qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
4,619,164 | <p>So one way to solve this ordinary differential equation is by computing the integral of both sides of <span class="math-container">$$\frac{y'}{y} = 1$$</span>
However, I did it another way and I think I made a mistake, but where is it? What I did is:
<span class="math-container">$$y' = y $$</span>
<span class="math-container">$$ \int y'dx = \int y dx $$</span>
<span class="math-container">$$ \int \frac{dy}{dx}dx = \int y dx $$</span>
<span class="math-container">$$ \int dy = \int y dx $$</span>
<span class="math-container">$$ y = yx + C $$</span>
<span class="math-container">$$ y = \frac{C}{1-x}$$</span></p>
| PrincessEev | 597,568 | <p>You're treating <span class="math-container">$y$</span> as a constant when you say</p>
<p><span class="math-container">$$\int y \, dx = xy+C$$</span></p>
<p>In reality, <span class="math-container">$y$</span> is a function of <span class="math-container">$x$</span>. For instance, this claims that</p>
<p><span class="math-container">$$\int e^x \, dx = xe^x + C$$</span></p>
<p>when we let <span class="math-container">$y(x)=e^x$</span>, but this claim is obviously problematic.</p>
|
3,546,274 | <p>I am trying out the first probability problem at this <a href="https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041-probabilistic-systems-analysis-and-applied-probability-fall-2010/assignments/MIT6_041F10_assn04.pdf" rel="nofollow noreferrer">link</a> which looks something like this:</p>
<p><a href="https://i.stack.imgur.com/O75nj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O75nj.png" alt="img"></a></p>
<p>I would like to ask what is the difference between <span class="math-container">$\mathbf P(...)$</span> and <span class="math-container">$p_{...}(...)$</span>? Aren't they both just probability? Why bother using different symbols? Thanks.</p>
| Yanirmr | 750,776 | <p>The probability of an event A is written as <span class="math-container">$P(A)$</span>, <span class="math-container">$p(A)$</span>, or <span class="math-container">$Pr(A)$</span>.</p>
<p>I'm recommending on this answers for your question: </p>
<ul>
<li><a href="https://stats.stackexchange.com/questions/260711/upper-case-p-or-lower-case-p-to-denote-p-values-and-probabilities-in-frequen">Upper case (P) or lower case (p) to denote p-values and probabilities in frequentist and Bayesian statistics</a></li>
<li><a href="https://stats.stackexchange.com/questions/108441/which-notation-and-why-textp-pr-textprob-or-mathbbp">Which notation and why: P(), Pr(), Prob(), or P()</a></li>
</ul>
|
3,546,274 | <p>I am trying out the first probability problem at this <a href="https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041-probabilistic-systems-analysis-and-applied-probability-fall-2010/assignments/MIT6_041F10_assn04.pdf" rel="nofollow noreferrer">link</a> which looks something like this:</p>
<p><a href="https://i.stack.imgur.com/O75nj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O75nj.png" alt="img"></a></p>
<p>I would like to ask what is the difference between <span class="math-container">$\mathbf P(...)$</span> and <span class="math-container">$p_{...}(...)$</span>? Aren't they both just probability? Why bother using different symbols? Thanks.</p>
| Graham Kemp | 135,106 | <p><span class="math-container">$\mathsf P(X=x)$</span> is the probability for the event of <span class="math-container">$X=x$</span>.</p>
<p><span class="math-container">$p_{\small X\!}(x)$</span> is the probability mass function of random variable <span class="math-container">$X$</span>, measured at <span class="math-container">$x$</span>.</p>
<p>Okay, yes, these are the same thing.</p>
<p>However, while we may generally measure the probability of <em>various</em> types of events, the probability mass function (pmf) of a random variable is something of special interest. So we have a special representation for it.</p>
<p>It describes the distribution of the random variable and can be used to derive various properties of the distribution. The mean, the variance, et cetera.</p>
<p>[Also <span class="math-container">$p_{\small X\!}(x)$</span> takes up <em>slightly</em> less typesetting on the page than <span class="math-container">$\mathsf P(X=x)$</span> , so can save space in long formulae.]</p>
|
1,294,792 | <p>Can a 2-dimensional sphere be decomposed into two disjoint homeomorphic subspaces? If yes, can these subspaces be non-discrete / connected / have some other good properties?</p>
| Timbuc | 118,527 | <p>Some ideas. There's quite a few things to justify, among them taking the Principal Value of the series after the logarithmic differentiation.</p>
<p>Take the infinite product for the sine function:</p>
<p>$$\sin \pi z=\pi z\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)=\pi z\prod_{0\neq n=-\infty}^\infty\left(1+\frac zn\right)\implies$$</p>
<p>$$\log\sin\pi z=\log\pi z+\sum_{0\neq n=-\infty}^\infty\log\left(1+\frac zn\right)$$</p>
<p>and now differentiate:</p>
<p>$$\pi\cot\pi z=\frac1z+\sum_{0\neq n=-\infty}^\infty\frac1{n+z}=\frac1z+\sum_{n=1}^\infty\left(\frac1{z+n}+\frac1{z-n}\right)=$$</p>
<p>$$=\frac1z+\sum_{n=1}^\infty\frac{2z}{z^2-n^2}$$</p>
|
712 | <p>I read <a href="https://math.stackexchange.com/questions/625/why-is-the-derivative-of-a-circles-area-its-perimeter-and-similarly-for-spheres">this question</a> the other day and it got me thinking: the area of a circle is $\pi r^2$, which differentiates to $2 \pi r$, which is just the perimeter of the circle. </p>
<blockquote>
<p>Why doesn't the same thing happen for squares? </p>
</blockquote>
<p>If we start with the area formula for squares, $l^2$, this differentiates to $2l$ which is sort of right but only <em>half</em> the perimeter. I asked my calculus teacher and he couldn't tell me why. Can anyone explain???</p>
| Isaac | 72 | <p>If you use the formula you describe for squares, your measurements are coming from one corner of the square. Imagine those measurements growing slowly. The square will grow, but only along the two sides opposite the corner from which you measured, so the derivative of the area formula is only the perimeter on those two sides.</p>
<p>Alternately, consider measuring the size of a square by the distance d from its center to the midpoint of a side. This would make the side length 2d, the perimeter 8d and the area 4d^2. Now, the derivative of the area is the perimeter. (Also, if you imagine growing the square slowly with this measurement, it grows from the center outward, growing on all four sides.)</p>
|
3,841 | <p>Taking tori in symmetric products and "miraculously" proving that the Floer homology is independent of choices always seemed, well, miraculous. Some time ago Max Lipyanski explained to me the origins of this construction from gauge theory on surfaces, a la Atiyah-Floer conjecture, which I have then forgotten. What is the origin of Heegard Floer?</p>
| Ben Webster | 66 | <p>I think the crude answer is that there is (or maybe just should be) an extended 4 dimensional TQFT that assigns the Fukaya category of a symmetric product to a surface, and the usual Heegard-Floer Lagrangian to a 3 manifold. So, the usual definition of Heegard-Floer is the gluing formula for a Heegard splitting, and invariance is no miracle at all.</p>
|
941,596 | <p>Assuming I'm flipping $M$ biased coins with different probability for heads $p_i, i=\{1,...,M\}$. What is the probability of having $k$ times head? Is there a distribution function known for this?</p>
| mathse | 136,490 | <p>You may want to look for "Poisson binomial distribution", e.g., <a href="http://en.wikipedia.org/wiki/Poisson_binomial_distribution" rel="nofollow">http://en.wikipedia.org/wiki/Poisson_binomial_distribution</a></p>
|
4,340,560 | <p>I have to teach limits to infinity of real functions of one variable.
I would like to start my course with a beautiful example, not simply a basic function like <span class="math-container">$1/x.$</span> For instance, I thought of using the functions linked to the propagation of covid-19 and show that, under the basic model, the number of contaminations will go to <span class="math-container">$0$</span> when time goes to <span class="math-container">$+\infty.$</span> However, this is a bad idea because the model is not so easy to explain and moreover students are sick of covid-subjects.</p>
<p>Hence, I ask you some help to find interesting examples from physics, geography, etc ... I suppose that an example with "time" going to <span class="math-container">$+\infty$</span> would be nice.</p>
| Chilote | 113,061 | <p><strong>Application:</strong> A limit can be interpreted as a prediction of the result of an experiment that can be held (hypothetically) for eternity without having to do it for eternity.</p>
<p>For example, if you throw a 6-side dice the probability of getting a 1 is <span class="math-container">$\frac{1}{6}.$</span>
If you throw the dice <span class="math-container">$k$</span> times, the probability that the output is <span class="math-container">$k$</span>-times 1 is <span class="math-container">$\left(\frac{1}{6}\right)^k$</span>.</p>
<p>The limit <span class="math-container">$\displaystyle\lim_{k\to\infty}\dfrac{1}{6^k}=0$</span> can be interpreted as: the more times we throw the dice, the probability of always getting the same output will be closer to zero. However, never will be zero.</p>
<p>Similarly, the limit <span class="math-container">$\displaystyle\lim_{k\to\infty}\left(1-\dfrac{1}{6^k}\right)=1$</span> can be interpreted as: the more times we throw the dice, the probability of not getting the same output in all the throws will be closer to one. However, never will be 1.</p>
<p>Here the limit assures us certain predictions, a tendency, without having to throw a dice many times.</p>
|
2,133,769 | <p>Asymptotic notations are a little vague for me at the moment. Here I have a problem that asks me if two equations are equal to each other, and it involves the big O notation. I wrote my question in the image in magenta color.</p>
<p><a href="https://i.stack.imgur.com/hxDx3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hxDx3.jpg" alt="enter image description here"></a></p>
| Joffan | 206,402 | <p>Your suggested solution assumes $x_4$ and $x_5$ are zero. The given solution acknowledges that there will be two remaining unknowns and set those to be $s=x_4$, $t=x_5$ (as you can see from the last two elements of the tuples they multiply) and solves for those.</p>
<p>[You can think of this as two more equations:<br>
$0x_1+0x_2+0x_3+1x_4+0x_5=s$<br>
$0x_1+0x_2+0x_3+0x_4+1x_5=t$<br>
]</p>
|
240,139 | <p>Does a quadratic form always come from symmetric bilinear form ?
We know when $q(x)=b(x,x)$ where $q$ is a quadratic form and $b$ is a symmetric bilinear form.
But when we just take a bilinear form and $b(x,y)$ and write $x$ instead of $y$,does it give us a quadratic form ?</p>
| Christopher A. Wong | 22,059 | <p>Yes, every quadratic form (over a finite-dimensional $\mathbb{R}$-space) can be expressed in terms of a symmetric bilinear form, because if your quadratic form $Q(x)$ (for $x \in \mathbb{R}^n$) is written as
$$ Q(x) = \sum_{i\le j} c_{ij} x_i x_j $$
then $Q(x) = x^T Ax$, where $A$ is a symmetric matrix given by $A = (a_{ij})$ with
$$a_{ij} = a_{ji} = \frac{c_{ij}}{2}, i < j, \quad a_{ii} = c_{ii}$$</p>
|
2,167,541 | <p>Is this expression true?$$y(x) =\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h \rightarrow 0}\frac{f(x)-f(x-h)}{h}$$</p>
<p>I'm taking the limit of h goes to 0 for y(x) and I got the latter equation. Can I just write y(x) = df(x)/dx?</p>
| Yusuf Kanat | 669,262 | <p><span class="math-container">$=\dfrac{1}{(2k+1)!}\displaystyle\sum_{i=0}^k \frac{(2k+1)!}{(2i+1)!(2k)(2i)!}$</span></p>
<p><span class="math-container">$=\dfrac{1}{(2k+1)!} \displaystyle\sum_{i=0}^k \begin{pmatrix} 2k+1 \\ 2i+1 \end{pmatrix}$</span></p>
<p><span class="math-container">$=\dfrac{1}{(2k+1)!}\begin{pmatrix} 2k+1 \\ 1 \end{pmatrix} + \begin{pmatrix} 2k+1 \\ 3 \end{pmatrix} +....$</span></p>
<p><span class="math-container">$=\dfrac{1}{(2k+1)!}(2^{2k+1-1})$</span></p>
<p><span class="math-container">$=\dfrac{1}{(2k+1)!}2^{2k}$</span></p>
<p><span class="math-container">$=2\displaystyle\sum_{k=0}^\infty (-1)^k x^{2k+1}\dfrac{1}{(2k+1)!}2^{2k}$</span></p>
<p><span class="math-container">$=\displaystyle\sum_{k=0}^\infty (-1)^k\dfrac{(2x)^{2k+1}}{(2k+1)!}$</span></p>
<p><span class="math-container">$=\sin2x$</span> </p>
|
4,387,141 | <p>I have the following instance of a semidefinite program (SDP):</p>
<p><span class="math-container">$$
d + f \to \max
$$</span></p>
<p><span class="math-container">$$
a + b + c = 1
$$</span></p>
<p><span class="math-container">$$
\begin{pmatrix}
a & d & e\\
d & b & f\\
e & f & c
\end{pmatrix} \succeq 0
$$</span></p>
<p>If I use Sylvester's criterion for this symmetric matrix, I get the following system:</p>
<p><span class="math-container">$$
\begin{cases}
a, b, c \geq 0 \\
ab - d^2 \geq 0\\
bc - f^2 \geq 0\\
ac - e^2 \geq 0\\
abc - af^2 - be^2-cd^2 + 2def \geq 0
\end{cases}
$$</span></p>
<p>From the first three inequalites the following evaluation holds</p>
<p><span class="math-container">$$
d \leq \sqrt{ab}, \qquad f \leq \sqrt{bc} \implies d + f \leq \sqrt{b}(\sqrt{a} + \sqrt{c})
$$</span></p>
<p>How can I maximize the right part of this inequality, or should I try to go another way to solve SDP?</p>
<p>I solved this problem in Mathematica and got that the following optimum</p>
<p><a href="https://i.stack.imgur.com/PgjiI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PgjiI.png" alt="enter image description here" /></a></p>
| Ben Grossmann | 81,360 | <p>I will assume that <span class="math-container">$r \leq n$</span>. Otherwise, there's some extra "bookkeeping" to be done regarding the singular values of <span class="math-container">$A$</span>.</p>
<p>It's easy to show that taking <span class="math-container">$X$</span> to be the leading left singular vectors of <span class="math-container">$A$</span> (i.e. the leading columns of <span class="math-container">$U$</span>) yields <span class="math-container">$\|X^TA\| = \sigma_1^2 + \cdots + \sigma_r^2$</span>, where <span class="math-container">$\sigma_1 \geq \cdots \geq \sigma_n$</span> denote the singular values of <span class="math-container">$A$</span>. To show that this is the maximal possible value of <span class="math-container">$\|X^TA\|$</span>, it suffices to show that <span class="math-container">$\sigma_i(A) \geq \sigma_i(X^TA)$</span> for all <span class="math-container">$i = 1,\dots,r$</span>.</p>
<p>One strategy is to use the <a href="https://en.wikipedia.org/wiki/Min-max_theorem#Min-max_principle_for_singular_values" rel="nofollow noreferrer">Courant-Fischer (min-max) theorem</a>, noting that <span class="math-container">$\|X^Tv\| \leq \|v\|$</span> holds for all vectors <span class="math-container">$v$</span>. Another approach is to note that for any such <span class="math-container">$X$</span>, there exists a <span class="math-container">$Y$</span> such that <span class="math-container">$U = [X\ \ Y]$</span> is square with orthonormal columns and hence unitary. Note that <span class="math-container">$A$</span> has the same singular values as the matrix
<span class="math-container">$$
U^TA = \pmatrix{X^TA\\ Y^TA}.
$$</span>
Now, the singular values of <span class="math-container">$U^TA$</span> are the positive eigenvalues of the block-matrix
<span class="math-container">$$
\pmatrix{0 & (U^TA)^T\\ U^TA & 0} = \pmatrix{0 & A^TX & A^TY\\X^TA & 0 & 0\\Y^TA & 0 & 0}.
$$</span>
By <a href="https://en.wikipedia.org/wiki/Min-max_theorem#Cauchy_interlacing_theorem" rel="nofollow noreferrer">Cauchy's interlacing theorem</a>, these eigenvalues interlace the positive eigenvalues of the principal submatrix
<span class="math-container">$$
\pmatrix{0 & (X^TA)^T\\ X^TA & 0},
$$</span>
whose positive eigenvalues are the singular values of <span class="math-container">$X^TA$</span>.</p>
|
4,387,141 | <p>I have the following instance of a semidefinite program (SDP):</p>
<p><span class="math-container">$$
d + f \to \max
$$</span></p>
<p><span class="math-container">$$
a + b + c = 1
$$</span></p>
<p><span class="math-container">$$
\begin{pmatrix}
a & d & e\\
d & b & f\\
e & f & c
\end{pmatrix} \succeq 0
$$</span></p>
<p>If I use Sylvester's criterion for this symmetric matrix, I get the following system:</p>
<p><span class="math-container">$$
\begin{cases}
a, b, c \geq 0 \\
ab - d^2 \geq 0\\
bc - f^2 \geq 0\\
ac - e^2 \geq 0\\
abc - af^2 - be^2-cd^2 + 2def \geq 0
\end{cases}
$$</span></p>
<p>From the first three inequalites the following evaluation holds</p>
<p><span class="math-container">$$
d \leq \sqrt{ab}, \qquad f \leq \sqrt{bc} \implies d + f \leq \sqrt{b}(\sqrt{a} + \sqrt{c})
$$</span></p>
<p>How can I maximize the right part of this inequality, or should I try to go another way to solve SDP?</p>
<p>I solved this problem in Mathematica and got that the following optimum</p>
<p><a href="https://i.stack.imgur.com/PgjiI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PgjiI.png" alt="enter image description here" /></a></p>
| KBS | 532,783 | <p>I can propose an alternative solution. The objective is to find a rank <span class="math-container">$r$</span> matrix <span class="math-container">$X\in\mathbb{R}^{m\times r}$</span> that mimimizes <span class="math-container">$||X^TA||_F$</span> and that satisfies <span class="math-container">$X^TX=I$</span>.</p>
<p>It is known that <span class="math-container">$$||X^TA||_F^2=\mathrm{trace}(A^TXX^TA)=\mathrm{trace}(AA^TXX^T).$$</span></p>
<p>If we let <span class="math-container">$X=(x_1,\ldots,x_r)$</span>, we obtain that</p>
<p><span class="math-container">$$\mathrm{trace}(AA^TXX^T)=\sum_{i=1}^r\mathrm{trace}(A^TAx_ix_i^T)=\sum_{i=1}^rx_i^TA^TAx_i$$</span> where <span class="math-container">$x_i^Tx_i=1$</span> and <span class="math-container">$x_i^Tx_j=0$</span>, <span class="math-container">$i\ne j$</span>.</p>
<p>Now, using the fact that the matrix <span class="math-container">$A^TA$</span> is positive semidefinite, we can order its (real and nonnegative) eigenvalues as <span class="math-container">$\lambda_1\ge\lambda_2\ge\ldots\lambda_n$</span>. We also note that eigenvectors of symmetric matrices are orthogonal.</p>
<p>Using this information, we can can see that to maximize the above sum, it is enough to pick the <span class="math-container">$x_i$</span>'s as the eigenvectors associated with the <span class="math-container">$r$</span> largest eigenvalues of <span class="math-container">$A^TA$</span>.</p>
<p>This can be done recursively. First, we choose <span class="math-container">$x_1$</span> to the eigenvector associated with <span class="math-container">$\lambda_1$</span>, as this is the best we can do. Now, we need to choose <span class="math-container">$x_2$</span>. Since, it has to be orthogonal to <span class="math-container">$x_1$</span>, the best choice is to pick <span class="math-container">$x_2$</span> to be eigenvector associated with <span class="math-container">$\lambda_2$</span>. And so on..</p>
<p>This leads to the result that the maximum of <span class="math-container">$||X^TA||_F^2$</span> over all rank-<span class="math-container">$r$</span> matrices <span class="math-container">$X$</span> such that <span class="math-container">$X^TX=I$</span> is given by <span class="math-container">$\sum_{i=1}^r\lambda_i$</span> or, equivalently, to the sum of the <span class="math-container">$r$</span> largest singular values of <span class="math-container">$A$</span>.</p>
<p>This leads to the final result that the maximum of <span class="math-container">$||X^TA||_F$</span> over all rank-<span class="math-container">$r$</span> matrices <span class="math-container">$X$</span> such that <span class="math-container">$X^TX=I$</span> is equal to the square root of the sum of the <span class="math-container">$r$</span> largest singular values of <span class="math-container">$A$</span>.</p>
|
1,326,652 | <p>What would be the nature of the roots of the equation $$2x^2 - 2\sqrt{6} x + 3 = 0$$</p>
<p>My book says that as the discriminant is 0 so the roots are rational and equal.
But discriminant can be used for determining the nature of roots only when the roots are rational numbers. Is the answer in the book wrong because actually the nature of roots should be irrational?</p>
| Bernard | 202,857 | <p>If the discriminant of $ax^2+bx+c$ is $0$, there is a double root, i. e. the quadratic can be written as $\,a(x-\xi)^2\,$ for some real number $\xi$ (if the coefficients $a,b,c$ are real, of course). Furthermore, this double root is equal to:
$$\xi=-\frac b{2a}.$$
So here the double root is actually <em>irrational</em>, equal to $\,\dfrac{\sqrt6}2$.</p>
|
3,014,843 | <p>My calculusbook simply states that:
<span class="math-container">$$\sin^2\left(\frac{\sqrt{x}}{2}\right)\le\left(\frac{\sqrt{x}}{2}\right)^2$$</span>
...but I don't immediately see why this is true. </p>
<p>What (probably) simple trick am I missing? Apparently it is "obvious"...</p>
| Matthew C | 25,959 | <p>This will follow if you show that <span class="math-container">$\sin(t) \leq t$</span> for all <span class="math-container">$t\geq 0$</span>: To do that, you can try to show that if <span class="math-container">$f(0) = g(0)$</span> and <span class="math-container">$f'(t) \leq g'(t)$</span> for all <span class="math-container">$t\geq 0$</span>, then <span class="math-container">$f(t) \leq g(t)$</span> for all <span class="math-container">$t\geq 0$</span>.</p>
|
3,014,843 | <p>My calculusbook simply states that:
<span class="math-container">$$\sin^2\left(\frac{\sqrt{x}}{2}\right)\le\left(\frac{\sqrt{x}}{2}\right)^2$$</span>
...but I don't immediately see why this is true. </p>
<p>What (probably) simple trick am I missing? Apparently it is "obvious"...</p>
| hamam_Abdallah | 369,188 | <p>By MVT</p>
<p><span class="math-container">$$\sin(X)-\sin(0)=X\cos(c)$$</span></p>
<p>thus</p>
<p><span class="math-container">$$|\sin(X)|\le |X|$$</span></p>
<p>and</p>
<p><span class="math-container">$$\sin^2(X)\le X^2$$</span></p>
<p>now apply to
<span class="math-container">$$X=\frac{\sqrt{x}}{2}$$</span></p>
|
303,944 | <p>Interesting problem I spotted while learning:</p>
<blockquote>
<p>Let <span class="math-container">$X=\left\{1,..,n\right\}$</span>. We randomly select subset of <span class="math-container">$X$</span> and name it <span class="math-container">$A$</span>. Each subset if equally likely.</p>
<p>a) Find the expected value of the sum of elements of A.</p>
<p>b) Find the expected value of the sum of elements of A, on condition that it has <span class="math-container">$k$</span> elements.</p>
</blockquote>
<p>a) I think I know how to solve a). If each subset is selected with the same probability then I think it is equivalent to selecting each element of <span class="math-container">$X$</span> with probability <span class="math-container">$\frac{1}{2}$</span>. So, using indicators, we got that expected value we are looking for is <span class="math-container">$\frac{n(n+1)}{4}$</span>. But I can't find any rigorous argument why it is equivalent to selecting each element with probability <span class="math-container">$1/2$</span>.</p>
<p>b) Small observation with <span class="math-container">$k=1$</span> (each element selected with probability <span class="math-container">$1/n$</span>) and <span class="math-container">$k=n$</span> (each element selected with probability <span class="math-container">$1$</span>) gives me feeling that approach from a) can be used with probability <span class="math-container">$k/n$</span> and then the result is <span class="math-container">$\frac{k(n+1)}{2}$</span>. But it is much less intuitive than observation in a). No idea, how to prove this. Can anyone help?</p>
| Ross Millikan | 1,827 | <p>For a) you can pair each subset with its complement to show that a given element is in half the subsets, so is in the chosen subset half the time.</p>
<p>For b) each element has the same chance to be chosen. If you want to get $k$ of them, you need that chance to be $\frac kn$.</p>
<p>Good thinking on both.</p>
|
432,223 | <p>I wanted to know, how can I factor $x^6 +5x^3 +8$, I have no idea. Is there any method to know if a polynomial is factored. Just some advice will do.</p>
<p>Help appreciated.</p>
<p>Thanks.</p>
| Eric Tressler | 26,785 | <p>Let $y = x^3$ to obtain $y^2 + 5y + 8 = 0$. This factors as $y = \frac{-5 \pm \sqrt{-7}}{2} = \frac{-5}{2} \pm \frac{\sqrt{7}}{2}i$. These have modulus $r = \sqrt{25/4 + 7/4} = 2\sqrt{2}$. Now solve $5/2 = 2\sqrt{2} \cos \theta$ to find the angle $\theta = \cos^{-1}(5/4\sqrt{2})$. Our two roots correspond to $re^{\pi-\theta}$ and $re^{\pi+\theta}$.</p>
<p>Now we have $x^3 = re^{\pi-\theta}$ and $x^3 = re^{\pi+\theta}$ to contend with. For the former, one root is $x_1 = \sqrt[3]{r}e^{(\pi-\theta)/3}$, so the other two are $x_2 = \sqrt[3]{r}\exp(\frac{\pi-\theta}{3} + 2\pi/3)$ and $x_3 = \sqrt[3]{r}\exp(\frac{\pi-\theta}{3} + 4\pi/3)$.</p>
<p>Similarly, the roots of the other equation are $x_4 = \sqrt[3]{r}e^{(\pi+\theta)/3}$, $x_5 = \sqrt[3]{r}\exp(\frac{\pi+\theta}{3} + 2\pi/3)$, and $x_6 = \sqrt[3]{r}\exp(\frac{\pi-\theta}{3} + 4\pi/3)$.</p>
|
17,885 | <p>In most education systems, Mathematics is a compulsory subject from primary school all the way to the start of university. A common reason given is that essential concepts like addition and multiplication are taught to the children. </p>
<p>But for many high school students, especially those who are keen on pursuing the humanities, they do not see any point in studying Mathematics for the rest of their schooling life, or how any concept in Mathematics could possibly be applied in their future work.</p>
<p>Why is Mathematics a compulsory subject for high school students, especially those who are clearly studying in Humanities streams?</p>
| JRN | 77 | <p>Underwood Dudley answers the question "What is mathematics education for?" in <a href="https://www.ams.org/notices/201005/rtx100500608p.pdf" rel="noreferrer">this article</a> from 2010 (<em>Notices of the American Mathematical Society</em>, vol. 57, no. 5, pp. 608-613) (even though the title of the article is "What Is Mathematics For?").</p>
<blockquote>
<p>So that there is no confusion, let me say that by “mathematics” I mean algebra, trigonometry, calculus, linear algebra, and so on: all those subjects beyond arithmetic. There is no question about what arithmetic is for or why it is supported. Society cannot proceed without it. Addition, subtraction, multiplication, division, percentages: though not all citizens can deal fluently with all of them, we make the assumption that they can when necessary. Those who cannot are sometimes at a disadvantage.</p>
<p>Algebra, though, is another matter. Almost all citizens can and do get through life very well without it, after their schooling is over. Nevertheless it becomes more and more pervasive, seeping down into more and more eighth-grade classrooms and being required by more and more states for graduation from high school. There is unspoken agreement that everyone should be exposed to algebra. We live in an era of universal mathematical education.</p>
</blockquote>
<p>He concludes:</p>
<blockquote>
<p>What mathematics education is for is not for jobs. It is to teach the race to reason. It does not, heaven knows, always succeed, but it is the best method that we have. It is not the only road to the goal, but there is none better.</p>
</blockquote>
|
17,885 | <p>In most education systems, Mathematics is a compulsory subject from primary school all the way to the start of university. A common reason given is that essential concepts like addition and multiplication are taught to the children. </p>
<p>But for many high school students, especially those who are keen on pursuing the humanities, they do not see any point in studying Mathematics for the rest of their schooling life, or how any concept in Mathematics could possibly be applied in their future work.</p>
<p>Why is Mathematics a compulsory subject for high school students, especially those who are clearly studying in Humanities streams?</p>
| WGroleau | 13,472 | <p>Because</p>
<blockquote>
<p>many high school students, especially those who are keen on pursuing
the humanities, they do not see any point in studying Mathematics for
the rest of their schooling life, or how any concept in Mathematics
could possibly be applied in their future work.</p>
</blockquote>
<p>shows they have no clue what mathematics is worth, therefore it needs to be compulsory to protect them from themselves.</p>
|
17,885 | <p>In most education systems, Mathematics is a compulsory subject from primary school all the way to the start of university. A common reason given is that essential concepts like addition and multiplication are taught to the children. </p>
<p>But for many high school students, especially those who are keen on pursuing the humanities, they do not see any point in studying Mathematics for the rest of their schooling life, or how any concept in Mathematics could possibly be applied in their future work.</p>
<p>Why is Mathematics a compulsory subject for high school students, especially those who are clearly studying in Humanities streams?</p>
| Aaron Peterson | 13,494 | <p>Mathematic notation and numbers are fundamental to basic communication and problem solving.</p>
<p>Our existence depends on it.</p>
<p>It is also very fun. </p>
<p>I think this question would come up less often if we presented kids with problems to solve, and then they realize that they need tools. </p>
<p>My physics teacher derived formulas from observation and testing. </p>
<p>Memorising tables sucks.
recognizing a pattern can be fun. </p>
|
372,181 | <p>The theory of real closed fields is decidable. The <a href="https://en.wikipedia.org/wiki/Hyperreal_number" rel="noreferrer">hyperreals</a> satisfy that theory, so we can interpret statements in the theory of real closed fields as being about hyperreals.</p>
<p>If we add a unary predicate for "is a standard real number" to the language, is the theory still decidable?</p>
| Emil Jeřábek | 12,705 | <p>Yes, the theory is decidable.</p>
<p>If <span class="math-container">$F$</span> is an ordered field and <span class="math-container">$R\subseteq F$</span> a non-cofinal subfield, then
<span class="math-container">$$O=\{x\in F:\exists u\in R\:(-u\le x\le u)\}$$</span>
is a convex valuation ring of <span class="math-container">$F$</span>, with maximal ideal
<span class="math-container">$$I=\{x\in F:\forall u\in R_{>0}\:(-u\le x\le u)\}.$$</span>
<span class="math-container">$R$</span> embeds as a cofinal subfield in the residue field <span class="math-container">$O/I$</span>; in general, the embedding may be proper, but if <span class="math-container">$R=\mathbb R$</span>, then <span class="math-container">$R=O/I$</span>, as <span class="math-container">$\mathbb R$</span> is complete.</p>
<p>Thus, let <span class="math-container">$T$</span> be the theory of structures <span class="math-container">$(F,R,+,\cdot,<)$</span> such that</p>
<ol>
<li><p><span class="math-container">$F$</span> is a real-closed field,</p>
</li>
<li><p><span class="math-container">$R$</span> is a non-cofinal subfield of <span class="math-container">$F$</span>, and</p>
</li>
<li><p>the canonical embedding of <span class="math-container">$R$</span> into the residue field <span class="math-container">$O/I$</span> as defined above is surjective (and therefore an isomorphism).</p>
</li>
</ol>
<p>Then <span class="math-container">$T$</span> is a recursively axiomatized theory, it is valid in the hyperreal structures <span class="math-container">$({}^*\mathbb R,\mathbb R)$</span>, and it is complete (see below), hence it is decidable, and axiomatizes the first-order theory of <span class="math-container">$({}^*\mathbb R,\mathbb R)$</span>.</p>
<p>As pointed out in a comment by Erik Walsberg (thanks!), the completeness of <span class="math-container">$T$</span> is a special case of a more general result on tame elementary extensions of o-minimal structures due to Van den Dries and Lewenberg [1]. (Here, tameness is basically the axiom 3 above.) Their results also show that <span class="math-container">$T$</span> is model-complete, and in fact, that it has quantifier elimination in a language expanded with function symbols for roots of polynomials (which make the theory of real-closed fields universally axiomatized) and for the “standard part” map <span class="math-container">$\mathrm{st}\colon O\to R$</span> such that <span class="math-container">$x-\mathrm{st}(x)\in I$</span>.</p>
<p>Let me indicate how to prove a weaker result: the theory <span class="math-container">$T_0$</span> of structures <span class="math-container">$(F,O)$</span> such that</p>
<ol>
<li><p><span class="math-container">$F$</span> is a real-closed field,</p>
</li>
<li><p><span class="math-container">$O$</span> is a proper convex subring of <span class="math-container">$F$</span>,</p>
</li>
</ol>
<p>is complete and decidable. This follows from the Ax–Kochen–Ershov principle, which states that two henselian valued fields of residue characteristic <span class="math-container">$0$</span> with elementarily equivalent residue fields and value groups are elementarily equivalent.</p>
<p>First, it is an easy consequence of basic facts about valued fields that if <span class="math-container">$F$</span> is a real-closed field with a convex valuation ring <span class="math-container">$O$</span>, then the valued field <span class="math-container">$(F,O)$</span> is henselian, the residue field <span class="math-container">$O/I$</span> is real-closed, and the value group <span class="math-container">$F^\times/O^\times$</span> is divisible.</p>
<p>Thus, if <span class="math-container">$(F,O)$</span> and <span class="math-container">$(F',O')$</span> are two models of <span class="math-container">$T_0$</span>, their residue fields are elementarily equivalent by completeness of the theory of real-closed fields, and their value groups are elementarily equivalent by completeness of the theory of divisible totally ordered abelian groups, hence the valued fields <span class="math-container">$(F,O)$</span> and <span class="math-container">$(F',O')$</span> are elementarily equivalent by the AKE principle.</p>
<p>As shown by Cherlin and Dickman [2], <span class="math-container">$T_0$</span> has quantifier elimination in the language of ordered rings expanded with the predicate
<span class="math-container">$$x\mid y\iff y\in xO.$$</span></p>
<p><strong>References:</strong></p>
<p>[1] Lou van den Dries, Adam H. Lewenberg: <em><span class="math-container">$T$</span>-convexity and tame extensions</em>, Journal of Symbolic Logic 60 (1995), no. 1, pp. 74–102, doi: <a href="https://doi.org/10.2307/2275510" rel="nofollow noreferrer">10.2307/2275510</a>. On <a href="https://www.jstor.org/stable/2275510" rel="nofollow noreferrer">JSTOR</a>.</p>
<p>[2] Gregory Cherlin, Max A. Dickmann: <em>Real closed rings II. Model theory</em>, Annals of Pure and Applied Logic 25 (1983), no. 3, pp. 213–231, doi: <a href="https://doi.org/10.1016/0168-0072(83)90019-2" rel="nofollow noreferrer">10.1016/0168-0072(83)90019-2</a>.</p>
|
1,313,216 | <p>Its my first time on here and my maths is poor so please be kind. I am working on a Masters dissertation focused on document clustering methods in which I would like to apply a weight based on the time interval between two documents. </p>
<p>I am looking for some help coming up with a function to express a time interval with results between 0 and 1. The reason I want to map the results to a maximum value of zero is that this is being applied as a weight to a cosine similarity metric where identical articles would receive a cosine measurement of 1 etc.</p>
<p>Example 1, the date difference between 31/05/2015 and 20/06/2015 is 9 days.
Example 2, the date difference between 31/05/2015 and 20/01/2015 is 129 days.</p>
<p>I would like to apply a function whereby example 1 has a higher value (towards the 1 end of the scale) and example 2 has a lower value (towards the 0 end of the scale). If the date difference was only 1, the value of 1 should apply.</p>
<p>I hope this makes sense. Any help anyone can offer me would be greatly appreciated. </p>
<p>Thank You</p>
<p>Claire</p>
| Hrodelbert | 133,762 | <p>The simplest approach in my opinion is to consider the function
$$
w(d) =\frac{1}{d},
$$
where $d$ is the number of days difference. It is obvious that $w(1) = 1$ and that $0<w(d)<1$ for all $d\geq1$, as you wanted, and by coincidence, this function also has the property that $w(9)=\frac{129}{9}w(129)$, or more generally
$$
w(d_1) = \frac{d_2}{d_1} w(d_2).
$$
You can generalize this function if you want:
$$
w_A(d) = \frac{A+1}{d+A}
$$
also obeys all your properties for a positive constant $A$. Note that this is not the most general function satisfying your constraints. Even after replacing $d$ with some monotonically ascending function of $d$, $w$ will still not be the most general function. </p>
|
4,349,748 | <p>I would appreciate some help with this.</p>
<p>Is:
<span class="math-container">$$
\nabla \cdot \left( \nabla \vec{v}\right)^T= \nabla \left( \nabla \cdot \vec{v}\right)
$$</span></p>
<p>How can I show this? Is the gradient of a vector mathematically defined?</p>
<p>Best regards</p>
| Son Gohan | 865,323 | <p>To me it is ok, but since you are in a metric space you could use sequences as well to characterize the closure of your set.</p>
<p>If instead you wanted to do it because you wanted to exercise with nets (whose convergence is sometimes called Moore-Smith convergence), then all you have to do is to mimick the sequence proof as you did.</p>
|
64,022 | <p><strong>Edit</strong></p>
<p>(As Robert pointed out, what I was trying to prove is incorrect. So now I ask the right question here, to avoid duplicate question)</p>
<p>For infinite independent Bernoulli trials with probability $p$ to success, define a random variable N which equals to the number of successful trial. Intuitively, we know if $p > 0$, $\Pr \{N < \infty \} = 0$, in other word $N \rightarrow \infty$. But I got stuck when I try to prove it mathematically.</p>
<p>\begin{aligned}
\Pr \{ N < \infty \}
& = \Pr \{ \cup_{n=1}^{\infty} [N \le n] \} \\
& = \lim_{n \rightarrow \infty} \Pr \{ N \le n \} \\
& = \lim_{n \rightarrow \infty}\sum_{i=1}^{n} b(i; \infty, p) \\
& = \sum_{i=1}^{\infty} b(i; \infty, p) \\
\end{aligned}</p>
<p>I've totally no idea how to calculate the last expression.</p>
<hr>
<p>(Original Question)</p>
<p>For infinite independent Bernoulli trials with probability $p$ to success, define a random variable N which equals to the number of successful trial. Can we prove that $\Pr \{N < \infty \} = 1$ by:</p>
<p>\begin{aligned}
\Pr \{ N < \infty \}
& = \Pr \{ \cup_{n=1}^{\infty} [N \le n] \} \\
& = \lim_{n \rightarrow \infty} \Pr \{ N \le n \} \\
& = \lim_{n \rightarrow \infty}\sum_{i=1}^{n} b(i; \infty, p) \\
& = \sum_{i=1}^{\infty} b(i; \infty, p) \\
& = \lim_{m \rightarrow \infty}\sum_{i=1}^{m} b(i; m, p) \\
& = \lim_{m \rightarrow \infty}[p + (1 - p)]^m \\
& = \lim_{m \rightarrow \infty} 1^m \\
& = 1
\end{aligned}</p>
<p>I know there must be some mistake in the process because if $p = 1$, N must infinite. So the equation only holds when $ p < 1 $. Which step is wrong?</p>
| Srivatsan | 13,425 | <p>You want to compute the probability of $s$ successes for $s = 0, 1, 2, \ldots$. Here the crucial point is that $s$ is fixed first, and then you compute the probability that you get $s$ successes when you throw infinitely many coins (each of success probability $p$). In other words, we want
$$
\lim_{m \to \infty} b(s; m, p) = \lim_{m \to \infty} \binom{m}{s} p^s (1-p)^{m-s} = (\frac{p}{1-p})^s \lim_{m \to \infty} \binom{m}{s} (1-p)^m.
$$
You can intuitively see that this answer should come out to be $0$ (since you are throwing infinitely many coins). How can we justify that rigorously? By upper bounding the function of $m$ suitably, and then using the sandwich theorem.</p>
<p>When $s$ is fixed, the first term $\binom{m}{s}$ is at most a polynomial in $s$, since we can upper bound it loosely by $\binom{m}{s} \leq m^s$. On the other hand, $(1-p)^m$ goes to zero exponentially fast. Can you use this to finish the proof?</p>
|
64,022 | <p><strong>Edit</strong></p>
<p>(As Robert pointed out, what I was trying to prove is incorrect. So now I ask the right question here, to avoid duplicate question)</p>
<p>For infinite independent Bernoulli trials with probability $p$ to success, define a random variable N which equals to the number of successful trial. Intuitively, we know if $p > 0$, $\Pr \{N < \infty \} = 0$, in other word $N \rightarrow \infty$. But I got stuck when I try to prove it mathematically.</p>
<p>\begin{aligned}
\Pr \{ N < \infty \}
& = \Pr \{ \cup_{n=1}^{\infty} [N \le n] \} \\
& = \lim_{n \rightarrow \infty} \Pr \{ N \le n \} \\
& = \lim_{n \rightarrow \infty}\sum_{i=1}^{n} b(i; \infty, p) \\
& = \sum_{i=1}^{\infty} b(i; \infty, p) \\
\end{aligned}</p>
<p>I've totally no idea how to calculate the last expression.</p>
<hr>
<p>(Original Question)</p>
<p>For infinite independent Bernoulli trials with probability $p$ to success, define a random variable N which equals to the number of successful trial. Can we prove that $\Pr \{N < \infty \} = 1$ by:</p>
<p>\begin{aligned}
\Pr \{ N < \infty \}
& = \Pr \{ \cup_{n=1}^{\infty} [N \le n] \} \\
& = \lim_{n \rightarrow \infty} \Pr \{ N \le n \} \\
& = \lim_{n \rightarrow \infty}\sum_{i=1}^{n} b(i; \infty, p) \\
& = \sum_{i=1}^{\infty} b(i; \infty, p) \\
& = \lim_{m \rightarrow \infty}\sum_{i=1}^{m} b(i; m, p) \\
& = \lim_{m \rightarrow \infty}[p + (1 - p)]^m \\
& = \lim_{m \rightarrow \infty} 1^m \\
& = 1
\end{aligned}</p>
<p>I know there must be some mistake in the process because if $p = 1$, N must infinite. So the equation only holds when $ p < 1 $. Which step is wrong?</p>
| uforoboa | 15,453 | <p>Let us call $E_{k,n}:=$ probability of winning exactly $k$ times after $n$ trials. Let now $$E_k=\lim_{n\to+\infty}E_{k,n}.$$</p>
<p>It holds
$$P(E_k)=\lim_{n\to\infty}P(E_{k,n})=\lim_{n\to+\infty}\binom{n}{k}p^k(1-p)^{n-k}$$
Because $E_{k,n}\subseteq E_{k,n+1}$ and of course one has</p>
<p>$$0\leq P(E_k)= \lim_{n\to+\infty}\left(\frac{p}{1-p}\right)^k\binom{n}{k}(1-p)^n\leq C(p,k)\lim_{n\to+\infty}n^k(1-p)^n=0.$$ </p>
<p>Now, the probability you are asking to find is clearly contained in the event $$\bigcup_{k=0}^{+\infty}E_k,$$
hence, by monotonicity and subadditivity of the probability measure, one has that the probability of winning a finite number of times in an infinite sequence of trials lesser or equal than
$$\lim_{i\to+\infty}\sum_{k=1}^iP(E_k)=0,$$ and so it is $0$.</p>
|
823,641 | <p>Find and classify the critical points of this function: $f(x,y)= (x^y)-(xy)$ in the domain $x>0, y>0$.</p>
<p>I am having trouble treating x and y as constants when taking partial derivatives.</p>
| Claude Leibovici | 82,404 | <p><strong>Hint</strong></p>
<p>If the names make trouble to you because they are names of variables, consider first $f_1(x)=x^k-kx$ and compute its derivative with respect to $x$ as usual; this will give you $$\frac {df_1(x)}{dx}=k x^{k-1}-k$$ so, replacing $k$ by $y$, $$f'_x(x,y)=y x^{y-1}-y$$ Do a similar thing using $f_2(y)=k^y-ky$; this will give you $$\frac {df_2(y)}{dy}=k^y \log(k)-k$$ so, replacing now $k$ by $x$ $$f'_y(x,y)=x^y \log (x)-x$$</p>
<p>Is this making things clearer ? When you take the partial derivative with respect to one variable the other variables are considered as constants.</p>
|
2,534,369 | <p>I am trying to work my through the exercises in Spivak's <em>Calculus on Manifolds.</em> I am currently working on the exercises in Chapter 3 which deals with Integration. I am having trouble with the following question:</p>
<blockquote>
<p>Let:</p>
<p>\begin{equation}
f(x,y)=\begin{cases}
0, & \text{if $x$ is irrational}.\\
0, & \text{if $x$ is rational, $y$ is irrational}. \\
1/q, & \text{if $x$ is rational, $y=p/q$ in lowest terms}.
\end{cases}
\end{equation}</p>
<p>Show that $f$ is integrable on $A = [0,1] \times [0,1]$ and $\int_A f = 0$.</p>
</blockquote>
<p>I was thinking of trying to prove that this set is Jordan Measurable and that it's Jordan measure is zero and that it is therefore Riemann Integrable but I am not sure how to do this or if it is even the best way to solve this problem.</p>
<p>If I could show that $f$ is continuous on $A$ up to a set of Jordan Measure $0$, then $f$ would be integrable but again, I'm not sure I can do this or if its even appropriate for this problem.</p>
<p>Any assistance that anyone could provide would be greatly appreciated.</p>
<p>Thank you.</p>
| RRL | 148,510 | <p>Hint: For any partition $P$ of $A$ the lower sum $L(P,f) = 0$ since any rectangle must contain a point $(x,y)$ where $x$ is irrational and $f(x,y) = 0.$ Next show that the upper sum $U(P,f)$ can be arbitrarily close to zero if the partition is sufficiently fine. Just extend the proof for the one-dimensional case given <a href="https://math.stackexchange.com/a/1427790/148510">here</a>.</p>
<p><strong>Aside</strong></p>
<p>This function is peculiar in that it is Riemann integrable on $[0,1]^2$, but for fixed rational $y$, the function $f(\cdot,y)$ is a non-Riemann-integrable Dirichlet function and $\int_0^1 f(x,y) \, dx$ does not exist as a Riemann integral.</p>
<p>In this case, the iterated integral</p>
<p>$$\int_0^1 \left(\int_0^1 f(x,y) \, dx \right) \, dy$$</p>
<p>does not exist.</p>
|
123,675 | <p>What are some "applications to" / "connections with" topology that one could hope to reasonably cover in a first course on topos theory (for master students)? I have an idea of what parts of the theory I would like to cover, however, I would love some more nice examples and applications. Of course, I have some ideas of my own, but am open to suggestions. Thank you!</p>
<p>P.S.</p>
<p>I am also interested in perhaps learning about some new connections myself, which would be out of reach for such a course, so feel free to leave these as well, qualified as such.</p>
| Ivan Di Liberti | 104,432 | <p>I will copy and paste the description of chapter 2 and 3 of my master thesis.</p>
<p>Chapter 2 plays an important motivational role in the thesis and is aimed at pointing out geometric characteristics of a Grothendieck topos. For a topological space $X$ one can consider the category of its open sets $\mathcal{O}(X)$ that is a complete Heyting algebra, so that we have a functor $$ \text{Spaces} \to \text{cHa}^{\text{op}}. $$ We study properties of this functor concluding that there is a huge subcategory of spaces (sober ones) that embeds into $\text{cHa}^{\text{op}}$ via this functor. For a notational motivation we call Locales the opposite of cHa,
$$\text{SobSpaces} \hookrightarrow \text{Locales} $$
So locales naturally are generalized (sober) spaces. </p>
<p>In Chapter 2 we present the notion of localic topos that is a Grothendieck topos on a locale and we prove that there is an equivalence of category between the category of locales and the category of localic toposes</p>
<p>\begin{matrix} \text{Locales} & \leftrightarrows & \text{LocToposes.} & \end{matrix} </p>
<p>This equivalence is the precise sense in which a localic topos is a generalized topological space, that is the same in which its associated locale is a generalized topological space.</p>
<p>A generic Grothendieck topos has not this fascinating property, there is not a topological space from which it comes from, but precisely in this rift one can collocate Barr's theorem.</p>
<p>Chapter 3 is devoted to proving Barr's theorem that we can formulate right now:</p>
<blockquote>
<p>Any Grothendieck topos is covered by a localic boolean one.</p>
</blockquote>
<p>This theorem states that not any Grothendieck topos is geometric but not far from it there is an other Grothendieck topos that is not only geometric (better say localic) but it is also boolean.
This is the first and most naive interpretation of Barr's theorem.</p>
|
2,008,437 | <p>Given four points in the plane, there exists a one-dimensional family of conics through these, often called a pencil of conics. The locus of the centers of symmetry for all of these conics is again a conic. What's the most elegant way of computing it?</p>
<p>I know I could choose five arbitrary elements from the pencil, compute their centers and then take the conic defined by these. I can also do so on a symbolic level, to obtain a general formula. But that formula is at the coordinate level, and my CAS is still struggeling with the size of the polynomials involved here. There has to be a better way.</p>
<p>Bonus points if you know a name for this conic. Or – as the center is the pole of the line at infinity – a name for the more general locus of the pole of an arbitrary line with respect to a given pencil of conics.</p>
| MvG | 35,416 | <p>Based on the fundamental theorem of projective geometry, it must be possible to take any projectively invariant property which can be described as a polynomial in the coordinates of the involved points and turn that into an expression in determinants of these homogeneous coordinates of these points instead. Using randomized evaluation of polynomial expressions, one can find valid combinations of relevant determinants and similar terms. I've used this technique before while formulating <a href="https://math.stackexchange.com/q/1848710/35416">this question of mine</a>, and now remembered to do it here as well.</p>
<p>The linear dependencies identified by my randomized evaluation look like this:</p>
<pre><code>goal 1
[B,C,D] [A,C,D] (l,AB) 1
[B,C,D] [A,B,D] (l,AC) 1
[B,C,D] [A,B,C] (l,AD) 1
[A,C,D] [A,B,D] (l,BC) 1
[A,C,D] [A,B,C] (l,BD) 1
[A,B,D] [A,B,C] (l,CD) 1
<l,D> [B,C,D] (AB,AC) 1
<l,C> [B,C,D] (AB,AD) 1
<l,D> [A,C,D] (AB,BC) 1
<l,C> [A,C,D] (AB,BD) 1
<l,D> [A,B,C] (AB,CD) 1
<l,C> [A,B,D] (AB,CD) 1
<l,B> [A,C,D] (AB,CD) 1
<l,A> [B,C,D] (AB,CD) 1
<l,B> [B,C,D] (AC,AD) 1
<l,D> [A,B,D] (AC,BC) -1 -1 -1 -1 -1 -1
<l,D> [A,B,C] (AC,BD) 1
<l,C> [A,B,D] (AC,BD) 1 1 -1 -1 1 -1 -1 -1 -1
<l,B> [A,C,D] (AC,BD) 1
<l,A> [B,C,D] (AC,BD) 1
<l,B> [A,B,D] (AC,CD) -1 -1 1 1 -1 1 -1 1 -1
<l,D> [A,B,C] (AD,BC) 1
<l,C> [A,B,D] (AD,BC) 1 -1 1 1 -1 1 1 1 -1
<l,B> [A,C,D] (AD,BC) -1 -1 1 1 -1 -1 -1 1 -1 1
<l,A> [B,C,D] (AD,BC) 1
<l,C> [A,B,C] (AD,BD) 1 -1 1 1 1 1
<l,B> [A,B,C] (AD,CD) 1 1 -1 -1 1 -1 1 1 -1 1
<l,A> [A,C,D] (BC,BD) 1 1 -1 -1 1 1 -1 -1 -1 -1
<l,A> [A,B,D] (BC,CD) -1 1 -1 -1 -1 -1 -1 1 1
<l,A> [A,B,C] (BD,CD) -1 1 1 1 1
</code></pre>
<p>The first column contains a non-zero entry in the row titled <code>goal</code>, so ignoring that entry, the rest is a formula for the conic I'm after. The other columns are just fancy descriptions of the zero matrix. Adding any of these to the first column <em>may</em> lead to a simpler formula in some way.</p>
<p>Here is how to read the notation of the terms. The letters <code>A</code>, <code>B</code>, <code>C</code> and <code>D</code> are the four points defining the pencil of conics. The letter <code>l</code> gives the line at infinity, which is required to make this a projectively invariant concept. The square brackets like <code>[A,B,C]</code> are determinants. Angle brackets like <code><l,D></code> are scalar products. If homogeneous coordinates are formed using a one in the last coordinate, and the line at infinity is described as $[0:0:1]$, then these will be just one. (As the line at infinity might be spanned by two points, this is essentially like a three-point determinant as well.) Parentheses like <code>(AB,CD)</code> are matrices of degenerate conics. $A\times B$ is one line joining two of the given points, and $C\times D$ is the other. So $(A\times B)(C\times D)^T$ will be a rank one matrix describing this pair of lines. Adding its transppose makes the matrix symmetric without changing the quadratic form. <code>(l,CD)</code> is the same thing with the line at infinity taking the role of one of the lines connecting two points. (Since the matrix constitutes a quadratic form, plugging a point <code>X</code> into said form would turn <code>(AB,CD)</code> into <code>[ABX][CDX]</code>, i.e. again a product of determinants.)</p>
<p>While this makes for a shorter formula than doing this at the coordinate level, the approach does convey little geometric intuition. But perhaps playing around with the space of possible formulas will allow someone to come up with a more elegant formula which can be interpreted geometrically again.</p>
|
2,008,437 | <p>Given four points in the plane, there exists a one-dimensional family of conics through these, often called a pencil of conics. The locus of the centers of symmetry for all of these conics is again a conic. What's the most elegant way of computing it?</p>
<p>I know I could choose five arbitrary elements from the pencil, compute their centers and then take the conic defined by these. I can also do so on a symbolic level, to obtain a general formula. But that formula is at the coordinate level, and my CAS is still struggeling with the size of the polynomials involved here. There has to be a better way.</p>
<p>Bonus points if you know a name for this conic. Or – as the center is the pole of the line at infinity – a name for the more general locus of the pole of an arbitrary line with respect to a given pencil of conics.</p>
| MvG | 35,416 | <p><em>While there were many good answers here, I feel that two comments were even more useful to me, so I'll combine them to a CW answer.</em></p>
<p><a href="https://math.stackexchange.com/users/305862/jeanmarie">@JeanMarie</a> quoted <a href="http://en.wikipedia.org/wiki/Nine-point_conic" rel="nofollow noreferrer">Wikipedia</a>:</p>
<blockquote>
<p>In 1912 Maud Minthorn showed that the nine-point conic is the locus of the center of a conic through four given points.</p>
</blockquote>
<p>So this gives a name to the conic I described. It also characterizes it as a conic passing through the centers of the sides as well as through the points of intersections of opposite sides.</p>
<p>The former of these properties is something <a href="https://math.stackexchange.com/users/59379/achille-hui">@achille hui</a> deduced from my specification of the conic:</p>
<blockquote>
<p>Given four points $P_1,P_2,P_3,P_4$ in general position. If you take $P_5=P_i+P_j-P_k$ where $i\neq j\neq k\in\{1,2,3,4\}$, the center of the conic passing through $P_1,\dots,P_5$ will be $\frac12(P_i+P_j)$. If you know the locus of center is a conic, this gives you $6$ easy to compute points for constructing it.</p>
</blockquote>
<p>So yes, taking the midpoint for every pair of points, dropping one of these and then constructing the conic through the other five is an elegant approach.</p>
<p>It is not perfect in terms of simplicity of the resulting formulation, though. Labeling the points $A,B,C,D$ and taking all pairwise midpoints except for $\frac12(C+D)$ (or rather $D_z\cdot C + C_z\cdot D$ in homogeneous coordinates), I found the following removable common factor in a homogeneous formulation of this construction:</p>
<p>$$A_z^3\cdot B_z^3\cdot C_z\cdot D_z\cdot\begin{vmatrix}
A_zB_x-B_zA_x & C_zD_x-D_zC_x \\
A_zB_y-B_zA_y & C_zD_y-D_zC_y
\end{vmatrix}$$</p>
<p>The first four factors indicate that the construction outlined above has a removable singularity (resulting in a null matrix instead of some conic) if one of the defining points is at infinity. This should not be a problem in non-projective scenarios where all inputs are guaranteed to be finite. It's also not surprising as the midpoint of a segment with one endpoint at infinity will be said point at infinity, so three of the six pairwise midpoints will be identical.</p>
<p>The last term, the determinant, characterizes a situation where the line $AB$ and the line $CD$ are parallel. In that case, four of the five defining points would lie on the parallel halfway between $AB$ and $CD$, and the remaining midpoint of $AB$ is not enough to define the second line of this degenerate conic. So be sure that the midpoint you omit corresponds to a line which is not parallel to its counterpart.</p>
<p><em>My other answer discusses a formulation which avoids these removable singularities.</em></p>
|
1,085,164 | <p><strong>I am currently taking a course on Numerical PDE. The course covers the following topics listed below.</strong> </p>
<p><strong>Chapter 1: Solutions to Partial Dierential Equations:</strong></p>
<p><strong>Chapter 2: Introduction to Finite Elements:</strong></p>
| Zaheer kiyani | 964,430 | <p>If any one tell me the basic and easy book with examples to understand the course of Numerical methods of PDE's.
Contents are given:
Boundary and initial conditions, Polynomial approximations in higher dimensions. Finite Difference Method: Finite Difference approximation. Finite Element Method: The Galerkin method in one and more dimension, Error bound on the Galarkin method, The Method of Collocation, Error bounds on the Collocation method, Comparison of efficiency of the finite difference and finite element method. Application to the solution of Linear and non-linear Partial Differential Equations appearing in Physical and engineering Problems.</p>
|
3,024,456 | <p>I am working on a problem and I am confused if exponents can be split up in the manner below. </p>
<p><span class="math-container">$$x^{k+2} = x^{k+1}+x^{k+1} = 2x^{k+1} $$</span></p>
<p>I apologize if this is a simple question I tried to look up exponent rules but I couldn't find exponents + a number. </p>
| PrincessEev | 597,568 | <p>No, they definitely cannot be split up like that. To my knowledge there isn't any general law for <span class="math-container">$x^a + x^b = \text{something}$</span>.</p>
<p>A few handy exponent laws can be found at <a href="http://mathworld.wolfram.com/ExponentLaws.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/ExponentLaws.html</a></p>
|
2,315,599 | <p>I am trying to evaluate the integral<br>
$\int (\frac{1}{z(\exp(z)-1)}dz$ $\mathbb{C}:|z|=1\}$</p>
<p>using Laurent theorem. $0$ is a singularity point and both denominators becomes $0$ when $z=0$. So it's hard to find a method to evaluate this using Residue theorem. Could someone please show me how to solve this.
Thank you.</p>
| Saketh Malyala | 250,220 | <p><strong>I would appreciate if someone could check this over for me:</strong></p>
<p>Consider $d$, the distance between her and the tram. </p>
<p>Also consider $r_0$ and $r_1$, the speeds of her and the tram.</p>
<p>We have (1) --> $d=(r_0+r_1)t$.</p>
<p>We also have (2) --> $d+\frac{1}{5}r_1=\frac{1}{5}r_0+(r_0+r_1)(t-\frac{1}{20})$</p>
<p>We know that $r_0=5.5$.</p>
<p>Now we subtract the equations from each other and plug in known values to get $r_1$.</p>
<p>Simplify the second equation a bit first.</p>
<p>$d=\frac{1}{5}(r_0-r_1)+(r_0+r_1)t-\frac{1}{20}({r_0+r_1})$.</p>
<p>Subtracting (1) from (2) gives $0=\frac{1}{5}(5.5-r_1)-\frac{1}{20}(5.5+r_1).$</p>
<p>Finally, we get $r_1,$ the speed of the tram, $\boxed{3.3}$ km/h.</p>
|
118,658 | <p>I have set of files in the format 20160615-1-0.asc, 20160615-2-0.asc,... 20160615-5-0.asc etc. The following commands can List the file names at once which can be used to import later. </p>
<pre><code> files = Range[1, 5];
c = "D:\\20160615\\20160615-" <> ToString[#] <> "-0.asc" & /@files
</code></pre>
<p>I want to use "0" part of the "-0.asc" string as a variable so that I can make different lists of multiple files like {-1-0.asc, -2-0.asc, ...-5-0.asc}, {-1-1.asc, -2-1.asc, ...-5-1.asc},{-1-3.asc, -2-3.asc, ...-5-3.asc}..and so on.</p>
| george2079 | 2,079 | <p>a good use for the "new in 10" <code>StringTemplate</code> : </p>
<pre><code> Table[
StringTemplate["D:\\20160615\\20160615-`1`-`2`.asc"][i, j], {i, 3}, {j, 3}]
</code></pre>
<blockquote>
<p><a href="https://i.stack.imgur.com/x637Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x637Z.png" alt="enter image description here"></a></p>
</blockquote>
<p>incedentally, if you need to zero pad your numbers as they often occur in file names you can use this mess:</p>
<pre><code>Table[
StringTemplate[
"D:\\20160615\\20160615-<*TemplateExpression[IntegerString[`1`,10,2]]*>-`2`.asc"
][i, j], {i, 3}, {j, 3}]
</code></pre>
<p><a href="https://i.stack.imgur.com/LayXA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LayXA.png" alt="enter image description here"></a></p>
|
3,556,109 | <p>Suppose that <span class="math-container">$f : (0, 2] \to \mathbb{R}$</span> is a continuous function and that <span class="math-container">$\lim_{x \to 0}f(x) = L$</span> for some <span class="math-container">$L \in \mathbb{R}$</span>.
Define <span class="math-container">$g : [0, 2] \to \mathbb{R}$</span> via <span class="math-container">$g(0)=L$</span> and <span class="math-container">$g(x)= f(x)$</span> for <span class="math-container">$x \in (0,2]$</span>.</p>
<p>(i) Show that <span class="math-container">$g$</span> is continuous on <span class="math-container">$[0, 2]$</span>.</p>
<p>(ii) Show that <span class="math-container">$f$</span> is uniformly continuous on <span class="math-container">$(0, 2]$</span>.</p>
<p>I'm struggling to know where to start! I understand that <span class="math-container">$|g(y)-g(x)|< \varepsilon$</span> for all <span class="math-container">$y \in [0,2]$</span> with <span class="math-container">$|y-x|<\delta$</span>.</p>
<p>I just can't get the formal proof started.</p>
| Rezha Adrian Tanuharja | 751,970 | <p>Actually Holder’s inequality for <span class="math-container">$p=q=2$</span> is stronger than CS inequality.</p>
<p>Triangle inequality:</p>
<p><span class="math-container">$\left|\sum{x_{i}y_{i}}\right|\leq\sum{\left|{x_{i}y{i}}\right|}$</span></p>
<p>Therefore:</p>
<p><span class="math-container">$\left|\sum{x_{i}y_{i}}\right|\leq\sum{\left|{x_{i}y{i}}\right|}\leq\sqrt{\sum{x_{i}^{2}}}\sqrt{\sum{y_{i}^{2}}}$</span></p>
|
40,669 | <p><a href="http://en.wikipedia.org/wiki/Wang_tile">Wang tiles</a> are (by Wikipedia): "equal-sized squares with a color on each edge which can be arranged side by side (on a regular square grid) so that abutting edges of adjacent tiles have the same color; the tiles cannot be rotated or reflected."</p>
<p>The usual decision problem associated with them is: given a set of tiles, can they tile the plane?</p>
<p>This problem is known to be undecidable and has a nice history (Wang originally gave a decision procedure, but it works only if the tiles form a periodic tiling, while there exists sets which give rise only to nonperiodic tilings). It has fascinated me personally for a very long time</p>
<p>The state-of-the-art proof of undecidability I'm familiar with is that of Raphael Robinson ("Undecidability and Nonperiodicity for Tilings of the Plane," Inventiones Mathematicae, 12(3), 1971 pp. 177–209). It is quite a difficult proof, and I have never seen it in textbooks.</p>
<p>I'm familiar with a much easier version - this time the set only needs to tile a quadrant of the plane, and the corner tile is already given. This problem is much easier to handle - the given corner tile enables us to "run a Turing machine" in the tiling (each row is a configuration) and undecidability follows. However, this result is primarily a "folklore" version - I don't recall it in textbooks at all.</p>
<p>This leads to my question: has Robinson's method been improved but remains folklore, and so not shown in textbooks? Is there a relatively simple proof of the undecidability of the general tiling problem I'm missing?</p>
| Ben Standeven | 29,318 | <p>There's a newer proof given in "Two-by-Two Substitution Systems and the Undecidability of the Domino Problem", by Nicolas Ollinger. It seems to be available online at: <a href="http://hal.inria.fr/docs/00/26/01/12/PDF/sutica.pdf" rel="noreferrer">http://hal.inria.fr/docs/00/26/01/12/PDF/sutica.pdf</a>. Hopefully, it is easier to understand than Robinson's proof.</p>
|
2,204,834 | <p>I have to show by induction that this function is a multiple of 8. I have tried everything but I can only show that is multiple of 4, some hints? The function is
$$5^{n+1}+2\cdot 3^n+1 \hspace{1cm}\forall n\ge 0$$, because it is a multiple of 8, you can say that$$5^{n+1}+2\cdot 3^n+1=8\cdot m \hspace{1cm}\forall m\in\mathbb{N}$$.</p>
| User8976 | 98,414 | <p>By induction:</p>
<p>It is true for case $n=1$, Let it be true for $n=k$ then </p>
<p>$5^{k+2} +2.3^{k+1} +1 = 5.5^{k+1}+3.2.3^k+1 = 2(5^{k+1} -1) + 3(5^{k+1}+2.3^k+1)$</p>
<p>The second part is a multiple of $8$ and one can easily show that $(5^{k+1} -1)$ is a multiple of $4$.</p>
<p>Hint for showing $(5^{k+1} -1)$ is a multiple of $4$:</p>
<p>$5^{k+2}-1 = 4.5^{k+1} + 5^{k+1} -1$</p>
|
4,263,836 | <blockquote>
<p>I would like to compute <span class="math-container">$\mathrm{Ext}^i_{\mathbb Z}(\mathbb Q, \mathbb Z/2\mathbb Z)$</span>.</p>
</blockquote>
<p>So from the definition I learned in my class, I need to find a free resolution of <span class="math-container">$\mathbb Q$</span> over <span class="math-container">$\mathbb Z$</span>, the first step is to find a surjection of <span class="math-container">$P^0\to\mathbb Q$</span> where <span class="math-container">$P^0$</span> is a free <span class="math-container">$\mathbb Z$</span>-module. Then apply <span class="math-container">$\text{Hom}_\mathbb Z(-,\mathbb Z/2\mathbb Z)$</span> and compute the cohomology. However, I find it hard to come up with an easy surjection onto <span class="math-container">$\mathbb Q$</span>. Since <span class="math-container">$\mathbb Q$</span> is countable, we can find a bijection <span class="math-container">$\phi:\mathbb Z\to\mathbb Q$</span> and consider the infinite direct sum <span class="math-container">$\oplus_{i=1}^{\infty} \mathbb Z$</span> and a map <span class="math-container">$$\psi:\oplus_{i=1}^{\infty}\mathbb Z\to \mathbb Q$$</span> given by <span class="math-container">$\psi((a_1, a_2,\ldots))=\sum_{i=1}^\infty a_i\phi(i).$</span> This is clearly a surjection and a morphism. However, it is hard to compute the cohomology coming from this map since I do not know how to characterize its kernel.</p>
<p>So I am just wondering is there a "better" resolution which can helps me compute Ext functor easily? Any hint is appreciated.</p>
| Jun Koizumi | 966,250 | <p>Taking an explicit free resolution of <span class="math-container">$\mathbb{Q}$</span> is not impossible but difficult.
Instead we can use the functoriality of <span class="math-container">$\operatorname{Ext}^i$</span> to determine <span class="math-container">$\operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2)$</span>.
Consider the map
<span class="math-container">$$
\varphi\colon \operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2)\to \operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2);\quad x\mapsto 2x.
$$</span>
This map is induced by the isomorphism <span class="math-container">$\mathbb{Q}\to \mathbb{Q};\;x\mapsto 2x$</span>, so <span class="math-container">$\varphi$</span> is an isomorphism.
However, it is also induced by the zero map <span class="math-container">$\mathbb{Z}/2\to \mathbb{Z}/2;\;x\mapsto 2x$</span>, so <span class="math-container">$\varphi$</span> is a zero map.
Putting these together we get
<span class="math-container">$$
\operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2)=0
$$</span>
for all <span class="math-container">$i\geq 0$</span>.</p>
|
2,561,245 | <p>Here is the question:
Let $f(x)$ be an integrable function on $[-1,1]$. $f(x)$ is continuous at $x=0$. Define $g(x)=\int_0^{\int_0^xf(n)dn}f(y)dy$ and prove $g'(0)=f(0)^2$.</p>
<p>My initial thought was to look at $g(0)$ and say when you take an integral from $a$ to $b$ and $a=b$ then the integral is $0$, but I do not know how to show $f(0)^2=0$ from that point.</p>
<p>My next thought was that I will need to use the Fundamental Theorem of Calculus II. Here is the definition from the book I am studying from:</p>
<p>Let $f$ be an integrable function on $[a,b]$. For x in $[a,b]$, let $F(x)=\int_a^xf(t)dt$. Then $F$ is continuous on $[a,b]$. If $f$ is continuous at $x_0$ in $(a,b)$, then $F$ is differentiable at $x_0$ and $F'(x_0)=f(x_0)$.</p>
<p>My work using this Theorem:</p>
<p>Let $h(x)=\int_0^xf(n)dn$. Since $f$ is integrable on $[-1,1]$, for $x$ in $[-1,1]$ we have that $h(x)$ is continuous. Since $f$ is continuous at $x=0$, then $h$ is differentiable at $x=0$ and $h'(0)=f(0)$.</p>
<p>Now relating $g$ to $h$, we know $g(x)=\int_0^{h(x)}f(y)dy$. This looks very similar to $h$, so we can say $g(x)=h(h(x))$. Then $g'(0)=h'(h(0))=f(h(0))=f(0)$, since $h(0)=0$ by definition of an integral with parameters that are equal. Now from here, we have $g'(0)=f(0)$, but I need to show $g'(0)=f(0)^2$. This is only true if they are both equal to either $0$ or $1$ I believe, but I am not sure how to do this or if I am even going in the right direction for this proof.</p>
<p>If anyone can tell me if I amm onto something or if I am way off base it would be much appreciated.</p>
| Dylan | 135,643 | <p>Here's what I would do. Let
$$h(x) = \int_0^x f(n)\ dn $$</p>
<p>then we immediately have $h'(x) = f(x)$ by the FTOC, and $h(0) = 0$</p>
<p>By definition
$$ g(x) = \int_0^{h(x)} f(y)\ dy $$</p>
<p>Using the FTOC again we have
$$ g'(x) = f\big(h(x)\big)\cdot h'(x) = f\big(h(x)\big) \cdot f(x) $$</p>
<p>Therefore
$$ g'(0) = f\big(h(0)\big) \cdot f(0) = f(0)\cdot f(0) = \big(f(0)\big)^2 $$</p>
<p><strong>EDIT:</strong> I read through your work more carefully. It's true that
$$ g(x) = h(h(x)) $$</p>
<p>However your derivative is incorrect
$$ g'(x) = h'\big(h(x)\big)\cdot h'(x) = f(h(x)) \cdot f(x) $$</p>
<p>As a consequence of the derivative chain rule. You can then proceed as above.</p>
|
3,048,340 | <blockquote>
<p>Suppose <span class="math-container">$f$</span> is a convex function and differentiable everywhere on <span class="math-container">$(0, \infty)$</span> and satisfies
<span class="math-container">$$\lim_{x \rightarrow\infty}{f(x)}=A$$</span>
where <span class="math-container">$A$</span> is an arbitrary real number. Show that <span class="math-container">$$\lim_{x\rightarrow\infty}{f'(x)}=0$$</span></p>
</blockquote>
<p><strong>My attempt:</strong> Suppose <span class="math-container">$f'(x_0)>0$</span> for <span class="math-container">$x_0 \in (0, \infty)$</span> such that
<span class="math-container">$$f(x) \geq f(x_0)+f'(x_0)(x-x_0)$$</span>
Taking limit that tends to infinity on both sides obtain
<span class="math-container">$$A=\lim_{x \rightarrow \infty}{f(x)} \geq \lim_{x \rightarrow \infty}{[f(x_0)+f'(x_0)(x-x_0)]}=\infty$$</span>
which contradicts the fact that <span class="math-container">$A$</span> is finite. Similar to the case when <span class="math-container">$f'(x_0)<0$</span>, Hence, <span class="math-container">$f'(x_0)=0$</span> for <span class="math-container">$x_0 \in (0, \infty)$</span>. Take <span class="math-container">$x_0 \rightarrow \infty$</span> and it follows the conclusion.</p>
<p><strong>Note:</strong> Any constant function <span class="math-container">$f$</span> is a trivial case on this question. </p>
<p>In this question, I illustrated an example <span class="math-container">$f(x)=e^{-x}$</span> and referring to its geometry interpretation of derivative, but I hope to know that if my attempt is wrong as I make <span class="math-container">$x_0\rightarrow \infty$</span> in my last progess, which is weird and correct it.</p>
| Kavi Rama Murthy | 142,385 | <p>For a convex differentiable function we canwrite <span class="math-container">$f(x)=f(0)+\int_0^{x} f'(t)\, dt$</span>. The hypothesis implies that <span class="math-container">$\int_0^{x} f'(t)\, dt$</span> remains bounded as <span class="math-container">$ x\to \infty$</span>. Note that <span class="math-container">$f'$</span> is increasing, so it has a (finite or infinite) limit as <span class="math-container">$ x\to \infty$</span>. If the limit is not <span class="math-container">$0$</span> you get a contradiction to the fact that <span class="math-container">$\int_0^{x} f'(t)\, dt$</span> remains bounded as <span class="math-container">$ x\to \infty$</span>. (The integral goes to <span class="math-container">$+\infty$</span> or <span class="math-container">$-\infty$</span> according as <span class="math-container">$\lim f'(x) >0$</span> or <span class="math-container">$\lim f'(x) <0$</span>).</p>
|
356,217 | <p>Find the following infinite sum : $$q\sin a+q^2\sin 2a+\ldots+q^n\sin na+\ldots$$ where $|q|<1$ .It would be good if you could find it without the help of any auxiliary sequences using only trigonometric formulas.</p>
| Community | -1 | <p><strong>Hint</strong>
We calculate the sum of geometric series
$$\sum_{n=1}^\infty q^ne^{ina}$$
then we take the imaginary part.</p>
|
2,362,485 | <blockquote>
<p>Evaluate: $$\lim_\limits{x\to \pi/4} \frac {2-\csc^2 x}{1-\cot x}$$</p>
</blockquote>
<p>My Attempt:
\begin{align}\lim_\limits{x\to \pi/4} \frac {2-\csc^2 x}{1-\cot x}&=\lim_{x\to \pi/4} \frac {2-\csc^2 x}{1-\cot x}\\\\
&=\lim_{x\to \pi/4} \frac {2-\frac {1}{\sin^2 x}}{1-\frac {\cos x}{\sin x}}\\\\
&=\lim_{x\to \pi/4} \frac {2\sin^2 x - 1}{\sin^2 x - \sin x\cdot\cos x}\end{align}</p>
| Claude Leibovici | 82,404 | <p>Considering
$$ A=\dfrac {2-\csc^2( x)}{1-\cot(x)}$$ let $x=y+\frac \pi 4$ to get (after simplification) $$A=\frac{2 \cos (y)}{\sin (y)+\cos (y)}$$ and now, look for the limit when $y\to 0$.</p>
|
1,842,093 | <p><a href="https://i.stack.imgur.com/A5U3D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A5U3D.png" alt="enter image description here"></a></p>
<p>$ABCD$ forms a square. $CDE$ forms a triangle. Given $\measuredangle AED=15^{\circ}$ and $DE=CE$, prove $\triangle CDE$ is equilateral. </p>
<p>The question is surprising hard, the problem is basically proving $\measuredangle BAE=\measuredangle AEB$. Can I have some hint. </p>
| Amritanshu | 349,576 | <p>Ok ! Here we go ..</p>
<p>$ABCD$ is a square.
That means -> $AB=BC=CD=AD$</p>
<p>Now if you carefully notice, $CD$ is the base of the $\Delta CDE$</p>
<p>Given that $DC = DE$</p>
<p>We can conclude that $\Delta ADE$ is an isosceles triangle.
Therefore ,
$\angle AED = \angle EAD$</p>
<p>So, $\angle EAD = 15º$</p>
<p>Now, By Angle-Sum Property in $\Delta ADE$</p>
<p>$\angle AED + \angle EAD + \angle ADE = 180º $</p>
<p>=> $\angle ADE = 150º $</p>
<p>=> $90º + \angle CDE = 150º$</p>
<p>=> $\angle CDE = 60º$</p>
<p>Also you can notice that $\Delta DCE$ isosceles.
Therefore , </p>
<p>$\angle DEC = \angle DCE$</p>
<p>By Angle-Sum Property in $\Delta DEC$ </p>
<p>$\angle CDE + \angle DCE + \angle CED = 180º$</p>
<p>=> $60º + \angle DCE + \angle DCE = 180º$</p>
<p>=> $2\angle DCE = 120º$</p>
<p>=> $\angle DCE = 60º$</p>
<p>Therefore,</p>
<p>$\angle CDE = \angle DCE = \angle CED = 60º$</p>
<p>So $\Delta DCE$ is equilateral !</p>
<p>Hence Proved </p>
|
7,900 | <p>This semesters I am teaching a course to 12-13 year old kids in which they are supposed to learn the basics of spreadsheet usage.</p>
<p>I am having difficult in coming up with fun / interesting exercises to teach them the basic functions, such as average, max, minimum and the like. One of the biggest problems is having only a 50 minutes long lesson every two weeks.</p>
<p>I've tried so far:</p>
<ul>
<li>Teaching them to plot relations using a dispersion graphic: they would produce the axes and plot relations such as linear and quadratic formulas, exponential, etc.</li>
<li>Introduced them to simple recurrences such as the Fibonacci sequence and the discrete logistic map.</li>
<li>Using stock data to teach them some descriptive statistics.</li>
</ul>
<p>What else would be proper, and preferably fun, to teach them? </p>
| celeriko | 3,237 | <p>Based on your unfortunate lack of time my answer is going to be a list of suggestions that I believe could be completed and meaningful given the amount of time you have with your students. Not all of them pertain directly with math (calculations, computations, formulas, etc), but I believe that 1) the real power of spreadsheets is their ability to organize, structure, and access data and 2) that the thinking behind these concepts is very mathematical. </p>
<ul>
<li><p>You mentioned that they have already used the technology to explore fitting various functions to graphs. I think this can open up into a great lesson on growth of functions, having them write basic formulas to model linear, log, quadratic, cubic, exponential, factorial, etc. and then graph them and see how the growth differs between the types of functions</p></li>
<li><p>You mentioned some recursive sequences/series that you looked at, how about Newton's method, calculating square roots or logarithms "by hand", or approximating pi by one of the many ways that are out there. Very cool things to be able to do by hand because it takes a lot of the "magic" out of pressing a button on a calculator and lets the students know that there is a reason why these things work and are the way they are</p></li>
<li><p>Have them do more with descriptive data but let them find their own datasets for data that might be meaningful to them (possibly video game sales, sports records, music sales, etc). It will reinforce their prior learnings and give them a sense of agency and ownership. I would have them research during their time out of the classroom and come with a dataset ready to go. Alternatively, you could spend one class finding datasets, discussing different biases, sample sizes, correlation, causation, etc. and then the next class analyze it</p></li>
<li><p>A large part of being able to effectively use a spreadsheet is organization and performing tasks like sort and merge. If you have not already talked about this I would suggest doing so and possibly using a classroom dataset of say, everyones height, eye color, length of name in letters, etc and either start with a poorly organized structure of the data and discuss why it is poor and how to fix it or start from scratch and discuss along the way meaningful ways to structure data</p></li>
<li><p>Use whichever spreadsheet software (i'm assuming google sheets or excel) as a means to talk about software, coding, and computer languages, possibly having them write a very simple macro or two after a lesson or two. I think that the real power in these software is their extensibility, if you know how to work it. I know you said that you are tasked with the basics, but it sounds like they are catching on so there is no reason not to push them to explore the tool</p></li>
</ul>
<p>I hope this is helpful and I wish you the best of luck :)</p>
|
7,900 | <p>This semesters I am teaching a course to 12-13 year old kids in which they are supposed to learn the basics of spreadsheet usage.</p>
<p>I am having difficult in coming up with fun / interesting exercises to teach them the basic functions, such as average, max, minimum and the like. One of the biggest problems is having only a 50 minutes long lesson every two weeks.</p>
<p>I've tried so far:</p>
<ul>
<li>Teaching them to plot relations using a dispersion graphic: they would produce the axes and plot relations such as linear and quadratic formulas, exponential, etc.</li>
<li>Introduced them to simple recurrences such as the Fibonacci sequence and the discrete logistic map.</li>
<li>Using stock data to teach them some descriptive statistics.</li>
</ul>
<p>What else would be proper, and preferably fun, to teach them? </p>
| Amir Asghari | 1,217 | <p>you might find this paper "<a href="http://link.springer.com/article/10.1007/s10758-005-8420-9" rel="nofollow">Designing Spreadsheet-Based Tasks for Purposeful Algebra</a>" (and a couple of some others written by the same authors and about the same year) useful. </p>
<p>PS. Originally, I meant to put this answer as a comment. But, I realized that I don't know how to hyperlink inside a comment. Could someone who knows please let me know how?</p>
|
7,900 | <p>This semesters I am teaching a course to 12-13 year old kids in which they are supposed to learn the basics of spreadsheet usage.</p>
<p>I am having difficult in coming up with fun / interesting exercises to teach them the basic functions, such as average, max, minimum and the like. One of the biggest problems is having only a 50 minutes long lesson every two weeks.</p>
<p>I've tried so far:</p>
<ul>
<li>Teaching them to plot relations using a dispersion graphic: they would produce the axes and plot relations such as linear and quadratic formulas, exponential, etc.</li>
<li>Introduced them to simple recurrences such as the Fibonacci sequence and the discrete logistic map.</li>
<li>Using stock data to teach them some descriptive statistics.</li>
</ul>
<p>What else would be proper, and preferably fun, to teach them? </p>
| Jasper | 1,147 | <p><em><a href="http://rads.stackoverflow.com/amzn/click/0393310728" rel="nofollow">How to Lie with Statistics</a></em>. Many graphs in newspapers are misleading because the authors used Excel's default graph settings, instead of making sure to set the range of the y-axis to start at zero.</p>
<p>Are you using a recent version of Microsoft Excel, such as Excel 2013? Excel 2013's PowerPivot and PivotChart features are amazing. [Note: I currently work as a contractor for Microsoft using these tools.] Bill Jelen (Mr. Excel) has written a book on <em><a href="http://rads.stackoverflow.com/amzn/click/0789748754" rel="nofollow">Excel 2013 Pivot Table Data Crunching</a></em>.</p>
<p>If you have a suitable data set (like the populations of various states over time, or baseball statistics, or...) you could have the students graph interesting data changes (like the population trend of a city or state, or a batter's on-base percentage <em>versus</em> time, or a batter's home-run rate with runners in scoring position, or...)</p>
<p>A graph of the population of a chosen city or state over time is nice. A PivotChart of the same thing is amazing. You can graph how the number of teenage girls in a city has changed over time. You can then graph how the number of 13 year-old girls and 13 year-old boys in the city has changed. Then, you can change a single filter to get a similar graph for a different city.</p>
<p>In my job, I try to make graphs that are very easy to understand. That usually means:</p>
<ul>
<li>Limiting the number of things being graphed.</li>
<li>Avoiding having lots of lines cross each other in the plot area.</li>
<li>A chart title that describes the point of the chart.</li>
<li>Either an easy-to-read axis scale, or a data table</li>
<li>Descriptive axis titles (unless the axis scale shows that the units are times)</li>
<li>Consistently coloring matching things in different graphs.</li>
<li>Unless I have a really good reason not to (like for date/time axes), starting axis scales at zero.</li>
<li>Where practical, making the axis scales match in corresponding graphs.</li>
<li>Avoiding complicated definitions for what is being graphed. It is easier to say "At-bats in May" or "At-bats before injury" than "At-bats from May 4 through June 6".</li>
<li>Making sure that the field names (as shown on the chart) are easy to understand.</li>
<li>Hiding redundant features (like a legend that says "Total" when there is only one variable being graphed).</li>
</ul>
|
2,760,113 | <p>Suppose $f'$ is continuous on $[a,b]$ and $\epsilon >0$ is given.
Prove that $\exists\; \delta>0$ such that $$\left\lvert \frac{f(t)-f(x)}{t-x}-f'(x)\right\rvert<\epsilon$$</p>
<p>$\forall \;0<|t-x|<\delta, a\le x \le b ,a\le t\le b$.</p>
<p>How to solve this problem since using compactness argument ?</p>
| Christian Blatter | 1,303 | <p>Since $f'$ is continuous we may write
$$f(t)-f(x)=\int_x^tf'(\xi)\>d\xi=(t-x)\int_0^1f'\bigl(x+\tau(t-x) \bigr)\>d\tau\ ,$$
and therefore
$${f(t)-f(x)\over t-x}-f'(x)=\int_0^1\bigl(f'\bigl(x+\tau(t-x)\bigr) -f'(x)\bigr)\>d\tau\qquad(t\ne x)\ .\tag{1}$$
Now $f'$ is automatically uniformly continuous on $[a,b]$. Given an $\epsilon>0$ we therefore can find a $\delta>0$ such that $0<|h|<\delta$ implies $\bigl|f'(x+h) -f'(x)\bigr|<\epsilon$ for arbitrary $x$, $x+h\in[a,b]$. From $(1)$ it then follows that
$$\left|{f(t)-f(x)\over t-x}-f'(x)\right|<\epsilon$$
whenever $0<|t-x|<\delta$, and both $x$ and $t$ are in $[a,b]$.</p>
|
1,207,134 | <p>Given the image:
<img src="https://i.stack.imgur.com/EJ3ax.jpg" alt="enter image description here"></p>
<p>and that $x_0 = 1, y_0=0$ and $\text{angles} \space θ_i
, i = 1, 2, 3, · · ·$ can be arbitrarily picked.</p>
<p>How can I derive a recurrence relationship for $x_{n+1}$ and $x_n$?</p>
<p>I actually know what the relationship is, however, don't know how to derive it.</p>
<p>Although it is obvious to see that $x_0 = x_1$</p>
<p>so for $n = 0$</p>
<p>We have $x_{1} = x_{0}+0 $</p>
<p>The recurrence relations are</p>
<p>$x_{n+1}=x_n−y_ntan(θ_{n+1})$</p>
<p>and </p>
<p>$y_{n+1}=y_n+x_ntan(θ_{n+1})$
But I can't get the derivation.</p>
<p><img src="https://i.stack.imgur.com/ZBzWJ.jpg" alt="enter image description here"></p>
<p>I tried taking some arbitrary right angle triangle and constructing two vectors.</p>
<p>$\mathop r_{\sim} = \langle x_n,y_n \rangle $</p>
<p>and a vector perpendicular to $\mathop r_\sim$, $\mathop d_{\sim} = \langle a,b\rangle$ such that</p>
<p>$\mathop r_{\sim} \cdot \mathop d_{\sim} = 0 $</p>
<p>Then we can let construct a unit vector for $\mathop d_\sim$ and eventually construct a line through a point?</p>
| marty cohen | 13,079 | <p>This is a generalization of the
"Spiral of Theodorus"
which is discussed in many places.
Here is the obvious one:
<a href="http://en.wikipedia.org/wiki/Spiral_of_Theodorus" rel="nofollow">http://en.wikipedia.org/wiki/Spiral_of_Theodorus</a></p>
<p>Using the methods there
as a guide.</p>
<p>(I guess this should be a comment
rather than an answer -
I was too impulsive.)</p>
|
1,207,134 | <p>Given the image:
<img src="https://i.stack.imgur.com/EJ3ax.jpg" alt="enter image description here"></p>
<p>and that $x_0 = 1, y_0=0$ and $\text{angles} \space θ_i
, i = 1, 2, 3, · · ·$ can be arbitrarily picked.</p>
<p>How can I derive a recurrence relationship for $x_{n+1}$ and $x_n$?</p>
<p>I actually know what the relationship is, however, don't know how to derive it.</p>
<p>Although it is obvious to see that $x_0 = x_1$</p>
<p>so for $n = 0$</p>
<p>We have $x_{1} = x_{0}+0 $</p>
<p>The recurrence relations are</p>
<p>$x_{n+1}=x_n−y_ntan(θ_{n+1})$</p>
<p>and </p>
<p>$y_{n+1}=y_n+x_ntan(θ_{n+1})$
But I can't get the derivation.</p>
<p><img src="https://i.stack.imgur.com/ZBzWJ.jpg" alt="enter image description here"></p>
<p>I tried taking some arbitrary right angle triangle and constructing two vectors.</p>
<p>$\mathop r_{\sim} = \langle x_n,y_n \rangle $</p>
<p>and a vector perpendicular to $\mathop r_\sim$, $\mathop d_{\sim} = \langle a,b\rangle$ such that</p>
<p>$\mathop r_{\sim} \cdot \mathop d_{\sim} = 0 $</p>
<p>Then we can let construct a unit vector for $\mathop d_\sim$ and eventually construct a line through a point?</p>
| Narasimham | 95,860 | <p>The problem is not well-posed. There are several solutions to the given problem depending on discretization choice. One more condition is needed to solve the finite "spiral". I am assuming that the real aim here is to find the spiral which is described by the tip of rotating vector.</p>
<p>An extreme choice of discretization is when the different $(x_i,y_i),(x_{i+1},y_{i+1}) $ are spaced infinitesimally close to each respecting the given relations among the discrete increments.</p>
<p>We have the same scenario that marks difference between <em>finite difference calculus</em> and the <em>infinitesimal calculus</em>.</p>
<p>Eliminating $ \tan \theta_{n+1}$ between the two given relations which is quite a straight forward thing,</p>
<p>$$ \dfrac{y_{n+1} - y_n}{x_{n+1} - x_n}=- \dfrac{x_n}{y_n} \tag{1}$$ </p>
<p>$$ \dfrac{\Delta y} {\Delta x}=- \dfrac{x}{y}\tag{2} $$ </p>
<p>which passes to the limit by differential calculus:</p>
<p>$$ \dfrac{d y} {d x}=- \dfrac{x}{y} \rightarrow x \;{d x} + y\;{d y}=0 \tag{3} $$ </p>
<p>which integrates to $$ x^2 + y^2 = R^2 \tag{4} $$ where R depends on starting $ x_i,y_i $ values, $ x_i^2 + y_i^2 = R^2 $; if $y_i=0$ ,then $ x_i = R.$</p>
<p>It is a Circle, the limiting case among a set of infinite solutions.</p>
|
2,338,797 | <p>A bag contains $3$ red balls and $4$ blue balls. Find the probability of picking, at random, two balls of </p>
<p>a) the same color </p>
<p>b) different colors</p>
| Mandar Kulkarni | 458,156 | <p>I think it is important to note here that the order in which balls are chosen does not matter.</p>
<p>The total number of ways of drawing 2 balls out of 7 is 7C2 i.e. 21</p>
<p>For first case we either draw 2 red (3C2) or 2 blue balls (4C2), for total favorable cases of 9</p>
<p>Hence probability of drawing same color balls is $\frac{9}{21}$</p>
<p>The probability of drawing different color balls is therefore $1-\frac{9}{21}=\frac{12}{21}$</p>
|
1,455,098 | <p>I am reading the book Foundation of Machine Learning, and the author has many proofs for different theorems. Here is one part of a proof about Perceptron algorithm which I don't quite understand</p>
<p>Note: $w_t$ is a vector</p>
<p>Let
\begin{equation}
w_{t+1} = \begin{cases}
w_t + ny_t x_t & \text{if } y_t(w.x_t) < 0 \\
w_t & \text{if } y_t(w.x_t) > 0 \\
w_t & \text{otherwise}
\end{cases}
\end{equation}</p>
<p>and in the proof, he has
\begin{equation}
\|w_{T+1}\| = \sqrt{\sum_{t \in T} \|w_{t+1}\|^2 - \| w_t \|^2}
\end{equation}</p>
<p>he mentioned that above equation uses (telescoping sum, $w_0 = 0$). I have no idea why lhs is equal to rhs using telescoping sum, can you help me explain this?</p>
<p>Thanks in advance!</p>
| Greg Martin | 16,078 | <p>This is an algebraic identity that has very little to do with the specific function: for <em>any</em> $f\colon \mathbb Z_{\ge0} \to \mathbb R_{\ge0}$ be any function satisfying $f(0)=0$, we have
$$
f(T+1) = \sqrt{\sum_{0\le t\le T} \big( f(t+1)^2-f(t)^2 \big)}.
$$
To see this, it's equivalent to show that
$$
f(T+1)^2 = \sum_{0\le t\le T} \big( f(t+1)^2-f(t)^2 \big).
$$
And this is easy to show by induction, or indeed by recognizing that the right-hand side is a telescoping sum. Taking $T=3$ for example:
$$
f(4)^2 = (f(1)^2-f(0)^2) + (f(2)^2-f(1)^2) + (f(3)^2-f(2)^2) + (f(4)^2-f(3)^2).
$$</p>
|
1,472,901 | <h2>Code first :</h2>
<pre><code>sum = 0
for(i=1; i<=n; i++){
for(j=1; j<=i; j++){
if(j%i == 0){
for(k=1; k<=n; k++){
sum = sum + k;
}
}
}
}
</code></pre>
<hr>
<p>Total no. of iterations in $j = 1+2+3+4+ \dots +n
\\= \frac{n. (n+1)}{2} \\= \Theta(n^2)$
total no. of times $k$ loop iterates = $n\times n = \Theta(n^2)$</p>
<p>So, time complexity $= \Theta(n^2) + \Theta(n^2)$
$= (n^2).$</p>
<h2>Code second :</h2>
<pre><code>for(int i =0 ; i < =n ; i++) // runs n times
for(int j =1; j<= i * i; j++) // same reasoning as 1. n^2
if (j % i == 0)
for(int k = 0; k<j; k++) // runs n^2 times? <- same reasoning as above.
sum++;
</code></pre>
<hr>
<p>Correct Answer$: n×n^2×n = O(n^4)$</p>
<hr>
<blockquote>
<p>Please check whether my solution is correct ?</p>
</blockquote>
| Sujith Sizon | 262,120 | <h2>Critical points</h2>
<p>refers to the set of all points which satisfies at least one of the following conditions :</p>
<p>(A) $f'(x)=0$ ie: turning points.</p>
<p>(B) $f'(x)$ does not exist.</p>
<p>(C) sign changes of $f'(x)$ in nbd of $x$ (not a completely necessary condition as these will be included in (B))</p>
<p>Whereas </p>
<h2>Inflection points</h2>
<p>are those points which satisfies atleast one of the following conditions :</p>
<p>(A) sign changes of $f''(x)$ in the neighbourhood of $x$ .</p>
|
4,164,405 | <p>Grothendieck has proven that whenever <span class="math-container">$X\longrightarrow\operatorname{Spec}(A)$</span> is a proper morphism of Noetherian schemes, <span class="math-container">$F$</span> is coherent over <span class="math-container">$X$</span> and flat over <span class="math-container">$\operatorname{Spec}(A)$</span>, then there exists a finite complex of finitely generated projective modules over <span class="math-container">$A$</span>
<span class="math-container">\begin{equation*}
0\longrightarrow K^{0}\longrightarrow...\longrightarrow K^{n}\longrightarrow 0
\end{equation*}</span>
such that for any <span class="math-container">$A$</span>-module <span class="math-container">$M$</span> there exists an isomorphism of <span class="math-container">$A$</span>-modules
<span class="math-container">\begin{equation*}
H^{p}(X,F\otimes_{A}M)\cong H^{p}(K\otimes_{A}M)\text{.}
\end{equation*}</span>
So far, so good. Is not the Čech complex precisely one example of such a Grothendieck complex? By construction, the Čech complex consists of projective modules, and it satisfies the isomorphism condition.</p>
| Aphelli | 556,825 | <p>The answer is that the groups of the Cech complex aren’t finitely generated except when <span class="math-container">$X$</span> is finite over <span class="math-container">$A$</span>. I’m studying the cohomology of <span class="math-container">$\mathcal{O}_X$</span>.</p>
<p>Indeed, let <span class="math-container">$X=\cup_i{U_i}$</span> be a finite affine cover. Thus every finite intersection of the <span class="math-container">$U_i$</span> is affine and the Cech complex computes the sheaf cohomology.</p>
<p><span class="math-container">$\bigoplus_i{\mathcal{O}_X(U_i)}$</span> is the first nonzero group in the Cech complex, so if it is finitely generated, then all the <span class="math-container">$\mathcal{O}_X(U_i)$</span> are finitely generated <span class="math-container">$A$</span>-modules, thus they are finite (hence proper) over <span class="math-container">$A$</span>, and thus are closed in <span class="math-container">$X$</span>, so that <span class="math-container">$X$</span> is the disjoint reunion of the <span class="math-container">$U_i$</span> which are finite.</p>
|
650,291 | <p>I'm not so good at combinatorics, but I want to know if my answer for this question is right. Originally this question is written in spanish and it says:</p>
<blockquote>
<p>Se dispone de una colección de 30 pelotas divididas en 5 tamaños distintos y 6 colores diferentes de tal manera que en cada tamaño hay los seis colores.¿Cuántas colecciones de 4 pelotas tienen exactamente 2 pares de pelotas del mismo tamaño (que no sean las 4 del mismo tamaño)?.</p>
</blockquote>
<p>And here's a translation made by me:</p>
<blockquote>
<p>A collection of 30 balls is available, separated in 5 different sizes and 6 different colors in a way that in each size there are six colors. How many collections of 4 balls have exactly 2 pairs of balls of same size (which those 4 balls aren't of same size)?</p>
</blockquote>
<p>I first wrote this table:</p>
<p>\begin{array}{c|c|c|c|c}
\cdot & Size 1 & Size 2 & Size 3 & Size 4 & Size 5 \\
\hline
Color 1 & ① & ❶ & ⒈ & ⑴ & ⓵ \\
Color 2 & ② & ❷ & ⒉ & ⑵ & ⓶ \\
Color 3 & ③ & ❸ & ⒊ & ⑶ & ⓷ \\
Color 4 & ④ & ❹ & ⒋ & ⑷ & ⓸ \\
Color 5 & ⑤ & ❺ & ⒌ & ⑸ & ⓹ \\
Color 6 & ⑥ & ❻ & ⒍ & ⑹ & ⓺
\end{array}</p>
<p>So, an example of a collection of 4 balls that have exactly 2 pairs of balls of the same size is:</p>
<blockquote>
<p>①②❶❷</p>
</blockquote>
<p>So, for the first column (Size 1) there are 15 combinations of having 2 balls:</p>
<pre><code>①② ②③ ③④ ④⑤ ⑤⑥
①③ ②④ ③⑤ ④⑥
①④ ②⑤ ③⑥
①⑤ ②⑥
①⑥
</code></pre>
<p>Which is the same as: </p>
<p>$$C_{6}^{2} = \frac{6!}{(6-2)!2!} = 15$$</p>
<p>Or the same as:</p>
<p>$$\sum_{k=1}^{5}k = 15$$</p>
<p>Then, for each row we have 10 combinations:</p>
<pre><code>①❶ ❶⒈ ⒈⑴ ⑴⓵
①⒈ ❶⑴ ⒈⓵
①⑴ ❶⓵
①⓵
</code></pre>
<p>Which is the same as:</p>
<p>$$C_{5}^{2} = \frac{5!}{(5-2)!2!} = 10$$</p>
<p>Or the same as:</p>
<p>$$\sum_{k=1}^{4}k = 10$$</p>
<p>And so, by the <a href="http://en.wikipedia.org/wiki/Rule_of_product" rel="nofollow">rule of product</a> I say that the number of collections of 4 balls having exactly 2 pairs of the same size is:</p>
<p>$$C_{6}^{2} C_{5}^{2} = 150$$</p>
<p>I'll be grateful if someone check my answer and give me further details :D</p>
| Woria | 119,309 | <p>In this kind of problems, the key idea is to transform it into steps and cases in a way that you can use "Addition law" and "Multiplication law". Here you need tow different pairs of balls of the same size, so</p>
<p><strong>Step 1:</strong> Select 2 different sizes from those existing 5 sizes;</p>
<p><strong>Step 2:</strong> Select 2 balls in each of those 2 selected sizes;</p>
<p><strong>Step 3:</strong> Use multiplication law.</p>
<p>And don't forget that "select" means "combination"!</p>
|
1,712,933 | <p>I am slightly ashamed to be asking this, but I have been recently reflecting on changing variables in very simple problems. If I missed a question that already discusses this please point it out to me and I will delete this one.
Anyhow writing this will probably be a learning experience.</p>
<p>Directly from the Wikipedia page on the argument I take as an example the equation:</p>
<p>$$x^6 - 9 x^3 + 8 = 0. \, $$</p>
<p>I quickly recognize this as a high school problem and use the methods that were taught to me, namely I set $x^3 = u$ so $x = u^{1/3}$.</p>
<p>Then I proceed to solve quadratic equation that results from this substitution, and only at the end I apply the reverse transformation $x^3 = u$ to get an answer for my starting variable. With not much imagination I always thought that the function used when changing variables (in the above case $f(x) = x^3)$ should be bijective in the domain of interest of the starting equation. This is because I need the inverse to return to my "starting variable".</p>
<p>But I notice on Wikipedia that a bit more is required; the change of variable function should be a diffeomorphism, we need differentiability (and even smooth manifolds for the domain and the image).</p>
<p>This is where I realized that I was never taught a proof of why the change of variables method work or how it works but I was just applying these substitutions blindly.</p>
<p>So could someone kindly point me to a source where I can improve my understanding on this very powerful method by adding rigour to what I am doing and possibly even a geometric interpretation.</p>
| Jasper | 316,202 | <p>This answer will give an idea on why change of variables works/is allowed when <em>integrating</em>.</p>
<p>I will give you an idea (heuristic) on why these requirements for $u$ are needed. The basic idea behind substition of variables is that you choose a different basis over which you know the solution. </p>
<p>Basically you want to solve $\int_a^b f(x) dx$.</p>
<p>If you think about a two-dimensional Euclidean $(x,y)$-grid (actually: manifold), then you can think of $dx$ as a vector (actually: covector) that defines the direction-step in the $x$-direction.</p>
<p>In the integral expression, $x$ is just a dummy, so you can choose it as anything you would like it to be, but then you need to change it at any place.</p>
<p>You've chosen $u = x^{\frac{1}{3}}$. This function is "sufficiently nice" in the sense that you can invert it, differentiate it infinitely many times and that it's continuous over $\mathbb{R}$ and everything else.</p>
<p>You can replace $dx$ now by $[\text{Something}]du$. You know $x = u^3$, so $x'(u) = \frac{dx}{du} = 3 u^2$. Multiply both sides by the differential of $u$ to find $dx = 3 u^2 du$.</p>
<p>Now you can transform everything from the basis in $x$ to a basis in $u$, so $$\int_a^b f(x) dx = \int_{a^{1/3}}^{b^{1/3}} f(u) (3 u^2 du)$$</p>
<p>Why do you need differentiability? For instance, consider $u = \frac{1}{x}$ and suppose that the point $x=0$ is within the interval $\langle a,b\rangle$. What is the value of $u$ when we consider $x = 0$?</p>
<p>The same way diffeomorphism. Suppose that the basis transformation is not injective. For instance consider $u = x$ if $x<0$ and $u = x+2$ if $x\geq 0$. There is no value of $x$ that maps to $u=1$. Now try to integrate over $x$ from -1 to 1. Then $\int_{-1}^1 f(x) dx = \int_{-1}^{3} f(?) du$. What would be the value of the latter integral?</p>
|
13,230 | <p>Introduction:</p>
<p>Let A be a subset of the naturals such that <span class="math-container">$\sum_{n\in A}\frac{1}{n}=\infty$</span>. The <a href="https://en.wikipedia.org/wiki/Erd%C5%91s_conjecture_on_arithmetic_progressions" rel="nofollow noreferrer">Erdos Conjecture</a> states that A must have arithmetic progressions of arbitrary length. </p>
<p>Question:</p>
<p>I was wondering how one might go about <em>categorizing</em> or <em>generating</em> the divergent series of the form in the introduction above. I'm interested in some particular techniques and I list some examples below:</p>
<p>If we let <span class="math-container">$S$</span> be the set of such divergent series: <span class="math-container">$S=\left[ A: \sum_{n\in A}\frac{1}{n}=\infty, \ A\in\mathbb{N} \right]$</span>, what kind of operations are there that would make S a group, or at the very least a semigroup? I'm rather vague on what the operatons should be for a reason, because although I presume trivial operations exist, their usefulness in understanding the members of <span class="math-container">$S$</span> would be questionable. </p>
<p>Alternately, can one look at these divergent sums through the technique of Ramanujan summation (think: <span class="math-container">$1+2+3+\ldots =^R -\frac{1}{12}$</span>, <span class="math-container">$R$</span> emphasizing Ramanujan summation)? The generalizations of Ramanujan summation (a good reference <a href="http://algo.inria.fr/seminars/sem01-02/delabaere2.pdf" rel="nofollow noreferrer"> here </a>) allow one to assign values to some of these series and give some measure of what kind of divergence is occurring. Moreover, basic series manipulations that hold for convergent series tend to carry over to Ramanujan summation, so can one perhaps look at the set <span class="math-container">$S$</span> above as a set of equivalence classes in the sense of two elements being equivalent if they share the same Ramanujan summation constant. </p>
<p>Thanks in advance for any input!</p>
| Thomas | 20,974 | <p>An idea that I came up with is that if the conjecture is true, then there would be a finite upper bound on how big the sum could be if the set had no arithmetic progression of length k, lets call this upper bound f(k). Obviously f(1)=0, as if there are any members then there is an arithmetic progression of that length. f(2)=1 (using the set {1}), as adding more numbers would result in an arithmetic progression of length two or more. I am currently working on f(3)</p>
|
2,226,307 | <p>Let <span class="math-container">$c\in \mathbb{R}$</span> and <span class="math-container">$f:\mathbb{R}\to \mathbb{R}$</span> is continuous at <span class="math-container">$c$</span>. If for every positive <span class="math-container">$\delta$</span>, there is a point <span class="math-container">$y\in (c-\delta, c+\delta)$</span> such that <span class="math-container">$f(y)=0$</span>, show that <span class="math-container">$f(c)=0.$</span></p>
<p>I am unable to solve the problem. I don't know how to start the problem. I need a help.</p>
<p>Edit:</p>
<blockquote>
<p>From the definition of continuity of <span class="math-container">$f(x)$</span> at <span class="math-container">$x=c$</span>. For any <span class="math-container">$\epsilon > 0$</span>, <span class="math-container">$\exists \delta > 0$</span>:</p>
<p><span class="math-container">$|f(x) - f(c)| < \epsilon \, \, \, \, \mbox{whenever} \, \, \, \, |x-c| < \delta$</span>.</p>
<p>Let <span class="math-container">$x=y\in (c-\delta, c+\delta)$</span> st <span class="math-container">$f(y)=0$</span> then <span class="math-container">$|f(c)|<\epsilon$</span> whenever <span class="math-container">$ |x-c| < \delta$</span></p>
<p>then <span class="math-container">$f(c)=0$</span> (proved)</p>
<p><strong>Whether the proof by <span class="math-container">$\epsilon-\delta$</span> method is correct</strong>?</p>
</blockquote>
| L.F. Cavenaghi | 248,387 | <p>Hint: $\delta_n := 1/n$. Then there exists $y_n$ such that $f(y_n) = 0.$ But $0 = \lim f(y_n) = f(\lim y_n)$ when $n\to \infty.$ But $\lim y_n = c.$</p>
|
322,598 | <p><a href="https://en.wikipedia.org/wiki/Partially_ordered_set" rel="noreferrer">Partially ordered sets</a> (<em>posets</em>) are important objects in combinatorics (with <a href="https://gilkalai.wordpress.com/2019/02/05/extremal-combinatorics-v-posets/" rel="noreferrer">basic connections to extremal combinatorics</a> and to algebraic combinatorics) and also in other areas of mathematics. They are also related to <em>sorting</em> and to other questions in the theory of computing. I am asking for a list of open questions and conjectures about posets.</p>
| Tri | 51,389 | <p>Let <span class="math-container">$P,Q$</span> be posets. Let <span class="math-container">$Q^P$</span> denote the poset of order-preserving maps from <span class="math-container">$P$</span> to <span class="math-container">$Q$</span>, where <span class="math-container">$f\le g$</span> if <span class="math-container">$f(p)\le g(p)$</span> for all <span class="math-container">$p\in P$</span>.</p>
<p>A poset has the <em>fixed point property</em> if for all <span class="math-container">$f\in P^P$</span>, there exists <span class="math-container">$p\in P$</span> such that <span class="math-container">$f(p)=p$</span>.</p>
<p><strong>If <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> have the fixed point property, does <span class="math-container">$P\times Q$</span>?</strong></p>
<p>Roddy, Rutkowski, and Schroeder proved the answer is "yes" if <span class="math-container">$P$</span> is finite.</p>
<p><strong>If <span class="math-container">$P,Q$</span> are finite and non-empty, does <span class="math-container">$P^P\cong Q^Q$</span> imply <span class="math-container">$P\cong Q$</span>?</strong> Duffus and Wille proved that the answer is "yes" if <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> are connected. </p>
<p>Dwight Duffus and Rudolf Wille, "A theorem on partially ordered sets of order-preserving mappings," <em>Proceedings of the American Mathematical Society</em> <strong>76</strong> (1979), 14-16 </p>
<p><strong>Conjecture.</strong> One of these questions is important. The other may not be of interest to anyone but myself.</p>
|
244,875 | <p>\begin{align}
\min_{x} c^Tx \\
s.t.~Ax=b
\end{align}
Note that here $x$ is unrestricted. I need to prove that the dual of this program is given by
\begin{align}
\max_{\lambda} \lambda^Tb \\
s.t.~\lambda^TA\leq c^T
\end{align}</p>
<p>But in the constraint, I always get an equality (using what I learnt)
\begin{align}
\max_{\lambda} \lambda^Tb \\
s.t.~\lambda^TA = c^T
\end{align}
Please give some explanation also. </p>
| littleO | 40,119 | <p>You wrote the dual problem correctly. Perhaps whoever wrote your assignment forgot to include the constraint $x \geq 0$ in the primal problem.</p>
<p>Edit: here's how I derived the dual problem. The Lagrangian is
\begin{align}
L(x,\nu) &= \langle c, x \rangle + \langle \nu, b - Ax \rangle \\
&= \langle c, x \rangle - \langle \nu, Ax \rangle + \langle \nu, b \rangle \\
&= \langle c, x \rangle - \langle A^T \nu, x \rangle + \langle \nu, b \rangle \\
&= \langle c - A^T \nu, x \rangle + \langle \nu, b \rangle.
\end{align}</p>
<p>The dual function is
\begin{equation}
g(\nu) =
\begin{cases}
\langle \nu,b \rangle & \quad \text{if } A^T \nu = c \\
-\infty & \quad \text{otherwise}.
\end{cases}
\end{equation}</p>
<p>So the dual problem is
\begin{align}
\text{maximize} & \quad \langle \nu, b \rangle \\
\text{subject to} & \quad A^T \nu = c.
\end{align}</p>
|
2,944 | <p>Let $f$ be a diffeomorphism, say from $\mathbb R^n$ to $\mathbb R^n$ , such as the transition map between two coordinate charts on a differentiable manifold.</p>
<p>A differential $n$-form (or rather its coefficient function which is obtained by using the canonical one-chart atlas on $\mathbb R^n$) then transforms essentially by multiplication with $\mathrm{det}(Df)$, while integrals transform essentially by multiplication of the integrand with $\lvert\mathrm{det}(Df)\rvert$.</p>
<p>(This is the reason for the necessity to choose an orientation in order to define the integral of a top form on a differentiable manifold.)</p>
<p><em>Question: What is an intuitive or conceptional reason for these different transformation behaviours of forms and integrands?</em></p>
| J. M. ain't a mathematician | 498 | <p>Briefly, $p\rightarrow q$ can be read as "if $p$, then $q$"; "if not p then not q" would be the "inverse", and "if not q then not p" is the "contrapositive", which has the same truth value as the original proposal.</p>
|
1,363,407 | <p>If $x,y,z$ are elements of a group such that $xyz=1,$ then which of the following are true?</p>
<ol>
<li>$yzx=1$</li>
<li>$yxz=1$</li>
<li>$zxy=1$</li>
<li>$zyx=1$</li>
</ol>
<p>I have found options 1 and 3 to be correct, but how to prove that options 2 and 4 are wrong (that is, what the given answers say)?</p>
| egreg | 62,967 | <p>You have
$$
xyzz^{-1}=1z^{-1}
$$
so
$$
xy=z^{-1}
$$
hence
$$
zxy=zz^{-1}=1
$$</p>
<p>Similarly, you can show that $yzx=1$. So you're right in saying that 1 and 3 hold *without any other assumption on the group.</p>
<p>Since 2 and 4 obviously hold when the group is abelian, you can find a counterexample only in a non abelian group, the simplest one is $S_3$.</p>
<p>Consider $x=(12)$, $y=(123)$, $z=(23)$ and do the computations.</p>
|
1,363,407 | <p>If $x,y,z$ are elements of a group such that $xyz=1,$ then which of the following are true?</p>
<ol>
<li>$yzx=1$</li>
<li>$yxz=1$</li>
<li>$zxy=1$</li>
<li>$zyx=1$</li>
</ol>
<p>I have found options 1 and 3 to be correct, but how to prove that options 2 and 4 are wrong (that is, what the given answers say)?</p>
| lisyarus | 135,314 | <p>1 and 3 are true, which follows from a little more general result: if $x_1 x_2 \dots x_n = 1$ and $\pi$ is a cyclic permutation, then $x_{\pi(1)}x_{\pi(2)}\dots x_{\pi(n)} = 1$.</p>
<p>To prove that 2 and 4 are false, it is sufficient to provide counterexamples.</p>
<p>2) Let $z = 1$, then $xyz$ = $xy$ and $yxz$ = $yx$. Take $x$ and $y$ to be non-commuting elements of some group. Then $xy \neq yx$ and $xyz \neq yxz$.</p>
<p>4) Let $y = 1$, then $xyz$ = $xz$ and $zyx$ = $zx$. Take $x$ and $z$ to be non-commuting elements of some group. Then $xz \neq zx$ and $xyz \neq zyx$.</p>
<p>For an example of non-commuting pairs of elements, you can use $\mathbb{S}_3$ (group of permutations on 3-element set) and take transpositions $(12)$ and $(13)$, which do not commute: $(12)(13) = (231)$ and $(13)(12) = (312)$.</p>
<p>Note that all 4 statements hold in an abelian group.</p>
|
367,638 | <p>Reading over a book on computability, it asserts that in P.C., if A is a theorem, then A has arbitrarily many proofs. I can't see how that would work, would you do an infinite loop in the sequence of well-formed-formulae?</p>
| xyzzyz | 23,439 | <p>A proof is a sequence of steps that leads to desired conclusion at the end. Nobody says that all these steps have to be relevant. For instance, just before concluding the proof you can put lots of logical tautologies. For a human reader, they will obviously be irrelevant to the proof, but nevertheless, the new proof, the bigger part of which is pointless, will still be a correct proof.</p>
<p>For instance, here's a proof that all numbers divisible by 4 are also divisible by 2:</p>
<blockquote>
<p>Let $n$ be a number divisible by $4$. It means there's some $k$ such that $n = 4k$. But $4k = 2\cdot 2k$. Putting $l = 2k$, we have $n = 2l$. Thus $n$ is divisble by 2.</p>
</blockquote>
<p>Here's another proof.</p>
<blockquote>
<p>Let $n$ be a number divisible by $4$. It means there's some $k$ such that $n = 4k$. But $4k = 2\cdot 2k$. Putting $l = 2k$, we have $n = 2l$. We see that $n = 2l$. We also have $4k = 2 \cdot 2k$. We know that $n$ is divisible by $4$. We see that $n = 2l$. Thus $n$ is divisble by 2.</p>
</blockquote>
<p>We obviously can make even longer (or arbitrarily long) proof.</p>
|
1,511,753 | <p>Using the concept of self-similarity, it's possible to encode the decimal expansion of a number as a sort of 'fractal' object. For instance, consider the sequence,</p>
<p>$$(1) \quad C_0=0.1, \ C_1=0.101, \ C_2=0.101000101, 0.101000101000000000101000101,...,C_n$$</p>
<p>The astute reader will notice this is analogous to the construction of the <a href="https://en.wikipedia.org/wiki/Cantor_set" rel="nofollow">Cantor Set</a>. The number I'd assume is irrational. However there is a fairly simple way to construct the number, and thus find it's decimal expansion. In fact, the number $C_n$ satisfies,</p>
<p>$$(2) \quad C_{n+1}=C_n+C_n \cdot 10^{-2 \cdot 3^{n}}$$</p>
<blockquote>
<p>Do similar methods exist for other reals such as $\sqrt{2}$ or $\pi$? If they do, how are these methods developed?</p>
</blockquote>
| BrianO | 277,043 | <p>It depends on the definition of the real in question. Some, like $\sqrt 2$ and $\pi$, are defined precisely enough that their decimal expansions can be computed to arbitrary precision. For the two reals just mentioned, the methods used aren't the same, and each takes advantage of aspects of the defined real peculiar to that real (or, a class of analogous reals). </p>
<p>However, there are clearly-defined reals whose decimal expansions are <em>not</em> computable: <a href="https://en.wikipedia.org/wiki/Chaitin's_constant" rel="nofollow">Chaitin's constants</a>. The wiki article just cited summarizes these as defining a "halting probability" — "a real number that informally represents the probability that a randomly constructed program [<em>in a particular formalism, e.g. Turing machines, systems of equations, etc.</em>] will halt". Each formalism for computation has its own particular halting probability, i.e. Chaitin constant. It goes on to mention that "each halting probability is a normal and transcendental real number that is not computable, which means that there is no algorithm to compute its digits".</p>
<p>In general, there are (many, many) more reals than there are algorithms, and for the "typical" irrational, there is no procedure for computing its decimal expansion.</p>
|
2,957,440 | <p>Knowing <span class="math-container">$f(1) = 2, f(4)=7,f'(1) = 5,f'(4)=3$</span></p>
<p>Find
<span class="math-container">$\int_1^4xf''(x)dx$</span></p>
<p>It is obvious that if there was no <span class="math-container">$x$</span> the answer would be <span class="math-container">$f'(4) - f'(1)$</span> from the fundamental theorem of calculus. I'm not sure what I do with that <span class="math-container">$x$</span> though.Is it treated like a constant in this case? Do I need to use integration by parts or a u-substitution? If so I can't seem to find the correct substitution. I feel like I'm overlooking something obvious...</p>
| Robert Lewis | 67,071 | <p>To evaluate, according to the given data, the integral</p>
<p><span class="math-container">$\displaystyle \int_1^4 xf''(x) \; dx, \tag 1$</span></p>
<p>set</p>
<p><span class="math-container">$u(x) = x, \; v(x) = f'(x); \tag 2$</span></p>
<p>then</p>
<p><span class="math-container">$du = dx, \; dv = f''(x) dx; \tag 3$</span></p>
<p>now integrate by parts, using</p>
<p><span class="math-container">$u \; dv = d(uv) - v \; du; \tag 4$</span></p>
<p>one finds</p>
<p><span class="math-container">$\displaystyle \int_1^4 xf''(x) \; dx = ( x f'(x) \vert_1^4 - \int_1^4 f'(x) \; dx$</span>
<span class="math-container">$= ( x f'(x) \vert_1^4 - (f(4) - f(1)) = (4(3) - 1(5)) - (7 - 2) = 2, \tag 5$</span></p>
<p>in agreement with the result of our colleague Parcly Taxel.</p>
|
1,350,733 | <p>The following (multiple choice) problem is from a test review.</p>
<blockquote>
<p>For the given matrix $A$, find a basis for the corresponding eigenspace for the given eigenvalue.</p>
<p>$$A = \begin{bmatrix}1 & 6 & 6 \\ 6 & 1 & -6 \\ -6 & 6 & 13\end{bmatrix};\quad \lambda = 7.$$</p>
</blockquote>
<p>The four given options are</p>
<blockquote>
<p><strong>A)</strong> $\left\{ \begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix} \right\}$<br>
<strong>B)</strong> $\left\{ \begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix} \right\}$<br>
<strong>C)</strong> $\left\{ \begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix} \right\}$<br>
<strong>D)</strong> $\left\{ \begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix} \right\}$</p>
</blockquote>
<p>The correct answer is <strong>A</strong>.</p>
<hr>
<p>My answer is not among the answer choices. What am I doing wrong?</p>
<p>First, reduce $(A-\lambda I)$:</p>
<p>$$\begin{align}
A - \lambda I &=
\begin{bmatrix}1 & 6 & 6 \\ 6 & 1 & -6 \\ -6 & 6 & 13\end{bmatrix} - \begin{bmatrix}7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7\end{bmatrix} \\
&= \begin{bmatrix}-6 & 6 & 6 \\ 6 & -6 & -6 \\ -6 & 6 & 6\end{bmatrix} \\
&\sim \begin{bmatrix}-1 & 1 & 1 \\ 1 & -1 & -1 \\ -1 & 1 & 1\end{bmatrix} \\
&\sim \begin{bmatrix}1 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
\end{align}$$</p>
<p>Parametric form solving for $(A-\lambda I) = \vec 0$:</p>
<p>$$\begin{align}
x_1 &= x_2 + x_3 \\
x_2 &= x_2 \\
x_3 &= x_3
\end{align}$$
has solutions $x_2(1,1,0) + x_3(1,0,1)$ for all $x_2 , x_3$.
So a basis for the null-space of $(A-\lambda I)$ consists of the vectors $\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$ and $\begin{bmatrix}1 \\0 \\ 1\end{bmatrix}$.</p>
<p>So, am I wrong or is the solution wrong?</p>
| Rory Daulton | 161,807 | <p>As @GTonyJacobs correctly said in a comment, your answer is correct, but there are infinitely many answers to the question and the one you found is not the one given as one of the multiple choice answers.</p>
<p>You asked @GTonyJacobs in a comment of your own what to do in such a case. The first thing is for your to recognize if a question has multiple answers. Any question asking for a basis of a non-trivial vector space over the rationals or reals will have infinitely many answers. If your given basis is not one of the choices given, you then need to see which of the given answers spans the same space as your basis.</p>
<p>In this particular case, this is easily done. One of your vectors, $\begin{bmatrix}1\\0\\1\end{bmatrix}$, is one of the vectors in choice (A). If you add the two vectors in choice (A) you get your other basis vector, $\begin{bmatrix}1\\1\\0\end{bmatrix}$. Adding two vectors is an elementary operation, so the two bases cover the same subspace.</p>
<p>In a more difficult case, write each basis as a matrix of row vectors then calculate the reduced row echelon form. The two subspaces covered by the bases are equal if and only if the two reduced row echelon forms are equal. In this particular case, you want to reduce the two matrices</p>
<p>$$\begin{bmatrix}1&0&1\\0&1&-1\end{bmatrix},\quad \begin{bmatrix}1&1&0\\1&0&1\end{bmatrix}$$</p>
<p>(Choice (A) is first, yours is second.) These both reduce to</p>
<p>$$\begin{bmatrix}1&0&1\\0&1&-1\end{bmatrix}$$</p>
<p>which is choice (A), of course, though that is not relevant. This shows that your basis and choice (A) cover the same subspace, so you choose (A).</p>
<p>By the way, choices (B) and (C) can be rejected immediately, since they have the wrong number of dimensions. You could find the reduced row echelon form of choice (D) and see that it differs from yours.</p>
|
3,336,592 | <p>I could prove the following result from my Real Analysis course:</p>
<blockquote>
<p>Let <span class="math-container">$f:[0,1] \rightarrow [0,1]$</span> be an increasing mapping. Then it has a fixed point.</p>
</blockquote>
<p>I understand that this is a very baby version of Tarski’s Fixed Point Theorem. Now, I wish to generalize this a little bit and get the following:</p>
<blockquote>
<p>Let <span class="math-container">$f:[0,1]^n \rightarrow [0,1]^n$</span> in which <span class="math-container">$f$</span> is increasing in the sense that if <span class="math-container">$y \geq x$</span> coordinate wise then <span class="math-container">$f(y) \geq f(x)$</span> coordinate wise. Then, f has a fixed point.</p>
</blockquote>
<p>From my point of view, we could just pick a point <span class="math-container">$x_0 \in [0,1]^n$</span>, fix all coordinates but one and apply the above lemma to that coordinate. Then, when the first coordinate of the fixed point is found, we do the same for the second and so on.</p>
<p>However, I am not sure this route would be successful and even if it is, I can’t write the extension formally. Any ideas? Thanks a lot in advance!</p>
| Noah Schweber | 28,111 | <p>As Chris Eagle said, your example for (1) is wrong. Removing the characteristic specification does the trick (as they observe), but there are also far simpler examples. For instance, take the empty language <span class="math-container">$\{\}$</span> (so only "<span class="math-container">$=$</span>" allowed, besides the pure logical grammar) and consider the theory <span class="math-container">$$T=\{\exists x,y\forall z(x=z\vee y=z)\}.$$</span> This theory has exactly two models up to isomorphism, a one-element set <span class="math-container">$M_1$</span> and a two-element set <span class="math-container">$M_2$</span>. These aren't elementarily equivalent, so <span class="math-container">$T$</span> isn't complete, but it is decidable since we have <span class="math-container">$$T\vdash\varphi\quad\iff M_1\models\varphi\mbox{ and }M_2\models\varphi,$$</span> and checking whether a sentence holds in a finite structure is computable.</p>
|
103,358 | <p>I am trying to illustrate some simple ideas with exponents. I can manually express something like $5^4$ as $5 \cdot 5 \cdot 5 \cdot 5$, but wondered how to get <em>Mathematica</em> to do that for me.</p>
<p>I found the example below in the documentation, but can't figure out how to "massage" it to work with a number that only has one factor, for example I would like $625$ to be represented as $5 \cdot 5 \cdot 5 \cdot 5$:</p>
<pre><code>CenterDot @@ (Superscript @@@ FactorInteger[20!])
</code></pre>
<p>Any ideas would be appreciated.</p>
| Searke | 144 | <p>First, try resetting Mathematica by following this article:</p>
<p><a href="http://support.wolfram.com/kb/12464" rel="nofollow">http://support.wolfram.com/kb/12464</a></p>
<p>Even if there's no reason why it should work, please try it. </p>
<hr>
<p>If that doesn't work, then try running the Mathematica kernel:</p>
<p><a href="http://support.wolfram.com/kb/12414" rel="nofollow">http://support.wolfram.com/kb/12414</a></p>
<p>You should be able to evaluate simple things like 2+2. Can you? If not, are there error messages? Try to evaluate:</p>
<pre><code>SystemInformation["Small"]
</code></pre>
<p>Does that work? What is the output?</p>
<hr>
<p>If none of this works, you may want to contact Wolfram Technical Support by email using this form:</p>
<p><a href="http://www.wolfram.com/support/contact/email/?topic=Technical" rel="nofollow">http://www.wolfram.com/support/contact/email/?topic=Technical</a></p>
|
3,619,835 | <blockquote>
<p>Find the maximum of the function <span class="math-container">$f(x,y) = (a + x)(b + y)$</span> under the constraint <span class="math-container">$d = x + y$</span>, where <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, <span class="math-container">$d$</span> are known.</p>
</blockquote>
<p>It seems "obvious" to me that you'd want to split it between the two so that both sides of the product are as equal as possible but no idea how to prove the result.</p>
<p>Sorry I don't know what kind of math this is so I tagged a few.</p>
| jobseeker_68141 | 771,246 | <p>It's labeled as pre-calculus problem so I'll solve it without calculus.</p>
<p><span class="math-container">$f(x,y)=(a+x)(b+y)=(a+x)(b+d-x)=-x^2+(b+d-a)x+(ab+ad)=-(x-(b+d-a)/2)^2+(ab+ad+(b+d-a)^2/4)$</span></p>
<p>The maximum can be obtained at <span class="math-container">$x=(b+d-a)/2$</span> and the maximum value is <span class="math-container">$ab+ad+(b+d-a)^2/4$</span></p>
|
547,784 | <p>Given random variable $F_{x}(n)=P(X=n)=\frac{c}{n(n+1)}$ calculate $c$ and $P(X>m)$ for $m=1,2,3...$.</p>
<p>First of all $\lim\sum\limits_{n \in \mathbb{N}} F_{x}(n)=1 $, so $$\lim\sum\limits_{n \in \mathbb{N}} F_{x}(n)=\frac{c}{n(n+1)} + \frac{c}{(n+1)(n+2)} ... \leq c(\frac{1}{2}+\frac{1}{4} ...)\leq c$$ </p>
<p>So $c$ has to be less than $1$. But how do I go about calculating $P(X>m)$?</p>
| Community | -1 | <p>$2^{n+1}=2^n.2=2.\sum_{v=0}^{n} \binom{n}{v}= \binom{n}{0}+\big\{\binom{n}{0}+ \binom{n}{1}\big\} +\big\{ \binom{n}{1}+\binom{n}{2}\big\}+\dots + \binom{n}{n}+ \binom{n}{n}$</p>
<p>It would be necessary to use :$$\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$$</p>
|
547,784 | <p>Given random variable $F_{x}(n)=P(X=n)=\frac{c}{n(n+1)}$ calculate $c$ and $P(X>m)$ for $m=1,2,3...$.</p>
<p>First of all $\lim\sum\limits_{n \in \mathbb{N}} F_{x}(n)=1 $, so $$\lim\sum\limits_{n \in \mathbb{N}} F_{x}(n)=\frac{c}{n(n+1)} + \frac{c}{(n+1)(n+2)} ... \leq c(\frac{1}{2}+\frac{1}{4} ...)\leq c$$ </p>
<p>So $c$ has to be less than $1$. But how do I go about calculating $P(X>m)$?</p>
| Haha | 94,689 | <p>You can use the Newton's binomial that states that $(a+b)^n=\sum_{k=0}^n \binom{n}{k}a^{k}b^{n-k}$. Thus you have that $2^n=(1+1)^n=\sum_{k=0}^n \binom{n}{k}1^{k}1^{n-k}=\sum_{k=0}^n \binom{n}{k}$</p>
|
483,602 | <p>$$
\large\sqrt{8+\sqrt{5}-\sqrt{6-2\sqrt{5}}}
$$</p>
<p>I am asked to write the following as a ratio of two integers (i.e.,) show that this value is rational
(Do not use calculator)</p>
<p>What I tried doing was putting everything to the exponent of $.5$ and then was unsure how to proceed </p>
| Josephine Moeller | 950 | <p>Andre's hint is great, but more generally, if you need to show that a quantity containing a square root is rational, try completing the square on the quantity under the root. </p>
<p>The most inside root is best to start with because it's easier to analyze and things might cancel.</p>
<p>For example, to do that on $6 - 2\sqrt{5}$, you square $\sqrt{5}$ to get $5$, so $6 - 2\sqrt{5} + 5 - 5 = 5 - 2\sqrt{5} + 1 = (\sqrt{5}-1)^2$.</p>
|
2,654,702 | <p>I'm working on generating a counter example that shows projections from $X \times Y$ do not always map closed sets to closed sets. I'm working backwards, that is, from the decompositions to the product space. My attempt is to see if I might generate a closed product space whose projection maps to clopen sets. </p>
<p>I don't know if my example is on the right track, but in any case I'm trying to gain some intuition into the kind of product space generated by these intervals. For example, in the lower left hand corner of the product, I believe the point in $\mathbb{E}^2$ looks something like $(0, 1/n)$. Is this right? If so, is there an "opening" in the product space at $(0,0)$? Is this even the right way to think about this, or is my intuition misguided? </p>
| fleablood | 280,126 | <p>You are missing the point of the exercise.</p>
<p>The exercise is to <em>extend</em> the definition of $b^n; n \in \mathbb N$ to $b^x; x \in \mathbb R$.</p>
<p>At this point $b^n$ is defined to mean "$b$ multiplied by itself $n$ times". This fine but it's not a very useful definition. What if $x \not \in \mathbb N$. What does $b^x$ mean then?</p>
<p>So we <em>extend</em> the definition. Well we defined $b^{\frac 1n}$ to mean the positive number $c$ so that $c^n = b$. This had <em>NOTHING</em> whatsoever to do with $b^n$ equaling $b*b*b... *b$ and the fact that they both looked like $b^{something}$ was entirely coincidental.</p>
<p>So we have $b^n = b*b*...*b$ by definition. And $b^x$ is utterly undefined. We'll define $b^{\frac nm}$ as $c^n$ where $c^m = b$. Now, wait, you ought to be saying. That's an <em>ENTIRELY</em> different definition and has <em>nothing</em> to do with the old definition $b^n = b*b*b....*b$.</p>
<p>But that's okay, because we are <em>extending</em> a definition. We just have to prove that with the <em>new</em> definition of $b^{\frac nm} = c^n; c^m =b$ that <em>IF</em> $\frac nm = k \in \mathbb N$ then $b^{\frac nm} = c^n$ will <em>also</em> be so that $b^{\frac nm} = b*b*b*...*b$ $k$ times.</p>
<p>If so, the <em>new</em> definition does 1) agrees with the old one and 2) allows for the term $b^x$ to be defined for more cases of $x$.</p>
<p>It does.</p>
<p>Okay. So $b^r; r\in \mathbb Q$ is defined. But $b^x; x \not \in \mathbb Q$ is <em>NOT</em> defined.</p>
<p>We need to <em>extend</em> the definition.</p>
<p>So we do that by DEFINING $b^x := \sup \{b^r| r\in \mathbb Q; r \le x\}$.</p>
<p>That will be our <em>new</em> definition and it will be a good one if it does the two things:</p>
<p>1) agrees with the old one and 2) allows for the term $b^x$ to be defined for more cases of $x$.</p>
<p>Well, it certainly does number 2) but does it do number 1)? That's what we have to prove.</p>
<p>If $r \in \mathbb Q$ does $b^r = \sup B(r)$. If so then the new definition agrees with the old one.</p>
<p>So you prove it for $r\in \mathbb Q$.</p>
<p>!!!!YOU DO <strong><em>NOT</em></strong> HAVE TO PROVE $b^x = \sup B(x)$ if $x \not \in \mathbb Q$. You CAN'T prove it even if you wanted to.... because that is the DEFINITION of $b^x; x \not \in \mathbb Q$.</p>
|
2,790,910 | <blockquote>
<p>Let <span class="math-container">$X \sim N (0, 1)$</span> and <span class="math-container">$Y ∼ N (0, 1)$</span> be two independent random variables, and define <span class="math-container">$Z = \min(X, Y )$</span>. Prove that <span class="math-container">$Z^2\sim\chi^2(1),$</span> i.e. Chi-Squared with degree of freedom <span class="math-container">$1.$</span></p>
</blockquote>
<p>I found the density functions of <span class="math-container">$X$</span> and <span class="math-container">$Y,$</span> as they are normally distributed. How would one use the fact that <span class="math-container">$Z = \min(X,Y)$</span> to answer the question? Thanks!</p>
| honeybadger | 471,078 | <p>Since $X$ and $Y$ are independent, we can see that $f(x,y) = f(x)f(y)$. </p>
<p>Now, if we transform our axis from $X,Y$ to $X_1,Y_1$, where $X_1 = \frac{1}{\sqrt{2}}(X+Y)$ and $X_2 = \frac{1}{\sqrt{2}}(X-Y)$ - essentially, we are rotating the $XY$ space by a +45 degrees. </p>
<p>Since the distribution $f(x,y)$ has radial symmetry, we can see that $f(x_1,x_2) =f(x,y) $. From this we can see that, even $X_2$ $∼ N (0, 1)$. </p>
<p>Now, the variable $Z = min(X,Y)$ lies to the space right of the line $\frac{1}{\sqrt{2}}(X-Y)$ in the $XY$ space. </p>
<p>Therefore, distribution of $Z^2$ is the same as the distribution of $X_2^{2}$ - which is essentially a Chi-square distribution with degrees of freedom = 1. </p>
|
4,115,308 | <p>Hi I have a problem where I need to solve the following set of equations:</p>
<p><span class="math-container">$$
v = U u
$$</span></p>
<p><span class="math-container">$$
u = 1 -Uv
$$</span></p>
<p><span class="math-container">$$
U^2 = u^2 + v^2
$$</span></p>
<p>I have tried subbing <span class="math-container">$u$</span> and <span class="math-container">$v$</span> into the expression for <span class="math-container">$U^2$</span> but it seems to get very messy very quickly.</p>
<p>Any help solving for <span class="math-container">$u$</span>,<span class="math-container">$v$</span> and <span class="math-container">$U$</span> would be greatly appreciated.</p>
| David | 919,460 | <p>Use that rank = row rank = column rank. Let <span class="math-container">$r$</span> be the rank of the original matrix <span class="math-container">$A.$</span> Form <span class="math-container">$B,$</span> by choosing <span class="math-container">$r$</span> linearly independent columns of <span class="math-container">$A.$</span> Form <span class="math-container">$C$</span> by choosing <span class="math-container">$r$</span> linearly independent rows of <span class="math-container">$B.$</span> Since the <span class="math-container">$r\times r$</span> matrix <span class="math-container">$C$</span> has rank <span class="math-container">$r,$</span> it has a nonzero determinant. (Observe that a submatrix cannot have larger rank than the original matrix.)</p>
|
2,808,445 | <p>I should mention this up front: this is a question about semantics. It is extremely formal, and a bit pedantic. Most people won't care.</p>
<p>Is det, i.e the determinant function, a single one? Or is it multiple functions, one for each field?</p>
<p>Let $\mbox{Mat}(K)$ denote the set of square matrices over the field $K$ (This is nonstandard notation). Then $\det(x) \in K$ for $x\in \mbox{Mat}(K)$. So we can view $\det$ in this context as a function $\det: \mbox{Mat}(K) \rightarrow K$. If you take this view, then you are forced to acknowledge that 'det' actually refers to infinitely many functions, one for each field $K$, and that which 'det' we are referring to in a particular expression depends entirely on context.</p>
<p>I should note that here, we take the view that a function MUST possess a domain and codomain, that it is IS NOT just a collection of (input, output) pairs with some sort of implicit domain and codomain.</p>
<p>On the other hand, we may view $\det$ as a single function, taking in ANY square matrix, and returning... something. This is fine, but what would the codomain be? It would have to consist of... everything, and that is not a good codomain for a single function object. Ideally, the codomain would be a bit more expressive. However, this, formally, is not a problem. </p>
<p>So, which is it? The former or the latter?</p>
| Leandro | 633 | <p>What about define it as a function with following domain and codomain
$$
\mathrm{det}: \bigcup_{n=1}^{\infty}\ \bigcup_{K:\ K\ \text{is a field}} M_{n\times n}(K)\to \bigcup_{K:\ K\ \text{is a field}} K
$$</p>
|
2,426,450 | <p>Today when I was solving problems from GRE Manhattan I ran into a strange word problem.</p>
<p><a href="https://i.stack.imgur.com/nqzes.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nqzes.png" alt="enter image description here"></a></p>
<p>The first line of the problem already seems weird since $\frac{3}{8}$ of $420$ is not an integer, namely it is equal to $\frac{3}{8}\cdot 420=157,5$. Am I right?</p>
| Ken Draco | 431,886 | <p>Yes, you are absolutely right. A typo has crept in or something like that. It happens both in actual tests and in various textbooks.</p>
<p>Students who took French: $3/8\cdot420=157.5$</p>
<p>Students who took geography: $2/5\cdot 420=168$</p>
<p>Students who took both: $1/4\cdot 420=105$</p>
<p>Students who took geography but not French: $168-105=63$</p>
<p>Students who took French but not geography: $157.5-105=52.5$</p>
<p>Students who took neither: $420-63-52.5-105=199.5$</p>
<p>Therefore answer C should be correct. A Venn diagram is the best way to illustrate the solution.</p>
<p>Please note that unfortunately this is normal for a test. There might be $1\mbox{-}2\%$ of questions, which are either badly prepared, or allow for some ambiguity, or as in your case something like a fractional number of persons pops up, or etc. The verbal section allows ten times as much ambiguity which is not good but not a deal breaker. I mean a student can still pass his/her test with flying colors. It’s just that mathematical questions are much more precise and have no ambiguity and no mistakes with some rare exceptions as was your question. My advice is not to waste time on such things. Such aberrations will pop up very, very rarely in math. So, there’s nothing wrong with Manhattan, Princeton, Barron’s, Kaplan or the actual exam. They are all good books in my opinion.</p>
|
370,151 | <p>Let $f: \Bbb R → \Bbb R$ be a continuous function such that $f(x)=x$ has no real solution . Then is it true that $f(f(x))=x$ also has no real solution ? </p>
| Marc van Leeuwen | 18,880 | <p>Just a variation of the proof. Suppose $f(f(x))=x$. Define $g:x\mapsto f(x)-x$, which is continuous, and has $g(f(x))=f(f(x))-f(x)=-(f(x)-x)=-g(x)$. Then by the intermediate value theorem $g(y)=0$ for some $y$ in the closed interval bounded by $x$ and $f(x)$, and then of course $f(y)=y$.</p>
|
1,789,033 | <p>Let $X$ be an irreducible affine variety. Let $U \subset X$ be a nonempty open subset. Show that dim $U=$ dim $X$.</p>
<p>Since $U \subset X$, dim $U \leq$ dim $X$ is immediate. I also know that the result is not true if $X$ is any irreducible topological space, so somehow the properties of an affine variety have to come in. I have tried assuming $U=X$ \ $V(f_1,...,f_k)$ but I don't know how to continue on.</p>
<p>Any help is appreciated!</p>
| Slade | 33,433 | <p>Let $X=\operatorname{Spec} A$, and pick some nonzero $f\in A$ with $\operatorname{Spec} A_f = D(f) \subset U$. Then we can lift chains of closed sets to see that $\dim D(f) \leq \dim U
\leq \dim X$.</p>
<p>But as shown <a href="https://math.stackexchange.com/questions/1784246/a-be-an-affine-k-algebra-and-f-be-a-non-zero-divisor-of-a-then-can-one-say-that">here</a>, we have $\dim D(f) = \dim A_f = \dim A = \dim X$. We can also see this quite quickly by noting that $A$ and $A_f$ have the same fraction fields, and therefore the same transcendence degree over $k$.</p>
<hr>
<p>Note that this is false for general rings $A$, even if $A$ is assumed to be a noetherian domain. In fact, any DVR is a counterexample.</p>
|
3,720,856 | <p><strong>Question:</strong>
Prove that <span class="math-container">$\sin(nx) \cos((n+1)x)-\sin((n-1)x)\cos(nx) = \sin(x) \cos(2nx)$</span> for <span class="math-container">$n \in \mathbb{R}$</span>.</p>
<p><strong>My attempts:</strong></p>
<blockquote>
<p>I initially began messing around with the product to sum identities, but I couldn't find any way to actually use them.<br />
I also tried compound angles to expand the expression, but it became too difficult to work with.</p>
</blockquote>
<p>Any help or guidance would be greatly appreciated</p>
| robjohn | 13,854 | <p><span class="math-container">$$
\begin{align}
\sin(nx)\cos((n+1)x)
&=\frac{\sin(nx+(n+1)x)+\sin(nx-(n+1)x)}2\tag1\\
&=\frac{\sin((2n+1)x)-\sin(x)}2\tag2\\
\sin((n-1)x)\cos(nx)
&=\frac{\sin((2n-1)x)-\sin(x)}2\tag3
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$(1)$</span>: identity: <span class="math-container">$\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}2$</span><br />
<span class="math-container">$(2)$</span>: simplify<br />
<span class="math-container">$(3)$</span>: apply <span class="math-container">$(2)$</span> for <span class="math-container">$n-1$</span></p>
<p>Therefore,
<span class="math-container">$$
\begin{align}
\sin(nx)\cos((n+1)x)-\sin((n-1)x)\cos(nx)
&=\frac{\sin((2n+1)x)-\sin((2n-1)x)}2\tag4\\
&=\sin(x)\cos(2nx)\tag5
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$(4)$</span>: subtract <span class="math-container">$(3)$</span> from <span class="math-container">$(2)$</span><br />
<span class="math-container">$(5)$</span>: identity: <span class="math-container">$\sin(a)-\sin(b)=2\sin\left(\frac{a-b}2\right)\cos\left(\frac{a+b}2\right)$</span></p>
|
1,526,653 | <p>I am having troubles isolating and solving for $\frac{dy}{dx}$,</p>
<p>$\sin(x+y)=xy$</p>
<p>$\cos(x+y)(1+\frac{dy}{dx})= x\frac{dy}{dx} + y$</p>
<p>What would be the next few step to solve for $\frac{dy}{dx}$?</p>
| Michael Hardy | 11,667 | <p>$$
\cos(x+y)\left(1+ \frac{dy}{dx}\right)= x\frac{dy}{dx} + y \qquad \longleftarrow \text{ error: You need $y$ here, not $x$.}
$$
Expand:
$$
\cos(x+y)\cdot1 + \cos(x+y)\frac{dy}{dx} = x\frac{dy}{dx} + y
$$
Transpose:
$$
\cos(x+y)\frac{dy}{dx} - x\frac{dy}{dx} = \cos(x+y)\cdot1 + y
$$
Factor
$$
(\cos(x+y) - x) \frac{dy}{dx} = \cos(x+y)\cdot1 + y
$$
Then divide both sides by $\cos(x+y) - x$.</p>
|
1,526,653 | <p>I am having troubles isolating and solving for $\frac{dy}{dx}$,</p>
<p>$\sin(x+y)=xy$</p>
<p>$\cos(x+y)(1+\frac{dy}{dx})= x\frac{dy}{dx} + y$</p>
<p>What would be the next few step to solve for $\frac{dy}{dx}$?</p>
| Claude Leibovici | 82,404 | <p>Hoping that you already know it, the <a href="https://en.wikipedia.org/wiki/Implicit_function_theorem" rel="nofollow">implicit function theorem</a> is very useful. </p>
<p>In you case, consider the function $$F=\sin(x+y)-xy=0$$ Computing each derivative, considering that the other variable is a constant, gives $$F'_x=\cos(x+y)-y$$ $$F'_y=\cos(x+y)-x$$ $$y'=\frac{dy}{dx}=-\frac{F'_x}{F'_y}=-\frac{\cos(x+y)-y}{\cos(x+y)-x}$$</p>
|
2,744,172 | <p>(Note: This question is tangentially related to this <a href="https://math.stackexchange.com/q/3609304">later one</a>.)</p>
<p>Let <span class="math-container">$$\sigma(x) = \sum_{d \mid x}{d}$$</span> denote the <em>sum of divisors</em> of <span class="math-container">$x \in \mathbb{N}$</span>, where <span class="math-container">$\mathbb{N}$</span> is the set of natural numbers or positive integers.</p>
<p>Recall that a <em>Descartes number</em> is an odd number <span class="math-container">$n = km$</span>, with <span class="math-container">$1 < k$</span>, <span class="math-container">$1 < m$</span>, satisfying <span class="math-container">$$\sigma(k)(m+1)=2km.$$</span> (<span class="math-container">$m$</span> is called the quasi-Euler prime of <span class="math-container">$n$</span>.) Note that we define <span class="math-container">$\sigma(m) := m + 1$</span> even when <span class="math-container">$m$</span> is composite (that is, we <em>pretend</em> that <span class="math-container">$m$</span> is prime).</p>
<p>Notice that the lone Descartes number that is known is
<span class="math-container">$$\mathscr{D} = k'm' = {{3003}^2}\cdot{22021}.$$</span></p>
<p>In particular, note that:</p>
<p><strong>(1)</strong> <span class="math-container">$k$</span> is a square.</p>
<p><strong>(2)</strong> <span class="math-container">$\sigma(k)/m = 2k - \sigma(k)$</span></p>
<p><strong>(3)</strong> <span class="math-container">$m \equiv 1 \pmod 4$</span></p>
<p>I want to prove that it must necessarily be the case that <span class="math-container">$m < k$</span>, for a Descartes number <span class="math-container">$n = km$</span>.</p>
<p><strong>Lemma 1</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span>, then <span class="math-container">$k \neq m$</span>.</p>
<p>To this end, suppose that <span class="math-container">$m = k$</span>. Then we have
<span class="math-container">$$\frac{\sigma(k)}{k} = \frac{2m}{m + 1} = \frac{2k}{k + 1},$$</span>
from which it follows that
<span class="math-container">$$\sigma(k) = \frac{2k^2}{k + 1} = \frac{2k^2 - 2}{k + 1} + \frac{2}{k + 1} = \frac{2(k - 1)(k + 1)}{k + 1} + \frac{2}{k + 1} = 2(k - 1) + \frac{2}{k + 1}.$$</span></p>
<p>Since <span class="math-container">$\sigma(k)$</span> and <span class="math-container">$2(k - 1)$</span> are integers, it follows that <span class="math-container">$2/(k+1)$</span> is also an integer, which means that <span class="math-container">$(k + 1) \mid 2$</span>. This implies that <span class="math-container">$k + 1 \leq 2$</span>, from which wet get <span class="math-container">$m = k \leq 1$</span>. This last inequality contradicts the condition <span class="math-container">$1 < k$</span>, <span class="math-container">$1 < m$</span>.</p>
<p><strong>Lemma 2</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m,k)=1$</span>, then
<span class="math-container">$$\frac{\sigma(m)}{k} \neq \frac{\sigma(k)}{m}.$$</span></p>
<p>Suppose to the contrary that <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m,k)=1$</span>, and that <span class="math-container">$\sigma(m)/k = \sigma(k)/m$</span>. Then it follows that
<span class="math-container">$$\frac{\sigma(m)}{k} = \frac{\sigma(k)}{m} = r \in \mathbb{N},$$</span>
from which we obtain
<span class="math-container">$$\frac{\sigma(m)}{k}\cdot\frac{\sigma(k)}{m} = r^2 \in \mathbb{N},$$</span>
contradicting
<span class="math-container">$$\sigma(k)\sigma(m) = \sigma(k)(m+1) = 2km,$$</span>
since the last two equations imply that <span class="math-container">$r^2 = 2$</span>.</p>
<p><strong>Lemma 3</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m, k) = 1$</span>, then we have</p>
<p><strong>(a)</strong> <span class="math-container">$\sigma(m) \neq \sigma(k)$</span></p>
<p><strong>(b)</strong> <span class="math-container">$\sigma(m) \neq k$</span></p>
<p><strong>(c)</strong> <span class="math-container">$\sigma(k) \neq m$</span></p>
<p>Proof of <strong>(a)</strong>: Suppose that <span class="math-container">$\sigma(k) = \sigma(m) = m + 1 \equiv 2 \pmod 4$</span>. This contradicts the fact that <span class="math-container">$k$</span> is a square, since then <span class="math-container">$\sigma(k) \equiv 1 \pmod 2$</span>.</p>
<p>Proof of <strong>(b)</strong>: Suppose that <span class="math-container">$\sigma(m) = k$</span>. Then the even number <span class="math-container">$m + 1 = \sigma(m)$</span> is equal to the odd number <span class="math-container">$k$</span>, which is a clear contradiction.</p>
<p>Proof of <strong>(c)</strong>: Suppose to the contrary that <span class="math-container">$\sigma(k) = m$</span>. Then we obtain the estimate
<span class="math-container">$$\frac{\sigma(k)}{m} + \frac{\sigma(m)}{k} = 1 + 2 = 3,$$</span>
which contradicts the known quantity
<span class="math-container">$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002.$$</span></p>
<p><strong>Lemma 4</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m, k)=1$</span>, then the following biconditionals hold:
<span class="math-container">$$m < k \iff \sigma(m) < \sigma(k) \iff \frac{\sigma(m)}{k} < \frac{\sigma(k)}{m}$$</span></p>
<p>We consider three different cases:</p>
<p><strong>Case (1)</strong>:
<span class="math-container">$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} = \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k}$$</span></p>
<p>Case (1) is equivalent to <span class="math-container">$\sigma(m) = \sigma(k)$</span> (which is ruled out by Lemma 3 (a)) or <span class="math-container">$k = m$</span> (which is ruled out by Lemma 1).</p>
<p><strong>Case (2)</strong>:
<span class="math-container">$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} < \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k}$$</span></p>
<p>Case (2) implies the estimate
<span class="math-container">$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} < \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k} < \frac{9 + 1}{9} + 2 = \frac{28}{9},$$</span>
which again contradicts the known quantity
<span class="math-container">$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002.$$</span></p>
<p><strong>Case (3)</strong>:
<span class="math-container">$$\frac{\sigma(m)}{m} + \frac{\sigma(k)}{k} < \frac{\sigma(m)}{k} + \frac{\sigma(k)}{m}$$</span></p>
<p>Case (3) is equivalent to the truth of the biconditional <span class="math-container">$m < k \iff \sigma(m) < \sigma(k)$</span> (by virtue of Lemma 1, Lemma 2, and Lemma 3), which in turn is equivalent to the truth of the biconditional
<span class="math-container">$$m < k \iff \sigma(m) < \sigma(k) \iff \frac{\sigma(m)}{k} < \frac{\sigma(k)}{m}.$$</span></p>
<p>By Lemma 4, we have the following possibilities:</p>
<p><strong>(A)</strong> <span class="math-container">$k < \sigma(k) < m < \sigma(m)$</span></p>
<p><strong>(B)</strong> <span class="math-container">$m < \sigma(m) < k < \sigma(k)$</span></p>
<p>Note that Case <strong>(A)</strong> implies that
<span class="math-container">$$\frac{\sigma(k)}{m} = 2k - \sigma(k) < 1$$</span>
forcing <span class="math-container">$2k - \sigma(k) = 0$</span> (i.e. <span class="math-container">$k$</span> must be perfect). This contradicts the fact that <span class="math-container">$k$</span> is a square.</p>
<p>Hence we necessarily have Case <strong>(B)</strong>, and a proof for the following theorem:</p>
<p><strong>THEOREM</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m, k) = 1$</span>, then <span class="math-container">$k$</span> is <em>not</em> an odd almost perfect number.</p>
<p><strong>QUESTIONS</strong></p>
<p><em>(I)</em> Can we remove the reliance of the proof on the condition
<span class="math-container">$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002?$$</span></p>
<p><em>(II)</em> To what extent can we relax the condition <span class="math-container">$\gcd(m, k)=1$</span> in the <strong>THEOREM</strong>?</p>
| mathlove | 78,967 | <p>This is a partial answer.</p>
<p><em>(II)</em> : I think that we can prove <span class="math-container">$m\lt k$</span> without assuming that <span class="math-container">$\gcd(k,m)=1$</span>. </p>
<hr>
<p>(1) Your proof for Lemma 1 is correct.</p>
<p>(2) We can prove <span class="math-container">$\frac{\sigma(m)}{k} \neq \frac{\sigma(k)}{m}$</span> without assuming <span class="math-container">$\gcd(k,m)=1$</span>. </p>
<p>Supposing that <span class="math-container">$\sigma(m)/k = \sigma(k)/m$</span> gives
<span class="math-container">$$\frac{m+1}{k}= \frac{2k}{m+1}\implies \bigg(\frac{m+1}{k}\bigg)^2=2\implies \frac{m+1}{k}=\sqrt 2$$</span>
which is a contradiction since LHS is rational while RHS isn't.</p>
<p>(3) You can prove <strong>(a)</strong> <span class="math-container">$\sigma(m) \neq \sigma(k)$</span> <strong>(b)</strong> <span class="math-container">$\sigma(m) \neq k$</span> <strong>(c)</strong> <span class="math-container">$\sigma(k) \neq m$</span> without assuming <span class="math-container">$\gcd(k,m)=1$</span> since you haven't used <span class="math-container">$\gcd(k,m)=1$</span> in your proof for Lemma 3.</p>
<p>(4) Lemma 4 is true without assuming <span class="math-container">$\gcd(k,m)=1$</span>. The proof for <span class="math-container">$m\lt k\iff \sigma(m)\lt\sigma(k)$</span> is written in the question. The proof for <span class="math-container">$m\lt k\iff\frac{\sigma(m)}{k}\lt \frac{\sigma(m)}{k}$</span> is written in the comments below.</p>
<p>(5) In <strong>(A)</strong>, another way to get a contradiction : We have <span class="math-container">$0\lt 2k-\sigma(k)\lt 1$</span> which contradicts that <span class="math-container">$2k-\sigma(k)$</span> is an integer. </p>
<hr>
<p>In conclusion, we can prove <span class="math-container">$m\lt k$</span> without assuming that <span class="math-container">$\gcd(k,m)=1$</span>. </p>
<p>(Moreover, I think that you can prove <span class="math-container">$m\lt k$</span> without using Lemma 2.)</p>
|
4,005,381 | <p>Why the probability of a single element event is "zero" on the continuous model?, the explanation given is based on probability additivity axioms. but how? some more explanation with the source would be helpful.</p>
<p>cool thanks!</p>
| YiFan | 496,634 | <p>Suppose otherwise, so that the chance of picking any element is positive, say <span class="math-container">$p$</span>. Assuming uniform distribution, this probability is equal for all values in the interval by definition. Then for each of the infinitely many values in the interval, it is chosen with probability <span class="math-container">$p$</span>, so the total probability would be infinite, which is absurd.</p>
<p>Of course, you don't need a uniform distribution either. You can easily convince yourself that summing uncountably many positive values is going to give you an infinite result.</p>
|
1,363,967 | <p>A group $G$ acts on a set $X$ transitively and a normal subgroup $H$ fixes a point $x_{0} \in X$, i.e. $h \cdot x_{0}=x_{0}$ for all $h \in H$. Show that $h \cdot x = x$ for all $h \in H$ and $x \in X$.</p>
<p>Since the action is transitive $\mathcal{O}_{x}=\{g\cdot x : g \in G,~ x \in X\}=X$ for any $x \in G$. </p>
<p>I've been fooling around with the fact that $g\cdot x = x_{0}$ for some $g \in G$ and using the fact that $H$ is a normal subgroup but haven't really gotten anywhere. I feel like this should follow straight from the axioms of group action and definition of normal subgroup, or am I missing something?</p>
| P Vanchinathan | 28,915 | <p>This follows from the following: Let $G$ act on a set $X$,(need not be transitive) and let $Y$ be the subset consisting of all the fixed points for a normal subgroup $H\subset G$. Then $g.y\in Y$ for all $y\in Y\ g\in G$.</p>
|
299,263 | <p>What is the set of polynomials with integer coefficients look like? Help!</p>
<p>Thanks!</p>
| Brian M. Scott | 12,042 | <p>Here, as an example of the sort of thing that’s wanted, is a recursive definition of the set $T$ of positive integers that multiples of $3$:</p>
<blockquote>
<ol>
<li>$3\in T$. </li>
<li>If $m,n\in T$, then $m+n\in T$. </li>
<li>$T$ contains only those integers required by the first two clauses.</li>
</ol>
</blockquote>
<p>To see that $12\in T$, for instance, we note that $3\in T$ by (1), so $3+3=6\in T$ by (2), and then $6+6\in T$ by a second application of (2).</p>
<p>Now let $T'=\{3n:n\in\Bbb Z^+\}$; we’d like to prove that $T=T'$, i.e., that we really did define the set of positive multiples of $3$. We first prove by induction that $T\subseteq T'$. </p>
<p>The base case is easy: $3=3\cdot1\in T'$. That is, the one member of $T$ that we’re explicitly given in (1) is in $T'$. Now we show that the constructor clause (2) preserves membership in $T'$: if $m,n\in T'$, then there are positive integers $a$ and $b$ such that $m=3a$ and $n=3b$, and therefore $$m+n=3a+3b=3(a+b)\in T'$$ as well. In short, we start with an element of $T'$, namely $3$, and we add new elements to $T$ <em>via</em> (2) in a way that can only give us multiples of $3$ if we start with multiples of $3$, so end up with nothing but multiples of $3$: $T\subseteq T'$.</p>
<p>To show that $T'\subseteq T$ (and hence that $T=T'$), suppose that $T'\setminus T\ne\varnothing$, and let $m$ be the smallest member of $T'\setminus T$. Since $m\in T'$, $m=3k$ for some positive integer $k$. We know that $k\ne 1$, since $3\cdot1=3\in T$, so $k>1$. But then $k-1$ is still a positive integer, so $3(k-1)\in T'$. $3(k-1)<m$, and $m$ is the <strong>smallest</strong> member of $T'\setminus T$, so $3(k-1)\notin T'\setminus T$. Certainly $3(k-1)$ <strong>is</strong> in $T'$, so this implies that $3(k-1)\in T$ as well. But then $$m=3k=(3k-3)+3=3(k-1)+3$$ is the sum of two members of $T$, so $m\in T$ by clause (2) of the definition of $T$. And this is a contradiction, since we assumed that $m\notin T$. Thus, the supposition that $T'\setminus T\ne\varnothing$ must be false, i.e., we must have $T'\subseteq T$.</p>
<p>You want a recursive definition of the set of polynomials in one variable $x$ with integer coefficients. (I’m assuming that polynomials in one variable were intended; if that’s not the case, the definition is a little more complicated, but the ideas are similar.) The simplest ones are the constant ones; take those as the base elements, analogous to $3$ in my example. If $P$ is the set of polynomials that you’re trying to define, you could start with this:</p>
<blockquote>
<ol>
<li>Every integer belongs to $P$.</li>
</ol>
</blockquote>
<p>How do you build up more complicated polynomials from simpler ones? You can multiply by $x$, or you can add two polynomials that you already have. For instance, if want $3x^2-2x+4$, you could get it from $3,-2$, and $4$ by the following steps: multiply $3$ by $x$ to get $3x$; add $-2$ to get $3x-2$; multiply by $x$ to get $3x^2-2x$; and add $4$. Thus, you’ll have two growth clauses instead of the one in my example. One is will say that if $p(x)$ is in $P$, so is $xp(x)$; what will the other say?</p>
<p>Your question doesn’t say so, but you may well be expected to prove that your definition really does define the intended set of polynomials. Such proofs are typically similar in general outline to the one that I gave with my example.</p>
|
3,827,650 | <p>Does the generalised integral</p>
<p><span class="math-container">$\int_{0}^{\pi}\frac{\sqrt x}{\sin x}dx$</span></p>
<p>converge or diverge?</p>
<p>The first thing I would do here is split it into two integrals</p>
<p><span class="math-container">$$\int_0^\pi \frac{\sqrt x}{\sin{x}}dx=\int_0^{\frac{\pi}{2}} \frac{\sqrt x}{\sin{x}}dx+\int_{\frac{\pi}{2}}^\pi \frac{\sqrt x}{\sin{x}}dx$$</span></p>
<p>But then I am a bit stuck. I don't know if I now should compare it to something (and in that case what?), or if I should expand it with Taylor or something.</p>
| Oliver Díaz | 121,671 | <p>The first integral in the split is convergent; the second intergral in the split howeverdiverges since</p>
<p><span class="math-container">$$\int^{\pi}_{\pi/2}\frac{\sqrt{x}}{\sin x}\,dx\geq \sqrt{\pi/2}\int^\pi_{\pi/2}\frac{dx}{\sin x}$$</span>
Since
<span class="math-container">$$\frac{2}{\pi}\leq \frac{\cos x}{\pi/2-x}\leq 1,$$</span></p>
<p><span class="math-container">$$
\begin{align}
\int^{\pi}_{\pi/2}\frac{1}{\sin x}\,dx &=\int^{\pi/2}_0\frac{dx}{\sin(x+\pi/2)}=\int^{\pi/2}_0\frac{\pi/2 -x}{(\pi/2-x)\cos x}\,dx\\
&\geq \int^{\pi/2}_0\frac{dx}{\pi/2-x}=\int^{\pi/2}_0\frac{du}{u}=\infty
\end{align}
$$</span></p>
|
232,387 | <p>A discrete random variable $X$ of values in $\mathbb N$ verifies the property that
$$P(X=k)=\cfrac 23 (k+1)P(X=k+1)$$
What is the distribution of $X$?</p>
<p>I found that
$$P(X\ge 0)=\sum_{k=0}^\infty P(X=k)=\sum_{k=0}^\infty\cfrac 23 (k+1)P(X=k+1)=\cfrac 23\sum_{k=1}^\infty kP(X=k)=\cfrac 23\text E(X)=1$$
$\ \ \ \ \ \ \ \ \text E(X) = 1.5$</p>
<p>I also found that $$P(X=k)=\cfrac{3^k}{2^k\cdot k!}\cdot P(X=0)$$
That is the only thing I could get out of the given property, I couldn't find the expression for $P(X=k)$ which is the actual question.</p>
| André Nicolas | 6,312 | <p>For a counterexample, use $(1,0)$ and $(0,1)$.${}{}{}{}{}{}{}{}$</p>
|
530 | <p>Tau ($\tau = 2 \pi$) has more merits in its application, but pi is the established standard in industry and education. Is the trade-off of teach-ability of circle concepts worth the subsequent confusion due to pi's omnipresence?</p>
<p>How can both be most effectively taught in order to maximize student's understanding of the use of the circle constant?</p>
| quid | 143 | <p>One should teach $\pi$. One might discuss that there is a choice that is made that is somewhat arbitrary, and there are also reasons for a different choice but I do not see this as that relevant to make much ado about this. </p>
<p>It should perhaps also be noticed that there are two conflicting proposals for $\tau$. Eagle (1958) proposed $\pi/2$ and Palais (2001) proposed $2 \pi$ (for detailed references see <a href="http://en.wikipedia.org/wiki/Tau_%282%CF%80%29#tau">the Wikipedia page on Tau</a>) and chance are there are others (at least others that proposed the same if not still other values). This further illustrates that the situation is not that clear, though <a href="http://www.math.utah.edu/~palais/pi.html">"$\pi$ is wrong!"</a> certainly makes interesting points. </p>
<p>Moreover, it is not quite clear <em>when</em> a choice differing from the mainstream one should happen in the teaching. To have somewhat good reasons to motivate a choice different than $\pi$ one needs more advanced mathematics than one typically has available when first needing a cricle constant. </p>
<p>Purely 'geometrically' the standard choice is at least as good, in my opinion. On the one hand, one can make an argument that area is the more straight forward notion than length, leading rather to $\pi$; second; if one goes with length to compare to the diameter does also no harm in that context and again has some merit, it might even be more natural.</p>
|
530 | <p>Tau ($\tau = 2 \pi$) has more merits in its application, but pi is the established standard in industry and education. Is the trade-off of teach-ability of circle concepts worth the subsequent confusion due to pi's omnipresence?</p>
<p>How can both be most effectively taught in order to maximize student's understanding of the use of the circle constant?</p>
| P. Williams | 221 | <p>I cannot imagine a context where it would benefit a student to know about $\tau$ over and above knowing about $\pi$. Every text book a student will ever encounter will exclusively talk about $\pi$. Every calculator a student will ever use will have the constant $\pi$ pre-programmed in.</p>
<p>Although there are perfectly lovely arguments to show that $\tau$ is a nicer constant, it is just $2\pi$, so it really doesn't matter.</p>
<p>At most I would introduce $\tau$ to higher ability students as an interesting aside, but I cannot see the value in actually teaching it as a useful constant.</p>
|
530 | <p>Tau ($\tau = 2 \pi$) has more merits in its application, but pi is the established standard in industry and education. Is the trade-off of teach-ability of circle concepts worth the subsequent confusion due to pi's omnipresence?</p>
<p>How can both be most effectively taught in order to maximize student's understanding of the use of the circle constant?</p>
| vonbrand | 123 | <p>There is a simple solution to this debate:</p>
<p><a href="http://xkcd.com/1292/" rel="noreferrer" title="Conveniently approximated as e+2, Pau is commonly known as the Devil's Ratio (because in the octal expansion, '666' appears four times in the first 200 digits while no other run of 3+ digits appears more than once.)"><img src="https://i.stack.imgur.com/hXQjd.png" alt="1.5 pi = pau"></a></p>
|
530 | <p>Tau ($\tau = 2 \pi$) has more merits in its application, but pi is the established standard in industry and education. Is the trade-off of teach-ability of circle concepts worth the subsequent confusion due to pi's omnipresence?</p>
<p>How can both be most effectively taught in order to maximize student's understanding of the use of the circle constant?</p>
| Garrett | 933 | <h1>$\tau$ should be taught in schools</h1>
<p>There's plenty of material arguing why $\tau$ is a much more intuitive and easier to teach concept (some of my favorites: <a href="http://www.maa.org/sites/default/files/pdf/Mathhorizons/apr12_aftermath.pdf">1</a>,<a href="https://www.youtube.com/watch?v=jG7vhMMXagQ">2</a>,<a href="http://tauday.com/tau-manifesto">3</a>) and I don't want to rehash their arguments, but if you think that $\tau$ is just to make equations look nicer, please check out those resources. </p>
<p>The question at hand is whether math educators should teach $\tau$. <strong>I will assume that $\tau$ is a more intuitive and didactically nicer concept than $\pi$ for the sake of argument - if you disagree, you can stop reading here.</strong></p>
<h1>Many arguments against $\tau$ are appeals to tradition</h1>
<p>There seems to be a lot of: </p>
<blockquote>
<p>$\pi$ is what we've always used, it is what we currently use, so we should continue using it. </p>
</blockquote>
<p>In fact, many of the arguments are generic enough to be able to apply to any proposed change. For example, imagine when the first few countries were switching to the metric system, I'm sure exactly the same arguments were used. </p>
<p>Here is part of EuYu's answer, but with $\tau$ and $\pi$ replaced with "the metric system of units" and "the British system", respectively:</p>
<blockquote>
<p>I feel that it is perhaps a little irresponsible to teach the metric system of units instead of British units. As a first introduction, it is the norm which should be taught: teaching a rare alternative to the British units only serves to confuse students, especially when almost all available resources use British units instead of metric units. Imagine a student's confusion when they see meters in class and feet everywhere else.</p>
</blockquote>
<h1>Thinking long-term</h1>
<p>Some of the arguments for the status-quo have seemed a bit short-sighted. Many changes have a switching cost, but pay perpetual benefits after the switch is over, as in switching to metric units. I think we should be asking ourselves: "Do we really want to be using a didactally and intuively inferior circle constant for the next 500 years?" instead of "Do we want to have to rewrite our textbooks?"</p>
<ul>
<li>Yes, textbooks would need to be rewritten, but textbooks get updated all the time with new, modern notation. This is a good thing.</li>
<li>Yes, "the students would not be able to read older literature" as Markus Klein points out in his answer. True, but when I read a very old journal paper, I already have a hard time. Again, it's because notation changes, usually for the better. I'm glad that people are no longer using the same exact math/physics notation that was used 100 years ago.</li>
</ul>
<h1>Educators can make this change happen</h1>
<p>Perhaps more people would agree to switch to $\tau$ if everyone were to do it at once - say, starting in 2015, all professors, all publishers, all researchers, etc. would switch to $\tau$ in one fell swoop. I'm down with that, but that's not going to happen if I've understood human nature correctly. </p>
<p>It has to start somewhere and educators are the gatekeepers of knowledge - educators teach the youngsters and write the textbooks. If the change is going to happen, it will happen thanks to them.</p>
<h1>My Personal experience of teaching with $\tau$</h1>
<p>I'm the TA for a Math Methods for Physics course at an American univeristy, and I use $\tau$ in my discussions. I wasn't sure how the students would react, but being nimble-minded, young students, they caught on pretty quickly.</p>
<p>I also felt it wasn't a huge burden on any students who were opposed to $\tau$ - I reminded them that they could always just write down "$2\pi$" in their notes every time they see me write "$\tau$". And I reminded them that they were allowed to ask questions using $\pi$ if they preferred (although most of them just asked questions using $\tau$).</p>
<p>Why don't you give $\tau$ a try in your class? See if your students like it. You may find yourself pleasantly surprised with how easily your students become fluent in $\tau$.</p>
|
2,122,149 | <p>Using the standard topology on $\mathbb{R}^d$, interior of a subset $S$ of topological space $M$ is defined as the set of all elements of $S$ except the boundary of $S$. </p>
<p>However, I only understand the boundary in this sense to be defined with the standard topology on $\mathbb{R}^d$, so how would we define boundary and "interior" in topological spaces in general?</p>
| TheGeekGreek | 359,887 | <p>Let $(X,\mathcal{T})$ be a topological space. For $A \subseteq X$ we define the <strong>interior of A</strong> as $$\operatorname{Int}A:=\bigcup\{C \subseteq X : C \subseteq A, C \in \mathcal{T}\}$$ Furthermore, the <strong>exterior of A</strong> is $$\operatorname{Ext}A := X \setminus \overline{A}$$ where $\overline{A}$ is the <strong>closure of A</strong> defined by $$\overline{A}:=\bigcap\{B \subseteq X : B \supseteq A, B^c \in \mathcal{T}\}$$ Then we define the <strong>boundary of A</strong> as $$\partial A := X \setminus (\operatorname{Int}A \cup \operatorname{Ext}A)$$ Sure, this notions are very abstract. But they reduce to the familiar in for example metric spaces like $\mathbb{R}^n$. Howevery, many properties can be in fact proved in this general setting. Like for example if a sequence $x_n$ in $A$ converges to $x$, then $x \in \overline{A}$.</p>
|
451,326 | <p>I am having trouble understanding a certain part of the proof on why a function cannot approach two different limits near $a$, so I will just list the relevant parts. If this is not enough/ambiguous then please tell me and I will type out the whole proof.</p>
<p>So, suppose we now have:</p>
<p>$$ \text{if } 0<|x-a|<\delta_1, \text{ then } |f(x)-l|<\epsilon \hspace{5cm} (1)$$</p>
<p>and</p>
<p>$$ \text{if } 0<|x-a|<\delta_2, \text{then} |f(x)-m|<\epsilon \hspace{5cm} (2)$$</p>
<p>and here's a quote from the text:</p>
<blockquote>
<p>We have had to use two numbers, $\delta_1$ and $\delta_2$, since there is no guarantee that the $\delta$ which works in one definition will work in the other. But, in fact, it is now easy to conclude that for any $\epsilon>0$ there is some $\delta>0$ such that, for all $x$,
$$ \text{if } 0<|x-a| < \delta, \text{then } |f(x)-l| < \epsilon \text{ and } |f(x)-m| \lt \epsilon$$
we simply chose $\delta=\text{min}(\delta_1,\delta_2)$</p>
</blockquote>
<p>I understand the need to use two distinct $\delta$. What I don't get is why selecting a $\delta$ that is the minimum of $\delta_1$ and $\delta_2$ will make that $\delta$ work in both (1) and (2). I mean, the limits are different so why would I expect that the delta that is the minimum of the two equations will satisfy both equations?</p>
<p>Thank you in advance for any help provided.</p>
| Ben Grossmann | 81,360 | <p>Hint:</p>
<p>Let's begin with the expression
$$
\frac{f(t) - f(t_0)}{t - t_0} = \frac{1}{t-t_0}\left( \frac{{t_0}^n \displaystyle\int_{B(y,t)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n} - \frac{{t}^n \displaystyle\int_{B(y,t_0)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n}\right)
$$
And rewrite that, saying $t=t_0+\Delta t$, as
$$
\frac{f(t_0+\Delta t) - f(t_0)}{\Delta t} = \frac{1}{\Delta t}\left( \frac{{t_0}^n \displaystyle\int_{B(y,t)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n} - \frac{(t_0+\Delta t)^n \displaystyle\int_{B(y,t_0)} u(x) \ dx}{|B(0,1)| t^n {t_0}^n}\right)
$$</p>
<p>Expand the second term and you can cancel the ${t_0}^n$ term, then use Lebesgue's theorem to evaluate the limit.</p>
|
14,167 | <p>I randomly pick a natural number <em>n</em>. Assuming that I would have picked each number with the same probability, what was the probability for me to pick <em>n</em> before I did it?</p>
| Gregory Fenn | 389,331 | <p>Hi :) I've always had a nagging suspicion about drawing (uniform) random numbers from an infinite set. I'm not convinced it's possible: here's an intuition about why I'm sceptical: it would be nice if someone could explain where I go wrong. My counterargument is structured as follows: if it's possible to draw random natural numbers (i.e. from the whole infinite set) then there must be a probability of selecting an EVEN number: I argue that <span class="math-container">$Pr(even) = \frac{1}{2}$</span> AND <span class="math-container">$Pr(even) = 1$</span>, causing a contradiction. </p>
<p>First we need to define:
URD: A "uniform random draw" for a set <span class="math-container">$S$</span>, (<span class="math-container">$S = \mathbb{N} = \{1, 2, 3, ... \}$</span> in this case) is a random selection <span class="math-container">$x$</span> from <span class="math-container">$S$</span> such that every <span class="math-container">$y$</span> in <span class="math-container">$S$</span> has an equal chance of being picked. This is what we're doing in the video, and it's what we usually mean when we just say "random". </p>
<p>Consider the statement '<span class="math-container">$Pr(\texttt{n even}) = 0.5$</span>', where <span class="math-container">$n$</span> is a random element from <span class="math-container">$\mathbb{N}$</span> -- this is what I assume most people would agree with. I disagree that this is true, I think it depends on arbitrary selection rules. </p>
<p>Firstly the intuition we normally have in mine: if you represent a natural number as a rounded scaler from 1 to infinity or a finite string of bits defining a binary number (like your computer does), then it's true that <span class="math-container">$Pr(\texttt{n even}) = 0.5$</span> : because it boils down to whether the last bit is 0 (even) or 1 (odd), and since both 0 and 1 bits are equally likely, it follows that Pr(even) = Pr(not even), which must sum to 1 so Pr(even) = 1/2. You can jiggle this for <span class="math-container">$Pr(\texttt{n is a multiple of m})$</span> to get <span class="math-container">$\frac{1}{m}$</span>. It's easy to see that this way of drawing number satisfies URD. </p>
<p>There is another way to uniquely represent numbers, not just as a line or scale: we can represent n as a product of primes via the fundamental theorem of arithmetic. </p>
<p>Consider <span class="math-container">$n = 2^{a1} * 3^{a2} * 5^{a3} * 7^{a4} * 11^{a5} * ..... $</span></p>
<p>If we start by drawing random URD numbers a1, a2, a3, ..., over <span class="math-container">$\{0, 1, 2, ....\}$</span> and then defining <span class="math-container">$n$</span> as above, we have drawn a random <span class="math-container">$n$</span>. In fact, since every different sequence <span class="math-container">$\{a1, a2, a3 ... \}$</span> defines one, and EXACTLY one, natural number <span class="math-container">$n$</span>, it also follows that this method of drawing <span class="math-container">$n$</span> should also satisfy URD. </p>
<p>But under this last method, <span class="math-container">$prob(\texttt{n is even}) = prob(\texttt{n not odd}) = 1 - prob(a1 = 0) = 1 - 0 = 1$</span></p>
<p>If you want to be more formal, you can note the following observation: suppose you have <span class="math-container">$N$</span> (non-empty) sets <span class="math-container">$S_1, S_2, ...., S_N$</span>. We want to choose a random element <span class="math-container">$X$</span> from the cartesian product set <span class="math-container">$S = S_1 \times S_2 \times S_3 .... \times S_N$</span>. Then drawing <span class="math-container">$X$</span> is equivalent to drawing <span class="math-container">$N$</span> elements <span class="math-container">$s_i$</span> from <span class="math-container">$S_i$</span>, and then setting <span class="math-container">$X = (s_1, s_2, ..., s_N)$</span>.
In effect, I've used this result in my argument above that <span class="math-container">$Pr(even) = 1$</span>, because I've quoted the fundamental theorem of arithmetic and defined <span class="math-container">$S_i = \{\texttt{powers of the }$</span>i<span class="math-container">$\texttt{th prime}\}$</span>, e.g. <span class="math-container">$S_2 = \{1, 3, 9, 27, 81, ....\}$</span>. </p>
<p>Since I've shown that Pr(even) = 1/2 and shown that Pr(even) = 1, it follows that it makes no sense to even assign a probability to drawing random numbers from the infinite set of natural numbers.</p>
<p>I assume I'm wrong somewhere, but I don't know where I've gone wrong?</p>
|
978,985 | <p>Why must every vector in V belongs to one of the generalised eigenspaces of $T: V \to V?$ Is there a simple proof for this? Can someone provide me with an intuition behind it?</p>
<p>Note that V is an algebraically closed field.</p>
| Yiorgos S. Smyrlis | 57,021 | <p>We have that
$$
S+T=T(I+T^{-1}S).
$$
Define
$$
K=\left(\sum_{n=0}^\infty (-1)^n (T^{-1}S)^n\right)T^{-1}.
$$
Then, the series above converges as $\|T^{-1}S\|<1$, and hence
$K$ is a bounded linear operator, and is readily shown that
$$
K(S+T)=I.$$</p>
|
978,985 | <p>Why must every vector in V belongs to one of the generalised eigenspaces of $T: V \to V?$ Is there a simple proof for this? Can someone provide me with an intuition behind it?</p>
<p>Note that V is an algebraically closed field.</p>
| BBVM | 291,058 | <p>If $X=Y$ in your question, we have
\begin{equation*}
S+T: X \rightarrow X.
\end{equation*}</p>
<p>$S+T$ is a injective map, because $(S+T)(x) = 0 \Rightarrow x= 0 $.</p>
<p>Now, we want to show that for all $y \in X$, exists $x \in X$ such that $(S+T)(x)=y$. See that,
\begin{equation*}
(S+T)(x) = y \Leftrightarrow S(x) +T(x) =y \Leftrightarrow T^{-1}(y) - T^{-1}S(x) = x
\end{equation*}
Set, $A(x) = T^{-1}(y) - T^{-1}S(x)$. It's easy to show that $A$ is a contraction, then $S+T$ is surjective. </p>
|
1,496,676 | <p>let's $F(x)= x'Ax + a'x $</p>
<p>where $x'Ax$ is a quadratic form and $a'$ is defined as a vector. </p>
<p>$$A:= \left[ \begin{matrix} 6 &1&1 \\ 1&2&0 \\ 1&0&4\end{matrix}\right] $$</p>
<p>Does there exist a global minimum point (absolute min) for this function $F(x)$?</p>
<p>note: I know that if the function is given as a linear structure, whether this is convex or concave is not important. we only exemine $x'Ax$. </p>
<p>but i cannot perfectly prove this question. thank you for helping. </p>
| xjtein | 241,759 | <p>A is symmetric and we can make it diagonal using an orthogonal matrix. Then we convert our problem into the form:
$$f(x)=\lambda_1 x^2+\lambda_2 y^2+\lambda_3 z^2+ a x+by +c z$$, which is high school math.
We find the eigenvalues of A are about $\{6.6,3.6,1.7\}$, therefore there exists minimum value of f.</p>
|
1,857,630 | <p>Let $\mathcal{A}$ be a non (necessarily) unital commutative Banach algebra, and let
$$ M_{\mathcal{A}} = \{ \phi:\mathcal{A} \to \mathbb{C} : \phi \mbox{ is multiplicative and not trivial}\} $$
and
$$ \mathrm{Max}(\mathcal{A})=\{ I \lhd \mathcal{A} : I \mbox{ maximal} \}.$$
If $\mathcal{A}$ is unital, it is well known that there is a bijection between $M_{\mathcal{A}}$ and $\mathrm{Max}(\mathcal{A})$ sending each functional to its kernel (the inverse is given by the quotient and the Gelfand-Mazur theorem). </p>
<p>My question is, </p>
<blockquote>
<p>is this still a bijection in the non-unital case?</p>
</blockquote>
<p>I'm aware that if $\mathcal{A}$ is a commutative C*-algebra it is still a bijection. Also that the restriction gives a bijection from $M_{\tilde{\mathcal{A}}} \setminus \{ \pi:\tilde{\mathcal{A}} \to \mathbb{C} \}$ to $M_{\mathcal{A}}$; but this fact don't seem enough to conclude the result. I haven't been able to find a source for this. </p>
<p>Thanks in advance.</p>
| Jonas Meyer | 1,424 | <p>The kernels of nonzero homomorphisms to $\mathbb C$ are <em>modular</em> ideals, terminology that might help you find more references.</p>
<p>Without any further restriction on the algebras, using the zero product is a way to provide trivial counterexamples. E.g., take $\mathbb C$ with the $0$ product, which has maximal ideal $\{0\}$ and no nonzero homomorphisms to $\mathbb C$.</p>
<p>Googling led me to the following maybe more interesting example, Example 1.3 in <a href="http://arxiv.org/abs/math/9909149" rel="noreferrer">this Feinstein and Somerset article</a>: Take $C[0,1]$ with its usual Banach space structure but with multiplication $(f\diamond g)(t) = f(t)g(t)t$. Then the ideal of functions vanishing at $0$ is maximal but not the kernel of a nonzero homomorphism to $\mathbb C$. </p>
|
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