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 |
|---|---|---|---|---|
101,929 | <p>I am trying to do the following :-</p>
<pre><code>f[x_] := {x, 2};
Do[f[x_] :=Append[f[x], {x, 4}],{3}] % runs into recursion limit
</code></pre>
<p>because I want to grow my function (which is a list of functions) in a loop ? What is the right way to do this ?</p>
| djphd | 27,205 | <p>This can be done pretty neatly with <code>Nest</code> in which case everything is built before assigning it to your function.</p>
<pre><code>f[x_,n1_,n2_,n3_]:=Nest[Append[#,{x,n2}]&,{x,n1},n3]
f[x, 2, 4, 3]
{x, 2, {x, 4}, {x, 4}, {x, 4}}
</code></pre>
<p>Now you can set the constants and number of repetitions as you like each time. If you only want one or two things to change then just fix the <code>n</code>'s to constants.</p>
|
1,901,170 | <blockquote>
<p>Consider the region $D$ bounded by the positive $y$-axis, the line $y=8$ and the line $y=x^3$. Evaluate the following integral between $x=0$ and $x=2$.
$$\iint_D \frac{x^3y^2}4\ dx\ dy$$</p>
</blockquote>
<p>What are the limits and what is the solution? I've asked people and they all give conflicting ideas!</p>
| Dave L. Renfro | 13,130 | <p>Andres Mejia has recommended a couple of books (<strong>Div, Grad, Curl</strong> is especially nice), but his books are for vector <strong>calculus</strong> and from your question it sounds like what you want (now, at least) are books for vector <strong>geometry</strong>.</p>
<p>Consider <a href="http://rads.stackoverflow.com/amzn/click/0486466728" rel="nofollow noreferrer"><strong>Elementary Vector Geometry</strong></a> by Seymour Schuster.</p>
<p>I got a copy of this (the original printing; it's been reprinted by Dover Publications) several years ago, and it seemed to me that it would serve nicely as supplementary reading for just what you are taking. In fact, a few years ago I was tutoring someone in an engineering multivariable calculus course, and I found this book helpful.</p>
<p>Also, see <a href="http://rads.stackoverflow.com/amzn/click/0387964053" rel="nofollow noreferrer"><strong>Calculus of Several Variables</strong></a> by Serge Lang.</p>
<p>Lang has a reputation for writing books that are concise with a lot left to the reader (for others reading, in 1978-79 I had a couple of graduate algebra out of Lang's <a href="http://rads.stackoverflow.com/amzn/click/038795385X" rel="nofollow noreferrer"><strong>Algebra</strong></a>, so I'm also speaking from personal observation), but this particular book is quite the opposite in that it is much more straightforward and reader friendly than the typical multivariable calculus text. Lang's book also has a very gentle treatment of vectors. For comments about Lang's book, see <a href="http://mathforum.org/kb/message.jspa?messageID=7365390" rel="nofollow noreferrer">here</a> (21 January 2011 ap-calculus post of mine archived at Math Forum) and <a href="https://math.stackexchange.com/questions/31733/would-it-be-fine-to-use-serge-langs-two-calculus-books-as-textbooks-for-freshma">Would it be fine to use Serge Lang's two Calculus books as textbooks for freshman as Maths major?</a>.</p>
|
2,919,096 | <p>I am trying to solve a system of two second-order ODEs. After separating them, I obtained a fourth-order independent ODE as illustrated below. I wonder if there is a specific technique to solve it.</p>
<p>$$y^{(4)}+\frac{a_1}{x} y^{(3)}+\frac{a_2}{x^2}y^{(2)}+a_3y^{(2)}+\frac{a_4}{x}y^{(1)}+a_5y=0$$</p>
| Przemo | 99,778 | <p>This is a follow-up to the answer given by doraemonpaul above.Let us express the unknown function through its Laplace transform as follows:
<span class="math-container">\begin{equation}
y(x)=\int\limits_0^\infty e^{-x s} K(s) ds
\end{equation}</span>
By inserting the above into the ODE and then by integrating by parts in order to eleiminate the powers of <span class="math-container">$x$</span> from the integrand in the <span class="math-container">$s$</span>-domain we obtain the following equation:
<span class="math-container">\begin{eqnarray}
&&\frac{d^2}{d s^2} \left[s^4 K(s) \right] - a_1 \frac{d}{d s} \left[s^3 K(s)\right]+(a_2+a_3 \frac{d^2}{d s^2}) \left[s^2 K(s) \right] - a_4 \frac{d}{d s}\left[ s K(s) \right] + a_5 \frac{d^2}{d s^2} \left[ K(s) \right] =\\
&& -a_5 x K(0) - a_5 K^{'}(0)
\end{eqnarray}</span></p>
<p>Now let us for simplicity assume that <span class="math-container">$K(0)=K^{'}(0)=0$</span> which means that <span class="math-container">$y(x) = O(1/x^3)$</span> when <span class="math-container">$x\rightarrow \infty$</span>. Then we have:
<span class="math-container">\begin{eqnarray}
&&\frac{a_2 s^2 K(s) + \frac{d}{d s}\left( \frac{d}{d s}\left((a_5+a_3 s^2+s^4)K(s)\right) - (a_1 s^3 + a_4 s)K(s)\right)}{a_5+a_3 s^2+s^4}=\\
&&
\frac{K[s] \left(s^2 (-3 a_1+a_2+12)+2 a_3-a_4\right)}{a_3 s^2+a_5+s^4}-\frac{s K'(s) \left((a_1-8) s^2-4 a_3+a_4\right)}{a_3 s^2+a_5+s^4}+K''(s) =0
\end{eqnarray}</span></p>
<p>Then by substituting for <span class="math-container">$u=s^2$</span>, <span class="math-container">$d/d s = 2 \sqrt{u} d/d u$</span> and <span class="math-container">$d^2/d u^2 = 2 d/d u + 4 u d^2/d u^2$</span> we get a following ODE satisfied by the Laplace transform:
<span class="math-container">\begin{equation}
\frac{K[u] (u (-3 a_1+a_2+12)+2 a_3-a_4)}{4 u (u (a_3+u)+a_5)}+\frac{K'(u) (a_5-u ((a_1-9) u-5 a_3+a_4))}{2 u (u (a_3+u)+a_5)}+K''(u)=0
\end{equation}</span>
Now we use the common technique to eliminate the coefficient at the first derivative. We write <span class="math-container">$K(u) = M(u) \cdot V(u)$</span> where <span class="math-container">$M(u)$</span> is the exponential from minus one half times the antiderivative of the coefficient in question. In this case it reads:
<span class="math-container">\begin{equation}
M(u):=\frac{(u (a_3+u)+a_5)^{\frac{a_1-8}{8}} \exp \left(-\frac{(a_1 a_3-2 a_4) \tan ^{-1}\left(\frac{a_3+2 u}{\sqrt{4 a_5-a_3^2}}\right)}{4 \sqrt{4
a_5-a_3^2}}\right)}{\sqrt[4]{u}}
\end{equation}</span>
Now define:
<span class="math-container">\begin{eqnarray}
P^{(4)}(u)&:=&u^4 \left(-a_1^2+2 a_1+4 a_2+3\right)+\\
&&u^3 (a_3 (-2 a_1+4 a_2+6)-2 (a_1-3)a_4)+\\
&&u^2 \left(2 a_5 (-3 a_1+2 a_2+3)+3 a_3^2+2 a_3 a_4-a_4^2\right)+\\
&&u (6 a_3 a_5-2 a_4 a_5)+\\
&&3 a_5^2
\end{eqnarray}</span>
and then the function <span class="math-container">$V(u)$</span> satisfies the following ODE:
<span class="math-container">\begin{equation}
V^{''}(u)+\frac{P^{(4)}(u)}{16 u^2(a_5+a_3 u + u^2)^2}V(u)=0
\end{equation}</span>
which maps to the Heun equation <a href="https://en.wikipedia.org/wiki/Heun_function" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Heun_function</a> .</p>
|
910,552 | <blockquote>
<p>$\sin(nx)$ does not contain Cauchy subsequence in $L^p([0,2\pi]) $ for $1\leq p < \infty$</p>
</blockquote>
<p><strong>My attempt:</strong> </p>
<p>Set $f_n(x) = \sin(nx)$.</p>
<p>Argue by contradiction, suppose there exists a Cauchy subsequence $f_{n_k}$ in $L^p$ for some fixed $p$, then $f_{n_k}$ converges strongly to some $f \in L^p$, this $f$ must be the zero function since we know that $f_{n_k}$ converges weakly to zero in $L^p$. </p>
<p>But $f\equiv 0$ is impossible since $||f_{n_k}||_p$ is bounded below by a positive number. To see this, we look at the preimage
$$\{{|f_{n_k}|}^p \geq \big( \frac{\sqrt 2}{2}\big)^p \} = \{{|f_{n_k}|}\geq \frac{\sqrt 2}{2} \} $$
the measure of this set is bounded below by $\pi$. By Chebyshev's Inequality ${||f_{n_k}||_p}^p$ is bounded below by $\pi\big( \frac{\sqrt 2}{2}\big)^p $.</p>
<p>Is this okay? thank you very much!</p>
| Davide Giraudo | 9,849 | <p>The given arguments are good. I agree with Golbez, the fact that if $(f_{n_k})_k$ converge weakly to something, then it converges to $0$ has to be detailed, for example noticing that $\int_{[0,2\pi]} f_{n_k}g\mathrm d\lambda\to 0$ for each continuous function $g$ and concluding by a density argument. </p>
|
4,159,337 | <p>How would you show
<span class="math-container">$$\log_2 3 + \log_3 4 + \log_4 5 + \log_5 6 > 5?$$</span></p>
<p>After trying to represent the mentioned expression in the same base, a messy expression is created.
The hint mentioned that the proof can take help of quadratic equations.</p>
<p>Could you provide some input?
Is it a good idea to think of some graphical solution?</p>
| Boshu | 257,404 | <p><span class="math-container">$\log_2 3 + \log_3 4 + \log_4 5 + \log_5 6 > 4\times\sqrt[4]{\frac{\log 3}{\log 2}\times\frac{\log 4}{\log 3}\times\frac{\log 5}{\log 4}\times\frac{\log 6}{\log 5}}=4\times\sqrt[4]{\log_2 6}.$</span></p>
<p>Now, <span class="math-container">$5/4=1.25$</span>. So we are done if we can show that <span class="math-container">$1.25^4<2.45<\log_2 6$</span>. To see this, observe that <span class="math-container">$2^{2.5}=\sqrt{2^5}=\sqrt 32<6$</span>. That gives you the result.</p>
|
626,920 | <p>$a_n=\sum_{k=1}^{n} \frac{1}{n+k}=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}$</p>
<p>How to find $\lim a_n$?</p>
| WLOG | 21,024 | <p>With Eulero-Mascheroni : $$\sum_{k = 1}^{n}\frac{1}{k} - \log{n} \rightarrow \gamma$$ $$\sum_{k = 1}^{2n}\frac{1}{k} - \log{2n} \rightarrow \gamma$$ so $$\sum_{k = n+1}^{2n}\frac{1}{k} - \log{2n} +\log{n} \rightarrow 0$$ and then$$\sum_{k = n+1}^{2n}\frac{1}{k} \rightarrow \log{2}$$</p>
|
3,805,452 | <p>Given a stick and break it randomly at two places, what is the probability that you can form a triangle from the pieces?</p>
<p>Here is my attempt and the answer does not match, so I am confused what went wrong with this argument.</p>
<p>I first denote the two randomly chosen positions by <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>, and let <span class="math-container">$A=\max(X,Y)$</span>, <span class="math-container">$B=\min(X,Y)$</span>. We are interested in the probability of the event <span class="math-container">$\{A>\frac{1}{2}, B>A-\frac{1}{2}\}$</span>. Thus, we want the joint distribution of <span class="math-container">$A$</span> and <span class="math-container">$B$</span>. To compute that, I computed
<span class="math-container">$$F_{A,B}(w,z)=\mathbb{P}(A\leq w, B\leq z)=\mathbb{P}(A\leq w)-\mathbb{P}(A\leq w, B>z)=\mathbb{P}(X\leq w,Y\leq w)-\mathbb{P}(X\leq w, Y\leq w, X>z, Y>z)$$</span>
Therefore, we have if <span class="math-container">$z\leq w$</span>
<span class="math-container">$$F_{A,B}(w,z)=w^2-(w-z)^2$$</span>
otherwise
<span class="math-container">$$F_{A,B}(w,z)=w^2$$</span>
Then the joint density of <span class="math-container">$A$</span> and <span class="math-container">$B$</span> is
<span class="math-container">$$f_{A,B}(w,z)=\frac{\partial^2 F}{\partial w\partial z}(w,z)=2$$</span>
if <span class="math-container">$z\leq w$</span> and <span class="math-container">$0$</span> otherwise.<br />
Finally
<span class="math-container">$$\mathbb{P}(A>\frac{1}{2},B>A-\frac{1}{2})=\int_{\frac{1}{2}}^1\int_{w-\frac{1}{2}}^w2dzdw=\frac{1}{2}$$</span>
The answer is <span class="math-container">$\frac{1}{4}$</span> instead, but I can't figure out what went wrong with this argument.</p>
| Draditate | 819,566 | <p>Here is my view of this problem, not a debug of your procedure(sorry)</p>
<p>Actually I figure this 'forming triangle' problem by 'forming triangles'.</p>
<p><a href="https://i.stack.imgur.com/LxTB7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LxTB7.png" alt="enter image description here" /></a></p>
<p>Following your definition of X,Y, the joint choice of X,Y are uniform in the unit cell (density 1). The task remained is to find the cutting position of X and Y such that resulting 3 segments can form a triangle.</p>
<p>Consider case when Y > X (so three segments length are X, Y-X, 1-Y). Inserting the fact that sum of any two sides is greater than the third. The constraints are: <span class="math-container">$X + (Y-X) > 1- Y , X + (1-Y) > Y-X$</span> and <span class="math-container">$Y-X + 1-Y > X$</span>, resulting in <span class="math-container">$Y>0.5, X > Y - 0.5$</span> and <span class="math-container">$X < 0.5$</span>, which the boundary of Area A1 in my figure. With similar argument, you can sketch Area A2.</p>
<p>The geometric area of A1 and A2 are 1/8 respectively, sum these two, that is how I get 1/4.</p>
|
3,816,140 | <p>The points A, B, C, D have position vectors <strong>a</strong>, <strong>b</strong>, <strong>c</strong> and <strong>d</strong> respectively. Prove that the lines joining the midpoints of opposite edges of the tetrahedron ABCD bisect each other and give the position vector of the point of intersection.</p>
<p>I have started by working out the position vectors of E and F:</p>
<p><strong>b</strong> + <span class="math-container">$\vec BA$</span> = <strong>a</strong></p>
<p>so <span class="math-container">$\vec BA$</span> = <strong>a</strong> - <strong>b</strong> and position vector of E is <span class="math-container">$\frac{(a - b)}{2}$</span></p>
<p>and that for F is <span class="math-container">$\frac{d - c}{2}$</span>.</p>
<p>But I can proceed no further.</p>
<p><a href="https://i.stack.imgur.com/D7zGU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D7zGU.png" alt="enter image description here" /></a></p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</span>
<span class="math-container">\begin{align}
&\bbox[#ffd,5px]{\sum_{i = 1}^{n - 3}{{n - 3 \choose i}{n + i - 1 \choose i} \over i + 1}\,\pars{-1}^{i + 1}} =
\sum_{i = 1}^{n - 3}{{n - 3 \choose i} \over i + 1}\,\pars{-1}^{i + 1}\bracks{z^{i}}\pars{1 + z}^{n + i - 1}
\\[5mm] = &\
-\bracks{z^{0}}\pars{1 + z}^{n - 1}\sum_{i = 1}^{n - 3}{{n - 3 \choose i} \over i + 1}\pars{-\,{1 \over z} - 1}^{i}
\\[5mm] = &\
-\bracks{z^{0}}\pars{1 + z}^{n - 1}\int_{0}^{1}\sum_{i = 1}^{n - 3}{n - 3 \choose i}\bracks{-\pars{1 + {1 \over z}}t}^{i}\,\dd t
\\[5mm] = &\
-\bracks{z^{0}}\pars{1 + z}^{n - 1}\int_{0}^{1}
\braces{\bracks{1 - \pars{1 + {1 \over z}}t}^{n - 3}
- 1}\,\dd t
\\[5mm] = &\
-\bracks{z^{0}}\pars{1 + z}^{n - 1}\bracks{%
{z^{n + 1} - \pars{-1}^{n}z^{3} \over \pars{n - 2}z^{n}\pars{1 + z}} - 1}
\\[5mm] = &\
1 -
{1 \over n - 2}\bracks{z^{n}}\pars{1 + z}^{n - 2}
\bracks{z^{n + 1} - \pars{-1}^{n}z^{3}}
\\[5mm] = &\
1 + {\pars{-1}^{n} \over n - 2}{n - 2 \choose n - 3}
= \bbx{\large 1 + \pars{-1}^{n}\,,\quad n \geq 4} \\ &
\end{align}</span></p>
|
3,713,206 | <p>Spivak (3rd edition) proposes solving the integral <span class="math-container">$$\int \frac{1+e^x}{1-e^x} dx$$</span> by letting <span class="math-container">$u=e^x$</span>, <span class="math-container">$x=\ln(u)$</span>, and <span class="math-container">$dx=\frac{1}{u}du$</span>. This results in the integral <span class="math-container">$$\int \frac{1+u}{1-u}\frac{1}{u}du\\=\int \frac{2}{1-u}+\frac{1}{u}du=-2\ln(1-u)+\ln(u)=-2\ln(1-e^x)+x$$</span></p>
<p>From this example, Spivak argues that a similar method will work on any integral of the form <span class="math-container">$\int f(g(x))dx$</span> whenever <span class="math-container">$g(x)$</span> is invertible in the appropriate interval. Because this method is not a simple application of the substitution theorem, Spivak provides the following justification for his claim.</p>
<p>Consider continuous <span class="math-container">$f$</span> and <span class="math-container">$g$</span> where <span class="math-container">$g$</span> is invertible on the appropriate interval. Applying the above method to the arbitrary case, we let <span class="math-container">$u=g(x)$</span>, <span class="math-container">$x=g^{−1}(u)$</span>, and <span class="math-container">$dx=(g^{−1})′(u)du$</span>. Thus, we need to show that <span class="math-container">$$∫f(g(x))dx=∫f(u)(g^{−1})′(u)du$$</span> To prove this equality Spivak uses a more typical substitution <span class="math-container">$u=g(x)$</span>, <span class="math-container">$du=g′(x)dx$</span> and applies it by noting that <span class="math-container">$$∫f(g(x))dx=∫f(g(x))g′(x)\frac{1}{g′(x)}dx$$</span> Presumably using the substitution theorem, which roughly states that <span class="math-container">$∫f(g(x))g'(x)dx=∫f(u)du$</span>, Spivak asserts that <span class="math-container">$$∫f(g(x))g′(x)\frac{1}{g′(x)}dx=∫f(u)\frac{1}{g′(g^{−1}(u))}du$$</span> Then, because <span class="math-container">$(g^{-1})'(u)=\frac{1}{g'(g^{-1}(u))}$</span> Spivak concludes <span class="math-container">$$∫f(u)\frac{1}{g′(g^{−1}(u))}du=∫f(u)(g^{−1})′(u)du$$</span></p>
<p>I lose track of the argument when Spivak argues that <span class="math-container">$$∫f(g(x))g′(x)\frac{1}{g′(x)}dx=∫f(u)\frac{1}{g′(g^{−1}(u))}du$$</span> In the original example, it was clear to me how we could apply the substitution theorem to make this equality true because <span class="math-container">$\frac{1}{g'(x)}$</span> was in fact a function of <span class="math-container">$g(x)$</span> as <span class="math-container">$g'(x)=g(x)$</span>. But this is not necessarily true in all cases, or so it seems. How do we know that <span class="math-container">$f(g(x))\frac{1}{g'x}$</span> can be written in the form <span class="math-container">$h(g(x))$</span> for some continuous function <span class="math-container">$h$</span>? To sum up, my main questions is, how do we use the substitution theorem to justify the equality <span class="math-container">$$∫f(g(x))g′(x)\frac{1}{g′(x)}dx=∫f(u)\frac{1}{g′(g^{−1}(u))}du$$</span></p>
| Joe | 623,665 | <p>Believe it or not,
<span class="math-container">$$
\int f(g(x)) \, dx
$$</span>
is still of the correct form to admit substitution, provided that <span class="math-container">$g$</span> is invertible. To begin with, consider what we are aiming for: we know that any integral of the form
<span class="math-container">$$
\int \phi'(\mu(x))\mu'(x) \, dx
$$</span>
is equal to <span class="math-container">$\phi(\mu(x))+C$</span>, and making the substitution <span class="math-container">$u=\mu(x)$</span> works out just the same (hence why integration by substitution is valid in the first place). This means that if <span class="math-container">$f(g(x))$</span> can be written as <span class="math-container">$\phi'(\mu(x))\mu'(x)$</span> for some functions <span class="math-container">$\phi'$</span> and <span class="math-container">$\mu$</span>, then substitution is applicable. First, rewrite the integral as
<span class="math-container">$$
\int \frac{f(g(x))}{g'(x)}g'(x) \, dx
$$</span>
Now, the derivative of the inner function <span class="math-container">$g$</span> appears elsewhere, which is what we wanted. From here, it is simply a matter of rearrangement:
<span class="math-container">\begin{align}
\int \frac{f(g(x))}{g'(x)}g'(x) \, dx &= \int \frac{f(g(x))}{g'(g^{-1}(g(x)))}g'(x) \, dx \\
&= \int \phi'(g(x))g'(x) \, dx
\end{align}</span>
where <span class="math-container">$\phi'$</span> is the function defined by
<span class="math-container">$$
\phi'(x) = \frac{f(x)}{g'(g^{-1}(x))} \, .
$$</span>
(It should now be clear why it is necessary that <span class="math-container">$g$</span> is invertible.) We find that the integral is equal to
<span class="math-container">$$
\int f(g(x)) \, dx = \int \phi'(g(x))g'(x) \, dx = \phi(g(x)) + C
$$</span>
Compare this method with how we usually make substitutions. When we set <span class="math-container">$u=g(x)$</span> so that <span class="math-container">$x = g^{-1}(u)$</span> and <span class="math-container">$dx = (g^{-1})'(u) \, du$</span>, it may seem that we are doing something completely different to the method shown above. However, if we realise that
<span class="math-container">$$
(g^{-1})'(u) = \frac{1}{g'(g^{-1}(u))}
$$</span>
then it becomes clear that the two methods are virtually the same:
<span class="math-container">\begin{align}
\int f(g(x)) \, dx &= \int f(u) (g^{-1})'(u) \, du \\
&= \int \frac{f(u)}{(g'(g^{-1}(u))} \, du \\
&= \int \phi'(u) \, du \\
&= \phi(u) + C \\
&= \phi(g(x)) + C \, .
\end{align}</span></p>
|
764,437 | <p>I am trying to prove that $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$.</p>
<p>My initial plan was to use the first isomorphism theorem. I showed that there is a map $\phi: \mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$, given by $\phi(f)=f(i).$ This map is onto and homorphic. The part I have a question on is showing that the $ker(\phi) = (x^{2}+1)$. </p>
<p>One containment is trivial, $(x^2+1)\subset ker(\phi)$. To show $ker(\phi)\subset (x^2+1)$, let $f \in ker(\phi)$, then f has either $i$ or $-i$ as a root. Sot $f=g(x-i)(x+i)=g(x^2+1).$
How can I prove that $f \in \mathbb{Z}[x]\rightarrow g \in \mathbb{Z}[x]$? </p>
| Christoph | 86,801 | <p>Let me elaborate on Daniel Fischers comment. You have a ring homomorphism $\phi: \mathbb Z[x]\to\mathbb Z[i]$ given by $x\mapsto i$. Take $f \in \ker \phi$. By the division algorithm,
$$ f = q\cdot(x^2+1) + r,$$
where $\deg r < 2$ and $q,r\in\mathbb Z[x]$, since $x^2+1$ is monic. Applying $\phi$ to this equation yields
$$ 0 = \underbrace{\phi(q)\cdot(i^2+1)}_0 + \phi(r).$$
Since $r$ is of degree $<2$, we can write $r = ax+b$ for some $a,b\in\mathbb Z$. Then $\phi(r)=0$ gives
$$ai+b=0.$$
This equation in $\mathbb Z[i]$ implies $a=b=0$, so we have $r=ax+b=0\in\mathbb Z[x]$ and therefore
$$
f = q\cdot (x^2+1) \in \langle x^2+1\rangle.
$$
We conclude that $\ker \phi \subseteq \langle x^2+1\rangle$.</p>
|
3,984,322 | <p>I have the 2nd order homogeneous ODE:
<span class="math-container">$$y(x)''+\frac{2}{x}y'(x)-By=0 $$</span>
Where <span class="math-container">$B$</span> is a constant. Since I am a bit rusty with analytical solutions, I plugged it in MatLab's symbolic solver which found the following solution.
<span class="math-container">$$y(x)=C_1\frac{e^{\sqrt{B}x}}{x}-C_2\frac{e^{-\sqrt{B}x}}{2\sqrt{B}x}$$</span>
Unofrtunately, the constant B takes both positive and negative values, but since the ODE is derived from a physical problem I'd like the solution to always be real.
Any suggestion/comment is highly appreciated.</p>
| user577215664 | 475,762 | <p><span class="math-container">$$y''+\frac{2}{x}y'-\beta y=0$$</span>
<span class="math-container">$$xy''+{2}y'-\beta xy=0$$</span>
Substitute <span class="math-container">$z=\sqrt \beta x ( \beta >0)$</span>:
<span class="math-container">$$x{\beta} y''(z)+{2}{\sqrt {\beta}}y'(z)-\beta xy(z)=0$$</span>
<span class="math-container">$$z y''+2y'-zy=0$$</span>
<span class="math-container">$$(zy)''-zy=0$$</span>
Charactristic polynomial is:
<span class="math-container">$$r^2-1=0 \implies r= \pm 1$$</span>
<span class="math-container">$$\implies zy=c_1e^{-z}+c_2e^{z}$$</span>
<span class="math-container">$$\boxed {y(x)=\dfrac {1}{x} \left (k_1e^{-\sqrt {\beta}x}+k_2e^{\sqrt {\beta}x} \right)}$$</span></p>
<hr />
<p>For <span class="math-container">$\beta <0 $</span> :
<span class="math-container">$$xy''+{2}y'+\alpha xy=0$$</span>
Where <span class="math-container">$\alpha=-\beta $</span>.
Substitute <span class="math-container">$z=\sqrt \alpha x $</span>:
<span class="math-container">$$x{\alpha} y''(z)+{2}{\sqrt {\alpha}}y'(z)+\alpha xy(z)=0$$</span>
<span class="math-container">$$z y''+2y'+zy=0$$</span>
<span class="math-container">$$(zy)''+zy=0$$</span>
<span class="math-container">$$ \implies r^2+1=0 \implies r=\pm i$$</span>
<span class="math-container">$$\implies zy=c_1\cos z+c_2 \sin z$$</span>
Unsubstitute <span class="math-container">$z= \sqrt {\alpha } x$</span>.
<span class="math-container">$$\boxed {y(x)=\dfrac {1}{x} \left (k_1\cos {(\sqrt {\alpha}x)}+k_2 \sin ({\sqrt {\alpha}x}) \right)}$$</span></p>
|
99,617 | <p>How can I animate a point on a polar curve? I have used <code>Animate</code> and <code>Show</code> together before in order to get the curve and the moving point together on the same plot, but combining the polar plot and the point doesn't seem to be working because point only works with Cartesian coordinates.</p>
<p>Here is the code I used before to animate a point on a parametric curve. For higher values of <code>a</code> and <code>theta</code>, you can see the point moving along the curve better (I was required to animate all three parameters).</p>
<pre><code>Animate[
Show[
ParametricPlot[{a Cos[θ] t, a Sin[θ] t - 4.9 t^2}, {t, 0, 15}, AxesLabel -> {"x", "y"},
PlotRange -> {{0, 50}, {0, 30}}],
Graphics[{Red, PointSize[.05], Point[{a Cos[θ] t, a Sin[θ] t - 4.9 t^2}]}]
],
{t, 0, 5, Appearance -> "Labeled"},
{a, 1, 20, Appearance -> "Labeled"},
{θ, 0, Pi/2, Appearance -> "Labeled"},
AnimationRunning -> False
]
</code></pre>
<p>Here is the code I tried to use to animate a point on a polar curve, but the point does not even show up.</p>
<pre><code>Animate[
Show[
PolarPlot[2 Sin[4*θ], {θ, 0, 2 Pi}],
Graphics[Red, PointSize[Large], Point[{2 Sin[4*θ] Cos[θ], 2 Sin[4*θ] Sin[θ]}]]
],
{θ, 0, 2 Pi},
AnimationRunning -> False
]
</code></pre>
| m_goldberg | 3,066 | <p>It is not necessary to resort to <code>Show</code>. Your animation can easily be accomplished with <code>Epilog</code>.</p>
<pre><code>Animate[
PolarPlot[2 Sin[4 θ], {θ, 0, 2 Pi},
Epilog ->
{Red, PointSize[Large], Point[{2 Sin[4 θ] Cos[θ], 2 Sin[4*θ] Sin[θ]}]}],
{θ, 0, 2 Pi},
AnimationRunning -> False]
</code></pre>
|
99,617 | <p>How can I animate a point on a polar curve? I have used <code>Animate</code> and <code>Show</code> together before in order to get the curve and the moving point together on the same plot, but combining the polar plot and the point doesn't seem to be working because point only works with Cartesian coordinates.</p>
<p>Here is the code I used before to animate a point on a parametric curve. For higher values of <code>a</code> and <code>theta</code>, you can see the point moving along the curve better (I was required to animate all three parameters).</p>
<pre><code>Animate[
Show[
ParametricPlot[{a Cos[θ] t, a Sin[θ] t - 4.9 t^2}, {t, 0, 15}, AxesLabel -> {"x", "y"},
PlotRange -> {{0, 50}, {0, 30}}],
Graphics[{Red, PointSize[.05], Point[{a Cos[θ] t, a Sin[θ] t - 4.9 t^2}]}]
],
{t, 0, 5, Appearance -> "Labeled"},
{a, 1, 20, Appearance -> "Labeled"},
{θ, 0, Pi/2, Appearance -> "Labeled"},
AnimationRunning -> False
]
</code></pre>
<p>Here is the code I tried to use to animate a point on a polar curve, but the point does not even show up.</p>
<pre><code>Animate[
Show[
PolarPlot[2 Sin[4*θ], {θ, 0, 2 Pi}],
Graphics[Red, PointSize[Large], Point[{2 Sin[4*θ] Cos[θ], 2 Sin[4*θ] Sin[θ]}]]
],
{θ, 0, 2 Pi},
AnimationRunning -> False
]
</code></pre>
| MarcoB | 27,951 | <p>Conceptually you are there already; you only had a small syntax error in your code, i.e. <strong>a missing set of braces</strong> around the items in the <code>Graphics</code> expression:</p>
<pre><code>Animate[
Show[
PolarPlot[2 Sin[4*t], {t, 0, 2 Pi}],
Graphics[{Red, PointSize[0.02], Point[{2 Cos[θ] Sin[4 θ], 2 Sin[θ] Sin[4 θ]}]}]
],
{θ, 0, 2 Pi},
AnimationRunning -> False
]
</code></pre>
<p>The problem is more visible if you temporarily comment out the <code>PolarPlot</code> expression from your code; the resulting <code>Graphics</code> will then complain of an incorrect syntax issue, which is what led me to discover the problem.</p>
<p>As a small improvement, I would suggest that you write your transformed point as:</p>
<pre><code>Point[Sin[4 θ] {Cos[θ], Sin[θ]}]
</code></pre>
<p>This way you can avoid having to rewrite your polar function twice.</p>
|
3,786,797 | <p>I have two variables <span class="math-container">$x$</span> and <span class="math-container">$y$</span> which are measured at two time steps <span class="math-container">$0$</span> and <span class="math-container">$1$</span>. Available to me is the percent change from the first timestep to the second, relative to the first i.e. <span class="math-container">$\frac{x_1 - x_0}{x_0}$</span> and <span class="math-container">$\frac{y_1 - y_0}{y_0}$</span>. I would like to recover the ratio <span class="math-container">$\frac{x_1}{y_1}$</span>. I've tried adding 1 = <span class="math-container">$\frac{x_0}{x_0} = \frac{y_0}{y_0}$</span> to the two percent changes and then dividing them but this only gets me to <span class="math-container">$\frac{x_1}{y_1} \cdot \frac{y_0}{x_0}$</span>. It seems to me that my problem has 2 equations and 4 unknowns, even though 2 of the unknowns don't appear in the desired ratio, making it have infinite solutions. Is that correct? Otherwise, is there another manipulation I can make to get my desired ratio?</p>
| N. F. Taussig | 173,070 | <p><strong>Method 1:</strong> Arrange the five women in a row. This can be done in <span class="math-container">$5!$</span> ways. Doing so creates six spaces in which we can place the men, four between successive pairs of women and two at the ends of the row.</p>
<p><span class="math-container">$$\square W \square W \square W \square W \square W \square$$</span></p>
<p>To ensure that no two of the men are adjacent, we must select five of these six spaces in which to place the men, which can be done in <span class="math-container">$\binom{6}{5}$</span> ways. The men can be arranged in the selected spaces in <span class="math-container">$5!$</span> ways. Hence, the number of seating arrangements of five men and five women in which no two of the men are seated next to each other is
<span class="math-container">$$5!\binom{6}{5}5! = 86400$$</span></p>
<p><strong>Method 2:</strong> We use the <a href="https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle" rel="nofollow noreferrer">Inclusion-Exclusion Principle</a> to correct your attempt, which is incorrect since you are subtracting cases in which there is more than one pair of adjacent men multiple times.</p>
<p>Rather than arranging men and women, we will count the number of arrangements of five Ms and five Ws in a row in which no pair of Ms are consecutive, then multiply our result by the number of ways of arranging five men in the positions occupied by the Ms and the number of ways of arranging five women in the positions occupied by the the Ws.</p>
<p>An arrangement of five Ms and five Ws is completely determined by choosing which five of the ten positions are occupied by the Ms since the remaining five positions must be occupied by Ws. Hence, there are
<span class="math-container">$$\binom{10}{5}$$</span>
possible arrangements of five Ms and five Ws. From these, we must subtract those arrangements in which one or more pairs of Ms are adjacent.</p>
<p><em>A pair of Ms is adjacent</em>: We have nine objects to arrange: MM, M, M, M, W, W, W, W, W. Choose one of the nine positions for MM and three of the remaining eight positions for the three single Ms. This can be done in
<span class="math-container">$$\binom{9}{1}\binom{8}{3}$$</span></p>
<p>However, if we subtract this from the total, we will have subtracted too much since we will have subtracted each arrangement in which there are two pairs of adjacent Ms twice, once for each way of designating one of those pairs as the pair of adjacent Ms. Since we only want to subtract these cases once, we must add them to the total.</p>
<p><em>Two pairs in which a pair of Ms is adjacent</em>: This can occur in two ways. The pairs may be overlapping (three consecutive Ms) or disjoint.</p>
<p>Two overlapping pairs of adjacent Ms: We have eight objects to arrange: MMM, M, M, W, W, W, W, W. Choose one of the eight positions for MMM and two of the remaining six positions for the two single Ms, which can be done in
<span class="math-container">$$\binom{8}{1}\binom{6}{2}$$</span>
ways.</p>
<p>Two disjoint pairs of adjacent Ms: We again have eight objects to arrange: MM, MM, M, W, W, W, W, W. Choose two of the eight positions for the MMs and one of the remaining six positions for the single M, which can be done in
<span class="math-container">$$\binom{8}{2}\binom{6}{1}$$</span>
ways.</p>
<p>However, if we add these cases to the total, we will not have subtracted those cases in which there are three pairs of adjacent Ms. This is because we subtracted them three times, once for each of the three ways of designating one of the pairs of adjacent Ms as the pair of adjacent Ms, and added them three times, once for each of the <span class="math-container">$\binom{3}{2}$</span> ways of designating two of the three pairs of adjacent Ms as the two pairs of adjacent Ms. Thus, we must subtract them from the total.</p>
<p><em>Three pairs in which a pair of Ms is adjacent</em>: This can occur in two ways. The three pairs overlap (in which case there are four consecutive Ms) or there are two overlapping pairs and another pair which is disjoint from the overlapping pairs.</p>
<p>Three overlapping pairs: We have seven objects to arrange: MMMM, M, W, W, W, W, W. Choose one of the seven positions for the MMMM and one of the remaining six positions for the M, which can be done in
<span class="math-container">$$\binom{7}{1}\binom{6}{1}$$</span>
ways.</p>
<p>Two overlapping pairs and a pair disjoint from the overlapping pairs: We again have seven objects to arrange: MMM, MM, W, W, W, W, W. Choose one of the seven positions for the MMM and one of the remaining six positions for the MM, which can be done in
<span class="math-container">$$\binom{7}{1}\binom{6}{1}$$</span>
ways.</p>
<p>However, if we subtract these cases from the total, we will have subtracted too much. This is because we have subtracted each case in which there are four pairs of consecutive Ms twice. This is because we first subtracted such cases four times, once for each ways of designating one of those pairs as the pairs of adjacent Ms; then added them six times, once for each of the <span class="math-container">$\binom{4}{2}$</span> ways of designating two of those four pairs as the two pairs of adjacent Ms; then subtracted them four times, once for each of the <span class="math-container">$\binom{4}{3}$</span> ways of designating three of those four pairs as the three pairs of adjacent Ms. Since we only want to subtract the cases in which there are four pairs of adjacent Ms once, we must add them to the total.</p>
<p><em>Four pairs of adjacent Ms</em>: Since there are a total of five Ms, this can only occur if all five Ms are consecutive. We have six objects to arrange: MMMMM, W, W, W, W, W. Choose which of the six positions is occupied by MMMMM, which can be done in
<span class="math-container">$$\binom{6}{1}$$</span>
ways.</p>
<p>By the Inclusion-Exclusion Principle, the number of arrangements of M, M, M, M, M, W, W, W, W, W in which no two Ms are consecutive is
<span class="math-container">$$\binom{10}{5} - \binom{9}{1}\binom{8}{3} + \binom{8}{1}\binom{6}{2} + \binom{8}{2}\binom{6}{1} - \binom{7}{1}\binom{6}{1} - \binom{7}{1}\binom{6}{1} + \binom{6}{1} = 6$$</span></p>
<p>Multiplying by the <span class="math-container">$5!$</span> ways we can arrange the men in the positions occupied by Ms and the <span class="math-container">$5!$</span> ways we can arrange the women in the positions occupied by Ws gives
<span class="math-container">$$\left[\binom{10}{5} - \binom{9}{1}\binom{8}{3} + \binom{8}{1}\binom{6}{2} + \binom{8}{2}\binom{6}{1} - \binom{7}{1}\binom{6}{1} - \binom{7}{1}\binom{6}{1} + \binom{6}{1}\right]5!5! = 86,400$$</span>
as above.</p>
|
3,585,111 | <p>Given that <span class="math-container">$x^{\frac{2}{3}}=({x^2})^{\frac{1}{3}}$</span>, I thought that <span class="math-container">$f(x)$</span> is continuous in <span class="math-container">$\mathbb{R}$</span> but when I plot this function, I get something like this in wolfram.</p>
<p><a href="https://i.stack.imgur.com/pwL67.png" rel="nofollow noreferrer">enter image description here</a></p>
<p>What happens to <span class="math-container">$f(-5)$</span>? is not <span class="math-container">$f(-5)=25^{\frac{1}{3}}$</span>. I find this plot odd. There is nothing for negative <span class="math-container">$x$</span> values. I thought the graph should be symmetrical about the <span class="math-container">$y$</span> axis.</p>
<p>So does it mean <span class="math-container">$f$</span> is not continuous on all of <span class="math-container">$\mathbb{R}$</span>?</p>
| Deepak | 151,732 | <p>Not sure if this is the issue but, allowing for complex answers, there are actually three cube roots of every real number (except <span class="math-container">$0$</span>). For negative real numbers, the principal cube root (one with the smallest argument) is a complex number with non-zero imaginary part. Some computer and calculator implementations are programmed to return this principal root when a cube root function is called. Since it's not real, it cannot be plotted on the real axis.</p>
<p>Also, when you take the exponent of <span class="math-container">$\frac 23$</span>, there may be differing implementations of how you execute that. There is no reason to expect the value to be squared first (giving non-negative values) before taking the cube root, which is why the above "issue" manifests.</p>
|
33,125 | <p>For some fixed $n$ define the quadratic form $$Q(x,y) = x^2 + n y^2.$$</p>
<p>I think that if $Q$ represents $m$ in two different ways then $m$ is composite.</p>
<p>I can prove this for $n$ prime. I was hoping someone could give me a hint towards proving this result for general $n$? Also would be interested in generalizations if any are known! Thanks a lot.</p>
| Gerry Myerson | 8,269 | <p>John Brillhart published a paper about this in the American Mathematical Monthly some time in the past year. </p>
|
381,313 | <p>Solve the limits of below:</p>
<p>(1) $\lim\limits_{n \to \infty} \int_0^n (1+\frac{x}{n})^n e^{-2x}dx$.</p>
<p>(2) $\lim\limits_{n \to \infty} \int_0^n (1-\frac{x}{n})^n e^{\frac{x}{2}}dx$. </p>
<p>(3) $\int_0^{\infty} \frac{e^{-x}\sin^2 x}{x}dx$</p>
<p>(Hint: show $f(x,y)=e^{-x}\sin2xy$ is integrable on $[0,\infty) \times [0,1]$)</p>
| Henry T. Horton | 24,934 | <p>Definitely not.</p>
<p>Here's a simple example. Depending on the choice of transition functions, a bundle of the form
$$[-1,1] \hookrightarrow E \to S^1$$
and structure group $\Bbb Z/2 = \{-1,1\}$ (acting on the fiber $[-1,1]$ via multiplication) can have total space a cylinder or a Möbius band.</p>
<p><strong>Edit:</strong> Let $[-1,1] \hookrightarrow E \to S^1$ be a fiber bundle with structure group $\Bbb Z/2 = \{-1, 1\}$ as above. I'll construct the cases where $E$ is the cylinder $S^1 \times [-1,1]$ and where $E$ is the Möbius band using transition functions.</p>
<p>Suppose that $E$ is trivialized over $U_1 = (-\varepsilon, \pi +\varepsilon)$ and $U_2 = (\pi-\varepsilon, \varepsilon)$ (here $\varepsilon$ is just a small positive number and we are thinking of $S^1$ as $\Bbb R/2\pi\Bbb Z$). We have that
$$U_1 \cap U_2 = (-\varepsilon, \varepsilon) \cup (\pi-\varepsilon, \pi+\varepsilon).$$
Assume that on the component $(\pi-\varepsilon, \pi+\varepsilon)$ of $U_1 \cap U_2$, the transition function is given by
$$\theta_{12}(x) = 1 \text{ for all } x \in (\pi-\varepsilon, \pi+\varepsilon).$$
Since a transition function must be continuous and $\Bbb Z/2$ is disconnected, we see that the only possibilities for $\theta_{12}|_{(-\varepsilon,\varepsilon)}$ are
\begin{align*}
\text{(a)} & \quad \theta_{12}(x) = 1 \text{ for all } x \in (-\varepsilon,\varepsilon), \\
\text{(b)} & \quad \theta_{12}(x) = -1 \text{ for all } x \in (-\varepsilon,\varepsilon).
\end{align*}
Case (a) gives the trivial bundle over $S^1$. In case (b), we have that the total space $E$ is the Möbius band.</p>
<p>To visualize the two cases (a) and (b) above, note that based on our choices so far, we have a strip $[-\varepsilon,2\pi+\varepsilon] \times [-1,1]$ and we need to glue the two ends $[-\varepsilon,0] \times [-1,1]$ and $[2\pi, 2\pi+\varepsilon] \times [-1,1]$ together to obtain $E$. Choice (a) just identifies the two ends trivially and gives us the cylinder $S^1 \times [-1,1]$. Choice (b) makes us rotate one of the ends before gluing, introducing a twist and hence giving us a Möbius band.</p>
<p>If we fix the base space $B$, fiber $F$, structure group $G$, and transition functions $\{\theta_{\beta\alpha}\}$, then the total space $E$ <strong>is</strong> determined. Essentially, $E$ is made up of pieces $U_\alpha \times F$ where $\{U_\alpha\}$ is an open cover of $B$, and the transition functions $\{\theta_{\beta\alpha}\}$ tell us how to glue together these pieces along overlaps where $U_\alpha \cap U_\beta \neq \varnothing$. The transition functions are just a recipe for gluing together the locally trivial pieces of the bundle to obtain $E$.</p>
|
2,348,641 | <p><span class="math-container">$f$</span> is a periodic continuous function of period <span class="math-container">$T > 0 $</span>.</p>
<ol>
<li><p>Let <span class="math-container">$ a > 0$</span>, prove that: <span class="math-container">$ \int_{0}^{+ \infty} f(t)e^{-at}dt $</span> is absolutely convergent.</p>
</li>
<li><p>Find a positive constant <span class="math-container">$ \lambda $</span> dependent only on <span class="math-container">$a$</span> and <span class="math-container">$T$</span> such that:
<span class="math-container">$$ \int_{0}^{+ \infty} f(t)e^{-at}dt = \lambda \int_{0}^{T}f(t)e^{-at}dt $$</span></p>
</li>
<li><p>Find the expression of <span class="math-container">$ \mu $</span> that is dependent only on <span class="math-container">$ a$</span> and <span class="math-container">$T$</span> such that:</p>
</li>
</ol>
<p><span class="math-container">$$\left|\int_{0}^{T}f(t)e^{-at}\,dt\right| \leq \mu\sqrt{\int_{0}^{T}f^{2}(t)e^{-at}\,dt} $$</span></p>
<ol start="4">
<li>Find <span class="math-container">$ \lim_{a \to + \infty } f(t) e^{-at} dt$</span>.</li>
</ol>
<hr />
<ol>
<li><p><span class="math-container">$f$</span> is continuous on the closed interval <span class="math-container">$[0, T]$</span>, there is an <span class="math-container">$M \in \mathbb{R}$</span> such that:</p>
<p><span class="math-container">$|f(t)| < M \implies |f(t) e^{-at}| < Me^{-at} $</span></p>
</li>
</ol>
<p><span class="math-container">$ \int_{0}^{+ \infty}Me^{-at}$</span> is convergent, by comparison test, so does <span class="math-container">$\int_{0}^{+ \infty}|f(t) e^{-at}|$</span>, which proves that <span class="math-container">$ \int_{0}^{+ \infty} f(t)e^{-at}dt $</span> is absolutely convergent. Is this correct?</p>
<p>I also need help with question 2, 3 and 4.</p>
<p>Thank you for your help.</p>
| Jack D'Aurizio | 44,121 | <p>$4$. Since $f$ is a bounded function the given limit equals $0$ for any $t>0$;</p>
<p>$3$. By the Cauchy-Schwarz inequality</p>
<p>$$ \left|\int_{0}^{T}f(t)\,e^{-at}\,dt\right|\leq \sqrt{\int_{0}^{T}e^{-at}\,dt\int_{0}^{T}f(t)^2 e^{-at}\,dt} $$
hence we may take $\mu$ as $\sqrt{\frac{1-e^{-aT}}{a}}$;</p>
<p>$2$. $$\int_{0}^{+\infty}f(t)e^{-at}\,dt = \int_{0}^{T}f(t)\left(e^{-at}+e^{-a(t+T)}+\ldots\right)\,dt $$
hence we may take $\lambda$ as $\frac{1}{1-e^{-aT}}$.</p>
|
1,540,082 | <p>I need to prove that the following sequence converges as $n\to\infty$:
$\sum\limits_{k=n+1}^{2n} 1/k$</p>
<p>The problem is that I've only ever seen sums from i to n for example. I'm confused because not only is 2n alien to me, but also k and n are related.</p>
<p>So far I've tried substituting k for n+1 to simplify the expression, but it led nowhere. I also tried to enter this into spreadsheet calculation in hopes to get an idea where the border might be, but the result just grows and grows. I almost expected this because the sequence is reminiscent of the harmonic series which diverges. The key difference is obviously the beginning and end of the sum, but I can't figure out how to tackle this.</p>
<p>Any ideas or pointers are appreciated</p>
| Eric Wofsey | 86,856 | <p>Let us write $a_n=\sum_{k=n+1}^{2n} 1/k$, and let's compare $a_n$ to $a_{n+1}$. They have almost the same terms, except $a_n$ has one more term at the beginning and $a_{n+1}$ has two more terms at the end. We get that $$a_{n+1}-a_n=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}>\frac{1}{2n+2}+\frac{1}{2n+2}-\frac{1}{n+1}=0.$$</p>
<p>Thus the numbers $a_n$ are increasing. Now note that for each $n$, every term in the sum for $a_n$ is at most $1/(n+1)$, and there are $n$ terms, so $a_n\leq n/(n+1)<1$. So the sequence $a_n$ is monotone increasing and bounded above by $1$. It follows that it must converge.</p>
|
3,279,965 | <p>Let <span class="math-container">$A$</span> be a non-symmetric <span class="math-container">$n\times n$</span> real matrix. Assume that all eigenvalues of <span class="math-container">$A$</span> have positive real parts. Could you please show some simple conditions such that <span class="math-container">$A+A^T$</span> is positive definite? Thank you so much.</p>
| Community | -1 | <p>As the OP writes, that follows is a class of <span class="math-container">$A$</span> s.t. <span class="math-container">$A+A^T>0$</span>.</p>
<p>Choose <span class="math-container">$A=[a_{i,j}]$</span> s.t. for every <span class="math-container">$i$</span>, </p>
<p><span class="math-container">$2a_{i,i}>\sum_{j\not= i}(|a_{i,j}+a_{j,i}|)$</span>;</p>
|
3,779,674 | <p>I've been working on this question for some time now, and have not had any significant progress towards a solution:</p>
<blockquote>
<p>Let <span class="math-container">$\Omega = \{z = x +iy \in \mathbb{C} : |y| \leq 1\}.$</span> If <span class="math-container">$f(z) = z^2 +2$</span>, then draw a sketch of
<span class="math-container">$$f(\Omega) = \{ f(z) : z \in \Omega\}.$$</span>
Justify your answer.</p>
</blockquote>
<p>Naturally I first determined that the set <span class="math-container">$\Omega$</span> essentially refers to the strip of complex numbers in the plane which have their imaginary part between <span class="math-container">$i$</span> and <span class="math-container">$-i$</span>, so I then set about attempting to plot the particualar set. However, I'm not sure how to go about drawing this — I tried 'mapping' random points on the curve by drawing arrows from <span class="math-container">$z$</span> to <span class="math-container">$f(z)$</span>, and found some success:</p>
<ul>
<li>The purely imaginary numbers 'map' to the region <span class="math-container">$[1, 2)$</span> on the real line.</li>
<li>The purely real numbers 'map' to the region <span class="math-container">$[0, \infty)$</span> on the real line.</li>
<li>The complex numbers of the form <span class="math-container">$x+i$</span> trace out the path of the curve <span class="math-container">$y=2\sqrt{x+1}$</span>, as would be expected.</li>
</ul>
<p>After this, though, I'm having a really hard time trying to find any sort of pattern between these observations, and I'm nowhere near trying to find the drawing of the curve as I'd like to be. The closest I've come is this crude sketch on Desmos.</p>
<p> <a href="https://i.stack.imgur.com/jVIa4l.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jVIa4l.png" alt="Sketch" /></a></p>
<hr />
<p>Just as an add-on, this question is expected to be solved on a normal sheet of paper. It appeared in <a href="https://www.isical.ac.in/%7Eadmission/IsiAdmission2017/PreviousQuestion/BStat-BMath-UGB-2019.pdf" rel="nofollow noreferrer">this test</a> last year which was expected to be solved by high-schoolers.</p>
| Ben Grossmann | 81,360 | <p>I am not sure how fruitful your approach will be. One method that will work is as follows: with polynomial long-division, we find that
<span class="math-container">$$
(t + 1)^{10}(t - 1)^2 + t = q(t)(t^3 - t^2 - 2t + 1) + 4631 t^2 + 3721 t - 2560.
$$</span>
Thus, we have
<span class="math-container">$$
(A + I)^{10}(A - I)^2 + A = 4631 A^2 + 3721A - 2560 I.
$$</span></p>
|
2,232,677 | <p>A parallellogram $ABCD$ is given, where $E$ is the midpoint of $BC$ and $F$ is a point on $AD$ so that $|FD| = 3|AF|$. $G$ is the point where $BF$ and $AE$ intersects.</p>
<p>Express the vector $AG$ in terms of vectors $AB$ and $AD$.</p>
<p>My solution to the problem is the following: Impose an <em>affine transformation</em> so that $ABCD$ becomes a unit square with point $A$ residing on origo. Then $BF$ is on the line $y = \frac{1}{4} - \frac{1}{4}x$ and $AE$ on $y = \frac{1}{2}x$. The lines' intersection gives me the required coefficients to express $AG$ in terms of $AB$ and $AD$. </p>
<p>My question is how would you solve this problem without the affine transformation? The reason I'm asking is because this problem was given at an early stage of the course before affine transformations was introduced. So I want to know if there is a simpler or "more intuitive" way to solve it which I haven't learned.</p>
| Ted Shifrin | 71,348 | <p>Sure: My standard way of teaching this material is to set $\overrightarrow{AB}=\vec x$ and $\overrightarrow{AD} = \vec y$ and write everything appropriate in terms of those vectors. In particular, $\overrightarrow{AE} = \vec x + \frac12\vec y$ and $\overrightarrow{AF} = \frac14\vec y$. We want to find $t$ so that $\vec x+t(\frac14 y-\vec x)$ is a scalar multiple of $\vec x+\frac12\vec y$.
Setting $$\vec x+t(\tfrac14 y-\vec x)=c(\vec x+\tfrac12\vec y)$$
and using linear independence of $\vec x$ and $\vec y$, it is easy to finish.</p>
|
1,575,920 | <p>The global error of $\int f(x) \mathrm{d}x$, between two
$x$-values by the trapezoidal rule is
$-(1/12)h^3f''(ξ)$</p>
<p>a) $f(x) = x^3, x := [0.2,0.5]$</p>
<p>Find the value for $ξ$. Not really sure where to start with this problem. We never did any examples in class.</p>
| marty cohen | 13,079 | <p>Do this:</p>
<p>Find the actual value of the integral.</p>
<p>Find the trapezoidal rule estimate.</p>
<p>Get the difference between the two.
That is the error.</p>
<p>Since the error is of the form
$-(1/12)h^3f''(ξ)$,
and you know $h$ and $f$
(and therefore $f''$),
find $ξ$
from the error,
$h$,
and $f''$.</p>
|
2,911,849 | <p>Let $S_i$ be a countable family of sets and $f$ a continuous map (say everything is in the complex plane).
Is it true that
$$
f( \overline{ \cup_i S_i } ) =
\overline{ \cup_i f( S_i ) }
$$
or perhaps the inclusion $\subseteq$?</p>
| Community | -1 | <p>The identity is not even true for <em>one</em> set, because some continuous functions simply do not send closed sets to closed sets. Take, for instance, $f(z)=i\Im z+\arctan\Re z$, which sends $\Bbb C$ to $\left\{z\in\Bbb C\,:\, \lvert\Re z\rvert<\frac\pi2\right\}$. The $\subseteq $ is true, because $\overline{\bigcup_i f[S_i]}$ is a closed set containing $f\left[\bigcup_i S_i\right]=\bigcup_i f[S_i]$, and so by continuity $f^{-1}\left[\overline{\bigcup_i f[S_i]}\right]$ must be a closed set. Since $f^{-1}[A]\subseteq f^{-1}[B]$ whenever $A\subseteq B$, it must also contain $f^{-1}\left[f\left[\bigcup_i S_i\right]\right]=\bigcup_i S_i$. Finally, since $f^{-1}\left[\overline{\bigcup_i f[S_i]}\right]\supseteq \bigcup_i S_i$, $$\bigcup_i f[S_i]=f\left[\bigcup_i S_i\right]\subseteq f\left[f^{-1}\left[\overline{\bigcup_i f[S_i]}\right]\right]=\operatorname{im}f\cap\overline{\bigcup_i f[S_i]}\subseteq \overline{\bigcup_i f[S_i]}$$</p>
|
77,980 | <p>How do I check whether a give function is 'algebraic' or not? I have a function $m(z) = 2\pi i z^n$ where $z \in \mathbb{C} \backslash \mathbb{R}$ and $n \in \mathbb{Z}$. I can write this as $w - 2\pi i z^n = 0$ where $w = m(z)$ and it is a polynomial. I think this is an algebraic function? Is my understanding correct?</p>
<p>What is the general approach to find whether a given function is algebraic or not?</p>
| Cyril Banderier | 85,924 | <p>If $F(z)=z^\gamma$ is an algebraic function (over $\mathbb Q$),
then $F'(z)$ is also an algebraic function (over $\mathbb Q$),
and thus $F'(1)$ is an algebraic number (over $\mathbb Q$).
Therefore, $z^\gamma$ algebraic implies $\gamma$ algebraic.
Equivalently, if $\gamma$ is not algebraic, then $z^\gamma$ is not algebraic.</p>
<p>In some cases, one knows a Puiseux series expansions of a function $F(z)$;
such an expansion is unique and is of the type
$$F(z) = \sum_{k\geq k_0} f_k \times \big((z-z_0)^{r} \big)^k \,.$$
What is more, the theory of Newton polygon for algebraic functions implies that the exponent $r$ in this Puiseux expansion is then a rational number. This gives a nice way to test if $F(z)$ is algebraic or not!</p>
<p>Therefore, one can say that $z^\gamma$ is an algebraic function if and only if $\gamma$ is a rational number (the same reasoning proves that $z^{\sqrt 2}$ is not an algebraic function). [It just requires a little lemma showing that $z^\gamma$ cannot be rewritten as an infinite sum of $z^{rk}$, which follows easily by considering the limit when $z\rightarrow 0$.]</p>
|
145,868 | <p>I have a data set like below:</p>
<pre><code>368 150 0.31895877
369 150 0.32141133
370 150 0.3190167
371 150 0.33890757
372 150 0.3649395
373 150 0.37585051
374 150 0.37916244
375 150 0.37619812
376 150 0.37287654
377 150 0.37249031
378 150 0.37102263
379 150 0.34839918
380 150 No Data
381 150 No Data
382 150 No Data
383 150 No Data
384 150 No Data
385 150 No Data
386 150 No Data
387 150 No Data
388 150 No Data
389 150 No Data
390 150 No Data
391 150 No Data
392 150 No Data
</code></pre>
<p>This data set sometimes contains "No Data" and I want to ListPointPlot3D with this data.</p>
<p>Therefore, I imported this data and plotted it:</p>
<pre><code>data=Import["datafile.txt","Table"];
ListPointPlot3D[data]
</code></pre>
<p>As I expected, I got error about the invalid list.</p>
<p>How can I plot the graph with these No Data rows eliminated?</p>
<hr>
<p>This is a sample file which I want to plot with.</p>
<p><a href="https://www.taiki-home.net/nextcloud/index.php/s/iyy60PCz5niXBTW" rel="nofollow noreferrer">datafile.txt</a></p>
| halirutan | 187 | <p>As you said in the comments, the data for your histograms did contain negatives and you squared all of them. I suggest to consider looking at your values before you square them</p>
<p><img src="https://i.stack.imgur.com/0HVJp.png" alt="Mathematica graphics"></p>
<p>Although I have no insight into the underlying process that created the data, one could assume that you have 4 mixed normal distributions here.
It seem the left peak is again divide into two distributions.
Let us use this for a start and define a mixture distribution:</p>
<pre><code>dist = MixtureDistribution[{a, b, c,
d}, {NormalDistribution[μ1, σ1],
NormalDistribution[μ2, σ2],
NormalDistribution[μ3, σ3],
NormalDistribution[μ4, σ4]}]
</code></pre>
<p>Now we can estimate the parameters by using <code>FindDistributionParameters</code>. I have roughly estimated the initial the initial conditions by just looking where the peaks are, and how high and wide they are:</p>
<pre><code>sol =
FindDistributionParameters[data,
dist, {{a,
1}, {μ1, -.4}, {σ1, .1}, {b, .3}, {μ2, .1}, {\
σ2, .05}, {c, .5}, {μ3, .4}, {σ3, .1}, {d, .4}, {\
μ4, -.2}, {σ4, .1}}]
(* {a -> 0.376702, b -> 0.125485, c -> 0.275036,
d -> 0.222777, μ1 -> -0.395739, σ1 -> 0.0586256, μ2 ->
0.103749, σ2 -> 0.0496838, μ3 -> 0.337538, σ3 ->
0.0866439, μ4 -> -0.217508, σ4 -> 0.0911891} *)
</code></pre>
<p>With 250.000 data values, Mathematica has no problem finding a good fit. Looking at it reveals:</p>
<pre><code>pdf1 = PDF[dist, x] /. sol;
Show[
Histogram[data, {0.03}, "PDF"],
Plot[PDF[dist, x] /. sol, {x, -.6, .6}]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/8gNez.png" alt="Mathematica graphics"></p>
<p>Now, since you squared all your values, we need to transform the found mixture distribution as well and you will end up with a distribution that has only values for positive x and fits your data very nicely:</p>
<pre><code>pdf = PDF[TransformedDistribution[u^2, u \[Distributed] dist], x] /. sol;
Show[
Histogram[data^2, {0.007}, "PDF"],
Plot[pdf, {x, 0, .6}]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/IaxSv.png" alt="Mathematica graphics"></p>
|
1,977,345 | <p>Queston;</p>
<p>Given that 2 is a generator of cyclic group U(25), find all generators.</p>
<p>I am only conversant with the finding the mod which is very long with this question. Pls can someone enlighten me on how to get it done faster.</p>
<p>I am new to the cyclic group.</p>
<p>solution
U(25) = {1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19,21,22,23,24}</p>
<p>2^20 = 1 (mod 25)</p>
<p>Thanks</p>
| Andreas Caranti | 58,401 | <p>The formula you need is the one that tells you that if $g$ is a generator of the cyclic group $G$ of order $n$, then $g^{k}$ is a generator if and only if $\gcd(n, k) = 1$.</p>
|
18,564 | <p>I want to get the eight points of intersection from the equations <code>2 Abs[x] + Abs[y] == 1</code> and <code>Abs[x] + 2 Abs[y] == 1</code>. To solve these equations, I tried</p>
<pre><code>Solve[{2 Abs[x] + Abs[y] == 1, Abs[x] + 2 Abs[y] == 1}, {x, y}]
</code></pre>
<p>but could only get the four points. I then tried </p>
<pre><code>y /. Quiet@Solve[#, y] /.
Abs[x_] -> {x, -x} & /@ {2 Abs[x] + Abs[y] == 1, Abs[x] + 2 Abs[y] == 1}
p = ({x, y} /. Solve[y == #] & /@ {{1 - 2 x, (1 - x)/2}, {(1 - x)/2, (
1 + x)/2}, {(1 + x)/2, 1 + 2 x}, {1 + 2 x, -1 - 2 x}, {-1 - 2 x,
1/2 (-1 - x)}, {1/2 (-1 - x),
1/2 (-1 + x)}, {1/2 (-1 + x), -1 + 2 x}, {-1 + 2 x, 1 - 2 x}})~
Flatten~1
</code></pre>
<blockquote>
<p>{{1/3, 1/3}, {0, 1/2}, {-(1/3), 1/3}, {-(1/2), 0}, {-(1/3), -(1/3)}, {0, -(1/2)}, {1/3, -(1/3)}, {1/2, 0}}</p>
</blockquote>
<p>It works, but I don't like it. Could you recommend a better method?</p>
<p><img src="https://i.stack.imgur.com/0GPLC.jpg" alt="enter image description here"></p>
| kglr | 125 | <p>Using <code>MeshFunctions</code>:</p>
<pre><code>ContourPlot[{2 Abs[x] + Abs[y] == 1, Abs[x] + 2 Abs[y] == 1}, {x, -1, 1}, {y, -1, 1},
MeshFunctions -> {ConditionalExpression[#2, Abs[#] < 1] &,
ConditionalExpression[#, Abs[#2] < 1] &, Abs[#2] - Abs[#] &},
Mesh -> {{0}, {0}, {0}}, MeshStyle -> Directive[Red, PointSize[Large]]]
</code></pre>
<p><a href="https://i.stack.imgur.com/C46R4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C46R4.png" alt="enter image description here"></a></p>
|
918,693 | <p>I've been at this for a few hours now, and it's frighteningly similar to the problem stated here: </p>
<p><a href="https://math.stackexchange.com/questions/388356/how-to-prove-int-0-infty-e-x2cos2bx-dx-frac-sqrt-pi2-e-b2">How to prove $\int_0^\infty e^{-x^2}cos(2bx) dx = \frac{\sqrt{\pi}}{2} e^{-b^2}$</a></p>
<p>but with enough change that it's still proving problematic. I also had some parts of the solution fly right over my head! I've been trying to differentiate w.r.t. b and find a clever u-sub to no avail. I'm also curious of using Euler's identity to exchange the cos for some exponentials, but I keep hitting a wall there as well. Help on either front would be great.</p>
<p>Edit: a>0 and b>0 are arbitrary constants just to be complete about this.</p>
<p>Update: Solved thanks to voldemort's solution and the Differential Equation solution from the linked problem, thanks all!</p>
| voldemort | 118,052 | <p>Just substitute $u^2=ax^2$, and use the result of the previous integration.</p>
<p>So, we must have $a >0$ (convergence issues), and thus, $u=\sqrt a x$, and $du=\sqrt a dx$.</p>
<p>After this substution, your integrand becomes $\frac{1}{\sqrt a} (e^{-u^2}\cos(\frac{b}{\sqrt a}u) du)$, and this is just the integral you linked to, with $\frac{b}{\sqrt a}$ instead of $2b$.</p>
<p>Remember that $x,u,b$ etc are just "variables".</p>
<p>Previous integral is also same as $\int_{0}^{\infty}e^{-y^2}\cos(b_1y)dy=\frac{\sqrt{\pi}}{2}e^{-b_1^2/4}$- hopefully it will be easier to make the connection now, by taking $b_1=\frac{b}{\sqrt a}$.</p>
|
962,301 | <p>$$(\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+(\sqrt 5-\sqrt 4)…$$</p>
<p>I have found partial sums equal to natural numbers. The first 3 addends sum to 1. The first 8 sum to 2. The first 15 sum to 3. When the minuend in an addend is the square root of a perfect square, the partial sum is a natural number. So I believe this series to be divergent.</p>
<p>Am I right? Have I used correct terminology? How would this be expressed using sigma notation? Is there a proof that this series diverges?</p>
| please delete me | 168,166 | <p>By telescoping, we see that the $n$th partial sum is $\sqrt {n+1}-1$, so this series diverges.</p>
|
2,679,525 | <p>Find the $\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$ over $\mathbb{Q}[x]$. Then find two polynomials $a(x),b(x) \in \mathbb{Q}[x]$ such that, $$a(x)(x^3-6x^2+14x-15) + b(x)(x^3-8x^2+21x-18)=\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$$ </p>
<p>I have managed to find,
$$x^3-6x^2+14x-15=(x-3)(x^2-3x+5)$$
$$x^3-8x^2+21x-18=(x-3)(x-3)(x-2)$$</p>
<p>Now since $x^2-3x+5$ is irreducible over $\mathbb{Q}[x]$ and so the greatest common divisor is $(x-3)$. Now to find $a(x)$ and $b(x)$ I have no clue how to do that. I have looked online and it seems there is extended euclidean algorithm for polynomials but I haven't formally learned it in my class yet, so I was wondering if there is another efficient way to find these polynomials. Any help is appreciated, thanks!</p>
| Community | -1 | <p>If I interpreted your question correctly and you want to do something like this, where in:</p>
<p>$a(x)(p)+b(x)(q)=gcd(p,q)$, $a=1$, and $b=1$</p>
<p>You can restate your expression as:</p>
<p>$$a(x)(x-3)(x^2-3x+5)+b(x)(x-3)(x-3)(x-2)$$
$$(x-3)\left(a(x)(x^2-3x+5)+b(x)(x-3)(x-2)\right)$$</p>
<p>Since this has to equal the $gcd$, which is $x-3$:</p>
<p>$$(x-3)\left(a(x)(x^2-3x+5)+b(x)(x-3)(x-2)\right)=x-3$$</p>
<p>$$\left(a(x)(x^2-3x+5)+b(x)(x-3)(x-2)\right)=1$$</p>
<p>Note: There are infinitely many solutions for $a(x)$ and $b(x)$ (but might not satisfy the definition of polynomials), since $0\le a(x)\le 1$ and $0\le b(x)\le 1$</p>
<p>Here, $a(x)(x^2-3x+5)$ and $b(x)(x-3)(x-2)$ sum up to $1$, so one instance is that they <em>can</em> individually sum up to $\dfrac 12$, since $\dfrac 12+\dfrac 12=1$.</p>
<p>So:</p>
<p>$$a(x)(x^2-3x+5)=\dfrac 12$$</p>
<p>$$b(x)(x-3)(x-2)=\dfrac 12$$</p>
<p>Can you figure this out?</p>
|
13,478 | <p>This may be a silly question - but are there interesting results about the invariant: the minimal size of an open affine cover? For example, can it be expressed in a nice way? Maybe under some additional hypotheses?</p>
| Dan Petersen | 1,310 | <p>The minimal number of open affine sets needed to cover (the coarse moduli space of) $M_{g,n}$ is a famous open problem. The conjecture, which is due to Looijenga if I recall correctly, is that the minimum is given by $g - \delta_{n,0}$ if $g > 1$. Proving this would in one swoop give a proof of Diaz's theorem and several different vanishing theorems on intersection numbers.</p>
|
1,343,554 | <p>This is exercise 6.10 in Resnick's book "A Probability Path".</p>
<p>We're given a sequence of random variables $(X_n)$ and an increasing function
$f: [0, \infty) \rightarrow [0, \infty)$ such that </p>
<p>$$
\frac{f(x)}{x} \rightarrow \infty, \qquad x \rightarrow \infty,
$$</p>
<p>and</p>
<p>$$
\sup_{n \geq 1} E(f(|X_n|)) < \infty,
$$</p>
<p>and the goal is to conclude that $\{X_n\}$ is uniformly integrable.
I know I'm supposed to show my work so far but honestly I find the problem hard to even get a start on, mostly because the existence of $f$ is abstract and I just don't see how to get from A to B.
No, I am not a student. Any help is appreciated.</p>
| David C. Ullrich | 248,223 | <p>Hmm. Writing integrals as integrals: Say $\epsilon>0$. You need to show there exists $\delta>0$ so $P(A)<\delta$ implies $\int_A|X_n|\,dP<\epsilon$.
You know that $\int f(|X_n|)\,dP\le K$ for some fixed $K$. Hmm.</p>
<p>Ok. Say $L>0$ is a large number. Choose $M$ so $$\frac{f(t)}t>L\quad(t>M).$$ Now write $A=B\cup C$, where $B=\{t\in A:|X_n(t)|>M\}$ and $C=A\setminus B$. Now
$$\int_C|X_n|\,dP\le MP(C)\le MP(A).$$So $\int_C|X_n|<\epsilon/2$ if $M\delta<\epsilon/2$. Progress...</p>
<p>Ah. On the other hand, on $B$ we have $|X_n|>M$, hence $f(|X_n|)/|X_n|>L$, or
$|X_n|<f(|X_n|)/L$. So $$\int_B|X_n|\,dP\le\frac1L\int_Bf(|X_n|)\,dP\le\frac KL.$$</p>
<p>There we are. First choose $L$ so that $K/L<\epsilon/2$. Now $L$ determines $M$; choose $\delta$ so $M\delta<\epsilon/2$ and you're done.</p>
|
1,463,887 | <p>$f(x,y)=x^2-y^2$ constraint to $x^2+y^2=1$</p>
<p>$f_x=2x$ and $f_y=-2y$ $\implies$ the critical point is at $(0,0)$</p>
<p>However, $(0,0)$ does not occur in the constraint. does that mean i don't have to consider it?</p>
<p>So if we work of the boundary curves $y=\pm\sqrt{1-x^2}$</p>
<p>we get a min @ $(0,1)$ with the value $-1$ and a max @ either $(1,0) or (-1,0)$ with the value $1$</p>
<p>did i do it correctly?</p>
| juantheron | 14,311 | <p>Given $x^2+y^2 = 1\;,$ Let $x=\cos \phi$ and $y=\sin \phi\;,$ Then $x^2-y^2 = \cos^2 \phi-\sin^2 \phi = \cos 2\phi$</p>
<p>Now Using $\bullet\; -1\leq \cos 2\phi \leq 1$</p>
<p>So we get $x^2-y^2\in \left[-1,1\right]$</p>
|
1,836,998 | <p>$$
\lim_{x\to 0^{+}}\left[\left(1+\frac{1}{x}\right)^x+\left(\frac{1}{x}\right)^x+\left(\tan(x)\right)^{\frac{1}{x}}\right]$$</p>
<p>Attempt: I used $\tan(x)\approx x$ also $(1+n)^{1/n}=e$ so I let $x=0+h$ and lim changes to lim h tending to $0$. But that gives me $e+\infty+h^{1/h}$. While the answer is an integer between $0-9$ . Where is my mistake?.Thanks</p>
| egreg | 62,967 | <p>Since you seem to be losing yourself among infinitesimals, I can suggest a safer way; when exponentials are involved, passing to the logarithm is often simpler.</p>
<p>Besides, the substitution $x=0+h$ is of no consequence.</p>
<p>Consider
$$
\lim_{x\to0^+}\log\left(\left(1+\frac{1}{x}\right)^{x}\right)=
\lim_{x\to0^+}x\log\left(1+\frac{1}{x}\right)=
\lim_{t\to\infty}\frac{\log(1+t)}{t}=0
$$
Moreover
$$
\lim_{x\to0^+}\log\bigl((1/x)^x\bigr)=
\lim_{x\to0^+}-x\log x=0
$$
Similarly,
$$
\lim_{x\to0^+}\log\bigl((\tan x)^{1/x}\bigr)=
\lim_{x\to0^+}\frac{\log\tan x}{x}=-\infty
$$
Thus your limit is
$$
1+1+0
$$</p>
|
1,911,601 | <p>Alright so I had to find the area of the common region determined by $y>=x^{0.5}$ and $x^2+y^2<2$. And I proceeded like this --
<a href="https://i.stack.imgur.com/R19kg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R19kg.jpg" alt="enter image description here"></a></p>
<p>But I'm not getting the expected answer, are there any flaws in my logic or approach? Please let me know friends. </p>
| heropup | 118,193 | <p>Your calculation is not correct, because the upper boundary of the integral $$\int_{x=0}^{\sqrt{2}} \sqrt{x} \, dx$$ is a vertical line at $x = \sqrt{2}$, not the arc of the circle centered at the origin with radius $\sqrt{2}$. Therefore, you cannot use this method to perform the area calculation; you have subtracted off an additional sliver-shaped area (shown in orange) outside the circle but underneath the parabola.</p>
<p><a href="https://i.stack.imgur.com/OM8Kq.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OM8Kq.gif" alt="enter image description here"></a></p>
<p>To get the correct area, you would need to locate the point of intersection of the circle and the parabola. This corresponds to the solution of the system $$x^2 + y^2 = 2, \quad y^2 = x$$ in the first quadrant. Then you can either split up the area of the shaded region, or you can express the shaded region as a difference of suitable areas, but in any case, you <strong>must</strong> use the fact that the two curves intersect at this point.</p>
<p>Fortunately, this point is simply $(1,1)$, so one direct method of calculation is to write $$A = \int_{x=0}^1 \sqrt{2-x^2} - \sqrt{x} \, dx.$$ Alternatively, we can write it as $$A = \frac{\pi}{4} - \int_{x=0}^1 \sqrt{x} - x \, dx,$$ if we recognize that the desired area is one eighth of the area of the circle, minus a sliver-shaped region whose boundary is the parabola from $x = 0$ to $x = 1$, and the line $y = x$ on this same interval. See the following picture:</p>
<p><a href="https://i.stack.imgur.com/bLkQT.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bLkQT.gif" alt="enter image description here"></a></p>
<p>Personally, I think this method is easiest from a computational perspective.</p>
<p>A third method is to do the integration along the $y$-axis: $$A = \int_{y=0}^1 y^2 \, dy + \int_{y=1}^\sqrt{2} \sqrt{2-y^2} \, dy.$$ There are quite a number of different ways to go about the calculation, but all should yield $$\pi/4 - 1/6.$$</p>
|
204,780 | <p>Let $X$ and $Y$ be independent discrete random variables, each taking values $1$ and $2$ with probability $1/2$ each. How do I calculate the covariance between $max(X,Y)$ and $min(X,Y)$?</p>
| robjohn | 13,854 | <p>First,
$$
\begin{array}{}
\mathrm{P}(\max(X,Y)=2)=\frac34\qquad\text{and}\qquad\mathrm{P}(\max(X,Y)=1)=\frac14\\
\mathrm{P}(\min(X,Y)=2)=\frac14\qquad\text{and}\qquad\mathrm{P}(\min(X,Y)=1)=\frac34\\
\end{array}
$$
Therefore,
$$
\mathrm{E}(\max(X,Y))=\frac74\qquad\text{and}\qquad\mathrm{E}(\min(X,Y))=\frac54
$$
Furthermore,
$$
\begin{align}
\mathrm{P}(\max(X,Y)\min(X,Y)=4)=\mathrm{P}(\min(X,Y)=2)&=\frac14\\
\mathrm{P}(\max(X,Y)\min(X,Y)=2)\hphantom{\ =\mathrm{P}(\min(X,Y)=2)}&=\frac12\\
\mathrm{P}(\max(X,Y)\min(X,Y)=1)=\mathrm{P}(\max(X,Y)=1)&=\frac14\\
\end{align}
$$
Therefore,
$$
\mathrm{E}(\max(X,Y)\min(X,Y))=\frac94
$$
Thus,
$$
\begin{align}
&\mathrm{Cov}(\max(X,Y),\min(X,Y))\\
&=\mathrm{E}(\max(X,Y)\min(X,Y))-\mathrm{E}(\max(X,Y))\mathrm{E}(\min(X,Y))\\
&=\frac1{16}
\end{align}
$$</p>
|
1,798,710 | <blockquote>
<p>Prove that any continuous bijection $f:X \rightarrow Y$ from a compact space $X$ to a Hausdorff space $Y$ is a homeomorphism</p>
</blockquote>
<p>Requirements for a homeomorphism $f:X \rightarrow Y$:</p>
<ol>
<li>$f$ is continuous</li>
<li>$f$ is bijective</li>
<li>$f^{-1}$ is continuous</li>
</ol>
<p>The first two properties are given in the question, so we just need to show that the inverse is continuous.</p>
<p>So $f^{-1}(Y)$ is the preimage of a Hausdorff space to a compact space. Why is this continuous?</p>
| Community | -1 | <p>Since the $f$ is continuous and bijective, it is a homeomorphism if and only if it is closed. But closed subset of a compact space are compact, image of compact is compact, and compact subsets of a Hausdorff space are closed.</p>
|
732,548 | <p>The proof for this using Dependent choice, is easy. I forgot how to prove this with only AC$_\omega$. Could someone help?</p>
<p>I feel really sorry that i keep asking these basic questions.. This would be my last one. Thank you in advance</p>
| Felix Marin | 85,343 | <p>$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\lim_{x \to 0^{+}}\sum_{n = 1}^{\infty}{2x \over n^{2}x^{2} + 1} = \pi:
\ {\large ?}}$.</p>
<blockquote>
<p>\begin{align}
\lim_{x \to 0^{+}}\sum_{n = 1}^{\infty}{2x \over n^{2}x^{2} + 1}
&=2\lim_{x \to 0^{+}}\bracks{{1 \over x}
\sum_{n = 0}^{\infty}{1 \over \pars{n + 1 + \ic/x}\pars{n + 1 - \ic/x}}}
\\[3mm]&=2\lim_{x \to 0^{+}}\bracks{{1 \over x}
\,{\Psi\pars{1 + \ic/x} - \Psi\pars{1 - \ic/x}\over
\pars{1 + \ic/x} - \pars{1 - \ic/x}}}
\end{align}
where $\ds{\Psi\pars{z}}$ is the
<a href="http://people.math.sfu.ca/~cbm/aands/page_258.htm" rel="nofollow">Digamma Function</a> ${\bf 6.3.1}$
and we used the <a href="http://people.math.sfu.ca/~cbm/aands/page_259.htm" rel="nofollow">identity</a>
${\bf 6.3.16}$.</p>
</blockquote>
<p>\begin{align}
\lim_{x \to 0^{+}}\sum_{n = 1}^{\infty}{2x \over n^{2}x^{2} + 1}
&=2\,\lim_{x \to 0^{+}}\Im\Psi\pars{1 + {\ic \over x}}
\end{align}</p>
<blockquote>
<p>With the <a href="http://people.math.sfu.ca/~cbm/aands/page_259.htm" rel="nofollow">identity</a>
${\bf 6.3.13}$:
\begin{align}
\color{#66f}{\large\lim_{x \to 0^{+}}
\sum_{n = 1}^{\infty}{2x \over n^{2}x^{2} + 1}}
&=\lim_{x \to 0^{+}}\bracks{-x + \pi\coth\pars{\pi \over x}}
=\color{#66f}{\LARGE\pi}
\end{align}
since $\ds{\lim_{x \to 0^{+}}\coth\pars{\pi \over x} = 1}$.</p>
</blockquote>
|
3,629,790 | <p>So I am a bit stuck on applying the limit definition for this problem. This is what I have so far: </p>
<p><span class="math-container">$\lvert x - c \rvert < \delta $</span> then <span class="math-container">$\lvert f(x) - l \rvert < \epsilon$$ $$\lvert x + 2 \rvert < \delta$</span> <span class="math-container">$$\lvert x^2 + x -2 \rvert < \epsilon $$</span> </p>
<p>which means <span class="math-container">$\lvert x -1 \rvert \lvert x + 2 \rvert < \epsilon $</span> </p>
<p><span class="math-container">$\lvert x + 2 \rvert < {\epsilon\over \lvert x -1 \rvert} $</span></p>
<p>let <span class="math-container">$\delta = 1 $</span> then: <span class="math-container">$\lvert x + 2 \rvert < \delta < 1 $</span> which means
<span class="math-container">$-1 < x -1 < 1 $</span>
<span class="math-container">$-4 < x -1 < -2 $</span>
Thus the upper bound of <span class="math-container">$\lvert x -1 \rvert < -2 $</span> is <span class="math-container">$-2$</span></p>
<p>let <span class="math-container">$$\delta = \min(1, {\epsilon\over -2}) $$</span></p>
<p>But what is confusing me here is the fact that, how is <span class="math-container">$ \lvert x + 2 \rvert < {\epsilon\over -2} $</span></p>
<p>Because if <span class="math-container">$\epsilon$</span> is any number more than <span class="math-container">$0$</span>, then <span class="math-container">${\epsilon\over -2} $</span> will be negative and it can't be greater than <span class="math-container">$ \lvert x + 2 \rvert $</span>. </p>
| John Titor | 770,201 | <p>If you choose ε so that <span class="math-container">$0<ε<2$</span> then from <span class="math-container">$|f(x)-l|<ε$</span> <span class="math-container">$\Rightarrow$</span> <span class="math-container">$|x^2+x-5+3|<ε$</span> <span class="math-container">$\Rightarrow$</span> <span class="math-container">$-ε < x^2+x-2 < ε$</span> <span class="math-container">$\Rightarrow$</span><br>
<span class="math-container">$( 0 < x ^2 + x -(2- ε ) ) \land ( x^2 + x- (2+ε) < 0 )$</span> <span class="math-container">$\Rightarrow$</span> ( x < ( -1 - √(1+4(2 - ε) ) ) / 2 , ( -1 + √ (1 + 4( 2- ε) ) ) /2 < x ) ∧ ( (- 1-√(1+4(2+ε) ))/2 < x < (-1+√(1+4(2+ε) ))/2)<br>
By combing the inequalities we will acquire 2 intervals containing 1 and -2 respectively. We consider the interval that always contains -2, and then you are done. This is not complete but you can fill in the rest of the details.</p>
|
298,121 | <p>I was watching a video on functional programming and the speaker was introducing functional programming notation for functions:</p>
<p>". . . if <strong>g(a)</strong> is a function <strong>g</strong> with variable <strong>a</strong>, we can write it also as <strong>g a</strong> . . . now, this looks like <strong>g</strong> <em>times</em> <strong>a</strong> and this would still be true if <strong>g</strong> were a linear equation . . ."</p>
<p>What does he mean by this? Why is <strong>g(a)</strong> the same as <strong>g</strong> times <strong>a</strong> if <strong>g</strong> is a linear equation? Or am I misinterpreting him? Here's the link: <a href="http://youtu.be/ZhuHCtR3xq8" rel="nofollow">http://youtu.be/ZhuHCtR3xq8</a><a href="http://youtu.be/ZhuHCtR3xq8" rel="nofollow"></a> (Starting at ~12:30)</p>
| Albert | 19,331 | <p>Recall that a function $f$ is called linear if it verifies $f(x+y)=f(x)+f(y)$ and $f(\lambda x)=\lambda f(x)$ for all $x,y$ in its domain of definition and $\lambda \in \mathbb{R}$.</p>
<p>It turns out that the only linear functions on $\mathbb{R}$ are the $f(x)=ax$, where $a$ is a constant : in other words, in one dimension, linear functions are multiplication functions. </p>
<p>In higher dimensions though, there are a lot of different linear functions, but we tend to still use the notation $fx$ instead of $f(x)$. For one thing, it is coherent with the definition of linearity : $f(x+y)=fx+fy$ and $f(\lambda x)=\lambda (fx)$. So in a way you can think of it as a sort of "generalized multiplication", the great difference being that division in in general not allowed.</p>
<p>Addendum : the term of "equation" in inappropriate in this context, I guess the author meant function, or maybe expression.</p>
|
84,255 | <p>It is just like the linear algebra over commutative ring (maybe advanced linear algebra), that is a nature extension and can make the structure of Lie algebra more algebraic, but I find little book discussing this topic.</p>
<p>I just know Weibel’s book “Homological Algebra” discuss something, but that is just a form, do not pay more attention to the Lie algebra and show some difference between field coefficient and ring coefficient. </p>
<p>Does anybody know something else？</p>
| mathphysicist | 2,149 | <p>Actually, it is possible to go even further and define Lie algebras over <em>noncommutative</em> rings, see the paper <a href="http://math.uoregon.edu/~arkadiy/noncomloopalg.pdf">Lie algebras and Lie groups over noncommutative rings</a> by Berenstein and Retakh.</p>
|
84,255 | <p>It is just like the linear algebra over commutative ring (maybe advanced linear algebra), that is a nature extension and can make the structure of Lie algebra more algebraic, but I find little book discussing this topic.</p>
<p>I just know Weibel’s book “Homological Algebra” discuss something, but that is just a form, do not pay more attention to the Lie algebra and show some difference between field coefficient and ring coefficient. </p>
<p>Does anybody know something else？</p>
| MTS | 703 | <p>One interesting fact about Lie algebras over commutative rings is that the PBW theorem can fail. I highly recommend the paper <a href="http://www.ams.org/mathscinet-getitem?mr=148717" rel="nofollow">A remark on the Birkhoff-Witt theorem</a> by P.M. Cohn. The fact that PBW can fail is probably addressed already in darij's comment to your original question, but I wanted to cite this paper separately because it is short (7 pages) and very well-written. In it, Cohn proves PBW for Lie algebras (over arbitrary commutative rings) whose additive groups are torsion-free, and also gives counterexamples where the additive group of the Lie algebra is a $p$-group.</p>
<p>Happy reading!</p>
|
352,501 | <p>There are several things that confuse me about this proof, so I was wondering if anybody could clarify them for me.</p>
<blockquote>
<p><strong>Lemma</strong> Let <span class="math-container">$G$</span> be a group of order <span class="math-container">$30$</span>. Then the <span class="math-container">$5$</span>-Sylow subgroup of <span class="math-container">$G$</span> is
normal.</p>
</blockquote>
<blockquote>
<p><strong>Proof</strong> We argue by contradiction. Let <span class="math-container">$P_5$</span> be a <span class="math-container">$5$</span>-Sylow subgroup of <span class="math-container">$G$</span>. Then the
number of conjugates of <span class="math-container">$P_5$</span> is congruent to <span class="math-container">$1 \bmod 5$</span> and divides <span class="math-container">$6$</span>. Thus, there must be six conjugates of <span class="math-container">$P_5$</span>. Since the number of conjugates is the index of the normalizer, we see that <span class="math-container">$N_G(P_5) = P_5$</span>.</p>
</blockquote>
<p>Why does the fact that the order of <span class="math-container">$N_G(P_5)$</span> is 5 mean that it is equal to <span class="math-container">$P_5$</span>?</p>
<blockquote>
<p>Since the <span class="math-container">$5$</span>-Sylow subgroups of <span class="math-container">$G$</span> have order <span class="math-container">$5$</span>, any two of them intersect in the
identity element only. Thus, there are <span class="math-container">$6\cdot4 = 24$</span> elements in <span class="math-container">$G$</span> of order <span class="math-container">$5$</span>. This leaves <span class="math-container">$6$</span> elements whose order does not equal <span class="math-container">$5$</span>. We claim now that the <span class="math-container">$3$</span>-Sylow subgroup, <span class="math-container">$P_3$</span>, must be normal in <span class="math-container">$G$</span>.</p>
</blockquote>
<blockquote>
<p>The number of conjugates of <span class="math-container">$P_3$</span> is congruent to <span class="math-container">$1 \bmod 3$</span> and divides <span class="math-container">$10$</span>. Thus, if
<span class="math-container">$P_3$</span> is not normal, it must have <span class="math-container">$10$</span> conjugates. But this would give <span class="math-container">$20$</span> elements of order <span class="math-container">$3$</span> when there cannot be more than <span class="math-container">$6$</span> elements of order unequal to <span class="math-container">$5$</span> so that <span class="math-container">$P_3$</span> must indeed be normal.</p>
</blockquote>
<blockquote>
<p>But then <span class="math-container">$P_5$</span> normalizes <span class="math-container">$P_3$</span>, and hence <span class="math-container">$P_5P_3$</span> is a subgroup of <span class="math-container">$G$</span>. Moreover, the Second Noether Theorem gives</p>
</blockquote>
<blockquote>
<p><span class="math-container">$(P_5P_3)/P_3 \cong P_5/(P_5 \cap P_3)$</span>.</p>
</blockquote>
<blockquote>
<p>But since <span class="math-container">$|P_5|$</span> and <span class="math-container">$|P_3|$</span> are relatively prime, <span class="math-container">$P_5 \cap P_3 = 1$</span>, and hence <span class="math-container">$P_5P_3$</span> must have order <span class="math-container">$15$</span>.</p>
</blockquote>
<p>Why do we need to use the second Noether theorem? Why can't we just use the formula <span class="math-container">$\frac{|P_5||P_3|}{|P_3 \cap P_5|}$</span> to compute the order?</p>
<blockquote>
<p>Thus, <span class="math-container">$P_5P_3 \cong Z_{15}$</span>, by Corollary <span class="math-container">$5.3.17$</span>. But then <span class="math-container">$P_5P_3$</span> normalizes <span class="math-container">$P_5$</span>, which contradicts the earlier statement that <span class="math-container">$N_G(P_5)$</span> has order <span class="math-container">$5$</span>.</p>
</blockquote>
<p>Why do we have to realize that <span class="math-container">$P_5P_3$</span> is isomorphic to <span class="math-container">$Z_{15}$</span>? Also, how can we conclude that <span class="math-container">$P_5P_3$</span> normalizes <span class="math-container">$P_5$</span>?</p>
<p>Thanks in advance.</p>
| DonAntonio | 31,254 | <p>Short Answers:</p>
<p>1.- Because $\,P_5\le N_G(P_5)\,$ and $\,|N_G(P_5)|=|P_5|\,$</p>
<p>2.- The "2nd Noether Theorem" seems to be what others (like me) call one (the second or the third, usually) of the isomorphism theorems, and the formula you want to use is precisely the order of the group </p>
<p>$$P_5P_3/P_3\cong P_5/(P_5\cap P_3)$$</p>
<p>otherwise you wouldn't be able to deduce $\,|P_3P_5|=15\,$ ...</p>
<p>3.- The group $\,\Bbb Z_{15}-\,$ the cyclic group of order $\,15\,$ is abelian , so it trivially normalizes its own subgroups...(this means $\,P_5\,$ is normal in $\,P_3P_5\,$ )</p>
|
2,007,123 | <p>Suppose $R=\mathbb{Q}[x]$ is a ring and define $T$ to be $T=\{f(x) \in R : f(0) \neq 0\}$ </p>
<p>Prove that $\frac{x}{1}$ is the only irreducible element in $T^{-1}R$ (disregarding associates).</p>
<hr>
<p>My approach: assume $\frac{x}{1}$ is reducible. So there exist non-unit elements $\frac{a}{b}, \frac{c}{d}$ in $T^{-1}R $ such that $\frac{x}{1}=\frac{a}{b} \frac{c}{d}$. So $x=ac$ and since the degree of $x$ is $1$ it follows that the degree of $a$ or $c$ needs to be $0$. So one of the elements $\frac{a}{b}, \frac{c}{d}$ is a unit and therefore our assumption that $\frac{x}{1}$ is reducible can't be true. Is my reasoning thus far correct?</p>
<p>Now I need to show that $\frac{x}{1}$ is the <strong>only</strong> irreducible element. But how can one prove that?</p>
| Han de Bruijn | 96,057 | <blockquote>
<p>Why a bilinear interpolation with a quadrilateral?</p>
</blockquote>
<p>As pointed out in the EDIT of the question, this issue is a bit more subtle.
<BR>
At first the comment by Rahul is applied here as a heuristics.
Rotate our $[-1,+1]×[-1,+1]$ square over $45^o$, with $xy \to \frac{1}{2}(y^2-x^2)$ as a consequence.
Instead of the basic polynomial $\,xy\,$ one gets two basic polynomials $x^2$ and $y^2$, five in total :
$1,x,y,x^2,y^2$ . For the basic shape this would imply <I>five</I> nodal points instead of four.
Now take a look at the picture below: an extra nodal point $(0)$ in the middle has been provided.
<BR>
<a href="https://i.stack.imgur.com/xgHF2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xgHF2.jpg" alt="enter image description here"></a>
<BR>
The shape on the right side is known in Finite Difference circles as a <B>five point star</B>.
It will be demonstrated here that it is possible to treat this Finite Difference pencil as if
it were a Finite Element. Let the coordinates of the <I>parent</I> five point star be given by:
$$
(0) = (0,0) \quad ; \quad \begin{cases} (1) = (-1,0) \quad ; \quad (2) = (+1,0) \\
(3) = (0,-1) \quad ; \quad (4) = (0,+1)\end{cases}
$$
Let function behaviour "inside" the five point star be approximated by a quadratic interpolation
between the function values at the vertices or nodal points, let $T$ be such a function and use
its Taylor expansion around the origin $(0)$:
$$
T(\xi,\eta) = T(0) + \frac{\partial T}{\partial \xi}(0).\xi
+ \frac{\partial T}{\partial \eta}(0).\eta
+ \frac{1}{2} \frac{\partial^2 T}{\partial \xi^2}(0).\xi^2
+ \frac{1}{2} \frac{\partial^2 T}{\partial \eta^2}(0).\eta^2
$$
Specify $T$ for the vertices with the basic polynomials of the five point star:
$$
T_0 = T(0)\\
T_1 = T(0) - \frac{\partial T}{\partial \xi}(0) + \frac{1}{2} \frac{\partial^2 T}{\partial \xi^2}(0)\\
T_2 = T(0) + \frac{\partial T}{\partial \xi}(0) + \frac{1}{2} \frac{\partial^2 T}{\partial \xi^2}(0)\\
T_3 = T(0) - \frac{\partial T}{\partial \eta}(0) + \frac{1}{2} \frac{\partial^2 T}{\partial \eta^2}(0)\\
T_4 = T(0) + \frac{\partial T}{\partial \eta}(0) + \frac{1}{2} \frac{\partial^2 T}{\partial \eta^2}(0)\\
\quad \mbox{ F.E. } \leftarrow \mbox{ F.D. }
$$
Solving these equations is not much of a problem
and well-known Finite Difference schemes are recognized:
$$
T(0) = T_0 \\
\frac{\partial T}{\partial \xi}(0) = \frac{T_2-T_1}{2}\\
\frac{\partial T}{\partial \eta}(0) = \frac{T_4-T_3}{2} \\
\frac{\partial^2 T}{\partial \xi^2}(0) = T_1-2T_0+T_2 \\
\frac{\partial^2 T}{\partial \eta^2}(0) = T_3-2T_0+T_4 \\
\quad \mbox{ F.D. } \leftarrow \mbox{ F.E. }
$$
Finite Element shape functions may be constructed as follows:
$$
T = N_0.T_0 + N_1.T_1 + N_2.T_2 + N_3.T_3 + N_4.T_4 = \\
T_0 + \frac{T_2-T_1}{2}\xi + \frac{T_4-T_3}{2}\eta
+ \frac{T_1-2T_0+T_2}{2}\xi^2 + \frac{T_3-2T_0+T_4}{2}\eta^2 =\\
(1-\xi^2-\eta^2)T_0 + \frac{1}{2}(-\xi+\xi^2)T_1 + \frac{1}{2}(+\xi+\xi^2)T_2
+\frac{1}{2}(-\eta+\eta^2)T_3 + \frac{1}{2}(+\eta+\eta^2)T_4 \\ \Longrightarrow \quad
\begin{cases}
N_0 = 1-\xi^2-\eta^2 \\
N_1 = (-\xi+\xi^2)/2\\
N_2 = (+\xi+\xi^2)/2\\
N_3 = (-\eta+\eta^2)/2\\
N_4 = (+\eta+\eta^2)/2
\end{cases}
$$
It is assumed that the <I>same parameters</I> $(\xi,\eta)$ are employed for the function $T$
as well as for the (global Cartesian) coordinates $x$ and $y$. Herewith it is expressed that
we have, as with the linear triangle and the bilinear quadrilateral, an <I>isoparametric
transformation</I>:
$$
\begin{cases}
x = N_0.x_0 + N_1.x_1 + N_2.x_2 + N_3.x_3 + N_4.x_4 \\
y = N_0.y_0 + N_1.y_1 + N_2.y_2 + N_3.y_3 + N_4.y_4
\end{cases} \quad \Longleftrightarrow \\
\begin{cases}
x =& x_0 + (x_2-x_1)/2\cdot\xi + (x_4-x_3)/2\cdot\eta\\
&+ \left[(x_1+x_2)/2-x_0\right]\cdot\xi^2 + \left[(x_3+x_4)/2-x_0\right]\cdot\eta^2 \\
y =& y_0 + (y_2-y_1)/2\cdot\xi + (y_4-y_3)/2\cdot\eta\\
&+ \left[(y_1+y_2)/2-y_0\right]\cdot\xi^2 + \left[(y_3+y_4)/2-y_0\right]\cdot\eta^2
\end{cases}
$$
Now take a look at the picture below and let attention be shifted from the original quadrilateral
to the quadrilateral that joins the midpoints of the edges of the original. The latter is known as
the <A HREF="https://en.wikipedia.org/wiki/Varignon%27s_theorem" rel="nofollow noreferrer"><B>Varignon parallelogram</B></A>
and it may be associated with our five point star.
<BR>
<a href="https://i.stack.imgur.com/8XJJH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8XJJH.jpg" alt="enter image description here"></a>
<BR>
When doing so, the diagonals of the parallelogram become the local coordinate axes of the star and
by a well-known property of the diagonals of a parallelogram we have:
$$
\begin{cases} x_0 = (x_1+x_2)/2 \\ x_0 = (x_3+x_4)/2 \end{cases} \quad \mbox{and} \quad
\begin{cases} y_0 = (y_1+y_2)/2 \\ y_0 = (y_3+y_4)/2 \end{cases} \quad \Longrightarrow \\
\begin{cases}
x = x_0 + (x_2-x_1)/2.\xi + (x_4-x_3)/2.\eta \\
y = y_0 + (y_2-y_1)/2.\xi + (y_4-y_3)/2.\eta
\end{cases}
$$
Swithing <A HREF="http://math.stackexchange.com/questions/2007116/quadrilateral-interpolation/2012594#2012594"><B>back</B></A> to the (numbering of) the original quadrilateral (on the left in the picture) we have:
$$\require{cancel}
\begin{array}{l}
x(\xi,\eta) = A_x + B_x.\xi + C_x.\eta\cancel{+ D_x.\xi.\eta} \\
y(\xi,\eta) = A_y + B_y.\xi + C_y.\eta\cancel{+ D_y.\xi.\eta}
\end{array} \qquad \mbox{ where: } \\
\begin{array}{ll}
A_x = \frac{1}{4} ( x_1 + x_2 + x_3 + x_4 ) & ; \quad
A_y = \frac{1}{4} ( y_1 + y_2 + y_3 + y_4 ) \\
B_x = \frac{1}{4} \left[(x_2 + x_4) - (x_1 + x_3)\right] & ; \quad
B_y = \frac{1}{4} \left[(y_2 + y_4) - (y_1 + y_3)\right] \\
C_x = \frac{1}{4} \left[(x_3 + x_4) - (x_1 + x_2)\right] & ; \quad
C_y = \frac{1}{4} \left[(y_3 + y_4) - (y_1 + y_2)\right] \\
\cancel{D_x = \frac{1}{4} ( + x_1 - x_2 - x_3 + x_4 )} & ; \quad
\cancel{D_y = \frac{1}{4} ( + y_1 - y_2 - y_3 + y_4 )}
\end{array}
$$
Which is precisely the original bilinear interpolation,
where the non-linear $\,\xi.\eta\,$ terms simply have been erased.
Due to the linearity achieved, the local parameters $(\xi,\eta)$ can easily be expressed now in the global coordinates $(x,y)$ :
$$
\begin{bmatrix} \xi \\ \eta \end{bmatrix} =
\begin{bmatrix} B_x & C_x \\ B_y & C_y \end{bmatrix}^{-1}
\begin{bmatrix} x-A_x \\ y-A_y \end{bmatrix}
$$
Because of the <I>isoparametrics</I>, exactly the same interpolation is applicable to any other
function $T$. Substitute $\,\xi(x,y),\eta(x,y)\,$ in:
$$
T(\xi,\eta) = A_T + B_T.\xi + C_T.\eta
\quad \mbox{ where: } \quad
\begin{cases}
A_T = \frac{1}{4} ( T_1 + T_2 + T_3 + T_4 ) \\
B_T = \frac{1}{4} \left[(T_2 + T_4) - (T_1 + T_3)\right] \\
C_T = \frac{1}{4} \left[(T_3 + T_4) - (T_1 + T_2)\right]
\end{cases}
$$
The idea is to evaluate function values at the midpoints of the edges of the bilinear quadrilateral.
Joining these midpoints together gives the Varignon parallelogram. Then employ the now <I>linear</I> interpolation of
this parallelogram as an extrapolation for points inside the original quadrilateral (under the assumption
that there is no problem with determining if a point is inside/outside an arbitrary convex quadrilateral
within an unstructured grid). Here is a visualization of the substitute interpolation. The original
quadrilateral is in black (with red vertices), the Varignon parallelogram is in blue, the $(\xi,\eta)$
coordinate axes are in yellow, the area covered by the substitute interpolation, with $-1 < \xi+\eta < +1$ and $-1 < \xi-\eta < +1$ , is in grey. There are four
<A HREF="http://math.stackexchange.com/questions/541926/converting-triangles-to-isosceles-equilateral-or-right/542684#542684"><B>triangles</B></A> remaining.
<BR>
<a href="https://i.stack.imgur.com/BX6hx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BX6hx.jpg" alt="enter image description here"></a></p>
<p>LATE EDIT. Continuing story at:
<H3><LI><A HREF="https://math.stackexchange.com/questions/2023382/any-employment-for-the-varignon-parallelogram">Any employment for the Varignon parallelogram?</A></LI></H3></p>
|
2,943,637 | <p>I want to prove that if <span class="math-container">$|z|=1 $</span> then <span class="math-container">$z^8-3z^2+1 \neq 0$</span>. I tried to prove the reciprocal by taking norms in <span class="math-container">$z^8-3z^2+1= 0$</span> and then solving for <span class="math-container">$ |z|$</span> but it does not work. I also assume <span class="math-container">$| z|=1 $</span> and then trying to see that <span class="math-container">$| z^8-3z^2+1 |> 0 $</span> but it did not work neither.</p>
<p>Any ideas on this?</p>
| Umberto P. | 67,536 | <p>Try the reverse triange inequality:
<span class="math-container">$$|z^8 - 3z^2 + 1| \ge |3z^2| - |z^8| - 1.$$</span></p>
|
2,943,637 | <p>I want to prove that if <span class="math-container">$|z|=1 $</span> then <span class="math-container">$z^8-3z^2+1 \neq 0$</span>. I tried to prove the reciprocal by taking norms in <span class="math-container">$z^8-3z^2+1= 0$</span> and then solving for <span class="math-container">$ |z|$</span> but it does not work. I also assume <span class="math-container">$| z|=1 $</span> and then trying to see that <span class="math-container">$| z^8-3z^2+1 |> 0 $</span> but it did not work neither.</p>
<p>Any ideas on this?</p>
| Mohammad Riazi-Kermani | 514,496 | <p><span class="math-container">$$|z|=1 \implies |z^8|=1 $$</span></p>
<p>Therefore, <span class="math-container">$|3z^2-1|=|z^8|=1$</span></p>
<p>Note that <span class="math-container">$|3z^2-1| \ge ||3z^2|-1| =2$</span></p>
<p>Which is not consistent with <span class="math-container">$|3z^2-1|=1$</span></p>
|
3,401 | <p>Consider an idealized classical particle confined to a two-dimensional surface that is frictionless. The particle's initial position on the surface is randomly selected, a nonzero velocity vector is randomly assigned to it, and the direction of the particle's movement changes only at the surface boundaries where perfectly elastic collisions occur (i.e. there is no information loss over time).</p>
<p>My question is - Does there exist such a bounded surface where the probability of the particle visiting any given position at some time 't', P(x,y,t), becomes equal to unity at infinite time? In other words, no matter where we initialize the particle, and no matter the velocity vector assigned to it, are there surfaces that will always be 'everywhere accessible'?</p>
<p>(Once again, I welcome any help asking this question in a more appropriate manner...)</p>
| Andy Putman | 317 | <p>I'll interpret you question to be asking about whether the particle paths are "equidistributed" in the sense of dynamical systems. There is a large literature on this sort of thing, though usually instead of "particles" the authors talk about "billiards". While I don't know the answer to your question as stated, I do know that there are many examples where the paths become equidistributed for "generic" choices of positions and initial directions (in other words, the "bad" choices form a set of measure zero). </p>
<p>Many examples and results of this form can be found in the wonderful survey "Rational billiards and flat structures" by Masur and Tabachnikov, which is available on Masur's <a href="http://math.uchicago.edu/~masur/" rel="noreferrer">web page</a>.</p>
<p>EDIT : I forgot a nice reference! Serge Tabachnikov has written a very accessible book entitled "Geometry and Billiards" which is available on his webpage <a href="http://www.math.psu.edu/tabachni/Books/books.html" rel="noreferrer">here</a>.</p>
|
1,535,137 | <p>I recently used the method of least norms to solve an underdetermined system of linear equations for a problem at work. This got me thinking, if I were to think about this more generally, does such a solution exits?</p>
<p>That is, if $T$ is an operator on an inner-product space $V$, and if $x,y\in V$, is it possible to find a solution to
$$Tx=y$$
such that $\|x\|$ is minimal?</p>
<p>I know that this doesn't exactly line up with the problem at work because there I was going from $\mathbb{R}^n$ to $\mathbb{R}^m$, but I thought it may be simpler to consider the case of an operator on an inner product space than a linear map between two different spaces.</p>
<p>I want to write this up to present it to a class, so if you know of any good sources where I might read up on this in depth, I would appreciate that as well.</p>
| user219967 | 219,963 | <p>When $n>m$, we have $rank(AB)\le min\{rank(A),rank(B)\}\Rightarrow rank(AB)\le m\Rightarrow Det(AB)=0$</p>
|
1,535,137 | <p>I recently used the method of least norms to solve an underdetermined system of linear equations for a problem at work. This got me thinking, if I were to think about this more generally, does such a solution exits?</p>
<p>That is, if $T$ is an operator on an inner-product space $V$, and if $x,y\in V$, is it possible to find a solution to
$$Tx=y$$
such that $\|x\|$ is minimal?</p>
<p>I know that this doesn't exactly line up with the problem at work because there I was going from $\mathbb{R}^n$ to $\mathbb{R}^m$, but I thought it may be simpler to consider the case of an operator on an inner product space than a linear map between two different spaces.</p>
<p>I want to write this up to present it to a class, so if you know of any good sources where I might read up on this in depth, I would appreciate that as well.</p>
| Archis Welankar | 275,884 | <p>Let A be an m×n matrix and B an n×m matrix. Write [n] for the set { 1, ..., n }, and $\tbinom{[n]}m$ for the set of m-combinations of [n] (i.e., subsets of size m; there are $\tbinom nm of them)$. The Cauchy–Binet formula then states</p>
<p>$\det(AB) = \sum_{S\in\tbinom{[n]}m} \det(A_{[m],S})\det(B_{S,[m]})$ .a good source would be Wikipedia</p>
|
2,828,023 | <p><strong>Four cards are dealt off the top of a well-shuffled deck.
Find the chance that:</strong></p>
<p><strong>You get a queen or a king.</strong> </p>
<p>The solution shows $ 1- \frac { \binom {44}{4}}{ \binom {52}{4}}$ </p>
<p>I get that in this case it's easier to use compliment rule, however I'm trying another method and I got the following:</p>
<p>$ \binom {4}{1} \cdot\ \frac {4}{52} \cdot\ \frac {48}{51} \cdot\ \frac {47}{50} \cdot\ \frac {46}{49} $ + $ \binom {4}{1} \cdot\ \frac {4}{52} \cdot\ \frac {48}{51} \cdot\ \frac {47}{50} \cdot\ \frac {46}{49} $ - [ ($ \binom {4}{1} \cdot\ \frac {4}{52} \cdot\ \frac {48}{51} \cdot\ \frac {47}{50} \cdot\ \frac {46}{49} $)^2]</p>
<p>My reasoning is P(1 Queen or 1 King) = P(1 Queen) + P(1 King) - P[P(1 queen) x P (1 king)]</p>
<p>The answer that I got is about 0.2499 which is different from the solution obtained from the compliument method, what did I miss here?</p>
<p>Thank you!</p>
| SlipEternal | 156,808 | <p>$\dfrac{\dbinom{8}{1}\dbinom{44}{3}+\dbinom{8}{2}\dbinom{44}{2}+\dbinom{8}{3}\dbinom{44}{1}+\dbinom{8}{4}\dbinom{44}{0}}{\dbinom{52}{4}} = 1-\dfrac{\dbinom{8}{0}\dbinom{44}{4}}{\dbinom{52}{4}}$</p>
|
2,493,802 | <p>If $A \subset \mathbb R$, we have the distance between a point in $\mathbb R$ and $A$ as</p>
<p>$$d(x,A)=\inf\{|x-a| \mid a \in A\}.$$</p>
<p><strong>For $\varepsilon>0$, if we define $A(\varepsilon) = \{x \in \mathbb R \mid d(x,A) < \varepsilon \}$, how do we show this set is open?</strong></p>
<hr>
<p>I know that $A \subset A(\varepsilon)$ since for any point in $a \in A$ we will have $d(a, A)=0 < \varepsilon$ for any $\varepsilon >0$.</p>
<p>I can see that $A(\varepsilon)$ is sort of like a neighborhood of the set $A$ but I am having trouble formalizing this.</p>
| Arun | 399,173 | <p>Geometrically one can guess that for any $b\in A(\varepsilon)$, the ball of radius <strong>$\frac{\varepsilon -d(b,A)}{2}$</strong> and centered at $b$ will lie inside $A(\varepsilon)$. Of course you need to prove it mathematically.</p>
|
3,930,373 | <p>Hi so im working on a question about finding the <span class="math-container">$rank(A)$</span> and the <span class="math-container">$dim(Ker(A))$</span> of a 7x5 Matrix. Without being given an actual matrix to work from.</p>
<p>I have been told that that the homogeneous equation <span class="math-container">$A\vec x=\vec0$</span> has general solution <span class="math-container">$\vec x=\lambda \vec v$</span> for some non zero <span class="math-container">$\vec v$</span> in <span class="math-container">$R^{5}$</span>.</p>
<p>So my thinking so far is that I know for an <span class="math-container">$m*n$</span> matrix we know that:</p>
<p><span class="math-container">$rk(A)+dimker(A)=n$</span>
which must mean that <span class="math-container">$rk(A)+dimker(A)=5$</span></p>
<p>but this is where I get stuck and dont know how to proceed.</p>
<p>Any help is greately appreciated.</p>
<p><a href="https://i.stack.imgur.com/2kLWC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2kLWC.png" alt="enter image description here" /></a>
.</p>
<p>This is the exact question for the person who asked.</p>
| Vercassivelaunos | 803,179 | <p>We can say so with perfectly good conscience. Ignoring all the notational baggage, the function in question is
<span class="math-container">$$f:\{0\}\longrightarrow\mathbb R,~0\mapsto0.$$</span>
Here, <span class="math-container">$\{0\}$</span> is a metric space with the only possible metric on a singleton set: <span class="math-container">$d(0,0)=0$</span>. With this metric, the set <span class="math-container">$\{0\}$</span> gains the discrete topology, with respect to which <em>every</em> function to <em>any</em> other metric space (or even topological space) is continuous. It's not a particularly interesting topology, but it's a perfectly fine one.</p>
<p>Another argument why it's perfectly fine to allow such a function to be continuous: Let <span class="math-container">$f:X\longrightarrow Y$</span> be a constant map where <span class="math-container">$X,Y$</span> are metric spaces (or even topological ones). Then <span class="math-container">$f$</span> is continuous. This is a very intuitive statement which is also easy to prove in the context of topology. Your function <em>is</em> a constant function between metric spaces, so we'd have to make an exception for this theorem, which in my opinion would be way worse than calling your function continuous.</p>
|
2,972,611 | <p><strong>Q</strong>: Prove the given inequalities for positive a,b,c:<br><span class="math-container">$(i) \left[\frac{bc+ca+ab}{a+b+c}\right]^{a+b+c}>\sqrt{(bc)^a.(ca)^b.(ab)^c}$</span><br><span class="math-container">$(ii) \left(\frac{a+b+c}{3} \right)^{a+b+c}<a^ab^bc^c<\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c} $</span><br>I know that G.M<span class="math-container">$\le$</span>A.M and somehow i guess it must be used.But i really struggling to prove this kind of inequality.Any hints or solution will be appreciated.And i do apologize if this question is very basic.<br>Thanks in advance.</p>
| DeepSea | 101,504 | <p><span class="math-container">$a)$</span>We have by the weighted AM-GM inequality: <span class="math-container">$\sqrt{(ab)^{\frac{c}{a+b+c}}\cdot (bc)^{\frac{a}{a+b+c}}\cdot (ca)^{\frac{b}{a+b+c}}}\le \sqrt{\dfrac{3abc}{a+b+c}}\le \dfrac{ab+bc+ca}{a+b+c}\iff(ab+bc+ca)^2 \ge 3abc(a+b+c)\iff (ab)^2+(bc)^2+(ca)^2 \ge abc(a+b+c) $</span> which is true. </p>
<p><span class="math-container">$b)$</span> Consider <span class="math-container">$f(x) = x\ln x \implies f''(x) = \dfrac{1}{x} >0 $</span>. Thus <span class="math-container">$f$</span> is convex and using <span class="math-container">$f(a)+f(b)+f(c) \ge 3f\left(\frac{a+b+c}{3}\right)$</span>, the answer follows for the left inequality. For the right one, apply again the weighted AM-GM inequality <span class="math-container">$a^{\frac{a}{a+b+c}}\cdot b^{\frac{b}{a+b+c}}\cdot c^{\frac{c}{a+b+c}}\le \dfrac{a^2+b^2+c^2}{a+b+c}$</span> . QED.</p>
|
4,494,489 | <p>These are questions I’m stuck on:</p>
<p>(i) Prove for <span class="math-container">$k\in \mathbb{R}^+:\frac{2}{\sqrt{k+1}+\sqrt{k}}\lt\frac{1}{\sqrt{k}}$</span></p>
<p>(ii) Prove <span class="math-container">$16\lt\sum_{k=1}^{80}\frac{1}{\sqrt{k}}\lt17$</span></p>
<p>I did the first one but I just equated them until I got <span class="math-container">$\sqrt{k+1}\gt\sqrt{k}$</span>
Which I think I did something wrong because I assumed I was supposed to use induction.</p>
<p>For the second one I am just completely stuck, I tried to convert 16 and 17 into summations with the same limits and got</p>
<p><span class="math-container">$\sum_{k=1}^{80}\frac{16k}{3240}=16$</span></p>
<p><span class="math-container">$\sum_{k=1}^{80}\frac{17k}{3240}=17$</span></p>
<p>But after that I just didn’t know what to do.</p>
| true blue anil | 22,388 | <p>Why do you want to raise a hornet's nest when a simple solution exists ?</p>
<p>Anyway, rethink your formula considering that if event <span class="math-container">$E$</span> has occurred when <span class="math-container">$i=0$</span> (round <span class="math-container">$1$</span>), it can occur in the next round only if both <span class="math-container">$A$</span> and <span class="math-container">$B$</span> progress to that round</p>
|
1,124,417 | <p>I was working my way through some Propositional Logic and had the following doubt :</p>
<blockquote>
<p><strong>Why is this true :</strong></p>
<p>((p <span class="math-container">$\Rightarrow$</span> r) <span class="math-container">$\land$</span> (q <span class="math-container">$\Rightarrow$</span> r)) <span class="math-container">$\equiv$</span> ((p <span class="math-container">$\lor$</span> q) <span class="math-container">$\Rightarrow$</span> r)</p>
</blockquote>
<p><em><strong>Please provide an intuitive explanation and not one that uses a truth table or logic identities to simplify the expression . I have already done both of them :)</strong></em></p>
| Community | -1 | <p>One can get some information with basic geometry, using the fact that this equation corresponds to a <em>conic</em> plane curve. There are only two possible answers, and they are exact (no error terms); deciding which answer is correct seems harder. </p>
<p>Let me write my thoughts in case they are helpful to you.</p>
<p>Your $F$ is a homogeneous polynomial of degree 2, irreducible over the algebraic closure, so it defines a smooth conic $C$ in $\mathbf P^2(\mathbf F_q)$. The number of solutions of the equation will be $1+(q-1)\cdot |C(\mathbf F_q)|$, where $|C(\mathbf F_q)|$ is the number of points of $\mathbf P^2(\mathbf F_q)$ that lie on the curve. (The extra 1 is the trivial solution $(0,0,0)$.)</p>
<p>Now: a smooth conic $C$ over any field $k$ is $k$-isomorphic to $\mathbf P^1_k$ iff it has a rational point over $k$. (Proof: stereographic projection.) And we certainly know how many points $\mathbf P^1$ has over $\mathbf F_q$, namely $q+1$. So
$$|C(\mathbf F_q)|=\cases{0 \quad \text{ if $C$ has no $\mathbf F_q$-point}\\ q+1 \quad\text{ otherwise}}$$</p>
<p>So your equation either has 1 solution (the trivial one) or $1+(q-1)(q+1)=q^2$ solutions.</p>
<p>Deciding which is the case seems a little trickier. If $a^2=b$ it is trivial to find a solution, so assume that isn't the case. Then for fixed $x$ and $y$, the quadratic equation $F(x,y,z)=0$ has discriminant</p>
<p>$$\Delta(x,y) = 4a^2(x+y)^2-8(a^2-b)(x^2+y^2+xy)$$</p>
<p>and one has to decide whether there exist $x, y \in \mathbf F_q$ such that $\Delta(x,y)$ is a square in $\mathbf F_q$. I can't see a good way to do this at the moment — I am making my answer community wiki so that someone else can fill in the details.</p>
|
1,124,417 | <p>I was working my way through some Propositional Logic and had the following doubt :</p>
<blockquote>
<p><strong>Why is this true :</strong></p>
<p>((p <span class="math-container">$\Rightarrow$</span> r) <span class="math-container">$\land$</span> (q <span class="math-container">$\Rightarrow$</span> r)) <span class="math-container">$\equiv$</span> ((p <span class="math-container">$\lor$</span> q) <span class="math-container">$\Rightarrow$</span> r)</p>
</blockquote>
<p><em><strong>Please provide an intuitive explanation and not one that uses a truth table or logic identities to simplify the expression . I have already done both of them :)</strong></em></p>
| Sergio Da Silva | 130,769 | <p>I want to add a tool into the mix which might simplify the situation for future computations. This comes from the theory of Frobenius splittings. Define a near splitting of $\mathbb{F}_q[x_1,..,x_n]$ by the map called $Tr(*)$ (the trace map) defined on monomials $m$ (and extended linearly) by: </p>
<p>$$Tr(m)=\begin{cases}
\frac{\sqrt[q]{m\prod_i x_i}}{ \prod_i x_i}& \text{if $m\prod_i x_i$ is a $q^{th}$ power}\\
0 & \text{otherwise}\\
\end{cases}$$</p>
<p>Now let $f\in\mathbb{F}_q[x_1,...,x_n]$ be a polynomial of degree at most $n>0$. Then the number of points $v\in\mathbb{F}_q^n$ in the affine hypersurface defined by $f=0$ is congruent to $(-1)^{n-1}Tr(f^{q-1})$. The reference can be found in Allen Knutson's paper on page 3: <a href="http://arxiv.org/pdf/0911.4941v1.pdf" rel="nofollow">http://arxiv.org/pdf/0911.4941v1.pdf</a></p>
<p>Using the answers above, you can reduce to the case that there are either $1$ or $q^2$ solutions. By degree considerations, we can see that for the $f$ you defined, $f^{q-1}$ will not have any monomials of the form $(xyz)^{q-1}$, so that $Tr(f^{q-1}) = 0$. The splitting method shows that the answer has to be congruent to $0$, so the answer has to be $q^2$.</p>
|
1,203,116 | <p>I'm trying to prove that in such triangle, AB>AC, but I can't suceed. I tried to use some properties of isósceles triangle since the sides are equal, but couldn't find something nice. Also, whats the criteria needed to prove that a side in a triangle is greater than the other? Thank you so much!</p>
| user26486 | 107,671 | <p><strong>Hint:</strong></p>
<p>(the diagram is not meant to be to scale)</p>
<p><span class="math-container">$$\angle BAD\stackrel{\text{isosc}}=\angle B:=a$$</span>
<span class="math-container">$$\angle ADC\stackrel{\text{isosc}}=\angle C\stackrel{(1)}=2a$$</span><br />
<span class="math-container">$\angle C>\angle B\implies AB>AC$</span>. For a proof of this, see <a href="http://aleph0.clarku.edu/%7Edjoyce/java/elements/bookI/propI18.html" rel="nofollow noreferrer">this</a> and then <a href="http://aleph0.clarku.edu/%7Edjoyce/java/elements/bookI/propI19.html" rel="nofollow noreferrer">this</a>.</p>
<p><span class="math-container">$(1)$</span>: triangle angles add up to <span class="math-container">$180^{\circ}$</span>.
<img src="https://i.stack.imgur.com/CQ6OV.png" alt="enter image description here" /></p>
|
3,879,009 | <p>How to prove this lemma?</p>
<blockquote>
<p><span class="math-container">$\min(x,y,z) \leq ax+by+cz \leq \max(x,y,z)$</span>, with <span class="math-container">$a+b+c = 1$</span> for any real numbers <span class="math-container">$x,y,z$</span> and <span class="math-container">$a,b,c$</span> positive.</p>
</blockquote>
| Jean Marie | 305,862 | <p>Proof with geometric flavour:</p>
<p>Consider <span class="math-container">$ax+by+cz$</span> as a <strong>barycentric combination</strong> of points <span class="math-container">$x,y,z$</span> with positive coefficients <span class="math-container">$a,b,c$</span> whose sum is <span class="math-container">$1$</span>. The result is necessarily between the two extreme points, i.e., <span class="math-container">$\min(x,y,z)$</span> and <span class="math-container">$\max(x,y,z)$</span>.</p>
<p>(one could have also used a <strong>convexity</strong> argumentation).</p>
|
2,996,476 | <p>So I was asked to solve the following sum using Fourier Series:</p>
<p><span class="math-container">$$\sum_{n\in\Bbb N}\frac{(-1)^n}{4n^2-1}^{(1)}$$</span></p>
<p>I decided to try splitting it up into two sums (for even and odd <span class="math-container">$n$</span>) in the hopes that it would telescope and be easier to evaluate. It did not, but I was able to salvage my work by employing the arctangent Taylor Series. The purpose of this post is twofold: to post a rather unusual approach and to ask others how they might do this problem more easily/differently.</p>
<p>First I noted</p>
<p><span class="math-container">$$\sum_{n\in\Bbb N}\frac{(-1)^n}{4n^2-1}=\sum_{n\in\mathbb N}\frac{1}{16n^2-1}^{(2)}-\sum_{n\in\mathbb N}\frac{1}{4(2n-1)^2-1}^{(3)}$$</span></p>
<p>Using partial fractions of the first sum on the RHS:</p>
<p><span class="math-container">$$(2)\;\sum_{n\in\mathbb N}\frac{1}{16n^2-1}=\frac{1}{2}\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)$$</span></p>
<p>Using partial fractions on the second sum on the RHS:</p>
<p><span class="math-container">$$(3)\;\sum_{n\in\mathbb N}\frac{1}{4(2n-1)^2-1}=\frac{1}{2}\sum_{n\in\mathbb N}\left(\frac{1}{4n-3}-\frac{1}{4n-1}\right)$$</span></p>
<p><span class="math-container">$$=-\frac{1}{2}+\frac{1}{2}\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)$$</span></p>
<p>Combining results for the even and odd sums from the RHS:</p>
<p><span class="math-container">$$\sum_{n\in\mathbb N}\frac{1}{16n^2-1}-\sum_{n\in\mathbb N}\frac{1}{4(2n-1)^2-1}=-\frac{1}{2}+\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)^{(4)}$$</span></p>
<p>Then I saw that the rightmost sum was the following:</p>
<p><span class="math-container">$$\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)=\sum_{n=2}^\infty\frac{(-1)^{n}}{2n-1}^{(5)}$$</span></p>
<p>Note the Taylor series for arctangent:</p>
<p><span class="math-container">$$\text{atan}(x)=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{2n-1}$$</span></p>
<p>Letting <span class="math-container">$x=1$</span>:</p>
<p><span class="math-container">$$\text{atan}(1)=\frac{\pi}{4}=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{2n-1}$$</span></p>
<p>Accounting for index and sign difference, we see that</p>
<p><span class="math-container">$$(5)\;\sum_{n=2}^\infty\frac{(-1)^{n}}{2n-1}=1-\frac{\pi}{4}$$</span></p>
<p>And finally, from the <span class="math-container">$-1/2$</span> term in <span class="math-container">$(4)$</span>, we see that</p>
<p><span class="math-container">$$(1)\;\sum_{n\in\Bbb N}\frac{(-1)^n}{4n^2-1}=\frac{2-\pi}{4}$$</span></p>
| Batominovski | 72,152 | <p>Here is my attempt to follow the instruction to use Fourier series. First, as you did, I rewrite the sum as
<span class="math-container">$$\sum_{n=1}^\infty\,\frac{(-1)^n}{4n^2-1}=\sum_{n=1}^\infty\,\frac{(-1)^n}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}+\sum_{n=1}^\infty\,\frac{(-1)^n}{2n-1}\,.\tag{*}$$</span></p>
<p>Let <span class="math-container">$f:\mathbb{R}\to\mathbb{R}$</span> be the periodic function with period <span class="math-container">$2\pi$</span> defined by
<span class="math-container">$$f(x):=\begin{cases}+1&\text{if }x\in(0,+\pi)\,,\\0&\text{if }x\in\{-\pi,0,+\pi\}\,,\\-1&\text{if }x\in(-\pi,0)\,.\end{cases}$$</span>
Then, as <span class="math-container">$f$</span> is an odd function, <span class="math-container">$f$</span> has a Fourier sine-series:
<span class="math-container">$$f(x)=\sum_{k=1}^\infty\,b_k\,\sin(kx)$$</span>
for some real numbers <span class="math-container">$b_1,b_2,b_3,\ldots$</span>. We see that
<span class="math-container">$$\begin{align}b_k&=\frac{1}{\pi}\,\int_{-\pi}^{+\pi}\,\sin(kx)\,f(x)\,\text{d}x=\frac{2}{\pi}\,\int_0^\pi\,\sin(kx)\,\text{d}x
\\&=\frac{2}{\pi\,k}\,\Big.\big(-\cos(kx)\big)\Big|_{x=0}^{x=\pi}=\frac{2\,\big(1-(-1)^k\big)}{\pi\,k}
\end{align}$$</span>
for all <span class="math-container">$k=1,2,3,\ldots$</span>.
Consequently,
<span class="math-container">$$f(x)=\sum_{k=1}^\infty\,\frac{2\,\big(1-(-1)^k\big)}{\pi\,k}\,\sin(kx)=\frac{4}{\pi}\,\sum_{n=1}^\infty\,\frac{\sin\big((2n-1)x\big)}{2n-1}\,.$$</span>
Because <span class="math-container">$f\left(\dfrac{\pi}{4}\right)=1$</span>, we get
<span class="math-container">$$1=\frac{4}{\pi}\,\sum_{n=1}^\infty\,\frac{\sin\left(\frac{(2n-1)\,\pi}{4}\right)}{2n-1}=\frac{4}{\pi}\,\sum_{n=1}^\infty\,\frac{(-1)^{n-1}}{2n-1}\,.$$</span>
This shows that
<span class="math-container">$$\sum_{n=1}^\infty\,\frac{(-1)^n}{2n-1}=-\frac{\pi}{4}\,.$$</span>
By (*), we conclude that
<span class="math-container">$$\sum_{n=1}^\infty\,\frac{(-1)^n}{4n^2-1}=\frac{1}{2}-\frac{\pi}{4}=\frac{2-\pi}{4}\approx -0.28539816\,.$$</span></p>
|
2,996,476 | <p>So I was asked to solve the following sum using Fourier Series:</p>
<p><span class="math-container">$$\sum_{n\in\Bbb N}\frac{(-1)^n}{4n^2-1}^{(1)}$$</span></p>
<p>I decided to try splitting it up into two sums (for even and odd <span class="math-container">$n$</span>) in the hopes that it would telescope and be easier to evaluate. It did not, but I was able to salvage my work by employing the arctangent Taylor Series. The purpose of this post is twofold: to post a rather unusual approach and to ask others how they might do this problem more easily/differently.</p>
<p>First I noted</p>
<p><span class="math-container">$$\sum_{n\in\Bbb N}\frac{(-1)^n}{4n^2-1}=\sum_{n\in\mathbb N}\frac{1}{16n^2-1}^{(2)}-\sum_{n\in\mathbb N}\frac{1}{4(2n-1)^2-1}^{(3)}$$</span></p>
<p>Using partial fractions of the first sum on the RHS:</p>
<p><span class="math-container">$$(2)\;\sum_{n\in\mathbb N}\frac{1}{16n^2-1}=\frac{1}{2}\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)$$</span></p>
<p>Using partial fractions on the second sum on the RHS:</p>
<p><span class="math-container">$$(3)\;\sum_{n\in\mathbb N}\frac{1}{4(2n-1)^2-1}=\frac{1}{2}\sum_{n\in\mathbb N}\left(\frac{1}{4n-3}-\frac{1}{4n-1}\right)$$</span></p>
<p><span class="math-container">$$=-\frac{1}{2}+\frac{1}{2}\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)$$</span></p>
<p>Combining results for the even and odd sums from the RHS:</p>
<p><span class="math-container">$$\sum_{n\in\mathbb N}\frac{1}{16n^2-1}-\sum_{n\in\mathbb N}\frac{1}{4(2n-1)^2-1}=-\frac{1}{2}+\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)^{(4)}$$</span></p>
<p>Then I saw that the rightmost sum was the following:</p>
<p><span class="math-container">$$\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)=\sum_{n=2}^\infty\frac{(-1)^{n}}{2n-1}^{(5)}$$</span></p>
<p>Note the Taylor series for arctangent:</p>
<p><span class="math-container">$$\text{atan}(x)=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{2n-1}$$</span></p>
<p>Letting <span class="math-container">$x=1$</span>:</p>
<p><span class="math-container">$$\text{atan}(1)=\frac{\pi}{4}=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{2n-1}$$</span></p>
<p>Accounting for index and sign difference, we see that</p>
<p><span class="math-container">$$(5)\;\sum_{n=2}^\infty\frac{(-1)^{n}}{2n-1}=1-\frac{\pi}{4}$$</span></p>
<p>And finally, from the <span class="math-container">$-1/2$</span> term in <span class="math-container">$(4)$</span>, we see that</p>
<p><span class="math-container">$$(1)\;\sum_{n\in\Bbb N}\frac{(-1)^n}{4n^2-1}=\frac{2-\pi}{4}$$</span></p>
| Community | -1 | <p>Consider a <span class="math-container">$2\pi$</span>-periodic function <span class="math-container">$f:\Bbb{R}\to\Bbb{C}$</span> whose values on <span class="math-container">$(-\pi,\pi]$</span> are given by <span class="math-container">$f(x)=\operatorname{sign}x\sin\alpha x$</span>. Then, <span class="math-container">$f$</span> is even so it has a Fourier series of the form
<span class="math-container">$$f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos nx,$$</span>
where <span class="math-container">$a_n=\frac1\pi\int_{-\pi}^\pi f(x)\cos nx\ dx=\frac{2}{\pi}\int_0^\pi f(x)\cos nx\ dx$</span> (recalling that <span class="math-container">$f$</span> is even). Because <span class="math-container">$2\sin\alpha x\cos nx=\sin(n+\alpha)x-\sin(n-\alpha)x$</span>, we conclude that
<span class="math-container">$$a_n=\begin{cases}\frac{1-(-1)^n\cos\alpha \pi}{\pi}\left(\frac{1}{n+\alpha}-\frac{1}{n-\alpha}\right)&\text{if }n\neq \pm\alpha,\\ 0&\text{if }n=\pm\alpha.\end{cases}$$</span></p>
<p>Take <span class="math-container">$\alpha=1/2$</span>, so that
<span class="math-container">$$a_n=\frac{1-(-1)^n\cos\frac{\pi}{2}}{\pi}\left(\frac{1}{n+1/2}-\frac{1}{n-1/2}\right)=-\frac{4}{\pi(4n^2-1)}.$$</span>
That is
<span class="math-container">$$\operatorname{sign}x\sin\frac{x}{2}=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos nx=\frac2\pi-\frac4\pi\sum_{n=1}^\infty\frac{1}{4n^2-1}\cos nx$$</span>
for <span class="math-container">$-\pi< x\le \pi$</span>. Plugging in <span class="math-container">$x=\pi$</span> (note that <span class="math-container">$f$</span> is continuous at <span class="math-container">$x=\pi$</span>, so we don't have to worry about jumps over there), we have
<span class="math-container">$$1=\operatorname{sign}{\pi}\sin\frac{\pi}{2}=\frac2\pi-\frac4\pi\sum_{n=1}^\infty\frac{\cos n\pi}{4n^2-1}=\frac2\pi-\frac4\pi\sum_{n=1}^\infty\frac{(-1)^n}{4n^2-1}.$$</span>
So <span class="math-container">$\sum_{n=1}^\infty\frac{(-1)^n}{4n^2-1}=\frac{2-\pi}{4}$</span>.</p>
<hr>
<p>Similarly, consider a <span class="math-container">$2\pi$</span>-periodic function <span class="math-container">$g:\Bbb{R}\to\Bbb{C}$</span> whose values on <span class="math-container">$(-\pi,\pi]$</span> are given by <span class="math-container">$g(x)=\operatorname{sign}x\cos\alpha x$</span>. Then, <span class="math-container">$g$</span> is odd so it has a Fourier series of the form
<span class="math-container">$$g(x)=\sum_{n=1}^\infty b_n\sin nx,$$</span>
where <span class="math-container">$b_n=\frac1\pi\int_{-\pi}^\pi g(x)\sin nx\ dx=\frac{2}{\pi}\int_0^\pi g(x)\sin nx\ dx$</span> (recalling that <span class="math-container">$g$</span> is odd). Because <span class="math-container">$2\cos\alpha x\sin nx=\sin(n+\alpha)x+\sin(n-\alpha)x$</span>, we conclude that
<span class="math-container">$$b_n=\begin{cases}\frac{1-(-1)^n\cos\alpha \pi}{\pi}\left(\frac{1}{n+\alpha}+\frac{1}{n-\alpha}\right)&\text{if }n\neq \pm\alpha,\\ 0&\text{if }n=\pm\alpha.\end{cases}$$</span></p>
|
4,541,392 | <p>I have an integral that looks like the following:
<span class="math-container">$$
\int_a^b \frac{1}{\left(1 + cx^2\right)^{3/2}} \mathrm{d}x
$$</span>
I have seen a method of solving it being to substitute <span class="math-container">$x = \frac{\mathrm{tan}(u)}{\sqrt{c}}$</span>; however, this seems somewhat sloppy to me. Is there perhaps a better way of tackling this integral?</p>
| Accelerator | 1,065,595 | <p>Here's one idea that requires <span class="math-container">$c > 0$</span> and <span class="math-container">$a,b \in \mathbb{R}$</span>. Let <span class="math-container">$I$</span> be the given integral. If <span class="math-container">$x = \frac{\sinh{(u)}}{\sqrt{c}}$</span>, then the integral becomes</p>
<p><span class="math-container">$$
\eqalign{
I &= \frac{1}{\sqrt{c}}\int_{\operatorname{arcsinh}\left(a\sqrt{c}\right)}^{\operatorname{arcsinh}\left(b\sqrt{c}\right)}\operatorname{sech}^{2}\left(u\right)du \cr
&= \frac{1}{\sqrt{c}}\tanh{(u)}\Bigg|_{\operatorname{arcsinh}\left(a\sqrt{c}\right)}^{\operatorname{arcsinh}\left(b\sqrt{c}\right)} \cr
&= \frac{1}{\sqrt{c}}\left(\tanh\left(\operatorname{arcsinh}\left(b\sqrt{c}\right)\right)-\tanh\left(\operatorname{arcsinh}\left(a\sqrt{c}\right)\right)\right) \cr
&= \frac{b}{\sqrt{cb^{2}+1}}-\frac{a}{\sqrt{ca^{2}+1}}.
}
$$</span></p>
|
2,058,560 | <p>Can anyone please simplify this boolean expression ? My answer always reduces to a single variable i.e $x$ but my instructor reduced to three literals.
$$x'y'z+x'yz'+xy'z'+xyz$$ </p>
| gt6989b | 16,192 | <p>We have
$$
\begin{split}
x'y'z+x'yz'+xy'z'+xyz
&= x'(y'z+yz') + x(y'z' + yz) \\
&= x'(\overline{yz'} + yz') + x(\overline{yz} + yz)
\end{split}
$$
and now note that if $y=0=z$ then the expression comes out $x$ and in all other cases, it will depend of $x'$ - you cannot really simplify any further</p>
|
100,957 | <p>I have this picture of small particles in a polymer film. I want to count how many particles in the figure, so that I can have a rough estimation of the particle density. But the image quality is poor, I had a hard time to do it.</p>
<p><a href="https://i.stack.imgur.com/ez50R.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/ez50R.jpg" alt="particle in film - grayscale"></a>, <a href="https://i.stack.imgur.com/0OJd7.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/0OJd7.jpg" alt="particle in film - color"></a></p>
<p>I have tried several ways to do it, but failed. below is the code.
The first method I tried is:</p>
<pre><code>`SetDirectory["C:\\Users\\mayao\\documents"]
image = Import["Picture3.jpg"];
imag2 = Binarize[image, {0.0, 0.8}];
cells = SelectComponents[DeleteBorderComponents[imag2], "Count", -400];
circles = ComponentMeasurements[ImageMultiply[image,cells],"Centroid", "EquivalentDiskRadius"}][[All, 2]]; Show[image, Graphics[{Red, Thick, Circle @@ # & /@ circles}]]
</code></pre>
<p>Here is what I got:
<a href="https://i.stack.imgur.com/f7u8C.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/f7u8C.jpg" alt="enter image description here"></a></p>
<p>So it does not count all the particle. Plus, it sometimes take several particle as one.</p>
<p>I read another method from a thread here, the code is:</p>
<pre><code>obl[transit_Image] := (SelectComponents[
MorphologicalComponents[
DeleteSmallComponents@
ChanVeseBinarize[#, "TargetColor" -> Black],
Method -> "ConvexHull"], {"Count", "SemiAxes"},
Abs[Times @@ #2 Pi - #1] < #1/100 &]) &@
transit; GraphicsGrid[{#, obl@# // Colorize,
ImageMultiply[#,
Image@Unitize@
obl@#]} & /@ (Import /@ ("C:\\Users\\mayao\\documents\\" <> # \
& /@ {"Picture1.jpg", "Picture2.jpg", "Picture3.jpg",
"Picture1.jpg"}))]
</code></pre>
<p>But it does not recognize the single particles:</p>
<p><a href="https://i.stack.imgur.com/zav3Y.png" rel="noreferrer"><img src="https://i.stack.imgur.com/zav3Y.png" alt="enter image description here"></a></p>
<p>Is there any other method to do this task?
Thanks a lot for any suggestions.</p>
| M.R. | 403 | <p>In the post referenced in the comment, the key ideas is </p>
<blockquote>
<p>In order to separate the overlapping cells, I extract a set of markers
by finding local maximas of the DistanceTransform and use these as
markers in WatershedComponents</p>
</blockquote>
<p>However exactly following that example verbatim won't work (because there are hyperparameters) and would give something like this:</p>
<p><a href="https://i.stack.imgur.com/eusjf.png" rel="noreferrer"><img src="https://i.stack.imgur.com/eusjf.png" alt="enter image description here"></a></p>
<p>Here are the preprocessing steps I would take:</p>
<pre><code>s1 = GradientFilter[img, 5] // ImageAdjust;
s2 = Threshold[s1, {"Hard", "Cluster"}];
s3 = FillingTransform[s2];
s4 = DeleteSmallComponents[s3, 30];
</code></pre>
<p>The bottom line is that it is tough for a human to delineate the blobs, so it is for a computer. However you can play with the "minimum saliency" threshold as follows:</p>
<pre><code>marker = MaxDetect[DistanceTransform[s4, Padding -> 0] // ImageAdjust,
0.1];
Manipulate[
w = WatershedComponents[GradientFilter[s4, 3], marker,
Method -> {"MinimumSaliency", ms}];
cells = SelectComponents[w, {"Area", "EquivalentDiskRadius"},
10 < #1 < 1000 && #2 > 4 &];
measures =
ComponentMeasurements[{cells, img}, {"Centroid",
"EquivalentDiskRadius", "Area"}];
Show[img,
Graphics[{Thickness[0.003], Opacity[0.6, Yellow],
Circle @@ # & /@ (measures[[All, 2, 1 ;; 2]])}],
ImageSize -> 800], {ms, 0, 1}]
</code></pre>
<p>To manually to achieve to desired cell separations:</p>
<p><a href="https://i.stack.imgur.com/gEkla.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/gEkla.jpg" alt="enter image description here"></a></p>
<p>which will look much better:</p>
<p><a href="https://i.stack.imgur.com/X3s8z.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/X3s8z.jpg" alt="enter image description here"></a></p>
|
990,467 | <p>I am having some trouble understanding these two questions. Any help is appreciated. Scanned questions are included at the end.</p>
<p>6) We are given the function $ f(x) =\frac{1 - 2x} {2x^2 - 3x - 2} $</p>
<p>6 a) Find the equation of the vertical asymptotes. Explain how.</p>
<p>For the above question how did they first get the equation $ x =( 3 \pm √25 ) / 4, $ </p>
<p>and then get x = 2 and x = -1/2 out of it?</p>
<p>6 b) Find the equation of the horizontal asymptotes. Use a limit.</p>
<p>For this question I understand that when the degree of the numerator is less than the degree of the denominator it results in a horizontal asymptote. Thus here we get y = 0. Right? But I would still like to know if it is the same procedure they used in the answer sheet to get the answer 0/2.</p>
<p><img src="https://i.stack.imgur.com/hKhb4.jpg" alt="enter image description here"></p>
| robjohn | 13,854 | <p>Letting $u=\cos(x)$, then we need to have
$$
4^u=1-u^2
$$
which implies
$$
u^2+4^u=1
$$
that is, $(u,2^u)$ crosses the unit circle.</p>
<p>$\hspace{4.5cm}$<img src="https://i.stack.imgur.com/IoJWg.png" alt="enter image description here"></p>
<p>Now that the equation is $2^{\large \cos(x)}=|\!\sin(x)|$, we also need to consider the curve $(u,-2^u)$, in red.</p>
<p>The point $u=0$ is there, as is the point $u=-0.82560777817003350220$. Thus, we have solutions
$$
\color{#00A000}{\cos^{-1}(0)=\frac\pi2}
$$
and
$$
\color{#C00000}{-\cos^{-1}(0)=-\frac\pi2}
$$
and
$$
\color{#00A000}{\cos^{-1}(-0.82560777817003350220)=2.5420748334255680556}
$$
and
$$
\color{#C00000}{-\cos^{-1}(-0.82560777817003350220)=-2.5420748334255680556}
$$</p>
|
1,235,973 | <p>Im reaching a point in programing where I need to create basic shapes which I simply cant since my math skills are very bad. After finding out that the skills required are trigonometry I read a few books. The result was that I couldnt understand any word at all.</p>
<p>What is required in order to understand basic 2D and 3D trigonometry? Calculus? Algebra? Also how long in general would it take for an average person to learn the material. It would be great if you could give me a guide for someone like me who only knows adding, multiplication, division and substracting.</p>
| Community | -1 | <p>I am useing Khan academy to catch up on things I have hut the algebra 1 and 2 sections so I am going to learn liner and diffrential algebra before I go back to the algebra 1 and 2 sections, but when I am doing liner algebra I have noticed like the Pythagorean theorem is in the geometry section and some of the algebra sums deal ing geometry and trigometry sections, so I might just go sstudy geometry and trigomety then go back to liner, diffrential and algebra 1 and 2 sections before moving onto pre calculus and calculus sections because I whant to build a solid foundation of things before going to the calculus sections.</p>
|
2,248,303 | <p>Evaluate the integral $ \int_{0}^{\infty} {\frac{\sin ^3 (x)}{x} }\:dx$ </p>
| Jack D'Aurizio | 44,121 | <p>We have
$$ \int_{0}^{+\infty}\frac{\sin(mx)}{x}\,dx=\frac{\pi}{2} $$
for any $m\in\mathbb{N}^+$, hence it is enough to expand $\sin^3(x)$ as a Fourier sine series. Actually
$$ \sin^3(x) = \frac{3}{4}\sin(x)-\frac{1}{4}\sin(3x) $$
hence
$$ \int_{0}^{+\infty}\frac{\sin^3(x)}{x}\,dx = \left(\frac{3}{4}-\frac{1}{4}\right)\frac{\pi}{2} = \color{red}{\frac{\pi}{4}}.$$</p>
|
2,692,124 | <blockquote>
<p>A group <span class="math-container">$G$</span> is finite iff all its elements have the same finite order?</p>
</blockquote>
<p>One direction (when <span class="math-container">$G$</span> is finite) is trivial following from Lagrange's theorem. But if a each element of <span class="math-container">$G$</span> has the same finite order (says <span class="math-container">$n<\infty$</span>), i.e. <span class="math-container">$\forall g\in G$</span>, <span class="math-container">$g^n=e_G$</span>. Will that implies <span class="math-container">$G$</span> is finite? I try to find some counter examples like infinite group consists of only nilpotent elements (not nilpotent group), but it doesn't work. So the statement is true?</p>
<p>=========================</p>
<p>Edit: The statement should be "A group <span class="math-container">$G$</span> is finite iff the order of any elements divides a fixed finite number n?" So one direction should be: Given <span class="math-container">$n<\infty$</span>, <span class="math-container">$\forall g\in G$</span>, if <span class="math-container">$g^{m}=e_G\neq g^{m-1}$</span>, then <span class="math-container">$m\mid n\implies |G|<\infty$</span></p>
| Exodd | 161,426 | <p>Try
$$\mathbb Z /2\mathbb Z \times \mathbb Z /2\mathbb Z \times\mathbb Z /2\mathbb Z \times\dots.$$
Every element has order 2.</p>
|
1,808,183 | <p>Consider the Cauchy problem:
$$\left\{\begin{array}{lll}
x^2\partial_x u+y^2\partial_yu=u^2\\
u(x,2x)=1
\end{array}\right.$$
It is easy to show that the characteristic equations are given by:
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{px^2+qy^2}=\frac{dp}{2pu-2xp}=\frac{dq}{2qu-2yq}=dt$$
By the Lagrange-Charpit's method, we need a first integral $\Psi$ different to $\Phi=x^2p+y^2q-u^2$.</p>
<p>How I can find $\Psi$? Maybe we should get directly the characteristic curves of the previous system?</p>
<p>Many thanks!</p>
| izmirlig | 149,896 | <p>The explanation contains some errors so I will start at the beginning. <span class="math-container">$[X]$</span> is the distribution of the random variable, <span class="math-container">$X$</span>. <span class="math-container">$[X | Y]$</span> is the conditional distribution of <span class="math-container">$X$</span> given <span class="math-container">$Y$</span>.</p>
<p>Let <span class="math-container">$(W(t): t\geq 0 )$</span> be standard brownian motion. </p>
<p><span class="math-container">$[W(s) | W(t) = y] = P( W(s) \in dx, W(t) \in dy ) / P( W(t) \in dy)$</span></p>
<p><span class="math-container">$=P( W(t) \in dy | W(s) = x ) P(W(s) \in dx)/ P( W(t) \in dy)$</span></p>
<p><span class="math-container">$=\frac{1}{\sqrt{t-s}} \phi\left(\frac{y-x}{\sqrt{t-s}}\right) \frac{\frac{1}{\sqrt{s}}\phi(\frac{x}{\sqrt{s}})}{\frac{1}{\sqrt{t}}\phi(\frac{y}{\sqrt{t}})}$</span> </p>
<p>Completing the square in <span class="math-container">$x$</span> gives </p>
<p><span class="math-container">$\sqrt{\frac{t}{s(t-s)}} \phi\left(\frac{x - \frac{s}{t}y}{\sqrt{s(t-s)/t}}\right)$</span></p>
<p>Now the mean, variance and second moment are straightforward:</p>
<p><span class="math-container">$\mathrm{E}\left[ W(s) | W(t) = y\right] = \frac{s}{t} y$</span></p>
<p><span class="math-container">$\mathrm{var}\left[ W(s) | W(t) = y \right] = \frac{s(t-s)}{t}$</span></p>
<p><span class="math-container">$\mathrm{E}\left[ W(s)^2 | W(t) = y\right] = \frac{s(t-s)}{t} + \frac{s^2}{t^2} y^2$</span></p>
<p>Finally the covariance. Assume <span class="math-container">$ r < s < t $</span> :</p>
<p><span class="math-container">$\mathrm{cov}\left[ W(r), W(s) | W(t) = y\right]$</span></p>
<p><span class="math-container">$=\mathrm{E}\left[W(r)W(s) | W(t) = y\right] - \frac{r s y^2}{t^2}$</span></p>
<p>Continuing with the cross moment in the first term above,</p>
<p><span class="math-container">$\mathrm{E}\left[W(r)W(s) | W(t) = y\right]$</span></p>
<p><span class="math-container">$= \mathrm{E}\left[\mathrm{E}\left[W(r)W(s) | W(s), W(t) = y\right] | W(t)=y\right]$</span></p>
<p><span class="math-container">$= \mathrm{E}\left[\mathrm{E}\left[W(r)W(s) | W(s)\right] | W(t)=y\right]$</span></p>
<p><span class="math-container">$= \mathrm{E}\left[W(s) \mathrm{E}\left[W(r)| W(s)\right] | W(t)=y\right]$</span></p>
<p><span class="math-container">$= \mathrm{E}\left[W(s) \frac{r}{s} W(s) | W(t)=y\right]$</span></p>
<p><span class="math-container">$=\frac{r}{s}\left[\frac{s(t-s)}{t} + \frac{s^2}{t^2} y^2\right]$</span></p>
<p>In the above, the first equality follows, as someone pointed out above,
by the iterated expectations, e.g. <span class="math-container">$\mathrm{E}[ \mathrm{E}[ Y | X ] ] = \mathrm{E}[Y]$</span>. If the outer expectation is conditional as well, everything is conditional upon that: <span class="math-container">$\mathrm{E}[ \mathrm{E}[ Y | X, W ] | W] = \mathrm{E}[Y | W]$</span>. The next equality follows because <span class="math-container">$W(r)$</span> and <span class="math-container">$W(t)$</span> are independent given <span class="math-container">$W(s)$</span> because of the Markovian property of brownian motion. The next equality follows because we can pull <span class="math-container">$W(s)$</span> out of the inner conditional expectation, because, conditional upon it, it is a constant. The next equality is the above result regarding the mean, and the final equality follows from the above result regarding the second moment. </p>
<p>Having computed the cross moment, we can plug its value back into the above expression for the covariance. After simplifying, we obtain:</p>
<p><span class="math-container">$\mathrm{cov}\left[ W(r), W(s) | W(t) = y\right] = \frac{r(t-s)}{t}$</span></p>
<p>The mean, variance, and covariance function for brownian bridge now follow if we replace <span class="math-container">$(r,s,t,y)$</span> with <span class="math-container">$(s,t,1,0)$</span>.</p>
|
2,070,353 | <p>Let there be two arbitrary functions $f$ and $g$, and let $g \circ f$ be their composition.</p>
<ol>
<li><p>Suppose $g \circ f$ is one-one. Then prove that this implies $f$ is also one-one.</p></li>
<li><p>Suppose $g \circ f$ is onto. Then prove that this implies $g$ is also onto.</p></li>
</ol>
<p>I tried to do both of the proofs by contradiction method but got stuck in that approach. Please help me in this proof. Proofs in more than one way is welcomed.</p>
| Fawad | 369,983 | <p>$$a_n= 2n+1$$
$$a_1 = 3$$
$$a_n=7$$ $$2n+1=7$$ $$n=3$$ $$a_3=7$$</p>
<p>So series will become 3,5,7,9,…</p>
<p>So before 7 there are 2 terms which are not included in your series.
So there are total $n-2$ terms</p>
|
2,657,112 | <p>I am currently reading the paper "PRIMES is in p" and have come across some notation that I don't quite understand in this following sentence</p>
<blockquote>
<p>Consider a prime $q$ that is a factor of $n$ and let $q^k || n$. Then...</p>
</blockquote>
<p>What does the notation $q^k || n$ mean here?</p>
<p>The full paper can be found <a href="https://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf" rel="noreferrer">here</a> and the notation described above is used in the proof on page 2</p>
| Community | -1 | <p>It means that $k$ is the higher power that divides $n$. In other words:
$$q^k\|n\iff q^k|n\wedge q^{k+1}\nmid n$$</p>
|
1,755,107 | <p>Let $(X,d)$ be a metric space and let $A \subseteq X$. We define the distance from a point $x \in X$ to $A$ by $d(x,A)= \inf \{ d(x,a) : a \in A \} $.</p>
<p>What will be the value of $d(x, \emptyset )$? I am confused between $+ \infty $ and $- \infty$. Also, is it possible to find $ \min \{ d(x,a) : a \in \emptyset \} $? (I know $\emptyset$ is empty, just asking symbolically whether we can take $\min$ instead of $\inf$)</p>
<p>Thanks.</p>
| bof | 111,012 | <p>$d(x,\emptyset)=\inf\emptyset=\infty.$ That makes sense because in general, if $A\subseteq B,$ then $d(x,A)\ge d(x,B);$ so $d(x,\emptyset)\ge d(x,B)$ for all $B.$</p>
<p>If the infimum of a set $S$ is <em>attained</em>, that is, if $\inf S\in S,$ then $\min S=\inf S;$ otherwise, $\min S$ does not exist. Thus $\min\emptyset$ does not exist, just as $\min\{\frac1n:n\in\mathbb N\}$ does not exist. It should be no surprise that the empty set has no <em>least</em> element, seeing as it has no elements at all.</p>
|
1,599,502 | <p>Solve for $x$ in exact value:</p>
<p>$\\3^{2x}-3^{x+2}+8=0$</p>
<p>I have tried substituting $3^x$ $=a$ but I didn't get anywhere.</p>
<p>$\\a^2-a^{1+\frac{2}{x}}+8=0$</p>
| Nate | 91,364 | <p>You need to be a bit more specific about what you mean by an algebra.</p>
<p>If you are in the setting of finite dimensional associative algebras then the Artin-Wedderburn theorem tells us that any such simple algebra is a matrix ring $M_n(D)$ of a finite dimensional division algebra $D$ over the ground field $k$. In this case there is a unique irreducible module, namely $D^n$ (which is the unique simple in the category of $A$ modules).</p>
<p>Note that the converse is not quite true in this setting, the algebra $k[x]/(x^2)$ has a unique simple module $k$ where $x$ acts by zero. However it is not a simple algebra since it has the ideal generated by $x$, and its category of modules is not semisimple.</p>
<p>If you are talking about Lie algebras, then your example of $\mathfrak{sl}_2$ shows that a simple Lie algebra does not have a unique irreducible representation. Note that the universal enveloping algebra of $\mathfrak{sl}_2$ is neither finite dimensional nor simple so this doesn't violate the above discussion.</p>
<p>One more case to consider is that of an infinite dimensional simple associative algebra. One of the simplest such examples is the Weyl algebra $\mathbb{C}[x,y]/(xy-yx-1)$, which has no finite dimensional modules and infinitely many non-isomorphic irreducible infinite dimensional modules. </p>
|
1,262,444 | <p>I study hard throughout the year and I am able to solve most problems in the text assigned to us and I am frequently the only one who can solve the hardest problems in the assignments or the problem sets. However, I can rarely do exceptionally well at math tests in my University and I frequently run out of time while sitting for a test. My grades have suffered and I am unable to understand why.[I end up with 65% or something like that in my exams while others score like 80%.]</p>
<p>Since, I am able to understand every theorem in the book and then solve most of the problems in the book(and solve the hardest problems in any test), it seems to me that I should also have good grades on most tests I take.
I will appreciate it if people give me some advice on how to do well in the tests.
Thank you.</p>
| alkabary | 96,332 | <p>You know, Exams are like tournament or competitions. Some people are great at practice but when they enter a championship, and feel under pressure, they under perform. I am telling you this because I am student athlete , and I did track and field for 4 years while doing my double major in computer science and mathematics. </p>
<p>So with that being said, I highly suggest that you put yourself under exam situation while studying or throughout the semester. Like you said, you are solving hard problems. But maybe it takes you too long to solve it. You should time yourself, and try to do a given set of problems in a certain amount of time and see how well you do. And I have to say that being an athlete and competing at a high level made me handle exams quite well. I was mentally and physically trained to perform under stress.</p>
<p>So My other suggestion to you is to do activities that will help you handle stress better . I am not saying do track and field, maybe go to the gym, enrol in some sort of competitions in some hobby you like,... etc.</p>
<p>I actually feel some people are quite talented with handling exams, apart from the fact that they are hard working or not. </p>
<p>Good luck man. You will do better next time !</p>
|
851,112 | <p>My goal is to obtain a reasonable approximation of the Gini index of a company (UBS).
I need to obtain an estimate of the salaries distribution from publicly available data:</p>
<ul>
<li>Nuber of employees=60205 </li>
<li>total compensation paid=15.182E9 CHF</li>
<li>min salary=50000 CHF</li>
<li>median salary = 100000 CHF</li>
<li>max salary=11430000 CHF</li>
</ul>
<p>I know it's very underdeterminated, but what's the best that can be obtained from this ?</p>
| Matt L. | 70,664 | <p>Another nice way of seeing this equality is noting that</p>
<p>$$\sin\theta=\frac{1}{2i}(p-p^*)\quad\text{and}\quad\sin\phi=\frac{1}{2i}(q-q^*)$$</p>
<p>So the right-hand side is</p>
<p>$$\frac{p-p^*+q-q^*}{p-p^*-q+q^*}=\frac{(p+q)-(p+q)^*}{(p-q)-(p-q)^*}$$</p>
<p>Now note that $p^*=1/p$ and $q^*=1/q$, so we get</p>
<p>$$\frac{(p+q)-(\frac{1}{p}+\frac{1}{q})}{(p-q)-(\frac{1}{p}-\frac{1}{q})}=\frac{(p+q)-\frac{p+q}{pq}}{(p-q)+\frac{p-q}{pq}}=
\frac{(p+q)(1-\frac{1}{pq})}{(p-q)(1+\frac{1}{pq})}=\frac{(p+q)(pq-1)}{(p-q)(pq+1)}$$</p>
|
3,525,356 | <p><a href="https://i.stack.imgur.com/8OitL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8OitL.png" alt="enter image description here"></a></p>
<p>Can someone help me on this? Any help will be greatly appreciated. </p>
| ancient mathematician | 414,424 | <p>If <span class="math-container">$M$</span> has order <span class="math-container">$p$</span> then its minimal polynomial divides <span class="math-container">$X^p-1=(X-1)^p$</span>; and so is <span class="math-container">$(X-1)^2$</span>. So by the Jordan Canonical Form Theorem all elements of order <span class="math-container">$p$</span> are conjugate to <span class="math-container">$\begin{pmatrix}1&1\\0&1\end{pmatrix}$</span>.</p>
<p>To see how many conjugates this element has we must calculate the size of its centraliser: this is an easy calculation, the centraliser is just the set of all <span class="math-container">$\begin{pmatrix}a&b\\0&a\end{pmatrix}$</span>, where <span class="math-container">$a\not=0$</span>. </p>
<p>The whole group has order <span class="math-container">$(p^2-1)(p^2-p$</span>); the centraliser has order <span class="math-container">$(p-1)p$</span>.</p>
<p>There are therefore <span class="math-container">$(p^2-1)$</span> elements of order <span class="math-container">$p$</span>. </p>
|
3,525,356 | <p><a href="https://i.stack.imgur.com/8OitL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8OitL.png" alt="enter image description here"></a></p>
<p>Can someone help me on this? Any help will be greatly appreciated. </p>
| Nourr Mga | 707,002 | <p>Since <span class="math-container">$GL_{2}(\Bbb Z_{p})$</span> contains <span class="math-container">$p+1$</span> Sylow p-subgroups and each of the them contains <span class="math-container">$p-1$</span> elements of order <span class="math-container">$p$</span>, then the total number of elements of order <span class="math-container">$p$</span> in <span class="math-container">$GL_{2}(\Bbb Z_{p})$</span> is equal to <span class="math-container">$(p+1)(p-1)=p^{2}-1$</span>. </p>
|
2,137,269 | <p>I just sat in on a lecture on exponential generating functions in combinatorics (I have no formal education in combinatorics myself). It was quite interesting, but I'm afraid I don't actually understand what the generating function is/does. I've tried doing some minimal research online, but everything I've seen seems to be either too complex or too general to understand well. For example, I know how to find the generating function for permutations of a finite set, $\frac{1}{1-x}$. But what role does $x $ play here, and what does the generating function tell us? I don't see how it's at all related to the species of permutations itself.</p>
| Asinomás | 33,907 | <p>The generating function is a power series that is assigned to a sequence. If you have a sequence $\{a_n|n\in \mathbb N\}$ you can assign to this sequence a power series: $\sum\limits_{n=0}^\infty a_nx^n$.</p>
<p>Manipulating the power series using algebra (power series form a ring) can help to find a closed form for the sequence.</p>
|
2,137,269 | <p>I just sat in on a lecture on exponential generating functions in combinatorics (I have no formal education in combinatorics myself). It was quite interesting, but I'm afraid I don't actually understand what the generating function is/does. I've tried doing some minimal research online, but everything I've seen seems to be either too complex or too general to understand well. For example, I know how to find the generating function for permutations of a finite set, $\frac{1}{1-x}$. But what role does $x $ play here, and what does the generating function tell us? I don't see how it's at all related to the species of permutations itself.</p>
| Karl | 203,893 | <p>Suppose you have the sequence $2, 5, 8, 11,...$ </p>
<p>One idea might be to have the terms in the sequence as the coefficients of some polynomial. This polynomial is called a generating function of the sequence.</p>
<p>For example:</p>
<p>$$ g(x)=2x^1+5x^2+8x^3+11x^4+...$$</p>
<p>All seems fine but how do you recover the sequence if you have the generating function? In this example it is obvious but let's try out a couple of ideas to gain a greater insight.</p>
<p>Notice that differentiation of a power reduces its power by one. We are going to use this idea as a way of sliding the terms of the sequence forwards.</p>
<p>Suppose we want the first term we differentiate $g(x)$ once.</p>
<p>$$g'(x)=2+10x+24x^2+44x^3+....$$</p>
<p>Notice now we can evaluate the polynomial at $x=0$ and find the first term! $g'(0)=2$</p>
<p>The second term is more troublesome, can you see why?</p>
<p>Note the second derivative of the function is $g''(x)=10+48x+132x^2+...$ and that $g''(0)=10$ which is not the 2nd term. All is not lost as we can divide $g''(0)$ by $2$ to recover the second term (we multiplied by $2$ in differentiating). This problem will be exaggerated if we want higher terms in the sequence. The 4th term is found by computing $g^{iv}(0)$ and dividing by $4 \times 3 \times 2$ or $4!$</p>
<p>This situation is a pain and so we may choose to adapt the method.</p>
<p>One idea might be to preload the generating function with the dividing factorials first.</p>
<p>Let's define $$G_1(x)=2\frac{x^1}{1!}+5\frac{x^2}{2!}+8\frac{x^3}{3!}+11\frac{x^4}{4!}+...$$</p>
<p>Now we may differentiate the appropriate number of times and substitute $x=0$ with out hassle.</p>
<p>Finally we can observe the connection to the exponential function as $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$</p>
<p>Hope this helps to demystify the exponential at least in part.</p>
|
785,603 | <p>What do Sylow 2-subgroups of finite simple groups look like?</p>
<p>It'd be nice to have explanations of the Sylow 2-subgroups of finite simple groups. There are many aspects to the question, so I envision many answers. I tried to give an answer of "broad strokes with references." It would be reasonable to answer just for a single infinite family of finite simple groups too. Here are some specific aspects that could be addressed.</p>
<ul>
<li>Which finite simple groups have the same Sylow 2-subgroup?</li>
<li>Are there ways in which Sylow 2-subgroups of finite simple groups are vastly different from a "random" or "typical" 2-group? (For instance, are the nilpotency class and derived length related in some special way compared to the general 2-group.)</li>
<li>What do the normalizers of the Sylow 2-subgroups look like (in cases where we understand the Sylow itself well)?</li>
<li>What are its characteristic, normal in the normalizer, or normal subgroups?</li>
<li>What are the possible fusion systems?</li>
</ul>
| Jack Schmidt | 583 | <p>Nonabelian finite simple groups come in a few types:</p>
<ul>
<li>Alternating groups</li>
<li>Classical groups in odd characteristic</li>
<li>Exceptional groups in odd characteristic</li>
<li>Groups in even characteristic (classical or exceptional)</li>
<li>Sporadic groups</li>
</ul>
<p>In cases 1,2,3 the Sylow 2-subgroups are (slightly deformed versions of) direct products of wreath products $P_2 \wr C_2 \wr C_2 \wr \ldots \wr C_2$ where $P_2$ is the Sylow $2$-subgroup of a tiny group from the family. In case 4, the groups are best understood using linear algebra. In case 5, it would be nice to know which sporadics “borrow” a Sylow 2-subgroup and which have their own unique Sylow 2-subgroup.</p>
<h2>Alternating</h2>
<p>The Sylow 2-subgroups of the symmetric groups are direct products of wreath products of Sylow 2-subgroups of $S_2$ -- this was known in the 19th century. The Sylow 2-subgroups of the alternating groups are index 2 subgroups.</p>
<p>For $n=4m+2$ and $n=4m+3$, the copies of $S_{4m}$ inside $A_n$ have odd index $2m+1$ or $(4m+3)(2m+1)$, so the Sylow 2-subgroup of $S_{4m}$ is isomorphic to the Sylow 2-subgroups of $A_{4m+2}$ and $A_{4m+3}$.</p>
<p>Weisner (1925) computes the order of the normalizers of the Sylow $p$-subgroups of symmetric and alternating groups (so counts them). The main result for us is that Sylow 2-subgroups are self-normalizing in simple alternating groups (except $A_5$ with normalizer $A_4$).</p>
<p>Weir (1955) computes the characteristic subgroups of the Sylow $p$-subgroup of the symmetric groups, but only for odd $p$. Lewis (1968) modifies this to handle $p=2$ for both symmetric and alternating groups. Dmitruk–Suščanskʹkiĭ (1981) take the approach of Kaloujnine (1945-1948), again handling $p=2$ and alternating groups.</p>
<p>Harada–Lang (2005) observes that the Sylow 2-subgroups of $A_{4m}$ and $A_{4m+1}$ are directly indecomposable (while those of $A_{4m+2}$ and $A_{4m+3}$ are directly indecomposable iff $m$ is a power of $2$).</p>
<ul>
<li>Weisner, Louis;
<a href="http://dx.doi.org/10.2307/2370639">“On the Sylow Subgroups of the Symmetric and Alternating Groups.”</a>
Amer. J. Math. 47 (1925), no. 2, 121–124.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=1506549">1506549</a>
DOI:<a href="http://dx.doi.org/10.2307/2370639">10.2307/2370639</a></li>
<li>Kaloujnine, Léo
“La structure des p-groupes de Sylow des groupes symétriques finis.”
Ann. Sci. École Norm. Sup. (3) 65, (1948). 239–276.
<sup><sub>Also see: C. R. Acad. Sci. Paris
<a href="http://www.ams.org/mathscinet-getitem?mr=14087">221 (1945), 222–224</a>; ibid.
<a href="http://www.ams.org/mathscinet-getitem?mr=16416">222 (1946), 1424–1425</a>; ibid.
<a href="http://www.ams.org/mathscinet-getitem?mr=18178">223 (1946), 703–705</a>; ibid.
<a href="http://www.ams.org/mathscinet-getitem?mr=19089">224 (1947), 253–255</a>.</sub></sup></li>
<li>Weir, A. J.
<a href="http://dx.doi.org/10.2307/2033425">“The Sylow subgroups of the symmetric groups.”</a>
Proc. Amer. Math. Soc. 6 (1955), 534–541.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=72142">72142</a>
DOI:<a href="http://dx.doi.org/10.2307/2033425">10.2307/2033425</a></li>
<li>Lewis, Robert Edward.
<a href="http://gateway.proquest.com/openurl?url_ver=Z39.88-2004%26rft_val_fmt=info:ofi/fmt:kev:mtx:dissertation%26res_dat=xri:pqdiss%26rft_dat=xri:pqdiss:6910772">“On the Sylow two-subgroups of the alternating groups.”</a>
Thesis (Ph.D.)–University of Illinois at Urbana-Champaign. 1968. 48 pp.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=2617989">2617989</a></li>
<li>Dmitruk, Ju. V.; Suščanskʹkiĭ, V. Ī.
“Construction of Sylow 2-subgroups of alternating groups and normalizers of Sylow subgroups in symmetric and alternating groups.”
Ukrain. Mat. Zh. 33 (1981), no. 3, 304–312.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=621637">621637</a></li>
<li>Harada, Koichiro; Lang, Mong Lung.
<a href="http://dx.doi.org/10.1007/s10440-004-5618-0">“Indecomposable Sylow 2-subgroups of simple groups.”</a>
Acta Appl. Math. 85 (2005), no. 1-3, 161–194.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=2128910">2128910</a>
DOI<a href="http://dx.doi.org/10.1007/s10440-004-5618-0">10.1007/s10440-004-5618-0</a></li>
</ul>
<h2>Classical groups in odd characteristic</h2>
<p>There is a huge difference in Sylow $p$-subgroup structure depending on whether $p$ is the characteristic of the field. In this section we assume $p$ is not the characteristic of the field.</p>
<p>In case $p$ is not the characteristic, then Weir (1955) showed that symmetric groups and classical groups are very similar, but again $p=2$ was left out until Carter-Fong (1964), and then more uniformly in Wong (1967). Algorithms to handle all Sylow $p$-subgroups of classical groups are described in Stather (2008).</p>
<p>The gist is that in GL, GO, GU, and Sp, the Sylow $p$-subgroups are direct products of wreath products of cyclic groups of order $p$ with the Sylow $p$-subgroup of the two-dimensional groups. For SL, SO or $\Omega$, SU the answers are more complicated, but only because an easy to understand part has been chopped off the top.</p>
<ul>
<li>Weir, A. J.
<a href="http://dx.doi.org/10.2307/2033424">“Sylow $p$-subgroups of the classical groups over finite fields with characteristic prime to $p$.”</a>
Proc. Amer. Math. Soc. 6 (1955), 529–533.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=72143">72143</a>
DOI:<a href="http://dx.doi.org/10.2307/2033424">10.2307/2033424</a></li>
<li>Carter, Roger; Fong, Paul.
<a href="http://dx.doi.org/10.1016/0021-8693(64)90030-4">“The Sylow 2-subgroups of the finite classical groups.”</a>
J. Algebra 1 (1964) 139–151.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=166271">166271</a>
DOI:<a href="http://dx.doi.org/10.1016/0021-8693(64)90030-4">10.1016/0021-8693(64)90030-4</a></li>
<li>Wong, Warren J.
<a href="http://dx.doi.org/10.1007/BF01112816">“Twisted wreath products and Sylow 2−subgroups of classical simple groups.”</a>
Math. Z. 97 (1967) 406–424.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=212095">212095</a>
DOI:<a href="http://dx.doi.org/10.1007/BF01112816">10.1007/BF01112816</a></li>
<li>Stather, Mark.
<a href="http://dx.doi.org/10.1016/j.jalgebra.2007.03.013">“Constructive Sylow theorems for the classical groups.”</a>
J. Algebra 316 (2007), no. 2, 536–559.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=2356843">2356843</a>
DOI:<a href="http://dx.doi.org/10.1016/j.jalgebra.2007.03.013">10.1016/j.jalgebra.2007.03.013</a></li>
</ul>
<h2>Exceptional groups in odd characteristic</h2>
<p>Sylow 2-subgroups for finite groups of Lie type are similar to the classical case: there is a 2-dimensional group $P_2$ and a “top” group $X$ (which need not be $C_2 \wr C_2 \wr \ldots \wr C_2$, but that is probably the correct picture to have) such that the $X$-conjugates of $P_2$ are commute with each other, so that $X \ltimes P_2^n$ is a Sylow 2-subgroup. The $P_2$ are the Sylow 2-subgroups of the so-called “fundamental subgroups” of Aschbacher (1977), where we view groups of Lie type as built up from rank 1 groups, in this case commuting rank 1 subgroups isomorphic to SL2. These are used in Aschbacher (1980) to describe groups in which a Sylow 2-subgroup is contained in a unique maximal subgroup, and Harada–Lang (2005) describes which Sylow 2-subgroups are indecomposable. GLS I.A.4.10 covers Aschbacher's ideas as well.</p>
<ul>
<li>Aschbacher, Michael.
<a href="http://dx.doi.org/10.2307/1971100">“A characterization of Chevalley groups over fields of odd order.”</a>
Ann. of Math. (2) 106 (1977), no. 2, 353–398.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=498828">498828</a></li>
<li>Aschbacher, Michael.
<a href="http://dx.doi.org/10.2307/1971063">“A characterization of Chevalley groups over fields of odd order. II.”</a>
Ann. of Math. (2) 106 (1977), no. 3, 399–468.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=498829">498829</a></li>
<li>Aschbacher, Michael.
<a href="http://dx.doi.org/10.1016/0021-8693(80)90095-2">“On finite groups of Lie type and odd characteristic.”</a>
J. Algebra 66 (1980), no. 2, 400–424.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=593602">593602</a>
DOI:<a href="http://dx.doi.org/10.1016/0021-8693(80)90095-2">10.1016/0021-8693(80)90095-2</a></li>
<li>Gorenstein, Daniel; Lyons, Richard; Solomon, Ronald.
<em>The classification of the finite simple groups.</em> Number 3. Part I. Chapter A.
Almost simple K-groups. Mathematical Surveys and Monographs, 40.3. American Mathematical Society, Providence, RI, 1998. xvi+419 pp. ISBN: 0-8218-0391-3
MR<a href="http://www.ams.org/mathscinet-getitem?mr=1490581">1490581</a></li>
<li>Harada, Koichiro; Lang, Mong Lung.
<a href="http://dx.doi.org/10.1007/s10440-004-5618-0">“Indecomposable Sylow 2-subgroups of simple groups.”</a>
Acta Appl. Math. 85 (2005), no. 1-3, 161–194.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=2128910">2128910</a>
DOI<a href="http://dx.doi.org/10.1007/s10440-004-5618-0">10.1007/s10440-004-5618-0</a></li>
</ul>
<h2>Groups in characteristic 2</h2>
<p>Here the Sylow 2-subgroups are basically groups of upper triangular matrices and are often best understood in terms of linear algebra. Weir (1955) describes the characteristic subgroups and those normalized by important subgroups of GL. These general ideas work in all the groups of Lie type. The main description I know is Chevalley's commutator formula, as explained in Carter (1972).</p>
<p>XXX: Decent reference for the classical, and then the exceptional. Maybe specifically handle Suzuki and Ree.</p>
<ul>
<li>Weir, A. J.
<a href="http://dx.doi.org/10.2307/2032788">“Sylow $p$-subgroups of the general linear group over finite fields of characteristic $p$.”</a>
Proc. Amer. Math. Soc. 6, (1955). 454–464.
MR<a href="http://www.ams.org/mathscinet-getitem?mr=69175">69175</a>
DOI:<a href="http://dx.doi.org/10.2307/2032788">10.2307/2032788</a></li>
<li>Carter, Roger W.
<em>Simple groups of Lie type.</em>
Pure and Applied Mathematics, Vol. 28. John Wiley & Sons, London-New York-Sydney, 1972. viii+331 pp.
Mr<a href="http://www.ams.org/mathscinet-getitem?mr=407163">407163</a></li>
</ul>
<h2>Sporadic</h2>
<p>I think each one is a special snowflake. XXX: Lookup coincidences in Sylow structure.</p>
|
785,603 | <p>What do Sylow 2-subgroups of finite simple groups look like?</p>
<p>It'd be nice to have explanations of the Sylow 2-subgroups of finite simple groups. There are many aspects to the question, so I envision many answers. I tried to give an answer of "broad strokes with references." It would be reasonable to answer just for a single infinite family of finite simple groups too. Here are some specific aspects that could be addressed.</p>
<ul>
<li>Which finite simple groups have the same Sylow 2-subgroup?</li>
<li>Are there ways in which Sylow 2-subgroups of finite simple groups are vastly different from a "random" or "typical" 2-group? (For instance, are the nilpotency class and derived length related in some special way compared to the general 2-group.)</li>
<li>What do the normalizers of the Sylow 2-subgroups look like (in cases where we understand the Sylow itself well)?</li>
<li>What are its characteristic, normal in the normalizer, or normal subgroups?</li>
<li>What are the possible fusion systems?</li>
</ul>
| Jack Schmidt | 583 | <h2>Normalizers and top fusion</h2>
<p>Kondratʹev (2005) contains a very clear description of the normalizers of Sylow 2-subgroups $S$ of finite simple groups. Typically there is not much going on: $N_G(S) = S$ unless:</p>
<ul>
<li>$G$ is a group of Lie type in even characteristic (then $N_G(S)$ is a Borel subgroup and $N_G(S)/S$ is a maximally-split maximal torus, which is usually not the identity unless $q=2$)</li>
<li>$S$ is abelian, so described in <a href="http://en.wikipedia.org/wiki/Walter_theorem">Walter's theorem</a> (Janko-Ree or certain PSL2)</li>
<li>$N_G(S)=S C_G(S)$ as in PSL, PSU, E6 and <sup>2</sup>E6</li>
<li>$N_G(S)/S$ is an elementary abelian 3-group ($J_2,J_3, HN,PSp_{2m}$ for most values of $m$)</li>
</ul>
<p>So other than groups of even characteristic, the automorphisms of $S$ induced from $G$ are very limited, usually inner, but with an order 7 in the Janko–Ree case and order 3 in the Janko–Ree, three more sporadics, and most of the symplectic groups.</p>
<h2>Bibliography</h2>
<ul>
<li>Kondratʹev, A. S.
<a href="http://www.ams.org/leavingmsn?url=http://dx.doi.org/10.1007/s11006-005-0133-9">“Normalizers of Sylow 2-subgroups in finite simple groups.”</a> (<a href="http://dx.doi.org/10.4213/mzm2593">original</a>)
Mat. Zametki 78 (2005), no. 3, 368--376; translation in
Math. Notes 78 (2005), no. 3-4, 338–346
MR<a href="http://www.ams.org/mathscinet-getitem?mr=2227510">2227510</a></li>
</ul>
|
494,141 | <p>I'm slightly confused about representations of Simple functions. I take it that simple functions are of the form $\phi(x) = \sum_{k=1}^N a_k \chi_{A_k}(x)$ where $\chi$ is the indicator function and that $A_k$'s are measurable sets.</p>
<p>Now on building the theory of integration using simple functions, it appears that it is necessary to construct the canonical form of simple functions for a definition of the Lebesgue integral. (well that's how Stein does it).</p>
<p>But my question is, if $\phi$ is a simple function defined above where it is not in its canonical form, will writing it in its canonical form ensure that the resulting disjoint sets are measurable? Or is this tautological by definition??</p>
<p>I cannot seem to grasp this because subsets of measurable sets aren't necessarily measurable.</p>
| QED | 91,884 | <p>Assume $f:(\Omega,\mathcal{A})\rightarrow\mathbb{R}$ is a simple function. Then $f$ is a measurable function that takes only finitely many real values. Let $\{c_1,\cdots,c_n\}$ be the set of possible values of $f$. Define $A_i=\{\omega\in\Omega:f(\omega)=c_i\}=f^{-1}\left(\{c_i\}\right)$. Then $A_i$'s are measurable and
$$f=\sum_{i=1}^nc_i\chi_{A_i}$$
These $A_i$'s are disjoint and their union is $\Omega$. Thus for every simple function $f$ I can choose such a representation such that the $A_i$'s form a partition of $\Omega$.</p>
|
3,560,378 | <p>Let <span class="math-container">$f: \mathbb{R} \to \mathbb{R}$</span> be a convex function. Let <span class="math-container">$x,y,z \in \mathbb{R}$</span> and <span class="math-container">$a,b,c \geq 0$</span> such that <span class="math-container">$a+b+c=1$</span>. We want to show that <span class="math-container">$f(ax+by+cz) \leq af(x)+bf(y)+cf(z)$</span>.</p>
<p>Obviously, if <span class="math-container">$c=0$</span>, the result is equivalent to the definition of convexity. I think I should use the two variable case to prove the the three variable case (and ideally generalize to <span class="math-container">$n$</span> variables), but I don't see how to do it.</p>
| Donald Splutterwit | 404,247 | <p><span class="math-container">\begin{eqnarray*}
f(ax+by+cz) &=& f \left( ax +(b+c) \frac{by+cz}{b+c} \right) \\
& \leq & a f(x) +(b+c) f \left( \frac{by+cz}{b+c} \right) \\
\end{eqnarray*}</span>
Now
<span class="math-container">\begin{eqnarray*}
f \left( \frac{by+cz}{b+c} \right) & \leq & \frac{b}{b+c} f ( y) + \frac{c}{b+c} f ( z) \\
\end{eqnarray*}</span>
Put these together and you have the result.</p>
|
3,203,460 | <p>The whole question looks like-<br></p>
<blockquote>
<p><span class="math-container">$f$</span> is a holomorphic function on the interior of the open unit disc
(say <span class="math-container">$D$</span>) around <span class="math-container">$0$</span> and <span class="math-container">$f(z)\ne0\ \forall z\in D$</span>. Suppose,
<span class="math-container">$|f(z)|\to1$</span> as <span class="math-container">$|z|\to1$</span>. Is f a constant function? Now, if we remove
the non-vanishing condition then is <span class="math-container">$f$</span> constant?<br></p>
</blockquote>
<p>The 2nd part is easy if we take <span class="math-container">$f(z)=z$</span>, then <span class="math-container">$|f(z)|\to1$</span> as <span class="math-container">$|z|\to1$</span> but <span class="math-container">$f(0)=0$</span>.<br>
For the first part, I tried to show <span class="math-container">$f'(z)=0\ \forall z\in D$</span><br>
From application of Cauchy intrgral formula, we know that <span class="math-container">$f'(z)={1\over 2\pi i}\int_{C} \frac{f(\zeta)}{(\zeta-z)^2}d\zeta$</span> where <span class="math-container">$z\in D$</span> and <span class="math-container">$C$</span> is a circle around <span class="math-container">$z$</span> inside <span class="math-container">$D$</span>.<br>
So, <span class="math-container">$|f'(z)|\le\sup\{\frac{f(\zeta)}{(\zeta-z)^2}|\zeta\in C\}\operatorname{Radius}(C)$</span><br>
Can use these concepts and the condition <span class="math-container">$|f(z)|\to1$</span> as <span class="math-container">$|z|\to1$</span> to show <span class="math-container">$f$</span> is constant? I even don't know the answer, so may be the <span class="math-container">$f$</span> is non-constant. But I think it will be constant.<br>
Can anybody solve the problem? Thanks for assistance in advance.</p>
| Martin R | 42,969 | <p>You have to show that the given conditions imply that <span class="math-container">$|f(z)| \le 1$</span> in the unit disk <span class="math-container">$\Bbb D$</span>. Then the same reasoning can be applied to <span class="math-container">$1/f$</span> (since <span class="math-container">$f$</span> has no zeros). It follows that <span class="math-container">$|f(z)| = 1$</span> in <span class="math-container">$\Bbb D$</span>, and the maximum modulus principle (or open mapping principle) implies that <span class="math-container">$f$</span> is constant.</p>
<p>So it remains to show that:</p>
<blockquote>
<p>Let <span class="math-container">$f$</span> be holomorphic in the unit disk <span class="math-container">$\Bbb D$</span> with the property that <span class="math-container">$|f(z_n)| \to 1$</span> for each sequence <span class="math-container">$(z_n)$</span> in <span class="math-container">$\Bbb D$</span> with <span class="math-container">$|z_n| \to 1$</span>. Then <span class="math-container">$|f(z)| \le 1$</span> for all <span class="math-container">$z \in \Bbb D$</span>.</p>
</blockquote>
<p>If <span class="math-container">$f$</span> were continuous on the closure <span class="math-container">$\overline{\Bbb D}$</span> then this would be a simple application of the maximum modulus principle. For the general case we can consider the maximum modulus on an increasing sequence of disks, and proceed as follows:</p>
<p>For <span class="math-container">$0 \le r < 1$</span> let
<span class="math-container">$$
M(r) = \max \{ |f(z)| : |z| = r \}
$$</span>
be the maximal modulus of <span class="math-container">$f(z)$</span> on the circle <span class="math-container">$|z| = r$</span>. It follows from the maximum modulus principle that <span class="math-container">$M(r)$</span> is increasing in <span class="math-container">$r$</span>. Now choose a sequence of increasing radii converging to one:
<span class="math-container">$$
0 < r_1 < r_2 < r_3 < \ldots \, , \quad r_n \to 1 \, .
$$</span>
Then
<span class="math-container">$$
M(r_1) \le M(r_2) \le M(r_3) \le \ldots \, .
$$</span>
For each circle <span class="math-container">$|z| = r_n$</span>, <span class="math-container">$|f(z)|$</span> attains the maximum value <span class="math-container">$M(r_n)$</span> at a point <span class="math-container">$z_n$</span>, i.e. there is a sequence <span class="math-container">$(z_n)$</span> in <span class="math-container">$\Bbb D$</span> such hat
<span class="math-container">$$
|z_n| = r_n \text{ and } |f(z_n)| = M(r_n)
$$</span>
for all <span class="math-container">$n$</span>. In particular <span class="math-container">$|z_n| \to 1$</span>, which implies
<span class="math-container">$$
M(r_n) = |f(z_n)| \to 1
$$</span>
for <span class="math-container">$n \to \infty$</span>. It follows that <span class="math-container">$M(r_n) \le 1$</span> for all <span class="math-container">$n$</span>, and consequently <span class="math-container">$M(r) \le 1$</span> for all <span class="math-container">$0 \le r < 1$</span>.</p>
|
3,203,460 | <p>The whole question looks like-<br></p>
<blockquote>
<p><span class="math-container">$f$</span> is a holomorphic function on the interior of the open unit disc
(say <span class="math-container">$D$</span>) around <span class="math-container">$0$</span> and <span class="math-container">$f(z)\ne0\ \forall z\in D$</span>. Suppose,
<span class="math-container">$|f(z)|\to1$</span> as <span class="math-container">$|z|\to1$</span>. Is f a constant function? Now, if we remove
the non-vanishing condition then is <span class="math-container">$f$</span> constant?<br></p>
</blockquote>
<p>The 2nd part is easy if we take <span class="math-container">$f(z)=z$</span>, then <span class="math-container">$|f(z)|\to1$</span> as <span class="math-container">$|z|\to1$</span> but <span class="math-container">$f(0)=0$</span>.<br>
For the first part, I tried to show <span class="math-container">$f'(z)=0\ \forall z\in D$</span><br>
From application of Cauchy intrgral formula, we know that <span class="math-container">$f'(z)={1\over 2\pi i}\int_{C} \frac{f(\zeta)}{(\zeta-z)^2}d\zeta$</span> where <span class="math-container">$z\in D$</span> and <span class="math-container">$C$</span> is a circle around <span class="math-container">$z$</span> inside <span class="math-container">$D$</span>.<br>
So, <span class="math-container">$|f'(z)|\le\sup\{\frac{f(\zeta)}{(\zeta-z)^2}|\zeta\in C\}\operatorname{Radius}(C)$</span><br>
Can use these concepts and the condition <span class="math-container">$|f(z)|\to1$</span> as <span class="math-container">$|z|\to1$</span> to show <span class="math-container">$f$</span> is constant? I even don't know the answer, so may be the <span class="math-container">$f$</span> is non-constant. But I think it will be constant.<br>
Can anybody solve the problem? Thanks for assistance in advance.</p>
| Subham Jaiswal | 238,932 | <p>INCORRECT (1 does not imply 2, since in Cauchy's inequality <span class="math-container">$R$</span> gets smaller and smaller as we go close to the boundary. We do get uniform continuity on any sub disc of radius <span class="math-container">$r < 1$</span> for any <span class="math-container">$r$</span>).
I unable to add this as a comment so writing here:</p>
<ol>
<li><p>Show that <span class="math-container">$f$</span> is bounded.</p>
</li>
<li><p>Use 1. to show that <span class="math-container">$f$</span> is uniformly continuous.</p>
</li>
<li><p>So, <span class="math-container">$f$</span> can be extended to the closed unit disc and we have <span class="math-container">$|f(z)| = 1$</span> whenevar <span class="math-container">$|z| = 1$</span>.</p>
</li>
<li><p>Now extend <span class="math-container">$f$</span> to all of <span class="math-container">$\mathbb{C}$</span> by a reflection type construction.</p>
</li>
<li><p>Now use Louville's theorem.</p>
</li>
</ol>
|
406,252 | <p>How to define the amount of additions. E.g. $1+2+3+4+5+6+7+8+9$</p>
<p>Are there $9$ additions, because of the nine numbers that are added together.
Or can you also say that there are $8$ additions, because there are only $8$ '$+$' signs.</p>
| iostream007 | 76,954 | <p>Addition is a binary operator that uses two operands to execute.so if there are $n$ number of elements then number of addition will be $$\;n-1$$.In other answers people uses "addition of $0$",because $0$ is addition identity,it will give $n$ addition but it is only in case of certain operator like $+,-$ but there are other binary operator also like $\times ,/$ with these operator we can not use $0$ there.For these operator we have multiplicative identity $1$.But regardless we can say there are minimum $$n-1$$ operations.</p>
|
432,291 | <p>This question is related to <a href="https://mathoverflow.net/questions/421450/homotopy-type-theory-how-to-disprove-that-0-operatornamesucc0-in-the-ty">Homotopy type theory : how to disprove that <span class="math-container">$0=\mathrm{succ}(0)$</span>
in the type <span class="math-container">$\mathbb N$</span></a>.</p>
<p>Section 2.13 in <a href="https://homotopytypetheory.org/book/" rel="nofollow noreferrer">The HoTT Book</a> uses "the encode-decode method to characterize the path space of the natural numbers" by defining a type family:</p>
<p><span class="math-container">$$\mathrm{code} : \mathbb N \to \mathbb N \to \cal U$$</span></p>
<p>with</p>
<p><span class="math-container">$$\begin{array}{rcl}
\mathrm{code}(0,0) & :\equiv & \mathbf 1\newline
\mathrm{code}(\mathrm{succ} (m),0) & :\equiv & \mathbf 0\newline
\dots & :\equiv & \dots\newline
\dots & :\equiv & \dots
\end{array}$$</span></p>
<p>To my understanding, <span class="math-container">$\mathrm{code}$</span> is only well-defined, if (in particular) <span class="math-container">$0:\mathbb N$</span> and <span class="math-container">$\mathrm{succ}(0):\mathbb N$</span> are <strong>not</strong> judgementally equal. How can we be sure that this is the case?</p>
| Mike Shulman | 49 | <p>Daniel's answer is correct that the judgmental distinctness of <span class="math-container">$0$</span> and <span class="math-container">$\mathsf{succ}(m)$</span> is not what justifies a definition by pattern-matching. However, it is still a meaningful question of how to prove that <span class="math-container">$0$</span> and <span class="math-container">$\mathsf{succ}(m)$</span> are not judgmentally equal. (Stretching terminology a bit, Daniel answered your <a href="https://xyproblem.info/" rel="nofollow noreferrer">X</a> and I'm going to answer your Y.)</p>
<p>Note that this is a metatheoretic question, in contrast to the internal proof that <span class="math-container">$0\neq \mathsf{succ}(m)$</span>. Thus, even though judgmental equality implies typal equality, the answer to that question does not answer this one.</p>
<p>It's also worth noting that this is not a trivial question, even though <span class="math-container">$0$</span> and <span class="math-container">$\mathsf{succ}$</span> are distinct constructors. From an "algebraic" perspective on type theory, judgmental equality is just the smallest congruence generated by certain rules. Thus, two terms that don't "look" like they could possibly be equal might turn out to be equal by passing through some chain of forwards and backwards equalities to other terms.</p>
<p>For the same reason, it is a "global" question about the type theory, not one that can be answered by referring only to <span class="math-container">$\mathbb{N}$</span>. For instance, in a higher inductive type such as the interval <span class="math-container">$\mathsf{I}$</span>, with constructors <span class="math-container">$\mathsf{zero},\mathsf{one} : \mathsf{I}$</span> and <span class="math-container">$\mathsf{seg}:\mathsf{zero}=\mathsf{one}$</span>, there is no "local" reason for <span class="math-container">$\mathsf{zero}$</span> and <span class="math-container">$\mathsf{one}$</span> to be judgmentally equal; but if the type theory also includes the equality reflection rule, then <span class="math-container">$\mathsf{seg}$</span> implies that <span class="math-container">$\mathsf{zero}\equiv\mathsf{one}$</span>.</p>
<p>One answer to this question involves giving an algorithm for checking whether two terms are judgmentally equal. Probably the simplest such algorithm involves giving a "rewriting system" under which every term can be reduced to a "normal form", and then two terms are judgmentally equal if they reduce to the same normal form (up to <span class="math-container">$\alpha$</span>-equivalence). Roughly speaking, this rewriting system consists of what the HoTT Book calls "computation rules". One then has to prove that this rewriting system is terminating, and that the relation of "reducing to the same normal form" is a congruence; thus it coincides with judgmental equality. Then one can simply observe that <span class="math-container">$0$</span> and <span class="math-container">$\mathsf{succ}(m)$</span> are (when <span class="math-container">$m$</span> is a variable) distinct normal forms.</p>
<p>In practice, nowadays more complicated algorithms are used. Among other reasons, this is in order to also include what the HoTT Book calls "uniqueness rules". These algorithms often go by names like "normalization by evaluation". The crucial idea is that there are two "phases" of the algorithm, one which applies the computational rewrites, and another which applies the uniqueness rules in the <em>expansionary</em> direction (<span class="math-container">$f \mapsto (\lambda x. f(x))$</span> or <span class="math-container">$u \mapsto (\pi_1(u),\pi_2(u))$</span>) by inspecting their types. But the basic idea is the same: after proving that the algorithm is terminating and complete, we can run the algorithm on two terms to verify that they are unequal (or equal).</p>
<p>With that said, I don't know whether anyone has <em>actually</em> done this for Book HoTT. It's known for MLTT, on which Book HoTT is based, but Book HoTT adds not just axioms (which don't disrupt such an algorithm) to MLTT but new judgmental equalities (the computation rules for point-constructors of HITs), which then have to be incorporated in the algorithm. So (even laying aside the point that "Book HoTT" is not precisely specified because it is open-ended with respect to what HITs are definable) there may technically be an open question here. However, I think no type theorist who's familiar with such proofs would have much doubt that standard techniques ought to apply to Book HoTT (and it's possible that someone has already done it).</p>
<p><strong>Edit:</strong> Simon has pointed out in the comments what I should clearly have realized myself, namely that another way to deduce that two terms are not judgmentally equal is to give a model in which they are interpreted by distinct things. Thus, for instance, the set-theoretic model proves this for MLTT, and the simplicial set model proves it for Book HoTT (modulo details in the construction of that model that aren't yet in the literature, such as showing that the univalent universes are closed under a sufficiently large class of higher inductive types).</p>
|
2,179,012 | <p>Let $V_1$ be a Banach space and let $A_1 \subset V_1$ be a compact subset of $V_1$</p>
<p>Let $V_2$ be a Banach space and let $A_2 \subset V_2$ be a compact subset of $V_2$</p>
<p>Let $f: A_1 \to A_2 $ be a homeomorphism</p>
<p>Let $x_1,y_1 \in A_1$ such that $$\left\| x_1-y_1 \right\|=\sup_{v,u \in A_1} \left\| v-u \right\|$$</p>
<p>Let $x_2,y_2 \in A_2$ such that $$\left\| x_2-y_2 \right\|=\sup_{v,u \in A_2} \left\| v-u \right\|$$</p>
<p>I would like to know if is true that
$$ f(x_1)=x_2 \text{ and } f(y_1)=y_2$$</p>
<p>thanks</p>
| Mahmoud Hassan | 282,900 | <p>Here are the three books I study from:</p>
<ul>
<li>S. Ponnusamy, Herb Silverman - Complex Variables with Applications.</li>
<li>Ravi P. Agarwal - An Introduction to Complex Analysis.</li>
<li>Jerrold E. Marsden, Michael J. Hoffman - Basic Complex Analysis</li>
</ul>
|
567,428 | <p>Let $A \ne V$ be a subspace of $V$ and $B$ a linearly independent subset of $V$. Prove that $B$ can be completed to a basis of $V$ with vectors from $V \setminus A$.</p>
<p>OK, I started with: </p>
<p>$K=\operatorname{lin}(v_1,v_2,v_3,\dotsc)$</p>
<p>and if any of $v_1,v_2,\dotsc$ belongs to $A$ we can replace this vector by a linear combination of vectors which don't belong to $A$ </p>
<p>So $v_{x_1},v_{x_2},\dotsc \in A$</p>
<p>$v_{x_1}\in\operatorname{lin}(w_{1,1},w_{1,2},\dotsc)$<br>
$v_{x_2}\in\operatorname{lin}(w_{2,1},w_{2,2},\dotsc)$<br>
etc.</p>
<p>Hence $V=\operatorname{lin}(w_{1,1},w_{1,2},\dots,w_{2,1},\dots,w_{i,j})$</p>
<p>and I don't know what to do next, but what I know is that I should use the Steinitz exchange lemma.</p>
<p>Does anybody have an idea how to finish this prove?</p>
| rnrstopstraffic | 218,011 | <p>Assume $$Dim(V)=m,$$ $$Dim(A)=n<m,$$ $$\text{and } B=\{v_1,...,v_n\}.$$ </p>
<p>Then $$Span(B)=A\neq V,$$</p>
<p>so $\exists v_{n+1}\in V\backslash A$, which means that there is no $n$-tuple $(c_1,...,c_n)$ with $v_{n+1}=c_1v_1+...+c_nv_n$ , i.e. $B\cup \{v_{n+1}\}$ is linearly independent. If $n+1=m$, we are done because we have a linearly independent set of $m$ vectors, which must be a basis for $V$. Otherwise, repeat the process.</p>
|
6,056 | <p>Are there any good places for me to sell off my mathematics books online especially Springer and Dover books?</p>
<p>(I thought perhaps this was off topic, but then I thought everybody in math probably has the problem of having bought pricey books that they would prefer to get some money back for, at one point or another.)</p>
| Jeremy Batterson | 2,302 | <p>Amazon will buy back a lot of textbooks at pretty fair rates. Check <a href="http://www.amazon.com/Textbooks-Trade-In/b/ref=bhp_9p2_A_txbq410_05?ie=UTF8&node=2205237011&pf_rd_m=ATVPDKIKX0DER&pf_rd_s=center-6&pf_rd_r=0RH8PF985VSEMWFKG1SV&pf_rd_t=101&pf_rd_p=1275995182&pf_rd_i=283155" rel="nofollow">this</a> out.</p>
|
310,238 | <p>The <a href="https://en.wikipedia.org/wiki/Weyl_character_formula" rel="nofollow noreferrer">Weyl character formula</a> and <a href="https://en.wikipedia.org/wiki/Weyl_character_formula#Weyl_denominator_formula" rel="nofollow noreferrer">the denominator identity</a> play important roles in the representation theory of classical simple Lie algebras and Kac-Moody Lie algebras over $\mathbb{C}$</p>
<p>Can you suggest any reference for similar formulas for Lie super-algebras? I suppose there are such formulas, at least for classes of (say classical simple) Lie super-algebras.</p>
| Konstantinos Kanakoglou | 85,967 | <p>I agree with the suggestion in the comments for searching the front of the math arXiv (as an entry point), because this is a quite broad and active topic (and i am not sure it can be fully covered within a couple of references). </p>
<p>However, i think it might be particularly useful to mention: </p>
<ul>
<li>The <a href="https://arxiv.org/abs/hep-th/9607161v1" rel="nofollow noreferrer">Dictionary on Lie Superalgebras</a>, by L. Frappat, A. Sciarrino, P. Sorba, sect. 7, p.10-12, where character and supercharacter formulas are given for the classical, basic Lie superalgebras $A(m,n)$, $B(0,n)$, $C(n+1)$.</li>
</ul>
<p>and also the following papers by S.J Kang :</p>
<ul>
<li><a href="https://core.ac.uk/download/pdf/81208379.pdf" rel="nofollow noreferrer">S.J. Kang, Graded Lie Superalgebras and the Superdimension Formula, J. of Alg., v.204, 597-655, 1998</a>, </li>
<li><a href="https://www.cambridge.org/core/journals/proceedings-of-the-london-mathematical-society/article/graded-lie-superalgebras-supertrace-formula-and-orbit-lie-superalgebras/3B6060558C726FF852AAEBAFA0797AE4" rel="nofollow noreferrer"> S.J. Kang, J.H. Kwon, Graded Lie Superalgebras, Supertrace Formula, and Orbit Lie Superalgebras, Proc. of the London Math. Soc. v.81, 3, November 2000 , p. 675-724</a> (see <a href="https://arxiv.org/pdf/math/9809025.pdf" rel="nofollow noreferrer">here</a> for the arXiv version), where a generalized denominator identity is obtained for a<br>
$(\Gamma\times\mathcal{A})$-graded Lie superalgebra, where $\Gamma$ is a countable abelian semigroup and $\mathcal{A}$ is a countable abelian group satisfying a certain finiteness condition. </li>
</ul>
|
1,564,750 | <p>Imagine that your country's postal system only issues 2 cent and 5 cent stamps. Prove that it possible to pay for postage using only these stamps for any amount n cents, where n is at least 4.</p>
<hr>
<p>My attempt (using Strong induction, I know we can use induction but since they say you can can apply strong induction for generalized/weak induction cases):</p>
<p>Base case:
$$n=4: 2 \times2cents$$
$$n=5: 0 \times 5cents$$
$$n=6: 3 \times 2cents$$
$$n=7: 1 \times 2 cents + 1 \times 5 cents$$</p>
<p>You might ask why I have so many base cases, this is my reason why: The question states that we can pay for postage using 2 and 5 cents only. Hence, we have 3 general cases: </p>
<p>1: ONLY 2-cents stamps are used</p>
<p>2: ONLY 5-cents stamps are used</p>
<p>3: 2-cents and 5-cents stamps are used.</p>
<p>(Till now, are my base cases valid?)</p>
<p>Assume that for n=k, P(k) is true and that we need to show P(k+1)</p>
<p>$\textbf{Induction hypothesis:}$ P(k) is true when $4\le i \le k$ and $k \ge 7$</p>
<p>Since $4\le (k-3) \le k$ , P(k-3) or i is true by Induction hypothesis.</p>
<p>Now, k-3 cents can be formed using 2 and 5-cent stamps.</p>
<p>To get k+1 stamps, we can just replace it with $\textbf{four}$ 2-cent stamps?</p>
<hr>
<p>Thank you! Is my proof valid or no? Any alternatives for this question using strong induction? Also, if I have used strong mathematical induction wrong or any of the steps are incorrect, please explain why.</p>
| Mario Carneiro | 50,776 | <p>Here's another induction-free proof, using the fact that integers are even or odd:</p>
<p>If $n$ is even, then $n=2k$ and we are done ($k$ two-cent stamps).</p>
<p>If $n$ is odd, so $n=2k+1$, then since $n\ge4$ we have $k\ge\frac32$ whence $k\ge2$ because $k$ is an integer. Therefore $m=k-2\ge0$, and $n=2m+5$, so we can use $m$ two-cent stamps and one five-cent stamp.</p>
|
2,925,160 | <p>We know that <span class="math-container">$\mathbb{R}$</span> is a linear space over <span class="math-container">$\mathbb{Q}$</span>, denoted by <span class="math-container">$\mathbb{R_{\mathbb{Q}}}$</span>,where the vector addition is the real numbers and scalar multiplication is the number in <span class="math-container">$\mathbb{Q}$</span> times the number in <span class="math-container">$\mathbb{R}$</span>.
Then <span class="math-container">$ \forall n\in\mathbb{N^{+}}$</span>,<span class="math-container">$\{1,\sqrt[n]{2},\sqrt[n]{2^2},\cdots,\sqrt[n]{2^{n-1}}\}$</span> is a linearly independent set in <span class="math-container">$\mathbb{R_{\mathbb{Q}}}.$</span>
Further,we can get <span class="math-container">$\mathbb{R_{\mathbb{Q}}}$</span> is an infinite dimensional linear space.</p>
<p>Let <span class="math-container">$\mathbb{Q}(\sqrt[3]{2}):=\{a+b\sqrt[3]{2}+c\sqrt[3]{2^2} \Big|a,b,c\in \mathbb{Q}\}.$</span></p>
<p>Similarly，<span class="math-container">$\mathbb{R}$</span> is a linear space over <span class="math-container">$\mathbb{Q}(\sqrt[3]{2})$</span>, denoted by <span class="math-container">$\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$</span>, where the vector addition is the real numbers and scalar multiplication is the number in <span class="math-container">$\mathbb{Q}(\sqrt[3]{2})$</span> times the number in <span class="math-container">$\mathbb{R}$</span>. We can proof that <span class="math-container">$\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$</span> also is an infinite dimensional linear space. In fact, If we suppose <span class="math-container">$\dim(\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}})\in \mathbb{N^{+}},$</span> then <span class="math-container">$\dim(\mathbb{R_{\mathbb{Q}}})\in\mathbb{N^{+}}$</span> holds. But forall <span class="math-container">$n\in\mathbb{N^{+}},$</span> how can I find <span class="math-container">$n$</span> elements of <span class="math-container">$\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$</span> such that these are linearly independent in <span class="math-container">$\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$</span>?</p>
| Lubin | 17,760 | <p>Here’s a way of looking at your problem that makes use of ramification theory. It may be too advanced for your taste, but at least it does provide an answer to your question.</p>
<p>You’re looking for many real quantities, preferably algebraic, that will be linearly independent over the real field <span class="math-container">$\Bbb Q\left(2^{1/3}\right)$</span>. As an extension of <span class="math-container">$\Bbb Q$</span>, this field is ramified over only <span class="math-container">$2$</span> and <span class="math-container">$3$</span>. So if you get lots of real fields that are <em>un</em>ramified over <span class="math-container">$2$</span> and <span class="math-container">$3$</span>, you have a solution to your problem.</p>
<p>A quadratic extension <span class="math-container">$\Bbb Q(\sqrt n\,)$</span> of <span class="math-container">$\Bbb Q$</span> (with <span class="math-container">$n$</span> squarefree) is unramified above <span class="math-container">$2$</span> if <span class="math-container">$n\equiv1\pmod4$</span>. And unramified above <span class="math-container">$3$</span> if <span class="math-container">$n$</span> is prime to <span class="math-container">$3$</span>. So take positive values of <span class="math-container">$n$</span> satisfying these conditions, such as <span class="math-container">$5,13,17,41,77,\cdots$</span> As you see with <span class="math-container">$77$</span>, they don’t need to be prime.</p>
<p>Anyway, adjoin the square roots of <span class="math-container">$m$</span> of these numbers, and you should get a field of dimension <span class="math-container">$2^m$</span> over <span class="math-container">$\Bbb Q\left(2^{1/3}\right)$</span>.</p>
|
217,580 | <p>Consider a compact manifold $M$ with boundary and corner. As an example, we could have the cube $\{(x_1, x_2,..x_n) \in \mathbb{R}^n : x_i \in [0,1]\}$. We could very well define the Laplacian $\Delta$ on such an $M$ with Dirichlet or Neumann boundary conditions. In such situations, are the eigenfunctions of the Laplacian smooth? It seems to me that this requires some sort of elliptic regularity theory for singular spaces (maybe there is an elliptic regularity theory for manifolds with Lipschitz boundaries). A reference would be highly appreciated.</p>
| Michael Renardy | 12,120 | <p>There is an extensive literature on elliptic problems on domains with corners. Grisvard's book (Elliptic Problems in Nonsmooth Domains) is a good place to start. </p>
|
3,507,927 | <p>Is there any simplification by the way of standard properties of Laplace Transform for this? <span class="math-container">$$L_{s} [t f(t) f'(t)]$$</span></p>
<p>where <span class="math-container">$f'(t)$</span> is <span class="math-container">$\frac{d}{dt} f(t)$</span></p>
<p>Alternately, is there a simplification of <span class="math-container">$L_{s} [f(t) f'(t)]$</span>, so that I can compute my original expression by computing the derivative <span class="math-container">$-\frac{d}{ds} L_{s} [f(t) f'(t)]$</span> ?</p>
| user577215664 | 475,762 | <p>Since:
<span class="math-container">$$\mathcal {L}\{tg(t)\}=-\frac {d}{ds}\mathcal {L}\{g(t)\}$$</span>
We have:
<span class="math-container">$$\mathcal {L}[tf(t)f′(t)]=-\frac {d}{ds}\mathcal {L}\{f(t)f′(t)\}$$</span>
<span class="math-container">$$\mathcal {L}[tf(t)f′(t)]=-\frac 1 2\frac {d}{ds}\mathcal {L}\{(f^2(t))′\}$$</span>
Use the fact that :
<span class="math-container">$$ \mathcal{L}{(g'(t))}=sG(s)-g(0) $$</span>
<span class="math-container">$$
\begin{align}
\mathcal {L}[tf(t)f′(t)]&=-\frac 1 2\frac {d}{ds}(s\mathcal {L}\{f^2(t)\}(s)-f^2(0)) \\
\mathcal {L}[tf(t)f′(t)]&=-\frac 1 2\frac {d}{ds}(s\mathcal {L}\{f^2(t)\}(s)) \\
\end{align}
$$</span></p>
|
3,490,024 | <p>This is part of a physics problem I was doing yesterday. I am supposed to find the maximum value of
<span class="math-container">$$\mu^{-1}\cos\theta+\sin\theta$$</span>
This is is supposed to produce the result of <span class="math-container">$\sqrt{1+\mu^{-2}}$</span>. However I am not getting that. By taking the derivative of the equation, and setting it equal to zero I got
<span class="math-container">$$-\mu^{-1}\sin\theta+\cos\theta=0$$</span>
<span class="math-container">$$\mu\cos\theta=\sin\theta$$</span>
<span class="math-container">$$\mu=\tan\theta$$</span>
I don’t get the same result of <span class="math-container">$\sqrt{1+\mu^{-2}}$</span> after plugging <span class="math-container">$\tan^{-1}{(\mu)}$</span> back into our equation. Can anyone help?</p>
| Abhishek Vangipuram | 630,593 | <p>It is known that for <span class="math-container">$A\cos(\theta) + B\sin(\theta)$</span>, we can represent it as 1 trigonometric function: <span class="math-container">$\sqrt{A^2+B^2}\cos(\theta + \phi)$</span> or <span class="math-container">$\sqrt{A^2+B^2}\sin(\theta+\omega)$</span>. Clearly it can be <span class="math-container">$\sin$</span> or <span class="math-container">$\cos$</span> because they are just shifted waves of each other.</p>
<p>From these expressions of 1 sinusoid, it is clear the maximum value is <span class="math-container">$\sqrt{A^2+B^2}$</span>, giving your answer of <span class="math-container">$\sqrt{1+\mu^{-2}}$</span>.</p>
<p>Using the calculus approach, we get that a critical point (other than 0) is <span class="math-container">$\arctan(\mu)$</span>.</p>
<p>Plugging this in the original function we get <span class="math-container">$\frac{1}{\mu}\cos(\arctan(\mu))+\sin(\arctan(\mu))$</span>.</p>
<p>Now it will help to draw a right triangle with legs <span class="math-container">$\mu$</span>, <span class="math-container">$1$</span> and hypotenuse <span class="math-container">$\sqrt{1+\mu^2}$</span> because then there is an angle of the triangle equal to <span class="math-container">$\arctan(\mu)$</span>.</p>
<p>Using this triangle, <span class="math-container">$\sin(\arctan(\mu))=\mu/\sqrt{1+\mu^2}$</span> and <span class="math-container">$\cos(\arctan(\mu))=1/\sqrt{1+\mu^2}$</span>. </p>
<p><span class="math-container">$$\mu^{-1}/\sqrt{1+\mu^2} + \mu/\sqrt{1+\mu^2} = \mu^{-1}(1+\mu^2)/\sqrt{1+\mu^2} = \mu^{-1}\sqrt{1+\mu^2}=\sqrt{1+\mu^{-2}},$$</span>the desired result as the maximum value.</p>
|
1,836,236 | <p>For which $a$ and $b$ is this matrix diagonalizable?</p>
<p>$$A=\begin{pmatrix} a & 0 & b \\ 0 & b & 0 \\ b & 0 & a \end{pmatrix}$$</p>
<p>How to get those $a$ and $b$? I calculated eigenvalues and eigenvectors, but don't know what to do next?</p>
| DonAntonio | 31,254 | <p>That is a symmetric matrix: it is <strong>always</strong> diagonalizable over a field with characteristic$\;\neq2\;$. Not only that: it is <em>orthogonally</em> diagonalizable.</p>
|
3,676,003 | <p>I would like to find the parametric equation for a curve starting at radius of curvature 10 at angle 0 degrees and ending at radius of curvature 100 at 90 degrees. The equation for change in radius of curvature along the path will be specified. </p>
<p>One can imagine this as a 90 degree arc of a circle, except the radius of the circle is changing along the path. This is similar to the concept of Euler curves, except the curvature only changes linearly for Euler curves. </p>
<p>Any suggestions on how to approach this problem? </p>
<p>Edit: Thanks for the many replies! Actually my question is that the rate of change of radius of curvature can be any function. In my case, I'd like it to be a tanh function going from radius r1 to radius r2. The radii I mentioned above are also arbitrary values. I am looking for an analytical or numerical way to approach the generalized problem. </p>
| Narasimham | 95,860 | <p>Natural/intrinsic differential equations need to be used. The Whewell's equation binding tangent rotation <span class="math-container">$\phi$</span> and arc length <span class="math-container">$s$</span> is more suitable compared to Cesàro equation <span class="math-container">$(\kappa, s)$</span> and it can be used for a given variation of curvature thus :</p>
<p><span class="math-container">$$\frac{1}{\kappa} = R= \frac{ds}{d \phi} = a \tanh (\phi) \tag1 $$</span></p>
<p>Integrate using <em>arbitrary</em> boundary conditions ( does not matter because the entire track can be subjected to any Euclidean motion, i.e., rotation and shift in plane) arc length <span class="math-container">$ s=0 ,\, \phi= -\frac{\pi}{4}$</span></p>
<p><span class="math-container">$$s= a\,log\,\dfrac{\cosh \phi}{\cosh \pi/4} ;\; \cosh \phi= \cosh \pi/4 \cdot e^{s/a}\tag2 $$</span></p>
<p>and parametric coordinates</p>
<p><span class="math-container">$$ x= \int \cos\phi \;ds,\,y= \int \sin\phi \;ds,\,\tag3 $$</span></p>
<p>Designated constant <span class="math-container">$a$</span> is obtained (by numerical solution) of two simultaneous equations for<span class="math-container">$(\phi_i, a) $</span>
<span class="math-container">$$ R_1= a \tanh \phi_i,\; R_2= a\tanh (\phi_i+ \pi/2)\tag4 $$</span></p>
<p>Numerical integration gives the required bent curve.</p>
<p>EDIT1:</p>
<p>The unmodified <span class="math-container">$\tanh(x)$</span> has a maximum of <span class="math-container">$45^\circ$</span> against the required <span class="math-container">$90^\circ$</span> tangent rotation. It does not result in the required rotation, so is an unsuitable choice for specifying the curve. </p>
<p>An Ellipse rotates <span class="math-container">$90^\circ$</span> for curvature ratio <span class="math-container">$ (a/b)^3 $</span> in a quadrant. Also the <em>director circle</em> is the solution of perpendicular tangents in any ellipse.</p>
<p>Other examples can be suggested: the <em>Cornu spiral</em> and the <em>Involute</em> can accommodate any large rotation between perpendicular tangents.</p>
|
286,312 | <p>I'm starting to understand how induction works (with the whole $k \to k+1$ thing), but I'm not exactly sure how summations play a role. I'm a bit confused by this question specifically:</p>
<p>$$
\sum_{i=1}^n 3i-2 = \frac{n(3n-1)}{2}
$$</p>
<p>Any hints would be greatly appreciate</p>
| Sigur | 31,682 | <p>For $n=1$ you have $\sum_{i=1}^1(3i-2)=3-2=1=\frac{1(3\cdot 1-1)}{2}$.</p>
<p>Suppose that $$\sum_{i=1}^n(3i-2)=\frac{n(3n-1)}{2}$$ and prove that $$\sum_{i=1}^{n+1}(3i-2)=\frac{(n+1)(3(n+1)-1)}{2}.$$</p>
|
971,997 | <p>How do we find
$$\Re\left[\int_0^{\large\frac{\pi}{2}} e^{\Large e^{i\theta}}d\theta\right]$$</p>
<p>In the shortest and easiest possible manner?</p>
<p>I cannot think of anything good.</p>
| Leucippus | 148,155 | <p>Using
\begin{align}
e^{x} = \sum_{r=0}^{\infty} \frac{x^{r}}{r!}
\end{align}
then
\begin{align}
I &= \int_{0}^{\pi/2} e^{e^{i \theta}} \, d\theta = \sum_{r=0}^{\infty} \frac{1}{r!} \, \int_{0}^{\pi/2} e^{i \theta} \, d\theta \\
&= - i \,\sum_{r=0}^{\infty} \frac{e^{\pi i r/2} -1}{r \cdot r!} \\
&= -i \, \sum_{r=0}^{\infty} \frac{i^{r}-1}{r \cdot r!} \\
&= Si(1) + \frac{\pi}{2} + i \left( Ei(1) - Ci(1) \right)
\end{align}
where $Si(x)$, $Ci(x)$, and $Ei(x)$ are the Sine Integral, Cosine Integral, and Exponential Integral, respectively. This leads to
\begin{align}
\Re\int_{0}^{\pi/2} e^{e^{i\theta}} d\theta &= \int_{0}^{\pi/2} e^{\cos(\theta)} \, \cos(\sin(\theta)) \, d\theta = Si(1) + \frac{\pi}{2} \\
\Im \int_{0}^{\pi/2} e^{e^{i\theta}} d\theta &= \int_{0}^{\pi/2} e^{\cos(\theta)} \, \sin(\sin(\theta)) \, d\theta = Ei(1) - Ci(1)
\end{align}</p>
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