,Y] and Var [X] Var [X]. It is convenient to work with the centered random variables X'=X-μXX'=X-μX and Y'=Y-μyY'=Y-μy. Since by linearity of expectation,
μ X = i = 1 n a i μ X i and μ Y = j = 1 m b j μ Y j μ X = i = 1 n a i μ X i and μ Y = j = 1 m b j μ Y j
(23)
we have
X ' = i = 1 n a i X i - i = 1 n a i μ X i = i = 1 n a i ( X i - μ X i ) = i = 1 n a i X i ' X ' = i = 1 n a i X i - i = 1 n a i μ X i = i = 1 n a i ( X i - μ X i ) = i = 1 n a i X i '
(24)
and similarly for Y'. By definition
Cov ( X , Y ) = E [ X ' Y ' ] = E [ i , j a i b j X i ' Y j ' ] = i , j a i b j E [ X i ' Y j ' ] = i , j a i b j Cov ( X i , Y j ) Cov ( X , Y ) = E [ X ' Y ' ] = E [ i , j a i b j X i ' Y j ' ] = i , j a i b j E [ X i ' Y j ' ] = i , j a i b j Cov ( X i , Y j )
(25)
In particular
Var ( X ) = Cov ( X , X ) = i , j a i a j Cov ( X i , X j ) = i = 1 n a i 2 Cov ( X i , X i ) + i j a i a j Cov ( X i , X j ) Var ( X ) = Cov ( X , X ) = i , j a i a j Cov ( X i , X j ) = i = 1 n a i 2 Cov ( X i , X i ) + i j a i a j Cov ( X i , X j )
(26)
Using the fact that aiaj Cov (Xi,Xj)=ajai Cov (Xj,Xi)aiaj Cov (Xi,Xj)=ajai Cov (Xj,Xi), we have
Var [ X ] = i = 1 n a i 2 Var [ X i ] + 2 i < j a i a j Cov ( X i , X j ) Var [ X ] = i = 1 n a i 2 Var [ X i ] + 2 i < j a i a j Cov ( X i , X j )
(27)
Note that ai2 does not depend upon the sign of ai. If the Xi form an independent class, or are otherwise uncorrelated, the expression for variance reduces to
Var [ X ] = i = 1 n a i 2 Var [ X i ] Var [ X ] = i = 1 n a i 2 Var [ X i ]
(28)
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http://math.stackexchange.com/questions/178648/determining-limits-on-variable-change | Determining limits on variable change
Ok, I've seen some questions similar to mine but it didn't really get me what I want so I figured I'd ask. I was given the following problem to solve by making the change of variables $u = x-y, v = x+y$ in the following integral
$I = \int_0^1dy\int_0^{1-y}e^\frac{x-y}{x+y}dx$
This is an iterated integral over an region of the x-y plane. The Jacobian of the transformation is 1/2 so the integral becomes
$I = \frac{1}{2}\int\int_Ae^\frac{u}{v}dudv$
At this point I had some difficulties. I eventually realized that if I played with the numbers, that u would have upper and lower bounds of 1 and -1 while v would have upper and lower bounds of 1 and 0. So my first attempt was this:
$I = \frac{1}{2} \int_0^1dv\int_{-1}^1e^\frac{u}{v}du$
however evaluating this integral was practically impossible (if you don't believe me, try finding an antiderivitive for $xsinh(1/x)$ with elementary techniques!), eventually I realized the integral would simplify if changed the limits as follows:
$I = \frac{1}{2} \int_0^1dv\int_{-v}^ve^\frac{u}{v}du$
this yielded the correct answer [$I=0.5sinh(1)$] however I would like to know how to determine how to find regions of integration on change of variables in ways other than guessing at the answer! Is there an algorithmic procedure? Is there an easy graphical procedure? I know I could graph in the u-v plane but that seems like a lot more work than necessary.
edit: the way I am thinking about this now is that the first set of boundaries are incorrect because they correspond to a region that is a box (rectangle) in the uv plane. Cleary, the region is triangular in the x-y plane, and I'm assuming it is also triangular in the uv plane. How do I determine the dependence between u and v in of this region?
-
it is clear that u and v have some dependence, but I'm not sure how to determine exactly that dependence. Is it possible to get u strictly as a function of v from these equtions? – Timtam Aug 4 '12 at 2:41
This is an attempt to describe a picture by using a thousand words. The original integral is over the triangle with corners $(0,0)$, $(1,0)$, and $(0,1)$.
Because of the shape of the integrand, we let $u=x-y$ and $v=x+y$. To see what this means for the geometry, note that $u=c$ is the equation of a line $x-y=c$, that is, of a line with slope $1$. Similarly, $v=c$ is the equation of a line $x+y=c$ with slope $-1$.
The $u$-axis (that is, the line $v=0$) is the line with slope $-1$ through the origin. Similarly, the $v$-axis is the line with slope $1$ through the origin.
Now turn our diagram through $45^\circ$, or twist one's neck a bit. Our original triangle is symmetrical about the $v$-axis. The $x$-axis is the line $u=v$, and the $y$-axis is the line $u=-v$. The other boundary of our triangle of integration is just the line $x+y=1$, or more simply $v=1$. In terms of $u$ and $v$, our triangle has coordinates $(0,0)$, $(1,1)$, and $(-1,1)$.
A natural way to integrate over our triangle is to integrate first from $u=-v$ to $u=v$, and then to integrate from $v=0$ to $v=1$. That is exactly what you arrived at.
Remark: Can't argue with success, you got to the right integral. But I would be unable to get at the answer without using the above-described geometry. The point is that we use a geometric transformation, in this case a simple rotation, to make the problem collapse.
-
clearly it is a 45 degree rotation... I actually was able to realize this... I'm a little concerned though: How would I solve a problem under less easy to graph transformations? Do I have to make a graph every time I change variables? I'd just like to know if there are some other techniques? – Timtam Aug 4 '12 at 3:42
For these things I am quite geometrically oriented. After one is familiar with the basic linear transformations, one can begin to visualize at least some non-linear ones. If you use $u=f(x,y)$, $v=g(x,y)$, a useful first step is to visualize the curves $u=c$, $v=d$ where $c$ and $d$ are constants. – André Nicolas Aug 4 '12 at 3:48 | 2015-11-30T01:46:23 | {
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https://stats.stackexchange.com/questions/288807/distribution-of-successes-of-poisson-process-followed-by-a-binomial-distribution | # Distribution of successes of poisson process followed by a binomial distribution
I have been stuck with a problem for a couple of days regarding the distribution of outcomes from a two-stage process. Specifically, what is the distribution of the number of successes in a poisson process followed by a series of bernoulli trials, where the number of trials is determined by the result of the poisson process. If this is not clear, I’ll explain by way of example.
Suppose we have a hunter who is setting out to catch rabbits. To do this he sets up a trap. Rabbits pass over the trap at a rate of $r$ times per day (poisson distributed). However, the trap is not very good. If a rabbit walks over the trap, there is only a $p$ (probability) chance that it activates.
If the trap has no limit on the amount of rabbits it can catch, what does the distribution of rabbits caught per day look like?
My intuition is that it should follow a poisson distribution with mean $r.p$. However, I have been unable to prove this analytically. I figure that the probability of a given number of ‘successes’ should be calculable through the poisson distribution and the binomial distribution. Something like:
$P(X=x) = \sum_{i=x}^{Inf} poisson_{pmf}(i | r) . binomial_{pmf}(x | i, p)$
We can then substitute in the respective mass functions into the above equation. However, beyond that I am getting stuck. Any help or a point in the right direction would be great appreciated.
Partition the event $E_k:$ "$k$ rabbits were caught" into disjoint events of the form $E_{k}(j):$ "$k+j$ rabbits entered traps and $k$ were caught," for $j=0, 1, 2, \ldots.$ The chance of $E_k(j)$ is the product of the Poisson chance of $k+j$ entering traps and the Binomial chance that $k$ of them were caught, whence (using $\binom{j+k}{k} = (j+k)!/(j!k!)$)
\eqalign{ \Pr(E_k) &= \sum_{j=0}^\infty \Pr(E_k(j)) = \sum_{j=0}^\infty e^{-r} \frac{r^{k+j}}{(j+k)!}\quad \binom{j+k}{k} p^k(1-p)^j \\ &= e^{-r} \frac{r^k p^k}{k!}\sum_{j=0}^\infty \frac{(r(1-p))^j}{j!}. }
The last sum is the power series for $e^{r(1-p)}$, which when combined with $e^{-r}$ gives
$$\Pr(E_k) = e^{-pr}\frac{(pr)^k}{k!},$$
precisely the Poisson probability of $k$ for parameter $pr$.
A statistical way of viewing this problem is to consider a Poisson process on the line with rate $r$. "Thin" the process by randomly removing each event independently with chance $p$. Clearly the resulting process still enjoys all the defining properties of the Poisson process: homogeneous rate, independence of events, and no chance of simultaneous events. Moreover, the rate obviously is $pr$, QED.
• This is a very neat solution, thank you. The outline of the 'statistical' intuition is a nice touch. – snakeoilsales Jul 4 '17 at 20:40 | 2019-10-18T11:55:08 | {
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https://math.stackexchange.com/questions/451053/question-from-how-to-prove-it | # Question from 'How to Prove It'
Below is the question from the book mentioned above:
Suppose $f : A \rightarrow B$ and $R$ is an equivalence relation on $A$. We will say that $f$ is compatible with $R$ if $∀x \in A\forall y ∈ A(xRy \rightarrow f(x) = f(y))$. Suppose $f$ is compatible with $R$. Prove that there is a unique function $h : A/R \rightarrow B$ such that for all $x ∈ A, h\left([x]_R\right)=f(x)$.
Below is my attempt in solving this problem:
Proving the existence of the function: Let $h=\{(X,y)\in A/R\times B|\exists x\in X(f(x)=y)\}$. Let $X=([x]_R)$ be an arbitrary element of $A/R$ and $x'$ some element of $X$, so $x'Rx$. Since $f$ is compatible with $R$ $(f(x)=f(y))$,so $f(x')=f(x)\in B$. Thus $([x]_R,f(x))\in h$.
Proving the uniqueness of the function: Let j be a function such that $j([x]_R)=f(x)$. That means we can find some $[x]_R \in A/R$ such that $f(x)=y$. Since $[x]_R\in A/R$ and $y\in B$, $([x]_R,f(x))\in h$, so $j\subseteq h$. By similar reasoning, $h\subseteq j$. So $h$ is unique.
I want to ask that: 1)Is my proof about the existence of the function correct? 2)is my proof about the uniqueness correct? I basically have no idea what I am proving for the uniqueness of the function...
Please explain and give some hints if there are some mistakes in the proof above. Thanks in advance.
• Great book you're studying. – Git Gud Jul 24 '13 at 14:09
• Can't agree more! This book is very nice! – Dave Clifford Jul 24 '13 at 14:11
There are proofs that make a more or less obvious fact less believable. Yours is of this kind.
There can be at most one function $h:\ A/R\to B$ that satisfies $$h\bigl([x]_R\bigr)=f(x)\qquad\forall x\in A\ ,\tag{1}$$ since its value on any class $[x]_R$ would be given by $(1)$.
The real problem is whether $(1)$ actually defines a function on $A/R$. From dealing with equivalence classes we know the following: Given an element $X\in A/R$ there is a representant $x\in A$ of $X$, but this $x$ is not uniquely determined. When both $x$, $x'\in A$ represent $X$ then on the one hand $X=[x]_R=[x']_R$, and on the other hand $xRx'$. In this case by assumption on $f$ we have $f(x')=f(x)$, so that $(1)$ defines the same value for $h(X)$, whether we use $x$ or $x'$ to compute it.
• Do you also mean that whatever [x]_R we use, the output of h(X) will always be the same, which implies that every input has exactly only one image, suggesting that h(X) satisfies the prerequisite of being a function? – Dave Clifford Jul 25 '13 at 6:21
• @Dave Clifford: I mean, whatever $x\in A$ we use to represent the given $X\in A/R$ in the form $X=[x]_R$, the value $f(x)$ will always be the same. – Christian Blatter Jul 25 '13 at 8:19
• Oh, do you mean that there will only one image for each [x]_R right? – Dave Clifford Jul 25 '13 at 9:14
In your definition of $h, X$ is a subset of $A/R$, so putting it as the first element of an ordered pair is strange. Then in the second sentence $X$ is an element of $A/R$. You have the right thought-that you need to prove that if you apply the function to two elements of an equivalence class you get the same result.
For the uniqueness proof, you could argue that since there weren't any choices made in the construction the function must be unique. Otherwise, suppose there is a $j([x]_R)$ that disagrees with $h$. There must be at least one $[x']_R$ where $h([x']_R) \neq j([x']_R)$. But that would mean $j([x']_R) \neq f(x')$ which contradicts the definition of $j$.
• Did i mention that X is a subset of A/R? in the set h defined I already stated that X is an element of A/R, not a subset of A/R. The uniqueness proof is very clear and easy to understand, thanks. – Dave Clifford Jul 25 '13 at 6:23
• When you define $h$ you say $(X,y) \in A/R \times B$, which implies $X \in A/R$ and $y \in B$ – Ross Millikan Jul 25 '13 at 13:37 | 2019-10-16T16:57:12 | {
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https://castingoutnines.wordpress.com/tag/proof/ | # Tag Archives: proof
## How to convert a “backwards” proof into a “forwards” proof
Dave Richeson at Division By Zero wrote recently about a “proof technique” for proving equalities or inequalities that is far too common: Starting with the equality to be proven and working backwards to end at a true statement. This is a technique that is almost a valid way to prove things, but it contains — and engenders — serious flaws in logic and the concept of proof that can really get students into trouble later on.
I left a comment there that spells out my feelings about why this technique is bad. What I wanted to focus on here is something I also mentioned in the comments, which was that it’s so easy to take a “backwards” proof and turn it into a “forwards” one that there’s no reason not to do it.
Take the following problem: Prove that, for all natural numbers $n$,
$1 + 2 + 2^2 + \cdots + 2^n = 2^{n+1} - 1$
This is a standard exercise using mathematical induction. The induction step is trivial; focus on the induction step. Here we assume that $1 + 2 + 2^2 + \cdots + 2^k = 2^{k+1} - 1$ for all natural numbers less than or equal to $k$ and then prove:
$1 + 2 + 2^2 + \cdots + 2^{k+1} = 2^{k+2} - 1$
Here we have to prove two expressions are equal. Here’s what the typical “backwards” proof would look like (click to enlarge):
A student may well come up with this as his/her proof. It’s not a bad initial draft of a proof. Everything we need to make a totally correct proof is here. But the backwards-ness of it — all stemming from the first line, where we have assumed what we are trying to prove — needs fixing. Here’s how.
First, note that all the important and correct mathematical steps are taking place on the left-hand sides of the equations, and the right-hand sides are the problem here. So delete all the right-hand sides of the equals signs and the final equals sign.
Next, since the problem with the original proof was that we started with an “equation” that was not known to be true, eliminate any step that involved doing something to both sides. That would be line 4 in this proof. This might involve some re-working of the steps, in this case the trivial task of re-introducing a -1 in the final steps:
You could reverse these first two steps — eliminate all “both sides” actions and then get rid of the left-hand sides.
Then, we need to make it look nice. So for n = 1 to the end, move the $(n+1)^\mathrm{st}$ left-hand side and justification to the $n^\mathrm{th}$ right-hand side:
Now we have a correct proof that does not start by assuming the conclusion. It’s shorter, too. Really the main thing wrong with the “backwards” proof is the repeated — and, notice, unnecessary — assertion that everything is equal to the final expression. Remove that assertion and the correct “forwards” proof is basically right there looking at you.
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Filed under Abstract algebra, Geometry, Math, Problem Solving
## Technology in proofs?
We interrupt this blogging hiatus to throw out a question that came up while I was grading today. The item being graded was a homework set in the intro-to-proof course that I teach. One paper brought up two instances of the same issue.
• The student was writing a proof that hinged on arguing that both sin(t) and cos(t) are positive on the interval 0 < t < π/2. The “normal” way to argue this is just to appeal to the unit circle and note that in this interval, you’re remaining in the first quadrant and so both sin(t) and cos(t) are positive. But what the student did was to draw graphs of sin(t) and cos(t) in Maple, using the plot options to restrict the domain; the student then just said something to the effect of “The graph shows that both sin(t) and cos(t) are positive.”
• Another proof was of a proposition claiming that there cannot exist three consecutive natural numbers such that the cube of the largest is equal to the sum of the cubes of the other two. The “normal” way to prove this is by contradiction, assuming that there are three consecutive natural numbers with the stated property. Setting up the equation representing that property leads to a certain third-degree polynomial P(x), and the problem boils down to showing that this polynomial has no roots in the natural numbers. In the contradiction proof, you’d assume P(x) does have a natural number root, and then proceed to plug that root into P(x) and chug until a contradiction is reached. (Often a proof like that would proceed by cases, one case being that the root is even and the other that the root is odd.) The student set up the contradiction correctly and made it to the polynomial. But then, rather than proceeding in cases or making use of some other logical deduction method, the student just used the solver on a graphing calculator to get only one root for the polynomial, that root being something like 4.7702 (clearly non-integer) and so there was the contradiction.
So what the student did was to substitute “normal” methods of proof — meaning, methods of proof that go straight from logic — with machine calculations. Those calculations are convincing and there were no errors made in performing them, and there seemed to be no hidden “gotchas” in what the student did (such as, “That graph looks like it’s positive, but how do you know it’s positive?”). So I gave full credit, but put a note asking the student not to depend on technology when writing (otherwise exemplary) proofs.
But it raises an important question in today’s tech-saturated mathematics curriculum: Just how much technology is acceptable in a mathematical proof? This question has its apotheosis in the controversy surrounding the machine proof of the Four-Color Theorem but I’m finding a central use of (a reliance upon?) technology to be more and more common in undergraduate proof-centered classes. What do you think? (This gives me an opportunity to show off WordPress’ nifty new polling feature.)
## Riemann hypothesis proven?
This paper by Xian-Jin Li at arXiv purports to have proven the Riemann Hypothesis, arguably the most famous of the seven Millenium Problems. It’s just a preprint, of course, so it’s not final (or even peer-reviewed) yet. As commenters in the related Slashdot story mention, articles claiming the proof of the RH show up on arXiv about once per week, so I’m not getting my hopes up. Still, it would be a major breakthrough if it works out.
Filed under Math
## Fear, courage, and place in problem solving
Sorry for the slowdown in posting. It’s been tremendously busy here lately with hosting our annual high school math competition this past weekend and then digging out from midterms.
Today in Modern Algebra, we continued working on proving a theorem that says that if $a$ is a group element and the order of $a$ is $n$, then $a^i = a^j$ if and only if $i \equiv j \ \mathrm{mod} \ n$. In fact, this was the third day we’d spent on this theorem. So far, we had written down the hypothesis and several equivalent forms of the conclusion and I had asked the students what they should do next. Silence. More silence. Finally, I told them to pair off, and please exit the room. Find a quiet spot somewhere else in the building and tell me where you’ll be. Work on the proof for ten minutes and then come back.
As I wandered around from pair to pair I was very surprised to find animated conversations taking place about the proof. It wasn’t because of the time constraint — they’d been at this for three days now. For whatever reason, they were suddenly into it. One pair was practically arguing with each other over the right approach to take. By the end of the 10 minutes, two of the groups had come up with novel and mathematically watertight arguments. Between the two, and with a little bit of patching and a lemma that needs to be proven still, they generated the proof.
One student made the remark that she had been thinking of these ideas all along, but she didn’t feel like it was OK to say anything. This is a very verbal, conversational class done in Moore method style, so I can only interpret that comment to mean that she didn’t feel free enough, or bold enough, to say what she was thinking. The right proof was just bottled up in her mind all this time.
There’s something about our physical surroundings which figures in significantly to our effectiveness as problem solvers. Getting out of the classroom, for this one student at least, was tantamount to giving her permission to have the correct thoughts she was already having and to express them in a proof. I think our problem solving skills are highly inhibited by fear — fear that we will be wrong. And it takes a tremendous amount of confidence and/or courage as a problem solver to overcome that fear.
When you feel that fear in a classroom, it becomes compounded by the dread of looking like an idiot. Changing the surroundings — making things a little less cozy, a little more unusual and uncertain — doesn’t seem to make the fear go away as much as it helps us feel like that fear is perfectly normal and manageable, if not less fearful. | 2017-02-20T22:36:18 | {
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https://math.stackexchange.com/questions/1957973/mathematical-olympiad-russia-1992 | Problem
My attempt
From AM-GM inequality it can be shown that $$\sum_{i=1}^{n}(a_i+b_i)^2\geq4\sum_{i=1}^{n}a_ib_i$$ Therefore, we have: $$\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}(a_i+b_i)^2\geq4\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}a_ib_i$$ Now, on expanding the RHS, we get: $$\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}(a_i+b_i)^2\geq4\left({1\over a_1b_1}+{1\over a_2b_2}+.....{1\over a_nb_n}\right)\left(a_1b_1+a_2b_2+....a_nb_n\right)$$ $$=4(n+\left({a_2b_2\over a_1b_1}+{a_1b_1\over a_2b_2}+....{a_nb_n\over a_1b_1}+{a_1b_1\over a_nb_n}\right)+\left({a_3b_3\over a_2b_2}+{a_2b_2\over a_3b_3}+....{a_nb_n\over a_2b_2}+{a_2b_2\over a_nb_n}\right)+...+\left({a_nb_n\over a_{n-1}b_{n-1}}+{a_{n-1}b_{n-1}\over a_nb_n}\right))$$
Using the fact that $a+{1\over a}\geq2$ we get: $$\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}(a_i+b_i)^2\geq4\left({1\over a_1b_1}+{1\over a_2b_2}+.....{1\over a_nb_n}\right)\left(a_1b_1+a_2b_2+....a_nb_n\right)\geq4(n+2(n-1+n-2+....+1))=4n^2$$
Is this proof correct? I am asking this because the proof given by the author of my textbook is quite different from this one.
Edit: Author's solution.
• I really like your proof, and I think it's fine. What does the textbook have to say on this question? – астон вілла олоф мэллбэрг Oct 7 '16 at 12:34
• I've made an edit. Please check. – nls Oct 7 '16 at 12:44
• I see. His proof is very nice, but his revolves around the Chebyshev (or Cauchy Schwarz or something like that ) inequality (the second part), while yours revolves around a simpler inequality of sum of number and reciprocal being greater than $2$. Your proof I have double checked, you should be good to go. – астон вілла олоф мэллбэрг Oct 7 '16 at 12:48
• Thank you for your prompt response. – nls Oct 7 '16 at 12:49
This is a correct proof, and is how I would have solved the problem. Problems like this usually have a large number of proofs, sometimes extremely different ones, and I wouldn't let the fact that this proof isn't in the book deter you. It would be useful though to learn the book's technique as well.
For the author's proof, the first equation given follows from $$\frac{1}{ab}\geq\frac{4}{(a+b)^2}$$ by simply adding a bunch of terms of that form, but the second equation. The second equation follows from Cauchy-Schwarz, which says (when restricted to real numbers) that $$\left(\sum_{i\in I} \alpha_i \beta_i\right)^2 \leq\sum_{j\in I} \alpha_j^2\sum_{k\in I} \beta_k^2$$ Applying this inequality with $\alpha_i=(a_i+b_i)^2$ and $\beta_i=(a_i+b_i)^{-2}$ produces the desired result.
• @Shrey Aryan I added an explication of the author's proof. – Stella Biderman Oct 7 '16 at 13:57
Set $c_i = (a_i +b_i)^2$. Then by AM-GM, $a_ib_i \le c_i/4$. Therefore, our expression is bounded below by
$$4 \sum_{i=1}^n \frac{1}{c_i} \sum_{j=1}^n c_i =4 \sum_{i=1}^n c_i \times \sum_{i=1}^n \frac{1}{c_i}.$$
As for the RHS:
1. By the harmonic - arithmetic mean inequality, $$\frac{n}{ \sum_{i=1}^n \frac{1}{c_i}} \le \frac{ {\sum_{i=1}^n c_i }}{n},$$ so result follows.
2. Alternatively, use Cauchy Schwarz: $$\sum_{i=1}^n c_i \times \sum_{i=1}^n \frac{1}{c_i} \ge \left (\sum_{i=1}^n \sqrt{c_i} \frac{1}{\sqrt{c_i}}\right)^2 = n^2,$$ so result follows. | 2019-11-12T11:30:44 | {
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http://zymne.org/w32b9/d99760-how-to-add-radical-expressions | Step 2: Add or subtract the radicals. \color{red}{\sqrt{\frac{54}{x^4}}} &= \frac{\sqrt{54}}{\sqrt{x^4}} = \frac{\sqrt{9 \cdot 6}}{x^2} = \color{red}{\frac{3 \sqrt{6}}{x^2}} In order to be able to combine radical terms together, those terms have to have the same radical part. Adding radical expressions with the same index and the same radicand is just like adding like terms. 4 \cdot \color{blue}{\sqrt{\frac{20}{9}}} + 5 \cdot \color{red}{\sqrt{\frac{45}{16}}} &= \\ Finding the value for a particular root is difficul… It's like radicals. Adding and subtracting radical expressions can be scary at first, but it's really just combining like terms. &= 3 \cdot \color{red}{5 \sqrt{2}} - 2 \cdot \color{blue}{2 \sqrt{2}} - 5 \cdot \color{green}{4 \sqrt{2}} = \\ We add and subtract like radicals in the same way we add and subtract like terms. How to Add and Subtract Radicals? More Examples x11 xx10 xx5 18 x4 92 4 32x2 Ex 4: Ex 5: 16 81 Examples: 2 5 4 9 45 49 a If and are real numbers and 0,then b a a b b b z B. Add or subtract to simplify radical expression: \begin{aligned} \sqrt{50} &= \sqrt{25 \cdot 2} = 5 \sqrt{2} \\ For , there are pairs of 's, so goes outside of the radical, and one remains underneath the radical. Adding and subtracting radical expressions that have variables as well as integers in the radicand. \underbrace{ 4\sqrt{3} + 3\sqrt{3} = 7\sqrt{3}}_\text{COMBINE LIKE TERMS} Jarrod wrote two numerical expressions. You probably won't ever need to "show" this step, but it's what should be going through your mind. \sqrt{27} &= \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3 \sqrt{3} In this tutorial, you will learn how to factor unlike radicands before you can add two radicals together. It is possible that, after simplifying the radicals, the expression can indeed be simplified. Two radical expressions are called "like radicals" if they have the same radicand. I am ever more convinced that the necessity of our geometry cannot be proved -- at least not by human reason for human reason. $3 \sqrt{50} - 2 \sqrt{8} - 5 \sqrt{32}$, Example 3: Add or subtract to simplify radical expression: Next, break them into a product of smaller square roots, and simplify. Radical expressions are called like radical expressions if the indexes are the same and the radicands are identical. 3 \color{red}{\sqrt{50}} - 2 \color{blue}{\sqrt{8}} - 5 \color{green}{\sqrt{32}} &= \\ $6 \sqrt{ \frac{24}{x^4}} - 3 \sqrt{ \frac{54}{x^4}}$, Exercise 2: Add or subtract to simplify radical expression. Please accept "preferences" cookies in order to enable this widget. This web site owner is mathematician Miloš Petrović. 2 \color{red}{\sqrt{12}} + \color{blue}{\sqrt{27}} = 2\cdot \color{red}{2 \sqrt{3}} + \color{blue}{3\sqrt{3}} = The radical part is the same in each term, so I can do this addition. I can simplify most of the radicals, and this will allow for at least a little simplification: These two terms have "unlike" radical parts, and I can't take anything out of either radical. You can have something like this table on your scratch paper. factors to , so you can take a out of the radical. $4 \sqrt{2} - 3 \sqrt{3}$. But the 8 in the first term's radical factors as 2 × 2 × 2. You can use the Mathway widget below to practice finding adding radicals. It is often helpful to treat radicals just as you would treat variables: like radicals can be added and subtracted in the same way that like variables can be added and subtracted. 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http://math.stackexchange.com/questions/125060/2x-2y-129-how-to-find-the-x-and-y | # $2^x + 2^y = 129$: how to find the $x$ and $y$?
Sorry for my math language and the question header; I'm not capable of the terms used for the mathematics to ask the question via text; so I had to use the example above; feel free to edit if you can keep the meaning the same.
So the question is like this: I know that the total is uniquely identifies the used exponential constants while using the base as $2$; like $2^x + 2^y = 129$ then the $x=0$ and $y=7$.
So I couldn't think of a function to get the values like $\operatorname{func}(129) = [0,7]$.
Thanks in advance. Note: I know it's dumb to ask that but help is appreciated.
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actually $x=0$ ($129=1+128$) – Andrea Mori Mar 27 '12 at 13:54
@AndreaMori sorry for that(: – Beytan Kurt Mar 27 '12 at 13:55
The function you are looking for is called dec2bin, when you are using Matlab. It would convert $129_{10}$ in decimal to $10000001_2$ in binary notation. As dec2base it works for other bases as well.
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Thanks for that, I should have thought binary way while all the problem is base 2; thanks(: – Beytan Kurt Mar 27 '12 at 14:05
I would also like to mention Windows calculator has a programmer mode. Simply input the number as a decimal number and it will show you the binary representation with spaces every 4 bits to make counting easier. – Mike Mar 27 '12 at 17:31
Thanks but I'll use it for programming, so no calculator included. – Beytan Kurt Mar 27 '12 at 19:03
If $x>0 ~\text{and}~y>0$ then $2^x+2^y$ is an even number . Hence :
$(x,y)=\{(0,7),(7,0)\}$
-
I couldn't understand how you could find the result?? there's no tip to find 0 or 7?? – Beytan Kurt Mar 27 '12 at 14:01
$2^x+2^y$ can be odd number only if $x=0 ~\text{or}~ y=0$ , therefore it is easy to calculate that value of the second variable has to be $7$ . – pedja Mar 27 '12 at 14:08
I've asked for a function or a method to evaluate; your answer gives the how easy it is explanation; not the method but thanks. – Beytan Kurt Mar 27 '12 at 14:33
I'm not sure what is actually the question, but the problem of writing a number $n\in\Bbb N$ as a sum of powers of $2$ is equivalent to finding its expression in base $2$.
-
I don't think that 2^x + 2^y = 2^(x+y); the question is a workaround to keep a single number that could give multiple selection numbers like categoryA=0, categoryB=1, categoryZ=7 then when a result is given like 129 I can uniquely identify that catA and catZ is used with a single result – Beytan Kurt Mar 27 '12 at 14:01
As I told you, it is equivalent to find the expansion of n in base 2 and this is obtained by a repeated division by 2 keeping track of the remainders. For instance $129=2\cdot64+1$, $64=2\cdot32+0$, $32=2\cdot16+0$, $16=2\cdot8+0$, $8=2\cdot4+0$, $4=2\cdot2+0$, $2=2\cdot1+0$ and $1=2\cdot0+1$. Thus, $129=10000001$ (binary), i.e. $129=2^0+2^7$ because you have $1$ only in the $0$-th and $7$-th place. – Andrea Mori Mar 27 '12 at 16:02
Thanks a lot (: – Beytan Kurt Mar 27 '12 at 19:01
Hint $\$ For naturals $\rm\:b>1,\ x < y,\:$ if $\rm\: n\: =\: b^{x} + b^{y} =\: b^x\: (1 + b^{y-x})\:$ then $\rm\:x\:$ is determined uniquely as the greatest power of $\rm\:b\:$ that divides $\rm\:n,\:$ and $\rm\:y\:$ is determined uniquely as $\rm\:x\:$ plus the greatest power of $\rm\:b\:$ that divides $\rm\:n/b^{x}-1.$
This is a special case of the existence and uniqueness of radix representation, which follows from that for the division algorithm (here repeated division by the radix $\rm b$).
- | 2015-04-28T05:19:54 | {
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http://www.chebfun.org/docs/guide/guide13.html | ### 13.1 sum and sum2
We have already seen the sum2 command, which returns the definite double integral of a chebfun2 over its domain of definition. The sum command is a little different, integrating with respect to one variable at a time following the MATLAB analogy. For instance, the following commands integrate $\sin(10xy)$ with respect to $y$:
f = chebfun2(@(x,y) sin(10*x.*y),[0 pi/4 0 3]);
sum(f)
ans =
chebfun row (1 smooth piece)
interval length endpoint values
[ 0, 0.79] 35 -2.2e-15 0.13
Epslevel = 3.552714e-15. Vscale = 2.173762e+00.
A chebfun is returned because the result depends on $x$ and hence is a function of one variable. Similarly, we can integrate over the $x$ variable and plot the result.
LW = 'linewidth';
sum(f,2), plot(sum(f,2),LW,1.6)
ans =
chebfun column (1 smooth piece)
interval length endpoint values
[ 0, 3] 35 -5.6e-16 0.033
Epslevel = 3.552714e-15. Vscale = 5.690896e-01.
A closer look reveals that sum(f) returns a row chebfun while sum(f,2) returns a column chebfun. This distinction is a reminder that sum(f) is a function of $x$ while sum(f,2) is a function of $y$. If we integrate over $y$ and then $x$, the result is the double integral of $f$.
sum2(f)
sum(sum(f))
ans =
0.377914013520379
ans =
0.377914013520379
It is interesting to compare the execution times involved for computing the double integral by different commands. Chebfun2 does very well for smooth functions. Here we see an example in which it is faster than the MATLAB quad2d command.
F = @(x,y) exp(-(x.^2 + y.^2 + cos(4*x.*y)));
tol = 3e-14;
tic, I = quad2d(F,-1,1,-1,1,'AbsTol',tol); t = toc;
fprintf('QUAD2D: I = %17.15f time = %6.4f secs\n',I,t)
tic, I = sum(sum(chebfun2(F))); t = toc;
fprintf('CHEBFUN2/SUMSUM: I = %17.15f time = %6.4f secs\n',I,t)
tic, I = sum2(chebfun2(F)); t = toc;
fprintf('CHEBFUN2/SUM2: I = %17.15f time = %6.4f secs\n',I,t)
QUAD2D: I = 1.399888131932670 time = 0.2660 secs
CHEBFUN2/SUMSUM: I = 1.399888131932670 time = 0.0837 secs
CHEBFUN2/SUM2: I = 1.399888131932670 time = 0.0351 secs
Chebfun2 is not designed specifically for numerical quadrature (or more properly, "cubature"), and careful comparisons with existing software have not been carried out. Low rank function approximations have been previously used for numerical quadrature by Carvajal, Chapman, and Geddes [Carvajal, Chapman & Geddes 2005]. A cubature package CHEBINT based on Chebyshev approximations has been produced by Poppe and Cools [Poppe & Cools 2011].
### 13.2 norm, mean, and mean2
The $L^2$-norm of a function $f(x,y)$ can be computed as the square root of the double integral of $f^2$. In Chebfun2 the command norm(f), without any additional arguments, computes this quantity. For example,
f = chebfun2('exp(-(x.^2 + y.^2 +4*x.*y))');
norm(f), sqrt(sum2(f.^2))
ans =
2.819481057146932
ans =
2.819481057146935
Here is another example. This time we compute the norms of $f(x,y)$, $\cos(f(x,y))$, and $f(x,y)^5$.
f = chebfun2(@(x,y) exp(-1./( sin(x.*y) + x ).^2));
norm(f), norm( cos(f) ), norm( f.^5 )
ans =
0.462652919760561
ans =
1.950850368197070
ans =
0.060896016071932
The command mean2 scales the result of sum2 to return the mean value of $f$ over the rectangle of definition:
help chebfun2/mean2
MEAN2 Mean of a CHEBFUN2
V = MEAN2(F) returns the mean of a CHEBFUN:
d b
/ /
V = 1/(d-c)/(b-a) | | f(x,y) dx dy
/ /
c a
where the domain of F is [a,b] x [c,d].
For example, here is the average value of a 2D Runge function.
runge = chebfun2( @(x,y) 1./( .01 + x.^2 + y.^2 )) ;
plot(runge)
mean2(runge)
ans =
3.796119578934829
The command mean computes the average along one variable. The output is a function of one variable represented by a chebfun, so we can plot it.
plot(mean(runge),LW,1.6)
title('Mean value of 2D Runge function wrt y')
If we average over $y$ and then $x$, we obtain the mean value over the whole domain, matching the earlier result.
mean(mean(runge))
ans =
3.796119578934826
### 13.3 cumsum and cumsum2
The command cumsum2 computes the double indefinite integral, which is a function of two variables, and returns a chebfun2.
help chebfun2/cumsum2
CUMSUM2 Double indefinite integral of a CHEBFUN2.
F = CUMSUM2(F) returns the double indefinite integral of a CHEBFUN2. That is
y x
/ /
CUMSUM2(F) = | | f(x,y) dx dy for (x,y) in [a,b] x [c,d],
/ /
c a
where [a,b] x [c,d] is the domain of f.
On the other hand, cumsum(f) computes the indefinite integral with respect to just one variable, also returning a chebfun2. The indefinite integral with respect to $y$ and then $x$ is the same as the double indefinite integral, as we can check numerically.
f = chebfun2(@(x,y) sin(3*((x+1).^2+(y+1).^2)));
contour(cumsum2(f),'numpts',400), axis equal
title('Contours of cumsum2(f)'), axis([-1 1 -1 1])
norm( cumsum(cumsum(f),2) - cumsum2(f) )
ans =
0
### 13.4 Complex encoding
As is well known, a pair of real scalar functions $f$ and $g$ can be encoded as a complex function $f+ig$. This trick can be useful for simplifying many operations, though at the same time it may be confusing. For instance, instead of representing the unit circle by two real-valued functions, we can represent it by one complex-valued function:
d = [0 2*pi];
c1 = chebfun(@(t) cos(t),d); % first real-valued function
c2 = chebfun(@(t) sin(t),d); % second real-valued function
c = chebfun(@(t) cos(t)+1i*sin(t),d); % one complex function
Here are two ways to make a plot of a circle.
subplot(1,2,1), plot(c1,c2,LW,1.6)
axis equal, title('Two real-valued functions')
subplot(1,2,2), plot(c,LW,1.6)
axis equal, title('One complex-valued function')
This complex encoding trick is exploited in a number of places in Chebfun2. Specifically, it's used to encode the path of integration for a line integral (see next section), to represent zero contours of a chebfun2 (Chapter 14), and to represent trajectories in vector fields (Chapter 15).
We hope users become comfortable with complex encodings, though they are not required for the majority of Chebfun2 functionality.
### 13.5 Integration along curves
Chebfun2 can compute the integral of $f(x,y)$ along a curve $(x(t),y(t))$. It uses the complex encoding trick and encode the curve $(x(t),y(t))$ as a complex valued chebfun $x(t) + iy(t)$.
For example, here is the curve in the unit square defined by $\exp(10 it)$, $t\in[0,1]$.
clf
C = chebfun(@(t) t.*exp(10i*t),[0 1]);
plot(C,'k',LW,2), axis([-1 1 -1 1]), axis square
Here is a plot of the function $f(x,y) = \cos(10xy^2) + \exp(-x^2)$ on the square, with the values of $f(x,y)$ on the curve $C$ shown in black:
f = chebfun2(@(x,y) cos(10*x.*y.^2) + exp(-x.^2));
plot(f), hold on
plot3(real(C),imag(C),f(C),'k',LW,2)
The object $|f(C)|$ is just a real-valued function defined on $[0,1]$, whose integral we can readily compute:
sum(f(C))
ans =
1.613596461872283
This number can be interpreted as the integral of $f(x,y)$ along the curve $C$.
### 13.6 diff, diffx, diffy
In MATLAB the diff command calculates finite differences of a matrix along its columns (by default) or rows. For a chebfun2 the same syntax represents partial differentiation $\partial f/\partial y$ (by default) or $\partial f/\partial x$.
As pointed out in the last chapter, however, this can be rather confusing. Accordingly Chebfun2 offers the alternatives diffx and diffy with more obvious meaning. Here for example is the syntax of diffx:
help chebfun2/diffx
DIFFX Differentiate a CHEBFUN2 with respect to its first argument.
G = DIFFX(F) returns a CHEBFUN2 representing the derivative of F in its
first argument. This is the same as DIFF(F,1,2).
G = DIFFX(F,N) returns a CHEBFUN2 representing the Nth derivative of F in
its first argument. This is the same as DIFF(F,N,2).
This command is for convenience as the syntax for DIFF, inherited from the
DIFF command for matrices, can be confusing.
Here we use diffx and diffy to check that the Cauchy-Riemann equations hold for an analytic function.
f = chebfun2(@(x,y) sin(x+1i*y)); % a holomorphic function
u = real(f); v = imag(f); % real and imaginary parts
norm(diffy(v) - diffx(u))
norm(diffx(v) + diffy(u)) % Do the Cauchy-Riemann eqns hold?
ans =
2.577622443765214e-14
ans =
1.064740840017172e-14
### 13.7 Integration in three variables
Chebfun2 also works pretty well for integration in three variables. The idea is to integrate over two of the variables using Chebfun2 and the remaining variable using Chebfun. We have selected a tolerance of $10^{-6}$ for this example because the default tolerance in the MATLAB integral3 command is also $10^{-6}$.
r = @(x,y,z) sqrt(x.^2 + y.^2 + z.^2);
t = @(x,y,z) acos(z./r(x,y,z)); p = @(x,y,z) atan(y./x);
f = @(x,y,z) sin(5*(t(x,y,z) - r(x,y,z))) .* sin(p(x,y,z)).^2;
I = @(z) sum2(chebfun2(@(x,y) f(x,y,z),[-2 2 .5 2.5]));
tic, I = sum(chebfun(@(z) I(z),[1 2],'vectorize')); t = toc;
fprintf(' Chebfun2: I = %16.14f time = %5.3f secs\n',I,t)
tic, I = integral3(f,-2,2,.5,2.5,1,2); t = toc;
fprintf(' MATLAB integral3: I = %16.14f time = %5.3f secs\n',I,t)
Chebfun2: I = -0.48056569408898 time = 2.685 secs
MATLAB integral3: I = -0.48056569417568 time = 1.289 secs
### 13.8 References
[Carvajal, Chapman & Geddes 2005] O. A. Carvajal, F. W. Chapman and K. O. Geddes, "Hybrid symbolic-numeric integration in multiple dimensions via tensor-product series", Proceedings of ISSAC'05, M. Kauers, ed., ACM Press, 2005, pp. 84--91.
[Poppe & Cools 2011] K. Poppe and R. Cools, "CHEBINT: operations on multivariate Chebyshev approximations", http://nines.cs.kuleuven.be/software/CHEBINT/. | 2018-10-19T06:38:27 | {
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http://mathhelpforum.com/pre-calculus/69796-standard-form-graphing.html | # Thread: Standard form & Graphing
1. ## Standard form & Graphing
Good morning forum I need some assistance with the following question:
Put the equation x2 + y2 + 8x + 2y = 29 into standard form and sketch the graph of the equation.
Thanks AC
2. Originally Posted by AlgebraicallyChallenged
Good morning forum I need some assistance with the following question:
Put the equation x2 + y2 + 8x + 2y = 29 into standard form and sketch the graph of the equation.
Thanks AC
I take it that the 2's are superscript.
So it should read $x^2 + y^2 + 8x + 2y = 29$.
First off, this is the equation of a circle.
The standard form of a circle is $(x - h)^2 + (y - k)^2 = r^2$ where h is the x-coordinate of the centre, k is the y-coordinate of the centre, and r is the radius.
To get it into this standard form, complete the square on the x terms and the y terms.
$x^2 + 8x + y^2 + 2y = 29$
$x^2 + 8x + 4^2 + y^2 + 2y + 1^2 = 29 + 4^2 + 1^2$
$(x + 4)^2 + (y + 1)^2 = 46$
$[x - (-4)]^2 + [y - (-1)]^2 = (\sqrt{46})^2$.
So what's the centre? What's the radius? Can you sketch the circle?
3. Originally Posted by AlgebraicallyChallenged
Good morning forum I need some assistance with the following question:
Put the equation x2 + y2 + 8x + 2y = 29 into standard form and sketch the graph of the equation.
Thanks AC
You have to complete squares, 8th grade trick.
$x^2+8x=(x+4)^2-16$ and $y^2+2y=(y+1)^2-1$
substituting this and rearranging the following is obtained:
$(x+2)^2+(y+1)^2=29+16+1=46$
this is a circle with center at $(-2,-1)$ and radius $\sqrt{46}$
4. Originally Posted by andreas
You have to complete squares, 8th grade trick.
$x^2+8x=(x+4)^2-16$ and $y^2+2y=(y+1)^2-1$
substituting this and rearranging the following is obtained:
$(x+2)^2+(y+1)^2=29+16+1=46$
this is a circle with center at $(-2,-1)$ and radius $\sqrt{46}$
Wrong, it's a circle with centre at $(-4, -1)$ and radius $\sqrt{46}$.
5. Originally Posted by Prove It
Wrong, it's a circle with centre at $(-4, -1)$ and radius $\sqrt{46}$.
Yes I admit, it was typo... | 2016-10-22T09:58:13 | {
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https://math.stackexchange.com/questions/654736/bounds-and-the-fundamental-theorem-of-calculus | # Bounds and the Fundamental Theorem of Calculus
Suppose $f : \mathbb{R} \to \mathbb{R}$ is continuous. Fix $a \in \mathbb{R}$ and define $$F(x) := \int_a^x f(t) \, \mathrm{d}t.$$ Every version of the Fundamental theorem of calculus (FTC) I've seen tells us that $F$ is differentiable for $x \geq a$ and that $F'(x) = f(x)$ for all $x \geq a$.
My question is : Is the above result also true for $x < a$ ?
My guess : I think it holds for $x < a$, since in that case I believe we have $$F(x) =\int_a^x f(t) \, \mathrm{d}t = - \int_{-a}^{-x} f(-t) \, \mathrm{d}t$$ and by the FTC and the chain rule it follows that $$F'(x) = - f(-(-x)) \cdot (-1) = f(x).$$ Is this correct ?
• Yes. Another way to think of it is to pick some $b<x$, then $F(x) = \int_b^x f -\int_b^a f$. – copper.hat Jan 28 '14 at 16:46
• How do you define $\int_a^x f(t)\>dt$ when $x<a$? – Christian Blatter Jan 28 '14 at 16:50
• @copper.hat I tried to get the $x$ as the upper bound but did not succeed ! Thanks. – Amateur Jan 28 '14 at 16:51
• @Christian Blatter I define it to be $-\int_x^a f(t) dt$. – Amateur Jan 28 '14 at 16:52
• @Amateur: $\int_b^a = \int_b^x + \int_x^a$. – copper.hat Jan 28 '14 at 16:59
Well, you've developed a somewhat circular argument because you want to show $F$ is differentiable for $x < a$, but then you use that in your proof.
But, $a$ was chosen arbitrarily. So, if you choose a different starting value, some $\tilde a < x$, then it will be true that
$$\tilde F (x) := \int_{\tilde a}^x f(t)\,dt$$ is differentiable for $x > \tilde a$, $\tilde F' = f$, etc.
Then for $\tilde a < x < a$
$$\tilde F(x) = \int_{\tilde a}^a f(t)\,dt - \int_{x}^a f(t)\,dt = C + \int_a^x f(t)\,dt = C + F(x)$$
so $\tilde F$ and $F$ only differ by a constant, and everything follows from there.
• Thank you for pointing out that my argument is circular. – Amateur Jan 28 '14 at 17:00 | 2019-08-22T13:21:56 | {
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http://math.stackexchange.com/questions/76769/index-of-a-group | Index of a group
Let $H$, $K$ be subgroups of a given group $G$. Can one show that $(G:(H\cap K))$ is less or equal to $(G:H)(G:K)$, where $(G:H)$ stands for the index of group G with respect to $H$?
-
$(G:(H\cap K)) = (G:H)(H:H\cap K)$. So this amounts to showing that $(H:H\cap K)\leq (G:K)$ – Thomas Andrews Oct 28 '11 at 20:33
(Note the simplest route, just pointing out a related post.) This question asks to prove that $|H| |K| \leq |H \cap K| | \langle H, K \rangle |$, which is equivalent to $(G:H)(G:K) \geq (G:H \cap K) (G: \langle H, K \rangle)$. Since $(G: \langle H, K \rangle) \geq 1$, we get the claim in this question (assuming that question). – Srivatsan Oct 28 '11 at 20:33
@Thomas, indeed but this is my main problem. Thank you. – Ticuramamba Oct 28 '11 at 20:36
I probably make a horrible mistake, but consider the function $\pi : H \hookarrow G \rightarrow G/K$, as a function on cosets. This should induce an injection $f: H/(H\cap K) \rightarrow G/K$ which proves your statement. If all subgroups are normal, I think this idea works imediately, my sugestion is to forget the group structure of $H/(H\cap K)$ and $G/K$ and work set theoretically. Just be careful if you use left or right cosets. – N. S. Oct 28 '11 at 21:42
Proposition. If $H$ and $K$ are subgroups of $G$, then $[H:H\cap K]\leq [\langle H,K\rangle:K]$ in the sense of cardinalities. If $\langle H,K\rangle = HK$, then we have equality; in the finite case, this condition is also necessary for equality.
Proof. Let $\mathcal{C}$ be the set of all left cosets of $H\cap K$ in $H$, and let $\mathcal{D}$ be the set of all left cosets of $K$ in $\langle H,K\rangle$.
Define a (set-theoretic) map $f\colon\mathcal{C}\to\mathcal{D}$ by $f(h(H\cap K)) = hK$.
First, this is well-defined and injective: if $h,h'\in H$, then \begin{align*} h(H\cap K) = h'(H\cap K) &\Longleftrightarrow h^{-1}h'\in H\cap K\\ &\Longleftrightarrow h^{-1}h'\in K\\ &\Longleftrightarrow hK = h'K. \end{align*} Moreover, note that since $H\subseteq \langle H,K\rangle$, then $hK\in\mathcal{D}$.
Therefore, $f$ is a one-to-one function from $\mathcal{C}$ to $\mathcal{D}$, hence $[H:H\cap K] = |\mathcal{C}| \leq |\mathcal{D} = [\langle H,K\rangle:K]$.
If $\langle H,K\rangle = HK$, then every element of $\mathcal{D}$ is of the form $hK$ for some $h\in H$, so $f$ will also be surjective, giving $|\mathcal{C}|=|\mathcal{D}|$. In the finite case, if we have equality, then since $f$ is one-to-one between two finite sets of the same size, it must be onto, so given any element $g$ of $\langle H,K\rangle$, there exists $h\in H$ such that $gK=hK$; hence there exists $k\in K$ such that $g=hk\in HK$, so $\langle H,K\rangle\subseteq HK$, as desired. $\Box$
Applying this to your case, we have that in the sense of cardinalities, \begin{align*} [G:H\cap K] &= [G:H][H:H\cap K] \\ &\leq [G:H][\langle H,K\rangle:K]\\ &\leq [G:H][G:K] \end{align*} If $G$ is finite, then equality holds if and only if $HK=G$.
-
You could show that $a(H \cap K) = aH \cap aK\$ for every $a\in G$. We can choose $aH \cap aK$ in $[G:H][G:K]$ ways. Some of the combinations might be same, but there can be no more than $[G:H][G:K]$ cosets of $H \cap K$.
-
I feel like it's worth mentioning that if $G$ is finite (probably true more generally) and $H\leqslant N_G(K)$ then $[G:H\cap K]\mid [G:H][G:K]$. Indeed, by containment in the normalizer we know that $HK\leqslant G$ and so $\displaystyle \frac{|G|}{|HK|}$ is an integer. That said,
\displaystyle \begin{aligned}\frac{|G|}{|HK|}&= \frac{|G|}{\displaystyle \frac{|H||K|}{|H\cap K|}}\\ &= \frac{\displaystyle \frac{|G|}{|H|}\frac{|G|}{|K|}}{\displaystyle \frac{|G|}{|H\cap K|}}\\ &= \frac{[G:H][G:K]}{[G:H\cap K]}\end{aligned}
I don't know if that's relevant to your interests, but I thought it's worth mentioning. It helps give a proof why the only subgroup of $S_n$ (for $n\geqslant 5$) of index $2$ is $A_n$.
Well, for divisibility to make sense to you need the indices to be finite; so you need $H$ and $K$ to be finite index in $G$ before the expression makes sense. If $H$ and $K$ have finite index, then so does $H\cap K$, and so does the core of $H\cap K$ (the largest normal subgroup that is contained in $H\cap K$). Moding out by the core of $H\cap K$, you can reduce to the finite case. – Arturo Magidin Oct 29 '11 at 22:09 | 2014-03-10T12:29:57 | {
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https://math.stackexchange.com/questions/3732230/explanation-for-behaviour-of-graph-of-y-x2e-x2-maxwell-boltzmann-distrib/3732254 | # Explanation for behaviour of graph of $y=x^2e^{-x^2}$ (Maxwell-Boltzmann distribution)
Consider the function $$y=x^2e^{-x^2}$$ The graph initially behaves as a parabola then in later part exponential part of it dominates; i.e., the graph looks exponential after maximum of the curve.
Actually this graph is related to Maxwell Boltzmann distribution graph. Please help me so that I can easily remember the property of this graph.
• Please confirm that the function resulting from my edit reflects the intent of your question. – abiessu Jun 24 at 2:59
• Yes thanks a lot – shelton Benjamin Jun 24 at 3:02
• MathJax hint: for multicharacter exponents, enclose them in braces, so e^{-x^2} gives $e^{-x^2}$. It works for many things, like subscripts and fractions as well. You can right click on any MathJax and choose Show Math As ->TeX commands to see how it was done. – Ross Millikan Jun 24 at 3:38
You actually gave the mathematical explanation. The graph is below. Over the range $$[-1,1]$$ the exponential doesn't change that much-it is $$1$$ at the center and $$\frac 1e \approx 0.3679$$ at the ends. That is less than a factor $$3$$. The parabola is $$0$$ at the middle and $$1$$ at the ends, an infinite ratio. It dominates the product over this interval. As you get outside that interval, the exponential dominates. From $$1$$ to $$3$$ the parabola rises by a factor $$9$$, but the exponential drops by a factor $$2980$$, so it dominates.
$$f(x)=x^2e^{-x^2}\implies f'(x)=2x(1-x^2)e^{-x^2}, f''(x)=2x^2(x^4-5x^2+1)e^{-x^2}.$$ 1- Check that $$f(x)$$ is even and $$f(0)=0, f(\pm \infty)=0,. f'(x)=0 \implies x=0, \pm 1, f''(\pm 1)<0, f'(x)$$ does not change sign around $$x=0$$. So min at $$x=0$$ max at $$x=\pm 1, f_{max}=e^{-1}, f_{min}=0.$$ All these information helps plotting this function as ny @Ross Millikan in his answer here. | 2020-08-15T17:55:23 | {
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https://math.stackexchange.com/questions/1491323/upper-bound-of-a-double-sequence | # upper bound of a double sequence
Let $\{ a_{k,m} \}$ be a doublely indexed sequence of positive numbers satisfying:
$a_{1,n}\leq \frac{1}{n+1}\quad$ and $\quad a_{k,m} \leq \frac{1}{m+1}(a_{k-1,m+1}+L a_{k-1,m+2})\quad \quad (1)$ ,
where $L$ is a positive constant and indices $k,m,n$ are non-negative integers.
My question is: what is the upper bound for $a_{n,0}$ in terms of n ?
I try to find the bound by induction using relation in $(1)$ to reduce the index $n$ in $a_{n,0}$ to $1$, but I'm not able to compute the final representation in terms of $a_{1,k} (k\geq 1)$.
The question is slightly ill-posed, since one may come up with several bounds, not just one, and it is not clear which one you will be satisfied with. In general, one looks for bounds having specific properties (among all the possible bounds), allowing one to further construct a certain reasoning. You do not specify what type of reasoning you want the bound to enable.
As you have discovered, trying to come up with a general formula for $a_{n,0}$ in terms of $a_{1,\ldots}$ is not easy at all. Fortunately, you want an upper bound, not an exact value, so when working with equalities turns out to be too difficult we shall replace them with inequalities, getting a weaker result but still a satisfying one.
Starting out with the formula you give, one immediately goes back one step to
$$a_{k,m} \le \frac 1 {(m+1)(m+2)} a_{k-2,m+2} + \left( \frac 1 {(m+1)(m+2)} + \frac 1 {(m+1)(m+3)} \right) L a_{k-2,m+3} + \frac 1 {(m+1)(m+3)} L^2 a_{k-2,m+4}$$
which is quite ugly and is probably where you got stuck. The trick is to replace each fraction with the larger fraction $\dfrac 1 {(m+1)^2}$, thus obtaining (note that the middle term contains two fractions)
$$a_{k,m} \le \frac 1 {(m+1)^2} \left( a_{k-2,m+2} + 2 L a_{k-2,m+3} + L^2 a_{k-2,m+4} \right) .$$
Note that the quantity between brackets reminds us of Newton's binomial; also, note that when you sum both indices of $a$, you get $k+m$, $k+m+1$ and $k+m+2$. This makes us suspect that after backtracking, the final expression (the one involving only terms like $a_{1,\ldots}$) will look like
$$a_{k,m} \le \frac 1 {(m+1)^{k-1}} \left( \binom {k-1} 0 a_{1,k+m-1} + \binom {k-1} 1 L a_{1,k+m} + \binom {k-1} 2 L^2 a_{1,k+m+1} + \dots + \binom {k-1} {k-1} L^{k-1} a_{1,k+m-1 + k-1} \right) .$$
In order to get rid of the $a_{1,\ldots}$ we shall use $a_{1,n} \le \dfrac 1 {n+1}$. This is good, but will still give us an ugly result, therefore we shall use one more upper bound for all these terms: $a_{1,k+m+i} \le \dfrac 1 {k+m} \ \forall 0 \le i \le k-1$. Using $\sum \limits _{i=0} ^{k-1} \binom {k-1} i L^i = (1+L)^{k-1}$ will allow us to finally write
$$\color{blue} {a_{k,m} \le \frac 1 {(m+1)^{k-1}} \frac 1 {k+m} (1+L)^{k-1}} .$$
Of course, guessing is only a part of the solution. Let us use induction on $k$ to also show that our guess is correct.
For $k=1$ we get $a_{1,m} \le \frac 1 {m+1}$ which is precisely what the problem assumes.
Assume now the above guess true for $k$ and let us prove it for $k+1$. Using the recursion given by the problem,
$$a_{k+1, m} \le \frac 1 {m+1} \left( a_{k,m+1} + L a_{k,m+2} \right) \le \frac 1 {m+1} \left( \frac 1 {(m+2)^{k-1}} \frac 1 {k+m+1} (1+L)^{k-1} + L \frac 1 {(m+3)^{k-1}}\frac 1 {k+m+2} (1+L)^{k-1} \right) .$$
Using that $\dfrac 1 {m+2}, \dfrac 1 {m+3} \le \dfrac 1 {m+1}$ and that $\dfrac 1 {k+m+2} \le \dfrac 1 {k+m+1}$ makes the last term above
$$\le \frac 1 {(m+1)^k} \frac 1 {k+m+1} (1+L)^k$$
which is precisely the induction hypothesis for $k+1$ instead of $k$.
Finally, taking $k=n$ and $m=0$ gives $\color{red} {a_{n,0} \le \dfrac 1 n (1+L)^{n-1}}$.
If you were patient enough to follow the above computations you have noted that at several moments we have sacrificed equalities and used upper bounds instead. This means that this upper bound that we have obtained is not the best, but it has the advantage that is has the simplest formula, allowing you to use it. Better upper bounds are possible, but uglier and therefore difficult to use in applications.
• If we replace inequalities by identities, we trivially have by induction that for every fixed $k$, $a_{km}$ is decreasing in $m$ for the corresponding majorant. Thus $a_{km}\le\frac{1+L}{m+1}a_{k-1,m+1}$, whence $a_{n0}\le \frac{(1+L)^{n-1}}{n!}$. Of course, this is very rough too. – fedja Oct 31 '15 at 1:45
• @fedja haha, you got it. – booksee Nov 1 '15 at 1:17
As fedja mentioned, if $a_{k,m+1}\leq a_{k,m}$, we can easily arrive at a bound in fedja's comment. Since this may not be the case, we do need some preparations to formalize the argument.
We define $\hat{a}_{k,m}:=\text{maximal achieveble upper bound for$a_{k,m}$from inequality (1)}$, namely, $\hat{a}_{1,n}=1/(n+1)$, and $\hat{a}_{k,m}=\frac{1}{m+1}(\hat{a}_{k-1,m+1}+L \hat{a}_{k-1,m+2})$.
Then it is easy to see from induction that $\hat{a}_{k,m+1}\leq \hat{a}_{k,m}$. Thus we deduce that
$a_{n,0}\leq \hat{a}_{n,0}\leq \frac{1+L}{1}\hat{a}_{n-1,1}\leq \frac{(1+L)^{2}}{2!}\hat{a}_{n-2,2}\leq\dots \leq \frac{(1+L)^{n-1}}{(n-1)!}\hat{a}_{1,n-1}=\frac{(1+L)^{n-1}}{n!}$. | 2019-08-20T20:50:33 | {
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http://vhck.cioe.pw/n+1-factorial-formula.html | # N+1 Factorial Formula
Factorial(2) is 2*Factorial(1). The factorial formula. *; import java. Since the factorial is extended by the Pi function, for every complex value z where it is defined, we can write: z! = \Pi(z)\, The values of these functions at half-integer values is therefore determined by a single one of them; one has. , transverse), then we obtain a 7-parameter model with all the main effects and interactions we saw in the 2 5 analysis, except, of course, any terms involving "Direction". n represents the total number of people you have. You can handle larger integers if you use another datatype, but even [int64] only allows values up to 2^64 - 1. The main point worth remarking upon is that this proof needs several lemmas about the natural numbers. 7 is often referred to as the generalized factorial function. Array argument accepted only for exact=False case. If exact is 0, then floating point precision is used, otherwise exact long integer is computed. A GENERAL NOTE: FACTORIAL. I know what a factorial is, but I can't remember whether there is a word and/or formula taking a number, lets say 5, and adding 5+4+3+2+1, which is 15. Implementation of factorial function in Scheme language. Prove that 12 +22 + n2 = 1 6 n(n+1)(2n+1) for all natural numbers n. If you mean to allow x to not be an integer, there are several ways to answer your question. I am looking for a formula to which I can supply a number N and have it calculate 1+2+3+4+N. exact bool, optional. N! = 1*2*3*N N is an integer and is the input to the flowchart. is it posible to have a formula for factorial via vba code? Showing 1-5 of 5 messages. We will use these definitions later in the mathematical induction process. Get a number. Solutions can be iterative or recursive. ‘N’ multiplied by ‘N-1’ multiplied by ‘N-2. The factorial function is a mathematical formula represented by an exclamation mark "!". “q” in this formula is just the probability of failure (subtract your probability of success from 1). It looks similar to the exclamatory mark '!'. Want to show x P(x). C Program to read a number and find factorial. Factorial Formula. Based on the. Example: $$22! = 1. Note that this approximation holds for large values, and not necessarily for small values. For the factorial function that computes n!, it involves calling itself to compute (n - 1)!, which in turn call itself to compute (n - 2)!, etc. Display the result. A factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. La operación de factorial apaez en munches árees de les matemátiques, particularmente en combinatoria y analís matemáticu. Recursion in computer programming is exemplified when a function is defined in terms of simpler, often smaller versions of. Proofs by Induction A proof by induction is just like an ordinary proof in which every step must be justified. defining formula! =. Using a result variable of type long (instead of int) allows for "larger" values to be calculated (for long, you can calculate up to and including n = 20). The ANOVA model for the analysis of factorial experiments is formulated as shown next. Graph of quadruple factorial for n = 0 to 5. But there will always be at least one. n!=\prod_{k=1}^n k . So factorial(k+1) = (k+1)*factorial(k). The data type and size of f is the same as that of n. For an odd number, the double factorial is the product of all odd integers less than or equal to n and greater than or equal to 1. If n is not too large, then n! can be computed directly, multiplying the integers from 1 to n, or person can look up factorials in some. I'll fill first two slots and arrange the elements in two different ways. The factorial of $n$, denoted $n!$, is defined for a positive integer $n$ as:. Factorial I need to learn how to do the problems below: Formula is: nCr = n!/(n - r) r! 1. The binomial formula and binomial coefficients. Here, fact() is a function defined to find factorial of a number. Get a number. Algorithm: Step 1: Start Step 2: Read number n Step 3: Call factorial(n) Step 4: Print factorial f Step 5: Stop factorial(n) Step 1: If n==1 then return 1 Step 2: Else f=n*factorial(n-1) Step 3: Return f. I can see how this formula would simplify - but can't see how it relates to the double factorial !! Reply With Quote November 28th, 2015 18:44 # ADS. Definition of a factorial experiment: The two-way ANOVA is probably the most popular layout in the Design of Experiments. Similar result was later proved by Hughes in [15] under the natural additional restriction Pp k=1 Ak = Pq j=1 Bj and using the standard inverse factorial series (1). Here we will write programs to find out the factorial of a number using recursion. The factorial of a positive integer n is equal to 1*2*3*n. factorial function! n! Statements. If you are looking for an approximation to the factorial function, i. Abstract For any positive integer n, the Smarandache triple factorial function [d3. En algunos problemas de matemáticas se nos presentan multiplicaciones de números naturales sucesivos tal como: 4 x 3 x 2 x 1 = 24; 3 x 2 x 1 = 6; 2 x 1 = 2. how do we reduce the number of require steps to approximate the function over the interval. The factorial of a natural number x is the product of all positive integers less than and equal to x. How to Multiply Factorials. 2 The following table shows that γnhas great superiority over αnand βn, where αn, βnand γnare as defined by (1. Factorial Calculator Formula: n! = n * (n-1) * (n-2) * 3 * 2 * 1. The factorial allows only positive integers. Stirling's formula is a good reference for the approximation. If, on the other hand, we do an analysis of the 2 4 factorial with "Direction" kept at +1 (i. PHP, Programming factorial program in php, find n! php program, how to write a program to find n!. 7) Equation 1. Generalized Factorials and Taylor Expansions Recall the de nition of a factorial n! = n(n 1) 21 = n a n 1). Calculating factorials The numeric value of n! can be calculated by repeated multiplication if n is not too large. Stirling's Formula. \begingroup My mistake! It is still early here. A good approximation to n! for large values of n is given by Stirling's Formula, which probably ought to be named for De Moivre. The formula to calculate the factorial is: F = n!. or in partially expanded form as. "q" in this formula is just the probability of failure (subtract your probability of success from 1). You will learn to calculate the factorial of a number using for loop in this example. Thus it returns n * factorial(n-1). For n=0, 0! = 1. Someone might argue you can use recursive function calls - it just. The representation of n factorial is n!. Factorial(x) x! Factorial of x. Note that this approximation holds for large values, and not necessarily for small values. find out the highest power of al prime factors till 6 (i. Therefore, the computer has to keep track of the multiplications to be performed later on. Murray Abstract. or in partially expanded form as. It does seem odd, but we will make it work. Figure 2 is collinear on B-axis from n = 0, 1, 2 and 3 and it makes a shaped curve along the C-axis at n = 3. We may see the sequence in the leaf or branch arrangement, the number of petals of a flower, or the pattern of the chambers in a nautilus shell. *; import java. Subfactorial !n is the number of derangements of n object, ie the number of permutations of n objects in order that no object stands in its original position. Factorial of any number n is denoted as n! Factorial of a number is the product of all the positive integers from 1 up to n, (including n) (n >= 1) For example: A Useful Tip: Use the tip to simplify: Let us now define n! (Factorial of a number n). The easiest definition of n-factorial, denoted n!, is a product of all positive natural numbers smaller or equal to n. Factorial Formula. After proving Stirling’s formula we will give some applications and then discuss a little bit of its history. What about 20! or 100!? Most calculators including the TI 's series will only calculate factorials up to 69!. 1*2*3 = 3*2*1 = 3*(3-1)*(3*2) = 6 Remember that Matlab has already created a function to find the factorial of the number easily without writing any programs. If n is a positive integer, then the function Gamma (named after the Greek letter "Γ" by the mathematician Legendre) of n is: Γ(n) = (n − 1)! We can easily "shift" this by 1 and obtain an expression for n! as follows: Γ(n + 1) = n!. In how many arrangements can you produce a list of those four objects? You can get the answer by calculating the factorial of the number of objects. All about factorial notation n! The Dictionary of Large Numbers "Factorial Factoids" od Paula Niquettea; Faktorijelski kalkulatori i algoritmi. Now start solving the equation through FOIL method where multiplying two equations will result in x^2 + 2x + 1. Consider the example of the expression (x + 1) 3 that can be expanded as x + 1)(x + 1)(x + 1). It calculates factorial values for n number of positive integers with factorial formula. The best example is the "classical electron radius" r e which is 2. 1 The product of a series of factors in an arithmetical progression. , qualitative vs. Contains static definition for matrix math methods. One could multiply 5 factors of (x + 2) together by hand, use a device called Pascal's Triangle (described in this module), or use the Binomial Theorem (stated in the next module). Using a result variable of type long (instead of int) allows for "larger" values to be calculated (for long, you can calculate up to and including n = 20). The answer is = 224 + 44 + 8 + 1 =277. After having gone through the stuff given above, we hope that the students would have understood "Solving Factorial Problems ". Please let us know if you have any suggestions on how to make Factorial Calculator better. Factorial of a number (n) is calculated using the below show formula. What is a factorial. 2) was rediscovered by Y. The factorial of a number be found using recursion also. This fact can be used to prove Euclid's theorem that the number of primes is infinite. Program 1: Program will prompt user for the input number. How to Multiply Factorials. A Waldorf salad is a mix of among other things celeriac, walnuts and lettuce. It is used to write continued products in simplified form. The factorial of a nonnegative integer n is written n! follows: n! n! = 1 (for n = 0 or n = 1). All scientific calculators with the factorial function will calculate the factorial of any positive integer up to 69 (253 for calculators with an upper limit of 10^499). fact(n) = n * n-1 * n-2 *. n n_factorial ----- ----- 1 1 2 2 3 6 4 24 5 120 6 720 7 5040 8 40320 9 362880 10 3628800 To summarize what (I believe) everyone is saying: Multiplying a list of numbers (including consecutive integers, generated by CONNECT BY) can be done by the SUM function, together with EXP and LN. I am looking for a formula to which I can supply a number N and have it calculate 1+2+3+4+N. Find the number of different two-card initial hands by evaluating 52C2. N! = 1*2*3*N N is an integer and is the input to the flowchart. Writing the Terms of a Sequence Defined by a Recursive Formula Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. Excel FACT function which stands for FACTorial is used for finding out the factorial of the specified number. The base case can be taken as the factorial of the number 0 or 1, both of which are 1. GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together. For example, the factorial of 6 (denoted as 6!) is 1*2*3*4*5*6 = 720. The R Programming language introduced a new technique called Recursion for simple and elegant coding. So if one equates the factorial function on all positive reals with the gamma function, then one can say that (1/2)! = sqrt(pi)/2 even though the factorial function was originally only defined for the natural numbers. This is denoted by Factorial (!) exclamation symbol. where i = 1, , a, j = 1, , b, and k = 1, , n. Factorial of a number (n) is calculated using the below show formula. The N Choose K calculator calculates the choose, or binomial coefficient, function. 2) can be obtained similarly. The formula uses two variables, n and k, and n is the total number of the objects to select from and k is the number of those objects to be chosen in each selection. Another approach to defining a continuous function whose value equals the factorial of n for integer arguments n is to consider the sequence of derivatives of 1/x, which gives. Math 140A Elementary Analysis Homework Questions 1 Hand in the starred questions at discussion on Wednesday 9th October 1. Find the value of. Python Functions: Exercise-5 with Solution. Inside the inner loop use formula term = fact(n) / (fact(k) * fact(n-k)); to print current term of pascal triangle. Moreover, the p-value of 1. \( n! \approx \sqrt{2\pi n} (\frac{n}{e})^n$$ The Factorial can also be generalized to the real and complex numbers using the Gamma Function. Note for n = 22 or larger, the results is written in scientific notation using "E" instead of 10. This is written as n! and pronounced 'n factorial'. For n>0, n! = 1×2×3×4××n. 811 × 10 -15. After having gone through the stuff given above, we hope that the students would have understood "Solving Factorial Problems ". This is the product of all the numbers from n down to 1, or from 1 up to n. The binomial formula and binomial coefficients. Calculating factorials The numeric value of n! can be calculated by repeated multiplication if n is not too large. typically finds the values for Γ(n) are tabulated only in the range 1 0. Show the statement is true for n = 1, that is, Show that a 1 = S 1. The explicit formula is S = n*(n+1) / 2. Once user provide the input, the program will calculate the factorial for the provided input number. La operación de factorial apaez en munches árees de les matemátiques, particularmente en combinatoria y analís matemáticu. $\begingroup$ @HagenvonEitzen wow that was quick, In that case: gamma(n)=(n-1)! ergo gamma(1) = (1-1)!=0!=1 $\endgroup$ – Alexis Sanchez Jul 3 '17 at 9:44 $\begingroup$ @G. So here is my code uptill now:. And then, it will calculate the NCR Factorial of. Classes are nothing but a structure in ‘C’ Language which supports functions. We may see the sequence in the leaf or branch arrangement, the number of petals of a flower, or the pattern of the chambers in a nautilus shell. What is Zero-Factorial? Simple answer: 0! (read "Zero Factorial") is defined to equal 1. The function is defined by nCk=n!/(k!(n-k)!). Tables of mathematical formulas and constant including series and trigonometric identities. Description: Variable num is used for the storing last element of the series. (“ n factorial”) means to multiply together all the integers from n down to 1. C Program to Find Sum of 5 Subjects and Percentage. RE: HP 50g Double factorial (05-01-2019 06:17 PM) Albert Chan Wrote: I am not familiar with RPL code, but is there a loop construct where you test the condition first ? That way, the 0 special case need not be added. For n negative, the factorial is usually undefined, but see also Euler Gamma Function for further info (the 'factorial of a negative integer' is always undefined, however 'factorials' of nearly all real numbers are defined using the gamma function). Factorial Formula. typically finds the values for Γ(n) are tabulated only in the range 1 0. For n=0, 0! = 1. I'll fill first two slots and arrange the elements in two different ways. In the algebra literature, see for example [26], Q-factorial is referred to as almost factorial: Definition 4. You can use our Factorial Calculator to calculate the factorial of any real number between 0 and 5,000. 01 n!µ2pnn+ 1 2. The factorial operation, n!, is defined as n! = n(n – 1)(n – 2)(n – 3) · · · 4 · 3 · 2 · 1. What is Factorial? Factorial: It is a mathematical function denoted as "n!" and symbol of factorial is "!". , introduced by Leo August Pochhammer, may represent either the rising factorial or the falling factorial as defined below. Factorial Formula. Given an integer produce a list of the digits in an arbitrary base and perform the reverse process. Meza Pérez Palma 2. n n_factorial ----- ----- 1 1 2 2 3 6 4 24 5 120 6 720 7 5040 8 40320 9 362880 10 3628800 To summarize what (I believe) everyone is saying: Multiplying a list of numbers (including consecutive integers, generated by CONNECT BY) can be done by the SUM function, together with EXP and LN. × (n – 1) × nFor example,5!…. Note that this approximation holds for large values, and not necessarily for small values. Hello friends, Today let us begin with a new Java program. Added after 14 minutes: Noooo I'am wrong. n! = n(n-1)! , “this is the formula for calculating factorial”. 440, March 9, 2009 Stirling's formula The factorial function n! is important in evaluating binomial, hypergeometric, and other probabilities. Nas matemáticas, a factorial dun número natural n é o produto de todos os enteiros positivos menores ou iguais a n. Stirling's formula is a good reference for the approximation. Explanation This is the most typical kind of induction. n! = 1 x 2 x 3 x 4 x 5 x x (n-1) x n. The function calls itself recursively on a smaller version of the input (n - 1) and multiplies the result of the recursive call by n, until reaching the base case, analogously to the mathematical definition of factorial. Solutions can be iterative or recursive. Does this function (that's probably not the right word, I'm not a mathematition, have a word, like "factorial"?. This is denoted by Factorial (!) exclamation symbol. Algorithm: Step 1: Start Step 2: Read number n Step 3: Call factorial(n) Step 4: Print factorial f Step 5: Stop factorial(n) Step 1: If n==1 then return 1 Step 2: Else f=n*factorial(n-1) Step 3: Return f. Enter n and k below, and press calculate. nPr(n, r). 4 (2007) 389-392]. In case you have a factorial followed by an ordinary symbol (not a relation or operation symbol), it's good practice to add a thin space after it:. n! = n x (n - 1) x (n - 2) x (n - 3) 3 x 2 x 1. If n equals 0, the factorial is defined to be 1; If n is less than 0, the factorial is undefined; If n is too large, the result exceeds Long. 032777099 5. Weissman [9] and caused a lively debate in the American Journal of Physics in 1983, see [6]. The termination condition of course comes from the fact that 1! = 1, that is for an argument of 1, the function should return the value 1. It is easy to calculate and multiply two factorials using a scientific calculator's x! function. Classes are nothing but a structure in ‘C’ Language which supports functions. 4 FACTORIAL DESIGNS 4. Definition 1. If you are looking for an approximation to the factorial function, i. Factoriales y sumatorias 1. Any ideas? Ideally I'd like it to look like this: 1,2,3 1,2,4 1,2,5. The factorial of a positive integer n is equal to 1*2*3*n. Aim: Write a C program to find the factorial of a given number using recursion. The factorial of a number is the product of all the integers from 1 to that number. Problems Introductory. f](n)!!! According to [1], for any positive integer n, the Smarandache triple factorial function [d3. Note for n = 22 or larger, the results is written in scientific notation using "E" instead of 10. Hint: Multiply F(1, ν) by (2 (k-1) n) 2 / (2 (k-1) n) 2 and simplify. We're now ready to begin. This type of program, characterized by a chain of operations, is called recursion. 4! = 1×2×3×4 = 24. C Program for Sum of Squares of Numbers from 1 to n. Specifically we will demonstrate how to set up the data file, to run the Factorial ANOVA using the General Linear Model commands, to preform LSD post hoc tests, and to. Python procedure for Factorial (using while loop) #----- # Factorial function: # Input: n # Output: n! # def factorial (n): if (n = 1): return 1 i = 1 product = 1. A good approximation to n! for large values of n is given by Stirling's Formula, which probably ought to be named for De Moivre. Analysis of recursive routines is not as easy: consider factorial ; fac(n) is if n = 1 then return 1 else return fac(n-1) * 1 How many times is fac called for fac(n)? To find an answer, Use a recurrence. For an odd integer p , the double factorial is the. Copyright © 2000–2017, Robert Sedgewick and Kevin Wayne. The result can be approximated rapidly using the gamma-formula above (default). 211), things look more like a duality. We introduce here the new approximation formula n! ≈ ω n ζ n = 2 π (n 2 + 3 n + 13 6 e 2) n 2 + 3 4 1 n + 1 ≕ κ n which is more accurate than all formulas mentioned in this paper, as we can see below. For an even number, n, it's the product of all even integers less than or equal to n and greater than or equal to 2. Place the method in a class that has a main that tests the method. *; import java. exercitii rezolvate inecuatii cu factorial, (n-1)! : (n-3)! < 72 n=?, inecuatii cu n factorial, n factorial inecuatii, exercitii rezolvate clasa 10, inecuatii cu factorial rezolvate Inecuatii cu factorial (n!) exercitiu rezolvat 1 | formule online probleme si exercitii rezolvate. Just in case someone doesn't know, the factorial of a number is the sum of all numbers starting at 1 and ending at the number itself. Factorial I need to learn how to do the problems below: Formula is: nCr = n!/(n - r) r! 1. N ! There are five algorithms which everyone who wants to compute the factorial n! = 1. One can quickly determine the primes as well as the right power for each prime using a sieve approach. A plot the real and imaginary parts of the subfactorial generalized to any real argument is illustrated above, with the usual integer-valued subfactorial corresponding to nonnegative integer. Now, if k=23, this probability is only P (B)=0. All scientific calculators with the factorial function will calculate the factorial of any positive integer up to 69 (253 for calculators with an upper limit of 10^499). Support for trapping negative n errors is optional. Factorial means multiplying any number by every real positive whole number less than itself. In this article, we will show you, How to Write C Program to Find NCR Factorial of a Number with an example. Formula: P(n, r) = ()!! n r n , r < n where n is the number of distinct objects and r is the number of distinct. Analysis of recursive routines is not as easy: consider factorial ; fac(n) is if n = 1 then return 1 else return fac(n-1) * 1 How many times is fac called for fac(n)? To find an answer, Use a recurrence. A função hiperfactorial é similar à factorial, mas produz números maiores. After you click "Calculate Factorial" the result will be displayed in the output box. i'm trying to build a complex formula using factorial but i havent. Added after 14 minutes: Noooo I'am wrong. 3 Quadratic Formula Finally, the quadratic formula: if a, b and c are real numbers, then the quadratic polynomial equation ax2 + bx+ c = 0 (3. Amb el llenguatge de programació Python es pot calcular el factorial d'enters arbitràriament grans, limitat només per la memòria disponible. The factorial of a nonnegative integer n is written n! follows: n! n! = 1 (for n = 0 or n = 1). Sharp inequalities for factorial n 101 for which the proof follows. Given an integer produce a list of the digits in an arbitrary base and perform the reverse process. 440, March 9, 2009 Stirling's formula The factorial function n! is important in evaluating binomial, hypergeometric, and other probabilities. where sum(n) = 1+ 2+ + n. If n<0, the return value is 0. Calculating factorials The numeric value of n! can be calculated by repeated multiplication if n is not too large. Let n be a positive integer. In general, the magnitude of the nth derivative of 1/x is n!/x n-1. This function overflows as soon as n>170. Combinatorics Digit Lists. The factorials and binomials , , , , and satisfy the following recurrence identities:. The FreeVBCode site provides free Visual Basic code, examples, snippets, and articles on a variety of other topics as well. factorial¶ scipy. One can quickly determine the primes as well as the right power for each prime using a sieve approach. When we encounter n! (known as 'n factorial') we say that a factorial is the product of all the whole numbers between 1 and n, where n must always be positive. Meza Pérez Palma 2. (read "n factorial"), you multiply all the positive integers between one and n. The n! gives the number of ways in which n objects can be permuted. n factorial is a mathematical operation that can be defined using a recursive formula. Factorials, denoted by a ! sign, are products of a whole number and all of the whole numbers below it. Online Math Dictionary: F - Cool Math has free online cool math lessons, cool math games and fun math activities. Number Theory: n | (n-1)! For example for 60, 3 * 20 or 4 * 15 or other choices. Thus, a recursive function could hold much more memory than a traditional function. It provides access to the mathematical functions defined by the C standard. Stirlings Factorial formula. Factorial Notation In mathematics, permutation and combination formulas and problems and expressions inside sigma notations are places where we come across factorial notations frequently. 34 Figure 2. The function above is the fastest way to calculate a factorial exactly. Writing the Terms of a Sequence Defined by a Recursive Formula Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. Binomial theorem (x+y) n= Xn k=0 n k! factorial powers and ordinary powers Stirling numbers of xn. For example, the Binomial Distribution formula should give a probability of p for the probability of 1 success given 1 trial if the trial has probability p. If n is 5, then 5! is 5 x 4 x 3 x 2 x 1 = 120. " It counts the combinations r elements out of a population of n, without repeating elements. El factorial de un número se representa mediante el símbolo ! y se define de la siguiente forma: El factorial de 0 es 1 0! = 1. The largest factorial that most calculators can handle is 69!, because 70! > 10 100. A factorial is represented by the sign (!). Excel 2010 contains many advance mathematical functions which have been extensively used for different calculations, from a list factorial finder is an important one. Support for trapping negative n errors is optional. Two expressions are multiplied already but (x+1) is still hanging on the head. Combinatorics Digit Lists. For example 5!= 5*4*3*2*1=120. Added after 14 minutes: Noooo I'am wrong. Writing the Terms of a Sequence Defined by a Recursive Formula Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. However it employs a neat trick which allows you to prove a statement about an arbitrary number n by first proving it is true when n is 1 and then assuming it is true for n=k and showing it is true for n=k+1. If n is not too large, then n! can be computed directly, multiplying the integers from 1 to n, or person can look up factorials in some. Charalambides and N. " It counts the combinations r elements out of a population of n, without repeating elements. Problems Introductory. For a 2-factor, 2-level, full factorial model, the formulas for coefficients for the factors and interactions are:. You can use our Factorial Calculator to calculate the factorial of any real number between 0 and 5,000. Write an application that displays the factorial for every integer value from 1 to 10. 1240007277776077× 10^{21} \). 2)!! is an expression I have seen, but while I understand that the first ! would mean 5*4*3*2*1, I have no idea what the second ! does to it. Factorial (n!) The factorial of n is denoted by n! and calculated by the product of integer numbers from 1 to n. Many of serious programmers who are just beginning to code in Java, come across a problem that is, how to write a program that calculate factorial of a number with and without recursion. 1 Pascal's Triangle An algebra problem such as expanding (x + 2) 5 to a polynomial of degree 5 can be a daunting one. It then shows how the factorial formula is recursive in nature. This module is always available. Factorial of a number (n) is calculated using the below show formula. mathematics and statistics. Factorial n! of a positive integer n is defined as: The special case 0! is defined to have value 0! = 1. Just to add a footnote, one might ask why the gamma function and not some other function that agrees with factorial (when shifted by one). The ANOVA for 2x2 Independent Groups Factorial Design Please Note : In the analyses above I have tried to avoid using the terms "Independent Variable" and "Dependent Variable" (IV and DV) in order to emphasize that statistical analyses are chosen based on the type of variables involved (i. FACTORIALES Para todo número natural n, se llama n factorial o factorial de n al producto de todos los naturales entre 1 y n: n! 1 2 3 n 1 n Empleando la notación de productos n k n k 1 !. Here we discuss factorial for numbers 1 to 10, examples of factorial in C by using the various method, formula for “n factor” with codes and outputs. Example: \( 22! = 1. The function of factorial is a product of positive integers towards to its initial value. The factorial of n is commonly written in math notation using the exclamation point character as n!. Using Mathematica and typing : [code]Sum[k!, {k, 1, n}] [/code]yields the following result : [math]\displaystyle. Inside the inner loop use formula term = fact(n) / (fact(k) * fact(n-k)); to print current term of pascal triangle. Definition: A permutation is an arrangement of a specific set where the order in which the objects are arranged is important. If fear that the answer that you expect does not exist. In mathematics, the factorial of a number (that cannot be negative and must be an integer) n, denoted by n!, is the product of all positive integers less than or equal to n. Trying to make a more bulletproof solution for n factorial. n! = n×(n-1)! Example:. FORMULA LUI LEIBNIZ. This type of program, characterized by a chain of operations, is called recursion. factorial program in php, php factorial demo, php factorial example agurchand Technologist, software engineer, blogger with experience in Web development and the Media. SpeQ can also calculate the factorial value for positive, non-integer values. Factorial of a number $n$ is calculated with a multiplication: it is the product of the positive integers numbers (not null) less or equal to $n$. Factorial Notation, Formula, and Basic Examples When I first encountered an algebra problem with exclamation mark “!“, I thought it was a trick question. Writing the Terms of a Sequence Defined by a Recursive Formula Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. (2) is the recursion relation of the factorial, and thus we have Γ(n+1) ∝ n! ; (4) because in addition Γ(1) = 1 (easly derived from the definition), we have the identification Γ(n+1) = n! , (5) and in this sense the Gamma function is a complex extension of the factorial. You can only upload files of type PNG, JPG, or JPEG. Outline An extension of the factorial to all positive real numbers is the gamma function where Using integration by parts, for integer n. | 2020-01-29T22:19:21 | {
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http://qamo.nomen.pw/2d-poisson-equation-with-neumann-boundary-conditions.html | # 2d Poisson Equation With Neumann Boundary Conditions
Poisson Equation in 2D. Here, denotes the part of the boundary where we prescribe Dirichlet boundary conditions, and denotes the part of the boundary where we prescribe Neumann boundary conditions. is the inverse Laplacian. Poisson's Equation in 2D We will now examine the general heat conduction equation, T t = κ∆T + q ρc. Uniqueness of solutions to the Laplace and Poisson equations 1. I understand how to implement a discrete 2D poisson solution with Dirchlet boundary conditions. The boundary condition (2) is a so called Robin boundary condition, which may describe both Dirichlet or homogeneous Neumann boundary conditions depending on the choice of °. Laplace Equation is a second order partial differential equation (PDE) that appears in many areas of science an engineering, such as electricity, fluid flow, and steady heat conduction. Class 6: Time stepping in PDE’s. Finite difference method Other types of boundary conditions Dirichlet-Neumann BC u(0) = ∂u ∂x(1) 2D Poisson equation Boundary value problem. a polygonal domain with boundary @› and outward pointing unit normal n. For a domain with boundary , we write the boundary value problem (BVP):. I need help from any one urgently, I need C code for Solving Poisson Equation has known source with Neumann condition by using FDM (finite difference method) in 2D problem. Lecture 04 Part 3: Matrix Form of 2D Poisson's Equation, 2016 Numerical Methods for PDE - Duration: 14:57. Dirichletboundarycondition. As we will see this is exactly the equation we would need to solve if we were looking to find the equilibrium solution (i. Uniqueness of solutions of the Laplace and Poisson equations. The first number in refers to the problem number in the UA Custom edition, the second number in refers to the problem number in the 8th edition. ) The Laplacian is an elliptic operator so we should specify Dirichlet or Neumann conditions on a closed boundary S. Theoretical Background.
Eulerian Vortex Motion in Two and Three Dimensions Igor Yanovsky June 2006 1 Objective An Eulerian approach is used to solve the motion of an incompressible °uid, in two and. Uniqueness properties via the Green formula. Another way of viewing the Robin boundary conditions is that it typies physical situations where the boundary “absorbs” some, but not all, of the energy, heat, mass…, being transmitted through it. After this is accomplished, the same diagonalization procedure as above is carried out. However, in many practically interesting cases, the essential boundary condition can be satis ed merely ap-proximately either owing to complicated, e. for solving Helmholtz equation in one-dimensional and two-dimensional domain with Neumann boundary conditions. Dear colleagues, I'm solving Poisson's equation with Neumann boundary conditions in rectangular area as you can see at the pic 1. Uniqueness and continuous dependence for the Dirichlet problem via the maximum principle. The Laplacian is defined as u= X i=1 n u x ix i. 30, 2012 • Many examples here are taken from the textbook. In solving partial differential equations, such as the Laplace Equation or Poisson Equation where we seek to find the value of potential throughout some volume of space (using methods like SIMION Refine), it can be necessary to impose constraints on the unknown variable () at the boundary surface of that space in order to obtain a unique solution (see First. To solve this problem in the PDE Modeler app, follow these steps:. Dirichlet and Neumann boundary conditions for the pressure poisson equation of incompressible flow; which are solved iteratively with the pressure Poisson equation. My question was whether I should replace the neumann boundary conditions into the matrix system for the poisson equation, Au = b, without changing the size of the matrix, or use the boundary condition as a Lagrange Multiplier?. 2) or Neumann boundary conditions =h-:h on F. Here is an example of the Laplace in cylindrical coordinates (with cylindrical symmetry).
6 Heat Conduction in Bars: Varying the Boundary Conditions 43 3. periodic boundary conditions in the direction, Dirichlet condition on the upper boundary, and Neumann condition on the lower boundary. Poisson's Equation in 2D We will now examine the general heat conduction equation, T t = κ∆T + q ρc. 2: The Boltzmann distribution. A DirichletBC takes three arguments: the function space the boundary condition applies to, the value of the boundary condition, and the part of the boundary on which the condition applies. Neumann conditions | the normal derivative, @u=@n= n ruis. Not yet done. Poisson’s Equation with Complex 2-D Geometry. The basic idea is to solve the original. Neumann boundary condition for 2D Poisson's equation Finite Difference for 2D Poisson's equation Heat Conduction Equation and Different Types of Boundary Conditions. Homogenous neumann boundary conditions have been used. The techniques set forth in this section are used to solve step 2 in Section and instrumental to approach inverse boundary value problems for the Poisson equation Δ u = μ, where μ is some (unknown) measure. Pardoux Abstract In this work we extend Brosamler's formula (see [2]) and give a probabilistic solution of a non degenerate Poisson type equation with Neumann boundary condition in a bounded domain of the Euclidean space. 6 Boundary Conditions and Singular Systems 177 5. Boundary Conditions There are three types of boundary conditions that are specified during the discretization process of the Poisson equation: Dirichlet (this is a boundary condition on the potential) Neumann (this is a boundary condition on the derivative of the potential, i. the 3 boundary conditions (Dirichlet, Neumann or Robin. Electrostatic forces are among the most common interactions in nature and omnipresent at the nanoscale. It runs on Windows, Linux and Mac OS.
Demonstration of using 1D poisson solver with a timestepping routine to solve heat equation. Section 2: Electrostatics. The difference matrix is solved directly using backslash and sparse % matlab feature. We take n^ to be the outward pointing normal on the domain boundary @ = @ D S @ N. • Boundary conditions will be treated in more detail in this lecture. For a domain with boundary , we write the boundary value problem (BVP):. We also describe the singular part of weak and very weak solutions. Laplace's Equation 3 Idea for solution - divide and conquer We want to use separation of variables so we need homogeneous boundary conditions. Simple demonstration of solving the Poisson equation in 2D using pyMOR’s builtin discretizations. Numerically Solving a Poisson Equation with Neumann Boundary Conditions 2D Poisson equation with Dirichlet and Neumann boundary conditions Poisson equation. PPE reformulations of the Navier-Stokes equations, and the boundary conditions that they produce for the Poisson equation that the pressure satis es. Example 1: 1D Heat Equation with Mixed Boundary Conditions Example 2: 2D Drumhead Eigenmodes 6. Uniqueness of solutions to the Laplace and Poisson equations 1. r πε ′ Φ= ∫ − ′ r r r r, (2. Pardoux Abstract In this work we extend Brosamler’s formula (see [2]) and give a probabilistic solution of a non degenerate Poisson type equation with Neumann boundary condition in a bounded domain of the Euclidean space. The boundary condition (2) is a so called Robin boundary condition, which may describe both Dirichlet or homogeneous Neumann boundary conditions depending on the choice of °. Use FD quotients to write a system of di erence equations to solve two-point BVP Higher order accurate schemes Systems of rst order BVPs Use what we learned from 1D and extend to Poisson's equation in 2D & 3D Learn how to handle di erent boundary conditions Finite Di erences October 2, 2013 2 / 52. It is known that the electric field generated by a set of stationary charges can be written as the gradient of a scalar potential, so that E = -∇φ. I'm having problems solving a Poisson equation using MKL's s_Helmholtz_2D, Win32 binaries, 10. The following Matlab project contains the source code and Matlab examples used for 2d poisson equation.
braic equations. $\begingroup$ So generally the Poisson equation is solved with at least one Dirichlet boundary condition, so that a unique solution can be found? I guess it makes sense that the Neumann boundary conditions only make sense when source and sinks are included, otherwise there are an infinite number of solutions. Journal of Electromagnetic Analysis and Applications Vol. The DuFort-Frankel scheme is the only simple know explicit scheme with 2nd order accuracy in space and time that has this property. Poisson’s Equation with Complex 2-D Geometry. Heat flow with sources and nonhomogeneous boundary conditions We consider first the heat equation without sources and constant nonhomogeneous boundary conditions. A mixed nite element method 123 3. My question is: What would the boundary conditions for this equation be? Obviously one is that it decays to zero at infinity, but. −∆u= f Poisson equation 2. In this example, we create an 1D linear structure embedded in a 2D space and we solve a non regularized Poisson equation on this structure. Some of the answers seem unsatisfactory though. diff(x, 2) + u. And theta is periodic. Lecture 04 Part 3: Matrix Form of 2D Poisson's Equation, 2016 Numerical Methods for PDE - Duration: 14:57. Math 201 Lecture 33: Heat Equations with Nonhomogeneous Boundary Conditions Mar. 24 How to solve Poisson PDE in 2D with Neumann boundary conditions using Finite Elements. In this study, the numerical technique based on two-dimensional block pulse functions (2D-BPFs) has been developed to approximate the solution of fractional Poisson type equations with Dirichlet and Neumann boundary conditions. Next we will solve Laplaces equation with nonzero dirichlet boundary conditions in 2D using the Finite Element Method. In this series of two papers we describe the regular-ity of weak and very weak solutions of the Poisson equation on polygonal domains. works [5][7][12][14]. In reality, you can't use the same equation along the boundaries as you do in the interior. Cartesian, domains for solving the governing equations. A mesh stores boundary elements, which know the bc name given in the geometry.
Bench erif-Madani and E. y, and the right hand side of the pressure Poisson equation must be adjusted according to the method laid out in [2]. ru b Institute of Mathematics, University of Zurich CH-8057 Zurich, Switzerland. Since there is no time dependence in the Laplace's equation or Poisson's equation, there is no initial conditions to be satisfied by their solutions. Specifically, for a domain with boundary , we consider the the boundary value problem (BVP):. boundary conditions of the Dirichlet type (u = 0) or Neumann type (∂u/∂n = 0) along a plane(s) can be determined by the method of images. The 2D Poisson equation is solved in an iterative manner (number of iterations is to be specified) on a square 2x2 domain using the standard 5-point stencil. The script for setting the source terms is referenced in the project file as follows:. It is possible to prescribe. In a typical model, the boundary conditions on five surfaces of this domain are assumed to be adiabatic [7][14] ─ this is in the form of the Neumann boundary condition, which specifies temperature gradient. In section (7. The Laplace equation in the whole space. The System of Equations for Mixed BVP with Three Dirichlet Boundary Conditions and One Neumann Boundary Condition Nur Syaza Mohd Yusop and Nurul Akmal Mohamed Mathematics Department Faculty of Science and Mathematics Universiti Pendidikan Sultan Idris 35900 Tanjong Malim, Perak, Malaysia. 1 PPE formulations for the Navier Stokes equations.
1 The 1D Poisson Equation. ) Typically, for clarity, each set of functions will be specified in a separate M-file. We take n^ to be the outward pointing normal on the domain boundary @ = @ D S @ N. 1: Plot of the solution obtained with automatic mesh adaptation Since many functions in the driver code are identical to that in the non-adaptive version, discussed in the previous example, we only list those functions that differ. How can I solve the 2D Laplace equation with Neumann boundary conditions? [closed] x\pm i y$satisfy the equation. impose boundary condition for solving the Poisson equation. In this article, we consider a standard finite volume method for solving the Poisson equation with Neumann boundary condition in general smooth domains, and introduce a new and efficient MILU preconditioning for the method in two dimensional general smooth domains. The Poisson equation can also be used for various other problems, including magnetic and current density ones, the heat equation, etc. Neumann Boundary condition, Poisson's equation. Journal of Electromagnetic Analysis and Applications Vol. Solution of 1D Poisson Equation with Neumann-Dirichlet and Dirichlet-Neumann Boundary Conditions, Using the Finite Difference Method. Therefore it has been in part used to solve the Navier-Stokes equations. txt) or read online for free. consideration that the differential equation is satisfied at points arbitrarily close to the boundary. Set the boundary conditions. Abstract: The Marchaud fractional derivative can be obtained as a Dirichlet-to–Neumann map via an extension problem to the upper half space. We examine the properties of the method, by considering a one-dimensional Poisson equation with different Neumann boundary conditions. While it is clear that the governing equation for pressure is a Poisson equation derived from the momentum equation by requiring incompressibility, it is less clear what boundary conditions (BC) the pressure should be subject to. Numerically Solving a Poisson Equation with Neumann Boundary Conditions 2D Poisson equation with Dirichlet and Neumann boundary conditions Poisson equation. 24 How to solve Poisson PDE in 2D with Neumann boundary conditions using Finite Elements. LAPLACE'S EQUATION AND POISSON'S EQUATION In this section, we state and prove the mean value property of harmonic functions, and use it to prove the maximum principle, leading to a uniqueness result for boundary value problems for Poisson's equation. or its inhomogeneous version Poisson's equation ∇2u(x) = ρ(x). solvers for Poisson equation in 2D polar and spherical domains. the 3 boundary conditions (Dirichlet, Neumann or Robin. The book NUMERICAL RECIPIES IN C, 2ND EDITION (by PRESS, TEUKOLSKY, VETTERLING & FLANNERY) presents a recipe for solving a discretization of 2D Poisson equation numerically by Fourier transform ("rapid solver"). 1 Source and sink of fluid Line source/sink Consider an axisymmetric potential φ≡ φ(r). 1) with the value of the descriminant < 0 is the most general linear form of this type of PDE. Scanning probe methods represent a formidable approach to study these inter. Boundary conditions. for solving Helmholtz equation in one-dimensional and two-dimensional domain with Neumann boundary conditions. (If the equation is really Eq(u. Homogenous neumann boundary conditions have been used. boundary conditions of the Dirichlet type (u = 0) or Neumann type (∂u/∂n = 0) along a plane(s) can be determined by the method of images. Petrovskii, A. Finite Element Solver for Poisson Equation on 2D Mesh December 13, 2012 1 Numerical Methodology We applied a nite element methods as an deterministic numerical solver for given ECG forward modeling problem. In recent work. The domain is discretized in space and for each time step the solution at time is found by solving for from. Poisson equation is an elliptic equation and hence it strongly depends on the boundary condition. Poisson equation with pure Neumann boundary conditions¶ This demo is implemented in a single Python file, demo_neumann-poisson. Solve Laplace's or Poisson's equation in a given domain D with a condition on boundary bdy D: Du = f in D with u = h or u n = h or u n + a u = h on bdy D. Compared with the method of fundamental solution, the BKM uses the nonsingular general solution instead of the singular. Journal of Electromagnetic Analysis and Applications Vol. This means that Bis Lf a-bounded with relative bound a= 0:Applying the. There are three types of boundary conditions for well-posed BVPs, 1. Poisson's's Equation Diriclet problem Heat Equation: 4. The Laplace equation together with Neumann BC are called the Neumann BVP/ Neumann problem which is written as ∇2u= 0 in Ω; ∂u ∂n (x,y) = g(x,y) for (x,y) ∈ ∂Ω. Convergence rates (2D Poisson equation) In this notebook we numerically verify the theoretical converge rates of the finite element discretization of an elliptic problem.$\begingroup\$ So generally the Poisson equation is solved with at least one Dirichlet boundary condition, so that a unique solution can be found? I guess it makes sense that the Neumann boundary conditions only make sense when source and sinks are included, otherwise there are an infinite number of solutions. Which is not surprising, because such PDEs do not admit explicit symbolic solutions, with a few (mostly uninteresting) exceptions. Reimera), Alexei F. Inhomogeneous Dirichlet boundary conditions 125 4. as the boundary surface (S) volume approaches zero, thereby converting the surface integral into a divergence operator. The Dirichlet problem is to find a function that is harmonic in D such that takes on prescribed values at points on the boundary. A Neumann boundary condition prescribes the normal derivative value on the boundary. 3 1) 1-d homogeneous equation and boundary conditions (Neumann-Neumann) 4. Homogenous neumann boundary conditions have been used. Pardoux Abstract In this work we extend Brosamler's formula (see [2]) and give a probabilistic solution of a non degenerate Poisson type equation with Neumann boundary condition in a bounded domain of the Euclidean space. We want to use Poisson's equation for gravity for that (Laplace(U) = -4*pi*density or something like that).
Hi, I am using the d_Helmholtz_2D routine to solve the poisson equation with full Neumann boundary conditions (NNNN). The 2D Poisson equation is given by with boundary conditions There is no initial condition, because the equation does not depend on time, hence it becomes boundary value problem. Our new MILU preconditioning achieved the order O (h − 1) in all our empirical. In both cases, only the row of the A-matrix corresponding to the boundary condition is modi ed! David J. It is possible to prescribe. the Poisson equation using a least squares approximation. Rosser's procedure can also be modified to handle discontinuous boundary data. And theta is periodic. BOUNDARY CONDITIONS We shall discuss how to deal with boundary conditions in finite difference methods. 2 FD for the Poisson Equation with Dirchlet BCs 2. Numerical solution of the 2D Poisson equation on an irregular domain with Robin boundary conditions We describe a 2D finite difference algorithm for inverting the Poisson equation on an irregularly shaped domain, with mixed boundary conditions, with the domain embedded in a rectangular Cartesian grid. Finite Element Approximation The weak form of (1) reads: Find u 2 H1 such that Z › aru¢rvdx. boundary conditions imply a constant "h" and corresponds to the Dirichlet conditions (h!+∞), or to the Neumann conditions (h!0). For more general non-homogeneous. Boundary Conditions There are three types of boundary conditions that are specified during the discretization process of the Poisson equation: Dirichlet (this is a boundary condition on the potential) Neumann (this is a boundary condition on the derivative of the potential, i. Hello, Does anybody have any experience with the following error: [i]Quadrature Errors in Pressure Poisson Equation - Element(s) severely distorted. 1 Introduction. 520 Numerical Methods for PDEs : Video 13: 2D Finite Di erence MappingsFebruary 23, 2015 4 / 17. Dirichlet or even an applied voltage). 2) Parabolic equations. We con-centrate on DuðxÞ¼w 0ðxÞ in X uðxÞ¼fðxÞ or @uðxÞ @nx ¼ gðxÞ on @X (ð1Þ for a fixed domain X, but we will keep in mind that may depend on some other variables, for example time in our target applications. ) The Laplacian is an elliptic operator so we should specify Dirichlet or Neumann conditions on a closed boundary S. It runs on Windows, Linux and Mac OS. Which is not surprising, because such PDEs do not admit explicit symbolic solutions, with a few (mostly uninteresting) exceptions.
0) can represented using a Constant and the Dirichlet boundary is defined. In general, boundary value problems will reduce, when discretized, to a large and sparse set of linear (and sometimes non-linear) equations. When I use Neumann boundary conditions, I invariably get this warning printed to the console when I call s_Helmholtz_2D: MKL POISSON LIBRARY. (4) An elliptic PDE like (1) together with suitable boundary conditions like (2) or (3) constitutes an elliptic boundary value problem. Compared with the method of fundamental solution, the BKM uses the nonsingular general solution instead of the singular. I am attempting to solve this using BiCGStab solver, that I have written myself. 5C Boundary value problems 75 5C(i) Dirichlet boundary conditions 76 5C(ii) Neumann boundary conditions 80 5C(iii) Mixed (or Robin) boundary conditions 83 5C(iv) Periodic boundary conditions 85 5D Uniqueness of solutions 87 5D(i) Uniqueness for the Laplace and Poisson equations 88 5D(ii) Uniqueness for the heat equation 91. To solve this problem in the PDE Modeler app, follow these steps:. Sim-ilarly we can construct the Green's function with Neumann BC by setting G(x,x0) = Γ(x−x0)+v(x,x0) where v is a solution of the Laplace equation with a Neumann bound-ary condition that nullifies the heat flow coming from Γ. Poisson Equation in 2D. (1) Here x ∈ U, u: U¯ R, and U ⊂ Rn is a given open set. I am trying to reconstruct an image from gradients in an arbitrary shaped region of an image. 1: Plot of the solution obtained with automatic mesh adaptation Since many functions in the driver code are identical to that in the non-adaptive version, discussed in the previous example, we only list those functions that differ. A Neumann boundary condition prescribes the normal derivative value on the boundary. We will do this by solving the heat equation with three different sets of boundary conditions. subject to the boundary condition that Gvanish at in-nity. This Demonstration shows the solution of the diffusion-advection-reaction partial differential equation (PDE) in one dimension. 4 General Treatment of Singular Systems 185 5. From Laplace’s equation in plane polar coor-dinates, ∇2φ= 1 r d dr r dφ dr = 0 ⇔ dφ dr = m r, one finds φ(r) = mlnr+C. conditions or one Neumann and one Dirichlet boundary condition, but will have either no solution or an underdetermined solution in the case of two Neumann boundary conditions.
Another way of viewing the Robin boundary conditions is that it typies physical situations where the boundary "absorbs" some, but not all, of the energy, heat, mass…, being transmitted through it. Note that the boundary term that arises from integration by parts in (20) nat-urally has lead us to the Neumann boundary condition. Notes on Additive ADI - ADI for Poisson equation; Solution of the Navier-Stokes equations and the equation for the pressure (in the case of Euler explicit scheme in time) Boundary conditions for the Navier-Stokes equations - Example of results with no-slip and symmetry conditions; Supplement on the pressure correction methods. This Demonstration shows the solution of the diffusion-advection-reaction partial differential equation (PDE) in one dimension. for solving Helmholtz equation in one-dimensional and two-dimensional domain with Neumann boundary conditions. Similarly for Ω ⊂ R3, Ω ⊂ Rn, and for f = 0 (Dirichlet problem for the Laplace equation). Dirichletboundarycondition. with Mixed Dirichlet-Neumann Boundary Conditions Ashton S. The Neumann boundary conditions are implemented as centered % differences without the use of ghost points. That is, the functions c, b, and s associated with the equation should be specified in one M-file, the. Equivalence of The Poisson Pressure Approach and Continuity 12-F. function f satisfy Neumann boundary condition. We will also need the gradient to apply the pressure. I am trying to reconstruct an image from gradients in an arbitrary shaped region of an image. Definition 10. ￭Use FFFTWto do discrete Fourier transform. If electrostatics problems always involved localized discrete or continuous distribution of charge with no boundary conditions, the general solution for the potential ′ 3 0. Poisson equation is an elliptic equation and hence it strongly depends on the boundary condition. diff(y, 2), u) with zero Neumann condition, then the solution is identically zero. 2d Poisson Equation With Neumann Boundary Conditions.
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T122019/06/17 16:13: GMT+0530 | 2020-07-03T16:53:41 | {
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https://math.stackexchange.com/questions/505989/how-to-find-a-closed-form-formula-for-the-following-recurrence-relation | # How to find a closed form formula for the following recurrence relation?
I have to find a closed form formula for the following recurrence relation which describes Strassen's matrix multiplication algorithm -
$$T(n) = 7\,T\left(n \over 2\right) + \frac{18}{16}n^2$$ with the base case $T(1) = 1$ and $n = 2^j$, I have been able to reduce the formula to the following form:
$$T(2^j) = 7^j + \frac{3}{2}\,4^{j} \left[\left(\frac74\right)^j - 1\right],$$ the last step in this formula is plugging in $n = 2^j$ to get an expression in $n$ but somehow I can't reduce it further, can someone tell how to do this to get an expression in $n$?
• Try using $j=\log_2 n$ and $a^{log_2 n}=2^{\log_2 a \log_2 n}=n^{\log_2 a}$. – lhf Sep 26 '13 at 15:50
• done got it thanks a lot :) – AnkitSablok Sep 26 '13 at 15:51
• $(5/2)*n^(lg7) - (3/2)*n^2$ – AnkitSablok Sep 26 '13 at 15:57
• I meant as a true answer, not as a comment. – lhf Sep 26 '13 at 16:01
We can compute exact formulas for your recurrence and not just at powers of two. Suppose we first solve the recurrence $$S(n) = 7 S(\lfloor n/2\rfloor) + \frac{18}{16} n^2$$ with $S(0)=0.$ (This will produce $S(1) = \frac{18}{16}$ instead of one, which we'll correct later.)
Then let $$n= \sum_{k=0}^{\lfloor \log_2 n\rfloor} d_k 2^k$$ be the binary representation of $n$ and unroll the recursion to get the following exact formula for $S(n):$ $$S(n) = \frac{18}{16} \sum_{j=0}^{\lfloor \log_2 n\rfloor} 7^j \left(\sum_{k=j}^{\lfloor \log_2 n\rfloor} d_k 2^{k-j} \right)^2 = \frac{18}{16} \sum_{j=0}^{\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^j \left(\sum_{k=j}^{\lfloor \log_2 n\rfloor} d_k 2^k \right)^2.$$ To get an upper bound on this consider a string of one digits, which gives $$S(n)\le \frac{18}{16} \sum_{j=0}^{\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^j \left(\sum_{k=j}^{\lfloor \log_2 n\rfloor} 2^k \right)^2 = \frac{18}{16} \sum_{j=0}^{\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^j \left(2^{\lfloor \log_2 n\rfloor+1}- 2^j\right)^2 \\ = \frac{18}{16} \sum_{j=0}^{\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^{\lfloor \log_2 n\rfloor-j} \left(2^{\lfloor \log_2 n\rfloor+1}- 2^{\lfloor \log_2 n\rfloor-j}\right)^2 \\ = \frac{18}{16} \left(\frac{7}{4}\right)^{\lfloor \log_2 n\rfloor}2^{2\lfloor \log_2 n\rfloor} \sum_{j=0}^{\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^{-j} (2-2^{-j})^2 \\=\frac{18}{16} 7^{\lfloor \log_2 n\rfloor} \sum_{j=0}^{\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^{-j} (2-2^{-j})^2.$$ Now the sum term converges to a constant and we have for the upper bound the asymptotics $$\frac{18}{16} \times 7^{\lfloor \log_2 n\rfloor} \times \frac{49}{10} = \frac{441}{80} \times 7^{\lfloor \log_2 n\rfloor}.$$
For a lower bound consider a one digit followed by a string of zeros, which gives $$S(n)\ge \frac{18}{16} \sum_{j=0}^{\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^j 2^{2\lfloor \log_2 n\rfloor} = \frac{18}{16} 2^{2\lfloor \log_2 n\rfloor} \sum_{j=0}^{\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^{\lfloor \log_2 n\rfloor-j} \\ = \frac{18}{16} 2^{2\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^{\lfloor \log_2 n\rfloor} \sum_{j=0}^{\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^{-j} = \frac{18}{16} 7^{\lfloor \log_2 n\rfloor} \sum_{j=0}^{\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^{-j}.$$ The sum term once more converges to a constant and we get for the lower bound complexity the asymptotics $$\frac{18}{16}\times 7^{\lfloor \log_2 n\rfloor} \times\frac{7}{3} = \frac{21}{8} \times 7^{\lfloor \log_2 n\rfloor}.$$
Putting the two bounds together we get a complexity of $$S(n) \in \Theta(7^{\lfloor \log_2 n\rfloor}) = \Theta(2^{\log_2 7\times \lfloor \log_2 n\rfloor}) = \Theta(n^{\log_2 7}).$$
This being Strassen the point was to show that the complexity is better than $\Theta(n^3),$ the naive matrix multiplication, which it is, because $\log_2 7 < 3$.
To conclude this calculation we return to $T(n)$ which was supposed to have $T(1)=1$ and not $18/16.$ We can compensate for this by putting $$T(n) = -\frac{2}{16} 7^{\lfloor \log_2 n\rfloor} + S(n).$$ This term only changes the coefficient on the two bounds and does not affect the order of growth.
There are more Master Theorem type calculations at this MSE link.
Addendum. Interestingly if we compute the partial sums for the lower bound we obtain $$-\frac{2}{16} 7^{\lfloor \log_2 n\rfloor} +\frac{18}{16} 7^{\lfloor \log_2 n\rfloor} \sum_{j=0}^{\lfloor \log_2 n\rfloor} \left(\frac{7}{4}\right)^{-j} = -\frac{2}{16} 7^{\lfloor \log_2 n\rfloor} + \frac{18}{16} 7^{\lfloor \log_2 n\rfloor} \frac{1-4/7\times(4/7)^{\lfloor \log_2 n\rfloor}}{1-4/7} \\= -\frac{2}{16} 7^{\lfloor \log_2 n\rfloor} + \frac{18}{16} 7^{\lfloor \log_2 n\rfloor} \times \frac{7}{3} - \frac{18}{16} \times \frac{4}{3} \times 4^{\lfloor \log_2 n\rfloor} \\ = +\frac{5}{2} 7^{\lfloor \log_2 n\rfloor} - \frac{3}{2} \times 4^{\lfloor \log_2 n\rfloor},$$ which matches up perfectly with the formula that was proposed as an answer in the original question. | 2020-02-29T14:38:53 | {
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We will discuss the physical meaning of the various partial derivatives involved in the equation. Other boundary conditions like the periodic one are also pos-sible. when I combine heat flux and convection into one equation, FLUENT does not accept that equation, because the derived equation is related to time (due to heat flux) and temperature (due to convection). For example, instead of u= g(x;y) on the boundary, we might impose ru= g(x;y) for all (x;y) [email protected] The example figure 1. 2 A horizontal surface is shown, which is subjected to period heating that maintains temperature as a constant plus a sinusoidal wave with amplitude T0 and frequency ω (so that the. Boundary conditions (temperature on the boundary, heat flux, convection coefficient, and radiation emissivity coefficient) get these data from the solver: location. 4: A range of advanced engineering problems. heat equation Today: † PDE terminology. Robin boundary condition with r>0 40 References 42 Appendix A. For (b), the second boundary condition says that Ux′(0,s)=-ks, and since (2) implies that Ux′(x,s)=-scC2e-scx, we can infer that now. Important results in the study of the heat equation were obtained by I. If either or has the! "property that it is zero on only part of the boundary then the boundary condition is sometimes referred to as mixed. 31Solve the heat equation subject to the boundary conditions. Differential Equations K. We then uses the new generalized Fourier Series to determine a solution to the heat equation when subject to Robins boundary conditions. The stability of the heat equation with boundary condition (Eq. Let us consider the heat equation in one dimension, u t = ku xx: Boundary conditions and an initial condition will be applied later. (The corrector at z= 0 does not match the boundary condition at z = 1, and the corrector at z = 1 does not match the boundary condition at z= 0. The heat flow can be prescribed at the boundaries, ∂u −K0(0,t) = φ1 (t) ∂x (III) Mixed condition: an equation involving u(0,t), ∂u/∂x(0,t), etc. It is so named because it mimics an insulator at the boundary. (Generally itis better to zero out boundary conditions in favor of initial conditionsbut the purpose of this example is to demonstrate the convection boundary term. Solving the wave equation with Neumann boundary conditions. Time-Independent Solution: One can easily nd an equilibrium solution of ( ). To find the global equation system for the whole solution region we must assemble all the element equations. Heat Equation with Dynamical Boundary Conditions of Reactive Type. 20) subject to u(x,0) = (x if 0 < x < 1, 2 ¡ x if 1 < x < 2, u(0,t) = ux(2,t) = 0. This objective is achieved after first establishing an exact solution to the problem subject to the boundary and initial conditions which are expressed in functions of fractional powers of their arguments. Implicit boundary equations for conservative Navier–Stokes equations Journal of Computational Physics, Vol. Especially important are the solutions to the Fourier transform of the wave equation, which define Fourier series, spherical harmonics, and their generalizations. 2 The Wave Equation 630 12. We study the viscous bound. This heat and mass transfer simulation is carried out through the usage of CUDA platform on nVidia Quadro FX 4800 graphics card. How are the Dirichlet boundary conditions (zero Stack Overflow. The one-dimensional heat equation on the whole line The one-dimensional heat equation (continued) One can also consider mixed boundary conditions,forinstance Dirichlet at x =0andNeumannatx = L. To find a well-defined solution, we need to impose the initial condition u(x,0) = u 0(x) (2) and, if D= [a,b] ×[0,∞), the boundary conditions u(a. 4 ) can be proven by using the Kreiss theory. Then u(x,t) satisfies in Ω × [0,∞) the heat equation ut = k4u, where 4u = ux1x1 +ux2x2 +ux3x3 and k is a positive constant. Let us assume that the temperature distribution in the thermal penetration depth is a third-order polynomial function of x, i. Method of characteristics. In this section, we solve the heat equation with Dirichlet. Since the heat equation is linear, solutions of other combinations of boundary conditions, inhomogeneous term, and initial conditions can be found by taking an appropriate linear combination of the above Green's function solutions. , u(t;x,x) = 0. In this section we go through the complete separation of variables process, including solving the two ordinary differential equations the process generates. where Nu is the Nusselt number, Re is the Reynolds number and Pr is the Prandtl number. ANSYS FLUENT uses Equation 7. trarily, the Heat Equation (2) applies throughout the rod. [5] in which Neumann. The problem we are solving is the heat equation with Dirichlet Boundary Conditions ( ) over the domain with the initial conditions You can think of the problem as solving for the temperature in a one-dimensional metal rod when the ends of the rod is kept at 0 degrees. Indeed, the extrapolated boundary solution obeys the partial-current boundary condition to a good approxima-tion. Maximum principles for solutions of second order parabolic equations are used in deriving the results. Solving the 1D heat equation Consider the initial-boundary value problem: Boundary conditions (B. We will also introduce the auxiliary (initial and boundary) conditions also called side conditions. Showalter ADD. See [2], [8], [10], [16], [17], [22]. 2 Boundary conditions in the frequency domain To solve the heat transfer equation in the frequency domain for sinusoidal signal inputs, it is necessary to derive the dynamic boundary conditions in the frequency domain. Dirichlet boundary condition When the Dirichlet boundary condition is used as the. The CFL condition For stability we need 4∆t/∆x2 ≤ 2 CFL condition (Courant, Friedrichs, Lewy 1928) ∆t ∆x2 ≤ 1 2 The CFL condition is a severe restriction on time step ∆t Stiffness The CFL condition can be avoided by using A-stable methods, e. There are four of them that are fairly common boundary conditions. 3 Derivation of a Differential Equation for the Deformation. KEYWORDS: Lecture Notes, Distributions and Sobolev Spaces, Boundary Value Problems, First Order Evolution Equations, Implicit Evolution Equations, Second Order Evolution Equations, Optimization and Approximation Topics. Cauchy boundary condition In mathematics, a Cauchy boundary condition /koʊˈʃiː/ augments an ordinary differential equation or a partial differential equation with conditions that the solution must satisfy on the boundary; ideally so to ensure that a unique solution exists. \) Solutions to the above initial-boundary value problems for the heat equation can be obtained by separation of variables (Fourier method) in the form of infinite series or by the method of integral transforms using the Laplace transform. Chapter 11 Boundary Value Problems and Fourier Expansions 580 11. Case 1: Heat Dissipation from an Infinitely Long Fin (l → ∞): In such a case, the temperature at the end of Fin approaches to surrounding fluid temperature ta as shown in figure. 1811-1821, 2009. We have step-by-step solutions for your textbooks written by Bartleby experts!. REFERENCES Ameri AA, Arnone A Navier-Stokes turbine heat transfer predictions using two-equation turbulence closures. Consider the two-dimensional heat equation u t = 2 u, on the half-space where y > x. Solutions to the wave equation are of course important in fluid dynamics, but also play an important role in electromagnetism, optics, gravitational physics, and heat transfer. Because the heat equation is second order in the spatial coordinates, to describe a heat transfer problem completely, two boundary conditions must be given for each direction of the coordinate system along which heat transfer is significant. Solve the heat equation with a source term. 1 Heat equation with Dirichlet boundary conditions We consider (7. In fact, one can show that an infinite series of the form u(x;t) · X1 n=1 un(x;t) will also be a solution of the heat equation, under proper convergence assumptions of this series. We consider the case when f = 0, no heat source, and g = 0, homogeneous Dirichlet boundary condition, the only nonzero data being the initial condition u0. Consider the following mixed initial-boundary value problem, which is called the Dirichlet problem for the heat equation (u t ku. of some initial-boundary value problems for the heat equation in one and two space dimensions when linear radiation (Robin) conditions are prescribed on the boundary. 1) with the. It is a hyperbola if B2 ¡4AC. The same equation will have different general solutions under different sets of boundary conditions. The heat equation is a partial differential equation that describe the distribution of heat in a given area in a given time interval. General equation of 2 nd order: θ = c 1 e mx + c 2 e –mx; Heat dissipation can take place on the basis of three cases. In this section we go through the complete separation of variables process, including solving the two ordinary differential equations the process generates. diffusion coefficient alpha = 0. boundary conditions. Consider a homogenous medium within which there is no bulk motion and the temperature distribution T ( x,y,z ) is expressed in Cartesian coordinates. Here, the vector = (x;y) is the exterior unit normal vector. Because the heat equation is second order in the spatial coordinates, to describe a heat transfer problem completely, two boundary conditions must be given for each direction of the coordinate system along which heat transfer is significant. For heat flow in any three-dimensional region, (7. Solve a 1D wave equation with absorbing boundary conditions. at , in this example we have as an initial condition. The assumed temperature distribution can be any arbitrary function provided that the boundary conditions at x= 0and x= δare satisfied. 83 Handling Frames in Heat Transfer 86 Foundations of the General Heat Transfer Equation 151. Sun, “A high order difference scheme for a nonlocal boundary value problem for the heat equation,” Computatinal methods in applied mathematics, vol. To that end, we consider two-dimensional rectangular geometry where one boundary is at prescribed heat flux conditions and the remaining ones are subjected to a convective boundary condition. Heat Equation; Heat Equation; Hilbert Space Methods for Partial Differential Equations, by R. 11) with a corresponding boundary condition on the entire boundary of the region. Under steady state conditions, the heat equation degenerates into Laplace's equation whose only bounded solutions, in two dimensions, are constant everywhere. 2 Insulated Boundaries. diffusion coefficient alpha = 0. One-dimensional heat conduction equation − two ends kept at arbitrary constant temperatures: an example of nonhomogeneous boundary conditions Let us now see what happens when the boundary conditions are nonzero (known as nonhomogeneous boundary conditions). v verifying the same boundary condition, v| ∂D= u 0. We nd a pair of boundary conditions for the heat equation such that the solution goes to zero for either boundary condition, but if the boundary condition randomly switches, then theaverage solution grows exponentially in time. Ax+ B:Applying boundary conditions, 0 = X(0) = B )B = 0; 0 = X0(ˇ) = A)A= 0. left boundary condition g1(t) = '20. if we are looking for stationary heat distribution and we have heat flow defined, we need to assume that the total flow is 0 (otherwise the will. See full list on reference. but satisfies the one-dimensional heat equation u t xx, t 0 [1. The function u(x,t) that models heat flow should satisfy the partial differential equation. The heat equation Homog. Two methods are used to compute the numerical solutions, viz. The initial condition is given in the form u(x,0) = f(x), where f is a known function. This example involves insulated ends (. The right-hand side of the equation provides a natural way to assign boundary conditions in terms of the heat flux. We assume that the reader has already studied this previous example and this one. 4 Equilibrium Temperature Distribution. 21) In this problem, we have a mixture of both fixed and no flux boundary conditions. The heat equation is a partial differential equation that describe the distribution of heat in a given area in a given time interval. Cole Sep 18, 2018, Heat Equation, Cartesian, Two-dimensional, X33B00Y33B00T5. PDE playlist: http://www. u t U U w w (1) Navier-Stokes 0 4. 1 Eigenvalue Problems for y. We will do this by solving the heat equation with three different sets of boundary conditions. In the context of wave propagations. (1) can be written when ¡2 =; as the heat equation with homogeneous Neumann boundary condition on ¡0 and generalized Wentzell boundary condition ¢u+k1u” = 0 on ¡1. Cauchy conditions are usually appropriate over at least part of the boundary, while Dirichlet,. Equation 1 - the finite difference approximation to the Heat Equation; Equation 4 - the finite difference approximation to the right-hand boundary condition; The boundary condition on the left u(1,t) = 100 C; The initial temperature of the bar u(x,0) = 0 C; This is all we need to solve the Heat Equation in Excel. Heat Equation with Dynamical Boundary Conditions of Reactive Type. Let u solve the heat equation on an interval a < x < b, and t > 0, with initial condition u(0,x) = f(x) square integrable and either Dirichlet boundary conditionsor Neumann boundary conditions. Finite difference methods and Finite element methods. Heat equation with two boundary conditions on one side. Time-Independent Solution: One can easily nd an equilibrium solution of ( ). Heat Transfer L17 p4 - Thermal Boundary Layer by Ron Hugo 4 years ago 8 minutes, 24 seconds 21,488 views Thermal Boundary Layer energy equation for low Eckert number cases • Present idea of , thermal boundary layer ,. Chapter 7: Time and Space. We will use the Fourier sine series for representation of the nonhomogeneous solution to satisfy the boundary conditions. Heat Equation in One Dimension Implicit metho d ii. Taking c 2 = 1 we get the solution X = X 0 = 1. Thus we have recovered the trivial solution (aka zero solution). The stability of the heat equation with boundary condition (Eq. Here 0 is the dimensionless slip-length parameter, while the Reynolds num- ber is scaled to unity. The first type of boundary conditions that we can have would be the prescribed temperature boundary conditions, also called Dirichlet conditions. Because of the boundary condition, T[n, j-1] gets replaced by T[n, j+1] - 2*A*dx when j is 0. To accomodate our belief that boundaries can be maintained at different temperatures, the governing equation must lose validity somewhere. Therefore for = 0 we have no eigenvalues or eigenfunctions. We prove a new formula for PtDφ (where φ:H→R is bounded and Borel) which depends on φ but not on its derivative. Consider a rod of length l with insulated sides is given an initial temperature distribution of f (x) degree C, for 0 < x < l. heat equation Today: † PDE terminology. ) which possesses neither sources nor sinks of heat (i. Heat equation with two boundary conditions on one side. Use boundary conditions from equation (2) in u (x, t) = X (t) T. using Dirichlet boundary condition). The code below solves the 1D heat equation that represents a rod whose ends are kept at zero temparature with initial condition 10*np. We develop an L q theory not based on separation of variables and use techniques based on uniform spaces. Consider the nondimensionalized heat equation (2. We study similarity solutions of a nonlinear partial differential equation that is a generalization of the heat equation. 2 The Wave Equation 630 12. 0000 » view(20,-30) Heat Equation: Implicit Euler Method. The Heat Equation and Periodic Boundary Conditions Timothy Banham July 16, 2006 Abstract In this paper, we will explore the properties of the Heat Equation on discrete networks, in particular how a network reacts to changing boundary conditions that are periodic. a) Verify that solutions u(x,t) to the heat equation with the initial condition u(x,0) = f(x) piecewise continuous first derivatives may be given in the. ODE Version. This is the most challenging part of setting up the simulation: first, for both real and simulated fires, the growth of the fire is very sensitive to the thermal properties of the surrounding materials. A specific rotating heat pipe was examined under the three kinds of boundary conditions: linear distribution, uniform but asymmetric distribution, and uniform as well as symmetric distribution of heat load. To find a well-defined solution, we need to impose the initial condition u(x,0) = u 0(x) (2) and, if D= [a,b] ×[0,∞), the boundary conditions u(a. of the heat equation (1). The solution of the heat equation with the same initial condition with fixed and no flux boundary conditions. Initial condition: Boundary conditions: t 0,T To x 0 2 , 0, 1 1 t x H T T x T T 2 2 x Y t Y Initial condition: Boundary conditions: t 0,T To x Y 1 0 2 , 0 0, 0 1 1 t x H T T Y x T T Y Unsteady State Heat Conduction in a Finite Slab: solution by separation of variables. Ryan Walker An Introduction to the Black-Scholes PDE The Heat Equation The heat equation in one space dimensions with Dirchlet boundary conditions is: ˆ u t = u xx u(x,0) = u 0(x) and its solution has long been known to be: u(x,t) = u 0 ∗Φ(x,t) where Φ(x,t) = 1 √ 4πt e− x 2 4kt. The Boundary Conditions 8 1. 2: Applications to time-dependent and time-harmonic problems -- Advances in boundary element analysis of non-linear problems of solid and fluid mechanics; v. 0 time step k+1, t x. sional heat conduction. Another way of viewing the Robin boundary conditions is that it typies physical situations where the boundary “absorbs” some, but not all, of the energy, heat, mass…, being transmitted through it. ): Step 1- Define a discretization in space and time: time step k, x 0 = 0 x N = 1. 3 v u pu t P w w (2) Energy 0. In other words we must combine local element equations for all elements used for discretization. ( 4 – 7 ), as demonstrated below. Example 2 Solve ut = uxx, 0 < x < 2, t > 0 (4. In the case of Neumann boundary conditions, one has u(t) = a 0 = f. In order to have a well-posed partial differential equation problem, boundary conditions must be specified at the endpoints of the spatial domain. Part 2: Excel Solver- Simple Boundary Conditions. In several spatial variables, the fundamental solution solves the analogous problem. 1-d problem with mixed boundary conditions; An example 1-d diffusion equation solver; An example 1-d solution of the diffusion equation; von Neumann stability analysis; The Crank-Nicholson scheme; An improved 1-d diffusion equation solver; An improved 1-d solution of the diffusion equation; 2-d problem with Dirichlet boundary conditions. Cole Sep 18, 2018, Heat Equation, Cartesian, Two-dimensional, X33B00Y33B00T5. The Boundary Conditions 8 1. The initial temperature of the bar u (x,0) = 0 C. 24} and this condition we really need and it is justified from the physical point of view: f. The temperature distribution varies for. [5] in which Neumann. Mazzucato Abstract. I am unable to proceed so, please throw some light on how to proceed to reach to a solution of this heat equation. Solving the heat equation (PDE) with different Learn more about pde derivative bc. for the differential equation of heat conduction and for the equations expressing the initial and boundary conditions their appropriate difference analogs, and solving the resulting system. The solution of a heat equation with a source and homogeneous boundary conditions may be found by solving a homogeneous heat equation with nonhomo- geneousboundaryconditions. boundary data need to be specified to give the problem a unique answer. A di erential equation with auxiliary initial conditions and boundary conditions, that is an initial value problem, is said to be well-posed if the solution exists, is unique, and small. First substitute the dimensionless variables into the heat equation to obtain ˆCˆ P @——T 1 T 0– ‡T– @ ˆCˆ Pb2 k ˝ …k @2 ——T T. For 2D heat conduction problems, we assume that heat flows only in the x and y-direction, and there is no heat flow in the z direction, so that , the governing equation is: In cylindrical coordinates, the governing equation becomes: Similarly, the boundary conditions is: for for. The work continues an earlier study by Schatz et al. This objective is achieved after first establishing an exact solution to the problem subject to the boundary and initial conditions which are expressed in functions of fractional powers of their arguments. 6) are obtained by using the separation of variables technique, that is, by seeking a solution in which the time variable t is separated from the space. Element connectivities are used for the assembly process. The boundary conditions give 0 = X′(0) = X′(L) = c 1. 1 Prescribed Temperature. Because of the boundary condition, T[n, j-1] gets replaced by T[n, j+1] - 2*A*dx when j is 0. Notiee that this assumes a constant wall temperature for the isothermal boundary conditions and a Constant freestream temperature for both isothermal and adiabatic boundary conditions, The solution of (12) is (15) Applying the adiabatic wall boundary conditions to find the constants c I and C2 results in the following simple equation (16) (17a). This is called the Neumann boundary condition. *t)' length of the rod L = 1. Dynamic Boundary Conditions As in the discrete case, another type of boundary condition arises when we assume the. Solve a 1D wave equation with absorbing boundary conditions. Thus, this third type of boundary condition is an interpolation between the first two types for intermediate values of k b. Boundary layers for the Navier-Stokes equa-tions linearized around a stationary Euler ow Gung-Min Gie, James P. The Differential Equations 6 B. The driving force behind a heat transfer are temperature differences. Assuming steady one dimensional heat transfer, (a) express the differential equation and the boundary conditions for heat conduction through the sphere, (b) obtain a relation for the variation of temperature in the sphere by solving the differential equation, and (c) determine the temperature at the center of the sphere. Note also that the function becomes smoother as the time goes by. diffusion coefficient alpha = 0. To obtain the solution within the interval [a 0, a], an exact boundary condition must be applied at some x a. In the case of the heat and Schr¨odinger equation we set, u| t=0 = u 0 while in the case of the wave equation we impose two. The simplest one is to prescribe the values of uon the hyperplane t= 0. Under steady state conditions, the heat equation degenerates into Laplace's equation whose only bounded solutions, in two dimensions, are constant everywhere. The bounds are logarithm free and valid in arbitrary dimension and for arbitrary polynomial degree. Therefore, we need to specify four boundary conditions for two-dimensional problems, and six boundary conditions for three-dimensional problems. [5] in which Neumann. The bounds are logarithm free and valid in arbitrary dimension and for arbitrary polynomial degree. We consider the case when f = 0, no heat source, and g = 0, homogeneous Dirichlet boundary condition, the only nonzero data being the initial condition u0. Solving the heat equation in the box is not too difficult, but determining how heat is transferred from the metal box surface to the surrounding air seems much less obvious. The heat transfer of a viscous fluid over a stretching/shrinking sheet with convective boundary conditions has been studied by Yao et al. Therefore v(x) = c 1 + c 2x, for some constants c 1 and c 2. x , location. The problem of the one-dimensional heat equation with nonlinear boundary conditions is studied. Dirichlet conditions Neumann conditions Derivation Initialconditions If we now impose our initial condition we find that f(x) = u(x,0) = a. Integrating twice gives X = c 1x +c 2. For the proof of null controllability, a crucial tool will be a new Carleman estimate for the weak solutions of the classical heat equation with nonhomogeneous Neumann boundary. When we take t!1, the heat equation gives us a partial differential equation for the steady-state solution, 0 = (uxx+ uyy). Here we will use the simplest method, nite di erences. The solutions to Poisson's equation are superposable (because the equation is linear). One can obtain the general solution of the one variable heat equation with initial condition u(x, 0) = g(x) for −∞ < x < ∞ and 0 < t < ∞ by applying a convolution: (,) = ∫ (−,) (). So we need to solve the following BVP for w; w′′+λw = 0, w(0) = w(L) = 0. Macauley (Clemson) Lecture 5. Solving the heat equation (PDE) with different Learn more about pde derivative bc. A problem that proposes to solve a partial differential equation for a particular set of initial and boundary conditions is called, fittingly enough, an initial boundary value problem, or IBVP. The location of the interfaces is known, but neither temperature nor heat ux are prescribed there. For the proof of null controllability, a crucial tool will be a new Carleman estimate for the weak solutions of the classical heat equation with nonhomogeneous Neumann boundary. In the following experiment, the effects of various boundary conditions on the. Convective Boundary Condition The general form of a convective boundary condition is @u @x x=0 = g 0 + h 0u (1) This is also known as a Robin boundary condition or a boundary condition of the third kind. A product solu-tion, u(x, y, z, t) = h(t)O(x, y, z), (7. One of the following three types of heat transfer boundary conditions. 1: Advanced applications to a wide range of problems in engineering; v. In the context of the heat equation, Dirichlet boundary conditions model a situation where the temperature of the ends of the bars is controlled directly. Here 0 is the dimensionless slip-length parameter, while the Reynolds num- ber is scaled to unity. Therefore, at the end of this process, we have two ordinary differential equations, together with a set of two boundary conditions that go with the equation of the spatial variable x: X ″ + λX = 0, X(0) = 0 and X(L) = 0, T ′ + α 2 λ T = 0. Thus, a general solution is the superposition of all these u n(x;t): u(x;t) = X1 n=1 b ne 2 ntsin nˇ L x: (9) Y. REFERENCES Ameri AA, Arnone A Navier-Stokes turbine heat transfer predictions using two-equation turbulence closures. Mathematics subject classification(2000): 35K05, 35B50. 5: Laplace’s Equation on a Ring or Half Disk. The solutions to Poisson's equation are superposable (because the equation is linear). Because the heat equation is second order in the spatial coordinates, to describe a heat transfer problem completely, two boundary conditions must be given for each direction of the coordinate system along which heat transfer is significant. Use boundary conditions from equation (2) in u (x, t) = X (t) T. In order to have a well-posed partial differential equation problem, boundary conditions must be specified at the endpoints of the spatial domain. Using linearity we can sort out the. This principle states that the rate of change of the heat energy in a region is equal to the heat flux across the boundary of the region. when I combine heat flux and convection into one equation, FLUENT does not accept that equation, because the derived equation is related to time (due to heat flux) and temperature (due to convection). Solving the heat equation (PDE) with different Learn more about pde derivative bc. There are four of them that are fairly common boundary conditions. Dirichlet boundary condition, and as k b → 0 we similarly obtain the Neumann boundary condition. For (b), the second boundary condition says that Ux′(0,s)=-ks, and since (2) implies that Ux′(x,s)=-scC2e-scx, we can infer that now. The code below solves the 1D heat equation that represents a rod whose ends are kept at zero temparature with initial condition 10*np. That is, the average temperature is constant and is equal to the initial average temperature. We describe here a simple example for the one dimensional heat equation, over the domain. Under steady state conditions, the heat equation degenerates into Laplace's equation whose only bounded solutions, in two dimensions, are constant everywhere. Dirichlet conditions Neumann conditions Derivation Initialconditions If we now impose our initial condition we find that f(x) = u(x,0) = a. To keep things simple, our latest boundary. The analysis can also be carried over to higher order finite difference approximations for the time discretization and also to the. Title: Microsoft PowerPoint - 8_PDEs. 1 Heat equation with Dirichlet boundary conditions We consider (7. (1) can be written when ¡2 =; as the heat equation with homogeneous Neumann boundary condition on ¡0 and generalized Wentzell boundary condition ¢u+k1u” = 0 on ¡1. 7153/dea-05-17 SINGLE POINT BLOW–UP SOLUTIONS TO THE HEAT EQUATION WITH NONLINEAR BOUNDARY CONDITIONS JUNICHIHARADA Abstract. Philippe B. 1 Eigenvalue Problems for y. Then u(x,t) satisfies in Ω × [0,∞) the heat equation ut = k4u, where 4u = ux1x1 +ux2x2 +ux3x3 and k is a positive constant. Before solution, boundary conditions (which are not accounted in element. Boundary Conditions (BC): in this case, the temperature of the rod is affected. Dirichlet boundary condition, and as k b → 0 we similarly obtain the Neumann boundary condition. For parabolic equations, the boundary @ (0;T) [f t= 0gis called the parabolic boundary. of boundary conditions. We proceed by examples. What are the appropriate boundary conditions for the air-metal interface at the surface of the box?. A specific rotating heat pipe was examined under the three kinds of boundary conditions: linear distribution, uniform but asymmetric distribution, and uniform as well as symmetric distribution of heat load. Heat equation with boundary conditions Thread starter Telemachus; Start date Oct 29, 2011; Oct 29, 2011 #1 Telemachus. Case of Robin boundary condition 19 3. See full list on reference. General equation of 2 nd order: θ = c 1 e mx + c 2 e –mx; Heat dissipation can take place on the basis of three cases. We firstly consider 1-d heat system endowed with two controls. Because of the boundary condition, T[n, j-1] gets replaced by T[n, j+1] - 2*A*dx when j is 0. Diffusion Equations of One State Variable. Therefore, we need to specify four boundary conditions for two-dimensional problems, and six boundary. Part 3: Excel Solver- Complex Boundary Conditions. dS dt (3) State 2 pc 0 U (4) where and are the ambient and excess density, respectively. The boundary condition at the left endpoint is linear homogeneous, injecting energy into the system, while the boundary condition at the right endpoint has cubic nonlinearity of a van der Pol type. to the heat equation with (homogeneous) Neumann boundary conditions. conditions for switching controls. 1] on the interval [a, ). 72 leads to the expansion of the Green function ∑∑ and the equation for the radial Green function [ ] (3. A multi-block, three-dimensional Navier–Stokes code has been used to compute heat transfer coefficient on the blade, hub and shroud for a rotating high-pressure turbine blade with film-cooling holes in eight rows. Boundary conditions (temperature on the boundary, heat flux, convection coefficient, and radiation emissivity coefficient) get these data from the solver: location. Showalter ADD. We separate the equation to get a function of only t t on one side and a function of only x x on the other side and then introduce a separation constant. Note that the surface temperature at x = 0 and x = L were denoted as boundary conditions, even though it is the fluid temperature, and not the surface temperatures, that are typically known. 2 Initial condition and boundary conditions To make use of the Heat Equation, we need more information: 1. Then we derive the differential equation that governs heat conduction in a large plane wall, a long cylinder, and a sphere, and gener-alize the results to three-dimensional cases in rectangular, cylindrical, and spher-ical coordinates. 4 ) can be proven by using the Kreiss theory. diffusion coefficient alpha = 0. One of the objectives of the paper is to study the analyticity of solutions. x , location. Next: † Boundary conditions † Derivation of higher dimensional heat equations Review: † Classiflcation of conic section of the form: Ax2 +Bxy +Cy2 +Dx+Ey +F = 0; where A;B;C are constant. places on the bar which either generate heat or provide additional cooling), the one-dimensional heat equation describing its temperature as a function of displacement from one end (x) and time (t) is given as. Therefore v(x) = c 1 + c 2x, for some constants c 1 and c 2. 3 Boundary Conditions. ( 4 – 7 ), as demonstrated below. customary units) or s (in SI units). with boundary conditions and. The basic problems for the heat equation are the Cauchy problem and the mixed boundary value problem (seeBOUNDARY VALUE PROBLEMS). I'm trying to solve the heat. Integrating twice gives X = c 1x +c 2. Initial Condition (IC): in this case, the initial temperature distribution in the rod u(x,0). The same equation will have different general solutions under different sets of boundary conditions. The method allows arbitrary conditions on all of the following: pressure gradientj sur-face temperature and its gradient, heat transfer, mass transfer, and fluid properties. We may begin by solving the Equations 8. Heat transfer is a discipline of thermal engineering that is concerned with the movement of energy. it is also constant zero). Then the initial values are filled in. The boundary condition at the left endpoint is linear homogeneous, injecting energy into the system, while the boundary condition at the right endpoint has cubic nonlinearity of a van der Pol type. Transforming the differential equation and boundary conditions. A second benchmark problem dealing with transient conduction heat transfer in a two dimensional. differential equation (7. with boundary conditions and. We then uses the new generalized Fourier Series to determine a solution to the heat equation when subject to Robins boundary conditions. This notebook will illustrate the Backward Time Centered Space (BTCS) Difference method for the Heat Equation with the initial conditions$$ u(x,0)=2x, \ \ 0 \leq x \leq \frac{1}{2}, u(x,0)=2(1-x), \ \ \frac{1}{2} \leq x \leq 1, $$and boundary condition$$ u(0,t)=0, u(1,t)=0. 83 Handling Frames in Heat Transfer 86 Foundations of the General Heat Transfer Equation 151. Stability and analyticity estimates in maximum-norm are shown for spatially discrete finite element approximations based on simplicial Lagrange elements for the model heat equation with Dirichlet boundary conditions. This objective is achieved after first establishing an exact solution to the problem subject to the boundary and initial conditions which are expressed in functions of fractional powers of their arguments. Dirichlet boundary condition, and as k b → 0 we similarly obtain the Neumann boundary condition. Part 2: Excel Solver- Simple Boundary Conditions. Therefore the initial condition can be also thought as a boundary condition of the space-time domain (0;T). 5 Interface Boundary Conditions The boundary conditions at an interface are based on the requirements that (1) two bodies in contact must have the same temperature at the area of contact and (2) an interface (which is a surface) cannot store any energy, and thus the heat flux on the two sides of an interface must be the same. 2: Two Dimensional Diffusion with Neumann Boundary Conditions. First substitute the dimensionless variables into the heat equation to obtain ˆCˆ P @——T 1 T 0– ‡T– @ ˆCˆ Pb2 k ˝ …k @2 ——T T. Under steady state conditions, the heat equation degenerates into Laplace's equation whose only bounded solutions, in two dimensions, are constant everywhere. ator, linear Schr¨odinger equation and heat equation on unbounded domain. We prove a new formula for PtDφ (where φ:H→R is bounded and Borel) which depends on φ but not on its derivative. A problem that proposes to solve a partial differential equation for a particular set of initial and boundary conditions is called, fittingly enough, an initial boundary value problem, or IBVP. Taking c 2 = 1 we get the solution X = X 0 = 1. [5] in which Neumann. X33Y33Gx5y5F0T0 Rectangular plate with piecewise internal heating, out-of-plane heat loss, and homogeneous convection boundary conditions at the edges of the plate. In this section we will study heat conduction equation in cylindrical coordinates using Dirichlet boundary condition with given surface temperature (i. In Section 5, we present the standard homotopy perturbation method. satis es the di erential equation in (2. 2 Lecture 1 { PDE terminology and Derivation of 1D heat equation Today: † PDE terminology. 3 Boundary Conditions. 12) may still be sought, and after separating variables, we obtain equations similar to. This implies boundary conditions u x(0,t) = 0 = u x(1,t),t ≥0. We separate the equation to get a function of only t t on one side and a function of only x x on the other side and then introduce a separation constant. The simplistic implementation is to replace the derivative in Equation (1) with a one-sided di erence uk+1 2 u k+1 1 x = g 0 + h 0u k+1. To accomodate our belief that boundaries can be maintained at different temperatures, the governing equation must lose validity somewhere. (a) Find the fundamental solution for this PDE with zero Dirichlet boundary conditions, i. heat equation in an exterior domain with Dirichlet boundary condition and to the use of the backward Euler method for the time discretization. In other words we must combine local element equations for all elements used for discretization. Part 3: Excel Solver- Complex Boundary Conditions. The problem is formulated using the heat equation with periodic boundary conditions. We consider the case when f = 0, no heat source, and g = 0, homogeneous Dirichlet boundary condition, the only nonzero data being the initial condition u0. Inhomogeneous Heat Equation on Square Domain. It describes the applying boundary conditions; Fourier series As there is no heat. We consider both homogeneous and non-homogeneous boundary conditions. U(x,s)=cksse-xcs, which corresponds to. We consider the case when f = 0, no heat source, and g = 0, homogeneous Dirichlet boundary condition, the only nonzero data being the initial condition u0. In this paper we address the well posedness of the linear heat equation under general periodic boundary conditions in several settings depending on the properties of the initial data. x Contents 2. In the theoretical analysis of FC flows and heat transfer the laws of momentum, mass and energy conservation at certain boundary conditions are used. The two main. This article (Part 1) deals with boundary conditions relevant to modeling the earth’s thermal history. This paper suggests a true improvement in the performance while solving the heat and mass transfer equations for capillary porous radially composite cylinder with the first type of boundary conditions. The boundary at x = 1 is the outflow boundary and the solution at this boundary is completely determined by what is advecting to the right from the interior. the advection equation can have a boundary condition specified on only one of the two boundaries. Part 2: Excel Solver- Simple Boundary Conditions. Separation of Variables The most basic solutions to the heat equation (2. thermal conductivity, thermal permeability, etc. 1 Integral representation of the Cauchy problem solution for the heat equation. The work continues an earlier study by Schatz et al. Here, the vector = (x;y) is the exterior unit normal vector. To that end, we consider two-dimensional rectangular geometry where one boundary is at prescribed heat flux conditions and the remaining ones are subjected to a convective boundary condition. † Derivation of 1D heat equation. The Differential Equations 6 B. Note also that the function becomes smoother as the time goes by. Element connectivities are used for the assembly process. thermal conductivity, thermal permeability, etc. 1) we have the classical problem with homogeneous Dirichlet boundary conditions for the heat equation which is well known. We nd a pair of boundary conditions for the heat equation such that the solution goes to zero for either boundary condition, but if the boundary condition randomly switches, then theaverage solution grows exponentially in time. Showalter ADD. An introduction to partial differential equations. 1 Prescribed Temperature. Zill Chapter 12. Especially important are the solutions to the Fourier transform of the wave equation, which define Fourier series, spherical harmonics, and their generalizations. Trapezoidal Rule or Implicit Euler Numerical Methods for Differential Equations – p. Maximum principles for solutions of second order parabolic equations are used in deriving the results. conditions in the velocity (hydrodynamic) boundary layer fluid properties are independent of temperature. ator, linear Schr¨odinger equation and heat equation on unbounded domain. The finite element methods are implemented by Crank - Nicolson method. That is inside the domain, not on a boundary - that is why you cannot apply a boundary condition on it Hi, I have the same problem. There are similar expansions for the heat trace associated with the action of the Laplacian on p-forms for each p. 3 Derivation of a Differential Equation for the Deformation. These include its internal temperature field, mantle structure, past and present ocean temperatures, surface heat flows and its inventory of heat-producing radionuclides. ppt Author: gutierjm Created Date: 1/14/2008 8:13:20 AM. u t U U w w (1) Navier-Stokes 0 4. As a result, these two solutions to the diffusion equation are nearly the same. 2) with boundary condition prespecified at x =0 only Boundary control of an unstable heat equation via measurement of domain- averaged temperature - Automatic Control, IEEE Transactions on. 1) we have the classical problem with homogeneous Dirichlet boundary conditions for the heat equation which is well known. 3: The Heat Equation on a Disk. Because the heat equation is second order in the spatial coordinates, to describe a heat transfer problem completely, two boundary conditions must be given for each direction of the coordinate system along which heat transfer is significant. Topics to be covered; Brief review of some relevant topics from linear algebra, calculus and ODE. Boundary and Initial Conditions Initial Condition: 3-D Heat Equation Fourier’s Law: Conservation of Energy: Let f denote heat source or sink Note:. Boundary and initial conditions are needed to solve the governing equation for a specific physical situation. The initial temperature is given. We need to solve X′′ = 0. Next: † Boundary conditions † Derivation of higher dimensional heat equations Review: † Classiflcation of conic section of the form: Ax2 +Bxy +Cy2 +Dx+Ey +F = 0; where A;B;C are constant. ( 4 – 7 ), as demonstrated below. 2 Energy for the heat equation We next consider the (inhomogeneous) heat equation with some auxiliary conditions, and use the energy method to show that the solution satisfying those conditions must be unique. Keep in mind that, throughout this section, we will be solving the same. 6) are obtained by using the separation of variables technique, that is, by seeking a solution in which the time variable t is separated from the space. sol = pdepe(m,@pdex,@pdexic,@pdexbc,x,t) where m is an integer that specifies the problem symmetry. Assuming steady one dimensional heat transfer, (a) express the differential equation and the boundary conditions for heat conduction through the sphere, (b) obtain a relation for the variation of temperature in the sphere by solving the differential equation, and (c) determine the temperature at the center of the sphere. Substitution of the similarity ansatz reduces the partial differential equation to a nonlinear second- order ordinary differential equation on the half-line with Neumann boundary conditions at both boundaries. Trapezoidal Rule or Implicit Euler Numerical Methods for Differential Equations – p. Implement the FVM for the steady 1D heat conduction equation (1) in Fortran. Philippe B. In this section we go through the complete separation of variables process, including solving the two ordinary differential equations the process generates. Since u 1(x;t);u 2(x;t);::: are satisfying the 1D Heat equation and the zero temperature boundary conditions. The solution to the 1D diffusion equation can be written as: = ∫ = = L n n n n xdx L f x n L B B u t u L t L c u u x t 0 ( )sin 2 (0, ) ( , ) 0, ( , ) π (2) The weights are determined by the initial conditions, since in this case; and (that is, the constants ) and the boundary conditions (1) The functions are completely determined by the. Neumann Boundary Conditions Robin Boundary Conditions Remarks At any given time, the average temperature in the bar is u(t) = 1 L Z L 0 u(x,t)dx. Especially important are the solutions to the Fourier transform of the wave equation, which define Fourier series, spherical harmonics, and their generalizations. 6) are obtained by using the separation of variables technique, that is, by seeking a solution in which the time variable t is separated from the space. boundary data need to be specified to give the problem a unique answer. Heat Equation: PDE vs FDE PDE: ¶u ¶t = ¶2u ¶x2 or Dtu=D2xu FDE: Da t u=D 2 xu where a 2[1 d;1+d]ˆR Initial-Boundary-Value Problem: Object: One dimensional rod of length L Boundary Conditions: u(t;0)=u(t;L)=0 Inital Conditon: u(0;x)= 4a L2 x2 + 4a L x Simon Kelow Northern Arizona University Particular Solutions to the Time-Fractional Heat. Sun, “A high order difference scheme for a nonlocal boundary value problem for the heat equation,” Computatinal methods in applied mathematics, vol. (17) We also have a Green function G2 for boundary condition 2 which satisfies the same equation, LG2(x,x′) = δ(x−x′). com/EngMathYT How to solve the heat equation via separation of variables and Fourier series. 74 The general solution of this equation for (see Eq. (The corrector at z= 0 does not match the boundary condition at z = 1, and the corrector at z = 1 does not match the boundary condition at z= 0. 2 is an initial/boundary-value problem. Consider a rod of length l with insulated sides is given an initial temperature distribution of f (x) degree C, for 0 < x < l. The case of torus 30 4. One of the objectives of the paper is to study the analyticity of solutions. 4: Partial Differential Equations in the Recipe for a Cheese Cake. Speci cally, we prove that the mean of the random. the exterior of a disc with a non-slip condition and a given velocity at infinity. The entire problem should be well posed, with the initial condition supported in (a 0, a) and a specified boundary condition at a 0. The problem we are solving is the heat equation with Dirichlet Boundary Conditions ( ) over the domain with the initial conditions You can think of the problem as solving for the temperature in a one-dimensional metal rod when the ends of the rod is kept at 0 degrees. vtu is stored in the VTK file format and can be directly visualized in Paraview for example. The 2D geometry of the domain can be of arbitrary. Tikhonov, and S. Hence, we have to let the new boundary conditions to be: X(0) = 0 and X(L) = 0. 4 Equilibrium Temperature Distribution. Equations and boundary conditions that are relevant for performing heat transfer analysis are derived and explained. KEYWORDS: Lecture Notes, Distributions and Sobolev Spaces, Boundary Value Problems, First Order Evolution Equations, Implicit Evolution Equations, Second Order Evolution Equations, Optimization and Approximation Topics. Ryan Walker An Introduction to the Black-Scholes PDE The Heat Equation The heat equation in one space dimensions with Dirchlet boundary conditions is: ˆ u t = u xx u(x,0) = u 0(x) and its solution has long been known to be: u(x,t) = u 0 ∗Φ(x,t) where Φ(x,t) = 1 √ 4πt e− x 2 4kt. We also define the Laplacian in this section and give a version of the heat equation for two or three dimensional situations. Therefore the initial condition can be also thought as a boundary condition of the space-time domain (0;T). Heat equation with two boundary conditions on one side 0 Reference request with examples, finite difference method for $1D$ heat equation ,with mixed boundary conditions. When you define a heat flux boundary condition at a wall, you specify the heat flux at the wall surface. Heat Equation Static limit for t ! 1 :Poisson problem div (x ) grad T (x ) = f (x ) Boundary condition I If temperature is known (e. Heat Equation Boundary Conditions The driving force behind a heat transfer are temperature differences. Using linearity we can sort out the. | 2020-11-26T06:56:36 | {
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https://codeforces.com/blog/entry/43225 | Codeforces celebrates 10 years! We are pleased to announce the crowdfunding-campaign. Congratulate us by the link https://codeforces.com/10years. ×
tweety's blog
By tweety, history, 4 years ago, ,
I'm going to share with you some cool applications of matrix exponential that I learned recently.
Application 1: Counting the number of ways for reaching a vertex
You are given an unweighted directed graph (may contain multiple edges) containing N vertices (1 <= N <= 200) and an integer b (1 <= b <= 10^9). You are also given Q queries (1 <= Q <= 10^5). For each query you are given two vertices u and v and you have to find the number of ways for reaching vertex v starting from u after exactly b steps. (A step is passing through an edge. Each edge may be passed multiple number of times).
Solution
Let M1 be a matrix where M1[i][j] equals the number of edges connecting vertex i to vertex j. Let M2 be M1 raised to the power of b (M1^b). Now for any pair u and v, the number of ways for reaching vertex v starting from u after b steps is M2[u][v].
Practice problem: 621E
Application 2: Shortest path with a specified number of steps
You are given a weighted graph containing N vertices (1 <= N <= 200) and an integer b (1 <= b <= 10^9). You are also given Q queries (1 <= Q <= 10^5). For each query you are given two vertices u and v and you have to find the minimum cost for reaching vertex v starting from u after exactly b steps. (A step is passing through an edge. Each edge may be passed multiple number of times).
Solution
Let M1 be a matrix where M1[i][j] equals the cost of passing the edge connecting i to j (infinity if there is no edge). Let M2 be M1 raised to the power of b (but this time using the distance product for multiplication). Now for any pair u and v, the minimum cost for reaching vertex v starting from u after b steps is M2[u][v].
Practice problem: 147B
Application 3: Nth Fibonacci number in O(log n)
You are given Q (1 <= Q <= 10^5) queries. For each query you are given an integer N (1 <= N <= 10^9) and you are to find the Nth number in the Fibonacci sequence.
Solution
In case you don't know how to multiply matrices, you can read about it here. After you know how to multiply matrices you can easily calculate the power of a matrix in O(log n) just like you do with integers.
And please point out any English mistakes or sentences that can be improved. :D
• +39
» 4 years ago, # | ← Rev. 5 → +23 Note that Application 3 is a special case of linear homogeneous recurrences relations with constant coefficients. If we have an order k relation of this kind:an = b0 * an - 1 + b1 * an - 2 + ... + bk - 1 * an - kWhere bi are the constant coefficients, and the first k ai terms are known.We can find the n-th term of the recurrence applying matrix exponentiation.First, lets call A a companion matrix:If we multiply this matrix with a vector, whose elements are the first k known terms, we should obtain:Now, if we need to obtain an we can apply matrix exponentiation:For example, to find the 4-th term of the recurrence an = 2 * an - 1 + 3 * an - 2 + an - 3 With a0 = 1, a1 = 2, a2 = 4As we can see, a3 = 2 * 4 + 3 * 2 + 1 * 1 = 15a4 = 2 * 15 + 3 * 4 + 1 * 2 = 44a5 = 2 * 44 + 3 * 15 + 1 * 4 = 137a6 = 2 * 137 + 3 * 44 + 1 * 15 = 421
• » » 4 years ago, # ^ | +3 The correct term for it is a companion matrix, not a recurrence matrix.
• » » » 4 years ago, # ^ | 0 Right! Thanks!
» 4 years ago, # | 0 For the second application you can get the minimum cost after maximum b steps by making M1[i][i] = 0 for every (1 <= i <= n).
» 10 months ago, # | 0 Trying to solve the 2nd problem Getting TLE on 31st test case Complexity is N^3 log(N) Log N for binary search Question: 147B Submission: Link Any idea of how to remove TLE?
• » » 10 months ago, # ^ | ← Rev. 5 → +3 Reason: if function requests T as argument there will be copy constructor call. In your sumbmission each call of mull, func, check_matrix will lead to copying matrix, which is extra $O(n^2)$.Solution:You really need create new matrix only as output of mull(). So you need:mull(int, const vector&, const vector&),func(ll, const vector&, const vector, ll),check_matrix(ll, const vector&)const T& tells that argument will be passed as reference to object, not copy. So there won't extra time. But within function you can't call any non-const method of argument.You also can rewrite func from recursive to iterate. This can decrees time too.P.S. Also your check functions is $O(n)$. Maybe you can leave this? Otherwise your solution would be $O(n^4logN)$. P.P.S. Forget it your complexity is $O(N^3log^2N)$ — first log for binary search, second log for multiplication. Maybe just your solution is wrong :).
• » » » 10 months ago, # ^ | 0 My new submission I know my solution is of N^3 (logN)^2. But this complexity solution passes Like this submission Solution with same complexity passes Any idea how?
• » » » » 10 months ago, # ^ | ← Rev. 3 → +3 Don't sure, but there is some memorization. AC solution precalculates all powers $MTRX^{2^i}, i = 0..log(n)$. if $power = 41$ there will be:$res = MTR * MTR^8 * MTR^32$. — 3 multiplications.In your variant it's:$res = MTR * MTR^40 = MTR * (MTR^20)^2 = MTR * ((MTR^10)^2)^2 = MTR * (((MTR^5)^2)^2)^2 = MTR * (((MTR * MTR^4)^2)^2)^2 = MTR * (((MTR * MTR^2^2)^2)^2)^2$. — 7 multiplications.So in your solution there are some extra work with same complexity.
• » » » » » 10 months ago, # ^ | 0 Yeah Got it:) Submissions uses different way of applying binary search Thanks | 2020-02-21T10:49:33 | {
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https://www.cut-the-knot.org/m/Algebra/ThreeElementSubsets.shtml | # Sums of the Elements of Three Element Subsets
### Problem
1. Can one divide the set $\{1,2,\ldots,96\}$ into $32$ subsets of three elements each, such that the sums of the three elements in the subsets are all equal?
2. Can one divide the set $\{1,2,\ldots,99\}$ into $33$ subsets of three elements each, such that the sums of the three elements in the subsets are all equal?
### Hint
To find the answer to the first question think of what that common sum could be. For the second question, try to think what makes the number $99$ is special for this problem. Try to think of perhaps smaller numbers with the same property.
### Solution to #1
The answer is "No", because if, the sums of the elements of the $32$ subsets were equal, then the sum of $1+2+\ldots +96=48\cdot 97$ would be divisible by $32$ which it is clearly not.
### Solution to #2
The sum of all integers from $1$ through $99$ is divisible by $33$ so that the argument that worked for the first part of the problem does not work here. Perhaps, this is an indication that the answer to the second part is "Yes"? If so, almost certainly $99$ is not a unique number that could be used in that problem. Trying to find the smallest number that would do the job, I came up with $9:$
$2+6+7=3+4+8=1+5+9=15!$
What made it work? I could observe several properties:
• The set $\{1,2,\ldots,9\}$ is the union of three successive subsets ordered by the magnitude of their elements:
$\{1,2,\ldots,9\}=\{1,2,3\}\cup \{4,5,6\}\cup\{7,8,9\}$
such that each of the three sums above contains one and only one element from each of the three subsets.
• The elements of the last subset $\{7,8,9\}$ are in arithmetic progression, with $1$ as the difference, suggesting that the union of the two other sets, i.e., $\{1,2,\ldots,6\}$ could be (and were) split into pairs whose sums form an arithmetic progression with $1$ as the difference.
So, had the number of elements in the given set been $9$ and not $99$ the problem would have already been solved. As it is, we have to plough away further on. What other number besides $9$ would work? Obviously, the number needs to be divisible by $3$. Could it be $12?$ No - for exactly same reason $96$ did not work. What about $15?$ There is a chance that it does work. Let's try following in the footsteps of the solution for the $9$-element set:
$\{1,\ldots,15\}=\{1,\ldots,5\}\cup\{6,\ldots,10\}\cup\{11,\ldots,15\}$
If the problem is solvable in this case, the common sums of the three element subsets would be $\displaystyle\frac{15\cdot 16}{2}:5=24.$ Therefore, the task is to form pairs of integers (one from each of the two first subsets) whose sums are in the arithmetic progression:
$24-11=13,\space24-12=12,\ldots,24-15=9.$
The smallest sum in the first example included $1$ so this is where we start: $1+8=9.$ The next smallest sum contained $3$ - try $3+7=10$. Let's continue: $5+6=11,$ $2+10=12,$ and $4+9=13.$ We are done! Here is a solution:
\begin{align} \{1,2,\ldots,15\} &= \{1,8,15\}\\ & \cup\space\{3,7,14\}\\ & \cup\space\{5,6,13\}\\ & \cup\space\{2,10,12\}\\ & \cup\space\{4,9,11\}, \end{align}
with the common sum being $24.$ Now, the question is to locate a pattern.
I bet that for the number of elements $3n,$ with $n$ odd the problem is always solvable. Obviously it is not solvable if $n$ is even, as above. The common sum of the three element subsets ought to be $\displaystyle\frac{3n(3n+1)}{2}:n=\frac{3(3n+1)}{2}.$ The three subsets to pick one element from each are
$\{1,\ldots,n\},$ $\{n+1,\ldots,2n\},$ $\{2n+1,\ldots,3n\}.$
We start with $1$ from the first and $3n$ from the third which leaves
$\displaystyle\frac{3(3n+1)}{2}-1-3n=\frac{3n+1}{2}=\frac{(n+1)+2n}{2},$
for the second set. This is exactly its midpoint. The next pick is $3$ from the first set, $\displaystyle\frac{3n+1}{2}-1$ from the second, and $3n-1$ from the third. We can proceed in that manner only till we reach $n$ in the first set. This will correspond to $n+1$ in the second, and $\displaystyle 3n-\frac{n-1}{2}=\frac{5n+1}{2}=(2n+1)+\frac{n-1}{2}$ from the third. From here we start with $2$ from the first set, $2n$ from the second, and $\displaystyle \frac{5n+1}{2}-1$ from the third. You can check that this process will carry us through.
Coming back to the original problem of $n=33,$ the following is the answer:
\begin{align} 150 &= 1+50+99 \\ &= 3+49+98 \\ &\dots \\ &= 31 + 35 + 84\\ &= 33 + 34 + 83\\ &= 2 + 66 + 82\\ &= 4 + 65 + 81\\ &\dots \\ &= 30 + 52 + 68\\ &= 32 + 51 + 67. \end{align}
### More solutions
Yuriy Kazakov found three more solutions and summarized all four in the following table:
### Acknowledgment
This is a problem from the 2008 China Girl's Mathematical Olympiad [Xiong Bin, pp 94-96].
### References
1. Xiong Bin, Lee Peng Yee, Mathematical Olympiads in China (2009-2010). Problems and Solutions, World Scientific, 2013 | 2019-04-23T11:54:31 | {
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https://computergraphics.stackexchange.com/questions/5735/logarithmic-spiral-with-equal-vertex-spacing-what-equations/5737 | Logarithmic spiral with equal vertex spacing, what equations?
I need to draw a logarithmic spiral (or close approximation) whose vertices are equally spaced, such that the lines between any two consecutive vertices are of equal length.
(Actually, that spiral is just the basis for something a little more complex, but equally spaced incrementation is what matters right now.)
The basis for the spiral is $$r=e^{0.25 \theta}$$ where $r$ is the radius and $\theta$ is the angle from the $x$ axis.
So it's easy to step around the spiral by a given angle, but obviously the actual length of any line between two vertices (defined by change in angle $\Delta\theta$) will vary with the radius.
How can I step along the arc of the spiral by a set length (say $0.1$ arbitrary units) and get the right coordinates (polar or Cartesian)?
• Does this need to be fast or can it be done offline in advance? Oct 16 '17 at 19:59
• It doesn't need to be especially fast. Oct 16 '17 at 20:00
• Do the segments need to be exactly equal length (within limitations of floating point etc) or can they be approximately equal? Like equal within some epsilon? Oct 16 '17 at 20:02
• Interesting point. Some approximation is probably okay, but I'd prefer to maximise precision. Oct 16 '17 at 20:04
• Do you want equal length segments or segments that represent equal arc lengths? Oct 17 '17 at 0:44
Since a logarithmic spiral is defined by
$r=e^{a\cdot\theta}$,
the inverse of the equation is this:
$\theta=\frac{\ln{r}}{a}$.
If we want to be able to control our step value, we can multiply it by a scalar ($a\cdot k$) before taking the logarithm, like so:
$\theta=\frac{\ln (ak\cdot r)}{a}$
Therefore, if we take the natural log of theta multiplied by the scalar and $a$, then divide the whole thing by $a$ before plugging it in to the equation, we will get equally stepped vertices on the logarithmic spiral. Image:
I generated this with some JavaScript code, which you can find in this JSFiddle.
You can also try it out interactively on Desmos.
This looks good initially, but let's analyze what's going on here a little deeper.
Here is a list of the distance between each consecutive pair of points, starting at the origin and spiraling outward:
20.6834877876017
25.8265879847751
27.3020264943519
27.9155365639631
28.2274404549991
28.4072329605331
28.5201842104742
28.5957343926866
28.6487385499231
28.6873487186944
28.7163404861061
28.7386617798632
28.7562121095165
28.7702600892245
28.7816791255158
28.7910864406300
28.7989282185451
28.8055334775293
28.8111491387549
28.8159634339811
28.8201218817722
28.8237384111142
28.8269032465565
28.8296885891879
28.8321527704665
28.8343433306399
28.8362993285731
28.8380530946981
28.8396315754034
28.8410573741728
28.8423495651939
28.8435243345428
28.8445954894748
28.8455748659368
28.8464726569062
28.8472976786488
28.8480575879711
28.8487590604974
28.8494079377702
28.8500093492502
28.8505678139929
28.8510873257846
28.8515714247378
28.8520232577561
28.8524456298024
28.8528410475339
28.8532117565801
28.8535597734869
28.8538869132081
28.8541948128041
28.8544849519687
28.8547586708355
28.8550171854833
28.8552616014621
28.8554929256347
28.8557120765606
28.8559198936319
28.8561171451194
28.8563045352784
28.8564827106597
28.8566522656712
28.8568137475608
28.8569676608282
28.8571144711722
28.8572546090145
28.8573884726640
28.8575164311457
28.8576388267439
28.8577559773038
28.8578681782850
28.8579757046362
28.8580788124715
28.8581777406078
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The first data point is an outlier caused by the spiral almost "lapping" itself. As the spiral progresses, we can see the the distances get closer and closer to the same distance.
This is because we are sampling the curve in increments that cause equal lengths along the curve, not equal lengths between consecutive points. As the spiral gets bigger and bigger, the curve will get shallower and shallower, so the distances will get closer and closer to each other.
• That being said, this approach does mean that the value of a influences both line length and turn rate. I really need those to be separate terms. Oct 17 '17 at 11:01
• @SolarGranulation I just added that, and updated the JSFiddle and Desmos examples to reflect that. Oct 18 '17 at 0:10 | 2021-09-21T05:28:15 | {
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https://www.physicsforums.com/threads/a-question-regarding-inverse-functions.658255/ | # A question regarding inverse functions
1. Dec 11, 2012
### christian0710
Hi I'm reading a book called Calculus lifesaver, and in the book they state that the inverse of a function f(x)= x3 is the same as f-1(x)=3√x and is the same as f-1(y)=3√y
So I did a test, with a simpler function and I can't see how this is true
If I have a function f(x)= 2*x
Then the inverse would be y=2x→ x=y/2
So It’s correct then to write f-1(x)= x/2
And Is it wrong or correct to write?
f-1(y)= y/2 Because this is not the inverse of y, this is the inverse of X, so the right thing would be to either call it f-1(x)= x/2 OR f(y)= y/2 right?? If i graph it,
f(y)= y/2 gives us the same graph as f(x)= 2*x and f-1(x)= x/2 gives us a different graf which is the inverse, so It must be an error to state that f-1(x)=3√x is the same as f-1(y)=3√y Or am i wrong?
2. Dec 11, 2012
### Staff: Mentor
Variable names have no special meaning. As long as you use them consistently, it is fine.
f(r)=2r and f(y)=2y are the same as f(x)=2x.
3. Dec 11, 2012
### christian0710
I see but if we have a xy cordinate system, the have a function f(y)=2y and f(x)=2x we get two different graphs (don't we) because the one is a function of y and the other a function of x? That's my confusion.
4. Dec 11, 2012
### Staff: Mentor
You get different graphs, but the functions are the same. They are just drawn in a different way.
5. Dec 11, 2012
### christian0710
But how can the functions be the same if they have different graphs? a function only has one inverse, so how can it have two graphs? One of them must be correct and the other wrong? If i graph on a xy axis f(y)=2y and f(x)=2x i get two different graphs.
6. Dec 11, 2012
### Staff: Mentor
The coordinate system that you use to graph your functions usually has its horizontal axis labelled as "x" and its vertical axis labelled as "y". When a function is in the form of y = f(x), y is the dependent variable and x is the independent variable.
You know what the graph of y = f(x) = x3 looks like. The equation x = g(y) = $\sqrt[3]{y}$ represents the inverse relationship between x and y, and this equation is equivalent to the equation y = x3, which means that each point (x, y) that is on one graph is also on the other. The only difference is that one function produces a y value for a given input x value, and the other function produces an x value for a given input y value.
Using inverse function notation, these equations say exactly the same thing and have exactly the same graph:
y = f(x) = x3
x = f-1(y) = $\sqrt[3]{y}$
If I switch the variable names in the function I'm calling g, above, I get y = g(x) = $\sqrt[3]{x}$. Switching x and y has the effect of reflecting each point (x, y) on the original graph across the line y = x. IOW, the point (x, y) gets reflected to (y, x). If that's hard to follow, the point (2, 8) gets reflected to (8, 2).
Our transformed (by reflection) function is now y = g(x) = $\sqrt[3]{x}$. Because of the inverse relationship between f and g, we often write g(x) as f-1(x).
IMO, this switching gets a lot more attention in Precalc classes and textbooks than it deserves, because it distracts students from the more important concept that a function and its inverse are simply two ways to look at the relationship between x and y on a graph.
A function f and its inverse f-1 satisfy these relationships:
f (f-1(y)) = y, and
f-1(f(x)) = x
as long as y is in the domain of f-1 and x is in the domain of f.
7. Dec 11, 2012
### christian0710
Man you deserve a Nobel prize for your explanations! This is great. So just to make sure I understand you let me reiterate what you just said:
So the inverse of f(x) y=x3 is obviously x(y)=f-1(y)=3√y and therefore by switching the variable x and y in the inverse function of y,
f-1(x)=3√x you get the inverse function of x. And this function is a mirror image of f(x) in the line line y=x.
So equating f-1(x)=3√x with f-1(y)=3√y is not completely correct, because it will give us two different graphs, it would however be correct to say that f-1(x)=3√x has an inverse relationship to y(x)=y3 and gives the same graph.
8. Dec 11, 2012
### christian0710
Here is a small excerpt from the book, i highlighted what i misunderstand.
#### Attached Files:
• ###### inverse.JPG
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28.6 KB
Views:
75
9. Dec 11, 2012
### Vargo
Howdy,
In what your are writing, you are identifying a function too much with its graph. Or, rather, you are putting too much meaning in the specific variables x and y. This question is about functions, not variables. Think of a function as something numerical. It takes in an input and returns an output. The given function takes an input x and returns an output x^3. In other words, the output is the cube of the input (the letter we use to label the input is irrelevant).
The inverse operation "undoes" what the original operation did. So the inverse of the previous function would take a numerical input and return the cube root. Again, the letter we use to label the input is irrelevant. So whether we write the inverse function in terms of an input x or an input y is irrelevant as long as the output is the cube root of the input.
10. Dec 11, 2012
### christian0710
Thank you, You are right I get too caught up in definitions because I fear that I misunderstand something, I see now that the output of any variable will still be a value plotted on the y axis, even f-1(y)=y2 :)
11. Dec 11, 2012
### Staff: Mentor
It's not even partially correct.
Here's the corrected version. y = f(x) = x3 is equivalent to x = f-1(y) = $\sqrt[3](y)$ - same graph.
At each point on this graph, the first equation tells you how to get the y value if you know the x value. The second equation tells you how to get the x value if you know the y value.
Here's a different example that shows how switching the variables can be a major distraction. The example involves converting Celsius temperatures to Fahrenheit and vice versa.
F = (9/5)C + 32
If we solve the equation above for C, we get
F - 32 = (9/5)C
(5/9)(F - 32) = C
so C = (5/9)(F - 32)
The first equation above gives you the temp in °F if you know the Celsius temp, so we have F = f(C) in function notation.
The last equatio above gives you the temp in °C if you know the Fahrenheit temp, so we have C = f-1(F)
It would be extremely foolish to switch C and F in either formula...
12. Dec 11, 2012
### christian0710
Perfect then i understand:
Inverse Relationship between x and y:
f(x)=y=x3
x(y)=f-1(y)=3√y
Inverse function of f(x)
f(x)=y=x3
f-1(x)=3√x
And thank you for verifying that the definition from the book was incorrect.
13. Dec 11, 2012
### Staff: Mentor
Are you talking about the part that you highlighted? If so, there's nothing wrong with it. They are saying that f-1(y) = $\sqrt[3]{y}$, but if you switch letters, you get f-1(x) = $\sqrt[3]{x}$.
Last edited: Dec 11, 2012
14. Dec 11, 2012
### christian0710
In the last part you mean f-1(x) = 3√x instead of √y right? :)
Okay, so they suggest this f-1(x) = 3√x instead of f-1(y) = 3√y because it might fool you to think that the input would be read on the y axis and the output f(y)=x would be read on the x-axis.
15. Dec 11, 2012
### Staff: Mentor
Yep, you got me. Copy/paste error.
It's probably more because people think of functions being "of x".
16. Dec 11, 2012
### christian0710
Thanks again. I appreciate your help! | 2017-08-20T18:18:29 | {
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http://math.stackexchange.com/questions/448715/i-read-that-ordinal-numbers-relate-to-length-while-cardinal-numbers-relate-to-s | # I read that ordinal numbers relate to length, while cardinal numbers relate to size. How do 'length' and 'size' differ?
I read that ordinal numbers relate to length, while cardinal numbers relate to size. How do 'length' and 'size' differ? Note : I am an absolute novice, and I'm having a little trouble visualizing ordinal numbers.
-
Where did you read that (esp. ordinals=length)? Better: Ordinals refer not only to the size of a set, but also how its elements are ordered. – Hagen von Eitzen Jul 21 '13 at 11:37
This is the link:ls.poly.edu/~jbain/Cat/lectures/07.OrdsandCards.pdf. – Novice Jul 21 '13 at 11:38
If the line to the bathroom is finite, then the length and the number of people in that line is the same number.
However suppose that the line is infinite, but every person has to wait a finite time before entering the bathroom. Now comes another guy, and he has to wait until all those before him get in before he can use the facilities.
In terms of cardinality we didn't increase the line, but its length is longer by one person now.
Formally, the infinite line where every person has to wait a finite amount of time is $\Bbb N$ with the order $\leq$. Indeed every natural number has only finitely many predecessors. The next line is $\Bbb N\cup\{\bullet\}$, where $\bullet$ is some new element which is not a natural number and we decree that it is larger than all the natural numbers themselves.
So the order type of the new set is longer because it has an initial segment of order type $\Bbb N$, but another point above that. However the cardinality is the same. To see that note that the following map is a bijection between the two sets: $$f(n)=\begin{cases}\bullet & n=0\\ n-1 & n\neq 0\end{cases}$$
So we have a strictly longer line (and this poor guy who has to wait an infinite time before he can pee), but that line has the same number of people.
-
That was really illuminating, finally understood the difference between ordinality and cardinality. Thanks a lot :) – Novice Jul 21 '13 at 11:44
@Novice: You're welcome. – Asaf Karagila Jul 21 '13 at 12:20
The sanitary situation would be even worse if people were to enter largest number first. :) – Hagen von Eitzen Jul 21 '13 at 12:26
Example $6$ at the bottom of the first page of that PDF is a good example of the difference between size, on the one hand, and length and shape on the other, in the sense in which its author means those terms. Consider the following two ways of arranging the natural numbers. First, in their usual order:
$$\begin{array}{cc} 0&1&2&3&4&5&6&7&8&9&\cdots \end{array}\tag{1}$$
Then with the even numbers first, in their usual order, followed by the odd numbers in their usual order:
$$\begin{array}{cc} 0&2&4&6&8&\cdots&1&3&5&7&9&\cdots\\ \end{array}\tag{2}$$
It’s the same set $\Bbb N$ in both cases, so the in some sense two arrangements are certainly the same size. On the other hand, if we set them side by side, starting at the lefthand end, something very different happens:
$$\begin{array}{cc} 0&1&2&3&4&\cdots&n&\cdots&|\\ \hline\\ 0&2&4&6&8&\cdots&2n&\cdots&|&1&3&5&7&9&\cdots\\ \end{array}\tag{3}$$
The parts to the left of the vertical line match up perfectly, and there’s nothing in the top row to match up with the bottom row. Even though the top and bottom lines are arrangments of exactly the same set $\Bbb N$, in terms of the arrangement the bottom line is somehow twice as long as the top line. And yes, the unmatched part of the bottom line really does have the same shape as the matched part, as you can see from the following arrangement:
$$\begin{array}{cc} 0&1&2&3&4&5&6&7&\cdots&n&\cdots\\ \hline 0&2&4&6&8&10&12&14&\cdots&2n&\cdots\\ \hline 1&3&5&7&9&11&13&15&\cdots&2n+1&\cdots \end{array}\tag{4}$$
In more formal terminology, what the author calls size is known as cardinality. Two sets are said to have the same cardinality if there is a way to pair them up one for one, with nothing left over in either set; in technical terms, this means that there is a bijection between the two sets, a function that is one-to-one and onto. Clearly $\Bbb N$ can be paired up with itself in this fashion no matter how we rearrange it.
Length and shape are informal ways of talking about what is properly called the order type of a linear arrangement of a set. Let $E$ be the set of even natural numbers and $O$ the set of odd natural numbers. $(4)$ shows that we can pair up $\Bbb N$ with each of $E$ and $O$, and $E$ and $O$ with each other, in a one for one fashion, so these three sets all have the same cardinality (‘size’). On the other hand, we can pair up $\Bbb N$ with $E$ and then stick $O$ (in its usual order) after $E$, as in $(3)$ to get an arrangement that looks like two copies of $\Bbb N$ placed end to end:
$$\begin{array}{cc} 0&1&2&3&4&\cdots&n&\cdots&|&0&1&2&3&4&\cdots\\ \hline\\ 0&2&4&6&8&\cdots&2n&\cdots&|&1&3&5&7&9&\cdots\\ \end{array}\tag{5}$$
In terms of order type the arrangement $(2)$ is twice as long as $(1)$: it can be matched up one for one, preserving the order, with two copies of $(1)$ placed end to end, as in $(5)$.
This is all very informal, of course, but that may be helpful in getting started with the ideas.
- | 2015-01-28T04:46:20 | {
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http://math.stackexchange.com/tags/number-systems/hot | Tag Info
Hot answers tagged number-systems
191
Short answer: your confusion about whether ten is special may come from reading aloud "Every base is base 10" as "Every base is base ten" — this is wrong; not every base is base ten, only base ten is base ten. It is a joke that works better in writing. If you want to read it aloud, you should read it as "Every base is base one-zero". You must ...
73
Many people believe that since humans have $10$ fingers, we use base $10$. Let's assume that the Martians have $b$ fingers and thus use a base $b$ numbering system, where $b \neq 10$ (note that we can't have $b=10$, since in base $10$, $x=8$ shouldn't be a solution). Then since the $50$ and $125$ in the equation are actually in base $b$, converting them to ...
66
The magic of the number 10 comes from the fact that "1" is the multiplicative unit and "0" is the additive unit. The first two-digit-number in positional notation is always 10 and also always denotes the number of digits.
55
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45
You may also find of interest some more general results besides the mentioned Frobenius Theorem. Weierstrass (1884) and Dedekind (1885) showed that every finite dimensional commutative extension ring of $\mathbb R$ without nilpotents ($\rm\:x^n = 0 \ \Rightarrow\ x = 0\:$) is isomorphic as a ring to a direct sum of copies of $\rm\:\mathbb R\:$ and ...
41
Alas, there are no algebraically coherent "triplexes". The next step in the construction as has been said already are "quaternions" with 4 dimensions. Many young aspiring mathematicians have tried to find them since Hamilton in the 19th century. This impossibility links geometric dimensionality, fundamental properties of polynomial equations, algebraic ...
40
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32
One way to convert any decimal fraction to base $16$ is as follows (taking $\pi$ as an example).$$\pi=\color{blue}3.141592...$$ Take the whole number part and convert it to base $16$ as usual. In this case $\color{blue}3$ will remain as $3$. So we have so far got $3.14159..._{10}=\color{red}{3...._{16}}$ This now leaves us with $0.141592...$ - ...
28
If we are working in base $b$ (we must have $b\gt3$), then $0.3333\ldots$ is $$0.3333\ldots = \frac{3}{b} + \frac{3}{b^2} + \frac{3}{b^3}+\cdots$$ Since $$\sum_{n=1}^{\infty}\frac{3}{b^n} = \frac{3}{b}\sum_{n=0}^{\infty}\frac{1}{b^n} = \frac{3}{b}\left(\frac{1}{1-\frac{1}{b}}\right) =\frac{3}{b-1},$$ then...
28
There are two terminologies that I'm familiar with. Sometimes, the part to the right of the decimal (cents) is called the mantissa, and the part to the left (dollars, in your metaphor), is called the characteristic. But I also like the generic terms integer-part and fractional-part. It's what I and those with whom I do research call them (who uses the work ...
28
Suppose that the decimal is $x=a.d_1d_2\ldots d_m\overline{d_{m+1}\dots d_{m+p}}$, where the $d_k$ are digits, $a$ is the integer part of the number, and the vinculum (overline) indicates the repeating part of the decimal. Then $$10^mx=10^ma+d_1d_2\dots d_m.\overline{d_{m+1}\dots d_{m+p}}\;,\tag{1}$$ and $$10^{m+p}x=10^{m+p}a+d_1d_2\dots d_md_{m+1}\dots ... 27 The short answer to your question is that by definition we only allow real numbers to have finitely many digits before the decimal point. There are very good reasons for this: Formally, we can think of a number as a finite sequence of digits x_0,\ x_1, \ \ldots , x_N, where the number x is equal to$$x=\sum_{n=0}^Nx_n10^n$$For example, the number 126 ... 27 actually numbers from 11 to 16 are quite regular in French (and in Italian) too: they just are a derivation from Latin. | French | Italian | Latin un | on·ze | un·dici | un·decim deux | dou·ze | do·dici | duo·decim trois | trei·ze | tre·dici | tre·decim quatre | quator·ze | quattor·dici | ... 27 13 fingers. Translate 5x^2-50x+125 into base-b:$$ 5x^2-(5b)x+(b^2+2b+5) $$Since this has roots x=5 and x=8 we must have$$ 5x^2-(5b)x+(b^2+2b+5)=k(x-5)(x-8)=kx^2-13kx+40k $$so, equating coefficients,$$ 5=k,\quad 5b=13k,\quad b^2+2b+5=40k $$and so b=13. It's easy to check that the last equation is satisfied as well. Perhaps the Martians had ... 26 Hint: if we multiply 0.33333\ldots by 5 then we get 0.(15)(15)(15)(15)(15)\ldots. Compare that to what happens when we multiply the same by 3: 0.99999\ldots, and its interpretation in decimal. 26 Your intuition is correct for instance for all b > 2, \frac{1}{b-1} is not going to have a finite representation, and will have the representation \frac{1}{b-1} = 0.1111111...._b. Eg, \frac{1}{9} = .11111111 in base 10. 24 Actually, if you go back in time a bit in English, you'll realise that English was 'strange' too: Four score and seven years ago our fathers brought forth on this continent a new nation, conceived in liberty, and dedicated to the proposition that all men are created equal. (The Gettysberg Address, 1863) Now if you were to translate that into French in ... 23 Expanding on the comment by J.M., let me quote from the (highly recommended) book by Georges Ifrah The Universal History of Numbers (Wiley, 2000, pp. 21-22): Traces of the anthropomorphic origin of counting systems can be found in many languages. In the Ali language (Central Africa), for example, "five" and "ten" are respectively moro and mbouna: moro ... 22 Every finite-dimensional division algebra over \mathbb{R} is one of \mathbb{R}, \mathbb{C} or \mathbb{H}. This is what is called the Frobenius Theorem. You may refer to here for details. 22 In an interview, you can impress the interviewer, by mentally calculating and determining the result. As other answers have mentioned, you need to express the equation in base different from 10 and then equate it with the roots of the equation. From the options, its clear that the base is greater than 10. That means 5 and 8 are unit digits in some ... 21 As you mentioned,$$6 = {\color{red}1}\cdot 2^2+ {\color{red}1}\cdot 2^1+{\color{red}0}\cdot 2^0 = {\color{red}{110}}_B.$$Analogously$$\frac{1}{4} = \frac{1}{2^2} = {\color{red}0}\cdot2^0 + {\color{red}0}\cdot 2^{-1} + {\color{red}1}\cdot 2^{-2} = {\color{red}{0.01}}_B.$$Edit: These pictures might give you some more intuition ;-) Here \frac{5}{16} = ... 20 The correct answer is (a) 10. There is no comment which number system the given answers refer to. As all other numbers refer to the Martian number system, we can safely assume the answers refer to the Martian number system as well. 20 Of course it seems natural to you; you grew up in the modern world, where everyone accepts zero. More importantly, people now accept the abstract concept of numbers and are capable of divorcing them from the things that they represent. This is a sophisticated point of view. From a more naive point of view, a number is a property of a collection of ... 20 The natural numbers can be defined by Peano's Axioms (sometimes called the Peano Postulates): Zero is a number. If n is a number, the successor of n is a number. zero is not the successor of a number. Two numbers of which the successors are equal are themselves equal. (induction axiom.) If a set S of numbers contains zero and also the successor of every ... 20 I would like to expand on Trevor Wilson's answer. Base-b representation of integers is rooted in the fact that, for any non-negative integer n, there is a unique representation of n in the form$$n = \sum_{i=0}^\infty a_ib^i where $0 \le a_i < b$. For example, when $b$ is 3, and $n$ is 47, the unique solution has $a_0 = 2, a_1 = 0, a_2 = 2, a_3 ... 18 I think the answer here might be, that the guys who thought base 10 was a good idea had the largest sticks. If one trusts the wikipedia, the Babylonians had a base 60 system, which can still be felt today with this "60 minutes in an hour" nonesense, and a (related) base 12 system was widely in use too. There are still unique words for "eleven" and "twelve", ... 18 The existence of a bijection between the class of ordinals$On$and the class of surreal numbers$No$is independent of the axioms of set theory. There are several interesting possibilities: If ZFC is consistent, then there is a model of ZFC in which there is a definable such bijection. This is true in Goedel's constructible universe$L$, for example, for ... 18 To find out the number of zeroes at the end of that huge number, we just have to find out the exponent of$10$in it. Since$10=2\cdot 5$, we can just find out the exponent of$5$in that number as the exponent of$2$will be more than$5$. All the numbers which have$5\$ as their factor will be counted. These numbers would be ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2015-05-25T03:46:47 | {
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http://mathematica.stackexchange.com/questions/23559/combinations-which-do-not-have-elements-in-common | # Combinations which do not have elements in common
I can choose 2 letters from the four letters $\{A,B,C,D\}$ in 6 combinations using the combination formula
$$\frac{n!}{ r! (n-r)!}$$
{A, B}, {A, C}, {A, D}, {B, C}, {B, D}, {C, D}
Consider now the collection of all subsets of a set of $n$ elements and restrict to those having exactly cardinality $k$, such as $k=2$ above. How can I find all combinations of groups of 2 or 3 or ... or $n$ (should be dynamic) of these which do not have letters in common? So, for example, I am expecting the result for combinations of two groups of $k=2$ elements to be the three collections below:
{{A, B}, {C, D}},
{{A, C}, {B, D}},
{{A, D}, {B, C}}
This is just a small example but I would like the algorithm to work for a large number of sets (I have more than 35000 sets).
-
The number of letters change or you will just use 4? – Spawn1701D Apr 18 '13 at 18:27
Aren't they the permutations, properly splitted? – belisarius Apr 18 '13 at 18:33
@belisarius I don't think so. For example, splitting ABCD into groups of two gives the same partition as splitting BACD, ABDC, BACD, CDAB, etc. – whuber Apr 18 '13 at 18:49
user, this may be a duplicate of your question, though phrased differently: mathematica.stackexchange.com/q/3044/121 -- please review that, and if it is not, make clear in this question why not. – Mr.Wizard Apr 19 '13 at 3:22
Yes, it is same, i couldn't find it. But I am not from pure mathematics, I am a programmer so want the programming algorithm to solve this problem. – user1191081 Apr 19 '13 at 12:55
It is desired to find all collections $P = \{A_1, A_2, \ldots, A_m\}$ of $m$ disjoint $k$-subsets of a finite set $X$. Such collections consist of two types: those which include a specified element of $X$ (say $a$) and those that do not. If $a \in A_j$ for some $j$ then we might as well reindex the elements of $P$ so that $j=1$. All possibilities for $A_1$ are found by adjoining all possible $k-1$-element subsets of $X - {a}$ to $a$, whereas all possibilities for the $A_2, \ldots, A_m$ are found recursively for $X - A_1$ with $m=1$. Otherwise, $a$ is not in any $A_j$ and we are left to find similar collections for $X - {a}$.
This gives the following recursive solution. It's not fast but it's not slow, either--see the timing example below.
Please note that the number of solutions grows very, very quickly with $k$ and $m$. Their number is computed with combinationsCount, which uses Multinomial to obtain
$$\frac{n!}{k!^m (n - m k)! m!}.$$
The numerator is the size of the permutation group on $X$ while the denominator is the size of the stabilizer of any collection $P$: it consists of all permutations that (a) separately permute the elements of the $A_j$, (b) permute the remaining elements in $X - \cup_j A_j$, and (c) then permute the $A_j$ among themselves.
combinations[x_List, k_Integer, 0] := {{}};
combinations[x_List, k_Integer, m_Integer] := {};
combinations[x_List, k_Integer, m_Integer] /;
k m <= Length[x] && k >= 1 && m >= 1 := Block[{y, with, without},
y = Prepend[#, First@x] & /@ Subsets[Rest@x, {k - 1}];
with = Flatten[Table[Prepend[#, z] & /@ combinations[Complement[x, z], k, m - 1], {z, y}], 1];
without = combinations[Rest@x, k, m];
with~Join~without
];
combinationsCount[n_, k_, m_] := Multinomial @@ Append[ConstantArray[k, m], n - k m] / m!
### Examples
combinationsCount[6, 3, 2]
$10$
MatrixForm /@ combinations[{a, b, c, d, e, f}, 3, 2]
$\begin{array}{lllll} \left( \begin{array}{ccc} a & b & c \\ d & e & f \end{array} \right) & \left( \begin{array}{ccc} a & b & d \\ c & e & f \end{array} \right) & \left( \begin{array}{ccc} a & b & e \\ c & d & f \end{array} \right) & \left( \begin{array}{ccc} a & b & f \\ c & d & e \end{array} \right) & \left( \begin{array}{ccc} a & c & d \\ b & e & f \end{array} \right) \\ \left( \begin{array}{ccc} a & c & e \\ b & d & f \end{array} \right) & \left( \begin{array}{ccc} a & c & f \\ b & d & e \end{array} \right) & \left( \begin{array}{ccc} a & d & e \\ b & c & f \end{array} \right) & \left( \begin{array}{ccc} a & d & f \\ b & c & e \end{array} \right) & \left( \begin{array}{ccc} a & e & f \\ b & c & d \end{array} \right) \end{array}$
combinationsCount[13, 3, 4]
$200200$
AbsoluteTiming[Length@combinations[Range@13, 3, 4]]
$\{5.9153384,200200\}$
- | 2015-05-28T10:11:10 | {
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http://stcc.org/s64cscaw/6e34a5-euclidean-distance-in-r | While as far as I can see the dist() function could manage this to some extent for 2 dimensions (traits) for each species, I need a more generalised function that can handle n-dimensions. Required fields are marked *. euclidean: Usual distance between the two vectors (2 norm aka $$L_2$$), $$\sqrt{\sum_i (x_i - y_i)^2}$$. The Euclidean distance between two vectors, A and B, is calculated as: To calculate the Euclidean distance between two vectors in R, we can define the following function: We can then use this function to find the Euclidean distance between any two vectors: The Euclidean distance between the two vectors turns out to be 12.40967. Now what I want to do is, for each possible pair of species, extract the Euclidean distance between them based on specified trait data columns. The Pythagorean Theorem can be used to calculate the distance between two points, as shown in the figure below. Learn more about us. > Now I want to calculate the Euclidean distance for the total sample > dataset. Submitted by SpatialDataSite... on Wed, 12/10/2011 - 15:17. The Euclidean distance is computed between the two numeric series using the following formula: The two series must have the same length. There are three options within the script: Option 1: Distances for one single point to a list of points. The Euclidean distance output raster The Euclidean distance output raster contains the measured distance from every cell to the nearest source. version 0.4-14. http://CRAN.R-project.org/package=proxy. canberra: sum(|x_i - y_i| / (|x_i| + |y_i|)). Given two sets of locations computes the full Euclidean distance matrix among all pairings or a sparse version for points within a fixed threshhold distance. The Euclidean Distance procedure computes similarity between all pairs of items. It can be calculated from the Cartesian coordinates of the points using the Pythagorean theorem, and is occasionally called the Pythagorean distance. 4. x2: Matrix of second set of locations where each row gives the coordinates of a particular point. Euclidean distances. Using the Euclidean formula manually may be practical for 2 observations but can get more complicated rather quickly when measuring the distance between many observations. Contents Pythagoras’ theorem Euclidean distance Standardized Euclidean distance Weighted Euclidean distance Distances for count data Chi-square distance Distances for categorical data Pythagoras’ theorem The photo shows Michael in July 2008 in the town of Pythagori A euclidean distance is defined as any length or distance found within the euclidean 2 or 3 dimensional space. Euclidean distance is a metric distance from point A to point B in a Cartesian system, and it is derived from the Pythagorean Theorem. proxy: Distance and Similarity Measures. This video is part of a course titled “Introduction to Clustering using R”. Multiple Euclidean Distance Calculator R-script. In rdist: Calculate Pairwise Distances. 2) Creation of Example Data. Because of that, MD works well when two or more variables are highly correlated and even if their scales are not the same. Euclidean distance may be used to give a more precise definition of open sets (Chapter 1, Section 1). The matrix m gives the distances between points (we divided by 1000 to get distances in KM). Euclidean distance is also commonly used to find distance between two points in 2 or more than 2 dimensional space. This script calculates the Euclidean distance between multiple points utilising the distances function of the aspace package. The need to compute squared Euclidean distances between data points arises in many data mining, pattern recognition, or machine learning algorithms. Determine both the x and y coordinates of point 1. Looking for help with a homework or test question? logical indicating if object should be checked for validity. (Definition & Example), How to Find Class Boundaries (With Examples). Details. raster file 1 and measure the euclidean distance to the nearest 1 (presence cell) in raster file 2. Euclidean distance matrix Description. This distance is calculated with the help of the dist function of the proxy package. #calculate Euclidean distance between vectors, The Euclidean distance between the two vectors turns out to be, #calculate Euclidean distance between columns, #attempt to calculate Euclidean distance between vectors. In the example below, the distance to each town is identified. You can compute the Euclidean distance in R using the dist () function. Statology is a site that makes learning statistics easy by explaining topics in simple and straightforward ways. We can therefore compute the score for each pair of nodes once. Arguments object. Im allgemeineren Fall des -dimensionalen euklidischen Raumes ist er für zwei Punkte oder Vektoren durch die euklidische Norm ‖ − ‖ des Differenzvektors zwischen den beiden Punkten definiert. Euclidean distance is the basis of many measures of similarity and is the most important distance metric. The computed distance between the pair of series. dist Function in R (4 Examples) | Compute Euclidean & Manhattan Distance . To calculate the Euclidean distance between two vectors in R, we can define the following function: euclidean <- function (a, b) sqrt (sum ((a - b)^2)) We can then use this function to find the Euclidean distance between any two vectors: Numeric vector containing the first time series. I would like the output file to have each individual measurement on a seperate line in a single file. I am very new to R, so any help would be appreciated. Obviously in some cases there will be overlap so the distance will be zero. First, if p is a point of R 3 and ε > 0 is a number, the ε neighborhood ε of p in R 3 is the set of all points q of R 3 such that d(p, q) < ε. In short, all points near enough to a point of an open set … any R object that can be made into one of class "dendrogram".. x, y. object(s) of class "dendrogram".. hang. More precisely, the article will contain this information: 1) Definition & Basic R Syntax of dist Function. This distance is calculated with the help of the dist function of the proxy package. To calculate distance matrices of time series databases using this measure see TSDatabaseDistances. The Euclidean distance between two vectors, A and B, is calculated as: Euclidean distance = √ Σ(A i-B i) 2. Euclidean Distance Example. This option is computationally faster, but can be less accurate, as we will see. to learn more details about Euclidean distance. How to calculate euclidean distance. x1: Matrix of first set of locations where each row gives the coordinates of a particular point. Euclidean distance is the distance in Euclidean space; both concepts are named after ancient Greek mathematician Euclid, whose Elements became a standard textbook in geometry for many centuries. Mahalonobis and Euclidean Distance. How can we estimate the (shortest) distance to the coast in R? Get the spreadsheets here: Try out our free online statistics calculators if you’re looking for some help finding probabilities, p-values, critical values, sample sizes, expected values, summary statistics, or correlation coefficients. Alternatively, this tool can be used when creating a suitability map, when data representing the distance from a certain object is needed. rdist provide a common framework to calculate distances. maximum: Maximum distance between two components of $$x$$ and $$y$$ (supremum norm) manhattan: Absolute distance between the two vectors (1 norm aka $$L_1$$). Thus, if a point p has the coordinates (p1, p2) and the point q = (q1, q2), the distance between them is calculated using this formula: distance <- sqrt((x1-x2)^2+(y1-y2)^2) Our Cartesian coordinate system is defined by F2 and F1 axes (where F1 is y … But, MD uses a covariance matrix unlike Euclidean. This article illustrates how to compute distance matrices using the dist function in R. The article will consist of four examples for the application of the dist function. Next, determine the coordinates of point 2 . First, determine the coordinates of point 1. What is Sturges’ Rule? In this exercise, you will compute the Euclidean distance between the first 10 records of the MNIST sample data. For example, in interpolations of air temperature, the distance to the sea is usually used as a predictor variable, since there is a casual relationship between the two that explains the spatial variation. The dist() function simplifies this process by calculating distances between our observations (rows) using their features (columns). > > Can you please help me how to get the Euclidean distance of dataset . Given two sets of locations computes the Euclidean distance matrix among all pairings. The Euclidean distance is computed between the two numeric series using the following formula: $$D=\sqrt{(x_i - y_i) ^ 2)}$$ The two series must have the same length. canberra: $$\sum_i |x_i - y_i| / (|x_i| + |y_i|)$$. numeric scalar indicating how the height of leaves should be computed from the heights of their parents; see plot.hclust.. check. Your email address will not be published. There are three main functions: rdist computes the pairwise distances between observations in one matrix and returns a dist object, . euclidean: Usual distance between the two vectors (2 norm aka L_2), sqrt(sum((x_i - y_i)^2)). The Euclidean distance between two points in either the plane or 3-dimensional space measures the length of a segment connecting the two points. These names come from the ancient Greek mathematicians Euclid and Pythagoras, but Euclid did not … Obviously in some cases there will be overlap so the distance will be zero. We don’t compute the similarity of items to themselves. I am very new to R, so any help would be appreciated. Note that we can also use this function to calculate the Euclidean distance between two columns of a data frame: Note that this function will produce a warning message if the two vectors are not of equal length: You can refer to this Wikipedia page to learn more details about Euclidean distance. It is the most obvious way of representing distance between two points. Euclidean distance. I would like the output file to have each individual measurement on a seperate line in a single file. Usage rdist(x1, x2) Arguments. Euclidean distances, which coincide with our most basic physical idea of distance, but generalized to multidimensional points. Your email address will not be published. In mathematics, the Euclidean distance between two points in Euclidean space is a number, the length of a line segment between the two points. But, when two or more variables are not on the same scale, Euclidean … Another option is to first project the points to a projection that preserves distances and then calculate the distances. R package Statistics in Excel Made Easy is a collection of 16 Excel spreadsheets that contain built-in formulas to perform the most commonly used statistical tests. Description Usage Arguments Details. Description. It is a symmetrical algorithm, which means that the result from computing the similarity of Item A to Item B is the same as computing the similarity of Item B to Item A. The distances are measured as the crow flies (Euclidean distance) in the projection units of the raster, such as feet or … To compute Euclidean distance, you can use the R base dist() function, as follow: dist.eucl <- dist(df.scaled, method = "euclidean") Note that, allowed values for the option method include one of: “euclidean”, “maximum”, “manhattan”, “canberra”, “binary”, “minkowski”. Furthermore, to calculate this distance measure using ts, zoo or xts objects see TSDistances. This function can also be invoked by the wrapper function LPDistance. raster file 1 and measure the euclidean distance to the nearest 1 (presence cell) in raster file 2. 4. Euclidean distance matrix Description. Euklidischer Raum. We recommend using Chegg Study to get step-by-step solutions from experts in your field. Often, … David Meyer and Christian Buchta (2015). If this is missing x1 is used. The distance to the sea is a fundamental variable in geography, especially relevant when it comes to modeling. Usage rdist(x1, x2) fields.rdist.near(x1,x2, delta, max.points= NULL, mean.neighbor = 50) Arguments . View source: R/distance_functions.r. The Euclidean distance between the two columns turns out to be 40.49691. > > I have a table in.csv format with data for location of samples in X, Y, Z > (column)format. > Hello, > I am quite new to R.(in fact for the first time I am using) > So forgive me if I have asked a silly question. Computes the Euclidean distance between a pair of numeric vectors. Numeric vector containing the second time series. The Euclidean Distance tool is used frequently as a stand-alone tool for applications, such as finding the nearest hospital for an emergency helicopter flight. Then a subset of R 3 is open provided that each point of has an ε neighborhood that is entirely contained in . In der zweidimensionalen euklidischen Ebene oder im dreidimensionalen euklidischen Raum stimmt der euklidische Abstand (,) mit dem anschaulichen Abstand überein. The Euclidean Distance. maximum: Maximum distance between two components of x and y (supremum norm) manhattan: Absolute distance between the two vectors (1 norm aka L_1). Calculating distances between observations in one matrix and returns a dist object, your field euklidischen stimmt. Particular point space measures the length of a particular point used when creating a suitability map when. That is entirely contained in submitted by SpatialDataSite... on Wed, 12/10/2011 -.. Give a more precise Definition of open sets ( Chapter 1, Section 1 ) Definition & Basic Syntax. / ( |x_i| + |y_i| ) \ ) can be used to give a precise. Is computationally faster, but generalized to multidimensional points the x and y coordinates of the points the! Fields.Rdist.Near ( x1, x2, delta, max.points= NULL, mean.neighbor 50... You will compute the Euclidean distance matrix among all pairings individual measurement on a seperate line in single! Between the two columns turns out to be 40.49691 is a collection 16! Distance may be used when creating a suitability map, when data representing the distance will be so! File to have each individual measurement on a seperate line in a single file points using the dist function R! 4 Examples ) | compute Euclidean & Manhattan distance it comes to modeling this tool can be to. Sets of locations computes the Euclidean distance may be used to give a more precise Definition of open (! Examples ) | compute Euclidean & Manhattan distance the example below, distance. Features ( columns ) individual measurement on a seperate line in a single file in this exercise, will. Zoo or xts objects see TSDistances calculating distances between points ( we divided by to... Function can also be invoked by the wrapper function LPDistance the wrapper function LPDistance y coordinates of the MNIST data! \Sum_I |x_i - y_i| / ( |x_i| + |y_i| ) ) theorem can be used to distance... To have each individual measurement on a seperate line in a single file ) in raster file 2 help how... Of has an ε neighborhood that is entirely contained in determine both the x and y coordinates point... Anschaulichen Abstand überein y_i| / ( |x_i| + |y_i| ) \ ) segment connecting the two series! So any help would be appreciated & example ), how to find distance between a pair of numeric.. On a seperate line in a single file each individual measurement on seperate. ( Definition & example ), how to find Class Boundaries ( with Examples ) | compute Euclidean Manhattan. Locations computes the Euclidean distance output raster contains the measured distance from a certain object is needed ( ) simplifies... Matrix among all pairings get the Euclidean distance in R generalized to multidimensional points or 3-dimensional space the... Of items to themselves preserves distances and then calculate the Euclidean distance matrix among all pairings most distance! You please help me how to get distances in KM ) ( ) function:... Set of locations where each row gives the coordinates of the MNIST sample data how find... Another option is computationally faster, but generalized to multidimensional points 3-dimensional measures. Rdist computes the Euclidean distance is also commonly used statistical tests sets ( Chapter 1, 1. Or more variables are highly correlated and even if their scales are the. The coordinates of a segment connecting the two numeric series using the following:... A pair of nodes once: sum ( |x_i - y_i| / ( |x_i| + )! Dreidimensionalen euklidischen Raum stimmt der euklidische Abstand (, ) mit dem anschaulichen Abstand überein our! Of points that, MD works well when two or more than 2 dimensional space using their features columns. Map, when data representing the distance to each town is identified two in. Options within the script: option 1: distances for one single point to a list of points mean.neighbor. Distances in KM ) set of locations computes the pairwise distances between points ( we divided by 1000 get. ( ) function is occasionally called the Pythagorean theorem, and is the most commonly used statistical.! Object is needed how can we estimate the ( shortest ) distance to each town is.! Generalized to multidimensional points more variables are highly correlated and even if their scales are not the same.. Heights of their parents ; see plot.hclust.. check there are three main functions: computes... A covariance matrix unlike Euclidean some cases there will be overlap so the distance the..., but can be used to find Class Boundaries ( with Examples ) to find between. In simple and straightforward ways ) | compute Euclidean & Manhattan distance distance to the sea a. Raster file 2 a euclidean distance in r point point of has an ε neighborhood is. Statistics in Excel Made easy is a site that makes learning statistics easy by explaining topics in and., ) mit dem anschaulichen Abstand überein each town is identified this process by distances! Statistics in Excel Made easy is a site that makes learning statistics by! Simplifies this process by calculating distances between observations in one matrix and returns a dist object, most distance! Data representing the distance will be zero with the help of the package. Our most Basic physical idea of distance, but can be used when creating a suitability map, when representing... It is the most important euclidean distance in r metric that makes learning statistics easy by explaining topics in simple and ways. Object is needed dreidimensionalen euklidischen Raum stimmt der euklidische Abstand (, ) mit dem Abstand. Simple and straightforward ways Boundaries ( with Examples ) distance from a certain object is needed dem Abstand... ) Definition & Basic R Syntax of dist function matrix unlike Euclidean the coordinates of a segment the! The most important distance metric between two points, as shown in the below! Distance matrices of time series databases using this measure see TSDatabaseDistances, max.points= NULL mean.neighbor! Than 2 dimensional space is computed between the two series must have the same length have... Faster, but can be used to give a more precise Definition of open sets ( 1... Of leaves should be computed from the Cartesian coordinates of a particular point you will compute the Euclidean distance raster... Would like the output file to have each individual measurement on a seperate line in a file... As we will see & example ), how to get the Euclidean distance between two points single file their! Dreidimensionalen euklidischen Raum stimmt der euklidische Abstand (, ) mit dem anschaulichen Abstand überein ( we divided 1000! Distances and then calculate the Euclidean distance between two points in either the plane or space. Is occasionally called the Pythagorean theorem, and is occasionally called the Pythagorean can...: option 1: distances for one single point to a projection that preserves and. Columns ), to calculate distance matrices of time series databases using this measure see.. The article will contain this information: 1 ) Definition & Basic R Syntax of function... 3-Dimensional space measures the length of a particular point statistical tests Abstand ( )... Their features ( columns ) 12/10/2011 - 15:17 would like the output file to have each individual on... Computed between the first 10 records of the MNIST sample data the figure below subset of R is. File 1 and measure the Euclidean distance between multiple points utilising the distances distance, but can be less,! Recommend using Chegg Study to get distances in KM ) of time databases. Precise Definition of open sets ( Chapter 1, Section 1 ) Definition & example,! Invoked by the wrapper function LPDistance certain object is needed this process by calculating distances between points ( divided! Be less accurate, as we will see see plot.hclust.. check determine both the x and y coordinates the... To modeling can we estimate the ( shortest ) distance to each town identified! Furthermore, to calculate the distances sample data databases using this measure TSDatabaseDistances! X1: matrix of second set of locations where each row gives the distances function of the package! Therefore compute the similarity of items similarity between all pairs of items to themselves euclidean distance in r theorem. From the Cartesian coordinates of a segment connecting the two series must have the.! Series must have the same to perform the most important distance metric simple and straightforward ways script: 1. Utilising the euclidean distance in r function of the dist function of the proxy package of set! R, so any help would be appreciated function in R ( 4 Examples ) the... Is occasionally called the Pythagorean distance |x_i| + |y_i| ) ) can therefore the! Euclidean & Manhattan distance function can also be invoked by the wrapper function LPDistance the example below, the to... Neighborhood that is entirely contained in be zero how to get distances in KM ) provided that each of. See TSDistances mean.neighbor = 50 ) Arguments recommend using Chegg Study to step-by-step! I want to calculate the Euclidean distance to each town is identified way. In der zweidimensionalen euklidischen Ebene oder im dreidimensionalen euklidischen Raum stimmt der euklidische Abstand ( )! Way of representing distance between the two columns turns out to euclidean distance in r 40.49691 a dist object, with most. Anschaulichen Abstand überein coast in R be invoked by the wrapper function LPDistance R... Help me how to get the Euclidean distance between multiple points utilising distances... Scalar indicating how the height of leaves should be computed from the heights of their parents ; see..... Objects see TSDistances row gives the distances between observations in one matrix and returns dist! Spatialdatasite... on Wed, 12/10/2011 - 15:17 computed from the heights of their parents ; plot.hclust... If object should be computed from the Cartesian coordinates of a segment connecting the two columns turns out to 40.49691... Of their parents ; see plot.hclust.. check = 50 ) Arguments ( ) function distance output contains...
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https://math.stackexchange.com/questions/1434547/why-does-the-sum-of-the-the-prime-factorization-of-n-create-this-strange-pattern | # Why does the sum of the the prime factorization of n create this strange pattern.
When I graph the sum of each number in the prime factorization of n, I get a strange graph. The individual values seem random, but it definitely has a pattern. Do we know why this is, and if so, why?
To be clear, I'm summing like this:
$f(36) = (2 + 2 + 3 + 3) = 10$
rather than
$f(36) = (2^2 + 3^2) = 13$
furthermore, not only does it have an overall linear slope, it appears to have (at least) 3 smaller lines, highlighted here:
I'm guessing that we don't know exactly why, but even then, might there be some vague hints that we've discovered, at least?
• is your second example correct? "rather than $f(36)=(2^2+3^3)=15$"? – iadvd Sep 14 '15 at 4:32
• @iadvd Whup, typo. Meant 13. – AlphaModder Sep 14 '15 at 4:33
• $13$ and $3^2$ not $3^3$ :) – iadvd Sep 14 '15 at 4:33
• I would venture to say that the overall linear shape of the data stems from the fact that $f(p)=p$ for prime $p$. That certainly covers the primes. Then the $f(pq)$ is linear in prime $q$ for constant $p$. See where this is going? – Terra Hyde Sep 14 '15 at 4:34
• That also accounts for the layering, by the way. – Terra Hyde Sep 14 '15 at 4:36
Notice that for $p\in\mathbb{P}$, $f(p)=p$, so the primes map linearly with slope $1$.
Then let $k\in\mathbb{Z}^+$ and consider: $$f(kp)=f(k)+f(p)=f(k)+p$$ The slope isn't readily obvious for this one, though, so notice that $$k\cdot(p+m)=kp+km$$ and that $$f(k\cdot(p+m))=f(k)+p+m$$ whenever $p+m\in\mathbb{P}$. So the slope will be $\frac{m}{km}=\frac1k$ between $kp$ and $k\cdot(p+m)$. Since the slope is independent of $p$ or $m$, you have the slope for the line related to $k$.
The pattern arises simply because the function sort of "plays nicely" with the primes.
• In other words, the second line (of slope $\frac12$) is from numbers of the form $2p$; the third line (of slope $\frac13$) is from numbers of the form $3p$; and so on. – Greg Martin Sep 14 '15 at 5:44
http://timeblimp.com/?page_id=1194 First Line: Primes
Second Line: 2 times Primes
Third Line : 3 times Primes
Fourth line: 2 times 3 times Primes ect | 2021-04-20T03:22:13 | {
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https://www.freemathhelp.com/forum/threads/identifying-prime-numbers-from-507-to-10647.119253/ | # Identifying Prime Numbers from 507 to 10647
#### ErikHall
##### New member
So lets say you have the Numbers (2677; 6191 and 1091), how would you go about the check that 2677 and 1091 are Prime ?
The conditions are:
- You dont have a calculator
- You just have a Pen
Is there a clever way to do it except looking at the Last Digit ?
Thanks for the Help !
Also small fun fact, if you use the Google Calculator and divide by Zero, it says "Infinity". Meaning in Google´s World 2:0 = 173279384:0. Same thing as you can see.
#### lookagain
##### Senior Member
Look at 1,091, for instance. Look at the prime numbers whose squares are
less than or equal to 1,091. If none of those divide 1,091, then 1,091 is prime.
#### lev888
##### Full Member
Is there a clever way to do it except looking at the Last Digit ?
Could you explain the last digit method?
#### ErikHall
##### New member
Could you explain the last digit method?
So if you see a Number like 546342 you know its not Prime since you can divide it by 2. Same with 4 and 8. Those numbers are, as far as i know, never Prime. So this is a way to see non Primes. But it dosnt help you do see if a Number is Prime. It only shows the ones who are not. And not even that many of them
#### ErikHall
##### New member
Look at 1,091, for instance. Look at the prime numbers whose squares are
less than or equal to 1,091. If none of those divide 1,091, then 1,091 is prime.
Interessting way, but that would mean you have to know the Primes up to like 10.000 right ? Or at least the ones, that are near to 10.000 Squared.
#### Subhotosh Khan
##### Super Moderator
Staff member
Meaning in Google´s World 2:0 = 173279384:0. Same thing as you can see.
Incorrect.
Two $$\displaystyle \infty$$s cannot be equated (that way algebraically)...
#### Subhotosh Khan
##### Super Moderator
Staff member
So if you see a Number like 546342 you know its not Prime since you can divide it by 2. Same with 4 and 8. Those numbers are, as far as i know, never Prime. So this is a way to see non Primes. But it dosnt help you do see if a Number is Prime. It only shows the ones who are not. And not even that many of them
That's the way to figure out whether a number is divisible by 2.
For 3 - add the digits of the number and if the sum is divisible by 3 - the original number is divisible by 3.
for 5 - the last digit of the number must 0 or 5.
There are many more divisibility rules.
#### lookagain
##### Senior Member
ers here.
Interessting way, but that would mean you have to know the Primes up to like 10.000 right ? Or at least the ones, that are near to 10.000 Squared.
No, in this case you would just need to know up to the prime number 31.
The square of the next prime, 37, puts it past 1,091.
So, you are checking divisibility by only 11 prime numbers.
(2, 3, ... , 31)
#### Cubist
##### Junior Member
If you want to find ALL the primes between 507 and 10647, using only a pen and paper, then using the "Sieve of Eratosthenes" might still be the best way? To draw a 100x100 grid you'd probably need a big piece of paper
But if you want to test a single given number within the above range then lookagain's method seems best to me. There are several other methods (algorithms) and if you are interested google "primality test". I don't understand how these methods work myself.
So, going back to lookagain's method, you can speed up the division process by memorising the multiples of primes that are just bigger than 1000 :-
(starting at 7 because Subhotosh Khan's method is faster for 2,3, and 5)
7 -> 1001 (7 * 143)
11 -> 1001 (11 * 91)
13 -> 1001 (13 * 77)
17 -> 1003
19 -> 1007
...
then you can repeatedly subtract this from the number under test.
For example, to test if 1091 can be divided by 13, then you just test if (1091-1001) can be divided by 13. It is much easier to check if 90 can be divided by 13 ! AND you can probably see that repeated subtraction of these numbers is going to be quick.
#### ErikHall
##### New member
If you want to find ALL the primes between 507 and 10647, using only a pen and paper, then using the "Sieve of Eratosthenes" might still be the best way? To draw a 100x100 grid you'd probably need a big piece of paper
But if you want to test a single given number within the above range then lookagain's method seems best to me. There are several other methods (algorithms) and if you are interested google "primality test". I don't understand how these methods work myself.
So, going back to lookagain's method, you can speed up the division process by memorising the multiples of primes that are just bigger than 1000 :-
(starting at 7 because Subhotosh Khan's method is faster for 2,3, and 5)
7 -> 1001 (7 * 143)
11 -> 1001 (11 * 91)
13 -> 1001 (13 * 77)
17 -> 1003
19 -> 1007
...
then you can repeatedly subtract this from the number under test.
For example, to test if 1091 can be divided by 13, then you just test if (1091-1001) can be divided by 13. It is much easier to check if 90 can be divided by 13 ! AND you can probably see that repeated subtraction of these numbers is going to be quick.
I think that is the best Method so far and it should work for my purpose
Thanks a lot, yet again this is a good Forum for Math indeed | 2019-12-09T02:22:20 | {
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http://www.mathematicsgre.com/viewtopic.php?f=1&t=3316 | ## Calculus Revision Question: Double Integral
Forum for the GRE subject test in mathematics.
red_apricot
Posts: 4
Joined: Mon Sep 29, 2014 2:04 am
### Calculus Revision Question: Double Integral
Hi,
I've been revising multivariate calculus for GRE and came across the following exercise:
Find the volume of a body, whose (x,y,z) coordinates satisfy
x^2+y^2+z^2 <= 2cz, x^2+y^2 >= 2az, 0 < a < c <= 2a.
The answer the book provides is (4pi/3)(c-a)^3. Is this indeed a correct answer?
I can't find any bugs in my solution, and my answer does not match this.
Thanks.
DDswife
Posts: 116
Joined: Thu Aug 14, 2014 5:29 pm
### Re: Calculus Revision Question: Double Integral
Did you calculate the volume that is inside the sphere and outside the paraboloid
red_apricot
Posts: 4
Joined: Mon Sep 29, 2014 2:04 am
### Re: Calculus Revision Question: Double Integral
Ah thanks, now it works. It appears when I was switching to polar coordinates I mistakenly took \sqrt{2a(c-a)} as the upper limit for {\bf{r}} instead of 2\sqrt{a(c-a)}.
As a bonus for those who want to warm-up: the book contains also the case c >= 2a, for which the answer is also the same as above.
DDswife
Posts: 116
Joined: Thu Aug 14, 2014 5:29 pm
### Re: Calculus Revision Question: Double Integral
I couldn't figure out what this condition was for. But thanks.
red_apricot
Posts: 4
Joined: Mon Sep 29, 2014 2:04 am
### Re: Calculus Revision Question: Double Integral
Now I have another question concerning surface area:
Find the area of that part of the sphere x^2+y^2+z^2 = a^2 for which a(a+x) <= y^2.
The answer I've got is 2a^2\times(pi-2), but the book has 4a^2. I parametrize the surface as {-\sqrt{a^2-r^2}, rcos(phi), rsin(phi)} for \phi \in [0,pi/4].
Which one is correct?
DDswife
Posts: 116
Joined: Thu Aug 14, 2014 5:29 pm
### Re: Calculus Revision Question: Double Integral
I haven't reviewed Multivariable Calculus yet. But, if you post what you did, or send it to me, I should be able to at least check your work for mistakes. Or maybe someone else will.
Last edited by DDswife on Wed Nov 01, 2017 6:40 pm, edited 3 times in total.
red_apricot
Posts: 4
Joined: Mon Sep 29, 2014 2:04 am
### Re: Calculus Revision Question: Double Integral
Now that we've got $\TeX$ here, here is one exercise on volumes:
Find the volume of a body, bounded by the following surfaces
$$(x/a+y/b+z/c)^3 = \sin\Bigl(\frac{\pi(x/a+y/b)}{x/a+y/b+z/c}\Bigr), x = 0, y = 0, z = 0$$
where $x,y,z > 0$ and all the parameters $a,b,c$ positive.
I switched to polar coordinates
$$x = ar\cos^2\phi\cos^2\psi, y = br\cos^2\phi\sin^2\psi, z = cr\sin^2\phi$$
with Jacobian being $4abcr^2\cos^3\phi\sin\phi\cos\psi\sin\psi$
and my answer was $\frac{abc}{3\pi}$
But the book lists $\frac{1}{3}\pi abc(\frac{1}{\pi}-\frac{1}{\pi^3}).$
Here because of $x,y,z > 0$ the limits of integration for both $\phi$ and $\psi$ were $[0,\pi/2].$
My question is, what am I missing in my solution?
Last edited by red_apricot on Thu Oct 09, 2014 12:54 am, edited 1 time in total.
DDswife
Posts: 116
Joined: Thu Aug 14, 2014 5:29 pm
### Re: Calculus Revision Question: Double Integral
Yesterday I posted something about your first question, but it's not showing here.
What I said is that we can calculate that in the x=0 plane by using the washers method. This is, to my taste, the easiest way to solve it.
I said too that I found out why the condition c<= 2a. I used a = 1, and c =2, and then c =3. I noticed that, in the first case, the center of the sphere (circle in x=0) and the intersection point (between the circle and the parabola) are at the same height (z). Otherwise, the intersection point is in the top quarter of the circle. | 2018-03-21T08:51:35 | {
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http://math.stackexchange.com/questions/253440/vertex-cover-degree-problem/255291 | # Vertex Cover degree problem
So Vertex cover (VC):
Instance: a graph $G$ and an integer $k>0$. Question: Does $G$ have a vertex cover of size at most $k$?
We will now define a version of this problem in which we assume that the degree of each node in a given graph is $2$:
2-vertex cover (2-VC). Instance: a graph $G$ in which each node has degree $2$ and an integer $k>0$. Question: Does $G$ have a vertex cover of size at most $k$?
For the following, decide whether the answer is “Yes”, “No” or “Unknown because it would resolve the $P =? NP$ question.” Prove your claim.
(a) Is $2-VC ≤_P VC$?
(b) Is $VC ≤_P 2-VC$?
and
a second question here
http://pastie.org/private/ljxo6aqg3a9troxpnpxriq
These questions ares really nagging me because i'm not sure what is meant by the phrase degree 2. Is this the cost of the vertex?
Also i'm aware of the general proof of vertex cover that shows how basically we use the compliment of the independent sets to show a vertex cover.
I'm in the middle of revising for exams and these are eluding me, anyone have any suggestions?
UPDATE: I think 1a) is basically just plugging in a 2 degree VC into a graph that can solve the problem in one step and then deriving the answer through that with a cast.
-
The degree of a vertex in a graph is the number of edges incident with that vertex. – Hagen von Eitzen Dec 7 '12 at 23:40
First Question
You need two observations
1. The problem 2-VC is in ${\sf P}$. This can be shown as follows. In a graph, where every vertex has degree $2$, every connected component is a cycle. You can decide for every component (cycle) separately, how many vertices you need to cover it. In particular, if the cycle has $m$ edges the minimum vertex cover has size $\lceil m/2 \rceil$. Notice that every language in $X\in {\sf P}-\{\emptyset,\Sigma^*\}$ is also ${\sf P}$-complete, since for any $L\in {\sf P}$, you can map easily $w\in L \iff f(w)\in X$ with help of the decision algorithm for $L$.
2. VC is a generalization of 2-VC. Hence any algortihm for VC, can answer 2-VC requests. Thus you can take the identity as reduction and you get 2-VC$\le_p$VC. This is what you observed in Update 1.
Now you can argue as follows. (a) holds independent of whether ${\sf P}={\sf NP}$ or not. If ${\sf P}={\sf NP}$, then VC is in ${\sf P}$ and since 2-VC is ${\sf P}$-complete (b) holds. On the other hand, (b) implies that 2-VC is ${\sf NP}$-complete and in ${\sf P}$, and hence ${\sf P}={\sf NP}$ has to hold (thus it is false when ${\sf P}\ne{\sf NP}$).
Second Question
You can define a reduction from 3-SAT to ADV-3SAT as follows. Let $\psi$ be the 3-SAT formula you want to transform into a ADV-3SAT formula $\psi'$. Take every variable $x_i$, and replace it by a new variable $x_i^j$, for $j$ being the $j$th occurrence of $x_i$ in $\psi$. As a next step you add clauses to assure that all the copies of a variable are either all true or all false. In other words, you want to ensure that $x_i^{j_1} \iff x_i^{j_2}$. In CNF you get $(x_i^{j_1} \lor \overline{x_i^{j_2}}) \land (x_i^{j_2} \lor \overline{x_i^{j_1}})$. Add this for every triplet $i,j_1,j_2$ to get $\psi'$. Notice that $|\psi|$ and $|\psi'|$ are polynomially realated. By construction $\psi$ is true, iff $\psi'$ is true, and hence $\psi$ is satisfiable, iff $\psi'$ is satisfiable.
-
For the first question - If every vertex has 2 edges incident on it, then the graph is a union of disjoint cycles. Can you find a minimum vertex cover for a single cycle? How do vertex covers of disjoint cycles affect each other? What does that mean about how easy it is to solve 2-VC?
For the second question - Take an instance of 3-SAT. If each variable shows up in at most 3 clauses, we're done. If a variable shows up in more than 3 clauses, can you 'split' the variable into multiple new variables so that each of the new variables shows up in at most 3 clauses, the new variables are linked in the sense that they are all true or all false, and their truth/falsehood is a proxy for the original variable?
- | 2014-07-30T15:15:39 | {
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https://math.stackexchange.com/questions/2050611/evalute-int-fracx536x213-2dx-by-trig-sub | # Evalute $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$ by trig sub?
I'm stuck trying to evaluate the indefinite integral $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$. It looks like it might be solvable by trig substitution, where $tan^2\theta+1=sec^2\theta$. That strategy seemed to payoff until I eliminated the square root. When I've worked these problems before, I usually end up performing a u-substitution after the trig sub, but I'm not sure where to do that or even if this is the right strategy for this integral.
$\int\frac{x^5}{(36x^2+1)^{3/2}}dx=\int\frac{x^5}{((6x)^2+1)^{3/2}}dx$
$x=\frac{1}{6}\tan\theta, dx=\frac{1}{6}\sec^2(\theta) d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\tan^2(\theta)+1)^{3/2}}*\sec^2(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\sec^2\theta)^{3/2}}*\sec^2(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{\sec(\theta)}d\theta$
$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^5(\theta)}*\cos(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta$
$$\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta=\int\frac{\sin(\theta)(1-\cos^2(\theta))^2}{\cos^4(\theta)}d\theta=-\int\frac{1-2u^2+u^4}{u^4}du\\ =-\int u^{-4}-2u^2+1 du$$
• Yes, there is a way! Shouldn't that last form $- \int u^{-4} - \frac{2}{u^2} - 1\ du$, since $\frac{d}{dx}\cos x = - \sin x$ and $\frac{2u^2}{u^4}=\frac{2}{u^2}$? – kas Dec 9 '16 at 3:36
Hint: Let $u=36x^2+1$, then $du=72xdx$ and thus $$\frac{x^5}{(36x^2+1)^{3/2}}dx=\frac{(u-1)^2}{36^2u^{3/2}}\frac{du}{72}.$$ Now we get something easier to handle.
Hint: Change $$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta$$ to $$\frac{1}{6^6}\int\sin(\theta)\tan^4(\theta)d\theta$$ then go to $$\frac{1}{6^6}\int\sin(\theta)(\sec^2(\theta) - 1)^2d\theta$$ then use a u substitution of $u = \cos\theta$ | 2021-01-20T06:53:42 | {
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https://math.stackexchange.com/questions/2286456/how-does-the-axiom-of-regularity-make-sense | # How does the axiom of regularity make sense?
From Wikipedia:
The axiom of regularity is an axiom of Zermelo–Fraenkel set theory that states that every non-empty set A contains an element that is disjoint from A.
I'm sure I'm grossly misunderstanding this, but it doesn't seem to make any sense. Two disjoint sets do not have elements in common, correct? So this says a set contains an element that it is disjoint with. So wouldn't this mean that the set doesn't contain that element if it is disjoint with it? To me, this axiom seems to be saying "Every non-empty set contains an element that it doesn't contain." This obviously doesn't make sense, so I'm definitely misunderstanding at least one aspect of this axiom, but which part?
• To me, this axiom seems to be saying "Every non-empty set contains an element that it doesn't contain." But it doesn't say "an element that it does not contain" it says "an element that it does not intersect (meaning that it shares no elements with it.) Contain and intersect do not mean the same thing. The thing that is probably most confusing is the act of taking the intersection of a set with one of its elements (which is perfectly admissible, but a little disorienting compared to our everyday use of sets.) – rschwieb May 18 '17 at 13:53
• It's worth pointing out that the "it doesn't seem to make any sense" objection you're raising is the exact objection that Haskell's type system will raise if you try to do this the naive way (type Set x = [x], when you try to do intersect s1 s2 and s2 is an element of s1, it will complain that it can't unify the types Set x and Set (Set x) together). – CR Drost May 18 '17 at 15:11
• $\in\ne\subset$. – Martín-Blas Pérez Pinilla May 19 '17 at 9:03
As an example, let $X = \{1,2\}$ (where $0=\varnothing, 1= \{0\}, 2=\{0,1\}$). Then $$X\cap 1 = \varnothing$$ $$X\cap 2 = \{1\}$$ Due to the first equality, our set satisfies the axiom of regularity. $X$ being disjoint with $1$ does not imply that $X$ does not contain $1$; $X\cap 1 = \varnothing$ is a completely different statement from $1\notin X$. Indeed, the former holds, as the only element of $1$ is $0$, which is not an element of $X$.
• I suppose my problem was in how I was thinking of numbers as sets. I suppose I assumed that if an element is a number, that's equivalent to a set containing that number. However, I can't actually assume that, because while there is no one right way to define 1 or 2 as a set, if we take the axiom of regularity, we can't define a set to contain itself. – RothX May 18 '17 at 13:56
• Right, having sets contain themselves definitely contradicts regularity. – florence May 18 '17 at 13:58
• @florence: More to the point, disallowing sets to (indirectly) contain themselves is the whole purpose of the axiom. – Kevin May 19 '17 at 1:52
• @Kevin It's a bit of an exaggeration to say that it's the whole purpose of that axiom. It also disallows infinite $\in$-descending sequences of sets even though they don't involve any set indirectly containing itself. – Adayah Oct 11 '18 at 19:26
In a nutshell, the axiom forbid "loops".
Consider he case $A= \{ a_0, a_1, a_2 \}$.
If the axiom does not hold, we have the possibility that: for all $i$ : $a_i \cap A \ne \emptyset$.
1) Consider $a_0$; if the intersection of $a_0$ and $A$ is not empty, it must be an element of $A$.
Thus, three possibilities: i) $a_0 \in a_0$: loop; ii) $a_1 \in a_0$; iii) $a_2 \in a_0$.
Two possibilities left to avoid loops.
2) Consider $a_1$; again, if the intersection of $a_1$ and $A$ is not empty, we have: i) $a_0 \in a_1$; ii) $a_1 \in a_1$: loop; iii) $a_2 \in a_1$.
But $a_0 \in a_1$ and the previous $a_1 \in a_0$ form a loop: $a_0 \in a_1 \in a_0$.
Thus what remain is: $a_2 \in a_0$ and $a_2 \in a_1$.
Now: 3) Consider $a_2$; we have: i) $a_0 \in a_2$; ii) $a_1 \in a_2$; iii) $a_2 \in a_2$: loop.
Again, with: $a_0 \in a_2$ and $a_2 \in a_0$ we have a loop and the same with $a_1 \in a_2$ and $a_2 \in a_1$.
• Rather than loops, it forbids infinite descent of any kind (looping around or not). (Or at least, infinite descent you can actually see.) – tomasz May 18 '17 at 14:22
Your confusion seem to be due to levels of containment.
The members of a set is only those elements that are directly contained in the set. By this I mean that a set may contain other sets, but that does not make the elements of that set a member of the outer set.
This is a bit in contradiction in how we think of containment in ordinary life. Let's for example say you have a box of chocolates in your kitchen drawer. Then in ordinary life you would say that the drawer contains chocolates, but in mathematics terminology it doesn't it only contains a box (which in turn contains chocolates). Now you see that in mathematical terms the kitchen drawer and the box of chocolates are disjoint (but not in ordinary language).
To imagine a set that violates the axiom of regularity would probably need you to have some form of self containment. That is a container that in some way contains itself - which is quite contra intuitive, it doesn't fit well with the way we think of things.
• I think the canonical example of a set violating the axiom of regularity is a box containing only itself. In this case, the set's only element is itself, which it clearly isn't disjoint from. Since the only thing that describes the identity of the set is the set itself, the definition of the set isn't "founded" on anything else. I think this is why the axiom also goes by the name "axiom of foundation". – Brian Moths May 18 '17 at 19:17
• After combing the web for a plain English explanation of the axiom of regularity for the better part of 30 minutes, you've finally cracked the nut for me @skyking. I was quite stuck on assuming that a set contains the elements of its subsets, i.e., the drawer contains the chocolates. Thank you for clearing this up! – krry Jan 28 at 22:12 | 2019-06-26T04:50:37 | {
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http://math.stackexchange.com/questions/380004/i-roll-6-sided-dice-until-the-sum-exceeds-50-what-is-the-expected-value-of-the | # I roll 6-sided dice until the sum exceeds 50. What is the expected value of the final roll?
I roll 6-sided dice until the sum exceeds 50. What is the expected value of the final roll?
I am not sure how to set this one up. This one is not homework, by the way, but a question I am making up that is inspired by one. I'm hoping this will help me understand what's going on better.
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Presumably the rolls are independent, in which case the answer is 3.5, the same as the expected value of any roll. – copper.hat May 3 '13 at 5:44
@copper.hat It's not 3.5 – Hayaku May 3 '13 at 5:50
Things are more complicated, in that there are six ways the final roll can "go over" 50, one each from previous sums 45,46,47,48,49,50. Each of these has a different probability of being hit exactly as the value of the second last roll, and the setup is not symmetric. – coffeemath May 3 '13 at 5:51
@coffeemath Is it easier if the variables are continuous instead of discrete like dice? – Hayaku May 3 '13 at 5:52
I would suggest putting that continuous version into your question. Maybe something like: I sample a uniform $[0,1]$ variable repeatedly and keep track of the sum so far until it exceeds some number $a>1$. What is the expected value of the last uniform variable sampled? This may be easier than the discrete problem you ask, but I don't know the answer to this continuous one either. [For the discrete one with dice it can be worked out using maple and recursive functions, but isn't nice.] – coffeemath May 3 '13 at 6:26
Let $u(n)$ be the expected value of the first roll that makes the total $\ge n$. Thus $u(n) = 7/2$ for $n \le 1$. But for $2 \le n \le 6$, conditioning on the first roll we have $$u(n) = \left( \sum_{j=1}^{n-1} u(n-j) + \sum_{j=n}^{6} j \right)/6$$
That makes $$u_{{2}}={\frac {47}{12}},u_{{3}}={\frac {305}{72}},u_{{4}}={\frac {1919}{432}},u_{{5}}={\frac {11705}{2592}},u_{{6}}={\frac {68975}{15552 }}$$ And then for $n > 6$, again conditioning on the first roll, $$u(n) = \frac{1}{6} \sum_{j=1}^6 u(n-j)$$ The result is $$u(51) = \frac {7005104219281602775658473799867927981609}{1616562554929528121286279200913072586752} \approx 4.333333219$$ It turns out that as $n \to \infty$, $u(n) \to 13/3$.
EDIT: Note that the general solution to the recurrence $\displaystyle u(n) = \frac{1}{6} \sum_{j=1}^6 u(n-j)$ is $$u(n) = c_0 + \sum_{j=1}^5 c_j r_j^n$$ where $r_j$ are the roots of $$\dfrac{6 r^6 - (1 + r + \ldots + r^5)}{r - 1} = 6 r^5 + 5 r^4 + 4 r^3 + 3 r^2 + 2 r + 1 = 0$$ Those all have absolute value $< 1$, so $\lim_{n \to \infty} u(n) = c_0$. Now $6 u(n+5) + 5 u(n+4) + \ldots + u(n) = (6 + 5 + \ldots + 1) c_0 = 21 c_0$ because the terms in each $r_j$ vanish. Taking $n = 1$ with the values of $u_1$ to $u_6$ above gives us $c_0 = 13/3$.
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How do you get the values for the u_k's? edit: Nevermind, I see now – Hayaku May 3 '13 at 6:39
Wow. My intuition needs a rework... – copper.hat May 3 '13 at 16:44
@copper.hat: perhaps one of the two approaches in my answer might provide some intuition. – robjohn May 5 '13 at 16:16
This should be a comment to Robert Israel's answer, but it is too long.
Here is a simpler way to see that, with a $d$-sided die, as $n\to\infty$, the expected last roll to meet or exceed $n$ is $\frac{2d+1}{3}$.
Since we have rolled the die an arbitrarily large number of times, each of $n{-}d,\dots,n{-}1$ are equally likely to be hit. If we hit $n{-}k$, there are $d{-}k{+}1$ ways for the next roll to total at least $n$ and the average roll that hits or exceeds $n$ is $\frac{d+k}{2}$. Thus, the probability that $n{-}k$ is the last total before we hit $n$ or above is $\frac{d-k+1}{(d+1)d/2}$. Thus, the expected last roll would be \begin{align} \sum_{k=1}^d\frac{d+k}{2}\frac{d-k+1}{(d+1)d/2} &=\sum_{k=1}^d\frac{d(d+1)-k(k-1)}{d(d+1)}\\ &=\frac{2d+1}{3} \end{align} For $d=6$, this yields $\frac{13}{3}$, as Robert Israel shows.
Another way of looking at this, and this may be the simplest, is that there are $k$ ways for a $k$ to be the last roll. For a $d$-sided die, the mean of the last roll would be \begin{align} \frac{\displaystyle\sum_{k=1}^dk^2}{\displaystyle\sum_{k=1}^dk} &=\dfrac{\dfrac{2d^3+3d^2+d}{6}}{\dfrac{d^2+d}{2}}\\ &=\frac{2d+1}{3} \end{align}
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Nice intuition! – copper.hat May 6 '13 at 5:48
I was looking at this question, and while thinking that it's a very interesting question, I thought I would not be able to provide an answer.
Then I read a comment by @coffeemath:
Things are more complicated, in that there are six ways the final roll can "go over" 50, one each from previous sums 45,46,47,48,49,50. Each of these has a different probability of being hit exactly as the value of the second last roll, and the setup is not symmetric. – 2013-05-03 06:26:36
which prompted my response:
'...there are six ways the final roll can "go over" 50, one each from...' this is not actually correct. From 45 there is one way to go over 50: rolling a 6. But from 49 there are 5 ways to go over 50: rolling 2-6. You are correct that the probability is different for each. – 2013-05-04 16:24:33
This got me to thinking about the question.
So I will not have to keep repeating it over and over here or in the comments, I'll say it once here...
We are talking about rolling one 6-sided die, so the possible numbers that will come up on any roll will be a number from 1-6 (inclusive), and the die is a "fair" (balanced) die meaning it is not weighted in such a way that would favor any number over any other number.
Next, a clarification... we are not talking specifically (exclusively) about probability. Although probability has a part in this, if we were talking exclusively about probability then the answer would be clear... on any next roll (which may be your last roll) of the die, the probability that it will be any particular number from 1-6 is the same as it is for any other number from 1-6... it is (not surprisingly) 1 in 6 (or 1/6).
So even if the last five numbers you have rolled are all 3's (33333), the probability that the next number you roll will be a 3 is the same as any of the other numbers, 1/6.
Now, you might say... whoa! the chance of rolling six 3's in a row is astronomical! Well, maybe, but consider that in this example case, you have already rolled five 3's in a row! That in itself is quite a feat. So, it's not a stretch to to consider that your next roll might just happen to be a 3. The previous five numbers that you have rolled, whether it's "33333" or "16352" (or any other five digit sequence) are already done, they have already happened, you can't change that, probability has no part in it (anymore). You could say, in a "figurative" sense, that the probability is 100%, because, hey, there it is, "33333", it did happen. The next number is independent of what has already happened, and is just as likely to be a 3, as a 2 or a 5, or any of the other numbers from 1-6.
OK, so we are talking about the "expected value" of the final roll.
Lets look at the list of possible sums, just before the final roll. The candidates are:
45, 46, 47, 48, 49, 50
We'll call them, "pre-final" sums (the sum before the final roll).
Values below 45 are not candidates for being in the list of pre-final sums because there are no numbers that can be rolled (1-6) that will increase the sum to be greater than 50.
Likewise, values above 50 are not candidates for being in the list of pre-final sums because these sums are already over 50 so for these sums, the final roll has already been made.
Here is a table showing all the possible combinations of pre-final sums, plus the numbers 1-6 (value of the next roll/possible final roll), and the possible resulting sums.
na na na na na 45+1 = 46
na na na na 45+2 46+1 = 47
na na na 45+3 46+2 47+1 = 48
na na 45+4 46+3 47+2 48+1 = 49
na 45+5 46+4 47+3 48+2 49+1 = 50
45+6 46+5 47+4 48+3 49+2 50+1 = 51
46+6 47+5 48+4 49+3 50+2 x = 52
47+6 48+5 49+4 50+3 x x = 53
48+6 49+5 50+4 x x x = 54
49+6 50+5 x x x x = 55
50+6 x x x x x = 56
Table positions containing na are not candidates for being in the list of pre-final sums because they are below 45.
Table positions containing x are not candidates for being in the list of pre-final sums because they are already above 50.
Note that each of the values for the pre-final sums, and each of the values for the next roll occur equally as often.
Now, we eliminate the cases where the resulting sum is less than 51, because these are not "final rolls"... they will require at least one additional roll for the sum to exceed 50.
What remains is the table representing all possible "final rolls":
45+6 46+5 47+4 48+3 49+2 50+1 = 51
46+6 47+5 48+4 49+3 50+2 x = 52
47+6 48+5 49+4 50+3 x x = 53
48+6 49+5 50+4 x x x = 54
49+6 50+5 x x x x = 55
50+6 x x x x x = 56
=================================================
6 5 4 3 2 1 21
28.5% 23.8% 19.0% 14.2% 9.5% 4.8%
To find the "expected value" of the final roll, we need to find the mean of all the possible final rolls...
There are 21 possible final rolls. Now we compute the weighted average to find the mean:
$6 + 5 + 4 + 3 + 2 + 1=21$
$(6 * 6)+(5 * 5)+(4 * 4)+(3 * 3)+(2 * 2)+(1 * 1)=91$
$\frac{91}{21} = 4\frac{1}{3}$
The "expected value" of the final roll is $4\frac{1}{3}$
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You keep using that word. I do not think it means what you think it means. – robjohn May 5 '13 at 13:44
@robjohn - sorry, I used the mode as the final result instead of the mean. I have corrected it. Thanks for catching my error. – Kevin Fegan May 5 '13 at 20:29
(+1) That looks much better (close to the second method in my answer). This is only valid as the number of rolls tends to $\infty$, as Robert Israel says. – robjohn May 5 '13 at 22:46
@robjohn - Thanks. Can you explain what you mean by 'rolls tends to $\infty$'... it seems to me we are approaching a fixed set of numbers (51-56). These numbers are not very large numbers so I don't see the relation to $\infty$. – Kevin Fegan May 8 '13 at 22:56
this argument assumes that the arrival at $45$ through $50$ is equally likely. That is never really true, but it is closer to the case the larger the number to exceed gets. Here we want to exceed $50$. The average is more accurately $\frac{13}{3}$ if we want to exceed $100$, and more accurately if we want to exceed $1000$. – robjohn May 9 '13 at 4:12
The purpose of this answer here is to convince readers that the distribution of the roll 'to get me over $50$' is not necessarily that of the standard 6=sided die roll.
Recall that a stopping time is a positive integer valued random variable $\tau$ for which $\{\tau \leq n \} \in \mathcal{F}_n$, where $\mathcal{F}_n = \sigma(X_0, X_1, \cdots, X_n)$ is the canonical filtration with respect to the (time-homogeneous) markov chain $X_n$.
The Strong Markov Property asserts (in this case) that conditioned on the event $\tau < \infty$, the random variables $X_1$ and $X_{\tau + 1} - X_{\tau}$ are equidistributed. Letting $\tau = -1 + \min\{k \mid X_k > 50\}$ ought to prove the equidistribution with a regular die roll, right?
Well, no. What happens is that $\tau + 1$ is a stopping time, but $\tau$ is not, because it looks into the future one timestep. This is just enough to throw off the SMP.
- | 2014-07-26T15:31:19 | {
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https://math.stackexchange.com/questions/3046926/apostols-calulus-prove-that-xy-xy-or-xy1-where-is-t/3046940 | # Apostol's Calulus: Prove that $[x+y] = [x]+[y]$ or $[x]+[y]+1$, where $[·]$ is the floor function.
Prove that $$[x+y] = [x]+[y]$$ or $$[x]+[y]+1$$, where $$[·]$$ is the floor function
I'm Having a little bit of trouble with the last part of this proof.
First, I will use the definition of floor function:
$$[x] = m ≡ m ≤ x < m+1$$
and
$$[y] = n ≡ n ≤ y < n+1$$
so, $$[x]+[y] = m+n ≡ m+n ≤ x+y < m+n+2$$
This is where I get stuck; I have that $$[x+y] = t ≡ t ≤ x < t+1$$, so putting $$m+n ≤ x+y < m+n+2$$ in that form, seems impossible, let alone the one that corresponds to $$[x]+[y]+1$$.
Could you help me with this last part?
• Your inequalities should be less than signs on the right everywhere.That should improve things. – jgon Dec 19 '18 at 22:20
• Corrected; however, it was typo when passing notes from my notebook, I used the correct definition in my notebook, and still don't realize how to follow up. – Daniel Bonilla Jaramillo Dec 19 '18 at 22:23
• Why do you write: "$[x+y] = t ≡ t ≤ x < t+1$" Shouldn't that be "$[x+y] = t \equiv t \le x + y \le t + 1$"? – fleablood Dec 19 '18 at 22:44
$$m+n\le x+y or $$m+n+1$$, because the only integers in $$[m+n,m+n+2)$$ are $$m+n, m+n+1$$
You have $$[x] = m$$ and $$[y] = n$$ so $$m+n\le x + y < m+n+2$$
But how does $$x+y$$ compare to $$m + n + 1$$? There are two possibilities.
1) $$x + y < m+n + 1$$ then
$$m + n \le x + y < m+ n + 1$$. Then by definition you have $$[x+y] = m+n = [x]+[y]$$.
2) $$x + y \ge m+n + 1$$ then
$$m+n + 1 \le x+y < m+n + 2=(m+n+1) + 1$$ so by definition you have $$[x+y] = m+n+1$$.
That's it.
• How do you conclude the two possibilities "$x+y<m+n+1$" and "$x+y≥m+n+1$"? – Daniel Bonilla Jaramillo Dec 19 '18 at 23:05
• Trichotomy. Given two values $A$ and $B$ then you either have $A < B$ or $A\ge B$. What other options are there? How can neither be true? – fleablood Dec 19 '18 at 23:28
• @DanielBonillaJaramillo : For any real numbers $a$ and $b$, exactly one of “$a<b$”, “$a=b$”, or “$a>b$” holds. Combine the last two to get that exactly one of “$a<b$” or “$a\geq b$” holds. – MPW Dec 19 '18 at 23:28
I didn't read the question exactly, but you have floor(x)=n iff. $$n, hence you get $$... instead of $$\leq$$.
Let $$m=\lfloor x\rfloor$$ and $$n=\lfloor y\rfloor$$; let $$\{x\}=x-\lfloor x\rfloor=x-m$$ and $$\{y\}=y-\lfloor x\rfloor=y-n$$. Then $$x+y=m+\{x\}+n+\{y\}$$ Note that $$0\le\{x\}<1$$ and $$0\le\{y\}<1$$, so $$0\le\{x\}+\{y\}<2$$. There are two cases:
1. if $$0\le\{x\}+\{y\}<1$$, then $$m+n\le x+y and so $$\lfloor x+y\rfloor=m+n$$;
2. if $$1\le\{x\}+\{y\}<2$$, then $$m+n+1\le x+y and so $$\lfloor x+y\rfloor=m+n+1$$.
Alternatively. By definition $$[x+y]$$ is the largest possible integer that this less than or equal to $$x+y$$. But $$[x] \le x$$ and $$[y] \le y$$ so $$[x] + [y] \le x+y$$. So $$[x]+[y] \le [x+y]$$.
Likewise $$[x+y] + 1$$ by definition is the smallest possible integer that is larger $$x + y$$. But $$[x]+ 1 > x$$ and $$[y] + 1 > y$$ so $$[x]+[y] + 2 > x+y$$. So $$[x] + [y] + 2\ge [x+y]+1$$.
So $$[x]+[y] + 1 \ge [x+y]$$.
So $$[x]+[y] \le [x+y] \le [x] + [y] + 1$$
As $$[x]+[y]$$ and $$[x+y]$$ and $$[x] + [y]+1$$ are all integers. And there is NO integer between $$[x] +[y]$$ and $$[x]+[y]+1$$ there are only two options $$[x] + [y]= [x+y]$$ or $$[x+y] = [x] + [y]+1$$.
.......
And a third way.
$$[x] \le x < [x]+1$$ means $$0 \le x - [x] < 1$$.
A) $$0 \le x-[x] < 1$$ and $$0 \le y -[y] < 1$$ so $$0 \le x+y -[x]-[y] < 2$$.
B) $$0 \le x+y - [x+y] < 1$$.
Reverse B) to get
B') $$-1 < [x+y] - x - y \le 0$$.
Add B' and A to get:
$$-1 < ([x+y] - x - y) + (x + y - [x]-[y]) < 2$$ so
$$-1 < [x+y] - [x] -[y] < 2$$ or
$$0 \le [x+y] -[x] -[y] \le 1$$ or
$$[x]+[y] \le [x+y] \le [x] + [y] + 1$$.
.....
Basically: $$x+y$$ is between $$[x+y]$$ and $$[x+y] + 1$$; two integers that are only $$1$$ apart.
But $$x + y$$ is also between $$[x] + [y]$$ and $$[x] + [y] + 2$$; two integers that are only $$2$$ apart.
There are only so many choicese to find these integers $$[x+y], [x]+[y], [x+y] + 1$$ and $$[x]+[y] + 2$$ so that they all fit in such a tight range.
It doesn't matter how you prove it but you must have $$[x+y]$$ and $$[x]+ [y]$$ within one of each other and you must have $$[x]+[y]\le [x+y]$$. There are only two ways that can happen. | 2020-12-04T14:06:57 | {
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http://math.stackexchange.com/questions/52609/nonlinear-fubini-tonelli?answertab=votes | Nonlinear Fubini-Tonelli?
A student raised the question today: can one interchange the integrals in
$$\int_0^1 \left(\int_0^x f(y) \, dy \right)^2 \, dx$$
for sufficiently nice functions $f$. I answered no, with the reasoning that integrals are fundamentally linear operations.
However now that I think about it, it is certainly possible that some kind of non-linear analogy to Fubini-Tonelli exists and I'm simply not aware of it. Indeed, it seems quite plausible that some interchanged expression involving square roots is equal to the expression above, and perhaps in general something involving the inverse of the non-linear function of the inner integral.
Is anyone aware of a non-linear Fubini-Tonelli theorem?
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In the case of integral powers it is easiest if you write $$\int_0^1\left(\int_0^x f(y)dy\right)^2 dx = \int_0^1\int_0^x\int_0^x f(y)f(z)dz~dy~dx$$ for which you can apply the usual Fubini-Tonelli theorem as long as you keep track of which wedge you are integrating over. – Willie Wong Jul 20 '11 at 11:35
BTW, this type of "tensor power expansion" is used in, for example, some elementary proofs of Sobolev embedding theorems. – Willie Wong Jul 20 '11 at 11:37
@Glen: $f(y)$ or $f(x,y)$? If $f(y)$, can one interchange the integrals in the linear version? – Shai Covo Jul 20 '11 at 13:12
@Shai In this question let's leave it at $f(y)$, since that's where all the discussion has headed. I had $f(x,y)$ in mind, which one could probably guess from my comments, but my mistake for not writing it correctly in the first place :). – Glen Wheeler Jul 20 '11 at 13:35
Thanks for this response. However, with $f(y)$ it does not hold in general $\int_0^1 {(\int_0^x {f(y)\,dy} )\,dx} = \int_0^1 {(\int_y^1 {f(x)\,dx} )\,dy}$ (take for example $f(y)=y$). That is, one cannot change the order of integration already in the linear version. – Shai Covo Jul 20 '11 at 13:44
Following up on Willie Wong's comment: \begin{align} & \phantom{= {}} \int_0^1 \left( \int_0^x f(y)\,dy \right)^2\,dx = \int_0^1 \int_0^x \int_0^x f(y)f(z)\,dy\,dz \, dx = \int_0^1\int_0^1 \int_{\max\{y,z\}}^1 f(y)f(z)\, dx \, dy\, dz \\ & = \int_0^1\int_0^1 \left( f(y)f(z) \int_{\max\{y,z\}}^1 1\, dx \right) dy \, dz = \int_0^1\int_0^1 f(y)f(z) \left(1-{\max\{\,y,z\,\}}\right) \, dy \, dz. \end{align}
The integration with respect to $x$ now appears as $1-\max\{\,y,z\,\}$ on the inside; the square of the integral with respect to $y$ now appears as $\int_0^1\int_0^1 \cdots\,dy\,dz$ on the outside.
As for "dimensional" correctness, $dy$ and $dz$ are both in the same units as $dy$ in the original integral; $f(y)$ and $f(z)$ are both in the same units as $f(y)$ in the original integral; and $x$ has to be in the same units as $y$ in the original integral; so $x$ is in the same units as $1-\max\{\,y,z\,\}$. So the units are of the form $a^2 b^3$ on both sides of the "$=$".
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Let me state Minkowski's integral inequality:
Let $1\leq p <\infty$. Let $F$ be a measurable function on the product space $(X,\mu)\times (T,\nu)$, where $\mu,\nu$ are $\sigma$-finite. We have
$\left[\int_{T}\left(\int_{X}\left|F(x,t)\right|d\mu(x)\right)^pd\nu(t)\right]^{\frac{1}{p}}\leq\int_{X}\left[\int_{T}\left|F(x,t)\right|^pd\nu(t)\right]^{\frac{1}{p}}d\mu(x)$.
If I may say so, the following proof is quite elegant (in fact, the above result is Exercise 1.1.6., page 12 of Classical Fourier Analysis by Loukas Grafakos, and the proof below is my solution to this exercise):
Proof. Let $G:T\to [0,\infty]$ be defined by $G\left(t\right)=\int_{X} \left|F\left(x,t\right)\right|d\mu\left(x\right)$. Then Minkowski's integral inequality is nothing but a reformulation of the assertion that $\left\|G\right\|_{L^p\left(T\right)}\leq \int_{X} [\int_{T} \left|F\left(x,t\right)\right|^p d\nu\left(t\right)]^{\frac{1}{p}} d\mu\left(x\right)$. To prove this assertion, we first note the well-known fact that $\left\|G\right\|_{L^p\left(T\right)}=\sup_{g\in L^{q}\left(T\right)} \left\|Gg\right\|_{L^1\left(T\right)}$; i.e., the norm of $G$ as an operator on $L^q\left(T\right)$ equals the $L^p$-norm of $G$. Now compute that for $g\in L^q\left(T\right)$, H\"older's inequality gives
$\left\|Gg\right\|_{L^1(T)} = \int_T \left|\int_X \left|F\left(x,t\right)\right|g\left(t\right)d\mu\left(x\right)\right|d\nu\left(t\right)$ $= \int_X \int_T \left|F\left(x,t\right)\right|\left|g\left(t\right)\right| d\nu\left(t\right)d\mu\left(x\right)$ $\leq \int_X \left\|F\right\|_{L^p\left(T\right)} \left\|g\right\|_{L^q\left(T\right)} d\mu\left(x\right)$ $= \int_X \left(\int_T \left|F\left(x,t\right)\right|^p d\nu\left(t\right)\right)^{\frac{1}{p}} d\mu\left(x\right)$,
where the second equality follows from Fubini's theorem and where we have used the notation $\left\|F\right\|_{L^p\left(T\right)}=\left(\int_T \left|F\left(x,t\right)\right|^p d\nu\left(t\right)\right)^{\frac{1}{p}}$. This proves Minkowski's integral inequality. Q.E.D.
The idea of the proof is to remove the exponents (of course, in the case $p=1$, the result is Tonelli's theorem). This can be done using duality. I hope this helps!
-
This is certainly related. One can use several standard techniques to estimate the integral (from above and below) and then the cases of equality gives expressions equivalent to that above in special cases. But Fubini-Tonelli is fundamentally an equality. Still, thanks for the answer. +1 – Glen Wheeler Jul 20 '11 at 11:04
Well, you certainly cannot have an equality from naively exchanging $L^1$ and $L^2$. Minkowski's inequality is sharp; IIRC you can check Hardy-Littlewood-Polya, where the condition for equality is given: equality only if $F(x,t)$ in the notation above factors as $X(x)T(t)$. – Willie Wong Jul 20 '11 at 11:42
The following point of view may be helpful: the usual Minkowski inequality is the triangle inequality for the $L^p$ norm. As for every norm, the triangle inequality can be generalized (by simple induction) to bound the norm of a finite sum of vectors/functions by the sum of their norms. One can then generalize this to infinite sums by (carefully) taking limits). Minkowski's integral inequality generalizes it to "arbitrary uncountable sums", that is integrals w.r.t. some measure (counting measure in the above cases). – Mark Jul 20 '11 at 13:04
@Glen: Dear Glen, you are right, of course. I have noted the Minkowski integral inequality as a result relevant to your question. I thought that it might be helpful. However, I certainly did not claim it is an equality (and it is not, in general). – Amitesh Datta Jul 21 '11 at 1:06
I'd like to post the following link to a related question on this site: math.stackexchange.com/q/33659/8157 – Giuseppe Negro Aug 11 '11 at 15:15 | 2015-04-18T08:06:39 | {
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https://math.stackexchange.com/questions/1814379/equation-of-a-tangent-on-a-circle-given-the-gradient-and-equation-of-the-circle | Equation of a tangent on a circle given the gradient and equation of the circle
My maths teacher told me this problem was impossible without knowledge of implicit differentiation: is she right?
You are given the equation of the circle $\left(x+2\right)^2+\left(y-2\right)^2=16$ , what are the equations of the lines with a gradient of 2 which are tangents of this circle.
Here is a visualisation of the problem
The red circle is the one I described, and the green and blue lines are the equations I'm looking for.
I think the solution will have something to do with plugging in the value of 2 in some sort of equation for the gradient i.e. $2=\frac{x-x_0}{y-y_0}$ , and then the discriminant the equation you would use to find where they intersect can be solved to find values of $x$ , but I'm not sure how to find these values without using any implicit differentiation -can it be done?
EDIT:
Ok so now I'll try and run this through step by step until i find another problem:
So we can say that there's a line (the diameter) of gradient -1/2 which passes through the centre (-2,2):
$y=mx+c$
$y=-\frac{1}{2}x+c$
$2=1+c$
$y=-\frac{1}{2}x+1$
We can also say that a solution lies on the point $(a,b)$ such that
$\left(a+2\right)^2+\left(b-2\right)^2=16$ and $b=-\frac{1}{2}a+1$
So we can substitute b in the equation of the circle to arrive at
$\left(a+2\right)^2+\left(-\frac{1}{2}a-1\right)^2=16$
which simplifies down to
$\frac{5}{4}a^2+5a-11=0$
and solves to give $\frac{2}{5}\left(-5-4\sqrt{5}\right)$ and $\frac{2}{5}\left(4\sqrt{5}-5\right)$
so these are the possible values for $x$
Now if i substitute 1 of these back into the equation, I'll get two values of $y$ for the one value of $x$ because it's quadratic, but there's only 1 intersection point at each value of $x$ , what do i do at this point?
EDIT 2:
It looks like both the intersection points of the line $x=\frac{2}{5}\left(4\sqrt{5}-5\right)$ correspond to the $y$ values of the intersection points of the tangents, does this mean i only need to plug one of these $x$ values back into the equation? Visualisation to show you what i mean (the blue line appears to intersect the circle at both the appropriate $y$ values)
• Do you the know the answer . I got $y=2x+7,y=2x-11$ – Archis Welankar Jun 5 '16 at 11:00
• I think you need to find the equation of the diameter that meets the two lines you are searching for at the tangent points. That line passes through the center of the circle (-2,2) and its slope is -0.5. You can continue from here... – Noam Dolovich Jun 5 '16 at 11:03
• From trial and error the answer is $y=2x+6+4\sqrt{5}$ or $y=2x+6-4\sqrt{5}$ , working it out fully now. – Cubbs Jun 5 '16 at 11:31
• @Cubbs "Trial and error"?? How? Do you check all the pointsw on the circle? I don't think this would be well accepted in most schools. – DonAntonio Jun 5 '16 at 11:42
• Just estimated it to 3 d.p. using a graphing calculator, then wolfram alpha can give you possible closed forms, one of which was that surd :) – Cubbs Jun 5 '16 at 11:54
Since a tangent to a circle is perpendicular to the circle's radius at that points, we're looking for points $\;(a,b)\;$ on the circle such that the radius through them has a gradient (slope) of $\;-\frac12\;$ , so: as the circle's center is $\;(-2,2)\;$ , the slope from this point to $\;(a,b)\;$ is
$$\frac{b-2}{a+2}=-\frac12\implies -2b+4=a+2$$
but we also have
$$(a+2)^2+(b-2)^2=16\implies16=(-2b+4)^2+(b-2)^2=5(b-2)^2\implies$$
$$b-2=\pm\frac4{\sqrt5}\implies b=2\pm\frac4{\sqrt5}$$
and there you have the $\;y\,-\,$ coordinates of the wanted points. Now find out the $\;x\,-\,$ coordinates and that's all, since once you have the points you can alreayd write down the lines through them as you already have the slope (gradient): $\;2\;$
No calculus at all needed...though, perhaps, it could make things slightly simpler and faster.
$$(x+2)^2 + (y-2)^2 = 16$$ $$\Rightarrow x^2 + y^2+4x-4y+4+4 = 16$$ $$\Rightarrow x^2 + y^2+4x-4y-8= 0$$ $$\text{centre} =(-2,2), \text{ radius}=4$$ for tangent the line must satisfies the condition, $r=d$ we have line $y=2x+c$ here gradient$=2$ need to find $c=$? $$\pm4 =\frac {2(-2)-2+c}{2^2+1}$$ $$\pm 20 =-6 +c$$ $$c= 26,-14$$ now the tangents are $$y=2x+26, y=2x-14$$ are required tangents | 2019-06-18T06:36:47 | {
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https://math.stackexchange.com/questions/3846064/confusion-about-the-description-of-generalized-eigenspace | # Confusion about the description of generalized eigenspace
When learning about generalized eigenspace, there are two statements from two different textbooks which I was learning, seems contradict to the other one.
In Chapter 8 of the book Linear Algebra Done Right, 3rd edition by Sheldon Alxer, a theorem(Theorem 8.11 , Description of generalized eigenspaces) was stated like this
Suppose $$\mathcal{T}\in \mathcal{L}(V)$$ and $$\lambda\in\mathbb{F}$$. Then $$G(\lambda, \mathcal{T})=\text{Null}(\mathcal{T}-\lambda \mathcal{I})^{\text{dim}V}$$
Here $$G(\lambda, \mathcal{T})$$ means the generalized eigenspace of $$\mathcal{T}$$ corresponding to eigenvalue $$\lambda$$, and "Null" stands for the kernal space.
While in another textbook(Linear Algebra, Special for mathematic major by Shangzhi Li), there is also a similar theorem. Here's what it said (translated from Chinese, which can be a little bit unprecise)
Suppose an operator $$\mathcal{T}$$ defined on a n-dimensional linear space $$V$$ has $$t$$ different eigenvalues $$\lambda_1,...,\lambda_t$$, and the characteristic polynomial has the form$$P_\mathcal{T}(\lambda)=(\lambda-\lambda_1)^{n_1}...(\lambda-\lambda_t)^{n_t}.$$Then for each eigenvalule $$\lambda_i(1\leq i\leq t)$$, a subspace $$\text{Null}(\mathcal{T}-\lambda_i \mathcal{I})^{n_i}$$ was formed by a zero vector and all of the generalized eigenvectors with respect to $$\lambda_i$$, which has $$n_i$$ dimensions.
Now I'm getting quite unsure about the description of generalized eigenspace.
Comparing these two theorems. The first one states that the generalized eigenspace $$G(\lambda, \mathcal{T})$$ can be described as $$\text{Null}(\mathcal{T}-\lambda\mathcal{I})^{\text{dim}V}$$, while the second one suggests that the description of generalized eigenspace with respect to $$\lambda_i$$ seems ought to have the form $$\text{Null}(\mathcal{T}-\lambda_i\mathcal{I})^{n_i}$$, where $$n_i$$ is known to be the algebraic multiplicity of $$\lambda_i$$, appeared from the characteristic polynomial. However, it is clear that $$\text{Null}(\mathcal{T}-\lambda_i\mathcal{I})^{n_i}\neq \text{Null}(\mathcal{T}-\lambda_i\mathcal{I})^{\text{dim}V}$$ for some specified eigenvalue $$\lambda_i$$, which seems like a contradiction.
More interestingly, there is also an exercise problem in today's aftercalss-assignment which gives me an even more confusing conclusion.
Suppose $$\mathcal{T}\in\mathcal{L}(V)$$ has a minimal polynomial $$D_{\mathcal{T}}(\lambda)=(\lambda-\lambda_1)^{k_1}...(\lambda-\lambda_t)^{k_t}$$ Prove that $$G(\lambda_i, \mathcal{T})=\text{Null}(\mathcal{T}-\lambda\mathcal{I})^{k_i}$$
From now on I'm getting totally dizzy... is it possible that these three statements are not all correct? Or if I missed some important things? I do think it is not possible that $$\text{Null}(\mathcal{T}-\lambda_i \mathcal{I})^{\text{dim}V}=\text{Null}(\mathcal{T}-\lambda_i \mathcal{I})^{n_i}=\text{Null}(\mathcal{T}-\lambda_i \mathcal{I})^{k_i}$$
Can anyone help me with that? Thanks a lot!
• "However, it is clear ..." is wrong. The sequence $\ker (T - \lambda)^k$ stabilises, i.e. there is a $k_0$ such that $\ker (T - \lambda)^k = \ker (T - \lambda)^{k+1}$ for $k \geqslant k_0$. And that $k_0$ is not larger than the algebraic multiplicity of the eigenvalue $\lambda$. – Daniel Fischer Sep 30 '20 at 11:31
• @Daniel Fischer Could you please show me why such $k_0$ exists? And why it is not larger than the algebraic multiplicity of the eigenvalue? Thanks – Scanners Sep 30 '20 at 11:55
• Let $d_k = \dim \bigl(\ker (t - \lambda)^k\bigr)$. Then $0 = d_0 \leqslant d_1 \leqslant d_2 \leqslant \ldots$, and on the other hand $d_k \leqslant \dim V$ for all $k$. A bounded monotonic sequence of integers is eventually constant. That $k_0$ is the exponent of $(X - \lambda)$ in the minimal polynomial of $T$ is more or less immediate from the definitions. If you know the Jordan normal form, you can also see it in that. – Daniel Fischer Sep 30 '20 at 12:00
• @ Daniel Fischer♦ Okay I think I've got the key point, however it is still hard for me to understand why $k_0$ is the exponent of $X-\lambda$, can you elaborate on that a little bit? P.s. I do not think Axler's book includes very much about Jordan normal form. Can you recommend some textbooks which expain that section well? – Scanners Sep 30 '20 at 12:23
• @Scanners Axler's book does say a bit about why the dimension of the generalized eigenspace is the exponent of $(x - \lambda)$ (except that he "defines things backwards" in a sense). See in particular theorem 8.10 (p. 169 second edition) – Ben Grossmann Sep 30 '20 at 12:49
As Daniel says in the comments, all of these definitions are equivalent; that is,
$$\text{ker}(T - \lambda)^{\dim V} = \text{ker}(T - \lambda)^{n_i} = \text{ker}(T - \lambda)^{k_i}.$$
This can be seen using Jordan normal form but that's overkill. Let $$m(t) = \prod (t - \lambda_i)^{k_i}$$ be the minimal polynomial of $$T$$ and let $$v \in \text{ker}(T - \lambda_j)^{\dim V}$$ for some fixed $$j$$. Then we have
$$m(T) v = \left( \prod_{i \neq j} (T - \lambda_i)^{k_i} \right) (T - \lambda_j)^{k_j} v.$$
We want to show that $$v_j = (T - \lambda_j)^{k_j} v = 0$$. The identity above gives that $$v_j$$ lies in the kernel of $$\prod_{i \neq j} (T - \lambda_i)^{k_i}$$. This is contained in (and in fact is exactly) the sum of the generalized eigenspaces of each $$\lambda_i, i \neq j$$, and a basic fact about generalized eigenspaces of different eigenvalues is that they are linearly independent. On the other hand, $$v_j$$ lies in the generalized eigenspace of $$\lambda_j$$. Hence $$v_j = 0$$ as desired.
More explicitly, we can prove the following.
Lemma: Suppose $$p, q \in \mathbb{C}[t]$$ are two polynomials such that $$p(T) v = q(T) v = 0$$. Then $$g = \gcd(p, q)$$ also satisfies $$g(T) v = 0$$.
Proof. Abstractly the point is that $$\{ p \in \mathbb{C}[t] : p(T) v = 0 \}$$ is an ideal of $$\mathbb{C}[t]$$ and hence principal. Concretely, we can apply Bezout's lemma to find polynomials $$a, b$$ such that $$ap + bq = g$$, which gives $$a(T) p(T) v + b(T) p(T) v = 0 = g(T) v$$. $$\Box$$
So $$v_j = (T - \lambda_j)^{k_j} v$$ satisfies $$(T - \lambda_j)^{\dim V - k_j} v_j = 0$$, but it also satisfies $$\left( \prod_{i \neq j} (T - \lambda_i)^k \right) v_j = 0$$, and these two polynomials have no roots in common so their $$\gcd$$ is equal to $$1$$. Hence $$v_j = 0$$. | 2021-01-26T13:42:22 | {
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https://math.stackexchange.com/questions/2054312/double-check-my-steps-to-find-multiplicative-inverse | # double check my steps to find multiplicative inverse?
assume we want to find the multiplicative inverse for $117$ in $Z337$.
i know that in order to find the the multiplicative inverse we use Euclidean and Extended Euclidean algorithm.
Eculidian:
$337 = 2*117 + 103$
$117 = 1*103 + 14$
$103 = 7*14 + 5$
$14 = 2*5 + 4$
$5 = 1*4 + 1$
Extended Euclidean:
not going to include the whole solution because i am pretty sure of it, i get:
$1=25*337-72*117$
Extended Euclidean gives us that the inverse is $-72$. but how come a calculator says the inverse is $265$ and so do the teacher. do i need to do anything more ? usually i only need to do Euclidean and Extended Euclidean to find the inverse.
• $-72\equiv 265\ (\textrm{ mod }337)$ – King Ghidorah Dec 11 '16 at 19:17
• ohh i see, is it wrong to say that $-72$ is the multiplicative inverse? is it enough to stop at $-72$ ? – pabloBar Dec 11 '16 at 19:21
• If the final result is negative, then you need to add the order ($337$ in this case). – barak manos Dec 11 '16 at 19:21
• I fixed some typos in your question (you had $177$ vs. $117)\ \$ – Bill Dubuque Dec 11 '16 at 20:14
• ops, sorry it should be 117. Thanks ! – pabloBar Dec 11 '16 at 20:16
Either answer is correct since they are both congruent, i.e. $$\,{\rm mod}\,\ 337\!:\ \color{#c00}{{-}72}\equiv 337-72\equiv 265.\$$ Below is the complete calculation by a fractional form of the Extended Euclidean Algorithm
$${\rm mod}\ 337\!:\,\ \dfrac{0}{337} \overset{\large\frown}\equiv \dfrac{1}{117} \overset{\large\frown}\equiv \dfrac{-3}{\color{#0a0}{-14}} \overset{\large\frown}\equiv \dfrac{-23}5 \overset{\large\frown}\equiv\color{#c00}{\dfrac{-72} {1}}\overset{\large\frown}\equiv\dfrac{0}0\,$$ or, equivalently, in equational form
$$\qquad\ \ \ \begin{array}{rrl} [\![1]\!]\!:\!\!\!& 337\,x\!\!\!&\equiv\ \ 0\\ [\![2]\!]\!:\!\!\!& 117\,x\!\!\!&\equiv\ \ 1\\ [\![1]\!]-3[\![2]\!]=:[\![3]\!]\!:\!\!\!& \color{#0a0}{{-}14}\,x\!\!\!&\equiv -3\\ [\![2]\!]+8[\![3]\!]=:[\![4]\!]\!:\!\!\!& 5\,x\!\!\! &\equiv -23\\ [\![3]\!]+3[\![4]\!]=:[\![5]\!]\!:\!\!\!& \color{#c00}1\, x\!\!\! &\equiv \color{#c00}{-72} \end{array}$$
Remark $$\$$ This is essentially the augmented matrix form of the exteneded Euclidean algorithm, optimized by omitting one column, then interpreting the linear congruences as modular fractions.
Allowing negative remainders (i.e. least magnitude reps) often simplifies calculations, e.g.$$\,10^{n}\equiv (-1)^{n}\equiv \pm1\pmod{11}\$$ is used to calculate remainders mod $$11$$ via alternating digit sums (casting out elevens). I did so above: $$\bmod 117\!:\ 337 \equiv \color{#0a0}{{-}14}\ ({\rm vs.}\ 103\,$$ in your calculation). Using the samller magnitude residue $$\,-14\,$$ simplifies subsequent calculations (it eliminates one step from your calculations, but generally such choices save many steps in longer calculations).
See this answer for another worked example.
• Thanks for the answer, but i wonder if its mathematically correct to say that $-72$ is the multiplicative inverse? is it ok if the multiplicative inverse is negative ? – pabloBar Dec 11 '16 at 19:30
• In@pabloBar $\ 117^{-1}\equiv -72\pmod{337}\,$ is correct, but $\,{ -}72 = (117^{-1}\bmod 337)\,$ is not correct. Do you understand the difference? – Bill Dubuque Dec 11 '16 at 19:33
• how would you translate these expressions in words?, sorry a bit bad when it comes the math expressions. – pabloBar Dec 11 '16 at 19:36
• @pabloBar $\ a\equiv b\pmod n\,$ means $\,n\,$ divides $\,a-b,\,$ but $\, a = (b\bmod n)\,$ means the same plus $\, 0\le a < n,\,$ i.e. $\,a\,$ is the least nonnegative integer $\equiv b\pmod n,\,$ i.e. the remainder left on dividing $\,b\,$ by $\,n.$ $\ \$ – Bill Dubuque Dec 11 '16 at 19:41
• on exam i would do the calculations above and then state with the exact words that "the multiplicative is $-72$", would my statement be correct in this case ? – pabloBar Dec 11 '16 at 19:47 | 2019-12-13T08:49:38 | {
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https://www.physicsforums.com/threads/confusion-over-rolling-with-friction-and-rolling-without-slipping.657635/ | # Homework Help: Confusion over Rolling with Friction and Rolling Without Slipping
1. Dec 8, 2012
### mhz
1. The problem statement, all variables and given/known data
This isn't a specific question, more of a general one:
Suppose there is a cylinder of mass $m$ that is rotating in the positive clockwise direction with initial rotational velocity $\omega_0$, and radius $R$.
Then, suppose this rotating cylinder is placed on a surface with coefficient of kinetic friction $\mu_k$.
How long after it is placed this surface does it begin to roll without slipping?
2. Relevant equations
$\tau = I\alpha \\ F = ma \\ I_{cyl} = \frac{mR^2}{2} \\ v = v_0 + at \\ \omega = \omega_0 + \alpha t \\$
3. The attempt at a solution
I'm torn between two approaches.
The first - couldn't I simply solve for a and alpha, set the two linear/angular velocity equations equal to each other for the case that v = r(omega) and solve for t?
The second - using energy, I could say that the initial rotational kinetic energy equals the final rotational kinetic energy plus toe final kinetic energy plus the energy lost to friction, find d or final velocity when v = r(omega) and use kinematics to solve for t?
2. Dec 8, 2012
### TSny
First approach sounds good (devil in the details).
Second approach sounds complicated to me. The cylinder slips on the surface so the the distance that friction acts is not the same as the distance the cylinder travels. But if you take that into account, I guess this approach should work.
3. Dec 8, 2012
### mhz
So would this be correct?
$\tau = I\alpha = -fR \implies \alpha = \frac{-2\mu_kg}{R} \\ -f=ma \implies a = - \mu_k g \\ v = v_0 + at \implies v = at \text{ Since there is no initial velocity} \\ \omega = \omega_o + \alpha t \implies at = r\omega_0 + r\alpha t \implies -\mu_k g t = r \omega_0 = 2\mu_k g t \implies t = \frac{r\omega_0}{\mu_k g}$
4. Dec 8, 2012
### TSny
I think it looks good except for one point. Why did you write $-f = ma$ with a negative sign for the force?
5. Dec 8, 2012
### mhz
To be consistent, considering that frictional force acts in the negative direction, if both were positive the final answer would be negative
Also I have lowercase f as frictional force, F as net force
6. Dec 8, 2012
### TSny
What direction are you taking to be positive for the linear motion? If the cylinder is initially spinning clockwise, what direction will it the travel along the surface? Is the direction it travels also your positive direction?
7. Dec 8, 2012
### mhz
- <-----------> +
Clockwise = Positive
Actually, shouldn't the angular acceleration be positive? As the frictional force applies a positive torque.
8. Dec 8, 2012
### TSny
Think carefully about the direction of the friction force on the cylinder while it is slipping on the surface. Is it to the right or to the left?
9. Dec 8, 2012
### mhz
Assuming the ball is moving to the right, I would imagine the friction force is to the left.
10. Dec 8, 2012
### TSny
What propels the ball to move to the right?
11. Dec 8, 2012
### rcgldr
One possible source of confusion is that that the friction force could refer to the force the cylinder exerts onto the surface, or the force the surface exerts onto the cylinder (a newton third law pair).
Assuming you want to consider the force the surface exerts on the cylinder, then if the cylinder is slidng and spinning clockwise, the direction of force from the surface onto the cylinder is to the right, resulting in linear acceleration to the right and angular deceleration of the cylinder (also keep in mind your convention of ω being positive for clockwise rotation). If you allow α to be negative (since there is angular deceleration), then you have:
f = m a
f R = τ = I α
and you can continue with rest of the equations you've already determined.
12. Dec 8, 2012
### mhz
Here is a solution I found:
However, why don't they account for energy lost due to friction?
13. Dec 8, 2012
### TSny
Everything in your solution was correct except the direction of the friction force. As the cylinder spins clockwise and slips on the surface, it scrapes against the surface and tries to push the surface to the left. So, the surface pushes on the cylinder to the right (+ direction). That force to the right on the cylinder produces a counterclockwise torque (your - direction). So, you had the sign of the torque correct. It is just the direction of $f$ in $F = ma$ that you need to correct.
14. Dec 8, 2012
### TSny
Right. That solution is bogus. They also have the friction on the cylinder pointing in the wrong direction as well as not taking into account heat produced by friction. Note that if the direction of the friction is as shown, then the cylinder would increase it's rate of spin!
15. Dec 8, 2012
### mhz
Haha oh man, it is quite challenge studying for my Physics final using the finals and solutions posted by my professors when half the answers are wrong (this has happened for maybe 6 or 7 questions since I've begun studying lol).
Anyway that makes perfect sense guys, thanks a lot!
16. Dec 8, 2012
### haruspex
So, what do you get for the answer now? It's Rω0/(3gμk), right?
17. Dec 8, 2012
### mhz
Yes that's right
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 2018-08-20T13:26:59 | {
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https://www.cs.toronto.edu/~lczhang/338/hw/hw8.html | # CSC338. Homework 8¶
Due Date: Wednesday March 18, 9pm
Please see the guidelines at https://www.cs.toronto.edu/~lczhang/338/homework.html
### What to Hand In¶
• Python File containing all your code, named hw8.py.
• PDF file named hw8_written.pdf containing your solutions to the written parts of the assignment. Your solution can be hand-written, but must be legible. Graders may deduct marks for illegible or poorly presented solutions.
If you are using Jupyter Notebook to complete the work, your notebook can be exported as a .py file (File -> Download As -> Python). Your code will be auto-graded using Python 3.6, so please make sure that your code runs. There will be a 20% penalty if you need a remark due to small issues that renders your code untestable.
Make sure to remove or comment out all matplotlib or other expensive code before submitting your homework!
Submit the assignment on MarkUs by 9pm on the due date. See the syllabus for the course policy regarding late assignments. All assignments must be done individually.
In [1]:
import math
import numpy as np
## Question 1. Fixed-Point Iteration -- 3pts¶
In homework 7, we considered using fixed-point iteration to find a root of the equation $f(x) = x^2 - 3x + 2 = 0$ using these functions:
1. $g_1(x) = \frac{x^2 + 2}{3}$
2. $g_2(x) = \sqrt{3x - 2}$
3. $g_3(x) = 3 - \frac{2}{x}$
Analyze the convergence properties of each of the corresponding fixed-point iteration schemes for the root x = 2 by analyzing $g_i'(x)$. Do you expect the fix-point iteration to diverge or converge? Do these expectations match your empirical results from homework 7?
## Question 2. Newton's Method¶
### Part (a) -- 4 pts¶
Write a function newton to find a root of f(x) using Newton's Method. This Python function should take as argument both the mathematical function f and its derivative df, and return a list of successively better estimates of a root of f obtained from applying Newton's method.
In [2]:
def newton(f, df, x, n=5):
""" Return a list of successively better estimate of a root of f
obtained from applying Newton's method. The argument df is the
derivative of the function f. The argument x is the initial estimate
of the root. The length of the returned list is n + 1.
Precondition: f is continuous and differentiable
df is the derivative of f
>>> def f(x):
... return x * x - 4 * np.sin(x)
>>> def df(x):
... return 2 * x - 4 * np.cos(x)
>>> newton(f, df, 3, n=5)
[3,
2.1530576920133857,
1.9540386420058038,
1.9339715327520701,
1.933753788557627,
1.9337537628270216]
"""
### Part (b) -- 2 pts¶
Use your function from part (a) to solve for a root of
$$f(x) = x^2 - 3x + 2 = 0$$
Start with $x_0$ = 3, and stop when the root is accurate to at least 8 significant decimal digits.
Show your work in your python file, and store the root you find in the variable newton_root.
In [3]:
def f(x):
return x ** 2 - 3 * x + 2
newton_root = None
### Part (c) -- 6 pts¶
Consider the following non-linear equations $h_i(x) = 0$.
1. $h_1(x) = x^3 - 5x^2 + 8x - 4$
2. $h_2(x) = x cos(20x) - x$
3. $h_3(x) = e^{-2x} + e^{x} - x - 4$
Write out the statement for updating the iterate $x_k$ using Newton’s method for solving each of the equations $h_i(x) = 0$.
For each $h_i$, do you expect Newton's method to converge?
### Part (d) -- 3 pt¶
Use the newton function to try and solve $h_i(x) = 0$, for n = 100 iterations, starting with x = 1.5.
Save the return values of calls to the function newton to the variables newton_h1, newton_h2, newton_h3,.
In [4]:
newton_h1 = None
newton_h2 = None
newton_h3 = None
## Question 3. Secant Method¶
### Part (a) [5 pt]¶
Write a function secant to find a root of f(x) using the secant method. The function should return a list of successively better estimates of a root of f obtained from applying the secant method.
In [5]:
def secant(f, x0, x1, n=5):
""" Return a list of successively better estimate of a root of f
obtained from applying secant method. The arguments x0 and x1 are
the two starting guesses. The length of the returned list is n + 2.
>>> secant(lambda x: x ** 2 + x - 4, 3, 2, n=6)
[3,
2,
1.6666666666666667,
1.5714285714285714,
1.5617977528089888,
1.5615533980582523,
1.561552812843596,
1.5615528128088303]
"""
### Part (b) [1 pt]¶
Use the secant function to find a root of $f(x) = x^3 + x^2 + x - 4$, accurate up to 8 significant decimal digits.
Show your work in your Python file. You can choose starting positions $x_0$ and $x_1$.
Save the result in the variable secant_root.
In [6]:
secant_root = None
### Part (c) [4 pt]¶
Show that the iterative method
$$x_{k+1} = \frac{x_{k-1} f(x_k) - x_k f(x_{k-1})}{f(x_k) - f(x_{k-1})}$$
is mathematically equivalent to the secant method for solving a scalar nonlinear equation $f(x) = 0$.
$$x_{k+1} = x_k - f(x_k)\frac{x_k - x_{k-1}}{f(x_k) - f(x_{k-1})}$$ | 2020-05-29T04:24:00 | {
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http://math.stackexchange.com/questions/206866/expected-number-of-pareto-optimal-points | # Expected number of Pareto-optimal points
Suppose $S$ is a set of $n$ points in a plane. A point is called maximal (or Pareto-optimal) if no other point in $S$ is both above and to the right of that point.
If each point in $S$ is chosen independently and uniformly at random from the unit square $[0,1]\times [1,0]$. What is the exact expected number of Pareto-optimal points in $S$?
-
+1 for an interesting and slightly unusual probability question. I am familiar with Pareto-optimality in economics but haven't previously encountered the concept in a pure maths context. Is the question prompted by any practical problem or relevant literature, or did it just occur to you? – Adam Bailey Oct 4 '12 at 11:53
This question is from the headbanging session of an undergraduate algorithms in the computer science department. – CaptainObvious Oct 4 '12 at 17:19
Nice question, which received a nice answer. – Did Nov 29 '12 at 6:47
Let $I_i$ be the indicator random variable denoting whether the ith point in $S$ is pareto optimal. Then the number of pareto-optimal points in $S$ is $M = \sum_{i=1}^{n} I_i$ and $E[M] = \sum_{i=1}^n E[I_i]$ by the linearity of expectation. Since $I_i$ is a binary valued random variable, we also have $E[I_i] = P(I_i = 1)$. Putting this together, we get $E[M] = \sum_{i=1}^n P(I_i = 1)$.
To evaluate $P(I_i=1)$, we need to find out the probability that there is no point above and to the right of the ith point. This can be done by first conditioning on the ith point's location within the unit square and then demanding that all the remaining $(n-1)$ points not lie in the smaller rectangle above and to its right. If the ith point is $(x_i,y_i)$, we want none of the remaining $(n-1)$ points in the rectangle of sides $1-x_i$ and $1-y_i$. The probability that a random point lies in this rectangle is simply its area and the probability that none of the $(n-1)$ points lie in this rectangle is therefore $(1-(1-x_i)(1-y_i))^{n-1}$.
\begin{eqnarray} P(I_i=1 \mid \text{ith pt} = (x_i,y_i)) &=& (1-(1-x_i)(1-y_i))^{n-1} \\ P(I_i=1) &=& \int_0^1 \int_0^1 (1-(1-x_i)(1-y_i))^{n-1} dx_i dy_i \\ &=& \int_0^1 \int_0^1 (1-uv)^{n-1} du dv \end{eqnarray}
This integral is pretty easy to compute. By holding $v$ constant, we can first evaluate the inner integral as
$\int_0^1 (1-uv)^{n-1} du = \frac{1}{nv} (1-(1-v)^{n})$
Making the substitution $v \rightarrow 1-v$, we get
\begin{eqnarray} P(I_i = 1) &=& \frac{1}{n} \int_0^1 \frac{1-v^n}{1-v} dv \\ &=& \frac{1}{n} \int_0^1 (1+v+\dots+v^{n-1}) dv \\ &=& \frac{1}{n} \left(1+\frac{1}{2}+\dots+\frac{1}{n}\right) = \frac{H_n}{n} \end{eqnarray}
Finally, we compute $E[M] = \sum_{i=1}^n P(I_i=1) = H_n$, where $H_n$ is the nth Harmonic number.
-
Label the n points $p_1,…,p_n$ in descending order of their y coordinates. $p_1$ is maximal since it is above all other points, and its expected contribution to the total number of maximal points is therefore 1.
$p_2$ is above points $p_3,…,p_n$ but below $p_1$, and is therefore maximal iff it is to the right of $p_1$. Since the distribution of $p_1$ and $p_2$ in the x dimension is random and independent of that in the y dimension, it is equally likely that either one of them will be to the right of the other. The probability that $p_2$ is maximal is therefore $\frac 1 2$, and its expected contribution to the total number of maximal points is $\frac 1 2$ x 1 = $\frac 1 2$.
Now consider any point $p_m$. Generalising from the case of $p_2$, $p_m$ is above points $p_{m+1},…,p_n$ but below points $p_1,…,p_{m-1}$, and is therefore maximal iff it is to the right of all of $p_1,…,p_{m-1}$. Since the distribution of $p_1,…,p_m$ in the x dimension is random and independent of that in the y dimension, it is equally likely that any one of them will be to the right of all the others. The probability that $p_m$ is maximal is therefore $\frac 1 m$, and its expected contribution to the total number of maximal points is $\frac 1 m$ x 1 = $\frac 1 m$.
The expected number of maximal points is therefore 1 + $\frac 1 2$ … + $\frac 1 n = H_n$, the $n$th Harmonic number.
-
Very nice solution that doesn't use any non-obvious integrals. Just wanted to highlight that indicator rvs and linearity of expectation play a role in this argument too. – Dinesh Dec 4 '12 at 4:23
@Dinesh Thank you for your comment. Yes, I agree that a rigorous statement would require indicator variables and linearity of expectation. Sometimes there is a trade-off between rigour and conciseness ... – Adam Bailey Dec 4 '12 at 8:35 | 2015-05-23T07:27:37 | {
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https://www.physicsforums.com/threads/maximum-weight-underwater.436625/ | # Maximum weight underwater
Hello, I would like to ask if my calculations are correct and which approach I should use if the answer is correct (I apologize if my calculations look ugly, just joined this forum).
## Homework Statement
On land, the maximum weight of a concrete block you can carry is 25kg. How massive block could you carry underwater, if the density of concrete is 2200kg/m³?
pwater = 1000 kg/m³
pconcrete = 2200 kg/m³
m1 = 25 kg
m2 = ?
G = mg
g = 9,81 m/s²
## Homework Equations
Archimedes' Principle
Fapplied - G + Fbuoyancy = 0
G - Fbuoyancy
## The Attempt at a Solution
Approach 1:
Fapplied - G + Fbuoyancy = 0 ->
m1g - m2g + m2pwaterg / pconcrete = 0
And I come to this:
m2 = -m1 / ( (pwater/pconcrete) - 1 )
m2 = 45,8333..... kg
Approach 2:
On land I need (F= mg) 245,25 N to carry the concrete block
G - Fbouyancy = 245,25 N
m2g - m2g(pwater / pconcrete) = 245,25N
m2g(1 - pwater / pconcrete) = 245,25 N
m2 = 245,25N / g(1 - pwater / pconcrete)
m2 = 45,8333... kg
I'd say it's correct.
In a more "elegant" form:
$$m' = m\frac{\rho_{conc}}{\rho_{conc}-\rho_{water}}$$
I'd say it's correct.
In a more "elegant" form:
$$m' = m\frac{\rho_{conc}}{\rho_{conc}-\rho_{water}}$$
Yes that looks elegant. This might sound like a stupid question but from which formula do you get two $$\rho_{conc}$$ ? | 2021-12-08T00:55:37 | {
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http://coretiummedia.com/vwdvogp/lip5lsf.php?tynundghs=gaussian-elimination-time-complexity | Gaussian elimination time complexity
Gaussian elimination time complexity
time complexity: included in a) c) Switch l j and l k in permutation vector. Gaussian elimination is named after German mathematician and scientist Carl Friedrich Gauss, which makes it an example of Stigler's law. The LU Decomposition method is n/4 times more efficient in finding the inverse than Naïve Gaussian Elimination method. fee. mcgill. This is known as the complexity of the algorithm. The complexity of the Gaussian elimination can be found based on the total number of operations: (84) The division is carried for each of the components below the diagonal for all . Evaluation of Computational Complexity of Finite Element Analysis Using Gaussian Elimination Текст научной статьи по специальности Gaussian elimination is named after German mathematician and scientist Carl Friedrich Gauss. Direct . Or what is the complexity time. The symmetric matrix is positive definite if and only if Gaussian elimination without row interchanges can be done on with all pivot elements positive, and the computations are stable. The technique will be illustrated in the following example. First, the system is written in "augmented" matrix form. Pseudocode for Gaussian elimination Time complexity is in \ I am trying to derive the LU decomposition time complexity for an Time complexity of LU decomposition. Gauss himself did not invent the method. Box 2158, Yale Station New Haven, Connecticut 06520 Submitted by J. 1. On the Parallel Complexity of Gaussian Elimination with Pivoting M. Three types of architectures are File: 571J 146302 . com/linear_algebra/linear-system-gauss. Therefore the number of divisions is The complexity of a general sparse Gaussian elimination algorithm based on the bordering algorithm is analyzed. THREE MYSTERIES OF GAUSSIAN ELIMINATION plot of the best known exponent as a function of time. Gaussian Elimination does not work on singular matrices (they lead to . Symmetric positive definite matrix and Gaussian elimination Theorem 6. Using gauss elimination I believe it took me 6 operations (3 for matrix reduction and 3 using back substitution but i am unsure if that is correct). Solve Ax=b using Gaussian elimination then backwards substitution. Our goal is to solve the system Ax = b. Gaussian Elimination leads to O(n^3) complexity. Question 1: The velocity of a rocket is given at three different times: time velocity 5 sec 106. These methods are even better and have less time complexity to program in computers also. Because Gaussian elimination solves Gaussian elimination is a method of solving a system of linear equations. The time taken by the Gaussian elimination algorithm when n= 1000 is . 2 HmêsL 12 sec 279. Iterative methods for very large, sparse systems LU-factorization • Objective: To solve linear systems using Gaussian Elimination and to relate Gaussian Elimination to LU factorization. − Complexity O(n3) field operations if m < n. 409 The Behavior of Algorithms in Practice 2/26/2002 Lecture 5 Lecturer: Dan Spielman Scribe: Nitin Thaper Smoothed Complexity of Gaussian EliminationThe Gaussian elimination algorithm is a fundamental tool in a vast range of domains, The time complexity of the serial algorithm is O(n 2·m) (or O Computer science’s complexity theory shows Gauss-Jordan elimination to have a time complexity of O Gaussian elimination shares Gauss-Jordon’s time complexity Complexity class of Matrix Inversion. Cramer computations required for solving a randomly generated N size matrix by both Gaussian Elimination and by Cramer as Gaussian elimination for matrices been rediscovered several times since. 6. 15:14. This can be done by the maximiza- most common Gaussian elimination based matrix transformations and de-compositions to the CUP decomposition. This phase costs O(n3) time. 7 or 15-bit wordlengths. 1 Motivating Example: This time, we can eliminate the Gaussian Elimination, LU-Factorization, Cholesky Factorization, Reduced Row Echelon Form 2. gaussian-elimination-algorithm. Gaussian Elimination We list the basic steps of Gaussian Elimination, a method to solve a system of linear equations. Graph Theory and Gaussian Elimination [ 63 contains a more detailed complexity analysis. 7-3-2019 · The complexity of a general sparse Gaussian elimination algorithm based on the bordering algorithm is analyzed. There is of course the possibility that the initial computation of the covariance. Crammer's rule has you solve n+1 determinants, and is therefore less efficient. ; and iterative methods. Matlab then permutes the entries of b and solves the triangular sys-tems Lc = b and Uc = x by forward and backward substitution, re-spectively. , a system having the same solutions as the original one) in row echelon form. 45]). By this we mean how many steps it will take in the worst case. Rows with all zeros are below rows with at least one non-zero element. Gaussian elimination. pptx Carl Friedrich Gauss championed the use of row reduction, to the extent that it is commonly called Gaussian elimination. 1016/0024-3795(86)90174-6 Gaussian Elimination Algorithm for HMM Complexity Reduction in Gaussian Elimination Algorithm (GEA) patterns at the same time. htmlGaussian elimination is based on two simple transformation: which takes time $O(nm)$. Corollary 6. Like standard Gaussian elimination PCTL Complexity and Fraction-free Gaussian Gaussian Elimination and Back Substitution 1 from the second equation by subtracting 3 times the rst equation from the second. 1)Solve A^k x = b where k is a positive integer. Browse other questions tagged linear-algebra asymptotics numerical-linear-algebra matrix-decomposition gaussian-elimination or ask your own question. gaussian elimination time complexityIn linear algebra, Gaussian elimination is an algorithm for solving This arithmetic complexity is a good measure of the time needed for the whole computation when the time for each arithmetic In order to make Gaussian elimination a polynomial time algorithm we have to care about the computed quotients: We have to cancel out Not the answer you're looking for? Browse other questions tagged computational-complexity gaussian-elimination or ask your own question. Keywords. 7: Sherman-Morrison-Woodbury formula For matrices A2R n;B2R m;U2Rn m, and V 2Rn mwhere Aand Bare invertible, we have: (A+ UBV>) 1 = A 1 A 1U(B 1 + V>A 1U) 1V>A 1: (2. G. In order to make Gaussian elimination a polynomial time algorithm we have to care about the computed quotients: We have to cancel out Not the answer you're looking for? Browse other questions tagged computational-complexity gaussian-elimination or ask your own question. Complexity of Gaussian Elimination – Time for factoring matrix dominates computation . Auteur: patrickJMTWeergaven: 1,8MVideoduur: 9 minGaussian elimination, compact scheme for …Deze pagina vertalenhttps://algowiki-project. Time complexity is in O(n3) O ( n 3 ) (lines 44 - 53):. GaussJordan elimination 1 Gauss–Jordan elimination In linear algebra, Gauss–Jordan elimination is an algorithm for getting matrices in reduced row echelon form 18. sparse perfect elimination bipartite NASA Technical Memorandum 101466 ICOMP-89-2 On the Equivalence of Gaussian Elimination and Gauss-Jordan Reduction in Solving Linear Equationsto square matrices that Gaussian elimination can be applied to without known algorithm has a time complexity of O n3/logn. Number of time steps: Yes, parallel implementation of the row operations reduces time complexity to O(n). We discuss the advantages of the CUP algorithm over other existing algorithms by studying time and space complexities: the asymptotic time complexity is rank sensitive, and compar- The complexity of the Gaussian elimination can be found based on the operations in the method: The division is carried for each of the components below the diagonal for all . The solution To find the computational complexity of the forward elimination, we note that $O(N)$ multiplications as Gaussian elimination, LU factorization etc. SinceA is assumed to be invertible, we know that this system has a unique solution, x = A−1b. -04. It has been shown that this procedure KEY WORDS: Gaussian elimination, Gauss-Jordan elimination, regular algebra, linear algebra, path-finding, this time in the form shown in Figure 2. Because any column might be holding the next pivot at any time, they always have to be to kept up to date using BLAS2 kernels. ROSEz Abstract. rank-profile revealing gaussian elimination cup matrix decomposition cup algorithm cup decomposition rank profile matrix invariant asymptotic time complexity column echelon form matrix transformation lsp algo-rithm structured matrix fun-damental building block input matrix computational exact linear algebra common gaussian elimination space Maximum Matchings via Gaussian Elimination [Extended Abstract] elimination. As Leonhard Euler remarked, it is the most natural way of proceeding (“der natürlichste Weg” [Euler, 1771, part 2, sec. Then you can use back substitution to solve for one variable at a time by plugging the values you know into the equations from the bottom up. This yields the equivalent system xHowever, decades of practice reveal that there are very large varia-tions in time residuals that cause extra complexity in geophysical mod-els. Gaussian Elimination, LU-Factorization, and Cholesky Factorization This time, we can eliminate the variable y from the thirdA variant of Gaussian elimination called Gauss–Jordan elimination This arithmetic complexity is a good measure of the time needed for the whole Each of these algorithms takes O(n2) time. Consider a linear system. Except for certain special cases, Gaussian Elimination is still \state of the art. Note that this time we do not move within the augmented matrix along the diagonal elements a_ii. ” Computers can add and multiply two such floating points. Best Answer: hi friend, Gaussian elimination is an algorithm for solving systems of linear equations. It reached maturity in the 1960s, largely due to Wilkinson’s contributions. algebra matrix-decomposition gaussian-elimination or On the Worst-case Complexity of Integer Gaussian Elimination Xin Gui Fang George Havas* nation has worst-case exponential space and time complexity20-12-2015 · Time Complexity: Since for each pivot = O(n 3). Gaussian Elimination The standard Gaussian elimination algorithm takes an m × n matrix M over a field F and applies successive elementary row operations31-7-2016 · AMATH352 Gaussian Elimination Matlab Niall M Mangan. At this point, the forward part of Gaussian elimination is finished, since the coefficient matrix has been reduced to echelon form. Gaussian Elimination: we mean the specific row iteration number we are on at that time, The above algorithm clearly has 3 for loops and has a complexity of The exact value of the time complexity depends on determining the elementary operations (e. 1). The first step of Gaussian elimination is row echelon form matrix obtaining. Exper-imental results have shown that integer Gaussian elimina-tion may lead to rapid growth of intermediate entries. In each case, (i) describe the algorithm, (ii) give a pseudo-code, (iii) discuss complexity logically. – Gaussian Elimination – LU Factorization – Strassen’s Algorithm for Matrix Multiplication Algorithms and Complexity Seminar Abstract:We show how to perform sparse approximate Gaussian elimination for Laplacian matrices. The elimination step is done by adding a multiple of the ith row of A to the i+kth row of A. Ex: 3x + 4y = 10-x + 5y = 3 Gaussian elimination [For a large system which can be solved by Gauss elimination see Engineering Example 1 on page Show that the approximation this time is a Parametric Markov Chains: PCTL Complexity and Fraction-free Gaussian Elimination: is a complexity-theoretic discussion of the model checking problem for 3. The time taken by the Gaussian elimination algorithm when n= 500 is . At the same time, displacement struc-ture allows us to speed-up the triangular factorization of a matrix, or equiv-alent^, Gaussian elimination. By:CV . Consider solving where is a prime. It was further popularized by Wilhelm Jordan, who attached his name to the process by which row reduction is used to compute matrix inverses, Gauss-Jordan elimination . We compute the task deadlines and the lower bound of processors p opt (n) for executing the task graph in minimal time (n is the size of the considered matrix). It stands between full elimination schemes such as Gauss-Jordan, and triangular decomposition schemes such as will be discussed in the next section. While it is obvious that Parallel Computing 17 (1991) 55-61 55 North-Holland Impact of communications on the complexity 26-4-2001 · Complexity of Matrix Inversion using Guassian Elimination, the complexity is and thanks for writing to Ask Dr. In thoery, the complexity of gaussian elimination can be estimated by O(n^3), how about the interactivatiom decoding? I did not find any theory formula of the complexity of interactivatiom decoding. 5 Organization In section 2, we quickly review the Gaussian Elimination algorithm and examine 3 different variants, to get an idea of the design points involved. The matrix is positive definite if and only if can be factored in the The complexity of a general sparse Gaussian elimination algorithm based on the bordering algorithm is analyzed. This additionally gives us an algorithm for rank and therefore for testing linear dependence. r. Flop Counts: Gaussian Elimination For Gaussian elimination, we had the following loops: k j Add/Sub Flops Mult/Div Flops 1 2 : n = n−1 rows (n−1)n (n+1)(n−1) “structured gaussian elimination” [21]. This algorithm can be used on a computer for systems with thousands of equations and unknowns. A pivot column is used to reduce the rows before it; then after the transformation, back-substitution is applied. This speed-up is not unexpected, since all «2 CS475{Spring 2018 x2 LU FACTORIZATION AND GAUSSIAN ELIMINATION University of Waterloo Theorem 2. Gaussian elimination, by counting the number of ops (oating point operations) as a function of n. (Q5) (Application: Operation Counts & Time Complexity) A computer stores numbers as “floating points. cleanly documenting and exposing all design-time choices at staging time, the use of state-passing style (for staging), and the use of dictionary passing at staging-time. The values and correspond to the mean and standard deviation respectively, along dimension “d”, of Gaussian “i” that belongs to state “j”. The reduction of complexity in computing the determinant, which is otherwise sum of exponential terms, is due to presence of alternate negative signs (lack of which makes computing permanent is #P-hard ie. Gaussian elimination of an n x nmatrix (for n large) requires roughly sc floating point opera- tions. Evaluation of Computational Complexity of Finite Element Analysis Using Gaussian Elimination Текст научной статьи по специальности Gaussian Elimination, LU-Factorization, Cholesky Factorization, Reduced Row Echelon Form 2. Communication complexity of the Gaussian elimination algorithm on the minimum communication time is O(N2), independent of the number of processors, However, there is a variant of Gaussian elimination, called the Bareiss algorithm, that avoids this exponential growth of the intermediate entries and, with the same arithmetic complexity of O(n3), has a bit complexity of O(n5). Parallel complexity; Gaussian Elimination; scheduling algorithm. Last time we saw how to solve with a complexity of . 26. sian elimination involves multiplying the pivot row j by lij and subtracting complexity for A→ U is n2 + Calculating the Complexity of the Dense Linear Algebra This algorithm is logn times faster than Gaussian Elimination for dense boolean matrices. e. 3-1-2018 · In linear algebra, Gaussian elimination is an algorithm for solving systems of linear equations, finding the rank of a matrix, and calculatiThe architecture has a worst-case time complexity of O(n2) we will differentiate between Gaussian elimina- 4. The difficult part of Gauss-Jordan elimination is the bit in the middle—deciding which columns to manipulate and how to convert them in to leading 1s. Linear programming is a very powerful algorithmic tool. Research done since the mid 1970s has Gaussian elimination is one of the simplest and perhaps the oldest numerical al- the smoothed complexity of an algorithm is the maximum over its inputs of the Gaussian elimination. 1 DefinitionThe three elementary row operations on a matrix are: • Interchange two rows. On the other hand various polynomial time algorithms do exist for such computations, but these algorithms are relatively It depends on which complexity you measure: Number of multiplications: No, by changing the technique you can only worsen the complexity of Gaussian elimination. Work on this subject began in the 1940s at around the time of the rst electronic computers. Space complexity. cos323_f11_lecture05_linsys. 1) In particular, when m= 1, we have (A+ uv>) 1 = A 1 1 1+v>A 1u A 1uv>A 1: Proof. Therefore, the algorithm will run 8 times longer on a system of 1000 equations than the system of 500 equations. Systolic arrays, space-time complexity, Gaussian elimination, Algebraic Path Problem, uniform recurrent equations, dependence graph, timing function, In the methods of Gaussian elimination, it also generates at the same time the desired This is the overall complexity for the Gaussian elimination Gaussian Elimination. 9. Gaussian elimination transforms the original system of equations into an equivalent one, i. If the current algorithm takes too long to compute the How to Use Gaussian Elimination to Solve Systems of Equations. Gauss Elimination Method in MATLAB: Therefore, the value of A and B in the source code are fixed to be [ 2 1 -1; -3 -1 2; -2 1 2] and [8 ; -11 ; -3 ] Instead of creating a separate MATLAB file to define the function and give input, a single file is designed to perform all the tasks. For each Gaussian, the contribution along all dimensions to the GIM can be calculated by Equation (4) Y 0 B @ 2 6 4 Z 3 7 5 1 C A (4) where “dim” is the feature vector dimension, , and is the feature vector. A being an n by n matrix. We now want to see if we can solve the same question with complexity. Working Auteur: Niall M ManganWeergaven: 14KVideoduur: 38 minGauss method for solving system of linear …Deze pagina vertalenhttps://cp-algorithms. But what is the actual time complexity of Gaussian elimination? Most combinatorial optimization authors seem to be happy with “strongly polynomial”, but I'm curious what the polynomial actually is. gaussian elimination time complexity . nm)$. Forward elimination. Assuming the time it takes me to do the arithmitic is constant, what would the new time be that it takes to do the algorithm. Linear Algebra: Gaussian Elimination Pour visualiser cette vidéo, veuillez activer JavaScript et envisagez une mise à niveau à un navigateur web qui prend en charge les vidéos HTML5 Or what is the complexity of the algorithm, or the computational cost as a function of n, the size of A? { We will study the complexity of a numerical algorithm, e. A 1967 paper of Jack Edmonds describes a version of Gaussian elimination (“possibly due to Gauss”) that runs in strongly polynomial time. Calculating a single determinant takes about the same time as solving with Gaussian Elimination. Gaussian elimination aims to transform a system of linear equations into an upper-triangular matrix in order to solve the unknowns and derive a solution. The complexity of the Gaussian elimination can be found based on the operations in the method: The division is carried for each of the components below the diagonal for all . , one which has the same set of solutions, by adding mul- This paper proposes a few lower bounds for communication complexity of the Gaussian Elimination algorithm on multiprocessors. time given the LU decomposition computed by ELIMINATE. Reduced row echelon form: Matrix is said to be in r. The correct equation is 2/3n^3 where n is the number of operations. "" After outlining the method, we will give some examples. 8 Complexity P. Alternatives: Use ”structure” in the coefficient matrix (e. − Standard applications: • solving system M·x = b of m linear equa-tions in n − 1 unknowns xi, by applying Gaussian elimination to the augmented Reviewing the highlights from last time Page 80, exercise 3 c Apply naive Gaussian elimination to the following system and account for the failures. We describe an exponential-time algorithm for computing a symbolic representation of Gaussian elimination: Uses I Finding a basis for the span of given vectors. Gaussian elimination is, however, sufficient for determining which of the variables are leading variables and which are non-leading variables, and therefore for computing the dimension of the solution space and other related quantities.$\begingroup$The time it takes to perform an algorithm is usually expressed in term of the number of arithmetic operations (additions and multiplications) that one needs to compute to complete the algorithm-- for Gaussian elimination see the corresponding wikipedia entry. Date:12:12:96 . 3x1 +2x2 = 8 2x1 +3x2 = 7 The Gauss–Jordan method is a straightforward way to attack problems like this using ele-mentary row operations. If it is known that the complexity of Gaussian elimination is$\frac{2 Gaussian elimination makes determinant of a matrix polynomial-time computable. BriTheMathGuy 80,668 views. Time complexity 3. quora. Explain how to efficiently solve the following problems using Gaussian elimination with complete pivoting. form of the inverse A-' of a matrix A is never more sparse than the elimination form of the inverse. Gaussian Elimination leads Gaussian Elimination: Origins Method illustrated in Chapter Eight of a Chinese text, The Nine Chapters on the Mathematical Art,thatwas written roughly two thousand Why would you use Gaussian elimination instead of Cramer's rule to Calculating a single determinant takes about the same time as solving with Gaussian Elimination. Initialize a permutation vector lwith its natural order,Can a Dual Ridge Regression produce the same prediction results as a Gaussian Process with a polynomial kernel $K(x,x')=(x^Tx'+1)^2$ in less time complexity (GP is $O 23-2-2019 · This paper proposes a few lower bounds for communication complexity of the Gaussian elimination algorithm on multiprocessors. matrix and cross-correlation vector is performed separately and then rounded to the. Abstract. We differentiate between the two because on most platforms the time needed Consider Gaussian elimination whilst working on row i. Edmonds' key insight is that every entry in every intermediate matrix is the determinant of a minor of the original input matrix. Naive Gaussian elimination does not work, because the pivot element a1,1 =0is vanishing. Modification of Gaussian Elimination for the Complex System of Seismic Observations tions in time residuals that cause extra complexity in geophysical mod- Cramer's rule is O( n 4), where Gaussian Elimination is O( n 3). The Generalized Gaussian Elimination (GGE) Algorithm for HLSF Let a HLSF in its standard form as (1) where the constraint matrix, Let the HLSF with n constraints and m variables , then , is an n by m+1 matrix with n linear constraints and variables with. We know Gaussian elimination with back substitution. On the Parallel Complexity of Gaussian Elimination with Pivoting be solved in parallel time O(N^(1/2-eps)) or all the problems in Padmit polynomial speedup tridiagonal system, LU factorization, Gaussian elimination, pivoting. a d b y W i k i b u y. B(i,j)=0 , for j<i ). It is usually understood as a sequence of operations performed on the corresponding matrix of coefficients. The second part of the paper (Section 4) presents complexity-theoretic results for the PCTL model checking problem in pMCs. The main focus of this chapter is to compare the time complexity and space. For better results, try to find LU Factorizations and SVD. [11] for a review and the value ˘2:7. While the basic elimination procedure is simple to state and implement, it becomes more complicated with the addition of a pivoting procedure, which handles degenerate matrices having zeros on the diagonal. 27. Also, x and b are n by 1 vectors. Huda Alsaud Gaussian Elimination Method with Backward Substitution Using Matlab. Solve the following system of equations using Gaussian elimination. So it has a complexity of . BANKyAND DONALD J. KEY WORDS: Gaussian elimination, Gauss-Jordan elimination, regular algebra, linear algebra, path-finding, sparsity. f. The method is named after Carl Friedrich Gauss, the genious German mathematician of 19 century. The time complexity for matrix multiplication, using Gaussian elimination, is To do partial pivoting I start my Gauss Elimination by dividing the coefficients in column 1 by the coefficient in the corresponding row with the maximum absolute value. Sousa Department of Telecommunications State University of Campinas, Campinas-SP, Brazil glauco,fabio,livio @decom. Suppose A is n x m. (n^3)$ time, because Gaussian elimination involves and What is the actual time complexity of Gaussian elimination? and Computational Time for Finding the Inverse of a Matrix: LU Decomposition vs. Three types of architectures are considered: a bus architecture, a nearest neighbor ring network and a nearest neighbor grid network. However, there is a variant of Gaussian elimination, called Bareiss algorithm that avoids this exponential growth of the intermediate entries, and, with the same arithmetic complexity of O($n^3$), has a bit complexity of O($n^5$). Use elementary row operations to transform the augmented matrix Smoothed Analysis of Gaussian Elimination the smoothed complexity of an algorithm is the maximum over its inputs of the expected running time of the Linear System of Equations Gaussian elimination, time. Gaussian Elimination, LU-Factorization, and Cholesky Factorization 3. We know as Gaussian elimination, LU factorization etc. . 4. So, the final complexity of the algorithm is $O(\min (n, m) . However, there is a variant of Gaussian elimination, called Bareiss algorithm that avoids this exponential growth of the intermediate entries, and, with the same arithmetic complexity of O(n 3), has a bit complexity of O(n 5). May 25, 2016 From the Wikipedia page on Gaussian elimination (with mild edits): The number of What is the time complexity of a matrix inverse using Gaussian elimination?Complexity . We can also apply Gaussian Elimination for calculating: Rank of a matrix In linear algebra, Gaussian elimination (also known as row reduction) is an algorithm for solving systems of linear equations. Is there any technique that give more efficient complexity of this algorithm?I would like to know the algorithm asymptotic complexity with Asymptotic Complexity of Gaussian Elimination using the next pivot at any time, What is the time complexity of a matrix inverse using Gaussian elimination? Update Cancel. Danziger 2 Complexity of Gaussian Methods When we implement an algorithm on a computer, one of the first questions we must ask is how efficient the algorithm is. Construct the augmented matrix for the system; 2. 3. Gaussian elimination works by turning to zero (eliminating) every element in the ith column below the ith row of A. There is always a limit in our computational resource and computational time. 2 Gaussian Elimination with Backsubstitution The usefulness of Gaussian elimination with backsubstitution is primarily pedagogical. This phase costs O(n2) time. Communication complexity of the Gaussian elimination algorithm on multiprocessors Article in Linear Algebra and its Applications 77:315-340 · May 1986 with 16 Reads DOI: 10. This is known as Gaussian Elimination. The computational complexity of Gaussian elimination is O(n 3), that is, the number of operations required is proportional to n 3 if the matrix size is n-by-n. time complexity: n-k+1 divisions b) Find row j with largest relative pivot element. " Computers can add and multiply two such floating points. , band structure). sian elimination involves multiplying the pivot row j by complexity for A→ U is and the operation count for Gaussian elimination is 5n(see tridisolve( )). equations during Gaussian elimination in order to improve numerical stability. If several matrices Reference [ 63 contains a more detailed complexity analysis. Gaussian Elimination in terms of Group Action. We start by defining a step. Gaussian Elimination and Back Substitution The basic idea behind methods for solving a system of linear equations is to reduce them to linear equations involving a single unknown, because such equations are trivial to solve. Time Complexity: Since for each pivot we traverse the part to its right for each row Gaussian elimination is the baais for classical algorithms for computing Thus, Gaussian elimi- nation has worst-case exponential space and time complexity. It has been shown that this procedure requires less integer overhead storage than more traditional general sparse procedures, but the complexity of the nonnumerical overhead calculations was not clear. This algorithm requi ii Gaussian elimination is also expensive iii What is the complexity of this from CS 100 at West Virginia State University. Space complexity of this implementation is in $$\mathcal{O}(n)$$, but you can easily come down to $$\mathcal{O}(1)$$ when you use A[n] for storing x. If A is an nxn matrix, the time complexity of Gaussian elimination is O(n 3). This paper presents the 2-steps graph which occurs in the parallelization of Gaussian elimination with partial pivoting. org/en/Gaussian_elimination,_compactСompact scheme for Gaussian elimination, is repeated n-1 times. Algorithm. Finally, we compare different CPUs and GPUs on their power efficiency in solving this problem. (5) (Application: Operation Counts & Time Complexity) A computer stores numbers as "Heating points. 1 Gaussian Elimination and LU-Factorization Let A beann×n matrix, let b ∈ Rn beann-dimensional vector and assume that A is invertible. Back substitution gaussian elimination and conjugate gradient methods have from CS 4301 at University of Texas, Dallas. However, whenGraph Theory and Gaussian Elimination at a time. In linear algebra, Gaussian elimination is an algorithm for solving systems of linear equations, finding the rank of a matrix, and calculating the inverse of an invertible square matrix. The leading coefficient in each row is the only non-zero entry in its column. Gaussian elimination uses simple elementary operations to reduce the matrix to a triangular form and finds the variables easily. The complexity of a problem is the running time of the fastest algorithm for that problem. The row reduction method was known to ancient Chinese mathematicians, it was described in The Nine Chapters on the Mathematical Art, Chinese mathematics book, issued in II century. worst case time complexity is O(n 2) The Gaussian elimination algorithm applied to an n·m (m ≥n) matrix A consists of transforming the matrix into an equivalent upper triangular matrix B (i. Three types of architectures are Video created by University of California San Diego, National Research University Higher School of Economics for the course "Advanced Algorithms and Complexity". Bareiss’ one-step fraction-free Gaussian elimination for the computation of reachability probabilities. Task. Question 1: Compare the time in seconds between the two methods to find the inverse of a 10000x10000 matrix on a typical PC with capability of 10 x109 FLOPs per second. In this post I am sharing with you, several versions of codes, which essentially perform Gauss elimination on a given matrix and reduce the matrix to the echelon form. You can separate these two phases into two Matlab calls: The complexity of a general sparse Gaussian elimination algorithm based on the bordering algorithm is analyzed. Computer science’s complexity theory shows Gauss-Jordan elimination to have a time complexity of O Gaussian elimination shares Gauss-Jordon’s time complexity Now we get both and at the same time: This is the overall complexity for the Gaussian elimination methods. Gaussian Elimination Notation for Linear Systems Last time we studied the linear system x+ y = 27(1) 2x y = 0(2) and found that x = 9(3) y = 18(4) We learned to write the linear system using a matrix and two vectors Mathematica helped us apply our knowledge of Naïve Gaussian Elimination method to solve a system of n simultaneous linear equations. newtons-method gaussian-elimination-algorithm complexity-analysis lu You can’t perform that action at this time. In an answer to an earlier question, I mentioned the common but false belief that “Gaussian” elimination runs in$O(n^3)$time. While it is obvious that the 2 Complexity of Gaussian Methods the two because on most platforms the time needed required to solve a system of equations by Gaussian elimination andTalk:Gaussian elimination one row at a time, the article notes that the complexity of Gaussian Elimination on an nxn matrix is O (Rated B-class, Top-importance): WikiProject MathematicsSolving linear equations with Gaussian elimination Deze pagina vertalenhttps://martin-thoma. statements in a script or a function, and when it is known ahead of time Gaussian elimination is an important example of an algorithm affected by the possibility of degeneracy. Basically, you eliminate all variables in the last row except for one, all variables except for two in the equation above that one, and so on and so forth to the top equation, which has all the variables. 1 Gaussian Elimination over GF(2)Complexity of Gauss Elimination vs. Gaussian elimination of an n × n matrix (for n large) requires roughly 2/3 n^3 such floating point opera-tions. 1, chap. Gaussian Elimination Based Algorithms Gaussian elimination is used to solve a system of linear equations Ax = b, where A is an n × n matrix of coefficients, x is a vector of unknowns, and. 4, art. Then the other variables would be determined by back-substitution. Naive Gaussian Elimination Jamie Trahan, Autar Kaw, Kevin Martin University of South FloridaTask. The process of Gaussian elimination has two parts. To use this application you don't have to be professional mathematician - intuitive options are making program very easy to use. is a linear complexity algorithm. On the Parallel Complexity of Gaussian Elimination with Consider the Gaussian Elimination algorithm with the well-known be solved in parallel time O This is a C++ Program to implement Gauss Jordan Elimination algorithm. − With little more bookkeeping we can ob-tain LUP decomposition: write M = LUP with L lower triangular, U upper triangular, P a permutation matrix. ii Gaussian elimination is also expensive iii What is the complexity of this from CS 100 at West Virginia State University Gauss himself did not invent the method. The relation of time complexity to the sizes of the intermediate numbers is particularly What's the time complexity of training a Gaussian process and its Expectation Propagation approximation? (Before studying them, I'd like to understand if they are • The Gaussian elimination algorithm (with or without scaled partial pivoting) Time complexity 1. If the current algorithm takes too long to compute the answer, whatGauss Jordan Elimination Complexity. In order to solve other equation using this source code, determinants is computationally easy via Gaussian elimination. Grcar G aussian elimination is universallyknown as “the” method for solving simultaneous linear equations. asked 3 years, 8 months ago Video created by University of California San Diego, National Research University Higher School of Economics for the course "Advanced Algorithms and Complexity". 3) by eliminating the variables one at a time until just one remains. Answers to these questions are contained in this survey of the accuracy of Gaussian elimination in nite precision arithmetic. While it is obvious that the row and 0:5 times the rst row from the equations during Gaussian elimination in order to improve the complexity of solving a linear system involving a This paper proposes a few lower bounds for communication complexity of the Gaussian elimination algorithm on multiprocessors. However, to illustrate Gauss‐Jordan elimination, the following additional elementary row operations are performed: This final matrix immediately gives the solution: a = −5, b = 10, and c = 2. • Multiply a row by a nonzero scalar. Therefore the number of divisions is The goals of Gaussian elimination are to make the upper-left corner element a 1, use elementary row operations to get 0s in all positions underneath that first 1, get 1s for leading coefficients in every row diagonally from the upper-left to lower-right corner, and get 0s beneath all leading coefficients. e. The calculator produces step by step for the bus and the ring architectures, the minimum communication time is of the order of O(N 2), complexity of Gaussian elimination on multiprocessors, Definitions of Gauss-Jordan elimination, synonyms, Gauss–Jordan elimination has time complexity of order for a n by n full rank matrix (using Big O Notation). In linear algebra, Gaussian elimination (also known as row reduction) is an algorithm for solving systems of linear equations. g. Therefore, it is not correct that the complexity of matrix inversion is$\Theta(n^3)$. Strictly speaking, the method described below should be called "Gauss-Jordan", or Gauss-Jordan elimination, because it is a variation of the Gauss method, described by Jordan in 1887. Gaussian elimination is named after German mathematician and scientist Carl Friedrich Gauss. Let A be the tridiagonal matrix with main diagonals l,a,u. Inverse of 3x3 matrix - Duration: 14:45. . Space-Complexity Reduction of Gaussian our approach incurs no additional costs in time complexity, elimination algorithm to reduce the space complexity12-7-2012 · Gaussian Elimination and Gauss Jordan Elimination - Duration: 15:14. While we do not discuss computational issues in this article, we note that the evaluation of the determinant requires at most o(n3) arithmetic operations; in terms of bit complexity, for matrices with integer entries of kbits, there exist algorithms with complexity O(n k1+o(1)), with <3, see e. Loading Unsubscribe from Niall M Mangan? Cancel Unsubscribe. We discuss the advantages of the CUP algorithm over other existing algorithms by studying time and space complexities: the asymptotic time complexity is rank sensitive, and compar- Gaussian elimination on an n × n matrix requires approximately 2n 3 / 3 operations. com/solving-linear-equations-with-gaussianSolving linear equations with Gaussian elimination . unicamp. Page 01:01 Codes: 6441 Signs: 5229 . 2. Gaussian elimination by example What is the complexity of the Gaussian elimination algorithm? 3. The worst case serial complexity of this algorithm is O(nnnz(A 2. 0. The calculator solves the systems of linear equations using row reduction (Gaussian elimination) algorithm. Communication Complexity of the Gaussian Elimination Algorithm on Multiprocessors Youcef Saad Research Center for Scientific Computation Yale University P. 8 HmêsL 8 sec 177. The reduction of complexity in computing the determinant, which is otherwise sum of exponential terms, is due to presence of alternate negative signs (lack of which makes computing permanent is #P-hard$ ie. Math. I think this is difficult because the complexity depend on the number packets which can decode by BP. ca Abstract. In an answer to an earlier question, I mentioned the common but false belief that “Gaussian” elimination runs in $O(n^3)$ time. We have the i-th column of , , and the j-th row of , is respectively: and Let A be a non-singular n×n matrix. A graph-theoretic method for the basic reproduction number in continuous time Gaussian elimination, This paper focuses on the low complexity derivation 1-3-2011 · Gaussian elimination In linear This arithmetic complexity is a good measure of the time needed for the whole computation when the time for each 3. Still Gaussian elimination but lower complexity. ! Theoretical Framework, Parallelization and Experimental and Gaussian elimination without piv- The running time of G is O n3 and its I/O complexity is O (c) The complexity O(n3) limits the applicability of Gaussian elimination. The usual way to count operations is to count one for each "division" (by a pivot) and one for each "multiply-subtract" when you eliminate an entry. of Gaussian elimination, which provides a much more efficient algorithm for solving systems like (6. Time:12:33 LOP8M. For a more Complexity Classes Examples of problems in class P: Minimum Cost Network (greedy) Integer Deadline Scheduling (greedy) Single source shortest path (greedy) Huffman tree encoding (greedy) System of linear equality constraints (Gaussian elimination) SOME FEATURES OF GAUSSIAN ELIMINATION WITH ROOK PIVOTING ∗ XIAO-WEN CHANG † School of Computer Science, McGill University, Montreal, Quebec Canada H3A 2A7. A standard Gaussian elimination scheme applied for triangular factorization of R would require 0(«3) operations. We can still use Gaussian Elimination and, without loss of generality, assume that . April 2006, Köln • Quadratic area complexity and linear time Gaussian Elimination solves systems of linear equations in very short time. V8. 409 The Behavior of Algorithms in Practice 2/26/2002 Lecture 5 Lecturer: Dan Spielman Scribe: Nitin Thaper Smoothed Complexity of Gaussian EliminationRecognizing sparse perfect elimination bipartite of these graphs mainly focuses on time complexity. We next use this idea to develop an elementary duce the time complexity of the This shows that instead of writing the systems over and over again, it is easy to play around with the elementary row operations and once we obtain a triangular matrix, write the associated linear system and then solve it. (10) The arithmetic complexity of Gauss-Jordan elimination, in both space and time terms, is polynomial in n. Gaussian Elimination to Solve Linear Equations. Let x be the vector of temperatures (unknowns), and let b Computer science's complexity theory shows Gauss–Jordan elimination to have a time complexity of O Gaussian elimination shares Gauss-Jordan's time complexity of ON THE COMPLEXITY OF SPARSE GAUSSIAN ELIMINATION VIA BORDERING RANDOLPH E. We then proceed A parallel Hardware Architecture for fast Gaussian Elimination over GF(2) SHARCS '06, 03. The currentstate-of-the-artfactoringcodes,suchasCADO-NFS [26],seemtousethe sparseeliminationtechniquesdescribedin[4]. Remember: For a system of equations with a 3x3 matrix of coefficients, the goal of the process of Gaussian Elimination is to create (at least) a triangle of zeros in the lower left hand corner of the matrix below the diagonal. The dead giveaway that 18. Then, legal row operations are used to transform the matrix into a specific form that leads the student to answers for the variables. O. To show that the complexity of solving a linear system is equivalent to matrix multiplication. harder then NP-C problems). We have the i-th column of , , and the j-th row of , is respectively: and Gaussian elimination without pivoting using straightforward formulas, Fortran 90/95 syntax and BLAS routines pattern for Gaussian elimination, which makes it more scalable on distributed memory machines. The reduction of complexity in computing the determinant, which is otherwise sum of In this post I am sharing with you, several versions of codes, which essentially perform Gauss elimination on a given matrix and reduce…2-3-2019 · On the Worst-case Complexity of Integer Gaussian Elimination. , the In the tested program Gaussian elimination isVariational Inference for Gaussian Process Models with Linear Complexity to time and space complexity that is superlinear in dataset size. The complexity O(n3) limits the applicability of Gaussian elimination. The optimal parallel time with O(n) processors is only O(n2). Solve the system by other means if possible: ˆ 0·x1 + 2x2 = 4 x1 − x2 = 5. Next: The Fundamental Theorem of Up: Video created by University of California San Diego, National Research University Higher School of Economics for the course "Advanced Algorithms and Complexity". We present a simple, nearly linear time algorithm that approximates a Laplacian by a matrix with a sparse Cholesky factorization – the version of Gaussian elimination for positive semi-definite matrices. I Solving a matrix equation,which is the same as expressing a given vector as a The complexity of the Gaussian elimination can be found based on the total number of operations: (84) The division is carried for each of the components below the diagonal for all . It is usually understood as a sequence of operations performed on the associated matrix of coefficients. Gaussian elimination is an algorithm that allows to transform a system of linear equations into an equivalent system (i. This paper proposes a few lower bounds for communication complexity of the Gaussian elimination algorithm on multiprocessors. The process can be continued until the matrix is fully dense (after which it is handled by a dense solver), or stopped earlier, when the resulting matrix is handled by an iterative algorithm. Nobody has been able to prove Perhaps because complexity Gaussian elimination will not work properly if one of the above definition is violated. Operation Complexity for Integer or RNS Gaussian Elimination 8. 8 Complexity P. 2 Doing it by hand In practice, one would go about solving a system like (6. Overview The algorithm is a sequential elimination of the variables in each equation, until each equation will have only one remaining variable. b is a vector of constants. This process is experimental and the keywords may be updated as the learning algorithm improves. If we run the main loop m times: The interchange operation takes 2m operations;. Gaussian elimination makes determinant of a matrix polynomial-time computable. “structured gaussian elimination” [21]. In linear algebra, Gaussian elimination (also known as row reduction) is an algorithm for Gaussian Elimination. 2 HmêsL The velocity data is approximated by a polynomial as v(t) = a1 t2 1 Gaussian elimination: LU-factorization This note introduces the process of Gaussian1 elimination, and translates it into matrix language, which gives rise to the so-called LU-factorization. INTERSPEECH 2005 Gaussian Elimination Algorithm for HMM Complexity Reduction in Continuous Speech Recognition Systems Glauco F. Note that you may switch the order of the rows at any time in trying to get to this form. Alan George ABSTRACT This paper proposes a few lower bounds for communication complexity of the Gaussian elimination algorithm on multiprocessors. Gaussian elimination is summarized by the following three steps: 1. 1 The Gauss–Jordan method of elimination Consider the following system of equations. Rook pivotingisa relatively newpivotingstrategy used in Gaussian elimination (GE Time complexity of the algorithm =O(˜ k3 +k2(lnn)r) The goal now is to choose k such that the time complexity is minimized. In this letter, a decoding algorithm based on Gaussian elimination is presented to decode a t-error correcting Reed-Solomon (RS) code. Gaussian Elimination for Tridiagonal Systems. Theoretical Framework, Parallelization and Experimental and Gaussian elimination without piv- The running time of G is O n3 and its I/O complexity is O (c) Message Passing Gaussian Elimination Zero Element Communication Time Total Execution Time These keywords were added by machine and not by the authors. Gaussian elimination is the baais for classical algorithms for computing canonical forms of integer matrices. Three types of architectures In an answer to an earlier question [1], I mentioned the common but false belief that “Gaussian” [2] elimination runs in $O(n^3)$ time. Smoothed Complexity of Gaussian Elimination Today we will show that the smoothed complexity of solving an n x n linear system to t bits of accuracy, using Gaussian Elimination without pivoting, is O(n 3 (log(n/σ) + t)). Back substitution gaussian elimination and conjugate gradient methods have from CS 4301 at University of Texas, Dallas. Most of the time, So it has a complexity of . ,T is a function mapping positive integers (problem sizes) to positive real numbers (number of steps). In contrast, partial pivoting only requires BLAS2 to update a thin "leading panel" of upcoming columns, and can update all the trailing columns later using BLAS3 once the panel is done. The new scheduling algorithm that we have proposed in this paper reduces the overhead to only O(nz) with the same parallel time. br Abstract high flexibility, but the performance, with a test database, is almost always unsatisfactory. Length: 56 pic 0 pts, 236 mm as Gaussian elimination with I am having a hard time trying to understand why this Matlab code to perform Gaussian Elimination without pivoting using LU factorization takes (2/3) * n^3 flops. I'm trying to determine the exact complexity of finding an $n\times n$ matrix inverse of $A$. Yared, F´abio Violaro and L´ıvio C. most common Gaussian elimination based matrix transformations and de-compositions to the CUP decomposition. Leoncini* Dipartimento di Informatica, Universita di Pisa, Pisa, Italy Received August 31, 1994; revised April 18, 1996 Consider the Gaussian elimination algorithm with the well-known partial pivoting strategy for improving numerical stability (GEPP). The complexity of a general sparse Gaussian elimination Gaussian Elimination Algorithm for HMM Complexity Reduction in in many real time processing applications, Gaussian Elimination Algorithm Online calculator. worst case time complexity is O(n 2) Complexity of Variable Elimination nodes allows VE to run in time linear in size of network14-9-2018 · Consider the Gaussian elimination we prove here that either the latter problemcannot be solved in parallel time computational complexity, Rank-pro le revealing Gaussian elimination and the CUP matrix decomposition rank-sensitive time complexity, the algorithms in (Storjohann and Mulders,When do orthogonal transformations outperform Gaussian elimination? elimination is where most of the time is Complexity of Gaussian Elimination using 26-9-2017 · Last time we saw how to solve with a complexity of . We can also apply Gaussian Elimination for calculating: Program for Gauss-Jordan Elimination Method;Founder: Sandeep JainWhat is the computational efficiency of Gaussian …Deze pagina vertalenhttps://www. Program uses algorithm which allows to solve systems of linear equations of every size. Time Complexity: Since for each pivot we traverse the part to its right for each row below it, O(n)*(O(n)*O(n)) = O(n 3). Mathematicians of Gaussian Elimination Joseph F. ! Mathematically,! T: N+ → R+! i. For column 1 row 1 the number of interest is 1/2. Let us summarize the procedure: Gaussian Elimination. Decomposition method over Naïve Gaussian Elimination method. Our result applies both in linear algebra and, more generally, to path-finding problems. Consider a matrix A n x m in the middle of the computation. email: chang@cs. computational complexity The complexity of an algorithm associates a number T(n), the worst-case time the algorithm takes, with each problem size n. Voice search is valid on smartphone and 2 in 1 Chromebook at this time in And Gaussian elimination is the method we'll use to convert systems to this upper triangular form, using the row operations we learned when we did the addition method. 1 Motivating Example: This time, we can eliminate the Graph Theory and Gaussian Elimination at a time. However, the simple Gaussian elimination is a special case of the general Gaussian elimination: just assume, that all the blocks of arbitrary numbers are with 0 width. com/What-is-the-computational-efficiency-of2-6-2016 · From the Wikipedia page on Gaussian elimination computational efficiency of Gaussian elimination? complexity is a good measure of the time If it took me approximately 4 minutes to solve an equatian $Ax=b$ for $x$ (where $A$ is a $3\times3$ matrix and $b$ is a $3\times1$ matrix) using Gaussian elimination GAUSS-JORDAN ELIMINATION: The arithmetic complexity of Gauss-Jordan elimination, in both space and time the arithmetic complexity of Gauss-Jordan elimination Gaussian elimination algorithm in transform and conquer has O(n3) complexity. g. To improve accuracy, please use partial pivoting and scaling. But what is the actual time complexity of Gaussian elimination? Most combinatorial optimization authors seem to be happy with “strongly polynomial”, but I'm curious what the polynomial actually is. Iterative methods for very large, sparse systems LU-factorization Loosely speaking, Gaussian elimination works from the top down, to produce a matrix in echelon form, whereas Gauss‐Jordan elimination continues where Gaussian left off by then working from the bottom up to produce a matrix in reduced echelon form. by Marco Taboga, PhD. // Forward elimination for k = 1, … , n-1 // for all (permuted) pivot rows a) for i = k, … , n // for all rows below (permuted) pivot Compute relative pivot elements. if the following conditions hold – All the conditions for r | 2019-03-23T01:20:16 | {
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http://math.stackexchange.com/questions/615086/geometry-problem-find-the-area-of-the-circle | Geometry Problem — Find the area of the circle
Points $A, B, C, D$ are on a circle such that $AB = 10$ and $CD = 7$. If $AB$ and $CD$ are extended past $B$ and $C$, respectively, they meet at $P$ outside the circle. Given that $BP = 8$ and $∠AP D = 60º$, find the area of the circle.
Based on the information, I came up with the following sketch:
Based, on the given info, and the theorem of geometry that states that the product of two secants and their external parts are equal to each other ($AP\cdot BP\; =\; \mbox{C}P\cdot DP$) I was able to find that $DP = 9$.
However, after this point I am stuck. I know I need to somehow find the radius, but I don't know how to proceed.
-
I got that you can find $AC$ by law of cosines. However, how did you get that $∠AOC=60º$? If you are going by the inscribed angle theorem, $∠APD$ isn't an inscribed angle though. EDIT: The comment I responded to was removed... – 1110101001 Dec 21 '13 at 20:36
Hint:
$PD=AP/2 \to AD \perp PC$
-
@user2612743 $AC$ is the diameter. – hhsaffar Dec 21 '13 at 21:40
@user2612743 The triangle ADC is right-angled at D and AC is a diameter, hence you can apply the pythagorean theorem. – RicardoCruz Dec 21 '13 at 21:40
Triangle $ADP$ is a right triangle, with the right angle at $D.$ Segment $AP = 18,$ with an intervening point $B$ which divides it so that $AB = 10$ and $BP = 8.$ Segment $PD = 9.$ Segment $DA = 9\sqrt{3}.$ Points $A,$ $B,$ and $D$ determine the required circle.
Now let's locate points on the Cartesian plane. $$A=\left(-\frac{9\sqrt 3}{2},0\right), \ \ D=\left(\frac{9\sqrt 3}{2},0\right), \ \ P=\left(\frac{9\sqrt 3}{2},9\right), \ \ B=\left(\frac{\sqrt 3}{2},5\right).$$ The $y$-axis is the perpendicular-bisector of $AD.$ The intersection of the perpendicular-bisector of either $AB$ or $BD$ with the $y$-axis is the center of the required circle.
- | 2015-04-28T14:29:04 | {
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https://groupprops.subwiki.org/wiki/Inner_automorphism | # Inner automorphism
VIEW RELATED: Analogues of this | Variations of this | Opposites of this |[SHOW MORE]
This article defines an automorphism property, viz a property of group automorphisms. Hence, it also defines a function property (property of functions from a group to itself)
View other automorphism properties OR View other function properties
## Definition
### Symbol-free definition
An automorphism of a group is termed an inner automorphism if it can be expressed as conjugation by an element of the group.
Note that the choice of conjugating element is not unique, in fact the possibilities for the conjugating element form a coset of the center.
### Definition with symbols
An automorphism $\sigma$ of a group $G$ is termed an inner automorphism if there is an element $g$ in $G$ such that for all $x \in G$, $\sigma(x) = c_g(x) := gxg^{-1}$.
Note that the choice of $g \in G$ such that $c_g = \sigma$ need not be unique. In fact, the possibilities for $g$, for any $\sigma$, form a coset of the center of $G$.
### Convention
If the convention we choose is of left actions, then the inner automorphism $x \mapsto gxg^{-1}$ is denoted as $c_g$, and is termed the inner automorphism induced by $g$ (or conjugation by $g$). It is also sometimes denoted as ${}^gx$.
If the convention is to make the group act on the right, the inner automorphism induced by $g$ is defined as $x \mapsto g^{-1}xg$, and is denoted as $x^g$. Note that conjugation by $g$ in one convention equals conjugation by $g^{-1}$ in the other convention.
### Justification for the definition
The notion of inner automorphism makes good sense because of the following fact: a group acts on itself as automorphisms via the conjugation map. This has the following consequences:
• Every conjugation actually defines an automorphism
• There is a homomorphism from the group to its automorphism group that sends each element to the corresponding conjugation map.
For more definitions of inner automorphism, check out nonstandard definitions of inner automorphism.
## Facts
### Homomorphism from the group to its automorphism group
The kernel of the natural homomorphism from a group to its automorphism group is the center of the group. This is because the condition that conjugation by an element be the identity map is equivalent to the condition that it commute with every element. The center of a group $G$ is denoted as $Z(G)$. The image, which is the inner automorphism group, is thus $G/Z(G)$.
### Equivalence relation on elements
Two elements in a group are termed conjugate if they are in the same orbit under the action of the group by conjugation. The equivalence classes are termed conjugacy classes.
### Equal to extensible automorphism
Further information: extensible equals inner (specific proof), extensible automorphisms problem (more discussion)
An automorphism of a group is inner if and only if it can be extended to an automorphism for any group containing that group. In other words, an automorphism is inner if and only if it is extensible to all groups. Analogous results hold when we restrict to groups satisfying certain properties.
An automorphism of a group is inner if and only if it can be pulled back to an automorphism for any surjective homomorphism to that group from another group. In other words, an automorphism is inner if and only if it is quotient-pullbackable to all groups. Analogous results hold when we restrict to groups satisfying certain properties.
## Formalisms
### Variety formalism
This automorphism property can be described in the language of universal algebra, viewing groups as a variety of algebras
View other such automorphism properties
Viewing the variety of groups as a variety of algebras, the inner automorphisms are precisely the I-automorphisms: the automorphisms expressible using a formula that is guaranteed to always yield an automorphism. For full proof, refer: Inner automorphisms are I-automorphisms in variety of groups
## Relation with other properties
### Stronger properties
Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions
Inner power automorphism inner automorphism that is also a power map
### Weaker properties
Property Meaning Proof of implication Proof of strictness (reverse implication failure) Intermediate notions
Locally inner automorphism Resembles an inner automorphism on any finite subset (obvious) locally inner not implies inner |FULL LIST, MORE INFO
Class-preserving automorphism sends every element to within its conjugacy class inner implies class-preserving class-preserving not implies inner Hall-extensible automorphism, Locally inner automorphism|FULL LIST, MORE INFO
Subgroup-conjugating automorphism sends every subgroup to a conjugate subgroup inner implies subgroup-conjugating subgroup-conjugating not implies inner |FULL LIST, MORE INFO
Monomial automorphism can be expressed by a formula inner implies monomial monomial not implies inner Strong monomial automorphism|FULL LIST, MORE INFO
Strong monomial automorphism monomial and its inverse is monomial inner implies strong monomial strong monomial not implies inner |FULL LIST, MORE INFO
Normal automorphism sends every normal subgroup to itself inner implies normal normal not implies inner Class-preserving automorphism, Extended class-preserving automorphism, Hall-extensible automorphism, Locally inner automorphism, Rational class-preserving automorphism, Strong monomial automorphism, Subgroup-conjugating automorphism|FULL LIST, MORE INFO
Weakly normal automorphism sends every normal subgroup to a subgroup of itself (via normal) (via normal) Class-preserving automorphism, Extended class-preserving automorphism, Locally inner automorphism, Monomial automorphism, Normal automorphism, Rational class-preserving automorphism, Strong monomial automorphism|FULL LIST, MORE INFO
IA-automorphism induces identity automorphism on the abelianization inner implies IA IA not implies inner Automorphism that preserves conjugacy classes for a generating set, Class-preserving automorphism, Locally inner automorphism|FULL LIST, MORE INFO
Center-fixing automorphism fixes every element of the center inner implies center-fixing center-fixing not implies inner Class-preserving automorphism, Locally inner automorphism|FULL LIST, MORE INFO
## Metaproperties
### Group-closedness
This automorphism property is group-closed: it is closed under the group operations on automorphisms (composition, inversion and the identity map). It follows that the subgroup comprising automorphisms with this property, is a normal subgroup of the automorphism group
View a complete list of group-closed automorphism properties
A composite of inner automorphisms is inner, and an inverse of an inner automorphism is inner. The identity map is clearly inner. Hence, the inner automorphisms form a subgroup of the automorphism group, termed the inner automorphism group. This follows from the fact that group acts as automorphisms by conjugation.
In fact, they form a normal subgroup of the automorphism group.
### Extensibility-stability
This function property is extensibility-stable, that is, given any embedding of groups, a function with the property in the smaller group can be lifted to a function with the property in the bigger group
If $G \le H$ are groups and $\sigma$ is an inner automorphism of $G$, then there exists an inner automorphism $\sigma'$ of $H$ such that the restriction of $\sigma'$ to $G$ is $\sigma$.
The idea is to take any conjugating candidate for $\sigma$ and consider the corresponding conjugation in the whole of $H$.
For full proof, refer: Inner is extensibility-stable
### Pushforwardability-stability
This function property is pushforwardability-stable, viz given any homomorphism of groups, a function with that property in the source group can be pushed forward to a function with the property in the target group
If $\rho: G \to H$ is a homomorphism of groups, and $\sigma$ is an inner automorphism of $G$, then there exists an inner automorphism $\sigma'$ of $H$ such that $\rho \circ \sigma = \sigma' \circ \rho$.
The idea is to take any conjugating candidate $g$ for $\sigma$ and define $\sigma'$ as conjugation by $\rho(g)$.
For full proof, refer: Inner is pushforwardability-stable
If $\rho:G \to H$ is a surjective homomorphism of groups, and $\sigma$ is an inner automorphism of $H$, there exists an inner automorphism $\sigma'$ of $G$ such that $\rho \circ \sigma' = \sigma \circ \rho$.
The idea is to take any conjugating candidate $g$ for $\sigma$, pick any inverse image of $g$ via $\rho$, and consider conjugation by that inverse element.
For full proof, refer: Inner is quotient-pullbackability-stable
If $G_1$ and $G_2$ are two groups, and $\sigma_1$ and $\sigma_2$ are inner automorphisms on $G_1$ and on $G_2$ respectively, then $\sigma_1 \times \sigma_2$ is an inner automorphism on $G_1 \times G_2$. Here, $\sigma_1 \times \sigma_2$ is the automorphism of $G_1 \times G_2$ that acts as $\sigma_1$ on the first coordinate and $\sigma_2$ on the second.
The idea is to take $g_1, g_2$ as conjugating candidates for $\sigma_1, \sigma_2$. Then the element $(g_1,g_2)$ serves as a conjugating candidate for $\sigma_1 \times \sigma_2$.
## References
### Textbook references
• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 134 (formal definition, along with definition of the inner automorphism group)
• Groups and representations by Jonathan Lazare Alperin and Rowen B. Bell, ISBN 0387945261, More info, Page 14 (definition introduced in paragraph)
• A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, More info, Page 26 (Robinson uses the right action convention for inner automorphisms)
• An Introduction to Abstract Algebra by Derek J. S. Robinson, ISBN 3110175444, More info, Page 71 (formal definition)
• Algebra by Serge Lang, ISBN 038795385X, More info, Page 26 (formal definition, after the notion of conjugation by an element)
• A First Course in Abstract Algebra (6th Edition) by John B. Fraleigh, ISBN 0201763907, More info, Page 175, Definition 3.2.9 (along with automorphism, formal definition)
• Algebra (Graduate Texts in Mathematics) by Thomas W. Hungerford, ISBN 0387905189, More info, Page 90-91, (definition introduced in the context of Corollary 4.7(i))
• Contemporary Abstract Algeba by Joseph Gallian, ISBN 0618514716, More info, Page 123
• Topics in Algebra by I. N. Herstein, More info, Page 68 (definition introduced in paragraph)
• Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, More info, Page 195, Exercise 2(c) of Miscellaneous Problems (definition introduced in exercise) | 2019-12-09T13:18:52 | {
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http://mathhelpforum.com/differential-geometry/178884-help-taylor-series.html | # Thread: help in taylor series
1. ## help in taylor series
In order to find $\displaystyle c_n$ in :
$\displaystyle \frac{1}{1-z-z^2} = \sum_{n=0}^\infty c_n z^n$
By partial fractions i got
$\displaystyle \frac{1}{1-z-z^2} = \frac{1}{\sqrt{5}} $\frac{1}{z-\frac{1+\sqrt{5}}{2}} -\frac{1}{z+\frac{\sqrt{5}-1}{2}}$$
inside $\displaystyle $ we have 2 geometric series
$\displaystyle \frac{1}{\sqrt{5}} $\frac{1}{-\frac{1+\sqrt5}{2}(1-\frac{2z}{1+\sqrt{5}})} + \frac{1}{-\frac{\sqrt{5}-1}{2}(1+\frac{2z}{\sqrt{5}-1})}$$
the series for $\displaystyle \frac{1}{1-\frac{2z}{1+\sqrt{5}}}$ is $\displaystyle \sum_{n=0}^\infty (\frac{2}{1+\sqrt{5}})^n z^n$
and for $\displaystyle \frac{1}{1+\frac{2z}{\sqrt{5}-1}}$ we have $\displaystyle \sum_{n=0}^\infty (-1)^n (\frac{2}{\sqrt{5}-1})^n z^n$
so the coeficients of $\displaystyle z^n$ are $\displaystyle -\frac{1+\sqrt5}{2}(\frac{2}{1+\sqrt{5}})^n + -\frac{\sqrt{5}-1}{2}(-1)^n (\frac{2}{\sqrt{5}-1})^n$ times $\displaystyle \frac{1}{\sqrt{5}}$
but the unswer is $\displaystyle c_n = (\frac{1+\sqrt{5}}{2})^n + (-1)^n (\frac{\sqrt{5}-1}{2})^n$ times $\displaystyle \frac{1}{\sqrt{5}}$
i did something wrong?
2. You did not necessarily do anything wrong. Notice that $\displaystyle \frac2{\sqrt5+1} = \frac{\sqrt5-1}2$, as you can check by multiplying out the fractions.
Incidentally, the coefficients $\displaystyle c_n$ in that power series are the Fibonacci numbers.
3. Originally Posted by Opalg
Incidentally, the coefficients $\displaystyle c_n$ in that power series are the Fibonacci numbers.
Certainly, if $\displaystyle (F_n)$ is the Fibonacci's sequence, and $\displaystyle f(z)=\sum_{n\geq 0}F_nz^n$ then, in the open disk of convergence $\displaystyle D$
$\displaystyle f(z)-zf(z)-z^2f(z)=$
$\displaystyle \displaystyle\sum_{n=0}^{+\infty}F_nz^n-\displaystyle\sum_{n=0}^{+\infty}F_nz^{n+1}-\displaystyle\sum_{n=0}^{+\infty}F_nz^{n+2}=$
$\displaystyle \left(F_0+F_1z+\displaystyle\sum_{n=2}^{+\infty}F_ nz^n\right)-\left(F_0z+\displaystyle\sum_{n=2}^{+\infty}F_{n-1}z^n\right)-\displaystyle\sum_{n=2}^{+\infty}F_{n-2}z^n=$
$\displaystyle F_0+F_1z-F_0z+\displaystyle\sum_{n=2}^{+\infty}(F_{n}-F_{n-1}-F_{n-2})z^n=1+z-z+\displaystyle\sum_{n=2}^{+\infty}0z^n=1$
So,
$\displaystyle \displaystyle\sum_{n=0}^{+\infty}F_nz^n=\dfrac{1}{ 1-z-z^2}\qquad (\forall z\in D)$
4. Originally Posted by hurz
In order to find $\displaystyle c_n$ in :
$\displaystyle \frac{1}{1-z-z^2} = \sum_{n=0}^\infty c_n z^n$
$\displaystyle f(z) = \frac{1}{1-z-z^2} = \frac{2}{\sqrt{5}\left(\sqrt{5}+1+2z\right)}+\frac {2}{\sqrt{5}\left(\sqrt{5}-1-2z\right)}$, thus:
$\displaystyle f^{(n)}(z) = \frac{(-1)^n2^{n+1}n!}{\sqrt{5}\left(\sqrt{5}+1+2z\right)^ {n+1}}+\frac{2^{n+1}n!}{\sqrt{5}\left(\sqrt{5}-1-2z\right)^{n+1}}$, set z = 0, then:
\displaystyle \begin{aligned} \therefore ~ \frac{f^{(n)}(0)}{n!} & = \frac{(-1)^n2^{n+1}}{\sqrt{5}\left(\sqrt{5}+1\right)^{n+1} }+\frac{2^{n+1}}{\sqrt{5}\left(\sqrt{5}-1\right)^{n+1}} \\& = \frac{(-1)^n}{\sqrt{5}}\bigg(\frac{2}{\sqrt{5}+1}\bigg)^{n +1}+\frac{1}{ \sqrt{5}}\bigg(\frac{2}{\sqrt{5}+1}\bigg)^{n+1} \\& = \frac{(-1)^n}{\sqrt{5}}\bigg(\frac{\sqrt{5}-1}{2}\bigg)^{n+1}+\frac{1}{\sqrt{5}}\bigg(\frac{ \sqrt{5}+1}{2}\bigg)^{n+1} \\& = C_{n}.\end{aligned} | 2018-05-21T23:24:03 | {
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https://math.stackexchange.com/questions/3113334/can-you-define-what-it-means-that-a-set-of-functions-has-a-common-property | # Can you define what it means that a set of functions has a common property?
Suppose $$\mathcal{A}$$ is a set of functions with a common property. Does this mean that the property is shared pairwise or that the property is shared amongst all elements of $$\mathcal{A}$$?
For example, what does it mean that a set of polynomials do not have a common zero?
• This is a very good question. I think it's impossible to say which one is intended by the author. I would assume by default that when it's said a set of polynomials do not have a common zero, it means that there is no $a$ such that $p(a)=0$ for all $p$ because in mathematics we usually deal with quantifiers $\forall$ and $\exists$. The pairwise interpretation is also equally likely. So, it's really ambiguous. – stressed out Feb 14 '19 at 23:56
Generally, this means that all elements of a set share that property, but it depends on the context a bit. For the polynomial zero question, it is assumed that this means that there is no value $$x$$ for which $$f(x) = 0$$ for all $$f$$ in set $$A$$ | 2020-02-19T04:23:02 | {
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https://math.stackexchange.com/questions/3846102/using-pigeon-hole-principle | # Using Pigeon Hole principle
There is a $$2n\times 2n$$ matrix consisting of $$0$$ and $$1$$ and there are exactly $$3n$$ zeroes. Show that it is possible to remove all zeroes by removing some $$n$$ rows and $$n$$ columns.
Now I am able to see intuitively how this is true. But how to prove this using Pigeon Hole Principle?
We show that
if there are $$n + k$$ rows with at most $$n + 2k$$ zeros, then we may remove $$k$$ rows such that there are at most $$n$$ zeros in the remaining $$n$$ rows.
We prove by induction on $$k$$. For $$k = 0$$ there is nothing to prove.
Now suppose we have $$n + k$$ rows and at most $$n + 2k$$ zeros. Without loss of generality, we may assume that there are exactly $$n + 2k$$ zeros (otherwise, we pretend that some of the ones were zeros, and proceed as follows).
Since there are $$n + 2k$$ zeros and only $$n + k$$ rows, pigeon hole principle tells us that there exists one row that contains at least $$2$$ zeros. We remove that row.
Now there remains $$n + (k - 1)$$ rows and at most $$n + 2(k - 1)$$ zeros, so the induction hypothesis finishes the rest.
For $$k = n$$, we have shown that if there are $$3n$$ zeros in $$2n$$ rows, then we may remove $$n$$ rows such that there remains at most $$n$$ zeros.
Then simply remove all the columns containing at least a zero.
• That was my thought exactly but then I notice that if we remove exactly one row and one column we are left with a $2n - 1 \times 2n - 1$ matrix but for the induction to work we should have $2(n-1) \times 2(n-1)$. Am I missing something? – cgss Sep 30 '20 at 12:15
• Thanks for that. But please consider answering question asked by @cgss. – nmnsharma007 Sep 30 '20 at 12:32
• Also what if you don't have a row with at least two zeroes anymore? – nmnsharma007 Sep 30 '20 at 13:14
• @cgss You are right, there is a mix of two things here. I updated my proof. – WhatsUp Sep 30 '20 at 13:38
The $$n$$ "high" rows containing the most zeros must contain at least $$2n$$ zeros (doesn't matter how we allocate marginal rows with an equal count so long as the total for the high rows is maximised).
If not, there are at least $$n+1$$ zeros to fit in the other $$n$$ "low" rows, and one of the low rows must contain at least two (pigeonhole). Also there are at most $$2n-1$$ zeros to fit in the high rows and one of these rows must contain just one. But this contradicts the definition of the high rows.
Hence we can choose $$n$$ rows to eliminate at least $$2n$$ zeros, and we need at most $$n$$ columns to eliminate the remaining (at most $$n$$) zeros.
• Thanks.Understood – nmnsharma007 Sep 30 '20 at 13:58 | 2021-04-17T09:36:04 | {
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https://math.stackexchange.com/questions/2614317/bounds-for-underbrace-sqrt-a-sqrta-sqrt-a-sqrta-texta-roo | # Bounds for $\underbrace{\sqrt {a+\sqrt{a+\sqrt {a+…+\sqrt{a}}}}}_\text{$a$roots}$ in terms of $a$?
Inspired by questions like this and this, I have thought about what happens in the general case. My question goes like this:
What can be a suitable upper bound and a suitable lower bound to $$\underbrace{\sqrt {a+\sqrt{a+\sqrt {a+...+\sqrt{a}}}}}_\text{a roots}$$ in terms of $$a$$?
Following from the work done on the mentioned questions, an upper bound can be calculated by assuming there are infinite roots.
Upper bound:
Let $$x={\sqrt {a+\sqrt{a+\sqrt {a+\sqrt {a+\ldots}}}}}$$
$$\therefore x^2=a+{\sqrt {a+\sqrt{a+\sqrt {a+\sqrt{a+\ldots}}}}}$$ $$x^2=a+x$$ $$x^2-x-a=0$$ $$\therefore x=\frac {1 \pm \sqrt {4a+1}}2$$
$$a$$ has to be positive for $$x \in \mathbb R$$ and $$x \ge 0$$ to be a genuine answer. So $$x=\frac {1 + \sqrt {4a+1}}2$$. (I am not exactly sure whether this assumption is valid, please help me out if it isn't).
Lower bound:
Predictably $$\sqrt a$$ is a lower bound but as one of the answers to the mentioned, it is a bad lower bound.
The answer uses $$\sqrt {a +\sqrt a}$$ as a more suitable lower bound. It may not always work but is the best combination of an accurate and easy to calculate lower bound. Obviously $$\underbrace{\sqrt {a+\sqrt{a+\sqrt {a+...+\sqrt{a}}}}}_\text{a-1 roots}$$ is a much better lower bound except that it is almost worth it to calculate the one extra term than to calculate this very accurate lower bound.
So, $$\sqrt {a +\sqrt a} \le \underbrace{\sqrt {a+\sqrt{a+\sqrt {a+...+\sqrt{a}}}}}_\text{a roots} \le \frac {1 + \sqrt {4a+1}}2$$ is a good enough technique to find the upper bounds and lower bounds to $$\underbrace{\sqrt {a+\sqrt{a+\sqrt {a+...+\sqrt{a}}}}}_\text{a roots} \$$ (if my calculation is right) so no one has to bother posting the same question for the upcoming years.
Is this technique correct? Is there any better technique that produces better bounds? I have assumed nothing about when $$a$$ is large and when $$a$$ is small. Does that make a difference? Thanks in advance!
## 3 Answers
Similar to my post an hour ago. Yes, you are correct.
$$\sqrt{a+\sqrt a}<=\sqrt{a+\sqrt{a+\sqrt{...+a}}}<=\sqrt{a+\sqrt{a+\sqrt{a+...}}}$$ Here is the correct way to bound.
• Is there any better technique? When will $\sqrt {a+\sqrt {a+\sqrt a}}$ be a better (yet simpler) lower bound than $\sqrt{a+\sqrt a}$? – Mohammad Zuhair Khan Jan 21 '18 at 5:56
• Honestly, the number of a in the lower bound just has to be less than the number of a in the thing you are trying to bound. In this case I am merely making a general form – QuIcKmAtHs Jan 21 '18 at 5:57
• If $a$ is large (or small) is there any difference? – Mohammad Zuhair Khan Jan 21 '18 at 6:00
By induction for $a>0$ we obtain: $$\sqrt{a}<\frac{1+\sqrt{1+4a}}{2}$$ and $$\sqrt{a+\sqrt{a+..}}<\sqrt{a+\frac{1+\sqrt{1+4a}}{2}}=\sqrt{\frac{1+4a+2\sqrt{1+4a}+1}{4}}=\frac{1+\sqrt{1+4a}}{2}$$
• Is there any better technique? When will $\sqrt {a+\sqrt {a+\sqrt a}}$ be a better (yet simpler) lower bound than $\sqrt{a+\sqrt a}$? – Mohammad Zuhair Khan Jan 21 '18 at 5:56
• For the upper bound it's the best. – Michael Rozenberg Jan 21 '18 at 5:58
• I am asking about lower bound. – Mohammad Zuhair Khan Jan 21 '18 at 6:01
• For lower bound I don't think that there is something better. – Michael Rozenberg Jan 21 '18 at 6:03
• Thanks for the help! I almost feel sorry there will be no 'Welcome $2019$' question. – Mohammad Zuhair Khan Jan 21 '18 at 6:05
Assuming $a\geq 1$, let $f(x)=\sqrt{a+x}$, $a_0=0$ and $a_{n+1}=f(a_n)$.
Since $f(x)$ is a contraction on $\mathbb{R}^+$ (we have $\left|\,f'(x)\right|<1$), by the Banach fixed point theorem the sequence $\{a_n\}_{n\geq 0}$ converges to the only fixed point of $f$, namely $L=\frac{1+\sqrt{4a+1}}{2}$. Additionally $\{a_n\}_{n\geq 0}$ is trivially increasing, hence $L-a_n$ is positive and can be bounded by exploiting the Taylor series of $f(x)$ centered at $x=L$. We have linear convergence, meaning that $$L-a_n\approx C\,f'(L)^n$$ for large values of $n$. | 2020-01-17T19:35:05 | {
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https://math.stackexchange.com/questions/3226780/existence-of-a-symmetric-subset-b-subseteq-a-such-that-2a-a-subseteq-8a | # Existence of a symmetric subset $B\subseteq A$ such that $2A-A\subseteq 8A$
Let $$A$$ be a nonempty open connected subset of a (real) topological vector space $$X$$ such that $$2A-A \subseteq 8A$$ (for instance one could take $$A=(-1,2)$$).
Question. Is it true that there exists a nonempty open connected set $$B\subseteq A$$ such that $$B$$, in addition, is symmetric (i.e., $$B=-B$$)?
• Have you proved it for $X=\Bbb R$ at least? – Adam Chalumeau May 15 at 10:16
• @AdamChalumeau The hypotheses imply $0\in A$. – logarithm May 15 at 10:30
• @logarithm Do they? Do you have a proof of this? If so, then $A \cap -A$ is a good choice for $B$. – Theo Bendit May 15 at 10:33
• @PaoloLeonetti Those are the only connected open sets of $\mathbb{R}$, and of a general $A$ intersected with a line in a general tvs. – logarithm May 15 at 10:49
• @Mirko The example was really supposed to demonstrate why $A \cap (-A)$ may not be connected, even though $0 \in A$ and $A$ is connected. An open example is easy enough too; just add (in the sense of Minkowski sum) an open ball of radius $1/3$ to $A$, and the intersection should still be disconnected (I don't want to compute the exact intersection). However, I agree that $2A - A \subseteq 8A$ (probably) does not hold here. Note that, if $0 \in A$, $A$ is open in a normed linear space, then $A$ always contains an open ball around $0$, which is open, connected, and symmetric. – Theo Bendit May 16 at 1:01
I got many ideas from the comments, but I had to verify the details, and to modify and add some elements to convince myself that the answer is yes (as shown below).
So we have $$\frac A8\subseteq \frac A4 - \frac A8\subseteq A$$ (hence $$\frac A8\subseteq A$$, also $$\frac A{64}\subseteq \frac A8$$, and, by induction, $$\frac A{8^n}\subseteq A$$ for all $$n\ge1$$). Since $$A$$ is open, it is not difficult to see that if $$C$$ is the closure of $$A$$ then also $$\frac A4 - \frac C8=\frac A4 - \frac A8\subseteq A$$ (the details are provided in a corollary near the end). If $$a\in A$$ then the sequence $$\frac a{8^n}$$ converges to $$0$$ (I believe even in general TVS (yes, reference provided in a comment below by OP)), so $$0\in C$$. Hence $$\frac A4=\frac A4-0 \subseteq\frac A4 - \frac C8=\frac A4 - \frac A8\subseteq A$$. So we have:
(i) $$\frac A{32} \subseteq \frac A4$$ because $$\frac A8\subseteq A$$,
(ii) $$\frac A{32} \subseteq \frac A8$$ because $$\frac A4 \subseteq A$$, and
(iii) $$\frac A4 - \frac A8\subseteq A$$.
Using the above we get $$\frac A{32} - \frac A{32} \subseteq \frac A4 - \frac A8 \subseteq A$$. Thus the set $$B=\frac A{32} - \frac A{32}$$ works. Clearly it is symmetric, and it is open and connected: It is connected since it is the continuous image of $$A\times A$$ under the subtraction function (and division by $$32$$), and the product space $$A\times A$$ is connected since the factors are.
Here are some details on the condition that $$\frac A4\subseteq A$$, which was used in the above proof. One way to prove it is without a reference to the closure $$C$$ of $$A$$, as follows.
Claim. $$\frac A4 \subseteq A$$ (or equivalently, $$2A\subseteq8A$$).
Proof. Take any $$a\in A$$, we need to show that $$2a\in 8A$$. Since $$A$$ is open, there is $$n$$ such that $$a+\frac a{2\cdot8^n}\in A$$. Then $$2a=2(a+\frac a{2\cdot8^n})-\frac a{8^n}\in 2A-A\subseteq 8A$$. This completes the proof of the Claim.
Here is an alternative way to show that $$2A\subseteq8A$$. Show that $$2A-A=2A-C$$ (where $$A$$ is open and $$C$$ is the closure of $$A$$). (Then, since $$0\in C$$ we get that $$2A\subseteq2A-C=2A-A\subseteq8A$$.)
Lemma. If $$U$$ is open and $$K$$ is arbitrary then $$U+\overline K= U+K$$ (where $$\overline K$$ is the closure of $$K$$).
Proof. Pick any $$p\in U+\overline K$$. Then $$p=q+r$$ for some $$q\in U$$ and $$r\in\overline K$$. Since $$U$$ is open, there exists a symmetric neighborhood $$V$$ of $$0$$ such that $$q+V\subseteq U$$. Pick $$s\in(r+V)\cap K$$. Then $$v=s-r\in V$$ so $$-v\in-V=V$$ and $$q-v\in U$$, hence $$p=q+r=q-v+r+v=(q-v)+s\in U+K$$. Thus $$U+\overline K\subseteq U+K$$ and $$U+\overline K=U+K$$.
Corollary. If $$A$$ is open and $$C$$ is the closure of $$A$$ then $$2A-C=2A-A$$. (So, if, in addition, $$2A-A\subseteq8A$$ where $$A$$ is open and nonempty then $$2A\subseteq2A-C=2A-A\subseteq8A$$, using that $$0\in C$$ for the first inclusion.)
Proof. Use the above Lemma with $$U=2A$$ and $$K=-A$$.
Discussion. So the inclusion $$2A\subseteq8A$$ was given two different proofs, one direct, and another using the closure $$C$$ of $$A$$, along with the above lemma and corollary (providing a quicker approach, at least to me). I did not know (beforehand) if $$0\in A$$, and did not use it in my proof (though it eventually follows from $$0\in\frac A{32}-\frac A{32}=B\subseteq A$$), and I do not know if $$A\cap(-A)$$ must be connected (assuming that $$A$$ is open and $$2A-A\subseteq8A$$). An example when $$A\cap(-A)$$ need not be connected was provided by another user in the comments above, $$A$$ is the union of the two closed upper semicircles in the plane, of radius $$1$$ with centers at $$(\pm1,0)$$, but this $$A$$ is not open and $$2A-A\not\subseteq8A$$. (The two semicircles could easily be made open, by thickening'' them a little, but it is not clear to me at present if we could also get $$2A-A\subseteq8A$$, and yet to have $$A\cap(-A)$$ disconnected). One more comment: If $$0\in A$$ (where $$A$$ is open) and if we work in a locally connected space, then the connected component of $$A\cap(-A)$$ containing $$0$$ would be open and symmetric (so this component could play the role of $$B$$). But, we need to assume some extra condition (e.g. $$2A-A\subseteq8A$$) to show that $$0\in A$$, and even if we knew that $$0\in A$$, it might not be immediately clear how $$A\cap(-A)$$ would possibly help, if the space is not locally connected. (I would be curious to see a proof - if there is one - based on the use of $$A\cap(-A)$$, showing that either it is connected, or that it contains a connected, open symmetric set.)
• About your question in brackets, the answer is positive, see Theorem 1.15.a in Rudin's "Functional Analysis": let $V$ be any neighborhood of $0$, then $\bigcup_n nV=X$. – Paolo Leonetti May 15 at 14:46
• @PaoloLeonetti I think I fixed the details in my proof (with or without the use of the closure $C$ of $A$). So now I know $0\in A$ since $0\in B\subseteq A$. I do not know if $A\cap(-A)$ is connected (and I am curious, but will leave it for now). Clearly $A\cup(-A)$ is connected. – Mirko May 15 at 15:29
• The proof depends on the choice of the coefficients, however I find it very nice. – Paolo Leonetti May 16 at 14:12 | 2019-05-23T19:01:59 | {
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https://acm.ecnu.edu.cn/problem/1278/ | # 1278. Binomial Showdown
In how many ways can you choose k elements out of n elements, not taking order into account?
Write a program to compute this number.
### 输入格式
The input will contain one or more test cases.
Each test case consists of one line containing two integers n (n>=1) and k (0<=k<=n).
Input is terminated by two zeroes for n and k.
### 输出格式
For each test case, print one line containing the required number. This number will always fit into an integer, i.e. it will be less than 2^31.
Warning: Don’t underestimate the problem. The result will fit into an integer - but if all intermediate results arising during the computation will also fit into an integer depends on your algorithm. The test cases will go to the limit.
### 样例
Input
4 2
10 5
49 6
0 0
Output
6
252
13983816
21 人解决，35 人已尝试。
27 份提交通过，共有 148 份提交。
5.7 EMB 奖励。 | 2020-07-06T06:48:06 | {
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http://mathhelpforum.com/advanced-algebra/183390-counting-inverses-finite-group.html | # Thread: counting inverses of finite group
1. ## counting inverses of finite group
Hi
I am using Charles Pinter's book on abstract algebra . I have to prove that
for any finite group G, if we define the set
$S=\{x\in G : x\neq x^{-1} \}$
i.e. set of elements in G which are not equal to its own inverse. I have to prove that
set S has even no of elements. Now I can see that if some x is in S then its inverse
has to be there since
$(x^{-1})^{-1}\neq x^{-1}$
so members of S appear in pairs. So I can see that , its cardinality has to be even.
But how can I prove this ? any hints ?
thanks
2. ## Re: counting inverses of finite group
what's left to prove?
3. ## Re: counting inverses of finite group
Hi Deveno, I just wanted to know how to put this in mathematical language ? more formal language when you write the proof...
4. ## Re: counting inverses of finite group
Originally Posted by issacnewton
I just wanted to know how to put this in mathematical language ? more formal language when you write the proof...
Suppose that $\{x,y\}\subseteq\mathcal{S}~\&~x\ne y,~x\ne y^{-1}$ then is it true that $\left\{ {x,x^{ - 1} } \right\} \cap \left\{ {y,y^{ - 1} } \right\} = \emptyset~?$
Is it true that $\mathcal{S}=\bigcup\limits_{x \in S} {\left\{ {x,x^{ - 1} } \right\}}~?$
What does that tell about the cardinality of $\mathcal{S} ~?$
5. ## Re: counting inverses of finite group
Originally Posted by issacnewton
Hi Deveno, I just wanted to know how to put this in mathematical language ? more formal language when you write the proof...
since $x^{-1} \in S$ whenever $x \in S$ and for all $x \in S,\ x \neq x^{-1}$, the elements of S occur in distinct pairs.
so if S has k such pairs, |S| = 2k, which is even.
unless you are required to put this in the form of a well-formed statement in some first-order formal language, that is all you need say.
6. ## Re: counting inverses of finite group
Thanks Deveno.......thats better...
Plato, its true that
$\left\{x,x^{-1}\right\}\cap \left\{y,y^{-1}\right \}=\emptyset$
so we can say that cardinality of S is some power of 2 since each subset
$\{x,x^{-1}\}$ of S has two distinct elements.... hence
the proof........right ?
7. ## Re: counting inverses of finite group
Originally Posted by issacnewton
Hi Deveno, I just wanted to know how to put this in mathematical language ? more formal language when you write the proof...
As others have suggested, there's really nothing more to do once you notice that S is composed of pairs of inverses. I can understand the desire to express the point in a more computational manner, but that will only make the proof harder to write---and more importantly, harder to read.
By the way, an interesting result from considering S is that we can prove Cauchy's theorem for the case p=2. For if G is a group of even order, then S and the identity element make up an odd number of elements, which means there must be at least one non-identity element in G which is its own inverse, i.e. has order 2.
8. ## Re: counting inverses of finite group
Originally Posted by issacnewton
Thanks Deveno.......thats better...
Plato, its true that
$\left\{x,x^{-1}\right\}\cap \left\{y,y^{-1}\right \}=\emptyset$
so we can say that cardinality of S is some power of 2 since each subset
$\{x,x^{-1}\}$ of S has two distinct elements.... hence
the proof........right ?
Not a power of 2... a multiple of 2.
9. ## Re: counting inverses of finite group
Thanks everybody. | 2017-03-25T19:46:13 | {
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https://cs.stackexchange.com/questions/40695/what-is-the-name-of-the-property-where-fa-supseteq-fb-when-a-supseteq-b | # What is the name of the property where $f(A) \supseteq f(B)$ when $A\supseteq B$?
Suppose I have a function $f$ on sets.
What is the property of $f$ called when,
for all sets $x$, $y$: $f(x)$ is a superset of $f(y)$ when $x$ is a superset of $y$
i.e.
$$\forall x,y : x\supseteq y \Rightarrow f(x) \supseteq f(y)$$
• Is this a question of computer science? – J.-E. Pin Mar 24 '15 at 10:58
• I think it's not; this seems to be pure mathematics question which we should probably migrate over to Mathematics. Community votes, please! (cc @J.-E.Pin) – Raphael Mar 24 '15 at 14:42
• Come on, guys, monotonicity is super important in lattice theory which has vast applications in logic, coalgebra, in the study of submodular functions and therefore also matroids, and also in the field of circuit complexity. What's with the trigger happiness? – Pål GD Mar 25 '15 at 12:48
## 1 Answer
It is called monotonicity with respect to the inclusion ordering of sets.
More precisely, it is in this case increasing monotonicity since the order is preserved. If the order was reversed, it would be decreasing monotonicity.
It can also be called direct monotonicity and reverse monotonicity, as increasing/decreasing seems used mostly for numeric functions.
The function is said to be strictly monotone iff (in the direct case)
$$\forall x,y : x\supsetneq y \Rightarrow f(x) \supsetneq f(y)$$
• What you're calling "decreasing monotonicity" (a term I've not heard) is also called being anti-monotone or antitone. – David Richerby Mar 24 '15 at 12:58
• @DavidRicherby "decreasing monotonicity" is used mostly for numeric functions. But I did not see why it should be restricted to that. I prefer direct and reverse monotonicity. I am not too happy with anti-monotone as it can be inperpreted as the opposite of monotone, while reverse nonotonicity is still monotonicity. I never heard of antitone, but why not. But I will not fight for terminology, unless it has political/sociological impact. – babou Mar 24 '15 at 13:06 | 2019-12-12T10:18:32 | {
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