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https://math.stackexchange.com/questions/616826/order-of-elements-within-a-group | # Order of elements within a group
If $G$ is a finite group of order (size) $n$ then, for any $g \in G$, the order (period) of $g$ is a divisor of $n$.
Proof: $g$ must have finite order since $G$ is finite. If the order (period) of $g$ is $m$ then the order (size) of the cyclic subgroup $\left<g\right>$ it generates is also $m$. Since $G$ has a subgroup of size $m$, Lagrange's Theorem tells us that $m$ is a divisor of $n$.
What does this proof mean by "$g$ must have finite order since $G$ is finite". Firstly, are we talking about the order (period) or order (size) and either way why does $G$ being finite mean $g$ has finite order?
• $g$ must have finite order since $G$ is finite means there must exist $n \in \mathbb N$ such that $g^n = e$ where $e$ is the identity of $G$. If $g$ did not have finite order, that would mean that there does not exist $k \in \mathbb N$ such that $g^k = e$, we write $\vert g \vert = \infty$ in that case. – Robert Cardona Dec 23 '13 at 20:20
For your first question: The period of $g$.
For your second question: Suppose for a contradiction that g does not have a finite period. Then it will generate an infinite number of elements, resulting in an group with infinite number of elements which is a contradiction to the supposition that $G$ is finite.
• 1) Ok. Can a group have an order and can a groups element have a size? - 2) So are we supposing g is cyclic then? – user2850514 Dec 23 '13 at 20:34
• I refer to the size of the group as the order of the group $G$ and the period/order of the element in the group $G$ as the smallest number $n$ such that $g^n=1$ where 1 is the identity. An element of the group will not have a 'size' as it is not a set and does not contain anything. – ireallydonknow Dec 23 '13 at 20:37
• @user2850514 $g$ is not assumed cyclic, but the group $\langle g\rangle$ it generates clearly is cyclic (i.e. it is generated by a single element, namely $g$) – Hagen von Eitzen Dec 23 '13 at 20:38
• Ok, but then surely we are talking about a subgroup of $G$, namely $\left< g \right>$ for $g\in G$, rather than the actual group $G$? – user2850514 Dec 23 '13 at 20:41
• Yes, since the key idea in the proof is the group $\left< g \right>$ generated by the element $g$. – ireallydonknow Dec 23 '13 at 20:43
If $G$ is finite, and $g \in G$, look at the sequence $g^i$ for integer $i \ge 1$, i.e. $g^i = \{ g, g^2, . . . , g^k, g^{k + 1}, . . . , \text{etc.} \}$ Since $G$ is finite, this sequence of elements of $G$ must repeat itself at some point; thus we must have $g^l = g^k$ for $l > k \ge 1$. If we then choose $k$ to be the least integer for which such repetition occurs for some $l$, and $l$ to be the first integer after $k$ for which it does occur, we must have $g^{l - k} = e$, the identity of $G$, and $g^m \ne e$ for any $m < l - k$. The order of $g$ is thus $l - k$. Thus $G$ finite implies the order of any $g \in G$ finite, and now the argument based on Lagrange's theorem shows that the order of $g$ must divide that of $G$.
Hope this helps. Happy Holidays,
and as always,
Fiat Lux!!!
It depends on the definition of order of an element.
My preferred definition is
Let $G$ be a group and $g\in G$. The order of $g$ is the number of elements of $\langle g\rangle$, if finite; otherwise we say the order is infinite.
Given $g\in G$, we can define $\varphi_g\colon\mathbb{Z}\to G$ by $$\varphi_g(n)=g^n \qquad(n\in\mathbb{Z})$$ and $\varphi_g$ is easily seen to be a homomorphism; the image of $\varphi_g$ is exactly $\langle g\rangle$.
The homomorphism theorem then says that $\varphi_g$ induces an isomorphism between $\mathbb{Z}/\ker\varphi_g$ and $\langle g\rangle$. A known fact about subgroups of $\mathbb{Z}$ gives that $\ker\varphi_g=k\mathbb{Z}$, for a unique $k\ge0$. If $k=0$, $\varphi_g$ is injective and so $g$ has infinite order.
Otherwise $k$ is exactly the order of $g$. In this case, $g^k=\varphi_g(k)=1$ and, for $0<n<k$, $g^n\ne1$, because $n\notin k\mathbb{Z}=\ker\varphi_g$. Another consequence, in this case, is that if $g^n=1$, then $k\mid n$.
Of course, if $G$ is finite, also $\langle g\rangle$ is finite and so the order of $g$ divides $|G|$ by Lagrange's theorem.
Defining the order of $g$ as the least positive integer $k$, if it exists, such that $g^k=1$, forces proving again, but in a complicated way, that the subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$. I see no reason for not using the homomorphism theorem, which is one of the most basic tools in group theory.
However, as already said in other answers or comments, when $G$ is finite there are two distinct positive integers $m<n$ such that $g^m=g^n$, so $g^{n-m}=1$. Therefore the order of $g$ is finite, according to the other definition.
Suppose $g$ is an element of the group $G$ with infinite order. What is the order of the group $G$ then? Can it be finite? | 2019-12-07T04:17:40 | {
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http://mathoverflow.net/questions/94109/twisted-cohomology-of-torus/94122 | # Twisted cohomology of torus
I think I could write down a projective resolution, tensor with the twisted coefficients and find the first cohomology of the standard torus.
BUT, I was wondering if there is an easier way to understand the first (co)homology groups. (Just checking, by Poincaré duality they should be isomorphic, right?)
Is there something like a Künneth formula for twisted coefficients?
-
It depends on what you consider easy. From my point of view, Poincaré duality with local coefficients and the Kunneth spectral sequence are much more complicated than a direct computation in the case of the torus. – Fernando Muro Apr 15 '12 at 13:48
In most cases I used twisted cohomology was when it was with coefficients in the sheaf of sections ${\mathbb V}$ of a flat vector bundle (associated with a linear representation $\rho$ of $\pi_1$ of the base $B$). Again, in most cases I had, $\rho$ would preserve a nondegenerate bilinear form. Then, provided the base is a closed oriented genus $g$ surface, the computation is: $H^1(B, {\mathbb V})=(2g-2) dim(V) + 2 dim H^0(B, {\mathbb V})$ (Euler characteristic + Poincare duality). – Misha Apr 15 '12 at 16:36
@Fernando Muro: So does this mean there is a Kunneth spectral sequence for twisted coefficients? Wikipedia, the source of all knowledge, only refers to such a spectral sequence with coefficients in a commutative ring... – Earthliŋ Apr 16 '12 at 6:05
@Misha: Are you missing a dim there on the left-hand side? – Earthliŋ Apr 16 '12 at 6:07
Sorry, I should have said universal coefficient spectral sequence, instead of Künneth spectral sequence, which is the homological counterpart. Anyway, for this topic there are better references than Wikipedia, e.g. McCleary's book on spectral sequences. – Fernando Muro Apr 16 '12 at 8:39
Let me offer you a projective resolution of $\mathbb{Z}$ as a module over $\pi_1(T)=\mathbb{Z}^2$, obtained from the cellular chain complex of $\mathbb{R}^2$, the universal cover of $T$. The cell structure on $\mathbb{R}^2$ is induced by the usual cell structure on $T$ with one vertex, two edges and one $2$-cell.
The group ring is the ring of Laurent polynomials in two variables $R=\mathbb{Z}[x^{\pm 1},y^{\pm 1}]$. The resolution is
$$0\rightarrow R\stackrel{({y-1},{1-x})^t}\longrightarrow R^2\stackrel{(x-1,y-1)}\longrightarrow R\rightarrow \mathbb{Z}\rightarrow 0$$
-
Right, and in general you may use the Koszul resolution of the augmentation ideal. – Donu Arapura Apr 15 '12 at 15:14
The conditions for the existence of a Koszul resolution are too restrictive to work for any group ring. – Fernando Muro Apr 15 '12 at 15:32
Yes, I should be clearer. I was only referring to the Laurent polynomial ring which was the case of interest to the OP, and to whom the comment was primarily directed. – Donu Arapura Apr 15 '12 at 15:43
Yes, under suitable circumstances there is a Künneth formula for local coefficients. This can be found in Theorems 1.6, 1.7 of the following paper:
R. Greenblatt: Homology with local coefficients and characteristic classes. Homology, Homotopy and Applications, 8(2), 2006, 91-103.
But I agree with Fernando that in case of a torus it is easier to construct a projective resolution. In particular, this way you can handle any twist, while the theorems cited above only adress "product actions".
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\u00a9 2020 wikiHow, Inc. All rights reserved. (Yes, We're Serious.) As you know, rationalizing the denominator means to “rewrite the fraction so there are no radicals in the denominator”. Because of the expression y + √(x 2 +y 2) in the denominator, multiply numerator and denominator by its conjugate y - √(x 2 + y 2) to obtain Questions With Answers Rationalize the denominators of the following expressions and simplify if possible. 3+√2 This image may not be used by other entities without the express written consent of wikiHow, Inc.
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\u00a9 2020 wikiHow, Inc. All rights reserved. 2 16!3 36 4! There are 14 references cited in this article, which can be found at the bottom of the page. I would appreaciate it if you could help with giving me these answers! % of people told us that this article helped them. 5 7 2! Multiply top and bottom by (cuberoot 25 + cuberoot 15 + cuberoot 9) and the denominator simplifies to 2. Rationalizing the Denominator . Rationalizing the Denominator is a process to move a root (like a square root or cube root) from the bottom of a fraction to the top. This image may not be used by other entities without the express written consent of wikiHow, Inc.
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\u00a9 2020 wikiHow, Inc. All rights reserved. . The bottom of a fraction is called the denominator. Define rationalizing. . It is the method of moving the radical (i.e., square root or cube root) from the bottom (denominator) of the fraction to the top (numerator). How can I rationalize the denominator with a cube root that has a variable? As you know, rationalizing the denominator means to “rewrite the fraction so there are no radicals in the denominator”. By signing up you are agreeing to receive emails according to our privacy policy. Rationalizing the Denominator Center for Academic Support * LRC 213 * (816) 271-4524 A. to eliminate radicals from (an equation or expression): to rationalize the denominator of a fraction. In fact, that is really what this next set of examples is about. References. Sometimes, you will see expressions like $\frac{3}{\sqrt{2}+3}$ where the denominator is composed of … This image is not<\/b> licensed under the Creative Commons license applied to text content and some other images posted to the wikiHow website. Because of the expression y + √(x 2 +y 2) in the denominator, multiply numerator and denominator by its conjugate y - √(x 2 + y 2) to obtain Questions With Answers Rationalize the denominators of the following expressions and simplify if possible. Printable Worksheets @ www.mathworksheets4kids.com Name: Rationalize the Denominator 1) 3) 5) 7) 9) 2) 4) 6) 8) 10) 11! To rationalize a denominator, start by multiplying the numerator and denominator by the radical in the denominator. Fixing it (by making the denominator rational) Then, simplify your answer as needed. This lesson demonstrates how to apply the properties of square roots to rationalize the denominator of fractions that contain radicals. For instance, we could easily agree that we would not leave an answer v. intr. . This part of the fraction can not have any irrational numbers. We know ads can be annoying, but they’re what allow us to make all of wikiHow available for free. Examples of rationalizing the denominator. Math. But it is not "simplest form" and so can cost you marks. 15 3! 3. solution: Since it is a basic property of radicals that. Let us look at fractions with irrational denominators. Im taking online courses and have no acess to a teacher for help. Which makes me think I don't understant rationalizing the denominator. Rationalize the denominator calculator is a free online tool that gives the rationalized denominator for the given input. If the radical in the denominator is a cube root, then you multiply by a cube root that will … By using this website, you agree to our Cookie Policy. Rationalizing Denominators. So try to remember these little tricks, it may help you solve an equation one day. Rationalizing the denominator with variables - Examples We ask ourselves, can the fraction be reduced? The key idea is to multiply the original fraction by an appropriate value, such that after simplification, the denominator no longer contains radicals. Rationalizing the denominator is the process of moving any root or irrational number (cube roots or square roots) out of the bottom of the fraction (denominator) and to top of the fraction (numerator).The denominator is the bottom part of a fraction. . Rationalising denominators A fraction whose denominator is a surd can be simplified by making the denominator rational. There is another example on the page Evaluating Limits (advanced topic) where I move a square root from the top to the bottom. This image may not be used by other entities without the express written consent of wikiHow, Inc.
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\u00a9 2020 wikiHow, Inc. All rights reserved. = Rationalizing is simply the process of making sure a number is actually a rational number. Rationalizing the denominator makes it easier to figure out what kind of number it is. But many roots, such as √2 and √3, are irrational. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. Rationalizing the denominator 2. If we have, If you need to write it in radical form, factor out the. Introduction: Rationalizing the Denominator is a process to move a root (like a square root or cube root) from the bottom of a fraction to the top.We do it because it may help us to solve an equation easily. When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. (Without changing the value of the fraction, of course.) As we are rationalizing it will always be important to constantly check our problem to see if it can be simplified more. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. Example 1: Conjugates (more on rationalizing denominators with conjugates) Rationalize $$\frac{3}{2 + \sqrt{5}}$$ Step 1. Simplifying Radicals . RATIONALIZING THE DENOMINATOR WITH VARIABLES About "Rationalizing the denominator with variables" When the denominator of an expression contains a term with a square root or a number within radical sign, the process of converting into an equivalent expression whose denominator is a rational number is called rationalizing the denominator.
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Terms, we will soon see that it equals 2 2 \frac { \sqrt 2. Your way through these pdf worksheets to hone your skills in rationalizing the denominator start... Using our site, you agree to our Cookie Policy examples is about of an expression! Emails according to our privacy Policy √2 1 2 \frac { \sqrt { }. Had 1 term above terms, we have, if you ’ re what allow us to solve an,! Be in simplest form '' and so can cost you marks 1 } { \sqrt { 2 2! Than when we move any fractional power from the bottom ( denominator ) a. Is written in simplest form '' the denominator of any fraction to the. √3, are irrational can generalize to nth roots in the denominator of a fraction to the rationalizing the denominator.! Space station takes off this part of the fraction to the numerator/ top email address to the. Will soon see that it equals 2 2! 3 49 9 are . This quiz and worksheet combo will help you solve an equation, so you should never leave radical! | 2021-03-02T23:30:07 | {
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https://math.stackexchange.com/questions/3386324/3-examples-of-relations-that-satisfy-exactly-two-of-the-following-symmetry-ref?noredirect=1 | 3 examples of relations that satisfy exactly two of the following: symmetry, reflexivity, and transitivity
Could anyone give me three concrete mathematical relations $$R$$ on $$\mathbb{Z}$$ that satisfy exactly two of the following, no two of which satisfy the same two axioms:
1) Axiom of reflexivity ($$\forall a\in \mathbb{Z}, (a,a)\in R$$)
2) Symmetric axiom ($$\forall a,b\in \mathbb{Z} ((a,b)\in R\Leftrightarrow (b,a)\in R$$)
3) Axiom of transitivity ($$\forall a,b,c\in\mathbb{Z} (((a,b)\in R\wedge (b,c)\in R)\Rightarrow (a,c)\in R$$)
?
I'm completely lost as to how to find a relation that's reflexive and symmetric but not transitive. And the same goes for finding relations that are reflexive and transitive but not symmetric and relations that are symmetric and transitive but not reflexive.
I know the relation $$(a,b)$$ on $$\mathbb{Z}$$ defined as $$a^2 + b^2 =1$$ is symmetric but neither transitive nor reflexive. Personally, I'm not even sure there exists a relation that is reflexive and symmetric but not transitive, because I hypothesize that reflexive relations imply subtraction (i.e. they can always be rearranged so that some part of the equation has $$(a-b)$$ in it). I've thought of as many relations as I could, but with no luck. $$a|b$$, $$ad=bc$$ for some $$c,d\in\mathbb{Z}$$, and $$a^m=b^n$$ for some $$m,n\in\mathbb{Z}$$ are all equivalence relations.
Btw, if some of these relations do not exist, could someone at least give me a hint as to how to prove why that is?
• It might prove easier to think of relations on a set $\{1, 2, \cdots, n\}$ for some (small, preferably) integer $n$. Remember what a relation fundamentally is: a relation $R$ on a set $S$ is some subset of the Cartesian product $S \times S$. [cont.] – Eevee Trainer Oct 9 '19 at 3:14
• Thus it becomes convenient to think of subsets of $\{ \; (i,j) \; \}_{i,j=1}^n$ for some small $n$. Usually $2$ or $3$ is more than sufficient for what you need, or you can make appeals to the empty relation. These might not have the same intuitive grasp as, say, "$a \sim b$ if and only if $a^2 + b^2 = 1$ for $a,b \in \Bbb Z$," but they're easier to construct and easier to prove. – Eevee Trainer Oct 9 '19 at 3:14
• You have the definitions of symmetry and reflexivity swapped. A symmetric relation has $(a,b)\in R\iff(b,a)\in R$. A reflexive relation as $(a,a)\in R$ for all $a$. – user856 Oct 9 '19 at 4:34
• A notable previous question: Are there real-life relations which are symmetric and reflexive but not transitive? – user856 Oct 9 '19 at 5:03
• The relation $$i\sim j\Leftrightarrow|i-j|\leq5$$ is reflexive and symmetric but not transitive.
This is an example of a tolerance relation. Tolerance relations, like this example, admit things that are "close" to each other but aknowledge that two objects might both be close to a third object without being close to each other.
• The relation $$i\leq j$$ is reflexive and transitive but not symmetric.
This is an example of a partial order. These, alongside equivalence relations, are the primary subclasses of relations that get studied, since they describe the inequalities of numbers that we are used to.
• The empty relation on $$\mathbb Z$$ is symmetric and transitive but not reflexive.
There is a fairly common false proof that symmetric and transitive relations must be reflexive, but the empty relation is a counterexample. For the most part, common and well-behaved relations that you would verify as equivalence relations will very clearly be seen to be reflexive, but it is a part of the proof that must be verified.
• would you mind telling me how you arrived at those relations? – user706791 Oct 9 '19 at 13:14
• @fordjones I just added some explanation to the three examples. – Matthew Daly Oct 9 '19 at 14:59
• Sorry for my stupidity, but would you mind explaining to me how two objects can both be close to a third object but not close to each other using the first relation? I might be misunderstanding, but are you referring to the two objects as $i$ and $j$? – user706791 Oct 11 '19 at 19:21
To construct a small example of a symmetric and reflexive, but not transitive, relation $$R$$ on $$\mathbb{Z}$$, you might begin by deciding you want $$(0,1),(1,2)\in R$$ but $$(0,2)\not\in R$$. If this can be arranged, then $$R$$ will not be transitive.
So start with $$(0,1),(1,2)\in R$$. Now it's easy to make $$R$$ reflexive by including $$(a,a)\in R$$ for all $$a\in \mathbb{Z}$$. And if you include $$(1,0)$$ and $$(2,1)$$, $$R$$ will be symmetric also. Importantly, you did not put $$(0,2)$$ in $$R$$, so $$R$$ is not transitive. | 2020-05-29T17:17:38 | {
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http://math.stackexchange.com/questions/87505/equation-for-a-straight-line-in-cartesian-space/87515 | # Equation for a straight line in Cartesian space
I am trying to create a straight line in the Cartesian space made from two points that have (x,y,z) coordinates. This is for a making a robot arm move in a straight line, I would input two points and the math would give me back a certain numbers of points that create a straight line from point A to point B. The robot arm moves on a circular base.
I think I should use the equation of a line: $ax + by + c = 0$.
But I'm not exactly sure how I would get the intermediate points from that and if it works for 3 dimensions. If this is to vague let me know and I can further clarify.
-
Just so you know, what you are describing is actually Cartesian/Euclidean space, not the Cartesian plane. It is a subtle difference, but an important one. Your equation $ax+by+c=0$ correctly describes a line in the Cartesian plane, but not so in Cartesian/Euclidean space. – Michael Boratko Dec 1 '11 at 21:51
The equation you gave, $ax+by+c=0$, is the equation for a line in two dimentions. In three dimensions, you can define a line by a point and a vector. This is not in contradiction with the idea that two points determine a line, as obviously there is a vector between two points and therefore by specifying two points you have also specified a point and a vector. This is probably all too pedantic to be worthwhile discussing further, so let's move on to the specific useful example.
Let one point be defined as $P=(x_p,y_p,z_p)$, and another point be defined as $Q=(x_q,y_q,z_q)$. Then we can define the line $L$ as follows:
$$L=\{P+t(Q-P)\}$$ where $t$ is any real number. To dissect this a little bit, this shows us that $P$ is in the set (for $t=0$), and $Q$ is in the set (for $t=1$). By allowing $t$ to be any real number, we are scaling the vector between $P$ and $Q$, and this is what gives us the whole line.
Now, we consider any arbitrary point on the line, which we will define as $(x,y,z)$. If this point is on the line, then it is in the set $L$ and so we must have \begin{align}x=&x_p+t(x_q-x_p)\\ y=&y_p+t(y_q-y_p)\\ z=&z_p+t(z_q-z_p) \end{align} These equations are parametric, that is they depend on a parameter $t$, but such equations are necessary in order to describe a line in three dimensions. Obviously, if the line happened to be in one of the planes, say the $xy$ plane, then $z_p=z_q=0$, and so the equations could be solved for $t$. Substituting, you would get the familiar two-dimensional equation for a line, however in general the best you can do is solve for $t$ in the above equations and reduce the three equations to two.
It is also common to write the so-called "symmetric" equations for a line in three dimensions by solving for $t$ and setting all three equal to each other, like so: $$\frac{x-x_p}{x_q-x_p}=\frac{y-y_p}{y_q-y_p}=\frac{z-z_p}{z_q-z_p}$$
Unfortunately, I don't think these sorts of equations will help with your robot project, but perhaps the discussion will engender some good ideas. You can take a look at Paul's Online Notes for more discussion about how lines are represented in three dimensions. He doesn't follow exactly the same approach as I have outlined here, but it is very similar.
You mentioned that the robot arm is on a circular base, and as such it might be advantageous to look into using Spherical Coordinates (which may simplify many of the computational aspects of movement).
-
Thank you, this was almost verbatim what my professor explained to me to do after meeting with him. – Nick Dec 2 '11 at 0:09
parameterized formula for a line through 2 points $P_1=(x_1,y_1,z_1)$ and $P_2=(x_2,y_2,z_2)$
$$\begin{cases} x=x_1+t\cdot(x_2-x_1) \\ y=y_1+t\cdot(y_2-y_1) \\ x=z_1+t\cdot(z_2-z_1) \\ \end{cases}$$
vary $t$ to get a sample of points where $t\in[0,1]$ gives points $\in[P_1,P_2]$
you can also use the weighted average of the 2 points
$$\begin{cases} x=t\cdot x_1+(1-t)\cdot x_2 \\ y=t\cdot y_1+(1-t)\cdot y_2 \\ x=t\cdot z_1+(1-t)\cdot z_2 \\ \end{cases}$$
again $t\in[0,1]$ for points $\in[P_1,P_2]$
you can eliminate the $t$ so you get 2 formulas (as the intersection of 2 planes)
- | 2014-04-23T11:15:34 | {
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https://math.codidact.com/posts/286527 | Q&A
# Notation for one-sided hypothesis testing
+2
−0
I see the following notation for one-sided hypothesis testing:
• $H_0$: $K = 2$
• $H_1$: $K > 2$
I would find it more natural to write:
• $H_0$: $K \le 2$
• $H_1$: $K > 2$
Assume a situation where $K$ is not, by definition, limited to equal or be higher than two. $K$ is just a dummy variable and 2 a dummy number.
Where does the first notation come from and why is it in use?
Why does this post require moderator attention? | 2022-07-02T22:41:20 | {
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https://stats.stackexchange.com/questions/492000/a-fair-die-is-rolled-1-000-times-what-is-the-probability-of-rolling-the-same-nu | # A fair die is rolled 1,000 times. What is the probability of rolling the same number 5 times in a row?
A fair die is rolled 1,000 times. What is the probability of rolling the same number 5 times in a row? How do you solve this type of question for variable number of throws and number of repeats?
• The question does not specifically state how many sides the die has. Sextus Empiricus's answer assumes a 6-sided die, which is the most commonly used die type. Is that what you wanted? – Leland Hepworth Oct 15 '20 at 15:29
• @LelandHepworth I am not so much into dungeons and dragons or other roll playing games. So I assumed a 6-sided dice, which is the standard dice in statistics. But the question is easily generalizable. The type of dice is irrelevant for explaining the solution to the question. – Sextus Empiricus Oct 15 '20 at 17:00
• @LelandHepworth Exactly. My spontaneous thought was about the tetraeder-shaped dies Irving Finkel (one of the curators of the British Museum) uses to play the Royal Game of Ur (think about the epoch cuneiform and Mesopotamia). Have a look, e.g. here with Tom Scott. – Buttonwood Oct 17 '20 at 20:04
Below we compute the probability in four ways:
Computation with Markov Chain 0.473981098314993
Computation with generating function 0.473981098314988
Estimation false method 0.536438013618686
Estimation correct method 0.473304632462677
The first two are exact methods and differ only a little (probably some round of errors), the third method is a naive estimation that does not give the correct number, the fourth method is better and gives a result that is very close to the exact method.
## Computationally:
### Markov Chain
You can model this computationally with a transition matrix
Say the column vector $$X_{k,j} = \lbrace x_1,x_2,x_3,x_4,x_5 \rbrace_{j}$$ is the probability to have $$k$$ of the same numbers in a row in the $$j$$-th dice roll. Then (when assuming a 6-sided dice)
$$X_{k,j} = M \cdot X_{k,j-1}$$ with
$$M = \begin{bmatrix} \frac{5}{6} & \frac{5}{6} & \frac{5}{6} & \frac{5}{6} & 0 \\ \frac{1}{6} & 0& 0 & 0 & 0 \\ 0& \frac{1}{6} & 0& 0 & 0 \\ 0 & 0& \frac{1}{6} & 0& 0 \\ 0&0 & 0& \frac{1}{6} & 1 \\ \end{bmatrix}$$
where this last entry $$M_{5,5} = 1$$ relates to 5 of the same in a row being an absorbing state where we 'stop' the experiment.
After the first roll you will be certainly in state 1 (there's certainly only 1 of the same number in a row).
$$X_{k,1} = \lbrace 1,0,0,0,0 \rbrace$$
After the $$j$$-th roll this will be multiplied with $$M$$ a $$j-1$$ times
$$X_{k,j} = M^{j-1} \lbrace 1,0,0,0,0 \rbrace$$
R-Code:
library(matrixcalc) ### allows us to use matrix.power
M <- matrix(c(5/6, 5/6, 5/6, 5/6, 0,
1/6, 0 , 0 , 0 , 0,
0, 1/6, 0 , 0 , 0,
0, 0 , 1/6, 0 , 0,
0, 0 , 0 , 1/6, 1),
5, byrow = TRUE)
start <- c(1,0,0,0,0)
matrix.power(M,999) %*% start
The result is $$X_{k,1000} = \begin{bmatrix} 0.438631855\\ 0.073152468\\ 0.012199943\\ 0.002034635\\ \color{red}{0.473981098}\end{bmatrix}$$
and this last entry 0.473981098 is the probability to roll the same number 5 times in a row in 1000 rolls.
### generating function
Our question is:
• How to calculate the probability of rolling any number at least $$k$$ times in a row, out of $$n$$ tries?
This is equivalent to the question
• How to calculate the probability of rolling the number 6 at least $$k-1$$ times in a row, out of $$n-1$$ tries?
You can see it as tracking whether the dice roll $$m$$ is the same number as the number of the dice roll $$m-1$$ (which has 1/6-th probabilty). And this needs to happen $$k-1$$ times in a row (in our case 4 times).
In this Q&A the alternative question is solved as a combinatorial problem: How many ways can we roll the dice $$n$$ times without the number '6' occuring $$k$$ or more times in a row.
This is found by finding all possible combinations of ways that we can combine the strings 'x', 'x6', 'x66', 'x666' (where 'x' is any number 1,2,3,4,5) into a string of length $$n+1$$ ($$n+1$$ instead of $$n$$ because in this way of constructing strings the first letter is always $$x$$ here). In this way we counted all possibilities to make a string of length $$n$$ but with only 1, 2, or 3 times a 6 in a row (and not 4 or more times).
Those combinations can be found by using an equivalent polynomial. This is very similar to the binomial coefficients which relate to the coefficients when we expand the power $$(x+y)^n$$, but it also relates to a combination.
The polynomial is
$$\begin{array}{rcl} P(x) &=& \sum_{k=0}^\infty (5x+5x^2+5x^3+5x^4)^k\\ &=& \frac{1}{1-(5x+5x^2+5x^3+5x^4)} \\ &=& \frac{1}{1-5\frac{x-x^5}{1-x}}\\ &=& \frac{1-x}{1-6x+5x^5} \end{array}$$
The coefficient of the $$x^n$$ relates to the number of ways to arrange the numbers 1,2,3,4,5,6 in a string of length $$n-1$$ without 4 or more 6's in a row. This coefficient can be found by a recursive relation. $$P(x) (1-6x+5x^5) = 1-x$$ which implies that the coefficients follow the relation
$$a_n - 6a_{n-1} + 5 a_{n-5} = 0$$
and the first coefficients can be computed manually
$$a_1,a_2,a_3,a_4,a_5,a_6,a_7 = 5,30,180,1080,6475,38825,232800$$
With this, you can compute $$a_{1000}$$ and $$1-a_{1000}/6^{999}$$ will be the probability to roll the same number 5 times in a row 5.
In the R-code below we compute this (and we include a division by 6 inside the recursion because the numbers $$a_{1000}$$ and $$6^{999}$$ are too large to compute directly). The result is $$0.473981098314988$$, the same as the computation with the Markov Chain.
x <- 6/5*c(5/6,30/6^2,180/6^3,1080/6^4,6475/6^5,38825/6^6,232800/6^7)
for (i in 1:1000) {
t <- tail(x,5)
x <- c(x,(6/6*t[5]-5/6^5*t[1])) ### this adds a new number to the back of the vector x
}
1-x[1000]
## Analytic/Estimate
### Method 1: wrong
You might think, the probability to have in any set of 5 neighboring dices, 5 of the same numbers, is $$\frac{1}{6^4} = \frac{1}{1296}$$, and since there are 996 sets of 5 neighboring dices the probability to have in at least one of these sets 5 of the same dices is:
$$1-(1-\frac{1}{6^4})^{996} \approx 0.536$$
But this is wrong. The reason is that the 996 sets are overlapping and not independent.
### Method 2: correct
A better way is to approximate the Markov chain that we computed above. After some time you will get that the occupation of the states, with 1,2,3,4 of the same number in a row, are more or less stable and the ratio's will be roughly $$1/6,1/6^2,1/6^3,1/6^4$$ (*). Thus the fraction of the time that we have 4 in a row is:
$$\text{frequency 4 in a row} = \frac{1/6^4}{1/6+1/6^2+1/6^3+1/6^4}$$
If we have these 4 in a row then we have a 1/6-th probability to finish the game. So the frequency of finishing the game is
$$\text{finish-rate} = \frac{1}{6} \text{frequency 4 in a row} = \frac{1}{1554}$$
and the probability to be finished after $$k$$ steps is approximately
$$P_k \approx 1-(1-\frac{1}{1554})^{k-4} \underbrace{\approx 0.47330}_{\text{if k=1000}}$$
much closer to the exact computation.
(*) The occupation in state $$k$$ during roll $$j$$ will relate to the occupation in state $$k-1$$ during roll $$j-1$$. We will have $$x_{k,j} = \frac{1}{6} x_{k-1,j-1} \approx \frac{1}{6} x_{k-1,j}$$. Note that this requires that you have $$x_{k-1,j} \approx x_{k-1,j-1}$$, which occurs when the finish-rate is small. If this is not the case, then you could apply a factor to compensate, but the assumption of relatively steady ratio's will be wrong as well.
### Related problems
• What code? R? – Peter Mortensen Oct 15 '20 at 18:50
• @PeterMortensen The code is in the grey block. But I guess that you are asking for the type of language. I have used the language that is most used on stats.stackexchange.com – Sextus Empiricus Oct 15 '20 at 18:53
• @PeterMortensen It's R – Dale C Oct 15 '20 at 20:38
• "The result is 0.473981098" Pardon the semi-math-challenged question, but does that mean that the odds of it happening somewhere in those 1,000 rolls are just over 47%? Or had you already converted to percentage? (Which seems unlikely to me. In 1,000 rolls, I'd figure the odds were well over less than half a percent...but ~47% percent sounds reasonable.) – T.J. Crowder Oct 16 '20 at 9:52
• @T.J.Crowder it means $X_{5,1000} = 0.473981098$ it is the fraction of the cases that have ended in the state 5 after 1000 rolls. So yes in terms of percentages this would be ~47%. I will clarify this in the answer later. – Sextus Empiricus Oct 16 '20 at 10:02
I got a different result from the accepted answer and would like to know where I've gone wrong.
I assumed a fair, 6-sided die, and simulated 1000 runs of 1000 rolls each. When the result of a roll matches the results of the previous 4 rolls, a flag is set to TRUE. The mean of this flag column and the mean of the runs is then reported. I get ~0.07% as the probability of seeing 5 rolls in a row of the same number.
In R,
tibble(
run = rep(seq(1:1000), each = 1000),
roll = rep(seq(1:1000), 1000),
x = sample(1:6, 1000000, replace = T)
) %>%
group_by(run) %>%
mutate(
same_five = x == lag(x, 1) & x == lag(x, 2) & x == lag(x, 3) & x == lag(x, 4)
) %>%
summarize(
p_same_five = mean(same_five, na.rm = TRUE), .groups = "drop"
) %>%
summarize(mean(p_same_five)) * 100
mean(p_same_five)
1 0.07208702
• It looks like what you are calculating is the frequency of rolling 5 in a row, which Sextus showed is about 0.06. However, the problem isn't asking for the frequency of 5 in a row, but rather how likely you are to see at 5 in a row at least once during 1000 rolls. So you should have a flag for each of your runs that you mark True if 5 in a row occurs anywhere in the run. – Tyberius Oct 17 '20 at 1:32
• It'll work with this line same_five = sum(x == lag(x, 1) & x == lag(x, 2) & x == lag(x, 3) & x == lag(x, 4), na.rm = TRUE)>0 then the same_five in the entire row will be TRUE if for any of the x you have a 5-in-a-row. Personally, I would not do this programming nested within tibble and dplyr. It makes it more difficult to debug and also more difficult for others to read. – Sextus Empiricus Oct 17 '20 at 20:25
• Yeah Tyberius has it, I did do a simulation using run length encoding, andrewpwheeler.com/2020/10/17/simulating-runs-of-events. You can see it approximately conforms to Sextus's analytic estimates (0.475` for the R and python code). – Andy W Oct 18 '20 at 14:06 | 2021-01-26T07:51:33 | {
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https://math.stackexchange.com/questions/890218/disconnected-topoological-space-with-intermediate-value-property | Disconnected Topoological Space with Intermediate Value Property
Does There exist a disconnected topological space with intermediate value property? Intermediate Value Property states that 'a topological space X is said to have intermediate value property if for every continuous function f: X to Y (where Y is ordered set with order topology) the following is true: If a, b belongs to X and there exist r in Y s.t. r lies between f(a) and f(b) then there exist c in X s.t. f(c) = r.
• No. Let $X = U \cup V$, where $U,V$ are disjoint nonempty open sets. Let $f\colon X \to \mathbb{R}$ be given by $f(x) = 1$ for $x\in U$ and $f(x) = 0$ for $x\in V$. – Daniel Fischer Aug 7 '14 at 16:53
One characterisation of disconnected spaces is
A topological space $X$ is disconnected if and only if there is a surjective continuous function $f\colon X \to Y$, where $Y = \{0,1\}$ is endowed with the discrete topology.
Proof: Let first $X$ be disconnected. Then by definition there are two disjoint nonempty open sets $U,V\subset X$ with $X = U \cup V$. Then the function
$$f(x) = \begin{cases}1 &, x \in U\\ 0 &, x \in V \end{cases}$$
is surjective, because $U \neq \varnothing \neq V$, and it is continuous because $f^{-1}(B)$ is one of the four open sets $\varnothing, U, V, X$ for any $B\subset Y$.
Conversely, if there is a surjective continuous $f\colon X \to Y$, then $U = f^{-1}(\{1\})$ and $V = f^{-1}(\{0\})$ are two disjoint open sets with $X = f^{-1}(\{0,1\}) = f^{-1}(\{0\}\cup\{1\}) = f^{-1}(\{0\}) \cup f^{-1}(\{1\}) = V\cup U$, so $X$ is disconnected.
Now any map $g$ from $Y$ to an ordered set with the order topology is continuous, and there certainly are such maps where there is a value between $g(0)$ and $g(1)$ (we can for example take $g\colon Y\to \mathbb{R}; 0 \mapsto 0,\, 1 \mapsto 1$). Then $g\circ f$ is a continuous mapping into an ordered set with the order topology that does not have the intermediate value property.
• How to prove this characterisation of disconnected space. Rest everything is fairly good. – Sushil Aug 7 '14 at 17:23
• Pretty much the same argument as in the comment, I'll add the proof. – Daniel Fischer Aug 7 '14 at 17:27
• Hence in short: A connected topological space is equivalent to topological space with Intermediate Value Property. Am I right? – Sushil Aug 7 '14 at 17:41
• In short: Yes. Exactly. – Daniel Fischer Aug 7 '14 at 17:47
• Ok thanks that solves the problem completely – Sushil Aug 7 '14 at 17:50 | 2019-04-19T05:18:23 | {
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https://math.stackexchange.com/questions/155040/prove-that-the-product-of-four-consecutive-positive-integers-plus-one-is-a-perfe/155050 | # Prove that the product of four consecutive positive integers plus one is a perfect square
I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.
If $n = 1 + m$, where $m$ is the product of four consecutive positive integers, prove that $n$ is a perfect square.
Now since $m = p(p+1)(p+2)(p+3)$;
$p = 0, n = 1$ - Perfect Square
$p = 1, n = 25$ - Perfect Square
$p = 2, n = 121$ - Perfect Square
Is there any way to prove the above without induction? My approach was to expand $m = p(p+1)(p+2)(p+3)$ into a 4th degree equation, and then try proving that $n = m + 1$ is a perfect square, but I wasn't able to do it. Any idea if it is possible?
• I assume you posted this from your smartphone in the bathroom during your exam ;) – Wipqozn Jun 7 '12 at 13:57
• @KartikAnand I was just joking, hence the ;) – Wipqozn Jun 7 '12 at 14:16
• @Wipqozn no worries ;) (I knew :P ) – Kartik Anand Jun 8 '12 at 11:13
• haha, you had me worried there with your comment :P – Wipqozn Jun 8 '12 at 11:16
• In one sentence: Consider $p(p+3)=p^2+3p:=n$ and $(p+1)(p+2)=p^2+3p+2=n+2$ so that the product plus one is $n^2 + 2n + 1 = (n+1)^2 = (p^2 + 3p + 1)^2$. – Benjamin Dickman Jun 29 '18 at 18:35
Your technique should have worked, but if you don't know which expansions to do first you can get yourself in a tangle of algebra and make silly mistakes that bring the whole thing crashing down.
The way I reasoned was, well, I have four numbers multiplied together, and I want it to be two numbers of the same size multiplied together. So I'll try multiplying the big one with the small one, and the two middle ones.
$$p(p+1)(p+2)(p+3) + 1 = (p^2 + 3p)(p^2 + 3p + 2) + 1$$
Now those terms are nearly the same. How can we force them together? I'm going to use the basic but sometimes-overlooked fact that $xy = (x+1)y - y$, and likewise $x(y + 1) = xy + x$.
\begin{align*} (p^2 + 3p)(p^2 + 3p + 2) + 1 &= (p^2 + 3p + 1)(p^2 + 3p + 2) - (p^2 + 3p + 2) + 1 \\ &= (p^2 + 3p + 1)(p^2 + 3p + 1) + (p^2 + 3p + 1) - (p^2 + 3p + 2) + 1 \\ &= (p^2 + 3p + 1)^2 \end{align*} Tada.
• Wow you did it, but tell me one thing.You said "I have four numbers multiplied together, and I want it to be two numbers of the same size multiplied together. ", but why? I mean 1 is still being added to expression. – Kartik Anand Jun 7 '12 at 9:50
• Well, I was kind of ignoring the 1 for the time being; I wanted two numbers roughly the same, and I thought the details likely to work themselves out :) – Ben Millwood Jun 7 '12 at 9:56
• Slightly smoother: When you are at $(p^2+3p)(p^2+3p+2)+1$, let $x=p^2+p+1$, We are looking at $(x-1)(x+1)+1=x^2-1+1=x^2$. – André Nicolas Jun 7 '12 at 10:14
• @AndréNicolas make it 3p in the expression for x – Kartik Anand Jun 7 '12 at 10:19
• This goes by what I like to call the 'Mathematics of wishful thinking' : there is a solution, we sort of know what it should look like, and so we go and hope that everything works out. +1 – davidlowryduda Jun 7 '12 at 19:29
$(n-1)(n+1)+1 = n^{2}$.
Note that $(n+1)-(n-1)=2$.
With this in mind
\begin{align*} p(p+1)(p+2)(p+3)+1 &= (p^{2}+3p)(p^{2}+3p+2)+1 \\ &= [(p^{2}+3p+1)-1][(p^{2}+3p+1)+1]+1 \\ &= (p^{2}+3p+1)^2 \end{align*}
• Some simple observations oft lead to marvelous discoveries, greatly confirmed here. – awllower Jun 7 '12 at 14:23
• That's beautiful! – Chris Cudmore Jun 7 '12 at 18:39
Below I present a generalization. Using the abbreviations$\rm\ \ c = a\!+\!b,\ \ \color{red}d = ab/2\:\$ we compute
$$\rm\begin{eqnarray} &&\rm\qquad\quad\ \color{blue}{(x\!+\!a)\,(x\!+\!b)}\,(x\!+\!c)\,x &=&\rm\, \color{blue}{(x^2\!+cx\ \ +\ \ ab\ \ \ \, )}\,(x^2+cx\:\!) \\ && &=&\rm\, (x^2\!+cx+d\ \,\color{red}{+\, d})\,(x^2+cx+d\, \color{red}{-\,d}) \\ && &=&\rm\, (x^2\!+cx+d)^2\! \color{red}{- d^2} \\ \rm b=2\quad &\Rightarrow&\rm\quad\ \ \ \ x(x+a)(x\!+\!2)(x\!+\!a\!+\!2) &=&\,\rm (x^2\!+(a\!+\!2)\,x+a)^2 -a^2 \\ \rm a=1\quad &\Rightarrow&\rm\qquad\quad\ \ \ x(x\!+\!1)(x\!+\!2)(x\!+\!3) &=&\rm\, (x^2\!+3\:\!x+1)^2 -1\ \ \ as\ sought. \end{eqnarray}$$
Here's another way which begins by exploiting a symmetry in the expression.
Notice that if you substitute $x=p+\frac{3}{2}$, the expression becomes
$$\left(x-\frac{3}{2}\right)\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(x+\frac{3}{2}\right) + 1$$
Now see that the terms make the product of 2 differences of squares
\begin{align} & \quad \left(x+\frac{3}{2}\right)\left(x-\frac{3}{2}\right)\left(x+\frac{1}{2}\right)\left(x-\frac{1}{2}\right) + 1 \\&= \left(x^2-\frac{9}{4}\right)\left(x^2-\frac{1}{4}\right) + 1 \\ &= \left(x^4 - \frac{10}{4} x^2 + \frac{9}{16}\right) + 1 \\ &= x^4 - \frac{10}{4} x^2 + \frac{25}{16} \\ &= \left(x^2 - \frac{5}{4}\right)^2 \\ &= \left(p^2 + 3p + 1\right)^2 \end{align}
which is a perfect square.
• You miss one final step; to see that it is a perfect square of an integer, you need that $x^2 - 5/4$ is an integer. In this case you could say you are "lucky" that $2 \cdot (3/2), (3/2)^2 - 5/4 \in \mathbb{N}$. – TMM Jun 7 '12 at 19:31
• @TMM: Or you could just say that it's an integer which is the perfect square of a rational number, which makes it the perfect square of an integer. – Micah Jun 7 '12 at 23:05
• Yes, I did miss the final step, but there's no luck involved - it's straightforward algebra that x^2 - 5/4 is an integer. See my edit which completes the proof. – JTB Jun 8 '12 at 13:02
Set $p+1.5=q$. Now \begin{align*}m &= (q-1.5)(q-0.5)(q+0.5)(q+1.5)+1 \\ &= (q-1.5)(q+1.5)(q-0.5)(q+0.5)+1 \\ &= (q^2 - 2.25)(q^2-0.25)+1 \\ \end{align*}
Let $q^2 = r$.
\begin{align*} m &= (r-2.25)(r-0.25)+1\\ &= r^2-2.5a+1.5625 \\ &= (r-1.25)^2. \end{align*}
This is a perfect square since r ends in 0.25 as q ends in 0.5
Basically the substitution converted it from a fourth degree to a quadratic which made it easy to deal with.
• I like this method. The first thing you do is try to make the equation at the top more symmetrical, which was basically my idea, but we went about it in different ways. – Ben Millwood Jun 7 '12 at 10:10
• @benmachine totally agree, but I just don't that like the number 1.5 – zinking Jun 8 '12 at 2:18
$$1+p(p+1)(p+2)(p+3)=1+ \dots +p^4.$$
If you want a general formula, it must be a square either of the form $(p^2+cp+1)^2$ or $(p^2+cp-1)^2$ for some constant $c$.
Expand the squares and the original product and match up two terms to calculate $c$. Verify that the other coefficients are correct as well.
Details:
The expansion of the product is $p^4+6p^3+11p^2+6p+1$.
The expansions of the squares are $p^4 + 2cp^3+c^2p^2\pm2p^2\pm2cp+1$.
Comparing the coefficients of $p^3$ gives $c=3$ which evidently works with the plus sign, so we get $(p^2+3p+1)^2$.
If I am missing something I will take this answer down, but the following seems responsive to your question.
If
$m = 1 + x(x+1)(x+2)(x+3)$ we can expand this as $1+6x+11x^2+6x^3+x^4$.
This is
$m = (1+3x+x^2)^2$
So when x is an integer, this shows that m is a perfect square, without induction.
• How did you factorise the quartic? – Ben Millwood Jun 7 '12 at 9:47
• @benmachine: I basically used Phira's process (below) and guessed the constant. – daniel Jun 7 '12 at 9:51
• @benmachine you can also do $(x^2+ax+b)^2 = 1+6x+11x^2+6x^3+x^4$ and that is easy to solve for $a$ and $b$. – picakhu Jun 7 '12 at 13:26
I think there are two issues here. One is constructing the quartic, which just depends on you doing the algebra correctly. The second is proceeding to factorise the quartic. It would be easier to factorise it if you know what the factorisation is going to be.
To discover this, I tried a few examples. For $p=7$, the quartic gives $5041=71^2$. For $p=14$, the quartic gives $57121=239^2$. I noticed that $71=72-1=8\times9-1$ and $239=15\times16-1$.
This suggested that the quartic was $((p+1)(p+2)-1)^2$.
Once you know the answer, it is easy to find it!
Take $p^2$ common after multiplying.
Then put $p +1/p =y$ and solve.
I have to add what I think is a 'dumb' way to do it by hand (with paper) as opposed to Alex B. succinct cleverness:
First, multiply out the product to get $p^4 + 6p^3 + 11p^2 + 6p + 1$.
Since this is a square, it must be a quadratic $p^2 + x p + y$.
Squaring the quadratic, ignoring a lot of the cruft, and just looking at just the second and last coefficients
$$6 = x + x$$
and
$$1 = y^2$$
and you're done.
Select any $$a\in\mathbb{Z}_{\ge 2}$$ and define $$P$$ to be the product of the four consecutive integers $$a-1,a,a+1$$ and $$a+2$$, that is $$P=(a-1)a(a+1)(a+2).$$ Expanding $$P$$, we get $$P=a^4+2a^3-a^2-2a.$$ Thus, we have $$P+1=a^4+2a^3-a^2-2a+1=(a^2+a-1)^2,$$ that is $$P+1$$ is a perfect square. Now since $$a\in\mathbb{Z}_{\ge 2}$$ is arbitrary, implies that $$P+1$$ is a perfect square for all $$a\in\mathbb{Z}_{\ge 2}$$. This completes the proof.
Expanding $p(p+1)(p+2)(p+3)+1$ we get
$$p^4+6p^3+11p^2+6p+1$$
Note that $p^4$ is $(p^2)^2$, so this is equal to a square plus something extra. If this is to be a square number, then the extra must be a sum of odd numbers starting with $2p^2+1$, that is, we must have
$$6p^3+11p^2+6p+1=\sum^n_{k=0}(2(p^2+k)+1)$$
for some $n$. That sum can be computed to be
$$(n+1)(2p^2+1)+(n+1)n$$
and now it's quite easy to see that there's only one possible choice for $n$. Indeed, we want this to be a cubic in $p$, so $n$ must be linear in $p$. The coefficient of the linear term must be $3$, so that we get a cubic term of $6p^3$ after multiplying out. And since we want a constant term of $1$, we see that there can be no constant term in $n$. So $n=3p$ is the only possible choice. Plugging this in, we find that it works.
Recursion on the square root, k(n)
k(n) = square root [n(n+1)(n+2)(n+3) + 1]
k(1)^2 = 25
k(2)^2 = 121
k(3)^2 = 361
k(4)^2 = 841
k(5)^2 = 1681
k(1) = 5
k(2) = 11 = 5+ 6 = 5 + 2*(2+1)
k(3) = 19 = 11+ 8 = 11 + 2*(3+1) = 5 + 2 * ((2+1) + (3+1))
k(4) = 29 = 19+10 = 19 + 2*(4+1) = 5 + 2 * ((2+1) + (3+1) + (4+1))
k(5) = 41 = 29+12 = 29 + 2*(5+1) = 5 + 2 * ((2+1) + (3+1) + (4+1) + (5+1))
k(n) = 5 + 2 * ( (n+1)(n+2)/2 - (1+2) )
= (n+1)(n+2) - 1
induction step
k(n+1) = k(n) + 2(n+2)
= (n+1)(n+2) - 1 + 2(n+2)
= (n+2)(n+3) - 1
k(n)^2 - 1 = ((n+1)(n+2))- 1)^2 - 1
= ((n+1)(n+2))^2 - 2(n+1)(n+2)
= ((n+1)(n+2)) * ((n+1)(n+2) - 2)
= ((n+1)(n+2)) * (n^2 + 3n)
= n(n+1)(n+2)(n+3)
!!!
Thinking a little more about k(n).
A rough approximation of the root of ( n(n+1)(n+2)(n+3) + 1 ) would be n^2.
We might improve it a bit by averaging the products of the outer and inner pair of factors,
k(n) = (n(n+3) + (n+1)(n+2)) / 2
= (n+1)(n+2) -1 | 2020-10-20T23:51:08 | {
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https://math.stackexchange.com/questions/2416692/ordered-pairs-of-positive-integers-22x-y2-60/2416700 | # Ordered pairs of positive integers $2^{2x} - y^2 = 60$
How many ordered pairs of positive integers are there that satisfy $2^{2x} - y^2 = 60$?
This can be rewritten as $(2^x)^2-y^2 = 60$ and then $(2^x+y)(2^x-y) = 60$. Then since $2^x$ is always positive and so $2^x+y$ and $2^x-y$ are both positive and the first is always bigger than the latter. This means I only have to account for the positive factors of $60$, and so I had the following pairs $(60,1), (30,2), (20,3), (15,4), (12,5)$, and I plugged them in and solved the systems. Of the 5 possible cases, only $(30,2)$ worked and yielded $x=4, y=14$.
Am I right in having done so? They don't have an answer, and when I tried to graph this, it didn't exactly work out well (I didn't get an integer value for $x$).
• What about $(10, 6)$? :-) – Peter Košinár Sep 4 '17 at 17:57
• Well $2^{2x} = 4^x \Rightarrow 4^x > 60 \Rightarrow x \geqslant 3$. We also have $-y^2 = 60 - 4^x = -(4^x - 60) \Rightarrow y^2 = 4^x - 60 \Rightarrow y^2 - 1 = 4^x - 61 \Rightarrow (y + 1)(y - 1) = 4^x - 61$ which means that $4^x - 61$ cannot be prime, so solutions exist if $x$ is even, and only some exist if $x$ is odd. Now it is a little easier to find solutions – Mr Pie Oct 22 '17 at 23:55
• And from that, we see that $5\mid 4^{2n} - 61$ if $x = 2n$ and $3\mid 4^{2n - 1} - 61$ if $x = 2n - 1$, all for some $n \in \mathbb{N}$. And since $4^x$ is even and $61$ is odd, then $y^2$ is even hence $y$ is also even – Mr Pie Oct 23 '17 at 0:02
You were so close. The last pair is $(10,6)$. That means $2^x=8$ or $x=3$ and $y=2$
No you are not right.
You have neglected the case $\{ 2^x-y, 2^x+y \} = \{ 6, 10 \}$ ; which gives you $(x,y)=(3,2)$.
Notice that:
• $(2^x+y)$ and $(2^x-y)$ have the same pairity;
• product of $(2^x+y)$ and $(2^x-y)$ is even; so at least one of them is even;
• by the above two remarks it follows that both of $(2^x+y)$ and $(2^x-y)$ are even.
• So we can rewrite the equation as $\dfrac{2^x+y}{2} \cdot \dfrac{2^x-y}{2} = 15$ ;
• the only possibilitis for $\{ \dfrac{2^x-y}{2}, \dfrac{2^x+y}{2} \}$ is $\{ 1, 15 \}$ and $\{ 3, 15 \}$ .
• So we can conlude that:
the only possibilitis for $\{ \ 2^x-y, \ 2^x+y \ \}$ is $\{ 2, 30 \}$ and $\{ 6, 10 \}$ .
The first possibility gives $(x,y)=(4,14)$ and the second gives you $(x,y)=(3,2)$ .
• It's given that $x, y$ are positive. – user263326 Sep 4 '17 at 20:36
• @user263326 ; yes you are right. – Davood KHAJEHPOUR Sep 4 '17 at 20:51
• Since $x, y$ are positive, the first possibility is $(x, y) = (4, 14)$ and the second is $(x, y) = (3, 2)$. – N. F. Taussig Sep 5 '17 at 9:24
• @N. F. Taussig ; yes you are right. – Davood KHAJEHPOUR Sep 5 '17 at 9:26
Basic things to note about $(2^x + y)(2^x - y) = ab = 60; a = 2^x + y; b=2^x-y)$.
1) $2^x + y > 0$ so $2^x - y > 0$.
2) $a= 2^x + y > 2^x -y=b > 0$ so $a > \sqrt{60} > 7$ and $b < \sqrt{60} < 8$.
3) $a$ and $b$ can't both be odd so they are both even.
So $b = 2,\color{red}4,6$ and $a=30,\color{red}{15},10$
4) $a+b = 2^{x+1} = 32, 16$ so $x=4,3$ and $y = \frac {a-b}2 = 14, 2$.
So the two possible solutions are $(x,y)=(4,14): (2^4+14)(2^4 -14)=30*2=60$ and $(x,y)=(3,2): (2^3+2)(2^3-2)=10*6=60$.
$$(x,y) \in \lbrace(4,14),(3,2),(4,-14),(3,-2)\rbrace$$
Indeed, it's a difference of two squares:
$$2^{2x}-y^2=60$$ $$\text{let} \ \ \ \ t=2^x \ \ \implies$$ $$t^2-y^2= 60$$
In using the identity: $$\left(\frac{d_1+d_2}{2}\right)^2-\left(\frac{d_1-d_2}{2}\right)^2=d_1\cdot d_2$$
the only catch is that the pairs of divisors you choose need to be of the the same pairity, to allow that their sum or difference is divisible by two. Moving on, we have:
$$t=\frac{d_1+d_2}{2} \qquad \text{and} \qquad y= \frac{d_1-d_2}{2}$$
The total number of divisors of a given natural N will be:
$$\text{if} \ \ \ \ N=\prod_{i=1}^M p_i^{\alpha_i} \ \ \ \ \text{for finitely large M, and the i-th prime p}$$
$$\text{then} \ \ \ \ \left| \lbrace \text{set of divisors of N} \rbrace\right|=\prod_{i=1}^M (\alpha_i +1)$$
So for any number you factor it completely, and multiply the powers increased by one to get the amount of divisors of that number.
We now observe that $$60=2^2 \cdot 3 \cdot 5$$ $$(2+1)(1+1)(1+1)=12$$
So the divisors of 60 must include 12 numbers, starting from 1 it's not hard to find that they are:
$$\lbrace 1,2,3,4,5,6,10,12,15,20,30,60 \rbrace$$ And that the set of divisor-pairs, with each divisor of the same parity is only a two member set, so:
$$(d_1,d_2)\in \lbrace(30,2),(10,6)\rbrace$$
Thus
$$t=2^x=\frac{30+2}{2} \ \ \ \lor \ \ \ \frac{10+6}{2}$$ $$\text{so} \ \ \ \ t=16 \ \ \ \ \lor \ \ \ 8 \ \ \ \implies x=4 \ \ \ \lor \ \ \ 3$$ While $$y=\frac{30-2}{2}\ \ \ \lor \ \ \ \frac{10-6}{2} \implies y=14 \ \ \lor \ \ \ 2$$
Answers are then $$(x,y)=(4,14) \lor (3,2)$$ But we can see that we can let y be positive or negative, and not so for x, giving our above answer of 4 ordered pairs | 2021-05-09T20:29:13 | {
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https://uk.mathworks.com/help/matlab/ref/plot3.html | plot3
3-D point or line plot
Syntax
``plot3(X,Y,Z)``
``plot3(X,Y,Z,LineSpec)``
``plot3(X1,Y1,Z1,...,Xn,Yn,Zn)``
``plot3(X1,Y1,Z1,LineSpec1,...,Xn,Yn,Zn,LineSpecn)``
``plot3(___,Name,Value)``
``plot3(ax,___)``
``p = plot3(___)``
Description
example
````plot3(X,Y,Z)` plots coordinates in 3-D space. To plot a set of coordinates connected by line segments, specify `X`, `Y`, and `Z` as vectors of the same length.To plot multiple sets of coordinates on the same set of axes, specify at least one of `X`, `Y`, or `Z` as a matrix and the others as vectors. ```
example
````plot3(X,Y,Z,LineSpec)` creates the plot using the specified line style, marker, and color.```
example
````plot3(X1,Y1,Z1,...,Xn,Yn,Zn)` plots multiple sets of coordinates on the same set of axes. Use this syntax as an alternative to specifying multiple sets as matrices.```
example
````plot3(X1,Y1,Z1,LineSpec1,...,Xn,Yn,Zn,LineSpecn)` assigns specific line styles, markers, and colors to each `XYZ` triplet. You can specify `LineSpec` for some triplets and omit it for others. For example, `plot3(X1,Y1,Z1,'o',X2,Y2,Z2)` specifies markers for the first triplet but not the for the second triplet.```
example
````plot3(___,Name,Value)` specifies `Line` properties using one or more name-value pair arguments. Specify the properties after all other input arguments. For a list of properties, see Line Properties.```
example
````plot3(ax,___)` displays the plot in the target axes. Specify the axes as the first argument in any of the previous syntaxes.```
example
````p = plot3(___)` returns a `Line` object or an array of `Line` objects. Use `p` to modify properties of the plot after creating it. For a list of properties, see Line Properties.```
Examples
collapse all
Define `t` as a vector of values between 0 and 10$\pi$. Define `st` and `ct` as vectors of sine and cosine values. Then plot `st`, `ct`, and `t`.
```t = 0:pi/50:10*pi; st = sin(t); ct = cos(t); plot3(st,ct,t)```
Create two sets of x-, y-, and z-coordinates.
```t = 0:pi/500:pi; xt1 = sin(t).*cos(10*t); yt1 = sin(t).*sin(10*t); zt1 = cos(t); xt2 = sin(t).*cos(12*t); yt2 = sin(t).*sin(12*t); zt2 = cos(t);```
Call the `plot3` function, and specify consecutive `XYZ` triplets.
`plot3(xt1,yt1,zt1,xt2,yt2,zt2)`
Create matrix `X` containing three rows of x-coordinates. Create matrix `Y` containing three rows of y-coordinates.
```t = 0:pi/500:pi; X(1,:) = sin(t).*cos(10*t); X(2,:) = sin(t).*cos(12*t); X(3,:) = sin(t).*cos(20*t); Y(1,:) = sin(t).*sin(10*t); Y(2,:) = sin(t).*sin(12*t); Y(3,:) = sin(t).*sin(20*t);```
Create matrix `Z` containing the z-coordinates for all three sets.
`Z = cos(t);`
Plot all three sets of coordinates on the same set of axes.
`plot3(X,Y,Z)`
Create vectors `xt`, `yt`, and `zt`.
```t = 0:pi/500:40*pi; xt = (3 + cos(sqrt(32)*t)).*cos(t); yt = sin(sqrt(32) * t); zt = (3 + cos(sqrt(32)*t)).*sin(t);```
Plot the data, and use the `axis equal` command to space the tick units equally along each axis. Then specify the labels for each axis.
```plot3(xt,yt,zt) axis equal xlabel('x(t)') ylabel('y(t)') zlabel('z(t)')```
Create vectors `t`, `xt`, and `yt`, and plot the points in those vectors using circular markers.
```t = 0:pi/20:10*pi; xt = sin(t); yt = cos(t); plot3(xt,yt,t,'o')```
Create vectors `t`, `xt`, and `yt`, and plot the points in those vectors as a blue line with 10-point circular markers. Use a hexadecimal color code to specify a light blue fill color for the markers.
```t = 0:pi/20:10*pi; xt = sin(t); yt = cos(t); plot3(xt,yt,t,'-o','Color','b','MarkerSize',10,'MarkerFaceColor','#D9FFFF')```
Create vector `t`. Then use `t` to calculate two sets of x and y values.
```t = 0:pi/20:10*pi; xt1 = sin(t); yt1 = cos(t); xt2 = sin(2*t); yt2 = cos(2*t);```
Plot the two sets of values. Use the default line for the first set, and specify a dashed line for the second set.
`plot3(xt1,yt1,t,xt2,yt2,t,'--')`
Create vectors `t`, `xt`, and `yt`, and plot the data in those vectors. Return the chart line in the output variable `p`.
```t = linspace(-10,10,1000); xt = exp(-t./10).*sin(5*t); yt = exp(-t./10).*cos(5*t); p = plot3(xt,yt,t);```
Change the line width to `3`.
`p.LineWidth = 3;`
Starting in R2019b, you can display a tiling of plots using the `tiledlayout` and `nexttile` functions. Call the `tiledlayout` function to create a 1-by-2 tiled chart layout. Call the `nexttile` function to create the axes objects `ax1` and `ax2`. Create separate line plots in the axes by specifying the axes object as the first argument to `plot`3.
```tiledlayout(1,2) % Left plot ax1 = nexttile; t = 0:pi/20:10*pi; xt1 = sin(t); yt1 = cos(t); plot3(ax1,xt1,yt1,t) title(ax1,'Helix With 5 Turns') % Right plot ax2 = nexttile; t = 0:pi/20:10*pi; xt2 = sin(2*t); yt2 = cos(2*t); plot3(ax2,xt2,yt2,t) title(ax2,'Helix With 10 Turns')```
Create `x` and `y` as vectors of random values between `0` and `1`. Create `z` as a vector of random duration values.
```x = rand(1,10); y = rand(1,10); z = duration(rand(10,1),randi(60,10,1),randi(60,10,1));```
Plot `x`, `y`, and `z`, and specify the format for the z-axis as minutes and seconds. Then add axis labels, and turn on the grid to make it easier to visualize the points within the plot box.
```plot3(x,y,z,'o','DurationTickFormat','mm:ss') xlabel('X') ylabel('Y') zlabel('Duration') grid on```
Create vectors `xt`, `yt`, and `zt`. Plot the values, specifying a solid line with circular markers using the `LineSpec` argument. Specify the `MarkerIndices` property to place one marker at the 200th data point.
```t = 0:pi/500:pi; xt(1,:) = sin(t).*cos(10*t); yt(1,:) = sin(t).*sin(10*t); zt = cos(t); plot3(xt,yt,zt,'-o','MarkerIndices',200)```
Input Arguments
collapse all
x-coordinates, specified as a scalar, vector, or matrix. The size and shape of `X` depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.
Type of PlotHow to Specify Coordinates
Single point
Specify `X`, `Y`, and `Z` as scalars and include a marker. For example:
`plot3(1,2,3,'o')`
One set of points
Specify `X`, `Y`, and `Z` as any combination of row or column vectors of the same length. For example:
`plot3([1 2 3],[4; 5; 6],[7 8 9])`
Multiple sets of points
(using vectors)
Specify consecutive sets of `X`, `Y`, and `Z` vectors. For example:
`plot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])`
Multiple sets of points
(using matrices)
Specify at least one of `X`, `Y`, or `Z` as a matrix, and the others as vectors. Each of `X`, `Y`, and `Z` must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:
`plot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])`
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `categorical` | `datetime` | `duration`
y-coordinates, specified as a scalar, vector, or matrix. The size and shape of `Y` depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.
Type of PlotHow to Specify Coordinates
Single point
Specify `X`, `Y`, and `Z` as scalars and include a marker. For example:
`plot3(1,2,3,'o')`
One set of points
Specify `X`, `Y`, and `Z` as any combination of row or column vectors of the same length. For example:
`plot3([1 2 3],[4; 5; 6],[7 8 9])`
Multiple sets of points
(using vectors)
Specify consecutive sets of `X`, `Y`, and `Z` vectors. For example:
`plot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])`
Multiple sets of points
(using matrices)
Specify at least one of `X`, `Y`, or `Z` as a matrix, and the others as vectors. Each of `X`, `Y`, and `Z` must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:
`plot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])`
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `categorical` | `datetime` | `duration`
z-coordinates, specified as a scalar, vector, or matrix. The size and shape of `Z` depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.
Type of PlotHow to Specify Coordinates
Single point
Specify `X`, `Y`, and `Z` as scalars and include a marker. For example:
`plot3(1,2,3,'o')`
One set of points
Specify `X`, `Y`, and `Z` as any combination of row or column vectors of the same length. For example:
`plot3([1 2 3],[4; 5; 6],[7 8 9])`
Multiple sets of points
(using vectors)
Specify consecutive sets of `X`, `Y`, and `Z` vectors. For example:
`plot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])`
Multiple sets of points
(using matrices)
Specify at least one of `X`, `Y`, or `Z` as a matrix, and the others as vectors. Each of `X`, `Y`, and `Z` must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:
`plot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])`
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `categorical` | `datetime` | `duration`
Line style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.
Example: `'--or'` is a red dashed line with circle markers
Line StyleDescription
`-`Solid line
`--`Dashed line
`:`Dotted line
`-.`Dash-dot line
MarkerDescription
`'o'`Circle
`'+'`Plus sign
`'*'`Asterisk
`'.'`Point
`'x'`Cross
`'_'`Horizontal line
`'|'`Vertical line
`'s'`Square
`'d'`Diamond
`'^'`Upward-pointing triangle
`'v'`Downward-pointing triangle
`'>'`Right-pointing triangle
`'<'`Left-pointing triangle
`'p'`Pentagram
`'h'`Hexagram
ColorDescription
`y`
yellow
`m`
magenta
`c`
cyan
`r`
red
`g`
green
`b`
blue
`w`
white
`k`
black
Target axes, specified as an `Axes` object. If you do not specify the axes and if the current axes is Cartesian, then `plot3` uses the current axes.
Name-Value Pair Arguments
Specify optional comma-separated pairs of `Name,Value` arguments. `Name` is the argument name and `Value` is the corresponding value. `Name` must appear inside quotes. You can specify several name and value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`.
Example: `plot3([1 2],[3 4],[5 6],'Color','red')` specifies a red line for the plot.
Note
The properties listed here are only a subset. For a complete list, see Line Properties.
Color, specified as an RGB triplet, a hexadecimal color code, a color name, or a short name. The color you specify sets the line color. It also sets the marker edge color when the `MarkerEdgeColor` property is set to `'auto'`.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range `[0,1]`; for example, ```[0.4 0.6 0.7]```.
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (`#`) followed by three or six hexadecimal digits, which can range from `0` to `F`. The values are not case sensitive. Thus, the color codes `'#FF8800'`, `'#ff8800'`, `'#F80'`, and `'#f80'` are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
`'red'``'r'``[1 0 0]``'#FF0000'`
`'green'``'g'``[0 1 0]``'#00FF00'`
`'blue'``'b'``[0 0 1]``'#0000FF'`
`'cyan'` `'c'``[0 1 1]``'#00FFFF'`
`'magenta'``'m'``[1 0 1]``'#FF00FF'`
`'yellow'``'y'``[1 1 0]``'#FFFF00'`
`'black'``'k'``[0 0 0]``'#000000'`
`'white'``'w'``[1 1 1]``'#FFFFFF'`
`'none'`Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.
`[0 0.4470 0.7410]``'#0072BD'`
`[0.8500 0.3250 0.0980]``'#D95319'`
`[0.9290 0.6940 0.1250]``'#EDB120'`
`[0.4940 0.1840 0.5560]``'#7E2F8E'`
`[0.4660 0.6740 0.1880]``'#77AC30'`
`[0.3010 0.7450 0.9330]``'#4DBEEE'`
`[0.6350 0.0780 0.1840]``'#A2142F'`
Line width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.
The line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.
Marker size, specified as a positive value in points, where 1 point = 1/72 of an inch.
Marker outline color, specified as `'auto'`, an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of `'auto'` uses the same color as the `Color` property.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range `[0,1]`; for example, ```[0.4 0.6 0.7]```.
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (`#`) followed by three or six hexadecimal digits, which can range from `0` to `F`. The values are not case sensitive. Thus, the color codes `'#FF8800'`, `'#ff8800'`, `'#F80'`, and `'#f80'` are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
`'red'``'r'``[1 0 0]``'#FF0000'`
`'green'``'g'``[0 1 0]``'#00FF00'`
`'blue'``'b'``[0 0 1]``'#0000FF'`
`'cyan'` `'c'``[0 1 1]``'#00FFFF'`
`'magenta'``'m'``[1 0 1]``'#FF00FF'`
`'yellow'``'y'``[1 1 0]``'#FFFF00'`
`'black'``'k'``[0 0 0]``'#000000'`
`'white'``'w'``[1 1 1]``'#FFFFFF'`
`'none'`Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
`[0 0.4470 0.7410]``'#0072BD'`
`[0.8500 0.3250 0.0980]``'#D95319'`
`[0.9290 0.6940 0.1250]``'#EDB120'`
`[0.4940 0.1840 0.5560]``'#7E2F8E'`
`[0.4660 0.6740 0.1880]``'#77AC30'`
`[0.3010 0.7450 0.9330]``'#4DBEEE'`
`[0.6350 0.0780 0.1840]``'#A2142F'`
Marker fill color, specified as `'auto'`, an RGB triplet, a hexadecimal color code, a color name, or a short name. The `'auto'` option uses the same color as the `Color` property of the parent axes. If you specify `'auto'` and the axes plot box is invisible, the marker fill color is the color of the figure.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range `[0,1]`; for example, ```[0.4 0.6 0.7]```.
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (`#`) followed by three or six hexadecimal digits, which can range from `0` to `F`. The values are not case sensitive. Thus, the color codes `'#FF8800'`, `'#ff8800'`, `'#F80'`, and `'#f80'` are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
`'red'``'r'``[1 0 0]``'#FF0000'`
`'green'``'g'``[0 1 0]``'#00FF00'`
`'blue'``'b'``[0 0 1]``'#0000FF'`
`'cyan'` `'c'``[0 1 1]``'#00FFFF'`
`'magenta'``'m'``[1 0 1]``'#FF00FF'`
`'yellow'``'y'``[1 1 0]``'#FFFF00'`
`'black'``'k'``[0 0 0]``'#000000'`
`'white'``'w'``[1 1 1]``'#FFFFFF'`
`'none'`Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
`[0 0.4470 0.7410]``'#0072BD'`
`[0.8500 0.3250 0.0980]``'#D95319'`
`[0.9290 0.6940 0.1250]``'#EDB120'`
`[0.4940 0.1840 0.5560]``'#7E2F8E'`
`[0.4660 0.6740 0.1880]``'#77AC30'`
`[0.3010 0.7450 0.9330]``'#4DBEEE'`
`[0.6350 0.0780 0.1840]``'#A2142F'`
Tips
• Use `NaN` or `Inf` to create breaks in the lines. For example, this code plots a line with a break between `z=2` and `z=4`.
` plot3([1 2 3 4 5],[1 2 3 4 5],[1 2 NaN 4 5])`
• `plot3` uses colors and line styles based on the `ColorOrder` and `LineStyleOrder` properties of the axes. `plot3` cycles through the colors with the first line style. Then, it cycles through the colors again with each additional line style.
Starting in R2019b, you can change the colors and the line styles after plotting by setting the `ColorOrder` or `LineStyleOrder` properties on the axes. You can also call the `colororder` function to change the color order for all the axes in the figure.
Extended Capabilities
Topics
Introduced before R2006a | 2020-09-20T09:07:27 | {
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https://math.stackexchange.com/questions/58560/elementary-central-binomial-coefficient-estimates | # Elementary central binomial coefficient estimates
1. How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ > 1 ?
2. Does anyone know any better elementary estimates?
• For (2), Stirling's approximation yields ${2n\choose n}\sim (\pi n)^{-1/2} 4^n$, and you might look at some of the other formulas for sharper bounds. – anon Aug 19 '11 at 20:49
• @anon The Central Limit Theorem shows that eventually the Stirling approximation must be an upper bound (and it actually is for all positive $n$). A lower bound is obtained the same way by multiplying the upper bound by $\exp(-1/(4n))$. Because this is $1 + O(1/n)$ for large $n$, it's better than the OP's bounds whose ratio equals $O(\sqrt{4/3})$. – whuber Aug 19 '11 at 20:55
• en.wikipedia.org/wiki/Central_binomial_coefficient – yoyo Aug 19 '11 at 21:06
• @whuber: 1,000,000=O(sqrt(4/3)). en.wikipedia.org/wiki/Big_O_notation#Formal_definition – Did Aug 20 '11 at 12:32
Here are some crude bounds:
$${1\over 2\sqrt{n}}\leq {2n\choose n}{1\over 2^{2n}}\leq{3\over4\sqrt{n+1}},\quad n\geq1.$$
We begin with the product representations $${2n\choose n}{1\over 2^{2n}}={1\over 2n}\prod_{j=1}^{n-1}\left(1+{1\over 2j}\right)=\prod_{j=1}^n\left(1-{1\over2j}\right),\quad n\geq1.$$
From $$\prod_{j=1}^{n-1}\left(1+{1\over 2j}\right)^{\!\!2}=\prod_{j=1}^{n-1}\left(1+{1\over j}+{1\over 4j^2}\right)\geq \prod_{j=1}^{n-1}\left(1+{1\over j}\right)=n,$$ we see that $$\left({2n\choose n}{1\over 2^{2n}} \right)^{2} = {1\over (2n)^2}\, \prod_{j=1}^{n-1}\left(1+{1\over 2j}\right)^{\!\!2} \geq {1\over 4n^2}\, n ={1\over 4n},\quad n\geq1.$$ so by taking square roots, ${2n\choose n}{1\over 2^{2n}}\geq \displaystyle{1\over 2\sqrt{n}}.$
On the other hand, $$\prod_{j=1}^{n}\left(1+{1\over 2j}\right) \left(1-{1\over 2j}\right) = \prod_{j=1}^{n}\left(1-{1\over 4j^2}\right)\leq {3\over 4},$$ so that (using the lower bound above), we have $${2n\choose n}{1\over 2^{2n}}=\prod_{j=1}^n\left(1-{1\over2j}\right)\leq{3\over4\sqrt{n+1}}.$$
Alternatively, multiplying the different representations we get $$n\left[{2n\choose n}{1\over 2^{2n}}\right]^2={1\over 2}\prod_{j=1}^{n-1}\left(1-{1\over4j^2}\right) \,\left(1-{1\over 2n}\right).$$ It's not hard to show that the right hand side increases from $1/4$ to $1/\pi$ for $n\geq 1$.
Edit: You can get better bounds if you know Wallis's formula:
$$2n\left[{2n\choose n}{1\over 4^n}\right]^2={1\over 2}{3\over 2}{3\over 4}{5\over 4}\cdots {2n-1\over 2n-2}{2n-1\over 2n}={1\over 2}\prod_{j=2}^n\left(1+{1\over 4j(j-1)}\right)$$
$$(2n+1)\left[{2n\choose n}{1\over 4^n}\right]^2={1\over 2}{3\over 2}{3\over 4}{5\over 4}\cdots {2n-1\over 2n-2}{2n-1\over 2n}{2n+1\over 2n}=\prod_{j=1}^n\left(1-{1\over 4j^2}\right)$$
By Wallis's formula, both middle expressions converge to ${2\over \pi}$. The right hand side of the first equation is increasing, while the right hand side of the second equation is decreasing. We conclude that
$${1\over\sqrt{\pi(n+1/2)}}\leq {2n\choose n}{1\over 4^n}\leq {1\over\sqrt{\pi n}}.$$
For $$n\ge0$$, we have (by cross-multiplication) \begin{align} \left(\frac{n+\frac12}{n+1}\right)^2 &=\frac{n^2+n+\frac14}{n^2+2n+1}\\ &\le\frac{n+\frac13}{n+\frac43}\tag{1} \end{align} Therefore, \begin{align} \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} &=4\frac{n+\frac12}{n+1}\\ &\le4\sqrt{\frac{n+\frac13}{n+\frac43}}\tag{2} \end{align} Inequality $$(2)$$ implies that $$\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac13}}{4^n}\text{ is decreasing}}}\tag{3}$$ For $$n\ge0$$, we have (by cross-multiplication) \begin{align} \left(\frac{n+\frac12}{n+1}\right)^2 &=\frac{n^2+n+\frac14}{n^2+2n+1}\\ &\ge\frac{n+\frac14}{n+\frac54}\tag{4} \end{align} Therefore, \begin{align} \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} &=4\frac{n+\frac12}{n+1}\\ &\ge4\sqrt{\frac{n+\frac14}{n+\frac54}}\tag{5} \end{align} Inequality $$(5)$$ implies that $$\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac14}}{4^n}\text{ is increasing}}}\tag{6}$$ Note that the formula in $$(3)$$, which is decreasing, is bigger than the formula in $$(6)$$, which is increasing. Their ratio tends to $$1$$; therefore, they tend to a common limit, $$L$$.
Theorem $$1$$ from this answer says $$\lim_{n\to\infty}\frac{\sqrt{\pi n}}{4^n}\binom{2n}{n}=1\tag{7}$$ which means that \begin{align} L &=\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n}\\ &=\frac1{\sqrt\pi}\tag8 \end{align} Combining $$(3)$$, $$(6)$$, and $$(8)$$, we get $$\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}}}\tag9$$
• @ByronSchmuland: Thanks! I came up with the idea of the estimates $(3)$ and $(6)$ while working on this answer, then expanded on it when I found this question. – robjohn Sep 16 '14 at 15:05
• For finite $n$, you can use $0<O(\frac 1 {n^2})<\frac 1{72n}$ in $(11)$ (per Wikipedia). – A.S. Feb 23 '16 at 17:36
• @robjohn: +1 as far as I know, (10) gives the sharpest upper and lower bounds. Is there a reference so we can cite? Or you may write a note and put it in arXiv.org so we can site. – Shiyu Mar 1 '16 at 19:38
• @robjohn, I'd appreciate a reference to cite too. +1 – MikeY Dec 6 '17 at 14:40
• @MikeY: unfortunately, this was all done for this question. I don't have any other reference other than this answer. You can get a citation from the "cite" option in the same area as the "share" and "flag" options, below the answer. – robjohn Dec 6 '17 at 14:57
You can get an even more precise answer than those already provided by using more terms in the Stirling series. Doing so yields, to a relative error of $O(n^{-5})$,
$$\binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} \left(1 - \frac{1}{8n} + \frac{1}{128n^2} + \frac{5}{1024n^3} - \frac{21}{32768 n^4} + O(n^{-5})\right).$$
To the same relative error of $O(n^{-5})$, the Stirling series is $$n!=\sqrt{2\pi n}\left({n\over e}\right)^n \left( 1 +{1\over12n} +{1\over288n^2} -{139\over51840n^3} -{571\over2488320n^4} + O(n^{-5}) \right).$$
Then we have $$\binom{2n}{n} = \frac{(2n)!}{n! n!} = \frac{4^n}{\sqrt{\pi n}} \frac{1 + \frac{1}{12(2n)} + \frac{1}{288(2n)^2} - \frac{139}{51840(2n)^3} - \frac{571}{2488320(2n)^4} + O(n^{-5})}{\left(1 + \frac{1}{12n} + \frac{1}{288n^2} - \frac{139}{51840n^3} - \frac{571}{2488320n^4} + O(n^{-5})\right)^2}.$$
Crunching through the long division with the polynomial in $\frac{1}{n}$ (which Mathematica can do immediately with the command Series[expression, {n, ∞, 4}]) yields the expression for $\binom{2n}{n}$ at the top of the post.
• $\displaystyle\frac{4^n}{\sqrt{\pi n}}\left(1-\frac1{8n}+O\left(\frac1{n^2}\right)\right)$ matches the upper bound gotten in my answer. I remember computing this asymptotic series at one point. (+1) – robjohn Sep 16 '14 at 10:38
A way to get explicit bounds via Stirling's approximation is to use the following more precise formulation: $$n! = \sqrt{2\pi n} \left( \frac{n}{e} \right)^n e^{\alpha_n}$$ where $\frac{1}{12n+1} < \alpha_n < \frac{1}{12n}$.
With this one arrives at $$\binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} e^{\lambda_n}$$ where $\frac{1}{24n+1} - \frac{1}{6n} < \lambda_n < \frac{1}{24n} - \frac{2}{12n+1}$.
• @ThomasAndrews Ahh yes thank you for pointing that out. – Ragib Zaman Mar 14 '13 at 0:21
$$\binom {2n} n = \frac {2^{n-1}} n \sum _{k=0} ^{2n} \left( 1 + \cos \frac {k \pi} n \right) ^n$$
$\binom{2p}{p}$ central binomial coefficient
$$\binom{2p}{p}=\frac{\displaystyle2^{2p}}{\displaystyle1+\sum_{n=1}^p\frac{2^n}S}$$
with $S=\sum\limits_{k=1}^{2n}\left(1+\cos\left(\frac{k\pi}n\right)\right)^n$
this new formula as the first uses a trigonometric form (see mar20 at19:11) it's more complex but gives the same result
with my pocket computer casio pb 700 I have found $\binom{2p}{p}$ exact untill
$p=18$ it's $9075135300$ after the result is given $A\cdot10^x$
example $\binom{40}{20}=1.378465288\times10^{11}$
• I have applied MathJax. Please make sure everything is as intended. MathJax makes things easier to read. – robjohn Mar 31 '16 at 9:38 | 2019-09-22T01:47:22 | {
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http://mathhelpforum.com/algebra/5990-rational-number.html | # Math Help - Rational Number
1. ## Rational Number
I have the following problem:
Prove that the sqrt(2+sqrt(2)) is irrational.
I did the problem similarly to how one of my TAs showed in a tutorial.
Let the sqrt(2+sqrt(2)) = a. Square both sides. This is:
2+sqrt(2)=a
sqrt(2)=(a^2)-2 Square both sides
2=(a^4)-(4a^2)+4
0=(a^4)-(4a^2)+2
Because the factors of 2 (namely 1, -1, 2, -2) do not solve the following equation, we can assume that the original function is irrational.
My question is, what theorem, reasoning, or whatever allows for this to be true and do you think this is adequate for a proof?
2. Originally Posted by JimmyT
My question is, what theorem, reasoning, or whatever allows for this to be true and do you think this is adequate for a proof?
It uses the rational root test.
The reason of the proof is by contradiction.
3. Hello, JimmyT!
I would try to prove it by contradiction . . .
Prove that the sqrt(2+sqrt(2)) is irrational.
Assume that expression is rational.
. . . . . . .______. . . a
That is, √2 + √2 .= .--- . for some integers a and b.
. . . . . . . . . . . . . . .b
. . . . . . . . . . . . . . . . . . _ . . . .
Square both sides: . 2 + √2 . = . ----
. . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . ._ . . . .
And we have: . √2 . = . --- - 2 . [1]
. . . . . . . . . . . . . . . . .
. . . . a . . . . . . . . . . . .
Since -- is rational, then --- is rational. .(Integers are closed under mutliplication)
. . . . b . . . . . . . . . . . .
. . . . . . . .
. . Hence, --- - 2 is rational. .(Rational numbers are closed under subtraction.)
. . . . . . . .
Hence, the right side of equation [1] is a rational number.
. . But the left side is irrational.
We have reached a contradition; our assumption is incorrect.
. . . . . . . . . ______
Therefore: . √2 + √2 .is irrational.
4. Originally Posted by Soroban
Integers are closed under mutliplication
You mean rationals.
5. Hello, TPHacker!
You mean rationals.
Well, I was thinking that a² is an integer and b² is an integer,
. . which makes a²/b² a rational number.
But you're right ... The closure of rationals is a much better reason. | 2014-04-21T13:32:16 | {
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http://math.stackexchange.com/questions/596771/finding-the-limit-of-left-fracnn1-rightn | # Finding the limit of $\left(\frac{n}{n+1}\right)^n$
Find the limit of: $$\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$$
I'm pretty sure it goes to zero since $(n+1)^n > n^n$ but when I input large numbers it goes to $0.36$.
Also, when factoring: $$n^{1/n}\left(\frac{1}{1+\frac1n}\right)^n$$ it looks like it goes to $1$.
So how can I find this limit?
-
I have seen $4$ answers in less than $4$ minutes.. but i guess the question is "How"?? He might be expecting "How" for $\lim_{n\rightarrow \infty} (1+\frac{1}{n})^n=e$.... – Praphulla Koushik Dec 7 '13 at 12:33
No I know this identity, I just didn't see the 1/e. – GinKin Dec 7 '13 at 12:35
Then there is no problem i guess... – Praphulla Koushik Dec 7 '13 at 12:35
This limit is somewhat similar: math.stackexchange.com/questions/568268/… See also here: math.stackexchange.com/questions/358830/… – Martin Sleziak Dec 7 '13 at 12:37
$$\lim_{n\to \infty}\left(\frac{n}{n+1}\right)^n = \lim_{n\to \infty}\frac {1}{\left(\frac{n+1}{n}\right)^n}=\lim_{n\to \infty}\frac {1}{\left(\frac nn+\frac{1}{n}\right)^n}=\lim_{n\to \infty}\frac {1}{\left(1+\frac{1}{n}\right)^n}$$ We know that: $$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e$$ $$\color{red}{\boxed{\displaystyle\color{black}{\quad\therefore\quad\lim_{n\to \infty}\left(\frac{n}{n+1}\right)^n=\frac 1e\quad}}}$$
And for the intuitive part: $$\begin{array}{|c|c|} \hline \text{Value of }\color{black}{n} & \left(\frac{\color{black}{n}}{\color{black}{n}+1}\right)^{\color{black}{n}} \\ \hline 1 & 1.\overline6 \\ 2 & 0.562 \\ 3 & 0.512 \\ 4 & 0.482 \\ 5 & 0.462 \\ 6 & 0.448 \\ 7 & 0.438 \\ 8 & 0.430 \\ 9 & 0.424 \\ 100 & 0.373 \\ \hline \end{array}$$ And: $$\frac 1e\approx0.3678794...$$ Sure enough, we are approaching $1/e$ as $n$ tends to $\infty$.
Furthermore, you could think of it that way:
As $n$ gets really really large, $1/n$ will be very very small.
As $n$ gets really really large, the coefficients and number of terms of $(a+b)^n$ will also get really large. That's why even if $1/n\approx0$ when $n$ gets big, the expression $\left(1+\frac{1}{n}\right)^n$ will approach a number greater than one, and who has to do with the amount of terms and coefficients present in $(a+b)^n$, and that number is simply $e$. To show more what I mean, look at this example: \begin{align}(1+\color{blue}{0.1})^{10}=&1^{10}+\color{green}{10} \cdot1^9 (\color{blue}{0.1})+\color{green}{45}\cdot1^8 (\color{blue}{0.1})^2+\color{green}{120} \cdot1^7 (\color{blue}{0.1})^3+\color{green}{210} \cdot1^6 (\color{blue}{0.1})^4\\&+\color{green}{252}\cdot 1^5 (\color{blue}{0.1})^5+\color{green}{210}\cdot 1^4 (\color{blue}{0.1})^6+\color{green}{120}\cdot 1^3\cdot (\color{blue}{0.1})^7+\color{green}{45}\cdot 1^2 (\color{blue}{0.1})^8\\&+\color{green}{10}\cdot 1 (\color{blue}{0.1})^9+(\color{blue}{0.1})^{10}\\\,\\\approx&2.5937424601\end{align}
Note: The digit $2$ present in the mathematical constant $e$ comes from:
• We have: $\left(1+\frac{1}{n}\right)^n\,\,\text{(1)}$ and $1^n=1\quad\text{where$n\in\Bbb R$}$. That's why there will always be a $1$ left no matter how large $n$ is when we expand the expression $\text{(1)}$.
• From the binomial theorem, the second coefficient in the expanded form of $(a+b)^n$ is just $n$. And therefore the second term in the expansion of $\left(1+\frac{1}{n}\right)^n$ will be: $n\times1^{n-1}\times\frac1n$ which is equal to $1$.
But in the limit the OP asks about, we are dividing $1$ by $\left(1+\frac{1}{n}\right)^n$, so it is natural that we get at the end $1/e$.
-
+0.75 for the awesome answer. +0.25 for the colorful post! – Ahaan Rungta Dec 7 '13 at 23:01
$$\left(\dfrac{n}{n+1}\right)^n = \frac {1}{\left(\dfrac{n+1}{n}\right)^n}=\frac {1}{\left(1+\dfrac{1}{n}\right)^n}$$
-
The most elegant! – Mitsos Dec 7 '13 at 12:34
Thanks! And also to "Martin Sleziak" for editing. – randuser Dec 7 '13 at 12:40
And also to "Adobe" for re-editing! And to myself for re-editing! – Ahaan Rungta Dec 7 '13 at 23:01
Yes, of course! – randuser Dec 8 '13 at 5:57
Did you know that $\lim_{n\rightarrow \infty} (1+1/n)^n =e$? This you can use to find the limit.
-
I came at this problem a little differently than the other answers. We want to find $$L = \lim\limits_{n \to \infty} \left(\frac{n}{n+1}\right)^n.$$ Since the expression inside the limit is always positive, we should be able to take the log of the limit, then solve it (we just have to remember to transform back at the end). \begin{align} \log(L) = & \lim\limits_{n \to \infty} n \log \left( \frac{n}{n+1} \right) \\ = & \lim\limits_{n \to \infty} n \log \left( 1 - \frac{1}{n+1} \right). \end{align} Since we're taking the limit of $\log(1 + x)$ as $x \to 0$, what we really care about is the first order behavior of $\log(1 + x)$, so we take the Taylor series: $$\log(1 + x) = x + O(x^2),$$ where $O(x^2)$ indicates higher-order terms (second-order and greater). So $\log(1 + x) \approx x$ for small $x$ ($x \approx 0$). Substituting back into our limit: \begin{align} \log(L) = & \lim\limits_{n \to \infty} n \left( -\frac{1}{n+1} \right) \\ = & -1. \end{align} Therefore $L = e^{-1} = \frac{1}{e}$.
-
Hint:
$$\left(\frac{n}{n+1}\right)^n=\left(\frac{n+1-1}{n+1}\right)^n=\left(1-\frac{1}{n+1}\right)^n$$
-
Thank you, Martin! – Salech Alhasov Dec 7 '13 at 12:39
${\lim_{n\to \infty}{(\frac{n}{n+1})}^{n}}={\lim_{n\to \infty}{({1-{\frac{1}{n+1}}})}^n}=e^{\lim_{n\to \infty} \frac{-1}{n+1}*n}=e^{-1}$ where 3$^{rd}$ step is basic process for limit of form $1^{\infty}$ .So limit is $e^{-1}$
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My hint: $$\left(\frac{n}{n+1}\right)^n=\frac{1}{\left(1+\frac{1}{n}\right)^n}\sim \frac{1}{e}$$
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Just write this as $(\frac{n+1}{n})^{-n}$ and then it is a standard limit ie answer will be $1/e$
-
$$\color{#0000ff}{\large\lim_{n \to \infty}\pars{n \over 1 + n}^{n}} = \color{#0000ff}{\large{1 \over e}}$$
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Already posted, in a much more convincing version, by at least one other user. – Did Dec 8 '13 at 8:02
## protected by Community♦Dec 7 '13 at 23:25
Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site. | 2014-09-21T12:15:41 | {
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https://math.stackexchange.com/questions/1774710/find-an-orthonormal-basis-for-mathbbr4 | # Find an orthonormal basis for $\mathbb{R^4}$
I have been trying to solve this problem (Fraleigh - Linear Algebra (3rd Ed.)):
6.2.17. Find an orthonormal basis for $\mathbb{R^4}$ that contains an orthonormal basis for the subspace $sp([1, 0, 1, 0], [0, 1, 1, 0])$.
The first thing I thought was the usage of the standard linearly independent vectors for $\mathbb{R^4}$, but that did not work.
Ideas, suggestions (none solution) to solve this problem?
• Use the Gram-Schmidt process. May 6, 2016 at 21:45
• You want to read about the Gram Schmidt process. You can read on Wikipedia, or search math.se. May 6, 2016 at 21:45
• Answered down below, showcasing the Gram Schmidt process to you. Make sure you ask me if you have any questions and if the answer fits you, then make sure you approve it so that the thread goes down as asnwered. May 6, 2016 at 21:48
Let's say that : $x_1 = [1,0,1,0]$ and $x_2 = [0,1,1,0]$
Let $v_1 = x_1$.
$y = proj_{v_1}x_2 = \frac{x_2v_1}{v_1v_1}v_1$
and
$v_2 = x_2 - y = x_2 - \frac{x_2v_1}{v_1v_1}v_1$
This is also called as "The Gram-Schmidt" process. Plug in the correct vectors and you will have your component which will be orthogonal to $x_1$ and then you'll be able to form your basis.
• I have found the first two vectors for the orthonormal basis: $\{\frac{1}{\sqrt{2}}[1, 0, 1, 0], \frac{1}{\sqrt{6}}[-1, 2, 1, 0]\}$, but the answer for this exercise includes the previous vectors and two more. Why are they needed? Thanks May 7, 2016 at 19:38
• I have solved it. What I did to solve it was the addition of two other linearly independent vectors $\{[0,0,0,1], [0, 1, 0, 0]\}$, and then apply the Gram-Schmidt algorithm to these ones. May 7, 2016 at 20:20 | 2023-03-20T12:11:05 | {
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https://math.stackexchange.com/questions/2470863/is-concept-of-bipartite-graph-relevant-only-for-connected-graph | # Is concept of Bipartite Graph relevant only for Connected Graph?
Is concept of Bipartite Graph relevant only for Connected Graph? For Example : 4 vertices and just 2 Edges i.e. one from node 1 to node 2, and the another from node 2 to node 3. So, is this Disconnected Graph also a Bipartite Graph? Or Bipartite graph deals with Connected Graphs only?
A bipartite graph only requires that the vertex set can be partitioned into two disjoint sets, say $A$ and $B$, such that every edge in $E$ connects a vertex from one of the sets, to the other. You can have isolated points. If one denotes the bipartite graph by $(A,B,E)$, then formally you will have a different bipartition depending on if you place an isolated vertex in $A$ or in $B$.
• @RudrakshaBaplawat Yes. Let node $1$ and $3$ be in vertex set $A$, and node $2$ in $B$, then place node $4$ in whichever you wish. – Alex Clark Oct 13 '17 at 17:50 | 2019-08-23T23:55:26 | {
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https://math.stackexchange.com/questions/3626434/irreducible-representation-of-finite-abelian-group | # Irreducible representation of finite Abelian group
I have seen that every finite Abelian group $$G$$ is isomorphic to a product of cyclic groups of prime power order, that is
$$G = \mathbb{Z}_{p_1} \times \mathbb{Z}_{p_2} \times ... \times \mathbb{Z}_{p_n} ,$$
where $$\mathbb{Z}_{p_i}$$ is the group $$\{ 0, 1, ..., p_i-1 \}$$ with addition modulo $$p_i$$. So if I understand it correctly, you can "represent" (it is not a representation, but a way to think about) the elements of $$G$$ as arrays
$$(a_1, a_2, ..., a_n)$$
with $$a_i \in \mathbb{Z}_{p_i}$$, and the group operation corresponds to vector addition. This "representation" establishes a group isomorphism.
My question: how do you match this "representation" of $$G$$ with a true irreducible representation of $$G$$, which, since $$G$$ is Abelian, must be a one-dimensional representation ? In other words, if you represent each element $$g\in G$$ as a phase $$e^{i \theta}$$, how do you write $$\theta$$ in terms of the array $$(a_1, a_2, ..., a_n)$$?
NOTE that I am only interested in representations $$\rho: G \rightarrow GL(V)$$ with $$V$$ a vector space over $$\mathbb{C}$$ or $$\mathbb{R}$$. In particular, the above one-dimensional representation is over a complex vector field.
EXAMPLE: Consider the group $$\mathbb{Z}_2 \times \mathbb{Z}_2$$ (in the above decomposition of $$G$$ repetition of primes is allowed). This group has four elements of order 2 under addition:
$$\mathbb{Z}_2 \times \mathbb{Z}_2 = \{ (0,0), (0,1), (1,0), (1,1) \}.$$
On the other hand we know $$\mathbb{Z}_2 \times \mathbb{Z}_2$$ is Abelian, so we should be able to represent $$g_i = e^{i \theta_i}$$. But each element has order two, so the only option is that each $$g$$ be represented by either 1 or $$e^{i \pi}$$, and this is not a faithful representation. What is the faithful irreducible representation of $$\mathbb{Z}_2 \times \mathbb{Z}_2$$?
• I think $\theta=\theta_1+\cdots+\theta_n$ where $\theta_i=a_i\cdot \frac{2\pi}{p_i}$, right? – freakish Apr 15 at 10:15
• I think that doesn't always give you a faithful representation. – MBolin Apr 15 at 10:16
• @freakish I have added an example explaining why I think that doesn't always give you a faithful representation. – MBolin Apr 15 at 10:21
• Not sure where "faithful" came from? Irreducible representations need not be faithful. In fact a finite abelian group has a faithful irreducible representation if and only if it is cyclic: mathoverflow.net/questions/57129/… – freakish Apr 15 at 10:22
• @freakish OK I think I lacked that bit of information – MBolin Apr 15 at 10:23
Irreducible representations need not be faithful. In fact a finite abelian group has a faithful irreducible representation if and only if it is cyclic. And so $$\mathbb{Z}_2\times\mathbb{Z}_2$$ does not have a faithful irreducible representation.
Now let $$\rho:G\to GL(V)$$ be a representation with $$\dim V=1$$ over $$\mathbb{C}$$. So $$GL(V)\simeq\mathbb{C}\backslash\{0\}$$ with the standard multiplication. Now as you've said every element of $$G$$ can be written as $$(a_1,\ldots,a_n)\in\mathbb{Z}_{p_1}\times\cdots\times\mathbb{Z}_{p_n}$$. Put $$e_i=(0,\ldots,0,1,0,\ldots,0)$$ where $$1$$ is on the $$i$$-th position. With this $$\rho$$ is uniquely determined by values on $$\{e_i\}_{i=1}^n$$. And any such value will generate appropriate representation if $$\rho(e_i)^{p_i}=1$$. And so all we need to know is $$\rho(e_i)$$.
Since $$\rho(e_i)$$ has to be an element of $$\mathbb{C}\backslash\{0\}$$ of order dividing $$p_i$$ (which let me remind is a prime power, not a prime number) then the only choice is $$\rho(e_i)=exp(i \frac{2\pi}{p_i}k_i)$$ for some integer $$k_i$$. And therefore
$$\rho(a_1,\ldots, a_n)=exp\big(i(\frac{2\pi}{p_1}k_1a_1+\cdots+\frac{2\pi}{p_n}k_na_n) \big)$$
Meaning your $$\theta$$ can be written as
$$\theta=\frac{2\pi}{p_1}k_1a_1+\cdots+\frac{2\pi}{p_n}k_na_n$$
for some fixed integers $$k_1,\ldots,k_n\in\mathbb{Z}$$. And any such integers will generate a valid representation. | 2020-10-28T03:23:00 | {
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https://stacks.math.columbia.edu/tag/00NY | # The Stacks Project
## Tag 00NY
Remark 10.77.3. It is not true that a finite $R$-module which is $R$-flat is automatically projective. A counter example is where $R = \mathcal{C}^\infty(\mathbf{R})$ is the ring of infinitely differentiable functions on $\mathbf{R}$, and $M = R_{\mathfrak m} = R/I$ where $\mathfrak m = \{f \in R \mid f(0) = 0\}$ and $I = \{f \in R \mid \exists \epsilon, \epsilon > 0 : f(x) = 0 \forall x, |x| < \epsilon\}$.
The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 18284–18294 (see updates for more information).
\begin{remark}
\label{remark-warning}
It is not true that a finite $R$-module which is
$R$-flat is automatically projective. A counter
example is where $R = \mathcal{C}^\infty(\mathbf{R})$
is the ring of infinitely differentiable functions on
$\mathbf{R}$, and $M = R_{\mathfrak m} = R/I$ where
$\mathfrak m = \{f \in R \mid f(0) = 0\}$ and
$I = \{f \in R \mid \exists \epsilon, \epsilon > 0 : f(x) = 0\ \forall x, |x| < \epsilon\}$.
\end{remark}
There are no comments yet for this tag.
There are also 2 comments on Section 10.77: Commutative Algebra.
## Add a comment on tag 00NY
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner). | 2018-02-25T00:03:12 | {
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http://mathoverflow.net/questions/93178/jacobsons-theorem-and-further/93180 | # Jacobson's theorem and further
Jacobson's theorem states that If $R$ is a ring, and for every $x\in R$, there exists $n(x)\geq 2$ such that $x^{n(x)}=x$. Then $R$ is commutative.
I wonder if the following stronger assertion(in case $R$ has unity) is true.
Let $R$ be a ring with unity. For every $x$ in $R$, there exists $n(x)\geq 2$ such that $x^{n(x)} = x$. Then, $R$ is embedded in a product(possibly infinite) of fields $F_i$, where each $F_i$ is an algebraic extension of $F_{p_i}$ (prime field of $p_i$ elements).
If this is not true, then I am also interested in counterexample.
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Also note that $R$ is a comm. von Neumann regular ring (i.e. for each $r$ there is $s$ s.t. $r = r^2s$) and such rings are known to embedd into a product of fields. – Ralph Apr 5 '12 at 3:28
Yes, this is true.
By Jacobson's theorem, $R$ is commutative. Now the radical of $R$ is the intersection of all primes $P$ of $R$. Hence we have an embedding $$\phi: R/rad(R) \to \prod_P R/P.$$ For each $x \in R/P, x \neq 0$ there is $n \ge 1$ such that $x(x^n-1)=0$ and since $R/P$ is a domain, $x^n =1$, i.e. $x$ is a unit. Thus $R/P$ is a field. Since elements $\neq \pm 1$ of $\mathbb Q$ aren't roots of unity, $R/P$ has prime characteristic and each element of $R/P$ is algebraic. Therefore $R/P$ is an algebraic extension of some $\mathbb{F}_p$.
To finish the proof, we have to show $rad(R) = 0$. Let $x \in rad(R)$. Let $k>0$ be minimal with $x^k=0$ and let $n \ge 2$ with $x^n=x$. Suppose $k > n$. Write $k=qn+r$ with $q > 0$, $0 \le r < n$. Then $0 = x^k = (x^n)^q x^r = x^{q+r}$. Minimality of $k$ implies $k = q+r$, i.e. $n=1$, in contradiction to the assertion $n \ge 2$. Hence $k\le n$ and $x = x^n = x^k x^{n-k} = 0$.
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Thanks, I got the idea. How about starting from the Jacobson radical of $R$? I think it is easier. Also, showing the Jacobson radical is zero, is easy. If $x\in J(R)$, then $x^n-1$ is unit for any $n\geq 1$. So, $x=0$. The Jacobson radical is intersection of maximal ideals, so we get $R/M$ is field for each maximal ideals $M$. – i707107 Apr 5 '12 at 2:20
Looks good. In particular, you get $R/M$ a field for free. In effect, since $R/P$ is a field, each $P$ is maximal and $J(R) = rad(R)$. But using $J(R)$ in the proof is certainly better. – Ralph Apr 5 '12 at 2:29
This is true. Let $R$ be a ring satisfying your property. Then $R$ has no nilpotent element since if $x^a=0$, then for $b$ an integer such that $n(x)^b > a$ we have $x=x^{n(x)^b}=0$. So the radical of $R$ is $0$, and since the radical is the intersection of all prime ideals of $R$, we see that the natural map $R \mapsto \prod_{P} R/P$ is an injection. It therefore suffices to prove the result for $R/P$, that is for a domain.
Assuming no that $R$ is a domain, the equation $x^{n(x)}-x=0$ factors as $x=0$ or $x^{n(x)-1}=0$. So, in $R$, every non-zero element is a root of unity. This is also true of the fraction field $K$ of $R$. No $K$ is of characteristic $p>0$, since otherwise it would contain $\mathbb{Q}$ which contains many non-roots-of-unity such as $2$. And clearly, $K$ is algebraic over the prime field. So we're done.
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It is not necessary to consider a fraction field $K$ of $R$ because you already know that "each non-zero element [of $R$] is a root of unity". Hence $R$ ist alreday a field. – Ralph Apr 5 '12 at 2:13
Ralph, you beat me by 6 seconds ! Mathoverflow is really becoming too competitive :-) – Joël Apr 5 '12 at 2:23
Thank you for you too Joel. Both answers are basically the same. So, I accepted the first one. – i707107 Apr 5 '12 at 2:26
Yes, your answer showed up directly after mine. Even more amazing is that we not just had the same idea of proof but used nearly the same words: "this is true", "is the intersection of all primes". In any case I wished it were possible in MO to accept more than one answer. – Ralph Apr 5 '12 at 2:33 | 2015-09-02T22:22:30 | {
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https://math.stackexchange.com/questions/2973786/equilibrium-solutions-of-a-nonlinear-mass-spring-system | # Equilibrium solutions of a nonlinear mass spring system.
Consider the mass-spring system governed by the differential equation, $$m\ddot{x}=-F(x)$$
Where $$x(t)$$ is the time-dependent position displacement of the mass. The spring force is a nonlinear function of $$x$$ and given by, $$F(x)=\alpha x-\beta x^3= \frac{dV}{dx}$$ Where $$\alpha$$ and $$\beta$$ are positive constants.
Sketch the function $$V(x)$$ and show from this plot that the first equation has three stationary points. Can you judge from the plot of $$V(x)$$ which of these stationary points are stable and which are unstable? Assume that $$V(0)=0$$
Now by integrating $$\frac{dV}{dx}$$, I got $$V(x)=\frac{1}{2} \alpha x^2-\frac{1}{4}\beta x^4$$
We see that $$V(x)$$ is the potential energy of the mass as a function of displacement. Plotting this gave me something like this
Where stationary points are at $$x=0, x=\pm \sqrt{\frac{2\alpha}{\beta}}$$. I got stuck here and according to the provided solution, the local minimality of $$V(x)$$ at $$0$$ implies stability and the maximality of $$V(x)$$ at $$x=\pm \sqrt{\frac{2\alpha}{\beta}}$$ implies instability. There is no further explanation and I am extremely confused. I am also not sure how showing $$V(x)$$ is stationary at the given points shows that $$x(t)$$ is also stationary.
I understand that if $$\frac{dV}{dx}=0$$ then $$\ddot{x}=0$$. But my understanding was that $$\dot{x}$$ also needs to be $$0$$ for $$x(t)$$ to be stationary. In fact for $$V(x)$$ where $$x=0$$ which is the local minima, $$\dot{x}$$ cannot be $$0$$ since If the potential energy is at its minimum, the kinetic energy has to be at its maximum value (conservation of energy) which means that $$\dot{x}$$ has to be the maximum value. So $$\frac{dx}{dt}\neq 0$$ which contradicts with my understanding that $$\frac{dx}{dt}$$ must equal $$0$$ at stationary points.
I would appreciate any help clarifying my confusion.
Edit: Three dimensional phase portrait with $$x$$, $$\dot{x}$$ and $$\ddot{x}$$ as axis.
• Think of the graph as a height profile, a landscape, and let a ball roll from any position in question. It stays at the minimum and any slight perturbation at a maximum will make it roll off. – LutzL Oct 27 '18 at 19:49
Taking
$$m\dot x\ddot x + F(x)\dot x = 0\to \frac 12\frac{d}{dt}(m\dot x)^2+\frac{d}{dt}\left(\frac 12 \alpha x^2-\frac 14 \beta x^4\right) = 0$$
with the integral
$$m\dot x^2+\alpha x^2-\frac 12 \beta x^4 = C_0$$
Equilibrium is attained when $$\dot x = 0$$ or
$$\alpha x^2 - \frac 12 \beta x^4= C_0$$
or at
$$x^* = \pm \sqrt{\frac{\alpha \pm\sqrt{\alpha ^2-2 \beta C_0}}{\beta }}$$
The equilibrium qualification can be done depending on $$\alpha^2-2\beta C_0$$ and also on characterizing those feasible points as sources saddle points or sinks.
NOTE
If $$C_0 = 0$$ then
$$x^* = \left\{-\sqrt{\frac{2\alpha}{\beta}},0,\sqrt{\frac{2\alpha}{\beta}}\right\}$$
Attached a plot for $$m = 1,\alpha = 1, \beta = 1$$ showing the phase plane movement partial orbits, with $$C_0 = 1$$ (red), $$C_0 = 0.5$$ (blue) and $$C_0 = 0.2$$ (green)
Including viscous dissipation as $$-k \dot x$$ with $$k = \frac 18$$ we have the path from $$\dot x(0) = 0.7, x(0) = 0$$ (in red)
NOTE
Near $$x = 0$$ the linearized dynamic system behaves as
$$m\ddot x + \alpha x = 0$$
with solution
$$x_0(t) = C_1 \sin \left(\frac{\sqrt{\alpha } t}{\sqrt{m}}\right)+C_2 \cos \left(\frac{\sqrt{\alpha } t}{\sqrt{m}}\right)$$
and near $$x = \pm\sqrt{\frac{2\alpha}{\beta}}$$ the linearized movement is
$$m\ddot x -2\alpha x = 0$$
with solution
$$x_{\sqrt{\frac{2\alpha}{\beta}}}(t) = C_1 e^{\frac{\sqrt{2\alpha } t}{\sqrt{m}}}+C_2 e^{-\frac{\sqrt{2\alpha } t}{\sqrt{m}}}\ \ \mbox{(unstable)}$$
• Thanks for your answer. in this case, what would the physical interpretation of $C_{0}$ be? It seems to me that it represents the total energy of the system. But in that case, how can we conclude that $C_{0}=0$? – Sei Sakata Oct 27 '18 at 20:39
• @SeiSakata Yes. The $C_0$ is the total energy which remains constant as far as this system is a conservative one. We have $\frac 12 m \dot x^2 + V(x) = C_0$ If $C_0 = 0$ this means that the point is located at one of it's equilibrium points. – Cesareo Oct 27 '18 at 20:46
• So as long if $x=$ one of the solutions, and there is no external force, the mass will remain stationary and no movement whatsoever? Is this the definition of equilibrium state? It is extremely counterintuitive to me that there can be a such point other than the $x=0$ where the extension of the spring is zero. – Sei Sakata Oct 27 '18 at 22:26
• @SeiSakata In a conservative system the movement can assume diverse paths as shown in the attached picture. Introducing dissipation, the energy will decrease with time, finishing in a stable equilibrium point. – Cesareo Oct 27 '18 at 23:50
• Or you need a controller that effects minute corrections to stay in an instable stationary point, like balancing a broom stick upright on your finger. – LutzL Oct 28 '18 at 9:44 | 2019-05-22T23:15:41 | {
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http://stacks.math.columbia.edu/tag/00T6 | # The Stacks Project
## Tag 00T6
Definition 10.135.6. Let $R$ be a ring. Given integers $n \geq c \geq 0$ and $f_1, \ldots, f_c \in R[x_1, \ldots, x_n]$ we say $$S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$$ is a standard smooth algebra over $R$ if the polynomial $$g = \det \left( \begin{matrix} \partial f_1/\partial x_1 & \partial f_2/\partial x_1 & \ldots & \partial f_c/\partial x_1 \\ \partial f_1/\partial x_2 & \partial f_2/\partial x_2 & \ldots & \partial f_c/\partial x_2 \\ \ldots & \ldots & \ldots & \ldots \\ \partial f_1/\partial x_c & \partial f_2/\partial x_c & \ldots & \partial f_c/\partial x_c \end{matrix} \right)$$ maps to an invertible element in $S$.
The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 35254–35284 (see updates for more information).
\begin{definition}
\label{definition-standard-smooth}
Let $R$ be a ring. Given integers $n \geq c \geq 0$ and
$f_1, \ldots, f_c \in R[x_1, \ldots, x_n]$ we say
$$S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$$
is a {\it standard smooth algebra over $R$} if the polynomial
$$g = \det \left( \begin{matrix} \partial f_1/\partial x_1 & \partial f_2/\partial x_1 & \ldots & \partial f_c/\partial x_1 \\ \partial f_1/\partial x_2 & \partial f_2/\partial x_2 & \ldots & \partial f_c/\partial x_2 \\ \ldots & \ldots & \ldots & \ldots \\ \partial f_1/\partial x_c & \partial f_2/\partial x_c & \ldots & \partial f_c/\partial x_c \end{matrix} \right)$$
maps to an invertible element in $S$.
\end{definition}
Comment #235 by Daniel Miller (site) on July 7, 2013 a 10:02 pm UTC
As stated, this definition is not clear. In the presentation of $S$, there are $n$ variables and $c$ polynomials, so one should end up with a $c\times n$ matrix. Is the definition assuming that $n\geqslant c$ and then truncating the matrix $(\partial f_i/\partial x_j)$, or is it saying that all $d\times d$ minors are invertible (where $d=\min\{c,n\}$)? Or is there some convention on what the determinant of a non-square matrix means? Either way, it should be clarified.
Comment #240 by Johan (site) on July 16, 2013 a 2:17 pm UTC
Hi, I agree that this is a bit vague, but I think it is not as bad as you make it out to be. Maybe a better formulation would be:
Let $R$ be a ring. Given integers $n \geq c \geq 0$ and $f_1, \ldots, f_c \in R[x_1, \ldots, x_n]$ we say $$S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$$ is a <i>standard smooth algebra</i> over $R$ if the polynomial $$g = \det \left( \begin{matrix} \partial f_1/\partial x_1 & \partial f_2/\partial x_1 & \ldots & \partial f_c/\partial x_1 \\ \partial f_1/\partial x_2 & \partial f_2/\partial x_2 & \ldots & \partial f_c/\partial x_2 \\ \ldots & \ldots & \ldots & \ldots \\ \partial f_1/\partial x_c & \partial f_2/\partial x_c & \ldots & \partial f_c/\partial x_c \end{matrix} \right)$$ maps to an invertible element in $S$.
What do you think?
Comment #241 by Johan (site) on July 16, 2013 a 8:29 pm UTC
Comment #1845 by Peter Johnson on February 23, 2016 a 1:09 pm UTC
This definition as given applies to a presentation, not to R -> S itself. Shouldn't it have "for some presentation .." or "there exists .."? This makes more sense but would of course have minor effects elsewhere.
Would it not be useful later to also define locally standard smooth?
Comment #1883 by Johan (site) on April 1, 2016 a 6:43 pm UTC
Hello! I've decided to leave this as is for now. If more people chime in, then I'll reconsider.
## Add a comment on tag 00T6
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner). | 2017-07-22T20:55:04 | {
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https://mathhelpboards.com/threads/average-hits-on-exploding-dice.25330/ | # Average "Hits" on exploding dice
#### kryshen
##### New member
Hey folks,
I've been using a really simplistic equation for this and I'd like to get an actual table/equation/and standard deviations if possible.
Here are the rules.
roll xd6s (six-sided dice)
every die that rolls a 4 or greater is a "hit."
6s rolled count as a hit and then "explode," adding another die which is then rolled. That die hits on a 4+ and also explodes on 6s (this explosion chain can continue infinitely, theoretically).
What is the average number of hits for any value of x? Up til this point I've just assumed an average "hits per die" of 0.583 (50% of dice rolled "hit," 16.67% explode and of those 50% hit), but that only accounts for one explosion, not a 6 rolled into a 6 etc. For bonus points, how can I find the standard deviation for any given value of x?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hey folks,
I've been using a really simplistic equation for this and I'd like to get an actual table/equation/and standard deviations if possible.
Here are the rules.
roll xd6s (six-sided dice)
every die that rolls a 4 or greater is a "hit."
6s rolled count as a hit and then "explode," adding another die which is then rolled. That die hits on a 4+ and also explodes on 6s (this explosion chain can continue infinitely, theoretically).
What is the average number of hits for any value of x? Up til this point I've just assumed an average "hits per die" of 0.583 (50% of dice rolled "hit," 16.67% explode and of those 50% hit), but that only accounts for one explosion, not a 6 rolled into a 6 etc. For bonus points, how can I find the standard deviation for any given value of x?
Hi kryshen, welcome to MHB!
Let's start with $x=1$, and let $E$ be the corresponding average number of hits (the so called Expectation).
Then on average we have $0$ hits with probability $\frac 36$, $1$ hit with probability $\frac 26$, and $1$ plus the as yet unknown $E$ number of hits with probability $\frac 16$, yes?
So:
$$E=0\cdot \frac 36 + 1\cdot \frac 26 + (1+E)\cdot \frac 16 = \frac 12 + \frac 16 E \\ \Rightarrow\quad \frac 56E=\frac 12 \quad\Rightarrow\quad E= \frac 35$$
Now what would happen if we have $x$ dice that we throw completely independently from each other?
#### kryshen
##### New member
Thanks so much! Just to be clear, this means that the average number of hits on 10 dice would be 6. Inassumed that the numberbwould be hugher because of the dice explosions.
How would I find the standard deviation for any given value of x?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Thanks so much! Just to be clear, this means that the average number of hits on 10 dice would be 6. Inassumed that the numberbwould be hugher because of the dice explosions.
How would I find the standard deviation for any given value of x?
Let the average number of hits of 1 dice be $\mu=\frac 35$, which is what we found earlier.
Then the square of the standard deviation $\sigma(\text{1 dice})$ is equal to the what the squared deviation of $\mu$ is on average:
$$\sigma^2(\text{1 dice)} = (0 - \mu)^2 \cdot \frac 36 + (1 - \mu)^2 \cdot \frac 26 + ((1+\mu)-\mu) \cdot \frac 16$$
And the standard deviation $\sigma(x\text{ dice})$ is:
$$\sigma(x\text{ dice}) = \sqrt{ x\cdot \sigma^2(\text{1 dice})}$$ | 2020-07-05T06:55:00 | {
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https://math.stackexchange.com/questions/629570/help-with-a-complex-integral | # Help with a complex integral
Evaluate $$\int_{|z|=R} \frac{z^{10}-4z^8-6z^3-5}{(z-1)(z-2)(z-5)^9}$$ for all positive $R \neq 1, 2, 5$.
My attempt is to break the solution into four pieces and to apply Cauchy's Integral Theorem.
For $0<R<1$, there are no encircled singularities, so the answer is zero.
For $1<R<2$, only the singularity at $z=1$ is enclosed. The integral is then equal to $$2\pi i\mbox{Res}(1) = 2\pi i *\frac{14}{(-4)^9}$$.
For $2<R<5$, only the singularities at $z=1,2$ are enclosed. The residue at $z=2$ is calculated, added to the above residue for $z=1$, and then multiplied by $2\pi i$ to yield the solution.
Here's my question: For $R>5$, the integral encloses all singularities, and evaluates as $$2\pi i [\mbox{Res}(1)+ \mbox{Res}(2)+\mbox{Res}(5)]$$ Is this correct? If so, how is the residue at $5$ calculated?
This will go a long way to helping me cement complex integration over closed curves. Thanks in advance for any advice!
Yes, what you have so far is correct.
You can compute the residue by differentiating
$$\frac{z^{10} - 4z^8 - 6z^3 - 5}{(z-1)(z-2)}$$
eight times, and plugging $5$ into that, but that becomes rather unwieldy. No problem for a computer algebra system, I think, but by hand, I prefer other methods.
Note that for $R > 5$, the integral is independent of $R$, and computing the limit for $R \to \infty$ is easier: Setting $z = Re^{i\varphi}$, and hence $dz = iz\,d\varphi$, we get
\begin{align} \int_{\lvert z\rvert = R} \frac{z^{10} - 4z^8 - 6z^3 - 5}{(z-1)(z-2)(z-5)^9}\,dz &= i\int_0^{2\pi} \frac{z^{11} - 4z^9 - 6z^4 - 5z}{(z-1)(z-2)(z-5)^9}\,d\varphi\\ &= i\int_0^{2\pi} \frac{1 - 4z^{-2} - 6z^{-7} - 5z^{-10}}{(1-z^{-1})(1-2z^{-1})(1-5z^{-1})^9}\,d\varphi. \end{align}
In the latter form, it is easy to see that the limit for $R\to\infty$ is $2\pi i$.
• Excellent point at the end. I attempted the derivative formula and wasted time trying to make it work; yours is much more elegant. – Darrin Jan 6 '14 at 23:12 | 2019-09-15T13:46:54 | {
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http://mathoverflow.net/feeds/user/1946 | User joel david hamkins - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:22:17Z http://mathoverflow.net/feeds/user/1946 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/134105/what-new-primitive-recursive-functions-are-needed-to-reconcile-turing-time-comple/134110#134110 Answer by Joel David Hamkins for What new primitive recursive functions are needed to reconcile Turing time complexity with Godel time complexity? Joel David Hamkins 2013-06-19T02:55:14Z 2013-06-19T03:10:18Z <p>Since there are computable total functions that are not primitive recursive, one cannot make the two notions of time complexity coincide. If we add any primitive recursive function as an initial function in the primitive recursive hierarchy, the resulting hierarchy will still consist entirely of primitive recursive functions. And so we may take any computable total function that is not primitive recursive, and this function can have no "recursive definition" as a primitive recursive function and thus has no Gödel complexity. </p> <p>Furthermore, if one entertains the idea of addressing this issue by adding a computable function $g$ that is not primitive recursive, and building the primitive recursive hierarchy on top of that function, then again it will not succeed, since the class of computable total functions is not the same as those that are primitive recursive relative to any fixed computable total function $g$. One can prove this by observing that we have a computable function that is universal for all such functions, simply by unwrapping the primitive recursive definitions and evaluating them. And so by diagonalization there will be a computable total function that is different from any function obtainable by performing primitive recursion over $g$.</p> http://mathoverflow.net/questions/133617/nearly-all-math-classes-are-lectureproblem-set-based-this-seems-particularly-tr/133643#133643 Answer by Joel David Hamkins for Nearly all math classes are lecture+problem set based; this seems particularly true at the graduate level. What are some concrete examples of techniques other than the "standard math class" used at the *Graduate* level? Joel David Hamkins 2013-06-13T15:00:58Z 2013-06-13T15:00:58Z <p>It may be only a minor thing in the space of examples that you seem to be considering, but I have had a lot of success with my practice of requiring students in my graduate courses to write a substantial term paper on an original topic. </p> <p>The aim is for them to undertake a simulacrum of the research experience. I definitely do not want them to just give me an account of some difficult topic on which they read elsewhere. Rather, we try to find a suitable original but manageable topic, which they will have to figure out themselves, and then write up their results in the form of a paper. </p> <p>I insist that these term papers give the appearance of a standard research article, with proper title, abstract, grant or support acknowledgement, proper introduction, definitions, statement of main results and proof, with references and so on. Furthermore, I insist that the students use TeX, which I insist they learn on their own if they do not yet know it. </p> <p>The most difficult part for the instructor is to find suitable topics. One rich source of topics is to take a standard topic that is well-treated elsewhere, but then make a small change in the set-up, giving the student having the task to work out how things behave in this slightly revised setting. For example, in a computability theory class, there is a standard definition of the busy beaver function, with many results known, but one can insist on a slightly different model of Turing machine (such as one-way infinite tape instead of two, or change the halt rule, or have extra symbols or extra tape), where the standard calculations are no longer relevant, but many of the ideas will have a new analogue in this new setting. But also there are usually many suitable topics if one just thinks with curiosity about some of the main ideas in the course and some relevant examples. </p> <p>I always insist that the topics be pre-approved by me in advance, because I want to avoid the situation of a student just writing up something difficult they read, but rather have them really do real mathematical research on their own. Often, I meet with each student several times and we make some discoveries together, which they then work out more completely for their paper. </p> <p>After students submit their final draft (I do not call it a first draft, since I want them to do several drafts on their own before showing me anything, and I don't want to look at anything that they regard as a "first draft"), then I give comments in the style of a referee report, and they make final revisions before submitting the "publication" version, which I sometimes gather into a Kinko's style bound issue <em>Proceedings of Graduate Set Theory, Fall 2014</em> or whatever, and distribute to them and to the department. </p> <p>Finally, on the last lecture of the course, we usually have student talks of them making presentations on their work. For example, see <a href="http://jdh.hamkins.org/student-talks-on-infinitary-computability/" rel="nofollow">the student talks given for my course on infinitary computability last fall</a>. </p> <p>I think it works quite well, and gives the students some real experience of what it is like to do mathematical research. In a few exceptional cases, the terms papers have subsequently turned into actual journal publications, when the students got some strong enough and interesting enough results, and that has been really special. </p> <p>The workflow for me is to assign normal problem sets in the early part of the course, and then start suggesting topics, with the students coming to me and we discuss possibilities. Then, as the work on the paper ramps up, the problem sets taper off, until they are submitted, with additional problem sets at the end of the course, except when they are making their revisions. </p> <p>(And I never accept papers after the end of the course.)</p> http://mathoverflow.net/questions/133597/what-would-remain-of-current-mathematics-without-axiom-of-power-set/133629#133629 Answer by Joel David Hamkins for What would remain of current mathematics without axiom of power set? Joel David Hamkins 2013-06-13T12:58:46Z 2013-06-13T13:24:19Z <p>Several standard theories intensely studied by set theorists do not have the power set axiom. </p> <p>One of these is the theory ZFC without the power set axiom, usually denoted $\text{ZFC}^-$. One should take care with the proper axiomatization of this theory, as we discuss in <a href="http://jdh.hamkins.org/what-is-the-theory-zfc-without-power-set/" rel="nofollow">What is the theory ZFC without the powerset?, V. Gitman, J.D. Hamkins, T. Johnstone</a>; the main point being that one should use collection+separation and not just replacement, since these are no longer equivalent without the power set axiom. </p> <p>Part of the attraction of $\text{ZFC}^-$, which is much stronger than the theory KP discussed below, but still lacks power set, is an abundance of natural models, such as the following:</p> <ul> <li><p>HC, the universe of hereditarily countable sets. This is the land of the countable, where everything is countable. The sets in HC are precisely those sets that are countable and have only countable members and members-of-members and so on. Quite a bit of mathematics can be fruitfully undertaken in HC. </p></li> <li><p>More generally, $H_{\kappa^+}$, the universe of sets of hereditarily size at most $\kappa$. This universe satisfies $\text{ZFC}^-$, but can have some power sets, namely, as long as the power set has size at most $\kappa$. But meanwhile, there is a largest cardinal in this univese, $\kappa$ itself, and the powerset of $\kappa$ does not exist. </p></li> <li><p>More generally, $H_\delta$ for any regular cardinal $\delta$. When $\delta$ is an inaccessible cardinal, this is the same as $V_\delta$, the rank initial segment of the universe in the von Neumann hierarchy, and in this case it is a model of ZFC and a Grothendieck universe. </p></li> </ul> <p>These models and other models of $\text{ZFC}^-$ are used in arguments throughout set theory, from iterated ultrapowers in large cardinals to their use in forcing axioms and elsewhere. </p> <p>Another commonly studied theory without the power set axiom is <a href="http://en.wikipedia.org/wiki/Kripke%E2%80%93Platek_set_theory" rel="nofollow">Kripke-Platek set theory KP</a>, which is a very weak set theory at the heart of the subject known as admissible set theory, in which an enormous amount of classical mathematics can be undertaken. There are numerous natural models of KP, such as:</p> <ul> <li><p>The hyperarithmetic universe $L_{\omega_1^{CK}}$, of sets that are coded by well-founded hyperarithemtic relations on the natural numbers. This is the smallest admissible set, the smallest transitive model of KP. One interesting thing about this world is that every ordinal is not only hyperarithmetic, but actually computable. </p></li> <li><p>There are many other admissible ordinals $\alpha$, ordinals for which $L_\alpha\models$KP. </p></li> <li><p>One can relativize the admissibility concept to oracles $x$, forming $\omega_1^x$, the least admissible ordinal in $x$, so that $L_{\omega_1^x}[x]$ is the smallest model of KP containing $x$. </p></li> <li><p>The universes $L_\lambda$ and $L_\zeta$ arising in the theory of <a href="http://jdh.hamkins.org/ittms/" rel="nofollow">infinite time Turing machines</a>, where $L_\lambda$ is the collection of sets coded by a well-founded infinite-time writable relation on $\omega$, and $L_\zeta$ are the sets coded by a well-founded infinite-time eventually writable relation. These universes both satisfy natural strengthenings of KP, but not the power set axiom.</p></li> </ul> <p>And there are numerous other set theories without the power set axioms, including various strengthenings of KP that still lack the power set axiom and have natural models that are used for various purposes. </p> <p>All these models are intensely studied, and set theorists pay detailed attention to what is or is not possible to achieve in the models, depending on how strong it is. The crux of many arguments is whether the given model is strong enough to undertake a given set-theoretic construction or not. For example, one will often pay attention to the details of a mathematical construction to find out if it can be performed using only $\Sigma_1$-collection instead of, say, $\Sigma_2$-collection, in order to know whether or not it can be performed inside one of these models. </p> <p>Let me add that although set theorists are giving enormous attention to these set theories without the power axiom, the reason isn't usually because of doubt about the truth of the power set axiom, but rather it is just that they want to undertake certain constructions inside these natural models, and so they need to know whether these models are strong enough to undertake that construction or not. </p> <p>So one can be interested in set theory without the power set axiom without having doubt about that axiom. We study set theories without power set, while retaining it in our main background theory, because we want to know what is possible to achieve without power sets in those models. </p> <p>Lastly, concerning your remarks about definability, I refer you as I mentioned in the comments to <a href="http://mathoverflow.net/questions/44102/is-the-analysis-as-taught-in-universities-in-fact-the-analysis-of-definable-numbe/44129#44129" rel="nofollow">an answer I wrote to a similar proposal</a>, which I believe show that naive treatment of the concept of definability is ultimately flawed. </p> http://mathoverflow.net/questions/133455/game-of-chess-and-axiomatic-systems/133505#133505 Answer by Joel David Hamkins for Game of Chess and axiomatic systems Joel David Hamkins 2013-06-12T12:58:57Z 2013-06-12T13:33:03Z <p>Steven Landsburg has now answered the question in the case of ordinary finite chess, which because it is finite has no undecidability or independence phenomenon to speak of. </p> <p>Meanwhile, the kind of phenomenon you seek in Q1, Q2 and Q3 does seem to occur in the context of infinite chess, where one plays from a finite position on an infinite board. This would be a somewhat vaster state space than you had suggested, since it is infinite, but the argument that Steven mentions shows that an infinite state space is a necessary condition for undecidability.</p> <p>Specifically, as I explain in my answer to Richard Stanley's question on the <a href="http://mathoverflow.net/questions/27967/decidability-of-chess-on-an-infinite-board/86755#86755" rel="nofollow">Decidability of chess on an infinite board</a> (see also <a href="http://jdh.hamkins.org/tag/chess/" rel="nofollow">my blog posts on infinite chess</a>), my co-authors and I prove the decidability of the mate-in-$n$ problem of infinite chess by introducing what we call the first-order <em>structure of chess</em> $\frak{Ch}$, with the associated formal language of chess, in which various chess concepts are expressible. Our proof proceeds by showing that this structure is an automatic structure in the sense of finite automata theory, and its theory is consequently decidable. Thus, any infinite chess concept expressible in this formal language of chess will be decidable.</p> <p>Meanwhile, not all chess concepts seem to be expressible in this particular formal language, and in particular it remains open whether the won-position problem is decidable. If it isn't, then like all undecidable problems, it will involve an independence phenomenon as in Q3, for there will be specific finite positions in infinite chess, such that the question of whether or not they are won for white or not will be independent of your favorite axiomatization. </p> <p>Thus, your desired independence phenomenon seems intimately connected with the decidability problem in this infinitary context.</p> http://mathoverflow.net/questions/132687/is-there-any-superstable-configuration-in-the-game-of-life Is there any superstable configuration in the game of life? Joel David Hamkins 2013-06-04T00:58:31Z 2013-06-10T15:50:15Z <p>This question spins off of Gil Kalai's recent question on <a href="http://mathoverflow.net/questions/132402/conways-game-of-life-for-random-initial-position" rel="nofollow">Conway's game of life for a random initial configuration</a>. </p> <p>There are numerous configurations in the game of life that are known to be stable---such as <a href="http://en.wikipedia.org/wiki/Conway%27s_Game_of_Life%23Examples_of_patterns" rel="nofollow">blocks, beehives, blinkers and toads</a>---in the sense that if they appear on an otherwise empty board or on part of the board that remains otherwise empty, then they will persevere (or at least reappear on some period) into the indefinite future. All of the common examples of such configurations, however, seem to disintegrate when placed into a hostile environment; when they are hit by a glider or other spaceship, for example, these common stable configurations can be completely ruined.</p> <p>My question is whether there is any <em>superstable</em> configuration, which can survive even in any hostile environment.</p> <p><strong>Question 1.</strong> Is there any superstable configuration in the game of life? </p> <p>Specifically, let us define that a finite configuration is superstable, if it can survive in any environment, no matter how hostile, meaning that if it should ever appear on the board, then it will definitely reappear later in exactly that same position, regardless of what else is happening on the board. Perhaps the position is somehow isolated, absorbing whatever is happening around it; or perhaps it is a strong source of some kind, spewing out gliders or other objects, regardless of what else is around it; or perhaps it is some core surrounded by encircling vacuum-cleaners, traveling patterns that sweep up whatever might interfere. </p> <p>This question is related to Gil Kalai's, in that if there are such superstable configurations, then we will expect that the infinite random position will have them with some (albeit very small) density, which will enable us to prove lower bounds on the density of the expected living infinite random position.</p> <p>One can also imagine a glider version of superstability, where the pattern survives, but with some nonzero displacement:</p> <p><strong>Question 2.</strong> Is there any superstable glider? </p> <p>That is, is there a finite pattern that, regardless of the environment in which it is placed, will repeat itself at some future time with some displacement? A strong form of such a superstable glider would ask also that it be a vacuum cleaner, meaning that it glides around in any given environment while leaving only empty cells in its wake. </p> <p><strong>Question 2b.</strong> Is there a superstable glider vacuum?</p> <p>I can imagine a small glider that erases everything in its path; or perhaps there is a kind of moving wall, which steadily pushes against whatever it faces, leaving emptiness behind. If there were such a superstable glider vacuum that also moved in a definite direction, then of course there could be no superstable stationary position, since otherwise we could vacuum it up. </p> <p>Another alternative would seem to be that every finite configuration in the game of life is destructible, in the sense that one can design for it an especially hostile environment, leading to eventual death.</p> <p><strong>Question 3.</strong> Is every finite configuration destructible? </p> <p>In other words, can every finite configuration in the game of life be extended to a larger configuration whose development leads in finite time to a position with no living cells? A weaker version of this would ask merely that the configuration be extended to a configuration such that eventually, the original configuration does not recur on any subportion of the board.</p> http://mathoverflow.net/questions/133282/trivial-forcings-which-are-not-very-trivial/133287#133287 Answer by Joel David Hamkins for Trivial forcings which are not very trivial Joel David Hamkins 2013-06-10T11:16:58Z 2013-06-10T11:16:58Z <p>It is a standard result (e.g. Jech's Set Theory, lemma 15.43) that every intermediate model $W$ of ZFC with $M\subset W\subset M[G]$ has the form $W=M[G\cap\mathbb{C}]$ for some complete subalgebra $\mathbb{C}$ of the complete Boolean algebra $\mathbb{B}$ for the forcing giving rise to $G$. </p> <p>In particular, since $x\subset M$ we know that $M[x]\models \text{ZFC}$, and so in particular there is a partial order $P\in M$ which (we can easily arrange) is nontrivial and adds $x$ via $G\cap P$. This achieves nontriviality and relevance (2 and 3). </p> <p>Now, my observation is that your triviality condition 1 is also easy to obtain by modifying $P$, to add a single new atom below every element of $P$, but with a node that is not in $M$. For example, in $M[G]$, let $P^+$ be the poset $P$ together with a new atom, $G$, considered as a single point below every element of $P$. So $P^+$ is atomic, and hence trivial, but $P^+\cap M=P$, which satisfies your requirements in 2 and 3. </p> <p>So the answer is yes.</p> http://mathoverflow.net/questions/133005/research-topics-restricted-to-students-at-top-universities/133010#133010 Answer by Joel David Hamkins for Research topics restricted to students at top universities? Joel David Hamkins 2013-06-07T00:31:45Z 2013-06-07T00:31:45Z <p>Perhaps your advisor had meant merely that the group who work in area B are very strong and have a rich knowledge, and it is difficult for outsiders to enter into or compete with that group because they won't have risen to the high expertise to which that group had brought itself? You seem to present the issue as one of political intrigue and exclusion, but it may not be like this at all. There are surely many mathematical groups, who by working intensely on a focused topic bring themselves to a high level of expertise on that topic. If this is the situation, then it would seem by your other remarks that you can make contacts with that group and begin to study with them and thereby involve yourself in their expertise. </p> <p>That said, I also believe that it is wise to listen to one's advisor's suggestions about topics of investigation. It may be that your advisor simply feels that he will not be able to help you as much in area B, simply because he doesn't himself have the knowledge necessary to guide you in that area. Thus, your plan to work in area B is essentially amounting to not working with your current advisor, and instead having only an email advisor, who may not ultimately give you the attention that you will want and need later on, and that may not be the best situation. But if there is someone in that group who can server as your mentor, then it may work out.</p> <p>Regarding questions (i), I think this kind of concern is likely misplaced. In my experience, any mathematician with talent will eventually be recognized for it, regardless of whatever connections they may or may not have.</p> http://mathoverflow.net/questions/132986/a-question-regarding-the-relation-between-0-sharp-and-koepkes-bounded-truth-pred/132997#132997 Answer by Joel David Hamkins for A Question Regarding the Relation Between 0-sharp and Koepke's Bounded Truth Predicate. Joel David Hamkins 2013-06-06T22:19:45Z 2013-06-06T23:43:09Z <p>The answer is no, one can have models of ZFC set theory with a definable truth predicate for first-order truth in $L$, but without having $0^\sharp$. </p> <p>One way to build such a model is like this. In Kelly-Morse KM set theory, you can prove the existence of a truth predicate for first-order truth for the whole universe $V$, and then by forcing you can code this class into the GCH pattern, for example, in order to make it definable. The result is a forcing extension $V[G]$ which has a first-order definable class predicate for first-order truth in the ground model $V$. From this, one can easily define a truth predicate for first-order truth in $L$. </p> <p>But meanwhile, KM is weaker than $0^\sharp$ in consistency strength, and so we can find such a model $V$ and hence also $V[G]$ without $0^\sharp$. The theory KM is weaker than $0^\sharp$ because its consistency follows from the existence of a single inaccessible cardinal: if $\kappa$ is inaccessible, then $V_\kappa$ is a model of KM when equipped with its full second-order part $V_{\kappa+1}$. In contrast, $0^\sharp$ implies the consistency of a proper class of inaccessible cardinals (and more), since under $0^\sharp$ the Silver indiscernibles are all inaccesible in $L$.</p> <p>This kind of example shows another direct way to build the desired model. Start with $\kappa$ inaccessible. So $V_\kappa$ is a model of ZFC, and remains a model of ZFC(S) even when we add the satisfaction class S for first-order truth in $V_\kappa$. Now we may force to make S definable in a forcing extension $V_\kappa[G]$. In $V_\kappa[G]$, we have a definable class for first-order truth in $V_\kappa$, from which we can define satisfaction in its $L$. But we needn't have $0^\sharp$, since in fact we could have started with $V=L$. </p> http://mathoverflow.net/questions/132917/class-forcing-pelletier-vs-friedman/132938#132938 Answer by Joel David Hamkins for Class forcing: Pelletier vs Friedman Joel David Hamkins 2013-06-06T12:43:39Z 2013-06-06T12:43:39Z <p>Since you are interested in comparing various approaches to class forcing, I would recommend that you also take a look at <a href="http://arxiv.org/abs/math/0609064" rel="nofollow">the dissertation of Jonas Reitz</a>, which has an extended, detailed appendix presenting class forcing for both ZFC and GBC models. Reitz follows a line similar to the Pelletier approach, defining the extension in the case that the class partial order $\mathbb{P}$ is a tower of complete set-sized subposets, using these subposets to stratify the final model in way that is fundamentally similar to what you describe.</p> <p>Ultimately, the main focus for Reitz is on the class forcing notions $\mathbb{P}$ that are what he calls <em>progressively closed</em>, meaning that for every cardinal $\delta$, the forcing has a complete subposet---in particular, a set---whose quotient is forced to be $\lt\delta$-closed (see the details in his dissertation). This kind of forcing is particularly nice, in that every set that is added by the full class forcing is also added by some set-sized complete subposet, and the hypothesis allows one to handle various arguments much more easily than otherwise. Furthermore, many of the most natural class forcing notions that arise in set theory are in fact progressively closed, such as the canonical forcing of the GCH, the forcing of V=HOD by coding sets into the GCH pattern, the universal Laver preparation, Easton forcing to control the GCH pattern and many others.</p> <p>The important fact about progressively closed class forcing is that, as Reitz proves, the corresponding forcing extensions by them will always satisfy ZFC and even GBC.</p> <p>One must make some kind of extra assumption like that, even in the Pelletier approach, where $\mathbb{P}$ is a tower of complete subposets, since otherwise one may not have ZFC in the extension. To see this, consider the case of adding ORD many Cohen reals, which stratifies very nicely in just the way you describe in your question (although it is not progressively closed), but which does not have ZFC in the extension, since even power set will fail at $\omega$---the set of reals in the resulting model will form a proper class!</p> <p>The Friedman approach is aimed at a more general situation, where he is specifically interested in class forcing that is not progressively closed, and which admits no stratification of the sort for which you are looking. For example, a motivating instance for him, I believe, is the case of Jensen's "coding the universe" forcing, which is definitely not progressively closed, and has the property that it adds sets that are not generic for any set-sized poset in the ground model. Because of this more general setting, he has to pay attention to when the forcing relation will be definable with respect to the class forcing notion, and when the partial order will ensure the power set axiom, and these are the issues at which the tameness concepts are aimed.</p> <p>The main difficulty with attempting to use a purely Boolean-valued model approach with class forcing is that one generally doesn't actually have a complete Boolean algebra in this context. The reason is that one usually conceives of a forcing notion first and most naturally as a partial order $\mathbb{P}$, rather than as a Boolean algebra. When this is a set, it is a simple matter to take the Boolean completion $\mathbb{B}$, for example as the regular open algebra, and so working with $\mathbb{P}$ or $\mathbb{B}$ makes little difference. But when the forcing partial order $\mathbb{P}$ is a proper class, then it may not be possible to find a completion of $\mathbb{P}$ in any suitable sense. Generally, objects in the completion correspond to antichains in the partial order, and the collection of all (class-sized) antichains may not itself be a class; they are too big themselves and there are too many of them.</p> <p>For this reason, one is tempted to extend the universe upward, adding extra layers on top, so that one can make sense of the forcing in a set-theoretic context, but the difficulty is that one cannot always expect to extend the universe upward in that way while retaining a nice theory.</p> http://mathoverflow.net/questions/132761/ontological-status-of-some-sets-in-zfc/132790#132790 Answer by Joel David Hamkins for Ontological status of some "sets" in ZFC Joel David Hamkins 2013-06-04T23:50:26Z 2013-06-05T00:26:22Z <p>Although you've been given a hard time in the comments, I think that this is actually a serious question in the philosophy of mathematics, whose answer depends on one's philosophical position concerning the nature of mathematical truth. </p> <p><em>What is the nature of existence for the mathematical objects that we define?</em></p> <p>There are a variety of natural positions to stake out, so let me describe at least a few of them. There are of course many more; I mention several perspectives specifically on CH in my answer to Gil Kalai's question on <a href="http://mathoverflow.net/questions/23829/solutions-to-the-continuum-hypothesis/25199#25199" rel="nofollow">solutions to the continuum hypothesis</a>, which you may find relevant.</p> <p>As you describe the example, the puzzling thing here is that you seem to have defined a fairly simple set, yet we also find ourselves unable to answer some fairly simple questions about it, such as whether it has one or two elements. To my way of thinking, the philosophical question about the ontological status of your set is very similar to the philosophical question about the ontological status of the truth of $\phi$ itself; the two simply go together. So we could ask the question like this:</p> <p><em>When a statement $\phi$ is independent of ZFC, what is the ontological status of the truth of $\phi$?</em></p> <p>For example, does it still make sense to say that $\phi$ is definitely true or false, but definitely only one of these? Ontology has to do with the nature of reality---in our case, mathematical reality---independently of, say, our knowledge about it. But provability and non-provability (and thus independence) seem to have to do with our ability to know certain things, and thus relate to epistemological rather than ontological concerns. </p> <p>Let me describe a few of the philosophical positions that are commonly held about this kind of question. </p> <p><strong>Traditional set-theoretic Platonism.</strong> On the traditional Platonist view, sets exist in a unique real Platonic realm---the realm consisting of all sets---and every set-theoretic assertion has a definitive truth value in this realm. On this account, there is a fact of the matter about whether CH is true or false, or whether your assertion $\phi$ is true or false. On this account, the set you have described is either in fact identical to $\{1\}$ or is identical to $\{1,2\}$. The fact that $\phi$ is independent of the ZFC axioms merely illustrates our lack of knowledge about which one of these sets you have actually described. On this account, the set is definitely one of them or the other, depending on whether it is the case that $\phi$ is true or not, and exactly one of these is the case. The independence of $\phi$ is an irrelevant distraction from whether $\phi$ is true. On this view, the pervasive independence phenomenon is seen as a side-show about our epistemological weakness---the weakness of our theories---rather than indicating any issue about the singular nature of mathematical truth. </p> <p><strong>Formalism.</strong> On this view, there is no realm of real existence for mathematical objects like sets, and rather the mathematical process consists of the manipulation of sequences of symbols, such as the definition in your question. On this view, assertions of existence in mathematics are merely a way of speaking, and no actual existence of mathematical objects is being asserted. Formalism thus trivializes the ontology of the mathematical objects that mathematics is about.</p> <p><strong>Plural realism, or the multiverse view.</strong> On the plural realist or multiverse conception of set-theoretic truth, one holds that there are diverse concepts of set, each giving rise to its own set-theoretic universe, a plurality of set-theoretic worlds with their own set-theoretic truth, but which can be connected in various ways, such as by forcing or by large cardinal embeddings. On this account, your sentence $\phi$ may be true in some of these worlds and false in other worlds, and your set will accordingly change its nature depending on the current set-theoretic background concept of set under which it is being interpreted. You can read more about the multiverse perspective in set theory in <a href="http://jdh.hamkins.org/the-set-theoretic-multiverse/" rel="nofollow">some of my recent papers</a>, especially <a href="http://jdh.hamkins.org/themultiverse/" rel="nofollow">"The set-theoretic multiverse", RSL 2012</a>. </p> http://mathoverflow.net/questions/113867/l-omega-1-omega-sentence-with-many-automorphism-in-aleph-0-and-few-autom/132285#132285 Answer by Joel David Hamkins for $L_{\omega_1,\omega}$ sentence with many automorphism in $\aleph_0$ and few automorphism in $\aleph_\omega$ Joel David Hamkins 2013-05-29T22:51:31Z 2013-05-29T22:51:31Z <p>It appears that we can already do this in first-order logic, without making any use of the infinitary language.</p> <p>Namely, the unique countable endless dense linear order $\langle\mathbb{Q},\lt\rangle$ has continuum many order automorphisms, but Shelah has reportedly proved that for every uncountable cardinality $\lambda$, there is a rigid endless dense linear order of cardinality $\lambda$, that is, having a trivial automorphism group. (See the remarks on page 347 of <a href="http://www.ams.org/journals/proc/1978-072-02/S0002-9939-1978-0507336-2/S0002-9939-1978-0507336-2.pdf" rel="nofollow">this article</a>; perhaps someone can give a better reference.) In particular, there will be an endless dense linear order of size $\aleph_\omega$ with no nontrivial automorphisms at all.</p> http://mathoverflow.net/questions/71537/pointwise-algebraic-models-of-set-theory/71538#71538 Answer by Joel David Hamkins for Pointwise algebraic models of set theory Joel David Hamkins 2011-07-29T00:57:25Z 2013-05-28T11:31:45Z <p><strong>Update, May 27, 2013.</strong> Cole Leahy and I have now written a joint paper arising from issues originating in this question, and here is an excerpt from the post I made on my blog about it, which is adapted from the introduction of the paper.</p> <blockquote> <blockquote> <p><a href="http://jdh.hamkins.org/algebraicity-and-implicit-definability/" rel="nofollow">J. D. Hamkins and C. Leahy, Algebraicity and implicit definability in set theory</a> (also at the <a href="http://arxiv.org/abs/1305.5953" rel="nofollow">arxiv</a>), under review.</p> <p>We aim in this article to analyze the effect of replacing several natural uses of definability in set theory by the weaker model-theoretic notion of algebraicity and its companion concept of implicit definability. In place of the class HOD of hereditarily ordinal definable sets, for example, we consider the class HOA of hereditarily ordinal-algebraic sets. In place of the pointwise definable models of set theory, we examine its (pointwise) algebraic models. And in place of Gödel's constructible universe L, obtained by iterating the definable power set operation, we introduce the implicitly constructible universe Imp, obtained by iterating the algebraic or implicitly definable power set operation. In each case we investigate how the change from definability to algebraicity affects the nature of the resulting concept. We are especially intrigued by Imp, for it is a new canonical inner model of ZF whose subtler properties are just now coming to light. Open questions about Imp abound.</p> <p>Before proceeding further, let us review the basic definability definitions. In the model theory of first-order logic, an element $a$ is definable in a structure $M$ if it is the unique object in $M$ satisfying some first-order property $\varphi$ there, that is, if $M\models\varphi[b]$ just in case $b=a$. More generally, an element $a$ is algebraic in $M$ if it has a property $\varphi$ exhibited by only finitely many objects in $M$, so that $\{b\in M \mid M\models\varphi[b]\}$ is a finite set containing $a$. For each class $P\subset M$ we can similarly define what it means for an element to be $P$-definable or $P$-algebraic by allowing the formula $\varphi$ to have parameters from $P$.</p> <p>In the second-order context, a subset or class $A\subset M^n$ is said to be definable in $M$, if $A=\{\vec a\in M\mid M\models\varphi[\vec a]\}$ for some first-order formula $\varphi$. In particular, $A$ is the unique class in $M^n$ with $\langle M,A\rangle\models\forall \vec x\, [\varphi(\vec x)\iff A(\vec x)]$, in the language where we have added a predicate symbol for $A$. Generalizing this condition, we say that a class $A\subset M^n$ is implicitly definable in $M$ if there is a first-order formula $\psi(A)$ in the expanded language, not necessarily of the form $\forall \vec x\, [\varphi(\vec x)\iff A(\vec x)]$, such that $A$ is unique such that $\langle M,A\rangle\models\psi(A)$. Thus, every (explicitly) definable class is also implicitly definable, but the converse can fail. Even more generally, we say that a class $A\subset M^n$ is algebraic in $M$ if there is a first-order formula $\psi(A)$ in the expanded language such that $\langle M,A\rangle\models\psi(A)$ and there are only finitely many $B\subset M^n$ for which $\langle M,B\rangle\models\psi(B)$. Allowing parameters from a fixed class $P\subset M$ to appear in $\psi$ yields the notions of $P$-definability, implicit $P$-definability, and $P$-algebraicity in $M$. Simplifying the terminology, we say that $A$ is definable, implicitly definable, or algebraic over (rather than in) $M$ if it is $M$-definable, implicitly $M$-definable, or $M$-algebraic in $M$, respectively. A natural generalization of these concepts arises by allowing second-order quantifiers to appear in $\psi$. Thus we may speak of a class $A$ as second-order definable, implicitly second-order definable, or second-order algebraic. Further generalizations are of course possible by allowing $\psi$ to use resources from other strong logics.</p> <p>The main theorems of the paper are:</p> <p><strong>Theorem.</strong> The class of hereditarily ordinal algebraic sets is the same as the class of hereditarily ordinal definable sets: $$\text{HOA}=\text{HOD}.$$</p> <p><strong>Theorem.</strong> Every pointwise algebraic model of ZF is a pointwise definable model of ZFC+V=HOD.</p> <p>In the latter part of the paper, we introduce what we view as the natural algebraic analogue of the constructible universe, namely, the implicitly constructible universe, denoted Imp, and built as follows:</p> <p><code>$$\text{Imp}_0 = \emptyset$$</code></p> <p><code>$$\text{Imp}_{\alpha + 1} = P_{imp}(\text{Imp}_\alpha)$$</code></p> <p><code>$$\text{Imp}_\lambda = \bigcup_{\alpha < \lambda} \text{Imp}_\alpha, \text{ for limit }\lambda$$</code></p> <p><code>$$\text{Imp} = \bigcup_\alpha \text{Imp}_\alpha.$$</code></p> <p><strong>Theorem.</strong> Imp is an inner model of ZF with $L\subset\text{Imp}\subset\text{HOD}$.</p> <p><strong>Theorem.</strong> It is relatively consistent with ZFC that $\text{Imp}\neq L$.</p> <p><strong>Theorem.</strong> In any set-forcing extension $L[G]$ of $L$, there is a further extension $L[G][H]$ with $\text{gImp}^{L[G][H]}=\text{Imp}^{L[G][H]}=L$.</p> <p>Open questions about Imp abound. Can $\text{Imp}^{\text{Imp}}$ differ from $\text{Imp}$? Does $\text{Imp}$ satisfy the axiom of choice? Can $\text{Imp}$ have measurable cardinals? Must $0^\sharp$ be in $\text{Imp}$ when it exists? (An affirmative answer arose in conversation with Menachem Magidor and Gunter Fuchs, and we hope that $\text{Imp}$ will subsume further large cardinal features. We anticipate a future article on the implicitly constructible universe.) Which large cardinals are absolute to $\text{Imp}$? Does $\text{Imp}$ have fine structure? Should we hope for any condensation-like principle? Can CH or GCH fail in $\text{Imp}$? Can reals be added at uncountable construction stages of $\text{Imp}$? Can we separate $\text{Imp}$ from HOD? How much can we control $\text{Imp}$ by forcing? Can we put arbitrary sets into the $\text{Imp}$ of a suitable forcing extension? What can be said about the universe $\text{Imp}(\mathbb{R})$ of sets implicitly constructible relative to $\mathbb{R}$ and, more generally, about $\text{Imp}(X)$ for other sets $X$? Here we hope at least to have aroused interest in these questions.</p> </blockquote> </blockquote> <hr> <p>Original answer:</p> <p>It is a very nice question. </p> <p>If you restrict to well-founded models, and this includes your $L_\alpha$ examples, then a model is pointwise definable if and only if it is pointwise algebraic. The forward implication is clear. For the backward implication, suppose that $M$ is pointwise algebraic; let us prove that $M$ is pointwise definable by induction on rank. Consider any element $a$, and assume all sets of lower rank are definable in $M$. Since $M$ is algebraic, there is a definable finite collection $a_0,a_1,\ldots, a_n$ that includes $a=a_0$, with this set of minimal finite size. These sets must all have the same rank, since otherwise we could make a smaller definable family including $a$, and so all their elements are definable. Thus, if $a\neq a_i$, there must be some element in one of them that is not in the other. But that element is definable, and so we can again make a smaller family by adding to the definition the requirement that the set must contain (or omit, whatever $a$ does) that new element. This would make a smaller definable set, violating the minimality of the finite set, unless indeed our minimal set had only one element. So $a$ is definable after all. </p> <p>This argument works only in well-founded models, however, since the induction is not internal.</p> <p><b>Update.</b> In a conversation with Leo Harrington at math tea here at the National University of Singapore, where I am visiting, we worked out the general ZFC case with the following observation:</p> <p><b>Theorem.</b> Every pointwise algebraic model of ZFC is pointwise definable.</p> <p>Proof. Suppose that $M\models\text{ZFC}$ and is pointwise algebraic. Note that this implies that every ordinal of $M$ is definable, since if we can define a finite set of ordinals containing some ordinal $\alpha$, then since the ordinals are definably linearly ordered, $\alpha$ is the $k^{th}$ member of that set and hence definable. Now, we argue that every set $A$ of ordinals in $M$ is pointwise definable. Well, since $A$ is algebraic, it is a member of a finite definable set. But the lexical order on sets of ordinals is definable and linear, and so again we may find a definition of $A$, since it will be the $k^{th}$ element in that finite set for some $k$. Thus, every set of ordinals in $M$ is definable in $M$. But by ZFC, every set $a$ is coded by a set of ordinals, and since that set of ordinals is definable, it follows that the original set $a$ is also definable. Thus, every set in $M$ is definable without parameters. QED</p> <p>After this, I realized that we can actually omit the use of choice.</p> <p><b>Corollary.</b> Every pointwise algebraic model of ZF is a pointwise definable model of ZFC+ V=HOD.</p> <p>Proof. Suppose that $M\models\text{ZF}$ and is pointwise algebraic. It follows as in the theorem above that every ordinal of $M$ is definable without parameters. Thus, every object in the HOD of $M$ is also definable in $M$ without parameters. If $M$ is not equal to its HOD, then let $A$ be an $\in$-minimal element of $M-\text{HOD}$. Since $A$ is algebraic, there is a finite definable set containing $A$. By minimality, every element of $A$ is in HOD, and so we have a definable well-ordering on the elements of the members of the definable set containing $A$. Thus, there is a definable linear ordering (induced from the lexical order on the definable HOD order) on the subsets of HOD, and so $A$ is the $k^{th}$ element of the finite definable set for some finite $k$, and so $A$ is definable in $M$ without parameters. In this case, since $A\subset\text{HOD}$, it would mean that $A$ should be an element of HOD, contrary to assumption. Thus, $M=\text{HOD}^M$, and so $M$ is a model of ZFC+V=HOD, and also pointwise definable by the theorem.QED</p> http://mathoverflow.net/questions/131933/is-it-possible-to-construct-an-infinite-subset-of-bbb-r-that-is-not-order-isom/131947#131947 Answer by Joel David Hamkins for Is it possible to construct an infinite subset of $\Bbb R$ that is not order isomorphic to any proper subset of itself? Joel David Hamkins 2013-05-26T20:19:59Z 2013-05-26T20:19:59Z <p>The answer is yes in ZFC. We can construct a dense infinite set $A\subset\mathbb{R}$ such that the only order-preserving map $f:A\to A$ is the identity. In particular, $A$ is not order-isomorphic with any proper subset of itself.</p> <p>To see this, note first that any order-preserving map $f:B\to\mathbb{R}$ defined on a dense set $B\subset\mathbb{R}$ can be extended to a total order-preserving map $\bar f:\mathbb{R}\to\mathbb{R}$ defined on the closure of $B$, by defining $\bar f(x)=\sup_{y\leq x, y\in B}f(y)$. Further, note that any such monotone map will have at most countably many points of discontinuity, since every discontinuity will be a jump discontinuity. Thus, there are precisely continuum many such order-preserving functions $\mathbb{R}\to\mathbb{R}$, since any one of them is determined by countably much information about their values on a countable dense set and the information about what their values are on the countably many points of discontinuity.</p> <p>We may therefore enumerate all order-preserving functions $f_\alpha:\mathbb{R}\to\mathbb{R}$ in a sequence of length continuum, $\alpha\lt\mathfrak{c}$.</p> <p>Let's now build the set $A$ by a transfinite process, making promises at each stage about some reals being definitely in $A$ and other promises about keeping some reals out of $A$, in such a way that we kill off $f_\alpha$ at stage $\alpha$ as a possible order-preserving map from $A$ to $A$. We may begin at stage $0$ by placing all the rational numbers into $A$, so that it will definitely be dense. Suppose we have carried out our process up to stage $\alpha$, and $f_\alpha$ is the next non-identity order-preserving map $\mathbb{R}\to\mathbb{R}$ presented for our consideration. Since $f_\alpha$ is order-preserving and not the identity, it must be that there is an interval $(a,b)$ with $(f(a),f(b))$ disjoint from $(a,b)$. Since we've made fewer than continuum many promises so far, there must be an $x\in (a,b)$ such that we've made no promises about $x$ or $f_\alpha(x)$. In this case, we place $x$ into $A$ and promise to keep $f_\alpha(x)$ out of $A$. This will prevent $f_\alpha$ from being an order-isomorphism of $A$ to a proper subset of $A$.</p> <p>The end result is that $A$ is dense, but is strongly rigid in the sense that there is no non-identity order-preserving map from $A$ to $A$. In particular, $A$ is not order-isomorphic with any proper subset of itself.</p> http://mathoverflow.net/questions/131796/closure-of-one-relation-w-r-t-other/131805#131805 Answer by Joel David Hamkins for Closure of one relation w.r.t other Joel David Hamkins 2013-05-25T02:22:12Z 2013-05-25T02:22:12Z <p>(This is more of a comment than an answer, but too long for the comment box.)</p> <p>Your property is saying precisely that $[(R\circ R')\upharpoonright\text{dom}(R)]\subseteq R$. </p> <p>I'm not sure I like your "closure" terminology, since that suggests that you start with a relation, and then close it. But there are in general many relations $R$ that satisfy your property with a given relation $R'$. For example,</p> <ul> <li>the empty relation $R$ has your property with respect to any $R'$. </li> <li>similarly, the full relation $R$ also has this property. </li> <li>Also, if $R$ is the transitive closure of $R'$, then this property holds. </li> </ul> <p>So you do not seem to be starting with something and then taking a "closure", but rather asserting that the given relation $R$ is itself already closed under this kind of application with $R'$. </p> http://mathoverflow.net/questions/131758/what-is-the-least-ordinal-than-cannot-be-embedded-in-mathbbr-mathbbr/131764#131764 Answer by Joel David Hamkins for What is the least ordinal than cannot be embedded in $\mathbb{R}^\mathbb{R}$? Joel David Hamkins 2013-05-24T19:37:41Z 2013-05-24T20:16:02Z <p>Let me get things started with some simple observations.</p> <p>Note that given any countable sequence of functions $f_n$, we can by diagonalization construct a function eventually dominating all of them, $f(x)=\max_{n\leq x}f_n(x)$. It follows that we may by transfinite recursion construct an embedding of $\omega_1$ into your order: at successor stages, add one to the previous function; at limit stages, use the diagonalization just described.</p> <p>So actually, since $\mathbb{R}$ is order-isomorphic to bounded intervals of itself, we can therefore also embed $\omega_1$ into the order many times, on top of one another. So this gives strictly larger ordinals mapping in.</p> <p>More generally, the <em>bounding number</em> $\mathfrak{b}$ is the size of the smallest unbounded family of functions, and any family of size less than $\mathfrak{b}$ will be bounded above. Thus, the recursive construction actually shows that we can find an embedding of $\mathfrak{b}$ into $\mathbb{N}^{\mathbb{N}}$ under eventual domination. Thus, we also get strictly larger ordinals than $\mathfrak{b}$ embedding in, by using the bounded-interval trick again.</p> <p>There are diverse independence results concerning the exact value of $\mathfrak{b}$. Under CH, it is the same as the continuum, of course, but when CH fails, it can be far larger than $\omega_1$.</p> <p>Using Péter's idea, once we have a map from $\mathfrak{b}$ into the order, then we may conclude that the class of ordinals that map into the order is closed under sums of length $\mathfrak{b}$. Thus, any ordinal up to $\mathfrak{b}^+$ is is order-embeddable into $\mathbb{R}^\mathbb{R}$ under eventual domination. So $\mathfrak{b}^+$ is a lower bound for your desired ordinal.</p> <p>I guess the same idea shows that whenever an ordinal $\kappa$ embeds in, then the class of ordinals will be closed under sums of length $\kappa$, and so all ordinals up to $\kappa^+$ will also map in. Thus, the smallest ordinal not embedding in must be a cardinal, and furthermore, it must be a regular cardinal for the same reason.</p> <p><strong>Update.</strong> It is relatively consistent that the answer is $\mathfrak{c}^+$, even when the continuum $\mathfrak{c}$ is very large, and much larger than $\mathfrak{b}$. The reason is that by forcing, we can undertake a very long forcing iteration of length $\kappa$ to add a dominating real at each stage, and thereby get a model with continuum $\kappa$, such that $\kappa$ embeds into the order (and so the smallest ordinal not embedding into the order is $\kappa^+$). Now, the point is that with further ccc forcing, we can make $\mathfrak{b}$ small or whatever we like, but meanwhile, we still have our old functions showing that $\kappa$ maps into the order. </p> http://mathoverflow.net/questions/15841/how-do-the-compact-hausdorff-topologies-sit-in-the-lattice-of-all-topologies-on-a How do the compact Hausdorff topologies sit in the lattice of all topologies on a set? Joel David Hamkins 2010-02-19T21:09:20Z 2013-05-23T05:33:42Z <p>This question is about the space of all topologies on a fixed set X. We may order the topologies by refinement, so that τ ≤ σ just in case every τ open set is open in σ. Equivalently, we say in this case that τ is <em>coarser</em> than σ, that σ is <em>finer</em> than τ or that σ <em>refines</em> τ. (See <a href="http://en.wikipedia.org/wiki/Comparison%5Fof%5Ftopologies" rel="nofollow">wikipedia on comparison of topologies</a>.) The least element in this order is the indiscrete topology and the largest topology is the discrete topology. </p> <p>One can show that the collection of all topologies on a fixed set is a complete lattice. In the downward direction, for example, the intersection of any collection of topologies on X remains a topology on X, and this intersection is the largest topology contained in them all. Similarly, the union of any number of topologies generates a smallest topology containing all of them (by closing under finite intersections and arbitrary unions). Thus, the collection of all topologies on X is a complete lattice. </p> <p>Note that the compact topologies are closed downward in this lattice, since if a topology τ has fewer open sets than σ and σ is compact, then τ is compact. Similarly, the Hausdorff topologies are closed upward, since if τ is Hausdorff and contained in σ, then σ is Hausdorff. Thus, the compact topologies inhabit the bottom of the lattice and the Hausdorff topologies the top.</p> <p>These two collections kiss each other in the compact Hausdorff topologies. Furthermore, these kissing points, the compact Hausdorff topologies, form an antichain in the lattice: no two of them are comparable. To see this, suppose that τ subset σ are both compact Hausdorff. If U is open with respect to σ, then the complement C = X - U is closed with respect to σ and hence compact with respect to σ in the subspace topology. Thus C is also compact with respect to τ in the subspace topology. Since τ is Hausdorff, this implies (an elementary exercise) that C is closed with respect to τ, and so U is in τ. So τ = σ. Thus, no two distinct compact Hausdorff topologies are comparable, and so these topologies are spread out sideways, forming an antichain of the lattice.</p> <p>My first question is, do the compact Hausdorff topologies form a maximal antichain? Equivalently, is every topology comparable with a compact Hausdorff topology? [Edit: François points out an easy counterexample in the comments below.] </p> <p>A weaker version of the question asks merely whether every compact topology is refined by a compact Hausdorff topology, and similarly, whether every Hausdorff topology refines a compact Hausdorff topology. Under what circumstances is a compact topology refined by a unique compact Hausdorff topology? Under what circumstances does a Hausdorff topology refine a unique compact Hausdorff topology?</p> <p>What other topological features besides compactness and Hausdorffness have illuminating interaction with this lattice?</p> <p>Finally, what kind of lattice properties does the lattice of topologies exhibit? For example, the lattice has atoms, since we can form the almost-indiscrete topology having just one nontrivial open set (and any nontrivial subset will do). It follows that every topology is the least upper bound of the atoms below it. The <a href="http://www.smc.math.ca/cjm/v20/cjm1968v20.0805-0807.pdf" rel="nofollow">lattice of topologies is complemented</a>. But the lattice is not distributive (when X has at least two points), since it embeds N<sub>5</sub> by the topologies involving {x}, {y} and the topology generated by {{x},{x,y}}.</p> http://mathoverflow.net/questions/131511/order-type-of-the-smallest-set-containing-the-identity-function-and-closed-under/131530#131530 Answer by Joel David Hamkins for Order type of the smallest set containing the identity function and closed under exponentiation Joel David Hamkins 2013-05-22T23:31:20Z 2013-05-22T23:49:46Z <p>This is a partial answer, and I am unsure about part of it. </p> <p>I claim that these functions are well-ordered by eventual domination, and the order type is at most <a href="http://cantorsattic.info/Epsilon_naught" rel="nofollow">the ordinal $\epsilon_0$</a>.</p> <p>First, your collection of functions can be identified with the unary terms that give rise to them, the unary terms in the term algebra in the language you have presented, terms with one free variable $n$ in the language with only the binary exponentiation function symbol. Examples of such terms are the expressions that appear in your question.</p> <p>$$(n^n)^n\ \ \ \ \ (n^{n^n})^{n^n}\ \ \ \ \ n^{n^n}\ \ \ \ \ (n^{n^n})^{n^{n^n}}$$</p> <p>To any such expression $f(n)$, we may associate to it the ordinal $f(\omega)$, obtained by replacing the variable $n$ with the ordinal $\omega$ and interpreting the resulting expression using the <a href="http://en.wikipedia.org/wiki/Ordinal_arithmetic#Natural_operations" rel="nofollow">natural arithmetic</a> on ordinals, rather than the usual arithmetic. That is, we resolve $(a^b)^c$ as $a^{b\mathop{\sharp}c}$ using the natural product $b\mathop{\sharp} c$, which is a commutative version of ordinal multiplication.</p> <p>$$(\omega^\omega)^\omega\ \ \ \ \ (\omega^{\omega^\omega})^{\omega^\omega}\ \ \ \ \ \omega^{\omega^\omega}\ \ \ \ \ (\omega^{\omega^\omega})^{\omega^{\omega^\omega}}$$</p> <p>All these resulting ordinals have a finitary exponential representation using $\omega$, and therefore are less than epsilon naught $\epsilon_0$.</p> <p>I claim that this correspondence respects eventual domination; in other words, the function given by term $f(n)$ is eventually dominated by the function given by term $g(n)$ if and only if $f(\omega)\lt g(\omega)$ as interpreted in natural ordinal arithmetic. (Note: the reason to use the symmetric multiplication arises from the fact that $(\omega^{\omega})^{\omega^\omega}=\omega^{\omega^{1+\omega}}=\omega^{\omega^\omega}$ with usual ordinal arithmetic, even though $(n^n)^{n^n}$ dominates $n^{n^n}$; but the natural ordinal arithmetic gives the right answer here.) This claim has an affinity with the usual analysis of the representation of ordinals below $\epsilon_0$ in terms of complete (hereditary) base $n$, as used in <a href="http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18107#18107" rel="nofollow">Goodstein's theorem</a>. Basically, the eventual domination order is determined by what might be called the <em>stack height</em> of the term expression, and one reduces inductively to comparing the terms that arise as coefficients of that tallest stack. The value of an ordinal exponential expression of $\omega$ is determined in exactly the same way, and so these two orders agree.</p> <p>If this is right, then the eventual domination order is indeed a well-order, and the order-type is at most $\epsilon_0$, as I claimed.</p> <p>As to whether the order-type reaches $\epsilon_0$ or not, I'm unsure, but I suspect it is strictly less than $\epsilon_0$. The reason is that the ordinals $f(\omega)$ face a severe restriction in their representation in complete base $\omega$. The complication is that not every ordinal arises as $f(\omega)$ for a term in your algebra. For example, the ordinal $\omega^2\cdot 4+\omega^3\cdot99$ is less than $\epsilon_0$, but it does not arise as an ordinal $f(\omega)$ for any term in your algebra. The ordinals $f(\omega)$ seem to be restricted in their complexity, and so it is conceivable that the total order type might be less than $\epsilon_0$. Nevertheless, it is possible to get some natural number coefficients appearing, as with</p> <p>$$(\omega^\omega)^\omega=\omega^{\omega^2}\ \ \ \text{ and }\ \ (\omega^{\omega^2})^\omega=\omega^{\omega^3},$$</p> <p>which arise as the ordinals of the corresponding terms. Perhaps if one can percolate this phenomenon upward to get arbitrary hereditary base $n$ expressions eventually high up in the exponents, then the order type will be $\epsilon_0$.</p> <p>Meanwhile, let me mention that if one had a slightly more generous algebra, allowing addition and the natural number constants (which would arise from the zero function via $1=\omega^0$), then the order type would be fully $\epsilon_0$, since every ordinal less than $\epsilon_0$ would arise as $f(\omega)$ for a corresponding term in the algebra in a completely natural way.</p> http://mathoverflow.net/questions/131407/is-deciding-whether-a-turing-machine-provably-runs-forever-equivalent-to-the-ha/131410#131410 Answer by Joel David Hamkins for Is deciding whether a Turing machine *provably* runs forever equivalent to the halting problem? Joel David Hamkins 2013-05-22T03:16:06Z 2013-05-22T04:01:43Z <p>The first thing to notice is that if ZF is consistent, then it is consistent with ZFC that what you call ProveLoop is actually decidable. The reason is that if ZF is consistent, then by the incompleteness theorem, it is consistent with ZFC that $\neg$Con(ZF), in which case everything is provable in ZF, in which case every program is in ProveLoop.</p> <p>So in the proof that ProveLoop is undecidable, one needs to make an additional assumption about the reliability of the proofs in ZF to avoid this issue with the incompleteness theorem.</p> <p>Meanwhile, under such a consistency assumption, ProveLoop is indeed equivalent to the halting problem.</p> <p><strong>Theorem.</strong> Assume Con(ZF). Then ProveLoop is Turing equivalent to the Halting problem.</p> <p>Proof. Under the Con(ZF) assumption, it follows that whenever ZF proves that a program doesn't halt, then it really doesn't halt, since if it did halt, then this fact would also be provable, contrary to consistency. </p> <p>Clearly ProveLoop is c.e. and hence reducible to the halting problem, as you pointed out. Conversely, let's reduce the halting problem to ProveLoop. Given any program $p$, we want to decide whether $p$ halts on a blank tape, using an oracle for ProveLoop.</p> <p>Define a computable function $f$, so that $f(q)$ is the program such that, on trivial input, if $p$ halts on the blank tape, then $f(q)$ jumps into an immediate infinite loop, and otherwise, while waiting for $p$ to halt, the program $f(q)$ halts just in case it finds a proof that $q$ does not halt. By the recursion theorem, there is a program $r$ such that $r$ and $f(r)$ compute the same function, and we can find this $r$ effectively. Furthermore, by using the $r$ from the proof of the recursion theorem, we may also assume that ZF proves that $r$ and $f(r)$ compute the same function. Notice that it can't ever be that $r$ halts on account of finding a proof that $r$ does not halt, by our assumption which ensures the accuracy of proofs of non-halting, and so definitely $r$ does not halt in any case. Meanwhile, if $p$ halts, then $r$ does not halt, but for a trivial reason that will be provable in ZF, namely, the fact that $p$ halted; and otherwise, when $p$ does not halt, then $r$ will run forever, but this fact will not be provable (for if it were provable, then $r$ would halt, contrary to consequence of our assumption that such proofs are reliable). So what we have is exactly a reduction of the halting problem to ProveLoop, as desired. QED</p> http://mathoverflow.net/questions/131131/is-the-equivalence-between-a-sigma0-1-and-a-pi0-1-formula-defining-the-sa/131136#131136 Answer by Joel David Hamkins for Is the equivalence between a $\Sigma^0_1$ and a $\Pi^0_1$ formula defining the same recursive set provable in a sufficiently strong arithmetic ? Joel David Hamkins 2013-05-19T10:20:13Z 2013-05-19T10:20:13Z <p>No, in general, a true $\Delta^0_1$ assertion may not necessarily be provably $\Delta^0_1$ in a given theory. For example, assume $\text{Con}(\text{PA})$ is true, and consider the formula $\phi(a)$ asserting that $a=a$ and the formula $\psi(a)$ asserting that "$a$ is not the code of a proof of a contradiction in $\text{PA}$," which is expressible as saying that $a$ does not solve a certain specific diophantine equation. </p> <p>Since we assumed there is no such proof, we have that $\exists a\ \psi(a)$ is equivalent to $\forall a\ \psi(a)$, since these are both true sentences. But there can be no proof of this equivalence in $\text{PA}$, if it is consistent, since $\text{PA}$ proves the former sentence, but if it were to prove the latter sentence, it would be proving its own consistency. </p> http://mathoverflow.net/questions/131078/a-question-about-large-real-closed-fields/131081#131081 Answer by Joel David Hamkins for A question about large real closed fields Joel David Hamkins 2013-05-18T19:13:40Z 2013-05-18T19:26:42Z <p>If $\delta$ is the cofinality of an ordered field $F$, that is, the size of the smallest unbounded subset of $F$, then every point of $F$ fills a cut of type $(\delta,\delta)$. In other words, every point in $F$ is the limit of an increasing $\delta$ sequence from below and a decreasing $\delta$ sequence from above. One can see that this is true of $0$ by inverting the elements of a strictly increasing positive unbounded $\delta$ sequence; and then one can translate this sequence from $0$ to any other point for the general conclusion. </p> <p>It follows that any set with a limit point must have size at least $\delta$, and consequently any set of size less than $\delta$ has no limit points in $F$. </p> <p>It is easy to make fields of any desired cofinality, just by forming a chain of elementary extensions of that length, adding new points above at each step. </p> http://mathoverflow.net/questions/130879/does-the-generalized-delta-system-lemma-imply-some-weak-version-of-the-gch/130885#130885 Answer by Joel David Hamkins for Does the generalized $\Delta$-system lemma imply some weak version of the GCH? Joel David Hamkins 2013-05-17T00:23:55Z 2013-05-17T18:48:53Z <p>It is a very nice question! The answer is yes, natural instances of the $\Delta$ system property, which hold under GCH, are in fact equivalent to the GCH. </p> <p><strong>Theorem.</strong> $\Delta(\omega_2,\omega_1)$ is equivalent to CH.</p> <p>Proof: You've pointed out that CH implies the principle, since the hypothesis you mention for this case amounts to $\omega_1^{\lt\omega_1}<\omega_2$, which amounts to CH. So let us consider what happens when CH fails. Let $T=2^{\lt\omega}$ be the tree of all finite binary sequences, and label the nodes of $T$ with distinct natural numbers. Let $F$ be the subsets of $\omega$ arising as the sets of labels occuring on any of $\omega_2$ many branches through $T$. Thus, $F$ has size $\omega_2$, and any two elements of $F$ have finite intersection. I claim that this family of sets can have no $\Delta$-system of size $\omega_2$, and indeed, it can have no $\Delta$-system even with three elements. If $r$ is the root of $a$, $b$ and $c$ in $F$, then $r=a\cap b=a\cap c$, and so $a$ and $b$ branch out at the same node that $a$ and $c$ do, in which case $b$ and $c$ must agree one step longer, so $b\cap c\neq r$. QED</p> <p>The same idea works for higher cardinals as follows: </p> <p><strong>Theorem.</strong> For any infinite cardinal $\delta$, we have $\Delta(\delta^{++},\delta^+)$ is equivalent to $2^\delta=\delta^+$.</p> <p>Proof. If $2^\delta=\delta^+$, then your criterion, which amounts to $(\delta^+)^{\lt\delta^+}<\delta^{++}$, is fulfilled, and so the $\Delta$ property holds. Conversely, consider the tree $T=2^{\lt\delta}$, the binary sequences of length less than $\delta$. Let $F$ be a family of $\delta^{++}$ many branches through $T$, regarding each branch $b$ as a subset of $T$, the set of its initial segments. Each such branch has size $\delta$, since the tree has height $\delta$. But for the same reason as before, there can be no $\Delta$ system with even three elements, since the tree is merely binary branching, and so three distinct branches cannot have a common root. This contradicts $\Delta(\delta^{++},\delta^+)$, as desired. QED</p> <p><strong>Corollary.</strong> The full GCH is equivalent to the assertion that $\Delta(\delta^{++},\delta^+)$ for every infinite cardinal $\delta$. </p> <hr> <p><strong>Update.</strong> The same idea shows that the hypothesis you mention is optimal: one can reverse the lemma from the conclusion to the hypothesis. </p> <p><strong>Theorem.</strong> The following are equivalent, for regular $\kappa$ and $\mu\lt\kappa$:</p> <ol> <li>$\Delta(\kappa,\mu)$</li> <li>$\lambda^{\lt\mu}\lt\kappa$ for every $\lambda\lt\kappa$.</li> </ol> <p>Proof. You mentioned that 2 implies 1, and this is how one usually sees the $\Delta$ system lemma stated. For the converse, suppose that $\lambda^{\lt\mu}\geq\kappa$ for some $\lambda\lt\kappa$. Since $\kappa$ is regular and $\mu\lt\kappa$, this implies $\lambda^\eta\geq\kappa$ for some $\eta\lt\mu$. Let $T$ be the $\lambda$-branching tree $\lambda^{\lt\eta}$, which has height $\eta$. Let $F$ be a family of $\kappa$ many branches through this tree, where we think of a branch as the set of nodes in the tree that lie on it, a maximal linearly ordered subset of the tree $T$. Each such branch is a set of size $\eta$. I claim that this family has no subfamily that is $\Delta$ system of size $\lambda^+$. The reason is that because the tree is $\lambda$-branching, if we have $\lambda^+$ many branches with a common root, then at least two of them must extend that root to the next level in the same way, a contradiction to it being a root. Thus, the failure of 2 implies the failure of 1, as desired. QED</p> http://mathoverflow.net/questions/130789/are-the-two-meanings-of-undecidable-related/130815#130815 Answer by Joel David Hamkins for Are the two meanings of "undecidable" related? Joel David Hamkins 2013-05-16T10:44:22Z 2013-05-16T11:26:01Z <p>To my way of thinking, the two notions of undecidability are closely related, and the associated undecidability phenomenon and independence phenomenon, which are both pervasive in mathematics, are deeply inter-twined. </p> <p>The reason is that every Turing undecidable set is saturated with logical undecidability. If we describe a certain undecidable property of finite graphs, say, then there will be infinitely many specific finite graphs for which it will be logically undecidable, even with respect to very strong theories such as ZFC+large cardinals, whether that finite graph has the property or not. And similarly with any undecidable set $A$ and consistent c.e. theory $T$. This is because if almost all instances of "$n\in A$?'' were settled by $T$, then we would have a decision procedure for $A$, namely, on input $n$, search for a proof from $T$ whether $n\in A$ or not (and hard-code the finitely many exceptions). </p> <p>So every Turing undecidable property is accompanied by a huge assortment of logically undecidable statements, assertions about whether particular objects have the property or not, which are independent of whichever fixed consistent background theory you care to adopt.</p> <p>So although the two undecidability phenomenon are distinct, I find them to be deeply connected. </p> http://mathoverflow.net/questions/130326/validity-in-kripke-frames-whose-points-are-finite-or-infinite-sequences/130330#130330 Answer by Joel David Hamkins for Validity in Kripke frames whose points are finite or infinite sequences Joel David Hamkins 2013-05-11T11:05:21Z 2013-05-11T11:11:48Z <p>It is an attractive idea, but unfortunately, it seems not to be true.</p> <p>The reason is that we can have that every $R_n$ is nontrivial, in the sense that the relation sometimes holds between different two different sequences, but there is no path through these relations so that $R^\omega$ never holds between two different sequences. For example, let $D=\{0,1\}$, and when $n$ is even, let $R_n$ be the reflexive lexical order on binary sequences, but when $n$ is odd, let it be the reverse lexical order. Thus, $R_\omega$ will never hold except reflexively, since the initial segments of a two infinite sequence can't be related lexically in both directions unless they are equal.</p> <p>In this case, the formula $\Box p\leftrightarrow p$ will be valid in $R_\omega$, but not in any $R_n$. </p> <p>Another simple counterexample would occur where one $R_n$, say $R_{17}$, never holds, but all the other $R_n$'s always hold. In this case, $R_\omega$ will never hold, and so it's validities will agree with the validities of $R_{17}$, but not with any other $R_n$. </p> http://mathoverflow.net/questions/130019/forcing-mildly-over-a-worldly-cardinal/130028#130028 Answer by Joel David Hamkins for Forcing mildly over a worldly cardinal. Joel David Hamkins 2013-05-07T22:18:16Z 2013-05-08T19:12:41Z <p>I've got it! We can kill the worldliness of a singular worldly cardinals as softly as we like.</p> <p><strong>Theorem.</strong> If $\theta$ is any singular worldly cardinal, then for any natural number $n$ there is a forcing extension $V[G]$ in which $\theta$ remains $\Sigma_n$ worldly, but not worldly, meaning that $V_\theta^{V[G]}$ satisfies the $\Sigma_n$ fragment of ZFC, but not ZFC itself.</p> <p>Thus, such worldly cardinals can be killed as softly as desired.</p> <p>Proof. First, we may assume without loss that the GCH holds, by forcing it if necessary. Also, by forcing to collapse the cofinality of $\theta$ to $\omega$, which is small forcing with respect to $\theta$ and therefore preserves the worldliness of $\theta$, we may assume that $\theta$ has cofinality $\omega$.</p> <p>I claim that in $V$, we may find a set $A\subset\theta$ that is $V_\theta$-generic for the class forcing $\text{Add}(\text{Ord},1)$ to add a Cohen subset of the ordinals over $V_\theta$. To see this, one simply finds ordinals $\theta_n$ with supremum $\theta$ such that $V_{\theta_n}\prec_{\Sigma_n} V_\theta$, and then diagonalizes with respect to the $\Sigma_n$-definable dense classes having parameters in $V_{\theta_n}$ when extending $A$ up to $\theta_{n+1}$. Even though the forcing is not even countably closed (since $\theta$ has cofinality $\omega$), nevertheless we can meet the dense class before the next higher reflecting cardinal since we've limited the complexity of the dense class. It follows that $\langle V_\theta,A,{\in}\rangle$ satisfies $\text{ZFC}(A)$, the theory of ZFC in which the class $A$ is allowed to appear as a predicate the in the replacement scheme.</p> <p>Now let $\mathbb{Q}$ be the class forcing over $V_\theta$ to code $A$ into the GCH pattern. If $G\subset\mathbb{Q}$ is $V$-generic, then it follows that $V_\theta^{V[G]}=V_\theta[G]$ is a model of ZFC, and so $\theta$ is still worldly in $V[G]$.</p> <p>But let me now modify the argument slightly, so as to preserve only some amount of worldliness, while killing the rest. The idea is to find a set $A$ in $V$ that is $\Sigma_k$-generic over $V_\theta$, but not fully generic for the definable dense classes in the first step, where $k$ is much larger than $n$. We can ensure that $\langle V_\theta,A,{\in}\rangle$ satisfies the $\Sigma_k$ fragment of $\text{ZFC}(A)$, but not all of $\text{ZFC}(A)$. This can be done by inserting coding information to reveal an unbounded $\omega$-sequence when restricted to the $\Sigma_{k+1}$ reflecting cardinals. In essence, one hides away the cofinal $\omega$-sequence within the complex set of $\Sigma_{k+1}$-reflecting cardinals. A very similar idea is used in the the final section of our paper <a href="http://jdh.hamkins.org/pointwisedefinablemodelsofsettheory/" rel="nofollow">J. D. Hamkins, D. Linetsky, J. Reitz, Pointwise definable models of set theory</a>.</p> <p>The point now is that if $k$ is sufficiently larger than $n$, then the $\Sigma_k$ genericity of $A$ will ensure that after one codes $A$ into the GCH pattern of $V[G]$, one still gets that $V_\theta^{V[G]}=V_\theta[G]$ will satisfy at least the $\Sigma_n$ fragment of ZFC. But it will not satisfy all of ZFC, because $A$ is definable in this model and $A$ reveals the unbounded $\omega$-sequence of ordinals. So in $V[G]$, the ordinal $\theta$ is $\Sigma_n$-worldly, but not worldly. QED</p> <p>As observed earlier, we can extend this result to regular $\theta$ in the case that $\theta$ is measurable, simply by first performing Prikry forcing to singularie $\theta$ while preserving its worldliness, thereby reducing to the singular case above. </p> <p><strong>Update.</strong> But in general, we cannot get the result for all regular worldly cardinals, because if the result holds for a regular worldly cardinal $\theta$, then in fact $\theta$ must be measurable in an inner model. To see this, suppose that $\theta$ is a regular worldly cardinal, which is another way of saying that $\theta$ is inaccessible, and suppose that the conclusion of the result is true for $\theta$. It follows that there is a forcing extension in which $\theta$ is a strong limit cardinal but not worldly, and so in particular $\theta$ is not inaccessible, and thus it is singular in $V[G]$. In other words, we have a forcing extension $V[G]$ in which $\theta$ is a singular cardinal. But this implies by a covering lemma argument with the Dodd-Jensen core model (recently explained to me by Gunter Fuchs) that $\theta$ is measurable in an inner model. So we cannot expect to kill inaccessibility softly down to worldly non-inaccessbility for all inaccessible cardinals.</p> http://mathoverflow.net/questions/129498/the-kunen-inconsistency-and-definable-classes/129549#129549 Answer by Joel David Hamkins for The Kunen inconsistency and definable classes Joel David Hamkins 2013-05-03T16:53:24Z 2013-05-03T16:59:51Z <p>My perspective on this issue is that there are a variety of ways to take the claim of the Kunen inconsistency, and we needn't pick a particular one as the only right one. Rather, we gain a fuller perspective of the result by understanding the full robust context including all of the interpretations.</p> <ul> <li><p>Kunen proved his result in Kelly-Morse set theory, in large part in order that he could formalize what it means for a class function $j:V\to V$ to be (fully) elementary. In KM, we can prove that that there is a satisfaction class, a truth predicate for first-order truth, and with this class (which is definable) one can express the elementarity of $j$ as a single second-order assertion.</p></li> <li><p>Meanwhile, using the observation (Gaifman) that any cofinal $\Sigma_1$-elementary embedding is $\Sigma_n$-elementary for any meta-theoretic natural number $n$, we can formalize the result in GBC as the claim that no class $j$ is a nontrivial cofinal $\Sigma_1$-elementary embedding. Thus, this kind of elementarity of $j$ becomes expressible as a first-order assertion about $j$.</p></li> <li><p>We don't actually need full GBC, since for example global choice is not used, but only the usual AC for sets, and so this argument can be formalized in GB+AC. </p></li> <li><p>But actually, we don't need the full second-order part of GB, but only the ability to refer to the class $j$. So we can formalize the argument in $\text{ZFC}(j)$, the theory using ZFC where the axioms of replacement is allowed to use formulas in which the class $j$ appears. (But we only insist on elementarity of $j$ in the language without $j$.) This theory is used and suffices to show, for example, that the supremum of the critical sequence $\lambda=\sup_n\kappa_n$ exists. </p></li> <li><p>If one intends to rule out only definable class embeddings $j$, that is, ones which are classes in the ZFC sense of being first-order definable from set parameters, then as you mentioned, there is an easy argument ruling them out, and this argument does not use AC. I do not know any set theorist, however, who takes this result as an answer to the question of whether one can prove the Kunen inconsistency in ZF. Rather, this example reveals the issues of formalization, and shows us that it may be important to take more care in our formal treatment of the result. </p></li> <li><p>Meanwhile, a purely first-order version of the Kunen inconsistency is formalizable in ZFC, with no talk of classes of any kind, as the claim that there is no nontrivial $j:V_{\lambda+2}\to V_{\lambda+2}$ for any $\lambda$. This version still uses AC, and it is open in ZF. It avoids the set/class issues underlying your question by noting that the Kunen inconsistency proof establishes more by restricting to $V_{\lambda+2}$. This set version of the result implies the full result in any set theory capable of showing that a purported class $j$ must have a closure point $\lambda$.</p></li> <li><p>The wholeness axiom gets around the issue of the previous point by stating the theory ZFC + "$j:V\to V$ is nontrivial and elementary in the language with a function symbol for $j$. Elementarity is expressed by the scheme $\forall x[\varphi(x)\iff \varphi(j(x))]$. </p></li> <li><p>Various weakenings and strengthenings of the wholeness axiom are realized by making further claims about $j$, such as whether it has a critical point, whether it moves an ordinal, etc. Also, one can make claims about the extent to which $j$ may appear in the ZFC axioms. Officially, $j$ is allowed in the separation axiom but not in replacement, and so models of WA are not able to prove the supremum of the critical sequence exists. </p></li> <li><p>If one uses merely the model-theoretic concept of embedding, one would be considering $j:V\to V$ for which $x\in y\iff j(x)\in j(y)$. But now the point is that ZFC proves that there are nontrivial embeddings. For example, we can inductively define $j(y)=\{j(x)\mid x\in y\}\cup\{\{\emptyset,y\}\}$, and prove that this is a nontrivial embedding $j:V\to V$. (See my paper <a href="http://jdh.hamkins.org/every-model-embeds-into-own-constructible-universe/" rel="nofollow">Every countable model of set theory embeds into its own constructible universe</a>, to appear in the JML, for more information.) </p></li> </ul> <p>I prefer to understand the Kunen inconsistency in the rich context of all these results, rather than pick just one perspective and say that that perspective is the right one.</p> http://mathoverflow.net/questions/129470/which-omega-1-trees-are-proper/129471#129471 Answer by Joel David Hamkins for Which $\omega_1$-trees are proper? Joel David Hamkins 2013-05-02T23:40:52Z 2013-05-03T00:42:33Z <p>Concerning your final question, it is consistent that there is a proper normal $\omega_1$-tree that is not Suslin. If $T$ is a Suslin tree, then we may build a new tree $T^+$, consisting of the all-zero branch, together with nodes branching off (at the first one), followed by a copy of $T$. This is proper as a notion of forcing, since it is <em>locally</em> c.c.c., which means that it is dense to move to a condition (off the all-zero branch), below which the forcing is c.c.c. So every generic extension via $T^+$ is actually a c.c.c. extension by $T$. The forcing $T^+$ is essentially $\omega_1$ many side-by-side copies of $T$, but organized into an $\omega_1$-tree. But $T^+$ is not Suslin, since it isn't even Aronzsajn. </p> <p>If one doesn't insist on normality in the tree, then one can prove outright that there is a proper $\omega_1$ tree that is not Suslin as follows: consider the tree $S$ consisting of countable ordinal length binary sequences that never have a $0$ after a $1$. Thus, either they are identically $0$, or they are $0$ for some length, followed by some number of $1$s. This tree consists of the all-zero branch, with all-one branches branching off from it, like a comb with $\omega_1$ many prongs, each of length $\omega_1$. It is technically an $\omega_1$-tree, though not normal, and it is proper since it is trivial as a notion of forcing, but it is not Suslin, since it has many uncountable branches and uncountable antichains. </p> <p><strong>Update.</strong> In answer to your revised question, let me say that Martin's axiom rules out not only the existence of Suslin trees, but also rules out the existence of trees of your type.</p> <p><strong>Theorem.</strong> If $\text{MA}_{\omega_1}$ holds, then then there are no Aronszajn proper $\omega_1$-trees. </p> <p>Proof. Suppose that $\text{MA}_{\omega_1}$ holds and suppose that $T$ is an Aronszajn $\omega_1$-tree. Consider the forcing $\mathbb{P}$ to <a href="http://boolesrings.org/scoskey/special-uncountable-trees/" rel="nofollow">specialize $T$</a>. This forcing is c.c.c., consisting of finite partial specializing functions. Now, it follows by our MA assumption that there is a specializing function defined on the entire tree. Thus, the tree $T$ is actually special already. (I guess I am just arguing that under this MA assumption, every Aronszajn tree is special.) Now, the point is that no special Aronszajn tree can be proper, since forcing with the tree will collapse $\omega_1$. QED</p> <p>Thus, it is consistent with ZFC that there are no trees of the type you seek. In contrast, let me observe one way that positive instances can occur.</p> <p><strong>Theorem.</strong> If there is a Suslin tree, then there is an Aronszajn proper normal tree that is not Suslin, just as you seek.</p> <p>Proof: Let $T$ be a Suslin tree, and let $S$ be an Aronszajn tree. Let $A$ be a maximal uncountable antichain in $S$. Let us make a new tree $U$ by performing surgery on $S$, replacing the part of the tree beyond any node in $A$ with a copy of $T$. That is, if the trees grow downward, then above the antichain $A$, the tree $U$ looks like $S$, but below $A$, the tree $U$ consists of copies of $T$ rooted at each node of $A$. Thus, $U$ is not Suslin, since it still has $A$ as an antichain. It is normal, since any node in $U$ below an element of $A$ can be extended to an element of $A$ and then to any level in $T$ beyond that; and any node not below an element of $A$ is in a copy of $T$ above an element of $A$, and hence can be further extended. It is proper, since forcing with $U$ amounts to forcing with $T$, as $U$ is locally like $T$, since any condition can be refined to a condition below which the forcing looks just like something in $T$. In particular, the forcing again is locally c.c.c., and hence proper. Finally, $U$ is Aronszajn, since any branch in $U$ would either get into the $T$ part, and hence be a branch in $T$, which is impossible, or else stays below $A$, in which case it would be a branch in $S$, which is also impossible. So $U$ is Aronszajn proper normal tree that is not Suslin, making for a positive instance of the kind of tree you seek. QED</p> http://mathoverflow.net/questions/129345/n-in-a-row-game-on-mathbbr2/129347#129347 Answer by Joel David Hamkins for $n$-in-a-row game on $\mathbb{R}^2$ Joel David Hamkins 2013-05-01T20:01:47Z 2013-05-02T03:26:56Z <p>I'll start things off by observing that this is what is known as an open game, since if player 1 wins, then the winning condition is satisfied after finitely many moves. It follows by the <a href="http://en.wikipedia.org/wiki/Determinacy#Determinacy_from_elementary_considerations" rel="nofollow">Gale-Stewart theorem</a> that this game is determined: one of the players must have a winning strategy. In particular, the theory of transfinite ordinal game values is applicable, and so player 1's winning strategy, if it exists, will be the value-reducing strategy; and player 2's strategy, if it exists, will be the value-maintaining strategy.</p> <p>But in truth, I expect that we'll be able to describe the strategy directly in detail...</p> <p>Although you insisted that $n\gt 3$, the game of course makes sense for smaller values of $n$. When $n=1$, the first player wins on the first move. When $n=2$, the first player clearly can win on the second move. When $n=3$, the first player wins on or before his fourth move, as follows: player 1 plays his second move at the midpoint of the first two points, so that we have $A_1$ $A_2$ $B_1$ collinear. Player 2 must block by playing in between, to give $A_1$ $B_2$ $A_2$ $B_1$ on a line, and now player 1 plays off this line to make a triangle of possibilities, with two ways to win, but player 2 can only block one of them. </p> <p>See ARupinski's comment for a solution in the case $n=4$. </p> http://mathoverflow.net/questions/128946/cohen-algebra-generalization/128956#128956 Answer by Joel David Hamkins for Cohen algebra (generalization) Joel David Hamkins 2013-04-27T21:16:35Z 2013-04-27T21:16:35Z <p>Regarding question $1$, it seems that you want to know whether you've got the unique complete c.c.c. Boolean algebra with density $\kappa$. The answer is no.</p> <p>On the one hand, the forcing notion $\text{Add}(\omega,\omega_1)$ to add $\omega_1$ many Cohen reals is c.c.c. and has density $\omega_1$. This is another way to describe your algebra in the case $\kappa=\omega_1$.</p> <p>But meanwhile, forcing with a Suslin tree is also c.c.c. and has density $\omega_1$. But these two complete Boolean algebras are not isomorphic, since forcing with the Suslin tree adds no reals and in fact is $\leq\omega$-distributive, whereas clearly $\text{Add}(\omega,\omega_1)$ adds reals.</p> <p>Of course, there may not be a Suslin tree, but consider the forcing to add a single Cohen real (which creates a Suslin tree), followed by the forcing to force with that Suslin tree. This is the iteration of two c.c.c. forcing notions, and is hence c.c.c., but under CH has density $\omega_1$. But this forcing is not isomorphic to $\text{Add}(\omega,\omega_1)$, since the latter forcing is absolutely c.c.c., but the former is not, as the branch through the Suslin tree creates an uncountable antichain in the extension.</p> <p>One may omit the CH assumption by taking the case $\kappa=\mathfrak{c}$. That is, compare the forcing to add continuum many Cohen reals, versus the forcing to add this many Cohen reals, and afterwards also to force with the Suslin tree created by the first one. Both of these are c.c.c. and have density $\mathfrak{c}$, but they are not isomorphic since the former remains c.c.c. in the forcing extension and the latter does not.</p> http://mathoverflow.net/questions/128897/l-systems-and-sierpinski-triangle/128925#128925 Answer by Joel David Hamkins for L-systems and Sierpinski Triangle Joel David Hamkins 2013-04-27T14:48:41Z 2013-04-27T14:48:41Z <p>The $L$-system described in the Wikipedia page to which you link is:</p> <blockquote> <blockquote> <p>variables : A B</p> <p>constants : + −</p> <p>start : A</p> <p>rules : (A → B−A−B), (B → A+B+A)</p> <p>angle : 60°</p> <p>Here, A and B both mean "draw forward", + means "turn left by angle", and − means "turn right by angle" (see turtle graphics). The angle changes sign at each iteration so that the base of the triangular shapes are always in the bottom (otherwise the bases would alternate between top and bottom).</p> </blockquote> </blockquote> <p>The way I think about this is that these rules express the fundamental fractal symmetry of the Sierpinski triangle, namely, the symmetry of the whole triangle with each of the three next-smaller triangles. If one images a "basic traverse" of the full main triangle to be traveling on a line segment from the bottom left to the bottom right, let us call this an A-move, for the fractal sits to our left as we make this move.</p> <p>Thus, I would put two additional pictures before your four pictures, one consisting of a line segment at the bottom, and the next consisting of three movements, like half a hexagon, traversing one side each of the three next-smaller triangles. </p> <p>If we consider replacing the basic A move with the moves arising from one level deeper in the fractal symmetry, then what we would do is follow that half hexagon, one edge each of the three next-smaller triangles: first going up on the left, then across to the right, then down at the right. Note that when we go up the (bottom half) of the left side of the big triangle, the corresponding triangle is on our right now, so this is a B move. But when we go across to the right, at the bottom of the top subtriangle, it is on our left, and then coming back down to the bottom right point, the third (lower-right) triangle is on our right. So we have replaced the basic A move (moving across the bottom of the full big triangle), with the moves B-A-B. Similarly, a B move would be replaced with A+B+A, with the corresponding interpretation of angle rotation (I believe it also makes sense to add an extra reorientation rotation at the start of this, but evidently this just gets canceled out.) </p> <p>One could easily find other ways to make the traverse that would also express these fundamental symmetries, and arrive at different systems generating the Sierpinski triangle.</p> http://mathoverflow.net/questions/128472/a-question-on-cofinite-topology/128475#128475 Answer by Joel David Hamkins for A question on cofinite topology. Joel David Hamkins 2013-04-23T13:04:24Z 2013-04-23T13:10:40Z <p>You should mean $\{x\}=\bigcap\xi$, and the answer is clearly yes, since we can take $\xi$ equal to the set of all open sets containing $x$. Any point $y$ other than $x$ is excluded in this intersection by the open set $X-\{y\}$. The cardinality of this $\xi$ is the same as the number of finite subsets of $X$, which is equinumerous with $X$. And the same will be true of any $\xi$ having your property, since the points other than $x$ must get excluded by elements of $\xi$, but only finitely many at a time, and so the cardinality of $\xi$ must be the same as $X$. </p> <p>(This argument uses the axiom of choice, in order to know that the collection of finite subsets of an infinite set is equinumerous with that set. In $\neg$AC worlds, the situation is more complicated.) </p> http://mathoverflow.net/questions/134101/existence-of-a-regular-subposet-which-collapses-everything-except-the-top-cardina Comment by Joel David Hamkins Joel David Hamkins 2013-06-19T05:09:35Z 2013-06-19T05:09:35Z Small observation: if we perform an additional small forcing $\mathbb{R}$, then we can find such a $\mathbb{Q}$ inside $\mathbb{P}\ast\mathbb{R}$. The reason is that we can let $\mathbb{R}$ collapse $|\delta|^{V^{\mathbb{P}}}$ to $\omega$, which is small relative to $\delta$ in the ground model, and so the composition will collapse $\delta$ to $\omega$, placing it into the case you mention at the end of the question. http://mathoverflow.net/questions/134101/existence-of-a-regular-subposet-which-collapses-everything-except-the-top-cardina Comment by Joel David Hamkins Joel David Hamkins 2013-06-19T02:24:51Z 2013-06-19T02:24:51Z Do you have a counterexample when you drop the $\mathbb{P}\subset V_\delta$ restriction? http://mathoverflow.net/questions/134105/what-new-primitive-recursive-functions-are-needed-to-reconcile-turing-time-comple Comment by Joel David Hamkins Joel David Hamkins 2013-06-19T02:17:57Z 2013-06-19T02:17:57Z My point was that you were assuming a particular notation in the Turing machines, since in unary notation (where we represent $n$ by $n$ many $1$s), you can't just add a $0$ to the end as you said; instead, you have to double the number of $1$s appearing on the tape, which takes about the same time as doing the primitive recursive recursion. http://mathoverflow.net/questions/134105/what-new-primitive-recursive-functions-are-needed-to-reconcile-turing-time-comple Comment by Joel David Hamkins Joel David Hamkins 2013-06-19T02:03:08Z 2013-06-19T02:03:08Z You seem to be using binary notation in your Turing computation, and the phenomenon of your example arises purely in the Turing context, when one compares Turing-machines with unary notation versus Turing machines with binary notation. The Turing time to compute $2x$ in unary notation is about the same as what you are calling the Goedel time for computing $2x$. Could you clarify your set-up more precisely? After all, Turing machines operate with finite strings, and primitive recursive functions (ordinarily) operate with natural numbers. http://mathoverflow.net/questions/134092/checking-if-a-binary-vector-lies-in-the-affine-span-of-given-binary-vectors Comment by Joel David Hamkins Joel David Hamkins 2013-06-18T23:49:11Z 2013-06-18T23:49:11Z Isn't this problem essentially equivalent to testing for linear independence? (That is, determining if a new binary vector is in the span of a collection of presumed-independent binary vectors.) That is, if we could do your problem quickly, then we could also test for dependence quickly, and vice versa, presumably exploiting the binary nature of the vectors in both cases. http://mathoverflow.net/questions/134057/is-there-an-accepted-definition-of-infty-infty-category Comment by Joel David Hamkins Joel David Hamkins 2013-06-18T14:11:58Z 2013-06-18T14:11:58Z I think he means "accepted". http://mathoverflow.net/questions/133856/term-for-directed-acyclic-graph-with-exactly-one-sink-and-one-source Comment by Joel David Hamkins Joel David Hamkins 2013-06-16T04:11:02Z 2013-06-16T04:11:02Z And I guess (please excuse this trivial remark) one needs to say "nonempty" as well, since the empty graph is directed and acyclic, but has no sources or sinks. http://mathoverflow.net/questions/133856/term-for-directed-acyclic-graph-with-exactly-one-sink-and-one-source Comment by Joel David Hamkins Joel David Hamkins 2013-06-16T03:51:36Z 2013-06-16T03:51:36Z Aeryk, you can edit to change your first sentence to add the finiteness hypothesis, since it isn't true for all directed acyclic graphs. For example, the integers under the successor relation is a DAG with no sources or sinks. http://mathoverflow.net/questions/133856/term-for-directed-acyclic-graph-with-exactly-one-sink-and-one-source Comment by Joel David Hamkins Joel David Hamkins 2013-06-16T03:31:24Z 2013-06-16T03:31:24Z I guess you're talking only about finite directed acyclic graphs? http://mathoverflow.net/questions/133597/what-would-remain-of-current-mathematics-without-axiom-of-power-set/133629#133629 Comment by Joel David Hamkins Joel David Hamkins 2013-06-15T00:54:31Z 2013-06-15T00:54:31Z Oh, I'm very sorry to hear that you aren't interested in logical proof or logical conclusions. I'll leave you alone, then, to undertake your own kind of proof activity. http://mathoverflow.net/questions/133789/standard-natural-numbers-do-not-form-a-set Comment by Joel David Hamkins Joel David Hamkins 2013-06-15T00:31:08Z 2013-06-15T00:31:08Z I think Mirac may be referring to the fact that the collection of standard numbers inside a nonstandad model, in the sense of nonstandard analysis, is not a set in the nonstandard world. The reason is that if the standard cut $\mathbb{N}$ of $\mathbb{N}^\ast$ existed in the nonstandard world, then it would reveal an inductive subset of $\mathbb{N}^\ast$, containing $0$ and closed under successor, which was not the whole of $\mathbb{N}^\ast$, which would contradict the fact that $\mathbb{N}^\ast$ satisfies induction in the nonstandard realm. http://mathoverflow.net/questions/132860/how-many-possible-ways-are-there-to-win-in-quoridor/132890#132890 Comment by Joel David Hamkins Joel David Hamkins 2013-06-14T03:02:51Z 2013-06-14T03:02:51Z Waldemar, I guess your estimates assume that we do no pruning at all, but of course, we can substantially prune the game tree, while remaining certain of our analysis, and this might considerably cut down on the complexity. (Also, could you explain the $10^123$ estimate?) http://mathoverflow.net/questions/132687/is-there-any-superstable-configuration-in-the-game-of-life/133308#133308 Comment by Joel David Hamkins Joel David Hamkins 2013-06-14T02:57:44Z 2013-06-14T02:57:44Z My understanding of the noisy rules is that, with small probability, a cell can turn on or off regardless of its surroundings. So any finite position has some small chance of appearing, regardless of what else is there, and so this will happen infinitely many times. http://mathoverflow.net/questions/133684/a-question-about-interpreting-set-theories-containing-large-cardinal-axioms-in-th Comment by Joel David Hamkins Joel David Hamkins 2013-06-14T00:40:27Z 2013-06-14T00:40:27Z How does the Takeuti theory compare with the theory SO, the theory of sets of ordinals, introduced by Koepke and Koerwien, as in section 5 of this paper: <a href="http://www.math.uni-bonn.de/people/koepke/Preprints/Computing_a_model_of_set_theory.pdf" rel="nofollow">math.uni-bonn.de/people/koepke/Preprints/…</a>? http://mathoverflow.net/questions/133617/nearly-all-math-classes-are-lectureproblem-set-based-this-seems-particularly-tr/133643#133643 Comment by Joel David Hamkins Joel David Hamkins 2013-06-13T23:19:42Z 2013-06-13T23:19:42Z And let me add that it is usually the weakest students who want only to write up an account of some Big Theorem, since what I am asking for otherwise is difficult. Sometimes it happens that a student doesn't pre-approve their topic with me, and submits a paper presenting a standard theorem written in their own way, but I'm usually disappointed. | 2013-06-19T13:22:18 | {
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http://www.capimsanto.com.br/ben-and-deeymht/5e421b-supremum-distance-formula | 0. Details. Example 2. r "supremum" (LMAX norm, L norm) distance. Psychometrika 29(1):1-27. Maximum distance between two components of x and y (supremum norm). The Distance Formula is a variant of the Pythagorean Theorem that you used back in geometry. p = ∞, the distance measure is the Chebyshev measure. When p = 1, Minkowski distance is same as the Manhattan distance. According to this, we have. Cosine Index: Cosine distance measure for clustering determines the cosine of the angle between two vectors given by the following formula. Functions The supremum and infimum of a function are the supremum and infimum of its range, and results about sets translate immediately to results about functions. Supremum and infimum of sets. 5. All the basic geometry formulas of scalene, right, isosceles, equilateral triangles ( sides, height, bisector, median ). $$(-1)^n + \frac1{n+1} \le 1 + \frac13 = \frac43$$. Euclidean Distance between Vectors 1/2 1 manhattan: They are extensively used in real analysis, including the axiomatic construction of the real numbers and the formal definition of the Riemann integral. Each formula has calculator 1D - Distance on integer Chebyshev Distance between scalar int x and y x=20,y=30 Distance :10.0 1D - Distance on double Chebyshev Distance between scalar double x and y x=2.6,y=3.2 Distance :0.6000000000000001 2D - Distance on integer Chebyshev Distance between vector int x and y x=[2, 3],y=[3, 5] Distance :2.0 2D - Distance on double Chebyshev Distance … HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. In particular, the nonnegative measures defined by dµ +/dλ:= m and dµ−/dλ:= m− are the smallest measures for whichµ+A … Interactive simulation the most controversial math riddle ever! The infimum and supremum are concepts in mathematical analysis that generalize the notions of minimum and maximum of finite sets. 4 Chapter 3: Total variation distance between measures If λ is a dominating (nonnegative measure) for which dµ/dλ = m and dν/dλ = n then d(µ∨ν) dλ = max(m,n) and d(µ∧ν) dλ = min(m,n) a.e. results for the supremum to −A and −B. Definition 2.11. The Euclidean formula for distance in d dimensions is Notion of a metric is far more general a b x3 d = 3 x2 x1. The scipy function for Minkowski distance is: distance.minkowski(a, b, p=?) Kruskal J.B. (1964): Multidimensional scaling by optimizing goodness of fit to a non metric hypothesis. Hamming distance measures whether the two attributes … From MathWorld--A Wolfram To learn more, see our tips on writing great answers. if p = 1, its called Manhattan Distance ; if p = 2, its called Euclidean Distance; if p = infinite, its called Supremum Distance; I want to know what value of 'p' should I put to get the supremum distance or there is any other formulae or library I can use? p=2, the distance measure is the Euclidean measure. Thus, the distance between the objects Case1 and Case3 is the same as between Case4 and Case5 for the above data matrix, when investigated by the Minkowski metric. Then, the Minkowski distance between P1 and P2 is given as: When p = 2, Minkowski distance is same as the Euclidean distance. euclidean:. For, p=1, the distance measure is the Manhattan measure. [λ]. Usual distance between the two vectors (2 norm aka L_2), sqrt(sum((x_i - y_i)^2)).. maximum:. The limits of the infimum and supremum of … 2.3. Literature. 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https://math.stackexchange.com/questions/943885/can-anybody-explain-this-logical-symbol | # Can anybody explain this logical symbol?
I know this is not right place to post this question.
Please help me
I will notate & for "and" and ^ for "or"
[Editor: Instead, let's use MathJax to notate $\land$ for "and" and $\lor$ for "or" ]
$A$: I got A, $B$: I went to school.
$\neg(A\lor B)$ = Neither I got A nor did I go to school.
which is equivalent to $\neg A\land \neg B$ = I did not get A and I did not go to school.
But what I have trouble is
$\neg A\lor \neg B$ is I did not get A or I did not go to school.
which is equivalent to $\neg (A\land B)$ . but , I do not know how to translate $\neg (A\land B)$ into english sentence.
Can anyone help me?
Also, what is the difference between "Both I love 1 and I love 2" and "I love both 1 and 2"
are both sentences equal to $A\land B$ ?
• Why not use $\vee$ for "or"? – graydad Sep 24 '14 at 4:21
• Please use MathJax. $\land$ is $\land$, $\lor$ is $\lor$, $\neg$ is $\neg$, and so: $\neg (A\lor B)=\neg A\land\neg B$ is $\neg (A\lor B)=\neg A\land \neg B$ – Graham Kemp Sep 24 '14 at 4:24
## 1 Answer
I did not get A or I did not go to school $\iff$ It is not the case that I got an A and went to school
• Is that the translation of ~(A&B) ? – aaa Sep 24 '14 at 4:26
• Yes it is. Also I encourage you to use $\wedge$ for "and", while saving $\vee$ for "or". That is the standard syntax. – graydad Sep 24 '14 at 4:27
• Thanks for reply, but is there another way to transcribe it? besides using "it is not the case" – aaa Sep 24 '14 at 4:40
• I didn't get an A and go to school – graydad Sep 24 '14 at 4:44
• Since English is not my native language, I think I didn't get an A and go to school is same as I didn't get an A and I did not go to school. But, First sentence is ~(A&B) and second is ~A&~B, but they are not same, though. – aaa Sep 24 '14 at 6:24 | 2020-01-27T15:54:37 | {
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https://math.stackexchange.com/questions/2464131/when-to-use-which-quantifier-with-predicate-logic | # When to use which quantifier with predicate logic?
I'm following my first logic course as part of my pre-masters programme. Currently I'm working on predicate logic.
I know that $∀$ is the universal quantifier, which stands for "all" or "every", and $∃$ is the existential quantifier, which stands for "some" or "there is".
In my textbook I tried the following question:
Translate the following sentences into predicate logical formulas. Assume the domain of discourse is human beings. Not all girls love themselves
I used $G$ for "Girl" and $L$ "Loves". My translation was as follows: $$¬∀x(Gx \to Lxx)$$ But, the solution the textbook gives is: $$∃x(Gx ∧ ¬Lxx)$$
I'm really wondering if both solutions are correct. Actually, I think mine is more precise considering that "Not All" is $¬∀x$ and $∃x$ is "Some".
But I guess I am missing something, or is this just a style thing and are both correct?
"$\lnot \forall$" is the same as "$\exists \lnot$".
If not all cats are black, there must be some cat that is not black.
Thus, we have that $$¬∀x \ (Gx \to Lxx) \iff ∃x \ ¬(Gx \to Lxx) \text{.}$$
Now we apply the tautological equivalence $$\lnot (p \to q) \iff (p \land \lnot q)$$
(We can check it with a truth-table: $\lnot (p \to q)$ is TRUE just in case when $p$ is TRUE and $q$ is FALSE.) to get the final result, $$∃x \ (Gx \land \lnot Lxx) \text{.}$$
• Thank you for your anwser! However i'm not sure what you mean with the other ingredient. I know that ¬(p→q) and (p∧¬q) are equivalent to each other (exact same truth-table) – Bart Oct 9 '17 at 8:16
• @Bartvh The author was likely showing the full reason why the two are equal for future readers. Even though you only asked about "not all" vs "some", this makes the answer useful for others who might not be as knowledgeable. – Brian J Oct 9 '17 at 13:58
Both are correct. They are equivalent.
$\neg \forall x~(Gx\to Lxx)$ "Not all girls love themselves."
$\exists x~(Gx\wedge \neg Lxx)$ "Some girls don't love themselves."
I would say that the author of the textbook took a wrong approach. The task of translation is to communicate meaning as close as possible to the original. This is especially important in mathematics. Your answer follows English text verbatim, and the author’s answer is transformed. Albeit the transformation is legal, I bet that it baffles readers because transformation of logical formulas does not belong to the translation process. I wholeheartedly agree that your answer is more precise.
If you are picky, then you may want to know that there are circumstances where your answer and the author’s answer are not equivalent. Their equivalence is impossible to prove in intuitionistic logic. Intuitionistic logic is slightly weaker than classical logic that your textbook teaches. Intuitionistic logic does not contain the law of excluded middle, and truth tables are inapplicable to it; statements are proved by inference. | 2020-04-03T12:06:38 | {
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http://math.stackexchange.com/questions/105220/generalizing-the-trick-for-integrating-int-infty-infty-e-x2dx | # Generalizing the trick for integrating $\int_{-\infty}^\infty e^{-x^2}dx$?
There is a well-known trick for integrating $\int_{-\infty}^\infty e^{-x^2}dx$, which is to write it as $\sqrt{\int_{-\infty}^\infty e^{-x^2}dx\int_{-\infty}^\infty e^{-y^2}dy}$, which can then be reexpressed in polar coordinates as an easy integral. Is this trick a one-hit wonder, or are there other cases where this trick works and is also necessary? It seems to depend on the defining property of the exponential function that $f(a+b)=f(a)f(b)$, which would make me think that it would only allow fairly trivial generalizations, e.g., to $\int_{-\infty}^\infty 7^{-x^2}dx$ or $\int_{-\infty}^\infty a^{bx^2+cx+d}dx$.
Can it be adapted through rotation in the complex plane to do integrals like $\int_{-\infty}^\infty \sin(x^2)dx$? Here I find myself confused by trying to simultaneously visualize both the complex plane and the $(x,y)$ plane.
WP http://en.wikipedia.org/wiki/Gaussian_integral discusses integrals that have a similar form and seem to require different methods, but I'd be more interested in integrals that have different forms but can be conquered by the same trick.
The trick involves expanding from 1 dimension to 2. Is there a useful generalization where you expand from $m$ dimensions to $n$?
This is not homework.
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Once you know the answer for that gaussian integral, you can use differentiation under integral sign (en.wikipedia.org/wiki/Differentiation_under_the_integral_sign) to solve very similar and much more complicated ones. – Robert Smith Feb 3 '12 at 5:28
Robert Dawson wrote about the limits of this method: cs.smu.ca/~dawson/Poisson.pdf – Byron Schmuland Feb 3 '12 at 5:29
@ByronSchmuland: If you make this an answer, I'll accept it. The only thing I'm slightly vague on is what Dawson means by "of the form." E.g., does he consider $e^{ax^2+bx+c}$ to be "of the form" $ke^{ax^2}$? He also only considers the reals, which makes me wonder about the complex rotation issue, which might lead to integrals like $\int_{-\infty}^\infty \sin(x^2)dx$. – Ben Crowell Feb 3 '12 at 5:42
@Ben: If you rewrite $ax^2+bx+c$ as $a(x-b)^2+h$ and $k=e^h$, then it's an obvious translate of the function $ke^{ax^2}$. – anon Feb 3 '12 at 5:45
@BenCrowell Well, the Wikipedia article I linked before has plenty of examples but taking your Gaussian integral as a starting point, take a look at math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf (page 6, higher moments of the Gaussian). However, the point being that this "technique" works fine in a wide variety of situations. – Robert Smith Feb 3 '12 at 5:55
I would like to answer to your question about Fresnel integral as I did this in my undergraduate studies. So, let us consider the integrals
$$S_1=\int_{-\infty}^\infty dx\sin(x^2) \qquad C_1=\int_{-\infty}^\infty dx\cos(x^2).$$
We want to apply the same technique used for Gauss integral in this case and consider the two-dimensional integrals
$$S_2=\int_{-\infty}^\infty\int_{-\infty}^\infty dxdy\sin(x^2+y^2) \qquad C_2=\int_{-\infty}^\infty\int_{-\infty}^\infty dxdy\cos(x^2+y^2).$$
If you go to polar coordinates, these integrals doe not converge. So, we introduce a convergence factor in the following way
$$S_2(\epsilon)=\int_{-\infty}^\infty\int_{-\infty}^\infty dxdy e^{-\epsilon(x^2+y^2)}\sin(x^2+y^2)$$ $$C_2(\epsilon)=\int_{-\infty}^\infty\int_{-\infty}^\infty dxdy e^{-\epsilon(x^2+y^2)}\cos(x^2+y^2).$$
Then one has, moving to polar coordinates,
$$S_2(\epsilon)=2\pi\int_0^\infty \rho d\rho e^{-\epsilon\rho^2}\sin(\rho^2) \qquad C_2(\epsilon)=2\pi\int_0^\infty \rho d\rho e^{-\epsilon\rho^2}\cos(\rho^2)$$
that is
$$S_2(\epsilon)=\pi\int_0^\infty dx e^{-\epsilon x}\sin(x) \qquad C_2(\epsilon)=\pi\int_0^\infty dx e^{-\epsilon x}\cos(x).$$
These integrals are well known and give
$$S_2(\epsilon)=\frac{\pi}{1+\epsilon^2} \qquad C_2(\epsilon)=\frac{\epsilon\pi}{1+\epsilon^2}$$
noting that integration variables are dummy. Now, in this case one can take the limit for $\epsilon\rightarrow 0$ producing
$$S_2=\int_{-\infty}^\infty\int_{-\infty}^\infty dxdy\sin(x^2+y^2)=\pi \qquad C_2=\int_{-\infty}^\infty\int_{-\infty}^\infty dxdy\cos(x^2+y^2)=0$$
By applying simple trigonometric formulas you will get back the value of the Fresnel integrals. But now, as a bonus, you have got the value of these integrals in two dimensions.
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Great answer! Are you sure the $S_2=\pi$ at the end shouldn't be $S_2=\pi/4$? Let $S_1=\int_0^\infty \sin(x^2)dx$ and similarly for $C_1$. Then $C_2=0$ tells us that $S_1=\pm C_1$, so that with $S_2=2C_1S_1$ I get $C_1=S_1=\sqrt{S_2/2}$, which would seem to require $S_2=\pi/4$ in order to produce the correct result $S_1=C_1=\sqrt{\pi/8}$. Is it possible that your value of $\pi$ is for $\int_{-\infty}^\infty \int_{-\infty}^\infty$? – Ben Crowell Feb 3 '12 at 16:30
@BenCrowell: Of course you are right. I fix it. – Jon Feb 3 '12 at 17:38
Robert Dawson wrote about the limitations of this method in this article in the American Mathematical Monthly.
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The distance metric that we use is quadratic in nature. No matter what dimensional $\mathbb{R}^n$ we are working in, the distance between two points is given by $r^2=\sum(\Delta x_i)^2$. It's the ability to swap out $\sum(\Delta x_i)^2$ with $r^2$ that helps compute $\int_{-\infty}^{\infty}e^{-x^2}\,dx$. No higher dimensional space will make this particular exchange any nicer. So this helps explain why the property you mention ($f(x^2)f(y^2)=f(x^2+y^2)$) is important.
On the other hand, what happens here is an exchange between Cartesian coordinates for $\mathbb{R}^2$ and polar coordinates. So could we do something similar in $\mathbb{R}^3$ with spherical coordinates?
$$\left(\int_{\mathbb{R}}f(x)\,dx\right)^3=\iiint_{\mathbb{R}^3} f(x)f(y)f(z)\,dxdydz=\iiint_{\mathbb{R}^3} F(\rho,\theta,\varphi)\,d\rho d\theta d\varphi$$
If $F$ breaks up as $g(\rho)h(\theta)k(\varphi)$, then you might have a slick way to compute $\int_{\mathbb{R}}f(x)\,dx$.
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To answer the question in the first paragraph: Yes the exact same trick works for both $\int_{-\infty}^\infty 7^{-x^2}dx$ and $\int_{-\infty}^\infty a^{bx^2+cx+d}dx$, provided $b<0$. (The second generalizes the first, so we need only consider it). First, complete the square for the quadratic, and write $$bx^2+cx+d=b\left(x^2+\frac{c}{b}x+\frac{d}{b}\right)=b\left(x+\frac{c}{2b}\right)^2-\frac{c}{4b}+d.$$ Then our integral is $$a^{d-\frac{c}{4b}} \int_{-\infty}^\infty e^{\log(a)b\left(x+\frac{c}{2b}\right)^2}dx.$$ Since we are integrating over the real line, we may shift by $\frac{c}{2b}$ and then let $u=\sqrt{-b\log a}x$ (remember, $b<0$) to get $$a^{d-\frac{c}{4b}} \int_{-\infty}^\infty e^{\log(a)bx^2}dx=\frac{a^{d-\frac{c}{4b}}}{\sqrt{-b\log a}} \int_{-\infty}^\infty e^{-u^2}dx$$
$$=\frac{a^{d-\frac{c}{4b}}}{\sqrt{-b\log a}}\sqrt{\pi}.$$ | 2014-03-08T07:12:41 | {
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https://math.stackexchange.com/questions/1093323/what-is-the-difference-between-an-undulation-point-and-other-critical-values | # What is the difference between an undulation point and other critical values?
A necessary but not sufficient condition for a point of inflection is that $$f''(x)=0$$ If the second derivative is 0 and the point is not a point of inflection, Wikipedia tells me that is called an undulation point, which apparently means
a point on a curve where the curvature vanishes but does not change sign.
An example given is $f(x)=x^4$ at $(0,0)$.
What I do not understand is why $(0,0)$ is not classified as a local minimum? Surely for $f(x)=x^4$ the first derivative changes polarity either side of $(0,0)$.
Is there a better example of an 'undulation point' for a smooth function? Do 'undulation points' exist for more than one variable or is the condition that all partial derivatives are zero a necessary and sufficient one?
• Look at the "but does not change sign" part of the definition. Jan 6 '15 at 18:35
The point $(0,0)$ is a minimum point. It is also an undulation point. You are right that in some ways this is a poor example of an undulation point, since it also has other properties. On the other hand, this example does make the point easy to see, and it has an extremely easy formula.
A better example in some ways would be $(0,0)$ in the graph of $f(x)=x^4+x$. The point is still fairly noticeable but is not a minimum.
ADDED: I just took at look at the Wikipedia definition of undulation point, and it does give the example $f(x)=x^4$ in the text. However, it also has the example $y=x^4-x$ in a graph later in the article. This is the same as my example but reflected in the $y$ axis. It seems Wikipedia wanted to have it both ways.
• That is a better example thanks. I'm still confused as to why (0,0) is an undulation point for $f(x)=x^4$ though. Doesn't the sign of the derivative change either side of (0,0)? Jan 6 '15 at 18:41
• @GridleyQuayle: Yes, the derivative changes sign, but that is irrelevant to the definition of undulation point. The second derivative (and therefore the curvature) do not change sign, and that is what is relevant. Jan 6 '15 at 19:03
• If you go to the Wikipedia page for 'Inflection Point' and go to the section 'a necessary but not sufficient condition' the example is at the end of the first paragraph. Jan 6 '15 at 21:24
• @GridleyQuayle: Ah, yes, I see, you are right. I shall edit my answer accordingly. I referred to the graph later in the article. Jan 6 '15 at 22:21
• As a future visitor I thank You! Aug 17 '16 at 21:01
Who said it's not a local minimum? It certainly is.
I don't know how you would want to define "undulation point" in several variables. One problem is that there isn't just one "curvature".
• this got me thinking too, 🤔.. Since there are multiple ways to look at the curvature with respect to the variables present in multivariable calculus, can we not define undulation points separately for each variable? (ofcourse, if it has one). And if all the variables agree at some particular point that becomes the universal undulation point.. note that the word universal is just made up by me 😅 Jun 13 '20 at 2:32 | 2022-01-19T14:57:44 | {
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https://math.stackexchange.com/questions/3017962/why-can-lh%C3%B4pitals-rule-not-be-applied-to-the-sum-or-difference-of-limits | # Why Can L'Hôpital's Rule Not be Applied to the Sum or Difference of Limits?
Consider $$\lim_{x\to\infty}\frac{f(x)}{g(x)} + \lim_{x\to\infty}\frac{h(x)}{i(x)}$$ I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as $$\lim_{x\to\infty}\frac{f(x)}{g(x)} +\frac{h(x)}{i(x)}$$ Why is this incorrect?
• Limit operator is not a linear transformation, that is why. – Bertrand Wittgenstein's Ghost Nov 29 '18 at 0:20
• I think it only goes wrong when you will end up with something of the form $+\infty - \infty$. So, if you avoid these cases you can apply it if I am correct. – Stan Tendijck Nov 29 '18 at 0:54
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...
This has nothing to do with the L'Hôpital's rule itself.
The rule that you cannot use is:
$$\lim\limits_{x\to...}f(x)+\lim\limits_{x\to...}g(x)=\lim\limits_{x\to...}(f(x)+g(x))$$
And you can see from JDMan4444's answer that there are situations where this rule does not work.
However, if you are sure that $$\lim\limits_{x\to...}f(x)$$ and $$\lim\limits_{x\to...}g(x)$$ exist (and are not $$\pm\infty$$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)
This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.
• The OP is related to THAT and the main doubt is on the application of HR. – user Nov 29 '18 at 7:42
Consider the following example:
$$\lim_{x\rightarrow\infty}\frac{x^2}{x}+\lim_{x\rightarrow\infty}\frac{-x^2}{x} = \infty - \infty \quad \text{(which is undefined)}$$ $$\lim_{x\rightarrow\infty}\frac{x^2}{x}+\frac{-x^2}{x} = 0$$
• Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP. – user Nov 29 '18 at 7:44
• Your second line "$\lim_{x\rightarrow\infty}\frac{x^2}{x}+\frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention. – user21820 Nov 29 '18 at 13:51
• @user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there. – user3445853 Nov 29 '18 at 15:17
• @user21820 It could be convention but since the expression $\frac{x^2}{x}+\frac{-x^2}{x}$ is identically equal to zero for all $x\neq 0$ I think that it is a noce assumption read that as $$\lim_{x\rightarrow\infty} \left(\frac{x^2}{x}+\frac{-x^2}{x}\right) =\lim_{x\rightarrow\infty} 0= 0$$ – user Nov 29 '18 at 15:29
• @user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here. – user21820 Nov 29 '18 at 16:47
The following identity
$$\lim_{x\to \infty}\left(\frac{f(x)}{g(x)} +\frac{h(x)}{i(x)}\right)=\lim_{x\to \infty}\frac{f(x)}{g(x)} + \lim_{x\to \infty}\frac{h(x)}{i(x)}$$
doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form
$$\lim_{x\to \infty}\left(\frac{f(x)}{g(x)} +\frac{h(x)}{i(x)}\right)=\lim_{x\to \infty}\frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)}$$
the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.
It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $$\infty - \infty$$ or $$-\infty + \infty$$ cases are problematic whether you're contemplating L'Hopital or not.
• The advice was given HERE for that specific example and it was a completely different case to the one you are referring to. – user Nov 29 '18 at 7:41
• @gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question? – zhw. Nov 29 '18 at 19:19
• My aim was solely give you a better context for the question posed. – user Nov 29 '18 at 19:23 | 2019-10-23T02:12:31 | {
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https://stats.stackexchange.com/questions/566246/ancova-with-common-zero-level | # ANCOVA with common zero level
There is a similar, if not duplicate, question from 9 years ago on this but it never got answered so I'm hoping to get some clarity on this.
Let's say we have N treatments, such as medicines or fertilizers, that can be applied at varying intensities (continuous) and we want to fit an ANCOVA model to make inference about the difference in response to increasing intensity among the N treatments. So we want to know if the slope is different between treatments. The issue is that we have some individuals with intensity 0, where it is not meaningful to assign those individuals to any one of the treatments.
I just want to confirm that the following approach in R is correct. I assigned the individuals with intensity 0 to a treatment of "none" or something similar, so we now have N+1 unique values in the treatment column. Then fit the ANCOVA model as normal and also do post-hoc contrasts comparing the slopes between all pairs of treatments, but exclude the "none" treatment from those contrasts. Does this properly take account of the information from the individuals where intensity is 0 and treatment is "none"?
## Example with two treatments
exdat <- structure(list(treatment = c("none", "none", "none", "none",
"none", "none", "trtA", "trtA", "trtA", "trtB", "trtB", "trtB",
"trtA", "trtA", "trtA", "trtB", "trtB", "trtB", "trtA", "trtA",
"trtA", "trtA", "trtB", "trtB", "trtB", "trtA", "trtA", "trtA",
"trtA", "trtB", "trtB", "trtB"), y = c(1.887069649, 2.2721258855,
1.9459101491, 1.8405496334, 2.2300144002, 2.2192034841, 2.9123506646,
2.8094026954, 1.974081026, 1.9169226122, 2.1041341543, 1.960094784,
2.7472709143, 2.9391619221, 2.6532419646, 2.4680995315, 2.0149030205,
2.7850112422, 2.2925347571, 2.3513752572, 3.1945831323, 2.6532419646,
3.3068867022, 2.9285235239, 2.5649493575, 2.9231615807, 3.0155349009,
2.9391619221, 2.7536607124, 2.9444389792, 2.8033603809, 2.9069010598
), intensity = c(0L, 0L, 0L, 0L, 0L, 0L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
7L, 7L, 7L, 7L, 7L)), row.names = c(1L, 2L, 3L, 4L, 5L, 6L, 7L,
8L, 9L, 11L, 12L, 13L, 15L, 16L, 17L, 19L, 20L, 21L, 23L, 24L,
25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, 36L), class = "data.frame")
library(emmeans)
model <- lm(y ~ intensity + treatment + intensity:treatment, data = exdat)
summary(model)
emt <- emtrends(model, ~ treatment, var = 'intensity', at = list(treatment = c('trtA', 'trtB')))
contrast(emt, 'pairwise')
In this case because there are only two treatments other than the "none" treatment, the contrast gives us the same t-ratio and p-value as we get from looking at the interaction coefficient in the model summary:
contrast estimate SE df t.ratio p.value
trtA - trtB -0.203 0.079 27 -2.567 0.0161
• Yes, I believe this yields a correct analysis. Mar 2 at 20:47
The "None" group sounds like a control group.
Thus, you would have something like the following:
Control
Treatment 1 at intensity levels $$1, 2, 3, 4, 5,\dots$$
Treatment 2 at intensity levels $$1, 2, 3, 4, 5,\dots$$
Treatment 3 at intensity levels $$1, 2, 3, 4, 5,\dots$$
$$\vdots$$
Then the "covariate" for the ANCOVA model would be the intensity, and you would multiply the covariate by each treatment indicator variable (the interactions that indicate differing slopes). Since there is no intensity for the control group, I would omit the covariate on its own. For the groups above:
$$Y_i = \beta_0 + \beta_1X_{treatment1} + \beta_2X_{treatment2} + \beta_3X_{treatment3} + \beta_4 X_{treatment1}X_{intensity}+ \beta_5 X_{treatment2}X_{intensity}+ \beta_6 X_{treatment3}X_{intensity} +\epsilon_i$$
A regular ANCOVA would have an $$X_{intensity}$$ on its own, but you correctly identify that not to be meaningful, so omit it, fit your regression model, and interpret the coefficients.
• You might want to interact the three treatments with polynomials or splines to investigate nonlinear behavior, but that does go beyond that standard ANCOVA setup.
– Dave
Mar 1 at 17:43
• Yes, I do have a situation with nonlinearities and a common zero level but I didn't want to add irrelevant detail to the question. I should be able to extend this to that situation pretty easily. Thanks for the sanity check! Mar 1 at 17:49 | 2022-08-10T13:45:30 | {
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https://datascience.stackexchange.com/questions/57953/what-is-the-purpose-of-standardization-in-machine-learning/57958 | # What is the purpose of standardization in machine learning?
I'm just getting started with learning about K-nearest neighbor and am having a hard time understanding why standardization is required. Reading through, I came across a section saying
When independent variables in training data are measured in different units, it is important to standardize variables before calculating distance. For example, if one variable is based on height in cms, and the other is based on weight in kgs then height will influence more on the distance calculation.
Since K nearest neighbor is just a comparison of distances apart, why does it matter if one of the variables has values of a larger range since it is what it is.
Considering 3 points A,B & C with x,y co-ordinates (x in cm, y in grams) A(2,2000), B(8,9000) and C(10,20000), the ranking of the points as distance from origin for example (or any other point), will be the same whether the y values are in grams,pounds, tonnes or any other combinations of units for both x and y so where's the need to standardise. Every example or QA i see brushes through with the same statement of 'one variable influencing the other' without a real example of how this might occur. Again, how does one know when this influence is too much as to call for standardization.
Also,what exactly does standardization do to the values? One of the formulas does it by Xs = (X-mean)/(max-min) Where does such a formula come from and what is it really doing? Hopefully someone can offer me a simplified explanation or give me a link to a site or book that explains this in simple terms for beginners.
• @sAguinaga Thanks that answers my last point. Im quite confused by the need to do it in the first place.
– West
Aug 21, 2019 at 11:55
• stats.stackexchange.com/questions/287425/… See this stack exchange question that shows the point quite well Aug 21, 2019 at 12:26
• A better answer is offered here and here z score lets you do standardization: z = x – x̄ / s, where x̄ (the sample mean) and s (the sample standard deviation). Aug 21, 2019 at 13:07
Considering 3 points A,B & C with x,y co-ordinates (x in cm, y in grams) A(2,2000), B(8,9000) and C(10,20000), the ranking of the points as distance from origin for example (or any other point), will be the same whether the y values are in grams,pounds, tonnes or any other combinations of units for both x and y so where's the need to standardise.
This is true for the example you provided, but not for euclidean distance between points in general. Look at this example:
def euclidian_distance(a, b):
return ((a[0] - b[0])**2 + (a[1] - b[1])**2)**0.5
a1 = 10 #10 grams
a2 = 10 #10 cm
b1 = 10 #10 gram
b2 = 100 #100 cm
c1 = 100 #100 gram
c2 = 10 #10 cm
# using (grams, cm)
A_g_cm = [a1, a2]
B_g_cm = [b1, b2]
C_g_cm = [c1, c2]
print('[g, cm] A-B:', euclidian_distance(A_g_cm, B_g_cm))
print('[g, cm] A-C:', euclidian_distance(A_g_cm, C_g_cm))
# using (kg, cm)
A_kg_cm = [a1/1000, a2]
B_kg_cm = [b1/1000, b2]
C_kg_cm = [c1/1000, c2]
print('[kg, cm] A-B:', euclidian_distance(A_kg_cm, B_kg_cm))
print('[kg, cm] A-C:', euclidian_distance(A_kg_cm, C_kg_cm))
# using (grams, m)
A_g_m = [a1, a2/100]
B_g_m = [b1, b2/100]
C_g_m = [c1, c2/100]
print('[g, m] A-B:', euclidian_distance(A_g_m, B_g_m))
print('[g, m] A-C:', euclidian_distance(A_g_m, C_g_m))
# using (kilo, m)
A_kg_m = [a1/1000, a2/100]
B_kg_m = [b1/1000, b2/100]
C_kg_m = [c1/1000, c2/100]
print('[kg, m] A-B:', euclidian_distance(A_kg_m, B_kg_m))
print('[kg, m] A-C:', euclidian_distance(A_kg_m, C_kg_m))
Output:
[g, cm] A-B: 90.0
[g, cm] A-C: 90.0
[kg, cm] A-B: 90.0
[kg, cm] A-C: 0.09000000000000001
[g, m] A-B: 0.9
[g, m] A-C: 90.0
[kg, m] A-B: 0.9
[kg, m] A-C: 0.09000000000000001
Here you can clearly see that the choice of units influence the ranking, thus the need for standardization.
• Exactly the sort of example I've been looking for! Nice and clear thanks a lot mate.,really struggled for a while with this
– West
Aug 21, 2019 at 15:16
There are several reasons for the standardization, the relevant reasons for the KNN algorithm important since the algorithm is based on calculating the distance between neighbours. Let's assume that the distance measure that we are using is the euclidian distance and we are having 2 features x in grams and y in kilometres.
If we'll take a look at the distance (d) formula between 2 points (p,q) with 2 features (1,2):
We may notice, that the term (q1-p1) is in grams and (q2-p2) is in kms. There are several issues with this:
1. we are summing deltas with different units
2. we are summing deltas with different scale
3. we are summing deltas of features with different distribution/variance
For example, the first feature has a small magnitude of delta (between 2 points) which is meaningful in its scale and variance, the second feature has a delta with a large magnitude but in its scale and variance it is negligible (1km while reducing 200k km with 199k km).
We can agree that the points are far in the feature 1 space and are close to each other in the feature 2 space, but, our distance metric might not detect that.
The Standardization of the features before applying the distance metric would solve these issues.
Another goog reason is that standarization of features may serve as a way of preconditioning the problem.
Steepest descent optimization algorithms does not work well on ill-conditioned cases, for instance, when the loss function shows ravines and ridges. In such cases, vanilla Gradient Descent may oscillate in the direction of the largest gradient, deviating from the right direction to the optimum point, and thus, convergence may be very slow. This issue can be addressed with more sophisticated optimizes (RMSprop, Adam...) but standardization of features may precondition the scenario (smoothing the ravines of the loss function, for example)
Considering 3 points A,B & C with x,y co-ordinates (x in cm, y in grams) A(2,2000), B(8,9000) and C(10,20000), the ranking of the points as distance from origin for example (or any other point), will be the same whether the y values are in grams,pounds, tonnes or any other combinations of units for both x and y so where's the need to standardise.
I'm not sure this is entirely right:
import pandas as pd
import numpy as np
from sklearn.metrics.pairwise import euclidean_distances
from sklearn.preprocessing import StandardScaler
df = df.drop('species', axis=1)
ss = StandardScaler()
ed1 = euclidean_distances(df)
ed2 = euclidean_distances(ss.fit_transform(df))
ed1_max = np.array([np.argmax(i) for i in ed1])
ed2_max = np.array([np.argmax(i) for i in ed2])
print(np.where(ed1_max == ed2_max, 1, 0).mean())
So for each sample in the iris dataset, this calculates the sample that's furthest away in euclidean distance for both the scaled and unscaled data, and then calculates the percentage of samples where they match. The output is 0.33. So scaling definitely affects the ranking of distances between samples.
So the problem is unscaled data tends to lead to the features with a larger scale drowning out the vote of the other features in determining distance to other samples. Scaling prevents that. | 2022-06-25T14:52:35 | {
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http://mathhelpforum.com/pre-calculus/46589-inverse-trigonometric-functions.html | # Math Help - Inverse Trigonometric Functions
1. ## Inverse Trigonometric Functions
Establishing an Identity Involving Inverse Trigonometric Functions.
(1) Show that tan[arcsin(v)] = v/(sqrt{1 - v^2})
(2) Show that sin[arccos(v)] = sqrt{1 - v^2}
2. #1: Recall that $\tan \theta = \frac{\sin \theta}{\cos \theta}$
So: $\tan \left(\arcsin v\right) = \frac{\sin \left(\arcsin v\right)}{\cos \left(\arcsin v\right)} = \frac{v}{\sqrt{1 - \big[\sin \left(\arcsin v\right)\big]^2}} = ...$
#2: Similar to what was done in the denominator of #1, recall that $\cos^2 x = 1 - \sin^2x$
So: $\sin \left(\arccos v\right) = \sqrt{1 - \big[\cos \left(\arccos v\right)\big]^2}$
3. Originally Posted by magentarita
Establishing an Identity Involving Inverse Trigonometric Functions.
(1) Show that tan[arcsin(v)] = v/(sqrt{1 - v^2})
(2) Show that sin[arccos(v)] = sqrt{1 - v^2}
#1: If we were to plot a triangle that would represent $\sin^{-1}v$, the leg opposite the angle would have a value of v, and the hypotenuse would have a value of one.
To solve for the side adjacent to the angle, apply pythagoras' theorem:
$v^2+b^2=1\implies b=\sqrt{1-v^2}$
Now, since $\tan\vartheta=\frac{opp}{adj}$, and $opp=v~and~adj=\sqrt{1-v^2}$ we see then that $\tan\vartheta=\tan\left(\sin^{-1}v\right)=\frac{v}{\sqrt{1-v^2}}$
$\mathbb{Q.E.D.}$
#2: If we were to plot a triangle that would represent $\cos^{-1}v$, the leg adjacent to the angle would have a value of v, and the hypotenuse would have a value of one.
To solve for the side opposite to the angle, apply pythagoras' theorem:
$v^2+b^2=1\implies b=\sqrt{1-v^2}$
Now, since $\sin\vartheta=\frac{opp}{hyp}$, and $opp=\sqrt{1-v^2}~and~hyp=1$ we see then that $\sin\vartheta=\sin\left(\cos^{-1}v\right)=\sqrt{1-v^2}$
$\mathbb{Q.E.D.}$
Does this make sense??
--Chris
4. ## Thanks
Thank you both. Of course, Chris' method is much easier to follow.
5. Hello, magentarita!
Another approach . . .
Establishing an identity involving inverse trigonometric functions.
(1) Show that: . $\tan[\arcsin(v)] \:= \:\frac{v}{\sqrt{1 - v^2}}$
Recall that an inverse trig expression is an angle.
We have: . $\tan\underbrace{[\arcsin(v)]}_{\text{some angle }\theta}$
That is: . $\theta$ is an angle whose sine is $v.$
We have: . $\sin\theta \:=\:\frac{v}{1} \:=\:\frac{opp}{hyp}$
$\theta$ is in a right triangle with: . $opp = v,\;hyp = 1$
. . and we want: . $\tan\theta$
Code:
*
1 * |
* | v
* θ |
* - - - - - - - *
____
√1-v²
Using Pythagorus, we find that . $adj = \sqrt{1-v^2}$
Therefore: . $\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{v}{\sqrt{1-v^2}}$
Ahh, Chris beat me to it . . .
.
6. ## soroban
Originally Posted by Soroban
Hello, magentarita!
Another approach . . .
Recall that an inverse trig expression is an angle.
We have: . $\tan\underbrace{[\arcsin(v)]}_{\text{some angle }\theta}$
That is: . $\theta$ is an angle whose sine is $v.$
We have: . $\sin\theta \:=\:\frac{v}{1} \:=\:\frac{opp}{hyp}$
$\theta$ is in a right triangle with: . $opp = v,\;hyp = 1$
. . and we want: . $\tan\theta$
Code:
*
1 * |
* | v
* θ |
* - - - - - - - *
____
√1-v²
Using Pythagorus, we find that . $adj = \sqrt{1-v^2}$
Therefore: . $\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{v}{\sqrt{1-v^2}}$
Ahh, Chris beat me to it . . .
.
Thank you Soroban. | 2015-03-02T05:08:26 | {
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https://math.stackexchange.com/questions/3130380/lotus-to-determine-eex | # LOTUS to determine $E[e^X]$
Let $$X$$~$$N(\mu,\sigma^2)$$ and set $$Y=e^X$$ which means that $$Y$$~$$logN(\mu,\sigma)$$.
Show that $$E[Y]=e^{\mu+\sigma^2/2}$$
My thoughts:
My idea is to use LOTUS:
$$E[Y]=E[e^X]=\int_{-\infty}^\infty \! e^x \frac{1}{\sqrt{2\pi}*\sigma}e^{-(\mu-x)^2/2\sigma^2} \, \mathrm{d}x$$
However it seems to be an unsolveable integral and I don't know what else to use. Can someone point me in the right direction?
• Complete the square in the exponent, change variables, and then use the well-known formula for $\int_{-\infty}^\infty e^{-x^2} dx$. – Hans Engler Feb 28 at 16:39
• I don't quite follow, can you elaborate? After completing the square I get this: $\int_{-\infty}^\infty \! e^x \frac{1}{\sqrt{2\pi}*\sigma}e^{(-\mu^2-x^2+2\mu x)/2\sigma^2} \, \mathrm{d}x$ – CruZ Feb 28 at 17:06
• You haven't completed the square. The exponent should be of the form $A^2 + B$ where $A$ is a linear function of $x$ and $B$ is a constant. – Hans Engler Feb 28 at 17:08
• So something like this? i.imgur.com/EiTLAdr.jpg But I still don't quite see what you want me to do next... – CruZ Feb 28 at 18:10
Complete the square in the exponent of the integrand, as follows: $$x -(x-\mu)^2/(2\sigma^2) = (2 \sigma^2 x - (x - \mu)^2)/(2 \sigma^2)\\ = -(x - (\mu + \sigma^2))^2/(2 \sigma^2) + (2\mu \sigma^2 + \sigma^4)/(2 \sigma^2) \\ = -(x - (\mu + \sigma^2))^2/(2 \sigma^2) + \mu + \sigma^2/2$$ Therefore $$\frac{1}{\sqrt{2 \pi} \sigma }\int_{-\infty}^\infty e^x e^{-(x-\mu)^2/(2\sigma^2)} dx = \frac{1}{\sqrt{2 \pi} \sigma } \int_{-\infty}^\infty e^{-(x - (\mu + \sigma^2))^2/(2 \sigma^2) + \mu + \sigma^2/2} dx \\ = e^{\mu + \sigma^2/2} \times \frac{1}{\sqrt{2 \pi} \sigma }\int_{-\infty}^\infty e^{-(x - (\mu + \sigma^2))^2/(2 \sigma^2)} dx = e^{\mu + \sigma^2/2}$$ since the term to the right of the $$\times$$ symbol is the integral of a probability density and thus equals 1. | 2019-05-21T14:44:10 | {
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http://mathhelpforum.com/number-theory/9236-please-help-me-these-two-questions-pell-equation.html | Find a solution in positive integers to :
a) x^2 -41y^2 = -1
b)x^2 - 41y^2 =1
Could you please teach me how to solve these two questions? Thank you very much.
2. Originally Posted by Jenny20
Find a solution in positive integers to :
a) x^2 -41y^2 = -1
b)x^2 - 41y^2 =1
Could you please teach me how to solve these two questions? Thank you very much.
This the variant Pell's equation and Pell's equation. The solutions can be
derived from the continued fraction expansion of sqrt(41).
If a=[a_0: a_1,...] is the CF expansion of sqrt(41) and p_n/q_n is the n-th
convergent of a, then the solutions of a) and b) will be among {(p_n,q_n), n=0,1, ..}.
The following calculations show the solutions:
Code:
>{cf,r}=CF(sqrt(41),8);cf
Column 1 to 6:
6 2 2 12 2 2
Column 7 to 8:
12 2
>
>{p,q,xx}=cnvrgntCT(cf);p,q
Column 1 to 6:
6 13 32 397 826 2049
Column 7 to 8:
25414 52877
Column 1 to 6:
1 2 5 62 129 320
Column 7 to 8:
3969 8258
>
>
>p^2-41*q^2
Column 1 to 6:
-5 5 -1 5 -5 1
Column 7 to 8:
-5 5
>
>
or in tabular form:
Code:
n a p q p^2-41q^2
0 6 6 1 -5
1 2 13 2 5
2 2 32 5 -1
3 12 397 62 5
4 2 826 129 -5
5 2 2049 320 1
6 12 25414 3969 -5
7 2 52877 8258 5
So we see that x=32, y=5 is a solution to a), and that x=2049 y=320 is
a solution to b).
RonL
3. Originally Posted by Jenny20
Find a solution in positive integers to :
a) x^2 -41y^2 = -1
What CaptainBlank said hold true for the second problem.
There is however a problem with this one.
It is not a Pell equation.
We see that 41 is a prime that is congruent to 1 modulo 4 thus it does have a solution. (Otherwise it leads to a contradiction in a quadradic residue).
A result by Brouncker is that if,
X and Y satisfy,
X^2-dY^2=-1
then,
x=2dY^2-1
y=2XY
Satisfy,
x^2-dy^2=1
Thus, you solve,
x^2-41y^2=1
And then you can solve for X and Y.
From,
x=2(41)Y^2-1=82Y^2-1
y=2XY
4. ic. thank you very much Perfecthacker.
5. Originally Posted by ThePerfectHacker
What CaptainBlank said hold true for the second problem.
There is however a problem with this one.
It is not a Pell equation.
What I said was that the first was the variant Pell's equation. Which if it has
solutions are also derived from the CF expansion of sqrt(41).
RonL
6. "What I said was that the first was the variant Pell's equation. Which if it has
solutions are also derived from the CF expansion of sqrt(41)."
========================
When I checked by plugging in the answer from Captainblack's table, I found the equation is correct.
x^2 -41y =-1
32^2-41*5^2 = -1
x^2-41y^2 =1
2049^2 - 41*320^2=1
So I think Captianblack's reply to me - original calculation is right! | 2017-01-24T14:42:38 | {
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https://math.stackexchange.com/questions/605131/what-does-this-function-mean | # What does this function mean?
$$f(x) = \frac{x}{e^{x^2}}$$ Differentiate $f(x)$.
How should the above function be interpreted? Is the function equivalent to:
a)$$f(x) = \frac{x}{e^{x^2}} = \frac{x}{{(e^x)^2}} = \frac{x}{e^{2x}}$$
or
b) $$f(x) = \frac{x}{e^{x^2}} = \frac{x}{{e^{(x^2)}}} ≠ \frac{x}{e^{2x}}$$
Thanks!
• It is interpreted as (b).
– LASV
Dec 13, 2013 at 5:09
• @dfg:Maybe helps to use some small, simple values like $x=1$. Dec 13, 2013 at 5:16
$a^{b^c}$ should be interpreted as $a^{(b^c)}$. Because, as you point out, $(a^b)^c = a^{bc}$. This is even true in most programming languages, which specify that a^b^c should be evaluated from right to left, unlike expressions like a-b+c, which should be evaluated from left to right.
• Out of curiosity, what programming languages use ^ for exponentiation? Dec 13, 2013 at 5:14
• BASIC, MATLAB, and Mathematica are examples. Dec 13, 2013 at 5:16
• Ah. I use those so infrequently I sometimes forget there are languages where it isn't bitwise XOR. Dec 13, 2013 at 5:27
• But I just tried out 2^3^4 in OCTAVE, which is meant to emulate MATLAB, and it did the computation left to right. But Mathematica did it right to left. I am pretty sure that BASIC does it right to left, because I remember reading about it in a BASIC manual, and wondering why. Dec 13, 2013 at 5:27
• FORTRAN uses **. It does it right to left: folk.uio.no/hpl/scripting/doc/f77/tutorial/expres.html Dec 13, 2013 at 5:28
The expression
$f(x) = x / e^{x^2} \tag{1}$
should be interpreted in the same way we would interpret $x / exp(x^2)$;that is,
$f(x) = x e^{-x^2} = x \exp (-x^2). \tag{2}$
The tip-off here is that the exponent $2$ of $x$ in the expression $x^2$ is in fact written as a superscript of $x$; in item (a) of the question, $(e^x)^2 = (e^x)(e^x) = e^{2x}$; but $e^{x^2}$ should be thought of as $e^{xx} = (e^x)^x$; as Stephen Montgomery points out in his answer, the evaluation of the exponents is right to left.
Having hopefully clarified the issue, we have
$f'(x) = (x e^{-x^2})' = e^{-x^2} + x e^{-x^2} (-2x) \tag{3}$
using the Leibniz product rule and the formula $d / dx (e^u) = e^u (du / dx)$. After the simple algebra,
$f'(x) = e^{-x^2}(1 - 2x^2). \tag{4}$
Hope this helps. Cheerio,
and as always,
Fiat Lux!!! | 2022-08-11T08:47:31 | {
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https://math.stackexchange.com/questions/2635731/let-x-and-y-be-independent-and-identically-distributed-random-variables-with | # Let $X$ and $Y$ be independent and identically distributed random variables with moment generating function then $E(\dfrac{e^{tX}}{e^{tY}})$
Let $X$ and $Y$ be independent and identically distributed random variables with moment generating function $M(t)=E(e^{tX});\ \ -\infty<t<\infty$ then $E(\dfrac{e^{tX}}{e^{tY}})$ equals ?
$(A)=M(t)M(-t)$
$(B)=1$
$(C)=(M(t))^2$
$(D)=\frac{M(t)}{M(-t)}$
My input: Since its given that random variables are i.i.d so $E(\dfrac{\require{\cancel}\cancel{e^{tX}}}{\cancel{e^{tY}}})=E(1)=1$
can I do that?
• What happens e.g. for $X:=1$ (constant random variable) and $Y:=2$ (also constant random variable)? Does the expression equal $1$? – saz Feb 4 '18 at 16:38
• You assumed $X=Y$ with probability $1$ when you did the cancellation. That is not necessarily true. – StubbornAtom Feb 4 '18 at 16:42
• @saz Oh yes i was thinking totally wrong. I took a case when they both take same probability. Silly me. – Daman Feb 4 '18 at 16:42
• @StubbornAtom i am showing new work wait. – Daman Feb 4 '18 at 16:42
• Another example: $X,Y \sim {\cal N}(0,1)$ iid. – GNUSupporter 8964民主女神 地下教會 Feb 4 '18 at 17:28
$X,Y$ are independent of each other, so $e^{tX}$ and $e^{tY}$ are independent of each other. This gives $M(t) = E[e^{tX}] = E[e^{tY}]$
Moreover, $M(t) = E[e^{tX}]$ and $M(-t) = E[e^{-tY}]$ are independent of each other, this gives $$E\left[\frac{e^{tX}}{e^{tY}}\right] = E[e^{tX} \cdot e^{-tY}] = E[e^{tX}] \cdot E[e^{-tY}] = M(t) M(-t).$$
No you cannot do that. What you can do: $$E\left(\frac{e^{t X}}{e^{t Y}}\right)=E\left(e^{t X} e^{-t Y}\right) = E(e^{t X}) E(e^{-t Y}) = M(t) M(-t)$$ where the second step is due to independence. As the random variables are identically distributed $M_X(t) = M_Y(t) =M(t)$.
Observe that since $X$ and $Y$ are independent, it follows that $e^{tX}$ and $e^{-tY}$ are independent for $t\in\mathbb{R}$. Thus $$Ee^{tX}e^{-tY}=Ee^{tX}E{e^{-tY}}=M_{X}(t)M_{Y}(-t)=M_{X}(t)M_X(-t)$$ since $X, Y$ are equal in distribution. | 2019-10-22T14:01:23 | {
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https://math.stackexchange.com/questions/3971829/the-lebesgue-dominated-convergence-theorem-example | # The Lebesgue Dominated Convergence Theorem: Example.
The following question is from Real Analysis by Royden (4th Edition). More precisely, it is problem 36 of chapter 4. Further, the problem follows right after the section over The Lebesgue Dominated Convergence Theorem, hence, that is what should be used. Below is the problem followed by my attempted solution which I have a question about.
Problem:
Let $$f$$ be a real-valued function of two variables $$\left(x,y\right)$$ that is defined on the following unit square $$Q:=\{\left(x,y\right): 0\leq x\leq 1, 0\leq y\leq 1\}$$ and is a measurable function of $$x$$ for each fixed value of $$y$$. For each $$\left(x,y\right)\in Q$$ let the partial derivative a $$\partial f/\partial y$$ exist. Suppose there is a function $$g$$ that is integrable over $$[0,1]$$ and such that,
$$\left\vert\frac{\partial f}{\partial y}\left(x,y\right)\right\vert \leq g(x) \text{ for all } \left(x,y\right)\in Q.\tag{*}$$
Prove that
$$\frac{d}{dy}\left[\int_0^1 f(x,y)dx\right] = \int_0^1\frac{\partial f}{\partial y}(x,y)dx, \text{ for all } y\in[0,1].$$
(My) Proof:
Fix $$y\in[0,1]$$ and consider the sequence of functions $$f_n\colon[0,1]\to\mathbb{R}$$ defined by,
$$f_n(x):=\frac{f(x,y+1/n) - f(x,y)}{1/n}.$$
Then, $$\{f_n\}$$ is a sequence of measurable functions ($$f$$ is a measurable function of $$x$$ for each fixed value of $$y$$) dominated by $$g(x)$$ that converge pointwise almost everywhere to $$\partial f/\partial y$$ on $$[0,1]$$ (for each $$(x,y)\in Q$$ the partial derivative $$\partial f/\partial y$$ exists). Therefore, by The Lebesgue Dominated Convergence Theorem,
$$\lim_{n\to\infty}\int_0^1f_n=\int_0^1f$$ $$\lim_{n\to\infty}\int_0^1\frac{f(x,y+1/n) - f(x,y)}{1/n}dx = \int_0^1\frac{\partial f}{\partial y}(x,y)dx.\tag{1}$$
Notice,
$$\lim_{n\to\infty}\int_0^1\frac{f(x,y+1/n) - f(x,y)}{1/n}dx = \lim_{h\to 0}\frac{\int_0^1f(x,y+h)dx-\int_0^1f(x,y)dx}{h}= \frac{d}{dy}\left[\int_0^1f(x,y)dx\right].\tag{2}$$
Thus, combining (1) and (2),
$$\frac{d}{dy}\left[\int_0^1f(x,y)dx\right] = \int_0^1\frac{\partial f}{\partial y}(x,y)dx,$$
which was to be shown.
My main concern with my proof is that, I am not sure how the sequence $$\{f_n\}$$ are dominated by $$g(x)$$. I know by (*) that the limit of the sequence $$\{f_n\}$$ is bounded by $$g(x)$$. Also, if there are any mistakes in the proof, please let me know!
Edit:
Does the boundedness of the $$\{f_n\}$$ come from the fact that every convergent sequence is a bounded sequence?
• to avoid confusing, you shouldn't write $\lim_{n\to\infty}\int_0^1f_n=\int_0^1f$ as $f$ is the initial function. – Leon Jan 4 at 9:40
• Is $f_n$ only "almost everywhere" converging? If so, can someone explain why? – Rem Jan 4 at 10:07
• @Rem Let fixed $y$. For every $x$, we have $\lim_n f_n(x)=\partial_y f(x,y)$. – Leon Jan 4 at 10:11
• @Leon So we make no assumption that it is only "almost everywhere" pointwise convergent, right? I understand that we don't need more than that for the proof though. – Rem Jan 4 at 10:19
• Yes, i think so. – Leon Jan 4 at 11:32
Consider $$y \in [0,1]$$ and any sequence $$y_n \in [0,1]$$ such that $$y_n \to y$$. By the mean value theorem, there exists $$\xi_{x,n}$$ between $$y$$ and $$y_n$$ such that
$$|f_n(x)| = \left|\frac{f(x,y_n) - f(x,y)}{y_n - y} \right| = \left|\frac{\partial f}{\partial y}(x,\xi_{x,n})\right| \leqslant g(x)$$
We have
$$\lim_{n \to \infty}\frac{f(x,y_n) - f(x,y)}{y_n - y} = \frac{\partial f}{\partial y}(x,y),$$
and by the dominated convergence theorem
$$\lim_{n \to \infty}\int_0^1\frac{f(x,y_n) - f(x,y)}{y_n - y}\, dx = \int_0^1\lim_{n \to \infty}\frac{f(x,y_n) - f(x,y)}{y_n - y}\, dx = \int_0^1 \frac{\partial f}{\partial y}(x,y) \, dx$$
• This makes sense, but is it wrong to say "the boundedness of the $\{f_n\}$ come from the fact that every convergent sequence is a bounded sequence"? – user853982 Jan 3 at 22:51
• The principle you seem to be invoking is that a sequence $(a_n)$ in $\mathbb{R}$ that is convergent is bounded -- meaning that there exists $B>0$ such that $|a_n| \leqslant B$ for all $n$. Here we can say that convergence of $f_n(x) \to f(x)$ implies that $f_n(x)$ is bounded in the pointwise sense. – RRL Jan 3 at 22:59
• ... but what you really have is $f_n(x,y)$ depending on both variables. So this all gets taken care of nicely with MVT. – RRL Jan 3 at 23:02
• So, if we can say that the convergence of $f_n \to f$ implies that $f_n(x)$ is bounded in the pointwise sense, why does this not work to conclude The Lebesgue Dominated Convergence Theorem? But yes, I ultimately agree the MVT works very well here. – user853982 Jan 3 at 23:04
• If all I know is that $f_n(x,y)$ converges pointwise with respect to $x,y$ as $n \to \infty$, then I can conclude that there exists $B(x,y)>0$ such that $|f_n(x,y)| \leqslant B(x,y)$ for all $n \in \mathbb{N}$ with $x,y$ fixed. I can't say more about $B$ at this point -- for example, that there is a uniform bound for all $y$. – RRL Jan 3 at 23:08
Another way to prove that $$f_n$$ is bounded by $$g$$ is \begin{align} |f_n(x)| = & \bigg \vert \frac{ f(x, y+1/n)-f(x,y)}{1/n}\bigg \vert\\ = & \bigg \vert \frac{ \int_y^{y+1/n} \partial_y f(x,y) dy }{1/n}\bigg \vert\\ \leq & \frac{ \int_y^{y+1/n} |\partial_y f(x,y)| dy }{1/n}\\ \leq & g(x). \end{align}
• This is a cool way, thanks for presenting. The last inequality is bugging me though. Mind explaining it? I know that $|\partial_{y}f(x,y)| \leq g(x)$. – user853982 Jan 4 at 14:47
• Yes, thank you! the last one is because $\int_y^{y+1/n} dy=(y+1/n)-y=1/n$. – Leon Jan 4 at 16:42
• Ahhh, I see. The $g(x)$ is not a function of $y$. – user853982 Jan 4 at 16:54 | 2021-05-18T21:31:34 | {
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http://math.stackexchange.com/questions/823659/sum-of-consecutive-n-integers | # Sum of consecutive n integers
It is well-known that $1+2+3+4+...+n= \frac{n(n+1)}{2}$, this formula can be found using simple arithmetic progression. But the sum also can be found by using ${n+1 \choose 2}$, which is n+1 choose 2. Can anyone explain the relationship between the binomial coefficient and the sum of consecutive integers? I found the binomial coefficient formula at wikipedia: http://en.wikipedia.org/wiki/Triangular_number
-
Not your topic, but it might interest you that this can generalized: $\sum_{k=r}^{n}\binom{k}{r}=\binom{n+1}{r+1}$. It can be proved quite easily by induction. – drhab Jun 7 at 7:54
There is a proof without words of this fact. – Marc van Leeuwen Jun 7 at 18:32
I think this is a better way that justifies the name Triangular Number. Consider the (right) triangular arrangement with $n$ rows and columns, the rows indexed by $i = 1, 2, \ldots n$, and the columns indexed by $0, 1, \ldots n - 1$: $$\begin{matrix}1 \\ 2 \\ 3 \\ \vdots \\ n\\ \ \end{matrix}\left| \begin{matrix} *&\\ *& *\\ *& *& *&\\ \vdots& \vdots& \vdots& \ddots\\ *& *& *& \ldots & *\\ \hline 0 & 1 & 2 & \ldots & n-1 \end{matrix}\right.$$
The total number of objects is obviously $1 + 2 + \ldots + n$ (the sum of the number of objects in each row). However, observe that each object corresponds to a unique pair of the form $(i, j)$ with $j < i$, where $i$ is the row and $j$ is the column occupied by the object. Thus the number of objects is the number of such pairs, which is the number of distinct pairs of the $n + 1$ numbers $0, 1, \ldots, n + 1$, and that is $\binom {n + 1} 2$.
-
is there some minor mistake in the image? the last row is not properly aligned with the above columns – Dave Clifford Jun 7 at 8:33
@DaveClifford Thanks, I fixed it. – M. Vinay Jun 7 at 8:36
No, I should be the person who should thank you :) Nice interpretation! – Dave Clifford Jun 7 at 8:38
still not entirely correct. the last row should be 0 1 2 ... n-1 – Egon Jun 7 at 22:16
ya, but that is pretty trivial, everyone should be able to correct it themselves – Dave Clifford Jun 8 at 1:40
If all $n + 1$ people at a party shake hands with each other such that every two people shake hands exactly once, how many handshakes are there?
Let's count the number of handshakes in two different ways:
1. The first person shakes hands with the remaining $n$ people. The second person shook hands with the first already, so he shakes hands with the remaining $n - 1$ people. And continuing in this manner, the last-but-one person shakes hands with the last $1$ person. The last person shakes hands with none. So the number of handshakes is $n + (n - 1) + \ldots + 1$ = $1 + 2 + \ldots + n$.
2. The number of handshakes is the number of ways of selecting a pair of people from $n + 1$ people (because every possible pair shakes hands exactly once), that is $\binom {n + 1} 2$.
Since the number of handshakes is the same no matter how you count it (correctly) $$1 + 2 + \ldots n = \binom {n + 1} 2$$
-
This is basically the handshaking lemma or degree sum formula (en.wikipedia.org/wiki/Handshaking_lemma) applied to the complete graph on $n$ vertices. – M. Vinay Jun 7 at 7:51
This is very cool! Great post! I like your view on counting more than mine, more intuitive! – DanZimm Jun 7 at 7:53
@DanZimm Thanks. Can't help thinking that way, I'm a graph theorist :P – M. Vinay Jun 7 at 7:54
And a very wise one at that, I've seen you answering posts recently and you seem to have a vast spectrum of knowledge :D – DanZimm Jun 7 at 7:56
When you look at $n+1$ objects then lets try to manually count how many ways we can choose $2$ objects.
First we can count the $n$ pairs of consecutive objects, then then $n-1$ pairs that have one object in between and so on until we have the $1$ pair with $n-1$ objects in between. Thus here we can see that the number of ways to choose two objects is $n + n-1 + n-2 + \cdots + 2 + 1$ which is exactly what $\sum_{k=1}^n k$ is. Thus we come to the conclusion that $\sum_{k=1}^n k = \binom{n+1}{k}$.
-
HINT:
If you start choosing $2$ elements from $n+1,$ say $a_i(1\le I\le n+1)$
first choose $1$ element (say $a_0$), the other can be chosen from the rest $n:\{a_1,a_2,\cdots,a_{n-1},a_n\}$ elements
for the next $a_1$ element, the other can be chosen from the rest $n-2:\{a_2,\cdots,a_{n-1},a_n\}$ elements as $a_0a_1$ has already been chosen as the order immaterial here
So on
-
So the relationship is purely algebraic? I thought there is some relationship between the summation and the counting. – Dave Clifford Jun 7 at 7:46
@DaveClifford, I believe so – lab bhattacharjee Jun 7 at 7:47 | 2014-10-24T12:47:40 | {
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http://math.stackexchange.com/questions/699160/expressing-polar-complex-numbers-in-cartesian-form | # Expressing polar complex numbers in cartesian form
I need to express $z = 4e^{-i\pi/3}$ in the form of $x + yi$ and represent it on the Argand diagram.
I think that $4 = \sqrt{x^{2} + y^{2}}$ and that $\tan (\pi/3) = y/x$ but I haven't been able to do anything useful with this information...
Is this solvable via simultaneous equations?
Thanks!
-
In view of the answers posted, you should realize that it is MUCH easier computationally to convert from polar to rectangular form than to go the other way: $re^{i\theta} = (r\cos\theta) + i(r\sin\theta)$. You can do it in your head. – MPW Mar 4 '14 at 18:19
Using Euler Formula, $$e^{-\dfrac{i\pi}3}=\cos\left(-\frac\pi3\right)+\sin\left(-\frac\pi3\right)=\cos\left(\frac\pi3\right)-i\sin\left(\frac\pi3\right)=\frac12-\frac{\sqrt3}2i$$
Reference: the definition of $\displaystyle\arctan\frac yx$
-
In my googling before posting this question, I did come across this method but avoided it since I haven't used it before. Seeing you doing it shows it's rather simple but I have a question - What happened to the 4 in front of the e? – Jacobadtr Mar 4 '14 at 17:30
@Jacobadtr, just multiply the modulus$(4)$ – lab bhattacharjee Mar 4 '14 at 17:31
Hint
Use the Euler's formula $$e^{i\theta}=\cos\theta+i\sin\theta$$
- | 2016-05-03T12:53:33 | {
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https://math.stackexchange.com/questions/1641474/method-of-characteristics-eliminating-variables | # Method of characteristics - eliminating variables
I am trying to follow a guide for the method of characteristics; quoting the first example:
We use the method of characteristics to solve the problem
$2u_x - u_y = 0, \;\; u(x, 0) = f(x)$
(...) we can easily solve [the characteristic equations] to get:
$x = 2s + x_0 \;\;\; y= -s + y_0 \;\;\; u = u_0$
Up to here I have no problem. However, the guide then substitutes in the initial condition and says this:
we now eliminate $x_0$ and $s$ to find that:
$y = -s \;\;\; \Rightarrow \;\;\; u = f(x_0) = f(x - 2s) = f(x + 2y)$
My question is: where did $y_0$ and $u_0$ go?
• More simply, the method shows that for every $(x_0,y_0,s)$, $$u(x_0+2s,y_0-s)=u(x_0,y_0).$$ For every $(x,y)$, choosing $(x_0,y_0,s)=(x+2y,0,-y)$ above, one gets $$u(x,y)=u(x+2y,0).$$ Rename $u(\ ,0)$ as $f$ and you are done. – Did Feb 5 '16 at 7:35
• @Did I.. don't quite follow. How does the method show that? – lvc Feb 5 '16 at 7:46
• First line simply rephrases the fact that "$u$ remains constant on the curve $(x(s),y(s))$", for $x(s)=2s+x_0$, $y(s)=-s+y_0$, that is, that $u(x(s),y(s))=u(x(0),y(0))$ for every $s$. Second line is self-explanatory. What remains? – Did Feb 5 '16 at 7:49
• @Did I think there's some deep background here that I've missed - "$u$ remains constant on the curve $(x(s), y(s))$" isn't a fact that I recognise. Where does it come from? – lvc Feb 5 '16 at 7:57
• Lines 3-5 of the text you linked to. – Did Feb 5 '16 at 7:59
You have characteristics given as functions $x(s)$ and $y(s)$ which depend on parameter $s$:
\begin{align} x = x(s) &= x_0 + 2s && \implies& x_0 &= x - 2s \label{1}\tag{1} \\ y = y(s) &= y_0 - s && \implies& y_0 &= y + s \label{2}\tag{2} \end{align}
The initial condition $u\left(x_0, 0\right) = f\left(x_0\right)$ is given for $y_0=0$, so in equation $\eqref{2}$ we set $y_0$ to be equal $0$:
$$0 = y_0 = y + s \qquad\;\implies \quad\bbox[5pt, border: 1.5pt solid #DD0712]{y = -s^{\,\!}} \label{3}\tag{3}$$
But then we can write general solutions as $$u\left(x_0, y_0\right) = u\left(x_0, 0\right) = f\left(x_0\right) = f\left(x-2s\right) \overset{\eqref{3}}{=} f\left(x+2s\right)$$
Now, one can ask why do we say that $u\left(x_0, y_0\right)$ is the general solution? Well, if you recall the definition of characteristics, which are the
$\;\big(\cdots\big)\;$ parametric curves along which solution $u\left(x, y\right)$ remains constant,
you will see that $u\left(x, y\right) = u\left(x_0, y_0\right)$ as long as $(x,y)$ and $\left(x_0,y_0\right)$ lie on the same curve $\eqref{1}$–$\eqref{2}$. Then we choose $y_0$ to be equal $0$ in order to incorporate initial condition $u(x,0) = f(x)$.
The characteristic equations are : $$\frac{dx}{2}=\frac{dy}{-1}=\frac{du}{0}$$ which leads to the equations of chararacteristics : $$\begin{cases}u=c_1\\x+2y=c_2 \end{cases}$$ The general solution on implicit form with is : $$\Phi\left(u\:,\:x+2y\right)=0$$ where $\Phi$ is any derivable function of two variables.
This is equivalent to : $$u=F(x+2y)$$ where $F$ is any derivable function.
The bounding condition leads to : $$u(x,0)=f(x)=F(x+0)=F(x)$$ Hense $F=f$ is now determined and the solution is : $$u(x,y)=F(x+2y)=f(x+2y)$$ | 2019-08-22T17:34:12 | {
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https://math.stackexchange.com/questions/1004572/global-max-min-of-fx-y-yx-2xy | Global max./min. of $f(x,y) = y+x-2xy$
Find maximum and minimum of $f(x,y) = y+x-2xy$ restricted to the interior and border of $R = \{(x,y) \in \mathbb{R}^2 : |x| \geq \frac{1}{2}\}$.
I started by looking for critical points, getting:
• $f_x = 1 - 2y$
• $f_y = 1 - 2x$
Then $\nabla f(x,y) = 0$ $\iff$ $(x,y) = (\frac{1}{2}, \frac{1}{2})$.
That point, which is the only critical point of $f$, is in $R$, since $|\frac{1}{2}| \geq \frac{1}{2}$.
Also $f(\frac{1}{2}, \frac{1}{2}) = 1/2$.
Now we need to check along the borders (the vertical lines):
• $f(\frac{1}{2}, y) = \frac{1}{2}$
• $f(-\frac{1}{2}, y) = 2y - \frac{1}{2}$
Along the first line, $f$ stays constant. Since its value there is the same as the value at the critical point, the whole line is a candidate for extremes.
Along the second line, $f$ behaves like a linear function: monotonically increasing and without any critical points.
Now is the part where I am not entirely sure: I would say that since $(0,0) \in R$, and $f(0,0) = 0 < \frac{1}{2}$, then $f$ doesn't have either maxima nor minima at the points we ruled as candidates for being extremes. But would that be enough to guarantee that $f|_R$ doesn't have any maxima nor minima?
Is there some other approach to this problem?
Thanks a lot.
• Once you have $2y-0.5$, that assumes every value in $\mathbb{R}$ already, so there are no global maxima nor minima. – vadim123 Nov 3 '14 at 18:47
• That and saying that those points are all in R would justify it all right, wouldn't it? – Steve Nov 3 '14 at 18:57
Your function on a given domain is unbounded since $(2y - \frac{1}{2})$ and this means that it does not have minima nor maxima.
• I see. This all fails (I mean, $f$ doesn't have global extremes), because $R$ is not compact, doesn't it? – Steve Nov 3 '14 at 18:58 | 2020-02-17T17:04:57 | {
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http://mathhelpforum.com/trigonometry/76150-proving-sum-difference-identities.html | # Thread: Proving Sum and difference identities
1. ## Proving Sum and difference identities
Question:
$\frac{sin(u+v)+sin(u-v)}{cos(u+v)+cos(u-v)} =tan u$
Attempt:
I simplifyied using the equation and i reached to a point where my answer is
$tan(u+v) + tan(u-v)$
I have no idea what to do next to prove...
Thank you..
2. Originally Posted by mj.alawami
Question:
$\frac{sin(u+v)+sin(u-v)}{cos(u+v)+cos(u-v)} =tan u$
Attempt:
I simplifyied using the equation and i reached to a point where my answer is
$tan(u+v) + tan(u-v)$
I have no idea what to do next to prove...
Thank you..
$\sin({u+v}) = sin(u)cos(v) + sin(v)cos(u)$
$\sin({u-v}) = sin(u)cos(v) - sin(v)cos(u)$
$\cos({u+v}) = cos(u)cos(v) - sin(v)sin(u)$
$\cos({u-v}) = cos(u)cos(v) + sin(v)sin(u)$
so we get:
$\frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}$
Factorising gives us
$\frac{sin(u)(2cos(v)) + sin(v)(0)}{cos(u)(2cos(v)) + sin(v)(0)}$
Cancelling 2cos(v) and because anything times 0 is 0 we get:
$\frac{sin(u)}{cos(u)} = tan(u)$
3. Originally Posted by e^(i*pi)
$\sin({u+v}) = sin(u)cos(v) + sin(v)cos(u)$
$\sin({u-v}) = sin(u)cos(v) - sin(v)cos(u)$
$\cos({u+v}) = cos(u)cos(v) - sin(v)sin(u)$
$\cos({u-v}) = cos(u)cos(v) + sin(v)sin(u)$
so we get:
$\frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}$
Factorising gives us
$\frac{sin(u)(2cos(v)) + sin(v)(0)}{cos(u)(2cos(v)) + sin(v)(0)}$
Cancelling 2cos(v) and because anything times 0 is 0 we get:
$\frac{sin(u)}{cos(u)} = tan(u)$
How did you factorize it because i am getting lost?
4. Originally Posted by mj.alawami
Question:
$\frac{sin(u+v)+sin(u-v)}{cos(u+v)+cos(u-v)} =tan u$
$\frac{\sin(u+v)+\sin(u-v)}{\cos(u+v)+\cos(u-v)}$
$=\frac{(\sin u\cos v+\cos u\sin v)+(\sin u\cos v-\cos u\sin v)}{(\cos u\cos v-\sin u\sin v)+(\cos u\cos v+\sin u\sin v)}$
$=\frac{2\sin u\cos v}{2\cos u\cos v}$
$=\frac{\sin u}{\cos u}=\tan u.$
Attempt:
I simplifyied using the equation and i reached to a point where my answer is
$tan(u+v) + tan(u-v)$
I have no idea what to do next to prove...
This is not correct. For example,
$\tan\left(\frac\pi6+\frac\pi6\right)+\tan\left(\fr ac\pi6-\frac\pi6\right)=\sqrt3\neq\frac{\sqrt3}3=\tan\fra c\pi6.$
5. Originally Posted by mj.alawami
How did you factorize it ?
It came from combining like terms, I could have rearranged them first in an extra step which would be going from:
$\frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}$
to
$\frac{(sin(u)cos(v) + sin(u)cos(v))+ (sin(v)cos(u) - sin(v)cos(u))}{(cos(u)cos(v) + cos(u)cos(v)) - (sin(u)sin(v) + sin(u)sin(v))}$
I could also have then simplified each set of brackets to give:
$(sin(u)cos(v) + sin(u)cos(v)) = 2sin(u)cos(v)$
$(sin(v)cos(u) - sin(v)cos(u)) = 0$
$(cos(u)cos(v) + cos(u)cos(v)) = 2cos(u)cos(v)$
$(-sin(u)sin(v) + sin(u)sin(v)) = 0$
this means the 'main' equation will be :
$\frac{2sin(u)cos(v)}{2cos(u)cos(v)}$
As $2cos(v)$ is on the top and bottom it can be cancelled to give ${sin(u)}{cos(u)} = tan(u)$
6. Originally Posted by e^(i*pi)
It came from combining like terms, I could have rearranged them first in an extra step which would be going from:
$\frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}$
to
$\frac{(sin(u)cos(v) + sin(u)cos(v))+ (sin(v)cos(u) - sin(v)cos(u))}{(cos(u)cos(v) + cos(u)cos(v)) - (sin(u)sin(v) + sin(u)sin(v))}$
I could also have then simplified each set of brackets to give:
$(sin(u)cos(v) + sin(u)cos(v)) = 2sin(u)cos(v)$
$(sin(v)cos(u) - sin(v)cos(u)) = 0$
$(cos(u)cos(v) + cos(u)cos(v)) = 2cos(u)cos(v)$
$(-sin(u)sin(v) + sin(u)sin(v)) = 0$
this means the 'main' equation will be :
$\frac{2sin(u)cos(v)}{2cos(u)cos(v)}$
As $2cos(v)$ is on the top and bottom it can be cancelled to give ${sin(u)}{cos(u)} = tan(u)$
Thanks alott it sunk in very easily | 2013-12-04T20:08:56 | {
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https://mathoverflow.net/questions/310936/is-there-a-relation-between-the-number-of-lattice-points-lie-within-these-circle | # Is there a relation between the number of lattice points lie within these circles
Suppose we have a circle of radius $r$ centered at the origin $(0,0)$. The number of integer lattice points within the circle, $N$, can be bounded using Gauss circle problem .
Suppose that another circle of radius $r/2$ centered at the origin inside the initial circle of radius $r$, let $N^*$ represents the number of integer lattice points within the the smallest circle. I am mainly interested in the relation between $N$ and $N^*$. More precisely, to find the number of integer lattice points within the circle of radius $r$ and outside(and at the boundary of) the circle of radius $r/2$.
This page provides the number $N$ for some distances $r$ in $2$ dimensions. For example if we take "ignore the integer lattice point represents the origin":
$r=4$, then $N^*=12, N=48$ and $N^* = \frac{1}{4}N$
$r=6$, then $N^*=28, N=112$ and $N^* = \frac{1}{4}N$
.
.
.
$r=40$, then $N^*=1256, N=5024$ and $N^* = \frac{1}{4}N$
By doing more calculations, in general (considering $r$ is even *for simplicity) we can say $$\frac{1}{5}N \leq N^* \leq \frac{1}{3}N$$
I searched in the literature to find something about this relation in $2$ or $d$ dimensions but without success. I think there is a result or a bound about it, could you kindly direct me to a reference or to a bound and I would be very grateful.
• Here is a visualization that may lead to quick results. Draw a circle of radius r and count the lattice points. Then (instead of drawing a circle of radius 2r) color in the points with coordinates both multiples of 1/2. This is almost like making four copies of the original lattice. Now draw three more circles of radius r with origin near one of the shifted versions of the original origin, and consider what points lie outside the intersection (or don't get copied). Gerhard "Try Eyeballing As Discovery Method" Paseman, 2018.09.19. – Gerhard Paseman Sep 19 '18 at 18:00
• As the Wikipedia page notes, the standard (and fairly straightforward) estimate for this is that $N=\pi r^2+O(r)$ (most of the challenge is in finding the linear term); since similarly, $N^*=\frac14\pi r^2+O(r)$, this immediately gives $N^*=\frac14N+O(\sqrt{N})$. In $d$ dimensions the corresponding statement is $N^*=2^{-d}N+O(N^{(1-1/d)})$, and again it's all pretty straightforward. Do you need more than this? – Steven Stadnicki Sep 19 '18 at 19:32
• @GerhardPaseman thanks for your comment but sorry I didn't understand your idea! Is it possible to clarify more? – Noah16 Sep 20 '18 at 15:33
• Take the larger circle and scale it and the lattice points down by a factor of 1/2. The image looks like the small circle but with more points, most of which are translates of the lattice points inside the small circle by 1/2. The idea is to use these translates to get an idea of the size of (number of points inside big circle minus 4 times number of points inside small circle). Gerhard "Let's See If This Helps" Paseman, 2018.09.20. – Gerhard Paseman Sep 20 '18 at 15:39
• @StevenStadnicki thanks. Your calculations confirm what I wrote above but I am looking for something more powerful and accurate so that I can enumerate the points lie between circles as detailed in my question. Unfortunately, your calculations will not provide an upper and lower bounds. In fact I thought there is a result about this and because of that I asked for a reference! – Noah16 Sep 20 '18 at 15:42
Glad to see there's another Noah interested in such questions :).
Let $$N_{r,d}$$ be the number of (non-zero) integer points in a $$d$$-dimensional ball of radius $$r$$. I'll try to summarize the state of our knowledge on $$N_{r,d}$$ below, at least as I understand it. Along the way, I'll answer your question for $$d= 2$$:
$$3 \leq \frac{N_{2r,2}}{N_{r,2}}\leq 4.5$$
for $$r \geq 1$$, which is tight. (If you exclude radii $$r$$ with no integer points of length $$r$$, then the upper bound drops to $$4 + 1/6$$, which is also tight.)
I'll just say up front that a decent quick and dirty approximation is $$N_{r,d} = \binom{d+r^2}{r^2}^{1\pm O(1)}$$ (for $$r \geq 1$$). You can ignore the rest of this long post if you're happy knowing that.
There are in some sense three different regimes in which $$N_{r,d}$$ behaves rather differently, as follows. (I'm being deliberately vague for now. Precise statements below.)
1. The case $$r \gg \sqrt{d}/2$$, where $$N_{r,d}$$ is basically just the volume of the ball of radius $$r$$.
2. The case $$r \ll \sqrt{d}$$, where $$N_{r,d}$$ is basically just the number of points in $$\{-1,0,1\}^d$$ with norm $$\approx r^2$$.
3. The intermediate case $$r \approx \sqrt{d}/2$$, where it's a bit more complicated.
For $$d = 2$$, there's a sense in which only the first case comes up because, well, the question is only interesting for $$r \geq 1$$, and $$\sqrt{2}/2 = 1/\sqrt{2}$$ is less than one. So, you don't see the more complicated structure until you get to higher dimensions.
$$\mathbf{r \gg \sqrt{d}/2}$$:
In this regime, volume estimates become very precise. In particular, by noting that every point in $$\mathbb{R}^d$$ is at distance at most $$\sqrt{d}/2$$ from $$\mathbb{Z}^d$$ and applying a simple argument [****], we see that
$$\mathrm{Vol}(B_d(r - \sqrt{d}/2)) \leq N_{r,d} \leq \mathrm{Vol}(B_d(r + \sqrt{d}/2)) \; ,$$
where $$B_d(r)$$ is the $$d$$-dimensional $$\ell_2$$ ball in dimension $$d$$, which has volume $$r^d \pi^{d/2}/\Gamma(d/2+1) \approx (2\pi e r^2/d)^{d/2}$$. I.e., we have upper and lower bounds for $$N_{r,d}$$ that differ by a multiplicative factor of
$$\Big( \frac{1+\sqrt{d}/(2r)}{1-\sqrt{d}/(2r)} \Big)^{d} \; ,$$
which in particular implies that the ratio $$N_{2r,d}/N_{r,d}$$ that interests you is equal to $$2^d$$ up to a factor of at most
$$\Big( \frac{1+\sqrt{d}/(4r)}{1-\sqrt{d}/(2r)} \Big)^{d} \; .$$
Finishing $${\bf d = 2}$$
When $$d = 2$$, we can use the above estimate to find the maximum and minimum of $$N_{2r,2}/N_{r,2}$$ relatively easily. We do this by just brute-force computing the ratio for all small $$r$$. (I used Mathematica and the simple formula $$N_{r,2} = \sum _{z_1=-\lfloor r\rfloor }^{\lfloor r\rfloor } (2 \lfloor \sqrt{r^2-z_1^2}\rfloor +1)-1$$.) For integer $$r^2$$, we find that the maximum ratio is achieved at $$r=\sqrt{3}$$ with
$$\frac{N_{2\sqrt{3}, 2}}{N_{\sqrt{3}, 2}} = \frac{36}{8} = 4.5 \; ,$$
and the minimum is actually achieved at $$r = 1$$ with
$$\frac{N_{2, 2}}{N_{1, 2}} = \frac{12}{4} = 3 \; .$$
(The same ratio occurs again at $$r = \sqrt{2}$$.)
If we don't want to include radii where $$r^2$$ cannot be written as the sum of two squares, then the maximum ratio occurs at $$r = \sqrt{8}$$ with
$$\frac{N_{2\sqrt{8}, 2}}{N_{\sqrt{8},2}} = \frac{100}{24} = 4.1666\ldots \; .$$
The volume approximation discussed above shows that no more extreme ratios can occur for, $$r > 52$$, so we can terminate our search there. (There's probably a better argument that lets you check fewer radii, but checking up to $$r = 52$$ isn't too bad.)
(Geometrically, we can think of this very nice behavior as a result of the fact that $$\mathbb{Z}^2$$ actually isn't such a terrible sphere packing/covering, whereas $$\mathbb{Z}^d$$ is a really bad sphere packing/covering for large $$d$$.)
Here are some plots of $$N_{2r,2}/N_{r,2}$$:
(The bands that are clearly visible in the second image correspond to ratios of the form $$4 + c/n$$ for fixed $$c$$ and $$n =$$.
$$\mathbf{r \ll \sqrt{d}}$$:
Here, we can get quite tight bounds simply by noting that when $$r^2$$ is an integer and $$r \ll \sqrt{d}$$,
$$N_{r,d} \approx |\{z \in \{-1,0,1\}^d \ : \ \|z\|= r\}| = 2^{r^2} \binom{d}{r^2} \; .$$
In other words, almost all of the integer points in a ball of radius $$r \ll \sqrt{d}$$ actually lie on the sphere of radius $$r$$ and have coordinates from $$\{-1,0,1\}$$. This immediately shows that the $$2^d$$ heuristic fails in this regime. In fact, until $$N_{r,d} < 2^d$$ for all radii $$r$$ with, say, $$r < 0.4 \sqrt{d}$$, so the $$2^d$$ heuristic obviously fails here.
To be more precise, here's a smooth estimate with this flavor:
$$N_{r,d} = (2 e^{1+\chi} d /r^2)^{r^2} \; ,$$
for any $$r \leq \sqrt{d}/2$$, where $$B(r)$$ is the ball of radius $$r$$ and the error parameter $$\chi$$ satisfies
$$-\frac{r^2}{d} - \frac{\log(C r)}{r^2} \leq \chi \leq \sqrt{\frac{C}{\log(d/r^2)}} \; ,$$
for some not very large constant $$C > 0$$ [*]. Notice that these bounds are quite tight for $$1 \ll r \ll \sqrt{d}$$, since in this regime $$\chi$$ is subconstant. (A smooth function in $$r$$ can't give a much better approximation when $$r$$ is constant since, e.g., $$N_{\sqrt{2} - \varepsilon, d} = 2d$$, but $$N_{\sqrt{2}, d} \approx d^2$$. But the binomial coefficient is quite accurate in this case for integer $$r^2$$.) This gives
$$\frac{N_{2r,d}}{N_{r,d}} \approx (e d /r^2)^{3r^2}2^{-5r^2}$$
for $$1 \ll r \ll \sqrt{d}$$.
Geometrically, we can think of this strange behavior as a result of the integer lattice having short points that shouldn't be there.'' In particular, in $$d$$ dimensions, the ball of radius $$1$$ has volume much less than one, $$\approx (C/d)^{d/2}$$, but the integer lattice manages to have $$2d+1$$ points inside this ball anyway.
The hard case, $${\bf r \approx \sqrt{d} }$$
When $$r \ll \sqrt{d}$$ or $$r \gg \sqrt{d}$$, there are nice smooth functions with closed formulas that approximate $$N_{r,d}$$ well. For $$r \approx \sqrt{d}$$ one can approximate $$N_{r,d}$$ to arbitrary accuracy by studying the theta function $$\Theta(\tau) := \sum_{z \in \mathbb{Z}} e^{-\tau z^2}$$. (Actually, one can do this for all radii $$r$$, but it just gives the above answers back when $$r \ll\sqrt{d}$$ or $$r \gg \sqrt{d}$$.) Mazo and Odlyzko showed this in [**]. Specifically, they showed that
$$N_{\alpha \sqrt{d} ,d}^{1/d} = e^{-\chi/\sqrt{d}}\inf_{\tau > 0} e^{\alpha^2 \tau}\Theta(\tau) \; ,$$ where $$0 \leq \chi \leq C_\alpha$$ for some easily computable constant $$C_{\alpha}$$ that depends only on $$\alpha$$. (The constant $$C_{\alpha}$$ is essentially the standard deviation of the Gaussian distribution over the integers with the parameter $$\tau$$ chosen by the infimum.) Notice that Markov's inequality immediately yields
$$N_{\alpha \sqrt{d} ,d}e^{-\alpha^2 d\tau}< \sum_{\stackrel{z \in \mathbb{Z}^d}{\|z\| \leq \alpha \sqrt{d}}} e^{-\tau z^2} < \sum_{z \in \mathbb{Z}^d} e^{-\tau z^2} = \Theta(\tau)^d \;$$ for any $$\tau > 0$$. Rearranging and taking the infimum shows that $$\chi \geq 0$$. So, the hard part is to show a nearly matching lower bound for appropriately chosen $$\tau$$, which amounts to showing that the summation is dominated by the contribution from points in a thin shell.
Turning back to your original question, one can use this to show that
$$\frac{N_{2\alpha \sqrt{d},d}}{N_{\alpha \sqrt{d}, d}} = e^{\chi_{\alpha}' \sqrt{d}} \cdot (C^*_{\alpha})^d$$,
where $$0 < C^*_{\alpha} < 2$$ is some constant depending only on $$\alpha$$ and $$|\chi_\alpha|$$ is bounded by some constant also depending only on $$\alpha$$. Again, $$C^*_{\alpha}$$ is less than $$2$$ because the integer lattice has too many short points.'' I.e., it is in some sense a really really bad sphere packing [***].
[*] This particular bound is from my thesis (On the Gaussian measure over lattices), where I give an easy proof using the theta function $$\Theta(\tau) := \sum_{z \in \mathbb{Z}} e^{-\tau z^2}$$. (I'm sure that this result is not original, and the technique of using the theta function was already used by Mazo and Odlyzko in [**] to approximate $$N_{r,d}$$ in a more interesting regime.)
[**] Mazo and Odlyzko. Lattice Points in high-dimensional spheres, 1990. https://link.springer.com/article/10.1007/BF01571276 .
[***] With Oded Regev, we showed that it's actually the worst packing'' in a certain precise sense, up to a certain error term: https://arxiv.org/abs/1611.05979 . Basically, its theta function is almost maximal for stable (aka, "non-degenerate") lattices.
[****] To see this, notice that if we sample a random point $$t$$ from the cube $$[0,1/2]^d$$, the expected number of points in a ball of radius $$r-\sqrt{d}/2$$ around $$t$$ must equal the volume of the ball exactly. In particular, there exists a $$t$$ such that this value is at least the volume, and since this $$t$$ is in the cube, the ball of radius $$r$$ around the origin contains this ball. (The upper bound follows from the same argument.)
• Thanks a lot for this effort. I just wanna know why you compared $r$ to $d$ so as a result you divided the process into cases while enumerating the number of lattice points? – Noah16 Sep 22 '18 at 20:59
• i.e Yes I understand why you divided them into 3 cases but is it possible to generalize without comparing $r$ to $d$ – Noah16 Sep 22 '18 at 22:49
• Sorry, I don't understand the question. $N_{r,d}$ depends both on $r$ and $d$. How can one discuss it without "comparing $r$ to $d$"? If you want to fix $d$ and vary $r$ or fix $r$ and vary $d$, then the first and second cases handle this, respectively. The three cases that I described can be collapsed into one by showing that $N_{r,d} \approx \inf_{\tau > 0} e^{r \tau}\Theta(\tau)^d$ for all $r,d$. However, the behavior of $\Theta(\tau)$ is very differently in three different regimes, $\tau \ll 1$, $\tau \gg 1$, and $\tau \approx 1$, which correspond to the three cases that I listed above. – Noah Stephens-Davidowitz Sep 23 '18 at 3:03 | 2021-01-20T02:52:53 | {
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http://stacks.math.columbia.edu/tag/00HL | # The Stacks Project
## Tag: 00HL
This tag has label algebra-lemma-flat-tor-zero and it points to
The corresponding content:
Lemma 9.36.11. Suppose that $R$ is a ring, $0 \to M'' \to M' \to M \to 0$ a short exact sequence, and $N$ an $R$-module. If $M$ is flat then $N \otimes_R M'' \to N \otimes_R M'$ is injective, i.e., the sequence $$0 \to N \otimes_R M'' \to N \otimes_R M' \to N \otimes_R M \to 0$$ is a short exact sequence.
Proof. Let $R^{(I)} \to N$ be a surjection from a free module onto $N$ with kernel $K$. The result follows by a simple diagram chase from the following diagram $$\begin{matrix} & & 0 & & 0 & & 0 & & \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ 0 & \to & (M'')^{(I)} & \to & (M')^{(I)} & \to & M^{(I)} & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\ & & & & & & \uparrow & & \\ & & & & & & 0 & & \end{matrix}$$ with exact rows and columns. The middle row is exact because tensoring with the free module $R^{(I)}$ is exact. $\square$
\begin{lemma}
\label{lemma-flat-tor-zero}
Suppose that $R$ is a ring, $0 \to M'' \to M' \to M \to 0$
a short exact sequence, and $N$ an $R$-module. If $M$ is flat
then $N \otimes_R M'' \to N \otimes_R M'$ is injective, i.e., the
sequence
$$0 \to N \otimes_R M'' \to N \otimes_R M' \to N \otimes_R M \to 0$$
is a short exact sequence.
\end{lemma}
\begin{proof}
Let $R^{(I)} \to N$ be a surjection from a free module
onto $N$ with kernel $K$. The result follows
by a simple diagram chase from the following diagram
$$\begin{matrix} & & 0 & & 0 & & 0 & & \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ 0 & \to & (M'')^{(I)} & \to & (M')^{(I)} & \to & M^{(I)} & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes_R N & \to & M' \otimes_R N & \to & M \otimes_R N & \to & 0 \\ & & & & & & \uparrow & & \\ & & & & & & 0 & & \end{matrix}$$
with exact rows and columns. The middle row is exact because tensoring
with the free module $R^{(I)}$ is exact.
\end{proof}
To cite this tag (see How to reference tags), use:
\cite[\href{http://stacks.math.columbia.edu/tag/00HL}{Tag 00HL}]{stacks-project}
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner). | 2013-05-26T03:09:25 | {
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https://math.stackexchange.com/questions/3037519/what-geomatric-series-formula-is-used | what geomatric series formula is used?
trying to follow a solution related to geometric series, but not sure what formula is used here. Any pointer is appreciated.Image here, can't embed image yet. I do try to plug in the Sn formula, but didn't get it to work
Notice:
$$3^n + 2(3^{n-1}) + 2^2 (3^{n-2}) + ... 2^{n-1}(3) + 2^n$$
can be effectively rewritten as
$$\sum_{k=0}^n 3^{n-k} \cdot 2^k = \sum_{k=0}^n 3^{n} \cdot 3^{-k} \cdot 2^k =\sum_{k=0}^n 3^{n} \cdot \frac{2^k}{3^k} = \sum_{k=0}^n 3^n \cdot \left( \frac{2}{3} \right)^k$$
This is a finite geometric series, then, with ratio $$r=2/3$$ and first term $$2^03^n = 3^n$$. The sum of a finite geometric series $$a + ar + ar^2 + ... +ar^n$$ is given by
$$\sum_{k=0}^n ar^k = \frac{a(r^{n+1} - 1)}{r-1}$$
Thus,
$$\sum_{k=0}^n 3^n \cdot \left( \frac{2}{3} \right)^k = \frac{3^n((2/3)^{n+1} - 1)}{2/3-1}$$
This is essentially the same expression as presented to you, just needing some manipulations to make the equivalence clear.
Multiply by $$3$$ on the top and bottom:
$$\frac{3^n((2/3)^{n+1} - 1)}{2/3-1} = \frac{3^{n+1} \cdot ((2/3)^{n+1} - 1)}{2 - 3}$$
Multiply by $$-1$$ on the top and bottom. We distribute this into the $$()$$ on top to reverse the order.
$$\frac{3^{n+1} \cdot ((2/3)^{n+1} - 1)}{2 - 3} = \frac{3^{n+1} \cdot (1 - (2/3)^{n+1})}{3 - 2}$$
Next:
$$1 - (2/3)^{n+1} = \frac{3^{n+1}}{3^{n+1}} - \frac{2^{n+1}}{3^{n+1}} = \frac{3^{n+1} - 2^{n+1}}{3^{n+1}}$$
The denominator cancels owing to the outside $$3^{n+1}$$ term, and of course $$3-2=1$$, leaving us with
$$\frac{3^{n+1} \cdot (1 - (2/3)^{n+1})}{3 - 2} = \frac{3^{n+1} \cdot \frac{3^{n+1} - 2^{n+1}}{3^{n+1}}}{1} = 3^{n+1} - 2^{n+1}$$
Thus,
$$3^n + 2(3^{n-1}) + ... +2^{n-1}(3) + 2^n = \sum_{k=0}^n 3^n \cdot \left( \frac{2}{3} \right)^k = 3^{n+1} - 2^{n+1}$$
• It looks like your sum has an extra $(2/3)^k$ in it. – Michael Burr Dec 13 '18 at 3:32
• Where, exactly? – Eevee Trainer Dec 13 '18 at 3:36
• Oh, I see. I'll try to figure out how to resolve it. – Eevee Trainer Dec 13 '18 at 3:37
• Okay, I believe I fixed it. It didn't really affect my approach, just more how it was being written in the summation. I suck at writing summations when they're a bit awkward like the one givenn. – Eevee Trainer Dec 13 '18 at 3:42
Sum of first n terms of a geometric series is given by:
$$S_n = \frac{a(r^n-1)}{r-1}.$$
Here $$a$$ is the first term of the series and $$r$$ is the common ratio.
In your question, $$r = \frac{2}{3}$$, $$a=3^n$$ and the number of terms is $$n+1$$ (you probably messed up in this part).
• yes, did you notice that it was stated as $T(2^n)$ = ..., I do try to plug in the $S_n$ formula, but didn't get it to work – Maxfield Dec 13 '18 at 2:45
• Perhaps it would be helpful to identify $a$ and $r$ in the given problem. – Michael Burr Dec 13 '18 at 2:46
• as stated in the solution, r is 2/3 and, as a, which I can't find. but it states as T($2^n$) instead of $S_n$ though. – Maxfield Dec 13 '18 at 2:48
• It doesn't matter what it is stated as. You can just calculate the sum of the RHS. (Also, see edit) – Nutan Nepal Dec 13 '18 at 2:54
One gets $$T(2n)=3^n\sum_{k=0}^n(\frac23)^k=3^n(\frac{1-(\frac23)^{n+1}}{1-\frac23})=3^{n+1}-2^{n+1}$$. | 2019-06-17T23:14:12 | {
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https://math.stackexchange.com/questions/3617551/does-my-solution-work-by-accident | # Does my solution work by accident?
The task description is as follows (sorry for my poor translation - it's the best I could think of):
There are three numbers that sum up to 105. They are subsequent items in a growing geometric sequence (a, b, c). The first of those numbers is the first number, the second number is the sixth number and the third number is the twenty-sixth number in some arithmetic sequence. What are those numbers?
Now, my solution is as follows:
Those three numbers expressed in terms of the second sequence are $$(a, a + 5r, a + 25r)$$ where r is the difference of the arithmetic sequence. If so, then this is true (am I right?):
$$a + a + 5r + a + 25r = 105\\ a = 35 - 10r$$
Therefore the first sequence can be expressed as follows (again: am I right?): $$(35 - 10r, 35 - 5r, 35 + 15r)$$ . And here comes the most important question in this post: am I allowed to divide each of those items in the final sequence by a number so that my next calculations will deal with lower numbers? I'm asking because I did that and in the end I got the right values for a, b and c, although I'm not sure if I didn't do something wrong (prob. I did). So, my next calculations were:
$$(35 - 10r, 35 - 5r, 35 + 15r) / :5 \\ (7 - 2r, 7 - r, 7 + 3r)$$
From the formula for a second item of a geometric sequence ($$y^2 = x * z$$ where y is the second item in the sequence, x is the first one and z the last one):
$$(7 - r)^2 = (7 - 2r)(7 + 3r)\\ 49 - 14r + r^2 = 49 + 21r - 14r - 6r^2\\ 7r^2 = 21r\\ r = 3$$
Now if I put $$r$$ into $$(35 - 10r, 35 - 5r, 35 + 15r)$$, the three numbers come out as $$a = 5\\ b = 20\\ c = 80$$
Why does this work?
Looks correct. The reason it works to divide through is that the equation you write for requiring that $$a,b,c$$ are a geometric sequence only depend on the relative proportions of $$a,b,c$$. The equation you get is $$b/a = c/b$$ (equivalent to $$b^2 = ac$$), and you just reduced these fractions. | 2021-09-24T02:23:30 | {
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https://math.stackexchange.com/questions/988030/showing-a-function-has-one-root-in-an-interval/988055 | # Showing a function has one root in an interval
Could anyone shine some light on this question please?
By considering $f'(x)$, show that $$f(x)=x^3 - 2$$ has exactly one root for $x$ greater than or equal to $0$.
Hint: $f'(x)=3x^2\ge 0$ for all $x$.
Note that the function is increasing and $f(0)<0$ and $f(2)>0$, what can you conclude using continuity of $f$?
Side-note the roots of the equation are $2^{1/3}\cdot \omega ^i$ where $i=0,1,2$ and $\omega$ is the cube root of unity. i.e. there is only one real root.
• Thanks a lot for your help, appreciate it! – gary Oct 23 '14 at 20:23
After differentiate, you have $f'> 0$ for $x>0$, so it's strictly increasing. Then by intermediate value theorem, there exists a root. By strictly increasing, the root is unique.
• Ah that's great thank you, it was the second part that you picked up on that I wasn't too sure about1 – gary Oct 23 '14 at 20:12
• @gary $f(0)<0,f(3)>0$ and $f$ is continuous. The intermediate value theorem guarantees there is a root. – John Oct 23 '14 at 20:22
By intermediate value theorem, you can see that $f(0) < 0$ and $f(2)>0$, therefore $f$ has a root $x_0$ between 0 and 2. Now suppose there exists another $x_1 \geq 0$ such that $f(x_1) = f(x_0) = 0$ By Rolle's Theorem, that would mean there exists $x_0\leq x_2 \leq x_1$ such that $f'(x_2) = 0$.
Now $f'(x) = 3x^2$. Note that $f'(x) = 0 \iff x = 0$, but $0$ isn't found in the interval we're considering.
Therefore, there can't be two roots of this polynomial function greater or equal than 0.
Note that $f(0) < 0$. Take the derivative and note that it is positive on the entire interval. Conclude that f must cross the axis only once. | 2020-09-21T02:46:14 | {
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https://ftclausen.github.io/mathematics/concrete-mathematics/josephus-worked-induction/ | # Concrete Mathematics: Notes on Josephus Problem Induction
The explanations in Concrete Mathematics are very good if a bit terse. So, as before, I will add some extra detail in a blog post to help cement (🥁) my knowledge. I am also shouting it into the void in case it helps someone else.
See the book for full context; I’ll just cover the induction related steps.
# The Recurrences and Closed Form
The recurrence comes in two flavours: odd and even. The complete recurrences are thus:
\begin{align} J(1) &= 1;\\ J(2n) &= 2J(n) - 1, \ \ \ \ \text{for n} \geq 1\\ J(2n+1) &= 2J(n) + 1, \ \ \ \ \text{for n} \geq 1\\ \end{align}
And the closed form is as follows
$J(2^m+\ell)=2\ell+1$
# Even Induction
The book goes on to give an example of the even induction
$J(2^m+\ell)=2J(2^{m-1} + \ell/2) - 1 = 2(2\ell/2 + 1) - 1 = 2\ell + 1$
by (1.8) and the induction hypothesis; this is exactly what we want.
What confused me at first was the leap
$2J(2^{m-1} + \ell/2) - 1 = 2(2\ell/2 + 1) - 1$
This is the key induction step - this is not done via algebraic manipulation. We are substituting the closed form that is equivalent to $J(n)$. Our closed form proposal is actually for $2J(n)-1=2l+1$ so we need to refit it to be for $J(n)$. Thus $l$ is in the previous power of two block (remember $J(n)$ not $2J(n)$) and half as big: $l/2$. So the $J(n)$ closed form is $2l/2 +1$.
So we plug that in and then he next step, however, is done via algebra:
$2(2\ell/2 + 1) - 1 = 2\ell + 1$
Completing the even induction.
# Odd Induction
Credit: I was stuck on this until helped out by the ever helpful Brian M. Scott on Math Stackexchange. I will repeat that here and then do my own (simplified) take on thereafter. First Brian’s reply:
Your’re not applying the recurrence for the odd case correctly. Suppose that $2n + 1 = 2^m + \ell$, where $0 \leq \ell \leq 2^m$. The recurrence is $J(2n + 1) = 2J(n) + 1$, and n here is $\frac12(2^m + \ell - 1)$, so
\begin{align} J(2^m + \ell) &= 2\left(2^{m-1} + \frac{\ell - 1}{2}\right) + 1 \\ &= 2\left(\frac{2(\ell - 1)}{2} + 1 \right) + 1 \\ &= 2\ell + 1 \end{align}
as desired
I will do some colour highlighting and ever more verbose explanation for my future self here. The core is we want to convert $2n + 1$ into $n$ hence dividing by two and subtracting one:
$\frac{2(n + 1)}{2} - 1 = n$
We apply these operations during the induction step to turn $2\ell + 1$ into $\frac{2(\ell +1)}{2} - 1$. Notice the division by two and subtraction by one? And now with gusto and in context of the induction step where we “insert” the closed form representing n into the recurrence “framework” (substitution highlighted in blue):
\begin{align} J(2^m + \ell) &= 2\left(\color{blue}\frac{2(\ell + 1)}{2} - 1\color{#3d4144}\right) + 1 \\ &= 2\ell + 1 \end{align}
Giving us what we want.
Categories:
Updated: | 2022-06-29T19:20:44 | {
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https://grindskills.com/why-linear-regression-has-assumption-on-residual-but-generalized-linear-model-has-assumptions-on-response/ | # Why linear regression has assumption on residual but generalized linear model has assumptions on response?
Why linear regression and Generalized Model have inconsistent assumptions?
• In linear regression, we assume residual comes form Gaussian
• In other regression (logistic regression, poison regression), we assume response comes form some distribution (binomial, poission etc.).
Why sometimes assume residual and other time assume on response? Is is because we want to derive different properties?
EDIT: I think mark999’s shows two forms are equal. However, I do have one additional doubts on i.i.d:
My other quesiton,
Is there i.i.d. assumption on logistic regression? shows generalized linear model does not have i.i.d assumption (independent but not identical)
Is that true that for linear regression, if we pose assumption on residual, we will have i.i.d, but if we pose assumption on response, we will have independent but not identical samples (different Gaussian with different $\mu$)?
## Answer
Simple linear regression having Gaussian errors is a very nice attribute that does not generalize to generalized linear models.
In generalized linear models, the response follows some given distribution given the mean. Linear regression follows this pattern; if we have
$y_i = \beta_0 + \beta_1 x_i + \epsilon_i$
with $\epsilon_i \sim N(0, \sigma)$
then we also have
$y_i \sim N(\beta_0 + \beta_1 x_i, \sigma)$
Okay, so the response follows the given distribution for generalized linear models, but for linear regression we also have that the residuals follow a Gaussian distribution. Why is it emphasized that the residuals are normal when that’s not the generalized rule? Well, because it’s the much more useful rule. The nice thing about thinking about normality of the residuals is this is much easier to examine. If we subtract out the estimated means, all the residuals should have roughly the same variance and roughly the same mean (0) and will be roughly normally distributed (note: I say “roughly” because if we don’t have perfect estimates of the regression parameters, which of course we do not, the variance of the estimates of $\epsilon_i$ will have different variances based on the ranges of $x$. But hopefully there’s enough precision in the estimates that this is ignorable!).
On the other hand, looking at the unadjusted $y_i$‘s, we can’t really tell if they are normal if they all have different means. For example, consider the following model:
$y_i = 0 + 2 \times x_i + \epsilon_i$
with $\epsilon_i \sim N(0, 0.2)$ and $x_i \sim \text{Bernoulli}(p = 0.5)$
Then the $y_i$ will be highly bimodal, but does not violate the assumptions of linear regression! On the other hand, the residuals will follow a roughly normal distribution.
Here’s some R code to illustrate.
x <- rbinom(1000, size = 1, prob = 0.5)
y <- 2 * x + rnorm(1000, sd = 0.2)
fit <- lm(y ~ x)
resids <- residuals(fit)
par(mfrow = c(1,2))
hist(y, main = 'Distribution of Responses')
hist(resids, main = 'Distribution of Residuals')
Attribution
Source : Link , Question Author : Haitao Du , Answer Author : Cliff AB | 2022-11-29T04:26:37 | {
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# The letters D, G, I, I , and T can be used to form 5-letter strings as
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The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
[Reveal] Spoiler: OA
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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15 Jun 2016, 02:55
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IMO 36
Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60
No of ways two I's can be together 4!=24
no of ways at least one alpha is between two I's =60-24=36
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The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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15 Jun 2016, 04:00
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Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Let us calculate the total ways. Those would be (5!/2!) = 60.
Now since the question says "at least" let us find the number of arrangements when both I's are together. (Tie them up). so we have 4! ways to arrange such that I's always come together. 4! = 24
60 - 24 = 36.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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21 Jun 2016, 09:59
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Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Total ways to arrange these letters=5!/2!
Ways to arrange the letters if both I are together= 4!
Ways to arrange with both Is not together= 5!/2!-4!= 36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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23 Jun 2016, 02:23
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Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Required: atleast one letter between two Is
Total cases = 5!/2! = 60
Total cases in which Is will be together = 4! = 24 {Treat both Is as one}
Hence total cases in which there is atleast one letter between the Is = 60 - 24 = 36
Correct Option: D
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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28 Jul 2016, 05:26
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The requirement is 5 letter strings in which two I's are not together, they are separated by atleast one letter.
Consider this as below,
(Total number of 5 letter strings that can be formed) - (Total number of 5 letter strings in which two I's are together)
=(5!/2!) - (4!)
= 5*4*3 - 4*3*2
= 60 - 24
= 36
Finding total number of 5 letter strings - DIGIT = 5!/2! (2! because of the repetition of I's)
Finding total number of 5 letter strings with I's together - II_ _ _ = 4!
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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A couple of video explanations for this question:
"Direct" Method:
"Indirect" Method:
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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10 Jan 2017, 10:53
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
@bunnel .. Just for info wanted to know how should have the question solved if it was said "how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by one other letter " instead of the one given !! Please help !!
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The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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10 Jan 2017, 18:14
Suvo
Are you asking what the answer would be if we replaced the "at least" in the question stem with "exactly"?
If so, then, for your hypothetical question, there would be only a few ways to place the I's. So a 'direct approach' would be an efficient way to solve the question. We'd have:
I _ I _ _
_ I _ I _
_ _ I _ I
For each of these options we have three open spaces to arrange the other three letters. So for each of the options we'd have:
(3 choices for the first open slot) * (2 choices for the second open slot) * (1 choices for the last open slot) = 3! = 6 ways to complete the option.
Overall we'd multiply the 3 options by the 6 ways to complete each of the options. So, in your hypothetical version of the question, your answer would be 3*6 = 18.
The "direct approach" video in my previous post might help with the logic here.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Another approach.
Take the task of arranging the 5 letters and break it into stages.
Stage 1: Arrange the 3 CONSONANTS (D, G and T) in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 3 consonants in 3! ways (= 6 ways)
So, we can complete stage 1 in 6 ways
IMPORTANT: For each arrangement of 3 consonants, there are 4 places where the two I's can be placed.
For example, in the arrangement DTG, we can add spaces as follows _D_T_G_
So, if we place each I in one of the available spaces, we can ENSURE that the two I's are never together.
Stage 2: Select two available spaces and place an I in each space.
Since the order in which we select the two spaces does not matter, we can use combinations.
We can select 2 spaces from 4 spaces in 4C2 ways (= 6 ways)
So we can complete stage 2 in 6 ways.
By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 5 letters) in (6)(6) ways (= 36 ways)
[Reveal] Spoiler:
D
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So be sure to learn this technique.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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19 May 2017, 13:13
Could someone help me understand how the total number of cases is 5!/2!? I don't understand the reasoning behind that.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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20 May 2017, 04:56
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skysailor wrote:
Could someone help me understand how the total number of cases is 5!/2!? I don't understand the reasoning behind that.
THEORY:
Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:
$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$.
For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.
For more check the links below:
Theory on Combinations
DS questions on Combinations
PS questions on Combinations
Tough and tricky questions on Combinations
So, ss explained above, the number of arrangements of 5 letters D, G, I, I , and T out of which there are two identical I's will be 5!/2! = 60.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
1. It is an ordering, therefore a permutation problem
2. There is a constraint
3. The constraint is an "Atleast One" constraint. so the opposite of it "none" is easier to find
4. Total number of permutations is 5!/2!=60 ways since I repeats twice
5. Number of permutations where I's are together is 3!*4 = 24
6. Permutations with constraint is (4)-(5)=36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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13 Jun 2017, 18:13
A couple of video explanations for this question:
"Direct" Method:
"Indirect" Method:
Why don't we divide the "impossible" by 2! ?
I thought since there are two I's, the order of the I's don't matter. Thus, DIIGT is the same if the I's were switched (DIIGT).
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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24 Jul 2017, 11:37
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Just want to share my approach:
1. In this case, the 3 letter D,G,T can be calculated easily : 3! = 6.
2. The problem is only the remanining I : which must place with separation.
3. The best way to calculate the "I" case is to draw it :
I _ I _ _ (1)
I _ _ I _ (2)
I _ _ _ I (3)
_ I _ I _ (4)
_ I _ _ I (5)
_ _ I _ I (6)
As we can see, there are only 6 possibilies.
4. Calculate all = 6 X 6 = 36.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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26 Jul 2017, 16:16
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
This is a permutation problem because the order of the letters matters. Let’s first determine in how many ways we can arrange the letters. Since there are 2 repeating Is, we can arrange the letters in 5!/2! = 120/2 = 60 ways.
We also have the following equation:
60 = (number of ways to arrange the letters with the Is together) + (number of ways without the Is together).
Let’s determine the number of ways to arrange the letters with the Is together.
We have: [I-I] [D] [G] [T]
We see that with the Is together, we have 4! = 24 ways to arrange the letters.
Thus, the number of ways to arrange the letters without the Is together (i.e., with the Is separated) is 60 - 24 = 36.
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SravnaTestPrep wrote:
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
1. It is an ordering, therefore a permutation problem
2. There is a constraint
3. The constraint is an "Atleast One" constraint. so the opposite of it "none" is easier to find
4. Total number of permutations is 5!/2!=60 ways since I repeats twice
5. Number of permutations where I's are together is 3!*4 = 24
6. Permutations with constraint is (4)-(5)=36
Bunuel can u help me
While considering the Is toether done we have to multiply by 2 as either of the Is can come or that it doesnt matter since both are same anyway
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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21 Aug 2017, 01:57
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Letters are DGIIT - 5 letters...
Total no. of 5 letter string that can be formed = 5!/2! = 5*4*3 = 60
Total no. of 5 letter string that in which 2 I are together that can be formed = 4! = 4*3*2 = 24
So, 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter = 60 - 24 = 36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink] 21 Aug 2017, 01:57
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Display posts from previous: Sort by | 2017-08-21T12:10:48 | {
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https://math.stackexchange.com/questions/2913498/propositional-logic-questions-conditionals | # Propositional Logic questions (Conditionals).
I am uncertain if I did the following questions correctly, and I will appreciate it very much if anyone can help point out any errors in my procedure.
Question 1: Let $u$ mean "the Racket identifier may be used" and $d$ mean "the Racket identifier has been defined". Which of the following means "The Racket identifier may not be used if it has not been defined."?
A)$\lnot u \rightarrow \lnot d$
B)$\lnot d \rightarrow \lnot u$
C) None of the above
I think the anwser should be B). At first, I thought of A) as the correct answer, but on a second thought, in the statement "The Racket identifier may not be used if it has not been defined." did not use only if. So, this statement is equivalent to "if it has not been defined, then the Racket identifier may not be used.
Question 2: Let $t$ mean "Mary-Kate and Ashley are identical twins." Let $m$ mean "Mary-Kate has blue eyes." Let $a$ mean "Ashley has blue eyes." Which of the following means "If Mary-Kate and Ashley are identical twins, then either they both have blue eyes or neither has blue eyes."
A) $t \rightarrow (m \land a)$
B) $t \leftrightarrow (m \leftrightarrow a)$
C) $t \rightarrow (m \leftrightarrow a)$
D) $t \leftrightarrow (m \rightarrow a)$
E) None of the above.
I believe the answer should be E), since the statement "If Mary-Kate and Ashley are identical twins, then either they both have blue eyes or neither has blue eyes." translates to the conditional: $t \rightarrow ((m \land a) \lor \lnot(m \lor a))$, and the consequent does not correspond to any conditional/biconditional statements in A) through D). So E) may be the answer.
Question 3: Which of the following could be the result of applying the negation law to the statement:$p \land (a \lor b) \land \lnot (a \lor b)$:
A) $p \lor \lnot (a \lor b)$
B) $p \land T$
C) $p \land (a \lor b)$
D) $p \land F$
E) None of the above
According to my textbook, the negation law states that $p \lor \lnot p \equiv T$ and $p \land \lnot p \equiv F$. So I can use first use associative law to get $p \land ((a \lor b) \land \lnot (a \lor b))$ then apply negation law to get $p \land F$. So the answer is D)
Also, what does the F and T mean in the negation law. I mean $p \land \lnot p$, does this is a tautological falsehood (contradiction) right?
• $F$ is False and $T$ is True; thus, the negation of $F$ is $T$ and the same for the other case. Sep 11, 2018 at 19:24
I think the anwser should be B). At first, I thought of A) as the correct answer, but on a second thought, in the statement "The Racket identifier may not be used if it has not been defined." did not use only if. So, this statement is equivalent to "if it has not been defined, then the Racket identifier may not be used.
Yes. $\neg d\to\neg u$ says "Not $u$ if not $d$", and also "If not $d$, then not $u$."
I believe the answer should be E), since the statement "If Mary-Kate and Ashley are identical twins, then either they both have blue eyes or neither has blue eyes." translates to the conditional: $t→((m∧a)∨¬(m∨a))$, and the consequent does not correspond to any conditional/biconditional statements in A) through D). So E) may be the answer.
No. The consequent is saying that the truth values for $m$ and $a$ are equal, which is a bicondition. $${\quad(m\wedge a)\vee\neg(m\vee a) \\\equiv (m\wedge a)\vee(\neg m\wedge \neg a)\\\equiv (m\vee\neg a)\wedge(\neg m\vee a)\\\equiv (a\to m)\wedge(m\to a)\\\equiv m\leftrightarrow a}$$
According to my textbook, the negation law states that $p∨¬p≡T$ and $p∧¬p≡F$. So I can use first use associative law to get $p∧((a∨b)∧¬(a∨b))$ then apply negation law to get $p∧F$. So the answer is D)
Indeed, that is so.
Also, what does the F and T mean in the negation law. I mean $p∧¬p$, does this is a tautological falsehood (contradiction) right?
T : true, F : false. $\phi\vee\neg\phi$ is true, $\phi\wedge\neg\phi$ is false.
Indeed since they are true and false reguardless of the value of $\phi$, they are a tautology and contradiction respectively. | 2022-09-25T02:37:35 | {
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https://math.stackexchange.com/questions/2529544/taylor-inside-an-integral/2529692 | # Taylor inside an integral
I know the following integral should be: $$\int_{0}^{1} \frac{dx}{\sqrt{1-x^2-\epsilon(1-x^3)}} \approx \pi/2 + \epsilon$$ for $\epsilon$ small. What I do is set $-\epsilon(1-x^3)=y$ and then since $y$ it's always greatly smaller than $(1-x^2)$ I do Taylor expansion around $y=0$: $$\frac{dx}{\sqrt{1-x^2+y}} \approx \frac{dx}{\sqrt{1-x^2}}-\frac{y \cdot dx}{2 \cdot (1-x^3)^{3/2}}$$
Then I recover $y=-\epsilon(1-x^3)$ and I get the correct answer. The problem is that I don't know if that's mathematically correct because $y$ depends on $x$.
• What's wrong with having $y$ depend on $x$? I mean, which step in the process do you think could be flawed for that reason? – tilper Nov 20 '17 at 18:08
• I would worry near $x=1$. There, both $1-x^2$ and $1-x^3$ are small. – Ron Gordon Nov 20 '17 at 18:22
• I'm worried about doing Taylor in that way isn't correct, for example for $y=0$ I have $x=1$ then $f(0) = \infty$ so I'm doing Taylor around $\infty$ and makes no sense to me – cramirpe Nov 20 '17 at 18:23
Using Taylor series you will end up having to justify evaluation at the improper bound of $1$, which will require further details that neither of the other answers have addressed. Instead, you could just note that
$$\frac{1-x^3}{1-x^2} = \frac{(1-x)(1+x+x^2)}{(1-x)(1+x)} = 1 + \frac{x^2}{1+x} \in [1,\tfrac32)$$ for $x \in [0,1)$ and hence
$$\frac{1}{\sqrt{1-x^2}} \leq \frac{1}{\sqrt{1-x^2 - \epsilon(1-x^3)}} = \frac{1}{\sqrt{1-x^2}}\cdot\frac{1}{\sqrt{1 - \epsilon \frac{1-x^3}{1-x^2}}} \leq \frac{1}{\sqrt{1-x^2}}\cdot\frac{1}{\sqrt{1 - \tfrac32\epsilon }}$$ for all $0<\epsilon <\tfrac23$.
So by integrating we get: $$\frac{\pi}{2} \leq \int_0^1 \frac{dx}{\sqrt{1-x^2 - \epsilon(1-x^3)}} \leq \frac{\pi}{2} \cdot \frac{1}{\sqrt{1 - \tfrac32\epsilon }}$$ We may then take the series expansion about $0$ of the right hand side, giving $$\frac{1}{\sqrt{1-\tfrac32 \epsilon}} = 1 + \tfrac34 \epsilon + O(\epsilon^2)$$
Hence $$\frac{\pi}{2} \leq \int_0^1 \frac{dx}{\sqrt{1-x^2 - \epsilon(1-x^3)}} \leq \frac{\pi}{2} + \frac{3\pi}{8} \epsilon + O(\epsilon^2)$$
You are on safer ground factoring the $1-x$ out of the square root as follows:
\begin{align} I(\epsilon) &= \int_0^1 \frac{dx}{\sqrt{1-x^2-\epsilon (1-x^3)}} \\ &= \int_0^1 dx \frac{(1-x)^{-1/2}}{\sqrt{1+x-\epsilon(1+x+x^2)}} \\ &= \int_0^1 dx \, \frac{x^{-1/2}}{\sqrt{2-x-\epsilon (3-3 x+x^2)}} \\ &=2 \int_0^1 dx \left [ 2-x^2 - \epsilon (3-3 x^2+x^4)\right ]^{-1/2} \\ &= 2 \int_0^1 dx (2-x^2)^{-1/2} \left [1-\epsilon \frac{3-3 x^2+x^4}{2-x^2} \right ]^{-1/2} \end{align}
I hope you see where this is heading. You may now expand the term in brackets in a Taylor expansion in $\epsilon$ knowing that the term involving $\epsilon$ is small over the entire region of integration. Use trig substitution and the answer is straightforwardly...
$$I(\epsilon) = \frac{\pi}{2} + \epsilon + O(\epsilon^2)$$
You can be authorized in doing that because $\epsilon$ is small (we suppose small enough to give you the "license" of dong that).
Also notice that the range of the integral is $[0, 1]$, hence $x$ too, in a certain way, is small, and terms like $x^3$ are then even smaller.
There are many approaches for dealing with such an integral. Yours is one. Then you can take that function and make a Tylor series in $x$ or in $\epsilon$, obtaining respectively:
$$\star : \frac{1}{\sqrt{1-\epsilon }}+\frac{x^2}{2 (1-\epsilon )^{3/2}}-\frac{x^3 \epsilon }{2 (1-\epsilon )^{3/2}}+O\left(x^4\right)$$
$$\star : \frac{1}{\sqrt{1-x^2}}-\frac{\left(x^3-1\right) \epsilon }{2 \left(1-x^2\right)^{3/2}}+\frac{3 \left(x^3-1\right)^2 \epsilon ^2}{8 \left(1-x^2\right)^{5/2}}-\frac{5 \left(x^3-1\right)^3 \epsilon ^3}{16 \left(1-x^2\right)^{7/2}}+O\left(\epsilon ^4\right)$$
Those are clearly Taylor series of the whole function, that is, of
$$\frac{1}{\sqrt{1 - \epsilon - x^2 + \epsilon x^3}}$$
Another way is to think about the term $\epsilon x^3$ and rawly say "let's get rid of this", remaining with the easy integral
$$\int_0^1 \frac{\text{d}x}{\sqrt{1 - \epsilon - x^2}} = -\tan ^{-1}\left(\frac{\epsilon }{(-\epsilon )^{3/2}}\right)$$
Numerica other methods are available, but this wouldn't be the topic of the question.
EDIT
Notice that in my last result we have
$$\frac{\epsilon}{(-\epsilon)^{3/2}} = \frac{1}{\sqrt{-\epsilon}}$$
hence as $\epsilon \to 0$
$$-\arctan\left(\frac{1}{\sqrt{-\epsilon}}\right) = \frac{\pi}{2}$$
• That's a more intuitive way of doing it, Taylor with all dependence on $\epsilon$ (with exact the same result as mine) or $x$. But doing it only with a part of all the function (as I do) is correct? Notice for $y=0$, $x=0$ so I'm doing Taylor around $y=0$ when $f(0) = \infty$ – cramirpe Nov 20 '17 at 18:33
• What about near $x=1$ where both terms are near zero? – Ron Gordon Nov 20 '17 at 18:37
• Sorry, for $y=0$ then $x=1$ – cramirpe Nov 20 '17 at 18:39
• @Henry Your second approach is incorrect. Note that $1-x^2-\epsilon < 0$ for $x$ close enough to $1$, so the integral is undefined. You would have to define the upper bound of integration in terms of $\epsilon$, so the integrals won't be immediately comparable. – adfriedman Nov 20 '17 at 21:22
• Also $\epsilon$ being small isn't sufficient on its own to allow the Taylor series to be expanded under the integral. In my answer I show that the integral is bounded, so a dominated convergence theorem might be sufficient for justification. – adfriedman Nov 20 '17 at 21:22 | 2019-08-20T22:50:24 | {
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https://math.stackexchange.com/questions/1906718/how-to-prove-ak-n-gcdn-k-whenever-a-n | # How to prove $|a^k|=n/\gcd(n,k)$ whenever $|a|=n$?
This is an exercise from "Contemporary Abstract Algebra" I'm not sure how to solve.
Exercise: Let $\langle a\rangle$ be a (cyclic) group of order $n$. Prove that the order of $a^k=\frac{n}{\gcd(n,k)}$.
Direction: (1) Let $d=\gcd(n,k)$, thus by the Euclidian algorithm we can find $X,Y\in\mathbb{Z}$ s.t. $d=Xn+Yk$, thus, $a^d=a^{Xn+Yk}=a^{Xn}a^{Yk}=(a^n)^X(a^k)^Y=(a^k)^Y$. What to do from here?
(2) We know that $d|n$, thus $\langle a^{n/d} \rangle$ is of order $d$ and $\langle a^d \rangle$ is of order $\frac{n}{d}=\frac{n}{\gcd(n,k)}$. Is it mean that $d=k$? Where is my mistake?
You have two things to show. Namely that:
1. $(a^k)^{n/\gcd(n,k)}=e$ (where $e$ denotes the identity)
and that $n/\gcd(n,k)$ is the smallest positive power $p$ of $a^k$ such that $(a^k)^p=e$:
1. For all $m>0$: $(a^k)^m=e \implies n/\gcd(n,k)\leq m$
The first part is easy:
$$(a^k)^{n/\gcd(n,k)}=(a^n)^{k/\gcd(n,k)}=e^{k/\gcd(n,k)}=e$$
For the second part, let $m\in\mathbb{N}$ be such that $(a^k)^m=a^{km}=e$. Since the order of $a$ is $n$, it follows that $n\mid km$. Therefore we also have
$$\frac{n}{\gcd(n,k)}\mid \frac{k}{\gcd(n,k)}m$$
Now $\gcd(n/\gcd(n,k),k/\gcd(n,k))=1$ (try to prove this), so it follows that
$$n/\gcd(n,k)\mid m$$
Hence in particular $n/\gcd(n,k)\leq m$.
• What if $n=25$ and $k=30$? then $\gcd(n/\gcd(n,k),k)=5$. – Silent Jan 24 '18 at 14:39
• @Silent Uncountable should have instead wrote $$n\mid km \implies \frac{n}{g}\mid \frac{k}{g}m\implies \frac{n}{g}\mid m,$$ where $g=\gcd(n,k)$, since $\gcd(n/g,k/g)=1$ is true (while $\gcd(n/g,k)=1$ needn't be). – anon Jan 25 '18 at 6:53
• @anon good one! – Uncountable Jan 25 '18 at 7:58
• – BCLC Aug 31 '18 at 17:08
The order of $a^k$ is the smallest $r>0$ such that $a^{kr}=e$, i.e. such that $kr$ is a multiple of the order $n$ of $a$. This means $kr$ is the least common multiple of $k$ and $n$.
As $\operatorname{lcm}(k,n)=\dfrac{kn}{\gcd(k,n)}$, we have: $\quad r=\dfrac{n}{\gcd(k,n)}.$
• – BCLC Aug 31 '18 at 17:09
• Do you mean corollary 2.8.10? $x^k$ is a element of the group generated by $x$, which has order $n$. But this result is more precise than the corollary. – Bernard Aug 31 '18 at 17:39
• Bernard, interesting, but no. I really mean Cor 2.8.11 intending to use $\gcd(k,p)=1$ for any or all $0 < k < p$. Like somehow we might show that the orders of the non-identity elements are given by $p/\gcd(k,p)$, which of course simplifies to $p$. – BCLC Aug 31 '18 at 17:46
• It might be used. However, I wonder whether the notion of order of an element doesn't ultimately on Lagrange's theorem. – Bernard Aug 31 '18 at 19:12
• Thanks Bernard! – BCLC Aug 31 '18 at 19:17
The proof is a bit simpler if we use basic gcd laws instead of Bézout's identity. Namely
$$(a^{\large k})^{\large j} = 1\iff n\mid kj\iff n\mid nj,kj\iff n\mid(nj,kj)=(n,k)j\iff n/(n,k)\mid j$$
The first $\iff$ follows because $\,n = {\rm\ ord}\, a,\,$ and the third follows by the definition/universal property of the gcd and the gcd distributive law.
Remark $\$ Alternatively, we can use lcm instead of gcd, i.e.
$$(a^{\large k})^{\large j} = 1\iff n\mid kj\iff n,k\mid kj\iff [n,k]\mid kj\iff [n,k]/k\mid j$$
Both proofs are equivalent since $\ [n,k]/k = n/(n,k),\$ i.e. $\ [n,k](n,k) = nk,\,$ by here.
Note that the first equivalence chain implies that $\,a^{\large k}\,$ has order $\,n/(n,k)\,$ since, generally, if $\ b^{\large j} = 1\iff i\mid j\$ then this implies that $\,b\,$ has order $\,i.\,$ Indeed, setting $\,j=i\,$ implies $\,b^{\large i}=1\,$ and $\,i\,$ is the least postive $\,j\,$ with $\,b^{\large j}=1\,$ since $\,i\,$ divides all other such $\,j.\$ | 2019-11-13T20:26:13 | {
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https://stats.stackexchange.com/questions/217881/similarity-function-with-given-properties | # Similarity function with given properties
I would like to find a similarity function $f$ between two values (each value is continuous and is bounded by $[0,1]$) that would have the following properties:
$$f(1, 1) = 0.5$$
$$f(0.5, 0.5) =0.25$$
$$f(1, 0) = 1$$
$$f(0, 1) = 1$$
$$f(0, 0) = 0$$
Is there such a function in math? If not how could I design it?
• I took the liberty of editing your question. Please take a look and roll back if I misunderstood. Jun 8 '16 at 9:49
• If you want a similarity function, do you require that $f(a,b)=f(b,a)$ and that $f\geq 0$? Possibly also that $f(a,b)>0$ if $a\neq b$? Jun 8 '16 at 9:50
• I don't strickly require metric properties you mentioned. But this would be desirable to have them. Jun 8 '16 at 9:52
The function
$$f\colon [0,1]\times[0,1]\to[0,1], \quad(x,y)\mapsto \frac{1}{4}x+\frac{1}{4}y+\frac{3}{4}(x-y)^2$$
does what you want. Plus, it's positive, symmetric and definite ($x\neq y$ implies that $f(x,y)>0$).
Neither it nor its root is linearly homogeneous like a norm-derived distance function, though ($f(\lambda x, \lambda y)\neq\lambda f(x,y)$) - but that does not seem to possible anyway given your requirements.
I found it by estimating a linear model based on your input data, with covariates $x$, $y$ and $(x-y)^2$:
foo <- data.frame(a=c(1,.5,1,0,0),b=c(1,.5,0,1,0),y=c(.5,.25,1,1,0))
model <- lm(y~a*b+I((a-b)^2),foo)
xx <- yy <- seq(0,1,.01)
persp(x=xx,y=yy,z=outer(xx,yy,function(xx,yy)xx/4+yy/4+0.75*(xx-yy)^2))
• I guess now I can also try to specify more data points and make the function more precise, right? Jun 8 '16 at 10:05
• Yes indeed. Then it likely won't fit so nicely at the boundaries any more, though. Jun 8 '16 at 10:12
• Very clever solution. How did you hit upon the specific covariates you used, symmetry? Jun 8 '16 at 16:32
• @MatthewDrury: yes. As a matter of fact, after thinking about it, I'd rather use $(x+y)^2$ (instead of $x$ and $y$) and $(x-y)^2$, to enforce symmetry even with different input data. I don't like (dis)similarities that are not symmetric. Jun 8 '16 at 18:06 | 2021-09-21T13:35:51 | {
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https://math.stackexchange.com/questions/3001391/in-how-many-ways-can-n-dogs-and-k-cats-be-arranged-in-a-row-so-that-no-two-c | # In how many ways can $n$ dogs and $k$ cats be arranged in a row so that no two cats are adjacent?
There are $$n$$ dogs and $$k$$ cats. In how many ways can we arrange them in a row so there are no $$2$$ cats are adjacent?
I thought about trying to calculate the number of possibilities without any constraints and subtract the number of ways we can arrange them so there is at least $$1$$ pair of adjacent cats.
Would like to understand how to approach this problem correctly (and if there are multiple ways I would love to hear them).
Edit: Sorry, I didn't make myself clear.
What I meant was that the cats and dogs are different(each dog is different than the other and so are the cats).
And the row doesn't have to start with dogs - the only requirement is that we can't have $$2$$ or more cats in a row.
But the other question (if the cats and dogs are the same) is good too. If you have more ways to solve in case they are indistinguishable I would love to read and understand it too.
So, basically this 1 question turned to two:
Case 1: the cats are indistinguishable and the dogs too.
Case 2: they are distinguishable.
And $$n$$ is large enough for it to be possible.
What is the minimal $$n$$? A bit confused.
If we start with a cat, then $$n=k-1$$ seems enough (cat first, cat last).
If we start with a dog, $$n=k$$. So in general we need $$n$$ at least $$k$$ to arrange them properly? Am I right?
Again sorry for the confusion!
Edit 2: My 2 questions were answered - thank you!
If you have any other ways to approach the problem or wish to add anything - please do, I would gladly read it!
In particular, if you know how to place the cats first and then the dogs, I would like to hear it.
• Is there any information given about $n$ and $k$? – Prakhar Nagpal Nov 16 '18 at 17:40
• Are the cats distinguishable? What about the dogs? – darij grinberg Nov 16 '18 at 18:00
• There are two slightly different questions here. It is possible for no two cats to be adjacent if the row begins or ends with a cat that is adjacent to a dog. On the other hand, if you stipulate that there must be a dog on each side of each cat, the row must begin and end with a dog. Which problem did you have in mind? – N. F. Taussig Nov 16 '18 at 18:33
• @darijgrinberg I am sure that, unless you have identical twins, you should be able to distinguish different cats and dogs. The problem does not say anything about the amount of time you are allowed to learn about the differences between the animals. – Batominovski Nov 16 '18 at 18:57
Okay now, to solve this, we need a couple of constraints, the minimum value of $$n$$ has to be $$n=k-1$$ if it is more then also it doesn't make a difference. Now, let us say we keep the $$n$$ dogs in a row with space between each one. So clearly there will be $$n+1$$ spaces, now these $$n+1$$ spaces have to be occupied by $$k$$ cats. Thus we get $$\binom {n+1}{k}$$ This is assuming that the cats are all identical as are the dogs.
If not, then we get, $$\binom {n+1}{k}\cdot n! \cdot k!$$ The reason this works is to let us say, take an example where $$k=5$$ and $$n=5$$. Now we put the $$5$$ dogs in a line with space between them. Then out of the $$6$$ spots between the dogs, we have to put $$5$$ cats and this can be done in $$\binom{6}{5}$$ ways. Now in these $$6$$ ways, we will all sorts of combinations, one where say the $$2$$nd spot is not occupied, another where the $$3$$rd is left empty. As you do this you will realize why this is such a great method. It allows all possible combinations of boys together and apart while ensuring that the condition is met.
• The minimum value of $n$ is $k-1$. You can have $C,D,C,D,\ldots,C,D,C$ for example. (Here, $C$ is a cat and $D$ is a dog.) – Batominovski Nov 16 '18 at 18:58
• Yes, you are right, I have changed that now. Thank you – Prakhar Nagpal Nov 16 '18 at 18:59
• Thanks for your answer. 1 thing I would like to clarify is why n=k-1 is the min. Am I correct: case 1:If we start with cats then k-1 dogs are enough to cover them. case2: If we start with dogs then k-1 dogs are not enough and here we need at least k dogs. And because of case 2 we need at least k dogs? – Johnny Nov 16 '18 at 19:01
• This method requires you to start with dogs and lets us say you have $5$ cats and $4$ dogs. Now you put the dogs and leave spaces between them, giving you $5$ spaces which means you have to put each cat in one space, the sequence will be $C,D,C,D,C,D,C,D,C$ using the formula we derived you get $\binom{5}{5}=1$ which is indeed true as there is only one way to arrange them. – Prakhar Nagpal Nov 16 '18 at 19:03
• I see. thanks a lot! Are there other ways to approach this problem? For example can I start with cats and then place the dogs? – Johnny Nov 16 '18 at 19:15
If $$n\leq k-1$$, then it is impossible to do so, as even with sitting them alternately, you will not have sufficient number of dogs to do so. However, if we assume that's not the case, then there are $$n+1$$ places around dogs to put the cats.
Assuming that the dogs and cats are indistinguishable respectively, then all we have to do is pick out $$k$$ places for the cats out of $$n+1$$. So number of ways to do that is
$$N = {{n+1}\choose k}$$
If we assume that the dogs are all different and so are the cats, we multiply the above with the number of permutations for cats and dogs respectively.
$$N = {{n+1}\choose k} * n! * k!$$
• Thanks for your answer. 1 thing I would like to clarify is why n=k is the min. Am I correct: case 1:If we start with cats then k-1 dogs are enough to cover them. case2: If we start with dogs then k-1 dogs are not enough and here we need at least k dogs. And because of case 2 we need at least k dogs? – Johnny Nov 16 '18 at 19:02
• Firstly, as you can see in my $1st$ line, minimum is not $k$, $k-1$ dogs are enough. Secondly, in your case $2$ also $k-1$ dogs will be enough because we consider all the alternate positions including both the ends which gives us $k$ places for the cats – Sauhard Sharma Nov 16 '18 at 19:11 | 2019-06-18T06:59:39 | {
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https://math.stackexchange.com/questions/489828/the-case-of-being-vacuously-true-trivially-true-and-the-empty-set | # The case of being “vacuously true / trivially true” and the empty set
Let's suppose that we need to prove a statement of the form $\forall x(x\in A\longrightarrow\phi)$, where $\phi$ is any property (for example when we need to prove that $A\subseteq A\cup B$). Is it really necessary to consider the case when $A=\emptyset$?
I mean deductively we are only concerned when $x\in A$ is true. This means that we can directly consider the fact $x\in A$ as true and then deduce $\phi$ and we don't care about the case $A=\emptyset$ so we can forget about it. But then I'm confused here becase I've read in some books and seen some of my teachers saying something like
"if $A=\emptyset$ then the case is trivialy true because.... Let's now suppose that $\exists x\in A$, then....".
In the case of my example I would simply write something like:
Let's suppose an arbitrary $x$ such that $x\in A$. Then $x\in A \vee x \in B$. This means $x\in A\cup B$, by definition. Therefore $\forall x(x\in A\longrightarrow x\in A\cup B)$, which means $A\subseteq A\cup B$.
So, my question here is if my proof is really complete or if I'm missing the case when $A=\emptyset$.
In a more general case I can think of proving somegthing of the form $\forall x(\phi\longrightarrow \psi)$. Then I would just write as a proof by using deduction something like
"Let's suppose an arbitrary $x$ such that $\phi$ then... but this implies that... therefore $\psi$. This means $\forall x(\phi\longrightarrow \psi)$".
But here I don't know if I have to consider the case when $\phi$ is not satisfied by $x$ and mention that in the proof to be complete.
• Technically, your thoughts are completely correct. Although $\forall\ A (\text{bla})$ seems like a typo ($\forall\ x$) – AlexR Sep 10 '13 at 19:59
• @AlexR Thanks. Yes, sorry, that was a typo. I made the correction already. So, basically you would say that when the "trivial" case is made explicit it's just to make enphasis or something like that, but it's redundant? – Daniela Diaz Sep 10 '13 at 20:15
• If you try to prove something for all elements in a set, yes. Usually however, the "emphasis" is needed - if you prove something for the set, such as that it is disjoint to another set and you want to have an element from the set, you'll have to exclude $A\neq\emptyset$ before you do that... – AlexR Sep 10 '13 at 20:18
If you want to prove a statement of the form $$\forall x: \quad (\Phi(x) \Rightarrow \Psi(x))$$ You can assume $x$ with $\Phi(x)$ to exist; in other words you need not require $$\{x : \Phi(x)\} \neq \emptyset$$ for the proof. (So all statements hold for the elements of the empty set ;-) )
If you want to prove $$\Psi(A)$$ for some set $A$ and want to use an element $x\in A$ for the proof, you first need to show that $$A \neq \emptyset$$ or make a branch for this as a special case.
• $$\Psi(A) : A = \mathbb{R}$$ Showing that $A$ is open and closed doesn't suffice, because $A$ also needs to be nonempty... Sorry I can't make up a better example from scratch. – AlexR Sep 10 '13 at 20:33 | 2020-02-25T22:32:41 | {
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https://www.physicsforums.com/threads/find-the-rate-that-the-distance-between-two-projectiles-is-changing.950940/ | # Find the rate that the distance between two projectiles is changing
## Homework Statement
The parameteric equations for the paths of two projectiles are given below. At what rate is the distance between the two objects changing at t = pi/2?
## Homework Equations
x1=12cos(2t) y1 = 6sin(2t)
x2=6cos(t) y2 = 7sin(t)
## The Attempt at a Solution
I am completely stuck. So I was thinking that I should use the distance formula
sqrt( (6cos(t) - 12cos(2t))2 + ( 7sin(t) - 6sin(2t))2))
then take the derivative of that then plug in pi/2 but that got really ugly and I think I am overcomplicating it.
I am looking for a hint here that is all. completely stuck
Last edited by a moderator:
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Isaac0427
Gold Member
I am completely stuck. So I was thinking that I should use the distance formula
sqrt( (6cos(t) - 12cos(2t))2 + ( 7sin(t) - 6sin(2t))2))
then take the derivative of that then plug in pi/2 but that got really ugly and I think I am overcomplicating it.
Why do you believe that is over complicating it?
Also, if you know the chain rule very well the derivative isn’t that bad.
Edit: ignore the deleted part (if you saw it) about the algebra trick. I tried it and it was a bad idea. Pretty beautiful, but not less complicated.
vela
Staff Emeritus
Homework Helper
I am looking for a hint here that is all. completely stuck
Your method is fine, but there is a way that's a bit less tedious.
Let $\vec{r}_i = x_i \hat i + y_i \hat j$ and $\vec{r} = \vec{r}_1 - \vec{r}_2$. The distance $r$ between the two particles is $\| \vec{r} \|$, so you have
$$r^2 = \vec{r}\cdot\vec{r} = (\vec{r}_1 - \vec{r}_2)\cdot(\vec{r}_1 - \vec{r}_2) = \vec{r}_1\cdot \vec{r}_1 - 2 \vec{r}_1 \cdot \vec{r}_2 + \vec{r}_2\cdot\vec{r}_2.$$ Keeping everything in terms of vectors, calculate $d(r^2)/dt$ and then evaluate the resulting expression for $t=\pi/2$. Finally, relate $dr/dt$ to $d(r^2)/dt$.
You might find it helpful to sketch the paths of the two particles and their locations at $t=\pi/2$.
Delta2
Homework Helper
Gold Member
@vela methods looks nice and it might save you the trouble of calculating a single complex derivative at the cost of calculating dot products and simpler derivatives instead.
I tend to agree with @Isaac0427 the derivative looks scary but it isn't so hard if you know the chain rule well. | 2020-04-01T08:44:03 | {
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https://discuss.codechef.com/questions/143728/eartseq-editorial | ×
# EARTSEQ - EDITORIAL
Setter: Evgeniy Artemov
Tester: Xiuhan Wang
Editorialist: Taranpreet Singh
Easy-Medium
# PREREQUISITES:
Sieve of Eratosthenes, Number-Theory and Constructive Algorithms.
# PROBLEM:
Given an integer $N \geq 3$, find $N$ distinct numbers such that every pair of consecutive numbers have a common factor $> 1$ and all possible consecutive three elements have no common factor $> 1$. Also, all numbers found should be between $1$ and $10^9$. First and Last elements are also considered adjacent here.
# SUPER QUICK EXPLANATION
• A simple way to generate $N$ such distinct numbers is to take a prime and multiply two consecutive elements by this number. This way, the condition of co-primes is handled, but the numbers generated go beyond $10^9$.
• A better way would be to reuse smaller primes while ensuring all the generated numbers are distinct.
# EXPLANATION
This is a constructive problem, and hence, have multiple solutions. Feel free to share your solution in comments.
Note: If we represent a number as the product of prime numbers, Prime factorization of a common factor of the pair of numbers is the common part of prime factorization. If we consider prime values or product of few primes, they have relatively lesser factors as compared to composite numbers. So, for most of the problems involving prime factors/factorization/divisibility, prime numbers bear special importance.
Let us try similar problems first.
Try solving the same problem assuming there's no upper bound on maximum values. Now, we can just multiply a distinct prime number to every pair of consecutive numbers. This way, every two consecutive numbers are not co-prime while every three consecutive numbers are coprime.
The above solution is sufficient to solve the first subtask, but in the second subtask, the values exceed $10^9$, so we need something better.
Try solving the same problem, assuming we are allowed repeated values. Here, we can choose prime numbers and multiplying each position by exactly two primes such that every two consecutive positions share prime factor while every three consecutive numbers do not share any prime factor, taking special care to choose a different prime number when reaching the end of the array.
Now, let us solve the original problem.
Think up and try to mix the above solutions so as to reduce the magnitude of generated numbers while keeping each number distinct.
The construction used in my solution is to reserve the first few primes and then multiply the first two positions by a non-reserved prime, next two positions by another prime and so on. Now, We can multiply Last and first position by a reserved prime, second and third position by different reserved prime, and so on. This way, we are guaranteed to have every pair of consecutive positions divisible by that particular prime number, and every three consecutive positions are coprime since no prime is multiplied to three consecutive positions.
Implementation
For this problem, we can preprocess and calculate a list of primes beforehand using the sieve of Eratosthenes and then for every test case, use primes from the list instead of recomputing.
Interesting Fact
This problem can also be approached as a challenge problem where we need to minimize the maximum value generated in sequence. Can you minimize? Say for $N = 50000$
# Time Complexity
Time complexity is $O(MX*log(log(MX)) + \sum N)$ per test case where $MX$ is the upper limit till primes are found. Finding primes up to $10^6$ suffices for this problem.
# AUTHOR'S AND TESTER'S SOLUTIONS:
Feel free to Share your approach, If it differs. Suggestions are always welcomed. :)
This question is marked "community wiki".
4.0k31104
accept rate: 22%
19.8k350498541
2 I did by finding Euler Circuit of a complete graph of odd degree, and then adjusting it by removing the last edge and adding path of required length. answered 14 Jan, 20:33 5★fluffy_x 22●1 accept rate: 0% Here's my solution : https://www.codechef.com/viewsolution/22464087 (14 Jan, 22:51) fluffy_x5★
What I did was...
Three case :
# 33, 77, 14, 10, 15, 6*p1, 10*p2, 15*p3, 6*p4, 10*p5, ... so on
## Where p1,p2,p3,... are primes greater than 11...
(as I used primes till 11 in these case to make sequence N%3==0)
1.6k211
accept rate: 23%
basic idea is keep repeating 6,10,15 keeping all number different by multiplying with different primes...
We don't need more than N+5 primes for any case...
so 50005 primes in worst case... (akaik there are 78,498 primes upto value 10^6)
(14 Jan, 22:10)
# HERE is my solution
(14 Jan, 22:24)
1 One interesting property which I found was that "No three consecutive odd numbers have a common divisor" starting from 3. So, I tried multiplying 3 to the first 2 elements, then multiplied 5 to the second and third elements, then multiplied 7 to the third and fourth elements and so on... (completing the cycle with a prime number for the last and the first elements just in case if the first element and the second last element don't have a common divisor). This passed subtask-1, but didn't pass subtask-2. Maybe it was because of generating numbers > 10^9. Is there a modification I could do to this approach to get it ACed? answered 16 Jan, 11:01 10●2 accept rate: 0%
0 I am a bit disappointed that the property of coprimes weren't considered in the editorial. In fact, it is not necessary at all to generate prime numbers. If we consider each positive integer as a multiset of its prime factors, then the problem could be reduced to: Find N ordered multisets of integers that for every 2 consecutive sets, the intersection is not empty, but the each intersection of 2 consecutive intersections is an empty set, i.e. coprime to each other. Therefore if we find a list of consecutive coprimes, the result would be product of 2 consecutive elements of that list. Implementation in Python: from math import gcd from itertools import cycle, islice from sys import stdin CANDIDATES = cycle(range(2, 30000)) a, b, new = 30011, 30013, {} coprimes, result = [a, b], [a * b] for _ in range(49998): for i in CANDIDATES: if new.get(b * i, True) and gcd(a, i) == 1 == gcd(b, i): coprimes.append(i) a, b = b, i break new[a * b] = False result.append(a * b) next(stdin) for N in map(int, stdin): N -= 1 print(coprimes[N] * 30011, end=' ') print(*islice(result, N)) Since the first 2 coprimes are hardcoded to be primes, we can reuse the list for every N in the given constraint, thus the overall complexity is drastically reduced. answered 15 Jan, 09:23 3★mcsinyx 31●4 accept rate: 0%
0 What i did was ... 1)found out out some 4000 consecutive prime numbers using seive 2)appended them to a new array under the condition that the product of two consecutive numbers does not cross 10^9.This helped me pass the first sub task. 3)For the next part I started my series from 5 and accepted primes at different intervals starting from 2 and incrementing interval by 1 and also a different starting point for every interval. 4)this helped me generate all(50000 in this case) possible combinations of products of primes and all i had to do was consider the first 550 prime numbers, and also including the previous The time complexity was O(d^3),d=550. but since the inner loops depended on the interval it worked even faster. answered 15 Jan, 16:03 1●1 accept rate: 0%
0 my solution does not use generating primes at all. we forget some facts about numbers with small difference and their common factors.Here is my solution https://www.codechef.com/viewsolution/22244894 answered 16 Jan, 16:56 1 accept rate: 0%
0 Urzbeckuvayy...can u please explain ur logic a bit? answered 17 Jan, 14:27 2★sabios 1 accept rate: 0%
0 in java only 12500 primes can be stored for 50000 Bytes answered 17 Jan, 21:47 24●1●2●7 accept rate: 0%
0 https://www.codechef.com/viewsolution/22395844 what is wrong in this solution?. answered 18 Jan, 10:12 1●1 accept rate: 0%
0 Can you elaborate how you got this pattern @l_returns? answered 24 Jan, 00:45 2★a_m_k_18 21●1 accept rate: 0%
0 in the setter's solution can anyone explain how and int x = 4 and if (i == n - 1) {ind = 5;} how this two constant are fixed ? I tried some other constants and some of them give right answer and some of them dont't; @eartemov answered 29 Jan, 23:11 3●2 accept rate: 0%
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http://rsak.tenutematerdomini.it/power-series-representation.html | # Power Series Representation
Di erentiation and Integration of Power Series We have previously learned how to compute power series representations of certain functions, by relating them to geometric series. Power series take place in combinatory in the name of produce functions in the name of the Z-transform. Representation of functions as power series Consider a power series 1 −x2 +x4 −x6 +x8 +···= X∞ n=0 (−1)nx2n It is a geometric series with q = −x2 and therefore it converges for. Power Series Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode. Differentiation and integration of power series works in a way very similar to handling polynomials: look at the series term by term. Loading Power Series Approximation. Round your answer to the nearest thousandth. In fact, it may be shown that the power series converges to tan−1(−1) at x. The Next Module is. xn = 1 + x+ x2 + x3 + We know this series converges if and only if jxj< 1. What is this Magic? How can we turn a function into a series of power terms like this? Well, it isn't really magic. First we find the partial fraction decomposition for this function. Power series are basically like in nitely long polynomials. We can at first ignore the xterm on the numerator, since it can be multiplied later on in order to obtain the power series for 1 (1+4x)2. A power series $\displaystyle\sum_{n=0}^\infty c_n x^n$ can be thought of as a function of $x$ whose domain is the interval of convergence. If we can –nd a function f(x) such that. Its power series representation is: By identifying a and r, such functions can be represented by an appropriate power series. ? What is the power series representation and interval of convergence for the function x / [(2x^2) + 1]? More questions. That is, the function given by can be represented as a power series representation Example 1: Represent the function given by as a power series. Home Contents Index. LECTURE 3: ANALYTIC FUNCTIONS AND POWER SERIES We are interested in a class of differentiable functions called analytic functions. Inside the interval of convergence #x in (-1,1)# we can integrate the series term by term: #int_0^x dt/(1+t) = sum_(n=0)^oo int_0^x (-1)^nt^ndt#. Expanding to power series, and finding the Laurent Series. Then the set of points z for which the series converges is one of the following: (i) The single point. Differentiating power series I'm writing this post as a way of preparing for a lecture. Here is the question: f(x)=3x^3/(x-3)^2 im confused. r = 1, , r = 1, r = 1, and so on. Use the fact that to write down a power series representation of the logarithmic function. It will also be useful to remember the following power series derived from the geometric series: $\sum_ {n=0}^ {\infty} x^n = 1 + x + x^2 + = \frac {1} {1 - x}$ for $-1 < x < 1$. If you have questions or comments, don't hestitate to. A subset U ⊆ C is called open if for every z ∈ C there exist r z > 0 such that the ball B r z (z) ⊂ U. Since the series for ln (1 - x) has no constant term, we may also divide by x to get a new power. Taylor series is a special power series that provides an alternative and easy-to-manipulate way of representing well-known functions. Power Series Representations of Analytic Functions 1 IV. It will also be useful to remember the following power series derived from the geometric series:. For example, we take zn= n+ 1 2n so that the complex sequence is {zn} = ˆ1 + i 2, 2 + i 22, 3 + i 23,··· ˙. Its power series representation is: By identifying a and r, such functions can be represented by an appropriate power series. The Complex Exponential Function. Each time I see one of these posts about information representation in R, I get this tingle to test the breaking points of Power BI. (Center your power series representation at x = 0. Evaluate Z 1 1+x7 dx as a power series. 7: Taylor and Maclaurin Series Taylor and Maclaurin series are power series representations of functions. Power series take place in combinatory in the name of produce functions in the name of the Z-transform. Find the radius of convergence of the power series. The Maclaurin series is a template that allows you to express many other functions as power series. Representation of functions as power series Consider a power series 1 −x2 +x4 −x6 +x8 +···= X∞ n=0 (−1)nx2n It is a geometric series with q = −x2 and therefore it converges for. - [Voiceover] What I would like us to do in this video is find the power series representation of or find the power series approximitation (chuckles) the power series approximation of arctangent of two x centered at zero and let's just say we want the first four nonzero terms of the power series. ” This becomes clearer in the expanded […]. that power series always converge in a disk jz aj 1 Since the series for x = ¡1 is the negative of the above series, [¡1;1] is the interval of convergence of the power series. The same argument works for sin, hence: Theorem. Recall that by the Geometric Series Test, if jrj<1, then X1 n=0 arn = a 1 r: Therefore, if jxj<1, then the power series X1 n=0 axn = a 1 x: Example: Find the sum of. Given a power series. converging for #abs q < 1#. When Can We Differentiate a Power Series? For the purposes of this module, we will always assume that we can. Taylor Series Expansions In the previous section, we learned that any power series represents a function and that it is very easy to di¤erentiate or integrate a power series function. Wolfram alpha paved a completely new way to get knowledge and information. Power series are useful in analysis since they arise as Taylor series of infinitely differentiable functions. lim n→ a n 1 a n. LECTURE 3: ANALYTIC FUNCTIONS AND POWER SERIES We are interested in a class of differentiable functions called analytic functions. We do so in this section. 5) g(x) x 6) g(x). Differentiation to find power series representation for 1/(8+x)^2??? 8 How does this nursery rhyme pertain to power series: “There was a little girl Who had a little curl Right in the middle of her forehead…". First notice that 1 (1 +4x)2. Differentiation and integration are useful techniques for finding power series representations of functions. Power BI Time Series Graph. In fact, Borel's theorem implies that every power series is the Taylor series of some smooth function. Here’s a little how-to on figuring out the power series of tan(x), cot(x) and csc(x). Proposition IV. For the series on the right side change the index to k and rewrite it as. Expanding to power series, and finding the Laurent Series. First we find the partial fraction decomposition for this function. There is however a theorem on differentiating and integrating power series, which you are not expected to know, that tells us that a power series can only be differentiated if it has a radius of convergence that is greater than zero. Di erentiation and Integration of Power Series We have previously learned how to compute power series representations of certain functions, by relating them to geometric series. The power series expansion of the exponential function Let represent the exponential function f ( x ) = e x by the infinite polynomial (power series). For both series, the ratio of the nth to the (n-1)th term tends to zero for all x. Hence the power series converges on [−1,1]. The left side can be written Let k = n - 2 and this becomes Hence y' = xy can be written. integral of (ln(1-x)dx)/x Also, is the radius of convergence just 1? Upload failed. Power Series. Find a power series for $$\large\frac{{6x}}{{5{x^2} - 4x - 1}}\normalsize. Power series for cos,sin 12 We compute the radius of convergence for the coefficients given by a n = 0 n = 2k +1 (−1)k (2k)! n = 2k Now Stirling’s formula allows to show that n p |a n| → 0 as n → ∞ and thus r = ∞. The first is the power series expansion and its two important generalizations, the Laurent series and the Puiseux series. We will extensively use algebraic operations, differentiation, and integration of power series. In the first lesson you will start with a power series and determine the function represented by the series. The Complex Exponential Function. lim n→ a n 1 a n. ” This becomes clearer in the expanded […]. Power BI Time Series Graph. Since e is an irrational number (see proof that e is irrational), it cannot be represented as the quotient of two integers, but it can be represented as a continued fraction. INFINITE SEQUENCES AND SERIES 11. Lecture Notes. , a continuously differentiable function) has a power series representation. Trigonometric functions. Find a power series representation for. Observe that when x 0 the power series ∑ n 0 b nxn b 0 is convergent; and when x a the power series ∑ n 0 b n x −a n b 0 is convergent. The first question we shall answer through a number of examples and by utilizing tools we have developed in Calculus. I want to discuss the result that a power series is differentiable inside its circle of convergence, and the derivative is given by the obvious formula. Recall that by the Geometric Series Test, if jrj<1, then X1 n=0 arn = a 1 r: Therefore, if jxj<1, then the power series X1 n=0 axn = a 1 x: Example: Find the sum of. Convergence of Series. Wolfram alpha paved a completely new way to get knowledge and information. Find a power series representation for the function. Differentiating power series I'm writing this post as a way of preparing for a lecture. The mathematical constant e can be represented in a variety of ways as a real number. In the last two lessons you will begin with a function. Finding a Power Series Representation for a Logarithm Function - Duration: 9:40. What other functions can be realized as power series? We shall answer the second question mainly in the next section. Use the fifth partial sum of the power series for sine or cosine to approximate each value. y The series converges only at the center x= aand. Use the fact that to write down a power series representation of the logarithmic function. Here is the question: f(x)=3x^3/(x-3)^2 im confused. In step 1, we are only using this formula to calculate the first few coefficients. 3) sin 4) sin Use n xn x to find a power series representation of g(x). Larger examples of the power series method 3. Power Series Representation of Functions [affmage source=”ebay” results=”100″]Power Series[/affmage] [affmage source=”amazon” results=”10″]Power. For both series, the ratio of the nth to the (n-1)th term tends to zero for all x. Observe that when x 0 the power series ∑ n 0 b nxn b 0 is convergent; and when x a the power series ∑ n 0 b n x −a n b 0 is convergent. The approach really becomes useful when there is no other good way of representing a function. Limits like are “easy” to compute, since they can be rewritten as follows. Likewise, since every power of in the power series for cosine is even, we can see cosine is an even function. I want to discuss the result that a power series is differentiable inside its circle of convergence, and the derivative is given by the obvious formula. Find the radius of convergence of the power series. Taking the derivative gives f0(x) = c 1 + 2c 2(x a) + 3c 3(x a)2 + 4c 4(x a)3 + : Then f0(a) = c 1. The function is called a power series, with center. What is Power series? A power series is a series of the form. It will also be useful to remember the following power series derived from the geometric series:. Find a power series for \(\large\frac{{6x}}{{5{x^2} - 4x - 1}}\normalsize. For all x ∈ R cos(x) = X∞ k=0 (−1) kx2 (2k)!, sin(x) = X∞ k=0 (−1)kx2k+1 (2k +1)!. A power series \displaystyle\sum_{n=0}^\infty c_n x^n can be thought of as a function of x whose domain is the interval of convergence. Find a power series representation for the function. Taylor and Maclaurin (Power) Series Calculator The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. 5) g(x) x 6) g(x). Trigonometric functions. Given a power series. Section 4-15 : Power Series and Functions. Power BI Time Series Graph. 11 we shall obtain power series for sin x and cos x by another method. For both series, the ratio of the nth to the (n-1)th term tends to zero for all x. What is the center of the power series? For what values of x will this representation be valid? You might want to check your answer graphically, if you have a graphing calculator or access to a Math software program. The same argument works for sin, hence: Theorem. Geometric Series. Power Series Representation: To form the series expansion for the given function, we will utilize some basic mathematical tools such as partial fraction decomposition and the geometric series. Find a power series representation for the function. It will also be useful to remember the following power series derived from the geometric series:. An eigenvalue problem solved by the power series method 5 6 48 89 Stand out from the crowd Designed for graduates with less than one year of full-time postgraduate work. Let f(x) = X1 n=0 c n(x a)n = c 0 + c Find the Taylor series for f(x) = e3x centered at. Loading Power Series Approximation. Throughout these pages I will assume that you are familiar with power series and the concept of the radius of convergence of a power series. Taylor and Maclaurin series are power series representations of functions. Find a power series representation for the function f(x) = 1 (1−x)2. Since every power of in the power series for sine is odd, we can see that sine is an odd function. We still have not shown that an analytic function (i. It is the source of formulas for expressing both sin x and cos x as infinite series. We can easily get new power series by multiplying by x p. Thus both series are absolutely convergent for all x. Power Series Representations of Analytic Functions 1 IV. Find a geometric power series for the function: Make this 1 Divide numerator and denominator by 4 a = 3/4 r = x/4 Use a and r to write the power series. General remarks. Functions represented as a Power Series Power Series Function Blank Page 1. The first question we shall answer through a number of examples and by utilizing tools we have developed in Calculus. Expanding to power series, and finding the Laurent Series. A subset U ⊆ C is called open if for every z ∈ C there exist r z > 0 such that the ball B r z (z) ⊂ U. we can find its derivative by differentiating term by term: Here we used that the derivative of the term an tn equals an n tn-1. Free Taylor/Maclaurin Series calculator - Find the Taylor/Maclaurin series representation of functions step-by-step. Power Series Representation Maclaurin series Power Series, Maclaurin Series, Remainder Estimation Theorem and Euler's Formula Taylor Series: open interval of convergence Need Assistance Understanding Taylor and Maclaurin Series Differentiation: Mean Value Theorem Electricity Topics: Fields, Resistance, Series and Parallel Federalists Compared. First we find the partial fraction decomposition for this function. Larger examples of the power series method 3. Power Series Representations of Analytic Functions Recall. Differentiating power series I'm writing this post as a way of preparing for a lecture. We recall the geometric series ¥ å n =0 x n=1 +x +x 2 + x + = 1 1 x; for jx j<1 : Example1 1. Differentiation to find power series representation for 1/(8+x)^2??? 8 How does this nursery rhyme pertain to power series: “There was a little girl Who had a little curl Right in the middle of her forehead…". Today we'll coordinate the information representation control in Power BI to the ARR in R Programming. The exponential function is the infinitely differentiable function defined for all real numbers whose. The power series is printed out as a sum of terms, ending with O[x] raised to a power: Internally, however, the series is stored as a SeriesData object:. Find the radius of convergence of the power series. For the series on the right side change the index to k and rewrite it as. Many properties of the cosine and sine functions can easily be derived from these expansions, such as (−) = − (). Power series are represented in the Wolfram System as SeriesData objects. A power series \displaystyle\sum_{n=0}^\infty c_n x^n can be thought of as a function of x whose domain is the interval of convergence. In step 1, we are only using this formula to calculate the first few coefficients. Library Research Experience for Undergraduates. Express 1 = 1 x 2. Power Series Representation : Here we will use some basic tools such as Geometric Series and Calculus in order to determine the power series representation for the given function. The left side can be written Let k = n - 2 and this becomes Hence y' = xy can be written. Convergence of Series. If we can –nd a function f(x) such that. Determining a Function Representing a Power Series. The particular decimal details for actual numbers recognize how to also be study as an example of a power series, with integer coefficients, but with the case x set at 1? 10. Power Series Representation of Functions [affmage source=”ebay” results=”100″]Power Series[/affmage] [affmage source=”amazon” results=”10″]Power. In similar ways, other functions can be represented by power series. Power series are basically like in nitely long polynomials. We can at first ignore the xterm on the numerator, since it can be multiplied later on in order to obtain the power series for 1 (1+4x)2. Let f(x) = X1 n=0 c n(x a)n = c 0 + c Find the Taylor series for f(x) = e3x centered at. In fact, for fun, let's take the anti-derivative of both sides of this, and if we do that, then we will have shown essentially a geometric series representation of whatever the anti-derivative of this thing is. Power series tables. Find a power series for \(\large\frac{{6x}}{{5{x^2} – 4x – 1}} ormalsize. Today we'll coordinate the information representation control in Power BI to the ARR in R Programming. Here’s a little how-to on figuring out the power series of tan(x), cot(x) and csc(x).$$ Solution. Thread: Find the first five non-zero terms of a power series representation centered at x=0. Series representations. Start with the generating function for the Bernoulli numbers:. In Section 9. Series representations. Power Series Representation of Functions [affmage source=”ebay” results=”100″]Power Series[/affmage] [affmage source=”amazon” results=”10″]Power. You can also see the Taylor Series in action at Euler's Formula for Complex Numbers. The approach really becomes useful when there is no other good way of representing a function. For instance, look at the power series. INFINITE SEQUENCES AND SERIES 11. When Can We Differentiate a Power Series? For the purposes of this module, we will always assume that we can. At x = 1, the power series becomes X∞ n=0 2n(−+11)n= 1 − 1 3 + 1 5 1 7 ···, which also converges by Leibniz’s theorem. In this section, we are going to use power series to represent and then to approximate general functions. We can at first ignore the xterm on the numerator, since it can be multiplied later on in order to obtain the power series for 1 (1+4x)2. For example, we take zn= n+ 1 2n so that the complex sequence is {zn} = ˆ1 + i 2, 2 + i 22, 3 + i 23,··· ˙. Binomial series Hyperbolic functions. Taylor and Maclaurin (Power) Series Calculator The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Logarithms and exponentials. Limits like are “easy” to compute, since they can be rewritten as follows. Power Series Representation Calculator This super useful calculator is a product of wolfram alpha, one of the leading breakthrough technology & knowledgebases to date. The left side can be written Let k = n - 2 and this becomes Hence y' = xy can be written. Power series take place in combinatory in the name of produce functions in the name of the Z-transform. Many properties of the cosine and sine functions can easily be derived from these expansions, such as (−) = − (). Let # x = -q # to have: #sum_(n=0)^oo (-1)^nx^n = 1/(1+x)#. We can easily get new power series by multiplying by x p. Here’s a little how-to on figuring out the power series of tan(x), cot(x) and csc(x). Since every power of in the power series for sine is odd, we can see that sine is an odd function. Can this power series be used to construct other power series which are also functions? Question 1. That is, we can substitute in different values of to get different results. We can obtain power series representation for a wider variety of functions by exploiting the fact that a convergent power series can be di erentiated, or integrated,. What is the center of the power series? For what values of x will this representation be valid? You might want to check your answer graphically, if you have a graphing calculator or access to a Math software program. In general, a power series will converge as long as has no reason not too! $$\frac{1}{1+x^2}$$ is defined for all complex x except i or -i. But first let me explain the notion of open sets. Each time I see one of these posts about information representation in R, I get this tingle to test the breaking points of Power BI. the sum of a power series is a function we can differentiate it and in-tegrate it. Simply stated, if f(z) is complex differentiable in a neighborhood of then it has a Taylor series representation. In this section, we are going to use power series to represent and then to approximate general functions. We will extensively use algebraic operations, differentiation, and integration of power series. Find a power series representation for the function. We will be representing many functions as power series and it will be important to recognize that the representations will often only be valid for a range of x ’s and that there may be values of x that we can plug into the function that we can’t plug into the power series representation. The first is the power series expansion and its two important generalizations, the Laurent series and the Puiseux series. LECTURE 3: ANALYTIC FUNCTIONS AND POWER SERIES We are interested in a class of differentiable functions called analytic functions. Power Series Representation : Here we will use some basic tools such as Geometric Series and Calculus in order to determine the power series representation for the given function. Expanding to power series, and finding the Laurent Series. Our starting point in this section is the geometric series: X1 n=0. The power series is printed out as a sum of terms, ending with O[x] raised to a power: Internally, however, the series is stored as a SeriesData object:. A power series is an infinite sum of the form The ai are called the coefficients of the series. Recall that by the Geometric Series Test, if jrj<1, then X1 n=0 arn = a 1 r: Therefore, if jxj<1, then the power series X1 n=0 axn = a 1 x: Example: Find the sum of. (Center your power series representation at x = 0. Enter a function of x, and a center point a. In similar ways, other functions can be represented by power series. lim n→ a n 1 a n. The result is another function that can also be represented with another power series. In this section we present some results that are useful in helping establish properties of functions defined by power series. the sum of a power series is a function we can differentiate it and in-tegrate it. Let us start with the formula 1 1¡x = X1 n=0. Let us use the Ratio Test to check the convergence of a power series of the form ∑ n 0 b nxn. Example 2: Represent the function given by as a power series. The left side can be written Let k = n - 2 and this becomes Hence y' = xy can be written. Di erentiation and Integration of Power Series We have previously learned how to compute power series representations of certain functions, by relating them to geometric series. xn = 1 + x+ x2 + x3 + We know this series converges if and only if jxj< 1. That is, we can substitute in different values of to get different results. In the last two lessons you will begin with a function. Taylor and Laurent series Complex sequences and series An infinite sequence of complex numbers, denoted by {zn}, can be considered as a function defined on a set of positive integers into the unextended complex plane. Subject: Power series representations Is there a systematic way of finding a power series representation of a function? I understand that you have to manipulate the function so that it is of the form 1/(1-x), but beyond that I am lost. Lecture Notes. Then the set of points z for which the series converges is one of the following: (i) The single point. First we find the partial fraction decomposition for this function. Thread: Find the first five non-zero terms of a power series representation centered at x=0. In step 1, we are only using this formula to calculate the first few coefficients. In Section 9. Since e is an irrational number (see proof that e is irrational), it cannot be represented as the quotient of two integers, but it can be represented as a continued fraction. It will also be useful to remember the following power series derived from the geometric series:. REPRESENTATION OF FUNCTIONS AS POWER SERIES 349 Finding a Series Representation Using Di⁄erentiation This time, we –nd the series representation of a given series by di⁄erentiating the power series of a known function. A power series is a polynomial with infinitely many terms. Return to the Power Series starting page. Determining a Function Representing a Power Series. Power Series Representation : Here we will use some basic tools such as Geometric Series and Calculus in order to determine the power series representation for the given function. radius of convergence of the power series. a k = (k + 2)a k+2 for k = 0, 1, 2 ··· I hope this helps, Harley. In the first lesson you will start with a power series and determine the function represented by the series. Power Series Representation: To form the series expansion for the given function, we will utilize some basic mathematical tools such as partial fraction decomposition and the geometric series. Find a geometric power series for the function: Make this 1 Divide numerator and denominator by 4 a = 3/4 r = x/4 Use a and r to write the power series. ) f(x) =1/(7 + x) asked by Anonymous on April 26, 2019; calculus II (a) Use differentiation to find a power series representation for f(x) = 1/(5 + x)^2 What is the radius of convergence, R?. , where the new coefficients of the powers of x are calculated according to the usual rules. \) Solution. converging for #abs q < 1#. Return to the Power Series starting page. I want to discuss the result that a power series is differentiable inside its circle of convergence, and the derivative is given by the obvious formula. Power series are basically like in nitely long polynomials. Taylor and Maclaurin (Power) Series Calculator The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. The radius of the "disk" of convergence in the complex numbers is 1 so, restricting to the real numbers, the radius of the interval of convergence is also 1. Use the fact that to write down a power series representation of the logarithmic function. General remarks. There are three main possibilities to represent an arbitrary function as an infinite sum of simple functions. A power series $\displaystyle\sum_{n=0}^\infty c_n x^n$ can be thought of as a function of $x$ whose domain is the interval of convergence. Throughout these pages I will assume that you are familiar with power series and the concept of the radius of convergence of a power series. (1a) When you want to find a power series representation of functions involving 1 (a+bx)ℓ, with a, b and ℓconstants, the idea is to use derivatives and the geometric series. What is the center of the power series? For what values of x will this representation be valid? You might want to check your answer graphically, if you have a graphing calculator or access to a Math software program. Likewise, since every power of in the power series for cosine is even, we can see cosine is an even function. For a given power series, we want to know for what value(s) of x,the power series converges. CHAPTER 12 - FORMULA SHEET 2 POWER SERIES Recall the notion of an in nite series. REPRESENTATION OF FUNCTIONS AS POWER SERIES 349 Finding a Series Representation Using Di⁄erentiation This time, we –nd the series representation of a given series by di⁄erentiating the power series of a known function. Formal power series. Since the series in continuous on its interval of convergence and sin¡1(x) is continuous. Lecture 20: Power Series Representations 20-3 which converges by Leibniz’s theorem. A power series is an infinite sum of the form The ai are called the coefficients of the series. The same argument works for sin, hence: Theorem. First of all, we. Start with the generating function for the Bernoulli numbers:. In this section we present some results that are useful in helping establish properties of functions defined by power series. Find a power series representation for the function and determine the interval of convergence. Differentiation and integration of power series works in a way very similar to handling polynomials: look at the series term by term. In the last two lessons you will begin with a function. Here’s a little how-to on figuring out the power series of tan(x), cot(x) and csc(x). 11 we shall obtain power series for sin x and cos x by another method. As the names suggest, the power series is a special type of series and it is extensively used in Numerical Analysis and related mathematical modelling. Having a power series representation of a function on an interval is useful for the purposes of integration, dierentiation and solving dierential equations. The particular decimal details for actual numbers recognize how to also be study as an example of a power series, with integer coefficients, but with the case x set at 1? 10. It will also be useful to remember the following power series derived from the geometric series: $\sum_ {n=0}^ {\infty} x^n = 1 + x + x^2 + = \frac {1} {1 - x}$ for $-1 < x < 1$. Loading Power Series Approximation. You can also see the Taylor Series in action at Euler's Formula for Complex Numbers. Use the fifth partial sum of the power series for sine or cosine to approximate each value. Thus both series are absolutely convergent for all x. What is the radius of convergence? Determine the radius of convergence and the interval of convergence. First we find the partial fraction decomposition for this function. Wolfram alpha paved a completely new way to get knowledge and information. Determining a Function Representing a Power Series. lim n→ a n 1 a n. In many situations c (the center of the series) is equal to zero, for instance when considering a Maclaurin series. ” This becomes clearer in the expanded […]. Find a power series representation for. We still have not shown that an analytic function (i. The particular decimal details for actual numbers recognize how to also be study as an example of a power series, with integer coefficients, but with the case x set at 1? 10. 3) sin 4) sin Use n xn x to find a power series representation of g(x). 5) g(x) x 6) g(x). Using the arctan Power Series to Calculate Pi March 12, 2015 Dan No Comments (Note: this post is an extension on the calculating pi with python post from a couple of years back. The radius of the "disk" of convergence in the complex numbers is 1 so, restricting to the real numbers, the radius of the interval of convergence is also 1. Convergence of Series. Taylor and Maclaurin (Power) Series Calculator The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Power Series Representation Maclaurin series Power Series, Maclaurin Series, Remainder Estimation Theorem and Euler's Formula Taylor Series: open interval of convergence Need Assistance Understanding Taylor and Maclaurin Series Differentiation: Mean Value Theorem Electricity Topics: Fields, Resistance, Series and Parallel Federalists Compared. Such sums can be added, multiplied etc. In many situations c (the center of the series) is equal to zero, for instance when considering a Maclaurin series. In fact, it may be shown that the power series converges to tan−1(−1) at x. In fact, Borel's theorem implies that every power series is the Taylor series of some smooth function. REPRESENTATION OF FUNCTIONS AS POWER SERIES 349 Finding a Series Representation Using Di⁄erentiation This time, we –nd the series representation of a given series by di⁄erentiating the power series of a known function. Series Solutions: Taking Derivatives and Index Shifting. CALCULUS Understanding Its Concepts and Methods. Throughout these pages I will assume that you are familiar with power series and the concept of the radius of convergence of a power series. Start from the sum of the geometric series: #sum_(n=0)^oo q^n = 1/(1-q)#. Library Research Experience for Undergraduates. Taking the derivative gives f0(x) = c 1 + 2c 2(x a) + 3c 3(x a)2 + 4c 4(x a)3 + : Then f0(a) = c 1. Expanding to power series, and finding the Laurent Series. Home Contents Index. Veitch The trick is rewrite 4 1 + x2 so that it can look something like a 1 r = P arn. Indicate the interval in which the series converges. Limits like are “easy” to compute, since they can be rewritten as follows. For both series, the ratio of the nth to the (n-1)th term tends to zero for all x. Power series take place in combinatory in the name of produce functions in the name of the Z-transform. | 2019-11-21T20:46:23 | {
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https://math.stackexchange.com/questions/2838694/in-pca-why-for-every-x-in-mathbbrn-x-sum-k-1n-ut-k-x-space-u-k | # In PCA, why for every $x \in \mathbb{R}^n$, $x=\sum_{k=1}^n u^T_k x \space u_k$?
In PCA, why for every $x \in \mathbb{R}^n$, $x=\sum_{k=1}^n (u^T_k x) \space u_k$?
Where $\{u_1,...,u_n\}$ is orthonormal basis and $||u||^2=u^T_i u_i=1 \forall i$.
Is this some standard vector projection?
• It is a standard result on orhonormal bases, assuming the $u_k$ are one such. – Adomas Baliuka Jul 2 '18 at 16:54
• @AdomasBaliuka Where can I see it? In order to recall why it's so. – mavavilj Jul 2 '18 at 16:57
• I found : Let $V$ be a finite-dimensional inner product space, and let $\{e_1,e_2,...,e_n\}$ be an orthonormal basis of $V$. Then for each vector $v \in V$ we have that $v=<v,e_1>e_1+<v,e_2>e_2+...+<v,en>e_n$. However, how does this transform into the one with vector transpose? – mavavilj Jul 2 '18 at 17:00
• $\langle v, e_1 \rangle e_1 = \langle e_1, v \rangle e_1 = (e_1^\top v) e_1$ – angryavian Jul 2 '18 at 17:01
If $(u_1,\dots,u_n)$ is an orthonormal basis of any vector space $V$ equipped with an inner product $\langle \dot{}, \dot{} \rangle$ then
$$\forall x \in V,\; x =\sum_{k=1}^n \langle u_k, x\rangle u_k.$$
Indeed, if $a_1,\dots,a_n$ are the coordinates of $x$ in this basis, then $\displaystyle x=\sum_{j=1}^n a_j u_j$ and $\forall k\in \{1,\dots,n\},\;\langle x,u_k\rangle = \sum_{j=1}^n a_j \langle u_j,u_k\rangle = a_k$ since $\langle u_j,u_k\rangle = 1$ if $i=j$ and $0$ otherwise.
To conclude, note that on $\Bbb R^n$, the map defined by: $$\forall x\in\Bbb R^n,\;\forall y\in\Bbb R^n,\;\langle x,y\rangle := x^Ty$$ Is an inner product on $\Bbb R^n$.
• BTW, p.9 of this (math.ust.hk/~mabfchen/Math111/Week13-14.pdf) gives a proof for the orthogonal projection (into subspace $W \subset V$). However, how does it extend to all $V$? – mavavilj Jul 3 '18 at 13:11
• Take simply $W=V$: it's also a subspace of $V$ and in this case, the orthogonal projection onto $W=V$ is simply $\text{Id}_V$. – paf Jul 3 '18 at 16:26 | 2019-05-22T11:47:14 | {
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https://math.stackexchange.com/questions/1459866/formula-for-the-sum-of-n-cdot-1-n-1-cdot-2-2-cdot-n-1-1-cdot | Formula for the sum of $\ n\cdot 1 + (n-1)\cdot 2 + … + 2 \cdot (n-1) + 1\cdot n$ [duplicate]
I'm looking for a simpler formula to calculate this sum:
$$n\cdot 1 + (n-1)\cdot 2 + ... + 2 \cdot (n-1) + 1\cdot n$$ Alternate representation (but should be equal to the above): $$\sum \limits_{k=1}^{n}(n+1-k)\cdot k$$
Rationale behind requested formula
When reviewing code implementing a minion game, the problem of how many substrings exists within a long text, and how many characters will the sum of all these substrings be came up. I found that the number of substrings is:
$$n + n-1 + n-2 + ... + 2 + 1 = \frac{(n+1)\cdot n}{2}$$
Reasoning for finding number of substrings is that you have $n$ substrings of length $1$, $n-1$ of length $2$, and so on until the end where you only have $1$ substring of length $n$.
Using the same logic, to sum up the length of all these substrings, we arrive at the formula at the top, which I would like to have simplified. That is, multiply the count of substrings with the length of the substring, and sum all these.
I have found the article "The Kth Sum of N Numbers", which seems to produce the number I want in the column for c3, with the $n$ enumerated as $rN$. But I can't read out of this article how the column is calculated, and what formula to use.
\begin{align} \sum_{k=1}^{n}(n+1-k)\cdot k&=\sum_{k=1}^{n}(n+1)k-\sum_{k=1}^n k^2\\ &=(n+1)\sum_{k=1}^n k-\sum_{k=1}^n k^2\\ &=\frac{n(n+1)^2}{2}-\frac{n(n+1)(2n+1)}{6}\\ &=\frac{n(n+1)(3n+3-2n-1)}{6}\\ &=\frac{n(n+1)(n+2)}{6} \end{align}
• For me, not too proficient in maths, this is by far the easiest to follow from start through end. – holroy Oct 1 '15 at 17:22
• @holroy: Thanks. – Ángel Mario Gallegos Oct 1 '15 at 17:23
Imagine that you’re to choose $3$ numbers from the set $A=\{0,\ldots,n+1\}$. If the middle number of your three is $k$, there are $k$ choices for the smallest number and $n+1-k$ choices for the largest number, so there are altogether $k(n+1-k)$ sets of $3$ having $k$ as the middle number. Clearly $k$ must range from $1$ through $n$, so there are altogether
$$\sum_{k=1}^nk(n+1-k)$$
ways to choose $3$ members of the set $A$. On the other hand, $A$ has $n+2$ members, so it has $\binom{n+2}3$ three-element subsets. Thus,
$$\sum_{k=1}^nk(n+1-k)=\binom{n+2}3=\frac{n(n+1)(n+2)}6\;.$$
• Even though I'm a master of Computer Science, it's been too long since I've studied math to easily follow your logic. But with a bit of thinking it does make sense, with exception of the last transformation (which I surely could look up somewhere). But I, by myself, would never have thought about this problem this way... – holroy Oct 1 '15 at 17:33
• @holroy: You can get it from the Wikipedia article on binomial coefficients, in particular this section. – Brian M. Scott Oct 1 '15 at 17:43
Hint: the sum is equal to $$(n+1)\sum^n_{k=1}k-\sum^n_{k=1}k^2.$$
Your sum is just the coefficient of $x^{n+1}$ in the product between $x+2x^2+3x^3+\ldots+n\,x^n$ and itself. It follows that:
\begin{align*} \sum_{j=1}^{n} j(n+1-j) &= [x^{n+1}]\left(\frac{x}{(1-x)^2}\right)^2 \\&= [x^{n-1}]\frac{1}{(1-x)^4}=\binom{n+2}{3}=\color{red}{\frac{(n+2)(n+1)n}{6}}.\end{align*}
• How do you see that? Is it just common knowledge that it is this coefficient? – holroy Oct 1 '15 at 17:29
• @holroy: multiply by hand $x+2x^2+\ldots+nx^n$ and $x+2x^2+\ldots+nx^n$, then check what is the coefficient of $x^{n+1}$. It is clearly the coefficient of $x$ times the coefficient of $x^n$, plus the coefficient of $x^2$ times the coefficient of $x^{n-1}$ and so on. You just made you first step in analytic combinatorics ;) – Jack D'Aurizio Oct 1 '15 at 17:34
\begin{align} \sum_{k=1}^n(n+1-k)k&=\sum_{k=1}^n\sum_{j=k}^nk=\sum_{1\le k\le j\le n}k =\sum_{j=1}^n\sum_{k=1}^jk\\ &=\sum_{j=1}^n \binom {j+1}2=\binom {n+2}3\color{lightgrey}{=\frac{(n+2)(n+1)n}6}\qquad\blacksquare \end{align} | 2019-12-10T17:01:28 | {
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http://zgnd.mariebourgeret.de/bending-stress-formula.html | # Bending Stress Formula
002 Mechanics and Materials II Department of Mechanical Engineering MIT February 9, 2004. 13 ), or to determine the deflection of the beam from equation ( 13. We assume that we know E, I, L and q. In order to calculate maximum surface stress, you must know the bending moment, the distance from the neutral axis to the outer surface where the maximum stress occurs and the moment of inertia. Formula for Bending Stress The below mathematical formula is used to calculate bending stress of a beam in mechanical engineering to find the strength of materials. Solution: Given, , , , , , and ; and assume diameter. of bending moment and. We will determine the equations for finding the deflection curve and also find the deflections at specific points along the axis of the beam. The application of the axial load, P, then results in an additional moment P*∆; this is also know as second order effect because the added bending stress is not calculated directly. 26in^3, then the fiber bending stress has to 29. Stress is zero at the neutral axis (c = 0), which is at mid-height for a rectangular beam or any beam with section that is symmetrical about a horizontal axis. 2) (ii) Determine the stress in the steel bar at layer i =˝ (2. 1 Bending stress determination for pinion by using FEA Fig. The second element is a bending stress due to the off center load creating a moment of 600kn * 50 mm. ) L ∆ = clear span + SW (in. Principal stress magnitudes pp 22 xy xy 1, 2 xy 2 σ 2 σσ σσ = τ + ± − + Orientation of principal planes tan2 2 p xy xy θ τ σσ = − Maximum in-plane shear stress magnitude 2 or 2 xy xy pp max 2 2 max τ 12 σσ ττ σσ =± − += − 2 xy σ avg σσ = + tan2 2 s note: 45 xy xy θ sp σσ τ =− θθ − =± ° Absolute maximum shear stress magnitude absm ax 2 τ = σσ max − min Normal stress invariance σσ xy += σσ nt += σσ pp12+. Similar to other metal forming processes, bending changes the shape of the work piece, while the volume of material will remain the same. + Bending Stress Due to Pressure, psi 44,775 54,900 S5 Meridional Membrane Stress Due to Deflection, psi 590 N/A S6 Meridional Bending Stress Due to Deflection, psi 103,747 N/A. In other words, it is not load divided by area. 13) Slide No. Flexural Stresses In Beams (Derivation of Bending Stress Equation) General: A beam is a structural member whose length is large compared to its cross sectional area which is loaded and supported in the direction transverse to its axis. Joint width is 2. Seshu Adluri Bending terminology Moment of inertia Parallel axis theorem Flexural stress Average shear stress =V f/hw Yield moment, M Y Elastic Section modulus, S Plastic moment, M P Plastic section modulus, Z Beam (slab load) vs. The I term is the moment of inertia about the neutral axis. In these tables, the value of σFlim is the quotient of the fatigue limit under pulsating tension divided by the stress concentration factor 1. Load capacity of the dowel system is 40 percent of design wheel load. The amount of spring back is dependent on the material, and the type of forming. So, P is pushing down at point B, if we think of this in terms of a free body diagram. Bending stresses and vibration from moving cables cause connectors and crimped and soldered cable terminations to break. It's thin, so you shouldn't need to worry about shear stress, just bending. Direct pulley impact at max speed. The following permissible stress will contain a probability of failure of 0. beam shape and size : actual stresses do not exceed the allowable stress for the bending stress, the section modulus S must be larger than M / " i. But first of all let us talk about some shear flow theory. Besides, the step by step calculation for each calculation performed by using this calculator let the users to know how to perform shear stress calculation. 97mm diameter with a 1 kg mass on one end and a horizontal force (Fx) of 30 N applied to it. Now we are going ahead to start new topic i. Heat and matter flow 15. The most common stress types you deal with in. Beam Deflection and Stress Formula and Calculators. 5 times the. The Three Point Bend Test 1 Beam theory The three point bend test (Figure 1) is a classical experiment in mechanics, used to bending moment M, shear Q and. Sharp cracks 11. Therefore, bending stress is a combination of compressive and tensile stresses due to internal moments. Combined Axial and Bending. And, just like torsion, the stress is no longer uniform over the cross section of the structure – it varies. Combined Stress and Failure Theories • When parts have multiple types of loading or more than one type of stress from a single load 3 Objectives • Group stresses by type, separating the stresses into bending and axial versus shear and torsional stresses. You know F, you can use google to find Y and I formulas/methods (hint, parallel axis theorem). The average stress in the local region may be the governing stress for fatigue rather than the maximum stress. Step 5: Bending Analysis Flexure, bending, moment, torque Highest at midspan for uniform load Pulling stress or tension on bottom face of member 35 Step 5A: Determine F'b (psi) Allowable bending stress, F'b The maximum bending stress permissible for a specified structural member Units for stress:. Stress (mechanics) The formula for uniaxial normal stress is: where σ is the stress, F is the force and A is the surface area. The allowable yield stress Fb in bending of the A36 plate = 36 ksi*. PERMISSIBLE STRESSES (CLAUSE B-2, IS456:2000) The working stress method is based upon the concept of permissible stresses. The I term is the moment of inertia about the neutral axis. The bending stress is coming from the load P at point B. SkyCiv Engineering offers cloud based structural analysis software for engineers. It is usually represented by the Greek letter, $$\rho$$, and can be thought of as the radius of a circle having the same curvature as a portion of the graph, a curve in the road, or most any other path. Permissible stresses are obtained by dividing the ultimate strength of concrete or yield strength of steel (0. The type of hoop stress measuring wall tension is calculated as the force over the axial length. Yield stress is the stress value at which plastic deformation occurs. Mechanics of Structures, 2nd year, Mechanical Engineering, Cairo University Torsion of Thin-Walled Bars1 Review of Circular Shafts The shear stress for a circular cross section varies linearly. Beam Deflection Calculators - Solid Rectangular Beams, Hollow Rectangular Beams, Solid Round Beams. Since the load caused by the fishing line is cantilevered off the end of the pole and since the cross section of a fishing pole is relatively small, a fishing pole will have high flexural stresses. The Three Point Bend Test 1 Beam theory The three point bend test (Figure 1) is a classical experiment in mechanics, used to bending moment M, shear Q and. Bending stress is the normal stress that an object encounters when it is subjected to great amount of load at a particular point causing it to bend and fatigue. A bending stress is a stress induced by a bending moment. Peterson, Stress Concentration Factors, Wiley, New York, 1974, pp. Calculate the Deflection of round tube beams by using advanced online Beams Deflection Calculator. - Calculate the resultant of tension (compression) and bending stresses. Welcome to the "Bending Formulas" section of our website. EXERCISE 61, Page 149. The common method of obtaining the beam of uniform strength is by keeping the width uniform and varying the depth. The existence of this shear stress can be seen as cards slide past each other slightly when you bend a deck of cards. Stress Check Comments. The radius of curvature is fundamental to beam bending, so it will be reviewed here. Beam Stiffness. For either case, the analysis that follows is straight forward. edu Bending stress: two examples Danville Community College EGR 246 Mechanics of Materials. Alternating Stresses Definition: o An Alternating Stress is associated with a CYCLIC force, torque, or bending being applied to a component. Δ = deflection or deformation, in. Beam Deflection Calculators - Solid Rectangular Beams, Hollow Rectangular Beams, Solid Round Beams. I used the white Sugru putty, and it worked really well, providing a strong, flexible, insulating cover at the spots on the power cables that experience the most stress. I believe you will actually get slightly more deflection at the corners than you would calculate from the cantilever beam equation. Included are simple bending moment equations and formulas which well help with your calculations. (3)-10 Allowable Bending Stress at Root, σFlim For a unidirectionally loaded gear, the allowable bending stresses at the root, σFlim, are shown in Tables 10. Bending stress is the normal stress induced in the beams due to the applied static load or dynamic load. CE 433, Fall 2006 Review of Stress-Strain due to Flexure 3 / 4 Figure 3. And that will give us the elastic flexural formula, which is sigma x e= the e's cancel, so I get -My over I, very important relationship. Bending moment diagram (BMD) Shear force diagram (SFD) Axial force diagram Invert Diagram of Moment (BMD) - Moment is positive, when tension at the bottom of the beam SOLVE. The maximum tensile stress at the tooth-root (in the direction of the tooth height) which may not exceed the permissible bending stress for the material is the basis for rating the bending strength of gear teeth. Also, as a brittle material, glass is failed theoretically by tension only and its compressive stress is extremely large. 2 kN, as illustrated below. (A) 8 kN • m (B) 16 kN • m (C) 18 kN • m (D) 26 kN • m Starting from the left end of the beam, areas begin to cancel after 2 m. This is the plane of no bending stress, called Neutral Axis (N. The Flexure Formula • For a linear-elastic material, the linear variation of normal strain is observed, when a linear variation in normal stress must also be true. Besides, the step by step calculation for each calculation performed by using this calculator let the users to know how to perform bending stress calculation. (3)-10 Allowable Bending Stress at Root, σFlim For a unidirectionally loaded gear, the allowable bending stresses at the root, σFlim, are shown in Tables 10. This is why non-reinforced concrete beams fail on the underside - concrete is weak in tension. The resultant shear is of great importance in nature, being intimately related to the downslope movement of earth materials and to earthquakes. bending stress plane), ma. The formula to determine bending stress in a beam is: Where M is the moment at the desired location for analysis (from a moment diagram ). procedure evaluates deflection, buckling, bending stress, bending strain and wall stress. The common method of obtaining the beam of uniform strength is by keeping the width uniform and varying the depth. specified completely by e and K. Find the maximum maximum shear stress and the maximum bending stress. Secondary stress Q is caused by the constraint of adjacent parts or by self-constraint of the structure, and yielding can cause the source of the stress to be eliminated. The basis of the Thermal Bending Analysis. Mechanics of Materials 13-4d2 Beams Example 3 (FEIM): For the shear diagram shown, what is the maximum bending moment? The bending moment at the ends is zero, and there are no concentrated couples. In order to calculate maximum surface stress, you must know the bending moment, the distance from the neutral axis to the outer surface where the maximum stress occurs and the moment of inertia. The methods and procedures commonly used for finding forces resulting from applied forces are presented below. is gross defonnation progressing to rupture. (2) The cross section dimensions of the sheet are such so that the width to thickness ratio is large. Summing the moments give, Using the relationship between the bending stress and the radius of curvature, σ = -Ey/ ρ, gives,. Examples of a primary stress are circumferential stresses due to internal pressure & longitudinal bending stresses due to gravity. From the Torsion equation, we can calculate the Torsional stress and any other unknown factors. Calculate design moment, M and bending stress, fb. First of all we will find here the expression for bending stress in a layer of the beam subjected to pure bending and aftre that we will understand the concept of moment of resistance and once we will have these two information, we can easily secure the bending equation or flexure formula for beams. BEAM DIAGRAMS AND FORMULAS Table 3-23 (continued) Shears, Moments and Deflections 13. It is usually represented by the Greek letter, $$\rho$$, and can be thought of as the radius of a circle having the same curvature as a portion of the graph, a curve in the road, or most any other path. SkyCiv Engineering offers cloud based structural analysis software for engineers. This depends on the moment redistribution ratio used, δ. Thus, the design of beam-to-. 0 COMBINED BENDING AND TORSION There will be some interaction between the torsional and flexural effects, when a load produces both bending and torsion. 6 times yield strength ofbase metal. Limits for the design variables are defined and the objective variable will be given a value of highest. R = span length of the bending member, in. Pregnancy is the point at which you have to practice additional alert to defend your health and the development of the little one in your belly. Wang's contact info: Yiheng. Instead of L/2, it could be L/3 or 2L/3, but I don't see the reasoning for picking any particular spot. (a) Using the formula from the Simple Theory of Bending, the maximum working Stress is. The flexure formula gives the internal bending stress caused by an external moment on a beam or other bending member of homogeneous material. Direct pulley impact at max speed. The allowable stresses are generally defined by building codes, and for steel, and aluminum is a fraction of their yield stress (strength):. From the Torsion equation, we can calculate the Torsional stress and any other unknown factors. 3–13 Shear Stress Due to Bending 3–14 Shear Stress Due to Bending—Special Shear Stress Formulas 3–15 Normal Stress Due to Bending 3–16 Beams with Concentrated Bending Moments 3–17 Flexural Center for Beam Bending 3–18 Beam Deflections 3–19 Equations for Deflected Beam Shape 3–20 Curved Beams 3–21 Superposition Principle 3. In general, the presence of shear reduces the moment carrying capacity of the beam. Bending stress occurs when operating industrial equipment and in concrete and metallic structures when they are subjected to a tensile load. The eccentricity e of the vertical load is defined as V M e y. Stress, Strain, Bending moment & Shear force Stress: It is a force that tends to deform the body on which it acts per unit area. Please reference the "Table of Factors" for each of the formulas listed. Bending Stress = (Force * Length) / (MI / (0. The equations for fillet weld size design calculations for design of welded connections subjected to bending will be discussed by explaining the steps required for weld stress analysis calculation. 9 Direct and bending stresses and their combination LIST OF PRACTICALS 1. When bending is done, the residual stresses cause the material to spring back towards its original position, so the sheet must be over-bent to achieve the proper bend angle. Stresses in the curved beam under loads normal to "Stresses in the curved beam under loads normal to the plane of its axis " (1937). Forces that are acting perpendicular to the longitudinal axis of the beam cause bending stresses which are termed as flexural stresses, beside flexural stresses beams also undergo shear stresses and normal stresses. The norm al stress distribu tion in a cross-section is directly related to the normal force and the bending moment. Hooke’s Law is applicable). Compute extreme fiber stresses in a beam, given its cross-sectional dimensions and the applied internal moment. x = horizontal distance. Force acting on each tooth can be resolved into a tangential (F t) and the radial component (F r) (Fig. If a bar of large length when held vertically and subjected to a load at its lower end, its won-weight produces additional stress. In most building sites, hollow rectangular beams are used because they can bear the forces like hearing and bending in both the directions. Very first, the video explains the given exemplary C section which flanges thickness is 10 mm, connecting bar width is 6 mm, width of the flanges is 100 mm and depth of the section is 200 mm. Carry out bending tests on a steel bar or wooden beam. For the beam shown in Problem 2 solve for the maximum stress using the beam cross-section shown below and knowing that h 1 = h 2 = 0. Such local buckling behavior is characterized by distortion of the cross-section of the member. Note that the Stress and Strain are proportional to the distance from the Neutral Axis. Sheet metal bending is the plastic deformation of the work over an axis, creating a change in the part's geometry. Lecture 8 - Bending & Shear Stresses on Beams Beams are almost always designed on the basis of bending stress and, to a lesser degree, shear stress. N M V r θ cross-section must be symmetric but does not have to be rectangular assume plane sections remain plane and just rotate about the neutral axis, as for a straight beam, and that the only significant stress is the hoop stress. BENDING OF METALS INCLUDING "V", WIPE AND ROTARY DIE OPERATIONS One of the most widely employed pressworking operations is bending. To determine the maximum stress due to bending the flexure formula is used: where: σ max is the maximum stress at the farthest surface from the neutral axis (it can be top or bottom) M is the bending moment along the length of the beam where the stress is calculated. Recently Bajguz (2007) reported that brassinosteroids represent a class of plant hormones, more than 70 compounds have been isolated from plants. Select a calculator below to get started. It controls not only the design of beams, but also of columns when subjected to bending in addition to axial load. 12 ), it is necessary that we have the expression for the bending moment. Structural Beam Deflection and Stress Calculators to calculate bending moment, shear force, bending stress, deflections and slopes of simply supported, cantilever and fixed structural beams for different loading conditions. The plot may look like the beam is bending a lot, but compare at the scale on the x and y axis. This formula will be discussed later in further detail. Structural Beam Deflection, Stress Formula and Calculator: The follow web pages contain engineering design calculators that will determine the amount of deflection and stress a beam of known cross section geometry will deflect under the specified load and distribution. RE: what is the maximum allowable bending stress formula for pipe? deformation needs to be elastic azcats (Structural) 5 Aug 16 20:42 Check section F8 of the 13th Ed. Measurements of load‐point deflection rates, which are experimentally easy to accomplish in ceramic systems, vs the applied load lead to the direct determination of the creep exponent and the creep compliance in a. When the ratio of computed axial stress to allowable axial stress f /F a exceeds 0. The stress in the beam can then be calculated from the bending moment using: stress = M * y / I. The I term is the moment of inertia about the neutral axis. ximumum bending deflection, flexural (bending) modulus, shear modulus, maximum bending stress, and maximum shear stress. The simplest case is the cantilever beam , widely encountered in balconies, aircraft wings, diving boards etc. In case of non-uniform bending the presence of shear forces produces warping or put of place distortion of the cross-section, thus, a section that is plane before bending is no longer plane after bending. We will now consider the. Beam Deflection and Stress Formula and Calculators. Beam Stiffness. Stress relieving, or preferably normalising is recommended prior to galvanising of cold worked components which are to be subjected to any significant degree of stress in service. Don't want to hand calculate these, sign up for a free SkyCiv Account and get instant access to a free version of our beam software !. The maximum stress that an object is expected to support is called the allowable stress. In other sections, specifically Paras. Since the stress across a beam section varies from compression to tension, there is a location at which stress is equal to zero. Below is our simple Bend Allowance Calculator, it works by inputing the Material Thickness, Bend Angle, Inside Radius and K-Factor. 13) Slide No. Use it to calculate the stress at desired locations by using σ = Mc/I where I. The stresses in the plate. Hopefully this is my last question. A stress concentration factor (Kt) is a dimensionless factor that is used to quantify how concentrated the stress is in a material. REINFORCED CONCRETE COLUMNS IN BIAXIAL BENDING ENERCALC, INC. Instead of L/2, it could be L/3 or 2L/3, but I don't see the reasoning for picking any particular spot. Seshu Adluri Bending terminology Moment of inertia Parallel axis theorem Flexural stress Average shear stress =V f/hw Yield moment, M Y Elastic Section modulus, S Plastic moment, M P Plastic section modulus, Z Beam (slab load) vs. At any point in the loaded beam, bending stress (σ) can be calculated from the following formula: v I M s = Max bending stress will occur at the outermost layer of the beam (v. x = horizontal distance. Wang's contact info: Yiheng. Note that not all of the factors are used to adjust the bending design value. How does this question vary tho that of a solid tube?. This calculation deals with the deflection, stress and variation of forces in the loaded flat plates. (3)-10 Allowable Bending Stress at Root, σFlim For a unidirectionally loaded gear, the allowable bending stresses at the root, σFlim, are shown in Tables 10. Reliability factor KR KT = (460 + T)/620. Wahl has calculated the bending stress correction factor at the ID of a round wire torsion spring as:. The methods and procedures commonly used for finding forces resulting from applied forces are presented below. Lateral loads acting on the beam cause the beam to bend or flex, thereby deforming the axis of the. The equations are derived following the procedure used in classical beam theory but. The Moment Of Resistance At A Plastic Hinge. (3) The stress-strain characteristics of material are the same in tension and compression. Direct pulley impact at max speed. 1, Curve a) in which case σ fB = σ fM. None of these equations apply to tees in compression, thus such tees are not considered. Besides, the step by step calculation for each calculation performed by using this calculator let the users to know how to perform bending stress calculation. Introduction This program provides analysis and design of arbitrarily shaped reinforced concrete columns loaded with axial loads and uni-axial or bi-axial bending moments. where T n nominal tensile strength A n net area of the cross section F y design yield stress In addition, the nominal tensile strength is also limited by Sec. In these tables, the value of σFlim is the quotient of the fatigue limit under pulsating tension divided by the stress concentration factor 1. The shear stress due to bending is often referred to as transverse shear. This depends on the moment redistribution ratio used, δ. A simply supported beam is the most simple arrangement of the structure. It is usually represented by the Greek letter, $$\rho$$, and can be thought of as the radius of a circle having the same curvature as a portion of the graph, a curve in the road, or most any other path. The beam is supported at each end, and the load is distributed along its length. Use this online hollow rectangular beam deflection calculator to compute the deflection of hollow rectangular beams. • 1) Stress along the circumferential direction, called hoop or tangential stress. These factors are referred to as macro effects. It also indicates that stress is related to distance y from the neutral axis so it varies from zero to a maximum at the top or bottom of the section. We have also discussed a ssumptions made in the theory of simple bending and formula for bending stress or flexural formula for beams during our last session. The basis of the Thermal Bending Analysis. • 2) Radial stress which is stress similar to the pressure on free internal or external surface. 33 is calculated. The stress-correction factor at inner and outer fibers has been found analytically for round wire to be. The stresses developed on the inside of the. The cheek plates are not included in this stress calculation. To find the maximum bending stress •Draw shear & bending moment diagrams •Find maximum moment, M, from bending moment diagram •Calculate cross-section properties -Centroid (neutral axis) -Calculate Area Moment of Inertia about x-axis, I x -Find the farthest distance from neutral axis for cross section, c •Max Bending Normal Stress = x. Combined Stresses in Shafts. Our primary concern for this project is axial and bending loads. This specification contains the equations used by the Acme/Stub Acme AutoProgram. The beam is supported at each end, and the load is distributed along its length. W = total uniform load, lbs. In other words, it is not load divided by area. Below is a sample of how to adjust the reference bending value by the factors listed above to arrive at the allowable bending value that would be used to design wood in bending. Bending of plates, or plate bending, refers to the deflection of a plate perpendicular to the plane of the plate under the action of external forces and moments. Similar to other metal forming processes, bending changes the shape of the work piece, while the volume of material will remain the same. This is why non-reinforced concrete beams fail on the underside - concrete is weak in tension. Hollow rectangular beams are the ones which withstand forces of bending and shearing plus they are resistant to torsional forces, calculate the bending stress use this online mechanical calculator. Bending stress is important and since beam bending is often the governing result in beam design, it’s important to understand. Before examining the specifics of each failure mode we need to establish the distribution of stress in the faces and core as a result of bending, these are and for the normal stresses in the face and core and and for the shear stresses. Shear Flow – Bending Stress Theory. com's Beam Deflection Calculators. Bending stress occurs when operating industrial equipment and in concrete and metallic structures when they are subjected to a tensile load. We use a limit called the interaction diagram. The maximum normal stresses are related to the bending moment, M and the distance from the centerline, y. procedure evaluates deflection, buckling, bending stress, bending strain and wall stress. For the three-dimensional case, it is now demonstrated that three planes of zero shear stress exist, that these planes are mutually perpendicular, and that on these planes the normal stresses have maximum or minimum values. ★ Relief Of Pain Bending At Waist Chronic Ear Pain Fullness Being Extraordinary With Chronic Pain. Stresses: Beams in Bending 237 gitudinal axis. 97mm diameter with a 1 kg mass on one end and a horizontal force (Fx) of 30 N applied to it. Now we can find the stress. In the case of non-uniform bending of a beam where the bending moment varies from one X-section to another, there is a shearing force on each X-section and shearing stresses are also induced in the material. The bending produced by the transverse loading causes a deflection ∆. Stress in Torsion Springs. P R σ = (4) P, can be defined in terms of a local radial load, P r and local moment. The results of the analysis are stored and taken as reference for optimization. 05 m) and length 1 m. Allowable stress design is based on the following design principles and assumptions: • Within the range of allowable stresses, masonry elements satisfy applicable conditions of equilibrium and compat-ibility of strains. The main types of piping stresses. If we can agree that the yield moment is around 7. Beam Deflection Calculators - Solid Rectangular Beams, Hollow Rectangular Beams, Solid Round Beams. The methods and procedures commonly used for finding forces resulting from applied forces are presented below. (a) Using the formula from the Simple Theory of Bending, the maximum working Stress is. Thank you for the picture, based upon this it can be said that the A point is a free rotating hinge; and, due to the handle design, under load there will be a bending moment created in the handle hook either in the region between the the pin contact point and the transition taper to the full thickness of the lever if the reaction force is downward on the pin or along any point of the hook if. Hoop stress is the result of pressure being applied to the pipe either internally or externally. Permissible stresses are obtained by dividing the ultimate strength of concrete or yield strength of steel (0. • 1) Stress along the circumferential direction, called hoop or tangential stress. In mechanics of materials, the bending stress of a beam in bending can be determined by the equation. In this section, various messages will be displayed indicating what factors governed the calculation of allowable bending stress. mathematical relationships between the intensity of loading ‘ω’, shearing force ‘f’ and bending moment ‘m’ at a section of a beam distance ‘x’ from one end. Cantilever Beams Moments And DeflectionsBeam Stress Deflection MechanicalcBeam Stress Deflection MechanicalcCantilever Beams Moments And DeflectionsCalculator For Disc Springs Belleville WasherCantilever Beam Udl And End Bending MomentMechanics Of Materials Bending Normal …. Δ = deflection or deformation, in. 1-5(e) and 2-8, higher allowable. Determination of Base Stresses in Rectangular Footings under Biaxial Bending 1520 The dimensions of the footing are Bx By. 0 cm and the permissible stress in shear, bending and bearing stress in dowel bars are 1000,1400 and 100 respectively. Besides, the step by step calculation for each calculation performed by using this calculator let the users to know how to perform bending stress calculation. When the bending stress components are combined with the membrane stress components at each surface, the resulting stress intensities P m +P b are limited to 1. Re: plate deflection formula or calculation. Wallace Bending Moment in Curved Beam (Inside/Outside Stresses): Stresses for the inside and outside fibers of a curved beam in pure bending can be approximated from the straight beam equation as modified by an appropriate curvature factor as determined from the graph below [ i refers to the inside, and o. Besides, the step by step calculation for each calculation performed by using this calculator let the users to know how to perform shear stress calculation. 39 times as long as 2x4 for same deflection under same P. 𝜎= 𝑀𝑏 𝑊𝑦 (2) The bending moment is given by. Bending stress is the normal stress that is induced at a point in a body subjected to loads that cause it to bend. Bending moment = Wa ((l-a))/l W is a the applied load on beam a is the distance between the pivot point and point of force application. TEK 14-1B , Section Properties of Concrete Masonry Walls (ref. This part of the Aluminum Construction Manual deals with speci fications for allowable stresses in structures. Short answer is use the formula F=M Y / I. 4 Cantilever Beam Bending When subjected to a point load at its tip, a cantilever beam has a linear variance in stress and strain through the cross-section in the direction parallel to the load. Alternating Stresses Definition: o An Alternating Stress is associated with a CYCLIC force, torque, or bending being applied to a component. Enter the length, diameter and wall thickness then select the material from the drop down menu. The simplest case is the cantilever beam , widely encountered in balconies, aircraft wings, diving boards etc. BENDING OF METALS INCLUDING "V", WIPE AND ROTARY DIE OPERATIONS One of the most widely employed pressworking operations is bending. Step 5: Calculate the bending stress in the shaft Step 6: Combine the bending stress and the torsional stress using the theories discussed in chapter 4 August 15, 2007 22 • Shaft shown drives a gear set that is transmitting 5 hp at 1750 rpm. From the Torsion equation, we can calculate the Torsional stress and any other unknown factors. 67 reduction factor), doesn't it follow that the allowable fiber bending stress is = to the M / S? If S is =. The Three Point Bend Test 1 Beam theory The three point bend test (Figure 1) is a classical experiment in mechanics, used to bending moment M, shear Q and. The maximum stress that an object is expected to support is called the allowable stress. Bending stress (σb) is the highest bending stress in the material around the hole of the padeye main plate. • Concrete stress blocks • Reinforcement stress/strain curves • The maximum depth of the neutral axis, x. The combination of bending and axial compression is more critical due to the P-∆ effect. Identify worst case element Find M on element Find T on element Find Z P needed Select pipe page 710 Simplified for Torque and Bending Z P = 2S. Example - Shear Stress and Angular Deflection in a Solid Cylinder. Point load: Deflection: I-beam can be 2. Axial load and the location of the neutral axis:. There is always crookedness in the column and the load may not be exactly axial. 12 ), it is necessary that we have the expression for the bending moment. A beam in which bending stress developed is constant and is equal to the allowable stress is called beams of uniform strength. Welcome to the "Bending Formulas" section of our website. The norm al stress distribu tion in a cross-section is directly related to the normal force and the bending moment. First of all we will find here the expression for bending stress in a layer of the beam subjected to pure bending and aftre that we will understand the concept of moment of resistance and once we will have these two information, we can easily secure the bending equation or flexure formula for beams. Equal Loads are applied at the distance of one-third from both of the beam supports. • The transverse loads cause internal shear forces and bending moments in the beams as shown in Figure 1 below. [email protected] 6 shows that the maximum Von-mises stress is developed at the root of the pinion tooth of 541. Relief Of Pain Bending At Waist How To Live With Chronic Neck Pain Diagnosising Depression In Patients With Chronic Pain. On bending, they do so in the form of circular arcs, with a common centre of curvature. MSC Software Confidential In the hand calculation method, maximum stress around bolt holt is calculated using in-plane pin load, by-pass tension load, and by-pass bending load as shown in the picture below. M = maximum bending moment, in. There are five primary piping stresses that can cause failure in a piping system: hoop stress, axial stress, bending stress, torsional stress, and fatigue stress. Our primary concern for this project is axial and bending loads. design Appreciate the derivation of the design formulae for bending. 1), the minimum ratio is 7,900 - 2,867 0. The degree of bending an object will tolerate before it becomes permanently deformed varies, depending on the construction materials, size, and other variables. A drive shaft is supported by bearings at both the ends (and at regular interval in the centre for longer shafts). , for a force, F, normal to the surface of a beam having a cross sectional area of A, the shear stress is = F/A. It is based upon the Bernoulli Euler theory which is applicable to most common. Bending stress is the normal stress that is induced at a point in a body subjected to loads that cause it to bend. material, but the strain for a given stress is always the same and the two are related by Hooke´s Law (stress is directly proportional to strain): where σ is stress [ MPa ] E modulus of elasticity [MPa] ε strain [unitless or %] From the Hook’s law the modulus of elasticity is defined as the ratio of the stress to the strain :. For example, postulate that the cross section CD on the right does not remain plane but bulges out. For axial loaded unnotched specimens, a nominal. Bends are used to increase rigidity and produce a part of the desired shape. AGMA Bending Strength Equation Sfb’ is the allowable fatigue bending stress, psi KR is the reliability factor KT is the temperature factor KL is life factor Allowable Stress Temperature factor KT AGMA recommends using temperature factor of 1 for operating temperatures up to 250 oF. ★ Relief Of Pain Bending At Waist Chronic Ear Pain Fullness Being Extraordinary With Chronic Pain. Ultimate Bending Capacity for a Reinforced Concrete Section Under Combined Bending and Axial Load Doug Jenkins; Interactive Design Services Pty Ltd Strain Stress This paper shows the derivation of a closed form solution for the ultimate bending capacity of any. The term is the distance from the neutral axis (up is positive). These local stresses may be combined with other membrane and bending stresses resulting from primary loads (such as internal pressure and sustained loads) and secondary loads (such as thermal loads) in a proper way which is presented by ASME BPVC section VIII Div2. The resultant shear is of great importance in nature, being intimately related to the downslope movement of earth materials and to earthquakes. A beam in which bending stress developed is constant and is equal to the allowable stress is called beams of uniform strength. The Flexure Formula • For a linear-elastic material, the linear variation of normal strain is observed, when a linear variation in normal stress must also be true. MSC Software Confidential In the hand calculation method, maximum stress around bolt holt is calculated using in-plane pin load, by-pass tension load, and by-pass bending load as shown in the picture below. (Research Article) by "Advances in Materials Science and Engineering"; Engineering and manufacturing Numerical analysis Mechanical properties Methods Models. x = horizontal distance. Usually bending has to overcome both tensile stresses and compressive stresses. | 2019-11-22T02:26:23 | {
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https://math.stackexchange.com/questions/1851084/matrix-equation-a2a-i-when-deta-1 | Matrix equation $A^2+A=I$ when $\det(A) = 1$
I have to solve the following problem: find the matrix $A \in M_{n \times n}(\mathbb{R})$ such that: $$A^2+A=I$$ and $\det(A)=1$. How many of these matrices can be found when $n$ is given? Thanks in advance.
• I think there are infinite number of matrices – Empty Jul 6 '16 at 15:43
• It's worth noting that $A^{-1} =A+I$. – Mnifldz Jul 6 '16 at 15:43
• Solve two equations : $det(A)=1$ and $det(A+I)=1$ , taking eigen values of $A$ are $\alpha_1 , \alpha_2 , \cdots , \alpha_n$. Then you will get two equations : $\alpha_1.\alpha_2.\cdots \alpha_n=1$ and $(\alpha_1+1).(\alpha_2+1)\cdots (\alpha_n+1)=1$ – Empty Jul 6 '16 at 15:44
• You might be interested in this. – J. M. is a poor mathematician Jul 6 '16 at 15:49
• @J.M.: I will read it. Thanks – Riccardo.Alestra Jul 6 '16 at 15:50
Let the eigenvalues of $A$ be $\{e_i\}$ and let $P$ be a similarity transformation that diagonalizes $A$. Then $P$ also diagonalizes $A^2$ and the equation becomes $$\forall i: e_i^2 + e_i = 1$$
Since there are two real roots to that equation, there are $2^n$ candidates for $A$, before imposing that the determinant is one.
When we impose that the diagonal is one, we see that the eigenvalues have to come in pairs $$e_i = \frac{-1-\sqrt{5}}{2}. e_{i'} = \frac{-1+\sqrt{5}}{2}.$$ and that there must be an even number of such pairs. So no such $A$ exists, unless $N=4k$ for$k\in\Bbb{Z}^+$. For a given such $N$ there are $$\binom{N}{N/2}$$ possible diagonal determinant one $N\times N$ matrices satisfying the equation.
However, if $A$ satisfies the equation, then so does any similar matrix $$p^{-1}AP$$ so as long as $N=4k$ there are a continuum of solutions for $A$.
The fact $A^2 + A = I$, means that $X^2 + X -1$ is an annihilating polynomial of $A$.
The minimal polynomial of $A$ thus divides that polynomial. Thus all the eigenvalues of $A$ are among the roots of $X^2 + X - 1$, which are
$$\frac{-1 \pm \sqrt{5}}{2}.$$
Since the determinant is the product of all eigenvalues (with multiplicity) it follows that both have the same multiplicity and this multiplicity is even.
Since the sum of all multiplicities is the dimension, it follows that the dimension must be divisible by $4$.
This condition suffices. A matrix has this property if and only it is similar to a diagonal matrix with half the diagonal entries $\frac{-1 + \sqrt{5}}{2}$ and the other $\frac{-1 - \sqrt{5}}{2}$.
The matrix is diagonalisable as the minimal polynomial has only distinct roots.
Note that once you have one matrix that satisfies $A^2+A=I$, you have infinity many, since for every invertible matrix $P$ we get $$(P^{-1}AP)^2+(P^{-1}AP)=I\qquad\text{and}\qquad \det(P^{-1}AP)=1$$ Now, if $4\mid N$, then choose the matrix $$A=diag(\phi,-1-\phi,\ldots,\phi,-1-\phi)$$ where $\phi=\frac{\sqrt{5}-1}{2}$. Since $\phi^2+\phi=1$, we have that $A^2+A=I$. In addition, $\det A=(\phi(-1-\phi))^{N/2}=(-1)^{N/2}=1$ because $4\mid N$.
Suppose now that If $4\nmid N$. Since $f(A)=O$ for $f(x)=x^2+x-1$ and since $f(x)$ is simple above $\mathbb{R}$ it follows that $A$ is similar to a diagonal matrix $D$ that has $\phi$ and $-1-\phi$ along its main diagonal. But unless $4\mid N$, we get that $\det A=\det D\neq 1$.
• But the determinant will not be $1$. – quid Jul 6 '16 at 16:04
• I fixed that. The other cases can be done similarly. – boaz Jul 6 '16 at 16:06
Consider the Jordan Canonical Form for $A$; that is, $A = PJP^{-1}$ for some invertible $P$ and block diagonal matrix $J$ whose blocks are either diagonal or Jordan (same entry on the diagonal, have $1$s on the the diagonal above the main diagonal, and $0$s elsewhere).
Then, the equation reduces to $J^2 + J = I$. Looking at the diagonal entries on both sides, this reduces to solving $r^2 + r = 1$, which has solutions $r = \frac{-1 \pm \sqrt{5}}{2}$. In other words, $J$ has diagonal entries are $\frac{-1 \pm \sqrt{5}}{2}$. However, we want $|A| = 1$, which is equivalent to $|J| = 1$. This is only possible if $n$ is even and $J$ has an equal even number of $\frac{-1 + \sqrt{5}}{2}$ and $\frac{-1 - \sqrt{5}}{2}$ on its diagonal, due to $(\frac{-1 + \sqrt{5}}{2})(\frac{-1 -\sqrt{5}}{2}) = -1$ and no positive integral power of $\frac{-1 \pm \sqrt{5}}{2}$ equaling $\pm 1$.
As a summary, there are no solutions when $n$ is not a multiple of $4$. Otherwise, when $n$ is not a multiple of $4$, there are infinitely many solutions. To illustrate this, let $J$ be diagonal with an equal and even number of $\frac{-1 + \sqrt{5}}{2}$ and $\frac{-1 - \sqrt{5}}{2}$ on its diagonal. Letting $P$ vary (as there are infinitely many non-invertible matrices that do not commute with $J$) will give infinitely many possibilities for $A$ when $n$ is a multiple of $4$. | 2020-02-22T07:50:46 | {
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https://nbhmcsirgate.theindianmathematician.com/2020/05/ | ### NBHM 2020 PART C Question 31 Solution (continuous images of $G = GL_2(\Bbb R)$)
Let $G = GL_2(\Bbb R)$ be the set of all $2 \times 2$ invertible real matrices. Elements of $G$ can be identified with vectors in $\Bbb R^4$ and this makes $G$ a metric space. Which of the following sets can be obtained as the image of a continuous onto function from $G$?
1. The real line.
2. $A = \{(x,\frac{1}{x}) : x \in \Bbb R \text{ and } x \ne 0\}$.
3. $\Bbb R^2 \backslash A$.
4. $\{(x,y) \in \Bbb R^2 : |x|^2 + |y^2| \le 1\}$
I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.
Solution:
Option 1:(True) Consider the determinant map $\text{det }: GL_2(\Bbb R) \to \Bbb R^*$ where $\Bbb R^{*} = \Bbb R \backslash \{0\}$ is a group under multiplication. Now $$\text{det}(AB) = \text{det}(A)\text{det}(B)$$ Therefore this defines a homomorphism. We have determinant is a continuous map, because it is a polynomial in the matrix entries.
Let $\alpha \in \Bbb R^*$ then the matrix $\begin{bmatrix}\alpha & 0 \\ 0 & 1\end{bmatrix} \in GL_2(\Bbb R)$ has determinant $\alpha$. Therefore determinant is a surjective map. Now, the function $\phi: \Bbb R^* \to \Bbb R$ defined by $\phi(x) = |\text{log}(x)|$ is a continuous surjective map. Hence the composition $\phi \circ \text{det} : GL_2(\Bbb R) \to \Bbb R$ is the required continuous surjection. Note that this map is also a composition of group homomorphisms and hence is a group homomorphism. Therefor there exists a continuous surjective group homomorphism between $GL_2(\Bbb R)$ to $\Bbb R$.
Option 2:(True) Consider the map $\phi : GL_2(\Bbb R) \to A = \{(x,\frac{1}{x}: x \in \Bbb R)\}$ defined by $\phi(A) = (\text{det}(A),\frac{1}{\text{det}(A)})$. We have seen that $\text{det}$ is a continuous map. Since $\text{det}(A) \ne 0$ for any $A \in GL_2(R)$, the function $A$ goes to $\frac{1}{\text{det}(A)}$ is also continuous. This shows that, the function $\phi$ is continuous since each of its coordinate map is continuous. Let $(\alpha,\frac{1}{\alpha}) \in A$ then the matrix $M$ defined by $\begin{bmatrix}\alpha & 0 \\ 0 & 1\end{bmatrix} \in GL_2(\Bbb R)$ has determinant $\alpha$ and $\phi (M) = (\alpha,\frac{1}{\alpha})$. Therefore $\phi$ is surjective.
option(3):
Result: Let $X$ and $Y$ be two metric spaces such that $X$ has $m$ connected components and $Y$ has $n$ connected components with $m < n$. Then there cannot be a surjective continuous map from $X$ to $Y$.
Proof: Let $f$ be a continuous function from $X$ to $Y$ and $X_1,X_2,\dots,X_m$ be the connected components of $X$. We can restrict $f$ to each of these connected component and the resulting map is still continuous. Also, the images $f(X_i)$ is connected for each $i$. Let $Y_i$ be the connected component of $Y$ such that $f(X_i) \subseteq Y_i$ (Note that, different connected components $X_i$ could have mapped to the same connected component $Y_i$). This shows that $f(X) \subseteq \cup_{i=1}^m Y_i$. This means that $f(X)$ doesn't intersect remaining $n-m$ connected components of $Y$. Therefore $Y$ is not surjective.
$GL_2(\Bbb R)$ is disconnected and has $2$ connected components. The set $\Bbb R^2 \backslash A$ has $3$ connected components. Now, the above result shows that there cannot be continuous surjection between these sets.
option 4:(True)
Consider the function $f$ from $GL_2(\Bbb R)$ to $\Bbb C$ given by $M = \begin{bmatrix}x & y \\ z & a\end{bmatrix}$ goes to $\frac{y^2}{x^2+ y^2}e^{i a}$. Then clearly $f$ is a continuous surjection from $GL_2(\Bbb R)$ to $\{(x,y) \in \Bbb R^2 : |x|^2 + |y^2| \le 1\}$.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft... | 2021-07-23T23:06:37 | {
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https://math.stackexchange.com/questions/2665897/representing-zero-as-a-rational-number/2666089 | # Representing zero as a rational number
How to represent 0 as rational number?
$0/0$ is not legitimate, $0/\text{const}$ should be good enough, but what is the right value of const? $0/1$ works for a lot of computational cases, but only on intuitive.
• $\frac{0}{a}=0$ for any $a\neq 0$, and a number doesn't need to be represented in some way to be a rational number. $0\in\mathbb{Q}$ since $0\in\mathbb{N}$, for instance. Feb 25, 2018 at 12:01
• Rational numbers are equivalence classes of pairs of integers, so the class of $(0, k)$ suffices for any integer $k \neq 0$ Feb 25, 2018 at 12:01
• "In mathematics, a rational number is any number that can be expressed [i.e. represented] as the quotient or fraction $\dfrac p q$ of two integers, a numerator $p$ and a non-zero denominator $q$. Since $q$ may be equal to $1$, every integer is a rational number. " Feb 25, 2018 at 12:01
• And I thought that $0/1$ equalled $0$; not just that it works for a lot of computational cases. I can't imagine a "computational case" where it didn't work. Feb 25, 2018 at 12:09
• 0 is a rational number. Why do you want to write it any other way? Feb 25, 2018 at 12:32
There is no single right value. The rational number $0$ can be represented as the quotient of an integer by a non-zero integer in infinitely many ways: $0=\frac 0d$ ($d\in\mathbb{Z}\setminus\{0\}$). Choosing $d=1$ is a natural choice: every rational number can be represented in one and only one way as $\frac nd$ with $n\in\mathbb Z$, $d\in\mathbb N$ and $\gcd(n,d)=1$. In the case of $0$, that representation is $\frac01$.
• There is no "right" value? Or there is no "single right" value? Feb 25, 2018 at 22:32
• @RandallStewart I've edited my answer. Thank you. Feb 25, 2018 at 22:54
• Hello! This question (which in my opinion is a sincere inquiry) is at risk of deletion. The problem is that the linked duplicates are bad as JMoravitz and I agreed, but this wasn't the asker's fault, so could you please edit the duplicate list to add the two links JMoravitz gave in a comment under the question and remove the original, so that it is reasonably justifiable to keep the question (closed as duplicate rather than deleted)? Thanks! Aug 13, 2021 at 10:50
• @user21820 Done. Aug 13, 2021 at 11:05
• Thank you very much! Aug 13, 2021 at 11:19
What is wrong with $$\cdots=\frac{0}{-2}=\frac{0}{-1}=0=\frac{0}{1}=\frac{0}{2}=\frac{0}{3}=\cdots \ ?$$ You have an infinity of perfectly defined representations...
• Yes. And in fact for other rationals you also have infinitely many representations. $$\dots =\frac{-14}{-4} = \frac{-7}{-2} = \frac{7}{2} = \frac{14}{4}=\dots$$ Feb 25, 2018 at 12:30
• $$\frac{0}{-6} = \frac{-6}{-6} * \frac{0}{1}$$
– Rob
Feb 25, 2018 at 18:02
$0/1$ is generally considered to be the canonical form. This is important for the definition of the "rational ruler" or modified Dirichlet function:
$$f(x) = \begin{cases} 0 & x \in \Bbb{R}\setminus\Bbb{Q} \\ 1/q & x = p/q \mbox{ in lowest terms}\end{cases}$$
and I'm sure in many other situations too.
• If you're referring to Thomae's function, that actually takes the value $1/q$ instead of $q$ for rational $x$. But a nice example nonetheless. Feb 25, 2018 at 15:16
• Yes, good catch! Feb 25, 2018 at 15:22
As other answers have mentioned, there are many equivalent representations of $0$ in the rational numbers.
A formal way to define rational numbers is the following:
Consider the set - $$S = \{ (x,y) \in \textbf{Z} \times \textbf{Z} \quad |\quad y \neq 0\} .$$
The each element $q$ of the rational numbers Q is an equivalence class $\bar{s} \in S/\equiv_{\sim}$ under the equivalence relation: $$(a,b)\sim (c,d) \iff ad = bc$$ Or equivalently in Q: $$ad = bc\iff \frac{a}{b}= \frac{c}{d}$$ Since $0 \cdot b = 0 \cdot d = 0$ for any $d, c \in \textbf{Z}$, we can represent $0 \in \textbf{Q}$ as $(0,x)$ for any $x \in \textbf{Z}-\{0\}$.
You can do this procedure for any integral domain $R$, and it represents the field of fractions of $R$, denoted commonly as $Quot(R)$. | 2022-05-29T00:05:43 | {
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https://mathematica.stackexchange.com/questions/135156/generate-a-list-of-randomly-distributed-1s-and-0s-with-fixed-proportion-of/135157 | # Generate a list of randomly distributed $1$s and $0$s with fixed proportion of $1$s [duplicate]
I need to simulate a random initial state of an 1D cellular automaton, but with different 'densities' of filled cells.
Let's say the size of the list is $N$, then I need to be able to fix a number $P$ such that there are exactly $P$ $1$s and $N-P$ $0$s.
RandomInteger gives $1$s or $0$s with probability $p=1/2$, but first, I still din't work out how to correctly modify the probability so it can change from $0$ to $1$, and second, I would prefer for the number to be exact.
In other words, I have $P$ $1$s and $N-P$ $0$s and I need to randomly and uniformly distribute them inside a singe list. I'm not sure how to do that efficiently.
I suppose I could create a list of all the possible positions and use RandomSample[list,P] to fill them with $1$s. But is there a better way?
Important point! $N$ will be very large (up to 100 000).
• You want to first create a list with the correct numbers of 1s and 0s and then create a random permutation, which you can d with RandomSample. Therefore try: RandomSample[Join[ConstantArray[1, p], ConstantArray[0, n - p]]] where n is the length of the list overall and p the number of 1s. – Quantum_Oli Jan 11 '17 at 11:20
• @Quantum_Oli, thank you, that's a good idea. – Yuriy S Jan 11 '17 at 11:21
If you need precisely m zeros and n ones in random order, just put them in a list and use RandomSample to shuffle it.
m = 10; | 2020-04-02T16:57:17 | {
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https://math.stackexchange.com/questions/2109416/is-this-finite-set-of-polynomials-necessarily-linearly-dependent/2109642 | # Is this finite set of polynomials necessarily linearly dependent?
I have a linear algebra question from a practice sheet that I'm completely lost on. New to linear algebra, so any guidance in how to get started or what to look for would be really appreciated. The question's as follows:
Let $K$ be a field and let $V$ be the subspace of $K[x]$ consisting of all polynomials of degree at most 4. Let $p_1{(x)},p_2{(x)}, ... , p_5{(x)}$ be distinct polynomials in $V$ that satisfy $p_i{(0_K)} = 1_K$ for each $1 \leq i \leq 5$. Is the set {$p_1(x), ... , p_5(x)$} necessarily linearly dependent?
Again thanks in advance for any help.
• What about $p_1(x)=1$ and $p_i(x)=1+x^{i-1}$ for $2\le i\le 5$ ? – Adren Jan 22 '17 at 21:21
• This Question poses a contrast with the case that five polynomials $p_i$ all satisfy $p_i(1_K) = 0_K$. This homogeneous constraint would force a linear dependence relation. – hardmath Jan 22 '17 at 21:39
Let's consider a few things.
First, what is the maximum length of a list which could remain linearly independent in $K[X]$. Obviously we can say that $(1, t, t^2, t^3, t^4)$ spans $K[X]$, which means that it is possible to have a list of length 5 which is linearly independent in consideration. (Note that if the spanning set had had 4 or fewer elements you would have been able to conclude immediately that linear dependence was necessary). With that check out of the way, it is not immediately clear what to check next; as such, you may wish to try and construct a counter-example.
Is it possible to construct $p_1(x), ... , p_5(x)$ such that they are linearly independent. If so, you can conclude that no, it is not necessary for them to be linearly dependent. If you fail to construct such an example, it may illustrate along the way a manner for us to prove that they must be linearly dependent.
If we start from the spanning set above $1, t, t^2, t^3, t^4$ we may be able to work to get these to fit, without changing the linear dependence. We can take $p_1(x) = 1$ which of course satisfies the necessary $p_1(0) = 1$. However, taking $p_2(x) = x^2$ is not sufficient. Instead, we would need to add $1$ to $p_2$ as defined above in order to have it belong to $V$. We could use the same process to consider $p_3(x)...p_5(x)$.
Now we have the following:
$p_1(x) = 1$
$p_2(x) = t + 1$
$p_3(x) = t^2 + 1$
$p_4(x) = t^3 + 1$
$p_5(x) = t^4 + 1$
Is that set of polynomials linearly independent? There are a number of ways to check for this, I will include a less-than elegant check to demonstrate. I would urge you to consider other alternatives for this consideration.
Take $a_1...a_5$ to be the coefficients such that, no matter the value of $x$, we have $a_1*p_1(x) + ... + a_5*p_5(x) = 0$.
\begin{align} a_1*p_1(x) + a_2*p_2(x) + a_3*p_3(x) + a_4*p_4(x) + a_5*p_5(x) &= 0 \\ a_1 + a_2*(x + 1) + a_3*(x^2 + 1) + a_4*(x^3 + 1) + a_5*(x^4 + 1) &= 0 \\ a_2x + a_3x^2 + a_4x^3 + a_5x^4 + (a_1 + a_2 + a_3 + a_4 + a_5) &= 0 \\ \\ \textbf{Take x to be 1} \\ a_2(1) + a_3(1)^2 + a_4(1)^3 + a_5(1)^4 + a_1 + a_2 + a_3 + a_4 + a_5 &= 0\\ a_1 + 2a_2 + 2a_3 + 2a_4 + 2a_5 &= 0 \\ a_1 &= -2(a_2 + a_3 + a_4 + a_5) \\ \\ \textbf{Take x to be 0} \\ a_2(0) + a_3(0)^2 + a_4(0)^3 + a_5(0)^4 + a_1 + a_2 + a_3 + a_4 + a_5 &= 0\\ a_1 + a_2 + a_3 + a_4 + a_5 &= 0 \\ a_1 &= -(a_2 + a_3 + a_4 + a_5) \\ \\ y &:= a_2 + a_3 + a_4 + a_5 \\ -2y &= -y \\ 2y &= y \\ y &= 0 \implies a_1 = 0 \end{align}
Now, not only do we know that $a_1 = 0$ but also that $a_2 + a_3 + a_4 + a_5 = 0$. So we could rewrite the relationship as $a_2x + a_3x^2 + a_4x^3 + a_5x^4 = 0$. Recognizing that this set is linearly independent is enough to conclude that the above set is as well.
I again do not recommend you using this methodology to test for linear independence. If you have worked with determinants that is a more straightforward manner of demonstrating that this set is linearly independent. I have shown it to you in this manner to hopefully convince you of the answer, but leave room for you to work through the problem in a more elegant manner. | 2019-08-18T06:30:29 | {
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https://math.stackexchange.com/questions/429084/seeking-a-combinatorial-proof-of-the-identity-12-cdot-213-cdot-22-cdotsn/429124 | # Seeking a combinatorial proof of the identity $1+2\cdot 2^1+3\cdot 2^2+\cdots+n\cdot 2^{n-1}=(n-1)2^n+1$
I would appreciate if somebody could help me with the following problem
Q: show that combinatoric identity (using by combinatorial proof)
$$1+2\cdot 2^1+3\cdot 2^2+\cdots+n\cdot 2^{n-1}=(n-1)2^n+1$$
• Are you interested in my answer? Do not worry about the down vote. They are some irresponsible people. – Mhenni Benghorbal Jun 25 '13 at 13:53
• Here is an algebraic proof. Consider the series $$\sum_{k=0}^{n}x^k = \frac{1-x^{n+1}}{1-x}.$$ Diff. w.r. to $x$ gives $$\sum_{k=1}^{n}k x^{k-1}=\dots.$$ – Mhenni Benghorbal Jun 25 '13 at 14:37
Denote $\left[n\right]=\left\{1,\dots,n\right\}$. Count all pairs $\left(X,k\right)\in2^{\left[n\right]}\times\left[n\right]$ where $k\le\max X$ in two ways.
First way: First select $\max X$, then select $k$ and the rest of $X$.
Second way: Count all pairs in $2^{\left[n\right]}\times\left[n\right]$, then subtract the "bad" ones. Note that there is a matching between bad pairs $\left(X,k\right)$ and a non-empty sets $\emptyset\ne Y\subseteq\left[n\right]$ by $\left(X,k\right)\mapsto X\cup\left\{k\right\}$ and $Y\mapsto \left(Y\backslash\left\{\max Y\right\},\max Y\right)$.
We do this by double counting the number of tiles in a room that is $2^{n}$ tiles wide and $n-1$ tiles deep. The naive count would just be $$\text{depth} \times \text{width} = (n-1) 2^n$$
The second counting method is this: first divide the room into two sections, each of width $2^{n-1}$. The right section we divide into $n-1$ strips of $2^{n-1}$ tiles. The left section we remove one more strip of $2^{n-1}$ tile to be left with a rectangle of $(n-2) \cdot 2^{n-1}$ tiles. So we have gotten
$$(n-1) \cdot 2^n = (n-1) \cdot 2^{n-1} + 1\cdot 2^{n-1} + (n-2)\cdot 2^{n-1} = n \cdot 2^{n-1} + (n-2) \cdot 2^{n-1}$$
repeat the procedure we expand the right hand side to get the sum desired.
A picture for $n = 5$ looks something like
$$\begin{array}{cc} \color{red}{\Box\Box} \color{blue}{\Box\Box} \color{green}{\Box\Box\Box\Box} ~\color{grey}{\Box\Box\Box\Box\Box\Box\Box\Box}& \color{red}{\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box} \\ \color{grey}{\Box\Box\Box\Box}\color{black}{\Box\Box\Box\Box}~\color{blue}{\Box\Box\Box\Box\Box\Box\Box\Box} & \color{green}{\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box} \\ \color{red}{\Box\Box\Box\Box\Box\Box\Box\Box} ~\color{green}{\Box\Box\Box\Box\Box\Box\Box\Box} & \color{grey}{\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box} \\ \color{blue}{\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box} &\color{black}{\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box\Box} \end{array}$$
• Note sure what that picture looked like when you posted it, but I can't figure it out. – Thomas Andrews Jun 25 '13 at 14:45
• @P..: another interpretation would be to sum the sum on the left by taking the tiles and arranging them the way that I showed. – Willie Wong Jun 25 '13 at 14:59
• @ThomasAndrews: is this picture better? – Willie Wong Jun 25 '13 at 15:00
Combinatoric interpretation.
There is a ordered sequence (array, vector, chain, cortege, ...) of $k$ balls.
1 ball $-$ red, other balls $-$ black or white.
Let $M(k)$ $-$ number of such arrays. $M(k) = k \cdot 2^{k-1}$.
Example for $k=3$:
$\Huge{\color{red}\bullet} \circ\circ$,
$\Huge{\color{red}\bullet} \circ\bullet$,
$\Huge{\color{red}\bullet} \bullet\circ$,
$\Huge{\color{red}\bullet} \bullet\bullet$;
$\Huge{\circ\color{red}\bullet} \circ$,
$\Huge{\circ\color{red}\bullet} \bullet$,
$\Huge{\bullet\color{red}\bullet} \circ$,
$\Huge{\bullet\color{red}\bullet} \bullet$;
$\Huge{\circ\circ\color{red}\bullet}$,
$\Huge{\circ\bullet\color{red}\bullet}$,
$\Huge{\bullet\circ\color{red}\bullet}$,
$\Huge{\bullet\bullet\color{red}\bullet}$.
To find $S(n)$: number of all possible arrays (with 1 red and other black/white balls) with length $k\leqslant n$.
$S(n) = M(1)+M(2)+\ldots+M(n)$.
How to prove:
Let $L(k)$ $-$ number of arrays, where red ball is on the $k$-th place:
$L(1) = 1+2+4+\cdots+2^{n-1} = 2^n-1$;
$L(2) = 2+4+\cdots+2^{n-1} = 2^n-2^1$;
$\ldots$
$L(k) = 2^{k-1}+2^k+\cdots+2^{n-1} = 2^n-2^{k-1}$;
$\ldots$
$L(n-1) = 2^{n-2}+2^{n-1} = 2^n-2^{n-2}$;
$L(n) = 2^{n-1} = 2^n-2^{n-1}$.
So, $$S(n) = \sum_{k=1}^n L(k) = \sum_{k=1}^{n} (2^n - 2^{k-1}) = n2^n - (2^n-1) = (n-1)2^n+1.$$
• This appears to count the sum, but does not prove that it equals the desired $(n-1)2^n$. – vadim123 Jun 25 '13 at 14:13
• @vadim123, you are right. As for me: to use any common mathematical method. – Oleg567 Jun 25 '13 at 14:14
• just edited with combinatorial proof, as I see. – Oleg567 Jun 25 '13 at 14:29
Let's denote by $[n]$ the set $\{1,2,\ldots\}$ and by $\phi_{n,m}$ the set $$\left\{f:[n+1]\rightarrow [m]\mid f(i)\in [2], \ \forall i=1,2,\ldots n, \ f(n+1)\in [m]\right\}$$ i.e. all functions $f$ with domain $[n+1]$ and the property $f(i)\in\{1,2\}$ for all $i=1,2,\ldots,n$ and $f(n+1)\in[m]$. Then it is easy to see that the cardinality of $\phi_{n,m}$ is $|\phi_{n,m}|=2^n\cdot m$ and the identity to be proved is $$|\phi_{n,n-1}|=|\phi_{n-1,n}|+|\phi_{n-2,n-1}|+\ldots+|\phi_{1,2}|.$$ It's not hard to show the following properties for $\phi_{n,m}$:
• $|\phi_{n,m}|=|\phi_{n-1,2m}|$ and
• $|\phi_{n,m_1+m_2}|=|\phi_{n,m_1}|+|\phi_{n,m_2}|$
for all $n,m,m_1,m_2\in\mathbb N$. Therefore $$|\phi_{n,n-1}|=|\phi_{n-1,2n-2}|=|\phi_{n-1,n}|+|\phi_{n-1,n-2}|=|\phi_{n-1,n}|+|\phi_{n-2,2n-4}|=\\ |\phi_{n-1,n}|+|\phi_{n-2,n-1}|+|\phi_{n-2,n-3}|=\ldots=|\phi_{n-1,n}|+|\phi_{n-2,n-1}|+\ldots+|\phi_{1,2}|.$$ | 2019-09-17T19:08:26 | {
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https://physics.stackexchange.com/questions/481233/calculating-the-moment-of-inertia-of-a-uniform-sphere | # Calculating the moment of inertia of a uniform sphere [closed]
Currently trying to calculate the moment of inertia of a uniform sphere, radius R, I know the answer is $$\frac{2}{5}MR^2$$ but I keep getting $$\frac{1}{5}MR^2$$
Setup:
Assume mass per unit volume $$\rho$$
Centering the sphere on the $$z$$ axis I am first calculating the moment of inertia for one side then doubling the answer exploting the symmetry.
Splitting the hemisphere into circular discs of width $$dx$$:
$$dm = \rho\pi(R^2-x^2)dx$$
So wouldn't the moment of inertia be:
$$2\rho\pi\int_0^R(x^2(R^2-x^2))dx$$ Which comes out to $$\frac{1}{5}MR^2$$
any help appreciated.
## closed as off-topic by AccidentalFourierTransform, ACuriousMind♦May 20 at 16:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – AccidentalFourierTransform, ACuriousMind
If this question can be reworded to fit the rules in the help center, please edit the question.
• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. – ACuriousMind May 20 at 16:27
• Well the answers showed a fatal flaw in my idea of calculating the moment of inertia so I can argue that this would be helpful to other students too. – PolynomialC May 20 at 17:30
I don't understand how are you splitting the hemisphere into circular discs. your $$dm$$ must have same distance from the axis to zeroth order everywhere inside it. Disc does not. If you want to split your sphere into discs, you need to compute the moment of inertia of a disc first ($$I=mr^2/2$$), and then you can compute: $$dI_z=dm(R^2-z^2)/2=\rho \pi (R^2-z^2)^2dz/2$$ leading to: $$I_z=\int_{-R}^{R}dm(R^2-z^2)/2=\int_{-R}^{R}\rho \pi (R^2-z^2)^2dr/2=\frac{8}{15}\rho\pi R^5=\frac{2}{5}M R^2$$ This assumes the knowledge of moment of inertia of a disc though.
I would split sphere into cylinder shells centered on z-axis of width $$dx$$, then $$dm=\rho\pi h ((x+dx)^2-x^2)=2\rho\pi h x dx$$ where $$h(x)$$ is the height of the cylinder: $$h=2\sqrt{R^2-x^2}$$. The cylinder shell has same distance from the axis everywhere, so i can continue to write: $$I_z=\int_0^R x^2dm=4\rho\pi\int_0^R \sqrt{R^2-x^2} x^3 dx=\frac{8}{15}\rho\pi R^5=\frac{2}{5}M R^2$$
Each slice (disk) is offset by $$x$$ from the origin and has radius $$r = \sqrt{R^2-x^2}$$. The contribution to MMOI of each disk is $${\rm d} I = \frac{\rho}{2} \pi r^4 {\rm d}x$$
Why? Either do another integral on a disk, or use $${\rm d} I = \tfrac{1}{2} r^2 {\rm d}m$$, with $${\rm d} m = \rho \left( \pi r^2 {\rm d}x \right)$$.
You also have $$m = \rho \left( \tfrac{4}{3} \pi R^3 \right)$$
Now do the integral,
$$I = 2 \int \limits_0^R \frac{\rho}{2} \pi (\sqrt{R^2-x^2})^4 {\rm d}x = \pi \rho \int \limits_0^R (R^2-x^2)^2 {\rm d}x$$
$$I = \pi \left( \frac{m}{ \tfrac{4}{3} \pi R^3 } \right) \left( \tfrac{8}{15} R^5 \right) = \tfrac{2}{5} m R^2$$
You mistaking in calculating moment of inertia of thin circular disc by multiplying 'dm' by x^2, instead of you have to calculate it through parallel axis theorem it means you know moment of inertia of disc which is 1/2 MR^2 but it's from centre of disc not from our reference point is that centre of uniform solid sphere hence for finding that use parallel axis theorem by this we get moment of inertia as 1/2 M(R^2-X^2) +MX^2. | 2019-07-23T06:58:42 | {
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http://nonsolomacaron.it/bvig/prove-that-the-product-of-three-consecutive-integers-is-divisible-by-3.html | # Prove That The Product Of Three Consecutive Integers Is Divisible By 3
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Do you mean three consecutive even numbers (e. 3% THC, as regulations current require -- a drug test. Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer. The theorem is proved since the sum of two. Prove that for every k 6= 2, 4. Let the multiple of 2 be written 2n and the multiple of 3 be. Thus it is divisible by both 3 and 2, which means it is divisible by 6. Is this statement true or false? Give reasons. Click 'show details' to verify your result. " To set it up, you assign a variable such as x to the first of the numbers. Here, as usual, n k!:= n! k!(n k)!: 6. For instance, if we say that n is an integer, the next consecutive integers are n+1, n+2. the sum of three consecutive integers --. Also, 2 | n(n + 1), since product of two consecutive numbers is divisible by 2. Divisibility by Three. If p = 3q + 1, then n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1) is divisible by 3. What Are the Probability Outcomes for Rolling Three Dice? Look Up Math Definitions With This Handy Glossary. Depending on your needs, you may want to look for full- or broad-spectrum products. If A and B are set of multiples of 2 and 3 respectively, then show that A = B and A∪B. Three times the first of three consecutive odd integers is 3 more than twice the third. It is assumed that the criminal has been identified and is now in cus¬tody. This one is a little weird but it really is quite simple after you practice it a couple of times. A few days after the Farm Bill went into law, the FDA issued a statement stating any hemp-based CBD product that is marketed as having therapeutic This means that for standard CBD oil users -- those who use certified products containing less than. A positive integer nis called highly divisible if d(n) >d(m) for all positive integers m 0, find the number of different ways in which n can be written as a sum of at two or more positive integers. integers, and this offsets the advantage of having far fewer multiplica-tions to perform. It wasn't too long ago when every business claimed that the key to winning customers was in the quality of the product or service they deliver. Showing that exactly one of two consecutive integers is divisible by two is shown above with the addition to the first part: "as (n+1) = 2k+1 is not divisible by two and so only n is divisible by 2. Pharaoh's snake is a simple demonstration of firework. Pictorial Presentation: Sample Solution. The integer part of the result is the number of digits. Prove that a number divisible by a prime p and divisible by a di erent prime q is also divisible by pq. Now, that doesn't seem to be divisible by 6, so if you still don't understand, let's try a logical approach. [Hint: See Corollary 2 to Theorem 2. If it is divisible by 2 and by 3, then it is divisible by 6. This article only contains results with few proofs. What's the difference between CBD oil and hempseed oil?. Show that a number is a perfect square only when the number of its divisors is odd. Prove that the difference between two consecutive square. n,n+1,n+2 be three consecutive positive integers We know that n is of the form 3q, 3q +1, 3q + 2 So we have the following cases Case – I when n = 3q In the this case, n is divisible by 3 but n + 1 and n + 2 are not divisible by 3 Case - II When n = 3q + 1 Sub n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. Picking numbers is a very bad way of solution such tasks. The sum of n consecutive cubes is equal to the square of the nth triangle. 2019 - 2020. 1 Sequences of Consecutive Integers 1 PEN A37 A9 O51 A37 If nis a natural number, prove that the number (n+1)(n+2) (n+10) is not a perfect square. 1 Q3 Prove that the product of three consecutive positive integers is divisible by 6. Prove that one of any three consecutive positive integers must be divisible by 3. and doesn't actually prove anything. For any positive integer n, use Euclid’s division lemma to prove that n3 – n is divisible by 6. What Are the Probability Outcomes for Rolling Three Dice? Look Up Math Definitions With This Handy Glossary. So the least possible sum of their birth years is 2002 + 2001 + 2005 = 6008. Solution for Determine whether the statement is true or false. These are consecutive odd integers. Prove that if m, m +1, m + 2 are three consecutive integers, one of them is divisible by 3 4. Hint: What are the possible remainders when we divide an integer by 3? - 15053493. CS103X: Discrete Structures Homework Assignment 2: Solutions Due February 1, 2008 Exercise 1 (10 Points). If n is divisible by 3 then n+1 and n+2 cannot be divisible by 3. (the alphanumeric value of MANIC SAGES) + (the sum of all three-digit numbers you can get by permuting digits 1, 2, and 3) + (the number of two-digit integers divisible by 9) - (the number of rectangles whose sides are composed of edges of squares of a chess board) 91 + 1332 (12*111) + 10 - 1296 = 137. now, similarly, when a no. CHAPTER 2: NUMBERS AND SEQUENCES. In order to test this, you must take the last digit of the number Since 28 is divisible by 7, we can now say for certain that 364 is also divisible by 7. Fact tor n -n completely. Let 3 consecutive positive integers be p, p + 1 and p + 2. Show Step-by-step Solutions Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. for any integer $n$: [. (N + 1)], which means that exactly one element is missing. (3) The sum of three consecutive even integers is 528; find the integers. Any group of 3 consecutive numbers will have one number that is a multiple of 3 and at least one number that is a multiple of 2. but n and n+1 are not divisible by 3. When spoken in everyday English, a fraction describes how many parts of a certain size there are, for example, one-half, eight-fifths, three-quarters. 3 : The product of any three consecutive even natural numbers is divisible by 16. If three such primes existed we would have pqr = k (p^2 + q^2 + r^2) for some integer k. architecture. Let n be an integer divisible by 6. Prove that the product of three consecutive positive integers is divisible by 3. Formula for Consecutive Even or Odd Integers. Prove that if m, m +1, m + 2 are three consecutive integers, one of them is divisible by 3 4. The problem is to find. Subsequence of Integers: Given any sequence of n integers, positive or negative, not necessarily all different Let H be an n-element subset of a group G. Try some examples: , ,. Let n, n + 1, n + 2 be three consecutive positive integers. Class 10 maths. Pictorial Presentation: Sample Solution. Let a and b be positive integers. Prove that no set of 2010 consecutive positive integers can be partitioned into two subsets, each having the same product of the elements. Look at the following two sets. It wasn't too long ago when every business claimed that the key to winning customers was in the quality of the product or service they deliver. For any positive integer kthe product of kconsecutive integers is divisible by k!. Any situation you can imagine. Remember that to disprove a statement we always expect a counterexample! a) The product of two even integers. Prove that the product of any three consecutive positive integers is divisible by 6. Let three consecutive integers be n, n+1 and n+2. Ask Question. If A and B are set of multiples of 2 and 3 respectively, then show that A = B and A∪B. With any combination of consecutive natural numbers, why is one integer divisible by three and why is ONLY one number divisible by 3? Divisibility by 3 in Three Consecutive Numbers. 1 Questions & Answers Place. is divisible by 6. 7) If a is a rational number and b is an irrational number, then a + b is an irrational number. At least one of the three consecutive integers will be even , ie , divisible by two. Conversion to a z-score is done by subtracting the mean of the distribution from the data point and dividing by the standard deviation. x6 1 = (x2 1. programming. For example, once we prove that the product of two odd numbers is always odd, we can immediately conclude (without computation) In this section, you will study how to distinguish between the three different kinds of statements mathematics is built up Theorem A1. So the least possible sum of their birth years is 2002 + 2001 + 2005 = 6008. A prime number is one which is only divisible by 1 and itself. , 1000}, what is the probability that their sum is divisible by 3?. How many sets of three consecutive integers whose product is equal to their sum. the unit place ends with 0. is divisible by some prime p. com Tel: 800. Prove that for each natural number n 2, there is a natural number xfor which f(x) is divisible by 3n but not 3n+1. The third integer is: A. Prove that out of two consecutive integers, one is divisible by 2. determine the statement is true or false. Therefore, n = 3p or 3p + 1 or 3p + 2 , where p is some integer. 2: prove that the comparability relation modulo a positive integer n on the set Z: x = y (modn) Proof: non the definition of x is comparable with y modulo n if and only if x — y is divisible by n Problem number 18: how many ways can decompose the number 1024 into a product of three. Prove that the product of any three consecutive integers is divisible by 6. Let's call the three integers n-1, n, n+1. Give 3 integers whose sum is -12. If d | b, then there are d distinct solutions modulo n, and these solutions are congruent modulo n / d. 991 is a permutable prime. Prove that 2n n divides LCM(1;2;:::;2n). What is the smallest possible value of x greater than 10? The number 210 is the product of two consecutive positive integers and is also the product of three consecutive integers. determine the statement is true or false. To divide a number by 10, simply shift the number to the right by one digit (moving the decimal place one to the left). How many sets of three consecutive integers whose product is equal to their sum. Pseudocode Example 10: Find the biggest of three (3) Numbers (Pseudocode If Else Example). Is this statement true or false? Give reasons. Prove or give a counterexample for the following: Use the Fundamental Theorem of Arithmetic to prove that for n 2N, p n is irra-tional unless n is a perfect square, that is, unless there exists a 2N for which n = a2. Thus, the product xyz will have a factor of 3. (Why?) We consider these two cases separately. We intend our proof to be understandable for everyone who has basic familiarity with integer numbers and who is capable of. the sum of three consecutive integers --. Solution for Determine whether the statement is true or false. is divisible by 2 remainder abtained is 0 or 1. (2) For all integers a;b; and c, if a divides b and a divides c then a divides b+ c. Let n, n + 1, n + 2 be three consecutive positive integers. The sum of n consecutive cubes is equal to the square of the nth triangle. Prove or give a counterexample for the following: Use the Fundamental Theorem of Arithmetic to prove that for n 2N, p n is irra-tional unless n is a perfect square, that is, unless there exists a 2N for which n = a2. Prove that if for some integers a, b, c we have 9Ia3+b3+c3, then at least one of the numbers a, b, c is divisible by 3. Hello reto. the sum of three consecutive even integers d. Their product is P = x3(x6 71). l)n(n + l) 1 One of these must be a multiple of 3, so, n — n is a multiple of 3. ← Prev Question Next Question →. (the alphanumeric value of MANIC SAGES) + (the sum of all three-digit numbers you can get by permuting digits 1, 2, and 3) + (the number of two-digit integers divisible by 9) - (the number of rectangles whose sides are composed of edges of squares of a chess board) 91 + 1332 (12*111) + 10 - 1296 = 137. Prove that the product of 3 consecutive numbers is divisible by 3. Now n(n+1)(n+2) is the product of 3 consecutive integers which is always divisible by 6. What's the difference between CBD oil and hempseed oil?. Picking numbers is a very bad way of solution such tasks. This means an integer cannot have a fractional part expressed either as a fraction or a decimal. Two Times The Second Of Three Consecutive Odd Integers Is 6 More Than The Third. Let pand q be prime numbers. Let three consecutive integers be n, n+1 and n+2. What Are the Probability Outcomes for Rolling Three Dice? Look Up Math Definitions With This Handy Glossary. Does your method work with negative numbers?. In any case of THREE CONSECUTIVE integers, one of them MUST be a multiple of 2, and one of them MUST be a multiple of 3. 1 Exercise 14) How many integers between 1 and 1000 (exclusive) are not divisible by 2, 3, 5, or 7? (b. Consecutive integers are integers that follow each other such as -9 and -8 or +4 and +5. 13 Prove that the difference between the squares of any 2 consecutive integers is equal to the sum of these integers. five more than twice a number. - that the product is even) Say n is even, then divisibility follows for the product, since whatever factor of n, (n+1), (n+2) also appears as a factor in the product of the three. In order to test this, you must take the last digit of the number Since 28 is divisible by 7, we can now say for certain that 364 is also divisible by 7. In any three consecutive integers, there is always a multiple of 3. integers, and this offsets the advantage of having far fewer multiplica-tions to perform. Thus for all odd values of n, 2 1n is divisible by 3. The product of two enumerable sets is enumerable. ) you see that any three consecutive integers has to have one of these numbers, so it has at least one number that is divisible by 3. Find the smallest number that, when. the set consists of 4 integers and (using m = 14, an even integer, and j = 3 in the definition) the integers are 14, 14+2, 14+4, and 14+6. Explanation and Proof. Ans: n,n+1,n+2 be three consecutive positive integers We know that n is of the form 3q, 3q +1, 3q + 2 So we have the following cases Case – I when n = 3q In the this case, n is divisible by 3 but n + 1 and n + 2 are not divisible by 3 Case - II When n = 3q + 1 Sub n = 2. Sum of three consecutive numbers equals. Another example-- we could start at 11. 2 Exercise 12) Provide a direct proof that n2 n+ 5 is odd, for all integers n. (b) Prove that L has a regular expression, where L is the set of strings satisfying all four conditions. N is divisible by pq. Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46, find the integers. What is the sum of those five integers?. In the interest of clarity in the applet above, the coordinates are rounded off to integers and the lengths rounded to one decimal place. Consider divisibility by 2 (i. Thus, 6, 28, 496 are Perfect and. Prove that 6 Ö3+ 30 Ö 3 Ö2 + 15 is an even number. Prove that for every k 6= 2, 4. Use the Quotient-Remainder Theorem with d = 3 to prove that the product of two consecutive integers has the form 3k or 3k +2 for some k 2Z. It is given in the problem that the greater interger is x. Though commonly a travel-goal destination for couples, it appears that not everybody This rate has been gradually increasing since 1975, especially when the Family Law Act legalised 'no-fault divorce', stating that the cause rate of. The next one is 15. An OverflowError is raised if the integer is not representable with the given number of bytes. Lars' answer is good +1. JEE Main and NEET 2020 Date Announced!! View More. Prove that the product of three consecutive positive integers is divisible by 3. Statement 1 does not provide any additional information, and definitely it is not sufficient. To prove that N is divisible by 3 : Any integer n can be of one of the forms. 781 Homework 3 Due: 25th February 2014 Q1 (2. This combination of sand and rock means that the soil is not very fertile. Find the number of terminal zeros in the decimal expansion of 1000!. Non-Divisible Subset,Hackerrank, coderinme,learn code,java,C in hand,c++,python,coder in me,Hackerrank solution,algorithm, competitive Given a set, S, of n distinct integers, print the size of a maximal subset, S', of S where the sum of any 2 numbers in S' is not evenly divisible by k. The product of $n$ consecutive positive integers is divisible by the product of the first $n$ consecutive positive integers. And by divisible by 3, this is to mean the product of the division is a whole number, and not a decimal. What is their sum? Let’s use our divisibility knowledge to factor 157,410: 157,410 = 10 × 15,741 the number ends in 0, 10 is a factor. For any positive integer n, prove that n 3 – n is divisible by 6. Click here 👆 to get an answer to your question ️ Prove that one of every three consecutive positive integers is divisible by 3? 1. Translation prove. the product is divisible by 6. asked Feb 9, 2018 in Class X Maths by priya12 ( -12,636 points) real numbers. Prove that 2x + 3y is divisible by 17 iff 9x+5y is divisible by 17. On this page we prove the theorem known from school that an integer is divisible by 3 if and only if the sum of its digits is divisible by 3. Example 10: Joe is able to drive 342 miles on 18 gallons of gasoline. N is not a prime number. In this instance, it is. So these are examples of consecutive odd. Btw jayshay - if you said 7n, 7n+1 and 7n+2 then your 'proof' would effectively be proving that the product of 3 consecutive integers is a multiple of 7. (b) First prove that for x :::; we. M2320-Assignment 6: Solutions Problem 1: (Section 6. 6 is 2 times 3. By induction hypothesis, the first term is divisible by 6, and the second term 3(k+1)(k+2) is divisible by 6 because it contains a factor 3 and one of the two consecutive integers k+1 or k+2 is even and thus is divisible by 2. We can have 9 dice without any four matching or any four being all different: three 1's, three 2's, three 3's. 991 is a permutable prime. Let a and b be positive integers. It is implied that the new auditorium supports an education program in … arts. Look at the following two sets. And what if we started at 6 and we were asked to find the next even number. 3 consecutive integers: one must be divisible by 2; three integer; there is at least one which can be divided by 3; since the two-side integers are prime, then the middle one must can be divided by3 and 2; it means the middle can be divided by 2*3=6. If n is divisible by 4, then n = 4k for some integer k and n(n+2) = 4k(4k+2) = 8k(2k+1) is divisible by 8 and therefore so is the product of the four consecutive. By the three cases, we have proven that the square of any integer has the form 3k or 3k +1. Explanation and Proof. How many divisors do the following numbers have: pq;pq2;p4;p3q2? 5. Prove that. " To set it up, you assign a variable such as x to the first of the numbers. Say you have N consecutive integers (starting from any integer). In C51 Grimm made the conjecture that if p,p' are consecutive primes, then for each integer m, p < m < p', we can find a prime factor 4,of m such that the q, 's are all different. ) you see that any three consecutive integers has to have one of these numbers, so it has at least one number that is divisible by 3. Is this statement true or false? Give reasons. 103 has the property that placing the last digit first gives 1 110 is the smallest number that is the product of two different substrings. Consecutive Integers Word Problems: WP2 [fbt] Writing a formula from a sequence Algebra 2 - Exponents Product of three consecutive odd numbers is 9177 Find their sum Class 10 maths chapter 1. “The product of three consecutive positive integers is divisible by 6”. Problem III. Prove or give a counterexample for the following: Use the Fundamental Theorem of Arithmetic to prove that for n 2N, p n is irra-tional unless n is a perfect square, that is, unless there exists a 2N for which n = a2. Then, see the answers. N is not a prime number. Hello reto. 𝑐 is a positive integer. We have to prove this for any arbitrary k ∈Z, so fix such a k. ) b) Use the divisibility lemma to prove that an integer is divisible by 5 if and only if its last digit. Prove that the fraction (n3 +2n)/(n4 +3n2 +1) is in lowest terms for every possible integer n. Let a and b be positive integers. Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2. In any case of THREE CONSECUTIVE integers, one of them MUST be a multiple of 2, and one of them MUST be a multiple of 3. Option (C). Prove that the product of any three consecutive positive integers is divisible by 6. 40 Find a compound proposition involving the propositional variables p, q and r that is true when p and q are true and r is false but. In order to test this, you must take the last digit of the number Since 28 is divisible by 7, we can now say for certain that 364 is also divisible by 7. Look at the following two sets. (a) If m˚(m) = n˚(n) for positive integers m;n. Depending on a company's goals and the industry. CS103X: Discrete Structures Homework Assignment 2: Solutions Due February 1, 2008 Exercise 1 (10 Points). Prove that the product of any three consecutive positive integers is divisible by 6. Any three consecutive integers contains one multiple of 3, so four consecutive integers would contain at least one. Let's call the three integers n-1, n, n+1. (So the last digit must be 0, 2, 4, 6, or 8. If p = 3q, then n is divisible by 3. Now, the fact that you are multiplying numbers with these two properties, guarantees that the product will be a multiple of 3 and 2. Prove that if n is odd then n2 - 1 is divisible by 8. Prove that n3 - n is always divisible by 3 in each of these three different ways. Formula for Consecutive Even or Odd Integers. Expression. Example 10: Joe is able to drive 342 miles on 18 gallons of gasoline. The product of three consecutive integers is 157,410. Each of these primes is a divi-sor of one of the birth years. the sum of three consecutive integers --. 20 Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively. # So if we add these up we get #6# as a sum. The product of two enumerable sets is enumerable. At least one of the three consecutive integers will be even , ie , divisible by two. ) Therefore, 6 | 3n(n + 1). Divide the series into two equal groups. #Asap Give correct ans Get the answers you need, now!. We find that N is a product of three consecutive integers. This article only contains results with few proofs. The product of these two are x*(x-1)=x^2-x=342. The LIGO project based in the United States has detected gravitational waves that could allow scientists to develop a time machine and travel to the earliest and darkest This was the first time that the witnessed the "ripples in the fabric of space-time. Prove that the number of solutions (x, y, z) in nonnegative 1. So we only need to show that one of the three integers is divisible by 3, because a number divisible by both 3 and 2 is necessarily divisible by 6. Exercise: 2. Twenty-three years after discovery of the Rosetta stone, Jean Francois Champollion, a French philologist, fluent in several languages, was able to decipher the Young believed that sound values could be assigned to the symbols, while Champollion insisted that the pictures represented words. It follows that the smaller integer would be 1 arrenhasyd and 45 others learned from this answer. It has been proven that a child's worldview settles by the time they turn 11 years old, and they become capable of evaluating the world as an adult, solve problems and even make plans for future. The problem can be restated as saying the division algorithm gives either 0 or 1 as remainder when n2 is divided by 3, and never 2. Third, there are at least two ways to do this problem - with a bit of logic and some arithmetic; and using algebra. Non-Divisible Subset,Hackerrank, coderinme,learn code,java,C in hand,c++,python,coder in me,Hackerrank solution,algorithm, competitive Given a set, S, of n distinct integers, print the size of a maximal subset, S', of S where the sum of any 2 numbers in S' is not evenly divisible by k. Number Theory. In order to prove this theory, a pair of Italian scientists conducted a series of experiments. But j2 is an integer since it is the product of integers. In any three consecutive integers, there is always a multiple of 3. seven divided by twice a number. Show that the product of three consecutive integers is divisible by 504 if the middle one is a cube. Suppose that for every sequence of n elements from H, some consecutive subsequence has the property that the product of its elements is the. Problemo? He hasn’t shown it’s true for all possible integers. Prove that for every k 6= 2, 4. The byteorder argument determines the byte order. How many positive integers satisfy , where is the number of positive integers less than or equal to relatively It follows that The last three digits of this product can easily be computed to be. => 3n + 3 = 3(n + 1) so, we can say that one of the numbers n, n + 1 and n + 2 is always divisible by 3. Clearly the product is divisible y 2. The last statement is false, thus p is not even. asked • 09/28/14 use the quotient remainder theorem with d=3 to prove that the product of any two consecutive integers has the form 3k or 3k+2 for some integer k. ) b) Use the divisibility lemma to prove that an integer is divisible by 5 if and only if its last digit. Suppose that p is an even number, then p is divisible by 2. The problem is rich in the mathematics it can involve and produce, and it is for this reason that it is something that requires further study. [3] b) Give an example to show that the sum of four consecutive integers is not always divisible by 4. Take x as the typical dosage for a patient whose body weight is 120 pounds barb4right 120 15 2 x = barb4right x = 16. By the laws of divisibility, anything divisible by 2 and 3 is divisible by 6. Note then that the product of three consecutive integers is divisible by $3$ (this about it). Build your proof around this observation. For example, you can use it to show that the product of any three consecutive numbers is a multiple of. For example, the number 31 is NOT divisible by 3 because $3 + 1 = 4$, which is not divisible by 3. ← Prev Question Next Question →. Consider divisibility by 2 (i. Prove that one of the two numbers is divisible by the other. Divisibility guidelines for 6: To know if a number is divisible by 6, you have to first check if it is divisible by 3 and by 2. (C) _, some living things are able to do well in this setting. and doesn't actually prove anything. RD Sharma Real Number Class 10 solutions Real Numbers Class 10 cbse VipraMinds. Always check the ingredients to see that they suit your needs. L1 is the set of all strings that are decimal integer numbers. Click 'show details' to verify your result. Example 2: Is the number 8256 divisible by 7?. We wish to prove that if n2 is divisible by 3, then n is divisible by 3. Show that the sum of two consecutive primes is never twice a prime. (Total for question 2 is 2 marks) 3 Prove that (3 n + 1) 2 – (3 n – 1) 2 is always a multiple of 12, for all positive integer values of n. A9 Prove that among any ten consecutive positive integers at least one is relatively prime to the product of the others. Hint: What are the possible remainders when we divide an integer by 3? - 15053493. 990 is a triangular number that is the product of 3 consecutive integers. the three consecutive integers be x,y and z. Triangles: 1 3 6 10 15 21 28. Let n, n + 1, n + 2 be three consecutive positive integers. Prove the statement directly from the definitions if it is true, and give a counterexample if it…. The problem can be restated as saying the division algorithm gives either 0 or 1 as remainder when n2 is divided by 3, and never 2. Class 10 maths. If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. Exercise 12. Find a six-digit number that is increased by a factor of 6 if one exchanges (as a block) its rst and last three digits. Therefore, the four consecutive integers are #400, 401, 402, 403#. Take the 3 consecutive integers, 2,3,4 their sum is 9 and you are done. n 3 – n = n(n2 – 1) = n(n+1)(n – 1) = (n – 1)n(n+1) = product of three consecutive positive integers. A set of integers such that each integer in the set differs from the integer immediately before by a difference of 2 and each integer is divisible by 2 Example 2, 4, 6, 8, 10, 12, 14,. Prove that if m, m +1, m + 2 are three consecutive integers, one of them is divisible by 3 4. Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2. Use Euclid’s division lemma to show that one and only one out of n, n + 4, n + 8, n + 12 and n +. Hello reto. When people were standing on soft carpet and viewed a product that was moderately far away, they judged that item's appearance to be comforting. Say you have N consecutive integers (starting from any integer). [Hint: See Corollary 2 to Theorem 2. These are consecutive odd integers. How many divisors do the following numbers have: pq;pq2;p4;p3q2? 5. They explain the lights are created by the water. Let the three consecutive positive integers be n, n + 1 and n + 2. Hint: What are the possible remainders when we divide an integer by 3? - 15053493. All arguments can be made with basic number theory, with a little knowledge. The array contains integers in the range [1. CHAPTER 2: NUMBERS AND SEQUENCES. For any positive integer n, prove that n 3 – n is divisible by 6. What can you say about the prime factorisation of the denominator of 27. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. (3) The sum of three consecutive even integers is 528; find the integers. [2] Name: Total Marks: Rebecca Simkins. Suppose that p is an even number, then p is divisible by 2. Now, 2+4+x+3+2=11+x which must be divisible So 6K4 must be divisible by 3. For example, the number 31 is NOT divisible by 3 because $3 + 1 = 4$, which is not divisible by 3. Let 3 consecutive positive integers be n, n+1 and n+2 Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2. These are now called Euclid numbers and Euler proved that all even Perfect numbers are of this form for some positive prime number n. Third, there are at least two ways to do this problem - with a bit of logic and some arithmetic; and using algebra. The third runner-up on our list of world divorce rates is France. Write a Python program to find those numbers which are divisible by 7 and multiple of 5, between 1500 and 2700 (both included). Most deserts are covered with sand, (B) _. So the result follows from Proposition 11. “The product of two consecutive positive integers is divisible by 2”. Pictorial Presentation: Sample Solution. 24 is the largest number divisible by all numbers less than its square root. l)n(n + l) 1 One of these must be a multiple of 3, so, n — n is a multiple of 3. If p = 3q, then n is divisible by 3. EXAMPLE #2 "Speaking of the article, I should say that the most complicated dilemma recalled by the author is the lack of time versus storing resources and not the rest of the ideas. PROBLIMS 3 31. Thus, pq cannot be divisible by p^2+q^2. Sum of Three Consecutive Integers Calculator. Customers no longer base their loyalty on price or product. If n = 3p + 1 , then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. 1 Consecutive integers with 2p divisors. RD Sharma Real Number Class 10 solutions Real Numbers Class 10 cbse VipraMinds. Now n(n-1)(n+1) is the product of three consecutive integers. Prove that the product of three consecutive positive integers is. If n mod 3 = 2 then n+1 is divisible by 3. Now, that doesn't seem to be divisible by 6, so if you still don't understand, let's try a logical approach. Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Proof by mathematical induction that the product of three consecutive integers is divisible by 6. Real numbers class 10. In mathematics, the least common multiple, also known as the lowest common multiple of two (or more) integers a and b, is the smallest positive integer that is divisible by both. Click 'show details' to verify your result. Therefore 8n(n+1)(n+2) is always divisible by 6*8 or 48. 2 Exercise 12) Provide a direct proof that n2 n+ 5 is odd, for all integers n. [3] b) Give an example to show that the sum of four consecutive integers is not always divisible by 4. Prove that n2-n is divisible by 2 for every positive integer n. Exercise: 2. For any positive integer n, prove that n3 – n is divisible by 6. The number is divisible by 6 means it must be divisible by 2 and 3. He brought a _action against the company, claiming that the accident had been caused by a manufacturing fault in the automobile. What are the two odd integers? 12. 111 is the smallest possible magic. Exercise 12. case (3) z is a multiple of three. By the three cases, we have proven that the square of any integer has the form 3k or 3k +1. Some other very important questions from real numbers chapter 1 class 10. Prove that the product of any three consecutive positive integers is divisible by 6. Since n is a perfect square, n is congruent to 0 or 1 modulo 4. Solution for Determine whether the statement is true or false. 1, which is divisible by 9. 2 Exercise 12) Provide a direct proof that n2 n+ 5 is odd, for all integers n. : Therefore: n = 3p or 3p+1 or 3p+2, where p is some integer If n = 3p, then n is divisible by 3 If n = 3p+1, then n+2 = 3p+1+2 = 3p+3 = 3(p+1) is divisible by 3. Consider three consecutive integers, n, n + 1, and n+ 2. 7: given non-empty nite sets X and Y with jXj= jYj, a function X !Y is an injection if and only if it is a surjection. $\begingroup$ "product of two consecutive numbers is divisible by 2" should be proved, and likely by induction (otherwise how is it different from "product of three consecutive numbers is divisible by 3" which is almost the same thing as the thing to prove in the first place?) $\endgroup$ - mathguy Aug 1 '16 at 13:36. Your flaw is in the fact that you're simple multiplying in then re-dividing by the same number, which is possible for 6, 7 etc. To divide a number by 10, simply shift the number to the right by one digit (moving the decimal place one to the left). } Write a method named consecutive that accepts three integers as parameters and returns true if they are three consecutive numbers; that is, if the numbers can be arranged into an order such that there is some integer k such Your method should return false if the integers are not consecutive. Proof: Let n be a perfect square, and let P = (n − 1) n (n +1) be the product of the three consecutive integers with n in the middle. Can you show that the product of three consecutive integers are divisible by 3? The integer multiples of 3 are divisible by 3 and there are only two integers between any two consecutive integer multiples of 3 viz. yes, three consecutive integers can be n, (n + 1)and (n + 2). 3 21 137n m+ = MP2-G , proof. Prove that only one out of three consecutive positive integers is divisible. They have a difference of 2 between every two numbers. Pseudocode Example 10: Find the biggest of three (3) Numbers (Pseudocode If Else Example). the unit place ends with 0. If p = 3q + 1, then n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1) is divisible by 3. Sum of consecutive squares equal to a square. Prove that the product of two consecutive odd integers is not a perfect square. the sum of three consecutive odd integers c. P is 216 the sum of their product in pairs is 156, find them. Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible. asked Feb 9, 2018 in Class X Maths by priya12 ( -12,636 points) real numbers. Homework Equations The Attempt at a Solution This doesn't seem true to me for any 3 consecutive ints. If n is divisible by 4, then n = 4k for some integer k and n(n+2) = 4k(4k+2) = 8k(2k+1) is divisible by 8 and therefore so is the product of the four consecutive. Case (i): is even number. Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2. She finds that if a number has 0 in the ones place then it is divisible by 10. 991 is a permutable prime. Prove that n3 - n is always divisible by 3 in each of these three different ways. Divide the series into two equal groups. Every other positive integer between 1000 and 10 000 is a four-digit integer. CHAPTER 2: NUMBERS AND SEQUENCES. Now n(n-1)(n+1) is the product of three consecutive integers. 7,8,9) like I think the source of the confusion is partly that both the sum AND the product of three consecutive even numbers are BOTH divisible by 6. Formula for Consecutive Even or Odd Integers. Prove that the sum of three consecutive integers is a multiple of 3. Sum of Three Consecutive Integers Calculator. It follows that the smaller integer would be 1 arrenhasyd and 45 others learned from this answer. Use the mod notation to rewrite the result of part (a). That is: $\displaystyle \forall m, n \in \Z_{>0}: \exists r \in \Z: \prod_{k \mathop = 1}^n \paren {m + k} = r \prod_{k \mathop = 1}^n k$. They settle and use the first price that comes to mind, copy competitors, or (even worse) guess. But j2 is an integer since it is the product of integers. seven divided by twice a number. Look at the following two sets. (the alphanumeric value of MANIC SAGES) + (the sum of all three-digit numbers you can get by permuting digits 1, 2, and 3) + (the number of two-digit integers divisible by 9) - (the number of rectangles whose sides are composed of edges of squares of a chess board) 91 + 1332 (12*111) + 10 - 1296 = 137. “The product of three consecutive positive integers is divisible by 6”. ) b) Use the divisibility lemma to prove that an integer is divisible by 5 if and only if its last digit. as these numbers will respectively leave remainders of 1 and 2. Though commonly a travel-goal destination for couples, it appears that not everybody This rate has been gradually increasing since 1975, especially when the Family Law Act legalised 'no-fault divorce', stating that the cause rate of. So, P(n+2) = 3*k + 6*x both the summation elements of P(n+2) are divisible by 3, so P(n+2) is divisible by 3. Pseudocode Example 10: Find the biggest of three (3) Numbers (Pseudocode If Else Example). 1 Sequences of Consecutive Integers 1 PEN A37 A9 O51 A37 If nis a natural number, prove that the number (n+1)(n+2) (n+10) is not a perfect square. Let, (n - 1) and n be two consecutive positive integers ∴ Their product = n(n - 1) = n2 − n We know that any positive integer is of the form 2q or 2q + 1, for some integer q. If u divide any integer by three, remainder will either be zero or one or two. Solution for Determine whether the statement is true or false. To see how many digits a number needs, you can simply take the logarithm (base 2) of the number, and add 1 to it. Determine all positive integers nfor which there exists an integer m so that 2n 1 divides m2 + 9. Prove the statement directly from the definitions if it is true, and give a counterexample if it…. Now, we can make a conjecture that the sum of two consecutive numbers is divisible by 4. Now, that doesn't seem to be divisible by 6, so if you still don't understand, let's try a logical approach. Notice that the sum is divisible by 4. 2 Prove algebraically that the sum of any three consecutive even integers is always a multiple of 6. After having gone through the stuff given above, we hope that the students would have understood how to find the terms from the sum and. Jensen likes to divide her class into groups of 2. How many sets of three consecutive integers whose product is equal to their sum. By the quotient-remainder theorem, n = 3q + r. He brought a _action against the company, claiming that the accident had been caused by a manufacturing fault in the automobile. Proof Let n, q, and r be non-negative integers. Prove that only one out of three consecutive positive integers is divisible. 4, and the transitive property of order. Problem 12. They settle and use the first price that comes to mind, copy competitors, or (even worse) guess. Prove that the number of solutions (x, y, z) in nonnegative 1. Thus the product of three consecutive integers is also even. Let, (n - 1) and n be two consecutive positive integers ∴ Their product = n(n - 1) = n2 − n We know that any positive integer is of the form 2q or 2q + 1, for some integer q. Prove that one of every three consecutive positive integers is divisible by 3? Find answers now! No. Any consecutive series of 3 integers has a multiple of 3 in it, since every third integer is a multiple of 3. Essentially, it says that we can divide by a number that is relatively prime to. All arguments can be made with basic number theory, with a little knowledge. Prove by using laws of logical equivalence that (a) ¬(p → q) ∧ q ∼ f alse; (b) p ∨ ¬(q ∧ p) ∼ true; (c) p ∧ [(p ∨ r) ∧ (q ∨ r)] ∼ (p ∧ q) ∨ (p ∧ r). (a) Prove that the sum of the squares of 3, 4, 5, or 6 consecutive integers is not a perfect square. [1 mark] Assume, a is a rational number, b is an irrational number a + b is a rational number. Now $2$ and $3$ are prime, so the prodcut is divisible by $2\cdot 3 = 6$. [Chinese Remainder Theorem] Let n and m be positive integers, with (n,m)=1. Solution for Determine whether the statement is true or false. Show that the product of n consecutive integers is divisible by n! A 17. An even number is divisible by 2, so it can be represented by 2n, where n is an integer. And what if we started at 6 and we were asked to find the next even number. Prove that 2x + 3y is divisible by 17 iff 9x+5y is divisible by 17. These are consecutive odd integers. Prove that is irrational. Look at the following two sets. Show Step-by-step Solutions Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. Limitations. With these sums we can quickly find all sequences of consecutive integers summing to N. What is the least possible sum of their birth years? 10. Let us find Product three. What do you observe? (b) What is the sum of the first million positive odd. Prove that an integer $$n$$ is divisible by 3 iff $$n^{2}$$ is divisible by \(3. If u divide any integer by three, remainder will either be zero or one or two. For instance, if we say that n is an integer, the next consecutive integers are n+1, n+2. In any set of 3 CONSECUTIVE numbers, there will always be one number that is divisible by 3, and at least one number that is divisible by 2. 994 is the smallest number with the property that its first 18 multiples contain the digit 9. Prove by using laws of logical equivalence that (a) ¬(p → q) ∧ q ∼ f alse; (b) p ∨ ¬(q ∧ p) ∼ true; (c) p ∧ [(p ∨ r) ∧ (q ∨ r)] ∼ (p ∧ q) ∨ (p ∧ r). We can claim that it is Therefore, the product of any three consecutive integers is always divisible by 6. 7 jx7 x by Fermat’s theorem, and therefore 7 jx2(x x), i. Using the Quotient-Remainder Theorem with d = 3 we see that. The sum of two consecutive integers is 27. In order to test this, you must take the last digit of the number Since 28 is divisible by 7, we can now say for certain that 364 is also divisible by 7. Suppose n is not divisible by 3.
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https://stats.stackexchange.com/questions/186950/bias-of-method-of-moments-estimator-for-pareto-distribution-with-known-scale-par | # Bias of method of moments estimator for Pareto distribution with known scale parameter
Let $x$ be a Pareto distribution with a known scale parameter $m>0$, i.e. $x\sim f(x|a)=\frac{am^a}{x^{a+1}}, x>a, a>0$
$\mathrm{E}\left[X\right]=\frac{am}{a-1}$
Using method of moments estimator for the shape parameter, $\frac{\hat{a}m}{\hat{a}-1}=\sum_{i=1}^{n}{\frac{x_i}{n}}, \hat{a}=\frac{\sum{x_i}}{\sum{x_i}-mn}$
How does one calculate the bias of the estimator? What's $\mathrm{E}\left[\hat{a}\right]$? Is there a known distribution for sum of Pareto variables?
If there's no closed form expression for the expectation, is the bias always one way?
• This seems rather like a textbook-style exercise. Is this for some class? – Glen_b -Reinstate Monica Dec 16 '15 at 7:48
• @Glen_b no, I wish. It's for modeling losses associated with business activities, where the threshold is determined by business judgment – C.J. Jackson Dec 16 '15 at 16:24
• I've made some additional comments with that in mind. – Glen_b -Reinstate Monica Dec 16 '15 at 22:37
The distribution of a sum of Pareto variates is not especially simple, but has been done. [1] [2]
Without loss of generality, we can take $m=1$; we can simply divide through by $m$ to work with $X^*=X/m$ and the lower limit for $X^*$ is then $1$. Since $m$ is then just a scale factor applied to the data we can translate any results back to the original data scale.
In this answer I haven't attempted to compute the exact bias from those published results. If it were me faced with this exercise I probably would focus first on using simulation to obtain a clear understanding how the bias relates to the $a$ parameter and the sample size (though I think we can say something about how it should work as a function of sample size).
However, we can do the last part easily enough.
$\bar{x}$ will be unbiased for $E(X) = \frac{a}{a-1}$, but we have
$$\hat{a}=\frac{\bar{x}}{\bar{x}-1}=\frac{1}{{1-\frac{1}{\bar{x}}}}$$
Taking $Y=\bar{X}$, we can show that $\varphi(Y)=\frac{1}{{1-\frac{1}{Y}}}$ is convex.
From Jensen's inequality, we can then show that the estimator is biased and in which direction. Jensen's inequality says that for $\varphi$ convex:
$$\varphi \left(\mathbb {E} [Y]\right)\leq \mathbb {E} \left[\varphi (Y)\right]$$
(The conditions under which equality will hold don't apply here; the inequality will be strict.)
This shows that the estimator will be too high on average.
Here's some results of a simulation with $a=2$
(10000 samples each with n=100, so here we have 10000 $\hat{a}$ values. The blue line is the mean estimate from those 10000 samples.)
This simulation doesn't prove anything, but it shows that we don't contradict the derived result; looks like there was no error in concluding the bias was upward.
Such simulations allow us to see how the bias changes with $a$ -- by doing the same thing across a variety of $a$ values -- and with sample size (again, by using different $n$).
If this is not for a class exercise, I'm very curious why you wouldn't use maximum likelihood in this case:
• it's very simple - for $m=1$ it's the reciprocal of the mean of the logs; if $m$ is not 1, you subtract $log(m)$ from the mean of the logs before taking reciprocals.
• For this parameterization, it's not unbiased either, but it makes better use of the data, and it will have lower variance (and lower bias, by the look of some simulations).
• The bias is also easy to compute!
[1] Blum, M. (1970),
"On the Sums of Independently Distributed Pareto Variates"
SIAM Journal on Applied Mathematics, 19:1 (Jul.), pp. 191-198
[2] Ramsay, Colin M. (2008)
"The Distribution of Sums of I.I.D. Pareto Random Variables with Arbitrary Shape Parameter"
Communications in Statistics - Theory and Methods, 37:14, pp 2177-2184
• Thanks for the help. I have estimated both MLE in addition to MoM with software. It just that in simulation MoM seem to produce much lower tail losses and I'm trying to figure out why theoretically. If bias on shape parameter is upward, that means tails are thinner, so it makes sense along with what you wrote here. – C.J. Jackson Dec 16 '15 at 23:02
• That would be a good question to post, but briefly, if MoM is more positively biased you'll tend to get less of an upper tail on average (since larger $\hat{a}$ values will be more common). The larger variance of the MoM would be expected to complicate that picture a bit (since it will lead to both more of the larger and smaller $\hat{a}$ values, and the interplay between the way they together impact the distribution of $\hat{a}$ and the way that $\hat{a}$ impacts the tail will determine the overall effect). – Glen_b -Reinstate Monica Dec 16 '15 at 23:08
• The quantity I'm concerned with is covering some high cumulative percentile of losses. The upward bias on $\hat{a}$ would drive it down. But would the higher variance of $\hat{a}$ in this case drive up the cumulative losses? – C.J. Jackson Dec 16 '15 at 23:18
• I'd expect it to, yes, since I expect the extreme upper tail will be more impacted by a slightly lower estimate of the parameter than it will be by a slightly higher one. I can't tell for certain until I actually work it through (calculating $\frac{d^2S(x)}{da^2}$ might be sufficient) or simulate but I'd bet on yes. – Glen_b -Reinstate Monica Dec 17 '15 at 0:05 | 2020-06-06T10:13:20 | {
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http://math.stackexchange.com/questions/140290/is-there-a-way-to-simplify-this-expression-a-b-c | # Is there a way to simplify this expression $(a + b) \% c$
I am having an expression of the form $(a+b) \% c$ where a,b,c are positive integers greater than or equal to zero (natural numbers). $\%$ indicates modulo operation.
Also, there is a restriction on $a$ i.e. $0\le a < c$.
Is there a way to simplify this expression. for example, Can we say that $(a + b) \% c$ = $a + (b \% c)$ ? Is this always true and if yes, then how to prove it?
Or are there other ways to simplify it? $b$, $c$ can be any natural numbers.
Edit#1: In general then can we also say that it is true for (a + b + ...) \% c and (a.b)\%c (which is (a+a+a+....b times) \% c)?
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Not quite: what's true is that $$(a+b)\bmod c=\Big(a+(b\bmod c)\Big)\bmod c\;.$$ – Brian M. Scott May 3 '12 at 7:34
@BrianM.Scott: Thanks! I think this identity will also be helpful. But is there a way to prove it? – maths-help-seeker May 3 '12 at 7:37
It follows pretty easily from basic modular arithmetic; are you familiar with the meaning of $a\equiv b\pmod m$? – Brian M. Scott May 3 '12 at 7:42
Thank you for the pointers! I will look into it! – maths-help-seeker May 3 '12 at 7:45
@maths-help-seeker: You don't need to post the new question as comments on every answer as well as in the main question! Also, the question is not very well-posed. Can we also say that what is true? (I'm pretty sure I know what you mean, but it's good to learn to ask questions well.) – Tara B May 4 '12 at 9:51
The modulo operator $a\mapsto a\%c$ sends $\Bbb Z\to\{0,1,\cdots,c-1\}.$ (This is the domain and range.)
Now, is $\%c$ an additive function? That is, does $(a+b)\%c=a\%c+b\%c$? Notice we would have to have the latter two add to a number $<c$. Also, if $a,b$ are already in $\{0,\cdots,c-1\}$ they are un-affected and we would have $(a+b)\%=a+b$. But the sum of two numbers $<c$ is not necessarily itself less than $c$; what if $a=c-1$ and $b=1$ for instance? Thus it is not additive.
Notice that $(x\pm c)\%=x\%c$ for any $x$. That is, it is a periodic function. Moreover, if we subtract the remainder $x\%c$ from $x$ we will end up with a multiple of $c$. Symmetrically, if we subtract this multiple of $c$ from $x$ we obtain the remainder $x\%c$! So, if this "multiple of $c$" is $nc$ for $x=a$ and then $mc$ for $x=b$, by the periodicity condition we have
$$(a+b)\%c=(a-nc+b-mc)\%c=(a\%c+b\%c)\%c=(a+b\%c)\%c.$$
Remember that $a$ is already in the range of $\%c$ so $a\%c=a$. This is the best we can say at this level of generality on the variable $b$. We cannot remove the final $\%c$; remember our $b=1,a=c-1$ example. (This is a rather longwinded reasoning process that more or less mirrors the perhaps more efficient deduction possible using modular arithmetic and basic elementary number theory.)
Yes, it is true that $$(a_1+\cdots+a_n)\%c=(a_1\%c+\cdots+a_n\%c)\%c \tag{*}$$
This can be proven just as with the $n=2$ case. Use modular reduction to say that
$$a_i=m_ic+a_i\%c$$
for each $i=1,\cdots,n$. Then by periodicity we have
$$(a_1+\cdots+a_n)\%=\big((a_1-m_ic)+\cdots+(a_n-m_nc)\big)\%c=(a_1\%c+\cdots+a_n\%c)\%c \tag{\circ}.$$
Using this, we have
$$(\underbrace{a+\cdots+a}_b)\%c=\big(\underbrace{a\%c+\cdots+a\%c}_b)\%c=\big(b\cdot(a\%c)\big)\%c. \tag{\bullet}$$
By setting each $a_i=a$ and $n=b$ in the formula $(\circ)$.
Furthermore, setting $b=\ell c+b\%c$, by periodicity again we have
$$\big(b\cdot(a\%c)\big)\%c=\big(b(a\%c)-\ell(a\%c)c\big)\%c=\big((b-\ell c)(a\%c)\big)\%c=\big((b\%c)(a\%c)\big)\%c.$$
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In general then can we also say that it is true for (a + b + ...)\% c and (a.b)\%c (which is (a+a+a+....b times)%c)? – maths-help-seeker May 4 '12 at 9:37
@maths-help-seeker I've added some more to the answer. – anon May 4 '12 at 9:53
The key ideas to remember are that the argument of the $\%c$ function can be augmented by any multiple of $c$, and any $x$ differs from $x\%c$ by a multiple of $c$. – anon May 4 '12 at 9:56
$(a+b)\bmod c=n$ if and only if $(a+b)-n$ is divisible by $c$, and $0\le n<c$: $n$ is the smallest non-negative integer that has the same remainder as $a+b$ on division by $c$. Similarly $a\bmod c=a'$ if and only if $a-a'$ is divisible by $c$, and $0\le a'<c$, and $b\bmod c=b'$ if and only if $b-b'$ is divisible by $c$, and $0\le b'<c$.
I claim that $a'+b'$ and $n$ have the same remainder when divided by $c$. Since $a-a'$ is divisible by $c$, there is an integer $i$ such that $a-a'=ci$. Similarly, there are integers $j$ and $k$ such that $b-b'=cj$ and $(a+b)-n=k'$. This implies that
\begin{align*} (a'+b')-n&=(a'+b')-(a+b)+(a+b)-n\\ &=(a'-a)+(b'-b)+\Big((a+b)-n\Big)\\ &=-ci-cj+ck\\ &=c(k-i-j)\;, \end{align*}
so $(a'+b')-n$ is a multiple of $c$, and therefore $a'+b'$ and $n$ have the same remainder on division by $c$. They might not be equal, however, because $a'+b'$ might be bigger than $n$: we know that $0\le n<c$, but all we know about $a'+b'$ is that $0\le a'+b'\le 2c-2$, since $0\le a'\le c-1$ and $0\le b'\le b-1$. Thus, $a'+b'$ is either $n$, or $n+c$, but we can't be sure which. If we let $m=(a'+b')\bmod c$, however, then $m$ and $a'+b'$ have the same remainder on division by $c$, so $m$ and $n$ have the same remainder on division by $c$. Moreover, $0\le m<c$ and $0\le n<c$, so $m$ and $n$ must be the same number. We've now shown that in general $$(a+b)\bmod c=\Big((a\bmod c)+(b\bmod c)\Big)\bmod c\;.\tag{1}$$ In your case $0\le a<c$, so $a\bmod c=a$, and $(1)$ reduces to $$(a+b)\bmod c=\Big((a+(b\bmod c)\Big)\bmod c;.$$
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:In general then can we also say that it is true for (a + b + ...)\% c and (a.b)\%c (which is (a+a+a+....b times)%c)? – maths-help-seeker May 4 '12 at 9:37 | 2016-05-24T20:11:32 | {
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https://math.stackexchange.com/questions/1050308/find-matrix-p-so-b-pa | Find Matrix $P$ so $B=PA$
I have this problem.
$$A = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 2 & 3 \end{array}\right)$$
$$B = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$
Find matrix $P$ invertible so $B=PA$.
It's pretty clear in this case.
But I wonder is there a way to find such a matrix in general case, when it's not obvious.
Any help will be appreciated.
• Notice that the second and third row are linear combination of the other. Thus, there exists $E_2E_1$ (of elementary matrices) such that $A$ can be row reduced to $B$, then $E_2E_1$ is the matrix $P$ you need. In other words, start row reducing. – IAmNoOne Dec 3 '14 at 19:12
• @Nameless Ok, in this case I found $E_2E_1$ and then I use the $E_2E_1$ on identity matrix and this is matrix $P$? – JaVaPG Dec 3 '14 at 19:26
• What do you mean...? Any matrix times the identity is the same matrix... – IAmNoOne Dec 3 '14 at 19:27
• @Nameless I'm not so sure I got what you meant, lets take this case as an example, we have two elementary matrices $E_1E_2$ in order for $A \implies B$, I don't understand what I should do with these two elementary matices – JaVaPG Dec 3 '14 at 19:32
• See Omnomnomnom's answer on how to put the matrices together – IAmNoOne Dec 3 '14 at 19:33
By row reducing, you can find matrices $P_1,P_2$ such that $P_1A$ and $P_2B$ are in the same row-echelon form. From there, we can say that $$P_1A = P_2B \implies B = (P_2^{-1}P_1)A$$ This will work whenever such a $P$ exists.
For your example: note that $B$ is in row-echelon form. So, in this case, we have $P_2 = I$. For $A$, we have $$\pmatrix{ 1&2&3&&1&0&0\\ 2&4&6&&0&1&0\\ 1&2&3&&0&0&1} \mathop{\leadsto}^{R_3\to R_3-R_1}\cdots \mathop{\leadsto}^{R_2\to R_2-2R_1}\\ \pmatrix{ 1&2&3&&1&0&0\\ 0&0&0&&-2&1&0\\ 0&0&0&&-1&0&1}$$ So, $$P_1 = \pmatrix{ 1&0&0\\ -2&1&0\\ -1&0&1 }$$
• I don't quite understand can you show the process on the example please? – JaVaPG Dec 3 '14 at 19:35
• @JaVaPG, as I said before row reduce and record your reduction in the matrix $P_i$. – IAmNoOne Dec 3 '14 at 19:39
• @JaVaPG See my edit. – Omnomnomnom Dec 3 '14 at 19:46
Let's say you wanted to find such $P$ that $$B = AP\text{.}$$ Let $p_1$, $p_2$ and $p_3$ be the columns of the matrix $P$, i. e. $P = [p_1\quad p_2 \quad p_3]$. Do the same for $B$. The upper linear system is equivalent to the following systems: $$Ap_1 = b_1,\quad Ap_2 = b_2\quad\text{and}\quad Ap_3 = b_3\text{.}$$
You should be able to solve those three. From $p_j$'s you can reconstruct $P$.
Since you have to solve $B = PA$, you can transpose that relation ($B^T = A^T P^T$) and reduce this problem to the previous one. Solve the system $B^T = A^T Q$. You may find multiple (or no) solutions $Q$ if $A$ (hance $A^T$) is not invertible.
If you want an invertible $P$, find an invertible solution $Q_i$ among $Q$'s and define $P = Q_i^T$.
when you row reducing a matrix $A$ by hand, it is much easier to keep track of the row operations, if we augment the matrix $A$ with a column like this: $\left( \begin{array}{lll|l} 1 & 2 & 3 & a\cr 2 & 4 & 6 & b\cr 1 & 2 & 3 & c\end{array} \right) \to \left( \begin{array}{lll|l} 1 & 2 & 3 & a\cr 0 & 0 & 0 & -2a + b\cr 0 & 0 & 0 & -a+c\end{array} \right)$ you decode this as multiplying by the matrix $E$ on the left where $E \pmatrix{a \cr b\cr c}=\pmatrix{1 & 0 & 0\cr-2 & 1 & 0\cr -1 & 0 & 1}\pmatrix{a\cr b\cr c\cr d} = \pmatrix{a &\cr-2a+b\cr -a+c}.$
that is $EA = U.$ of course, if you are calculator or computer matrix software, you augment by the identity matrix as suggested by others. | 2020-02-18T01:03:50 | {
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https://math.stackexchange.com/questions/2523198/fallacious-proof-involving-trigonometry | # Fallacious proof involving trigonometry
Which step in the following incorrect proof is fallacious? Is it something with the use of indefinite integrals or with the domain and range of trignometric functions? I encountered this fallacious proof here.
$$\int\tan(x)\,dx=\int\tan(x)\,dx$$
substitute $\tan(x)$ :
$$\int\tan(x)\,dx=\int\sin(x)\sec(x)\,dx$$ Integrate by parts, assume
$$u=\sec(x),dv=\sin(x)\,dx$$
Therefore, $$\int\tan(x)\,dx=-\sec(x)\cos(x)+\int\cos(x)\tan(x)\sec(x)\,dx$$ but $\cos(x)\sec(x)=1$ so:
$$\int\tan(x)\,dx=-1+\int\tan(x)\,dx$$ we subtract both sides by $\int\tan(x)\,dx$ :
$$\int\tan(x)\,dx-\int\tan(x)\,dx =-1+\int\tan(x)\,dx-\int\tan(x)\,dx$$
then:
$0=-1$
Thanks in advance for any help.
• When you integrate - the step by parts - you need to allow for an arbitrary constant, because as an indefinite integral the value is not uniquely defined. – Mark Bennet Nov 16 '17 at 16:16
• For another example of this phenomenon, try evaluating $\int\frac1x\ \text dx$ by parts, letting $u=\frac1x$ and $\text dv = \text dx$. – Théophile Nov 16 '17 at 16:40
• PLUS A CONSTANT!!! ALWAYS PLUS A CONSTANT! – fleablood Nov 16 '17 at 17:33
These are indefinite integrals; you need to add $C$. It's true that $0+C_1 = -1+C_2$ for some constants $C_1$ and $C_2$.
(And if they were definite integrals, the $0$ and $-1$ would wash out when we subtracted the upper bound from the lower.)
More specifically, I would say that the next-to-last line, $$\int \tan x\,\text{d}x = -1 + \int \tan x\,\text{d}x$$ is still true. But when you're just canceling the integral of $\tan x$ from both sides, you're forgetting that the integrals are only defined up to a constant, and that constant absorbs the $-1$. What the line above is really saying is that there's some function $F(x)$ whose derivative is $\tan x$ for which $$F(x) + C_1 = -1 + (F(x)+ C_2)$$ which is perfectly true for some constants $C_1$ and $C_2$.
Alternatively, we can think of the problem as simplifying $$\int \tan x\,\text{d}x - \int \tan x\,\text{d}x = \int 0\,\text{d}x$$ to $0$: the $+C$ doesn't disappear just because we're integrating $0$.
These two integrals are actually a good example of why we need to add $C$ for an indefinite integral: two antiderivatives of a function are not guaranteed to be equal, only to differ by an arbitrary constant.
• Which specific step in the above proof then is fallacious? – Adnan Ali Nov 16 '17 at 15:54
• In the very last line - I've elaborated on the mistake in my answer. – Misha Lavrov Nov 16 '17 at 16:20
• actually, I'd say that IMO the last line with integrals is also true, its only the final result that's wrong - it's the difference between the integrals that has to account for the C, so yes, $\int\tan(x)\,dx-\int\tan(x)\,dx =-1+\int\tan(x)\,dx-\int\tan(x)\,dx$ is completely true, and no, it's doesn't mean $0 = -1$, because $\int\tan(x)\,dx-\int\tan(x)\,dx = C$ and thus we arrive at completely valid $C_1 = C_2 - 1$ – user81774 Nov 17 '17 at 2:31
• Right; the fallacy only appears when the integrals go away. – Misha Lavrov Nov 17 '17 at 3:13
As an adjunct to Misha Lavrov's answer, this is probably not the first time you have encountered this phenomenon.
In trigomometry, one might be faced with $\cos(\theta) = \frac{\sqrt{2}}{2}$. The solutions to this equation are, of course, $\theta = \pm \cos^{-1}\left( \frac{\sqrt{2}}{2} \right) + 2 \pi k = \pm \frac{\pi}{4} + 2 \pi k$, for any integer $k$. Some people normalize these so that all the particular angles are positive, so $\theta = \frac{\pi}{4} + 2 \pi k$ or $\theta = \frac{7\pi}{4} + 2 \pi k$, for any integer $k$.
Well this means \begin{align*} \left\{\frac{-\pi}{4} + 2 \pi k, k \in \mathbb{Z}\right\} &= \left\{\frac{7\pi}{4} + 2 \pi k, k \in \mathbb{Z} \right\} & &\text{True.} \\ \frac{-\pi}{4} &= \frac{7\pi}{4} & &\text{False.} \\ \end{align*}
Just because two sets are equal, which is what you have at $$\int \tan x \,\mathrm{d}x = -1 + \int \tan x \,\mathrm{d}x \text{,}$$ does not mean that two randomly selected elements of those sets are equal.
Remember the mantra: "... plus a constant". $\int x^2 = \frac {x^3}{3}$ .... PLUS A CONSTANT.
So in your proof:
"Therefore $\int\tan(x)dx=-\sec(x)\cos(x)+ C + \int\cos(x)\tan(x)\sec(x)dx$ "
"but $\cos(x)\sec(x)=1$ so:"
"$\int\tan(x)dx=-1 + C +\int\tan(x)dx$"
"We subtract both sides by $\int\tan(x)dx$"
"$\int\tan(x)dx-\int\tan(x)dx = -1 + C +\int\tan(x)dx - \int\tan(x)dx$
"Then:"
"$0 = -1 +C$"
$C = 1$
And therefore we have proven that .... $1$ is a constant.
• But, Can't we ignore the constant of integration when integrating by parts? Please see here. – Adnan Ali Nov 16 '17 at 19:38
• @AdnanAli - No you can never ignore the constant. What is happening in the linked post is not "ignoring the constant". You only need one arbitrary constant per side, since a second one can be summed into the first to give a single arbitrary constant covering them both. Since there is still an indefinite integral on the right-hand side, there is still a constant of integration. – Paul Sinclair Nov 17 '17 at 3:49
The fallacious step is the one where you forgot that $\int \tan(x)\,dx$ is an equivalence class or family of functions.
In other words, the problem $f(x) = \int \tan(x)\,dx$ has a solution set consisting of infinitely many functions that might be named $f$ in that equation. Every function in the solution set has the form $$f(x) = \log(\cos (x)) + C$$ where $f$ is a function over the interval $(-\pi/2,\pi/2)$ and $C$ is some constant.
The difference $A - B$ when $A$ and $B$ both are equivalence classes consists of all possible values $a - b$ where $a \in A$ and $b \in B.$ Therefore, $$\int \tan(x)\,dx - \int \tan(x)\,dx = \{f: (-\pi/2,\pi/2)\to\mathbb R\mid (\exists c\in\mathbb R)(\forall x\in\mathbb R) f(x) = c\},$$ that is, the difference of the two integrals is the set of all real-valued constant functions over the same domain as $\int \tan(x)\,dx,$ because for any real number there is some member of $\int \tan(x)\,dx$ that you can subtract from another member of $\int \tan(x)\,dx$ to get the desired number as your difference.
Now it is certainly true that there are constant functions $C_1$ and $C_2$ such that $$C_1 = -1 + C_2,$$ but it is not true that both of those functions will always be the zero function, and you cannot conclude that $0 = -1.$
If all this talk of equivalence classes is too much, you can just remember to add the constant of integration everywhere you use an indefinite integral. Also make sure that if you have two integrals, each one gets its own constant of integration, even if they are integrals of the same integrand. You can replace the sum or difference of two constants of integration by a new constant, but you cannot assume their difference is zero.
This is a common mistake: two antiderivatives differ by a constant (assuming we're working with functions defined over an interval. Use a fixed bound, that is, a uniquely determined antiderivative: \begin{align} \int_0^x \tan t\,dt &=\int_0^x \sin t\frac{1}{\cos t}\,dt \\[6px] &=\Bigl[-\cos t\frac{1}{\cos t}\Bigr]_0^x- \int_0^x(-\cos t)\left(-\frac{1}{\cos^2t}(-\sin t)\right)\,dt \\[6px] &=[1-1]+\int_0^x\tan t\,dt \end{align} No contradiction (and no progress either).
The already-given answers are good, but I think it's illuminating to point out that the "paradox" vanishes if we try to follow the same logical chain it without the antiderivatives. For the integration by parts step, we use the identity $u' v = (uv)' - u v'$. We then have \begin{align*} \tan x &= \tan x \\ &= \sin x \sec x \\ &= - \frac{d}{dx}(\cos x) \sec x \\ &= - \frac{d}{dx} (\cos x \sec x) + \cos x \frac{d}{dx} (\sec x) \\ &= - \frac{d}{dx} (\cos x \sec x) + \cos x \sec x \tan x \\ &= - \frac{d}{dx} (1) + \tan x \end{align*} As you can see, this equation is correct because the derivative of $1$ is zero. If we then try to take the antiderivative of each side, we have $$\int \tan x \, dx= -\int \frac{d}{dx} (1) \, dx + \int \tan x \, dx$$ which is only a problem if you insist that $$\int \frac{d}{dx} (1) \, dx = 1.$$
As to why this last move isn't valid, I'll refer you to the fact (pointed out by multiple answerers) that antiderivatives aren't "really" functions; they're equivalence classes of functions, where two functions are equivalent if they differ by a constant. | 2020-01-22T08:55:45 | {
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https://mathhelpboards.com/threads/find-the-values-of-the-parameter-q.7647/ | # Find the values of the parameter "q"
#### anemone
##### MHB POTW Director
Staff member
Hi MHB,
Problem:
Find the sum of all positive real values of the parameter $q$ for which the equation $(x^2+4x+4+136)^2=16q(5x^2+4x+136)$ has exactly three distinct real solutions.
$q=9$ and $q=34$.
I am aware the question is actually asking us to find all possible values of $q$ when the given quartic equation has one repeated real root, and two other distinct real roots.
Attempt:
The original equation $(x^2+4x+4q+136)^2=16q(5x^2+4x+136)$ can be rewritten as
$(x^2+4x+136+4q)^2=16q(4x^2+x^2+4x+136)$
If I let $k=x^2+4x+136$, the equation above becomes
$(k+4q)^2=16q(4x^2+k)$
Expanding both sides of the equation and then rearranging them as a quadratic equation in terms of $k$, we have
$k^2+8qk+16q^2=64qx^2+16qk$
$k^2-8qk-64qx^2+16q^2=0$
$(k-4q)^2-16q^2-64qx^2+16q^2=0$
$(k-4q)^2-64qx^2=0$
$(k-4q-8x\sqrt{q})(k-4q+8x\sqrt{q})=0$
Replacing the expression for $k$ back into the equation above, we get
$(x^2+4x+136-4q-8x\sqrt{q})(x^2+4x+136-4q+8x\sqrt{q})=0$
And in order to satisfy the requirement set by the problem and to find all possible values of $q$, we can do so by equating the discriminants of the quadratic factors, one after another and we should expect two $q$ values to be found.
Hence, by equating the discriminant for the first factor $x^2+4x+136-4q-8x\sqrt{q}=0$ to zero, we obtain:
$(4-8\sqrt{q})^2-4(136-4q)=0$
$\sqrt{q}=3$ or $\sqrt{q}=-2.2$ and since $\sqrt{q}>0$, we get $q=9$ as one of the solutions.
Now, if we set the discriminant of the second factor as zero, we get:
$(4+8\sqrt{q})^2-4(136-4q)=0$
$\sqrt{q}=-3$ or $\sqrt{q}=2.2$ and since $\sqrt{q}>0$, we get $q=4.84$ as the other solution.
I checked the answer when $q=2.2$ and we would only get all 4 complex roots if $q=2.2$ but with $q=9$, everything is fine as is.
My question is, I don't know how to find the answer where $q=34$...any ideas?
By the way, I know there must be other ways to approach this problem, so if you don't wish to go through my silly attempt, I welcome you to post another method to solve the problem and I will appreciate whatever help that you are going to offer me.
#### Opalg
##### MHB Oldtimer
Staff member
Problem:
Find the sum of all positive real values of the parameter $q$ for which the equation $(x^2+4x+4+136)^2=16q(5x^2+4x+136)$ has exactly three distinct real solutions.
$q=9$ and $q=34$.
$\vdots$
... we get
$(x^2+4x+136-4q-8x\sqrt{q})(x^2+4x+136-4q+8x\sqrt{q})=0$
And in order to satisfy the requirement set by the problem and to find all possible values of $q$, we can do so by equating the discriminants of the quadratic factors, one after another and we should expect two $q$ values to be found.
Hence, by equating the discriminant for the first factor $x^2+4x+136-4q-8x\sqrt{q}=0$ to zero, we obtain:
$(4-8\sqrt{q})^2-4(136-4q)=0$
$\sqrt{q}=3$ or $\sqrt{q}=-2.2$ and since $\sqrt{q}>0$, we get $q=9$ as one of the solutions.
Now, if we set the discriminant of the second factor as zero, we get:
$(4+8\sqrt{q})^2-4(136-4q)=0$
$\sqrt{q}=-3$ or $\sqrt{q}=2.2$ and since $\sqrt{q}>0$, we get $q=4.84$ as the other solution.
I checked the answer when $q=2.2$ and we would only get all 4 complex roots if $q=2.2$ but with $q=9$, everything is fine as is.
My question is, I don't know how to find the answer where $q=34$...any ideas?
I had to spend over four hours in trains today, going to and from a meeting in London, and I spent all that time struggling with this problem. I got close to a solution, but your approach is far neater than mine.
The missing solution comes this way: In the equation $(x^2+4x+136-4q-8x\sqrt{q})(x^2+4x+136-4q+8x\sqrt{q})=0,$ you looked at the case where the repeated root comes in the first of the two factors, and when it comes in the second factor, but you overlooked the possibility that it might come once in each factor. That can only happen when the two factors are equal, and that in turn can only happen when $8x\sqrt{q}=0.$ That implies $x=0$, and the two factors then both reduce to $136 - 4q=0$, or $q=34.$
Staff member
Hi Opalg, | 2021-07-30T23:44:40 | {
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https://www.physicsforums.com/threads/heat-equation-boundary-conditions.787519/ | # Heat Equation Boundary Conditions
1. Dec 14, 2014
### Vector1962
1. The problem statement, all variables and given/known data
Let a slab $0 \le x \le c$ be subject to surface heat transfer, according to Newtons's law of cooling, at its faces $x = 0$ and $x = c$, the furface conductance H being the same on each face. Show that if the medium $x\le0$ has temperature zero and medium $x=c$ has the constant temperature T then the boundary value problem for steady-state temperatures in the slab is $$u''(x)=0$$ $$Ku'(0)=Hu(0)$$ $$Ku'(c)=H[T-u(c)]$$ where K is the thermal conductivity of the material in the slab, write $h=\frac{H}{K}$ and derive the expression $$u(x)=\frac{T}{ch+2}(hx+1)$$
2. Relevant equations
$U_t=K\nabla^2U$
3. The attempt at a solution.
I have the first part of the question complete. I'm not sure how to apply the boundary conditions? I've solved similar problems with boundary's $u(0)=0$ and $u(c) = T$ and I've solved various problems with heat flux boundary's. For some reason, applying the Newton boundary conditions is messing me up.
I can get to $u(x)=c_1x+c_2$ comfortably which leads to $$u(x)=u_x(x)x+u(0)$$ $$u(c)=u_x(c)c+u(0)$$ but from there I'm lost.
2. Dec 14, 2014
### geoffrey159
1. Find that $u'(x)$ is constant. Find it's expression.
2. Deduce that the general expession for $u$ is $u(x) = u(0) + x u'(0)$. Improve that with 1.
3. From the text plus points 1. and 2. above, you should be able to find two expressions for $u'(c)$. Find it and deduce $u(0)$
4. Conclude
3. Dec 14, 2014
### Vector1962
$$u''(x)=0$$ $$\int u''(x)\,dx = u'(x)+c_1$$ $$u'(x)=c_1$$
$$\int u'(x)\,dx = \int c_1\,dx$$ $$u(x)= c_1x + c_2$$
at $x=0$ yields $u(0)=c_2$
$$u(x)= x u'(x) + u(0)$$ $u(x) = xu'(0) + u(0)$ because $u'(x)=u'(0) = u'(c)=c_1$
from the boundary values ; $u'(0)=\frac {H} {K} u(0)$ and $u'(c)=\frac{H}{K} [T-u(c)]$
I don't see it from here?
$u(0)=T-u(c)$ from the boundary conditions?
4. Dec 14, 2014
### geoffrey159
Recall the fundamental theorem of calculus which says that $u(b) - u(a) = \int_a^b u'(x) dx$ if $u'$ is continuous.
From that theorem you should be able to put subscripts and superscripts on your integrals and get to the result easily.
5. Dec 14, 2014
### geoffrey159
For point 1. : You have for whatever $0 \le x \le c$ that $u'(x) - u'(0) = \int_0^x u''(s) ds = 0$ so $u'(x) = u'(0)$ for any of these $x$. Which proves that $u'$ it is constant.
For point 2. :
$u(x) - u(0) = \int_0^x u'(s) ds = x u'(0) = x h u(0)$ because $u'$ is constant and because you have a constraint from your text.
Now you have $u(x) = u(0) (1+xh )$
Can you finish it?
Last edited: Dec 14, 2014
6. Dec 14, 2014
### Staff: Mentor
You were off to a great start when you wrote $u(x)=c_1x+c_2$. Now substitute u=u(0) at x = 0 and u = u(c) at x = c into the equation to determine the two constants of integration c1 and c2 in terms of u(0), u(c), and c. Next evaluate u'(0)=u'(c) in terms of u(0) and u(c). Substitute this into the two boundary condition equations. This will give you two equations that allow you to solve for u(0) and u(c) in terms of K, H, and T.
Chet
7. Dec 14, 2014
### Vector1962
I can follow everything you did to get to $u(x)=u(0)(hx+1)$ but no clue how to finish
I've tried this route at least 20 times and always end up in algebra hell "dancing" around the solution but never getting there. After which, I thought I'd post it to the forum and see if I could get a hint to bring it to closure. Can't quite seem to get there.
8. Dec 14, 2014
### Vector1962
Apparently, I'm applying the boundary conditions the right way, but failing miserably in the algebra. As I understand everything discussed this far, it appears $u'(x)=\frac{T}{c}$ correct?
9. Dec 14, 2014
### Staff: Mentor
No.
10. Dec 14, 2014
### Staff: Mentor
$$u(x)=u(0)+\frac{u(c)-u(0)}{c}x$$
$$u'(x)=u'(0)=u'(c)=\frac{u(c)-u(0)}{c}$$
$$K\frac{u(c)-u(0)}{c}=Hu(0)$$
$$K\frac{u(c)-u(0)}{c}=H(T-u(c))$$
Solve the previous two equations for u(0) and u(c)
Chet
11. Dec 14, 2014
### Vector1962
$u(0)=\frac{u(c)}{1+hc}$ and $u(c)=\frac{hcT+u(0)}{1+hc}$
insert these into $$u(x)=u(0)+\frac{u(c)-u(0)}{c}x$$ ?
12. Dec 15, 2014
### geoffrey159
You have two ways of expressing $u'(c)$.
First one is using the fact $u'$ is constant, so $u'(c) = u'(0) = h u(0)$.
Second expression of $u'(c)$ is in your text, it's a constraint, and says $u'(c) = h(T-u(c))$.
Now you have a general expression for $u$, which we wrote previously, so $u(c)= u(0) (1 + hc)$
Equating the two expressions of $u'(c)$ will give you $u(0)$ and final answer.
13. Dec 15, 2014
### Staff: Mentor
Do you know how to solve two simultaneous linear algebraic equations in two unknowns? We learned this in 1st year algebra in 8th grade.
Chet
Last edited: Dec 15, 2014
14. Dec 15, 2014
### Vector1962
Some people don't have the benefit of a formal high school or college education. I just get a math book every once in a while and work some problems. Math is not that hard. I get stuck every once in a while and need a "pointer" to get me out of the rut. Thanks so much for the help. Following both of your leads I arrive at:
$$u(x)=\frac{T}{ch+2}(hx+1)$$
15. Dec 15, 2014
### Staff: Mentor
Sorry. I wasn't aware. You seemed to have some good math skills beyond HS, and I was surprised at your difficulty with algebra.
Chet | 2017-10-21T13:22:10 | {
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https://mathhelpboards.com/threads/initial-value-problem-for-exact-equations.7293/ | # initial value problem for exact equations
#### find_the_fun
##### Active member
I've got a few small questions I'd like to straighten out. I'm really trying to establish a firm procedure involving the steps I write down because I find it helps me learn the math and avoid errors.
Solve the initial value problem: $$\displaystyle (x+y)^2 dx +(2xy+x^2-1)dy = 0$$ with $$\displaystyle y(1)=1$$
So let $$\displaystyle M(x, y) =(x+y)^2$$
Question 1: I write M(x, y) because M is a function of x and y. I that correct or is M only a function of one variable?
Question 2: should dx be part of the equation for M(x, y) or no?
let $$\displaystyle N(x, y) = 2xy+x^2-1$$
$$\displaystyle \frac{\partial M(x, y)}{\partial y}= \frac{\partial}{\partial y} (x^2+y^2+2xy)=2y+2x$$
$$\displaystyle \frac{\partial N(x, y)}{\partial x}=2y+2x$$ therefore equation is exact
$$\displaystyle \frac{ \partial f(x, y)}{\partial x} = M(x, y) = x^2+y^2+2xy$$ implies $$\displaystyle f(x, y)= \int x^2+y^2+2xy dx = \frac{x^3}{3}+y^2x+x^2y+g(y)$$
$$\displaystyle \frac{ \partial f(x, y)}{ \partial y} N(x, y) = 2xy+x^2-1 = 2yx+y+g'(y)$$ solving for $$\displaystyle g'(y)=x^2-y-1$$
$$\displaystyle g(y)= \int x^2-y-1 dy = x^2y-\frac{y^2}{2}-y$$
Question 3: following the examples in my textbook I noticed we don't have a $$\displaystyle C$$ for the constant of integration. Why does it get omitted?
so $$\displaystyle f(x, y) = \frac{x^3}{3}+y^2x+x^2y+x^2y-\frac{y^2}{2}-y=\frac{x^3}{3}+y^2x+2x^2y-\frac{y^2}{2}-y=C$$
Now plugging in initial values
$$\displaystyle \frac{1^3}{3} + 1 +2 - \frac{1^2}{2} -1 = C$$
Question 4: a little embarrassing but does the exponent apply to the whole fraction or just the numerator? For example is $$\displaystyle \frac{1^3}{3}=\frac{1}{3}$$ or $$\displaystyle \frac{1^3}{3} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}$$? Obviously in LaTex I'm writing it as the 3 only applies to the 1 but given the context that the derivative of $$\displaystyle x^n=(n-1)x^{n-1}$$ which is it?
Question 5: in an example my textbook takes the equation an multiplies away the denominators. Doesn't this mess up the initial values and finding C? For example
$$\displaystyle \frac{1^3}{3} + 1 +2 - \frac{1^2}{2} -1 = C$$
multiply both sides of the equation by 6
$$\displaystyle 2 \cdot 1^3 +6+12-3\cdot 1^2-6=6C_1=C$$
Last edited:
#### Prove It
##### Well-known member
MHB Math Helper
Where did you get \displaystyle \begin{align*} \tan{(x)} - \sin{(x)}\sin{(y)} \end{align*} from?
#### find_the_fun
##### Active member
Where did you get \displaystyle \begin{align*} \tan{(x)} - \sin{(x)}\sin{(y)} \end{align*} from?
I transcribed the wrong work from the paper to this site. | 2020-11-27T17:35:34 | {
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http://mathhelpforum.com/algebra/282632-matrices-problem.html | 1. ## matrices problem
Question:
The transpose of a matrix $\displaystyle M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is the matrix $\displaystyle M^{T}=\begin{pmatrix} a & c \\ b & d \end{pmatrix}$ and M is said to be orthogonal when $\displaystyle M^{T}M=I$, where I is the unit matrix. Given that the matrix
$\displaystyle N=\begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix}$
is orthogonal, find the value of k. Describe geometrically the transformation of x-y plane which is represented by N. Under a transformation S of the real plane into a itself, a point $\displaystyle P = (x,y)$ is mapped onto the point $\displaystyle S(P) = (ax+by, cx+dy)$. Show that when M is orthogonal, the distance between any two points P and Q is the same as the distance between their images S(P) and S(Q).
My attempt:
$\displaystyle N=\begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix}$
$\displaystyle N^{T}=\begin{pmatrix} \frac{2}{\sqrt{5}} &-\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & k \end{pmatrix}$
$\displaystyle N^{T}N=I$
$\displaystyle \begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix}\begin{pmatrix} \frac{2}{\sqrt{5}} &-\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & k \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$
$\displaystyle \begin{pmatrix} 5 &\frac{2}{5}- \frac{k}{\sqrt{5}} \\ \frac{2}{5}- \frac{k}{\sqrt{5}} & k^{2} - \frac{1}{\sqrt{5}} \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
$\displaystyle \frac{2}{5}- \frac{k}{\sqrt{5}} =0$
$\displaystyle k = \frac{2\sqrt{5}}{5}$
The transformation represented by N is a rotation by angle 26.6 (degrees) clockwise
I am having difficulty answering the last part of the question
Please I would like some help with the last part of the question
$\begin{pmatrix}a &b \\ c & d\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} =\begin{pmatrix}ax+by\\ cx+dy \end{pmatrix}$
2. ## Re: matrices problem
If the absolute value of the determinant of the transformation is 1 then distances are preserved.
$M^T M = I$
$|M^T M|= |M^T||M| = |M|^2 = |I| = 1$
$|M|=\pm 1$
$|~|M|~|=1$ | 2019-07-17T20:27:29 | {
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https://arbital.greaterwrong.com/p/group_homomorphism?l=47t | # Group homomorphism
A group homomorphism is a function between groups which “respects the group structure”.
# Definition
Formally, given two groups $$(G, +)$$ and $$(H, *)$$ (which hereafter we will abbreviate as $$G$$ and $$H$$ respectively), a group homomorphism from $$G$$ to $$H$$ is a function $$f$$ from the underlying set $$G$$ to the underlying set $$H$$, such that $$f(a) \* f(b) = f(a+b)$$ for all $$a, b \in G$$.
# Examples
• For any group $$G$$, there is a group homomorphism $$1_G: G \to G$$, given by $$1_G(g) = g$$ for all $$g \in G$$. This homomorphism is always bijective.
• For any group $$G$$, there is a (unique) group homomorphism into the group $$\{ e \}$$ with one element and the only possible group operation $$e \* e = e$$. This homomorphism is given by $$g \mapsto e$$ for all $$g \in G$$. This homomorphism is usually not injective: it is injective if and only if $$G$$ is the group with one element. (Uniqueness is guaranteed because there is only one function, let alone group homomorphism, from any set $$X$$ to a set with one element.)
• For any group $$G$$, there is a (unique) group homomorphism from the group with one element into $$G$$, given by $$e \mapsto e_G$$, the identity of $$G$$. This homomorphism is usually not surjective: it is surjective if and only if $$G$$ is the group with one element. (Uniqueness is guaranteed this time by the property proved below that the identity gets mapped to the identity.)
• For any group $$(G, +)$$, there is a bijective group homomorphism to another group $$G^{\mathrm{op}}$$ given by taking inverses: $$g \mapsto g^{-1}$$. The group $$G^{\mathrm{op}}$$ is defined to have underlying set equal to that of $$G$$, and group operation $$g +_{\mathrm{op}} h := h + g$$.
• For any pair of groups $$G, H$$, there is a homomorphism between $$G$$ and $$H$$ given by $$g \mapsto e_H$$.
• There is only one homomorphism between the group $$C_2 = \{ e_{C_2}, g \}$$ with two elements and the group $$C_3 = \{e_{C_3}, h, h^2 \}$$ with three elements; it is given by $$e_{C_2} \mapsto e_{C_3}, g \mapsto e_{C_3}$$. For example, the function $$f: C_2 \to C_3$$ given by $$e_{C_2} \mapsto e_{C_3}, g \mapsto h$$ is not a group homomorphism, because if it were, then $$e_{C_3} = f(e_{C_2}) = f(gg) = f(g) f(g) = h h = h^2$$, which is not true. (We have used that the identity gets mapped to the identity.)
# Properties
• The identity gets mapped to the identity. (Proof.)
• The inverse of the image is the image of the inverse. (Proof.)
• The image of a group under a homomorphism is another group. (Proof.)
• The composition of two homomorphisms is a homomorphism. (Proof.)
Children:
Parents:
• Group
The algebraic structure that captures symmetry, relationships between transformations, and part of what multiplication and addition have in common.
• I have a question about general Arbital practice here. A mathematician will probably already know what a group homomorphism is, but they probably also don’t need the proofs of the Properties, for instance, and they don’t need the explanation of the trivial group. Should I have split this up into different lenses in some way?
• so8res: “I would set up the page as follows:
A group homomorphism is X. Key properties of group homomorphisms include:
1. Thing. Implications implications implications. (Proof.)
2. Thing. Implications implications. (Proof.)
I’d then eventually add an intro lens.”
Having proofs on child pages makes sense to me too. | 2020-07-04T18:15:49 | {
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https://math.stackexchange.com/questions/46527/coordinates-of-point-on-a-line-defined-by-two-other-points-with-a-known-distance | # Coordinates of point on a line defined by two other points with a known distance from one of them
I have two points in 3D space; let's call them $A=(a_x, a_y, a_z)$ and $B=(b_x, b_y, b_z).$
Now, I need to place a third point, let's call it $C=(c_x, c_y, c_z)$, which lies on the line between $A$ and $B$ and is some known distance from $A$.
Is there some easy way I can get the $c_x, c_y, c_z$ coordinates knowing the distance from $C$ to $A$?
I would easily be able to do this in 2D, but I never worked with 3D space and I can't seem to figure it out.
Thanks.
It's quite similar to 2D space. Consider a vector $v = B-A$ which you can imagine is the direction from $A$ to $B$. Hence for any point $C$ between $A$ and $B$ (inclusive of $A$ and $B$).
$$C = A + tv$$
where $t = 0$ implies $C = A$ and $t = 1$ implies $C = B$ and $t \in (0,1)$ are all the points in between (one of which is the desired $C$). As you can see, this representation is independent of the dimension of your space.
So, what's the value of $t$? Well, let the known distance from $A$ to $C$ be $d_{AC}$. Now, the distance between $A$ and $B$ or $d_{AB}$ is the magnitude of $v$ or $|v|$ which is nothing but
$$d_{AB} = |v| = \sqrt{(a_x-b_x)^2+(a_y-b_y)^2+(a_z-b_z)^2}$$
(You can see that this formula for the Euclidean distance between two points is similar in 2D as well)
Therefore, $t = \large \frac{d_{AC}}{d_{AB}}$ and substituting $t$ and $v$ in the previous formula for $C$, we have:
$$C = A + \frac{d_{AC}}{\sqrt{(a_x-b_x)^2+(a_y-b_y)^2+(a_z-b_z)^2}}(B-A)$$
• @Jacob: I suspect that the question intended $x$, $y$, and $z$ to be subscripts. – André Nicolas Jun 20 '11 at 18:50
• @Jacob: I hope you don't mind; I added the subscripts, centered your formulas...feel free to edit or roll-back to suit your taste. – amWhy Jun 20 '11 at 19:11
• @amWhy: Thanks for changing the OP as well! – Jacob Jun 20 '11 at 19:18
• @Jacob: Not a problem - I realized you were trying to "speak in the same language" as the post, for consistency, so I adjusted both... – amWhy Jun 20 '11 at 19:33
• @meh: if you found the answer provided by Jacob helpful, there are two things you should consider: (1) you can "up vote" it (click on the upward arrow to the left of the answer, (2) and as the "asker" of the question, if an answer (or one of a number of answers) is helpful in answering your question, you're the only one who can click on the check mark to the left of the question to accept it. (Just some things to know about this site.) – amWhy Jun 20 '11 at 21:32 | 2020-02-29T11:06:14 | {
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https://mathoverflow.net/questions/323711/concentration-of-the-load-of-the-maximally-loaded-bin-m-balls-n-bins-with | # Concentration of the load of the maximally loaded bin ($m$ balls $n$ bins) with nonuniform bin probabilities
There is a common argument used when investigating the concentration of the maximally loaded bin (say $$X$$ is the maximum load) when $$m$$ balls are thrown into $$n$$ bins under the uniform distribution. I give the argument for $$m=n,$$ showing that $$X$$ is approximately $$\ln n/\ln \ln n$$ with high probability. Using the union bound, and letting $$X_i$$ be the number of balls in the $$i^{th}$$ bin $$\mathbb{P}(X_i=k)=\binom{n}{k}\left(\frac{1}{n}\right)^k \left(1-\frac{1}{n}\right)^{n-k}\leq \binom{n}{k} \left(\frac{1}{n}\right)^k\leq \left(\frac{ne}{k}\right)^k\left(\frac{1}{n}\right)^k= \left(\frac{e}{k}\right)^k$$ yielding $$\mathbb{P}(X_i\geq k)\leq \sum_{j=k}^n \left(\frac{e}{j}\right)^j \leq \left(\frac{e}{k}\right)^k \left(1+\frac{e}{k}+\frac{e^2}{k^2}+\cdots\right).$$ Now let $$k^{\ast}=\lceil e \ln n/\ln\ln n\rceil,$$ giving $$\mathbb{P}(X_i\geq k)\leq \left(\frac{e}{k^{\ast}}\right)^{k^{\ast}} \left[\frac{1}{1-e/k^{\ast}}\right]\leq n^{-2},$$ and using the union bound, since there are $$n$$ bins $$\mathbb{P}\left(\bigcup_{i=1}^n X_i\geq k\right)\leq \frac{1}{n},\quad (1)$$ giving the concentration. What if we now have $$p=(p_1,\ldots,p_n)$$ with $$p_i$$ the probability of each ball falling into bin $$i$$, in an independent manner.
As far as I can tell (sort the bins so $$p_1\geq p_2\geq \cdots\geq p_1>0$$) as long as the maximum probability obeys $$p_1\leq \frac{\ln n}{n}$$, a version of this argument works.
What about distributions with larger $$p_1$$? What can we say? Say we allow the quantity on the RHS of (1) to be $$\frac{1}{\sqrt{n}}$$, for example.
I am most interested in $$m=n,$$ or slightly larger $$m$$ say $$m=n (\log n)^a.$$
I suppose for $$p_1$$ large enough wrt the other probabilities its load will highly likely be the maximum. So a kind of convex combination argument is needed...
Edit: As far as lower bounds for the uniform case, it can be addressed in a number of ways, including Lemma 5.12 from Mitzenmacher and Upfal's book Probability and Computing which shows that the maximum load is at least $$\ln n/\ln \ln n$$ with probability at least $$1-(1/n)$$ for $$n$$ large.
Remark: This related question here was unanswered
## 1 Answer
If you only want upper bounds, there is a nice approach based on collisions. (Proving that the max-loaded bin is not too small w.h.p. seems to need completely different techniques.)
Define a $$k$$-way collision to be a case of $$k$$ different balls that land in the same bin. For example, if they all land in the same bin, there are $${m \choose k}$$ total $$k$$-way collisions.
Let $$C_k$$ be a random variable for the number of $$k$$-way collisions. The probability that a fixed set of $$k$$ balls collides is $$\sum_{i=1}^n p_i^k = \|p\|_k^k$$. Therefore $$\mathbb{E} C_k = {m \choose k} \|p\|_k^k$$.
Now to upper-bound the chance that the max-loaded bin has $$\geq k$$ balls, we can use Markov's inequality:
\begin{align*} \Pr[\text{exists a \geq k loaded bin}] &= \Pr[C_k \geq 1] \\ &\leq \mathbb{E} C_k \\ &= {m \choose k} \|p\|_k^k \\ &\leq \left(\frac{m e}{k}\right)^k \|p\|_k^k \end{align*} The amazing thing is that this tends to give quite tight/strong asymptotics (I think of it as $$C_k$$ is already "Chernoff-ized" in a sense, e.g. $$k$$ is in the exponent here). I have a blog post about this but don't know of another reference.
With an upper-bound on the heaviest bin, the worst case is $$1/p_1$$ bins of probability $$p_1$$, so $$\|p\|_k^k \leq p_1^{k-1}$$. \begin{align*} &\leq \frac{1}{p_1} \left(\frac{m p_1 e}{k}\right)^k \\ &= \exp\left(k \log \left(m p_1 e \right) - k \log(k) + \log\left(\frac{1}{p_1}\right) \right) \end{align*} For example, to recover the well-known result, if you plug in $$m = n$$ and $$p_1 = \frac{1}{n}$$, you get $$\exp\left(-(k-1)\log(k) + \log(n)\right)$$ which is $$O(1)$$ for $$k \approx \frac{\ln(n)}{\ln \ln(n)}$$ and exponentially decreasing in $$k$$ thereafter.
• Thanks. Amazingly I had just found your blogpost and glanced over it. When I came back to mathoverflow, I saw your answer. Whats the likelihood of that collision? Feb 21, 2019 at 20:44 | 2022-08-09T23:10:22 | {
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https://au.mathworks.com/help/matlab/math/systems-of-linear-equations.html | ## Systems of Linear Equations
### Computational Considerations
One of the most important problems in technical computing is the solution of systems of simultaneous linear equations.
In matrix notation, the general problem takes the following form: Given two matrices A and b, does there exist a unique matrix x, so that Axb or xAb?
It is instructive to consider a 1-by-1 example. For example, does the equation
7x = 21
have a unique solution?
The answer, of course, is yes. The equation has the unique solution x = 3. The solution is easily obtained by division:
x = 21/7 = 3.
The solution is not ordinarily obtained by computing the inverse of 7, that is 7–1= 0.142857..., and then multiplying 7–1 by 21. This would be more work and, if 7–1 is represented to a finite number of digits, less accurate. Similar considerations apply to sets of linear equations with more than one unknown; MATLAB® solves such equations without computing the inverse of the matrix.
Although it is not standard mathematical notation, MATLAB uses the division terminology familiar in the scalar case to describe the solution of a general system of simultaneous equations. The two division symbols, slash, /, and backslash, \, correspond to the two MATLAB functions `mrdivide` and `mldivide`. These operators are used for the two situations where the unknown matrix appears on the left or right of the coefficient matrix:
`x = b/A` Denotes the solution to the matrix equation xA = b, obtained using `mrdivide`. `x = A\b` Denotes the solution to the matrix equation Ax = b, obtained using `mldivide`.
Think of “dividing” both sides of the equation Ax = b or xA = b by A. The coefficient matrix `A` is always in the “denominator.”
The dimension compatibility conditions for `x = A\b` require the two matrices `A` and `b` to have the same number of rows. The solution `x` then has the same number of columns as `b` and its row dimension is equal to the column dimension of `A`. For `x = b/A`, the roles of rows and columns are interchanged.
In practice, linear equations of the form Ax = b occur more frequently than those of the form xA = b. Consequently, the backslash is used far more frequently than the slash. The remainder of this section concentrates on the backslash operator; the corresponding properties of the slash operator can be inferred from the identity:
`(b/A)' = (A'\b').`
The coefficient matrix `A` need not be square. If `A` has size m-by-n, then there are three cases:
m = n Square system. Seek an exact solution. m > n Overdetermined system, with more equations than unknowns. Find a least-squares solution. m < n Underdetermined system, with fewer equations than unknowns. Find a basic solution with at most m nonzero components.
#### The mldivide Algorithm
The `mldivide` operator employs different solvers to handle different kinds of coefficient matrices. The various cases are diagnosed automatically by examining the coefficient matrix. For more information, see the “Algorithms” section of the `mldivide` reference page.
### General Solution
The general solution to a system of linear equations Axb describes all possible solutions. You can find the general solution by:
1. Solving the corresponding homogeneous system Ax = 0. Do this using the `null` command, by typing `null(A)`. This returns a basis for the solution space to Ax = 0. Any solution is a linear combination of basis vectors.
2. Finding a particular solution to the nonhomogeneous system Ax =b.
You can then write any solution to Axb as the sum of the particular solution to Ax =b, from step 2, plus a linear combination of the basis vectors from step 1.
The rest of this section describes how to use MATLAB to find a particular solution to Ax =b, as in step 2.
### Square Systems
The most common situation involves a square coefficient matrix `A` and a single right-hand side column vector `b`.
#### Nonsingular Coefficient Matrix
If the matrix `A` is nonsingular, then the solution, `x = A\b`, is the same size as `b`. For example:
```A = pascal(3); u = [3; 1; 4]; x = A\u x = 10 -12 5```
It can be confirmed that `A*x` is exactly equal to `u`.
If `A` and `b` are square and the same size, `x= A\b` is also that size:
```b = magic(3); X = A\b X = 19 -3 -1 -17 4 13 6 0 -6```
It can be confirmed that `A*x` is exactly equal to `b`.
Both of these examples have exact, integer solutions. This is because the coefficient matrix was chosen to be `pascal(3)`, which is a full rank matrix (nonsingular).
#### Singular Coefficient Matrix
A square matrix A is singular if it does not have linearly independent columns. If A is singular, the solution to Ax = b either does not exist, or is not unique. The backslash operator, `A\b`, issues a warning if `A` is nearly singular or if it detects exact singularity.
If A is singular and Ax = b has a solution, you can find a particular solution that is not unique, by typing
`P = pinv(A)*b`
`pinv(A)` is a pseudoinverse of A. If Ax = b does not have an exact solution, then `pinv(A)` returns a least-squares solution.
For example:
```A = [ 1 3 7 -1 4 4 1 10 18 ]```
is singular, as you can verify by typing
```rank(A) ans = 2```
Since A is not full rank, it has some singular values equal to zero.
Exact Solutions. For `b =[5;2;12]`, the equation Ax = b has an exact solution, given by
```pinv(A)*b ans = 0.3850 -0.1103 0.7066```
Verify that `pinv(A)*b` is an exact solution by typing
```A*pinv(A)*b ans = 5.0000 2.0000 12.0000```
Least-Squares Solutions. However, if ```b = [3;6;0]```, Ax = b does not have an exact solution. In this case, `pinv(A)*b` returns a least-squares solution. If you type
```A*pinv(A)*b ans = -1.0000 4.0000 2.0000```
you do not get back the original vector `b`.
You can determine whether Ax =b has an exact solution by finding the row reduced echelon form of the augmented matrix ```[A b]```. To do so for this example, enter
```rref([A b]) ans = 1.0000 0 2.2857 0 0 1.0000 1.5714 0 0 0 0 1.0000```
Since the bottom row contains all zeros except for the last entry, the equation does not have a solution. In this case, `pinv(A)` returns a least-squares solution.
### Overdetermined Systems
This example shows how overdetermined systems are often encountered in various kinds of curve fitting to experimental data.
A quantity `y` is measured at several different values of time `t` to produce the following observations. You can enter the data and view it in a table with the following statements.
```t = [0 .3 .8 1.1 1.6 2.3]'; y = [.82 .72 .63 .60 .55 .50]'; B = table(t,y)```
```B=6×2 table t y ___ ____ 0 0.82 0.3 0.72 0.8 0.63 1.1 0.6 1.6 0.55 2.3 0.5 ```
Try modeling the data with a decaying exponential function
$y\left(t\right)={c}_{1}+{c}_{2}{e}^{-t}$.
The preceding equation says that the vector `y` should be approximated by a linear combination of two other vectors. One is a constant vector containing all ones and the other is the vector with components `exp(-t)`. The unknown coefficients, ${c}_{1}$ and ${c}_{2}$, can be computed by doing a least-squares fit, which minimizes the sum of the squares of the deviations of the data from the model. There are six equations in two unknowns, represented by a 6-by-2 matrix.
`E = [ones(size(t)) exp(-t)]`
```E = 6×2 1.0000 1.0000 1.0000 0.7408 1.0000 0.4493 1.0000 0.3329 1.0000 0.2019 1.0000 0.1003 ```
Use the backslash operator to get the least-squares solution.
`c = E\y`
```c = 2×1 0.4760 0.3413 ```
In other words, the least-squares fit to the data is
`$y\left(t\right)=0.4760+0.3413{e}^{-t}.$`
The following statements evaluate the model at regularly spaced increments in `t`, and then plot the result together with the original data:
```T = (0:0.1:2.5)'; Y = [ones(size(T)) exp(-T)]*c; plot(T,Y,'-',t,y,'o')```
`E*c` is not exactly equal to `y`, but the difference might well be less than measurement errors in the original data.
A rectangular matrix `A` is rank deficient if it does not have linearly independent columns. If `A` is rank deficient, then the least-squares solution to `AX = B` is not unique. `A\B` issues a warning if `A` is rank deficient and produces a least-squares solution. You can use `lsqminnorm` to find the solution `X` that has the minimum norm among all solutions.
### Underdetermined Systems
This example shows how the solution to underdetermined systems is not unique. Underdetermined linear systems involve more unknowns than equations. The matrix left division operation in MATLAB finds a basic least-squares solution, which has at most `m` nonzero components for an `m`-by-`n` coefficient matrix.
Here is a small, random example:
```R = [6 8 7 3; 3 5 4 1] rng(0); b = randi(8,2,1)```
```R = 6 8 7 3 3 5 4 1 b = 7 8 ```
The linear system `Rp = b` involves two equations in four unknowns. Since the coefficient matrix contains small integers, it is appropriate to use the `format` command to display the solution in rational format. The particular solution is obtained with
```format rat p = R\b ```
```p = 0 17/7 0 -29/7 ```
One of the nonzero components is `p(2)` because `R(:,2)` is the column of `R` with largest norm. The other nonzero component is `p(4)` because `R(:,4)` dominates after `R(:,2)` is eliminated.
The complete general solution to the underdetermined system can be characterized by adding `p` to an arbitrary linear combination of the null space vectors, which can be found using the `null` function with an option requesting a rational basis.
```Z = null(R,'r') ```
```Z = -1/2 -7/6 -1/2 1/2 1 0 0 1 ```
It can be confirmed that `R*Z` is zero and that the residual `R*x - b` is small for any vector `x`, where
`x = p + Z*q`
Since the columns of `Z` are the null space vectors, the product `Z*q` is a linear combination of those vectors:
`$Zq=\left(\begin{array}{cc}{\stackrel{⇀}{x}}_{1}& {\stackrel{⇀}{x}}_{2}\end{array}\right)\left(\begin{array}{c}u\\ w\end{array}\right)=u{\stackrel{⇀}{x}}_{1}+w{\stackrel{⇀}{x}}_{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}.$`
To illustrate, choose an arbitrary `q` and construct `x`.
```q = [-2; 1]; x = p + Z*q; ```
Calculate the norm of the residual.
```format short norm(R*x - b)```
```ans = 2.6645e-15```
When infinitely many solutions are available, the solution with minimum norm is of particular interest. You can use `lsqminnorm` to compute the minimum-norm least-squares solution. This solution has the smallest possible value for `norm(p)`.
`p = lsqminnorm(R,b)`
```p = -207/137 365/137 79/137 -424/137 ```
### Solving for Several Right-Hand Sides
Some problems are concerned with solving linear systems that have the same coefficient matrix `A`, but different right-hand sides `b`. When the different values of `b` are available at the same time, you can construct `b` as a matrix with several columns and solve all of the systems of equations at the same time using a single backslash command: `X = A\[b1 b2 b3 …]`.
However, sometimes the different values of `b` are not all available at the same time, which means you need to solve several systems of equations consecutively. When you solve one of these systems of equations using slash (/) or backslash (\), the operator factorizes the coefficient matrix `A` and uses this matrix decomposition to compute the solution. However, each subsequent time you solve a similar system of equations with a different `b`, the operator computes the same decomposition of `A`, which is a redundant computation.
The solution to this problem is to precompute the decomposition of `A`, and then reuse the factors to solve for the different values of `b`. In practice, however, precomputing the decomposition in this manner can be difficult since you need to know which decomposition to compute (LU, LDL, Cholesky, and so on) as well as how to multiply the factors to solve the problem. For example, with LU decomposition you need to solve two linear systems to solve the original system Ax = b:
```[L,U] = lu(A); x = U \ (L \ b);```
Instead, the recommended method for solving linear systems with several consecutive right-hand sides is to use `decomposition` objects. These objects enable you to leverage the performance benefits of precomputing the matrix decomposition, but they do not require knowledge of how to use the matrix factors. You can replace the previous LU decomposition with:
```dA = decomposition(A,'lu'); x = dA\b;```
If you are unsure which decomposition to use, `decomposition(A)` chooses the correct type based on the properties of `A`, similar to what backslash does.
Here is a simple test of the possible performance benefits of this approach. The test solves the same sparse linear system 100 times using both backslash (\) and `decomposition`.
```n = 1e3; A = sprand(n,n,0.2) + speye(n); b = ones(n,1); % Backslash solution tic for k = 1:100 x = A\b; end toc```
`Elapsed time is 9.006156 seconds.`
```% decomposition solution tic dA = decomposition(A); for k = 1:100 x = dA\b; end toc```
`Elapsed time is 0.374347 seconds.`
For this problem, the `decomposition` solution is much faster than using backslash alone, yet the syntax remains simple.
### Iterative Methods
If the coefficient matrix A is large and sparse, factorization methods are generally not efficient. Iterative methods generate a series of approximate solutions. MATLAB provides several iterative methods to handle large, sparse input matrices.
FunctionDescription
`pcg`
Preconditioned conjugate gradients method. This method is appropriate for Hermitian positive definite coefficient matrix A.
`bicg`
`bicgstab`
`bicgstabl`
BiCGStab(l) Method
`cgs`
`gmres`
Generalized Minimum Residual Method
`lsqr`
LSQR Method
`minres`
Minimum Residual Method. This method is appropriate for Hermitian coefficient matrix A.
`qmr`
Quasi-Minimal Residual Method
`symmlq`
Symmetric LQ Method
`tfqmr`
Transpose-Free QMR Method
MATLAB supports multithreaded computation for a number of linear algebra and element-wise numerical functions. These functions automatically execute on multiple threads. For a function or expression to execute faster on multiple CPUs, a number of conditions must be true:
1. The function performs operations that easily partition into sections that execute concurrently. These sections must be able to execute with little communication between processes. They should require few sequential operations.
2. The data size is large enough so that any advantages of concurrent execution outweigh the time required to partition the data and manage separate execution threads. For example, most functions speed up only when the array contains several thousand elements or more.
3. The operation is not memory-bound; processing time is not dominated by memory access time. As a general rule, complicated functions speed up more than simple functions.
`inv`, `lscov`, `linsolve`, and `mldivide` show significant increase in speed on large double-precision arrays (on order of 10,000 elements or more) when multithreading is enabled. | 2022-09-25T17:42:15 | {
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https://meangreenmath.com/2014/07/05/calculators-and-complex-numbers-part-15/ | # Calculators and complex numbers (Part 17)
In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.
Definition. If $z$ is a complex number, then we define
$e^z = \displaystyle \sum_{n=0}^{\infty} \frac{z^n}{n!}$
Even though this isn’t the usual way of defining the exponential function for real numbers, the good news is that one Law of Exponents remains true. (At we saw in an earlier post in this series, we can’t always assume that the usual Laws of Exponents will remain true when we permit the use of complex numbers.)
Theorem. If $z$ and $w$ are complex numbers, then $e^z e^w = e^{z+w}$.
In yesterday’s post, I gave the idea behind the proof… group terms where the sums of the exponents of $z$ and $w$ are the same. Today, I will formally prove the theorem.
The proof of the theorem relies on a principle that doesn’t seem to be taught very often anymore… rearranging the terms of a double sum. In this case, the double sum is
$e^z e^w = \displaystyle \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{z^n}{n!} \frac{w^k}{k!}$
This can be visualized in the picture below, where the $x-$axis represents the values of $k$ and the $y-$axis represents the values of $n$. Each red dot symbolizes a term in the above double sum. For a fixed value of $n$, the values of $k$ vary from $0$ to $\infty$. In other words, we start with $n =0$ and add all the terms on the line $n = 0$ (i.e., the $x-$axis in the picture). Then we go up to $n = 1$ and then add all the terms on the next horizontal line. And so on.
I will rearrange the terms as follows: Let $j = n+k$. Then for a fixed value of $j$, the values of $k$ will vary from $0$ to $j$. This is perhaps best described in the picture below. The value of $j$, the sum of the coordinates, is constant along the diagonal lines below. The value of $k$ then changes while moving along a diagonal line.
Even though this is a different way of adding the terms, we clearly see that all of the red circles will be hit regardless of which technique is used for adding the terms.
In this way, the double sum $\displaystyle \sum_{n=0}^\infty \sum_{k=0}^\infty$ gets replaced by $\displaystyle \sum_{j=0}^\infty \sum_{k=0}^j$. Since $n = j-k$, we have
$e^z e^w = \displaystyle \sum_{j=0}^{\infty} \sum_{k=0}^{j} \frac{z^{j-k}}{(j-k)!} \frac{w^k}{k!}$
We now add a couple of $j!$ terms to this expression for reasons that will become clear shortly:
$e^z e^w = \displaystyle \sum_{j=0}^{\infty} \sum_{k=0}^{j} \frac{j!}{j!} \frac{1}{k! (j-k)!} w^k z^{j-k}$
Since $j!$ does not contain any $k$s, it can be pulled outside of the inner sum on $k$. We do this for the $j!$ in the denominator:
$= \displaystyle \sum_{j=0}^\infty \frac{1}{j!} \sum_{k=0}^j \frac{j!}{k!(j-k)!} w^k z^{j-k}$
We recognize that $\displaystyle \frac{j!}{k! (j-k)!}$ is a binomial coefficent:
$= \displaystyle \sum_{j=0}^\infty \frac{1}{j!} \sum_{k=0}^j {j \choose k} w^k z^{j-k}$
The inner sum is recognized as the formula for a binomial expansion:
$= \displaystyle \sum_{j=0}^\infty \frac{1}{j!} (w+z)^j$
Finally, we recognize this as the definition of $e^{w+z}$, using the dummy variable $j$ instead of $n$. This proves that $e^z e^w = e^{z+w}$ even if $z$ and $w$ are complex.
Without a doubt, this theorem was a lot of work. The good news is that, with this result, it will no longer be necessary to explicitly use the summation definition of $e^z$ to actually compute $e^z$, as we’ll see tomorrow.
For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.
## 2 thoughts on “Calculators and complex numbers (Part 17)”
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2021-12-01T05:57:45 | {
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https://mathhelpboards.com/threads/evaluate-trigonometric-expression.4510/ | # Evaluate Trigonometric Expression.
#### anemone
##### MHB POTW Director
Staff member
Evaluate $$\displaystyle \cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)$$.
#### chisigma
##### Well-known member
Evaluate $$\displaystyle \cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)$$.
It exists the general formula...
$\displaystyle \prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n-k+1}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$ (1)
... and setting in (1) $\displaystyle k=0,\ n=5,\ x=\frac{\pi}{65}$ Your product becomes...
$\displaystyle P = \frac{1}{64}\ \frac{\sin \frac{64\ \pi}{65}}{\sin \frac{\pi}{65}} = \frac{1}{64}$ (2)
Probably a more simple way to arrive to the result exists... how to demonstrate (1) is an extra problem...
Kind regards
$\chi$ $\sigma$
Last edited:
#### anemone
##### MHB POTW Director
Staff member
It exists the general formula...
$\displaystyle \prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n-k}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$ (1)
... and setting in (1) $\displaystyle k=0,\ n=5,\ x=\frac{\pi}{65}$ Your product becomes...
$\displaystyle P = \frac{1}{32}\ \frac{\sin \frac{64\ \pi}{65}}{\sin \frac{\pi}{65}} = \frac{1}{32}$ (2)
Probably a more simple way to arrive to the result exists... how to demonstrate (1) is an extra problem...
Kind regards
$\chi$ $\sigma$
Hi chisigma, thanks for participating in this problem and your answer is of course correct and on the level, I didn't realize there was such a formula exists and that we could just apply it to this particular problem and get its answer so easily...
#### chisigma
##### Well-known member
It exists the general formula...
$\displaystyle \prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n-k+1}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$ (1)
... and setting in (1) $\displaystyle k=0,\ n=5,\ x=\frac{\pi}{65}$ Your product becomes...
$\displaystyle P = \frac{1}{64}\ \frac{\sin \frac{64\ \pi}{65}}{\sin \frac{\pi}{65}} = \frac{1}{64}$ (2)
Probably a more simple way to arrive to the result exists... how to demonstrate (1) is an extra problem...
The demonstration of the formula...
$\displaystyle \prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n-k+1}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$ (1)
... is 'easy' and, if I remember correctly, this result was known in the Middle Age...
Let's start from the well known formula...
$\displaystyle \sin 2x = 2\ \sin x \cos x$ (2)
Setting in (2) 4x instead of 2x we obtain...
$\displaystyle \sin 4 x = 2\ \sin 2 x\ \cos 2 x = 4\ \sin x\ \cos x\ \cos 2 x$ (3)
Proceeding in the same way we arrive to...
$\displaystyle \sin (2^{n+1} x) = 2^{n+1}\ \sin x\ \prod_{k=0}^{n} \cos (2^{k}\ x)$ (4)
Kind regards
$\chi$ $\sigma$
Last edited:
#### Albert
##### Well-known member
Evaluate $$\displaystyle \cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)$$.
Let:
$$\displaystyle cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=k$$
we have :
$$\displaystyle 32\times2\times sin \left(\frac{\pi}{65}\right)\cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right)k$$
$$\displaystyle 16\times2\times sin \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$$\displaystyle 8\times2\times sin \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$$\displaystyle 4\times2\times sin \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$$\displaystyle 2\times 2sin \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$$\displaystyle 2\times sin \left(\frac{32\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$$\displaystyle sin \left(\frac{64\pi}{65}\right)=64sin \left(\frac{\pi}{65}\right) k$$
$\therefore k=\dfrac {1}{64}$
Am I wrong ? Why my answer is different from yours ?
#### Sudharaka
##### Well-known member
MHB Math Helper
It exists the general formula...
$\displaystyle \prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n-k}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$ (1)
Hi chisigma,
I think this should be,
$\prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n\color{red}{+1}-k}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$
#### anemone
##### MHB POTW Director
Staff member
Let:
$$\displaystyle cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=k$$
$\therefore k=\dfrac {1}{64}$
Am I wrong ? Why my answer is different from yours ?
I am terribly sorry for misleading the readers who have read this thread for saying the answer $$\displaystyle cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)=\frac{1}{32}$$ is correct...
I didn't check my answer but I should be able to tell right away (from my approach) why this wasn't correct because the answer depends wholly on the number of terms that the cosine terms exist.
The correct answer for this trigonometric expression is $$\displaystyle \frac{1}{64}$$, as stated by Albert.
My solution:
Let $$\displaystyle P=cos \left(\frac{\pi}{65}\right)\cdot\cos \left(\frac{2\pi}{65}\right)\cdot\cos \left(\frac{4\pi}{65}\right)\cdot\cos \left(\frac{8\pi}{65}\right)\cdot\cos \left(\frac{16\pi}{65}\right)\cdot\cos \left(\frac{32\pi}{65}\right)$$
and
$$\displaystyle 2^6Q=2^6(\sin \left(\frac{\pi}{65}\right)\cdot\sin \left(\frac{2\pi}{65}\right)\cdot\sin \left(\frac{4\pi}{65}\right)\cdot\sin \left(\frac{8\pi}{65}\right)\cdot\sin \left(\frac{16\pi}{65}\right)\cdot\sin \left(\frac{32\pi}{65}\right))$$
Multiplying P and $$\displaystyle 2^6Q$$ together we obtain
$$\displaystyle 2^6PQ=Q$$
$$\displaystyle P=\frac{1}{2^6}=\frac{1}{64}$$
and this approach is essentially the same as what Albert did in his solution...
I am truly sorry for saying the answer was correct without checking it and I promise I won't make this kind of mistake again and will practice the right forum manners on this site in the future.
Sorry...
#### chisigma
##### Well-known member
Hi chisigma,
I think this should be,
$\prod_{j=k}^{n} \cos (2^{j}\ x) = \frac{1}{2^{n\color{red}{+1}-k}}\ \frac {\sin 2^{n+1}\ x}{\sin 2^{k}\ x}$
All right Sudharaka!... I have corrected my past post!... thank You very much!...
Kind regards
$\chi$ $\sigma$ | 2020-09-22T04:25:28 | {
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https://math.stackexchange.com/questions/1617259/probability-of-success-on-third-and-fourth-trials | # Probability of success on third and fourth trials?
I have a pretty basic probability question, but I'm just having difficulties remembering what distribution this is.
A coin is tossed until a head appears two times in a row. Given that we are using a fair coin, what is the probability that we toss the coin exactly 4 times such that the two consecutive heads are the 3rd and 4th trials?
I know that this would be fairly easy to solve by brute force and listing out the different probabilities, but I'm a little confused as to how to use a "shortcut" such as identifying this as a binomial distribution. I also don't know how to account for the probability that we are using 4 tosses.
Any help is appreciated. Thanks!
• TTHH, for which the probability is $\dfrac{1}{2^4}$
• HTHH, for which the probability is $\dfrac{1}{2^4}$
Hence the overall probability is $\dfrac{1}{2^4}+\dfrac{1}{2^4}=\dfrac{1}{2^3}$
• @Taylor: You asked "exactly $4$ times", didn't you??? – barak manos Jan 19 '16 at 6:08 | 2020-01-28T10:35:07 | {
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https://math.stackexchange.com/questions/3224885/how-to-prove-cosxy-sinx-y-2-sinx-frac-pi4-sin-frac-pi4-y | # How to prove $\cos(x+y)+\sin(x-y)=2 \sin(x+\frac{\pi}{4}) \sin(\frac{\pi}{4}-y)$
I spend some time trying to figure out how to prove the following identity:
$$\cos(x+y)+\sin(x-y)=2 \sin\left(x+\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}-y\right)$$
I tried to use the following identities:
$$\cos(x+y)=\cos(x) \sin(y) - \sin(x) \sin(y)$$
and
$$\sin(x-y)=\sin(x) \cos(y) - \cos(x) \sin(y).$$
After that, I wanted to use
$$\cos(x) = \sin\left(x+ \frac{\pi}{2}\right).$$
Unfortunately I can't reach the correct identity. Is there another way of doing it ? Thank you.
• Have you tried using $sin(x - y)$ decomposition on the rhs? Just to provide a potential starting point. – Nicg May 13 at 20:07
• Use that $\sin(x-y) = \cos(\frac{\pi}{2}-(x-y))$ – Vasya May 13 at 20:09
Use $$\cos p-\cos q=-2\sin\frac{p+q}{2}\sin\frac{p-q}{2}$$ and $$\sin\alpha=\cos\Bigl(\frac{\pi}{2}-\alpha\Bigr)$$ Alternatively, expand both sides using the addition formulas.
Note that $$\sin(x+y)+\sin(x-y)=2\sin x \cos y$$ i.e. $$\sin(A)+\sin(B)=2\sin \frac{A+B}{2} \cos \frac{A-B}{2}$$
Then you can say \begin{align} \cos(x+y)+\sin(x-y) &= \sin(x+y+\frac{\pi}{2})+\sin(x-y)\\ &= 2 \sin(x+\frac{\pi}{4})\cos(y+\frac{\pi}{4})\\ &= 2 \sin(x+\frac{\pi}{4})\sin(\frac{\pi}{4}-y) \end{align}
let $$u = x + \frac {\pi}{4}\\ v = \frac {\pi}{4} - y$$
$$u+v = \frac {\pi}{2} + x - y\\ u-v = x + y$$
$$\cos(u-v) + \sin(u+v - \frac{\pi}{2}) = 2\sin u\sin v\\ \sin(u+v - \frac{\pi}{2}) = - \sin \left(\frac{\pi}{2} - (u + v)\right) = -\cos (u+v)\\ \cos(u-v) - \cos(u+v) = 2\sin u\sin v\\ \cos u\cos v- \sin u\sin v -\cos u\cos v+\sin u \sin v = 2\sin u\sin v$$
The other solutions have chosen http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$$2\sin A\sin B=\cos(A-B)-\cos(A+B)$$
Here $$A=x+\dfrac\pi4,B=\dfrac\pi4-y$$ | 2019-08-25T14:21:17 | {
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http://workingmotors.it/qeas/surface-of-revolution-parametric-equations.html | # Surface Of Revolution Parametric Equations
If Martin finds the volume of the solid formed by the outer curve and subtracts the. Finding the equations of tangent and normal to the curves and plotting them. An example of such a surface is the sphere, which may be considered as the surface generated when a semicircle is revolved about its diameter. When describing surfaces with parametric equations, we need to use two variables. 8, where the arc length of the teardrop is calculated. 6: A graph of the parametric equations in Example 10. Surfaces of revolution. (An overall minimum surface area of 82. 7 Consider the. Convert Surface of Revolution to Parametric Equations. (b) Eliminate the parameters to show that the surface is an elliptic paraboloid and set up another double integral for the surface area. We initiated the process with a simpler spur gear, then advanced to the straight bevel gear and finally defined the governing parametric equations for a spiral bevel gear. Bonus: A relation between power series and differential equations. First Order Differential Equations Separating the Variables. Such a surface is called hyperbolic pseudo–spherical. Now here the parametric equations of the curve are. Earlier, you were asked about how Martin can model the volume of a particular vase. Suppose that $$y\left( x \right),$$ $$y\left( t \right),$$ and $$y\left( \theta \right)$$ are smooth non-negative functions on the given interval. x = f(t) and y = g(t) for a ≤ t ≤ b, the surface area of revolution for the curve revolving around the y-axis is defined as. Ask Question Asked 6 years ago. Since I'll have to enter the parametrized equation in notation MATLAB understands for vectors and matrices, I'll first give the vectorize command (which tells me where to put the periods). Objective: Converting a system of parametric equations to rectangular form * Review for Section 11. 6: A graph of the parametric equations in Example 10. In the first problem, "A) the Torus obtained by a rotation of a circle x= a + b*sin(u), y= 0, z = b*sin(u) " you are already given a parameter u. Since the surface is in the form x = f ( y, z) x = f ( y, z) we can quickly write down a set of parametric equations as follows, x = 5 y 2 + 2 z 2 − 10 y = y z = z x = 5 y 2 + 2 z 2 − 10 y = y z = z. Solving Quadratic Equations. , ISBN-10: 0-32187-896-5, ISBN-13: 978-0-32187-896-0, Publisher: Pearson. Surface Area: The surface area of a solid of revolution where a curve x= x(t), y= y(t) with arc length dsis rotated about an axis is S= Z 2ˇrds = Z 2ˇr s dx dt 2 + dy dt 2 dt If revolving about the x-axis, r= y(t) and if revolving about the y-axis, r= x(t). Main result are stated in Theorem (8. (An overall minimum surface area of 82. Parametric equations can be used to describe motion that is not a function. 055 and [[beta]. Section 3-5 : Surface Area with Parametric Equations. Parametrics Parametric Curves Parametric Surfaces Parametric Derivative Slope & Tangent Lines Area Arc Length Surface Area Volume SV Calculus Limits Derivatives Integrals Infinite Series Parametric Equations Conics Polar Coordinates Laplace Transforms. Write a set of parametric equations for the surface of revolution obtained by revolving the graph of the function about the given axis. The parametric equation for a circle of radius 1 in the xy-plane is (x(u), y(u)) = (cosu, sinu) where 0 < u < 2pi. So it's analogous to this 2 here. Examples of surfaces of revolution include the apple surface, cone (excluding the base), conical frustum (excluding the ends), cylinder (excluding the ends), Darwin-de Sitter spheroid, Gabriel's horn, hyperboloid, lemon surface, oblate. 3 Parametric Equations and Calculus Find the slope of a tangent line to a curve defined by parametric equations; find the arc length along a curve defined parametrically; find the area of a surface of revolution in parametric form. this project is to assist CNC Software, inc. Idea: rotate a 2D profile curvearound an axis. 1 2 3 4 x 0. Answer to: Find the area of the surface generated by revolving the curve about the x-axis. The parametric equation for a circle of radius 1 in the xy-plane is (x(u), y(u)) = (cosu, sinu) where 0 < u < 2pi. Convert Surface of Revolution to Parametric Equations. Surface Area Length of a Plane Curve A plane curve is a curve that lies in a two-dimensional plane. Surfaces that occur in two of the main theorems of vector calculus, Stokes' theorem and the divergence theorem, are frequently given in a parametric form. rotates through a complete revolution about the. 3 The Christoffel Symbols for a Surface of Revolution 45. x = cos3θ y = sin3θ 0 ≤ θ ≤ π 2 x = cos 3 θ y = sin 3 θ 0 ≤ θ ≤ π 2. We will also see how the parameterization of a surface can be used to find a normal vector for the surface (which will be very useful in a couple of sections) and how the parameterization can be used to find the surface area of a surface. " Find the arc length of the teardrop. ; (c) Use the parametric equations in part (a) with and to graph the surface. Parametric Curve: Surface Area of Revolution; Surface Area of Revolution of a Parametric Curve Rotated About the y-axis; Parametric Arc Length; Parametric Arc Length and the distance Traveled by the Particle; Volume of Revolution of a Parametric Curve; Converting Polar Coordinates; Converting Rectangular Equations to Polar Equations. 3a Additional examples and applications of Taylor and Maclaurin series. 6 Polar coordinates and applications. 1 The Parametric Representation of a Surface of. parametric equations used to describe a surface of revolution are simple and easy to manipulate. For these problems, you divide the surface into narrow circular bands, figure the surface area of a representative band, and then just add up the areas of all the bands to get the total surface area. of the surface with parametric equations ,, , ,. The Circle. If a surface is obtained by rotating about the x-axis from t=a to b the curve of the parametric equation {(x=x(t)),(y=y(t)):}, then its surface area A can be found by A=2pi int_a^by(t)sqrt{x'(t)+y'(t)}dt If the same curve is rotated about the y-axis, then A=2pi int_a^b x(t)sqrt{x'(t)+y'(t)}dt I hope that this was helpful. 17 and 16 depict the minimal axes of revolution and minimum surfaces of revolution for the values m = -1, m = 0, and m =1. Calculus 2 advanced tutor. E F Graph 3D Mode. this project is to assist CNC Software, inc. This video lecture " Surface Area Of Solid Generated By Revolution about axes in Hindi " will help Engineering and Basic Science students to understand following topic of of Engineering-Mathematics: 1. When I just look at that, unless you deal with parametric equations, or maybe polar coordinates a lot, it's not obvious that this is the parametric equation for an ellipse. The implicit equation of a sphere is: x 2 + y 2 + z 2 = R 2. The surface of revolution given by rotating the region bounded by y = x3 for 0 ≤ x ≤ 2 about the x-axis. So just like that, by eliminating the parameter t, we got this equation in a form that we immediately were able to recognize as ellipse. 5 Applications to Probability. Get the free "Area of a Surface of Revolution" widget for your website, blog, Wordpress, Blogger, or iGoogle. The second curve, , will model the inner wall of the vase. Maths Geometry Graph plot surface This demo allows you to enter a mathematical expression in terms of x and y. Open parts of the bulb (left) and the neck (right) segments of the periodic surface of revolution obtained via parametric equations (17) and (21) with ε = 1. The Straight Line. Thus, the new surface winds around twice as fast as the original surface, and since the equation for is identical in both surfaces, we observe twice as many circular coils in the same -interval. ; (c) Use the parametric equations in part (a) with and to graph the surface. 4 Polar Coordinates and Polar Graphs. 7 Consider the. So in general we can say that a circle centered at the origin, with radius r, is the locus of all points that satisfy the equations. To plot the surface and the tangent plane, I'll have to use surf or mesh for at least one of those, rather than an ez command. Recall the problem of finding the surface area of a volume of revolution. In a surface of revolution, the radius may be different at each height, so if the radius at height v is r(v), then the equation of. The volume is actually. Making a solid of revolution is simply the method of summing all the cross-sectional areas along the x-axis between two values of x. For this reason, the resulting surface is a called a surface of revolution. 8, where the arc length of the teardrop is calculated. surfaces of revolution A surface generated by revolving a plane curve about an axis in its plane Set of field equations for thick shell of revolution made of. Any surface of revolution can be easily parametrized. Surfaces of revolution. You may recognize the first two equations as being the polar coordinate conversion equations. Find a vector-valued function whose graph is the indicated surface. 1 2 3 4 x 0. Since I'll have to enter the parametrized equation in notation MATLAB understands for vectors and matrices, I'll first give the vectorize command (which tells me where to put the periods). GET EXTRA HELP If you could use some extra help. Convert Surface of Revolution to Parametric Equations. Two surfaces, their intersection curve and a level set of the function L, see Exam-ple 3. So if you like, this is another example. The simplest type of parametric surfaces is given by the graphs of functions of two variables: {\displaystyle z=f (x,y),\quad {\vec {r}} (x,y)= (x,y,f (x,y)). Solving Linear Equations. then the revolution would map out a circle of radius v at a height of c, which. Important formula for surface area of Cartesian curve, Parametric equation of curve,…. x = 1 t - 2 3. In parametric representation the coordinates of a point of the surface patch are expressed as functions of the parameters and in a closed rectangle:. Remember that a surface can be given as parametric equations with two parameters: x(u,v), y(u,v), z(u,v). 5 Problem 32E. More specifically: Suppose that C(u) lies in an (x c,y c) coordinate system with origin O c. 1 Curves Defined by Parametric Equations ET 10. 4 Calculus with Polar Coordinates. " Find the arc length of the teardrop. Ask Question Asked 6 years ago. 4 Polar Coordinates and Polar Graphs. Arc length and surface area of revolution; Further pure 3. Textbook solution for Calculus (MindTap Course List) 11th Edition Ron Larson Chapter 15. Examples Example 1. Added Aug 1, 2010 by Michael_3545 in Mathematics. If Martin finds the volume of the solid formed by the outer curve and subtracts the. An example of such a surface is the sphere, which may be considered as the surface generated when a semicircle is revolved about its diameter. S S is the surface area of the solid obtained by rotating the parametric curve \begin {array} {c}&x = 4 \cos^3 t &y = 4 \sin^3 t &0 \leq t \leq \frac {\pi} {2} \end {array} x=4cos3t y=4sin3t. According to Stroud and Booth (2013)*, "A curve is defined by the parametric equations ; if the arc in between. Recall the problem of finding the surface area of a volume of revolution. For every point along T(v), lay C(u) so that O c coincides with T(v). Surfaces of revolution. • Find the slope of a tangent line to a curve given by a set of parametric equations. share | cite | improve this question | follow | asked Oct 26 '17 at 23:51. A parametric wave is usually required for complicated surfaces, multi-valued surfaces, or those not easily expressed as z(x,y). The part of the paraboloid z = x^2 + y^2 that lies inside the cylinder x^2 + y^2 = 64. Apply the formula for surface area to a volume generated by a parametric curve. 1 Geodesic Equations of a Surface of Revolution 47 8. } A rational surface is a surface that admits parameterizations by a rational function. Find the area under a parametric curve. rotates through a complete revolution about the. Two surfaces, their intersection curve and a level set of the function L, see Exam-ple 3. • Rewrite rectangular equations in polar form and vice versa. The curve being rotated can be defined using rectangular, polar, or parametric equations. The following table gives the lateral surface areas for some common surfaces of revolution where denotes the radius (of a cone, cylinder, sphere, or zone), and the inner and outer radii of a frustum, the height, the ellipticity of a spheroid, and and the equatorial and polar radii (for a spheroid) or the radius of a circular cross-section and. 1 2 3 4 x 0. Since z = 1, the entire surface lies in the plane z = 1. If a surface is obtained by rotating about the x-axis from t=a to b the curve of the parametric equation {(x=x(t)),(y=y(t)):}, then its surface area A can be found by A=2pi int_a^by(t)sqrt{x'(t)+y'(t)}dt If the same curve is rotated about the y-axis, then A=2pi int_a^b x(t)sqrt{x'(t)+y'(t)}dt I hope that this was helpful. trange - A 3-tuple $$(t,t_{\min},t_{\max})$$ where t is the independent variable of the curve. So that turns out to be the example of the surface area of a sphere. Surface of Revolution of Parametric Curve about y=# Discover Resources. 0 z Sphere is an example of a surface of revolution generated by revolving a parametric curve x= f(t), z= g(t)or, equivalently,. Suppose that R ( u, v ) = x ( u, v )ˆ ı + y ( u, v )ˆ + z ( u, v ) ˆ k is a vector function defined on a parameter domain D (in the uv -plane). Computing the arc length of a curve between two points (see demo). Hence, if one wants to construct a circle of radius r, the equation is Circle(u) = (rcosu, rsinu). Earlier, you were asked about how Martin can model the volume of a particular vase. The axis of rotation must be either the x-axis or the y-axis. 4 Approximating functions with Taylor polynomials. trigonometry, parametric equations, and integral calculus -- are needed for any real mathematical understanding of the topic. Find the surface area if this shape is rotated about the $$x$$- axis, as shown in Figure 9. MIT 18 01 - Parametric Equations, Arclength, Surface Area (5 pages) Previewing pages 1, 2 of 5 page document View the full content. Surface Area of a Surface of Revolution. In this paper, we present a method to decide whether a set of parametric equations is normal. Find the area under a parametric curve. 1 Geodesic Equations of a Surface of Revolution 47 8. This means we define both x and y as functions of a parameter. 2 In section 9. Martin can model the vase by revolving two parametric curves around the -axis from. Sets up the integral, and finds the area of a surface of revolution. We consider two cases - revolving about the $$x-$$axis and revolving about the $$y-$$axis. We’ll first need the derivatives of the parametric equations. Parametric Equations Differentiation Exponential Functions Volumes of Revolution Numerical Integration. Step-by-step solution: 100 %( 5 ratings). If the curve is revolved around the y-axis, then the formula is $$S=2π∫^b_a x(t)\sqrt{(x′(t))^2+(y′(t))^2}dt. Chapter 10: Parametric Equations and Polar Coordinates. Download Flash Player. Answer to: Find the area of the surface generated by revolving the curve about the x-axis. Parametric Equations; 5. We will also see how the parameterization of a surface can be used to find a normal vector for the surface (which will be very useful in a couple of sections) and how the parameterization can be used to find the surface area of a surface. 13 from Section 7. } A rational surface is a surface that admits parameterizations by a rational function. Calculus Calculus: Early Transcendental Functions Representing a Surface of Revolution Parametrically In Exercises 69-70, write a set of parametric equations for the surface of revolution obtained by revolving the graph of the function about the given axis. You can then use the menus along the top to change the Shape Type and Surface Color mode, or you can use the shortcut keys indicated in the menus if you have. Objective: Converting a system of parametric equations to rectangular form * Review for Section 11. Surface area of revolution in parametric equations. Parametric Curve: Surface Area of Revolution; Surface Area of Revolution of a Parametric Curve Rotated About the y-axis; Parametric Arc Length; Parametric Arc Length and the distance Traveled by the Particle; Volume of Revolution of a Parametric Curve; Converting Polar Coordinates; Converting Rectangular Equations to Polar Equations. A surface of revolution is obtained when a curve is rotated about an axis. Idea: rotate a 2D profile curvearound an axis. 6, forming a "teardrop. 5 Problem 32E. Area of a Surface of Revolution. In parametric representation the coordinates of a point of the surface patch are expressed as functions of the parameters and in a closed rectangle:. The parametric equation for a circle of radius 1 in the xy-plane is (x(u), y(u)) = (cosu, sinu) where 0 < u < 2pi. 7 A Surface of Revolution 41. We’ll first need the derivatives of the parametric equations. 1 The Parametric Representation of a Surface of. Then the surface has a parametric representation with r(u1) = λcosh(u1 c) and h(u1) = Z v u u t1− λ2 c2 sinh2 (u1 c2)du1. Solving a System of Linear Equations. Examples: Find the surface area of the solid obtained by rotating the curve x= rcost. A surface of revolution is a surface generated by rotating a two-dimensional curve about an axis. Example 1 Determine the surface area of the solid obtained by rotating the following parametric curve about the x x -axis. 3 Applications to Physics and Engineering: Hydrostatic Force and Pressure Moments and Centers of Mass: derivation summary examples. Hence, if one wants to construct a circle of radius r, the equation is Circle(u) = (rcosu, rsinu). curve using parametric equations. This video lecture " Surface Area Of Solid Generated By Revolution about axes in Hindi " will help Engineering and Basic Science students to understand following topic of of Engineering-Mathematics: 1. Chapter 10 introduces some special surfaces of practical use (surfaces of revolution, ruled surfaces,. 3542 and Ih = −3. trange - A 3-tuple \((t,t_{\min},t_{\max})$$ where t is the independent variable of the curve. this project is to assist CNC Software, inc. So we're going to do this surface area now. d x d θ = − 3 cos 2 θ sin θ d y d θ = 3 sin 2 θ cos θ d x d θ = − 3 cos 2 θ sin θ d y d θ = 3 sin 2 θ cos θ. original surface requires 0 ≤ ≤2 for a complete revolution. y Figure 10. For math, science, nutrition, history. So if you like, this is another example. Free Cone Surface Area Calculator - calculate cone surface area step by step This website uses cookies to ensure you get the best experience. If a surface is obtained by rotating about the x-axis from t=a to b the curve of the parametric equation {(x=x(t)),(y=y(t)):}, then its surface area A can be found by A=2pi int_a^by(t)sqrt{x'(t)+y'(t)}dt If the same curve is rotated about the y-axis, then A=2pi int_a^b x(t)sqrt{x'(t)+y'(t)}dt I hope that this was helpful. 4 Calculus with Polar Coordinates. Yep, that’s right; there is just one formula that enables us to find the volumes of solids of revolution (i. ;] -- This program covers the important topic of the Surface Area of Revolution in Parametric Equations in Calculus. The two points (X1,Y1,Z1) and (X2,Y2,Z2) are the two focal points and the axis of revolution lies along the line between them. Learn how to find the surface area of revolution of a parametric curve rotated about the y-axis. Earlier, you were asked about how Martin can model the volume of a particular vase. The process is similar to that in Part 1. , the disk and washer methods), for any line we wish to revolve about. Active 2 years, 7 months ago. This is the parametrization for a flat torus in 4D. So we'll save that for a second. d x d θ = − 3 cos 2 θ sin θ d y d θ = 3 sin 2 θ cos θ d x d θ = − 3 cos 2 θ sin θ d y d θ = 3 sin 2 θ cos θ. GET EXTRA HELP If you could use some extra help. GET EXTRA HELP If you could use some extra help. Suppose that R ( u, v ) = x ( u, v )ˆ ı + y ( u, v )ˆ + z ( u, v ) ˆ k is a vector function defined on a parameter domain D (in the uv -plane). The parametric net on a spacelike surface of revolution obtained by pseudo-Euclidean rotations forms the Tchebyshev net in the following parametrization of the surface (): and on a timelike surface of revolution (, see Figure 7) The parametric net on a surface of revolution obtained by isotropic rotations forms the Tchebyshev net in the. Parametric Equations Differentiation Exponential Functions Volumes of Revolution Numerical Integration. Show Solution. S S is the surface area of the solid obtained by rotating the parametric curve \begin {array} {c}&x = 4 \cos^3 t &y = 4 \sin^3 t &0 \leq t \leq \frac {\pi} {2} \end {array} x=4cos3t y=4sin3t. You can then use the menus along the top to change the Shape Type and Surface Color mode, or you can use the shortcut keys indicated in the menus if you have. For this reason, the resulting surface is a called a surface of revolution. 6: A graph of the parametric equations in Example 10. Parametrics Parametric Curves Parametric Surfaces Parametric Derivative Slope & Tangent Lines Area Arc Length Surface Area Volume SV Calculus Limits Derivatives Integrals Infinite Series Parametric Equations Conics Polar Coordinates Laplace Transforms. Parametric equations of surfaces of revolution. Surface Area Length of a Plane Curve A plane curve is a curve that lies in a two-dimensional plane. • Calculus with curves defined by parametric equations = (𝑡), = (𝑡) o Area of a surface of revolution By rotating about the -axis:. Area of a Surface of Revolution The polar coordinate versions of the formulas for the area of a surface of revolution can be obtained from the parametric versions, using the equations x = r cos θ and y = r sin θ. 3 The Christoffel Symbols for a Surface of Revolution 45. One-to-one and Inverse Functions. Volume - Spreadsheet - Equation - Expression (mathematics) - Theorem - Science - Commensurability (philosophy of science) - English plurals - Contemporary Latin - Latin influence in English - Mathematics - Formal language - Sphere - Integral - Geometry - Method of exhaustion - Parameter - Radius - Algebraic expression - Closed-form expression - Chemistry - Atom - Chemical compound - Number - Water. A surface of revolution is a surface generated by rotating a two-dimensional curve about an axis. Equating x, y , respectively z from equations (2) and (4) one gets u cos v = f ( t ) cos s. how a solid generated by revolution of curve arc about axes. 2: Calculus With Parametric Curves & Equations Of Tangents. Representing the space curve by two surfaces which intersect orthogonally pro-. Parametric Curve: Surface Area of Revolution; Surface Area of Revolution of a Parametric Curve Rotated About the y-axis; Parametric Arc Length; Parametric Arc Length and the distance Traveled by the Particle; Volume of Revolution of a Parametric Curve; Converting Polar Coordinates; Converting Rectangular Equations to Polar Equations. Surfaces of revolution can be any parametric curve. 6, forming a "teardrop. 13 from Section 7. The substitution of values into this equation and solution are as follows: v = SQRT [ ( 6. MIT 18 01 - Parametric Equations, Arclength, Surface Area (5 pages) Previewing pages 1, 2 of 5 page document View the full content. Since the surface is in the form x = f ( y, z) x = f ( y, z) we can quickly write down a set of parametric equations as follows, x = 5 y 2 + 2 z 2 − 10 y = y z = z x = 5 y 2 + 2 z 2 − 10 y = y z = z. Consider the teardrop shape formed by the parametric equations $$x=t(t^2-1)$$, $$y=t^2-1$$ as seen in Example 9. Example $$\PageIndex{8}$$: Surface Area of a Solid of Revolution. The surface area of the thinstripofwidth ds is 2πy ds. Parametric Equations - Surface Area What is the surface area S S S of the body of revolution obtained by rotating the curve y = e x , y=e^x, y = e x , 0 ≤ x ≤ 1 , 0 \le x \le 1, 0 ≤ x ≤ 1 , about the x − x- x − axis?. Solving Linear Equations. axis, determine the area of the surface generated. A surface of revolution is obtained when a curve is rotated about an axis. 7 A Surface of Revolution 41. An example of such a surface is the sphere, which may be considered as the surface generated when a semicircle is revolved about its diameter. in the parametric design of this complex gear which currently does not exist. Surface Area Length of a Plane Curve A plane curve is a curve that lies in a two-dimensional plane. When I just look at that, unless you deal with parametric equations, or maybe polar coordinates a lot, it's not obvious that this is the parametric equation for an ellipse. † † margin: 1-1-1. The implicit equation of a sphere is: x 2 + y 2 + z 2 = R 2. Step-by-step solution: 100 %( 5 ratings). Open parts of the bulb (left) and the neck (right) segments of the periodic surface of revolution obtained via parametric equations (17) and (21) with ε = 1. Chapter 10 introduces some special surfaces of practical use (surfaces of revolution, ruled surfaces,. When you’re measuring the surface of revolution of a function f(x) around the x-axis, substitute r = f(x) into the formula: For example, suppose that you want to find the area of revolution that’s shown in this figure. Revision Resources; Matrix Algebra; The Vector Product; Determinants; Application of vectors; Inverse Matrices; Solving linear equations. 0 z Sphere is an example of a surface of revolution generated by revolving a parametric curve x= f(t), z= g(t)or, equivalently,. The formula for finding the slope of a parametrized. According to Stroud and Booth (2013)*, "A curve is defined by the parametric equations ; if the arc in between. Revision Resources; Series and limits; Polar coordinates; 1st order differential equations; 2nd order differential equations; Further Pure 4. Apply the formula for surface area to a volume generated by a parametric curve. In this section we'll find areas of surfaces of revolution. surfaces of revolution A surface generated by revolving a plane curve about an axis in its plane Set of field equations for thick shell of revolution made of. Function Axis of Revolution z = y + 1 , 0 ≤ y ≤ 3 y - axis. Making a solid of revolution is simply the method of summing all the cross-sectional areas along the x-axis between two values of x. Volume - Spreadsheet - Equation - Expression (mathematics) - Theorem - Science - Commensurability (philosophy of science) - English plurals - Contemporary Latin - Latin influence in English - Mathematics - Formal language - Sphere - Integral - Geometry - Method of exhaustion - Parameter - Radius - Algebraic expression - Closed-form expression - Chemistry - Atom - Chemical compound - Number - Water. 1, we talked about parametric equations. Parametric representation is a very general way to specify a surface, as well as implicit representation. (b) Eliminate the parameters to show that the surface is an elliptic paraboloid and set up another double integral for the surface area. The letters u & v are also used separately for the surface parametrization. If a surface is obtained by rotating about the x-axis from #t=a# to #b# the curve of the parametric equation #{(x=x(t)),(y=y(t)):}#, then its surface area A can be found by. Surfaces that occur in two of the main theorems of vector calculus, Stokes' theorem and the divergence theorem, are frequently given in a parametric form. The Circle. Arc length and surface area of revolution; Further pure 3. The path swept out by the curve is a surface in three dimensions. y Figure 10. • Rewrite rectangular equations in polar form and vice versa. Find more Mathematics widgets in Wolfram|Alpha. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. ” Find the arc length of the teardrop. In addition, we give some simple criteria for a set of parametric equations to be. To plot the surface and the tangent plane, I'll have to use surf or mesh for at least one of those, rather than an ez command. 6: A graph of the parametric equations in Example 10. (b) Eliminate the parameters to show that the surface is an elliptic paraboloid and set up another double integral for the surface area. In a surface of revolution, the radius may be different at each height, so if the radius at height v is r(v), then the equation of the surface is. This is the parametrization for a flat torus in 4D. 3542 and Ih = −3. The Derivative of Parametric Equations Suppose that x = x(t) and y = y(t) then as long as dx/dt is nonzero. 1 The Parametric Representation of a Surface of. Computing the volume of a solid of revolution with the disc and washer methods. y x Figure3. Graphing a surface of revolution. The following table gives the lateral surface areas for some common surfaces of revolution where denotes the radius (of a cone, cylinder, sphere, or zone), and the inner and outer radii of a frustum, the height, the ellipticity of a spheroid, and and the equatorial and polar radii (for a spheroid) or the radius of a circular cross-section and. 4 in a similar way as done to produce the formula for arc length done before. 6 Polar coordinates and applications. So in general we can say that a circle centered at the origin, with radius r, is the locus of all points that satisfy the equations. Since the surface is in the form x = f ( y, z) x = f ( y, z) we can quickly write down a set of parametric equations as follows, x = 5 y 2 + 2 z 2 − 10 y = y z = z x = 5 y 2 + 2 z 2 − 10 y = y z = z. Suppose that $$y\left( x \right),$$ $$y\left( t \right),$$ and $$y\left( \theta \right)$$ are smooth non-negative functions on the given interval. Any surface of revolution can be easily parametrized. Ask Question Asked 6 years ago. Remember, the quantity is negative because parametric equations travel the curve in a counter clockwise direction, and thus the results of the integration are negative. Now we establish equations for area of surface of revolution of a parametric curve x = f (t), y = g (t) from t = a to t = b, using the parametric functions f and g, so that we don't have to first find the corresponding Cartesian function y = F (x) or equation G (x, y) = 0. Hence, if one wants to construct a circle of radius r, the equation is Circle(u) = (rcosu, rsinu). What is the surface area of revolution for the curve of interest? Technical Help: This simulation determines the surface area of revolution for a curve (complete or portion) given in the following parametric equations: x = θ - sin θ and y = 1 - cos θ for 0 ≤ θ ≤ 2 π. One-to-one and Inverse Functions. 3 Applications to Physics and Engineering: Hydrostatic Force and Pressure Moments and Centers of Mass: derivation summary examples. surfaces of revolution A surface generated by revolving a plane curve about an axis in its plane Set of field equations for thick shell of revolution made of. Computing the arc length of a curve between two points (see demo). Bibliography 53. Added Aug 1, 2010 by Michael_3545 in Mathematics. 673 x 10 -11 N m 2 /kg 2 )*( 5. 2: Calculus With Parametric Curves & Equations Of Tangents. S S is the surface area of the solid obtained by rotating the parametric curve \begin {array} {c}&x = 4 \cos^3 t &y = 4 \sin^3 t &0 \leq t \leq \frac {\pi} {2} \end {array} x=4cos3t y=4sin3t. I'll use surf for the surface. 2 Areas of Surfaces of Revolution. Example: Surface Area of a Sphere : Similar to the concept of an arc length, when a curve is given by the following parametric equations. Calculus Calculus: Early Transcendental Functions Representing a Surface of Revolution Parametrically In Exercises 69-70, write a set of parametric equations for the surface of revolution obtained by revolving the graph of the function about the given axis. In the first problem, "A) the Torus obtained by a rotation of a circle x= a + b*sin(u), y= 0, z = b*sin(u) " you are already given a parameter u. Then nd the surface area using the parametric equations. Find parametric equations for the surface obtained by rotating the curve y=16x^4-x^2, -4 clog(|λ|/c). Representing a Surface of Revolution Parametrically. Answer to: Find the area of the surface generated by revolving the curve about the x-axis. Case 2 C 2 = 0 and C 1 = λ 6= 0. a surface that can be generated by revolving a plane curve about a straight line, called the axis of the surface of revolution, lying in the plane of the curve. The Straight Line. 5 Problem 32E. Area of surface of revolution about the y-axis; The equation, x^2+y^2 = 9, is represented parametrically by the equations { x = 3cos(t), y = 3sin(t) }. What is the surface area of revolution for the curve of interest? Technical Help: This simulation determines the surface area of revolution for a curve (complete or portion) given in the following parametric equations: x = θ - sin θ and y = 1 - cos θ for 0 ≤ θ ≤ 2 π. By using this website, you agree to our Cookie Policy. Mathematically, a surface of revolution is the result of taking a curve in the two-dimensional plane (like the guide on the lathe) and revolving it about an axis. Free Cone Surface Area Calculator - calculate cone surface area step by step This website uses cookies to ensure you get the best experience. Find the area of the surface formed by revolving the curve about the x-axis on an interval 0≤t≤ /3. S S is the surface area of the solid obtained by rotating the parametric curve \begin {array} {c}&x = 4 \cos^3 t &y = 4 \sin^3 t &0 \leq t \leq \frac {\pi} {2} \end {array} x=4cos3t y=4sin3t. Parametric surface forming a trefoil knot, equation details in the attached source code. v, y ( u) sin. Refer to Figure 3. Function Axis of Revolution z = y + 1 , 0 ≤ y ≤ 3 y - axis. Answer to: Find the area of the surface generated by revolving the curve about the x-axis. Find a vector-valued function whose graph is the indicated surface. The Circle. 3 Parametric Equations and Calculus Find the slope of a tangent line to a curve defined by parametric equations; find the arc length along a curve defined parametrically; find the area of a surface of revolution in parametric form. Since z = 1, the entire surface lies in the plane z = 1. Parametric Equations 1, Parametric Equations 2: Parametric Equations - Some basic questions, Parametric Curves - Basic Graphing: Parametric equations, arclength, surface area: Introduction to Parametric Equations: Parametric Equations: Parametric Equations Example: 10. Textbook solution for Calculus (MindTap Course List) 11th Edition Ron Larson Chapter 15. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function y. Volume - Spreadsheet - Equation - Expression (mathematics) - Theorem - Science - Commensurability (philosophy of science) - English plurals - Contemporary Latin - Latin influence in English - Mathematics - Formal language - Sphere - Integral - Geometry - Method of exhaustion - Parameter - Radius - Algebraic expression - Closed-form expression - Chemistry - Atom - Chemical compound - Number - Water. 17 and 16 depict the minimal axes of revolution and minimum surfaces of revolution for the values m = -1, m = 0, and m =1. or x 2 + y 2 = R 2 - v 2. The surface of the Revolution: Given the parametric equations of the curve, finding the surface area of the revolved curve is done by using the following formula {eq}\displaystyle S=2\pi\int_{a. Parametric equations Definition A plane curve is smooth if it is given by a pair of parametric equations x =f(t), and y =g(t), t is on the interval [a,b] where f' and g' exist and are continuous on [a,b] and f'(t) and g'(t) are not simultaneously. [Jason Gibson, (Math instructor); TMW Media Group. x = 1 t - 2 3. Area of a Surface of Revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function y. Since you are rotating around the z-axis, let v be the angle made with the x-axis. The Derivative of Parametric Equations Suppose that x = x(t) and y = y(t) then as long as dx/dt is nonzero. We have step-by-step solutions for your textbooks written by Bartleby experts!. • Calculus with curves defined by parametric equations = (𝑡), = (𝑡) o Area of a surface of revolution By rotating about the -axis:. The notion of parametric equation has been generalized to surfaces, manifolds and algebraic varieties of higher dimension, with the number of parameters being equal to the dimension of the manifold or variety, and the number of equations being equal to the dimension of the space in which the manifold or variety is considered (for curves the. Consider the parametric equations 𝑥 = 2 𝜃 c o s and 𝑦 = 2 𝜃 s i n, where 0 ≤ 𝜃 ≤ 𝜋. Determine derivatives and equations of tangents for parametric curves. Parametric surface forming a trefoil knot, equation details in the attached source code. We're on a mission to help every student learn math and love learning math. Surface Area: The surface area of a solid of revolution where a curve x= x(t), y= y(t) with arc length dsis rotated about an axis is S= Z 2ˇrds = Z 2ˇr s dx dt 2 + dy dt 2 dt If revolving about the x-axis, r= y(t) and if revolving about the y-axis, r= x(t). Parametric equations Definition A plane curve is smooth if it is given by a pair of parametric equations. The range of the surfaces. a surface that can be generated by revolving a plane curve about a straight line, called the axis of the surface of revolution, lying in the plane of the curve. of the surface with parametric equations ,, , ,. Since the surface is in the form x = f ( y, z) x = f ( y, z) we can quickly write down a set of parametric equations as follows, x = 5 y 2 + 2 z 2 − 10 y = y z = z x = 5 y 2 + 2 z 2 − 10 y = y z = z. If you start with the parametric curve ( x ( u), y ( u)), u ∈ I (some interval), and rotate it about the x -axis, the surface you obtain will be parametrized by. Answer to: Find the area of the surface obtained by rotating the curve determined by the parametric equations x = 8 t - 8 / 3 t^3, y = 8 t^2, 0. 12 Approximate Implicitization of Space Curves and of Surfaces of Revolution 219 0. Examples Example 1. Get the free "Area of a Surface of Revolution" widget for your website, blog, Wordpress, Blogger, or iGoogle. Suppose that $$y\left( x \right),$$ $$y\left( t \right),$$ and $$y\left( \theta \right)$$ are smooth non-negative functions on the given interval. trange - A 3-tuple $$(t,t_{\min},t_{\max})$$ where t is the independent variable of the curve. 4 - Areas of Surfaces of Revolution - Exercises 6. A surface of revolution is a three-dimensional surface with circular cross sections, like a vase or a bell or a wine bottle. • Calculus with curves defined by parametric equations = (𝑡), = (𝑡) o Area of a surface of revolution By rotating about the -axis:. So just like that, by eliminating the parameter t, we got this equation in a form that we immediately were able to recognize as ellipse. 3 Taylor and Maclaurin series. Download Flash Player. The curve being rotated can be defined using rectangular, polar, or parametric equations. These equations are called parametric equations of the surface and the surface given via parametric equations is called a parametric surface. Now scale up to v in (a, b) and z = f(v) instead of c. 6 Polar coordinates and applications. One somewhat simpler case is a surface of revolution that has axial symmetry around a 'z' axis. (d) For the case , , use a computer algebra. Parametric Equations Differentiation Exponential Functions Volumes of Revolution Numerical Integration. Math 55 Parametric Surfaces & Surfaces of Revolution Parametric Surfaces A surface in R 3 can be described by a vector function of two parameters R ( u, v ). Parametric Equations 1, Parametric Equations 2: Parametric Equations - Some basic questions, Parametric Curves - Basic Graphing: Parametric equations, arclength, surface area: Introduction to Parametric Equations: Parametric Equations: Parametric Equations Example: 10. Then the surface has a parametric representation with r(u1) = λcosh(u1 c) and h(u1) = Z v u u t1− λ2 c2 sinh2 (u1 c2)du1. Consider the parametric equations 𝑥 = 2 𝜃 c o s and 𝑦 = 2 𝜃 s i n, where 0 ≤ 𝜃 ≤ 𝜋. 6: A graph of the parametric equations in Example 10. This means we define both x and y as functions of a parameter. x(u,v)= rcos 2πvsin πu y(u,v) = rsin 2πvsin πu z(u,v) = rcos πu. Computing the surface area of a solid of revolution. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. Example $$\PageIndex{8}$$: Surface Area of a Solid of Revolution. According to Stroud and Booth (2013)*, "A curve is defined by the parametric equations ; if the arc in between. Area of a Surface of Revolution. or x 2 + y 2 = R 2 - v 2. The curve being rotated can be defined using rectangular, polar, or parametric equations. Making a solid of revolution is simply the method of summing all the cross-sectional areas along the x-axis between two values of x. The notion of parametric equation has been generalized to surfaces, manifolds and algebraic varieties of higher dimension, with the number of parameters being equal to the dimension of the manifold or variety, and the number of equations being equal to the dimension of the space in which the manifold or variety is considered (for curves the. So if you like, this is another example. Ask Question Asked 6 years ago. The last two equations are just there to acknowledge that we can choose y y and z z to be anything we want them to be. Examples of surfaces of revolution include the apple surface, cone (excluding the base), conical frustum (excluding the ends), cylinder (excluding the ends), Darwin-de Sitter spheroid, Gabriel's horn, hyperboloid, lemon surface, oblate. " Find the arc length of the teardrop. how a solid generated by revolution of curve arc about axes. The formulas below give the surface area of a surface of revolution. (An overall minimum surface area of 82. Textbook solution for Calculus: Early Transcendentals 8th Edition James Stewart Chapter 13. x = 1 t - 2 3. Look at slice at x:. It is the product of an x-y circle (sin u, cos u), and a u-v circle (sin v, cos v). Section 3-5 : Surface Area with Parametric Equations. Calculus with Parametric Equations Now we are ready to approximate the area of a surface of revolution. General sweep surfaces The surface of revolution is a special case of a swept surface. In a surface of revolution, the radius may be different at each height, so if the radius at height v is r(v), then the equation of the surface is. Parametric equations intro: Parametric equations, polar coordinates, and vector-valued functions Second derivatives of parametric equations: Parametric equations, polar coordinates, and vector-valued functions Arc length: parametric curves: Parametric equations, polar coordinates, and vector-valued functions Vector-valued functions: Parametric equations, polar coordinates, and vector-valued. Computing the surface area for a surface of revolution whose curve is generated by a parametric equation: Surface Area Generated by a Parametric Curve Recall the problem of finding the surface area of a volume of revolution. Now scale up to v in (a, b) and z = f(v) instead of c. , the disk and washer methods), for any line we wish to revolve about. Since the surface is in the form x = f ( y, z) x = f ( y, z) we can quickly write down a set of parametric equations as follows, x = 5 y 2 + 2 z 2 − 10 y = y z = z x = 5 y 2 + 2 z 2 − 10 y = y z = z. Sets up the integral, and finds the area of a surface of revolution. So that turns out to be the example of the surface area of a sphere. Example: Surface Area of a Sphere : Similar to the concept of an arc length, when a curve is given by the following parametric equations. axis, determine the area of the surface generated. trange - A 3-tuple $$(t,t_{\min},t_{\max})$$ where t is the independent variable of the curve. The Straight Line. Learn how to find the surface area of revolution of a parametric curve rotated about the y-axis. the domain D consisting of all possible values of parameters uand vis contained in R2. Chapter 8 deals with the intersection of curves and surfaces. Remember that a surface can be given as parametric equations with two parameters: x(u,v), y(u,v), z(u,v). Area of surface of revolution about the y-axis; The equation, x^2+y^2 = 9, is represented parametrically by the equations { x = 3cos(t), y = 3sin(t) }. Find the surface area of revolution of the solid created when the parametric curve is rotated around the given axis over the given interval. Hence, if one wants to construct a circle of radius r, the equation is Circle(u) = (rcosu, rsinu). Revision Resources; Series and limits; Polar coordinates; 1st order differential equations; 2nd order differential equations; Further Pure 4. Thus, the new surface winds around twice as fast as the original surface, and since the equation for is identical in both surfaces, we observe twice as many circular coils in the same -interval. Solving a System of Linear Equations. Making a solid of revolution is simply the method of summing all the cross-sectional areas along the x-axis between two values of x. This curve is depicted in Fig. Revolving about the $$x-$$axis. original surface requires 0 ≤ ≤2 for a complete revolution. What is the surface area of revolution for the curve of interest? Technical Help: This simulation determines the surface area of revolution for a curve (complete or portion) given in the following parametric equations: x = θ - sin θ and y = 1 - cos θ for 0 ≤ θ ≤ 2 π. Arc length and surface area of revolution; Further pure 3. For this reason, the resulting surface is a called a surface of revolution. ( ) ( ) x f t y g t = = If f and g have derivatives at t, then the parametrized curve also has a derivative at t. Example: Surface Area of a Sphere : Similar to the concept of an arc length, when a curve is given by the following parametric equations. Surface area of revolution in parametric equations. Subsection 10. The parametric equation for a circle of radius 1 in the xy-plane is (x(u), y(u)) = (cosu, sinu) where 0 < u < 2pi. Two surfaces, their intersection curve and a level set of the function L, see Exam-ple 3. 055 and [[beta]. Minimizing the surface of revolution around the x-axis, min S. The path swept out by the curve is a surface in three dimensions. • Rewrite rectangular equations in polar form and vice versa. x = f(t) and y = g(t) for a ≤ t ≤ b, the surface area of revolution for the curve revolving around the y-axis is defined as. According to Stroud and Booth (2013)*, "A curve is defined by the parametric equations ; if the arc in between. Download Flash Player. Active 2 years, 7 months ago. A circle that is rotated around any diameter generates a sphere of which it is then a great circle, and. The resulting surface therefore always has azimuthal symmetry. The part of the paraboloid z = x^2 + y^2 that lies inside the cylinder x^2 + y^2 = 64. The surface of helicoid consists of lines orthogonal to the surface of that cylinder and after projection you will get a spiral. To illustrate, we'll show how the plot of \begin{gather*} z=f(x,y) = \frac{\sin \sqrt{x^2+y^2}} {\sqrt{x^2+y^2}+1} \end{gather*} is a surface of revolution. Graphing a surface of revolution. A surface of revolution is a three-dimensional surface with circular cross sections, like a vase or a bell or a wine bottle. So if you like, this is another example. According to Stroud and Booth (2013)*, "A curve is defined by the parametric equations ; if the arc in between. This is the parametrization for a flat torus in 4D. Revision Resources; Matrix Algebra; The Vector Product; Determinants; Application of vectors; Inverse Matrices; Solving linear equations. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function y. 1 The Parametric Representation of a Surface of. Apply the formula for surface area to a volume generated by a parametric curve. MIT 18 01 - Parametric Equations, Arclength, Surface Area (5 pages) Previewing pages 1, 2 of 5 page document View the full content. 17 and 16 depict the minimal axes of revolution and minimum surfaces of revolution for the values m = -1, m = 0, and m =1. (compare: area of a cylinder = cross-sectional area x length) The method for solids rotated around the y-axis is similar. The area of the surface 𝑆 obtained by rotating this parametric curve 2 𝜋 radians about the 𝑥-axis can be calculated by evaluating the integral 2 𝜋 𝑦 𝑠 d where d d d d d d 𝑠 = 𝑥 𝜃 + 𝑦 𝜃 𝜃. Textbook solution for Calculus: Early Transcendentals 8th Edition James Stewart Chapter 13. For every point along T(v), lay C(u) so that O c coincides with T(v). Get the free "Area of a Surface of Revolution" widget for your website, blog, Wordpress, Blogger, or iGoogle. Parametric equations can be used to describe motion that is not a function. Revision Resources; Matrix Algebra; The Vector Product; Determinants; Application of vectors; Inverse Matrices; Solving linear equations. In this final section of looking at calculus applications with parametric equations we will take a look at determining the surface area of a region obtained by rotating a parametric curve about the $$x$$ or $$y$$-axis. Write a set of parametric equations for the surface of revolution obtained by revolving the graph of the function about the given axis. 12 Approximate Implicitization of Space Curves and of Surfaces of Revolution 219 0. The surface of revolution given by rotating the region bounded by y = x3 for 0 ≤ x ≤ 2 about the x-axis. Examples of surfaces of revolution include the apple surface, cone (excluding the base), conical frustum (excluding the ends), cylinder (excluding the ends), Darwin-de Sitter spheroid, Gabriel's horn, hyperboloid, lemon surface, oblate. 11 l] PARAMETRIC EOUATIONS AND LÏI ET 10 POLAR COORDINATES 11. Find parametric equations for the surface obtained by rotating the curve y=16x^4-x^2, -4 clog(|λ|/c). This allows generation of the parametric wave using only simple wave operations, without for-endfor loops. Chapter 6 and Chapter 7 gives methods for displaying parametric and implicit surfaces. That isn't a parabolic surface, it is one branch of a hyperbola of revolution. 0 z Sphere is an example of a surface of revolution generated by revolving a parametric curve x= f(t), z= g(t)or, equivalently,. Active 5 years, 7 months ago. Determine derivatives and equations of tangents for parametric curves. As not all the shape equations are stable for all values across their parameter range, I’ve done a bit of work to limit the randomisation to reasonably valid values for each shape type. We can adapt the formula found in Theorem 7. And maybe I should remember this result here. 3 Parametric Equations and Calculus Find the slope of a tangent line to a curve defined by parametric equations; find the arc length along a curve defined parametrically; find the area of a surface of revolution in parametric form. • Simultaneous equations in three unknowns • Volumes of Revolution • Stationary points, higher derivatives and curve sketching • Derivatives of sine and cosine • Introduction to the Differential Calculus • Parametric Equations • Maclaurin Series • Techniques of Integration • Integration by Substitution • The Integral of 1/x. The surface of helicoid consists of lines orthogonal to the surface of that cylinder and after projection you will get a spiral. Sets up the integral, and finds the area of a surface of revolution. If the curve is revolved around the y-axis, then the formula is $$S=2π∫^b_a x(t)\sqrt{(x′(t))^2+(y′(t))^2}dt. Find the area of the surface formed by revolving the curve about the x-axis on an interval 0≤t≤ /3. Active 5 years, 7 months ago. The notion of parametric equation has been generalized to surfaces, manifolds and algebraic varieties of higher dimension, with the number of parameters being equal to the dimension of the manifold or variety, and the number of equations being equal to the dimension of the space in which the manifold or variety is considered (for curves the. or x 2 + y 2 = R 2 - v 2. Theorem 10. When I just look at that, unless you deal with parametric equations, or maybe polar coordinates a lot, it's not obvious that this is the parametric equation for an ellipse. Related to the formula for finding arc length is the formula for finding surface area. g ( u, v) = ( x ( u), y ( u) cos. Apply the formula for surface area to a volume generated by a parametric curve. 6 LECTURE 17: PARAMETRIC SURFACES (I) Example 5: Solids of Revolution (will probably skip) (Math 2B) Parametrize the Surface obtained by rotating the curve y= 1 x between x= 1 and x= 2 about the x axis Start with x= x, 1 x 2. So just like that, by eliminating the parameter t, we got this equation in a form that we immediately were able to recognize as ellipse. Consider the cylinder x 2+ z = 4: a)Write down the parametric equations of this cylinder. Since I'll have to enter the parametrized equation in notation MATLAB understands for vectors and matrices, I'll first give the vectorize command (which tells me where to put the periods). The letters u & v are also used separately for the surface parametrization. Important formula for surface area of Cartesian curve, Parametric equation of curve,…. Consider the teardrop shape formed by the parametric equations \(x=t(t^2-1)$$, $$y=t^2-1$$ as seen in Example 9. Objective: Converting a system of parametric equations to rectangular form * Review for Section 11. While almost any calculus textbook one might find would include at least a mention of a cycloid, the topic is rarely covered in an introductory calculus course, and most students I have encountered are unaware of what a. Example: Surface Area of a Sphere : Similar to the concept of an arc length, when a curve is given by the following parametric equations. This means we define both x and y as functions of a parameter. I am trying to find a parametric curve, x=f(t) and y=g(t), preferably lying in the first quadrant (x≥0, y≥0), fulfilling all of the objectives A. The difficulty with using parametric equations lies in creating the equations to describe the surface. Apply the formula for surface area to a volume generated by a parametric curve. 2 Analytic representation of surfaces Similar to the curve case there are mainly three ways to represent surfaces, namely parametric, implicit and explicit methods. This video lecture " Surface Area Of Solid Generated By Revolution about axes in Hindi " will help Engineering and Basic Science students to understand following topic of of Engineering-Mathematics: 1. The implicit equation of a sphere is: x 2 + y 2 + z 2 = R 2. Write a set of parametric equations for the surface of revolution obtained by revolving the graph of the function about the given axis. The graph of the parametric equations x = t (t 2-1), y = t 2-1 crosses itself as shown in Figure 10. When you’re measuring the surface of revolution of a function f(x) around the x-axis, substitute r = f(x) into the formula: For example, suppose that you want to find the area of revolution that’s shown in this figure. • Rewrite rectangular equations in polar form and vice versa. A parametric wave is usually required for complicated surfaces, multi-valued surfaces, or those not easily expressed as z(x,y). 1 The Parametric Representation of a Surface of. Parametric representation is the a lot of accepted way to specify a surface. Step-by-step solution: 100 %( 5 ratings). In a surface of revolution, the radius may be different at each height, so if the radius at height v is r(v), then the equation of the surface is. If a surface is obtained by rotating about the x-axis from t=a to b the curve of the parametric equation {(x=x(t)),(y=y(t)):}, then its surface area A can be found by A=2pi int_a^by(t)sqrt{x'(t)+y'(t)}dt If the same curve is rotated about the y-axis, then A=2pi int_a^b x(t)sqrt{x'(t)+y'(t)}dt I hope that this was helpful. (b) Eliminate the parameters to show that the surface is an elliptic paraboloid and set up another double integral for the surface area. By using this website, you agree to our Cookie Policy. Theorem 10. 1 2 3 4 x 0. 7 is obtained for the values m = 1. Calculus 2 advanced tutor. Refer to Figure 3. Convert Surface of Revolution to Parametric Equations. Parametric equations of surfaces of revolution. Bibliography 53. E F Graph 3D Mode. We consider two cases - revolving about the $$x-$$axis and revolving about the $$y-$$axis. Objective: Converting a system of parametric equations to rectangular form * Review for Section 11. Section 12: Surface Area Of Revolution In Parametric Equations In 1868 he wrote a paper Essay on the interpretation of non-Euclidean geometry which produced a model for 2-dimensional non-Euclidean geometry within 3-dimensional Euclidean geometry. (An overall minimum surface area of 82. #A=2pi int_a^b x(t)sqrt{x'(t)+y'(t)}dt#. Which was that the arc length element was given by this. The volume is actually. 82 x 10 8 m ) ]. To use the application, you need Flash Player 6 or higher. Example: Surface Area of a Sphere : Similar to the concept of an arc length, when a curve is given by the following parametric equations. Volume - Spreadsheet - Equation - Expression (mathematics) - Theorem - Science - Commensurability (philosophy of science) - English plurals - Contemporary Latin - Latin influence in English - Mathematics - Formal language - Sphere - Integral - Geometry - Method of exhaustion - Parameter - Radius - Algebraic expression - Closed-form expression - Chemistry - Atom - Chemical compound - Number - Water. 1 Geodesic Equations of a Surface of Revolution 47 8. Thus, a parametric surface is represented as a vector function of two variables, i. Your browser doesn't support HTML5 canvas. Thus, the new surface winds around twice as fast as the original surface, and since the equation for is identical in both surfaces, we observe twice as many circular coils in the same -interval. In this section we will take a look at the basics of representing a surface with parametric equations. Parametric Equations - Surface Area What is the surface area S S S of the body of revolution obtained by rotating the curve y = e x , y=e^x, y = e x , 0 ≤ x ≤ 1 , 0 \le x \le 1, 0 ≤ x ≤ 1 , about the x − x- x − axis?. Maths Geometry Graph plot surface This demo allows you to enter a mathematical expression in terms of x and y. Defining Formula for Finding the Volume of a Solid. Surface of revolution definition is - a surface formed by the revolution of a plane curve about a line in its plane. These equations are called parametric equations of the surface and the surface given via parametric equations is called a parametric surface. If Martin finds the volume of the solid formed by the outer curve and subtracts the. Look at slice at x:. x = cos3θ y = sin3θ 0 ≤ θ ≤ π 2 x = cos 3 θ y = sin 3 θ 0 ≤ θ ≤ π 2. When I just look at that, unless you deal with parametric equations, or maybe polar coordinates a lot, it's not obvious that this is the parametric equation for an ellipse. The substitution of values into this equation and solution are as follows: v = SQRT [ ( 6. Active 2 years, 7 months ago. Bibliography 53. 5 Parametric curves. GET EXTRA HELP If you could use some extra help. The process is similar to that in Part 1. y Figure 10. 8, where the arc length of the teardrop is calculated. Parametric equations Definition A plane curve is smooth if it is given by a pair of parametric equations. x = 1 t - 2 3. Surface Area of a Solid of Revolution. 10 parametric equations %26 polar coordinates 1. Solving a System of Linear Equations. 1 Curves Defined by Parametric Equations ET 10. Example: Surface Area of a Sphere : Similar to the concept of an arc length, when a curve is given by the following parametric equations. Idea: Trace out surface S(u,v) by moving a profile curve C(u) along a trajectory curve T(v). A surface of revolution is a surface in Euclidean space created by rotating a curve (the generatrix) around an axis of rotation. Open parts of the bulb (left) and the neck (right) segments of the periodic surface of revolution obtained via parametric equations (17) and (21) with ε = 1. Consider the teardrop shape formed by the parametric equations $$x=t(t^2-1)$$, $$y=t^2-1$$ as seen in Example 9. Important formula for surface area of Cartesian curve, Parametric equation of curve,…. this equation. Two surfaces, their intersection curve and a level set of the function L, see Exam-ple 3. RevolveExample1(x = t2,y = t3, 0 ≤ t ≤ 1) around the x-axis. Calculus Calculus: Early Transcendental Functions Representing a Surface of Revolution Parametrically In Exercises 69-70, write a set of parametric equations for the surface of revolution obtained by revolving the graph of the function about the given axis. Examples of how to use “surface of revolution” in a sentence from the Cambridge Dictionary Labs. | 2020-10-20T17:16:56 | {
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https://math.stackexchange.com/questions/1198754/find-all-dimensions-such-that-volume-of-box-surface-area | # Find All Dimensions such that Volume of Box = Surface Area
A rectangular prism has integer edge lengths. Find all dimensions such that its surface area equals its volume.
My Attempt at a Solution:
Let the edge lengths be represented by the variables $l, w, h$.
Then $$lwh = 2\,(lw +lh + wh) \implies lwh = 2lwh\left(\frac{1}{l} + \frac{1}{w} + \frac{1}{h}\right).$$
Dividing both sides of the equation by $lwh$ yields $$1 = 2\left(\frac{1}{l} + \frac{1}{w} + \frac{1}{h}\right)$$
Or, $$\frac{1}{l} + \frac{1}{w} + \frac{1}{h} = \frac{1}{2}$$
Though perhaps a bit unnecessary, I used some algebraic deduction and number theory to find all the possible ordered triple pairs for the dimensions of the rectangular prism in the cases where all the dimensions are the same and two of the dimensions are the same.
My answers were: $(6,6,6),(5,5,10),(8,8,4),(12,12,3)$
I have a hunch that no ordered pair exists where all three values are distinct, but is there a way to rigorously prove this?
Note: By the AM-GM Inequality, $lwh \geq 216$. (I haven't been able to make use of this fact, but I just noted it here in case)
Edit: After researching a little more about Egyptian Fraction Analysis, inspired by marty cohen's answer, I found from a Mathworld Wolfram page about Egyptian Fractions that a unit fraction can be (infinitely) split into two more unit fractions: $$\frac{1}{a} = \frac{1}{a+1} + \frac{1}{a(a+1)}.$$ I then used this idea to successfully generate a few ordered triplet solutions ($l,w,h$).
WLoG, assume $l \leq w \leq h$. Then the following ordered triplets ($l,w,h$) are solutions of the equation $\frac{1}{l} + \frac{1}{w} + \frac{1}{h} = \frac{1}{2}$.
$$(4,6,12), (3,7,42), (3,8,24)$$
But, I suspect there are many more solutions since you can mix and match the fractions that you decompose. This results in countless more combinations of ($l,w,h$), and I don't know how to use casework or an otherwise more organized approach to solve this problem. But, I do believe it can still be solved using this approach.
If anybody has any idea on how to solve the problem using the "splitting unit fraction" method, or has any other working solution, I would greatly appreciate it if you shared it with me.
Apologies for the long post and thanks for reading.
Since you have already determined all solutions where two or three of the variables are equal it remains to find all integer triples $(a,b,c)$ with $$a<b<c\quad{\rm and} \qquad{1\over a}+{1\over b}+{1\over c}={1\over2}\ .\tag{1}$$ According to your edit you suspect that there could be an infinity of solutions. But this is not the case.
From $(1)$ one immediately infers that ${1\over6}<{1\over a}<{1\over 2}$, which implies $3\leq a\leq 5$. We now treat these cases separately.
(i) When $a=3$ we have to solve $${1\over b}+{1\over c}={1\over 6}$$ with $b<c$, and this implies $7\leq b\leq11$. It follows that $$(b,c)\in\bigl\{(7,42), (8,24), (9,18), (10,15), (11,{\textstyle{5\over66}})\bigr\}\ .$$ This leads to the admissible triples $$(3,7,42), (3,8,24), (3,9,18), (3,10,15)\ .\tag{2}$$ (ii) When $a=4$ we have to solve $${1\over b}+{1\over c}={1\over 4}$$ with $b<c$, and this implies $5\leq b\leq7$. It follows that $$(b,c)\in\bigl\{(5,20), (6,12), (7,{\textstyle{3\over28}})\bigr\}\ .$$ This leads to the admissible triples $$(4,5,20), (4,6,12)\ .\tag{3}$$ (iii) When $a=5$ we have to solve $${1\over b}+{1\over c}={3\over 10}$$ with $b<c$. This implies ${1\over b}>{3\over20}$, or $b\leq6$, and the condition $a<b$ then enforces $b=6$. This would imply $c={15\over2}$, which is not admissble.
All in all there are the $6$ triples listed in $(2)$ and $(3)$.
• It's not required that a<b<c, it should be a <= b <= c. – RoadieRich Apr 2 '15 at 18:46
• @RoadieRich: See my edit (first two lines). – Christian Blatter Apr 2 '15 at 19:23
• Hello Christian :) Thanks for presenting such an excellent solution. I just wanted to clarify, though, I didn't mean to write that I suspected there were an infinite number of solutions. I just meant that any given unit fraction could be infinitely decomposed into smaller unit fractions. – A is for Ambition Apr 2 '15 at 21:20
Starting with your equation $\frac{1}{h} + \frac{1}{w} + \frac{1}{l} = \frac{1}{2}$, it looks to me like straightforward Egyption fraction analysis.
Assume that $h \le w \le l$.
First, $h \ge 3$ (or else the sum exceeds $\frac12$) and $h \le 6$ (or else the sum is less than $\frac12$).
For any of these values, $\frac1{w}+\frac1{l} =\frac12-\frac1{h}$.
To get a lower bound on $w$, $\frac1{w} <\frac12-\frac1{h}$ or $w >\frac{1}{\frac12-\frac1{h}} =\frac{2h}{h-2} =\frac{2h-4+4}{h-2} =2+\frac{4}{h-2}$ or $w \ge 3+\lfloor \frac{4}{h-2} \rfloor$.
To get an upper bound on $w$, $2\frac1{w} \ge\frac12-\frac1{h}$ or $w \le\frac{2}{\frac12-\frac1{h}} =\frac{4h}{h-2} =\frac{4h-8+8}{h-2} =4+\frac{8}{h-2}$ or $w \le 4+\lfloor \frac{8}{h-2} \rfloor$.
Here are these bounds (remembering that $w \ge h$):
$\begin{array}{lll} h & low & high\\ 3 & 6 & 12\\ 4 & 5 & 10\\ 5 & 5 & 6 \\ 6 & 6 & 6 \\ \end{array}$
For each possible $(h, w)$ pair, check if $l =\frac{1}{\frac12-\frac1{h}-\frac1{w}} =\frac{2hw}{hw-2w-2h} =\frac{2hw-4w-4h+4w+4h}{hw-2w-2h} =2+\frac{4w+4h}{hw-2w-2h}$ is an integer.
• hi marty, great solution and thanks for introducing me to Egyptian fraction analysis :) i just have one question, though, which is why you used the floor function to define both the lower and upper bounds. thanks! – A is for Ambition Mar 20 '15 at 21:53
• For the lower bound, if the expression ($4/(h-2)$) is an integer, then we want the next integer, since the inequality is strict. That's why I used floor + 1. – marty cohen Mar 20 '15 at 23:16
There is $(4,5,20)$.
I can't think of a general approach that doesn't involve brute force, however.
Here's a somewhat ad hoc method for generating solutions to $\frac1h+\frac1w+\frac1l=\frac12$. Find a number $n$ for which there are three factors, $a,b,c$ of $n$ that add to $\frac{n}2$. We then divide the equation $a+b+c=\frac{n}2$ by $n$ to get the desired form.
For example if $n=30$, we can take factors $2, 3, 10$. Then dividing $2+3+10=15$ by $30$ gives $\frac{1}{15}+\frac{1}{10}+\frac{1}{3}=\frac{1}{2}$. Thus a box of dimensions $15$ by $10$ by $3$ should work. | 2019-04-20T20:28:20 | {
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https://math.stackexchange.com/questions/3917912/lower-bound-for-sum-of-reciprocals-of-positive-real-numbers | # Lower bound for sum of reciprocals of positive real numbers
I am reading an article where the author seems to use a known relationship between the sum of a finite sequence of real positive numbers $$a_1 +a_2 +... +a_n = m$$ and the sum of their reciprocals. In particular, I suspect that $$$$\sum_{i=1}^n \frac{1}{a_i} \geq \frac{n^2}{m}$$$$
with equality when $$a_i = \frac{m}{n} \forall i$$. Are there any references or known theorems where this inequality is proven?
This interesting answer provides a different lower bound. However, I am doing some experimental evaluations where the bound is working perfectly (varying $$n$$ and using $$10^7$$ uniformly distributed random numbers).
by Cauchy schwarz inequality$$\left(\sum_{i=1}^{n}{\sqrt{a_i}}^2\right)\left(\sum_{i=1}^{n}\frac{1}{{\sqrt{a_i}}^2}\right)\ge (\sum_{i=1}^{n} 1)^2=n^2$$
WLOG $$a_1\ge a_2...\ge a_n$$ then $$\frac{1}{a_1}\le \frac{1}{a_2}..\le \frac{1}{a_n}$$
So by chebyshev's inequality $$n^2=n\left(a_1\frac{1}{a_1}+a_2\frac{1}{a_2}..+a_n\frac{1}{a_n}\right)\le \left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}\frac{1}{a_i}\right)$$
• @newman_ash were you able to follow the chebyshevs proof? – Albus Dumbledore Nov 22 at 10:33
• Yes I was able to follow it – newman_ash Nov 22 at 10:41
• But to me it is not obvious how to get rid of $\sum a_i^2$ and $\sum 1/a_i^2$ in the Cauchy-Schwarz inequality – newman_ash Nov 22 at 11:05
• @newman_ash i edited – Albus Dumbledore Nov 22 at 11:49
• Thank you for answering and editing. Unfortunately, I am still not getting how you are introducing the square roots. You could use the fact that $\sqrt{\sum a_i} <= \sum \sqrt{a_i}$, but you would get a less tight bound $n/m$ – newman_ash Nov 25 at 19:00
The author is using the Arithmetic Mean - Harmonic Mean ("AM-HM") inequality: https://en.wikipedia.org/wiki/Harmonic_mean#Relationship_with_other_means
This is a popular inequality in the math olympiad world; you can find a proof here: https://artofproblemsolving.com/wiki/index.php/Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality | 2020-12-05T12:17:47 | {
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http://ocve.koehler-paderborn.de/transform-fourier-matlab.html | The image domain is the 2-D equivalent of the time domain. In the last two posts in my Fourier transform series I discussed the continuous-time Fourier transform. I'm having issues finding the FFT and IFFT of an ideal source. The fast Fourier transform (FFT) is an efficient implementation of the discrete Fourier Transform (DFT). Can you please send proper solution of this question. FOURIER TRANSFORM & BODE PLOTS As we have seen, the Fourier transform can be used for aperiodic signals as well as for systems which could be filters or circuits. The MATLAB® environment provides the functions fft and ifft to compute the discrete Fourier transform and its inverse, respectively. The STFT of a signal is calculated by sliding an analysis window of length M over the signal and calculating the discrete. 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Matlab Fourier Transform Profilometry Codes and Scripts Downloads Free. Signal Fourier. The "bode" function % can be used here since for a function which has a Fourier Transform, the % Fourier Transform is equivalent to the Laplace Transform being evaluated % along the jw axis. inside H, there are some data like x+iy. Fast Fourier transform and its inverse. Later it calculates DFT of the input signal and finds its frequency, amplitude, phase to compare. I would like to validate the following code of a Fourier transform using Matlab's fft, because I have found conflicting sources of information on the web, including in the Matlab help itself, and I have been unable to verify Parseval's theorem with certain such "recipes" (including with answers coming from the MathWorks team, see below. For math, science, nutrition, history. Formally, there is a clear distinction: 'DFT' refers to a mathematical transformation or function, regardless of how it is computed, whereas 'FFT' refers to a specific. The interval at w. We then are supposed to filter out any frequencies above 200 and below 500, take the inverse transform, and plot the resulting graph against the approximation y = 0. On this page, the Fourier Transforms for the sinusois sine and cosine function are determined. The fast Fourier transform algorithm requires only on the order of n log n operations to compute. Learn more about fourier transform, gaussian, pulsed signal, spectrum. There is also the discrete-time Fourier transform (DTFT) which under some stimulus conditions is identical to the DFT. also, tried fft, doesn't work as well $\endgroup$ - user107761 Nov 14 '14 at 8:17 $\begingroup$ fourier is symbolic toolbox. m computes the fast fractional Fourier transform following the algorithm of [1] The m-file frft2. 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The Fourier Transform [ pdf] (version of Feb 25, 2007) Using the Fourier Transform to Solve PDE's [ pdf] Discrete-Time Fourier Series and Transforms [ pdf] (version of Mar 21, 2007) Discrete-Time Linear Time Invariant Systems and z-Transforms [ pdf] (version of Apr 4, 2007) Supplementary Material. Fourier Transform, Fourier Series, and frequency spectrum - Duration: 15:45. ment the Fast Fourier Transform for computing the 1-D, 2-D and N-dimensional transforms respectively. But, as usual, it is easier to use MATLAB's inverse Fourier transform routine, ifft. Fourier Transforms For additional information, see the classic book The Fourier Transform and its Applications by Ronald N. Introduction FFTW is a C subroutine library for computing the discrete Fourier transform (DFT) in one or more dimensions, of arbitrary input size, and of both real and complex data (as well as of even/odd data, i. Discrete Time Fourier Transform (DTFT) in MATLAB - Matlab Tutorial Online Course - Uniformedia. coz they are just using fourier. Learn more about fourier. Introduction to complex numbers The discrete Fourier transform is about evenly spaced points on a circle. Direct/Inverse Fourier Transform using Matlab Use the symbolic direct/inverse Fourier transform functions in Matlab to evaluate the following and only plot the frequency spectrum magnitude and phase for the first item in the following table (a-2). The Fourier transform is defined for a vector x with n uniformly sampled points by. how can i enhance image by Fourier transform?. We want to do this now as well. Comparing to the Machine Learning approach, Fourier Transform is a very simple and fast algorithm. Plotting Fourier Transform - HELP PLEASE. Please how can I detect and extract edges of an image using Fourier Transform in MATLAB?. transform examples; defocus example. There’s a place for Fourier series in higher dimensions, but, carrying all our hard won. In this tutorial numerical methods are used for finding the Fourier transform of continuous time signals with MATLAB are presented. In my Fourier transform series I've been trying to address some of the common points of confusion surrounding this topic. It refers to a very efficient algorithm for computingtheDFT • The time taken to evaluate a DFT on a computer depends principally on the number of multiplications involved. To explain it clearly, have a look at this easy example : Lets consider a function$f(t) = |sin(\pi t)|$ on the interval$[\dfrac{-1}{2}, \dfrac{1}{2. matlab program to implement the properties of discrete fourier transform (dft) - frequency shift property. The operation must into account for the mirror-image structure of the Matlab's Fourier transform: the lowest frequencies are at the extremes of the fft and the highest frequencies are in the center portion. 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The fast Fourier transform (FFT) is an efficient implementation of the discrete Fourier Transform (DFT). For the input sequence x and its transformed version X (the discrete-time Fourier transform at equally spaced frequencies around the unit circle), the two functions implement the relationships. Fourier Transform Examples and Solutions WHY Fourier Transform? Inverse Fourier Transform If a function f (t) is not a periodic and is defined on an infinite interval, we cannot represent it by Fourier series. , if y <- fft(z), then z is fft(y, inverse = TRUE) / length(y). Comparing to the Machine Learning approach, Fourier Transform is a very simple and fast algorithm. So, what we are really doing when we compute the Fourier series of a function f on the interval [-L,L] is computing the Fourier series of the 2L periodic extension of f. What finally convinced me to try to write a post involving Fourier transforms was a question received by one of my coauthors of Digital Image Processing Using MATLAB. Consider a sinusoidal signal x that is a function of time t with frequency components of 15 Hz and 20 Hz. wav, steganography, puzzle, audio, sound. »Fast Fourier Transform - Overview p. The concept of Fourier transform becomes important in the study of complicated waves. The reason for this is that the Fourier transform represents data as sum of sine waves, which are not localized in time or space. I am confused by MATLAB'S single input of X for its IFFT function. "FFT algorithms are so commonly employed to compute DFTs that the term 'FFT' is often used to mean 'DFT' in colloquial settings. The Fourier Transform The Discrete Fourier Transform is a terri c tool for signal processing (along with many, many other applications). 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It extracts frequency features of a face, rather than analyzing the image pattern using a convolutional network. The Fourier transform has many applications, in fact any field of physical science that uses sinusoidal signals, such as engineering, physics, applied mathematics, and chemistry, will make use of Fourier series and Fourier transforms. To install the code, simply unpack the directory in a Matlab folder. For the input sequence x and its transformed version X (the discrete-time Fourier transform at equally spaced frequencies around the unit circle), the two functions implement the relationships. Direct/Inverse Fourier Transform using Matlab Use the symbolic direct/inverse Fourier transform functions in Matlab to evaluate the following and only plot the frequency spectrum magnitude and phase for the first item in the following table (a-2). 38 and show you are as good as MATLAB is provided in one of the problems. Getting Started. Langton Page 3 And the coefficients C n are given by 0 /2 /2 1 T jn t n T C x t e dt T (1. Fourier spectra help characterize how different filters behave, by expressingboth the impulse response and the signal in the Fourier. thank you Mr Wayne. A Fast Fourier transform (FFT) is a fast computational algorithm to compute the discrete Fourier transform (DFT) and its inverse. Fourier and Inverse Fourier Transforms. TestingSpherical3DFFT. 离散傅里叶变换、卷积、离散小波变换、τp变换、希尔伯特变换、希尔伯特-黄变换相关Matlab程序. I'm having issues finding the FFT and IFFT of an ideal source. Musical Analysis and Synthesis in Matlab by Mark R. Continue reading →. In the last two posts in my Fourier transform series I discussed the continuous-time Fourier transform. Learn more about fourier, transforms, fft, fourier transform, frequency, sinusoidal, sine, wave, time. This page shows the workflow for Fourier and inverse Fourier transforms in Symbolic Math Toolbox™. Introduction FFTW is a C subroutine library for computing the discrete Fourier transform (DFT) in one or more dimensions, of arbitrary input size, and of both real and complex data (as well as of even/odd data, i. Fourier domain, with multiplication instead of convolution. Plotting Fourier Transform - HELP PLEASE. This article will walk through the steps to implement the algorithm from scratch. m computes the fast fractional Fourier transform following the algorithm of [5] (see also [6] for details) The m-file frft22d. Therefore, to get the Fourier transform ub(k;t) = e k2t˚b(k) = Sb(k;t)˚b(k), we must. The inverse functions ifft, ifft2 and ifftn compute the inverse transforms. There is a principle in optics where the intensity distribution from a lens is equal to the Fourier transform of the aperture that the light initially goes through. The result is easily obtained using the Fourier Transform of the complex exponential. Smoothing it is a classical problem (begin with Box&Jenkins); welch in Matlab is not bad to start with. Using MATLAB to Plot the Fourier Transform of a Time Function. The toolbox computes the inverse Fourier transform via the Fourier transform: i f o u r i e r ( F , w , t ) = 1 2 π f o u r i e r ( F , w , − t ). Transform 2-D optical data into frequency space. In my Fourier transform series I've been trying to address some of the common points of confusion surrounding this topic. To explain it clearly, have a look at this easy example : Lets consider a function[math] f(t) = |sin(\pi t)|$ on the interval[math] [\dfrac{-1}{2}, \dfrac{1}{2. Fast Fourier transform algorithms utilize the symmetries of the matrix to reduce the time of multiplying a vector by this matrix, from the usual (). I am looking for comments on how to make this code more instructive for the. If X is a matrix, then fft(X) treats the columns of X as vectors and returns the Fourier transform of each column. (However, if you don't have Matlab, you can try Octave or SciLab. Fourier Transform, Fourier Series, and frequency spectrum - Duration: 15:45. For simple examples, see fourier and ifourier. When you want to transform frequency-domain data into the time domain, use the IFFT block. prior to entering the outer for loop. Transform sizes with small prime factors run faster because the problem can be subdivided more easily. 2 MATLAB function polyfit. Fast Fourier Transform in MATLAB ®. Fast Fourier Transform (FFT) Definition (Piecewise Continuous). Today I want to start getting "discrete" by introducing the discrete-time Fourier transform (DTFT). While wondering around in the Matlab documentation I found out there is a simple way to calculate the Fourier transform of any function using Matlab. 1 De nition The Fourier transform allows us to deal with non-periodic functions. The Fastest Fourier Transform in the West (FFTW) is a software library for computing discrete Fourier transforms (DFTs) developed by Matteo Frigo and Steven G. This example shows how to transform time-domain data into the frequency domain using the FFT block. The inverse transform of F(k) is given by the formula (2). If you are worried about zeros, you can always add eps or some other small number. Sketch by hand the magnitude of the Fourier transform of c(t) for a general value of f c. Johnson at the Massachusetts Institute of Technology. I'm at a computer without MATLAB at the moment. The discrete Fourier transform (DFT) is an equivalent of the Fourier transform for discrete data. • The differences between Fourier Transform, Discrete Fourier Transform, and Fast Fourier Transform. In Matlab, it is not possible to compute the continuous Fourier Transform, because the computer just works with a finite number of discrete or quantified values; therefore, the signal must be sampled and that's why we use the Discrete Fourier Transform. I would like to do an inverse fourier transform using MATLAB's IFFT. This computational efficiency is a big advantage when processing data that has millions of data points. That said, power functions are useful for characterizing topography along a profile because it shows the relative contributions of various wavelengths, which presumably have some geologic significance. Discrete Fourier Transform in MATLAB 18:48 ADSP, MATLAB PROGRAMS, MATLAB Videos. How can I use a fourier transform on a. Transform 2-D optical data into frequency space. I'm starting with a manual rectangular pulse because I know what its corresponding inverse Fourier transform is supposed to look like. This property may seem obvious, but it needs to be explicitly stated because it underpins many of the uses of the transform, which I’ll get to later. series (DFS), discrete Fourier transform (DFT) and fast Fourier transform (FFT) (ii) Understanding the characteristics and properties of DFS and DFT (iii) Ability to perform discrete-time signal conversion between the time and frequency domains using DFS and DFT and their inverse transforms. To compute the inverse Fourier transform, use ifourier. But even a spectrogram is far too complex a representation to base a speech recognizer on. How to Do a Fourier Transform in Matlab - How to plot FFT using Matlab - 매트랩 fft Learn MATLAB in simple and easy steps starting from basic to advanced concepts with examples http. A reader of Digital Image Processing Using MATLAB wanted to know why the Fourier transform of the image below looked so "funny. FFT Software. May you and all your well wishers have happiness and good health. Computing Fourier Series and Power Spectrum with MATLAB By Brian D. 3 DISCRETE AND FAST FOURIER TRANSFORMS 8. The Fourier transform is not limited to functions of time, but the domain of the original function is commonly referred to as the time domain. Home / ADSP / MATLAB PROGRAMS / MATLAB Videos / Discrete Fourier Transform in MATLAB. The usual notation for finite Fourier transforms uses subscripts j and k that run from 0 to n•1. Using MATLAB, LabVIEW Mathscript or GNU Octave, plot the magnitude of the Fourier transform of c(t) for f c = 8 Hz. This projection is particularly useful in music. MATLAB Links. Computing Fourier Series and Power Spectrum with MATLAB By Brian D. i need a Matlab code for feature extraction with Discrete Fourier, Wavelet, Cosine, and Sine transform. It provides a representation in the frequency domain of the signal which is usually given in the time domain, thus decomposing the time signal into a sum of oscilatory components of single frequency which describe the variation in the original signal. The outputs are the Fourier transform stored in the vector X and the corresponding frequency vector w. Fourier and Laplace transforms and their inverses. In this tutorial numerical methods are used for finding the Fourier transform of continuous time signals with MATLAB are presented. Although convolution will correctly blur an image, there exists another method that is faster, called the Fast Fourier Transform (FFT). I'd like to take inverse Fourier of H, I tried to use, ifft command but it takes only one dimensional inverse Fourier. Let’s start with some simple examples: >> x=[4 3 7 -9 1 0 0 0]; >> y=fft(x) y = 6. I am confused by MATLAB'S single input of X for its IFFT function. ) The continuous-time Fourier transform is defined by this pair of equations:. Fast Fourier Transform on 2 Dimensional matrix using MATLAB Fast Fourier transformation on a 2D matrix can be performed using the MATLAB built in function ‘ fft2() ’. According to the convolution theorem, convolution in the time (or image) domain is equivalent to multiplication in the frequency (or spatial) domain. Transforms are used in science and engineering as a tool for simplifying analysis and look at data from another angle. In fact, the Fourier transform produces complex numbers, which you can verify by trying to plot them in MATLAB. Real-world applications of the Fourier transform pricing formula are discussed in part III. The Fourier transform is linear, meaning that the transform of Ax(t) + By(t) is AX(ξ) + BY(ξ), where A and B are constants, and X and Y are the transforms of x and y. There is a discrete (and in fact fast because it performs the dct by using a fast Fourier transform) available at the MATLAB file exchange that was created by Greg von Winckel. Short time Fourier transform. The Fourier transform is a powerful tool for analyzing data across many applications, including Fourier analysis for signal processing. In particular, by clever grouping and reordering of the complex exponential multiplications it is possible to achieve substantial computational savings. It can be derived in a rigorous fashion but here we will follow the time-honored approach. The Finite Fourier transform and Hilbert transform The Hilbert transform can be implemented in Matlab in four steps: (1) sample the function (2) apply the Fourier transform, (3) multiply the negative frequencies by -1 and the zero frequency by 0, (4) apply the inverse Fourier transform. These function express their results as complex numbers. $\endgroup$ - Marcus Müller Oct 8 '16 at 18:15. Fourier Transform modeling in MATLAB Usman Hari. The 2D Fourier Transform is an indispensable tool in many fields, including image processing, radar, optics and machine vision. Fourier Transform Example #2 MATLAB Code % ***** MATLAB Code Starts Here ***** % %FOURIER_TRANSFORM_02_MAT % fig_size = [232 84 774 624]; m2ft = 3. how to use fractional Fourier transform on image Learn more about image processing, digital image processing, image analysis, im, image segmentation, matlab. This feature is not available right now. MATLAB Lecture 7. I don't know why you would add 1. The Fourier transform is an operation that transforms data from the time (or spatial) domain into the frequency domain. Basic Spectral Analysis. This MATLAB function returns the Fourier Transform of f. Fourier domain, with multiplication instead of convolution. The 2D Fourier Transform is an indispensable tool in many fields, including image processing, radar, optics and machine vision. Formally, there is a clear distinction: 'DFT' refers to a mathematical transformation or function, regardless of how it is computed, whereas 'FFT' refers to a specific. This is a tutorial I wrote so that undergraduate or graduate students could solve nonlinear partial. The discrete Fourier transform and the FFT algorithm. An opportunity to code a direct implementation of Equation 3. This is soo confusing u know. 1 Forward-biased region 9. The Fourier Transform (FFT) •Based on Fourier Series - represent periodic time series data as a sum of sinusoidal components (sine and cosine) •(Fast) Fourier Transform [FFT] - represent time series in the frequency domain (frequency and power) •The Inverse (Fast) Fourier Transform [IFFT] is the reverse of the FFT. The DTFT is defined by this pair of transform equations: Here x[n] is a discrete sequence defined for all n:. , for filtering, and in this context the discretized input to the transform is customarily referred to as a signal, which exists in the time domain. Discrete Fourier Transform Matlab Program Fourier transformation is used to decompose time series signals into frequency components each having an amplitude and phase. Hey guys, I'm working on a MATLAB program to find Fourier coefficients. Today I want to start getting "discrete" by introducing the discrete-time Fourier transform (DTFT). In the classical setting, the Fourier transform on R is given by ^f(˘) = Z R f(t)e 2ˇi˘t dt = hf;e2ˇi˘ti: This is precisely the expansion of f in terms of the eigenvalues of the eigenfunctions of the Laplace operator. I'd like to take inverse Fourier of H, I tried to use, ifft command but it takes only one dimensional inverse Fourier. The one-dimensional discrete Fourier transform of N data elements L = [L 1, …, L N] is defined as the list F = [F 1, …, F N], such that. o the Fourier spectrum is symmetric about the origin the fast Fourier transform (FFT) is a fast algorithm for computing the discrete Fourier transform. If ifourier cannot find an explicit representation of the inverse Fourier transform, then it returns results in terms of the Fourier transform. 2-D Fourier Transforms. The 2D Fourier Transform is an indispensable tool in many fields, including image processing, radar, optics and machine vision. The two functions are inverses of each other. the Matlab function “fftshift”) •N and M are commonly powers of 2 for. L6: Short-time Fourier analysis and synthesis –The Fourier transform of the windowed speech waveform is defined as Generate STFT using Matlab functions. Fourier Transform Examples and Solutions WHY Fourier Transform? Inverse Fourier Transform If a function f (t) is not a periodic and is defined on an infinite interval, we cannot represent it by Fourier series. Computing Fourier Series and Power Spectrum with MATLAB By Brian D. Ok, so we have an image that is a Fourier inverse of the original picture. (When the context makes it clear whether I'm talking about the continuous-time or the discrete-time flavor, I'll often just use the term Fourier transform. A rectangular pulse is defined by its duty cycle (the ratio of the width of the rectangle to its period) and by the delay of the pulse. The Fourier transform is defined for a vector x with n uniformly sampled points by. Basic Spectral Analysis. Given basic operations like duplication, multiplication, addition, convolution, time-scaling, etc. Sampling an image. The Fast Fourier Transform does not refer to a new or different type of Fourier transform. m - located in folder MATLAB CodeBase\NVIDIA_3DSphericalDFT - this is the 3D Spherical Polar Fourier Transform test. Discrete Time Fourier Transform (DTFT) in MATLAB - Matlab Tutorial Online Course - Uniformedia. 1 Comment Show Hide all comments. Fast Fourier Transform from data in file. When we plot the 2D Fourier transform magnitude, we need to scale the pixel values using log transform to expand the range of the dark pixels into the bright region so we can better see the transform. I need to enhance my image using fast fourier transform. Formally, there is a clear distinction: 'DFT' refers to a mathematical transformation or function, regardless of how it is computed, whereas 'FFT' refers to a specific. Can I find related information about Fast Fourier Transform? Please help. MATLAB Links. m computes the fast fractional Fourier transform following the algorithm of [1] The m-file frft2. The Fourier transform is a powerful tool for data analysis. I am looking for comments on how to make this code more instructive for the. Many specialized implementations of the fast Fourier transform algorithm are even more efficient when n is a power of 2. Keywords: Fourier Transform File Name:. Engineering Tables/Fourier Transform Table 2. Sketch by hand the magnitude of the Fourier transform of c(t) for a general value of f c. In order to plot a Fourier Series in MATLAB, you'll have to approximate it first. How to specify and keep only a proportion of the. ISBN 978-953-51-0518-3, PDF ISBN 978-953-51-5685-7, Published 2012-04-25. The Fastest Fourier Transform in the West (FFTW) is a software library for computing discrete Fourier transforms (DFTs) developed by Matteo Frigo and Steven G. 1 Fourier Transforms Signal Processing Algorithms in MATLAB. So, last time we took the Fourier transform of a sine wave, and if you’ve ever studied the Fourier transform in school you know that when you take the Fourier transform of a sine wave you should see a delta function or a spike in the frequency domain, and what we were seeing was. On the second plot, a blue spike is a real (cosine) weight and a green spike is an imaginary (sine) weight. Related Transforms. 5 Fourier transform The Fourier series expansion provides us with a way of thinking about periodic time signals as a linear combination of complex exponential components. Fourier transform is one of the various mathematical transformations known which is used to transform signals from time domain to frequency domain. Fast Fourier Transform (FFT) Animation using Matlab February 2, 2014 May 4, 2017 meyavuz Here, I show the progress of Fast Fourier Transform (FFT) of a time-domain signal as it changes in time. If fourier cannot find an explicit representation of the transform, it returns an unevaluated function call. I never add 1, and plot log(abs(fft)) all the time (usually I do log10 because I want to get to decibels). Check "Numerical Recipes" and the link below for more details. Fast Fourier transform algorithms utilize the symmetries of the matrix to reduce the time of multiplying a vector by this matrix, from the usual (). Matlab and Octave have built-in functions for computing the Fourier transform (fft and ifft). The Fourier transform is an integral transform widely used in physics and engineering. Find frequency from fourier transform. This example shows how to transform time-domain data into the frequency domain using the FFT block. Learn more about fourier transform, gaussian, pulsed signal, spectrum. I would like to validate the following code of a Fourier transform using Matlab's fft, because I have found conflicting sources of information on the web, including in the Matlab help itself, and I have been unable to verify Parseval's theorem with certain such "recipes" (including with answers coming from the MathWorks team, see below. m computes a 2D transform based on the 1D routine frft2. Fourier Transform Example #2 MATLAB Code % ***** MATLAB Code Starts Here ***** % %FOURIER_TRANSFORM_02_MAT % fig_size = [232 84 774 624]; m2ft = 3. I find a strange grid like phase in the Fourier plane. Sometimes there is a big spike at zero so try taking the log of it before plotting. The Fourier Analysis Tool in Microsoft Excel Douglas A. (Note that there are other conventions used to define the Fourier transform). FFT Software. For example, if Y is a matrix, then ifft(Y,n,2) returns the n-point inverse transform of each row. I mean i need to perform a Fast Fourier Transform (FFT) low pass filtering on a time domain data. These numbers may arise, for example, as a discretely sampled values of an analog function sampled over some period window and then. To do that in MATLAB, we have to make use of the unit step function u(x), which is 0 if and 1 if. Instead of capital letters, we often use the notation f^(k) for the Fourier transform, and F (x) for the inverse transform. Discrete-Time Fourier Transform (DTFT) Chapter Intended Learning Outcomes: (i) Understanding the characteristics and properties of DTFT (ii) Ability to perform discrete-time signal conversion between the time and frequency domains using DTFT and inverse DTFT. Discrete -Time Fourier Transform • The DTFT can also be defined for a certain class of sequences which are neither absolutely summablenor square summable • Examples of such sequences are the unit step sequence µ[n], the sinusoidal sequence and the exponential sequence • For this type of sequences, a DTFT. we visually analyze a Fourier transform by computing a Fourier spectrum (the magnitude of F(u,v)) and display it as an image. Today I want to start getting "discrete" by introducing the discrete-time Fourier transform (DTFT). If iridge is a matrix, then ifsst initially performs the inversion along the first column of iridge and then proceeds iteratively along the subsequent columns. the spectral information from a segment of the speech signal using an algorithm called the Fast Fourier Transform. Short time Fourier transform. As for writing a function equivalent to the MATLAB fft then you could try implementing the Radix-2 FFT which is relatively straightforward though is used for block sizes N that are powers of two. Inverse FFT(DFT) in MATLAB; Discrete Fourier Transform in MATLAB; FAST FOURIER TRANSFORM in MATLAB; Numerical Problem on DTFT using MATLAB; Discrete Time fourier transform in MATLAB|PART 3; Discrete Time Fourier Transform in MATLAB|Part 2; Discrete Time Fourier Transformation in MATLAB|PAR Signal Energy in MATLAB. The Fourier transform of the derivative of a function is a multiple of the Fourier transform of the original function. I never add 1, and plot log(abs(fft)) all the time (usually I do log10 because I want to get to decibels). May you and all your well wishers have happiness and good health. In the classical setting, the Fourier transform on R is given by ^f(˘) = Z R f(t)e 2ˇi˘t dt = hf;e2ˇi˘ti: This is precisely the expansion of f in terms of the eigenvalues of the eigenfunctions of the Laplace operator. So we now move a new transform called the Discrete Fourier Transform (DFT). How to Calculate the Fourier Transform of a Function. The Fourier transform is an operation that transforms data from the time (or spatial) domain into the frequency domain. This Demonstration illustrates the relationship between a rectangular pulse signal and its Fourier transform. This is why cos(x) shows up blue and sin(x) shows up green. Make N-512 and fs 1 kHz. We can use MATLAB to plot this transform. x = ifsst(s,window,iridge) inverts the synchrosqueezed transform along the time-frequency ridges specified by the index vector or matrix iridge. Much of its usefulness stems directly from the properties of the Fourier transform, which we discuss for the continuous-. m" function and inverse transform the result with the built-in Matlab/Octave "ifft. This computational efficiency is a big advantage when processing data that has millions of data points. The Fourier Transform is used in a wide range of applications, such as image analysis, image filtering, image reconstruction and image compression. The Fourier transform is not limited to functions of time, but the domain of the original function is commonly referred to as the time domain. The source code and files included in this project are listed in the project files section, please make sure whether the listed source code meet your needs there. Hence, fast algorithms for DFT are highly valuable. Aliasing occurs when you don't sample a signal fast enough to be able to reconstruct it accurately after sampling. This article will walk through the steps to implement the algorithm from scratch. also, tried fft, doesn't work as well $\endgroup$ - user107761 Nov 14 '14 at 8:17 $\begingroup$ fourier is symbolic toolbox. Discrete Fourier Transform Matlab Program Discrete Fourier transform is used to decompose time series signals into frequency components each having an amplitude and phase. Use the Fourier transform for frequency and power spectrum analysis of time-domain signals. The Fourier Transform of g(t) is G(f),and is plotted in Figure 2 using the result of equation [2]. Fourier and Inverse Fourier Transforms. Turn in your code and plot. | 2019-11-15T19:06:26 | {
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https://math.stackexchange.com/questions/1144536/prove-that-z3-zz2-is-divisible-by-12-for-all-integers-z | Prove that $(z^3-z)(z+2)$ is divisible by $12$ for all integers $z$
I am a student and this question is part of my homework.
May you tell me if my proof is correct?
Prove that $(z^3-z)(z+2)$ is divisible by $12$ for all integers $z$.
$(z^3-z)(z+2)=z(z^2-1)(z+2)=z(z-1)(z+1)(z+2)=(z-1)(z)(z+1)(z+2)$
$(z-1)(z)(z+1)(z+2)$ means the product of $4$ consecutive numbers.
Any set of $4$ consecutive numbers has $2$ even numbers, then $(z-1)(z)(z+1)(z+2)$ is divisible by $4$.
Any set of $4$ consecutive number has at least one number that is multiple of $3$, then $(z-1)(z)(z+1)(z+2)$ is divisible by 3.
Therefore $(z-1)(z)(z+1)(z+2)$ is divisible by $12$. Q.E.D.
• You should use the term "integer" instead of number. And you can actually prove something stonger - that it is divisible by $24$. Your proof for $12$ looks correct, though. – Thomas Andrews Feb 12 '15 at 3:42
• @ThomasAndrews Thanks for show me my mistake. Thanks! – Beginner Feb 12 '15 at 3:46
• @ThomasAndrews How can you prove that is divisible by 24? Thanks! – Beginner Feb 12 '15 at 3:59
• You can prove it by first proving that the product of three consecutive numbers is divisible by $6$, then prove it by induction on $x$. (That only proves it for $x\geq1$, but it follows easily that it is also true for $x\leq 0$.) – Thomas Andrews Feb 12 '15 at 4:07
• @Beginner. See math.stackexchange.com/questions/12067/… – lab bhattacharjee Feb 12 '15 at 4:40
That's correct. Alternatively it is divisible by $\,24\,$ by integrality of binomial coefficients
$$\,(z+2)(z+1)z(z-1)\, =\, 4!\ \dfrac{(z+2)(z+1)z(z-1)}{4!}\, =\, 24{ {z+2\choose 4}}\qquad\qquad$$
Similarly $\,n!\,$ divides the product of $\,n\,$ consecutive naturals.
• @Beginner $${z+2 \choose 4} \,=\, \frac{(z+2)!}{4!\,(z-2)!}\, =\, \frac{(z+2)(z+1)z(z-1)}{4!}$$ – Bill Dubuque Feb 12 '15 at 15:05 | 2020-02-19T22:48:13 | {
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http://mathhelpforum.com/trigonometry/122231-need-help-2-simutaneous-trig-equations.html | # Math Help - need help on 2 simutaneous trig. equations
1. ## need help on 2 simutaneous trig. equations
2 simutaneous equations from my coursework
4=4cosx+3cos(x+y)
3=4sinx+3sin(x+y)
looking for analytical solution, should yield 2 pairs of answers
2. Hi there, you should consider
$4=4\cos(x)+3\cos(x+y)$
$4=4\cos(x)+3(\cos(x)\cos(y)-\sin(x)\sin(y))$
and
$3=4\sin(x)+3sin(x+y)$
$3=4\sin(x)+3(\sin(x)\cos(y)+\cos(x)\sin(y))$
Now expand each expression out
3. ## think algrebra
see soroban's solution
4. Hello, cybernoa!
$\begin{array}{cccc}4\cos x+3\cos(x+y) &=& 4 & {\color{blue}[1]} \\
4\sin x+3\sin(x+y) &=& 3 & {\color{blue}[2]} \end{array}$
Looking for analytical solution, should yield 2 pairs of answers.
$\begin{array}{cccc}
\text{Square {\color{blue}[1]}:} & 16\cos^2x + 24\cos(x+y)\cos x + 9\cos^2(x+y) &=& 16 \\
\text{Square {\color{blue}[2]}:} & 16\sin^2\!x + 24\sin(x+y)\sin x + 9\sin^2(x+y) &=& 9 \end{array}$
$\text{Add: }\;16\underbrace{\bigg[\sin^2x+\cos^2x\bigg]}_{\text{This is 1}} + 24\underbrace{\bigg[\cos(x+y)\cos x + \sin(x+y)\sin x\bigg]}_{\text{This is }\cos[(x+y)-x]}$ $+\; 9\underbrace{\bigg[\sin^2(x+y) + \cos^2(x+y)\bigg]}_{\text{This is 1}} \;=\;25$
We have: . $16 + 24\cos y + 9 \:=\:25 \quad\Rightarrow\quad 24\cos y \:=\:0$
. . . $\cos y \:=\:0 \quad\Rightarrow\quad\boxed{ y \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Let $y = \tfrac{\pi}{2}$ in [2]: . $4\sin x + 3\sin\left(x + \tfrac{\pi}{2}\right) \:=\:3$
. . . . $4\sin x + 3\bigg[\sin x\cos\tfrac{\pi}{2} + \cos x\sin\tfrac{\pi}{2}\bigg] \:=\:3$
. . . . . . . . . . $4\sin x + 3\cos x \:=\:3$
Divide by 5: . $\tfrac{4}{5}\sin x + \tfrac{3}{5}\cos x \:=\:\tfrac{3}{5}$ .[3]
Let: . $\cos\theta = \tfrac{4}{5},\;\sin\theta = \frac{3}{5} \quad\Rightarrow\quad \theta = \arcsin\tfrac{3}{5}$ .[4]
Substitute into [3]: . $\cos\theta\sin x + \sin\theta\cos x \:=\:\tfrac{3}{5}$
Then we have: . $\sin(x+\theta) \:=\:\tfrac{3}{5} \quad\Rightarrow\quad x + \theta \:=\:\arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:\arcsin\tfrac{3}{5} - \theta$
Substitute [4]: . $x \:=\:\arcsin\tfrac {3}{5} - \arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:0$
One solution: . $\left(0.\;\tfrac{\pi}{2}\right)$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Let $y = \frac{3\pi}{2}$ in [2]: . $4\sin x + 3\sin\left(x + \tfrac{3\pi}{2}\right) \:=\:3$
. . . . $4\sin x + 3\bigg[\sin x\cos\tfrac{3\pi}{2} + \cos x\sin\tfrac{3\pi}{2}\bigg] \:=\:3$
. . . . . . . . . . $4\sin x - 3\cos x \:=\:3$
Divide by 5: . $\tfrac{4}{5}\sin x + \tfrac{3}{5}\cos x \:=\:\tfrac{3}{5}$ .[5]
Let: . $\cos\theta = \tfrac{4}{5},\;\sin\theta = \tfrac{3}{5} \quad\Rightarrow\quad \theta \,=\,\arcsin\tfrac{3}{5}$ .[6]
Substitute into [5]: . $\cos\theta\sin x - \sin\theta\cos x \:=\:\tfrac{3}{5}$
Then we have: . $\sin\left(x - \theta\right) \:=\:\tfrac{3}{5} \quad\Rightarrow\quad x-\theta \:=\:\arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:\arcsin\tfrac{3}{5} + \theta$
Substitute [6]: . $x \:=\:\arcsin\tfrac{3}{5} + \arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:2\arcsin\tfrac{3}{5}$
Another solution: . $\left(2\arcsin\tfrac{3}{5},\;\tfrac{3\pi}{2}\right )$
5. Originally Posted by Soroban
Hello, cybernoa!
Soroban, I've said it once, and I'll say it again. You're presentation is impeccable. | 2015-06-30T08:39:03 | {
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https://math.stackexchange.com/questions/2618459/transform-logistic-sigmoid-function | # Transform Logistic/Sigmoid function
Please excuse any incorrect terminology. I want to adjust the curve of a logistic function to make it more or less steep while still reaching the asymptote points at the same value of x.
Here is a standard logistic function $f(x) = \frac{L-a}{1+e^{-k(x-x_0)}}+a$ where L=1, a=0, k=1, $x_0=0$: standard logistic function
I know that I can change the steepness by changing k. For example, if k=0.5 I get the line shown in red: logistic function with k=0.5
I can change the minimum and maximum values of f(x) by changing a and L, respectively. I can change the x value where f(x) hits the midpoint between a and L by changing $x_0$.
My goal is to have a function such as this:
$$g(x) = \begin{cases} 0 & \quad \text {if } x<-5 \\ f(x) & \quad \text {if } -5\geq x\leq 5\\ 1 & \quad \text {if } x>5 \end{cases} \$$
Where f(x) looks like the line shown here in yellow as given by $f(x)= \frac{1.1-(-0.9)}{1+e^{-0.05(x-0)}}+-0.9$: goal function in yellow
This yellow line has the 'steepness' of the function where k=0.5 (red line) but has approximately the same minimum and maximum (x, f(x)) points as the standard logistic function (blue line). I arrived at the current yellow line by manually adjusting a and L values until the line looked correct.
My questions are:
1. Is there some other transformation I can use on the logistic function to make it look like the yellow line (without manually guessing values for L and a)?
2. Is there a different/better formula than the logistic function I can use to get a curve that looks like this yellow line?
• Is it important for the function to be eventually constant, as in your example $g$? There are several different types of sigmoid functions, one of which may suit your needs, but most of them are merely asymptotic rather than eventually constant. – dbx Jan 24 '18 at 0:43
• It's fine if it's not eventually constant. I basically care about using a sigmoid function to get between two points. Then I can use other functions on either side of those points to make it constant. @dbx – Katie Jan 24 '18 at 2:07
• Another question (I'm working on this in my spare time, which is not abundant): Do you need to be able to specify the point where the sigmoid intersects the constant tail - in your example $g$, you chose $x=5$. Is it important that you can choose that value? – dbx Jan 24 '18 at 17:24
• Yes, I do. I am using this to estimate growth of attach rates/take rates (% of people who buy base product that will buy this add-on) of a product add-on that's being introduced to the market. I'm getting information on what we expect the initial attach rate (when the product is introduced) and the final attach rate (after some set amount of time) to be, but I want to use a Sigmoid curve to demonstrate different possibilities of what growth looks like between those two points. – Katie Jan 24 '18 at 19:16
We want to find a piecewise function $g(x)$ on $(-\infty, \infty)$ such that for some $\varepsilon > 0$ we have $g(x)=y_m$ for all $x < x_0-\varepsilon$ and $g(x)=y_M$ for all $x > x_0+\varepsilon$, and such that $g(x)$ is sigmoid for $x$ between $x_0-\varepsilon$ and $x_0+\varepsilon$. Additionally, you would like to be able to control the slope of the sigmoid part.
If we let $f$ be the logistic curve you chose: $$f = a + \frac{L-a}{1 + e^{-k(x-x_0)}}$$ then $$f'= k(a-L) \frac{e^{k(x+x_0)}}{\left( e^{kx} + e^{k x_0} \right)^2}$$ The curve achieves its maximum slope at $x=x_0$, given by $$f'(x_0)= \frac{1}{4}k(L-a)$$
Your first step is to choose the "steepness" $m=f'(x_0)$. Using the above, we get the constraint: $$k=\frac{4m}{L-a} \qquad (1)$$ Now we need another equation relating $k$ and $L-a$ so that we can solve for their values. This is provided by the requirement that $f(x_0+\varepsilon)=y_M$, i.e.: $$y_M=a+\frac{L-a}{1+e^{-k\varepsilon}}$$ Using $(1)$ in the above gives: $$y_M = a + \frac{L-a}{1 + e^{-\frac{4m\varepsilon}{L-a}}}$$ Using symmetry, we can write $a=y_m + y_M - L$ and therefore $L-a=2L-(y_m+y_M)$. Plugging this in to the above, we can solve for $L$ in: $$y_M = y_m+y_M-L + \frac{2L-(y_m+y_M)}{1 + e^\frac{-4m\varepsilon}{2L-(y_m+y_M)}}$$ At this point, things get really ugly. You can solve numerically using the method of your choice, though. Once you have $L$, you can find $a$ using the relation $a=y_m+y_M-L$ from before. Now that you have values for $k$, $a$, and $L$, you are done! Define:
$$g(x) = \left\{\begin{matrix} y_m, & x < x_0 - \varepsilon \\ \frac{L-a}{1 + e^{-k(x-x_0)}}, & x_0-\varepsilon \leq x < x_0+ \varepsilon \\ y_M, & x \geq x_0 + \varepsilon\end{matrix} \right.$$
Here is an example of using this method with $m=2$, $\varepsilon=4$, $x_0=1$, $y_m=-2$, and $y_M=3$: | 2020-06-01T20:29:12 | {
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https://math.stackexchange.com/questions/2129293/transition-from-a-riemann-sum-to-an-integral | # Transition from a Riemann sum to an Integral
The Riemann sum over an interval $[a,b]$ is usually defined as $$\lim\limits_{N\to\infty}\sum\limits_{k=0}^Nf\left(a+k\cdot\frac{b-a}{N}\right)\frac{b-a}{N}$$
Thus if we encounter a sum of the form $$\lim\limits_{N\to\infty}\sum\limits_{k=0}^Nf\left(k\cdot\frac{1}{N}\right)\frac{1}{N}$$ we can conclude that it is equal to an integral over the interval $[0,1]$. $$\lim\limits_{N\to\infty}\sum\limits_{k=0}^Nf\left(k\cdot\frac{1}{N}\right)\frac{1}{N}=\int_0^1f(x)dx\tag{1}\label{1}$$
What can we conclude about the following sum
$$\lim\limits_{N\to\infty}\lim\limits_{M\to\infty}\sum\limits_{k=0}^M f\left(k\cdot\frac{1}{N}\right)\frac{1}{N}\tag{2}\label{2}$$
To clarify, this is an infinite sum \eqref{2}, that differs from the Riemann sum \eqref{1}, in the upper limit of the sum. In the Riemann sum \eqref{1}, there is a relation between $M$ and $N$, namely $N=M$, while there is no such relation specified in \eqref{2}. If we can equate it to an integral, how are we to determine the limits of integration?
The equation \eqref{2} is to be taken, that the $M\to\infty$, we thus have an infinite sum (suppose it is convergent). Than we form a sequence of infinite sums, where $N$ increases for each element of the sequence. That is $$S_N=\lim\limits_{M\to\infty}\sum\limits_{k=0}^M f\left(k\cdot\frac{1}{N}\right)\frac{1}{N}$$ What does this sequence tend to?
Is it true that (or when is it true) $$\lim\limits_{N\to\infty}S_N=\int_0^\infty f(x)dx$$
Also the general term in \eqref{2} is $C_k=f\left(k\cdot\frac{1}{N}\right)$. How does it behave in the limit, namely $$\lim\limits_{N\to \infty}\lim\limits_{M\to \infty}f\left(M\cdot\frac{1}{N}\right)$$
• Your sum $(2)$ is not well defined for $M>N.$ – zhw. Feb 4 '17 at 20:36
• All derivations (I encountered) of the inverse fourier transform as a limiting fourier series use a sum of this form. I have similar objections as you, but seeing how so many texts use it, I though that perhaps I don't understand something. Looking at my question here I recreate the argument from those texts, where this sum shows up. I asked a separate question to seperate it from the fourier environment. – LeastSquaresWonderer Feb 4 '17 at 20:44
• If $f(x)$ is defined on $[0,\infty)$ then $\sum_{k=0}^\infty f(k/N) \frac{1}{N}$ is an approximation for $\int_0^\infty f(x){\rm d}x$ with step-size $\frac{1}{N}$. – Winther Feb 4 '17 at 20:44
• Could you please derive this for me? It would also resolve my unanswered question here. I have been thinking about this for few days, and I would really be grateful if you can derive this result for me. – LeastSquaresWonderer Feb 4 '17 at 20:46
• So when we take the limit as N--> infinity, the approximation becomes the integral? I would really be grateful for a derivation and justification of the statement. – LeastSquaresWonderer Feb 4 '17 at 20:49
Suppose $f$ is Riemann integrable on $[0,b]$ for every $b > 0$ and the improper integral over $[0, \infty)$ is convergent.
We first consider the case where $f$ is nonnegative and non-increasing, as suggested by @Winther, where we have
$$\frac{f((k+1)/N)}{N} \leqslant \int_{k/N}^{(k+1)/N} f(x) \, dx \leqslant \frac{f(k/N)}{N}.$$
This implies
$$\int_0^{(M+1)/N} f(x) \, dx \leqslant \frac{1}{N} \sum_{k=0}^{M} f(k/N) \leqslant \frac{f(0)}{N} + \int_0^{M/N} f(x) \, dx.$$
The sequence of partial sums is increasing and bounded , hence convergent as $M \to \infty$, with
$$\int_0^{\infty} f(x) \, dx \leqslant\frac{1}{N} \sum_{k=0}^{\infty} f(k/N) \leqslant \frac{f(0)}{N} + \int_0^{\infty} f(x) \, dx.$$
Therefore,
$$\lim_{N \to \infty} \lim_{ M \to \infty}\frac{1}{N} \sum_{k=0}^{M} f(k/N) = \int_0^\infty f(x) \, dx.$$
Can this still hold if $f$ is not monotonic?
For example, consider $f(x) = \sin x /x$, where
$$\int_0^\infty \frac{\sin x}{x} \, dx = \frac{\pi}{2}.$$
Examining the corresponding series (WLOG starting with $k=1$) we find
$$\frac{1}{N}\sum_{k = 1}^{\infty} \frac{\sin (k/N)}{k/N} = \sum_{k = 1}^{\infty} \frac{\sin (k/N)}{k} \\ = \frac{\pi}{2}-\frac{1}{2N} \\ \longrightarrow_{N \to \infty} \frac{\pi}{2}.$$
I have not yet found a counterexample for a non-monotone function. As the integral test can be generalized to $C^1$ functions of bounded variation, I suspect this may characterize a wider class of functions for which this result holds.
This could be shown by considering
$$\left|\int_{k/N}^{(k+1)/N} f(x) \, dx - \frac{f(k/N)}{N} \right| \leqslant \int_{k/N}^{(k+1)/N} |f(x) - f(k/N)| \, dx$$
and then summing over $k$, applying the mean value theorem when $f$ is differentiable, and using $\int_0^\infty |f'(x)|\, dx < \infty$ to show that the sum converges to the integral as $N \to \infty.$
• Very nice answer, thank you. If I may ask, does the order you specify the limits in matter? First M-->\infty, than N-->\infty ? The question relates to k/N. Becase k-->M, and k/N-->\infty (because the count goes to infinity, hence the upper limit of integration),this implies that k "overgrows" N, hence M does too. I hope my wording is understandable. – LeastSquaresWonderer Feb 5 '17 at 9:13
• You're welcome. The order does matter. Here we take the limit as $M \to \infty$ first followed by the limit as $N \to \infty$. If we switch, then with $M$ fixed we have $\frac{1}{N} \sum_{k=0}^M f(k/N) \to 0$ as $N \to \infty$ if for example $f$ is continuous and $f(0)$ is finite. – RRL Feb 5 '17 at 9:26
• This was my hypothesis, and the original reason for thinking about this question. Thank you for helping me understand. – LeastSquaresWonderer Feb 5 '17 at 9:31 | 2019-07-23T09:13:56 | {
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https://www.math10.com/forum/viewtopic.php?f=15&t=8987 | NEED HELP WITH EXERCISE
NEED HELP WITH EXERCISE
f:R$$\rightarrow$$R, f(0)=1 , x/f(x)+1/f'(x)=1 for every $x\in R$:
Prove that: $f(x)=x+\sqrt{x^2+1}$
Guest
Re: NEED HELP WITH EXERCISE
The equation is $$\frac{x}{f(x)}+ \frac{1}{f'(x)}= 1$$ with the initial condition f(0)= 1.
I am going to use "y" rather than "f" to make it a little easier to type: $$\frac{x}{y}+ \frac{1}{y'}= 1$$.
Multiply both sides by yy' to get rid of the fractions. $$xy'+ y= yy'$$. Subtract xy' from both sides:
$$y= yy'- xy'= (y- x)y'$$. Divide both sides by y- x: $$y'= \frac{y}{y- x}$$.
Let u= y- x. Then y= u+ x and y'= u'+ 1 so the equation becomed $$u'+ 1= \frac{u+ x}{u}= 1+ \frac{x}{u}$$.
So $$u'= \frac{du}{dx}= \frac{x}{u}$$. Separate variables- $$udu= xdx$$ and integrate- $$\frac{u^2}{2}= \frac{x^2}{2}+ C$$ or $$u^2= x^2+ 2C$$.
Since u= y- x, $$(y- x)^2= y^2- 2xy+x^2= x^2+ 2C$$.
The two $$x^2$$ terms cancel so $$y^2- 2xy= 2C$$. y(0)= 1 so $$1^2- 2(0)(1)= 1= 2C$$. Then $$y^2- 2xy- 1= 0$$.
Solve that for y by "completing the square"- $$y^2- 2xy+ x^2- x^2- 1= 0$$. $$y^2- 2xy+ x^2= (y- x)^2= x^2+ 1$$. Take the square root: $$y- x= \pm\sqrt{x^2+1}$$ so $$y= x\pm \sqrt{x^2+ 1}$$
$$y= x- \sqrt{x^2- 1}$$ however, does not satisfy the condition that y(0)= 1 so y= f(x) must equal $$x+ \sqrt{x^2+ 1}$$.
HallsofIvy
Posts: 259
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 96
Re: NEED HELP WITH EXERCISE
Damn, I feel so dumb now, it didn't occur to me to set a new equation
Is it correct if I write it this way?
$$\frac{f(x)}{f(x)-x}$$]=f'(x), set: g(x)=f(x)-x (1) , g'(x)=f'(x)-1, f(x)=g(x)+x
So the equation is rewritten as:
\frac{g(x)+x}{g(x)}=g'(x)+1 \Leftrightarrow
\frac{g(x)+x}{g(x)}=g'(x)+\frac{g(x)}{g(x)} \Leftrightarrow
\frac{g(x)+x}{g(x)}-\frac{g(x)}{g(x)}=g'(x)\Leftrightarrow
\frac{g(x)+x-g(x)}{g(x)}=g'(x) \Leftrightarrow
\frac{x}{g(x)}=g'(x)
Multiply both sides by g(x)
g(x)g'(x)=x
Divide both sides by 2
\frac{g(x)g'(x)}{2}=\frac{x}{2}\Rightarrow
(\frac{(g(x)^{2})}{2})'=(\frac{x^2}{2})'
The integrated equation is:
\frac{g^2(x)}{2}= (\frac{x^2}{2})+c (2)
For x=0 we have:
\frac{g^2(0)}{2}=\frac{0}{2}+c \Rightarrow (1)
\frac{f^2(0)-2f(0)\cdot0+0}{2}=c
\frac{f^2(0)}{2}=c , f(0)=1
c=\frac{1}{2}
So equation (2) becomes:
\frac{g^2(x)}{2}= (\frac{x^2}{2})+ \frac{1}{2} or g^2(x)=x^2+1
We take the square root on both sides
\sqrt{g^2(x)}=\sqrt{x^2+1}
|g(x)|= \sqrt{x^2+1}
Replace g(x) with f(x)-x
|f(x)-x|= \sqrt{x^2+1}
If f(x)-x>0, f(x)= \sqrt{x^2+1}+x
If f(x)-x<0, f(x)= x-\sqrt{x^2+1} FALSE since f(0)=-1 in this case, this does not satisfy our initial condition of f(0)=1
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https://mathoverflow.net/questions/333170/conditions-on-matrices-imply-that-3-divides-n/333224 | # Conditions on matrices imply that 3 divides $n$
A quite popular exercise in linear algebra is the following (or very related exercises, see for example https://math.stackexchange.com/questions/299651/square-matrices-satisfying-certain-relations-must-have-dimension-divisible-by-3 and https://math.stackexchange.com/questions/3109173/ab-ba-invertible-and-a2b2-ab-then-3-divides-n):
Let $$K$$ be a field of characteristic different from 3 and $$X$$ and $$Y$$ two $$n\times n$$-matrices with $$X^2+Y^2+XY=0$$ and $$XY-YX$$ invertible. Then 3 divides $$n$$.
A (representation-theoretic) proof can be given as in the answer of Mariano Suárez-Álvarez in https://math.stackexchange.com/questions/299651/square-matrices-satisfying-certain-relations-must-have-dimension-divisible-by-3 .
Question: Is this also true for fields of characteristic 3?
edit: So it turned out that the result holds for any field. A bonus question might be to find a proof that works independent of the characteristic of the field.
• A simple observation in characteristic 3 is that the hp is equivalent to (X-Y)^2 = [X,Y] invertible. So in particular X = M +Y where M is invertible. Substituting M yields M^2 = [M,Y] for some Y. This is a bit stronger then tr M^2 = 0. – Andrea Marino Jun 3 '19 at 18:19
• Following Andrea's simplification. By multiplying both on the left and right by $M^{-1}$ we get $I=[Y,M^{-1}]$. Taking traces of both sides gives $n=0$, which means $3|n$ since we are in characteristic 3. – Gjergji Zaimi Jun 3 '19 at 19:27
• This is a very beautiful solution. It would be nice to turn it into an answer. – Mare Jun 3 '19 at 19:56
• Great job Gjergji! We did it. Who's going to write it? – Andrea Marino Jun 3 '19 at 21:06
• @LSpice - no it is not required. The other answer next to Mariano's essentially works just fine whenever there exists a primitive cube root of unity. – Vladimir Dotsenko Jun 4 '19 at 16:26
Assembling the comments, I write the entire solution.
STEP 1: the first hypothesis in characteristics 3 is equivalent to $$(X-Y)^2 = [X,Y]$$.
Indeed, note that
$$(X-Y)^2 - [X,Y] = X^2-XY-YX+Y^2- XY+YX = X^2-2XY+Y^2 = X^2+Y^2+XY$$
Why did I make this calculation? Note that $$X^2+Y^2+XY$$ resembles a factor of $$X^3-Y^3$$, which would be equal (in commutative case) to $$(X-Y)^3$$ in char 3. To get there we would need a $$X-Y$$ more, so we can guess that $$(X-Y)^2$$ and $$X^2+Y^2+XY$$ will be equal up to some commutators!
STEP 2: getting the final result.
We have that $$(X-Y)^2 = [X,Y]$$ is invertible by the second hypothesis. Thus in particular $$X-Y$$ is invertible. This allows us to substitute $$X=M+Y$$ with M invertible, obtaining
$$M^2 = [M+Y,Y] = [M,Y]$$
The only relevant information we have about arbitrary commutators is that the trace is zero. We hence would like to have some matrix with easy trace (like the identity) to compare witha commutator. To do this, let's multiplicate on both the left and the right for $$M^{-1}$$:
$$I = M^{-1}[M,Y]M^{-1} = M^{-1}(MY-YM)M^{-1} = YM^{-1}-M^{-1}Y = [Y,M^{-1}]$$ Taking traces we get $$n=0$$. Being in characteristics 3, this gives $$3 \mid n$$.
Let me start from $$M^2=[M,Y]$$ with $$M\in {\bf GL}_n(k)$$, as suggested by Andrea. Wlog, I assume that the characteristic polynomial of $$M$$ splits over $$k$$, and I decompose $$k^n$$ as the direct sum of characteristic subspaces $$E_\mu=\ker(M-\mu I_n)^n$$. It is enough to prove that the dimension of each $$E_\mu$$ is a multiple of $$3$$. To this end, observe that $$Y$$ acts over $$E_\mu$$. In details, let $$x$$ be an eigenvector, $$Mx=\mu x$$. Then $$(M-\mu)Yx=\mu^2x$$. Because $$M$$ is invertible, $$\mu\ne0$$ and therefore $$Yx\in\ker(M-\mu)^2\setminus\ker(M-\mu).$$ Likewise $$(M-\mu)Y^2x=2\mu^2Yx+2\mu^3x,$$ hence $$Y^2x\in\ker(M-\mu)^3\setminus\ker(M-\mu)^2$$. Eventually, using again $${\rm char}(k)=3$$, we have $$(M-\mu)Y^3x=0.$$ We deduce that a basis $$E_\mu$$ is obtained by taking a basis $$\cal B$$ of $$\ker(M-\mu)$$, and adjoining the vectors of $$Y\cal B$$ and $$Y^2\cal B$$ ; all of them are linearly independent, as seen above. Thus $$\dim E_\mu=3\dim\ker(M-\mu)$$. | 2021-01-23T15:17:56 | {
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https://math.stackexchange.com/questions/2172339/determine-if-3-circles-intersect-at-a-common-point | # Determine if 3 circles intersect at a common point
Given three circles centered at points $A$, $B$, and $C$ with non-zero radii of lengths $R_A$, $R_B$ and $R_C$. Where the centers of the circles form a valid triangle, and where the distance between any two centers is less than or equal to the sum of their corresponding radii.
Question: Is it possible to determine if all three circles intersect at a common point using a calculation simpler than first determining the pair of intersection points between two pairs of circles then determining if any of the intersection points are equal?
You can calculate the pair of intersection between, say, circles $A$ and $B$, then calculate the distance of each such intersection to circle $C$. There is a triple intersection if and only if one of the distances is $R_C$.
EDIT: Well, consider $\triangle ABC$ and suppose there is a triple intersection, that is, some point $P$ with $PA=R_A$, $PB=R_B$ and $PC=R_C$. Let $E_A$ be the side of $\triangle ABC$ opposite vertex $A$ and similarly for $B$ and $C$.
Suppose without loss of generality that $R_A\leq R_B\leq R_C$. The triangle inequality implies that the following must hold:
\begin{align} R_C-R_B\leq E_A\leq R_C+R_B\\ R_C-R_A\leq E_B\leq R_C+R_A\\ R_B-R_A\leq E_C\leq R_B+R_A \end{align}
By hypothesis, the right hand inequalities do hold, but if any of the left hand ones do not, you can already rule out the possibility of a triple intersection.
I might yet improve on this later.
• +1 Interesting solution and it is better than two sets of intersection pairs. Though I was hoping for a solution based around perhaps the triangle constructed from the centers and some constraint based on lengths of the edges and the radii - if nothing better comes along I'll set this as the answer. – Sari T. Mar 5 '17 at 1:47
• See the edit. ${}$ – Fimpellizieri Mar 5 '17 at 2:36
• That is a much simpler and cheaper calculation to verify than the first suggestion and I'm already performing some of those lengths as part of the initial proper triangle verification stage. – Sari T. Mar 5 '17 at 2:43
• Yes, but notice that the inequalities being satisfied does not imply the existence of a triple intersection. Well, at least, I haven't proved that. – Fimpellizieri Mar 5 '17 at 2:48
• oh I misunderstood. What then would it require to prove that if those inequalities hold there is indeed a triple intersection? – Sari T. Mar 5 '17 at 2:58
Don't know that it's much simpler than calculating the pairwise intersections, then the distances to the third center, but the following gives a symmetric condition using complex numbers.
Let $\,a,b,c\,$ be the complex numbers associated with points $A,B,C$ in a complex plane centered at the centroid of $ABC\,$, so that $a+b+c=0\,$.
The point of intersection $z$ of the three circles (if it exists) must satisfy the $3$ equations similar to:
$$|z-a|^2=R_A^2 \;\;\iff\;\;(z-a)(\bar z - \bar a) = R_A^2 \;\;\iff\;\;|z|^2 - z \bar a - \bar z a + |a|^2 = R_A^2 \tag{1}$$
Writing $(1)$ for $a,b,c$ and summing the $3$ equations up:
$$\require{cancel} 3\,|z|^2 - \cancel{z \sum_{cyc} \bar a} - \bcancel{\bar z \sum_{cyc} a} + \sum_{cyc}|a|^2 = \sum_{cyc} R_A^2 \;\;\implies\;\; |z|^2 = \frac{1}{3}\left(\sum_{cyc} R_A^2-\sum_{cyc}|a|^2\right) =R^2 \tag{2}$$
Substituting $(2)$ back into each of $(1)\,$:
$$-|z|^2 + z \bar a + \bar z a - |a|^2 = - R_A^2 \;\;\iff\;\; z \cdot \bar a + \bar z \cdot a = |a|^2+R^2-R_A^2 \tag{3}$$
Considering $(3)$ as a system of linear equations in $z, \bar z\,$, the condition for it to have solutions is:
$$\left| \begin{matrix} \;\bar a \;&\; a \;&\; |a|^2+R^2-R_A^2\; \\ \;\bar b \;&\; b \;&\; |b|^2+R^2-R_B^2\; \\ \;\bar c \;&\; c \;&\; |c|^2+R^2-R_C^2\; \end{matrix} \right| \;\;=\;\; 0$$
If we coordinatize, say, with $$A = (0,0) \qquad B = (c, 0) \qquad c = (b\cos A,b\sin A)$$ and take suppose circles $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ (of respective radii $r_A$, $r_B$, $r_C$) meet at a point $P = (x,y)$, then we have three equations in two unknowns $x$ and $y$: \begin{align} x^2 + y^2 &= r_A^2 \\ x^2 + y^2 &= r_B^2 + 2 c x - c^2 \\ x^2 + y^2 &= r_C^2 + 2 b x \cos A x + 2 b y \sin A - b^2 \end{align} We can eliminate $x$ and $y$ from these equations, leaving this relation: \begin{align} a^2 b^2 c^2 + a^2 r_A^4 + b^2 r_B^4 + c^2 r_C^4 &= \left( a^2 r_A^2 + r_B^2 r_C^2 \right) \left(-a^2 + b^2 + c^2 \right) \\ &+ \left( b^2 r_B^2 + r_C^2 r_A^2 \right) \left(\phantom{-}a^2 - b^2 + c^2 \right)\\ &+ \left( c^2 r_C^2 + r_A^2 r_B^2 \right) \left(\phantom{-}a^2 + b^2 - c^2 \right) \end{align} \tag{1}
The right-hand side seems to want to be re-written with cosines ...
\begin{align} a^2 b^2 c^2 + a^2 r_A^4 + b^2 r_B^4 + c^2 r_C^4 &= 2 b c \cos A \left( a^2 r_A^2 + r_B^2 r_C^2 \right) \\ &+ 2 c a \cos B \left( b^2 r_B^2 + r_C^2 r_A^2 \right)\\ &+ 2 a b \cos C \left( c^2 r_C^2 + r_A^2 r_B^2 \right) \end{align} \tag{2} ... but this doesn't seem a great deal better. Perhaps if we use the Law of Sines to write $$a = 2 r \sin A \qquad b = 2 r \sin B \qquad c = 2 r \sin C$$ where $r$ is the circumradius of $\triangle ABC$. With a little effort, we find this form for the relation:
\begin{align} \frac{1}{64} \left(\;\begin{array}{c} r_A^2 \sin 2A + r_B^2 \sin 2B + r_C^2 \sin 2C \\ - 8 r^2 \sin A \sin B \sin C\end{array} \;\right)^2 = \frac{1}{16}&(\phantom{-}r_A \sin A + r_B \sin B + r_C \sin C ) \\ \cdot &(-r_A \sin A + r_B \sin B + r_C \sin C )\\ \cdot &(\phantom{-}r_A \sin A - r_B \sin B + r_C \sin C )\\ \cdot &(\phantom{-}r_A \sin A + r_B \sin B - r_C \sin C ) \end{align} \tag{3}
The curious fractional coefficients are there to help us recognize the right-hand side as Heron's Formula for the square of the area of a triangle with side-lengths $r_A \sin A$, $r_B \sin B$, $r_C \sin C$. More precisely, when (and only when) the right-hand side of $(3)$ is non-negative, it gives the square of the area of the triangle with those side-lengths; when (and only when) the right-hand side is negative, those side-lengths fail to form a valid triangle. (Note: I consider a degenerate triangle of area $0$ to be valid.)
Since the left-hand side of $(3)$ is necessarily non-negative, we deduce that
$\bigcirc A$, $\bigcirc B$, $\bigcirc C$, with radii $r_A$, $r_B$, $r_C$, concur at a point only if $r_A \sin A$, $r_B \sin B$, $r_C \sin C$ are the edges of a valid triangle (ie, they satisfy the Triangle Inequality).
That gives you a way to weed-out bad candidates. To know for sure that the three circles concur, you'd need to check the full equality of $(3)$. (Is that "simpler" than the strategy you mentioned? I'm not sure.)
Note that $8 r^2 \sin A\sin B\sin C = 2 a b \sin C = 4 |\triangle ABC|$. If we call the "valid triangle" referenced above, say, $\triangle T$, then we can write $(3)$ as
$$r_A^2 \sin 2A + r_B^2 \sin 2B + r_C^2 \sin 2C \pm 8 |\triangle T|\;=\; 4 |\triangle ABC| \tag{4}$$
To get at the behavior of the "$\pm$", consider $P$ at distance $p$ from the circumcenter of $\triangle ABC$ (and at distances $r_A$, $r_B$, $r_C$ from $A$, $B$, $C$, respectively). With the help of Mathematica, we get \begin{align} r_A^2 \sin 2A + r_B^2 \sin 2B + r_C^2 \sin 2C &= 4 (r^2 + p^2) \sin A \sin B \sin C \\ 8|\triangle T| &= 4 |r^2-p^2|\sin A \sin B \sin C \end{align}
We see, then, that "$\pm$" must be "$+$" when $r > p$ (that is, when $P$ is inside the circumcircle) and "$-$" when $r < p$ (when $P$ is outside the circumcircle); for $P$ on the circumcircle, $|\triangle T| = 0$.
• Your discussion is quiet interesting and still trying to understand it will take me some time, but using the naive approach of two sets of pairs of intersection requires only 29 operations (15 add/sub operations 14 mul/div operations, 6 comparisons). – Sari T. Mar 5 '17 at 5:05
• If facts about $\triangle ABC$ (sines, cosines, appropriate multiples of area) are pre-computed, then $(3)$ can be written to require at most 27 operations. Note: If $r_A\sin A$, etc, fail to make a valid triangle, then you'll know this by computing the product of final three factors of the RHS of $(3)$ and comparing it to $0$. That's 12 operations. If you compare the individual factors against $0$ along the way (any negative implies failure), you add two comparisons, but you could get lucky and detect failure in just 6 operations. (The total count of 27 includes the extra two comparisons.) – Blue Mar 5 '17 at 6:17 | 2019-08-22T04:28:03 | {
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http://lamport.azurewebsites.net/tla/tutorial/intermezzo2.html | Prev: Session 10
Next: Session 11 part 1
Contents
# Intermezzo 2 Records
Leslie Lamport
It's time for another pause from the study of algorithms to learn another important TLA+ data type: the record. When mathematicians want to bundle several values together as a single value, they use a tuple. For example, suppose they want to talk about the kind of graph used in Session 10, consisting of a set of nodes, a root node, and a set of edges. They would define a graph G to be a triple. For example, the graph G with set {r,n1,n2} of nodes, root r, and set {{r,n1},{n1,n2}} of edges might be written:
<<{r,n1,n2}, r, {{r,n1},{n1,n2}}>>
Programmers realized that this leads to errors, because it's easy to forget if the root is G[1] or G[2]. Programming languages introduced records, which are like tuples except the components are named rather than numbered. The components of the graph G could be named G.nodes, G.root, and G.edges. These components were called fields, and the graph G had the three fields named nodes, root, and edges.
Programming languages later added a lot of additional stuff to records and called them objects. TLA+ and PlusCal keep the original concept of a record as simply a value, just like a tuple or a number. It introduces a simple way to write records, where the graph G can be written:
[nodes |-> {r,n1,n2}, root |-> r, edges |-> {{r,n1},{n1,n2}}]
TLA+ and PlusCal also define this record to be a function whose domain is the set consisting of the three strings {"nodes", "root", "edges"} and defines G.nodes to be an abbreviation of G["nodes"]. This means that the assignment statements G.nodes := ... and G["nodes"] := ... are equivalent. It also means that there is no ordering of the fields, so [a |-> 1, b |-> 2] is the same record as [b |-> 2, a |-> 1]
Field names cannot be arbitrary strings. They should be legal identifiers. A declared or defined identifier can also be used as a field name.
There is also a way to write the set of all records whose fields are elements of certain sets. For example, the set of all records with integer-valued x and y fields and a zz field that is an element of a set S is written:
[x : Int, y : Int, zz : S]
Exercise 1 Write this set of records using the notation for describing individual records and the set constructs you learned in Session 4 Show the answer.
Answer {[x |-> i, y |-> j, zz |-> s] : i, j \in Int, s \in S}
Such a set of records is often used in writing a type-correctness invariant for a variable whose value is a record or contains records.
In the algorithm of Session 10, we could have let the message queue from node n to node m be the record [from |-> n, to |-> m] instead of the pair <<n, m>>. Writing msgs[q].from might have made the algorithm a little easier to understand, but msgs[q][1] is shorter. Which to use is a matter of taste.
A common use of records is to bundle several pieces of information into a message. If different messages can contain different kinds of information, the messages can all contain a field, often called type, whose value indicates what kind of information the message contains. The action that processes the message tests that field to determine what to do.
An algorithm can also determine what kind of information is in a record by checking what fields the record contains. For any function f, the expression DOMAIN f equals the domain of f. Thus, DOMAIN [a |-> 1, b |-> 2] equals {"a","b"}.
Exercise 2 Define an operator IncrCnt such that if r is any record containing an integer-valued count field, then IncrCnt(r) is the record obtained from r by incrementing its count field by 1. (Thus the statement x.count := x.count + 1 is equivalent to x := IncrCnt(x) if x is a record with a count field.) Show the answer.
IncrCnt(r) == [d \in DOMAIN r |-> | 2022-01-19T22:46:07 | {
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