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 |
|---|---|---|---|---|
2,976,613 | <p>I want to show, that <span class="math-container">$a:=\sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n$</span> is not converging, because <span class="math-container">$\lim \limits_{n \to \infty}(a)\neq 0 \; (*)$</span>. Therefore, the series can't be absolute converge too.</p>
<p>Firstly, I try to simplify the term. After that I want to find the limit.</p>
<p>Unfortunately, I can't seem to find any good equation with that I can clearly show <span class="math-container">$(*)$</span>.
<span class="math-container">\begin{align}
\sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n&=\left (\dfrac{\not{n}\cdot (2+n^2)}{\not n \cdot (\frac{3}{n}-4)}\right)^n\\
&=\left(\dfrac{2+n^2}{\frac 3n-4}\right)^n\\
&= \cdots
\end{align}</span></p>
<p>How to go on?</p>
| Doesbaddel | 587,094 | <p><span class="math-container">$\sum \limits_{n=0}^{\infty} \left(\underbrace{\dfrac{2n+n^3}{3-4n}}_{a_n}\right)^n$</span></p>
<p>Root test: Prove:<span class="math-container">$\lim \limits_{n \to \infty} \sqrt[n]{\mid a_n \mid}<1 \implies a_n \text{ is absolute converging.}$</span></p>
<p><span class="math-container">\begin{align}
&\lim \limits_{n \to \infty} \sqrt[n]{\left( \dfrac{2n+n^3}{3-4n}\right)^n} \\
=&\lim \limits_{n \to \infty}=\frac{2n+n^3}{3-4n}\\
=&\lim \limits_{n \to \infty}=\dfrac{n(n^2+2)}{n(\frac 3n-4)}\\
=&\lim \limits_{n \to \infty}=\dfrac{\overbrace{n^2+2}^{\infty}}{\underbrace{\frac 3n-4}_{0}}\\
=&\lim \limits_{n \to \infty}(a_n)=\infty\\&
\implies a_n \text{ and } \sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n \text{ are not absolute convergent.}
\end{align}</span></p>
<p>I just noticed, that this would only prove absolute convergence and not convergence.</p>
|
813,121 | <p>I have to calculate this : $$ \lim_{x\to 0}\frac{2-x}{x^3}e^{(x-1)/x^2} $$ Can somebody help me?</p>
| meh | 70,191 | <p>$$\lim_{x\to0}(f(x)g(x)) = \lim_{x\to0}(f(x)) \cdot \lim_{x\to0}(g(x)) $$</p>
<p>With that being said you can let $f(x) = (2-x)/x^3 $ and $ g(x) = e^{(x-1)/x^2} $</p>
<p>I hope this helps.</p>
|
183,881 | <p>I want to replace an element of a matrix by a function of its current value. The code I'm using is as follows</p>
<pre><code>T = {{2, 3}, {3, 2}, {1, 4}}
KNeeds = Transpose[T][[2]]
pos = Flatten[{Position[KNeeds, Min[KNeeds]], 2}]
ReplacePart[T, pos -> T[[pos[[1]]]] - Epsilon]]
</code></pre>
<p>I was hoping to obtain <code>{{2, 3}, {3, 2-Epsilon}, {1, 4}},</code> but I get <code>{{2, 3}, {3, {3 - Epsilon, 2 - Epsilon}}, {1, 4}}.</code> </p>
<p>I've also tried <code>ReplacePart[T, pos -> T[[pos]] - Epsilon]],</code> but then I get <code>{{2, 3}, {3, {{3 - Epsilon, 2 - Epsilon}, {3 - Epsilon,
2 - Epsilon}}}, {1, 4}}.</code></p>
| halirutan | 187 | <p>If I understood your question correctly, then you can rip out the inner part of your code above and use <code>ListLinePlot</code> by creating a table containing <code>{x, productY}</code> elements:</p>
<pre><code>pts = Table[{i, 0, 1}, {i, 0, 10}];
ListLinePlot[
Table[{x,
R1 = Cylinder[{{3, 0, 0}, {3, 0, 2}}, x];
R2 = Cylinder[{{7, 0, 0}, {7, 0, 2}}, x];
R3 = RegionIntersection[R1, R2];
Count[RegionMember[R3, pts], True]},
{x, 5}]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/YRyFW.png" alt="Mathematica graphics"></p>
|
1,149,685 | <p>Suppose that 4 guests check their hats when they arrive at a restaurant, and that these hats are returned to them in a random order when they leave. Determine the probability that no guest will receive the proper hat.</p>
<p>My attempt:
So I know that there are 4! ways of "arranging" the hats.
The probability of each receiving their hat is 1/4 and the probability that they will not receive the right hat is 3/4.</p>
<p>I think the answer should be something like:</p>
<p>$\frac{}{4!}$</p>
<p>where the 4! is from the total ways of arranging the hats, and the numerator is the number of "correct" ways to arrange the hats (so no one gets their own hat)But I can't seem to think of how to express that mathematically.</p>
<p>How should I think about this problem instead?</p>
<p>So the book gives the answer 3/8 and myclassmate said the answer is </p>
<p>$\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}=\frac{3}{8}$</p>
<p>But I don't understand why this is?</p>
| Alex | 38,873 | <p>Use inclusion-exclusion theorem/formula/principle. Total number of ways to distribute 4 hats amon 4 people is $4!$. Now subtract the number of ways 1 person leaves with his hat, add two people left with their hats, subtract three people left with their hats, add 4 ppl left with their hats (the last one is obviously 1). </p>
|
1,149,685 | <p>Suppose that 4 guests check their hats when they arrive at a restaurant, and that these hats are returned to them in a random order when they leave. Determine the probability that no guest will receive the proper hat.</p>
<p>My attempt:
So I know that there are 4! ways of "arranging" the hats.
The probability of each receiving their hat is 1/4 and the probability that they will not receive the right hat is 3/4.</p>
<p>I think the answer should be something like:</p>
<p>$\frac{}{4!}$</p>
<p>where the 4! is from the total ways of arranging the hats, and the numerator is the number of "correct" ways to arrange the hats (so no one gets their own hat)But I can't seem to think of how to express that mathematically.</p>
<p>How should I think about this problem instead?</p>
<p>So the book gives the answer 3/8 and myclassmate said the answer is </p>
<p>$\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}=\frac{3}{8}$</p>
<p>But I don't understand why this is?</p>
| Balthier TSK | 869,590 | <p>Sorry for the very late reply, lmao.
Anyhow, the answer is to use derangements: an arrangement such that no elements appear in their original position.
Basically, if you map a hat and its owner, you can find an arrangement such that no hat is mapped to its original owner again, which is just derangement.
It is also useful to know when the question asked it "If at least one person leave with their original hats", that would just be the complement of the derangement and be
(1 - Pr(x)), where x is the probability that no one leaves with their hats.</p>
<p>Using derangement, you can see that 9 arrangements fit the category and since there's 4! available permutation, the result is 9/24 = 3/8</p>
|
4,320,849 | <p>I had this problem in an exam I recently appeared for:</p>
<blockquote>
<p>Find the range of
<span class="math-container">$$y =\frac{x^2+2x+4}{2x^2+4x+9}$$</span></p>
</blockquote>
<p>By randomly assuming the value of <span class="math-container">$x$</span>, I got the lower range of this expression as <span class="math-container">$3/7$</span>. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.</p>
<p>Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.</p>
<p>Another guy I met outside the exam center, told me he used an approach of <span class="math-container">$x$</span> tending to infinity in both cases and got the maximum value of this expression as <span class="math-container">$1/2$</span>. But before I could ask him to explain more on this method, he had to leave for his work.</p>
<p>So, will someone please throw some light on this method of <span class="math-container">$x$</span> tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).</p>
| Nikola Alfredi | 670,977 | <p>The question can be easily solved by this technique:</p>
<p>As <span class="math-container">$\displaystyle y = \frac {x^2 + 2x + 4}{2x^2 + 4x + 9} \implies 2y = \frac {2x^2 + 4x + 9 - 1}{2x^2 + 4x + 9}$</span>.</p>
<p>Thus, <span class="math-container">$\displaystyle 2y = 1-\frac {1}{2(x + 1)^2 + 7} $</span></p>
<p>Squares can never be less than zero so the minimum value of the function : <span class="math-container">$\displaystyle 2(x + 1)^2 + 7 $</span> would be <span class="math-container">$7$</span> , or Maximum value of <span class="math-container">$\displaystyle \frac {1}{2(x + 1)^2 + 7} $</span> is <span class="math-container">$\displaystyle \frac {1}{7} $</span>.</p>
<p>This tells that minimum value of <span class="math-container">$y $</span> will be <span class="math-container">$\displaystyle \frac{3}{7}$</span>.</p>
<p>And so on.. check for <span class="math-container">$x \rightarrow \infty$</span>.</p>
<p>From here you can easily tell the maximum and minimum values : <span class="math-container">$\displaystyle y \in \left [ \frac {3}{7}, \frac {1}{2} \right ) $</span></p>
|
125,399 | <p>How can I solve the following integral?
$$\int_0^\pi{\frac{\cos{nx}}{5 + 4\cos{x}}}dx, n \in \mathbb{N}$$</p>
| Christian Blatter | 1,303 | <p>Here is a solution without residue calculus:</p>
<p>Via partial fractions and geometric series one arrives at
$$\eqalign{{1\over 5+4\cos x}&={1\over(2+e^{ix})(2+e^{-ix})}=\ldots\cr &={1\over 3}-{1\over3}{e^{ix}/2\over 1+e^{ix}/2}-{1\over3}{e^{-ix}/2\over 1+e^{-ix}/2}\cr &= {1\over3}+{1\over3}\sum_{k=1}^\infty(-1)^k{e^{ikx}\over 2^k} +{1\over3}\sum_{k=1}^\infty(-1)^k{e^{-ikx}\over 2^k}\cr &={1\over3}+{2\over3}\sum_{k=1}^\infty(-1)^k{\cos(k x)\over 2^k}\ .\cr}$$
Therefore
$$\int_0^\pi{\cos(n x)\over 5+4\cos x}\ dx={2(-1)^k\over3\cdot 2^n}{1\over2}\int_{-\pi}^\pi \cos^2(nx) dx={(-1)^n\over 3\cdot 2^n}\pi\qquad(n\geq0)\ .$$</p>
|
1,206,528 | <p>Find the matrix $A^{50}$ given</p>
<p>$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$</p>
<p>I was practicing some questions for my exam and I found questions of this form in a previous year's paper.</p>
<p>I don't know how to do such questions.</p>
<p>Please assist over this question.</p>
<p>Thank You</p>
| mvw | 86,776 | <p>And yet another approach: Splitting $A$ into a diagonal matrix $D$ and a pertubation matrix $P$.</p>
<p>We look at the first powers of $P$:</p>
<p>$$
P^0 =
\left(
\begin{matrix}
1 & 0\\
0 & 1
\end{matrix}
\right)
\quad
P^1 =
\left(
\begin{matrix}
0 & -1 \\
0 & 0
\end{matrix}
\right)
\quad
P^2 =
\left(
\begin{matrix}
0 & 0 \\
0 & 0
\end{matrix}
\right)
$$
and note that quadratic powers and higher of $P$ vanish.
Then
$$
DP = 2 P \quad PD = P
$$
Thus
\begin{align}
A^n
&= (D + P)^n \\
&= (D^2 + DP + PD + P^2) (D+P)^{n-2} \\
&= (D^2 + (1+2^1)P) (D+P)^{n-2} \\
&= (D^3 + (1+2^1)PD + D^2P + 3P^2)(D+P)^{n-3} \\
&= (D^3 + (1+2^1)P + 2^2 P)(D+P)^{n-3} \\
&= (D^3 + (1+2^1+2^2) P)(D+P)^{n-3} \\
&= D^n + (1+2^1+\cdots+2^{n-1})P \\
&= D^n + (2^n-1)P
\end{align}</p>
<p>This gives
$$
A^n
=
\left(
\begin{matrix}
2^n & 0 \\
0 & 1^n
\end{matrix}
\right)
+
\left(
\begin{matrix}
0 & 1-2^n \\
0 & 0
\end{matrix}
\right)
=
\left(
\begin{matrix}
2^n & 1-2^n \\
0 & 1
\end{matrix}
\right)
$$</p>
|
1,206,980 | <p>Here is the problem: if $v^1$, $v^2$, ..., $v^d$ is an orthonormal basis in $\mathbb{R}^d$, then show that
$$
||A - A\sum_{i = 1}^k v^i(v^i)^T ||^2_F = \sum_{i = k+1}^d||Av^i ||_2^2.
$$</p>
<p>I am having trouble understanding this step of a longer proof. The text that I got this from just says that this holds because $v^1$, $v^2$, ..., $v^d$ is an orthonormal basis of $\mathbb{R}^d$. </p>
| pjs36 | 120,540 | <p>The $x$ and $y$ are dummy variables, they don't really matter. For example, if $p(t) = t - 1$, we have $p(x) = x - 1$, so that</p>
<p>\begin{align*}\int_0^t\int_0^y (x - 1)\ dx\ dy &= \int_0^t\left[\frac{x^2}{2} - x\right]_0^y\ dy\\
&=\int_0^t\left(\frac{y^2}{2}-y\right)\ dy \\
&=\left[\frac{y^3}{6} - \frac{y^2}{2}\right]_0^t\\
&=\frac{1}{6}t^3 - \frac{1}{2}t^2.
\end{align*}</p>
<p>Perhaps somebody smarter than me knows how to compute the final transformation matrix from smaller matrices (ones involving integration), but I'd just bite the bullet and compute the integral for each basis polynomial in $P_3$.</p>
<p>You may be able to find the transformation matrix for a single integration, and apply that twice to your basis vectors, but I'm not positive it will work out.</p>
|
2,815,900 | <p>Consider the proposition $∀x (P (x) → Q (x))$, where $P (x)$ and $Q (x)$ are predicates on a domain $U$. Prove or disprove the following statement, justifying the answer.</p>
<p>If the statement $∀x (P (x) → Q (x))$ is FALSE then it's FALSE also $∀x P (x)$</p>
<p>How can I solve this exercise?</p>
| Ovi | 64,460 | <p>If $∀x (P (x) → Q (x))$ is false then it just means there exists some $x$ such that $P(x)$ is true and $Q(x)$ is false. It does not tell us anything about weather $P(x)$ is true for every $x$ or not ... can you take it from here?</p>
|
637,379 | <p>So I have a $10$ gallon aquarium slightly salty aquarium...</p>
<p>When I add water the water is at $.7$% salt. ($7$ part per thousand)</p>
<p>I let $1$ gallon evaporate, at which point I have a $\frac{7}{9}$% salinity
I then drain $1$ additional gallon. now I have $8$ gallons of $\frac{7}{9}$% salinity water.</p>
<p>I then replace the missing $2$ gallons with $.7$% salinity water again.</p>
<p>If I perform this over and over does the water continue to get ever saltier, or does it approach some limit.</p>
| kaine | 53,641 | <p>Hint the final concentration of one cycle will be: $$f(x) =.8(\frac{10x}{9})+.2(.7)$$ where $x$ is $.7$ initially. </p>
<p>Does $f(x)$ increase or decrease for each $x$? </p>
<p>Is there any case where $f(x)=x$? </p>
<p>If you can, look at the derivative of $f(x)$.</p>
|
1,571,083 | <p>Given the 2 terms
$$ \frac{k + a}{k + b}$$
and $$\frac{a}{b}$$
with $a, b, k \in \mathbb{R^+}$ and $a > b$</p>
<p>I want to show, that the first term is always bigger than the second one.</p>
<p><strong>My try</strong>
$$
\frac{k + a}{k + b} > \frac{a}{b} \\
\frac{k + b + (a - b)}{k + b} > \frac{b + (a - b)}{b} \\
1 + \frac{a - b}{k + b}> 1 + \frac{a - b}{b} \\
\frac{1}{k + b} > \frac{1}{b} \\
k + b > b
$$
the problem is, that I think the inequality sign has to change to $<$ in the last step because of calculating the inverse on both sides. That would mean
$$ k + b < b$$
which is nonsense.</p>
| costrom | 271,075 | <p>The odds of winning the first contest is:</p>
<p>$\frac{1}{31}+\frac{30}{31}\cdot\frac{1}{30} = \frac{2}{31}$ - either they win the first or second draw - and there is no replacement between draws</p>
<p>The odds of winning the second contest is $\frac{1}{25}$, as you said.</p>
<p>The odds of winning both is the product of these, or</p>
<p>$\frac{2}{775} \approx 0.26\%$</p>
<p>So it's unlikely, but it's not necessarily an indication of cheating.</p>
<p>If it happens any more times, I'd start to think about it though...</p>
|
1,887,395 | <p>Let $D=\{Z\in\mathbb{C}:|z|<1\}$. Suppose $f:\overline D\to\mathbb{C}$ is continuous, holomorphic in $D$, and $f(\partial D)\subset \mathbb{R}$. Define $h:\mathbb{C}\to\mathbb{C}$ by
$$
h(z)=\begin{cases}
f(z) &\text{if}\ z\in \overline D\\
\overline{f(1/\overline z)} &\text{if}\ z\in\mathbb{C}\setminus \overline{D}
\end{cases}$$
Show that $h$ is entire and $f$ is a constant function.</p>
<p>My work: I can show that $h$ is holomorphic in $\mathbb{C}\setminus \overline{D}$. </p>
<p>To show $h$ is entire, by using Morera's theorem, considering arbitrary triangular path, the integral of $h$ is zero if it does not cross the boundary of the unit disc, however, what if the triangular path cross the unit disc? How to show in this case the integral is also zero.</p>
<p>Could anyone kindly help? Thanks!</p>
| Robert Z | 299,698 | <p>According to Lemma 1 in this <a href="http://www.doiserbia.nb.rs/img/doi/1452-8630/2012/1452-86301200016S.pdf" rel="nofollow">article</a> by A. Sofo and D. Cvijovic, we have that
$$\sum_{n=1}^{\infty} H_{n}\left(\frac{1}{n}-\frac{1}{n+k}\right)=
k\sum_{n=1}^{\infty} \frac{H_{n}}{n(n+k)}=
\frac{\pi^2}{6}+\frac{H^2_{k-1}+H^{(2)}_{k-1}}{2}$$
where $H_n=\sum_{j=1}^n\frac{1}{j}$ and $H^{(2)}_n=\sum_{j=1}^n\frac{1}{j^2}$.</p>
|
4,295,652 | <p>In the integration formula <span class="math-container">$\int dx/x = log x + c$</span>, Is the log natural or log base 10? The formula appears in many problems and i just got a problem wrong for apparently using the wrong log. Could you please enlighten me about the right log to be used in integration.</p>
| Community | -1 | <p>In the formula
<span class="math-container">$$
\int \frac{1}{x}\; dx = \log(x)+C
$$</span>
the logarithm is with base <span class="math-container">$e$</span>. Most calculus books write
<span class="math-container">$$
\int \frac{1}{x}\; dx = \log|x|+C\tag{1}
$$</span>
with the absolute value on the right-hand side. The formula (1) is understood as <span class="math-container">$\log|x|$</span> is an antiderivative of <span class="math-container">$\frac1x$</span> on any <strong>open interval</strong> contained in the set <span class="math-container">$\mathbb{R}\setminus\{0\}$</span>.</p>
<p>[<em>If one wants to consider any domain for the function <span class="math-container">$\frac1x$</span> that is not connected, for instance, its <a href="https://en.wikipedia.org/wiki/Domain_of_a_function#Natural_domain" rel="nofollow noreferrer">natural domain</a> <span class="math-container">$(-\infty,0)\cup(0,\infty)$</span>, then one needs more than one "artibrary constants" and different cases for each connected component of the domain on the right-hand side. In other words, the general solutions to the differential equation
<span class="math-container">$$
\frac{dy}{dx}=\frac{1}{x}
$$</span>
on an open subset that is <em>not</em> connected has more than one "arbitrary constant" C's.</em>]</p>
<hr />
<p>In many other contexts, the base may not be important due to the fact that they only cause a factor of a constant. But when doing integrals, the natural log is the one that is an antiderivative of <span class="math-container">$\frac1x$</span>.</p>
|
4,295,652 | <p>In the integration formula <span class="math-container">$\int dx/x = log x + c$</span>, Is the log natural or log base 10? The formula appears in many problems and i just got a problem wrong for apparently using the wrong log. Could you please enlighten me about the right log to be used in integration.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>The formula is <span class="math-container">$$\int\,\frac{\mathrm{d}x}{x}=\ln(|x|)+C$$</span></p>
<p>For the definition of the natural logarithm as an integral we have</p>
<p><span class="math-container">$$ Ln (x) = \int _1^x \frac {1}{t} dt , x>0 $$</span></p>
|
2,349,312 | <blockquote>
<p>Let $F$ be a subfield of the complex numbers. Let $A,B\in F^{n\times n}$ and $X\in F^{n\times 1}.$ If $A\ne B,$ then $AX\ne BX$ for every non-zero $X.$</p>
</blockquote>
<p>I am NOT sure if the above statement is right. I did some computation with random matrices and it seems true to me. This is the argument I tried :</p>
<p>Let $A=[A_1,A_2,\dots,A_n]$ and $B=[B_1,B_2,\dots,B_n]$ where $A_i,B_i \in F^{n\times 1}$ are columns of $A$ and $B$ respectively. Let $X=\begin{bmatrix}x_1\\\vdots \\x_n\end{bmatrix}.$</p>
<p>Since $A\ne B,$ choose $j$ such that $A_j\ne B_j.$ Let $J_1$ be the set of all such $j$'s and $J_2=I$ \ $ J_1$ where $I=\{1,2,\dots ,n\}.$ Then, $AX=\sum_{i\in J_1} x_iA_i +\sum_{i\in J_2} x_iA_i \ne \sum_{i\in J_1} x_iB_i + \sum_{i\in J_2} x_iB_i= BX.$</p>
<p><strong>(1) Is my argument correct?<br>(2) Also, can $F$ be generalized?</strong> </p>
| Bernard | 202,857 | <p>It's not correct, as it would imply that, as soon as $A\ne B$, $A-B$ is non-singular. Why should it be true?</p>
<p>Actually, $A\ne B$ only implies that for <em>some</em> $X$, one has $AX\ne BX$.</p>
|
2,349,312 | <blockquote>
<p>Let $F$ be a subfield of the complex numbers. Let $A,B\in F^{n\times n}$ and $X\in F^{n\times 1}.$ If $A\ne B,$ then $AX\ne BX$ for every non-zero $X.$</p>
</blockquote>
<p>I am NOT sure if the above statement is right. I did some computation with random matrices and it seems true to me. This is the argument I tried :</p>
<p>Let $A=[A_1,A_2,\dots,A_n]$ and $B=[B_1,B_2,\dots,B_n]$ where $A_i,B_i \in F^{n\times 1}$ are columns of $A$ and $B$ respectively. Let $X=\begin{bmatrix}x_1\\\vdots \\x_n\end{bmatrix}.$</p>
<p>Since $A\ne B,$ choose $j$ such that $A_j\ne B_j.$ Let $J_1$ be the set of all such $j$'s and $J_2=I$ \ $ J_1$ where $I=\{1,2,\dots ,n\}.$ Then, $AX=\sum_{i\in J_1} x_iA_i +\sum_{i\in J_2} x_iA_i \ne \sum_{i\in J_1} x_iB_i + \sum_{i\in J_2} x_iB_i= BX.$</p>
<p><strong>(1) Is my argument correct?<br>(2) Also, can $F$ be generalized?</strong> </p>
| Fred | 380,717 | <p>Let $A = \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}, B= \begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}$ and $X=\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}$ and look what happens !</p>
|
762,964 | <blockquote>
<p>Let <span class="math-container">$X$</span> be from a normal distribution <span class="math-container">$N(\theta,1)$</span>.</p>
<p>a) Find a sufficient statistic for <span class="math-container">$\theta$</span>.<br />
b) Is <span class="math-container">$S_n^2$</span> a
sufficient statistic for <span class="math-container">$\theta$</span>?</p>
</blockquote>
<p><strong>My answers</strong></p>
<p>For part a)</p>
<p>Since the joint p.d.f is <span class="math-container">$1 \over (2\pi)^{n/2}$$e^{{-1 \over 2}\sum(x_i-\theta)^2}$</span> I can say that <span class="math-container">$\sum X_i$</span> is a sufficient statistic for <span class="math-container">$\theta$</span> because <span class="math-container">$e^{{-1 \over 2}\sum(x_i-\theta)^2}$</span> depends on X only through the values of <span class="math-container">$\sum X_i$</span> right? Because if I know the value of <span class="math-container">$\sum X_i$</span> then I know <span class="math-container">$\sum X_i^2$</span> as well.</p>
<p>For part b)</p>
<p>Expanding the joint p.d.f as <span class="math-container">$$\frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\sum(x_i-\theta)^2} = \frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\sum(x_i- \bar x + \bar x-\theta)^2} = \frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\Big[\sum(x_i- \bar x)^2+n(\bar x-\theta)^2\Big]} = \frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\Big[{\sum(x_i- \bar x)^2 \over n-1}n-1+n(\bar x-\theta)^2\Big]}$$</span></p>
<p>Now can I say <span class="math-container">$S_n^2$</span> is a sufficient statistic for <span class="math-container">$\theta$</span> . Is it a problem that I have <span class="math-container">$\bar x$</span> in the function <span class="math-container">$g(S_n^2,\theta)$</span>?.<br />
Because <span class="math-container">$\bar x$</span> is a particular value I thought <span class="math-container">$g(S_n^2,\theta)$</span> depends on <span class="math-container">$\theta $</span> only through the values of <span class="math-container">$S_n^2$</span>.</p>
| Henry | 6,460 | <p>(a) Taking your joint probability density of $\frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\sum(x_i-\theta)^2}$, you can expand this into $$\left(\frac{1}{(2\pi)^{n/2}}e^{-\sum x_i^2 /2}\right)\left(e^{-n\theta^2/2+\theta \sum x_i }\right)$$ where the left part does not depend on $\theta$ and the right part is a function of $\theta$ and $\sum x_i$, implying by Fisher's factorisation theorem that $\sum x_i$ is a sufficient statistic for $\theta$</p>
<p>(b) $S_n^2$ (you do not say, but presumably the sample variance, or possibly the sample second moment about $0$ or perhaps $\sum x_i^2$) is not a sufficient statistic for $\theta$. One way of seeing this is that multiplying all the $x_i$ observations by $-1$ would not change $S_n^2$, and so it cannot give any information to distinguish between the population mean of the original normal distribution being $\theta$ or being $-\theta$</p>
|
3,767,421 | <p>This is a rather soft question.</p>
<p><strong>My understanding:</strong></p>
<p>Suppose we have <span class="math-container">$x \in ℝ$</span> and <span class="math-container">$x^2 = -1$</span> [in the normal interpretation].</p>
<p>Then the statement "<em>there exists <span class="math-container">$r \in ℝ$</span> such that <span class="math-container">$r^2 = -1$</span></em>" is true.</p>
<p>This is because <span class="math-container">$x \in ℝ$</span> and <span class="math-container">$x^2 = -1$</span> form a contradiction, and under contradictory settings any statement follows. That is, in an inconsistent system, any statement is true. [UPDATE: this should be "any statement can be proved" as pointed out in the following answers/comments]</p>
<p><strong>My question:</strong></p>
<p>So, is that still valid to say that the negation of a true statement in an inconsistent system is false? If yes, then we would have any statement in an inconsistent system is simultaneously true and false. [UPDATE: this implication is actually wrong and has been corrected in the following answers/comments]</p>
<p>Or do we rather leave <em>false</em> to be <em>undefined</em> in inconsistent systems? (since I think the definition of <em>false</em> is to some extent redundant in such systems)</p>
<p><strong>Motivation:</strong></p>
<p>I am thinking about what does it actually mean when we say some statement is <em>true</em>.</p>
<p>In a vacuous implication, we say that the premise is <em>false</em>. However, for example, when we are using proof by contradiction to <em>test</em> if a statement is false, we actually treat the statement as if it is a true statement <em>until</em> we hit a contradiction, and then conclude that the statement is false, under the given settings. In other words, a statement is not necessary to be false if we don’t expect consistent system in the first place.</p>
| Mauro ALLEGRANZA | 108,274 | <p>There are different (equivalent) definitions of <a href="https://en.wikipedia.org/wiki/Consistency" rel="nofollow noreferrer">consistency</a>.</p>
<p>Basically, an <a href="https://en.wikipedia.org/wiki/Consistency#Definition" rel="nofollow noreferrer">inconsistent system</a> is a system that proves a sentence <span class="math-container">$\varphi$</span> and its negation <span class="math-container">$¬ \varphi$</span>.</p>
<p>If so, due to the fact that the negation of a True sentence is False, and vice versa, an inconsistent system is a system that proves True sentences as well as False ones.</p>
<p>Thus,</p>
<blockquote>
<p>YES, we have false statement in inconsistent systems.</p>
</blockquote>
<p>Regarding your example, we assume that we know <em>facts</em> about <em>real</em> numbers (i.e. mathematical objects whose collection is named with <span class="math-container">$\mathbb R$</span>), where for simplicity I'll equate a "mathematical fact" with the content expressed by a mathematical theorem.</p>
<p>It is a theorem that, for every real number <span class="math-container">$r : r^2 \ge 0$</span>.</p>
<p>This means that if we can prove that, for some real <span class="math-container">$x$</span>, we have <span class="math-container">$x^2=-1$</span>, this fact contradicts the above theorem.</p>
<p>This amounts to having found an inconsistency in the system we have used to prove it.</p>
<blockquote>
<p>Does it mean that every statement in an inconsistent system is simultaneously true and false?</p>
</blockquote>
<p>If we agree that there are mathematical objects called (real) numbers and there are objective facts regarding them that we can "discover" through proofs in a suitable system describing them, we accept the <a href="https://en.wikipedia.org/wiki/Principle_of_bivalence" rel="nofollow noreferrer">"classical" concept of Truth</a> and thus we cannot have statements that are both True and False.</p>
<p>Thus, if we have an inconsistent theory of real numbers, i.e. a system that proves both a statement <span class="math-container">$\varphi$</span> and its negation <span class="math-container">$\lnot \varphi$</span>, we have to conclude that the system is a wrong description of the reals and we have to fix it (as happened already in the past).</p>
<hr />
<p><em>References</em>:</p>
<ul>
<li><p>Jan Wolenski, <a href="https://books.google.it/books?id=L5PHDwAAQBAJ&pg=PA1" rel="nofollow noreferrer">Semantics and Truth (2019, Springer)</a></p>
</li>
<li><p>Stewart Shapiro, <a href="https://pdfs.semanticscholar.org/bd1c/6420eeb215a607f50a7ec165fd2cc66d85fb.pdf" rel="nofollow noreferrer">We hold these truths to be self-evident: But what do we mean by that? (<em>RevSL</em>,2009)</a></p>
</li>
</ul>
|
594,498 | <p>Let $U$ be a linear subspace of $V$. Show that $\dim_K(U)\leq \dim_K(V)$ and conclude, that $\dim_K(U)= \dim_K(V) \Leftrightarrow U = V$. Is the equivalence $\dim_K(U)= \dim_K(V) \Leftrightarrow U = V$ also true for $\dim_K(V)= \infty$?</p>
<p>I've proved so far that $\dim_K(U)= \dim_K(V) \Leftrightarrow U = V$,but i have no idea, if this is still true for $\dim_K(V)= \infty$. I have problems to show that $\dim_K(U)\leq \dim_K(V)$ (doesn't this follow by definition?)</p>
| vadim123 | 73,324 | <p>Let $B_U$ be a basis for $U$. Then $B_U$ is an independent subset of $V$, and hence (assuming that $V$ has a basis) may be extended to a basis for $V$, which has cardinality at least as large as $|B_U|=dim_K(U)$.</p>
<p>If both vector spaces are infinite-dimensional it need not be true that they are equal. For example, consider $V=P(t)$, the set of all polynomials in $t$, and $U=tP(t)$, the set of all polynomials in $t$ with constant term equal to zero.</p>
|
2,249,036 | <p>I was watching a video on PDEs and when arriving at the part of Fourier Series, the professor said:</p>
<blockquote>
<p>And one of the most fascinating reads I ever had was a paper by Riemann on the history of this [Fourier Series].</p>
</blockquote>
<p>I tried looking for it but didn't succeed, and I was wondering if anyone could provide a link or a reference to this paper.</p>
<p>Thank you very much.</p>
<p>In case you need it, the link of the video is: <a href="https://www.youtube.com/watch?v=cf8rgx60IKA" rel="nofollow noreferrer">https://www.youtube.com/watch?v=cf8rgx60IKA</a> on the minute 14:21 (The audio on that specific part is messed up, so be careful if you are using headphones).</p>
| angryavian | 43,949 | <p>If you have two vectors $v=(v_1,v_2)$ and $w=(w_1,w_2)$, then their dot product is $$v \cdot w = v_1 w_1 + v_2 w_2.$$ Additionally, we also have
$$v \cdot w = \|v\| \|w\| \cos \theta$$ where $\|v\|=\sqrt{v_1^2+v_2^2}$, $\|w\|=\sqrt{w_1^2+w_2^2}$, and $\theta$ is the angle between $v$ and $w$. <a href="https://proofwiki.org/wiki/Cosine_Formula_for_Dot_Product" rel="nofollow noreferrer">Proof here</a>.</p>
<p>Now, consider $v=(\cos \alpha, \sin \alpha)$ and $w=(\cos \beta, \sin \beta)$. Do you now see why the formula for $\cos(\alpha-\beta)$ then follows? Hint: we have $\|v\|=\|w\|=1$ and $\theta=\alpha-\beta$.</p>
<hr>
<p>For your second question,note $\cos(\alpha+\beta)=\cos(\alpha-(-\beta))$,
then apply your formula for the cosine of a difference of angles, and then use the fact that $\sin$ is an odd function, and $\cos$ is an even function. This has been done for you in the comments.</p>
|
839,764 | <p>I am wondering how to show the following proposition</p>
<p><strong>If A, B, and C are set and $A \subset B ,B \subset C \rightarrow A \subset C$</strong></p>
<p><strong>My proof</strong> </p>
<p>Let x be any integer.</p>
<p>If $x\in A$ then $x\in B$ because all element of A are in B. If $x \in B \rightarrow x\in C$ because all element of B are in C.</p>
<p>Since B contains A and C contains B, then C contains A logically.</p>
<p>the end</p>
| amWhy | 9,003 | <p>You don't need the last sentence. You simply draw the conclusion that $x \in A\implies x \in C $, and so $A\subset C$.</p>
<p>Polishing it up a tad:</p>
<p>Let $x$ be any element in $A$. Then, $x \in A\implies x\in B$ because $A \subset B$. Then $x \in B \implies x\in C$, since $B\subset C$.</p>
<p>Therefore, $$x\in A \implies x \in C$$ and by definition of a subset, $A\subset C$.</p>
|
839,764 | <p>I am wondering how to show the following proposition</p>
<p><strong>If A, B, and C are set and $A \subset B ,B \subset C \rightarrow A \subset C$</strong></p>
<p><strong>My proof</strong> </p>
<p>Let x be any integer.</p>
<p>If $x\in A$ then $x\in B$ because all element of A are in B. If $x \in B \rightarrow x\in C$ because all element of B are in C.</p>
<p>Since B contains A and C contains B, then C contains A logically.</p>
<p>the end</p>
| beep-boop | 127,192 | <p>$$\underbrace{x \in A \implies x \in B}_{\text{since} \ A \subset B} \implies x \in C \ \ (\text{since } B \subset C),$$ so ( by the transitive property of implications) $x \in A \implies x \in C$ for any $x$, so $A \subset C $.</p>
<p>$\square$</p>
<p>A couple of notes: $x$ does not have to be an integer; it doesn't even have to be a number. </p>
<p>$A$ could be the set of all people in the world and $x$ could be an arbitrary person.</p>
<p>This proof shows that, for any <em>element</em> $x$ which is in $A$, $x$ is automatically in $C$.</p>
<p>Also, don't finish with "the end"- it's not a fairy tale!</p>
<p>A nice, concise way to end a proof is with a $\square$.</p>
|
2,786,138 | <blockquote>
<p>Find all triples of prime numbers <span class="math-container">$(p,q,r)$</span> such that <span class="math-container">$$p^q+q^r=r^p.$$</span></p>
</blockquote>
<p>I proved that when <span class="math-container">$r=2$</span>, the equation becomes <span class="math-container">$$p^q+q^2=2^p.$$</span> Then I tried to use reciprocity laws and Fermat's little theorem. I could prove that <span class="math-container">$p\equiv 7\pmod 8$</span> and that <span class="math-container">$p>q$</span>.</p>
<p>The equation appeared in some olympiad . They asked to prove that <span class="math-container">$r=2$</span>. So I am trying to find at least one triple.</p>
| Servaes | 30,382 | <p>It is clear that precisely one of $p$, $q$ and $r$ must equal $2$. We'll first show that in fact $r=2$:</p>
<hr>
<p><strong>Observation 1:</strong> $r=2$.</p>
<p>If $q=2$ then reducing mod $3$ shows that
$$p^2+2\equiv r^p\pmod{3},$$
so either $p=3$ or $r=3$. Both are easily verified to be impossible.</p>
<p>If $p=2$ then
$$2^q+q^r=r^2,$$
but for all primes $q,r>2$ we have $q^r>r^2$, a contradiction. </p>
<hr>
<p>This leaves us with finding odd primes $p$ andd $q$ satisfying
$$p^q+q^2=2^p.$$
We first make a few more simple observations.</p>
<hr>
<p><strong>Observation 2:</strong> $q^2\equiv2\pmod{p}$.</p>
<p>By Fermat's little theorem
$$q^2\equiv p^q+q^2\equiv2^p\equiv2\pmod{p}.$$</p>
<p><strong>Observation 3:</strong> $q<p$.</p>
<p>As $p>2$ we have $2^q<p^q+q^2=2^p$, so $q<p$.</p>
<hr>
<p>The problem can be rephrased in the number ring $\Bbb{Z}[\sqrt{2}]$. This is a unique factorization domain with unit group $\{\pm1\}$. Setting $m:=\frac{p-1}{2}$ we can rewrite the equation above as
$$p^q=-(q+2^m\sqrt{2})(q-2^m\sqrt{2}).$$
The greatest common divisor of the two factors on the right hand side divides both $2q$ and $p^q$. Because $p$ and $q$ are distinct, these two factors are coprime and so both are $q$-th powers in $\Bbb{Z}[\sqrt{2}]$. Let $a,b\in\Bbb{Z}$ be such that
$$q+2^m\sqrt{2}=(a+b\sqrt{2})^q.$$
Then $q-2^m\sqrt{2}=(a-b\sqrt{2})^q$ and hence
$$p^q=-(a+b\sqrt{2})^q(a-b\sqrt{2})^q=(2b^2-a^2)^q,$$
which shows that $p=2b^2-a^2$. </p>
<hr>
<p><strong>Observation 4:</strong> $a=\pm q$ and $b\mid2^m$. <em>(Thanks to barto's comment)</em></p>
<p>Set $\alpha:=a+b\sqrt{2}$ and $\bar{\alpha}:=a-b\sqrt{2}$. Because
$$(\alpha+\bar{\alpha})\mid(\alpha^2+\bar{\alpha}^q)\qquad\text{ and }
\qquad(\alpha-\bar{\alpha})\mid(\alpha^2-\bar{\alpha}^q),$$
we see that $2a\mid2q$ and $2b\sqrt{2}\mid2^{m+1}\sqrt{2}$. The binomial expansion
$$q+2^m\sqrt{2}=(a+b\sqrt{2})^q,$$
shows that $q\mid a^q$, and hence that $q\mid a$ so $a=\pm q$.</p>
<p><strong>Observation 5:</strong> $b=\pm1$.</p>
<p>By observation 3 we have $a^2=q^2<p^2$, and so
$$p=2b^2-a^2>2b^2-p^2.$$
Because $p>q>2$ we have $p\geq5$ and so $b^2<\frac{p^2+p}{2}\leq(p-1)^2$.
Also, by observation 2
$$p=2b^2-a^2=2b^2-q^2\equiv2b^2-2\pmod{p},$$
and so $b^2\equiv1\pmod{p}$. This means $b=\pm1$.</p>
<hr>
<p>We have $a=\pm q$ and $b=\pm1$ and so
$$p=2b^2-a^2=2-q^2<0,$$
a contradiction.</p>
|
2,696,097 | <p>$ \lim_{n\to \infty} ( \lim_{x\to0} (1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} $</p>
<p>The answer should be $ {e}^\frac{1}{3} $</p>
<p>I haven't encountered problems like this before and I'm pretty confused, thank you.</p>
<p>I guess we must use the remarkable limit of $ \frac{\tan(x)}{x} $ when $ x $ approaches $ 0 $ by dividing and then multiplaying with $x$.</p>
| Nicky Hekster | 9,605 | <p>In general, if $H$ is a <em>proper</em> subgroup of a group $G$, then $\langle G\backslash H \rangle = G$, that is, the set-theoretic complement generates the whole group. This follows from the fact that a group cannot be the union of two proper subgroups (I leave the proof to you). </p>
|
43,640 | <p>Consider the following problem:</p>
<p>Let ${\mathbb Q} \subset A\subset {\mathbb R}$, which of the following must be true?</p>
<p>A. If $A$ is open, then $A={\mathbb R}$</p>
<p>B. If $A$ is closed, then $A={\mathbb R}$</p>
<p>Since $\overline{\mathbb Q}={\mathbb R}$, one can immediately get that B is the answer. </p>
<p>Here are my questions:</p>
<blockquote>
<p>Why A is not necessarily true? What can be a counterexample?</p>
</blockquote>
| Luboš Motl | 10,599 | <p>A counterexample for the rule A is
$$ {\mathbb R} \backslash F $$
where $F$ is any non-empty finite (or countable) set of irrational numbers. For example
$$ {\mathbb R} \backslash \{\pi\} $$
Note that if I remove the point $\pi$, the set is still open on both sides from $\pi$. Because $\pi$ isn't rational, the set above still contains all rational numbers.</p>
|
292,780 | <p>I have to realize an NFA that recognizes the language of strings on the alphabet {a, b} ending with: bb, ba, baa. I thought that there must be the following states:</p>
<p>$q_0$: the string ends with bb.</p>
<p>$q_1$: the string ends with ba.</p>
<p>$q_2$: the string ends with baa.</p>
<p>Is right the definition of the states? Or missing some state?</p>
<p>Thanks.</p>
| vonbrand | 43,946 | <p>You need at least a start state, which loops back on each character, An then you need states for, e.g., $q_2 \rightarrow q_2' \rightarrow q_2'' \rightarrow q_2'''$ ($q_2$ means "$baa$ still to go", $q_2'$ is "want $aa$", $q_2''$ means "need $a$", and $q_2'''$ is final). [Sorry, would put symbols over the arrows if I knew how. Perhaps some kind editor does?.]</p>
|
3,346,543 | <p>I am aware this is a pretty big topic, but the attempts at layman's explanations I have seen either barely provide commentary on the formal proofs, or fail to provide an explanation (e.g "it gets too complex" does not really say anything)</p>
<p>Is there a good intuitive explanation as to why we fail to obtain a general solution for a 5th+ degree polynomial, and why this happens at the 5th degree and not below or above?</p>
| Moishe Kohan | 84,907 | <p>See this detailed sketch of Arnold's proof of Abel-Ruffini theorem here: web.williams.edu/Mathematics/lg5/394/ArnoldQuintic.pdf. It is intended for a layman who can handle some basic group theory (and a bit of topology). It still takes about 5 pages.</p>
|
2,777,051 | <p>When talking about operator semigroups, some books use the notation $(e^{tA})_{t \geq 0}.$ Doesn't this create an ambiguity? Why are results proved for this semigroup instead of taking a general semigroup, say $(T(t))_{t \geq 0}$? </p>
| David C. Ullrich | 248,223 | <p>The two are equivalent under weak continuity assumptions. In particular if $(T(t))$ is a semigroup and $||T(t)-I||\to0$ as $t\to0$ then there exists a bounded $A$ such that $T(t)=e^{tA}$.</p>
<p><strong>Edit:</strong> Or so I'd read. Too stubborn to look at the proof - I think this works:</p>
<p>If $h>0$ is small enough then $$||I-\frac1h\int_0^h T(t)dt||<1,$$so $\frac1h\int_0^hT(t)dt$ is invertible.</p>
<p>Since $T$ is a semigroup it seems like the standard method of evaluating a finite geometric series should say something about that integral. Say $J=\int_0^h T(t)dt$, just because "$I$" is taken. For small $\epsilon>0$ we have $$ (T(\epsilon)-I)J=\int_h^{h+\epsilon}T-\int_0^\epsilon T$$. So $$\lim_{\epsilon\to0^+}\frac{T(\epsilon)-I}{\epsilon}=(T(h)-I)J^{-1}:=A.$$</p>
<p>Hence the semigroup property shows that $$D_RT(t)=AT(t),$$where $D_R$ denotes the right-hand derivative. It's easy to see that this implies that the two-sided derivative exists, but never mind that. The product rule shows that $$D_R(T(t)e^{-tA})=0,$$hence $T(t)e^{-tA}$ is constant, hence $$T(t)=e^{tA}.$$</p>
<p><strong>Heh:</strong> That has very little to do with operator theory; in fact the same argument works if $B$ is a Banach algebra with identity and $T:(0,\infty)\to B$ is a continuous semigroup.</p>
|
497,687 | <p>How would I integrate the following.</p>
<p>$$
\int_{\ln2}^{\ln3}\frac{e^{-x}}{\sqrt{1-e^{-2x}}}\,dx
$$ </p>
<p>I think I have to use the $\arcsin(x)$ formula.</p>
<p>Which means would I use $y=u^2$ with $u=e^{-x}$ but that does not seem to work.</p>
| Norbert | 19,538 | <p>It does work
$$
\int_{\ln 2}^{\ln 3}\frac{e^{-x}}{\sqrt{1-e^{-2x}}}dx
\overset{\quad u=e^{-x}}{=}
-\int_{1/2}^{1/3}\frac{du}{\sqrt{1-u^2}}dx
$$
$$
=-\arcsin u\Biggl|_{1/2}^{1/3}
=-\arcsin(1/3)+\frac{\pi}{6}
$$</p>
|
991,569 | <p>\begin{align}
x-\hphantom{2}y+2z+2t&=0 \\
2x-2y+4z+3t&=1 \\
3x-3y+6z+9t&=-3 \\
4x-4y+8z+8t&=0
\end{align}</p>
<p>Solve for x,y,z and t</p>
<p><img src="https://i.stack.imgur.com/y3R7x.jpg" alt="reduced row echelon form"></p>
| Community | -1 | <p>Then treat any two variables as constants and express other two in terms of those constants</p>
<p>For Example if you keep $x$ and $y$ as constants then
$$z=\frac{y-x+2}{2}$$
$$t=-1$$</p>
|
3,567,245 | <p>I am trying to compute <span class="math-container">$\int_{C}Log(z+3)$</span>, where C is a circle centered at the origin, with radius of 2, oriented once counterclockwise. </p>
<p>I kind of get the idea of how to compute <span class="math-container">$\int Log(z)$</span> on the unit circle. I know <span class="math-container">$z=e^{i \theta}$</span> and <span class="math-container">$dz=ie^{i\theta}d\theta$</span> gives</p>
<p><span class="math-container">$\int Log(e^{i\theta})ie^{i\theta}d\theta = \int -\theta e^{i \theta}d \theta$</span></p>
<p>and it follows from there, but I am not sure if it is the same approach for <span class="math-container">$Log(z+3)$</span> or for a circle of radius 2. So far I am thinking <span class="math-container">$z=2e^{i \theta}$</span> since the radius is 2 but that's as far as I got. Do I take the same approach? Computing </p>
<p><span class="math-container">$\int_{C}Log(2e^{i \theta}+3)2ie^{i \theta}d \theta$</span> </p>
<p>if so I have no idea where to go from there.</p>
<p>Thanks :)</p>
| David Melo | 662,316 | <p>You can use the same method, but from there you're gonna need to split your interval into injective portions of a substitution and then calculate it by substitution of variables noting that the exponential function is holomorphic.</p>
<p>Since you're computing a contour integral of a "well behaved" function you're just gonna end up at <span class="math-container">$0$</span>. (Cauchy's Integral Theorem), since the derivative <span class="math-container">$\frac{d}{dz}Log(z+3) = \frac{1}{z+3}$</span> with only pole at <span class="math-container">$z=-3$</span> which is not on the region delimited by your contour.</p>
<p>If you're a math student I suggest reading a bit on substitution of variables for complex integrals and Cauchy's Integral Theorem.</p>
|
2,745,436 | <p>For some reason I have been struggling with this problem for the past couple hours.</p>
<p><a href="https://i.stack.imgur.com/diFLe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/diFLe.png" alt="enter image description here"></a></p>
<p>I believe I have solved part a.</p>
<p>Since there are 6 states (assuming a standard die and the die is fair), then there is an equal chance of landing in any of the 6 states for every dice throw. Therefore I concluded that the transition matrix looks as such:</p>
<p><code>[1 0 0 0 0 0]
[0 1 0 0 0 0]
[0 0 1 0 0 0]
[0 0 0 1 0 0]
[0 0 0 0 1 0]
[0 0 0 0 0 1]</code></p>
<p>For part b I setup my starting matrix as:</p>
<p>s1 = </p>
<p><code>[1]
[0]
[0]
[0]
[0]
[0]</code></p>
<p>I multiplied my transition matrix by <code>s1</code> 5 times and arrived at the same vector <code>s1</code>. Arithmetically, this makes sense. However, logically it does not.</p>
<p>Can somebody help explain where I went wrong?</p>
<p><strong>Edit:</strong> Based on the answers I believe the transition matrix would be:</p>
<p><code>[1/6 0 0 0 0 0]
[1/6 2/6 0 0 0 0]
[1/6 1/6 3/6 0 0 0]
[1/6 1/6 1/6 4/6 0 0]
[1/6 1/6 1/6 1/6 5/6 0]
[1/6 1/6 1/6 1/6 1/6 1]</code></p>
| Arnaud Mortier | 480,423 | <p><strong>Hint.</strong> The entry $m_{i,j}$ in the transition matrix is the probability to go to state $i$ at the next throw given that you are currently in state $j$. Therefore it is lower triangular, but not quite the identity.</p>
<p>E.g. if you are in state $4$ you have a $4/6$ chance to stay in state $4$. Therefore $m_{4,4}=4/6$.</p>
|
2,343,216 | <p>Let $\operatorname{gd}(x)$ the Gudermannian function, defined as in this <a href="http://mathworld.wolfram.com/Gudermannian.html" rel="nofollow noreferrer">MathWorld's article</a>, and implemented in Wolfram Language as <em>Gudermannian[x]</em>. </p>
<p>This afternoon I've spent an hour playing with <a href="https://www.wolframalpha.com" rel="nofollow noreferrer">Wolfram Alpha online calculator</a> about integrals, for instance, like this $$I=\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx.$$
Notice that the plot in previous article of MathWorld tell us that for $0<x<1$ one has $0<\operatorname{gd}(x)<1$. </p>
<p>My calculations to get an approximation were $$I\approx\int_0^1 \operatorname{gd}(x)\,dx=-\frac{\pi}{2}+i\left(\operatorname{Li_2}(-ie)-\operatorname{Li_2}(ie)\right)-2C,$$ where I've combined with the indefinite integral of the Gudermannian functions that tell us MathWorld. Here $C$ is Catalan's constant. But this quantity is about $\approx 0.464065$.</p>
<p>But WA calculated $I$ as $0.365619$, play this code:
<code>integrate log(1+(gd(x)))dx, from x=0 to x=1</code></p>
<blockquote>
<p><strong>Question.</strong> Can you provide us a better approximation, than mine, for our integral $$\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx?$$
Please provide also the more important justifications in your calculations. Also if your approach uses the numerical analysis and/or calculations with your computer provide us some detail of your calculation/method. <strong>Thanks in advance.</strong></p>
</blockquote>
| wythagoras | 236,048 | <p>The Gudermannian function is lower-bounded by the line going to $(0,0)$ and $(1,\operatorname{gd}(1))$. However, it doesn't make a very large difference, and it hence gives a reasonable approximation. </p>
<p>See, for example, a plot here on <a href="https://www.wolframalpha.com/input/?i=(Gudermannian%5B1%5D-Gudermannian%5B0%5D)x,+Gudermannian%5Bx%5D+from+x%3D0+to+x%3D1%7D" rel="nofollow noreferrer">Wolfram Alpha</a>.</p>
<p>Now note that </p>
<p>\begin{align*} \int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx &> \int_0^1\log\left(1+\operatorname{gd}(1)x\right)\,dx \\ &= \left[ x \log(\operatorname{gd}(1) x + 1) + \frac{\log(\operatorname{gd}(1) x + 1)}{\operatorname{gd}(1)} - x\right]^1_0 \\ & = \log(\operatorname{gd}(1)+ 1) +\frac{\log(\operatorname{gd}(1)+ 1)}{\operatorname{gd}(1) } - 1 \\ &\approx 0.344043 \end{align*}</p>
<hr>
<p>Even more simply, one can approximate the integral by $$\frac{\log(1+\operatorname{gd}(1))+\log(1+\operatorname{gd}(0))}{2} = \frac{\log(1+\operatorname{gd}(1))}{2} \approx 0.31184$$</p>
|
2,343,216 | <p>Let $\operatorname{gd}(x)$ the Gudermannian function, defined as in this <a href="http://mathworld.wolfram.com/Gudermannian.html" rel="nofollow noreferrer">MathWorld's article</a>, and implemented in Wolfram Language as <em>Gudermannian[x]</em>. </p>
<p>This afternoon I've spent an hour playing with <a href="https://www.wolframalpha.com" rel="nofollow noreferrer">Wolfram Alpha online calculator</a> about integrals, for instance, like this $$I=\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx.$$
Notice that the plot in previous article of MathWorld tell us that for $0<x<1$ one has $0<\operatorname{gd}(x)<1$. </p>
<p>My calculations to get an approximation were $$I\approx\int_0^1 \operatorname{gd}(x)\,dx=-\frac{\pi}{2}+i\left(\operatorname{Li_2}(-ie)-\operatorname{Li_2}(ie)\right)-2C,$$ where I've combined with the indefinite integral of the Gudermannian functions that tell us MathWorld. Here $C$ is Catalan's constant. But this quantity is about $\approx 0.464065$.</p>
<p>But WA calculated $I$ as $0.365619$, play this code:
<code>integrate log(1+(gd(x)))dx, from x=0 to x=1</code></p>
<blockquote>
<p><strong>Question.</strong> Can you provide us a better approximation, than mine, for our integral $$\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx?$$
Please provide also the more important justifications in your calculations. Also if your approach uses the numerical analysis and/or calculations with your computer provide us some detail of your calculation/method. <strong>Thanks in advance.</strong></p>
</blockquote>
| Claude Leibovici | 82,404 | <p>For sure, you could use Taylor expansion
$$\log (1+\text{gd}(x))=x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{12}+\frac{3 x^5}{40}-\frac{x^6}{18}+\frac{169
x^7}{5040}-\frac{19 x^8}{840}+\frac{2171 x^9}{120960}+O\left(x^{10}\right)$$ and integrate which would lead to $$\int_0^1 \log (1+\text{gd}(x))\,dx=\frac{443161}{1209600}\approx 0.366370$$ Continuing the expansion to $O\left(x^{20}\right)$ would give $$\int_0^1 \log (1+\text{gd}(x))\,dx=\frac{35582823156912797}{97316080327065600}\approx 0.365642$$ May be, better would be to use Padé approximants; a very simple would be
$$\log (1+\text{gd}(x))\approx \frac{x-\frac{x^2}{2}}{1-\frac{x^2}{6}}$$ and the integration would lead to $$\int_0^1 \log (1+\text{gd}(x))\,dx\approx 3+3 \log \left(\frac{6}{5}\right)-3 \sqrt{6} \tanh ^{-1}\left(\frac{1}{\sqrt{6}}\right)\approx 0.361349$$ Using the next one
$$\log (1+\text{gd}(x))\approx \frac{-\frac{17 x^3}{60}+\frac{6 x^2}{5}+x}{\frac{2 x^2}{5}+\frac{17 x}{10}+1}$$ we can still integrate and get a bunch of logarithms for a value $\approx 0.364644$.</p>
<p>Just for the fun, I put in the next table the decimal value of the integral of the $[n,2]$ Padé approximant
$$\left(
\begin{array}{cc}
n & \text{result} \\
2 & 0.3613491513 \\
3 & 0.3646438085 \\
4 & 0.3661395229 \\
5 & 0.3657879040 \\
6 & 0.3654692466 \\
7 & 0.3655746142 \\
8 & 0.3656449849 \\
9 & 0.3656298359 \\
10 & 0.3656072339
\end{array}
\right)$$</p>
|
693,550 | <p>Consider the function $F(x) = x^2-2x+2$.</p>
<p>Find an interval in which the function is contractive and find the fixed point in this interval.</p>
<p>What is the convergence rate of the fixed point iteration: $x_{n+1} = F(x_n)$ in that interval?</p>
<p>I'm lost on fixed point iteration. I've watched a few youtube videos, but I'm still not understanding contractive intervals and methods for finding fixed points in functions.</p>
| Mhenni Benghorbal | 35,472 | <p>You need to consider the recurrence relation</p>
<p>$$ x_{n+1}= x_{n}^2-x_n+2 ,$$</p>
<p>which suggests considering the function</p>
<p>$$f(x) = x^2-x+2 \implies f'(x) = 2x-1. $$</p>
<p>Now, to find the interval, use the condition $|f'(x)|<1$. See a <a href="https://math.stackexchange.com/questions/267648/prove-that-a-function-having-a-derivative-bounded-by-0-49-has-a-unique-solution/267677#267677">related problem</a>.</p>
|
693,550 | <p>Consider the function $F(x) = x^2-2x+2$.</p>
<p>Find an interval in which the function is contractive and find the fixed point in this interval.</p>
<p>What is the convergence rate of the fixed point iteration: $x_{n+1} = F(x_n)$ in that interval?</p>
<p>I'm lost on fixed point iteration. I've watched a few youtube videos, but I'm still not understanding contractive intervals and methods for finding fixed points in functions.</p>
| Prasenjit D Wakode | 429,349 | <p>Try the form x(n+1)=(2x(n)-2)^(0.5) or x(n+1)=i(-2x(n)+2)^(0.5) of given equation. This converges to (1+i) which is one of the root of given equation. You note here that except infinities, any real or complex number as initial guess x(0) will converge x(n+1) to (1+i) .</p>
|
1,665,107 | <p>I must use a generating function to solve this question: </p>
<p>In how many ways can you collect six dollars from eight people if six people give either $0$ or $1$ dollars and the other two people each give $0$, $1$, or $5$ dollars?</p>
<p>Here is what I have so far.</p>
<p>The generating function is $(1+x)^6(1+x+x^5)^2$. Then let $f(x)= (1+x)^6$, and $g(x) = (1+x+x^5)^2$. We want the coefficient of $x^6$. Then, we note f(x) has the expansion:</p>
<p>$(1+x)^6$ = $1$ + $6 \choose 1$$x$ + $6 \choose 2$$x^2$ + ... + $6 \choose 6$$x^6$</p>
<p>But here is where I get stuck. We can let $h(x) = f(x)g(x)$ where $a_0b_0 + (a_1b_0+a_0b_1)x + (a_2b_0+ a_1b_1 + a_0b_2)x^2 + ... + (a_rb_0+ a_{r-1}b_1 + a_{r-2}b_2 + ... + a_0b_r)x^r$,</p>
<p>But I can't figure out the expansion of $g(x) = $$(1+x+x^5)^2$.</p>
<p>Can someone help me find the expansion of $g(x)$, and help me solve the problem?</p>
| vonbrand | 43,946 | <p>The generating function for each of the first class of people is <span class="math-container">$1 + z$</span>, the second class <span class="math-container">$1 + z + z^5$</span>. In all, the generating function of the number of ways to collect <span class="math-container">$n$</span> dollars in all is:</p>
<p><span class="math-container">$\begin{equation*}
(1 + z)^6 (1 + z + z^5)^2
\end{equation*}$</span></p>
<p>A computer algebra system (yes, I'm a lazy bum) tells you the coefficient of <span class="math-container">$z^8$</span> is 71.</p>
|
3,517,722 | <p>Let <span class="math-container">$\alpha:= \sqrt[7]{2},\omega:= e^{\frac{2\pi i }{7}}\in \Bbb C$</span>. We set <span class="math-container">$E:=\Bbb Q(\alpha,\omega)$</span> and <span class="math-container">$B:=\Bbb Q(\omega+\omega^2+\omega^4)\leq E$</span>. </p>
<p>Thus, we have the Tower of Fields
<span class="math-container">$$\Bbb Q \leq B \leq E.$$</span>
We can prove that <span class="math-container">$|\mathrm{Aut}(E/\Bbb Q)|=42$</span> and that <span class="math-container">$\theta,\sigma\in \mathrm{Aut}(E/\Bbb Q)$</span>, given by
<span class="math-container">$$\theta:\alpha \longmapsto \alpha \omega,\ \omega \longmapsto \omega$$</span>
<span class="math-container">$$\sigma: \alpha \longmapsto \alpha,\ \omega \longmapsto \omega^3.$$</span></p>
<p><strong><em>Question:</em></strong> How can we prove that <span class="math-container">$\mathrm{Aut}(E/B)=\langle \theta,\sigma^2\rangle$</span>?</p>
<p>A first thought is to compute the Galois group <span class="math-container">$\mathrm{Aut}(E/B)$</span>, but this seems to be extremely hard, because of <span class="math-container">$B$</span>. Another might be to use somehow the Fundamental Theorem of Galois Theory, so to exploit the relations <span class="math-container">$[E:B]=|\mathrm{Aut}(E/B)|$</span> and then find the order of the subgroup <span class="math-container">$\langle \theta,\sigma^2 \rangle$</span>.</p>
<p>Note that <span class="math-container">$\theta^i(\alpha)=\alpha\omega^i$</span> and <span class="math-container">$\sigma^j(\omega)=\omega^{3^j}$</span>.</p>
| Jyrki Lahtonen | 11,619 | <p>Below please find a few pieces that help you to onoe way of proving the claims.</p>
<ul>
<li>The calculation <a href="https://math.stackexchange.com/a/299417/11619">here</a> reveals that <span class="math-container">$B=\Bbb{Q}(\sqrt{-7})$</span>. Therefore <span class="math-container">$[B:\Bbb{Q}]=2$</span>, and you know that <span class="math-container">$[E:B]=21$</span>.</li>
<li>You can easily verify that <span class="math-container">$\theta$</span> and <span class="math-container">$\sigma^2$</span> leave <span class="math-container">$B$</span> fixed pointwise, so they belong to <span class="math-container">$Gal(E/B)$</span>.</li>
<li>The calculation showing that <span class="math-container">$G=Gal(E/\Bbb{Q})$</span> is generated by <span class="math-container">$\theta$</span> and <span class="math-container">$\sigma$</span> more specifically describes the Galois group as a semidirect product
<span class="math-container">$$G\simeq C_7\rtimes C_6,$$</span> where the normal subgroup <span class="math-container">$\simeq C_7$</span> is generated by <span class="math-container">$\theta$</span> and the non-normal factor by <span class="math-container">$\sigma$</span>. This follows also from the relation
<span class="math-container">$$\sigma\theta\sigma^{-1}=\theta^3$$</span> that you can verify by calculating their effects on <span class="math-container">$\alpha,\omega$</span>.</li>
<li>It is the easy to check that the subgroup generated by <span class="math-container">$\theta$</span> and <span class="math-container">$\sigma^2$</span> has order <span class="math-container">$21$</span> (see reuns's answer for a way of making that explicit), so it must be all of <span class="math-container">$Gal(E/B)$</span>.</li>
</ul>
<hr>
<p>Extras:</p>
<ul>
<li><span class="math-container">$\Bbb{Q}(\omega)/\Bbb{Q}$</span> is Galois with an <em>abelian</em> Galois group isomorphic to <span class="math-container">$\Bbb{Z}/7\Bbb{Z}^*\simeq C_6$</span> generated the restriction of <span class="math-container">$\sigma$</span>. Therefore the intermediate field <span class="math-container">$B$</span> must also be Galois over <span class="math-container">$\Bbb{Q}$</span>.</li>
<li>You see that
<span class="math-container">$$\sigma(\omega+\omega^2+\omega^4)=\omega^3+\omega^6+\omega^5=-1-(\omega+\omega^2+\omega^4)$$</span> implying that <span class="math-container">$\sigma$</span> does not fix <span class="math-container">$B$</span> element wise but <span class="math-container">$\sigma^2$</span> does. This also implies that <span class="math-container">$[B:\Bbb{Q}]=2$</span> without the calculation relating <span class="math-container">$B$</span> to <span class="math-container">$\sqrt{-7}$</span>.</li>
</ul>
|
3,517,722 | <p>Let <span class="math-container">$\alpha:= \sqrt[7]{2},\omega:= e^{\frac{2\pi i }{7}}\in \Bbb C$</span>. We set <span class="math-container">$E:=\Bbb Q(\alpha,\omega)$</span> and <span class="math-container">$B:=\Bbb Q(\omega+\omega^2+\omega^4)\leq E$</span>. </p>
<p>Thus, we have the Tower of Fields
<span class="math-container">$$\Bbb Q \leq B \leq E.$$</span>
We can prove that <span class="math-container">$|\mathrm{Aut}(E/\Bbb Q)|=42$</span> and that <span class="math-container">$\theta,\sigma\in \mathrm{Aut}(E/\Bbb Q)$</span>, given by
<span class="math-container">$$\theta:\alpha \longmapsto \alpha \omega,\ \omega \longmapsto \omega$$</span>
<span class="math-container">$$\sigma: \alpha \longmapsto \alpha,\ \omega \longmapsto \omega^3.$$</span></p>
<p><strong><em>Question:</em></strong> How can we prove that <span class="math-container">$\mathrm{Aut}(E/B)=\langle \theta,\sigma^2\rangle$</span>?</p>
<p>A first thought is to compute the Galois group <span class="math-container">$\mathrm{Aut}(E/B)$</span>, but this seems to be extremely hard, because of <span class="math-container">$B$</span>. Another might be to use somehow the Fundamental Theorem of Galois Theory, so to exploit the relations <span class="math-container">$[E:B]=|\mathrm{Aut}(E/B)|$</span> and then find the order of the subgroup <span class="math-container">$\langle \theta,\sigma^2 \rangle$</span>.</p>
<p>Note that <span class="math-container">$\theta^i(\alpha)=\alpha\omega^i$</span> and <span class="math-container">$\sigma^j(\omega)=\omega^{3^j}$</span>.</p>
| Chris | 326,865 | <p><em>Taking into account Jyrki's answer, I tried to collect the pieces. Please, have a look.</em></p>
<p>(1) <em>By the first lines of his answer:</em> If <span class="math-container">$[B:\Bbb{Q}]=2$</span>, then
<span class="math-container">$$[E:\Bbb Q]=[E:B][B:\Bbb Q] \implies 42=[E:B]\cdot 2 \iff [E:B]=21=|\mathrm{Aut}(E/B)|.$$</span> Since <span class="math-container">$B=\Bbb Q(i\sqrt 7)=\langle1,i\sqrt 7\rangle$</span>, we can see that <span class="math-container">$\theta(b)=\sigma^2(b)=b,\ \forall b\in B$</span>. Hence,
<span class="math-container">$$\theta,\sigma^2 \in \mathrm{Aut}(E/B)\implies \langle \theta,\sigma^2 \rangle \subseteq \mathrm{Aut}(E/B).$$</span> But also we have <span class="math-container">$|\langle \theta,\sigma^2\rangle|=|\langle \theta\rangle | \cdot |\langle \sigma^2\rangle|=7 \cdot 3 =21=|\mathrm{Aut}(E/B)|$</span>. Therefore, <span class="math-container">$$\mathrm{Aut}(E/B)=\langle \theta,\sigma^2\rangle.$$</span></p>
<hr>
<p>(2) <em>By the extras:</em> We observe that
<span class="math-container">$$\theta(\omega+\omega^2+\omega^4)=\omega+\omega^2+\omega^4,\\
\sigma^2(\omega+\omega^2+\omega^4)=\dotsb =\omega+\omega^2+\omega^4, $$</span>
so, <span class="math-container">$\omega+\omega^2+\omega^4 \in \mathrm{Fix}(\langle \theta,\sigma^2 \rangle) $</span>.
Then, we have the tower of fields
<span class="math-container">$$\Bbb Q \leq B=\Bbb Q (\omega+\omega^2+\omega^4) \leq \mathrm{Fix}(\langle \theta,\sigma^2 \rangle).$$</span>
But,
<span class="math-container">\begin{align*}
|G: \langle \theta,\sigma ^2 \rangle |=[\mathrm{Fix}(\langle \theta,\sigma^2\rangle):\Bbb Q] & \iff \frac{|G|}{|\langle \theta,\sigma^2\rangle|}=[\mathrm{Fix}(\langle \theta,\sigma^2\rangle):\Bbb Q] \\
& \iff \frac{42}{21}=[\mathrm{Fix}(\langle \theta, \sigma^2\rangle):\Bbb Q] \\
& \iff [\mathrm{Fix}(\langle \theta,\sigma^2\rangle):\Bbb Q]=2.
\end{align*}</span>
So, <span class="math-container">$B=\Bbb Q$</span> <em>or</em> <span class="math-container">$B=\mathrm{Fix}(\langle \theta,\sigma^2\rangle)$</span>. But <span class="math-container">$\omega+\omega^2+\omega^4 \notin \Bbb Q$</span>. So,
<span class="math-container">$$B=\mathrm{Fix}(\langle \theta,\sigma^2\rangle) \implies [B:\Bbb Q]=2.$$</span> Therefore,
<span class="math-container">$$[E:\Bbb Q]=[E:B][B:\Bbb Q] \iff [E:B]=21 \iff |\mathrm{Aut}(E/B)|=21.$$</span>
Now <span class="math-container">$[\underbrace{\Bbb Q(\omega+\omega^2+\omega^4)}_{B}:\Bbb Q]=2 < \infty \iff \omega+\omega^2+\omega^4$</span> is algebraic over <span class="math-container">$F$</span> and therefore it determines uniquely the basis of <span class="math-container">$B$</span>. So, for showing that <span class="math-container">$\theta,\sigma^2 \in \mathrm{Aut}(E/B)$</span>, it sufficies to show that <span class="math-container">$\theta,\sigma^2 $</span> leave <span class="math-container">$\omega+\omega^2+\omega^4$</span> fixed, and indeed that happens. So, <span class="math-container">$\langle \theta,\sigma^2 \rangle \subseteq \mathrm{Aut}(E/B)$</span>. Also, like (1), <span class="math-container">$|\langle \theta,\sigma^2 \rangle|=21=|\mathrm{Aut}(E/B)|$</span> and eventually
<span class="math-container">$$\mathrm{Aut}(E/B)=\langle \theta,\sigma^2 \rangle,$$</span>
as wanted.</p>
|
3,469,582 | <p>I have a feeling that <span class="math-container">$\lim_{x\rightarrow\infty}\frac{\log f(x^2)}{f(x)}=0$</span> for any positive monotone increasing function such that <span class="math-container">$\lim_{x\rightarrow\infty}f(x)=\infty$</span>, (in reality I am thinking mostly about super-exponential growth here but I do not think it might make a difference). The idea, intuitively, is that the logarithm takes away more growth than the square in the argument boosts. Though, I am not sure if this is true (maybe adding some more hypotheses) because I find it hard to prove neatly, having <span class="math-container">$f$</span> in between mixing these two contributions up. Tried L'Hospital but did not work and standard limit stuff neither, unsuccessfully. Anybody sees which way of approaching this could be productive? Don't need detailed answers, if you have a hunch that'll be enough.</p>
| WhatsUp | 256,378 | <p>Define a sequence <span class="math-container">$a_n$</span> with arbitrary <span class="math-container">$a_0>1$</span> and <span class="math-container">$a_{n + 1}=e^{a_n}$</span>.</p>
<p>Define a function <span class="math-container">$f$</span> such that <span class="math-container">$f(x)= a_0$</span> for <span class="math-container">$x< 2$</span> and <span class="math-container">$f(x) = a_n$</span> for <span class="math-container">$2^{2^{n - 1}}\leq x < 2^{2^n}$</span>.</p>
<p>Then <span class="math-container">$f$</span> is a counter example.</p>
<p>It can be modified (by "smoothing out" the jump points) to an example of continuous or even <span class="math-container">$\mathcal C^\infty$</span> function.</p>
|
223,509 | <p>Let $\ulcorner \cdot \urcorner$ be a fixed encoding of formulas by numbers, e.g.
let $\ulcorner \varphi \urcorner$ denote the <a href="https://en.wikipedia.org/wiki/G%C3%B6del_numbering" rel="nofollow">Godel number</a> of $\varphi$.
Let $T$ be a first-order arithmetic theory, e.g. PA.
Let $\Phi$ be a class of closed first-order formulas e.g. $\Sigma^0_1$.
Let $Tr$ be a first-order formula defining
the truth predicate for the formulas in $\Phi$ in the standard model:</p>
<blockquote>
<p>for all $\varphi \in \Phi$, $\mathbb{N} \vDash \varphi$ iff
$\mathbb{N} \vDash Tr(\ulcorner \varphi \urcorner)$</p>
</blockquote>
<p>E.g.
we have:
$\mathbb{N} \vDash Tr(\ulcorner \top \urcorner)$,
$\mathbb{N} \nvDash Tr(\ulcorner \bot \urcorner)$,
$\mathbb{N} \nvDash Tr(\ulcorner \exists x \ (x + 1 = 0)\urcorner)$,
etc.</p>
<p>Consider the following property: </p>
<blockquote>
<p>for all $\varphi \in \Phi$,
$T \vdash Tr(\ulcorner \varphi \urcorner) \to \varphi$.</p>
</blockquote>
<p><strong>Does this property have a standard name?</strong></p>
<hr>
<p>The reflection is similar to the property above
with <em>truth</em> replaced with <em>provability</em>:
let $\square$ denote the provability in theory $T'$.
Then reflection property is:</p>
<blockquote>
<p>for all $\varphi \in \Phi$, $T \vdash \square(\ulcorner \varphi \urcorner) \to \varphi$.</p>
</blockquote>
| Ali Enayat | 9,269 | <p>Biconditionals of the form $Tr(\varphi) \leftrightarrow \varphi$ are known as <em>T-sentences</em> in the literature of philosophical logic, and they date back to Tarski's groundbreaking work on the notion of truth in formalized languages. </p>
<p>However, I have not seen a special name for implications of the form mentioned in the question.</p>
|
2,554,380 | <p>With the metric space $(X,d) : X = \Bbb R$ and $d(x,y) = |x| + |y|$ for $x\neq y$, $d(x,x) = 0$ and $A = \{0\}$. Is interior of $A$ empty? </p>
<p>In usual metric it would be empty but with this metric I conclude that it is $A$ itself.</p>
<p>Def: $ a \in A $ is an interior point of $A$ if $\exists \epsilon > 0: B(a,\epsilon) \subset A$</p>
<p>$B(a,\epsilon) = \{x \in \Bbb R | |x| + |a| < \epsilon\} = \{0\}$ </p>
<p>For the condition $B(a,\epsilon) \subset A$ in Def, should it be proper subset or $\{0\} \subset \{0\}$ is okay?</p>
| Peter Szilas | 408,605 | <p>$f:D \rightarrow \mathbb{R} $ is uniformly continuous in </p>
<p>$D$, if</p>
<p>for every $\epsilon >0$ there is a $\delta$ such that </p>
<p>$|x-x'| \lt \delta$ implies </p>
<p>$|f(x)-f(x')| \lt \epsilon$ for all $x,x' \in D.$</p>
<p>Given: </p>
<p>$|f(x)-f(x')| \le c|x-x'|$, $c>0$, for $ x,x' \in D.$</p>
<p>Let $ \epsilon$ be given.</p>
<p>Choose $\delta =\epsilon/c.$</p>
<p>Then for $x,x' \in D:$</p>
<p>$|x-x'| \lt \delta$ implies</p>
<p>$|f(x)-f(x')| \le c|x-x'| \lt c\delta = \epsilon$.</p>
|
930,458 | <p>would somebody algebraic-geometry-savvy please help me with this problem, as I am pretty new to algebraic geometry:
I have to find smallest algebraic variety (irreducible algebraic set) in $\mathbb{C^{2}}$ that contains all the points with integer coordinates. I don't really know how to start, so at least an idea, hint or even better sketch of the solution would be much appreciated. Thank you very much.</p>
| Rene Schipperus | 149,912 | <p>The only such is the plane itself. Note that there can be no algebraic curve passing through all those points since then it would have an infinite number of intersections with a line, which is impossible.</p>
|
236,229 | <p>I have a list of 3 pairs of integers, for each pair <code>i</code> I want to replace the one having the smallest absolute value with <code>xi</code> and the other one with its sign. Like this:</p>
<pre><code>In: {{-5, 1}, {7, -3}, {4, 4}}
Out: {{-1, x2}, {1, x3}, {x4, 1}} (*for the last one {1, x4} would be fine as well*)
</code></pre>
<p>My attempt is this, I fear it's a bit too convoluted:</p>
<pre><code>list = {{-5,1},{7,-3},{4,4}}
list = MapThread[#1 /. #2 -> #3 &, {list, If[Total[#] > 0, Min[#], Max[#]] & /@ list, {x1, x2, x3}}];
list = Map[If[NumericQ[#], Sign[#], #] &, list, {2}];
</code></pre>
<p>I'm using this in a fractal related problem, so what's the most optimized way?</p>
| kglr | 125 | <pre><code>ClearAll[f]
f = MapIndexed[If[LessEqual @@ Abs @ #,
{Symbol["x" <> ToString[#2[[1]]]], Sign @ Last @ #},
{Sign @ First @ #, Symbol["x" <> ToString[#2[[1]]]]}] &];
list = {{-5, 1}, {7, -3}, {4, 4}};
f @ list
</code></pre>
<blockquote>
<pre><code>{{-1, x1}, {1, x2}, {x3, 1}}
</code></pre>
</blockquote>
<p>Also</p>
<pre><code>symbolpositions = MapIndexed[Flatten[{#2, #}] &][2 - Boole[LessEqual @@@ Abs[list]]];
symbols = Symbol["x" <> ToString[#]] & /@ Range[Length@list];
ReplacePart[Sign@list, Thread[symbolpositions -> symbols]]
</code></pre>
<blockquote>
<pre><code> {{-1, x1}, {1, x2}, {x3, 1}}
</code></pre>
</blockquote>
|
236,229 | <p>I have a list of 3 pairs of integers, for each pair <code>i</code> I want to replace the one having the smallest absolute value with <code>xi</code> and the other one with its sign. Like this:</p>
<pre><code>In: {{-5, 1}, {7, -3}, {4, 4}}
Out: {{-1, x2}, {1, x3}, {x4, 1}} (*for the last one {1, x4} would be fine as well*)
</code></pre>
<p>My attempt is this, I fear it's a bit too convoluted:</p>
<pre><code>list = {{-5,1},{7,-3},{4,4}}
list = MapThread[#1 /. #2 -> #3 &, {list, If[Total[#] > 0, Min[#], Max[#]] & /@ list, {x1, x2, x3}}];
list = Map[If[NumericQ[#], Sign[#], #] &, list, {2}];
</code></pre>
<p>I'm using this in a fractal related problem, so what's the most optimized way?</p>
| Αλέξανδρος Ζεγγ | 12,924 | <p><a href="https://reference.wolfram.com/language/ref/MapIndexed.html" rel="nofollow noreferrer"><code>MapIndexed</code></a> should be the suitable function working at the overall level. A version slightly different from that of @kglr 's, with <a href="https://reference.wolfram.com/language/ref/OrderingBy.html" rel="nofollow noreferrer"><code>OrderingBy</code></a>, <a href="https://reference.wolfram.com/language/ref/ToExpression.html" rel="nofollow noreferrer"><code>ToExpression</code></a>, and <a href="https://reference.wolfram.com/language/ref/StringTemplate.html" rel="nofollow noreferrer"><code>StringTemplate</code></a></p>
<pre><code>symbol = ToExpression[StringTemplate["x``"] @@ #] &;
func = If[Greater @@ OrderingBy[#, Abs],
{Sign[#[[1]]], symbol[#2]},
{symbol[#2], Sign[#[[2]]]}
] &;
list = {{-5, 1}, {7, -3}, {4, 4}};
MapIndexed[func][list]
</code></pre>
<blockquote>
<pre><code>{{-1, x1}, {1, x2}, {x3, 1}}
</code></pre>
</blockquote>
|
2,607,668 | <p>I am trying to prove/disprove $\operatorname{Arg}(zw)=\operatorname{Arg}(z)+\operatorname{Arg}(w)$. Apparently $\operatorname{Arg}(zw)=\operatorname{Arg}(z)+\operatorname{Arg}(w)+2k\pi$ where $k=0,1,\text{ or }-1$, but I have no idea why. I keep on finding that answer online. I am very lost on how to prove this statement, any help would be great. Thank you. </p>
| joeyg | 454,784 | <p>Denote $z=r_1\cdot e^{i\theta_1}$, $z=r_2\cdot e^{i\theta_2}$.</p>
<p>Then $zw=r_1\cdot e^{i\theta_1}\cdot r_2\cdot e^{i\theta_2}=r_1r_2e^{i(\theta_2+\theta_2)}$, and since the argument of a complex number $re^{i\theta}$ is $\theta$, we receive that $Arg(z)+Arg(w)=\theta_1+\theta_2=Arg(zw)$ up to $\pm 2\pi$.</p>
|
711,549 | <p>Consider the successor of the largest finite ordinal that will ever be considered alone. But then it wasn't the largest finite ordinal that will ever be considered alone. How do we get around this paradox? The largest finite ordinal that will ever be considered alone does exist, and yet we can consider its successor.</p>
| vadim123 | 73,324 | <p>This apparent paradox has nothing to do with ordinals, or metalanguage. It is a time travel/omniscience paradox. Let's strip away the unnecessary bits.</p>
<p>How many words will the longest sentence I personally will say tomorrow be? Maybe I predict seven, but that doesn't stop me from saying a sentence with eight words. No matter my prediction, I can break it.</p>
<p>You will note that the difficulty disappears if the time travel is taken away. If we ask about the largest ordinal that has <em>previously</em> been considered, that is a unique number (that periodically increases with time).</p>
|
1,461,484 | <p>For example, look at this sentence from Perko's text on dynamical system</p>
<p>"It follows from Cauchy Schwarz inequality that if $T \in L(R^n)$ is represented by the matrix $A$ with respect to the standard basis for $R^n$ $_\cdots$" pg 11</p>
<p>What does it mean for a $T: R^n \to R^n$ to be represented by a matrix? Isn't it by definition that $T: R^n \to R^n$ is equivalent to an n by n matrix? Can someone translate exactly what it means by "represented" versus "not represented"?</p>
| user2357112 | 91,416 | <p>The number after nine is represented in the usual base system by the sequence of digits $10$, but it isn't a sequence of digits. Properties like "two digits long" are properties of the representation, while properties like "has four factors" are properties of the number. We could represent the number in binary, as $1010$, and that would be just as valid.</p>
<p>Similarly, a linear transformation $T$ may be represented by a matrix $M$, but it isn't the matrix. Properties like "upper triangular" are properties of the representation, while properties like "injective" are properties of the linear transformation. In a different basis, $T$ might be represented by a matrix other than $M$, and that representation would be just as valid.</p>
|
3,848,517 | <p>I have a conjecture in my mind regarding Arithmetic Progressions, but I can't seem to prove it. I am quite sure that the conjecture is true though.</p>
<p>The conjecture is this: suppose you have an AP (arithmetic progression):
<span class="math-container">$$a[n] = a[1] + (n-1)d$$</span>
Now, suppose our AP satisfies the property that the sum of the first <span class="math-container">$n$</span> terms of our AP is equal to the sum of the first <span class="math-container">$m$</span> terms:
<span class="math-container">$$S[n] = S[m]$$</span>
but <span class="math-container">$n \neq m$</span>. I want to prove two theorems:</p>
<ul>
<li>The underlying AP <span class="math-container">$a[n]$</span> must be <strong>symmetric</strong> with respect to the point at which it becomes zero.</li>
<li><span class="math-container">$S[n + m] = 0$</span></li>
</ul>
<h2>A Numerical Example</h2>
<p>Consider the AP:
<span class="math-container">$$a[n] = 4 - n = (3, 2, 1, 0, -1, -2, -3)$$</span>
This is an AP with common difference <span class="math-container">$d = -1$</span> and first term <span class="math-container">$a[1] = 3$</span>:
<a href="https://i.stack.imgur.com/NKmZF.png" rel="nofollow noreferrer">Here is the MATLAB plot of this AP</a>. As you can see in the plot, our AP is <strong>symmetric</strong> with respect to the point <span class="math-container">$n = 4$</span>:
<span class="math-container">$$a[4-1] = -a[4+1] = 1$$</span>
<span class="math-container">$$a[4-2] = -a[4+2] = 2$$</span>
<span class="math-container">$$a[4-3] = -a[4+3] = 3$$</span></p>
<p>Now, here is the sum of our AP:
<span class="math-container">$$S[n] = (3,5,6,6,5,3,0)$$</span>
<a href="https://i.stack.imgur.com/Ep0aB.png" rel="nofollow noreferrer">Here is the MATLAB plot of the summation</a>. You can clearly see that:
<span class="math-container">$$S[1] = S[6] = 3$$</span>
<span class="math-container">$$S[2] = S[5] = 5$$</span>
<span class="math-container">$$S[3] = S[4] = 6$$</span></p>
<p>and you can also see that:
<span class="math-container">$$S[1 + 6] = S[7] = 0$$</span>
<span class="math-container">$$S[2 + 5] = S[7] = 0$$</span>
<span class="math-container">$$S[3 + 4] = S[7] = 0$$</span></p>
<p>Can you please help me out with this problem? Any guidance will be very welcome. I am actually an Engineering student, so my Pure Math skills are not that strong.</p>
<p>Thank you!</p>
| Kavi Rama Murthy | 142,385 | <p>If <span class="math-container">$f_n(x) =0$</span> for <span class="math-container">$x <0$</span> as well as for <span class="math-container">$x >\frac 1n$</span> and <span class="math-container">$f_n(x)=4n^{2}x(\frac 1n -x)$</span> for <span class="math-container">$0 <x<\frac 1n$</span> then you can check that <span class="math-container">$f_n \to f$</span> point-wise but supremum of <span class="math-container">$f_n(x)$</span> is <span class="math-container">$1$</span> for each <span class="math-container">$n$</span>.</p>
<p>Under uniform convergence the conclusion is true.</p>
|
3,391,118 | <p>let <span class="math-container">$x , y$</span> be a real numbers with <span class="math-container">$-3\leq x \leq5$</span> and <span class="math-container">$-2\leq y\leq -1$</span> , I ask if the range of <span class="math-container">$x-y$</span> is <span class="math-container">$[-1,6]$</span> or <span class="math-container">$[-1,1]$</span> using two cases negative <span class="math-container">$x$</span> and positive <span class="math-container">$x$</span> and takin g sup is <span class="math-container">$1$</span> ? </p>
| Bernard | 202,857 | <p>No, you cannot subtract inequalities, only add them, and changing signs in an inequality reverses it.</p>
<p>Here a way to go:
<span class="math-container">$$−2≤y≤−1\iff 1\le -y\le 2$$</span>
so, adding the latter to the inequalities with <span class="math-container">$x$</span> yields
<span class="math-container">$$-3+1\le x+(-y)=x-y\le 5+2,\quad\text{ i.e. }\quad -2\le x-y\le 7.$$</span></p>
|
3,391,118 | <p>let <span class="math-container">$x , y$</span> be a real numbers with <span class="math-container">$-3\leq x \leq5$</span> and <span class="math-container">$-2\leq y\leq -1$</span> , I ask if the range of <span class="math-container">$x-y$</span> is <span class="math-container">$[-1,6]$</span> or <span class="math-container">$[-1,1]$</span> using two cases negative <span class="math-container">$x$</span> and positive <span class="math-container">$x$</span> and takin g sup is <span class="math-container">$1$</span> ? </p>
| Z Ahmed | 671,540 | <p>Two in-equations in the same direction can be added but they can never be subtracted.
Let us take two parts of four given in-equations: <span class="math-container">$x \le 5~ and ~ -2 \le y~$</span> we can add the two to get <span class="math-container">$x-2 \le 5+y \implies x-y \le 7.$</span> Next, let us take other two parts as <span class="math-container">$-3 \le x, ~ and ~ y \le -1$</span>. These two can be added to get <span class="math-container">$y-3 \le x-1 \implies x-y \ge -2.$</span> So finally we have <span class="math-container">$$ -2\le x-y \le 7$$</span></p>
|
3,344,766 | <p>It is too be proved that <span class="math-container">$3(p^2)+61$</span> is a perfect square only for <span class="math-container">$$p=1 ; p=N$$</span>
The question arises from a problem in arithmatic progression with a certain soft constraint of natural numbers. At the end a quadratic is formed and since our constraint is of natural number The discriminants should be a perfect square.
If we take <span class="math-container">$p= 1$</span> we are able to solve the question. Our teacher left it to us to prove <span class="math-container">$p=1$</span> is the only solution.
I have not proceeded due to lack of ideas.</p>
| Will Jagy | 10,400 | <p>Given any solution with positive integers <span class="math-container">$q^2 - 3 p^2 = 61,$</span> we get an infinite string of (larger) positive solutions by
<span class="math-container">$$ (q,p) \mapsto (2q+3p, \; q+2p) $$</span></p>
<p>For example,
<span class="math-container">$$ (8,1) \mapsto (19, \; 10) \mapsto (68, \; 39) \; ... $$</span></p>
<p><span class="math-container">$$ (13,6) \mapsto (44, \; 25) \mapsto (163, \; 94) \; ... $$</span></p>
<p>By Cayley-Hamilton or direct calculation, these can be split into four one-variable recurrences, under the rule
<span class="math-container">$$ x_{n+2} = 4 x_{n+1} - x_n \; .$$</span></p>
<p><span class="math-container">$$ 8, 19, 68, 253, 944, 3523, 13148, 49069, 183128, 683443, 2550644,... $$</span>
<span class="math-container">$$ 1, 10, 39, 146, 545, 2034, 7591, 28330, 105729, 394586, 1472615, ... $$</span>
<span class="math-container">$$ 13, 44, 163, 608, 2269, 8468, 31603, 117944, 440173, 1642748, 6130819, ... $$</span>
<span class="math-container">$$ 6, 25, 94, 351, 1310, 4889, 18246, 68095, 254134, 948441, 3539630,... $$</span></p>
<pre><code>jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
Automorphism matrix:
2 3
1 2
Automorphism backwards:
2 -3
-1 2
2^2 - 3 1^2 = 1
w^2 - 3 v^2 = 61 = 61
Wed Sep 4 18:54:19 PDT 2019
w: 8 v: 1 SEED KEEP +-
w: 13 v: 6 SEED BACK ONE STEP 8 , -1
w: 19 v: 10
w: 44 v: 25
w: 68 v: 39
w: 163 v: 94
w: 253 v: 146
w: 608 v: 351
w: 944 v: 545
w: 2269 v: 1310
w: 3523 v: 2034
w: 8468 v: 4889
w: 13148 v: 7591
w: 31603 v: 18246
w: 49069 v: 28330
w: 117944 v: 68095
w: 183128 v: 105729
w: 440173 v: 254134
w: 683443 v: 394586
w: 1642748 v: 948441
w: 2550644 v: 1472615
w: 6130819 v: 3539630
w: 9519133 v: 5495874
w: 22880528 v: 13210079
w: 35525888 v: 20510881
w: 85391293 v: 49300686
Wed Sep 4 18:56:21 PDT 2019
w^2 - 3 v^2 = 61 = 61
jagy@phobeusjunior:~$
</code></pre>
|
231,403 | <p>Let $1 \le p < \infty$. For all $\epsilon > 0$, does there exist $C = C(\epsilon, q)$ such that$$\|u\|_{L^p(0, 1)} \le \epsilon \|u'\|_{L^1(0, 1)} + C\|u\|_{L^1(0, 1)} \text{ for all }u \in W^{1, 1}(0, 1)?$$</p>
| Jochen Wengenroth | 21,051 | <p>The inequality you ask for is more or less a restatement of the compactness of an appropriate embedding: Let $p\le q\le r$ be three norms on a vector space such that the ball $B_r(1)=\lbrace x\in X. r(x)\le 1\rbrace$ is relatively compact in $(X,q)$. Then, for every $\varepsilon>0$ there is $C>0$ such that
$$
q(x) \le \varepsilon r(x) + C p(x).
$$</p>
<p>Indeed, the topologies induced by $p$ and $q$ on $B_r(1)$ coincide because a (relatively) compact space does not admit strictly coarser Hausdorff topologies. For $\varepsilon >0$ there is thus $\delta>0$ such that
$$
B_p(\delta)\cap B_r(1) \subseteq B_q(\varepsilon).
$$
For $x\in X$ with $r(x)<\infty$ and $t=p(x)/\delta +r(x)$ you have
$\frac 1t x \in B_p(\delta)\cap B_r(1)$ and hence $q(x)\le t\varepsilon$ which proves the inequality for $C=\varepsilon/\delta$.</p>
<p>In your case, $r$ is the $W^{1,1}$-norm, $p$ the $L^1$-norm, and $q$ is the $L^p$-norm.</p>
|
2,234,478 | <p>Given 3 points, A (x1, y1), B (x2, y2) and C (x3, y3), what is the best way to tell if all 3 points lie within a circle of a given radius <strong>r</strong>?</p>
<p>The best I could come up with was to find the Fermat point F (x4, y4) for the triangle ABC, and then check if the distance from F to each A, B and C is less than <strong>r</strong>. Is there a way to do this more efficiently?</p>
| Mark Fischler | 150,362 | <p>Alpha only gives a numeric approximation which, unless you mis-typed your equation into alpha, should be about 1.56.</p>
<p>There is no simple closed for for the answer.</p>
<p>But there is something cute about this: Full Mathematica fails to give a numerical solution using its NSolve method, while Alpha does give a numerical solution -- even when I type it in using "NSolve[x^(2^(1/x))==2,x]" (!)</p>
|
2,234,478 | <p>Given 3 points, A (x1, y1), B (x2, y2) and C (x3, y3), what is the best way to tell if all 3 points lie within a circle of a given radius <strong>r</strong>?</p>
<p>The best I could come up with was to find the Fermat point F (x4, y4) for the triangle ABC, and then check if the distance from F to each A, B and C is less than <strong>r</strong>. Is there a way to do this more efficiently?</p>
| egreg | 62,967 | <p>The equation can be written
$$
\sqrt[x]{2}\log x=\log 2
$$
(natural logarithm); setting $t=1/x$, it becomes
$$
-2^t\log t=\log 2
$$
Consider the function $f(t)=2^t\log t+\log 2$. We see that
$$
\lim_{t\to0}f(t)=-\infty
\qquad
\lim_{t\to\infty}f(t)=\infty
$$
Moreover
$$
f'(t)=2^t\cdot\log 2\cdot\log t+\frac{2^t}{t}>0
$$
so the function is increasing. Then the equation $f(t)=0$ has exactly one solution.</p>
<p>Since $f(1)=\log2>0$ and
$$
f(1/2)=-\sqrt{2}\log2+\log2=(1-\sqrt{2})\log2<0,
$$
we know the solution is in the interval $(1/2,1)$, so the solution of the original equation is in $(1,2)$.</p>
|
1,149,907 | <p>What are some different methods to evaluate</p>
<p>$$ \int_{-\infty}^{\infty} x^4 e^{-ax^2} dx$$</p>
<p>for $a > 0$.</p>
<p>This integral arises in a number of contexts in Physics and was the original motivation for my asking. It also arises naturally in statistics as a higher moment of the normal distribution.</p>
<p><strong>I have given a few methods of evaluation below.</strong> Anyone know of others?</p>
| D. Thomine | 20,413 | <p>Generating functions, as often, provide a nice way to compute these integrals.</p>
<p>Let $I_n := \int_\mathbb{R} x^n e^{-ax^2} \ dx$, and for all complex $\lambda$:</p>
<p>$$G(\lambda) := \sum_{n=0}^\infty \frac{I_n}{n!}\lambda^n.$$</p>
<p>Then:</p>
<p>$$G(\lambda) = \int_\mathbb{R} \sum_{n=0}^\infty \frac{(\lambda x)^n}{n!} e^{-ax^2} \ dx = \int_\mathbb{R} e^{\lambda x -ax^2} \ dx = e^{\frac{\lambda^2}{4a}} \int_\mathbb{R} e^{-a\left(x-\frac{\lambda}{2a}\right)^2} \ dx = e^{\frac{\lambda^2}{4a}} \int_\mathbb{R} e^{-ax^2} \ dx.$$</p>
<p>Now, since $\int_\mathbb{R} e^{-ax^2} \ dx = \sqrt{\pi}/\sqrt{a}$, we finally get:</p>
<p>$$\sum_{n=0}^\infty \frac{I_n}{n!}\lambda^n = \frac{\sqrt{\pi} e^{\frac{\lambda^2}{4a}}}{\sqrt{a}} = \sum_{n=0}^\infty \frac{\sqrt{\pi}}{\sqrt{a}} \frac{1}{(4a)^n n!}\lambda^{2n}.$$</p>
<p>Identification of the coefficients yields $I_{2n+1} = 0$ for all $n$, and:</p>
<p>$$I_{2n} = \frac{\sqrt{\pi}}{\sqrt{a}} \frac{(2n)!}{(4a)^n n!}.$$</p>
<p>For $n=2$, this yields:</p>
<p>$$I_4 = \frac{\sqrt{\pi}}{\sqrt{a}} \frac{24}{32a^2} = \frac{3\sqrt{\pi}}{4a^{5/2}}.$$</p>
|
4,415,037 | <p>I would like to prove upper and lower bounds on <span class="math-container">$|\cos(x) - \cos(y)|$</span> in terms of <span class="math-container">$|x-y|$</span>. I was able to show that <span class="math-container">$|\cos(x) - \cos(y)| \leq |x - y|$</span>. I'm stuck on the lower bound. Does anyone know how to approach this?</p>
<p>Update: Over the interval <span class="math-container">$[0,\pi/2]$</span>, I was able to show that <span class="math-container">$|\cos(x) - \cos(y)| \geq \frac{2 \min(x,y)}{\pi}|x-y|$</span>. But I would like a lower bound that holds for any interval.</p>
| Swapnajit Das | 857,214 | <p><span class="math-container">$|\cos(x)-\cos(y)|\geq 0$</span> for all <span class="math-container">$x$</span>, <span class="math-container">$y \in$</span> <span class="math-container">$\mathbb{R}$</span>. It is possible for equality to hold, for example when <span class="math-container">$x=y$</span>.</p>
<p>By the <a href="https://en.wikipedia.org/wiki/Mean_value_theorem" rel="nofollow noreferrer">mean value theorem</a>,</p>
<p><span class="math-container">$\cos(x)-\cos(y) =-\sin(c)(x-y)$</span> for some <span class="math-container">$c \in (x, y)$</span>.</p>
<p>Since <span class="math-container">$\sin$</span> is bounded above and below by <span class="math-container">$\pm 1$</span>, we have <span class="math-container">$|\cos(x)-\cos(y)| \leq |x-y|$</span></p>
<p>So we can conclude that <span class="math-container">$0 \leq|\cos(x)-\cos(y)| \leq |x-y|$</span>.</p>
|
2,869,305 | <p>I need some help evaluating this limit... Wolfram says it blows up to infinity but I don't think so. I just can't prove it yet.</p>
<p>$$ \lim_{x\to\infty}(x!)^{1/x}-\frac{x}{e} $$</p>
| pancini | 252,495 | <p>It diverges, albeit slowly.</p>
<p>By <a href="https://en.wikipedia.org/wiki/Stirling%27s_approximation" rel="nofollow noreferrer">Stirling's Approximation</a>, the function approaches</p>
<p>$$f(x)=(2\pi x)^{1/x}\left( \frac xe\right)-\frac xe=\left(\frac xe\right)\left( (2\pi x)^{1/x}-1\right).$$</p>
<p>Now it is true that $(2\pi x)^{1/x}$ goes to $1$ as $x\to\infty$, but we need more information. Using L'Hospital's rule,</p>
<p>\begin{align}
\lim_{x\to\infty}\frac{(2\pi x)^{1/x}-1}{e/x}&=\lim_{x\to\infty}\frac{-(2 \pi x)^{1/x} x^{-2} (\log (2 \pi x)-1)}{-ex^{-2}}\\
&=\lim_{x\to\infty}e^{-1}(2 \pi x)^{1/x} (\log (2 \pi x)-1)\\
&=\infty
\end{align}
since the factor $e^{-1}(2 \pi x)^{1/x}$ is bounded and nonzero and $\log(2\pi x)-1\to\infty$.</p>
<hr>
<p>As an aside, check out the harmonic prime series $1/2+1/3+1/5+1/7+\cdots$. This diverges, but so slowly that the first million primes don't even make it past $3$.</p>
|
57,232 | <p>I have huge matrices in the form of</p>
<pre><code>mtx1 = {{24+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8]},{24+24 FF[5,10] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[7,8] GG[7,8]},{24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]}};
</code></pre>
<p>but the matrices I use are much bigger. Now I want to get rid of each term that contains <code>FF[___]</code> or <code>GG[___]</code>. Both always come together, therefore I used</p>
<pre><code>mtx2 = mtx1 /. FF[___] -> 0;
(* mtx2={{24, 24, 24}, {24, 24, 24}, {24, 24, 24}} *)
</code></pre>
<p>and got the desired result in mtx2. Unfortunatly it turns out that this zeroing is extremly time-consuming. For my huge matrices, it takes on the order of 100 seconds. </p>
<p><strong>Question:</strong></p>
<p>Is there a more time-efficient way to zero all <code>FF[___]</code>-terms in <code>mtx1</code>?</p>
<p><strong>Comparison:</strong></p>
<p>I compare several approaches, for a big 3 big test-matrix. The approaches also include the construction of the matrix.</p>
<p><em>my original approach</em></p>
<ul>
<li>{174.8417751, 65.4913582, 25.3878123} seconds</li>
</ul>
<p><em>Coefficient-Creation of Matrix</em></p>
<ul>
<li>{134.4920621, 51.4260521sec, 19.6079772} seconds</li>
</ul>
<p><em>belisarius' Block-evaluation methode</em></p>
<ul>
<li>{82.3675688, 31.5639078, 12.3822025} seconds</li>
</ul>
<p><em>eldo's Join/Partition</em></p>
<ul>
<li>{77.8615328, 29.0973367, 11.2742769} seconds</li>
</ul>
<p><em>kguler's Block-evaluation</em></p>
<ul>
<li>{75.8906436, 29.1315892, 11.6544345} seconds</li>
</ul>
<p><em>Mr.Wizard's mtx1[[All, All, 1]]</em></p>
<ul>
<li>{75.5589726, 29.0378220, 11.9491954} seconds</li>
</ul>
<p><strong>Edit</strong></p>
<p>The full problem, including the matrix-creation is posted here: <a href="https://mathematica.stackexchange.com/questions/57241/time-efficient-creation-of-matrix">
Time-efficient creation of matrix
</a></p>
| eldo | 14,254 | <p>Given the structure of your SQUARE example matrix this should be fast:</p>
<pre><code>dim = First @ Dimensions @ mtx1
</code></pre>
<blockquote>
<p>3</p>
</blockquote>
<pre><code>tup = Join [#, {1}] & /@ Tuples[Range@dim, 2]
</code></pre>
<blockquote>
<p>{{1, 1, 1}, {1, 2, 1}, {1, 3, 1}, {2, 1, 1}, {2, 2, 1}, {2, 3, 1},
{3,1, 1}, {3, 2, 1}, {3, 3, 1}}</p>
</blockquote>
<pre><code>Partition[mtx1[[#1, #2, #3]] & @@@ tup, dim]
</code></pre>
<blockquote>
<p>{{24, 24, 24}, {24, 24, 24}, {24, 24, 24}}</p>
</blockquote>
<p>Oneliner:</p>
<pre><code>With[{dim = First@Dimensions@mtx1},
Partition[mtx1[[#1, #2, #3]] & @@@ #, dim] &[
Join [#, {1}] & /@ Tuples[Range@dim, 2]]]
</code></pre>
|
4,580,484 | <p>It is often said that "differential forms are used for integration".</p>
<p>Typically people like to talk about the integral <span class="math-container">$\int_M \omega$</span> of a differential form <span class="math-container">$\omega$</span>, and exterior derivative, one of the most important operation of differential form, is defined such that Stokes' theorem <span class="math-container">$\int_{\partial M} \omega = \int_M \mathrm{d} \omega$</span> can hold.</p>
<p>Symplectic manifold is a manifold equipped with a symplectic 2-form <span class="math-container">$\omega$</span>. Therefore, according to the former consideration, it should be interesting to talk about integral of the symplectic form on a two-dimensional submanifold of a symplectic manifold.</p>
<p>However, it seems that people in the field of symplectic geometry is quite indifferent on this. Why?</p>
| Quaere Verum | 484,350 | <p>This is only true if you are very pedantic about what you mean by "symplectic geometry". Namely, if you include symplectic topology, then one of the most important invariants of a symplectic manifold requires integrating the symplectic form over <span class="math-container">$2$</span>-dimensional submanifolds. To be more precise, one considers maps <span class="math-container">$u:\Sigma\to M$</span> for <span class="math-container">$\Sigma$</span> a Riemann surface, and then integrates <span class="math-container">$\int_\Sigma u^*\omega$</span>. If <span class="math-container">$u$</span> is an embedding, then this is clearly the same as integrating over the submanifold <span class="math-container">$u(\Sigma)\subseteq M$</span>.</p>
<p>The reason why this is done, is because one wants to "count" the number of pseudo-holomorphic maps <span class="math-container">$\Sigma\to M$</span>, up to reparameterisation. To be able to do this, one has to restrict to maps with finite energy, i.e. <span class="math-container">$\int_\Sigma u^*\omega<\infty$</span>. Then the "moduli space" of solutions can be constructed, and compactified (finite energy is crucial to ensure compactification). If this moduli space has expected dimension <span class="math-container">$0$</span>, then the compactification results in a finite number of points, which can be counted in some appropriate sense. The numbers which are obtained in this way are called the Gromov-Witten invariants of the symplectic manifold.</p>
|
2,249,929 | <p>Is there a more general form for the answer to this <a href="https://math.stackexchange.com/q/1314460/438622">question</a> where a random number within any range can be generated from a source with any range, while preserving uniform distribution? </p>
<p><a href="https://stackoverflow.com/q/137783/866502">This</a> question for example looks familiar and is changing a range of 1-5 to 1-7</p>
| Marcus Andrews | 97,648 | <p>I wrote a Python implementation a while back that did this:</p>
<pre><code>def uniform_generator(m, n): #mimics m-sided die using an n-sided die
"""
Expected number of rolls
E = r * n^r / m
where r = int(ceil(log(m, n)))
"""
r = int(ceil(log(m, n)))
while True:
candidate = sum(n**power * randint(0, n-1) for power in range(r)) + 1
if candidate <= m:
return candidate
</code></pre>
<p>Not optimized or anything, but gets the job done.</p>
<p>How this works (using an $n$-sided die to mimic a $m$-sided die), starting off with two facts:</p>
<ol>
<li><p>Generating a number from something like $\{1, 2, 3, ..., n\}$ is the same as generating a number from $\{0, 1, 2, ..., n-1\}$ and adding $1$. We'll focus on this latter version.</p></li>
<li><p>If we had $m < n$ (using a bigger die to mimic a smaller one), we could simply continue rolling the $n$-sided die until we got something $\leq m$ and then just take that result. Let's say we wanted to know the probability of rolling a $1$ using this strategy. There is a $1/n$ chance we roll a $1$, and a $(n-m)/n$ chance we roll too high and have to start again. This implies $p = 1/n + (n-m)p/n$, and we get $p = 1/m$, the same as if we had used a $m$-sided die to begin with. The point of this paragraph is to show that you can mimic a smaller uniform distribution by using a larger one and simply trying again if the result is too large.</p></li>
</ol>
<p>Moving on: Instead of generating a number from $\{0, 1, 2, ..., n-1\}$ directly, we're actually building a base-$n$ number that is capable of returning any number in this set (and possibly a few more numbers, which we would ignore).</p>
<p>For instance, if you're familiar with binary (base-$2$), we could write the number $5$ as $(1 \cdot 2^2) + (0 \cdot 2^1) + (1 \cdot 2^0)$. Each term in parentheses you can think of as a "bit column" -- we're generating each digit of the expansion independently using our $n$-sided die (any digit in base $n$ ranges from $0$ to $n-1$). </p>
<p>In base $n$, the smallest number that can be generated is $0$, and the largest is $n^r-1$ given $r$ columns. Each result is equally likely since each digit of that result is being generated independently.</p>
<p>We just need to ensure we have enough columns such that we can generate any number from $0$ to $m-1$, i.e. where $n^r \geq m$, and this occurs at $r = \lceil \log_n(m) \rceil$. Then we use the $n$-sided die to generate a digit for each column and then add up the result from the expansion. </p>
<p>If the result plus $1$ (see fact $1$ from earlier) happens to be greater than $m$, we start over (see fact $2$ from earlier).</p>
|
1,955,509 | <p>There's this exercise in Hubbard's book:</p>
<blockquote>
<p>Let $ h:\Bbb R \to \Bbb R $ be a $C^1$ function, periodic of period $2\pi$, and define the function $ f:\Bbb R^2 \to \Bbb R $ by
$$f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=rh(\theta)$$</p>
<p>a. Show that $f$ is a continuous real-valued function on $\Bbb R^2$.</p>
<p>b. Show that $f$ is differentiable on $\Bbb R^2 - \{\mathbf 0\}$.</p>
<p>c. Show that all directional derivatives of $f$ exist at $\mathbf 0$ if and only if</p>
<p>$$ h(\theta) = -h(\theta + \pi) \ \text{ for all } \theta $$</p>
<p>d. Show that $f$ is differentiable at $ \mathbf 0 $ if an only if $h(\theta)=a \cos \theta + b \sin \theta$ for some number $a$ and $b$. </p>
</blockquote>
<p>I can't find how to prove $ f $ is continuous, I tried to prove
$$ \lim_{\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix} \to \begin{pmatrix}s\cos\phi\\s\sin\phi \end{pmatrix}} f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=s\ h(\phi) $$ for all $s$ and $\phi$.
But I can't do much else.</p>
| E.H.E | 187,799 | <p>Hint:
$${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\quad {\text{ for }}|x|<1\!$$
differentiate it and multiply by $x$</p>
<p>$${\frac {x}{(1-x)^{2}}}=\sum _{n=0}^{\infty }nx^{n}$$
differentiate it again and multiply by $x$
$${\frac {x(x+1)}{(1-x)^{3}}}=\sum _{n=0}^{\infty }n^2x^{n}$$</p>
|
2,253,768 | <p>I am currently working on a small optimization problem in which I need to find an optimal number of servers for an objective function that incorporates the Erlang Loss formula. To this end, I have been searching for an expression for the first order difference of the Erlang Loss formula with respect to the number of servers, i.e. $B(E,m+1)-B(E,m)$, where <em>m</em> is the number of servers and $B(E,m)$ is given by:</p>
<p>$B(E,m)={\dfrac {{\dfrac {E^{m}}{m!}}}{\sum _{{i=0}}^{m}{\dfrac {E^{i}}{i!}}}}$</p>
<p>Unfortunately, until now I can't derive or find such an expression (if it exists) and was wondering whether one on this forum could help me out? </p>
<p>Many thanks in advance! </p>
| quasi | 400,434 | <p>First we show $(M,x)$ is a proper ideal of $R[x]$.
<p>
Suppose instead that $(M,x) = (1)$.
<p>
Then $1 \in (M,x)$ implies $1 = m + xg(x)\;$for some $m \in M$, some $g \in R[x]$.
<p>
Since $M$ is maximal in $R,\, M$ is a proper ideal of $R$, hence $m \ne 1$.
<p>
Then </p>
<p>\begin{align*}&1 = m + xg(x)\\[4pt]
\implies\;&1-m = xg(x)
\end{align*}</p>
<p>contradiction, since $xg(x)$ can't be a nonzero constant.</p>
<p>It follows that $(M,x)$ is a proper ideal of $R[x]$, as claimed.</p>
<p>Let $f \in R[x]\setminus(M,x)$.
<p>
Then to show $(M,x)$ is maximal in $R[x]$, it suffices to show $(M,x,f) = (1)$.
<p>
Letting $f(x) = r + xg(x)$, for some $r \in R$, some $g \in R[x]$, we can't have $r \in M$, else $f \in (M,x)$.
<p>
Since $M$ is maximal in $R$, it follows that as an ideal of $R,\,(M,r) =(1),],$ hence also $(M,r) = (1)$ as an ideal of $R[x]$.
<p>
Then
\begin{align*}
&f(x) = r + xg(x)\\[4pt]
\implies\;&r \in (x,f)\\[4pt]
\implies\;&r \in (M,x,f)\\[4pt]
\implies\;&(M,r) \subseteq (M,x,f)\\[4pt]
\implies\;&(M,x,f) = (1)
\end{align*}</p>
<p>It follows that $(M,x)$ is maximal in $R[x]$, as was to be shown.</p>
|
3,440,732 | <p>How can I show that <span class="math-container">$P=\{\{2k-1,2k\},k\in \mathbb {N}\}$</span> is the basis of <span class="math-container">$\mathbb N$</span>.
Obviously we have to show that basis's orders do work.
But the problem is if I take two subsets that belong to <span class="math-container">$P$</span> , for example <span class="math-container">$P_1=\{1,2\}$</span> , <span class="math-container">$P_2=\{3,4\}$</span> , I get <span class="math-container">$P_1 \cap P_2=\emptyset$</span> , then how can I show that for every <span class="math-container">$n\in\mathbb N$</span> , there is <span class="math-container">$P_3\in P$</span> such that <span class="math-container">$n\in P_3$</span>?</p>
| Henno Brandsma | 4,280 | <p>Any <em>partition</em> <span class="math-container">$P_i, i \in I$</span> of a set <span class="math-container">$X$</span> is a base for a topology on <span class="math-container">$X$</span>. A partition obeys:</p>
<ol>
<li><span class="math-container">$\bigcup_{i \in I} P_i = X$</span>.</li>
<li><span class="math-container">$\forall i \neq j: P_i \cap P_j = \emptyset$</span>.</li>
</ol>
<p>This translates nicely into the two conditions for definitng a base for some topology: 1. is exactly the first : every point of <span class="math-container">$X$</span> lies in some base element.</p>
<p>The other one says that whenever we have base elements <span class="math-container">$B_1,B_2$</span> and for any <span class="math-container">$z \in B_1 \cap B_2$</span>, there must be an base element <span class="math-container">$B_3$</span> such that <span class="math-container">$z \in B_3 \subseteq B_1 \cap B_2$</span>, see <a href="https://en.wikipedia.org/wiki/Base_(topology)" rel="nofollow noreferrer">Wikipedia</a> or Munkres' text book.</p>
<p>This latter condition is easy to check, as the only way we can be in this situation of two basic sets <span class="math-container">$P_i$</span> and <span class="math-container">$P_j$</span> and a <span class="math-container">$z$</span> in their intersection if when <span class="math-container">$P_i=P_j$</span> by 2. and so we can take the required base element to be that same <span class="math-container">$P_i$</span>!. So the other condition is fulfilled. QED.</p>
|
263,961 | <p>Some of the more organic theories considered in model theory (other than set theory, which, from what I've seen, seems to be quite distinct from "mainstream" model theory) are those which arise from algebraic structures (theories of abstract groups, rings, fields) and real and complex analysis (theories of expansions of real and complex fields, and sometimes both).</p>
<p>While relationships with algebra seem quite apparent, I wonder what are some interesting results in real and complex analysis that have nice model-theoretical proofs (or better yet, only model-theoretical proofs are known!)? </p>
<p>Of course, there's nonstandard analysis, but I hope to see some different examples. That said, I wouldn't mind seeing a particularly interesting application of nonstandard analysis. :)</p>
<p>I hope the question is at least a little interesting. I have only the very basic knowledge of model theory of that type (and the same applies to nonstandard analysis), so it may seem a little naive, but I got curious, hence the question.</p>
| Haskell Curry | 39,362 | <p>There is a result in functional analysis whose first known proof uses non-standard techniques:</p>
<blockquote>
<p><strong>Theorem</strong> If a bounded linear operator $ T $ on a Hilbert space $ \mathcal{H} $ is polynomially compact, i.e., $ P(T) $ is compact for some non-zero polynomial $ P $, then $ T $ has an invariant subspace. This means that there is a non-trivial proper subspace $ W $ of $ \mathcal{H} $ such that $ p(T)[W] \subseteq W $.</p>
</blockquote>
<p>The proof was given by Allen Bernstein and Abraham Robinson. Their result is significant because it is related to the so-called <em>Invariant-Subspace Conjecture</em>, an important unsolved problem in functional analysis. Paul Halmos, a staunch critic of non-standard analysis, supplied a standard proof of the result almost immediately after reading the pre-print of the Bernstein-Robinson paper. In fact, both proofs were published in the same issue of the <em>Pacific Journal of Mathematics</em>!</p>
|
263,961 | <p>Some of the more organic theories considered in model theory (other than set theory, which, from what I've seen, seems to be quite distinct from "mainstream" model theory) are those which arise from algebraic structures (theories of abstract groups, rings, fields) and real and complex analysis (theories of expansions of real and complex fields, and sometimes both).</p>
<p>While relationships with algebra seem quite apparent, I wonder what are some interesting results in real and complex analysis that have nice model-theoretical proofs (or better yet, only model-theoretical proofs are known!)? </p>
<p>Of course, there's nonstandard analysis, but I hope to see some different examples. That said, I wouldn't mind seeing a particularly interesting application of nonstandard analysis. :)</p>
<p>I hope the question is at least a little interesting. I have only the very basic knowledge of model theory of that type (and the same applies to nonstandard analysis), so it may seem a little naive, but I got curious, hence the question.</p>
| Tomasz Kania | 17,929 | <p>Using <em>continuous model theory</em>, Ilijas Farah, Brad Hart and David Sherman proved <a href="http://arxiv.org/abs/1111.0998" rel="nofollow">`blind man's version' of the Connes Embedding Problem</a>:</p>
<blockquote>
<p>There exists a separable ${\rm II}_1$-factor $M$ such that every other separable ${\rm II}_1$-factor embeds into an ultrapower of $M$.</p>
</blockquote>
<p>I find this result pretty amusing. On the C*-algebra side, I like results à la the Löwenheim–Skolem theorem: build a non-separable C*-algebra with your favourite axiomatisable properties and go to a separable C*-algebra which inherits those properties.</p>
|
4,007,450 | <p>Let <span class="math-container">$y=f(x)$</span> the graph of a real-valued function. We define its curvature by : <span class="math-container">$$curv(f) = \frac{|f''|}{(1+(f')^2)^{3/2}}$$</span></p>
<p>I would like to know if there is any function (apart from the trivial anwser <span class="math-container">$f(x)=0$</span>) whose curvature is itself. So what is the fixed point of <span class="math-container">$curv$</span> ?</p>
| Community | -1 | <p>The hypothesis implies an exponent of two for the group. But then it is abelian, since <span class="math-container">$ab=(ab)^{-1}=b^{-1}a^{-1}=ba$</span>, for any two elements of the group <span class="math-container">$a$</span> and <span class="math-container">$b$</span>.</p>
|
3,194,169 | <p>I am not a mathematician but I have a question.</p>
<p>I found </p>
<ul>
<li><span class="math-container">$1 / 3 = 0.33333....$</span></li>
<li><span class="math-container">$2 / 3 = 0.666666...$</span></li>
<li><span class="math-container">$3 / 3 = 1$</span> while it should be <span class="math-container">$0.999999....$</span> </li>
</ul>
<p>Same as dividing by <span class="math-container">$9$</span></p>
<ul>
<li><span class="math-container">$1/9=0.11111..$</span></li>
<li><span class="math-container">$2/9= 0.2222..$</span></li>
<li><span class="math-container">$3/9=0.3333..$</span></li>
<li><span class="math-container">$9/9=1$</span> while it should be <span class="math-container">$0.99999..$</span></li>
</ul>
<p>Divided by <span class="math-container">$7$</span> is little bit different. As </p>
<ul>
<li><span class="math-container">$1/7= 0.142857 \, 142857 \, 142857$</span></li>
<li><span class="math-container">$2/7= 0.285714 \, 285714\, 285714$</span></li>
<li><span class="math-container">$3/7= 0.428571\, 428571\, 428571$</span></li>
</ul>
<p>The same pattern of number repeated in the same arrangement. But with <span class="math-container">$7/7$</span> the answer is <span class="math-container">$1$</span> </p>
<p>I can understand the approximation but with the number increases the error increased.</p>
<p>To explain it:</p>
<pre><code> Should be
1/3=0.3333. 1/3=0.3333
2/3=0.6666. 2/3=0.6666
3/3=1. 3/3=0.9999
4/3=1.333. 4/3=1.2222
5/3=1.666. 5/3=1.55555
6/3=2 6/3=1.88888
</code></pre>
<p>Explanation is appreciated </p>
| Bernard | 202,857 | <p>There is a <em>bijection</em> between infinite ultimately periodic decimal expansions and rational numbers, <em>except</em> when the period consists only of the digit <span class="math-container">$9$</span>. In this case,, one proves that deleting the final <span class="math-container">$9$</span>s and replacing last digit <span class="math-container">$d\ne 9$</span> with the digit <span class="math-container">$d+1$</span> defines a decimal number which is equal to the given infinite decimal expansion. </p>
|
3,409,342 | <p>Let <span class="math-container">$X$</span> be a set containing <span class="math-container">$A$</span>.</p>
<p>Proof:
<span class="math-container">$y\in A \cup (X \setminus A) \Rightarrow y\in A$</span> or <span class="math-container">$y \in (X \setminus A)$</span></p>
<p>If <span class="math-container">$y \in A$</span>, Then <span class="math-container">$y \in X$</span> because <span class="math-container">$A \subset X$</span>. </p>
<p>If <span class="math-container">$y \in (X \setminus A)$</span>, Then <span class="math-container">$y \in X$</span> and <span class="math-container">$y \notin A$</span>. So <span class="math-container">$y \in X$</span>.</p>
<p>Therefore <span class="math-container">$y \in A \cup (A \setminus X) \Rightarrow y \in X$</span>.</p>
<p>Now I've proved that every element of <span class="math-container">$A \cup (A \setminus X)$</span> is also an element of <span class="math-container">$X$</span>.
But how to prove vice versa??</p>
| Manx | 483,923 | <p>Let <span class="math-container">$A\subseteq X$</span></p>
<p><span class="math-container">$\Leftarrow:$</span></p>
<p>Show</p>
<p><span class="math-container">$$X\subseteq A\cup(X-A)$$</span></p>
<p>Let <span class="math-container">$\phi_1,\phi_2$</span> be some formula</p>
<p>write <span class="math-container">$A=\{a:\phi_1(a)\}, X=\{x:\phi_2(x)\}$</span></p>
<p>Since <span class="math-container">$A\subseteq X$</span> we have for any <span class="math-container">$t$</span> that <span class="math-container">$\phi_1(t)$</span> implies <span class="math-container">$\phi_2(t)$</span></p>
<p>Note that <span class="math-container">$$A\cup(X-A)=\{t:\phi_1(t)\vee(\phi_2(t)\wedge\neg\phi_1(t))\}$$</span></p>
<p>To show <span class="math-container">$X$</span> is a subset of <span class="math-container">$A\cup(X-A)$</span> is same as to show that</p>
<p><span class="math-container">$$\forall t,\phi_2(t)\rightarrow\phi_1(t)\vee(\phi_2(t)\wedge\neg\phi_1(t))$$</span></p>
<p>Which is clearly a tautology</p>
<p>Hence proved</p>
<p><span class="math-container">$\Rightarrow:$</span></p>
<p>Similarly, for another direction we also need the condition <span class="math-container">$A\subseteq X$</span>, just consider</p>
<p><span class="math-container">$$\forall t,\big((\phi_1(t)\rightarrow\phi_2(t))\wedge(\phi_1(t)\vee(\phi_2(t)\wedge\neg\phi_1(t)))\big)\rightarrow\phi_2(t)$$</span></p>
<p>Not hard to see, it's also a tautology</p>
<p>Which proved <span class="math-container">$A\cup(X-A)\subseteq X\tag*{$\square$}$</span></p>
|
877,477 | <p>I'm stuck with the following question, which looks quite innocent.</p>
<p>I'd like to show that if a covering space map $f:\tilde{X}\to X$ between cell complexes is null-homotopic, then the covering space $\tilde{X}$ must be contractible.</p>
<p>Since $f$ is null-homotopic there exists a homotopy $H_t:\tilde{X}\to X$ from $H_0=x_0$ to $H_1=f$ and I would like to use it to construct another homotopy $G:\tilde{X}\to \tilde{X}$ from $G_0=\tilde{x}_0$ to $G_1=Id_{\tilde{X}}$.</p>
<p>By the homotopy lifting property, $H_t$ lifts to a homotopy $\tilde{H}_t:\tilde{X}\rightarrow \tilde{X}$ such that $H_t(x)=f(\tilde{H}_t(x))$ and $\tilde{H}_0(x)=\tilde{x}_0$</p>
<p>So we have a homotopy $\tilde{H}_t:\tilde{X}\rightarrow \tilde{X}$ from $\tilde{H}_0(x)= \tilde{x}_0$ to $\tilde{H}_1(x)$ and besides $f(x)=H_1(x)=f(\tilde{H}_1(x))$.</p>
<p>If $f$ was injective we would be done, but in principle $\tilde{H}_1(x)$ could be any point in $f^{-1}(x_0)$ right?</p>
| Kyle | 153,841 | <p>As you said, we're given a covering $f: \tilde X \to X$ and a homotopy $H_t : \tilde X \to X$ such that $H_0(\tilde x)=f(\tilde x)$ and $H_1(\tilde x)=x_0$ for some fixed $x_0 \in X$. Since $\operatorname{id}_{\tilde X}$ is a lift of $H_0=f$, there is a unique lift $\tilde H_t : \tilde X \to \tilde X$ of the homotopy $H_t$ such that $\tilde H_0= \operatorname{id}_{\tilde X}$. Since $f\circ \tilde H_1=H_1$ is a constant function and $f$ is a local homeomorphism, $\tilde H_1$ is locally constant. Since $\tilde X$ is connected, $\tilde H_1$ is constant, i.e. $\tilde X$ is contractible.</p>
<p><strong>Note:</strong> We must assume that $\tilde X$ is connected. To see this, consider any covering $\tilde X= X \sqcup X \to X$, where $X$ is contractible.</p>
|
2,319,119 | <p>Been given a question and find it to be too vague to know what's going on. </p>
<p>The question is: </p>
<p>$f(x) = 2x + 2$. Define $f(x)$ recursively.</p>
<p>I'm just quite puzzled as there is no $f(0)$, $f(1)$ or $f(x-1)$ function to go by other than the original function.</p>
<p>Supposed to be in the form of $f(x-1)$.</p>
<p>Any help appreciated thanks.</p>
| N. S. | 9,176 | <p><strong>Hint:</strong> If $\| f-g\|_\infty <\epsilon$ then
$$\left| \int_0^1 f(t) dt -\int_0^1 g(t) dt \right| <\epsilon$$</p>
|
492,392 | <p>I found this math test very interesting. I would like to know how the answer is being calculate?</p>
| Mufasa | 49,003 | <p>HINT: $49a+49b=6272$ can be written as $49(a+b)=6272$ and the average of $a$ and $b$ can be written as $\large\frac{a+b}{2}$</p>
|
2,563,300 | <p>If we put the Cartesian coordinates of a point in a 2 dimensional locus' equation then we get zero as the value if the point lies on the locus.
On putting coordinates of all other points in the equation which do not lie on the locus, we get a numerical value. Does this numerical value convey any information about the position of the point with respect to the locus?</p>
<p>Example:
Suppose you have the equation of a straight line as 2x+y=0. If you put the coordinates of the points lying on the line, you get R.H.S. as 0,i.e., the condition is satisfied. However if we put in the coordinates of points(x,y) which lie lower than the line in the L.H.S. (2x+y) , we get R.H.S.<0 and on putting the values of of points that lie above the line, we get R.H.S.>0. Can we derive any relationship between the position of points and the value if R.H.S. we get by putting their coordinates in the equation's L.H.S. ?</p>
| Karn Watcharasupat | 501,685 | <p>Let's look at it this way.</p>
<hr>
<p><strong>'At least one' case</strong></p>
<p>If at least one succeeds then we have
\begin{align}
P(\text{at least one success})&=P(A\cap B)+P(A\cap B')+P(A'\cap B)\\
&=\frac{1}{3}\cdot\frac{1}{5}+\frac{1}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{1}{5}\\
&=\frac{7}{15}
\end{align}</p>
<p>Now <strong>given</strong> that this is true, the cases where the success is B are $A\cap B$ and $A'\cap B$ (which is in fact, just $B$).</p>
<p>So we have
\begin{align}
P(B\mid\text{at least one success})&=\frac{P(B\cap\text{at least one success})}{P(\text{at least one success})}\\
&=\frac{P(A\cap B)+P(A'\cap B)}{P(A\cap B)+P(A\cap B')+P(A'\cap B)}\\
&=\frac{\frac{1}{3}\cdot\frac{1}{5}+\frac{2}{3}\cdot\frac{1}{5}}{\frac{1}{3}\cdot\frac{1}{5}+\frac{1}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{1}{5}}\\
&=\frac{\frac{3}{15}}{\frac{7}{15}}\\
&=\frac{3}{7}
\end{align}</p>
<p>Now can you do the rest in a similar manner? I will put the full solution down below in a while.</p>
<hr>
<p><strong>'At most one' case</strong></p>
<p>We have
\begin{align}
P(\text{at most one success})&=P(A'\cap B')+P(A\cap B')+P(A'\cap B)\\
&=\frac{2}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{1}{5}\\
&=\frac{14}{15}
\end{align}</p>
<p>Now <strong>given</strong> that this is true, the only case where the success is A is $A\cap B'$.</p>
<p>So we have
\begin{align}
P(A\mid\text{at most one success})&=\frac{P(A\cap\text{at most one success})}{P(\text{at most one success})}\\
&=\frac{P(A\cap B')}{P(A'\cap B')+P(A\cap B')+P(A'\cap B)}\\
&=\frac{\frac{1}{3}\cdot\frac{4}{5}}{\frac{14}{15}}\\
&=\frac{2}{7}
\end{align}</p>
<hr>
<p><strong>'Exactly one' case</strong></p>
<p>We have
\begin{align}
P(\text{exactly one success})&=P(A\cap B')+P(A'\cap B)\\
&=\frac{1}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{1}{5}\\
&=\frac{6}{15}
\end{align}</p>
<p>Now <strong>given</strong> that this is true, the only case where the success is A is $A\cap B'$.</p>
<p>So we have
\begin{align}
P(A\mid\text{exactly one success})&=\frac{P(A\cap\text{exactly one success})}{P(\text{exactly one success})}\\
&=\frac{P(A\cap B')}{P(A\cap B')+P(A'\cap B)}\\
&=\frac{\frac{1}{3}\cdot\frac{4}{5}}{\frac{6}{15}}\\
&=\frac{2}{3}
\end{align}</p>
|
4,095,715 | <p>I know how to do these in a very tedious way using a binomial distribution, but is there a clever way to solve this without doing 31 binomial coefficients (with some equivalents)?</p>
| Sid | 428,996 | <p><span class="math-container">$(1+x^2+x^4+x^6)^n = (1+x^4)^n(1+x^2)^n$</span></p>
<p>the <span class="math-container">$r^{th}$</span> term in first series: <span class="math-container">${n \choose r} x^{4r}$</span></p>
<p>the <span class="math-container">$l^{th}$</span> term in second series: <span class="math-container">${n \choose l} x^{2l}$</span></p>
<p>so we have <span class="math-container">$4r+2l = 12 \implies r=0, l=6; r=1, l=4; r=2, l=2; r=3, l=0$</span></p>
<p>So we have coefficient: <span class="math-container">${n \choose 0}{n \choose 6} +{n \choose 1}{n \choose 4} + {n \choose 2}{n \choose 2} + {n \choose 3}{n \choose 0}$</span></p>
|
4,064,353 | <blockquote>
<p>The number <span class="math-container">$1.5$</span> is special because it is equal to one quarter of the sum of its digits, as <span class="math-container">$1+5=6$</span> and <span class="math-container">$\frac{6}{4}=1.5$</span> .Find all the numbers that are equal to one quarter of the sum of their own digits.</p>
</blockquote>
<p>I was puzzling over this question for a while, but I couldn't find a formula without using the <span class="math-container">$\sum$</span> , but I can't really solve generalizations, only come up with them. The only thing I could come up with was to brute-force it, but I can't really come up with any 'special' numbers. Any help?</p>
| fleablood | 280,126 | <p>My first draft was messy.</p>
<p>With hindsight.</p>
<p>Let <span class="math-container">$N = \frac {K}4 = M\frac i4$</span> where <span class="math-container">$K$</span> is the sum of the digits. <span class="math-container">$M$</span> is the quotient integer of dividing <span class="math-container">$K$</span> by <span class="math-container">$4$</span> and <span class="math-container">$i=0,1,2,3$</span> is the fractional remainder.</p>
<p>Note: <span class="math-container">$K$</span> is the sum of the digits of <span class="math-container">$N$</span> which is also the sum of the digits of <span class="math-container">$M$</span> plus the sum of the digits of <span class="math-container">$\frac i4$</span>.</p>
<p>So <span class="math-container">$4N = 4M + i =K$</span>. If the digits of <span class="math-container">$M$</span> are <span class="math-container">$d_k$</span> then</p>
<p><span class="math-container">$4N = 4M + i=4d_m \times 10^m + ..... + 4d_1\times 10 + 4d_0 + \begin{cases}0\\ 1\\ 2\\ 3\end{cases}$</span></p>
<p>While <span class="math-container">$K =d_m + ..... + d_1 + d_0 +\text{sum of the digits of }\frac i4$</span> so</p>
<p><span class="math-container">$4d_m \times 10^m + ..... + 4d_1\times 10 + 4d_0 + \begin{cases}0\\ 1\\ 2\\ 3\end{cases}=d_m + ..... + d_1 + d_0 +\text{sum of the digits of }\frac i4$</span></p>
<p><span class="math-container">$d_m(4\cdot 10^m-1) + .... d_1(40 -1) + 3d_0 + \begin{cases}0\\ 1\\ 2\\ 3\end{cases}= \text{sum of the digits of }\frac i4$</span></p>
<p>But <span class="math-container">$\text{sum of the digits of }\frac i4 =\begin{cases}0\\2+5=7\\5\\7+5=12\end{cases}$</span></p>
<p>So we have <span class="math-container">$d_m(4\cdot 10^m-1) + .... d_1(40 -1) + 3d_0 = \begin{cases}0-0=0\\2+5-1=6\\5-2=3\\7+5-3=9\end{cases}$</span></p>
<p>But the RHS is less than <span class="math-container">$10$</span> so none of the <span class="math-container">$d_{k_{k> 0}}$</span> can be none zero and <span class="math-container">$M$</span> is a single digit, <span class="math-container">$d_0$</span>.</p>
<p>And we have</p>
<p><span class="math-container">$3d_0 = \begin{cases}0\\6\\3\\9\end{cases}$</span></p>
<p>So <span class="math-container">$M=d_0 = \begin{cases}0\\2\\1\\3\end{cases}$</span></p>
<p>ANd <span class="math-container">$N = M +\frac i4 = \begin{cases}0\\2.25\\1.5\\3.75\end{cases}$</span></p>
<p>==== first answer below (more thought and scrabble and not as slick-- a "rough draft")=====</p>
<p>So the sum of the digits is an integer. And and integer divided by <span class="math-container">$4$</span> will result in four possible cases.</p>
<p><span class="math-container">$\frac {integer}4 = \begin{cases}n.00\\n.25\\n.5\\n.75\end{cases}$</span>.</p>
<p>So the least power of <span class="math-container">$10$</span> in the number can be <span class="math-container">$-2$</span>.</p>
<p>So let <span class="math-container">$N = \sum_{k=-2}^m a_k\times 10^m$</span> (where <span class="math-container">$a_m \ne 0$</span> but the other <span class="math-container">$a_k$</span> may be).</p>
<p><span class="math-container">$ \sum_{k=-2}^m a_k\times 10^m = \frac {\sum_{k=-2}a_k}4< \frac{9\times (m+2)}4$</span> but even more so <span class="math-container">$a_{-1}+ a_{-2} \le 7+5 = 12$</span> so <span class="math-container">$N < 9m + 12$</span> and <span class="math-container">$a10^m \le N < \frac {9m + 12}4<3m + 3$</span> can give us an upper limit of <span class="math-container">$m$</span>.</p>
<p>If <span class="math-container">$m \ge 2$</span> then <span class="math-container">$3m + 3 < 10^m$</span> obviously so <span class="math-container">$m\le 1$</span>. If <span class="math-container">$m =1$</span> then <span class="math-container">$10a_1 < 6$</span> is impossible. SO <span class="math-container">$m\le 0$</span></p>
<p>So we have four cases:</p>
<p><span class="math-container">$N = a.00$</span> and so <span class="math-container">$a = \frac 14 a$</span> and <span class="math-container">$a = 0$</span> and <span class="math-container">$N = 0$</span>. That's a solution.</p>
<p><span class="math-container">$N = a.25$</span> and <span class="math-container">$a +\frac 14 = \frac 14(a + 7)$</span> so <span class="math-container">$4a+1=a+7$</span> so <span class="math-container">$3a=6$</span> so <span class="math-container">$a=2$</span> and <span class="math-container">$N = 2.25$</span> yield <span class="math-container">$2.25 =\frac 14 \times 9$</span>. That's a solution.</p>
<p><span class="math-container">$N = a.5$</span> and <span class="math-container">$a + \frac 12 = \frac 14(a+5)$</span> so <span class="math-container">$4a + 2 = a+5$</span> so <span class="math-container">$3a = 3$</span> and <span class="math-container">$a = 1$</span> and <span class="math-container">$N=1.5$</span> yields your solution.</p>
<p><span class="math-container">$N = a.75$</span> and <span class="math-container">$a + \frac 34 =\frac 14(a+12)$</span> so <span class="math-container">$4a + 3=(a+12)$</span> so <span class="math-container">$3a =9$</span> and <span class="math-container">$a = 3$</span> and <span class="math-container">$N=3.75 = 3\frac 34 = \frac {15}4 = \frac {3+7+5}4$</span> is a solution.</p>
<p>Pretty cool and pretty cute.</p>
|
2,808,512 | <p>Show that if an $n\times n$ matrix $A$ satisfies $A^T=-A$, then $x^TAx=0$ for any $n\times 1$ vector $x$.</p>
<p>My attempt: Since matrix transpose won't affect the diagonal entries, so matrix $A$ has only zeros on its diagonal. </p>
<p>Then I tried to write $x$ in the form of $\begin{pmatrix}
x_1\\
x_2\\
\vdots\\
x_n
\end{pmatrix}$ and $A$ in the form of $\begin{pmatrix}
0 & a_{11} & \cdots & a_{1n}\\
a_{21} & 0 & \cdots & \cdots \\
\vdots &\ddots &\ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & 0
\end{pmatrix}$ and multiply them in this form, but it doesn't make any sense to me.</p>
| Lord KK | 446,929 | <p>What you can see is that $x^TAx$ is of type $(1\times1)$, hence symmetric, i.e.,</p>
<p>$$(x^TAx)^T = x^TAx.$$
On the other hand we have
$$x^TA^Tx= x^TAx.$$</p>
<p>From $A^T=-A$ we get $$-x^TAx = x^TAx\iff 2x^TAx =0\iff x^TAx=0.$$</p>
|
313,298 | <p>For example, let's say that I have some sequence $$\left\{c_n\right\} = \left\{\frac{n^2 + 10}{2n^2}\right\}$$ How can I prove that $\{c_n\}$ approaches $\frac{1}{2}$ as $n\rightarrow\infty$?</p>
<p>I'm using the Buchanan textbook, but I'm not understanding their proofs at all.</p>
| Kaster | 49,333 | <p>$$
\lim_{n \rightarrow \infty} \frac {n^2+10}{2n^2} = \lim_{n \rightarrow\infty} \frac {1+\frac {10}{n^2}}{2} = \lim_{n \rightarrow \infty} \left (\frac 12+\frac{10}{2n^2} \right) = \frac 12+5\lim_{n \rightarrow \infty}\frac 1{n^2}=\frac 12
$$</p>
|
1,130,645 | <p>I am asked to prove: if <span class="math-container">$|a|<\epsilon$</span> for all <span class="math-container">$\epsilon>0$</span>, then <span class="math-container">$a=0$</span></p>
<p>I can prove this as follows.</p>
<p>Assume <span class="math-container">$a \not= 0$</span></p>
<p>I want to show then that <span class="math-container">$|a| \geq \epsilon$</span> for some <span class="math-container">$\epsilon$</span></p>
<p>We let <span class="math-container">$\epsilon = \frac{|a|}{2}$</span> and we are done.</p>
<p>However, I am curious if you can use the idea of infinitesimals to prove this directly.
Can I let <span class="math-container">$\epsilon$</span>=an (or maybe the?) <a href="http://mathworld.wolfram.com/Infinitesimal.html" rel="nofollow noreferrer">infinitesimal value</a> and then just show that:</p>
<p><span class="math-container">$|a|<\epsilon$</span> <span class="math-container">$\implies$</span> <span class="math-container">$-\epsilon<a<\epsilon$</span> <span class="math-container">$\implies$</span> <span class="math-container">$|a|=a=0$</span></p>
<p>I don't know much about infinitesimals but wanted to try and prove this directly. This was the first thing that occurred to me for some reason. Can anyone shed some light on infinitesimals and whether or not I can use them this way? this might be completely off base because I really have almost no knowledge about how infinitesimals fit into mathematics and mathematical thinking.</p>
| Hagen von Eitzen | 39,174 | <p>You can do so if you do work with infinitesimals (say Robinson's hyperreal numbers) and, to really make the proof this short, make sure that $a$ is a standard number. One needs to know that application of the $\operatorname{st}()$ part to both sides of a strict inequality produces a corresponding un-stricted inequality:
$$\begin{align} \forall\epsilon>0\colon |a|<\epsilon&\implies \forall\epsilon>0\colon \operatorname{st}(|a|)\le\operatorname{st}(\epsilon)\\&\implies 0\le |a|=\operatorname{st}(|a|)\le \operatorname{st}(\epsilon)=0\text{ for any positive $\epsilon\approx 0$}\end{align}$$</p>
|
351,534 | <p>It was already known to Weil that a sufficiently reasonable cohomology theory for algebraic varieties over <span class="math-container">$\mathbb{F}_p$</span> would allow for a possible solution to the Weil conjectures. </p>
<p>It was also understood that such a cohomology theory could not take values in vector spaces over either the rational numbers <span class="math-container">$\mathbb{Q}$</span> or the <span class="math-container">$p$</span>-adic numbers <span class="math-container">$\mathbb{Q}_p$</span>. The classical argument that I know is that the cohomology of a supersingular elliptic curve should admit an action by a quaternion algebra structure of the endomorphism group of the elliptic curve, and representation theory rules out such an action. (Side question: Due to whom is this reasoning, and was it known to Weil's generation?)</p>
<p>This reasoning does not, however, rule out the possibility of cohomology with coefficients in <span class="math-container">$\mathbb{Q}_\ell$</span> with <span class="math-container">$\ell \neq p$</span>, and indeed <span class="math-container">$\ell$</span>-adic cohomology would end up being found. </p>
<p><strong>Question.</strong> Were there any other historical reasons to believe that looking at cohomology valued in <span class="math-container">$\mathbb{Q}_\ell$</span>-vector spaces would lead to something fruitful?</p>
<p>N.B. This question was posted on MathSE, but I got advised to post it here instead.</p>
| S. Carnahan | 121 | <p>One historical reason for considering <span class="math-container">$\ell$</span>-adic cohomology, not completely disconnected from the example you introduce, is that for a curve over a field, we get a natural Galois representation by taking the <span class="math-container">$\ell$</span>-adic Tate module of the Jacobian (i.e., the projective limit of <span class="math-container">$\ell$</span>-power torsion). Furthermore, if such a curve is defined over a subfield of the complex numbers, then the rank of the Tate module is equal to the rank of the classical degree 1 cohomology of the complexified curve. We now know that the Tate module is naturally dual to étale cohomology in degree 1.</p>
<p>One might then reasonably hope for a similar relationship between the higher degree classical cohomology of higher dimensional varieties and certain <span class="math-container">$\ell$</span>-adic Galois representations naturally attached to the variety.</p>
|
3,203,577 | <p>What is the computational complexity of solving a linear program with <span class="math-container">$m$</span> constraints in <span class="math-container">$n$</span> variables?</p>
| Zhao Song | 324,173 | <p>Brand's 2020 result derandomized the Cohen, Lee and Song result.
Here is the link
<a href="https://arxiv.org/pdf/1910.11957.pdf" rel="noreferrer">https://arxiv.org/pdf/1910.11957.pdf</a></p>
|
2,512,396 | <p>i am trying to compute inverse Laplace transform of function = $\tan ^{-1} (2/s)$</p>
| Nosrati | 108,128 | <p>$${\cal L}\Big(\frac{\sin2t}{t}\Big)=\int_s^\infty\frac{2}{s^2+4}~ds=\arctan\frac{s}{2}\Big|_s^\infty=\frac{\pi}{2}-\arctan\frac{s}{2}=\arctan\frac{2}{s}$$</p>
|
3,565,727 | <blockquote>
<p>Show that the antiderivatives of <span class="math-container">$x \mapsto e^{-x^2}$</span> are uniformly continuous in <span class="math-container">$\mathbb{R}$</span>.</p>
</blockquote>
<p>So we know that for a function to be uniformly continuous there has to exists <span class="math-container">$\varepsilon$</span> s.t when <span class="math-container">$|x-y| < \delta$</span> implies <span class="math-container">$|f(x)-f(y)| < \varepsilon$</span>.</p>
<p>But how do we go about this since we cannot really integrate <span class="math-container">$e^{-x^2}$</span>and we need the antiderivative?</p>
| Gary | 83,800 | <p>Any antiderivative can be written in the form
<span class="math-container">$$
\int_0^x {e^{ - t^2 } dt} + C,
$$</span>
where <span class="math-container">$C$</span> is a constant. Thus
<span class="math-container">$$
\left| {\int_0^x {e^{ - t^2 } dt} + C - \left( {\int_0^y {e^{ - t^2 } dt} + C} \right)} \right| = \left| {\int_y^x {e^{ - t^2 } dt} } \right| \le \left| {\int_y^x {1dt} } \right| = \left| {x - y} \right|.
$$</span>
Can you finish from here?</p>
|
3,565,727 | <blockquote>
<p>Show that the antiderivatives of <span class="math-container">$x \mapsto e^{-x^2}$</span> are uniformly continuous in <span class="math-container">$\mathbb{R}$</span>.</p>
</blockquote>
<p>So we know that for a function to be uniformly continuous there has to exists <span class="math-container">$\varepsilon$</span> s.t when <span class="math-container">$|x-y| < \delta$</span> implies <span class="math-container">$|f(x)-f(y)| < \varepsilon$</span>.</p>
<p>But how do we go about this since we cannot really integrate <span class="math-container">$e^{-x^2}$</span>and we need the antiderivative?</p>
| egreg | 62,967 | <p>Every function <span class="math-container">$f$</span> defined and differentiable over an interval, with a bounded derivative, is uniformly continuous. Indeed, by the mean value theorem, if <span class="math-container">$x\ne y$</span>, we can say that
<span class="math-container">$$
\frac{f(x)-f(y)}{x-y}=f'(c)
$$</span>
with <span class="math-container">$c$</span> between <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. Therefore, if <span class="math-container">$|f'(x)|\le L$</span> for every <span class="math-container">$x$</span>, we get
<span class="math-container">$$
|f(x)-f(y)|\le L|x-y|
$$</span>
Thus <span class="math-container">$f$</span> is not only uniformly continuous, but also Lipschitzian, which is a stronger property.</p>
<p>Since <span class="math-container">$0\le e^{-x^2}\le 1$</span>, we're done.</p>
|
2,197,065 | <p>Baire category theorem is usually proved in the setting of a complete metric space or a locally compact Hausdorff space.</p>
<p>Is there a version of Baire category Theorem for complete topological vector spaces? What other hypotheses might be required?</p>
| Michael Hardy | 11,667 | <p>You can take an average of several distances, as with several times.</p>
<p>One place where "average time" occurs is in statistical things where, for example, a phone call arriving at a switchboard has no effect on the timing of another phone call arriving, and one speaks of the average time from the arrival of one phone call to the next.</p>
<p>"Average speed" arises when say a car passed a certain point at 1:00 PM and reached a point $35$ miles from there at 2:00 PM, so that you can say its average speed during that hour was 35 miles per hour.</p>
<p>Perhaps "average distance" does not occur in this way in standard examples in math textbooks, but there is no reason why you can't take an average of several distances. And astronomers do speak of such things as the average distance from the earth to the sun (since the distance undergoes seasonal variations).</p>
|
391,020 | <p>Does there exist a non-zero homomorphism from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}$ ? If yes, state the mapping. How is this map exactly? </p>
| Billy | 39,970 | <p>Suppose there is a homomorphism $f: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}$. What could $f(1)$ be? Let's call it $a$. Then $f(2) = f(1+1) = f(1)+f(1) = a+a = 2a$, and likewise $f(3) = 3a$, $\dots,$ $f(\underbrace{1+\cdots+1}_{n\text{ times}}) = na$. But $\underbrace{1+\cdots+1}_{n\text{ times}} = 0$, and $f(0) = 0$. So what can $a$ be?</p>
<p>Once you've worked out which value(s) of $a$ is/are allowed, can you check that the associated map $f$ is a homomorphism? (We've already done this, really, but make sure you're sure of that.)</p>
|
738,435 | <p>I am having a problem understanding how to determine if a function is one to one.</p>
<p>The problem is: Show that the function f(x) = 3x+4 is one-to-one.</p>
<p>Also, I'm being thrown off by the notation x[subset 1] = x[subset 2], what does that mean, loosely speaking? "In my eyes" that would mean two different values are being represented. </p>
| A. Wong | 100,535 | <p>In general, to show that a function is one to one you would suppose that $f(x)=f(y)$ and show that it must be the case that $x=y$.</p>
<p>In this case you start by assuming that $3x+4=3y+4$ and then proceed to show that $x=y$.</p>
<p>I believe when you say x[subset 1]=x[subset 2] you are actually saying $x_{1}=x_{2}$ which in my answer is just the $x$ and $y$.</p>
<p><img src="https://i.stack.imgur.com/o1zBE.jpg" alt="enter image description here"></p>
<p>Consider the graph of your function to be more specific. To see that it is a one to one function, you draw a horizontal line and see that the horizontal line never intersects the graph more than once. This tells you that for each value of $x$ your function only takes one value of $f(x)$. </p>
<p>By using the method of proof I described above you are showing that for every value on the $y$ axis you have only one value on the $x$ axis that gives you that value.</p>
|
1,758,194 | <p>Consider the function f: {-1, +1} -> R defined by</p>
<p>$f(x)= \arcsin (\frac{1+x^2}{2x})$.</p>
<p>Due to the following two inequalities :</p>
<p>(i) $1+x^2 \geq 2x$</p>
<p>(ii)$1+x^2 \geq -2x$ , </p>
<p>the function can only be defined at $x=1$ and $x=-1$. I have learnt that the epsilon delta definition only includes those values of $x$ which are in the domain of $f(x)$. But in this case, the function isn't defined on <em>either</em> side of x=1.</p>
<p>So this is my question: Is it correct to say that the limit as $x$ approaches $1$ of $f(x$) is $\frac{\pi}{2}$ ?</p>
<p>Can the above question be given a <strong>definitive</strong> "yes" or "no" answer, or must it(unfortunately) vary from person to person? </p>
<p>If the latter, is the "precise" definition of a limit not precise enough?</p>
<p>How can the answer be proved or disproved using the epsilon delta definition?</p>
<p>I have also read that functions are by default continuous at isolated points. Can I conclude from the definition of continuity (the limit equals the value of the function evaluated at the point) that the limit must exist?</p>
<p><strong>Note :</strong> Forgive my ignorance but I do not know a thing about topology. I'm looking for a detailed answer but in simple terms, preferably written in the language of calculus.</p>
<p>Thanks for the help. </p>
| eyeballfrog | 395,748 | <p>Usually when talking about limits of functions $f:D\subseteq \mathbb R \rightarrow \mathbb R$ in calculus, we work with functions whose domain is not a discrete set. Because of this, an extra condition on the limit is left out: the limit can only be taken at limit points of the domain. A point $c$ is a limit point of the domain $D$ if, for all $\delta > 0$, there exists $x\in D$ with $0 < |x-c| < \delta$. Note that $c \in D$ is not required. It should be fairly clear why this condition is desired in the definition of the limit. So the epsilon-delta definition of the limit at $c$ should be augmented with the condition $\forall(\delta > 0) \exists (x \in \mathrm{dom}[f]): 0 < |x-c| < \delta$.</p>
<p>The domain of your function $f$ is $\{-1, 1\}$. Like all discrete sets, it has no limit points, and so the limit is not defined anywhere.</p>
|
1,668,653 | <blockquote>
<p>If $\sum a_n$ is convergent and $a_n>0$, then $\sum (-1)^n a_n$ is convergent.</p>
</blockquote>
<p>So far I've tried convergence/divergence tests and also I tried to prove this using partial sums. But tests do not work because the latter series <em>includes negative terms</em>. Maybe the Alternating Series Test could have worked but "$a_n$ is decreasing" is not an initial condition. I know the proof will be very easy but I am stuck here. Could you provide me with a little <strong>hint</strong>?</p>
| Nick | 272,448 | <p>HINT: any absolutely convergent series is convergent.Let me know if you want the full solution</p>
|
527,328 | <p>$$\lim_{x\rightarrow\infty}\left(\frac{x+1}{x-2}\right)^{2x-1}$$</p>
<p>What are the steps to solve it? Probably the division should be multiplied by some expression.</p>
| DonAntonio | 31,254 | <p>$$\left(\frac{1+x}{-2+x}\right)^{-1+2x}=\left[\left(1+\frac3{x-2}\right)^{x-2}\right]^{2}\left(1+\frac3{x-2}\right)^3$$</p>
<p>And now use the basic lemma</p>
<p>$$\lim_{x\to\infty}\left(1+\frac a{f(x)}\right)^{f(x)}=e^a$$</p>
<p>where $\;a\in\Bbb R\;$ and $\;f\;$ is a funcion s.t.</p>
<p>$$\lim_{x\to\infty}f(x)=\infty$$</p>
|
3,540,593 | <p>Are <span class="math-container">$n$</span> vectors are orthogonal if performing the inner product of all <span class="math-container">$n$</span> vectors at once yields zero?</p>
<p>In other words, could I say that <span class="math-container">$\hat{i} \perp \hat{j} \perp \hat{k}$</span>?</p>
<p>For example, suppose I have the three vectors <span class="math-container">$\hat{i} = <2, 0, 4>; \hat{j}= <0, 1, 0>; \hat{k} = <2, 0, 1>$</span>. Their inner product would be <span class="math-container">$2(0)(2) + 0(1)(0) + 4(0)(1) = 0$</span>. Can I say that the three vectors <span class="math-container">$\hat{i}, \hat{j}, $</span> and <span class="math-container">$ \hat{k}$</span> are orthogonal, or do I have to multiply out each pair?</p>
| lab bhattacharjee | 33,337 | <p>Hint</p>
<p>We need <span class="math-container">$p$</span> to divide <span class="math-container">$k^2-1=(k+1)(k-1)$</span></p>
<p>But as <span class="math-container">$k+1-(k-1)=2,p$</span> must divide exactly one of <span class="math-container">$k+1,k-1$</span></p>
<p>Now we need <span class="math-container">$1<k<p$</span></p>
|
162,863 | <p>if I have a quaternion which describes an arbitrary rotation, how can I get for example only the half rotation or something like 30% of this rotation?</p>
<p>Thanks in advance!</p>
| Christian Blatter | 1,303 | <p>See here: <a href="http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation" rel="nofollow">Wikipedia on quaternions and rotations</a>.</p>
<p>To the benefit of all: One can encode a rotation $T\in SO(3)$ as a quaternion $q+ q_1 i+ q_2 j + q_3 k$ of norm $1$. Here $q=\cos{\theta\over2}$ and $(q_1,q_2,q_3)=\sin{\theta\over2}{\bf a}$, where ${\bf a}$ denotes the unit vector on the axis of $T$ and $\theta$ the rotation angle of $T$. The essential point is that to the composition of two rotations $T_1$, $T_2\in SO(3)$ corresponds the product of the two associated norm-$1$ quaternions.</p>
<p>As to your question: From the given quaternion $(q,{\bf q})$ you can read off the rotation angle $\theta$ (there are some sign conventions to observe) and the axis ${\bf a}$. For $\sqrt{T}$ (or another fraction of $T$) the axis is the same, but you will need a pocket calculator to compute $\cos\theta'$ and $\sin\theta'$, where $\theta'$ denotes the desired fraction of $\theta$.</p>
|
4,079,842 | <p>Consider the below function.
<span class="math-container">$$f(x,y) =
\begin{cases}
1 & xy \neq 0 \\
0 & xy = 0
\end{cases}$$</span>
Suppose i want to calculate the below repeated limit.
<span class="math-container">$$\lim_{x\to0}\lim_{y\to0}f(x,y)$$</span>.
In general textbooks in India for BS courses,the above repeated limit computation is given to be <span class="math-container">$1$</span> which is fine if we know that the point <span class="math-container">$(x,y)\neq 0$</span>. However , if we be really precise, then ,this limit computation is<span class="math-container">$$\lim_{x\to0}[\lim_{y\to0}f(x,y)]$$</span>.
The inner limit is unknown to us as we donot know about the nature of <span class="math-container">$x$</span>. So my question is , in repeated limit calculation , do we assume that there is some distance left that we still have to move some in <span class="math-container">$x$</span> and <span class="math-container">$y$</span> to approach a given point?</p>
| Adam Rubinson | 29,156 | <p><span class="math-container">$$\lim_{x\to0}[\lim_{y\to0}f(x,y)]$$</span></p>
<blockquote>
<p>The inner limit is unknown to us
as we donot know about the nature of <span class="math-container">$x$</span>.</p>
</blockquote>
<p>This is false. We do know the nature of <span class="math-container">$x$</span>: it is either approaching <span class="math-container">$0$</span> from above or from below.</p>
<p>To calculate the double limit <span class="math-container">$$\large{\lim_{x\to0^-}[\lim_{y\to0}f(x,y)]},$$</span></p>
<p>fix <span class="math-container">$x<0$</span> and <strong>then</strong> calculate <span class="math-container">$\large{L_x = \lim_{y\to0}f(x,y).}\quad (1)$</span></p>
<p>The value of <span class="math-container">$\large{L_x}$</span> depends on the value of <span class="math-container">$x$</span> you fixed before you calculated <span class="math-container">$\lim_{y\to0}f(x,y).$</span></p>
<p>Now repeat process <span class="math-container">$(1)$</span> whilst taking values of <span class="math-container">$x<0$</span> closer and closer to <span class="math-container">$0.$</span></p>
<p>In other words,</p>
<p><span class="math-container">$$\large{\lim_{x\to0^-}[\lim_{y\to0}f(x,y)] = \lim_{x\to0^-}L_x}.$$</span>
<span class="math-container">$$$$</span>
Similarly, To calculate the double limit <span class="math-container">$$\large{\lim_{x\to0^+}[\lim_{y\to0}f(x,y)]},$$</span></p>
<p>fix <span class="math-container">$x>0$</span> and <strong>then</strong> calculate <span class="math-container">$\large{L_x = \lim_{y\to0}f(x,y).}\quad (2)$</span></p>
<p>The value of <span class="math-container">$\large{L_x}$</span> depends on the value of <span class="math-container">$x$</span> you fixed before you calculated <span class="math-container">$\lim_{y\to0}f(x,y).$</span></p>
<p>Now repeat process <span class="math-container">$(2)$</span> whilst taking values of <span class="math-container">$x>0$</span> closer and closer to <span class="math-container">$0.$</span></p>
<p>In other words,</p>
<p><span class="math-container">$$\large{\lim_{x\to0^+}[\lim_{y\to0}f(x,y)] = \lim_{x\to0^+}L_x}.$$</span></p>
<p>If <span class="math-container">$\large{\lim_{x\to 0^+}L_x}$</span> and <span class="math-container">$\large{\lim_{x\to 0^+}L_x}$</span> both exist and are equal, then by definition they are equal to <span class="math-container">$\lim_{x\to0}[\lim_{y\to0}f(x,y)]$</span>.</p>
|
492,125 | <p>How many different equivalence relations can be defined on a set of five elements?</p>
| user248070 | 248,070 | <p>The formula is $2^n-n$ where $n$ denotes the number of elements. So here, the answer would be $2^5-5= 32-5 = 27$.</p>
|
4,106,575 | <p>Consider a number drawn from <span class="math-container">$U(1,100)$</span>. When we make an incorrect guess, we are told whether the target number is smaller or larger. So we employ a binary search approach, where the first guess is 50. If that is not the target, then that means our next guess is going to be 25 or 75.</p>
<p>What is the probability that binary search ends in 2 guesses?</p>
<p>At first, I thought it was <span class="math-container">$P(B|A)P(A)$</span> where <span class="math-container">$A$</span> is the event that the first guess is wrong and <span class="math-container">$B$</span> is the event that the second guess is right. We know <span class="math-container">$P(A) = 99/100$</span>. When <span class="math-container">$A$</span> occurs, that means we would either search in the interval <span class="math-container">$[1,49]$</span> or <span class="math-container">$[51, 100]$</span>, which has 49 and 50, elements, respectively.
So I believe <span class="math-container">$P(B | A) = 49/100 * 1/49 + 50/100 * 1/50 = 2/100$</span>.</p>
<p>So <span class="math-container">$P(B|A)P(A) = 99/100 * 2/100 = 0.198$</span>.</p>
<p>Then I second guessed myself and started wondering why is it not <span class="math-container">$2/100$</span>? If it ends in 2 guesses, it means the target number is 25 or 75. Since the target number is uniformly chosen, the probability that it being 25 or 75 is <span class="math-container">$2/100=1/50 = 0.2$</span>.</p>
<p>Both of these answers seem right to me, but 1 one of them must be wrong. Which one is wrong?</p>
| Ahmad Beirami | 915,095 | <p>The correct answer is <span class="math-container">$0.02$</span>, and your second approach is correct.</p>
<p>Here is how you can make your first approach work:
You should multiply 2/99 (probability of ending at the second round given you did not end in the first round -- see below) by 99/100 (probability of not ending in the first round) and you will get the same answer.
<span class="math-container">$$
P(B|A) = \frac{49}{99} \frac{1}{49} + \frac{50}{99}\frac{1}{50} = \frac{2}{99}.
$$</span></p>
|
1,422,859 | <p>$$\sqrt{1000}-30.0047 \approx \varphi $$
$$[(\sqrt{1000}-30.0047)^2-(\sqrt{1000}-30.0047)]^{5050.3535}\approx \varphi $$
Simplifying Above expression we get<br>
$$1.0000952872327798^{5050.3535}=1.1618033..... $$
Is this really true that
$$[\varphi^2-\varphi]^{5050.3535}=\varphi $$</p>
| Vincenzo Oliva | 170,489 | <p>No, not at all. Factoring and then using a well known property of $\varphi$: $$\left(\varphi(\varphi-1)\right)^{5050.3535}=\left(\frac{\varphi}{\varphi}\right)^{5050.3535}=1\ne\varphi. $$</p>
<p>The argument needs to be a bit larger than $1$. You're dealing with an approximation.</p>
|
1,037,621 | <p>There are 20 people at a chess club on a certain day. They each find opponents and start playing. How many possibilities are there for how they are matched up, assuming that in each game it does matter who has the white pieces (and who has the black ones). </p>
<p>I thought it might be $$\large2^{\frac{20(20-1)}2}$$ is this correct?</p>
| Suzu Hirose | 190,784 | <p>If we arrange the players in some order, say of height, and let the shortest player choose first, then there are 19 people for him to choose from. Then let the next shortest remaining player choose, and there are 17 people left to choose, and so on. Thus the number of ways to choose pairs is
$$
19\times17\times15\times13\times ...\times3\times1
$$
This is equal to
$$
{20!\over 2^{10}10!}
$$
If we care about which is white and which is black, once we have chosen the pairs, the number of ways to choose who is white and who is black among the ten pairs is $2^{10}$, so we get
$$
{20!\over2^{10}10!}\times2^{10}={20!\over10!}
$$</p>
|
2,387,435 | <p><a href="https://i.stack.imgur.com/RxHmS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RxHmS.png" alt="enter image description here"></a></p>
<p>We wish to construct a rectangular auditorium with a stage shaped as a semicircle of radius $r$, as shown in the diagram below (white is the stage and green is the seating area). For safety reasons, light strips must be placed on the perimeter of the seating area. If we have $45\pi + 60$ meters of light strips, what should $r$ be so that the seating area is maximized?</p>
<p>My issue is that I tried to solve it the "geometric way",
figuring that the perimeter (in r) for the seating area is going to be $(3+\pi)r$. I set it equal to $45\pi+60$ to try and find the closest number to equal the value of $r$. Is this the way I should approach the problem, or is there a "calculus" style method I should use to come to a solution?</p>
<p>Edit: </p>
| Siong Thye Goh | 306,553 | <p>I will leave $R_y(\theta)$ as an exercise for you. </p>
<p>Below is a verification of $R_x(\theta)$ is a rotation about the $x$-axis.</p>
<p>Denote the Pauli matrices as
$$X=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, Y=\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}, Z=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$</p>
<p>We have the following expression for $R_x(\theta)$
$$R_x(\theta) = \cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X$$</p>
<p>Consider $$\rho=\frac12(I+r_xX+r_yY+r_zZ)$$</p>
<p>$\begin{align}\rho' &= R_x(\theta)\rho R_x(\theta)^\dagger \\
&=\frac12(I+r_xR_x(\theta)XR_x(\theta)^\dagger+r_yR_x(\theta)YR_x(\theta)^\dagger+r_zR_x(\theta)ZR_x(\theta)^\dagger)\end{align}\tag{1}$</p>
<p>Now, let's analyze each separate term:</p>
<p>\begin{align}R_x(\theta)XR_x(\theta)^\dagger &= \left(\cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X\right)X\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\left(\cos \left( \frac{\theta}{2} \right)X-i\sin\left( \frac{\theta}{2}\right)I\right)\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\cos^2 \left( \frac{\theta}{2}\right)X-i\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)I+i\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)I+\sin^2 \left( \frac{\theta}{2}\right)X\\
&= X\end{align}</p>
<p>\begin{align}R_x(\theta)YR_x(\theta)^\dagger &= \left(\cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X\right)Y\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\left(\cos \left( \frac{\theta}{2} \right)Y+\sin\left( \frac{\theta}{2}\right)Z\right)\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\cos^2 \left( \frac{\theta}{2}\right)Y+\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Z+\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Z-\sin^2 \left( \frac{\theta}{2}\right)Y\\
&= \cos (\theta) Y + \sin(\theta)Z\end{align}</p>
<p>\begin{align}R_x(\theta)ZR_x(\theta)^\dagger &= \left(\cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X\right)Z\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\left(\cos \left( \frac{\theta}{2} \right)Z-\sin\left( \frac{\theta}{2}\right)Y\right)\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\cos^2 \left( \frac{\theta}{2}\right)Z-\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Y-\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Y-\sin^2 \left( \frac{\theta}{2}\right)Z\\
&= -\sin (\theta) Y + \cos(\theta)Z\end{align}</p>
<p>\begin{align}\rho'&=\frac12 \left(I+r_xX+r_y(\cos (\theta) Y + \sin(\theta)Z)+r_z(-\sin (\theta) Y + \cos(\theta)Z)\right)\\
&= \frac12 \left(I+r_xX+(r_y(\cos (\theta) - r_z \sin (\theta)) Y + (r_y\sin(\theta) + r_z\cos(\theta))Z\right)\\\end{align}
Hence if we write</p>
<p>$$\rho'=\frac12 \left(I+r_x'X+r_y'Y+r_z'Z\right)$$</p>
<p>$$\begin{bmatrix}r_x' \\ r_y' \\ r_z'\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos \theta & - \sin \theta\\ 0 & \sin \theta & \cos \theta\end{bmatrix}\begin{bmatrix}r_x \\ r_y \\ r_z\end{bmatrix}$$</p>
|
2,387,435 | <p><a href="https://i.stack.imgur.com/RxHmS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RxHmS.png" alt="enter image description here"></a></p>
<p>We wish to construct a rectangular auditorium with a stage shaped as a semicircle of radius $r$, as shown in the diagram below (white is the stage and green is the seating area). For safety reasons, light strips must be placed on the perimeter of the seating area. If we have $45\pi + 60$ meters of light strips, what should $r$ be so that the seating area is maximized?</p>
<p>My issue is that I tried to solve it the "geometric way",
figuring that the perimeter (in r) for the seating area is going to be $(3+\pi)r$. I set it equal to $45\pi+60$ to try and find the closest number to equal the value of $r$. Is this the way I should approach the problem, or is there a "calculus" style method I should use to come to a solution?</p>
<p>Edit: </p>
| Frunobulax | 93,252 | <p>I'm a bit late to the party, but here's another way to prove that <span class="math-container">$R_x(\alpha)$</span> is a rotation about the <span class="math-container">$x$</span> axis on the Bloch sphere.</p>
<p>First, consider the following way to map every vector <span class="math-container">$\mathbf v$</span> of <span class="math-container">$\mathbb C^2$</span> to <span class="math-container">$\mathbb{R}^3$</span>: Normalize <span class="math-container">$\mathbf v$</span> to <span class="math-container">$\mathbf v_0$</span> and compute the point <span class="math-container">$\mathbf p_0$</span> on the Bloch sphere to which <span class="math-container">$\mathbf v_0$</span> would be mapped. Then map <span class="math-container">$\mathbf v$</span> to <span class="math-container">$\|\mathbf v\|\cdot \mathbf p_0$</span>. Using this mapping, every function from <span class="math-container">$\mathbb C^2$</span> to <span class="math-container">$\mathbb C^2$</span> induces a function from <span class="math-container">$\mathbb{R}^3$</span> to <span class="math-container">$\mathbb{R}^3$</span>.</p>
<p>Now, the function described by the Pauli matrix <span class="math-container">$X$</span> maps <span class="math-container">$\bigl(\cos(\theta/2),\mathrm{e}^{\mathrm{i}\phi}\sin(\theta/2)\bigr)^T$</span> to <span class="math-container">$\bigl(\mathrm{e}^{\mathrm{i}\phi}\sin(\theta/2),\cos(\theta/2)\bigr)^T$</span> which is physically indistinguishable from <span class="math-container">$\bigl(\cos(\pi/2-\theta/2),\mathrm{e}^{-\mathrm{i}\phi}\sin(\pi/2-\theta/2)\bigr)^T$</span>. Therefore, the induced map on <span class="math-container">$\mathbb{R}^3$</span> maps the point with spherical coordinates <span class="math-container">$(r,\theta,\phi)$</span> to the one with <span class="math-container">$(r,\pi-\theta,-\phi)$</span>. This is obviously a linear map.</p>
<p>Because <span class="math-container">$R_x(\alpha)$</span> is <span class="math-container">$\cos(\alpha/2)I-\mathrm{i}\sin(\alpha/2)X$</span>, the map induced by <span class="math-container">$R_x(\alpha)$</span> is also linear and to see what it does we only need to check what it does to the three base vectors of <span class="math-container">$\mathbb{R}^3$</span>.</p>
<p>The base vector in <span class="math-container">$z$</span> direction has <span class="math-container">$\phi=0$</span> and <span class="math-container">$\theta=0$</span> and corresponds to <span class="math-container">$(1,0)^T$</span> in <span class="math-container">$\mathbb{C}^2$</span>. Its image under <span class="math-container">$R_x(\alpha)$</span> is <span class="math-container">$\bigl(\cos(\alpha/2),\mathrm{e}^{-\mathrm{i}\pi/2}\sin(\alpha/2)\bigr)^T$</span> and the image of the base vector under the induced map is thus <span class="math-container">$\phi'=-\pi/2$</span> and <span class="math-container">$\theta'=\alpha$</span>.</p>
<p>The base vector in <span class="math-container">$y$</span> direction has <span class="math-container">$\phi=\pi/2$</span> and <span class="math-container">$\theta=\pi/2$</span> and corresponds to <span class="math-container">$\bigl(\cos(\pi/4),\mathrm{i}\sin(\pi/4)\bigr)^T$</span> in <span class="math-container">$\mathbb{C}^2$</span>. Its image under <span class="math-container">$R_x(\alpha)$</span> is
<span class="math-container">\begin{align*}
&\begin{pmatrix}
\cos(\alpha/2)\cos(\pi/4)+\sin(\alpha/2)\sin(\pi/4) \\
-\mathrm{i}\sin(\alpha/2)\cos(\pi/4)+\mathrm{i}\cos(\alpha/2)\sin(\pi/4)
\end{pmatrix} \\ =
&\begin{pmatrix}
\cos(-\alpha/2)\cos(\pi/4)-\sin(-\alpha/2)\sin(\pi/4) \\
\mathrm{i}\sin(-\alpha/2)\cos(\pi/4)+\mathrm{i}\cos(-\alpha/2)\sin(\pi/4)
\end{pmatrix} =
\begin{pmatrix}
\cos(\pi/4-\alpha/2) \\
\mathrm{e}^{\pi/2} \sin(\pi/4-\alpha/2)
\end{pmatrix}
\end{align*}</span>
and the image of the base vector under the induced map is thus <span class="math-container">$\phi'=\pi/2$</span> and <span class="math-container">$\theta'=\pi/2-\alpha$</span>.</p>
<p>Finally, the base vector in <span class="math-container">$x$</span> direction has <span class="math-container">$\phi=0$</span> and <span class="math-container">$\theta=\pi/2$</span> and corresponds to <span class="math-container">$\bigl(\cos(\pi/4),\sin(\pi/4)\bigr)^T$</span> in <span class="math-container">$\mathbb{C}^2$</span>. Its image under <span class="math-container">$R_x(\alpha)$</span> is
<span class="math-container">\begin{align*}
&\begin{pmatrix}
\cos(\alpha/2)\cos(\pi/4)-\mathrm{i}\sin(\alpha/2)\sin(\pi/4) \\
-\mathrm{i}\sin(\alpha/2)\cos(\pi/4)+\cos(\alpha/2)\sin(\pi/4)
\end{pmatrix} \\ =
&\frac1{\sqrt2}\begin{pmatrix}
\cos(-\alpha/2)+\mathrm{i}\sin(-\alpha/2) \\
\cos(-\alpha/2)+\mathrm{i}\sin(-\alpha/2)
\end{pmatrix} =
\frac1{\sqrt2}\begin{pmatrix}
\mathrm{e}^{-\mathrm{i}\alpha/2} \\
\mathrm{e}^{-\mathrm{i}\alpha/2}
\end{pmatrix} = \mathrm{e}^{-\mathrm{i}\alpha/2}
\begin{pmatrix}
\cos(\pi/4) \\
\sin(\pi/4)
\end{pmatrix}
\end{align*}</span>
which is physically indistinguishable from <span class="math-container">$\bigl(\cos(\pi/4),\sin(\pi/4)\bigr)^T$</span>. Thus, the image of this base vector under the induced map is <span class="math-container">$\phi'=0$</span> and <span class="math-container">$\theta'=\pi/2$</span>, i.e., the vector doesn't move.</p>
<p>All three base vector are mapped to where they should be mapped by a rotation by <span class="math-container">$\alpha$</span> about the <span class="math-container">$x$</span> axis.</p>
|
3,397,834 | <blockquote>
<h2><span class="math-container">$$49y″−98y′+48y= 0 \quad\quad\quad \,\, y(2)=3,y′(2)=9.$$</span></h2>
</blockquote>
<p>When I solved, I got that my <span class="math-container">$r_1= \frac67$</span> and <span class="math-container">$r_2= \frac87.$</span> Then I got that <span class="math-container">$y= C_1e^{(\frac67)}t +C_2e^{\frac87}t.$</span> I’m not sure how to proceed from here in regards to substituting <span class="math-container">$y(2)= 3$</span> and <span class="math-container">$y’(3)=9$</span> to get <span class="math-container">$C_1$</span> and <span class="math-container">$C_2$</span> values. </p>
| Dr. Sonnhard Graubner | 175,066 | <p>we get for <span class="math-container">$$y(x)\neq 0$$</span> <span class="math-container">$$y'(x)=-\frac{x^4}{y^4}$$</span> so
<span class="math-container">$$y''(x)=-\frac{4x^3y^4-x^44y^3y'}{y^8}$$</span>
plugging <span class="math-container">$$y'=-\frac{x^4}{y^4}$$</span> we get
<span class="math-container">$$y''=-\frac{4(x^3y^5+x^8)}{y^{9}}$$</span></p>
|
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