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 |
|---|---|---|---|---|
342,491 | <p>How to prove the following:</p>
<p>$a_n = \left\{\left(1+\frac{1}{n}\right)^n\right\}$ is bounded sequence, $ n\in\mathbb{N}$</p>
| André Nicolas | 6,312 | <p>Below is the same answer as the one by Ju'x, in slightly different language. We want to show that $\left(1+\dfrac{1}{n}\right)^n \lt e$, or equivalently that $1+\frac{1}{n}\lt e^{1/n}$. The series expansion of $e^{1/n}$ is a sum of positive terms, the first two of which are $1$ and $\dfrac{1}{n}$.</p>
<p>The downside is that this approach is not available if we intend to <em>define</em> $e$ as the limit of $\left(1+\dfrac{1}{n}\right)^n$.</p>
|
3,932,279 | <p>Here, I am taking Lebesgue measure. There is a notion that a set <span class="math-container">$E$</span> is measruable iff <span class="math-container">$m_*(E) = m^*(E)$</span>. Let <span class="math-container">$E' = [0,1] \cap \mathbb I$</span>, where <span class="math-container">$\mathbb I$</span> denote the set of irrational numbers.</p>
<p>By definition, <span class="math-container">$m_*(E) = \sup_{F}\{|F| \mid F \subset E, F \text{ compact}\}$</span>. Then, <span class="math-container">$m_*(E') = \sup_{F}\{m^*(F) \mid F \subset E', F \text{ compact}\}$</span>. Take some compact <span class="math-container">$F \subset E'$</span> such that <span class="math-container">$m^*(E') < m^*(F) + \epsilon$</span>. Then,</p>
<ol>
<li><span class="math-container">$F$</span> must not contain a closed interval.</li>
<li><span class="math-container">$F$</span> is nowhere dense in <span class="math-container">$[0,1]$</span> since if it is dense in some <span class="math-container">$O \subset [0,1]$</span>, then compactness <span class="math-container">$F$</span> implies that <span class="math-container">$\mathbb Q \cap F \neq \varnothing$</span>.</li>
<li><span class="math-container">$F$</span> has strictly positive measure.</li>
</ol>
<p>How is this possible?</p>
| Measure me | 854,564 | <p>You actually know that there exists by other routes, in fact you know that <span class="math-container">$|[0,1]\cap \mathbb{Q}|=0$</span> so it is measurable in the Lebesgue complete measure and it implies that <span class="math-container">$|[0,1]\cap \mathbb{I}|=1$</span>.</p>
<p>So you know it is possible, the only thing is "how?"; well, we know there exist some ugly sets out there, like the Cantor set, actually you may be able to use a generalized Cantor set.</p>
<p>An other idea is to use the fact that the sets <span class="math-container">$\{ x \}$</span> with <span class="math-container">$x\in [0,1]\cap \mathbb{I}$</span> are closed and bounded, so compact, and also every finite union of these singletons. It is true that their measure is <span class="math-container">$0$</span>, but we are talking about their <span class="math-container">$\sup$</span>, so there is hope.</p>
|
2,321,850 | <p>The base is the semicircle $$y=\sqrt{16−x^2},$$
where -4 $\le$ $x$ $\le$ 4. The cross-sections perpendicular to the $x$-axis are squares.
$$\\$$
So far this is what I have:</p>
<p>$\int (Area)\,dx$</p>
<p>$\implies$ $\int \frac{π}{2} ($$r^2$$)\ dx$.</p>
<p>$r$ = $\frac{\sqrt{16−x^2}}{2}$</p>
<p>$\implies$ $ \frac {π}{8}$ $\int 16- $$x^2$$\ dx$.
$$\\$$
I'm confused with what I have to do with the information given about the squares.</p>
| BBot | 454,930 | <p>You must use the area of a square.</p>
<p>Remember that the area of a square is $s^2$.</p>
<p>For this problem, your side length is $y=\sqrt{16−x^2}$, which means your area for the square is $y=\sqrt{16−x^2}^2.$
$$\\$$ </p>
<p>$$\int_{-4}^{4}{16 +x^2 } \,dx$$
$$\\$$
Can you continue from there?</p>
|
1,341,771 | <p>I'm trying to show that given three distinct points $z_1,z_2,z_3\in\mathbb C$, the rational function
$$
f(z) = \frac{(z-z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} = \frac{(z_2 - z_3)z + (z_1z_3 - z_1z_2)}{(z_2 - z_1)z + (z_1z_3 - z_2z_3)} = \frac{az + b}{cz + d}
$$
is a Möbius transformation. That is, I must show that $ad - bc \neq 0$. I worked out that
\begin{align*}
ad - bc & = (z_1z_2z_3 - z_2^2z_3 - z_1z_3^2 + z_2z_3^2) - (z_1z_2z_3 - z_1^2z_3 - z_1z_2^2 + z_1^2z_2) \\
& = - z_2^2z_3 - z_1z_3^2 + z_2z_3^2 + z_1^2z_3 + z_1z_2^2 - z_1^2z_2 \\
& = (z_2z_3^2 - z_2^2z_3) + (z_1^2z_3 - z_1z_3^2) + (z_1z_2^2 - z_1^2z_2) \\
& = z_1^2(z_3 - z_2) + z_2^2(z_1 - z_3) + z_3^2(z_2 - z_1),
\end{align*}
but I'm unsure of where to go from here to show that this expression is nonzero.</p>
| Travis Willse | 155,629 | <p>We can factor $z_3$ from the expression for $d$ and $z_1$ from the expression for $c$ and then group:
\begin{align}
ad-bc
&= (z_2 - z_3) (z_1 z_3 - z_2 z_3) - (z_2 - z_1) (z_1 z_3 - z_1 z_2) \\
&= (z_1 - z_2) (z_2 - z_3) z_3 - (z_1 - z_2) (z_2 - z_3) z_1 \\
&= (z_1 - z_2) (z_2 - z_3) (z_3 - z_1) .
\end{align}
Since $z_1, z_2, z_3$ are distinct, $ad - bc$ is a product of three nonzero expressions and hence nonzero.</p>
|
11,172 | <p>As a TA who led calculus* 1 and 2 discussion section and holds office hour** in the previous year, I heard the following (wrong) arguments several times.</p>
<blockquote>
<ol>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} \sqrt{x+1}-\sqrt{x}=0$</span> because <span class="math-container">$\infty-\infty=0$</span>.</p>
</li>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} x^{1/x}=1$</span> because <span class="math-container">$\infty^0=1$</span>.</p>
</li>
<li><p><span class="math-container">$\int_1^{\infty}f(x)dx$</span> and <span class="math-container">$\int_1^{\infty}g(x)dx$</span> both diverge so <span class="math-container">$\int_1^{\infty}f(x)+g(x)dx$</span> diverge.</p>
</li>
</ol>
</blockquote>
<p>I usually explain the arguments are not true in general by providing a (very trivial) counter example, for example,</p>
<blockquote>
<ol>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)=\infty$</span> and <span class="math-container">$\displaystyle \lim_{x\to \infty} g(x)=\infty$</span> does not guarantee <span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)-g(x)=0$</span>, for example, <span class="math-container">$f(x)=x+1$</span> and <span class="math-container">$g(x)=x$</span>.</p>
</li>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)=\infty$</span> and <span class="math-container">$\displaystyle \lim_{x\to \infty} g(x)=0$</span> does not guarantee <span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)^{g(x)}=1$</span>, for example, <span class="math-container">$f(x)=2^x$</span> and <span class="math-container">$g(x)=1/x$</span>.</p>
</li>
<li><p>False in general, for example <span class="math-container">$f(x)=-g(x)=1$</span></p>
</li>
</ol>
</blockquote>
<p>After giving explanations like that I sometime heard "But in your examples you can cancel the expression/formula..." and I was not sure how to continue. I tried the following methods, non of them seem to work very well.</p>
<p>a. Provide a much more complicated counter example which requires a few minutes of calculation to get the answer. This often leads to further confusion.</p>
<p>b. Just say that is the wrong way to do it. It sounds like "I'm the teacher so believe me." and doesn't do too much.</p>
<p>c. Show them the correct way to do their problems. This is almost like b (Why is your way the right way and mine is the wrong way?).</p>
<p>I'm looking for a better way to deal with questions like these.</p>
<p>*<span class="math-container">$\epsilon-\delta$</span> definition is not introduced.
** Office hour is in tutoring center where I'm also responsible for students take the class from the professors I'm not TA'ing for.</p>
| Daniel R. Collins | 5,563 | <p>The first thing I would say is that principally you want to stick with strategy (a), and not veer into (b) or (c). You definitely want to dig into, and spend time on, these trouble spots with proper mathematical reasoning; certainly not "Because I said to do it this way", as that doesn't actually count as math at all. Counterexamples are logically the fundamental way one knocks down conjectures like those, so students should exercise and be trained in that method of dialogue. </p>
<p>What I do in a situation like this is try to jot down those trouble spots and turn them into quizzes for the whole class. Usually if I can turn it into (a) a binary "true or false" question, and (b) get the whole class in on the discussion, then we very quickly come to a consensus on the correct answer. </p>
<p>If you're solely working in a tutorial/one-on-one situation, then this becomes trickier to implement. Perhaps write down the statement as a "true or false" question and try to pin the student down to the one correct response (no "but this case was different" allowed). Possibly leverage one or two other nearby students to get their input on the discussion. </p>
|
467,279 | <p>I'm reading Intro to Topology by Mendelson.</p>
<p>I'm in the section titled "Compact Metric Spaces".</p>
<p>The problem is in the title.</p>
<p>My attempt at the proof is as follows:</p>
<p>Let $\{a_n\}_{n=1}^\infty$ be a Cauchy sequence in $X$. We will show that $\{a_n\}_{n=1}^\infty$ converges to a point in $X$. Consider the set $S=\{a_n:n\in\mathbb{N}\}$. Then there are two cases to consider, $S$ finite and $S$ infinite. If $S$ is finite then there exists some $N\in\mathbb{N}$ such that $a_n=a$ for some $a\in S$ and so $\{a_n\}_{n=1}^\infty\to a$. Suppose now that $S$ is infinite. Then $S$ has at least one accumulation point in $X$, call it $a$. Thus, the neighborhood $B(a;\frac{1}{n})$ contains a point $a_n\in S$ and $\lim\limits_{n\to\infty} a_n=a$.</p>
<p>My concern with this proof is no where did I use the fact that the sequence was Cauchy, other than supposing it was. I know this is a flaw in my proof since I have to use the hypothesis some where. </p>
<p>I was also considering looking at the $\text{sup} S$, but I'm not sure how to go about using that fact or whether or not that's the right approach.</p>
<p>Thanks for any help or feedback!</p>
| Henno Brandsma | 4,280 | <p>In the cases you handle you only can construct a subsequence of the original sequence that converges to some $a$. In the case where $S$ is finite, so finitely many values $a_n$ occur, we can conclude (pigeon hole principle) that there exists $a \in S$ and infinitely many $n$ (say all $n \in M \subset \mathbb{N}$ that have $a_n = a$. This gives us a constant subsequence (all with value $a$) and thus trivially a convergent subsequence. But not yet convergence of the <em>whole</em> sequence (without using Cauchy).</p>
<p>Also, when $S$ is infinite, it has some limit point $a$, and then again <em>all</em> you can do
at first is construct a subsequence of $a_n$ that converges to $a$: pick $n_1$ such that $d(a_{n_1}, a) < 1$, and having picked $n_1 < n_2 < \ldots < n_k$ such that $d(a_{n_i}, a) < \frac{1}{i}$ for all $i \le k$, we then pick $n_{k+1} > n_k$ such that $d(a_{n_{k+1}}, a) < \frac{1}{k+1}$, which can be done as there infinitely many points of the sequence in any open ball around $a$. And then $a_{n_m} \to a$ as $m \to \infty$.</p>
<p>Now where Cauchy is used is in the lemma: let $a_n$ be a Cauchy sequence in $(X,d)$ and let $a_{n_k}$ be a subsequence that converges to some $a \in X$. Then $a_n$ converges to $a$ as well.</p>
<p>Proof: let $\epsilon>0$. Pick $N \in \mathbb{N}$ such that for all $n,m \ge N$ we have
$d(a_n, a_m) < \frac{\epsilon}{2}$, by Cauchyness. Also, pick $k$ such that $n_k > N$ and
$d(a_{n_k}, a) < \frac{\epsilon}{2}$, by convergence of the subsequence to $a$.</p>
<p>Now for any $n \ge N$: $d(a_n, a) \le d(a_n, a_{n_k}) + d(a_{n_k}, a) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. So having found $N$ for all $\epsilon>0$, $a_n \to a$ as $n \to \infty$, as required.</p>
|
2,895,923 | <p>I am trying to solve this simple geometry problem but I am always tangled in so many equations it makes my head spin. I tried solving it via similar triangles but i cant seem to eliminate all the unwanted variables. Please help.</p>
<p>I have to prove $ r_1\times r_3=(r_2)^2$</p>
<p><a href="https://i.stack.imgur.com/5KQru.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5KQru.png" alt="Question Image"></a></p>
<p>Thank you</p>
| Narasimham | 95,860 | <p><a href="https://i.stack.imgur.com/xZxzS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xZxzS.png" alt="Contacting Circles"></a></p>
<p>We have $h'$s of contacting circles from common apex A, the series of similar triangles:</p>
<p>$$ h_2-h_1=2r_1, h_3-h_2=2r_2, h_4-h_3=2r_3 \tag1$$</p>
<p>$$ \sin \alpha = \frac {(h_2-h_1)/2}{(h_2+h_1)/2}= \frac {h_2-h_1}{h_2+h_1} =\frac {h_3-h_2}{h_3+h_2}\tag2$$</p>
<p>Apply componendo/dividendo to each pair</p>
<p>$$\frac{1+\sin \alpha}{1-\sin \alpha}= \frac{h_2}{h_1} = \frac{h_3}{h_2} =\frac{h_4}{h_3} = = k \tag3$$</p>
<p>which gives successive geometric means that are also called <em>powers of circle</em> corresponding to extreme $h$ values product of each pair:</p>
<p>$$ r_1=\sqrt{h_1h_2},\, r_2=\sqrt{h_2h_3},\, r_3=\sqrt{h_3h_4} \, \tag4$$</p>
<p>$$ \frac{r_2^2 }{r_1 r_3}=\frac{h_2 h_3}{\sqrt{h_1h_2h_3h_4}}=\sqrt{\frac{h_2 h_3}{h_1 h_4}}=\sqrt{k/k}=1 \tag5$$</p>
|
830,755 | <p>I would like to compute the following,
$$
\int_0^{\infty}\int_0^{\infty}e^{-x^2-2xy-y^2}\ dx\,dy
$$
It is obvious that we can rewrite the integral above to,
$$
\int_0^{\infty}\int_0^{\infty}e^{-(x+y)^2}\ dx\,dy
$$
so we are ending up with something looking like a gaussian integral. I think that a smart substitution would help but all I tried ended up to be something I am not able to compute...</p>
<p>I really would appreciate any hint.</p>
<p>Thanks in advance!</p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
\newcommand{\wt}[1]{\widetilde{#1}}$</span>
<span class="math-container">$\ds{\int_{0}^{\infty}\int_{0}^{\infty}\expo{-\pars{x + y}^{2}}\,\dd y\,\dd x:\
{\large ?}}$</span></p>
<blockquote>
<p><span class="math-container">\begin{align}
&\color{#66f}{\large%
\int_{0}^{\infty}\int_{0}^{\infty}\expo{-\pars{x + y}^{2}}\,\dd y\,\dd x}
\\[5mm] = &\
\int_{0}^{\infty}\int_{x}^{\infty}\expo{-y^{2}}\,\dd y\,\dd x
=\left.\int_{0}^{\infty}\int_{0}^{\infty}\expo{-y^{2}}\,\dd y\,\dd x
\right\vert_{y\ >\ x}
\\[3mm]&=\left.\int_{0}^{\infty}\expo{-y^{2}}\int_{0}^{\infty}\,\dd x\,\dd y
\right\vert_{x\ <\ y}
=
\int_{0}^{\infty}\expo{-y^{2}}\int_{0}^{y}\,\dd x\,\dd y
=\int_{0}^{\infty}y\expo{-y^{2}}\,\dd y
\\[5mm] = &\
\left.-\,\half\,\expo{-y^{2}}\right\vert_{0}^{\infty} =
\color{#66f}{\Large\half}
\end{align}</span></p>
</blockquote>
|
2,190,508 | <p>Let $f:[a,b] \to \mathbb R$ be nice differentiable function. By Fundamental Theorem of Calculus (FTC), we have $$\int_a^{b} f'(t) dt = f(b) -f(a).$$</p>
<p>By Cauchy-Schwartz inequality, we have, $\int_{a}^{b} |f(t)| dt \leq (b-a)^{1/2}(\int_a^b |f(t)|^2 dt)^{1/2} .$ Using this, and FTC, we have $\int_{a}^{b} (f'(t))^2 dt \geq (b-a)^{-1} (f(b)-f(a))^2.$ </p>
<blockquote>
<p>Question: (1) Can compute $\int_a^{b}(f'(t))^2 dt$ precisely? (2) Can we improve inequality $\int_{a}^{b} (f'(t))^2 dt \geq (b-a)^{-1} (f(b)-f(a))^2$ ?</p>
</blockquote>
| Andrew D. Hwang | 86,418 | <p>In reverse order: If
$$
f(x) = f(a) + \underbrace{\frac{f(b) - f(a)}{b - a}}_{m}(x - a),\quad a \leq x \leq b,
$$
then $f'(t) \equiv m$, so
$$
\int_{a}^{b} f'(t)^{2}\, dt = m^{2} (b - a)
= \frac{\bigl(f(b) - f(a)\bigr)^{2}}{b - a}.
$$
In words, the stated inequality is saturated for affine functions (as would be expected from Cauchy-Schwarz), so without further constraints on $f$ there is no tighter lower bound.</p>
<p>As for evaluating the integral of $(f')^{2}$, presumably that means something like "in terms of $a$, $b$, $f(a)$, $f(b)$, $f'(a)$, $f'(b)$". The answer is (tentatively, pending your exact meaning, but pretty firmly) no: Letting $\phi$ be a function that vanishes to second order at $a$ and $b$ (e.g., $(x - a)^{2} (x - b)^{2}$ multiplied by a smooth function), and replacing $f$ by $f + \phi$ does not change the value of any functional of the endpoint values, but does ("usually") change the integral of the derivative squared.</p>
|
35,327 | <p>This is a bit of a follow-up to a previous question: <a href="https://mathematica.stackexchange.com/questions/35256/how-can-i-merge-multiple-sets-of-morphological-components-perhaps-selected-usin">How can I merge multiple sets of morphological components (perhaps selected using different metrics)?</a></p>
<p>I've run into a few problems recently where it's actually rather difficult to select morphological components in an image based on size or geometry metrics, and really it would be much easier to just click on them, or to select them based on a coordinate in their interior. </p>
<p>Consider the task of selecting an arbitrary subset of morphological components in this image:</p>
<pre><code>Import["http://i.stack.imgur.com/gSXIj.png"]
</code></pre>
<p>Is something like this possible?</p>
<hr>
<p>Here's an update based on nikie's comment, where I believe he's suggesting we can do this:</p>
<pre><code>image = Binarize[Import["http://i.stack.imgur.com/gSXIj.png"]];
m = MorphologicalComponents[image];
m // Colorize
exMorphologicalComponentNumOne = PixelValue[Image[m], {50, 214}]
exMorphologicalComponentNumOneTEST = PixelValue[Image[m], {49, 213}]
exMorphologicalComponentNumTwo = PixelValue[Image[m], {206, 146}]
exMorphologicalComponentNumTwoTEST = PixelValue[Image[m], {203, 142}]
</code></pre>
<p>This, I believe, is telling us the index value for the morphological components containing the pixels at {50, 214} and at {206, 146} in the image. Here, <code>PixelValue</code> simply takes a pixel coordinate and returns whatever is sitting at this index in <code>ImageData[image]</code>. So if you look at the output for <code>MorphologicalComponents[image]</code>, you'll notice that the matrix is the same size as the output from <code>ImageData[image]</code> and that the positions in the image corresponding to a morphological component carry the value of the component's index.</p>
<p>This is a very good start (thank you nikie!), but it still isn't clear to me how to quickly select a subset of morphological components based on their index. It would also be really nice to be able to to the click-based selection I mention in the title, since here, we have to write down and retype coordinates using the locator pane. This becomes kind of time consuming if we need to select a large subset of morphological components in multiple images.</p>
| Dr. belisarius | 193 | <p>Based on nikie's comment:</p>
<pre><code>i = Import["http://i.stack.imgur.com/gSXIj.png"];
k = Image[MorphologicalComponents[i]];
DynamicModule[{pts = {{-1, 1}/2}},
{LocatorPane[Dynamic[pts], Image[k, ImageSize -> 300], LocatorAutoCreate -> True,
Appearance -> Graphics[{Red, Disk[]}, ImageSize -> 10]],
Dynamic@SelectComponents[k, "Label", MemberQ[PixelValue[k, #] & /@ Round@pts, 1. #] &]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/JkEDe.gif" alt="enter image description here"></p>
<p><strong>Previous answer</strong></p>
<pre><code>i = Import["http://i.stack.imgur.com/gSXIj.png"];
k = Image[MorphologicalComponents[i]];
DynamicModule[{pts = {{-1, 1}/2}},
{LocatorPane[Dynamic[pts], Image[k, ImageSize -> 300], LocatorAutoCreate -> True,
Appearance -> Graphics[{Red, Disk[]}, ImageSize -> 10]],
PixelValue[k, #] & /@ Dynamic@Round@pts}]
</code></pre>
<p><img src="https://i.stack.imgur.com/6nfJR.png" alt="Mathematica graphics"></p>
|
2,764,447 | <p>In the following question:</p>
<p><a href="https://i.stack.imgur.com/Rgzjr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rgzjr.jpg" alt="enter image description here" /></a>
I'm getting the answer as LHL=-infinity while RHL= infinity</p>
<p>The answer given to this question is D.</p>
<p>However as far as I know when a limit tends to infinity it does exist. I don't know if I'm wrong, could someone please help.</p>
| Dylan Zammit | 400,332 | <p>The formal definition of the existence of a limit at a point $a$:
$$\forall\epsilon>0\exists\delta>0, |x-a|<\delta\implies|f(x)-L|<\epsilon \text{ where } L\in\mathbb{R}$$</p>
<p>So no, the limit must be a constant.</p>
|
3,266,705 | <p>I know that if <span class="math-container">$f(x)$</span> is irreducible then <span class="math-container">$\langle f(x) \rangle$</span> is a prime ideal. Then I thought: is it maximal? Then I search about it, I find that it is not maximal ideal but cannot find any proof. </p>
<p>Any help is appreciated. Thanks.</p>
| Daniel Schepler | 337,888 | <p>If <span class="math-container">$f$</span> has degree 0, i.e. <span class="math-container">$f$</span> is a constant polynomial with a prime value <span class="math-container">$p$</span>, then <span class="math-container">$\langle p \rangle \subsetneqq \langle p, x \rangle \subsetneqq \langle 1 \rangle$</span>.</p>
<p>If <span class="math-container">$f$</span> has positive degree, let <span class="math-container">$p$</span> be a prime number not dividing the leading coefficient of <span class="math-container">$f$</span>. Then <span class="math-container">$\langle f \rangle \subsetneqq \langle f, p \rangle \subsetneqq \langle 1 \rangle$</span>.</p>
|
322,448 | <p>I'm supposed to show that: </p>
<p>$$y=\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$ </p>
<p>and the required answers are: $$ P=5, Q=4, R=-4 $$ </p>
<p>I tried to solve this with partial fractions like so: </p>
<p>$$5(x-1)(x+2) = A(x+3) + B(x-2)$$ </p>
<p>$\implies$ $A$=4, $B$=-4<br>
$\implies$ $Q$=4, R=-4 </p>
<p>But where does $P$=5 come from?</p>
<p>Or should I have first multiplied out the numerator and denominator and then used long division to solve?</p>
| Archy Will He 何魏奇 | 65,082 | <p>You are given the wrong expression. After some calculation I figure out that in order for these two expressions to be equivalent,</p>
<p>$$\frac{5(x-1)(x-2)}{(x-2)(x+3)} = 5 + \frac{4}{(x-2)} + \frac{-4}{(x+3)}$$ </p>
<p>instead of having $\frac{5(x-1)(x-2)}{(x-2)(x+3)}$, it should be</p>
<p>$$\frac{5(x-1)(x+2)}{(x-2)(x+3)}$$</p>
<p>Now solve it again it would work well.</p>
<p>My approach to solving it is </p>
<p>$$\frac{5(x-1)(x+2)}{(x-2)(x+3)} = \frac{5x^2+5-10}{(x-2)(x+3)} =\frac{\frac{5x^2+5-10}{(x-2)}}{(x+3)}$$</p>
<p>$$\frac{\frac{5x^2+5-10}{(x-2)}}{(x+3)} = \frac{5x+15+\frac{40}{(x-2)}}{(x+3)}$$</p>
<p>$$\frac{5x+15+\frac{40}{(x-2)}}{(x+3)} = 5+\frac{40}{(x-2)(x+3)}$$</p>
<p>$$5+\frac{40}{(x-2)(x+3)}=5 + \frac{4}{(x-2)} + \frac{-4}{(x+3)}$$</p>
|
2,207,260 | <p>I have to find the asymptotics of the following integral</p>
<p>$$\int^1_0 \frac{\sin(x)}{x(1+x)^n} dx$$</p>
<p>as $n\to\infty$.</p>
<p>I know I am supposed to use the lebesgue dominated convergence theorem and create a sequence $b_n$ and aim for $\lim_{n\to\infty} a_n/b_n =1$ but I can't seem to get my head round it.</p>
<p>Any help I would be very much grateful for</p>
| Mark Viola | 218,419 | <p>Let $I_n$ be given by</p>
<p>$$\begin{align}
I_n&=\int_0^1 \frac{\sin(x)}{x(1+x)^n}\,dx\\\\
&=\int_0^1 (1+x)^{-n}\,dx+\sum_{k=1}^\infty\frac{(-1)^{k}}{(2k+1)!}\int_0^1x^{2k}(1+x)^{-n}\,dx\\\\
&=\frac{1}{n-1}-\frac{2}{2^n(n-1)}+\sum_{k=1}^\infty\frac{(-1)^{k}}{(2k+1)!}\int_0^1 x^{2k}(1+x)^{-n}\,dx\tag 1
\end{align}$$</p>
<p>We can evaluate the integral on the right-hand side of $(1)$ be making the substitution $x\to x-1$ and using the binomial theorem to write $(x-1)^{2k}=\sum_{\ell = 0}^{2k}\binom{2k}{\ell}(-1)^{2k-\ell}x^\ell$. Proceeding we find</p>
<p>$$\int_0^1x^{2k}(1+x)^{-n}\,dx=\sum_{\ell = 0}^{2k}\binom{2k}{\ell}(-1)^{2k-\ell}\frac{1-2^{-n+\ell+1}}{n-(\ell+1)}$$</p>
<p>which shows that $\int_0^1x^{2k}(1+x)^{-n}\,dx=O\left(\frac{1}{n^{2k+1}}\right)$</p>
<p>Finally, we have </p>
<p>$$I_n=\frac{1}{n-1}-\frac{1}{3(n-1)(n-2)(n-3)}+O\left(\frac1{n^5}\right)$$</p>
|
3,374,024 | <p>Basicly, i am bad at mathematics very hard to me got score 70. I try learn trigonometry now, starting from the very basic.
I practice to find the value of <span class="math-container">$$\sin(135^\circ)$$</span> using my knowing about <span class="math-container">$\sin(45^\circ)={\sqrt2\over2}$</span>, <span class="math-container">$\cos(45^\circ)={\sqrt2 \over 2}$</span>, <span class="math-container">$\sin(90^\circ)=1$</span>, <span class="math-container">$\cos(90^\circ)=0$</span>, the quadrant rule and sum angle formula. I hope the expert can help me to correcting and giving a tips.</p>
<p>Using the quadrant rule :
Since <span class="math-container">$135^\circ$</span> on second quadrant the sine sign is positive</p>
<p>Using sum formula of sin :
<span class="math-container">$$\sin(135^\circ)=\sin(90^\circ+45^\circ)$$</span>
<span class="math-container">$$=\sin(90^\circ)×\cos(45^\circ)+\cos(90^\circ)×\sin(45^\circ)$$</span>
<span class="math-container">$$=1×\cos(45^\circ)+0×\sin(45^\circ)$$</span>
<span class="math-container">$$=\cos(45^\circ)+0$$</span>
<span class="math-container">$$=\cos(45^\circ)$$</span>
<span class="math-container">$$={\sqrt2\over2}$$</span>
For your informatiom, i am just economic faculty at second semester, and my college tutor giving us a calculus, that contain trigonometry. I need to learn this because 2 weeks again i got calculus examination. Thank you for your support and tips and explanation.</p>
<p>Is my practice is correct ? Can i get the simple way to remember the sum formula of sine and cosine ? What suitable practice for me after i practicing with <span class="math-container">$135^\circ$</span> ?</p>
| ComFreek | 85,341 | <p>I think the authors just want to make a difference between the theory of sequents and the meta level. The meta level here is talking about the theory of sequents.</p>
<p>I previously had a paragraph in this answer elaborating on possible differences in connotations, but I've come to the conclusion that I cannot separate those terms in a clear-cut fashion and thus deleted the paragraph.</p>
|
4,527,455 | <p>My goal is to find all values of "a" so that the circle <span class="math-container">$x^2 - ax + y^2 + 2y = a$</span> has the radius 2</p>
<p>The correct answer is: <span class="math-container">$a = -6$</span> and <span class="math-container">$a = 2$</span></p>
<p>I tried solving it by doing this:<br/>
<span class="math-container">$x^2 - ax + y^2 +2y=a$</span><br/>
<span class="math-container">$x^2 - ax + (y+1)^2-1=a$</span><br/>
<span class="math-container">$(x - \frac a2)^2 - (\frac a2)^2 + (y+1)^2-1=a$</span><br/>
<span class="math-container">$(x - \frac a2)^2 - {a^2\over 4} + (y+1)^2-1=a$</span><br/>
<span class="math-container">$(x - \frac a2)^2 + (y+1)^2=a + {a^2\over 4} + 1$</span><br/>
<span class="math-container">$(x - \frac a2)^2 + (y+1)^2={a^2+4a + 4\over 4}$</span><br/></p>
<p>We want the radius to be 2 so set this <span class="math-container">${a^2+4a + 4\over 4}$</span> equal to 2<br/>
<span class="math-container">${a^2+4a + 4\over 4}=2$</span><br/>
<span class="math-container">$a^2+4a + 4=8$</span><br/>
<span class="math-container">$a^2+4a -4=0$</span><br/><br/>
Solve for a:<br/>
<span class="math-container">$a=-2 \pm \sqrt{4+4}$</span><br/>
<span class="math-container">$a=-2 \pm \sqrt{8}$</span><br/><br/>
This is not correct as you can see. I don't understand what I do wrong, I'm not sure if there is one of those tiny mistakes somewhere in my solving process or if I'm completely wrong from the beginning. Thanks in advance.</p>
| eMathHelp | 166,193 | <p><span class="math-container">$\frac{a^2+4a+4}{4}$</span> should equal <span class="math-container">$2^2=4$</span>, not <span class="math-container">$2$</span>.</p>
|
56,804 | <p>I know the Galois group is $S_3$. And obviously we can swap the imaginary cube roots. I just can't figure out a convincing, "constructive" argument to show that I can swap the "real" cube root with one of the imaginary cube roots. </p>
<p>I know that if you have a 3-cycle and a 2-cycle operating on three elements, you get $S_3$. I have a general idea that based on the order of the group there's supposed to be at least a 3-cycle. But this doesn't feel very "constructive" to me. </p>
<p>I wonder if I've made myself understood in terms of what kind of argument I'd like to see?</p>
| Adam Karlson | 301,192 | <p>I am not sure if this is too late. But the symmetry can be seen by plotting on the complex plane. The roots form the vertices of the equilateral triangle. So, you can rotate all of them and preserve symmetry. Hand-wavy but I guess it works.</p>
|
1,939,107 | <p>An unbiased coin is tossed repeatedly and outcomes are recorded. What is the expected no of toss to get HT ( one head and one tail consecutively) ?</p>
<p>I tried to solve above question using the following probability tree diagram.</p>
<p><a href="https://i.stack.imgur.com/10cO3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/10cO3.png" alt="enter image description here"></a></p>
<p>Probability in each branch is = $0.5$. I double circled the satisfying toss events.
While observing the diagram I noticed that, from 2nd toss onward our required event starts showing up. Additionally,</p>
<ol>
<li>in the $\text{2nd}$ toss (or the 3rd level) we have one satisfying case.</li>
<li>in the $\text{3rd}$ toss (or the 4th level) we have two satisfying case.</li>
<li>in the $\text{4th}$ toss (or the 5th level) we have three satisfying case.</li>
<li>in the $\text{5th}$ toss (or the 6th level) we have four satisfying case.</li>
<li>etc.</li>
</ol>
<p>i.e. in the $\text{kth}$ toss we would have $(k-1)$ satisfying case.
So,</p>
<p>$\\ E = \sum_{k=2}^{\infty } k.P(k)\
\\ E = \sum_{k=2}^{\infty } k.\left \{ (k-1)*(0.5)^k \right \}\\
\\ E = \sum_{k=2}^{\infty } \left \{ (k^2-k)*(0.5)^k \right \}\\$</p>
<p>I understand the series summation would converge because we have the $0.5^{k}$ term which eventually becomes zero.</p>
<p>But could not solve this polynomial and power series summation.</p>
<p>Please suggest how to solve this and or any other method.</p>
| epi163sqrt | 132,007 | <p>We can describe all valid tosses as follows:</p>
<blockquote>
<p>The valid tosses are</p>
<ul>
<li>zero or more tosses of T followed by </li>
<li>zero or more tossed of H followed by</li>
<li>HT</li>
</ul>
<p>This corresponds to the tree diagram where we can see following valid paths starting from the root</p>
<p>\begin{array}{lllllllllll}
&HT\qquad&H^2T\qquad&H^3T\qquad&\color{blue}{H^4T}\qquad&H^5T\qquad&\cdots\\
&THT&TH^2T&\color{blue}{TH^3T}&TH^4T&TH^5T&\cdots\\
&T^2HT&\color{blue}{T^2H^2T}&T^2H^3T&T^2H^4T&T^2H^5T&\cdots\\
&\color{blue}{T^3HT}&T^3H^2T&T^3H^3T&T^3H^4T&T^3H^5T&\cdots\\
&T^4HT&T^4H^2T&T^4H^3T&T^4H^4T&T^4H^5T&\cdots\\
&\quad\vdots&\quad\vdots&\quad\vdots&\quad\vdots&\quad\vdots&\\
\end{array}</p>
<p>Note the entries in a diagonal have the same length, starting with the first node $HT$ of length $2$. The entries in the diagonal with length $5$ are marked in blue.</p>
</blockquote>
<p>From this scheme we can derive the expectation value. We obtain following the same arrangement</p>
<p>\begin{align*}
E[X]=\sum_{n=2}^\infty x_n p_n&=2p(HT)+3p(H^2T)+4p(H^3T)+\color{blue}{5p(H^4T)}+\cdots\\
&\qquad+3p(THT)+4p(TH^2T)+\color{blue}{5p(TH^3T)}+6p(TH^4T)+\cdots\\
&\qquad+4p(T^2HT)+\color{blue}{5p(T^2H^2T)}+6p(T^2H^3T)+7p(T^2H^4T)+\cdots\\
&\qquad+\color{blue}{5p(T^3HT)}+6p(T^3H^2T)+7p(T^3H^3T)+8p(T^3H^4T)+\cdots\\
&\qquad+\cdots\\
\end{align*}</p>
<p>We can add up these values along the diagonals. Note, the diagonal marked in blue has $4$ entries</p>
<p>$$\{HHHHT,THHHT,TTHHT,TTTHT\}$$
of length $5$. </p>
<blockquote>
<p>We obtain
\begin{align*}
E[X]&=\sum_{n=1}^\infty n(n+1)\left(\frac{1}{2}\right)^{n+1}\\
&=\left.\sum_{n=1}^\infty n(n+1)x^{n+1}\right|_{x=\frac{1}{2}}\\
&=x^2\left.\sum_{n=1}^\infty n(n+1)x^{n-1}\right|_{x=\frac{1}{2}}\\
&=x^2\left.\sum_{n=2}^\infty (n-1)nx^{n-2}\right|_{x=\frac{1}{2}}\\
&=\left.\left(x^2D_x^2\sum_{n=0}^\infty x^n\right)\right|_{x=\frac{1}{2}}\\
&=\left.\left(x^2D_x^2\frac{1}{1-x}\right)\right|_{x=\frac{1}{2}}\\
&=\left.\frac{2x^2}{(1-x)^3}\right|_{x=\frac{1}{2}}\\
&=4
\end{align*}</p>
</blockquote>
<p>Here we use the differential operator $D_x:=\frac{d}{dx}$ and apply it to the <em><a href="https://en.wikipedia.org/wiki/Geometric_series#Geometric_power_series" rel="nofollow">geometric series</a></em>. </p>
|
3,238,306 | <p>Hello I have the differential equation <span class="math-container">$$x'=\frac{8t+10x}{17t+x}$$</span>
Brought it in a eqation where I can substitute <span class="math-container">$$ u=\frac{x}{t}$$</span>
and after some transformations I got the equation
<span class="math-container">$$ \frac{17+u}{8-7u-u^2}du=\frac{1}{t}dt $$</span>
with the following equation
<span class="math-container">$$ -2ln(u-1)+ln(u+8)=ln(t)+c$$</span>
Then I tried to multiply it with <span class="math-container">$$e$$</span> to make the <span class="math-container">$$ln()$$</span> disappear
then I got the equation <span class="math-container">$$ \frac{u+8}{(u-1)^2}=te^c$$</span> but now I dont know how to resubstitute or to break the fracture.Thanks,Ciwan.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>I would write <span class="math-container">$$a(b-1)\geq b$$</span> and this is <span class="math-container">$$a(b-1)\geq b-1+1$$</span> or
<span class="math-container">$$(b-1)(a-1)\geq 1$$</span> this is true since <span class="math-container">$$a-1\geq 1$$</span> and <span class="math-container">$$b-1\geq 1$$</span></p>
|
2,845,001 | <p>If I know the length of a chord of a circle and the length of the corresponding arc, but do not know the circle's radius, is there a formula by which I can calculate the length of the sagitta?</p>
| Claude Leibovici | 82,404 | <p>Starting from Ted Shifrin's answer, I think that we can have a reasonable estimate of $\theta$ using the Padé approximant built at $\theta=0$
$$\dfrac{\sin(\theta)}{\theta}=\frac{1-\frac{53 }{396}\theta^2+\frac{551 }{166320}\theta^4 } {1+\frac{13 }{396}\theta^2+\frac{5 }{11088}\theta^4 }$$ which is quite good for $0 \le \theta \le \pi/2$ (the maximum error is $2\times 10^{-6}$ at the upper bound).</p>
<p>Using it,solving
$$\dfrac{\sin(\theta)}{\theta} =a$$ reduces to a quadratic equation in $\theta^2$ the solution of which being
$$\theta=\sqrt{6\,\frac{455 a+1855- \sqrt{35} \sqrt{-3985 a^2+130862 a+25583}}{551-75
a} }$$</p>
<p>If you look <a href="https://math.stackexchange.com/questions/871516/how-could-i-improve-this-approximation">here</a>, you would find some interesting numerical approaches for the solution of $\sin(\theta)=a{\theta} $.</p>
<p><strong>Edit</strong></p>
<p>If you want a much better approximation, use
$$\dfrac{\sin(\theta)}{\theta}=\frac{ 1-\frac{9168 }{68821}\theta^2+\frac{433 }{133573}\theta^4} {1+\frac{11374 }{340015}\theta^2+\frac{85 }{175694}\theta^4 }$$</p>
|
2,871,084 | <p>Consider that we have a circle drawn a round the <strong>origin (0,0)</strong>. That circle has some points drawn on its circumference. Each of those points has range and azimuth $(r,\theta)$, the <strong>r</strong> and <strong>$\theta$</strong> values of these points are calculate with responding to the <strong>origin (0,0)</strong>. </p>
<p>I want to move or translate that circle and its (on-circumference) points to a new center (x,y), the new center can be in any quadrant. Here is an image for more demonstration (consider that it was translated to the first quadrant):<a href="https://i.stack.imgur.com/slaiD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/slaiD.png" alt="enter image description here"></a></p>
<p><strong>The question is what is the new $(r, \theta)$ of each point with responding to the origin (0,0) after translation to the new center?</strong></p>
| zhw. | 228,045 | <p>It's easy to see every function in the range is Lipschitz, but there are plenty of functions in $C[a,b]$ that are not Lipschitz. For example, $g(x)=\sqrt {x-a}.$ Hence the integral operator is not surjective.</p>
|
1,050,141 | <p>Find the roots of the equation $3^{x+2}$+$3^{-x}$=10 . By inspection the roots are $x=0$ and $x=-2$. But how can I solve this equation otherwise? </p>
| Dr. Sonnhard Graubner | 175,066 | <p>your equation is equivalent to $3^{2x}-\frac{10}{9}3^x+\frac{1}{9}=0$ solve this as a quadratic equation.</p>
|
85,622 | <p>I'm trying to understand how this works in terms of Galois theory and local class field theory. Assume we have an extension of local fields $E/L/K$ s.t. $E/L$ and $L/K$ are abelian. I'm interested in recognizing when $E/K$ is Galois. Clearly, $E/K$ is Galois if and only if $E$ is always fixed by an extension of an $L/K$ automorphism to $E$, but this is tricky to compute.</p>
<p>I overheard a brief conversation that this can be done through Galois groups and some group actions that occur in the tower, but I haven't found anything explicit through google. We should be able to see from how $Gal(L/K)$ acts on something whether or not the extension is Galois. I'm having trouble seeing what the action should be. I hope someone who knows what I'm talking about could write it down explicitly. Since the Galois groups should correspond through local class field theory to very concrete objects which are quotients of $E^\times$, $L^\times$ and $K^\times$. I was wondering how this action on the Galois side is expressed on the local field side?</p>
<p>I'm interested in this since it clearly would provide a tool for constructing some solvable extensions of e.g. $\mathbb{Q}_p$. I apologize for being fuzzy, but I don't know how to be more explicit.</p>
| Franz Lemmermeyer | 3,503 | <p>If $L/K$ is abelian and $K/k$ cyclic, then $L/k$ is going to be normal if the Galois group at the gottom acts on the class group attached to $L/K$ via class field theory, and it will be abelian if action fixes the classes. This is due to Hasse and follows easily from the standard results in class field theory. This is Prop. 1.2.8 in my <a href="http://www.rzuser.uni-heidelberg.de/~hb3/publ/pcft.pdf" rel="nofollow">survey</a> on class field towers (I apologize for the outdated content. All of this needs to be rewritten).</p>
<p>There are similar constructions in Kummer theory going back to Kummer; this can be used in several proofs of the Kronecker-Weber theorem and should be e.g. in Washington's book; see also Prop. 2.2.1 in the survey</p>
<p>If the base extension is abelian and not cyclic, stuff happens. There are many articles investigating this problem, but nothing as simple as in the cyclic case.</p>
|
380,696 | <p>Show that for any odd $n$ it follows that $n^2 \equiv 1 \mod 8$ and for uneven primes $p\neq 3$ we have $p^2 \equiv 1\mod 24$.</p>
<p>My workings so far: I proceeded by induction. Obviously $1^2 \equiv 1 \mod 8$. Then assume that for a certain uneven number $k$ we have $k^2 \equiv 1 \mod 8$. Then the the next uneven number is $k+2$ and $(k+2)^2 = k^2 + 4k + 4 \equiv 4k + 5 \mod 8$. by the induction hypothesis. Now since $k$ is uneven we can write it as $k=2j+1$ and thus $4k+5 =8j+9 \equiv 1 \mod 8$ and we have shown what was asked for by induction.</p>
<p>However, in the case of a prime number $n$ I am not so certain how to proceed because I can't use induction. It comes down to proving that $24|(p^2-1)$. This is certainly the case for the first uneven prime $p \neq 3$, namely $p=5$ such that $p^2-1 =24$. How would I proceed from here, or how should I approach the problem differently? Many thanks in advance!</p>
| lab bhattacharjee | 33,337 | <p>For the second question, any prime $>3,$ can be expressed as $6a\pm1$ where $a$ is a positive integer</p>
<p>So, $p^2=(6a\pm1)^2=36a^2\pm12a+1=24a^2+24\cdot\frac{a(a\pm1)}2+1\equiv1\pmod{24}$ as $a(a\pm1)$ is divisible by $2$</p>
<p>Evidently, we don't need prime, any odd number not divisible by $3$ will satisfy this</p>
|
306,178 | <p>Given
$$
y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1.
$$
Show that $\lbrace y_n \rbrace$ is a decreasing sequence. Anyone can help ? I consider the ratio $\frac{y_{n+1}}{y_n}$ but I got stuck.</p>
| robjohn | 13,854 | <p>Note that
<span class="math-container">$$
\begin{align}
\frac{\left(1+\frac1n\right)^{n+1}}{\left(1+\frac1{n+1}\right)^{n+2}}
&=\left(\frac{n+1}{n}\right)^{n+1}\left(\frac{n+1}{n+2}\right)^{n+2}\\
&=\frac{n}{n+1}\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+2}\\
&=\frac{n}{n+1}\left(1+\frac1{n(n+2)}\right)^{n+2}\\
&\ge\frac{n}{n+1}\left(1+\frac{n+2}{n(n+2)}\right)\\
&=1
\end{align}
$$</span>
Therefore,
<span class="math-container">$$
\left(1+\frac1{n+1}\right)^{n+2}\le\left(1+\frac1n\right)^{n+1}
$$</span></p>
<hr />
<p>Similarly,
<span class="math-container">$$
\begin{align}
\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n}
&=\left(\frac{n+2}{n+1}\right)^{n+1}\left(\frac{n}{n+1}\right)^n\\
&=\frac{n+1}{n}\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}\\
&=\frac{n+1}{n}\left(1-\frac1{(n+1)^2}\right)^{n+1}\\
&\ge\frac{n+1}{n}\left(1-\frac{n+1}{(n+1)^2}\right)\\
&=1
\end{align}
$$</span>
Therefore,
<span class="math-container">$$
\left(1+\frac1{n+1}\right)^{n+1}\ge\left(1+\frac1n\right)^n
$$</span></p>
<hr />
<p><strong>Bernoulli's Inequality</strong></p>
<p>In the preceding, we used <a href="http://en.wikipedia.org/wiki/Bernoulli%27s_inequality" rel="nofollow noreferrer">Bernoulli's Inequality</a>: for all <span class="math-container">$x\ge-1$</span> and non-negative integer <span class="math-container">$n$</span>,
<span class="math-container">$$
(1+x)^n\ge1+nx
$$</span>
This can be proven by induction:</p>
<p>Note that the inequality above is true for <span class="math-container">$n=0$</span>.</p>
<p>Suppose that <span class="math-container">$x\ge-1$</span> and for a non-negative integer <span class="math-container">$n$</span>, we have
<span class="math-container">$$
(1+x)^n-nx\ge1
$$</span>
Then
<span class="math-container">$$
\begin{align}
(1+x)^{n+1}-(n+1)x
&=(1+x)^n-nx+x(1+x)^n-x\\
&\ge1+x((1+x)^n-1)\\
&\ge1
\end{align}
$$</span>
If <span class="math-container">$-1\le x\le0$</span>, then both <span class="math-container">$x$</span> and <span class="math-container">$(1+x)^n-1$</span> are negative. If <span class="math-container">$x\ge0$</span>, then both both <span class="math-container">$x$</span> and <span class="math-container">$(1+x)^n-1$</span> are positive. Therefore, if <span class="math-container">$x\ge-1$</span>, <span class="math-container">$x((1+x)^n-1)\ge0$</span>. This justifies the last inequality above.</p>
<p>Note that if <span class="math-container">$x\ne0$</span> and <span class="math-container">$n\ge1$</span>, the last inequality is strict. Thus, for <span class="math-container">$x\ne0$</span> and <span class="math-container">$n\ge2$</span>, we have
<span class="math-container">$$
(1+x)^n\gt1+nx
$$</span></p>
<hr />
<p><strong>Negative Exponents</strong></p>
<p>Bernoulli's Inequality is also true for negative integer exponents. That is, for <span class="math-container">$x\gt-1$</span> and non-negative <span class="math-container">$n\in\mathbb{Z}$</span>,
<span class="math-container">$$
1-nx\le(1+x)^{-n}
$$</span>
Suppose that
<span class="math-container">$$
(1-nx)(1+x)^n\le1
$$</span>
which is trivially true for <span class="math-container">$n=0$</span>, and strictly true for <span class="math-container">$x\ne0$</span> and <span class="math-container">$n=1$</span>. Then
<span class="math-container">$$
\begin{align}
(1-(n+1)x)(1+x)^{n+1}
&=(1-nx)(1+x)^n-(n+1)x^2(1+x)^n\\
&\le1
\end{align}
$$</span>
Therefore, for all non-negative <span class="math-container">$n\in\mathbb{Z}$</span>,
<span class="math-container">$$
(1-nx)\le(1+x)^{-n}
$$</span>
where the inequality is strict for <span class="math-container">$x\ne0$</span> and <span class="math-container">$n\ge1$</span>.</p>
|
2,213,047 | <p>Let $F$ be the set of all functions $f : \mathbb{R} \to \mathbb{R}$. A relation $c$ is defined on $F$ by
$f c g$ if and only if $f(x) ≤ g(x)$ for all $x ∈ \mathbb{R} $.
Prove that '$c$' is a partial order.</p>
<p>I have proved that the relation is <em>reflexive</em>. Now I must prove that it is <em>antisymmetric</em> (and later <em>transitive</em>). </p>
<p>My working thus far: </p>
<p>Suppose $a, b ∈ F$. We must prove that if $a c b$ and $b c a$ then $a = b$. Now, if $a c b$ and $b c a$ then $f(a) ≤ f(b)$ and $f(b) ≤ f(a)$. Therefore, $f(a) = f(b)$. </p>
<p><em>Now, how to prove that $a = b$? There is no indication that $f$ is an injective function.</em> </p>
| R.W | 253,359 | <p>I think you are confuse about your own notation. You define the relation as</p>
<blockquote>
<p>Let <span class="math-container">$\mathcal{F}(\mathbb{R},\mathbb{R})$</span> be the set of all functions <span class="math-container">$f: \mathbb{R}\to \mathbb{R}$</span>. So define <span class="math-container">$\color{blue}{c}$</span> as the relation defined on <span class="math-container">$\mathcal{F}(\mathbb{R},\mathbb{R})$</span> as, for <span class="math-container">$f,g \in \mathcal{F}(\mathbb{R},\mathbb{R})$</span> then <span class="math-container">$f \color{blue} c g \Leftrightarrow f(x)\leq g(x) \, \forall x \in \mathbb{R}$</span>.</p>
</blockquote>
<p>Now take a look in your proof: You say that <span class="math-container">$a,b \in \mathcal{F}(\mathbb{R},\mathbb{R})$</span> and then in your proof you use <span class="math-container">$a,b$</span> as variables for <span class="math-container">$f,g$</span>! But your answer is not all wrong. Let's see it</p>
<p>Suppose <span class="math-container">$a,b \in \mathcal{F}(\mathbb{R},\mathbb{R})$</span>. <a href="https://en.wikipedia.org/wiki/Antisymmetric_relation" rel="nofollow noreferrer">We want to prove that</a> <span class="math-container">$(a \color{blue }c b $</span> and <span class="math-container">$b \color{blue}c a )$</span> implies <span class="math-container">$a = b$</span>. Now what you've done is </p>
<ol>
<li><span class="math-container">$a \color{blue}c b \implies \color{red}\forall x \in \mathbb{R}, a(x)\color{red}\leq b(x)$</span></li>
<li><span class="math-container">$b \color{blue}ca \implies \color{red}\forall x \in \mathbb{R}, b(x)\color{red}\leq a(x)$</span></li>
</ol>
<p>And then <span class="math-container">$\color{red}{(1)\text{ and } (2)\text{ together}}$</span> implies that <span class="math-container">$\forall x \in \mathbb{R}, a(x) = b(x)$</span> wich implies that <span class="math-container">$a = b \in \mathcal{F}(\mathbb{R}, \mathbb{R})$</span>. In blue is the relation and in red is important things that should be in your argument. Now you should try to answer transitive. But that should follow directly from arguments of inequality as the one above.</p>
|
2,673,078 | <p>Generalize the Monty Hall problem where there are $n \geq 3$ doors, of
which Monty opens $m$ goat doors, with $1 \leq m \leq n$.<br><strong>Original Monty Hall Problem:</strong> There are $3$ doors, behind one of which there is a car (which you want), and behind the other two of which there are goats (which you don’t want). Initially, all possibilities are equally likely for where the car is. You choose a door, which for concreteness we assume is Door $1$. Monty Hall then opens a door to reveal a goat, and offers you the option of switching. Assume that Monty Hall knows which door has the car, will always open a goat door and offer the option of switching, and as above assume that if Monty Hall has a choice between opening Door $2$and Door $3$, he chooses Door $2$ and Door $3$ with equal probability .<br>Find the probability that the strategy of always switching succeeds, given that Monty opens Door $2$.</p>
<p><strong>My approach:</strong><br>
Let $C_i$ be the event that car is behind the door $i$,
$O_i$ be the event that Monty opened door $i$ and $X_i$ be the event that intially I chose door $i$. Here $i=1,2,3,...,n$.<br>
Let's start with case where I chose $X_1$. Then:<br>
$P(O_{j_1, j_2, ..., j_m}|C_1, X_1) = {{n-1}\choose{m}}(\frac{1}{n-1})^m$, here $j \in$ {$m$ doors out of $n-1$, i.e., exclude Door$1$ }<br>
$P(O_{k_1, k_2, ..., k_m}|C_t, X_1) = {{n-2}\choose{m}}(\frac{1}{n-2})^m$,
here $k \in$ {$m$ doors out of $n-2$, i.e., exclude Door$1$ & Door$t$}, $t \in$ {$2,3, ..., n$}<br>
Also, $P(C_r|X_s) = \frac{1}{n}$, here $r,s \in$ {$1,2,...,n$}</p>
<p>Probability of winning by switching is,</p>
<p>$$P(C_3 | O_{k_1, k_2, ..., k_m}, X_1) = \frac{P(O_{k_1, k_2, ..., k_m}|C_3, X_1).P(C_3|X_1)}{P(O_{m-doors}|X_1)}$$</p>
<p>$$= \frac{P(O_{k_1, k_2, ..., k_m}|C_3, X_1).P(C_3|X_1)}{P(O_{j_1, j_2, ..., j_m}|C_1, X_1).P(C_1|X_1) + \sum_{t=2}^n(P(O_{k_1, k_2, ..., k_m}|C_t, X_1).P(C_t|X_1))}$$</p>
<p>$$ = \frac{{{n-2}\choose{m}}(\frac{1}{n-2})^m.\frac{1}{n}}{(\frac{1}{n}).({{n-1}\choose{m}}(\frac{1}{n-1})^m + {{n-2}\choose{m}}(\frac{1}{n-2})^m.(n-1))}$$</p>
<p>$$= \frac{(n-m-1)(n-1)^{m-1}}{(n-2)^m + (n-m-1)(n-1)^m}$$</p>
<p>However, the correct answer is $\frac{(n-1)}{(n-m-1).n}$. Any insights to what I have done wrong.</p>
| manisar | 843,108 | <p>While the original question has been answered, I'll post another solution to the actual problem at hand.</p>
<p>It will greatly simplify the calculations if we assume that the host momentarily identified a set <span class="math-container">$y > m$</span> doors (just in his mind) and opened <span class="math-container">$m$</span> doors from this <span class="math-container">$y$</span> set. Later we'll just make <span class="math-container">$y=n-1$</span>.</p>
<p>In the beginning, the winning probability of group <span class="math-container">$y$</span> as a whole is simply <span class="math-container">$y \over n$</span>. And, if <span class="math-container">$y$</span> wins, the winning probability of each door inside it is <span class="math-container">$1 \over y$</span>. With <span class="math-container">$m$</span> non-winning doors now opened from <span class="math-container">$y$</span>, the winning probability of each door in <span class="math-container">$y$</span> becomes <span class="math-container">$1 \over y-m$</span>, provided <span class="math-container">$y$</span> wins. Hence the absolute winning probability of each door within <span class="math-container">$y$</span> is simply <span class="math-container">${y \over n} \times {1 \over y-m}$</span>, or <span class="math-container">${1 \over n} \times {y \over y-m}$</span>. Since <span class="math-container">${y \over y-m} >1$</span>, <strong>switching to any door within <span class="math-container">$\boldsymbol{y}$</span> is always advantageous.</strong></p>
<p>Finally, on getting rid of <span class="math-container">$y$</span> by making it as large as <span class="math-container">$n-1$</span>, i.e. it having every door except the player's original choice, we get the winning probability of switching which is <span class="math-container">$$ {1 \over n} \times {y \over y-m} = {1 \over n} \times {n-1 \over n-1-m} $$</span></p>
<p><strong>Have a look at these two pages for comparing the real world results with the predictions from the equation</strong> - <a href="https://randompearls.com/reference/tools/monty-hall-paradox-tool/" rel="nofollow noreferrer">Monty Hall Paradox Test Tool</a> and <a href="https://randompearls.com/science-and-technology/mathematics/monty-hall-paradox-probability-equations/" rel="nofollow noreferrer">The Monty Hall Paradox Probability Equations</a>.</p>
|
2,462,852 | <p>I want to prove that $f(x)=x^3+x+2$, $f: \mathbb R \rightarrow \mathbb R$ is bijective without calculus. My attempts at showing to prove that it' injective and surjective are written below:</p>
<p>$1)$ Injectivity: </p>
<p>I want to show that $\forall a,b \in \mathbb R$ $f(a)=f(b) \implies a=b$.</p>
<p>I started like this:
$$f(a)=f(b) \implies a^3+a+2=b^3+b+2$$
$$\implies a(a^2+1)=b(b^2+1)$$
$$\implies \frac{a}{b}=\frac{b^2+1}{a^2+1}$$
Then I said since $\frac{b^2+1}{a^2+1}>0$ $\forall a,b \in \mathbb R$ then either $a \land b < 0$ or $a \land b > 0$. (For the case when $b=0 \land a \in \mathbb R$ it would be easy to prove that $a=b$.) From there it seemed pretty obvious that $\frac{a}{b}=\frac{b^2+1}{a^2+1} \implies a=b$ so I couldn't really draw a logical argument to show that $a=b$.</p>
<p>$2)$ Surjectivity: </p>
<p>I want to show that $\forall b \in \mathbb R$ $\exists a \in \mathbb R$ s.t. $f(a)=b$. </p>
<p>I started like this:</p>
<p>Let $b \in \mathbb R$ and set $f(a)=b$ then we have:</p>
<p>$$a^3+a+2=b$$
$$\implies a^3+a=b-2$$
$$\implies a(a^2+1)=b-2$$</p>
<p>But then I couldn't find an expression for $a \in \mathbb R$ in terms of $b$.
So I'm wondering if anyone can tell me how I can proceed with my surjectivity and injectivity proofs.</p>
| Guy | 206,544 | <p>As for the injectivity, if $a>b$, then $a^3>b^3$, so $f(a)=a^3+a+2>b^3+b+2=f(b)$.</p>
<p>As for the surjectivity, there is a formula to find a real solution of an equation of degree 3; it is not very nice, but you can use it. You can find it <a href="https://en.wikipedia.org/wiki/Cubic_function#Roots:_Nature_and_derivation" rel="nofollow noreferrer">here</a></p>
|
688,711 | <p>Assume the desity of air <span class="math-container">$\rho$</span> is given by</p>
<p><span class="math-container">$\rho(r)=\rho_0$$e^{-(r-R_0)/h_0}$</span> for <span class="math-container">$r\ge R_0$</span></p>
<p>where <span class="math-container">$r$</span> is the distance from the centre of the earth, <span class="math-container">$R_0$</span> is the radius of the earth in meters, <span class="math-container">$\rho_0=1.2kg/m^3$</span> and <span class="math-container">$h_0=10^4m$</span></p>
<p>Assuming the atmosphere extends to infinity, calculate the mass of the portion of the earth's atmosphere north of the equator and south of <span class="math-container">$30^\circ$</span>N latitude.</p>
<p>How do I even start this problem? Do I need to convert it into spherical coordinates? But then what limits do I use for the integration?</p>
| cffk | 36,388 | <p>You don't have to do integrals! Divide atmospheric pressure A = 101.3 kPa by g = 9.8 m/s<sup>2</sup> to give the mass per unit area (kg/m<sup>2</sup>). Multiply this by the area of the earth and you're done. (Assumptions: g is a constant over the height of the atmosphere; g independent of latitude; neglect the mass of the air displaced by the volume of the land about sea level.)</p>
<p><b>ADDENDUM:</b> Also you can use the fact that 1 atmosphere = 760 Torr = 15 lb-force/in<sup>2</sup> to estimate the mass of the atmosphere per unit area as 0.76m ρ<sub>Hg</sub> or 15 lb/in<sup>2</sup> (ρ<sub>Hg</sub> = density of mercury = 13.53 metric tonnes/m<sup>3</sup>).</p>
|
165,235 | <p>I have a weighted, directed graph with 100 vertices and the maximal number of edges, 9900. Are there <em>Mathematica</em> tools or packages available to visualize which edges have large weights? (If you know of non-<em>Mathematica</em> software, that's fine, too.)</p>
<p>Background: My institution has 100 majors, and I'm interested in studying how students change majors, where they start compared to where they end up.</p>
| Szabolcs | 12 | <p>I will use <a href="http://szhorvat.net/mathematica/IGraphM" rel="noreferrer">IGraph/M</a> for the following answer.</p>
<p>Here's a complete directed graph on 100 with weighted edges. There are lots of edges with small weights and a few with large weights.</p>
<pre><code>graph = CompleteGraph[100, DirectedEdges -> True,
EdgeWeight -> RandomVariate[ExponentialDistribution[5], 100 99]];
</code></pre>
<p>We will:</p>
<ul>
<li>remove edges below a certain weight threshold</li>
<li>style the remaining edges based on weight by adjusting their opacity and thickness</li>
</ul>
<p>First, I need to utility functions.</p>
<p>Threshold an array (replace values below the threshold by zero). <a href="http://reference.wolfram.com/language/ref/Threshold.html" rel="noreferrer"><code>Threshold</code></a> does not seem to work on sparse arrays.</p>
<pre><code>threshold[arr_, th_] := arr UnitStep[arr - th]
</code></pre>
<p>Scale the elements of an array so that the largest is 1.</p>
<pre><code>scale[arr_] := arr/Max@Abs[arr]
</code></pre>
<p>Now we convert the graph to a weighted adjacency matrix, threshold, convert back, then style it.</p>
<pre><code>IGWeightedAdjacencyGraph[
VertexList[graph],
threshold[WeightedAdjacencyMatrix[graph], 0.9],
GraphLayout -> "CircularEmbedding", ImageSize -> Large
] //
IGEdgeMap[
Directive[AbsoluteThickness[5 #], Opacity[#], Arrowheads[0.05 #]] &,
EdgeStyle -> scale@*IGEdgeProp[EdgeWeight]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/e584f.png" rel="noreferrer"><img src="https://i.stack.imgur.com/e584f.png" alt="enter image description here"></a></p>
<p>We could use a smaller threshold to include more edges, but compute the opacity based on a power of the weight, to emphasize strong edges and fade out weak ones.</p>
<pre><code>IGWeightedAdjacencyGraph[VertexList[graph],
threshold[WeightedAdjacencyMatrix[graph], 0.7],
GraphLayout -> "CircularEmbedding", ImageSize -> Large] //
IGEdgeMap[
Directive[AbsoluteThickness[5 #], Opacity[#^3],
Arrowheads[0.05 #]] &, EdgeStyle -> scale@*IGEdgeProp[EdgeWeight]]
</code></pre>
<p><a href="https://i.stack.imgur.com/YkrQI.png" rel="noreferrer"><img src="https://i.stack.imgur.com/YkrQI.png" alt="enter image description here"></a></p>
<p>These choices about visualization are yours to make.</p>
|
213,533 | <p>I am trying to play a sequence of sounds in Mathematica using <code>Table</code>. My real example is much more complicated so that is why I got to using <code>Table</code>, </p>
<p>Why doesn't the follow code work? </p>
<pre><code>frequencies = {{100, 200}, {200, 300}, {400, 600}};
Sound[
Table[
Sound[Play[Total[Sin[# 2 Pi t] & /@ frequencies[[i]]], {t, 0, 0.5}]],
{i, 1, 3, 1}]]
</code></pre>
<p>If I run the sounds as a list it works fine, but when using table it does not, I just get the following:</p>
<blockquote>
<pre><code>Sound[{Sound[
Sound[SampledSoundFunction[
Function[{Play`Time494},
Block[{t =
0. + 0.000125 Play`Time494}, (Total[(Sin[#1 2 \[Pi] t] &) /@
frequencies[[i]]] - 4.44089*10^-16) 0.568158]], 4000,
8000]]],
Sound[Sound[
SampledSoundFunction[
Function[{Play`Time495},
Block[{t =
0. + 0.000125 Play`Time495}, (Total[(Sin[#1 2 \[Pi] t] &) /@
frequencies[[i]]] + 1.88738*10^-15) 0.525731]], 4000,
8000]]],
Sound[Sound[
SampledSoundFunction[
Function[{Play`Time496},
Block[{t =
0. + 0.000125 Play`Time496}, (Total[(Sin[#1 2 \[Pi] t] &) /@
frequencies[[i]]] + 8.88178*10^-16) 0.525731]], 4000,
8000]]]}]
</code></pre>
</blockquote>
<p>Any help appreciated</p>
| A little mouse on the pampas | 42,417 | <p>This code is written by Shenlan.</p>
<pre><code> Block[{s1 = "C", s2 = "D", s3 = "E", s4 = "F", s5 = "G", s6 = "A",
s7 = "B", t1 = 0.25, t2 = 0.5},
Sound[SoundNote @@@ {{s3, t1}, {s3, t1}, {s4, t1}, {s5,
t1},(*||*){s5, t1}, {s4, t1}, {s3, t1}, {s2, t1}, {s1, t1}, {s1,
t1}, {s2, t1}, {s3, t1},(*||*){s3, t2}, {None, t1/2}, {s2,
t1/2}, {s2, t2},(*||*){s3, t1}, {s3, t1}, {s4, t1}, {s5,
t1},(*||*){s5, t1}, {s4, t1}, {s3, t1}, {s2, t1},(*||*){s1,
t1}, {s1, t1}, {s2, t1}, {s3, t1},(*||*){s2, t1}, {None,
t1/2}(*Rest*), {s1, t1/2}, {s1, 2 t1},(*||||*){s2, t1}, {s2,
t1}, {s3, t1}, {s1, t1},(*|*){s2, t1}, {s3, t1/2}, {s4,
t1/2}, {s3, t1}, {s1, t1},(*|*){s2, t1}, {s3, t1/2}, {s4,
t1/2}, {s3, t1}, {s2, t1},(*|*){s1, t1/2}, {s2, t1/2}, {s5,
t1}, {s3, t1/2}, {s3, t1/2}, {s3, t1}, {s4, t1}, {s5, t1}, {s5,
t1}, {s4, t1}, {s3, t1/2}, {s4, t1/2}, {s2, t1/2}, {s1, t1}, {s1,
t1}, {s2, t1}, {s3, t1}, {s2, t1}, {None, t1/2}(*Rest*), {s1,
t1/2}, {s1, 2 t1}}]]
</code></pre>
|
1,212,177 | <p>Does $i\arg(e^{2z})=2iy?$ If it does I have solved my problem, and hence it seems like it must be the case, but I don't see it.</p>
<p>$$i\arg(e^{2z})=i\arg(e^{2x+2iy})=i\arg(e^{2x}e^{2iy})\implies \theta=2y(?)$$ Why does the $2x$ get 'ignored'?</p>
| AlexR | 86,940 | <p>Maybe it becomes a little clearer when looking at the principal branch of the logarithm:</p>
<p>$$\log z = \ln |z| + i\arg z$$
So $\arg (e^{2z}) = \Im \log(e^{2z}) = \Im 2z = 2\Im z = 2y$. Note that we have assumed $y\in (-\pi, \pi)$ and your equality is only valid in this case. What you can always say is that
$$i\arg (e^{2z}) = 2yi + 2\pi i k$$
for some $k\in\mathbb Z$.</p>
|
1,997,341 | <p>We know that the binomial theorem and expansion extends to powers which are non-integers. </p>
<p>For integer powers the expansion can be proven easily as the expansion is finite. However what is the proof that the expansion also holds for fractional powers? </p>
<p>A simple an intuitive approach would be appreciated.</p>
| Cheerful Parsnip | 2,941 | <p>You could calculate, for example, $(1+x)^{1/2}=a_0+a_1x+a_2x^2+\cdots$ by squaring both sides and comparing coefficients. For example we can get the first three coefficients by ignoring all degree $3$ terms and higher:
$$1+x=a_0^2+2a_0a_1x+2a_0a_2x^2+a_1^2x^2+\cdots$$</p>
<p>From here we can conclude that $a_0=\pm1$ (we'll take $+1$ to match what happens when $x=0$). Then comparing coefficients of $x$ we have $2a_1=1$, so $a_1=1/2$. Finally, comparing coefficients of $x^2$, we have $2a_0a_2+a_1^2=0$, so $2a_2+1/4=0$ and $a_2=-1/8$.</p>
<p>You can definitely get as many coefficients as you want this way, and I trust that you can even derive the binomial coefficient formula. However, this is not any easier than the Taylor series, where you take $(1+x)^{1/2}=a_0+a_1x+a_2x^{2}+\cdots$ and find the coefficients by saying the $n$th derivatives on both sides have to be equal at $0$. </p>
<p>For example, plugging in $0$ on both sides we conclude $a_0=1$. Calculating the first derivative of both sides, we have
$$\frac{1}{2}(x+1)^{-1/2}=a_1+2a_2x+\cdots$$
Plugging in $0$, we get $a_1=1/2$. Taking the derivative one more time, we see
$$(-1/2)(1/2)(1+x)^{-3/2}=2a_2+\cdots$$
Plugging in $x=0$, we have $(-1/2)(1/2)=2a_2$, or $a_2=-1/8$. The advantage to this way, is that it is much easier to see the pattern of coefficients!</p>
<p>Unfortunately, there is a big hole in both arguments. They will give you what the coefficients have to be, but they won't prove that the series expansion converges in the first place. We started off by assuming you could write $1+x$ as an infinite power series, but there is no guarentee that this exists, and actually it doesn't converge unless $|x|<1$, which we never used. So you need to estimate the error in Taylor's formula to complete the proof rigorously.</p>
|
1,883,765 | <p>In the following situation I know how to find the <code>red</code> distance as <code>(diagonal - diamater) / 2</code> however I'm not sure how to find the <code>yellow</code> and <code>green</code> distances.</p>
<p><a href="https://i.stack.imgur.com/RMNLh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RMNLh.png" alt="enter image description here"></a></p>
| Joshua Lochner | 351,844 | <p>In the above diagram: </p>
<p>Construct a square, such that the diagonal (as shown by the red line) connects the outside square and the circle in the shortest length...</p>
<p>Now, we can see that the green and yellow lines are equal (sides of a square)... Let the red line $= red$, Therefore the green and yellow line = $\frac{red}{\sqrt2}$</p>
<hr>
<p>To answer this in terms of the radius, $r$:</p>
<ul>
<li>Side length of square = $2r$</li>
<li>Diagonal length = $2r\sqrt2$</li>
<li>red section = $\frac{1}{2}$diagonal - $r$ $= r\sqrt2 - r$</li>
</ul>
<p>Therefore Green/Yellow section = $\frac{r\sqrt2 - r}{\sqrt2}$</p>
<p>$$=r - \frac{r}{\sqrt2}$$
$$=r - \frac{r\sqrt2}{2}$$</p>
|
2,939,605 | <p>The given task goes as follows:</p>
<blockquote>
<p>Show that <span class="math-container">$ f: \mathbb{R} \longrightarrow \mathbb{R}$</span> defined by <span class="math-container">$f(x) = \sqrt{1 + x^2} $</span> is not a polynomial function.</p>
</blockquote>
<p>I tried this approach - if <span class="math-container">$f(x)$</span> is a <span class="math-container">$n$</span>-degree polynomial function, then the <span class="math-container">$(n+1)$</span>-st derivative equals to 0 and I was trying to determine the <span class="math-container">$k$</span>-th derivative of <span class="math-container">$f(x)$</span> (and show it differs from 0 for any <span class="math-container">$k$</span>) but without success. Since <span class="math-container">$f(x)$</span> is continuous and defined over whole R domain, I have no idea how to carry on. Any ideas? </p>
| Chinnapparaj R | 378,881 | <p>Suppose <span class="math-container">$f$</span> is a polynomial,namely <span class="math-container">$$f(x)=\sqrt{1+x^2}={a_0+a_1x+\cdots+a_nx^n}=p(x)$$</span> for some <span class="math-container">$n$</span>. Then <span class="math-container">$$\frac{f(x)}{x}=\frac{p(x)}{x}$$</span> for all non zero <span class="math-container">$x$</span>. Hence <span class="math-container">$$\lim_{x \rightarrow \infty} \frac{p(x)}{x}=\lim_{x \rightarrow \infty}\frac{f(x)}{x}=\lim_{x \rightarrow \infty} \frac{\sqrt{1+x^2}}{x}=1$$</span></p>
<p>which means <span class="math-container">$p(x)$</span> and <span class="math-container">$x$</span> are polynomials of same degree, a contradiction!</p>
|
2,939,605 | <p>The given task goes as follows:</p>
<blockquote>
<p>Show that <span class="math-container">$ f: \mathbb{R} \longrightarrow \mathbb{R}$</span> defined by <span class="math-container">$f(x) = \sqrt{1 + x^2} $</span> is not a polynomial function.</p>
</blockquote>
<p>I tried this approach - if <span class="math-container">$f(x)$</span> is a <span class="math-container">$n$</span>-degree polynomial function, then the <span class="math-container">$(n+1)$</span>-st derivative equals to 0 and I was trying to determine the <span class="math-container">$k$</span>-th derivative of <span class="math-container">$f(x)$</span> (and show it differs from 0 for any <span class="math-container">$k$</span>) but without success. Since <span class="math-container">$f(x)$</span> is continuous and defined over whole R domain, I have no idea how to carry on. Any ideas? </p>
| Angel Moreno | 327,493 | <p>The limit of the function <span class="math-container">$f(x)$</span> when <span class="math-container">$x$</span> tends to infinity is <span class="math-container">$x$</span>.
<span class="math-container">$$\lim_{x\to\infty} f(x) = x$$</span></p>
<p>Therefore, either the function <span class="math-container">$f(x)$</span> is a first degree polynomial or it is not a polynomial.</p>
<p>Also, <span class="math-container">$f ^ 2$</span> must be <span class="math-container">$x ^ 2 + 1$</span>.</p>
<p>The polynomials of the first degree squared are either simply <span class="math-container">$x ^ 2$</span> or have a component <span class="math-container">$x$</span> other than zero:
<span class="math-container">$$ (x+a)^2 = x^2 + 2ax +a^2 $$</span></p>
<p>Therefore <span class="math-container">$f (x)$</span> is not a first degree polynomial (<span class="math-container">$f^2$</span> it has no <span class="math-container">$x$</span> component, and is not <span class="math-container">$ x^2 $</span>).</p>
<p>And since it can not have more degrees, <span class="math-container">$f (x)$</span> is not a polynomial.</p>
|
2,448,745 | <p>I am trying to derive the general form of the equation of the circle given two points $(x_1,y_1)$, $(x_2,y_2)$and the angle $\theta$ subtended by the chord joining the two points. </p>
<p>So, $\theta$ is basically the angle between two lines
$\implies \tan \theta = |\dfrac{\dfrac{y-y_1}{x-x_1}-\dfrac{y-y_2}{x-x_2}}{1+( \dfrac{y-y_1}{x-x_1}\dfrac{y-y_2}{x-x_2})}| \\ \implies(x-x_1) (x-x_2)+(y-y_1)(y-y_2)= \pm \cot\theta[(y-y_1)(x-x_2)-(x-x_1)(y-y_2)]$</p>
<p>But this is the equation of the major arc right? How do I derive the equation of the circle from it?</p>
| Tobias Kildetoft | 2,538 | <p>If we do this in two dimensions, like you suggest, then we get something that starts looking like a $2$-category (possibly with just one object, i.e. a monoidal category).</p>
<p>A common way to describe "nice" monoidal categories is in terms of diagrams, and these diagrams can then be composed both horizontally, by putting them next to each other (this is the "tensor product" in the category), or vertically, by placing one on top of the other (the "usual composition" in the category). Of course, this has the issue of which objects can be composed, since not all of them need to be composable.</p>
<p>For examples of this, see for example the paper "Soergel Calculus" by Elias and Williamson (this forms the basis of their Annals paper in which they prove the Soergel conjecture).</p>
|
72,098 | <p>This is my first time using Stack Exchange, but it looks like a good resource. I am here to ask a couple questions about my homework. We're working from the latest edition of <em>Abstract Algebra</em> by Herstein. This is problem #3 from p. 73 of that book, and it reads as follows:</p>
<blockquote>
<p>Let $G$ be any group and $A(G)$ the set of all 1-1 mappings of $G$, as a set, onto itself. Define $L_{a} \colon G \to G$ by $L_{a}(x)=xa^{-1}$. Prove that:<br>
(a) $L_{a} = A(G)$.<br>
(b) $L_{a}L_{b} = L_{ab}$.<br>
(c) The mapping $\psi:G \rightarrow A(G)$ defined by $\psi(a) = L_{a}$ is a monomorphism of G into A(G).</p>
</blockquote>
<p>I believe that I have answered parts b & c correctly, but have some concerns about the rigorousness of my approaches to all three problems. I will start out here by including my proposed solution to part a. Any advice or comments would be greatly appreciated.</p>
<p>(a) $A(G)$, as the set of all 1-1 mappings of $G$ onto itself, can be represented as the set of all operations on an element $x \in G$ such that the result is also in $G$. Since $G$ is a group, we know that this set can be represented as the set of all group multiplications $yx \mid x \in G, y \in G$ for a given element $x$. This is because any $x$ element can be fixed as the first parameter, while the $y$ elements are taken over every element of G. We have then that $yx \in G$, due to G's closure under group multiplication. Furthermore, any element $e \in G$ has an inverse element $e^{-1} \in G$, since $G$ is a group. This means that we can consider any element in $G$ as the inverse of its inverse: $e = (e^{-1})^{-1}$. This, when plugged in for our variable $x \in G$ above, gives us that $\forall x \in G, \forall y \in G, yx^{-1} \in G$. This is exactly our given mapping $L_{a}$ above, with the labels rearranged. This shows that $L_{a}$ is a mapping from $G \rightarrow G$, as required. In order to show why this set contains every such possible mapping, we will assume that there is a mapping $M_{a}(x) : G \rightarrow G$ such that $M_{a}(x) \notin L_{a}$. This mapping, as a mapping from G to G, must take the form of a group multiplication: $M_{a}(x) = x*a, x\in G$. However, since G is a group, we have again that $a = (a^{-1})^{-1}$, or, if we let $m = a^{-1}$, that $a = m^{-1}$, and so our mapping can be rewritten as $M_{a}(x) = x*m^{-1} m \in G$. However, this is exactly the same mapping as our above $L_{a}(x)$, showing that every mapping in $A(G)$ can indeed be written as a multiplication between some element $x \in G$ and another element's inverse $a^{-1} \in G$, and thus that $L_{a} \in A(G)$. $\blacksquare$</p>
<p>Thanks for taking the time to check this out, even if you don't feel you can offer any help.
<strong>EDIT</strong>: Thanks to those who pointed out that I had used =, not $\in$, above by mistake. Also appreciated is the edit to italicize my variables. Now I know how to as well. :) </p>
<p><strong>EDIT</strong>: The responses so far have been so helpful, I'd like to put the other two parts of my solution up to solicit feedback on them as well. Hopefully they aren't as muddled as the first part's was.<br>
<strong>changed a bit in response to feedback</strong><br>
(b)
$L_{a}(x) = xa^{-1}$<br>
$L_{b}(x) = xb^{-1}$<br>
$(L_{a}L_{b})(x) = L_{a}(L_{b}(x))$<br>
$L_{a}(L_{b}(x)) = L_{a}(xb^-1)$<br>
$L_{a}(xb^-1) = xb^{-1}a^{-1}$<br>
$xb^{-1}a^{-1} = x(ab)^{-1} = L_{ab}(x)$<br>
$L_{a}(x)L_{b}(x) = L_{ab}(x) \quad\blacksquare$ </p>
<p>(c)
$\psi(a) = L_{a}(x) = xa^{-1}$. To show that this mapping is a monomorphism of $G$ into $A(G)$, we will first rely on part (a) to state that $\psi(a) = L_{a}$ is indeed a mapping from $G$ to $A(G)$. Now we must show that $\psi$ is a monomorphism of $G$ into $A(G)$. First we will show that $\psi$ is a homomorphism of $G$ into $A(G)$. To this end, we will appeal to the results of our calculations in part (b) to state that $L_{a}L_{b} = L_{ab}$, which of course implies that $\psi(a)\psi(b) = L_{a}L_{b} = L_{ab} = \psi(ab)$, which proves that $\psi$ is a homomorphism. In order to continue and show that $\psi$ is a monomorphism, we must show that it is an injective (1-1) mapping. Let $\psi(a) = Z = \psi(b)$. This can be written as:
$\psi(a) = L_{a}(e) = ea^{-1} = Z$<br>
$\psi(b) = L_{b}(e) = eb^{-1} = Z$<br>
$Z = ea^{-1} = eb^{-1}$<br>
$a^{-1} = b^{-1} \Rightarrow a = b$<br>
The last line of the above follows from the uniqueness of inverse elements in $G$. This shows that the only way for two output values of $\psi$ to be equal is for their inputs to be equal as well, and thus $\psi$ is an injective homomorphism, or a monomorphism from $G$ to $A(G)$. $\blacksquare$</p>
| Arturo Magidin | 742 | <p>I'm somewhat concerned with what you write, as I am not sure what it is you are trying to say.</p>
<p>You seem to be trying to show that if $f\in A(G)$, then $f=L_a$ for some $a$; first, <em>nobody is asking you to prove that</em>. And second, you don't seem to be doing that. And third, what you write, $L_a=A(G)$, does not even make sense! $L_a$ is a function from $G$ to itself; $A(G)$ is a set of functions from $G$ to itself. You are trying to show that a single function is equal to a set of functions. That is going to be rather hard to do...</p>
<p>Suppose that $f\colon G\to G$ is a (set-theoretic) 1-1 function.</p>
<p>It <em>is</em> true that for <em>each</em> $y\in G$, since $f(y)\in G$ by hypothesis, and since in a group we can always solve any equation of the form $ay=b$, we can find some $x$, *which depends on $y$ *, such that $f(y) = yx$. <strong>However</strong>, in general, different $y$'s will require different $x$s <em>with the same function $f$</em>.</p>
<p>So I do not see how you can simply state, as you do when you write:</p>
<blockquote>
<p>Since G is a group, we know that this set can be represented as the set of all group multiplications $yx | x\in G,y\in G$ for a given element $x$.</p>
</blockquote>
<p>In fact, this assertion is wrong: suppose that every bijection $f\colon G\to G$ is indeed of the form $f(y)=yx$ for some $x$. Then $f(e) = ex = x$, so for every $y$ we would have $f(y)=yf(e)$. It is very easy to see that this cannot be true for most groups, because if you have any bijection $G\to G$, you can compose it with a function that transposes $e$ and $x\neq e$ that does not have order $2$, and still get a bijection. This composition will be a bijection, but will map $e$ to $xf(e)$, but $x$ will not be mapped to $x^2f(e)$, but to $f(e)$. So not every function in $A(G)$ can be of the form $L_x$ for some $x$. The rest of the paragraph is, in my opinion, a big muddle.</p>
<p>In any case, you are not being asked to prove that. You are being asked to show that if $a\in G$, then the map $L_a\colon G\to G$ is an element of $A(G)$. That is, you need to show that $L_a$ is a bijection of $G$ onto itself. You are <em>given</em> that $L_a$ is a function from $G$ to $G$, so what you need to show is that $L_a$ is one-to-one and onto. </p>
<p>(As an aside, your attempt to argue about a function $M_a$ which is not one of the $L_a$ also gets off on the wrong foot; if you wanted to show by contradiction that every element of $A(G)$ is <em>some</em> $L_a$, you would need to start by assuming there is a function $M$ in $A(G)$ such that *<em>for every $b\in G$ *</em> we have $M\neq L_b$; you only assume that $M\neq L_a$ for a particular $a$, and that's no good). </p>
<hr/>
<p><strong>Added.</strong> The penultimate line for part (b) should have $L_{ab}(x)$, not $L_{ab}$. Otherwise, it is correct.</p>
<p>The first line of (c) is incorrect. $L_a(x)$ is an element of $G$ (namely, $xa^{-1}$). But $\psi(a)$ is an element of $A(G)$ (namely, $L_a$). $\psi(a)$ does not equal $L_a(x)$. </p>
<p>You don't need to show that $\psi$ is a map from $G$ to $A(G)$: it is <strong>defined</strong> to be a map from $G$ to $A(G)$. This follows from (a), since you know, if you do (a) correctly, that $L_a\in A(G)$ for any $a\in G$.</p>
<p>Several times you confuse the function $L_a$, with the value of the function at $x$, $L_a(x)$. Remember that "$x$" is one of the names for elements of $G$: don't use it as if this were calculus and you call the function "$f(x)$"! The name of the function is just $L_a$, not $L_a(x)$. </p>
<p>You correctly show it is a homomorphism.</p>
<p>To prove it is 1-1, you are not really doing a proof by contradiction, you are doing a direct proof: you are showing that if $\psi(a)=\psi(b)$, then $a=b$. This is a <em>direct proof</em>, do it as a direct proof. But the line right after that you again commit the <em>faux pas</em> of confusing the function $\psi(a)$ with the particular value of $\psi(a)$ at the element $x$. $\psi(a)$ <strong>is not</strong> equal to $xa^{-1}$, and $\psi(b)$ <strong>is not</strong> equal to $xb^{-1}$. </p>
<p>Instead, you have to assume that $L_a=L_b$ <em>as functions</em>, meaning that <em>for every</em> $x\in G$ you have $L_a(x)=L_b(x)$. At <em>that</em> point you can plug in some values for $x$ and see what you can conclude. Might I suggest using the fact that $L_a(e) = L_b(e)$ to conclude that $a=b$?</p>
|
2,735,854 | <p>How many permutations of the word $STRESSLESSNESS$ begin OR end with an $E$? </p>
<p>Correct me if I'm wrong, but you would have to subtract the permutations where $E$ begins AND ends the permutation?</p>
| Michael Hardy | 11,667 | <p>One way of doing it is this:</p>
<p>number that begins with an E, plus number than ends with an E, minus number that both begins and ends with an E.</p>
<p>Another way is this:</p>
<p>First find the total number of permutations, then subtract the number that begin and end with a non-E.</p>
|
104,328 | <p>This question is related to <a href="https://mathematica.stackexchange.com/questions/104310/export-bug-in-combination-with-paralleltable">Export bug in combination with ParallelTable?</a>.</p>
<p>The following code shows an even stranger bug than in the upper question.</p>
<p><strong>The y-axis should always be logarithmic, but in the resulting plot files it jumps randomly between logarithmic and linear</strong>. The linear axis numbers are wrong.</p>
<pre><code>ChoiceDialog[{FileNameSetter[Dynamic[outputDir], "Directory"], Dynamic[outputDir]}];
SetDirectory[outputDir];
image = ColorConvert[ExampleData[{"TestImage", "Lena"}], "Grayscale"];
levels = ImageLevels[image, "Byte"];
hist = Histogram[WeightedData @@ Transpose[levels], 256,
ScalingFunctions -> "Log", ImageSize -> 600];
ParallelTable[
fileName = StringJoin[outputDir, "\\histogram_parallel_table_", ToString[i], ".png"];
Export[fileName, hist, "PNG"],
{i, 1, 10}
];
</code></pre>
<p><strong>This error does not occur when <code>Table</code> is used or when logarithmic scaling is removed.</strong></p>
<p>For one run I got e.g. the following two different plots (there are more):</p>
<p><code>i=1</code>:</p>
<p><a href="https://i.stack.imgur.com/CNwd7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CNwd7.png" alt="enter image description here"></a></p>
<p><code>i=6</code>:</p>
<p><a href="https://i.stack.imgur.com/EHAcy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EHAcy.png" alt="enter image description here"></a></p>
<p>I am programming with Mathematica 10.3.1.0 on Windows 10 Professional 64 Bit and have an i7-4940-MX 3,1 GHz processor (4 cores).</p>
| mrz | 19,892 | <p>This is a workaround, which I found at </p>
<blockquote>
<p>example/OutOfCoreImageHistogramComputation</p>
</blockquote>
<pre><code>counts = ConstantArray[0, 256];
f[x_] := Block[{b = Clip[Floor[255*x + .5], {0, 255}]}, counts[[b + 1]]++];
image = ColorConvert[ExampleData[{"TestImage", "Lena"}], "Grayscale"];
ImageScan[f, image];
BarChart[counts, BarSpacing -> 0, ImageSize -> 600, ScalingFunctions -> "Log"]
ChoiceDialog[{FileNameSetter[Dynamic[outputDir], "Directory"], Dynamic[outputDir]}];
SetDirectory[outputDir];
ParallelTable[
fileName = StringJoin[outputDir, "\\histogram_parallel_table_", ToString[i], ".png"];
Export[fileName, plot, "PNG"],
{i, 1, 10}
];
</code></pre>
<p>All plots look the same and the vertical scale is always logarithmic.</p>
<p>For all <code>i</code>:</p>
<p><a href="https://i.stack.imgur.com/vqeKW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vqeKW.png" alt="enter image description here"></a></p>
|
2,698,284 | <p>Here is the problem: </p>
<p>Let A be a set with n elements. Find an expression for S(n, 2), the number of partitions of A into exactly two subsets. You can either start with the general recurrence for S(n, k), or count S(n, 2) directly. </p>
<p>I'm having trouble understanding what exactly it wants me to do. So far I know that it wants me to find an expression that gives the number of combinations of A into sets like (x, y). I got the equation: (n - 1) * 2 for number of partitions of A into two parts, but I don't know understand what the problem means by creating two subsets. What am I doing wrong?</p>
| Michael Rees | 881,353 | <p>I think that you need S(n, 2)=2^(n-1)+1, assuming that both sub-sets are required to not be empty.</p>
<p>For example n=3 gives 3 subsets (i.e {{1,2}, {3}}, {{1,3}, {2}}, and {{2,3}, {1}}).</p>
<p>There is a recurrence relation: S(n+1,2)=2*S(n,2)+1</p>
<p>This can be interpreted as saying that when you and a new element, then this element can be put in any of the previous sub-sets, or in a sub-set by itself. So, for S(4,2) you would have:
{{1,2,4}, {3}}, {{1,3,4}, {2}}, and {{2,3,4}, {1}}) as well as
{{1,2}, {3,4}}, {{1,3}, {2,4}}, and {{2,3}, {1,4}}), and as well as
{{1,2,3}, {4}}</p>
<p>So S(4,2)=2*3+1=7 etc.</p>
<p>In general, there is a recurrence relation: S(n+1,k)=kS(n,k)+S(n,k-1) for k between 1 and n. This can be interpreted in the same way as the example where k=2. [and the sum over all k for any fixed n is the number of partitions, which is known as Bell's number]</p>
|
2,655,178 | <p>I am asked to find the equation of a cubic function that passes through the origin. It also passes through the points $(1, 3), (2, 6),$ and $(-1, 10)$. </p>
<p>I have walked through many answers for similar questions that suggest to use a substitution method by subbing in all the points and writing in terms of variables. I have tried that but I don't really know where to take it from there or what variables to write it as. </p>
<p>If anyone could provide their working out for this problem it would be extremely enlightening. </p>
| Narasimham | 95,860 | <p>To determine a conic we need to solve 5 equations with 5 given points.</p>
<p>Likewise here we are given 4 points and 4 simultaneous equations, not involving any $xy $ term.. so solve it by Cramer's determinants. </p>
|
2,329,003 | <p>Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3 $ in the plane $ x+y+z=1 $ with positive orientation . $$ $$ I have thought the parametriation: </p>
<p>\begin{align} x(t)=1+ 3 \cos (t) \hat j +3 \sin (t) \hat k \\ y(t)=1+3 \cos (t) \hat i+3 \sin (t) \hat k \\ z(t)=-1+3 \cos (t) \hat i +3 \sin (t) \hat j , \ \ 0 \leq t \leq 2 \pi \end{align} But I am not sure . <strong>Any help is there ?</strong></p>
| Donald Splutterwit | 404,247 | <p>The plane can be parameterised as follows
\begin{eqnarray*}
x=1+t \\
y=1+s \\
z=-1-t-s.
\end{eqnarray*}
Now substitute this into the equation for the sphere $(x-1)^2+(y-1)^2+(z+1)^2=9$. We have
\begin{eqnarray*}
t^2+s^2+(t+s)^2=9 \\
\end{eqnarray*}
Rearrange this to</p>
<p>\begin{eqnarray*}
(2t+s)^2+3s^2=18
\end{eqnarray*}
This can be paramterised </p>
<p>\begin{eqnarray*}
s=\sqrt{6} \sin(\theta) \\
2t+s= \sqrt{18} \cos(\theta)
\end{eqnarray*}
Now substitute back into the parameterisation of the plane</p>
<p>\begin{eqnarray*}
x=1+ \frac{\sqrt{18} \cos(\theta) - \sqrt{6} \sin(\theta)}{2} \\
y=1+\sqrt{6} \sin(\theta) \\
z=-1-\frac{\sqrt{18} \cos(\theta)+\sqrt{6} \sin(\theta)}{2}
\end{eqnarray*}</p>
|
212,466 | <p>Consider the list <code>list={1,2,3,4,5,6,7,8}</code> and imagine I wanted to change it to <code>{{1,2,3},{4,5,6},{7,8}}</code>. How can I do this?</p>
<p>Using <code>ArrayReshape</code> in the usual manner</p>
<pre><code>list = {1, 2, 3, 4, 5, 6, 7, 8};
ArrayReshape[list, {3, 3}]
</code></pre>
<p>yields <code>{{1, 2, 3}, {4, 5, 6}, {7, 8, 0}}</code>. Is it possible to use <code>ArrayReshape</code> and avoid that extra 0 at the end? Essentially, I want to reshape my list according to its size.</p>
| OkkesDulgerci | 23,291 | <pre><code>list = {1, 2, 3, 4, 5, 6, 7, 8};
Partition[list, UpTo[3]]
</code></pre>
<blockquote>
<p>{{1, 2, 3}, {4, 5, 6}, {7, 8}}</p>
</blockquote>
|
4,585,589 | <p>In the 4th edition of "Matrix Computations", Golub and Van Loan present "Algorithm 5.1.1 (Householder Vector)". The first couple of lines (translated into MATLAB-syntax) read:</p>
<pre><code>m = length(x); sigma = x(2:m)'*x(2:m); v = [1; x(2:m)];
if sigma == 0 && x(1) >= 0
beta = 0;
elseif sigma == 0 && x(1) < 0
beta = -2;
else
...
</code></pre>
<p>The <code>else</code> clause handles the case where <code>sigma</code> is nonzero and no code after the provided snippet modifies <code>v</code> if <code>sigma</code> is zero. The matrix form of the resulting Householder transformation is <span class="math-container">$I-\beta v v^T$</span>.</p>
<p>The <code>elseif</code> clause is a little strange. It doesn't appear in the algorithm listing in the
3rd edition, so it was added for the 4th, presumably for numerical stability. However, it seems to me to generate a <span class="math-container">$(\beta, v)$</span> pair that doesn't map to an orthogonal matrix. For example, if <code>x = [-1; 0; 0]</code> then <code>sigma == 0</code> and <code>x(1) < 0</code> so we get <code>beta = -2</code> and <code>v = [1; 0; 0]</code> and a Householder transformation of <code>[3, 0, 0; 0, 1, 0; 0, 0, 1]</code> which is not an orthogonal matrix.</p>
<p>So my questions are:</p>
<ul>
<li>What benefit is there to handling that case separately, rather than just setting <code>beta = 0</code> if <code>sigma = 0</code>?</li>
<li>Does the resulting Householder transformation need to be applied differently?</li>
</ul>
| Lutz Lehmann | 115,115 | <p>You are correct, this is another error in this algorithm (complete code in <a href="https://github.com/Jiangtang/Matrix-Computations/blob/master/Archives/matlab_GVL/Chapter5/Functions/House.m" rel="nofollow noreferrer">a github archive</a>, are there more official online sources?). It is clearly stated in the header comments that</p>
<pre><code>% v is a column m-vector with v(1) = 1 and beta is a scalar such
% that P = I - beta*v*v' is orthogonal and Px = norm(x)*e1.
</code></pre>
<p>and in the general case <code>beta</code> is computed as</p>
<pre><code> beta = 2*v(1)^2/(sigma+v(1)^2);
v = v/v(1);
</code></pre>
<p>according to the documented intent. When testing this there apparently were no tests for trivial cases. On the other hand, getting <code>sigma == 0</code> in the usual situation of a call from the middle of some numerical algorithm has probability zero.</p>
<hr />
<p>The more serious error is in the algorithm design, in the choice of the construction of the reflection normal <code>v</code> as a continuous vector field on the punctured sphere, excepting the "north" pole, as done with the first special case.</p>
<p>The question of course remains if that makes sense at all. Comparing a floating point value with <span class="math-container">$0$</span> in a numerical algorithm smells of bad design. If one has to avoid a singularity in one point, then it is to be expected that results close to that singularity are ill-behaved, with a dramatic loss in accuracy or the magnification of floating-point noise to the level of the intended results.</p>
<p>The stated aim is to get the first component of the reflected vector to be positive. The general construction is to find one of the two bisectors of the lines along the given vectors <span class="math-container">$\vec x$</span>, and some unit vector <span class="math-container">$\vec e$</span>, typically <span class="math-container">$\vec e=\vec e_1=(1,0,…,0)^T$</span>,
<span class="math-container">$$
\vec v = \frac{\pm\|\vec x\|\vec e+\vec x}{\bigl\|\pm\|\vec x\|\vec e+\vec x\bigr\|}
=\frac{\pm\|\vec x\|\vec e+\vec x}{\sqrt{2\|\vec x\|(\|\vec x\|\pm\vec e^T\vec x)}}.
$$</span>
The recommendation for numerical stability is to take the variant that makes the denominator largest. This switches the variants at the equator (normal plane to <span class="math-container">$\vec e$</span>), this switch is a jump by <span class="math-container">$90°$</span> in the direction.</p>
<p>With <span class="math-container">$\vec e=\vec e_1$</span>, the reflected vector has as first component
<span class="math-container">\begin{align}
x_{1,\rm reflect}
&=x_1-2\frac{(\pm\|\vec x\|+x_1)\|\vec x\|(\pm x_1+\|\vec x\|)}{2\|\vec x\|(\|\vec x\|\pm x_1)}
\\&=\mp \|\vec x\|
\end{align}</span></p>
<hr />
<p>So to always end up with a positive number one would have to chose the lower sign variant, with the special case where that does not work. This can even be done in a numerically stable way when <span class="math-container">$\vec x\approx \|\vec x\|\vec e_1$</span>, as by binomial extension
<span class="math-container">$$
x_1-\|\vec x\|=\frac{-(\|\vec x\|^2-x_1^2)}{\|\vec x\|+x_1}
=-\frac{\|\vec x_{2:n}\|^2}{\|\vec x\|+x_1}
$$</span></p>
<p>Why would that be bad? Explore the reflector construction for <span class="math-container">$\vec x=\pmatrix{1\\\bar x}$</span> with <span class="math-container">$\|\bar x\|\sim\mu=10^{-16}$</span> some quasi-random accumulation of rounding errors from previous computational steps. Then using the "bad choice" the reflector unit normal will be close to <span class="math-container">$\pmatrix{-\|\bar x\|/2\\\bar x/\|\bar x\|}$</span>. Thus a small error in the input leads to a large difference in the result, a random point on the equator. This kind of unpredictability, "catastrophic" error magnification, is usually unwanted in stable numerical algorithms.</p>
<hr />
<p>The "good choice" would always result in a reflector normal close to the north unit vector <span class="math-container">$\pmatrix{1\\0}$</span>, this kind of dependency is what is actually desired.</p>
<p>Now one could ask what about points on the equator under the "good choice", then both the midpoint to the north pole and the midpoint to the south pole are equally good. While it is true that under slight perturbations one of the two is selected quasi randomly, the result still stays close to one of these two possibilities, this is still numerically stable under a slight extension of this idea.</p>
|
1,098,587 | <blockquote>
<p>Prove that for a bipartite graph $G$ on $n$ vertices the number of edges in $G$ is at most $\frac{n^2}{4}$. </p>
</blockquote>
<p>I used induction on $n$. </p>
<p>Induction hypothesis: Suppose for a bipartite graph with less than $n$ vertices the result holds true. </p>
<p>Now take a bipartite graph on $n$ vertices.Let $x,y$ be two vertices in $G$ where an edge exist between $x$ and $y$. Now remove these two vertices from $G$ and consider this graph $G'$. $G'$ has at most ${(n-2)^2}\over4$. Add these two vertices back. Then the number of edges $G$ can have is at most </p>
<p>$$|E(G')|+d(x)+d(y)-1$$ </p>
<p>My question is in my proof I took $d(x) + d(y) \le n$, where $d(x)$ denotes the degree of vertex $x$. Can I consider $d(x)+d(y) \leq n$? I thought the maximum number of edges is obtained at the situation $K_{\frac n 2,\frac n 2}$</p>
| Matt Samuel | 187,867 | <p>There is no need to use induction here. A bipartite graph is divided into two pieces, say of size $p$ and $q$, where $p+q=n$. Then the maximum number of edges is $pq$. Using calculus we can deduce that this product is maximal when $p=q$, in which case it is equal to $n^2/4$.</p>
<p>To show the product is maximal when $p=q$, set $q=n-p$. Then we are trying to maximize $f(p)=p(n-p)$ on $[0,n]$. We have $f'(p)=n-2p$, and this is zero when $p=n/2$. After checking the end points we conclude that the maximum is $n^2/4$ at $p=n/2$.</p>
|
1,050,664 | <p>Can someone explain why the inverse $4$ modulo $9$ is $7$? What am I missing?
$$9 = 2\cdot4 + 1$$
$$1 = 9-4\cdot2$$
$$1 = -2\cdot4 + 1\cdot9$$</p>
<p>Isn't then $-2$ inverse of $4$ modulo $9$?</p>
| John Butnor | 185,327 | <p>To find the modular multiplicative inverse x of 4(mod 9), we must solve the equation </p>
<p>x = 4^(-1)(mod 9)</p>
<p>4x = 1(mod 9)</p>
<p>4x = 28(mod 9)</p>
<p>x = 7(mod 9)</p>
<p>The multiplicative inverse is 7. If you multiply a number by its inverse, you get 1.</p>
<p>7*4 = 28, but 28(mod 9) = 1.</p>
|
2,100,793 | <p>Let us say that I have a set $A=\{1,2, 3\}$. Now I need to access, say, the element $3$ of $A$. How do I achieve this?</p>
<p>I know that sets are unordered list of elements but I need to access the elements of a set. Can I achieve this with a tuple? Like $A=(1, 2, 3)$, should I write $A(i)$ to access the i-th element of $A$? Or is there any other notation? </p>
<blockquote>
<p>If I have a list of elements, what is the best mathematical object to represent it so that I can freely access its elements and how? In programming, I would use <code>arrays</code>.</p>
</blockquote>
| Milo Brandt | 174,927 | <p>You are speaking of the index operator one often finds in programming when using arrays. Given that sets are unordered, it does not make sense to do this with sets, as $\{1,2\}=\{2,1\}$. </p>
<p>You probably want to use a <em>tuple</em>, which is just a (finite) sequence of elements, and is denoted as you wrote $A$. If $A=(1,2,3)$ is a tuple, then I think the most common way to index it is to use subscripts - so $A_1=1$ and $A_2=2$ and so on, though one sometimes sees $A(1)$ and $A(2)$ as well, since one can think of $A$ as a function from the set $\{1,2,3\}$ to $\mathbb R$. Occasionally, one even sees $A^1$ and $A^2$. Really, what matters is that you are consistent and clear in how you write indexing.</p>
|
356,279 | <p><strong>Definitions</strong> (<a href="https://www.ias.ac.in/article/fulltext/pmsc/122/03/0459-0467" rel="nofollow noreferrer">https://www.ias.ac.in/article/fulltext/pmsc/122/03/0459-0467</a>):</p>
<p>Given a planar convex region <span class="math-container">$C$</span> (could be smooth or polygonal), an <strong>area bisector</strong> of <span class="math-container">$C$</span> is any line that partitions <span class="math-container">$C$</span> into 2 pieces of equal area. A <strong>'fair bisector'</strong> is a line that partitions <span class="math-container">$C$</span> into 2 pieces of <em>equal area and equal perimeter</em>.</p>
<p>Thru every point on the boundary of <span class="math-container">$C$</span>, an area bisector can be drawn (for a description of their properties, please see 'Mathematical Omnibus' by Fuchs and Tabachnikov, Lecture 11). But it can be seen that a convex planar region can have just a single fair bisector (eg. for a thin isosceles triangle, the only fair bisector is the bisector of its apex angle) or a finite number of them (in which case, their number is necessarily odd as can be seen from simple continuity arguments; see reference at the top) or infinitely many.</p>
<p><strong>Observations:</strong> For regions with a center of symmetry such as a circular disk or ellipse or regular polygon with even number of sides, all fair bisectors are concurrent. But, numerically, we see that for a general convex region <span class="math-container">$C$</span> with finitely many fair bisectors, the fair bisectors are not necessarily concurrent but usually very close to being so. Clearly,for a general <span class="math-container">$C$</span> with exactly 3 fair bisectors, they determine a small triangular region deep in the interior of <span class="math-container">$C$</span>. For <span class="math-container">$C$</span>s with more fair bisectors, their many possible intersections will divide the interior of <span class="math-container">$C$</span> into many regions. Let us refer to the union of those regions which do not share the outer boundary of <span class="math-container">$C$</span> as the 'core' of <span class="math-container">$C$</span>.
The core must lie deep inside <span class="math-container">$C$</span>.</p>
<p><strong>Questions:</strong></p>
<ol>
<li><p>For which convex shape of <span class="math-container">$C$</span> is the area of the 'core' of <span class="math-container">$C$</span> the largest as a fraction of the area of <span class="math-container">$C$</span>? Intuitively, a relatively large core is a measure of the asymmetry of <span class="math-container">$C$</span>. Can one say (say) that such a shape is always one with exactly 3 fair bisectors?</p>
</li>
<li><p>Generalizing a bit, what about lines that break off the same fraction <span class="math-container">$t$</span> of the area and outer boundary length of <span class="math-container">$C$</span>? For a circular disk, it appears that only for <span class="math-container">$t=1/2$</span>, we have such lines (any diameter). Are there <span class="math-container">$C$</span>'s for which such lines exist for several (maybe even arbitrarily many) different values of <span class="math-container">$t$</span>?</p>
<p>Guess: All centrally symmetric convex regions (rectangles, ellipses,...) appear to give such only one single partitioning line that divides both area and outer perimeter in same ration - only for <span class="math-container">$t=1/2$</span>. But general convex regions with no symmetry might give infinitely many such lines - one such partitioning line for each orientation - and a different value <span class="math-container">$t$</span> for each orientation. And the set of these lines might even have interesting envelopes.</p>
</li>
</ol>
<p>These questions have obvious higher dimensional analogs.</p>
| Nandakumar R | 142,600 | <p>An answer to the question 2 (written with K Sheshadri)</p>
<p>The guess made above that needs to be proved: For a general convex polygonal region with no symmetries, for every orientation, we have a unique line with that orientation that separates out the same fraction of both area and outer boundary length. The value of this common fraction of area and perimeter separated out will vary continuously with orientation.</p>
<p><strong>Proof :</strong>
Consider the convex polygonal region <span class="math-container">$C$</span> and a given orientation (direction). Draw both tangents to <span class="math-container">$C$</span> in that orientation. We assume both these tangents to touch <span class="math-container">$C$</span> at a single vertex (coincidences of tangents with entire edges of <span class="math-container">$C$</span> can be dealt with by small perturbations). Let these parallel tangents be distance <span class="math-container">$D$</span> apart. By sliding a line coincident with one of the tangents perpendicular to itself until it coincides with the other tangent to <span class="math-container">$C$</span>, we get a continuous range of cutting lines. Let these cutting lines be parametrized by <span class="math-container">$d$</span>, the perpendicular distance from the tangent from which we began sliding the cutting line. </p>
<p>For each value of <span class="math-container">$d$</span>, we have a line that cuts <span class="math-container">$C$</span>. Plot against <span class="math-container">$d$</span>, the fraction of area (call this fraction Af) of the full <span class="math-container">$C$</span> that the piece separated from <span class="math-container">$C$</span> has and also the fraction of perimeter (call this fraction Pf) for the same piece. Obviously, as <span class="math-container">$d$</span> goes from 0 to <span class="math-container">$D$</span> both Af and Pf go from 0 to 1.</p>
<p>Now, we observe that the plot of Af against d has a quadratic behavior at both ends. Its plot will be continuous and made of several parabolic segments - beginning with an upward parabolic piece (where, as <span class="math-container">$d$</span> starts from 0, Af also starts from 0) and ending with a downward parabolic piece (when Af tends to 1 as <span class="math-container">$d$</span> approaches <span class="math-container">$D$</span>). Moreover, due to convexity of <span class="math-container">$C$</span>, the curve of Af rises monotonically. </p>
<p>On the other hand, Pf has a linear behaviour throughout including at ends. This graph is a continuous polyline and also rises monotonically. </p>
<p>From the above observations, as <span class="math-container">$d$</span> is increased from 0, the Af curve (quadratic) begins <strong>lower</strong> than Pf (linear) curve and as <span class="math-container">$d$</span> tends to <span class="math-container">$D$</span>, Af approaches 1 from <strong>above</strong> the Pf curve. This plus the monotonically rising nature of both graphs plus their start values both being 0 and end values both being 1 guarantee that they have to necessarily intersect at some intermediate value of <span class="math-container">$d$</span>; at these intersections, obviously, Af = Pf. It appears that convexity of <span class="math-container">$C$</span> also guarantees there will be only one such intersection. </p>
<p>Thus we have, for every orientation, a value of <span class="math-container">$d$</span> for which Af and Pf have same value - as claimed. If <span class="math-container">$C$</span> is centrally symmetric (circle, ellipse, rectangle, regular polygon with even number of sides...), the only such value of <span class="math-container">$d$</span> is <span class="math-container">$D$</span>/2 and the common value of the fractions is 1/2 for all orientations. This will not be the case for asymmetric convex polygonal <span class="math-container">$C$</span>'s - we have different common Af and Pf values for different orientations. This fraction should change continuously with orientation. </p>
<p>We guess that the envelope etc. of the cutting lines with common Af and Pf for each orientation might have interesting properties. </p>
|
4,232,796 | <p>Consider <span class="math-container">$\sqrt{x^2+y^2}+2\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-6x-8y+25}$</span>. I need to find global or local minima of this function, but Wolfram Alpha doesn't seem to find one; the answer is that <span class="math-container">$1 + 2\sqrt{5}$</span> is its global minimum</p>
<p>Am I doing something wrong? I use <a href="https://www.wolframalpha.com/input/?i=minimum+sqrt%28x%5E2%2By%5E2%29%2B2sqrt%28x%5E2%2By%5E2-2x%2B1%29%2Bsqrt%28x%5E2%2By%5E2-6x-8y%2B25%29" rel="nofollow noreferrer">this</a> input for the global minimum, and <a href="https://www.wolframalpha.com/input/?i=local+minimum+sqrt%28x%5E2%2By%5E2%29%2B2sqrt%28x%5E2%2By%5E2-2x%2B1%29%2Bsqrt%28x%5E2%2By%5E2-6x-8y%2B25%29" rel="nofollow noreferrer">this</a> input for local minima.</p>
| Giorgos Giapitzakis | 907,711 | <p>Consider the set <span class="math-container">$$D = \left\{\frac{1}{k+1}: k \in \mathbb{N}\right\}$$</span>
Since <span class="math-container">$f^{(n)}|_D\equiv 0$</span>, <span class="math-container">$f^{(n)}$</span> is holomorphic and <span class="math-container">$0$</span> is an accumulation point of <span class="math-container">$D$</span>, frοm the identity theorem we conclude that <span class="math-container">$f^{(n)}\equiv 0$</span> on <span class="math-container">$\mathbb{C}$</span>. From that it's easy to conclude that <span class="math-container">$f$</span> is a polynomial of degree at most <span class="math-container">$n-1$</span>.</p>
|
3,575,015 | <p>This question is taken out of the text book High Dimensional Probability by Roman. I am unsure from the context if it means locally or globally Lipschitz. But the question is as follows:</p>
<p>Every differentiable function <span class="math-container">$f:\mathbb{R}^n\rightarrow{\mathbb{R}}$</span> is Lipschitz, and </p>
<p><span class="math-container">$\|f\|_{Lip}\le{∥∇f∥_∞}$</span> .</p>
<p>I have shown that if a function is continuously differentiable, it is locally Lipchitz. But since the question uses <span class="math-container">$\|f\|_{Lip}$</span> I believe I am to show that if a function is only differentiable, it is globally Lipchitz.</p>
<p>Any hints / clues?</p>
| lonza leggiera | 632,373 | <p>From the comments, I gather that if <span class="math-container">$\ f_t\ $</span> is a solution to the initial value problem
<span class="math-container">$$
a_tf''(x) + b_tf'(x) + c_tf(x)=0\ ,\\
f(0)=k, f'(0)=0
$$</span>
where <span class="math-container">$\ a_t, b_t, c_t>0\ $</span> and <span class="math-container">$\ \frac{b_t}{2\sqrt{a_tc_t}}=t\ $</span>, then the problem is to find the range of values of <span class="math-container">$\ b_t\ $</span> for any <span class="math-container">$\ t>1\ $</span> such that the ratio
<span class="math-container">$$
\frac{f_1(x)}{f_t(x)}
$$</span>
converges as <span class="math-container">$\ x\rightarrow\infty\ $</span>. It is not clear to me from the statement of the question, or the OP's comments, exactly which of the quantities <span class="math-container">$\ t\ $</span>, <span class="math-container">$\ a_t\ $</span>, <span class="math-container">$\ b_t\ $</span>, <span class="math-container">$\ c_t\ $</span> are to be considered fixed and which variable, or precisely what relations are assumed to hold between them, apart from the the fact that they're all positive, <span class="math-container">$\ t\ge1\ $</span>, and <span class="math-container">$\ b_t=2t\sqrt{a_tc_t}\ $</span>. Here, I shall assume that <span class="math-container">$\ a_1$</span>, <span class="math-container">$\ c_1\ $</span> and <span class="math-container">$\ a_t$</span>, <span class="math-container">$\ b_t$</span>, and <span class="math-container">$\ c_t$</span> for <span class="math-container">$\ t>1\ $</span> can all vary independently of each other, subject only to the inequality <span class="math-container">$\ b_t>2\sqrt{a_tc_t}\ $</span> for <span class="math-container">$\ t>1\ $</span> and the positivity restrictions. Once these values have been given, the values of <span class="math-container">$\ b_1= 2\sqrt{a_1c_1}\ $</span> and <span class="math-container">$\ t= \frac{b_t}{2\sqrt{a_tc_t}}\ $</span> are determined.</p>
<p>For <span class="math-container">$\ t>1\ $</span>,
<span class="math-container">$$
f_t(x)=\frac{k\left(b_t+r_t\right)e^{-\frac{\left(b_t-r_t\right)x}{2a_t}}}{2r_t}-\frac{k\left(b_t-r_t\right)e^{-\frac{\left(b_t+r_t\right)x}{2a_t}}}{2r_t}
$$</span>
where <span class="math-container">$\ r_t=\sqrt{ b_t^2-4a_tc_t}\ $</span>, and for <span class="math-container">$\ t=1\ $</span>,
<span class="math-container">$$
f_1(x)=ke^{-x\sqrt{\frac{c_1}{a_1}}}\left(1+ x\sqrt{\frac{c_1}{a_1}}\right)\ .
$$</span>
Therefore,
<span class="math-container">\begin{align}
\frac{f_1(x)}{f_t(x)}&=\frac{e^{\left(\frac{\left(b_t-r_t\right)}{2a_t}-\sqrt{\frac{c_1}{a_1}}\right)x}\left(1+ x\sqrt{\frac{c_1}{a_1}}\right)}{\left(\frac{k\left(b_t+r_t\right)}{2r_t}-\frac{k\left(b_t-r_t\right) e^{-\frac{r_tx}{a_t}}}{2r_t}\right)}\ .
\end{align}</span>
Since <span class="math-container">$\ a_t, r_t > 0\ $</span>, this ratio converges as <span class="math-container">$\ x\rightarrow\infty\ $</span> if and only if
<span class="math-container">$$
\frac{\left(b_t-r_t\right)}{2a_t}-\sqrt{\frac{c_1}{a_1}}<0\ ,
$$</span>
or
<span class="math-container">$$
b_t-2a_t \sqrt{\frac{c_1}{a_1}}< \sqrt{ b_t^2-4a_tc_t}
$$</span>
when it converges to <span class="math-container">$0$</span>. This inequality is automatically satisfied if <span class="math-container">$\ b_t\le 2a_t \sqrt{\frac{c_1}{a_1}}\ $</span>. On the other hand, if <span class="math-container">$\ b_t> 2a_t \sqrt{\frac{c_1}{a_1}}\ $</span>, both sides of the inequality are positive, and it is therefore equivalent to the one obtained by squaring both sides:
<span class="math-container">$$
b_t^2-4a_tb_t \sqrt{\frac{c_1}{a_1}}+4a_t^2\left(\frac{c_1}{a_1}\right)<b_t^2-4a_tc_t\ ,
$$</span>
which reduces to
<span class="math-container">$$
b_t>c_t \sqrt{\frac{a_1}{c_1}}+a_t \sqrt{\frac{c_1}{a_1}}\ .
$$</span>
If <span class="math-container">$\ a_t \sqrt{\frac{c_1}{a_1}}\ge c_t \sqrt{\frac{a_1}{c_1}}\ $</span>(i.e. if <span class="math-container">$\ \frac{a_t}{c_t}\ge \frac{a_1}{c_1}\ $</span>, and, in particular, if <span class="math-container">$\ \frac{a_t}{c_t}= \frac{a_1}{c_1}\ $</span>), then <span class="math-container">$\ b_t> 2a_t \sqrt{\frac{c_1}{a_1}}\ $</span> implies that <span class="math-container">$ b_t>$$c_t \sqrt{\frac{a_1}{c_1}}+$$a_t \sqrt{\frac{c_1}{a_1}}\ $</span>, and so <span class="math-container">$\ \lim_\limits{x\rightarrow\infty}\frac{f_1(x)}{f_t(x)}=0\ $</span> for all <span class="math-container">$\ b_t>2\sqrt{a_tc_t}\ $</span> in this case.</p>
<p>On the other hand, if <span class="math-container">$\ \frac{a_t}{c_t}<\frac{a_1}{c_1}\ $</span>, then <span class="math-container">$\ \frac{f_1(x)}{f_t(x)}=0\ $</span> does not converge to a finite limit whenever <span class="math-container">$\ 2a_t \sqrt{\frac{c_1}{a_1}}<$$b_t\le$$ c_t \sqrt{\frac{a_1}{c_1}}+$$a_t \sqrt{\frac{c_1}{a_1}}\ $</span>.</p>
|
1,041,623 | <p>I have been suggested to read the <a href="http://press.princeton.edu/chapters/gowers/gowers_VIII_6.pdf"><em>Advice to a Young Mathematician</em> section </a> of the <em>Princeton Companion to Mathematics</em>, the short paper <a href="http://alumni.media.mit.edu/~cahn/life/gian-carlo-rota-10-lessons.html"><em>Ten Lessons I wish I had been Taught</em></a> by Gian-Carlo Rota, and the <a href="http://terrytao.wordpress.com/career-advice/"><em>Career Advice</em></a> section of Terence Tao's blog, and I am amazed by the intelligence of the pieces of advice given in these pages. </p>
<p>Now, I ask to the many accomplished mathematicians who are active on this website if they would mind adding some of their own contributions to these already rich set of advice to novice mathematicians. </p>
<p>I realize that this question may be seen as extremely opinion-based. However, I hope that it will be well-received (and well-answered) because, as Timothy Gowers put it,</p>
<blockquote>
<p>"The most important thing that a young mathematician needs to learn is
of course mathematics. However, it can also be very valuable to learn
from the experiences of other mathematicians. The five contributors to
this article were asked to draw on their experiences of mathematical
life and research, and to offer advice that they might have liked to
receive when they were just setting out on their careers."</p>
</blockquote>
| David | 117,084 | <p>$ \clubsuit$ Try to view every thing to one dimension or two dimension where we can see geometrically.. </p>
<p>$\quad$ It is quite often helps to realize the things and to generalize it.</p>
<p>$\clubsuit$ Aim to guess the answer first. If we realize that the answer is correct then proving </p>
<p>$\quad$ is not that difficult.</p>
<p>$\clubsuit$ If you are finding something wrong. Try to give counter examples or prove that </p>
<p>$\quad$ why this is wrong.</p>
<p>$\clubsuit $ Teaching and explaining is the best way to understand the mathematics...</p>
<p>$\clubsuit$ It is better to view the old things in new ways.</p>
<p>$\quad$ For, example nowadays we are thinking widely through all subjects. </p>
<p>$\quad$ If we wish to study about groups try to give examples in analysis like $\mathcal C[a,b]$ etc.</p>
|
1,041,623 | <p>I have been suggested to read the <a href="http://press.princeton.edu/chapters/gowers/gowers_VIII_6.pdf"><em>Advice to a Young Mathematician</em> section </a> of the <em>Princeton Companion to Mathematics</em>, the short paper <a href="http://alumni.media.mit.edu/~cahn/life/gian-carlo-rota-10-lessons.html"><em>Ten Lessons I wish I had been Taught</em></a> by Gian-Carlo Rota, and the <a href="http://terrytao.wordpress.com/career-advice/"><em>Career Advice</em></a> section of Terence Tao's blog, and I am amazed by the intelligence of the pieces of advice given in these pages. </p>
<p>Now, I ask to the many accomplished mathematicians who are active on this website if they would mind adding some of their own contributions to these already rich set of advice to novice mathematicians. </p>
<p>I realize that this question may be seen as extremely opinion-based. However, I hope that it will be well-received (and well-answered) because, as Timothy Gowers put it,</p>
<blockquote>
<p>"The most important thing that a young mathematician needs to learn is
of course mathematics. However, it can also be very valuable to learn
from the experiences of other mathematicians. The five contributors to
this article were asked to draw on their experiences of mathematical
life and research, and to offer advice that they might have liked to
receive when they were just setting out on their careers."</p>
</blockquote>
| Michael | 155,065 | <p>Here is a late answer (several years late it seems): </p>
<p>1) One minor piece of advice is to have respect for others. (Well, perhaps that is a major piece of advice.) </p>
<p>Here is a minor example:
People on stackexchange that
answer a question two hours after a major hint
has been given, using the basic idea of the hint,
but who provide no mention of the hint or the
person who gave it, do not demonstrate respect for
others. To me, this behavior suggests the person is
rude, unsafe, and may have other work that
is heavily borrowed from others (without sufficient
referencing). </p>
<p>An example on the flipside: If this kind of behavior does happen, <em>give the offender the benefit of the
doubt</em>: Maybe that person accidentally left a reference off,
or was somehow unaware. </p>
<p>2) I resonate with not being afraid to "ask dumb questions" from your link here:
<a href="https://terrytao.wordpress.com/career-advice/" rel="noreferrer">https://terrytao.wordpress.com/career-advice/</a></p>
<p>If you strive to always be right, and to always ask "smart questions," you likely will not get very many new results. To make progress you must also make mistakes. </p>
<p>3) Do not put arbitrary constraints on yourself. </p>
<p>I give some examples on that last point in these slides I made on "thinking outside the box." A simple example that I repeatedly observed when tutoring middle school and high school students is when they set up their math homework by first boxing in the amount of space needed for each problem. [These slides are from a powerpoint talk I gave several years ago (as a second part of a larger seminar at the University of Southern California). These slides are supposed to change when clicked, though some PDF viewers only seem to allow scrolling.]</p>
<p><a href="http://ee.usc.edu/assets/008/61688.pdf" rel="noreferrer">http://ee.usc.edu/assets/008/61688.pdf</a></p>
|
549,411 | <p>How would I go about generalizing the product rule to the product of $n$ functions $\psi_1(x), \ \psi_2(x), ..., \ \psi_n(x)$? That is, I'm hoping to obtain an expression for</p>
<p>$$
\frac{d}{dx} \prod_{j = 1}^n \psi_j(x)
$$</p>
| Steven Stadnicki | 785 | <p>These are known as <a href="http://en.wikipedia.org/wiki/Beatty_sequence" rel="nofollow"><em>Beatty sequences</em></a> and they've been quite thoroughly studied (you can find one of your two sequences at <a href="http://oeis.org/A001951" rel="nofollow">http://oeis.org/A001951</a> ). The most interesting property (and most relevant here) is that if $r$ and $s$ are irrational numbers with $r,s \gt 1$ and $\frac1r+\frac1s=1$, then the sequences $R=\left\{\lfloor (nr)\rfloor\right\}$ and $S=\left\{\lfloor (ns)\rfloor\right\}$ are complementary: $R\cap S=\emptyset$ and $R\cup S=\mathbb{N}$.</p>
|
488,287 | <p>I'm trying to solve the following problem : </p>
<p>$ABCD$ is a square of side $4$ units. Find the area of the shaded region as shown in the figure.</p>
<p><img src="https://i.stack.imgur.com/S095z.png" alt="Figure"></p>
<p>The area of square is obviously $16$ , but what after that?</p>
| Hagen von Eitzen | 39,174 | <p>Let $P(x)\equiv (x=1)$ and $Q(x)\equiv (x=0)$ and use the fact that $1\ne 0$</p>
|
488,287 | <p>I'm trying to solve the following problem : </p>
<p>$ABCD$ is a square of side $4$ units. Find the area of the shaded region as shown in the figure.</p>
<p><img src="https://i.stack.imgur.com/S095z.png" alt="Figure"></p>
<p>The area of square is obviously $16$ , but what after that?</p>
| Ittay Weiss | 30,953 | <p>Let $P(x)$ be the claim that $x$ is an odd integer. Let $Q(x)$ be the claim that $x$ is an even integer. Then $\exists x(P(x)\wedge Q(x))$ is false while $\exists x P(X) \wedge \exists x Q(x)$ is true. </p>
|
278,637 | <p>After spending more than 1 hr on this, and looking at many questions, I give up as I am not able to figure a solution.</p>
<p>I have my current package in the standard location given by <code>FileNameJoin[{$UserBaseDirectory, "Applications"}]</code> which on windows is</p>
<pre><code>C:\Users\Owner\AppData\Roaming\Mathematica\Applications
</code></pre>
<p>My package is <code>nma</code>, so I have in the above the standard set up of <code>nma.m</code> and <code>kernel\init.m</code> where init.m loads all the files using <code>Get</code>.</p>
<pre><code>C:\Users\Owner\AppData\Roaming\Mathematica\Applications\nma\nma.m
C:\Users\Owner\AppData\Roaming\Mathematica\Applications\nma\A.m
C:\Users\Owner\AppData\Roaming\Mathematica\Applications\nma\B.m
C:\Users\Owner\AppData\Roaming\Mathematica\Applications\nma\kernel\init.m
</code></pre>
<p>The above is all working fine. I can do</p>
<pre><code> <<nma`
</code></pre>
<p>From any notebook, and all the files in the package are loaded just fine.</p>
<p>But I do not like to keep my software on the C drive. My backup software only backups another drive. So I wanted to move <code>C:\Users\Owner\AppData\Roaming\Mathematica\Applications\nma\</code> to say <code>G:\</code> drive and the whole tree to say <code>G:\mathematica_version\code\nma\</code>. But when I did this, and made sure to open the notebook in same folder and made sure to do <code>SetDirectory[NotebookDirectory[]]</code> so that current directory is the above where all the files including <code>kernel</code> folder are, now when I do</p>
<pre><code> <<nma`
</code></pre>
<p>It no longer loads all the files in the package, it only loads <code>nma.m</code>, because I assume it does not use <code>init.m</code> in the <code>kernel</code> folder anymore, where <code>init.m</code> was loading all the other .m files.</p>
<p>It seems <code>application_name/kernel/init.m</code> is only used when the package lives in the standard location <code>C:\Users\Owner\AppData\Roaming\Mathematica\Applications</code> ? Is this true?</p>
<p><strong>My question is</strong>, how to copy <code>Applications\nma\...</code> from C drive to any other location and have</p>
<pre><code><<nma`
</code></pre>
<p>work the same as before? i.e. as if the package was in the standard location?</p>
<p>I tried my other things, like <code>AppendTo[$Path,"G:\\mathematica_version\\code\\nma"]</code> but this had no effect.</p>
<p>So currently after moving the package to the different location, I have to now manually do a <code>Get</code> on each file. This is something that <code>init.m</code> was doing before. i.e. I am doing this now</p>
<pre><code> SetDirectory[NotebookDirectory[]]
Get["nma.m"]
Get["A.m"]
Get["B.m"]
etc...
</code></pre>
<p>for each .m file. I would prefer to have init.m do this as before if possible.</p>
<p><strong>Update</strong></p>
<p>I also tried the following. Changed <code>$UserBaseDirectory</code> to point to the location in the other disk. Removed all the <code>Application\</code> folder and moved it to the new location. But this did not work. When I did</p>
<pre><code><<nma`
</code></pre>
<p>It did not load the package. To change <code>$UserBaseDirectory</code> I had to do the following</p>
<pre><code> Unprotect[<span class="math-container">$UserBaseDirectory]
$</span>UserBaseDirectory="new path here"
Protect[$UserBaseDirectory]
</code></pre>
<p>So I copied the Application folder back to where it was. I thought may be by moving the whole Application tree and changing <code>$UserBaseDirectory</code> will make it work. May be I did not do it right, I looked at option inspector and did not see where <code>$UserBaseDirectory</code> is defined. Is it ok to change <code>$UserBaseDirectory</code> to new location? If so, what is the correct way to do it?</p>
| Michael E2 | 4,999 | <p>Just a refactoring of the OP's idea:</p>
<pre><code>scalarize // ClearAll;
scalarize[x_List] /; MatchQ[MinMax@Dimensions@x, {1, 1}] :=
First@Flatten[x];
scalarize[x_] := x;
</code></pre>
<p>Another way:</p>
<pre><code>scalarize // ClearAll;
scalarize[x_List] :=
CheckAbort[
x /. {Except[{_}, _List] :> Abort[]}; First@Flatten@x,
x];
scalarize[x_] := x;
</code></pre>
|
718,609 | <p>This theorem is the converse of Wilson's theorem:</p>
<blockquote>
<p>If $n$ is composite and $n>4$, then $(n-1)! \equiv 0 \pmod n$</p>
</blockquote>
<p>The question holds up for all the composites I have tried but I'm struggling to form a proof for all composites greater than $4$.</p>
| André Nicolas | 6,312 | <p>Suppose first that there exist integers $a$ and $b$, such that $1\lt a\lt b\le n-1$ and $ab=n$. Then since $b\le n-1$, it follows that $ab$ divides $(n-1)!$. For each of $a$ and $b$ appears among the numbers from $1$ to $n-1$.</p>
<p>But $n$ can be composite without meeting the above condition: It could be that the <strong>only</strong> proper factorizations of $n$ have the shape $n=a^2$. (This only happens if $a$ is prime.) In that case, if $a\ne 2$, then $a$ and $2a$ are both $\le n-1$. So $2a^2$ divides $(n-1)!$, and therefore $a^2$ divides $(n-1)!$. </p>
|
47,753 | <p>What are the Method options Solve command accepts? Solve has the Method option, however documentation contains no methods that it accepts...</p>
| János Tóth | 13,216 | <p>With <code>Method -> Reduce</code>, <code>Solve</code> uses only equivalent transformations and finds all solutions.</p>
<pre><code>Solve[x E^x == 1/2, x, Method -> Reduce]
</code></pre>
<blockquote>
<pre><code>{{x -> ConditionalExpression[ProductLog[C[1], 1/2], C[1] ∈ Integers]}}
</code></pre>
</blockquote>
|
2,226,386 | <p>Solve $$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dz}{x+y}.$$ Solve the similtaneous equation by using method of multipliers. How can we choose these multipliers? Is there any specific method to know these multipliers?</p>
| Andrei | 331,661 | <p>It all comes from the invariance with respect to the reference frame. What I mean is that if you draw the same triangle a little to the left, or right, or up, or down, the problem should not change. In fact you can even rotate the whole figure, and the answer should be the same. That means that you can choose any origin, and any two perpendicular directions through that origin to be your coordinate frame, without changing the answer. In this case they choose the bottom of the triangle (let's call this $B$). Then $2a$ is the coordinate of point $C$, the left corner. $2b$ is the coordinate of point $A$ (top corner of your triangle) <strong>with respect to point $B$</strong>. So in that reference frame, the absolute coordinate of point $A$ is $2a+2b$. Similarly, the lower corner $B$ coordinate <strong>with respect to $A$</strong> is $2c$. So in any reference frame, we can call coordinates of points $A, B, C$ as $z_A, z_B, z_C$. Then The side $BA$ is going to be given by $z_B-z_A$ and so on. In this case they just called $z_B-z_A=2c$</p>
|
2,996,775 | <p><span class="math-container">$$ \frac {1}{\log_2(x-2)^2} + \frac{1}{\log_2(x+2) ^2} =\frac5{12}.$$</span></p>
<p>I made the graph using wolfram alpha it is giving answer as
6. But how to solve it algebraically?
base of logarithm is 2.</p>
<p>Tried using taking Lcm but then two different log terms
are getting formed.</p>
| Batominovski | 72,152 | <p>Define <span class="math-container">$f:\mathbb{R}\setminus\{\pm1,\pm3\}\to\mathbb{R}$</span> by <span class="math-container">$$f(x):=\frac{1}{\log_2\big((x-2)^2\big)}+\frac{1}{\log_2\big((x+2)^2\big)}$$</span> for all real numbers <span class="math-container">$x\neq \pm1,\pm2,\pm3$</span>, and <span class="math-container">$f(\pm 2)$</span> is defined to be <span class="math-container">$\dfrac14$</span>. Note that <span class="math-container">$f$</span> is a continuous even function (i.e., <span class="math-container">$f(-x)=f(x)$</span>). Therefore, it suffices to solve <span class="math-container">$f(x)=y$</span> for <span class="math-container">$x\geq 0$</span>. </p>
<p>Observe that <span class="math-container">$f$</span> is strictly increasing on <span class="math-container">$[0,1)$</span> with local minimum <span class="math-container">$f(0)=1$</span> and without upper bound. Also, <span class="math-container">$f$</span> is strictly increasing on <span class="math-container">$(1,2]$</span>, with local maximum <span class="math-container">$f(2)=\dfrac14$</span> and without lower bound. On <span class="math-container">$[2,3)$</span>, <span class="math-container">$f$</span> is strictly decreasing with local maximum <span class="math-container">$f(2)=\dfrac14$</span> and without lower bound. Then, <span class="math-container">$f$</span> is strictly decreasing on <span class="math-container">$(3,\infty)$</span> with local infimum <span class="math-container">$\lim\limits_{x\to\infty}\,f(x)=0$</span> and without upper bound. I leave the proof of the above analysis to you. </p>
<p>Hence, for any real number <span class="math-container">$y$</span>, the number <span class="math-container">$n(y)$</span> of <span class="math-container">$x\in\mathbb{R}$</span> such that <span class="math-container">$f(x)=y$</span> is
<span class="math-container">$$n(y)=\left\{\begin{array}{ll}
4&\text{if }y>1\,,\\
3&\text{if }y=1\,,\\
2&\text{if }\frac14< y<1\,,\\
4&\text{if }y=\frac14\,,\\
6&\text{if }0<y<\frac14\,,\\
4&\text{if }y\leq 0\,.
\end{array}\right.$$</span>
In particular, <span class="math-container">$n\left(\dfrac{5}{12}\right)=2$</span>. Since <span class="math-container">$x=\pm6$</span> are solutions to <span class="math-container">$f(x)=\dfrac{5}{12}$</span>, they are the only solutions.</p>
|
3,370,061 | <p>Could anyone clarify why this Boolean expression AB'+AB'AC' = AB'? I did not understand what happened to the C'</p>
| Bram28 | 256,001 | <p><span class="math-container">$$AB'+AB'AC'=AB'(1+AC')=AB'1=AB'$$</span></p>
|
1,463,258 | <p>Given a cumulative distribution function of the form $P(X\leq x) = 1- e^{-\lambda x^3}$, is there any way to represent it in terms of an exponential or any other distribution? I've thought about exponentials, but don't know how to deal with the cubed term. Thanks!</p>
| Clarinetist | 81,560 | <p>The cumulative distribution function $$F_{X}(x) = 1-e^{-\lambda x^3}\text{, }x \geq 0$$
implies that $X$ follows a (two-parameter) <a href="https://en.wikipedia.org/wiki/Weibull_distribution#Distribution_function" rel="nofollow">Weibull</a> distribution. Using the parametrization in the link, the <a href="https://en.wikipedia.org/wiki/Weibull_distribution#Definition" rel="nofollow">density function</a> is $f\left(x, \dfrac{1}{\sqrt[3]{\lambda}},3\right)$. </p>
<p>As Andre has shown, if $Y$ is exponential with mean $\dfrac{1}{\lambda}$, $X = Y^{1/3}$ has this distribution function. See also <a href="http://www.math.wm.edu/~leemis/chart/UDR/PDFs/ExponentialWeibull.pdf" rel="nofollow">this link</a>.</p>
|
947 | <p>I'm looking for the algorithm that efficiently locates the "loneliest person on the planet", where "loneliest" is defined as:</p>
<p>Maximum minimum distance to another person — that is, the person for whom the closest other person is farthest away.</p>
<p>Assume a (admittedly miraculous) input of the list of the exact latitude/longitude of every person on Earth at a particular time.</p>
<p>Also take as provided a function <span class="math-container">$d(p_1, p_2)$</span> that returns the distance on the surface of the earth between <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span> - I know this is not trivial, but it's "just spherical geometry" and not the important (to me) part of the question.</p>
<p>What's the most efficient way to find the loneliest person?</p>
<p>Certainly one solution is to calculate <span class="math-container">$d(\ldots)$</span> for every pair of people on the globe, then sort every person's list of distances in ascending order, take the first item from every list and sort those in descending order and take the largest. But that involves <span class="math-container">$n(n-1)$</span> invocations of <span class="math-container">$d(\ldots)$</span>, <span class="math-container">$n$</span> sorts of <span class="math-container">$n-1$</span> items and one last sort of <span class="math-container">$n$</span> items. Last I checked, <span class="math-container">$n$</span> in this case is somewhere north of six billion, right? So we can do better?</p>
| j.c. | 353 | <p>It may be possible to reformulate your problem so that "<a href="https://doi.org/10.1007/s007780050006" rel="nofollow noreferrer">distance-based outlier</a>" algorithms will apply.</p>
|
3,790,932 | <p>If <span class="math-container">$f : \bar{\mathbb{Q}} \to \mathbb{Q}$</span> is a continuous function, where <span class="math-container">$\bar{\mathbb{Q}}$</span> denotes the set of algebraic numbers, does such function have to be constant?</p>
| freakish | 340,986 | <blockquote>
<p>can such function be constant?</p>
</blockquote>
<p>I assume you mean "does it have to be constant?". No, it doesn't.</p>
<blockquote>
<p><strong>Lemma.</strong> There is a non-constant continuous function <span class="math-container">$f:X\to Y$</span> between any countable metrizable space <span class="math-container">$X$</span> with at least two points <em>and</em> any topological space <span class="math-container">$Y$</span> with at least two points.</p>
</blockquote>
<p><em>Proof.</em> If <span class="math-container">$X$</span> has an isolated point, say <span class="math-container">$x_0$</span>, then we can simply glue any value on <span class="math-container">$\{x_0\}$</span> with any other constant value on <span class="math-container">$X\backslash\{x_0\}$</span>. The "glueing" is done via <a href="https://en.wikipedia.org/wiki/Pasting_lemma" rel="nofollow noreferrer">pasting lemma</a>.</p>
<p>If <span class="math-container">$X$</span> does not have an isolated point then by the famous Sierpiński theorem <span class="math-container">$X$</span> is homeomorphic to the standard <span class="math-container">$\mathbb{Q}$</span>, and thus there is a non-constant continuous function to <span class="math-container">$Y$</span> because we can partition <span class="math-container">$\mathbb{Q}$</span> into two disjoint closed subsets and again apply the pasting lemma on constant pieces. <span class="math-container">$\Box$</span></p>
<p>So assuming you deal with the standard <span class="math-container">$\overline{\mathbb{Q}}$</span>, which is a countable subspace of complex numbers (and thus metric), then the answer is that there are such non-constant continuous functions.</p>
<p>Of course the Sierpiński theorem I used is highly non-trivial, but on the other hand it covers a very wide range of cases.</p>
|
1,160,699 | <p>In my discrete math book, I was tasked with finding a counterexample for this:</p>
<blockquote>
<p>If $n$ is prime, then $2^n-1$ is prime.</p>
</blockquote>
<p>Does there exist a counterexample for such a statement? Also, am I wrong in thinking that when something asks for a counterexample, is it looking for some logic that proves the original statement to be false?</p>
<p>Any help is appreciated, as I've got a test on subjects like this tomorrow.</p>
| Tim Raczkowski | 192,581 | <p>The statement says $2^n-1$ is prime for <strong>all</strong> primes $n$. A counterexample would be a prime number $n$ such that $2^n-1$ is not prime.</p>
|
1,696,713 | <p>I am solving exact differential equation, but I am stuck on the step on how to simplify this term or how to rewrite it. </p>
<p>$e^{-2\ln{\sin{x}}}$</p>
| Marco Disce | 17,010 | <p>I would say the best mathematical setting to represent physical causation are <a href="https://en.wikipedia.org/wiki/Dynamical_system_(definition)" rel="nofollow">dynamical systems</a>:</p>
<ul>
<li>You have a global "state" $x(t)$ (a vector with many coordinates) with flour, milk and eggs represented by set of variables changing in the time variabe $t$, </li>
<li>you have physical laws represented by a rule linking $x(t)$ to $x(t+dt)$ (maybe a differential equation),</li>
<li>you have a a "macro"-state $X(t)$ that is represented by a set of "micro"-state $x(t)$ that we identify (for example there are several micro-state that we would call "bread" and several micro-states that we would call "unmixed flour, milk and eggs"),</li>
<li>you have causation if the macro-state $A$ (in any given possible micro-state configuration) always evolves to the macro-state $B$ (in some micro-state configuration), then you would say that "$A$ causes $B$".</li>
</ul>
|
484,550 | <p>The problem is as follows: let $n_1, n_2,..., n_t$ be positive integers. Prove that if $n_1+n_2+...+n_t-t+1$ objects are placed into $t$ boxes, then for some $i, i=1, 2, ..., t$, the $i$th box contains at least $n_i$ objects. </p>
<p>I'm having difficulty getting started in developing a proof because I have no intuition as to why this should be true or whether or not it actually is. Could someone help get me started?</p>
| Barry Cipra | 86,747 | <p>A way to think about it is to rewrite $n_1+n_2+\cdots n_t - t+1$ as </p>
<p>$$(n_1-1)+(n_2-1)+\cdots+(n_t-1)+1$$</p>
<p>and then think that if you want try to avoid putting $n_i$ balls into the $i$th box, you can put at most $n_1-1$ balls in the first box, $n_2-1$ in the second, etc., which will leave $1$ ball left, which will have to go somewhere.</p>
|
209,760 | <p>Is it possible to transform this equation to give <span class="math-container">$R$</span>?
<span class="math-container">$$y=x\left[\frac{\left(1+\frac{R}{12}\right)^{12\times{25}}}{\frac{R}{12}}-1\right]$$</span> </p>
| draks ... | 19,341 | <p>You could rewrite like $\frac{R}{12}\left(\frac yx+1\right)-\sum_{k=0}^{12\times 25}\binom{12\times 25}{k}\left(\frac R{12}\right)^k=0$, but to my knownledge, there is no general way to solve that other than numerical.</p>
|
2,489,429 | <p>$\int \frac{x}{x+1}\,dx$ so I set substitution to $x+1=t$ then I differentiate to get $dx=dt$ and then proceed to get $\int 1\,dt - \int \frac{1\,dt}{t}$ to get $t-\ln(|t|)+C$ but when I replace back the substitution I get $x+1$ in the final answer instead of just the $x$ , so my question is why was it a mistake to integrate $\int 1\,dt$ instead to switch back to x and then integrate $\int 1\,dx$</p>
| Jaideep Khare | 421,580 | <p>There is nothing wrong in your solution .</p>
<p>$x+1-\ln|x+1| + C$</p>
<p>Now let $C+1=C'$</p>
<p>$x-\ln|x+1|+C'$</p>
<p>You can always adjust the constant of integration.</p>
|
689,921 | <p>There are two teams.Two games were played.There are three possible outcomes which are win, lose or draw. how many permutations are there?</p>
| Guy | 127,574 | <p>The first outcome can either be a win, lose or draw. Let's assume that it was a win. In this case, the second outcome can also have $3$ possibilities. If the first outcome was instead either a lose or draw, even then in both cases, the second outcome can either be win, lose or draw. </p>
<p>Thus the total permuations is $3+3+3=9$</p>
<p>In general, if there were $n$ games played, then the number of permutations would be $3^n$.</p>
|
1,012,985 | <p>Let $A = \left[ \begin{matrix} 3 & 2 & 1\\ 5 & 0 &1\end{matrix}\right]$, </p>
<p>how can I know if there is a matrix $N$ , st. $AN=0$ (N is not a zero matrix) </p>
| Robert Israel | 8,508 | <p>Hint: what could you say about the columns of such a matrix?</p>
|
1,552,747 | <p>While we were introduced to integration, we were told about some basic concepts that, as we were told, could not be proved based on our level of sophistication. They are as follows:</p>
<ol>
<li>$$\int_a^b \! f(x) \, \mathrm{d}x=\phi(b)-\phi(a),$$ where $\phi$ is a primitive of $f$ in $[a,b]$</li>
<li>$$\int_b^a \! f(x) \, \mathrm{d}x=-\int_a^b \! f(x) \, \mathrm{d}x$$</li>
</ol>
<p>When I learned Riemann Integral, I thought I would be able to prove them, as they didn't seem to be too out-of-the-earth type. So, my question is what is the concept of a primitive according to Riemann Integration? And how can it be used to prove $(1)$? If there isn't any, then where should I look?</p>
<p>Also, how can I even conceptualise $(2)$? I mean, Riemann Integration is defined using Sums. How can I Sum in the opposite way? Does it even matter? And how does it become negative?</p>
<p>I'm just a curious high school student, familiar with the basic concepts of Riemann Integration. So, spare me if my questions are too dumb. And it'd be great if you can suggest some study material where I should look for these type of concepts. Thank you in advance.</p>
| User2956 | 283,118 | <p>$\int_a^b \! f(x) \, \mathrm{d}x=g(b)-g(a)$ is a result obtained directly from the fundamental theorem of calculus.</p>
<p>Say there's a function $f(x)$</p>
<p>Now the area under the curve from $0$ up til $x$ can be found using Reimann sums. Let A(x) = $\int_0^x f(t) dt$ where A(x) is the area function that gives you the area under the curve of $f(x)$ from $0$ up to an $x$ value</p>
<p>It can be easily shown that $A '(x) = f(x)$ (<a href="https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_meaning" rel="nofollow">https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_meaning</a>)</p>
<p>Which means, $A(x)$ is <strong>an antiderivative</strong> of $f(x)$
Using this result, we can compute areas under the curve of a function easily.
Depending on your choice of antiderivative, you can calculate the area from any point $a$ up to $x$ under the curve. For e.g if you want to find the area from $0$ up to $x$, you will choose an antiderivative g(x) that satisfies the relation g(0) = 0.</p>
<p>Now say you want to find the area from $a$ up to $b$.
To make things easier for us, we can choose any antiderivative g(x) we like, evaluate the value of $g(b)$ ie. find the area under the curve up till $b$ and from it subtract $g(a)$ ie the area under the curve up to $a$ which will leave us with the area from $a$ to $b$</p>
<p>The proof for the above can be found here -<a href="https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Proof_of_the_second_part" rel="nofollow">https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Proof_of_the_second_part</a></p>
|
1,037,530 | <p>what is the basic difference between a mapping and a function? many say they are same but the opposite views are also seen. is mapping a restricted version of a function?</p>
| Ittay Weiss | 30,953 | <p>Probably most modern texts would agree that there is no mathematical difference between function and mapping. Historically, and not a very long time ago, and in difference disciplines the two were used to mean different things. For instance, in topology and metric space theory sometimes 'mapping' stands for 'continuous function'. In analysis it is common to speak of functionals, which are linear operators to the ground field. In that context 'function' may stand for a functional from the ground field (necessarily to the ground field), while 'mapping' stands for a continuous function. So, I'm afraid historically things are far from clear-cut. But again, the modern approach is very clear: both mean the exact same thing. </p>
|
1,928,892 | <p>I'm solving a graduate entrance examination problem.
We are required to establish the inequality using the following result:</p>
<p>for $x,y > 0$, $\frac{x}{y} + \frac{y}{x} > 2$ (1), which is easy to prove as it is equivalent to $(x - y)^2 > 0$.</p>
<p>But when it comes to an inequality combining $x, y, z$, I got stuck as I've tried to develop the expression into one single fraction and obtain something irreducible.</p>
<p>Any hints ? My intuition tells me that for $x,y,z >0$, any fraction of the form $\frac{x}{y+z}$ is greater than 1/2. As there are three fractions of this kind with mute variables playing symmetrical roles, we get: $1/2 + 1/2 + 1/2 = 3/2$.</p>
<p>I just don't figure out how to play with the result (1).</p>
| Ian Miller | 278,461 | <p>Apply your formula three times:</p>
<p>$$\frac{x+y}{x+z}+\frac{x+z}{x+y}>2$$</p>
<p>$$\frac{y+x}{y+z}+\frac{y+z}{y+x}>2$$</p>
<p>$$\frac{z+x}{z+y}+\frac{z+y}{z+x}>2$$</p>
<p>Combining gives:</p>
<p>$$\frac{x+y}{x+z}+\frac{x+z}{x+y}+\frac{y+x}{y+z}+\frac{y+z}{y+x}+\frac{z+x}{z+y}+\frac{z+y}{z+x}>6$$</p>
<p>$$\frac{x+y+2z}{x+y}+\frac{x+2y+z}{x+z}+\frac{2x+y+z}{y+z}>6$$</p>
<p>$$1+\frac{2z}{x+y}+1+\frac{2y}{x+z}+1+\frac{2x}{y+z}>6$$</p>
<p>$$2\left(\frac{z}{x+y}+\frac{y}{x+z}+\frac{x}{y+z}\right)>3$$</p>
<p>$$\frac{z}{x+y}+\frac{y}{x+z}+\frac{x}{y+z}>\frac{3}{2}$$</p>
|
2,238,165 | <p>Can anyone help with this limit?</p>
<p>\begin{equation*}
\lim_{x \rightarrow 4}
\frac{16\sqrt{x-\sqrt{x}}-3\sqrt{2}x-4\sqrt{2}}{16(x-4)^2}
\end{equation*}</p>
<p>I've tried a variable change of \begin{equation*} y=\sqrt{x} \end{equation*} but this didn't help.</p>
| Kanwaljit Singh | 401,635 | <p>Hint -</p>
<p>You have to take cases where 5 repeated at least twice and 7 repeated at least thrice.</p>
<p>Alternate way is find the number of ways number with length N can be arranged and then subtract cases with at most 5 repeating only once and 7 repeating at most twice.</p>
|
2,661,004 | <p>I have the following given as an answer to a questions, but I don't understand how I get $1+ \log n$ in the second step. I think I am to use L'Hospital's, but deriving the logarithm is $1/(n\ln(b))$, so I don't know why I'm keeping $\log(n)$. I know it's probably something basic, but what am I missing about that step? </p>
<p><a href="https://i.stack.imgur.com/rQY1W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rQY1W.png" alt="enter image description here"></a></p>
| The Phenotype | 514,183 | <p>An open ball in $(1,7]$ is of the form $(1,7]\cap (a,b)$ by definition of <a href="https://en.wikipedia.org/wiki/Subspace_topology" rel="nofollow noreferrer">subspace topology</a>. So for example the open ball $(1,7]\cap (2,8)=(2,7]$ in $(1,7]$ (a subspace of $\mathbb{R}$) includes $7$.</p>
|
131,583 | <p>My question is: usually, a partial differential equation, for example, those coming from physics, is written in a language of vector calculus in a local coordinate. Is there any way (or any <strong>algorithm</strong>) that we can use to rewrite it using language of differential forms, tensor, exterior calculus, Hodge star and other operators which are coordinate independent? An example, the <strong>Grad f</strong> can be rewritten as a geometric form: (df)#, where # is a sharp operator turning a one-form into a vector. I am currently facing this problem to turn a partial differential equation into its coordinate-independent form, which involves forms, tensors, exterior calculus and other operators. </p>
<p>Thank you for anyone who help me about this problem!</p>
| Igor Khavkine | 2,622 | <p>This is rather standard, though, unfortunately this knowledge is confined to a rather narrow segment of mathematicians working on PDEs. The main reason is that, most of the time, this is not necessary for practical problems. In fact there are multiple ways converting a PDE into invariant form.</p>
<p>In mathematical physics, this question comes up when trying to generalized equations from flat space to curved space. The main technical device is to introduce explicit dependence on a metric $g$ and use covariant differentiation, with respect to $g$, to replace ordinary derivatives. The invariant form of the original flat space equations is then obtained by simply replacing $g$ by the Euclidean metric. This method is sometimes colloquially known as the <em>comma goes to semicolon rule</em>, where the punctuation signs represent notation for ordinary and covariant derivatives. More on this can be found in textbooks on GR. For instance, <a href="http://books.google.com/books?id=w4Gigq3tY1kC">Gravitation</a> by Misner, Thorne & Wheeler would definitely discuss this.</p>
<p>I would guess that you might be happy to stop here. But if you are looking for more generality and mathematical sophistication, read on.</p>
<p>Another method is to repeatedly introduce auxiliary dependent variables, until the PDE system becomes first order and this can be done in such a way that all dependent variables are actually differential forms and differentiations are only effected through the exterior derivative. Once in this form, the PDE system is manifestly invariant. This approach was pioneered by Cartan and is explained in depth in the well known book <a href="http://books.google.com/books?id=pPmHMAEACAAJ">Exterior Differential Systems</a> by Bryant et al.</p>
<p>Yet another way, which does not involve reduction to a first order system, is to notice that derivatives of arbitrary order are mathematically described in an invariant way by <a href="http://en.wikipedia.org/wiki/Jet_%2528mathematics%2529">jets</a>. An arbitrary PDE can then be described in terms of jets. Once one unwinds all the relevant definitions, this kind of description is manifestly invariant, but not necessarily always helpful, since all it really does is bundle up the complicated rules for transforming higher derivatives under coordinate changes into the language of jets. Nevertheless, I think it is important to learn about it, as it provides the tools to study PDE systems in a very deep way. An elementary treatment of PDEs in the context of jets can be found in the book <a href="http://books.google.com/books?id=sI2bAxgLMXYC">Applications of Lie Groups to Differential Equations</a> by Olver. A much more sophisticated treatment can also be found in the aforementioned book by Bryant et al. Another very helpful reference is <a href="http://www.mat.univie.ac.at/~michor/kmsbookh.pdf">Natural Operations in Differential Geometry</a> by Kolar, Michor & Slovak.</p>
|
1,348,763 | <p>Using the comparison test I am supposed to figure out whether this integral converges or diverges. what other function should I use? Also, the inequality stating that $1/\sqrt{e^x+1}$ is larger or smaller than another function must be proven.</p>
| Community | -1 | <p>First, why is there any question of convergence or divergence? Check that the integrand is continuous and defined for all real numbers, so there's no vertical asymptotes to be concerned about; that leaves only the upper bound of $\infty$.</p>
<p>Next, let's guess whether it converges or not. For this, we try to understand what happens to the integrand as $x\to\infty$. Then $e^x\to\infty$, but 1 is just 1 ($e^x$ is "big", and 1 is "regular size"). So $1+e^x\approx e^x$, and
$$ \frac1{\sqrt{1+e^x}}\approx\frac1{\sqrt{e^x}} = e^{-x/2} $$
That's an integrand you could just integrate directly; so do that, and verify that $\int_1^\infty e^{-x/2}\,dx$ is finite.</p>
<p>At this point we suspect that the original integral converges. To prove that, we'll use the comparison theorem, as you said. The integrand is positive, and we expect to show that it converges, so we're looking for an upper bound:
$$ \frac1{\sqrt{1+e^x}} \le g(x) $$
where $\int_1^\infty g(x)\,dx$ is finite. Now, if we're lucky, the same approximation we did above will work as an inequality in the direction we need. It turns out that in this case, we're lucky: since $1\ge 0$, we have $\sqrt{1+e^x}\ge\sqrt{0+e^x}=e^{x/2}$, and so
$$ \frac1{\sqrt{1+e^x}} \le e^{-x/2} $$
and off we go.</p>
<p>(If it had turned out that we wanted the opposite inequality, which will happen in other problems of this type, then you could get one as follows: for $x\ge0$ we have $1\le e^x$, so $\sqrt{1+e^x}\le\sqrt{e^x+e^x} = \sqrt2 e^{x/2}$, and
$$ \frac1{\sqrt{1+e^x}} \ge \frac1{\sqrt2} \,e^{-x/2} $$
This would be good enough for the purposes of the comparison test, since the $\frac1{\sqrt2}$ will just pop out in front of the integral and not affect whether it comes out to be finite or infinite. This is a common trick: when we were just approximating, we threw away the 1 as negligibly small in context; but if doing that yields an inequality in the wrong direction, then rather than throw it away, replace it with a copy of the nearby large terms and combine them, perhaps at the cost of a constant factor.)</p>
|
194,128 | <p>I am trying to solve this problem by induction. The sad part is that I don't have a very strong grasp on solving by inductive proving methods. I understand that there is a base case and that I need an inductive step that will set $k = n$ and then one more step that basically sets $k = n + 1$. </p>
<p>Here is the problem I am trying to solve: </p>
<blockquote>
<p>If $f(n) = \sum_{i = 0}^n X_{i}$, then show by induction that $f(n) = f(n - 1) + X_{n-1}$.</p>
</blockquote>
<p>Can I have someone please try to point me in the right direction? </p>
<p>*EDIT: I update the formula to the correct one. I wasn't sure how to typeset it correctly and left errors in my math. Thank you for those that helped. I'm still having the problem but now I have the proper formula posted. </p>
| Raul Gonzalez | 89,717 | <p>To prove by induction, you need to prove two things. First, you need to prove that your statement is valid for $n=1$. Second, you have to show that the validity of the statement for $n=k$ implies the validity of the statement for $n=k+1$. Putting these two bits of information together, you effectively show that your statement is valid for any value of $n$, since starting from $n=1$, the second bit that you proved above shows that the statement is also valid for $n=2$, and then from the validity of the statement for $n=2$ you know it will also be valid for $n=3$ and so on. </p>
<p>As for your problem, we proceed as follows:</p>
<p>(Part 1) Given the nature of your problem, it is safe to assume $f(0)=0$ (trivial sum with no elements equals zero). In this case it is trivial to see that $f(1)=f(0)+X_{1}$. </p>
<p>(Part 2) Assume the proposition is true for $n=k$, so $$f(k)=\sum_{i=1}^kX_i=f(k-1)+X_k$$
Adding $X_{k+1}$ to both sides of the equation above we get
$$f(k)+X_{k+1}=\sum_{i=1}^kX_i=f(k+1)$$
So, the truth of the statement for $n=k$ implies the truth of the statement for $n=k+1$, and so the result is proved for all $n$. As Mr. Newstead said above, the induction step is not really necessary because of the nature of the problem, but it's important to have this principle in mind for more complicated proofs.</p>
|
27,865 | <p>Both the Laplace transform and the Fourier transform in some sense decode the "spectrum" of a function. The Laplace transform gives a power-series decomposition whereas the Fourier transform gives a harmonic (or loop-based) decomposition.</p>
<p>Are there deep connections between these two transforms? <a href="https://math.stackexchange.com/questions/7301/connection-between-fourier-transform-and-taylor-series">The formulaic connection</a> is clear, but is there something deeper?</p>
<p>(Maybe the answer will involve spectral theory?)</p>
| JAYSHREE RATHOD | 307,432 | <p>Fourier transform does not exist for every signal application.So by introducing the region of convergence in Fourier transform which is known as Laplace Transform one may have indirectly the Fourier transform of signal. </p>
|
2,254,694 | <p>I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:</p>
<p>Attempt 1:</p>
<p>Write out the terms of $b_n$:</p>
<p>$$b_1=a_{2}-\frac{a_{1}}{2}$$
$$b_2=a_{3}-\frac{a_{2}}{2}$$
$$b_3=a_{4}-\frac{a_{3}}{2}$$
$$\cdots$$
$$b_n=a_{n+1}-\frac{a_{n}}{2}$$</p>
<p>Adding up the terms you get:</p>
<p>$$\sum_{i = 1}^n b_i=a_{n+1}+\frac{a_n}{2}+\frac{a_{n-1}}{2}+\cdots+\frac{a_2}{2}-\frac{a_1}{2}.$$</p>
<p>But a dead end here.</p>
<p>Attempt 2:</p>
<p>For $ε=\dfrac{1}{2}$, $\exists K$ such that $\forall n>K$, $$\left|a_{n+1}-\frac{a_n}{2}\right|<\frac{1}{2}.$$</p>
<p>Now I attempt to prove $\{a_n\}$ is Cauchy and hence converges. </p>
<p>For $m>n>K$,
\begin{align*}
|a_m-a_n|&=\left|a_m-\frac{a_{m-1}}{2}+\frac{a_{m-1}}{2}-\frac{a_{m-2}}{2^2}+\cdots -+\frac{a_{n+1}}{2^{m-n-1}}-a_n\right|\\
&\leq \left|a_m-\frac{a_{m-1}}{2}\right|+\frac{1}{2}\left|a_{m-1}-\frac{a_{m-2}}{2}\right|+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&\leq \frac{1}{2}+\frac{1}{2} × \frac{1}{2}+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&<1+\left|\frac{a_n}{2^{m-n}}-a_n\right|,
\end{align*}
and a dead end. </p>
| mercio | 17,445 | <p>Let $\epsilon > 0$.</p>
<p>Since $a_{n+1}-a_n/2$ converges to $0$, there is an integer $m$ such that
for any $n \ge m$, $|a_{n+1}-a_n/2| \le \epsilon/4$. </p>
<p>For such an $n$, you have<br>
$|a_{n+1}| - \epsilon/2 \\
\le |a_{n+1} - a_n/2| + |a_n/2| - \epsilon/2 \\
\le \epsilon/4 + |a_n|/2 - \epsilon/2 \\
= |a_n|/2 - \epsilon/4 \\
= (|a_n| - \epsilon/2)/2$</p>
<p>Intuitively you can interpret this as something saying that $|a_n|$ has to decrease somewhat exponentially at least until $|a_n|$ gets too close to $\epsilon/2$.</p>
<p>Then let us show there is an $m' \ge m$ such that $|a_{m'}| \le \epsilon$. </p>
<p>If $|a_m| \le \epsilon$ then we are already done by picking $m'=m$, so suppose $|a_m| > \epsilon$.<br>
Now, $(|a_m|- \epsilon/2) / (\epsilon/2) > 1 > 0$ so there is an integer $k$ such that $2^k \ge (|a_m|- \epsilon/2) / (\epsilon/2)$.<br>
Looking at $m' = m+k$ we get<br>
$|a_{m+k}| - \epsilon/2 \le (|a_m| - \epsilon/2) 2^{-k} \le \epsilon/2$,<br>
and so $|a_{m'}| \le \epsilon$. </p>
<p>Then we can prove by induction that for any $n \ge m'$, $|a_n| \le \epsilon$ : </p>
<p>This is true for $n=m'$.<br>
Suppose $n \ge m'$ and $|a_n| \le \epsilon$.<br>
Then $|a_n|-\epsilon/2 \le \epsilon/2$, and so because $n \ge m$,<br>
$|a_{n+1}| - \epsilon/2 \le (|a_n| - \epsilon/2)/2 \le \epsilon/4 < \epsilon/2$, and finally $|a_{n+1}| \le \epsilon$. </p>
<p>Therefore, for all $n \ge m', |a_n| \le \epsilon$, and we have shown that the sequence $a_n$ converges to $0$.</p>
|
4,556,193 | <p>I meet a series of the form</p>
<p><span class="math-container">$$\sum_{n=0}^{\infty} \frac{x^n}{(2n-1)!!}$$</span>
where <span class="math-container">$(-1)!! = 1$</span>.</p>
<p>I guess it is a Taylor expansion of a function but I don't know what it is. Could anyone here help me?</p>
<p>Remark: The problem comes from calculating a renewal process. Assume <span class="math-container">$N(t)$</span> is a renewal process with interarrival time <span class="math-container">$X_i$</span> where <span class="math-container">$X_i$</span> i.i.d. follow <span class="math-container">$\chi^2_1$</span>. Then the arrival time of the <span class="math-container">$k$</span>th event is <span class="math-container">$S_k \sim \chi^2_k$</span>. Then the renewal function is</p>
<p><span class="math-container">$$m(t) = \mathbb{E}N(t) =\sum_{k=1}^\infty Pr(S_k \leq t)$$</span></p>
<p>which is</p>
<p><span class="math-container">$$\sum_{k=1}^\infty \int_0^t \frac{x^{k/2-1}e^{-x/2}}{2^{k/2}\Gamma(k/2)}dx.$$</span></p>
<p>We can exchange the summation and the integral and divide the summation into two parts according to <span class="math-container">$k$</span> is even or odd.</p>
<p>The part for <span class="math-container">$k$</span> is even is easy. But for <span class="math-container">$k$</span> is odd, I think we need to deal the series in the beginning of the problem.</p>
| robjohn | 13,854 | <p>To compute
<span class="math-container">$$
f(x)=\sum_{n=0}^\infty\frac{x^n}{(2n-1)!!}\tag1
$$</span>
consider
<span class="math-container">$$
\begin{align}
g(x)
&=\frac{f(2x)}{\sqrt{2x}}\tag{2a}\\
&=\sum_{n=0}^\infty\frac{(2x)^{n-1/2}}{(2n-1)!!}\tag{2b}
\end{align}
$$</span>
where
<span class="math-container">$$
\begin{align}
g'(x)
&=\sum_{n=0}^\infty\frac{(2x)^{n-3/2}}{(2n-3)!!}\tag{3a}\\
&=g(x)-(2x)^{-3/2}\tag{3b}
\end{align}
$$</span>
since <span class="math-container">$(-3)!!=-1$</span>.</p>
<p>Applying an <a href="https://en.wikipedia.org/wiki/Integrating_factor" rel="nofollow noreferrer">integrating factor</a> of <span class="math-container">$e^{-x}$</span> to <span class="math-container">$(3)$</span> yields
<span class="math-container">$$
(e^{-x}g(x))'=-e^{-x}(2x)^{-3/2}\tag4
$$</span>
and thus,
<span class="math-container">$$
\begin{align}\newcommand{\erf}{\operatorname{erf}}
g(x)
&=\frac{e^x}{2\sqrt2}\int_x^\infty e^{-t}t^{-3/2}\,\mathrm{d}t+ce^x\tag{5a}\\
&=-\frac{e^x}{\sqrt2}\int_x^\infty e^{-t}\,\mathrm{d}t^{-1/2}+ce^x\tag{5b}\\
&=\frac1{\sqrt{2x}}-\frac{e^x}{\sqrt2}\int_x^\infty e^{-t}t^{-1/2}\,\mathrm{d}t+ce^x\tag{5c}\\
&=\frac1{\sqrt{2x}}-\sqrt2e^x\int_{\sqrt{x}}^\infty e^{-t^2}\,\mathrm{d}t+ce^x\tag{5d}\\
&=\frac1{\sqrt{2x}}+\sqrt2e^x\int_0^{\sqrt{x}}e^{-t^2}\,\mathrm{d}t\tag{5e}\\
&=\frac1{\sqrt{2x}}+e^x\sqrt{\frac\pi2}\erf\left(\sqrt{x}\right)\tag{5f}
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$\text{(5a):}$</span> solve <span class="math-container">$(4)$</span> for <span class="math-container">$g(x)$</span> where <span class="math-container">$c$</span> is a constant TBD<br />
<span class="math-container">$\text{(5b):}$</span> prepare to integrate by parts<br />
<span class="math-container">$\text{(5c):}$</span> integrate by parts<br />
<span class="math-container">$\text{(5d):}$</span> substitute <span class="math-container">$t\mapsto t^2$</span><br />
<span class="math-container">$\text{(5e):}$</span> set <span class="math-container">$c=\sqrt{\pi/2}=\sqrt2\int_0^\infty e^{-t^2}\mathrm{d}t$</span><br />
<span class="math-container">$\phantom{\text{(5e):}}$</span> so that <span class="math-container">$g$</span> is an odd function of <span class="math-container">$\sqrt{x}$</span><br />
<span class="math-container">$\text{(5f):}$</span> <span class="math-container">$\int_0^xe^{-t^2}\,\mathrm{d}t=\frac{\sqrt\pi}2\erf(x)$</span></p>
<p>Reversing <span class="math-container">$\text{(2a)}$</span> by setting <span class="math-container">$f(x)=\sqrt{x}\,g(x/2)$</span>, we get
<span class="math-container">$$
f(x)=1+e^{x/2}\sqrt{\frac{\pi x}2}\erf\left(\sqrt{x/2}\right)\tag6
$$</span></p>
|
1,517,245 | <p>I would appreciate if somebody could help me with the following problem</p>
<p>Q: Arrangements of $1,2,..,81$ in an $9 \times 9$ matrix such that each row and each column is increasing. How many possible $X$ ?</p>
<p><a href="https://i.stack.imgur.com/Te1TB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Te1TB.png" alt="enter image description here"></a></p>
| J.-E. Pin | 89,374 | <p>A monoid with an distinguished absorbing element is called a <em>monoid with zero</em> in the literature.
A monoid with zero in which the units are exactly the nonzero elements is called a <em>group with zero</em> [1, p. 5; 2, Def 1.3.1, p. 34; 4] or a <em>$0$-group</em> [3].</p>
<p>[1] A. H. Clifford, G. B. Preston, The algebraic theory of semigroups. Vol. I. Mathematical Surveys, No. 7 American Mathematical Society, Providence, R.I. 1961 xv+224 pp. </p>
<p>[2] P.M. Higgins, Techniques of semigroup theory. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1992. x+258 pp. ISBN: 0-19-853577-5</p>
<p>[3] J.M. Howie, Fundamentals of semigroup theory. London Mathematical Society Monographs. New Series, 12. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1995. x+351 pp. ISBN: 0-19-851194-9</p>
<p>[4] Lallement, Gérard. Semigroups and combinatorial applications. Pure and Applied Mathematics. A Wiley-Interscience Publication. John Wiley & Sons, New York-Chichester-Brisbane, 1979. xi+376 pp. ISBN: 0-471-04379-6</p>
|
2,303,339 | <p>How does this proof work?</p>
<p><strong>Theorem.</strong>$\quad$Let $G$ be a group. Then $G$ has a unique identity. </p>
<p><strong>Proof.</strong>$\quad$Assume that $e$ and $f$ are two identities in $G$. Since $e$ is
an identity, $ef=f$; and since $f$ is an identity, $ef=e$. Thus $e=ef=f$. </p>
<p>I think need to get my understanding of variables sorted out, because when I read the first line of the proof I picture $e$ as an object different from $f$ and it's confusing to then read the conclusion that $e$ and $f$ are equal. Also, how does this show that $G$ has a unique identity?</p>
| Kyle Miller | 172,988 | <p>In general, a matrix <span class="math-container">$X\in \mathbb{M}_n(\mathbb{F})$</span> is determined by the linear map <span class="math-container">$\mathbb{M}_n(\mathbb{F})\to \mathbb{F}$</span> defined by <span class="math-container">$Y\mapsto \operatorname{tr}(XY)$</span>. This is because if <span class="math-container">$E_{ij}$</span> is the <span class="math-container">$n\times n$</span> elementary matrix that's all <span class="math-container">$0$</span> except for a <span class="math-container">$1$</span> in the <span class="math-container">$(i,j)$</span> entry, then <span class="math-container">$\operatorname{tr}(XE_{ij})$</span> is the <span class="math-container">$(i,j)$</span> entry of <span class="math-container">$X$</span>.</p>
<p>Thus, it suffices to show that for all <span class="math-container">$A,B,C\in \mathbb{M}_2(\mathbb{F})$</span> that
<span class="math-container">$$
\operatorname{tr}(ABC+BAC)=\operatorname{tr}((\operatorname{tr}B)AC + (\operatorname{tr}A)BC +(\operatorname{tr}(AB) - \operatorname{tr}A\operatorname{tr}B)C ).
$$</span>
Applying linearity of trace, this is equivalently
<span class="math-container">$$
\operatorname{tr}ABC+\operatorname{tr}BAC=\operatorname{tr}B \operatorname{tr}(AC) + \operatorname{tr}A \operatorname{tr}(BC) +\operatorname{tr}(AB)\operatorname{tr}C - \operatorname{tr}A\operatorname{tr}B\operatorname{tr}C
$$</span>
and just for sake of putting things in order to show how symmetric the situation is, this is
<span class="math-container">$$
\operatorname{tr}ABC+\operatorname{tr}CBA= \operatorname{tr}A \operatorname{tr}(BC) + \operatorname{tr}B \operatorname{tr}(CA) + \operatorname{tr}C\operatorname{tr}(AB) - \operatorname{tr}A\operatorname{tr}B\operatorname{tr}C.
$$</span>
(The theoretical context here is that we have these two trilinear forms on <span class="math-container">$\mathbb{M}_2(\mathbb{F})$</span> that we want to show are equal. In fact, these are trilinear forms invariant under the action of <span class="math-container">$GL(2)$</span> on <span class="math-container">$\mathbb{M}_2(\mathbb{F})$</span> by conjugation. The vector space of such invariant trilinear forms for general <span class="math-container">$GL(n)$</span> and <span class="math-container">$\mathbb{M}_n(\mathbb{F})$</span> happens to be spanned by the six terms present in the above equation, and what we're wanting to show is that when <span class="math-container">$n=2$</span> that this spanning set is linearly dependent in this way. Linear dependence occurs exactly when <span class="math-container">$n=0,1,2$</span>, and, fancifully, when <span class="math-container">$n=-1,-2$</span>, otherwise the vector space of invariant trilinear forms is indeed 6-dimensional.)</p>
<p>Recall that the third exterior power of a two-dimensional vector space is zero-dimensional, i.e. <span class="math-container">$\bigwedge^3\mathbb{F}^2\approx 0$</span>. Regarding the exterior power as a subspace of <span class="math-container">$\mathbb{F}^2\otimes \mathbb{F}^2\otimes \mathbb{F}^2$</span>, then the projection <span class="math-container">$p$</span> onto the third exterior power is given by
<span class="math-container">$$v_1\otimes v_2\otimes v_3 \mapsto \frac{1}{6}\sum_{\sigma\in S_3}(-1)^\sigma v_{\sigma(1)}\otimes v_{\sigma(2)}\otimes v_{\sigma(3)}$$</span>
where <span class="math-container">$(-1)^\sigma\in\{-1,1\}$</span> is the sign of the permutation <span class="math-container">$\sigma$</span> in the symmetric group <span class="math-container">$S_3$</span> on three elements. (If you care about finite-characteristic fields, you can get rid of the <span class="math-container">$\frac{1}{6}$</span>, and it's ok using a scale multiple of a projection for the following.) Since the subspace is zero-dimensional, this projection function is identially zero, hence we get the identity
<span class="math-container">$$
v_1\otimes v_2 \otimes v_3 - v_1\otimes v_3 \otimes v_2 + v_2\otimes v_3 \otimes v_1 - v_2\otimes v_1 \otimes v_3 + v_3\otimes v_1 \otimes v_2 - v_3\otimes v_2 \otimes v_1 = 0
$$</span>
for all <span class="math-container">$v_1,v_2,v_3\in\mathbb{F}^2$</span>.</p>
<p>This next part is tricky to explain. I'm staring at some <a href="https://en.wikipedia.org/wiki/Penrose_graphical_notation" rel="nofollow noreferrer">Penrose notation</a> that convinced me of it, but I'll try to do it symbolically. The idea is that we can take the three matrices as linear operators <span class="math-container">$\mathbb{F}^2\to\mathbb{F}^2$</span>, tensor them together to get an operator <span class="math-container">$A\otimes B\otimes C:\mathbb{F}^2\otimes \mathbb{F}^2\otimes \mathbb{F}^2\to \mathbb{F}^2\otimes \mathbb{F}^2\otimes \mathbb{F}^2$</span>, composing this with <span class="math-container">$p$</span>, and then taking the trace. Considering that <span class="math-container">$p=0$</span>, we have
<span class="math-container">$$\operatorname{tr}(p\circ(A\otimes B\otimes C))=0.$$</span>
Perhaps you can convince yourself that, using the expansion of <span class="math-container">$p$</span> into six permutations, that this is
<span class="math-container">$$
\operatorname{tr}A\operatorname{tr}B\operatorname{tr}C
- \operatorname{tr}A\operatorname{tr}(BC)
+ \operatorname{tr}(ABC)
- \operatorname{tr}C\operatorname{tr}(AB)
+ \operatorname{tr}(CBA)
- \operatorname{tr}B\operatorname{tr}(CA)
= 0
$$</span>
which after moving some terms to the other side is exactly the equality we wanted to establish. Sorry for the lack of details in this last step -- if anyone wants a Penrose diagram proof I can add it.</p>
|
3,971,220 | <p>Let <span class="math-container">$G$</span> be a group (not necessary abelian) generated by <span class="math-container">$a, b$</span> that has a group presentation: <span class="math-container">$$\langle a, b \mid a^m = 1, b^n = 1, ba = a^rb\rangle. $$</span>Then how can I prove that <span class="math-container">$(a^sb^t)(a^ub^v) = a^xb^y$</span>, where <span class="math-container">$x$</span> is the remainder of <span class="math-container">$s + u(r^t)$</span> when divided by <span class="math-container">$m$</span>, and <span class="math-container">$y$</span> is the remainder of <span class="math-container">$t + v$</span> when divided by <span class="math-container">$n$</span>? It seems that it can be solved by changing <span class="math-container">$b^{...}a^{...}$</span> form into <span class="math-container">$a^{...}b^{...}$</span> form successively using the condition <span class="math-container">$ba = a^rb$</span>, but I don't understand precisely how to do this.</p>
| José Carlos Santos | 446,262 | <p>No, it is not correct. If <span class="math-container">$G$</span> is generated by <span class="math-container">$a$</span>, then its quotient groups are:</p>
<ul>
<li><span class="math-container">$G/\{e\}\simeq G$</span>;</li>
<li><span class="math-container">$G/\{e,a^5\}\simeq\Bbb Z_5$</span>;</li>
<li><span class="math-container">$G/\{e,a^2,a^4,a^6,a^8\}\simeq\Bbb Z_2$</span>;</li>
<li><span class="math-container">$G/G$</span>, which is the trivial group.</li>
</ul>
|
3,971,220 | <p>Let <span class="math-container">$G$</span> be a group (not necessary abelian) generated by <span class="math-container">$a, b$</span> that has a group presentation: <span class="math-container">$$\langle a, b \mid a^m = 1, b^n = 1, ba = a^rb\rangle. $$</span>Then how can I prove that <span class="math-container">$(a^sb^t)(a^ub^v) = a^xb^y$</span>, where <span class="math-container">$x$</span> is the remainder of <span class="math-container">$s + u(r^t)$</span> when divided by <span class="math-container">$m$</span>, and <span class="math-container">$y$</span> is the remainder of <span class="math-container">$t + v$</span> when divided by <span class="math-container">$n$</span>? It seems that it can be solved by changing <span class="math-container">$b^{...}a^{...}$</span> form into <span class="math-container">$a^{...}b^{...}$</span> form successively using the condition <span class="math-container">$ba = a^rb$</span>, but I don't understand precisely how to do this.</p>
| Community | -1 | <p>Any quotient of a cyclic group is cyclic. One way to see it is that it is a homomorphic image (of the cyclic group).</p>
<p>Furthermore, any homomorphic image has order dividing the order of the group (first isomorphism theorem).</p>
<p>Finally it is easy to see that all the cyclic groups of orders dividing, in this case <span class="math-container">$10$</span>, do occur as quotients.</p>
<hr />
<p>As far as what you did, a bunch of them are the same. So there are only four.</p>
|
1,279,947 | <p>In physics, 1m is defined 'the length which light travels in certain time'.
And i've thought 2m has twice many elements than 1m.
But today i've realized intervals (0,1) and (0,2) are equipotent.
I was shocked because i've thought length and cardinality are similar concepts.
Cardinality is clearly defined in set theory. But what is length? </p>
| Simone | 77,622 | <p>Just talking in an heuristic way, you can think that the idea of length is similar to the idea of area or volume, which is generalized in mathematics to measure theory. In some sense, you'd like to associate a positive real number to a subset of a set such that if you consider the union of disjoint pieces you get the sum of measures of pieces, if you compute the measure of the empty set you get $0$. This is the closest idea to length. In fact you can define such a measure, the Lebesgue one, in a way that the measure of an interval $(a,b)$ is $a-b$. But don't forget that you are doing mathematics, so don't leave everything to usual intuition. In fact, also in measure theory strange things can happen. If you want a beautiful example look, e.g. on Wikipedia, at Banach-Tarski paradox. Notice also that cardinality has nothing to do with length, also if you think to finite sets. In fact you'd agree that in Euclidean Geometry points have no dimension. In particular a finite set of discrete points e.g. in $\mathbb{R}^{2}$ is supposed to have no "dimension", or "length" if you prefer, instead it's cardinality is different from $0$, so also in finite case cardinality is not useful to find a concept of length.</p>
|
1,279,947 | <p>In physics, 1m is defined 'the length which light travels in certain time'.
And i've thought 2m has twice many elements than 1m.
But today i've realized intervals (0,1) and (0,2) are equipotent.
I was shocked because i've thought length and cardinality are similar concepts.
Cardinality is clearly defined in set theory. But what is length? </p>
| Chris Culter | 87,023 | <p>See 5xum's answer! Regarding this part:</p>
<blockquote>
<p>Cardinality is clearly defined in set theory. But what is length?</p>
</blockquote>
<p>Cardinality is meaningful for all sets, which is why it's typically encountered in a first course on set theory. Measure is a more complicated concept, which is typically first defined for subsets of $\mathbb R$ in a real analysis course. Just because $(0,1)$ and $(0,2)$ are sets doesn't mean that set theory tells us everything about them!</p>
|
3,556,334 | <p>Given a rectangle with dimensions <span class="math-container">$10$</span>cm and <span class="math-container">$6$</span>cm, show that for every <span class="math-container">$3$</span> points in the interior of the rectangle, the area of the triangle is less than <span class="math-container">$30$</span> cm<span class="math-container">$^2$</span>.</p>
<p>I draw the diagonals but now I am stuck.</p>
| Andrei | 331,661 | <p>Let's have the rectangle <span class="math-container">$ABCD$</span> and <span class="math-container">$P$</span>, <span class="math-container">$Q$</span>, and <span class="math-container">$R$</span> three points inside. Let's use <span class="math-container">$A=(0,0)$</span>, <span class="math-container">$B=(a,0)$</span>, <span class="math-container">$C=(a,b)$</span>, and <span class="math-container">$D=(0,b)$</span>. The <span class="math-container">$x$</span> coordinates of the three interior points are in the interval<span class="math-container">$(0,a)$</span> and the <span class="math-container">$y$</span> coordinates are in the <span class="math-container">$(0,b)$</span>. From <span class="math-container">$P,\ Q,\ R$</span>, choose the point with minimum <span class="math-container">$x$</span> and draw a parallel to <span class="math-container">$y$</span> axis. Repeat the procedure for the point with maximum <span class="math-container">$x$</span> coordinate. Now do the same for the <span class="math-container">$y$</span> coordinate. You get a rectangle <span class="math-container">$A'B'C'D'$</span> that is in the interior of <span class="math-container">$ABCD$</span> (so the area is smaller). At least one of <span class="math-container">$P,\ Q,\ R$</span> is in the corner of this rectangle. Without loss of generality, assume <span class="math-container">$P=A'$</span>, <span class="math-container">$Q$</span> is on <span class="math-container">$B'C'$</span>, and <span class="math-container">$R$</span> is on <span class="math-container">$C'D'$</span>. The lengths of the rectangle sides are <span class="math-container">$a'$</span> and <span class="math-container">$b'$</span>. The coordinates of <span class="math-container">$Q$</span> with respect to <span class="math-container">$P$</span> are <span class="math-container">$(a', y)$</span>, and <span class="math-container">$R$</span> is at <span class="math-container">$(x,b')$</span>. We have <span class="math-container">$0\le x\le a'$</span> and <span class="math-container">$0\le y\le b'$</span>. Now let's write the area of the <span class="math-container">$PQR$</span> triangle by subtracting from the area of <span class="math-container">$A'B'C'D'$</span> the triangles that are formed with the sides:
<span class="math-container">$$\begin{align}A_{PQR}&=A_{A'B'C'D'}-A_{QA'B'}-A_{RA'D'}-A_{QRC'}\\&=a'b'-\frac12a'y-\frac12b'x-\frac12(a'-x)(b'-y)\\&=\frac12 a'b'-\frac12xy\le\frac12a'b'\lt \frac12 ab\end{align}$$</span></p>
|
2,924,549 | <p>Given that the three fundemental geometries are euclidean geometry (zero curvature), riemannian geometry (positive curvature) and lobachevskian geometry (negative curvature), I am curious as to what the link is between these curvatures and the conic sections of an ellipse, an hyperbole and a circle. </p>
<p>I have seen euclidean geometry be called parabolic geometry, riemannian geometry be called elliptical geometry and lobachevskian geometry be called hyperbolic geometry but I have not found any further explanation. How can the 2D shapes be mathematically linked to the 3D surfaces? Why is a parabola associated with flat space while a hyperbole is associated with a saddle shaped 3D space and an ellipse is associated with spherical space? </p>
<p>I have tried to look at curvature for a connection, however, I have only found explanations of curvature and Gaussian curvature seperately but not anything connecting the curvature of 3D space with 2D space. </p>
<p>I also read that gaussian curvature = 1 - eccentricity. This would theoretically make sense as for example a parabola has an eccentricity of 1 giving a curvature of zero which would relate it to flat space. However, I can only imagine this being true for some special cases as I have struggled to find a published explenation. Is there any proof for this formula? </p>
| Zeno Rogue | 265,219 | <p>AFAIK these names come from far-fetched analogies, and other meanings of these three Greek words, rather than any specific similarities to the similarly named curves.</p>
<p>The name "elliptic geometry" is especially confusing to people who hear it for the first time, because it sounds like it would describe ellipsoids, while it actually deals with perfect spheres. At least it makes some sense if you consider circles/spheres to be special cases of ellipses/ellipsoids (but why not name based on the special case then?).</p>
<p>The name "hyperbolic geometry" feels quite appropriate though if you think of the Minkowski hyperboloid model, and also as you would use hyperbolic functions (sinh, cosh) for most distance-related computations in this geometry.</p>
<p>One quite far-fetched relation between Euclidean geometry and a shape of parabola is that horospheres in Minkowski hyperboloid model are represented as paraboloids, and the geometry on a horosphere is Euclidean.</p>
|
2,924,549 | <p>Given that the three fundemental geometries are euclidean geometry (zero curvature), riemannian geometry (positive curvature) and lobachevskian geometry (negative curvature), I am curious as to what the link is between these curvatures and the conic sections of an ellipse, an hyperbole and a circle. </p>
<p>I have seen euclidean geometry be called parabolic geometry, riemannian geometry be called elliptical geometry and lobachevskian geometry be called hyperbolic geometry but I have not found any further explanation. How can the 2D shapes be mathematically linked to the 3D surfaces? Why is a parabola associated with flat space while a hyperbole is associated with a saddle shaped 3D space and an ellipse is associated with spherical space? </p>
<p>I have tried to look at curvature for a connection, however, I have only found explanations of curvature and Gaussian curvature seperately but not anything connecting the curvature of 3D space with 2D space. </p>
<p>I also read that gaussian curvature = 1 - eccentricity. This would theoretically make sense as for example a parabola has an eccentricity of 1 giving a curvature of zero which would relate it to flat space. However, I can only imagine this being true for some special cases as I have struggled to find a published explenation. Is there any proof for this formula? </p>
| brainjam | 1,257 | <p>From Greenberg's <em>Euclidean and Non-Euclidean Geometries</em>:</p>
<blockquote>
<p>[I]n the early nineteenth century two alternative geometries were
proposed. In hyperbolic geometry (from the Greek <em>hyperballein</em>, "to
exceed") the distance between the rays increases. In elliptic geometry
(from the Greek <em>elleipein</em>, "to fall short") the distance decreases and
the rays eventually meet.</p>
</blockquote>
<p>So the terms are not directly based on the names of conic curves, but come from the same roots. The Greek <em>parabole</em> had connotations of juxtaposition, comparison of one thing with another, likeness, similitude. So it was the "just right" that separated <em>hyperballein</em> and <em>elleipein</em>.</p>
<p>Some more background at Perisho, <a href="https://www.jstor.org/stable/24338341" rel="nofollow noreferrer"><em>The Etymology Of Mathematical Terms</em></a></p>
<p><strong>Update:</strong> I was leafing through Klein's <a href="https://archive.org/details/elementarymathem0002-K67klei/page/182/mode/2up" rel="nofollow noreferrer"><em>Elementary mathematics from an advanced standpoint : geometry</em></a> and noticed his claim to naming these geometries.</p>
<p><a href="https://i.stack.imgur.com/2tw9K.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2tw9K.jpg" alt="enter image description here" /></a></p>
<p>Interested parties can follow the URL for context, and the article he refers to is <a href="https://archive.org/details/mathematischean22behngoog/page/n594/mode/2up" rel="nofollow noreferrer">here</a> (in German). (See also Campo and Papadopoulos,
<a href="https://arxiv.org/pdf/1406.7309.pdf" rel="nofollow noreferrer"><em>On Klein’s So-called Non-Euclidean geometry</em></a>, page 18).</p>
<p>His reference to asymptotes of conics may sound strange because we are taught that only the hyperbola has asymptotes. But if asymptotes are defined as tangents to ideal points (points on the line at infinity) on the conic, then the ellipse has these but they are imaginary.</p>
|
2,390,804 | <p>How does one compute the following integral?
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$$</p>
<hr>
<p>I have tried extending $x$ to the complex plane then evaluating the following contour integral
$$\oint_C \frac{\sqrt{x}e^{ix}}{1+x^2} dx$$
with the contour $C$ running along the whole real axis and then upper semicircle. I obtain
$$\int_{0}^{\infty} \frac{\sqrt{x}\big(\cos(x)+\sin(x)\big)}{1+x^2}\,\mathrm dx=\frac\pi{e\sqrt2}$$
but not the original integral.</p>
| Hans | 64,809 | <p>My goal is to prove
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =\frac{\pi}{2\sqrt{2}\:e}\left(-e^2\text{erfc}(1)+\text{erf}(1) +1\right)$$
as stated in @JJacquelin's answer. My following attempt falls short of that. But it does provide an equivalent integral with much faster convergence.</p>
<p>Take the contour integral $$2\oint_C \frac{z^2e^{iz^2}}{1+z^4} dz$$
with contour $C$ run along the real axis from $0$ to $R$, trace the circle $Re^{i\theta}$ with $\theta$ running from $0$ to $\frac\pi 4$, then roll back to the origin along $xe^{i\frac\pi 4}$ with $x$ running from $R$ to $0$ while circumventing around the point $e^{i\frac\pi 4}$ clockwise with radius $\delta>0$. The integral on the octal circle vanishes as $R\to\infty$. We have
$$\oint_C \frac{z^2e^{iz^2}}{1+z^4} dz=\int_{0}^{\infty} \frac{x^2e^{ix^2}}{1+x^4} dx-e^{i\frac34\pi}\int_0^\infty \frac{x^2e^{-x^2}}{1-x^4}dx+\cdots$$</p>
<p>(to be continued) </p>
|
13,467 | <p>I will present two problems alongside solutions, student is doing problems of type I like a cakewalk but has several issues with the problems of type 2;</p>
<p><strong>Type I</strong></p>
<blockquote>
<p>Consider an experiment of rolling two dice:</p>
<p>Sample space $$ S = \{(1,1),(1,2),(1,3), \cdots, (6,6) \} $$</p>
<p>Let $A$ be the event of getting 6 as sum on two dice:</p>
<p>Event $$A = \{(1,5),(2,4),(3,3),(4,2),(5,1)\}$$ </p>
<p>Let $B$ be the event of getting 4 on first die:</p>
<p>Event $$B = \{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\}$$ </p>
<p>Now, the probability of getting sum of 6 on two dice given that first
die appears as 4 is given by</p>
<p>$$p(A \mid B) = \dfrac{p(A\cap B)}{p(B)} = \dfrac{p\{(4,2)\}}{p(B)} =
\dfrac{1}{6}$$</p>
</blockquote>
<p><strong>Type II</strong></p>
<blockquote>
<p>Let $C$ be the event of having cancer and $p(C) = 1/2$</p>
<p>The probability of having both tumor and cancer is $p(C\cap T) = 1/6$</p>
<p>Then the probability of having tumor given cancer is given by</p>
<p>$$p(T\mid C) = \dfrac{p(T\cap C)}{p(C)} = \dfrac{1}{3}$$</p>
</blockquote>
<p>Students are understanding type I problems and type 2 problems well, but some students are asking to present $C, T, C\cap T$ in terms of sets. I tried to convince them using Venn diagrams. But they are asking for either roster form (for discrete sets) or set builder form(for any set).</p>
<p>I am trying to do like follows </p>
<p>\begin{align}
\text{Sample space} = S &= \{x \mid \text{$x$ is a living being} \} \\
C &= \{x \mid \text{$x$ has cancer}\} \\
C\cap T &= \{x \mid \text{$x$ has both tumor and cancer}\} \\
\end{align}</p>
<p>Is it right way to do? Students are facing difficulty and they are asking every event inform of set, since definition of event is that it is a subset of sample space (which is a set).</p>
| Nate Bade | 9,019 | <p>To use set builder notation you need to carefully state to which set every element you want to discuss belongs. In general it goes </p>
<p>$P = \{x\in\text{some set} : x\text{ satisfies some truth evaluable proposition}\}$. </p>
<p>Here's how I would do it in your case:</p>
<p>Let $P$ be the set of all humans, let $C=\{x\in P: x\text{ has cancer}\}$ and let $T=\{x\in P: x\text{ has a tumor}\}$. Then $C\cap P$ is the set of humans that both have cancer and a tumor, or $C\cap P = \{x\in P: (x\text{ has a tumor})\wedge (x\text{ has cancer})\}$. In the previous, the wedge notation is the logical 'and'. </p>
|
2,993,958 | <blockquote>
<p>Do there exist three non null vectors <span class="math-container">$a,b,c$</span> with <span class="math-container">$$a.b=a.c$$</span> such that <span class="math-container">$b\ne c$</span>?</p>
</blockquote>
<p>My attempt:</p>
<p>a.(b-c)=0</p>
<p>But is it possible to say b-c=0 since a is non null?</p>
| edm | 356,114 | <p>Consider <span class="math-container">$$a=\begin{bmatrix}1\\0\end{bmatrix}, b=\begin{bmatrix}0\\1\end{bmatrix},c=\begin{bmatrix}0\\2\end{bmatrix}.$$</span> As long as <span class="math-container">$a$</span> is perpendicular to <span class="math-container">$b-c$</span>, you will get the equality <span class="math-container">$a\cdot b=a\cdot c$</span>.</p>
|
3,879,366 | <p>This might or might not be author specific, but in Algebraic Statistics by Seth Sullivant the author uses a notation of <span class="math-container">$S_{K, K}$</span> for example. So far I have not been able to find out what it means. Here is an <a href="https://i.stack.imgur.com/oEJHw.png" rel="nofollow noreferrer">example</a>. For reference, <a href="https://i.stack.imgur.com/43Wn6.png" rel="nofollow noreferrer">here</a> is the syntax the author uses for an induced subgraphs.</p>
| Claude Leibovici | 82,404 | <p><span class="math-container">$$g(x) = \Phi(x) = {\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{x}e^{-t^{2}/2}\,dt=\frac{1}{2} \left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)+1\right)$$</span>
<span class="math-container">$$f(x,g(x)) = x(1-g(x)-g(x)^2)=-\frac{1}{4} x \left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)
\left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)+4\right)-1\right)$$</span>
<span class="math-container">$$f'(x,g(x))= \frac{e^{-\frac{x^2}{2}} x
\left(\text{erfc}\left(\frac{x}{\sqrt{2}}\right)-3\right)}{\sqrt{2 \pi
}}+\frac{1}{4} \left(-\left(\text{erfc}\left(\frac{x}{\sqrt{2}}\right)-6\right)
\text{erfc}\left(\frac{x}{\sqrt{2}}\right)-4\right)$$</span>
<span class="math-container">$$f''(x,g(x))= \frac{e^{-\frac{x^2}{2}} \left(x^2-2\right)
\left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)+2\right)}{\sqrt{2 \pi
}}-\frac{e^{-x^2} x}{\pi }$$</span></p>
<p>Starting with <span class="math-container">$x_0$</span>, Newton methods solves pretty fast <span class="math-container">$f'(x,g(x))=0$</span> :
<span class="math-container">$$\left(
\begin{array}{cc}
n & x_n \\
0 & 0 \\
1 & 0.15666426716443753140 \\
2 & 0.15106897311919179609 \\
3 & 0.15106533761909824900 \\
4 & 0.15106533761750598442
\end{array}
\right)$$</span> At this point, <span class="math-container">$f''(x,g(x))$</span> is negative <span class="math-container">$(\approx -1.7003)$</span>; so, this point is a maximum of <span class="math-container">$f(x,g(x))$</span>.</p>
<p>So, the maximum of <span class="math-container">$f(x,g(x))$</span> is <span class="math-container">$\sim 0.0190825$</span>.</p>
<p><strong>Edit</strong></p>
<p>If we plot <span class="math-container">$f(x,g(x))$</span> and notice that the maximum is close to <span class="math-container">$x=0$</span>, we could avoid all these calculations building its Taylor expansion around <span class="math-container">$x=0$</span>.</p>
<p><span class="math-container">$$f(x,g(x))=\frac{x}{4}-\sqrt{\frac{2}{\pi }} x^2-\frac{x^3}{2 \pi }+O\left(x^4\right)$$</span> This would give the maximum
<span class="math-container">$$x_*=\frac{\sqrt{19}-4}{3} \sqrt{\frac{\pi }{2}}\approx 0.149938 \quad \text{and} \quad f(x_*,g(x_*))=\frac{19 \sqrt{19}-82}{54} \sqrt{\frac{\pi }{2}}\approx 0.0190105 $$</span> which is quite decent.</p>
<p>Similarly, for more accuracy, we could use one more term for the expansion
<span class="math-container">$$f(x,g(x))=\frac{x}{4}-\sqrt{\frac{2}{\pi }} x^2-\frac{x^3}{2 \pi }+\frac{x^4}{3 \sqrt{2 \pi}}+O\left(x^5\right)$$</span> and build the <span class="math-container">$[1,2]$</span> Padé approximant of <span class="math-container">$f'(x,g(x))$</span>. This would give
<span class="math-container">$$x_*=\frac{201 }{4 (420-\pi)}\sqrt{\frac{\pi }{2}}\approx 0.151080$$</span>
<span class="math-container">$$ f(x_*,g(x_*))=\frac{201 }{1024 (420-\pi)^4}\sqrt{\frac{\pi }{2}}\left(-2404548720+20194569 \pi -67776 \pi ^2+64 \pi ^3\right)$$</span> which is <span class="math-container">$\approx 0.0190786$</span>.</p>
|
4,141,812 | <p>Considering the vector space <span class="math-container">$\mathbb{R}^3$</span>, find all values of <span class="math-container">$k$</span> such that
<span class="math-container">\begin{align*}
\begin{bmatrix}
2k^2 \\
-3k \\
1
\end{bmatrix} \in \text{ span }
\left\{
\begin{bmatrix}
1 \\
1 \\
3
\end{bmatrix},
\begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix},
\begin{bmatrix}
1 \\
2 \\
4
\end{bmatrix}
\right\}.
\end{align*}</span></p>
<p>Here is my approach:<br />
By creating the system of equations
<span class="math-container">\begin{align*}
\begin{cases}
x_1 + x_3 &= 2k^2\\
x_1 + x_2 + 2x_3 &= -3k\\
3x_1 + x_2 + 4x_3 &= 1
\end{cases}
\end{align*}</span>
And then, I acquired its RREF by performing elementary row operations.
<span class="math-container">\begin{align*}
\begin{bmatrix}
1 & 0 & 1 & 2k^2 \\
0 & 1 & 1 & -2k^2-3k \\
0 & 0 & 0 & -4k^2 + 3k + 1
\end{bmatrix}
\end{align*}</span>
Since we get a pivot point at the augmented column,
<span class="math-container">\begin{align*}
-4k^2+3k+1 &= 0\\
k &= 1, -\dfrac{1}{4}
\end{align*}</span>
I'm not really sure with what I did in the last part. Is this right?</p>
| nir shahar | 890,149 | <p>Notice that the function <span class="math-container">$f(x):=e^x+x$</span> is continuous, and <span class="math-container">$\lim_{x\rightarrow \infty} f(x) = \infty$</span>, but also <span class="math-container">$\lim_{x\rightarrow -\infty} f(x)=-\infty$</span>.</p>
<p>In fact, we only need to know that there are two points <span class="math-container">$x_1,x_2$</span> such that <span class="math-container">$f(x_1)>0>f(x_2)$</span>. Now, since <span class="math-container">$f$</span> is continous, it must pass through <span class="math-container">$0$</span> at some point. Let this point be called <span class="math-container">$a$</span>. We will have: <span class="math-container">$\lim_{x\rightarrow a}\frac{1}{e^x+x}=\lim_{x\rightarrow a}\frac{1}{f(x)}=\pm\infty$</span> (the partial limits are either plus or negative infinity, or both of them. Nothing else)</p>
<p>Hence at that point <span class="math-container">$a$</span>, there is a vertical asymptote.</p>
|
216,040 | <p>A model for the movement of a stock price supposes that if the present price is S then after one period, say one second, it will either go up to uS with probability p or go down to dS with probability q = 1 - p. Assuming that successive movements are independent,approximate the probability that the stock will be up by at least 5% after the next 1000 periods for u = 1.02, d = 0.95 and p = 0.6</p>
| Ben | 27,132 | <p>The probability to win when $k$ = 3:
$$p = 6/6^3 = 1/36$$</p>
<p>The probability to win when $k$ = 2:</p>
<p>There are 3 ways to select 2 dice from a set of 3 dice($_3C_2$). For each of these, there is a 1/6 chance that the 2 dice will show the same number:
$$p = 3*{1 \over\ 6} = 1/2$$</p>
<p>To calculate our average winnings($w$), we multiply the probability of the event by the payoff if the even occurs:
$$w = 3*{1 \over 36} + 4*{1 \over 2} = {25 \over 12} \approx 2.08 $$</p>
<p>The average winnings per round will be about 2.08.</p>
|
1,148,674 | <p>I have always asked myself why this happens. </p>
<p>If $x = 4$, then $\sqrt{x} = 2$, but if I search for the $\sqrt{4}$, I get $2$ & $-2$.</p>
| Vladimir Vargas | 187,578 | <p>$\sqrt{\cdot}$ is a function. As a function it cannot return $2$ values. Actually mathematicians defined that function to show the positive root. What does it mean?</p>
<p>Suppose $x^2=b$. Then $x$ could take two vales: $\sqrt b$ and $-\sqrt b$. Both of them satisfy the equation. Meanwhile, if we have $x=\sqrt c$, then there is only $1$ value of $x$ that satisfy the equation, namely $\sqrt c$, which is a <strong>positive</strong> number.</p>
<p>If we put this more formally: Let $f:[0,\infty) \longrightarrow \mathbb R$ be a function defined by $x\longmapsto f(x)=\sqrt{x}$. We know that $f\subseteq \mathbb R\times \mathbb R:[(x,z),(x,y)\in f\Longrightarrow z=y]\wedge [(x,f(x))\in f\,\forall x \in [0,\infty)]$. Also $(x,\sqrt x)\in f$ iff $\sqrt x>0\wedge (\sqrt x)^2=x$. If we accept $(x,-\sqrt x)\in f$, then $-\sqrt x = \sqrt x$, and that's nonsense, the square root wouldn't be a function. However IT IS a function, and basically, it's just a definition that you take the positive root always.</p>
|
1,148,674 | <p>I have always asked myself why this happens. </p>
<p>If $x = 4$, then $\sqrt{x} = 2$, but if I search for the $\sqrt{4}$, I get $2$ & $-2$.</p>
| justsomeguy90 | 174,221 | <p>The map z->z^2 is taken to be ramification of order 2. Setting w=z^2 we have that the inverse function z=sqrt(w) has a branch point at zero, so that there is nontrivial monodromy about the origin in the sense of riemann surface theory. In this sense and modern algebraic senses, the square root fuction is a multivalued function. The convention is to choose the positive solution as the principle branch.</p>
<p>when talking about real numbers this means that sqrt(x^2)=±x, but we have made it a convention to select +x as the principal solution because of the most important applications, where physical measurements are positive.</p>
<p>In general, we have z^n=|w|*e^(2*i*pi) -> z=|w|^(1/n)*e^(2*i*(k/n)*pi) for k=0,1,...,n-1.</p>
<p>So, w=z^2 -> z=sqrt(|w|)<em>e^(i</em>k*pi) for k=0,1, which simplifies to z=±sqrt(|w|)</p>
<p>Also important, in the theory of generalized functions,</p>
<p>sqrt(z^2)=z*csgn(z)</p>
|
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