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 |
|---|---|---|---|---|
1,456,224 | <p>I've been asked to compute the Euler-Lagrange equation and second variation of the functional $$I[y]=\int_{a}^{b}(y'^2+y^4)dx$$
with boundary conditions $y(a)=\alpha$, $y(b)=\beta$. It's easy to see that $$I[y+\delta y]=I[y]+\int_{a}^{b}\delta y(4y^{3}-2y'') dx+\int_{a}^{b}(6y^{2}\delta y^{2}+\delta y'^{2})dx$$
So the Euler-Lagrange equation integrates to give $y^{4}-y'^{2}=k$, where $k$ is a constant of integration. We're then asked to solve this equation when $\alpha=\beta=0$. The equation is separable, but to my shame I can't do the integration (I think it involves special functions), so I looked for a different way. Completing the square on $I$ gives $$\int_{a}^{b}(y'^2+y^4)dx=\int_{a}^{b}(y'+y^2)^{2}dx-\int_{a}^{b}2y^{2}y'dx$$
But the final term is just $\left[\frac{2}{3}y^{3}\right]^{y=0}_{y=0}=0$, and the other two integrals are non-negative. The only way to extremise the RHS (I think) is to minimise it, and besides the second variation is non-negative, so we make the RHS zero by allowing $y'=-y^{2}$. But then the LHS forces both $y'=0$ and $y=0$ on all of $[a,b]$. Does this mean that the only solution is the zero function? Then, since $\delta^{2}I=\int_{a}^{b}\delta y'^{2}dx$, and this is positive unless $\delta y$ is constant (and hence $0$, by the boundary condition), do we have the zero function as the actual solution? </p>
<p>Sorry if this sounds a little incoherent, when I started writing the answer I forgot to consider $y(x)=0$ and found the other solutions to $y'=-y^{2}$, which can't possibly satisfy the boundary conditions. In fact, looking at it now, I'm starting to think that even completing the square was unnecessary.</p>
| Community | -1 | <p>The Euler-Lagrange equation </p>
<p>$$y^4 - (y')^2 = k$$</p>
<p>is actually not too hard to solve once you know $y(a) = y(b) = 0$. Note that $y(a) = 0$ imply $k \le 0$. If $k=0$, then $(y')^2 = y^4$ and you can check that the only solution (using $y(a) =0$ and $y'' - 4y^3 = 0$) is $y(t) = 0$. </p>
<p>If $k<0$, then
$$(y')^2 = \sqrt{y^4 - k} \ge \sqrt{-k} >0 . $$</p>
<p>Then $y' \ge (-k)^{\frac 14} $ or $y' \le -(-k)^{\frac 14}$. In both cases, $y(b) = 0$ is not possible. Thus $y(t) = 0$ is the only solution to the Euler-Lagrange equation. </p>
|
2,503,306 | <p>Suppose $g{^n}$=e. Show the order of $g$ divides $n$.</p>
<p>Would I use Eulers Theorem???;</p>
<p>$a{^{\phi p}}$ $\equiv1 \pmod p$</p>
<p>$a{^{p-1}}\equiv1 \pmod p$</p>
<p>$a{^p}\equiv a\pmod p$</p>
<p>So then I would have </p>
<p>$g{^n}\equiv g\pmod n$</p>
<p>then I think you use the $\gcd$, which states $\gcd(a,b) = 1$</p>
<p>or </p>
<p>$a=nq+r$ and $b=nq+r$</p>
<p>which is $a\equiv b\pmod n$??</p>
| lhf | 589 | <p>Consider $A = \{ k \in \mathbb Z : g^k =e \}$. Then $A$ is subgroup of $\mathbb Z$ and so $A=m\mathbb Z$, where $m$ is the smallest positive element of $A$. By definition, $m$ is the order of $g$. Thus, if $n\in A$, then $n$ is a multiple of $m$.</p>
<p>This approach may work for you if you see the subgroups of $\mathbb Z$ before talking about abstract groups.</p>
|
1,537,648 | <p>For example let's say we have a password combination of (a,b,c,d), if the password length was 1 then we'll have 4 possible passwords (a,b,c,d), now if the length was 2 then we'll have 20 possible passwords (a,b,...,dc,dd), I calculated this manually, I want the rule of calculating probability?</p>
| Iceman | 91,561 | <p>Suppose you want a password of length up to $n$. Suppose also that you have $m$ possible digits to pick from...</p>
<p>The total possible 1-passwords would just be $m$ (m letters)</p>
<p>The total possible 2-passwords would just be $m^2$ (m letters times m letters)</p>
<p>...</p>
<p>The total possible n-passwords would just be $m^n$ (m letters times m letters times...times m letters) n times</p>
<p>So the total possible amount of passwords would be</p>
<p>$$m+m^2+m^3+...+m^n$$</p>
<p>This is basically the geometric series $-1$. So</p>
<p>$$\sum_{k=1}^n{m^k}=\frac{m^{n+1}-1}{m-1}-1=\frac{m^{n+1}-m}{m-1}$$</p>
|
138,708 | <p>I'm dealing with derivatives of scalar functions of matrices and wondering if Mathematica can help me here.</p>
<p>The standard approach of expanding it in terms of components is cumbersome. As an motivating example, I want to minimize the following function, where $X$ is a matrix</p>
<p>$$f(X) = \text{tr}(X'X)$$</p>
<p>I can use <a href="https://tminka.github.io/papers/matrix/" rel="noreferrer">matrix differential calculus</a> to derive that one step of gradient descent to minimize this function is:</p>
<p>$$X^* = X - 2 X$$</p>
<p>On other hand, suppose $X$ is square, and my function is
$$g(X)=\text{tr}(X^2)$$</p>
<p>Now, a single step of gradient descent looks as follows
$$X^* = X - 2 X'$$</p>
<p>This can get complicated to do by hand, as an example, Exercise 1 of <a href="http://www.janmagnus.nl/misc/mdc2007-3rdedition" rel="noreferrer">Magnus</a> 9.10 asks to show that gradient descent step of the following function</p>
<p>$$h(X) = \det A X B$$</p>
<p>is the following</p>
<p>$$X^* = X-\det(AXB)(A'(B'X'A')^{-1}B')'$$</p>
<p>now take
$$h^*(X) = \text{tr}(AX'BXC)$$
formula to do a single step of gradient descent is
$$ \text{probably something simple} $$</p>
<p>Is there a way to get help deriving/checking expressions above in Mathematica?</p>
<p>(note, I'm using "gradient descent step" instead of "derivative" because there are multiple notations for derivative which differ in shape, but reformulating it as gradient descent removes ambiguity)</p>
| Carl Woll | 45,431 | <p>In this post I discuss a function <code>MatrixD</code> which attempts to take a matrix derivative following the guidelines given in the <a href="http://www2.imm.dtu.dk/pubdb/views/edoc_download.php/3274/pdf/imm3274.pdf" rel="noreferrer">The Matrix Cookbook</a>.</p>
<p>I still want to take advantage of the normal partial derivative function D, but I need to override the default handling of matrix functions. The basic approach is the following:</p>
<ol>
<li><p>Add <a href="http://reference.wolfram.com/language/ref/Inverse" rel="noreferrer"><code>Inverse</code></a>, <a href="http://reference.wolfram.com/language/ref/Tr" rel="noreferrer"><code>Tr</code></a>, <a href="http://reference.wolfram.com/language/ref/Transpose" rel="noreferrer"><code>Transpose</code></a>, <a href="http://reference.wolfram.com/language/ref/Det" rel="noreferrer"><code>Det</code></a>, <a href="http://reference.wolfram.com/language/ref/MatrixPower" rel="noreferrer"><code>MatrixPower</code></a>, and <a href="http://reference.wolfram.com/language/ref/MatrixFunction" rel="noreferrer"><code>MatrixFunction</code></a> to the list of "ExcludedFunctions" in <code>SystemOptions[]</code>. The default handling of derivatives of these symbols needs to be avoided (<code>D[Tr[X], X]</code> evaluating to <code>Tr'[X]</code> is not useful).</p></li>
<li><p>The matrix derivative of a matrix is not 1, but rather is given by the following:
$$\frac{\partial M_{\text{ij}}}{\partial M_{\text{kl}}}=\mathbb{J}_{\text{ijkl}}$$
where $\mathbb{J}_{\text{ijkl}}\equiv \delta _{\text{ik}} \delta _{\text{jl}}$.
Since <code>D[X, X]</code> invariably returns 1, I will instead use the following construct:</p>
<pre><code>MatrixD /: D[X, MatrixD, NonConstants->{X}] := $SingleEntryMatrix
</code></pre>
<p><code>MatrixD</code> here acts as a variable, and the <a href="http://reference.wolfram.com/language/ref/NonConstants" rel="noreferrer"><code>NonConstants</code></a> option prevents the derivative from being 0. This DownValue for <a href="http://reference.wolfram.com/language/ref/D" rel="noreferrer"><code>D</code></a> (as well as all of the others) are hidden inside the implementation of <code>MatrixD</code>.</p></li>
<li><p>The result of <code>MatrixD</code> will need to be simplified further. For this purpose, I wrote a <code>MatrixReduce</code> function. Some of the simplifications performed by <code>MatrixReduce</code> depend on whether the matrix variable is invertible or not. This is controlled by the "Invertible" option of <code>MatrixD</code> and <code>MatrixReduce</code>.</p></li>
<li><p>Scalar functions of a matrix (e.g., <code>Log[X]</code>) are not supported. Instead, one must use <a href="http://reference.wolfram.com/language/ref/MatrixFunction" rel="noreferrer"><code>MatrixFunction</code></a> (or <a href="http://reference.wolfram.com/language/ref/MatrixPower" rel="noreferrer"><code>MatrixPower</code></a>), e.g., <code>MatrixFunction[Log, X]</code> instead of <code>Log[X]</code>.</p></li>
<li><p><a href="http://reference.wolfram.com/language/ref/MatrixFunction" rel="noreferrer"><code>MatrixFunction</code></a> has a simple differentiation rule when it is inside of a <a href="http://reference.wolfram.com/language/ref/Tr" rel="noreferrer"><code>Tr</code></a>/<a href="http://reference.wolfram.com/language/ref/Det" rel="noreferrer"><code>Det</code></a>, but much more complicated rules if it is outside of a <a href="http://reference.wolfram.com/language/ref/Tr" rel="noreferrer"><code>Tr</code></a>/<a href="http://reference.wolfram.com/language/ref/Det" rel="noreferrer"><code>Det</code></a>. Hence, derivatives of <a href="http://reference.wolfram.com/language/ref/MatrixFunction" rel="noreferrer"><code>MatrixFunction</code></a> are only supported when they occur inside of a <a href="http://reference.wolfram.com/language/ref/Tr" rel="noreferrer"><code>Tr</code></a>/<a href="http://reference.wolfram.com/language/ref/Det" rel="noreferrer"><code>Det</code></a> wrapper.</p></li>
<li><p>For test purposes, I also created a <code>TestMatrixD</code> function. This function tests the <code>MatrixD</code> output by performing the matrix derivative using ordinary <a href="http://reference.wolfram.com/language/ref/D" rel="noreferrer"><code>D</code></a> rules as well as with the <code>MatrixD</code>, and comparing the output after substituting random real matrices for the variables and matrix constants.</p></li>
<li><p>There is at least one bug in the <code>MatrixD</code> implementation, when the argument to one of the matrix functions <a href="http://reference.wolfram.com/language/ref/Tr" rel="noreferrer"><code>Tr</code></a>, <a href="http://reference.wolfram.com/language/ref/Det" rel="noreferrer"><code>Det</code></a>, <a href="http://reference.wolfram.com/language/ref/Inverse" rel="noreferrer"><code>Inverse</code></a> and <a href="http://reference.wolfram.com/language/ref/Transpose" rel="noreferrer"><code>Transpose</code></a> is actually a scalar and not a matrix (e.g., <code>Tr[Det[X]]</code>). There may be more.</p></li>
</ol>
<p>The code is a little lengthy, so I put it on GitHub. You can download it at:</p>
<p><a href="https://github.com/carlwoll/MatrixD/releases" rel="noreferrer">https://github.com/carlwoll/MatrixD/releases</a></p>
<p>The source code can be viewed there as well. Download the file "MatrixD-1.0.paclet" and then install it with:</p>
<pre><code>PacletInstall[file]
</code></pre>
<p>Load the package with:</p>
<pre><code><<MatrixD`
</code></pre>
<p>Now, for your first couple elementary examples:</p>
<pre><code>MatrixD[Tr[X\[Transpose].X], X]//TeXForm
</code></pre>
<blockquote>
<p>$2 X$</p>
</blockquote>
<pre><code>MatrixD[Tr[X.X], X]//TeXForm
</code></pre>
<blockquote>
<p>$2 X^T$</p>
</blockquote>
<p>Your slightly more complicated <a href="http://reference.wolfram.com/language/ref/Det" rel="noreferrer"><code>Det</code></a> example:</p>
<pre><code>MatrixD[Det[A.X.B], X, "Invertible"->False]//TeXForm
</code></pre>
<blockquote>
<p>$\left| A.X.B\right| A^T.\left(B^T.X^T.A^T\right)^{-1}.B^T$</p>
</blockquote>
<p>And the output for your last example:</p>
<pre><code>MatrixD[Tr[A.X.B.X\[Transpose].C], X, "Invertible"->False]//TeXForm
</code></pre>
<blockquote>
<p>$A^T.C^T.X.B^T+C.A.X.B$</p>
</blockquote>
<p>I'm planning on adding support for vectors as well as matrices in the future.</p>
|
1,267,395 | <p>Julie is required to pay a 2 percent tax on all income over 3,000. She also has to pay 2.5 percent on all income over 20,000. She earned more than 20,000 and paid 992.50 what was her total income</p>
| user70160 | 228,922 | <p>She earns more than $20,000$</p>
<p>So she pays $2\%$ on her her income between $3,000$ and $20,000$, i.e. she pays $2\%$ on $17,000$ which gives:</p>
<p>$17,000\times2\% = 340$</p>
<p>Subtract that from the given total tax burden of $992.5$ to get the tax she paid for her income above $20,000$ as follows:</p>
<p>$992.5-340 = 652.5$</p>
<p>Define her total income as $x$ and her income above $20,000$ as $y$ so that:</p>
<p>$x = 20,000 + y$</p>
<p>The tax she pays on $y$ was calculated above as $625.5$ which according to the question is $2.5\%$ of $y$ (her income above $20,000$) so that:</p>
<p>$y\times 2.5\% = 652.5$</p>
<p>solving this for $y$ gives:</p>
<p>$y=\cfrac{652.5}{2.5\%}= \cfrac{652.5\times4}{10\%} = 6,525\times4 = 13,050\times2= 26,100.$</p>
<p>So $y$ (her income above $20,000$) was $26,100$ and her total income is hence:</p>
<p>$x= 20,000 + y =$</p>
<p>$20,000+26,100=$</p>
<p>$46,100.00$</p>
|
1,457,063 | <p>I am utterly confused on how to solve this problem. I found a lemma that says $|A\cup B|=|A|+|B|$ is true if the two sets are disjoint which makes sense, but how do I prove the entire statement. </p>
| Sak | 5,209 | <p>Consider the function $f:A\cup B\to A\sqcup \{0\}$ and $g:A\cup B\to B\sqcup\{0\}$ given by $f(x)=x$ if $x\in A$ and $f(x)=0$ in other case; $g(y)=y$ if $y\in B$ and $g(0)=0$ in other case. </p>
<p>These functions are surjective. Then, $|A\cup B|\leq |A|$ and $|A\cup B|\leq |B|$, and therefore $|A\cup B|\leq |A|+|B|$. </p>
<p>Since $A$ and $B$ can be included into $A\cup B$ via inclusions with disjoint images (since $A$ and $B$ are disjoint) you also have that $|A|+|B|\leq |A\cup B|$.</p>
|
1,174,359 | <p>I have the following problem where n is a positive integer $(n >= 1)$:</p>
<p>Prove that $\frac{1}{2n}\le\frac{1*3*5*...*(2n-1)}{2*4*...*2n}$</p>
<p>I know that I must start with the basic step showing that $P(1)$ is true as follows:
$1/(2*1) = 1/2$ so $P(1)$ is true.</p>
<p>Now follows the induction step where I must show that "if $P(k)$ then $P(k+1)$" where
$P(k)$ is $\frac{1}{2k}\le\frac{1*3*5*...*(2k-1)}{2*4*...*2k}$
and $P(k+1)$ is $\frac{1}{2(k+1)}\le\frac{1*3*5*...*(2k+1)}{2*4*...*2(k+1)}$</p>
<p>I will very much appreciate any help about how to continue.</p>
| Hippalectryon | 150,347 | <p>This is wrong. Because $f'(a)=b$ doesn't mean at all that $(f^{-1})'(a)=\frac{1}b$</p>
<p>(quick counterexample on $\Bbb{R}^+_*$ for $x=1$ : take $f(x)=e^x,f^{-1}(x)=\ln x$,$f'(1)=e,(f^{-1})'(1)=1$)</p>
<p>What you should do : find the inverse function of $f$, then compute its derivative the given point.</p>
<p>Hint to find $f^{-1}$ : the inverse function of $x^2$ is $\sqrt{x}$</p>
|
1,174,359 | <p>I have the following problem where n is a positive integer $(n >= 1)$:</p>
<p>Prove that $\frac{1}{2n}\le\frac{1*3*5*...*(2n-1)}{2*4*...*2n}$</p>
<p>I know that I must start with the basic step showing that $P(1)$ is true as follows:
$1/(2*1) = 1/2$ so $P(1)$ is true.</p>
<p>Now follows the induction step where I must show that "if $P(k)$ then $P(k+1)$" where
$P(k)$ is $\frac{1}{2k}\le\frac{1*3*5*...*(2k-1)}{2*4*...*2k}$
and $P(k+1)$ is $\frac{1}{2(k+1)}\le\frac{1*3*5*...*(2k+1)}{2*4*...*2(k+1)}$</p>
<p>I will very much appreciate any help about how to continue.</p>
| Mathemagician1234 | 7,012 | <p>y= f(x) = $\sqrt{5x}$ $\rightarrow$ $y^2$ = $5x$ $\rightarrow$ $\frac{1}{5}y^2$ = x </p>
<p>So $f^{-1}(x)$ = $\frac{1}{5}x^2$ , which is easily verified by taking $f^{-1}(f(x))$= x. </p>
<p>Now:
$(f^{-1})'(x)$ = $\frac{2}{5}x$. The slope of the tangent line of $f^{-1}$ at (4,$\frac{16}{5}$) is $(f^{-1})'(4)$ = $\frac{2}{5}*4$ =$\frac{8}{5}$. </p>
<p>So the tangent line to the curve of $(f^{-1})'(x)$ at ( 4,$\frac{16}{5}$) is y-$\frac{16}{5}$ = $\frac{8}{5}$ (x - 4) $\rightarrow$ y = $\frac{8}{5}x$ -$\frac{16}{5}$. </p>
<p>A graph of this situation is as follows: </p>
<p><img src="https://i.stack.imgur.com/GREGV.png" alt="enter image description here"></p>
<p>This is the point slope form of the tangent line. </p>
|
56,394 | <p>Hi!</p>
<p>While studying C*-algebras I found 2 different definitions for non degenerate representations (<em>-homomorphisms $\pi:\mathcal{A} \rightarrow B(\mathcal{h})$ where $\mathcal{A}$ is a C</em>-algebra and $B(\mathcal{h})$ is the space of bounded linear operators on some Hilbert space $\mathcal{h}$):</p>
<p>1) For every non-zero $\xi \in \mathcal{h}$ there exists $a \in \mathcal{A}$ such that $\pi(a)\xi \neq 0$;</p>
<p>2) The set $\{\pi(a)\xi \quad a \in \mathcal{A}, \xi \in \mathcal{h}\}$ is dense in $\mathcal{h}$.</p>
<p>Are they equivalent?</p>
<p>Thanks,
Alessandro</p>
| Stefan Waldmann | 12,482 | <p>In fact, for unital $C^*$-algebras non-degeneracy just means $\pi(1) = 1$. In the non-unital case there is even a sharper statement than your item (2): One can find for every $\phi$ and every $\epsilon > 0$ another vector $\psi$ and a <em>positive</em> algebra element $a \in \mathcal{A}^+$ with
\begin{equation}
\phi = \pi(a)\psi
\quad
\textrm{and}
\quad
\|\phi - \psi\| < \epsilon.
\end{equation}
This is nice as it shows that we do not just get a dense subspace and we get in some sense as close as possible to $\pi(1) = 1$. I found this in Blackadars encyclopedia book in Theorem II.5.3.7 and in II.6.1.5. Might be worth a look :)</p>
|
2,439,111 | <p>By definition, a function $ f: \mathbb{R} \rightarrow\mathbb{R}$ is linear iff</p>
<ol>
<li>$f(x+y)=f(x)+f(y)$ $ \forall x,y \in \mathbb{R}$ </li>
<li>$ f(bx) = bf(x)$ $ \forall b,x \in \mathbb{R}$</li>
</ol>
<p>I am trying to prove the following statement: </p>
<p>If $ f $ is a linear map defined above then $f$ has the following form: $f(x)=ax$ $ \forall x \in\mathbb{R}$.</p>
<p>Could you give a suggestion about where to start from? </p>
| Community | -1 | <ul>
<li><p>if part: $a(x+y) = (ax) + (ay)$ and $a(bx) = b(ax)$.</p></li>
<li><p>only if part: $f(x)=f(x\cdot1)=xf(1)=ax$.</p></li>
</ul>
|
2,426,892 | <blockquote>
<p>Between which two integers does <span class="math-container">$\sqrt{2017}$</span> fall? </p>
</blockquote>
<p>Since <span class="math-container">$2017$</span> is a prime, there's not much I can do with it. However, <span class="math-container">$2016$</span> (the number before it) and <span class="math-container">$2018$</span> (the one after) are not, so I tried to factorise them. But that didn't work so well either, because they too are not perfect squares, so if I multiply them by a number to make them perfect squares, they're no longer close to <span class="math-container">$2017.$</span> How can I solve this problem?</p>
<p>Update:
Okay, since <span class="math-container">$40^2 = 1600$</span> and <span class="math-container">$50^2 = 2500$</span>, I just tried <span class="math-container">$45$</span> and <span class="math-container">$44$</span> and they happened to be the answer - but I want to be more mathematical than that... </p>
| Jack D'Aurizio | 44,121 | <p>$\sqrt{2017}\approx\sqrt{2000}=20\sqrt{5}\approx 20\cdot 2.236 \approx 45$ and
$$44^2 = 1936,\qquad 45^2=2025$$
hence $\sqrt{2017}\in\color{red}{\left(44,45\right)}$.</p>
|
2,426,892 | <blockquote>
<p>Between which two integers does <span class="math-container">$\sqrt{2017}$</span> fall? </p>
</blockquote>
<p>Since <span class="math-container">$2017$</span> is a prime, there's not much I can do with it. However, <span class="math-container">$2016$</span> (the number before it) and <span class="math-container">$2018$</span> (the one after) are not, so I tried to factorise them. But that didn't work so well either, because they too are not perfect squares, so if I multiply them by a number to make them perfect squares, they're no longer close to <span class="math-container">$2017.$</span> How can I solve this problem?</p>
<p>Update:
Okay, since <span class="math-container">$40^2 = 1600$</span> and <span class="math-container">$50^2 = 2500$</span>, I just tried <span class="math-container">$45$</span> and <span class="math-container">$44$</span> and they happened to be the answer - but I want to be more mathematical than that... </p>
| Bram28 | 256,001 | <p>For a rough estimate, I'd first divide by $100$ and think about $20.17$, for $\sqrt{2017} = 10 \sqrt{20.17}$. In fact, I would just consider $\sqrt{20}$: </p>
<p>You know $4^2=16$ and $5^2 =25$, so $\sqrt{20}$ is between $4$ and $5$, and is in fact close to the middle of them, i.e close to $4.5$. Hence, $\sqrt{2017}$ will be close to $45$.</p>
<p>So, see what $45^2$ is ... and proceed from there.</p>
|
2,426,892 | <blockquote>
<p>Between which two integers does <span class="math-container">$\sqrt{2017}$</span> fall? </p>
</blockquote>
<p>Since <span class="math-container">$2017$</span> is a prime, there's not much I can do with it. However, <span class="math-container">$2016$</span> (the number before it) and <span class="math-container">$2018$</span> (the one after) are not, so I tried to factorise them. But that didn't work so well either, because they too are not perfect squares, so if I multiply them by a number to make them perfect squares, they're no longer close to <span class="math-container">$2017.$</span> How can I solve this problem?</p>
<p>Update:
Okay, since <span class="math-container">$40^2 = 1600$</span> and <span class="math-container">$50^2 = 2500$</span>, I just tried <span class="math-container">$45$</span> and <span class="math-container">$44$</span> and they happened to be the answer - but I want to be more mathematical than that... </p>
| Jam | 161,490 | <p>There are well enough good answers here, but no one's suggested this method yet, so I'll add it.</p>
<h3>Method <span class="math-container">$1$</span> - (Secant Approximation)</h3>
<p>We can take two simple perfect squares that straddle <span class="math-container">$2017$</span>. We're just getting a rough estimate, so we'd prefer numbers to be easy to work with than near <span class="math-container">$2017$</span>. For example, <span class="math-container">$1600\leq 2017\leq 3600$</span>. The secant line passing <span class="math-container">$(1600,40)$</span> and <span class="math-container">$(3600,60)$</span> should approximate <span class="math-container">$\sqrt{x}$</span> in the interval <span class="math-container">$[1600,3600]$</span>. This gives us: <span class="math-container">$\sqrt{x}=24+\frac{x}{100}+\varepsilon_1(x)$</span>. *
So, <span class="math-container">$\sqrt{2017}\approx44$</span>. Finally, we can check the squares of <span class="math-container">$43,44,45,\ldots$</span> and find our answer.</p>
<h3>Method <span class="math-container">$2$</span> - (Mean of <span class="math-container">$2$</span>nd Degree Taylor Polynomials)</h3>
<p>We can make method <span class="math-container">$1$</span> slightly more accurate, though it's not really vital to do so. The <span class="math-container">$2$</span>nd degree Taylor expansion of <span class="math-container">$\sqrt{x}$</span> at <span class="math-container">$x=a$</span> is <span class="math-container">$T_a(x)=\sqrt{a}+\frac{x-a}{80}+\frac{(x-a)^2}{8a\sqrt{a}}$</span>. Then the mean of <span class="math-container">$T_{1600}$</span> and <span class="math-container">$T_{3600}(x)$</span> should be a good estimate for <span class="math-container">$\sqrt{x}$</span> near the centre of the interval <span class="math-container">$[1600,3600]$</span>. Then <span class="math-container">$\sqrt{x}=60+\frac{x-2600}{80}-\frac{1}{2\cdot8}\left(\frac{(x-1600)^2}{40^3}+\frac{(x-3600)^2}{60^3}\right)+\varepsilon_2(x)$</span>.** Hence, <span class="math-container">$\sqrt{2017}\approx45$</span> and we can check the squares of nearby integers, as in method <span class="math-container">$1$</span>.</p>
<p>These methods would also work for any <span class="math-container">$2000<x<3000$</span> and could easily be adapted for other values of <span class="math-container">$x$</span>. They also give a convenient way of finding an initial guess, for methods such as Newton-Raphson (detailed in <em>@mathreadler</em>'s answer).</p>
<hr />
<h3>Accuracy of Methods</h3>
<p>* The error term reaches its maximum at <span class="math-container">$\varepsilon_1(2500)= 1$</span>. So, in the worst case scenario, we'd need to check the <span class="math-container">$3$</span> numbers <span class="math-container">$(y-1),y,(y+1)$</span>, where <span class="math-container">$y$</span> is the estimate of <span class="math-container">$\sqrt{x}$</span> from method <span class="math-container">$1$</span> and where <span class="math-container">$x\in[1600,3600]$</span>.</p>
<p>** For <span class="math-container">$x\in(1769,3110)$</span>, we have <span class="math-container">$\varepsilon_2(x)<0.67<\varepsilon_1(x)$</span> but for <span class="math-container">$x<1769$</span> or <span class="math-container">$x>3110$</span>, we have <span class="math-container">$\varepsilon_2(x)>\varepsilon_1(x)$</span>. In other words, method <span class="math-container">$1$</span> is more accurate than method <span class="math-container">$2$</span> in the centre of <span class="math-container">$[1600,3600]$</span>, but the opposite is true near the bounds of the interval. However, since we're interested in the centre of the interval, this is good. The error term, <span class="math-container">$\varepsilon_2(x)$</span>, reaches a minimum of <span class="math-container">$0$</span> around <span class="math-container">$x=2351$</span>.</p>
|
930,949 | <p>Given that the circle C has center $(a,b)$ where $a$ and $b$ are positive constants and that C touches the $x$-axis and that the line $y=x$ is a tangent to C show that $a = (1 + \sqrt{2})b$</p>
| Mauro ALLEGRANZA | 108,274 | <p>The language of <em>first-order</em> logic is made of :</p>
<ul>
<li><p>sentential <em>connectives</em></p></li>
<li><p><em>quantifiers</em></p></li>
<li><p>the <em>equality</em> symbol</p></li>
<li><p>countable many (individual) <em>variables</em> : $x_1, x_2, \ldots$</p></li>
<li><p>a set (possibly empty) of <em>constants</em></p></li>
<li><p>for each positive integer $n$, a set (possibly empty) of predicate variable : $P_1^n, \ldots$ .</p></li>
</ul>
<p>Thus we can build a formula like :</p>
<blockquote>
<p>$\exists x_2 \forall x_1 \lnot P_1^2(x_1,x_2)$.</p>
</blockquote>
<p>In order to give meaning to formulae, we need an <em>interpretation</em>, i.e. we have to choose a <em>domain</em> (usually not-empty) of objects : we will use it for interpreting the quantifiers, and a mapping of the constant symbols on "distinguished" objects of the domain and of the predicate symbols on <em>relations</em> (of suitable <em>arity</em>) on the domain.</p>
<p><em>Note</em> : you can see <a href="https://math.stackexchange.com/questions/567598/interpreting-quantified-predicate-formulas">here</a> some more details.</p>
<hr>
<p>We can consider some formalized language for mathematical theories :</p>
<p><em>(i)</em> Language of set theory</p>
<blockquote>
<p>Equality: $=$; one binary predicate symbol : $\in$; a constant symbol : $\emptyset$.</p>
</blockquote>
<p>Thus, using $\in$ in place of $P_1^2$ and introducing the abbreviations : $x \in y$ for $\in(x,y)$ and $x \notin y$ for $\lnot (x \in y)$, the above formula becomes :</p>
<blockquote>
<blockquote>
<p>$\exists x_2 \forall x_1 (x_1 \notin x_2)$</p>
</blockquote>
</blockquote>
<p>which is true in the <em>domain</em> of sets, because the domain includes the emptyset.</p>
<p><em>(ii)</em> Language of elementary number theory</p>
<blockquote>
<p>Equality: $=$; one binary predicate symbol : $<$; a constant symbol : $0$.</p>
</blockquote>
<p>Thus, using $<$ in place of $P_1^2$ and introducing the abbreviations : $x < y$ for $<(x,y)$ and $x \ge y$ for $\lnot (x < y)$, the above formula becomes :</p>
<blockquote>
<blockquote>
<p>$\exists x_2 \forall x_1 (x_1 \ge x_2)$</p>
</blockquote>
</blockquote>
<p>which is true in the <em>domain</em> of natural numbers, because the domain includes the number $0$.</p>
<hr>
<p>The <a href="http://en.wikipedia.org/wiki/Bounded_quantifier" rel="nofollow noreferrer">bounded quantifiers</a> are not part of the "usual" language of first-order logic.</p>
<p>In the language of set theory</p>
<blockquote>
<p>there are two bounded quantifiers: $\forall x \in t$ and $\exists x \in t$. These quantifiers bind the set variable $x$ and contain a term $t$ (which may not mention $x$ but which may have other free variables).</p>
<p>The semantics of these quantifiers is determined by the following rules:</p>
<blockquote>
<p>$\exists x \in t\ (\varphi) \Leftrightarrow \exists x ( x \in t \land \varphi)$</p>
<p>$\forall x \in t\ (\varphi) \Leftrightarrow \forall x ( x \in t \rightarrow \varphi)$</p>
</blockquote>
</blockquote>
<p>Their use is clear; they "bound" or "restrict" the "range of application" of the quantifier to a specific set "named" by the <em>term</em> $t$; of course, $t$ must be a primitive or defined term of the lenguage, like $\omega$ for the set of natural numbers.</p>
<hr>
<p>Consider now the expression :</p>
<blockquote>
<p>for every $x$ in the set of $D, P(x)$.</p>
</blockquote>
<p>We can translate it as : $\forall x \in D (P(x))$.</p>
<p>The question is : what is the meaning of $D$, outside set theory ?</p>
<p>For a formula in first-order language we do not have to specify the domain of the interpretation of the formula: we choose a domain when we interpret it and the quantifiers and variables will receive a meaning through the interpretation. </p>
<p>Consider now the traditional example : "all men are mortal" and translate it in first-order logic in the usual way; we have :</p>
<p>$\forall x (Man(x) \rightarrow Mortal(x))$.</p>
<p>If we note that a subset of the domain of the interpretation is the denotation of a (unary) predicate symbol, we can use the set $Men$ for the intended meaning of the predicate $Man(x)$, and rewrite the above formula as :</p>
<p>$\forall x \in Men (Mortal(x))$.</p>
<p>But in first-order logic there are <strong>no</strong> variables ranging over the subsets of the domain; thus, $D$ cannot be a variable: it must be an "abbreviation" for some suitable unary predicate.</p>
<p>Thus, outside of set theory, the above formula is of little significance.</p>
<hr>
<p>We can find bounded quantifiers used also in semi-formal mathematical contexts, like :</p>
<blockquote>
<p>$\forall x \in \mathbb R^+ \exists y (y^2=x)$.</p>
</blockquote>
<p>In mathematical analysis, where the domain is the set of <em>real</em> numbers, can be useful to use this symbol to specify different subsets of $\mathbb R$, like in the example above, which is an abbreviation for :</p>
<blockquote>
<p>$\forall x (x \ge 0 \rightarrow \exists y (y^2=x))$.</p>
</blockquote>
|
2,805,975 | <p>For matrices $A, B$, I would like to show and understand the intuition behind the following identity
$$
(A+B)^{-1} = A^{-1} - (A+B)^{-1} B A^{-1}
$$
assuming the inverses exist.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>$$(A+B)\big [ A^{-1} - (A+B)^{-1} B A^{-1}\big ] =I + BA^{-1}-BA^{-1}=I$$</p>
<p>Thus $$(A+B)^{-1} = A^{-1} - (A+B)^{-1} B A^{-1}$$</p>
<p>The intuition behind it is probably $$ \frac {1}{a+b} = \frac {1}{a} -\frac {1}{a+b}\frac {b}{a}$$</p>
<p>where a and b are real numbers. </p>
|
1,278,719 | <p>This is a problem from Artin's book "Algebra". In the fifth miscellaneous problem of the chapter "Vector spaces", he has asked to prove that:</p>
<p>If $\alpha$ is a cube root of $2$, then the real numbers $a+b\alpha +c\alpha ^2$ with $a,b,c \in \mathbb{Q}$ form a field.</p>
<p>I am stuck at proving this. For example, what would be the inverse of $1+\alpha +\alpha^2$? In the previous subpart, we were asked to prove that $1,\alpha, \alpha^2$ are linearly independent over $\mathbb{Q}$, and that went well. Using this result seems to give me there is no inverse of $1+\alpha + \alpha^2$ in the above set, which can't happen in a field.</p>
<p>Am I missing something? Thanks.</p>
| Trevor J Richards | 5,952 | <p>This series converges <strong>conditionally</strong> for every $x$. To see this, fix some $x\geq0$. Then once $n>x$, $\dfrac{1}{n+x}\geq\dfrac{1}{2n}$, so by the comparison test, the series $\displaystyle\sum_n\dfrac{1}{n+x}$ diverges. Similarly if $x<0$.</p>
|
4,572,505 | <p>There are many approximations of <span class="math-container">$\pi$</span> using trigonometric and rational numbers. But I created this one: <span class="math-container">$$\pi \approx \sqrt[11]{294204}$$</span> Which is correct to almost <span class="math-container">$8$</span> decimal places. Are there any other approximations of <span class="math-container">$\pi$</span> using radicals? I know of <span class="math-container">$\sqrt{10}$</span>, <span class="math-container">$\sqrt[3]{31}$</span>, <span class="math-container">$\sqrt[4]{97}$</span>, and so on. But are there more beautiful ways (like using <span class="math-container">$\varphi$</span> since it is composed of radicals)?</p>
| Tito Piezas III | 4,781 | <p>The well-known <span class="math-container">$\pi \approx \frac{\ln\left(5280^3\,+\,744\right)}{\sqrt{67}}$</span> and <span class="math-container">$\pi \approx \frac{\ln\left(640320^3\,+\,744\right)}{\sqrt{163}}$</span> involves the <em>j-function</em> and integers. But we can also use the <em>Dedekind eta function</em> and radicals,
<span class="math-container">$$\begin{align}
\pi &\approx \frac{\ln\left(2^6\phi^6-24\right)}{\sqrt{5}}\\
\pi &\approx \frac{\ln\left(2^6\phi^{12}+24\right)}{\sqrt{10}}\\
\pi &\approx \frac{\ln\left(2^{12}\phi^8-24\right)}{\sqrt{15}}\\
\pi &\approx \frac{\ln\left(2^6\phi^{24}-24\right)}{\sqrt{25}}
\end{align}$$</span>
with golden ratio <span class="math-container">$\phi =\frac{1+\sqrt5}2$</span>. The last two are correct to <span class="math-container">$9$</span> and <span class="math-container">$12$</span> decimal places, respectively.</p>
<p><strong>P.S.</strong> The consistent appearance of <span class="math-container">$24$</span>, the powers that are factors of <span class="math-container">$24$</span>, and the radicands that are multiples of <span class="math-container">$5$</span> are clues the above are <em><strong>not</strong></em> coincidences and in fact have a mathematical reason behind them. Lovely, aren't they?</p>
|
1,949,966 | <h2>Q 1a</h2>
<p>Is it possible to define a number $x$ such that $|x|=-1$, where $|\cdot|$ means absolute value, in the same manner that we define $i^2=-1$?</p>
<p>I have no idea if it makes sense, but then again, $\sqrt{-1}$ used to not be a thing either.</p>
<p>To be more explicit, I want as many properties to hold as possible, e.g. $|a|\times|b|=|a\times b|$ and $|a|=|-a|$, as some properties that seem to hold for all different types of numbers (or in some analogous way).</p>
<hr>
<h2>Q 1b</h2>
<p>If we let the solution to $|x|=-1$ be $x=z_1$, and we allow the multiplicativeness property,</p>
<p>$$|(z_1)^2|=1$$</p>
<p>Or, further,</p>
<p>$$|(z_1)^{2n}|=1\tag{$n\in\mathbb N$}$$</p>
<p>Note that this does not mean $z_1$ is any such real, complex, or any other type of number. We used to think $|x|=1$ had two solutions, $x=1,-1$, but now we can give it the solution $x=e^{i\theta}$ for $\theta\in[0,2\pi)$. Adding in the solution $(z_1)^{2n}$ is no problem as far as I can see.</p>
<p>However, there result in some problems I simply cannot quite see so clearly, for example,</p>
<p>$$|z_1+3|=?$$</p>
<p>There exists no such way to define such values at the moment.</p>
<p>Similarly, let $z_2$ be the number that satisfies the following:</p>
<p>$$|z_2|=z_1$$</p>
<p>As far as I see it, it is not possible to create $z_2$, given $z_1$ and $z_0\in\mathbb C$.</p>
<p>The following has a solution, in case you were wondering.</p>
<p>$$|\sqrt{z_1}|=i$$</p>
<p>so no, I did not forget to consider such cases.</p>
<p>But, more generally, I wish to define the following numbers in a recursive sort of way.</p>
<p>$$|z_{n+1}|=z_n$$</p>
<p>since, as far as I can tell, $z_{n+1}$ is not representable using $z_k$ for $k\le n$. In this way, the nature of $z_n$ goes on forever, unlike $i$, which has the solution $\sqrt i=\frac1{\sqrt2}(1+i)$.</p>
<p>So, my second question is to ask if anyone can discern some properties about $z_n$, defining them as we did above? And what is $|z_1+3|=?$</p>
<hr>
<h2>Q 2a</h2>
<p>This part is important, so I truly want you guys (and girls) to consider this:</p>
<blockquote>
<p>Can you construct a problem such that $|x|=-1$ will be required in a step as you solve the problem, but such that the final solution is a real/complex/anything already well known. This is similar to <em><a href="https://en.wikipedia.org/wiki/Casus_irreducibilis" rel="nofollow noreferrer">Casus irreducibilis</a></em>, which basically forced $i$ to exist by establishing its need to exist.</p>
</blockquote>
<p>I am willing to give a large rep bounty for anyone able to create such a scenario/problem. </p>
<hr>
<h2>Q 2b</h2>
<p>And if it is truly impossible, why? Why is it not possible to define some 'thing' the solution to the problem, keep a basic set of properties of the absolute value, and carry on? What's so different between $|x|=-1$ and $x^2=-1$, for example?</p>
<hr>
<h2>Thoughts to consider:</h2>
<p>Now, <a href="https://math.stackexchange.com/a/1345391/272831">Lucian</a> has pointed out that there are plenty of things we do not yet understand, like $z_i\in\mathbb R^a$ for $a\in\mathbb Q_+^\star\setminus\mathbb N$. There may very well exist such a number, but in a field we fail to understand so far.</p>
<p>Similarly, the triangle inequality clearly cannot coexist with such numbers as it is. For the triangle inequality to exist, someone has to figure out how to make triangles with non-positive/real lengths.</p>
<p>As for the <a href="https://en.wikipedia.org/wiki/Norm_(mathematics)#Definition" rel="nofollow noreferrer">properties/axioms of the norm</a> I want:</p>
<p>$$p(v)=0\implies v=0$$</p>
<p>$$p(av)=|a|p(v)$$</p>
| Dirk | 3,148 | <p>The absolute value is quite a different thing than a square. A square simply comes from multiplication and nothing else. Especially, a square does not need an order on the underlying structure. However, the absolute value can only be defined after an order in defined by setting
$$
|x| = \begin{cases} x & x\geq 0\\ -x & x < 0\end{cases}.
$$
So, it is indeed <em>defined</em> to be non-negative. It is not that you may have some algebraic structure with an absolute value and then ask yourself "What if $|x|$ is negative?" in the same way you ask about squares… Put differently:</p>
<p><strong>You can't deduce form the field axioms that $x^2 = -1$ has no solutions, but you can deduce from the axioms of the ordering that $|x|=-1$ has no solutions.</strong></p>
<p>To answer the actual question: I haven't seen variant of absolute values (or norms, or metrics) to take negative values and doubt that such a thing has been studied.</p>
|
1,706,939 | <p>Can anyone share an easy way to approximate $\log_2(x)$, given $x$ is between $0$ and 1?</p>
<p>I'm trying to solve this using an old fashioned calculator (i.e. no logs)</p>
<p>Thanks!</p>
<p>EDIT: I realize that I stepped a bit ahead. The x comes in the form of a fraction, e.g. 3/8, which is indeed between 0 and 1, but could also be written as log2(3) - log2(8). I am hoping there is a quick way to approximate this calculation to let's say 2 decimals</p>
| Community | -1 | <p>First normalize the value to the range $[1,2)$, multiplying by $2$ as long as necessary (the number of multiplies will form the integer part of the logarithm).</p>
<p>Then use the formula</p>
<p>$$\log_2\left(\frac{1+t}{1-t}\right)=\frac2{\ln(2)}\left(t+\frac{t^3}3+\frac{t^5}5\cdots\right)$$ evaluating for </p>
<p>$$t=\frac{x-1}{x+1}$$which will be in range $[0,\dfrac13)$. It will converge reasonably quickly, about one correct decimal per term. </p>
<hr>
<p>Another option is to keep a tabulated list of constants such as $1.5,1.25,1.125\cdots$ together with their logarithms, and use</p>
<p>$$x>C_i\to \log_2(x)=\log_2\left(\frac x{C_i}\right)+\log(C_i).$$</p>
<p>At the same time as you divide $x$ by the constants (but keeping $x>1$), you accumulate the logarithms of these constants. When $x\approx1$, you have it. You can choose the set of constants that suits you best.</p>
|
288,340 | <p>I am having difficulty understanding the recursive definition of a language. The problem asked how to write this non recursively. But I want to understand just how a recursive definition of a language works.</p>
<p>Recursive definition of a subset of L of $\{a,b\}^*$.</p>
<p>Basis : $a\in L$ </p>
<p>Recursive Definition : for any $x\in L$, $ax$ and $xb$ are in $L$.</p>
<p>Below is my attempt at explaining the recursive definition.</p>
<p>Starting with the basis $a$ saying that $ax$ is in it means that all strings formed such as $\{a,aa,aaa,...\}$ are present.</p>
<p>Defining all $xb$ represents $\{ab,abb,abbbb,...\}$</p>
<p>The answer I have so far is $\{a\}^*\{b\}^*$ but again it is understanding it that I am after.</p>
| Community | -1 | <pre><code> a
/ \
aa ab
/ \ / \
aaa aab aab abb
/ \ / \ / \ / \
aaaa aaab aaab aabb aaab aabb aabb abbb
</code></pre>
|
2,041,251 | <p>Suppose $f$ is a twice-differentiable function with $f(0) = 0$, $f\left(\frac12\right) = \frac12$ and $f'(0) = 0$. Prove that $|f''(x)| \ge 4$ for some $x \in \left[0,\frac12\right]$.</p>
<p>I've been stuck on this question for a while now without any idea on how to get started. Is it possible for someone to help me?</p>
<p>Thanks!</p>
| Paramanand Singh | 72,031 | <p>Use Taylor's theorem to get $$f\left(\frac{1}{2}\right)=f(0)+\frac{1}{2}\cdot f'(0)+\frac{1}{8}\cdot f''(c)$$ for some $c\in(0,1/2)$ and your answer follows. </p>
<p>Note that the result can not be obtained by mean value theorem directly, but rather one should observe that proofs of mean value theorem and Taylor's theorem depend on Rolle's theorem. </p>
|
1,079,262 | <p>How many permutations exist in the string $ABCDEFG$, starting from the smallest possible combination if the only direction allowed is forward? For example, B is the smallest possible combination in the string $BDEF$. The only direction being forward, $BD, BE, BF$ are larger permutations, etc.</p>
<p>NOTE: You can't input a character more than once in the same permutation (e.g. $ABCDF$ is allowed but $ABCBDF$ is not. $CDEF$ is allowed on it's own, but $CDEFC$ is not).</p>
<p>Plus, how many key permutations will exist in the same string $A,B,C,D,E,F,G$ if you can move forward and backwards but aren't allowed to repeat a character in the same permutation (e.g. $ABCDF$ is allowed but $ABCBDF$ is not. $CDEF$ is allowed on it's own, but $CDEFC$ is not)?</p>
| Felix Marin | 85,343 | <p>$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,{\rm Li}_{#1}}
\newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
There are, at least, two ways to find the residue:</p>
<ul><li>
\begin{align}
{\expo{z} \over \sin^{2}\pars{z}}&=
{\expo{k\pi} \over \sin^{2}\pars{z}}
+\dsc{\expo{k\pi}\pars{z - k\pi} \over \sin^{2}\pars{z}}
+{\expo{k\pi}\pars{z - k\pi}^{2} \over 2\sin^{2}\pars{z}} + \cdots
\\[5mm]&=
{\expo{k\pi} \over \sin^{2}\pars{z}}
+\dsc{\color{#00f}{\expo{k\pi}}
\bracks{{z - k\pi \over \sin\pars{z - k\pi}}}^{2}\,\color{magenta}{1 \over z - k\pi}}
+{\expo{k\pi}\pars{z - k\pi}^{2} \over 2\sin^{2}\pars{z}} + \cdots
\end{align}
</li>
<li><hr></li>
<li>
\begin{align}
&\lim_{z\ \to\ k\pi}\totald{}{z}
\bracks{\pars{z - k\pi}^{2}\,{\expo{z} \over \sin^{2}\pars{z}}}
=\color{#00f}{\expo{k\pi}}\ \underbrace{\lim_{z\ \to\ 0}\totald{}{z}
\bracks{z^{2}\expo{z} \over \sin^{2}\pars{z}}}_{\dsc{1}}
\end{align}
</li>
</ul>
<p>$$
\color{#66f}{\large\,{\rm Res}_{\bracks{z\ =\ k\pi}}\bracks{%
{\expo{z} \over \sin^{2}\pars{z}}}} = \color{#66f}{\large\expo{k\pi}}
$$</p>
|
1,079,262 | <p>How many permutations exist in the string $ABCDEFG$, starting from the smallest possible combination if the only direction allowed is forward? For example, B is the smallest possible combination in the string $BDEF$. The only direction being forward, $BD, BE, BF$ are larger permutations, etc.</p>
<p>NOTE: You can't input a character more than once in the same permutation (e.g. $ABCDF$ is allowed but $ABCBDF$ is not. $CDEF$ is allowed on it's own, but $CDEFC$ is not).</p>
<p>Plus, how many key permutations will exist in the same string $A,B,C,D,E,F,G$ if you can move forward and backwards but aren't allowed to repeat a character in the same permutation (e.g. $ABCDF$ is allowed but $ABCBDF$ is not. $CDEF$ is allowed on it's own, but $CDEFC$ is not)?</p>
| robjohn | 13,854 | <p>Your approach is correct. Perhaps more direct is
$$
\begin{align}
\frac{e^{k\pi}(1+z+z^2/2+O(z^3))}{z^2(1-z^2/3+O(z^4))}
&=\frac{e^{k\pi}(1+z+5z^2/6+O(z^3))}{z^2}\\
&=\color{#C00000}{e^{k\pi}}\left(\frac1{z^2}+\color{#C00000}{\frac1z}+\frac56+O(z)\right)
\end{align}
$$
Thus, the residue at $z=k\in\mathbb{Z}$ is $e^{k\pi}$.</p>
<p>In the end, everything is about computing the coefficient of $z^{-1}$, which is what you've done.</p>
|
106,031 | <p>I need to "monochromize" a large amount of plots (mostly coming from <code>ListPlot</code>) and export them to PDF. The problem is that I no longer have the data used to generate the plots, I only have notebooks that contain the plots. I attempted copy-pasting one plot and then something along the lines of <code>Show[plot,PlotTheme->"Monochrome"]</code> but that obviously did not work. I also tried <code>ColorConvert[plot,"Grayscale"]</code> but then the exported PDF was of extremely poor quality although I used the <code>ImageResolution</code> option of <code>Export</code>. One could, I think, extract the points from <code>FullForm</code> of the plots and simply plot them once more but this seems to be quite complicated given the quantity of the plots I have and their various structure (some of the <code>ListPlot</code> outputs contain two lists, some of them three, I also have a couple of simple graphics coming from <code>Plot</code>). Method of last resort is using some other software like Photoshop, but that makes the PDF many times larger - I need to use the plots with LaTeX so this would be the last option. </p>
| Sjoerd C. de Vries | 57 | <p>Example plot:</p>
<pre><code>Plot[{Cos[x], Sin[x], Tan[x]}, {x, 0, 2 π}]
</code></pre>
<p><img src="https://i.stack.imgur.com/DBYMS.png" alt="Mathematica graphics"></p>
<p>Now, just copy this plot from the notebook (using ctrl-C) and paste it in the front of following expression:</p>
<pre><code>/. a : RGBColor[__] :> ColorConvert[a, "Grayscale"]
</code></pre>
<p><img src="https://i.stack.imgur.com/a3Afw.png" alt="Mathematica graphics"></p>
<p>The result:</p>
<p><img src="https://i.stack.imgur.com/tYWdt.png" alt="Mathematica graphics"></p>
|
1,595,118 | <p>What is value of $a+b+c+d+e$? If given :</p>
<p>$$abcde=45$$</p>
<p>And $a,b, c, d, e$ all are distinct integer.</p>
<hr>
<p>My attempt :</p>
<p>I calculated, $45 = 3^2 \times 5$.</p>
<blockquote>
<p>Can you explain, how do I find the distinct values of $a,b, c, d, e$ ?</p>
</blockquote>
| Panglossian Oporopolist | 218,322 | <p>Indeed... the values would be $5,3,-3,-1,1$. So adding them should give you $5$.</p>
|
185,061 | <p>Let $p$ be a prime number. For which finite $p$-groups $H$ is there a finite $p$-group $G$ such that $[G,G] \cong H$?</p>
| Dietrich Burde | 32,332 | <p>There are some results for special cases. Burnside has proved in $1912$ that,
if $G$ is a non-metabelian $p$-group, then the centre of the derived group of $G$ cannot be
cyclic. In particular, a non-Abelian group of order $p^3$ cannot be the derived group of
a $p$-group. Blackburn later described the 2-generator groups that arise as commutator subgroups of $2$-generator $p$-groups, see <a href="http://journals.cambridge.org/article_S0305004100031959" rel="nofollow">here</a>. </p>
|
2,065,254 | <p>Let $f: \mathbb{R} \to \mathbb{R}$ be a function that is twice differentiable.</p>
<p>We know that:
$$\lim_{x\to-\infty}\ f(x) = 1$$</p>
<p>$$\lim_{x\to\infty}\ f(x) = 0$$</p>
<p>$$f(0) = \pi$$</p>
<p>We have to prove that there exist at least two points of the function in which $f''(x) = 0$.</p>
<p>How could we do it in a rigorous way? It is pretty intuitive, but in a rigorous way it isn't that simple for me...</p>
| Barry Cipra | 86,747 | <p>For the heck of it, let's do a proof that invokes only the Intermediate and Mean Value Theorems. In particular, let's eschew any explicit use of the Extreme Value Theorem.</p>
<p>If $f$ ever takes the same value infinitely often, the Mean Value (or Rolle's) Theorem applied twice shows there are two values of $x$ for which $f''(x)=0$. Indeed, it suffices for $f$ to take the same value <em>four</em> times: If $f(a)=f(b)=f(c)=f(d)$ with $a\lt b\lt c\lt d$, then there are points $a'\in(a,b)$, $b'\in(b,c)$, and $c'\in(c,d)$ for which $f'(a')=f'(b')=f'(c')=0$, and this in turn implies there are points $a''\in(a',b')$ and $b''\in(b',c')$ for which $f''(a'')=f''(b'')=0$.</p>
<p>We henceforth assume $f$ never takes the same value infinitely often. We now argue that there is a $c$ such that $f(c)\ge f(0)$ and $f'(c)=0$. (If we allowed ourselves the Extreme Value Theorem, this conclusion would be almost automatic and in fact we could have skipped the previous paragraph as well.)</p>
<p>If $f'(0)=0$, we simply let $c=0$. If $f'(0)\gt0$, then there exists $a\gt0$ for which $f(x)\gt f(0)$ for all $x\in(0,a)$. Since $f(0)\gt\lim_{x\to\infty}f(x)$, the Intermediate Value Theorem gives us a $b\in[a,\infty)$ for which $f(b)=f(0)$. Our new assumption on $f$ tells us there is a <em>smallest</em> such $b$, and the Intermediate Value Theorem guarantees $f(x)\gt f(b)=f(0)$ for all $x\in(0,b)$. The Mean Value (or Rolle's) Theorem now says there is a $c\in(0,b)$ for which $f'(c)=0$. A similar argument applies if $f'(0)\lt0$, with a <em>largest</em> $b\le a\lt0$ and $c\in(b,0)$.</p>
<p>Now let $B$ be a bound such that $f(x)\lt2$ for $|x|\ge B$, the existence of which is guaranteed by the limiting behavior of $f$. (Actually, $2$ is just a convenient number between between $\pi$ and $1$; any number between $f(c)$ and the larger of the two limiting values will do.) Let $g_+(x)=m_+(x-c)+f(c)$ and $g_-(x)=m_-(x-c)+f(c)$ be linear functions with $g_+(B)=g_-(-B)=2$, and consider the functions $f_+(x)=f(x)-g_+(x)$ for $x\ge c$ and $f_-(x)=f(x)-g_-(x)$ for $x\le c$. Note that $f_+(c)=f_-(c)=0$ and observe that</p>
<p>$$m_+={g_+(B)-g_+(c)\over B-c}={2-f(c)\over B-c}\lt0\quad\text{and}\quad m_-={g_-(c)-g_-(-B)\over c+B}={f(c)-2\over c+B}\gt0$$</p>
<p>The sign of the slope $m_+$ implies $f_+(x)\gt0$ for $x\in(0,a)$ and also in the limit as $x\to\infty$. But $f_+(B)=f(B)-g_+(B)\lt2-2=0$. The Intermediate Value Theorem tells us that $f_+(c)=f_+(b)=f_+(b')=0$ for two values $b_1$ and $b_2$ with $c\lt b_1\lt B\lt b_2$. Applying the Mean Value Theorem to $f_+(c)=f_+(b_1)$ and $f_+(b_1)=f_+(b_2)$ gives two values $b_1'$ and $b_2'$ with $c\lt b_1'\lt b_1\lt b_2'\lt b_2$ and $f_+'(b_1)=f_+'(b_2)=0$. And applied to $f_+'(b_1)=f_+'(b_2)=0$, MVT coughs up a point $d\in(b_1,b_2)$ for which $f_+''(d)=0$. But $g_+''(x)=0$ for <em>all</em> $x$, hence we have a point $d_+\gt c$ for which $f''(d_+)=0$.</p>
<p>The same argument, just with some of the signs reversed, shows there is a point $d_-\lt c$ for which $f''(d_-)=f_-''(d_-)=0$, and now we are truly done. We have $f''(x)=0$ at two distinct points, $x=d_+$ and $x=d_-$.</p>
|
3,043,296 | <p>Prop: For sets A and B, say A ~ B iff there exists a bijection from A to B. Then ~ is an equivalence relation on sets.</p>
<p>I understand that an equivalence relation holds the properties of reflexive, symmetric, and transitive. I am also aware of their definitions, however, I am struggling to write a proof for this proposition.</p>
<p>I would assume we can suppose there is a bijection between A and B, as this would imply there is a bijection between the two. This would also mean that the two sets have equal cardinality but from this point on I am completely lost, the direction of the proof seems very unclear.</p>
<p>A hidden answer (written proof) would be great with some visible guidance or hints so enhance my understanding.</p>
<p>Thank you.</p>
| Community | -1 | <p>We only need to show three properties: </p>
<p><strong>Reflexiveness:</strong></p>
<p>Does there exist for ANY set <span class="math-container">$A$</span> a bijective function between the set and itself?
If so, we can say that <span class="math-container">$A \sim A$</span>.</p>
<p>Hint, what if there were a function that assigns <span class="math-container">$f(a)=a$</span> for all <span class="math-container">$a \in A$</span>, how do we call this function?</p>
<p><strong>Symmetry:</strong>
If <span class="math-container">$A \sim B$</span>, there exists a bijection between these two sets. Any bijective function is also invertible, so does there exist a bijection between <span class="math-container">$B$</span> and <span class="math-container">$A$</span>? if so, we can say that <span class="math-container">$B \sim A$</span> </p>
<p><strong>Transivity:</strong></p>
<p>Finally, if <span class="math-container">$A\sim B$</span> and <span class="math-container">$B \sim C$</span>, we know there are bijective functions <span class="math-container">$f$</span> and <span class="math-container">$g$</span>, what do we know about the composition of two bijective functions?</p>
<p>Good luck!</p>
|
2,645,611 | <blockquote>
<p>Prove that:
<span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span></p>
</blockquote>
<h3>My work so far:</h3>
<p><span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span>
<span class="math-container">$$\frac{(n+0)!}{n!0!}+\frac{(n+1)!}{n!1!}+...+\frac{(n+n)!}{n!n!}=\frac{(2n+1)!}{n!(n+1)!}$$</span>
<span class="math-container">$$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+...+\binom{2n}{n}=\binom{2n+1}{n}$$</span></p>
<p><em>How to prove the last equality?</em></p>
| Robert Z | 299,698 | <p>Combinatorial proof of
$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$
that is, after dividing by $n!$,
$$\binom{n}{n}+\binom{n+1}{n}+\binom{n+2}{n}+...+\binom{2n}{n}=\binom{2n+1}{n+1}.$$
The RHS counts the number of $\{0,1\}$-strings of length $2n+1$ with $(n+1)$ $1$s. The LHS enumerates the same set according to the position $k$ of the last $1$ on the right: for $k=n+1,n+2,\dots, 2n+1$ they are
$$\binom{k-1}{n}.$$</p>
|
4,280,894 | <blockquote>
<p>Given
<span class="math-container">$$f(x)=1-x^2+x^3 \qquad x\in[0,1]$$</span>
calculate
<span class="math-container">$$
\lim_{n\rightarrow\infty}\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}
$$</span>
where <span class="math-container">$f^n(x)=\underbrace{f(x)·f(x)·\dots\text{·}f(x)}_{n\ \text{times}}$</span>.</p>
</blockquote>
<p>This is a question from CMC(Mathematics competition of Chinese)in <span class="math-container">$2017$</span>. The solution provides an idea: given <span class="math-container">$s∈(0,\frac{1}{2}),$</span> prove:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$</span> The final result is <span class="math-container">$\ln2.$</span></p>
<hr />
<p><strong>My approach</strong><br />
For this:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$</span>
I want to do piecewise calculation:<span class="math-container">$$\int_{s}^{1-s}f^n(x)dx+\int_{1-s}^{1}f^n(x)dx.$$</span>For this:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$</span>Here is the proof: when<span class="math-container">$\ \ n≥\frac{1}{s^2}$</span>, <span class="math-container">$$\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=\frac{\int_{0}^{s}(1-x^2(1-x))^ndx}{\int_{0}^{s}(1-x(1-x)^2)^ndx}\\\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{s}(1-x^2)^ndx}\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{1/\sqrt{n}}(1-\frac{x}{\sqrt{n}})^ndx}\\=\frac{\frac{4}{n+1}(1-(1-\frac{s}{4})^{n+1})}{\frac{\sqrt{n}}{n+1}(1-(1-\frac{1}{n})^{n+1})}\sim\frac{4}{\sqrt{n}(1-\frac{1}{e})}\rightarrow0.\\$$</span>For this:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$</span>Here is the proof: given <span class="math-container">$t,0<t<s<\frac{1}{2},$</span>then<span class="math-container">$$f(t)>f(s)>f(1-s).$$</span>Define <span class="math-container">$m_t=\min_{x\in[0,t]}f(x),M_s=\max_{x\in[s,1-s]}f(x),$</span> so<span class="math-container">$$m_t=f(t)>f(1-s)=M_s.$$</span><span class="math-container">$$\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{t}f^n(x)dx}$$</span><span class="math-container">$$\leq\frac{(1-2s)M_s ^n}{tm_t ^n}=\frac{1-2s}{t}(\frac{M_s}{m_t})^n\rightarrow0.\\$$</span> In conclusion,we can get:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.$$</span></p>
| River Li | 584,414 | <p><em>Alternative proof</em>:</p>
<p>Let
<span class="math-container">\begin{align*}
J_n &= \int_0^1 (1 - x^2 + x^3)^n \,\mathrm{d} x, \\
K_n &= \int_0^1 (1 - x^2 + x^3)^n \ln (x + 2)\,\mathrm{d} x.
\end{align*}</span></p>
<p>First, we have
<span class="math-container">$$K_n \ge \int_0^1 (1 - x^2 + x^3)^n \ln 2\,\mathrm{d} x = \ln 2 \cdot J_n. \tag{1}$$</span></p>
<p>Second, we have
<span class="math-container">\begin{align*}
J_n &\ge \int_0^{1/\sqrt{n}} (1 - x^2 + x^3)^n \,\mathrm{d} x \\
&\ge \int_0^{1/\sqrt{n}} (1 - 1/n)^n \,\mathrm{d} x \\
&= \frac{1}{\sqrt{n}}(1 - 1/n)^n. \tag{2}
\end{align*}</span></p>
<p>Third, we have
<span class="math-container">\begin{align*}
K_n - \ln 2 \cdot J_n &= \int_0^1 (1 - x^2 + x^3)^n \ln (1 + x/2)\,\mathrm{d} x \\
&\le \int_0^1 (1 - x^2 + x^3)^n \, \frac{x}{2}\,\mathrm{d} x \\
&= \int_0^{1/2} (1 - x^2 + x^3)^n \, \frac{x}{2}\,\mathrm{d} x
+ \int_{1/2}^{5/6} (1 - x^2 + x^3)^n \, \frac{x}{2}\,\mathrm{d} x\\
&\qquad + \int_{5/6}^1 (1 - x^2 + x^3)^n \, \frac{x}{2}\,\mathrm{d} x \\
&\le \int_0^{1/2} (1 - x^2 + x^3)^n \, (2x - 3x^2)\,\mathrm{d} x
+ \int_{1/2}^{5/6} (191/216)^n \, \frac{x}{2}\,\mathrm{d} x\\
&\qquad + \int_{5/6}^1 (1 - x^2 + x^3)^n \, (-2x + 3x^2)\,\mathrm{d} x \\
&= \frac{-(1 - x^2 + x^3)^{n + 1}}{n + 1}\Big\vert_0^{1/2}
+ \frac19(191/216)^n
+ \frac{(1 - x^2 + x^3)^{n + 1}}{n + 1}\Big\vert_{5/6}^1\\
&= \frac{24n - 167}{216n + 216}(191/216)^n
- \frac{7}{8n + 8}(7/8)^n + \frac{2}{n + 1} \tag{3}
\end{align*}</span>
where we have used:<br />
(i) <span class="math-container">$\ln(1 + u) \le u$</span> for all <span class="math-container">$u \ge 0$</span>;<br />
(ii) <span class="math-container">$x/2 \le 2x - 3x^2$</span> for all <span class="math-container">$x\in [0, 1/2]$</span>;<br />
(iii) <span class="math-container">$1 - x^2 + x^3 \le 191/216$</span> for all <span class="math-container">$x\in [1/2, 5/6]$</span>;<br />
(iv) <span class="math-container">$x/2 \le -2x + 3x^2$</span> for all <span class="math-container">$x\in [5/6, 1]$</span>.</p>
<p>Fourth, from (1), we have
<span class="math-container">$$\frac{K_n}{J_n} \ge \ln 2.$$</span>
From (1), (2) and (3), we have
<span class="math-container">$$\frac{K_n}{J_n} \le \ln 2 + \frac{\frac{24n - 167}{216n + 216}(191/216)^n
- \frac{7}{8n + 8}(7/8)^n + \frac{2}{n + 1}}{\frac{1}{\sqrt{n}}(1 - 1/n)^n}.$$</span>
Note that
<span class="math-container">$$\lim_{n\to \infty} \frac{\frac{24n - 167}{216n + 216}(191/216)^n
- \frac{7}{8n + 8}(7/8)^n + \frac{2}{n + 1}}{\frac{1}{\sqrt{n}}(1 - 1/n)^n} = 0.$$</span>
Thus, using the squeeze theorem, we have
<span class="math-container">$$\lim_{n\to \infty} \frac{K_n}{J_n} = \ln 2.$$</span></p>
<p>We are done.</p>
|
4,280,894 | <blockquote>
<p>Given
<span class="math-container">$$f(x)=1-x^2+x^3 \qquad x\in[0,1]$$</span>
calculate
<span class="math-container">$$
\lim_{n\rightarrow\infty}\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}
$$</span>
where <span class="math-container">$f^n(x)=\underbrace{f(x)·f(x)·\dots\text{·}f(x)}_{n\ \text{times}}$</span>.</p>
</blockquote>
<p>This is a question from CMC(Mathematics competition of Chinese)in <span class="math-container">$2017$</span>. The solution provides an idea: given <span class="math-container">$s∈(0,\frac{1}{2}),$</span> prove:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$</span> The final result is <span class="math-container">$\ln2.$</span></p>
<hr />
<p><strong>My approach</strong><br />
For this:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$</span>
I want to do piecewise calculation:<span class="math-container">$$\int_{s}^{1-s}f^n(x)dx+\int_{1-s}^{1}f^n(x)dx.$$</span>For this:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$</span>Here is the proof: when<span class="math-container">$\ \ n≥\frac{1}{s^2}$</span>, <span class="math-container">$$\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=\frac{\int_{0}^{s}(1-x^2(1-x))^ndx}{\int_{0}^{s}(1-x(1-x)^2)^ndx}\\\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{s}(1-x^2)^ndx}\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{1/\sqrt{n}}(1-\frac{x}{\sqrt{n}})^ndx}\\=\frac{\frac{4}{n+1}(1-(1-\frac{s}{4})^{n+1})}{\frac{\sqrt{n}}{n+1}(1-(1-\frac{1}{n})^{n+1})}\sim\frac{4}{\sqrt{n}(1-\frac{1}{e})}\rightarrow0.\\$$</span>For this:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$</span>Here is the proof: given <span class="math-container">$t,0<t<s<\frac{1}{2},$</span>then<span class="math-container">$$f(t)>f(s)>f(1-s).$$</span>Define <span class="math-container">$m_t=\min_{x\in[0,t]}f(x),M_s=\max_{x\in[s,1-s]}f(x),$</span> so<span class="math-container">$$m_t=f(t)>f(1-s)=M_s.$$</span><span class="math-container">$$\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{t}f^n(x)dx}$$</span><span class="math-container">$$\leq\frac{(1-2s)M_s ^n}{tm_t ^n}=\frac{1-2s}{t}(\frac{M_s}{m_t})^n\rightarrow0.\\$$</span> In conclusion,we can get:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.$$</span></p>
| Piquancy | 979,182 | <p>Fisrtly,we prove that:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$</span>
Perform piecewise calculation:<span class="math-container">$$\int_{s}^{1-s}f^n(x)dx+\int_{1-s}^{1}f^n(x)dx.$$</span>For this:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$</span>When<span class="math-container">$\ \ n≥\frac{1}{s^2}$</span>, <span class="math-container">$$\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=\frac{\int_{0}^{s}(1-x^2(1-x))^ndx}{\int_{0}^{s}(1-x(1-x)^2)^ndx}\\\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{s}(1-x^2)^ndx}\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{1/\sqrt{n}}(1-\frac{x}{\sqrt{n}})^ndx}\\=\frac{\frac{4}{n+1}(1-(1-\frac{s}{4})^{n+1})}{\frac{\sqrt{n}}{n+1}(1-(1-\frac{1}{n})^{n+1})}\sim\frac{4}{\sqrt{n}(1-\frac{1}{e})}\rightarrow0.\\$$</span>For this:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$</span>Given <span class="math-container">$t,0<t<s<\frac{1}{2},$</span>then<span class="math-container">$$f(t)>f(s)>f(1-s).$$</span>Define <span class="math-container">$m_t=\min_{x\in[0,t]}f(x),M_s=\max_{x\in[s,1-s]}f(x),$</span> so<span class="math-container">$$m_t=f(t)>f(1-s)=M_s.$$</span><span class="math-container">$$\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{t}f^n(x)dx}$$</span><span class="math-container">$$\leq\frac{(1-2s)M_s ^n}{tm_t ^n}=\frac{1-2s}{t}(\frac{M_s}{m_t})^n\rightarrow0.\\$$</span> In conclusion,we can get:<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.$$</span>
Secondly,we calculate the result:<span class="math-container">$\ln2$</span>.<br />
For <span class="math-container">$\varepsilon\in(0,\ln\frac{5}{4})$</span>,take <span class="math-container">$s=2(e^\varepsilon-1)$</span>,than <span class="math-container">$s\in(0,\frac{1}{2}),\ln\frac{2+s}{2}=\varepsilon.$</span><br />
From the above conclusion,we can know:<br />
<span class="math-container">$\exists N\geq1,s.t.$</span>,when <span class="math-container">$n\geq N$</span>,<span class="math-container">$$\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq \varepsilon.$$</span>So<span class="math-container">$$\lvert\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}-\ln2\rvert=\frac{\int_{0}^{1}f^n(x)\ln\frac{x+2}{2}dx}{\int_{0}^{1}f^n(x)dx}$$</span><span class="math-container">$$\leq\frac{\int_{0}^{s}f^n(x)\ln\frac{x+2}{2}dx}{\int_{0}^{s}f^n(x)dx}+\frac{\int_{s}^{1}f^n(x)\ln\frac{x+2}{2}dx}{\int_{0}^{s}f^n(x)dx}$$</span><span class="math-container">$$\le\ln\frac{s+2}{2}+\frac{\ln\frac{3}{2}\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq\varepsilon(1+\ln\frac{3}{2}).\\$$</span>Let <span class="math-container">$\varepsilon\rightarrow0$</span>,we can get<span class="math-container">$$\lim_{n\rightarrow\infty}\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}=\ln2.\\$$</span>That's all.</p>
|
4,180,750 | <p>Let's consider inner product space with vectors <span class="math-container">$x, y, z$</span> which satisfies:</p>
<p><span class="math-container">$$\|x+y+z\|^2 = 14$$</span></p>
<p><span class="math-container">$$\|x+y-z\|^2 = 2$$</span></p>
<p><span class="math-container">$$\|x-y+z\|^2 = 6$$</span></p>
<p><span class="math-container">$$\|x-y-z\|^2 = 10$$</span></p>
<p>I want to prove that <span class="math-container">$x$</span> is perpendicular to <span class="math-container">$y$</span>.</p>
<p><strong>My work so far</strong></p>
<p>In other words we want to prove that <span class="math-container">$\langle x, y \rangle = 0$</span>.</p>
<p>My first idea was to use Cauchy Schwarz inequality <span class="math-container">$ | \langle x , y \rangle |\le \|x\| \|y\|$</span>
and to show that our conditions force that <span class="math-container">$\|x\|\cdot\|y\| = 0$</span>. However I didn't manage to do anything sensible. Also I tried to somehow prove that under this conditions our norm has bo inducted by inner product - in other words we have that: <span class="math-container">$$2(\|x\|^2 + \|y\|^2) = \|x+y\|^2 + \|x-y\|^2$$</span></p>
<p>but also I didn't end up with something rational. Could you please give me a hand, what's the correct approach to this problem?</p>
| José Carlos Santos | 446,262 | <p>Assuming that you are working over <span class="math-container">$\Bbb R$</span>, let:</p>
<ul>
<li><span class="math-container">$s=\|x\|^2+\|y\|^2+\|z\|^2$</span>;</li>
<li><span class="math-container">$a=2\langle x,y\rangle$</span>;</li>
<li><span class="math-container">$b=2\langle x,z\rangle$</span>;</li>
<li><span class="math-container">$c=2\langle y,z\rangle$</span>.</li>
</ul>
<p>Then those four equalities tell you that<span class="math-container">$$\left\{\begin{array}{l}s+a+b+c=14\\s+a-b-c=2\\s-a+b-c=6\\s-a-b+c=10.\end{array}\right.$$</span>In particular,<span class="math-container">$$\begin{split}(s+a+b+c)+(s+a-b-c)-(s-a+b-c)-(s-a-b+c)
&=14+2-6-10 \\
&=0
\end{split}$$</span>In other words, <span class="math-container">$4a=0$</span>. But <span class="math-container">$a=2\langle x,y\rangle$</span>.</p>
|
1,705,081 | <p>It's a matrix solved with least squares equations (probaly). I used some calculator but can't get his outcome. If you have a way how to get to this please explain how.</p>
<p>[The math on that image is:
$$A = \left[\matrix{4&3&1&0&1\cr 5&2&1&0&1\cr 4&2&1&1&1\cr 3&1&0&1&1\cr 1&1&0&1&1\cr}\right], \quad\vec b = \left[\matrix{4\cr 6\cr 6\cr 3\cr 1\cr}\right]$$
"Least Squares" of $A\vec x = \vec b$ is $\vec x = \left[\matrix{1\cr -1\cr 2\cr 1\cr 1\cr}\right]$.]</p>
<p>Pic related:
<a href="http://postimg.org/image/7ellc2pnl/" rel="nofollow">http://postimg.org/image/7ellc2pnl/</a></p>
| iiivooo | 218,986 | <p>Your answer is correct. To compare the logarithm you should notice that
$$\lim_{x->\infty}\frac{(\log(x))^p}{x}=0$$
for any positive constant $p$ and from this you immediately see that $(\log n)^3$ has smaller rate that $\sqrt n\log n$ and both have smaller order than all polynomials. </p>
|
21,752 | <blockquote>
<p>"Let $P$ be the change-of-basis matrix
from a basis $S$ to a basis $S'$ in a
vector space $V$. Then, for any vector
$v \in V$, we have $$P[v]_{S'}=[v]_{S}
\text{ and hence, } P^{-1}[v]_{S} =
[v]_{S'}$$</p>
<p>Namely, if we multiply the coordinates
of $v$ in the original basis $S$ by
$P^{-1}$, we get the coordinates of
$v$ in the new basis $S'$." - Schaum's
Outlines: Linear Algebra. 4th Ed.</p>
</blockquote>
<p>I am having a lot of difficulty keeping these matrices straight. Could someone please help me understand the reasoning behind (what appears to me as) the counter-intuitive naming of $P$ as the change of basis matrix from $S$ to $S'$? It seems like $P^{-1}$ is the matrix which actually changes a coordinate vector in terms of the 'old' basis $S$ to a coordinate vector in terms of the 'new' basis $S'$...</p>
<p><em>Added:</em></p>
<blockquote>
<p>"Consider a basis $S =
\{u_1,u_2,...,u_n\}$ of a vector space
$V$ over a field $K$. For any vector
$v\in V$, suppose $v = a_1u_1
+a_2u_2+...+a_nu_n$</p>
<p>Then the coordinate vector of $v$
relative to the basis $S$, which we
assume to be a column vector (unless
otherwise stated or implied), is
denoted and defined by $[v]_S =
[a_1,a_2,...,a_n]^{T}$. "</p>
<p>"Let $S = \{ u_1,u_2,...,u_n\}$ be a
basis of a vector space $V$, and let
$S'=\{v_1,v_2,...,v_n\}$ be another
basis. (For reference, we will call
$S$ the 'old' basis and $S'$ the 'new'
basis.) Because $S$ is a basis, each
vector in the 'new' basis $S'$ can be
written uniquely as a linear
combination of the vectors in S; say,</p>
<p>$\begin{array}{c} v_1 = a_{11}u_1 +
a_{12}u_2 + \cdots +a_{1n}u_n \\ v_2 =
a_{21}u_1 + a_{22}u_2 + \cdots
+a_{2n}u_n \\ \cdots \cdots \cdots \\ v_n = a_{n1}u_1 + a_{n2}u_2 + \cdots
+a_{nn}u_n \end{array}$</p>
<p>Let $P$ be the transpose of the above
matrix of coefficients; that is, let
$P = [p_{ij}]$, where $p_{ij} =
a_{ij}$. Then $P$ is called the
\textit{change-of-basis matrix} from
the 'old' basis $S$ to the 'new' basis
$S'$." - Schaum's Outline: Linear Algebra 4th Ed.</p>
</blockquote>
<p>I am trying to understand the above definitions with this example:</p>
<p>Basis vectors of $\mathbb{R}^{2}: S= \{u_1,u_2\}=\{(1,-2),(3,-4)\}$ and $S' = \{v_1,v_2\}= \{(1,3), (3,8)\}$ the change of basis matrix from $S$ to $S'$ is $P = \left( \begin{array}{cc} -\frac{13}{2} & -18 \\ \frac{5}{2} & 7 \end{array} \right)$.</p>
<p>My current understanding is the following: normally vectors such as $u_1, u_2$ are written under the assumption of the usual basis that is $u_1 = (1,-2) = e_1 - 2e_2 = [u_1]_E$. So actually $[u_1]_S = (1,0)$ and I guess this would be true in general... But I am not really understanding what effect if any $P$ is supposed to have on the basis vectors themselves (I think I understand the effect on the coordinates relative to a basis). I guess I could calculate a matrix $P'$ which has the effect $P'u_1, P'u_2,...,P'u_n = v_1, v_2,..., v_n$ but would this be anything?</p>
| Agustí Roig | 664 | <p>Everybody studying the change of basis affair should work out some simple examples like the following. Consider this basis in $\mathbb{R}^2$:</p>
<p>$$
v_1 = (1,1) \qquad \text{and} \qquad v_2 = (1,-1) \ .
$$</p>
<p>Or, since we are going to stress the bases and coordinates thing, we could write it this way</p>
<p>$$
v_1 = (1,1)_e \qquad \text{and} \qquad v_2 = (1,-1)_e \ ,
$$</p>
<p>since these are coordinates in the standard basis</p>
<p>$$
e_1 = (1,0) \qquad \text{and} \qquad e_2 = (0,1) \ .
$$</p>
<p>The change of basis matrix from $v$ to $e$ is</p>
<p>$$
P =
\begin{pmatrix}
1 & 1 \\\
1 & -1
\end{pmatrix} \ .
$$</p>
<p>Now, take the vector </p>
<p>$$
u = 2v_1 - 3v_2 \ .
$$</p>
<p>Its coordinates in the $v$ basis are:</p>
<p>$$
u = (2,-3)_v \ .
$$</p>
<p>If you want to obtain its coordinates in the $e$ (standard) basis, you can do it by hand:</p>
<p>$$
u = 2v_1 - 3v_2 = 2(1,1)_e -3(1,-1)_e = (2-3, 2+3)_e = (-1, 5)_e \ .
$$</p>
<p>Now, you realise that these are <em>exactly the same</em> operations that you do when performing this matrix multiplication:</p>
<p>$$
P
\begin{pmatrix}
2 \\\
-3
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 \\\
1 & -1
\end{pmatrix}
\begin{pmatrix}
2 \\\
-3
\end{pmatrix}
=
\begin{pmatrix}
2 - 3 \\\
2 + 3
\end{pmatrix}
=
\begin{pmatrix}
-1 \\\
5
\end{pmatrix} \ .
$$</p>
<p><strong>Exercise.</strong> Maybe now you could redo yourself the proof of the change of basis theorem: take two arbitrary bases $v$ and $e$ in no matter which vector space, related by</p>
<p>$$
v_i = a^1_i e_1 + \cdots + a^n_i e_n \ , \qquad i = 1, \dots , n \ .
$$</p>
<p>Write down the change of basis matrix from $v$ to $e$ (that is, put the coordinates of the $v$ vectors as columns, like in the previous example):</p>
<p>$$
P =
\begin{pmatrix}
a^1_1 & \dots & a^1_n \\\
\vdots & \ddots & \vdots \\\
a^n_1 & \dots & a^n_n
\end{pmatrix} \ ,
$$</p>
<p>take any vector</p>
<p>$$
u = b^1v_1 + \cdots + b^nv_n \ ,
$$</p>
<p>and write down its coordinates in the $v$ basis. Finally, find out its coordinates in the $e$ basis (by hand and with the help of the matrix $P$).</p>
|
1,651,991 | <p>Let $p(x)$ be an odd degree polynomial and let $q(x)=(p(x))^2+ 2p(x)-2$ </p>
<p>a) The equation $q(x)=p(x)$ admits atleast two distinct real solutions.</p>
<p>b) The equation $q(x)=0$ admits atleast two distinct real solutions.</p>
<p>c) The equation $p(x)q(x)=4$ admits atleast two distinct real solutions.</p>
<p>which of the following are true?</p>
<p>i know that all the three are true but donot know how to prove them</p>
| GoodDeeds | 307,825 | <p>The straight line joining $(-2,-0.5)$ and $(0.25,0.5)$ is given by
$$y-0.5=\frac{0.5-(-0.5)}{0.25-(-2)}(x-0.25)$$
$$y-0.5=\frac{1}{2.25}(x-0.25)$$
$$2.25y-1.125=x-0.25$$
$$2.25y-x=0.875$$
$$18y-8x=7$$</p>
<p>Now, substituting $x=-1$, we get,
$$18y+8=7$$
$$y=-\frac{1}{18}$$</p>
|
1,634,807 | <p>The full question:</p>
<p>Having an equivalence relation $\sim$ on $\Bbb N$ defined by: $a \sim b$ meaning $a,b\in\Bbb N$ such that $a=b*10^k,$ for some $k\in\Bbb Z$, give a complete set of equivalence class representatives.</p>
<p>I am having trouble visualising these. I'm thinking you would need the whole set $\Bbb Z$. Does anyone have any ideas?</p>
<p>Help would be greatly appreciated! (this is my first post as well, so please say if anything is unclear)</p>
| Asinomás | 33,907 | <p>The numbers that are not multiples of $10$ give us a complete set of representatives, and in fact we have exactly one representative for each class. Let $X$ be the set of all positive numbers not divisible by $10$</p>
<p>To see why it is a complete set of representatives take $n\in\mathbb N$, suppose $10^k$ is the largest power of $10$ dividing $n$ ($k$ might be $0$), notice $n\equiv \frac{n}{10}\equiv\dots\equiv \frac{n}{10^k}\in X$.</p>
<p>To see why it does not contain two elements in the same class notice if $a\equiv b$ then one of $a$ and $b$ is a multiple of $10$.</p>
<hr>
<p>Edit: I assume $\mathbb N$ does not contain $0$, if we want $0\in\mathbb N$ just notice $0$ would be in a class by itself.</p>
|
4,398,864 | <p>How many ways are there to select a three digit number <span class="math-container">$\underline{A}\ \underline{B}\ \underline{C}$</span> so that <span class="math-container">$A \neq B$</span>, <span class="math-container">$B \leq C$</span>, and <span class="math-container">$A < C$</span>?</p>
<hr />
<p>I found the answer as <span class="math-container">$\displaystyle\sum_{k=2}^{9} (k^2-k)=240$</span> by evaluating some big summations. I was wondering if anyone had a solution that doesn't require evaluating big sums?</p>
<p>Thanks in advance.</p>
| lulu | 252,071 | <p>If the three digits are distinct and non-zero, then there are two ways to arrange any selection of three digits from <span class="math-container">$\{1,\cdots, 9\}$</span>. Thus this gives us <span class="math-container">$2\times \binom 93=168$</span> cases.</p>
<p>If <span class="math-container">$B=0$</span>, there are <span class="math-container">$\binom 92=36$</span> choices.</p>
<p>If <span class="math-container">$B=C>0$</span> there are <span class="math-container">$\binom 92=36$</span> choices.</p>
<p>And we are done, with <span class="math-container">$168+36+36=240$</span> cases.</p>
|
2,418,440 | <p><strong>Defn:</strong> Let $f$ be a function from $\mathbb{R}$ into a set $X$. We say that $f$ is <em>periodic</em> if there exists $p>0$ such that for all $x\in \mathbb{R}$, we have $f(x+p)=f(x)$. </p>
<p><strong>Prove</strong>: If $f$ is a continuous periodic function from $\mathbb{R}$ into a metric space $M$, then $f$ is uniformly continuous on $\mathbb{R}$.</p>
<p><strong>Attempt:</strong> I think I can use the fact that for all $x \in \mathbb{R}$, $[x,x+p]$ is a closed and bounded interval. Then $f$ is compact and hence uniformly continuous on the interval. </p>
<p>I also tried considering $[0,p]$. In that case, $x = np+\alpha$ and $y = mp + \beta$ for some $m,p \in \mathbb{Z}$ and $\alpha,\beta \in \mathbb{R}$. If $n<0$, then we can choose $\alpha \in \mathbb{R^\mathbf{-}}$, so that $f(np+\alpha)=f(|\alpha|)$ and not $f(1-\alpha)$. Hope that makes sense.</p>
<p>Then $\alpha,\beta \in [0,p]$, and I can find a $\delta$ that works for all $\alpha, \beta$ in $[0,p]$. The problem is, I need to constrain $|x-y|$ and somehow get that $|\alpha-\beta| < \delta$. So far I haven't figured out how to do this.</p>
<p>Been furrowing my brow at this for a while.. any hints very welcomed...</p>
<p>Thanks</p>
| DanielWainfleet | 254,665 | <p>For $r>0$ let $s_r\in (0,p)$ such that $\forall x,y\in [0,2p]\;(|x-y|<s_r\implies |f(x)-f(y)|<r). $</p>
<p>For any $x,y\in \mathbb R$ with $|x-y|<s_r$ there exists $n\in \mathbb Z$ such that $\{x,y\}\subset [np, (n+2)p].$ Because if $n=\max \{m\in \mathbb Z: mp\leq \min (x,y)\}$ then $\min (x,y)<(n+1)p,$ so $$np\leq \min (x,y)\leq\max (x,y)<\min (x,y)+s_r<(n+1)p+s_r<(n+2)p .$$</p>
<p>Now $\{x-np,y-np\}\subset [0,2p]$ and $|(x-np)-(y-np)|=|x-y|<s_r.\;$ Therefore $$|f(x)-f(y)|=|f(x-np)-f(y-np)|<r.$$ </p>
|
2,523,660 | <p>I'm trying to solve the next question:</p>
<p>For all $m\in I=(0,1)$ there is a subset $A_m \subseteq \mathbb{R}$ that $A_{m} = \{ a\in \mathbb{R} : a-\lfloor a \rfloor < m \} $. Find $\bigcap\limits_{m\in I} A_{m}$</p>
<p>So I think that the solution is $\bigcap\limits_{m\in I} A_m=\mathbb{Z}$, and I tried to prove it like this:</p>
<p>let $x\in\mathbb{Z}$ then $-x\leq-\lfloor x\rfloor<1-x$.</p>
<p>Therefore $0=-x+x\leq x-\lfloor x\rfloor<x+1-x=1$, so for every $m\in I, x\in A_{m} $</p>
<p>and then $x\in\bigcap\limits_{m\in I} A_{m}$, so $\mathbb{Z}\subseteq \bigcap\limits_{m\in I} A_{m}$.</p>
<p>Now I need to prove that $\bigcap\limits_{m\in I} A_{m} \subseteq \mathbb{Z}$, but how?</p>
| William Elliot | 426,203 | <p>$\pi$ is continuous. Thus its inverse image of an open set is open and a closed set closed. No, the inverse is not an open nor a closed map because it is not a map from X to X×Y. </p>
<p>It is a function from the power set of X to the power set of X×Y and those two power sets do not have a topology. Thus it is meaningless to ask if its open or closed.</p>
|
3,328,737 | <p>For any rational number, <span class="math-container">$\frac{p}{q}$</span> , <span class="math-container">$p$</span> and <span class="math-container">$q$</span> should be integers, <span class="math-container">$q\neq0$</span> and <span class="math-container">$p,q$</span> should not have any common factors.
Now, if we have two even numbers, say <span class="math-container">$2m$</span> and <span class="math-container">$2n$</span> where <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are integers.
<span class="math-container">$$\frac{\text{even}}{\text{even}}=\frac{2m}{2n}=\frac{m}{n}$$</span>
where(<span class="math-container">$\frac{m}{n}$</span>) nature is still unknown.
So, what nature does <span class="math-container">$\frac{\text{even}}{\text{even}}$</span> have, rational or irrational?</p>
<p>additional reference: <a href="https://www.quora.com/Is-2-4-rational" rel="nofollow noreferrer">https://www.quora.com/Is-2-4-rational</a></p>
| Ahmed Hossam | 430,756 | <p><a href="https://artofproblemsolving.com/wiki/index.php/Proof_by_contradiction" rel="nofollow noreferrer">https://artofproblemsolving.com/wiki/index.php/Proof_by_contradiction</a></p>
<p>Proof (by contradiction) that <span class="math-container">$\sqrt{2}$</span> is irrational.</p>
<p><strong>Assume for a minute, that</strong> <span class="math-container">$\sqrt{2}$</span> <strong>is rational</strong>, that is: it can be written as <span class="math-container">$\sqrt{2}=p/q$</span> for two specific integer numbers <span class="math-container">$p$</span> and <span class="math-container">$q$</span>, <strong>which don't share any common divisors greater than</strong> <span class="math-container">$1$</span>. </p>
<p>If now <span class="math-container">$\sqrt{2}=p/q$</span>, then we can square to get <span class="math-container">$2=p^2/q^2$</span> or multiply both sides by <span class="math-container">$q^2$</span> and get <span class="math-container">$2q^2 = p^2$</span>. Now the number <span class="math-container">$q^2$</span> is an integer, because <span class="math-container">$q$</span> is an integer, and the number <span class="math-container">$2q^2$</span> is an even number, because we are multiplying the integer <span class="math-container">$q^2$</span> by <span class="math-container">$2$</span> and the result of multiplication of any integer number by <span class="math-container">$2$</span> gives an even number.</p>
<p>If now <span class="math-container">$2q^2$</span> is an even number, then <span class="math-container">$p^2$</span> is an even number, because <span class="math-container">$2q^2=p^2$</span>. Thus the number <span class="math-container">$p$</span> <strong>is an even number</strong>, because <span class="math-container">$p^2$</span> is an even number.</p>
<p>If now <span class="math-container">$p$</span> is an even number, then we can express it as <span class="math-container">$p=2k$</span> for some integer <span class="math-container">$k$</span> and if <span class="math-container">$p=2k$</span>, then <span class="math-container">$p^2 = 4k^2$</span>.</p>
<p>So, we get <span class="math-container">$q^2=2k^2$</span>, because <span class="math-container">$2q^2=p^2$</span> and <span class="math-container">$p^2=4k^2$</span>. If we have <span class="math-container">$q^2=2k^2$</span>, then <span class="math-container">$q^2$</span> is an even number. If <span class="math-container">$q^2$</span> is an even number, then <span class="math-container">$q$</span> <strong>is an even number</strong>.</p>
<p>If <span class="math-container">$p$</span> is an even number and <span class="math-container">$q$</span> is an even number, then <span class="math-container">$p/q$</span> can be further simplified. This is a contradiction to the assumption, that <span class="math-container">$p$</span> and <span class="math-container">$q$</span> should not share any common divisors greater than <span class="math-container">$1$</span>. Thus, the assumption is false, it's opposite is true. This means <span class="math-container">$\sqrt{2}$</span> must be irrational.</p>
|
1,646,135 | <p>I have the subset $\left[0,1\right] \backslash \mathbb{Q}$ in $\mathbb{R} \backslash \mathbb{Q}$. </p>
<p>Am I right in thinking that this set is open and not closed in the space given?</p>
<p>Also, how do I go about finding the interior, closure and boundary?</p>
| lulu | 252,071 | <p>No such function exists. </p>
<p>To see that, suppose you had such an $f$. Let's compute a few examples.</p>
<p>Case I: $$A=\{1,2,3\},\,B=\{3,4,5\},C=\{1,4,6\}$$</p>
<p>Then it is easy to compute everything in your expression. We get $$0=f(3,3,3,1,1,1)$$</p>
<p>Case II: $$A=\{1,2,3\},\,B=\{1,4,5\},C=\{1,6,7\}$$</p>
<p>Then we get $$1=f(3,3,3,1,1,1)$$</p>
|
3,446,084 | <p>Suppose we have coprime integers <span class="math-container">$(a,b)$</span> and let <span class="math-container">$\ell \in \mathbb{Z}$</span> be arbitrary. The general solution to the linear Diophantine equation <span class="math-container">$ax+by=\ell$</span> is given by <span class="math-container">$x=\ell x' + bt$</span> and <span class="math-container">$y=\ell y' - at$</span> where <span class="math-container">$x', y'$</span> satisfy <span class="math-container">$ax'+by'=1$</span> and are found by the extended Euclidean algorithm, and <span class="math-container">$t$</span> ranges over <span class="math-container">$\mathbb{Z}$</span>.</p>
<p>I'm interested in minimal solutions to the equation, in the sense of the <span class="math-container">$L^{\infty}$</span> norm. That is to say, we're looking for solutions <span class="math-container">$(x,y)$</span> to the equation which minimize <span class="math-container">$\text{max}(|x|, |y|)$</span>. </p>
<p>If <span class="math-container">$\ell=1$</span>, then the minimal solution to <span class="math-container">$ax+by=1$</span> is exactly <span class="math-container">$(x', y')$</span>, where <span class="math-container">$(x', y')$</span> are defined as above, and it satisfies <span class="math-container">$|x'| \leq |b|$</span> and <span class="math-container">$|y'|\leq |a|$</span>. This is just because the extended Euclidean algorithm always produces minimal solutions. If <span class="math-container">$\ell \neq 1$</span>, then we might think that <span class="math-container">$(\ell x', \ell y')$</span> are the minimal solutions, but this is not correct. A simple example is <span class="math-container">$7x+3y=10$</span>. The optimal solution here is of course <span class="math-container">$(x,y)=(1,1)$</span>. This is better than any solution of the form <span class="math-container">$(10x', 10y')$</span>, where <span class="math-container">$(x',y')$</span> are minimal solutions to <span class="math-container">$7x+3y=1$</span>. </p>
<p>Is there, in general, a decent algorithm for the minimal solution? Alternatively, is there a (decent) estimate for the minimal solution? For coprime <span class="math-container">$(a,b)$</span>, if we define a function <span class="math-container">$f_{(a,b)}(n)$</span> given by <span class="math-container">$$f_{(a,b)}(n) = \min_{\{(x,y) \in \mathbb{Z}^2 : ax+by=n\}} \max (|x|, |y|)$$</span> is there anything that can be said to how <span class="math-container">$f$</span> behaves as <span class="math-container">$n \to \infty$</span>? Certainly, <span class="math-container">$f$</span> would tend to <span class="math-container">$\infty$</span>, but are there precise asymptotic estimates or concrete inequalities? If I had to guess, I would reckon that it would be <span class="math-container">$O(n)$</span>. </p>
<p>Note that, in more general terms, we can ask this question for any nonempty set <span class="math-container">$S$</span> of coprime integers, and linear Diophantine equations with coefficients in <span class="math-container">$S$</span>. Here, we've restricted ourselves to the case where <span class="math-container">$|S| = 2$</span>. A general solution would also be great. </p>
| A.J. | 654,406 | <p>Suppose <span class="math-container">$\,(x',y')\,$</span> is the minimal solution to the <span class="math-container">$\,ax+by=1\,$</span> generated by the extended Euclidean algorithm. Then as you pointed out, <span class="math-container">$\,(nx',ny')\,$</span> will be a solution to <span class="math-container">$\,ax+by=n\,$</span> but not necessarily minimal. However, note that <span class="math-container">$\,(nx'-kb,ny'+ka)\,$</span> will also be a solution for any <span class="math-container">$\,k \in \Bbb Z.\,$</span></p>
<p>To find the minimal solution to <span class="math-container">$\,ax+by=n\,$</span>, we simply need to find <span class="math-container">$\,k\,$</span> such that <span class="math-container">$\,nx'-kb\,$</span> and <span class="math-container">$\,ny'+ka\,$</span> are as close as possible [see note below]. To do that, we solve the equation <span class="math-container">$\,nx'-kb=ny'+ka\,$</span> for <span class="math-container">$\,k\,$</span>; if the equation yields an integral <span class="math-container">$\,k\,$</span>, then we will have our minimal solution, otherwise either <span class="math-container">$\,\lfloor k \rfloor\,$</span> or <span class="math-container">$\,\lceil k \rceil\,$</span> (whichever is closer to <span class="math-container">$\,k\,$</span>) will generate the minimal solution.</p>
<p>In your particular example, the extended Euclidean algorithm should generate the solution <span class="math-container">$\,(1,-2)\,$</span> as the minimal solution to <span class="math-container">$\,7x+3y=1,\,$</span> so our starting solution to <span class="math-container">$\,7x+3y=10,\,$</span> would be <span class="math-container">$\,(10,-20).\,$</span> Then solving the equation <span class="math-container">$\,10-3k=-20+7k\,$</span> gives <span class="math-container">$\,k=3\,$</span> which yields the minimal solution <span class="math-container">$\,(1,1).\,$</span></p>
<p>[Note: this assumes that <span class="math-container">$\,ab \gt 0;\,$</span> if not then one can just solve <span class="math-container">$\,ax-by=n\,$</span> and then use the solution <span class="math-container">$\,(nx'-kb,-[ny'+ka]).\,$</span>]</p>
|
2,922,881 | <p>I'm trying to evaluate the following complex integral using the residue method. $$\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz$$</p>
<p>The point $z_0=0$ seems to be a singularity. I'm not sure but I think it's also a non-removable one. I tried using the Taylor expansion of $e^x$ and $\cos{x}$ as that usually helps. </p>
<p>$$e^{\frac{1}{z}}\cos{\frac{1}{z}}=(1+\frac{1}z+\frac{1}{2!z^2}+\frac{1}{3!z^3}+...)(1-\frac{1}{2!z^2}+\frac{1}{4!z^4}-...)\\=>e^{\frac{1}{z}}\cos{\frac{1}{z}}=(1+\frac{1}{z}+...)$$</p>
<p>It seems like the negative power terms are infinite showing that $z_0=0$ is no pole. If I'm correct, the coefficient of $1/z$, which is $1$, is the residue of the singularity and this leads to the result:$$ \int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz=2\pi i$$ </p>
<p>I don't think I've evaluated other integrals with non-removable singularities and I'm not sure about the whole process..</p>
| eloiprime | 180,579 | <p>Assuming that $|z|=1$ is positively oriented (which is the convention), then your working looks fine to me. The residue theorem does not care about the type of singularities within the simply-connected domain around which you're integrating, just that there are finitely many such singularities. Furthermore, the method you used to compute the required residue does not require the implied singularity to be of a certain type; it may be taken as the definition of the residue.</p>
|
1,926,839 | <p>I have to find a function $f\in C^\infty(\mathbb{R})$ which, for fixed $a,b\in \mathbb{R}$, $a<b$, is identically 1 for $x\le a$, identically $0$ for $x\ge b$ and decreases in $a\le x\le b$. I've tried many times to write $f$ in $a\le x\le b$ using an exponential, but it didn't work. Can you help me?</p>
| Ian | 83,396 | <p>The function</p>
<p>$$f (x)=\begin {cases} e^{-1/x^2} & x>0 \\ 0 & otherwise \end {cases} $$
is smooth, increases to 1 from 0. Subtracting it from 1 gives a function that decreases from 1 to 0. Pick your favorite smooth increasing homeomorphism $h $ from $(a,b) $ to $(0,\infty)$. Then define $g(x)=1-f (h (x))$ on $(a,b) $ and extend from there.</p>
|
95,741 | <p>I wonder if there is any difference between mapping and a function. Somebody told me that the only difference is that mapping can be from any set to any set, but function must be from $\mathbb R$ to $\mathbb R$. But I am not ok with this answer. I need a simple way to explain the differences between mapping and function to a lay man together with some illustration (if possible).</p>
<p>Thanks for any help.</p>
| Alex Becker | 8,173 | <p>I'm afraid the person who told you that was wrong. There is no difference between a mapping and a function, they are just different terms used for the same mathematical object. Generally, I say "mapping" when I want to emphasize that what I am talking about pairing elements in one set with elements in another set, and "function" when I want to emphasize that the thing I am talking about takes input and returns output. But that's just a personal preference, and there is no convention I'm aware of.</p>
|
95,741 | <p>I wonder if there is any difference between mapping and a function. Somebody told me that the only difference is that mapping can be from any set to any set, but function must be from $\mathbb R$ to $\mathbb R$. But I am not ok with this answer. I need a simple way to explain the differences between mapping and function to a lay man together with some illustration (if possible).</p>
<p>Thanks for any help.</p>
| Will Jagy | 10,400 | <p>Not that much difference in the long run. When I use the word function I generally mean that a point maps to a single point. So, if a point might map to several points, I am not going to use that word, more likely mapping or transformation. In a recent article I had one of these, each point went to several points, and each point in the image probably had several pre-images, so I emphasized, in a traditional phrase, that the mapping was "many-to-many." Now, both primage and image were equivalence classes under a weaker equivalence, so the mapping did induce a function from "genus" to "genus," but was not well-defined on the level of isometry classes of quadratic forms.</p>
<p>Anyway, if a point goes to only a single point, you are allowed to call it a function. </p>
<p>EDIT: I see, you have finished college and are just asking about preferences. I've got to think about popularity in English... Function is used for $\mathbb C \mapsto \mathbb C,$ also maps from any smooth manifold to the reals. I might use function for almost any map into $\mathbb R^n$ from almost anything, but would be less likely to use function for a mapping between two other manifolds. Various kinds of mappings in algebra are unlikely to be called function. </p>
|
20,224 | <p>I was rereading an answer to an old question of mine and it included a reference to the fact that $2^{\omega_1}$ was separable. I'm having a hard time finding a reference for this fact, and the proof is not immediately obvious to me. Can anyone provide me with a cite and/or a proof? </p>
| David R. MacIver | 4,959 | <p>Should have searched a bit harder before asking this one. This is an immediate consequence of the Hewitt-Marczewski-Pondiczery theorem:</p>
<p>Let $m \geq \aleph_0$. If $\{X_s : s \in S\}$ are topological spaces with $d(X_s) \leq m$ and $|S| \leq 2^m$ then $d(\prod_s X_s) \leq m$.</p>
|
20,224 | <p>I was rereading an answer to an old question of mine and it included a reference to the fact that $2^{\omega_1}$ was separable. I'm having a hard time finding a reference for this fact, and the proof is not immediately obvious to me. Can anyone provide me with a cite and/or a proof? </p>
| Henno Brandsma | 2,060 | <p>This is indeed the Hewitt-Marczewski-Pondiczery theorem. My proof, following Engelking, is <a href="http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2010;task=show_msg;msg=0487.0001" rel="nofollow">here</a>. It's in fact not that hard, the fact for a product of copies of 2 point discrete spaces already implies the general theorem pretty quickly.</p>
|
3,299,469 | <p>Consider the set <span class="math-container">$A=\{n\ a \}$</span> where <span class="math-container">$a>0$</span> is a constant and <span class="math-container">$n \in \mathbb{N}$</span></p>
<p><strong>How shall we write this set <span class="math-container">$A$</span> in set theory?</strong></p>
<p>If we write it as <span class="math-container">$A=\{n\ a\ \backslash n \in \mathbb{N}, a>0 \}$</span> or <span class="math-container">$A=\{n\ a\ / n \in \mathbb{N}, a>0 \}$</span> will it mean just one set or a set of infinite sets?</p>
| J.G. | 56,861 | <p>You can also write it as <span class="math-container">$a\Bbb N$</span>.</p>
|
4,049,293 | <p>I am learning about the cross entropy, defined by Wikipedia as
<span class="math-container">$$H(P,Q)=-\text{E}_P[\log Q]$$</span>
for distributions <span class="math-container">$P,Q$</span>.</p>
<p>I'm not happy with that notation, because it implies symmetry, <span class="math-container">$H(X,Y)$</span> is often used for the joint entropy and lastly, I want to use a notation which is consistent with the notation for entropy:
<span class="math-container">$$H(X)=-\text{E}_P[\log P(X)]$$</span></p>
<p>When dealing with multiple distributions, I like to write <span class="math-container">$H_P(X)$</span> so it's clear with respect to which distribution I'm taking the entropy. When dealing with multiple random variables, it thinks it's sensible to make precise the random variable with respect to which the expectation is taken by using the subscript <span class="math-container">$_{X\sim P}$</span>. My notation for entropy thus becomes
<span class="math-container">$$H_{X\sim P}(X)=-\text{E}_{X\sim P}[\log P(X)]$$</span></p>
<p>Now comes the point I don't understand about the definition of cross entropy: Why doesn't it reference a random variable <span class="math-container">$X$</span>? Applying analogous reasoning as above, I would assume that cross entropy has the form <span class="math-container">\begin{equation}H_{X\sim P}(Q(X))=-\text{E}_{X\sim P}[\log Q(X)]\tag{1}\end{equation}</span>
however, Wikipedia makes no mention of any such random variable <span class="math-container">$X$</span> in the article on cross entropy. It speaks of</p>
<blockquote>
<p>the cross-entropy between two probability distributions <span class="math-container">$p$</span> and <span class="math-container">$q$</span></p>
</blockquote>
<p>which, like the notation <span class="math-container">$H(P,Q)$</span>, implies a function whose argument is a pair of distributions, whereas entropy <span class="math-container">$H(X)$</span> is said to be a function of a random variable. In any case, to take an expected value I need a (function of) a random variable, which <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> are not.</p>
<p>Comparing the definitions for the discrete case:
<span class="math-container">$$H(p,q)=-\sum_{x\in\mathcal{X}}p(x)\log q(x)$$</span>
and
<span class="math-container">$$H(X)=-\sum_{i=1}^n P(x_i)\log P(x_i)$$</span></p>
<p>where <span class="math-container">$\mathcal{X}$</span> is the support of <span class="math-container">$P$</span> and <span class="math-container">$Q$</span>, there would only be a qualitative difference if the events <span class="math-container">$x_i$</span> didn't cover the whole support (though I could just choose an <span class="math-container">$X$</span> which does).</p>
<p>My questions boil down to the following:</p>
<ol>
<li><p>Where is the random variable necessary to take the expected value which is used to define the cross entropy <span class="math-container">$H(P,Q)=-\text{E}_{P}[\log Q]$</span></p>
</li>
<li><p>If I am correct in my assumption that one needs to choose a random variable <span class="math-container">$X$</span> to compute the cross entropy, is the notation I used for (1) free of ambiguities.</p>
</li>
</ol>
| Hagen von Eitzen | 39,174 | <p><span class="math-container">$$ \begin{align}f(n+1)-f(n)&=\frac{(n+1)^5-n^5}5+\frac{(n+1)^3-n^3}3+\frac7{15}
\\&=\frac{5n^4+10n^3+10n^2+5n+1}5+\frac{3n^2+3n+1}3+\frac7{15}\\&=\text{integer}+\frac15+\frac13+\frac 7{15}\end{align}$$</span>
is an integer for all integers <span class="math-container">$n$</span> and so is <span class="math-container">$f(0)$</span>. The claim follows by induction.</p>
<hr />
<p>Here's a generalization for you: Suppose <span class="math-container">$p_1,p_2,\ldots,p_m$</span> are primes, <span class="math-container">$a_1,a_2,\ldots,a_m$</span> are integers and <span class="math-container">$c$</span> is real. Let
<span class="math-container">$$ f(n)=\frac{a_1n^{p_1}}{p_1}+\frac{a_2n^{p_2}}{p_2}+\cdots +\frac{a_mn^{p_m}}{p_m}+cn.$$</span>
If <span class="math-container">$f(1)$</span> is an integer, then <span class="math-container">$f(n)$</span> is an integer for all integers <span class="math-container">$n$</span>.</p>
|
194,671 | <p>I'm searching for two symbols - considering they exist - (1) unknown value; (2) unknown probability.</p>
<p><strong>Note</strong>: I thought that $x$ was used in a temporary context, whenever I see it, it remains unknown until an evaluation is made. I was thinking in a "unknown and impossible to be known" context. I'm not sure if this
context exists or if $x$ also express it.</p>
| Aang | 33,989 | <p>Generally used Symbol for unknown value is $x$ and other letters.</p>
<p>and</p>
<p>unknown probability of some event $A$ is denoted as $P(A)$ or $P_A$.</p>
|
194,671 | <p>I'm searching for two symbols - considering they exist - (1) unknown value; (2) unknown probability.</p>
<p><strong>Note</strong>: I thought that $x$ was used in a temporary context, whenever I see it, it remains unknown until an evaluation is made. I was thinking in a "unknown and impossible to be known" context. I'm not sure if this
context exists or if $x$ also express it.</p>
| Trevor Wilson | 39,378 | <p>One convention of notation is to use symbols from near the end of the alphabet (e.g., $x$, $y$, $z$, $t$ as others have answered) for "unknown" values and symbols from near the beginning of the alphabet (e.g., $a$, $b$, $c$) for "known" or "fixed" values. Of course conventions vary between fields of mathematics.</p>
|
3,174,982 | <p>I tried for few primes but they do not satisfy Eisenstein criterion.Also is there any approach other than brute force with the help of which we can find that prime.</p>
| Clive Newstead | 19,542 | <p><strong>Hint:</strong> If <span class="math-container">$f(x) = x^3 + x^2 - 2x - 1$</span>, then what is <span class="math-container">$f(x+2)$</span>?</p>
|
3,174,982 | <p>I tried for few primes but they do not satisfy Eisenstein criterion.Also is there any approach other than brute force with the help of which we can find that prime.</p>
| José Carlos Santos | 446,262 | <p>Apply Eisenstein's criterion to <span class="math-container">$(x+2)^3+(x+2)^2-2(x+2)-1$</span>, with <span class="math-container">$p=7$</span>.</p>
|
3,174,982 | <p>I tried for few primes but they do not satisfy Eisenstein criterion.Also is there any approach other than brute force with the help of which we can find that prime.</p>
| Bill Dubuque | 242 | <p>We don't need to pull the Eisenstein shift out of a hat like magic. I explain how to find it below.</p>
<p><strong>Hint</strong> <span class="math-container">$\, \bmod\color{#c00}7\!:\,\ f(x) \equiv (x\!-\!2)^{3} \ $</span> is a <em>prime power</em>. </p>
<p>So Eisenstein works on <span class="math-container">$\,g(x) = f(x\!+\!2) \equiv x^3\ $</span> by
<span class="math-container">$\,g(0) = f(2)\equiv 7\not\equiv 0\pmod{\!49}$</span></p>
<p><strong>Remark</strong> <span class="math-container">$\ $</span> Recall that the <a href="https://math.stackexchange.com/a/23877/242">key idea</a> behind the Eisenstein criteria is that polynomials satisfying the criterion are, mod <span class="math-container">$\,p,\,$</span> powers of a prime, viz. <span class="math-container">$\,\equiv x^n,\,$</span> and products of primes always factor <em>uniquely</em>. The same works for its shift <span class="math-container">$\,(x-c)^n,\,$</span> so we seek primes <span class="math-container">$\,p\,$</span> such that, mod <span class="math-container">$\,p,\,$</span> the polynomial is congruent to such a power (e.g. for motivation: <a href="https://math.stackexchange.com/a/93628/242">cyclotomic case</a>). The only primes <span class="math-container">$\,p\,$</span> that can yield such powers are those dividing the discriminant (here <a href="https://www.wolframalpha.com/input/?i=factor+PolynomialDiscriminant%5Bx%5E3%2Bx%5E2-2x-1,x%5D" rel="nofollow noreferrer">by Alpha</a> = <span class="math-container">$\,\color{#c00}7^2).\,$</span> Indeed, if <span class="math-container">$\,f\equiv a (x-c)^n,\,\ n> 1\,$</span> then <span class="math-container">$\,f\,$</span> and <span class="math-container">$\,f'\,$</span> have a common root <span class="math-container">$\,x\equiv c,\,$</span> hence their <a href="http://en.wikipedia.org/wiki/Resultant" rel="nofollow noreferrer">resultant</a> <span class="math-container">$\, R(f,f')\equiv 0.\,$</span> But this is, up to sign, the <a href="http://en.wikipedia.org/wiki/Discriminant#Discriminant_of_a_polynomial" rel="nofollow noreferrer">discriminant</a> of <span class="math-container">$\,f\,$</span> (presumed monic).</p>
|
14,583 | <p>Given a <code>Graph</code> with an automatically computed layout (i.e. not explicitly given <code>VertexCoordinates</code>, but using a <code>GraphLayout</code> method), how can we extract the coordinates of the vertices?</p>
<pre><code>In[]:= g = RandomGraph[{10, 20}, GraphLayout -> "SpringEmbedding"]
Out[]= << picture of graph >>
In[]:= PropertyValue[g, VertexCoordinates]
Out[]= Automatic (* <-- I'd like to have a list of coordinates here *)
</code></pre>
<p>It's possible to convert the graph into a graphics object using <code>Show</code> and extract the coordinates from there. Is there a less hacky, more direct/robust way?</p>
| ssch | 1,517 | <p>In version 8, you can use:</p>
<pre><code>VertexCoordinates /. AbsoluteOptions[g, VertexCoordinates]
</code></pre>
<p><code>AbsoluteOptions</code> is usually a good bet when other things just return <code>Automatic</code></p>
<p>In version 9, there's the <a href="http://reference.wolfram.com/mathematica/ref/GraphEmbedding.html"><code>GraphEmbedding</code></a> function:</p>
<pre><code>GraphEmbedding[g]
</code></pre>
|
1,631,505 | <p>Give the precise meaning of the limit- $$\lim_{x\to-\infty} f(x) = +\infty$$ (x is going to negative infinity, the symbol is hard to see)</p>
<p>I know that as $x$ gets smaller and smaller, $f(x)$ gets larger and larger, but how do I put that in terms of a precise definition?</p>
| Zhanxiong | 192,408 | <p>The definition is standard:</p>
<blockquote>
<p>For any $M > 0$, there exists $K < 0$ (usually sufficiently small) such that for all $x < K$, it holds $f(x) > M$.</p>
</blockquote>
<p>To make analogy to the normal function limit, you might also treat $-\infty$ as a special point, whose neighbors are $(-\infty, x)$ for any $x \in \mathbb{R}^1$.</p>
|
3,637,785 | <p>Prove that
<span class="math-container">$${2n \choose n} 2^{-2n} = (-1)^n {-\frac12 \choose n},$$</span></p>
<p><span class="math-container">$$\frac{1}n {2n -2 \choose n-1} 2^{-2n +1} = (-1)^{n-1} {\frac12 \choose n}.$$</span></p>
<p>The second part can be proved by replacing <span class="math-container">$n$</span> by <span class="math-container">$n-1$</span> in the first part. For the first part, I found that the right side is equal to <span class="math-container">${n-1/2 \choose n}$</span>, but when I expand the left side, I get something like
<span class="math-container">$$\frac{2n/4}{n}\frac{(2n-1)/4}{n-1}...\frac{(2n-n+1)/4}{1}$$</span>
which does not look similar with <span class="math-container">${n-1/2 \choose n}$</span>.</p>
<p>I appreciate if you give some help. </p>
| Olivier Roche | 649,615 | <p>Assume for contradiction that there are non zero natural numbers <span class="math-container">$a, b$</span> such that <span class="math-container">$a^2 = 2 b^2$</span>. Consider <span class="math-container">$a \neq 0 \in \mathbb{N}$</span> minimal such that there is <span class="math-container">$b \in \mathbb{N}$</span> (wlog) such that <span class="math-container">$a^2 = 2 b^2$</span>. As observed, the last digit of <span class="math-container">$a^2$</span> and <span class="math-container">$2b^2$</span> are <span class="math-container">$0$</span>, so that both are divisible by <span class="math-container">$5$</span>. </p>
<p>Hence, both <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are divisible by <span class="math-container">$\textbf{5}$</span> (and not necessarily by <span class="math-container">$10$</span>, as noted by @jamie in his excellent answer).</p>
<p>Then <span class="math-container">$a' := \frac{a}{5}$</span> and <span class="math-container">$b' := \frac{b}{5}$</span> also satisfy <span class="math-container">$a'^2 = 2 b'^2$</span>. It is a contradiction since <span class="math-container">$a' < a$</span>.</p>
<p>Btw I skiped the "infinite descent" but it amounts to this : if any <span class="math-container">$a, b$</span> satisfy <span class="math-container">$a^2 = 2 b^2$</span> then <span class="math-container">$a' := \frac{a}{5}$</span> and <span class="math-container">$b' := \frac{b}{5}$</span> also satisfy <span class="math-container">$a'^2 = 2 b'^2$</span> and eventualy you can't divide by <span class="math-container">$5$</span> anymore.</p>
|
2,573,492 | <p>In my course "Introduction To Algebraic Topology" I had following test problem:</p>
<blockquote>
<p>Exemplify a topological space with fundamental group $\mathbb{Z}/3\mathbb{Z}$.</p>
</blockquote>
<p>I was supposed to use this theorem:</p>
<blockquote>
<p>Let $Y$ be a simply connected topological space. If a group $G$ (finite or countable) acts on $Y$ freely and properly discontinuously, then fundamental group of the quotient space $Y/G$ is naturally isomorphic to $G$.</p>
</blockquote>
<p>So problem is I just wasn't able to come up with simply connected space such that $\mathbb{Z}/3\mathbb{Z}$ acts on it freely and properly discontinuously. </p>
<p>Any ideas? Thanks!</p>
| Andres Mejia | 297,998 | <p>There is an explicit construction:</p>
<p>Consider the CW complex given by three $0$ cells, three $1$ cells and a two cell pasted in (this is a filled in triangle.)</p>
<p>Now, considering the circle with a base point, and pasting in this $CW$-complex by identifying all three vertices with the base point (in a three fold way.)</p>
<p>The fundamental group of the resulting space is exactly $\mathbb Z_3$ by construction, since the two cell gives the relation $a^3=1$ on the fundamental group.</p>
<p>This is called the <a href="https://en.wikipedia.org/wiki/Presentation_complex" rel="nofollow noreferrer">presentation complex</a> by the way.</p>
<hr>
<p>If you want to construct the Cayley complex (the universal cover of the presentation complex), you will need to take three disks and paste them to the $1$ skeleton of the first construction. $\mathbb Z_3$ will act by cyclic permutation of the $2$-cells, and taking the quotient will give you exactly the presentation complex. Note that this is a simply connected space and that $\mathbb Z_3$ acts freely on the two cells.</p>
|
47,974 | <p>I am interested in the following question:</p>
<p>Is it known that <span class="math-container">$2$</span> is a primitive root modulo <span class="math-container">$p$</span> for infinitely many primes <span class="math-container">$p$</span>?</p>
<p>There is some information about Artin's conjecture in <a href="https://en.wikipedia.org/w/index.php?title=Artin%27s_conjecture_on_primitive_roots&oldid=374854868" rel="nofollow noreferrer">Wikipedia</a>.
I need to know if it is up-to-date and if one can say something about the case <span class="math-container">$n=2$</span>.</p>
| Misha | 21,684 | <p>Igor, the following is not an answer but, I think, is as close to an answer as one can get at this time. First of all, Bill Thurston is making two good points:</p>
<p>(a) There is no algorithm in the context of hyperbolic groups, as it follows from the Rips' 1982 construction. </p>
<p>(b) At this stage, there is no good notion of geometrically-finite discrete group actions on higher rank symmetric spaces, but people are working on it. For instance, "Anosov actions" of Guichard and Wienhard is a stab at defining analogues of "convex cocompact" actions (although, I think, their definition is too restrictive, but it is the best that we have at this point). In the case of discrete subgroups of $SL(n, {\mathbb R})$ one can (very) tentatively define geometrically finite groups as ones which have finitely-sided "Dirichlet-Selberg" fundamental domains (the definition is too limited even in rank 1 case but is good enough for the purposes of what follows). These domains were introduced in Selberg's famous 1960 paper (and, I think, deserve to be better known) and have the (algorithmic) advantage of being defined by linear inequalities in the space of symmetric matrices. With this definition, Thurston's suggestion goes through as one can algorithmically construct such fundamental domains using Jorgensen's algorithm (that you probably know from the real-hyperbolic case): Each time you add a bisector you check if Poincare's fundamental domain conditions hold. Of course, if there is no finitely-sided domain, this algorithm runs forever. </p>
<p>(c) On the other hand, I think, already for $n=4$, there is no algorithm. The idea is to exploit Mikhailova-type subgroups $\Lambda$ in products $\Gamma=F_2\times F_2\subset SL(4, {\mathbb Z})$ (see Mark's comments). Because of such subgroups, there is no algorithm to determine if a finite subset $P\subset \Gamma$ generates $\Gamma$. The problem, of course, is that $\Gamma$ is too small (not even Zariski dense in $SL(4, {\mathbb R})$). The idea is to look for finite subsets $Q\subset SL(4, {\mathbb Z})$ so that $\langle \Gamma \cup Q\rangle$ generates $SL(4, {\mathbb Z})$ (or, at least, a finite-index subgroup), while $\langle \Lambda \cup Q\rangle$ generates an infinite-index subgroup in $SL(4, {\mathbb Z})$. This is easier said than done, but, in principle, there is no reason to think that such sets do not exist. For instance, this idea does work for some lattices in $SO(4,1)$, where instead of $F_2\times F_2$ you take a lattice from $SO(3,1)$ (uniformizing a hyperbolic 3-manifold fibered over the circle) and as a subgroup $\Lambda$ you take a normal surface subgroup in $\Gamma$. This does not prove anything, of course, for $SL(4, {\mathbb Z})$ since "life is hard" in higher rank.</p>
<p>(d) Situation could be radically different for $SL(3, {\mathbb Z})$. Suppose that $\Lambda$ is an infinite index fg torsion-free sugroup of $SL(3, {\mathbb Z})$ which is Zariski dense in $SL(3, {\mathbb R})$. Sadly, at this point we know exactly two constructions of such subgroups: (1) free subgroups (Tits), (2) closed surface subgroups (originally due to Kac and Vinberg but, now, there are more constructions due to Alan Reid and maybe others). Both constructions are very "tame" and, I think, lead to geometrically finite subgroups. It is then not impossible that every fg subgroup of $SL(3, {\mathbb Z})$ is geometrically finite. </p>
|
912,426 | <p>A bag contains six chips, numbered 1 through 6. If two chips are chosen at random without replacement and the values on those two chips are multiplied, what is the probability that this product will be greater than 20?</p>
<p>I tried to solve by counting the total possibilities (36) and solving for 6 choices that worked, e.g. $4x6, 5x5, 5x6, 6x4,6x5,6x6$... so I thought the probability would be $1/6$. </p>
<p>How is my method incorrect?</p>
| amcalde | 168,694 | <p>$$\prod_{n=1}^N \frac{2n-1}{2n} = \exp\left\{ \log \left( \prod_{n=1}^N \frac{2n-1}{2n} \right)\right\} = \exp\left\{ \sum_{n=1}^N \log \left(\frac{2n-1}{2n} \right)\right\}$$</p>
<p>$$ = \exp\left\{ \sum_{n=1}^N \log(2n-1) - \log(2n)\right\} = \exp\left\{ \sum_{n=1}^N \log(2n)+\log(1-\frac{1}{2n}) - \log(2n)\right\} = \exp\left\{ \sum_{n=1}^N \log(1-\frac{1}{2n})\right\} $$</p>
<p>For large $n$ the log approximates as $\log(1-\frac{1}{2n}) \approx \frac{-1}{2n}$. In fact, keeping the higher order terms only makes it more negative so that the sum in the exponent diverges to negative infinity (as the harmonic sequence does).</p>
<p>Thus</p>
<p>$$\prod_{n=1}^\infty \frac{2n-1}{2n} = \exp( -\infty ) = 0$$</p>
|
478,566 | <p>I'm reading a book about combinatorics. Even though the book is about combinatorics there is a problem in the book that I can think of no solutions to it except by using number theory.</p>
<p>Problem: Is it possible to put $+$ or $-$ signs in such a way that $\pm 1 \pm 2 \pm \cdots \pm 100 = 101$?</p>
<p>My proof is kinda simple. Let's work in mod $2$. We'll have:</p>
<p>$\pm 1 \pm 2 \pm \cdots \pm 100 \equiv 101 \mod 2$
but since $+1 \equiv -1 \mod 2$ and there are exactly $50$ odd numbers and $50$ even numbers from $1$ to $100$ we can write:</p>
<p>$(1 + 0 + \cdots + 1 + 0 \equiv 50\times 1 \equiv 0) \not\equiv (101\equiv 1) \mod 2$ which is contradictory. </p>
<p>Therefore, it's not possible to choose $+$ or $-$ signs in any way to make them equal.</p>
<p>Now is there a combinatorial proof of that fact except what I have in mind?</p>
| Community | -1 | <p>Another answer that use almost the same idea: the sum or subtraction of two even or odd number is an even number. How many odd number we have?</p>
|
38,193 | <p>For simplicity, let me pick a particular instance of Gödel's Second Incompleteness
Theorem:</p>
<p>ZFC (Zermelo-Fraenkel Set Theory plus the Axiom of Choice, the usual foundation of mathematics) does not prove Con(ZFC), where Con(ZFC) is a formula that expresses that
ZFC is consistent.</p>
<p>(Here ZFC can be replaced by any other sufficiently good, sufficiently strong set of axioms,
but this is not the issue here.)</p>
<p>This theorem has been interpreted by many as saying "we can never know whether mathematics is consistent" and has encouraged many people to try and prove that ZFC (or even PA) is in fact inconsistent. I think a mainstream opinion in mathematics (at least among mathematician who think about foundations) is that we believe that there is no problem with
ZFC, we just can't prove the consistency of it.</p>
<p>A comment that comes up every now and then (also on mathoverflow), which I tend to agree with, is this:</p>
<p>(*) "What do we gain if we could prove the consistency of (say ZFC) inside ZFC? If ZFC were inconsistent, it would prove its consistency just as well."</p>
<p>In other words, there is no point in proving the consistency of mathematics by a mathematical proof, since if mathematics were flawed, it would prove anything, for instance its own non-flawedness.
Hence such a proof would not actually improve our trust in mathematics (or ZFC, following the particular instance).</p>
<p>Now here is my question: Does the observation (*) imply that the only advantage of the Second Incompleteness Theorem over the first one is that we now have a specific sentence
(in this case Con(ZFC)) that is undecidable, which can be used to prove theorems like
"the existence of an inaccessible cardinal is not provable in ZFC"?
In other words, does this reduce the Second Incompleteness Theorem to a mere technicality
without any philosophical implication that goes beyond the First Incompleteness Theorem
(which states that there is some sentence <span class="math-container">$\phi$</span> such that neither <span class="math-container">$\phi$</span> nor <span class="math-container">$\neg\phi$</span> follow from ZFC)?</p>
| Nik Weaver | 23,141 | <p>Yudkowsky and Herreshoff have a (messy but) <a href="https://intelligence.org/files/TilingAgentsDraft.pdf" rel="nofollow noreferrer">great paper</a> which relates the second incompleteness theorem to issues in theoretical artificial intelligence. (<a href="https://arxiv.org/pdf/1312.3626.pdf" rel="nofollow noreferrer">This paper</a> of mine might be a more accessible introduction to the subject.) In principle, one way an intelligent agent <span class="math-container">$M$</span> might achieve a goal is by building an auxiliary agent <span class="math-container">$M'$</span> and tasking it with the goal. But presumably <span class="math-container">$M$</span> cannot satisfy its criterion for action unless it can prove that <span class="math-container">$M'$</span> reasons consistently --- otherwise it could be building an agent who might fail because it reasons incorrectly. But by the second incompleteness theorem, <span class="math-container">$M$</span> cannot prove that the system within which it itself reasons is sound, which means that <span class="math-container">$M'$</span> would have to reason within a weaker system.</p>
<p>It's especially a problem for the idea of self-modifying AI. A sufficiently advanced AI ought to be better at designing AI's than we are. So we might want to design an AI which is capable of improving itself by modifying its own source code. But the incompleteness obstacle seems to imply that it could only do this at the cost of weakening the formal system in which it reasons. Since proof-theoretic strength is gauged by ordinals, after finitely many iterations it would reach imbecility.</p>
<p>At first sight it seems like there should be a trivial resolution, but the more you think about it, the more serious you realize the problem is.</p>
|
233,367 | <p>A set of $m$ non-zero <strong><em>rationals</em></strong> {$a_1, a_2, ... , a_m$} is called a <em><a href="https://web.math.pmf.unizg.hr/~duje/intro.html" rel="nofollow">rational Diophantine $m$-tuple</a></em> if $a_i a_j+1$ is a square. It turns out an $m$-tuple can be extended to $m+2$ if it has certain properties. The problem is to generalize the relations below to $m=5$.</p>
<blockquote>
<p><strong>I.</strong> $m=2$. Given $a,b$:</p>
</blockquote>
<p>$$ax_i+1 = \big(a\pm\sqrt{ab+1}\big)^2\tag1$$</p>
<p>then {$a,b,x_1$} is a triple for any choice of $x_i$ . However, {$a,b,x_1,x_2$} is a quadruple if,</p>
<p>$$2(a^2+b^2)-(a+b)^2-3 = y^2$$</p>
<p>Ex. From $a,b = \frac{1}{16},\frac{17}{4}$, we get $x_1,x_2 = \frac{33}{16},\frac{105}{16}$, a quadruple first found by Diophantus.</p>
<blockquote>
<p><strong>II.</strong> $m=3$. Given $a,b,c$:</p>
</blockquote>
<p>$$ax_i+1 = \big(a\sqrt{bc+1}\pm\sqrt{(ab+1)(ac+1)}\big)^2\tag2$$</p>
<p>then {$a,b,c,x_1$} is a quadruple. However, {$a,b,c,x_1,x_2$} is a quintuple if,</p>
<p>$$2(a^2+b^2+c^2)-(a+b+c)^2-3 = y^2\,^{\color{red}\dagger}$$</p>
<p>Ex. From $a,b,c = \frac{28}{5},\frac{55}{16},\frac{1683}{80}$, we get $x_1,x_2 = \frac{3}{80},1680$. </p>
<blockquote>
<p><strong>III.</strong> $m=4$. Given $a,b,c,d$:</p>
</blockquote>
<p>$$\small(ax_i+1)(abcd-1)^2 = \big(a\sqrt{(bc+1)(bd+1)(cd+1)}\pm\sqrt{(ab+1)(ac+1)(ad+1)}\big)^2\tag3$$</p>
<p>then {$a,b,c,d,x_1$} is a quintuple. However, {$a,b,c,d,x_1,x_2$} is a sextuple if,</p>
<p>$$2(a^2+b^2+c^2+d^2)-(a+b+c+d)^2-3-6abcd+(abcd)^2 = y^2\,^{\color{red}\dagger}$$</p>
<p>Ex. From $a,b,c,d = \frac{5}{4},\;\frac{5}{36},\;\frac{32}{9},\;\frac{189}{4}$, we get $x_1,x_2 = \frac{3213}{676},\;\frac{665}{1521}$, one of first sextuples found by Gibbs in 1999.</p>
<p>$^{\color{red}\dagger}$ These two can be satisfied by the parametric <a href="https://web.math.pmf.unizg.hr/~duje/ratio.html" rel="nofollow">example</a> in the variable $t$ in Dujella's website. </p>
<blockquote>
<p><strong>IV. Notes:</strong> </p>
</blockquote>
<p>In general, an $n$-tuple can be extended to a $n+1$ (unconditional) and $n+2$ (conditional) for $n=2,3,4$. Also, one root $x_i$ is equal to zero if,</p>
<p>$$(a-b)^2 = 4\\
(a+b-c)^2 = 4(ab+1)\\
(a+b-c-d)^2 = 4(ab+1)(cd+1)$$</p>
<p>for relations $(1), (2), (3)$, respectively.</p>
<blockquote>
<p><strong>V. Question:</strong></p>
</blockquote>
<p>For $m=5$, given $a,b,c,d,e$:</p>
<p>$$\text{LHS}? = \text{RHS}?\tag4$$</p>
<ol>
<li><strong><em>Can we find $(4)$, analogous to the first three?</em></strong> If yes, then maybe we can use known $5$-tuples or $6$-tuples to generate $7$-tuples, of which there is yet no known example. </li>
<li>The pattern is suggestive. But, like quintics, is there a Galois-theoretic restriction on five variables $a,b,c,d,e$ that prevent generalization for $m>4$? </li>
</ol>
| Tito Piezas III | 12,905 | <p>(<em>Too long for a comment, but may help in a generalization.</em>)</p>
<p>After some sleuthing around, it turns out $(1),(2),(3)$ can be encapsulated in the single equation,</p>
<p>$$(a b c d e + 2a b c + a + b + c - d - e)^2 = 4(a b + 1)(a c + 1)(b c + 1)(d e + 1)\tag1$$</p>
<p>which I think is by Dujella. For example,</p>
<ul>
<li>Let $a,b,c,d =1,3,0,0,\,$ yields $e_1, e_2 = 0,8$. </li>
<li>Let $a,b,c,d =1,3,8,0,\,$ yields $e_1, e_2 = 0,120$.</li>
<li>Let $a,b,c,d =1,3,8,120,\,$ yields $e_1, e_2 = 0,\frac{777480}{8288641}$.</li>
</ul>
<p>The last was also found by Euler, so he must have a version of $(1)$. Thus, in general, an $m$-tuple can be extended to a quintuple. However, if $e_1 e_2+1 =\square$, then it yields a sextuple as in the example above,</p>
<ul>
<li>Let $a,b,c,d = \frac{5}{4},\;\frac{5}{36},\;\frac{32}{9},\;\frac{189}{4},\,$ yields $e_1,e_2 = \frac{3213}{676},\;\frac{665}{1521}$.</li>
</ul>
<p>a solution unnoticed by Euler and only found in 1999. </p>
<p>Tinkering with $(1)$, it can be expressed by the <strong><em>elementary symmetric polynomials</em></strong> $\alpha_i$ in a much simpler form,</p>
<p>$$(\alpha_1-\alpha_5)^2=4(\alpha_2+\alpha_4+1)\tag2$$</p>
<p>where,</p>
<p>$$\begin{aligned}
\alpha_1 &=a + b + c + d + e\\
\alpha_2 &=a b + a c + b c + a d + b d + c d + a e + b e + c e + d e\\
\alpha_4 &=a b c d + a b c e + a b d e + a c d e + b c d e\\
\alpha_5 &=abcde\\
\end{aligned}$$</p>
<p>So if $(2)$ can be generalized, then the question can be rephrased as: <strong><em>Is there a version of $(2)$ using the elementary symmetric polynomials $\alpha_i$ for $a,b,c,d,e,f$?</em></strong></p>
|
3,977,531 | <p>The Cantor-Bendixson theorem implies that any closed subset of the Cantor set <span class="math-container">$\mathcal{C}$</span> can be described as a disjoint union of a set <span class="math-container">$\mathcal{C}_c$</span> that is homeomorphic to the original Cantor set, and a countable open set <span class="math-container">$\mathcal{C}_o$</span>.</p>
<p>The following <a href="https://math.stackexchange.com/questions/57260/subsets-of-the-cantor-set">answer</a>, and the referenced work by Schoenberg & Grunhage therein, implies that all noncompact open subsets of the Cantor set are homeomorphic to the Cantor set minus a point—say, <span class="math-container">$\mathcal{C}/ \{ 0\}$</span>. But this would mean that <span class="math-container">$\mathcal{C}_o$</span> is homeomorphic to <span class="math-container">$\mathcal{C}/ \{ 0\}$</span>, which would imply the Cantor set minus a point is countable, which seems strange.</p>
<p>Is this true, or am I missing something?</p>
| Mees de Vries | 75,429 | <p>The "countable open set <span class="math-container">$\mathcal C_o$</span>" is not open as a subset of the cantor set <span class="math-container">$\mathcal C$</span>, it is open as a subset of the set that can be written as <span class="math-container">$\mathcal C_c \sqcup \mathcal C_o$</span>. Non-empty open subsets of Cantor space <span class="math-container">$\mathcal C$</span> are never countable.</p>
|
3,977,531 | <p>The Cantor-Bendixson theorem implies that any closed subset of the Cantor set <span class="math-container">$\mathcal{C}$</span> can be described as a disjoint union of a set <span class="math-container">$\mathcal{C}_c$</span> that is homeomorphic to the original Cantor set, and a countable open set <span class="math-container">$\mathcal{C}_o$</span>.</p>
<p>The following <a href="https://math.stackexchange.com/questions/57260/subsets-of-the-cantor-set">answer</a>, and the referenced work by Schoenberg & Grunhage therein, implies that all noncompact open subsets of the Cantor set are homeomorphic to the Cantor set minus a point—say, <span class="math-container">$\mathcal{C}/ \{ 0\}$</span>. But this would mean that <span class="math-container">$\mathcal{C}_o$</span> is homeomorphic to <span class="math-container">$\mathcal{C}/ \{ 0\}$</span>, which would imply the Cantor set minus a point is countable, which seems strange.</p>
<p>Is this true, or am I missing something?</p>
| Henno Brandsma | 4,280 | <p>The Cantor-Bendixsson theorem says that <span class="math-container">$C$</span> is a union of it's scattered part (open) and its perfect part (closed). The scattered part here is empty (as <span class="math-container">$C$</span> has no isolated points, and <span class="math-container">$C'=\emptyset$</span>, and there is no scattering part) and it's already perfect. The theorem is valid but void here.</p>
<p>If <span class="math-container">$O \subseteq C$</span> is an open non-compact subset, it's locally compact (being open in compact) so it has a one-point compactification <span class="math-container">$\alpha O = O \cup \{\infty\}$</span>. This is compact metric totally disconnected and has no isolated points so is homeomorphic to <span class="math-container">$C$</span> by Brouwer's theorem and so <span class="math-container">$O$</span> is homeomorphic to <span class="math-container">$C$</span> minus (any) point (by homogeneity).</p>
|
1,884,958 | <p>How do I show that <span class="math-container">$$\text{Var}(aX+b)=a^2\text{Var}(X).$$</span>
Since I am reading statistics for the first time, I don't have any idea how to start.</p>
<p>Thanks for helping me.</p>
| Sachchidanand Prasad | 249,258 | <p>See the solution is easy but at least you have to try once. Just applying the definition of variance you will get the desired result. Although I am writing the solution but please try by yourself.</p>
<blockquote class="spoiler">
<p> $$\begin{align*}\mathsf{Var}(X)& =\mathsf E[X^2]-\mathsf E[X]^2\\[2ex]\implies \mathsf{Var}(aX+b) &=\mathsf E[(aX+b)^2]-\mathsf E[aX+b]^2\\[1ex]& = \mathsf E[a^2X^2+2abX+b^2]-(a\mathsf E[X]+b)^2 \quad \text{as } \mathsf E[aX]=a\mathsf E[X]\ \text{and}\ \mathsf E[b]=b\\[1ex]& = a^2\mathsf E[X^2]+2ab\mathsf E[X]+b^2-a^2\mathsf E[X]^2-2ab\mathsf E[X]-b^2\\[1ex]& = a^2(\mathsf E[X^2]-\mathsf E[X]^2)\\[1ex]& = a^2\mathsf {Var}(X)\end{align*}$$</p>
</blockquote>
|
1,736,376 | <p>Let $R$ be a ring and $I$ the set of non-invertible elements of $R$. </p>
<p>If $(I,+)$ is an additive subgroup of $(R,+)$, then show that $I$ is an ideal of $R$ and so $R$ is local. </p>
<p>$$$$ </p>
<p>I have done the following: </p>
<p>Since $(I,+)$ is an additive subgroup of $(R,+)$, we have that $\forall a,b \in I$ : $ab\in I$. </p>
<p>But how can we show that it holds that $ax\in I, \forall a\in I, \forall x\in R$ ? </p>
| user1043065 | 34,162 | <blockquote>
<p>Since $(I,+)$ is an additive subgroup of $(R,+)$, we have that $∀a,b∈I
> : ab∈I.$</p>
</blockquote>
<p>This reasoning is wrong. Moreover, you don't need to prove this statement. The product of a non-invertible element and any other element is non-invertible. Therefore it is immediate that $IR\subset R$. The set $I$ is by assumption a subgroup w.r.t. addition, hence an ideal.</p>
|
1,237,528 | <p>$$ \displaystyle {\int_{0}^{z}} \sqrt {1 + \tan^2(\dfrac{\pi}{4} \dfrac{z}{H} )} dz $$</p>
<p>_</p>
<p>$$ gives $$ </p>
<p>_</p>
<p>$$ \dfrac{4H}{\pi} {\sinh^{-1}} ( {\tan \dfrac{\pi}{4} \dfrac{z}{H} } ) $$</p>
<p>Please advise solution</p>
<p>edit:- </p>
<p>I can get to </p>
<p>$$\dfrac{4H}{\pi} \displaystyle {\int_{0}^{\dfrac{\pi z}{4H}}} \sec {u} {du}$$</p>
<p>Please help after this step ?</p>
| GFauxPas | 173,170 | <p>Use the identity:</p>
<p>$$\tan^2 \theta + 1 = \sec^2 \theta$$</p>
<p>Though the answer won't necessarily look like the answer given. Does it have to be in terms of $\sinh^{-1}$?</p>
|
118,873 | <p>I understand that the Mellin transform of a modular form is expected to satisfy RH when it is an eigenform of all Hecke operators, in which case it has an Euler product. Now about when the form is not an eigenform: Is it known a case where the zeros are all in the critical strip?</p>
| Marc Palm | 10,400 | <p>Perhaps I don't understand the question, but in its current form the answer is no.</p>
<p>A general modular form of fixed weight will be a linear combination of Hecke eigenforms of that weight. The Gamma factors will imply that there are trivial zeros outside the critical stripe.</p>
<p>But I guess you have an L-function with symmetric functional equation in mind? The Mellin transform is a linear operator, so the result is a linear combination of Hecke L functions times a Gamma factors. Assuming that you find some combination non-vanishing out the complement of closure of the critical stripe via some estimates I don't know, you will need a good argument for non-vanishing on the boundary of the critical stripe, where the function is expected to be universal. </p>
<p>What is the motivation for the question?</p>
|
2,916,099 | <p>Find a Mobius transformation $T$ from the unit disk to the right half plane with condition $T(0)=3$.</p>
<p>First, the transformation from the unit circle to the upper half plane is $T_1(z)=(1-i)\frac{z-i}{z-1}$.</p>
<p>So from the unit circle to the right half plane, $T_2(z)=-i(1-i)\frac{z-i}{z-1}$</p>
<p>How can I introduce the condition $T(0)=3$ ?</p>
<p>$T(0)=1-i\neq3$</p>
| Donald Splutterwit | 404,247 | <p>It is well worth knowing that the conformal map $z \rightarrow \frac{1+z}{1-z}$ permutes the regions shown below $(1234)$ and act similarly on the lower half plane.</p>
<p><a href="https://i.stack.imgur.com/zfRg2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zfRg2.png" alt="enter image description here"></a></p>
<p>In particular the unit circle is mapped to the right half plane, indeed one can check that the unit circle ($z=e^{i \theta} =\cos( \theta)+i \sin( \theta)$ is mapped to the imaginary axis
\begin{eqnarray*}
\frac{1+z}{1-z}= \frac{1+\cos( \theta)+i \sin( \theta)}{1-\cos( \theta)-i \sin( \theta)} \times \frac{1-\cos( \theta)+i \sin( \theta)}{1-\cos( \theta)+i \sin( \theta)} = \cdots
\end{eqnarray*}
Now in order to fulfill the requirement that $T(0)=3$ simply multiply this function by $3$ (and note that this will still map the unit circle to the imaginary axis) & so ... </p>
<p>\begin{eqnarray*}
T(z)=3 \frac{1+z}{1-z}
\end{eqnarray*}
will do the trick.</p>
|
2,178,318 | <p>On a test I wrote an implication arrow "$\implies$" to show that I deduced one statement from the previous one, but I didn't get full score since it was more accurate to use an equivalence arrow "$\iff$".
For example:
$$ 2x = 4 \implies x = 2 $$
but it's also true the other way around:
$$ 2x = 4 \impliedby x = 2$$
so it is more correct to write equivalence arrow:
$$ 2x = 4 \iff x = 2$$
Given this i would assume that if $Q \implies P$ is <strong>true</strong>, then $Q \impliedby P$ is <strong>false</strong>. <br><strong>Is this correct?</strong></p>
<p><br>I don't want to check whether a statement only implies or is equivalent to another every time I do some operations to it.
<br>So my second question is then: is there some other more loosely defined implication arrow that allows me to show that implication in one direction is true, without saying that implication the other direction is false? I also came across <a href="https://i.stack.imgur.com/FpyJi.png" rel="noreferrer">this picture</a>, but i'm not entirely sure what the difference between those two definitions are.</p>
| Bram28 | 256,001 | <p>As your own very example shows: just because the implication goes one way doesn't mean that it doesn't go the other way as well. In your case, it goes from left to right <em>and</em> from right to left, so we can write $P \Leftrightarrow Q$. But this does <em>not</em> mean that one of $P \Rightarrow Q$ or $Q \Rightarrow P$ is false. In fact, <em>both</em> would be true!</p>
<p>There is no commonly used symbol to say that you only have a one-way implication ... You'd have to explicitly say "$P \Rightarrow Q$ but not $Q \Rightarrow P$" ... Or $P \Rightarrow Q$ and $Q \not \Rightarrow P$</p>
|
1,321,544 | <p>How do you evaluate the following?</p>
<p>$$\cos\left[\cos^{-1}\left(\frac{3}{4}\right)\right]$$</p>
<p>To me the cosine of an arc cosine is just the value, which would be $3/4$.</p>
| frosh | 211,697 | <p>It is a fact that $f \circ f^{-1}(x)=x$ which is <a href="http://en.wikipedia.org/wiki/Identity_function" rel="nofollow">Identity Function</a>.</p>
|
3,203,282 | <p>Given that <span class="math-container">$C[-\pi,\pi]$</span> is complete:
How can we prove, by using the supremum norm, that the space:</p>
<p><span class="math-container">$$C_p[-\pi,\pi]=\{f\in C[-\pi,\pi]\mid f(-\pi)=f(\pi)\}$$</span></p>
<p>is also complete? thank you!</p>
| Siong Thye Goh | 306,553 | <p>We know that <span class="math-container">$$\cos (y) =\sum_{n=0}^\infty \frac{(-1)^n y^{2n}}{(2n)!}= 1-\frac{y^2}{2!}+\frac{y^4}{4!}+\ldots$$</span></p>
<p>from Taylor series of cosine. </p>
<p>We just replace <span class="math-container">$y=2x$</span>.</p>
|
2,358,838 | <p>I can see the answer to this in my textbook; however, I am not quite sure how to solve this for myself . . . the book has the following:</p>
<blockquote>
<p>To take advantage of the inductive hypothesis, we use these steps:</p>
<p>$ 7^{(k+1)+2} + 8^{2(k+1)+1} = 7^{k+3} + 8^{2k+3} $</p>
<p>$$
= 7\cdot7^{k+2} + 8^{2}\cdot8^{2k+1}\\
= 7\cdot7^{k+2} + 64\cdot8^{2k+1}\\
= 7(7^{k+2}+8^{2k+1})+57\cdot8^{2k+1}\\ $$</p>
</blockquote>
<p>While the answer is apparent to me <em>now</em>; how exactly would I go about figuring out a similar algebraic manipulation if I were to see something like this on a test? Is there an algorithm or a way of thinking about how to break this down that I'm missing? I think I'm most lost regarding the move from the second to last and last equations.</p>
<p><em>Source: Discrete Mathematics and its Applications (7th ed), Kenneth H. Rosen (p.322)</em></p>
| Paolo Leonetti | 45,736 | <p>\begin{align}
7^{n+2}+8^{2n+1}&=49\cdot 7^n+8\cdot 64^n \\
&\equiv -8\cdot 7^n+8\cdot 7^n\equiv 0\bmod{57}.
\end{align}</p>
|
65,304 | <p>I have a plane curve $C$ described by parametric equations $x(t)$ and $y(t)$ and a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$. The line integral of $f$ along $C$ is the area of the "fence" whose path is governed by $C$ and height is governed by $f$.</p>
<p><img src="https://i.stack.imgur.com/4rmZy.png" alt="enter image description here"></p>
<p>How can I generate a picture of the "fence" in Mathematica?</p>
<p>For the sake of a concrete example, let's borrow from Stewart (since I already borrowed his picture). For $0 \leq t \leq \pi$, define
$$
\begin{align*}
x(t) &= \cos t\\
y(t) &= \sin t\\
f(x,y) &= 2 + x^2y
\end{align*}
$$
so that
$$
\begin{align*}
f(x(t),y(t)) &= 2 + \cos^2 t \sin t.
\end{align*}
$$</p>
| Verbeia | 8 | <p>This is just a small tweak of Belisarius' answer, using <code>MeshFunctions</code> to get the vertical lines and <code>BoundaryStyle</code> to get the "fence".</p>
<pre><code>ParametricPlot3D[{Cos[t], Sin[t], z (2 + Sin[t] Cos[t]^2)}, {t,
0, \[Pi]/2}, {z, 0, 1}, MeshFunctions -> {#2 &}, MeshStyle -> {Red},
BoundaryStyle -> Directive[Thick, Blue], BoxRatios -> {1, 1, 1/2},
PlotStyle -> Opacity[.5]]
</code></pre>
<p><img src="https://i.stack.imgur.com/6fmJz.png" alt="enter image description here"></p>
<p>An alternative that gives more "even" vertical lines is to specify the mesh in terms of <code>t</code> and <code>z</code>. </p>
<pre><code>ParametricPlot3D[{Cos[t], Sin[t], z (2 + Sin[t] Cos[t]^2)}, {t,
0, \[Pi]/2}, {z, 0, 1},
Mesh -> {Range[0, 1, 0.1], Range[0, 2, 0.5], Range[0, 1, 0.1]},
MeshStyle -> {Directive[Red, Thick], Directive[Green, Thick]},
BoundaryStyle -> Directive[Blue, Thick], BoxRatios -> {1, 1, 1/2},
PlotStyle -> Opacity[.5]]
</code></pre>
<p><img src="https://i.stack.imgur.com/mobft.png" alt="enter image description here"></p>
<p>And as Belisarius points out in comments (I'm going to have to CW this one!), <code>MeshFunctions -> {#4 &}</code> is even evener :)</p>
<pre><code>ParametricPlot3D[{Cos[t], Sin[t], z (2 + Sin[t] Cos[t]^2)}, {t,
0, \[Pi]/2}, {z, 0, 1}, MeshFunctions -> {#4 &}, MeshStyle -> {Red},
BoundaryStyle -> Directive[Thick, Blue], BoxRatios -> {1, 1, 1/2},
PlotStyle -> Opacity[.5]]
</code></pre>
<p><img src="https://i.stack.imgur.com/ls9eu.png" alt="enter image description here"></p>
|
3,452,707 | <p>It is well known that <span class="math-container">$\sum_{k=0}^n{n\choose k} =2^n$</span>.</p>
<p><strong>My question:</strong> If <span class="math-container">$z$</span> is the limit point of an infinite sequence of real numbers <span class="math-container">$\{ a_n \}$</span>, then does <span class="math-container">$$\frac{{n \choose 1} a_1 + {n \choose 2} a_2+ \cdots+ {n \choose n} a_n}{2^n}$$</span> converge to <span class="math-container">$z$</span> as <span class="math-container">$n\ \to \infty$</span>?</p>
| Calvin Khor | 80,734 | <p>Something far more general is true. Define the doubly half-infinite matrix <span class="math-container">$ c_{mn}$</span>,
<span class="math-container">$$ c_{mn} = \frac1{2^m}\binom{m}{n}\mathbb 1_{n\le m} $$</span>
then you're asking if
<span class="math-container">$$ t_m := \sum_{n=0}^\infty c_{mn} a_n \to z?$$</span>
According to Hardy's Divergent Series, Theorem 2 (page 43) (apparently due to Toeplitz and Schur):</p>
<blockquote>
<p><strong>Theorem 2.</strong> (Paraphrased) For any such infinite matrix <span class="math-container">$(c_{mn})_{m,n\ge 0}$</span>, the statement
<span class="math-container">$$\text{“$a_n \to z$ implies $t_m \to z$''}$$</span>
is true iff the following conditions for <span class="math-container">$c_{mn}$</span> are satisfied:</p>
<ol>
<li><span class="math-container">$\sum_{n=0}^\infty |c_{mn} | < H $</span> where <span class="math-container">$H<\infty$</span> is uniform in <span class="math-container">$m$</span>,</li>
<li>for each <span class="math-container">$n$</span>, <span class="math-container">$c_{mn} \xrightarrow[m\to\infty]{} 0$</span>,</li>
<li><span class="math-container">$\sum_{n=0}^\infty c_{mn} \xrightarrow[m\to\infty]{} 1$</span>.</li>
</ol>
</blockquote>
<p>For your specific <span class="math-container">$c_{mn}$</span>, for each of these conditions:</p>
<ol>
<li><span class="math-container">$\sum_{n=0}^\infty |c_{mn} | =\sum_{n=0}^\infty c_{mn} = 2^{-m}\sum_{n=0}^m \binom{m}{n} = 1$</span>. </li>
<li><span class="math-container">$c_{mn} = \frac{m!}{2^m n!(m-n)!} \le \frac{m^n}{n!2^m} \xrightarrow[m\to\infty]{} 0.$</span></li>
<li>see 1.</li>
</ol>
<p>So the assumptions for Theorem 2 are verified, and the result holds.</p>
<p>(This type of result is a "regularity" result for a summation method, and such methods are the main focus of Hardy's book.)</p>
|
3,451,629 | <p>Let <span class="math-container">$f : \mathbb{R} \to \mathbb{R}$</span> be a continous map then which of the following cannot be the image of </p>
<p><span class="math-container">$[0,1)$</span> under <span class="math-container">$f$</span> ?</p>
<p>(a) <span class="math-container">$0$</span></p>
<p>(b) <span class="math-container">$(0,1)$</span></p>
<p>(c) <span class="math-container">$[0,1)$</span></p>
<p>(d) <span class="math-container">$[0,1]$</span></p>
<p>Now , I know the following theorem if <span class="math-container">$f$</span> is a continuous map iff inverse image of every closed set is a closed set.</p>
<p>So, (a) and (d) must be the correct choices,
But for option (a) I can easily set <span class="math-container">$f(x) = 0$</span>, hence it must be incorrect. </p>
<p>But I don't really understand why isn't option (a) the correct choice, According to the theorem all the hypothesis are satisfied hence <span class="math-container">$0$</span> should be one of the answer.</p>
<p>Can someone please clear my doubt as why is option (a) Not the correct choice ?</p>
| WoolierThanThou | 686,397 | <p>I think the only good way to go about this really is to apply the property that <span class="math-container">$f$</span> maps compacts to compacts. So we see that <span class="math-container">$f([0,1])$</span> is compact, and <span class="math-container">$f([0,1])=f([0,1))\cup f(\{1\})$</span>, but the latter is a single point, and adding a single point to <span class="math-container">$(0,1)$</span> can never make it closed (hence, not compact). Hence, <span class="math-container">$(0,1)$</span> must be wrong.</p>
<p>As for the other options:</p>
<p><span class="math-container">$\{0\}=f([0,1))$</span> if <span class="math-container">$f(x)=0$</span> for all <span class="math-container">$x$</span>.</p>
<p><span class="math-container">$[0,1)=f([0,1))$</span> if <span class="math-container">$f(x)=x$</span></p>
<p><span class="math-container">$[0,1]=f([0,1))$</span> if <span class="math-container">$f(x)=2\cdot|\frac{1}{2}-x|$</span></p>
|
1,453,067 | <p>My friend's professor raised this question in a coaching and he and I tried everything we could think of. But later I thought that since $\sin (2x) $ can have values only between -1 & +1 and anything but +1 makes the equation complex ( keeping in mind that the integral is meant to be non-complex), there is no solution to the integral. Am I correct?</p>
<p>$\int \sqrt {\sin(2x)-1}$</p>
| Adesh Tamrakar | 266,411 | <p>yes. its solvable
let $$\sin2x-1=t$$
then $$\cos2x= {\sqrt{t^2+2t}}$$
then after all simplification you'll get $$ \int \frac{2}{\sqrt{2-t}}.dt$$
now you can integrate it easily.</p>
|
1,478,142 | <p>Evaluate these limits by relating them to a derivative. </p>
<p>$\lim\limits_{x \to 0} \frac{\sqrt{\cos{x}}-1}{x}$</p>
| Dr. Sonnhard Graubner | 175,066 | <p>HINT: rewrite it in the form
$$\frac{(\sqrt{\cos x}-1)(\sqrt{\cos x}+1)}{x(\sqrt{\cos x}+1)(\sqrt{\cos x}+1)}$$</p>
|
748,371 | <p>Could you give me some hint how to deal with this question:</p>
<p>Suppose $a_n\le b_n \le c_n$ for almost all n, $b_n\to L$, $c_n-a_n\to 0$.
Prove: $a_n \to L,b_n \to L$.</p>
<p>Well, if $a_n\to a, b_n \to b$ and $c_n-a_n\to 0$, than $a=b$ and from Squeeze Theorem we can conclude a=b=L.</p>
<p>So, we need to prove that this is only possible case.</p>
<p>$a_n$ is bounded above by L, so it could not converge to $+\infty$,
$b_n$ is bounded below by L, so it could not converge to $-\infty$.</p>
<p>if $c_n-a_n\to 0$ it is not possible that $a_n\to $ some finite number and $b_n \to +\infty$
or $b_n\to $ some finite number and $a_n \to -\infty$.</p>
<p>I failed to eliminate the possibility of $a_n \to -\infty$ and $b_n \to +\infty$,
and the possibility of $a_n,b_n$ do not have any limit at all.</p>
<p>Thanks.</p>
| Indrayudh Roy | 70,140 | <p>The right hand side is Binomial theorem for we have $$(1+\frac{2}{n})^{n}=1+n.\frac{2}{n}+{n \choose 2}(\frac{2}{n})^{2}+\cdots \geq1+2=3$$ and hence $3^{\frac{1}{n}}\leq (1+\frac{2}{n})$, with equality holding only when $n=1$.(Since the $n$-th root fuction is increasing).</p>
|
748,371 | <p>Could you give me some hint how to deal with this question:</p>
<p>Suppose $a_n\le b_n \le c_n$ for almost all n, $b_n\to L$, $c_n-a_n\to 0$.
Prove: $a_n \to L,b_n \to L$.</p>
<p>Well, if $a_n\to a, b_n \to b$ and $c_n-a_n\to 0$, than $a=b$ and from Squeeze Theorem we can conclude a=b=L.</p>
<p>So, we need to prove that this is only possible case.</p>
<p>$a_n$ is bounded above by L, so it could not converge to $+\infty$,
$b_n$ is bounded below by L, so it could not converge to $-\infty$.</p>
<p>if $c_n-a_n\to 0$ it is not possible that $a_n\to $ some finite number and $b_n \to +\infty$
or $b_n\to $ some finite number and $a_n \to -\infty$.</p>
<p>I failed to eliminate the possibility of $a_n \to -\infty$ and $b_n \to +\infty$,
and the possibility of $a_n,b_n$ do not have any limit at all.</p>
<p>Thanks.</p>
| Sandeep Thilakan | 124,957 | <p>The left inequality (rather both) is straightforward using calculus. Consider $f:(\frac{2}{3}, \infty) \to \mathbb{R}$ $$f(x)=3-(1+ \frac{2}{3-2x})^x$$</p>
<p>Clearly $f(1) = 0$. Differentiating,
$$f'(x)=6 \left( 1+\frac{3}{3x-2}\right)^x \log \left( 1+\frac{2}{3x-2}\right)\frac{1}{(3x-2)^2}$$</p>
<p>which is positive for $x \geq 1$. Hence the result.</p>
|
748,371 | <p>Could you give me some hint how to deal with this question:</p>
<p>Suppose $a_n\le b_n \le c_n$ for almost all n, $b_n\to L$, $c_n-a_n\to 0$.
Prove: $a_n \to L,b_n \to L$.</p>
<p>Well, if $a_n\to a, b_n \to b$ and $c_n-a_n\to 0$, than $a=b$ and from Squeeze Theorem we can conclude a=b=L.</p>
<p>So, we need to prove that this is only possible case.</p>
<p>$a_n$ is bounded above by L, so it could not converge to $+\infty$,
$b_n$ is bounded below by L, so it could not converge to $-\infty$.</p>
<p>if $c_n-a_n\to 0$ it is not possible that $a_n\to $ some finite number and $b_n \to +\infty$
or $b_n\to $ some finite number and $a_n \to -\infty$.</p>
<p>I failed to eliminate the possibility of $a_n \to -\infty$ and $b_n \to +\infty$,
and the possibility of $a_n,b_n$ do not have any limit at all.</p>
<p>Thanks.</p>
| achille hui | 59,379 | <p>For the LHS, apply $\text{AM} \ge \text{GM}$ to $n$ numbers with $n-1$ copies of $1$ and one copy of $\frac13$, we get</p>
<p>$$\begin{align}
&\left(1 + \frac{2}{3n-2}\right)^{-1}= 1 - \frac{2}{3n} = \frac{1}{n}\left( (n-1)\times 1 + \frac{1}{3}\right) \ge \frac{1}{\sqrt[n]{3}}\\
\implies &
1 + \frac{2}{3n-2} \le \sqrt[n]{3}
\end{align}$$
For the RHS, apply $\text{AM} \ge \text{GM}$ to $n$ numbers with $n-1$ copies of $1$ and one copy of $3$, we get</p>
<p>$$\sqrt[n]{3} \le \frac{1}{n}\left( (n-1)\times 1 + 3 \right) = 1 + \frac{2}{n}$$</p>
|
1,784,912 | <p>In this question, I know that $\text{C},\text{R},\text{T},\text{A}\in\mathbb{R}^+$</p>
<p>I've this circuit (the bottom of the resitor is connected to earth ($0$)):
<a href="https://i.stack.imgur.com/hfKGJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hfKGJ.jpg" alt="enter image description here"></a></p>
<p>When I use Laplace transform I can find that:</p>
<ul>
<li>$$\text{V}_{\text{out}}(s)=\frac{\text{R}}{\text{R}+\frac{1}{\text{C}s}}\cdot\text{V}_{\text{in}}(s)$$</li>
</ul>
<p>My input function $\text{V}_{\text{in}}(t)$ is:</p>
<p><a href="https://i.stack.imgur.com/KDFT5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KDFT5.jpg" alt="enter image description here"></a></p>
<p>When I use Laplace transform, I can find that:</p>
<ul>
<li>$$\text{V}_{\text{in}}(s)=\frac{\text{A}\tanh\left(\frac{\text{T}s}{4}\right)}{s}$$</li>
</ul>
<p>Now, when I substitute that in, I get that:</p>
<ul>
<li>$$\text{V}_{\text{out}}(s)=\frac{\text{R}}{\text{R}+\frac{1}{\text{C}s}}\cdot\frac{\text{A}\tanh\left(\frac{\text{T}s}{4}\right)}{s}$$</li>
</ul>
<p>So, when I solved the inverse Laplace transform. I got ($\text{H}(x)$ is the Heaviside stepfunction):</p>
<ul>
<li>$$\text{V}_{\text{out}}(t)=\text{A}\exp\left[-\frac{t}{\text{CR}}\right]-2\text{A}\sum_{n=0}^{\infty}\text{H}\left(t-\frac{\text{T}n}{2}\right)\exp\left[-\frac{\left(t-\frac{\text{T}n}{2}\right)}{\text{CR}}\right]$$</li>
</ul>
<p>Now, when I choose values $\text{T}=\frac{1}{50},\text{R}=1980,\text{A}=6,\text{C}=\frac{47\times10^{-6}}{10}$</p>
<p>I got a graph that looks like:</p>
<p><a href="https://i.stack.imgur.com/MWobO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MWobO.jpg" alt="enter image description here"></a></p>
<blockquote>
<p>Q: When I build it I looked at the scope and that told me that the graph I should get looks somehing like the picture down here, where is my mistake?:</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/keQ13.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/keQ13.jpg" alt="enter image description here"></a></p>
<p>I noticed, when I took out the floor part of my $\text{V}_{\text{out}}(t)$ function (I get that with mathematica) and set $\text{A}=-6$ I got a graph that look more like the thing I expected, but here I dont understand why it oscillates around $18$, it should me around $0$:
<a href="https://i.stack.imgur.com/kyNbi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kyNbi.jpg" alt="enter image description here"></a></p>
| alexjo | 103,399 | <p>The transfer function is
$$
H(s)=\frac{V_{\text{out}}(s)}{V_{\text{in}}(s)}=\frac{s}{s+\frac{1}{\tau}}
$$
where $\tau=RC$. The input voltage is
$$
V_{\text{in}}(s)=\frac{A}{s}\tanh\left(\frac{sT}{4}\right)=\frac{A}{s}\left(1-\mathrm e^{-\frac{Ts}{2}}\right)^2\sum_{k=0}^{\infty} \mathrm e^{-kTs}
$$
and then
$$
V_{\text{out}}(s)=\frac{A}{s+\frac{1}{\tau}}\left(1-2\mathrm e^{-\frac{Ts}{2}}+\mathrm e^{-Ts}\right)\sum_{k=0}^{\infty} \mathrm e^{-kTs}=F(s)\sum_{k=0}^{\infty} \mathrm e^{-kTs}
$$
The inverse Laplace transform gives
$$
v_{\text{out}}(t)=\sum_{k=0}^{\infty} f(t-kT)
$$
where
$$
f(t)=A\left[\mathrm e^{-\frac{t}{\tau}} u\left(t\right)-
2\mathrm e^{-\frac{1}{\tau}\left(t-\frac{T}{2}\right)} u\left(t-\tfrac{T}{2}\right)+\mathrm e^{-\frac{1}{\tau}\left(t-T\right)} u\left(t-T\right)
\right]
$$
and $u(t)$ is the Heaviside function.</p>
<p>Check <a href="http://www4b.wolframalpha.com/input/?i=Plot[6+e%5E%28-1.07458+t%29+%281.0666+theta%28t-3%2F50%29-2.1104+theta%28t-1%2F20%29%2B2.08784+theta%28t-1%2F25%29-2.06553+theta%28t-3%2F100%29%2B2.04345+theta%28t-1%2F50%29-2.02161+theta%28t-1%2F100%29%2Btheta%28t%29%29,%7Bt,0,4%2F50%7D]" rel="nofollow">here</a> for the first 3 periods with your values.</p>
<p>If for simplicity we put $\tau=2$ and $T=1$ we have the plot <a href="http://www.wolframalpha.com/input/?i=sum_%7Bk%3D0%7D%5E%7B10%7D+6%28e%5E%28-%28%28t-kT%29%2Ftau%29%29+theta%28%28t-kT%29%29+-+2+e%5E%28-%281%2Ftau%29%28%28t-kT%29-T%2F2%29%29+theta%28%28t-kT%29-T%2F2%29%2Be%5E%28-%281%2Ftau%29%28%28t-kT%29-T%29%29+theta%28%28t-kT%29-T%29%29+where+tau%3D2+and+T%3D1" rel="nofollow">here</a></p>
|
71,117 | <p>I have this assertion: if $p$ is a prime such that $p\equiv 11 \pmod{56}$, then $p$ splits in $\mathbb{Z}[\sqrt{14}]$ (the discriminant of $\mathbb{Z}[\sqrt{14}]$ is $56$.)</p>
<p>Why? Does $p\equiv 11\pmod{56}$ imply $14$ is a quadratic residue mod $p$?</p>
| Eric Naslund | 6,075 | <p>$11$ is not a square modulo $56$. Suppose $11\equiv a^2 \pmod{56}$. Then reducing modulo $8$ we see that $$3\equiv a^2 \pmod{8}$$ which is impossible, since $1$ and $4$ are the only squares modulo $8$.</p>
<p>Here is one way to prove that $p\equiv 11 \pmod{56}$ does not factor. Since the norm is multiplicative, we would need the factors to have norm 11. But then $B^2-14A^2=11$ which implies that $3$ is a square modulo $7$, a contradiction.</p>
|
3,791,936 | <p>An advanced sum <a href="https://www.facebook.com/photo.php?fbid=3190290677734375&set=a.222846247812181&type=3&theater" rel="noreferrer">proposed</a> by Cornel Valean:</p>
<blockquote>
<p><span class="math-container">$$S=\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}$$</span>
<span class="math-container">$$=4\text{Li}_4\left(\frac12\right)-\frac12\zeta(4)+\frac72\zeta(3)-4\ln^22\zeta(2)+6\ln2\zeta(2)+\frac16\ln^42-1$$</span></p>
</blockquote>
<p>I managed to find the integral representation of <span class="math-container">$\ \displaystyle\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^2{2n\choose n}}\ $</span> but not <span class="math-container">$S$</span>:</p>
<p><a href="https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Reihenentwicklungen#Potenzen_des_Arkussinus" rel="noreferrer">Since</a></p>
<p><span class="math-container">$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n\choose n}}$$</span></p>
<p>we can write</p>
<p><span class="math-container">$$\frac{2\sqrt{x}\arcsin \sqrt{x}}{\sqrt{1-x}}=\sum_{n=1}^\infty\frac{2^{2n}x^{n}}{n{2n\choose n}}$$</span></p>
<p>now multiply both sides by <span class="math-container">$-\frac{\ln(1-x)}{x}$</span> then <span class="math-container">$\int_0^1$</span> and use that <span class="math-container">$-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$</span> we have</p>
<p><span class="math-container">$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^2{2n\choose n}}=-2\int_0^1 \frac{\arcsin \sqrt{x}\ln(1-x)}{\sqrt{x}\sqrt{1-x}}dx\tag1$$</span></p>
<p>But I could not get the integral representation of <span class="math-container">$S$</span>. Any idea?</p>
<p>In case you find the integral, I prefer solutions that do not use contour integration or you can leave it to me to give it a try. Thank you.</p>
<p>In case the reader is curious about computing the integral in <span class="math-container">$(1)$</span>, set <span class="math-container">$x=\sin^2\theta$</span> then use the Fourier series of <span class="math-container">$\ln(\cos \theta)$</span>.</p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</span>
<span class="math-container">\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{2^{2n}H_{n + 1} \over
\pars{n + 1}^{2}{2n \choose n}}} =
\sum_{n = 2}^{\infty}H_{n}\,{4^{n - 1} \over
n^{2}{2n - 2 \choose n - 1}} =
-1 + \sum_{n = 1}^{\infty}
H_{n}\,{4^{n - 1} \over n^{2}}\,{\Gamma\pars{n}\Gamma\pars{n} \over
\Gamma\pars{2n - 1}}
\\[5mm] = &\
-1 + \sum_{n = 1}^{\infty}
H_{n}\, 4^{n - 1}\pars{{2 \over n} - {1 \over n^{2}}}\,{\Gamma\pars{n}\Gamma\pars{n} \over\Gamma\pars{2n}}
\\[5mm] = &\
-1 +
2\sum_{n = 1}^{\infty}H_{n}\, 4^{n - 1}
\pars{\int_{0}^{1}x^{n - 1}\,\dd x}
\int_{0}^{1}y^{n - 1}\pars{1 - y}^{n - 1}\,\dd y
\\[2mm] &\
-\sum_{n = 1}^{\infty}H_{n}\, 4^{n - 1}
\bracks{-\int_{0}^{1}\ln\pars{x}x^{n - 1}\,\dd x}
\int_{0}^{1}y^{n - 1}\pars{1 - y}^{n - 1}\,\dd y
\\[5mm] = &\
-1 +
2\int_{0}^{1}\int_{0}^{1}
\sum_{n = 1}^{\infty}H_{n}\pars{4xy \over 1 - y}^{n - 1}
\,\dd x\,\dd y
\\[2mm] &\
+ \int_{0}^{1}\ln\pars{y}\int_{0}^{1}
\sum_{n = 1}^{\infty}H_{n}\, \pars{4xy \over 1 - y}^{n - 1}
\,\dd x\,\dd y
\\[5mm] = &\
-1 +
2\int_{0}^{1}\int_{0}^{4y/\pars{1 - y}}
\sum_{n = 1}^{\infty}H_{n}x^{n - 1}\,
{1 - y \over 4y}\,\dd x\,\dd y
\\[2mm] &\
+ \int_{0}^{1}\ln\pars{y}\int_{0}^{4y}
\sum_{n = 1}^{\infty}H_{n}\, x^{n - 1}\,{y - 1 \over 4y}
\,\dd x\,\dd y
\\[5mm] = &\
-1 +
{1 \over 2}\int_{0}^{1}{1 - y \over y}\int_{0}^{4y/\pars{1 - y}}
\bracks{-\,{\ln\pars{1 - x} \over 1 - x}}
\,{\dd x \over x}\,\dd y
\\[2mm] &\
+ {1 \over 4}\int_{0}^{1}{\pars{1 - y}\ln\pars{y} \over y}\int_{0}^{4y/\pars{1 - y}}
\bracks{-\,{\ln\pars{1 - x} \over 1 - x}}
\,{\dd x \over x}\,\dd y
\\[5mm] = &\
-1 - {1 \over 2}\int_{0}^{1}{\ln\pars{1 - x} \over x\pars{1 - x}}
\int_{0}^{x/\pars{x + 4}}{1 - y \over y}\,\dd y\,\dd x
\\[2mm] &\
- {1 \over 4}\int_{0}^{1}{\ln\pars{1 - x} \over x\pars{1 - x}}
\int_{0}^{x/\pars{x + 4}}{\pars{1 - y}\ln\pars{y} \over y}
\,\dd y\,\dd x
\\[5mm] = &\
-1 - {1 \over 4}\int_{0}^{1}{\ln\pars{1 - x} \over x\pars{1 - x}}
\int_{0}^{x/\pars{x + 4}}
{\pars{1 - y}\bracks{2 + \ln\pars{y}} \over y}\,\dd y\,\dd x
\end{align}</span>
The <span class="math-container">$\ds{y}$</span>-integration becomes:
<span class="math-container">$$
-2\ln\pars{x \over 4 + x} -
{1 \over 2}\ln^{2}\pars{x \over 4 + x} -
{4 \over 4 + x} - {x \over 4 + x}\ln\pars{4 + x \over x}
$$</span>
It seems to be a nasty job !!!. I hope somebody else can take it from here.</p>
|
3,891,551 | <p>Let <span class="math-container">$S$</span> be the unit sphere <span class="math-container">$x^2+y^2+z^2=1$</span> with the outward pointing normal vector <strong>n</strong>. Calculate the flux for the vector field <span class="math-container">$\mathbf{f}(\mathbf{r})=4\mathbf{r}$</span> through <span class="math-container">$S$</span>.</p>
<p><strong>What I have done so far:</strong></p>
<p>I have rearranged the equation to get <span class="math-container">$z=\sqrt{1-x^2-y^2}$</span> and thus <span class="math-container">$N(x,y)=(\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1)$</span></p>
<p>Now to calculate the flux the equation is <span class="math-container">$\iint\limits_{S}\mathbf{f}\cdot\mathbf{n}dS$</span>.</p>
<p>Hence I now have <span class="math-container">$$4\iint\limits_{S}(x,y, \sqrt{1-x^2-y^2} )\cdot (\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1)dS$$</span></p>
<p><span class="math-container">$$=4\iint\limits_{S}\frac{1}{\sqrt{-x^2-y^2+1}}dxdy$$</span></p>
<p>However I don't know how to get the limits for the integral.</p>
| Math Lover | 801,574 | <p>In spherical coordinates,</p>
<p><span class="math-container">$x = r \cos \theta \sin \phi, y = r \sin \theta \sin \phi, z = r \cos \phi$</span></p>
<p>Surface area element <span class="math-container">$dS = r^2 \sin \phi \ d \theta d \phi = \sin \phi \ d \theta d \phi \, $</span> (as <span class="math-container">$r = 1$</span>)</p>
<p>Please note the outward normal vector should be a unit vector pointing directly away from the origin for this surface. So using, <span class="math-container">$\vec{r} = x \hat{i} + y\hat{j} + z\hat{k} = \cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}$</span></p>
<p><span class="math-container">$\hat{n} = \frac{\vec{r}}{||r||} = \cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}$</span></p>
<p><span class="math-container">$\vec{F} = 4r = 4(\cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k})$</span></p>
<p><span class="math-container">$Flux = \displaystyle \iint_S \vec{F} \cdot \hat{n} dS$</span></p>
<p><span class="math-container">$ \displaystyle = 4 \int_0^{\pi} \int_0^{2\pi} (\cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}) \cdot (\cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}) \sin \phi \, d\theta d\phi$</span></p>
<p><span class="math-container">$ \displaystyle = 4 \int_0^{\pi} \int_0^{2\pi} \sin \phi d\theta d\phi = 16 \pi$</span></p>
|
3,891,551 | <p>Let <span class="math-container">$S$</span> be the unit sphere <span class="math-container">$x^2+y^2+z^2=1$</span> with the outward pointing normal vector <strong>n</strong>. Calculate the flux for the vector field <span class="math-container">$\mathbf{f}(\mathbf{r})=4\mathbf{r}$</span> through <span class="math-container">$S$</span>.</p>
<p><strong>What I have done so far:</strong></p>
<p>I have rearranged the equation to get <span class="math-container">$z=\sqrt{1-x^2-y^2}$</span> and thus <span class="math-container">$N(x,y)=(\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1)$</span></p>
<p>Now to calculate the flux the equation is <span class="math-container">$\iint\limits_{S}\mathbf{f}\cdot\mathbf{n}dS$</span>.</p>
<p>Hence I now have <span class="math-container">$$4\iint\limits_{S}(x,y, \sqrt{1-x^2-y^2} )\cdot (\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1)dS$$</span></p>
<p><span class="math-container">$$=4\iint\limits_{S}\frac{1}{\sqrt{-x^2-y^2+1}}dxdy$$</span></p>
<p>However I don't know how to get the limits for the integral.</p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\on}[1]{\operatorname{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</span>
<span class="math-container">$\ds{\bbox[5px,#ffd]{}}$</span></p>
<hr>
<span class="math-container">\begin{align}
&\bbox[5px,#ffd]{\left.\int 4\vec{r}\cdot\dd\vec{S}
\,\right\vert_{\ r\ =\ 1}} =
4\int r\,\hat{r}\cdot\pars{\hat{r}\verts{\dd\vec{S}} \over r^{2}} =
4\int\overbrace{r}^{\ds{1}}\,\,
\overbrace{\hat{r}\cdot\hat{r}}^{\ds{1}}\
\overbrace{{\verts{\dd\vec{S}} \over r^{2}}}^{\ds{\dd\Omega_{\,\vec{r}}}}
\\[5mm] = &\
4\ \underbrace{\int\dd\Omega_{\,\vec{r}}}_{\ds{4\pi}}\
=\ \bbx{16\,\pi} \\ &
\end{align}</span>
|
263,650 | <p>As proposed by Quillen, Drinfeld, and Deligne and other important mathematicians, there is supposed to be a philosophy that, at least over a field of characteristic zero, assigns to every "deformation problem" a differential graded Lie algebra or $L_{\infty}$-algebra that controls it. </p>
<p>I've seen this idea realized in various situations, like for example the deformation theory of a compact complex manifold, which is "controlled" by the Kodaira-Spencer DGLA, or the deformation theory of Dirac structures in exact Courant algebroids. However, from my naive point of view as an outsider, I see this set of techniques completely disconnected from a different type of moduli problems of more "analytic" character. I refer for example to the moduli of Einstein metrics or the moduli of instantons in Donaldson's theory. It looks like the DGLA-philosophy has played virtually no role in the study of the moduli spaces of Einstein metrics and instantons. Why is this so? Does the DGLA-principle still applies to these problems but it does not add anything interesting? What is the DGLA controlling the deformation theory of these problems? It looks like the more analysis requires a moduli space problem, the less of a relevant role is played by the DGLA-philosophy, which seems to be more "algebraically inclined". I wonder because sometimes it looks like the moduli-literature is very polarized in different communities using different techniques to study moduli problems, so I would like to know to which extent the methods of one community apply to the problem considered by a different community.</p>
<p>Thanks.</p>
| Igor Khavkine | 2,622 | <p>With Domenico's clear explanation, I can actually write down more or less explicitly the DGLA describing the deformations of Einstein metrics.</p>
<p>First, some notation. Let $\bar{g}_{ab}$ denote a given (background) Einstein metric, with corresponding Levi-Civita connection $\bar{\nabla}_a$, Riemann tensor $\bar{R}_{abc}{}^d$, and Ricci tensor $\bar{R}_{ac} = \bar{R}_{abc}{}^b$. It is given that $\bar{\nabla}_a \bar{g}_{bc} = 0$ and $\bar{R}_{ac} = k \bar{g}_{ac}$ for a fixed constant $k$.</p>
<p>Let $\nabla_a$ denote an arbitrary symmetric (torsion free) affine connection, which differs from the background Levi-Civita connection as $\nabla_a v^b = \bar{\nabla}_a v^b + C^b_{ac} v^c$. Let me call $C^b_{ac} = C^b_{(ac)}$ the corresponding Christoffel tensor (or connection coefficients). The equation for an Einstein metric $g_{ab} = \bar{g}_{ab} + h_{ab}$ (with the same constant $k$) can be written in the form
\begin{align*}
\nabla_a g_{bc}
&= \bar{\nabla}_a h_{bc}
- C^d_{ab} \bar{g}_{dc} - C^d_{ac} \bar{g}_{bd}
- C^d_{ab} h_{dc} - C^d_{ac} h_{bd} ,
\\
R_{ac}[C] - k h_{ac}
&= -kh_{ac} - \bar{\nabla}_a C^b_{bc} + \bar{\nabla}_b C^b_{ac}
+ C^b_{ac} C^d_{db} - C^b_{ad} C^d_{cb} ,
\end{align*}
where $R_{ac}[C] = R_{abc}{}^b[C]$ is the usual Ricci contraction of the curvature tensor $R_{abc}{}^d[C]$ of $\nabla_a$. The first of the above equations is the $g$-compatibility condition for $\nabla_a$. Solving it for the Christoffel tensors identifies $\nabla_a$ with the Levi-Civita connection of $g_{ab}$ and plugging that solution into the second equation gives the equation $R_{abc}[g] - k g_{ac} = 0$ for an Einstein metric, due to the identity $R_{ab}[g] = \bar{R}_{ab} + R_{ab}[C]$.</p>
<p>Now the equations are precisely in the form that Domenico described, $f[h,C] + Q[h,C;h,C] = 0$, with $f$ linear and $Q$ quadratic
\begin{align*}
\begin{pmatrix}
f_{abc}[h,C] \\ f_{ac}[h,C]
\end{pmatrix}
&= \begin{pmatrix}
\bar{\nabla}_a h_{bc}
- C^d_{ab} \bar{g}_{dc} - C^d_{ac} \bar{g}_{bd} \\
-kh_{ac} - \bar{\nabla}_a C^b_{bc} + \bar{\nabla}_b C^b_{ac}
\end{pmatrix}
\\
\begin{pmatrix}
Q_{abc}[h,C;h',C'] \\ Q_{ac}[h,C;h',C']
\end{pmatrix}
&= \frac{1}{2} \begin{pmatrix}
- C^d_{ab} h'_{dc} - C^d_{ac} h'_{bd}
- C'^d_{ab} h_{dc} - C'^d_{ac} h_{bd} \\
+ C^b_{ac} C'^d_{db} - C^b_{ad} C'^d_{cb}
+ C'^b_{ac} C^d_{db} - C'^b_{ad} C^d_{cb}
\end{pmatrix}
\end{align*}
It remains to describe how infinitesimal symmetries (diffeomorphisms) act on the $h_{ab}$ and $C^b_{ac}$ tensor fields, as well as on $f$ and $Q$. They are generated by vector fields $u^a$. They act on tensors via the usual Lie derivative, which we find convenient to express via $\bar{\nabla}_a$, so that $\mathcal{L}_u X^a = u^b \bar{\nabla}_b X^a - X^b \bar{\nabla}_b u^a$ and $\mathcal{L}_u Y_a = u^b \bar{\nabla}_b Y_a + Y_b \bar{\nabla}_a u^b$, and they act on each other via the usual Lie bracket $[u,u'] = \mathcal{L}_u u' = -\mathcal{L}_{u'} u$. But special attention must be paid to the identities
\begin{align*}
\mathcal{L}_u \bar{\nabla}_b X^a - \bar{\nabla}_b \mathcal{L}_u X^a
& = u^c \bar{\nabla}_c \bar{\nabla}_b X^a
- (\bar{\nabla}_b X^c) \bar{\nabla}_c u^a
+ (\bar{\nabla}_c X^a) \bar{\nabla}_b u^c \\
& \quad {}
-\bar{\nabla}_b (u^c \bar{\nabla}_c X^a - X^c \bar{\nabla}_c u^a) \\
&= u^c \bar{R}_{bcd}{}^{a} X^d + X^c \bar{\nabla}_b \bar{\nabla}_c u^a
= (\bar{\nabla}_{(b} \bar{\nabla}_{c)} u^a - u^d \bar{R}_{d(bc)}{}^a) X^c , \\
% &= u^c \bar{R}_{bcd}{}^{a} X^d - X^c \bar{R}_{bcd}{}^a u^d + X^c \bar{\nabla}_c \bar{\nabla}_b u^a \\
% &= X^c \bar{R}_{cdb}{}^a u^d + X^c \bar{\nabla}_c \bar{\nabla}_b u^a , \\
\mathcal{L}_u \bar{\nabla}_b Y_a - \bar{\nabla}_b \mathcal{L}_u Y_a
& = u^c \bar{\nabla}_c \bar{\nabla}_b Y_a
+ (\bar{\nabla}_b Y_c) \bar{\nabla}_a u^c
+ (\bar{\nabla}_c Y_a) \bar{\nabla}_b u^c \\
& \quad {}
-\bar{\nabla}_b (u^c \bar{\nabla}_c Y_a + Y_c \bar{\nabla}_a u^c) \\
&= -u^c \bar{R}_{bca}{}^{d} Y_d - Y_c \bar{\nabla}_b \bar{\nabla}_a u^c
= (\bar{\nabla}_{(b} \bar{\nabla}_{a)} u^c - u^d \bar{R}_{d(ba)}{}^c) Y_c ,
\end{align*}
which identify the action of $[\mathcal{L}_u, \bar{\nabla}_b]$ as a derivation on the algebra of tensors. This means that infinitesimal diffeomorphisms generate the infinitesimal transformation $(h,C) \mapsto (h,C) + \epsilon K[u;h,C] + O(\epsilon^2)$, where
\begin{equation*}
\begin{pmatrix}
K_{ab}[u;h,C] \\ K_{ab}^c[u;h,C]
\end{pmatrix}
= \begin{pmatrix}
\bar{g}_{ac} \bar{\nabla}_b u^c + \bar{g}_{cb} \bar{\nabla}_a u^c \\
\bar{\nabla}_{(a} \bar{\nabla}_{b)} u^c - u^d \bar{R}_{d(ab)}{}^c
\end{pmatrix} .
\end{equation*}</p>
<p>Finally, we can put these formulas together in the definition of a DLGA $(L,d,[-,-])$. $L$ itself will break down into a sum of sections of certain tensor bundles. For simplicity, I will write $T$ to denote the space of sections of the bundle of vectors, $S^2T^*$ for symmetric covariant 2-tensors, etc. The breakdown by degree is
\begin{gather*}
\begin{array}{c|ccccc}
& 0 && 1 && 2 \\
\hline
L & T &\to& S^2 T^*\oplus S^2T^*\otimes T &\to& S^2T^* \otimes T^* \oplus S^2 T^* \\
d & & K & & f &
\end{array} , \\
\begin{array}{c|ccc}
[-,-] & 0 & 1 & 2 \\
\hline
0 & [u,u'] & K[u;h',C'] & \mathcal{L}_u \\
1 & -K[u';h,C] & 2Q[h,C;h',C'] & 0 \\
2 & -\mathcal{L}_{u'} & 0 & 0
\end{array}
\end{gather*}
I believe that this DGLA could be extended by one more degree to take the Bianchi identities into account. But I will stop here.</p>
<p>There are of course other ways to present the same DGLA and one can find explicit attempts in the literature of writing it down. Here's one that uses a somewhat different presentation:</p>
<blockquote>
<p>Michael Reiterer, Eugene Trubowitz <em>The graded Lie algebra of general relativity</em> <a href="https://arxiv.org/abs/1412.5561" rel="nofollow noreferrer">arXiv:1412.5561</a></p>
</blockquote>
|
4,122,425 | <p>Let’s say a corona test is correct with <code>p=0.8</code>. If I now take two tests. What’s the probability that I get a correct result?</p>
<p>I think thought of <code>0.8*0.8</code>, but that makes now sense, since it should not decrease and <code>0.8+0.8</code> gives a probability over 1, which makes no sense either. Or maybe that Bayes probability example?</p>
<p>Edit:
I would like to extend my questions: What’s the probability that I am negativ with one and with two tests? There the probability with two should increase if I am actually negative?
Thanks for the answers.</p>
| Célio Augusto | 600,009 | <p>The probability that both tests are wrong is <span class="math-container">$(0.2)(0.2)=0.04$</span>. Therefore, the probability that <em>at least</em> one of them is right is <span class="math-container">$1-0.04=0.96$</span>.</p>
<p>This, if course, assumes that both tests are independent.</p>
|
2,420,727 | <p>I'm trying to evaluate </p>
<blockquote>
<p>$$\lim _{ x\to -\infty } \frac { 2x-3 }{ \sqrt { x^{ 2 }+7x-2 } } $$</p>
</blockquote>
<p>by rationalizing the denominator, but I am not getting anywhere. Can someone please help me with this?</p>
<p>Thanks</p>
| haqnatural | 247,767 | <p>$$\lim _{ x\to -\infty } \frac { 2x-3 }{ \sqrt { x^{ 2 }+7x-2 } } =\lim _{ x\to -\infty } \frac { x\left( 2-\frac { 3 }{ x } \right) }{ \left| x \right| \sqrt { \left( 1+\frac { 7 }{ x } -\frac { 2 }{ { x }^{ 2 } } \right) } } =\\ =\lim _{ x\to -\infty } \frac { x\left( 2-\frac { 3 }{ x } \right) }{ -x\sqrt { 1+\frac { 7 }{ x } -\frac { 2 }{ { x }^{ 2 } } } } =-2\\ $$</p>
|
2,807,611 | <p>I know the answer is $n=6$, but can't figure out how to solve.
I tried dividing by $n!$, but didn't work because there isn't one in RHS to simplify... also tried using Gamma function properties, but didn't work either... </p>
<p>Any help would be appreciated.</p>
<p>Thanks.</p>
| David Taylor | 538,979 | <p>Because of $a_n = (5!)^{-1}((n+2)! - n!)$ in $\mathbb{N}$, so $n \geq 5$, and $a_n \equiv 0\pmod{n}$ for $n \geq 6$. However, $330 = 6 \cdot 55 \not \equiv 0 \pmod{7}$ therefore $a_n = 330$ for some of $n \leq 6$. Therefore $a_n = 330$ at $n = 6$. </p>
|
3,954,865 | <p>I am trying to solve a question but stuck with the steps. I can not find any similar questions. With help of some online resources to calculate some parts of the question but I can see that is not enough. I know my approach has lack of information but, this is the only thing I have reached, I was covid ill at the class hours and can not follow the class examples, I thought someone can help me to solve and learn the subject.</p>
<p>With help of the answers from here I try to give an answer. Still need some improvements but tried to do my best. I still do not have answer for question D and confused about CL(the part C) and Significance level(part B)</p>
<p><strong>My answers:</strong></p>
<p><span class="math-container">$N\ =\ 9\ \ \ \ \ \ \ \ Sum\ of\ x\ =\ 3970\ \ \ \ \ \ \ \ Mean,\ µ = 441.1111 \ Variance,σ^2 = 161.1111$</span>
<span class="math-container">$ \sigma\ =\ \sqrt{161.1111} = 12.6929$</span></p>
<p><span class="math-container">$t\ =\ \frac{m\ -\ \mu}{s\ /\ \sqrt n}$</span></p>
<p><span class="math-container">$t\ =\ \frac{500\ -\ 441.1111}{12.6929\ /\ \sqrt9} = 13.918545$</span></p>
<p>We subtract 1 to get degrees free 9 - 1 = 8</p>
<p>Degrees of freedom = n – 1 = 8</p>
<p><span class="math-container">$Probability: P( T ≤ 13.918545) = 0.00000069 $</span> So, this is the p-Value</p>
<p><span class="math-container">$We\ will\ reject\ H_0\ at \ \alpha = 1% $</span> and also any > 1%</p>
<p><span class="math-container">$$ (i) 0.10\
The\ information\ from\ the\ first\ question\, the\ critical\ t-value\ for\ α = 0.10\ and\ df = 8,\ t_c=\ 1.86
\\ CI\ =\ (\bar{X}\ -\ \frac{t_c\ \times\ s}{\sqrt n},\ \bar{\ X}\ +\ \frac{t_c\ \times\ s}{\sqrt n})
\\CI\ =\ (441.1111\ -\ \frac{1.86\ \times\ 12.6929}{\sqrt9},\ \ 441.1111+\frac{1.86\ \times\ 12.6929}{\sqrt9})
= (441.1111 – 7.868, 441.1111 + 7.868)
= (433.243, 448.979)
$$</span></p>
<p><span class="math-container">$For\ the\ other\ t_c\ values:$</span></p>
<blockquote>
<p>(ii) 0.05 <span class="math-container">$t_c=\ 2.306$</span> CL = (431.354, 450.868) (iii) 0.01 <span class="math-container">$t_c=\
> 3.355$</span>
= (426.915, 455.308)</p>
</blockquote>
<p>Based on the answers in part 2 for (i) = 0.90, (ii) = 0.95, (iii) = 0.99, none of the confidence intervals contain 500.</p>
<p><strong>The Question:</strong></p>
<p>The worker says that the mean purchasing cost is 500 USD. We decide to test this.</p>
<p>For a random sample of 9 purchases drawn from a normally distributed population with unknown variance, the costs are:</p>
<pre><code>430, 450, 450, 440, 460, 420, 430, 450, 440.
</code></pre>
<p>A) Conduct a hypothesis test of whether the population mean purchasing equals 500 USD. Include all assumptions, the hypotheses, test statistic, and P-value and interpret the result in context.</p>
<p>B) For which significance levels can you reject <span class="math-container">$H_0?$</span> (i) 0.10, (ii) 0.05, or (iii)
0.01.</p>
<p>C) Based on the answers in part B), for which confidence levels would the confidence interval contain 500? (i) 0.90, (ii) 0.95, or (iii) 0.99.</p>
<p>D) Use part B) and part C) to illustrate the correspondence between results
of significance tests and results of confidence intervals.</p>
| tommik | 791,458 | <p>As population variance is unknown you have a t-score, not z-score.</p>
<p>I did not check your calculation but if the score is so high (in Absolute value) you will reject the null hypothesis for any significance level.</p>
<p>Thus you reject <span class="math-container">$H_0$</span> at <span class="math-container">$\alpha =1\%$</span> and also at any <span class="math-container">$\alpha>1\%$</span></p>
<p>The rest of the exercise follows as a consequence</p>
|
3,954,865 | <p>I am trying to solve a question but stuck with the steps. I can not find any similar questions. With help of some online resources to calculate some parts of the question but I can see that is not enough. I know my approach has lack of information but, this is the only thing I have reached, I was covid ill at the class hours and can not follow the class examples, I thought someone can help me to solve and learn the subject.</p>
<p>With help of the answers from here I try to give an answer. Still need some improvements but tried to do my best. I still do not have answer for question D and confused about CL(the part C) and Significance level(part B)</p>
<p><strong>My answers:</strong></p>
<p><span class="math-container">$N\ =\ 9\ \ \ \ \ \ \ \ Sum\ of\ x\ =\ 3970\ \ \ \ \ \ \ \ Mean,\ µ = 441.1111 \ Variance,σ^2 = 161.1111$</span>
<span class="math-container">$ \sigma\ =\ \sqrt{161.1111} = 12.6929$</span></p>
<p><span class="math-container">$t\ =\ \frac{m\ -\ \mu}{s\ /\ \sqrt n}$</span></p>
<p><span class="math-container">$t\ =\ \frac{500\ -\ 441.1111}{12.6929\ /\ \sqrt9} = 13.918545$</span></p>
<p>We subtract 1 to get degrees free 9 - 1 = 8</p>
<p>Degrees of freedom = n – 1 = 8</p>
<p><span class="math-container">$Probability: P( T ≤ 13.918545) = 0.00000069 $</span> So, this is the p-Value</p>
<p><span class="math-container">$We\ will\ reject\ H_0\ at \ \alpha = 1% $</span> and also any > 1%</p>
<p><span class="math-container">$$ (i) 0.10\
The\ information\ from\ the\ first\ question\, the\ critical\ t-value\ for\ α = 0.10\ and\ df = 8,\ t_c=\ 1.86
\\ CI\ =\ (\bar{X}\ -\ \frac{t_c\ \times\ s}{\sqrt n},\ \bar{\ X}\ +\ \frac{t_c\ \times\ s}{\sqrt n})
\\CI\ =\ (441.1111\ -\ \frac{1.86\ \times\ 12.6929}{\sqrt9},\ \ 441.1111+\frac{1.86\ \times\ 12.6929}{\sqrt9})
= (441.1111 – 7.868, 441.1111 + 7.868)
= (433.243, 448.979)
$$</span></p>
<p><span class="math-container">$For\ the\ other\ t_c\ values:$</span></p>
<blockquote>
<p>(ii) 0.05 <span class="math-container">$t_c=\ 2.306$</span> CL = (431.354, 450.868) (iii) 0.01 <span class="math-container">$t_c=\
> 3.355$</span>
= (426.915, 455.308)</p>
</blockquote>
<p>Based on the answers in part 2 for (i) = 0.90, (ii) = 0.95, (iii) = 0.99, none of the confidence intervals contain 500.</p>
<p><strong>The Question:</strong></p>
<p>The worker says that the mean purchasing cost is 500 USD. We decide to test this.</p>
<p>For a random sample of 9 purchases drawn from a normally distributed population with unknown variance, the costs are:</p>
<pre><code>430, 450, 450, 440, 460, 420, 430, 450, 440.
</code></pre>
<p>A) Conduct a hypothesis test of whether the population mean purchasing equals 500 USD. Include all assumptions, the hypotheses, test statistic, and P-value and interpret the result in context.</p>
<p>B) For which significance levels can you reject <span class="math-container">$H_0?$</span> (i) 0.10, (ii) 0.05, or (iii)
0.01.</p>
<p>C) Based on the answers in part B), for which confidence levels would the confidence interval contain 500? (i) 0.90, (ii) 0.95, or (iii) 0.99.</p>
<p>D) Use part B) and part C) to illustrate the correspondence between results
of significance tests and results of confidence intervals.</p>
| Botnakov N. | 452,350 | <p>I will give some formulas, which may be proved by standard methods.</p>
<p>We have <span class="math-container">$H_0: X_i \sim N(a, \sigma^2)$</span>,</p>
<p><span class="math-container">$a= 500$</span>, <span class="math-container">$n=9$</span>, <span class="math-container">$\overline{X} = \frac{\sum_{i=1}^n X_i}{n} = 441.1111...$</span>, <span class="math-container">$s^2 = \frac{\sum_{i=1}^n (X_i - \overline{X})^2}{n} = \hat{\sigma^2}\frac{n-1}{n}$</span>, <span class="math-container">$ \hat{\sigma^2} =\frac{\sum_{i=1}^n (X_i - \overline{X})^2}{n-1} = 161.1111...$</span>, <span class="math-container">$\hat{\sigma} = 12.69296...$</span>.</p>
<p>We know that <span class="math-container">$\xi_n = \frac{\sqrt{n}(\overline{X} - a)}{\sigma} \sim N(0,1)$</span>, <span class="math-container">$\eta_n = \frac{ns^2}{\sigma^2} \sim \chi_{n-1}^2$</span>, <span class="math-container">$\xi_n$</span> and <span class="math-container">$\eta_n$</span> are independent,
<span class="math-container">$$t = \frac{\xi_n}{\sqrt{\frac{\eta_n}{n-1} }} = \sqrt{n-1}\frac{ \overline{X} - a}{s} = \sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} \sim T_{n-1}.$$</span>
and also <span class="math-container">$-t \sim T_{n-1}$</span>, because <span class="math-container">$T_{n-1}$</span> is symmetric. Here <span class="math-container">$T_{n-1}$</span> is Student's distribution.</p>
<p>We have <span class="math-container">$t =\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} = -13.9185...$</span>.</p>
<p>If <span class="math-container">$H_0$</span> is true then <span class="math-container">$|t| < u_{\frac{1+\gamma}2}$</span> with probability <span class="math-container">$\gamma$</span>, where <span class="math-container">$u_c$</span> is a quantile function at <span class="math-container">$c$</span>.</p>
<p>Put <span class="math-container">$\gamma = 0.99$</span>. Hence <span class="math-container">$\frac{1+\gamma}2 = 0.995$</span>. We know that <span class="math-container">$u_{0.995} = 5.04305...$</span> and hence <span class="math-container">$|t| < 5.04305...$</span> with probability <span class="math-container">$0.99$</span>, if <span class="math-container">$H_0$</span> is true.</p>
<p>Hence we reject <span class="math-container">$H_0$</span> for all significance levels in <span class="math-container">$B$</span>.</p>
<p>The confidence interval for <span class="math-container">$a$</span> we get form a condition <span class="math-container">$$|\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}.$$</span></p>
<p>It has the form:
<span class="math-container">$$ a \in (\overline{X} - \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma} , \overline{X} + \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma}).$$</span></p>
<p>Even in case <span class="math-container">$\gamma = 0.99$</span>, when the confidence interval is wider, than in cases <span class="math-container">$\gamma = 0.95$</span> and <span class="math-container">$\gamma = 0.90$</span>, we saw that the condition <span class="math-container">$$|\sqrt{n}\frac{ \overline{X} - 500}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}$$</span> doesn't hold and hence the confinedce interval doesn't contain <span class="math-container">$a=500$</span>.</p>
<p>So in <span class="math-container">$C)$</span> we get that <span class="math-container">$500$</span> is not contained is confinedce interval in any case.</p>
<p>D) We see that <span class="math-container">$H_0$</span> is accepted (significance level is fixed) if and only if <span class="math-container">$a$</span> is contained in a confidence interval of correspoding confidence level. The correspondence is illustrated.</p>
<p><strong>Addition about D</strong>.</p>
<p><span class="math-container">$H_0: X_i \sim N(a, \sigma^2)$</span>.</p>
<p>Test has significanse level <span class="math-container">$1-\gamma$</span> (in other words, test has level of confidence <span class="math-container">$\gamma$</span>). Test says that we shoud accept <span class="math-container">$H_0$</span> if and only if <span class="math-container">$$|\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}.$$</span></p>
<p>The confidence interval has the form:
<span class="math-container">$$ a \in (\overline{X} - \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma} , \overline{X} + \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma}).$$</span></p>
<p>So the condition "<span class="math-container">$a$</span> belongs to the <span class="math-container">$\gamma$</span>-confidence interval" is equivalent to condition <span class="math-container">$$|\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}$$</span></p>
<p>which is a necessary and sufficient condition to accept <span class="math-container">$H_0: X_i \sim N(a, \sigma^2)$</span> with significanse level <span class="math-container">$1-\gamma$</span> .</p>
<p><strong>Conclusion.</strong> "<span class="math-container">$a$</span> belongs to the confidence interval (with level of condidence <span class="math-container">$\gamma$</span>)" if and only if we accept <span class="math-container">$H_0: X_i \sim N(a, \sigma^2)$</span>, using the test with significanse level <span class="math-container">$1-\gamma$</span> .</p>
|
21,238 | <p>Would someone be able to point me to a good resource explaining step by step the process for solving inhomogenous recurrence relations? (ie something of the form $ a_n = \sum{{b_i}{a_{n-i}}} + f(n)$ )</p>
| Heinrich Hartmann | 5,714 | <p>A topological remark:</p>
<p>If <span class="math-container">$E \subset Y$</span> is a closed analytic subspace of a smooth space <span class="math-container">$X$</span>, then the boundary
of a tubular neighbourhood is an (odd dimensional, real) sphere bundle over <span class="math-container">$E$</span>.</p>
<p>Thus if the blowup of <span class="math-container">$Z \subset X$</span> is smooth then the boundary of a tubular
neighbourhood of <span class="math-container">$Z \subset X$</span> can be written as a sphere bundle over the exceptional
divisor <span class="math-container">$E = P(N_Z X)$</span>.</p>
<p>One might use this to arrive at a contradiction e.g. by calculating cohomology.</p>
<p>Similar ideas were used by Mumford in his study of normal surface singularities.
<a href="https://doi.org/10.1007/BF02698717" rel="nofollow noreferrer">Mumford: The topology of normal singularities of an algebraic surface and a criterion for simplicity.</a></p>
|
2,604,844 | <p>I have the folowing induction :</p>
<p>"Every graph with n vertices and zero edges is connected"</p>
<ul>
<li><p>Base:</p>
<p>For $n=1$ graph with one vertice is connected, hence a graph with $1$ vertex and zero edge.</p></li>
<li><p>Assumpution:</p>
<p>Every graph with $n-1$ vertices and zero edges is connected.</p></li>
</ul>
<p>For every graph with $n$ vertices and zero edges lets remove one vertice hence we get a graph with $n-1$ vertices and zero edges, by the assumpution the graph is connected, therefore the original graph is connected.</p>
<p>I feel that when we remove one vertice from the graph with $n$ vertice it not must be a connected graph, Yet I'm not sure if thats the flaw, or how to proof it.</p>
<p>I read about the "all horses has the same color" yet I don't find how to use the method of it here.</p>
<p>Any ideas?</p>
<p>Thanks!</p>
| ArsenBerk | 505,611 | <p>The problem is here:</p>
<blockquote>
<p>For every graph with $n$ vertices and zero edges lets remove one vertice hence we get a graph with $n−1$ vertices and zero edges, by the assumpution the graph is connected, therefore the original graph is connected. </p>
</blockquote>
<p>Because first you supposed argument holds for $n-1$ vertices. Then you start with $n$ vertices and removed one to make your assumption true. But this doesn't imply that the graph with $n$ vertices is connected. For example, if you remove a vertex, that is not a leaf, from a tree you get a disconnected graph even though tree is connected itself. Instead, you should use induction as this one (after the base case):</p>
<p>Suppose the argument holds for $n-1$. Then if we <em>add</em> a vertex to have a graph with $n$ vertices, ...</p>
<p>However, when we add an isolated vertex (not connected to any other $n-1$ vertices), then by definition of connectivity, the graph with $n$ vertices will not be connected. So argument doesn't hold for all $n$.</p>
|
4,194,511 | <p><span class="math-container">$$ \frac {d^2y}{dx^2} + 5\frac{dy}{dx} = 15x^2 $$</span></p>
<p>It's solution is</p>
<p><span class="math-container">$$y = y_{h} + y_{p}$$</span></p>
<p>where <span class="math-container">$y_{h} $</span> is the solution for homogenous equation and <span class="math-container">$y_{p} $</span> is the particular solution</p>
<p>For homogenous solution:
It's auxiliary equation is</p>
<p><span class="math-container">$$m^2 + 5m = 0$$</span></p>
<p>on solving for m,</p>
<p><span class="math-container">$$m = 0, -5$$</span></p>
<p>so, <span class="math-container">$$y_{h} = C_{1} + C_{2} e^{-5x}$$</span></p>
<p>here</p>
<p><span class="math-container">$$R = 15x^2$$</span></p>
<p>so it's particular solution should be :</p>
<p><span class="math-container">$$y_{p} = K_{2}x^2 + k_{1}x + k_{0}$$</span></p>
<p>on differentiating with respect to x</p>
<p><span class="math-container">$$y\prime _{p}= 2xk_{2} + k_{1}$$</span></p>
<p>again</p>
<p><span class="math-container">$$y\prime\prime_{p} = 2k_{2}$$</span></p>
<p>using all above values in the initial equation gives</p>
<p><span class="math-container">$$2k_{2} + 10xk_{2} + 10k_{1} = 15x^2$$</span></p>
<p>But using this method I am unable to proceed for the value of all unknown especially <span class="math-container">$k_{0}$</span></p>
<p>The given answer is:</p>
<p><span class="math-container">$$y = C_{1} + C_{2}e^{-5x} + x^3 + \frac{3x^2}{5} - \frac{6}{25}x$$</span></p>
| jjagmath | 571,433 | <p>You are right. It's a common mistake that high school teachers make when they teach to solve even the most simple equations.</p>
<p>As José Carlos Santos says in his answer, you can track which of the steps of a solution are only implications or are actually equivalences.</p>
<p>Here's an easy example where not all the obtained possible solutions are solutions:</p>
<p>Take <span class="math-container">\begin{align}
& x = \sqrt{2-x^2}\\
\implies \quad & x^2 = 2-x^2\\
\implies \quad & 2x^2 = 2\\
\implies \quad & x^2 = 1\\
\implies \quad & x = 1 \text{ or } x=-1
\end{align}</span></p>
<p>But <span class="math-container">$-1$</span> is obviously not a solution.</p>
<p>For a slightly more interesting example try to solve (in <span class="math-container">$\Bbb C$</span>) the equation <span class="math-container">$$\sqrt{1+z}+\sqrt{1-z} = 1$$</span></p>
|
863,167 | <p>How can the signed area be 0? For example if you have 3 on positive x side and 3 on the negative x side then you get the signed area of 0? How can area be 0?</p>
| Ant | 66,711 | <p>You can interpret the integral of a function as "area" only if the function is positive.</p>
<p>Otherwise, you may interpret the result as the area under the positive part of the function minus the area above the negative part of the function. </p>
<p>(N.B. I'm guessing you are talking about integrals, but your post is not very clear)</p>
|
1,498,048 | <p>I can´t prove this problem. Can you help me? The problem says:</p>
<p><em>If $\{X_n\}$ is a sequence of identically distributed random variables with finite mean, then
$$lim_{n\to\infty}\frac{1}{n}\mathbb{E}\Big[\max_{1\leq j\leq n} |Xj|\Big] = 0$$
[HINT: Use Exercise 17 to express the mean of the maximum.]</em></p>
<p>In problem 17 show that if X is a positive random variable, then we have
$$\mathbb{E}[X]=\int_0^\infty \!\! P(X > x)\, dx=\int_0^\infty \!\! P(X \geq x)\, dx$$</p>
<p>I prove the problem 19 assuming independence and for the general case I find some upper bounds for the maximum of dependent variables on the internet but none worked.</p>
<p>Thank you for help me.</p>
| Kavi Rama Murthy | 142,385 | <p>Use Dominated convergence theorem after applying Exercise 17. The dominating function is P(|X_1|>x). </p>
|
1,242,570 | <p>I want to use the standard definition $x_n \rightarrow x$ if for all $\epsilon>0$ there exists $N$ such that if $n>N$ then $|x_N-x|<\epsilon$. </p>
<p>So my claim is $x_n\rightarrow 0$ If I set $N=\epsilon^2,$ then the following expression $|\sqrt{n^2+1}-n-0|<\epsilon$ will hold true. I solved for $N$ by squaring both sides $n^2+1-n^2<\epsilon^2$ Does this work? </p>
<p>Edit: Made a dumb algebraic mistake, thanks everyone.</p>
| Adhvaitha | 228,265 | <p><strong>HINT</strong> We have
$$x_n = \sqrt{n^2+1} - n \implies x_n = \dfrac1{\sqrt{n^2+1}+n} < \dfrac1{2n}$$</p>
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