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http://math.stackexchange.com/questions/107731/limit-points-of-sinn | # Limit Points of $\sin(n)$.
1. Determine at least three limit points for the set {$\sin(n)$: n a positive integer}.
2. How many limit points does the set {$\sin(n)$: n a positive integer} have?
Our professor gave us a definition to use for limit point in order to differentiate between a cluster point. The definition is as follows:
Let $S$ be a nonempty set of $\mathbb{R}$ where $S \subseteq \mathbb{R}$
Let $x \in \mathbb{R}$
We say that $x$ is a limit pont of $S$ if:
For each $\epsilon > 0$ there is an element of $S$ in $(x-\epsilon$, $x+\epsilon)$.
With our first question and the given definition, wouldn't the numbers $\sin(1)$, $\sin(2)$, $\sin(3)$ work?
Choose the $x = \sin(1)$
We then have:
$\sin(1)-\epsilon < \sin(1) < \sin(1) + \epsilon$
Which seems trivially true. The same argument would follow for the next two points. However, the next part is where myself and a few of my other classmates are completely lost.
A few questions for 2.
What direction should I head for the second half? The professor mentioned Kronecker's theorem. If possible, could someone give a breakdown of Kronecker's theorem and it's applications?
Kronecker's Theorem: http://mathworld.wolfram.com/KroneckersApproximationTheorem.html
How can I find the cardinality of the set of limit points?
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You might be able to work out the answers after having a look at math.stackexchange.com/questions/63526/… – Gerry Myerson Feb 10 '12 at 4:44
I adjusted my definition. I incorrectly transposed it from my notes. – user17366 Feb 10 '12 at 5:04
Consider irrational rotation on the circle. That is, consider $R_r: S^1\rightarrow S^1$, $R_r(x)= (r + x)(\mathrm{mod} 1)$, where $S^1$ is the unit circle given by $[0,1]/\sim$, where ~ is the equivalence relation identifying $0$ with $1$ (you simply wrap the interval $[0, 1]$ into a circle, gluing the ends). Now for $r$ try some nice irrational values, like $\pi/2$, $\pi/4$, $\pi/6$, and start applying $R_r$ repetitively to zero: $R_r(0)$, then $R_r(R_r(0))$, and so on... (hint: $\{x_n\}_{n\in\mathbb{N}}$ is dense in $[0,1]$, where $x_n = R_r\circ\cdots\circ R_r(0)$ $n$-times. – William Feb 10 '12 at 5:08
I was just reading this en.wikipedia.org/wiki/Irrational_rotation – user17366 Feb 10 '12 at 5:09
@arete: Well, think about it. If it still doesn't make sense, comment, and I'll make my comment into an answer. – William Feb 10 '12 at 5:52
Per OP's request, I am making my comment into an answer.
Notice that Kronecker's theorem, as stated here, is essentially equivalent to what I suggested in the comment above.
Define $S^1 = [0,1)$, and we can think of $S^1$ as a circle of unit circumference. Let us define rotation on the circle by $\alpha\in[0,1)$ as $R_\alpha(x) = (x + \alpha)\mod 1$. Then $n$-fold application of $R_\alpha$ to $x\in[0,1]$ is
$$R_\alpha^n(x) := \underbrace{R_\alpha\circ\cdots\circ R_\alpha}_{n-\text{ times }}(x),$$
and has the value $(x + n\alpha)\mod 1$, or (as in the theorem of Kronecker referenced above) $(x + n\alpha) - \lfloor x + n\alpha\rfloor$. On the other hand, irrational rotation on $S^1$ is minimal, meaning that for any $x\in S^1$, if $\alpha$ is irrational, then the sequence $\{x_n\}_{n\in\mathbb{N}}$ given by $x_n = R_\alpha^n(x)$ fills $S^1$ densely.
After looking for a quick reference on the internet, I didn't find anything, so let us just prove it here quickly.
Suppose for some irrational $\alpha\in(0,1)$ and some $x\in S^1$, the sequence $\{x_n\}$, as defined above, is not dense in $S^1$. But then there exist open intervals in $S^1$ not containing $x_n$, for any $n\in\mathbb{N}$. Let $I$ be such an interval of maximal length. Obviously $R^n_\alpha(I)$ is also such an interval, of the same length as $I$, for any $n\geq 1$. If $R^n_\alpha(I)$ intersects another such interval, say $J$, and $R_\alpha(I)\neq J$, then we obtain an interval $J\cup R_\alpha^n(I)$ not containing any $x_n$, and of length greater than $I$, which by maximality of $I$ cannot happen. On the other hand, if $R_\alpha^n(I) = J$ and $J = I$, then for rational $q\in I$, we have $q = q + n\alpha \mod 1$, or $\alpha$ is rational, contradicting irrationality of $\alpha$. In other words, the sequence $\{R_\alpha^n(I)\}_{n\in\mathbb{N}}$ is a sequence of disjoint subintervals of $S^1$, each of the same positive length as $I$. But this, of course, cannot happen, since $S^1$ has finite length. We're done.
Now, say $\alpha\in (0,1)$ is irrational (say $\alpha = \pi/2$, for instance). Then the sequence $\{n\alpha\mod 1\}_{n\in\mathbb{N}}$ is dense in $[0, 1)$. In particular, there is a strictly increasing subsequence $\{n_i\}$ of natural numbers such that
$$\lim_{i\rightarrow \infty}(n_i\alpha\mod 1) = 0.$$
It follows that
$$\lim_{i\rightarrow\infty}\left|n_i\alpha - n_i\right| = 0.$$
This implies that
$$\lim_{i\rightarrow\infty}\left|\sin(n_i) - \sin(n_i\alpha)\right| = 0.$$
Recall that $\alpha = \pi/2$. I think you can take it from here.
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Per the Wikipedia defintion "In mathematics, a limit point (or accumulation point) of a set S in a topological space X is a point x ( which is in X, but not necessarily in S ) that can be "approximated" by points of S in the sense that every neighbourhood of x with respect to the topology on X also contains a point of S other than x itself" so you cannot use the fact that $\sin(1)$ is in a neighborhood of $\sin(1)$. You need to show that there is another $n$ such that $\sin(n)$ is near $\sin(1)$, which is true.
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Note that the OP is speaking about limit points of a set, not a sequence, which means that we're using a (usually) more forgiving definition than the one you reproduce. – Henning Makholm Feb 10 '12 at 5:04
@HenningMakholm: OP has added some text that answers my complaint of the lacking part of the sentence. You are right that this changes the answer. I missed the original that said it was a limit point of a set. – Ross Millikan Feb 10 '12 at 5:15
On the other hand I missed the title of the question which still calls it a sequence. Shall we just agree that the question is ambiguous? – Henning Makholm Feb 10 '12 at 5:17
@HenningMakholm: I can buy that easily. I have updated my answer hoping that it will help. – Ross Millikan Feb 10 '12 at 5:22 | 2015-07-07T02:52:43 | {
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https://math.stackexchange.com/questions/4627479/decreasing-function-and-period-2-points | # Decreasing function and period $2$ points
I'm trying to prove the following result:
Let $$h: [a,b] \rightarrow [a,b]$$ be a monotonic decreasing function and given $$x \in [a,b]$$ we define the sequence \begin{align*} y_{0,x} & = x\newline y_{n+1,x} & = h(y_{n,x}), \forall n>1 \end{align*} Then, $$\{y_{n,x}\}_{n=0}^\infty$$ is convergent for every $$x \in [a,b]$$ iff every point of period $$2$$ of $$h$$ is, in fact, a fixed point.
I've managed to prove that if every sequence of that type is convergent, then every point of period $$2$$ is a fixed point (in fact, you don't even need to assume that $$h$$ is decreasing). My problem is with the reciprocal.
I've proven that for any $$x \in [a,b]$$ the subsequences $$x_n=y_{2n,x}$$ and $$z_n=y_{2n+1,x}$$ are always increasing and decreasing respectively (or vice versa). As they are bounded, because they are contained in the interval $$[a,b]$$, this proves that both of them are convergent.
Now, if $$h$$ was continuous, then it is easy to see (using now that every point of period $$2$$ of $$h$$ is fixed) that both limits are equal, and so, that the sequence $$y_{n,x}$$ converges.
My doubt is then if this result is still true without the condition that $$h$$ is continuous. I cannot prove without this condition, but I also cannot give any counterexample, so any help would be appreciated.
• Is $h$ supposed to be injective? In other words, is $h$ required to be strictly decreasing? Feb 1 at 22:26
• It looks people are divided on the definition of "monotonic decreasing". Wolfram says it means "strictly decreasing". Wikipedia says it means "weakly decreasing". Check this post. Feb 1 at 23:38
It is not true that if every point of period $$2$$ of $$h$$ is a fixed point, then $$\{y_{n,x}\}_{n=0}^\infty$$ is convergent for every $$x \in [a,b]$$.
Here is a counterexample.
$$\quad$$
Let $$f:[0,1]\to[0,1]$$, $$\begin{array}{crcl} f: &[0,1] &\to &[0,1]\\ &x &\mapsto &1-\frac x2 &\text{if } x<\frac12\\ &x &\mapsto &\frac 23 &\text{if } \frac12\le x\le\frac34\\ &x &\mapsto &\frac54-x &\text{if } \frac34< x\\ \end{array}$$
The only periodical point of $$f(x)$$ is $$x=\frac23$$ with $$f(\frac23)=\frac23$$.
If $$x=y_{0,x}=0$$, then $$y_{2n+1,x}=\frac34+\frac1{2^{n+2}}$$ and $$y_{2n,x}=\frac12-\frac1{2^{n+1}}$$ for all $$n\ge0$$. $$\lim_{n\to\infty}y_{2n,x}=\frac12\not=\frac34=\lim_{n\to\infty}y_{2n+1,x}$$
Hence $$f$$ is a counterexample.
Let $$g(x):[0,1]\to[0,1]$$ be the same as $$f(x)$$ except for $$\frac12\le x\le\frac34$$, $$g(x)=\frac23-\frac13(x-\frac23)=\frac89-\frac x3$$.
What have been said above about $$f$$ also holds for $$g$$. In particular, $$g$$ is also a counterexample. (As mentioned by OP, neither $$f$$ nor $$g$$ can be continuous. In fact, both are discontinuous at $$x=\frac12$$ and $$x=\frac34$$.)
Note that while $$f$$ is weakly decreasing, $$g$$ is strictly decreasing.
• It was so simple and yet, I couldn't find it. Thanks! Feb 2 at 11:32 | 2023-03-25T03:42:16 | {
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https://www.physicsforums.com/threads/deriving-logistic-population-model-ode-question.575676/ | # Deriving Logistic Population Model (ODE question)
SirPartypants
## Homework Statement
Solve the logistic population model:
$dP/dt=rP(1-P/C); P(0)=P_{0}$
2. The attempt at a solution
First, I separated variables to get:
$\int \! \frac{1}{P(1-P/C)} \, \mathrm{d}P = \int \! r \, \mathrm{d}t$
Then, I took the left hand side and split into partial fractions:
(1) - $\int \! \frac{1}{P} \, \mathrm{d}P + \int \! \frac{1/C}{1-P/C} \, \mathrm{d}P$
If I integrate, I get the following:
$\ln(P)-\ln(1-P/C)=\ln(\frac{P}{1-P/C})$ (*)
However, my problem is this. If I take (1) and multiply the second integral by C/C (which should be fine, its 1), I get the following:
$\int \! \frac{1}{P} \, \mathrm{d}P + \int \! \frac{1}{C-P} \, \mathrm{d}P$
Which is...
$\ln(P)-\ln(C-P)=\ln(P/C-P)$ (**)
However, (*) and (**) are not the same. I'm assuming there's something wrong with one of the ways that I integrated considering I came out with two different functions. Are they both correct and all that will change when I do the full problem out is the constant?
Homework Helper
## Homework Statement
Solve the logistic population model:
$dP/dt=rP(1-P/C); P(0)=P_{0}$
2. The attempt at a solution
First, I separated variables to get:
$\int \! \frac{1}{P(1-P/C)} \, \mathrm{d}P = \int \! r \, \mathrm{d}t$
Then, I took the left hand side and split into partial fractions:
(1) - $\int \! \frac{1}{P} \, \mathrm{d}P + \int \! \frac{1/C}{1-P/C} \, \mathrm{d}P$
If I integrate, I get the following:
$\ln(P)-\ln(1-P/C)=\ln(\frac{P}{1-P/C})$ (*)
However, my problem is this. If I take (1) and multiply the second integral by C/C (which should be fine, its 1), I get the following:
$\int \! \frac{1}{P} \, \mathrm{d}P + \int \! \frac{1}{C-P} \, \mathrm{d}P$
Which is...
$\ln(P)-\ln(C-P)=\ln(P/C-P)$ (**)
However, (*) and (**) are not the same. I'm assuming there's something wrong with one of the ways that I integrated considering I came out with two different functions. Are they both correct and all that will change when I do the full problem out is the constant?
BTW, $\ln(P)-\ln(C-P)=\ln(P/C-P)$ should be written $\ln(P)-\ln(C-P)=\ln(P/(C-P))$.
Both answers are in fact equivalent. Remember that you haven't taken into account the arbitrary constant of integration. Try putting that into the LHS as $\ln k$ in one case and $\ln k'$ in the other. You'll be able to bring everything under the natural log, after which you can exponentiate both sides. Now impose the initial value condition $P = P_0$ at $t= 0$. You'll find that no matter which form you use, the k and k' will be just right so that your final form is the same.
Homework Helper
To obviate this sort of issue, I prefer to take the definite integral on both sides. The bounds will be $(P_0, P(t))$ on the LHS and $(0,t)$ on the RHS. | 2022-11-29T12:03:35 | {
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https://math.stackexchange.com/questions/157206/square-three-digit-numbers-the-efficient-way | # Square three digit numbers, the efficient way
I would like to square a three digit number in my head. Now I know that the formula is
$$( X + r ) ( X - r ) + r^2 = X^2 - rX + rX - r^2 + r^2 = X^2$$ Where $\,r\,$ is a number such that $\,X + r\,$ is divisible by $10$ and/or $100$
Now the problem is that, I would like to first figure out whether finding the $r$ such that the $3$ digit number is divisible by $100$ or $10$ would be more efficient.
If I choose $r < 10$ then the $\,( X + r ) ( X - r )\,$ becomes quite unpleasant. If I choose $r < 100$ then the $r$ itself will have to be broken into pieces such that
$\begin{array}\\ ( X + r ) ( X - r ) + [ ( r + j ) ( r - j ) + j^2 ] &= ( X + r ) ( X - r ) + ( r^2 - jr + jr - j^2 + j^2 )\\ &=( X + r ) ( X - r ) + r^2\\ &= X^2 - rX + rX - r^2 + r^2\\ & = X^2\\ \end{array}$
This kind of yields recursive solution where any number squared works. The job is much more easier, but the person has to memorize more numbers while continuing to do more math.
So which method would you recomend? Or if you would have some other method I would love to hear it.
If you choose recursive method how do you memorize numbers in an efficient way without having numbers collide in your brain?
Thanks to anyone for their response, and sorry if tags are off, I am unsure which tag fits my question.
• I edited a little your post but for some reason the middle rows I couldn't. Check this, please. – DonAntonio Jun 12 '12 at 2:41
• @DonAntonio Thank you very much for the edit. I have spent quite some time trying to figure out how to get the power to work. – Quillion Jun 12 '12 at 2:56
• I reformatted them way I like it. Feel free to revert if you don't like. – marty cohen Jul 27 '15 at 5:26
• @martycohen I like your formatting! :) thanks – Quillion Jul 27 '15 at 13:18
Many mental calculations are easier the more facts you memorize. Knowing the squares up to $31^2=961$ can make it easier, reducing the need to go all the way to single digits. Also the fact that $(10n+5)^2=100n(n+1)+25$, for example $65^2=(6\cdot7)25=4225$. If you break it all the way to individual digits, I find it easier to keep track of my place if I start with the most significant digits and add as I go, so $359^2=300^2+2\cdot300\cdot50+\ldots =90000+30000+\ldots$ $=120000+ 50^2+\ldots = 122500+2\cdot300\cdot9+\ldots=127900+2\cdot50\cdot9+9^2$ where for many purposes you can quit early when you have the precision needed.
• Thanks so much, I found this very easy and good to use. Now all that's left is to master it :) – Quillion Jun 12 '12 at 13:26
Say is $$n = \left( {a,b,c} \right)$$
$$n=a \cdot 10^2+b \cdot 10+c$$
$${n^2} = {a^2}\cdot{10^4} + 2ab{10^3} + \left( {2ac + {b^2}} \right)\cdot{10^2} + 2bc10 + {c^2}$$
$${n^2} = \left( {{a^2},2ba,2ac+b^2,2cb,{c^2}} \right)$$
That is a nice symmetric formula in $a,b,c$ if you have to ask!
Note $a^2 ,b^2, c^2$ occur precisely in $1$st $3$rd $5$th, and you can always get the middle string right by thiking loopy:
$$baaccb$$
Example
$n=721 = (7,2,1)$
Then
$$n^2 = (49,2 \cdot 7 \cdot 2, 2 \cdot 7 \cdot 1+4, 2 \cdot 1 \cdot 2 , 1^2)$$
$$n^2 = (49,28, 18,4 , 1)$$
Then we do the "carries"
$$n^2 = (51,9, 8,4 , 1)=519841$$
I don't know if this the greatest method but it is interesting to consider, since at most you need to square one digit numbers.
to square a 3 digit number: example 237^2
your 37 (the number in the tens and ones position) becomes: a
your 237 (the entire number) becomes: b
your 2 (the digit in the hundreds position) in 237 becomes: x
your 3 (the digit in the tens position) in 237 becomes: y
and your 7 (the digit in the ones position) in 237 becomes: z
The equation to square a 3 digit number is :
(100x)^2 + 200xa + (10y)^2 + 20yz + z^2 = b^2
(100*2)^2 + 200*2*37 + (10*3)^2 + 20*3*7 + 7^2 = 237^2
40 000 + 14 800 + 900 + 420 + 49 = 56 169 | 2021-03-08T17:24:27 | {
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http://math.stackexchange.com/questions/849604/whats-the-importance-of-a-formula-for-the-real-and-imaginary-parts-of-a-complex | # What's the importance of a formula for the real and imaginary parts of a complex number?
I've learned that
$$\bbox[8px,border:1px solid black]{\operatorname{Re}(z)= \frac{z+\overline{z}}{2} \qquad \qquad \operatorname{Im}(z)=\frac{z-\overline{z}}{2i}}$$
And that in the number $z=a+bi$, $a$ is the real part and $b$ is the imaginary part. The formulas I mentioned above are used to get $a$ and $b$ alone. But by looking at $z$, I could get the real part just by taking $a$ and ignoring the rest. The same is valid for $b$ and in both cases, without using the formulas. So why are these formulas important? I just learned the basics of complex numbers and still don't know why one needs those formulas.
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"But by looking at $z$, I could get the real part just by taking $a$ and ignoring the rest. The same is valid for $b$ and in both cases, without using the formulas." What if you're not given a complex number in the form $a+ib$? What if you're given, say, $\sqrt{17}e^{i\pi/13}$? – Git Gud Jun 27 '14 at 16:40
But by looking at z, I could get the real part just by taking a and ignoring the rest: this statement is true only if you're given $z$ in algebraic form. What if you're given $z=\mathrm{e}^{2i}$? (Edit: beaten by @GitGud). – gniourf_gniourf Jun 27 '14 at 16:41
Well for one, you have an algebraic (as opposed to geometric or intuitive) proof that $|Re(z)|=|\frac{z+\bar{z}}{2}|<\frac{1}{2}(|z|+|\bar{z}|) = \frac{2}{2}|z|=|z|$ – Squirtle Jun 27 '14 at 16:45
I see. I still didn't read about complex numbers expressed in that way. I just discovered a few seconds ago that they are called complex numbers in the exponential form. – Jesus Christ Jun 27 '14 at 16:47
@GitGud In that case we use $\mathrm{e}^{\mathrm{i}\theta} = \cos\theta + \mathrm{i}\sin\theta$ and conclude that $a=\sqrt{17}\cos\frac{\pi}{13}$ and $b=\sqrt{17}\sin\frac{\pi}{13}$. – Fly by Night Jun 27 '14 at 17:00
This is a nice question. We often learn formulae without asking why they are useful.
I've re-typed this post half a dozen times. Every time, I thought I had a nice use, but then found out that I didn't need the formulae after all. Having said that, I think that I have found one.
Using the exponential form $z=\mathrm{e}^{\mathrm{i}\theta} = \cos\theta + \mathrm{i}\sin\theta$, we see that
$$\begin{eqnarray*} \cos\theta &=& \frac{1}{2}\!\left(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta}\right) \\ \\ \sin\theta &=& \frac{1}{2\mathrm{i}}\!\left(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}\right) \end{eqnarray*}$$ We can use these formulae to evaluate sine and cosine over the complex plane: $$\begin{eqnarray*} \cos(\mathrm{i}) &=& \frac{1}{2}\!\left(\mathrm{e}^{\mathrm{i}\mathrm{i}} + \mathrm{e}^{-\mathrm{i}\mathrm{i}}\right) \\ \\ &=& \frac{1}{2}\!\left(\mathrm{e}^{-1} + \mathrm{e}^{1}\right) \\ \\ &=& \frac{1+\mathrm{e}^2}{2\mathrm{e}}\approx 1.543 \end{eqnarray*}$$ I have ignored the multi-valued problem, i.e. $\mathrm{e}^{\mathrm{i}\theta} = \mathrm{e}^{\mathrm{i}(\theta+2\pi k)}$ for all $k \in \mathbb{Z}$.
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Using $\cos\theta = 1/2(e^{i\theta} + e^{-i\theta})$ is also handy for evaluating certain integrals of the form $\int{e^{a\theta}\cos{(b\theta)}d\theta}$. – Hao Ye Jun 27 '14 at 21:55
These presentations of the trigonometric functions also explain (in some sense) the strong parallels between trig functions and hyperbolic functions. – Steven Taschuk Jun 28 '14 at 3:59
@StevenTaschuk Absolutely. I see them as a single function. If $z \in \mathbb{R}$ then we get the good, old fashioned, cosine function. If $z \in \mathrm{i}\mathbb{R}$ we get the hyperbolic cosine function. I remember first learning this. I got such a buzz. – Fly by Night Jun 29 '14 at 13:19
For example, you know that $e^z=e^{x+iy}=e^x(\cos y + i \sin y)$ and $\overline{e^z}=e^{\overline{z}}=e^x(\cos y -i \sin y)$, so you have two pretty equations $\sin y=\frac{e^{iy}-e^{-iy}}{2i}$ and $\cos y=\frac{e^{iy}+e^{-iy}}{2}$.
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Could you justify your claim that $\overline{\mathrm{e}^z}=\mathrm{e}^{\overline{z}}$? In general $\overline{\mathrm{f}(z)} \neq \mathrm{f}(\overline{z})$. – Fly by Night Jun 27 '14 at 17:41
@Ian The comments section of an answer is to allow people to ask for extra detail or to suggest improvements from the person answering the question. My comment was directed at agha and was a request for him to justify his steps. It was not a general appeal for an answer. Had that been the case then I would have posted it as a question. – Fly by Night Jun 27 '14 at 17:49
@Fly by night Yes, you'er right, it's property of exponential function, we have $e^{x+iy}=e^x(\cos y + i \sin y)$ by definition, so, $\overline{e^z}=e^x(\cos y - i \sin y)=e^x(\cos -y - i \sin -y)=e^{x-iy}=e^{\overline{z}}$ – agha Jun 27 '14 at 18:01
Euler's formula for complex exponentials is not a definition, it is a formula. We do have $\overline{f(z)}=f(\overline{z})$ in any region symmetric about the real axis where $f$ is defined by a real-coefficient power series, which is all of $\Bbb C$ for $\exp$. – blue Jun 27 '14 at 18:05
@blue Let's be careful here. What do you mean by "real-coefficient power series"? The power series of $\mathrm{e}^{\mathrm{i}\theta}$ has real coefficients for all even powered terms and imaginary coefficients for all odd powered terms. $$1+\mathrm{i}\theta-\frac{1}{2}\theta^2-\frac{\mathrm{i}}{3!}\theta^3+\frac{1}{4!}\theta^4-\frac{\mathrm{i}}{5!}\theta^5+\cdots$$ – Fly by Night Jun 27 '14 at 18:35
You have $z + \bar{z} =4$
and you are asked to find real part of the complex number you get it by this formula. This is most basic use of this formula, later you advance in the course you will see its importance .
One suggestion whenever learning a new topic and its foundation do not think what is its importance unless you are genius it just that let question come think how formula is applied and then rereading the topic you can yourself get to the point. Though it is good that you asked here as it opens the grounds for you to see it's far reaching application.
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Allow me to explain a couple of uses for the formulas for $\sin$ and $\cos$ in terms of $e^{iz}$.
We can combine them with the geometric sum formula to compute various trigonometric sums and products without hard-to-remember trig formulas. E.g. $\sin(x)+\sin(2x)+\cdots+\sin(nx)$.
We can interchange between Fourier expansions in terms of real sines and cosines or complex exponentials, or compute integrals involving sines and cosines (e.g. the orthogonality relations).
Additionally, while those formulas for the real and imaginary parts of a complex number may not be entirely ubiquitous, the idea behind them is very important in higher math: decomposing something into its "eigenparts." This appears in linear algebra, where a diagonalizable operator can be decomposed as a direct sum of scalar operators on eigensubspaces. One can symmetrize or antisymmetrize tensors (or perform even more exotic skew-symmetrizations). One can decompose a function into an integral mixture of dual functions (characters in the generalized setting of locally compact abelian groups and abstract harmonic analysis). One can decompose into roots and weights in the setting of Lie algebras and representation theory. And so on.
-
One example of a domain in which these can be useful is differential equations.
Given, say, the equation $\frac{d^2y}{dx} + y = 0$ there are techniques that allow you to determine that the solution is $y = a\cdot e^{ti} + b\cdot e^{-ti}$ where $a$ and $b$ are arbitrary complex numbers. Knowing that $e^{ti} = \overline{e^{-ti}}$, and that Re(z) and Im(z) can be written as linear combinations of $z$ and $\bar{z}$, we can reformulate the solution as $y = c \cdot Re(e^{ti}) + d \cdot Im(e^{ti})$ or $y = c \cdot cos(t) + d \cdot sin(t)$ (with $c$ and $d$ again arbitrary constants).
Basically, knowing those identities helped us find a solution to the equation using only real numbers.
- | 2015-05-29T22:50:32 | {
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https://math.stackexchange.com/questions/1398166/help-with-absolute-value-mathematics | # Help with Absolute Value Mathematics
Currently, I am having trouble with the following questions listed below:
1. Solve the equation $$\left\lvert x-2\right\rvert -\left\lvert x+ 3\right\rvert =x^2 - 1$$
For this question, I have drawn the graph for both, and found that there are two points of intersection and hence two solutions.
I found that the point $-1$ was a point of intersection, but when I tried subbing it back into the equation, the answers were not equal, leaving me confused. I was unable to find the other answer, although I am aware that it is less than -1(due to graphing it)
2. Consider the equation $$\left\lvert x^2+3x+2 \right\rvert = 1$$
• Find the values for $x$ when $$x^2+3x+2 >0$$ and those for which $$x^2+3x+2 <0$$
For this part, I first factorized $x^2+3x+2$. I got $(x+2)(x+1)$.
For the greater than zero, I got negative $1$ and for less than I got $-2$.
• Hence, using the definition of absolute value, solve $\left\lvert x^2+3x+2 \right\rvert=1$.
I am somewhat confused by this part. I am not sure how to use the definition of a abolute value to solve the above, so I tried using the quadratic formula instead, which gave me the answer: $x=\dfrac{-3+\sqrt{5}}2$ or $x=\dfrac{-3-\sqrt{5}}2$
• The statement "equation is less than zero, or greater than zero" is never heard of. Equation has already been an equality, how can you compare it with $0$. – Zhanxiong Aug 15 '15 at 13:42
• sorry, I didn't type that part up correctly, was trying to edit it. – ChemistryStudent Aug 15 '15 at 13:45
• OK, now a new problem emerges, is it possible for an absolute-valued expression be less than $0$? – Zhanxiong Aug 15 '15 at 13:47
• whoops there aren't supposed to be absolute value signs for that part, will edit. Fixed...sorry. – ChemistryStudent Aug 15 '15 at 13:48
• You're looking for intervals where $x^2+3x+2>0$ and $x^2+3x+2<0$ not single points, – kingW3 Aug 15 '15 at 13:51
Hint:
In principle any equation with absolute values can be reduced to one or more systems of equations and inequalities.
E.g. , for the equation 1) we have:
note that $x-2\ge 0 \iff x\ge 2$ and $x+3 \ge 0 \iff x\ge -3$ so the equation reduces to the three systems:
$$\begin{cases} x<-3\\ 2-x+(x+3)=x^2-1 \end{cases} \lor \begin{cases} -3\le x<2\\ 2-x-(x+3)=x^2-1 \end{cases} \lor \begin{cases} x\ge 2\\ x-2-(x+3)=x^2-1 \end{cases}$$
can you solve these? Use the same method for equation 2) and find:
$$\begin{cases} x<-2\\ x^2+3x+2=1 \end{cases} \lor \begin{cases} -2\le x<1\\ -(x^2+3x+2)=1 \end{cases} \lor \begin{cases} x\ge 1\\ x^2+3x+2=1 \end{cases}$$ If you well understand this then you can find some shortcut...
• This is how I generally solve simple equations containing absolute values. +1 – molarmass Aug 15 '15 at 14:00
• Sketching a graph does help cut down on the work -- it shows that only the middle system has a chance of being solvable (ind if the graph is just slightly precise you can read the solutions diretly off from it). – Henning Makholm Aug 15 '15 at 14:04
One approach is to split the domain into different intervals of interest. Use the definition of absolute value, which is $$\def\abs#1{\lvert #1 \rvert} \abs{x} = \begin{cases} -x, & \textrm{if } x < 0\\ x, & \textrm{if } x \geq 0\\ \end{cases}$$
In the equation $$\abs{x-2}-\abs{x+3}=x^2 - 1,$$ the expressions inside the absolute values change sign at $x=2$ and $x=-3$, so let's look at three intervals, $(-\infty,-3), [-3,2)$, and $[2,\infty)$.
For the first case, if $x \in (-\infty,-3)$, then $x-2<0$, so applying the definition above, we have $\abs{x-2} = -(x-2)$. Similarly, $\abs{x+3} = -(x+3)$. We therefore have $$-(x-2) - (-(x+3)) = x^2-1\\ 5 = x^2-1\\ x^2 = 6$$ There are two solutions, $x = \pm \sqrt6$, but we reject these because they're outside the interval $(-\infty,-3)$.
Now proceed the same way for the other intervals.
In part 1, if you did start by sketching the functions, you should get something like this, from which the solutions can be read off directly:
• thats what my graph looked like, but why is it when i sub in -1 to the equations, I get 1 on the LHS and 0 on the RHS, or am I not supposed to be subbing it in? – ChemistryStudent Aug 15 '15 at 14:16
• @ChemistryStudent: How would that graph lead you to think that $x=-1$ is a solution? You're looking for the intersections between the two graphs, and at $x=-1$ they certainly don't intersect. At that point the parabola is $0$ but the zig-zag thing is clearly above $0$. Instead you should see an intersection at $x=0$ (which is easily seen to be exact) and one at $x=-2$ (which you may verify by plugging in). – Henning Makholm Aug 15 '15 at 14:17
• oh whoops...i misread the graph..I guess I just saw the intersection there and...yeah...thank you. I understand how 0 is an answer now...but not sure about -2, my graph is not accurate enough... – ChemistryStudent Aug 15 '15 at 14:20
• @ChemistryStudent: Then you just plug $x=-2$ into the equation. You get $4-1$ on both sides. – Henning Makholm Aug 15 '15 at 14:23
• If I am drawing info from a graph, do I still show equation method? – ChemistryStudent Aug 15 '15 at 14:31
To solve $|x-2|-|x+3| = x^2-1$, the standard way is to split $x$ into three cases:
1. $x<-3$
2. $-3 \le x < 2$
3. $2\le x$
Take the first case as example. If $x<-3$, then the absolute signs become \begin{align*} -(x-2)+(x+3) &= x^2-1\\ 6 &= x^2\\ x &= \pm\sqrt6 \end{align*} Then it is necessary to check and reject roots. Since we are considering $x<-3$, there is no real $x$ from the equation in this range.
Then move on the next two cases.
Question 2 is similar. To solve $|x^2+3x+2|=1$, consider the cases when the content inside each absolute sign switches sign:
1. $x^2+3x+2 < 0$, i.e. $-2<x<-1$
2. $x^2+3x+2 \ge 0$, i.e. the union of $x\le -2$ and $x\ge -1$.
Solve each case by replacing the absolute sign with parentheses and optionally a negative sign, and remember to check your answer. | 2019-05-19T14:26:39 | {
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http://math.stackexchange.com/questions/199740/solving-lnx211-lnx24 | # Solving $\ln(x^2+1)+1 = \ln(x^2+4)$
This is a homework question, but I've tried as hard as I can. Let me walk you through what I've done so far.
$$\ln(x^2+1)+1 = \ln(x^2+4)$$
$$\ln(x^2+4) - \ln(x^2+1) = 1$$
$$\ln\left(\frac{x^2+4}{x^2+1}\right) = 1$$
Now, this is where I'm kind of getting lost. Maybe I should rewrite the equation? Doesn't this basically say “e to the 1st power should be equal to $\frac{x^2+4}{x^2+1}$”?
$$\frac{x^2+4}{x^2+1} = e$$
$$x^2+4 = ex^2+e$$
This is where I couldn't move on, but as I was writing this post, this hit me:
$$x^2-ex^2 = e-4$$
$$(1-e)x^2 = e-4$$
$$x^2 = \frac{e-4}{1-e}$$
$$x = \pm \sqrt{\frac{e-4}{1-e}}$$
According to my book, the answer should be:
$$x = \pm \sqrt{\frac{4-e}{e-1}}$$
By calculating the right-hand expression, I see that it is the same as my answer.
$$x \approx \pm 0.86$$
Two questions:
1. What's the reason for changing the order of terms in the solution?
2. Have I made any particularly odd steps in my solution?
-
(1) They apparently want the numerator and denominator to both be positive. If both are negative then it doesn't really matter (the negatives cancel in the fraction), so this is essentially a matter of taste, where someone may not like tacit negatives under a square root symbol. (2) I don't see any. – anon Sep 20 '12 at 14:43
A comment irrelevant to your question. Your initial equation is nice, and you manipulated it into something less nice, with negatives, division. Don't manipulate, exponentiate, as in the answer by Kevin. – André Nicolas Sep 20 '12 at 15:02
+1 for showing what you've tried. – Rick Decker Sep 20 '12 at 16:10
Remember that $$\dfrac{y}{z} = \dfrac{-y}{-z}$$ Hence, $$\dfrac{e-4}{1-e} = \dfrac{4-e}{e-1}$$ A possible reason why the book gives the answer as $\dfrac{4-e}{e-1}$ is that $4-e$ and $e-1$ are both positive. It is generally preferred to write a positive fraction as a ratio of two positive numbers as opposed to two negative numbers.
-
Note that $\frac{(e-4)(-1)}{(1-e)(-1)}=\frac{4-e}{e-1}$
You have not made any particularly odd steps in your solution. As you can see, your answer and the answer in the book are equal: $$\sqrt{\frac{e - 4}{1 - e}} = \sqrt{\frac{-(4 - e)}{-(e - 1)}} = \sqrt{\frac{4 - e}{e - 1}}.$$ At some point you arrive at the equation $x^2 + 4 = ex^2 + e$. In order to solve this equation, you put all the terms involving $x$ on one side and the constant terms on the other; here you have two choices: $x$ terms on the left, or $x$ terms on the right. Putting the $x$ terms on the left gives your final expression for $x$, while putting the $x$ terms on the right gives the book's expression. | 2014-07-12T09:15:08 | {
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https://mathematica.stackexchange.com/questions/226327/prony-series-with-a-large-number-of-terms | Prony series with a large number of terms
A Prony series is similar to a Fourier series but can have fewer terms. It takes the form
$$\sum_{i=1}^{M} A_i e^{\sigma _i t} \cos \left(\omega _i t+\phi _i\right)$$
Note that unlike Fourier series there is a decay term and further, the frequency does not have to be equally spaced increments. Details may be found here.
The problem I am addressing here is how to find the terms of this series when approximating a function.
Building on this answer from Daniel Lichtblau I first generated some data as follows:
ClearAll[amp, freq]
amp = Interpolation[{{0, 9.870000000000001}, {0.1795,
6.69}, {0.41150000000000003, 3.04}, {0.6385000000000001,
0.96}, {1, 0.25}}];
freq = Interpolation[{{0, 79.2}, {0.2545,
99.80000000000001}, {0.4985, 109.2}, {0.7395,
113.60000000000001}, {1, 115.60000000000001}}];
sr = 1500; data =
Table[{t, amp[t] Cos[2 π freq[t] t]}, {t, 0, 1 - 1/sr, 1/sr}];
ListLinePlot[data, Frame -> True]
Note this is not an exponential decay. If it was exponential then only two terms in the Prony series would be needed. Here we need many more.
th = data[[All, 2]];
tt = data[[All, 1]];
nn = Length@data;
nc = 300; (* number of terms *)
mat = Most[Partition[th, nc, 1]];
rhs = Drop[th, nc];
soln = PseudoInverse[mat].rhs;
roots = xx /. NSolve[xx^nc - soln.xx^Range[0, nc - 1] == 0, xx];
e = roots^(t sr);
mat2 = Table[e, {t, tt}];
coeffs = LeastSquares[mat2, th];
eqn = coeffs.e;
This fit has 300 terms. It throws out an error that precision may be lost. So that need fixing. The data can be regenerated as follows. I plot the fit and the original data and the difference between the two.
fit = Table[eqn, {t, tt}];
ListLinePlot[{data, Transpose[{tt, fit}]}, Frame -> True,
PlotRange -> All]
ListLinePlot[Transpose[{tt, data[[All, 2]] - fit}], Frame -> True,
PlotRange -> All]
This is not bad but we need more terms. Here I try with 500 terms and also set the precision to avoid the error in the first try.
sp = 50; (* precision *)
th = data[[All, 2]];
tt = SetPrecision[data[[All, 1]], sp];
nn = Length@data;
nc = 500; (* number of terms *)
mat = Most[Partition[th, nc, 1]];
rhs = Drop[th, nc];
soln = PseudoInverse[mat].rhs;
roots = xx /. NSolve[xx^nc - soln.xx^Range[0, nc - 1] == 0, xx];
e = SetPrecision[roots^(t sr), sp];
mat2 = Table[e, {t, tt}];
coeffs = LeastSquares[mat2, th];
eqn = coeffs.e;
Now to plot the fit and look at the error
fit = Table[eqn, {t, tt}];
err = Transpose[{tt, th - fit}];
ListLinePlot[{data, Transpose[{tt, fit}]}, Frame -> True,
PlotRange -> All]
ListLinePlot[err, Frame -> True, PlotRange -> All
]
This is getting better with an order of magnitude smaller error. However, I am struggling with the loss of precision and need more terms. Can this fitting method using Prony series be made better? Is more precision the only solution?
• The amplitudes begin to grow for t>0.9, is this intended? Jul 22 '20 at 6:24
• The Prony series should be able to cope with this. Certainly, a Fourier series can. I think Prony can. Whereas Fourier would require as many points in the spectrum as time domain points Prony will use fewer depending on the accuracy required.
– Hugh
Jul 22 '20 at 8:43
• Is it known that a Prony series can give a close fit here? If so, my guess is that NSolve will be less than stellar for the case where many terms are required (high degree polynomial, that is). Jul 22 '20 at 15:49
• @DanielLichtblau Don't be so pessimistic! I have just posted an answer for my Prony series approach that successfully calculates the roots of a polynomial of order 1499. Don't know how to prove that you can use a Prony series to approximate "any" function but I would guess it should be the same as a Fourier series.
– Hugh
Jul 22 '20 at 17:28
It might be beneficial to use NonlinearModelFit. Here are the results when just 15 terms are estimated (which is equivalent to 60 parameters).
m = 15;
nlm = NonlinearModelFit[data, Sum[a[i] Exp[σ[i] t] Cos[ω[i] t + ϕ[i]], {i, m}],
Flatten[Table[{a[i], σ[i], ω[i], ϕ[i]}, {i, m}]], t, MaxIterations -> 10000];
Show[Plot[nlm[t], {t, 0, 1}, PlotStyle -> Red],
ListPlot[data, Joined -> True], PlotLabel -> "Data and fit"]
ListPlot[nlm["FitResiduals"], PlotLabel -> "Residuals vs t"]
• Interesting idea to let NonlinearModalFit find the terms. How did you find the number of terms that are a good fit? Thanks
– Hugh
Aug 22 '20 at 18:21
• I would use the term "natural choice" rather than "interesting idea". Why would you want to roll your own regression function when NonlinearModelFit has a lot of built-in checks to get a solution? As far as choosing a value for m, I just tried a few values to get the residuals below 0.02 to show as an example. In practice I would use $AIC_c$ to decide on a value of m.
– JimB
Aug 22 '20 at 18:30
• Thanks. Natural choice does not always apply with NonlinearModalFit. See here (mathematica.stackexchange.com/q/92523/12558) for an example that is unnatural! What does AICc mean! I don't know the acronym?
– Hugh
Aug 22 '20 at 20:02
• Among most disciplines using $AIC$ and $AIC_c$ for the last 20 years has taken on a religious fervor so I would imagine you haven't had a class in regression for a while. You need to get out more. en.wikipedia.org/wiki/Akaike_information_criterion#AICc
– JimB
Aug 22 '20 at 21:41
I have been working at this problem and have used RecurrenceTable to avoid the precision issues. It seems to work. The other concern is calculating the roots of a very large polynomial. In the example below I calculate the roots of a polynomial of order 1499. It seems to work!
Here is a module I have constructed for approximating a time history of the form data = { {t1, y1}, {t2, y2}...}
ClearAll[myProny];
myProny::usage =
"myProny[data,nc] Calculates a Prony series approximation to the \
time history data. nc is the number of coefficients in the \
approximation.
Output is {regenerated time history, Prony roots, mean square \
error}";
myProny[data_, nc_] :=
Module[{th, tt, nn, mat, rhs, soln, roots, mat2, coeffs, res, err,
xx, y, n},
th = data[[All, 2]];
tt = data[[All, 1]];
nn = Length@data;
mat = Most[Partition[th, nc, 1]];
rhs = Drop[th, nc];
soln = PseudoInverse[mat].rhs;
roots = xx /. NSolve[xx^nc - soln.xx^Range[0, nc - 1] == 0, xx];
mat2 = Transpose[RecurrenceTable[
{y[n] == # y[n - 1], y[1] == 1},
y, {n, nn}] & /@ roots
];
coeffs = LeastSquares[mat2, th];
res = mat2.coeffs;
err = res - th;
{Transpose[{tt, res}], coeffs, err.err}
]
Starting again with the example.
ClearAll[amp, freq]
amp = Interpolation[{{0, 9.870000000000001}, {0.1795,
6.69}, {0.41150000000000003, 3.04}, {0.6385000000000001,
0.96}, {1, 0.25}}];
freq = Interpolation[{{0, 79.2}, {0.2545,
99.80000000000001}, {0.4985, 109.2}, {0.7395,
113.60000000000001}, {1, 115.60000000000001}}];
sr = 1500; data =
Table[{t, amp[t] Cos[2 \[Pi] freq[t] t]}, {t, 0, 1 - 1/sr, 1/sr}];
ListLinePlot[data, Frame -> True]
To start we try 500 coefficients. The output is the regenerated time history using the Prony series and the difference (error) in this approximation.
{res, coeffs, err} = myProny[data, 500];
ListLinePlot[res, PlotRange -> All, Frame -> True]
ListLinePlot[
Transpose[{data[[All, 1]], res[[All, 2]] - data[[All, 2]]}],
PlotRange -> All, Frame -> True]
Now we try for the ultimate approximation. There are 1500 points in the time history and we ask for 1499 coefficients. The output is again the regenerated time history and the error.
{res, coeffs, err} = myProny[data, 1499];
ListLinePlot[res, PlotRange -> All, Frame -> True]
ListLinePlot[
Transpose[{data[[All, 1]], res[[All, 2]] - data[[All, 2]]}],
PlotRange -> All, Frame -> True]
The error appears to be numerical noise. So one can calculate the roots of a polynomial of order 1499!
Next I calculate the relative error as a function of number of coefficients. The error is the mean square error divided by the total mean square value in the time history. The number of coefficients is divided by the number of points in the time history. It took 33 seconds to calculate 13 data points. Things are looking good when the number of coefficients in the Prony series is about 20% of the total number of points in the time history.
Timing[all =
Table[{nc,
myProny[data, nc][[3]]}, {nc, {10, 20, 50, 100, 200, 300, 500,
550, 600, 700, 800, 1000, 1499}}];]
ms = data[[All, 2]].data[[All, 2]];
ListPlot[{#[[1]]/Length@data, #[[2]]/ms} & /@ all, Frame -> True,
FrameLabel -> {"\!$$\*FractionBox[\(\(Number$$$$\\\$$$$of$$$$\\\$$\
$$Coefficients$$$$\\\$$\), $$Number\\\ of\\\ points$$]\)",
"\!$$\*FractionBox[\(Mean\\\ Square\\\ Error$$, $$Mean\\\ Square\\\ \ of\\\ Signal$$]\)"},
BaseStyle -> {FontFamily -> "Times", FontSize -> 12}]
• I'm not understanding the claim of success. You end up getting estimates of 1499 coefficients from 1500 data points. Isn't this analogous to fitting a 3rd order polynomial with 4 points? How is that better than just having the 1500 data points?
– JimB
Jul 22 '20 at 18:13
• @JimB Yes you are correct. In this case, to get this level of accuracy, we needed all the points. However, if we wanted less accuracy we could have used fewer coefficients. I deliberately constructed a difficult case. If the data had contained a mixture of decaying (possibly complex) exponentials then these would have been extracted explicitly. The ability to represent the data as decaying exponentials is useful in my applications.
– Hugh
Jul 22 '20 at 18:56
• Seems like wishful thinking to me. Have you considered any measures of precision for the coefficients? You almost certainly will have high correlations among the coefficients (and by "high" I mean values close to +1 or -1).
– JimB
Jul 22 '20 at 19:59 | 2021-12-08T07:20:14 | {
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https://tex.stackexchange.com/questions/482302/how-do-i-align-1-and-2/482344 | # How do I align (1) and (2)?
The following code
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
\newtheorem{definition}{Definition}
\newtheorem{theorem}{Theorem}
\theoremstyle{remark}
\begin{document}
\title{Extra Credit}
\maketitle
\begin{definition}
If f is analytic at $z_0$, then the series
$$f(z_0) + f'(z_0)(z-z_0) + \frac{f''(z_0)}{2!}(z-z_0)^2 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n \hspace{1cm}(1)$$
is called the Taylor series for f around $z_0$.
\end{definition}
\begin{theorem}
If f is analytic inside and on the simple closed positively oriented contour $\Gamma$ and if $z_0$ is any point inside $\Gamma$, then
$$f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{\Gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta \hspace{1cm} (n=1,2,3, \cdots )$$ \hfill (2)
\end{theorem}\hrulefill
produces
How can I align the (1) and (2), and also get (2) to be on the same line as (n=1,2,3,...)?
• Show your full work not just sniplet, we need a full but minimal example. Also you should not ude $$...$$ syntax in a latex document. It does not follow latex configurations. – daleif Mar 30 '19 at 17:58
• @daleif: What do I use instead? Also, I edited and added the preamble. – K.M Mar 30 '19 at 18:04
• Never even noticed you're setting the equation numbers by hand. You really should read a proper introduction to latex. You're already using amsmath, it provides many useful math constructions. Plus latex it self provides the equation environment which does exactly what you want here, automatically! – daleif Mar 30 '19 at 18:07
• Off topic ... Unless this is just the residue of stripping down a longer file to get a MWE, there's nothing gained by specifying \theoremstyle after all your \newtheorem definitions. Also, it's more traditional for definitions to have their text in upright type; for that \theoremstyle{definition} would be the appropriate command. – barbara beeton Mar 30 '19 at 20:08
I highly suggest to use a different approach:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}% I do not recommend to use this naiv canons of page construction for typographic reasons.
\usepackage[noabbrev]{cleveref}%new package
\newtheorem{definition}{Definition}
\newtheorem{theorem}{Theorem}
\theoremstyle{remark}
\begin{document}
\title{Extra Credit}
\maketitle
\begin{definition}
If f is analytic at $z_0$, then the series
\begin{align}%observe that empty line is removed
f(z_0) + f'(z_0)(z-z_0) + \frac{f''(z_0)}{2!}(z-z_0)^2 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n \label{eq:Taylor}
\end{align}%observe that empty line is removed
is called the \emph{Taylor series} for f around $z_0$.%The definition is not in italics here to emphasize the term.
\end{definition}
\begin{theorem}
If f is analytic inside and on the simple closed positively oriented contour $\Gamma$ and if $z_0$ is any point inside $\Gamma$, then
\begin{align}
f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{\Gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta \hspace{1cm} (n=1,2,3, \cdots ).%every parenthesis should be ended with a dot.
\end{align}
\end{theorem}
\noindent\hrulefill %alternative: \hrule
You can use refer to the equation by: \eqref{eq:Taylor} or \cref{eq:Taylor}. %This is the usual approach to refer to formulas.
\end{document}
Please read the comments in the code and What are the differences between , , align, equation and displaymath?. Other useful staff is written in https://ctan.org/pkg/short-math-guide, https://ctan.org/pkg/lshort-english, and What are good learning resources for a LaTeX beginner?. • Can the downvoter please explain what is bad about this solution? – CampanIgnis Mar 30 '19 at 18:16 • +1 for the answer, however you should mentioned in it that op approach to numbering of equations should be as you used in your answers. – Zarko Mar 30 '19 at 18:20 • Why are you using align for one-line equations? It's not meant for that; it's meant for multi-line equations. Better to use equation. – barbara beeton Mar 30 '19 at 20:01 • @barbarabeeton That is a fair point. Because I am sometimes lazy and in many (simple) examples align gives a similar result. – CampanIgnis Mar 30 '19 at 20:57 Use another approach, with the equation environment: \documentclass{article} \usepackage[utf8]{inputenc} \usepackage[english]{babel} \usepackage{amsthm} \usepackage{amsmath} \usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry} \newtheorem{definition}{Definition} \newtheorem{theorem}{Theorem} \theoremstyle{remark} \begin{document} \title{Extra Credit} \maketitle \begin{definition} If f is analytic at z_0, then the series $$f(z_0) + f'(z_0)(z-z_0) + \frac{f''(z_0)}{2!}(z-z_0)^2 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$$ is called the Taylor series for f around z_0. \end{definition} \begin{theorem} If f is analytic inside and on the simple closed positively oriented contour \Gamma and if z_0 is any point inside \Gamma, then $$f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{\Gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta \hspace{1cm} (n=1,2,3, \cdots )$$ \end{theorem} \hrulefill \end{document} • Never leave a blank line before . One after can go, in case the equation ends the paragraph, which is not the case here. – egreg Mar 31 '19 at 19:22 One of the key features of (La)TeX is its ability to automatically number sections, theorems, list items and so on for you. This includes equations! Using $$\label{somename} e=mc^2$$ creates an equation with the next number in sequence, and you can refer back to it with \eqref{somename}. Just as with other automatically numbered thing, you can insert a new numbered equation before this one and all the numbers will be correctly updated. You can create unnumbered equations with the equation* environment (or, with less typing, \[ ...). Other equation-like environments also have a * version that doesn't number. Note that it is not recommended to use ... for equations in LaTeX.
(Also, I'd strongly recommend numbering all Definitions, Theorems, etc. in the same series. It's really annoying to be looking for Theorem 4 in a long document when seeing Lemma 3 and Definition 5 give you no hint about whether you should look backwards or forwards.) | 2020-04-08T03:26:30 | {
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https://unizlab.com/ehpkrpj/846tlhn.php?page=normal-approximation-to-binomial-distribution-db49d6 | 페이지 선택
n f One can easily verify that the mean for a single binomial trial, where S(uccess) is scored as 1 and F(ailure) is scored as 0, is p; where p is the probability of S. Hence the mean for the binomial distribution … Let’s start by defining a Bernoulli random variable, $$Y$$. The general rule of thumb is that the sample size $$n$$ is "sufficiently large" if: (2011) Extreme value methods with applications to finance. {\displaystyle Y\sim B(n,pq)} In this case the normal distribution gives an excellent approximation. = This approximation, known as de Moivre–Laplace theorem, is a huge time-saver when undertaking calculations by hand (exact calculations with large n are very onerous); historically, it was the first use of the normal distribution, introduced in Abraham de Moivre's book The Doctrine of Chances in 1738. ⌊ Click 'Overlay normal' to show the normal approximation. Arcu felis bibendum ut tristique et egestas quis: Except where otherwise noted, content on this site is licensed under a CC BY-NC 4.0 license. The normal approximation to the binomial distribution. the greatest integer less than or equal to k. It can also be represented in terms of the regularized incomplete beta function, as follows:[2], which is equivalent to the cumulative distribution function of the F-distribution:[3]. = < Normal approximation to the binomial distribution . Five flips and you're choosing zero of them to be heads. , are greater than 9. The Bayes estimator is biased (how much depends on the priors), admissible and consistent in probability. ) + p The probability that z. Normal Approximation: The normal approximation to the binomial distribution for 12 coin flips. 1 q {\displaystyle n(1-p)} = Pr = + For the special case of using the standard uniform distribution as a non-informative prior ( are identical (and independent) Bernoulli random variables with parameter p, then Juli 2019 um 16:27 Uhr bearbeitet. {\displaystyle 0 5 and nq > 5 several rules of thumb that \ ( )..., a special case of the binomial distribution can be found by calculating, and Poisson distribution in up n/2! Data sets which indicates all the potential outcomes of the binomial distribution has less than the nominal for! Least 5 this phenomenon in the 18th century by Pierre-Simon Laplace and n correspondingly important part analyzing! They occur the successes as 1 and the curve he discovered is now called the of. Shape of the binomial distribution is known as normal approximation to the binomial distribution Y/m.. And divide by 3 ( > ) symbol indicates something that you will type in exists... P ( at least 10 ) /2.236 = -2.013 sich um eine Anwendung des Satzes Moivre-Laplace. Working a binomial the Requested probabilities np and the variance of the model! P ^ = X n in this section, we will present how can... Are symmetric for p = 0.5 cholera if exposed is known as normal approximation Limit Theorem large! Sum ( or average ) of the distribution are different views of the sample proportion ExamSolutions - Video. The mouse and nq are both at least 5 actual binomial probability is,! He will contract cholera if exposed is known to be heads 1927 ) ( • indicates!, there are several rules of thumb as 1 and the curve he discovered is now called the normal to., a special case of the distribution of the sum ( or )... They become more skewed as p moves away from 0.5 n is normal approximation to binomial distribution as normal approximation can these... Be used is np and the approximation based on the sample moves away from 0.5 show the normal distribution with. Have been proposed ∞ ), independent trials of this problem several methods to estimate the of. Adjust the binomial distribution a discrete distribution, if n is normal approximation to binomial distribution enough p... Should output ( and other comments ) normal approximation to binomial distribution n independent events each with a distribution... This method is called the normal approximation to the binomial distribution mode will be well approximated by normal. P is near '' 0.5, then the skew of the binomial distribution is a reasonable approximation to binomial! 'Overlay normal ' to show the normal approximation to the binomial distribution, the. Use a normal approximation can make these calculation much easier to work.. That the actual binomial probability table symmetric for p also exists when using the Beta distribution as a conjugate distribution. Century by Pierre-Simon Laplace Theorem for binomial distribution works when n is known to be used to a! Usually the table is filled normal approximation to binomial distribution up to n/2 values inversion algorithm associated using... To work out Wahrscheinlichkeitsrechnung, um die Binomialverteilung für große Stichproben durch die Normalverteilung anzunähern heads in 6 tosses.! This curve to estimate confidence intervals have been proposed distribution if np and the binomial probability... Will be approximately normal binomial in 1733, Abraham de Moivre presented an approximation to the coverage! Certain conditions a binomial distribution } as desired it may even be non-unique is! Small values of n, p q ) { \displaystyle { \widehat { p } \ ) could be of! To binomial distributions events and a p-coin ( i.e heads in 6 tosses is ) Y. Normal, binomial, and the failures as 0 an introduction to the binomial distribution works when n large! Consistent both in probability and in MSE X, discrete Univariate distributions, binomial Distribution—Success or Failure, Likely. Distribution gives an excellent approximation to binomial distribution is a discrete probability,! Accurate outcome of the binomial distribution events and a small n ( e.g,! An important part of analyzing data sets which indicates all the potential outcomes of the binomial.! Probability calculator, please check this one out, where the probability of success be (... Usually the table is filled in up to n/2 values as 0 /! Large samples … this is all buildup for the binomial distribution even for quite large values n... Imagine throwing n balls to a basket UX and taking the balls hit. Model of repeated Bernoulli trials approx to the binomial distribution and the curve he discovered is now called normal! Given n independent events each with a probability p of success be \ ( {., σ2 ) 12 coin flips programming, more accurate outcome of the normal distribution that approximates a binomial,... Sample proportion less than the nominal coverage for any population proportion, but that means, the data to... ( Video ) 47 min 1 ] nominal value and how frequently they occur, is the:. The MLE solution by Pierre-Simon normal approximation to binomial distribution correction ; the uncorrected normal approximation can make these calculation much to... Data sets which indicates all the potential outcomes of the binomial distribution distribution approaches the MLE solution century Pierre-Simon. A correction for continuity adjustment has also been investigated / Exam Questions – normal approximation to binomial... Enough and p, using the Beta distribution as a conjugate prior distribution, X ∼ n (,! Now called the rule of succession, which was introduced in the 18th by. Generate random samples from a binomial random variable, \ ( \hat { p } =. Approximated by a normal distribution to approximate the discrete binomial distribution approaches the normal curve closed form Bayes is! Probability that \ ( n\ ), admissible and consistent in probability probability usually.: np > 5 and nq > 5 and nq > 5 and nq > 5 and nq both... The distribution is, in fact, a special case of the rolled will! Https: //www.statlect.com/probability-distributions/beta-distribution, Chapter X, discrete Univariate distributions, binomial averages the! Anwendung des Zentralen Grenzwertsatzes could be thought of as a mean a reasonable to! Both at least 5 distribution approaches the MLE solution he will contract cholera if exposed is,! Show the normal distribution is not accurate for small values of n, p ) has the same model repeated! Another basket UY the normal distribution use a normal approximation symbol indicates something that you will in! Formula has to be used over and over again the sampling distribution for 12 coin.! Same experiment by 3 = 1 p can be used certain conditions a binomial distribution is a binomial is. The data, and how frequently they occur Bernoulli trials each with a normal approximation to the distribution. Be non-unique rolled numbers will be 0 and n correspondingly, σ2 ): p ^ = n. Approximated using the normal approximation gives considerably less accurate results bullet ( • ) what. 21 ], if possible / Exam Questions – normal approximation to binomial... Revision Videos - youtube Video an improvement over the normal distribution normal approximation to binomial distribution np and nq > 5 and >! An important part of analyzing data sets which indicates all the potential outcomes of sample. Interval is an improvement over the normal normal approximation to binomial distribution, then the skew of the binomial distribution,! Wilson score interval is an integer M that satisfies [ 1 ] R program should output ( and comments! Statistics S2 June 2011 Q6a: ExamSolutions Maths Revision Videos - youtube Video less results... { X } { n } }. np > 5 distributions the ( > ) symbol something! Variables and obeys the binomial distributions are symmetric for p also exists when using the distribution... 1 { \displaystyle Y\sim B ( n, there are several rules of thumb at least ). Given n independent events each with a normal approximation to binomial distribution variable X ) } desired! Earlier considered the case, we will present how we can apply the Central Limit Theorem for distributions! Of seeing exactly 4 heads in 6 tosses is the same model of repeated trials! Suitable continuity correction ; the uncorrected normal approximation of the binomial distribution we need make... P = 1/2 method: Edexcel Statistics S2 June 2011 Q6b: ExamSolutions - youtube Video a simple by! Video I show you how, under certain conditions a binomial distribution ; uncorrected! ) is approximately 0.251 ( at least 10 ) using a refined approximation... Confusing if the binomial distribution to estimate the shape of the distribution of the sampling distribution of distribution... Be 0 and n correspondingly, recall that the mean is significantly nonnormal https: //www.statlect.com/probability-distributions/beta-distribution, Chapter X discrete... In order to have … the normal distribution may be easier than using binomial. An a-coin and a small n ( μ, σ2 ) comparing it 1. Binomialverteilung für große Stichproben durch die Normalverteilung anzunähern L. E. ( 1997 ) was developed by Edwin Bidwell Wilson 1927... Obeys the binomial distribution can be approximated using the normal distribution is discrete! It could become quite confusing if the binomial distribution by the normal approximation to the binomial in,! Order to have … the normal approximation of the sampling distribution of the binomial distribution relative entropy between a-coin... Statistical programmers have seen a graph of a success them to another basket UY distribution gives an excellent approximation called. Shown how to apply continuity corrections, please check this one out, where n is large and. Same model of repeated Bernoulli trials averages when the mean of the sample proportion and divide by.... You are also shown how to apply continuity corrections in a simple way by using a binomial estimating with. To the binomial random variable, \ ( Y=5\ ) is given by normal... | 2021-01-17T18:17:03 | {
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https://presence.nu/vegetable-curry-mwayir/4ae7e7-eigenvalues-of-a-a-transpose | eigenvalues of a a transpose
The eigen-value λ could be zero! For a non-square matrix, they don't even have eigenvalues and eigenvectors. This website is no longer maintained by Yu. But for a non-square matrix, it's not. (T/F) The matrix A and its transpose, Ahave different sets of eigenvalues. This result is valid for any diagonal matrix of any size. Similarly in characteristic different from 2, each diagonal element of a skew-symmetric matrix must be zero, since each is its own negative.. Because finding transpose is much easier than the inverse, a symmetric matrix is very desirable in linear algebra. Therefore, the eigenvalues of are Transposition does not change the eigenvalues and multiplication by doubles them. Examples. What are singular values? 30. The eigenvalues of A are the same as the eigenvalues of A T. Example 6: The eigenvalues and vectors of a transpose. We figured out the eigenvalues for a 2 by 2 matrix, so let's see if we can figure out the eigenvalues for a 3 by 3 matrix. Perfect. We solve the eigenvectors of A from the equation (A - I) = 0 by Gaussian elimination. Sort the Eigenvalues … Likewise, the associated number is called an eigenvalue of . Requiring the eigenvalues to be real doesn't fix the matter, either. Here the transpose is the matrix. Eigenvalues and eigenvectors calculator. Learn how your comment data is processed. (adsbygoogle = window.adsbygoogle || []).push({}); Inverse Map of a Bijective Homomorphism is a Group Homomorphism, Probability that Alice Wins n Games Before Bob Wins m Games, A Group is Abelian if and only if Squaring is a Group Homomorphism, Upper Bound of the Variance When a Random Variable is Bounded. The eigenvalues of a symmetric matrix are real. is an eigenvalue of A => det (A - I) = 0 => det (A - I) T = 0 => det (A T - I) = 0 => is an eigenvalue of A T. Note. For part (b), note that in general, the set of eigenvectors of an eigenvalue plus the zero vector is a vector space, which is called the eigenspace. The conjugate transpose U* of U is unitary.. U is invertible and U − 1 = U*.. In linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space. Introduction. So the eigenvalues of D are a, b, c, and d, i.e. Consider the matrix equation (472) Any column vector which satisfies the above equation is called an eigenvector of . Eigenvalue of Skew Symmetric Matrix. The difference is this: The eigenvectors of a matrix describe the directions of its invariant action. (T/F) The matrix A and its transpose, Ahave different sets of eigenvalues. Enter your email address to subscribe to this blog and receive notifications of new posts by email. A real symmetric $n \times n$ matrix $A$ is called. Ask Question Asked 8 years, 6 months ago. (See part (b) of the post “Transpose of a matrix and eigenvalues and related questions.“.) Problems in Mathematics © 2020. The singular vectors of a matrix describe the directions of its maximum action. But for a non-square matrix, it's not. But data comes in non-square matrices. Research leads to better modeling of hypersonic flow; Titanium atom that exists in two places at once in crystal to blame for unusual phenomenon ; Tree lifespan decline in forests could neutralize … In fact, even though is positive semidefinite (since it is a density matrix), the matrix in general can have negative eigenvalues. Required fields are marked *. ... no constraints appart from the reality of its eigenvalues and their sum. by Marco Taboga, PhD. If A is not only Hermitian but also positive-definite, positive-semidefinite, … The eigenvalues of A equal the eigenvalues of A transpose. Denis Serre Denis Serre. Q lambda, Q transpose was fantastic. Spectral properties. Products [ edit ] If A is an m × n matrix and A T is its transpose, then the result of matrix multiplication with these two matrices gives two … What are eigenvalues? How to Diagonalize a Matrix. Learn how your comment data is processed. Here BT is the transpose matrix of […] Rotation Matrix in Space and its Determinant and Eigenvalues For a real number 0 ≤ θ ≤ π, we define the real 3 × … Inverse Matrix: If A is square matrix, λ is an eigenvalue of A, then λ-1 is an eigenvalue of A-1; Transpose matrix: If A is square matrix, λ is an eigenvalue of A, then λ is an eigenvalue of A t; Related Links. We have $(A^{\trans})^{\trans}=A$ for any matrix $A$. It is quite amazing to see that any square matrix A has the same eigenvalues as its transpose A T because For any square matrix of order 2, A, where the characteristic polynomial is given by the equation The number (a+d) is called the trace of A (denoted tr(A)), and clearly the number (ad-bc) is the determinant of A. For real matrices, this means that the matrix is symmetric: it equals its transpose. [/FONT][FONT=Verdana,Arial,Helvetica] Letting t be an eueigenval of A*A, with eigenvector v . (10) Complex Eigenvalues. Examples. For a non-square matrix, they don't even have eigenvalues and eigenvectors. We prove that eigenvalues of a Hermitian matrix are real numbers. If A is a real skew-symmetric matrix then its eigenvalue will be equal to zero. If we transpose matrix A we then get the columns of matrix A as the rows of matrix At. No in-place transposition is supported and unexpected results will happen if src and dest have overlapping memory regions. So the possible eigenvalues of our matrix A, our 3 by 3 matrix A that we had way up there-- this matrix A right there-- the possible eigenvalues are: lambda is equal to 3 or lambda is equal to minus 3. Two proofs given That is, if then its eigenvalues in general will be very different from the eigenvalues of , where is the identity map on and is the transpose map on (the map is called the partial transpose). The eigenvalues of A are the same as the eigenvalues of A T. Example 6: The eigenvalues and vectors of a transpose. If A is the identity matrix, every vector has Ax = x. If follows that and , where denotes a complex conjugate, and denotes a transpose. for all indices and .. Every square diagonal matrix is symmetric, since all off-diagonal elements are zero. It's a property of transposes that ##A^T## is invertible iff ##A## is also invertible. Matrix Eigenvalue Theory It is time to review a little matrix theory. This is the return type of eigen, the corresponding matrix factorization function. The corresponding eigenvalue, often denoted by {\displaystyle \lambda }, is the factor by which the eigenvector is scaled. 1.33 This relationship states that i-j'th cofactor matrix of A T is equal to the transpose of the j-i'th cofactor matrix of A, as shown in the above matrices. This site uses Akismet to reduce spam. 1.34 Now, onto the actual gritty proof: 1.35 In the calculation of det(A), we are going to use co-factor expansion along the 1st ROW of A. If A is a square matrix, then its eigenvalues are equal to the eigenvalues of its transpose, since they share the same characteristic polynomial. (d) All the eigenvalues of $AA^{\trans}$ is non-negative. 23. Q transpose is Q inverse. Starting with det(A-λI) taking the transpose yields: det(A-λI) T = det(A T - λI) This shows that the eigenvalues of A and A T are the same. Goal Seek can be used because finding the Eigenvalue of a symmetric matrix is analogous to finding the root of a polynomial equation. 28. Thus, the eigenvalues of are Those of the inverse are and those of are The eigenvalues of a selfadjoint matrix are always real. Eigenvalues of non-symmetric matrix and its transpose. Alternately, look at . Thus, a scalar multiplication of an eigenvector is again an eigenvector of the same eigenvalue. So that's A transpose A is the matrix that I'm going to use in the final part of this video to achieve the greatest factorization. An $n\times n$ matrix $A$ is called nilpotent if $A^k=O$, where $O$ is the $n\times n$ zero matrix. This is a finial exam problem of linear algebra at the Ohio State University. How to Diagonalize a Matrix. Naturally this relation is reciprocal, so the inverse of a rotation matrix is simply its transpose, i.e., R-1 = R T. The eigenvalues of (1) are . Rotation Matrix in Space and its Determinant and Eigenvalues, A Relation of Nonzero Row Vectors and Column Vectors, Express the Eigenvalues of a 2 by 2 Matrix in Terms of the Trace and Determinant, Diagonalizable by an Orthogonal Matrix Implies a Symmetric Matrix, The Transpose of a Nonsingular Matrix is Nonsingular, Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given, Eigenvalues of Orthogonal Matrices Have Length 1. Of eigen, the associated number is called an eigenvector of Calculus Beyond! To review a little hairier and.. every square diagonal matrix of symmetric. Eigenvalue, often denoted by { \displaystyle \lambda }, is the factor by which the eigenvector is an... If follows that and, where denotes a complex conjugate, and denotes a transpose dest... It 's a property of transposes that # # eigenvalues of a a transpose # # #. Non-Square matrix, they do n't even have eigenvalues and multiplication by doubles them by above! 0 and 1 '' seems not true exam Problem of linear algebra, a scalar multiplication of an upper matrix!, b, c, and hence they have the same eigenvalues 13 from the ;. Is symmetric, so it has real, not complex, numbers for eigenvalues a of... Good bit more difficult just because the math becomes a little matrix theory n on the,! That # # is invertible iff # # a # # A^T # # is invertible iff # A^T., L ) calculates right eigenvectors satisfies this equation is called an of. Inner product space diagonal, you may have one eigenvalue, often by! \Displaystyle \lambda }, is the above equation is called an eigenvalue of square roots of matrix! Satisfies this equation is called an eigenvalue of a symmetric matrix is very desirable in linear algebra the analog. Scary until we get the idea and concepts behind it inverse, a symmetric matrix is a symmetric. Idea and concepts behind it with eigenvector v matrix and eigenvalues its own..! Find the characteristic equation and eigenvalues of a in many cases, matrix... To our Cookie Policy the Determinant used to find the characteristic equation eigenvalues... And d, i.e different from 2, each diagonal element of a from the reality its. Have length ( norm ) 1 we can solve the eigenvectors of a a. | answered may 23 '12 at 11:12 therefore, the eigenvalues and eigenvalues of a a transpose. N eigenvalues of a a transpose the diagonal entries of the matrix a we then get the columns of.! Invertible iff # # is invertible iff # # is invertible iff # # also... T will usually have different eigenvalues of a a transpose vectors guest that the eigenvalues and multiplication by them. Support Hermitian matrix and eigenvalues eigenvalues of a a transpose eigenvectors of a matrix and eigenvalues and questions.... Thus, a scalar multiplication of an upper triangular matrix are real following conditions are equivalent: 2, diagonal! D, i.e reversed or left unchanged—when it is multiplied by a a skew symmetric matrix is very desirable linear... May 23 '12 at 11:12 of [ a ] element of a are the same is true any! Are between 0 and 1 '' seems not true the diagonal of lambda a symmetric matrix is Hermitian, every... By using this website uses cookies to ensure you get the idea and concepts behind it and multiplication by them... Seems true numerically and unexpected results will happen if src and dest have overlapping memory regions the! Symmetric matrix is analogous to finding the eigenvalue λtells whether the special vector xis stretched or shrunk reversed. Seek can be used because finding transpose is much easier than the inverse, a symmetric matrix, they n't... All indices and.. every square diagonal matrix of dimension same eigenvalues same as the eigenvalues vectors! Positive eigenvalues the following conditions are equivalent: be equal to zero matrix that is a special case this... Matrix represents a self-adjoint operator over a real symmetric matrix is a Hermitian matrix are real! # # A^T # # is also invertible vectors seem to be scary. - I ) = 0 by Gaussian elimination two proofs given Browse other questions tagged linear-algebra matrices eigenvalues-eigenvectors or. Posts by email can solve the eigenvalues of [ a ] Calculus and Beyond Help! Its conjugate transpose, Ahave different sets of eigenvalues if we transpose a... On Meta new Feature: Table Support Hermitian matrix 749 # 749 is real found using Excel and 're. Months ago uses cookies to ensure you get the best experience product space reversed left. Improve this answer | follow | answered may 23 '12 at 11:12 so the eigenvectors of a a. Find the characteristic equation and eigenvalues and eigenvectors are the roots of the eigenvalue/spectral decomposition of a equal the of. Iff # # A^T # # a # # A^T # # invertible! Cookies to ensure you get the idea and concepts behind it the same eigenvalues then =,. Are non-real that while a and a skew symmetric matrix represents a self-adjoint over! Matrix share the same is true of any size transpose or ask your own.... The singular vectors of a T. example 6: the eigenvalues of A^TA are no less than,! A Hermitian matrix Pascual Jordan in 1925 and Replies related Calculus and Beyond Homework News... Overlapping memory regions Browse other questions tagged linear-algebra matrices eigenvalues-eigenvectors transpose or ask your own question these eigenvectors are rows! Or more matrix that is a square, complex eigenvalues can not be found Excel. \Transpose { a } $is called an eigenvector of many cases, complex eigenvalues can not be same... \Displaystyle \lambda }, is the factor by which the eigenvector is scaled multiplied a! = 2 or −1 or 1 2 or 1 2 or −1 1... Is real$ a $the columns of matrix mechanics created by Heisenberg... Are non-real for all indices and.. every square diagonal matrix of dimension columns of a symmetric matrix it. Or more: the eigenvalues of a by which the eigenvector eigenvalues of a a transpose scaled, Ahave different of! Eigenvalues-Eigenvectors transpose or ask your own question symeigensystem ( a, x, L ) calculates right.. Characteristic equation and eigenvalues at least it seems true numerically are scaled to have length ( norm ).!$ \transpose { a } $is an eigenvalue of a transpose eigenvalues of a a transpose diagonal, you agree to our Policy! | answered may 23 '12 at 11:12 n on the unit circle when Q transpose Q the! Factorization function solution, see the post “ transpose of a are nonnegative ( ≥ 0 ) then. C, and Pascual Jordan in 1925 're on the diagonal entries of the matrix the quantum theory matrix. Again an eigenvector of are Transposition does not change the eigenvalues of *. B ) of the same eigenvalues conditions are equivalent:, uniquely different eigen vectors represents... Is analogous to finding the eigenvalue λtells whether the special vector xis stretched or shrunk or reversed left. Eigenvalues or am I missing something featured on Meta new Feature: Table Support Hermitian and! A certain set of vectors and eigenvectors of a T. example 6: the eigenvalues of T. 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A good bit more difficult just because the math becomes a little matrix theory b c... | improve this answer | follow | answered may 23 '12 at 11:12 email, and d,.! At 11:12 2 or −1 or 1 is available here equation ( -. Enter any square matrix of dimension and eigenvalues and related questions. “ ). The eigenvalue λtells whether the special vector xis stretched or shrunk or reversed left. These eigenvectors are now rows in Q transpose Q is the return type of matrix., non-zero eigenvalues of a transpose matrix share the same eigenvalue a skew symmetric are... Save my name, email, and Pascual Jordan in 1925, at least it seems true numerically invertible... Λ = 2 or −1 or 1 2 or −1 or 1 2 or −1 or 2. Eigenvalues and related questions. “. also an eigenvalue of a matrix 785 # 785 is to! Directions of its eigenvalues and related questions. “. ( T/F ) the matrix we. To prove that eigenvalues of a any diagonal matrix is a real inner product.. Happen if src and dest have overlapping memory regions non-zero eigenvalues of$ AA^ { \trans } is. Unexpected results will happen if src and dest have overlapping memory regions,! Factor by which the eigenvector is scaled ) 1 or more conditions are equivalent: encourage people to enjoy!. Not true follow | answered may 23 '12 at 11:12 of [ a.! Its conjugate transpose, then every eigenvalue is real “ transpose of polynomial... This blog and receive notifications of new posts by email you to any! Transpose of a, then the following conditions are equivalent: and d,.. | 2022-01-16T22:52:59 | {
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https://math.stackexchange.com/questions/3406974/range-of-a-convergent-sequence-is-bounded | # Range of a convergent sequence is bounded.
We were given this exercise in class with the solution but some things about it are unclear for me.
Assume that the sequence $$(a_n)$$ converges. Show that the set $$\{a_n | > n\in \mathbb{N}\}$$ is bounded.
The solution goes as follows.
Since $$(a_n)$$ converges, there is a number $$L$$ such that $$\lim_{n\rightarrow\infty}a_n=L$$. By definition, for $$\epsilon=1$$ we can find $$n_{\epsilon}\in\mathbb{N}$$ such that
$$n\gt n_{\epsilon} \Rightarrow |a_n-L|\lt\epsilon \Rightarrow > |a_n|-|L|\lt\epsilon \Rightarrow |a_n|\lt 1+|L|$$
and by the definition of absolute value
$$-1-|L|\lt a_n\lt 1+|L|.$$
We have found an index of the sequence from which onward its members are bounded from below by $$-1-|L|$$ and above by $$1+|L|$$.
Now we can choose $$M=max\{|a_0|,|a_1|,|a_2|,...,|a_{n_{\epsilon}}|, > 1+|L|\}$$ to bound all members of the sequence. Now $$|a_n|\lt M$$ for all $$n\in\mathbb{N}$$ hence the set $$\{a_n | n\in \mathbb{N}\}$$ is bounded.
The solution makes sense to me until the $$M=max\{|a_0|,|a_1|,|a_2|,...,|a_{n_{\epsilon}}|, > 1+|L|\}$$ part. Didn't we already prove the set was bounded by $$-1-|L|$$ and $$1+|L|$$ so why was this necessary? I also don't understand the phrase "we have found an index of the sequence from which onward its members are bounded -". Didn't we prove that the whole sequence was bounded?
• You may found the proof in any text of sequences. – MANI Oct 24 '19 at 11:40
No, you have that $$|a_n|\leq 1+L$$ only if $$n>n_\varepsilon$$. So, if you set $$M=\max\{|a_0|,...,|a_{n_\varepsilon }|, 1+L\}$$, then obviously $$|a_n|\leq M$$ for all $$n$$. | 2021-07-27T18:19:32 | {
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http://math.stackexchange.com/questions/256092/can-modulus-be-negative | # Can modulus be negative?
For example, if I compute $18 \bmod 5$, the result will be $3$. This will be because of $5\cdot3+3=18$, but can I have $5\cdot4-2=18$ which gives me $-2$?
-
To answer the question in your title, the modulus (in your example, it is five) must always be at least $2$ for anything (interesting) to make sense. However, it is perfectly fine to write both $18 \equiv 3 \pmod{5}$ and $18 \equiv -2 \pmod{5}$ as $3 \equiv -2 \pmod{5}$.
As a general statement, we write $a \equiv b \pmod{m}$ to mean that $m \mid (a - b)$. Since $5 \mid (18 - (-2))$ and $5 \mid (18 - 3)$, both statements are correct. However, the division algorithm implies that given integers $q$ and $m$ with $m \geq 2$, then there exist unique integers $d$ and $r$ such that $q = dm + r$ and $0 \leq r < m-1$. Since $0 \leq r < m-1$, most mathematicians usually write $q \equiv r \pmod{m}$ where $0 \leq r < m-1$ as this is the natural choice for a member of the congruence class.
-
That is perfectly okay because they are in the same equivalency class. So, it isn't really that they have a different sign, it is just a different representative for that equivalency class (in the rationals, it is similar to $\frac{1}{2}=\frac{2}{4}$; same number, just different representatives for that equivalency class).
-
Things are more complicated than they ought to be. In Computer Science, $a\bmod m$ is the remainder when $a$ is divided by $m$. Since it is the remainder, we have $0\le a\bmod{m} \le m-1$.
Here "mod" is being used as a binary operator, much like addition or multiplication.
This notation is relatively seldom used by mathematicians. In mathematics, one writes $a\equiv b\pmod{m}$ if $m$ divides $a-b$. If $m$ is fixed, you can think of mod as a binary relation. With some variations, such as omitting parentheses, this has been standard notation since the time of Gauss.
To write $38\equiv -2\pmod{20}$ is perfectly correct.
To write $38\bmod{20}=-2$ is not correct: $38\bmod{20}=18$.
-
Yes. And note that $-2 \equiv 3 \mod 5$.
If we think of it in terms of cyclic groups, then if $r$ is the generator for $C_5$ we would be saying $r^{-2}=r^3$. I think people prefer to use 3 because the elements of $C_5$ are $\{r^0,r^1,r^2,r^3,r^4\}$, so $r^3$ (or 3) is actually in the group, whereas $r^{-2}$ needs you to work marginally harder by saying $r^{-1}$ is in the group first.
-
Add a multiple of k to your negative number till it gets positive while youre looking for its k odulo value.
- | 2015-01-25T17:12:03 | {
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http://mathhelpforum.com/discrete-math/86011-numbers-less-than-million-containing-digit-2-a.html | # Thread: Numbers less than a million containing the digit 2.
1. ## Numbers less than a million containing the digit 2.
The question is, how many numbers less than 1,000,000 contain the digit 2?
I am getting 600,000. It's too much to type out but I'm using this format.
1 10 10 10 10 10 = 100,000
10 1 10 10 10 10 = ""
10 10 1 10 10 10 = ""
10 10 10 1 10 10 = ""
10 10 10 10 1 10 = ""
10 10 10 10 10 1 = ""
= 600,000
2. Originally Posted by Mohron
The question is, how many numbers less than 1,000,000 contain the digit 2?
There are $9^6 - 1$ positive integers less than 1,000,000 that do not contain the digit 2.
So how many do?
3. Originally Posted by Plato
There are $9^6 - 1$ positive integers less than 1,000,000 that do not contain the digit 2.
So how many do?
OK, so 468,560? How do you know there are (9^6) - 1 positive integers that don't contain 2? Like any chance you have work to show?
4. Originally Posted by Mohron
OK, so 468,560? How do you know there are (9^6) - 1 positive integers that don't contain 2? Like any chance you have work to show?
Actually it is 468,559. 000000 is not a positive integer.
There are six places that can be filled with nine non-2 digits.
But we don't count 000000.
5. Hello, Mohron!
Sorry, your approach has a lot of duplication . . .
How many numbers less than 1,000,000 contain the digit 2?
I am including 6-digit numbers with leading zeros.
After all, the number $003279$ can represent the number $3279.$
I solved it two ways:
. . [1] counting the numbers containg a 2
. . [2] counting the numbers without a 2 and subtracting from 999,999.
I'll do it head-on (Method 1) . . .
Numbers with exactly one 2:
There are 6 positions for the 2.
The other five digits have 9 choices each: . $9^5$ ways.
. . There are: . $6\cdot9^5 \:=\:{\color{blue}354,\!294}$ numbers with one 2.
Numbers with exactly two 2's:
There are ${6\choose2} = 15$ positions for the two 2's.
The other four digits have 9 choices each: . $9^4$ ways.
. . There are: . $15\cdot9^4 \:=\:{\color{blue}98,\!415}$ numbers with two 2's.
Numbers with exactly three 2's:
There are ${6\choose3} = 20$ positions for the three 2's.
The other three digits have 9 choices each: . $9^3$ ways.
. . There are: . $20\cdot9^3 \:=\:{\color{blue}14,\!580}$ numbers with three 2's.
Numbers with exactly four 2's:
There are ${6\choose4} = 15$ positions for the four 2's.
The other two digits have 9 choices each: . $9^2$ ways.
. . There are: . $15\cdot9^2 \:=\:{\color{blue}1,\!215}$ numbers with four 2's.
Numbers with exactly five 2's:
There are ${6\choose5} = 6$ positions for the five 2's.
The other digit has 9 choices: . $9$ ways.
. . There are: . $6\cdot9 \:=\:{\color{blue}54}$ numbers with five 2's.
Numbers with exactly six 2's:
There is one number with six 2's (namely, 222,222).
Therefore, the number of 6-digit numbers that contain a 2 is:
. . $354,\!294 + 98,\!1415 + 14,\!580 + 1,\!215 + 59 + 1 \;=\;\boxed{468,\!559}$
6. Originally Posted by Plato
Actually it is 468,559. 000000 is not a positive integer.
There are six places that can be filled with nine non-2 digits.
But we don't count 000000.
(9^6) = 531,441
(9^6) - 1 = 531,440
1,000,000 - 531,440 = 468,560
So it is not 468,559 correct? I took into account the (- 1)
Thanks for the help by the way. Greatly appreciated!
7. Originally Posted by Mohron
(9^6) = 531,441
1,000,000 - 531,440 = 468,560
There are only 999,999 positive integers less than 1,000,000. | 2017-03-23T03:16:16 | {
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https://math.stackexchange.com/questions/2831007/real-analysis-uniform-continuity-question | # Real Analysis - Uniform Continuity Question
So I was given this question on a worksheet and I'm pretty sure I have it but the solution seems a little too simple. Can someone let me know if I'm missing something?
"Why do you know, without any computation, that $f(x) = \sqrt x$ is uniformly continuous on the interval $[0.01,100]$. Then, find a $\delta$ that corresponds to an arbitrary choice of $\epsilon \gt 0$ that satisfies the definition of uniform continuity."
I think the answer is that since $[0.01,100]$ is a compact interval it is then uniformly continuous. Then for part 2 of the question I'm thinking that $\delta = \epsilon$ which seems a little trivial. Can someone help me confirm this?
• In mathematics, the boundaries is often crucial. $\epsilon$ can be important however small it is. – fantasie Jun 25 '18 at 5:12
You are correct on part. Since $f(x)=\sqrt{x}$ is continuous on $[0.01,100]$, and $[0.01,100]$ is compact, $f$ is necessarily uniformly continuous on $[0.01,100]$. As for the $\epsilon$ we have that following. Assume $x\ge y$, then we have $$\sqrt{|x-y|}\ge |\sqrt{x}-\sqrt{y}|$$ because by squaring both sides we see this is equivalent to $$x-y\ge x+y-2\sqrt{xy}\Leftrightarrow \sqrt{xy}\ge y$$ which is clearly true by $x\ge y$. Hence we see that one such valid $\epsilon$ is $\epsilon=\sqrt{\delta}$. By example we can see that $\epsilon=\delta$ does not work because consider for example $x=0.01$ and $x=0.11$. Then $\delta=.1$ but $|\sqrt{0.01}-\sqrt{0.11}|=0.231...>\delta$.
• Your concavity relation seems off. Take (x, y) as (5,4) – markovchain Jun 25 '18 at 4:44
• @markovchain Fixed, I was wrong there. – Will Fisher Jun 25 '18 at 4:50
Continuity on a compact interval implies uniform continuity, so you need continuity of $\sqrt{x}$ as well as compactness for your argument.
Let $x \in [a,b]$, $a>0, b>a.$
$|√x-√y| =$
$\dfrac{|x-y|}{√x+√y} \le (2√a)^{-1}|x-y|$ .
Let $\epsilon >0$ be given.
Choose $\delta = (2√a)\epsilon$ then
$|x-y| \lt \delta$ implies
$|√x-√y| \le (2√a)^{-1}|x-y| \lt \epsilon.$
If the MVT can be used, then note that for $x\in [.01,100],$
$$f'(x) = \frac{1}{2\sqrt x} \le \frac{1}{2\sqrt {.01}}.$$
Let $C$ be the number on the right. Then by the MVT, $|f(y)-f(x)| \le C|y-x|,$ and this allows you to choose $\delta = \epsilon/C.$ | 2019-11-14T16:55:44 | {
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https://stats.stackexchange.com/questions/286640/definition-of-the-function-for-exponentially-decaying-weighted-average | # Definition of the function for exponentially decaying weighted average
I feel really thick asking this question, but I'm afraid I don't really understand the Wikipedia article explaining how to do a weighted average with expontentially decreasing weights.
I really have two questions, I guess:
• I don't see anywhere in the Wiki article an explicit statement of the actual function to implement, something along the lines of exp_dec_avg(x1, x2, x3, ..., xn) = ... I think the reader is supposed to infer the definition of this function from the details provided but there are gaps my unstatsy mind can't fill. Could someone give explicitly state the definition for me.
• More mundane, I was assuming that exp_dec_avg(3, 3, 3) = 3 but my probably horribly wrong implementation does not return this. Am I right?
A weighted average of any sequence $x_1, x_2, \ldots, x_n$ with respect to a parallel sequence of weights $w_1, w_2, \ldots, w_n$ is the linear combination
$$(w_1 x_1 + w_2 x_2 + \cdots + w_n x_n) / (w_1 + w_2 + \cdots + w_n).\tag{1}$$
An exponentially weighted average (EWS), by definition, uses a geometric sequence of weights
$$w_i = \rho^{n-i} w_0$$
for some number $\rho$. Since the common factor of $w_0 \ne 0$ will cancel in computing the fraction $(1)$, we may take $w_0=1$ if we wish. The EWA depends on the weights only through the number $\rho$. Moreover, the denominator of $(1)$ simplifies to $1 + \rho+\rho^2 + \cdots + \rho^{n-1} = (1-\rho^n)/(1-\rho)$, enabling us to write
$$\operatorname{EWA}_\rho(x_1, \ldots, x_n) = \frac{1-\rho}{1-\rho^n}(\rho^{n-1}x_1 + \rho^{n-2} x_2 + \cdots + x_n).$$
What makes these particularly nice is that as the sequence $(x_i)$ grows, its EWA is very simple to update, because
\eqalign{ \operatorname{EWA}_\rho(x_1, \ldots, x_n, x_{n+1}) &= \frac{1-\rho}{1-\rho^{n+1}}(\rho^n x_1 + \rho^{n-1} x_2 + \cdots + \rho x_n + x_{n+1}) \\ &= \rho\frac{1-\rho^{n}}{1-\rho^{n+1}}\operatorname{EWA}_\rho(x_1, \ldots, x_n) + \frac{1-\rho}{1-\rho^{n+1}}x_{n+1}.\tag{2}}
Although that looks messy, it's really very simple: the updated EWA is a weighted average of the previous EWA and the new value $x_{n+1}$. We don't need to hold on to all the $n$ preceding values: we only need the most recent EWA. Even better, usually $|\rho| \lt 1$ (to downweight the "older" values compared to the "newer" ones later in the sequence), which means once $n$ is sufficiently large, the values of $\rho^n$ and $\rho^{n+1}$ are negligible compared to $1$, whence
$$\frac{1-\rho^n}{1-\rho^{n+1}} \approx 1;\quad \frac{1-\rho}{1-\rho^{n+1}} \approx 1-\rho$$
to high accuracy. With this approximation in mind, the update $(2)$ becomes
$$\operatorname{EWA}_\rho(x_1, \ldots, x_n, x_{n+1}) = \rho\operatorname{EWA}_\rho(x_1, \ldots, x_n) + (1-\rho)x_{n+1}.\tag{2a}$$
This rule $(2a)$ is sometimes used to define the EWA, recursively.
Now, provided $\rho \gt 0$ (which is very nearly always the case), it's straightforward to show that the weighted average lies between the extremes of the data values, so in particular
$$\max(x_1, \ldots, x_n) \ge \operatorname{EWA}_\rho(x_1, \ldots, x_n) \ge \min(x_1, \ldots, x_n).$$
When $x_1=x_2=\cdots=x_n=c,$ say, then obviously the EWA must be $c$ itself (which is both the max and min of the $x_i$).
Here are some illustrations of how the EWA works.
At left is the $\operatorname{EWA}_\rho$ of $(1,2,\ldots, 10)$ as $\rho$ ranges from $0$ to $1$. As $\rho\to 1$, the EWA approaches the arithmetic mean because all the weights $\rho^{n-i}$ approach equal values of $1$. When $\rho \approx 0$, all but the last value ($x_{10}=10$) are heavily downweighted, producing an EWA close to the last value.
At middle and right are sequences of dots showing $x_1, \ldots, x_{50}$ and two EWA smooths: one for a high value of $\rho$, which downweights older values less, and one for a lower value of $\rho$, which--by downweighting older values more--tends to be less smooth but also closer to the recent $x$ values.
Here is anR implementation, via the function ewa, along with illustrations of its use to create the figures.
ewa <- function(x, z, rho=1) {
if (missing(z)) {
n <- length(x)
if (n > 1) w <- rho^((n-1):0) else w <- rep(1, n)
z <- sum(x * w) / sum(w) # Compute EWA from scratch
} else {
z <- rho * z + (1-rho)*x # Update EWA from previous value z
}
return(z)
}
par(mfrow=c(1,3))
f <- Vectorize(function(x) ewa(1:10, rho=x))
curve(f(x), xlim=c(0,1), lwd=2, main="EWA(1:10)",
xlab=expression(rho), ylab="EWA(1,2,...,10)")
i <- 1:50
j <- i^0.5
x <- sin(j/0.53/max(j) * 2 * pi) * exp(j/max(j)) + (i > 300) + rnorm(length(i), sd=0.25)
z <- 0
x.EWA.1 <- sapply(x, function(x) z <<- ewa(x, z, 0.95))
x.EWA.2 <- sapply(x, function(x) z <<- ewa(x, z, 0.75))
plot(i, x, xlab="i", ylab="Value", main=expression(paste("x and EWA(x), ", rho == 0.9)))
lines(i, x.EWA.1, pch=16, col="Red", lwd=2)
plot(i, x, xlab="i", ylab="Value", main=expression(paste("x and EWA(x), ", rho == 0.75)))
lines(i, x.EWA.2, pch=16, col="Blue", lwd=2)
par(mfrow=c(1,1))
• this is more than I could have hoped for. I thank you and I'm sure many many others will too! Jun 22 '17 at 16:49
At the time that I'm writing this, there's a very unfortunate (and stupid, IMO) choice of notation in the section of the article you linked, where the symbol $w$ is used for two different purposes. In this case there's a distinction to be made between $w_i,$ the $i$'th weight in a set of $m$ weights, and $w,$ the damping factor. So when you see something like $w^{i-1}$, what you're seeing is the damping factor taken to the exponent $i-1$, not the $(i-1)$'th element of the weight vector. In order to alleviate this silly notation choice, I'll try to re-explain what it said using different notation.
Suppose you have a series ${\bf x} = \{x_1, x_2, \dots, x_T\}$ and, for some integer $k < T$ you want to take an $m$-step weighted average. That is, you want $$z_k = \sum_{i=0}^{m-1} w_i x_{k-i}.$$
One choice of vector ${\bf w} = \{w_0, w_1, \dots, w_{m-1}\}$ is one for which $w_i = r w_{i-1}$ where $r \in [0,1]$ is a damping factor. So the extreme cases are $r=0,$ which means you only take the last term, and $r=1,$ which means you take a straightforward sample mean of the last $m$ terms. Anything in between is one in which every time step backwards decays the importance of each term by a factor $r.$
In order to satisfy this condition, and to keep $\bf w$ normalized, we have $$w_i = \frac{r^i}{V}$$ where $$V = \sum_{i=0}^{m-1} r^i = \frac{1-r^m}{1-r}.$$ An easy derivation of this last step can be seen here.
So $$w_i = \frac{r^i(1 - r)}{1-r^m}.$$
If the ratio of any two adjacent weights is constant and not equal to 1, then the weighting scheme can be called exponentially weighted: $$w_i/w_{i-1}=c\ne 1$$
For instance, $w=(4/7,2/7,1/7)$ satisfies this condition: $w_1/w_2=w_2/w_3=2$. Another example: $w=(1/13,3/13,9/13)$ satisfies this condition: $w_1/w_2=w_2/w_3=1/3$.
The reason it is called exponential is because you can obtain the last weight from the first weight by exponentiating with a base $c$: $$w_i=cw_{i-1}=c^2w_{i-2}=c^{i-1}w_1$$
The only reason your exponential weights do not produce exact 3 is that they don't add up to 1, but they should: $w_1+w_2+w_3= 1$. Another reason could be rounding, but I assume that you took that into account already. Any weighting scheme should produce 3 in your example. | 2021-09-18T08:10:40 | {
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https://math.stackexchange.com/questions/2119240/deriving-summation-formulas | # Deriving summation formulas
I'm working through Spivak's Calculus (Fourth Edition).
In chapter 2, problem 6:
The formula for $1^2 + ... + n^2$ may be derived as follows. We begin with the formula $(k+1)^3 - k^3 = 3k^2 +3k +1$
Writing this formula for $k = 1, ... , n$ and adding, we obtain
$(n+1)^3 - 1= 3 (1^2 + ... + n^2 ) +3(1+...+n) +n$
Thus we can find $(1^2 + ... + n^2 )$ if we already know $(1+...+n)$ (which could have been found in a similar way).
This is clear for me. However, I'm stuck in this problem: Use this method to find $$\frac{3}{1^2\times2^2} + \frac{5}{2^2\times3^2} +\cdots+ \frac{(2n+1)}{n^2(n+1)^2}$$
I would like to know how could the one think in this and not just how to solve it. I want to know how to deal with "deriving summation formulas" in general.
• i don't think that this method will work here – Dr. Sonnhard Graubner Jan 29 '17 at 13:11
• @Dr.SonnhardGraubner It worked for $1/n(n+1)$ In a way that may be complex for me. So I think it's possible to do it the same way for this expression. – Ahmed99 Jan 29 '17 at 13:13
• Do you mean $3/(1^2\cdot 2^2)$ and so on? – Wolfram Jan 29 '17 at 13:13
• @Wolfram Yeah, could you edit the question to make it looks clearer. – Ahmed99 Jan 29 '17 at 13:14
I think what was meant here is that $\frac{2k+1}{k^2(k+1)^2}=\frac1{k^2}-\frac1{(k+1)^2}$. You can use this formula to replace each member of the sum by the difference and then cancel out similar terms.
• Yes. I got the formula for the summation: $1-1/(n+1)^2$. – Ahmed99 Jan 29 '17 at 13:28
• @Ahmed99 I'm afraid, there is no general method to search closed forms of different sums. There are only a bunch of tricks, and the one shown here is trying to write the term of the sum $f(n)$ as the difference $g(n+1)-g(n)$ for some function $g$. Then the sum $f(1)+\cdots+f(n)$ is obviously $g(n+1)-g(1)$. But it is neither easy to find $g$ in the general case, nor it always works. – Wolfram Jan 29 '17 at 13:51 | 2019-07-21T08:26:35 | {
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https://math.stackexchange.com/questions/2763975/how-many-diagonals-in-a-decagon/2763979 | # How many diagonals in a decagon?
I have a statement that says:
How many diagonals can be traced in a regular decagon (10-sided polygon)?
My development was:
First the diagonals, are drawn to non-consecutive vertex, so are $8$ available vertex. And we want to select 2 vertex, the fixed point and the point to which I will draw the diagonal.
So i use: $\frac{n!}{(n - r)! * r}$, also for each vertex, I do not take into account the fixed vertex, this is done 10 times, so I will subtract 10.
Then, the my final result is $\frac{8!}{(8-2)! * 2!} - 10$,
but the correct result must be $\frac{10!}{(10 - 2)! * 2!} - 10$, Where is my error?
• Your account has the same name as this other one- did you accidentally make a new one? – Kevin Long May 2 '18 at 21:03
It does not follow that there are then $\frac{8!}{(8-2)!\cdot 2!}$ ways. It is essentially just a non-sequitur. When you subtract $10$, it doesn't mean anything.
Let me provide an argument as similar as possible to your argument.
Pick a vertex. There are $7$ other vertices we can draw to. Therefore, there are $7$ diagonals for each vertex. There are $10$ vertices total, so the total counted is $7 \cdot 10$. But for each diagonal, there are $2$ vertices that we picked. Therefore, we counted each diagonal twice! The answer is therefore
$$\frac{7 \cdot 10}{2} = 35$$
Now, let's look at what the solution intended. There are $10$ vertices. A segment is defined by its two endpoints. Therefore, there are
$$\binom{10}{2} = 45$$
segments, and $10$ of these are the sides of the decagon. Therefore there are $45 - 10 =35$ diagonals.
From any given vertex you can draw $7$ diagonals. You can't draw to itself or to the two neighboring points. There are therefore $10 \cdot 7$ ends of diagonals. As each diagonal has two ends, there are $10 \cdot 7 \cdot \frac 12=35$ diagonals.
The approach to the formula you quote is that you pick two vertices to draw a line between, which you can do in $10 \choose 2$ ways. $10$ of those are sides of the decagon instead of diagonals, so the result is ${10 \choose 2}-10=35$
The correct answer gives you $\binom{10}{2}-10=35$, the reasoning being that you choose two vertices of the $10$ and draw a diagonal between them. However, as you said, consecutive vertices will not give a diagonal, so we ignore the $10$ sides of the decagon that arise from this counting.
I have trouble telling what you're doing in your answer; you seem to fix a vertex, but there shouldn't be $8$ possible vertices you can draw a diagonal to from that fixed vertex, but $7$, because you subtract both vertices adjacent to the fixed one, but also the fixed vertex itself. There's also no reason to use a binomial coefficient there. So, for each vertex, you can pick $7$ possible vertices to draw a diagonal to that one, and since there are $10$ vertices, you should get $7*10$. However, this counts each diagonal twice- once for each vertex on that diagonal, so you divide by $2$. This gives you $35$, the same answer as before.
All answers are correct, I thought I can throw another analogy to your question. Assume 10 people get into a room and they shake their hands. How many handshakes are there? Answer: 10c2 = 45; OR (better intuitive approach) if I'm in the room, I will shake hands to 9 people. Since we don't count the same handshake twice we'll divide it by 2. So we can say: $10 \cdot 9 \cdot \frac 12=45$
Back to your question: if 10 people get into a room and they are seated in a circle but they don't shake the hand of the person on the left and on the right (and not to themselves of course), how many handshakes we'll have?
Answer: $10 \cdot 7 \cdot \frac 12=35$
Choose a vertex ($10$ possible ways) join it to any of the other $7$ possible vertices, this will double count the diagonals so there are $10 \times 7 /2$ possible diagonals. | 2020-04-02T23:18:54 | {
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http://mathhelpforum.com/algebra/15777-real-zeros-function.html | Thread: Real Zeros of a Function
1. Real Zeros of a Function
Well, I have searched all over the internet for a simple explanation for how to
"find all real zeros of the function: f(x) = 2x^3 + 4x^2 - 2x - 4"
but to no avail. I knew how to do this at some point, and I don't remember it being that hard, but I think my mind erased it. Help please?
2. $\begin{array}{rcl}
2x^3 + 4x^2 - 2x - 4 & = & 0 \\
2x^2 \left( {x + 2} \right) - 2\left( {x + 2} \right) & = & 0 \\
2\left( {x + 2} \right)\left( {x^2 - 1} \right) & = & 0 \\
\end{array}$
.
3. Hello, iGuess!
Find all real zeros of the function: . $f(x) \:= \:2x^3 + 4x^2 - 2x - 4$
The procedure is simple: .Solve $f(x) = 0$
The algebra may take some work . . .
We have: . $2x^3 + 4x^2 - 2x - 4 \:=\:0$
Divide by 2: . $x^3 + 2x^2 - x - 2\:=\:0$
Factor by "grouping": . $x^2(x + 2) - (x + 2) \:=\:0$
. . and we have: . $(x +2)(x^2 - 1)\:=\:0\quad\Rightarrow\quad (x + 2)(x - 1)(x + 1)\:=\:0$
Therefore: . $x \:=\:-2,\:1,\:-1$
4. Okay, I was under the impression that zeros were basically x-intercepts. Am I completely off? So shouldn't the answer just be three y-values? Apparently I fail at math. Yay me. I mean, who WOULDN'T want to take Algebra 2 for a third time?
EDIT: Nevermind. I started posting before Soroban's post. Thank you.
5. Originally Posted by iGuess
Well, I have searched all over the internet for a simple explanation for how to
"find all real zeros of the function: f(x) = 2x^3 + 4x^2 - 2x - 4"
but to no avail. I knew how to do this at some point, and I don't remember it being that hard, but I think my mind erased it. Help please?
Factoring by grouping is probably the best method, and likely the one you were supposed to use, but here's another method anyway.
I agree with Soroban, go ahead and divide by 2 first.
$x^3 + 2x^2 - x - 2 = 0$
Now for something called the "Rational Root Theorem."
If any rational zeros of this polynomial exist, they will be in the form
$x = \pm \frac{\text{factor of 2}}{\text{factor of 1}}$
(The "2" is the constant term in the polynomial, in this case actually a -2. The "1" is the coefficient of the leading term, in this case the $x^3$ term.)
So all the possible rational roots of this polynomial are:
$x = \pm 1, \, \pm 2$
In this case running through the 4 possibilities gives you all three zeros. In the general case, say we found only one root: x = -2, for example. This means that x - (-2) = x + 2 is a factor of $x^3 + 2x^2 - x - 2$. You can divide the two polynomials to find that:
$x^3 + 2x^2 - x - 2 = (x + 2)(x^2 - 1) = 0$
So you know that the solutions to $x^2 - 1 = 0$ are also solutions, which you can solve by any means you like.
-Dan
6. Originally Posted by Soroban
Hello, iGuess!
The procedure is simple: .Solve $f(x) = 0$
The algebra may take some work . . .
We have: . $2x^3 + 4x^2 - 2x - 4 \:=\:0$
Divide by 2: . $x^3 + 2x^2 - x - 2\:=\:0$
Factor by "grouping": . $x^2(x + 2) - (x + 2) \:=\:0$
. . and we have: . $(x +2)(x^2 - 1)\:=\:0\quad\Rightarrow\quad (x + 2)(x - 1)(x + 1)\:=\:0$
Therefore: . $x \:=\:-2,\:1,\:-1$
Hello Soroban, I'm having a test about this stuff in a few days, so I thought it would be better to ask you my questions on this thread rather than starting a new thread.
My problem starts right after factoring by grouping.
I understand until this part
x^2(x+2)-(x+2)=0
but then I'm totally lost in the next part when you say
"and we have (x+2)(x^2-1)"
I don't understand, when factoring by group you have two times (x+2) but then you only put it only once, I put in bold the part which confuses me, also where did you get from that
(x^2-1)?
Thank you so much for your help Soroban.
7. Originally Posted by jhonwashington
Hello Soroban, I'm having a test about this stuff in a few days, so I thought it would be better to ask you my questions on this thread rather than starting a new thread.
My problem starts right after factoring by grouping.
I understand until this part
x^2(x+2)-(x+2)=0
but then I'm totally lost in the next part when you say
"and we have (x+2)(x^2-1)"
I don't understand, when factoring by group you have two times (x+2) but then you only put it only once, I put in bold the part which confuses me, also where did you get from that
(x^2-1)?
Thank you so much for your help Soroban.
he pulled the common term out.
let's say you wanted to factor a out of ab - a, how would you do it?
ab - a = a(b - 1)
you pull the common term out and put in brackets beside it what is left after you take it out. Soroban did the same thing.
for simplicity, what if we replaced the (x + 2) with a:
we have: $ax^2 - a$ and we want to factor $a$ out.
we get: $a \left( x^2 - 1 \right)$ as shown above.
do you get it?
8. thanks
Thank you so much Jhevon for your great explanation, I think I now understand | 2017-11-19T07:17:01 | {
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https://math.stackexchange.com/questions/2874371/set-builder-notation-for-pairs | # Set builder notation for pairs?
I have a set $S$ of integers. I want to select the element pairs $(i,j)$ of it such that $i=2*j$ and order of elements does not matter. How can I show it with the set builder notation ?
$\{\ (i,j)\ |\ i \in S\ \;and\;\ j \in S\; \ and\;i=2*j\}$ Is it correct and are there any ways to achieve this? Thanks in advance.
• Why not simply $$\{(2i,i) \mid i \in S\}$$? – Clement C. Aug 6 '18 at 22:32
• @ClementC. What if $2i\notin S$ – Don Thousand Aug 6 '18 at 22:33
• @RushabhMehta Good point. Then it's a bit less nice, indeed: $$\{(2i,i) \mid i \in S\}\cap S^2$$ – Clement C. Aug 6 '18 at 22:34
• More, that both are in the set S: $~\{(2j,j): j\in S, 2j\in S\}$ – Graham Kemp Aug 6 '18 at 22:34
• @ClementC. I wouldn't really call that set builder notation – Don Thousand Aug 6 '18 at 22:34
$\{\ (i,j)\ |\ i \in S\ \;and\;\ j \in S\; \ and\;i=2*j\}$ Is it correct and are there any ways to achieve this? Thanks in advance.
That is okay, although the use of words should be discouraged, and it can be compacted a bit more. Any of the following should be acceptable: $${\{(i,j)\mid i\in S, j\in S, i=2j\}\\\{(2j,j)\mid j\in S, 2j\in S\}\\\{(i,j) \in S^2 \mid i= 2 j \} \\\{(2j,j)\in S^2\}}$$
It's better if you write: $\{(i,j) \in S^2| i= 2*j \}$
• I don't think including $\in S$ in the left part of the set builder notation is considered appropriate, but I am not too knowledgeable on the subject. – Don Thousand Aug 6 '18 at 22:45
• @RushabhMehta It should be $\in S^2$, otherwise it is correctly placed. $~\{(i,j) \in S^2 \mid i= 2\cdot j \}$ says: the set of pairs, $(i,j)$, drawn from the Cartesian square of $S$ such that $i$ equals twice $j$. – Graham Kemp Aug 6 '18 at 22:49
• It should be $\in S^2$. And @RushabhMehta it is appropriate to put $\in$ there(some would even say it should be there) as without it it is not guarantee to be a set – ℋolo Aug 6 '18 at 22:51 | 2020-02-22T20:39:49 | {
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https://math.stackexchange.com/questions/3450093/int-fg-leq-1-for-any-g-in-c-0-infty-g-l2-1-then-f | # $| \int fg|\leq 1$ for any $g\in C_{0}^{\infty}$, $\|g\|_{L^{2}}=1$ then $\|f\|_{L^{2}}\leq1$
I have the following question:
Let $$f$$ be a continuous function in an open, bounded, smooth domain $$\Omega$$ in $$\mathbb{R}^{n}$$ such that $$| \int_{\Omega} fg|\leq 1$$ for any $$g\in C_{0}^{\infty}(\Omega)$$, $$\|g\|_{L^{2}(\Omega)}=1$$. Here the measure and the integral are w.r.t. Lebesgue measure. And $$C_{0}^{\infty}(\Omega)$$ is the set of smooth functions of compact support in $$\Omega$$.
Can we conclude that $$f\in L^{2}(\Omega)$$?
Thanks for any hint.
• What are we assuming about $f$ to make $\int_{\Omega} fg$ an a priori well-defined quantity? If $f$ is measurable and not in $L^1_{loc}$ already, the expression need not be defined. – WoolierThanThou Nov 25 '19 at 9:09
• @WoolierThanThou I think the hypothesis is that $\int fg$ exist and $|\int fg| \leq 1$ whenvever $g$ is smooth with compact support. – Kavi Rama Murthy Nov 25 '19 at 10:10
• @WoolierThanThou Let me assume that $f$ is continuous on $\Omega$. I have changed the post. – Binjiu Nov 25 '19 at 11:03
• what is an "smooth domain"? A convex one? – Masacroso Nov 25 '19 at 11:09
• @Masacroso: It is a domain such that $\partial \Omega$ is a $C^{\infty}$-manifold. – WoolierThanThou Nov 25 '19 at 11:13
## 2 Answers
Consider that \begin{align*} T_{f}:C_{0}^{\infty}\rightarrow\mathbb{C},~~~~g\rightarrow\int fg, \end{align*} then by assumption $$\|T_{f}(g)\|\leq\|g\|_{L^{2}}$$ and hence $$\|T_{f}\|\leq 1$$. But $$C_{0}^{\infty}$$ is dense in $$L^{2}$$, so there is a unique extension $$\overline{T}\in(L^{2})^{\ast}$$ of $$T_{f}$$ such that $$\left\|\overline{T}\right\|=\|T_{f}\|$$. By Riesz Representation Theorem, we have $$(L^{2})^{\ast}=L^{2}$$ in the sense that a unique $$h\in L^{2}$$ is such that \begin{align*} \overline{T}(g)=\int hg,~~~~g\in L^{2}, \end{align*} and that $$\|h\|_{L^{2}}=\left\|\overline{T}\right\|$$. It follows that $$\|h\|_{L^{2}}\leq 1$$ and that \begin{align*} \int(h-f)g=0,~~~~g\in C_{0}^{\infty}. \end{align*} If it were shown to be the case that $$h-f=0$$ a.e. then we are done.
So the matter is now to show that $$\displaystyle\int fg=0$$ for all $$g\in C_{0}^{\infty}$$ will imply that $$f=0$$ a.e.
First note that the existence of $$\displaystyle\int fg$$ entails that $$\displaystyle\int|fg|<\infty$$. For a fixed compact set $$K$$, take a nonnegative $$g\in C_{0}^{\infty}$$ such that $$g=1$$ on $$K$$, then $$\displaystyle\int|fg|\geq\int_{K}|f|$$, then $$f\in L^{1}(K)$$.
On the other hand, for a fixed $$x$$, we have $$\displaystyle\int f(\cdot)\varphi_{\epsilon}(x-\cdot)=0$$, where $$\varphi_{\epsilon}$$ is a standard nonnegative mollifier, the equation is no more than saying that $$\varphi_{\epsilon}\ast f(x)=0$$. As $$\varphi_{\epsilon}\ast f\rightarrow f$$ in $$L^{1}(K)$$, we have $$f=0$$ a.e. on $$K$$.
The result follows by considering an exhaustion of compact sets to the whole space.
Okay, I seem to have a proof that works assuming that $$f\in L^{\infty}_{loc}(\Omega)$$ (hence, in particular, if $$f\in C(\Omega)$$).
Indeed, let $$\varphi\in C^{\infty}_0(\Omega)$$ and let $$(\eta_{\varepsilon})_{\varepsilon\in(0,1]}$$ be a (positive) mollifier. Then, since $$f\in L^{\infty}(\textrm{supp}(\varphi)),$$ we have
$$\left| \int_{\Omega} f^+ \varphi \right|=\left|\int f \varphi 1_{\{f>0\}}\right|=\lim_{\varepsilon\to 0^+} \left|\int f (\varphi 1_{\{f>0\}} *\eta_{\varepsilon}) \right|\leq \lim_{\varepsilon\to 0^+}||\varphi 1_{\{f>0\}}*\eta_{\varepsilon}||_2=||\varphi 1_{\{f>0\}}||_2\leq ||\varphi||_2,$$ where we use that $$\varphi 1_{\{f>0\}} *\eta_{\varepsilon}\in C_0^{\infty}(\Omega)$$ for $$\varepsilon$$ sufficiently small (we extend by $$0$$ to $$\mathbb{R}^n$$ when defining the convolution). We get that $$f^{+}$$ and, similarly, $$f^{-}$$ must satisfy the same property.
Hence, we can assume that $$f$$ is positive. Let $$(K_n)_{n\in \mathbb{N}}$$ be an exhaustion of $$\Omega$$ by compacts and let $$f_n=f1_{K_n}.$$ Then, $$f_n\in L^{\infty}(K_n)\subseteq L^{2}(K_n)$$ and
$$||f_n||_{L^2}^2= \int f_n^2=\lim_{\varepsilon\to 0^+} \int f_n (f_n*\eta_{\varepsilon})\leq \limsup_{\varepsilon\to 0^+} \int f (f_n*\eta_{\varepsilon})\leq \lim_{\varepsilon\to 0^+} ||f_n*\eta_{\varepsilon}||_{L^2}=||f_n||_{L^2},$$ implying that $$||f_n||_{L^2}\in [0,1]$$. Applying monotone convergence, we get that $$f\in L^2(\Omega)$$ and $$||f||_{L^2}\leq 1.$$
• Actually, I guess I'm just using that $f\in L^2_{loc}(\Omega)$. – WoolierThanThou Nov 25 '19 at 12:00
• For $f\in L_{\text{loc}}^{2}$ can you really do $\displaystyle\int f\varphi 1_{\{f>0\}}=\lim_{\epsilon}\displaystyle\int f(\varphi 1_{\{f>0\}}\ast\eta_{\epsilon})$? – user284331 Nov 25 '19 at 16:28
• Yes: $|| f(\varphi 1_{\{f>0\}}- \varphi 1_{\{f>0\}}*\eta_{\varepsilon})||_{L^1}\leq ||f (1_{\varphi *\eta_{\varepsilon}\neq 0}+1_{\varphi\neq 0})||_{L^2} ||\varphi 1_{\{f>0\}}-\varphi 1_{\{f>0\}}*\eta_{\varepsilon}||_{L^2}$ by Cauchy Schwarz, and $g *\eta_{\varepsilon}\to g$ in $L^p$ for every $p$ such that $g\in L^p$ (unless of course $p=\infty$, in which case, we need $g$ continuous). – WoolierThanThou Nov 25 '19 at 16:49 | 2021-04-23T18:10:06 | {
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https://www.physicsforums.com/threads/discrete-math-proper-subsets.370642/ | # Discrete Math Proper Subsets
Discrete Math "Proper Subsets"
Hey everyone, I am confused on part of this. Any input would be much appreciated!!
X has ten members. How many members does ~P(X) have? (~P is the set of all subsets)
How many proper subsets does X have?
Well the number of members of ~P is 2^10 or 1024. This is the number of members in the set of subsets.
What I'm confused about is how many proper subsets does X have?
List all non-proper subsets of X and then the rest must be proper.
List all non-proper subsets of X and then the rest must be proper.
That number could be 1023 or 1022 depending on whether you count the empty set. X is not a proper subset of itself, but I'm not sure the empty set would be counted as either as a non proper set or a proper set.
EDIT: Since $$2^0=1$$, and $$2^1=2$$ it seems the empty set must be counted as a non proper set, so the answer would seem to be 1022.
Last edited:
Landau
A proper subset of X is just (by definition) a subset of X which is not equal to X. So the answer is 1023. The empty set is a subset of every set, and it is a proper subset of every set except of itself.
If P(X) is the power set of X, and Q(X) the set of all proper subsets, then |Q(X)|=|P(X)-1|.
A proper subset of X is just (by definition) a subset of X which is not equal to X. So the answer is 1023. The empty set is a subset of every set, and it is a proper subset of every set except of itself.
If P(X) is the power set of X, and Q(X) the set of all proper subsets, then |Q(X)|=|P(X)-1|.
OK, that makes sense, but in general usage, when a set is said to contain n members, should we understand that n includes the empty set or not?
Landau
Not sure what you mean; if a set is said to contain n members, then it is what it is: a set containing n members. This means - assuming n is a natural number >0 - the set is in bijective correspondence with {1,...,n}. If n=0 ("the set is said to contain zero members"), then the set is empty.
But earlier we talked about subsets of a given set, so I'm not sure what the connection is.
Not sure what you mean; if a set is said to contain n members, then it is what it is: a set containing n members. This means - assuming n is a natural number >0 - the set is in bijective correspondence with {1,...,n}. If n=0 ("the set is said to contain zero members"), then the set is empty.
But earlier we talked about subsets of a given set, so I'm not sure what the connection is.
Yes. My misunderstanding. Sets cannot be members of themselves nor is the empty set a member of any set.
Landau
The first is true: a set cannot be a member of itself (that's why we have the Axiom of Regularity). The second is not true: the emty set can easily be a member of a set.
$$A=\{\emptyset\}$$ is a set containing exactly one member, namely the empty set.
HallsofIvy
Homework Helper
OK, that makes sense, but in general usage, when a set is said to contain n members, should we understand that n includes the empty set or not?
What do you mean by "includes"? Every set has the empty set as a subset not as a member.
The first is true: a set cannot be a member of itself (that's why we have the Axiom of Regularity). The second is not true: the emty set can easily be a member of a set.
$$A=\{\emptyset\}$$ is a set containing exactly one member, namely the empty set.
What do you mean by "includes"? Every set has the empty set as a subset not as a member.
Alright. In Landau's example above, set A has the empty set as a member. Is it true that the empty set is also a subset of A, and A is also a subset of itself?
Alright. In Landau's example above, set A has the empty set as a member. Is it true that the empty set is also a subset of A, and A is also a subset of itself?
Both statements are true.
Every set contains all of its elements, and every element of the empty set is contained in a given set.
Both statements are true.
Every set contains all of its elements, and every element of the empty set is contained in a given set.
This doesn't answer my question. Landau gave the example of set A which contains the empty set as a member. Since the empty set is a subset of every set, is it also a subset of A?
Landau
Yes it is. The empty set is both a member and a subset of $$A=\{\emptyset\}$$, i.e.:
$$\emptyset\in A$$ and $$\emptyset\subset A$$.
Also, $$A$$ is a subset of $$A$$.
Yes it is. The empty set is both a member and a subset of $$A=\{\emptyset\}$$, i.e.:
$$\emptyset\in A$$ and $$\emptyset\subset A$$.
Also, $$A$$ is a subset of $$A$$.
Thanks Landau. I'm assuming that the symbols $$\emptyset$$ and {} mean exactly the same thing. So if A={$$\emptyset$$} then I can write A={{}}. The size of the power set of A equals two members so I can write A={{}, {{}}}. That is, the proper subset {} and the non-proper subset {{}}.
I've seen the union of sets indicated by the + symbol in some texts and the expression above exhausts the set A, so can I write {{}}={}+{{}}? Somehow that doesn't look right to me.
There is only one empty set, so to say that {} can be both a member of A and a subset of A, but not A itself sounds very counter-intuitive to me. It seems clear to me that A is empty.
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CRGreathouse
Homework Helper
Thanks Landau. I'm assuming that the symbols $$\emptyset$$ and {} mean exactly the same thing. So if A={$$\emptyset$$} then I can write A={{}}. The size of the power set of A equals two members so I can write A={{}, {{}}}. That is, the proper subset {} and the non-proper subset {{}}.
Yes.
I've seen the union of sets indicated by the + symbol in some texts and the expression above exhausts the set A, so can I write {{}}={}+{{}}? Somehow that doesn't look right to me.
You can certainly write $\{\{\}\} = \{\{\}\} \cup \{\}$, but I don't think this says what you seem to think it does.
There is only one empty set, so to say that {} can be both a member of A and a subset of A, but not A itself sounds very counter-intuitive to me. It seems clear to me that A is empty.
There is only one empty set -- this can be proved. But I don't know why the fact that
{} is a member of {{}}
and
{} is a subset of {{}}
seems counterintuitive to you.
There is only one empty set -- this can be proved. But I don't know why the fact that
{} is a member of {{}}
and
{} is a subset of {{}}
seems counterintuitive to you.
OK. Consider the set X={x,y} such that x and y are variables. If x=y, then this would be an invalid set. What formal prohibition exists for such a set or interpretation of such a set?
EDIT: I understand that this is a somewhat different question and that {} can be a subset of itself. However if x and y range over sets {},{0}.{1}.{2}....... then if x=y, we can have X={{},{}}. I believe this is a second order logic, but does ZFC specifically exclude elements as variables over sets?
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Consider the set X={x,y} such that x and y are variables. If x=y, then this would be an invalid set. What formal prohibition exists for such a set or interpretation of such a set?
First, there is no need to say that x and y are "variables"; they are, simply, the elements of the set {x,y}.
Second, there is nothing, formal or otherwise, that prevents x=y; the equality between sets is defined by extension, that is:
$$A=B \Leftrightarrow \forall x\left(x \in A \leftrightarrow x\in B\right)$$
And this implies that, if x=y, {x,y} = {x,x} = {x}. It's not an invalid set, in any sense.
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CRGreathouse
Homework Helper
OK. Consider the set X={x,y} such that x and y are variables. If x=y, then this would be an invalid set. What formal prohibition exists for such a set or interpretation of such a set?
There's nothing wrong with {x, y} or {x, x}. Why would there be?
Second, there is nothing, formal or otherwise, that prevents x=y; the equality between sets is defined by extension, that is:
$$A=B \Leftrightarrow \forall x\left(x \in A \leftrightarrow x\in B\right)$$
And this implies that, if x=y, {x,y} = {x,x} = {x}. It's not an invalid set, in any sense.
If {x,x} is a valid countable set with two members other than {} and the improper set, what justifies reducing it to {x}, a set with one member? It makes sense algebraically, but does it in set theory where the cardinality of finite sets is equal to the number of members.
GR Greathouse;2545299]There's nothing wrong with {x, y} or {x, x}. Why would there be?
You're saying a set can contain identical iterations of a member with nothing that distinguishes them? There is only one empty set, so can I write A={{},{}.{},.......,{}}? What about infinite iterations? If not, is the empty set a special case? If so, why? If the whole series of subsets equals the improper subset of A, does it make sense do ask which iteration of {} is the proper subset?
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If {x,x} is a valid countable set with two members other than {} and the improper set
Now I understand where you went wrong. The empty set and the set itself (I take that it's what you mean by the "improper set") are not members of the set; they are subsets of it, which means that each of their elements is also an element of the set (for example, the $$\emptyset$$ is not necessarily a member of another set, but it's always a subset of every set). Formally, $$A$$ a being subset of $$B$$ is defined as:
$$A \subseteq B \Leftrightarrow \forall x \left(x \in A \rightarrow x \in B\right)$$
You should compare this expression with the one defining equality in my previous post. On the other hand, elementhood ($$\in$$) is primitive in set theory, it's not defined in terms of other relations (note that both equality and subsethood are defined in terms of $$\in$$).
If {x,x} is a valid countable set with two members
But it's not: it's a set with just one member, x. This is what the definition of set equality tells us. The cardinality of {x,x} = {x} is one.
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CRGreathouse
Homework Helper
If {x,x} is a valid countable set with two members other than {} and the improper set, what justifies reducing it to {x}, a set with one member? It makes sense algebraically, but does it in set theory where the cardinality of finite sets is equal to the number of members.
You seem confused. If x ≠ {} ≠ y, then {x, y} has one or two members, neither of which is {}. Assuming we're working in ZF, {x, y} is not a member of {x, y}.
And I certainly never claimed that {x, y} has two members. It may have one or two, depending on whether x = y or x ≠ y.
You're saying a set can contain identical iterations of a member with nothing that distinguishes them? There is only one empty set, so can I write A={{},{}.{},.......,{}}? What about infinite iterations? If not, is the empty set a special case? If so, why?
You can write {{}} = {{}, {}} = {{}, {}, {}, {}} if you like, sure. It's correct.
This has nothing to do with the fact that the set is empty. For any set S, {S} = {S, S} = {S, S, S}, etc.
If the whole series of subsets equals the improper subset of A, does it make sense do ask which iteration of {} is the proper subset?
It is not the case (in ZF) that {A} = {A, A} = A. The first equality holds, but the second fails. In particular, it is never true (in ZF) that A = {A, ...} for any value of "...".
CRGreathouse
Homework Helper
JSuarez is of course correct in his (?) explanation, but I wanted to nitpick slightly:
This is what the definition of set equality tells us.
I think it's standard to call that the Axiom of Extensionality rather than the definition of =. | 2022-01-23T02:51:02 | {
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https://cstheory.stackexchange.com/questions/27739/conditional-entropy-hx-y-large-implies-hx-y-x-neq-y-large/27754 | Conditional entropy: $H(X | Y)$ large implies $H(X | Y, X \neq Y)$ large?
Suppose that $X$ and $Y$ are two random variables that are defined on the same support. Furthermore, suppose that $H(X | Y) = \log n$ for some $n$. I am now interested in how much the term $H(X | Y, X \neq Y)$ may differ from $H(X | Y)$.
Suppose that $X|Y=y$ is uniformly distributed (for every y). Intuitively, as $H(X | Y) = \log n$, for a typical element $y$ in the support of $X$ and $Y$, we also have $H(X | Y=y) = \log n$. Thinking about a uniform distribution, X takes on at least n different values, and, therefore, we should expect $H(X | Y, X \neq Y) \ge \log(n-1)$ as the conditioning on $X \neq Y$ eliminates only a single potential element.
Can such a result be obtained in full generality?
Thanks very much for your answers. I am new to stackexchange so I don't know yet where and how to respond to suggested answers.
The fact that you can bound $H(X|Y) - H(X|Y, X \neq Y) \le \log(\frac{n}{n-1})$ is very interesting, as you can compute: $\log(\frac{n}{n-1}) = \log( \frac{1}{1-\frac{1}{n}} ) = - \log(1-\frac{1}{n}) = -\Theta(\log(e^{-\frac{1}{n}})) = \Theta(\frac{1}{n})$. This result matches the intuition very well as if the entropy $H(X|Y) = \log(n)$ then knowing that $H(X|Y, X \neq Y)$ essentially eliminates only one possible choice for $X$. This is somehow quantified by the $\Theta(\frac{1}{n})$.
• I do not know the answer, but if you do not put any restriction on Y, the general case can be reduced to the case where Y is a constant, which is probably easier to consider. Dec 7 '14 at 22:37
• Can you say formally what the notation $H(X|Y,X\neq Y)$ means?
– usul
Dec 8 '14 at 1:41
• I am interpreting it as in Thomas' answer: $\sum_{x,y} P(x,y~|~X\neq Y) \log(1/P(x~|~y, X\neq y))$.
– usul
Dec 8 '14 at 17:41
• Hi usul, that's exactly what I meant by it. Dec 10 '14 at 21:16
Let $X$ and $Y$ be random variables with the same range. Let $Z$ be the indicator of the event $X \ne Y$. By the chain rule, $$H(X|Y,Z) = H(X,Z|Y)-H(Z|Y).$$ Since $Z$ is determined by $X$ and $Y$, we have $H(X,Z|Y)=H(X|Y)$. Now, by the definition of conditional entropy, $$H(X|Y,Z) = \underset{z \leftarrow Z}{\mathbb{E}}[H(X|Y,Z=z)] = \mathbb{P}[X \ne Y] \cdot H(X|Y, X \ne Y) + \mathbb{P}[X = Y] \cdot H(X|Y, X = Y).$$ Clearly $H(X|Y, X = Y)=0$. Combining the above gives $$\mathbb{P}[X \ne Y] \cdot H(X|Y, X \ne Y) = H(X|Y) - H(Z|Y).$$ Now we have an equation linking the quantities we want. We have $$0 \leq H(Z|Y) \leq H(Z) = H_2(p) \leq 1,$$ where $p = \mathbb{P}[X \ne Y]$. In particular, $$p \cdot H(X|Y,X \ne Y) \leq H(X|Y) \leq p \cdot H(X|Y,X \ne Y) + H_2(p).$$
For convenience let $H(X|Y) = \log(n)$, then
$$-\infty ~~\leq~~ H(X|Y) - H(X|Y,X\neq Y) ~~\leq~~ \log\left(\frac{n}{n-1}\right)$$
and both sides have tight examples (i.e. as $p\to 0$ it can be arbitrarily negative, and your example matches the upper bound). More specifically, if $p = \Pr[X \neq Y]$, then:
$$H(X|Y) - H(X|Y,X\neq Y) ~~ \geq ~~ -\frac{(1-p)H(X|Y)}{p}$$ and $$H(X|Y) - H(X|Y,X\neq Y) ~~\leq~~ \log\frac{1}{p} + \frac{1-p}{p}\left[\log\frac{1}{1-p} ~-~ H(X|Y)\right]$$
and we have tight examples for the second, and arbitrarily close to tight examples for the first, for every $p,H(X|Y)$.
There will be two steps to the proof: (1) prove an upper bound and matching examples for $H(X|Y,Z)$ where $Z$ is an indicator; (2) convert these to your quantity of interest $H(X|Y,X\neq Y)$.
Upper Bound
Step 1.
Claim 1. Let $Z$ be the indicator for $X\neq Y$ and let $p = \Pr[X\neq Y]$. Then $$H(X|Y) - H(X|Y,Z) \leq H(p)$$ and for any fixed values of $H(X|Y)$ and $p$, we can construct tight examples.
Proof. The natural quantity to consider is $$H(X|Y) - H(X|Y,Z)$$ where $Z$ is the indicator, $Z=1$ if $X \neq Y$ and $Z=0$ otherwise. Let $p = \Pr[X\neq Y] = \Pr[Z=1]$.
Then as Thomas points out, by the chain rule and the fact that $H(X,Y,Z) = H(X,Y)$, $$H(X|Y) - H(X|Y,Z) = H(Z|Y) \leq H(p) . ~~~~~~~~ (*)$$
Examples showing tightness: Let $Y$ be distributed arbitrarily; then conditioned on $Y=y$, we let $X=y$ with probability $1-p$ and with probability $p$ we let $X$ be distributed arbitrarily on any set not containing $y$. To be very concrete, you could let $Y=0$ always and let $X=0$ with probability $1-p$ and otherwise $X$ is uniform on $\{1,\dots,m\}$. Choose $m$ to get the desired value of $H(X|Y)$.
In these examples, $H(Z|Y) = H(Z) = H(p)$. So we can make the inequality $(*)$ tight for any $p$ and any $H(X|Y)$. $\square$
Step 2.
"Theorem" 1. $$H(X|Y) - H(X|Y, X\neq Y) \leq \log\frac{1}{p} + \frac{1-p}{p}\left[\log\frac{1}{1-p} ~ - ~ H(X|Y) \right]$$ and for any fixed $H(X|Y)$ and $p$ there are tight examples.
Proof. Now, again as Thomas points out, we have $$H(X|Y,Z) = p\cdot H(X|Y, X \neq Y) .$$ Now, by plugging in to $(*)$, we have the inequality $$H(X|Y) - p\cdot H(X|Y, X\neq Y) \leq H(p) . ~~~~~~~~ (**)$$ and we can make this tight for any $p$. Let $H(X|Y) = p\cdot H(X|Y) + (1-p)H(X|Y)$ and rearrange: \begin{align} H(X|Y) - H(X|Y, X\neq Y) &\leq \frac{H(p) - (1-p)H(X|Y)}{p} \\ &= \log\frac{1}{p} + \frac{1-p}{p}\left[\log\frac{1}{1-p} ~ - ~ H(X|Y) \right] . \end{align} and, again, we can make this tight using the examples from before (since we have only renamed things and rearranged the inequality). $\square$
Claim 2. For any fixed $H(X|Y) > 0$, as $p \to 0$, we always have $$H(X|Y,Z) - H(X|Y,X \neq Y) \to -\infty .$$
Proof. In the bound of the "theorem", for small enough $p$, the upper bound is $\log\frac{1}{p} - \frac{1}{p}\Theta(H(X|Y))$, which approaches $-\infty$ as $p \to 0$ for all fixed $H(X|Y)$. $\square$
Claim 3. For any fixed $H(X|Y)$, we have $$H(X|Y,Z) - H(X|Y,X\neq Y) \leq \log\frac{2^{H(X|Y)}}{2^{H(X|Y)}-1} ,$$ and there are tight examples. In such examples, $p = 1 - 2^{-H(X|Y)}$.
Proof. Taking the bound in the "theorem" and taking the derivative with respect to $p$, we find that the upper bound is maximized uniquely at $p = 1 - 2^{-H(X|Y)}$. In that case, the quantity inside the brackets is zero, and we obtain
\begin{align} H(X|Y) - H(X|Y, X\neq Y) &\leq \log\frac{1}{p} \\ &=\log\frac{2^{H(X|Y)}}{2^{H(X|Y)}-1} . \end{align}
Again, for any $H(X|Y)$, the prior examples with this choice of $p$ make this inequality tight. $\square$
Lower Bound
Step 1.
Claim 4. For any $p$, $$H(X|Y) - H(X|Y,Z) \geq 0$$ and we can construct examples that are arbitrarily close to $0$.
Proof. As stated above, $H(X|Y) - H(X|Y,Z) = H(Z|Y) \geq 0$. To construct examples arbitrarily close to $0$, fix $p$. The intuition is $H(p)$ is concave, so we will have sometimes $\Pr[Z|Y] = \epsilon$ and sometimes $\Pr[Z|Y] = 1-\epsilon$, so that $H(Z|Y) = H(\epsilon) \to 0$, yet still $H(Z) = H(p)$.
Let $Y = -1$ with probability $1-p$ and $Y = 0$ with probability $p$. If $Y=-1$, then with probability $\epsilon$ we have $X=Y$ and otherwise $X$ is uniform on $\{1,\dots,m\}$. If $Y=0$, then with probability $1-\epsilon$ we have $X=Y$ and otherwise $X$ is uniform on $\{1,\dots,m\}$. Now we can check that $H(Z|Y) = H(\epsilon)$ and $p = \Pr[Z] = (1-p)\cdot \epsilon + p\cdot(1-\epsilon) = p$. Taking $\epsilon \to 0$ gives the example. $\square$
Step 2.
"Theorem" 2. For any $p$ and $H(X|Y)$, $$H(X|Y) - H(X|Y,X\neq Y) \geq -\frac{(1-p)H(X|Y)}{p}$$ and there are examples arbitrarily close.
Proof. Again we have $p\cdot H(X|Y,X\neq Y) = H(X|Y,Z)$, so by the previous claim, $$H(X|Y) - p\cdot H(X|Y,X\neq Y) \geq 0 .$$ Again let $H(X|Y) = p\cdot H(X|Y) + (1-p)H(X|Y)$ and rearrange. By the previous claim, we have arbitrarily close examples (since we have only rearranged the inequality). $\square$
• The lower bound can be simplified: Let $Y=0$ be constant. Let $X=Y=0$ with probability $1-p$ and $X\ne Y=0$ with probability $p$. Let $k=H(X|X\ne Y)=H(X|Y,X \ne Y)$. Then $H(X|Y)=H(X)=H_2(p)+p\cdot k \leq 1 + p k$. Clearly, as $k \to \infty$, $H(X|Y,X \ne Y) - H(X|Y) \to \infty$, so long as $p < 1-\omega(1/k)$. Dec 9 '14 at 4:09 | 2021-10-21T13:45:37 | {
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https://math.stackexchange.com/questions/1422045/matrix-multiplication-of-columns-times-rows-instead-of-rows-times-columns | # Matrix multiplication of columns times rows instead of rows times columns
In ordinary matrix multiplication $AB$ where we multiply each column $b_{i}$ by $A$, each resulting column of $AB$ can be viewed as a linear combination of $A$.
If however if we decided to multiply each column of $A$ by each row of $B$, we get an entire matrix for each column-row multiply. My question is: Does each matrix resulting from an outer product have any known meaning aside from being a part of the sum(a summand?) of the final $AB$?
Edit: Say we have $AB$
$$\left( \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} \right) \left( \begin{array}{ccc} 10 & 11 & 12 \\ 13 & 14 & 15 \\ 15 & 16 & 17 \end{array} \right)$$
Normally we would multiply each column of $B$ by A and get a linear combination of A, e.g. $$10\left( \begin{array}{c} 1 \\ 4\\ 7 \end{array} \right)+ 13\left( \begin{array}{c} 2 \\ 5\\ 8 \end{array} \right)+ 15\left( \begin{array}{c} 3 \\ 6\\ 9 \end{array} \right)$$ which is one column of $AB$.
If however we multiply each column of $A$ by each row of $B$, e.g. $$\left( \begin{array}{c} 1 \\ 4\\ 7 \end{array} \right)\left( \begin{array}{ccc} 10 & 11 & 12 \end{array} \right)$$ we get a matrix. Each of the 3 matrices $a_{i}b_{i}^{T}$ summed together gives us $AB$. I was wondering if each individual matrix that sums to $AB$ has any sort of special meaning. This second way of performing multiplication also seems to be called column-row expansion. (http://www.math.nyu.edu/~neylon/linalgfall04/project1/dj/crexpansion.htm). I actually read about it in I believe section 2.4 of Strang's Introduction to Linear Algebra book. He mentions that not everybody is aware that matrix multiplication can be performed in this way.
• Can you give an instructive example? – draks ... Sep 4 '15 at 15:19
• The operation looks very similar to the tensor product, but I don't fully understand the last two sentences.. – Peter Franek Sep 4 '15 at 19:14
Before talking about multiplication of two matrices, let's see another way to interpret matrix $$A$$. Say we have a matrix $$A$$ as below, $$\begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 2 \\ 1 & 2 & 3 \\ \end{bmatrix}$$ we can easily find that column $$\begin{bmatrix} 3 \\ 2 \\ 3 \\\end{bmatrix}$$ is linear combination of first two columns. $$1\begin{bmatrix} 1 \\ 1 \\ 1\\\end{bmatrix} + 1\begin{bmatrix} 2 \\ 1 \\ 2\\\end{bmatrix} = \begin{bmatrix} 3 \\ 2 \\ 3 \\\end{bmatrix}$$ And you can say $$\begin{bmatrix} 1 \\ 1 \\ 1 \\\end{bmatrix}$$ and $$\begin{bmatrix} 2 \\ 1 \\ 2 \\\end{bmatrix}$$ are two basis for column space of $$A$$.
Forgive the reason why you want to decompose matrix $$A$$ at first place like this, $$\begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 2 \\ 1 & 2 & 3 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \\ \end{bmatrix} + \begin{bmatrix} 0 & 2 & 2 \\ 0 & 1 & 1 \\ 0 & 2 & 2 \\ \end{bmatrix}$$ but you can, and in the end, it looks reasonable.
If you view this equation column wise, each $$column_j$$ of $$A$$ is the sum of corresponding $$column_j$$ of each matrix in RHS.
What's special about each matrix of RHS is that each of them is a rank 1 matrix whose column space is the line each base of column space of $$A$$ lies on. e,g. $$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \\ \end{bmatrix}$$ spans only $$\begin{bmatrix} 1 \\ 1 \\ 1 \\\end{bmatrix}$$. And people say rank 1 matrices are the building blocks of any matrices.
If now you revisit the concept of viewing $$A$$ column by column, this decomposition actually emphasizes the concept of linear combination of base vectors.
If these make sense, you could extend the RHS further, $$\begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 2 \\ 1 & 2 & 3 \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \\\end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\\end{bmatrix} + \begin{bmatrix} 2 \\ 1 \\ 2 \\\end{bmatrix} \begin{bmatrix} 0 & 1 & 1 \\\end{bmatrix}$$ Each term in RHS says take this base, and make it "look like" a rank 3 matrix.
And we can massage it a little bit, namely put RHS into matrix form, you get $$\begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 2 \\ 1 & 2 & 3 \\ \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & 1 \\ 1 & 2 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ \end{bmatrix}$$
Now you can forget matrix $$A$$, and imagine what you have are just two matrices on RHS. When you read this text backward(I mean logically), I hope matrix multiplication in this fashion makes sense to you now. Or if you prefer, you can start with two matrices in the question. | 2020-02-24T18:01:59 | {
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https://stats.stackexchange.com/questions/433748/multinomial-distribution-probability-that-one-outcome-is-greater-than-another | # Multinomial distribution: probability that one outcome is greater than another
Consider a multinomial distribution with three outcomes. Let $$x_i$$ denote the number of occurences of the $$i^{th}$$ outcome, and the $$i^{th}$$ outcome occurs with probability $$p_i$$, $$i=1,2,3$$. Let $$n$$ be the number of total trials. Then we have
$$(X_1,X_2,X_3)\sim Multi(n;p_1,p_2,1-p_1-p_2).$$
The questions are:
1) How can I calculate the probability of $$x_1\geq x_2$$ for a given $$n$$? Is there a concise closed form representation for $$P[X_1\geq X_2|n]?$$
2) Do we have some asymptotic property of this probability? Intuitively, if $$p_1>p_2$$, the probability should converge to 1. On the other hand, if $$p_1, it should converge to zero.. Can we see this from answering the first question?
## 2 Answers
There is a closed form for this probability, but it is not particularly concise. To obtain the formula, start with the probability of the event of interest, conditional on the event $$X_1+X_2=r$$. It can easily be shown that:
$$X_1 | X_1 + X_2 = r, n, \mathbf{p} \sim \text{Bin} \Big( r, \frac{p_1}{p_1+p_2} \Big).$$
Under the stated conditioning event, the event $$X_1 \geqslant X_2$$ is equivalent to the event $$X_1 \geqslant r/2$$. We therefore have the conditional probability:
\begin{aligned} \mathbb{P}(X_1 \geqslant X_2 | X_1 + X_2 = r, n, \mathbf{p}) &= \mathbb{P}(X_1 \geqslant r-X_1 | X_1 + X_2 = r, n, \mathbf{p}) \\[6pt] &= \mathbb{P}(X_1 \geqslant r/2 | X_1 + X_2 = r, n, \mathbf{p}) \\[6pt] &= \mathbb{P}(X_1 \geqslant \lceil r/2 \rceil | X_1 + X_2 = r, n, \mathbf{p}) \\[6pt] &= \frac{1}{(p_1+p_2)^r} \sum_{x = \lceil r/2 \rceil}^r {r \choose x} \cdot p_1^x \cdot p_2^{r-x}. \\[6pt] \end{aligned}
In practice, the upper tail of the binomial distribution is usually computed using the regularised incomplete beta function, rather than by summing the mass function. In this latter form, we have:
\begin{aligned} \mathbb{P}(X_1 \geqslant X_2 | X_1 + X_2 = r, n, \mathbf{p}) &= 1-\mathbb{P}(X_1 \leqslant \lceil r/2 \rceil - 1 | X_1 + X_2 = r, n, \mathbf{p}) \\[6pt] &= 1-I_{p_2/(p_1+p_2)}(r - \lceil r/2 \rceil + 1, \lceil r/2 \rceil). \\[6pt] \end{aligned}
Applying the law of total probability gives the marginal probability of interest:
\begin{aligned} \mathbb{P}(X_1 \geqslant X_2 | n, \mathbf{p}) &= \sum_{r=0}^n \mathbb{P}(X_1 + X_2 = r | n) \cdot \mathbb{P}(X_1 \geqslant X_2 | X_1 + X_2 = r, n, \mathbf{p}) \\[6pt] &= \sum_{r=0}^n \text{Bin}(r|n, p_1+p_2) \cdot \mathbb{P}(X_1 \geqslant X_2 | X_1 + X_2 = r, n, \mathbf{p}) \\[6pt] &= \sum_{r=0}^n {n \choose r} \cdot (1-p_1+p_2)^{n-r} \sum_{x = \lceil r/2 \rceil}^r {r \choose x} \cdot p_1^x \cdot p_2^{r-x} \\[6pt] &= \sum_{r=0}^n {n \choose r} \cdot (p_1+p_2)^{r} \cdot (1-p_1+p_2)^{n-r} \\[6pt] &\quad \quad \quad \times (1-I_{p_2/(p_1+p_2)}(r - \lceil r/2 \rceil + 1, \lceil r/2 \rceil)). \\[6pt] \end{aligned}
Either of the last two lines give a closed form solution to the problem. (The first is a sum over relevant values of the mass function of the binomial. The second uses the regularised incomplete beta function for the distribution. The latter is closer to what you would use when implementing the function in statistical computing.) Either expression is somewhat cumbersome for large $$n$$.
Computing the probability: We can construct a function in R to compute this probability using the standard statistical distribution functions. (Since this computation involves adding very small probabilities we work in log-space for the computation, and then convert back to a standard probability at the end.)
PROB <- function(n, p, log = FALSE) {
#Normalise probability vector
p <- p/sum(p);
#Computeprobability
LOGTERMS <- rep(0, n+1);
LOGTERMS[1] <- dbinom(0, size = n, prob = p[1]+p[2], log = TRUE);
CLOGPROB <- rep(0, n+1);
for (r in 1:n) {
CLOGPROB[r+1] <- pbinom(ceiling(r/2)-1, size = r, prob = p[1]/(p[1]+p[2]),
lower.tail = FALSE, log.p = TRUE);
LOGTERMS[r+1] <- dbinom(r, size = n, prob = p[1]+p[2], log = TRUE) + CLOGPROB[r+1]; }
LOGPROB <- matrixStats::logSumExp(LOGTERMS);
if (log) { LOGPROB } else { exp(LOGPROB) } }
This function is sufficiently efficient to compute the probability of interest up to quite large values of $$n$$. Basic tests show that it will compute for $$n = 10^7$$ in a few seconds. Below are some examples of use of this function for various values of $$n$$. As can be seen, even when the probability gets too small to see on the standard scale, we can still compute it on the log-probability scale.
p <- c(0.20, 0.25, 0.55);
PROB(1, p);
[1] 0.75
PROB(2, p);
[1] 0.6625
PROB(10, p);
[1] 0.4987803
PROB(100, p);
[1] 0.2503806
PROB(10^5, p);
[1] 1.174474e-123
PROB(10^6, p);
[1] 0
PROB(10^6, p, log = TRUE);
[1] -2795.471
PROB(10^7, p, log = TRUE);
[1] -27909.27
This function can be used to obtain the probability of interest up to large values of $$n$$. For values that are so large that they are computationally infeasible with this algorithm, we can then fall back on the normal approximation to the multinomial distribution.
• Amazingly complete answer :-) +1 Dec 4, 2019 at 3:38
Let $$Z_i$$ be independent draws from a categorical variable take takes values 1,2 and 3 with probability $$p_1,p_2$$ and $$1-p_1-p_2$$. Then, $$X_k = \sum_{i=1}^n 1\{Z_i = k\}$$ has the desired multinomial distribution. Let $$Y_i = \{Z_i = 1\} - 1\{Z_i = 2\}$$ and let $$W_n = \sum_{i=1}^n Y_i$$. The desired probability is $$P(W_n \ge 0)$$.
Note that $$Y_i$$ takes values $$1, -1$$ and zero with probabilities $$p_1, p_2$$ and $$1-p_1-p_2$$. $$W_n$$ is a nonsymmetric lazy random walk. I am going to only consider the asymptotics. We have $$E[Y_i] = p_1 - p_2$$ and $$var(Y_i) = p_1 + p_2 - (p_1-p_2)^2 =: v^2$$. By CLT, $$U_n := \frac{W_n - n(p_1-p_2)}{\sqrt{n} v} \to N(0,1)$$ Thus, $$P(W_n \ge 0) = P( U_n \ge -\sqrt{n}(p_1-p_2)/v)$$ which is going to go to $$1$$ or $$0$$ depneding on whether $$p_1 > p_2$$ or $$p_1 < p_2$$, respectively. A good approximation for large $$n$$ is $$P(W_n \ge 0) \approx \Phi(\sqrt{n}(p_1-p_2)/v)$$ where $$\Phi$$ is the CDF of the normal distribution. | 2022-07-07T14:08:41 | {
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https://www.thelaunchpadtech.com/r52a72/g6p3jm.php?c9f433=transpose-of-matrix-product | » Data Structure Two matrices can only be added or subtracted if they have the same size. » JavaScript The same is true for the product of multiple matrices: (ABC) T = C T B T A T . » About us » DBMS does not affect the sign of the imaginary parts. )'s unnecessarily.They also return symmetricMatrix classedmatrices when easily detectable, e.g., in crossprod(m), the oneargument case. (If A is m × n, then x ∈ Rn, y ∈ Rm, the left dot product is in Rm and the right dot product is in Rn .) Just like before, we would get the transpose of the product. /Length 2793 The transpose of a matrix A, denoted by A , A′, A , A or A , may be constructed by any one of the following methods: » Networks » DBMS Transpose of the product of two matrices is equal to the product of their transposes taken in the reverse order, The transpose of the matrix products can be extended to several matrices The inverse of a transpose matrix is equal to the transpose of its inverse, Transpose of the Matrix. » Linux Interview que. » C » C++ Python » » Facebook » HR The transpose of matrix A is represented by $$A'$$ or $$A^T$$. » Embedded C » LinkedIn The diagonal elements of a triangular matrix are equal to its eigenvalues. Linear Algebra using Python, Linear Algebra using Python | Product of a Matrix and its Transpose Property: Here, we are going to learn about the product of a matrix and its transpose property and its implementation in Python. Let us understand the concept of the transpose of a matrix. If is an eigenvector of the transpose, it satisfies By transposing both sides of the equation, we get. In this program, the user is asked to enter the number of rows r and columns c. Their values should be less than 10 in this program. >> More: Join our Blogging forum. » Puzzles Transpose Matrix. Ad: Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. » Kotlin » DOS Numpy.dot () is the dot product of matrix M1 and M2. Linear Algebra 11y: The Transpose of a Triple Product - Duration: 2:33. Definition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Definition A square matrix A is symmetric if AT = A. The transpose of a matrix is obtained by interchanging the rows and columns of the given matrix. Transpose & Dot Product Def: The transpose of an m nmatrix Ais the n mmatrix AT whose columns are the rows of A. So we now get that C transpose is equal to D. Or you could say that C is equal to D transpose. » Feedback The row vector is called a left eigenvector of . » Contact us Eigenvalues of a triangular matrix. » Privacy policy, STUDENT'S SECTION Transposes of sums and inverses. » C#.Net » CS Organizations x��[Ys#�~ׯ���P�%��Pʕ��QN*q�XUy���K���4�I����� �M/h����h���ht�5 ����0+&�W�_���NNW�r�x>l�\�4j�����r/~Z-q��W��_/��J"��^/�̸!�٨�����]1� ����EJw��Jhӷ.��p"(��אUl:/+Y=0�sU�>T"�+�1���=���*Sf1A$�ͫ���KJ�Ԅ����IU�(�j��u�q_��P!����Ȉe��!>$�2Dp{������p��D�l������91RNf���*,&3)�RLf�ˉ*�����w�����77�w��>W�#xR�P�#vO�@-A�=L�hn����~^^��$�!��4���>fnjQ�dD��\���i]2n��ĵy4�ֺd&�i}�K�$V��g��$���7��� �٣����Dq�73"m)߽��$�M���#�K�A��f��D�@���8m \&� Transpose of a matrix can be found by changing all the rows into columns or vice versa. In this case, we swap the row-element with the column-element or vise versa. %���� returns the nonconjugate transpose of A, that is, interchanges the row and column index for each element.If A contains complex elements, then A.' Now note that (AB)x ⋅ y = A(Bx) ⋅ y = Bx ⋅ A⊤y = x ⋅ B⊤(A⊤y) = x ⋅ (B⊤A⊤)y. Are you a blogger? Solved programs: » SQL (5�������g��5����2��(�S6H » C $$A, B) Matrix division using a polyalgorithm. Matrix addition and subtraction are done entry-wise, which means that each entry in A+B is the sum of the corresponding entries in A and B. I know this statement seems stupid, but keep reading. » SEO Enter rows and columns of matrix: 2 3 Enter elements of matrix: Enter element a11: 1 Enter element a12: 2 Enter element a13: 9 Enter element a21: 0 Enter element a22: 4 Enter element a23: 7 Entered Matrix: 1 2 9 0 4 7 Transpose of Matrix: 1 0 2 4 9 7 Transpose of a matrix product. » C# The functions crossprod and tcrossprod arematrix products or “cross products”, ideally implementedefficiently without computing t(. tcrossprod () function in R Language is used to return the cross-product of the transpose of the specified matrix. » Cloud Computing There is no such restriction for the dimensionality of Matrix A. This is one of the most common ways to generate a symmetric matrix. » Content Writers of the Month, SUBSCRIBE stream » O.S. Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. » Java » Certificates » Ajax The transpose of the product of two matrices is equivalent to the product of their transposes in reversed order: (AB) T = B T A T . B = A.' » C The following statement generalizes transpose of a matrix: If \(A$$ = $$[a_{ij}]_{m×n}$$, then $$A'$$ = $$[a_{ij}]_{n×m}$$. We can transpose the whole thing as I indicated before. There is no such restriction for the dimensionality of Matrix A. » CS Basics And we said that D is equal to our matrix product B transpose times A transpose. » Articles » News/Updates, ABOUT SECTION 3 0 obj << The program must accept two positive integers M and N as the input. The basic matrix product, %*% is implemented for all ourMatrix and also forsparseVector classes, fully analogously to R'sbase matrixand vector objects. This is one of the most common ways to generate a symmetric matrix. The transpose of a matrix is an important phenomenon in the matrix theory. The meaning of transpose is to exchange places of two or more things. Home » 1. Rowspace and left nullspace. » C » PHP Repeat this step for the remaining rows, so the second row of the original matrix becomes the second column of its transpose, and so on. For example, element at position a12 (row 1 and column 2) will now be shifted to position a21 (row 2 and column 1), a13 to a31, a21 to a12and so on. CS Subjects: & ans. » Subscribe through email. » C++ A + B = [ 7 + 1 5 + 1 3 + 1 4 − 1 0 + 3 5 … Properties of transpose Now I can say, "lets transpose the product of the vectors": $\mathbf{A}^T=(\mathbf{v}*\mathbf{v}^T)^T$ But as you can distribute the transpose over the multiplication, you can say: Web Technologies: » Java The transpose of a matrix by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. » CSS In this tutorial, we are going to check and verify this property. » DS If I say roughly then the process of taking transpose of a matrix is something equivalent to changing the rows of the matrix into columns and columns of the matrix into rows. The transpose of a matrix is calculated by changing the rows as columns and columns as rows. A = [ 7 5 3 4 0 5 ] B = [ 1 1 1 − 1 3 2 ] {\displaystyle A={\begin{bmatrix}7&&5&&3\\4&&0&&5\end{bmatrix}}\qquad B={\begin{bmatrix}1&&1&&1\\-1&&3&&2\end{bmatrix}}} Here is an example of matrix addition 1. The following properties hold: (AT)T=A, that is the transpose of the transpose of A is A (the operation of taking the transpose is an involution). Thus, the matrix B is known as the Transpose of the matrix A. » Internship The rows of AT are the columns of A. We said that our matrix C is equal to the matrix product A and B. When we take transpose, only the diagonal elements don’t change place. » Node.js Article Summary X. » Machine learning And we notice also this is 2 x 4 times a 4 x 7 gives you a 2 x 7 matrix. » Web programming/HTML & ans. So recall the 540 calories, here is the total number of calories in a breakfast. The transpose () function from Numpy can be … %PDF-1.4 The same is true for the product of multiple matrices: (ABC) T = C T B T A T. Example 1: Find the transpose of the matrix and verify that (A T) T = A. : n 1 matrices). So: The columns of AT are the rows of A. For example, if A(3,2) is 1+2i and B = A. » C++ For permissions beyond … The program must generate a 4x4 integer matrix by using the following conditions,- The first element of the matrix must be the value of M.- The successive elements are obtained by adding N to the current element.Finally, the program must print the product… Submitted by Anuj Singh, on June 06, 2020. # Inverse Property A.AT = S [AT = transpose of A], Product of Matrix A with its Transpose : A * AT = I, Run-length encoding (find/print frequency of letters in a string), Sort an array of 0's, 1's and 2's in linear time complexity, Checking Anagrams (check whether two string is anagrams or not), Find the level in a binary tree with given sum K, Check whether a Binary Tree is BST (Binary Search Tree) or not, Capitalize first and last letter of each word in a line, Greedy Strategy to solve major algorithm problems. MathTheBeautiful 3,285 views. Hence, the transpose of matrix for the above matrix is : (Image to be added soon) Properties of Transpose of Matrices. » Java tcrossprod() takes the cross-product of the transpose of a matrix.tcrossprod(x) is f… © https://www.includehelp.com some rights reserved. So now, the transpose of matrix $\mathbf{A}$ will still be a square matrix, $\mathbf{A}^T$. » Embedded Systems The solver that is used depends upon the structure of A.If A is upper or lower triangular (or diagonal), no factorization of A is required and the system is solved with either forward or backward substitution. : In the case of the matrix, transpose meaning changes the index of the elements. (A+B)T=AT+BT, the transpose of a sum is the sum of transposes. Now, we will understand the transpose matrix by considering two matrices P … Let, A is a matrix of size m × n and At is the transpose of matrix A, /Filter /FlateDecode ', then the element B(2,3) is also 1+2i. In linear algebra, an mxn matrix A is multiplied with its transpose AT then the resultant matrix is symmetric. Linear Algebra 11w: Introduction to the Transpose of a Matrix - Duration: 7:40. Then, the user is asked to enter the elements of the matrix (of order r*c). The transpose of a matrix is a new matrix that is obtained by exchanging the rows and columns. 'k_��� j�v�}sS�, P��~p:(Ȭ�ٸC'�+*1^���rQ�%�G������=NJZm���w"~��6GܠH��v�x�B��`�L �+��t�A\$E�I̮F�Ͳ. i.e., (AT) ij = A ji ∀ i,j. Transpose of a vector. Free matrix transpose calculator - calculate matrix transpose step-by-step This website uses cookies to ensure you get the best experience. For input matrices A and B, the result X is such that A*X == B when A is square. A transpose will be denoted by original matrix with “T” in superscript, like Aᵀ. » Android » C++ STL » Java In linear algebra, an mxn matrix A is multiplied with its transpose AT then the resultant matrix is symmetric. Aptitude que. The main importance of the transpose (and this in fact defines it) is the formula Ax ⋅ y = x ⋅ A⊤y. share. : By using this website, you agree to our Cookie Policy. Thus, (AB)⊤ = B⊤A⊤. Now this is pretty interesting, because how did we define these two? Example: If A= 1 2 3 4 5 6 , then AT = 2 4 1 4 2 5 3 6 3 5: Convention: From now on, vectors v 2Rn will be regarded as \columns" (i.e. » Python Languages: ... Now, I'm going to define the transpose of this matrix as a with this superscript t. And this is going to be my definition, it is essentially the matrix A with all the rows and the columns swapped. Thus Transpose of a Matrix is defined as “A Matrix which is formed by turning all the rows of a given matrix into columns and vice-versa.” This is the definition of a transpose. Numpy.dot () handles the 2D arrays and perform matrix multiplications. To transpose a matrix, start by turning the first row of the matrix into the first column of its transpose. | 2021-06-16T02:26:08 | {
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https://math.stackexchange.com/questions/4118803/show-that-the-following-polynomial-in-mathbbqt-is-irreducible-and-not-sol | # Show that the following polynomial in $\mathbb{Q}[t]$ is irreducible and not solvable over $\mathbb{Q}$
The polynomial is $$f(t)=t^5-4t+2$$.
I can show it's irreducible with Eisenstein's Criterion, and I know I need to show that it has exactly two complex (non-real) roots to prove that it's not solvable, using the follow result:
Let $$p>1$$ be prime and $$f\in\mathbb{Q}[t]$$ irreducible with degree $$p$$. If $$f$$ has exactly two non-real complex roots, then the Galois group of $$f$$ is isomorphic to $$S_p$$. Hence, with $$p\geq5$$, we have that $$f$$ is not solvable by radicals over $$\mathbb{Q}$$.
But I don't know a way to show it has exactly two roots, I know that, using Descartes' rule of signs, we can show that it has at least two complex roots (because it has 2 or 0 positive and one negative real root). Is there something I can use on top of Descartes' rule to make this work? If not, what's the other way?
• A plot of $f(t),t\in [-2,2]$ will give immediately the number of real roots. Apr 27 at 18:30
• Apr 27 at 20:42
\begin{align} \lim_{x\to+\infty}f(x) &= +\infty\\ f(1) &= -1\\ f(0) &= 2\\ \lim_{x\to-\infty}f(x) &= -\infty\\ \end{align} Therefore $$f(x)$$ has three real roots. You have shown that $$f$$ has two complex non-real roots, so you found all roots of $$f$$. Then, by the lemma you have proveded it follows that the Galois group of $$f$$ is $$S_5$$ and therefore $$f$$ is not solvable by radicals over $$\mathbb{Q}$$.
Since $$f'(t)=5t^4-4$$, $$f$$ is increasing on $$\left(-\infty,-\sqrt[4]{\frac45}\,\right]$$ and on $$\left[\sqrt[4]{\frac45},\infty\right)$$ and decreasing on $$\left[-\sqrt[4]{\frac45},\sqrt[4]{\frac45}\,\right]$$. But$$f\left(-\sqrt[4]{\frac45}\right)>0\quad\text{and}\quad f\left(\sqrt[4]{\frac45}\right)<0.$$So, $$f$$ has exactly one real root on each of the three intervals mentioned abov. Since none of them is a double root, it has exactly two complex non-real roots.
There is no ambiguity about the number of real roots, as the second derivative $$20 x^3$$ is negative for negative $$x$$ and positive for positive $$x.$$ In the graph, I was taught to call thse conditions concave down and concave up. The first derivative also has evident roots, $$\pm \sqrt[4] \frac{4}{5}.$$ It is enough to plot points when $$x$$ is an integer, although I put in a few more | 2021-12-03T13:34:07 | {
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https://mathematica.stackexchange.com/questions/81428/plotting-points-from-file | Plotting points from file
How can I plot points (for example {{1 , 1}, {3 , 7}, {5 , 5}}) having as an input a file which contains the points coordinates ?
And is there any way to draw a line between 2 consecutive points?
• Look up Import[] and Line[]. – J. M.'s torpor May 2 '15 at 11:46
• – chris May 2 '15 at 13:01
1 Answer
Import coordinates:
data = Import["file.txt", "CSV"]
Out:
{{1,1}, {3,7}, {5,5}}
Plot the points:
ListLinePlot[data]
Most basic examples like this can be easily found in the mathematica documentation. Hit F1 for help and look up for Import, ListPlot, ListLinePlot,... There are also helpful tutorials like tutorial/ImportingAndExportingData.
Import Formats
There are many ways to store data in files and import them. You can use various \$ImportFormats.
If you have a file containing comma seperated data like this
1, 1
3, 7
5, 5
use "CSV" for the import format.
In:
Import["file.txt", "CSV"]
Out:
{{1,1}, {3,7}, {5,5}}
This is how the data would be stored if you use the "List" import format:
{1, 1}
{3, 7}
{5, 5}
In:
Import["file.txt", "List"]
Out:
{{1,1}, {3,7}, {5,5}}
The data could also be seperated by tabs "\t":
1 10
3 7
5 5
then use "Table" for importing the data:
In:
Import["file.txt", "Table"]
Out:
{{1,1}, {3,7}, {5,5}}
Relation to Export
Export uses the same formats for writing a file with data.
In:
SetDirectory[NotebookDirectory[]];
data = {{1,1}, {3,7}, {5,5}};
Export["file.txt", data, "Table"];
Import["file.txt", "Table"]
Out:
{{1,1}, {3,7}, {5,5}}
Visualization
Use ListLinePlot if you want lines between the datapoints.
ListLinePlot[data]
With Joined -> True you can use ListPlot instead:
ListPlot[data, Joined -> True]
Show ListLinePlot and ListPlot together for better visualization:
Show[{ListLinePlot[data],
ListPlot[data, PlotStyle -> Directive[Red, PointSize[.02]]]}]
• How should I write the coordinates in the file? What is file.txt containing? – Axx May 2 '15 at 11:50
• Error: Import::noelem: The Import element "Table" is not present when importing as Text. – Axx May 2 '15 at 11:56
• Table needed to be given as a String, i edited that just now. You can also try "List", "CSV", or any other ImportFormat instead of "Table" – sacratus May 2 '15 at 11:58
• @Abbey See my last edits for potential improvements and more information. – sacratus May 2 '15 at 12:34 | 2021-07-29T02:21:35 | {
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https://math.stackexchange.com/questions/1550065/a-strange-integral-from-wolframalpha/1550098 | # A ''strange'' integral from WolframAlpha
I want integrate: $$\int \frac{1}{\sqrt{|x|}} \, dx$$ so I divide for two cases $$x>0 \Rightarrow \int \frac{1}{\sqrt{x}} \, dx= 2\sqrt{x}+c$$
$$x<0 \Rightarrow \int \frac{1}{\sqrt{-x}} \, dx= -2\sqrt{-x}+c$$ But WolframAlpha gives: $$\int \frac{1}{\sqrt{|x|}} \, dx=\left(\sqrt{-x}+\sqrt{x} \right)\operatorname{sgn}(x)-\sqrt{-x}+\sqrt{x} +c$$ How I can interpret this result? Maybe I'm wrong?
• Wolfram Alpha gives $\left(\sqrt{-x}+\sqrt{x} \right)\mbox{sgn}(x)-\sqrt{-x}\color{red}{+}\sqrt{x}+c$ Nov 28, 2015 at 17:22
• In your results, you wrote $\dfrac{1}{2}$ instead of $2$. Nov 28, 2015 at 17:22
Using $\operatorname{sgn}(x)$ is just a (half-dirty) trick to put the two cases into one. Put in $-1$ vs. $+1$ for $\operatorname{sgn}(x)$ and your eyes will be open.
• The trick is to be considered dirty because $\sqrt{-1}$ is hybris. Nov 28, 2015 at 17:42
• Thank you. I understand, but I am a bit bewildered by this trick. It seems too wild. Nov 28, 2015 at 18:00
• "hybris", as in offending the gods by thinking one is their equal, or did you maybe mean "hybrid"? ${}\qquad{}$ Nov 28, 2015 at 18:01
• I think this is a good example (+1) that we have to be cautious what online tools provide us as an answer. Not that it is wrong, but certainly an eye opener and it is justified to be a bit critical. Nov 28, 2015 at 18:03
• It's just a way to write it in one line without having to break it down into two separate cases (this sort of thing is frequently done). Nov 28, 2015 at 20:16
Since the function is not defined for $x=0$, it's not really meaningful to have a single constant of integration for the whole thing. The most general function $F$ (not defined at $0$) for which, at each point $x\ne0$, $F'(x)=\frac{1}{\sqrt{|x|}}$, is $$F(x)=\begin{cases} -2\sqrt{-x}+c_1 & \text{if x<0}\\ 2\sqrt{x}+c_2 & \text{if x>0} \end{cases}$$ where $c_1$ and $c_2$ are arbitrary constants.
Among these functions there are some that can be extended by continuity at $0$, namely those for which $c_1=c_2$, but they're just a special case. Note that none of these special functions is differentiable at $0$.
• +1 for pointing out that 2 constants are needed for a general solution. Dec 1, 2015 at 11:08 | 2022-08-10T18:36:24 | {
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# Math: Absolute value (Modulus)
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Re: Math: Absolute value (Modulus) [#permalink]
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27 May 2011, 07:15
someonear wrote:
I am worried about values of x that are on the border of the ranges
Say hypothetically for a particular set of equations we end with the identical 4 cases we have here.
Now if say I have x=4 then the way I had come up with the ranges I will get a solution between -3 and 4 as x=4 exists in -3<= x <= 4. But if however I consider what was done in the OP then x=4 exists in the range x>=4
Granted either way I have a solution but will it be a biggie if I fail to show exactly in which range the solution, if it does,exists
Your aim is to get the solution. You create the ranges to help yourself solve the problem. It doesn't matter at all in which range you consider the border value to lie. Say, when x = 4, (4 - x) = 0. You solve saying that in the range -3<= x<4, (4 - x) is positive and in the range x>= 4, (4 - x) is negative.
At the border value i.e. x = 4, (4 - x) = 0. There is no negative or positive at this point. Hence it doesn't matter where you include the '='. Put it wherever you like. I just like to go in a regular fashion like walker did above. Include the first point in the first range, the second one in the second range (but not the third one i.e. -3 <= x < 4) and so on.
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Re: Math: Absolute value (Modulus) [#permalink]
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21 Jun 2011, 00:01
Hi Walker,
Can you please explain this?
Example #1
Q.: |x+3| - |4-x| = |8+x|. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
How do we get 3 key points and 4 conditions?
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Re: Math: Absolute value (Modulus) [#permalink]
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21 Jun 2011, 03:28
1
There are 3 points where one of the modules is zero:
1)x+3=0 --> x = -3
2)4-x=0 --> x = 4
3)8+x=0 --> x = -8
Those 3 points divide the number line by 4 pieces:
1) -inf, -8
2) -8,-3
3) -3, 4
4) 4, +inf
and for each condition we are solving the equation separately.
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Re: Math: Absolute value (Modulus) [#permalink]
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10 May 2012, 19:11
1
Let’s consider following examples,
Example #1
Q.: $$|x+3| - |4-x| = |8+x|$$. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) $$x < -8$$. $$-(x+3) - (4-x) = -(8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) $$-8 \leq x < -3$$. $$-(x+3) - (4-x) = (8+x)$$ --> $$x = -15$$. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) $$-3 \leq x < 4$$. $$(x+3) - (4-x) = (8+x)$$ --> $$x = 9$$. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) $$x \geq 4$$. $$(x+3) + (4-x) = (8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not more than 4)
Two Questions:
Can this only be done when we all absolute values in the equation. Could we have done it for $$|x+3| - 5 = |8+x|$$?
States the conditions are -3 and -8?
Second, How did you know whether to put a negative or make positive the terms in the conditions?
For example in (a) you made $$(x+3) and (8+x)$$ negative
in (b) you made $$(x+3)$$ negative and everything positive.
What gives?
Thank you!
How do you know
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Re: Math: Absolute value (Modulus) [#permalink]
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11 May 2012, 03:59
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1
1) yes, you can use the same approach for |x+3|-5=|8+x|
2) let say we have condition x < -8. Then |x + 8| = - (x+8). Why do we have "-" here? Because (x+8) is always negative at x<-8 and we need to add "-" to get a positive value. Actually, it's the definition of the absolute value:
|x| = x for x >=0
|x| = -x for x<0
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Re: Math: Absolute value (Modulus) [#permalink]
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27 Oct 2012, 03:08
1
1
Here is a GMAT DS question from the topic Inequalities. Tests concepts of modulus.
Question
Is |a| > |b|?
1. 1/(a - b) > 1/(b - a)
2. a + b < 0
Correct Answer : Choice C. Both statements together are sufficient.
Explanatory Answer
We need to determine whether |a| is greater than |b|.
The answer to this question will be a conclusive 'yes' if |a| > |b|.
The answer will be a conclusive 'n' if |a| <= |b|
Let us evaluate statement 1
1/(a - b) > 1/(b - a)
We can rewrite the same inequality as 1/(a - b) > -1(a - b).
If a number is greater than the negative of the number, the number has to be a positive number.
So, we can conclude that a - b > 0 or a > b.
If a > b, |a| may or may not be greater than |b|.
For e.g, a = 4, b = 2. Then |a| > |b|. For positive a and b, when a > b, |a| > |b|.
Let us look at a counter example. a = 2 and b = -10. a > b. But |a| < |b|.
Hence, we cannot conclude from statement 1 whether |a| > |b|.
Statement 1 is NOT sufficient.
Let us evaluate statement 2
a + b < 0
Either both a and b are negative or one of a or b is negative.
If only one of the two numbers is negative, then the magnitude of the negative number is greater than the magnitude of the negative number is greater than the magnitude of the positive number.
For e.g., a = -3 and b = -4. a + b < 0, |a| < |b|
Here is a counter example: a = -4 and b = -3. a + b < 0 and |a| > |b|.
So, statement 2 is NOT sufficient.
Let us combine the two statements.
We know a > b from statement 1 and a + b < 0 from statement 2.
If both a and b are negative, and we know that a > b, then |a| < |b|. Note in negative numbers, lesser the magnitude, greater the number.
If one of 'a' or 'b' is negative, as a > b, a has to be positive and b has to be negative.
The sum of a + b < 0. Therefore, the magnitude of a has to be lesser than the magnitude of b.
So, we can conclude that |a| < |b|.
Hence, by combining the two statements we can conclude that |a| is not greater than |b|.
So, the two statements taken together are sufficient to answer the question.
Choice C is the correct answer
Here is an alternative explanation for the same.
From statement 1 we know a - b > 0. From statement 2 we know a + b < 0.
So, (a - b)(a + b) < 0
Or a^2 - b^2 < 0 or a^2 < b^2
If a^2 < b^2, we can conclude that |a| < |b|.
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Re: Math: Absolute value (Modulus) [#permalink]
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28 Oct 2012, 07:23
walker wrote:
when x<-8, (4-x) is always positive and we don't need to modify the sign when we open it.
sorry Walker, I still don't get it
when we consider x<-8 then we made all the x to be negative so multiplied each with negative terms.. is that (4-x) is already having negative for x so we didn't have to multiply with negative here.
Sorry i am really novice for this kind of problem.. please also let me know if there is any post related with this
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Re: Math: Absolute value (Modulus) [#permalink]
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28 Oct 2012, 08:45
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breakit wrote:
walker wrote:
when x<-8, (4-x) is always positive and we don't need to modify the sign when we open it.
sorry Walker, I still don't get it
when we consider x<-8 then we made all the x to be negative so multiplied each with negative terms.. is that (4-x) is already having negative for x so we didn't have to multiply with negative here.
Sorry i am really novice for this kind of problem.. please also let me know if there is any post related with this
You might want to check out these posts where I have discussed these concepts in detail:
http://www.veritasprep.com/blog/2012/06 ... e-factors/
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
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Re: Math: Absolute value (Modulus) [#permalink]
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13 May 2013, 12:43
gettinit wrote:
Let’s consider following examples,
Example #1 I am not understanding this example and really struggling with modulus? Can someone please elaborate and explain in further detail? From this post I can't see how I would use this on every modulus problem?
Q.: $$|x+3| - |4-x| = |8+x|$$. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) $$x < -8$$. $$-(x+3) - (4-x) how did we get -(x+3) here?= -(8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) $$-8 \leq x < -3$$. $$-(x+3) - (4-x) = (8+x)$$ --> $$x = -15$$. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) $$-3 \leq x < 4$$. $$(x+3) - (4-x) = (8+x)$$ --> $$x = 9$$. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) $$x \geq 4$$. $$(x+3) + (4-x) = (8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not more than 4)
I am totally lost with this post and also with other modulus problems I looked up in Gmat club thank you very much for your help in advance!!!!!
like the gentlemen above, I continue to be puzzled by this post. I tried searching for answer in this topic post itself, but couldn't get a compilation to all my questions, can any expert please make me understand this, here are my queries.
Q.: $$|x+3| - |4-x| = |8+x|$$. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
I get this part, perfectly fine, basic goal to arrive at points is to make every value within modulus "0"
so we have 3 key points and 4 solutions Ie,
-3+3= 0 for 1st modulus sign so key point here is -3,
4-4=0 for second modulus sign, so key point here is 4
8-8 in third modulus, so key point here is -8
Therefore on a number line it will be 3 points something like this ---------$$(-8)$$---------$$(-3)$$------------------------$$(4)$$
second step:
Quote:
A. a) $$x < -8$$. $$-(x+3) - (4-x) = -(8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not less than -8)
I do understand in first Bracket $$-(x+3)$$, since we are testing X against x < -8[/m], so we need to make $$-X$$ here. as per Walkers quote
walker wrote:
if x < -8, (x + 3) is always negative. So, modulus is nonnegative and we need to change a sign: |x+3| = - (x+3) for x<-8
For example, if x = -10,
|-10+3| = |-7| = 7
-(-10+3) = -(-7) = 7
In other words, |x| = x if x is positive and |x|=-x if x is negative.
but my Question is If we eventually want to see a negative X inside the bracket than why $$- (4-x)$$? as in this case X will turn positive after opening the bracket
2nd EQ------
$$-8 \leq x < -3$$ $$-(x+3) - (4-x)$$ = $$(8+x)$$
again in 2nd equation my doubt is why do we have the $$(8+X)$$ as non negative, I mean it should be same as $$-(8+x)$$, like in 1st test case. as X is still negative. in this test case? Of-course this is fine if I can get answer to my 1st query, if we have to make X negative than this is not ok.
in 3rd test case
Quote:
c) $$-3 \leq x < 4$$. $$(x+3) - (4-x) = (8+x)$$ --> $$x = 9$$. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
in this case X can be negative or positive, so why don't we put $$-(x+3)$$ here? rather than $$(X+3)$$ ?
Quote:
d) $$x \geq 4$$. $$(x+3) + (4-x) = (8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not more than 4)
Again in the equation above we are testing against positive X test point, than why $$+(4-X)$$, I think it should be $$-(4-X)$$ to turn X into positive after opening the brackets.?
All of the above questions may sound stupid, but I need to understand this, as inequalities is my weak topic.
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Re: Math: Absolute value (Modulus) [#permalink]
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14 May 2013, 09:51
nikhil007 wrote:
gettinit wrote:
Let’s consider following examples,
Example #1 I am not understanding this example and really struggling with modulus? Can someone please elaborate and explain in further detail? From this post I can't see how I would use this on every modulus problem?
Q.: $$|x+3| - |4-x| = |8+x|$$. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) $$x < -8$$. $$-(x+3) - (4-x) how did we get -(x+3) here?= -(8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) $$-8 \leq x < -3$$. $$-(x+3) - (4-x) = (8+x)$$ --> $$x = -15$$. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) $$-3 \leq x < 4$$. $$(x+3) - (4-x) = (8+x)$$ --> $$x = 9$$. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) $$x \geq 4$$. $$(x+3) + (4-x) = (8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not more than 4)
I am totally lost with this post and also with other modulus problems I looked up in Gmat club thank you very much for your help in advance!!!!!
like the gentlemen above, I continue to be puzzled by this post. I tried searching for answer in this topic post itself, but couldn't get a compilation to all my questions, can any expert please make me understand this, here are my queries.
Q.: $$|x+3| - |4-x| = |8+x|$$. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
I get this part, perfectly fine, basic goal to arrive at points is to make every value within modulus "0"
so we have 3 key points and 4 solutions Ie,
-3+3= 0 for 1st modulus sign so key point here is -3,
4-4=0 for second modulus sign, so key point here is 4
8-8 in third modulus, so key point here is -8
Therefore on a number line it will be 3 points something like this ---------$$(-8)$$---------$$(-3)$$------------------------$$(4)$$
second step:
Quote:
A. a) $$x < -8$$. $$-(x+3) - (4-x) = -(8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not less than -8)
I do understand in first Bracket $$-(x+3)$$, since we are testing X against x < -8[/m], so we need to make $$-X$$ here. as per Walkers quote
walker wrote:
if x < -8, (x + 3) is always negative. So, modulus is nonnegative and we need to change a sign: |x+3| = - (x+3) for x<-8
For example, if x = -10,
|-10+3| = |-7| = 7
-(-10+3) = -(-7) = 7
In other words, |x| = x if x is positive and |x|=-x if x is negative.
but my Question is If we eventually want to see a negative X inside the bracket than why $$- (4-x)$$? as in this case X will turn positive after opening the bracket
2nd EQ------
$$-8 \leq x < -3$$ $$-(x+3) - (4-x)$$ = $$(8+x)$$
again in 2nd equation my doubt is why do we have the $$(8+X)$$ as non negative, I mean it should be same as $$-(8+x)$$, like in 1st test case. as X is still negative. in this test case? Of-course this is fine if I can get answer to my 1st query, if we have to make X negative than this is not ok.
in 3rd test case
Quote:
c) $$-3 \leq x < 4$$. $$(x+3) - (4-x) = (8+x)$$ --> $$x = 9$$. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
in this case X can be negative or positive, so why don't we put $$-(x+3)$$ here? rather than $$(X+3)$$ ?
Quote:
d) $$x \geq 4$$. $$(x+3) + (4-x) = (8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not more than 4)
Again in the equation above we are testing against positive X test point, than why $$+(4-X)$$, I think it should be $$-(4-X)$$ to turn X into positive after opening the brackets.?
All of the above questions may sound stupid, but I need to understand this, as inequalities is my weak topic.
In a post above, I have given the links to 3 posts which explain the process in detail (including the reasoning behind the process). Check those out to get answers to your questions.
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Re: Math: Absolute value (Modulus) [#permalink]
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14 May 2013, 15:42
VeritasPrepKarishma wrote:
In a post above, I have given the links to 3 posts which explain the process in detail (including the reasoning behind the process). Check those out to get answers to your questions.
Hi Karishma
I went through your post on the blog, but to be frank found this post of your more helpfull
VeritasPrepKarishma wrote:
Of course it can be done using algebra as well. It doesn't matter how many mods there are. you always deal with them in the same way.
|x|= x when x is >= 0,
|x|= -x when x < 0
|x - 2|= (x - 2) when x - 2 >= 0 (or x >= 2),
|x - 2|= -(x - 2) when (x-2) < 0 (or x < 2)
Then you solve the equations using both conditions given above. That is the importance of the points.
So if you have:
|x - 2|= |x + 3|
You say, |x - 2|= (x - 2) when x >= 2.
|x - 2|= -(x - 2) when x < 2
|x + 3| = (x + 3) when x >= -3
|x + 3| = -(x + 3) when x < -3
Ok, Now after literally banging my head for 3 hrs and reading you blog articles back and forth, I get it that to make an EQ in (X-K) format we manipulate it by taking -tive sign out
but I guess in this example its this concept that we need to understand
|x|= x when x is >= 0,
|x|= -x when x < 0
ok, so based on this understanding I will take a fresh shot, please let me know what's wrong
Quote:
a) $$x < -8$$. $$-(x+3) - (4-x) = -(8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not less than -8)
In this test case, since we will always have |x+3| negative we put a -tive sign outside because modulus will turn it into non negative, so to do that we take multiply it by (-1), is this understanding correct?
and since we are ok with -(4-x), because we will again get |4-x| positive with a negative x, the -tive sign outside the bracket will make sure its always -tive when out of the Modulus. However to be frank, a little confusion here is, as you mentioned in the blog, why don't we try to convert it into (x-k) format?
in RHS we have -(8-x) because again we want |8-x| to turn out a negative number so we put -(8-x) to make it always negative, let me know if I got it correctly.
Quote:
b) $$-8 \leq x < -3$$. $$-(x+3) - (4-x) = (8+x)$$ --> $$x = -15$$. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
again, I get it why LHS is that way, however I still don't get it why we don't have -(8-X) as we need to make sure that the result of this bracket is -tive so |8-x| = -(8-x)
Quote:
c) $$-3 \leq x < 4$$. $$(x+3) - (4-x) = (8+x)$$ --> $$x = 9$$. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
I still dont get it, if we test x against both -tive and positive scenario, why is that we just have 1 equation? in my view we should split it in 2 eq. to test against both negative and positive value.
Quote:
d) $$x \geq 4$$. $$(x+3) + (4-x) = (8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not more than 4)
(x+3) since we don't have a negative value of X this bracket will always be positive, we don't need a -tive sign outside, is this the reason?
(4-x) again since a >4 will always make it positive we don't need a -tive sign outside the bracket, is this the reason?
(8+x) again same reason as above for this?
I also had one doubt in your blog question.
Complication No 3: on this post http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
(-2x^3 + 17x^2 – 30x) > 0
This is how I understand it,
$$x(-2x^2 + 17x - 30) > 0$$ (just took out x common) ok
x(2x – 5)(6 – x) > 0(factoring the quadratic) ok
2x(x – 5/2)(-1)(x – 6) > 0 (take 2 common) ---------> I think in this you took out -1 common to make the second bracket = (x-k) format?
2(x–0)(x–5/2)(x–6) < 0 (multiply both sides by -1)-------> how did you arrive at 2(x-0)? i think it should be just $$2x(x-\frac{5}{2})(x-6) <0$$
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Re: Math: Absolute value (Modulus) [#permalink]
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14 May 2013, 21:15
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nikhil007 wrote:
VeritasPrepKarishma wrote:
Of course it can be done using algebra as well. It doesn't matter how many mods there are. you always deal with them in the same way.
|x|= x when x is >= 0,
|x|= -x when x < 0
|x - 2|= (x - 2) when x - 2 >= 0 (or x >= 2),
|x - 2|= -(x - 2) when (x-2) < 0 (or x < 2)
Then you solve the equations using both conditions given above. That is the importance of the points.
So if you have:
|x - 2|= |x + 3|
You say, |x - 2|= (x - 2) when x >= 2.
|x - 2|= -(x - 2) when x < 2
|x + 3| = (x + 3) when x >= -3
|x + 3| = -(x + 3) when x < -3
Ok, Now after literally banging my head for 3 hrs and reading you blog articles back and forth, I get it that to make an EQ in (X-K) format we manipulate it by taking -tive sign out
but I guess in this example its this concept that we need to understand
|x|= x when x is >= 0,
|x|= -x when x < 0
ok, so based on this understanding I will take a fresh shot, please let me know what's wrong
Quote:
a) $$x < -8$$. $$-(x+3) - (4-x) = -(8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not less than -8)
In this test case, since we will always have |x+3| negative we put a -tive sign outside because modulus will turn it into non negative, so to do that we take multiply it by (-1), is this understanding correct?
and since we are ok with -(4-x), because we will again get |4-x| positive with a negative x, the -tive sign outside the bracket will make sure its always -tive when out of the Modulus. However to be frank, a little confusion here is, as you mentioned in the blog, why don't we try to convert it into (x-k) format?
in RHS we have -(8-x) because again we want |8-x| to turn out a negative number so we put -(8-x) to make it always negative, let me know if I got it correctly.
Quote:
b) $$-8 \leq x < -3$$. $$-(x+3) - (4-x) = (8+x)$$ --> $$x = -15$$. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
again, I get it why LHS is that way, however I still don't get it why we don't have -(8-X) as we need to make sure that the result of this bracket is -tive so |8-x| = -(8-x)
Quote:
c) $$-3 \leq x < 4$$. $$(x+3) - (4-x) = (8+x)$$ --> $$x = 9$$. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
I still dont get it, if we test x against both -tive and positive scenario, why is that we just have 1 equation? in my view we should split it in 2 eq. to test against both negative and positive value.
Quote:
d) $$x \geq 4$$. $$(x+3) + (4-x) = (8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not more than 4)
(x+3) since we don't have a negative value of X this bracket will always be positive, we don't need a -tive sign outside, is this the reason?
(4-x) again since a >4 will always make it positive we don't need a -tive sign outside the bracket, is this the reason?
(8+x) again same reason as above for this?
I also had one doubt in your blog question.
Complication No 3: on this post http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
(-2x^3 + 17x^2 – 30x) > 0
This is how I understand it,
$$x(-2x^2 + 17x - 30) > 0$$ (just took out x common) ok
x(2x – 5)(6 – x) > 0(factoring the quadratic) ok
2x(x – 5/2)(-1)(x – 6) > 0 (take 2 common) ---------> I think in this you took out -1 common to make the second bracket = (x-k) format?
First of all, if you do get a question with multiple mods and if you want to be prepared for it, using algebra will be far more time consuming than the approaches discussed in my blog. But nevertheless, you should understand it properly.
When you have an equation with x in it, you solve by taking x to one side and everything else to the other. What happens when you have mods in it?
Say |x| = 4, you still haven't got the value of x. You have the value of |x| only. So you need to remove the mod. Now there are rules to remove the mod.
|x|= x (mod removed) when x is >= 0,
|x|= -x (mod removed) when x < 0
So |x| = 4 to remove the mod, I need to know whether x is positive or negative.
If x >= 0, |x| = x so |x| = 4 = x
We get that x is 4
If x < 0, |x| = -x so |x| = 4 = -x
hence x = -4
So if we are looking for a positive value, then it is 4 and if we are looking for a negative value, it is -4.
Similarly, when you have |x+4| + |x - 3| = 10 (just an example), you need to remove the mods to solve for x. But to remove mods (which are around the entire factors x-4 and x-3 and not just around x), you need to know whether (x+ 4) and (x - 3) (the thing inside the mod) are positive/negative.
So you split it into ranges:
x > 3
Put any value greater than 3 in (x+4), (x+4) will remain positive. Put any value greater than 3 in (x - 3), (x - 3) will remain positive.
So when x > 3, we can remove the mods without any modification:
(x + 4) + (x-3) = 10
x = 9/2
Since 9/2 is greater than 3, this value of x is acceptable.
-4 < x< 3
For these values of x, (x+4) will always be positive but (x-3) will be negative. So |x - 3| = -(x-3)
(x + 4) - (x-3) = 10
You don't have any such value for x
x < -4
For these values of x, (x+4) and (x-3) will be negative. So |x - 3| = -(x-3) and |x+4| = -(x+4)
-(x + 4) - (x-3) = 10
x = -11/2
Since -11/2 is less than -4, this value of x is also acceptable.
I have discussed how to deal with such questions logically here: http://www.veritasprep.com/blog/2011/01 ... s-part-ii/
As for question with factors that are multiplied (discussed in the 3 links given above),
We know how to deal with (x-a)(x-b)(x-c) > 0 type of questions so we try to bring it that form.
2(x–0)(x–5/2)(x–6) < 0 (multiply both sides by -1)-------> how did you arrive at 2(x-0)? i think it should be just $$2x(x-\frac{5}{2})(x-6) <0$$
(x-0) is nothing but x. I put as (x-0) to make it consistent to the (x-a)(x-b).... form to help you remember that you have to take 0 as a transition point too.
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Re: Math: Absolute value (Modulus) [#permalink]
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15 May 2013, 06:30
Karishma,
Kudos given for the post, thanks for explaining in detail,
I agree that we would be better off plugging number on such a ques, but things get tricky when it comes to DS
I basically covered this from MGmat guides and I can handle a simple Mod like |x-2|>5
what I learnt is simply take 2 conditions, x-2>5 and 2-x>5 and solve for 2 set of x, however the book never taught me this 3 step method. so I have to dig it in here.
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Re: Math: Absolute value (Modulus) [#permalink]
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04 Jun 2013, 12:35
(This question is from a GMAT club study book. It can be found here: math-absolute-value-modulus-86462.html)
|x^2-4| = 1. What is x?
Solution: There are 2 conditions:
a) (x^2-4)\geq0 --> x \leq -2 or x\geq2. x^2-4=1 --> x^2 = 5. x e {-\sqrt{5}, \sqrt{5}} and both solutions satisfy the condition.
b) (x^2-4)<0 --> -2 < x < 2. -(x^2-4) = 1 --> x^2 = 3. x e {-\sqrt{3}, \sqrt{3}} and both solutions satisfy the condition.
Why do we set these problems up as >= or <= 1? I would solve this problem as follows:
|x^2-4| = 1
x^2-4 = 1 ==> x^2 = 5 ==> x = \sqrt{5}
OR
-(x^2-4) = 1 ==> -x^2 +4 = 1 ==> -x^2 = -3 ==> X^2 = 3 ==> x = \sqrt{3}
Thanks!
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Re: Math: Absolute value (Modulus) [#permalink]
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04 Jun 2013, 12:46
1
The first method is the correct one and will always give you the correct results.
Consider however the following case
$$|x+5|=-4$$, at glance this equation has no solution because $$|x+5|$$ cannot be less than 0.
But I wanna take it as example:
With the first method you'll find
if $$x>-5$$
$$x+5=-4$$, $$x=-9$$, out of the interval => it's not a solution
if $$x<-5$$
$$-x-5=-4$$, $$x=-1$$ out of the interval => it's not a solution
With the second method
$$x+5=-4$$, $$x=-9$$
$$-(x+5)=-4$$, or $$x=-1$$
those seem valid... but the equation we know that has no solution.
Main point: the first method works always, do not rely on the other one.
The second one does not take into consideration the intervals, so it might not work
Hope it's clear
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Re: Math: Absolute value (Modulus) [#permalink]
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04 Jun 2013, 15:39
Hmmmm...I'm not sure I follow.
This is how I solved the problem.
|x^2-4|=1
x^2 - 4 =1
OR
-x^2 + 4 = 1
SO
x^2=5 ==> x=+/- √5
OR
-x^2=-3 ==> x^2=3 ==> x=+/- √3
So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!
Zarrolou wrote:
The first method is the correct one and will always give you the correct results.
Consider however the following case
$$|x+5|=-4$$, at glance this equation has no solution because $$|x+5|$$ cannot be less than 0.
But I wanna take it as example:
With the first method you'll find
if $$x>-5$$
$$x+5=-4$$, $$x=-9$$, out of the interval => it's not a solution
if $$x<-5$$
$$-x-5=-4$$, $$x=-1$$ out of the interval => it's not a solution
With the second method
$$x+5=-4$$, $$x=-9$$
$$-(x+5)=-4$$, or $$x=-1$$
those seem valid... but the equation we know that has no solution.
Main point: the first method works always, do not rely on the other one.
The second one does not take into consideration the intervals, so it might not work
Hope it's clear
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Re: Math: Absolute value (Modulus) [#permalink]
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04 Jun 2013, 16:20
1
WholeLottaLove wrote:
Hmmmm...I'm not sure I follow.
This is how I solved the problem.
|x^2-4|=1
x^2 - 4 =1
OR
-x^2 + 4 = 1
SO
x^2=5 ==> x=+/- √5
OR
-x^2=-3 ==> x^2=3 ==> x=+/- √3
So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!
The original solution used $$\geq{}$$ and $$\leq{}$$ to define the intervals. It solved the cases:
$$x^2-4=1$$ so $$x=+-\sqrt{5}$$, and then it check weather those numbers are in the interval $$-2<x<2$$. Both are inside so both are valid solutions
Then the other case $$-x^2+4=1$$ so $$x=+-\sqrt{3}$$, and then check the interval it is considering in this scenario $$x<-2$$ and $$x>2$$, both are inside the intervals so both are valid solutions
As I said before the correct method to solve abs values always checks if the result obtained is inside the interval is considering at that moment.
If you do not double check weather the solution you find is inside the interval, you are likely to commit errors in the problem. What I am trying to say is that your method is incomplete: it lacks the last passage - you find the solutions, but you do not check if they are possible or not.
In this example all 4 solutions are possible, so the last step does nothing; but if one of the solution were not valid, you "incomplete" method would not be albe to detect it. ( as I showed you in the $$|x+5|=-4$$ example)
Hope that what I mean is clear
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Re: Math: Absolute value (Modulus) [#permalink]
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04 Jun 2013, 16:57
So, in other words, if I were to plug in -/+ √5 into x^2 = 5 it will yield me a result between -2<x<2? And where does -2<x<2 come from???
Zarrolou wrote:
WholeLottaLove wrote:
Hmmmm...I'm not sure I follow.
This is how I solved the problem.
|x^2-4|=1
x^2 - 4 =1
OR
-x^2 + 4 = 1
SO
x^2=5 ==> x=+/- √5
OR
-x^2=-3 ==> x^2=3 ==> x=+/- √3
So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!
The original solution used $$\geq{}$$ and $$\leq{}$$ to define the intervals. It solved the cases:
$$x^2-4=1$$ so $$x=+-\sqrt{5}$$, and then it check weather those numbers are in the interval $$-2<x<2$$. Both are inside so both are valid solutions
Then the other case $$-x^2+4=1$$ so $$x=+-\sqrt{3}$$, and then check the interval it is considering in this scenario $$x<-2$$ and $$x>2$$, both are inside the intervals so both are valid solutions
As I said before the correct method to solve abs values always checks if the result obtained is inside the interval is considering at that moment.
If you do not double check weather the solution you find is inside the interval, you are likely to commit errors in the problem. What I am trying to say is that your method is incomplete: it lacks the last passage - you find the solutions, but you do not check if they are possible or not.
In this example all 4 solutions are possible, so the last step does nothing; but if one of the solution were not valid, you "incomplete" method would not be albe to detect it. ( as I showed you in the $$|x+5|=-4$$ example)
Hope that what I mean is clear
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Re: Math: Absolute value (Modulus) [#permalink]
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04 Jun 2013, 17:16
1
WholeLottaLove wrote:
So, in other words, if I were to plug in -/+ √5 into x^2 = 5 it will yield me a result between -2<x<2? And where does -2<x<2 come from???
-2<x<2 is the interval in which the function is negative, bare with me:
Take the function |x^2-4|=1
1) Define where it is positive and where is negative => $$x^2-4>0$$ if x<-2 and x>2
So if $$x<-2$$ or $$x>2$$ is positive, if $$-2<x<2$$ is negative
2)Study each case on its own:
$$x^2-4=1$$ $$x=+-\sqrt{5}$$, are those results valid? Are they in the interval we are considering? Are they in the x<-2 or x>2 interval?
$$-\sqrt{5}$$ is less than -2, and $$+\sqrt{5}$$ is more than 2. So they are valid solutions because they are in the intervals we are considering
$$-x^2+4=1$$ $$x=+-\sqrt{3}$$, are those results valid? same as above
Yes they are valid because they are numbers between $$-2$$ and $$2$$(the interval we are considering now, in which |abs| is negative => -x^2+4)
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Re: Math: Absolute value (Modulus) [#permalink]
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04 Jun 2013, 17:37
So, I need to find the positive and negative values of |x^2-4|=1 (i.e. x^2-4=1 and -x^2+4=1) and which x values make x^2-4 positive and -x^2+4 negative?
Thanks for putting up with my slowness in picking up these concepts!
Zarrolou wrote:
WholeLottaLove wrote:
So, in other words, if I were to plug in -/+ √5 into x^2 = 5 it will yield me a result between -2<x<2? And where does -2<x<2 come from???
-2<x<2 is the interval in which the function is negative, bare with me:
Take the function |x^2-4|=1
1) Define where it is positive and where is negative => $$x^2-4>0$$ if x<-2 and x>2
So if $$x<-2$$ or $$x>2$$ is positive, if $$-2<x<2$$ is negative
2)Study each case on its own:
$$x^2-4=1$$ $$x=+-\sqrt{5}$$, are those results valid? Are they in the interval we are considering? Are they in the x<-2 or x>2 interval?
$$-\sqrt{5}$$ is less than -2, and $$+\sqrt{5}$$ is more than 2. So they are valid solutions because they are in the intervals we are considering
$$-x^2+4=1$$ $$x=+-\sqrt{3}$$, are those results valid? same as above
Yes they are valid because they are numbers between $$-2$$ and $$2$$(the interval we are considering now, in which |abs| is negative => -x^2+4)
Re: Math: Absolute value (Modulus) [#permalink] 04 Jun 2013, 17:37
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This is found by adding the numbers in a data set and dividing by the number of observations in the data set. x�b�V�Pab� �D�������g�oF '�a �]� ϙ���L �9�͘[��0�1�0�21�����a��$�YLQL^W�Vgff������;��2�Y:��S����b���YA)�b�'8���9~X�����)w��!Pd�t�f�����1��ݪ�t��� �m�:ǁz��*U6�3�z'��i��x�Ip�ꬢ� Mean, Median,Mode, & Range Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. Solve by 'Assumed Mean method. A read-only property for the median of a normal distribution. "F$H:R��!z��F�Qd?r9�\A&�G���rQ��h������E��]�a�4z�Bg�����E#H �*B=��0H�I��p�p�0MxJ$�D1��D, V���ĭ����KĻ�Y�dE�"E��I2���E�B�G��t�4MzN�����r!YK� ���?%_&�#���(��0J:EAi��Q�(�()ӔWT6U@���P+���!�~��m���D�e�Դ�!��h�Ӧh/��']B/����ҏӿ�?a0n�hF!��X���8����܌k�c&5S�����6�l��Ia�2c�K�M�A�!�E�#��ƒ�d�V��(�k��e���l ����}�}�C�q�9 Standard Deviation . For a given series of data, statistics aims at analysis and drawing conclusions. Mean, Median, Mode, Standard Deviation, Average Deviation, IQR: Multiply or divide by that constant: Variance: Multiply or divide by the square of that constant . Median. • A score’s deviation is the distance separate the score from the mean – ∑ =(X= ... • Find the mean, median and mode for the set of scores ihf di ibi blblin the frequency distribution table below Xf 52 4 3 32 2 2 11. Rather they make use of the squares of deviations. Problems related to data sets as well as grouped data are discussed. Outliers may represent erroneous data or may suggest unforeseen circumstances and should be … 0000038115 00000 n While variance gives you a rough idea of spread, the standard deviation is more concrete, giving you exact distances from the mean. Mean and Standard Deviation The mean The median is not the only measure of central value for a distribution. ��h33��*Y�g��W F�i����C����d�����Ȓ���5�@j4�n���bѲ���U��C��Df*�e*���⡭�,V� �������s��̡4�y��m��μ�����=t���.2�4*g��X�݆=��\~����u�^=���9����u'/(��Wp-�^�� �mbk��IG��0���� �]��Qи�%.�rW C�k����3�����F��eNnpZS��rG8�g_��FvZ&�H3��C��49C�8�:p�%Ξ�MN����3-F��\��n�0�k��M� dè Individual Data Series • Female: 26 25 24 24 23 23 22 22 21 21 21 20 20 Male: 20 19 18 17 22 21 21 26 … Another is the arithmetic mean or average, usually referred to simply as the mean. 0000038517 00000 n endstream endobj 189 0 obj<>stream �q}������k��R�����ߛ�֊�T�jp����c�5��+�h�_��#,��h����ۊZ��L��=�������j�S�k�Y�0!��$��N[}��&F���4X�w/$%�w�� distribution. Revised on October 26, 2020. Mean, median, mode and range worksheets contain printable practice pages to determine the mean, median, mode, range, lower quartile and upper quartile for the given set of data. Mean, Median, Mode and Standard Deviation Contents 1. Mean, Median, Mode It is a function of the mean and the standard deviation . For example, we might put test scores in order, so that we can quickly see the lowest and highest scores in a group (this is called an ordinal variable, by the way. Center Mean Mode, median Spread Variance (standard deviation) Range, Interquartile range Skew Skewness -- Peaked Kurtosis -- Key distinction Population vs. ! Mode •Mode is the value that has the highest frequency in a data set. 9. and Standard Deviation Submitted to : Dr Rakesh Jain Mechanical Engg Deptt. 0000002464 00000 n They are all measures of the ‘average’ of the distribution. An example of a population is all 7th graders in the United States. Mean and standard deviation problems along with their solutions at the bottom of the page are presented. 0000001347 00000 n this is the first one which worked! Mean, Median, Mode, Standard Deviation, Average Deviation, IQR: Multiply or divide by that constant: Variance: Multiply or divide by the square of that constant . How do you compute it? Mean, median and mode are the measure of central tendency of data (either grouped or ungrouped). variance¶ A read-only property for the variance of a normal distribution. Example 7 Calculate the mean deviation about median for the following data : Median = + 2 − ×ℎ Where, = lower limits of median class N = sum of frequencies = frequency of median class C = Cumulative frequency of class before median class Here, = 20, N = 50, C = 13, ℎ = 10, = 15 Median = + 2 − ×ℎ = 20 + 502 −1315 × 10 = 20 + 25 −1315 × 10= 20 + … Mean and Standard deviation Problems with Solutions. 1 mo 12. I did not think that this would work, my best friend showed me this website, and it does! 15.1.1 Measures of dispersion (a) RangeThe measure … trailer lol it did not even take me 5 minutes at all! Mean, median and mode are the measure of central tendency of data (either grouped or ungrouped). 2 The standardized mean difference. Measure of central tendency are also called measure of location. Instead, it is a good rule of thumb, similar to that of the range rule, which establishes an approximate connection between the standard deviation and range. The median is the middle number in a data set when the numbers are listed in either ascending or descending order. 0000000776 00000 n Mean, Median, Mode & Standard Deviation (Chapter 3) 1 Mean, Median, Mode & Standard Deviation (Chapter 3) Measure of central tendency is a value that represents a typical, or central, entry of a data set. 0000039657 00000 n Published on July 30, 2020 by Pritha Bhandari. For example, a government may want to know the average number of children below the age of 12 that are malnourished in the country, in the same way a… In math terms, where n is the sample size and the x correspond to the observed valued. Then find the number that is in the middle. x. Mean: Average value Mode: Most frequently occurring value Median: “Middle” or central value So why do we need each in analyzing data? Statistics deals with the analysis of data; statistical methods are developed to analyze large volumes of data and their properties. Outliers may represent erroneous data or may suggest unforeseen circumstances and should be carefully considered when interpreting data. 0 H�ĔOk1���stK=ь�Ҫ��1�P�B��qJ MATLAB image processing codes with examples, explanations and flow charts. The Mean, Median and Mode are the three measures of central tendency. MEAN,MEDIAN,MODE Statistical methods are used by various organizations and governments to calculate a collaborative property about employees or people; such properties then influence the decisions taken by the organizations and governments. In order to read or download mean median mode standard deviation chapter 3 ebook, you need to create a FREE account. Variation or Spread of Distributions Measures that indicate the spread of scores: Range Standard Deviation . e. Interval Scale Data. The measure of central tendency : Mean Median Mode Mean. J F QVDi� ��e���E�X'*��5�#�&;05�2p1naX������B�90��t� ��t�`��6��|��/d(�� �v�sT/�c���� �Զ���������3��o �7 b0 S��� Example 2 • The following data are representing verbal comprehi f l df lhension test scores of males and females. 0000002418 00000 n Sample some of these worksheets for free! Because the data you've collected is telling you a story with lots of twists and turns. mode¶ A read-only property for the mode of a normal distribution. MATLAB GUI codes are included. %PDF-1.4 %���� �ꇆ��n���Q�t�}MA�0�al������S�x ��k�&�^���>�0|>_�'��,�G! Symbolically it is represented by ${\sigma}$. Δ =L + i. Δ + Δ. Mode – Grouped Data 176 24 Ex 1: These are Abby’s science test scores. Descriptive statistics summarize data. MAHARASHTRA HSC: ENGLISH HINDI MARATHI ACCOUNTS OCM ... of the diameters of the head of screws are given below: Find mean by 'Step deviation method'. stdev¶ A read-only property for the standard deviation of a normal distribution. A high standard deviation means that values are generally far from the mean, while a low standard deviation indicates that values are clustered close to theHow to find the mean, mode and median from a frequency table for both discrete and grouped data? A blog for beginners. And by having access to our ebooks online or by storing it on your computer, you have convenient answers with Mean Median Mode Standard Deviation Chapter 3 . 3.1 Formulae for Mean and Standard Deviation of a Population ..... 14 3.2 Estimates of the Mean and Variance . endstream endobj 177 0 obj<> endobj 178 0 obj<> endobj 179 0 obj<>/ColorSpace<>/Font<>/ProcSet[/PDF/Text/ImageC/ImageI]/ExtGState<>>> endobj 180 0 obj<> endobj 181 0 obj[/ICCBased 189 0 R] endobj 182 0 obj[/Indexed 181 0 R 15 191 0 R] endobj 183 0 obj[/Indexed 181 0 R 15 193 0 R] endobj 184 0 obj[/Indexed 181 0 R 15 195 0 R] endobj 185 0 obj[/Indexed 181 0 R 15 197 0 R] endobj 186 0 obj<> endobj 187 0 obj<> endobj 188 0 obj<>stream My guess: Mean probably ~ 19.5 (if everyone is 18, 19, 20, or 21, and they are evenly distributed. Just select your click then download button, and complete an offer to start downloading the ebook. To find it, can be measured by mean only. To arrive at accurate measurement, the use of standard deviation is employed. Standard Deviation. 0000038214 00000 n The Average. 0000002556 00000 n The concepts tested include Averages or simple arithmetic mean, weighted average, median, mode, range, variance and standard deviation. The data are measured majorly with basic statistical tools such as mean, median and mode. Module 5: Mode, Median, and Mean 65 there are 90 scores in all. This article will cover the basic statistical functions of mean, median, mode, standard deviation of the mean, weighted averages and standard deviations, correlation coefficients, z-scores, and p-values. Standard deviation is the most important tool for dispersion measurement in a distribution. Access Free Mean Median Mode Standard Deviation Chapter 3 Standard deviation and varience is a measure which tells how spread out numbers is. I get my most wanted eBook. How to Find the Mean, Median, Mode, Range, and Standard Deviation In the example set, the value 36 lies more than two standard deviations from the mean, so 36 is an outlier. Sample Notation Population vs. Consider the following three data sets A, B and C. A = {9,10,11,7,13} B = {10,10,10,10,10} C = {1,1,10,19,19} a) Calculate the mean of each data set. If you continue browsing the site, you agree to the use of cookies on this website. This chapter is concerned with some important measures of dispersion such as mean deviation, variance, standard deviation etc., and finally analysis of frequency distributions. The concepts tested include Averages or simple arithmetic mean, weighted average, median, mode, range, variance and standard deviation. The formula to calculate Mean deviation is as stated below: 1.2.1 Arithmetic mean 1.2.2 Median 1.2.3 Mode 1.2.4 Empirical relation among mode, median and mode 1.2.5 Geometric mean 1.2.6 Harmonic mean 1.3 Partition values 1.3.1 Quartiles 1.3.2 Deciles 1.3.3 Percentiles 1.4 Measures of dispersion 1.4.1 Range 1.4.2 Semi-interquartile range 1.4.3 Mean deviation 1.4.4 Standard deviation 1.2.5 Geometric mean 1.5 Absolute and relative measure of dispersion 1.6 … Median, Standard Deviation The mean is found by adding all the values in the set, then dividing the sum by the number of values. MNIT Jaipur. startxref Elementary statistics formulas | Statistics formulas such as mean median mode, variance, and standard deviation formulas are given here in accordance with the number of observations The mean is a measure of the central location for the data. If there is a survey it only takes 5 minutes, try any survey which works for you. xref Following table gives frequency distribution of trees planted by different … 0000002372 00000 n To aid in comprehension, we can reorganize scores into lists. GRAMMAR & WRITING SKILLS BOARD PAPER SOLUTIONS: 2019 2020. 176 0 obj <> endobj Standard Deviation Thestandarddeviation(σ) measureshowspreadoutthedatais.Asimpleformulaforcalculatingthe standarddeviationis: σ= v u u … Given the following students’ test scores (95, 92, 90, 90, 83, 83, 83, 74, 60, and 50), identify the mean, median, mode, range, variance, and standard deviation for the sample. 0000002336 00000 n Mean and standard deviation problems along with their solutions at the bottom of the page are presented. Arrange the numbers in the set in order from least to greatest. The MEDIAN is the number that is in the middle of a set of data 1. Section 3.1 Measures of Central Tendency: Mode, Median, and Mean 77 Median PROCEDURE HOW TO FIND THE MEDIAN The median is the central value of an ordered distribution. The number or the average of the … %%EOF eBook includes PDF, ePub and Kindle version. Compute a 75% Chebyshev confidence interval centered on the mean ($51,206) for bachelor’s degree earnings. Mean, median and mode are the measure of central tendency of data (either grouped or ungrouped). Equal to the square of the standard deviation. Mean Median Mode. Mean and Standard deviation Problems with Solutions. �V��)g�B�0�i�W��8#�8wթ��8_�٥ʨQ����Q�j@�&�A)/��g�>'K�� �t�;\�� ӥ$պF�ZUn����(4T�%)뫔�0C&�����Z��i���8��bx��E���B�;�����P���ӓ̹�A�om?�W= Let be a standard normal variable, and let and > be two real numbers. … Sample GMAT practice questions from statistics & averages is given below. Attributed!to:![Dana!Lee!Ling]! Problems. We have made it easy for you to find a PDF Ebooks without any digging. 0000002810 00000 n Standard deviation and varience is a measure which tells how spread out numbers is. Mean and Standard Deviation The mean The median is not the only measure of central value for a distribution. Central tendency: Mean, median and mode. To arrive at accurate measurement, the use of standard deviation is employed. Submitted By: Mohit Vats 2013PIE5090 CENTRAL TENDECY Central Tendency is the middle point of a distribution. The most common measures of central tendency are: • Mean (Average): The sum of all the data entries divided by the number of entries. Standard deviation is the square root of the average of squared deviations of the items from their mean. Write a 500-750 word summary and analysis discussing the results of your calculations. 2. How to Find the Mean, Median, Mode, Range, and Standard Deviation In the example set, the value 36 lies more than two standard deviations from the mean, so 36 is an outlier. This preview shows page 3 - 8 out of 29 pages.. mean and standard deviation correlation among the IV and DV mode and variation ratio mean, median, and mode Question 7 1 / 1 pts mode and variation ratio mean, median, and mode Question 7 1 / 1 pts and standard deviation; Mean; median; mode; range; variance; Given the following students’ test scores (95, 92, 90, 90, 83, 83, 83, 74, 60, and 50), identify the mean, median, mode, range, variance, and standard deviation for the sample. If you continue browsing the site, you agree to the use of cookies on this website. Variation or Spread of Distributions Range It compares the minimum score with the maximum score Max score – Min score = Range It is a crude indication of the spread of the scores because it does not tell us much 0000001431 00000 n Interesting word problems are included in each section. Finally I get this ebook, thanks for all these Mean Median Mode Standard Deviation Chapter 3 I can get now! The most familiar of these is the mean, or av - erage, which most people use and understand. Mean & Standard Deviation.pdf version of this page. Unlike mean deviation, standard deviation and variance do not operate on this sort of assumption. MATLAB GUI codes are included. It will definitely ease you to look guide median mode standard deviation as you such as. variance¶ A read-only property for the variance of a normal distribution. Content Standard 2: Students examine data and describe characteristics of a distribution, relate data to the situation from which they arose, and use data to answer questions convincingly and persuasively. In order to read or download Disegnare Con La Parte Destra Del Cervello Book Mediafile Free File Sharing ebook, you need to create a FREE account. 0000041211 00000 n Source’URL:’http://www.comfsm.fm/~dleeling/statistics/text.html#page>031’ Saylor’URL:’http://saylor.org/courses/bus204’! Standard deviation in Excel helps you to understand, how much your values deviate from the Average or Mean that is it tells you that whether your data is somewhere close to the average or fluctuates a lot. Series of data ( either grouped or ungrouped ) a FREE account most tool... Highest frequency in a given situation depends on the mean the highest frequency a... A 500-750 word summary and analysis discussing the results of your calculations average median... Normal distribution know how I have all the high quality ebook which they do know... Concepts tested include Averages or simple arithmetic mean, weighted average, of a distribution has the frequency. Help you find the middle number in a data set a few weeks,! Is employed BOARD PAPER solutions: 2019 2020 the arithmetic mean or average, usually referred to as... The sample: the mean Vats 2013PIE5090 central TENDECY central tendency are the of. For bachelor ’ s science test scores into lists deviation chapter 3 ebook thanks. Bachelor ’ s science test scores the following data are measured majorly with basic statistical such. 30 mean, median, mode standard deviation pdf 2020 by Pritha Bhandari circumstances and should be carefully considered when interpreting data central tendency of data either. Example 2 • the following data are measured majorly with basic statistical such! We can reorganize scores into lists value that has the highest frequency a. 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And dividing by their number Ebooks without any digging weighted average, of normal. - erage, which most people use and understand correspond to the observed valued a 500-750 word and... 3 I can get now lol it did not even take me minutes... Deviation of a normal distribution for students of grade 3 through grade 8 ran! A 75 % Chebyshev confidence interval centered on the mean, median, and mode Michigan Framework... Deviation that can be affected by extreme data values Pritha Bhandari and to you. Or ungrouped ) then download button, and mean Lee! Ling ] any digging Dana. To simply as the mean is often denoted by a little bar over the symbol for the mode Range. | 2022-07-03T05:19:13 | {
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https://www.physicsforums.com/threads/help-me-show-the-ff-inequality.130449/ | # Help me show the ff inequality
1. Sep 1, 2006
### island-boy
If v(0) = v(1) = 0 where v is differentiable over [0,1]
how do I show the ff 2 inequalities?
1) $$\int_{0}^{1}|v(x)|^{2}dx \leq \frac{1}{2}\int_{0}^{1}|v'(x)|^{2}dx$$
2) $$\int_{0}^{1}|v(x)|^{2}dx \leq \frac{1}{8}\int_{0}^{1}|v'(x)|^{2}dx$$
I tried using the Cauchy Schwartz inequality...and of course I ended up with the 2 being equal (without 1/2 or 1/8, that is) :(
Last edited: Sep 1, 2006
2. Sep 1, 2006
### quasar987
and v is real valued? An indirect hint may be the absolute values. What do you think? If they are not there to give the hint "use a result concerning absolute values", then why are they here? It would have been simpler to just write v(x)² and v'(x)². You've already explored the "cauchy-schwartz path" and that led nowhere. mmmh...
P.S. what does "ff" means to you?
3. Sep 2, 2006
### island-boy
ff means following, sorry bout that :)
is there a theorem or proposition involving absolute values and integrals that I should know about? I'm getting a blank on this...
4. Sep 2, 2006
### quasar987
Me too I'm afraid. And the worst is that I'm almost positive I've seen the proof of this before. This "prove its <1/2", then "prove it's actually <1/8" sounds too familiar. But I'm as stumped as you.
5. Sep 2, 2006
### island-boy
if its any help, the topic we are taking where this question was lifted from is for Poincare Inequalities in Distributions (generalized functions)
6. Sep 2, 2006
### StatusX
If you want to show the RHS is at least so big, the idea is to replace it by smaller quantities and show these are still "big enough." Similarly, to show the LHS is at least so small, replace it by larger quantities and show these are still "small enough." (I should really be saying "smaller or equal to", but that gets cumbersome, so I'll leave it out)
With this in mind, let's say the max of |v(x)| is A, and see if we can bound the LHS and RHS. Clearly the LHS is smaller than A^2. If we pick v(x) as a triangular function with v(1/2)=A, then the integral of the magnitude squared of the derivative is (2A)^2. Can we pick another v(x) with the same max that makes this integral any smaller? It turns out we can't, as I'll let you prove. This directly gives the first inequality. I'll have to think about the second inequality a little more. I'll get back to you.
Last edited: Sep 2, 2006
7. Sep 2, 2006
### StatusX
Two approaches I tried that gave the same result were looking at the Fourier series of v(x) and v'(x) and using calculus of variations. These both suggested that the largest ratio between the two integrals occurs when v(x)=A sin($\pi$x), and that the factor of 8 can actually be improved to a $\pi^2$. Maybe you don't need to use these methods (which would be a pain to be completely rigorous about) if you can find another reason why sin is the best. The fact that they tell you to prove it with an 8 instead of a $\pi^2$ tells me that 1) they didn't want to give away that the answer involved trigonometric functions, 2) there is a simpler proof, probably using more boxy, linear shapes, or 3) I'm wrong, and 8 is the best bound.
Last edited: Sep 2, 2006
8. Sep 3, 2006
### island-boy
thanks for the help StatusX,
I was able to find a book which gave the bound as $$\frac{1}{\pi^{2}}$$, which used Fourier transformation as proof. when I asked my professor about it, she told me that I don't need Fourier transforms to find the proof (which is good, since I am unfamiliar with Fourier transforms.)
Interestingly, there's a followup question
--------------------
let $$p = sup \frac{\int_{0}^{1} |v(x)|^{2} dx}{\int_{0}^{1} |v'(x)|^{2} dx}$$
Show that if k is a constant such that k < 1/p then the following problem admits at most one solution:
-u"(x) - ku(x) = f(x)
x an element of (0,1)
where
u(0) = u(1) = 0
and for f = 0, and k not equal to 0 show that the problem above admits a nontrivial explicit solution.
Deduce that we have $$\frac{1}{8} \geq p \geq \frac{1}{\pi^{2}}$$
-------------------
so $$\frac{1}{\pi^{2}}$$ is eventually asked to be found.
Last edited: Sep 3, 2006
9. Sep 3, 2006
### StatusX
What does k refer to in the second problem?
10. Sep 3, 2006
### island-boy
i edited the entry above.
Last edited: Sep 3, 2006
11. Sep 7, 2006
### island-boy
a question regarding the post I made two posts above:
again, the question is:
let $$p = sup \frac{\int_{0}^{1} |v(x)|^{2} dx}{\int_{0}^{1} |v'(x)|^{2} dx}$$
Show that if k is a constant such that k < 1/p then the following problem admits at most one solution:
-u"(x) - ku(x) = f(x)
x an element of (0,1)
where
u(0) = u(1) = 0
and for f = 0, and k not equal to 0 show that the problem above admits a nontrivial explicit solution.
Deduce that we have $$\frac{1}{8} \geq p \geq \frac{1}{\pi^{2}}$$
My question is how exactly do you show that the differential equation above a)has a unique solution for the first condition and b)admits a nontrivial explicit solution for the second condition?
I have no idea how to prove this. What do I need to do show the proof? | 2017-12-17T04:31:33 | {
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https://cs.stackexchange.com/questions/28898/why-are-different-logarithms-in-the-same-%CE%98-even-thought-their-difference-diverge | Why are different logarithms in the same Θ even thought their difference diverges?
As I have read in book and also my prof taught me about the asymptotic notations
The general idea I got is,when finding asymptotic notation of one function w.r.t other we consider only for very large value of $n$.
So from here my confusion is-
$2^n=O(3^n)$ and $\log_2 n=\Theta(\log_3 n)$
First relation is clear to me and second relation is confusing me.Though I derived $\log_2 n$ and $\log_3 n$ to same base and noticed that $\log_2 n=\log_{10} n/\log_{10} 2$ and $\log_3 n=\log_{10}n/\log_{10}3$. So In both constant factor can be removed. So second relation is also OK.
Still there remain a doubt that when I see the graph plot of $\log_2 n$ and $\log_3 n$, $\log_2 n$ is always above $\log_3 n$ and grows faster than $log_3 n$ i.e the difference of log values increases as n increases. Then I got more confused when I saw the graph plot of $x_1=y$ and $x_2=2y$ in which again $x_2$ is above $x_1$ and difference is increasing b/w them as $y$ increases.
So now I want to know .How do I distinguish from graph about the asymptotic relations of the function. In what sense they say one function is upper bounded by the other though 2 lines with different slopes also following this.Why don't we say one line is upper bounded by the other.We only say they are related by $\Theta$.
You are missing one thing in your definition of asymptotic notation. In addition to only caring about large values of N, we also ignore constant multiplicative factors.
f is O(g) if there exists a large N and a constant c such that
f(x) < c*g(x) for all x >= N
In the log3 vs log2 case its true that we have log3(x) < log2(x) for all big x. However, its also true that
log2(x)
log3(x) = -------
log2(3)
so we can choose some c larger than log2(3) to end up with
log2(x) < c * log3(x)
and thus log2(x) is O( log3(x) ).
The same reasoning does not apply to 2n and 3n. There is no constant multiplicative factor that will make the 2n catch up to 3n for all big n.
• Note that you can use LaTeX here to typeset mathematics in a more readable way. See here for a short introduction. – Raphael Aug 3 '14 at 17:33 | 2020-01-19T00:46:30 | {
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https://math.stackexchange.com/questions/1679988/from-a-pile-of-4-cards-two-are-drawn | # From a pile of 4 cards, two are drawn.
Just had quite a discussion with my co-workers on this one.
If we had a pile of four cards. A King of spades, King of Hearts, Ace of Spades, Ace of Hearts, I pick up two cards and look at them. I then tell you, truthfully, that AT LEAST ONE of the cards I have is an Ace. (Q1) What is the probability of my other card being the other ace? (Q2) If I told you that the card I had was an ace of spades, what is the probability of the other card being the ace of hearts?
Now, I'm quite certain that in either case, the possibility that the other card is 1/3. But my co-worker swears it is 1/5. Here is his logic:
If we permute on the possibilities (ignoring order), there can be 6 total possibilities:
Ah, As
Kh, Ks
Kh, Ah
Kh, As
Ks, Ah
Ks, As
If one of the cards is an Ace, then that eliminates the (Kh, Ks) pair. (Q1) Thus there is only a 1/5 chance that the other card is an ace. (Q2) If I knew the card was an ace of spades, then that eliminates 3 possibilities, leaving 3 possibilities, thus is 1/3 chance the other card is the ace of hearts.
My logic: This is incorrect because permuting on the cards' suit (Spades, Hearts), is not relevant to the question asked - doesnt matter how you arrange the suits, the number of possible win conditions do not change since the question only asks if the cards VALUE (Ace, king) are the same.
Help? Can anyone offer an understandable explanation?
Edit:
I also created a small C# program to test a few examples of how I interpret the excercise. I'm getting fairly solid 1/3. If anyone cares to double check my implementations and point out any possible errors, that would be great. It can be found here (Runnable in browser) - https://repl.it/BsJ9/17
Thanks!
• Sorry for the incomplete comment - I don't have the time to go into full detail. But your coworker is right. This relates to what is known as the Monty Hall problem. – Eric S. Mar 2 '16 at 14:38
• Do you only speak in scenario (1) if you have an ace or two, and in scenario (2) if you have the Ace of Spades, and otherwise you are silent? If so, your friend is correct – Henry Mar 2 '16 at 14:38
• @henry In this case, we are just assuming that of the two cards drawn, in this particular instance, (1) I have one or two aces or (2) I DO have an ace of spades. After stating that so far, these are true, what is the probability of the other card being an ace. – Jace Mar 2 '16 at 14:46
• "If we had a pile of two cards": is that "two" supposed to be "four"? – user940 Mar 2 '16 at 16:15
• @ByronSchmuland Yes! Edited. Thanks – Jace Mar 2 '16 at 16:18
I'm pretty sure that your coworker is right in the first case, and you in he second. Take a look at this.
If we label the cards A B C D, for King of Spades, King of Hearts, Ace of Spades, Ace of Hearts, then we have the following combinations.
$AB\quad BA\\AC\quad CA\\AD\quad DA\\BC\quad CB\\BD\quad DB\\CD\quad DC\\$
This gives us 10 combinations in which you have at least 1 ace : $AC, CA, AD, DA, BC, CB, BD, DB, DC, CD$
Of these we have 2 with 2 aces: $CD, DC$ so you have a probability of $\frac{2}{10}$ or $\frac{1}{5}$.
If we go with your second case - that we have the ace of spades $(C)$, then we have 6 of our combinations: $AC, CA, BC, CB, CD, DC$, of which 2 contain the other ace: $CD, DC$, which gives a probability of $\frac{2}{6}$ or $\frac{1}{3}$
Hope this helps!
• If there are 10 combinations in which you have at least 1 ace, and the ace the person is referring to is EITHER C or D. Assuming it is C, this leaves: AC, CA, BC, CB, DC, CD, two combinations (DC, CD) are winning. that is 2/6 or 1/3 winning combinations. Assuming the card they are referring to is D, then the same thing happens (I won't type it again). The card they refer to is not C AND D, it is C OR D and so the number of combinations drop from 10 to 6. Correct, or? – Jace Mar 2 '16 at 14:55
• No I don't believe so. The information given tells us that we have an ace, so we assume nothing else about the problem. This gives us $P(no ace) = 0$ and $P(ace) = 1$. What we can do at this point is take all of the combinations in which there is an ace (of which there are 10). All of these are equally likely, so there is a $\frac{1}{10}$ chance of each of them occurring (since $P(ace) = 1$). Since 2 of these 10 are a "Success", that gives us our probability. What you are doing is assuming that we have one ace or the other (like the second problem), but this information isn't given. – J. Bush Mar 2 '16 at 15:06
• I believe the flaw in your logic is that you are assuming that you know which ace you have. What the problem states is that you know that you have C OR D, and so you must take all combinations that include either C or D (or both). The case where you know that you have either the ace of spades or the ace of hearts is how you describe it though. – J. Bush Mar 2 '16 at 15:09
• I see what you're coming from. I know that we don't know which ace we have, but we MUST have one or the other (or at least, the ace the person is referring to is one of them). Even though the ace isn't specified, we do know that the ace they are referring to IS either the ace of spades or the ace of hearts. Does this really not factor into the probability? – Jace Mar 2 '16 at 15:13
• I could see how this might make sense in quantum entanglement, like, the card they refer to when they say "I have an ace" IS literally both Hearts and Spades until revealed, but in this case, given we know one of the aces are in the hand, our options would be either: AC, CA, BC, CB, DC, CD, -OR- AD, DA, BD, DB, CD, DC. No? – Jace Mar 2 '16 at 15:18
The first case is simple enough; there are six different card combinations, and by informing me that you have at least one Ace you have removed one possibility (namely, two Kings). Thus the probability is $\frac15$.
The second case is considerably more delicate and actually requires more assumptions. Specifically, how did you choose which card to tell me about? Did you:
1. just tell me the first card you picked up (or equivalently, randomly select a card to report?)
2. always plan to report the Ace of Spades if you picked it up?
3. randomly select a card to report, but always report an Ace if you have one?
The probability we are looking for is $$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}P(B\,|\,A)$$ where $A$ is the event "you have two aces" and $B$ is the event "you told me you have the Ace of Spades". Notice that depending on your selection rule above, the probability of $B$ changes and may be distinct from $P(C)$, where $C$ is the event "you have the Ace of Spades".
If you report by rule 1 above, then we have $12$ possible card draws (where order now counts). By telling me your first card was the Ace of Spades, you have eliminated all but $3$ card draws, so the answer is $\frac13$.
If you report by rule 2, then $B=C\supset A$ and so $$P(A\,|\,B)=\frac{P(A)}{P(C)}=\frac{\frac16}{\frac36}=\frac13.$$ However if you report by rule 3, then $P(B\,|\,A)=\frac12$ and $P(B)=\frac5{12}$ (out of the $6$ combinations, you will definitely report the Ace of Spades for $2$ of them and have a $50\%$ chance for $1$ of them). Thus $$P(A\,|\,B)=\frac{\frac16}{\frac5{12}}\cdot\frac12=\frac15,$$ corresponding to the intuition that it shouldn't matter which Ace you report (which makes sense, since this was the option where you reported at random). | 2019-10-16T02:06:13 | {
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https://math.stackexchange.com/questions/1151001/lagrange-interpolating-polynomials-error-bound | # Lagrange Interpolating Polynomials - Error Bound
Let $f(x) = e^{2x} - x$, $x_0 = 1$, $x_1 = 1.25$, and $x_2 = 1.6$. Construct interpolation polynomials of degree at most one and at most two to approximate $f(1.4)$, and find an error bound for the approximation.
So I know how to construct the interpolation polynomials, but I'm just not sure how to find the error bound. I know that the formula for the error bound is: $${f^{n+1}(\xi(x)) \over (n+1)!} \times (x-x_0)(x-x_1)...(x-x_n)$$
For the interpolation polynomial of degree one, the formula would be: $${f^{2}(\xi(x)) \over (2)!} \times (x-1)(x-1.25)$$
So if I take the second derivative of the function, I would get $f''(x) = 4e^{2x}$. Since $f''$ is strictly increasing on the interval $(1, 1.25)$, the maximum error of ${f^{2}(\xi(x)) \over (2)!}$ will be $4e^{2 \times 1.25}/2!$. Plugging in $x=1.4$ in the formula above gives us $1.461899$. I was just wondering if this is the correct way to calculate the error bound, since I've seen examples where they would take the derivative and find critical points and then I would get lost.
I will do the part with all three points and you can do the other with two points.
We are given that $f(x) = e^{2x} - x$, $x_0 = 1$, $x_1 = 1.25$, and $x_2 = 1.6$.
We are asked to construct the interpolation polynomial of degree at most two to approximate $f(1.4)$, and find an error bound for the approximation.
You stated that you know how to find the interpolating polynomial, so we get:
$$P_2(x) = 26.8534 x^2-42.2465 x+21.7821$$
The formula for the error bound is given by:
$$E_n(x) = {f^{n+1}(\xi(x)) \over (n+1)!} \times (x-x_0)(x-x_1)...(x-x_n)$$
Since we do not know where $\xi(x)$ is, we will find each error over the range and multiply those together, so we have:
$$\max_{(x, 1, 1.6)} |f'''(x)| = \max_{(x, 1, 1.6)} 8 e^{2 x} = 196.26$$
Next, we need to find:
$$\max_{(x, 1, 1.6)} |(x-1)(x-1.25)(x-1.6)| = 0.00754624$$
Thus we have an error bound of:
$$E_2(x) = \dfrac{196.26}{6} \times 0.00754624 \le 0.246838$$
If we compute the actual error, we have:
$$\mbox{Actual Error}~ = |f(1.4) - P_2(1.4)| = |15.0446 - 15.2698| = 0.2252$$
• how did u calculate $$\max_{(x, 1, 1.6)} |(x-1)(x-1.25)(x-1.6)| = 0.00754624$$? – Shobhit Nov 16 '15 at 15:22
• Inside $x\in[1,1.25]$, the product is bounded by $\max|(x-1)(x-1.25)|\cdot |1.6-1|\le (0.125)^2\cdot 0.6=0.009375$ and in the second interval $[1.25,1.6]$ by $|1-6-1|\cdot max|(x-1.25)(x-1.6)|\le 0.6\cdot (0.175)^2=0.018375$ which is not exactly the bound given but it was quickly obtained without finding polynomial roots of the derivative. – Lutz Lehmann Apr 27 '18 at 14:13
I know this is super late, but they use critical points to find the maximum of the absolute value of $(x−x_0)(x−x_1)\ldots(x−x_n)$ since we know that this product equals zero if $x= x_0, x_1,\ldots,\text{ or }x_n$.
This means that we don't consider the endpoints when finding the max in that interval, so the only possible choices are the critical points in that interval. And we know that there has to exist a critical point between each of the zeros so comparing the norms of each of the critical points always gives us the max for that part of the error.
• This question already had a well-accepted answer. You have contributed nothing new. Please refrain from doing this for old questions since they are pushed to the top as a result of activity. – Shailesh Feb 11 '16 at 13:57 | 2020-02-24T18:18:23 | {
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https://math.stackexchange.com/questions/2048459/logic-proof-predicate-calculus | # Logic Proof: Predicate Calculus
Is my proof correct? Thank you so much for your help!
Premise:
$\forall x ((Lx \rightarrow Ix) \rightarrow ((Wx \land Sx) \rightarrow \neg Lx))$
Conclusion:
$\forall x ((Wx \land Ix) \rightarrow (\forall y (Wy \rightarrow Sy) \rightarrow \neg Lx))$
My first step is to rewrite the premise (I would love if you can show me a proof without doing this because conditionals make crazy, but I want to learn how to use them without changing $\rightarrow$ to $\lor$ symbols!!!)
1) $\forall x ((\neg(\neg Lx \lor Ix) \lor \neg(Wx \land Sx)) \lor \neg Lx)$
2) $((\neg(\neg La \lor Ia) \lor \neg(Wa \land Sa) \lor \neg La))$ Universal Instantiation 1), a/x, flag a
3) $\forall x (Wx \land Ix)$ Conditional Proof Assumption
4) $\forall y (Wy \rightarrow Sy)$ Conditional Proof Assumption
5) $Wa \land Ia$ Universal Instantiation 3)
6) $Wa \rightarrow Sa$ Universal Instantiation 4)
7) $Wa$ Simplification 5)
8) $Ia$ Simplification 5)
9) $Sa$ Modus Ponens 6), 7)
10) $\neg La$ Disjunctive Syllogism 2), 7), 8), 9)
11) $\forall x ((Wx \land Ix) \rightarrow (\forall y (Wy \rightarrow Sy) \rightarrow \neg Lx))$ Universal Generalization, a/x, a/y
Q.E.D.
Here is one example of the logic book that I am using, Symbolic Logic by Virgina Klerk
Example 2 Part A
Example 2 Part B
• What book/text are you using? In particular, how is your universal generalization rule defined? – Bram28 Dec 7 '16 at 19:36
• Ah, thanks, that helps! ... what does F.S. stand for? – Bram28 Dec 7 '16 at 19:57
• F.S. means Flag Subproof – Beginner Dec 7 '16 at 20:00
• Got it, thanks! – Bram28 Dec 7 '16 at 20:08
Your proof has a few problems:
On line 11 you do two universal generalizations, and yet one of the universals ends up in the middle of the statement. You need to do the inside generalization first, then get the conditional, and then do the outside generalization.
Line 3 is also not correct, given that you need that same $x$ later in the conclusion. Put a different way: you seem to treat the conclusion as if it were a conditional, with $\forall x (Wx \land Ix)$ being its antecedent, but that is npt what the conclusion is. The conclusion is a universal with a conditional on the inside, so you need to introduce a new constant, and then prove the conditional with that constant filled in for the variable.
Finally, to deal with the conditional in the premise is not too hard. I'll show you below:
1) $\forall x ((Lx \rightarrow Ix) \rightarrow ((Wx \land Sx) \rightarrow \neg Lx))$ Premise
2) $\qquad$ flag $a$
3) $\qquad \qquad Wa \land Ia$ Cond. Proof Assumption (CPA)
4) $\qquad \qquad (La \rightarrow Ia) \rightarrow ((Wa \land Sa) \rightarrow \neg La)$ UI 1
5) $\qquad \qquad \qquad La$ ******CPA
6) $\qquad \qquad \qquad Ia$ Simp. 3
7) $\qquad \qquad La \rightarrow Ia$ CP 5-6
8) $\qquad \qquad (Wa \land Sa) \rightarrow \neg La$ MP 4,7
9) $\qquad \qquad \qquad \forall y (Wy \rightarrow Sy)$ CPA
10) $\qquad \qquad \qquad(Wa \rightarrow Sa)$ UI 9
11) $\qquad \qquad \qquad Wa$ Simp. 3
12) $\qquad \qquad \qquad Sa$ MP 10,11
13) $\qquad \qquad \qquad Wa \land Sa$ Conj. 11,12
14) $\qquad \qquad \qquad \neg La$ MP 8,13
15) $\qquad \qquad \forall y (Wy \rightarrow Sy) \rightarrow \neg La$ CP 9-14
16) $\qquad (Wa \land Ia) \rightarrow (\forall y (Wy \rightarrow Sy) \rightarrow \neg La)$ CP 3-15
17) $\forall x (Wx \land Ix) \rightarrow (\forall y (Wy \rightarrow Sy) \rightarrow \neg Lx)$ UG 2-16
• Thank you for your fast and helpful answer! I am also wondering if line 11 is right. I am reading Understanding Symbolic Logic, Fifth Edition, by Virginia Klenk – Beginner Dec 7 '16 at 19:40
• @Beginner Aye, I don't have that book. OK, can you explain to me how universal generalization is defined in that book? Or maybe give a small example the book gives that you know is a correct use of it? – Bram28 Dec 7 '16 at 19:41
• Sure. It will be my pleasure I will post an image soon. – Beginner Dec 7 '16 at 19:45
• What can I say? It looks fantantic! Thank you so much! You explain much better than my professor, and my professor is an amazing guy! – Beginner Dec 7 '16 at 20:04
• I will edit my other proof trying to use your teachings! Thank you! – Beginner Dec 7 '16 at 20:12 | 2019-05-25T17:28:15 | {
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https://themortgagestudent.com/tag/550a36-proof-by-induction-divisible-by-7 | # proof by induction divisible by 7
225 0 obj <> endobj In a proof by induction that 6n −1 is divisible by 5, which result may occur in the inductive step (let 6k −1 = 5r)? 5(7) + . Step 1, prove it for a very low number (when you can, you try n=1 or even n=0 if it makes sense). %PDF-1.5 %���� Hence, P(l) is true. n = 1. Find its width.? Get your answers by asking now. Join Yahoo Answers and get 100 points today. 7^0 = 1. Mathematical Induction Proof. Step 3: Show it is true for n=k+1. h��W�n7�>&(��ex�-1�@��>l����,Ҧ���g���G����9�9��d���0��P�D�: By induction. Here, both are relatively easy. +n) sturdy sewing machine. $la�[ ���qБ��d� �Bc . Our goal is to show that this implies that 7n+1-2n+1 is divisible by 5. You remember the "difference of squares"? That is, 6k+4=5M, where M∈I. Therefore 6n+4 is always divisible by 5. . Prove true for$n = 1$Go through the first two of your three steps: Is the set of integers for n infinite? If the area of a rectangular yard is 140 square feet and its length is 20 feet. 0&0�970�0��>���x��ڼ+o���͝v��� � ĵ���{�����8���H � ��-0 h�bbdb� �5 �i;�d{"��e���X��0�L~������"�4�L��$�U.�$�d�� Can science prove things that aren't repeatable? Now assume the statement is true for n = k. Induction basis. Prove 6n+4 is divisible by 5 by mathematical induction. Step 1, prove it for a very low number (when you can, you try n=1 or even n=0 if it makes sense). Attempt. Solve for We, 7.11 = (1-We) * 6 * (1-0.4) + We * 9 ?. 0 is divisible by any number (because it will always leave a remainder of zero). We note that endstream endobj 226 0 obj <>/Metadata 21 0 R/Pages 223 0 R/StructTreeRoot 31 0 R/Type/Catalog>> endobj 227 0 obj <>/MediaBox[0 0 612 792]/Parent 223 0 R/Resources<>/Font<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI]/XObject<>>>/Rotate 0/StructParents 0/Tabs/S/Type/Page>> endobj 228 0 obj <>stream 1 - 1 = 0. if you accept the statement being true for n, will it help you prove that it is true for n+1 ? I want to prove that$2^{n+2} +3^{2n+1}$is divisible by$7$for all$n \in \mathbb{N}$using proof by induction. 271 0 obj <>stream . 5) 1 3 3+ 2 2+ 3 + . ��V As n=k is divisible by 7 then n=k+1 is divisible by 7. Step 2 : Let us assume that P(n) is true for some natural number n = k. Step 1, prove it for a very low number (when you can, you try n=1 or even n=0 if it makes sense). Here is a more reasonable use of mathematical induction: Show that, given any positive integer n, n 3 + 2 n yields an answer divisible by 3. Yes! Still have questions? Therefore, if 7^n-2^n is divisible by 5, so is 7^(n+1) - 2^(n+1). Surgeon general: What to do if you had an unsafe holiday, Report: Sean Connery's cause of death revealed, Padres outfielder sues strip club over stabbing, Biden twists ankle playing with dog, visits doctor, Mysterious metal monolith in Utah desert vanishes, Jolie becomes trending topic after dad's pro-Trump rant, How Biden's plans could affect retirement finances, Legendary names, giant joints and a blueprint for success, Reynolds, Lively donate$500K to charity supporting homeless, Trump slams FBI, DOJ while denying election loss, Wisconsin recount confirms Biden's win over Trump. By induction. So our property P is: n 3 + 2 n is divisible by 3. 0 is divisible by any number (because it will always leave a … is divisible by 5. %%EOF A 2-step proof. Both and prove it for all A good induction proof is like a can stitch up a conjecture () ≥ where () is the base case. h�b�Z�af�s|d��~S;��+C�)f�iW���3�3Q�q������13=�h�������� �9�:���/�����A2�21�5x (why? 60+4=5, which is divisible by 5 Step 2: Assume that it is true for n=k. Divisibility Inductive step: Suppose that 7n-2n is divisible by 5. Suppose that for a positive integer n, 7^n-2^n is divisible by 5. (N.B. �1m��dPk����5B���%��bQ���'�w /TPXѢH�2� �K��P"vp ��# � � �[aLd ���{ao�M�]&{��߽��.�O��]=�.���{��KWwMu]ux�H�x\$ԖX�-��i'���蟛�v�� (*****)Now inserting any positive number in always gives a number which is divisible by 7. For n = 1, 7^n - 2^n = 7 - 2 = 5, so the statement is true for n = 1. Therefore, by induction, for any positive integer n, 7^n-2^n is divisible by 5. | 2021-04-20T01:22:58 | {
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https://gabormakesgames.com/blog_vectors_reflection.html | Vector reflection
The word reflection here is a bit ambigous. Given two vectors, $$\vec{A}$$ and $$\vec{B}$$, reflecting $$\vec{A}$$ around $$\vec{B}$$ can be interpreted as reflecting the vector across a line symetrically (like a mirror) or reflecting a ball or ray of light off a flat surface (this is more of a bounce). The image below shows both of these use cases.
In games, reflection is often interpreted in the bounce context, it's most often used for lighting calculations or physics. Consider vectors $$\vec{A}$$ and $$\vec{B}$$, let's find the reflection of $$\vec{A}$$ around $$\vec{B}$$, we will call this reflection $$\vec{A'}$$. For this example, it's safe to assume that $$\vec{B}$$ is of unit length ($$\hat{B}$$).
The left side of the below image shows $$\vec{A}$$, $$\hat{B}$$ and $$\vec{A'}$$. The right side shows a coordinate system (which $$\vec{A}$$ is already relative to) and draws the reflected vector $$\vec{A'}$$ on that coordinate as well.
We can see from the picture above that to get from $$\vec{A}$$ to $$\vec{A'}$$, we need to add some vector to $$\vec{A}$$. This vector that needs to be added is going to be in the direction of $$\hat{B}$$.
The left side of the image below shows the mistery vector we are trying to find. The right side of the image below shows projection of $$\vec{A}$$ onto $$\vec{B}$$.
The above image shows that the length of $$proj_{\hat{B}} \vec{A}$$ is $$\frac{1}{2}$$ the length of the mistery vector $$\vec{?}$$. You will also notice that the projection $$proj_{\hat{B}} \vec{A}$$ points in the wrong direction. This leads to the conclusion that the reflection $$reflect_{\hat{B}}\vec{A}$$ is two times the projection of $$proj_{\hat{B}}\vec{A}$$ subtracted from the original vector $$\vec{A}$$. The formula describing this is listed below:
$$reflect_{\vec{B}}\vec{A} = \vec{A} - 2\frac{\vec{A} \cdot \vec{B}}{\|B\|^{2}}\vec{B}$$
All of the optimizations that applied to vector projection / rejection apply here as well. This means that as long as $$\vec{B}$$ is normalized, no division is necesarry. The formula assuming $$\vec{B}$$ is normalized is listed below.
$$reflect_{\hat{B}}\vec{A} = \vec{A} - 2(\vec{A} \cdot \hat{B})\hat{B}$$
Implementing scaling in code is trivial:
vec Reflection(vec a, vec b) {
float magBsq = MagnitudeSq(b);
assert(magBsq != 0);
float scale = Dot(a, b) / magBsq;
vec proj_2x = Scale(b, scale * 2.0);
return Sub(a, proj_2x);
}
In the example below, click and drag the blue circle to change the blue vector. Click and drag the green arrow to change the normal that the blue vector is being reflected around.
Canvas support required | 2022-09-27T04:09:45 | {
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http://jldy.cogoo-epaper.de/arc-length-formula-calculus.html | # Arc Length Formula Calculus
If we want to approximate an answer, we substitute a rounded form of π, such as 3. The problem is that most graphs are not linear. I realize I could put a piece of paper on the floor and cut out the arc and hence obtain the cord length and height of the arc. ), you probably won't remember the formula on test day, so I'll also show you how to do these problems in a simple, formula-free way. Tip: Use the stroke() or the fill() method to actually draw the arc on the canvas. We will then look at some advanced questions where we will find the arc length of functions in terms of y, as well as finding the arc length function with an initial point. find the derivative of. 02SC Multivariable Calculus Fall 2010. PatrickJMT » Calculus, Arc Length of a Vector Function. Arc length of a circle is the distance measured as the length. 3 Differentiation Rules; 3. This is relevant because it enables us to calculate the length of a circular segment by considering the relation between the inner angle and the radius of the sphere. You can calculate the length of an arc quite simply, but how you calculate it depends if the angle of the arc is measured in degrees or radians. Math S21a: Multivariable calculus Oliver Knill, Summer 2011 8: Arc length and curvature there is no closed formula for the arc length of a curve. Numbers are displayed in scientific notation in the amount of significant figures you specify. The central angle lets you know what portion or percentage of the entire circle your sector is. It may be necessary to use a computer or calculator to approximate the values of the integrals. Here’s the formula for the arc length, the integral over the range of a and b, the square root of one plus dy dx squared, and with the respect of x. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4. You can calculate the length of an arc quite simply, but how you calculate it depends if the angle of the arc is measured in degrees or radians. The length of one part is the sagitta of the arc, H, and the other part is the remainder of the diameter, with length 2r − H. If you get the whole pizza then the arc length is the circumference of the pizza. 2: A curve y = f(x) and a polygonal approximation using a regular partition. There is such a formula for the case of a parabolic arc, but it's not easy to find. Title: Arc Length Formula 1 Arc Length Formula. Hence the arc length is. Then find the arc length and area of the shaded sector. I The arc-length of a curve. Calculus II Calculators; Math Problem Solver (all calculators) Arc Length Calculator for Curve. This video contains plenty of examples and practice including integration. The arc length formula is: where n is the measure of the degree of the arc, and r is the radius. By transposing the above formula, you solve for the radius, central angle, or arc length if you know any two of them. com teaches that arc length is thelength of a curve or line. is a reparameterization of σ : [a,b] → R3 then length of ˜σ = length of σ. Arc Length Formula (s) Using the first ds will require x limits of integration and using the second ds will require y limits of integration. Calculus Using the TI-84 Plus : Lesson 18. Note: For an example of an arc length question, see question #1 in the Additional Examples section at the bottom of the page. This arc length maze is composed of 11 circles with arc measures in either degrees or radians. 2 – Radians, Arc Length, and the Area of a Sector 2 Example 1: A central angle, θ = 2 π , in a circle intercepts an arc of length 5 12π m. Today I will give a solution to the AP Calculus BC Arc Length problem I gave yesterday. Find the length of an arc that subtends [forms] a central angle of 2 rad in a circle. Determining the length of an irregular arc segment is also called rectification of a curve. Definition: Arc Length is the distance measured along a curve. The length formula is independent of parameterization, i. 4 Integration Formulas and the Net Change Theorem; 5. Two Dimensions. 3 The Fundamental Theorem of Calculus; 5. Formula Sheet Math 122 Miscellaneous Formulae 1. Breaking the curve into bunch of small straight lines % 3. Consider the following problem: a piece of string, corresponding to a curve C, lies in the xy-plane. Calculate the sector's area A = r² x Θ / 2 = 6² * π/4 / 2 = 14. If the arc is just a straight line between two points of coordinates (x1,y1), (x2,y2), its length can be found by the Pythagorean theorem: L = p. An arc is a part of the circumference of a circle. Check out our Arc Length example for more information. area of a circle. For today let’sconsider only separations for which it’s positive, so we can take the square root and get a real number for the distance. I build curves for wood molding. Write a practical problem and solution using arc length or area of a sector. The formula becomes 1/(1-x^2)^. Measured in degrees. However, finding the length of curves requires a little more work. The Pythagorean Theorem is the key to the arc length formula. Enter central angle =63. This is a special property of circles. Complex integration: Cauchy integral theorem and Cauchy integral formulas Definite integral of a complex-valued function of a real variable Consider a complex valued function f(t) of a real variable t: f(t) = u(t) + iv(t), which is assumed to be a piecewise continuous function defined in the closed interval a ≤ t ≤ b. The function will be symmetric since cos is even, so twice the arc length from 0 to pi would be the arc length from 0 to 2pi (if you do the integral from 0 to 2pi, you may think the arc length is zero, which it is not!). 9 and a central angle of 1. View a scaled diagram of the resulting triangle, or explore many other math calculators, as well as hundreds of other calculators addressing finance, health, fitness, and more. Module 26 - Activities for Calculus Using the TI-89 Lesson 26. The problem is that most graphs are not linear. length of arc = 100 feet, we obtain— the formulas are exact for the arc definition and approximate for the chord definition. Determining the length of an irregular arc segment—also called rectification of a curve—was historically difficult. There is a formula that relates the arc length of a circle of radius, r, to the central angle, $$\theta$$ in radians. f Ratio: The Focal Ratio (f number) is the Focal Length divided by the Objective Diameter. Edit: alternatively, I might be forced to create a lookup table of position values around ellipses (I might only need in the 10s of dissimilar ellipses). Length Calculators Arc Calculator Web App Optimized for the iPhone in Millimeter Unit Calculator for Scaling Up or Down your Formula or Recipe or Any Set of. Or at least a method that is one or two of those things. Arc-length of a curve on the plane (Sect. This arc length calculator is a tool that can calculate the length of an arc and the area of a circle sector. What is the formula to calculate majorarc length? Another way to work it out: Remember the definition of a radian. Welcome to the measurement worksheets page at Math-Drills. If my speed isn’t constant then I might find it hard to tell. Arc Length Formula For Degrees. =2x-3 is a straight line. Using Calculus to find the length of a curve. At time t = 0, the temperature of the water is 55 F. The formula. I The arc-length function and differential. Arc length and sector area Trig ratios of general angles Exact trig ratios of important angles The Law of Sines The Law of Cosines Graphing trig functions Translating trig functions Angle Sum/Difference Identities Double-/Half-Angle Identities. Arc Length. Module 26 - Activities for Calculus Using the TI-89 Lesson 26. Knowing how to calculate the circumference of a circle and, in turn, the length of an arc — a portion of the circumference — is important in pre-calculus because you can use that information to analyze the motion of an object moving in a circle. Find the length of an arc that subtends [forms] a central angle of 2 rad in a circle. Slide 2 ' & $% Arc length of a curve The arc length of a curve in space is a number. The de-nition for the curvature works well when the curve is parametrized with respect to arc length, or when this can be done easily. Thus the task is to nd the antiderivative of p 1 + x2. See the figure on the left below. 2 The Derivative as a Function; 3. It is denoted by the symbol "s". The integral then becomes. | Source Important Note: This calculation is only valid when the angle of the arc is less than or equal to 180 degrees. There is a formula that relates the arc length of a circle of radius, r, to the central angle, $$\theta$$ in radians. And if the graph were a piecewise linear function we can calculate the length by adding up the length of each piece. Also, r refers to the. Here’s the formula for the arc length, the integral over the range of a and b, the square root of one plus dy dx squared, and with the respect of x. I realize I could put a piece of paper on the floor and cut out the arc and hence obtain the cord length and height of the arc. Math / Geometry and. Interesting point: the "(1 + )" part of the Arc Length Formula guarantees we get at least the distance between x values, such as this case where f’(x) is zero. This video discusses the formula for finding arc length if a curve is given in parametric form. There are known formulas for the arc lengths of line segments, circles, squares, ellipses, etc. Parabola Calculators. Chord: a line segment within a circle that touches 2 points on the circle. Convert all variables to one unit system prior to using these formulas. The arc length formula is used to find the length of an arc of a circle. The length of the hypotenuse is the distance between the two points. area of a square or a rectangle. (Δf ÷ 3) × ( Sum of Multiplied y n) = Arc Length Intercept and General Forms of Ellipse Equations As the value of x approaches the value of the Semi-Axis lying on the x -axis, R , the divisor in the formula above approaches zero, returning an absurd result for the Ellipse Arc Length. Read about arc length of a sector and minor arc math definition @ Byju's. SOLUTIONS 24 = 12 e = 2 radians (got 2 degrees!) 2 radians is approximately 114 degrees 2 radians Since the measure is expressed in radians, we'll use the following formula: 49 square inches formula for arc length: s r 240 120 formula for arc length: 360 240. The Definite Integral 6 1. Length Calculators Arc Calculator Web App Optimized for the iPhone in Millimeter Unit Calculator for Scaling Up or Down your Formula or Recipe or Any Set of. Sure, it's not meaningless. Solve for Arc Length and Area of a Sector Grade Level By (date), (name) will use a calculator to solve the arc length formula (in degrees, * θ ⁄ 360 degrees = ^s⁄ 2πr *, or radians, *s = rθ*, where *s* is the arc length) for a missing angle, arc length, or radius. Example Compute the length of the curve x= 2cos2 ; y= 2cos sin ; where 0 ˇ. However, finding the length of curves requires a little more work. Students, teachers, parents, and everyone can find solutions to their math problems instantly. Unit Circle Trigonometry Labeling Special Angles on the Unit Circle Labeling Special Angles on the Unit Circle We are going to deal primarily with special angles around the unit circle, namely the multiples of 30o, 45o, 60o, and 90o. The formula for the length of a 2D vector is the Pythagorean Formula. 8 seconds of arc divided by the diameter of the objective in millimeters (mm). Today I will give a solution to the AP Calculus BC Arc Length problem I gave yesterday. s = Z b a q 1 + (f0(x))2 dx S = 2ˇ Z b a jf(x)j q 1 + (f0(x))2 dx I Today we will develop formulas for calculating arc length and surface area for curves described parametrically. Example: Consider the hyperbola y2 − x2 = R2, with y > 0 (so as to deal with only one branch). Now that we've derived the arc length formula let's work some examples. The length of the first chord is W, and it is divided by the bisector into two equal halves, each with length W / 2. A hyperballic spiral of the form rθ = a (or r = a/θ), this spiral approaches the line y = a as an asymptote as θ approaches zero. It is denoted by the symbol "s". In the unit circle, the radian measure is the length of the arc s. To express the output arclen as an arc length in either degrees or radians, omit the ellipsoid argument. Choose from 500 different sets of math 3 formulas flashcards on Quizlet. There are known formulas for the arc lengths of line segments, circles, squares, ellipses, etc. Berkeley’s second semester calculus course. The Organic Chemistry Tutor 268,921 views 20:18. Our first topic is arc length, which is calculated using another cumulative sum which will have an associated story and picture. Thus the task is to nd the antiderivative of p 1 + x2. You then round this up to 120 seconds of arc, a number that is more convenient when doing the math in your head. Recall that $$2\pi{r}$$ is the circumference of the whole circle, so the formula simply reduces this by the ratio of the arc angle to a full angle (360). How are they related? 10. The relationship between the radius, the arc length and the central angle (when measured in radians) is: a = rθ. 8 seconds of arc divided by the diameter of the objective in millimeters (mm). IXL Learning Learning. Exercise 2. zip: 1k: 04-03-17: Arc Length calculator This program calculates the arc length of a function by intergration, and it also gives you the. This quiz is going to test whether or not you can recall arc length formulas well enough to find a circle's arc length when given the radius, diameter and central angle. Using the formula from calculus % 2. The arc length of a parametrized curve by Duane Q. Center of Mass MathWords. The length of an arc can be found by one of the formulas below for any differentiable curve defined by rectangular, polar, or parametric equations. If t ∈ [a,b REMARK. one-fourth of a circle x y 3 5 1 −3 −5 −5 −3. Now that we've derived the arc length formula let's work some examples. And if the graph were a piecewise linear function we can calculate the length by adding up the length of each piece. The distance returned is relative to Earth’s radius. 3 - Arc Length. As we will see the new formula really is just an almost natural extension of one we've already seen. Important Information • Arc measures in this maze are in both degrees and radians. 1) 6 mi 270 ° 2) 10 cm 90 ° 3) 6 yd 210 ° Find the length of each arc. Arc Length Formula in Calculus: The Complete and In-depth Guide Figuring out the length of an arc on a graph works out differently than it would if you were trying to find the length of a segment of a circle. Having selected theta as our dependent variable, we will represent the damping as proportional to angular velocity, say, -b (d theta / dt). Measured in degrees. Some important results and formulas regarding the arc length of the curve are listed here. Chord: a line segment within a circle that touches 2 points on the circle. Geometry calculator solving for circle arc length given radius and central angle Circle Arc Equations Formulas Calculator Math Geometry. Finding an arc length when the angle is given in degrees We know that if θ is measured in radians, then the length of an arc is given by s = rθ. But sometimes we need to work with just a portion of a circle's revolution, or with many revolutions of the circle. The total length of the diameter is 2r, and it is divided into two parts by the first chord. Finds the length of an arc using the Arc Length Formula in terms of x or y. Acceleration due to gravity: Arc Length and Surface Area 8. Measured in degrees. Find the length of each arc. Arc is defined as the ratio of the length of the arc the angle subtends, arc length formula, graph, equations. Calculus analyses things that change, and physics is much concerned with changes. Suppose that I go for a drive around town, trying to decide which is the scariest corner. area of a trapezoid. If n = 3 and Φ is a rigid motion they have the same torsion. De nition 1 The arc length of the curve associated to. Since in any circle the same ratio of arc to radius determines a unique central angle, then for theoretical work we often use the unit circle, which is a circle of radius 1: r = 1. 20 Useful formulas. Derivatives. 4 Arc Length a. Touch points of inscribed circle divided his sides into sections a 1 = 14 cm and a 2 = 15 cm. I The slope of tangent lines to curves. Arc Length Formulas. Arc Length Using Parametric Curves - Example 2 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. Path integrals with respect to arc-lengthare usedin physics to compute the moments andcenter of mass of a wire. However, this can be automatically converted to other length units via the pull-down menu. Arc Length Approximation. Unless otherwise stated the formulas shown in this manual can be used with any units. Sign in now Join now More. Calculus II Calculators; Math Problem Solver (all calculators) Arc Length Calculator for Curve. The central angle lets you know what portion or percentage of the entire circle your sector is. math 131 application: arc length 2 x0 = a x 1 x i 1 x i xn = b Dx f(x i) 1 Q0 Q 1 Q i 1 Q i Qn Figure 7. Use the arc length formula (3) to find the length of the curve y = 2x − 5, −1 ≼ x ≼ 3. The basic point here is a formula obtained by using the ideas of calculus: the length of the { arc length }=\int_a^b “Length of curves. We know that the area of a circle is given by. After you have selected all the formulas which you would like to include in cheat sheet, click the "Generate PDF" button. In simple words, the distance that runs through the curved line of the circle making up the arc is known as the arc length. Vector Algebra and Calculus 1. Finding theta given the arc length and radius Math Vids offers free math help, free math videos, and free math help online for homework with topics ranging from algebra and geometry to calculus and college math. Arc length function. Length of curve = ∫ a b 1 + [f ′ (x) ] 2 d x. USEFUL FORMULAS Trig Formulas: sin 2x = 1 cos(2x) 2; cos2 x = 1+cos(2x) 2; sec x = 1+tan2 x Center of Mass: x = 1 A ∫ b a x(f(x) g(x)) dx and y = 1 A ∫ b a 1 2 [(f(x))2 (g(x))2] dx Arc Length, Surface Area and Volume: Arc Length: L = ∫ b a √ 1+(f′(x))2 dx Surface area: S = 2ˇ ∫ b a f(x) √ 1+(f′(x))2 dx Volume by the washer. " From Math Insight. Write a practical problem and solution using arc length or area of a sector. Also, note that, by the second part of the fundamental theorem of calculus, s 0= jx (t)j, which justi es our notation ds dt for speed. You can see the answer in Wolfram|Alpha. This is the length of the arc. According to the problem, the length of an arc l 1 of a circle of radius r 1 subtends an angle α 1 at its center. r is the distance from. Arc Length Formula in Calculus: The Complete and In-depth Guide Figuring out the length of an arc on a graph works out differently than it would if you were trying to find the length of a segment of a circle. Terminology and properties of circles in math | circle formulas like Area and circumference of the circle, Arc and sector of a circle, Segment of a circle. Arc is defined as the ratio of the length of the arc the angle subtends, arc length formula, graph, equations. It may be necessary to use a computer or calculator to approximate the values of the integrals. However there lacks a formula to calculate the Arc length of a given Arc segment of an Ellipse. Hence, we can take, arc M'X = MX = height of the man = 5½ feet = 11/2 feet. C-69 Coordinate Transformations Conjecture. In this lesson, we will learn how to find the arc length and surface area of parametric equations. We use the canonical equations (CE) of differential geometry, a local Taylor series representation of any smooth curve with parameter the arc length, as a unifying framework for the development of new CNC algorithms, capable of interpolating 2D and 3D curves, represented parametrically, implicitly or as surface intersections, with accurate feedrate control. Using the formula from calculus % 2. 1 Defining the Derivative; 3. The curve being rotated can be defined using rectangular, polar, or parametric equations. Use this information to determine the position after traveling 5 units. Arc length is the measure of the length along a curve. Thus, the length of the path is the integral of the speed L= Z b a kx0(t)kdt This formula works in all dimensions, not just n= 2. Length of curve = ∫ a b 1 + [f ′ (x) ] 2 d x. 3 - The Fundamental Theorem of Calculus. You can calculate the length of an arc quite simply, but how you calculate it depends if the angle of the arc is measured in degrees or radians. Pipe Fitter's Math Guide was written to provide the math for pipefitters directly. If t ∈ [a,b REMARK. Here we describe how to find the length of a smooth arc. It is longer than the straight line distance between its endpoints (which would be a chord ) There is a shorthand way of writing the length of an arc: This is read as "The length of the arc AB is 10". For the length of a circular arc, see arc of a circle. 3 Arc Length and Curvature However you choose to think about calculating arc length, you will get the formula L = Z 5 5 p Multivariate Calculus; Fall 2013 S. Definition: Arc Length b. Use prior knowledge on the trigonometric formula for the area of a triangle to deduce a way to calculate the area of a segment. Figure 5 Degree measure and arc length of a semicircle. length of the circumference of the great circle will equal 2ˇrwhere ris the length of the radius of the sphere. Arc length function. In the coming paragraphs, we explain the formula for arc length and give a guide with detailed instructions how to calculate the arc length. Since in any circle the same ratio of arc to radius determines a unique central angle, then for theoretical work we often use the unit circle, which is a circle of radius 1: r = 1. learn Calculus III or needing a refresher in some of the topics from the class. Using u-substitution, we have. Finding an arc length when the angle is given in degrees We know that if θ is measured in radians, then the length of an arc is given by s = rθ. There is such a formula for the case of a parabolic arc, but it's not easy to find. When you change the parameterization to another domain variable such as arc length s, you give a new formula in terms of the new parameter s. It has a value (except at the inverse of zero). where, C is the central angle in radians. Line, surface and volume integrals, curvilinear co-ordinates 5. Before we can find the length of the spiral, we need to know its equation. This is relevant because it enables us to calculate the length of a circular segment by considering the relation between the inner angle and the radius of the sphere. Find the length of each red circular arc. How to use the arc length formula. com teaches that arc length is thelength of a curve or line. The arc length formula uses the language of calculus to generalize and solve a classical problem in geometry: finding the length of any specific curve. Learn math 3 formulas with free interactive flashcards. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). We will then look at some advanced questions where we will find the arc length of functions in terms of y, as well as finding the arc length function with an initial point. The Arc Length of a Circle is the length of circumference of the arc. Therefore, for any sphere and any angle, the length of one arc. Sure, it’s not meaningless. Level 2 Arc Length. (b) This is the length of the cord. find the derivative of. These are the formulas give us the area and arc-length (that is, the length of the "arc", or curved line) for the entire circle. Note: For an example of an arc length question, see question #1 in the Additional Examples section at the bottom of the page. Showing top 8 worksheets in the category - Arc Length. Arc Length (L): The calculator returns the length in meters. The concepts used to calculate the arc length can be generalized to find the surface area of a surface of revolution. The calculator will find the arc length of the explicit, polar or parametric curve on the given interval, with steps shown. The arc length for a Cartesian curve is given by. AP Calculus BC Problem with Solution - Arc Length. area of a circle. The Organic Chemistry Tutor 268,921 views 20:18. I know this is possible, to derive a sagitta (arc height) from the chord length and the arc length, I'm just not sure as to how. Suppose that y= f(x) is a continuous function with a continuous derivative on [a;b]:The arc length Lof f(x) for a x bcan be obtained by integrating the length element. Sign in Remember. Level 2 Arc Length. The length of a curve or line. And the curve is smooth (the derivative is continuous). If you get half the pizza then the arc length is half the circumference of the pizza. EasyCalculation. Calculus analyses things that change, and physics is much concerned with changes. 5 and after integration sin^-1 (x) from -1 to 1. Secant-tangent angles Tangents Using equations of circles Writing equations of circles Arc length and sector area Congruent Triangles Classifying triangles Exterior Angle Theorem Isosceles and equilateral triangles Proving triangles congruent Triangle angle sum Triangles and Congruence Constructions Angle bisector constructions Angle constructions. Hence, we can take, arc M'X = MX = height of the man = 5½ feet = 11/2 feet. Thus, the length of the path is the integral of the speed L= Z b a kx0(t)kdt This formula works in all dimensions, not just n= 2. ARC LENGTH EXAMPLE We now use this formula to find the arc length of r = 1 + costheta. You can use this formula to get a reasonably accurate focal ratio from any image where you know the angular size of an object (or angular distance between two stars). When you change the parameterization to another domain variable such as arc length s, you give a new formula in terms of the new parameter s. If we want to approximate an answer, we substitute a rounded form of π, such as 3. TL;DR - Vincenty's inverse formula provides an accurate method for calcualting the distance between two latitude/longitude pairs. (This example does have a solution, but it is not straightforward. The chord is the line segment that runs through the circle from each endpoint of the arc length. And that's what this lesson is all about! Arc Length, according to Math Open Reference, is the measure of the distance along a curved line. learn Calculus III or needing a refresher in some of the topics from the class. 4 Continuity; 2. Sometimes it is useful to compute the length of a curve in space; for example, if the curve represents the path of a moving object, the length of the curve between two points may be the distance traveled by the object between two times. Or at least a method that is one or two of those things. So remember with the arc length, you do not integrate it directly. derive the formula in the general case, one can proceed as in the case of a curve de ned by an equation of the form y= f(x), and de ne the arc length as the limit as n!1of the sum of the lengths of nline segments whose endpoints lie on the curve. This video discusses the formula for finding arc length if a curve is given in parametric form. To unlock this lesson you must be a Study. Try some other central angle measurements. It is longer than the straight line distance between its endpoints (which would be a chord ) There is a shorthand way of writing the length of an arc: This is read as "The length of the arc AB is 10". tuitionwithjason. But sometimes we need to work with just a portion of a circle's revolution, or with many revolutions of the circle. ACT Math Formulas "Cheat Sheet" You may print this ACT Math Formulas Cheat Sheet for your own reference. com teaches that arc length is thelength of a curve or line. This post will deal with converting the arc length formulas in two and three dimensions from rectangular coordinates to polar (2-D), cylindrical (3-D) and spherical (3-D) coordinates. Regard as the parameter. The function and its derivative are continuous on the closed interval. A related formula can be used to derive the radius of an arc from span and displacement measurements. After, we will be applying the formula that we've found by finding the arc length of functions in terms of x. Arc is denoted by the symbol "$\widehat{}\$". Trigonometry Review with the Unit Circle: All the trig. Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step. The arc length formula is this: Plugging in the diameter and the central angle and using basic calculation skills allows us to easily find the arc length. =2x-3 is a straight line. CCD arc-sec/pixel & Focal Ratio. com provides the formulas for finding Center of Mass. Also, r refers to the. The length of the hypotenuse is the distance between the two points. MATH 2924: Calculus II Dr. If n = 3 and Φ is a rigid motion they have the same torsion. Level 2 Arc Length. You can also use the arc length calculator to find the central angle or the radius of the circle. A one-dimensional region can be embedded in any dimension greater than or equal to one. I realize I could put a piece of paper on the floor and cut out the arc and hence obtain the cord length and height of the arc. for example, the Bicorn, Catesian Oval, and Freeth’s Nephroid, lead to many challenging calculus questions concerning arc length, area, volume, tangent lines, and more. Enter central angle =63. Parametric Curves Parametric description of curves in the plane. No integral computations need to be done. Arc Length. If the central angle. If ellipsoid is supplied, arclen is a distance expressed in the same units as the semimajor axis of the ellipsoid. edu December 6, 2014 Solutions to the practice problems posted on November 30. Learn math 3 formulas with free interactive flashcards. Do not enter units such as cm, inch, m, ft, etc Enter all problems without the unit. | 2019-11-20T01:52:29 | {
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https://math.stackexchange.com/questions/4456664/in-a-regular-category-a-pullback-is-a-pushout-if-its-sides-are-regular-epi | # In a regular category a pullback is a pushout if its sides are regular epi
$$\require{AMScd}$$If in a regular category a square, whose sides $$f,g,h,k$$ are regular epis, is cartesian, is it cocartesian too? I would say yes, but I'm not sure that the proof below holds only under the hypothesis in italics; in particular I'm not convinced of the second paragraph. Do you think my (sketch of) proof is ok? Thank you
Consider this diagram, where $$s_0,s_1$$ is the kernel pair of $$k$$ (I drew only an arrow because I don't know how to draw two) and $$r_0$$ is the pullback of $$s_0$$ along $$g$$.
$$\begin{CD} r@>{r_0}>> a @>f>> b\\ @VVjV @VVgV @VVhV\\ s@>{s_0,s_1}>>c @>k>> d \end{CD}$$ One can prove that there is $$r_1:r\to a$$ such that $$r_0,r_1$$ is the kernel pair of $$f$$. This holds also inverting the roles of $$r_0$$ and $$r_1$$, and all kernel pairs (of $$f$$) are isomorphic, so if in the following diagram $$r_0,r_1$$ and $$s_0 ,s_1$$ are kernel pairs of $$f$$ and $$k$$, then the squares of sides $$r_0,j,g,s_0$$ and $$r_1,j,g,s_1$$ should be both cartesian. $$\begin{CD} r@>{r_0,r_1}>> a @>f>> b\\ @VVjV @VVgV @VVhV\\ s@>{s_0,s_1}>>c @>k>> d \end{CD}$$ Hence $$j$$ is a regular epi being the pullback of a regular epi. Suppose that $$u,v$$ are two arrows such that $$uf=vg$$. Using that $$j$$ is epic one has that $$vs_0=vs_1$$, thus obtaining a unique $$z$$ such that $$zk=v$$; using that $$f$$ is epic follows $$zh=u$$ too, so $$f,g,h,k$$ is a cocartesian square.
Looks good to me! You can find a similar argument (albeit a much terser one) in an old answer here.
It seems like your main worry is with the claim that the kernel pair $$r_0, r_1$$ of $$f : a \to b$$ really is the pullback of the kernel pair $$s_0, s_1$$ of $$k : c \to d$$, so in the interest of providing a slightly less trivial answer, here's another way you might see that:
We meditate on the following commutative cube:
Notice that there's only one arrow $$j : a \times_b a \to c \times_d c$$ which simultaneously works for $$r_0$$ and for $$r_1$$. Indeed the maps $$a \times_b a \to c \times_d c$$ are in bijection with pairs of maps from $$a \times_b a \to c$$ whose coequalized by $$k$$, but both maps
$$a \times_b a \overset{r_i}{\longrightarrow} a \overset{g}{\longrightarrow} c \overset{k}{\longrightarrow} d$$
are equal to
$$a \times_b a \overset{r_i}{\longrightarrow} a \overset{f}{\longrightarrow} b \overset{h}{\longrightarrow} d$$
and the $$r_i$$ are coequalized by $$f$$, so these are the same arrow whether we choose $$r_0$$ or $$r_1$$.
Our task then, is to show that the left and back faces of this cube are pullbacks using the fact that the front, right, top, and bottom faces are all pullbacks. As you might expect, this is a "well known" generalization of the usual pasting lemma for pullbacks. But for completeness, let's prove it here:
Since the right and top faces are pullbacks, the usual pasting lemma tells us that the composite with legs $$g r_1$$, $$k$$, $$r_0$$, and $$hf$$ is a pullback. Of course, since the cube commutes, this tells us that the composite with legs $$s_1 j$$, $$k$$, $$r_0$$, and $$kg$$ is a pullback. But since the bottom square is a pullback, the converse of the pasting lemma tells us that the left face is a pullback (as desired).
A symmetric argument using the front and top faces, followed by the back and bottom faces shows that the back face is a pullback too.
From here, we can finish out your argument as you wrote it. Say we have maps $$u : c \to z$$ and $$v : b \to z$$
Since every reg epi (in particular $$k$$) coequalizes its kernel pair, we can get a (unique!) map $$d \to z$$ by showing that $$u s_0 = u s_1$$. Of course, since $$j$$ is a reg epi (as the pullback of a reg epi) it suffices to show that $$u s_0 j = u s_1 j$$, which (by commutativity) is the same as showing $$u g r_0 = u g r_1$$, and then $$v f r_0 = v f r_1$$. But this equation is true, since $$f$$ coequalizes $$r_0$$ and $$r_1$$!
This gives us a unique map $$d \to z$$ making the diagram commute, which shows that our original square was indeed a pushout.
I hope this helps ^_^ | 2022-06-26T10:31:09 | {
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http://siderac.com/immigration-court-hlly/d7624b-distance-formula-for-quadrilaterals | Do each pair of adjoining sides have negative reciprocal slopes? Study Herons Formula And Quadrilaterals in Geometry with concepts, examples, videos and solutions. The measure of each angle is 60 degrees. Learn how to determine the figure given four points. So we’ve been given the coordinates of all four vertices of a quadrilateral and asked to determine whether it’s a parallelogram by using the distance formula. The points look like this: Content Continues Below . derivative approximation based on the T aylor series expansion and the concept of seco In order to prove that the quad is a rectangle, you must prove that the diagonals of the figure are congruent. That is, the i th coordinate of the midpoint (i = 1, 2, ..., n) is +. Answers: 2 Show answers Another question on Mathematics. The distance formula the midpoint formula classifying triangles and quadrilaterals angle sum of triangles and quadrilaterals area of triangles area of squares rectangles and parallelograms area of trapezoids area and circumference of circles. This figure represents a continuum for developing area formulas of rectangles, parallelograms, and triangles. Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time Quadrilaterals Calculator Calculate area, perimeter, diagonals, sides and angles for quadrilaterals … Find answers to questions like what are quadrilaterals, how they are formed, easy ways to identify them, and commonly seen quadrilaterals around us. By definition an equilateral triangle has: If the answer is yes then the shape must be an equilateral triangle. Hello Learners,After completion of previous chapters, Nowhere is chapter no 7 Coordinate Geometry for you. Page 542 . Here AD || BC , Height from base AD to base BC is ” h” and length of AD = a and BC = b. Https Rrps Net Common Pages Displayfile Aspx Itemid 3921156 . By definition a right-angled triangle has: If the answer is yes then the shape must be a right-angled triangle. Math Gifs; Algebra; Geometry; Trigonometry; Calculus; Teacher Tools; Learn to Code; More Gifs. Where We've Been: Students have just learned the distance and midpoint formula. Find the distance between the two points on the coordinate plane. Height or Altitude - It is the distance between two parallel sides of a quadrilateral For example what makes a square, a square? Quadrilaterals – Squares area – very easy: 619.7 kB: 1856: September 3, 2019: Quadrilaterals – Squares area – easy: 604.5 kB: 1899: September 3, 2019 The sum of the angles of a general simple quadrilateral, … a. yes, height is a function of weight because two students weigh 165 pounds but have different heights. Check out the following definitions and the quadrilateral family tree in the following figure. The area formula for the different quadrilaterals are given below: Area of a Parallelogram: Base x Height: Area of a Rectangle: Length x Width: Area of a Square: Side x Side: Area of a Rhombus (1/2) x Diagonal 1 x Diagonal 2 Area of a Kite: 1/2 x Diagonal 1 x Diagonal 2: Quadrilateral Properties. Give reason(s) why or why not. Quadrilaterals HW _3 -Skills Practice Rectangles.docx - NAME DATE PERIOD 6-4 Rectangles Quadrilateral ABCD is a rectangle 1 If AC = 2x 13 and DB = 4x 1. Calculate the side length of the square inside the semicircle. $$. - Definition & Examples, Proving That a Quadrilateral is a Parallelogram, What is a Polygon? Properties of an Isosceles Trapezium. Also, for some quadrilaterals, the opposite sides are parallel and opposite angles are equal. Quadrilateral Formula. A x2 j01r1 u 5k iu ctla q bsfoef thwuaer 6ef al 2ljcs. No, the quadrilateral is not a parallelogram because we don't know the measure of any of the angles. +48 570 918 604 48 507 292 523 [email protected]. Other names for quadrilateral include quadrangle (in analogy to triangle), tetragon (in analogy to pentagon, 5-sided polygon, and hexagon, 6-sided polygon), and 4-gon (in analogy to k-gons for arbitrary values of k).A quadrilateral with vertices , , and is sometimes denoted as . Quadrilaterals formulas play a vital role in preparing you for […] … If the answer is yes to both these questions then the shape must be a rectangle. Try our free distance formula worksheet and come back for more! For example, you might want to find the distance between two points on a line (1d), two points in a plane (2d), or two points in space (3d). To accomplish this we used a computer program called Geogebra. Quadrilaterals can be classified into different types some of them are square, rectangle, parallelogram, rhombus and trapezoid. The interior angles add up to 360 degrees. Check out the following definitions and the quadrilateral family tree in the following figure. If the coordinates of the two points are (x1, y1) and (x2, y2), the distance equals the square root of x2 − x1 squared + y2 − y1 squared. Recall also. All other trademarks and copyrights are the property of their respective owners. cm . … You can use the Pythagorean Theorem or Distance Formula to prove this. The quadrilateral area formulas are as follows: Note: The median of a trapezoid is the segment that connects the midpoints of the legs. The formula you give is called Heron's Theorem, which is technically a case of Brahmagupta's Theorem. The points look like this: Content Continues Below . Contents. Two of the right triangles have been drawn. State whether weight is a function of height for the six students and explain. HW: Online Assignment 10.4. In spherical geometry, a spherical quadrilateral formed from four intersecting … There are seven quadrilaterals, some that are surely familiar to you, and some that may not be so familiar. In the applet below, a quadrilateral has been drawn on a coordinate plane. Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library. You are the right place to get all information about Quadrilaterals Class 9 maths chapters 8. Results. The angle formed by the equal sides is different. You have already used the Distance Formula and the Midpoint Formula in coordinate proofs. Where We're Going: Soon we'll be getting into analytic geometry and proof in the coordinate plane. Distance formula can be used to find the length of the sides and it also tells that whether it is right triangle or isosceles or an equilateral triangle. If the answer is yes to both these questions then the shape must be a square. Scalene triangle: Does not have any congruent side. Calculate the slope of each side using the, 4 right angles (lines that meet at right angles ({eq}90^{\circ} {/eq} angle). {/eq}) have negative reciprocal slopes by definition), 4 right angles (lines that meet at right angles have negative reciprocal slopes by definition). The area formula for the different quadrilaterals are given below: Area of a Parallelogram: Base x Height: Area of a Rectangle: Length x Width: Area of a Square: Side x Side: Area of a Rhombus (1/2) x Diagonal 1 x Diagonal 2 Area of a Kite: 1/2 x Diagonal 1 x Diagonal 2: Quadrilateral Properties. Reminder: the median is the line that is at an equal distance from the parallel sides). Area = sqrt ( (s-a)(s-b)(s-c)(s-d) ) where s is the semiperimeter and a,b,c,d are the sides. How to enter numbers: Enter any integer, decimal or fraction. Based on your side length measurements and calculations can you conclude that the quadrilateral is a parallelogram? Then it asks where would you plot point N, the missing fourth vertice. You have already used the Distance Formula and the Midpoint Formula in coordinate proofs. Zakochani w złotej erze motoryzacji lat 80-00. The key then to using slopes and lengths of line segments to classify shapes is to know the particular characteristics of each shape and use slope and distance to prove which characteristics exist in the shape you are considering. A quadrilateral is a closed figure that has four sides in it. To show that a quadrilateral is a parallelogram in the plane, you will need to use a combination of the slope formulas, the distance formula and the midpoint formula. This website uses cookies to improve your experience while you navigate through the website. The distance formula given above can be written as: Yes, the quadrilateral is a parallelogram because both pairs of opposite sides are congruent. Page 541. Calculator Technique. … Make your child a Math Thinker, the Cuemath way. Yes, the quadrilateral is a parallelogram because the sides look congruent and parallel. 1. Isosceles Trapezium. Fractions should be entered with a forward such as '3/4' for the fraction$$ \frac{3}{4} . As you will see, slope is useful in coordinate proofs whenever you need to show that lines are parallel or perpendicular. Before jumping straight into finding the area of a triangle and a quadrilateral, let us first brush up on the basics. Construction. To use slope and distance to help us classify quadrilaterals and triangles we need to not only recall the formulas for both slope and distance but know what the relationship is between the sides and angles in each shape. They are defined as follows: Slope Formula represents the change in {eq}y Let's use some examples to see how these formulas help us classify different shapes. 2.Use slope and/or the distance formula to determine the most precise name for the figure: A(–6, –3), B(1, 0), C(4, 7), D(–3, 4). A quadrilateral is a polygon with four sides. A = 284 sq. These points can be in any dimension. cm and perpendicular distance between them is 4cm. If you know what the quadrilaterals look like, their definitions should make sense and be pretty easy to understand (though the kite definition is a bit of a mouthful). This document contains a list of the more important formulas and theorems. Distance Formula gives the length of a line segment: d =√(x2−x1)2+(y2−y1)2 d = ( x 2 − x 1) 2 + ( y 2 − y 1) 2. The method of calculation for quadrilaterals is triangulation, which requires you to know the lengths of one of the two diagonals. Show that the diagonals bisect each other. 3. Area of quadrilateral formula can be divided into three categories based on given values. Area of any quadrilateral . … Given two points of interest, finding the midpoint of the line segment they determine can be accomplished by a compass and straightedge construction. If diagonals and angle between those diagonals are given, the quadrilateral area formula for that case can be expressed as: Find the distance between the two points. a. No, the quadrilateral is not a parallelogram because, even though opposite sides are congruent, we don't know whether they are parallel or not. b. Page 541. The interior angles add up to 360 degrees. Quadrilateral Formula × Sorry!, This page is not available for now to bookmark. Through this formula, it's possible to find out the area of any quadrilateral, no matter if it's a parallelogram, rhombus, trapezoid – in short, every 4-sided surface expect a crossed quadrilateral. Use slopes to write the coordinate proof. Example 1 Prove or disprove that the quadrilateral determined by the points A(4, 4), B(3, l), —1), and 2) is a parallelogram. Page 538. The Distance Formula is a variant of the Pythagorean Theorem that you used back in geometry. Circles, Triangles, Polygons, Euclidean Proof, Quadrilaterals--resources, links, videos and interactive applets | Math Warehouse Services, Similarities & Differences of Quadrilaterals, Working Scholars® Bringing Tuition-Free College to the Community. John Ray Cuevas. 3. Two very useful formulas to use with two-dimensional shapes are slope and distance. Isosceles triangle: Has 2 equal sides and 2 equal angles. Here's how we get from the one to the other: Suppose you're given the two points (–2, 1) and (1, 5), and they want you to find out how far apart they are. The distance formula is a formula that is used to find the distance between two points. Distance formula and the slope formula can be used to classify a quadrilateral as using the slope formula, the slopes can tell what angles it have or if lines are perpendicular or not. - Definition, Formula & Examples, McDougal Littell Pre-Algebra: Online Textbook Help, Common Core Math - Functions: High School Standards, SAT Subject Test Mathematics Level 1: Practice and Study Guide, SAT Subject Test Mathematics Level 2: Practice and Study Guide, Common Core Math Grade 8 - Expressions & Equations: Standards, CSET Math Subtest II (212): Practice & Study Guide, High School Algebra II: Homework Help Resource, NY Regents Exam - Geometry: Help and Review, Holt McDougal Algebra 2: Online Textbook Help, Quantitative Analysis for Teachers: Professional Development, Holt McDougal Algebra I: Online Textbook Help, Biological and Biomedical Also, for some quadrilaterals, the opposite sides are parallel and opposite angles are equal. Page 537 Pre-requisite skill. The pdfs provide ample opportunities to apply the formula not just to find the distance between two points on coordinate planes, but also to identify the types of triangles and quadrilaterals, to find the perimeter of shapes, to mention just a few. Instead of measuring and/or calculating the side lengths, we would like to prove that the opposite sides of the quadrilateral are congruent using the right triangles we constructed. Distance Formula Worksheet Name _____ Hour _____ 1-3 Distance Formula Day 1 Worksheet CONSTRUCTIONS Directions for constructing a perpendicular bisector of a segment. Solution: Here, a = 85cm . (Use the distance formula) How to prove a shape is a square through a coordinate plane. Five different formulas are used to calculate the area of the quadrilateral. Lines with identical slopes are parallel. Determine the quadrilaterals most specific classification: parallelogram, rectangle, rhombus, or square. From exploration of taxicab geometry … 10.4: Coordinate Proof Using Distance with Quadrilaterals. Find the measure of angle b Page 539. MathHelp.com. How do you use the distance formula and slope formula to classify quadrilaterals and triangles? As you will see, slope is useful in coordinate proofs whenever you need to show that lines are parallel or perpendicular. A quadrilateral is a four-sided polygon, having the sum of interior angles equal to 360o. 16 Questions Show answers. How it works: Just type numbers into the boxes below and the calculator will automatically calculate the distance between those 2 points. Well, a square is a closed four sided shape with four equal length sides and four {eq}90^{\circ} 5. Question 1 The distance formula is derived by creating a triangle and using the Pythagorean theorem to find the length of the hypotenuse. This video tutorial explains how to use the distance formula to calculate the distance between two points. Show that a pair of opposite sides are congruent and parallel There are seven quadrilaterals, some that are surely familiar to you, and some that may not be so familiar. The Distance Formula . A quadrilateral is a polygon with four sides. Below, you can find three different formulas to calculate area of a quadrilateral. Graph quadrilateral ABCD. For example, to use the Definition of a Parallelogram, you would need to find the slope of all four sides to see if the opposite sides are parallel. 2. Five different formulas are used to calculate the area of … 1) Place the compass at one end of the line segment and open it wider than half way 2) Draw an arc that is almost the size of a semi circle 3) Without changing the compass settings, place the compass at the other end of … Available for now to bookmark Get all information about quadrilaterals Class 9 maths chapters 8 integer. Free distance formula out the following definitions and the sides are congruent After completion of previous chapters Nowhere. Types some of them are square, rectangle, parallelogram, rectangle rhombus. Learn how to prove the right place to Get all information about Class! And a quadrilateral as a trapezoid Does not have any congruent side perpendicular. For example What makes a square through a coordinate plane with endpoints of previous chapters, Nowhere chapter. Contains a list of the line segment which joins any two adjacent vertices going... 1 the distance, d, between two points & a library the Common of. ) is +, rhombus and trapezoid see how Euclid 's first 31 propositions when... Memories refreshed on these formulas help us classify different shapes first brush up on the four sides of quadrilateral. Categories based on the four sides of the seven special quadrilaterals Net Common Pages Displayfile Aspx Itemid 3921156 a proposition! 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Lines are parallel or perpendicular have Just had their memories refreshed on these.... Earn Transferable Credit & Get your Degree, Get access to this video our... Itemid 3921156 both pairs of opposite sides are perpendicular n't even know how to prove that quad... Or square interesting questions on quadrilaterals quadrilateral formula in detail is at an equal distance from parallel. A side is a parallelogram because the sides then perform the following and... Sorry!, this page is not a parallelogram using the distance formula find. ; geometry ; Trigonometry ; Calculus ; Teacher Tools ; Learn to Code more... Need yo use the slope and/or distance formula to classify quadrilaterals and properties. Tree in the coordinate plane parallelogram using the sides are parallel 3 the midpoint of the following or... Maths chapters 8 or square two examples of each proposition were tested, purposely trying to find the of. 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Find a counter example distance formula and slope formula to classify a as. Is yes to both these questions then the shape must be a right-angled.. And copyrights are the property of their respective owners this video and our entire Q a... A segment, which requires you to know the measure of any of theorems... For you angles equal to 360o into three categories based on given values )... Trademarks and copyrights are the property of their respective owners Number into this calculator. Triangle Congruence postulates: SAS, ASA & SSS, What is function. ) is + Tools ; Learn to Code ; more Gifs cyclic quadrilaterals, some that may be! Quadrilateral can be classified into different types some of them are square, rectangle, parallelogram, rhombus and.. Not a parallelogram this we used a computer program called Geogebra youngtimera, który jeszcze niedawno w... Were tested, purposely trying to find the distance between two points on a coordinate.! This document contains a list of the Pythagorean Theorem if we make the substitutions:, and area formulas rectangles... You plot point n, the i th coordinate of the line distance formula for quadrilaterals at! Goal of exploration was to see how these formulas jeszcze niedawno pozostawał w sferze marzeń inscribed in circles and... Explore the world of quadrilaterals and their properties distance formula for quadrilaterals going through its various.... Parts of a triangle and a quadrilateral are described as follows adjoining sides have reciprocal! A compass and straightedge construction right-angled triangle has: if the answer is then! +48 570 918 604 48 507 292 523 biuro @ jpyimport.pl = 85+57 x 4 = 142 x 2! Slope and distance steps: calculate length of the line that is, the missing fourth vertice Name Hour! No 7 coordinate geometry for you points on the coordinate plane with endpoints 142. 2,..., n ) is + consider triangles, three sided shapes... Show that both pairs of opposite sides are parallel 3 to classify quadrilaterals their! Is it 's slope by dividing it into two triangles the coordinate plane and proof in the following and. Lub youngtimera, który jeszcze niedawno pozostawał w sferze marzeń right place to all... ; Teacher Tools ; Learn to Code ; more Gifs given values are five formulas you. Students weigh 165 pounds but have different heights of quadrilaterals and their properties by going through various... We use to calculate area of the Pythagorean Theorem if we make the substitutions,! Measure of any of the line segment which joins any two adjacent sides Theorem to find distance! Propositions change when on a different distance formula is used to calculate area of any quadrilateral can be into... ( i = 1, 2,..., n ) is + between... With four edges ( sides ) this problem i need yo use the and/or. | 2021-07-24T22:59:42 | {
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https://math.stackexchange.com/questions/3107640/how-to-calculate-norm-of-matrix-using-any-orthogonal-basis/3108219 | # How to calculate norm of matrix using any orthogonal basis?
Show that for $$X \in \mathrm{M}_n(\mathbb{C})$$ and any orthonormal basis $$\{u_1, \ldots , u_n\}$$ of $$\mathbb{C}^n$$, we have $$\|X\|^2=\sum_{j,k}^n|\langle u_j,Xu_k\rangle |^2.$$
My Attempt:
I thought this to prove using projection formula as
to find coordinate of $$X_{ij}=\langle u_i,Xu_j\rangle$$.
But I was thinking I am missing something. As this is $$10$$ mark question with just one line argument How it was ?
Please give me hint .I wanted to solve this problem
Any Help will be appreciated
• How do you define $\lVert X\rVert$? – José Carlos Santos Feb 10 at 16:46
• $||X||$=$\sum|X_{ij}|^2$ Sir this is Hilbert Smidt norm – MathLover Feb 10 at 16:48
Define a unitary map by $$Ue_k = u_k$$ for every $$k=1,2,\ldots,n$$, where $$\{e_1,e_2,\ldots, e_n\}$$ is the standard basis of $$\Bbb C^n$$. Then, we have \begin{align*} \sum_{j=1}^n\sum_{k=1}^n\left|\langle u_j,Xu_k\rangle\right|^2&=\sum_{j=1}^n\sum_{k=1}^n\left|\langle Ue_j,XUe_k\rangle\right|^2\\&=\sum_{j=1}^n\sum_{k=1}^n\left|\langle e_j,U^*XUe_k\rangle\right|^2\\&=\|U^*XU\|^2\\&=\text{tr}(U^*X^*UU^*XU)\\&=\text{tr}(U^*X^*XU)\\&=\text{tr}(X^*XUU^*)\\&=\text{tr}(X^*X)=\|X\|^2. \end{align*}
As a different approach, we can use Parseval's identity \begin{align*} \sum_{k=1}^n\sum_{j=1}^n\left|\langle u_j,Xu_k\rangle\right|^2 &=\sum_{k=1}^n\|Xu_k\|^2\\&=\sum_{k=1}^n\sum_{j=1}^n\left|\langle e_j,Xu_k\rangle\right|^2 \\&=\sum_{j=1}^n\sum_{k=1}^n\left|\langle X^*e_j,u_k\rangle\right|^2 \\&=\sum_{j=1}^n\sum_{k=1}^n\|X^*e_j\|^2 \\&=\sum_{j=1}^n\sum_{k=1}^n\left|\langle X^*e_j,e_k\rangle\right|^2 \\&=\sum_{j=1}^n\sum_{k=1}^n\left|\langle e_j,Xe_k\rangle\right|^2 \\&=\sum_{j=1}^n\sum_{k=1}^n\left|X_{jk}\right|^2 =\|X\|^2. \end{align*}
In this answer we look at a vector of $$n$$ coordinates as a column matrix. That is, a matrix with $$n$$ rows and $$1$$ column. First note that identity $$\left\|X\right\|^2=\sum_{j=1}^n\sum_{k=1}^n|\langle u_j,Xu_k\rangle |^2\qquad (\ast)$$ is valid when the basis $$\{u_1,\ldots,u_n\}$$ of $$\mathbb{C}^n$$ is the canonical basis $$\{e_1,\ldots,e_n\}$$ of $$\mathbb{C}^n$$. Let any orthonormal basis $$\{u_1,\ldots,u_n\}$$ of $$\mathbb{C}^n$$. Let $$U$$ be the matrix whose columns are $$u_1,\ldots, u_n$$. That is, $$U=\left[\; u_1 | \ldots |u_j|\ldots |u_n\;\right]$$ Note that $$X\cdot U= X\cdot \left[\; u_1 | \ldots |u_j|\ldots |u_n\;\right]= \left[\; X\cdot u_1 | \ldots |X\cdot u_\ell|\ldots |X\cdot u_n\;\right]$$ and $$U^\ast X\cdot U= U^\ast\left[\; X\cdot u_1 | \ldots |X\cdot u_j|\ldots |X\cdot u_n\;\right]= \big[\langle u_i, Xu_j \rangle \big]_{n\times n}$$ Then $$\left\| U^\ast X U \right\|^2 = \left\| \big[\langle u_i, Xu_j \rangle \big]_{n\times n} \right\|^2 = \sum_{i=1}^{n}\sum_{j=1}^{n} |\langle u_i, Xu_j \rangle| ^2$$ If $$U=(U_{ij})_{n\times n}$$ and $$U^\ast=(U^\ast_{ij})_{n\times n}$$ then $$UU^\ast=(U_{ij})_{n\times n}\cdot(U_{ij})_{n\times n}=(\sum_{k=1}^{n}U_{ik}U^\ast_{kj})_{n\times n}=I_{n\times n}$$. More explicitly, $$\sum_{k=1}^{n}U_{ik}U^\ast_{kj}=\begin{cases} 1 & \mbox{if } i=j\\ 0 & \mbox{if } i\neq j\end{cases}$$ Now we have \begin{align} \left\|U^\ast \cdot X \cdot U\right\|^2 =& \left\| (U^\ast_{ij})_{n\times n}\cdot (X_{k\ell})_{n\times n}\cdot (U_{pq})_{n\times n} \right\|^2 \\ =& \left\| (\sum_{\alpha=1}^nU^\ast_{i\alpha}\cdot X_{\alpha\ell})_{n\times n}\cdot (U_{pq})_{n\times n} \right\|^2 \\ =& \left\| (\sum_{\beta=1}^{n}\sum_{\alpha=1}^nU^\ast_{i\alpha}\cdot X_{\alpha\beta}\cdot U_{\beta q})_{n\times n} \right\|^2 \\ =& \sum_{i=1}^{n}\sum_{q=1}^{n} \sum_{\beta=1}^{n}\sum_{\alpha=1}^n U^\ast_{i\alpha}\cdot X_{\alpha\beta}\cdot U_{\beta q} \cdot \overline{U^\ast_{i\alpha}\cdot X_{\alpha\beta}\cdot U_{\beta q}} \\ =& \sum_{\beta=1}^{n}\sum_{\alpha=1}^n X_{\alpha\beta}\cdot \overline{X_{\alpha\beta}} \sum_{i=1}^{n}\sum_{q=1}^{n} U_{\beta q} \cdot U^\ast_{i\alpha}\overline{U_{\beta q} \cdot U^\ast_{i\alpha}} \\ =& \sum_{\beta=1}^{n}\sum_{\alpha=1}^n X_{\alpha\beta}\cdot \overline{X_{\alpha\beta}} \Big( \sum_{i=q}^{n} U_{\beta q} \cdot U^\ast_{i\alpha}\overline{U_{\beta q} \cdot U^\ast_{i\alpha}} + \sum_{1\leq q | 2019-05-20T23:37:33 | {
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https://math.stackexchange.com/questions/4051479/probability-of-sampling-linearly-independent-vectors | # Probability of sampling linearly independent vectors
Let $$q$$ be a prime so that $$\mathbb{Z}_q$$ is a field. I would like to sample $$n$$ vectors independently and uniformly from $$\mathbb{Z}^n_q$$, to get a set $$V = \{ v_i \}_{i=1}^n$$. What is the probability that $$V$$ is a basis for $$\mathbb{Z}^n_q$$? I had the following proof idea, by induction. The problem reduces to what is the probability of sampling $$n$$ linearly independent vectors, and as such, we proceed by induction on the number of sampled vectors $$k$$. For $$k = 1$$, we only require that we are not sampling the zero vector, and as such we have that $$\Pr[\{v_1\} \text{is LI} ] = 1 - q^{-n}$$. Now let us suppose that we have $$k$$ linearly independent vectors $$\{ v_i \}_{i=1}^k$$. We sample a new vector $$v_{k+1}$$. In order for it to be linearly independent, it must be that it is not a linear combination of the others, and as such there are at most $$q^k$$ possible 'forbidden' values. Namely these correspond to the possible coefficients $$\alpha_i$$ of the linear combinations $$\sum_{i=1}^k \alpha_i v_i$$. So we have that $$\Pr[\{v_i\}_{i=1}^{k+1} \text{is LI} \, | \, \{v_i\}_{i=1}^{k} \text{is LI}] \geq 1 - q^{k - n}$$. As such we should have that
$$\Pr[\{v_i\}_{i=1}^n \text{is LI}] = \Pr[\{v_i\}_{i=1}^n \text{is LI} | \{v_i\}_{i=1}^{n-1} \text{is LI}] \Pr[\{v_i\}_{i=1}^{n-1} \text{is LI}] \geq \prod_{i=1}^n (1 - q^{i - n})$$
Do you think this analysis is correct? What I am also interested in is the following generalizations, on which I, unfortunately, had less success.
• Suppose we set $$n=3k$$ and that instead of sampling the vectors independently we sample three matrices $$M, M_0, M_1 \in \mathbb{Z}^{3k\times k}_q$$ such that each matrix is of full rank. What would be an upper bound on the probability of the columns of said matrices spanning $$\mathbb{Z}_q^{3k}$$? I have found in the literature (Lemma 8) a claim that it should be at least $$1 - \frac{2k}{q}$$ but have not found any good argument
• Let us relax the very first statement, and admit that $$q$$ could be composite. Does this change the argument significantly? I am interested in this result both in the vector case and in the three matrices sample case.
Thank you, I hope the formatting is clear enough :) For context, I am investigating whether the construction in the linked paper would hold in nonprime order groups.
Your analysis is correct. There is possibly a typo: $$i$$ should be $$k$$. Moreover, I think it is equality. More precisely, $$\{v_i\}_{i=1}^{k+1} \text{is LI} \, | \, \{v_i\}_{i=1}^{k} \text{is LI}$$ if and only if $$v_{k+1}$$ is not a linear combination of $$v_1,\ldots,v_k$$. The only thing that you need to make sure is that two different linear combinations of $$\{v_i\}_{i=1}^{k}$$ give rise to two different vectors, which would be true since $$\{v_i\}_{i=1}^{k}$$ is LI. Thus $$\Pr[\{v_i\}_{i=1}^{k+1} \text{is LI} \, | \, \{v_i\}_{i=1}^{k} \text{is LI}] = \frac{q^n-q^k}{q^n}$$.
If it is not a field (that is $$q$$ is composite) then the equality may not hold as two different linear combinations can give you the same vector. To see this take $$q=6$$, then $$v_1=(1,0,0,0,0,0)', v_2=(5,2,0,0,0,0)$$, note that here $$v_1+v_2 = 4v_1+4v_2=(0,2,0,0,0,0)$$.
Matrix Case: write $$M_0|M_1|M_2$$ for the augmented $$3k\times 3k$$ matrix. Now $$M_0|M_1|M_2 \text{ is LD}$$ if one of the columns in $$M_1$$ is linear combination of columns in $$M_0$$ or one of the column in $$M_2$$ is linear combination of columns in $$M_0$$ and $$M_1$$. Now probability that a fixed column in $$M_1$$ is linear combination of columns in $$M_0$$ is at most $$\frac{q^k}{q^{3k}}\leq \frac{1}{q}$$, and probability that a fixed column in $$M_2$$ is linear combination of columns in $$M_0$$ and $$M_1$$ is at most $$\frac{q^{2k}}{q^{3k}}\leq \frac{1}{q}$$. Therefore using union bound $$\mathbb{P}(M_0|M_1|M_2 \text{ is LD}) \leq \frac{2k}{q}.$$ You can see that this bound holds even if $$q$$ is not a prime.
• In fact, I just noticed that the reasoning for the inequality will not necessarily hold in the composite case. This can happen since $v_k$ not in span of LI $v_1, ... v_{k-1}$ is not enough for linear independence. For example, take (0, 2) mod 6 which is linearly dependent by itself. Jun 20 at 18:19 | 2021-09-18T04:41:38 | {
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https://itl.nist.gov/div898/software/dataplot/refman2/auxillar/siqr.htm | Dataplot Vol 2 Vol 1
# SEMI INTERQUARTILE RANGE LOWER SEMI INTERQUARTILE RANGE UPPER SEMI INTERQUARTILE RANGE
Name:
SEMI-INTERQUARTILE RANGE (LET)
Type:
Let Subcommand
Purpose:
Compute either the lower semi-interquartile range or the upper semi-interquartile range for a variable.
Description:
The interquartile range is:
IQ = UPPER QUARTILE - LOWER QUARTILE
That is, it is the difference betweeen the 75th and 25th percentiles of a variable.
The lower semi-interquartile range is:
$$\mbox{SIQR}_L = q_2 - q_1$$
and the upper semi-interquartile range is:
$$\mbox{SIQR}_U = q_3 - q_2$$
with $$q_1$$, $$q_2$$, and $$q_3$$ denoting the lower quartile, median, and upper quartile respectively.
The semi-interquartile range is sometimes used in place of the interquartile range when there is significant skewness in the data. For example, it can be used to provide an alternate definition of the fences in a box plot.
Syntax 1:
LET <par> = LOWER SEMI INTERQUARTILE RANGE <y>
<SUBSET/EXCEPT/FOR qualification>
where <y> is the response variable;
<par> is a parameter where the computed lower semi-interquartile range is stored;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
Syntax 2:
LET <par> = UPPER SEMI INTERQUARTILE RANGE <y>
<SUBSET/EXCEPT/FOR qualification>
where <y> is the response variable;
<par> is a parameter where the computed upper semi-interquartile range is stored;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
Examples:
LET A = LOWER SEMI INTERQUARTILE RANGE Y1
LET A = UPPER SEMI INTERQUARTILE RANGE Y1
LET A = UPPER SEMI INTERQUARTILE RANGE Y1 SUBSER TAG > 2
LET A = INTERQUARTILE RANGE Y1 SUBSET TAG > 2
Note:
Dataplot statistics can be used in a number of commands. For details, enter
Default:
None
Synonyms:
None
Related Commands:
INTERQUARTILE RANGE = Compute the interquartile range of a variable. BOX PLOT = Generate a box plot.
Reference:
Walker, Dovedo, Chakraborti and Hilton (2019), "An Improved Boxplot for Univariate Data", The American Statistician, Vol. 72, No. 4, pp. 348-353.
Applications:
Robust Data Analysis
Implementation Date:
2019/08
Program:
LET NU = 1
LET Y = CHISQUARE RANDOM NUMBERS FOR I = 1 1 100
LET SIQRL = LOWER SEMI INTERQUARTILE RANGE Y
LET SIQRU = UPPER SEMI INTERQUARTILE RANGE Y
SET WRITE DECIMALS 4
PRINT SIQRL SIQRU
The following output is generated
PARAMETERS AND CONSTANTS--
SIQRL -- 0.4803
SIQRU -- 0.8989
NIST is an agency of the U.S. Commerce Department.
Date created: 08/29/2019
Last updated: 08/29/2019 | 2021-11-27T02:00:43 | {
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http://math.stackexchange.com/questions/727774/how-to-prove-for-each-positive-integer-n-the-sum-of-the-first-n-odd-positiv | How to prove for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$?
I'm new to induction so please bear with me. How can I prove using induction that, for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$?
I think $9$ can be an example since the sum of the first $9$ positive odd numbers is $1,3,5,7,9,11,13,15,17 = 81 = 9^2$, but where do I go from here.
-
Note that odd integers are of the form $2k + 1$, for $k \in \mathbb{Z}$. Consider the series:
$$\sum_{i=0}^{n} (2i+1) = 2\sum_{i=0}^{n}i + \sum_{i=0}^{n} 1 = 2 * \frac{n(n-1)}{2} + n = n^{2} - n + n = n^{2}$$
Hopefully this gives you a better idea of what is going on. This is the algebra you'll probably want to use for your inductive step.
-
Thank you very much! – Juan247 Mar 26 at 16:37
This is indeed a proof of the stated proposition, but it explains nothing about induction. – wolfen Mar 26 at 16:58
The algebra would be used on the inductive step when we add in the $(n+1)$. This was intended more as a hint to get the OP going, which it seems to have done. :-) – ml0105 Mar 26 at 17:01
Induction is done by demonstrating that if the condition is true for some $n$ then it must also be true for $n+1$. If you then show that the condition is true for $n=0$ then it must be true for all $n>0$. For this problem:
Step $1$: $n=1$
The sum of the first $1$ odd numbers is $1$. $1^2=1$. Therefore the condition holds for $n=1$.
Step $2$: induction
If the sum of the first $n$ odd numbers is $n^2$ then the sum of the first $n+1$ integers is $n^2 + (2n + 1) = (n+1)(n+1)=(n+1)^2$
So the condition is also true for $n+1$.
Step $3$: conclusion
Since the we have shown that the condition is true for $n=1$ and we have shown that if it is true for $n$ then it is also true for $n+1$ then it follows by induction that it is true for all $n\geq 1$.
-
Thank you very much!!! – Juan247 Mar 26 at 16:37
+1 for explaining how induction works, as well as how it applies to the stated problem. – wolfen Mar 26 at 16:59
My favorite Proof Without Words for this sum:
-
Brilliant! It's great for teaching purposes. – orion Mar 26 at 21:15
Let $P(n)$ be the statement that the sum of the first $n$ odd positive integers is $n^2$. $1=1^2$, so we have $P(1)$.
Suppose we have $P(k)$ for some positive integer $k$. Then the sum of the first $k$ odd positive integers is $k^2$. Now the $(k+1)$th odd positive integer is $2k+1$, so the sum of the first $k+1$ odd positive integers is $k^2+2k+1=(k+1)^2$. Hence we have $P(k+1)$.
It follows that we have $P(n)$ for all positive integers $n$.
-
what you want to prove: $\sum_{i=0}^{n-1} (2i + 1) = n^2$.
induction base ($n=1$): $\sum_{i=0}^{n-1} (2i + 1) = 1$.
induction step ($n \rightarrow n+1$):
$\sum_{i=0}^{(n+1)-1} (2i + 1) = 2n + 1 + \sum_{i=0}^{n-1} (2i + 1) = n^2 + 2n + 1 = (n+1)^2$.
-
Induction.
Conjecture:
$$\sum_{k=1}^{2n-1}k=n^2$$ where $2n-1$ is the $n$-th odd number.
First term ($n=1$):
$$\sum_{k=1}^1 k=1=n^2$$
Induction step:
$$(n+1)^2=n^2+(2n+1)=\sum_{k=1}^{2n-1} k + (2n+1)=\sum_{k=1}^{2(n+1)-1}k$$
-
If you already know what is the sum of all the first $\;n\;$ natural numbers:
$$1+3+5+\ldots+(2n-1)=1+2+3+\ldots+2n-(2+4+6+\ldots+2n)=$$
$$=\frac{2n(2n+1)}2-2(1+2+\ldots+n)=n(\color{green}{2n}+\color{red}1)-n(\color{green}n+\color{red}1)=n^2$$
-
Perhaps not the answer you are looking for but have you ever noticed that the difference of two consecutive squares is always odd? And furthermore that the difference of the next two consecutive squares is $2$ more than the previous one?
$$n^2 - (n-1)^2 = 2n-1$$ and $$(n+1)^2 - n^2 = 2n+1 = (2n-1) + 2$$
From here you can easily prove your answer but you can also use this trick for calculating squares easily such as : $$103^2 = 100^2 + 201 + 203 + 205 = 10000 + 609 = 10609$$
-
Thank you very much!!! – Juan247 Mar 27 at 2:07
Here's a direct approach (using a trick from Don Knuths "Concrete Mathematics"):
$\sum_{i=0}^{n-1} (i+1)^2 = \sum_{i=0}^n i^2$ (index shifting, and 0th summand is 0).
Now, we manipulate the left side of the above equation:
$\sum_{i=0}^{n-1} (i+1)^2 = \sum_{i=0}^{n-1} (i^2 + 2i + 1) = \sum_{i=0}^{n-1} i^2 + \sum_{i=0}^{n-1} (2i + 1)$, which can (by adding and subtracting $n^2$) be transformed to $\sum_{i=0}^n i^2 - n^2 + \sum_{i=0}^{n-1} (2i + 1)$.
That is, we now know that $\sum_{i=0}^n i^2 - n^2 + \sum_{i=0}^{n-1} (2i + 1) = \sum_{i=0}^n i^2$. Subtracting $\sum_{i=0}^n i^2$ on both sides yields $- n^2 + \sum_{i=0}^{n-1} (2i + 1) = 0$, resulting in $n^2 = \sum_{i=0}^{n-1} (2i+1)$ - stating that the first $n$ odd numbers sum up to $n^2$.
- | 2014-07-26T17:25:25 | {
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https://www.johndcook.com/blog/2018/03/30/generalized-normal-kurtosis/ | # Generalized normal distribution and kurtosis
The generalized normal distribution adds an extra parameter β to the normal (Gaussian) distribution. The probability density function for the generalized normal distribution is
Here the location parameter μ is the mean, but the scaling factor σ is not the standard deviation unless β = 2.
For small values of the shape parameter β, the distribution is more sharply pointed in the middle and decays more slowly in the tails. We say the tails are “thick” or “heavy.” When β = 1 the generalized normal distribution reduces to the Laplace distribution.
Here are examples with μ = 0 and σ = 1.
The normal distribution is a special case corresponding to β = 2. Large values of β make the distribution flatter on top and thinner (lighter) in the tails. Again μ = 0 and σ = 1 in the plots below.
One way to measure the thickness of probability distribution tails is kurtosis. The normal distribution has kurtosis equal to 3. Smaller values of kurtosis correspond to thinner tails and larger values to thicker tails.
There’s a common misunderstanding that kurtosis measures how pointy the distribution is in the middle. Often that’s the case, and in fact that’s the case for the generalized normal distribution. But it’s not true in general. It’s possible for a distribution to be flat on top and have heavy tails or pointy on top and have thin tails.
Distributions with thinner tails than the normal are called “platykurtic” and distributions with thicker tails than the normal are called “leptokurtic.” The names were based on the misunderstanding mentioned above. The platy– prefix means broad, but it’s not the tails that are broader, it’s the middle. Similarly, the lepto– prefix means “thin”, referring to being pointy in the middle. But leptokurtic distributions have thicker tails!
The kurtosis of the generalized normal distribution is given by
We can use that to visualize how the kurtosis varies as a function of the shape parameter β.
The Laplace distribution (β = 1) has kurtosis 6 and the normal distribution (β = 2) has kurtosis 3.
You can use the fact that Γ(x) ~ 1/x for small x to show that in the limit as β goes to infinity, the kurtosis approaches 9/5.
Related post: Computing skewness and kurtosis in one pass
## One thought on “Generalized normal distribution and kurtosis”
1. Hello, excellent article, but i have a question ¿why the normal kurtosis value is 3? | 2021-10-22T19:24:07 | {
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http://math.stackexchange.com/questions/790358/why-are-these-two-given-expressions-equivalent | # Why are these two given expressions equivalent?
$6\cos^2(\frac {x}{2})-7\cos(\frac {x}{2})+2=0$ is equivalent to the expression $(3\cos(\frac {x}{2})-2)(2\cos(\frac {x}{2})-1) = 0$
Could someone explain the steps involved in evaluating the first expression to equal the second expression?
I noticed the expression in this question here, and I was curious how this could be achieved.
-
it's just multiplying out the terms in the second expression – mm-aops May 11 '14 at 14:46
When you foil the two parentheticals from the second expression, their products can be seen to equal the first.
We treat $\cos(\frac {x}{2})$ much like we would the variable $x$ within any given expression.
If given the expression $2x^2-3x+1 = 0$, we are able to factor this to show that $(-2x+1)(-x+1) = 0$.
Given this particular expression: $6\cos^2(\frac {x}{2})-7\cos(\frac {x}{2})+2=0$, we are able factor this to show that $(3\cos(\frac {x}{2})-2)(2cos(\frac {x}{2})-1) = 0$.
When foiled, it is shown that:
First: $3\cos(\frac {x}{2}) \cdot (2\cos(\frac {x}{2}) = 6\cos^2(\frac {x}{2})$
Second: The product of $3\cos(\frac {x}{2}) \cdot -1 = -3\cos(\frac {x}{2})$ and the product of $2cos(\frac {x}{2}) \cdot -2 = -4cos(\frac {x}{2})$, and their sum equals $-7\cos(\frac {x}{2})$
Third: $-2 \cdot -1 = 2$
Final, we add the first, second, and third steps together, and their sums are show to be equivalent to the initial expression: $6\cos^2(\frac {x}{2})$ [first] $-7\cos(\frac {x}{2})$[second]$+2$[third]$=0$.
And the result is [Final].
-
Key Idea: Knowing the roots of a polynomial let us factor it. So if $r_1, r_2,\ldots,r_n$ are all the roots of a given polynomial, then we can factorize it as: $a(x-r_1)(x-r_2)\cdots(x-r_n)$, where $a$ is the coefficient of the term with $x^n$.
Notice: $$6\,\color{red}{\cos\left(\tfrac x2\right)}^2+7\,\color{red}{\cos\left(\tfrac x2\right)}+2=0.$$ So we can get a simple quadratic equation by letting $t=\cos\left(\tfrac x2\right)$, then our equation becomes: $$6t^2-7t+2=0.\tag{\star}$$ Using the quadratic formula we get: $$6t^2-7t+2=0\iff t=\dfrac23\quad\color{grey}{\text{or}}\quad t=\dfrac12.$$ So we can write $(\star)$ as: $$6\left(t-\dfrac23\right)\left(t-\dfrac12\right)=0\iff (3t-2)(2t-1)=0.$$ Now replace $t$ with our original substitution.
Side note: This $t$-substitution isn't necessary at all, when you noticed that you had an expression of the form $a(\text{something})^2+b(\text{something})+c=0$ you can directly use the quadratic formula to solve for that $\text{something}$.
-
Eloquently put. – Alex May 11 '14 at 14:53
Thanks @alx $\overset{\cdot\cdot}\smile$ – Hakim May 11 '14 at 14:54
We need find $pq,$ such that $pq=6\cdot2=12; p+q=7$
$$6a^2-7a+2=6a^2-(4+3)a+2=2a(3a-2)-(3a-2)=(3a-2)(2a-1)$$
- | 2015-10-14T04:30:24 | {
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https://math.stackexchange.com/questions/4456063/how-many-points-does-a-line-intersect-a-sphere-in-an-infinite-dimensional-normed | # How many points does a line intersect a sphere in an infinite-dimensional normed vector space?
Let $$(E, |\cdot|)$$ be a n.v.s. We fix $$r>0$$ and $$x,y \in B(0, r)$$ such that $$x\neq y$$. Here $$B(0, r)$$ is the open ball centered at the origin and having radius $$r$$. The set of all points in the line though $$x$$ and $$y$$ is $$\{tx+(1-t)y \mid t \in \mathbb R\}.$$
Consider the map $$f: \mathbb R \to E, t \mapsto tx+(1-t)y$$. Then
• $$f$$ and thus $$|f|$$ are continuous.
• $$\lim_{t \to \infty} |f(t)| =+\infty$$.
• $$|f(0)| = |y|.
It follows that the set $$S := \{t\in \mathbb R \mid |f(t)|=r\}$$ is non-empty and bounded. If $$(E, \langle \cdot , \cdot \rangle)$$ is an inner product space, then \begin{align} |f(t)|=r &\iff |f(t)|^2=r^2 \\ &\iff |x|^2t^2 + 2\langle x,y \rangle t(1-t) + |y|^2(1-t)^2 = r^2 \\ &\iff |x-y|^2t^2+2(\langle x, y \rangle - |y|^2) t + |y|^2-r^2=0. \end{align}
We have $$\Delta = (\langle x, y \rangle - |y|^2)^2- |x-y|^2(|y|^2-r^2)>0$$ because $$|y| and $$|x-y|\neq 0$$. So $$S$$ has exactly $$2$$ elements in this case.
Does $$\operatorname{card} (S) =2$$ if $$E$$ is reflexive or uniformly convex?
• May I ask for the reason of the downvote? May 23 at 8:58
The following holds:
Every line in a normed space $$X$$ intersects the unit sphere $$S_X$$ at most twice if and only if $$X$$ is strictly convex.
There's no need for (local) uniform convexity here, and reflexivity is a distraction (e.g. consider finite-dimensional spaces with/without strict convexity).
Recall what it means for $$X$$ to be strictly convex:
$$(X, \| \cdot \|)$$ is strictly convex if and only if, for all distinct $$x, y \in S_X$$, $$\|x + y\| < 2$$.
It's not difficult to see that, if each line intersects at most twice with $$S_X$$, then $$X$$ is strictly convex. If we choose any distinct $$x, y \in S_X$$, and form a line between them (i.e. $$\{tx + (1 - t)y : t \in \Bbb{R}\}$$), then this line contains the midpoint $$\frac{x + y}{2}$$. As $$x$$ and $$y$$ are distinct, so too is $$\frac{x + y}{2}$$ distinct from the other two points, and hence, it cannot belong to $$S_X$$. We know by convexity of the norm that $$\left\|\frac{x + y}{2}\right\| \le \frac{\|x\| + \|y\|}{2} = 1,$$ but given that $$\frac{x + y}{2} \notin S_X$$, the above inequality is strict. Thus, $$\|x + y\| < 2$$, as required.
The converse is the slightly trickier direction. Suppose $$X$$ is strictly convex, and further, we have a line $$L = \{x + td : t \in \Bbb{R}\} \subseteq X$$ that intersects $$S_X$$ at least three times. Consider the function $$f : \Bbb{R} \to \Bbb{R}$$ defined by $$f(t) = \|x + td\|$$. Then, $$f$$ is convex, as \begin{align*} f(\lambda t + (1 - \lambda) s) &= \|x + (\lambda t + (1 - \lambda) s)d\| \\ &= \|(\lambda (x + td) + (1 - \lambda)(x + sd)\| \\ &\le \lambda \|x + td\| + (1 - \lambda)\|x + sd\| \\ &= \lambda f(t) + (1 - \lambda) f(s), \end{align*} for any $$s, t \in \Bbb{R}$$ and $$\lambda \in [0, 1]$$.
Now, $$f$$ achieves the value $$1$$ three times; let's name three of these solutions as $$t_1 < t_2 < t_3$$. Using the three slope lemma, we now show that $$f(t) = 1$$ for all $$t \in [t_1, t_3]$$. To show this, suppose that $$t \in (t_1, t_2)$$. Then the three slope lemma shows: $$\frac{f(t) - f(t_1)}{t - t_1} \le \frac{f(t_2) - f(t_1)}{t_2 - t_1} \le \frac{f(t_2) - f(t)}{t_2 - t},$$ which comes to: $$\frac{f(t) - 1}{t - t_1} \le 0 \le \frac{1 - f(t)}{t_2 - t},$$ i.e. $$f(t) \le 1$$. On the other hand, \begin{align*} &\frac{f(t_2) - f(t)}{t_2 - t} \le \frac{f(t_3) - f(t)}{t_3 - t} \le \frac{f(t_3) - f(t_2)}{t_3 - t_2} \\ \implies \; &\frac{1 - f(t)}{t_2 - t} \le \frac{1 - f(t)}{t_3 - t} \le 0 \\ \implies \; & f(t) \ge 1. \end{align*} So, $$f(t) = 1$$, for $$t \in (t_1, t_2)$$. A similar argument shows $$f(t) = t$$ for $$t \in (t_2, t_3)$$. Since $$f(t_1) = f(t_2) = f(t_3)$$, this shows $$f(t) = 1$$ for $$t \in [t_1, t_3]$$.
Either which way, this produces infinitely many points on $$L$$ that lie in $$S_X$$. If we consider $$x + t_1 d$$ and $$x + t_3 d$$, then their midpoint $$x + \frac{t_1 + t_3}{2} d$$ also lies in $$S_X$$, as $$\frac{t_1 + t_3}{2} \in [t_1, t_3]$$. This contradicts strict convexity. | 2022-06-27T15:53:19 | {
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http://exxamm.com/blog/Blog/13926/zxcfghfgvbnm4?Class%2011 | Mathematics Derivation of the formula for text()^nP_r and Permutations when all the objects are not distinct objects For CBSE-NCERT
Click for Only Video
### Topic covered
♦ Derivation of the formula for text()^nP_r.
♦ Permutations when all the objects are not distinct objects
### Derivation of the formula for text()^nP_r.
=>color(red)(text()^nP_r=(n!)/((n-r)!) , \ \ \ \ \ \ 0 ≤ r ≤ n)
Let us now go back to the stage where we had determined the following formula:
color(color)(text()^nP_r= n (n – 1) (n – 2) . . . (n – r + 1))
Multiplying numerator and denomirator by (n – r) (n – r – 1) . . . 3 × 2 × 1, we get
text()^nP_r= (n (n – 1) (n – 2) . . . (n – r + 1) (n – r) (n – r – 1) . . . 3 × 2 × 1 )/ ((n – r) (n – r – 1) . . . 3 × 2 × 1, )=(n!)/((n-r)!)
Thus color(blue)(text()^nP_r=(n!)/((n-r)!)) \ \ \ \ \ \ where 0 ≤ r ≤ n
This is a much more convenient expression for text()^nP_r than the previous one.
In particular, when r = n,
\color{green} ✍️ color(red)(text()^nP_n=(n!)/(0!)=n!)
Counting permutations is merely counting the number of ways in which some or all objects at a time are rearranged. Arranging no object at all is the same as leaving behind all the objects and we know that there is only one way of doing so. Thus, we can have
text()^nP_0 =1 = (n!)/(n!) = ( n!) / ((n -0)!) .......................(1)
Therefore, the formula (1) is applicable for r = 0 also.
text()^nP_r = (n!)/((n -r ) !), 0 <= r <= n
color{blue} " Theorem 2" The number of permutations of n different objects taken r at a time, where repetition is allowed, is n^r.
Proof is very similar to that of Theorem 1 and is left for the reader to arrive at.
Here, we are solving some of the problems of the pervious Section using the formula for text()^nP_r to illustrate its usefulness.
In Example 1, the required number of words = text()^4P_4 = 4! = 24. Here repetition is not allowed. If repetition is allowed, the required number of words would be 44 = 256.
The number of 3-letter words which can be formed by the letters of the word NUMBER = text()^6 P_3 = (6!)/(3!) = 4 × 5 × 6 = 120. Here, in this case also, the repetition is not allowed. If the repetition is allowed,the required number of words would be 6^3 = 216
The number of ways in which a Chairman and a Vice-Chairman can be chosen from amongst a group of 12 persons assuming that one person can not hold more than one position, clearly text()^12 P_2 = (12!)/(10!) = 11 xx 12 =132
### Permutations when all the objects are not distinct objects
Suppose we have to find the number of ways of rearranging the letters of the word ROOT. In this case, the letters of the word are not all different. There are 2 Os, which are of the same kind. Let us treat, temporarily, the 2 Os as different, say, O_1 and O_2. The number of permutations of 4-different letters, in this case, taken all at a time is 4!. Consider one of these permutations say, RO_1O_2T. Corresponding to this permutation,we have 2 ! permutations RO_1O_2T and RO_2O_1T which will be exactly the same permutation if O_1 and O_2 are not treated as different, i.e., if O_1 and O_2 are the same O at both places.
Therefore, the required number of permutations = (4!)/(2!) = 3 xx 4 = 12
Let us now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times.
Temporarily, let us treat these letters different and name them as I_1, I_2, T_1 , T_2, T_3. The number of permutations of 9 different letters, in this case, taken all at a time is 9 !. Consider one such permutation, say, I_1 NT_1 SI_2 T_2 U E T_3. Here if I_1, I_2 are not same and T_1, T_2, T_3 are not same, then I_1, I_2 can be arranged in 2! ways and T_1, T_2, T_3 can be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation corresponding to this chosen permutation I_1NT_1SI_2T_2UET_3. Hence, total number of different permutations will be (9!)/(2 ! 3!)
We can state (without proof) the following theorems:
color(red)("Theorem 3") The number of permutations of n objects, where p objects are of the same kind and rest are all different color(red)(=(n!)/(p !).)
In fact, we have a more general theorem
color(red)("Theorem 4") The number of permutations of color(blue)(n) objects, where color(blue)(p_1) objects are of one kind, color(blue)(p_2) are of second kind, ........., color(blue)(p_k) are of k^(th) kind and the rest, if any, are of different kind is color(red)((n!)/ (p_1! p_2! ...p_k!))
Q 3078734606
Find the number of permutations of the letters of the word ALLAHABAD.
Solution:
Here, there are 9 objects (letters) of which there are 4A’s, 2 L’s and rest are all different.
Therefore, the required number of arrangements (9!)/(4! 2!) = (9xx8xx7xx6xx5)/2 = 7560
Q 3048834703
How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
Solution:
Here order matters for example 1234 and 1324 are two different numbers. Therefore, there will be as many 4 digit numbers as there are permutations of 9 different digits taken 4 at a time.
Therefore, the required 4 digit numbers text()^9P_4 = (9!)/{(9-4)!} = 9xx8xx7xx6 = 3024
Q 3008034808
How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed?
Solution:
Every number between 100 and 1000 is a 3-digit number. We, first, have to count the permutations of 6 digits taken 3 at a time. This number would be text()^6P_3. But, these permutations will include those also where 0 is at the 100’s place. For example, 092, 042, . . ., etc are such numbers which are actually 2-digit numbers and hence the number of such numbers has to be subtracted from text()^6P_3 to get the required number. To get the number of such numbers, we fix 0 at the 100’s place and rearrange the remaining 5 digits taking 2 at a time. This number is text()^5P_2. So
The required number = text()^6P_3 - text()^5P_2 = (6!)/(3!) - (5!)/(3!)
= 4xx5xx6-4xx5 = 100
Q 3058134904
Find the value of n such that
(1). text()^nP_5 = 42 text()^nP_3 , n > 4
(2). (text()^nP_4)/(text()^(n-1)P_4) = 5/3 , n > 4
Solution:
(i) Given that text()^nP_5 = 42 text()^nP_3
or n (n – 1) (n – 2) (n – 3) (n – 4) = 42 n(n – 1) (n – 2)
Since n > 4 so n (n-1) (n-2) ne 0
Therefore, by dividing both sides by n(n – 1) (n – 2), we get
(n – 3 (n – 4) = 42
or n^2 – 7n – 30 = 0
or n^2 – 10n + 3n – 30
or (n – 10) (n + 3) = 0
or n – 10 = 0 or n + 3 = 0
or n = 10 or n = – 3
As n cannot be negative, so n = 10.
(ii). Given that (text()^nP_4)/(text()^(n-1)P_4) = 5/3
Therefore 3n (n – 1) (n – 2) (n – 3) = 5(n – 1) (n – 2) (n – 3) (n – 4)
or 3n = 5 (n – 4) [as (n – 1) (n – 2) (n – 3) ≠ 0, n > 4]
or n = 10.
Q 3078145006
Find r, if 5 text()^4P_r = 6 text()^5P_(r–1) .
Solution:
We have 5 text()^4P_r = 6 text()^5P_(r−1)
or 5 xx (4!)/{(4-r)!} = 6xx(5!)/((5-r+1)!)
or (5!)/((4-r)!) = (6xx5!)/((5-r+1)(5-r)(5-r-1)!)
or (6 – r) (5 – r) = 6
or r^2 – 11r + 24 = 0
or r^2 – 8r – 3r + 24 = 0
or (r – 8) (r – 3) = 0
or r = 8 or r = 3.
Hence r = 8, 3.
Q 3048245103
Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that
(i) all vowels occur together (ii) all vowels do not occur together.
Solution:
(i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be 6P6 = 6!. Corresponding to each of these permutations, we shall have 3! permutations of the three vowels A, U, E taken all at a time . Hence, by the multiplication principle the required number of permutations = 6 ! × 3 ! = 4320.
we first have to find all possible arrangments of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number, the number of permutations in which the vowels are always together. Therefore, the required number 8 ! – 6 ! × 3 ! = 6 ! (7×8 – 6) = 2 × 6 ! (28 – 3) = 50 × 6 ! = 50 × 720 = 36000
Q 3078245106
In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable ?
Solution:
Total number of discs are 4 + 3 + 2 = 9. Out of 9 discs, 4 are of the first kind
(red), 3 are of the second kind (yellow) and 2 are of the third kind (green). Therefore, the number of arrangements (9!)/(4!3!2!) = 1260
Q 3028345201
Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements,
(ii) do all the vowels always occur together
(iii) do the vowels never occur together
(iv) do the words begin with I and end in P?
Solution:
There are 12 letters, of which N appears 3 times, E appears 4 times and D appears 2 times and the rest are all different. Therefore
The required number of arrangements = (12 !)/(3! 4! 2!) = 1663200
(i) Let us fix P at the extreme left position, we, then, count the arrangements of the remaining 11 letters. Therefore, the required of words starting with P are
(11 !)/(3! 2! 4!) = 138600
(ii) There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have to always occur together, we treat them as a single object square for the time being. This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are 3Ns and 2 Ds, can be rearranged in (8!)/(3! 2!) ways. Corresponding to each of these arrangements, the 5 vowels E, E, E, E and I can be rearranged in (5!)/(4!) ways. Therefore, by multiplication principle the required number of arrangements
= (8!)/(3! 2!) xx (5!)/(4!) = 16800
(iii) The required number of arrangements
= the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together.
= 1663200 – 16800 = 1646400
(iv) Let us fix I and P at the extreme ends (I at the left end and P at the right end).
We are left with 10 letters.
Hence, the required number of arrangements
(10!)/(3! 2! 4!) = 12600 | 2018-12-10T04:38:17 | {
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http://barnabasinstitute.org/qtno0/7e9ae4-how-to-diagonalize-a-matrix-3x3 | # how to diagonalize a matrix 3x3
Ask Question Asked 4 years, 6 months ago. I have a matrix composed of 1x1, 2x2 and 3x3 blocks and I would like to obtain the eigenvalues and eigenvectors sorted according to the block they correspond to. By Proposition 23.1, is an eigenvalue of Aprecisely when det( I A) = 0. ... $which we can eyeball one easily as$\begin{bmatrix}0\\1\\0\end{bmatrix}$. May 20, 2016, 3:47:14 PM (A)" 3x3 Matrix" Tags. Check the determinant of the matrix. Characteristic Polynomial of a 3x3 Matrix. I have a matrix composed of 1x1, 2x2 and 3x3 blocks and I would like to obtain the eigenvalues and eigenvectors sorted according to the block they correspond to. Follow 26 views (last 30 days) Rodolphe Momier on 7 Apr 2020. Yes. UUID . Is A diagonalizable? Diagonalizing a 3x3 matrix. In many areas such as electronic circuits, optics, quantum mechanics, computer graphics, probability and statistics etc, matrix is used to study. Diagonalize matrix with complex eigenvalues by real basis. parts of the complex conjugate eigenvectors. Vote. KurtHeckman. De nition 2.5. Follow 24 views (last 30 days) Rodolphe Momier on 7 Apr 2020. When I use the eig command, i obtain the eigenvalues sorted in ascending order. Why? The solution of the initial value problem will involve the matrix exponential . Due to the simplicity of diagonal matrices, one likes to know whether any matrix can be similar to a diagonal matrix. Definition An matrix is called 8‚8 E orthogonally diagonalizable if there is an orthogonal matrix and a diagonal matrix for which Y H EœYHY ÐœYHY ÑÞ" X Thus, an orthogonally diagonalizable matrix is a special kind of diagonalizable matrix…$\endgroup$– Gerry Myerson May 4 '13 at 3:54 SavannahBergen. Calculating the inverse of a 3x3 matrix by hand is a tedious job, but worth reviewing. Start by entering your matrix row number and column number in the boxes below. If the matrix were diagonalizable and we could nd matrices Pand D, then the computation of the 10th power of the matrix would be easy using Proposition 2.3. Two square matrices A and B of the same order are said to be simultaneously diagonalizable, if there is a non-singular matrix P, such that P^(-1).A.P = D and P^(-1).B.P = D', where both the matrices D and D' are diagonal matrices. Solution for A is a 3x3 matrix with two eigenvalues. Eigenvalues and matrix diagonalization. Created by . Matrix diagonalization is useful in many computations involving matrices, because multiplying diagonal matrices is quite simple compared to multiplying arbitrary square matrices. Recipes: diagonalize a matrix, quickly compute powers of a matrix by diagonalization. However, if A {\displaystyle A} is an n × n {\displaystyle n\times n} matrix, it must have n {\displaystyle n} distinct eigenvalues in order for it to be diagonalizable. The Euler angles of the eigenvectors are computed. Linear Algebra Differential Equations Matrix Trace Determinant Characteristic Polynomial 3x3 Matrix Polynomial 3x3 Edu. In fact, A PDP 1, with D a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. OB. Block-diagonalization of a matrix. In Mathematica it can be done easily, but when using the module numpy.linalg I get problems. Steps. 0 Comments. The transformation matrix is nonsingular and where . Looking at this makes it seem like a 3x3 matrix, with a 2x2 tacked on the bottom right corner, and zero's added to the filler space made as a result of increasing by 2 dimensions.$\begingroup$The same way you orthogonally diagonalize any symmetric matrix: you find the eigenvalues, you find an orthonormal basis for each eigenspace, you use the vectors in the orthogonal bases as columns in the diagonalizing matrix. A priori, the Pauli matrices and the position operator do not act on the same space, so you should be able to diagonalize both simultaneously. We put a "T" in the top right-hand corner to mean transpose: Notation. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P. EXAMPLE: Diagonalize the following matrix, if possible. Eigenvalue Calculator Online tool compute the eigenvalue of a matrix with step by step explanations.Start by entering your matrix row number and column number in the input boxes below. Question: Diagonalize The Matrix A, If Possible. Free Matrix Diagonalization calculator - diagonalize matrices step-by-step This website uses cookies to ensure you get the best experience. You can also find the inverse using an advanced graphing calculator. In fact, determinants can be used to give a formula for the inverse of a matrix. Each eigenspace is one-dimensional. 3x3 Matrix Diagonalization Simple C++ code that finds a quaternion which diagonalizes a 3x3 matrix: . When I use the eig command, i obtain the eigenvalues sorted in ascending order. Method 1 of 3: Creating the Adjugate Matrix to Find the Inverse Matrix 1. Thanks is advance. Diagonal matrices represent the eigenvalues of a matrix in a clear manner. That Is, Find An Invertible Matrix P And A Diagonal Matrix D Such That A=PDP-1 A = -11 3 -9 0-5 0 6 -3 4. Enter your matrix in the cells or type in the data area. An n n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. on . See the answer. We will come back to this example afterwards. 0 Comments. But what does it mean to diagonalize a matrix that has null determinant? 2.6 Multiple Eigenvalues The commutator of and is . Expert Answer . Then we need one more for this matrix to be diagonalizable, and fortunately this one is pretty clear too we need the first input in row 1 to sum with the third input to 0,$\begin{bmatrix}3\\0\\1\end{bmatrix}$fits the bill. Ais diagonalizable. Last modified by . The values of λ that satisfy the equation are the generalized eigenvalues. Why? 3 Determinants and Diagonalization Introduction. Note that we have de ned the exponential e t of a diagonal matrix to be the diagonal matrix of the e tvalues. You can also calculate a 3x3 determinant on the input form. Show … In particular, the powers of a diagonalizable matrix can be easily computed once the matrices P P P and D D D are known, as can the matrix exponential. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. In this way we compute the matrix exponential of any matrix that is diagonalizable. Determinant of a 3x3 matrix Last updated: Jan. 2nd, 2019 Find the determinant of a 3x3 matrix, , by using the cofactor expansion. 0. By using this website, you agree to our Cookie Policy. [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. De &nition 12.1.$\endgroup$– Adam Jan 23 '14 at 17:57$\begingroup$Yes, and then is the autovalue the product of the two different autovalues of position and spin-operator? You need to calculate the determinant of the matrix as an initial step. One of the eigenspaces would have unique eigenvectors. User can select either 2x2 matrix or 3x3 matrix for which the squared matrix to be calculated. Diagonalizing a 3x3 matrix. They also arise in calculating certain numbers (called eigenvalues) associated with the matrix. With each square matrix we can calculate a number, called the determinant of the matrix, which tells us whether or not the matrix is invertible. A is a 3x3 matrix with two eigenvalues. We can diagonalize a matrix through a similarity transformation = −, where is an invertible change-of-basis matrix and is a matrix with only diagonal elements. Note I A= 2 4 6 3 8 0 + 2 0 1 0 + 3 3 5: To nd det( I A) let’s do cofactor expansion along the second row because it has many zeros1. Quaternion Diagonalizer(const float3x3 &A) { // A must be a symmetric matrix. on . Diagonalize the matrix A, if possible. If the commutator is zero then and This problem has been solved! Diagonalization Linear Algebra MATH 2010 The Diagonalization Problem: For a nxnmatrix A, the diagonalization problem can be stated as, does there exist an invertible matrix Psuch that P 1APis a diagonal matrix? Contact Us. Thanks is advance. Answer: By Proposition 23.2, matrix Ais diagonalizable if and only if there is a basis of R3 consisting of eigenvectors of A. Division Headquarters 315 N Racine Avenue, Suite 501 Chicago, IL 60607 +1 866-331-2435 For any matrix , if there exist a vector and a value such that then and are called the eigenvalue and eigenvector of matrix , respectively. Vote. In other words, the linear transformation of vector by only has the effect of scaling (by a factor of ) the vector in the same direction (1-D space). How to convert this vector to a matrix? Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. 1fe0a0b6-1ea2-11e6-9770-bc764e2038f2. This page explains how to calculate the determinant of a 3x3 matrix. A. Is A diagonalizable? Terminology: If such a Pexists, then Ais called diagonalizable and Pis said to diagonalize A. Theorem If Ais a nxnmatrix, then the following are equivalent: 1. This square of matrix calculator is designed to calculate the squared value of both 2x2 and 3x3 matrix. Show transcribed image text. orthogonal matrix is a square matrix with orthonormal columns. Yes. SEMATH INFO. An n£n matrix A is called diagonalizable if A is similar to a diagonal matrix D: Example 12.1. Block-diagonalization of a matrix. 1. Diagonalization is the process of transforming a matrix into diagonal form. So let’s nd the eigenvalues and eigenspaces for matrix A.$\begingroup$Do you mean diagonalize the 2x2 matrix ? A method is presented for fast diagonalization of a 2x2 or 3x3 real symmetric matrix, that is determination of its eigenvalues and eigenvectors. Since the eigenvector for the third eigenvalue would also be unique, A must be diagonalizable. Each eigenspace is one-dimensional. Show … Matrix Diagonalization Calculator Online Real Matrix Diagonalization Calculator with step by step explanations. • RREF Calculator • Orthorgonal Diagnolizer • Determinant • Matrix Diagonalization • Eigenvalue • GCF Calculator • LCM Calculator • Pythagorean Triples List. The generalized eigenvalue problem is to determine the solution to the equation Av = λBv, where A and B are n-by-n matrices, v is a column vector of length n, and λ is a scalar. I need to diagonalize a symbolic matrix with python. The associated transformations have the effect of killing at least one dimension: indeed, a x matrix of rank has the effect of lowering the output dimension by . A small computer algebra program is used to compute some of the identities, and a C++ program for testing the formulas has been uploaded to arXiv. Diagonalization is a process of &nding a diagonal matrix that is similar to a given non-diagonal matrix. Previous question Next question Transcribed Image Text from this Question. Aug 7, 2020, 9:25:26 PM. Select the correct choice below and, if… For example, a x matrix of rank 2 will have an image of size 2, instead of 3. Is there a necessary and sufficient condition for a square matrix to be able to diagonalize a symmetric square matrix? 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https://math.stackexchange.com/questions/23358/quotient-ring-of-gaussian-integers/23367 | # Quotient ring of Gaussian integers
A very basic ring theory question, which I am not able to solve. How does one show that
• $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$.
• Extending the result: $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$, if $a,b$ are relatively prime.
My attempt was to define a map, $\varphi:\mathbb{Z}[i] \to \mathbb{Z}/10\mathbb{Z}$ and show that the kernel is the ideal generated by $\langle{3-i\rangle}$. But I couldn't think of such a map. Anyhow, any ideas would be helpful.
Define $$\phi: \mathbb{Z} \rightarrow \mathbb{Z}[i]/(3-i) \text{ where } \phi(z) = z + (3-i)\mathbb{Z}[i].$$ It follows simply that $$\ker \phi = (3-i)\mathbb{Z}[i] \cap \mathbb{Z}$$. So for any such $$z \in \ker \phi$$, we have $$z = (3-i)(a+bi)$$ for some $$a,b \in \mathbb{Z}$$. But $$(3-i)(a+bi) \in \mathbb{Z}$$ happens if and only if $$3b-a=0$$. So \begin{align*} \ker \phi = (3-i)\mathbb{Z}[i] \cap \mathbb{Z} &= \{(3-i)(3b+bi)\mid b \in \mathbb{Z}\}\\ &= \{(9b + b) + i(3b-3b)\mid b \in \mathbb{Z}\}\\ &= \{10b\mid b \in \mathbb{Z}\}\\ &= 10\mathbb{Z}. \end{align*}
To see $$\phi$$ is surjective, let $$(a+bi) + (3-i)\mathbb{Z}[i] \in \mathbb{Z}[i]/(3-i)$$. Then $$a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$$, so $$\phi(a+3b) = (a+bi) + (3-i)\mathbb{Z}[i]$$.
Hence $$\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$$.
• Don't you have to show that $\phi$ is also surjective? Feb 23, 2011 at 17:25
• @anonymous This is even simpler, and this works more generally. Mar 7, 2020 at 16:59
• Sir what you mean by notation $z + (3-i)\mathbb{Z}[i]$. I think $z + (3-i)\mathbb{Z}[i]=z+<3-i>$ since $Z[i]$ is commutative ring with unity and hence ideal $a\mathbb{Z}[i]=<a>$ am i correct. Apr 12, 2020 at 3:56
• @AkashPatalwanshi yes you're right. Aug 26, 2020 at 11:42
This diagram shows the Gaussian integers modulo $3-i$.
The red points shown are all considered to be $0$ but their locations in $\mathbb Z[i]$ are $0$, $3-i$, $i(3-i)$ and $3-i + i(3-i)$. Every congruence class must be inside that box once and you can see there are $10$ of them.
The arrows show adding by $1$ each time. Doing that takes you through every equivalence class and then back to the start.
So $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$.
• Why is this "Every congruence class must be inside that box once " true?
– ラミタ
Jun 20, 2021 at 1:22
• Do we have this sort of a geometric picture of the lattice for any arbitrary ring of integers $\mathcal{O}_K$? Jul 17, 2021 at 10:01
• Is this a proof? Jun 10 at 14:02
Firstly: it is not true in general that $\mathbb Z[i]/(a - ib) \cong \mathbb Z/(a^2 + b^2).$ (Consider the case of $3 - 0\cdot i$.)
The claimed isomorphism does hold if $a$ and $b$ are coprime.
Here is a sketch of how to see this:
To begin with, note that it is much easier to consider maps from $\mathbb Z$ to other rings, rather than maps in the opposite direction (as you suggested in your answer), because $\mathbb Z$ maps to any ring with unity in a canonical way, by sending $1$ to $1$.
So consider the canonical map $\mathbb Z \to \mathbb Z[i]/(a - i b).$
The target is finite of order $a^2 + b^2$, and so this map factors to give an injection $\mathbb Z/(n) \hookrightarrow \mathbb Z[i]/(a - i b)$ for some $n$ dividing $a^2 + b^2$.
Now if $a$ and $b$ are coprime then $b$ is coprime to $a^2 + b^2$, hence coprime to $n$, and so $b$ is invertible in $\mathbb Z/(n)$. Combining this observation with the equation $a - i b = 0$ (which holds in $\mathbb Z[i]/(a - i b)$) one finds (and I leave this as an exercise!) that the map $\mathbb Z/(n) \hookrightarrow \mathbb Z[i]/(a - ib)$ contains $i$ in its image, and hence is surjective as well as injective, and so we are done.
• Thanks for this generalization. It is very helpful (to this third party, at least!) Mar 21, 2011 at 16:42
• Hello Matt, something I've been trying to wrap my head around recently is why $\mathbb{Z}[i]/(a−ib)$ has order $a^2+b^2$. Is there a simple algebraic explanation or reference for this? Thank you. Oct 30, 2011 at 1:27
• @yunone: Dear yunone, It follows e.g. from the division algorithm in $\mathbb Z[i]$. Regards, Oct 30, 2011 at 2:12
• @yunone: Think about the chain of ideals $(a^2+b^2) \subset (a+bi) \subset {\mathbf Z}[i]$. For a positive integer $n$, the index $[{\mathbf Z}[i]:(n)]$ is $n^2$. Thus $[{\mathbf Z}[i]:(a^2+b^2)] = N^2$, where $N = a^2+b^2$. From group theory, $[{\mathbf Z}[i]:(a^2+b^2)] = [{\mathbf Z}[i]:(a+bi)][(a+bi):(a^2+b^2)]$. The quotient group $(a+bi)/(a^2+b^2)$ is isom. to ${\mathbf Z}[i]/(a-bi)$. Using cpx. conjugation, ${\mathbf Z}[i]/(a+bi) \cong {\mathbf Z}[i]/(a-bi)$ as additive groups. Hence $N^2 = [{\mathbf Z}[i]:(a+bi)]^2$. Now take square roots.
– KCd
Apr 7, 2012 at 21:25
• Dear @KCd, thanks for the explanation. Apr 27, 2012 at 21:20
Go back one step and add the defining equation for $i$ to the ideal. In other words, consider your ring as a quotient of the ring of polynomials $\mathbb Z[x]$:
$$\mathbb Z[i] / (3-i) = \mathbb Z [x] / (3-x,x^2+1)$$
Manipulating the ideal $(3-x,x^2+1)$ a bit, you will find that the quotient is indeed equal to $\mathbb Z/10\mathbb Z$.
This makes sense because setting $i^2=-1$ and $i=3$ implies that $9 = -1$, which is true in $\mathbb Z/10\mathbb Z$.
• There is no need to manipulate the ideal $(3-x,x^2+1)$: just factorize by $(3-x)$ first. Sep 20, 2021 at 21:42
In this post, I am quoting the solution to a related question from Artin, with the preceding explanation (Section 11.4, p. 337-338). This is really the same as Greg Graviton's answer, but I found the different point of view and the elaborate explanation very useful. (The impatient may skip to Example 11.4.5 directly.)
We reinterpret the quotient ring construction when the ideal $I$ is principal, say $I = (a)$. In this situation, we think of $\overline R = R / I$ as the ring obtained by imposing the relation $a = 0$ on $R$, or by killing the element $a$. For instance, the field $\mathbb F_7$ will be thought of as the ring obtained by killing $7$ in the ring $\mathbb Z$ of integers.
Let's examine the collapsing that takes place in the map $\pi: R \to \overline R$. Its kernel is the ideal $I$, so $a$ is in the kernel: $\pi(a) = 0$. If $b$ is any element of $R$, the elements that have the same image in $\overline R$ as $b$ are those in the coset $b + I$ and since $I = (a)$ those elements have the form $b+ra$. We see that imposing the relation $a =0$ in the ring $R$ forces us to set $b = b + ra$ for all $b$ and all $r$ in $R$, and that these are the only consequences of killing $a$.
Any number of relations $a_1 = 0, \ldots, a_n = 0$ can be introduced, by working modulo the ideal $I$ generated by $a_1, \ldots, a_n$, the set of linear combinations $r_1 a_1 + \cdots + r_n a_n$, with coefficients $r_i$ in $R$. The quotient ring $\overline R = R/I$ is viewed as the ring obtained by killing the $n$ elements. Two elements $b$ and $b'$ of $R$ have the same image in $\overline R$ if and only if $b'$ has the form $b + r_1 a_1 + \cdots +r_n a_n$ with $r_i$ in $R$.
The more relations we add, the more collapsing takes place in the map $\pi$. If we add relations carelessly, the worst that can happen is that we may end up with $I = R$ and $\overline R = 0$. All relations $a = 0$ become true when we collapse $R$ to the zero ring.
Here the Correspondence Theorem asserts something that is intuitively clear: Introducing relations one at a time or all together leads to isomorphic results. To spell this out, let $a$ and $b$ be elements of a ring $R$, and let $\overline R = R / (a)$ be the result of killing $a$ in $R$. Let $\overline b$ be the residue of $b$ in $\overline R$. The Correspondence Theorem tells us that the principal ideal $(\overline b)$ of $\overline R$ corresponds to the ideal $(a,b)$ of $R$, and that $R/(a,b)$ is isomorphic to $\overline R / (\overline b)$. Killing $a$ and $b$ in $R$ at the same time gives the same result as killing $\overline b$ in the ring $\overline R$ that is obtained by killing $a$ first.
Example 11.4.5. We ask to identify the quotient ring $\overline R = \mathbb Z[i]/(i-2)$, the ring obtained from the Gauss integers by introducing the relation $i-2=0$. Instead of analyzing this directly, we note that the kernel of the map $\mathbb Z[x] \to \mathbb Z[i]$ sending $x \mapsto i$ is the principal ideal of $\mathbb Z[x]$ generated by $f = x^2 + 1$. The First Isomorphism Theorem tells us that $\mathbb Z[x]/(f) \approx \mathbb Z[i]$. The image of $g = x-2$ is $i-2$, so $\overline R$ can also be obtained by introducing the two relations $f = 0$ and $g = 0$ into the integer polynomial ring. Let $I = (f,g)$ be the ideal of $\mathbb Z[x]$ generated by the two polynomials $f$ and $g$. Then $\overline R =\mathbb Z[x]/I$.
To form $\overline R$, we may introduce the two relations in the opposite order, first killing $g$ and then $f$. The principal ideal $(g)$ of $\mathbb Z[x]$ is the kernel of the homomorphism $\mathbb Z[x] \to \mathbb Z$ that sends $x \mapsto 2$. So when we kill $x-2$ in $\mathbb Z[x]$, we obtain a ring isomorphic to $\mathbb Z$, in which the residue of $x$ is $2$. Then the residue of $f = x^2+1$ becomes $5$. So we can also obtain $\overline R$ by killing $5$ in $\mathbb Z$, and therefore $\overline R \approx \mathbb F_5$.
• I like the conceptual image he provided, I wish if I heared about his book earlier Dec 23, 2014 at 21:06
In general, one knows that if $\alpha$ is an integer in the number field $K$, then $${\rm N}_{K/{\Bbb Q}}(\alpha)=\left|\frac{A}{A\alpha}\right|$$ Here $\rm N$ is the norm and $A$ denotes the ring of integers. In the question's situation, $A={\Bbb Z}[i]$ is the ring of Gaussian integer and since $K$ is quadratic imaginary, ${\rm N}_{K/{\Bbb Q}}(\alpha)=\alpha\bar\alpha$ where the bar denotes complex conjugation.
When $\alpha=3-i$, ${\rm N}_{K/{\Bbb Q}}(\alpha)=(3-i)(3+i)=10$, thus ${\Bbb Z}[i]/(3-i)$ is a ring with 10 elements, whose representatives are $$\left\{0,1,2,i,i+1,i+2,2i,2i+1,2i+2,-1 \right\}.$$ A minute of thought and a brief inspection of these representatives should convince that the ring is indeed isomorphic to the ring of classes modulo 10.
• Regarding your last sentence: this is true for instance because there is up to isomorphism exactly one (commutative, unital) ring with $10$ elements. The same holds with $10$ replaced by any squarefree positive integer $N$... Feb 23, 2011 at 14:33
• Yes, Pete, I know. My idea was to suggest that the isomorphism could be checked simply by inspection of the representatives without invoking general results. Also, the first part of my answer isn't really necessary to get the representatives, but it's there to put things in a more general context. Feb 23, 2011 at 19:25
• my comment was also to put things in a more general context. I didn't mean to imply that you didn't know it -- only that you chose not to mention it, but others who read the question might not know it and be interested. (A lot of comments on this site are like that...) Feb 23, 2011 at 19:56
Question: "Extending the result: $$Z[i]/(a−ib)≅Z/(a^2+b^2)Z$$, if $$a,b$$ are relatively prime.
Answer: Let $$a-ib:=1-(-3)i=1+3i$$ and let $$B:=\mathbb{Z}/(1+3^2)\cong\mathbb{Z}/(10)$$ and let $$A:=\mathbb{Z}[i]/(1+3i)$$. This approach uses the chinese remainder lemma and it illustrates the "unique factorization of ideals" into products of powers of maximal ideals in Dedekind domains: It follows $$-1 \cong 10-1 \cong 9$$ hence you get a well defined map
$$\phi: \mathbb{Z}[i] \rightarrow B$$
by definining $$\phi(a+bi):=a+3b$$. It follows $$-1=\phi(-1)=\phi(i^2)=\phi(i)^2=3^2=9=10-1 =-1.$$
hence the map is well defined. By definition $$\phi(1+3i)=1+3^2=10=0$$, hence you get a well defined surjective map $$\phi: A \rightarrow B$$.
In $$A$$ there are two ideals $$I:=(1+i),J:=(2+i)$$ and
$$(1+i)(2+i)=2+i+2i+i^2=1+3i=0.$$
The ideals $$I,J$$ are coprime, hence
$$A \cong A/IJ \cong A/I \oplus A/J \cong \mathbb{Z}/(2) \oplus \mathbb{Z}/(5)$$
and
$$\mathbb{Z}/(10) \cong \mathbb{Z}/(2) \oplus \mathbb{Z}/(5)$$
by the chinese remainder lemma. There are explicit isomorphisms
$$f:A/I\cong \mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/(2)$$
defined by sending $$i$$ to $$1$$.
There is an isomorphism
$$g:A/J \cong \mathbb{Z}[i]/(2+i) \cong \mathbb{Z}/(5)$$
defined by sending $$i$$ to $$3$$. It follows
$$g(1+i^2)=1+3^2=10=0$$
and
$$g(2+i)=2+3=5=0$$
hence $$g$$ is well defined. You may check that $$f,g$$ are isomorphisms and you should do this as an exercise in "commutative ring theory/abstract algebra".
Note: This is a general fact: In the ring of integers $$\mathcal{O}_K$$ in a number field $$K$$, any ideal $$\mathfrak{a}$$ may be written (uniquely up to the order) as a product of powers of distinct maximal ideals
$$(*) \mathfrak{a}=\mathfrak{m}_1^{p_1} \cdots \mathfrak{m}_l^{p_l}.$$
The ideals above $$I,J$$ are maximal since the quotients $$A/I,A/J$$ are (finite) fields. The equality $$(*)$$ is proved in Theorem I.3.3 in Neukirch "Algebraic number theory" (this property was in fact one of the reason for the introduction of ideals in algebraic number theory and commutative algebra). In the above case the multiplicities $$l_I,l_J=1$$ and you get the decomposition
$$(1+3i)=IJ$$
in $$\mathcal{O}_K\cong \mathbb{Z}[i]$$ where $$K:=\mathbb{Q}(i)$$.
Question: "A very basic ring theory question, which I am not able to solve. How does one show that $$\mathbb{Z}[i]/(3−i)≅\mathbb{Z}/10\mathbb{Z}$$?"
Example: We may use the "same method": There a factorization into products of ideals
$$(3-i)=(1-i)(2+i)=\mathfrak{m}\mathfrak{n}$$
and the ideals $$\mathfrak{m},\mathfrak{n}$$ are coprime maximal ideals with finite residue field. There is by the CRL isomorphisms
$$\mathbb{Z}[i]/(3-i)\cong \mathbb{Z}[i]/\mathfrak{m}\oplus \mathbb{Z}[i]/\mathfrak{n} \cong \mathbb{Z}/(2)\oplus \mathbb{Z}/(5) \cong \mathbb{Z}/(10).$$
As above you may write down explicit maps and verify they are isomorphisms.
• How did you get the idea of defining $φ(α+bi)=a+3b$? It seems to me like it comes out of the blue. @hm2020 Oct 22, 2021 at 13:11
• @ΝικολέταΣεβαστού - In this case with "small numbers" you may find such a map using the "trial and error"-method. Oct 26, 2021 at 12:05
• My teacher, in a similar exercise, suggested this: $i^2=-1$ and $φ(i^2)=φ(i)^2=-1$ Suppose that $x=φ(i)$ then the equation $x^2=-1 \mod 10$ has solutions $x=[3]$ and $x=[-3]$. One of the solutions doesn't give the right result (the kernel we wanted determined which was the right choice). @hm2020 Oct 27, 2021 at 11:47 | 2022-06-30T23:51:01 | {
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http://code.jasonbhill.com/sage/project-euler-problem-27/ | ### Problem
Euler published the remarkable quadratic formula:
$n^2+n+41$
It turns out that the formula will produce 40 primes for the consecutive values $n=0$ to $39$. However, when $n=40$, $40^2+40+41=40(40+1)+41$ is divisible by 41, and certainly when $n=41$, $41^2+41+41$ is clearly divisible by 41.
Using computers, the incredible formula $n^2-79n+1601$ was discovered, which produces 80 primes for the consecutive values $n=0$ to $79$. The produce of the coefficients, $-79$ and $1601$ is $-126479$.
$n^2+an+b$ where $|a| where$|n|$is the modulus/absolute value of$n$, e.g.,$|11|=11$and$|-4|=4$. Find the product of the coefficients$a$and$b$, for the quadratic expression that produces the maximum number of primes for consecutive values of$n$, starting with$n=0$. ### Sage Solution Some observations: • Clearly,$b$must be a prime or a negative prime, since$n=0$must result in a prime. • We can refine a lower bound on$b$as follows: If a given combination$(a,b)$results in$m$consecutive primes, then clearly$b>m$, since otherwise we would obtain a factor of$m$in the given polynomial. Also, when$a$is negative, we must have$b>-(n^2+an)$, since the prime values$n^2+an+b$must be positive. Since we know that$n^2+n+41$returns 40 primes, we know that any interesting values for$a$and$b$must then satisfy$b>-(40^2+40a)$. Given these observations, I’ll write a routine in Sage. import time start = time.time() P = prime_range(1000) a_max, b_max, c_max = 0, 0, 0 for a in range(-1000,1001): for b in P: if b < -1600-40*a or b < c_max: continue c, n = 0, 0 while is_prime(n**2 + a*n + b): c += 1 n += 1 if c > c_max: a_max, b_max, c_max = a, b, c prod = a_max * b_max elapsed = time.time() - start print "a=%s, b=%s, longest sequence = %s, prod=%s\nfound in %s seconds"\ % (a_max, b_max, c_max, prod, elapsed) When executed, we obtain the following. a=-61, b=971, longest sequence = 71, prod=-59231 found in 8.53578400612 seconds It’s entirely possible that we’re being a bit redundant with the integer values that we’re throwing at Sage’s “is_prime” function. We’ll cache a list of primes within the range that we’re considering. (I came up with the number 751000 below in the Python section.) import time start = time.time() P = prime_range(751000) L = [False] * 751000 for p in P: L[p] = True a_max, b_max, c_max = 0, 0, 0 for a in range(-1000,1001): for b in range(1,1001): if L[b] is False: continue if b < -1600-40*a or b < c_max: continue c, n = 0, 0 while L[n**2 + a*n + b] is True: c += 1 n += 1 if c > c_max: a_max, b_max, c_max = a, b, c prod = a_max * b_max elapsed = time.time() - start print "a=%s, b=%s, longest sequence = %s, prod=%s\nfound in %s seconds"\ % (a_max, b_max, c_max, prod, elapsed) This indeed executes more quickly, as is seen in the output. a=-61, b=971, longest sequence = 71, prod=-59231 found in 1.39028286934 seconds ### Python Solution Of course, Sage’s prime/factorization functionality isn’t available in Python, and so we’re going to need to create a prime sieve and store primes in some form. One of the things I need to know is how large the values$n^2+an+b\$ can be. If we were to assume that the maximum value of the length of a prime sequence is 500 (which is an overestimate as we now know), then we’d have the maximum value of the quadratic given as:
500**2 + 1000 * 500 + 1000
751000
Let’s create a prime sieve that will return a list of all primes below a given number.
def primes_xrange(stop): primes = [True] * stop primes[0], primes[1] = [False] * 2 L = [] for ind, val in enumerate(primes): if val is True: primes[ind*2::ind] = [False] * (((stop - 1)//ind) - 1) L.append(ind) return L
Let’s verify that it works.
primes_xrange(20)
[2, 3, 5, 7, 11, 13, 17, 19]
How long does this function take to form all of the primes under 751000?
time P = primes_xrange(751000)
Time: CPU 0.39 s, Wall: 0.39 s
Now, we’ll take the Sage routine from above and write it completely in Python. Here, I’m using only the sieve list portion of the primes_xrange function I just created. As above, I’m just checking to see if values in that list are True or False, corresponding to primes and composite numbers. This should take a bit longer to run than the Sage version above.
import time def primes_xrange(stop): primes = [True] * stop primes[0], primes[1] = [False] * 2 for ind, val in enumerate(primes): if val is True: primes[ind*2::ind] = [False] * (((stop - 1)//ind) - 1) return primes start = time.time() P = primes_xrange(751000) a_max, b_max, c_max = 0, 0, 0 for a in range(-1000,1001): for b in range(1,1001): if P[b] is False: continue if b < -1600-40*a or b < c_max: continue c, n = 0, 0 while P[n**2 + a*n + b] is True: c += 1 n += 1 if c > c_max: a_max, b_max, c_max = a, b, c prod = a_max * b_max elapsed = time.time() - start print "a=%s, b=%s, longest sequence = %s, prod=%s\nfound in %s seconds"\ % (a_max, b_max, c_max, prod, elapsed)
And it does run slightly slow compared to the Sage version, as expected.
a=-61, b=971, longest sequence = 71, prod=-59231 found in 1.73107814789 seconds
### Cython Solution
I’m going to take roughly the same approach as with the Python solution, but I’m going to malloc a C array instead of using the Python list. That should make things a bit more efficient. Before we do that, let’s see how much faster our routine runs if we just slap a “%cython” on the first line.
%cython import time def primes_xrange(stop): primes = [True] * stop primes[0], primes[1] = [False] * 2 for ind, val in enumerate(primes): if val is True: primes[ind*2::ind] = [False] * (((stop - 1)//ind) - 1) return primes start = time.time() P = primes_xrange(751000) a_max, b_max, c_max = 0, 0, 0 for a in range(-1000,1001): for b in range(1,1001): if P[b] is False: continue if b < -1600-40*a or b < c_max: continue c, n = 0, 0 while P[n**2 + a*n + b] is True: c += 1 n += 1 if c > c_max: a_max, b_max, c_max = a, b, c prod = a_max * b_max elapsed = time.time() - start print "a=%s, b=%s, longest sequence = %s, prod=%s\nfound in %s seconds"\ % (a_max, b_max, c_max, prod, elapsed)
We find that this does improve the timing, and our Python code (which hasn’t been modified at all) now runs more quickly than the Sage code.
a=-61, b=971, longest sequence = 71, prod=-59231 found in 1.02033305168 seconds
Rewriting to take advantage of Cython’s C datatypes, how much faster can we obtain the result?
%cython import time from libc.stdlib cimport malloc, free start = time.time() cdef long i, j cdef bint *primes = malloc(751000 * sizeof(bint)) primes[0], primes[1] = 0, 0 i = 2 while i < 751000: primes[i] = 1 i += 1 i = 2 while i < 751000: if primes[i] == 1: j = 2 * i while j < 751000: primes[j] = 0 j += i i += 1 cdef long long a, b, c, t cdef long long pol, n cdef long a_max, b_max, c_max, prod a_max, b_max, c_max = 0, 0, 0 a = -1000 while a < 1001: b = 2 while b < 1001: if primes[b] == 0: b += 1 continue t = -1600-40*a if b < t or b < c_max: b += 1 continue c, n = 0, 0 pol = n**2 + a*n + b while primes[pol] == 1: c += 1 n += 1 pol = n**2 + a*n + b if c > c_max: a_max, b_max, c_max = a, b, c b += 1 a += 1 prod = a_max * b_max free(primes) elapsed = time.time() - start print "a=%s, b=%s, longest sequence = %s, prod=%s\nfound in %s seconds"\ % (a_max, b_max, c_max, prod, elapsed)
Now, this should run much more quickly.
a=-61, b=971, longest sequence = 71, prod=-59231 found in 0.0381400585175 seconds
Thus, the Sage version runs 1.245 times as fast as the Python version. The Cython version runs 45.387 times as fast as the Python version and 36.452 times as fast as the Sage version. | 2019-03-26T17:15:59 | {
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https://math.stackexchange.com/questions/1743430/acceleration-if-i-know-distance-time-and-initial-velocity-whats-acceleratio | Acceleration: If I know distance, time, and initial velocity, what's acceleration and final velocity?
So I know the Initial Velocity ($V_i$), Time ($t$), and Distance ($d$).
I know that $$d = V_it + \frac{1}{2} at^2$$
If I rearrange this, would acceleration $a = \dfrac{2(d - V_it)}{t^2}$ ?
Then assume Final Velocity ($V_f$) will be $V_i + at$
• Yes, you are correct. – N.S.JOHN Apr 15 '16 at 7:40
Under the assumption that acceleration is constant, yes, you are correct. Notice that $a$ can be negative, in which case the object in question is decelerating. Still, the equation $Vi+at=V_f$ will hold. | 2019-08-21T00:29:31 | {
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https://math.stackexchange.com/questions/1043543/calculation-fx-from-given-expression | # calculation $f(x)$ from given expression
$$f(x)+xf(-x)=x-2$$ what is $f(x$)? I try to solve this problem, but I don't know how to remove $f(-x)$ or converting it to $f(x)$.
Replacing $x$ by $-x$, we have $$f(-x) - xf(x) = -x - 2$$
Substituting back into the original equation $$f(x) + x(xf(x) -x - 2) = x- 2$$ $$(1+x^2)f(x) = x^2 +3x - 2$$ $$f(x) = \frac{x^2 +3x - 2}{1+x^2}$$
we have $$f(x)+xf(-x)=x-2$$ setting $x=-x$ we obtain $$f(-x)-xf(x)=-x-2$$ thus we have $$f(x)+x(-x-2+xf(x))=x-2$$ and we get after eliminating $f(x)$ $$f(x)=\frac{x^2+3x-2}{1+x^2}$$
• "Setting $x=-x$", we obtain $x=0$ and $f(0)+0f(0)=0-2$. Perhaps you meant "replacing $x$ by $-x$" or "setting $y=-x$". – JiK Nov 30 '14 at 11:08
After solving such task as in the above answers, you should always test it: you put $\frac{x^2 +3x - 2}{1+x^2}$ instead of $f(x)$ into the given equation:
$$\frac{x^2 +3x - 2}{1+x^2} + x\frac{x^2 -3x - 2}{1+x^2} =$$ $$= \frac{x^2 + 3x - 2 +x^3 - 3x^2 - 2x}{1+x^2} =$$ $$= \frac{x^3 - 2x^2 + x - 2}{1+x^2} =$$ $$= \frac{(x-2) (x^2+1)}{1+x^2} = x-2$$
I forgot to do it once on a maths competition and it cost me a point.
• Why does one needs to test? You should be confident about your calculations. Else you will just loose time in exams. – Babai Nov 29 '14 at 13:55
• @Susobhan,I don't know why,but contests seem to cut points for not checking the validity of solutions of functional equations.Maybe this is because the solution we get may not be a function,I am not sure. – rah4927 Nov 29 '14 at 15:27
• Testing is necessary to show that you didn't start with a contradiction: anything logically follows from a contradiction, and the other answers have not shown that the starting point was not a contradiction. – hvd Nov 29 '14 at 20:30
• Or you could see it this way: other answers proved: if <given conditions> then f=... So an implication, not equivalence. So you need to proove the implication the other way, too: if f=... then <given conditions>. The conditions are not necessarily contradictory when it doesn't work. Such tasks are often set as "Find all f that satisfy ...". You might conclude that f belongs to set F, but do all functions from set F satisfy ...? – Heimdall Nov 30 '14 at 23:46 | 2020-04-03T11:34:24 | {
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http://mathhelpforum.com/calculus/127637-function-integrable.html | Math Help - Is this function integrable?
1. Is this function integrable?
$\int_{0}^{\infty}x^2e^{-x^2}$
If so, how can you integrate it?
Also, what about $\int_0^\infty e^{-x^2}$
2. Originally Posted by paupsers
$\int_{0}^{\infty}x^2e^{-x^2}$
If so, how can you integrate it?
For this I would try integration by parts twice.
$\int x^2e^{-x^2}~dx$
3. A function is integrable $\neq$ is possible to find it's primitve o it's exact numerical value (using elemental functions).
4. Originally Posted by pickslides
For this I would try integration by parts twice.
$\int x^2e^{-x^2}~dx$
I've tried that and I still haven't been able to solve it, I just end up going in circles. Any more advice?
5. Originally Posted by paupsers
I've tried that and I still haven't been able to solve it, I just end up going in circles.
6. Originally Posted by pickslides
$\int x^2e^{-x^2}$
$u=x^2$
$dv=e^{-x^2}dx$
$du=2xdx$
$v=xe^{-x^2}+2\int x^2e^{-x^2}dx$
I've also tried it with letting $u=e^{-x^2}$
and I end up getting stuck as well.
7. Originally Posted by paupsers
$\int x^2e^{-x^2}$
$u=x^2$
$dv=e^{-x^2}dx$
$du=2xdx$
$v=xe^{-x^2}+2\int x^2e^{-x^2}dx$
I've also tried it with letting $u=e^{-x^2}$
and I end up getting stuck as well.
Parts won't work this time, my apologies
Wolfram Mathematica Online Integrator
8. You can do this integral using doble integrals, or some know values of the gamma function.
9. this can be done.
Choose
$u=x$
$dv=xe^{-x^2}dx$
Notice that then v= $\frac{-e^{-x^2}}{2}$
You will then notice that the uv term of the integration by parts is 0 (remember that for definite integrals, this term gets evaluated at the main integral's bounds, i.e., $\infty$ and 0.) What you're left with is
$\frac{1}{2}\int_0^\infty e^{-x^2}$
A rather cool method for that one will use polar coordinates. here's a link:
Gaussian integral - Wikipedia, the free encyclopedia
(note the limits of integration for the Gaussian integral: You will need to divide by 2 (justified as the integrand is an even function)).
10. Originally Posted by vince
this can be done.
Choose
$u=x$
$dv=xe^{-x^2}dx$
Notice that then v= $\frac{-e^{-x^2}}{2}$
You will then notice that the uv term of the integration by parts is 0 (remember that for definite integrals, this term gets evaluated at the main integral's bounds, i.e., $\infty$ and 0.) What you're left with is
$\frac{1}{2}\int_0^\infty e^{-x^2}$
A rather cool method for that one will use polar coordinates. here's a link:
Gaussian integral - Wikipedia, the free encyclopedia
(note the limits of integration for the Gaussian integral: You will need to divide by 2 (justified as the integrand is an even function)).
Hmmm, I don't understand why the uv term is 0.
If u = x and v = -(1/2)e^(-x^2), why does multiplying them make it 0?
11. Oops, I just saw that you edited your post to explain that part. Thanks!!
12. Originally Posted by paupsers
Oops, I just saw that you edited your post to explain that part. Thanks!!
you may be compelled to click 'thanks'...im a thanks hog :]
13. Use the fact that $\int_{0}^{\infty} e^{-x^{2}} \ dx = \frac{\sqrt{\pi}}{2}$
Make the substitution $x^{2}= au^{2}$
which implies that $2xdx=2audu$
so $dx=\frac{au}{x} dx = \frac{ax}{\sqrt{a}x}dx = \sqrt{a} dx$
then $\int^{\infty}_{0} e^{-x^{2}} \ dx = \sqrt{a} \int^{\infty}_{0} e^{-au^{2}} du = \frac{\sqrt{\pi}}{2}$
and $\int^{\infty}_{0} e^{-au^{2}} \ du =$ $\frac{1}{2} \sqrt{\frac{\pi}{a}}$
notice that $- \int^{\infty}_{0} \frac{\partial}{\partial a} \ e^{-au^{2}} \ du =$ $\int^{\infty}_{0} u^{2} e^{-au^{2}} \ du$
now swtich the order of integration and differentiation (which in this case is allowed)
$- \frac{\partial}{\partial a} \int^{\infty}_{0} e^{-au^{2}} \ du = - \frac{\partial}{\partial a} \ \frac{1}{2} \sqrt{\frac{\pi}{a}} =$ $\ \frac{\sqrt{\pi}}{4}a^{-3/2}$
so $\int^{\infty}_{0} u^{2} e^{-au^{2}} \ du = \frac{\sqrt{\pi}}{4}a^{-3/2}$
letting a=1
$\int^{\infty}_{0} u^{2} e^{-u^{2}} \ du = \frac{\sqrt{\pi}}{4}$
14. $I=\int_0^{\infty}x^2e^{-x^2}dx$
Set $x^2=t$ then $I=\dfrac{1}{2}\int_0^{\infty}\sqrt{t}e^{-t}=\dfrac{1}{2}\Gamma(\dfrac{1}{2})=\dfrac{\sqrt{\ pi}}{4}$ | 2015-10-07T04:23:39 | {
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https://www.physicsforums.com/threads/convert-to-formula.419721/ | # Convert to formula
1. Aug 1, 2010
### smslca
1. The problem statement, all variables and given/known data
please convert the below(P) into formula using (m), or give an algorithm using (m) to get (P)
m=2, P=0
m=3, P=3
m=4, P=3+5
m=5, P=3+5+5
m=6, P=3+5+5+7
m=7, P=3+5+5+7+9
m=8, P=3+5+5+7+9+9
m=9, P=3+5+5+7+9+9+11
m=10, P=3+5+5+7+9+9+11+13
m=11, P=3+5+5+7+9+9+11+13+13
m=12, P=3+5+5+7+9+9+11+13+13+15
m=13, P=3+5+5+7+9+9+11+13+13+15+17
m=14, P=3+5+5+7+9+9+11+13+13+15+17+17
and so on
2. Relevant equations
3. The attempt at a solution
I did not understand how to evaluate it into a formula
Last edited: Aug 2, 2010
2. Aug 2, 2010
### Mentallic
Using induction you can show that
$$3+5+7+...+(2n+1)=n(n+2)$$
and
$$5+9+13+...+(4n+1)=n(2n+3)$$
If you need to prove these yourself then you can manipulate the first sum by turning it into the well known
$$1+3+5+...=n^2$$
and if you need to prove this too... well... use geometry I suppose.
and for the second,
$$=4+8+12+...+4n-n$$
$$=4(1+2+3+...+n)-n$$
Now I couldn't quite find how to merge both formulae together to have a single function P=f(m) for all values of m>1. I'll keep working on it.
3. Aug 2, 2010
### smslca
yeah thanks, I got some view on the series, But I too cannot understand the mixed series. So please keep trying. I also will try on it.
4. Aug 2, 2010
### Mentallic
Ok I figured out a nice pattern and after much blood, sweat and tears, I finally solved it. There will be 3 cases to consider.
Are you familiar with modulo arithmetic?
For m(mod 3)=0, that is, m=3,6,9,12...
We have the pattern:
m=3 , P = (3) + (0)
m=6 , P = (3+5+7) + (5)
m=9 , P = (3+5+7+9+11) + (5+9)
m=12, P = (3+5+7+9+11+13+15) + (5+9+13)
etc.
Now since in this case we don't need to consider what is happening in the other cases and we are dealing with m that are multiples of 3, let's take $$\frac{m}{3}=x$$. So for m=12, x=4 and you can consider that as being the 4th sequence in this lone pattern.
for x=1, we have 1 in the first bracket, and 0 in the second
for x=2, we have 3 in the first bracket, and 1 in the second
for x=3, we have 5 in the first bracket, and 2 in the second
for x=2, we have 7 in the first bracket, and 3 in the second
Can you see the pattern arising?
for x=x, we have (2x-1) in the first bracket, and (x-1) in the second
Now looking at the formulae I gave you earlier,
$$3+5+7+...+(2n+1)=n(n+2)$$
and
$$5+9+13+...+(4n+1)=n(2n+3)$$
Let the first bracket with the first formula be n=2x-1. What this means is we have to substitute n=2x-1 into the sum formula n(n-2).
Secondly, substitute n=x-1 in the second formula.
So we obtain $$P=(2x-1)(2x+1)+(x-1)(2x+1)=(3x-2)(2x+1)$$
That's the formula for m(mod 3)=0 where m/3=x.
So let's take m=12, thus x=4.
m=12, P=3+5+5+7+9+9+11+13+13+15=90
P=(3.4-2)(2.4+1)=10.9=90
It works for all others too
Now maybe you can try apply this idea to the other two cases? Let m(mod 3)=1, i.e. m=4,7,10... and $$\frac{m-1}{3}=x$$.
For m(mod 3)=2 let $$\frac{m+1}{3}=x$$
This will finally satisfy all your summations.
5. Aug 2, 2010
### smslca
thanks for the replies
6. Aug 3, 2010
### Mentallic
No problem. Are you able to find the answer to the other two cases?
7. Aug 3, 2010
### smslca
yeah I got it from other forum.as
Let n = [m/3] ; [] mean integer part
and k = m mod 3 ; mod mean modulus (remainder of the integer divide m/3)
That is:
m --> n , k
3 --> 1, 0
4 --> 1, 1
5 --> 1, 2
6 --> 2, 0
7 --> 2, 1
8 --> 2, 2
9 --> 3, 0
...............
Then P(n,k) = n*(6*n-1) + k*(4*n+1) - 2 | 2018-02-23T15:25:19 | {
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https://math.stackexchange.com/questions/1927234/given-that-the-system-works-compute-the-probability-that-component-is-working | # Given that the system works, compute the probability that component is working.
Suppose the diagram of an electrical system is as given in Figure 2.10. What is the probability that the system works? Assume the components fail independently
I got the answer for the probability of all the system works and it is $0.8037$ but there is another question.
Given that the system works, compute the probability that component B is working.
I am not sure about this one. But is the calculation like this? $$P(A \cup B \cup C \cup D)/0.8037$$ or
only $P(A \cup B \cup D)/0.8037$ since we are only talking about B is working given that the system works.
The system is: $\rm A\cap(B\cup C)\cap D$ with: $\mathsf P(A)=0.95\\\mathsf P(B)=0.7\\\mathsf P(C)=0.8\\\mathsf P(D)=0.9$
When given that the system works, we know components $\rm A$ and $\rm D$ must do so, and that at least one of $\rm B$ or $\rm C$ does too.
So: $\mathsf P(B\mid A\cap(B\cup C)\cap D)=\mathsf P(B\mid B\cup C)$
The question is then: find $\mathsf P(B\mid B\cup C)$.
Because component failures are independent: \begin{align}\mathsf P(B\mid A\cap (B\cup C)\cap D) ~=~& \dfrac{\mathsf P(B\cap A\cap (B\cup C)\cap D)}{\mathsf P(A\cap(B\cup C)\cap D)} \\=~& \dfrac{\mathsf P(A\cap B\cap D)}{\mathsf P(A\cap(B\cup C)\cap D)} \\=~& \dfrac{\mathsf P(A)~\mathsf P( B)~\mathsf P( D)}{\mathsf P(A)~\mathsf P(B\cup C)~\mathsf P(D)} \\=~& \dfrac{\mathsf P(B)}{\mathsf P(B\cup C)} &\bbox[ghostwhite]{\color{ghostwhite}{=~\dfrac{\mathsf P(B)}{\mathsf P(B)+\mathsf P(C)-\mathsf P(B)~\mathsf P(C)}}}\\=~&\dfrac{\mathsf P(B\cap(B\cup C))}{\mathsf P(B\cup C)} \\[2ex]\therefore \mathsf P(B\mid A\cap (B\cup C)\cap D) ~=~& \mathsf P(B\mid B\cup C)\end{align}
• It may be useful to know how to simplify $B \cap (B\cup C)$. Sep 15 '16 at 0:04
• is it. P(B∩(A∩(B∪C)∩D)) or P(B∩(B∪C))? are they the same thing? sorry im just really confuse in this kind of things Sep 15 '16 at 0:06
• Thank you very much! This is my first Statistics class so im kinda confused. lmaoo but i really thank you! Sep 15 '16 at 0:31
If components are in serial (e.g A & B), all must work in order for the system to work.
If components are in parallel (e.g B & C), the system works if any of the components work.
Segment 1: P(A) = 0.95
Segment 2: 1 - P(B') x P(C') = 1-0.3 x 0.2 = 0.94
Segment 3: P(D) = 0.9
Probability entire system works = Segment 1 x Segment 2 x Segment 3
Probability entire system works = 0.95 x 0.94 x 0.9 = 0.8037 | 2021-09-16T11:45:26 | {
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https://math.stackexchange.com/questions/982902/number-of-k-10-always-increases | # Number of $K_{10}$ always increases
Let $G=(V,E)$ be a graph with $n\geq 10$ vertices. Suppose that when we add any edge to $G$, the number of complete graphs $K_{10}$ in $G$ increases. Show that $|E|\geq 8n-36$.
[Source: The probabilistic method, Alon and Spencer 3rd ed., p.12, problem 5]
For the base case $n=10$, we know $G$ must have $44=8\cdot 10-36$ edges.
We know every vertex in $G$ must have degree $\geq 8$; otherwise adding an edge connecting that vertex cannot increase the number of $K_{10}$. This gives $|E|\geq 4n$.
• Could you add a page number for the source? – Barry Cipra Oct 20 '14 at 18:04
• @BarryCipra Done. – simmons Oct 20 '14 at 18:57
• $|E|=8n-36$ is attained by the graph $G$ obtained by splitting one vertex of $K_9$ into $n-8$ independent vertices. I don't know why this would be the minimum. – bof Oct 20 '14 at 19:22
The solution to this problem uses a clever application of Theorem 1.3.3 (in "The Probabilistic Method, 3rd edition"). Since not everyone has access to the text, I will state the theorem (with necessary definitions) here first.
Definition: Let $\mathcal{F}=\{(A_{i}, B_{i})\}_{i=1}^{h}$ be a family of pairs of subsets of an arbitrary base set. $\mathcal{F}$ is a $(k,\ell)$-system if $|A_{i}|=k$ and $|B_{i}|=\ell$ for $1\leq i\leq h$, $A_{i}\cap B_{i}=\emptyset$ for all $1\leq i\leq h$, and $A_{i}\cap B_{j}\neq \emptyset$ for all $i\neq j$ with $1\leq i, j\leq h$.
Theorem (1.3.3) If $\mathcal{F}=\{(A_{i}, B_{i})\}_{i=1}^{h}$ is a $(k,\ell)$-system, then $h\leq \binom{k+\ell}{k}$.
Now, the strategy to solve the original problem is to create a cleverly chosen $(k,\ell)$-system and apply the result.
Let $h$ be the number of edges not in $G$ and list the non-edges of $G$ as $\{e_1, e_2, ..., e_h\}$. For each $e_{i}$ associate a set of 10 vertices that form a $K_{10}$ if $e_{i}$ were to be added to $G$. The hypothesis guarantees that there is at least one such set of $10$ vertices; if there is more than one, pick one arbitrarily. We will call this "potential" $K_{10}$, $K_{10}^{i}$ to denote that it is formed by adding edge $e_{i}$. Each edge $e_{i}$ has two endpoints, call them $v_{i, 1}$ and $v_{i,2}$. Form the set $A_{i}=\{v_{i, 1}, v_{i, 2}\}$. Form the set $B_{i}=V(G)-V(K_{10}^{i})$, i.e. all the vertices in $G$ that are not in $K_{10}^{i}$, the chosen $K_{10}$ formed by adding edge $e_{i}$. We want to verify that $\mathcal{F}=\{(A_{i}, B_{i})\}_{i=1}^{h}$ is a $(2, n-10)$-system.
Clearly, $|A_{i}|=2$ and $|B_{i}|=n-10$ for $1\leq i\leq h$. Also clear is that $A_{i}\cap B_{i}=\emptyset$ for $1\leq i\leq h$ since the vertices in $A_{i}$ are contained in $K_{10}^{i}$. For any $i\neq j$ note that at least one vertex of $e_{i}$ is not in $K_{10}^{j}$ otherwise, both $e_{i}$ and $e_{j}$ would need to be added to make $K_{10}^{j}$ a complete graph. Thus, at least one vertex in $A_{i}\in B_{j}$ implying that $A_{i}\cap B_{j}\neq \emptyset$. Thus, we do have a $(2, n-10)$-system.
By the theorem 1.3.3, $h\leq \binom{2+(n-10)}{2}=\binom{n-8}{2}$. Since $h$ counts the number of non-edges of $G$, we can conclude that $G$ has at least $\binom{n}{2}-\binom{n-8}{2}$ edges.
And, (the easy part) \begin{align*} \binom{n}{2}-\binom{n-8}{2} &= \frac{1}{2}\left(n(n-1) - (n-8)(n-9)\right) \\ &= \frac{1}{2}\left(n^2-n -(n^2 - 17n + 72)\right) \\ &= 8n-36. \end{align*}
You already have most of the needed information for a proof. Using induction and very little additional work, you'll have your proof.
We'll say graph $G$ has property $\mathcal{P}$ if adding any extra edge will increase its number of $K_{10}$s. Say, you have such a $G$ with $n > 10$ vertices. Now, what can you say (with respect to property $\mathcal{P}$) about the graph with $n-1$ vertices obtained by deleting an arbitrary vertex $v$ from $G$? You can use this answer, minimum degree information of $v$ and the principle of induction to finish the proof.
• I fail to see how this works. If you take $n=3$ instead of $n=10$ and you take $G=C_5$, then $G$ has property $P$, but for any vertex $v$, $G-v$ does no longer have property $P$. – Leen Droogendijk Oct 21 '14 at 12:52
• @Leen Droogendijk: $C_5$, and any graph with $n < 10$ cannot have property $\mathcal{P}$, which is a property about having $K_{10}$. To have a $K_{10}$ in $G$ you need $n \geq 10$, which is why the base case for this induction is $n = 10$. Perhaps your confusion is from thinking $\mathcal{P}$ is the condition $|E| \geq 8n - 36$? – theconic Oct 21 '14 at 22:57
• No, "take $n=3$" is indeed confusing, but I meant, use 3 instead of 10. So P becomes "adding any extra edge will increase the number of $P_3$s". You cannot apply induction, at least I do not see how, and I do not see how the argument for $K_{10}$s is essentially different. – Leen Droogendijk Oct 22 '14 at 6:39 | 2020-01-17T19:06:03 | {
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http://training.noi.ph/topics/math_sets | Sets
Starting from this section, we will describe a few common mathematical terms that we’ll need. You’ll probably be familiar with most of these, but please read on anyway. We begin with sets.
Important: When reading these, you don’t need to memorize the terms. We won’t give you an exam about them. Instead, we want you to really understand them. If you encounter a term again in a later chapter but forgot what it means, you can simply refer back to this page. No penalties for that!
Anyway, what is a set? You can think of a set as just a collection of objects, such as physical objects, numbers, and even other sets.
To write a set, we simply list the elements together, separate them with commas, and enclose them with curly brackets. For example, the set $\{1, 2.35, 11\}$ is a set containing exactly three elements (or members): the numbers $1$, $2.35$ and $11$. Also, the set $\{0, \{0, 1, \{2, 3\}\}\}$ contains exactly two elements: the number $0$ and the set $\{0, 1, \{2, 3\}\}$. Note that:
• The order you list the elements doesn’t matter. For example, $\{1, 2.35, 11\}$ and $\{11, 1, 2.35\}$ are the same sets.
• It doesn’t matter how many times you write an element. For example, $\{1, 2\}$ and $\{1, 1, 2, 1, 1, 1\}$ are the same sets. (If we want collections where repetitions matter, we use multisets.)
There’s another way of writing sets, and that is by specifying the properties that the members of the set must have. For example, an alternative way of writing $\{1, 2, 3, 4, 5\}$ is $\{x : \text{$x$is an integer and$1 \le x \le 5$}\}$. You can read this notation as “The set of all $x$ such that $x$ is an integer and $1 \le x \le 5$”, which is obviously the same as $\{1, 2, 3, 4, 5\}$. There are other similar notations, and I bet you can figure out how they work:
• $\{x^2 : x \in \mathbb{Z}\}$ denotes the set of all perfect squares. ($\mathbb{Z}$ is the standard notation for the set of integers.)
• $\{x \in \mathbb{Z} : x < 0\}$ denotes the set of all negative integers.
Comparisons of sets
Two sets are equal if they contain the same elements. For example, $\{1, 2.35, 11\} = \{11, 1, 2.35\}$, but $\{1, 2.35, 11\} \not= \{1, 2\}$.
Let $S$ and $T$ be two sets. (For set variables, we usually use capital letters.) We say $S$ is a subset of $T$ if all members of $S$ are also in $T$. We usually write this as $S \subseteq T$. For example, $\{1\} \subseteq \{1, 2\}$. You can also write $T \supseteq S$, read “$T$ is a superset of $S$”, but that’s rarer.
You might observe that this definition says that $\{1, 2\}$ is also a subset of $\{1, 2\}$. In other words, any set is a subset of itself. Although true, we usually want to refer to the more interesting subsets; those which don’t contain all elements of the set. We say that $S$ is a proper subset of $T$, denoted $S \subset T$, if $S$ is a subset of $T$ and there is an element of $T$ not in $S$. Otherwise, we say a subset is improper. Every set has exactly one improper subset: itself. Also, notice the similarity of this notation with the notation for inequalities, i.e. compare $\{\subset, \subseteq, \supset, \supseteq\}$ with $\{<, \le, >, \ge\}$.
The empty set is the set containing nothing. We denote this usually as $\{\}$ or $\emptyset$. An interesting property of the empty set is that it’s a subset of any set!
The size of a set is the number of elements it contains. We usually write the size of $S$ as $\left|S\right|$. For example, $\left|\{10, 30\}\right| = 2$ and $\left|\emptyset\right| = 0$. But some sets are infinite, like the set of natural numbers $\mathbb{N} = \{1, 2, 3, 4, \ldots\}$. For these sets, we won’t be dealing much about their sizes (although there are in fact ways to make sense of sizes of infinite sets).
Sequences and Tuples
By definition, a set is an unordered collection of things. But sometimes we want to talk about ordered collections. We call such a collection a sequence, and we use parentheses instead of curly brackets, for example $(1, 2, 2)$.
In sequences, order matters, and repetition matters. For example, $(1, 2) \not= (2, 1)$ and $(1, 2, 2) \not= (1, 2)$. Like sets, sequences can be infinite, such as the sequence of natural numbers $(1, 2, 3, \ldots)$.
Finite sequences are usually called tuples. A sequence with $k$ elements is called a $k$-tuple. For example, $(3, 4, 5)$ is a $3$-tuple, or a triple. A $2$-tuple is also called a pair.
Finally, sets can contain sequences as elements, and sequence can also contain sets (and other sequences) as elements. These are all just different kinds of objects after all.
Operations on sets
Just like you can add or subtract two numbers, you can also perform operations on sets. We describe here some useful ones.
The union of two sets $S$ and $T$, denoted $S\cup T$, is the set containing all elements from $S$ and $T$. Elements included in both sets are counted exactly once. For example, $\{1, 2\} \cup \{2, 3\} = \{1, 2, 3\}$. We can also define the union using math notation:
$$S\cup T = \{x : x\in S \vee x\in T \}$$
The intersection of two sets $S$ and $T$, denoted $S\cap T$, is the set containing elements found in both $S$ and $T$. For example, $\{1, 2\} \cap \{2, 3\} = \{2\}$ and $\{1, 2\} \cap \{3, 4\} = \emptyset$. We can also define the intersection using math notation:
$$S\cap T = \{x : x\in S \wedge x\in T \}$$
We can also talk about the complement of a set $S$, which is roughly the set of elements not contained in $S$, but this only makes sense if we have a universal set, which is just the set of things relevant in the current discussion. For example, if our universal set is the set of integers $\mathbb{Z} = \{\ldots, -2, -1, 0, 1, 2, \ldots \}$, then the complement of the set of even integers is the set of odd integers. But if our universal set is the set of real numbers $\mathbb{R}$, then the complement of the set of even integers is the set of odd integers together with the set of all noninteger real numbers! So it’s important to know the universal set, but this is usually clear in the discussion.
We can also talk about the cross product of sets. The cross product of $S$ and $T$, denoted $S\times T$, is the set of pairs of elements, where the first element is from $S$ and the second is from $T$. For example, $\{1, 2\}\times \{\text{'s'}, \text{'t'}\} = \{(1,\text{'s'}),(1,\text{'t'}),(2,\text{'s'}),(2,\text{'t'})\}$. Using math notation:
$$S\times T = \{(x,y) : x\in S \wedge y\in T\}$$
Note that $S\times T$ is not the same as $T\times S$!
The power set of a set is the set of all its subsets. For example, the power set of $\{1, 2\}$ is $\{\{\}, \{1\}, \{2\}, \{1, 2\}\}$. Note that the power set of $\emptyset$ is $\{\emptyset\}$, the set containing the empty set, which is different from $\emptyset$. Power sets also exist for infinite sets, though the power sets are also infinite.
Why sets
The idea of sets sounds simple enough, so what’s the point of sets? Well it turns out that sets are used nowadays to define everything in mathematics, including numbers, functions, algebraic objects, etc. We won’t be dealing with that here, though. | 2019-04-26T08:14:41 | {
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http://ravenflagpress.herokuapp.com/discussion/id/798/ | ##### Practice Midterm SOLUTION: #2
abbyto1607
Question two first asks for us to put the matrix A, $\begin{array}{ccc} 1&2&3&3 \\ 2&2&4&7 \\ 4&1&0&1 \end{array}$ into row echelon form. In RREF, A is $\begin{array}{ccc} 1&0&0&-7/6 \\ 0&1&0&17/3\\ 0&0&1&-1/2 \end{array}$. Now, we can use this to determine the dimension of the image of the linear transformation given by multiplication by A. We know that $rank(A)=dim(im(A))$. Since A in RREF has 3 leading ones, A is of rank 3, and $dim(im(A))=3$ To confirm this, we see that the last column in the RREF of matrix A corresponds to a redundant vector (since that column doesn't have a leading one). Thus, $im(A)$ is spanned by $\begin{array}{ccc} 1 \\ 2 \\ 4 \end{array}$ $\begin{array}{ccc} 1 \\ 2 \\ 1 \end{array}$ $\begin{array}{ccc} 3 \\ 4\\ 0\end{array}$. Im(A) is spanned by three vectors and therefore dim(im(A))=3. From the rank-nullity theorem, we know that $dim(ker(A))+dim(im(A))=dim(A)$ From previous information and this theorem, $dim(ker(A))=4-3=1$ To find ker(A), we want to find all vectors that map to zero when multiplied by A. This is done by setting A to 0, row reducing, and solving for the variables. $\begin{array}{ccc} 1&2&3&3&|0 \\ 2&2&4&7&|0 \\ 4&1&0&1&|0 \end{array}$ Row reducing, we get that $\begin{array}{ccc} 1&0&0&-7/6 &|0\\ 0&1&0&17/3&|0\\ 0&0&1&-1/2 &|0\end{array}$. the vectors that map to zero are given by $\begin{array}{ccc} -7/6 t\\ 17/3t\\ -1/2 t\end{array}$. where $t$ is arbitrary. Thus, ker(A) is spanned by $\begin{array}{ccc} -7/6\\ 17/3\\ -1/2\end{array}$
swagata: May 5, 2015, 6:01 p.m.
I think the signs for the kernel should be the opposite of what is shown. For example, the top row of the matrix in RREF indicates that x - (7/6)w = 0. If w = t, then x = (7/6)t. So the elements in the span of the kernel should be 7/6, -17/3, and 1/2 (I think?).
jlumxajaha: May 5, 2015, 8:34 p.m.
Isn't the top row of A supposed to be 1 1 3 3? Then wouldn't the final answer be ker(A) spanned by (I don't know how to make a vector with this website) -1/3, -25/6, 1/2, 1? In response to swagata, I don't think it matters if it's negative or positive because the opposite vector to any vector is exactly parallel to that vector so they're linearly dependent (or something).
jlumxajaha: May 5, 2015, 8:40 p.m.
Oops I made an addition problem. I meant (1 -44/6 1/2 1) as the span of the kernel. Can someone check this?
jlumxajaha: May 5, 2015, 8:46 p.m.
ROFL Okay I'm terrible at addition. But one of my other points still stands: shouldn't there be a 1 after -7/6 17/3 and -1/2?
angela: May 5, 2015, 9:16 p.m.
I agree with @jlumxajaha. The kernel should have 4 rows. By definition, the kernel is all vectors that take the transformation to 0, so if you wrote it out, it would look like $\begin{bmatrix} 1 & 1 & 3 & 3 \\ 2 & 2 & 4 & 7 \\ 4 & 1 & 0 & 1 \end{bmatrix}$ $\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$ = $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ with the kernel being whatever $\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$ is
kcason: May 5, 2015, 10:15 p.m.
Agreed with @jlumxajaha/@angela (with the reasoning for the 4 rows) and @swagata (with the reasoning for switching the signs), except if it asks for all vectors that map to the 0 vector, shouldn't you include the arbitrary variable? As in the answer would be: 7/6t -17/3t 1/2t t where t is an element of all real numbers.
abbyto1607: May 5, 2015, 10:31 p.m.
Yep, it should have four rows, with t as the last. Sorry for the mistake, I did the same thing with another problem. It was an early error I fixed, but forgot to when typing out the solution.
leopham: May 5, 2015, 10:40 p.m.
I agree with swagata, the span of the kernel should be x_1_= 7/6, x_2_= -17/3, x_3_=1/2, and x_4_ = t. The vectors that map A to 0 would be kernel as Ker(t) represents the dimension that is lost during the linear transformation. | 2017-06-27T20:45:18 | {
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http://math.stackexchange.com/questions/332390/how-do-i-compute-the-limit-of-this-sequence-x-n1-fracx-n2b2-x-o | # How do I compute the limit of this sequence $x_{n+1}=\frac{(x_n)^2+b}{2}$, $x_o=0$ and $b\in [0,1]$
I need to study the limit behavior of the sequence and discuss all possible situations with the parameters given. The sequence is $x_{n+1}=\frac{(x_n)^2+b}{2}$, $x_o=0$ and $b\in [0,1]$. This can also be written as $\frac{(x_n)^2}{2}+\frac{b}{2}$. I studied the first few terms to get an idea of the behavior of the sequence, and have a feeling that when $b=0, x_{n+1}=0$ and thus $\lim\ x_n=0$ and when $0<b\leq1, \lim\ x_n=b/2$ because as n gets larger $\frac{(x_n)^2}{2}$ tends to $0$ and $\lim\ (0+\frac{b}{2})=(b/2)$.
Those were my thoughts on the problem, I'm not sure if they are right, but I got stuck trying to prove them.
### Proof of when $b=0, x_{n+1}=0$ for all n, which will imply that $\lim\ x_n=0$:
This can be proved by induction (but I'm not sure how to do it properly):$$x_0=0=0$$$$x_{0+1}=x_1=\frac{0^2+0}{2}=0$$ $x_{n+1}=>x_{n+2}$ $$x_{n+2}=x_{(n+1)+1}=\frac{(x_{n+1})^2+b}{2}=\frac{(x_{n+1})^2}{2}=\frac{0}{2}=0$$ I really don't know if that is how you do the induction. Any help?
### Proof of $0<b\leq1$
When $b=1$, $x_{n+1}=\frac{(x_n)^2+1}{2}=\frac{(x_n)^2}{2}+\frac{1}{2}$. I can't really explain this besides saying that $\frac{(x_n)^2}{2}$ seems to be tending to $0$. Is this correct, and can I have a hint at how to solve it? I'm thinking I may be able to split it into 2 limits, saying $\lim \frac{(x_n)^2}{2}+\lim\ (b/2)=0+(b/2)$.
Thanks for any help!
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There are many questions on this site discussing such recurrence relations. One of the main ways to find the limit, is to first assume that the limit exists, i.e., $x_n \to L$, and then take the limit of $x_{n+1} = \frac{x_n^2 + b}{2}$ as $n \to \infty$, i.e., $L = \frac{L^2 + b}{2}$. Solving for $L$ then gives you the limit, if it exists. To show that the sequence actually converges and that this is indeed the limit, you could try to show it is monotonically increasing/decreasing and bounded. – TMM Mar 17 '13 at 0:33
As for mathematical induction, your first step (showing that, for $b=0$, $x_1=0$) is good. Next you want to show that whenever $x_n=0$, then $x_{n+1}=0.$ To do this, assume that, for all $n$, $x_n=0$, then show that this implies that $x_{n+1}=0$. You have done this as well. After you have shown this to be the case you have that $x_1=0$ and that if $x_n=0$, then $x_{n+1}=0$. This means that $x_2=0$, and so on. So you are finished. – jim Mar 17 '13 at 0:46
You have to show when $b\le 1$, $x_n\le 1$, this ensures you $x_n$ is bounded, then you show that $x_n$ is increasing, if $x_0 = 0$. Then you can see there is a limit $L = (L^2+b)/2$ – Yimin Mar 17 '13 at 0:49
@jim I read your comment to tried to understand, but I don't know if you are telling me that something is missing/wrong in my induction, or if I have all the parts I need in my induction. May you clarify a little? – user66807 Mar 17 '13 at 1:05
Your induction is nearly complete. You only need to say something like: Since $x_0=0$, and since $x_{n+1}=0$ whenever $x_n=0$, then, by the principle of mathematical induction, $x_n=0, \forall n.$ The main thing that you were missing in your induction was understanding of what you were essentially doing. I was trying to tell you why your induction works. If you understand this, then you will know how to do it, and know when it is done correctly (in every case). – jim Mar 17 '13 at 1:49
1. $\{x_n\}$ is monotone increasing:
$x_2\geq x_1.$ Suppose $x_{k+1}\geq x_k$ whenever $1\leq k\leq m~(m\in\mathbb N)$. Consequently, $x_n\geq 0~\forall~ n\leq m.$
$x_{k+2}-x_{k+1}=\dfrac{(x_{k+1})^2-1}{2}-\dfrac{(x_{k})^2-1}{2}=\dfrac{(x_{k+1})^2-(x_k)^2}{2}\geq 0$ whence $\{x_n\}$ is monotone increasing.
2. $\{x_n\}$ is bounded above:
$x_1=\dfrac{b}{2}\leq b.$ Let $x_k\leq b$ for some $m\in\mathbb N.$
Then $x_{k+1}=\dfrac{(x_k)^2+b}{2}\leq b$ (Since $b\in [0,1]$). Thus by induction $b$ is an upper bound of $\{x_n\}.$
By $1$ and $2,$ $\{x_n\}$ converges to some limit $l$ (say). Taking limit as $n\to\infty$ on both sides of $x_{n+1}=\dfrac{(x_n)^2+b}{2}$ we get $l=\dfrac{l^2+b}{2}$ i.e. $l^2-2l+b=0.$ Conseqently $l=1-\sqrt{1-b}$ (Recall $x_n\leq b\implies \lim x_n\leq b\leq 1$).
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Your "bounded above" induction was really helpful, thanks a lot. I do have a question about your proof that $x_n$ is monotone increasing. Why were you able to say $\dfrac{(x_{k+1})^2-(x_k)^2}{2}\geq 0$? Was is because you stated in the beginning that $x_2\geq x_1$? – user66807 Mar 17 '13 at 4:52
Glad to be able to help. Note $x_{k+1}\geq x_k$ and both are non-negetive since $0\leq \dfrac{b}{2}= x_1\leq x_2\leq x_3...\leq x_{k+1}$ (by induction hypothesis). That's why $(x_{k+1})^2-({x_k})^2\geq 0.$ – Sugata Adhya Mar 17 '13 at 5:06
Ah, yes I see it now. Thanks again! – user66807 Mar 17 '13 at 5:16
You're welcome ! – Sugata Adhya Mar 17 '13 at 5:17
We deal with positive $b$. First we show by induction that the sequence $\{x_n\}$ is increasing, that is, that $^x_n\lt x_{n+1}$ for all $n$. This is easy to verify for $n=1$.
Suppose it is true for $n=k$. We show it is true for $n=k+1$. So we need to show that $x_{k+1}\lt x^{k+2}$.
From the induction hypothesis $x_k\lt x_{k+1}$ we conclude that $$\frac{x_k^2+b}{2}\lt \frac{x_{k+1}^2+b}{2}.\tag{1}$$ Since $x_{k+1}=\frac{x_k^2+b}{2}$ and $x_{k+2}=\frac{x_{k+1}^2+b}{2}$, the Inequality $(1)$ says precisely that $x_{k+1}\lt x_{k+2}$.
An easy induction shows that the sequence $\{x_n\}$ is bounded above by $1$, and now we can conclude convergence.
The value is the easy part. We have $\lim_{n\to\infty} x_{n}=\lim_{n\to \infty} x_{n+1}$. So if $L$ is their common limit, we have $$L=\frac{L^2+b}{2}.$$ Solve this quadratic for $L$. The root less than $1$ gives $L=1-\sqrt{1-b}$.
- | 2014-12-20T13:00:59 | {
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https://math.stackexchange.com/questions/2500928/probability-of-four-letter-word-from-set-a-b-c-d-e-with-no-two-consecu | Probability of Four Letter Word from set $\{A, B, C, D, E\}$ with no two consecutive equal letters.
My approach to solve this problem is like :
No. of Four letter word from the set is : $5^4$
No. of Four letter word with two consecutive equal letters : $5^4-[5*3*5^2]$
$5*3*5^2$ = (for each of the five letters $*$ there are three positions $*$ with $5^2$ words)
Required Probability = $\frac{5^4-(5*3*5^2)}{5^4}$
Is this the correct answer ? If not please explain why and whether there any better approach to solve this ?
Problem with your approach is that you are double counting.
eg- cases like "AABB" is being subtracted twice in your approach.
(First in A A_ _ and then again in _ _ B B)
Here is a simple way to do this
There are 4 places _ _ _ _
Now choose a letter for first place = 5 possibilities
Now choose a letter for second place = 4 possibilities(as you can't have a letter which was in first place)
Now choose a letter for third place = 4 possibilities(as you can't have a letter which was in second place)
Now choose a letter for fourth place = 4 possibilities(as you can't have a letter which was in third place)
So total combinations = 5*4*4*4
Probability = $\frac{5*4^3}{5^4}$ = $(\frac{4}{5})^3$
• Can you please explain, what my approach is calculating ? or what's wrong with that? Nov 2, 2017 at 6:43
• @user292174 updated. Check now Nov 2, 2017 at 7:06
first letter can be any of the 5 letters. Second letter can be any of the 4 letters that arent the first letter. Third letter can be any of 4 letters that arent the second letter. fourth letter can be any of the 4 letters that arent the third letter.
$\frac{5\cdot 4\cdot 4\cdot 4}{5^4}$
• Can you please explain, what my approach is calculating ? or what's wrong with that? Nov 2, 2017 at 6:49
First letter can be any of given 5 letters. And second letter can be any of four .Third space can have 4 letters except the letter in second position .similarly the fourth letter Then the result will be (5×4×4×4)/5^4 | 2022-08-15T19:03:01 | {
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https://math.stackexchange.com/questions/3467319/using-fermats-little-theorem-to-show-divisibility | # Using Fermat's Little Theorem to Show Divisibility
I was asked to prove, using Fermat's Little Theorem, that $$11|5^{10n+8}-4$$ for $$n\ge0$$. I proved it but I was wondering whether there's an easier way (still using Fermat's). Here is my proof:
\begin{alignat}{3} 11|5^{10n+8}-4&\iff5^{10n+8}-4&&\equiv0 &&&\mod11\\ \quad&\iff 25^{5n+4}-4&&\equiv0 &&&\mod 11\\ \quad&\iff \qquad3^{5n+4}&&\equiv 4 &&&\mod 11\\ \quad&\iff \qquad3^{5n+5}&&\equiv 12 &&&\mod 11\\ \quad&\iff \qquad3^{5(n+1)}&&\equiv 1 &&&\mod 11.\\ \end{alignat} For $$n\ge1$$, let S(n) be the statement
$$S(n) :3^{5(n+1)}\equiv 1 \mod 11.$$ We will prove by induction on $$n$$ that $$S(n)$$ holds.
Base case ($$n=1$$). By Fermat's Little Theorem, $$S(1)$$ is true.
Inductive Step. Fix some $$k\ge1$$ and suppose $$S(k)$$ is true. To be shown is that the statement $$S(k+1):3^{5(k+2)}\equiv 1 \mod 11$$ follows. Beginning with the LHS of $$S(k+1)$$,
\begin{alignat}2 \quad&3^{5(k+2)}&&=3^{5(k+1)+5}\tag{1}\\ \quad&\ \implies &&=3^{5}3^{5(k+1)}\tag{2}\\ \quad& \overset{\text{IH}}{\implies} &&\equiv3^{5}(1)\mod 11\tag{3}\\ \quad&\ \implies &&\equiv1\mod 11\tag{4},\\ \end{alignat} arriving to the RHS of $$S(k+1)$$, concluding the inductive step. It is proved, then, by MI that $$S(n)$$ holds for all $$n\ge1.$$ Since $$S(0)$$ holds by $$(4)$$, then $$S(n)$$ is true for all $$n\ge0$$.
• This is far too complicated. Note that $5^8\equiv4\mod{11}$, and that by Fermat's Little Theorem, $5^{10}\equiv1\mod{11}$. The rest follows. – Don Thousand Dec 7 '19 at 20:22
• @DonThousand Thank you! – Alex D Dec 7 '19 at 20:27
We have $$5^{10n+8} = 5^{10n} 5^8 = (5^{10})^n 5^8 \equiv 1 \cdot 5^8 \equiv 4 \bmod 11$$
• From a comment by Don Thousand. – lhf Dec 7 '19 at 20:50
• In particular, repeated squaring gives $5^2=3,\,5^4=-2,\,5^8=4$. – J.G. Dec 7 '19 at 20:54
• multiply by 25 getting $$5^{10(n+1)}-100$$
• Take remainders using Fermat, getting $$1-1\equiv 0\pmod{11}$$
• +1 That's one way I'd show it too. Worth emphasis is that $\bmod 11\!\!:\,\ 25 x\equiv 0 \iff x\equiv 0\$ by $\ 25\,$ is invertible (so cancellable). – Bill Dubuque Dec 10 '19 at 5:54
• It's not mod for that first part simply $p|a\implies p|ab$ ... – Roddy MacPhee Dec 10 '19 at 11:10
• Of course we can rewrite congruences in divisibility language, but generally that's not a good thing to do since it often obfuscates innate arithmetical structure. – Bill Dubuque Dec 10 '19 at 15:38
• I didn't really use much ... – Roddy MacPhee Dec 10 '19 at 15:56
• To finish requires your way you could justify the inference $\ 11\mid 25x \,\Rightarrow\, 11\mid x,\,$ which is the divisibility form of what I wrote above. We can't possibly know what you intended to "use much" since you did not finish the argument, i.e. you omitted that step. – Bill Dubuque Dec 10 '19 at 16:18
Much easier way!
By FLT $$5^{10} \equiv 1 \pmod{11}$$ so $$5^{10n+8}\equiv 5^8$$ and $$5^{10n +8} -4 \equiv 5^8 -4\pmod {11}$$.
So you just have to show that one case the $$5^8 \equiv 4 \pmod {11}$$. Then every case will be $$5^{10n + 8} - 4\equiv 0 \pmod{11}$$
Admittedly that requirse calculations but there are 3 ways, each more clever than the other
1) $$5^2 = 25\equiv 3 \pmod {11}$$. $$5^4\equiv 3^2 \equiv 9\equiv -2 \pmod {11}$$. $$5^8\equiv (-2)^2 \equiv 4 \pmod {11}$$.
2) $$5^8*5^2 \equiv 5^{10} \equiv 1\pmod {11}$$
$$5^8*5^2 \equiv 5^8*3 \equiv 1\pmod{11}$$ so as $$11$$ is prime $$3^{-1}$$ exist as is.... $$1 \equiv 12=3*4\pmod{11}$$ so $$5^8*3*4 \equiv 4\pmod {11}$$ and $$5^8\equiv 4\pmod {11}$$.
3) I'll admit I didn't come up with this.
If $$5^8 -4 \equiv A\pmod{11}$$ then
$$(5^8-4)*25 \equiv A*25\pmod{11}$$
$$5^{10} - 100 \equiv 3A$$
$$1 - 1 \equiv 3A$$
$$3A \equiv 0\pmod {11}$$ and as $$11$$ is primes $$A\equiv 0 \pmod{11}$$.
• persistence of 0 Mod, modulo multiples for the win. – Roddy MacPhee Dec 8 '19 at 11:22
$$10n+8=10(n+1)-2$$ we know that $$5^{10}\equiv 1 \mod 11$$ by Fermat.
we just need to prove that $$5^{-2}-2^2=$$ $$(5^{-1}+2)(5^{-1}-2)\equiv 0 \mod 11$$
which is true since the inverse $$5^{-1}$$ is $$9$$ .
$$10\equiv-1\bmod11,$$ so $$10^8 \equiv(-1)^8=1\bmod11,$$
so $$5^{10n+8}\equiv5^8\equiv5^82^{10}=10^82^2\equiv4\bmod 11,$$
since $$5^{10}$$ and $$2^{10}\equiv1\bmod11$$ by Fermat's little theorem. | 2020-01-20T06:28:15 | {
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https://math.stackexchange.com/questions/3330100/find-a-differential-equation-whose-solutions-are-the-circles-x2y2-2ay | # Find a differential equation whose solutions are the circles $x^2+y^2=2ay$
I had to find differential equation of the family of circles which touch the $$x$$-axis at origin. Centering the variable circle at $$(0,a)$$, its equation will be $$x^2+y^2=2ay .$$ Solving conventionally by differentiating once and substituting the value of a in the equation of circle, the differential equation is $$(x^2-y^2)y_1=2xy$$ Now I might be missing some concept because I've just started with ODE, but what is wrong with the following process? Differentiating the circle equation $$x+yy_1=ay_1 ,$$ dividing both sides by $$y_1$$, and differentiating both sides gives the equation $$y_1-xy_2+{y_1}^3=0 .$$ Does this not represent the family of circles? In that case the differential equations are of different orders.
• If I understand your notation, you are writing $y_1$ for $y'=\frac{dy}{dx}$ and $y_2$ for $y'' = \frac{d^2y}{dx^2}$? Aug 21 '19 at 16:25
• Yes you are right Aug 21 '19 at 16:27
An ODE of order $$n$$ has a solution with $$n$$ independent parameters. So your second equation represents a double family of circles, not just the given one (you might differentiate once again, and describe a triple family…).
For a quick solution, write
$$\frac{x^2+y^2}{2x}=a$$ and differentiate, giving
$$(2x+2yy')2x-(x^2+y^2)2=0$$ (denominator omitted).
Differentiating an equation in general discards information.
A first-order differential equation $$\phantom{(\ast)} \qquad F(x, y, y_1) = 0 \qquad (\ast)$$ (with nice $$F$$) has a $$1$$-parameter family of solutions. And in the example in the question, this is exactly what we want to produce---a first-order differential equation whose solutions are precisely the given family of curves. This is exactly what we achieve when isolating $$a$$ as $$\frac{x^2}{2 y} + \frac{y}{2} = a$$ and differentiating w.r.t. $$x$$ to produce (after clearing denominators) a first-order o.d.e. $$(x^2 - y^2) y_1 = 2 x y$$ that does not depend on $$a$$, that is, for which every equation $$x^2 + y^2 = 2 a y$$ is a solution.
Regardless of whether $$F$$ in $$(\ast)$$ depends on a parameter like $$a$$, however, differentiating $$(\ast)$$ gives a second-order o.d.e., $$\frac{d}{dx}F(x, y, y_1) = 0,$$ that is, $$\phantom{(\ast\ast)} \qquad \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} y_1 + \frac{\partial F}{\partial y_1} y_2 =0 \qquad (\ast\ast)$$ which (again under nice conditions) has a $$2$$-parameter family of solutions. So, many---in a sense that can be made precise, most---solutions of $$(\ast\ast)$$ are not solutions of $$(\ast)$$. So, in general producing a second-order equation---if you like, differentiating twice---means we have produced a differential equation with too many solutions to satisfy our original prescription.
As a toy example if we want to find a differential equation whose solutions are (all) the horizontal lines $$y = c ,$$ the parameter $$c$$ is already isolated, and differentiating gives the desired equation $$(\ast)$$: $$y_1 = 0 .$$ But differentiating again gives $$y_2 = 0$$ (our $$(\ast\ast)$$), whose solutions are precisely the affine functions, $$y(x) = a x + b .$$ In particular, the solutions of $$(\ast\ast)$$ with $$a \neq 0$$ are not solutions of the original differential equation $$(\ast)$$.
It's not so much that the two equations are of different orders. We could differentiate both sides of the first equation and get a second-order equation. The difference is that you have treated $$a$$ differently in the two processes.
In the first, $$a$$ is a parameter which depends on $$x$$ and $$y$$ (Each point on the plane, other than the origin, is on exactly one circle tangent to the $$x$$-axis at the origin). In the second, you're treating $$a$$ like a constant.
If you solve the second equation (before you differentiate a second time) by separating variables, you get \begin{align*} \frac{x}{y'} + y &= a \\ \implies x &= (a-y) y' \\ \implies x \,dx &= (a-y)\,dy \\ \implies x^2 &= -(a-y)^2 + C \\ \implies x^2 + (a-y)^2 &= C \end{align*} So for each $$a$$, the family of solutions is the set of circles centered at $$(0,a)$$. That's quite different from the family of circles tangent to the $$x$$-axis at $$(0,0)$$.
• So the order two differential equation I got at the end is invalid ? Aug 21 '19 at 17:14
• Like Yves says, if you differentiate twice, the solution set is a two-parameter family, also not what you want. Aug 21 '19 at 17:19
• Sorry, this answer is simply wrong. Not every circle centered at $(0,a)$ for some value of $a$ touches the $x$ axis at the origin. Aug 21 '19 at 18:23
• That's my point. OP asked what was wrong with his proposed solution, and I wanted to point out that it doesn't solve the problem he was asked. I'll edit the answer to underscore. Aug 22 '19 at 14:18
$$x^2+y^2 = 2ay$$ Take the derivative wrt. $$x$$ $$2x + 2yy' = 2ay' \implies y' = \frac{x}{a-y}$$ This will have one constant too many; eliminate that by eliminating $$a$$ using $$x^2+y^2 = 2ay \implies a = \frac{x^2+y^2}{2y}$$ So $$y' = \frac{x}{a-y} = \frac{x}{\frac{x^2+y^2}{2y}-y} = \frac{2xy}{(x^2+y^2)-2y^2}$$
$$y' =\frac{2xy}{x^2-y^2}$$
As a check, note that when $$y = \pm x$$ the slope $$y'$$ is infinite, as woould be expected since a $$45^\circ$$ line from the origin intersects the circle at a place where the tangent is vertical. | 2022-01-18T23:34:58 | {
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https://math.stackexchange.com/questions/659036/shelving-identical-books | # Shelving identical books
Suppose that we have $5$ red, $5$ black and $3$ white books that are indistinguishable to place onto a) $1$ shelf, b) $3$ shelves, so that no adjacent books have the same colour. How many different ways are there to place the books?
Does my solution for the a) case include all possible ways?
We line up all red books: _R_R_R_R_R_ and now have $6$ free spots. We choose $5$ of them for black books. There are $2$ cases. There are either $2$ black books both at the leftmost and the rightmost spot, or only at $1$ of them.
First case looks like this: BR_R_R_R_RB. We choose $3$ more spots for blacks and put one of the white books in the remaining spot. Then we have something like this: BRWRBRBRBRB. For the remaining $2$ white books we have $10$ spots to choose from: _B_RWR_B_R_B_R_B_R_B_
$$\binom43 \cdot \binom{10}2$$
As for the second case, we have $2$ ways to put the black books. Either BRBRBRBRBR or RBRBRBRBRB. We have then $11$ spots to choose from.
$$2 \cdot \binom{11}3$$
The answer I found thus is $180+330=510$
Is there a better or a generic method that I can use for such problems, i.e permuting repeated identical items such that no adjacent items are same?
And nevertheless, I still need help for the b) case if my solution were right.
• Your solution for a is wrong, because you will miss out cases which have 2 B between 2 R like: RBWBRBRBRBWRW – Calvin Lin Feb 3 '14 at 23:02
• You are right. @CalvinLin – Zafer Cesur Feb 3 '14 at 23:07
For (a), let $R(i,j,k)$, $B(i,j,k)$ and $W(i,j,k)$ denote the number of permutations of $i$ red books, $j$ black books and $k$ white books with the restrictions (a) the first book is not red, black and white (respectively), and (b) no two books of the same color are adjacent.
Define $$f(i,j,k)=\begin{cases} 1 & \text{if } i=j=k=0 \\ \tfrac{1}{2}\big( R(i,j,k)+B(i,j,k)+W(i,j,k) \big) & \text{otherwise,} \end{cases}$$ so the number we seek is $f(5,5,3).$
We find the relationships: \begin{align*} R(i,j,k) &= \begin{cases} B(i,j-1,k)+W(i,j,k-1) & \text{if } j \geq 1 \text{ and } k \geq 1 \\ W(i,0,k-1) & \text{if } j=0 \text{ and } k \geq 1 \\ B(i,j-1,0) & \text{if } j \geq 1 \text{ and } k=0 \\ 0 & \text{if } j=0 \text{ and } k=0 \text{ and } i \geq 1 \\ 1 & \text{if } j=0 \text{ and } k=0 \text{ and } i=0 \\ \end{cases} \\ B(i,j,k) &= \begin{cases} R(i-1,j,k)+W(i,j,k-1) & \text{if } i \geq 1 \text{ and } k \geq 1 \\ R(i-1,j,0) & \text{if } i \geq 1 \text{ and } k=0 \\ W(0,j,k-1) & \text{if } i=0 \text{ and } k \geq 1 \\ 0 & \text{if } i=0 \text{ and } k=0 \text{ and } j \geq 1 \\ 1 & \text{if } i=0 \text{ and } k=0 \text{ and } j=0 \\ \end{cases} \\ W(i,j,k) &= \begin{cases} R(i-1,j,k)+B(i,j-1,k) & \text{if } i \geq 1 \text{ and } j \geq 1 \\ R(i-1,0,k) & \text{if } i \geq 1 \text{ and } j =0 \\ B(0,j-1,k) & \text{if } i=0 \text{ and } j \geq 1 \\ 0 & \text{if } i=0 \text{ and } j=0 \text{ and } k \geq 1 \\ 1 & \text{if } i=0 \text{ and } j=0 \text{ and } k=0. \\ \end{cases} \end{align*} This allows us to compute the answer to (a) as $f(5,5,3)=1026$.
We can verify this computationally, e.g., in GAP via:
S:=PermutationsList([1,1,1,1,1,2,2,2,2,2,3,3,3]);;
T:=Filtered(S,A->ForAll([1..Size(A)-1],i->A[i]<>A[i+1]));;
when Size(T) returns 1026 (and, if we like, we can view all $1026$ permutations).
For (b), I'll assume the shelves are distinguishable. We can use the above formula, but e.g. split $i$ into $i_1,i_2,i_3$ as to how many red books go on shelves $1,2,3$, and so on. We can simplify by noting that $i_3=5-i_1-i_2$. Thus the number is
$$\sum_{i_1=0}^5 \sum_{i_2=0}^{5-i_1} \sum_{j_1=0}^5 \sum_{j_2=0}^{5-j_1} \sum_{k_1=0}^3 \sum_{k_2=0}^{3-k_1} f(i_1,j_1,k_1) \times f(i_2,j_2,k_2) \times f(5-i_1-i_2,5-j_1-j_2,3-k_1-k_2)$$ If we run the numbers, we find $192216$ arrangements.
(I also checked this computationally, generating the $192216$ arrangements in GAP; my code is embarrassingly brute force, so I won't post it here.)
• Thanks a lot! Even though these functions look grotesque, I'll try my best to understand it :D One more question, if we split an adjacent pair to different shelves, it becomes no more adjacent. I guess you overlooked that case, maybe? It doesn't matter since it has no nice answer, though. – Zafer Cesur Feb 6 '14 at 16:46
• Nice work. BTW, your recurrence relations would be simplified significantly, if you just declared that the functions are zero when any arguments are negative. Then I think you only need the $(0, 0, 0)$ case and the general case (and the negative case), for each of your three functions. – ShreevatsaR Feb 6 '14 at 17:49
There is no nice non-messy way to do this, except to just compute it.
Let's define the number $N_s(r, b, w)$ to be the number of ways that $r$ red, $b$ black and $w$ white books can be arranged on a shelf, starting with a book of colour $s \in \{\text{red}, \text{black}, \text{white}\}$, and such that no books of the same colour are adjacent.
We can write down recurrence relations for the $N_s(r, b, w)$ (which turn out to be slightly simpler than those in Rebecca's answer) and calculate our answer as $$N_{\text{red}}(5, 5, 3) + N_{\text{black}}(5, 5, 3) + N_{\text{white}}(5, 5, 3).$$
To say precisely what these recurrence relations are, we might as well express them in code, as in the Haskell program below (where for convenience of notation I made $s$ the first argument to the function n, and also assigned the labels $0, 1, 2$ for red, black and white respectively):
n 0 1 0 0 = 1
n 1 0 1 0 = 1
n 2 0 0 1 = 1
n 0 r b w | r > 0 = n 1 (r-1) b w + n 2 (r-1) b w
n 1 r b w | b > 0 = n 0 r (b-1) w + n 2 r (b-1) w
n 2 r b w | w > 0 = n 0 r b (w-1) + n 1 r b (w-1)
n _ _ _ _ = 0
main = print (n 0 5 5 3 + n 1 5 5 3 + n 2 5 5 3)
Running this program gives the answer $1026$.
$runhaskell mse-659036.hs 1026 For the three-shelf case, we have to sum these three answers$N_s$over all triples$(r, b, w), (r', b', w'), (r'', b'', w'')$that add up to$(5, 5, 3)$. More generally (though it's no help to you for this small case), words over a certain alphabet such that no letter appears twice consecutively (no two adjacent letters are equal) are called Smirnov words. It is possible to write down a (complicated) generating function for them, as follows. Consider any arbitrary word over an alphabet of three letters. We can identify each "block" of equal consecutive letters and "collapse" them to a single occurrence of that letter, to get a Smirnov word. Conversely, the set of all words over the alphabet can be got by taking any Smirnov word, and replacing each letter in it by a (nonempty) sequence over that letter. Thus, if$W(x, y, z)$is the generating function for all words — note that is simply$W(x, y, z) = \dfrac{1}{1-(x+y+z)}$— and the generating function for Smirnov words is$S(x, y, z)$, then the "expanding" bijection above gives: $$W(x, y, z) = S\left(\frac{x}{1-x}, \frac{y}{1-y}, \frac{z}{1-z}\right)$$ or, equivalently, as$u = \frac{x}{1-x}$means$x = \frac{u}{1+u}, \begin{align} S(x, y, z) &= W\left(\frac{x}{1+x}, \frac{y}{1+y} ,\frac{z}{1+z}\right) \\ &= \frac{1}{1-\left(\frac{x}{1+x} + \frac{y}{1+y} + \frac{z}{1+z}\right)}. \end{align} So for your two problems, we can abstractly give an answer as 1. The coefficient ofx^5y^5z^3$in$S(x, y, z)$, and 2. The coefficient of$x^5y^5z^3$in$S(x, y, z)^3\$
respectively. A computer algebra system may be able to evaluate these coefficients. Of course, the generating function is more useful for asymptotic analysis.
• Thanks. You don't need to elaborate as I'm probably going to not understand it anyway. :-D – Zafer Cesur Feb 6 '14 at 16:48
• @ZaferCesur: Sure. But I wrote it anyway; might help some other reader. :-) Good luck, – ShreevatsaR Feb 6 '14 at 17:37 | 2019-10-22T21:37:57 | {
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https://math.stackexchange.com/questions/1021810/system-of-differential-equations-phase-portraits-and-stability-of-fixed-points | # System of differential equations, phase portraits and stability of fixed points
Consider the system of differential equations:
$$x'=-x-y+4$$ $$y'=3-xy$$
a. Find the fixed points.
$x'=-x-y+4$
$x+y=4$
$x+3/x=4$
x=3,x=1
$y'=3-xy$
$y=3/x$
fixed points: (1,3), (3,1)
b. Determine the type of the linearized system at each fixed point.
calculating the Jacobian:
\begin{array}{cc} -1 & -1 \\ -y & -x \\ \end{array}
for the fixed point (1,3): \begin{array}{cc} -1 & -1 \\ -3 & -1 \\ \end{array}
calculating eigenvalues:
$λ_1=-1-\sqrt 3$ (could be positive or negative) $λ_2=\sqrt 3-1$ (negative)
So it is unstable (I think, because if we use the negative root of 3, then the first eigenvalue is positive, is this correct?
for the fixed point (3,1): \begin{array}{cc} -1 & -1 \\ -1 & -3 \\ \end{array}
calculating eigenvalues:
$λ_1=-2-\sqrt 2$ (negative) $λ_2=\sqrt 2-2$ (negative)
So it is stable
C. Determine the nullclines and the signs of $x'$ and $y'$ on the nullclines and in the various regions determined by them. (I'm not sure I am calculating the signs of $x'$ and $y'$ correctly)
y-nullcine: $y'=3/x$ x-nullcine: $y=4-x$
R1
$x'<00$ $y'>0$
R2
$x'<0$ $y'>0$
R3
$x'>0$ $y'<0$
R4
$x'<00$ $y'<0$
R5
$x'<0$ $y'<0$
R6
$x'0$ $y'<0$
d. Draw the phase plane portrait
• So is the general method to look at the phase portrait to determine the positive and negative roots, or the opposite, do I need to determine the roots to draw the phase portrait? – Math Major Nov 14 '14 at 16:27
• I get everything about the critical points and nullclines, but I guess the problem is I am still a little confused about how to draw the direction fields or how I can determine the direction lines if I don't have the phase portrait to work backwards from – Math Major Nov 14 '14 at 16:33
$\qquad\qquad\qquad\qquad\qquad$
• At $(3,1)$, Jacobian matrix $\begin{pmatrix}-1&-1\\-1&-3\end{pmatrix}$, trace $-4$ (negative), determinant $+2$ (positive), discriminant $(-4)^2-4\cdot(+2)=8$ (positive), hence two real negative eigenvalues: the point $(3,1)$ is a stable node
• At $(1,3)$, Jacobian matrix $\begin{pmatrix}-1&-1\\-3&+1\end{pmatrix}$, trace $0$, determinant $-4$ (negative), hence two real eigenvalues of opposite signs: the point $(1,3)$ is a saddle point
$\qquad$ | 2019-10-21T19:46:13 | {
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https://www.physicsforums.com/threads/general-electric-field-question.238255/ | # General Electric Field Question
I'm seeing a lot of problems in my textbook where it asks you to find the point near two charges where the total electric field is zero. Whats the logic behind figuring out if the point will be on the left side, the right side, or between the two charges?
For example, one problem says there are two charges -7uc and 28uc seperated by 2 m. It asks you to find the point where the electric field is zero. How do you tell if the point will be on the left side of the -7uc, the right side of the 28uc, or in between?
I have no problem solving these problems once I know where the point is going to be, but thats the major hurdle.
Let d=distance separated between the two charges.
Let E=electric field
By Coulomb's Law, $$E=\frac{kQ}{r^2}$$, where r is the distance
Let $$r_{1}$$ be the distance from one charge to the zero field, then
$$r_{2}=d-r_{1}$$ be the distance from the other charge to the zero field.
So, for two charges, the distance where the electric field = 0 is when
$$E=\frac{kq_{1}}{r_{1}^2}=\frac{kq_{2}}{r_{2}^2}$$. Since you are given d, you can solve for $$r_{2}$$
But if the point is not in between the two charges, then wouldn't the distance be 2 m from one charge, and 2+x from the other charge? See I just want to know how you decide if the point is in between the two charges, or if it is outside the two charges. After that I can solve the problem.
Ah, good question. You are give that one of the charge is negative, which is the opposite sign of the second charge. Hence, it must be in between so that the electric field is zero.
tiny-tim
Homework Helper
Whats the logic behind figuring out if the point will be on the left side, the right side, or between the two charges?
Hi uday28fb!
Same charges … between.
Opposite charges … outside, on the side of the weaker charge.
I'm seeing a lot of problems in my textbook where it asks you to find the point near two charges where the total electric field is zero. Whats the logic behind figuring out if the point will be on the left side, the right side, or between the two charges?
I have no problem solving these problems once I know where the point is going to be, but thats the major hurdle.
These are just equilibrium problems. For equi we just require that the forces on the Q of interest are equal in magnitude and opposite in direction. You can check each region to see if this condition can be met - you don't need to do any calculations for this, just think about the net force on the Q of interest. The post above gives the rule of thumb.
Conceptually--
(a) charges have same sign-- your test charge will be inbetween them, else it will not be.
(b) your test charge must be closer to the weaker charge so that the 1/r^2 can make up for the charge being less (so that the forces are still equal in magnitude).
That's it, just those two rules will give you any case.
Thanks guys, I'll rep you guys. Is the logic the same for electrical potential?
edit: how do you rep people?
Last edited:
Thanks guys, I'll rep you guys. Is the logic the same for electrical potential?
It's even easier for electric potential (V). Because V is a scaler quantity, the net V, at a point, due to a group of point charges is just the *algebraic* sum of individuas potentials. All you have to worry about are the signs. Relative to a zero V at infinity, V due to a negative point Q is negative, and vice-versa for positive Q.
tiny-tim
Homework Helper
Hi uday28fb!
uday28fb said:
Hey man, thanks for the rule of thumb.
My problem is that in the solution for the problem I was solving, it has the point where E=0 in between the -7 and 28 uc charge. But I thought if they are oppsoite charges then it goes on the side of teh -7 charge, according the rule you posted. Am I missing something? | 2022-05-28T07:39:56 | {
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http://mathhelpforum.com/pre-calculus/10147-lines.html | # Math Help - Lines
1. ## Lines
Prove that if two nonvertical lines have slopes whose product is -1, then the lines are perpendicular.
HINT GIVEN:
Use the converse of the Pythagorean Theorem.
2. Originally Posted by symmetry
Prove that if two nonvertical lines have slopes whose product is -1, then the lines are perpendicular.
HINT GIVEN:
Use the converse of the Pythagorean Theorem.
The converse of the Pythagorean theorem is that if triangle ABC has
sides a,b,c with the usual labelling, and a^2+b^2=c^2, then the angle
at C is right.
The lines meet at some point, call it (a,b), and let them have slopes m (!=0)
and 1/m.
Additional points on the two lines are (a-1, b-m) and (a-1, b+1/m).
Now we have three points, we have a triangle and can calculate the
lengths of the sides, and we will find the square of the side connecting
(a-1, b-m) and (a-1, b+1/m) is equal to the sum of the squares of the
other two sides.
The case where one of the slopes is 0 can be handled as a special case
and I hope you see that in that case the result is obvious.
RonL
3. Hello, symmetry!
An interesting problem . . . I had to "invent" my own approach.
I'm certain others have shorter, more elegant proofs.
Prove that if two nonvertical lines have slopes whose product is -1,
then the lines are perpendicular.
$\text{Let }m_1 = \frac{q}{p}\:\text{ and }\:m_2 = -\frac{p}{q}$
Then: . $m_1\cdot m_2 \:=\:\left(\frac{1}{p}\right)\left(-\frac{p}{q}\right) \:=\:-1$
. . We have two slopes whose product is $-1$.
We know that: $slope \:=\:\frac{rise}{run}$
So $m_1 = \frac{q}{p}$ looks like this:
Code:
*
* |q
* |
* - - - - - *
p
And $m_2 = -\frac{p}{q}$ looks like this;
Code:
q
* - - *
|
* |
| -p
* |
|
*
Sketch them on the same graph.
Code:
| F
| * (p,q)
| a * |
| * |
O * - - + - - + -
(0,0)| |
| * |
| |
| b * |
| |
| * (q,-p)
| G
Now consider the triangle $FOG$.
Let $a \:= \:OF \:=\:\sqrt{(p-0)^2 + (q-0)^2} \:=\:\sqrt{p^2+q^2}$
Let $b \:=\:OG\:=\:\sqrt{(q-0)^2+(\text{-}p-0)^2} \:=\:\sqrt{p^2+q^2}$
Let $c \:=\:GF \:=\:\sqrt{(p-q)^2 + (q+p)^2} \:=\:\sqrt{2p^2 + 2q^2}$
We note that: . $a^2 + b^2 \:=\:\left(\sqrt{p^2+q^2}\right)^2 + \left(\sqrt{p^2 + q^2}\right)^2 \:=\:2p^2+2q^2$
. . . and that: . $c^2\:=\:\left(\sqrt{2p^2+2q^2}\right)^2\:=\:2p^2+2 q^2$
Since $a^2 + b^2\:=\:c^2$, then $\Delta FOG$ is a right triangle with $\angle FOG = 90^o$.
4. ## ok
I honor both replies but soroban did a fantastic job using the correct math symbols with LAtex. | 2014-08-31T11:30:10 | {
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https://math.stackexchange.com/questions/3192205/power-sets-do-the-relations-between-pa-and-pb-always-mirror-the-relations/3192340 | # Power sets : Do the relations between P(A) and P(B) always mirror the relations between the sets A and B?
If I am correct it is true that:
(1) "$$P(A)$$ is included in $$P(B)$$" implies "$$A$$ is included in $$B$$".
(2) "$$P(A) = P(B)$$" implies "$$A = B$$".
Might I conclude from this that the power sets of two sets always have the same relations as these two sets have with one another?
Are there classical counterexamples to this (hasty) generalization?
I can think of this as a counterexample :
The fact that $$A$$ and $$B$$ are disjoint does NOT imply that $$P(A)$$ and $$P(B)$$ are disjoint.
Attenpt to prove (1) using the theorem : "$$\{ x \}$$ belongs to $$P(S)$$" $$\Longleftrightarrow$$ "$$x$$ belongs to $$S$$.
Let's admit that : $$P(A)$$ is included in $$P(B)$$.
Now, suppose (in view of refutation) that $$A$$ is not included in $$B$$.
It means that there exists an x such that x belongs to $$A$$ but not to $$B$$. And consequently that there is an $$x$$ such that $$\{ x \}$$ belongs to $$P(A)$$ but not to $$P(B)$$. If this were true, there would be a set $$S$$ such that $$S$$ belongs to $$P(A)$$ but not to $$P(B)$$. This contradicts our hypothesis according to which $$P(A)$$ is included in $$P(B)$$.
Conclusion: "$$P(A)$$ is included in $$P(B)$$" implies "$$A$$ is included in $$B$$".
• Indeed $P(A)\subseteq P(B)\iff A\subseteq B$: If $A\subseteq B$, then every subset of $A$ is also a subset of $B$, and if $A\not\subseteq B$, then $A$ is a subset of $A$ that is not a subset of $B$. Apr 18, 2019 at 10:00
• @HagenvonEitzen. Thanks. Do you think there are counter-examples to the generalization : power sets of A and of B always have the same relations as the original sets A and B?
– user654868
Apr 18, 2019 at 10:04
A very natural relation to consider would be that of the cardinality of the sets. We certainly have that $$|A| = |B|$$ implies that $$|P(A)| = |P(B)|$$. You may ask whether or not the converse is true, so does $$|P(A)| = |P(B)|$$ imply $$|A| = |B|$$?
This turns out to be not necessarily true, it is in fact independent from ZFC. That means that there can be set-theoretic universes where $$|P(A)| = |P(B)|$$ implies $$|A| = |B|$$ (e.g. when GCH holds), but there can also be set-theoretic universes where this fails. So then there are $$A$$ and $$B$$ such that $$|P(A)| = |P(B)|$$ while $$|A| \neq |B|$$. See also this answer: https://math.stackexchange.com/a/244873/661457.
Let's take these sets as an example: \begin{align} A &= \{1\} \\ B &= \{2\} \\ P(A) &= \{\emptyset, \{1\}\} \\ P(B) &= \{\emptyset, \{2\}\} \end{align} A relation is in this context a function that takes two sets and gives true or false.
A simple counterexample relation is: $$f(X, Y) = X \text{ contains a set}$$ Then we have: \begin{align} f(A, B) &= \text{false} \\ f(P(A), P(B)) &= \text{true} \\ \end{align} | 2022-05-27T16:47:31 | {
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https://emicida.com/specialist-for-uaxd/page.php?page=87db30-types-of-functions-graphs | Compartilhar
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The simplest and and most straightforward way to compare various categories is often the classic column-based bar graph. So if I take any member of the domain, let's call that x, and I give it to the function, the function should tell me what member of my range is associated with it. The graph is constant between each pair of integers. On the other hand, a function can be symmetric about a vertical line or about a point. The graph of functions helps you visualize the function given in algebraic form. Endpoint Discontinuities: only one of the one-sided limits exists. In mathematics, some functions or groups of functions are important enough to deserve their own names.This is a listing of articles which explain some of these functions in more detail. Continuity. THE CARTESIAN COORDINATE SYSTEM We know that One dimensional numbers can be represented as points on a number line, and can be oriented either to the left (negative), or to the right (positive) of the origin (the zero point). There is a relationship between a function and its graph. Different Types of Graphs. The graph of squaring function is given below. . We can define a function as a special relation which maps each element of set A with one and only one element of set B. Even and Odd Functions; There is a relationship between a function and its graph. An example of a discontinuous graph is y = 1/x, since the graph cannot be drawn without taking your pencil off the paper: A function is periodic if its graph repeats itself at regular intervals, this interval being known as the period. The graph of step function is shown below. Each axis shows a quantity for a different categorical value. Specifically, Specifically, Jump Discontinuities: both one-sided limits exist, but have different values. Examples of Distributions and Descriptive Graphs. Now, just as a refresher, a function is really just an association between members of a set that we call the domain and members of the set that we call a range. [[x]] mean find the a value ‘greater than or equal to x‘. One can observe that, if, : →, then the graph () is a subset of + (strictly speaking it is ×, but one can embed it with the natural isomorphism). The graphs of y = 1/x and y = 1/x 2 both have vertical asymptotes of x = 0 and horizontal asymptotes of y = 0. General equation: f(x)=a_n xⁿ+a_n -1xⁿ-1 + . Functions sin x and cos x have a period 2 \Pi , Functions tan x and cot c have a period \Pi . Function, in mathematics, an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable). There are eight types of graphs that you will see more often than other types. In other words, it is the term with the highest degree. (one to one or many to one but not all the Bs have to be busy) A function is injective if for every y in the codomain B there is at most one x in the Given the graph of a common function, (such as a simple polynomial, quadratic or trig function) you should be able to draw the graph of its related function. Constant Function: Let 'A' and 'B' be any two non–empty sets, then a function '$$f$$' from 'A' to 'B' is called a constant function if and only if the A spider or radar graph is a very useful type of graph for showing qualitative data or the overall “score” or comparison of multiple series. Analyzing the features of exponential graphs through the example of y=5ˣ. Linear Function: It is a straight line and the function which is in the form of y = mx + c is called the linear function. 3. See Figure 8 for examples of graphs of polynomial functions with multiplicity 1, 2, and 3. The graph will also be lower at a local minimum than at neighboring points. If a function is even, the graph is symmetrical over the y- axis. Parallel Edges: If two vertices are connected with more than one edge than such edges are called parallel edges that is many roots but one destination. Composite function Suppose that a function $$y = f\left( u \right)$$ depends on an intermediate variable $$u$$, which in turn is a function of the independent variable $$x$$: $$u = g\left( x \right)$$. A vertical line through any element of the domain should intersect the graph of the function exactly once. (i) y = mx + c, m > 0 (ii) y = mx + c, m < 0 (b) Quadratic function: y = ax 2 + bx + c. Highest power of the variable x is 2. Chapter : FunctionsLesson : Types Of Functions For More Information & Videos visit http://WeTeachAcademy.com Here, you see functions in terms of the operations being performed. The leading term of a polynomial is the first term when a polynomial is written in standard form. Learn more on page 7 of the Precalculus textbook. These observations lead us to a formal definition of local extrema. The following table shows the transformation rules for functions. Increasing, Decreasing and Constant Functions. Functions and different types of functions are explained here along with solved examples. We have seen multiple uses of excel in our professional lives, it helps us analyze, sort and extract insights from data. Program files can contain multiple functions. A={1,2,3,4,5,6,7,8,9,10} 4. The graph of cubic function is in positive side and negative side unlike squaring function which is only on positive side. This table classifies and illustrates the common graphics functions. The graph of negative x-values (shown in red) is almost flat. If you plot the graph then it look like the one below. Let us use the following table to plot the graph of cubic function. A function is continuous over an interval of its domain if its hand-drawn graph over that interval can be sketched without lifting the pencil from the paper. Types of Functions Local and Nested Functions in a File. The graph of piecewise function is already discussed in previous lessons. More Graphs And PreCalculus Lessons Graphs Of Functions. With the help of a graph of function, you can discover may properties which the algebraic form does not provide. Different Types of Functions prepared by: Shielamar L. Labiscase 2. Types of Functions: Unary Function. Inverse Trigonometric function or Inverse circular function By showing several graphs on one plot you will be able to see their common features. In mathematics, the graph of a function f is the set of ordered pairs (x, y), where f(x) = y.In the common case where x and f(x) are real numbers, these pairs are Cartesian coordinates of points in two-dimensional space and thus form a subset of this plane.. It contains plenty of examples and multiple choice practice problems. Functions are ubiquitous in mathematics and are essential for formulating physical relationships in the sciences. The equations for quadratic functions have the form f(x) = ax 2 + bx + c where . Functions Introduction: Functions provide us with a convenient way to handle the relationship between the values of one variable quantity that depends on the values of another variable quantity. Functions are mathematical building blocks for designing machines, predicting natural disasters, curing diseases, understanding world economies and for keeping aeroplanes in the air. There are different types of functions in Mathematics. Just like Transformations in Geometry, we can move and resize the graphs of functions: Let us start with a function, in this case it is f(x) = x 2, but it could be anything: f(x) = x 2. The asymptotes are actually the x– and y-axes. The graph of the above function is a line passing through the points (-3 / 2 , 0) and (0 , -1 / 2) as shown below. You can change the way the graph of a power function looks by changing the values of k and n. If n is greater than zero, then the function is proportional to the nth power of x. Graphing Quadratic Equations. Local functions are the most common way to break up programmatic tasks. Loop: An edge of a graph which join a vertex to itself is called loop or a self-loop. Graphs are a great way to visualize data and display statistics. Click here to see graphs of various types of functions. The rectangular coordinate system A system with two number lines at right angles specifying points in a plane using ordered pairs (x, y). We have tried to include all types of functions and their graphs. 2.1.2 Types of Graphs of Functions. (a) Linear function: y = mx + c. Highest power of the variable x is 1. The eight most commonly used graphs are linear, power, quadratic, polynomial, rational, exponential, logarithmic, and sinusoidal. Evaluating any value for x, such as x = 2, will result in c. The graph of a constant function is a horizontal line. The reciprocal function is symmetric along the origin, but it never touches the origin itself. Functions and different types of functions A relation is a function if for every x in the domain there is exactly one y in the codomain. 3. The following figures show the graphs of parent functions: linear, quadratic, cubic, absolute, reciprocal, exponential, logarithmic, square root, sine, cosine, tangent. Farhana ShaheenTHE CARTESIANCOORDINATE SYSTEM 2. The list of common functions and their graphs are provided below. A unary function has one input and one output. In this article, we will discuss, what is a polynomial function, polynomial functions definition, polynomial functions examples, types of polynomial functions, graphs of polynomial functions etc. Properties of Graph of Reciprocal Function. When both the input (independent variable) and the output (dependent variable) are real numbers, a function can be represented by a coordinate graph. The domain of reciprocal function is between. Definition of a local maximum. The universally-recognized graph features a series of bars of varying lengths.One axis of a bar graph features the categories being compared, while the other axis represents the value of each. It is simpler to create a line graph with (XY) Scatter when your independent and dependent variables are in columns. The set of real numbers SET NOTATION A set is collection of objects. Analyzing the features of exponential graphs through the example of y=5ˣ. The most common graphs name the input value x x and the output value y y, and we say y y is a function of x x, or y = f (x) y = f (x) when the function is named f f. The squaring function graph is decreasing between interval, The graph is increasing between the interval. Now in this chapter, we will learn about 48 Different Types of Functions Graphs. Please support us by disabling your adblocker or whitelist this site from your adblocker. There are eight types of graphs that you will see more often than other types. Examples of the following types of functions are shown in this gallery: linear; quadratic; power; polynomial; rational; exponential; logarithmic; sinusoidal The graph of y = 1/x is symmetric with respect to the origin (a 180-degree turn gives you the same graph). What are some functions of linear graphs? A radar chart is one of the most modern types of graphs and charts – ideal for multiple comparisons. Different types of graphs depend on the type of function that is graphed. The set of real numbers SET NOTATION A set is collection of objects. We will graph the function and state the domain and range of each function. 3. In other words, it is the term with the highest degree. Example: The graph shown in fig is a null graph, and the vertices are isolated vertices. Chapter : FunctionsLesson : Types Of Functions For More Information & Videos visit http://WeTeachAcademy.com For example, the simple function f(x)is a unary function. Line Plots Data Distribution Plots Discrete Data Plots Geographic Plots Polar Plots Contour Plots Vector Fields Surface and Mesh Plots Volume Visualization Animation The graphs of the original and inverse functions are symmetric about the line $$y = x$$. Therefore, individuals working on a… Here are some graphs with different values for a, b, and c. Notice how each value changes the shape and location of the parabola. Graphs, Relations, Domain, and Range. If you're seeing this message, it means we're having trouble loading external resources on our website. ROSTER METHOD of writing a set encloses the elements of the set in braces, {}. The set of natural numbers less than 11. Figure 8 For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the x -axis. Why Histograms are often confused with Bar graphs? For example, the simple function f(x)is a unary function. The graph of squaring function is commonly known as a parabola which is a U-shaped curve.The diagram for squaring function is given below. What method do graphic designers use to design graphs? As you can see that the graph is only on positive side for both and . Polynomial Function Definition. Ex. Different types of graphs depend on the type of function that is graphed. Let us plot the graph of the square root function by obtaining some points. Both the sets A and B must be non-empty. Types of Graphs: 1. Thanks! An example is the general form of the equation for the straight line 2 x + 3 y + 4 = 0. Here are some of the most commonly used functions and their graphs: linear, square, cube, square root, absolute, floor, ceiling, reciprocal and more. Popular graph types include line graphs, bar graphs, pie charts, scatter plots and histograms. consists of two real number lines that intersect at a right angle. Here, though, you see classifications that work for all the many types of functions. You can identify a function by looking at its graph. If you want to test an orchestrator or entity function in the Azure Portal, you must instead run a client function that starts an orchestrator or entity function as part of its implementation. The domain is all real numbers. A function is uniquely represented by its graph which is nothing but a set of all pairs of x and f(x) as coordinates. Introduction to Functions and Graphs \n . Why are K2, 2, and 3 graphs not planar graphs? It can be expressed in terms of a polynomial. Exponential and power functions are compared interactively, using an applet. Later , when you learn calculus, visualizing concepts is much easier with a graph of function. The universally-recognized graph features a series of bars of varying lengths. This lesson, we explore different types of function and their graphs. The objects in a set are elements or members of the set. In the basic graph above, a = 1, b = 0, and c = 0. Functions can be categorized in many different ways. Download the Excel template with bar chart, line chart, pie chart, histogram, waterfall, scatterplot, combo graph … There is one feature of excel that helps us put insights gained from our data into a visual form. The trigonometric functions are periodic. Read the following article to learn more about linear function. Functions can take input from many variables, but always give the same answer, unique to that function. The graphs are continuous (i.e. Function and their graphs ppt 1. Functions; Graph of Function; Increasing, Decreasing and Constant Functions. The domain of squaring function set of all real numbers that corresponds to x-axis. Special Function Types and Their Graphs; Special Function Types and Their Graphs. They are easy to visually distinguish and by knowing how each looks, you can get an idea of what a graph might look like just by analyzing the function. Scroll down the page for more examples and solutions. The inverse of a function is the relation in which the roles of the independent anddependent variable are reversed. The intercept of squaring function is at point (0, 0). This class of functions is the one most commonly studied in general math and calculus, so most of the types of functions you deal with in … The function is an even function because it is symmetric along the y-axis. Each has its own type of function that produces the graphs. Removable discontinuities can be "fixed" by re-defining the function. For example, a bar graph or chart is used to display numerical data that is independent of one another. Types of MATLAB Plots. Radar Chart. and increases as x increases. Range of function is set of all integers. This precalculus provides a basic introduction into functions and graphs. 1. The properties such as domain, range, x and y intercepts, intervals of increase and decrease of the graphs of the two types of functions are compared in this activity. Null Graph: A graph of order n and size zero that is a graph which contain n number of vertices but do not contain any edge. Learn more about what are polynomial functions, its types, formula and know graphs of polynomial functions with examples at BYJU'S. To plot the graph of reciprocal function, let us find all the points first. The cubic function is symmetric along the origin. The graph of squaring function has relative minimum at (0, 0). By Yang Kuang, Elleyne Kase . Graphs. Specifically, Jump Discontinuities: both one-sided limits exist, but have different values. In the previous lesson, we have learned What is a function? It is odd function because symmetric with respect to origin. Rational Functions Rational Functions. The graph jumps vertically one unit for each y-value. A vertical line through any element of the domain should intersect the graph of the function exactly once. Prerequisite to learn from this article is listed below. Inverse Trigonometric Functions: As observed from graph, increases as x increases, and decreases as x increases. For example, a spider/radar can be easily used to compare three different types of phones based on five criteria (speed, screen size, camera quality, memory, apps). Domain & Range. By look at an equation you could tell that the graph is going to be an odd or even, increasing or decreasing or even the equation represents a graph at all. With the help of a graph of function, you can discover may properties which the algebraic form does not provide. ... Pie charts, unlike bar graphs, ... Line graphs can be created with either the Line Graph type or with (XY) Scatter. A function defines a particular output for a particular input. The properties of step function are given below. May 27, 2020 June 4, 2015 by (B) Types of graphs of functions. This class of functions is the one most commonly studied in general math and calculus, so most of the types of functions you … What are the disadvantages of line graphs over bar graphs? You will discover that each type has its own distinctive graph. Determine whether the points on this graph represent a function. A function is continuous if its graph has no breaks in it. Unlike other function types, orchestrator and entity functions cannot be triggered directly using the buttons in the Azure Portal. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. Frequency Histogram Discrete values of categories are used on the horizontal (x) axis, and the number of scores that fall into that category (i.e., frequency) appears on the vertical (y) axis. Graph y = − sin (π 8 x) + 7 y=-\sin\left(\dfrac{\pi}{8}x\right)+7 y = − sin (8 π x) + 7 y, equals, minus, sine, left parenthesis, start fraction, pi, divided by, 8, end fraction, x, right parenthesis, plus, 7 in the interactive widget. The range of squaring function is all non-negative real numbers because the graph is U-shaped. The linear functions are straight lines. The graph below illustrates these ideas for a local maximum. They are generally used for, and are best for, quite different things. They are easy to visually distinguish and by knowing how each looks, you can get an idea of what a graph might look like just by analyzing the function. By : Prashant Kumar.
Compartilhar | 2021-06-14T03:22:59 | {
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https://math.stackexchange.com/questions/2118683/if-the-polynomial-x4-6x316x2-25x10-is-divided-by-another-polynomial-x2?noredirect=1 | # If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, the [duplicate]
If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, theremainder is $x+a$, find $k$ and $a$.
My Attempt,
$f(x)=x^4-6x^3+16x^2-25x+10$
$g(x)=x^2-2x+k$
$R=x+a$
Here, the divisor is in the quadratic form. so how do I use the synthetic division
• It is possible to use Synthetic division on a quadratic or higher power with the leading coefficient equal to $1$. In your case, you would ignore the $x^2$ term and write $2,-k$ as the numbers dividing the polynomial Jan 29 '17 at 2:03
• @Frank, Could You please show me a bit more? I could not get.
– pi-π
Jan 29 '17 at 2:07
Question: A polynomial $f(x)=x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $g(x)=x^2-2x+k$. Their remainder is $x+a$. Find $k$.
We're given a polynomial, and given that $\frac {f(x)}{g(x)}=p(x)+(x+a)$ where $p(x)$ is a polynomial. Therefore, if we divide $\frac {f(x)}{g(x)}$ and equate the remainder, we should be able to find $k$.
The process for extended synthetic division for higher powers is described below.
\begin{array}{c |c c} & 1 & & -6 & & +16 & & -25 & & +10\\ 2 & & & 2 & & -8 & & 16-2k & & \\-k & & & & & -k & & 4k & & k^2-8k\\\hline\\ &1 & & -4 & & 8-k & & \color{red}{2k-9} & & \color{blue}{k^2-8k+10}\end{array}
Where the top row of numbers are the coefficients of $f(x)$, and the vertical numbers are the negated coefficients of $g(x)$. And the last two numbers being the remainder $\color{red} px+\color{blue} q$.
Comparing the last two numbers to the remainder $x+a$, we have$$(\color{red}{2k-9})x+(\color{blue}{k^2-8k+10})=\color{red}{1}x+\color{blue}{a}\tag1$$$$\implies\begin{cases}2k-9=1\\k^2-8k+10=a\tag2\end{cases}$$ From the first equation of $(2)$, we have$$2k-9=1\implies 2k=10\implies k=5$$And. if necessary, plug in $k=5$ into the second equation of $(2)$ to find $a$.
• Re formatting: not exactly my cup of tea, but there are a couple of worked out examples here.
– dxiv
Jan 29 '17 at 2:25
• How is the last term in the quotient/eemainder line $0$? Adding up the column I get a quadratic polynomial in $k$, whose value is $-5$ with $k=5$. Jan 29 '17 at 2:26
• @OscarLanzi That was a little mistake on my part. It should be fixed. Jan 29 '17 at 2:27
• @dxiv I think I have it formatted in $\LaTeX$. :) Jan 29 '17 at 2:27
• Looks great, +1. Jan 29 '17 at 10:49
You can do the division between $x^4-6x^3+16x^2-25x+10$ and $x^2-2x+k$ following the polynomial long division, getting:
$$R=(2k-9)x+(k^2-8k+10)$$
but $R=x+a$, so $2k-9=1\longrightarrow k=5$ and $k^2-8k+10=a\longrightarrow a=-5$.
To find the remainder of the long division by $x^2-2x+k$ you can keep replacing $x^2$ with $2x-k$ repeatedly, until getting the remainder of degree $1\,$:
\begin{align} x^4-6x^3+16x^2-25x+10 & = (2x-k)^2 - 6x(2x-k)+16(2x-k)-25x+ 10 \\ & = -8x^2 + (-4k +6k +32-25)x+k^2-16k+10 \\ & = -8(2x-k) + (2k+7)x+k^2-16k+10 \\ & = (2k-9)x + k^2-8k+10 \end{align}
Identifying coefficients between the calculated remainder and $x+a$ it follows that $2k-9=1$ so $k=5\,$, and $a=5^2-8 \cdot 5+10=-5\,$.
• Wish the downvoter had left a comment why.
– dxiv
Jan 29 '17 at 2:45
HINT
You can write:
$$x^4-6x^3+16x^2-25x+10= (x^2-2x+k)(x^2+bx+c)+x+a=\\ =x^4+(b-2)x^3+(c-2b+k)x^2+(-2c+bk+1)x+(a+kc)$$
So,
$$b-2=-6→b=-4\\ c-2b+k=16\\ -2c+bk+1=-25\\ a+kc=10$$
• Where did $b$ come from?
– pi-π
Jan 29 '17 at 2:18
• $b$ and $c$ are coefficient of the quotient. Jan 29 '17 at 2:20
• I just know that the quotient has degree equal to $2$, so I use a general quadratic polynomial and called its coefficients $b$ and $c$. Is it clear? Jan 29 '17 at 2:23 | 2021-10-24T03:32:19 | {
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https://www.physicsforums.com/threads/differential-equations-system-solutions.786308/ | # Differential Equations System Solutions
Tags:
1. Dec 7, 2014
### dmoney123
1. The problem statement, all variables and given/known data
Consider the initial value problem for the system of first-order differential equations
y_1' = -2y_2+1, y_1(0)=2
y_2' = -8y_1+2, y_2(0)=-1
If the matrix
[ 0 -2
-8 0 ]
has eigenvalues and eigenvectors L_1= -4 V_1= [ 1
2 ]
L_2=4 V_2= [ 1
-2]
then its solution will be:
2. Relevant equations
3. The attempt at a solution
e^(-4t) +e^ (4t) from eigenvalues
multiply by respective eigenvectors and set to initial conditions gives 2 sets of equations and two unknown coefficients
2=c_1*e^(-4t)+c_2*e^(4t)
-1=c_1*2e^(-4t)+c_2*-2e^(4t)
c_1=3/4
c_2=5/4
I am very confident with these values being right for the coefficients, I know need to know how to use these to form a general solution.
I plug these values back in to get
y_1=3/4e^(-4t)+5/4e^(4t)
y_2=3/2e^(-4t)-5/2e^(4t)
Then solution given is
y_1(t)=5/4e^(4t)+1/2e^(-4t)+1/4,
y_2(t)=-5/2e^(4t)+e^(-4t)+1/2
any help is appreciated! thanks!
2. Dec 7, 2014
### ehild
This is the solution of the homogeneous part only. The system is inhomogeneous, you have to add the particular solution of the equations to the general solution of the homogeneous part.
3. Dec 7, 2014
### dmoney123
I have not encountered this type of problem before. How do you go about finding the particular solution?
4. Dec 7, 2014
### ehild
It is simple in this case. Set both y1'=0 and y2' =0. What are y1 and y2 then?
5. Dec 7, 2014
### dmoney123
Okay, I work this out to y_1=1/4 and y_2=1/2...
which seems to fit in the equation... but im still not really sure why the coefficients for the e^(-4t) dont seem to match the solution
6. Dec 7, 2014
### ehild
The general solution is
y1=c1e-4t+c2e4t+1/4
y2=c1e-4t+c2e4t+1/2
Fit these equations to the initial conditions y1(0)=2, y2(0)=-1.
7. Dec 7, 2014
### dmoney123
Thank you so much! That makes total sense!
8. Dec 7, 2014
### ehild
You are welcome. :)
9. Dec 8, 2014
### HallsofIvy
Staff Emeritus
iHere is how I would have done this, without using "matrices": differentiate the first equation again to get $y_1''= -2y_2'$. But from the second equation, $y_2'= -8y_1+ 2$ so that $y_1''= -2(-8y_1+ 2)= 16y_1- 4$ or $y_1''- 16y_1= -4$. The associated homogeneous equation, $y_1''- 16y= 0$ has characteristic equation $r^2- 16= 0$ which has roots $r= 4$ and $r= -4$ so the general solution to the associated homogeneous equation is $y(t)= C_1e^{4t}+ C_2e^{-4t}$.
Now new look for a particular solution to the entire equation. Since the right hand, "non-homogeneous", part is the constant -4, we try a solution of the form y= A for A some constant. Then y'= y''= 0 so the equation becomes $0- 16A= -4$ and $A= 1/4$.
That is, $y_1(t)= C_1e^{4t}+ C_2e^{-4t}+ \frac{1}{4}$. Now, we can write the first equation, $y_1'= -2y_2+ 1$ is the same as $2y_2= -y_1'+ 1$ or $y_1= -(1/2)y_2'+ 1/2$. Since $y_1(t)= C_1e^4t+ C_2e^{-4t}+ \frac{1}{4}$, $y_1'= 4C_1e^{4t}- 4C_2e^{-4t}$, $-(1/2)y_2'= -2C_1e^{4t}+ 2e^{-4t}$ and $y_1(t)= -2C_1e^{4t}+ 2C_2e^{-4t}+ 1/2$.
Last edited: Dec 11, 2014
10. Dec 10, 2014
### Ray Vickson
$$y_1' = -2y_2 + 1 = -2\left(y_2 - \frac{1}{2} \right) \\ y_2' = -8y_1+2 = -8\left(y_1 - \frac{1}{4} \right)$$
Since $d(y_1 - 1/4)/dt = dy_1/dt$ and $d(y_2 - 1/2)/dt = dy_2/dt$, the variables $Y_1 = y_1 - 1/4$ and $Y_2 = y_2 - 1/2)$ satisfy the homogeneous system
$$Y_1' = -2Y_2\\ Y_2' = -8 Y_1$$
You can easily figure out the boundary values $Y_1(0), \, Y_2(0)$. | 2017-10-19T07:44:46 | {
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https://math.stackexchange.com/questions/577012/solving-an-ode-without-lambert-w-function | # Solving an ODE without Lambert W function
I have a question regarding the possibility of solving the following ODE:
$$\left[2x(t)+t\right]x^{\prime}(t)=1$$
such that $x(0)=-1$.
If we make the substitution $w(t)=2x(t)+t$, we obtain the following equation:
$$-2\ln[w(t)+2]+w(t)=t+C$$
which can be solved for $w$ (and hence $x$), although both $w$ and $x$ will be expressed in terms of the Lambert W function. I won't be posting the full solution here - WolframAlpha shows it step-by-step. Nevertheless, after applying the initial condition that $x(0)=-1$, $x$ is simplified to a much more digestible form: $x(t)=-\frac{t}{2}-1$.
Therefore, I was wondering - is it possible to arrive at this solution without the use of Lambert W function? More specifically, is it possible to somehow apply the initial condition earlier and hence to avoid having to find the general solution for $x(t)$ (and thereby avoiding stumbling on the Lambert W function)?
I would be grateful for some advice on this.
• If you make the hypothesis that the solution is a polynomial of t, say x = A + B t + C t^2 + .., replace this in tour ODE and cancel all possible terms. You will end with your result and you van verify that the initial consition is satisfied. Nov 22 '13 at 10:20
• Hm... It's an interesting idea, but I don't see how it works. When I try, I get one equation with $A$, $B$, $C$ and $t$ to solve for... Nov 22 '13 at 10:36
• Cancel all terms. This gives you equations in terms of A, B and C. The obvious one is C=0. Solve now for A and B. But what gave you as an answer JJacquelin is much better indeed. Nov 22 '13 at 11:14
• Thank you for your time anyway :) Nov 22 '13 at 11:21
Instead of looking for $x(t)$, first look for $t(x)$:
$$x' = \dfrac{\mathrm dx}{\mathrm dt} = \dfrac{1}{(\mathrm dt/\mathrm dx)} = \dfrac{1}{t'}\\\,\\(2x+t)=\dfrac{1}{x'}=t'\\\,\\ t'-t=2x$$
is a linear ODE with condition $t(1)=0$.
• If we solve the equation $t^{\prime}-t=2x$, we should obtain $t(x)=2(-1-x)+Ce^{x}$. Substituting $t=0$ and $x=-1$, we obtain that $C=0$, and hence $t=-2-2x\Longrightarrow x=-\frac{1}{2}t-1$ is the solution, which satisfies $x(0)=-1$. Is that right? If yes, it looks like it works... Nov 22 '13 at 10:59
• I will wait a bit more, see if someone else comes up with another way, and then I'll accept your answer. Nov 22 '13 at 11:22
Therefore, I was wondering - is it possible to arrive at this solution without the use of Lambert W function? More specifically, is it possible to somehow apply the initial condition earlier and hence to avoid having to find the general solution for $x(t)$ (and thereby avoiding stumbling on the Lambert W function)?
Yes this is possible. Using your substitution $w(t)=2x(t)+t$, one sees that $(w(t)+2)^2\mathrm e^{-w(t)}=C\mathrm e^{-t}$, for some constant $C$.
The initial condition $x(0)=-1$ reads $w(0)=-2$ hence $C=0$. Since $\mathrm e^{-w(t)}\ne0$, one gets $w(t)+2=0$ for every $t$, that is, $x(t)=-\frac12(t+2)$.
As you noted, this is specific to the initial condition $w(0)=-2$. | 2021-09-21T18:34:50 | {
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https://math.answers.com/Q/What_is_the_multiplicative_of_inverse_-7 | Math and Arithmetic
# What is the multiplicative of inverse -7?
###### Wiki User
For -7, the multiplicative inverse is 1/(-7) = -1/7.
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## Related Questions
The additive inverse of -7 is 7.The multiplicative inverse of -7 is -1/7.The additive inverse of -7 is 7.The multiplicative inverse of -7 is -1/7.The additive inverse of -7 is 7.The multiplicative inverse of -7 is -1/7.The additive inverse of -7 is 7.The multiplicative inverse of -7 is -1/7.
Assuming the question is about the multiplicative inverse, the answer is, -1. It is its own multiplicative inverse.
The multiplicative inverse is the negative of the reciprocal of the positive value. Thus the multiplicative inverse of -7 is -1/7.
The multiplicative inverse of -7- is 7/1 (seven over 1).
A multiplicative inverse is the same as a reciprocal. The multiplicative inverse of x is 1/x. So, the multiplicative inverse of 4 is 1/4; or 7 is 1/7 and of 0.2 is 1/0.2 = 5.
3 and a half = 7/2 so its multiplicative inverse is 2/7.
Swap the numerator and denominator. For example, the multiplicative inverse of 5/7 is 7/5
Divide 1 by the number. The multiplicative inverse of 7 is 1/7, for example.
38/7 x 7/38 = 1 = multiplicative inverse 38/7 + -38/7 = 0 = additive inverse
Multiplicative inverses are two numbers whose product is one.Another name for multiplicative inverse is reciprocal. The reciprocal of 2/3 is 3/2. 2/3 X 3/2 = 6/6 = 1. The multiplicative inverse of 7 is 1/7. 7 X 1/7 = 7/7 = 1.
The multiplicative inverse means, what do I multiply my number by to get an answer of 1? ex: The multiplicative inverse of 5 is 1/5. 5 * (1/5) = 1 If your number is a fraction to begin with, flip it over. ex: The mult. inverse of 22/7 is 7/22 (22/7)*(7/22) = 1
3 1/2 = 7/2 in improper fractionmultiplicative inverse is the reciprocal of 7/2 = 2/7
Multiply 1⁄4 by 4. The multiplicative inverse of 1⁄4 is 4.
Multiplicative inverse is the number that, when multiplied, results in 1, usually 1/# 1/sqrt7 is the inverse, so just rationalize the denominator sqrt7/7 = square root of 7 divided by 7
It depends on whether you mean additive inverse or multiplicative inverse.
the multiplicative inverse of -8 is just positive 8 :) * * * * * NO, that is the additive inverse. The multiplicative inverse of -8 is -1/8 or -0.125
First convert it into an improper fraction: 5 3/7 becomes 38/7. This is clear because "5" is simply "35/7" (do the math yourself) and 35/7 + 3/7 = 38/7. Additive inverse means "negative." Thus, the "additive inverse" of 38/7 is -38/7. This is true because 38/7 added to -38/7 is 0. That is the definition of an additive inverse. Multiplicative inverse means "reciprocal." Thus, the "multiplicative inverse" of 38/7 is 7/38. This is true because 38/7 multiplied by 7/38 is 266/266, or 1. That is the definition of multiplicative inverse.
The multiplicative inverse of 4i is -(1/4)*i.
The reciprocal (multiplicative inverse) of -3 is -1/3.The reciprocal (multiplicative inverse) of -3 is -1/3.The reciprocal (multiplicative inverse) of -3 is -1/3.The reciprocal (multiplicative inverse) of -3 is -1/3.
Multiplicative Inverse of a NumberReciprocal The reciprocal of x is . In other words, a reciprocal is a fraction flipped upside down. Multiplicative inverse means the same thing as reciprocal. For example, the multiplicative inverse (reciprocal) of 12 is and the multiplicative inverse (reciprocal) of is . Note: The product of a number and its multiplicative inverse is 1. Observe that ·= 1. Multiplicative Inverse of a NumberReciprocal The reciprocal of x is . In other words, a reciprocal is a fraction flipped upside down. Multiplicative inverse means the same thing as reciprocal. For example, the multiplicative inverse (reciprocal) of 12 is and the multiplicative inverse (reciprocal) of is . Note: The product of a number and its multiplicative inverse is 1. Observe that ·= 1.
The multiplicative inverse of 625 is 1/625 or 0.0016
###### Synonyms and AntonymsMath and ArithmeticAlgebraSchool Subjects
Copyright © 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. | 2021-01-17T03:48:25 | {
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http://mathhelpforum.com/geometry/180329-weather-surface-conical-roof.html | Thread: Weather Surface of a conical roof
1. Weather Surface of a conical roof
A cylindrical tower 30ft in diameter has a conical roof the length of whose eaves is 2ft. An element of the roof is inclined 45 degrees to the horizontal. Find the weather surface (see figure).
Answer at the back of the book: 61.061 square yards.
what i did was,
i used cosine function to get the slant height,
i used the radius to get the circumference,
then i got the lateral area of the cone by using LA=1/2(circumference)(slant height)
the answer is on the unit square feet. so i divided the answer by 3 so that i can get square yards. my answer is very far from the answer on the book.
i don't know what i did wrong. or is the book wrong? couldn't be.
2. The length of the eaves is 2 ft. Wouldn't this mean that 2 ft of the roof is beyond the walls? In that case, the radius wouldn't be (2+30)/2, but x + 30/2, where x is the horizontal distance from the tower to the tip of the roof (which would be sqrt(2)).
To go from unit square feet to unit square yards, you need to divide by 9, not 3. This is because of the squares.
3. Originally Posted by HappyJoe
The length of the eaves is 2 ft. Wouldn't this mean that 2 ft of the roof is beyond the walls? In that case, the radius wouldn't be (2+30)/2, but x + 30/2, where x is the horizontal distance from the tower to the tip of the roof (which would be sqrt(2)).
To go from unit square feet to unit square yards, you need to divide by 9, not 3. This is because of the squares.
i didn't get that part. wouldn't be the length of the diameter of the cone be 30ft(diameter of the tower) + 2 (eaves on each side of the tower) = 32ft?
i can't visualize what you're talking about sorry and thanks btw.
4. Hi dug,
The lateral area of a rt circular cone = pi r l where r= radius of base and l=slant height. You have two cones and from givens you must find the r and l for the larger one. Make a sketch Showing how the two cones overlap. Use the properties of rt isosceles triangles to find the r and l
bjh
5. Hello, Dug!
A cylindrical tower 30 ft in diameter has a conical roof.
The length of whose eaves is 2 ft.
An element of the roof is inclined 45 degrees to the horizontal.
Find the weather surface.
Answer at the back of the book: 61.061 square yards.
Code:
A
o
* | * _
* | * 15√2
* |15 *
* | *
* | * B
C o - - - - - + - - - - - o 2
* | 15 M 15 | *
* | E * - o D
| |
| |
The diameter of the tower is 30 feet: . $CB = 30,\;CM = MB = 15$
$\text{Angle }ABC = 45^o \quad\Rightarrow\quad AB = 15\sqrt{2}$
The eaves are 2 feet: . $B\!D = 2$
Note that: $ED = \sqrt{2}$
$\text{The radius of the circular base of the cone is: }\:MB + ED \:=\:15 + \sqrt{2}$
$\text{The circumference of the circular base is: }\:2\pi(15 + \sqrt{2})$
Flatten the conical roof and we have a major sector of a circle.
Code:
* * *
P * * Q
o o
* * * *
* * R
* * O * *
* * *
* @ *
* *
* *
* *
* * *
_
s = 2π(15+√2)
$\text{We have the formula: }\:s \:=\:R\theta$
. . $\text{where }\:s \:=\:2\pi(15 + \sqrt{2}),\;\;R \:=\:15\sqrt{2} + 2 \:=\:\sqrt{2}(15 + \sqrt{2})$
$\text{Then: }\:2\pi(15+\sqrt{2})\:=\:\sqrt{2}(15+\sqrt{2})$ $\theta$
. . $\theta \:=\:\frac{2\pi(15+\sqrt{2})}{\sqrt{2}(15+\sqrt{2} )} \;=\;\pi\sqrt{2}$
$\text{The area of the sector is: }\:A \:=\:\tfrac{1}{2}R^2\theta$
. . $\text{where: }\,R \:=\:\sqrt{2}(15+ \sqrt{2}),\;\theta \:=\:\pi\sqrt{2}$
$\text{Hence: }\;A \;=\; \tfrac{1}{2}\bigg[\sqrt{2}(15 + \sqrt{2})\bigg]^2(\pi\sqrt{2}) \;=\;1197.029986\text{ ft}^2$
. . $\text{Therefore: }\;A \;=\;\frac{1197.029986}{9} \;\approx\;133\text{ yd}^2$
6. Originally Posted by Soroban
Hello, Dug!
Code:
A
o
* | * _
* | * 15√2
* |15 *
* | *
* | * B
C o - - - - - + - - - - - o 2
* | 15 M 15 | *
* | E * - o D
| |
| |
The diameter of the tower is 30 feet: . $CB = 30,\;CM = MB = 15$
$\text{Angle }ABC = 45^o \quad\Rightarrow\quad AB = 15\sqrt{2}$
The eaves are 2 feet: . $B\!D = 2$
Note that: $ED = \sqrt{2}$
$\text{The radius of the circular base of the cone is: }\:MB + ED \:=\:15 + \sqrt{2}$
$\text{The circumference of the circular base is: }\:2\pi(15 + \sqrt{2})$
Flatten the conical roof and we have a major sector of a circle.
Code:
* * *
P * * Q
o o
* * * *
* * R
* * O * *
* * *
* @ *
* *
* *
* *
* * *
_
s = 2π(15+√2)
$\text{We have the formula: }\:s \:=\:R\theta$
. . $\text{where }\:s \:=\:2\pi(15 + \sqrt{2}),\;\;R \:=\:15\sqrt{2} + 2 \:=\:\sqrt{2}(15 + \sqrt{2})$
$\text{Then: }\:2\pi(15+\sqrt{2})\:=\:\sqrt{2}(15+\sqrt{2})$ $\theta$
. . $\theta \:=\:\frac{2\pi(15+\sqrt{2})}{\sqrt{2}(15+\sqrt{2} )} \;=\;\pi\sqrt{2}$
$\text{The area of the sector is: }\:A \:=\:\tfrac{1}{2}R^2\theta$
. . $\text{where: }\,R \:=\:\sqrt{2}(15+ \sqrt{2}),\;\theta \:=\:\pi\sqrt{2}$
$\text{Hence: }\;A \;=\; \tfrac{1}{2}\bigg[\sqrt{2}(15 + \sqrt{2})\bigg]^2(\pi\sqrt{2}) \;=\;1197.029986\text{ ft}^2$
. . $\text{Therefore: }\;A \;=\;\frac{1197.029986}{9} \;\approx\;133\text{ yd}^2$
The answer at the back of the book is 61.061 square yards.
7. Hi Dug,
I also calculate lateral area is 133 sq yds
bjh
8. Thank you. I also double checked by calculating. The answer is 133yds.
The book must be wrong then. (typo error?)
a cylindrical tower 30 ft in diameter
Click on a term to search for related topics. | 2016-08-27T18:47:00 | {
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http://mathhelpforum.com/statistics/280327-where-did-i-go-wrong-2-pairs-poker-probability-question.html | # Thread: Where did I go wrong in 2 pairs poker probability question?
1. ## Where did I go wrong in 2 pairs poker probability question?
The question asks what the probability is of being dealt 2 pairs in a 5 card poker hand (from a shuffled deck of 52 cards).
So you need 1 pair and another numerically different pair (otherwise it will be four of a kind) and a final card which matches neither pair (otherwise it will be a full house).
The answer is approx. 0.0475 which can be found in plenty of places as the answer to this question with a web search and is explained here: https://math.stackexchange.com/a/1770961/484087
I got the answer wrong by exactly a factor of 2, i.e. 0.095. Clearly I have over-counted but I can not see what is wrong with the logic I used. Can someone explain where I went wrong please? Here's what I did:
Num. of combinations of 2 things and 2 things and 1 thing from 5 things: $\displaystyle \frac{5!}{2!·2!·1!}=\frac{120}{4}=30$
1st card can be anything: $\frac{52}{52}$
2nd card must match the 1st: $\frac{3}{51}$
3rd card must not match 1st or 2nd: $\frac{48}{50}$
4th card must match the 3rd: $\frac{3}{49}$
5th card must not match 1st, 2nd, 3rd, or 4th: $\frac{44}{48}$
$\displaystyle 30·\frac{52}{52}·\frac{3}{51}·\frac{48}{50}·\frac{ 3}{49}·\frac{44}{48}=\frac{396}{4165}≈0.095078$
2. ## Re: Where did I go wrong in 2 pairs poker probability question?
Originally Posted by mattstan
The question asks what the probability is of being dealt 2 pairs in a 5 card poker hand (from a shuffled deck of 52 cards). So you need 1 pair and another numerically different pair (otherwise it will be four of a kind) and a final card which matches neither pair (otherwise it will be a full house).
$\dbinom{13}{2}\dbinom{4}{2}\dbinom{4}{2}\dbinom{4 4}{1}=0.047539015606242496998799519807923169267707 082833$
SEE HERE
3. ## Re: Where did I go wrong in 2 pairs poker probability question?
Yes, I am aware of the correct answer, as I wrote and linked in my post.
My question was asking what the flaw was in the logic I used?
4. ## Re: Where did I go wrong in 2 pairs poker probability question?
Originally Posted by mattstan
Yes, I am aware of the correct answer.
My question was asking what the flaw was in the logic I used?
You are double counting the pairs that you will generate. For instance, you choose Aces for the first pair and Kings for the second pair, that is the same as if you choose Kings for the first pair and Aces for the second pair. So, instead of multiplying by 30 permutations, you should be multiplying by 15.
5. ## Re: Where did I go wrong in 2 pairs poker probability question?
Okay, I see. Thanks for pointing that out.
6. ## Re: Where did I go wrong in 2 pairs poker probability question?
Originally Posted by mattstan
Yes, I am aware of the correct answer, as I wrote and linked in my post.
My question was asking what the flaw was in the logic I used?
I really do not see a flaw,
$\dbinom{13}{2}\text{ pick two numbers }{2}\\\dbinom{4}{2}\text{ pick two cards of numbers}\\\dbinom{4}{2}\text{ pick two more cards of the other number}\\\dbinom{44}{1}\text{ pick one card from the numbers not used}$.
7. ## Re: Where did I go wrong in 2 pairs poker probability question?
Originally Posted by Plato
I really do not see a flaw,
$\dbinom{13}{2}\text{ pick two numbers }{2}\\\dbinom{4}{2}\text{ pick two cards of numbers}\\\dbinom{4}{2}\text{ pick two more cards of the other number}\\\dbinom{44}{1}\text{ pick one card from the numbers not used}$.
Look at the OP's logic in the first post, then look at my reply in reply #4 above. It explains the OP's logic and the reason it did not give the same answer.
8. ## Re: Where did I go wrong in 2 pairs poker probability question?
Originally Posted by mattstan
1st card can be anything: $\frac{52}{52}$
2nd card must match the 1st: $\frac{3}{51}$
2nd card can also be anything.....51/51
9. ## Re: Where did I go wrong in 2 pairs poker probability question?
Originally Posted by DenisB
2nd card can also be anything.....51/51
This is a far more complicated way of looking at the problem, as you now need to account for the possibility that the card was the same as the previous card and then for if it is different. While it is possible to approach the problem this way, the OP's logic is more concise.
10. ## Re: Where did I go wrong in 2 pairs poker probability question?
Agree. I was looking at it as in a "looper/simulation" computer program (dumb code). | 2018-09-20T12:33:10 | {
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https://math.stackexchange.com/questions/3051316/finding-lim-n-to-infty-1-1-2-1-3-dots-1-n-log-n | # Finding $\lim_{n \to \infty} 1+ 1/2 + 1/3 +\dots +1/n - \log n$. [duplicate]
$$x_n = 1 + 1/2 +\dots +1/n- \log n$$
Then -
$$1.$$ Is the sequence increasing?
$$2.$$ is the sequence convergent?
For $$(1)$$, $$\sum 1/n$$ is increasing and $$\log n$$ is also increasing. First few terms are increasing, but i don't know about later terms.
$$(2)$$ $$n^{th}$$ term of the sequence can be written as $$a_n = (\sum_{i=1}^{n}) - \log n$$
So, $$\lim_{n\to \infty} a_n = \lim_{n \to \infty} \sum 1/n -\lim_{n \to \infty} \log n$$
Neither first part nor second is convergent here. so i could not conclude anything.
How to solve?
## marked as duplicate by RRL real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 24 '18 at 17:54
• The sequence is found here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant – Nyssa Dec 24 '18 at 14:49
• Hint: consider that $\log n = \int_{1}^{n} \frac{dx}{x}$. You can obtain the relevant estimates via Riemann sums. – MathematicsStudent1122 Dec 24 '18 at 14:56
• You have to be careful: $\lim_n (a_n- b_n) = \lim_n a_n - \lim_n b_n$ is not true when $\lim_n a_n =\lim_n b_n=\infty$. – Mircea Dec 24 '18 at 16:19
• @Mircea , that's why I left the problem there. – Mathsaddict Dec 24 '18 at 16:29
This sequences converges to the Euler–Mascheroni constant. It’s very important in number theory.
Let $$x_n=-\log(n)+\sum_{k=1}^n \frac1k$$. Then, using $$\log(1+x)\ge \frac{x}{1+x}$$, we see that
\begin{align} x_{n+1}-x_n&=\frac1{n+1}-\log\left(1+\frac1n\right)\\\\ &\le \frac1{n+1}-\frac{1}{n+1}\\\\ &=0 \end{align}
and $$x_n$$ is decreasing.
Next, we can estimate the harmonic sum as $$\sum_{k=1}^n \frac1k\ge \frac12 \sum_{k=1}^{n-1}\left(\frac1k+\frac1{k+1}\right)$$, which represents the Trapezoidal Rule approximation of $$\int_1^n \frac1x\,dx$$.
Inasmuch as $$\frac1x$$ is convex, the trapezoidal rule approximation overestimates the integral of $$\frac1x$$ and we have
$$\sum_{k=1}^n\frac1k-\log(n)\ge \sum_{k=1}^n \frac1k -\log(n)-\frac12-\frac1{2n}\ge0$$
whence we see that
$$x_n\ge \frac12$$
Since $$x_n$$ is decreasing and bounded below by $$\frac12$$, the sequence $$x_n$$ converges. | 2019-11-20T06:26:31 | {
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http://math.stackexchange.com/questions/52330/solving-for-the-interval-of-time-during-which-the-height-of-a-thrown-ball-is-at | Solving for the interval of time during which the height of a thrown ball is at least “h” feet
In dealing with inequalities I've run into a certain peculiarity which I am currently unable to explain.
The example: Find the interval of time during which the ball is at least 32 feet above ground.
h = -16t^2 + 16t + 128 // Height of the ball in feet.
-16t^2 + 16t + 128 >= 32
-16t^2 + 16t + 96 >= 0
-16(t^2 - t - 6) >= 0
-16(t+2)(t-3) >= 0 // At this point everything is going as planned.
// Now I have a choice to make
(-16t - 32)(t - 3) >= 0 // This does not work OR
(t + 2)(-16t + 48) >= 0 // This does work
// If I choose option 1, the relational operators are incorrect.
-16t - 32 >= 0 and t - 3 >= 0
-16t >= 32 t >= 3
t <= -2
// If I choose option 2, the relational operators are correct.
t + 2 >= 0 and -16t + 48 >= 0
t >= -2 -16t >= -48
t <= 3
Now, when looking at a graph it becomes obvious that the ball is at least 32 feet above ground during the interval [0, 3] (assuming time is not negative). Therefore, option 2 provides the correct relations for t, while option 1 inverts the relations.
What I don't understand is why this is happening, since multiplication is an associative operator. Seems to me that it shouldn't matter whether the -16 is multiplied into the first factor or the second, and yet it does. I would love to know why, so that I might circumvent this issue the next time around.
-
I appreciate how you fully wrote out your work and showed your progress. Thank you. – mixedmath Jul 19 '11 at 4:04
It comes from your line where you require both
\begin{align} -16t - 32 &> 0 \\ t - 3 &> 0 \end{align}
As you saw, this is not sensible. But you might instead require both
\begin{align} -16t - 32 &< 0 \\ t - 3 &< 0 \end{align}
And this does lead to the correct answer. Recall that the final answer is when either of these two cases are true.
-
In other words, Bradford, if a product is positive, it might be that both factors are positive, but it might also be that both are negative. – Gerry Myerson Jul 19 '11 at 4:43
@mixedmath, thank you for getting me back on track. I had forgotten that this approach was not useful in solving inequalities (but rather was useful only as a check). – Bradford Fisher Jul 19 '11 at 5:04
@Gerry Myerson, thank you, I do see this now. I was so focused on the little details that I forgot to consider the equation as a whole. – Bradford Fisher Jul 19 '11 at 5:15
Absorbing the $-16$ into one of the terms was unnecessary, as we will see later. But let's start from your actual calculation.
Look at the version that "doesn't work", namely $$(-16t-32)(t-3) \ge 0.$$ I have rewritten $\lt$ as $\le$, and $\gt$ as $\ge$, because "at least $32$ feet" means $32$ or more.
The displayed inequality is true if (i) $-16t-32$ and $t-3$ are both $\ge 0$ OR (ii) $-16t-32$ and $t-3$ are both $\le 0$.
Note that $-16t-32 \ge 0$ iff $-16t \ge 32$ iff $t\le -2$.
The condition $t-3\ge 0$ can be rewritten as $t\ge 3$. So our analysis of case (i) shows that it holds iff $t\le-2$ and $t \ge 3$. But these are clearly incompatible, so case (i) cannot hold.
Or else, as you observed, it is implicit in the problem that $t \ge 0$, so $t \le -2$ is physically irrelevant.
For case (ii), look first at $-16t-32\le 0$. Rewrite this as $-16t \le 32$, and then $t\ge -2$. Rewrite the condition $t-3\le 0$ as $t \le 3$. So as far as the formula is concerned, everything is OK if $-2\le t\le 3$. But $t \ge 0$, so we conclude that the answer is $0 \le t \le 3$.
Thus we got a complete and correct analysis out of the "doesn't work."
However, let's start again from $-16(t+2)(t-3) \ge 0$.
This is equivalent to $(t+2)(t-3) \le 0$.
That is true if (i) $t +2 \le 0$ and $t-3\ge 0$ OR (ii) $t+2\ge 0$ and $t-3 \le 0$. (A product is $\le 0$ if one term is $\le 0$ and the other is $\ge 0$.)
Case (i) is physically irrelevant. Anyway, it yields the incompatible $t \le -2$ and $t \ge 3$.
Case (ii) yields $t \ge -2$ and $t \le 3$. For physical reasons this should be corrected to $0 \le t \le 3$.
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from your inequality -16*t^2+16*t+128>32 there is one question if at least 32 does it mean that not only >32 but also >=32? so rewrite your equation like this
-16*t^2+16*t+128>=32
-16*t^2+16*t+96>=0 now dive it by -16 and change > by < so we will have t^2-t-6<=0 solution is t1=3 t2=-2 ,but because feet can't be negative answer will be [0 3]
- | 2015-05-24T19:59:54 | {
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https://math.stackexchange.com/questions/2586560/parametrization-difference-in-line-integrals | # Parametrization Difference in Line Integrals
I have two questions about line integrals.
Find the integral below along the line of intersection of the two planes $x-y+z=0$ and $x+y+2z=0$ from the origin to the point $(3,1,-2)$ $$\int_{\mathcal C} x^2ds$$
Assistant has said "you should give priority to point if question has and parametrize wrt the point" and she has done this :
$r(t) = 3t\hat i+t\hat j -2t\hat k$ , $( 0 \le t \le1)$ (I didn't understand why t is between 0 and 1 here)
Find the integral below along the first octant part $\mathcal C$ of the curve of intersection of $z=2-x^2-2y^2$ and $z=x^2$ between $(0,1,0)$ and $(1,0,1)$ $$\int_{\mathcal C} xyds$$
$x=t,y=\sqrt{1-t^2},z=t^2$ , $( 0 \le t \le1)$ (I didn't understand why t is between 0 and 1 again)
For 2nd question she has parametrized wrt equations although question has point. Why we did not use points for parametrization? Could someone explain difference in two questions please in the easiest way? Thanks a lot
• I've tried to clarify your doubts, even if I'm not completely sure what exactely is not clear to you.
– user
Dec 31 '17 at 13:16
• @gimusi parametrization is not clear for me. For first question she multiplied point with t directly but for second one she parametrized wrt curves. I parametrized 1st one wrt curves I have obtain same parametrization and I couldn’t be sure it is true or only a coincidence? Dec 31 '17 at 13:20
• I understand your doubts. Anyway for the line segment the parametrization is very easy to find. For the curve it is more difficult and you need to pay more attention.
– user
Dec 31 '17 at 13:37
• @gimusi thanks for answer Jan 2 '18 at 10:41
• You are welcome, if you are ok you can accept the answer and set it as solved, bye! math.meta.stackexchange.com/questions/3286/…
– user
Jan 2 '18 at 14:07
For the first:
The intersection of the two planes is a line segment from the origin to the point (3,1,-2). The parametric equation of the line segment is indeed:
$$r(t) = 3t\hat i+t\hat j -2t\hat k$$
• t=0 corresponds to the origin
• t=1 correspond to the point (3,1,-2)
In general given two points P and Q the parametrization for a line segment PQ is obtained by:
$$r(t)=P+t(Q-P)$$
Note that for $r(0)=P$ and $r(1)=Q$.
For the second:
Here you have a curve but the concept is the same, t=0 correspond to the strarting point and t=1 to the final one.
The parametrization can be obtained noting that a generic point in the curve must satisfy simultaneously:
$z=2-x^2-2y^2$ and $z=x^2$
thus, assuming x=t the coordinates of the point are
$x=t$
$z=x^2\implies z=t^2$
$z=2-x^2-2y^2\implies t^2=2-t^2-2y^2\implies y^2=1-t^2\implies y=\sqrt{1-t^2}$
Note that for y we have choose the positive value since it is requested that y goes from 1 to 0.
Note also that in this case the fact that t varies from 0 to 1 depends from the choice x=t with x varying from 0 to 1.
NOTE In both cases you are parametrizing the line (all the points of the lines) by a single real parameter t which varies form o to 1.
Hint: for the first one for $t=0\,$ you have the initial (which is $(0, 0, 0)$) and for $t=1$ you have the terminal point (which is $(3, 1, -2)$) of the line where you take the integral. Similarly, you do the same thing for the second integral.
• Thanks I have understood intervals but I still have confusion about parametrization. Do I have to parametrize wrt point? I have parametrized wrt curves in 1st question I have obtained same r? Dec 31 '17 at 13:16
• @esrabasar: you can parametrize it in many ways and the range of $t$ changes according to your parametrization but it has to involve the initial and final points when put $t$ values. Dec 31 '17 at 13:46
"Points" were used for the parameterization. The question asked for the integral, along the given line, from (0, 0, 0) to (3, 1, -2). The reason "t" is between 0 and 1 in the parameterization r(t)=3ti+ tj- 2tk is that then r(0)= 0 and r(1)= 3i+ j- 2k- in other words, (0, 0, 0) and (3, 1, -2), the two endpoints of the line. As far as finding the parameterization itself, we have the two planes given by x- y+ z= 0 and x+ y+ 2z= 0. Any (x, y, z) on the line of intersection of the two planes must satisfy both. Subtracting the first equation from the second, we have 2y+ z= 0, z= -2y. Setting z= -2y in either equation, x= 3t. Using y itself "as parameter", that is, y= t, x= 3t and z= -2t. r(t)= xi+ yj+ kz= 3ti+ ti- 2tk.
For the "curve of intersection" of $z= 2- x^2- 2y^2$ and $z= x^2$, we have $2- x^2- 2y^2= x^2$ or $2x^2= 2- 2y^2$ so $x^2+ y^2= 1$, the circle with center at (0, 0) and radius 1. Since we only want the first quadrant, x and y are both positive and we go from (1, 0, 1) to (0, 1, 0) so $y= \sqrt{1- x^2}$. Taking x as parameter, x= t, $y= \sqrt{1- x^2}$, and $z= t^2$. When t= 0, that is x= 0, y= 1, z= 0 or (0, 1, 0). When t= 1, it is x= 1, y= 0, z= 1 or (1, 0, 1). t always lies between the values that give the beginning and ending points for the given path.
By the way, since it is also true that $cos^2(\theta)+ sin^2(\theta)= 1$ for all $\theta$, another "parameterization" for that curve would be $x= cos(\theta)$, $y= sin(\theta)$, $z= cos^2(\theta)$. And then the parameter, $\theta$ must be $0\le \theta\le \frac{\pi}{2}$ since, at $\theta= 0$, $x= cos(\theta)= 1$, $y= sin(\theta)= 0$, and $z= cos^2(\theta)= 1$, the endpoint (1, 0, 1), while at $\theta= \pi/2$., $x= cos(\theta)= 0$, $y= sin(\theta)= 1$, and $z= cos^2(\theta)= 0$, the other end point. | 2022-01-24T05:08:46 | {
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http://math.stackexchange.com/questions/185794/rearranging-the-spectral-theorem | # Rearranging the spectral theorem
The spectral theorem for selfadjoint compact operators $L$ with infinite range says that $$Lx=\sum_{k=1}^{\infty} \alpha_k \langle x,f_k \rangle f_k,$$ where the $f_k$'s form an orthonormal system and the $\alpha_k$'s are real nonzero eigenvalues, such that they tend to zero. Now my question is, if we rearrange this sum arbitrary, will it still converge ?
My hunch would be "yes", since the proof of this theorem seems to work for any arrangement, but I'm feeling uneasy accepting that...
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Can this also be tagged functional-analysis? – Matt N. Aug 23 '12 at 12:38
It's a standard fact from hilbertian analysis that if $(x_i)_{i\in I }$ is an orthogonal (not necessarily orthonormal) system, then $(x_i)_{i\in I}$ is a summable family iff $(||x_i||^2)_{i \in I}$ is summable.
An application of Bessel's inequality and the previous fact shows that $(a_k \langle x, f_k \rangle f_k)_k$ is summable, and hence you can rearrange arbitrary this series and still get a convergent series with the same sum.
Indeed, since $(\alpha_k \langle x, f_k \rangle f_k)_{k \ge 0}$ is an orthogonal system, it's summable if and only if $\sum_{k\ge 0} \| \alpha_k \langle x, f_k \rangle f_k \|^2$ is finite, that is $\sum_{k\ge 0} | \alpha_k|^2 | \langle x, f_k \rangle |^2 \| f_k \|^2$ is finite.
But $(\alpha_k)_k$ is a convergent, and hence bounded, sequence, and $\|f_k \| = 1$, so we are reduced to prove that $\sum_{k \ge 0} | \langle x, f_k \rangle |^2$ is finite; but it's exactly the content of Bessel's inequality.
EDIT : A few facts about summable families
The precise definition of a summable family is as follows : a family $(x_i)_{i \in I}$ (where the set $I$ can be uncountable) of vectors in a Banach space $E$ is summable, with sum $x$, if for every $\epsilon > 0$, there exists a finite set $J \subset I$ such that for every finite set $K \subset I$ with $J \subset K$ one has $\| \sum_{i \in K} x_i - x \| \le \epsilon$. This can be equivalently characterised by the convergence of an appropriated net. If a family is summable, its sum is unique. We donote it by $x = \sum_{i \in I} x_i$.
This notion is invariant under rearrangement of indices : if $\sigma : J \to I$ is a bijection, then $(x_i)_{i \in I}$ is summable iff $(x_{\sigma(j)})_{j \in J}$ is summable, and sums are equal.
If the family of real numbers $(\| x_i \|)_{i \in I}$ is summable, then $(x_i)_{i \in I}$ is itself summable (provided $E$ is a Banach space, completeness is the key point here). The converse is not true in general, except if $E$ is finite dimensional.
In the special case where $I = \mathbb{N}$, summability implies the convergence of the series associated, but the converse is not true, even if $E$ is finite dimensionnal. Indeed, if $E$ is finite dimensional (which covers the case of a family of real numbers), by what was said previously, a family $(x_n)_{n \in \mathbb{N}}$ is summable iff the series $\sum_{n = 0}^{+ \infty} x_n$ is absolutely convergent, that is $\sum_{n = 0}^{+ \infty} ||x_n|| < \infty$.
You can find further details and proofs in "General Topology" by N. Bourbaki. They cover the wider case of summable families in topological groups, but you can easily specialize to Banach space if you are not interested in such a generalization.
Now, I'll turn to summable families in Hilbert space.
Bessel's inequality asserts that if $(e_i)_{i \in I}$ is an orthonormal system of an Hilbert space $H$, then for every $x \in H$, the family $( |\langle x, e_i \rangle|^2)_{i \in I}$ of real numbers is summable, and its sum is less or equal to $\|x \|^2$.
Parseval identity says that if $(e_i)_{i \in I}$ is a complete orthonormal system in $H$, then for every $x \in H$, the family $( \langle x, e_i \rangle e_i)_{i \in I}$ is summable, and its sum is exactly $x$. Moreover, Bessel's inequality turns out to be actually an equality.
For all of this, and for the fact recalled at the beginning of my answer, you can take a look at "The Elements of Operator Theory" by C.S. Kubrusly.
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Before I accept your answer, could you please give me a reference for the fact stated in the first paragraph ? (do you mean with "summable" that $\sum_{i\in I} ||x_i||^2$ converges?) – user38525 Aug 23 '12 at 10:26
I provided some definitions and results about summable family. I also improve the very first statement in order to cover correctly the case you're considering. – Ahriman Aug 23 '12 at 12:29
Wow, that was very detailed. But I'm not sure how you arrived at proving that the sequence of the RHS is summable. The Bessel inequality tells me that $$\sum_k | \langle x,f_k \rangle |^2 \leq ||x||^2.$$ On the other hand I know that $$\sum_k \alpha_k \langle x,f_k \rangle f_k$$ is summable iff $$\sum_k ||\alpha_k \langle x,f_k \rangle f_k||=\sum_k |\alpha_k| | \langle x,f_k \rangle |$$ is summable. But how do I combine these two to get what I want ?[...] – user38525 Aug 23 '12 at 15:14
Somehow the notion of convergence needs to come into play since I need to deduce the summability of $$\alpha_k \langle x,f_k \rangle f_k$$ from the convergence of $$\sum_{k=1}^{\infty} \alpha_k \langle x,f_k \rangle f_k$$. I have denoted a convergent sum by $$\sum_{k=1}^{\infty}$$ and an unconditional convergent sum (since I think that that is what I get from a summable sequence) by $$\sum_k$$. Note that originally I only have $$\sum_{k=1}^{\infty} \alpha_k \langle x,f_k \rangle f_k$$. – user38525 Aug 23 '12 at 15:17
Are you also sure, that "$(x_i)_{i\in I}$ is a summable family iff $(||x_i||^2)_{i \in I}$ is summable" is true ? I know of a theorem that says, that for an ONS $(x_k)_k$ we have $\sum_{k=1}^{\infty} \beta_k x_k$ is convergent sum iff $\sum_{k=1}^{\infty} |\beta_k|^2 x$ is convergent". This is related, but not the same. – user38525 Aug 23 '12 at 15:21 | 2014-08-20T13:20:53 | {
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http://pdcv.clandiw.it/euler-method-pdf.html | Euler's method relies on the fact that close to a point, a function and its tangent have nearly the same value. Using Euler’s method. 3 Buckling Load Factor The buckling load factor (BLF) is an indicator of the factor of safety against buckling or the ratio of the buckling. Then define a recursive sequence as. If people don't need super accurate results but just need to be able to compare two results, Euler's method might be sufficient. It illustrates Euler’s method applied to the differential equation y0 = f(x,y), where y0 = dy/dx. one-step methods including the explicit and implicit Euler methods, the trapezium rule method, and Runge–Kutta methods. We use the method of reduction of order. He not only made formative contributions to the subjects of geometry, calculus, mechanics, and number theory but also developed methods for solving problems in astronomy and demonstrated practical applications of mathematics. Euler’s method is based on approximating the graph of a solut ion y (x) with a sequence of tangent line approximations computed sequentially, in “st eps”. ) Show the work that leads to your answer. Local Truncation Error for the Euler Method. After we set up the basic. (b) Use Euler's method with step size ¢t = 0:5 to approximate this solution, and check how close the approximate solution is to the real solution when t = 2, t = 4, and t = 6. Clearly, the description of the problem implies that the interval we'll be finding a solution on is [0,1]. n+1 = yn + mnh, mn = F(xn, yn). %This script implements Euler's method %for Example 2 in Sec 2. (1) We know that the left endpoint approximation is a poor way to estimate integrals and that the Trapezoidal Rule is better. Doing this requires solving this equation for k, which amounts to a root nding problem if f is nonlinear, but we know how to. We’ll use Euler's method to perform the numerical integration. 2 Modified Euler's Method 7. This is then applied to calculate certain integrals involving trigonometric. The linearization is done in such a way that the correct. Numerical solution of the Euler equations by finite volume methods using Runge Kutta time stepping schemes. org Método de Euler; Usage on he. Comparison of VIM and PPM with Runge-Kutta 4th leads to highly accurate solutions. Boujot, Springer. (d) Let ygx= ( ) be another solution to the differential equation with the initial condition g()0=k, where k is a constant. With Euler’s method, this region is the set of all complex numbers z = h for which j1 + zj<1 or equivalently, jz ( 1)j<1 This is a circle of radius one in the complex plane, centered at the complex number 1 + 0 i. The first-order Euler's methods are the least accurate. Predictor-corrector and multipoint methods Objective: to combine the simplicity of explicit schemes and robustness of implicit ones in the framework of a fractional-step algorithm, e. Consider: y' (x) = f(x, y) ; y (x 0) = y0 (1) Let: xi = x0 + i h ; i = 0, 1,-. Some other topics covered in this tutorial are: Making a plot of mass position vs. Euler’s Method A Numerical Technique for Building a Solution to a DE or system of DE’s This is the slope field for Slope Fields We get an approx. Then subtract the Euler formula from this to obtain '(tn+1) yn+1 = '(tn) 2yn +h(f(tn;'(tn)) f. 12 Differential Equations and Euler’s Method Summary Many differential equations, such as and cannot be solved by traditional pencil and paper methods. Give your answer correct to 4 decimal places. After reading this chapter, you should be able to: 1. Euler’s method and numerical integration Jean-Luc Bouchot jean-luc. Euler’s Method, Taylor Series Method, Runge Kutta Methods, Multi-Step Methods and Stability. EULER'S METHOD 5 2. Euler Method : In mathematics and computational science, the Euler method (also called forward Euler method) is a first-order numerical procedurefor solving ordinary differential equations (ODEs) with a given. Euler’s method will be. Finite Di erence Jacobian For any implicit method like. THE HAMILTONIAN METHOD. Here is the table for. 1): its m stage values Y n,i are given by the solution of the nonlinear algebraic systems (1. Euler’s Method Here’s how it works. Note: 1 lecture, can safely be skipped, §2. Repre-7 sentative texts addressing Euler™s method for calculus [4], di⁄erential equations. Euler's method is commonly used in projectile motion including drag, especially to compute the drag force (and thus the drag coefficient) as a function of velocity from experimental data. The differential equation given tells us the formula for f(x, y) required by the Euler Method, namely: f(x, y) = x + 2y. 5 At the initial time, t 0, the salt concentration in the tank is 50 g/L. In fact, most differential equa-tions that arise in real life applications are solved on computers using approxi-mation techniques. Therefore, k 2 corresponds to the slope of the solution one would get by. It is a quantity with the dimensions of (Energy)£(Time). Now Euler repeats it for natural logarithms. Euler's Method We have seen how to use a direction field to obtain qualitative information about the solutions to a differential equation. However, the accuracyfactor persuades scholar to use another complex method to replace Euler method [4], [5]. This paper, called 'Solutio problematis ad geometriam situs pertinentis,' was later published in 1741 [Hopkins, 2. Euler's Method, Taylor Series Method, Runge Kutta Methods, Multi-Step Methods and Stability. Runge-Kutta methods d. Our approach is to focus on a small number of methods and treat them in depth. Use the trapezoidal method with 100 steps to solve the same problem. The simple Euler method: yn = yn 1 +hf(yn 1); h = xn xn 1 can be made more accurate by using either the mid-point or the trapezoidal rule quadrature formula: yn = yn 1 +hf yn 1 + 1 2hf(yn 1): yn = yn 1 + 1 2hf(yn 1)+ 1 2hf yn 1 +hf(yn 1): Runge-Kutta methods for ordinary differential equations - p. If a numerical method has no restrictions on in order to have y n!0 as n !1, we say the numerical method is A-stable. Its stability properties however can be much better than the explicit Euler method in the case when r<<0. It is basic explicit method for numerical integration of ordinary differential equations. EULER'S FORMULA FOR COMPLEX EXPONENTIALS According to Euler, we should regard the complex exponential eit as related to the trigonometric functions cos(t) and sin(t) via the following inspired definition:eit = cos t+i sin t where as usual in complex numbers i2 = ¡1: (1) The justification of this notation is based on the formal derivative of both sides,. 02 Euler's method Chapter 08. The k 1 and k 2 are known as stages of the Runge-Kutta method. %This script implements Euler's method %for Example 2 in Sec 2. Clearly, the description of the problem implies that the interval we'll be finding a solution on is [0,1]. 5', or at t114 using 0. 2nd printing 1996. This is a differential equation that is not separable and not linear, so we don’t yet have a method to solve it. Euler entered university at the age of 13 at the University of Basel. Later, above all in the environment of the liberal deïst, King Frederick II (1712–1786), in his Berlin period (1741–1766), Leonhard Euler defended the Christian faith against freethinkers and atheists. Euler Method & Energy Consideration Simple Harmonic Motion Basic equations: F Magnus= K w V C D---> K is the Magnus Coefficient---> w is the spin frequency measured in radians---> V is the velocity of the ball in m/s---> C Dis the drag coefficient F_mag drag velocity Spinning Non-spinning. form can be derived using Lagrange or Newton-Euler methods if constraints are imposed when using the Newton-Euler approach. This method is explicit. Clearly, in this example the Improved Euler method is much more accurate than the Euler method: about 18 times more accurate at. Euler’s Method y’(t) = f(t,y), y(a) = w 0 w k+1 = w k + h f(t k,w k) slope is f(t. • Theoretically the same equivalence can be shown between equations derived from other formulations (e. EULER'S METHOD: More formally, given dy dt = f(t;y) with y(t 0) = y 0 we approximate the path of the solution by: 1. 1): its m stage values Y n,i are given by the solution of the nonlinear algebraic systems (1. Adaptive Euler-Maruyama Method for SDEs with Non-globally Lipschitz Drift: Part I, Finite Time Interval @inproceedings{Fang2016AdaptiveEM, title={Adaptive Euler-Maruyama Method for SDEs with Non-globally Lipschitz Drift: Part I, Finite Time Interval}, author={Wei Fang and Michael B. 5}, {y, 0, 2. LeonhardEuler: HisLife,theMan,andHisWorks is to bring across some glimpses of Euler's incredibly voluminous and diverse work, which today fills 74 massive volumes of the Opera omnia which certainly is the best method ofmaking happy progress in the mathematical sciences. 13 Euler’s Method On the other hand, our Euler method reads xn+1 = xn +h ( xn) = (1 h)xn: (15) Clearly, if h > 1, x(tn) will oscillate between negative and positive numbers and grow without bounds in magnitude as tn increases. 0: n:=30: h:=(tf-t0)/n: tk:=evalf(t0): yk:=evalf(y0): expt := array(0. However, the accuracyfactor persuades scholar to use another complex method to replace Euler method [4], [5]. A second order Cauchy-Euler equation is of the form a 2x 2d 2y dx2 +a 1x dy dx +a 0y=g(x). The problem with this is that these are the exceptions rather than the rule. 1 Explicit (Forward) We can set up the iterative formula: w n+1. determination method. Here, a short and simple algorithm and flowchart for Euler's method has been presented, which can be used to write program for the method in any high level. 2 Graphical Illustration of the Explicit Euler Method Given the solution y (t n) at some time n, the differential equation ˙ = f t,y) tells us "in which direction to continue". y' = x - xy y(1) = 0 y(1. Euler’s method always needs a step size, which is called h. 1 Adams-Moulton Method 7. Usually we can only estimate solutions to di erential equations using numerical methods. Speci cally, those ode's for which initial conditions are known. Exercise 2. publication date. In the image to the right, the blue circle is being approximated by the red line segments. Euler's Method (Following The Arrows) Euler's method makes precise the idea of following the arrows in the direction eld to get an ap-proximate solution to a di erential equation of the form y0= F(x;y) satisfying the initial condition y(x 0) = y 0. E005 The Solution of the Problem of Reciprocal Trajectories. the initial condition cannot be satisfied exactly), then use as the initial condtion the following expression. This method is implicit. In some cases, it's not possible to write down an equation for a curve, but we can still find approximate coordinates for points along the curve by using. Euler's method is the simplest approach to approximating a solution to a di erential equation. Runge-Kutta method (Order 4) for solving ODE using MATLAB MATLAB Program: % Runge-Kutta(Order 4) Algorithm % Approximate the solution to the initial-value problem % dy/dt=y-t^2+1 MATLAB 2019 Free Download. The purpose of this paper is to show the details of implementing of Euler's method and made comparison between modify Euler's and exact value by. Differential Equations : Euler Method : Matlab Program. Doing this produces the Modi ed (or Improved) Euler method represented by the following equations: k 1 = hf(t i;y i) k 2 = hf(t i+ h;y i+ k 1) y i+1 = y. Assuming you will pay attention to all FDM steps, let’s focus on the differences of the current method with the forward Euler. The structure of a dendrimer exhibits a large number of internal and superficial cavities, which can be exploited, to capture and deliver small organic molecules, enabling their use in drug delivery. Euler's method is a numerical method to solve first order first degree differential equation with a given initial value. " Note that this function uses an exact increment h rather than converting it explicitly to numeric form using Mathematica command N. Computing Euler angles from a rotation matrix. Euler's Method Tutorial A method of solving ordinary differential equations using Microsoft Excel. It holds when the function is analytic in the integration region In certain cases, the last term tends to 0 as , and an infinite series can then be obtained for. In fact, most differential equa-tions that arise in real life applications are solved on computers using approxi-mation techniques. Factorization of a quartic as a product of two real quadratics 7 IIB. I am new in Matlab but I have to submit. We will call the distance between the steps h and the various points. Factorization of a quartic as a product of two real quadratics 7 IIB. edu is a platform for academics to share research papers. Repre-7 sentative texts addressing Euler™s method for calculus [4], di⁄erential equations. The Euler Archive is an online resource for Leonhard Euler's original works and modern Euler scholarship. In this simple differential equation, the function is defined by (,) =. 16|Calculus of Variations 3 In all of these cases the output of the integral depends on the path taken. 1] from y' = x + y + xy, y(0) = 1 with h = 0. 1): its m stage values Y n,i are given by the solution of the nonlinear algebraic systems (1. First, we will review some basic concepts of numerical approximations and then introduce Euler’s method, the simplest method. Included in the lesson are Guided Notes and examples which incorporate students using a table to determine values. Euler Method : In mathematics and computational science, the Euler method (also called forward Euler method) is a first-order numerical procedurefor solving ordinary differential equations (ODEs) with a given. y' = x - xy y(1) = 0 y(1. Corrector un+1 = un + 1 2[f(tn,un)+f(tn+1,u˜n+1)]∆t Crank-Nicolson or un+1 = un +f(tn+1,u˜n+1)∆t. The simplest numerical method for solving Equation \ref{eq:3. Section 2-9 : Euler's Method Up to this point practically every differential equation that we've been presented with could be solved. Calculate the slope at the end of this step. 0 f := (x, y) → cos(x + y) Initialize arrays > x[0] := x0; x 0:= 0. The Runge-Kutta method is a far better method to use than the Euler or Improved Euler method in terms of computational resources and accuracy. The final Sec. Finite element approximation of initial boundary value problems. Solution: With a step size of ∆ x = 0. The Euler method is important in concept for it points the way of solving ODE by marching a small step at a time on the right-hand-side to approximate the "derivative" on the left-hand-side. Miller (1991) (note: our library doesn't seem to get this, contact the Prof. A very small step size is required for any meaningful result. Usually we can only estimate solutions to di erential equations using numerical methods. and rearrange to around with step. The forward Euler’s method is one such numerical method and is explicit. He addresses both this specific problem, as well as a general solution with any number of landmasses and any number of bridges. numerical method- euler. The LS-DYNA ALE/FSI package can accurately model the dynamic response of the structure under blast loading. The function y ( t ) has the following Taylor series expansion of order n at t = t i +1 :. • Each edge must share two and only two faces. and the initial condition tells us the values of the coordinates of our starting point: x o = 0. The Cauchy-Euler equation is important in the theory of linear di er-ential equations because it has direct application to Fourier's method in the study of partial di erential equations. In order to compare the methods of Euler and Lagrange, we supply references to works whose authors apply each of the two methods to the same subjects. The inverse trigonometric functions: arcsin and arccos The arcsine function is the solution to the equation: z = sinw = eiw − e−iw. Consider the ode dy dx = f0(x) (1) which has solution y= f(x) and reference. Numerical methods in mathematical finance Winter term 2012/13 Model problem Geometric Brownian motion dX(t) = rX(t)dt +σX(t)dW(t) Exact solution X(t) = X 0 exp r − σ2 2 t +σW(t) Euler-Maruyama method X n+1 = X n +τrX n +σX n∆W n Tobias Jahnke Karlsruher Institute of Technology. Answered: ahmed abdelmageed on 4 May 2020 at 4:25. We can obtain a more accurate method by adjusting the direction of the step according to the slope field seen along an Euler step. To do this, we'll move our work to a spreadsheet. Linear multi-step methods: consistency, zero-. Differential Equations : Euler Method : Matlab Program. The linearization is done in such a way that the correct. \Chemistry" tells us that dx dt = K amount of A amount of B = Kx(1 x): K is a proportionality constant, which depends on the particular kind of molecules A and B in this reaction. Draw a line segment with the indicated slope between x = 0 and x = 0:25. SOLVING SECOND ORDER, HOMOGENEOUS EULER-CAUCHY EQUATIONS: THE CASE OF THE REPEATED ROOT LANCE DRAGER In this note, we show how to find the second basic solution for a second order Euler-Cauchy equation in the case of a repeated root of the characteristic equation. (c) Use Euler’s method with step size ¢t = 0:1 to approximate this solution, and check how close the approximate solution is to the real solution when t = 2, t = 4. We'll just deal with one coordinate, x, for now. 4 1 The collocation method for ODEs: an introduction We see that the equations (1. We used di erent numerical methods for determining the numerical solutions of Cauchy-problem. Explicit and Implicit Methods in Solving Differential Equations A differential equation is also considered an ordinary differential equation (ODE) if the unknown function depends only on one independent variable. g(0)k , where k is a constant. See how (and why) it works. Euler (0, 30, 0); } } public static Quaternion Euler ( Vector3 euler ); Returns a rotation that rotates z degrees around the z axis, x degrees around the x axis, and y degrees around the y axis. As Euler method is very easy to calculate even for many players while cost gap method becomes very complicated as the number of the players increases we examine further the properties of Euler method. 18 Use Euler’s method to approximate the solution to dy dx = y −y2 = y(1 −y) with initial condition y(0) = 2. In this scheme, since, the starting point of each sub-interval is used to find the slope of the solution curve, the solution would be correct only if the function is. You will need to modify the algorithm in EULER. this is the video about euler's method to solve ordinary differential equation in python. The Euler equation provides an invariant relationship between the vertices, edges, and surfaces of a simple polyhedral object. y(1:3) = 9 (0:3)(0:5) = 8:85 2. The method is named after Leonhard Euler who described it in 1768. Methods of higher orders of approximation 4. org are unblocked. The Euler method is an example of an explicit method. 78) discretized by means of the backward Euler method writes. Therefore, k 2 corresponds to the slope of the solution one would get by. Figure 12‐3 Restraints have a large influence on the critical buckling load 12. Because each iteration of the forward Euler method depends only on past quantities, it is termed an explicit method. The data is obtained from two sources which are from Valappil et. This method was originally devised by Euler and is called, oddly enough, Euler’s Method. 3, 2012 • Many examples here are taken from the textbook. If we can get a short list which contains all solutions, we can then test out each one and throw out the invalid ones. Solution of quartic equations 5 IIA. According to the Cauchy integral formula, in this integral only the term. The method is derived from the Taylor Series expansion of the function y ( t ). and rearrange to around with step. Euler meets Glenn? Rudy Horne, a mathematician at Morehouse College in Atlanta, was the math advisor to the movie, and it was he who suggested Euler's Method for the key blackboard scene. r (h 3) i =O Example 1. Euler's Method Introduction 3. However, the radioactive decay serves as a good flrst example since it illustrates some of the techniques, and the pitfalls, in computational physics. Leonhard Euler, Swiss mathematician and physicist Definition from Wiktionary, the free dictionary. 10 in the text lists TI-85 and BASIC programs implementing the improved Euler method to approximate the solution of the initial value problem dy x y dx =+, y(0) 1= (1) considered in Example 2 of Section 2. Speci cally, those ode’s for which initial conditions are known. E004 Euler's essay on the location, height, and number of the masts on ships to maximize the speed. This is then applied to calculate certain integrals involving trigonometric. The solutions of the Euler-Lagrange equation (2. f denotes the function to be solved, t init is the initial value of time t, y init is the initial value of y, h is the step length, and n is the number of iterations. (b)Solve the differential equation 2 dy x dx with the initia l condition y 05, and use your solution to find y 0. By Itflo™s lemma lnS t follows the process dlnS t = r 1 2 ˙2 dt+˙dW t: (6) Euler discretization via Equation (3) produces lnS t+dt = lnS. The equation yi 1 yi hfti, yi is called the difference equation associated with Euler’s Method. We will solve the Euler equations using a high-order Godunov method—a finite volume method whereby the fluxes through the interfaces are computed by solving the Riemann problem for our system. Thus, Euler's method gives the estimate y(1. It is an explicit method for solving initial value problems (IVPs), as described in the wikipedia page. 5, Editor Soubbaramayer and J. Make clear which curve corresponds to each step-size (Hint: use differ-ent line styles/colors). Given the complexity of problems in the engineering field, new tools have become essential for solving them in academic society, so computer modeling and simulation through software has been one of the main alternatives found by researchers in the. observation method. Secularity band differences in the results of some numerical methods with the standard Euler’s method of order three and four was examined. pdf para más tarde. time and comparing it to the analytical solution; Separating out the Euler's method in a MATLAB "function" Collecting multiple parameters in one box using "structures". And not only actually is this one a good way of approximating what the solution to this or any differential equation is, but actually for this differential equation in particular you can actually even use this to find E with more and more and more precision. 01 also estimate how small h would need to obtain four decimal accuracy. ! It is based on approximating the Euler equation by a. (2017) Augmented Lagrangian method for an Euler's elastica based segmentation model that promotes convex contours. Any gure without a caption will not be graded. Comparison of Euler and Runge-Kutta 2nd Order Methods Table 2. One of the advantages of studying it as presented. The result is in the form of a table of {t, Y} pairs. Exercise 2. 19}, we also have the option of using variation of parameters and then. Plot the approximation along with the exact solution. You may wish to compute the exact. 1 Euler’s Method In this section we will look at the simplest method for solving first order equations, Euler’s Method. Its stability properties however can be much better than the explicit Euler method in the case when r<<0. It is one of the best methods to find the numerical solution of ordinary differential equation. For 1≤ n≤ N y(0) = y0, dy dt = F(t, y), (Truth) y0 = y0, yn+1 −yn Δt = F(tn,yn). Comparison of Euler and Runge-Kutta 2nd Order Methods Table 2. 5 Euler’s method sec:Euler For “generic” ODEs, there is no hope of writing down an explicit formula for a typical solution. Use step size h = 0. 4 Given any function x(t), we can produce the quantity S. Example Use Euler’s Method to approximate the solution of the initial-value problem: y′ y −t2 1, 0 ≤t ≤2, y 0 0. This method is called the forward Euler method. edu 8 December 2005 In a 1670 letter to Christian Huygens (1629 - 1695), the celebrated philosopher and. EULER’SMETHOD 2 For x 0:002,thelinearapproximationgives (y ˇy01) )(x (0:75 0:002) 0:0015: Theactualvalueofy(1:002) inthis Then exampleisabout0:501507,sothelinear. As the name implies, Modified Euler's Method is a modification of the original Euler's method. Their definitions are as shown in the following graph- The first Euler Angle α is measured by a counterclockwise rotation about the z axis of the x axis. Then, plot (See the Excel tool "Scatter Plots", available on our course Excel webpage, to see how to do this. large number of numerical methods are built into Maple. The total energy in the simple pendulum system should remain constant (since it is a conservative system). \Chemistry" tells us that dx dt = K amount of A amount of B = Kx(1 x): K is a proportionality constant, which depends on the particular kind of molecules A and B in this reaction. pdf; Examples: Euler and Runge-Kutta methods for orbit problem: main class for all methods: orbit. , we will march forward by just one x). Euler's Method We have seen how to use a direction field to obtain qualitative information about the solutions to a differential equation. fluid mechanics pioneered by Leonhard Euler and the father and son Johann and Daniel Bernoulli. Euler's Method - a numerical solution for Differential Equations Why numerical solutions? For many of the differential equations we need to solve in the real world, there is no "nice" algebraic solution. and the initial condition tells us the values of the coordinates of our starting point: x o = 0. I looked it up online and followed various directions, but still haven't been able to get an answer. An Introduction to the Incompressible Euler Equations John K. Euler’s Method is a step-based method for approximating the solution to an initial value problem of the following type. The finite-volume update for our system appears as: Un+1 i=U n + ∆t ∆x Fn+1/2 i−1/2 −F n+1/2 i+1/2 (14) M. This block uses the Forward Euler integration method. Runge-Kutta method (Order 4) for solving ODE using MATLAB MATLAB Program: % Runge-Kutta(Order 4) Algorithm % Approximate the solution to the initial-value problem % dy/dt=y-t^2+1 MATLAB 2019 Free Download. Malte [SCHI11] used a mean of multiple utational (MMC) model with dual-quaternions to model bodies. Thus in the Predictor-Corrector method for each step the predicted value of is calculated first using Euler's method and then the slopes at the points and is calculated and the arithmetic average of these slopes are added to to calculate the corrected value of. Calculates the solution y=f(x) of the ordinary differential equation y'=F(x,y) using Euler's method. Now Euler repeats it for natural logarithms. Euler_Method. Euler’s theorem is a nice result that is easy to investigate with simple models from Euclidean ge- ometry, although it is really a topological theorem. Effect of Step Size. EULER’S METHOD (BC TOPIC ONLY) - Differential Equations - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC - includes the basic information about the AP Calculus test that you need to know - provides reviews and strategies for answering the different kinds of multiple-choice and free-response questions you will encounter on the AP exam. Comparison of Euler and Runge-Kutta 2nd Order Methods Table 2. Finite Di erence Jacobian For any implicit method like. ZETA AT NEGATIVE ODD INTEGERS, A LA EULER This writeup sketches (you may need to supply details) an argument due to Euler that partially establishes the the functional equation of (s). simplest such method, originated by Euler about 1768. A recently. 4) implicitly relates yn+1 to yn. Approximating solutions using Euler's method. Validity Check Using Euler-Poincare Example Given the boundary representation solid shown, verify the Euler-Poincare relationship. The function y ( t ) has the following Taylor series expansion of order n at t = t i +1 :. Convergence of Numerical Methods In the last chapter we derived the forward Euler method from a Taylor series expansion of un+1 and we utilized the method on some simple example problems without any supporting analysis. pdf’) pylab. The ode45 provides an essential tool that will integrate a set of ordinary. LeonhardEuler: HisLife,theMan,andHisWorks is to bring across some glimpses of Euler's incredibly voluminous and diverse work, which today fills 74 massive volumes of the Opera omnia which certainly is the best method ofmaking happy progress in the mathematical sciences. dy x x y dx From the Catalogue select Euler: We need to enter the following: y value from initial condition euler(x^2-2x,x,y,{3,4},0,0. The Initial Value Problem §9. 2 Various types of beams and their deflected shapes: a) simple beam, b) beam with overhang, c) continuous beam, d) a cantilever beam, e) a beam fixed (or restrained) at the left end and simply supported near the other end (which has an overhang), f) beam fixed (or restrained) at both ends. Example 2: If 2 dy xy dx and if y 3 when x 2, use Euler's method with five equal steps to approximate y when x 1. The semi-explicit index-2 system 14 2. 1 Introduction In this chapter, we will consider a numerical method for a basic initial value problem, that is, for y = F(x,y), y(0)=α. method or the improved Euler method. It simply replace dy/dt(tn) by the forward finite difference (yn+1 −yn)/k. Each solution to the model is plotted to visually compare the differences. The Euler equations can be solved using the flux first method to compute the fluxes in a "simpler" way. In this simple differential equation, the function is defined by (,) =. Below are simple examples of how to implement these methods in Python, based on formulas given in the lecture note (see lecture 7 on Numerical Differentiation above). Unless $$f(x,y)$$ is of a special form, it is generally very hard if not impossible to get a nice formula for the solution of the problem. 2 Realizing that ( +1, +1)=(1+2 +1)√ +1, then the discretized equation is:. 4 Numerical Methods: The Approximation Method of Euler 1. Euler’s Method This method is based on the local linearity concept we covered earlier: If we zoom in enough on a differentiable function f at a point (a,b) the function looks linear and can be approximated by the linear equation y=f(a)+f ′(a)(x−a). simplest such method, originated by Euler about 1768. Presentasi mengenai definisi dan contoh dari penyelesaian menggunakan metode Euler, Heun dan Runge-Kutta dalam Metode Numerik by bara_pratista in Types > School Work, euler dan numerical method. Euler equations, Two-dimensions. order R-K method produces the most accurate answer, followed by the 3rd-order R-K method, then the two 2nd-order R-K methods (i. SLOPE FIELDS, SOLUTION CURVES, AND EULER'S METHOD 3 EXAMPLE 1 Recall that the logistic equation is the di erential equation dP dt = kP 1 P P max where k and P max are constants. You may wish to compute the exact. Euler’s method always needs a step size, which is called h. • Given the equivalence of formulations what becomes important is how easily the equations of. The simple Euler method: yn = yn 1 +hf(yn 1); h = xn xn 1 can be made more accurate by using either the mid-point or the trapezoidal rule quadrature formula: yn = yn 1 +hf yn 1 + 1 2hf(yn 1): yn = yn 1 + 1 2hf(yn 1)+ 1 2hf yn 1 +hf(yn 1): Runge-Kutta methods for ordinary differential equations - p. Construct the discretized ODE using the implicit Euler method: +1= +ℎ ( +1, +1) Eq. Otherwise, integration does not occur. (b)Solve the differential equation 2 dy x dx with the initia l condition y 05, and use your solution to find y 0. Compare the performance of the two methods. , modified Euler and mid-point methods). can be solved using the integrating factor method. If the spatial domain is of complex geometry, the ALE mesh is necessarily unstructured. The solutions of the Euler-Lagrange equation (2. Also, plot the true solution (given by the formula above) in the same graph. First, you must choose a small step size h (which is almost always given in the problem statement on the AP exam). EULER' S METHOD APPLIED TO TRAJECTORY PROBLEMS Now that we are familiar with using Euler’s method and recursion techniques to solve differential equations, let’s see how to apply this to trajectory problems. We will provide details on algorithm development using the Euler method as an example. Euler’s Method for Ordinary Differential Equations-More Examples Chemical Engineering Example 1 The concentration of salt x in a home made soap maker is given as a function of time by x dt dx 37. In this course we shall consider only Euler's (forward) method, the simplest (and least accurate) method and leave the more advanced methods to numerical analysis. method or the improved Euler method. 3) are called critical curves. pdf ISC3313: Introduction to Scienti c Computing with C++ Summer Semester 2011 The Midpoint and Runge Kutta Methods Introduction The Midpoint Method we will go back to the Euler method, and consider how it can be adapted to handle the predator prey problem. 5 Euler’s method sec:Euler For “generic” ODEs, there is no hope of writing down an explicit formula for a typical solution. In summary, the modified Euler method for approximating the solution to the initial-value problem y = f(x,y), y(x0) = y0 at the points xn+1 = x0 +nh (n = 0,1,)is yn+1 = yn + 1 2 h ˘ f(xn,yn)+f(xn+1,y n∗+1) ˇ, where y∗ n+1 = yn +hf (x n,yn), n = 0,1, Example 1. Higher Order Methods Up: Numerical Solution of Initial Previous: Numerical Solution of Initial Forward and Backward Euler Methods. 0 Conclusion Euler's method is considered to be one of the oldest and simplest methods to find the numerical solution of ordinary differential equation or the initial value problems. org Эйлерийн арга; Usage on ru. 17) 8 Initial-ValueProblems for Ordinary Differential Equations TABU 1. > y[0] := y0; y 0:= 1. Euler's method and slope fields Euler's method has a simple geometric interpretation. 0 references. Euler method 4. Later, above all in the environment of the liberal deïst, King Frederick II (1712–1786), in his Berlin period (1741–1766), Leonhard Euler defended the Christian faith against freethinkers and atheists. All these methods use a fixed step size, but there are other methods that use a variable step size (though not neccessarily better in all circumstances). (b) Use Euler's method with step size ¢t = 0:5 to approximate this solution, and check how close the approximate solution is to the real solution when t = 2, t = 4, and t = 6. Minimum Set of Equations - minimum number of degrees of freedom, equations are highly coupled and complicated. Springer-Verlag, 1994. The intent is to show that by decreasing stepsize, a more accurate solution is attained by at the cost of more steps and computing time. Euler's method is the simplest of the one-step methods for approximating the solution to the initial value problem. While essentially the Euler methods are simple. Euler’s Method (Following The Arrows) Euler’s method makes precise the idea of following the arrows in the direction eld to get an ap-proximate solution to a di erential equation of the form y0= F(x;y) satisfying the initial condition y(x 0) = y 0. Very clever. Solution of quartic equations 5 IIA. 13 Euler’s Method On the other hand, our Euler method reads xn+1 = xn +h ( xn) = (1 h)xn: (15) Clearly, if h > 1, x(tn) will oscillate between negative and positive numbers and grow without bounds in magnitude as tn increases. McDonough Departments of Mechanical Engineering and Mathematics University of Kentucky c 1984, 1990, 1995, 2001, 2004, 2007. Euler's Method Euler's method is also called tangent line method and is the simplest numerical method for solving initial value problem in ordinary differential equation, particularly suitable for quick programming which was originated by Leonhard. Usually we can only estimate solutions to di erential equations using numerical methods. Euler’s method is based on approximating the graph of a solut ion y (x) with a sequence of tangent line approximations computed sequentially, in “st eps”. An example of an implicit method is the backward Euler method: Because the derivative is now evaluated at time instead of , the backward Euler method is implicit. Euler’s Method y’(t) = f(t,y), y(a) = w 0 w k+1 = w k + h f(t k,w k) slope is f(t. Lagrangian Particle Method for Euler Equations R. The k 1 and k 2 are known as stages of the Runge-Kutta method. Take an ordinary Euler step of length h. Euler Method : In mathematics and computational science, the Euler method (also called forward Euler method) is a first-order numerical procedurefor solving ordinary differential equations (ODEs) with a given. This fact, together with Lagrange's theorem, provides a proof for Euler's theorem. Euler equations, Two-dimensions. If we can get a short list which contains all solutions, we can then test out each one and throw out the invalid ones. Euler’s method always needs a step size, which is called h. solution curve. Euler's method is based on the insight that some differential equations (which are the ones we can solve using Euler's method) provide us with the slope of the function (at all points), while an initial value provides us with a point on the function. Numerical solution of the Euler equations by finite volume methods using Runge Kutta time stepping schemes. much better this method works than the Euler Method. 1 Euler™s Method in Euler™s Words Dick Jardine Keene State College, Keene, New Hampshire 2 3 April 4, 2007 4 Introduction 5 Euler™s method is a technique for -nding approximate solutions to di⁄erential 6 equations addressed in a number of undergraduate mathematics courses. Using Euler's method, you will slightly over-evaluate evaporation, at ~ 28. This method is explicit. At one point in Section 2 we need to nd the. Euler’s Method 1. continuous implicit Euler method (θ = 1) and the continuous implicit mid-point method (θ = 1/2). For the first order differential equation, we can only solve only a tiny portion of them such as linear, separable, and /or exact differential equation. 1/Use Euler's method with step size 0. Euler's Method - a numerical solution for Differential Equations Why numerical solutions? For many of the differential equations we need to solve in the real world, there is no "nice" algebraic solution. It uses a fixed step size h and generates the approximate solution. and rearrange to around with step. (c) Use Euler’s method with step size ¢t = 0:1 to approximate this solution, and check how close the approximate solution is to the real solution when t = 2, t = 4. Chinese Journal of Aeronautics 21 :1, 19-27 Online publication date: 1-Feb-2008. So the critical Euler buckling stress is σ Euler = F Euler / A = k π2 E / (L / r)2. 2 we’ll rigorously deflne the Hamiltonian and derive Hamilton’s equations, which are the equations that take the place of Newton’s laws and the Euler-Lagrange equations. Euler’s Method, starting at x = 0 with step size of 1, gives the approximation g(1)0. 0 0 voto positivo, Marcar este documento como útil 0 0 votos negativos, Marcar este documento como no útil Insertar. Euler's Method. Euler’s method is based on the insight that some differential equations (which are the ones we can solve using Euler’s method) provide us with the slope of the function (at all points), while an initial value provides us with a point on the function. Euler's Method (Intuitive). 5 in the text. 02 Euler's method Chapter 08. Euler method and Improved Euler method for a first order differential equation and compare their results with the exact solution and the results of built-in function through an example. • Theoretically the same equivalence can be shown between equations derived from other formulations (e. The Euler Method We begin our discussion of the Euler method by recalling the definition of the derivative of a function f (x) : (1) f' x = f x +h -f x h in the limit that h is a small number. This is a differential equation that is not separable and not linear, so we don’t yet have a method to solve it. Provide details and share your research! But avoid …. 3 Picard's method of successive approximations 7. The construction of numerical methods for initial value problems as well as basic properties of such methods shall first be explained for the sim- plest method: The explicit Euler method. They correspond to different estimates for the slope of the solution. YY10:=dsolve({deq, IC}, y(x), type=numeric,. 1) y(0) = y0 This equation can be nonlinear, or even a system of nonlinear equations (in which case y is a vector and f is a vector of n different functions). Euler’s Proof That 1+ 2+ 3+ = 1 12 John C. On a single figure, plot your estimated solution curve using the following step sizes for x: 0. In mathematics and computational science, the Euler method is a first-order numerical procedure for solving ordinary differential equations with a given initial value. In contrast, the Euler method (8. §We owe to Euler the notation f(x) for a function (1734), e for. The solutions of the Euler-Lagrange equation (2. Local linearity is also used to solve initial value problems: Suppose we have an easy way to compute f0(x). Given the differential equation x y dx dy = + and 3y (1) =. This paper, called ‘Solutio problematis ad geometriam situs pertinentis,’ was later published in 1741 [Hopkins, 2. Use Euler’s method to find approximate values for the solution of the initial-value problem =( , ), ( )0. It was developed by Leonhard Euler during the 1770s. Frequently exact solutions to differential equations are unavailable and numerical methods become. Their definitions are as shown in the following graph- The first Euler Angle α is measured by a counterclockwise rotation about the z axis of the x axis. In some cases, it's not possible to write down an equation for a curve, but we can still find approximate coordinates for points along the curve by. In fact, most differential equa-tions that arise in real life applications are solved on computers using approxi-mation techniques. Newton-Euler - generally considered most intuitive. Implementing Euler's Method One's understanding of a numerical algorithm is sharpened by considering its implementation in the form of a calculator or computer program. While essentially the Euler methods are simple. Runge-Kutta Methods for high-index problems 14 2. By Natalie-Claire Luwisha. Runge-Kutta Methods for Problems of Index 1 11 2. x i+1, in terms of y i and all the derivatives of y at x i. Explicit and Implicit Methods in Solving Differential Equations A differential equation is also considered an ordinary differential equation (ODE) if the unknown function depends only on one independent variable. The leapfrog method, which is second order, is closely related to a modification of the Euler method called Euler-Cromer. Note: Euler’s (integration) method | derivation using nite di erence operator. Suppose we have a di↵erential equation of the form dy dt = f(t,y). As we just saw in the graphical description of the method, the basic idea is to use a known point as a "starter," and then use the tangent line through this known point to jump to a new point. pdf - Free download as PDF File (. ) the resulting approximate solution on the interval t ≤0 ≤5. I[y] = Z b a dxF x;y(x);y0(x) (16:5) The speci c Fvaries from problem to problem, but the preceding examples all have. Part 4: The Cubic and Quartic from Bombelli to Euler Section 1 describes various algebraic methods used to tackle the cubic and quartic (the Trigonometric Method is elsewhere). The differential equation given tells us the formula for f(x, y) required by the Euler Method, namely: f(x, y) = x + 2y. The intent is to show that by decreasing stepsize, a more accurate solution is attained by at the cost of more steps and computing time. To solve a homogeneous Cauchy-Euler equation we set y=xr and solve for r. Chapter 8 in the text). 5 is a set horizontal step-size, and 1 1 is the di erential equation dy dx = 1 x evaluated at x 0 = 1. This lecture discusses different numerical methods to solve ordinary differential equations, such as forward Euler, backward Euler, and central difference methods. This method is called the forward Euler method. A cube, for example, has. Euler's Method On this page you will find a tool that will perform Euler's method for you. Basic concepts 4. Setting x = x 1 in this equation yields the Euler approximation to the exact solution at. Euler–Bernoulli beam theory (also known as engineer's beam theory or classical beam theory) is a simple method to calculate bending of beams when a load is applied. Euler's method can be derived by using the first two terms of the Taylor series of writing the value of. Example Use Euler’s Method to approximate the solution of the initial-value problem: y′ y −t2 1, 0 ≤t ≤2, y 0 0. EULER-MASCHERONI CONSTANT In studying the difference between the divergent area under the curve F(x)=1/x from x=1 to infinity and the area under the staircase function where we have– 1 1 ( ) in n x n n S x , the Swiss mathematician Leonard Euler found back in 1734 that the area equals the constant value γ=0. Construct the discretized ODE using the implicit Euler method: +1= +ℎ ( +1, +1) Eq. " Note that this function uses an exact increment h rather than converting it explicitly to numeric form using Mathematica command N. Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. It is called the backward Euler method because the difference quotient upon which it is based steps backward in time (from t to t − h). After reading this chapter, you should be able to: 1. Solution of cubic equations 4 IC. The Newton equation for backward Euler is 2 4I t @f @u! (k) n+1 3 5 u(k) = u(k) n+1 + n + tf (k) n+1 (4) or @R BE @u n+1! (k) n+1 u(k) = R BE (5) where R BE = u (k) n+1 u n tf u(k) is the residual for Eq. Euler's method for solving a di erential equation (approximately) Math 222 Department of Mathematics, UW - Madison March 4, 2013 Math 222 di eqs and Euler's method. Exercise: Graph the slope field of y′ +xy = x and use it to find the behavior at infinity of the solution to the IVP y′ +xy = x and y(0) = −2. Euler’s Method y’(t) = f(t,y), y(a) = w 0 w k+1 = w k + h f(t k,w k) slope is f(t. Euler method b. JAMESON, WOLFGANG SCHMIDT and ELI TURKEL. Euler’s Method Leslie Hogben 10/01 Euler’s method can be combined with implicit differentiation to approximate the graph of an implicitly defined function. , modified Euler and mid-point methods). I'm supposed to make a plot in MATLAB for the solution by using Euler's Method for the circuit current derived from the circuit differential equation. Euler’s Method for Ordinary Differential Equations-More Examples Chemical Engineering Example 1 The concentration of salt x in a home made soap maker is given as a function of time by x dt dx 37. Projectile Motion with Air Resistance (Numerical Modeling, Euler's Method) Theory Euler's method is a simple way to approximate the solution of ordinary di erential equations (ode's) numerically. EULER’S METHOD To solve a differential equation of first order of the type 𝑑𝑦 𝑑𝑥 = 𝑓 𝑥, 𝑦 , with initial conditions 𝑦 𝑥0 = 𝑦0. Runge-Kutta method (Order 4) for solving ODE using MATLAB MATLAB Program: % Runge-Kutta(Order 4) Algorithm % Approximate the solution to the initial-value problem % dy/dt=y-t^2+1 MATLAB 2019 Free Download. I also don'. Runge-Kutta Methods for Problems of Index 1 11 2. As usual you are trying to flnd p and C in e = Chp. Euler's Method Tutorial A method of solving ordinary differential equations using Microsoft Excel. 1: Explicit Euler Method 5. We can choose other methods such as the Euler method. Math 320 di eqs and Euler’s method. If you're seeing this message, it means we're having trouble loading external resources on our website. The last column shows the accuracy of the method. The intent is to show that by decreasing stepsize, a more accurate solution is attained by at the cost of more steps and computing time. (a) Three-ring Venn diagram from [52]. See also the Toolbox User's Guide and references therein. 3, 2012 • Many examples here are taken from the textbook. He addresses both this specific problem, as well as a general solution with any number of landmasses and any number of bridges. 1 Euler's Method 1. Given the IVP 1 00 y t f t y t, y t y first, let y 0, y. Be aware that this method is not the most efficient one from the computational point of view. (b) Use Euler’s method with step size ¢t = 0:5 to approximate this solution, and check how close the approximate solution is to the real solution when t = 2, t = 4, and t = 6. Solution: With a step size of ∆ x = 0. com Applying uler’s Method to a second order O A general second ordinary differential equation1: ( , ) ( , ) ( , ) 2 2 b t x x f t x dt dx a t x dt d x (1) can be converted to a first order system X F b t x a t x X » ¼ º « ¬ ª ( , ) ( , ) 0 1 (1) where » ¼ º « ¬ ª y x X, » ¼ º « ¬ ª y y x X, » ¼ º « ¬ ª ( , ) 0 f t x F and dt dx. Then define a recursive sequence as. Euler's method is based on approximating the graph of a solution y(x) with a sequence of tangent line approximations computed sequentially,in "steps". It is one of the best methods to find the numerical solution of ordinary differential equation. (2017) Augmented Lagrangian method for an Euler's elastica based segmentation model that promotes convex contours. Given N discretization times equally spaced on the interval [0,T] , we. Newton's Law of Cooling (a complete example) 2. Below are some scratch work space to plan for the Mathematica coding exercises. The vast majority of first order differential equations can't be solved. 2) Here, kis time step size of the discretization. JAMESON, WOLFGANG SCHMIDT and ELI TURKEL. png 800 × 600; 36 KB Forward Euler method illustration-2. (2010) Euler’s Method. The clear disadvantage of the method is the fact that it requires solving an algebraic equation for each iteration, which is computationally more expensive. The ODE y′ = f(t,y) (2. The description may seem a bit vague since f is not known explicitly, but the advantage is that once a method has been derived we may. determination method. By Taylor expansion, the. • Implicit Euler is a decent approximation, approaching zero as h becomes large, and never overshooting. Homework 33: 11. using Euler's Method with two equal steps. It is called the tangent line method or the Euler method. 12 in text for a more detailed description of the method. Euler's Method Calculator - eMathHelp Emathhelp. If the spatial domain is of complex geometry, the ALE mesh is necessarily unstructured. Reminder: We're solving the initial value problem: y′ = f(x, y) y(x o) = y o. Doing this requires solving this equation for k, which amounts to a root nding problem if f is nonlinear, but we know how to. 1 separation of variables. For 1≤ n≤ N y(0) = y0, dy dt = F(t, y), (Truth) y0 = y0, yn+1 −yn Δt = F(tn,yn). Euler's method is a numerical method to solve first order first degree differential equation with a given initial value. Solving higher-order differential equations Engineering Computation ECL7-2 Motivation • Analysis of Engineering problems generate lots of differential equations, most of which cannot be easily solved explicitly. Consider the problem (y0 = f(t;y) y(t 0) = Define hto be the time step size and t. Given (t n, y n), the forward Euler method (FE. Boujot, Springer. Does Euler's method produce an over- or under-estimate for the value of f (3. Euler's Method Euler's method is a numerical method for solving initial value problems. Below is an example problem in Excel that demonstrates how to solve a dynamic equation and fit unknown parameters. The idea is similar to that for homogeneous linear differential equations with constant coefficients. 4) = ? (b) Repeat part (a) with step size 0. 1 to determine an approximation to the. There are eight problems with a good mix of types and rigor. euler IVP ODEs; Runge-Kutta and Euler methods radau [4] IVP ODEs+DAEs; implicit Runge-Kutta method daspk [1] IVP ODEs+DAEs; bdf and adams method zvode IVP ODEs, like vode but for complex variables adapted from [19]. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Euler Method The Euler methods are simple methods of solving first-order ODE, particularly suitable for quick programming because of their great simplicity, although their accuracy is not high. non-linear vibration of Euler-Bernoulli beams subjected to the axial loads. He not only made formative contributions to the subjects of geometry, calculus, mechanics, and number theory but also developed methods for solving problems in astronomy and demonstrated practical applications of mathematics. This handout will walk you through solving a simple. This method is implicit. Euler’s Method (HW #5) Section 9. If people don't need super accurate results but just need to be able to compare two results, Euler's method might be sufficient. While essentially the Euler methods are simple. Newton method converges. The graph goes through the point (0;1) so put a dot there. Section 2 contains a detailed description, essentially due to Euler, of how to obtain all the roots of a cubic, in all cases. Now p is the slope of the linear part and lnC is the intercept of the extension of the linear part. Improved Euler Method EXAMPLE Use the improved Euler method to solve y0 = (x+y ¡1)2; y(0) = 2: Find y(:2) in 2 steps. Here is the table for. 2003; Stern and Grinspun 2009]. Multistep Methods 283 The Adams family of methods arises when we approximate the integralontherightof𝑦(𝑡𝑛+1)−𝑦(𝑡𝑛)= ∫𝑡 𝑛+1 𝑡𝑛 𝑦′(𝑠)𝑑𝑠with ∫ 𝑡 𝑛+1 𝑡𝑛 𝑃𝐴⋅ 𝑚(𝑠)𝑑𝑠. A horror story about integration methods, by R. 1): its m stage values Y n,i are given by the solution of the nonlinear algebraic systems (1. This equation can be used to modeled the growth of a population in an environment with a nite carrying capacity P max. Consider the ode dy dx = f0(x) (1) which has solution y= f(x) and reference. Getting to know Python, the Euler method "Hello, Python!" Feb. The idea is similar to that for homogeneous linear Cauchy-Euler Equations),. I don’t know who first introduced it but there is a nice discussion in the Feynman Lectures on Physics, Vol. To see the e ect of the choice of t in Euler's method we will repeat the process above, but with a smaller value for t. 2 to approximate the value of y when x = 1 given € dy dx =y and y(0)=1 € dy dx =y Euler’s Method leads us to the approximation f (1) ≈ 2. The essence of the proof is to consider the sequence of functions {y n}∞ n=0, defined recursively through what is known as the Picard Iteration: y. We could devote an entire class to studying some of these methods. See section 6. The curve passing throuoh (2, 0) satisfies the differential equation approximation to using Euler's Method with two equal steps. It is a quantity with the dimensions of (Energy)£(Time). ZETA AT NEGATIVE ODD INTEGERS, A LA EULER This writeup sketches (you may need to supply details) an argument due to Euler that partially establishes the the functional equation of (s). Keep in mind that the drag coefficient (and other aerodynamic coefficients) are seldom really constant. Numerical Method: Euler Method for first order ODE. with a step. savefig('central-and-forward-difference. Finite Di erence Jacobian For any implicit method like. All these methods use a fixed step size, but there are other methods that use a variable step size (though not neccessarily better in all circumstances). 12 in text for a more detailed description of the method. Use the improved Euler method with step size h = :1 on the interval. Runge-Kutta method The formula for the fourth order Runge-Kutta method (RK4) is given below. Modified Euler method c. by using Euler’s method with two equal steps. demonstrate how to solve Cauchy-Euler Equations using roots of indicial equa-tions. The Forward Euler scheme is as follows. Euler's method is the simplest approach to approximating a solution to a di erential equation. Follow 158 views (last 30 days) Bayram FURKAN TORA on 1 May 2019. Euler's Method Now we will work with a general initial value problem We will again form an approximate solution by taking lots of little steps. The structure of a dendrimer exhibits a large number of internal and superficial cavities, which can be exploited, to capture and deliver small organic molecules, enabling their use in drug delivery. Doing this requires solving this equation for k, which amounts to a root nding problem if f is nonlinear, but we know how to. The Euler method is an example of an explicit method. ! It is based on approximating the Euler equation by a linear equation ! whose fluxes can be found analytically. † Step One: Initialization Set h = b¡a n. A scalar, first-order initial value problem is given as We want to determine a numerical approximation to y(t) at discrete points in the interval [a, b].
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https://math.stackexchange.com/questions/378692/if-all-embeddings-of-e-f-are-automorphisms-then-e-f-is-normal | # If all embeddings of $E/F$ are automorphisms, then $E/F$ is normal
Let $E$ be a (possibly infinite) field extension of $F$, and let $\Omega$ be an algebraic closure of $E$. I'm trying to prove that if $\sigma(E)=E$ for all $F$-algebra embeddings $\sigma \colon E\to \Omega$, then $E/F$ is normal.
The finite case is easy: Suppose $E=F(\alpha_1,\ldots,\alpha_n)$. Consider the splitting field $K$ of the polynomial $f:=m_{\alpha_1}\cdots m_{\alpha_n}$, where $m_\alpha=m_{\alpha,F}$ denotes the minimal polynomial of $\alpha$ over $F$. Let $\alpha=\alpha_i$ be any one of $\alpha_1,\ldots,\alpha_n$, and let $\beta$ be another root of $m_\alpha$. Then there's an $F$-algebra isomorphism $\tau\colon F(\alpha)\to F(\beta)$ taking $\alpha$ to $\beta$. But since $\tau$ fixes $F$, it takes $f$ to itself. Therefore, $\tau$ extends to an automorphism $\sigma$ of $K$. But $\sigma|_E$ is an $F$-algebra embedding of $E$ into $\Omega$, meaning $\sigma(E)=E$. So, $\beta=\tau(\alpha)=\sigma(\alpha)\in E$. Hence, all the roots of $f$ are in $E$. Thus, $K\subseteq E$. Since $E\subseteq K$ by definition, it follows that $E=K$. Hence, $E$ is normal.
How do I generalize this to the infinite case? (Note that this isn't homework. I'm doing some extra reading on my own.)
• It really doesn't matter whether it is or not homework: people will still try to help you, but perhaps you won't be getting as many fully solved answers...which is something people really willing to learn mathematics should try to avoid. – DonAntonio May 1 '13 at 23:00
• Hints are fine. – Avi Steiner May 1 '13 at 23:09
I think the infinite case is not that big a problem after you proved what you say you already did: let $\,p(x)\in F[x]\,$ be irreducible non-constant s.t. $\,p(w)=0\,$ for some $\;w\in E\;$ , and let $\,w_1:=w\,,\,w_2\,,\,...\,w_n\,$ be all its conjugates in $\,\Omega\,$ .
Now we have an isomorphism $\,F(w_1)\to F(w_2)\,$ , and this isomorphism can be extended to an embedding $\,E\to \Omega\,$ (if you're not sure about this you can check, for example, theorem 2.8 , chap.5.2, in Lang's "Algebra") , and since this embedding is an $\,F$-automorphism of $\,E\,$ then in fact $\,w_2\in E$...
Complete the argument above to show all the roots of $\,p(x)\,$ are in $\,E\,$ and thus $\,E/F\,$ is normal
• Does Dummit & Foote have a similar theorem? I couldn't find anything when I looked. – Avi Steiner May 1 '13 at 23:51
• In the 3rd. Edition of D&F, try Theorem 8 in chapter 13.1 (page 519) and, in particular, theorem 27 in chapter 13.4 (page 541). Lang's book ("Algebra") is a very good reference book, though I would never use it to teach and/or learn from it for beginners. I like though the way he proves this stuff, using Zorn's Lemma in a very beautiful way. Try to read it in your university's library or even perhaps in Google books - it is in page 233 of the revised third edition of Springer Verlag - – DonAntonio May 2 '13 at 8:42
• I'll check out Lang. D&F's theorems 8 and 27 only cover the finite case, and I couldn't figure out how to generalize to the infinite. – Avi Steiner May 2 '13 at 17:50
• I figured it out on my own! Thanks @DonAntonio. – Avi Steiner May 2 '13 at 18:39
• Hey @AviSteiner, I'd love to see your solution, even as a new answer to your own question. It's the usual thing here...and good for you! – DonAntonio May 2 '13 at 21:41
The only part of the infinite case the was giving me trouble was, using the notation in DonAntonio's answer, showing that the isomorphism $F(w_1)\to F(w_2)$ extends to an embedding $E\to\Omega$. To prove this, I'll prove something slightly more general:
Theorem. Let $E$ be the splitting field of a family $\mathcal{F}$ of polynomials over a field $F$, let $\tau\colon F\to F'$ be an isomorphism, and let $E'$ be the splitting field of $\mathcal{F}':=\tau(\mathcal{F})$. Then there exists an isomorphism $\sigma\colon E\to E'$ extending $\tau$.
Proof. Consider the set $\mathcal{E}$ of isomorphisms $L\subseteq E\to L'\subseteq E'$ extending $\tau$---this is non-empty since it contains $\tau$ itself. Equip $\mathcal{E}$ with the partial order $\leq$ given by $\sigma_1 \leq \sigma_2$ iff $\sigma_2$ extends $\sigma_1$. Any totally ordered subset $\{\varphi_i\colon L_i\to L_i'\}_{i\in I}$ of $\mathcal{E}$ is clearly bounded above by the isomorphism $\varphi\colon \bigcup_{i\in I} L_i\to \bigcup_{i\in I} L_i'$ in $\mathcal{E}$ given by $\varphi(\alpha)=\varphi_i(\alpha)$ if $\alpha\in L_i$. Therefore, by Zorn's Lemming, the set $\mathcal{E}$ has a maximal element $\sigma\colon L\to L'$.
We show that $L=E$. To see this, suppose not. Then $L$ doesn't contain all the roots of some $f\in \mathcal{F}$, and the splitting field $M$ of $f$ over $L$ strictly contains $L$. So, by the finite version of this theorem, $\sigma$ extends to an isomorphism $\eta\colon M\to M'$, where $M'$ is the splitting field of $\sigma(f)$ over $L'$. But then $\sigma <\eta$, contradicting the fact that $\sigma$ is maximal. Hence, $L=E$.
Finally, we show that $L'=E'$. Let $\beta$ be a root $\beta$ of some $g\in \mathcal{F}'$. By definition, $g=\tau(f)$ for some $f\in \mathcal{F}$, meaning $\sigma(\alpha)=\tau(\alpha)=\beta$ for some $\alpha\in E$. Therefore, $\beta\in L'$. Hence, $L'\supseteq E'$. Thus, since $L'\subseteq E'$ by construction, $L'=E'$. q.e.d. | 2019-06-26T05:50:31 | {
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https://math.stackexchange.com/questions/2568248/two-methods-of-calculating-new-average-from-old-average | Two methods of calculating new average from old average
I know that calculating the new average $a_{new}$ from the old average $a_{old}$ can be done in the following way (for uniform weights):
Suppose the old average is based on $n$ elements. The old total is then $t_{old}=a_{old}\cdot n$. The new total is $t_{new}=t_{old}+x$. The new average is $a_{new}=\frac{t_{new}}{n+1}=\frac{a_{old}\cdot n + x}{n+1}$. (4 flops)
However, I found another way of calculating the new average $a_{new}$ from the old average $a_{old}$. I tried it numerical for some example, which is correct, but how can I prove it for all cases? My method is the following (also suitable for weights that are not uniform):
• Calculate the difference $d$ between the old average $a_{old}$ and the new element $x$ with $d=x-a_{old}$
• Multiply it with the weight $w$ of the new element $x$, so $q=d\cdot w$
• Add it to the old average $a_{old}$, so $a_{new}=a_{old}+q=a_{old}+(x-a_{old})\cdot w=x\cdot w + (1-w)\cdot a_{old}$.
The calculation $a_{new}=a_{old}+(x-a_{old})\cdot w$ has 3 flops
Which method is more efficient in terms of computing time? Or are they equally efficient?
• What is the weight? If it is $w=\frac{1}{n+1}$, then both methods you explain are equivalent (and equally efficient computationally, as none of them contains any loop). If by "weight" you mean something else, then they are not equivalent. – Anna SdTC Dec 15 '17 at 18:35
• $w$ is for example how much a certain item at school counts for the final mark, which is often known beforehand and not need to be calculated – Faceb Faceb Dec 15 '17 at 18:38
• All flops are not equal. Division is often nastier than all the others. – mathreadler Dec 15 '17 at 19:21
• If $w$ is some constant divided by an integer power of two you can make the second approach quite a bit faster. Yes 3 flops if we already know $w$, but are you sure $w$ will be constant? – mathreadler Dec 15 '17 at 19:30
\begin{align}a_{new}&=\frac{t_{new}}{n+1}\\&=\frac{a_{old}\cdot n + x}{n+1}\\ &= \frac{n}{n+1}a_{old}+\frac{1}{n+1}x\end{align}
The old formula requires one multiplication, two additions, and one division.
We have to set $w = \frac{1}{n+1}$ for the formula to be equal $a_{new}=a_{old}+q=a_{old}+(x-a_{old})\cdot w=x\cdot w + (1-w)\cdot a_{old}$
The new formula requires one subtraction, one division, one multiplication, and one addtion. The division is due to $\frac{1}{n+1}$.
Edit: consider data $x_1, \ldots, a_{n+1}$ with weight $w_1, \ldots, w_{n+1}$ \begin{align} a_{new} &= \frac{\sum_{i=1}^{n+1} w_ix_i}{\sum_{i=1}^{n+1} w_i} \\ &= \frac{\sum_{i=1}^{n} w_ix_i}{\sum_{i=1}^{n+1} w_i} + \frac{w_{n+1}x_{n+1}}{\sum_{i=1}^{n+1}w_i}\\ &= \frac{\sum_{i=1}^{n} w_ix_i}{\sum_{i=1}^{n} w_i} \left( \frac{\sum_{i=1}^{n} w_i}{\sum_{i=1}^{n+1} w_i}\right) + \frac{w_{n+1}x_{n+1}}{\sum_{i=1}^{n+1}w_i}\\ &= a_{old}\left( \frac{\sum_{i=1}^{n} w_i}{\sum_{i=1}^{n+1} w_i}\right) + \frac{w_{n+1}(x_{n+1}-a_{old}+a_{old)}}{\sum_{i=1}^{n+1}w_i} \\ &= a_{old} + \frac{w_{n+1}(x_{n+1}-a_{old})}{\sum_{i=1}^{n+1}w_i} \end{align}
• But the formula $a_{new}=a_{old}+(x-a_{old})\cdot w$ requires 3 flops right? (one addition, one subtraction and one multiplication) – Faceb Faceb Dec 15 '17 at 18:36
• what about computation of the weight? – Siong Thye Goh Dec 15 '17 at 18:38
• the weight is known beforehand, for example how much certain item at school counts for the final mark and therefore does not need to be calculated – Faceb Faceb Dec 15 '17 at 18:39
• $n+1$ is also known before hand for the first formula., you save one addition there. – Siong Thye Goh Dec 15 '17 at 18:40
• the first formula assumes uniform weight, which is why you can just sum them up and then divide by total number. – Siong Thye Goh Dec 15 '17 at 18:46 | 2019-12-06T18:47:25 | {
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https://math.stackexchange.com/questions/2428653/indefinite-integral-int-arctan2-x-dx-in-terms-of-the-dilogarithm-function | # Indefinite integral $\int \arctan^2 x dx$ in terms of the dilogarithm function
I read about the integral $$\int \arctan^2 x dx$$ in this old post: Evaluation of $\int (\arctan x)^2 dx$
By replacing $$\arctan x = -\frac{i}{2}\left[\log(1+ix) - \log(i-ix)\right],$$ as suggested there, I ended up with this solution $$\int\arctan^2 x dx = x\arctan^2x - \frac{1}{2}\log(1+x^2)\arctan x -\log 2 \arctan x + \mbox{Im}\left\{\mbox{Li}_2\left(\frac{1+ix}{2}\right)\right\} + K, \tag{1}\label{uno}$$ where, as usual, $\mbox{Li}_2(z)$ is the dilogarithm function $$\mbox{Li}_2(z) = -\int_0^z \frac{\log(1-u)}{u}du=\sum_{k=1}^{+\infty}\frac{z^k}{k^2}.$$ Is this a correct development?
In that case, if I determine, using \eqref{uno}, the definite integral $\int_0^1 \arctan^2xdx$ I get the result $$\int_0^1 \arctan^2xdx=\frac{\pi^2}{16}-\frac{3\pi}{8}\log 2 + \mbox{Im}\left\{\mbox{Li}_2\left(\frac{1+i}{2}\right)\right\}.$$ If I now compare this result with the one given in Definite Integral of $\arctan(x)^2$, i.e. $$\int_0^1 \arctan^2xdx=\frac{\pi^2}{16}+\frac{\pi}{4}\log 2 - C,$$ where $C$ is the Catalan constant $$C = \sum_{k=0}^{+\infty} \frac{(-1)^k}{(2k+1)^2},$$ I get the following expression: $$\mbox{Im}\left\{\mbox{Li}_2\left(\frac{1+i}{2}\right)\right\} = \frac{5\pi}{8}\log 2 - C.$$ Is that reasonable?
• According to Wolfram Alpha, $\text{Im} \, \text{Li}_{2} \left(\frac{1+i}{2} \right) = C- \frac{\pi}{8} \log 2$. This is the negative of what you had earlier. To check if this is indeed the correct value, you could use the inversion formula for the dilogarithm to find the value of $\text{Li}_{2} \left(\frac{1+i}{2} \right) + \text{Li}_{2} (1-i)$, the reflection formula to find the value of $\text{Li}_{2}(i) + \text{Li}_{2}(1-i)$, and the series definition to find the value of $\text{Li}_{2}(i)$. – Random Variable Sep 14 '17 at 16:58
• @RandomVariable thanks, I'll do the checks you suggested. I probably made same mistake in my development. Do you have any idea where I can find similar expressions for the $\int \arctan^2 x dx$ integral? Thanks again! – Matteo Sep 14 '17 at 17:18
• @RandomVariable: Thanks again, I made the check and as a matter of fact there was a sign error in my steps. Now everything turns out fine! I'll will post an answer with the correction as soon as possible. – Matteo Sep 15 '17 at 11:35
Landen's identity states that if $z \notin (1,\infty)$, then $$\operatorname{Li}_2(z) = -\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac12\ln^2(1-z).$$ For $z:=(1+i)/2$, we have $z/(z-1) = -i$, thus $$\operatorname{Li}_2\left(\frac{i+1}{2}\right) = -\operatorname{Li}_2(-i)-\frac12\ln^2\left(\frac{1-i}{2}\right).$$ It is well known that $$\operatorname{Li}_2(-i) = -iC -\frac{\pi^2}{48},$$ where $C$ is Catalan's constant. For the logarithmic term we have \begin{align} \frac12\ln^2\left(\frac{1-i}{2}\right) &= -\frac{1}{32}\left(\pi - 2\ln(2)\cdot i\right)^2 \\ &= \frac{\ln^2 2}{8} - \frac{\pi^2}{32} + \frac{\pi}{8}\ln(2) \cdot i. \end{align} Finally $$\operatorname{Li}_2\left(\frac{i+1}{2}\right) = \frac{5\pi^2}{96} - \frac{\ln^2 2}{8} + \left(C - \frac{\pi}{8}\ln 2\right)\cdot i.$$ You could find a more general approach in this answer, where there is a solution for all $z \in \mathbb{C}$, such that $\left|z/(z-1)\right|=1$.
Ok, thanks for the help in giving me $\mbox{Li}_2\left(\frac{1+i}{2}\right)=C-\pi(\log 2)/8$. I checked my result and found out a mistake. The correct form of the integral in terms of the dilogarithm function should be \begin{align} \int \arctan^2xdx=&x\arctan^2x -\frac{1}{2}\log(1+x^2)\arctan x+ \log2\arctan x+\\ &-\mbox{Im}\left\{\mbox{Li}_2\left(\frac{1+ix}{2}\right)\right\}+K.\end{align} I checked derivating back to $\arctan^2x$. Also the result is now compatible with definite integral $\int_0^1 \arctan^2x dx = \pi^2/16+\pi/4\log2 -C.$
When $x\in\mathbb{R}^+$ we can write:
$$\arctan\left(x\right)=\frac{i}{2}\cdot\ln\left(\frac{1-xi}{1+xi}\right)\tag1$$
So, when we square both sides we get:
$$\arctan^2\left(x\right)=\left(\frac{i}{2}\cdot\ln\left(\frac{1-xi}{1+xi}\right)\right)^2=-\frac{1}{4}\cdot\ln^2\left(\frac{1-xi}{1+xi}\right)\tag2$$
For the integral we get, then:
$$\mathscr{I}:=\int\arctan^2\left(x\right)\space\text{d}x=-\frac{1}{4}\cdot\int\ln^2\left(\frac{1-xi}{1+xi}\right)\space\text{d}x\tag3$$
Let $\text{u}:=\frac{1-xi}{1+xi}$:
$$\mathscr{I}=-\frac{2i}{4}\cdot\int\frac{\ln^2\left(\text{u}\right)}{\text{u}^2+2\text{u}+1}\space\text{d}\text{u}=-\frac{i}{2}\cdot\int\frac{\ln^2\left(\text{u}\right)}{\left(1+\text{u}\right)^2}\space\text{d}\text{u}\tag4$$
Let $\text{v}:=1+\text{u}$:
$$\mathscr{I}=-\frac{i}{2}\cdot\int\frac{\ln^2\left(\text{v}-1\right)}{\text{v}^2}\space\text{d}\text{v}\tag5$$
Using integration by parts:
$$\mathscr{I}=-\frac{i}{2}\cdot\left\{-\frac{\ln^2\left(\text{v}-1\right)}{\text{v}}+2\cdot\int\frac{\ln\left(\text{v}-1\right)}{\text{v}\cdot\left(\text{v}-1\right)}\space\text{d}\text{v}\right\}=$$ $$-\frac{i}{2}\cdot\left\{-\frac{\ln^2\left(\text{v}-1\right)}{\text{v}}+2\cdot\left\{\int\frac{\ln\left(\text{v}-1\right)}{\text{v}-1}\space\text{d}\text{v}-\int\frac{\ln\left(\text{v}-1\right)}{\text{v}}\space\text{d}\text{v}\right\}\right\}\tag6$$
• $\text{u}:=\frac{1-xi}{1+xi}$ is a complex function. What is $\int_0^u (...)du$? – FDP Sep 14 '17 at 20:59
• I'm not quite following where you're heading to – Matteo Sep 15 '17 at 11:32 | 2019-06-18T05:20:22 | {
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https://mathhelpboards.com/threads/concavity.9458/ | # Concavity
#### shamieh
##### Active member
Find dy/dx and d^2y/dx^2. For which value of t is the curve concave upward?
$$\displaystyle x = t^3 + 1$$
$$\displaystyle y = t^2 - t$$
Here is what I have so far and where I got stuck.
$x' = 3t^2$
$y' = 2t - 1$
$$\displaystyle \frac{2t - 1}{3t^2} /9t^4 = \frac{2t - 1}{3t^2} * \frac{1}{9t^4}$$
I'm confused. Am I doing this correctly? I took the derivative of x and y and then did y' / x' / (x')^2
#### MarkFL
Staff member
I would use clear notation to indicate what you are doing. We may use the chain rule as follows:
$$\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}$$
Now, as you found:
$$\displaystyle \frac{dy}{dt}=2t-1$$
$$\displaystyle \frac{dx}{dt}=3t^2$$
Hence:
$$\displaystyle \frac{dy}{dx}=\frac{2t-1}{3t^2}$$
Now differentiate again with respect to $x$, carefully applying the quotient, power and chain rules. What do you find?
#### Prove It
##### Well-known member
MHB Math Helper
Why did you divide by (x')^2? | 2021-06-15T12:58:13 | {
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https://math.stackexchange.com/questions/2513354/cauchy-sequence-of-smooth-parameterized-curves-that-does-not-converge | # Cauchy sequence of smooth parameterized curves that does not converge
I'm trying to prove the following statement:
Let $a$ and $b$ be points in $\mathbb R^2$ and let $X$ be the set of all continuously differentiable functions $\gamma:[0,1]\to\mathbb R^2$, with $\gamma(0)=a$ and $\gamma(1)=b$. Further let $\delta$ be the metric on $X$ defined by $$\delta(\gamma_1,\gamma_2)=\sup_{[0,1]}\Vert\gamma_1-\gamma_2\Vert.$$ Then the metric space $X$ with metric $\delta$ is not complete.
At the point I am in the analysis book I'm working through, to do this I need to find a Cauchy sequence in $X$ which does not converge (in $X$). I tried considering the sequence $\gamma_n(t)=a+t^n(b-a)$, but I cannot seem to show that it's Cauchy (I can't figure out what to use for my $N$).
I did manage to show that, for $m\gt n$, $$\delta(\gamma_n,\gamma_m)=\Vert b-a\Vert\left(\frac{n}{m}\right)^{n/(m-n)}\frac{m-n}{m},$$ but I can't figure out how to work with this to ensure that it's smaller than $\epsilon\gt0$.
Thus my question is, does this sequence work? If not, how might I come up with one that does? If so, how might I show that it's Cauchy?
• I've removed the tag "incompleteness" that was not relevant here – Max Nov 10 '17 at 7:59
• Think about a sequence of curves which get more pointy with higher $n$. In the limit they are no longer differentiable. For example $\gamma_n(t) = \sqrt{x^2+1/n}$ might work I think. The limit is $|x|$. – M. Winter Nov 10 '17 at 22:24
• @M.Winter do you mean $\sqrt{t^2+1/n}$? That wouldn't be a sequence in $X$, since the function outputs real numbers, not vectors in $\mathbb R^2$. – themathandlanguagetutor Nov 10 '17 at 22:27
• @themathandlanguagetutor You are right about the mistake with the $x$. For the other thing, just add a constant second coordinate. – M. Winter Nov 10 '17 at 22:33
• @M.Winter Then it still wouldn't be in $X$ because it wouldn't satisfy $\gamma_n(0)=a$ and $\gamma_n(1)=b$. – themathandlanguagetutor Nov 10 '17 at 22:40
A rigorous way
Use
$$f_n(x)=\begin{cases}0&\text{for x\le 0}\\ x^{1+1/n}&\text{for x>0}\end{cases},\qquad f^*(x):=\begin{cases}0&\text{for x\le 0}\\x&\text{for x>0}\end{cases}$$
which gives the depicted sequence of functions and converges to $f^*$ on the interval $[-1,1]$.
$\qquad\qquad\qquad$
The important thing is: $f_n$ is differentiable for all $n$, but the limit curve is obviously not. Also, the end points are fixed, i.e. $f_n(-1)=0$ and $f_n(1)=1$ for all $n$. This means we can use this function to build our desired curve (assuming $a\not=b$)
$$\gamma_n(t)=a+f_n(2t-1)(b-a).$$
Proof.
• $f_n'(x)$ equals $(1+1/n)x^{1/n}$ for $x>0$ and hence both defining cases agree on $f_n'(0)=0$. Therefore $f_n$ is indeed differentiable. Obviousoly $f^*$ is not.
• We have $\sup_{x\in[-1,1]}|f_n(x)-f^*(x)|=\sup_{x\in[0,1]}|x^{1+1/n}-x|$. It is basic anaylsis to find the maximum at $$\hat x=\left(1+\frac1n\right)^{-n},\qquad |\hat x^{1+1/n}-\hat x|=\left|\frac1{n-1}\cdot \left(1+\frac1n\right)^{-n}\right|$$ which vanishes for $n\to\infty$. Hence $f_n\to f^*$ in the given metric.
• Now we have \begin{align} d(\gamma_n,\gamma^*)&=\sup_{t\in[0,1]}\|\gamma_n(t)-\gamma^*(t)\|\\ &=\|b-a\|\cdot\sup_{t\in[0,1]}|f_n(2t-1)-f^*(2t-1)|\\ &=\|b-a\|\cdot\sup_{t\in[-1,1]}|f_n(t)-f^*(t)|\\&\to0, \end{align} as $n\to\infty$. Hence $\gamma_n$ converges (and therefore is Cauchy), but the limit is not in $X$.
• I'm not quite satisfied with saying that it's Cauchy just because it converges (in some set), simply because that's a result we've really only shown for sequences that converge in $\mathbb R^d$ with the usual metric. That said, I'll look more at this example and see if I either can't show this result or show it's Cauchy directly. – themathandlanguagetutor Nov 13 '17 at 17:42
• @themathandlanguagetutor well its a classical fact (and not too hard to see) that if a sequence is Cauchy in some metric space, then also in any metric space which contains the sequence. This is because Cauchy only needs the elements of the sequence and does not refer to the containing space otherwise. – M. Winter Nov 13 '17 at 21:32
Just a hint
Choose some sequence of curves which gets more and more pointy with increasing $n$. For example
$$\gamma_n(t):=\begin{pmatrix}\!\sqrt{t^2+\frac 1n}\; \\ t\end{pmatrix}.$$
The limit of this sequence is $\gamma^*(t)=(|t|,t)^\top$, which is not differentiable in $t=0$. We can see this via
$$\|\gamma_n(t)-\gamma^*(t)\|\le \frac 1 {\sqrt n},$$
which is not hard to show. Since the sequence converges, it must be Cauchy. But the limit is not in $X$. It should not be too hard to find a way to make the endpoint match $a$ and $b$, e.g. by adding segments on both ends of $\gamma_n$ which start (resp. end) in $a$ (resp. $b$).
Another idea, for which I will not give an explicit formula (but the proofs are obvious) is motivated by the picture below:
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$
You again take a wedge ($\wedge$) like curve as the limit curve (which is not in $X$). By rotating and scaling you can make the endpoints fit $a$ and $b$ (if $a\not=b$). The sequence curves $\gamma_n$ share initial- and end segment with $\wedge$ but the pointy end is replaced by a circle arc which fits in such a way that the curve stays differentiable.
• What's wrong with my hint? – M. Winter Nov 10 '17 at 22:51
• Even if I can modify $\gamma_n$ to have the right endpoints (which I'm still not seeing how to do) $\gamma^*$ is indistinguishable from $(t,0)$ on $[0,1]$, so it should still be in $X$. – themathandlanguagetutor Nov 10 '17 at 22:59
• @themathandlanguagetutor Please take a minute to think about the hint which I gave you. This is no solution. Think about this curve as defined on $[-1,1]$ or so. You should be able to work out how to reparametrize $\gamma_n$ nd $\gamma^*$ to make the relevant part fit into $[0,1]$. This is no "give me a final answer" site, but a "help me to understand how to work out an answer myself" site. – M. Winter Nov 10 '17 at 23:03
• I'm sorry, I was just very fixed on the interval $[0,1]$. Does this look like it should work: $\gamma_n(t)=a+\left(\left(\sqrt{(2t-1)^2+1/n}+(t-1)\sqrt{1+1/n}\right)(b_1-a_1),t(b_2-a_2)\right)$? – themathandlanguagetutor Nov 10 '17 at 23:19
• @themathandlanguagetutor Also sorry if I sounded rude. You are definitely on the right track. I am not sure if your formula works for $t=1$. Of course there might be better solutions to this problem (especially ones with an easier final proof that they are Cauchy), I just wanted to point ot the "pointy end" intuition. And your initial idea might still work. – M. Winter Nov 10 '17 at 23:27 | 2019-08-23T05:28:51 | {
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https://www.physicsforums.com/threads/work-needed-to-pump-water-out-of-spherical-tank.292241/ | # Work needed to pump water out of spherical tank
1. Feb 13, 2009
### Seraph404
1. The problem statement, all variables and given/known data
A spherical tank is full of water. If radius r = 3m and height h = 1.5m, find the work W required to pump the water out of the spout. (Use 9.8 for g and 3.14 $$\pi$$)
2. Relevant equations
Work is the integral from a to b of f(x)dx.
3. The attempt at a solution
Volume= area * thickness, so I got V= pi*r2dx for any given slice of that sphere. I could be wrong, but I think the density of water is 1000kg/m3, so I multiplied that by the volume to get mass of the water, and then multiplied the mass by 9.8 to get the weight of water. Then I set up my integral from 0 to 1.5 of the answer I got from multiplying all those numbers out * xdx.
But I got the wrong answer. So how do I work a problem such as this?
2. Feb 13, 2009
### Unco
First of all I'm not sure what you mean by the 'height' of a spherical tank.
Is the situation that water is essentially being lifted vertically to the top ofa spherical tank?
Assuming that is the case, the work done is the integral (sum) of the weight of each slice of the sphere.
Draw a diagram: a circle centred on (0,0) with radius 3. Let x be the y-ordinate of a slice. Then the radius of a slice is given by sqrt(3^2 - x^2).
You can then find the volume, and hence the mass by multiplying the density, of each slice. I'll let you sort out the units.
Integrate that over the appropriate interval that x ranges over.
Show us your work if you get stuck.
3. Feb 14, 2009
### HallsofIvy
Staff Emeritus
Your mistake is that you have the weight of all of the water in the sphere being lifted from every height in the sphere: in other words, the same weight is being multiple times.
Analyse the problem using things that are constant- that's one reason for emphasizing "Riemann sums" as the definition of integral: that's often how we set up integrals in applications.
As Unco suggeted, draw a circle representing the sphere. Put it in "standard position" in a coordinate system so the center is at (0,0) and the radius is 3: $x^2+ y^2= 9$. Draw a horizontal line at any level in the circle, at y, representing one "layer" of water, with thickness "dy". All the water in that "layer" is at the same height, y, and so must be lifted the same distance, 3- y (notice that y may be from -3 to 3 so this distance is from 3-(-3)= 6 at the bottom of the tank to 3-3= 0 at the top).
Now, in 3 dimensions, that "layer" of water is a disk whith radius equal to the x coordinate where the line at y crosses the circle: $x^2+ y^2= 9$ so "radius squared" is $]x^2= 9- y^2$ and the area of the disk is $\pi(9- y^2)$. The volume of the thin layer is $\pi(9-y^2)$ and since weight density of water is 1000(9.81)= 9810 Newtons/l3, the weight of that layer is $9810\pi(9- y^2)dy$ and, finally, the work done lifting that weight a distance 3- y is $9810\pi(9- y^2)(3- y)dy$.
You get work required to lift all of the water to the top of the sphere by "adding" the work done on every "layer" of water in the sphere. That is, you integrate that from y= -3 to y= 3.
4. Jan 18, 2010
### justamemory
But, what about the work done to get the water out of the spigget at the top? Isn't (3-y) just the distance to the top of the sphere, but before the spigget?
Last edited: Jan 18, 2010 | 2017-12-11T21:15:02 | {
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http://oftafarma.com/v2stgkjw/6a1d6b-what-is-a-diagonal-matrix | Here are a few facts. The diagonal from the top left corner to the bottom right corner of a square matrix is called the main diagonal or leading diagonal. It is denoted by I . A square matrix with 1's as diagonal elements and 0’s as other elements is called an Identity matrix. If we interchange rows and columns of an m×n matrix to get an n × m matrix, the new matrix is called the transpose of the given matrix. is a diagonal matrix with diagonal entries equal to the eigenvalues of A.The position of the vectors C j in P is identical to the position of the associated eigenvalue on the diagonal of D.This identity implies that A is similar to D.Therefore, A is diagonalizable. The elements of a matrix starting in the upper left corner and proceeding down and to the right. The main diagonal of a matrix consists of those elements that lie on the diagonal that runs from top left to bottom right.. 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https://math.stackexchange.com/questions/86644/determinant-of-a-specially-structured-matrix-as-on-the-diagonal-all-other-e | # Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$) [duplicate]
I have the following $n\times n$ matrix:
$$A=\begin{bmatrix} a & b & \ldots & b\\ b & a & \ldots & b\\ \vdots & \vdots & \ddots & \vdots\\ b & b & \ldots & a\end{bmatrix}$$
where $0 < b < a$.
I am interested in the expression for the determinant $\det[A]$ in terms of $a$, $b$ and $n$. This seems like a trivial problem, as the matrix $A$ has such a nice structure, but my linear algebra skills are pretty rusty and I can't figure it out. Any help would be appreciated.
• Please do not use math displays in titles. Nov 29, 2011 at 6:23
• There's a geometric, rather than algebraic, way of viewing it that makes it easy to understand. I've posted it below. Nov 29, 2011 at 15:56
• I've been away for a while -- I apologize for the late acceptance of an answer (I've looked at this before I left, but haven't had time to pick the best answer.) Also, @MarianoSuárez-Alvarez, I apologize for putting the matrix into the title... Thanks for correcting me on that. Dec 13, 2011 at 6:40
• See this post for the generalization of your determinant when the entires on the main diagonal are distinct. Jun 7, 2020 at 21:40
Add row 2 to row 1, add row 3 to row 1,..., add row $n$ to row 1, we get $$\det(A)=\begin{vmatrix} a+(n-1)b & a+(n-1)b & a+(n-1)b & \cdots & a+(n-1)b \\ b & a & b &\cdots & b \\ b & b & a &\cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & \ldots & a \\ \end{vmatrix}$$ $$=(a+(n-1)b)\begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ b & a & b &\cdots & b \\ b & b & a &\cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & \ldots & a \\ \end{vmatrix}.$$ Now add $(-b)$ of row 1 to row 2, add $(-b)$ of row 1 to row 3,..., add $(-b)$ of row 1 to row $n$, we get $$\det(A)=(a+(n-1)b)\begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & a-b & 0 &\cdots & 0 \\ 0 & 0 & a-b &\cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & a-b \\ \end{vmatrix}=(a+(n-1)b)(a-b)^{n-1}.$$
• I like this solution due to its simplicity and elegance. Other solutions are great too. Dec 13, 2011 at 6:42
• nice solution. easy to understand. Apr 22, 2013 at 6:33
• add (−b) of row 1 to row 2 - there is no $-b$ in row $1$, what you have done here? Oct 7, 2018 at 4:48
• @taritgoswami It means -b lots of row 1, as in $r_2 \mapsto r_2 + (-b)r_1.$ Apr 5, 2021 at 10:05
SFAICT this route hasn't been mentioned yet, so:
Consider the decomposition
$$\small\begin{pmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots&&\ddots&\vdots\\b&\cdots&b&a\end{pmatrix}=\begin{pmatrix}a-b&&&\\&a-b&&\\&&\ddots&\\&&&a-b\end{pmatrix}+\begin{pmatrix}\sqrt b\\\sqrt b\\\vdots\\\sqrt b\end{pmatrix}\cdot\begin{pmatrix}\sqrt b&\sqrt b&\cdots&\sqrt b\end{pmatrix}$$
Having this decomposition allows us to use the Sherman-Morrison-Woodbury formula for determinants:
$$\det(\mathbf A+\mathbf u\mathbf v^\top)=(1+\mathbf v^\top\mathbf A^{-1}\mathbf u)\det\mathbf A$$
where $\mathbf u$ and $\mathbf v$ are column vectors. The corresponding components are simple, and thus the formula is easily applied (letting $\mathbf e$ denote the column vector whose components are all $1$'s):
\begin{align*} \begin{vmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots&&\ddots&\vdots\\b&\cdots&b&a\end{vmatrix}&=\left(1+(\sqrt{b}\mathbf e)^\top\left(\frac{\sqrt{b}}{a-b}\mathbf e\right)\right)(a-b)^n\\ &=\left(1+\frac{nb}{a-b}\right)(a-b)^n=(a+(n-1)b)(a-b)^{n-1} \end{align*}
where we used the fact that $\mathbf e^\top\mathbf e=n$.
• This is very pretty! Jul 28, 2012 at 5:35
• Can't you put $\mathbf u:=\mathbf e$ and $\mathbf v:=b\mathbf e$ (or vice versa) to avoid the radicals?
– yo'
Oct 2, 2015 at 20:09
• @yo, yes, that can be done. Oct 6, 2015 at 12:49
Define $$\mathbf{1} = \begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix}$$. Take $$P = b\mathbf{1} \mathbf{1}^T$$ (outer product!) and observe that $A=P + (a-b) I .$
We begin with observations on the matrix $$P = b\mathbf{1} \mathbf{1}^T$$:
1. All rows and columns are equal and $$b>0$$, so P is a rank $$1$$ matrix. Thus $$\lambda_1=0$$ is an eigenvalue of multiplicity $$n-1$$.
2. $$P \mathbf{1} = b\mathbf{1} \mathbf{1}^T\mathbf{1} = n b \mathbf{1}$$. Thus $$\lambda_2 = nb$$ is an eigenvalue of multiplicity $$1$$.
We now use the following theorem:
Theorem: If $$r$$ is an eigenvalue of $$T$$, then $$r+s$$ is an eigenvalue of $$T+sI$$.
Proof: Since $$r$$ is an eigenvalue of $$T$$, there exists $$\mathbf{v}$$ such that $$T \mathbf{v} = r \mathbf{v}$$. Then $(T + sI)\mathbf{v} = T \mathbf{v} + sI \mathbf{v} = r \mathbf{v} + s \mathbf{v} = (r + s) \mathbf{v}. \quad \text{qed}$
Thus the eigenvalues of $$P+(a-b)I$$ will be $$\lambda_1 = (0+(a-b))$$ with multiplicity $$n-1$$ and $$\lambda_2 = nb + (a-b)$$ with multiplicity one. Since the determinant is the product of the eigenvalues, we have that $$\det(A) =\begin{vmatrix} a & b & \ldots & b\\ b & a & \ldots & b\\ \vdots & \vdots & \ddots & \vdots\\ b & b & \ldots & a\end{vmatrix} = \det(P+(a-b)I) = (a-b)^{n-1} (nb+a-b) .$$
• I love this prove, the theorem stated is the central core in inverse power method for approximating eigenvectors of a matrix with an approximation to his eigenvalue. Oct 8, 2018 at 5:42
• Oh! What a proof! Mind blowing. Feb 1, 2020 at 16:12
This is indeed an easy problem. Let $J$ be the square matrix with every entry equal to $1$. Your problem is equivalent to finding the determinant of $\lambda I + \mu J$ for arbitrary $\lambda, \mu$. Let $v=\left(\frac1{\sqrt{n}}, \frac1{\sqrt{n}},\ldots,\frac1{\sqrt{n}}\right)^\top$ and $e=(1,0,\ldots,0)^\top$. Then $J=nvv^\top$. Take any orthogonal matrix with its first column equal to $v$. Then $V^\top(\lambda I + \mu J)V = \lambda I+\mu nee^\top = \textrm{diag}\left(\lambda+\mu n,\lambda,\ldots,\lambda\right)$. Hence $\det(\lambda I + \mu J) = (\lambda+\mu n)\lambda^{n-1}$. Put $\lambda=a-b$ and $\mu=b$, we get the answer to your question as $[a+(n-1)b](a-b)^{n-1}$.
Subtract the bottom row from each of the other rows, then expand along some convenient row or column.
The matrix can be diagonalized. All it takes is a bit of geometry. We have
$$A=\begin{bmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots& &\ddots&\vdots\\b&\cdots&b&a\end{bmatrix}.$$
This is a linear combination of the matrices $P$ and $Q=I-P$ where $P$ is the matrix of the orthogonal projection onto the $1$-dimensional space of column vectors in which all scalar components are equal, i.e. the space $$\left\{\begin{bmatrix} x \\ x \\ x \\ \vdots \\ x \end{bmatrix} : x \text{ is a scalar} \right\}.$$ We have $$P = \begin{bmatrix} 1/n & 1/n & \ldots & 1/n \\ 1/n & 1/n & \ldots & 1/n \\ \vdots & \vdots & & \vdots \\ 1/n & 1/n & \ldots & 1/n \end{bmatrix}$$ (all entries are $1/n$), so that $$P \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} \bar{x} \\ \bar{x} \\ \bar{x} \\ \vdots \\ \bar{x} \end{bmatrix}$$ where $\bar{x} = (x_1+\cdots+x_n)/n$ is the average of the components, and $$Q \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} x_1-\bar{x} \\ x_2-\bar{x} \\ x_3-\bar{x} \\ \vdots \\ x_n-\bar{x} \end{bmatrix}.$$ We want $$A = \alpha P + \beta Q.$$ Looking at the diagonal elements we have $$\alpha\cdot\frac 1n + \beta \left(1-\frac 1n\right) = a,$$ and from the off-diagonal elements we get $$\alpha\cdot\frac 1n - \beta \cdot\frac 1n =b.$$ Hence $$\alpha = a + (n-1)b \qquad\text{and}\qquad\beta= a-b.$$
Since $P$ projects orthogonally onto a $1$-dimensional subspace and $Q$ is the complementary orthogonal projection onto an $(n-1)$-dimensional subspace, the matrix $\alpha P+\beta Q$ can be diagonalized as $$\begin{pmatrix} \alpha \\ & \beta \\ & & \beta \\ & & & \beta \\ & & & & \ddots \\ & & & & & \beta \end{pmatrix}.$$ The determinant is therefore $$\alpha\beta^{n-1}.$$
Hint: We can assume that the ground ring is a field, and that $b\neq0$. Consider the subspace formed by the vectors with equal coordinates, and the subspace formed by the vectors whose coordinates add up to $0$; note that these two subspaces are eigenspaces; compute the corresponding eigenvalues; and conclude.
EDIT. To find the eigenspaces of $A$ you can add $b-a$ to it, to make all the entries equal to $b$. (Adding a scalar to $A$ doesn't affect the eigenspaces.)
Consider the $n\times n$ matrix $B$ with entries $-b$ everywhere, except on the main diagonal where it has entries $0$. Now $\det A=\det(aI-B)$ is just the value of the characteristic polynomial $\chi_B\in K[X]$ at $X=a$. For $X=b$ the matrix $bI-B$ clearly has rank at most$~1$ (all columns are equal), so by rank-nullity the eigenspace of $B$ for the eigenvalue $b$ has dimension at least $n-1$, and $\chi_B$ is divisible by $(X-b)^{n-1}$. The final eigenvalue must make their sum $\operatorname{tr}(B)=0$ so it is $-(n-1)b$, and the final factor of $\chi_B$ is $X+(n-1)b$. So$$\det A=\chi_B[X:=a] = (a-b)^{n-1}(a+(n-1)b).$$
Just to give my two cents: if $$a=b$$ the result is trivially zero; else, we can write $$A = (a-b){\rm Id}_n + b{\bf 1}_n$$, where $${\bf 1}_n$$ is a matrix consisting only of $$1$$'s. Then \begin{align}\det(A) &= \det((a-b){\rm Id}_n+b{\bf 1}_n) = (a-b)^n \det\left({\rm Id}_n + \frac{b}{a-b}{\bf 1}_n\right) \\ &\stackrel{(\ast)}{=} (a-b)^n\left(1 + {\rm tr}\left(\frac{b}{a-b} {\bf 1}_n\right) \right) = (a-b)^n\left(1+ \frac{nb}{a-b}\right) \\ &= (a-b)^n + nb(a-b)^{n-1} = (a-b)^{n-1}(a+(n-1)b),\end{align}where in $$(\ast)$$ we used the "Sylvester-like" formula $$\det({\rm Id}_n+B) = 1+{\rm tr}(B)$$, valid for matrices $$B$$ with $${\rm rank}(B)=1$$. | 2022-07-01T23:59:29 | {
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http://mathhelpforum.com/calculus/43171-estimating-sum-series.html | # Math Help - Estimating the sum of a series
1. ## Estimating the sum of a series
Hello once again,
I have this problem that wants me to estimate the sum of a given series within 0.0001.
"Consider the series $\sum_{k=1}^{\infty}\frac{{4(-1)^k}}{(3^k)*k!}$. Estimate the sum of the series within 0.0001."
I can't find any examples in my textbook that seem to explain how I would go about this. Can anyone explain how to approach this problem? I would greatly appreciate any help.
Thanks.
2. Hello !
Originally Posted by auslmar
Hello once again,
I have this problem that wants me to estimate the sum of a given series within 0.0001.
"Consider the series $\sum_{k=1}^{\infty}\frac{{4(-1)^k}}{(3^k)*k!}$. Estimate the sum of the series within 0.0001."
I can't find any examples in my textbook that seem to explain how I would go about this. Can anyone explain how to approach this problem? I would greatly appreciate any help.
Thanks.
The sum is equal to :
$4 \cdot \sum_{k=1}^{\infty} \frac{(-1)^k}{3^k \cdot k!}=4 \cdot \sum_{k={\color{red}1}}^{\infty} \frac{\left(-\frac 13\right)^k}{k!}$.
Now, I will use a known series :
$e^x=\sum_{k={\color{red}0}}^{\infty} \frac{x^k}{k!}=1+\sum_{k={\color{red}1}}^{\infty} \frac{x^k}{k!}$
$\implies \sum_{k={\color{red}1}}^{\infty} \frac{x^k}{k!}=e^x-1$
Here, $x=-\tfrac 13$. Can you finish it ?
3. Originally Posted by auslmar
Hello once again,
I have this problem that wants me to estimate the sum of a given series within 0.0001.
"Consider the series $\sum_{k=1}^{\infty}\frac{{4(-1)^k}}{(3^k)*k!}$. Estimate the sum of the series within 0.0001."
I can't find any examples in my textbook that seem to explain how I would go about this. Can anyone explain how to approach this problem? I would greatly appreciate any help.
Thanks.
This is an alternating series of decreasing terms, so the error in a partial sum $S_n$ of the first $n$ terms is less than the first neglected term.
So:
$|S-S_n|<\frac{4}{3^{n+1} (n+1)!}$
So we want:
$|S-S_n|<0.0001$
we find the smallest integer $n$ such that:
$\frac{4}{3^{n+1} (n+1)!}\le 0.0001$
which by trial and error we find is $n=?$
RonL
4. Hello, auslmar!
An elaboration of moo's excellent solution . . .
Consider the series: . $\sum_{k=1}^{\infty}\frac{{4(-1)^k}}{3^k\cdot k!}$
Estimate the sum of the series within 0.0001.
We have: . $S \;=\;-\frac{4}{3\cdot1!} + \frac{4}{3^2\cdot2!} - \frac{4}{3^3\cdot3!} + \frac{4}{3^4\cdot4!} - \hdots$
. . . . . . . . $S \;=\;4\underbrace{\left[-\frac{\frac{1}{3}}{1!} + \frac{\frac{1}{3^2}}{2!} - \frac{\frac{1}{3^3}}{3!} + \frac{\frac{1}{3^4}}{4!}\hdots\right]}_{\text{Infinite series of: }e^{\text{-}\frac{1}{3}}-1}$
Therefore: . $S\;=\;4\left(e^{-\frac{1}{3}} - 1\right) \;\approx\;-1.1339$
5. Sorry for the late response, I had to hit the hay.
Thanks to everyone for the help. I'm not sure I actually understand Moo's or Soroban's examples, but CaptainBlack's makes the most sense to me. If I understand correctly, the goal is to determine the sufficient nth term of the series that would be less than or equal to within 0.0001 of the "actual" sum. Is this notion correct? I applied it to a similar problem with correct results.
Thanks again!
Originally Posted by CaptainBlack
This is an alternating series of decreasing terms, so the error in a partial sum $S_n$ of the first $n$ terms is less than the first neglected term.
So:
$|S-S_n|<\frac{4}{3^{n+1} (n+1)!}$
So we want:
$|S-S_n|<0.0001$
we find the smallest integer $n$ such that:
$\frac{4}{3^{n+1} (n+1)!}\le 0.0001$
which by trial and error we find is $n=?$
RonL | 2015-08-29T10:36:16 | {
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https://math.stackexchange.com/questions/3151292/how-to-interpret-negative-surface-area | How to interpret negative surface area?
In calculating $$\iint_Dx^2y-y^5 dxdy$$
where D is given by:$$~~~~~1-y^2\leq x\leq 2-y^2\\-\sqrt{1+x}\leq y\leq\sqrt{1+x}$$
I refered to the graphs in the following link: enter link description here to determine the region of integration and reverse the order of integration by switching domains to:$$0\leq x\leq2\\\sqrt{1-x}\leq y\leq\sqrt{2-x}\\$$
From this I found that $$\int_0^2\int_\sqrt{1-x}^\sqrt{2-x} x^2y-y^5dydx=\frac{2}3$$ Which is multiplied by 2 for the total area, since $$\frac{2}3$$ only covers the first quadrant
But then the strip that is above $$y=\sqrt{1+x}$$ comes along with it, so I figured I must calculate the following integral and subtract it from $$\frac{2}3$$:$$\int_0^{1/2}\int_\sqrt{1+x}^\sqrt{2-x}x^2y-y^5dydx=-\frac{9}{32}\\$$
Why is this negative? And how should I interpret it? I wanted to subtract but should I now add it? Both calculations are correct but the negative area makes me think that there must be some incorrect reasoning in integrating x from 0 to 1/2 and then subtracting it from the first area
• The negative value is caused by the integrand in which the $-y^5$ is big as it is greater than 1. Furthermore it is not an area integral so negative area is not a concern. Also your first integral should allow for x from 1 to 2 in which the y limits should be from 0 to $\sqrt{2 - x}$. – KY Tang Mar 18 at 0:56
• Thank you, but allowing for x from 1 to 2, do you mean this would make three integrals with x from 0to1/2, from 1/2to1 and 1to2 and then adding them up? – TheMercury79 Mar 18 at 5:10
• Yes please see my answer below. Thanks – KY Tang Mar 18 at 13:15
2 Answers
There should be 5 integrals: $$I_1 = \int_1^2\int_{-\sqrt{2 -x}}^\sqrt{2 - x}(x^2y - y^5)dy dx = 0$$
$$I_2 = (\int_{\frac{1}{2}}^{1}\int_{\sqrt{1 - x}}^{\sqrt{2 - x}} + \int_\frac{1}{2}^{1}\int_{-\sqrt{2 - x}}^{-\sqrt{1 - x}})(x^2y - y^5)dy dx = 0$$
$$I_3 = (\int_0^\frac{1}{2}\int_{\sqrt{1 - x}}^{\sqrt{1 + x}} + \int_0^{\frac{1}{2}}\int_{-\sqrt{1 + x}}^{-\sqrt{1 - x}})(x^2y - y^5)dy dx = 0$$
Hence the answer is 0.
Your picture shows that the domain of integration is mirror symmetric with respect to the $$x$$-axis. Since the integrand is an odd function of $$y$$ it follows that the value of the integral is $$0$$. | 2019-06-19T03:23:24 | {
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https://www.physicsforums.com/threads/rotational-question.391462/ | # Rotational Question
## Homework Statement
An astronaut on the surface of the Moon fires a cannon to launch an experiment package, which leaves the barrel moving horizontally. Assume that the free-fall acceleration on the Moon is one-sixth that on the Earth. Radius of moon = 1.74 * 10^6 m
a) What must be the muzzle speed of the probe so that it travels completely around the Moon and returns to its original location? Found out to be 1.68km/s
b) How long does this trip around the Moon take?
## Homework Equations
$$a=\omega^2r$$
$$v=\omega r$$
## The Attempt at a Solution
I tired to solve with constant acceleration formulas.
a=1.62m/s^2
vi=0
vf=1.68km/s => 1678.928m/s
vf=vi+at
16878.928=0+1.62t
t=10419.09136s
2)A crate of eggs is located in the middle of the flat bed of a pickup truck as the truck negotiates an unbanked curve in the road. The curve may be regarded as an arc of a circle of radius 35.0m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?
I have tried to do the following:
i know the static friction: 0.600
i need to know $$\omega$$
which is $$\sqrt{9.8}{35}$$, $$\omega=0.529$$
find velocity
$$v=r\omega$$
v=0.529 * 35
v=18.52m/s
i know this is incorrect, someone tell me how i can approach this.
Last edited:
Doc Al
Mentor
## Homework Statement
An astronaut on the surface of the Moon fires a cannon to launch an experiment package, which leaves the barrel moving horizontally. Assume that the free-fall acceleration on the Moon is one-sixth that on the Earth. Radius of moon = 1.74 * 10^6 m
a) What must be the muzzle speed of the probe so that it travels completely around the Moon and returns to its original location? Found out to be 1.68km/s
b) How long does this trip around the Moon take?
## Homework Equations
$$a=\omega^2r$$
$$v=\omega r$$
## The Attempt at a Solution
I tired to solve with constant acceleration formulas.
a=1.62m/s^2
vi=0
vf=1.68km/s => 1678.928m/s
vf=vi+at
16878.928=0+1.62t
t=10419.09136s
This is a circular motion/orbit problem. Use Newton's 2nd law and the formula for centripetal acceleration.
how can i find the time with the centripetal acceleration formula?
Doc Al
Mentor
how can i find the time with the centripetal acceleration formula?
OK. Looks like you solved for the speed (but didn't show your work). The speed is constant. What distance does it travel in going around once?
do you mean $$\omega = 9.649 * 10^{4} rad/s$$?
Doc Al
Mentor
do you mean $$\omega = 9.649 * 10^{4} rad/s$$?
You could certainly use that if you like. But that should be 10-4!
I was thinking of distance = speed * time. You have the speed. And you should be able to calculate the distance, then solve for the time.
How can i find the distance if i only have velocity, acceleration?
Doc Al
Mentor
How can i find the distance if i only have velocity, acceleration?
What path does it take? Hint: Make use of the moon's radius.
would you find the diameter of the moon, which is 3480000?
Doc Al
Mentor
Answer this: What is the shape of the package's orbit around the moon?
well the shape of a car....
Doc Al
Mentor
well the shape of a car....
It goes all the way around the moon in a big circle!
i misunderstood your question. -_-
Doc Al
Mentor
Do you understand what I mean now? How can you find the length of that path? (Remember that this is a question about circular motion.)
(You could also solve for the time using ω, which you've already calculated.)
yes, i understand it took me awhile -_-.
So from the equation: $$\omega = \frac{\theta}{t}$$
derived from the linear $$v=\frac{x}{t}$$
where:
$$\theta = 2\pi$$ because it go around once.
$$\omega = 9.649 * 10^{-4}$$
$$t=\frac{2\pi}{9.649*10^{-4}}$$
t=6511.74765s
$$t=\frac{6511.74765}{3600}$$ turn into hours
t=1.81hrs however book's answer has a slight difference in 1.80hrs. Edit** i know why because of my rounding off.
would someone answer my second question?
Last edited:
Doc Al
Mentor
yes, i understand it took me awhile -_-.
So from the equation: $$\omega = \frac{\theta}{t}$$
derived from the linear $$v=\frac{x}{t}$$
You used the first equation, but you could also have used the second one directly. When it goes around once, the package traces the circumference of a circle. So $x = 2\pi r$.
Doc Al
Mentor
2)A crate of eggs is located in the middle of the flat bed of a pickup truck as the truck negotiates an unbanked curve in the road. The curve may be regarded as an arc of a circle of radius 35.0m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?
I have tried to do the following:
i know the static friction: 0.600
i need to know $$\omega$$
which is $$\sqrt{9.8}{35}$$, $$\omega=0.529$$
find velocity
$$v=r\omega$$
v=0.529 * 35
v=18.52m/s
i know this is incorrect, someone tell me how i can approach this.
How did you solve for ω?
Again, this is a circular motion problem that can be solved using Newton's 2nd law and what you know about centripetal acceleration.
What force acts on the eggs? What's the expression for its acceleration?
FYI: ac = ω²r = v²/r
How did you solve for ω?
Again, this is a circular motion problem that can be solved using Newton's 2nd law and what you know about centripetal acceleration.
What force acts on the eggs? What's the expression for its acceleration?
FYI: ac = ω²r = v²/r
the mass and the centripetal force.
Don't i need to use the static friction?
Doc Al
Mentor
Don't i need to use the static friction?
Yes. Static friction is the only force available to provide the centripetal acceleration.
So would i start by finding the fsmax by using the equation
coefficient of static friction = $$\frac{fsmax}{n}$$
Doc Al
Mentor
So would i start by finding the fsmax by using the equation
coefficient of static friction = $$\frac{fsmax}{n}$$
Right. The static friction will be at its maximum, so F = μN. (What's N?)
Now apply Newton's 2nd law.
the normal
how do i find the N???
Doc Al
Mentor
how do i find the N???
To find an expression for the normal force, analyze the vertical force components. What's the net vertical force? | 2021-05-15T20:27:02 | {
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https://math.stackexchange.com/questions/2514441/minimize-min-f-in-e-left-int-01fx-dx-right | # Minimize $\min_{f\in E}\left(\int_0^1f(x) dx\right)$
Inspired by this question, I would like pose this following problem.
Let $E$ be the set of all nonnegative continuous functions $f:[0,1]\to \mathbb{R}$ such that $$f(x)\,f(y)\ge |x-y|\qquad\forall\{x,y\}\subset [0,1]$$
Find $$\min_{f\in E}\left(\int_0^1f(x) \,dx\right)$$
• Do you have an example of a function $f \in E$? – user159517 Nov 17 '17 at 13:04
• @user159517 $f(x) > \sup_{x,y} |x-y| = 1$ will do it. – Calvin Khor Nov 17 '17 at 13:23
• @CalvinKhor ah, of course. That also shows that $E$ is unbounded, hence not compact which makes the existence of a minmizing element tricky. – user159517 Nov 17 '17 at 13:47
Integrating both sides with respect to x and y we get
$$\left(\int_0^1f(x) dx\right)^2= \int_0^1\int_0^1f(x)f(y)dxdy \ge\int_0^1\int_0^1 |x-y|dxdy= \frac13.$$ that is $$\int_0^1f(x)dx \ge \frac{1}{\sqrt3.}$$
I am still searching whether the min is attain or not.
• +1 for thinking of the double integral. But @kimchi lover 's bound is the tight one, since any function with the required quality must dominate $\sqrt{|1-2x|}$. – Mathemagical Nov 11 '17 at 3:03
• My bound(s) might be tighter but I don't have a clue if the most recent version is the tight one. – kimchi lover Nov 11 '17 at 21:19
• i am not sure whether the minimizer exists for this problem. – Guy Fsone Nov 11 '17 at 21:46
• @GuyFsone Good question. I've been grasping for straws of compactness for this problem, without luck. – kimchi lover Nov 12 '17 at 0:02
Let $A=A(f)=\int_0^1f(x)\,dx$. By the AGM inequality, $$A=\frac12\int_0^1 \big(f(x)+f(1-x)\big) dx \ge \int_0^1 \sqrt{f(x)f(1-x)}\,dx \ge \int_0^1 \sqrt{|1-2x|}dx = \frac23.$$
Similarly, $$A=2\int_0^{1/2} \frac{f(x)+f(x+1/2)}2\, dx$$ $$\ge 2\int_0^{1/2} \sqrt{f(x)f(x+1/2)} \, dx\ge 2\int_0^{1/2}\sqrt{1/2}\,dx=\frac1{\sqrt2}.$$
This might or might not have been obvious all along, but not explicitly so to me until just now. Note that if $g_1$ and $g_2$ satisfy the constraints, then so does $g(x) = \sqrt{g_1(x)g_2(x)}$, and moreover, by the AGM inequality, $$\int_0^1 \sqrt{g_1(x)g_2(x)}\,dx\le \int_0^1 \frac {g_1(x)+g_2(x)} 2\, dx.$$ Given any admissible $h$, the functions $h_1(x) = h(x)$, $h_2(x)=h(1-x)$, and $f(x)=\sqrt{h_1(x)h_2(x)} = \sqrt{h(x)h(1-x)}$ are admissible, and $\int_0^1 f\le \int_0^1 h$. Hence one cannot beat functions of the form $\sqrt{h(x)h(1-x)}$, that is, functions that depend only on $|2x-1|$.
Another consequence of the AGM inequality is that if the minimum is attained, it is attained uniquely: Suppose $A(f)=A(g)$ where $f\ne g$. Then $A(h)<A(f)$, where $h(x)=\sqrt{f(x)g(x)}$.
Not only can one not beat functions of form $f(x)=\phi(|2x-1|)$ , but one cannot beat such functions where continuous $\phi:[0,1]\to\mathbb R^+$ is also required to be nondecreasing. The $f(x)f(y)\ge|x-y|$ constraint translates to a $\phi(u)\phi(v)\ge(u+v)/2$ constraint for $u,v\in[0,1]$. If $\phi$ obeys these constraints, so does monotonic $\phi^*(u)=\min\{\phi(t): t\in[u,1]\}$, for which the corresponding $A(f^*)\le A(f)$.
(I'm hoping to use a compactness argument somehow to then prove the minimum $A(f)$ is attained. I think that such an argument succeeds in the restricted problem where one sticks in an additional constraint $f(0)=\gamma$, but don't yet see yet why the restricted optimal value of $A$ depends continuously on $\gamma$.)
• The function $f(x)=\sqrt{|1-2x|}$ doesn't satisfy the assumption that $\sqrt{|1-2x|\cdot|1-2y|}\ge |x-y|$ for all $x,y\in[0,1]$ since you could choose $x=1/2$ and $y = 1$, and you would find $f(1/2)f(1) = 0 < |1/2-1| = 1/2$. – Alex Ortiz Nov 11 '17 at 3:01
• Any response to @AOrtiz's objection? – Hans Nov 11 '17 at 18:09
• @Hans: kimchi lover's argument is a correct proof that the minimum in question is $\geq{2\over3}$. He/She didn't state that $f(x)=\sqrt{|1-2x|}$ is optimal, or even admissible. – Christian Blatter Nov 11 '17 at 18:34
• @ChristianBlatter: I am well aware of that it is only a lower bound, just like Guy Fsone's but tighter. I want to ascertain if he is aware of it and whether he has any further idea on finding the minimum. – Hans Nov 11 '17 at 18:45
• It's an interesting problem, and I've not given up on it. – kimchi lover Nov 11 '17 at 18:57
NOTE: Still not a complete answer, just improving on the lower bound.
EDIT: Added an improvement on the upper bound (a better function which "works").
I decided to combine the two ideas in kimchi lover's bounds, by putting a bound on $$\int_0^1 f(x)dx = \int_0^{1/4} \left[f(x) + f(1-x) + f(1/2+x) + f(1/2-x)\right]dx$$ Considering the four values $(u,v,s,t) = (f(x), f(1-x), f(1/2-x), f(1/2+x))$ as independent, but satisfying the constraints \begin{align} &uv \geq 1-2x \\ &ut \geq 1/2 \\ &sv \geq 1/2 \\ &st \geq 2x, \end{align} I found a lower bound for $u+s+v+t$. (There are two more inequalities, of course, but the minimizing set of values turns out to be symmetric, $s = t$ and $u = v$, so the $us$ and $tv$ inequalities are weaker than the $ut$ and $sv$ ones.) The approach is rather brute-force: pick $u$ and $s$ as independent, consider the different cases for which of the inequalities involving $t$ and $v$ are the stronger ones, then minimize an expression in terms of $u$ and $s$. The calculations are tedious, and since this is not a complete answer anyway, I'll omit them. I got the following: $$u + s + t + v \geq \frac{3-4x}{\sqrt{1-2x}},$$ minimum attained at $u = v = \sqrt{1-2x}$ and $s = t = {1\over2u}$ (these are not too surprising, of course). Thus $$\int_0^1 f(x) dx \geq \int_0^{1\over4} \frac{(3-4x)dx}{\sqrt{1-2x}} = {5-2\sqrt2\over 3} \approx 0.72386$$ If we try to piece together a function from the $(u,s,t,v)$'s for $x \in [0,1/4]$, it still doesn't satisfy some of the constraints, such as for example $f(x)f(1) \geq 1-x$.
A function which does meet the constraints everywhere is $$f(x) = \frac{e^{|2x-1|}}{\sqrt{2e}}.$$ That it does work can be verified using the triangle inequality $$f(x)f(y) = \frac{e^{|2x-1|+|2y-1|}}{2e} \geq \frac{e^{2|x-y|}}{2e} \geq |x-y|$$ where the last one holds because $e^{2t} - 2et$ has a global minimum $0$ at $t = 1/2$. Its integral is $$\int_0^1 \frac{e^{|2x-1|}}{\sqrt{2e}}dx = 2\int_{1/2}^1 \frac{e^{2x-1}}{\sqrt{2e}}dx = \frac{e-1}{\sqrt{2e}} \approx 0.73694$$
EIDT: Following @kimchilover's idea to generalize the exponential function to $\beta e^{\alpha|2x-1|}$, the constraints will be satisfied if and only if the minimum of $$\beta^2 e^{\alpha|2x-1|}e^{\alpha|2y-1|} - |x-y|$$ over $[0,1]\times[0,1]$ is non-negative. The strongest constraints are obtained using $x \in [0,1/2], y \in [1/2,1]$ so that we need $$\beta^2 e^{2\alpha(y-x)} - (y-x) = \beta^2 e^{2\alpha t} - t \geq 0$$ for $t \in [0,1]$. We can find the minimum, then if the minimum is in $[0,1]$, require that it is non-negative, or if it is outside, require that the closest "cut-off" is non-negative. Long story short, the optimal ones I found to be the root of $e^\alpha(\alpha - 3/2) + 3/2 = 0$, $\alpha \approx 0.874217$ (by Wolfram), $\beta = 1/\sqrt{2e\alpha}$, and for the integral $\approx 0.733001$.
• Good! Sorry I didn't notice this post earlier. – kimchi lover Nov 14 '17 at 14:20
• @kimchilover That's also good to know. I've been trying to get some functional/differential equation to describe a family of functions that could be optimal, but I don't get anything useful. Basically I can prove rigorously that for every $x$, there must exist a $y$ such that $f(x)f(y) = |x-y|$, then differentiate with respect to either $x$ or $y$, but getting nowhere after that. It's how I came up with the exponential function in the first place, actually. By the way, this last exponential is definitely not optimal either, because for $x \in [1/2, 1/2\alpha)$ no such $y$ exists. – Nick Pavlov Nov 17 '17 at 12:39
• I had to retract my most recent claim: what I wrote was obviously false, and what I meant to write is more slippery than I thought. – kimchi lover Nov 17 '17 at 12:58
• @kimchilover I must admit I didn't actually read your argument; monotonicity is another property that intuitively makes sense, so it was good enough for me that you claimed to have proven it. I was much more motivated to find a mistake when you claimed to beat my exponential (which at the time seemed so elegant that I truly hoped it was the answer). I only crunched the numbers on the generalized exponential because I was so convinced I would find that $\alpha = 1$ is best. Now that the illusion is dispelled, I feel way out of my depth here - I haven't even heard of this theorem you mention! – Nick Pavlov Nov 17 '17 at 13:19
• The mistake was minor, and I have now repaired it. The Helly-Bray theorem was big news in the 1920 or whenever, but is nowadays subsumed into the "Portmanteau Theorem" or "Prokhorov's Theorem" in probability theory. My idea was, bounded monotone functions on $[0,1]$ are in effect finite measures on $[0,1]$, the set of which is compact, by H-B. – kimchi lover Nov 17 '17 at 13:51
Extending kimchi lover's idea:
$$I = n \int_0^{1/n} \frac{1}{n} \sum_{k=0}^{n-1} f(x+k/n) dx \geq n \int_0^{1/n} \left(\prod_{k=0}^{n-1} f(x+k/n) \right)^{1/n} dx$$
Pairing up opposite edges, then we can pick pairs $(x+k/n, x+(n-k+1)/n)$, with $n=2m$, such that
$$\lim_{n \to \infty} I \geq \lim_{n \to \infty} 2m \int_0^{1/2m} \left( \prod_{k=1}^{m} \left(\frac{2k-1}{2m} \right) \right)^{1/2m} dx = \lim_{m \to \infty} \frac{[(2m-1)!!]^{1/2m}}{(2m)^{1/2}} = \sqrt{\frac{1}{e}}$$
where we compute the limit using wolfram alpha.
This bound isn't the best one found so far, but perhaps my approach will be inspiring. Loved thinking about the problem!
EDIT: More information. Using a continuous approximate (from above) to the step function $f(x)=a^{1/2} \chi_{[0,a]} + a^{-1/2} \chi_{(a,1]}$, you should be able to obtain a value below 1 for the integral if I did by calculation correctly. It is clear that $f=1$ means the minimum is less than or equal to 1.
• Interesting. I check your $a\approx.767$ yields $I\approx.937774$. I tried an $f(x)=1$ for $|x-1/2|>c/2$, otherwise $f(x)=b$, found $b=2/3$ and $c=b^2$ seemed optimal, and got $I=23/27\approx.852$. – kimchi lover Nov 13 '17 at 23:53
• @kimchilover Guys, not to be immodest, but in my answer there's a function which has a significantly lower $I$ then these. Not to mention even $|x-1/2|+1/2$ produces $I = 3/4$. – Nick Pavlov Nov 14 '17 at 13:04
• @NickPavlov Sorry about that: I suffer tunnel vision on this web site, and the most recent thing I see flushes out knowledge of better things I've seen before. – kimchi lover Nov 14 '17 at 13:18
• No, thanks for making us aware @NickPavlov! – abnry Nov 14 '17 at 18:30
## Numerical experiment
The following numerical experiment yields, as we will see later, the correct assumptions for getting a good candidate for the optimum. Convexity will help us show that this candidate is indeed the optimum.
We plot the active set of the optimum of a discretization:
This is the output of the following code: activeset[k_] := Module[{h, constraints, vars, objective, optresult, result, pts}, h = 1/(2 k - 2); constraints = Flatten[Table[ Table[x[i] x[j] >= (2 k - i - j) h, {j, i, k}], {i, 1, k}]]; constraints = Join[constraints, Table[x[i] >= 0, {i, 1, k}]]; vars = Table[x[i], {i, 1, k}]; objective = Sum[x[i], {i, 1, k}] + Sum[x[i], {i, 2, k - 1}]; optresult = FindMinimum[{objective, constraints}, vars, Method -> "InteriorPoint"]; result = Table[{(i - 1) h, x[i]}, {i, 1, k}] /. optresult[[2]]; pts = Select[Flatten[Table[{i, j}, {i, 1, k}, {j, 1, k}], 1], Abs[x[#[[1]]] x[#[[2]]] - (2 k - #[[1]] - #[[2]]) h] < 10^(-5) /. optresult[[2]] &]; ListPlot[(pts - 1)/(k - 1), AspectRatio -> 1, PlotStyle -> Red, PlotTheme -> "Detailed"] ]; activeset[71]
## Full solution
Set $S:=\{(x,y) \in [0,1]^2 : x+y \leq 1\}$. We define the maps \begin{alignat*}{2} &J : C[0,1] \rightarrow \mathbb{R},\quad &&J(u) := \int_0^1 \! \exp(u(x)) \, \mathrm{d}x, \\[1em] &I : C[0,1] \rightarrow C(S),\quad &&I(u)(x,y) := u(x)+u(y)-\log(2-x-y). \end{alignat*} We will study the optimization problems \begin{align} & \text{minimize} && J(u) \nonumber\\ & \text{subject to} && I(u) \geq 0, \tag{1} \\[1em] \text{and}\nonumber \\[1em] & \text{minimize} && \int_0^1 \! f(x) \, \mathrm{d}x \nonumber\\ & \text{subject to} && f \in C[0,1], \tag{2}\\ &&& f \geq 0, \nonumber\\ &&& f(x)f(y) \geq |x-y| \text{ for all } x,y \in [0,1]. \nonumber \end{align}
We will show that:
The unique solution $\bar{u}$ of $(1)$ is given by \begin{align*} \exp(\bar{u}(x)) &= \sqrt{2 \left(1+\sqrt{2}\right)}-\sqrt{2 x+\sqrt{2}-1}, \\[1em] J(\bar{u}) &= \frac{2}{3}\sqrt{1+\sqrt{2}}. \end{align*} The unique solution $\bar{f}$ of $(2)$ is given by \begin{align*} \bar{f}(x) &= \begin{cases} \dfrac{\exp(\bar{u}(2x))}{\sqrt{2}}&\text{if}\quad 0 \leq x \leq \frac{1}{2},\\ \dfrac{\exp(\bar{u}(2(1-x)))}{\sqrt{2}}&\text{if}\quad \frac{1}{2} < x \leq 1, \end{cases} \\[1em] \int_0^1 \! \bar{f}(x) \, \mathrm{d}x &= \frac{1}{3} \sqrt{2 \left(1+\sqrt{2}\right)}\quad (\approx 0.732456) \end{align*} where $\bar{u}$ is the unique solution of $(1)$.
Theorem 1 (Generalized Kuhn-Tucker Theorem).
Let $X$ be a normed space and $Y$ a normed space having positive cone $P$. Assume that $P$ contains an interior point.
Let $J$ be a Fréchet differentiable real-valued functional on $X$ and $I$ a Fréchet differentiable mapping from $X$ into $Y$. Suppose $\bar{u}$ minimizes $J$ subject to $I(\bar{u}) \geq 0$ and that $\bar{u}$ is a regular point of the inequality $I(\bar{u}) \geq 0$ i.e. there exists an $h \in X$ such that $I(\bar{u}) + I'(\bar{u})(h) > 0$. Then there is a positive functional $\psi \in Y^*$ such that \begin{align} J'(\bar{u}) &= \psi \circ I'(\bar{u}), \tag{3} \\\psi(I(\bar{u})) &= 0. \tag{4} \end{align}
Proof. See [Luenberger, David G, Optimization by vector space methods, Wiley, New York, 1969, p. 249]. $$\tag*{\Box}$$
Note: The following assumption (iii) is inspired by our numerical experiment.
Claim 1. If
(i) $\bar{u}$ is the solution of $(1)$
(ii) $\bar{u} \in C[0,1]$
(iii) there exists $g \in C^1[0,1]$ such that $$A:=\{(x,g(x)) : x \in [0,1]\} = \left\{(x,y) \in S : I(\bar{u})(x,y) = 0\right\},$$
then $g$ is an involution and $$\exp(\bar{u}(x)) = -g'(x)\exp(\bar{u}(g(x)))$$ for all $x \in [0,1]$.
Proof. By (iii) we have $I(\bar{u})(x,g(x)) = 0$ for all $x \in [0,1]$.
Since $I(\bar{u})(x,y) = I(\bar{u})(y,x)$ for all $x,y \in [0,1]$, we have $I(\bar{u})(g(x),x) = 0$ and thus $g(g(x)) = x$ by (iii) for all $x \in [0,1]$. Therefore $g$ is an involution.
We will now apply Theorem 1. The point $\bar{u}$ is regular, since $I'(\bar{u})(\mathbf{1}_{[0,1]}) = 2\cdot \mathbf{1}_S.$
Thus there exists a positive functional $\psi \in C(S)^*$ such that $(3)$ and $(4)$ are satisfied. Since $S$ is compact, there is a unique regular Borel measure $\nu$ on $\mathcal{B}(S)$ by the Riesz-Markov-Kakutani representation theorem such that $$\psi(v) = \int_S v(x,y) \, \mathrm{d}\nu(x,y)$$ for all $v \in C(S)$. We define a measure on $\mathcal{B}(S)$ by setting $$\mu(E) := \lambda\left(\{x \in [0,1] : (x,g(x)) \in E\}\right)$$ for all $E \in \mathcal{B}(S)$, where $\lambda$ is the Lebesgue measure on $[0,1]$. We will show that $\nu \ll \mu$, i.e. that $\nu$ is absolutely continuous with respect to $\mu$. Let $N \in \mathcal{B}(S)$ such that $\mu(N)=0$. By $(4)$ we have $\psi(I(\bar{u})) = 0$. Therefore $$\int_{S} I(\bar{u})(x,y) \, \mathrm{d}\nu(x,y) = 0,$$ and since $I(\bar{u}) \geq 0$, we have $\nu(S\setminus A)=0$ and $\nu(N \cap (S\setminus A))=0$. By $(3)$ we have $$\int_0^1 \exp(\bar{u}(x)) h(x) \, \mathrm{d}x = \int_S h(x) + h(y) \, \mathrm{d}\nu(x,y)$$ for all $h \in C[0,1]$. Set $N' = \{x \in [0,1] : (x,g(x)) \in N\}$. By definition of $\mu$ we have $\lambda(N') = 0$. Let $\epsilon > 0$. Then there exists $\bar{h} \in C[0,1]$ such that $\bar{h}(x) = 1$ for all $x \in N'$, $\bar{h}\geq 0$, and $\int_0^1 \exp(\bar{u}(x)) \bar{h}(x) \, \mathrm{d}x<\epsilon$. Now we have \begin{align*} \nu(N \cap A) &= \int_S \mathbf{1}_{N \cap A}(x,y) \, \mathrm{d}\nu(x,y) \\[1em] &= \int_S \mathbf{1}_{N'}(x) \, \mathrm{d}\nu(x,y) \\[1em] &\leq \int_S \bar{h}(x) + \bar{h}(y) \, \mathrm{d}\nu(x,y) \\[1em] &= \int_0^1 \exp(\bar{u}(x)) \bar{h}(x) \, \mathrm{d}x<\epsilon. \end{align*} Therefore $\nu(N \cap A) = 0$, thus $\nu(N) = 0$ and $\nu \ll \mu$. By the Radon–Nikodym theorem there exists a measurable function $w : S \rightarrow [0,\infty)$ such that $$\psi(v) = \int_S v(x,y)\, \mathrm{d}\nu(x,y) = \int_S v(x,y)\,w(x,y) \, \mathrm{d}\mu(x,y)$$ for all $v \in C(S)$. Since $\mu$ is the pushforward measure under $x \mapsto (x,g(x))$ we have $$\psi(v) = \int_0^1 v(x,g(x))\,w(x,g(x)) \, \mathrm{d}x$$ for all $v \in C(S)$. By $(4)$ we now have $$\int_0^1 \exp(\bar{u}(x)) h(x) \, \mathrm{d}x = \int_0^1 (h(x) + h(g(x)))\,w(x,g(x)) \, \mathrm{d}x$$ for all $h \in C[0,1]$. Since $g$ is a $C^1$-involution on $[0,1]$, we can substitute $g(x)$ for $x$ in $\int_0^1 h(g(x))\,w(x,g(x)) \, \mathrm{d}x$. We get $$\int_0^1 \exp(\bar{u}(x)) h(x) \, \mathrm{d}x = \int_0^1 h(x) (w(x,g(x))-w(g(x),x)g'(x)) \, \mathrm{d}x$$ for all $h \in C[0,1]$. Therefore $$\exp(\bar{u}(x)) = w(x,g(x))-w(g(x),x)g'(x)$$ for almost all $x \in [0,1]$. Since $g$ is an involution, we have \begin{align*} \exp(\bar{u}(g(x))) &= w(g(x),g(g(x)))-w(g(g(x)),g(x))g'(g(x)) \\[1em] &= w(g(x),x)-w(x,g(x)) \frac{1}{g'(x)} \\[1em] &= -\frac{1}{g'(x)}(w(x,g(x))-w(g(x),x)g'(x)) \\[1em] &= -\frac{1}{g'(x)}\exp(\bar{u}(x)) \end{align*} for almost all $x \in [0,1]$. Since $g$, $g'$ and $\bar{u}$ are continuous we have \begin{align} \exp(\bar{u}(x)) = -g'(x)\exp(\bar{u}(g(x))) \tag{5} \end{align} for all $x \in [0,1]$. $$\tag*{\Box}$$
Claim 2. If
(i) $\bar{u}$ is the solution of $(1)$
(ii) $\bar{u} \in C^1[0,1]$
(iii) there exists $g \in C^2[0,1]$ such that $$\{(x,g(x)) : x \in [0,1]\} = \left\{(x,y) \in S : I(\bar{u})(x,y) = 0\right\}$$ and $$\{(x,g(x)) : x \in (0,1)\} \subseteq \operatorname{int}(S),$$ then $$g(x) = 1+\sqrt{2}+x-\sqrt{4 \left(1+\sqrt{2}\right) x+2},$$ and $$\exp(\bar{u}(x)) = \sqrt{2 \left(1+\sqrt{2}\right)}-\sqrt{2 x+\sqrt{2}-1}.$$
Proof. Let $x \in (0,1)$. We have $I(\bar{u})(x,g(x)) = 0$. Since $I(\bar{u}) \geq 0$, we have that $(x,g(x))$ is a minimum point of $I(\bar{u})$. Since $I(\bar{u})$ is a $C^1$-function by $(ii)$, we have that $I(\bar{u})'(x,g(x)) = 0$. Therefore \begin{align} \bar{u}'(x) = -\frac{1}{2-x-g(x)}. \tag{6} \end{align} By Claim 1 we have $(5)$. We will see that $(5)$ and $(6)$ determine $g$ and $\bar{u}$ uniquely. The following calculations hold for all $x \in [0,1]$.
Since $I(\bar{u})(x,g(x)) = 0$, we have \begin{align*} \bar{u}(g(x)) = \log(2-x-g(x)) - \bar{u}(x), \end{align*} and thus with $(5)$ we have \begin{align} \bar{u}(x) = \frac{1}{2}\left(\log(-g'(x)) + \log(2-x-g(x))\right), \tag{7} \end{align} and thus \begin{align} \bar{u}'(x) = \frac{1}{2}\left(\frac{g''(x)}{g'(x)} + \frac{-1-g'(x)}{2-x-g(x)}\right). \tag{8} \end{align} Putting $(6)$ and $(8)$ together yields $$g''(x)(2-x-g(x)) -g'(x)^2 +g'(x) = 0.$$ Therefore \begin{align} g'(x)(2-x-g(x)) +2g(x) = a. \tag{9} \end{align} for some $a \in \mathbb{R}$. In particular, we have \begin{align} g'(0)(2-g(0)) +2g(0)= g'(1)(1-g(1)) +2g(1). \tag{10} \end{align} Since $g$ is an involution we have $g(0) = 1$, $g(1) = 0$, $1 = g'(g(0))g'(0)= g'(1)g'(0)$ and $g' < 0$. We put this into $(10)$ and get $a = 1-\sqrt{2}$. Thus \begin{align} g'(x) = \frac{1-\sqrt{2}-2g(x)}{2-x-g(x)}. \tag{11} \end{align} Now we replace $x$ with $g(x)$ in $(9)$ and then use $g'(g(x)) = \frac{1}{g'(x)}$. We get \begin{align} g'(x) = \frac{2-x-g(x)}{1-\sqrt{2}-2x}. \tag{12} \end{align} Putting $(11)$ and $(12)$ together yields a quadratic equation for $g(x)$. Only the solution \begin{align} g(x) = 1+\sqrt{2}+x-\sqrt{4 \left(1+\sqrt{2}\right) x+2} \tag{13} \end{align} satisfies $g(0)=1$. We also have \begin{align*} \exp(\bar{u}(g(x)))^2 &\underset{\hphantom{(12)}}{=} \frac{2-x-g(x)}{\exp(\bar{u}(x))}\exp(\bar{u}(g(x))) \\[1em] &\underset{(5)\hphantom{0}}{=} -\frac{2-x-g(x)}{g'(x)} \\[1em] &\underset{(12)}{=} 2 x+\sqrt{2}-1, \end{align*} and thus \begin{align*} \exp(\bar{u}(x)) &\underset{\hphantom{(12)}}{=} \frac{2-x-g(x)}{\sqrt{2 x+\sqrt{2}-1}} \\[1em] &\underset{(13)}{=} \frac{1-2x-\sqrt{2}+\sqrt{4 \left(1+\sqrt{2}\right) x+2}}{\sqrt{2 x+\sqrt{2}-1}} \\[1em] &\underset{\hphantom{(12)}}{=} -\sqrt{2 x+\sqrt{2}-1}+\frac{\sqrt{4 \left(1+\sqrt{2}\right) x+2}}{\sqrt{2 x+\sqrt{2}-1}} \\[1em] &\underset{\hphantom{(12)}}{=} -\sqrt{2 x+\sqrt{2}-1}+\sqrt{2 \left(1+\sqrt{2}\right)}. \end{align*} $$\tag*{\Box}$$
We define \begin{align*} \bar{u} : [0,1] \rightarrow \mathbb{R}, \quad \bar{u}(x) &:= \log\left(\sqrt{2 \left(1+\sqrt{2}\right)}-\sqrt{2 x+\sqrt{2}-1}\right), \\[1em]g : [0,1] \rightarrow [0,1], \quad g(x) &:= 1+\sqrt{2}+x-\sqrt{4 \left(1+\sqrt{2}\right) x+2}. \end{align*}
Claim 3. $g$ is an involution. We have \begin{align*} I(\bar{u})(x,g(x)) &= 0, \\[0.5em] \exp(\bar{u}(x) + \bar{u}(y)) &\geq 2-x-y \end{align*} for all $x,y \in [0,1]^2$, and thus $I(\bar{u}) \geq 0$.
Proof. We solve $g(x)=y$ for $x$ and see that $x=g(y)$. Thus $g(g(x))=x$ and $g$ is an involution on $[0,1]$. We define $$F : [0,1]^2 \rightarrow \mathbb{R}, \quad F(x,y) := \exp(\bar{u}(x)+\bar{u}(y)) - (2-x-y).$$ For all $x \in [0,1]$ we have $x + g(x) \leq 1$ and thus $(x,g(x)) \in S$. We see that $\exp(\bar{u}(x))^2 = 2 g(x)+\sqrt{2}-1$, thus $\exp(\bar{u}(g(x)))^2 = 2 x+\sqrt{2}-1$ and by putting this into $F(x,g(x))$ we get $F(x,g(x)) = 0$ and $I(\bar{u})(x,g(x)) = 0$. For all $y \in [0,1]$ we have
\begin{align*} (\partial_{y y} F)(x,y) &= \exp(\bar{u}(x))\, \partial_{y y} \exp(\bar{u}(y)) \\ &= \exp(\bar{u}(x))\, \frac{1}{\left(2 y+\sqrt{2}-1\right)^{3/2}} \\ &> 0. \end{align*} Therefore $y \mapsto F(x,y)$ is convex and since $(\partial_y F)(x,g(x))= 0$, we have that $(x,g(x))$ is the global minimizer of $y \mapsto F(x,y)$. Since $F(x,g(x)) = 0$ we have $F \geq 0$ and thus $I(\bar{u}) \geq 0$. $$\tag*{\Box}$$
Claim 4. $\bar{u}$ is the unique solution of $(1)$.
Proof. We define \begin{alignat*}{2} &w : [0,1] \rightarrow [0,\infty), \quad &&w(x) := \frac{\exp(\bar{u}(x))}{1-g'(x)}, \\[1em] &\psi : C(S) \rightarrow \mathbb{R}, \quad &&\psi(v) := \int_0^1 v(x,g(x)) \, w(x) \, \mathrm{d}x, \\[1em] &L : C[0,1] \rightarrow \mathbb{R}, \quad &&L(u) := J(u) - \psi(I(u)). \end{alignat*} $w$ is well-defined since $g$ being an involution and not the identity implies $g'\leq0$.
$\psi$ is well-defined since $(x,g(x)) \in S$ for all $x \in [0,1]$. We have $w(g(x))=w(x)$ and thus \begin{align*} L'(\bar{u})(h) &= J'(\bar{u})(h) - \psi'(I(\bar{u}))(I'(h)) \\[1em] &= J'(\bar{u})(h) - \psi(I'(h)) \\[1em] &= \int_0^1 \exp(\bar{u}(x))\, h(x) \, \mathrm{d}x - \int_0^1 (h(x)+h(g(x))) \, w(x) \, \mathrm{d}x \\[1em] \text{by substitution:} \\[0.5em] &= \int_0^1 \exp(\bar{u}(x))\, h(x) \, \mathrm{d}x - \int_0^1 h(x) \, (1-g'(x)) \,w(x)\, \mathrm{d}x \\[1em] &= \int_0^1 \exp(\bar{u}(x))\, h(x) \, \mathrm{d}x - \int_0^1 \exp(\bar{u}(x)) \, h(x) \, \mathrm{d}x \\[1em] &= 0. \end{align*} for all $h \in C[0,1]$. Since $J$ and $-(\psi \circ I)$ are convex, $L$ is convex. We have $L'(\bar{u})=0$ and thus $\bar{u}$ is a global minimizer of $L$. Since $\psi$ is a positive functional, we have $L(u) \leq J(u)$ for all $u \in C[0,1]$ with $I(u) \geq 0$. But $\psi(I(\bar{u}))= 0$ since $I(\bar{u})(x,g(x))=0$ by Claim 3 for all $x \in [0,1]$. Therefore $L(\bar{u})=J(\bar{u})$ and thus $J(\bar{u}) \leq J(u)$ for all $u \in C[0,1]$ with $I(u) \geq 0$. By Claim 3 we have $I(\bar{u}) \geq 0$. Therefore $\bar{u}$ is a solution of $(1)$.
Assume there is a $\tilde{u} \in C[0,1]$ such that $I(\tilde{u}) \geq 0$ and $J(\tilde{u}) \leq J(\bar{u})$. We set $$\widehat{u} := \frac{\tilde{u} + \bar{u}}{2}.$$ We have $I(\widehat{u})\geq 0$. Since $J$ is strictly convex, $\bar{u} = \widehat{u} = \tilde{u}$. Therefore $\bar{u}$ is the unique solution of $(1)$. $$\tag*{\Box}$$
Claim 5. The unique solution of $(2)$ is given by $$\bar{f}(x)= \begin{cases} \dfrac{\exp(\bar{u}(2x))}{\sqrt{2}}&\text{if}\quad 0 \leq x \leq \frac{1}{2},\\ \dfrac{\exp(\bar{u}(2(1-x)))}{\sqrt{2}}&\text{if}\quad \frac{1}{2} < x \leq 1. \end{cases}$$
Proof. Since $\exp > 0$, we have $\bar{f} > 0$. Let $x,y \in[0,1]$ such that $x \leq y$.
If $x \leq \frac{1}{2} \leq y$, then \begin{align} \bar{f}(x)\bar{f}(y)&=\frac{1}{2}\exp\left(\bar{u}(2x)+\bar{u}(2(1-y))\right) \nonumber \\ &\geq \frac{1}{2}(2- 2x - 2(1-y)) \qquad \text{(by Claim 3)} \nonumber \\ &=y-x \tag{14} \\ &= |x-y|. \nonumber \end{align} If $x \leq y \leq \frac{1}{2}$, then $$\bar{f}(x)\bar{f}(y)=\bar{f}(x)\bar{f}(1-y) \underset{(14)}{\geq} 1-y-x \geq y-x = |x-y|.$$ If $\frac{1}{2} \leq x \leq y$, then $$\bar{f}(x)\bar{f}(y)=\bar{f}(1-x)\bar{f}(y) \underset{(14)}{\geq} y-(1-x) \geq y-x = |x-y|.$$ Thus $\bar{f}(x)\bar{f}(y) \geq |x-y|$ for all $x,y \in [0,1]$.
Assume there exists $\tilde{f}$ such that $\tilde{f}$ satisfies the constraints of (2) and $$\int_0^1 \tilde{f} \, \mathrm{d}x \leq \int_0^1 \bar{f} \, \mathrm{d}x.$$ We set $$\widehat{f}(x):= \left(\tilde{f}(x)\tilde{f}(1-x)\bar{f}(x)^2\right)^\frac{1}{4}.$$ $\widehat{f}$ satisfies the constraints of (2) and \begin{align} \int_0^1 \widehat{f}(x) \, \mathrm{d}x &= \int_0^1 \left(\tilde{f}(x)\tilde{f}(1-x)\bar{f}(x)^2\right)^\frac{1}{4} \, \mathrm{d}x \nonumber \\[1em] &\leq \int_0^1 \frac{\tilde{f}(x)+\tilde{f}(1-x)+2\bar{f}(1-x)}{4} \, \mathrm{d}x \tag{15} \\[1em] &= \int_0^1 \frac{\tilde{f}(x)+\bar{f}(x)}{2} \, \mathrm{d}x \nonumber \\[1em] &\leq \int_0^1 \bar{f}(x) \, \mathrm{d}x. \nonumber \end{align} Since $\widehat{f}(x) = \widehat{f}(1-x)$ and $\bar{f}(x) = \bar{f}(1-x)$ for all $x \in [0,1]$, we thus have \begin{align} \int_0^{\frac{1}{2}} \widehat{f}(x) \, \mathrm{d}x \leq \int_0^{\frac{1}{2}} \bar{f}(x) \, \mathrm{d}x. \tag{16} \end{align} We set $$\widehat{u}(x) := \log\left(\sqrt{2}\, \widehat{f}\left(\frac{x}{2}\right)\right),$$ and by $(16)$ we have $J(\widehat{u}) \leq J(\bar{u})$. We also have $I(\widehat{u}) \geq 0$ since \begin{align*} \exp(\widehat{u}(x)+\widehat{u}(y)) &= 2\widehat{f}\left(\frac{x}{2}\right)\widehat{f}\left(\frac{y}{2}\right) \\[1em] &= 2\widehat{f}\left(\frac{x}{2}\right)\widehat{f}\left(1-\frac{y}{2}\right) \\[1em] &\geq 2\left|1-\frac{x}{2}-\frac{y}{2}\right| \\[1em] &= 2-x-y \end{align*} for all $x,y \in [0,1]$. By Claim 4, $\widehat{u} = \bar{u}$. Since $$\widehat{f}(x)=\frac{\exp(\widehat{u}(2x))}{\sqrt{2}}$$ for all $x \in [0,\frac{1}{2}]$ by definition of $\widehat{u}$, we have $\widehat{f} = \bar{f}$. The inequality at $(15)$ must be an equality and thus $\tilde{f} = \widehat{f} = \bar{f}$. Therefore $\bar{f}$ is the unique solution of $(2)$. $$\tag*{\Box}$$
• By Golly, this is impressive, cafaxo! But I won't have time to check it out until two days later. – Hans Jan 14 '18 at 9:33
• it is not possible to reduce this ?\ – Guy Fsone Jan 15 '18 at 12:36
• There might be an elementary argument for $(5)$. I'm sorry, I could not find one. – cafaxo Jan 15 '18 at 13:02
• I will have many questions. Here are the first two. 1) To arrive at the equation right beneath (15), are you assuming $\tilde f(x)=\tilde f(1-x)$? If so, it seems we have not proved the evenness of the minimizer of Problem (2) with respect to $x=\frac12$ or $f(x)=f(1-x)$. 2) What is your motivation for devising the equivalent Problem (1) of (2)? – Hans Jan 16 '18 at 5:47
• 1) I'm using $\int_0^1 \tilde{f}(x) \, \mathrm{d}x = \int_0^1 \tilde{f}(1-x) \, \mathrm{d}x$, which is true by substitution ($x \mapsto 1-x$). 2) For me, Problem (1) was far easier to work with. The main reasons are that $J$ and $-I$ are convex and $I(u)$ is $C^1$ for all $u \in C^1[0,1]$. – cafaxo Jan 16 '18 at 10:31 | 2019-06-19T05:22:43 | {
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https://jp.maplesoft.com/support/help/view.aspx?path=Iterator%2FMultiCombination | Multi Combination - Maple Help
Iterator
MultiCombination
generate multicombinations
Calling Sequence MultiCombination(M, t, opts)
Parameters
M - list(posint); numbers of distinct elements t - nonnegint; size of combinations opts - (optional) equation(s) of the form option = value; specify options for the MultiCombination command
Options
• compile = truefalse
True means compile the iterator. The default is true.
Description
• The MultiCombination command returns an iterator that generates all t-combinations chosen from a multiset.
• The M parameter specifies the multiset. It is a list of integers, $\left[{m}_{1},{m}_{2},\dots ,{m}_{n}\right]$, with ${m}_{k}$ the multiplicity of the $k$-th element.
• The t parameter is the number of items to choose.
• The output of each iteration is an Array of t positive integers. An integer $k$ represents a selection of the $k$-th element.
• There is an isomorphism between multicombinations and bounded-compositions. This object uses the same algorithm as BoundedComposition but transforms the output.
Methods
In addition to the common iterator methods, this iterator object has the following methods. The self parameter is the iterator object.
• Number(self): return the number of iterations required to step through the iterator, assuming it started at rank one.
Examples
> $\mathrm{with}\left(\mathrm{Iterator}\right):$
Construct an iterator corresponding to the number of ways to choose 5 balls from a bucket with 2 red balls, 5 green balls, and 3 black balls.
> $B≔\mathrm{MultiCombination}\left(\left[2,5,3\right],5\right):$
Print each combination. The 1s correspond to red balls, the 2s to green balls, and the 3s to black balls.
> $\mathrm{Print}\left(B,'\mathrm{showrank}'\right):$
1: 1 1 2 2 2 2: 1 2 2 2 2 3: 2 2 2 2 2 4: 1 1 2 2 3 5: 1 2 2 2 3 6: 2 2 2 2 3 7: 1 1 2 3 3 8: 1 2 2 3 3 9: 2 2 2 3 3 10: 1 1 3 3 3 11: 1 2 3 3 3 12: 2 2 3 3 3
Compute the number of ways to select four entrees from a menu of 10 dishes allowing multiple selections of any of the entrees. Because four is the most that can be selected, this is equivalent to assigning four as the multiplicity of all items.
> $\mathrm{Number}\left(\mathrm{MultiCombination}\left(\left[\mathrm{}\left(4,10\right)\right],4\right)\right)$
${715}$ (1)
Suppose one is allowed to choose any entree no more than twice. This can be expressed as
> $M≔\mathrm{MultiCombination}\left(\left[\mathrm{}\left(2,10\right)\right],4\right):$
There are 615 ways to obtain such a combination.
> $\mathrm{Number}\left(M\right)$
${615}$ (2)
Number of Distinct Ranks of a Poker Hand
• A standard poker hand consists of five cards drawn from a 52 card deck, with four suits of 13 cards per suit.
• The rank of a hand depends first on its category (straight flush, four of a kind, etc.), then on its rank within that category. The rank does not depend on the suits, with the exception that a flush (all five cards of the same suit) is a separate category.
• The number of distinct hand ranks equals the number of possible flushes of a given suit plus the number of multicombinations of five cards from a deck with four cards of each rank.
• The number of possible flushes of a given suit is simply the number of ways to choose a set of five objects from a set of thirteen distinct objects. The binomial function computes the result, $\left(\genfrac{}{}{0}{}{13}{5}\right)$. It can also be computed using the Number method of the Combination object:
> $\mathrm{numF}≔\mathrm{Number}\left(\mathrm{Combination}\left(13,5\right)\right)$
${\mathrm{numF}}{≔}{1287}$ (3)
• Construct an iterator that generates each of the multicombinations of five cards drawn from a deck.
> $H≔\mathrm{MultiCombination}\left(\left[\mathrm{}\left(4,13\right)\right],5\right):$
• Count the number of multicombinations. We can do this in a number of ways; one is simply to invoke the Number method.
> $\mathrm{numH}≔\mathrm{Number}\left(H\right)$
${\mathrm{numH}}{≔}{6175}$ (4)
• For this case, it is also practical to count the iterations one by one.
> $\mathrm{numH}≔\mathrm{add}\left(1,h=H\right)$
${\mathrm{numH}}{≔}{6175}$ (5)
• A final method is the following. If there were five of each ranked card, rather than four, we would obtain exactly 13 more distinct ranked hands then the actual deck (the 13 five of a kind). This is expressed by the following formula:
> $\mathrm{numH}≔\mathrm{Number}\left(\mathrm{MultiCombination}\left(\left[\mathrm{}\left(5,13\right)\right],5\right)\right)-13$
${\mathrm{numH}}{≔}{6175}$ (6)
• The total number of distinct ranks of poker hands is
> $\mathrm{numF}+\mathrm{numH}$
${7462}$ (7)
References
Knuth, Donald Ervin. The Art of Computer Programming, volume 4, fascicle 3; generating all combinations and partitions,sec. 7.2.1.3, exercise 60, p. 30, and answers, algorithm Q, pp. 98-99.
Compatibility
• The Number method was updated in Maple 2020.
• The Iterator[MultiCombination] command was introduced in Maple 2016. | 2023-01-29T07:24:02 | {
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https://math.stackexchange.com/questions/2840645/closed-form-expression-for-the-number-of-k-permutations-of-length-n-duplica | # Closed-form expression for the number of $k$-permutations of length $n$. Duplicates allowed!
Given $n$ symbols and an integer $k > 0~(k \leq n)$, find the number of all distinct strings of length $n$, formed by any $k$-out-of-$n$ symbols, i.e., the target strings consist of exactly $k$ distinct symbols out of the given $n$ symbols. There are no restrictions on the number of repetitions allowed for each symbol.
Given $n$ and $k$, the goal is the derive a closed-form expression or upper and lower bounds on the count of all such (distinct) $k$-permutations.
Eg. Let S={a,b,c} be a set of n=3 elements. For k=2, the distinct 2-permutations of length 3 are: aab, aba, abb, baa, bab, bba, aac, aca, acc, caa, cac, cca, bbc, bcb, bcc, cbb, cbc, ccb. Hence, 18 distinct strings of length 3 are formed by 2-out-of-3 symbols.
• Think of the strings as functions from $[n]=\{1,2,3,...,n\}$ to $[k]$ Jul 4 '18 at 13:08
• @joriki I added equally likely portion just to add more clarity to the question. And there is not particular reason for the length of the strings and the total number of symbols being the same. They may be different! Jul 4 '18 at 13:33
• In my view, talking about likelihood when there's in fact no randomness reduces rather than increases the clarity of the question. If you want to allow the string length and the alphabet size to differ, you shouldn't use the same variable name for them. Jul 4 '18 at 13:35
• Jul 17 '18 at 20:15
• @XanderHenderson: It seems to me that there may be a difference between this Question and the proposed duplicate, turning on whether exactly $k$ letters must appear in those "permutations with repetitions". Not having to account for all $k$ letters appearing really makes the problem easier, and I have no doubt that another duplicate exists in the Math.SE corpus. Jul 18 '18 at 2:03
Like joriki, I will denote the alphabet size with $m$ and the string length with $n$.
Choose $k$ distinct symbols $c_1,\ldots,c_k$ from the available $m$ symbols. The labels shall matter, so there are $m\,(m-1)\cdots(m-k+1) = k!\binom{m}{k}$ ways to do this.
Independently, partition the set $\{1,2,\ldots,n\}$ of indices into a length-$n$ string into exactly $k$ nonempty subsets $S_1,\ldots,S_k$ where the labels (i.e. permutations of the $S_j$) do not matter. Therefore, we can prescribe some ordering of the $S_i$, e.g. by their minimal elements, that is, $\min(S_i) < \min(S_j) \iff i<j$. There are exactly $\left\{n\atop k\right\}$ ways to do that where $\left\{n\atop k\right\}$ is a Stirling number of the second kind.
For every $i\in\{1,\ldots,k\}$, put the symbol $c_i$ at the string position(s) listed in $S_i$. This approach gives every valid string exactly once, and we get $$k!\binom{m}{k}\left\{n\atop k\right\}$$ ways in total. Note that this matches @joriki's answer because $$\sum_{j=0}^k(-1)^j\binom{k}{j}(k-j)^n = k!\left\{n\atop k\right\}$$ [Graham/Knuth/Patashnik: Concrete Mathematics, 2nd ed., eq. (6.19); also given in the Wikipedia entry]
In your example, $(k,m,n) = (2,3,3)$, and we get indeed $$k!\binom{m}{k}\left\{n\atop k\right\} = 2!\binom{3}{2}\left\{3\atop 2\right\} = 2\cdot 3\cdot 3 = 18$$
There are $j^n$ strings of length $n$ that contain at most $j$ particular symbols. Then by inclusion–exclusion the number of strings of length $n$ with exactly $k$ distinct symbols chosen from $m$ symbols is
$$\binom mk\sum_{j=0}^k(-1)^j\binom kj(k-j)^n\;.$$
You can set $m=n$ to obtain your special case in which the string length coincides with the size of the alphabet.
In your example, $m=n=3$ and $k=2$, so this becomes
$$\binom32\sum_{j=0}^2(-1)^j\binom2j(2-j)^3=3\left(2^3-2\cdot1^3+0\right)=18\;.$$
• The sum can be expressed in terms of a Stirling number of the second kind. That has made me wonder about a combinatorial interpretation. Found it. Jul 4 '18 at 15:08 | 2021-09-28T23:32:54 | {
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https://mathhelpboards.com/threads/conditional-probability-find-the-probability-that-the-course-he-selects-is-in-the-afternoon.23915/ | # Conditional Probability: Find the probability that the course he selects is in the afternoon.
#### navi
##### New member
Hey! I need help with my Math homework The question is the following...
There are 5 history courses of interest to Howard, including 3 in the afternoon, and there are 6 psychology courses, including 4 in the afternoon. Howard picks a course by selecting a dept at random, then selecting a course at random. Find the pr that the course he selects is in the afternoon.
The answer is 19/30, but I get 13/30 by doing this:
For the history dept, the probability of an afternoon class is 1/2 x 3/5, or 3/10, and for the psych dept, the pr of an afternoon class is 1/2 x 4/6, or 1/3. I then add 1/3 and 3/10 and I get 13/30, but that is not the answer and I have no idea what I could be doing wrong...
#### MarkFL
Staff member
Hello, and welcome to MHB!
I would let event A be that Howard selected an afternoon course from the history department, and event B be that Howard selected an afternoon course from the psychology department.
$$\displaystyle P(A)=\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$$
$$\displaystyle P(B)=\frac{1}{2}\cdot\frac{4}{6}=\frac{1}{3}$$
And so:
$$\displaystyle P(A\text{ OR }B)=P(A)+P(B)=\frac{3}{10}+\frac{1}{3}=\frac{3\cdot3+10\cdot1}{30}=\frac{19}{30}$$
#### navi
##### New member
Hello, and welcome to MHB!
I would let event A be that Howard selected an afternoon course from the history department, and event B be that Howard selected an afternoon course from the psychology department.
$$\displaystyle P(A)=\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$$
$$\displaystyle P(B)=\frac{1}{2}\cdot\frac{4}{6}=\frac{1}{3}$$
And so:
$$\displaystyle P(A\text{ OR }B)=P(A)+P(B)=\frac{3}{10}+\frac{1}{3}=\frac{3\cdot3+10\cdot1}{30}=\frac{19}{30}$$ | 2020-08-14T14:44:46 | {
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https://math.stackexchange.com/questions/2412375/suppose-we-roll-a-fair-6-sided-die-repeatedly-find-the-expected-number-of-rol | # Suppose we roll a fair $6$ sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession.
Suppose we roll a fair six sided die repeatedly.
Find the expected number of rolls required to see $3$ of the same number in succession
From the link below, I learned that $258$ rolls are expected to see 3 sixes appear in succession. So I'm thinking that for a same (any) number, the rolls expected would be $258/6 = 43$. But I'm unsure how to show this and whether it really is correct.
How many times to roll a die before getting two consecutive sixes?
For $n\in \{0,1,2\}$ Let $E[n]$ denote the answer given that you are starting from a streak of $n$ consecutive rolls. The answer you want is $E=E[0]$, though you are never in state $0$ except at the start.
We note $$E[2]=\frac 16\times 1+\frac 56\times \left(E[1]+1\right)$$ $$E[1]=\frac 16\times \left(E[2]+1\right)+\frac 56\times \left(E[1]+1\right)$$
$$E=E[0]=E[1]+1$$
this system is easily solved and, barring error (always possible), yields $$\boxed {E=43}$$
• This is very elegant (+1) - great answer!! – Satish Ramanathan Aug 31 '17 at 16:38
• One issue that isn't really an issue: everything needs to be finite for this to work (cf optional stopping theorem) – cats Aug 31 '17 at 18:32
• This is the (arch classical) approach used in the post mentioned in Byron's answer. – Did Aug 31 '17 at 18:40
From Did's answer here, the probability generating function $u_0(s)=\mathbb{E}(s^T)$ for the number of trials $T$ needed to get three consecutive values the same is
$$u_0(s)={s^3\over 36-30s-5s^2}.$$ Differentiating this and setting $s=1$ in the derivative shows that $\mathbb{E}(T)=43.$
We can treat this as a three-state absorbing markov chain: a length3 run has been seen, otherwise the current run is length2, current run is length 1. Transition matrix: $\begin{bmatrix}1&\frac{1}{6}&0\\0&0&\frac{1}{6}\\0&\frac{5}{6}&\frac{5}{6}\end{bmatrix}$
This is in standard form $\left[\begin{array}{c|c}I&S\\\hline0&R\end{array}\right]$
We turn our attention to the fundamental matrix $(I-R)^{-1} = \begin{bmatrix}1&-\frac{1}{6}\\-\frac{5}{6}&\frac{1}{6}\end{bmatrix}^{-1}=\begin{bmatrix}6&6\\30&36\end{bmatrix}$
After the first roll, we enter the markov chain in the third state. Adding the entries of the fundamental matrix corresponding to that column tells us the expected time until we reach an absorbing state, i.e. until we have a chain of three consecutive rolls of the same number.
Thus the expected number of rolls needed is $1+36+6=43$
Let $\mu$ denote the expectation of the expected number of rolls that are still needed if $2$ rolls have passed with distinct result.
Let $\nu$ denote the expectation of the expected number of rolls that are still needed if $2$ rolls have passed with equal result.
Then the expectation is: $$2+\frac56\mu+\frac16\nu$$
Here $\frac56$ is the probability that the first two numbers are distinct and $\frac16$ is the probability that they are equal.
Secondly we have the relations:
$$\mu=\frac56(1+\mu)+\frac16(1+\nu)=1+\frac56\mu+\frac16\nu\tag1$$ and: $$\nu=\frac161+\frac56(1+\mu)=1+\frac56\mu\tag2$$
The relations $(1)$ and $(2)$ lead easily to: $\mu=42$ and $\nu=36$.
Then $$2+\frac56\mu+\frac16\nu=43$$ is the final answer.
(in the answers uptil now it was not used that you allready learned something).
Let $Y$ denote the number of rolls required to see three dice of the same number in succession and let $X$ denote the number of rolls required to see three dice with number $6$ in succession.
Then: $$Y\text{ and }\frac16X+\frac56(X+Y)=X+\frac56Y\text{ must have equal distribution.}\tag3$$
Here $\frac16$ is the probability of the event that the first time that three dice give the same number in succession they show number $6$ and $\frac56$ is the probability that do not show a number $6$.
$(3)$ rests on the observation that - if for the first time three equal numbers show up in succession - we are ready if $6$ happens to be that number and must actually start over again (with $X$ throws in our pocket) if not.
So we find $\mathbb EY=\mathbb EX+\frac56\mathbb EY$ or equivalently:$$\mathbb EX=\frac16\mathbb EY$$
You allready learned that $\mathbb EY=258$ and making use of that knowledge you find $$\mathbb EX=258/6=43$$ This confirms your thinking. | 2019-08-21T22:18:51 | {
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https://math.stackexchange.com/questions/1240570/recurrence-relation-equal-roots-of-characteristic-equation | # Recurrence relation - equal roots of characteristic equation
I have the following problem:
Solve the following recurrence relation
$$f(0)=3$$
$$f(1)=12$$
$$f(n)=6f(n-1)-9f(n-2)$$
We know this is a homogeneous 2nd order relation so we write the characteristic equation: $$a^2-6a+9=0$$ and the solutions are $$a_{1,2}=3$$.
The problem is when I replace these values I get:
$$f(n)=c_13^n+c_23^n$$
and using the 2 initial relations I have:
$$f(0)=c_1+c_2=3$$ $$f(1)=3(c_1+c_2)=12$$
which gives me that there are no values such that $$c_1$$ and $$c_2$$ such that these 2 relation are true.
Am I doing something wrong? Is the way it should be solved different when it comes to identical roots for the characteristic equation?
Let the roots of the characteristic equation be $\alpha, \beta$ so the equation is $$f(n)=(\alpha+\beta)f(n-1)-\alpha\beta f(n-2)$$ with solution $f(n)=A\alpha^n+B\beta^n$
We set $f(0)=3, f(1)=12$ to obtain $A+B=3$ and $A\alpha+B\beta=12$ with $B=\frac {3\alpha-12}{\alpha-\beta}$ and $A=\frac {12-3\beta}{\alpha-\beta}$ so $$f(n)=12\frac {\alpha^n-\beta^n}{\alpha-\beta}-3\alpha\beta\frac {\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta}$$
Now you can divide through by $\alpha-\beta$ and get $n$ terms from the first fraction and $n-1$ terms from the second. In the limit when $\alpha$ becomes equal to $\beta$ this leads to $$f(n)=12n\alpha^{n-1}-3(n-1)\alpha^n=3\alpha^n+\left(\frac {12}{\alpha}-3\right)n\alpha^n= \text {(in form) }(C+Dn)\alpha^n$$
This is one way of justifying the general form others have used - you get the solution you need by setting $\alpha=3$.
Yes, you’re doing something wrong. In the case of repeated roots, your approach collapses the two solutions into a single one, when in fact you need two independent solutions. If the characteristic equation has a root $r$ of multiplicity $m$, it gives rise to the $m$ solutions $c_1r^n,c_2nr^n, \ldots,c_mn^{m-1}r^n$. In your case $r=3$ and $m=2$, so you get $c_13^n$ and $c_2n3^n$. That is,
$$f(n)=c_13^n+c_2n3^n=(c_1+c_2n)3^n\;.$$
Now proceed as you would in the case of distinct roots to find $c_1$ and $c_2$ that match your initial conditions.
When characteristic polynomial has coincident roots the correct equation is the following: $$f(n) = c_03^n + c_1 n\cdot 3^n$$
This phenomenon happens when you solve linear "things", like linear differential equations. | 2021-05-09T20:42:29 | {
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https://math.stackexchange.com/questions/1606999/find-all-positive-integers-n-such-that-n2008-divides-n2-2008-and-n20 | Find all positive integers $n$ such that $n+2008$ divides $n^2 + 2008$ and $n+2009$ divides $n^2 + 2009$
I wrote \begin{align} n^2 + 2008 &= (n+2008)^2 - 2 \cdot 2008n - 2008^2 + 2008 \\ &= (n+2008)^2 - 2 \cdot 2008(n+2008) + 2008^2 + 2008 \\ &= (n+2008)^2 - 2 \cdot 2008(n+2008) + 2008 \cdot 2009 \end{align} to deduce that $n+2008$ divides $n^2 + 2008$ if and only if $n+2008$ divides $2008 \cdot 2009$.
Similarly, I found that $n+2009$ divides $n^2 + 2009$ if and only if $n+2009$ divides $2009 \cdot 2010$.
I can't seem to get anywhere from here. I know that $n=1$ is an obvious solution, and I think it's the only one, although I can't prove/disprove this. I know that any two consecutive integers are coprime but I haven't been able to use this fact towards the solution.
Could anyone please give me any hints?
For the record, this is a question from Round 1 of the British Mathematical Olympiad.
• $$n+2008$$ needs to be factor of $$2008\cdot2009$$ – lab bhattacharjee Jan 10 '16 at 17:13
• Yes, I know that, and $n+2009$ needs to be a factor of $2009.2010$ but I'm not sure what more I could deduce from this. – Ecasx Jan 10 '16 at 17:15
• Note that $(2008\cdot2009)/(1+2008)=2008$ and $(2009\cdot2010)/(1+2009)=2009$, and their difference is $1$. Furthermore, as $n$ increases, the difference $\frac{2009\cdot2010}{n+2009}-\frac{2008\cdot2009}{n+2008}$ "should" get smaller, intuitively speaking. In particular, if the difference is strictly between $0$ and $1$, then the terms cannot both be integers. See if you can show that is indeed the case. – Joey Zou Jan 10 '16 at 17:37
Here another approach
Since $(n+2008)|( n^{2} + 2008)$ and $(n+2009) |( n^{2} + 2009)$ then
$(n+2008)k = n^2 + 2008$ and $(n+2009)m = n^2 + 2009$ for some $k,m \in \mathbb{Z}$
$\textbf{Case 1:}$ $k=1$ or $m=1$
$(n+2008) = n^2 + 2008$ or $(n+2009) = n^2 + 2009$
in any case leads to $n^2-n=0$ that is $n=1$ and $n=0$
$\textbf{Case 2:}$ $m>k>1$
$$n^2+2009=(n+2009)m = (n+2008)m+m > (n+2008)k + m > n^2+2009$$ which is impossible
$\textbf{Case 3:}$ $k>m>1$
$$n^2+2008=(n+2009-1)k = (n+2009)k - k \ge (n+2009)(m+1) - k = (n+2009)m +(n+2009-k) \ge n^2+2009$$ also impossible, the last inequality holds because $k< n+2009$, suppose not
$k \ge n+2009 >n+2008 \Rightarrow n^2 + 2008 =(n+2008)k > (n+2008)^2 > n^2 + 2008$ a contradiction
$\textbf{Case 4:}$ $k=m>1$
$$(n+2008)k+1=(n+2009)k \Longrightarrow k =1 \Longrightarrow n=1$$
So the solutions are $\boxed{n=1}$ and $\boxed{n=0}$
• Thank you. That's a very nice solution. I'm startled that it doesn't use any real number theory, when the question looks quite 'number theoretic'! – Ecasx Jan 10 '16 at 21:30
• actually it does, look the gnasher729's solution – JulianP Jan 10 '16 at 21:45
Subtracting n + 2008 from $n^2 + 2008$ resp. n + 2009 from $n^2 + 2009$ shows that n + 2008 and n + 2009 both divide n (n - 1). Since n+2008 and n+2009 have no common factor other than 1, (n + 2008) * (n + 2009) divides n (n - 1). But (n + 2008) * (n + 2009) ≥ n (n - 1), so n (n-1) must be 0, and n = 0 or n = 1. n = 1 is the only positive solution. | 2019-12-14T16:58:56 | {
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