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 |
|---|---|---|---|---|
366,311 | <blockquote>
<p>Show that the sequence $\displaystyle (x_n)=\left( \sum_{i=1}^n\frac 1 i\right)$ diverge by epsilon delta definition.</p>
</blockquote>
<p>I'm not familiar with proving divergent sequence. Do anyone have any des? Thank you.</p>
| Warren Moore | 63,412 | <p>Are you familiar with the comparison test?</p>
<blockquote>
<p>Let $\sum x_n$ and $\sum y_n$ be series with $x_n\ge y_n\ge 0$ for sufficiently large $n$. Then if $\sum y_n$ diverges, $\sum x_n$ diverges.</p>
</blockquote>
<p>So you just need to compare it with a divergent series from below. </p>
|
34,215 | <p>How do professional mathematicians learn new things? How do they expand their comfort zone? By talking to colleagues? </p>
| Peter Shor | 2,294 | <p>Try to solve a famous open problem in the new field. Even though you'll almost certainly fail, you'll learn a lot of new mathematics on the way.</p>
|
188,087 | <p>Is there a function that can extract a list of variables in an expression?
For example, assume we have an expression</p>
<pre><code>x^2+y^3+z
</code></pre>
<p>This expression has variables x, y and z. The result should be</p>
<pre><code>{x, y, z}
</code></pre>
<p>. Is there a way to get this?</p>
| user49047 | 49,047 | <p>Listing the expression might be an alternative for this simple example</p>
<pre><code>List @@ (x^8 + y^3 + z) /. a_Symbol^n_ -> a
</code></pre>
|
1,858,529 | <p>I know, there are some threads dealing with this sum but I want to solve it with the integral test for convergence(<a href="https://en.wikipedia.org/wiki/Integral_test_for_convergence" rel="nofollow">more</a>)</p>
<blockquote>
<p>$$\sum\limits_{n=3}^{\infty} \frac{1}{n\log(n)\log(\log(n))}$$</p>
</blockquote>
<p>I can't find the right substitution here:
$$\int\limits_3^{\infty} \frac{1}{x\log(x)\log(\log(x))}dx$$ I used $t=\log(x)$ but it doesn't work. Any hints?</p>
| Jack D'Aurizio | 44,121 | <p>Just apply <a href="https://en.wikipedia.org/wiki/Cauchy_condensation_test" rel="nofollow noreferrer">Cauchy's condensation test</a> twice to get that the series is divergent.</p>
<p>As an alternative, the substitution <span class="math-container">$x=\exp(\exp t)$</span> gives:
<span class="math-container">$$ \int_{e^e}^{\exp(e^M)}\frac{dx}{x\log(x)\log\log(x)} = \log M,$$</span></p>
<p>hence <span class="math-container">$\sum_{n=3}^{N}\frac{1}{n\log(n)\log\log(n)}$</span> diverges like <span class="math-container">$\log\log\log N$</span>, i.e. extremely slow, but still.</p>
|
1,319,476 | <p>This is a question related to another posted question:</p>
<p>The answer to the following question "Find all solutions to: $e^{ix}=i$" is as follows: </p>
<p>"Euler's formula: $e^{ix}=\cos(x)+i\sin(x)$,</p>
<p>so: $ \cos x+i\sin x=0+1⋅i$</p>
<p>compare real and imaginary parts
$\sin(x)=1$
and
$\cos(x)=0$</p>
<p>$x=\frac{(4n+1)π}2$, $n∈$
(W stands for set of whole number W={0,1,2,3,.......,n})."</p>
<p>My question: Where does $x=\frac{(4n+1)π}2$, $n∈$ come from? </p>
<p>My steps: </p>
<ol>
<li><p>$\cos(x) + i\sin(x) = 0 + i(1)$</p></li>
<li><p>$\cos(x) = i(1 - \sin(x))$</p></li>
<li><p>... </p></li>
<li><p>how does $x=\frac{(4n+1)π}2$ follow? </p></li>
</ol>
| Community | -1 | <p>You said it, you compare real and imaginary parts. $\cos(x)=0$ and $\sin(x)=1$. When is this true?</p>
|
1,272,499 | <p>Definite integral of $$\int_0^{2\pi} \frac{1}{2+\cos x}$$ without using improper integral, I want to solve this without having to use $-\infty$ and $\infty$ on the integrals limits. Is that possible?</p>
<p>The only way I can think of solving that is by using Weierstrass. $u = \tan \frac{x}{2}$, don't you have to modify the lower and upper limits with that substitution? The only way I can think of to progress in this is to change the limits to $-\pi$ and $\pi$, but when you'll get $-\infty$ and $\infty$ upper and lower limits.</p>
| Mark Viola | 218,419 | <p>Another way of evaluating this integral is to use contour integration. </p>
<p>Let $z=e^{i\theta}$ so that $d\theta =dz/(iz)$, $\cos \theta =\frac12 (z+z^{-1})$ where $0\le \theta \le 2\pi$, and let $C$ be the unit circle $|z|=1$. </p>
<p>Then, </p>
<p>$$\begin{align}
I&=\int_0^{2\pi}\frac{1}{2+\cos \theta}d\theta\\\\
&=\oint_C \frac{1}{2+\frac12(z+z^{-1})}\frac{1}{iz}dz\\\\
&=\frac{2}{i}\oint_C \frac{1}{z^2+4z+1}dz\\\\
&=\frac{2}{i}2\pi i \text{Res}\left(\frac{1}{z^2+4z+1},z=-2+\sqrt
{3}\right)\\\\
&=2\pi\sqrt{3}/3
\end{align}$$</p>
|
207,418 | <p>The TAs in my department are stuck in assisting an undergraduate with the following problem:</p>
<p>$$\sum^{2k}_{i=0} C^{4k}_{2i}(-1)^{i}=2^{2k}(-1)^{k}.$$</p>
<p>We tried to solve this via induction (obviously failed), via various combinatorial identities, via generating functions, etc. Aside from the fact that nothing works, we also do not know how to solve this nicely by using some combinatorial interpretation given this is some undergraudate's HW. I venture to ask in here for I want to see how it can be properly understood. </p>
| Qiaochu Yuan | 232 | <p>I'll rewrite the identity as</p>
<p>$$\sum_{k=0}^{2n} {4n \choose 2k} (-1)^k = 2^{2n} (-1)^n$$</p>
<p>because I'm about to use $i$ for something else. Write this as</p>
<p>$$\sum_{k=0}^{4n} {4n \choose k} a_k = 4^n (-1)^n$$</p>
<p>where $a_k$ is the sequence of period $4$ which is equal to $0$ when $k$ is odd and which satisfies $a_{2k} = (-1)^k$.</p>
<p><strong>Claim:</strong> $$a_k = \frac{i^k + (-i)^k}{2}.$$</p>
<p>This is essentially an application of the <a href="http://en.wikipedia.org/wiki/Discrete_Fourier_transform">discrete Fourier transform</a>. In the high school competition circuit it is sometimes known as the "roots of unity filter." </p>
<p><strong>Corollary:</strong> The LHS of the identity evaluates to </p>
<p>$$\frac{(1 + i)^{4n} + (1 - i)^{4n}}{2}.$$</p>
<p>Since $(1 \pm i)^2 = \pm 2i$ we have $(1 \pm i)^4 = -4$, hence the LHS of the identity evaluates to $(-4)^n$ as desired. </p>
|
207,418 | <p>The TAs in my department are stuck in assisting an undergraduate with the following problem:</p>
<p>$$\sum^{2k}_{i=0} C^{4k}_{2i}(-1)^{i}=2^{2k}(-1)^{k}.$$</p>
<p>We tried to solve this via induction (obviously failed), via various combinatorial identities, via generating functions, etc. Aside from the fact that nothing works, we also do not know how to solve this nicely by using some combinatorial interpretation given this is some undergraudate's HW. I venture to ask in here for I want to see how it can be properly understood. </p>
| Brian M. Scott | 12,042 | <p>It can be proved by induction on $k$:</p>
<p>$$\begin{align*}
\sum^{2k}_{i=0}\binom{4k}{2i}(-1)^{i}&=\sum_{i=0}^{2k}\left(\sum_{j=0}^4\binom4j\binom{4k-4}{2i-j}\right)(-1)^i\\
&=\sum_{j=0}^4\binom4j\sum_{i=0}^{2k}\binom{4k-4}{2i-j}(-1)^i\\
&=\sum_{i=0}^{2k}\binom{4k-4}{2i}(-1)^i+4\sum_{i=0}^{2k}\binom{4k-4}{2i-1}(-1)^i+6\sum_{i=0}^{2k}\binom{4k-4}{2i-2}(-1)^i\\
&\qquad\qquad+4\sum_{i=0}^{2k}\binom{4k-4}{2i-3}(-1)^i+\sum_{i=0}^{2k}\binom{4k-4}{2i-4}(-1)^i\\
&=\sum_{i=0}^{2k-2}\binom{4k-4}{2i}(-1)^i+6\sum_{i=1}^{2k-1}\binom{4k-4}{2i-2}(-1)^i+\sum_{i=2}^{2k}\binom{4k-4}{2i-4}(-1)^i\\
&\qquad\qquad+4\sum_{i=1}^{2k-2}\binom{4k-4}{2i-1}(-1)^i+4\sum_{i=2}^{2k-1}\binom{4k-4}{2i-3}(-1)^i\\
&=\sum_{i=0}^{2k-2}\binom{4k-4}{2i}(-1)^i+6\sum_{i=0}^{2k-2}\binom{4k-4}{2i}(-1)^{i+1}+\sum_{i=0}^{2k-2}\binom{4k-4}{2i}(-1)^i\\
&\qquad\qquad+4\sum_{i=1}^{2k-2}\binom{4k-4}{2i-1}(-1)^i+4\sum_{i=1}^{2k-2}\binom{4k-4}{2i-1}(-1)^{i+1}\\
&=-4\sum_{i=0}^{2k-2}\binom{4k-4}{2i}(-1)^i\\
&=-4(-4)^{k-1}\\
&=(-4)^k\;.
\end{align*}$$</p>
|
207,418 | <p>The TAs in my department are stuck in assisting an undergraduate with the following problem:</p>
<p>$$\sum^{2k}_{i=0} C^{4k}_{2i}(-1)^{i}=2^{2k}(-1)^{k}.$$</p>
<p>We tried to solve this via induction (obviously failed), via various combinatorial identities, via generating functions, etc. Aside from the fact that nothing works, we also do not know how to solve this nicely by using some combinatorial interpretation given this is some undergraudate's HW. I venture to ask in here for I want to see how it can be properly understood. </p>
| Mike Spivey | 2,370 | <p>Here's a short proof.</p>
<p>$$\sum_{k=0}^{4n} \binom{4n}{k} i^k = (1+i)^{4n} = \left(\sqrt{2} e^{i \pi /4}\right)^{4n} = 4^n e^{i n \pi} = 4^n \left(\cos (n \pi) + i\sin (n \pi)\right) = (-4)^n.$$</p>
<p>Equating real and imaginary parts yields both the OP's identity</p>
<p>$$\sum_{k=0}^{2n} \binom{4n}{2k} (-1)^k = (-4)^n$$
as well as the bonus identity
$$\sum_{k=0}^{2n-1} \binom{4n}{2k+1} (-1)^k = 0.$$</p>
<p>The idea here is of course similar to that in Qiaochu Yuan's answer.</p>
|
23,020 | <p>I am in the process of learning about Mapping class groups. At this point it seems like most of what I've read involves very low dimensional (surfaces and 3-manifolds) applications.</p>
<p>I was wondering if they were studied (or arise naturally) in higher dimensional settings?</p>
<p>In particular, any references to their uses in homotopy theory would be appreciated.</p>
| Zoltan Zimboras | 7,317 | <p>As you mentioned, people have been studying the mapping class groups mostly in connection with (two-dimensional) surfaces, and also some work have been done for 3-manifolds - here the most notable result is Kojima's theorem which states that every finite group is the mapping class group of a compact hyperbolic 3-manifold.</p>
<p>I know only a very few cases, where mapping class groups appear in higher dimensions:</p>
<p>One is a recent generalization of Kojima's theorem for higher dimensional hyperbolic manifolds by Belolipetsky and Lubotsky. See: M. Belolipetsky and A. Lubotzky, Finite Groups and Hyperbolic Manifolds, Invent. Math. 162 (2005) 459–472.</p>
<p>Another one is the use of mapping class groups (of 2d surfaces) in the study of symplectic 4-manifolds (and even 6-manifolds) in <a href="http://math.berkeley.edu/~auroux/papers/mcg-farb.pdf" rel="noreferrer">http://math.berkeley.edu/~auroux/papers/mcg-farb.pdf</a></p>
<p>All the best,
Zoltan</p>
|
3,242,553 | <p>I got two sequences of stochastic process <span class="math-container">$(X_n(t))_{t \in [0,1]}$</span> and <span class="math-container">$(Y_{n}(t))_{t \in [0,1]}$</span>, defined on a probability space <span class="math-container">$(\Omega, \mathcal{F},P)$</span>, and know that their distance in the sup-norm on <span class="math-container">$[0,1]$</span> converges to <span class="math-container">$0$</span> almost surely, i.e.</p>
<p><span class="math-container">$\sup \limits_{t \in [0,1]} \vert X_n(t) - Y_n(t) \vert \to 0 \quad P-a.s., \quad n \to \infty$</span>, </p>
<p>or equivalently</p>
<p><span class="math-container">$P \left (\lim \limits_{n \to \infty} \sup \limits_{t \in [0,1]} \vert X_n(t) - Y_n(t) \vert = 0 \right ) = 1$</span>.</p>
<p>Now I'm wondering if this also implies the (pointwise) convergence in distribution of <span class="math-container">$X_n$</span> to <span class="math-container">$Y_n$</span>. This result seems very intuitive, but how does one formally show this?</p>
<p>Thanks!</p>
| gourab das | 582,064 | <p>One way of interpreting the convergence of a sequence Xn to X is to say that the ''distance'' between X and Xn is getting smaller and smaller. For example, if we define the distance between Xn and X as P(|Xn−X|≥ϵ), we have convergence in probability. One way to define the distance between Xn and X is</p>
<p>E(|Xn−X|r),
where r≥1 is a fixed number. This refers to convergence in mean. (Note: for convergence in mean, it is usually required that E|Xrn|<∞.) The most common choice is r=2, in which case it is called the mean-square convergence. (Note: Some authors refer to the case r=1 as convergence in mean.)</p>
<p>Convergence in Mean</p>
<p>Let r≥1 be a fixed number. A sequence of random variables X1, X2, X3, ⋯ converges in the rth mean or in the Lr norm to a random variable X, shown by Xn −→Lr X, if
limn→∞E(|Xn−X|r)=0.
If r=2, it is called the mean-square convergence, and it is shown by Xn −→−m.s. X.</p>
|
2,624,837 | <p>What is the integral of $$\int a^{x-1}dx?$$</p>
<p>is it $$\frac{a^{x-1}}{\log(a)} + c?$$</p>
<p>How can we derive the proper integral? Also can you please tell me the definite integral with limits, say b to c?</p>
| Peter Szilas | 408,605 | <p>Let $a>0$, $a,x$ real:</p>
<p>$a^{x-1} =\exp(\ln(a^{x-1}))= $</p>
<p>$\exp((x-1)\ln(a))= \exp(c(x-1))$, </p>
<p>where $c :=\ln(a).$</p>
<p>Use substitution: </p>
<p>$y=c (x-1)$ to integrate $\exp(y).$</p>
|
2,476,181 | <p>Where Ω = {1,2,...,p}, all Ω are equally likely and p is prime how would I show that if A and B are independent events then at least one of A and B is either ∅ or Ω?</p>
| XRBtoTheMOON | 478,087 | <p>You've got 13 letters. Choose 9 of them to make your consonants. Then you've got 21 consonants for the first, 20 for second, etc. Then 5 vowels to put in each of the other spots. So $\binom{13}{9}\cdot 21^{\underline{9}} \cdot 5^4$</p>
|
2,197,790 | <h3>Question</h3>
<blockquote>
<p>A sequence $\{a_n\}$ of real numbers is said to be a Cauchy sequence of for
each $\epsilon$ > 0 there exists a number $N > 0$ such that m, $n > N$ implies
that $|a_n − a_m| <\epsilon$.</p>
<p>Prove that every convergent sequence is a Cauchy sequence</p>
</blockquote>
<hr>
<h3>Attempt</h3>
<p>This is my first time hearing what a cauchy sequence is. I have no idea how to even start this. I googled cauchy sequence and I think its when $a_n$ converges to $a_{n+1}$? </p>
<p>Attempt:</p>
<p>WTS: $\exists a_m \in \mathbb R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in \mathbb N$, if $n > N$, then $|a_n - a_m| < \epsilon$</p>
<p>Let $\epsilon > 0$ be arbitrary</p>
<p>Choose N such that for $n > N$ we have $|a_n - a_m| < \epsilon$</p>
<p>Suppose $n > N$, then </p>
<p>??</p>
<p>Could someone point me to the right direction? Thx.</p>
| Jonathan Barkey | 414,649 | <p>A way to do this is to notice that since the sequence $(x_n)$ is convergent, then a tail of that sequence, let's call it $(x_m)$ is also convergent. That is, for all $\epsilon_1\gt0$, there exists $N\in\mathbb{N}$ such that for all $n\ge N$, $|x_n-x|\lt\epsilon_1$. Similarly, for all $\epsilon_2\gt0$, there exists $N\in\mathbb{N}$ such that for all $m\ge N$, $|x_m-x|\lt\epsilon_2$ for all $m\ge N\in\mathbb{N}$.</p>
<p>Do you know the triangle inequality? If so, you can relate $|x_n-x_m|$, $|x_n-x|\lt\epsilon_1$, and $|x_m-x|\lt\epsilon_2$ such that $\epsilon_1+\epsilon_2=\epsilon$ (a lot of examples you may see in the future use $\epsilon/2$ for these), showing that all convergent sequences are Cauchy.</p>
|
1,957,453 | <p>Can you please help me with this question?</p>
<p>Question: Find the radius of curvature, and the equation of the osculating circle, for the following curve for <span class="math-container">$t\geq0$</span>.</p>
<p><span class="math-container">$r(t) = \sin(\sqrt{e^t+1}) \hat{i} - \cos(\sqrt{e^t+1}) \hat{j} + 0 \hat{k}$</span></p>
<p>Attempt:
v = (e^t cos(sqrt(e^t+1))/(2sqrt(e^t+1) i + (e^t sin(sqrt(e^t+1))/(2sqrt(e^t+1)j
|v| = 0.5e^t sqrt(1/e^t+1)</p>
<p>I am not able to calculate the curvature and the equation of the osculating circle</p>
| June Mar Fajardo | 210,883 | <p>the curvature $\kappa(t)$ is $$\dfrac{||\vec{r}'(t)\times\vec{r}''(t)||}{||\vec{r}'(t)||^3}$$ so the radius $\rho$ is $1/\kappa.$ For the equation of osculating circle, note that the vector, with terminal point equal to the center of the osculating circle and initial point equal to the point $\vec{r}(t)$, is parallel to the unit normal vector $\vec{N}(t)$ of $\vec{r}(t).$ In other words, if $(a,b,c)$ is the center of the osculating circle, then $\langle a,b,c\rangle - \vec{r}(t)=\rho\vec{N}(t).$ Thus, once you know the center and the radius of the circle, you may construct the parametric equations of the osculating circle.</p>
|
4,638,170 | <p>I'm trying to write a proof to show that a tree structure of finite nodes terminate.</p>
<p>Suppose we can say that either a node is a parent of another node (<span class="math-container">$Pqp$</span>: <span class="math-container">$q$</span> is the parent of <span class="math-container">$p$</span>), or it is a terminal node (<span class="math-container">$Tp$</span>: <span class="math-container">$p$</span> does not have a parent).</p>
<ol>
<li><span class="math-container">$\forall_p \space Tp \otimes \exists_q Pqp$</span></li>
</ol>
<p>Every node that has a parent only has one parent.</p>
<ol start="2">
<li><span class="math-container">$\forall_{pqr} \space (Pqp \space \land \space Prp) \rightarrow q=r$</span></li>
</ol>
<p>We say that if node <span class="math-container">$q$</span> comes before node <span class="math-container">$p$</span> (<span class="math-container">$Bqp$</span>), then any node that comes before <span class="math-container">$q$</span> also comes before <span class="math-container">$p$</span>. Additionally, we deny that <span class="math-container">$p$</span> comes before <span class="math-container">$q$</span>. This avoids cyclical chains.</p>
<ol start="3">
<li><span class="math-container">$\forall_{pqr} \space Bqp \rightarrow ((Brq \rightarrow Brp) \space \land \space \lnot Bpq)$</span></li>
</ol>
<p>Any node that is the parent of a nother node must also come before it.</p>
<ol start="4">
<li><span class="math-container">$\forall_{pq} \space Ppq \rightarrow Bpq$</span></li>
</ol>
<p>Suppose we know that at least one node exists.</p>
<ol start="5">
<li><span class="math-container">$\exists p$</span></li>
</ol>
<p>I want to conclude that there must be a terminal node <span class="math-container">$\therefore \exists_p \space Tp$</span> however, one can always find a counter-example using premise 1 in that <span class="math-container">$p$</span> has a parent -- resulting in an infinite regress.</p>
<p>(tl;dr)</p>
<p>So my struggle is, I want to formalise the assumption that there are finite nodes. In all of the definitions of a finite set that I've seen, you need to specify some maximum N. In this case, I want to be intentionally vague about it -- only claiming that it's finite; not how it's finite.</p>
<p>An idea I had was to assume that the domain of discourse is positive integers with something like <span class="math-container">$N\in \mathbb{N_1}$</span> and <span class="math-container">$\forall_p \space p < N$</span> but something about this solution feels kind of hacky.</p>
<p>So is there a better way to do this and, if so what is it?</p>
| Paul Frost | 349,785 | <p>If we know that <span class="math-container">$U$</span> is both open and closed, we also know that <span class="math-container">$V = X \setminus U$</span> is open. You have <span class="math-container">$a \in U$</span>, thus <span class="math-container">$U \ne \emptyset$</span>. Since <span class="math-container">$U \cup V = X$</span> and <span class="math-container">$U \cap V = \emptyset$</span>, we conclude that <span class="math-container">$V = \emptyset$</span> because otherwise <span class="math-container">$X$</span> would not be connected. Thus <span class="math-container">$U = X$</span> which means that <span class="math-container">$X$</span> is path-connected because any point <span class="math-container">$x \in X$</span> admits a path connecting it with <span class="math-container">$a$</span>.</p>
<p>Note that the above argument shows that a space <span class="math-container">$X$</span> is connected iff the only clopen (= open and closed) subsets of <span class="math-container">$X$</span> are <span class="math-container">$X, \emptyset$</span>.</p>
<p>Let us now prove that <span class="math-container">$U$</span> is clopen.</p>
<ol>
<li><p>Let <span class="math-container">$x \in U$</span>. There exists a path-connected open neigborhood <span class="math-container">$U(x)$</span> of <span class="math-container">$x$</span> in <span class="math-container">$X$</span>. Clearly <span class="math-container">$U(x) \subset U$</span> since for each <span class="math-container">$y \in U(x)$</span> there exists a path in <span class="math-container">$X$</span> from <span class="math-container">$a$</span> to <span class="math-container">$x$</span> and a path in <span class="math-container">$U(x)$</span> from <span class="math-container">$x$</span> to <span class="math-container">$y$</span>, thus a path in <span class="math-container">$X$</span> from <span class="math-container">$a$</span> to <span class="math-container">$y$</span>.</p>
</li>
<li><p>Let <span class="math-container">$x \in X \setminus U$</span>. Again there exists a path-connected open neigborhood <span class="math-container">$U(x)$</span> of <span class="math-container">$x$</span> in <span class="math-container">$X$</span>. Assume <span class="math-container">$U(x) \cap U \ne \emptyset$</span>. Pick <span class="math-container">$y \in U(x) \cap U$</span>. There exists a path in <span class="math-container">$X$</span> from <span class="math-container">$a$</span> to <span class="math-container">$y$</span> and a path in <span class="math-container">$U(x)$</span> from <span class="math-container">$y$</span> to <span class="math-container">$x$</span>, thus a path in <span class="math-container">$X$</span> from <span class="math-container">$a$</span> to <span class="math-container">$x$</span> which means <span class="math-container">$x \in U$</span>, a contradiction. Thus <span class="math-container">$U(x) \subset X \setminus U$</span>.</p>
</li>
</ol>
<p>Note that the assumption of local path-connectedness can be weakened to prove that <span class="math-container">$X$</span> is path connected. It suffices to require that each <span class="math-container">$x \in X$</span> has an open neigborhood <span class="math-container">$U(x)$</span> such that for each <span class="math-container">$y \in U(x)$</span> there exists a path in <span class="math-container">$X$</span> from <span class="math-container">$x$</span> to <span class="math-container">$y$</span>.</p>
|
3,853,723 | <p>While brushing up on my old discrete mathematics skills I stumbled upon this problem that I can't solve.</p>
<p>In <span class="math-container">$\mathbb{R^2}$</span> the middle point of two coordinates is <span class="math-container">$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$</span>. Show that given five points in <span class="math-container">$\mathbb{Z^2}$</span> (points with intetger coordinates) there are at least one pair of them whose middle point also lays in <span class="math-container">$\mathbb{Z^2}$</span>. Let us consider the corresponding question in <span class="math-container">$\mathbb{R^3}$</span>. How many points in <span class="math-container">$\mathbb{Z^3}$</span> would you need to be sure that at least one pair of them has a middle point in <span class="math-container">$\mathbb{Z^3}$</span>?</p>
<p>I am thinking that using the <strong>pigeonhole principle</strong> is appropriate, how I would use it is unclear to me though.</p>
<p><a href="https://i.stack.imgur.com/g4WxU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g4WxU.png" alt="points in <span class="math-container">$\mathbb{Z^2}$</span>" /></a></p>
| Giovanni Barbarani | 444,892 | <p>EDIT: as pointed out in the comments the approch is wrong but I still think it is a bit valuable so I'm not deleting it for now.</p>
<p>I'll try. <span class="math-container">$B$</span> is a <span class="math-container">$k$</span>-algebra hence there exists a ring homorphism <span class="math-container">$T:k\rightarrow B$</span> such that <span class="math-container">$T(k)$</span> is in the center of <span class="math-container">$B$</span>. Given a matrix <span class="math-container">$M \in M_n(k)$</span> let me extend the notation naming <span class="math-container">$T(M) \in M_n(B)$</span> the matrix obtained applying <span class="math-container">$T$</span> elementwise.</p>
<p>Consider the following map
<span class="math-container">\begin{gather*}
\phi:M_n(k) \otimes_kB \longrightarrow M_n(B)\\
\phi(M \otimes b) = b \ T(M)
\end{gather*}</span></p>
<p>The map such defined is a <span class="math-container">$k$</span>-module homomorphism. Considering that the elements <span class="math-container">$E_{ij}\otimes b$</span> are mapped into the elements <span class="math-container">$bE_{ij}$</span> of <span class="math-container">$M_n(B)$</span> we can conclude that the map is surjective.</p>
<p>Now suppose two elements have the same image, so</p>
<p><span class="math-container">\begin{gather*}
\phi(M_1 \otimes b_1) = \phi(M_2 \otimes b_2) \\
b_1 T(M_1) = b_2 T(M_2)
\end{gather*}</span></p>
<p>This means</p>
<p><span class="math-container">\begin{gather*}
\forall(i,j) \ \ m_{ij}^1 b_1 = m_{ij}^2 b_2 \\
\forall(i,j) \ \ b_2 = (m_{ij}^2)^{-1}m^1_{ij} b_1 \\
\end{gather*}</span></p>
<p>So for each <span class="math-container">$(i,j)$</span> the ratio <span class="math-container">$r=(m_{ij}^2)^{-1}m^1_{ij}$</span> is constant, and we have <span class="math-container">$b_2 = rb_1$</span>, so</p>
<p><span class="math-container">\begin{gather*}
M_2 \otimes b_2 = M_2 \otimes rb_1 = rM_2 \otimes b_1 = M_1 \otimes b_1
\end{gather*}</span></p>
<p>demonstrating that <span class="math-container">$\phi$</span> is an injection on simple tensor.</p>
|
1,497,495 | <p>How would I solve this linear system equation?</p>
<p>$$\begin{cases}
2w+x-y=4\\
3z-x=6\\
-2y-x+9z+4w=7
\end{cases}$$</p>
<p>First I arranged them so I could make the metric and then I was stuck and I don't realy know how to continue, please help me.</p>
| Jon | 273,053 | <p>Try multiplying the first equation by $2$, and the second one by $3$:
\begin{cases}
4w+2x-2y=8\\
9z-3x=18\\
-2y-x+9z+4w=7
\end{cases}
Then,
\begin{cases}
4w-2y=8-2x\\
9z=18+3x\\
4w-2y+9z=7+x
\end{cases}</p>
<p>And the rest is by positioning the elements from the first two equations in the third one</p>
|
1,497,495 | <p>How would I solve this linear system equation?</p>
<p>$$\begin{cases}
2w+x-y=4\\
3z-x=6\\
-2y-x+9z+4w=7
\end{cases}$$</p>
<p>First I arranged them so I could make the metric and then I was stuck and I don't realy know how to continue, please help me.</p>
| egreg | 62,967 | <p>Write your system into normal form:
\begin{cases}
x-y+2w=4\\
x-3z=-6\\
x+2y-9z-4w=-7
\end{cases}</p>
<p>Now, depending on the tools you have available, there are several possibilities. The most efficient is Gaussian elimination: the matrix of the system is
\begin{align}
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
1 & 0 & -3 & 0 & -6 \\
1 & 2 & -9 & -4 & -7
\end{array}\right]
&\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
0 & 1 & -3 & -2 & -10 \\
1 & 2 & -9 & -4 & -7
\end{array}\right] && R_2\gets R_2-R_1 \\[6px]
&\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
0 & 1 & -3 & -2 & -10 \\
0 & 3 & -9 & -6 & -11
\end{array}\right] && R_3\gets R_3-R_1 \\[6px]
&\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
0 & 1 & -3 & -2 & -10 \\
0 & 0 & 0 & 0 & 19
\end{array}\right] && R_3\gets R_3-3R_2 \\[6px]
\end{align}
which shows the system has no solution.</p>
<p>Another possibility is to get $x$ from the second equation: $x=3z-6$. Substitute in the third equation to get
$$
3z-6+2y-9z-4w=-7
$$
or
$$
2y-6z-4w=-1
$$
Substitute also in the first equation to find $3z-6-y+2w=4$, so you get
$$
y=3z+2w-10
$$
and now back in the other equation
$$
6z+4w-6z-4w-20=-1
$$
or
$$
-20=-1
$$
that is false.</p>
|
1,497,495 | <p>How would I solve this linear system equation?</p>
<p>$$\begin{cases}
2w+x-y=4\\
3z-x=6\\
-2y-x+9z+4w=7
\end{cases}$$</p>
<p>First I arranged them so I could make the metric and then I was stuck and I don't realy know how to continue, please help me.</p>
| k170 | 161,538 | <p>$$2w+x-y=4$$
$$3z-x=6$$
$$-2y-x+9z+4w=7$$
Can be rewritten to
$$2w+x-y+0z=4$$
$$0w-x+0y+3z=6$$
$$4w-x-2y+9z=7$$
$$0w+0x+0y+0z=0$$
So now we have
$$\left[\begin{array}{cccc|c}
2 & 1 & -1 & 0 & 4 \\
0 & -1 & 0 & 3 & 6 \\
4 & -1 & -2 & 9 & 7 \\
0 & 0 & 0 & 0 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{cccc|c}
2 & 1 & -1 & 0 & 4 \\
0 & -1 & 0 & 3 & 6 \\
0 & 3 & 0 & -9 & 1 \\
0 & 0 & 0 & 0 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{cccc|c}
2 & 1 & -1 & 0 & 4 \\
0 & -1 & 0 & 3 & 6 \\
0 & 0 & 0 & 0 & 19 \\
0 & 0 & 0 & 0 & 0 \\
\end{array}\right]$$
This system is inconsistent because we have
$$0w+0x+0y+0z=19$$
Therefore this system has no solutions.</p>
|
3,099,815 | <p>I need some help on how to approach this problem. I can't seem to find any examples that help me understand this, so if anyone has an approach example to post I would be very grateful:</p>
<p>"Consider a relation <span class="math-container">$R$</span> defined on the set of integers. Determine for the following if the relation is reflexive, symmetric, and transitive: <span class="math-container">$R = \{(x, y)|x = 2y \}.$</span>"</p>
| user247327 | 247,327 | <p>You need to think about the basic definitions. Any relation, x~ y, is "reflexive" if and only if x~ x is true for any x. It is "symmetric" if and only if x~ y implies y~ x. It is "transitive" if and only if x~ y and y~ z implies x~ z.</p>
<p>Reflexive: is it true that x~ x, that is, x= 2x, for any real number x?</p>
<p>Symmetric: Suppose x~ y. That is, x= 2y. Does it the follow that y~ x, that y= 2x?</p>
<p>Transitive: Suppose x~ y and y~ z. That is, x= 2y and y= 2z. Does it follow that x~ z, that x= 2z?</p>
|
3,099,815 | <p>I need some help on how to approach this problem. I can't seem to find any examples that help me understand this, so if anyone has an approach example to post I would be very grateful:</p>
<p>"Consider a relation <span class="math-container">$R$</span> defined on the set of integers. Determine for the following if the relation is reflexive, symmetric, and transitive: <span class="math-container">$R = \{(x, y)|x = 2y \}.$</span>"</p>
| Sujit Bhattacharyya | 524,692 | <p>I assume you have a clear concept about the definition of Relations.</p>
<p>As given, <span class="math-container">$R=\{(x,y) : x=2y\}$</span> i.e. in the set of integers the relation <span class="math-container">$R$</span> is defined as,</p>
<p>for <span class="math-container">$x,y\in\mathbb{Z}$</span>, <span class="math-container">$x R y \iff x=2y$</span>.</p>
<p>Now <span class="math-container">$R$</span> is not reflexive, as <span class="math-container">$1\ne2$</span> so, <span class="math-container">$1\not R 1$</span> i.e. <span class="math-container">$(1,1)\notin R$</span>.</p>
<p><span class="math-container">$R$</span> is not symmetric, as <span class="math-container">$2R1\iff 2=2\times1$</span> but <span class="math-container">$1\not R 2$</span> as <span class="math-container">$\exists$</span> no integer <span class="math-container">$n$</span> s.t. <span class="math-container">$1=2n$</span>. </p>
<p><span class="math-container">$R$</span> is not transitive, as <span class="math-container">$4R2$</span> and <span class="math-container">$8R4$</span> but <span class="math-container">$8\not R 2$</span>.</p>
|
42,301 | <p>everyone! I am sorry, but I am an abcolute novice of Mathematica (to be more precise this is my first day of using it) and even after surfing the web and all documents I am not able to solve the following system: </p>
<pre><code>Solve[{y*(((y*x)/(beta*b))^(1/(beta - 1)) - v) - c*alpha ==
0, ((x/alpha))*(((y*x)/(alpha*beta*b))^(1/(beta - 1)) -
v) + (((y*x)/alpha) -
2*alpha*((yx)/(2*beta*b))^(1/(beta - 1)))*(1/(beta -
1))*(x/(alpha*beta*
b))*((y*x)/(alpha*beta*b))^((2 - beta)/(1 - beta)) == 0}, {x,
y}]
</code></pre>
<p>What I need is to solve following systems, getting x and y expressed through all these symbols. Is it even possible? Thank you in advance. </p>
| MathBiolGuy | 12,187 | <p>Thanks everyone for the reply. Didn't expect such an overwhelming response. I did a quick check on the speed of each of the solutions by making a random list of 2x10^7 elements and compared the timing (given in bold) using the 4 solutions given by Yi Wang, halirutan and sakra: </p>
<pre><code>a = RandomInteger[1000, {2*10^7, 3}];
</code></pre>
<p>Method 1:</p>
<pre><code>findLastMin[mat_] := Cases[mat, {__, Min@Last@Transpose@mat}]
findLastMin[a] // Timing
</code></pre>
<blockquote>
<p>{<strong>8.020000</strong>, {{710, 337, 0}, {347, 509, 0}, <<19744>>, {609, 151, 0}, {553, 806, 0}}}</p>
</blockquote>
<p>Method 2:</p>
<pre><code>First[SortBy[a, Last]] // Timing
</code></pre>
<blockquote>
<p>{<strong>18.216000</strong>, {0, 28, 0}}</p>
</blockquote>
<p>Method 3:</p>
<pre><code>Block[{min = Min[Last[Transpose[a]]]},
Do[If[Last[elm] === min, Return[elm]], {elm, a}]] // Timing
</code></pre>
<blockquote>
<p>{<strong>2.536000</strong>, {710, 337, 0}}</p>
</blockquote>
<p>Method 4:</p>
<pre><code>Fold[If[Last[#2] < Last[#1], #2, #1] &, {0, 0, Infinity}, a] // Timing
</code></pre>
<blockquote>
<p>{<strong>29.132000</strong>, {710, 337, 0}}</p>
</blockquote>
<p>Method 1 gives all solutions and is fairly quick. Once again thanks for all the solutions.</p>
|
212,949 | <p>A simple question:</p>
<p>I have this equation:</p>
<pre><code>eq1=Derivative[0, 1][T1][x, t] - Derivative[1, 0][T0][x, t]^2 -
T0[x, t]*Derivative[2, 0][T0][x, t] - Derivative[2, 0][T1][x, t] == 0;
</code></pre>
<p>I want only to select terms that contain T0 or its derivatives only, that is:</p>
<pre><code>-Derivative[1, 0][T0][x, t]^2 - T0[x, t]*Derivative[2, 0][T0][x, t]
</code></pre>
<p>Thanks in anticipation.</p>
| kcr | 49,048 | <p>Not the most elegant solution, but you can use the <code>Collect</code>command in the following manner</p>
<pre><code>eq1 = Derivative[0, 1][T1][x, t] -
Derivative[1, 0][T0][x, t]^2 -
T0[x, t]*Derivative[2, 0][T0][x, t] -
Derivative[2, 0][T1][x, t];
(Coefficient[#1, {T0[x, t], Derivative[1, 0][T0][x,
t], Derivative[1, 0][T0][x, t]^2}] & )[eq1]
</code></pre>
<p>This is telling you the coefficient in front of <code>T0[x, t]</code> and so on. </p>
|
4,380,475 | <p>I'm trying to differentiate <span class="math-container">$x\sqrt{4-x^2}$</span> using the definition of derivative.</p>
<p>So it would be something like</p>
<p><span class="math-container">$$\underset{h\to 0}{\text{lim}}\frac{(h+x) \sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}$$</span></p>
<p>I was trying to solve and I just can end up with something like</p>
<p><span class="math-container">$$\underset{h\to 0}{\text{lim}}\frac{(x+h)\sqrt{4-x^2-2xh-h^2}-x\sqrt{4-x^2}}h \cdot \frac{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$</span></p>
<p><span class="math-container">$$\underset{h\to 0}{\text{lim}}\frac{-3x^2h-3xh^2+4h-h^3+\sqrt{4-x^2}-\sqrt{4-x^2-2xh+h^2}}{h\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$</span></p>
<p>Now if I group on h, I will have some tricky 3 instead of 2.
The idea is I should have something like <span class="math-container">$h(2x^2+4)$</span> that would cancel up.</p>
<p>I'm quite stuck can I ask a little of help? I know wolframalpha exists but it refuses to create the step by step solution with the error "Ops we don't have a step by step solution for this query".</p>
<p>The final result shall be
<span class="math-container">$$-\frac{2 \left(x^2-2\right)}{\sqrt{4-x^2}}$$</span></p>
| 2'5 9'2 | 11,123 | <p>I think your algebra could look more like:</p>
<p><span class="math-container">$$\begin{align}
&\frac{(x+h)\sqrt{4-(x+h)^2}-x\sqrt{4-x^2}}{h}\cdot\frac{(x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}}{(x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}}\\
&=\frac{(x+h)^2(4-(x+h)^2)-x^2(4-x^2)}{h\left((x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}\right)}\\
\end{align}$$</span></p>
<p>This leaves no radicals in the numerator.</p>
<p>In the numerator, once this is multiplied out, all <span class="math-container">$h$</span>-free terms will have canceled out through adding terms to their negatives.</p>
<p>Then you can factor <span class="math-container">$h$</span> from the top and cancel the <span class="math-container">$h$</span> in the denominator. Then it will be OK to just let <span class="math-container">$h\to0$</span>.</p>
|
3,966 | <p>This type of answer is what I'm looking for:</p>
<pre><code>In[58]:= ArcTan @ 1
Out[58]= π/4
</code></pre>
<p>This is what mathematica gives me:</p>
<pre><code>In[59]:= ArcTan@2
Out[59]= ArcTan[2]
</code></pre>
<p>Is it possible to express <code>ArcTan</code> in terms of $\pi$? I understand some fractions would be hairy.</p>
<p>I am using <code>Mathematica 8</code>.</p>
| F'x | 700 | <p>Short answer: <strong>no</strong>, <code>ArcTan[2]</code> is not fraction of $\pi$. But this is more of a mathematics question than pertaining to <em>Mathematica</em>.</p>
<p>If you want to “check” that the result is not expressable as a fraction of $\pi$, you can check for the continued fraction reprentation of $\arctan(2)/\pi$, and see that it does not seem to converge:</p>
<pre><code>Table[FromContinuedFraction@ContinuedFraction[ArcTan[2]/\[Pi], n], {n, 20}]
</code></pre>
<p><img src="https://i.stack.imgur.com/0bRHS.png" alt="enter image description here"></p>
|
14,458 | <p>I want to be able to plot several numerical solutions of an ODE, plus its analytical solution in one plot, in order to see how the numerical solutions converge towards the analytical one with respect to the number of steps. The method I'm using is Euler's method for the equation
$ y'(t) = 1-t +4y(t), y(0)=1$</p>
<p>The code I have so far is: </p>
<pre><code>y[0]=1;
Do[y[n+1]=y[n]+0.01(1-0.01n+4y[n]), {n,0,99}]
y[100]
</code></pre>
<p>Is this doable? Thanks in advance for any help :)</p>
| Sascha | 4,597 | <p>Instead of implementing an explicit Euler method on your own, you could as well use the built-in option of <code>NDSolve</code>: (the value following <code>StartingStepSize</code> is your actual step size for the whole method since <code>"ExplicitEuler"</code> has no adaptive step size):</p>
<pre><code>AnalyticalSolution = DSolve[{y'[t] == 1 - t + 4*y[t], y[0] == 1}, y, t]
NumericalSolution = NDSolve[{y'[t] == 1 - t + 4*y[t], y[0] == 1}, y, {t, 0, 10},
Method -> "ExplicitEuler", StartingStepSize -> 0.01]
Plot[{y[t] /. NumericalSolution, y[t] /. AnalyticalSolution}, {t, 0, 10}]
</code></pre>
|
469,485 | <p>Here's the simple question:</p>
<p>Devon has a piece of poster board 45 cm by 20 cm.
His teacher challenges him to cut the board into parts, then rearrange</p>
<p>the parts to form a square.
a) What is the side length of the</p>
<p>square?
b) What are the fewest cuts</p>
<p>Devon could have made? Explain.</p>
<p>I understand part a (the answer is 30 or the square root of 900) but how many cuts would he have made to make it a square? 45+20 = 65, 65 doesn't divide in 30 and if I try taking 45-15 and moving the 15 to the 20, I now get 35*30 which is not = 900 (1050)? How come then, 45*20 = 90 but when you displace 15, you get a greater answer? I'm sure the error is somewhere in my conversion between the rectangle and the square.</p>
| Ross Millikan | 1,827 | <p>Part b doesn't have a nice algorithmic solution that I know of. Clearly you can cut the board int $5 \times 5$ squares and rearrange them to make a $30 \times 30$ square. That is a lot of cuts. Many times the answer is a cut (not a single straight line) that is a stairstep, then you move the stairs one notch. It doesn't work here, so I would endorse qalpha's solution.</p>
|
469,485 | <p>Here's the simple question:</p>
<p>Devon has a piece of poster board 45 cm by 20 cm.
His teacher challenges him to cut the board into parts, then rearrange</p>
<p>the parts to form a square.
a) What is the side length of the</p>
<p>square?
b) What are the fewest cuts</p>
<p>Devon could have made? Explain.</p>
<p>I understand part a (the answer is 30 or the square root of 900) but how many cuts would he have made to make it a square? 45+20 = 65, 65 doesn't divide in 30 and if I try taking 45-15 and moving the 15 to the 20, I now get 35*30 which is not = 900 (1050)? How come then, 45*20 = 90 but when you displace 15, you get a greater answer? I'm sure the error is somewhere in my conversion between the rectangle and the square.</p>
| André Nicolas | 6,312 | <p>We describe a nice way to do it, unfortunately in words. It really needs a picture. </p>
<p>Put down your cardboard rectangle, one corner at the origin, the long side along the positive $x$-axis. So the corners of your cardboard rectangle are at $(0,0)$, $(0,45)$, $(45,20)$, and $(0,20)$.</p>
<p>Draw a $30\times 30$ square, with corners $(0,0)$, $(30,0)$, $(30,30)$, and $(0,30)$.</p>
<p>Draw the line that joins $(0,30)$ to $(45,0)$.</p>
<p>This line will meet the top side of your cardboard rectangle at $P=(15,20)$, and will meet the right side of the square at $Q=(30,10)$. Let $R=(30,0)$ and $S=(45,0)$.</p>
<p>All set up! Use a <strong>razor knife</strong> to cut along the line $PS$. That will slice a substantial triangle from the cardboard. Leave it in place for now.</p>
<p>Use the razor knife to cut straight down along $QR$. This slices off a smallish triangle from the cardboard.</p>
<p><strong>Slide</strong> the big triangle upward until its top side agrees with the top line of the square. It will.</p>
<p>Slide the little triangle way up so that it fills in the top left corner of the square. It will.</p>
<p>Done, two cuts.</p>
<p>It is a very pretty construction, works uniformly for <strong>all rectangles</strong> that are not too skinny. If the rectangle is very skinny, a not too hard adjustment can be made.</p>
<p>You will have to <em>prove</em> that this works. Straight coordinate or similar triangle geometry. </p>
<p><strong>Remark:</strong> This construction is one of the steps in the proof of the Bolyai-Gerwien Theorem (which, as is so often the case, was proved a number of years earlier by at least two other people). The result is that if $A$ and $B$ are <strong>any</strong> polygonal regions with the same area, then $A$ can be cut into a finite number of polygonal pieces that can be reassembled to maske $B$.</p>
|
96,289 | <p>In 1995 (if I'm not mistaken) Taylor and Wiles proved that all semistable elliptic curves over $\mathbb{Q}$ are modular. This result was extended to all elliptic curves in 2001 by Breuil, Conrad, Diamond, and Taylor.</p>
<p>I'm asking this as a matter of interest. Are there any other fields over which elliptic curves are known to be modular? Are there any known fields for which this is not true for? </p>
<p>Also, is much research being conducted on this matter?</p>
| David Loeffler | 2,481 | <p>Yes, this is a <em>very</em> active area -- one of the major themes of current research in number theory. </p>
<p>Much of the recent work has focussed on proving something slightly weaker, but easier to get at, than modularity. An elliptic curve $E$ over a number field $K$ is said to be <em>potentially modular</em> if there is a finite extension $L / K$ such that $E$ becomes modular over $L$. This notion of potential modularity has been much studied by Richard Taylor and his coauthors, and turns out to be almost as good for most purposes as knowing modularity over $K$. </p>
<p>It's now known, for instance, that any elliptic curve over a totally real number field $K$ becomes modular over some totally real extension $L / K$; a bit of googling turns up <a href="http://www2.math.kyushu-u.ac.jp/~virdol/basechange2.pdf" rel="nofollow">http://www2.math.kyushu-u.ac.jp/~virdol/basechange2.pdf</a> (which shows that one can choose $L$ in a rather specific way, using work of Taylor and Skinner-Wiles to do the heavy lifting).</p>
<p>I'm not an expert in the area, but my impression from talking to genuine experts is that current methods are very much limited to the case where the elliptic curve is defined over a field which is either totally real or CM -- outside these situations modularity is much less well understood.</p>
<p>(EDIT: I should add that there are some totally real fields for which one can show modularity, rather than just potential modularity; Jarvis and Manoharmayum have shown, for instance, that every semistable elliptic curve over $\mathbb{Q}(\sqrt{2})$ is modular.) </p>
|
2,485,261 | <blockquote>
<p>$\displaystyle \sum_{k=0}^n k {n \choose k} p^k (1-p)^{n-k}$ with $0<p<1$</p>
</blockquote>
<p>I know of one way to evaluate it (from statistics) but I was wondering if there are any other ways. </p>
<p>This is the way I know:</p>
<p>Let </p>
<p>$$M(t)=\displaystyle \sum_{k=0}^n e^{kt} {n \choose k} p^k (1-p)^{n-k}$$</p>
<p>Then $$M(t)=\displaystyle \sum_{k=0}^n {n \choose k} (pe^t)^k (1-p)^{n-k}=(pe^t+1-p)^n$$</p>
<p>$$M'(t)=\displaystyle \sum_{k=0}^n ke^{kt} {n \choose k} p^k (1-p)^{n-k}=pe^tn(pe^t+1-p)^{n-1}$$</p>
<p>$$M'(0)=\displaystyle \sum_{k=0}^n k {n \choose k} p^k (1-p)^{n-k}=np$$</p>
| epi163sqrt | 132,007 | <p>A slightly different variation of an answer already given which might also be convenient.</p>
<blockquote>
<p>We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k}}
&=np(1-p)^{n-1}\sum_{k=1}^n\binom{n-1}{k-1}\left(\frac{p}{1-p}\right)^{k-1}\tag{1}\\
&=np(1-p)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}{k}\left(\frac{p}{1-p}\right)^{k}\tag{2}\\
&=np(1-p)^{n-1}\left(1+\frac{p}{1-p}\right)^{n-1}\tag{3}\\
&\color{blue}{=np}\tag{4}
\end{align*}</p>
</blockquote>
<p><em>Comment:</em></p>
<ul>
<li><p>In (1) we use the binomial identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ and we factor out $p(1-p)^{n-1}$ to prepare the index shift of the next line.</p></li>
<li><p>In (2) we shift the index to start from $k=0$.</p></li>
<li><p>In (3) we use the binomial summation formula.</p></li>
<li><p>In (4) we do some final simplifications.</p></li>
</ul>
|
4,284,803 | <p>I am solving a question and I can't get over this step: proving <span class="math-container">$$\sin \frac{1}{k} > \frac{1}{k} - \frac{1}{k^2}$$</span>
where <span class="math-container">$k$</span> is a positive integer.</p>
<p>I tried using induction, but I failed. One of my friends managed to prove it using derivatives but I am searching for a solution which does not involve calculus or series. Proving this would help me solve a convergence problem.</p>
| Albus Dumbledore | 769,226 | <p>well if you don't want to use calculus ...</p>
<p>from <a href="https://math.stackexchange.com/questions/98998/why-x-tanx-while-0x-frac-pi2">this</a> geometric proof by David Mitra we have <span class="math-container">$$\sin t\le t\le \tan t$$</span> Now <span class="math-container">$$\tan t/2\ge t/2 $$</span> <span class="math-container">$$\sin (t/2)\cos(t/2)\ge \frac{t(1-\sin^2(t/2))}{2}$$</span> <span class="math-container">$$\to \sin t\ge t(1-\sin^2 (t/2)\ge t(1-t^2/4)=t-t^3/4$$</span> Using this result we get <span class="math-container">$$\sin 1/k\ge \frac{1}{k}-\frac{1}{4k^3}$$</span> But <span class="math-container">$\frac{1}{k}\le 1$</span> so obviously <span class="math-container">$$\frac{1}{k}-\frac{1}{4k^3}\ge \frac{1}{k}-\frac{1}{k^2}$$</span> we are done!</p>
|
207,185 | <p>How would I go about proving this without a truth table?</p>
<p>$[(p \lor q) \implies r ] \implies [ \neg r \implies (\neg p \land \neg q)]$</p>
| RicardoCruz | 36,340 | <p>Let's consider only two different settings in our proof, as shown in figure 1 and 2. You can complete the proof for other settings.</p>
<p>(1) <strong>Case</strong> ($\angle ERM$ and $\angle MRF$ are adjacent angles).</p>
<p>See figure 1.</p>
<p><img src="https://i.stack.imgur.com/Z2J22.png" alt="MiqueSimi"></p>
<p>We know that $\triangle MRE$ is isosceles and $QR$ is bisector of $\angle MRE$. We also know that $\triangle MRF$ is isosceles and $PR$ is bisector of $\angle MRF$.</p>
<p>Let $m(\angle MRE) = 2 \alpha$ and $m(\angle MRF) = 2 \beta$, then we can conclude that $m(\angle ECF)= \alpha + \beta$, since $\angle ECF$ is a inscribed angle and its measure is half the measure of the central angle $\angle ERF$. ($m(\angle ERF) = 2\alpha + 2 \beta$).</p>
<p>So it follows that
$$\angle ECF = \angle QRP.$$</p>
<p>By a similar approach we can conclude that</p>
<p>$$\angle DAF = \angle QPR.$$</p>
<p>Therefore $\triangle ABC \sim \triangle QPR$.</p>
<p>(2) <strong>Case</strong> ($\angle ERM$ and $\angle MRF$ are not adjacent angles).</p>
<p>See figure 2.</p>
<p>In that setting there is a superposition of $\triangle MRE$ and $\triangle MRF$, so $m(\angle ERF) = 2 \beta - 2 \alpha$, and no more $2 \alpha + 2 \beta$.</p>
<p>Note that $m(\angle QRF)= \beta - \alpha$ and that $m(\angle ECF)= \beta - \alpha$, since $m(\angle ECF)$ is still half the measure of $\angle ERF$.</p>
<p>Therefore
$$\angle ECF = \angle QFR.$$</p>
<p>As in case 1 (by a similar approach), we can conclude that</p>
<p>$$\angle DAF = \angle QPR.$$</p>
<p>Therefore $\triangle ABC \sim \triangle QPR$, as in case 1.</p>
|
1,451,301 | <p>We're given a function $P_n(x)$ for $-1\leq x\leq1$ as follows :</p>
<p>$$P_n(x) = \int \limits_0^\pi \dfrac{1}{\pi}(x+i\sqrt{1-x^2} \cos\theta)^n \, d\theta$$</p>
<p>for $n=(0,1,2,3,\ldots)$, we need to prove that $|P_n(x)| \leq 1$.</p>
<p>I tried the following :</p>
<p>Let $z=x+i\sqrt{1-x^2}\cos\theta$ </p>
<p>I somehow want to prove that $|z|=|x+i\sqrt{1-x^2}\cos\theta|\leq1$, as that would imply that $|z^n|\leq1$, as $|z^n|=|z|^n$.</p>
<p>$$|z| = \sqrt{x^2+(1-x^2)\cos^2 \theta} \Longrightarrow \sqrt{x^2 \sin^2\theta + \cos^2\theta}.$$</p>
<p>Now, $x^2\leq1$ and $\sin^2\theta \leq1 \Longrightarrow x^2\sin^2\theta \leq1$.</p>
<p>Also, $\cos^2\theta \leq1 \Longrightarrow \sqrt{x^2\sin^2\theta+\cos^2\theta} \leq \sqrt{2}$, </p>
<p>But " $\leq1$" condition is required...</p>
<p>That's the point where I am stuck, could anyone help me with this?</p>
| Community | -1 | <p>why not do it like:
for the above to be true we assume the condition to be true and then equating both we will get $x^2\le 1$ which is always true for the given case. Hence, by true logic.</p>
|
1,451,301 | <p>We're given a function $P_n(x)$ for $-1\leq x\leq1$ as follows :</p>
<p>$$P_n(x) = \int \limits_0^\pi \dfrac{1}{\pi}(x+i\sqrt{1-x^2} \cos\theta)^n \, d\theta$$</p>
<p>for $n=(0,1,2,3,\ldots)$, we need to prove that $|P_n(x)| \leq 1$.</p>
<p>I tried the following :</p>
<p>Let $z=x+i\sqrt{1-x^2}\cos\theta$ </p>
<p>I somehow want to prove that $|z|=|x+i\sqrt{1-x^2}\cos\theta|\leq1$, as that would imply that $|z^n|\leq1$, as $|z^n|=|z|^n$.</p>
<p>$$|z| = \sqrt{x^2+(1-x^2)\cos^2 \theta} \Longrightarrow \sqrt{x^2 \sin^2\theta + \cos^2\theta}.$$</p>
<p>Now, $x^2\leq1$ and $\sin^2\theta \leq1 \Longrightarrow x^2\sin^2\theta \leq1$.</p>
<p>Also, $\cos^2\theta \leq1 \Longrightarrow \sqrt{x^2\sin^2\theta+\cos^2\theta} \leq \sqrt{2}$, </p>
<p>But " $\leq1$" condition is required...</p>
<p>That's the point where I am stuck, could anyone help me with this?</p>
| marty cohen | 13,079 | <p>\begin{align}
x^2\sin^2 t+\cos^2 t
&=x^2(1-\cos^2 t+\cos^2 t\\
&=(1-x^2)\cos^2t+x^2\\
\text{so}\\
|x^2\sin^2t+\cos^2t|
&\le|(1-x^2)\cos^2t+x^2|\\
&\le|(1-x^2)\cos^2t|+|x^2|\\
&\le|1-x^2|+|x^2|\\
&\le(1-x^2)+x^2\\
&=1
\end{align}</p>
|
24,305 | <p>I have several functions, let's assume they are:</p>
<pre><code>func1[x_]=x;
func2[x_]=3*x-5;
func3[x_]=0.1*x^2;
</code></pre>
<p>and a lot more like these.</p>
<p>For each and every one of these I want to do the following</p>
<pre><code>xvalues = Range[0, 500, 2.5];
points1 = Map[func1, xvalues];
Do[If points1[[[i]] < 0, points1[[i]] = 0, points1[[i]] = points1[[i]],
{i, 1, Length[points1]}]
table1 = Transpose[{xvalues, points1}];
</code></pre>
<p>Now seeing as I have a lot of these functiones, is there any way to automate this in some kind of routine? </p>
<p>While answering, please be aware that I don't really have any extensive Mathematica knowledge.</p>
| rcollyer | 52 | <p>I would simplify your code a bit, merging everything into the <code>Map</code> statement, and move everything into a function, as follows:</p>
<pre><code>process[func_, xvals_] :=
Block[{points},
points = Map[ With[{val = func@#}, UnitStep[val] val]&, xvals];
Transpose[{xvals, points}]
]
</code></pre>
<p>and then for your functions, you can simply run</p>
<pre><code>process[func1, Range[0, 500, 2.5]]
</code></pre>
<p>Or, if you prefer to bury your <code>xvals</code> inside your function, just do this, instead:</p>
<pre><code>process[func_, xvals_:Range[0, 500, 2.5]] :=
Block[{points},
points = Map[ With[{val = func@#}, UnitStep[val] val]&, xvals];
Transpose[{xvals, points}]
]
</code></pre>
|
24,305 | <p>I have several functions, let's assume they are:</p>
<pre><code>func1[x_]=x;
func2[x_]=3*x-5;
func3[x_]=0.1*x^2;
</code></pre>
<p>and a lot more like these.</p>
<p>For each and every one of these I want to do the following</p>
<pre><code>xvalues = Range[0, 500, 2.5];
points1 = Map[func1, xvalues];
Do[If points1[[[i]] < 0, points1[[i]] = 0, points1[[i]] = points1[[i]],
{i, 1, Length[points1]}]
table1 = Transpose[{xvalues, points1}];
</code></pre>
<p>Now seeing as I have a lot of these functiones, is there any way to automate this in some kind of routine? </p>
<p>While answering, please be aware that I don't really have any extensive Mathematica knowledge.</p>
| Dr. belisarius | 193 | <pre><code>f = {# &, 3*# - 5 &, 0.1*#^2 &};
xvalues = Range[0, 500, 2.5];
t1 = Through[f[xvalues]] /. x_ /; x < 0 -> 0;
ListPlot[t1, DataRange -> {0, 500}]
</code></pre>
<p><img src="https://i.stack.imgur.com/kh38S.png" alt="enter image description here"></p>
|
4,570,329 | <p>In the textbook that I am working through, it is left as an exercise to prove the following claim</p>
<blockquote>
<p>Consider two linear maps <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> from <span class="math-container">$\mathbb{R}^n$</span> to <span class="math-container">$\mathbb{R}^m$</span> (for <span class="math-container">$m \le n$</span>) such that <span class="math-container">$P$</span> is an injective linear map. If we assume that the first <span class="math-container">$m$</span> columns of P are linearly independent, then we need to show that <span class="math-container">$\exists \delta >0$</span> such that <span class="math-container">$$ \lvert \lvert Q-P \rvert \rvert < \delta \implies rank(Q) = m$$</span> where our choice of norm is the operator norm.</p>
</blockquote>
<p>This exercise is supposed to be equivalent to showing that the set of matrices in <span class="math-container">$\mathbb{R}^{mn}$</span> that have full rank form an open subset of <span class="math-container">$\mathbb{R}^{mn}$</span></p>
<p>As someone that doesn't have a math background, I find these types (epsilon-delta) of proofs particularly challenging, as I often find them difficult to build up an intuition for. As there are no solutions to the exercises, I would be grateful for any guidance here on how to prove the statement.</p>
| belkacem abderrahmane | 660,639 | <p>I will only assume that <span class="math-container">$m$</span> columns of <span class="math-container">$P$</span> are linearly independent, i.e <span class="math-container">$P$</span> is surjective (Since the columns of <span class="math-container">$P$</span> spans its image), So there's some linear map <span class="math-container">$T:\mathbb{R} ^{m} \to \mathbb{R} ^{n} $</span> such that <span class="math-container">$PT=I$</span>, So <span class="math-container">$P+Q^{*} =P(I+TQ^{*} )$</span>, Now we apply the following :<span class="math-container">$$ $$</span> lemma :if <span class="math-container">$S:\mathbb{R}^{n} \to \mathbb{R} ^{n} $</span> is a Linear map, such that <span class="math-container">$\lvert \lvert S\rvert \rvert <1$</span>,then <span class="math-container">$I+S$</span> is invertible (an isomorphism). <span class="math-container">$$ $$</span> if we choose <span class="math-container">$\lvert \lvert Q^{*} \rvert \rvert <\lvert \lvert T \rvert \rvert <\frac{1}{\lvert \lvert T\rvert \rvert} $</span>, then <span class="math-container">$\lvert \lvert TQ* \rvert \rvert <1$</span>,So (by our lemma), <span class="math-container">$I+TQ^{*} $</span> is invertible, hence <span class="math-container">$rank(P+Q) =m$</span>, now the proof is finished by putting <span class="math-container">$Q=P+Q^{*} $</span>(remark that <span class="math-container">$||P-Q||=||Q{*} ||$</span>, with <span class="math-container">$\delta=\frac{1}{\lvert \lvert T\rvert \rvert}$</span>
This shows that the set of linear maps having full rank is open, i. e if <span class="math-container">$P$</span> has a full rank, and <span class="math-container">$Q$</span> is close enough <span class="math-container">$(\lvert \lvert P-Q\rvert \rvert <\delta $</span>) , then <span class="math-container">$Q$</span> still have full rank.</p>
|
2,352,821 | <p>So, in order to obtain the required answer, I tried to apply some Taylor expansions, which led me to nowhere actually.
After a while I tried to use the summation theorem </p>
<p>$\sum_{n=-\infty}^{+\infty}{f\left(n\right)}=-\sum_{i=1}^{m}{Res_{z=z_i}{\pi\cot\left(\pi z\right)f\left(z\right)}}$ at $f\left(z\right)$'s poles.</p>
<p>Residue at $z=-\frac{1}{2}$ equals to residue at $z=\frac{1}{2}$, both of them are $-\frac{\pi ^2}{16}$.
What I got seems to be not the correct answer after all:</p>
<p>$-\sum_{i=1}^{m}{Res_{z=z_i}{\pi\cot\left(\pi z\right)f\left(z\right)}} = -\frac{\pi ^2}{8}$, so that $\sum_{n=-\infty}^{+\infty}{f\left(n\right)}=\frac{\pi ^2}{8}$
Since the $f\left(z\right)$ is even, I get $\sum_{n=0}^{+\infty}{f\left(n\right)}=\frac{\pi ^2}{16}$</p>
<p>However, the initial task was to find the sum of $\sum_{n=1}^{+\infty}{f\left(n\right)}$. I supposed that $\sum_{n=1}^{+\infty}{f\left(n\right)}=\sum_{n=0}^{+\infty}{f\left(n\right)}-f\left(0\right)=\frac{\pi ^2}{16}-1$, which is incorrect, as according to Mathematica, I should get $\frac{\pi ^2}{16}-\frac{1}{2}$. I don't know where is my mistake at this point. Perhaps, the whole approach is not well executed. </p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{n = 1}^{\infty}{1 \over \pars{4n^{2} - 1}^{2}} & =
\left.{1 \over 16}\,\partiald{}{a}\sum_{n = 1}^{\infty}{1 \over n^{2} - a^{2}}
\right\vert_{\ a\ =\ 1/2} =
\left.{1 \over 16}\,\partiald{}{a}\sum_{n = 0}^{\infty}\pars{%
{1 \over n + 1 - a} - {1 \over n + 1 + a}}{1 \over 2a}
\right\vert_{\ a\ =\ 1/2}
\\[5mm] & =
\left.{1 \over 32}\,\partiald{}{a}\pars{H_{a} - H_{-a} \over a}
\right\vert_{\ a\ =\ 1/2}\qquad\qquad\quad\pars{~H_{z}:\ Harmonic\ Number~}
\\[5mm] & =
\left.{1 \over 32}\,\partiald{}{a}\pars{H_{a - 1} + 1/a - H_{-a} \over a}
\right\vert_{\ a\ =\ 1/2}\qquad\pars{~H_{z}\ Recursive\ Property~}
\\[5mm] & =
\left.{1 \over 32}\,\partiald{}{a}\pars{{1 \over a^{2}} -
{H_{-a} - H_{a - 1} \over a}}\right\vert_{\ a\ =\ 1/2}
\\[5mm] & =
\left.{1 \over 32}\,\partiald{}{a}\pars{{1 \over a^{2}} -
{\pi\cot\pars{\pi a} \over a}}\right\vert_{\ a\ =\ 1/2}
\qquad\pars{~H_{z}\ Euler\ Reflection\ Formula~}
\\[5mm] & =
\left.{1 \over 32}\pars{-\,{2 \over a^{3}} +
{\pi\cot\pars{\pi a} \over a^{2}} +
{\pi^{2}\csc^{2}\pars{\pi a} \over a}}\right\vert_{\ a\ =\ 1/2} =
\bbx{\pi^{2} - 8 \over 16} \approx 0.1169
\end{align}</p>
|
4,291,880 | <p>I don't understand how you would take the conjugate of a quadratic equation and how it would be useful to solve this question.</p>
<p>I would normally show it by saying if <span class="math-container">$b$</span> is real, then it is equal to <span class="math-container">$\alpha$</span> times <span class="math-container">$\beta$</span>, say <span class="math-container">$a$</span> is equal to <span class="math-container">$-(\alpha + \beta)$</span> and then just show it, but this doesn't involve conjugates as far as I can tell.</p>
| Deepak | 151,732 | <p>You have that <span class="math-container">$z^2 + az + b = 0$</span>. The given proposition only holds for real <span class="math-container">$a$</span> and real <span class="math-container">$b$</span>. This needs to be stated.</p>
<p>You need to either show or assume the following properties of complex conjugation:</p>
<p><span class="math-container">$\overline {z + w} = \overline z + \overline w$</span> (conjugate of sum is sum of conjugates),</p>
<p><span class="math-container">$\overline {zw} = \overline z\cdot \overline w$</span> (conjugate of product is product of conjugates),</p>
<p>and</p>
<p><span class="math-container">$z = w \implies \overline z = \overline w$</span> (equality implies equality of conjugates)</p>
<p>All of these are trivial to show (if you need to) by letting <span class="math-container">$z = x + yi$</span> and <span class="math-container">$w = a + bi$</span>. The second one needs some algebraic expansion.</p>
<p>A trivial corollary of the second (when <span class="math-container">$z = w$</span>) is that <span class="math-container">$\overline z^2 = (\overline z)^2$</span></p>
<p>Also, the first and second one can be extended to the sum and product (respectively) of an arbitrary number of terms. You'll be doing that with the first. Justifying this is easy with the associative property of addition.</p>
<p>Anyway, once all that preliminary work done and established as lemmas (or assumed), you can simply invoke the third lemma to take the conjugate of both sides of the quadratic. Then invoke the first two lemmas to immediately give <span class="math-container">$(\overline z)^2 + a\overline z + b = 0$</span></p>
<p>From that, you can immediately conclude that <span class="math-container">$\overline z$</span> is a solution to the quadratic as well. This leaves two possibilities: <span class="math-container">$z = \overline z$</span> which implies real root(s) or <span class="math-container">$z \neq \overline z$</span> which implies conjugate non-real complex roots. Thus the proof is done.</p>
|
3,867,834 | <p>I gotta find the value of <span class="math-container">$x+y$</span> in the following image</p>
<p><a href="https://i.stack.imgur.com/j9RPH.png" rel="noreferrer"><img src="https://i.stack.imgur.com/j9RPH.png" alt="enter image description here" /></a></p>
<p>I have no info about if a point is the middle point of a length or even the figure is a square. I can prove that
<span class="math-container">$$x-y=30°$$</span>
And I tried to use similarity between triangles (and so, parallelism) but I got no clue at all.</p>
| nmasanta | 623,924 | <p>This is nothing but the <em><strong>Trapezoidal Rule with Error Formula</strong></em>, where <span class="math-container">$~-\frac{(b-a)^{3}}{12} f^{\prime \prime}(\eta)~$</span> is the error term <span class="math-container">$(a < η < b)$</span>.</p>
<p><strong>Derivation:</strong> Consider the integral <span class="math-container">$~\int^b_a f(x)~dx~.$</span> Divide the interval <span class="math-container">$~[a,b]~$</span> such that <span class="math-container">$~a=x_0,~x_1,~x_2,~\cdots,x_{n-1},~x_n=b~$</span> with spacing <span class="math-container">$~h~.$</span> For trapezoidal rule <span class="math-container">$~n=1~,$</span> i.e., in this case <span class="math-container">$~a=x_0,~b=x_1=x_0+h\implies h=b-a~.$</span><br>
Let <span class="math-container">$~I~$</span> be the analytical integration and <span class="math-container">$~I_T~$</span> be the trapezoidal integration, then
<span class="math-container">$$I=I_T+~\text{error term} \implies \text{error term} =I-I_T~$$</span>
Now by the first fundamental theorem of calculus,<span class="math-container">$$I=\int_a^bf(x)~dx=\int_{x_0}^{x_1}f(x)~dx=F(x_1)-F(x_0)=F(x_0+h)-F(x_0)$$</span>where <span class="math-container">$F(x)$</span> is the anti-derivative of <span class="math-container">$f(x)$</span> i.e., <span class="math-container">$~F'(x)=f(x)~.$</span><br>
Also from the trapezoidal rule, <span class="math-container">$$I_T=\int_a^bf(x)~dx=\dfrac {b-a}2\left[f(a)+f(b)\right]=\dfrac h2\left[f(x_0)+f(x_1)\right]=\dfrac h2\left[f(x_0)+f(x_0+h)\right]$$</span>
Therefore,<br> Error
<span class="math-container">$=\left[F(x_0+h)-F(x_0)\right]-\dfrac h2\left[f(x_0)+f(x_0+h)\right]$</span><br>
<span class="math-container">$=\left\{hF'(x_0)+\dfrac{h^2}{2!}F''(x_0)+\dfrac{h^3}{3!}F'''(x_0)+\cdots\right\}-\dfrac h2\left\{2f(x_0)+hf'(x_0)++\dfrac{h^2}{2!}f''(x_0)++\dfrac{h^3}{3!}f'''(x_0)+\cdots\right\}~,~~\text{(using Taylor series expansion)}$</span><br>
<span class="math-container">$=\left\{hf(x_0)+\dfrac{h^2}{2!}f'(x_0)+\dfrac{h^3}{3!}f''(x_0)+\cdots\right\}-\dfrac h2\left\{2f(x_0)+hf'(x_0)++\dfrac{h^2}{2!}f''(x_0)++\dfrac{h^3}{3!}f'''(x_0)+\cdots\right\}$</span><br>
<span class="math-container">$=\left(\dfrac 13-\dfrac 12\right)\dfrac{h^3}{2}f''(x_0)+O(h^4)$</span><br>
<span class="math-container">$=-\dfrac{h^3}{12}f''(x_0)~,~~\text{(neglecting the higher order)}$</span><br>
<span class="math-container">$\approx-\dfrac{(b-a)^3}{12}f''(\eta), \quad \eta \in(a, b)$</span></p>
<p>Therefore <span class="math-container">$$I=\int_{a}^{b} f(x) \mathrm{d} x = \frac{b-a}{2}[f(a)+f(b)]-\frac{(b-a)^{3}}{12} f^{\prime \prime}(\eta), \quad \eta \in(a, b)$$</span></p>
|
1,862,108 | <p>This question is related to maths, so I post here. Actually it's a computer science question and I am facing this type of question while learning Design and Analysis of Algorithms, but we all know that computer science has complete relation with maths. </p>
<p>Arrange the following functions in increasing order of growth rate (with $g(n)$ following $f(n)$ in your list if and only if $f(n)=O(g(n)))$.</p>
<p>a) $2^{log(n)}$<br>
b) $2^{2log(n)}$<br>
c) $n^5/2$<br>
d) $2^{n^2}$<br>
e) $n^2 log(n)$ </p>
<p>So i think the answer, in increasing order, is CEDAB.</p>
<p>Is it correct? I have confusion in option A and B. I think option A should be first... the one with the lower growth rate I mean. Please help me solve this. I faced this question in algorithm course part 1 assignment (Coursera).</p>
| Eduardo Sebastian | 451,521 | <p>Take the limit to the infinity of quotient of each combination and if the result is <span class="math-container">$0$</span> or more generally a constant, what it means? and if the result is <span class="math-container">$\infty$</span>?</p>
<p>That is the way to check what function grows faster that other when <span class="math-container">$n$</span> grows.</p>
|
2,791,914 | <ul>
<li>$\displaystyle \int_0^\infty \frac{\arctan\frac{x^3}{1+x^2}}{x^2} \, dx$ </li>
</ul>
<hr>
<p>So i know that this one converge from 1 to infinity (by Dirichlet rule), but i'm not sure about the 0 to 1 segment. I kind of think that it converge as well, but can't prove it myself. Any suggestions?</p>
| Andrei | 331,661 | <p><strong>HINT:</strong> The function is positive in that interval. Find the maximum (not at 0). It is a finite value. The integral from $0$ to $1$ must be less than the maximum multiplied by $1-0$</p>
|
1,579,371 | <p>I'm studying for my number theory test tomorrow, and these are the last questions in my study guide. I think I understand Fermat's factorization, however, I can't tell how my professor wants us to answer these questions. One of them is going to be on the exam.</p>
<ol>
<li><p>Set <span class="math-container">$n= 87463$</span> and <span class="math-container">$q(x) = x^2 - n$</span>. Explain how to use the congruences
<span class="math-container">\begin{eqnarray*}
q(265) &=& -2\times3\times13^2\times17,\\
q(278) &=& -3^3\times13\times29,\\
q(296) &=& 3^2\times17,\\
q(299) &=& 2\times3\times17\times19,\\
q(307) &=& 2\times3^2\times13\times29,\\
q(316) &=& 3^6\times17,
\end{eqnarray*}</span>
to factor <span class="math-container">$n$</span>.</p>
</li>
<li><p>Explain how the congruences below prove that <span class="math-container">$n = 2821$</span> is composite
<span class="math-container">\begin{eqnarray*}
2^{705} &=& 2605 \pmod n,\\
2^{1410} &=& 1520 \pmod n,\\
2^{2820} &=& 1 \pmod n.
\end{eqnarray*}</span>
I'm not so sure if these questions are related or not. In the second one, it is easy to see <span class="math-container">$705 = \frac{n-1}4$</span>, <span class="math-container">$1410 = \frac{n-1}{2}$</span> and <span class="math-container">$2820 = n-1$</span> however I'm not sure on which property to use here.</p>
</li>
</ol>
| Tito Piezas III | 4,781 | <p>(<em>Too long for a comment</em>.)</p>
<p>You may have missed the small even powers $k=2,4,6$
$$\begin{align}
&32\,\Re\operatorname{Li}_3\left(i\,\phi^2\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-8}\right)-4\operatorname{Li}_3\left(\phi^{-4}\right)+20\operatorname{Li}_3\left(\phi^{-2}\right)-7\ln^3(\phi^2)-16\,\zeta(3)\\
\,\\
&32\,\Re\operatorname{Li}_3\left(i\,\phi^4\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-16}\right)-4\operatorname{Li}_3\left(\phi^{-8}\right)+40\operatorname{Li}_3\left(\phi^{-2}\right)-46\ln^3(\phi^2)-32\,\zeta(3)\\
\,\\
&32\,\Re\operatorname{Li}_3\left(i\,\phi^6\right)\stackrel{\color{blue}?}=18\operatorname{Li}_3\left(\phi^{-8}\right)-72\operatorname{Li}_3\left(\phi^{-4}\right)+64\operatorname{Li}_3\left(\phi^{-3}\right)+90\operatorname{Li}_3\left(\phi^{-2}\right)-48\operatorname{Li}_3\left(\phi^{-1}\right)-1196\ln^3\phi-35\,\zeta(3)
\end{align}$$
More generally, for any <em>real</em> number $k$ it seems,
$$32\,\Re\operatorname{Li}_3\left(i\,\phi^k\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-4k}\right)-4\operatorname{Li}_3\left(\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$
though I have no proof yet.</p>
|
413,778 | <p>Let <span class="math-container">$G$</span> be a finite abelian group, <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> be two non-empty subsets of <span class="math-container">$G$</span> of equal size. Suppose that for each irreducible character <span class="math-container">$\chi$</span> of <span class="math-container">$G$</span> we have <span class="math-container">$\sum_{x\in X}\chi(x)=\sum_{y\in Y}\chi(y)$</span>. Is it true that <span class="math-container">$X=Y$</span> in general?</p>
| Geoff Robinson | 14,450 | <p>More generally, for any finite group <span class="math-container">$G$</span>, Abelian or not, it is true that <span class="math-container">$Z(\mathbb{C}G)$</span>, the center of the complex group algebra <span class="math-container">$\mathbb{C}G$</span>, has a <span class="math-container">$\mathbb{C}$</span>-basis of mutually orthogonal idempotents, <span class="math-container">$\{e_{\chi} : \chi \in {\rm Irr}(G) \},$</span> indexed by the complex irreducible characters of <span class="math-container">$G$</span>. In other words,we have <span class="math-container">$e_{\chi}^{2} = e_{\chi}$</span> for each <span class="math-container">$\chi$</span>, and <span class="math-container">$e_{\chi}e_{\mu} = 0$</span> when <span class="math-container">$\chi \neq \mu.$</span> This property ensures that the idempotents <span class="math-container">$\{ e_{\chi} \}$</span> are linearly independent</p>
<p>Given an element <span class="math-container">$X \in Z(\mathbb{C}G),$</span> we may write <span class="math-container">$X$</span> uniquely in the form
<span class="math-container">$X = \sum_{\chi} a_{\chi}(X) e_{\chi}$</span> where the <span class="math-container">$a_{\chi}(X)$</span> are complex numbers, and we may check that for each <span class="math-container">$\chi$</span>, the map <span class="math-container">$X \to a_{\chi}(X)$</span> is an algebra homomorphism from <span class="math-container">$Z(\mathbb{C}G)$</span> to <span class="math-container">$\mathbb{C}$</span>, using the orthogonality of the idempotents <span class="math-container">$e_{\chi}$</span>.</p>
<p>Notice that <span class="math-container">$a_{\chi}(e_{\mu}) = \delta_{\chi, \mu}$</span> by definition, and that <span class="math-container">$a_{\chi}(1_{G}) = 1$</span> for each <span class="math-container">$\chi.$</span> By Schur's Lemma, we see that <span class="math-container">$e_{\chi}$</span> is represented by a scalar matrix in any complex representation of <span class="math-container">$G$</span> affording irreducible character <span class="math-container">$\chi$</span>, and this matrix must be idempotent, so that <span class="math-container">$\chi(e_{\chi}) = \chi(1)$</span></p>
<p>Thus we have <span class="math-container">$a_{\chi}(e_{\mu}) = \frac{\chi(e_{\mu})}{\chi(1)}$</span> for all irreducible characters <span class="math-container">$\chi, \mu.$</span> Hence we have <span class="math-container">$a_{\chi}(X) = \frac{\chi(X)}{\chi(1)}$</span> for all <span class="math-container">$X \in Z(\mathbb{C}G)$</span>, since <span class="math-container">$\{e_{\mu} \}$</span> is a <span class="math-container">$\mathbb{C}$</span>-basis for <span class="math-container">$Z(\mathbb{C}G).$</span></p>
<p>Consequently, we may conclude that two elements <span class="math-container">$X,Y \in Z(\mathbb{C}G)$</span> are equal if and only if <span class="math-container">$\chi(X) = \chi(Y)$</span> for each complex irreducible character <span class="math-container">$\chi$</span> of <span class="math-container">$G$</span>.</p>
<p>In case <span class="math-container">$G$</span> is Abelian, the group algebra <span class="math-container">$\mathbb{C}G$</span> is commutative, and is equal to its center <span class="math-container">$Z(\mathbb{C}G)$</span>. Furthermore, <span class="math-container">$G$</span> has <span class="math-container">$|G|$</span> complex irreducible characters all of degree <span class="math-container">$1$</span>, so we obtain that <span class="math-container">$X,Y \in \mathbb{C}G$</span> are equal if and only if <span class="math-container">$\lambda(X) = \lambda(Y)$</span> for each irreducible (linear) complex character <span class="math-container">$\lambda$</span> of <span class="math-container">$G$</span>.</p>
|
2,543,169 | <p>The question is pretty self explanatory, but I’ve encountered situations where, for the length of some vector $\vec{a}$, to denote the length (or magnitude, which ever you prefer) as either $\| \vec{a}\|= \sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ or $|\vec{a}|= \sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ and I was wondering which notation is more widely accepted, per say? I’ve tried researching this and different websites actually use different notation. </p>
<p>Any help is appreciated, thank you.</p>
| GEdgar | 442 | <p>Definitions are not judged "correct" or "incorrect". Both of your notations are used in various contexts in mathematics. There is a problem only if the writer and the reader do not understand each other.</p>
|
919,572 | <p>Do you know any nice way of expressing </p>
<p>$$\sum_{k=0}^{n} \frac{H_{k+1}}{n-k+1}$$
?</p>
<p>Some simple manipulations involving the integrals lead to an expression that also uses<br>
the hypergeometric series. Is there any way of getting a form that doesn't use the HG function?</p>
| Jack D'Aurizio | 44,121 | <p>Yes:
$$\begin{eqnarray*}\color{red}{\sum_{k=1}^{n+1}\frac{H_k}{n+2-k}}=\sum_{k=1}^{n+1}\sum_{r+s=k}\frac{1}{rs}=\color{red}{H_{n+2}^2-H_{n+2}^{(2)}.}\end{eqnarray*}$$
For the proof, see <a href="https://math.stackexchange.com/questions/627936/how-find-this-sum-i-n-sum-k-0n-frach-k1h-n-k1k2">this other</a> question. It is line $(4)$ in my second answer.</p>
|
1,307,069 | <p>Let's look at $f(x)=\cos(x)$ defined on the interval $[0,\pi]$.</p>
<p>We know that for any function $g$ defined on $[0,\pi]$ we have:</p>
<p>$g(x)=\sum_{k=1}^{\infty}B_k\sin(kx)$ where $B_k=\frac{2}{\pi}\int_{0}^{\pi}g(x)\sin(kx)dx$. And $f$ is no different. So in our case:</p>
<p>$B_k=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(kx)dx=\frac{2}{\pi}\int_{0}^{\pi}\cos(x)\sin(kx)dx$</p>
<p>It can be shown that $\int\cos(x)\sin(kx)dx=\frac{\sin(x)\sin(kx)+k\cos(x)\cos(kx)}{1-k^2}$, so:</p>
<p>$B_k=\frac{2}{\pi}\int_{0}^{\pi}\cos(x)\sin(kx)dx=\frac{2}{\pi}\frac{-k\cos(k\pi)-k}{1-k^2}=\frac{2}{\pi}\frac{(-1)^{k+1}k-k}{1-k^2}$</p>
<p>So overall we should have $f(x)=\cos(x)=\sum_{k=1}^{\infty}\frac{2}{\pi}\frac{(-1)^{k+1}k-k}{1-k^2}\sin(kx)$ But that clearly can't be true, because $\cos(0)=1$ but that sum is equal to $0$ at $x=0$ since $\sin(0)=0$.</p>
<p>Where is the mistake? and not only that, we seem to have a big problem when $k=1$</p>
| Hans Lundmark | 1,242 | <p>Your formula after "it can be shown that" is clearly not valid for $k=1$,
so you simply have to compute $B_1$ using some different method. (For example, $\int \cos x \sin x \, dx= \frac12 \sin^2 x + C$.)</p>
|
4,118 | <p>I've recently dipped my toes into the world of number theory; and I've bought a book that to me is quite unconventional: R. P. Burn, <em>A Pathway into Number Theory</em>. I've yet to put the book through its paces, but it seems agreeable enough to me. The book is unique in that it poses a sequence of questions to you in the hope that you'll be able to answer them and by thus doing so, begin to discover the fundamentals of number theory.</p>
<p>This is a style of learning that I find agreeable as the knowledge I gain this way is assimilated and retained better. I like being able to discover for myself however most times I don't have the necessary direction (I am self-studying) but that's where this textbook comes in. I feel that this book in the process of nudging you in the right direction also helps you think more like a mathematician (from my very limited experience with it). </p>
<p>I enjoy books that give you a "pathway", although I guess this is the aim of all textbooks. Is it possible for anyone to recommend texts that take a similar aided discovery/inquiry based approach? </p>
| Jon Bannon | 354 | <p>Check out <a href="http://jiblm.org">http://jiblm.org</a>. There are lots of scripts here, some better than others. </p>
<p>A nice book in this style is <a href="http://www.maa.org/publications/ebooks/distilling-ideas">"Distilling Ideas" by Brain Katz and Michael Starbird</a>.</p>
<p>I also recommend the following method: Take a reputable text on a topic, and try to prove all the theorems for yourself. If you get stuck for a long time, take a quick peek to get unstuck…or find a new thing to assume by looking at the proof. You'll find that if you do this honestly that you can find out which ideas are novel and which are routine. I learned this approach from a Halmos student, and it really is enjoyable to do. The proofs in the book are just regarded as the answer key.</p>
<p>Have fun! </p>
|
4,338,190 | <p>Its required to prove that <span class="math-container">$|x^{1/n} -1| \lt \epsilon$</span> for <span class="math-container">$\epsilon \gt 0$</span> and <span class="math-container">$n \ge N$</span> where <span class="math-container">$N \in \mathbb N$</span>.<br>
Let <span class="math-container">$x^{1/n} -1 = h$</span> for <span class="math-container">$h \gt 0$</span>. So <span class="math-container">$x = (1+h)^{n}$</span>, and we now need to prove that <span class="math-container">$h \lt \epsilon$</span>. The binomial theorem gives <span class="math-container">$x \gt nh $</span> so <span class="math-container">$h \lt \frac{x}{n}$</span>. Now its sufficient to choose <span class="math-container">$n $</span> such that <span class="math-container">$\frac{x}{n} \lt \epsilon$</span>, i.e. <span class="math-container">$n \gt\frac{x}{\epsilon}$</span>. So the given condition is satisfied for all <span class="math-container">$ N \ge [\frac{x}{\epsilon}] + 1.$</span>(The square brackets stand for the floor function.)<Br>
Is this in form and content correct?</p>
| user2661923 | 464,411 | <blockquote>
<p>Let <span class="math-container">$x^{(1/n)} - 1 = h$</span> for <span class="math-container">$h>0.~~$</span> So <span class="math-container">$x=(1+h)^{(1/n)}$</span>.</p>
</blockquote>
<p>No, this is wrong. The intermediate conclusion is that <br>
<span class="math-container">$x = (1 + h)^n$</span>.</p>
<hr />
<p>An easier approach, if permitted, is to let <span class="math-container">$y = \log(x)$</span>.</p>
<p>Then, regardless of whether <span class="math-container">$y$</span> is positive, negative, or zero, you have that</p>
<p><span class="math-container">$$\lim_{n\to\infty} \frac{y}{n} = 0.$$</span></p>
<p>Therefore, it is routine to set up an <span class="math-container">$\varepsilon$</span> demonstration, choosing <span class="math-container">$N$</span> large enough so that for all <span class="math-container">$n \geq N$</span>, you have that <span class="math-container">$\displaystyle e^{(y/n)}$</span> is in the neighborhood of <span class="math-container">$(1 - \varepsilon, 1 + \varepsilon).$</span></p>
<p>That is, you want to choose <span class="math-container">$N$</span> large enough so that</p>
<p><span class="math-container">$\displaystyle \left|\frac{y}{N}\right| < \log(1 + \varepsilon)$</span></p>
<p>and</p>
<p><span class="math-container">$\displaystyle \left|\frac{y}{N}\right| < \left|\log(1 - \varepsilon)\right|.$</span></p>
<p><strong>Note</strong><br>
You can assume, without loss of generality, that <span class="math-container">$\varepsilon < 1$</span>. That is, if the demonstration holds for a specific <span class="math-container">$\varepsilon < 1$</span>, then the demonstration holds for any larger <span class="math-container">$\varepsilon_1$</span>.</p>
<p>This is because <br>
<span class="math-container">$0 < \varepsilon < \varepsilon_1 \implies (1 - \varepsilon, 1 + \varepsilon) \subseteq (1 - \varepsilon_1, 1 + \varepsilon_1).$</span></p>
|
2,390,077 | <p>is it possible to find a matrix $B$ which fulfills:</p>
<p>$BAA^TB^T=I$, where $I$ is identity matrix and $A$ strictly lower triangular?</p>
<p>Thank you very much in advance!</p>
| Dietrich Burde | 83,966 | <p>No, this is not possible. It is useful to write it down explicitly for $n=2$. Then either $A=0$, which has no solution, or we may assume that
$$
A=\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix},\quad
B=\begin{pmatrix} a & c \cr b & d \end{pmatrix}.
$$
Then the condition $BAA^ TB^ T=I$ is equivalent to the equations
\begin{align*}
a^2 & = 1 ,\\
b^2 & = 1,\\
ab & = 0.
\end{align*}
Clearly this has no solution.</p>
|
707,317 | <p>Let $g: R\rightarrow R$ be a twice differentiable function satisfying $g(0)=1, g'(0)=0$ and $ g''(x)-g(x)=0$, for all $x$ in R</p>
<p>Fix $x$ in R. Show that there exists $M>0$ such that for all natural number n and all θ from 0 to 1 $$ |g^{(n)}(θx)|\leq M$$</p>
<p>Also, find the coefficients of the Taylor expansion of $g$ about $0$, and prove that this expansion converges to $g(x)$ for all $x$ in R</p>
<p>p.s.
My idea is to start from proving that $g$ has derivatives of all orders, but I am not sure whether it is a correct start and how I can proceed. Any suggestion or attempt is appreciated. </p>
| Community | -1 | <p><strong>First Part:</strong></p>
<p>Since <span class="math-container">$g$</span> is differentiable on <span class="math-container">$\mathbb{R}$</span>, it is also continuous on <span class="math-container">$\mathbb{R}$</span>. Note that every continuous function must attain a maximum and a minimum on a closed interval, so <span class="math-container">$g$</span> must attain a minimum and maximum on <span class="math-container">$[0, x]$</span>. Let <span class="math-container">$g$</span> have a minimum at <span class="math-container">$x_m$</span> and maximum at <span class="math-container">$x_M$</span> for <span class="math-container">$x_m, x_M \in [0, x]$</span>, and let <span class="math-container">$M_1 = max(|g(x_m)|, |g(x_M)|)$</span>. Similarly, since <span class="math-container">$g'$</span> is continuous on <span class="math-container">$[0, x]$</span>, let <span class="math-container">$M_2 = max(|g'(x_m)|, |g'(x_M)|)$</span>, where <span class="math-container">$x_m, x_M \in [0, x]$</span> and <span class="math-container">$g'$</span> has its minimum at <span class="math-container">$x_m$</span> and maximum at <span class="math-container">$x_M$</span>. </p>
<p>Now let <span class="math-container">$M = max(M_1, M_2)$</span>. Then both <span class="math-container">$|g(\theta x)| \leq M$</span> and <span class="math-container">$|g'(\theta x)| \leq M$</span> for every <span class="math-container">$\theta \in [0, 1]$</span>. Also, since <span class="math-container">$g''(x) = g(x)$</span>, we have <span class="math-container">$|g''(\theta x)| \leq M$</span> and <span class="math-container">$|g^{(3)}(\theta x)| \leq M$</span> (since <span class="math-container">$g''(x) = g(x)$</span> implies <span class="math-container">$g^{(3)}(x) = g'(x)$</span>).</p>
<p>More generally, <span class="math-container">$g^{(2n + 1)}(x) = g'(x)$</span> and <span class="math-container">$g^{(2n)}(x) = g(x)$</span> for every <span class="math-container">$n\geq 0$</span>, which means <span class="math-container">$|g^{(n)}(\theta x)| \leq M$</span></p>
<p><strong>Second Part:</strong></p>
<p>IV_ does a good job explaining the Taylor series expansion part.</p>
|
773,880 | <p>What approach would be ideal in finding the integral $\int4^{-x}dx$?</p>
| Ellya | 135,305 | <p>Firstly, the most Important thing here is that $(4^{-x})'=(-\ln 4)4^{-x}$ </p>
<p>So we rewrite our integral as follows:</p>
<p>$\int 4^{-x}\,dx=\frac{-1}{\ln 4}\int -\ln 4\cdot 4^{-x}\,dx=\frac{-1}{\ln 4}\int (4^{-x})'\,dx=-\frac{1}{\ln 4}4^{-x}+C$.</p>
|
248,733 | <p>Assume the following matrix
$$
C_p^{(a,b)}:=\left(
\begin{array}{cccccc}
a &a &0 &\cdots &\cdots &0 \\
0 &0 &a &\ddots &\ddots &\vdots \\
\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\
\vdots &\ddots &\ddots &\ddots &\ddots &0 \\
0 &\cdots &\cdots &0 &0 &a \\
b &b &\cdots &\cdots &b &b \\
\end{array}
\right)_{p \times p}\, .
$$
Where $a$ and $b$ are any integer number. With the numerical simulation, i found that the $n$th power of the matrix $C_p^{(a,b)}$, has
the following form
$$
{(C_p^{(a,b)})}^n:=\left(
\begin{array}{cccccc}
{g_1^{a,b}}(n) &{g_1^{a,b}}(n) &\cdots &\cdots &{g_1^{a,b}}(n) \\
\\
{g_2^{a,b}}(n) &{g_2^{a,b}}(n) &\cdots &\cdots &{g_2^{a,b}}(n) \\
\\
\vdots &\cdots &\cdots &\cdots &\vdots \\
\vdots &\cdots &\cdots &\cdots &\vdots \\
\\
{g_p^{a,b}}(n) &{g_p^{a,b}}(n) &\cdots &\cdots &{g_p^{a,b}}(n) \\
\end{array}
\right)_{p \times p}\, .
$$
Where ${g_i^{a,b}}(n)$, $1\leq i \leq p$, are expressions based on the parameters $a$, $b$ and $n$. For example, two consecutive power
of the matrix $C_7^{(2,3)}$, are as follows
$$
{(C_7^{(2,3)})}^9:=
\left( \begin {array}{ccccccc} 8000&8000&8000&8000&8000&8000&8000
\\ 12000&12000&12000&12000&12000&12000&12000
\\ 30000&30000&30000&30000&30000&30000&30000
\\ 75000&75000&75000&75000&75000&75000&75000
\\ 187500&187500&187500&187500&187500&187500&187500
\\ 468750&468750&468750&468750&468750&468750&468750
\\ 1171875&1171875&1171875&1171875&1171875&1171875&
1171875\end {array} \right)\, .
$$</p>
<p>$$
{(C_7^{(2,3)})}^{10}:=
\left( \begin {array}{ccccccc} 40000&40000&40000&40000&40000&40000&
40000\\ 60000&60000&60000&60000&60000&60000&60000
\\ 150000&150000&150000&150000&150000&150000&150000
\\375000&375000&375000&375000&375000&375000&375000
\\ 937500&937500&937500&937500&937500&937500&937500
\\ 2343750&2343750&2343750&2343750&2343750&2343750&
2343750\\ 5859375&5859375&5859375&5859375&5859375&
5859375&5859375\end {array} \right)\, .
$$
Is there a way to find an explicit formula for ${g_i^{a,b}}(n)$, $1\leq i \leq p$ in general. The matrix $C_p^{(a,b)}$ is so interesting.
If $a=-b$ then
$$
\forall n\geq p \qquad {(C_p^{(a,b)})}^n=O_p\, .
$$
Where $O_p$ is a zero matrix of order $p$. In some cases, ${g_i^{a,b}}(n)$, $1\leq i \leq p$, are fixed. For example,
if $[a=-(d\pm1) \, \& \, b=d]$ or $[b=-(d\pm1) \, \& \, a=d]$ where $d$ is an integer number, then we have
$$
\forall n\geq p-1 \qquad {(C_p^{(a,b)})}^n=\pm F_p\, .
$$
Where $F_p$ is a fixed matrix of order $p$. For example, by using $C_5^{(-3,2)}$ and $C_4^{(3,-4)}$, we can see that
$$
C_5^{(-3,2)}=
\left( \begin {array}{ccccc} -3&-3&0&0&0\\0&0&-3&0
&0\\ 0&0&0&-3&0\\ 0&0&0&0&-3
\\ 2&2&2&2&2\end {array} \right) \Rightarrow
\forall n\geq 4 \quad {(C_5^{(-3,2)})}^n=
\left( \begin {array}{ccccc} 81&81&81&81&81\\ -54&-
54&-54&-54&-54\\ -18&-18&-18&-18&-18
\\ -6&-6&-6&-6&-6\\ -2&-2&-2&-2&-2
\end {array} \right)\, .
$$
$$
C_4^{(3,-4)}=
\left( \begin {array}{cccc} 3&3&0&0\\0&0&3&0
\\ 0&0&0&3\\ -4&-4&-4&-4
\end {array} \right)
\Rightarrow
\forall n\geq 3 \quad {(C_4^{(3,-4)})}^n=\pm
\left( \begin {array}{cccc} 27&27&27&27\\ -36&-36&-
36&-36\\ 12&12&12&12\\ -4&-4&-4&-4
\end {array} \right)\, .
$$
In some especial cases, i found an expression for ${g_i^{a,b}}(n)$, $1\leq i \leq p$. Assume $C_p^{(a,b)}$, for $a=b=1$, as follows
$$
C_p^{(1,1)}:=\left(
\begin{array}{cccccc}
1 &1 &0 &\cdots &\cdots &0 \\
0 &0 &1&\ddots &\ddots &\vdots \\
\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\
\vdots &\ddots &\ddots &\ddots &\ddots &0 \\
0 &\cdots &\cdots &0 &0 &1 \\
1 &1 &\cdots &\cdots &1 &1 \\
\end{array}
\right)_{p \times p}\, .
$$
With the induction on $n$, we can prove that for $n\geq p-1$, we have </p>
<p>$$
{(C_p^{(1,1)})}^n:=\left(
\begin{array}{cccccc}
2^{n-(p-1)} &2^{n-(p-1)} &\cdots &\cdots &2^{n-(p-1)} \\
\\
2^{n-(p-1)} &2^{n-(p-1)} &\cdots & \cdots & 2^{n-(p-1)} \\
\\
2^{n-(p-2)} &2^{n-(p-2)} &\cdots & \cdots & 2^{n-(p-2)} \\
\\
2^{n-(p-3)} &2^{n-(p-3)} &\cdots & \cdots & 2^{n-(p-3)} \\
\\
\vdots &\cdots &\cdots &\cdots &\vdots \\
\vdots &\cdots &\cdots &\cdots &\vdots \\
\\
2^{n-1} &2^{n-1} &\cdots &\cdots & 2^{n-1} \\
\end{array}
\right)_{p \times p}\, .
$$
Is there a method to find a general expression for ${g_i^{a,b}}(n)$, $1\leq i \leq p$? I would greatly appreciate for any suggestions.</p>
| Meysam Ghahramani | 84,430 | <p>Let $\left(
\begin{array}{c}
{g_1^{(a,b)}}(1) \\
\\
{g_2^{(a,b)}}(1) \\
\\
\vdots \\
\vdots \\
\\
{g_p^{(a,b)}}(1) \\
\end{array}
\right)$
be the first column of $C_p^{(a,b)}$. From equation ${(C_p^{(a,b)})}^n=C_p^{(a,b)}\cdot {(C_p^{(a,b)})}^{n-1}$ we conclude:</p>
<p>$$\left\{
\begin{array}{c}
g_1^{(a,b)}(n)=a\cdot (g_1^{(a,b)}(n-1)+g_2^{(a,b)}(n-1)) \\g_i^{(a,b)}(n)=a\cdot g_{i+1}^{(a,b)}(n-1) , 2\le i\le p-1\\g_p^{(a,b)}(n)=b \cdot \sum_1^p g_i^{(a,b)}(n-1) \\
\end{array}
\right.$$</p>
|
385,537 | <p>How would you go about proving the following?</p>
<p>$${1- \cos A \over \sin A } + { \sin A \over 1- \cos A} = 2 \operatorname{cosec} A $$</p>
<p>This is what I've done so far:</p>
<p>$$LHS = {1+\cos^2 A -2\cos A + 1 - \cos^2A \over \sin A(1-\cos A)}$$</p>
<p>....no idea how to proceed .... X_X</p>
| amWhy | 9,003 | <p>You did everything thus far correctly, I just pick up with where you left off in the second line:</p>
<p>$$\begin{align}(1 - \cos A)^2 + \sin^2 A \over \sin A(1 - \cos A)
& = \dfrac{1 - 2 \cos A + \cos^2 A + \sin^2 A}{\sin A(1 - \cos A)} \\ \\
& = {1 \color{blue}{\bf + \cos^2 A} -2\cos A + 1 \color{blue}{\bf - \cos^2A} \over \sin A(1-\cos A)} \\ \\
& = \dfrac{2 - 2\cos A}{\sin A(1 - \cos A)}\\ \\
& = \dfrac{2\color{red}{\bf (1-\cos A)}}{\sin A\color{red}{\bf (1 - \cos A)}}\\ \\
& = \frac{2}{\sin A} \\ \\
& = 2 \csc A
\end{align}$$</p>
|
1,958,491 | <p>Let $t^k$ act as the $k$-th derivative operator on the set of polynomials. So</p>
<p>$$t^k(x^n)=t^k x^n=(n)_kx^{n-k}$$</p>
<p>where $(n)_k=n(n-1)(n-2)...(n-k+1)$ is the falling factorial. Then with a formal power series, $f(t)=\sum_{k\ge 0}a_k\frac{t^k}{k!}$, the linear operator $f(t)$ acts as such that</p>
<p>$$f(t)(x^n)=f(t)x^n=\sum_{k=0}^n\binom{n}{k}a_k x^{n-k}$$</p>
<p>Therefore, depending on the coefficients of the power series, we can get some interesting binomial identites. For example, if $f(t)=e^{yt}$, since the coefficients $a_n=y^n$, we get</p>
<p>$$e^{yt}x^n=\sum_{k=0}^n\binom{n}{k}y^k x^{n-k}=(x+y)^n$$</p>
<p>by linearity, </p>
<p>$$(e^{yt}-1)x^n=(x+y)^n-x^n=\sum_{k=1}^{n}\binom{n}{k}y^k x^{n-k}$$</p>
<p>and perhaps not as obvious</p>
<p>$$\left(\frac{e^{yt}-1}{t}\right)x^n=\int_{x}^{x+y}u^ndu$$</p>
<p>Now suppose that $f(t)=e^{yt}-1-yt$. Then</p>
<p>$$(e^{yt}-1-yt)x^n=(x+y)^n-x^n-ynx^{n-1}=\sum_{k=2}^{n}\binom{n}{k}y^k x^{n-k}$$</p>
<p>Obviously there is a nice formed forward difference equation in the previous case that is not happening here. But there is a relationship with subtracted terms of the binomial expansion. What i would really like help understanding is whether or not a possible analogous integral representation exists for the following operator:</p>
<p>$$\left(\frac{e^{yt}-1-yt}{t^2}\right)x^n=\left(\sum_{k=0}^\infty\frac{y^{k+2}}{(k+2)(k+1)}\frac{t^k}{k!}\right)x^n=\sum_{k=0}^n\binom{n}{k}\frac{y^{k+2}}{(k+1)(k+2)}x^{n-k}$$ </p>
<p>$$=\sum_{k=0}^n\binom{n+2}{k+2}\frac{y^{k+2}}{(n+1)(n+2)}x^{n-k}=\frac{1}{(n+1)(n+2)}\sum_{k=2}^{n+2}\binom{n+2}{k}y^kx^{n+2-k}$$ </p>
<p>It is not as simple. Clearly $\frac{d^2}{dx^2}\frac{x^{n+2}}{((n+2)(n+1)}$. If I integrated below I think the math is correct</p>
<p>$$\int_x^{x+y}{\frac{u^{n+1}}{n+1}}du=\frac{1}{(n+1)(n+2)}\sum_{k=1}^{n+2}\binom{n+2}{k}y^kx^{n+2-k}$$ </p>
<p>Which is really close, but the lower bound on the summation is $1$, not $2$. Does any one have any insight in how i can fix this, if possible?</p>
| Han de Bruijn | 96,057 | <p>Introductionary reference:
<UL><LI>
<A HREF="http://www.alternatievewiskunde.nl/jaar2015/examples.htm" rel="nofollow">Operator Calculus</A>
</LI></UL>
So why not replace $\,t\,$ by the differential operator $\,d/dx\,$ everywhere it occurs? Next reference:</p>
<p><UL><LI>
<A HREF="http://math.stackexchange.com/questions/719487/exponential-of-a-function-times-derivative">Exponential of a function times derivative</A>
</LI></UL>
Therefore we have in general and especially:
$$
e^{y\,d/dx} f(x) = f(x+y) \quad \Longrightarrow \quad e^{y\,d/dx} x^n = (x+y)^n
$$
Indeed the operator $1/(d/dx)$ is the inverse of differentiation, which is integration. So let's take a look at your <I>perhaps not as obvious</I> expression:
$$
\left(\frac{e^{y\,d/dx}-1}{d/dx}\right)x^n=\frac{1}{d/dx}\left(e^{y\,d/dx}-1\right)x^n\\
=\int\left[(x+y)^n - x^n\right] dx = \frac{(x+y)^{n+1}-x^{n+1}}{n+1} + C
$$
with $C$ an arbitrary constant. Evaluation is not ambiguous, because the operators $1/(d/dx)$ and $\left(e^{y\,d/dx}-1\right)$ do indeed <I>commute</I>. Which
means that we can do algebra with <I>these</I> operators as with common numbers. The same in, last but not least:
$$
\left(\frac{e^{y\,d/dx}-1-y\,d/dx}{(d/dx)^2}\right)x^n=\int\left\{\int\left[(x+y)^n-x^n\right]dx\right\}dx-\int y\,x^n\,dx
\\=\frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} + Cx + D
$$
with $C$ and $D$ arbitrary constants. This result is deviant from your last formula. It is noticed that:
$$
\frac{d^2}{dx^2} \left[\frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} + C\,x + D\right]
= (x+y)^n - x^n - y\frac{dx^n}{dx}
$$
EDIT. I almost forgot to mention the stuff where it's all about: the <I>formal power series</I>
(<A HREF="https://en.wikipedia.org/wiki/Binomial_theorem" rel="nofollow">Wikipedia</A>).
$$
(y+x)^{n}=\sum _{k=0}^{n}{n \choose k}y^{n-k}x^{k} \quad \Longrightarrow \\
\frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} =\\
\frac{1}{(n+1)(n+2)}\left[\sum _{k=0}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k} - x^{n+2} - (n+2)y\,x^{n+1}\right]=\\
\frac{1}{(n+1)(n+2)}\left[\sum _{k=2}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k}
+ {n+2 \choose 0}y^0 x^{n+2} - x^{n+2} + {n+2 \choose 1}y^1 x^{n+1} - (n+2)y\,x^{n+1} \right]\\
=\frac{1}{(n+1)(n+2)}\left[\sum _{k=2}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k}\right]
$$
So here is your fix on the lower bound of the summation.</p>
|
1,484,838 | <p>Ok, so I just learned trig identities and I come across this problem that had it's answer to it, and I have no idea how they got to that answer.</p>
<p>Here is the problem:</p>
<p>$$
\frac{-\sec\theta}{1-\cos\theta}=\frac{-1-\sec\theta}{\sin^2\theta}
$$</p>
<p>Now the problem calls for the left side to be adjusted. Here's where it came to first:</p>
<p>$$
\frac{-1-\sec\theta}{1-\cos^2\theta}
$$</p>
<p>After that, it then came to the solution which was:</p>
<p>$$
\frac{-1-\sec\theta}{\sin^2\theta}
$$</p>
<p>I'm stumped on how they got to the second step. What am I missing?</p>
| Narasimham | 95,860 | <p>Multiply numerator and denominator by $ (1+ \cos \theta) $ and simplify.</p>
|
2,327,273 | <p>If a tree has 5 vertices of degree 2, 3 vertices of degree 3, 4 vertices of degree 4, then how many leaves are there in that tree? </p>
<p>I know the tree has at least 12 vertices and so it must have at least 11 edges. Also the number of leaves must be odd but I could not proceed further. </p>
| Donald Splutterwit | 404,247 | <p><a href="https://i.stack.imgur.com/PEdCG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PEdCG.png" alt="enter image description here"></a></p>
<p>Note that the number of leaves will be independent of how the verticies of valency $2,3$ & $4$ are joined. We reckon $\color{red}{13}$.</p>
|
181,702 | <p>I am working on getting the hang of proofs by induction, and I was hoping the community could give me feedback on how to format a proof of this nature:</p>
<p>Let $x > -1$ and $n$ be a positive integer. Prove Bernoulli's inequality:
$$ (1+x)^n \ge 1+nx$$</p>
<p><strong>Proof</strong>: </p>
<p>Base Case: For $n=1$, $1+x = 1+x$ so the inequality holds.</p>
<p>Induction Assumption: Assume that for some integer $k\ge1$, $(1+x)^k \ge 1+kx$. </p>
<p>Inductive Step: We must show that $(1+x)^{k+1} \ge 1+(k+1)x$</p>
<p><em>Proof of Inductive Step</em>:
$$\begin{align*}
(1+x)^k &\ge 1+kx \\
(1+x)(1+x)^k &\ge (1+x)(1+kx)\\
(1+x)^{k+1} &\ge 1 + (k+1)x + kx^2 \\
1 + (k+1)x + kx^2 &> 1+(k+1)x \quad (kx^2 >0) \\
\Rightarrow (1+x)^{k+1} &\ge 1 + (k+1)x \qquad \qquad \qquad \square
\end{align*}$$ </p>
| timur | 2,473 | <p>This looks fine to me. Just a small note on formatting of the inequalities: I would combine the third and fourth inequalities as
$$
(1+x)^{k+1} \geq 1+(k+1)x+kx^2>1+(k+1)x,
$$
so there is no need of the fifth line. Or even
$$
(1+x)^{k+1} = (1+x)(1+x)^{k} \geq (1+x)(1+kx)=1+(k+1)x+kx^2>1+(k+1)x.
$$</p>
|
189,650 | <p>let $S=\{s_1, s_2, s_3 \}$, if $s_1$ can be represented as a linear combination of $s_2$ and $s_3$, $s_2$ can be represented as a linear combination of $s_1$ and $s_3$ but $s_3$ can not be represented as a linear combination of $s_1$ or $s_2$ or $s_1$ and $s_2$, can we call $S$ a linearly dependent set? </p>
| Godot | 38,875 | <p>You said that </p>
<p>$$s_1=as_2+bs_3.$$</p>
<p>Therefore</p>
<p>$$s_1-as_2-bs_3=0$$</p>
<p>and your set is not linearly independent.</p>
|
116,394 | <p>After importing a sound file, how can I add an echo to it?</p>
<pre><code> sound = Import["test.wav", "SampleRate"]
</code></pre>
<p>It needs to be apply after time specified by user. This is as far as I have got:</p>
<pre><code> addEcho[sound_, time_] :=
Module[{tmp = sound, channels, samples, duration},
{channels, samples} = Dimensions[Import["test.wav", "Data"]];
duration = samples/tmp // N;
result]
</code></pre>
| demm | 30,122 | <p>The idea is to delay the sound (sometimes also reduce its amplitude) and add it to your original signal. Assuming that </p>
<p><code>Import["yourWavfile.wav", "Elements"]</code>
yields:
<code>{AudioChannels,AudioEncoding,Data,SampledSoundList,SampleRate,Sound}</code></p>
<p>the duration (in seconds) of your sound file is the length of your Data over the SampleRate (e.g. 20000). Thus to delay your signal by e.g. 0.2 seconds simply drop the first 0.2*20000=4000 points and add 4000 zeros at the end of the resulting list. In some cases you may also wish to reduce the intensity of the shifted measurements by multiplying by a given factor 0.5, 0.8 etc. (initially set it to 1). Finally, add the shifted signal to the original one and you should have an echo effect. I am sure there are more elegant ways to do it, here is my attempt (note that snd[[5,2]] is the SampleRate and snd[[3,2,1]] are your raw data):</p>
<pre><code>echo[sound_, delay_, intens_]:=Module[{snd, shift, secondSnd, echoSound},snd = Import[sound, "Rules"];shift = delay*snd[[5,2]];secondSnd=PadRight[Drop[snd[[3, 2, 1]], shift], Length[snd[[3, 2, 1]]]];echoSound = intens*secondSnd + snd[[3, 2, 1]];Sound[SampledSoundList[echoSound, snd[[5, 2]]]]]
</code></pre>
<p>Try <code>echo["MyWavFile.wav", 0.2, 1]</code> in which 0.2 is the delay in seconds and 1 is the amplitude of the shifted sound.</p>
|
3,298,412 | <blockquote>
<p>For an n-dimensional vector space <span class="math-container">$V$</span> and an ordered basis <span class="math-container">$B$</span> of <span class="math-container">$V$</span>
, the mapping <span class="math-container">$\Phi : \mathbb{R}^n → V , \Phi(e_i) = b_i, i = 1,...,n$</span>
is linear , where <span class="math-container">$E = (e_1,...,e_n)$</span> is the standard basis of
<span class="math-container">$\mathbb{R}^n$</span>.</p>
</blockquote>
<p>This is a paragraph from my textbook, I am trying to figure out why <span class="math-container">$\Phi$</span> is a linear map. </p>
<p>For an example, If I set <span class="math-container">$n=2$</span>, I can show that there exist a linear combination of <span class="math-container">$B$</span> vectors for each vector of <span class="math-container">$E$</span> where in this case <span class="math-container">$E = (e_1, e_2)$</span> (Cartesian basis) and <span class="math-container">$B = \{(1, -1)^T, (1, 1)^T\}$</span>. But I don't know what to do after this.</p>
<p>Accroding to my book, the mapping is also isomorphic since <span class="math-container">$\Phi$</span> is linear and <span class="math-container">$\dim(\mathbb{R}^n) = \dim(V)$</span>, I am just stuck at seeing the linearity part. Any pointers?</p>
| ancient mathematician | 414,424 | <p>As stated this is false unless all the <span class="math-container">$b_i=0$</span> . Let <span class="math-container">$\Phi$</span> be as stated on the <span class="math-container">$e_i$</span> and let <span class="math-container">$\Phi(x)=0$</span> otherwise. </p>
<p>Then <span class="math-container">$\Phi(2e_i)=0$</span> but <span class="math-container">$2\Phi(e_i)=2 b_i$</span> so for linearity we need <span class="math-container">$b_i=0$</span>.</p>
|
3,362,000 | <p>From listing the first few terms, I suspect that the sequence is increasing, so I wanted to use mathematical induction to verify my suspicion.</p>
<p>I have assumed that <span class="math-container">$a_k<a_{k+1}$</span>, I don't see how I can obtain <span class="math-container">$a_{k+1}<a_{k+2}$</span> because <span class="math-container">$\frac{1}{a_k}>\frac{1}{a_{k+1}}$</span></p>
| orlp | 5,558 | <p>Hint, prove the following theorems in order:</p>
<blockquote>
<p><strong>Theorem 1.</strong> <span class="math-container">$a_n$</span> is positive.</p>
<p><strong>Theorem 2.</strong> <span class="math-container">$a_n > a_{n-1}$</span></p>
</blockquote>
|
185,112 | <p>Does there exist $m,n\ge1$, an $m \times n$ matrix $A$, and a vector $x \in \mathbb{R}^n$ such that:</p>
<ul>
<li>The entries of $A$ are $\in \{0, 1\}$.</li>
<li>For all pairs of columns $u, v$ of $A$ the entries of $u - v$ are never either all non-negative or all non-positive (i.e. there is a positive entry and a negative entry in $u - v$).</li>
<li>$\sum_i x_i = 0$.</li>
<li>The entries of $Ax$ are all non-negative with at least one entry being strictly positive.</li>
</ul>
<p>Edit: It turns out this is not true via a concrete counterexample found by my collaborator. I would list it, but it's rather large.</p>
| Per Alexandersson | 1,056 | <p>Not if $n$ is too large compared to $m$:
For a fixed number of rows, there is only a finite set of possible columns, $2^m$.
Thus, if $n > 2^n$, some columns are identical. This contradicts property 2.</p>
<p>Using a finer reasoning about the second constraint, it should be easy to strengthen this observation.</p>
<p><em>As a related and easier problem:</em>
What is the maximal size of a subset of $\{0,1\}$-vectors of length $n$,
such that all pairs satisfy property 2?</p>
|
2,780,597 | <p>The <strong>definition</strong> of a <em>convex set</em> is geometrically intuitive. But the definition of <em>convex function</em> doesn't seem so intuitive: $S \subset \mathbb{R}^n$ is convex if given $x,y\in S$ the line segment joining $x,y$ is in $S$. </p>
<p>Let $f$ be a real valued function from an open interval $I$. Consider the graph of $f$ in plane: set theoretically it is
$$\{(x,f(x))\,|\, x\in I\}.$$
For any $a,b\in I$, consider the line joining $(a,f(a))$ and $(b,f(b))$. Then one of the following happens:</p>
<ol>
<li><p>The line lies above the graph of $f$.</p></li>
<li><p>The line lies below the graph of $f$.</p></li>
<li><p>None of these hold. </p></li>
</ol>
<blockquote>
<p><strong>Q.</strong> Suppose you know the definition of <em>convex set</em> and let $f:I\rightarrow \mathbb{R}$ be a function which does not satisfy (3). This means $f$ satisfies (1) or (2). What is <em>intuition way</em> to define function to be convex or concave?</p>
</blockquote>
<p>For example, in Wikipedia, it says that <em>a function is convex if graph above $f$ is convex</em>. But if we are trying to give intuitive definition of convex function based on convex set, according to convexity of region above $f$ or below $f$, what is intuitive way to decide one of them? </p>
<hr>
<p>Since $f$ is satisfying (1) or (2), so the words <em>convex function</em> and <em>concave function</em> are reserved for it; we can assign any word to any case without intuition. But, considering definition of <em>convex set</em> can we get intuition to define convex function? Note that almost everyone knows geometric explanation of the standard definition of convex function. </p>
| Masacroso | 173,262 | <p>In the <a href="https://en.wikipedia.org/wiki/Convex_function" rel="nofollow noreferrer">article of wikipedia about convex functions</a> the relation between the two concepts is stated in this way:</p>
<blockquote>
<p>A function $f:\Bbb R^n\to\Bbb R$ is convex if its <a href="https://en.wikipedia.org/wiki/Epigraph_(mathematics)" rel="nofollow noreferrer">epigraph</a> (the set of points on or above the graph of the function) is a convex set.</p>
</blockquote>
<p>where the epigraph of a function as the above is defined formally as follow:</p>
<p>$$\operatorname{epi}(f):=\left\{(x,\mu)\in\Bbb R^n\times\Bbb R: \mu\ge f(x)\right\}\tag1$$</p>
|
3,953,681 | <p>I have a basic question but I have failed in solving it. I have the equation of a cylinder which is <span class="math-container">$y^2 + z^2 = r^2$</span> (centered in the x-axis). The parametric equation (dependent on <span class="math-container">$L$</span> and <span class="math-container">$s$</span>) is <span class="math-container">$(x,y,z) = (L, r\cos(s), r\sin(s))$</span>.</p>
<p>I would like to rotate it certain angle <span class="math-container">$\theta$</span> (anticlockwise). Thus I have the new axis from the rotation as: <span class="math-container">$x=x'*\cos\theta + z'*\sin\theta$</span>, <span class="math-container">$y=y'$</span> and <span class="math-container">$z=r*\sin\theta$</span>. However, when rewriting the equation of the cylinder as <span class="math-container">$(y')^2 + (-x'*\sin\theta + z'*\cos\theta)^2 = r^2$</span> and parametrizing, I get: <span class="math-container">$(x,y,z) = (L, r*\cos(s), z+x'*\tan\theta)$</span>, with <span class="math-container">$z=r*\sin\theta$</span>. When I plot this, I get a elliptic cylinder.
Does anyone know what am I doing wrong? I need such equation because I will generate multiple cylinders later computationally.</p>
<p>I have followed previous posts such as <a href="https://math.stackexchange.com/questions/2733090/if-i-have-an-oblique-cylinder-can-i-trim-it-in-to-a-rectilinear-cylinder">If I have an oblique cylinder can I trim it in to a rectilinear cylinder?</a> but they actually obtain the elliptic cylinder.</p>
<p>Many thanks!</p>
| Travis Willse | 155,629 | <p><strong>Hint</strong> Your method seems to work, provided anyway that <span class="math-container">$a, b > 0$</span>: The substitution <span class="math-container">$x = \sqrt{\frac{a}{b}} \sin \theta$</span>, <span class="math-container">$dx = \sqrt{\frac{a}{b}} \cos \theta \,d\theta$</span> transforms the integral to
<span class="math-container">$$\frac{1}{\sqrt{a}} \int \csc \theta \,d\theta .$$</span></p>
<p>Another option avoids trigonometric functions altogether: Rewrite the integral as
<span class="math-container">$$\int \frac{x \,dx}{x^2 \sqrt{a - b x^2}} .$$</span></p>
<p>What substitution does this form of the integrand suggest?</p>
<blockquote class="spoiler">
<p><strong>Additional hint</strong> Consider the substitution <span class="math-container">$u^2 = a - b x^2$</span>, <span class="math-container">$du = -2 b x \,dx$</span>.</p>
</blockquote>
|
4,204,133 | <p>Trying to solve <a href="https://math.stackexchange.com/questions/4201328/how-can-i-arrange-a-group-of-people-at-tables-and-switch-them-around-so-that-no">this problem</a> led me to consider the following generalization.</p>
<p>Let <span class="math-container">$g$</span> and <span class="math-container">$p$</span> be positive integers. Imagine that you own <span class="math-container">$g$</span> distinct board games, where each game requires exactly <span class="math-container">$p$</span> people to play. You plan to host a game night, and invite exactly <span class="math-container">$p\times g$</span> friends over to play your games. You plan to do this over the course of <span class="math-container">$g$</span> rounds, where in each round, the group divides itself into <span class="math-container">$g$</span> groups of <span class="math-container">$p$</span> people, and each group plays a game. The goal is to do this in such that way that</p>
<ol>
<li><p>Everyone plays every game exactly once, and</p>
</li>
<li><p>No pair of people play together more than once.</p>
</li>
</ol>
<p>This is equivalent to finding a <span class="math-container">$g\times (gp)$</span> matrix such that each column is a permutation of <span class="math-container">$\{1,\dots,g\}$</span>, each number in <span class="math-container">$\{1,\dots,g\}$</span> appears <span class="math-container">$p$</span> times in each row, and where any two columns agree in at most one place (each column is a player, each row is a round, the entry says which game that player plays in that round). When <span class="math-container">$p=1$</span>, this is just a Latin square of order <span class="math-container">$g$</span>.</p>
<p>My question is, <strong>for what values of <span class="math-container">$g$</span> and <span class="math-container">$p$</span> is this possible</strong>? Also, I am curious if anyone has seen this problem in the literature. This certainly bears a resemblance to the social golfer problem. Both this and the SGP involve dividing <span class="math-container">$pg$</span> people into <span class="math-container">$g$</span> groups of <span class="math-container">$p$</span> over the course of several rounds, such that no pair of people play together twice. The difference is that here, people are playing distinct games instead of splitting into unlabeled groups, so constraint <span class="math-container">$(1)$</span> does not make sense in the context of SGP and makes this problem distinct.</p>
<p>This is possible in the trivial cases where <span class="math-container">$g=1$</span> or <span class="math-container">$p=1$</span>. Besides these, I know that <span class="math-container">$$g\ge p+1$$</span> is a necessary condition for a schedule to exist. Consider the <span class="math-container">$p$</span> people who play Monopoly (say) in the first round. In the next round, they must all play different games, so there must be at least <span class="math-container">$p$</span> other games besides Monopoly.</p>
<p>I suspect that this is also a sufficient condition, since I have only found positive results in this regime:</p>
<ul>
<li><p>You can succeed with <span class="math-container">$p=2$</span>, as long as <span class="math-container">$g$</span> is odd. For each <span class="math-container">$x,y\in \{0,1,\dots,g-1\}$</span>, persons numbered <span class="math-container">$2x$</span> and <span class="math-container">$2y+1$</span> will play game number <span class="math-container">$x+y\pmod g$</span> during round number <span class="math-container">$x-y\pmod g$</span>.</p>
</li>
<li><p>It is also possible when <span class="math-container">$(g,p)=(4,3)$</span> and <span class="math-container">$(5,4)$</span>. I found the schedule for the former case by hand, and for the latter case by computer. However, I cannot see how the <a href="https://math.stackexchange.com/a/4203659/177399">schedule</a> I found would generalize. Here is the solution for <span class="math-container">$(g,p)=(4,3)$</span>:
<span class="math-container">$$
\begin{array}{|cccccccccccc|}\hline
1&1&1&2&2&2&3&3&3&4&4&4\\
2&3&4&1&3&4&1&2&4&1&2&3\\
3&4&2&4&1&3&2&4&1&3&1&2\\
4&2&3&3&4&1&4&1&2&2&3&1\\\hline
\end{array}
$$</span></p>
</li>
</ul>
<p>Therefore, I further ask <strong>is it indeed true that a schedule exists if and only if <span class="math-container">$g\ge p+1$</span></strong>?</p>
| Mark Saving | 798,694 | <blockquote>
<p>Then there is <span class="math-container">$\xi_1 \in U$</span> with <span class="math-container">$\xi_1 = z$</span>.</p>
</blockquote>
<p>No. <span class="math-container">$z \in f(U) \subseteq \mathbb{R}^2$</span>, so <span class="math-container">$z$</span> is an ordered pair, not a single number <span class="math-container">$\xi_1$</span>.</p>
<blockquote>
<p>Conclude that <span class="math-container">$(z - r, z + r) \subseteq f(U)$</span>.</p>
</blockquote>
<p>What does this mean? <span class="math-container">$(z - r, z + r) \in \mathbb{R}^2$</span> is an ordered pair of numbers, so it doesn't make sense to say it's a subset of <span class="math-container">$f(U)$</span>.</p>
<p>Do you mean <span class="math-container">$(z - r, z + r) \in f(U)$</span>? If so, you should prove this by showing there exists <span class="math-container">$a, b$</span> such that <span class="math-container">$f(a, b) = (z - r, z + r)$</span>.</p>
|
1,511,246 | <blockquote>
<p>What is the value of $0.7\overline{54}$ +$0.69\overline2$?</p>
<p>(a) $\frac{1813}{900}$ (b) $\frac{1783}{910}$ (c) $\frac{14323}{9900} (d) \frac{13243}{9900}$</p>
</blockquote>
<p>I get</p>
<p>@edit</p>
<p>$$754-7/990 + 692-69/900$$=$747$/$990$ + $623$/$900$=$1$/$90$($747$/$11$ + $623$/$10$)</p>
<p>=($7470$/$11$ + $623$ . $11$/$10$)=($7470$ + $6853$)=$14323$/$9900$</p>
<p>Thankx for help I did not multiplied by $90$</p>
<p>Can anyone guide me how to solve the problem?</p>
| SchrodingersCat | 278,967 | <p>The question is not $0.754+0.692=?$ <br>
The question asks $0.7 \overline{54}+0.69\overline2=?$</p>
<p>So the answer will be $0.7 \overline{54}+0.69\overline2=\frac{754-7}{990}+\frac{692-69}{900}=\frac{747}{990}+\frac{623}{900}=1.44\overline{67}=\frac{14323}{9900}$</p>
|
1,511,246 | <blockquote>
<p>What is the value of $0.7\overline{54}$ +$0.69\overline2$?</p>
<p>(a) $\frac{1813}{900}$ (b) $\frac{1783}{910}$ (c) $\frac{14323}{9900} (d) \frac{13243}{9900}$</p>
</blockquote>
<p>I get</p>
<p>@edit</p>
<p>$$754-7/990 + 692-69/900$$=$747$/$990$ + $623$/$900$=$1$/$90$($747$/$11$ + $623$/$10$)</p>
<p>=($7470$/$11$ + $623$ . $11$/$10$)=($7470$ + $6853$)=$14323$/$9900$</p>
<p>Thankx for help I did not multiplied by $90$</p>
<p>Can anyone guide me how to solve the problem?</p>
| fleablood | 280,126 | <p>(754 - 7)/990 + (692 - 69)/900 = 747/990 + 623/900 = 7470/9900 + 6853/9900 = 14323/9900</p>
|
4,416,063 | <p>How to solve <span class="math-container">$\int\frac{\ln(x \ln(x))}{x} dx$</span>?</p>
<p>My work:<br />
Let <span class="math-container">$t = \ln(x) \implies x= e^t ; dt = \dfrac{dx}{x}$</span></p>
<p>So above integral changes to,
<span class="math-container">$$\int t ( e^t t) dt$$</span>
<span class="math-container">$$\int t^2 e^t dt$$</span></p>
<p>Using IBP to get:
<span class="math-container">$$t^2e^t - 2te^t + 2e^t + C$$</span></p>
<p>Undoing the substitution to get,
<span class="math-container">$$x\log^2(x) - 2x\log(x) + 2x + C$$</span></p>
<p>But differentiating this doesn't give the original integral. I've definitely done anything wrong which I'm unable to understand. Can anyone help me to solve it?</p>
<p>Also I'm wondering if We could solve it without using by parts.</p>
<hr />
<p><strong>Edit</strong><br />
I tried again as suggested in comments.
<span class="math-container">$$\int\dfrac{\log(x \log(x))}{x}dx \overset{t\to\log(x)}= \int\log(e^t t) dt = \int t + \log(t) = \dfrac{t^2}{2} + t\log(t) - t + C$$</span></p>
<p>By undoing the substitution,
<span class="math-container">$$\boxed{\dfrac{\log^2(x)}{2} + \log(x)\log(\log x) - \log(x)+ C}$$</span></p>
<p>What's wrong with this?</p>
| Lai | 732,917 | <p><span class="math-container">$$
\begin{aligned}
I&= \int \ln (x \ln x) d(\ln x) \\
=& \ln x \cdot \ln (x \ln x)-\int \ln x \cdot \frac{1}{x \ln x}\left(x \cdot \frac{1}{x}+\ln x\right) d x \\
=& \ln x \cdot\ln (x \ln x)-\int \frac{1+\ln x}{x} d x \\
=& \ln x \cdot\ln (x \ln x)-\int(1+\ln x) d(\ln x) \\
=& \ln x \cdot \ln (x \ln x)-\ln x-\frac{\ln ^{2} x}{2}+C .
\end{aligned}
$$</span></p>
|
181,499 | <p>In many of the classes that I teach, I require students to learn the basics of Mathematica which we use throughout the semester to do computations and to submit homeworks (in notebook form). Some students really like this and some... not so much. </p>
<p>Since I teach in an engineering department, almost everyone already knows some programming language: <em>Matlab</em>, <em>python</em>, <em>java</em>, or <em>C</em> are the most common, though there is quite a variety. One thing that I have found pretty effective is to try and relate Mathematica formalisms, structures, and ideas to those that students already know. For example:</p>
<p><span class="math-container">$-$</span> When talking about using the <a href="https://reference.wolfram.com/language/ref/Listable.html" rel="noreferrer"><code>Listable</code></a> Attribute of functions, I compare this to Matlab's <a href="https://www.mathworks.com/help/matlab/matlab_prog/vectorization.html" rel="noreferrer">vectorization</a></p>
<p><span class="math-container">$-$</span> When talking about alternatives for loops, Mathematica's <a href="https://reference.wolfram.com/language/ref/Table.html" rel="noreferrer"><code>Table</code></a> function is analogous to python's <a href="https://docs.python.org/3/tutorial/datastructures.html#list-comprehensions" rel="noreferrer">List Comprehensions</a>, for example, observe the similarity between</p>
<pre><code>squares = [x**2 for x in range(10)]
</code></pre>
<p>and</p>
<pre><code>squares = Table[x^2, {x, Range[10]}]
</code></pre>
<p><span class="math-container">$-$</span> Mathematica's Notebook format is analogous to <a href="https://jupyter.org/index.html" rel="noreferrer">Jupyter notebooks</a> which merge word processing, computation, and interactive presentations.</p>
<p>My question is this: What are some other analogies between Mathematica functions, expressions, and structures that might be helpful to new users in understanding "what Mathematica is thinking" or "why it works that way"?</p>
<p>Update: It seems that we have some very good answers for Matlab and for python. How about other languages? Any nice analogies for/with other popular languages?</p>
| sunt05 | 260 | <p>For those with experience in python, WRI has already provided <a href="http://www.wolfram.com/language/fast-introduction-for-programmers/en/" rel="noreferrer">a nice introductory tutorial</a> along with many analogies.</p>
<p>However, for the class intended for image processing as mentioned by the OP, pure python is for certain not enough: <a href="https://www.numpy.org" rel="noreferrer"><code>numpy</code></a>, <a href="https://pandas.pydata.org" rel="noreferrer"><code>pandas</code></a>, <a href="https://www.scipy.org" rel="noreferrer"><code>scipy</code></a> and <a href="https://pillow.readthedocs.io" rel="noreferrer"><code>pillow</code></a> are some of the essential packages to go with.</p>
|
213,872 | <p>I'm learning probability theory and I see the half-open intervals $(a,b]$ appear many times. One of theorems about Borel $\sigma$-algebra is that</p>
<blockquote>
<p>The Borel $\sigma$-algebra of ${\mathbb R}$ is generated by inervals of the form $(-\infty,a]$, where $a\in{\mathbb Q}$. </p>
</blockquote>
<p>Also, the distribution function induced by a probability $P$ on $({\mathbb R},{\mathcal B})$ is defined as
$$
F(x)=P((-\infty,x])
$$</p>
<p>Is it because for some theoretical convenience that the half-open intervals are used often in probability theory or are they of special interest?</p>
| shaman | 823,495 | <p>I would say that the answer is best presented in the reverse order of your question:</p>
<p>The (cumulative) distribution function is defined the way it is, because that is a natural way to think about accumulating probabilities as you observe more events in the sample space. So in a sense, the half-open intervals are the most fundamental events to consider. This is, in essence, what Hagen mentioned.</p>
<p>Also, sigma-algebras define what combinations of sets (i.e. the fundamental events) are also considered as events and thus can be assigned probability. The set of all such events on the real line, generated by the half-open intervals are the Borel sets.</p>
|
2,648,626 | <p>Is the set $(e_n)_{n>0}$ a (vector space) basis for the sequence Hilbert space $l^2$? It is a Hilbert space basis anyway.</p>
<p>I would say no, because the sequence $\left(\frac{1}{n}\right)_{n>0}$ is in $l^2$ but it can't be written as a finite linear combination of $e_i$'s.</p>
<p>Is that right?</p>
| José Carlos Santos | 446,262 | <p>Yes, it is entirely correct. Actually, it can be proved that a Hilbert space is infinite-dimensional if and only if no Hilbert basis is a basis in the Linear Algebra sense.</p>
|
112,432 | <p>a) Is true the following statement. Let $h$ be analytic in the unit disk such that $$|h(z)|\le \frac{|z|^2}{1-|z|^2},$$ then $$|h'(z)|\le \frac{2}{(1-|z|^2)^2}.$$
a') Is true the following statement. Let $h$ be analytic in the unit disk such that $$|h(z)|\le \frac{|z|^2}{1-|z|^2},$$ then the inequality $$|h'(z)|\le \frac{8}{\pi(1-|z|^2)^2}$$ is sharp. The inequality can be proved by using Schur test, and Riesz-Thorin convexity type theorem (Dunford & Schwartz 1958, §VI.10.11).</p>
<p>b) If $$|h(z)|\le \frac{|z|^2}{|1-z^2|}$$ then we have better conclusion $$|h'|\le \frac{2|z|}{(1-|z|^2)|1-z^2|}$$ and this follows by using Schwarz lemma. Namely in this case $$|H(z)|=|(1-z^2) h(z)/z^2|\le 1.$$ Then $$|H'(z)|\le \frac{1-|H(z)|^2}{1-|z|^2}.$$</p>
<p>As $$H'(z)=(1-z^2) h'(z)/z^2-2/z^3 h(z),$$ it follows that $$|(1-z^2) h'(z)/z^2|\le \frac{2(1-|z|^2)/|z|^3 h(z)+1-|H(z)|^2}{1-|z|^2}$$ $$\le \frac{2|H(z)|/|z| +1-|H(z)|^2}{1-|z|^2}\le \frac{2|z|^{-1}}{1-|z|^2}.$$</p>
<p>The question a) is related to precise estimation of norm of a Bergman projection into Bloch space and is far for being a homework.</p>
| Alexandre Eremenko | 25,510 | <p>Let $|z|=r$, apply the Cauchy estimate to the disc $|\zeta-z|<(1-r)/2$.
We obtain
$$|f'(z)|\leq \frac{2}{1-r}\frac{(1+r)^2}{(1-r)(3+r)}.$$
Maximizing the factor $(1+r)^2/(3+r)$ by Calculus, we obtain that is it at most $1$.</p>
<p>This gives
$$|f'(z)|\leq\frac{2}{(1-|z|)^2}$$
which is worse than conjectured only by a factor of $(1+|z|)^2$, which is at most $4$.
Perhaps one can improve the constant by applying Cauchy to a disc of radius $t\in(0,1-r)$,
and then optimizing in $t$, which leads to solving a cubic equation.</p>
<p>It is not likely that a simple extremal function exists, and probably for each $z$
there will be a different extremal function.</p>
|
3,041,656 | <p>I need some help in a proof:
Prove that for any integer <span class="math-container">$n>6$</span> can be written as a sum of two co-prime integers <span class="math-container">$a,b$</span> s.t. <span class="math-container">$\gcd(a,b)=1$</span>.</p>
<p>I tried to go around with "Dirichlet's theorem on arithmetic progressions" but didn't had any luck to come to an actual proof.
I mainly used arithmetic progression of <span class="math-container">$4$</span>, <span class="math-container">$(4n,4n+1,4n+2,4n+3)$</span>, but got not much, only to the extent of specific examples and even than sometimes <span class="math-container">$a,b$</span> weren't always co-prime (and <span class="math-container">$n$</span> was also playing a role so it wasn't <span class="math-container">$a+b$</span> it was <span class="math-container">$an+b$</span>).</p>
<p>I would appriciate it a lot if someone could give a hand here.</p>
| templatetypedef | 8,955 | <p>Here's another route you can take to solve this problem. For any <span class="math-container">$n \ge 7$</span>, you want to show that there is a number <span class="math-container">$a$</span> where</p>
<ol>
<li><span class="math-container">$gcd(a, n - a) = 1$</span>,</li>
<li><span class="math-container">$1 < a < n$</span>, and</li>
<li><span class="math-container">$1 < n - a < n$</span>.</li>
</ol>
<p>One option would be to choose <span class="math-container">$a$</span> to be the smallest prime number that doesn't divide <span class="math-container">$n$</span>. In that case, <span class="math-container">$gcd(a, n - a) = 1$</span> because otherwise you'd have <span class="math-container">$gcd(a, n - a) = a$</span>, meaning that <span class="math-container">$a$</span> divides <span class="math-container">$a + (n - a) = n$</span>, contradicting the fact that <span class="math-container">$a$</span> doesn't divide <span class="math-container">$n$</span>.</p>
<p>What you'll need to then show is that if you pick <span class="math-container">$n \ge 7$</span> that the smallest prime number that doesn't divide <span class="math-container">$n$</span> happens to be less than <span class="math-container">$n - 1$</span>. I'll leave that as an exercise to the reader. :-)</p>
|
2,516,942 | <p>Trying to find all solutions on (-infinity,+infinity) for :
$y''+4y = 0$</p>
<p>I know that the discriminant of the characteristic equation is -16 so the roots are complex. so $k=0.5 \cdot \sqrt{-16} = 2i$</p>
<p>$f_1(x) = e^{(2ix)} = \cos(2x) + i\sin(2x)$</p>
<p>$f_2(x) = e^{(-2ix)} = \cos(2x) - i\sin(2x)$</p>
<p>and so the general solution therefore is </p>
<p>$y=c_1(\cos(2x) + i\sin(2x)) + c_2(\cos(2x) - i\sin(2x))$</p>
<p>but the answers say that it is </p>
<p>$y=C_1\cos(2x) + C_2\sin(2x)$</p>
<p>So I am having trouble interpreting the real parts of the complex roots. Could someone please explain how to get to the answer from here?</p>
| fleablood | 280,126 | <p>Let's see:</p>
<p>$(2n!) = \prod\limits_{k=1..2n;n \text{even}}k*\prod\limits_{k=1...2; n\text{ odd}}k=\prod\limits_{k=1..n}2k\prod\limits_{k=1...2n; n\text{ odd}}k=2^n*n!*\prod\limits_{k=1...2n; n\text{ odd}}k$</p>
<p>So $\frac {(2n)!}{(n!)^2}\cdot \frac 1{2^{2n}} = \frac {\prod\limits_{k=1...2n; n\text{ odd}}k}{n!}*\frac 1{2^n}=\frac {\prod\limits_{k=1...2n; n\text{ odd}}k}{\prod\limits_{k=1...2n; n\text{ even}}k}=\prod\limits_{k=1}^n \frac {2k-1}{2k}$</p>
<p>This becomes $\frac 1{2\sqrt{n}} \le \prod\limits_{k=1}^n \frac {2k-1}{2k} \le \frac 1{\sqrt{2n+1}}$ or $\sqrt{2n+1} \le \prod\limits_{k=1}^n \frac {2k}{2k-1} \le 2\sqrt{n}$. Or $2n+1 \le \prod\limits_{k=1}^n \frac {4k^2}{(2k-1)^2}\le 4n$... which we can probably do by induction.</p>
<p>For $n=1$ then $3 \le 4 \le 4$</p>
<p>If true for $n=k$ then</p>
<p>$2k+1\le \prod\limits_{j=1}^k\frac {4j^2}{(2j-1)^2} \le4k$</p>
<p>$(2k+1)*\frac{4(k+1)^2}{(2k+1)^2}\le \prod\limits_{j=1}^{k+1}\frac {4j^2}{(2j-1)^2} \le 4k*\frac{4(k+1)^2}{(2k+1)^2}$</p>
<p>And $(2k+1)*\frac{4(k+1)^2}{(2k+1)^2}= (2k+1)*\frac {4k^2 + 8k + 4}{4k^2 + 4k+1}=(2k+1)(1 + \frac {4k+3}{4k^2 + 4k + 1})$</p>
<p>$= 2k+ 1 + \frac {2k(4k+3) + 4k+3}{4k^2 + 4k + 1}=2k+1 + \frac {8k^2 + 10k + 3}{4k^2 + 4k + 1}>2k+1 + 2 = 2(k+1) + 1$</p>
<p>while $4k*\frac{4(k+1)^2}{(2k+1)^2} =4k*\frac {4k^2 + 8k + 4}{4k^2 + 4k + 1} = 4k(1 + \frac {4k+3}{4k^2 + 4k + 1})=4k + \frac {16k^2 + 12k}{4k^2 + 4k +1} < 4k + \frac {16k^2 + 16k + 4}{4k^2 + 4k + 1} = 4(k+1)$</p>
<p>So... it is true (with equality only holding on the left side for $n=1$... which makes me a little uneasy).</p>
<p>Obviously this is not what they intended. But it works. I assume they wanted something to do with $\sqrt{2n + 1} = \sqrt{(n+1)^2 - n^2}$ but ... I couldn't seem to work that in.</p>
<p>Oh.... $\sqrt{2n + 1}=\sqrt{(n+1)^2 - n^2} = \sqrt{(n-1)(n+1)}< \sqrt{n^2} = n$. Maybe you can work that in. Well... I don't know. </p>
|
642,631 | <p>What is $[\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}]$?</p>
<p>On the one hand, we have $[\mathbb{Q}(i,\sqrt{2},\sqrt{3}):\mathbb{Q}(i,\sqrt{2})]\cdot[\mathbb{Q}(i,\sqrt{2}):\mathbb{Q}(i)]\cdot[\mathbb{Q}(i):\mathbb{Q}]=2^3=8.$</p>
<p>On the other hand, the minimum polynomial in $\mathbb{Q}[x]$ containing $i,\sqrt{2},\sqrt{3}$ as roots is $(x^2+1)(x^2-2)(x^2-3)$, which is of degree $6$.</p>
<p>What am I misunderstanding?</p>
| initial_D | 105,162 | <p>8 sounds right, and your reasoning is right as well. $\mathbb{Q}[i,\sqrt{2},\sqrt{3}]$ is the splitting field of the polynomial you gave. The only thing you might be missing is the degree of the splitting field is not necessarily the degree of the polynomial itself.</p>
|
3,910,053 | <p>The function
<span class="math-container">$u(x,t) = \frac{2}{\sqrt{\pi}}$$\int_{0}^\frac{x}{\sqrt{t}} e^{-s^2}ds$</span>
satisfies the partial differential equation</p>
<p><span class="math-container">$$\frac{\partial u}{\partial t} = K\frac{\partial^2u}{\partial x^2}$$</span></p>
<p>where <span class="math-container">$K$</span> is a positive constant number. Find the value of <span class="math-container">$K$</span>.</p>
<p>The value of <span class="math-container">$K$</span> that I calculated isn't a constant. Can anyone please help with this question?
Thank you so much!</p>
| Nevzat Eren Akkaya | 611,386 | <p>Let <span class="math-container">$c_1,...,c_n$</span> be eigenvalues of <span class="math-container">$A$</span> then</p>
<p><span class="math-container">$t=c_i$</span> <span class="math-container">$\Rightarrow$</span> <span class="math-container">$\det(A-tI)=0$</span></p>
<p><span class="math-container">$\det((A+2I)-tI)=\begin{vmatrix} 5-t & 2 \\ 5 & 2-t \end{vmatrix}=(t-5)(t-2)-10=t(t-7)$</span></p>
<p>roots of determinants are <span class="math-container">$7$</span> and <span class="math-container">$0$</span> so our eigenvalues are <span class="math-container">$0$</span> and <span class="math-container">$7$</span>.</p>
|
3,130,939 | <p>Suppose the following function with pi notation, with the pi denoting the iterated product, multiplying from <span class="math-container">$i = 0$</span> to <span class="math-container">$i = n$</span>:</p>
<p><span class="math-container">$$\prod_{i=0}^n \ln(y_i^{x - 1})$$</span></p>
<p>That is, the natural logarithm of <span class="math-container">$y$</span>, subscripted by <span class="math-container">$i$</span>, to the power of <span class="math-container">$x - 1$</span>.</p>
<p>What is the derivative of this product - to be clear, its derivative with respect to <span class="math-container">$x$</span>, not <span class="math-container">$y$</span>? </p>
| Paras Khosla | 478,779 | <p>Define <span class="math-container">$\alpha=\prod_{i=0}^n \ln(y_i)$</span>. Note that <span class="math-container">$\alpha$</span> is simply a constant. Use logarithm properties, use the power rule for derivatives and you're done.</p>
<p><span class="math-container">$$f(x)=\prod_{i=0}^{n}\ln{y_i}^{x-1}=\prod_{i=0}^{n}(x-1)\ln{y_i}$$</span><span class="math-container">$$\alpha(x-1)^{n+1}\implies \dfrac{\mathrm df}{\mathrm dx}=\alpha(n+1)(x-1)^n$$</span>
<span class="math-container">$$\boxed{\dfrac{\mathrm d}{\mathrm dx}\prod_{k=0}^n \ln{y_i}^{x-1}=(n+1)(x-1)^n\prod_{k=0}^n\ln y_i}$$</span></p>
|
634,127 | <p>How to prove this (true or not)?</p>
<blockquote>
<p>$f(a,b) = f(a,c)$ must hold if $b = c$</p>
</blockquote>
<p><b>Note:</b> <i><b>f(a,b)</b> is a function with <b>a</b> & <b>b</b></i> parameters</p>
<p>thanks</p>
| Doug Spoonwood | 11,300 | <p>If we assume that $\forall$x(x=x), then yes.</p>
<p>Suppose that b=c.</p>
<p>f(a,b)=f(a,b) since we can derive it from substitution and that $\forall$x(x=x).</p>
<p>Now since b=c we can replace just the second "b" by "c" and obtain</p>
<p>f(a,b)=f(a,c).</p>
|
203,456 | <p>Please help me proof $\log_b a\cdot\log_c b\cdot\log_a c=1$, where $a,b,c$ positive number different for 1.</p>
| DonAntonio | 31,254 | <p>Change all to the natural logarithm $\log\,$:</p>
<p>$$\log_ba\cdot\log_cb\cdot\log_ac=\frac{\log a}{\log b}\frac{\log b}{\log c}\frac{\log c}{\log a}$$</p>
<p>and voila.</p>
|
627,444 | <p>I guess that I'm quite familiar with the basic "everyday algebraic structures" such as groups, rings, modules and algebras and Lie algebras. Of course, I also heard of magmas, semi-groups and monoids, but they seem to be way to general notions as to admit a really interesting theory.</p>
<p>Thus, I'm wondering whether there are also other interesting algebraic structure (here, this means mainly some set $S$ together with a bunch of functions $f_i:S^n\to S$ satisfying some laws) which behave somewhat differently, i. e. satisfy some unusual relations like $(ab)c=(ca)(cb)$ or $ba=(aa)(bb)$, but in such a way that there is a decent amount of theory about them (some kind of nontrivial classification or representation theorem would be truly fascinating).</p>
<p>Bonus points if these structures arise naturally in some areas of mathematics.</p>
| Steven Gubkin | 34,287 | <p>Groupoids seem somewhat seem somewhat more exotic than groups, but are actually a lot more natural in many ways. For example the fundamental groupoid is really more "fundamental" than the fundamental group in many ways. </p>
<p><a href="http://en.wikipedia.org/wiki/Operad_theory" rel="nofollow">Operads</a> are also quite handy when doing algebraic topology. I recommend looking at Tom Leinster's book "Higher operads, higher categories".</p>
|
3,576,008 | <p><strong>Question:</strong></p>
<p>In acute <span class="math-container">$\Delta ABC$</span>, let <span class="math-container">$D$</span> be the foot of the altitude from <span class="math-container">$A$</span> to <span class="math-container">$BC$</span>, and let <span class="math-container">$\overline{AD}$</span>
intersect the circumcircle of <span class="math-container">$\Delta ABC$</span> at <span class="math-container">$E$</span>. </p>
<p>Let the circle with diameter <span class="math-container">$AE$</span> intersect lines <span class="math-container">$AB$</span> and <span class="math-container">$AC$</span> at <span class="math-container">$N$</span> and <span class="math-container">$M$</span>, respectively. Given that <span class="math-container">$DB=3NB$</span> and <span class="math-container">$MA=5NA$</span>, </p>
<p>then the value of <span class="math-container">$\displaystyle \frac{DC}{MC} $</span> can be written in simplest form as <span class="math-container">$\displaystyle \frac{a}{b}$</span>. What is the value of <span class="math-container">$a-b$</span>?</p>
<p><a href="https://brilliant.org/problems/interesting-circles/" rel="nofollow noreferrer">SOURCE</a></p>
<p><strong>My attempt to draw:</strong> Please guide me.</p>
<p><a href="https://i.stack.imgur.com/cRZK0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cRZK0.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/iV7vu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iV7vu.jpg" alt="enter image description here"></a></p>
| Matt | 263,495 | <p>It does indeed appear to indicate that there is a typo in the question as posed. The circle with the diameter <span class="math-container">$AE$</span> does not have to intersect <span class="math-container">$AB$</span> and <span class="math-container">$AC$</span> at all, and so in particular the point <span class="math-container">$M$</span> does not exist.</p>
|
1,380,508 | <p>Is there such a proof that states that the Runge Phenomena will always occur when interpolating with higher order polynomials or is this just observed empirically?</p>
| Community | -1 | <p>The Runge Phenomenon does not always occur. You can interpolate, say, $f(x)= e^x$ using equally spaced nodes on any interval $[a,b]$, and the interpolating polynomials will converge to $f$ uniformly. (The same holds for any function whose Taylor series has infinite radius of convergence.)</p>
<p>When the Runge phenomenon does occur, it is a lot easier to observe empirically than to write down a rigorous proof that it happens. This is why the textbooks on numerical analysis tend to do the former and not the latter. </p>
<p>For the classical example of intepolating $f(x)=1/(1+x^2)$ on $[-5,5]$ by equally spaced nodes, David Speyer <a href="https://math.stackexchange.com/a/807784">posted a complete proof</a> of divergence at $x=4$.</p>
|
14,552 | <p>What are good examples of proofs by induction that are relatively low on algebra? Examples might include simple results about graphs.</p>
<p>My aim is to help students get a sense of the logical form of an induction proof (in particular proving a statement of the form 'if $P(k)$ then $P(k+1)$'), independent of the way one might show that in a proof about series formulae specifically.</p>
| Aeryk | 401 | <p>How about: A tree with $n\ge 1$ vertices has $n-1$ edges.</p>
|
14,552 | <p>What are good examples of proofs by induction that are relatively low on algebra? Examples might include simple results about graphs.</p>
<p>My aim is to help students get a sense of the logical form of an induction proof (in particular proving a statement of the form 'if $P(k)$ then $P(k+1)$'), independent of the way one might show that in a proof about series formulae specifically.</p>
| ncr | 1,537 | <p>I think tiling problems are good for this kind of thing. See, for example, <a href="https://www.math.hmc.edu/funfacts/ffiles/20002.4.shtml" rel="nofollow noreferrer">this</a>. There they describe how to prove the statement "if you have a $2^n\times 2^n$ chessboard with one square missing, then you can tile it with L-shaped trominoes." There are other tiling questions, as well, such as the ones <a href="http://www.mathcircles.org/wp-content/uploads/2017/10/tiling.pdf" rel="nofollow noreferrer">here</a> that deal with triangular chessboards and trominoes. Yet others ask students to count the number of ways to tile something (e.g., <a href="http://www.algorithmist.com/index.php/UVa_10918" rel="nofollow noreferrer">here</a>) via linear recurrences (which usually can easily be proved inductively).</p>
|
292,651 | <blockquote>
<p>Does an integer $9<n<100$ exist such that the last 2 digits of $n^2$ is $n$? If yes, how to find them? If no, prove it.</p>
</blockquote>
<p>This problem puzzled me for a day, but I'm not making much progress. Please help. Thanks.</p>
| robjohn | 13,854 | <p>We are solving $n(n-1)=n^2-n\equiv0\pmod{100}$. Since $\gcd(n,n-1)=1$, one of $n$ or $n-1$ must be a multiple of $4$ while the other must be a multiple of $25$.This leads to the equations
$$
\begin{align}
4x-25y=+1\tag{1}\\
4x-25y=-1\tag{2}
\end{align}
$$
For $(1)$, $n=4x$ and $n-1=25y$. For $(2)$, $n=25y$ and $n-1=4x$.</p>
<p>Using the Euclidean algorithm, $(1)$ has solutions $(x,y)=(-6+25k,-1+4k)$ and $(2)$ has solutions $(6+25k,1+4k)$. The two solutions that give $4x$ and $25y$ between $9$ and $99$ are $(19,3)$ and $(6,1)$.</p>
<p>$(19,3)$ solves $(1)$ so $n=4x=76$ and $76^2=5776\equiv76\pmod{100}$</p>
<p>$(6,1)$ solves $(2)$ so $n=25y=25$ and $25^2=625\equiv25\pmod{100}$</p>
<p>Thus, the two integers that satisfy the given condition are $25$ and $76$.</p>
|
86,762 | <p>The other day, my teacher was talking infinite-dimensional vector spaces and complications that arise when trying to find a basis for those. He mentioned that it's been proven that some (or all, do not quite remember) infinite-dimensional vector spaces have a basis (the result uses an Axiom of Choice, if I remember correctly), that is, an infinite list of linearly independent vectors, such that any element in the space can be written as a finite linear combination of them. However, my teacher mentioned that actually finding one is really complicated, and I got a sense that it was basically impossible, which reminded me of Banach-Tarski paradox, where it's technically 'possible' to decompose the sphere in a given paradoxical way, but this cannot be actually exhibited. So my question is, is the basis situation analogous to that, or is it actually possible to explicitly find a basis for infinite-dimensional vector spaces?</p>
| David Wheeler | 23,285 | <p>The "hard case" is essentially equivalent to this one:</p>
<p>Find a basis for the real numbers <span class="math-container">$\mathbb{R}$</span> over the field of the rational numbers <span class="math-container">$\mathbb{Q}$</span>.</p>
<p>The reals are obviously an extension field of the rationals, so they form a vector space over <span class="math-container">$\mathbb{Q}$</span>. It should be clear that such a basis has to be uncountable (for if it were countable, the reals would likewise also be countable).</p>
<p>It should also be clear that such a basis is a subset of <span class="math-container">$\{1\} \cup \mathbb{R} \setminus \mathbb{Q}$</span>. The trouble is, that the power set of the reals is "so big" that it's not even clear how to name the sets we need to apply the axiom of choice TO. Linearly independent subsets however, DO satisfy the requirements for Zorn's Lemma, a form of the Axiom of Choice.</p>
<p>A relatively easy-to-follow proof of the existence of a basis for any vector space using Zorn's Lemma can be found here: <a href="https://planetmath.org/EveryVectorSpaceHasABasis" rel="nofollow noreferrer">Link</a></p>
|
611,529 | <p>$$i^3=iii=\sqrt{-1}\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)(-1)}=\sqrt{-1}=i
$$</p>
<p>Please take a look at the equation above. What am I doing wrong to understand $i^3 = i$, not $-i$?</p>
| Brian | 114,928 | <p>We cannot say that $\sqrt{a}\sqrt{b}=\sqrt{ab}$ for negative $a$ and $b$. If this were true, then $1=\sqrt{1}=\sqrt{\left(-1\right)\cdot\left(-1\right)} = \sqrt{-1}\sqrt{-1}=i\cdot i=-1$. Since this is false, we have to say that $\sqrt{a}\sqrt{b}\neq\sqrt{ab}$ in general when we extend it to accept negative numbers.</p>
|
611,529 | <p>$$i^3=iii=\sqrt{-1}\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)(-1)}=\sqrt{-1}=i
$$</p>
<p>Please take a look at the equation above. What am I doing wrong to understand $i^3 = i$, not $-i$?</p>
| Abramo | 3,416 | <p>Multiplying by $i$ in the complex plane amounts to a counterclockwise turn of $90$ degrees. Hence it is geometrically clear that $i^3 = -i$, as the following drawing shows</p>
<p><img src="https://i.stack.imgur.com/W8UJo.png" alt="enter image description here"></p>
<p>Regarding your equations, as remarked already, $\sqrt{a}\sqrt{b}=\sqrt{ab}$ holds only for positive real $a$ and $b$, so you cannot use it with complex numbers.</p>
|
455,230 | <p>I found this proposition and don't see exactly as to why it is true and even more so, why the converse is false:</p>
<p>Proposition 1. The equivalence between the proposition $z \in D$ and the proposition $(\exists x \in D)x = z$ is provable from the definitory equations of the existential quantifier and of the equality relation. If $D = \{t_{1},t_{2},...t_{n}\}$, the sequent
$z = t_{1} \vee z = t_{2} \vee z \vee ... z = t_{n} \vdash z \in D$ is provable from the definition of the additive disjunction $\vee$.</p>
<p>The converse sequent $z \in D \vdash z = t_{1} \vee z = t_{2} \vee ... z = t_{n}$ is not provable.</p>
<p>The author goes on to say: "We adopt the intuitionistic interpretation of disjunction. With respect to it, one can characterize a particular class of finite sets"</p>
<p>On the first part of the proposition, well I am not sure what the point is as if we take an element z in D, then we could just call this element x and hence this x = z. Is there something more to this? On the second part of Proposition 1 since $z = \text{ some } t \in D$ since t is in the set of axioms and $t = z$, then $t \vdash z$ as z is derivable from t.</p>
<p>For the converse if $z \in D$ then why wouldn't z = some $t \in D$? Is this because of the Incompleteness theorem? That perhaps there D as a set of axioms has some consequence which can not be proven by the set of axioms in D? Or perhaps I am way off here.</p>
<p>Any ideas?</p>
<p>Thanks,</p>
<p>Brian</p>
| Jeremy Rickard | 88,262 | <p>For an Artinian example, let $A$ be the algebra of upper triangular $2\times 2$ matrices over a field $K$, let $J$ be the left ideal consisting of all matrices with bottom row zero, and let $S$ be the simple left $A$-module $A/J$. Then it's easy to check that $Ja\neq0$ for every non-zero element $a$ of $A$, but $Js=0$ for every $s\in S$, so $S$ can't be isomorphic to a submodule of the regular module.</p>
|
3,715,484 | <p>As the title saying , the question is how to find the radius <span class="math-container">$R$</span> of convergence of <span class="math-container">$\sum_{n=1}^{\infty}\frac{\sin n}{n} x^n$</span>. My method is as the following:</p>
<p>When <span class="math-container">$x=1$</span>, it is well known that the series <span class="math-container">$\sum_{n=1}^{\infty}\frac{\sin n}{n}$</span> is convergent by Dirichlet's test, and so is <span class="math-container">$\sum_{n=1}^{\infty}(-1)^n \frac{\sin n}{n}$</span> . when <span class="math-container">$x>1$</span>, the limit <span class="math-container">$\lim_{n\to \infty} \frac{\sin n}{n} x^n $</span> does not exist. Therefore, if <span class="math-container">$x>1$</span>, the series <span class="math-container">$\sum_{n=1}^{\infty}\frac{\sin n}{n} x^n$</span> is not convergent. So <span class="math-container">$R=1$</span>. Is this solution right? or is there any other method to calculate the radius?</p>
<p>I would appreciate if someone could give some suggestions and comments.</p>
| José Carlos Santos | 446,262 | <p>By definition, the partial derivative at <span class="math-container">$(0,0)$</span> with respect to the direction <span class="math-container">$v$</span> is the limit<span class="math-container">$$\lim_{h\to0}\frac{f(hv)-f(0,0)}h.$$</span>So, compute this limit.</p>
|
4,203,079 | <p>I’m trying to grasp the idea behind quotient spaces and reading <a href="https://en.m.wikipedia.org/wiki/Quotient_space_(topology)" rel="nofollow noreferrer">this</a> wikipedia article. In the section ”Examples” they have the unit square <span class="math-container">$S^2$</span> homeomorphism example, which I tought would be something I could use to start building my understading of these. I’ve gone through abstract algebra course and now what an equivalence relation is however I still cannot understand the idea here.</p>
<blockquote>
<p>Consider the unit square <span class="math-container">$I^2 = [0,1] × [0,1]$</span> and the equivalence relation <span class="math-container">$\sim$</span> generated by the requirement that all boundary points be equivalent, thus identifying all boundary points to a single equivalence class. Then <span class="math-container">$I^2/ \sim$</span> is homeomorphic to the sphere <span class="math-container">$S^2$</span>.</p>
</blockquote>
<p>The problem is with the sentence.</p>
<blockquote>
<p>equivalence relation <span class="math-container">$\sim$</span> generated by the requirement that all boundary points be equivalent, thus identifying all boundary points to a single equivalence class.</p>
</blockquote>
<p>What does this exactly mean? All the boundary points of <span class="math-container">$I^2$</span> are the set <span class="math-container">$\partial I^2$</span> which I guess could be denoted as <span class="math-container">$[0,1] \times \{0\} \cup \{0\} \times [0,1] \cup [0,1] \times \{1\} \cup \{1\} \times [0,1]$</span>? Also the sentence ”thus identifying all boundary points to a single equivalence class.” is somewhat confusing. Any clarification for this would be greatly appreciated.</p>
| Moe Sarah | 787,944 | <p>Let A be the subset consisting of boundary points. Then define an equivalence relation on <span class="math-container">$I^2$</span> by declaring <span class="math-container">$x\sim y$</span> if and only if <span class="math-container">$x=y$</span> or <span class="math-container">$x,y \in A$</span> then the resulting quotient space is the space youre talking about and you can geometrically see is <span class="math-container">$S^2$</span>.</p>
<p>The whole concept of quotient space is constructed so that you can treat points not equal in the original space as equal in the quotient space (identifying points)</p>
|
451,131 | <blockquote>
<p><strong>Problem</strong>: If <span class="math-container">$\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $</span>. Find <span class="math-container">$f(x)$</span></p>
<p><strong>Solution</strong>: <span class="math-container">$\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $</span></p>
<p>Differenting both sides,we get</p>
<p><span class="math-container">$ f(x) \sin{x} \cos{x} = \frac {f'(x)}{2(b^2 - a^2)f(x)} $</span></p>
</blockquote>
<p>Am I doing right ?</p>
| Community | -1 | <p>Yes and to complete: we have</p>
<p>$$\int f'(x)(f(x))^{-2}\;dx=C\int\sin(2x)\;dx$$
where $C=b^2-a^2$ so
$$-\frac{1}{f(x)}=-\frac{C}{2}\cos(2x)+C'$$
and you can take $f(x)$ from it.</p>
|
2,856,373 | <blockquote>
<p>If <span class="math-container">$z_{1},z_{2}$</span> are two complex numbers and <span class="math-container">$c>0.$</span> Then prove that</p>
<p><span class="math-container">$\displaystyle |z_{1}+z_{2}|^2\leq (1+c)|z_{1}|^2+\bigg(1+\frac{1}{c}\bigg)|z_{2}|^2$</span></p>
</blockquote>
<p>Try: put <span class="math-container">$z_{1}=x_{1}+iy_{1}$</span> and <span class="math-container">$z_{2}=x_{2}+iy_{2}.$</span> Then
from left side</p>
<p><span class="math-container">$$(x_{1}+x_{2})^2+(y_{1}+y_{2})^2=x^2_{1}+x^2_{2}+2x_{1}x_{2}+y^2_{1}+y^2_{2}+2y_{1}y_{2}$$</span></p>
<p>Could some help me how to solve it further, Thanks in Advance.</p>
| Robert Z | 299,698 | <p>You are on the right track. Following your approach the inequality becomes
$$x^2_{1}+x^2_{2}+2x_{1}x_{2}+y^2_{1}+y^2_{2}+2y_{1}y_{2}\leq (1+c)(x_1^2+y_1^2)+\bigg(1+\frac{1}{c}\bigg)(x_2^2+y_2^2)$$
that is
$$2x_{1}x_{2}+2y_{1}y_{2}\leq c(x_1^2+y_1^2)+\frac{1}{c}(x_2^2+y_2^2).$$
Is it true that $2uv\leq cu^2+\frac{v^2}{c}$? (Hint. use AM-GM inequality).</p>
|
2,548,353 | <p><strong>Find the number of $4\times4$ matrices such that $|a_{ij}| = 1 \forall i,j\in[1,4]$ , and sum of every row and column is zero.</strong></p>
<p>I tried 'counting' the number of matrices that satisfy the above conditions, that is, elements are $1$ or $-1$ and sum of every row and column is zero.</p>
<p>In the attempt to generate a recursion I started off with a $2\times2$ matrix, for which case the answer is $2$. (First element is 1 or -1, other elements are decided accordingly)
However, this method becomes cumbersome and mathematically disappointing for $3x3$ and larger matrices.</p>
<p>Could someone please explain the method, or post a solution to the problem?
Is it possible to generalise the result to an nxn matrix? </p>
| Isaac Browne | 429,987 | <p>This is quite difficult to generalize to a $n \times n$ matrix. I would suggest generalizing to having two $1$'s in each row and column in an $n \times n$ matrix, as is done is problem $8$ of combinatorics in <a href="https://www.hmmt.co/archive/february/problems/2012/" rel="nofollow noreferrer">2012 February HMMT</a>. The recursion developed in the official solution solves your problem for $n=4$.</p>
<p>There is also a slightly simpler way of constructively counting it for the $4 \times 4$ problem. First we split the types into two cases.</p>
<p>The first is where we have two pairs of matching columns, as
$$\begin{bmatrix}
1 & -1 & 1 & -1 \\
1 & -1 & 1 & -1 \\
-1 & 1 & -1 & 1 \\
-1 & 1 & -1 & 1
\end{bmatrix}$$
We can count how many of these there are by first choosing two $1$'s out of $4$ spaces for the first column, and then $1$ from the next $3$ columns to replicate the first column.
$$\binom{4}{2} \binom{3}{1} = 18$$</p>
<p>The second case is where we have no repeated columns, as
$$\begin{bmatrix}
1 & -1 & \color{red}1 & -1 \\
1 & \color{red}1 & -1 & -1\\
-1 & \color{red}1 & -1 & 1\\
-1 & -1 & \color{red}1 & 1
\end{bmatrix}$$
Again we can start counting these by choosing two $1$'s out of $4$ spaces for the first column. Then we have two choices of pairing the $1$'s in the columns that share a $1$ row with the first column, which is represented in red to make it easier to see. Finally we have $3!$ ways of arranging the second three columns. Multiplying these together we have
$$\binom{4}{2} \cdot 2 \cdot 3! = 72$$
And finally we add the two cases to get $18+72 = \color{red}{90}$.</p>
|
2,548,353 | <p><strong>Find the number of $4\times4$ matrices such that $|a_{ij}| = 1 \forall i,j\in[1,4]$ , and sum of every row and column is zero.</strong></p>
<p>I tried 'counting' the number of matrices that satisfy the above conditions, that is, elements are $1$ or $-1$ and sum of every row and column is zero.</p>
<p>In the attempt to generate a recursion I started off with a $2\times2$ matrix, for which case the answer is $2$. (First element is 1 or -1, other elements are decided accordingly)
However, this method becomes cumbersome and mathematically disappointing for $3x3$ and larger matrices.</p>
<p>Could someone please explain the method, or post a solution to the problem?
Is it possible to generalise the result to an nxn matrix? </p>
| Isaac Browne | 429,987 | <p>And here is a tour of the HMMT solution for the generalized version, for posterity and for understanding.</p>
<p>The problem statement is now as follows: "Find the amount of $n \times n$ matrices do there exist for which all the columns and rows contain exactly two $1$'s and the rest of numbers are $-1$'s"</p>
<p>The first paragraph explains an important concept vital to the solution of the problem which involves cycles. A cycle of length $n$ is a permutation of objects $a_i$ where $a_{k_1}$ maps to $a_{k_2}$, $a_{k_2}$ maps to $a_{k_3}$, and so on until we have $a_{k_n}$ mapping back to $a_{k_1}$ where all $n$ $k_i$'s are distinct.</p>
<p>Now, when we look at the $n \times n$ matrix colored with cycles in mind. We can see that there exist parts of the matrix which trace out a path which maps to a cycle. As an example, this $5 \times 5$ matrix contains two cycles. One of length $2$ and one of length $3$.
$$\begin{bmatrix}
\color{blue}1 & \color{blue}1 & -1 & -1 & -1 \\
-1 & \color{blue}1 & -1 & \color{blue}1 & -1 \\
-1 & -1 & \color{red}1 & -1 & \color{red}1 \\
\color{blue}1 & -1 & -1 & \color{blue}1 & -1 \\
-1 & -1 & \color{red}1 & -1 & \color{red}1
\end{bmatrix}$$
For a $n \times n$ matrix, we can count the number of ways to create the matrix by creating a $k$ sized cycle and then filling in the rest with an $(n-k) \times (n-k)$matrix. If we denote the number of matrices of size $n \times n$ as $f(n)$ as is done in the HMMT solution, we can sum over all the different length cycles (of length at least $2$!) to get
$$f(n) = \sum_{k=2}^n a_kf(n-k) \tag{1}$$
Where $a_k$ is the number of ways to create a $k$ cycle in an $n \times n $ matrix. So now we must find $a_n$.
The easiest way to do this is as is done in the HMMT solution. Looking at the first column, we have $\binom{n}{2}$ ways of choosing the first two squares. Now we choose the second column. There are $n-1$ ways of choosing the second column, and $n-2$ ways of choosing the second $1$ in that column (if $n>2$). Then $n-2$ ways of choosing the third column and $n-3$ ways of choosing the second $1$ in that column (if $n>3$). Note that we start out from only one of the two squares in the first column to avoid over-counting.</p>
<p>We can see that this pattern countinues on until we choose the final column from $n-k+1$ remaining columns, and the second square must align with the other of the first two squares. Thus we have the formula
$$a_k = \frac{n(n-1)}{2} \times (n-1)(n-2) \times ... \times (n-k+1) =
\frac{n!(n-1)!}{2(n-k)!(n-k)!} \tag{2}$$
Combining $(1)$ and $(2)$, we get the formula from the official solution
$$f(n) = \sum_{k=2}^{n}\frac{1}{2} f(n-k) \frac{n!(n-1)!}{(n-k)!(n-k)!}$$
Then some rearranging to get
$$\frac{2nf(n)}{n!n!} = \sum_{k=2}^{n} \frac{f(n-k)}{(n-k)!(n-k)!}$$
And finally a trick to get rid of the summation
$$\frac{2nf(n)}{n!n!} - \frac{2(n-1)f(n-1)}{(n-1)!(n-1)!} = \sum_{k=2}^{n} \frac{f(n-k)}{(n-k)!(n-k)!} - \sum_{k=2}^{n-1} \frac{f(n-k)}{(n-k)!(n-k)!} \\
\frac{2nf(n)}{n!n!} - \frac{2(n-1)f(n-1)}{(n-1)!(n-1)!}=\frac{f(n-2)}{(n-2)!(n-2)!}$$
Finally, multiplying each side by $n!n!/2n$, we get our recursion for the solution to the generalized problem.
$$f(n) = (n)(n-1)f(n-1) + \frac{n(n-1)^2}{2}f(n-2)$$
And applying this, with $f(1) = 0$ and $f(2) = 1$, we get the following results, including $90$ for $n=4$, as expected.
$$f(3) = 6$$
$$\color{red}{f(4) = 90}$$
$$f(5) = 2040$$
$$f(6) = 67950$$
And here is the OEIS sequence <a href="https://oeis.org/A001499" rel="nofollow noreferrer">https://oeis.org/A001499</a></p>
|
443,736 | <p>Find the point on the parabola $3x^2+4x-8$ that is closest to the point $(-2,-3)$.</p>
<p>My plan for this problem was to use the distance formula and then that the derivative to get my answer. I'm having a little trouble along the way.</p>
<p>$$ d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$$</p>
| pre-kidney | 34,662 | <p>Suppose the closest point is at $p=(x_0,y_0)$, and set $q=(-2,-3)$. Then the tangent to the parabola at $p$ is perpendicular to $\ell$, the line through $p,q$.</p>
<p>Since $y'=6x+4$, the slope of the tangent is $6x_0+4$, so the slope of $\ell$ is $-\frac{1}{6x_0+4}$. Since it passes through $q$, we see the equation for $\ell$ is
$$y+3 = -\frac{x+2}{6x_0+4}$$</p>
<p>Combining with $y=3x^2+4x-8$ shows that $x_0$ is a root of
$$
(6x+4)(3x^2+4x-5)+(x+2)=18x^3+36x^2-13x-18=0
$$
and we end up numerically computing the roots from here.</p>
|
4,380,992 | <p>I'm trying to do the following exercise:</p>
<p><em>Find a non-homogeneous recurrence relation for the sequence whose general term is</em></p>
<p><span class="math-container">$$a_n = \frac{1}{2}3^n - \frac{2}{5} 7^n$$</span></p>
<p>From this expression we can obtain the roots of the characteristic polynomial <span class="math-container">$P(x)$</span>, which are <span class="math-container">$3$</span> and <span class="math-container">$7$</span>, so <span class="math-container">$P(x) = x^2 - 10x + 21$</span> and <span class="math-container">$a_n = 10a_{n-1} - 21a_{n-2} \; \forall \; n \ge 2, \; a_0 = \frac {1}{10}, \; a_1 = -\frac{13}{10}$</span>.</p>
<p>Now I don't know how to obtain a non-homogeneous recurrence relation given this homogeneous recurrence relation.</p>
| dxiv | 291,201 | <p>Write the relations for two consecutive terms:</p>
<p><span class="math-container">$$
\begin{align}
\begin{cases}
a_n &=\, \dfrac{1}{2}\,3^n - \dfrac{2}{5}\, 7^n
\\ a_{n+1} &=\, \dfrac{1}{2}\,3^{n+1} - \dfrac{2}{5}\, 7^{n+1} \,=\, \dfrac{3}{2}\,3^n-\dfrac{14}{5}\,7^n
\end{cases}
\end{align}
$$</span></p>
<p>Eliminate (for example) <span class="math-container">$\,7^n\,$</span> between the two:</p>
<p><span class="math-container">$$
\require{cancel}
a_{n+1} - 7 a_n = \left(\dfrac{3}{2}\,3^n-\cancel{\dfrac{14}{5}\,7^n}\right) - \left(\dfrac{7}{2}\,3^n - \cancel{\dfrac{14}{5}\, 7^n}\right) \;\iff\; a_{n+1} = 7a_n-2\cdot 3^n
$$</span></p>
<p>Note that the non-homogeneous recurrence is not unique. If you chose to eliminate the other power <span class="math-container">$\,3^n\,$</span>, for example, you would get <span class="math-container">$\,a_{n+1} = 3a_n-\dfrac{8}{5}\,7^n\,$</span>, which is equally valid, as are many others.</p>
|
1,621,347 | <p>Is there a closed-form expression for the following definite integral?
\begin{equation}
\mathcal{I} = \int_{\delta_1}^{\delta_2}(1+Ax)^{-L}x^{L}\exp\left(-Bx\right)dx,
\end{equation}
where $A$, $B$, $\delta_1$, and $\delta_2$ are positive constant. $L$ is a positive integer.</p>
<p>I am facing problem due to finite limits $\delta_1$ and $\delta_2$. I know the answer when "$\delta_1 = 0$ and $\delta_2 = \infty$." </p>
| Pierpaolo Vivo | 302,446 | <p>Use the identity
$$
\frac{1}{(1+A x)^{L}}=\frac{1}{\Gamma(L)}\int_0^\infty dy\ y^{L-1}e^{-y}e^{-y A x}
$$
and swapping the two integrals (using $\partial_B^{k} e^{-B x}=(-1)^k x^k e^{-B x}$) you get
$$
\mathcal{I}=\frac{(-1)^L}{\Gamma(L)}\partial_B^{(L)}\int_0^\infty dy\ y^{L-1}e^{-y}\underbrace{\int_{\delta_1}^{\delta_2}dx\ e^{-x (B+A y)}}_{\frac{e^{\text{$\delta_1 $} (-(A y+B))}-e^{\text{$\delta_2 $} (-(A y+B))}}{A y+B}}\ ,
$$
therefore
$$
\mathcal{I}=\frac{(-1)^L}{\Gamma(L)}\partial_B^{(L)}\left[\frac{e^{B/A} \Gamma (L) \left(\frac{A}{B}\right)^{-L} \left(\Gamma \left(1-L,B \left(\text{$\delta_1 $}+\frac{1}{A}\right)\right)-\Gamma \left(1-L,B \left(\text{$\delta_2 $}+\frac{1}{A}\right)\right)\right)}{B}\right]\ .
$$</p>
|
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