qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,998,938 | <p>How can I solve \begin{cases}
u_t-u_{xx}=0,&\text{if $0<x<1, t>0$}\\
u(0,t)=u(1,t)=0, & \text{if $t>0$}\\u(x,0)=u_0(x), &\text{if $x\in(0,1)$} \end{cases}</p>
<p>where $$u_0=min(x,1-x)$$</p>
| robjohn | 13,854 | <p>We can derive a recursion for $f_n$ which is valid for $n\ge2$ and $1\lt s\lt n$:
$$
\begin{align}
f_n(s)
&=\int_0^\infty\left\{\frac1t\right\}^nt^{s-1}\,\mathrm{d}t\\
&=\int_0^\infty\{t\}^nt^{-s-1}\,\mathrm{d}t\\
&=-\frac1s\int_0^\infty\{t\}^n\,\mathrm{d}t^{-s}\\
&=\frac ns\int_0^\infty\{t\}^{n-1}t^{-s}\,\mathrm{d}t-\frac{\zeta(s)}s\\
&=\frac nsf_{n-1}(s-1)-\frac{\zeta(s)}s\tag{1}
\end{align}
$$
We can explicitly compute $f_1(s)$ for $0\lt s\lt1$ using $(1)$-$(4)$ of <a href="https://math.stackexchange.com/a/162426">this answer</a>:
$$
\begin{align}
f_1(s)
&=\int_0^\infty\{t\}t^{-s-1}\,\mathrm{d}t\\
&=\lim_{L\to\infty}\int_0^L\{t\}t^{-s-1}\,\mathrm{d}t\\
&=\lim_{L\to\infty}-\frac1s\int_0^L\{t\}\,\mathrm{d}t^{-s}\\
&=\lim_{L\to\infty}\left(\frac1s\int_0^Lt^{-s}\,\mathrm{d}t-\frac1s\sum_{k=1}^Lk^{-s}\right)\\
&=\lim_{L\to\infty}\left(\frac1s\frac{L^{1-s}}{1-s}-\frac1s\sum_{k=1}^Lk^{-s}\right)\\
&=-\frac{\zeta(s)}s\tag{2}
\end{align}
$$
Multiply $(1)$ by $\frac{\Gamma(s+1)}{n!}$ and rearrange to get
$$
\frac{\Gamma(s+1)}{n!}f_n(s)-\frac{\Gamma(s)}{(n-1)!}f_{n-1}(s-1)=-\frac{\Gamma(s)\zeta(s)}{n!}\tag{3}
$$
Summing $(3)$, we get
$$
\frac{\Gamma(s+1)}{n!}f_n(s)-\frac{\Gamma(s-n+2)}{1!}f_1(s-n+1)=-\sum_{k=0}^{n-2}\frac{\Gamma(s-k)\zeta(s-k)}{(n-k)!}\tag{4}
$$
Applying $(2)$ to $(4)$, we get
$$
\frac{\Gamma(s+1)}{n!}f_n(s)=-\sum_{k=0}^{n-1}\frac{\Gamma(s-k)\zeta(s-k)}{(n-k)!}\tag{5}
$$
Simplifying $(5)$ gives
$$
f_n(s)=-\sum_{k=0}^{n-1}\frac{\binom{n}{k}}{\binom{s}{k}}\frac{\zeta(s-k)}{s-k}\tag{6}
$$</p>
<hr>
<p><strong>Extension by Analytic Continuation</strong></p>
<p>The recursion $(1)$ computes $f_n(s)$ from $f_1(s-n+1)$ and the integral for $f_1(s-n+1)$ in $(2)$ converges when $n-1\lt\mathrm{Re}(s)\lt n$. The integral for $f_n(s)$ in $(1)$ converges for $0\lt\mathrm{Re}(s)\lt n$, is analytic, and agrees with $(6)$ for $n-1\lt\mathrm{Re}(s)\lt n$. By analytic continuation, $(6)$ holds for $0\lt\mathrm{Re}(s)\lt n$.</p>
|
3,913,732 | <p>The following was asked by a high school student which I could not answer. Please help</p>
<p>In the figure below, show that the bisector of <span class="math-container">$\angle AEB$</span> and <span class="math-container">$\angle AFD$</span> intersect at perpendicular <img src="https://i.stack.imgur.com/76tDO.jpg" alt="enter image description here" /></p>
| Aman Kumar | 663,536 | <p>One way to solve it is as follows:</p>
<p><span class="math-container">$$\cos{(1.3\pi\cos{\theta})} = \cos{1.3\pi}$$</span></p>
<p>Using the identity I gave above,</p>
<p><span class="math-container">$$1.3\pi\cos{\theta} = 2n\pi\pm\ 1.3\pi.....n\in Z$$</span></p>
<p>Dividing both sides by <span class="math-container">$1.3\pi$</span>, we get,</p>
<p><span class="math-container">$$\cos{\theta} = \frac{20n}{13}\pm1$$</span></p>
<p>Solve for inequalities
<span class="math-container">$$-1<\frac{20n}{13}+1<1$$</span></p>
<p>and
<span class="math-container">$$-1<\frac{20n}{13}-1<1$$</span></p>
<p>since <span class="math-container">$\cos{\theta} \in [-1,1]$</span>. Then take inverse cosine for the solutions you get.</p>
|
132,862 | <p>Is it true that given a matrix $A_{m\times n}$, $A$ is regular / invertible if and only if $m=n$ and $A$ is a basis in $\mathbb{R}^n$?</p>
<p>Seems so to me, but I haven't seen anything in my book yet that says it directly.</p>
| Henry | 6,460 | <p>Informally, $\{0,1,2,3,\ldots\}\subseteq\mathbb{N}$ comes from the first and second axioms.</p>
<p>Of course you would need to define what $\{0,1,2,3,\ldots\}$ is. Perhaps writing it $\{0,S(0),S(S(0)),S(S(S(0))),\ldots\}$ makes it clearer. Then you can take a particular member of this set and use the first and second axioms to show it is in $\mathbb{N}$. </p>
|
33,817 | <p>It is an open problem to prove that $\pi$ and $e$ are algebraically independent over $\mathbb{Q}$.</p>
<ul>
<li>What are some of the important results leading toward proving this?</li>
<li>What are the most promising theories and approaches for this problem?</li>
</ul>
| muad | 4,361 | <p>There is a proof of the algebraic independence of $\pi$ and $e^\pi$ in <a href="http://www.springer.com/mathematics/numbers/book/978-3-540-41496-4">Introduction to Algebraic Independence Theory</a> and <em>a detailed exposition of methods created in last the 25 years</em> although I have not read it.</p>
|
1,665,833 | <p>Given that A $\in$ M $_{mxn}$ (<strong>R</strong>). Assume that {$v_1$...$v_n$} is a basis for $R^n$ such that {$v_1$...$v_k$} is a basis for Null(A). </p>
<p>How would I prove that {A$v_{k+1}$...A$v_n$} spans Col(A)?</p>
| Michael James Kali Galarnyk | 522,520 | <p>Here is some python code to make your own z-table (in case you find it useful)</p>
<pre><code>from scipy.integrate import quad
import numpy as np
import pandas as pd
def normalProbabilityDensity(x):
constant = 1.0 / np.sqrt(2*np.pi)
return(constant * np.exp((-x**2) / 2.0) )
standard_normal_table = pd.DataFrame(data = [],
index = np.round(np.arange(0, 3.5, .1),2),
columns = np.round(np.arange(0.00, .1, .01), 2))
for index in standard_normal_table.index:
for column in standard_normal_table.columns:
z = np.round(index + column, 2)
value, _ = quad(normalProbabilityDensity, np.NINF, z)
standard_normal_table.loc[index, column] = value
# Formatting to make the table look like a z-table
standard_normal_table.index = standard_normal_table.index.astype(str)
standard_normal_table.columns = [str(column).ljust(4,'0') for column in standard_normal_table.columns]
</code></pre>
<p>I also derived the math explaining how to make it <a href="https://towardsdatascience.com/how-to-use-and-create-a-z-table-standard-normal-table-240e21f36e53" rel="nofollow noreferrer">here</a>. </p>
|
3,620,375 | <p>I am asked to calculate the integral <span class="math-container">$$\int_C \frac{1}{z-a}dz$$</span> where <span class="math-container">$C$</span> is the circle centered at the origin with radius <span class="math-container">$r$</span> and <span class="math-container">$|a|\neq r$</span></p>
<p>I parametrized the circle and got <span class="math-container">$$\int_0^{2\pi}\frac{ire^{it}}{re^{it}-a}dt=\text{log}(re^{2\pi i}-a)-\text{log}(r-a)=\text{log}(r-a)-\text{log}(r-a)=0$$</span></p>
<p>Because of how the complex logarithm is defined, I am pretty sure that the first equaility is wrong, it it is, what is the correct solution?</p>
| Surb | 154,545 | <p>If <span class="math-container">$r<|a|$</span> the integral is indeed <span class="math-container">$0$</span>, but if <span class="math-container">$r>|a|$</span>, then using <a href="https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula" rel="nofollow noreferrer">Cauchy integral formula</a> (or more generally, Residue theorem), you get <span class="math-container">$2i\pi$</span>.</p>
<hr>
<p>You can also do what you did, but :</p>
<p>Notice that if <span class="math-container">$a\in \mathbb R$</span>, <span class="math-container">$a>0$</span>, then :</p>
<ul>
<li><p>If <span class="math-container">$r>a$</span>, then <span class="math-container">$$\ln(re^{2i\pi}-a)=\ln(e^{2i\pi}(r-a))=2i\pi+\ln(r-a),$$</span></p></li>
<li><p>If <span class="math-container">$r<a$</span>, then <span class="math-container">$$\ln(re^{2i\theta }-a)-\ln(r-a)=\ln(e^{i\pi}(a-r))-\ln(e^{i\pi}(a-r))=0.$$</span></p></li>
</ul>
<p>You can generalize this to the case <span class="math-container">$a\in \mathbb C$</span>, and distinguish the case <span class="math-container">$r>|a|$</span> or <span class="math-container">$r<|a|$</span>.</p>
|
3,079,493 | <p>Let <span class="math-container">$$D_6=\langle a,b| a^6=b^2=1, ab=ba^{-1}\rangle$$</span> <span class="math-container">$$D_6=\{1,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b\}$$</span></p>
<p>I would like to compute its character table and its irreducible representations.</p>
<p>I will explain what I have done so far and I will add some doubts I had while doing this.</p>
<p><strong>MY ATTEMPT</strong></p>
<ol>
<li><p>Compute conjugacy classes.
<span class="math-container">$$C_1=\{e\}, C_2=\{a,a^5\},C_3=\{a^2,a^4\}$$</span><span class="math-container">$$C_4=\{a^3\},C_5=\{b,a^2b,a^4b\},C_6=\{ab,a^3b,a^5b\}$$</span></p></li>
<li><p>Find <span class="math-container">$1$</span>-dimensional representations. Since <span class="math-container">$D_6/\{a,a^5\}\cong \mathbb{Z}_2$</span>, we have one more representation apart from <span class="math-container">$\alpha_1=id$</span>. That is <span class="math-container">$$\alpha_2: G \longrightarrow \mathbb{C}: a \mapsto 1, b\mapsto -1$$</span>
Again using irreducible representations from quotient group by normal subgroup, I considered <span class="math-container">$G/\{\overline{1},\overline{a},\overline{b},\overline{ab}\}\cong \mathbb{Z}_2\times\mathbb{Z}_2$</span> (since it is abelian). Then from here I obtained
<span class="math-container">$$\alpha_3:G\longrightarrow \mathbb{C}: a\mapsto -1, b\mapsto 1$$</span>
<span class="math-container">$$\alpha_4:G\longrightarrow \mathbb{C}: a\mapsto -1, b\mapsto -1$$</span></p></li>
<li><p>Find <span class="math-container">$2$</span>-dimensional representations. I have seen in my notes that for <span class="math-container">$D_n$</span> we can define <span class="math-container">$2$</span>-dimensional representations:
<span class="math-container">$$\alpha_5: G\longrightarrow GL_2(\mathbb{C}): a\mapsto \begin{bmatrix}cos(\frac{2\pi}{n}) & -sin(\frac{2\pi}{n})\\ sin(\frac{2\pi}{n}) &cos(\frac{2\pi}{n})\end{bmatrix}, b\mapsto \begin{bmatrix}1 & 0\\ 0 &-1\end{bmatrix}$$</span>
Hence my <span class="math-container">$$\alpha_5: G\longrightarrow GL_2(\mathbb{C}): a\mapsto \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} &\frac{1}{2}\end{bmatrix}, b\mapsto \begin{bmatrix}1 & 0\\ 0 &-1\end{bmatrix}$$</span></p></li>
<li><p>Build my character table.
<span class="math-container">\begin{array}{|c|c|c|c|}
\hline
& C_1 & C_2 &C_3 &C_4 &C_5 &C_6 \\ \hline
\chi_1& 1 & 1 &1 &1 &1&1 \\ \hline
\chi_2& 1 & 1 &1 &1 &-1 &-1 \\ \hline
\chi_3& 1 & -1 &1 &-1 &1 &-1 \\ \hline
\chi_4& 1 & -1 &1 &-1 &-1 &1 \\ \hline
\chi_5& 2 & 1 &-1 &-2 &0 &0 \\ \hline
\chi_6& 2 & -1 &-1 &2 &0 &0 \\ \hline
\end{array}</span></p></li>
</ol>
<p>where I have computed <span class="math-container">$\chi_6$</span> by the orthogonality formula <span class="math-container">$(\chi_6|\chi_j)=\delta_{6,j}$</span>.</p>
<p><strong>QUESTIONS</strong></p>
<ol>
<li>Is <span class="math-container">$D_6/\{a^2,a^4\}$</span> really abelian? I can not see it clearly.</li>
<li><p>My first question comes when I have to find <span class="math-container">$2$</span>-dimensional irreducible representations. I have find them because I have seen it in my notes. But how could I get <span class="math-container">$\alpha_5$</span> and <span class="math-container">$\alpha_6$</span> without knowing the special case of <span class="math-container">$D_n$</span>. I know that I also could get it from <span class="math-container">$S_3$</span> (one of them). But I have again the same problem, if you are looking for <span class="math-container">$2$</span>-dimensional irreducible representations of <span class="math-container">$S_3$</span>, how do you find them? (Both).</p></li>
<li><p>Now consider <span class="math-container">$X$</span> to be the set of the vertices of a regular <span class="math-container">$6$</span>-gon and consider the action of <span class="math-container">$D_6$</span> on the set <span class="math-container">$X$</span> by restricting the usual action of <span class="math-container">$D_6$</span> on the <span class="math-container">$6$</span>-gon to the set of vertices <span class="math-container">$X$</span>. Let <span class="math-container">$\phi$</span> be the induced permutation representation (over <span class="math-container">$\mathbb{C}$</span>) of <span class="math-container">$D_6$</span>. I would like to write it as a sum of irreducible representations by computing the in-product of the irreducible characters with <span class="math-container">$\chi_{\phi}$</span>. What should I do? I do not understand this induced permutation representation. Any help?</p></li>
</ol>
| Angina Seng | 436,618 | <p>In general, <span class="math-container">$D_n$</span> is a group of order <span class="math-container">$2n$</span> with a cyclic subgroup <span class="math-container">$C_n$</span>
of order <span class="math-container">$n$</span> generated by <span class="math-container">$a$</span> say. Also <span class="math-container">$D_n$</span> contains <span class="math-container">$b$</span> with <span class="math-container">$b^2=1$</span>
and <span class="math-container">$bab^{-1}=a^{-1}$</span>.</p>
<p>One can obtain degree <span class="math-container">$2$</span> characters of <span class="math-container">$D_n$</span> by inducing from degree
<span class="math-container">$1$</span> characters of <span class="math-container">$C_n$</span>. For each <span class="math-container">$j$</span> there is a representation <span class="math-container">$\rho_j$</span>
of <span class="math-container">$C_n$</span> taking <span class="math-container">$a^k$</span> to <span class="math-container">$\zeta^{jk}$</span> where <span class="math-container">$\zeta=\exp(2\pi i/n)$</span>. This
induces to a degree <span class="math-container">$2$</span> character of <span class="math-container">$D_n$</span> via
<span class="math-container">$$\chi_j(g)=\rho_j(g)+\rho_j(bgb^{-1})$$</span>
where we set <span class="math-container">$\rho_j(g)=0$</span> for <span class="math-container">$g$</span> outside <span class="math-container">$C_n$</span>. Then
<span class="math-container">$$\chi_j(a^k)=\zeta^{jk}+\zeta^{-jk}=2\cos\frac{2\pi jk}n$$</span>
and
<span class="math-container">$$\chi_j(a^kb)=0.$$</span></p>
<p>This character <span class="math-container">$\chi_j$</span> is irreducible unless <span class="math-container">$\zeta^j=\pm1$</span>. Together
with the degree <span class="math-container">$1$</span> characters of <span class="math-container">$D_n$</span>, the irreducible <span class="math-container">$\chi_j$</span> exhaust the
characters of <span class="math-container">$D_n$</span>.</p>
|
4,486,594 | <p>Let <span class="math-container">$X$</span> be the Riemann surface of <span class="math-container">$w^{2} \ =\text{sin} \ z$</span> in <span class="math-container">$ \mathbb{C}^{2}$</span>, i.e. let <span class="math-container">$X = \{(z,w): w^2 = \text{sin} \ z\}$</span>.</p>
<p>The Riemann surface structure on <span class="math-container">$X$</span> is obtained by paramertizing by <span class="math-container">$z$</span> at all places where <span class="math-container">$w \ne 0$</span> and parametrizing by <span class="math-container">$w$</span> at places where <span class="math-container">$w=0$</span>.</p>
<p>The question is to show is not the interior of a compact surface-with-boundary.</p>
<hr />
<p>My attempt:</p>
<p>Consider the holomorphic function <span class="math-container">$\displaystyle F:X\rightarrow \mathbb{C}$</span> given by <span class="math-container">$\displaystyle F:( z,w)\rightarrow w$</span>.
Now assume there is a compact surface-with-boundary <span class="math-container">$\displaystyle \tilde{X}$</span> such that interior of <span class="math-container">$\displaystyle \tilde{X}$</span> is <span class="math-container">$\displaystyle X$</span>.
It is natural to expect that <span class="math-container">$\displaystyle F$</span> extends to an holomorphic function <span class="math-container">$\displaystyle \tilde{F} :\tilde{X} \ \rightarrow \mathbb{C} P^{1}$</span>.</p>
<p>Now observe <span class="math-container">$\displaystyle \tilde{F}^{-1}( 0)$</span> has a limit point as it contains an infinite sequence <span class="math-container">$\displaystyle \{( n\pi ,0)\}_{n=1}^{\infty }$</span> in <span class="math-container">$\displaystyle \tilde{X}$</span>. Holomorphicity of <span class="math-container">$\displaystyle \tilde{F}$</span> forces <span class="math-container">$\displaystyle \tilde{F} $</span> to be a constant.
<br />
This is a contradiction as <span class="math-container">$\displaystyle F$</span> is not constant.</p>
<p>The issue with this proof is that there is no reason to believe that <span class="math-container">$\displaystyle F$</span> should have an extension <span class="math-container">$\displaystyle \tilde{F}$</span>.</p>
| Moishe Kohan | 84,907 | <p>The thing to prove is that the surface <span class="math-container">$X$</span> has infinite topological type. Instead of your function <span class="math-container">$F$</span>, I will consider the function <span class="math-container">$g(z,w)=z$</span> on <span class="math-container">$X$</span>, which is (generically) 2-to-1. Critical values of this function are <span class="math-container">$z_n=\pi n$</span>, <span class="math-container">$n\in {\mathbb Z}$</span> (the zeroes of the function <span class="math-container">$\sin(z)$</span>).</p>
<p>Consider a sequence of closed disks
<span class="math-container">$$
D_n=\bar{D}(0, \pi n + \frac{\pi}{2})\subset {\mathbb C},
$$</span>
each containing <span class="math-container">$2n+1$</span> critical values of the function <span class="math-container">$g$</span> and no critical values on the boundary. Let <span class="math-container">$X_n:= g^{-1}(D_n)$</span>. Then each <span class="math-container">$X_n$</span> has one boundary component (see below). Each critical value of <span class="math-container">$g$</span> will have exactly one preimage in <span class="math-container">$X$</span>. By the Riemann-Hurwitz formula,
<span class="math-container">$$
\chi(X_n)= 2\chi(D_n) - 2n-1= 1-2n.
$$</span>
Since <span class="math-container">$\partial X_n$</span> is connected, it follows that the genus of <span class="math-container">$X_n$</span> is <span class="math-container">$n$</span>. (This calculation also shows that <span class="math-container">$\partial X_n$</span> cannot consist of 2 components and, since <span class="math-container">$g$</span> is 2-to-1, the higher number of boundary components is impossible.) Hence, <span class="math-container">$X$</span> has infinite genus and, therefore, is not homeomorphic to the interior of a compact surface.</p>
|
704,680 | <p>We have $$\sqrt{x -2} = 3 -2\sqrt{x}$$.</p>
<p>I am to find whether a real number exists for this relation, and the real number that satisfies.</p>
<p>I start by squaring both sides, which yields: </p>
<p>$$x - 2 = 4x - 12\sqrt{x} + 9$$.</p>
<p>Whence:</p>
<p>$$ -3x = -12\sqrt{x} + 11 \\
\sqrt{x} = \frac{x}{4} + \frac{11}{12}.
$$</p>
<p>But once i get here i am stuck. How can i find whether a solution exists for x from here?</p>
| TheBridge | 4,437 | <p>Hi I think it is true that the operator $T$ that you defined is such that the image of any measurable set of $\mathcal{B}([0,+\infty))$ is a measurable set or as claimed that $T$ preserves measurability. </p>
<p>The idea is to view things in a "topological" way first. If you "examine" carefully $T$ I think that you would shortly realize that it is a continuous linear application from one Banach spaces to another and that moreover it is surjective, from this, every ingredients are here for the Banach-Schauder theorem to apply unless mistaken. </p>
<p>This proves that the image of every open sets of the space of continuous functions equipped with the sup norm (on every compacts) are open sets of the space at arrival which is naturally embedded in the very same space (you could refine the argument by defining the space of continuous functions that start at 0 but this doesn't really matter). </p>
<p>Now let's reconcile topological and meausre theory, it is proven in Karatzas and Shreve "Brownian Motion and Stochastic Calculus" (may be as an exercise left to the reader I can't remember) that the $\sigma$-algebra defined the way you did by projection restricted to the space of continuous functions is the same $\sigma$-algebra that the Borelian $\sigma$-algebra of the space of continuous functions equipped with the sup norm (on every compacts). </p>
<p>Hence I think we are done because $T$ preserves open sets which is a collection of sets the generated our $\sigma$-algebra $\mathcal{B}([0,+\infty))$, an application of the Monotone Class Theorem should be enough to conclude I think and I leave this part to you. </p>
<p>Best regards</p>
|
704,680 | <p>We have $$\sqrt{x -2} = 3 -2\sqrt{x}$$.</p>
<p>I am to find whether a real number exists for this relation, and the real number that satisfies.</p>
<p>I start by squaring both sides, which yields: </p>
<p>$$x - 2 = 4x - 12\sqrt{x} + 9$$.</p>
<p>Whence:</p>
<p>$$ -3x = -12\sqrt{x} + 11 \\
\sqrt{x} = \frac{x}{4} + \frac{11}{12}.
$$</p>
<p>But once i get here i am stuck. How can i find whether a solution exists for x from here?</p>
| Evan Aad | 37,058 | <p>The following is an addendum to <a href="https://math.stackexchange.com/users/442/gedgar">GEdgar</a>'s <a href="https://math.stackexchange.com/a/705320/37058">answer</a>, aimed to clarify some points for my future reference, as well as for the sake of other readers who, like me, do not find these points self evident. The principal results are propositions 6 and 7 below.</p>
<p><strong><em>Notation</em></strong></p>
<p>We denote the borel field on the real line by $\mathcal{B}$, the Borel field on $\mathbb{R}^2$ by $\mathcal{B}_2$ and in general for any $n \in \mathbb{N}_1$ the Borel field on $\mathbb{R}^n$ by $\mathcal{B}_n$. When convenient, we take $\mathbb{R}^2$'s elements to be column vectors.</p>
<hr>
<p><strong><em>Lemma 1</em></strong></p>
<p>A plane Borel set exists, whose projection on the $x$ coordinate is not a Borel set.</p>
<p><strong><em>Proof</em></strong></p>
<p>This is known as <a href="https://www.encyclopediaofmath.org/index.php/Suslin_theorem" rel="nofollow noreferrer">Suslin's Theorem</a>. Q.E.D.</p>
<hr>
<p><strong><em>Lemma 2</em></strong></p>
<p>If $A \in \mathcal{B}_2$ and $y \in \mathbb{R}$, then the section $A_y := \left\{x \in \mathbb{R} \mid: (x,y) \in A\right\}$ is Borel (i.e. $A_y \in \mathcal{B}$).</p>
<p><strong><em>Proof</em></strong></p>
<p>See Theorem A (p. 141) in Chapter 34 ("Sections") of Halmos, Paul R., "Measure Theory", Springer, 1974. Q.E.D.</p>
<hr>
<p><strong><em>Lemma 3</em></strong></p>
<p>a. If $T:\mathbb{R}^m \rightarrow \mathbb{R}^n$ for some $m,n \in \mathbb{N}_1$ is linear, then for all $A \in \mathcal{B}_n$, $T^{-1}(A) \in \mathcal{B}_m$.</p>
<p>b. If $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ for some $n \in \mathbb{N}_1$ is linear and invertible, then for all $A \in \mathbb{R}^n$, $A \in \mathcal{B}_n \iff T(A) \in \mathcal{B}_n$.</p>
<p><strong><em>Proof</em></strong></p>
<p>A linear transformation between finite dimensional linear spaces is continuous, and continuous functions are Borel. Q.E.D.</p>
<hr>
<p><strong><em>Lemma 4</em></strong></p>
<p>Let $A \subseteq \mathbb{R}$. Then $A \in \mathcal{B}$ iff $\left\{(x,0) :\mid x \in A\right\} \in \mathcal{B}_2$.</p>
<p><strong><em>Proof</em></strong></p>
<p>Suppose $A \in \mathcal{B}$. Denote by $T:\mathbb{R}^2 \rightarrow \mathbb{R}$ the projection of $\mathbb{R}^2$ on the $x$-coordinate: $T(v) := Pv$ with
$$
P := \left[\begin{array}{cc}
1 & 0
\end{array}\right]
$$
Then $T^{-1}(A) \in \mathcal{B}_2$ (Lemma 3a). Then $\left\{(x,0) :\mid x \in A\right\} \in \mathcal{B}_2$, as the intersection of $T^{-1}(A)$ with the $x$-axis, which is $\in \mathcal{B}_2$.</p>
<p>Conversely, suppose that $\left\{(x,0) :\mid x \in A\right\} \in \mathcal{B}_2$. Define the linear transformation $F:\mathbb{R}\rightarrow\mathbb{R}^2$ thus:
$$
F(x) := (x, 0)
$$
Then $A = F^{-1}\left(\left\{(x,0) :\mid x \in A\right\}\right) \in \mathcal{B}$ (Lemma 3a). Q.E.D.</p>
<hr>
<p><strong><em>Lemma 5</em></strong></p>
<p>If $A \in \mathcal{B}$, then $\left\{(x,-x) :\mid x \in A\right\} \in \mathcal{B}_2$.</p>
<p><strong><em>Proof</em></strong></p>
<p>Define $B := \left\{(x, 0) :\mid x \in A\right\}$. Then $B \in \mathcal{B}_2$ (Lemma 4). Denote by $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ the linear transformation $T(v) := Rv$ with
$$
R := \left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]
$$
$T$ is invertible. Hence $\left\{\left(x,-x\right) :\ x \in A\right\} = T(A) \in \mathcal{B}_2$ (Lemma 3b). Q.E.D.</p>
<hr>
<p><strong><em>Proposition 6</em></strong></p>
<p>There is a Borel set $E$ in $\mathbb{R}^2$ such that its difference set $F := \{x-y :\mid \left(x,y\right) \in E\}$ is not a Borel set.</p>
<p><strong><em>Proof</em></strong></p>
<p>Let $B \in \mathcal{B}_2$ be such that $B$'s projection on the $x$-coordinate is not Borel (Lemma 1). Consider the projection of $B$ on the $x$-axis, $PB$ with
$$
P = \left[\begin{array}{cc}
1 & 0 \\
0 & 0
\end{array}\right]
$$
Then $PB \notin \mathcal{B}_2$ (Lemma 4).</p>
<p>Define $E$ to be the rotation of $B$ $45^\circ$ clockwise, i.e. $E := AB$, where
$$
A := \left[\begin{array}{cc}
\sqrt{2} & \sqrt{2} \\
-\sqrt{2} & \sqrt{2}
\end{array}\right]
$$
We have $E \in \mathcal{B}_2$ and $\frac{1}{2}APA^{-1}E \notin \mathcal{B}_2$ (Lemma 3b). But
$$
\frac{1}{2}APA^{-1} = \left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]
$$
Hence
$$
\frac{1}{2}APA^{-1}E = \left\{\left(x - y, -(x - y)\right) :\mid \left(x,y\right) \in E\right\} = \left\{(x,-x) :\mid x \in F\right\}
$$
So $F \notin \mathcal{B}$ (Lemma 5). Q.E.D.</p>
<hr>
<p><strong><em>Proposition 7</em></strong></p>
<p>Using the definitions used in GEdgar's answer, $T(A) \notin \mathcal{B}_{[0,\infty)}$.</p>
<p><strong><em>Proof</em></strong></p>
<p>Define a function $g:\mathbb{R}\rightarrow\mathbf{C}_{[0,\infty)}$ thus: for every $t \in \mathbb{R}$ and every $x \in \left[0,\infty\right)$, $g(t)(x) := tx$.</p>
<p>$g$ is $\mathcal{B}/\mathcal{B}_{[0,\infty)}$-measurable. Therefore, had $T(A)$ been in $\mathcal{B}_{[0,\infty)}$, we would have $g^{-1}\left(T(A)\right) \in \mathcal{B}$. But $g^{-1}\left(T(A)\right) = F$. Q.E.D.</p>
|
704,680 | <p>We have $$\sqrt{x -2} = 3 -2\sqrt{x}$$.</p>
<p>I am to find whether a real number exists for this relation, and the real number that satisfies.</p>
<p>I start by squaring both sides, which yields: </p>
<p>$$x - 2 = 4x - 12\sqrt{x} + 9$$.</p>
<p>Whence:</p>
<p>$$ -3x = -12\sqrt{x} + 11 \\
\sqrt{x} = \frac{x}{4} + \frac{11}{12}.
$$</p>
<p>But once i get here i am stuck. How can i find whether a solution exists for x from here?</p>
| GEdgar | 442 | <p>Let's try this for the tail field.</p>
<p>Notation<br>
$\mathcal{B}_1$ the Borel sets for $\mathbb R$, a Polish space,<br>
$\mathcal{B}_2$ the Borel sets for $\mathbb R^2$, a Polish space,<br>
$\mathcal{B}_{\left[0,\infty\right)} = \sigma\big(\pi_t \colon t \in [0,\infty)\big)$ the Borel sets for $\mathbf{C}_{\left[0,\infty\right)}$, a Polish space,<br>
for $s > 0$, $\mathcal{T}_s = \sigma\big(\pi_t \colon t \ge s\big)$,<br>
$\mathcal{T} = \bigcap_{s>0}\mathcal{T}_s$, the tail field.</p>
<p>There is $E \in \mathcal{B}_2$ such that $F := \{x-y\colon (x,y) \in E\} \notin \mathcal{B}_1$. </p>
<p>Let $\phi$ be a sawtooth function, $\phi(t) = t$ for $0 \le t \le 1$, $\phi(t) = 2-t$ for $1 < t \le 2$, and periodic with period $2$. Let $Z = \{a\phi \colon a \in \mathbb R\}$, a closed set in $\mathbf{C}_{\left[0,\infty\right)}$ (one-dimensional subspace), and thus $Z \in \mathcal{B}_{\left[0,\infty\right)}$. </p>
<p>For $n \in \mathbb N$, let
$$
A_n := \left\{f \in \mathbf{C}_{\left[0,\infty\right)} \colon
(f(2n+1),f(2n)) \in E, \text{ and for all } t \ge 2n, f(t+2)=f(t)\right\} .
$$
Because we can check the last part (periodic from $2n$ on) just using rational $t$, we have $A_n \in \mathcal{T}_{2n}$. Also note $A_1 \subseteq A_2 \subseteq A_3 \subseteq \cdots$. Let $A = \bigcup_n A_n$. Then $A$ is a tail event, $A \in \mathcal{T}$. </p>
<p>We claim $T(A) \notin \mathcal{B}_{\left[0,\infty\right)}$. Suppose (for purposes of contradiction) that $T(A) \in \mathcal{B}_{\left[0,\infty\right)}$. Then also its complement $T(A)^c \in \mathcal{B}_{\left[0,\infty\right)}$. Therefore $Z \cap T(A) \in \mathcal{B}_{\left[0,\infty\right)}$ and $Z \cap T(A)^c \in \mathcal{B}_{\left[0,\infty\right)}$. But, in fact,
$$
T(A) \cap Z = \{a\phi\colon a \in F\},\qquad
T(A)^c \cap Z = \{a\phi\colon a \in F^c\} .
$$
Now (in Polish spaces) the continuous image of a Borel set is analytic, so the two sets sets
$$
\pi_1\big(T(A)\cap Z\big) = F\qquad\text{and}\qquad
\pi_1\big(T(A)^c\cap Z\big) = F^c
$$
are both analytic in $\mathbb R$, and therefore $F$ is Borel. This contradiction completes our proof.</p>
|
3,872,750 | <p>Suppose we have a series <span class="math-container">$$\sum_{n=2}^\infty (-1)^n \frac{n^2}{10^n} = \sum_{n=2}^\infty (-1)^n b_n$$</span>.</p>
<p>I want to apply the alternating series test to see if it converges.</p>
<p>I need to show that:</p>
<p><span class="math-container">$$\lim_{n \rightarrow \infty} \frac{n^2}{10^n} = 0$$</span></p>
<p><span class="math-container">$$\frac{(n+1)^2}{10^{n+1}} \leq \frac{n^2}{10^n}$$</span></p>
<p>To see that <span class="math-container">$\lim_{n \rightarrow \infty} \frac{n^2}{10^n} =0$</span>, we note that the denominator is going to grow way faster than the the numerator. If you want to show it using calculus, then you can do L'hoptials rule twice to get:</p>
<p><span class="math-container">$\lim_{n \rightarrow \infty} \frac{n^2}{10^n} = \lim_{n \rightarrow \infty} \frac{2}{10^xlog^2(10)}=0$</span></p>
<p>Next we want to know if</p>
<p><span class="math-container">$$\frac{(n+1)^2}{10^{n+1}} \leq \frac{n^2}{10^n}$$</span></p>
<p>is true for all <span class="math-container">$n$</span>. Rearranging, we can get:</p>
<p><span class="math-container">$$\frac{10^n}{10^{n+1}} \leq \frac{n^2}{(n+1)^2}$$</span></p>
<p><span class="math-container">$$\frac{1}{10} \leq \frac{n^2}{(n+1)^2}= (\frac{n}{n+1})^2$$</span></p>
<p>By plugging in values <span class="math-container">$n=2,3,...$</span> we can see that this inequality only gets more and more true as <span class="math-container">$n$</span> gets bigger since <span class="math-container">$\lim_{n \rightarrow \infty} (\frac{n}{n+1})^2 = 1$</span></p>
<p>To show this with calculus we consider the function <span class="math-container">$f(x) = \frac{x^2}{10^x}$</span> and take it's derivative and find when it is <span class="math-container">$\leq 0$</span> on our domain of interest <span class="math-container">$[2, \infty)$</span>:</p>
<p><span class="math-container">$f'(x) = -10^{-x}x(xlog(10)-2) \leq 0$</span></p>
<p>This will be true as long as <span class="math-container">$(xlog(10)-2)$</span> is not negative, so <span class="math-container">$x \geq \frac{2}{log(10)}$</span>, whicn includes our domain since we only care about <span class="math-container">$[2, \infty)$</span></p>
<p>Thus the series converges by the alternating series test</p>
| Arthur | 15,500 | <p>For an equation
<span class="math-container">$$
[a]x=[b]
$$</span>
(where <span class="math-container">$a,b\in\Bbb Z$</span> and <span class="math-container">$x\in\Bbb Z_n$</span> for some natural <span class="math-container">$n$</span>), there are either <span class="math-container">$\gcd(a,n)$</span> solutions, or no solutions. These solutions exist iff <span class="math-container">$b$</span> is a multiple of <span class="math-container">$\gcd(a,n)$</span>.</p>
<p>First of all, the smallest positive integer <span class="math-container">$k$</span> such that <span class="math-container">$[a]x=[k]$</span> has a solution is <span class="math-container">$\gcd(a,n)$</span> (the extended Euclidean algorithm or Bézout's identity can tell you this). Further, any multiple of <span class="math-container">$k$</span> also clearly gives solutions. Finally, if <span class="math-container">$\ell$</span> is an integer which is not a multiple of <span class="math-container">$k$</span>, then there cannot be any solutions to <span class="math-container">$[a]x=[\ell]$</span>, as that would imply a solution to <span class="math-container">$[a]x=[\gcd(k,\ell)]$</span>, where <span class="math-container">$0<\gcd(k,\ell)<k$</span>, which can't happen by definition of <span class="math-container">$k$</span>.</p>
<p>And why are there <span class="math-container">$\gcd(a,n)$</span> solutions? Because by the above characterisation of which conjugacy classes <span class="math-container">$[b]$</span> give solutions, there are <span class="math-container">$\frac{n}{\gcd(a,n)}$</span> such classes. And each of them have an equal number of solutions.</p>
<p>So in your case, for instance, the equations <span class="math-container">$[3]x=[b]$</span> will have <span class="math-container">$\gcd(3,6)$</span> solutions iff <span class="math-container">$b$</span> is a multiple of <span class="math-container">$\gcd(3,6)$</span>, and no solutions otherwise.</p>
|
73,785 | <p>I am new to Mathematica, and I have read this <a href="https://mathematica.stackexchange.com/questions/29203/determine-the-2d-fourier-transform-of-an-image">post</a> to understand how to perform Fourier transform on an image. My mission is to extract information on the typical distance between the black patches in the image I have attached here. The code that I attach here gives me the Fourier transform, but I don't know how to take out from the Fourier transform the values of the wavenumbers.</p>
<p>I have used this piece of code</p>
<pre><code>img = Import["example.jpg"];
Image[img, ImageSize -> 300]
data = ImageData[img];(*get data*)
{nRow, nCol, nChannel} = Dimensions[data];
d = data[[All, All, 2]];
d = d*(-1)^Table[i + j, {i, nRow}, {j, nCol}];
fw = Fourier[d, FourierParameters -> {1, 1}];
(*adjust for better viewing as needed*)
fudgeFactor = 100;
abs = fudgeFactor*Log[1 + Abs@fw];
Labeled[Image[abs/Max[abs], ImageSize -> 300],Style["Magnitude spectrum", 18]]
</code></pre>
<p>I have the following image on which I would like to perform this analysis - <img src="https://i.stack.imgur.com/hALsH.jpg" alt="enter image description here"></p>
| bill s | 1,783 | <p>It looks like random blobs, and that's what the FFT suggests...</p>
<pre><code>img = Import["http://i.stack.imgur.com/hALsH.jpg"];
imgBW = ImageData@ColorConvert[img, "Grayscale"];
imgZ = imgBW - Mean@Mean[imgBW];
xf = Abs[Fourier[imgZ, FourierParameters -> {1, -1}]];
{d1, d2} = Ceiling[Dimensions[xf]/2];
xCentered = RotateLeft[xf, {d1, d2}];
ArrayPlot[xCentered]
</code></pre>
<p><a href="https://i.stack.imgur.com/FtH3b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FtH3b.png" alt="enter image description here"></a></p>
<p>Here we zoom into the center where we can see it looks a lot like a randomized Gaussian blob:</p>
<pre><code>zoom = 50;
ArrayPlot[xCentered, PlotRange -> {{d1 - zoom, d1 + zoom}, {d2 - zoom, d2 + zoom}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/OIQi6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OIQi6.png" alt="enter image description here"></a></p>
<p>I think this means you are going to have to approach this a different way.</p>
|
73,785 | <p>I am new to Mathematica, and I have read this <a href="https://mathematica.stackexchange.com/questions/29203/determine-the-2d-fourier-transform-of-an-image">post</a> to understand how to perform Fourier transform on an image. My mission is to extract information on the typical distance between the black patches in the image I have attached here. The code that I attach here gives me the Fourier transform, but I don't know how to take out from the Fourier transform the values of the wavenumbers.</p>
<p>I have used this piece of code</p>
<pre><code>img = Import["example.jpg"];
Image[img, ImageSize -> 300]
data = ImageData[img];(*get data*)
{nRow, nCol, nChannel} = Dimensions[data];
d = data[[All, All, 2]];
d = d*(-1)^Table[i + j, {i, nRow}, {j, nCol}];
fw = Fourier[d, FourierParameters -> {1, 1}];
(*adjust for better viewing as needed*)
fudgeFactor = 100;
abs = fudgeFactor*Log[1 + Abs@fw];
Labeled[Image[abs/Max[abs], ImageSize -> 300],Style["Magnitude spectrum", 18]]
</code></pre>
<p>I have the following image on which I would like to perform this analysis - <img src="https://i.stack.imgur.com/hALsH.jpg" alt="enter image description here"></p>
| bill s | 1,783 | <p>One approach is to locate the black components and then measure some properties of them. Here we locate them using <code>MorphologicalComponents</code>, find the centroids using <code>ComponentMeasurements</code> and then calculate the distance between the centroids using <code>Nearest</code>.</p>
<pre><code>img = Import["http://i.stack.imgur.com/hALsH.jpg"];
imgBW = Binarize@ColorConvert[img, "Grayscale"];
comp = MorphologicalComponents[ColorNegate[Dilation[imgBW, 5]]] // Colorize
cent = ComponentMeasurements[comp, "Centroid"];
allNears = Nearest[cent[[All, 2]], #, 2] & /@ cent[[All, 2]];
Total[Norm[allNears[[#, 1]] - allNears[[#, 2]]] & /@
Range[Length[allNears]]]/Length[allNears]
</code></pre>
<p>As you can see, the average distance form the center of each blob to the nearest adjacent blob is about 33 pixels.</p>
<p><a href="https://i.stack.imgur.com/Xukai.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xukai.png" alt="enter image description here"></a></p>
|
1,617,698 | <p>While I was trying to find the formula of something by my own means I came across this sum which I need to solve, however I don't know if there is a solution for it, maybe it doesn't mean anything and I made a mistake. However if there's an equation which can replace this sum I will appreciate it a lot if you show me which one and how did you find the answer!</p>
| JimmyK4542 | 155,509 | <p>Let $S = \displaystyle\sum_{k = 1}^{n}\sin\dfrac{k\pi}{2n}$. (The $k = 0$ term is $0$, so we can ignore it). Then, by using the product to difference identity $\sin A \sin B = \dfrac{1}{2}\left(\cos(A-B)-\cos(A+B)\right)$, we have:</p>
<p>$S\sin \dfrac{\pi}{4n}$ $= \displaystyle\sum_{k = 1}^{n}\sin\dfrac{k\pi}{2n}\sin\dfrac{\pi}{4n}$ $= \dfrac{1}{2}\displaystyle\sum_{k = 1}^{n}\left(\cos\left(\dfrac{k\pi}{2n}-\dfrac{\pi}{4n}\right)-\cos\left(\dfrac{k\pi}{2n}-\dfrac{\pi}{4n}\right)\right)$ </p>
<p>$= \dfrac{1}{2}\displaystyle\sum_{k = 1}^{n}\left(\cos\dfrac{(2k-1)\pi}{4n} - \cos\dfrac{(2k+1)\pi}{4n}\right)$ $= \dfrac{1}{2}\left(\cos\dfrac{\pi}{4n} - \cos\dfrac{(2n+1)\pi}{4n}\right)$. </p>
<p>Now, divide both sides by $\sin\dfrac{\pi}{4n}$ to get $S = \dfrac{\cos\tfrac{\pi}{4n} - \cos\tfrac{(2n+1)\pi}{4n}}{2\sin\tfrac{\pi}{4n}}$. </p>
|
2,921,439 | <p>I got this summation from the book <a href="https://rads.stackoverflow.com/amzn/click/0201558025" rel="nofollow noreferrer">Concrete Mathematics</a> which I didn't exactly understand:</p>
<p>$$
\begin{align}
Sn &= \sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant j \lt k} {\frac{1}{k-j}} \\
&= \sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant k-j \lt k} {\frac{1}{j}} \\
&= \sum_{1 \leqslant k \leqslant n} \sum_{0 \lt j \leqslant k-1} {\frac{1}{j}} \\
\end{align}
$$</p>
<p>I didn't understant why $1 \leqslant j \lt k$ became $1 \leqslant k-j \lt k$ in the second line and why $1 \leqslant k-j \lt k$ became $0 \lt j \leqslant k-1$ in the third line.</p>
<p>Can you guys help me understanding that?</p>
| mvw | 86,776 | <p>There is an algorithm to handle such integrals: <a href="https://en.wikipedia.org/wiki/Partial_fraction_decomposition" rel="nofollow noreferrer">Partial fraction decomposition</a></p>
|
2,921,439 | <p>I got this summation from the book <a href="https://rads.stackoverflow.com/amzn/click/0201558025" rel="nofollow noreferrer">Concrete Mathematics</a> which I didn't exactly understand:</p>
<p>$$
\begin{align}
Sn &= \sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant j \lt k} {\frac{1}{k-j}} \\
&= \sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant k-j \lt k} {\frac{1}{j}} \\
&= \sum_{1 \leqslant k \leqslant n} \sum_{0 \lt j \leqslant k-1} {\frac{1}{j}} \\
\end{align}
$$</p>
<p>I didn't understant why $1 \leqslant j \lt k$ became $1 \leqslant k-j \lt k$ in the second line and why $1 \leqslant k-j \lt k$ became $0 \lt j \leqslant k-1$ in the third line.</p>
<p>Can you guys help me understanding that?</p>
| 5xum | 112,884 | <p>$$-x^3 - x^2 + 4x + 4 = -x^2(x+1) +4(x+1) = (x+1)(4-x^2)$$</p>
<p>meaning that</p>
<p>$$\frac{x+1}{-x^2-x^2+4x+4} = \frac{x+1}{(x+1)(-x^2+4)} = \frac{1}{-x^2+4}$$</p>
<p>which is also the derivative you are looking for</p>
|
2,921,439 | <p>I got this summation from the book <a href="https://rads.stackoverflow.com/amzn/click/0201558025" rel="nofollow noreferrer">Concrete Mathematics</a> which I didn't exactly understand:</p>
<p>$$
\begin{align}
Sn &= \sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant j \lt k} {\frac{1}{k-j}} \\
&= \sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant k-j \lt k} {\frac{1}{j}} \\
&= \sum_{1 \leqslant k \leqslant n} \sum_{0 \lt j \leqslant k-1} {\frac{1}{j}} \\
\end{align}
$$</p>
<p>I didn't understant why $1 \leqslant j \lt k$ became $1 \leqslant k-j \lt k$ in the second line and why $1 \leqslant k-j \lt k$ became $0 \lt j \leqslant k-1$ in the third line.</p>
<p>Can you guys help me understanding that?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Use that$$\frac{x+1}{-x^3-x^2+4x-4}=\frac{1}{4(x+2)}-\frac{1}{4(x-2)}$$</p>
|
132,226 | <p>After edit:</p>
<p>How do we show that for every (holomorphic) vector bundle over a curve, is it possible to deform it to another one which is decomposable (into line bundles)? </p>
<p>Before edit:</p>
<p>I am not sure how much obvious or wrong is the following question:</p>
<p>For every (holomorphic) vector bundle over a complex projective variety, is it possible to deform it to another one which is decomposable (into line bundles)?
I feel the answer is yes (since obstruction lies in the ext group and you can deform any element of ext group to zero) but I don't know how easy is a precise proof! </p>
<p>Is this true at least over curves?</p>
| David E Speyer | 297 | <p>Francesco, in comments, shows that any vector bundle on a curve degenerates to a direct sum of a line bundles. (By the way, I observe the convention that "degeneration" means moving towards the special fiber and "deformation" means moving away from it; you are doing degeneration.)</p>
<p>In the comments, the OP asks whether the moduli space of vector bundles on a curve with fixed rank and Chern character is connected. The answer is yes. Actually, there are a lot of subtleties in talking about this moduli space (stability issues), so I'll just directly answer the question about what can be connected to what in families over connected bases.</p>
<p>By Francesco's argument, we can degenerate from any vector bundle to a direct sum of line bundles. We need to show that, if $r=s$ and $\sum_{i=1}^r \deg L_i = \sum_{j=1}^s \deg M_j$, then we can build a path from $\bigoplus_{i=1}^r L_i$ to $\bigoplus_{j=1}^s M_j$. </p>
<p><b>Step 1</b> On a curve, for any two ample line bundles $L_1$ and $L_2$, there is a degeneration from $L_1 \oplus L_2$ to $\mathcal{O} \oplus (L_1 \otimes L_2)$. </p>
<p><b>Proof</b> Let $f_1$ and $f_2$ be sections of $L_1$ and $L_2$ with disjoint zero locus. Then
$$0 \to \mathcal{O} \stackrel{\begin{pmatrix} f \\ g \end{pmatrix}}{\longrightarrow} L_1 \oplus L_2 \stackrel{\begin{pmatrix} \otimes g & -f \otimes \end{pmatrix}}{\longrightarrow} L_1 \otimes L_2 \to 0$$
is exact. As in Francesco's argument, this shows we can degenerate from $L_1 \oplus L_2$ to $\mathcal{O} \oplus (L_1 \otimes L_2)$.</p>
<p><b>Step 2</b> On a curve, for any $r$ line bundles $L_1$, $L_2$ ..., $L_r$, there is a path connecting $\bigoplus L_i$ to $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes L_i$</p>
<p><b>Proof</b> Choose $D$ large enough that $\mathcal{O}(D)$ and $(\bigotimes L_i)(D)$ and all the $L_i(D)$ are ample. Recursively using Step 1 gives a family from $\bigoplus L_i(D)$ to $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes \left( L_i(D) \right)$. Step 1 also gives a path to this point from $\mathcal{O}(D)^{\oplus (r-1)} \oplus \left( \bigoplus L_i \right)(D)$.
So there is a path connecting $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes \left( L_i(D) \right)$ and $\mathcal{O}(D)^{\oplus (r-1)} \oplus \left( \bigoplus L_i \right)(D)$. Tensor that path with $\mathcal{O}(-D)$ to get a path joining $\bigoplus L_i$ to $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes L_i$.</p>
<p>So we can join $\bigoplus_{i=1}^r L_i$ to $\mathcal{O}^{\oplus (r-1)} \oplus \bigotimes L_i$ and we can do similarly with the $M$'s. If $\sum \deg L_i = \sum \deg M_i =d$ then, since $\mathrm{Pic}^d(X)$ is connected, we can find a path from $\bigotimes L_i$ to $\bigotimes M_i$.</p>
|
2,087,724 | <p>Let $\Omega $ a smooth domain of $\mathbb R^d$ ($d\geq 2$), $f\in \mathcal C(\overline{\Omega })$. Let $u\in \mathcal C^2(\overline{\Omega })$ solution of $$-\Delta u(x)+f(x)u(x)=0\ \ in\ \ \Omega .$$
Assume that $f(x)\geq 0$ for $x\in \Omega $. Prove that $$\int_{B(x,r)}|\nabla u|^2\leq \frac{C}{r^2}\int_{B(x,2r)}|u|^2,$$
for all $x\in \Omega $ and $r>0$, $B(x,2r)\subset \subset \Omega $ for somme $C\geq 0$ independant of $u,f,x$ and $r$.</p>
<p><strong>My attempts</strong></p>
<p>Using divergence theorem and that $\Delta u=fu$ in $\Omega $, I have that $$\int_{B(x,r)}|\nabla u|^2=\int_{B(x,r)}\text{div}(u\nabla u)-\int_{B(x,r)}u\Delta u=\int_{\partial B(x,r)}u\nabla u\cdot \nu-\int_{B(x,r)}fu^2.$$</p>
<p>But I can't do better. Any help would be welcome.</p>
| Glitch | 74,045 | <p>Suppose that $\varphi : \bar{B}(x,2r) \to [0,\infty)$ is Lipschitz and $\varphi =0$ on $\partial B(x,2r)$. Note that since $\varphi$ is Lipschitz, it is differentiable almost everywhere by Rademacher's theorem. Then
$$
0 \le \int_{B(x,2r)} \varphi^2 f u^2 = \int_{B(x,2r)} \varphi^2 u \Delta u = \int_{B(x,2r)} - \nabla (\varphi^2 u) \cdot \nabla u
$$<br>
since $\varphi$ vanishes on $\partial B(x,2r)$. Since
$$
\nabla(\varphi^2 u) = 2 \varphi \nabla \varphi u + \varphi^2 \nabla u
$$
we find, upon plugging in above, that
$$
0 \le -\int_{B(x,2r)}2 \varphi u \nabla \varphi \cdot \nabla u + \varphi^2 |\nabla u|^2,
$$
and so
$$
\int_{B(x,2r)} \varphi^2 |\nabla u|^2 \le \int_{B(x,2r)} -2 \varphi u \nabla \varphi \cdot \nabla u \le 2 \left(\int_{B(x,2r)} \varphi^2 |\nabla u|^2\right)^{1/2} \left(\int_{B(x,2r)} u^2 |\nabla \varphi|^2 \right)^{1/2}.
$$
From this we then see that
$$
\int_{B(x,2r)} \varphi^2 |\nabla u|^2 \le 4 \int_{B(x,2r)} u^2 |\nabla \varphi|^2.
$$</p>
<p>With the last inequality in hand we can prove the result. Set
$$
\varphi(y) =
\begin{cases}
1 & \text{if } |x-y| \le r \\
2 - |x-y|/r & \text{if } r < |x-y| \le 2r.
\end{cases}
$$
It's easy to see that $\varphi \ge 0$, $\varphi$ vanishes on the boundary, and $\varphi$ is Lipschitz. Also
$$
|\nabla \varphi(y)| =
\begin{cases}
0 & \text{if } |x-y| \le r \\
1/r & \text{if } r < |x-y| \le 2r.
\end{cases}
$$
Plugging this in above then gives the inequality
$$
\int_{B(x,r)} |\nabla u|^2 \le \int_{B(x,2r)} \varphi^2 |\nabla u|^2 \le 4 \int_{B(x,2r)} u^2 |\nabla \varphi|^2 = \frac{4}{r^2} \int_{B(x,2r) \backslash B(x,r)} u^2.
$$
This is actually stronger than the desired inequality.</p>
|
1,002,777 | <p>I want to convert this polynomoial to partial fraction.</p>
<p>$$
\frac{x^2-2x+2}{x(x-1)}
$$</p>
<p>I proceed like this:
$$
\frac{x^2-2x+2}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1}
$$
Solving,
$$
A=-2,B=1
$$
But this does not make sense. What is going wrong?</p>
| lab bhattacharjee | 33,337 | <p>As the highest power of $x$ in the numerator & the denominator are same,</p>
<p>using <a href="http://en.wikipedia.org/wiki/Partial_fraction_decomposition" rel="nofollow">Partial Fraction Decomposition</a>,
$$\frac{x^2-2x+2}{x(x-1)}=1+\frac Ax+\frac B{x-1}$$</p>
<p>$1$ is found by $$\frac{\text{The coefficient of the highest power of }x\text{ in the numerator}}{\text{The coefficient of the highest power of }x\text{ in the denominator}}$$</p>
|
2,293,162 | <p>$$f_n(x)=\begin{cases} 1-nx,&\text{for }x\in[0,1/n]\\
0 ,&\text{for }x \in [1/n,1]
\end{cases}$$ </p>
<p>Then which is correct option?</p>
<p>1.$\lim\limits_{n\to\infty }f_n(x)$ defines a continuous function on $[0,1]$. </p>
<p>2.$\lim\limits_{n\to\infty }f_n(x)$ exists for all $x\in [0,1]$. $f_n(0)=1$ and $f_n(1/n)=0$ </p>
<p>I think first option is correct but I have no proper justification for that just by looking at the graph of function, $f$ is continuous but again there is a question arises if $f$ is continuous then why $2$ is wrong? Please correct me and help me to understand this problem.</p>
| Community | -1 | <p>The function cannot be continuous at $x=0$, because its value is $1$ and the values at all $x>0$ are $0$. (Because for $x>0$, there is always an $n$ such that $1/n<x$).</p>
<p>The option 2. is correct because the limit is defined for all $x$ (as above) and the last two equalities come from the definition.</p>
|
3,948,418 | <p>Ok so on doing a whole lot of Geometry Problems, since I am weak at Trigonometry, I am now focused on <span class="math-container">$2$</span> main questions :-</p>
<p><span class="math-container">$1)$</span> <strong>How to calculate the <span class="math-container">$\sin,\cos,\tan$</span> of any angle?</strong></p>
<p><strong>Some Information</strong> :- This site :- <a href="https://www.intmath.com/blog/mathematics/how-do-you-find-exact-values-for-the-sine-of-all-angles-6212" rel="noreferrer">https://www.intmath.com/blog/mathematics/how-do-you-find-exact-values-for-the-sine-of-all-angles-6212</a> , produces a clear understanding and a detailed approach of finding the <span class="math-container">$\sin$</span> of any angle from <span class="math-container">$1$</span> to <span class="math-container">$90^\circ$</span> , and I found it very interesting. But now the Questions arise :-</p>
<p>Can you find the <span class="math-container">$\sin$</span>, <span class="math-container">$\cos$</span> or <span class="math-container">$\tan$</span> of any fraction angles, like <span class="math-container">$39.67$</span>? <br/>
Can you find the <span class="math-container">$\sin$</span>, <span class="math-container">$\cos$</span> or <span class="math-container">$\tan$</span> of recurring fractions like <span class="math-container">$\frac{47}{9}$</span>? <br/>
Can you find the <span class="math-container">$\sin$</span>, <span class="math-container">$\cos$</span> or <span class="math-container">$\tan$</span> of irrationals, like <span class="math-container">$\sqrt{2}?$</span></p>
<p>Since I am a bit new to Trigonometry, I will be asking if there is a formula to find the <span class="math-container">$\sin$</span> of fractions, or even recurring fractions. I can use the calculator to find them obviously, but I have another Question :-</p>
<p><span class="math-container">$2)$</span> <strong>How to calculate the trigonometric ratios of every angle in fractional form?</strong></p>
<p>We all know <span class="math-container">$\sin 45^\circ = \frac{1}{\sqrt{2}}$</span> , but what will be <span class="math-container">$\sin 46^\circ$</span> in fractions? I can use a calculator to calculate the decimal of it, but it is hard to deduce the fraction out of the value, especially because the decimal will be irrational. I know how to convert recurring decimals to fractions, but this is not the case. Right now I am focused on a particular problem, which asks me to find the <span class="math-container">$\sin$</span> of a recurring fraction, in a fraction form. I am struggling to do this unless I clear up the ideas.</p>
<p><strong>Edit</strong>: My problem is to find the <span class="math-container">$\sin$</span> of <span class="math-container">$\frac{143}{3}^\circ$</span> . I do not have any specific formula to find this, and I am mainly stuck here. I need a formula which shows how this can be done.</p>
<p>Can anyone help me? Thank You.</p>
| Tyma Gaidash | 905,886 | <p>This will be my attempt at answering your question about finding <span class="math-container">$\sin(\frac{143°}{3})=\sin(\frac x3)$</span>.</p>
<p>Let θ=<span class="math-container">$\frac x3$</span> and using <a href="https://mathworld.wolfram.com/Multiple-AngleFormulas.html" rel="nofollow noreferrer">this website</a></p>
<p><span class="math-container">$\sin(3θ)=3\sinθ-4\sin^3θ$</span> Therefore, we need to solve this equation for sin(θ):</p>
<p><span class="math-container">$4\sin^3θ-3\sinθ+ \sin(3θ)=0⇔ \sin^3θ -\frac 34 \sinθ+ \frac 14 \sin(3θ)=0$</span></p>
<p>Using Cardano’s Depressed Cubic Formula gets us the first root of the cubic equation gets us:</p>
<p><span class="math-container">$\sqrt[3]{ \frac {-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$</span>+ <span class="math-container">$\sqrt[3]{ \frac {-q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$</span> where q=<span class="math-container">$\frac 14 \sin(3θ)$</span>= <span class="math-container">$\frac 14 \sin(x)$</span>= <span class="math-container">$\frac 14 \sin(143°)$</span>= <span class="math-container">$\frac 14 \sin(37°)$</span> which is the constant value of the equation and p=-<span class="math-container">$\frac 34$</span> which is the coefficient of the linear value of degree one in the equation above.</p>
<p>Plug these into the formula and simplify to get an answer. We will use the fact that <span class="math-container">$\sin143°=\sin37°$</span>:</p>
<p><span class="math-container">$\sqrt[3]{ \frac {-(1/4)\sin37°}{2}+\sqrt{\frac{((1/4)\sin37°)^2}{4}+\frac{(-3/4)^3}{27}}}$</span>+ <span class="math-container">$\sqrt[3]{ \frac {-1/4)\sin37°}{2}-\sqrt{\frac{((1/4)\sin37°)^2}{4}+\frac{(-3/4)^3}{27}}}$</span>=</p>
<p><span class="math-container">$\sqrt[3]{ \frac {-\sin37°}{8}+\frac 18 \sqrt{\sin^2 37°-1}}$</span>+<span class="math-container">$\sqrt[3]{ \frac {-\sin37°}{8}-\frac 18 \sqrt{\sin^2 37°-1}}$</span>=<span class="math-container">$\frac 12 (\sqrt[3]{i\cos37°- \sin37°}-\sqrt[3]{\sin37°+i\cos37°})$</span>.</p>
<p>Unfortunately, <span class="math-container">$\sin37°=\cos53°$</span> and <span class="math-container">$\cos37°=\sin53°$</span> do not have easily solvable forms, but <a href="https://www.intmath.com/blog/wp-content/images/2011/06/exact-values-sin-degrees.pdf" rel="nofollow noreferrer">this website</a> has the exact values for sine. However, <span class="math-container">$\sin37°=\sin143°=\sin\frac{37π}{180}$</span> so here are the steps for finding this value:</p>
<p>1.Use the same technique but sin(5θ)=sinx,x=π. Then, <span class="math-container">$θ=\frac {x}{5}$</span> and using the multiple angle formulas for sin in the above website to get sinπ=0=<span class="math-container">$5y-20y^3-16y^6$</span>,and solve for <a href="https://mathworld.wolfram.com/TrigonometryAnglesPi5.html" rel="nofollow noreferrer"><span class="math-container">$y=\sinθ=\sin\frac π5$</span></a></p>
<p>2.Use the cubic technique on <span class="math-container">$sin\frac π5$</span> to get <span class="math-container">$\sin\frac{π}{15}$</span></p>
<p>3.Use the half angle formula twice to get <span class="math-container">$\sin\frac{π}{60}$</span></p>
<p>4.Use the cubic technique again on <span class="math-container">$sin\frac π{60}$</span> to get <span class="math-container">$\sin\frac{π}{180}$</span></p>
<p>5.Finally use the multiple angle formula for <span class="math-container">$\sin(37a)=\sin\frac{37π}{180}$</span></p>
<p>6.Evaluate <span class="math-container">$\sqrt{1-\sin^2\frac{37π}{180}}$</span>= <span class="math-container">$\cos\frac{37π}{180}$</span></p>
<p>This means the final answer is:
<span class="math-container">$\sin\frac{143°}{3}$</span>= <span class="math-container">$\frac 12 \bigg(\sqrt[3]{\bigg[}\bigg($</span><a href="https://i.stack.imgur.com/E8M4I.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E8M4I.jpg" alt="[" /></a><span class="math-container">$\bigg)i-\bigg($</span>
<a href="https://i.stack.imgur.com/nJTsO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nJTsO.jpg" alt="enter image description here" /></a><span class="math-container">$\bigg)-\sqrt[3]{\bigg[}$</span><img src="https://i.stack.imgur.com/W1WLE.jpg" alt="enter image description here" /><span class="math-container">$+\bigg($</span><a href="https://i.stack.imgur.com/Ong1c.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ong1c.jpg" alt="enter image description here" /></a><span class="math-container">$\bigg)i\bigg]\bigg)$</span></p>
<p>Here is <a href="https://www.wolframalpha.com/input/?i=%281%2F2%29%28cbrt%28-sin37%2Bicos37%29%2Bcbrt%28-sin37-icos37%29%29%29-sin%28143.01%2F3%29+&assumption=%22TrigRD%22+-%3E+%22D%22&assumption=%22i%22+-%3E+%22ImaginaryI%22&assumption=%7B%22FunClash%22%2C+%22cbrt%22%7D+-%3E+%7B%22CubeRoot%22%7D" rel="nofollow noreferrer">proof</a> of my <a href="https://www.wolframalpha.com/input/?i=cbrt%28-a-b%29%3D-cbrt%28a%2Bb%29" rel="nofollow noreferrer">answer</a>. Please correct me if I am wrong or give me feedback!</p>
|
2,612,794 | <p>I have a very elementar question but I do not see where my mistake is. </p>
<p>Suppose we have a sequence $(x_n)$ with
$\lim_{n\to\infty}x_n=1$. Moreover, suppose that the sequence $({x_n}^c)$ for some constant $c>1$ has limit $\lim_{n\to\infty}{x_{n}}^c=c$.</p>
<p>Then
$$
\lim_{n\to\infty}\log({x_n}^c)=\log(c).
$$</p>
<p>But since $\log({x_n}^c)=c\log(x_n)$, I also have</p>
<p>$$
\lim_{n\to\infty}\log({x_n}^c)=c\lim_{n\to\infty}\log(x_n)=0.
$$</p>
<p>Where is my mistake? Maybe in the assumptions of the sequences.</p>
| Michael Hardy | 11,667 | <p>If $\lim\limits_{n\,\to\,\infty} x_n =1$ then $\lim\limits_{n\,\to\,\infty} x_n^c = 1.$
\begin{align}
\lim_{n\,\to\,\infty} (x_n^c) & = \left( \lim_{n\,\to\,\infty} x_n \right)^c & & \text{because } x \mapsto x^c \text{ is a continuous function} \\[10pt]
& = 1^c = 1.
\end{align}</p>
|
2,165,213 | <p><strong>The Problem</strong></p>
<p>Let $V=k^3$ for some field $k$. Let $W$ be the subspace spanned by $(1,0,0)$ and let $U$ be the subspace spanned by $(1,1,0)$ and $(0,1,1)$. Show that $V= W \oplus U$. Explain your argument in detail.</p>
<hr>
<p><strong>What I Know</strong></p>
<ol>
<li><p>I know that a field $k^n=n$-tuples of elements of $k$.</p></li>
<li><p>I know that a subset $W$ of a vector space $V$ over a field $k$ is a <em>subspace</em> if the operations of $V$ make $W$ into a vector space over $k$.</p></li>
<li><p>I know that if span$(S)=V$ for a set $S$ in a vector space $V$, where $S$ is linearly independent, then $S$ is a <em>basis</em> for $V$.</p></li>
<li><p>I know that the <em>external direct sum</em> $V \oplus W$ for vector spaces $V$ and $W$ over a field $k$ is defined as the set of all ordered pairs $(v,w)$ such that $v\in V$ and $w \in W$.</p>
<hr></li>
</ol>
<p><strong>What I Don't Know</strong></p>
<ol>
<li><p>How to <em>apply</em> what I listed above to help me solve the problem. I am absolutely atrocious at this material and struggle so much in simply starting these problems.</p></li>
<li><p>If everything I listed above is even relevant to the problem at hand.</p></li>
<li><p>If what I listed above is insufficient to complete the problem. </p>
<hr></li>
</ol>
<p>Text: <em>Abstract Linear Algebra</em> by Curtis</p>
| InsideOut | 235,392 | <p>Let $q(x)$ be an element in $\Bbb R[x]$, and let $n=\deg q(x)$. Then
$$q(x)=(x-5)g(x)+r(x),$$ but $r(x)\in \Bbb R$, so it is a constant because $\deg g(x)=n-1$. </p>
<p>Thus $\Bbb R[x]/( x-5 ) \cong \Bbb R$, because $\ker\phi=\{p(x) : (x-5)/ p(x)\}\cong \Bbb R_{\ge 1}[x]$.</p>
<p>EDIT: the last isomorphism is the following. First of all choose $\mathcal{B}=\{1,x-5,x^2-25,\dots,x^n-5^n\dots\}$ as basis for $\Bbb R[x]$ and $\ker\phi=\langle \mathcal{B}\setminus \{1\}\rangle$. Then the isomorphism is given by a map sending $x^n-5^n$ to $x^n$.</p>
|
3,060,742 | <p><span class="math-container">$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = 1.644934$</span> or <span class="math-container">$\frac{\pi^2}{6}$</span></p>
<p>What if we take every 3rd term and add them up? </p>
<p>A = <span class="math-container">$ \frac{1}{3^2} + \frac{1}{6^2} + \frac{1}{9^2} + \cdots = ??$</span></p>
<p>How to take every 3rd-1 term and add them up?</p>
<p>B = <span class="math-container">$ \frac{1}{2^2} + \frac{1}{5^2} + \frac{1}{8^2} + \cdots = ??$</span></p>
<p>How to take every 3rd-2 term and add them up?</p>
<p>C = <span class="math-container">$ \frac{1}{1^2} + \frac{1}{4^2} + \frac{1}{7^2} + \cdots = ??$</span></p>
<p>I am not sure how to adapt Eulers methods as he used the power series of sin for his arguments: <a href="https://en.wikipedia.org/wiki/Basel_problem" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Basel_problem</a></p>
| robjohn | 13,854 | <p><strong>Polylogarithms</strong></p>
<p>A useful formula that can be applied here is
<span class="math-container">$$
\frac13\sum_{k=0}^2e^{2\pi ijk/3}=[3\mid j]\tag1
$$</span>
So
<span class="math-container">$$
\begin{align}\newcommand{\Li}{\operatorname{Li}}
\sum_{j=0}^\infty\frac1{(3j+1)^2}
&=\frac13\sum_{k=0}^2\sum_{j=1}^\infty e^{2\pi i(j-1)k/3}\frac1{j^2}\\
&=\frac13\left(\frac{\pi^2}6+e^{-2\pi i/3}\Li_2\left(e^{2\pi i/3}\right)+e^{2\pi i/3}\Li_2\left(e^{-2\pi i/3}\right)\right)\tag2
\end{align}
$$</span>
Mathematica gives <span class="math-container">$1.12173301393634378687$</span> using<br>
<code>N[1/3(Pi^2/6 + Exp[-2Pi I/3]PolyLog[2,Exp[2Pi I/3]]+
Exp[2Pi I/3] PolyLog[2,Exp[-2Pi I/3]]),20]</code></p>
<hr>
<p><strong>Extended Harmonic Numbers</strong></p>
<p>Another approach is to use the Extended Harmonic Numbers.
<span class="math-container">$$
H(x)=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\tag3
$$</span>
where
<span class="math-container">$$
H'(x)=\sum_{k=1}^\infty\frac1{(k+x)^2}\tag5
$$</span>
Giving
<span class="math-container">$$
\frac19H'\!\left(-\frac23\right)=\sum_{k=0}^\infty\frac1{(3k+1)^2}\tag6
$$</span>
Mathematica gives <span class="math-container">$1.1217330139363437869$</span> using<br>
<code>N[1/9HarmonicNumber'[-2/3],20]</code></p>
<hr>
<p><strong>Euler-Maclaurin Sum Formula</strong></p>
<p>Although it does not give a closed form, the Euler-Maclaurin Sum Formula allows us to accelerate the computation of the sum.
<span class="math-container">$$
\begin{align}
\sum_{k=0}^n\frac1{(3k+1)^2}
&\sim C-\frac1{3(3n+1)}+\frac1{2(3n+1)^2}-\frac1{2(3n+1)^3}+\frac9{10(3n+1)^5}\\
&-\frac{81}{14(3n+1)^7}+\frac{729}{10(3n+1)^9}-\frac{32805}{22(3n+1)^{11}}\tag7
\end{align}
$$</span>
Using <span class="math-container">$n=100$</span> in <span class="math-container">$(7)$</span>, we get
<span class="math-container">$$
\sum_{k=0}^\infty\frac1{(3k+1)^2}=1.1217330139363437868657782\tag8
$$</span></p>
|
2,154,960 | <p>Here is the question and i dont really understand</p>
<p>Point $(a,b)$ is on the function $f(x)=\frac{2}{x}$ $x>0$. Show that the area of the triangle formed by the tangent line at $(a,b)$ , the $x$ axis and $y$ axis is equals to $4$.</p>
<p>What is the question asking?</p>
<p>I used the first principle to find the derivative $f'(x)=\frac{-2}{x^2}$</p>
| Paul Sundheim | 88,038 | <ol>
<li>Find $b$ in terms of $a$ using $f(x)=2/x$. This gives you the point $(a,b)$ in terms of $a$.</li>
<li>Find the slope of the tangent line at $(a,b)$ using the derivative of $f$, in terms of $a$.</li>
<li>Find the $y$-intercept of the line using the point and the slope.</li>
<li>Use the equation of the line to get the $x$-intercept.</li>
<li>You now have two sides of a right triangle using the $x$ and $y$ intercepts. Use the area formula for a triangle to show that the area must be 4.</li>
</ol>
|
2,154,960 | <p>Here is the question and i dont really understand</p>
<p>Point $(a,b)$ is on the function $f(x)=\frac{2}{x}$ $x>0$. Show that the area of the triangle formed by the tangent line at $(a,b)$ , the $x$ axis and $y$ axis is equals to $4$.</p>
<p>What is the question asking?</p>
<p>I used the first principle to find the derivative $f'(x)=\frac{-2}{x^2}$</p>
| David G. Stork | 210,401 | <p>The questioner asked for help understanding the question itself, and here a picture is worth $10^3$ words:</p>
<p><a href="https://i.stack.imgur.com/RibfO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RibfO.png" alt="enter image description here"></a></p>
<p>The goal is then to find the value of $a$ (and then $b$) such that the area of the triangle defined by the tangent and the axes has the desired value. The figure above is to illustrate a <em>candidate</em> solution. The goal is to use simple algebra to find the actual solution values. Paul Sundheim's approach is straightforward indeed.</p>
|
355,296 | <p>How can we evaluate $$\displaystyle\int \frac{x^2 + x+3}{x^2+2x+5} dx$$ </p>
<p>To be honest, I'm embarrassed. I decomposed it and know what the answer should be but<br>
I can't get the right answer. </p>
| Christopher A. Wong | 22,059 | <p>You can decompose your integrand as follows:</p>
<p>$$ \frac{ x^2 + 2x + 5 - x - 1 - 1}{x^2 + 2x + 5} = 1 - \frac{x + 1}{x^2 + 2x + 5} - \frac{1}{(x+1)^2 + 4}$$</p>
<p>You can integrate the first term directly, the second term after the substitution $u = x^2 + 2x + 5$, and the third term by recalling that $(\arctan{x})' = 1/(x^2 + 1)$, and then using another substitution to make the expression look like the derivative of $\arctan$.</p>
|
355,296 | <p>How can we evaluate $$\displaystyle\int \frac{x^2 + x+3}{x^2+2x+5} dx$$ </p>
<p>To be honest, I'm embarrassed. I decomposed it and know what the answer should be but<br>
I can't get the right answer. </p>
| Community | -1 | <p><strong>Hint</strong> Use the decomposition
$$\frac{x^2 + x+3}{x^2+2x+5}=1-\frac{ x+2}{x^2+2x+5}=1-\frac{1}{2}\frac{ 2x+2}{x^2+2x+5}-\frac{ 1}{x^2+2x+5}$$
and
$$\frac{ 1}{x^2+2x+5}=\frac{ 1}{(x+1)^2+4}=\frac{1}{4}\frac{ 1}{(\frac{x+1}{2})^2+1}$$
the first fraction is on the form $\frac{f'}{f}$ and the second have the form $\frac{1}{u^2+1}$ by change of variable.</p>
|
642,443 | <p>Let $\{a_n\}_{n\ge1}^{\infty}=\bigg\{\cfrac{1}{1\cdot3}+\cfrac{1}{2\cdot4}+\dots+\cfrac{1}{n\cdot(n+2)}\bigg\}$. Find $\lim_{n\to \infty}{a_n}$.</p>
<p>I write:
$$\lim_{n\to \infty}{a_n}=\sum_{n=1}^{\infty}{\frac{1}{n\cdot(n+2)}}=\sum_{n=1}^{\infty}{\frac{1}{n^2+2n}}\approx\sum_{n=1}^{\infty}{\cfrac{1}{n^2}}$$</p>
<p>I put some values of $n$ for finding a pattern:
$$
\begin{array}{c|lcr}
n & \text{1}&\text{2}&\text{3}&\text{4}\\
\hline
\sum &\cfrac{1}{3}&\cfrac{11}{24}&\cfrac{21}{40}&\cfrac{17}{30}
\end{array}
$$</p>
<p>... but no hope.
I know that the limit/series converges, because $\forall n\in\mathbb N^*$:</p>
<ol>
<li><p>$\{a_n\}$ is increasing by the test of monothony : $a_{n+1}-a_{n}=\cfrac{1}{(n+1)(n+3)}>0$</p></li>
<li><p>$\{a_n\}$ is bounded : $0\le a_1=\frac{1}{3}\le a_2=\frac{11}{24}\le \dots \le a_n \le1$</p></li>
</ol>
<p>Wolfram says that the summation can be written as follows:
$$\cfrac{3}{4}-\cfrac{2n+3}{2(n+1)(n+2)}$$
How did it end up at this formula? Thank you.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\down}{\downarrow}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\fermi}{\,{\rm f}}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\half}{{1 \over 2}}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\ol}[1]{\overline{#1}}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\sum_{n = 1}^{\infty}{1 \over n\pars{n + 2}}&=
\sum_{n = 0}^{\infty}{1 \over \pars{n + 1}\pars{n + 3}}
={\Psi\pars{3} - \Psi\pars{1} \over 3 - 1}\tag{1}
\end{align}
where $\Psi\pars{z}$ is the $\it digamma$ function. By using the identity
$\Psi\pars{z} = 1/\pars{z - 1} + \Psi\pars{z - 1}$ we'll get:
$$
\Psi\pars{3} = \half + \Psi\pars{2} = \half + 1 + \Psi\pars{1} = {3 \over 2} + \Psi\pars{1}
$$
We replace this result in $\pars{1}$:
$$
\color{#00f}{\large\sum_{n = 1}^{\infty}{1 \over n\pars{n + 2}} = {3 \over 4}}
$$</p>
|
8,997 | <p>I have a set of data points in two columns in a spreadsheet (OpenOffice Calc):</p>
<p><img src="https://i.stack.imgur.com/IPNz9.png" alt="enter image description here"></p>
<p>I would like to get these into <em>Mathematica</em> in this format:</p>
<pre><code>data = {{1, 3.3}, {2, 5.6}, {3, 7.1}, {4, 11.4}, {5, 14.8}, {6, 18.3}}
</code></pre>
<p>I have Googled for this, but what I find is about importing the entire document, which seems like overkill. Is there a way to kind of cut and paste those two columns into <em>Mathematica</em>? </p>
| WReach | 142 | <p>Here is a manual method using copy-and-paste that is suitable for small volumes of data on an ad hoc basis...</p>
<p>1) Enter the following expression into a notebook, but don't evaluate it:</p>
<pre><code>data = ImportString["", "TSV"]
</code></pre>
<p>2) Copy the cells from the source spreadsheet onto the clipboard.</p>
<p>3) Paste the contents of the clipboard between the empty pair of quotes in the expression from <em>1)</em>. If Mathematica brings up a dialog asking whether you want "escapes inserted", select <em>Yes</em>. The result should look something like this:</p>
<pre>data = ImportString["<em>1.0\t2.0\t3.0
0.4\t0.5\t0.6
7.0\t8.0\t9.0
</em>","TSV"]</pre>
<p>4) Press <kbd>SHIFT</kbd>-<kbd>ENTER</kbd> to evaluate the expression.</p>
<p>5) <code>data</code> now contains the pasted cell data, interpreted as Tab-Separated Values (TSV).</p>
<p><strong>Alternate Approach</strong></p>
<p>Follows steps <em>1)</em> through <em>3)</em> as before, then...</p>
<p>4a) Triple-click on <code>ImportString</code> in the expression from <em>1)</em>.</p>
<p>5a) Press <kbd>CTRL</kbd>-<kbd>SHIFT</kbd>-<kbd>ENTER</kbd> to evaluate the highlighted expression in place.</p>
<p>6a) The notebook now contains an expression that looks like this:</p>
<pre><code>data = {{1., 2., 3.}, {0.4, 0.5, 0.6}, {7., 8., 9.}}
</code></pre>
<p>... but that expression has not been evaluated yet.</p>
|
481,167 | <p>Let $V$ be a $\mathbb{R}$-vector space.
Let $\Phi:V^n\to\mathbb{R}$ a multilinear symmetric operator.</p>
<p>Is it true and how do we show that for any $v_1,\ldots,v_n\in V$, we have:</p>
<p>$$\Phi[v_1,\ldots,v_n]=\frac{1}{n!} \sum_{k=1}^n \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-k}\phi (v_{j_1}+\cdots+v_{j_k}),$$
where $\phi(v)=\Phi(v,\ldots,v)$.</p>
<p>My question come from that, I have seen this formula when I was reading about mixed volume, and also when I was reading about mixed Monge-Ampère measure. The setting was not exactly the one of a vector space $V$ but I think the formula is true here and I am interested by having this property shown out of the specific context of Monge-Ampère measures or volumes. I have done some work in the other direction, <em>i.e.</em> starting from an operator $\phi:V\to\mathbb{R}$ satisfying some condition and obtaining a multilinear operator $\Phi$ ; bellow are the results I have seen in this direction.</p>
<p>I already know that if $\phi':V\to\mathbb{R}$ is such that for any $v_1,\ldots,v_n\in V$, $\phi'(\lambda_1 v_1+\ldots+\lambda_n v_n)$ is a homogeneous polynomial of degree $n$ in the variables $\lambda_i$, then there exists a unique multilinear symmetric operator $\Phi':V^n\to\mathbb{R}$ such that $\Phi'(v,\ldots,v)=\phi'(v)$ for any $v\in V$. Moreover $\Phi'(v_1,\ldots,v_n)$ is the coefficient of the symmetric monomial $\lambda_1\cdots\lambda_n$ in $\phi'(\lambda_1 v_1+\ldots+\lambda_n v_n)$ (see <a href="https://math.stackexchange.com/questions/469342/symmetric-multilinear-form-from-an-homogenous-form">Symmetric multilinear form from an homogenous form.</a>).</p>
<p>I also know that if $\phi'(\lambda v)=\lambda^n \phi'(v)$ and we define
$$\Phi''(v_1,\ldots,v_n)=\frac{1}{n!} \sum_{k=1}^n \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-k}\phi' (v_{j_1}+\cdots+v_{j_k}),$$
then $\Phi''(v,\ldots,v)=\frac{1}{n!} \sum_{k=1}^n (-1)^{n-k} \binom{n}{k} k^n \phi'(v)=\phi'(v)$ (see <a href="https://math.stackexchange.com/questions/465172/show-this-equality-the-factorial-as-an-alternate-sum-with-binomial-coefficients">Show this equality (The factorial as an alternate sum with binomial coefficients).</a>). It is clear that $\Phi''$ is symmetric, but I don't know if $\Phi''$ is multilinear.</p>
<p>Formula for $n=2$:
$$\Phi[v_1,v_2]=\frac12 [\phi(v_1+v_2)-\phi(v_1)-\phi(v_2)].$$</p>
<p>Formula for $n=3$:
$$\Phi[v_1,v_2,v_3]=\frac16 [\phi(v_1+v_2+v_3)-\phi(v_1+v_2)-\phi(v_1+v_3)-\phi(v_2+v_3)+\phi(v_1)+\phi(v_2)+\phi(v_3)].$$</p>
| Gilles Bonnet | 60,457 | <p><strong>This in not an answer</strong>, but an incomplete attempt of induction proof.</p>
<p>First, we will consider the following notation:
$$\Phi_v[v_1,\ldots,v_{n-1}]=\Phi[v_1,\ldots,v_{n-1},v],$$ so $\Phi_v:V^{n-1}\to\mathbb{R}$ is the multinear symmetric operator we obtain when we fix a variable in $\Phi$.
We then of course note $\phi_w(v)=\Phi_w[v,\ldots,v]=\Phi[v,\ldots,v,w]$.</p>
<p>In this 'answer', I show that the formula is proved if
$$\phi(v_1+\cdots+v_n)
=\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-1-k}\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k}),$$
which is probably not more easy then the polarization formula itself, but that where lead me my attempt of induction.</p>
<p>We also write
$$\Phi_v[v_1,\ldots,\hat{v_i},\ldots,v_n]=\Phi_v[v_1,\ldots,v_{i-1},v_{i+1},\ldots,v_n].$$</p>
<p>Let assume the formula is true for multilinear symmetric operator $V^{n-1}\to\mathbb{R}$.
Since $\Phi[v_1,\ldots,v_n]=\Phi_{v_i}[v_1,\ldots,\hat{v_i},\ldots,v_n]$ by symmetry, we have:
$$\Phi[v_1,\ldots,v_n]=\frac1n \sum_{i=1}^n \Phi_{v_i}[v_1,\ldots,\hat{v_i},\ldots,v_n].$$
By the induction we have
$$\Phi_{v_i}[v_1,\ldots,\hat{v_i},\ldots,v_n]
=\frac{1}{(n-1)!} \sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n\ ;\ j_l\neq i} (-1)^{n-1-k}\phi_{v_i} (v_{j_1}+\cdots+v_{j_k}).$$
So
$$\Phi[v_1,\ldots,v_n]
=\frac1{n!}\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-1-k}
\sum_{\{i\mid i\neq j_l \forall l\leq k\}}\phi_{v_i} (v_{j_1}+\cdots+v_{j_k}).$$
But
\begin{align}
\sum_{\{i\mid i\neq j_l \forall l\leq k\}}\phi_{v_i} (v_{j_1}+\cdots+v_{j_k})
&={\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k})-\phi_{v_{j_1}+\cdots+v_{j_k}}(v_{j_1}+\cdots+v_{j_k})}\\
&={\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k})-\phi(v_{j_1}+\cdots+v_{j_k})}.
\end{align}
So
\begin{align}
&\Phi[v_1,\ldots,v_n]\\
&=\frac1{n!}\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-1-k}
(\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k})-\phi(v_{j_1}+\cdots+v_{j_k}))\\
&=\frac1{n!}\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-k}
\phi(v_{j_1}+\cdots+v_{j_k})\\
&\qquad +\frac1{n!}\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-1-k}\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k}).\\
\end{align}
In this last expression the first part is almost the R.H.S. of the polarisation formula ; the sum goes until $n-1$ instead of $n$. But when $k=n$,
$\sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-k} \phi(v_{j_1}+\cdots+v_{j_k})
=\phi(v_1+\cdots+v_n)$.
Hence we will have prove the polarization formula if
$$\phi(v_1+\cdots+v_n)
=\sum_{k=1}^{n-1} \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-1-k}\phi_{v_1+\cdots+v_n}(v_{j_1}+\cdots+v_{j_k}).$$</p>
<p>Unfortunately I don't see how to show that, it is maybe as difficult as the polarization formula. </p>
|
204,592 | <p>The matrix exponential is a well know thing but when I see online it is provided for matrices. Does it the same expansion for a linear operator? That is if $A$ is a linear operator then $$e^A=I+A+\frac{1}{2}A^2+\cdots+\frac{1}{k!}A^k+\cdots$$</p>
| kalvotom | 38,469 | <p>Yes, you can define an exponential of any linear BOUNDED operator by this series. If the operator is unbounded then it is not always possible. </p>
|
1,285,014 | <p>Let $R,S$ be commutative rings with identity.</p>
<p>Proving that $X \sqcup Y$ is an affine scheme is the same as proving that $Spec(R) \sqcup Spec(S) = Spec(R \times S)$.</p>
<p>I proved that if $R,S$ are rings, then the ideals of $R \times S$ are exactly of the form $P \times Q$, where $P$ is an ideal of $R$ and $Q$ is an ideal of $S$.</p>
<p>However, for prime ideals this is not true in general.</p>
<p>If $I$ is a prime ideal of $R \times S$, then $I = \mathfrak{p} \times \mathfrak{q}$, where $\mathfrak{p}$ is a prime ideal of $R$ and $\mathfrak{q}$ is a prime ideal of $S$.</p>
<p>But if $\mathfrak{p}$ is a prime ideal of $R$ and $\mathfrak{q}$ is a prime ideal of $S$, it is not true in general that $\mathfrak{p} \times \mathfrak{q}$ is a prime ideal of $R \times S$.</p>
<p>Then, $Spec(R \times S) \subseteq Spec(R) \times Spec(S)$ and the reverse inclusion is false in general.</p>
<p>My question is, what is $Spec(R) \sqcup Spec(S)$ set-theoretically, in order to use what I proved above?</p>
| Demosthene | 163,662 | <p>Note that you can rewrite $p\to\neg(q\lor r)$ as $\neg p\lor(\neg(q\lor r))$, i.e. $\neg p\lor(\neg q\land\neg r)$. This fits exactly the truth table (as you would expect), and shows that the proposition is true whenever $p$ is false or both $q$ and $r$ are false.</p>
<p>This would usually enough be enough to answer your question, but we can also give a full disjunctive normal form of all the cases making the proposition true, as follows:
\begin{align}\neg p&\land q\land r\\\lor\ \neg p&\land q\land\neg r\\\lor\ \neg p&\land\neg q\land r\\\lor\ \neg p&\land\neg q\land\neg r\\\lor\quad p&\land\neg q\land\neg r\end{align}</p>
|
1,285,014 | <p>Let $R,S$ be commutative rings with identity.</p>
<p>Proving that $X \sqcup Y$ is an affine scheme is the same as proving that $Spec(R) \sqcup Spec(S) = Spec(R \times S)$.</p>
<p>I proved that if $R,S$ are rings, then the ideals of $R \times S$ are exactly of the form $P \times Q$, where $P$ is an ideal of $R$ and $Q$ is an ideal of $S$.</p>
<p>However, for prime ideals this is not true in general.</p>
<p>If $I$ is a prime ideal of $R \times S$, then $I = \mathfrak{p} \times \mathfrak{q}$, where $\mathfrak{p}$ is a prime ideal of $R$ and $\mathfrak{q}$ is a prime ideal of $S$.</p>
<p>But if $\mathfrak{p}$ is a prime ideal of $R$ and $\mathfrak{q}$ is a prime ideal of $S$, it is not true in general that $\mathfrak{p} \times \mathfrak{q}$ is a prime ideal of $R \times S$.</p>
<p>Then, $Spec(R \times S) \subseteq Spec(R) \times Spec(S)$ and the reverse inclusion is false in general.</p>
<p>My question is, what is $Spec(R) \sqcup Spec(S)$ set-theoretically, in order to use what I proved above?</p>
| Theuth | 240,068 | <p>See also K-map:
<a href="http://en.wikipedia.org/wiki/Karnaugh_map" rel="nofollow">http://en.wikipedia.org/wiki/Karnaugh_map</a>
It is a very clever and fast method to derive DNF and other useful things</p>
|
1,828,097 | <p>If we contruct two strainght lines as shown:<a href="https://i.stack.imgur.com/8K5Eo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8K5Eo.png" alt="enter image description here"></a></p>
<p>Then join them such that to complete a triangle. <a href="https://i.stack.imgur.com/Uvtnw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Uvtnw.png" alt="enter image description here"></a></p>
<p>It is taught that we can find infinity points on straight line. So there are infinity points on $DE$ and $BC$. </p>
<p>If we will join $A$ with $BC$ as shown:<a href="https://i.stack.imgur.com/HuWos.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HuWos.png" alt="enter image description here"></a>
We can find one point on $DE$ and corresponding point on $BC$. So point on $DE$ and $BC$ are same.</p>
<p>Hence, can we say that $\infty=\infty$, But why $\infty - \infty \neq0$</p>
<p>I'm not sure does this make any sense or not,your suggestions are appreciated.</p>
| abcabc123 | 347,408 | <p>Well first of all it is because $\infty$ is not a number, you can't do what you usually do with numbers but let's suppose here that it is, let's suppose $\infty$ is a number, the bigest number. Then you can imagine a finite constant $k$ added to $\infty$ which has to be $\infty$ also i.e. $k+\infty=\infty$. That then implies that
$$\infty-\infty=k+\infty-\infty=k$$
This would be true if $\infty$ was a defined number in the first place which it isn't but we assumed it was here and we found that $\infty-\infty=k$, which remember $k$ can be any number you want. </p>
<p>We say then $\infty-\infty$ is an undefined expression. Not because it is infinite, actually, not mainly for that reason. It is actually due to the fact that $\infty$ is not a defined number like $2$ or $\pi$ or other numbers like them.</p>
<p>We could define an infinite number. Consider the product of all natural numbers from 1 and give it a name, say $I$. Then $I-I$ really equals $0$ because I really is a mathematical object in this case and really equals the same thing. It would be surely weird to work with such infinite numbers, but it already has been done.</p>
<p>So say your last line is twice the length of the first, you say "well there are twice as many points in that second line", but consider this: it is well know that there is the same "infinity" of natural numbers than the "infinity" of even numbers, is because there exists a one-to-one correspondance between the sets, even if you may think, at first sight, that there are twice as many naturals than there are even numbers. Infinity is tricky and is treated differently whether you are talking about an amount, in sets of numbers or points for example, or more like a number.</p>
|
2,227,027 | <p>If $f(x)$ is defined everywhere except at $x=x_0$, would $f'(x_0)$ be undefined at $x=x_0$ as well?</p>
<p>One example is: $$f(x)=\ln(x)\rightarrow f'(x)=\frac{1}{x}$$</p>
<p>In this particular case, both $f(x)$ and $f'(x)$ are undefined at $x=0$. I wonder if this always holds true.</p>
<p>Thank you.</p>
| The Count | 348,072 | <p>This depends on some conventions, but the typical answer is <em>yes</em>, because if a function is not defined somewhere, it cannot have a slope there! In other words, we certainly don't have a slope where there is no function value!</p>
<p>We need a value $f(x_0)$ to plug into the limit definition of the derivative, after all.</p>
|
3,273,756 | <blockquote>
<p>I am supposed to give a 9-dimensional irreducible representation of <span class="math-container">$\mathfrak{so}(4)$</span>.</p>
</blockquote>
<p>I know that <span class="math-container">$\mathfrak{so}(4)\cong\mathfrak{so}(3)\oplus\mathfrak{so}(3)$</span> and hence I have a 6-dimensional reducible representation of <span class="math-container">$\mathfrak{so}(4)$</span>. But how do I get a irreducible 9-dimensional representation?
This may be a stupid question but I couldn't find anything.</p>
| user10354138 | 592,552 | <p><strong>Hint</strong>: The space of <span class="math-container">$4\times 4$</span> symmetric traceless matrices is of dimension <span class="math-container">$9$</span>.</p>
|
1,861,890 | <p>Given that $$s_{n}=\frac{(-1)^{n}}{n},$$ I want to show $$\lim_{n\to\infty}{s_{n}}=0$$ in the metric space $X=\mathbb{C}.$ However, it seems to me that <strong>Archimedean Property is not applicable to the case above</strong>, because $s_{n}$ is not always positive for each $n$. Then, how can I do that?</p>
| George Law | 141,584 | <p>You don’t need $s_n$ to be always positive; all you need is to show that for every real $\varepsilon>0$, $|s_n-0|<\varepsilon$ for all $n$ greater than some $N$.</p>
|
3,260,530 | <p>I read from wikipedia that a neighbourhood of a point <span class="math-container">$p$</span> is a subset <span class="math-container">$V$</span> of a topological space <span class="math-container">$\{X,\tau\}$</span> that includes an open set <span class="math-container">$U$</span> such that <span class="math-container">$p \in U$</span>. </p>
<p>I would like to clarify, suppose that <span class="math-container">$p \in U$</span>, but <span class="math-container">$p \not\in W$</span>, if <span class="math-container">$V=\{U,W\}$</span>, does <span class="math-container">$V$</span> qualify as a neighborhood of <span class="math-container">$p$</span> even if it includes the set <span class="math-container">$W$</span> which does not contain <span class="math-container">$p$</span>?</p>
<p>Does this mean that if the topological space is <span class="math-container">$\mathbb{R}$</span>, is it accurate to say that the set <span class="math-container">$\{(-1,1),(2,3)\}$</span> is also a neighborhood of <span class="math-container">$0$</span>?</p>
| Theo Bendit | 248,286 | <p>A subset <span class="math-container">$V$</span> of a topological space <span class="math-container">$X$</span> is a neighbourhood of point <span class="math-container">$p \in X$</span> if some open set <span class="math-container">$U$</span> exists such that
<span class="math-container">$$p \in U \subseteq V.$$</span>
The point <span class="math-container">$p$</span> is not usually a subset of <span class="math-container">$U$</span>, nor is <span class="math-container">$V$</span> typically a set containing two sets, one open. Your question doesn't make a whole lot of sense as stated.</p>
<p>What I will say is that <span class="math-container">$(-1, 1) \cup (2, 3) \subseteq \Bbb{R}$</span> is a neighbourhood of <span class="math-container">$0$</span> (and <span class="math-container">$e$</span> for that matter).</p>
|
2,400,336 | <p>My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$</p>
<p>Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$</p>
<p>But I still end up with an ugly radical expression.</p>
| Frank | 332,250 | <p>There's actually a general formula for these kinds of expressions. Namely$$\sqrt{X\pm Y}=\sqrt{\frac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}2}$$
Where $X,Y$ are real numbers. Simply substituting $X=6$ and $Y=\sqrt{20}$ gives the proper denesting. The proof of this is quite simple. Assume that$$X\pm Y=\left(\sqrt A\pm \sqrt B\right)^2$$and expand via binomial theorem. Collecting terms, you will end up with two equations from which you can solve for $A$ and $B$ in terms of $X$ and $Y$.</p>
|
1,261,504 | <p>I am trying to proof $ab = \gcd(a,b)\mathrm{lcm}(a,b)$.</p>
<p>The definition of $\mathrm{lcm}(a,b)$ is as follows:</p>
<p>$t$ is the lowest common multiple of $a$ and $b$ if it satisfies the following:</p>
<p>i) $a | t$ and $b | t$ </p>
<p>ii) If $a | c$ and $b | c$, then $t | c$.</p>
<p>Similiarly for the $\gcd(a,b)$.</p>
<p>Here is my proof:</p>
<p>Case I: $\gcd(a,b)\neq 1$</p>
<p>Suppose $\gcd(a,b) = d$.</p>
<p>Then $ab = dq_1b = dbq_1 = d(dq_1q_2)$</p>
<p>Claim: $\mathrm{lcm}(a,b) = dq_1q_2$</p>
<p>$a = dq_1 | dq_1q_2$ </p>
<p>$b = dq_2 | dq_2q_1$.</p>
<p>Supppose $\mathrm{lcm}(a,b) = c$.
Hence $c \leq dq_1q_2$ .</p>
<p>To get the other inequality we have $dq_1 | a$ and $dq_2 | b$. Hence $dq_1 \leq a \leq c \leq dq_1q_2$ similarly for $dq_2$.</p>
<p>Suppose that c is strictly less than $dq_1q_2$, so we have $dq_1q_2 < cq_2$ and $dq_1q_2 < cq_1$.</p>
<p>So $dq_1q_2 < c < cq_2 < dq_2^2q_1$ and $dq_1q_2 < c < cq_2 < dq_1^2q_2$, but $dq_1^2q_2 > dq_1q_2$ so $c < dq_1q_2$ and </p>
<p>$c > dq_1q_2$ contradiction. Hence $c = dq_1q_2$ </p>
<p>Notice that the case where $\gcd(a,b) = 1$ we can just set $q_1 = a$ and $q_2 = b$, and the proof will be the same.</p>
| BruceET | 221,800 | <p>(a) If you are using at a normal table, first determine whether it
gives areas $P(Z \le z)$ [or $P(0 < Z \le z)].$ In the first
instance look in the <em>body</em> of the table to find the closest
probability to .9500; in this case, probably a tie between .9496 and .0505. Then find the corresponding value of $z$ by looking at the
<em>margins</em>; in this case, 1.64 or 1.65, so split the difference
and say $z = 1.645.$ [In the second case, you'll look for the
closest probability to .4500, eventually obtaining the same result.]</p>
<p>(b) By symmetry, it is $z = -1.645.$</p>
<p>If you are using statistical software, then you can get these using the
'inverse CDF' or 'quantile' function. You will get more places of accuracy than from normal tables. In R, it is:</p>
<pre><code> qnorm(.95) # (a)
## 1.644854
qnorm(.05) # (b) Same as P(Z <= z) = .05
## -1.644854
</code></pre>
<p>In Minitab:</p>
<pre><code> MTB > invcdf .95;
SUBC> norm.
Inverse Cumulative Distribution Function
Normal with mean = 0 and standard deviation = 1
P( X <= x ) x
0.95 1.64485
</code></pre>
<p>Almost all statistical software packages have something like this,
but each has its own syntax for input and formatting for output.</p>
<p><em>Addendum:</em> I just noticed the nice example from @calculus. In software, you
can sometimes work such problems without 'standardizing'; that is, without subtracting the mean and dividing the difference by the SD. The answer is a little different because software doesn't need to round to two places before table look-up. (Not that the
small difference matters for most practical purposes.)</p>
<pre><code> pnorm(5, 4, sqrt(9))
## 0.6305587
</code></pre>
|
163,917 | <p>Suppose $B_{\epsilon}$ are closed subsets of a compact space and $B_{\epsilon} \supset B_{\epsilon'} \quad \forall \epsilon > \epsilon'$. Furthermore, $B_0 = \bigcap_{\epsilon>0} B_{\epsilon}$. For a continuous function $f$ can we conclude that
$$f(B_0) = \bigcap_{\epsilon>0} f(B_{\epsilon})?$$</p>
<p>I believe the answer to be yes. It seems this should be a well-known property---I'm having trouble finding a reference.</p>
| kaba | 51,161 | <p>The following answer is a generalization of the one given by Asaf Karagila.</p>
<h4>Theorem</h4>
<p>Let <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> be spaces, <span class="math-container">$I$</span> be an ordinal, <span class="math-container">$V : I \to \mathcal{P}(X)$</span> be decreasing, <span class="math-container">$V_i$</span> be closed, and <span class="math-container">$f : X \to Y$</span> have compact fibers. Then</p>
<p><span class="math-container">$$f(\bigcap V_I) = \bigcap_{i \in I} f(V_i).$$</span></p>
<h4>Proof</h4>
<p>The <span class="math-container">$\subset$</span> direction holds by set theory; i.e. without any requirements. We will show the <span class="math-container">$\supset$</span> direction.</p>
<p>Let <span class="math-container">$y \in \bigcap_{i \in I} f[V_i]$</span>, <span class="math-container">$C = f^{-1}[\{y\}]$</span>, and <span class="math-container">$V'_i = V_i \cap C$</span>. Then
<span class="math-container">\begin{aligned}
f[V'_i] & = f[V_i \cap C] \\
{} & = f[V_i \cap f^{-1}[\{y\}]] \\
{} & = f[V_i] \cap \{y\} \\
{} & = \{y\}.
\end{aligned}</span>
Therefore <span class="math-container">$\bigcap_{i \in I} f[V'_i] = \{y\} \neq \emptyset$</span>. By subspace topology, <span class="math-container">$V'_i$</span> is closed in <span class="math-container">$C$</span>. By assumption, <span class="math-container">$C$</span> is compact. Since closed subsets of compact subsets are compact, <span class="math-container">$V'_i$</span> is compact. By the lemma below, <span class="math-container">$f[\bigcap V'_I] \neq \emptyset$</span>. Therefore <span class="math-container">$f[\bigcap V'_I] = \{y\}$</span>; i.e. <span class="math-container">$y \in f[\bigcap V_I]$</span>. Therefore <span class="math-container">$f[\bigcap V_I] \supset \bigcap_{i \in I} f[V_i]$</span>.</p>
<h4>Special case</h4>
<p>Suppose <span class="math-container">$X$</span> is compact, <span class="math-container">$Y$</span> is <span class="math-container">$T_1$</span>, and <span class="math-container">$f$</span> is continuous. Then <span class="math-container">$f$</span> has compact fibers, since a closed subspace of a compact space is compact.</p>
<h4>Lemma</h4>
<p>Let <span class="math-container">$X, Y$</span> be spaces, <span class="math-container">$I$</span> be an ordinal, <span class="math-container">$V : I \to \mathcal{P}(X)$</span> be decreasing, <span class="math-container">$V_i$</span> be closed and compact, and <span class="math-container">$f : X \to Y$</span>. Then</p>
<p><span class="math-container">$$f[\bigcap V_I] = \emptyset \iff \bigcap_{i \in i} f[V_i] = \emptyset.$$</span></p>
<h4>Proof</h4>
<p>The direction <span class="math-container">$\impliedby$</span> holds by set theory without any restrictions. We will show <span class="math-container">$\implies$</span>.</p>
<p>Suppose <span class="math-container">$f[\bigcap V_I] = \emptyset$</span>. Then <span class="math-container">$\bigcap V_I = \emptyset$</span>, and so <span class="math-container">$\bigcup_{i \in I} (V_0 \setminus V_i) = V_0$</span>. By assumption, <span class="math-container">$V_i$</span> is closed. By subspace topology, <span class="math-container">$V_0 \setminus V_i$</span> is open in <span class="math-container">$V_0$</span>. Since <span class="math-container">$V_0$</span> is compact, there exists finite <span class="math-container">$I' \subset I$</span> such that <span class="math-container">$\bigcup_{i \in I'} (V_0 \setminus V_i) = V_0$</span>; i.e. <span class="math-container">$\bigcap V_{I'} = \emptyset$</span>. Since <span class="math-container">$V$</span> is decreasing, <span class="math-container">$V_{\max(I')} = \bigcap V_{I'} = \emptyset$</span>. Therefore <span class="math-container">$\bigcap_{i \in I} f[V_i] = \emptyset$</span>.</p>
<h4>Comments</h4>
<p>The theorem does not generalize to locally compact fibers: a counter-example is given by <span class="math-container">$I = \mathbb{N}$</span>, <span class="math-container">$V_i = \mathbb{R}^{\geq i}$</span>, <span class="math-container">$f : \mathbb{R} \to \mathbb{R}$</span>, and <span class="math-container">$f(x) = 0$</span>. Then <span class="math-container">$f(\bigcap V_I) = \emptyset$</span> and <span class="math-container">$\bigcap_{i \in I} f(V_i) = \{0\}$</span>.</p>
|
2,631,284 | <p>I'm trying to find all $n \in \mathbb{N}$ such that</p>
<p>$(n+2) \mid (n^2+5)$ </p>
<p>as the title says, I've tried numbers up to $20$ and found that $1, 7$ are solutions and I suspect that those are the only $2$ solutions, however I have no idea how to show that.</p>
<p>I've done nothing but basic transformations:</p>
<p>$(n+2) \mid (n^2+5)$ </p>
<p>$\iff n^2+5 = k(n+2)$</p>
<p>$\iff n^2+5 \mod(n+2) = 0$</p>
<p>$\iff (n^2 \mod(n+2) + 5 \mod(n+2)) \mod(n+2) = 0$</p>
<p>Now I suspect the next step is to find all possible solutions for</p>
<p>$n^2 \mod(n+2)$, which I have no idea how to do.</p>
| PNT | 873,280 | <p><span class="math-container">$(n+2) \mid (n^2+5)\implies \exists m \in \mathbb N$</span> such that :
<span class="math-container">$$\frac{n^2+5}{n+2}= m \iff n^2-mn+(5-2m)=0$$</span></p>
<p>By Solving the quadratic equation in <span class="math-container">$n$</span>, We get:
<span class="math-container">$$n = \frac{m\pm \sqrt{m^2-20+12m}}{2}$$</span>
Now since <span class="math-container">$n\in \mathbb N \implies m^2-20+12m$</span>, Must be a perfect square:
<span class="math-container">$$m^2+12m-20 = x^2 \iff (m+6)^2-x^2=56$$</span>
<span class="math-container">$$(m+6-x)(m+6+x)=56$$</span>
You can look at the Factors of <span class="math-container">$56$</span> and try to solve the systems of equations, but you will end up with :
<span class="math-container">$$(m,x) \in \{(9,13), (3,5)\}$$</span>
By Plugging the values of <span class="math-container">$m$</span> to the quadratic equation you will get :
<span class="math-container">$$n \in \{1,7\}$$</span></p>
<p>I think that solving the quadratic equation is extremely useful in questions about perfect squares or divisibility.</p>
|
104,626 | <p>I encountered the following differential equation when I tried to derive the equation of motion of a simple pendulum:</p>
<p>$\frac{\mathrm d^2 \theta}{\mathrm dt^2}+g\sin\theta=0$</p>
<p>How can I solve the above equation?</p>
| yoyo | 6,925 | <p>replacing $\sin\theta$ by $\theta$ (physically assuming small angle deflection) gives you a homogeneous second order linear differential equation with constant coefficients, whose general solution can be found in most introductory diff eq texts (or a google search). this new equation represents a simple harmonic oscillator (acceleration proportional to displacement, like a spring force).
$$
\theta''+g\theta=0
$$
has solutions $A\cos(\sqrt{g}t)+B\sin(\sqrt{g}t)$.
so, for example, if the initial displacement is $\theta_0$ and initial angular velocity is $0$ then the solution is
$$
\theta_0\cos(\sqrt{g}t)
$$</p>
|
2,403,404 | <p>I would be thankful if anyone can answer my question. This is a very basic question. Let's say we wish to minimise the quantity</p>
<p>$$\hat{h}= \|h-h_i\|+\lambda\|h-u\|,$$</p>
<p>where:</p>
<p>$$h=[13,17,20, 17, 20, 14, 17, 18, 16, 15, 15, 12, 19, 13, 17, 13]^\top,\\ h_i=[18, 17, 14, 13, 17, 15, 17, 19, 12, 20, 15, 13, 16, 17, 20, 13]^\top, \\u = [16, 16,16,16,16,16,16,16,16,16,16,16,16,16,16,16]^\top,\\
\text{with }\lambda \in [0,100].$$</p>
<p>I know this is a very basic question, but please help me to understand. Also, please suggest me any book where I can start from zero to learn to solve these kinds of problems.</p>
| Vasili | 469,083 | <p>Let's consider this triangle on the coordinate plane. Let A has coordinates (0,0) and B has coordinates (6,0). Point C will belong either to line y=4 or y=-4 but it's irrelevant, both cases will produce the same result. Let x be the abscissa of point C. We can define CA+CB as the following function $f(x)=\sqrt{x^2+4}+\sqrt{(6-x)^2+4}$. Now all we need to do is to find the minimum of this function when $x$ is between 0 and 6. $\frac{df}{dx}=\frac{x}{\sqrt{x^2+4}} - \frac{x-6}{\sqrt{(6-x)^2+4}}$. Solving $\frac{df}{dx}=0$ we find that $x$=3. Now all we need to do is to check the values of $f(0), f(3), f(6)$ to prove that the minimal value is $f(3)$. </p>
|
4,048,785 | <p>Show that <span class="math-container">$2r^2-3$</span> is never a square, <span class="math-container">$r=2,3,...$</span></p>
<p>I know that no perfect square can have <span class="math-container">$2, 3, 7$</span>, or <span class="math-container">$8$</span> as its last digit. I'm not sure how to do this with congruence/mod notation. Any hints or solutions are greatly appreciated. I also tried to assume it was equal to a perfect square and through some algebraic manipulations arrive at a contradiction, but that proved to be difficult. I'm curious to know if it can be done algebraically.</p>
| abiessu | 86,846 | <p><span class="math-container">$2r^2-3$</span> is odd, so let <span class="math-container">$(2y+1)^2=4y^2+4y+1=2r^2-3$</span>, then <span class="math-container">$4y^2+4y=2r^2-4$</span> so <span class="math-container">$2\mid r$</span> giving <span class="math-container">$r=2s\to y^2+y=2s^2-1$</span>. But <span class="math-container">$y^2+y$</span> is always even and <span class="math-container">$2s^2-1$</span> is always odd, so there is no solution in integers <span class="math-container">$y,r$</span>.</p>
|
3,893,908 | <p>I want to compute <span class="math-container">$$\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx$$</span></p>
<p>My approach is <span class="math-container">$$\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx=\int_{-\infty}^{\infty} \frac{1}{(x -\pi)^2}dx+\int_{-\infty}^{\infty} \frac{\cos(x)}{(x -\pi)^2}dx.$$</span></p>
<p>However, the first term on the RHS does not exist since there is a singularity <span class="math-container">$\pi$</span>.</p>
<p>How to overcome this problem?</p>
<p>Now follow the hint, I get <span class="math-container">$$\int_{-\infty}^{\infty} \frac{\sin x}{x}dx$$</span></p>
<p>BTW, if I want to use <strong>residue calculus</strong> to do this problem, how to do it? I am confused the pole at origin. Thanks!</p>
| Franklin Pezzuti Dyer | 438,055 | <p><strong>HINT:</strong> With the substitution <span class="math-container">$x\to x+\pi$</span>, we get</p>
<p><span class="math-container">$$\int_{-\infty}^\infty \frac{1-\cos x}{x^2}dx$$</span></p>
<p>then integrate by parts with <span class="math-container">$u=1-\cos x$</span> and <span class="math-container">$dv=dx/x^2$</span>. You should end up with a familiar integral...</p>
|
3,893,908 | <p>I want to compute <span class="math-container">$$\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx$$</span></p>
<p>My approach is <span class="math-container">$$\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx=\int_{-\infty}^{\infty} \frac{1}{(x -\pi)^2}dx+\int_{-\infty}^{\infty} \frac{\cos(x)}{(x -\pi)^2}dx.$$</span></p>
<p>However, the first term on the RHS does not exist since there is a singularity <span class="math-container">$\pi$</span>.</p>
<p>How to overcome this problem?</p>
<p>Now follow the hint, I get <span class="math-container">$$\int_{-\infty}^{\infty} \frac{\sin x}{x}dx$$</span></p>
<p>BTW, if I want to use <strong>residue calculus</strong> to do this problem, how to do it? I am confused the pole at origin. Thanks!</p>
| Henry Lee | 541,220 | <p><span class="math-container">$u=x-\pi,du=dx$</span>
<span class="math-container">$$\int_{-\infty}^\infty\frac{1+\cos(x)}{(x-\pi)^2}dx=\int_{-\infty}^\infty\frac{1+\cos(u+\pi)}{u^2}du=\int_{-\infty}^\infty\frac{1-\cos(u)}{u^2}du$$</span>
and now if we look at this for <span class="math-container">$u\to\infty$</span>:
<span class="math-container">$$\frac{1-\cos(u)}{u^2}\approx\frac 1{u^2}$$</span>
and we can also calculate that:
<span class="math-container">$$\lim_{u\to 0}\frac{1-\cos(u)}{u^2}=\frac12$$</span>
so we can determine that the function is finite for the domain its in and is convergent due to the comparison test.</p>
<p>You can also use the fact that:
<span class="math-container">$$1-\cos(x)=2\sin^2\left(\frac x2\right)$$</span></p>
|
1,638,051 | <p>$$\int\frac{dx}{(x^{2}-36)^{3/2}}$$</p>
<p>My attempt:</p>
<p>the factor in the denominator implies</p>
<p>$$x^{2}-36=x^{2}-6^{2}$$</p>
<p>substituting $x=6\sec\theta$, noting that $dx=6\tan\theta \sec\theta$ </p>
<p>$$x^{2}-6^{2}=6^{2}\sec^{2}\theta-6^{2}=6^{2}\tan^{2}\theta$$</p>
<p>$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\int\frac{6\tan\theta \sec\theta}{36\tan^{2}\theta}=\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}$$</p>
<p>using trig identities:
$$\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}=\frac{1}{6}\int \sin^{-1}\theta$$</p>
<p>now using integration by parts:
$$\frac{1}{6}\int \sin^{-1}\theta$$
$$u=\sin^{-1}\theta, du=\frac{1}{\sqrt{1-\theta^{2}}}, dv=1, v=\theta$$
using $uv-\int{vdu}$</p>
<p>$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\int{\frac{\theta}{\sqrt{1-\theta^{2}}}}d\theta\bigg)$$</p>
<p>now using simple substitution:$$z=1-\theta^{2}, dz=-2\theta d\theta, -\frac{1}{2}du=\theta d\theta$$</p>
<p>it is apparent that</p>
<p>$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}\int{\frac{dz}{\sqrt{z}}}\bigg)\bigg)$$</p>
<p>$$=\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}(2\sqrt{z})\bigg)\bigg)=\frac{1}{6}\bigg(\theta \sin^{-1}\theta+\sqrt{1+\theta^{2}}\bigg)$$</p>
<p>$$=\frac{1}{6}\theta \sin^{-1}\theta+\frac{1}{6}\sqrt{1+\theta^{2}}+C$$</p>
<p>I have the following questions:</p>
<p>1.This integral seems tricky and drawn out to me, is there another method that reduces the steps/ methods of integration? I had to use trig substitution, integration by parts, and substitution in order to solve the integral, what can I do to find easier ways to complete integrals of this type?</p>
<p>2.Is this solution even correct? wolfram alpha says the solution to this integral is $-\frac{x}{36\sqrt{x^{2}-36}}+C$ how can i determine equivalence?</p>
| helpmeh | 200,346 | <p>attempt 2:</p>
<p>$$\int\frac{dx}{(x^{2}-36)^{3/2}}$$</p>
<p>the factor in the denominator implies</p>
<p>$$(x^{2}-36)^{3/2}=({x^{2}-6^{2}})^{3/2}$$</p>
<p>substituting $x=6sec\theta$, noting that $dx=6tan\theta sec\theta d\theta$ </p>
<p>$$(x^{2}-6^{2})^{3/2}=(6^{2}sec^{2}\theta-6^{2})^{3/2}d\theta=(36tan^{2}\theta)^{3/2}=216tan^3\theta$$</p>
<p>$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\int\frac{6tan\theta sec\theta}{216tan^{3}\theta}d\theta=\frac{1}{36}\int\frac{sec\theta}{tan^{2}\theta}d\theta$$</p>
<p>using trig identities:
$$\frac{1}{36}\int\frac{sec\theta}{tan^{2}\theta}d\theta=\frac{1}{36}\int{cot(\theta)csc(\theta) d\theta}$$</p>
<p>the integral cot(x)csc(x) is know</p>
<p>$$\int cot(x)sec(x)dx=-csc(x)+C$$</p>
<p>so </p>
<p>$$\frac{1}{36}\int cot(\theta)csc(\theta)d\theta=\frac{1}{36}\bigg(-csc(\theta)\bigg)+C=-\frac{1}{36}csc(\theta)+C$$</p>
<p>Question 2 still applies:</p>
<p>2.Is this solution even correct? wolfram alpha says the solution to this integral is $-\frac{x}{36\sqrt{x^{2}-36}}+C$ how can i determine equivalence?</p>
|
23,994 | <p>For which values of m does the equation:
$$3 \ln x+m x^3 = 17$$
have $1$ solution? $2$ solutions? $0$ solution?</p>
<p>Thanks.</p>
| I. J. Kennedy | 130 | <p>Can't improve on Jonas Meyer's answer, but playing with <a href="http://omnium-gatherum.appspot.com/pages/so23994.html" rel="nofollow">this</a> geogebra applet might help you get a feel for the equation. Move the slider to change $m$ in increments of 1/1000000.</p>
|
2,683,326 | <p>I have a function $f(x)$ whose second order Taylor expansion is represented by $f_2(x)$. Is it true that $$f(x)>f_2(x)$$ for all $x$? Any help in this regard will be much appreciated. Thanks in advance.</p>
| BruceET | 221,800 | <p>For the case where the breaks are five independent observations from
$\mathsf{Unif}(0,1),$ an argument similar to @GrahamKemp's can be used
to show that the longest piece is longer than $1/2$ with probability $6/32 = 3/16 = 0.1875.$ </p>
<p>If the intervals are arranged in a circle, the probability that the clockwise
180-degree gap following any one initial point is empty is $1/32,$ so the
total probability of a piece greater than .5 in length is $6/32.$</p>
<hr>
<p>In the following simulation, the vector <code>x</code> contains points $0, 1,$ and the
five breakpoints, sorted in order. Taking the maximum difference, we get
the length <code>big</code> of the longest piece. The result <code>0.187218</code> of <code>mwan(big > .5)</code>
is consistent with $6/32$ within simulation error. </p>
<pre><code>set.seed(318) # retain this statement for exact same simulation; delete for frexh run
m = 10^6; cuts=5; big = numeric(m)
for(i in 1:m) {
x = c(0, sort(runif(cuts)), 1)
big[i] = max(diff(x)) }
mean(big); sd(big); mean(big > .5)
## 0.4082853
## 0.1081648
## 0.187218
</code></pre>
<p>A histogram of the lengths of the one million such longest pieces simulated
is shown below:</p>
<p><a href="https://i.stack.imgur.com/1wEGN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1wEGN.png" alt="enter image description here"></a></p>
<p><em>Note:</em> By contrast, if the bathtub-shaped distribution $\mathsf{BETA}(.5,.5)$ is used
to make the breaks, then there will be relatively few cuts near the middle of $(0,1),$
and the probability that the length of the biggest piece exceeds $1/2$ increases
to about 0.41. (This illustrates @hardmath's Comment.)</p>
<p><a href="https://i.stack.imgur.com/U7phu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U7phu.png" alt="enter image description here"></a></p>
|
2,837,683 | <p>I have to solve the integral $$\int_D \sqrt{x^2+y^2} dx dy$$ where $D=\{(x,y)\in\mathbb{R^2}: x^{2/3}+y^{2/3}\le1\}$.</p>
<p>I am not able to find a parameterization that suits the integrand.
I tried with $$\cases{x=(r\cos t)^3\\y=(r\sin t)^3}$$ in order to reduce the domain to a circle but then the integral becomes $$\int_0^1\int_0^{2\pi} r^3\sqrt{(\cos t)^6+(\sin t)^6}\cdot9r^5\cos^2 t\sin^2 t dt dr$$ and then I cannot proceed to solve the integral.</p>
<p>Should I change the parametrization? </p>
| Jack D'Aurizio | 44,121 | <p>We may exhibit a parametrization of the integration domain through $x=\rho\cos^3\theta, y=\rho\sin^3\theta$ with $\rho\in[0,1]$ and $\theta\in[0,2\pi)$. By considering the Jacobian of this transformation and symmetry, we get that the original integral equals</p>
<p>$$ 4 \int_{0}^{1}\int_{0}^{\pi/2}3\rho^2\sin^2\theta\cos^2\theta\sqrt{\sin^6\theta+\cos^6\theta}\,d\theta\,d\rho $$
or
$$ 4\int_{0}^{1}x^2\sqrt{1-x^2}\sqrt{x^6+(1-x^2)^3}\,dx\stackrel{x\mapsto\sqrt{x}}{=}2\int_{0}^{1}\sqrt{x(1-x)} \sqrt{1-3x(1-x)}\,dx $$
and by setting $x=\frac{1+t}{2}$ the last integral is converted into
$$ \frac{1}{2}\int_{-1}^{1}\sqrt{1-t^2}\sqrt{1-\frac{3}{4}(1-t^2)}\,dt=\color{blue}{\frac{1}{2}\int_{0}^{1}\sqrt{1-t^2}\sqrt{1+3t^2}\,dt} $$
which can be represented in terms of complete elliptic integrals of the first and <a href="http://mathworld.wolfram.com/EllipticIntegraloftheSecondKind.html" rel="nofollow noreferrer">second kind</a>:
$$\boxed{ \iint_{D}\sqrt{x^2+y^2}\,dx\,dy = \tfrac{2}{9}\, E\!\left(\tfrac{3}{4}\right)+\tfrac{1}{9}\,K\!\left(\tfrac{3}{4}\right)}$$
<sup>Here I am using Mathematica's convention: the argument of $E$ or $K$ is the elliptic modulus and not the elliptic parameter.</sup></p>
<p>Numerically, this is approximately $0.5087364114$. An explicit series representation is
$$ I=\frac{\pi}{18}\sum_{n\geq 0}\frac{3^n}{64^n}\binom{2n}{n}^2\frac{2n-3}{2n-1}.$$
The inequality $I\leq \frac{1}{\sqrt{3}}$ can be simply derived by applying the Cauchy-Schwarz inequality to the blue integral above.</p>
|
2,828,472 | <p>This question is regarding property of little o notation given in Apostol Calculus. The property is given on page 288 and stated as:</p>
<blockquote>
<p>Theorem 7.8 (c) As $x\to a$ we have $f(x)\cdot o (g(x)) = o(f(x)g(x))$.</p>
</blockquote>
<p>Here say $h(x) = o(g(x))$ then we have $f(x) \lim_{x\to a} \frac{h(x)}{g(x)} = 0$, and on the right, we have $j(x) = o(f(x)g(x)) $ then $\lim_{x\to a} \frac{j(x)}{f(x)g(x)} = 0$ .. I am confused to approach the proof. </p>
| epi163sqrt | 132,007 | <p>This answer is based upon Apostol's settings. First of all we should be aware that the equality sign $=$ here is used to indicate a subset relation $\subseteq$ between two sets
\begin{align*}
f(x)o(g(x))=o(f(x)g(x)\qquad\qquad \text{as }x\to a
\end{align*}
meaning that whenever a function $h=h(x)$ is an element of the left-hand side $f(x)o(g(x))$ it is also an element of the right-hand side $o(f(x)g(x))$ as $x\to a$.</p>
<blockquote>
<p>We obtain
\begin{align*}
\color{blue}{h(x)}&\color{blue}{=f(x)o(g(x))\qquad\text{as }x\to a}\\
&\quad:\Longleftrightarrow\ \exists h_1: h(x)=f(x)h_1(x)\quad\text{ and }\quad\lim_{x\to a}\frac{h_1(x)}{g(x)}=0\\
&\quad\ \ \Longrightarrow\ \exists h_1: h(x)=f(x)h_1(x)\quad\text{ and }\quad\lim_{x\to a}\frac{f(x)h_1(x)}{f(x)g(x)}=0\\
&\quad\ \ \Longrightarrow\ \ \color{blue}{h(x)=o(f(x)g(x))\qquad \text{as }x\to a}
\end{align*}
and the claim follows.</p>
</blockquote>
|
2,952,392 | <p>Revisit the following discussion: </p>
<p><a href="https://math.stackexchange.com/questions/843909/prove-that-the-inverse-image-of-an-open-set-is-open">Prove that the inverse image of an open set is open</a></p>
<p>Obviously, the above discussion is based on Euclidean space (which is also a metric space, so the proof is based on the open ball). Can we say the following:</p>
<p>Let <span class="math-container">$X,Y$</span> be any two topological spaces, </p>
<p><span class="math-container">$f: X \rightarrow Y$</span> be a continuous function. The inverse image of an open set is open under <span class="math-container">$f$</span>. </p>
<p>Can this famous theorem apply to any topological space? for example, Zariski space? </p>
| Robert Lewis | 67,071 | <p>Look at</p>
<p><span class="math-container">$p(x, y) = x^2 + y^2 - 1 \in \Bbb R[x, y]; \tag 1$</span></p>
<p>this polynomial has an uncountable infinity of zeroes</p>
<p><span class="math-container">$(x, y) = (\cos \theta, \sin \theta) \in \Bbb R^2, \; \theta \in [0, 2\pi). \tag 2$</span></p>
<p>The problem is that, though for each value of <span class="math-container">$x \in [-1, 1]$</span> there are <em>precisely</em> two values of <span class="math-container">$y$</span> such that <span class="math-container">$p(x, y) = 0$</span>, there are ucountable such <span class="math-container">$x$</span>, so we are really looking at an uncountable collection of real polynomials <span class="math-container">$p(x, y) \in \Bbb R[x][y]$</span>.</p>
<p>Similar situations occur in higher dimensions.</p>
|
2,952,392 | <p>Revisit the following discussion: </p>
<p><a href="https://math.stackexchange.com/questions/843909/prove-that-the-inverse-image-of-an-open-set-is-open">Prove that the inverse image of an open set is open</a></p>
<p>Obviously, the above discussion is based on Euclidean space (which is also a metric space, so the proof is based on the open ball). Can we say the following:</p>
<p>Let <span class="math-container">$X,Y$</span> be any two topological spaces, </p>
<p><span class="math-container">$f: X \rightarrow Y$</span> be a continuous function. The inverse image of an open set is open under <span class="math-container">$f$</span>. </p>
<p>Can this famous theorem apply to any topological space? for example, Zariski space? </p>
| Wuestenfux | 417,848 | <p>There is a characterization of zero-dimensional ideals in the polynomial ring <span class="math-container">$R={\Bbb K}[x_1,\ldots,x_n]$</span> using Gröbner bases. An ideal <span class="math-container">$I$</span> is zero-dimensional if its zero locus <span class="math-container">$V_{\Bbb K}(I)$</span> is finite.
Equivalently, let <span class="math-container">$G=\{g_1,\ldots,g_m\}$</span> be a reduced Gröbner basis for <span class="math-container">$I$</span> w.r.t. the lex order <span class="math-container">$>$</span> with <span class="math-container">$x_1>\ldots>x_n$</span> such that <span class="math-container">${\rm LT}_>(g_i)>{\rm LT}_>(g_{i+1})$</span> for all <span class="math-container">$i$</span>. Then for each <span class="math-container">$1\leq i\leq n$</span>, there exists <span class="math-container">$j=j_i$</span> such that <span class="math-container">${\rm LT}_>(g_{j_i})=x_i^{e_i}$</span> for some <span class="math-container">$e_i\geq 1$</span>; i.e., the initial term is a pure power of <span class="math-container">$x_i$</span>. This can be directly decided by inspection!</p>
|
10,601 | <p>It sometimes happens that the same user posts <strong>exactly</strong> the same question twice in a row.</p>
<p>Examples: </p>
<ul>
<li><a href="https://math.stackexchange.com/questions/446622/drawing-at-least-90-of-colors-from-urn-with-large-populations">1</a> <a href="https://math.stackexchange.com/questions/446631/number-of-draws-required-for-ensuring-90-of-different-colors-in-the-urn-with-la">2</a>. user asked twice for the probability of choosing 90% of the colors from a collection of $10^{10}$ balls of $10^7$ different colors</li>
<li><a href="https://math.stackexchange.com/questions/463610/distance-between-two-polyhedra">1</a> <a href="https://math.stackexchange.com/questions/463536/distance-between-two-disjoint-polyhedrons">2</a>. user asked twice for a proof that disjoint polyhedra must lie at positive distance from one another</li>
<li><a href="https://math.stackexchange.com/questions/328368/matrix-factorization-in-upper-and-lower-triangular-matrix">1</a> <a href="https://math.stackexchange.com/questions/328326/guassian-elimination-triangular-factorization">2</a>. user asked twice for the LU decomposition of the same 3×4 matrix</li>
<li><a href="https://math.stackexchange.com/questions/315841/aronszajns-criterion-for-euclidean-space-again">1</a> <a href="https://math.stackexchange.com/questions/315733/aronszajns-criterion-for-euclidean-space/">2</a>. user asked twice for clarification of lemma 3 from a certain paper of Arthan </li>
<li><a href="https://math.stackexchange.com/questions/451268/children-balls-homework">1</a> <a href="https://math.stackexchange.com/questions/449079/children-balls-choosability">2</a>, both deleted but visible to 10K users. User asked twice for proofs of the same claim about $2n$ children choosing from sets of $n$ colored balls</li>
<li><a href="https://math.stackexchange.com/questions/391489/sum-of-squared-cubed-combinations">1</a> <a href="https://math.stackexchange.com/questions/391640/sum-of-squared-cube-combinations">2</a>, second one deleted. user asked twice for a closed form for $\sum {n \choose k}^3$</li>
<li><a href="https://math.stackexchange.com/questions/233892/directed-graph-dijkstras-algorithm">1</a> <a href="https://math.stackexchange.com/questions/233770/directed-graph-max-flow-dijkstras-algorithm">2</a>, both deleted. user asked twice for a proof that a certain max-flow problem could be solved with Dijkstra's algorithm</li>
<li><a href="https://math.stackexchange.com/questions/209331/prove-that-every-problem-in-p-is-reducible">1</a> <a href="https://math.stackexchange.com/questions/208848/for-two-problems-a-and-b-if-a-is-in-p-then-a-is-reducible-to-b">2</a>. user asked twice for a proof that any problem $A$ in $\mathcal P$ is polytime-reducible to any other problem $B$</li>
</ul>
<p>It seems to me that the best way to handle these is to flag them for moderator attention, so that the moderators can immediately merge the questions or close or delete the second one. I can vote to close one as a duplicate of the other, but it sometimes takes along while to gather five votes to close, and in that time the following discussion, which should be happening in one place, is split between two. If it is, the moderators could merge the two questions and their answers, which the site members have no way to do, so I think the flag is required anyway.</p>
<p>Sometimes my flags have been accepted, other times rejected. It appears that some moderators see the matter the way I do, but others don't. I would like to hear other members' opinions on this, and if possible I would like a clear statement from the moderators about whether I should raise a flag in this situation.</p>
<p>Related: <a href="http://meta.math.stackexchange.com/questions/10359/closing-duplicate-questions-by-the-same-poster">Closing duplicate questions by the same poster</a>.</p>
| robjohn | 13,854 | <p>I think the first thing to do is to vote to close if you have the reputation to do so.</p>
<p>Flagging the moderators about a duplicate is okay if it is definitely a duplicate. If there is some question, then it would be better to allow the community to decide than have a moderator unilaterally decide for the community.</p>
|
10,601 | <p>It sometimes happens that the same user posts <strong>exactly</strong> the same question twice in a row.</p>
<p>Examples: </p>
<ul>
<li><a href="https://math.stackexchange.com/questions/446622/drawing-at-least-90-of-colors-from-urn-with-large-populations">1</a> <a href="https://math.stackexchange.com/questions/446631/number-of-draws-required-for-ensuring-90-of-different-colors-in-the-urn-with-la">2</a>. user asked twice for the probability of choosing 90% of the colors from a collection of $10^{10}$ balls of $10^7$ different colors</li>
<li><a href="https://math.stackexchange.com/questions/463610/distance-between-two-polyhedra">1</a> <a href="https://math.stackexchange.com/questions/463536/distance-between-two-disjoint-polyhedrons">2</a>. user asked twice for a proof that disjoint polyhedra must lie at positive distance from one another</li>
<li><a href="https://math.stackexchange.com/questions/328368/matrix-factorization-in-upper-and-lower-triangular-matrix">1</a> <a href="https://math.stackexchange.com/questions/328326/guassian-elimination-triangular-factorization">2</a>. user asked twice for the LU decomposition of the same 3×4 matrix</li>
<li><a href="https://math.stackexchange.com/questions/315841/aronszajns-criterion-for-euclidean-space-again">1</a> <a href="https://math.stackexchange.com/questions/315733/aronszajns-criterion-for-euclidean-space/">2</a>. user asked twice for clarification of lemma 3 from a certain paper of Arthan </li>
<li><a href="https://math.stackexchange.com/questions/451268/children-balls-homework">1</a> <a href="https://math.stackexchange.com/questions/449079/children-balls-choosability">2</a>, both deleted but visible to 10K users. User asked twice for proofs of the same claim about $2n$ children choosing from sets of $n$ colored balls</li>
<li><a href="https://math.stackexchange.com/questions/391489/sum-of-squared-cubed-combinations">1</a> <a href="https://math.stackexchange.com/questions/391640/sum-of-squared-cube-combinations">2</a>, second one deleted. user asked twice for a closed form for $\sum {n \choose k}^3$</li>
<li><a href="https://math.stackexchange.com/questions/233892/directed-graph-dijkstras-algorithm">1</a> <a href="https://math.stackexchange.com/questions/233770/directed-graph-max-flow-dijkstras-algorithm">2</a>, both deleted. user asked twice for a proof that a certain max-flow problem could be solved with Dijkstra's algorithm</li>
<li><a href="https://math.stackexchange.com/questions/209331/prove-that-every-problem-in-p-is-reducible">1</a> <a href="https://math.stackexchange.com/questions/208848/for-two-problems-a-and-b-if-a-is-in-p-then-a-is-reducible-to-b">2</a>. user asked twice for a proof that any problem $A$ in $\mathcal P$ is polytime-reducible to any other problem $B$</li>
</ul>
<p>It seems to me that the best way to handle these is to flag them for moderator attention, so that the moderators can immediately merge the questions or close or delete the second one. I can vote to close one as a duplicate of the other, but it sometimes takes along while to gather five votes to close, and in that time the following discussion, which should be happening in one place, is split between two. If it is, the moderators could merge the two questions and their answers, which the site members have no way to do, so I think the flag is required anyway.</p>
<p>Sometimes my flags have been accepted, other times rejected. It appears that some moderators see the matter the way I do, but others don't. I would like to hear other members' opinions on this, and if possible I would like a clear statement from the moderators about whether I should raise a flag in this situation.</p>
<p>Related: <a href="http://meta.math.stackexchange.com/questions/10359/closing-duplicate-questions-by-the-same-poster">Closing duplicate questions by the same poster</a>.</p>
| Willie Wong | 1,543 | <p>I look through the few cases where your flag was declined. In the ones that I saw, my instinct would have been to decline too: the OP generally have rephrased the question so that at first glance it is not entire obvious that the questions are exact duplicates. (Moving sentences around and such.) </p>
<p>It would help, I think, if you flagged using the free form field for those type of questions, and specify that (a) the questions are asked by the same user and (b) the two questions are the same up to permuting the sentences or that one question is the refinement/clarification of the other or something like that. </p>
<p>Basically, to add to <a href="http://meta.math.stackexchange.com/a/10602/1543">robjohn's answer</a>, moderators will close exactly duplicated posts; but sometimes it is not that easy to see that two posts are exact duplicates of each other. A little justification in your flag will help a lot. </p>
|
2,912,376 | <p>I understood why he chose the positive square root in the sin but why the tan is also positive ? Isn't the tan positive and negative in this interval ?
<a href="https://i.stack.imgur.com/2rBFs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2rBFs.png" alt="enter image description here"></a></p>
| Math Lover | 348,257 | <p>Note that when $-1 <x < 0$, $$\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x} < 0.$$</p>
|
340,575 | <p>I got my exam on Thursday, and just got a few questions left. Anyway I would aprreciate help a lot! Can anyone please help me to solve this task? You can see the picture below. The need is to finde the size of the two radius. I thought about working with cords, like the cord AC is the same size like another one. Still couldn´t really find something usefull. <img src="https://i.stack.imgur.com/nOGlt.png" alt="enter image description here"></p>
| Vincent Tjeng | 52,208 | <p>The key insight is that $\angle ACB=90°$.</p>
<p>To show this, we draw a line passing through $C$ that is tangent to both the circles $k_1,k_2$ at $C$. This is possible since the two circles are tangent to each other. Let this tangent intersect the line $AB$ at $D$.</p>
<p>Now, we have $DA=DC$ since $DA,DC$ are lines tangent to circle $k_1$ at $A,C$ respectively. Similarly, $DB=DC$. Combining these two equations, we have $DA=DC=DB$; therefore, $D$ is the center of the circle passing through points $A,B,C$, that is, the circumcircle of triangle $ABC$. Furthermore, $AB$ must be the diameter of this circle, and thus $\angle ACB=90°$ (by <a href="http://en.wikipedia.org/wiki/Thales%27_theorem" rel="nofollow">Thales' Theorem</a>).</p>
<p>Once we have shown that $\angle ACB=90°$, the rest of the problem can be solved by trigonometry. However, for a more elegant approach, you can consider the following.</p>
<p>Let the midpoint of $AC$ be $P_1$ and the midpoint of $CB$ be $P_2$. Then $P_1M_1C$ is similar to $CAB$, and $P_2M_2C$ is similar to $CBA$. (This can be shown by labeling one of the angles - say $\angle CAB=\alpha$ and then finding the size of the rest of the angles). Once you have these relations, you can compare the lengths via similar triangles to obtain the results</p>
<p>$$r_2=M_2C=P_2C\times \frac {BA}{CA}=\frac{1}{2}BC\times\frac {BA}{CA}=5.625$$</p>
<p>and similarly</p>
<p>$$r_1=M_1C=P_1C\times \frac {AB}{CB}=\frac{1}{2}AC\times\frac {AB}{CB}=10$$</p>
|
4,061,536 | <p>We know that in a finite group of order say <span class="math-container">$g$</span>, an element of the group will have order of element <span class="math-container">$m\leq g$</span>. However, is it necessarily true that at least one element in the group <span class="math-container">$\textbf{must}$</span> have order of element <span class="math-container">$g$</span>?</p>
| José Carlos Santos | 446,262 | <p>No. If <span class="math-container">$G$</span> has order <span class="math-container">$m$</span> and <span class="math-container">$H$</span> has order <span class="math-container">$n$</span>, then <span class="math-container">$G\times H$</span> has order <span class="math-container">$mn$</span>. But if <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are not relatively prime, then no element of <span class="math-container">$G\times H$</span> has order <span class="math-container">$mn$</span>.</p>
|
4,061,536 | <p>We know that in a finite group of order say <span class="math-container">$g$</span>, an element of the group will have order of element <span class="math-container">$m\leq g$</span>. However, is it necessarily true that at least one element in the group <span class="math-container">$\textbf{must}$</span> have order of element <span class="math-container">$g$</span>?</p>
| kabenyuk | 528,593 | <p>Here is the complete answer to your question.
A group of order <span class="math-container">$n$</span> has an element of order <span class="math-container">$n$</span> if and only if the group is cyclic. This statement is an easy exercise.</p>
|
1,116,022 | <p>I've always had this doubt.
It's perfectly reasonable to say that, for example, 9 is bigger than 2.</p>
<p>But does it ever make sense to compare a real number and a complex/imaginary one?</p>
<p>For example, could one say that $5+2i> 3$ because the real part of $5+2i
$ is bigger than the real part of $3$? Or is it just a senseless statement?</p>
<p>Can it be stated that, say, $20000i$ is bigger than $6$ or does the fact that one is imaginary and the other is natural make it impossible to compare their 'sizes'?</p>
<p>It would seem that the 'sizes' of numbers of any type (real, rational, integer, natural, irrational) can be compared, but once imaginary and complex numbers come into the picture, it becomes a bit counter-intuitive for me.</p>
<p>So, does it ever make sense to talk about a real number being 'more than' or 'less than' a complex/imaginary one?</p>
| JMP | 210,189 | <p>If $s>t$ then we have $s-t>0$. If $s$ and $t$ are complex, and $s-t=u$ (u<>0), then we need $u>0$. However as u lies on a circle with +ve radius $r, u$ is always greater than $0$. This means that $-u=t-s>0$, implying $t>s$, a contradiction, so there is no order over the complex numbers.</p>
|
1,116,022 | <p>I've always had this doubt.
It's perfectly reasonable to say that, for example, 9 is bigger than 2.</p>
<p>But does it ever make sense to compare a real number and a complex/imaginary one?</p>
<p>For example, could one say that $5+2i> 3$ because the real part of $5+2i
$ is bigger than the real part of $3$? Or is it just a senseless statement?</p>
<p>Can it be stated that, say, $20000i$ is bigger than $6$ or does the fact that one is imaginary and the other is natural make it impossible to compare their 'sizes'?</p>
<p>It would seem that the 'sizes' of numbers of any type (real, rational, integer, natural, irrational) can be compared, but once imaginary and complex numbers come into the picture, it becomes a bit counter-intuitive for me.</p>
<p>So, does it ever make sense to talk about a real number being 'more than' or 'less than' a complex/imaginary one?</p>
| Daniel W. Farlow | 191,378 | <p><strong>Observation:</strong> Many of the properties of the real number system $\mathbb{R}$ hold in the complex number system $\mathbb{C}$, but there are some rather interesting differences as well--one of them is the concept of <em>order</em>. The concept of order used in $\mathbb{R}$ does not carry over to $\mathbb{C}$. That is, we cannot compare two complex numbers $z_1=a_1+ib_1,b_1\neq0$, $z_2=a_2+ib_2,b_2\neq0$, by means of inequalities. Statements such as $z_1<z_2$ or $z_2\geq z_1$ have no meaning in $\mathbb{C}$ except in the special case when the two numbers $z_1$ and $z_2$ are real. Thus, if you see a statement such as $z_1=\alpha z_2, \alpha>0$, it is implicit from the use of the inequality $\alpha>0$ that the symbol $\alpha$ represents a real number. </p>
<p>A number system is said to be an <em>ordered system</em> provided it contains a subset $P$ with the following two properties:</p>
<ol>
<li>For any nonzero number $x$ in the system, either $x$ or $-x$ is (but not both) in $P$.</li>
<li>If $x$ and $y$ are numbers in $P$, then both $xy$ and $x+y$ are in $P$.</li>
</ol>
<p><strong>Question:</strong> In the real number system the set $P$ is the set of <em>positive</em> numbers. In the real number system we say $x$ is greater than $y$, written $x>y$, if and only if $x-y$ is in $P$. Can you see why the complex number system has no such subset $P$?</p>
<p><strong>Answer:</strong> By the conditions given for an ordered system, if $i\in P$, then $i\cdot i=-1\in P$. Thus, we have that $(-1)\cdot i=-i\in P$, which is a contradiction ($i$ and $-i$ cannot both be in $P$). If $-i\in P$, then $(-i)(-i)=-1\in P$. Thus, $(-1)(-i)=i\in P$, and this is also a contradiction. Consequently, no such subset $P$ exists.</p>
|
4,554,231 | <p>How do you evaluate this limit? I can't manage to do it, even after manipulating the limit expression in several different ways, and using L'Hôpital's rule.</p>
<p><span class="math-container">$$
\lim_{h\,\to\, 0^{+}}\,
\left(\frac{{\rm e}^{-1/h^{2}}\,}{h}\right)
$$</span></p>
| Robin | 602,386 | <p>We wish to evaluate</p>
<p><span class="math-container">$$L =\lim _{h \to 0^+} \left(\frac{e^{-{1/h^2}}}{h}\right).$$</span></p>
<p>Substitute <span class="math-container">$u = \frac{1}{h}$</span> to get</p>
<p><span class="math-container">$$L =\lim _{u \to \infty} (ue^{-u^2}).$$</span></p>
<p>We have <span class="math-container">$0 < ue^{-u^2} < ue^{-{u}}$</span> for all <span class="math-container">$u > 1$</span>. (*)</p>
<p>It is a standard result that <span class="math-container">$\lim _{u \to \infty} (ue^{-u}) = 0.$</span> (To prove this, consider the Taylor series of <span class="math-container">$e^{-u}$</span>.)</p>
<p>Thus by taking <span class="math-container">$u \to \infty$</span> in (*) we get <span class="math-container">$0 \leq L \leq 0$</span>.</p>
<p>Hence <span class="math-container">$L=0$</span>.</p>
|
2,651,394 | <p>I am attempting to create a function in Matlab which turns all matrix elements in a matrix to '0' if the element is not symmetrical. However, the element appears to not be reassigning.</p>
<pre><code>function [output_ting] = maker(a)
[i,j] = size(a);
if i ~= j
disp('improper input!')
else
end
c = 1;
b = a.';
while c < length(a) + 1
if a(c) == b(c)
c = c + 1;
continue
else
a(c) = 0;
c = c + 1;
end
end
disp(a)
end
</code></pre>
| lab bhattacharjee | 33,337 | <p>$$\ln\dfrac{1+\dfrac{\sin(x+h)-\sin x}{\sin x}}h$$</p>
<p>$$=\ln\dfrac{\left(1+\dfrac{\sin(x+h)-\sin x}{\sin x}\right)}{\dfrac{\sin(x+h)-\sin x}{\sin x}}\cdot\dfrac{\sin(x+h)-\sin x}{h\sin x}$$</p>
<p>Now $\lim_{u\to0}\dfrac{\ln(1+u)}u=1$</p>
<p>and $$\lim_{h\to0}\dfrac{\sin(x+h)-\sin x}h=\lim_{h\to0}\dfrac{\sin\dfrac h2\cos\left(x+\dfrac h2\right)}{\dfrac h2}=?$$</p>
|
3,005,208 | <p>I want to solve this polynomial analytically. I know the useful answer is between 0 and 1. Is there any way I can write the answer based on a, b, and c?
<span class="math-container">$$
6\cdot a \cdot x^4 + 2 \cdot b \cdot x^3-b \cdot c=0
$$</span>
Also, an approximate answer is acceptable, for example, an answer with 2% error.
I will appreciate if someone can help me on this subject.</p>
| Bayesian guy | 594,523 | <p>You could use Ferrari's method for solve in general.
<a href="https://proofwiki.org/wiki/Ferrari%27s_Method" rel="nofollow noreferrer">https://proofwiki.org/wiki/Ferrari%27s_Method</a> this is an easy algorithmic way to do it.</p>
|
86,067 | <p>So I am having an issue using <code>NDSolve</code> and plotting the function. So I have two different <code>NDSolve</code> calls in my plotting function. (They are technically the same, just have different names; but that can be changed back if at all possible because I want them to be the same.) But the second one is not working. </p>
<p>When I remove the plotting code from the <code>Manipulate</code> command, <code>Plot</code> works fine and outputs an answer. I just need the expanded form so that I can manipulate the variables (if there is a way around this, that would be great too!)</p>
<p>Here is what I have so far, any help would be appreciated as I have no idea why I am getting this error.</p>
<pre><code>Manipulate[
Plot[{
(Evaluate[
ReplaceAll[
Paorta[t],
NDSolve[
{Paorta'[t] ==
1/Caorta ((1/2*k*(1 + Cos[ω t]) + 10 - Paorta[t])/
Piecewise[{{Ro,
1/2*k*(1 + Cos[ω t]) + 10 - Paorta[t] > 0}},
x*Ro] - Paorta[t]/Rsystemic), Paorta[0] == 90},
{Paorta[t]},
{t, 0, 10}
]
]
]), (1/2*k*(1 + Cos[ω t]) +
10), (((1/2*k*(1 + Cos[ω t]) + 10) -
Evaluate[
ReplaceAll[Pao[t],
NDSolve[{Pao'[t] ==
1/Caorta ((1/2*k*(1 + Cos[ω t]) + 10 - Pao[t])/
Piecewise[{{Ro,
1/2*k*(1 + Cos[ω t]) + 10 - Pao[t] > 0}},
x*Ro] - Pao[t]/Rsystemic), Pao[0] == 90}, {Pao[t]}, {t,
0, 10}
]
]
])/Ro)},
{t, 0, 10},
ImageSize -> Large,
PlotRange -> Full,
PlotLegends -> {"Aortic Pressure", "Pressure in Left Ventricle",
"Flow"}
],
{{Caorta, 1/.48}, 1, 6},
{{Rsystemic, 3.1}, .1, 6},
{{x, 8000}, 1, 10000},
{{ω, 2 π}, π, 3 π},
{{k, 110}, 60, 200},
{{Ro, .01}, .007, .05}
]
</code></pre>
<p>Thanks in advance!</p>
| BlacKow | 8,114 | <p>Not a real answer. But you can try to put your plot in <code>Inset[]</code>, then add another <code>Inset[]</code> for x-axis and yet another <code>Inset[]</code> for y-axis
and then stitch all scales together…
Something like this (nothing is stitched)</p>
<pre><code>Graphics[{Transparent, Rectangle[],
Inset[ListPlot[{1, 2, 3, 4, 0}, Axes -> False, Joined -> True,
InterpolationOrder -> 0, Filling -> Bottom], {0, 0}, {0.1, -0.1},
1], Inset[
ListPlot[{}, AxesStyle -> Red, Axes -> {False, True}], {0,
0.05}, {0, 0}, 1],
Inset[ListPlot[{}, AxesStyle -> Blue, Axes -> {True, False}], {0.05,
0}, {0, 0}, 1]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/xGXyy.png" alt="enter image description here"></p>
<p>This <a href="https://mathematica.stackexchange.com/questions/83636/how-to-align-coordinate-systems-of-inset-and-enclosing-graphics">post</a> is potentially useful for aligning three <code>Inset</code>s</p>
|
246,817 | <p>We have the succession and its formula:
$$
1^2+4^2+\cdots+ (3k-2)^2 = \dfrac{k(6k^2-3k-1)}{2}
$$</p>
<p>Now we need to apply it for $k+1$:
$$
1^2+4^2+\cdots+ (3n-2)^2 +(3(k+1)-2)^2 = \\
\dfrac{k(6k^2-3k-1)}{2} + (3(k+1)-2)^2
$$</p>
<p>I know that the result must be $\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)$ but I wasn't able to find the elegant solution during the test. I can't even see from where to factor out $k+1$.</p>
<p>Care to give a hint?</p>
| Brian M. Scott | 12,042 | <p>First combine everything into a single fraction:</p>
<p>$$\begin{align*}
\frac{k(6k^2-3k-1)}{2} + \big(3(k+1)-2\big)^2&=\frac{k(6k^2-3k-1)}2+(3k+1)^2\\
&=\frac{k(6k^2-3k-1)+2(3k+1)^2}2\\
&=\frac{6k^3-3k^2-k+18k^2+12k+2}2\\
&=\frac{6k^3+15k^2+11k+2}2\;.\tag{1}
\end{align*}$$</p>
<p>At this point the easiest thing to do is to multiply out the desired expression,</p>
<p>$$\frac{(k+1)\left(6(k+1)^2-3(k+1)-1\right)}2\;,$$</p>
<p>and verify that it’s equal to $(1)$. If you want to keep working forward from $(1)$, however, you can. We know that $k+1$ should be a factor of the numerator, so we could simply try to divide it out, but we can easily confirm this: by inspection $k=-1$ is a zero of the numerator, so $k-(-1)=k+1$ is a factor. Ordinary polynomial long division or synthetic division quickly yields the factorization</p>
<p>$$6k^3+15k^2+11k+2=(k+1)(6k^2+9k+2)\;.$$</p>
<p>$6(k+1)^2=6k^2+12k+6$, so $6k^2+9k+2=6(k+1)^2-3k-4=6(k+1)^2-3(k+1)-1$, as desired.</p>
|
1,860,267 | <blockquote>
<p>Prove the convergence of</p>
<p><span class="math-container">$$\int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x$$</span></p>
</blockquote>
<p>First I thought the integral does not converge because</p>
<p><span class="math-container">$$\int\limits_1^{\infty} -\frac{1}{x} \,\mathrm{d}x \le \int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x$$</span></p>
<p>But in this case</p>
<p><span class="math-container">$$\int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x \le \int\limits_1^{\infty} \frac{1}{x^2} \, \mathrm{d}x$$</span></p>
<p>it converges concerning the majorant criterion. What's the right way?</p>
| robjohn | 13,854 | <p><span class="math-container">$$
\begin{align}
\left|\,\int_1^\infty\frac{\cos(x)}{x}\,\mathrm{d}x\,\right|
&\le\left|\,\int_1^\pi\frac{\cos(x)}{x}\,\mathrm{d}x\,\right|+\sum_{k=1}^\infty\left|\,\int_{(2k-1)\pi}^{(2k+1)\pi}\frac{\cos(x)}{x}\,\mathrm{d}x\,\right|\\
&=\left|\,\int_1^\pi\frac{\cos(x)}{x}\,\mathrm{d}x\,\right|+\sum_{k=1}^\infty\left|\,\int_{(2k-1)\pi}^{(2k+1)\pi}\cos(x)\left(\frac1{x}-\frac1{2k\pi}\right)\,\mathrm{d}x\,\right|\\
&\le\log(\pi)+\sum_{k=1}^\infty2\pi\left(\frac1{(2k-1)\pi}-\frac1{2k\pi}\right)\\[6pt]
&=\log(4\pi)
\end{align}
$$</span></p>
|
135,159 | <p>Slow morning.
Can someone help me figure it out? I have a feeling it is trivially easy and not worthy of a thread.
$$
3^{n+1} + 3^n = 4\cdot3^n
$$</p>
<p>Thanks.</p>
| Community | -1 | <p><strong>Answers this version of the question</strong></p>
<p>The previous version of the question claimed that for all <span class="math-container">$n \in \Bbb N$</span>, <span class="math-container">$$3^{n+1}+3n=4\cdot 3n \tag{1}$$</span></p>
<p>However <span class="math-container">$(1)$</span> is not true for all <span class="math-container">$n$</span> as noted in the next part of the answer.</p>
<blockquote>
<p>However what is true is: <span class="math-container">$$3^{n+1}+3^n=4\cdot3^n$$</span></p>
<p>To see this, note that <span class="math-container">$3^{n+1}=3^n \cdot 3$</span> and factor the <span class="math-container">$3^n$</span> out.</p>
</blockquote>
<p><strong>Answers the previous version of the question</strong></p>
<p>Your claim is simply not true. For <span class="math-container">$n=2$</span>, LHS equals <span class="math-container">$ 33$</span> while RHS evaluates to <span class="math-container">$24$</span>.</p>
|
4,469,583 | <p>How can I construct/define arbitrary semi-computable (but not computable) sets?</p>
<p>Recall that a set A is semi-computable if it is domain of a computable function f. Recall also that a set A is computable if and only if both A and the complement set (A<sup>c</sup>) are semi-computable.</p>
<p>In particular, I am looking for a semi-computable (but not computable) set A, such that A is a proper subset of E, where E is the set of even numbers. E = { 2x | x € N }</p>
| Wuestenfux | 417,848 | <p>Let <span class="math-container">$\phi_0,\phi_1,\phi_2,\ldots$</span> be an enumeration of all <span class="math-container">$n$</span>-ary partial recursive functions.</p>
<p>It is well-known that the set <span class="math-container">$K=\{x\mid x\in dom\,\phi_x\}$</span> is semi-computable. Moreover, the set <span class="math-container">$K\cap E$</span> is semi-computable which you can easily check.</p>
<p>Add: Suppose <span class="math-container">$K\cap E$</span> is decidable. Then the complement must be semi-decidable. Otherwise, <span class="math-container">$K$</span> would be decidable. Now change the encoding <span class="math-container">$\phi_n$</span> of the partial recursive function by exchanging even and odd.</p>
|
1,904,767 | <p>I'm trying to understand regularization in machine learning. one way of regularization is adding a l1 norm to the error function. This is said to produce sparsity. But I can't understand.</p>
<p>sparsity is defined as "only few out of all parameters are non-zero". But if you look at the l1 norm equation, it is the summation of parameters' absolute value. </p>
<p>Sure, a small l1 norm could mean fewer non-zero parameters. but it could also mean that many parameters are non-zero, only the values of them are close to zero.</p>
<p>So why adding l1 regularization will for sure guarantee the first case? not the second case?</p>
<p>to give a concrete example, we have 2 vectors, A and B.</p>
<p>A is [0.1, 0.1, 0.1], which is not sparse</p>
<p>B is [1000, 0, 0], which is sparse.</p>
<p>clearly l1 norm of A is smaller than that of B.</p>
<p>so why do we use l1 norm to ensure sparsity? they seem to be unrelated?</p>
| Robert Israel | 8,508 | <p>Perhaps this is discussing a situation where the possible parameter values are a discrete set. If every nonzero parameter has absolute value at least $\epsilon$, the number of nonzero parameters is at most $1/\epsilon$ times the $\ell^1$ norm of the parameter vector.</p>
|
70,728 | <p>I've started taking an <a href="http://www.ml-class.org/" rel="noreferrer">online machine learning class</a>, and the first learning algorithm that we are going to be using is a form of linear regression using gradient descent. I don't have much of a background in high level math, but here is what I understand so far.</p>
<p>Given <span class="math-container">$m$</span> number of items in our learning set, with <span class="math-container">$x$</span> and <span class="math-container">$y$</span> values, we must find the best fit line <span class="math-container">$h_\theta(x) = \theta_0+\theta_1x$</span> . The cost function for any guess of <span class="math-container">$\theta_0,\theta_1$</span> can be computed as:</p>
<p><span class="math-container">$$J(\theta_0,\theta_1) = \frac{1}{2m}\sum_{i=1}^m(h_\theta(x^{(i)}) - y^{(i)})^2$$</span></p>
<p>where <span class="math-container">$x^{(i)}$</span> and <span class="math-container">$y^{(i)}$</span> are the <span class="math-container">$x$</span> and <span class="math-container">$y$</span> values for the <span class="math-container">$i^{th}$</span> component in the learning set. If we substitute for <span class="math-container">$h_\theta(x)$</span>,</p>
<p><span class="math-container">$$J(\theta_0,\theta_1) = \frac{1}{2m}\sum_{i=1}^m(\theta_0 + \theta_1x^{(i)} - y^{(i)})^2$$</span></p>
<p>Then, the goal of gradient descent can be expressed as</p>
<p><span class="math-container">$$\min_{\theta_0, \theta_1}\;J(\theta_0, \theta_1)$$</span></p>
<p>Finally, each step in the gradient descent can be described as:</p>
<p><span class="math-container">$$\theta_j := \theta_j - \alpha\frac{\partial}{\partial\theta_j} J(\theta_0,\theta_1)$$</span></p>
<p>for <span class="math-container">$j = 0$</span> and <span class="math-container">$j = 1$</span> with <span class="math-container">$\alpha$</span> being a constant representing the rate of step. </p>
<p>I have no idea how to do the partial derivative. I have never taken calculus, but conceptually I understand what a derivative represents. The instructor gives us the partial derivatives for both <span class="math-container">$\theta_0$</span> and <span class="math-container">$\theta_1$</span> and says not to worry if we don't know how it was derived. (I suppose, technically, it is a computer class, not a mathematics class) However, I would very much like to understand this if possible. Could someone show how the partial derivative could be taken, or link to some resource that I could use to learn more? I apologize if I haven't used the correct terminology in my question; I'm very new to this subject.</p>
| Christian Sykes | 322,386 | <p>Despite the popularity of the top answer, it has some <strong>major</strong> errors. The most fundamental problem is that $g(f^{(i)}(\theta_0, \theta_1))$ isn't even <em>defined</em>, much less equal to the original function. The focus on the <a href="https://en.wikipedia.org/wiki/Chain_rule" rel="noreferrer">chain rule</a> as a crucial component is correct, but the actual derivation is not right at all.</p>
<p>So I'll give a correct derivation, followed by my own attempt to get across some intuition about what's going on with partial derivatives, and ending with a brief mention of a cleaner derivation using more sophisticated methods. That said, if you don't know some basic differential calculus already (at least through the chain rule), you realistically aren't going to be able to truly follow <em>any</em> derivation; go learn that first, from literally any calculus resource you can find, if you really want to know.</p>
<p>For completeness, the properties of the derivative that we need are that for any constant $c$ and functions $f(x)$ and $g(x)$,
$$\frac{d}{dx} c = 0, \ \frac{d}{dx} x = 1,$$
$$\frac{d}{dx} [c\cdot f(x)] = c\cdot\frac{df}{dx} \ \ \ \text{(linearity)},$$
$$\frac{d}{dx}[f(x)+g(x)] = \frac{df}{dx} + \frac{dg}{dx} \ \ \ \text{(linearity)},$$
$$\frac{d}{dx}[f(x)]^2 = 2f(x)\cdot\frac{df}{dx} \ \ \ \text{(chain rule)}.$$</p>
<p>Taking partial derivatives works essentially the same way, except that the notation $\frac{\partial}{\partial x}f(x,y)$ means we we take the derivative by treating $x$ as a variable and $y$ as a constant using the same rules listed above (and vice versa for $\frac{\partial}{\partial y}f(x,y)$).</p>
<hr>
<h2>Derivation</h2>
<p>We have</p>
<p>$$h_\theta(x_i) = \theta_0 + \theta_1 x_i$$</p>
<p>and</p>
<p>$$\begin{equation} J(\theta_0, \theta_1) = \frac{1}{2m} \sum_{i=1}^m (h_\theta(x_i)-y_i)^2\end{equation}.$$</p>
<p>We first compute</p>
<p>$$\frac{\partial}{\partial\theta_0}h_\theta(x_i)=\frac{\partial}{\partial\theta_0}(\theta_0 + \theta_1 x_i)=\frac{\partial}{\partial\theta_0}\theta_0 + \frac{\partial}{\partial\theta_0}\theta_1 x_i =1+0=1,$$</p>
<p>$$\frac{\partial}{\partial\theta_1}h_\theta(x_i) =\frac{\partial}{\partial\theta_1}(\theta_0 + \theta_1 x_i)=\frac{\partial}{\partial\theta_1}\theta_0 + \frac{\partial}{\partial\theta_1}\theta_1 x_i =0+x_i=x_i,$$</p>
<p>which we will use later. Now we want to compute the partial derivatives of $J(\theta_0, \theta_1)$. We can actually do both at once since, for $j = 0, 1,$</p>
<p>$$\frac{\partial}{\partial\theta_j} J(\theta_0, \theta_1) = \frac{\partial}{\partial\theta_j}\left[\frac{1}{2m} \sum_{i=1}^m (h_\theta(x_i)-y_i)^2\right]$$</p>
<p>$$= \frac{1}{2m} \sum_{i=1}^m \frac{\partial}{\partial\theta_j}(h_\theta(x_i)-y_i)^2 \ \text{(by linearity of the derivative)}$$</p>
<p>$$= \frac{1}{2m} \sum_{i=1}^m 2(h_\theta(x_i)-y_i)\frac{\partial}{\partial\theta_j}(h_\theta(x_i)-y_i) \ \text{(by the chain rule)}$$ </p>
<p>$$= \frac{1}{2m}\cdot 2\sum_{i=1}^m (h_\theta(x_i)-y_i)\left[\frac{\partial}{\partial\theta_j}h_\theta(x_i)-\frac{\partial}{\partial\theta_j}y_i\right]$$</p>
<p>$$= \frac{1}{m}\sum_{i=1}^m (h_\theta(x_i)-y_i)\left[\frac{\partial}{\partial\theta_j}h_\theta(x_i)-0\right]$$</p>
<p>$$=\frac{1}{m} \sum_{i=1}^m (h_\theta(x_i)-y_i)\frac{\partial}{\partial\theta_j}h_\theta(x_i).$$</p>
<p>Finally substituting for $\frac{\partial}{\partial\theta_j}h_\theta(x_i)$ gives us</p>
<p>$$\frac{\partial}{\partial\theta_0} J(\theta_0, \theta_1) = \frac{1}{m} \sum_{i=1}^m (h_\theta(x_i)-y_i),$$
$$\frac{\partial}{\partial\theta_1} J(\theta_0, \theta_1) = \frac{1}{m} \sum_{i=1}^m (h_\theta(x_i)-y_i)x_i.$$</p>
<hr>
<h2>Intuition for partial derivatives</h2>
<p>So what are partial derivatives anyway? In one variable, we can assign a single number to a function $f(x)$ to best describe the rate at which that function is changing at a given value of $x$; this is precisely the derivative $\frac{df}{dx}$of $f$ at that point. We would like to do something similar with functions of several variables, say $g(x,y)$, but we immediately run into a problem. In one variable, we can only change the independent variable in two directions, forward and backwards, and the change in $f$ is equal and opposite in these two cases. (For example, if $f$ is increasing at a rate of 2 per unit increase in $x$, then it's decreasing at a rate of 2 per unit decrease in $x$.)</p>
<p>With more variables we suddenly have infinitely many different directions in which we can move from a given point and we may have different rates of change depending on which direction we choose. So a single number will no longer capture how a multi-variable function is changing at a given point. However, there are certain specific directions that are easy (well, easier) and natural to work with: the ones that run parallel to the coordinate axes of our independent variables. These resulting rates of change are called partial derivatives. (For example, $g(x,y)$ has partial derivatives $\frac{\partial g}{\partial x}$ and $\frac{\partial g}{\partial y}$ from moving parallel to the x and y axes, respectively.) Even though there are infinitely many different directions one can go in, it turns out that these partial derivatives give us enough information to compute the rate of change for any other direction. (Strictly speaking, this is a slight white lie. There are functions where the all the partial derivatives exist at a point, but the function is not considered differentiable at that point. This happens when the graph is not sufficiently "smooth" there.)</p>
<p>In particular, the gradient $\nabla g = (\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y})$ specifies the direction in which g increases most rapidly at a given point and $-\nabla g = (-\frac{\partial g}{\partial x}, -\frac{\partial g}{\partial y})$ gives the direction in which g decreases most rapidly; this latter direction is the one we want for gradient descent. This makes sense for this context, because we want to decrease the cost and ideally as quickly as possible.</p>
<hr>
<h2>A higher level approach</h2>
<p>For the interested, there <em>is</em> a way to view $J$ as a simple composition, namely</p>
<p>$$J(\mathbf{\theta}) = \frac{1}{2m} \|\mathbf{h_\theta}(\mathbf{x})-\mathbf{y}\|^2 = \frac{1}{2m} \|X\mathbf{\theta}-\mathbf{y}\|^2.$$</p>
<p>Note that $\mathbf{\theta}$, $\mathbf{h_\theta}(\mathbf{x})$, $\mathbf{x}$, and $\mathbf{y}$, are now <em>vectors</em>. Using more advanced notions of the derivative (i.e. the total derivative or Jacobian), the multivariable chain rule, and a tiny bit of linear algebra, one can actually differentiate this directly to get</p>
<p>$$\frac{\partial J}{\partial\mathbf{\theta}} = \frac{1}{m}(X\mathbf{\theta}-\mathbf{y})^\top X.$$</p>
<p>The transpose of this is the gradient $\nabla_\theta J = \frac{1}{m}X^\top (X\mathbf{\theta}-\mathbf{y})$. Setting this gradient equal to $\mathbf{0}$ and solving for $\mathbf{\theta}$ is in fact exactly how one derives the explicit formula for linear regression.</p>
|
2,702,060 | <p>A little confusion on my part. Study of multi variable calculus and we are using the formula for length of a parameterized curve. The equation makes intuitive sense and I can work it OK. But I also recall using the same integral with out the parameterizing to find the length of a curve where the first term of the square root in just one. The former formula is the general case.</p>
<p>Now for the question: I had just previously used the integral for completing the quadrature i.e. Find the area under a curve. Is the single integral used for finding both area and length ? I guess I am trying to unify the concepts in my mind to understand the context of how they are used and know the difference. Thank you.</p>
| Arian | 172,588 | <p>For $n$ even i.e. $n=2m$ for some $m\in\mathbb{N}$. Let $S:=1+2+...+2m$ then</p>
<p>$$2S=S+S=(1+2+...+(2m-1)+2m)+(2m+(2m-1)+...+2+1)=((1+2m)+(2+(2m-1))+...+((2m-1)+2)+(2m+1))=\underbrace{((2m+1)+...+(2m+1))}_{2m-times}=(2m+1)\cdot 2m$$
Therefore $$S=\frac{(2m+1)2m}{2}=m(2m+1)$$
Analogue for $n$ odd. The formula is the same as in general for any $n$. You indeed just substitute directly into it. For instance
$$\frac{n(n+1)}{2}\Rightarrow \frac{2m(2m+1)}{2}=m(2m+1)$$</p>
|
2,702,060 | <p>A little confusion on my part. Study of multi variable calculus and we are using the formula for length of a parameterized curve. The equation makes intuitive sense and I can work it OK. But I also recall using the same integral with out the parameterizing to find the length of a curve where the first term of the square root in just one. The former formula is the general case.</p>
<p>Now for the question: I had just previously used the integral for completing the quadrature i.e. Find the area under a curve. Is the single integral used for finding both area and length ? I guess I am trying to unify the concepts in my mind to understand the context of how they are used and know the difference. Thank you.</p>
| user | 505,767 | <p>This classical result can be also easily proved by the following trick</p>
<p><a href="https://i.stack.imgur.com/3mlDm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3mlDm.png" alt="enter image description here"></a></p>
<p>An extended discussion also for more general cases here <a href="https://math.stackexchange.com/questions/2664218/how-are-the-solutions-for-finite-sums-of-natural-numbers-derived/2664224#2664224">How Are the Solutions for Finite Sums of Natural Numbers Derived?</a></p>
|
9,111 | <p>What function can I use to evaluate $(x+y)^2$ to $x^2 + 2xy + y^2$? </p>
<p>I want to evaluate It and I've tried to use the most obvious way: simply typing and evaluating $(x+y)^2$, But it gives me only $(x+y)^2$ as output. I've been searching for it in the last minutes but I still got no clue, can you help me?</p>
| image_doctor | 776 | <p>You might also try:</p>
<pre><code>Apart[(x + y)^2]
</code></pre>
<blockquote>
<p>x^2 + 2 x y + y^2</p>
</blockquote>
|
4,014,554 | <p>A simple heuristic of the first million primes shows that no prime number can be bigger than the sum of adding the previous twin primes.</p>
<p>Massive update:
@mathlove made a comment that leaves me completely embarrassed. <span class="math-container">$13 > 7 + 5$</span> I don’t know how I missed it and I deeply apologize to everyone!</p>
<p>I ask anyone qualified to suggest any edits for the question.</p>
<p><span class="math-container">$7 < 5 + 3$</span></p>
<p><span class="math-container">$11 < 7 + 5$</span></p>
<p><span class="math-container">$17 < 11 + 13$</span></p>
<p><span class="math-container">$23 < 17 + 19$</span></p>
<p>At larger numbers:</p>
<p><span class="math-container">$4886639 < 4886489 + 4886491$</span></p>
<p><span class="math-container">$5389451 < 5388869 + 5388871$</span></p>
<p><span class="math-container">$3155597 < 3154757 + 3154759$</span></p>
<p>I assume that if it could be proved, it would prove the twin prime conjecture of whether twin primes exist forever.</p>
<p>So I am not exactly seeking for a proof, but rather for possible explanations or references for why it is assumed true (or not)?</p>
<p>Also as the list grows, there seems to be a range for how small or big can a prime be in comparison to the sum of adding the previous twin primes.</p>
<p>As the list grows, a prime is usually never bigger or smaller than slightly above <span class="math-container">$50\%$</span> of the sum of the previous twin primes. Any references for such a range will be appreciated too.</p>
<p>*Update: When mentioning "the previous twin primes", I am implying to: <span class="math-container">$(107, 109), 113, 127, 131, (137, 139)$</span>.</p>
<p><span class="math-container">$131 < 107 +109$</span></p>
| Barry Cipra | 86,747 | <p>Let's modify the OP's observation as follows:</p>
<blockquote>
<p>Let <span class="math-container">$(p,p+2)$</span> and <span class="math-container">$(q,q+2)$</span> be <em>consecutive</em> pairs of twin primes,
e.g., <span class="math-container">$(107,109)$</span> and <span class="math-container">$(137,139)$</span>. Then (conjecturally) <span class="math-container">$q\lt2p+2$</span>.</p>
</blockquote>
<p>This is, essentially, a Bertrand's Postulate for twin primes, and it's not hard to confirm that it holds for entries at the outset for the sequence <a href="https://oeis.org/A001359" rel="nofollow noreferrer"><span class="math-container">$3,5,11,17,29,41,59,\ldots$</span></a>. The basic explanation can be found in the <a href="https://en.wikipedia.org/wiki/Twin_prime#First_Hardy%E2%80%93Littlewood_conjecture" rel="nofollow noreferrer">heuristic twin-prime "theorem"</a> <span class="math-container">$\pi_2(x)\sim2C_2x/(\ln x)^2$</span>, although arguing that it (conjecturally) holds for <em>all</em> twin primes, not just for ones that are sufficiently large -- i.e., giving a (heuristic) twin-prime analog of the <a href="https://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate" rel="nofollow noreferrer">proof of Bertrand's Postulate</a> -- seems problematic.</p>
<p>Remark: Amusingly, the modification proposed above of the OP's observation is technically agnostic with regard to the twin prime conjecture. Indeed, it would be easiest, in principle, to prove (or disprove) if there were an identifiable <em>last</em> pair of twin primes.</p>
|
644,057 | <p>I am having trouble with this problem:</p>
<p>Let $a_n$ be sequence of positive terms with $$\frac{a_{n+1}}{a_n}\lt \frac{n^2}{(n+1)^2}.$$
Then is the series $\sum a_n$ convergent?</p>
<p>Thanks for any help.</p>
| Yiorgos S. Smyrlis | 57,021 | <p>Note that
$$
a_n=a_1\prod_{k=1}^{n-1}\frac{a_{k+1}}{a_k}\le a_1\prod_{k=1}^{n-1}\frac{k^2}{(k+1)^2}=\frac{a_1}{n^2},
$$
and as the series $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}$ converges so does the series $\displaystyle\sum_{n=1}^\infty a_n$, as a consequence of the comparison test.</p>
|
2,336,988 | <blockquote>
<p>Let $a,b,c>0 ,2b+2c-a\ge 0,2c+2a-b\ge 0,2a+2b-c\ge 0$ show that
$$\sqrt{\dfrac{2b+2c}{a}-1}+\sqrt{\dfrac{2c+2a}{b}-1}+\sqrt{\dfrac{2a+2b}{c}-1}\ge 3\sqrt{3}$$</p>
</blockquote>
<p>I try use AM-GM and Cauchy-Schwarz inequality and from here I don't see what to do</p>
| Michael Rozenberg | 190,319 | <p>By AM-GM
$$\sum_{cyc}\sqrt{\frac{2b+2c}{a}-1}=\sum_{cyc}\frac{2\sqrt3(2b+2c-a)}{2\sqrt{3a(2b+2c-a)}}\geq$$
$$\geq\sum_{cyc}\frac{2\sqrt3(2b+2c-a)}{3a+2b+2c-a}=\sum_{cyc}\frac{2\sqrt3(2b+2c-a)}{2(a+b+c)}=3\sqrt3$$</p>
|
2,379,405 | <blockquote>
<p>Determine convergence or divergence of $$ \int_0^{\infty} \frac{1 + \cos^2x}{\sqrt{1+x^2}} dx$$</p>
</blockquote>
<p>As the graph of the function suggests convergence, Let's find an upper bound that converges.</p>
<p>$$ \int_0^{\infty} \frac{1 + \cos^2x}{\sqrt{1+x^2}} dx \leq
\int_0^{\infty} \frac{2}{\sqrt{1+x^2}} dx \leq
\int_0^{\infty} \frac{2}{x} dx$$</p>
<p>Given that $\int_0^{\infty} \frac{2}{x} dx $ diverges, i wont be able to show that the original converges.</p>
<p>What is my error? What upper bound could I choose?</p>
| user0102 | 322,814 | <p>Here it is an alternative approach:
\begin{align*}
\int_{0}^{\infty}\frac{1+\cos^{2}(x)}{\sqrt{1+x^{2}}}dx \geq \int_{0}^{\infty}\frac{dx}{\sqrt{1+x^{2}}} = \int_{0}^{\infty}\frac{\cosh(u)}{\sqrt{1+\sinh^{2}(u)}}du = \int_{0}^{\infty}du = +\infty
\end{align*}</p>
|
149,872 | <p>How would I show that $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$? </p>
<p>Im not sure how to begin, does it involve using $\sinh z=\frac{e^{z}-e^{-z}}{2}$ and $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$?</p>
| Javier | 2,757 | <p>If you use the sine addition formula, the pythagorean identities, and the fact that $\sin(ix)=i\sinh (x)$ and $\cos(ix) = \cosh(x)$, then you get this:</p>
<p>$$
\begin{align}
\sin(x+iy) &= \sin x \cos (iy)+\cos x \sin(iy) \\
&= \sin x \cosh y + i \cos x \sinh y
\end{align}
$$</p>
<p>$$
\begin{align}
|\sin(x+iy)|^2 &= (\sin x \cosh y)^2 + (\cos x \sinh y)^2
\end{align}
$$</p>
<p>Now you can get rid of the cosines knowing that $\cos^2 x + \sin^2 x = 1$ and that $\cosh^2 x - \sinh ^2 x = 1$. You can take it from there.</p>
<p>By the way, to get the sine addition formula and the sine and cosine of imaginary numbers, convert them to exponential form:</p>
<p>$$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$$
$$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$
$$\sinh x = \frac{e^x-e^{-x}}{2}$$
$$\cosh x = \frac{e^x+e^{-x}}{2}$$</p>
<p>Plug in what you want to find out; the derivation of the identities is straightforward.</p>
|
1,610,055 | <p>I feel rather silly for having to ask this question in specific and am by no means looking for a flat out step by step answer. I understand the definition for the euclidean norm in an n-dimensional space (as defined <a href="https://en.wikipedia.org/wiki/Norm_(mathematics)#Euclidean_norm" rel="nofollow">here</a>). I can't figure out how to apply it however to even a simple problem like this one:</p>
<blockquote>
<p>If $\| x - z \| \lt 2$ and $\| y - z \| \lt 3$, prove $\| x - y \| \lt 5$.</p>
</blockquote>
<p>Sorry in advance for lack of formatting, I'm new to math exchange and there's really nothing complicated to format. Again, I am not looking for a straight answer. My proof breaks down after I add the two assumptions and attempt to square both sides. Hopefully someone can point out the simple first step here. Thanks. </p>
| Nicolás Giossa | 303,222 | <p>First observe that $\ \|x-z\| < 2$ and $\ \|y-z\| < 3 \Rightarrow \ \|x-z\| + \ \|y-z\| < 5$. </p>
<p>So, using triangle inequality, we have $\ \|x-y\| = \ \|x-z+z-y\| \leq \ \|x-z\| + \ \|y-z\| < 5 $.</p>
|
2,436,167 | <p>I appear to be misunderstanding a basic probability concept. The question is: you flip four coins. At least two are tails. What is the probability that exactly three are tails? </p>
<p>I know the answer isn't 1/2, but I don't know why that's so. Isn't the probability of just getting 1 tail in the remaining two coins 1/2?</p>
<p>Thanks</p>
| RideTheWavelet | 394,393 | <p>I believe the source of your confusion is arising from the fact that the specific tosses on which the two tails took place are not specified. So if we compute the conditional probability which is being described, letting $A$ be the event that we get at least $2$ tails, and $B$ be the event that we get exactly $3$ tails, we get: $$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{P(B)}{P(A)}=\frac{4(1/2)^{4}}{(6+4+1)(1/2)^{4}}=\frac{4}{11}.$$ You can see quite clearly the importance of the number of ways these different sequences can happen: there are $6$ sequences of heads and tails that result in $2$ tails, but only $4$ sequences of heads and tails that result in $1$ tail, and this results in the probability being smaller than your original attempt.</p>
|
3,121,103 | <p>The integral surface of the first order partial differential equation <span class="math-container">$$2y(z-3)\frac{\partial z}{\partial x}+(2x-z)\frac{\partial z}{\partial y} = y(2x-3)$$</span> passing through the curve <span class="math-container">$x^2+y^2=2x, z = 0$</span> is</p>
<ol>
<li><span class="math-container">$x^2+y^2-z^2-2x+4z=0$</span> </li>
<li><span class="math-container">$x^2+y^2-z^2-2x+8z=0$</span> </li>
<li><span class="math-container">$x^2+y^2+z^2-2x+16z=0$</span> </li>
<li><span class="math-container">$x^2+y^2+z^2-2x+8z=0$</span></li>
</ol>
<p>My effort:</p>
<p>I find the mulipliers <span class="math-container">$x, 3y, -z$</span> and get the solution <span class="math-container">$x^2+3y^2-z^2=c_1$</span>. How to proceed further? Please help.</p>
| Aleksas Domarkas | 562,074 | <p>All surfaces passing through the curve <span class="math-container">$x^2+y^2=2x, z = 0$</span>. Bat only
first surface <span class="math-container">$$x^2+y^2-z^2-2x+4z=0\tag{1}$$</span> is solution of pde. We check it</p>
<p>Let <span class="math-container">$z=z(x,y)$</span>. We differentiate equation <span class="math-container">$(1)$</span> respect <span class="math-container">$x$</span> and <span class="math-container">$y$</span>:
<span class="math-container">$$2x-2zz'_x-2+4z'_x=0,\\
2y-2zz'_y+4z'_y=0.
$$</span>
Solving this system we get
<span class="math-container">$$z'_x=\frac{x-1}{z-2},\quad z'_y=\frac{y}{z-2}$$</span>
Next we substitute this in to pde. We get
<span class="math-container">$$\frac{2 \left( x-1\right) y\, \left( z-3\right) }{z-2}+\frac{y\, \left( 2 x-z\right) }{z-2}=\left( 2 x-3\right) \, y,$$</span>
<span class="math-container">$$(2x-3)y=(2x-3)y.$$</span>
Then <span class="math-container">$(1)$</span> is solution.</p>
|
2,524,487 | <p>I can not find a way to prove that the abelian group ($\mathbb{Q}_{>0}$,*) is a free abelian group with countable basis. Is is even true?</p>
| lisyarus | 135,314 | <p>By the fundamental theorem of arithmetic, any natural number can be expressed as $p_1^{k_1}\cdot \dots \cdot p_n^{k_n}$, where $p_i$ are primes and $k_i \in \mathbb{N}$.</p>
<p>Extending this to positive rationals, any positive rational can be expressed in the form $p_1^{k_1}\cdot \dots \cdot p_n^{k_n}$, where $p_i$ are primes and $k_i \in \mathbb{Z}$. This makes it into a free abelian group with $p_i$ as a basis.</p>
|
1,619,292 | <p>Let $\mathbf C$ be an abelian category containing arbitrary direct sums and let $\{X_i\}_{i\in I}$ be a collection of objects of $\mathbf C$. </p>
<p>Consider a subobject $Y\subseteq \bigoplus_{i\in I}X_i$ and put $Y_i:=p_i(Y)$ where $p_i:\bigoplus_{i\in I}X_i\longrightarrow X$
is the obvious projection. </p>
<p>Is $Y$ a subobject of $\bigoplus_{i\in I}Y_i$?</p>
<p>This seems so obvious, but I can't seem to be able to prove it. </p>
| matt | 118,474 | <p>Here are my suggestions:</p>
<ul>
<li>Stephen Strogatz <b>Nonlinear Dynamics and Chaos</b></li>
<li>Online SOOC <b>chaosbook.org</b></li>
<li>Edward Ott <b>Chaos in Dynamical Systems</b></li>
</ul>
<p>If you know nothing about nonlinear dynamics, then Strogatz is the best place to start. If you want to jump straight into chaos, then go with Edward Ott's book. I recently discovered the online SOOC--just started, but it seems very promising!</p>
|
1,346,073 | <p>$$100\frac{dy^2}{dx^2} + y = 0$$</p>
<p>Is this worked out by using the auxillary equation such that:</p>
<p>$$100m^2 + 1 = 0$$</p>
<p>so $m = \pm i\sqrt{1/100}$ ?</p>
<p>So the general solution would be
$y(x) = A cos (1/10) + B sin(1/10)$?</p>
<p>I am not sure if I've gone about this the right way.</p>
| orangeskid | 168,051 | <p>Also see <a href="http://www.jstor.org/stable/2372705?" rel="nofollow">Horn's theorem</a>. It says that two real $n$ vectors $\bf{x}$, $\bf{y}$ are the eigenvalues and the diagonal of a (real) symmetric matrix if
and only if $\bf{x}\succ \bf{y}$ in the <a href="https://en.wikipedia.org/wiki/Majorization" rel="nofollow">Schur order</a>, that is
\begin{eqnarray}x'_1 &\le &y'_1\\
x'_1 + x'_2 &\le& y'_1 + y'_2 \\
\ldots \\
x'_1 + \ldots + x'_n &=& y'_1 + \ldots + y'_n
\end{eqnarray}
where $\bf{x}'$ is the increasing rearrangement of $\bf{x}$ and same for $\bf{y}$' of $\bf{y}$.</p>
<p>Note that the only if part is not too hard, using reduction to principal axes, which shows that $\bf{y}$ is a mixture of $\bf{x}$ ( a sequence with all entries equal is small).</p>
|
4,520,506 | <p>I know that <span class="math-container">$ x \gt 0 $</span> because of logarithm precondition, and I can see that <span class="math-container">$ x \neq 1 $</span> because otherwise it would lead to <span class="math-container">$ 0^0$</span> which is problematic, but when I checked the graph of the function I have discovered that it started from <span class="math-container">$ 1 $</span> on <span class="math-container">$ x $</span> axis and <span class="math-container">$ (0,1) $</span> interval is not considered.</p>
<p>In sum, I thought that the domain should have been <span class="math-container">$ (0, \infty) - \{1\}$</span> but it seems to be <span class="math-container">$(1,\infty)$</span> and I can't figure out where I am wrong.</p>
| José Carlos Santos | 446,262 | <p>The expression <span class="math-container">$(k+1)(k+2)\ldots n$</span> is the product of the numbers <span class="math-container">$k+1$</span>, <span class="math-container">$k+2$</span>, …, <span class="math-container">$n$</span>. When <span class="math-container">$n=3$</span> and <span class="math-container">$k=2$</span> , there is only one such number: <span class="math-container">$2+1(=3)$</span>. So, the product is <span class="math-container">$3$</span>.</p>
<p>If <span class="math-container">$n=4$</span> and <span class="math-container">$k=2$</span>, there are two such numbers: <span class="math-container">$2+1(=3)$</span> and <span class="math-container">$2+2(=4)$</span>. And, indeed,<span class="math-container">$$\frac{4!}{2!}=12=3\times4.$$</span></p>
|
2,179,289 | <p>Every valuation ring is an integrally closed local domain, and the integral closure of a local ring is the intersection of all valuation rings containing it. It would be useful for me to know when integrally closed local domains are valuation rings.</p>
<p>To be more specific,</p>
<blockquote>
<p>is there a property $P$ of unitary commutative rings that is strictly weaker than being a valuation ring, such that an integrally closed local domain is a valuation ring iff it satisfies the property $P$.</p>
</blockquote>
| Badam Baplan | 164,860 | <p>A very late answer, but I think there are some other connections and references that merit mention and maybe will interest someone down the line.</p>
<p>As mentioned by M. Zafrullah, in the presence of the local condition we can reduce the search to a property making integrally closed domains Prüfer.</p>
<p>The first connection I'll point out was explained by Zafrullah in his post, but here is another way of thinking about it.</p>
<blockquote>
<p>Let <span class="math-container">$D$</span> be a domain. For a polynomial <span class="math-container">$f \in D[x]$</span>, let <span class="math-container">$c(f)$</span> denote the ideal generated by its coefficients<span class="math-container">$^1$</span>.</p>
<p><span class="math-container">$(1)$</span> <span class="math-container">$D$</span> is Prüfer iff all polynomials <span class="math-container">$f,g \in D[x]$</span> satisfy <span class="math-container">$c(f)c(g) = c(fg)$</span></p>
<p><span class="math-container">$(2)$</span> <span class="math-container">$D$</span> is integrally closed iff all polynomials <span class="math-container">$f,g \in D[x]$</span> satisfy <span class="math-container">$[c(f)c(g)]_v = c(fg)_v$</span></p>
</blockquote>
<p>A simple property that bridges (1) and (2) is the requirement that <strong>every finitely generated ideal is divisorial</strong>, so that's one possible answer to the question.</p>
<p>Another approach is to just jack all the results about when the integral closure of a domain is Prüfer, and there are many nice characterizations of that. First to lay out a couple of definitions.... A prime <span class="math-container">$P$</span> of <span class="math-container">$D[x]$</span> such that <span class="math-container">$P \cap D = 0$</span> is called an <em>upper to zero</em>. A domain <span class="math-container">$D$</span> is called a <em>UMt</em> if every upper to zero contains a polynomial <span class="math-container">$f$</span> such that <span class="math-container">$c(f)_v = D$</span>, or equivalently if every upper to zero is a maximal <span class="math-container">$t$</span>-ideal (the terminology UMt comes from <strong>U</strong>ppers to zero are <strong>M</strong>aximal <strong>t</strong>-ideals). Now here is a sample of results on Prüfer integral closure.</p>
<blockquote>
<p>Let <span class="math-container">$D$</span> be a domain with field of fractions <span class="math-container">$K$</span>, and let <span class="math-container">$D'$</span> be the integral closure. The following are equivalent<span class="math-container">$^2$</span>.</p>
<p><span class="math-container">$(a)$</span> <span class="math-container">$D'$</span> is Prüfer.</p>
<p><span class="math-container">$(b)$</span> <span class="math-container">$D'$</span> is a QQR-domain, i.e. every overring of <span class="math-container">$D'$</span> is the intersection of localizations of <span class="math-container">$D'$</span>.</p>
<p><span class="math-container">$(c)$</span> Every overring of <span class="math-container">$D$</span> is a UMt.</p>
<p><span class="math-container">$(c')$</span> <span class="math-container">$D$</span> is a UMt and the maximal ideals of <span class="math-container">$D$</span> are <span class="math-container">$t$</span>-ideals.</p>
<p><span class="math-container">$(c'')$</span> Every upper to zero contains a polynomial <span class="math-container">$f$</span> having <span class="math-container">$c(f) = D$</span>.</p>
<p><span class="math-container">$(d)$</span> Every <span class="math-container">$k \in K$</span> is the root of a polynomial <span class="math-container">$f \in D[x]$</span> having <span class="math-container">$c(f) = D$</span></p>
<p><span class="math-container">$(e)$</span> For every <span class="math-container">$k \in K$</span>, the extension <span class="math-container">$D \subseteq D[k]$</span> satisfies INC, i.e. if <span class="math-container">$P,Q$</span> are primes of <span class="math-container">$D[k]$</span> such that <span class="math-container">$P \cap D = Q \cap D$</span>, then <span class="math-container">$P,Q$</span> are incomparable.</p>
</blockquote>
<p>These give some nice non-trivial answers to the question of a when a local integrally closed domain is a valuation ring. For example, (d) says that <strong>a local integrally integrally domain is a valuation ring iff every element of the fraction field is the root of some polynomial having a unit coefficient</strong>
and <span class="math-container">$(b)$</span> enables us to check if the domain is a valuation ring by checking to see if there are 'complicated' overrings that can't be achieved by intersecting localizations.</p>
<p>Another approach we could take is related to the last one. Note that valuation rings have the property that every ideal is a <span class="math-container">$t$</span>-ideal. So we might as well check that the maximal ideal of our local ring is a <span class="math-container">$t$</span>-ideal right off the bat. If not, then we don't have a valuation ring at hand. But if yes, then it suffices to show that our ring is a <em>PvMD</em>, more generally than a Prüfer domain, and the analog of the above results for PvMDs then becomes available. For example, <strong>a local integrally closed domain is a valuation ring iff its maximal ideal is a <span class="math-container">$t$</span>-ideal and every element of the fraction field is the root of a polynomial <span class="math-container">$f$</span> having <span class="math-container">$c(f)_v = D$</span></strong>, etc. etc.</p>
<p><span class="math-container">$^1$</span> A standard reference is Gilmer's <em>Multiplicative Ideal Theory</em>. (1) is covered in 28.1-28.6, and (2) is 34.8. The main ingredients in these characterizations are the Dedekind-Mertens content formula for polynomials and the fact that the integral closure is the intersection of valuation overrings.</p>
<p><span class="math-container">$^2$</span> <span class="math-container">$(a) \iff (b)$</span> Easy using the characterization of Prüfer domains as having every overring integrally closed, and again using that integrally closed domains are the intersection of their valuation overrings (which for Prüfer rings are localizations at primes). <span class="math-container">$(a) \iff (c,c',c'')$</span> See <a href="https://www.tandfonline.com/doi/abs/10.1080/00927879808826181" rel="nofollow noreferrer">here for the originating work</a> and also section <span class="math-container">$1$</span> of <a href="https://pdfs.semanticscholar.org/03ce/a7fc4a4e147506c89b599cb36ef1c643f021.pdf" rel="nofollow noreferrer">this paper</a> which also mentions most of the results in this post. Note especially in <span class="math-container">$c'$</span> how the assumption that maximal ideals are <span class="math-container">$t$</span>-ideals implies that <span class="math-container">$I_t = D \implies I = D$</span>, which interacts with the UMt property by forcing every upper to zero to contain a polynomial with a unit coefficient. This has a similar effect to our first approach of requiring f.g. ideals to be divisorial. <span class="math-container">$(a) \iff (d)$</span> See <a href="https://projecteuclid.org/download/pdf_1/euclid.pjm/1102868625" rel="nofollow noreferrer">Theorem 5 here</a> <span class="math-container">$(d) \iff (e)$</span> Actually this could be stated more strongly for individual elements, which are usually called <em>primitive</em>, and for arbitrary ring extensions. See <a href="https://www.cambridge.org/core/journals/canadian-mathematical-bulletin/article/on-incextensions-and-polynomials-with-unit-content/C9BE6787DDACAC8778FFE8C7D0DE53F7" rel="nofollow noreferrer">the first theorem here</a> for details.</p>
|
276,987 | <p>I want to visualize the following set in Maple:</p>
<blockquote>
<p>$\lbrace (x+y,x-y) \vert (x,y)\in (-\frac{1}{2},\frac{1}{2})^{2} \rbrace$ </p>
</blockquote>
<p>Which commands should I use? Is it even possible?</p>
| Michael Greinecker | 21,674 | <p>Let $X$ be a set with at leas two elements $x$ and $y$. Let the topology consist of arbitrary unions of singletons from $X\backslash \{y\}$ and the doubleton $\{x,y\}$. The result is a fairly boring topology such that any finer topology is discrete.</p>
|
1,618,411 | <p>I'm learning the fundamentals of <em>discrete mathematics</em>, and I have been requested to solve this problem:</p>
<p>According to the set of natural numbers</p>
<p>$$
\mathbb{N} = {0, 1, 2, 3, ...}
$$</p>
<p>write a definition for the less than relation.</p>
<p>I wrote this:</p>
<p>$a < b$ if $a + 1 < b + 1$</p>
<p>Is it correct?</p>
| Andres Mejia | 297,998 | <p>$a<b \iff \exists p \in \mathbb{N_{>0}}$: $b=a+p$.</p>
|
1,618,411 | <p>I'm learning the fundamentals of <em>discrete mathematics</em>, and I have been requested to solve this problem:</p>
<p>According to the set of natural numbers</p>
<p>$$
\mathbb{N} = {0, 1, 2, 3, ...}
$$</p>
<p>write a definition for the less than relation.</p>
<p>I wrote this:</p>
<p>$a < b$ if $a + 1 < b + 1$</p>
<p>Is it correct?</p>
| fleablood | 280,126 | <p>You can either have a direct definition or a recursive definition. If you have a recursive definition you need a base case from which all cases arrive.</p>
<p>Your function appears to be recursive but it has no base case.</p>
<p>a < b if a + 1< b + 1 which raises the question what is the definition of a + 1 < b + 1 to which a + 1 < b+a if a + 2 < b +2, and final verification is pushed further and further away.</p>
<p>So if you are going to do recursion, you need a base case involving 0</p>
<ol>
<li>$0 < b$ if $b \ne 0$</li>
</ol>
<p>Now your definition $a < b$ if $a + 1 < b + 1$ ... isn't good because it is taking you <em>away</em> from the base case. We need a definition that either a) takes you <em>from</em> the base case to $a < b$ or b) takes you from $a < b$ <em>to</em> the base case.</p>
<p>Either</p>
<p>2a. $a < b $ if $a - 1 < b -1$ <em>(allows the user to start at $a<b$ and work down to $0 < b'$)</em></p>
<p>Or </p>
<p>2b. if $a < b$ then $a + 1 < b + 1$ <em>(allows the user to start at $0<b'$ annd work up to $a < b$)</em></p>
<p>will do. Which one you like is a matter of taste.</p>
<p>====</p>
<p>Then there is a <em>direct</em> definition. This is less obvious to see but more "powerful" and ,ahem, direct to use. When is $a < b$ true? It's true when $0 < b- a$ which, as these are natural numbers rather than integers, is true whenever $b - a \ne 0$ and $b - a$ is a legitimate natural number.</p>
<p>So</p>
<ul>
<li>$a < b$ if $b - a \ne 0$ and $b - a \in \mathbb N$.</li>
</ul>
|
4,403,081 | <p><span class="math-container">$$\int \dfrac{dx}{x\sqrt{x^4-1}}$$</span></p>
<p>I need to solve this integration.
I solved and got <span class="math-container">$\dfrac12\tan^{-1}(\sqrt{x^4-1}) + C$</span>, however the answer given in my textbook is <span class="math-container">$\dfrac12\sec^{-1}(x^2) + C$</span></p>
<p>How can I prove that both quantities are equal? Is there something wrong with my answer?</p>
<p><strong>EDIT:</strong></p>
<p>Here's my work:
<span class="math-container">$$\int\dfrac{dx}{x\sqrt{x^4-1}}= \dfrac{1}{4}\int\dfrac{4x^3 dx}{x^4\sqrt{x^4-1}}$$</span></p>
<p>Let <span class="math-container">$x^4 - 1 = t^2$</span>
<span class="math-container">$$\dfrac{1}{2}\int\dfrac{dx}{1 + t^2}$$</span></p>
<p><span class="math-container">$$\dfrac12 \tan^{-1}(\sqrt{x^4 -1 }) + C$$</span></p>
| Zaragosa | 691,503 | <p>The answer was already given by B. Goddard, here I try to give a visual answer:</p>
<p><a href="https://i.stack.imgur.com/jQJGX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jQJGX.png" alt="enter image description here" /></a></p>
|
4,403,081 | <p><span class="math-container">$$\int \dfrac{dx}{x\sqrt{x^4-1}}$$</span></p>
<p>I need to solve this integration.
I solved and got <span class="math-container">$\dfrac12\tan^{-1}(\sqrt{x^4-1}) + C$</span>, however the answer given in my textbook is <span class="math-container">$\dfrac12\sec^{-1}(x^2) + C$</span></p>
<p>How can I prove that both quantities are equal? Is there something wrong with my answer?</p>
<p><strong>EDIT:</strong></p>
<p>Here's my work:
<span class="math-container">$$\int\dfrac{dx}{x\sqrt{x^4-1}}= \dfrac{1}{4}\int\dfrac{4x^3 dx}{x^4\sqrt{x^4-1}}$$</span></p>
<p>Let <span class="math-container">$x^4 - 1 = t^2$</span>
<span class="math-container">$$\dfrac{1}{2}\int\dfrac{dx}{1 + t^2}$$</span></p>
<p><span class="math-container">$$\dfrac12 \tan^{-1}(\sqrt{x^4 -1 }) + C$$</span></p>
| tryingtobeastoic | 541,298 | <p>There is nothing wrong with your answer. The antiderivative you found is <a href="https://www.integral-calculator.com/#expr=%5Cfrac%7B1%7D%7Bx%5Csqrt%7Bx%5E4-1%7D%7D" rel="nofollow noreferrer">correct</a>.</p>
<p>Your book's answer is also <a href="https://www.integral-calculator.com/#expr=%5Cfrac%7B1%7D%7Bx%5Csqrt%7Bx%5E4-1%7D%7D" rel="nofollow noreferrer">correct</a>. The thing is that</p>
<p><span class="math-container">$$\frac{1}{2}\tan^{-1}(\sqrt{x^4-1})=\frac{1}{2}\sec^{-1}(x^2)$$</span></p>
<p>If you don't believe me, see their <a href="https://www.desmos.com/calculator/8hcmfzkbcv" rel="nofollow noreferrer">graphs</a>. The graphs of <span class="math-container">$\frac{1}{2}\tan^{-1}(\sqrt{x^4-1})$</span> and <span class="math-container">$\frac{1}{2}\sec^{-1}(x^2)$</span> are exactly the same.</p>
<p><strong>Edit:</strong></p>
<p>In order to prove that the quantities are indeed equal see <a href="https://math.stackexchange.com/a/4403111/768162">@Zaragosa's answer</a>.</p>
<hr />
<p><strong>PS:</strong> Whenever in doubt about calculus, use the <a href="https://www.derivative-calculator.net/" rel="nofollow noreferrer">derivative-calculator</a> or the <a href="https://www.integral-calculator.com/" rel="nofollow noreferrer">integral-calculator</a>.</p>
|
1,046,066 | <p>Is this series $$\sum_{n\geq 1}\left(\prod_{k=1}^{n}k^k\right)^{\!-\frac{4}{n^2}} $$ convergent or divergent?</p>
<p>My attempt was to use the comparison test, but I'm stuck at finding the behaviour of $\displaystyle \prod_1^n k^k$ as $n$ goes to infinity. Thanks in advance.</p>
| Milly | 182,459 | <p>Hint:
$$\log\left(\prod_{k=1}^n k^k\right) = \sum_{k=1}^n k\log k\cdots $$</p>
|
1,296,420 | <p>I was trying to find an example such that $G \cong G \times G$, but I am not getting anywhere. Obviously no finite group satisfies it. What is such group?</p>
| wlad | 228,274 | <p>Let $G = \mathbb Z ^ \mathbb N$ (with pointwise addition as the product). Then let $f:G \times G \longrightarrow G$ be $$f(g,h)(n) = \begin{cases} g(k), &n = 2k \\ h(k), &n = 2k+1 \end{cases}$$
You can verify $f$ is an isomorphism.</p>
|
1,296,420 | <p>I was trying to find an example such that $G \cong G \times G$, but I am not getting anywhere. Obviously no finite group satisfies it. What is such group?</p>
| Cameron Buie | 28,900 | <p>Let $G$ be the trivial group, for the only finite example.</p>
|
3,986,831 | <p>the question is:
true or false: if <span class="math-container">$f_n(x)'$</span> converges uniformly to <span class="math-container">$f(x)'$</span> then <span class="math-container">$f_n(x)$</span> converges uniformly to <span class="math-container">$f(x)$</span>. I tried many examples and they all confirmed the statement.</p>
| Yuval Peres | 360,408 | <p>True on a finite interval as long as convergence holds at one point. Simply integrate.</p>
|
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