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 |
|---|---|---|---|---|
3,449,826 | <p><span class="math-container">$C$</span> is any closed curve encompassing the whole branch cut.
The approach to this problem would involve using the residue theorem:</p>
<p>1) We first want to find the residues at infinity so we change the form to where we are able to perform a series expansion.</p>
<p>2) Obtain th... | Pedro Juan Soto | 601,282 | <p>We have that <span class="math-container">$$\sqrt{\frac{z-a}{z-b}} = 1 + \frac{b-a}{2}\frac{1}{z} + O\bigg(\frac{1}{z^2}\bigg)$$</span> therefore using Cauchy's integral formula we have that <span class="math-container">$$\oint \sqrt{\frac{z-a}{z-b}}dz = 2 \pi i \frac{b-a}{2} = (b-a)\pi i .$$</span></p>
|
1,210,018 | <p>$$ \begin{bmatrix}
1 & 1 \\
1 & 1 \\
\end{bmatrix}
\begin{Bmatrix}
v_1 \\
v_2 \\
\end{Bmatrix}=
\begin{Bmatrix}
0 \\
0 \\
\end{Bmatrix}$$</p>
<p>How can i solve this ?</p>
<p>I found it $$v_1+v_2=0$$ $$v_1+v_2=0$$ .</p>
<p>... | Jordan Glen | 225,803 | <p>You essentially have $1$ equation in two unknowns. Let $v_2= t$, where $t\in \mathbb R$. Then $v_1 = -v_2 = -t$.</p>
<p>So there are infinitely many solutions, all of which can be represented by $$V = \begin{pmatrix} -t\\t\end{pmatrix} = t\begin{pmatrix} -1 \\ 1\end{pmatrix}$$ again, where $t$ can take any value in... |
3,047,670 | <blockquote>
<p>Prove or disprove that <span class="math-container">$$\lim_{n\to \infty} \left(\frac{x_{n+1}-l}{x_n-l}\right)=\lim_{n\to\infty}\left(\frac{x_{n+1}}{x_n}\right)$$</span> where <span class="math-container">$l=\lim_{n\to \infty} x_n$</span></p>
</blockquote>
<p>I think that the above result is true,but ... | Love Invariants | 551,019 | <p>Take <span class="math-container">$x_n=n$</span><br>
Then LHS=<span class="math-container">$1\over0$</span> and RHS=<span class="math-container">$1+{1\over\infty}=1$</span><br>
Both aren't equal.</p>
|
3,587,891 | <blockquote>
<p>Suppose <span class="math-container">$f(x, y, z): \mathbb{R}^{3} \rightarrow \mathbb{R}$</span> is a <span class="math-container">$C^{2}$</span> harmonic function, that is, it satisfies <span class="math-container">$f_{x x}+f_{y y}+f_{z z}=0 .$</span> Let <span class="math-container">$E \subset \mathbb{... | Juan L. | 646,880 | <p>Let <span class="math-container">$(s_p)_{p \in U} \in \prod_{p \in U} \mathscr F_p$</span> be a compatible germ. By definition, there exists a cover <span class="math-container">$\{U_i\}_{i \in I}$</span> for <span class="math-container">$U$</span>, and elements <span class="math-container">$f_i \in \mathscr F(U_i)$... |
1,159 | <p>A lot of times I see theorems stated for local rings, but usually they are also true for "graded local rings", i.e., graded rings with a unique homogeneous maximal ideal (like the polynomial ring). For example, the Hilbert syzygy theorem, the Auslander-Buchsbaum formula, statements related to local cohomology, etc.<... | Greg Stevenson | 310 | <p>One small thing I know of which changes is that if one has a Z-graded-commutative noetherian ring (where Z is the integers) Matlis' classification of indecomposable injective modules goes through but with one small hiccup.</p>
<p>Every indecomposable injective is isomorphic to E(R/p)[n] for some unique homogeneous ... |
2,937,671 | <p>Definition <span class="math-container">$\{A_i\}_{i\in I}$</span> be an indexed family of classes; Let
<span class="math-container">$$A=\bigcup_{i\in I} A_i.$$</span></p>
<p>The <span class="math-container">$product$</span> of the classes <span class="math-container">$A_i$</span> is defined to be the class
<span cl... | Paul Frost | 349,785 | <p><span class="math-container">$I$</span> and <span class="math-container">$J$</span> are sets. Define a family <span class="math-container">$\{ X_k \}_{k \in \{ 1, 2 \}}$</span> by <span class="math-container">$X_1 = I, X_2 = J$</span> and form the product <span class="math-container">$P = \prod_{k \in \{ 1, 2 \}} X_... |
704,073 | <p>I encountered something interesting when trying to differentiate $F(x) = c$.</p>
<p>Consider: $\lim_{x→0}\frac0x$. </p>
<p>I understand that for any $x$, no matter how incredibly small, we will have $0$ as the quotient. But don't things change when one takes matters to infinitesimals?
I.e. why is the function $\fr... | AlexR | 86,940 | <p>Note that writing $f(x) = \frac0x$ results in $f(0) = \frac00$ wich <em>is</em> undefined. However, the singularity of $f$ is nice in the way that is can be continuously defined by $f(0) := 0$ (note the colon for <em>defining</em> the value). A limit is exactly this concept: What is the value of $f(x)$ when $x$ come... |
3,046,083 | <p>Is it true that the intersection of the closures of sets <span class="math-container">$A$</span> and <span class="math-container">$B$</span> is equal to the closure of their intersection?
<span class="math-container">$ cl(A)\cap{cl(B)}=cl(A\cap{B})$</span> ?</p>
| Henno Brandsma | 4,280 | <p>No, the rationals and the irrationals (in the reals) are disjoint so <span class="math-container">$\operatorname{cl}(A \cap B) = \operatorname{cl}{\emptyset}=\emptyset$</span> while <span class="math-container">$\operatorname{cl}(A) = \operatorname{cl}(B) = \mathbb{R}$</span></p>
|
865,598 | <p>How can I calculate this value?</p>
<p>$$\cot\left(\sin^{-1}\left(-\frac12\right)\right)$$</p>
| Rene Schipperus | 149,912 | <p>$$\cot x =\frac{\cos x}{\sin x}=\frac{\sqrt{1-\sin^2x}}{\sin x}$$
so we have $$\frac{\sqrt{\frac{3}{4}}}{-\frac{1}{2}}=\pm \sqrt{3}$$
There is not enough information in the problem to determine the sign.</p>
|
3,773,856 | <p>I'm having trouble with part of a question on Cardano's method for solving cubic polynomial equations. This is a multi-part question, and I have been able to answer most of it. But I am having trouble with the last part. I think I'll just post here the part of the question that I'm having trouble with.</p>
<p>We ha... | Community | -1 | <p>You can "brute force" the solution of the quadratic,</p>
<p><span class="math-container">$$x^2+qx-\frac{p^3}{27}=0,$$</span></p>
<p>giving two roots</p>
<p><span class="math-container">$$u^3,v^3=\frac{-q\pm\sqrt{q^2+\dfrac{4p^3}{27}}}2$$</span> which are complex. In polar form,</p>
<p><span class="math-con... |
3,051,176 | <blockquote>
<p>True of False: If <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are odd positive integers, then <span class="math-container">$n^2+m^2$</span> is not a perfect square.</p>
</blockquote>
<p>Anyway it is already appear <a href="https://math.stackexchange.com/ques... | José Carlos Santos | 446,262 | <p>The line that you mentioned is the line defined by the two points at which the tangent lines touch the circle. So, if you want to actually get those two points, compute the intersection between the line and the circle. Then, the tangent lines will be the lines passing through <span class="math-container">$(x_1,y_1)$... |
3,051,176 | <blockquote>
<p>True of False: If <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are odd positive integers, then <span class="math-container">$n^2+m^2$</span> is not a perfect square.</p>
</blockquote>
<p>Anyway it is already appear <a href="https://math.stackexchange.com/ques... | Mick | 42,351 | <p>I think what you have found in the text should be:-</p>
<p>If <span class="math-container">$P(x_1, y_1)$</span> is a point <strong>on</strong> the circle <span class="math-container">$C: x^2 + y^2 +2gx + 2fy + c = 0$</span>, then the equation of the tangent t<strong>ouching circle C at P</strong> is<br>
<span clas... |
2,412,783 | <p>I'm very new to linear algebra, and I have a homework problem that hasn't been covered in the book or by the professor. It seems like I have a fundamental misunderstanding of what matrices represent, but I can't find a good article or answer.</p>
<blockquote>
<p>Do the three lines $x_1 - 4x_2 = 1$, $2x_1 - x_2 = ... | Randall | 464,495 | <p>You can reduce it further. When you do, the first row is $(1,0,-13/7)$ which says that your simultaneous solution is $(-13/7, -5/7)$. So, there is one point of intersection.</p>
<p>Since you have only two variables but three equations, you would need an extra 0-row in order to get infinitely many solutions, and t... |
2,412,783 | <p>I'm very new to linear algebra, and I have a homework problem that hasn't been covered in the book or by the professor. It seems like I have a fundamental misunderstanding of what matrices represent, but I can't find a good article or answer.</p>
<blockquote>
<p>Do the three lines $x_1 - 4x_2 = 1$, $2x_1 - x_2 = ... | Atique M | 686,251 | <p>If your augmented matrix is of the form <span class="math-container">$[A | C]$</span> then system has unique solution if <span class="math-container">$rank(A)=rank(C)$</span>=number of unknowns.
Here, <span class="math-container">$ank(A)=rank(C)=$</span>number of unknowns=2
So, unique soon i.e a single intersection... |
3,795,234 | <p>(disclaimer: I am not well versed in mathematics so please excuse my poor notation / explanation)</p>
<p>Given a hexagon grid that defines it's "neighbours" via offsets on the axis' <span class="math-container">$q$</span> & <span class="math-container">$r$</span> like this :
<a href="https://i.stack.im... | roookeee | 817,286 | <p>(disclaimer: I had a really hard time to explain how this works and will describe some stuff that just "comes into existance" without explaining how I got there because I really can't explain it)</p>
<p>Given the following image:</p>
<p><a href="https://i.stack.imgur.com/I0dJE.png" rel="nofollow noreferrer... |
1,767,682 | <p>I was thinking about sequences, and my mind came to one defined like this:</p>
<p>-1, 1, -1, 1, 1, -1, 1, 1, 1, -1, 1, 1, 1, 1, ...</p>
<p>Where the first term is -1, and after the nth occurrence of -1 in the sequence, the next n terms of the sequence are 1, followed by -1, and so on. Which led me to perhaps a str... | Eric Towers | 123,905 | <p>Most methods of <a href="https://en.wikipedia.org/wiki/Divergent_series" rel="nofollow">summing divergent series</a> can be adapted to making divergent sequences converge. For instance, Cesaro summation replaces the $n^\text{th}$ term of a series with the mean of the first $n$ terms. You could do the same: replace... |
2,810,008 | <p>Can I investigate this limit and if yes, how? $${i^∞}$$</p>
<p>I am at a loss of ideas and maybe it is undefined?</p>
| ncmathsadist | 4,154 | <p>You can't. The powers of $i$ form a periodic progression among four values. Convergence means you have some complex number for which the following is true. For any neighborhood of the number, the sequence eventually enters that neighborhood and never re-exits.</p>
|
3,815,898 | <p>In my stats lecture, my professor introduced this theorem, however I don't quite understand what the theorem means and how he got from step 3 to 4. Also what does the prime symbol mean? Could someone paraphrase what the theorem means or point me to some online resource where I could learn more about this theorem? Th... | tommik | 791,458 | <p>This theorem is named "integral transformation theorem "</p>
<p>As you can see after few passages you get</p>
<p><span class="math-container">$$F_Y(y)=F_X[F_X^{-1}(y)]$$</span></p>
<p>now it is evident that applying both <span class="math-container">$F$</span> and <span class="math-container">$F^{-1}$</spa... |
4,250,845 | <p>Considering a Quadrilateral <span class="math-container">$ABCD$</span> where <span class="math-container">$A(0,0,0), B(2,0,2), C(2,2\sqrt 2,2), D(0,2\sqrt2,0)$</span>. Basically I have to find the <strong>Area</strong> of <strong>projection</strong> of quadrilateral <span class="math-container">$ABCD$</span> on the ... | Glorious Nathalie | 948,761 | <p>As indicated by Intelligenti pauca in the above comments, the best way to go is to find the area of the quadrilateral, then multiply the area found by the cosine of the angle between the two planes which is the same angle between the normals to the planes (or its supplement).</p>
<p><span class="math-container">$\be... |
1,778,098 | <p>Let $f,g: M^{k} \to N$ ($M$ and $N$ with out boundary ) such that they are homotopic then for $\omega$ a $k$-form on $N$ do we have that </p>
<p>$$ \int_M f^{\ast} \omega = \int_M g^{\ast} \omega$$ </p>
<p>as conclusion? I can't figure out a proof so I am starting to think that it is not true. I can't use Homolog... | Ted Shifrin | 71,348 | <p>You had better assume $M$ is a compact manifold without boundary and $\omega$ is a closed $k$-form on $N$. (The latter is automatic if $\dim N = k$ as well.) Then you can consider the homotopy mapping $H\colon M\times [0,1]\to N$ and apply Stokes's Theorem to $H^*\omega$. </p>
|
655,261 | <p>Let meagre subsets be defined as:<br>
$A\text{ meagre}\iff A=\bigcup_{\lvert K\rvert\leq\aleph_0} A_k\text{ with }\overline{A_k}°=\varnothing$<br>
Then it satisfies:<br>
$B\subseteq A\text{ meagre}\Rightarrow B\text{ meagre}$<br>
$A_k\text{ meagre}\Rightarrow\bigcup_{\lvert K\rvert\leq\aleph_0} A_k\text{ meagre}$</p... | Martín-Blas Pérez Pinilla | 98,199 | <p>The property "being a set" (not very demanding!) verifies:</p>
<p>$B\subset A$ is a set $\Rightarrow$ $B$ is a set.</p>
<p>$A_k\text{ is a set}\Rightarrow\bigcup_{\lvert K\rvert\leq\aleph_0} A_k\text{ is a set}$</p>
|
655,261 | <p>Let meagre subsets be defined as:<br>
$A\text{ meagre}\iff A=\bigcup_{\lvert K\rvert\leq\aleph_0} A_k\text{ with }\overline{A_k}°=\varnothing$<br>
Then it satisfies:<br>
$B\subseteq A\text{ meagre}\Rightarrow B\text{ meagre}$<br>
$A_k\text{ meagre}\Rightarrow\bigcup_{\lvert K\rvert\leq\aleph_0} A_k\text{ meagre}$</p... | Community | -1 | <p>If $A$ is meager, then there is a sequence of nowhere dense subsets $\{A_n\}$ such that $A=\bigcup_{n=1}^\infty A_n$. Now $B=B\cap A=\bigcup_{n=1}^\infty(B\cap A_n)$. We just need show that each $B\cap A_n$ is nowhere dense, and this will finish the proof for the first.</p>
<p>But we can show something a little mor... |
2,278,991 | <p>(NOTE: I am new to proof construction. Don't panic if your heart beat increase.)</p>
<p>Proof 1:</p>
<p>suppose $x$ is even and prime, then there is $k$ in $\mathbb N$ such that </p>
<p>$x = 2k$</p>
<p>But x is only divisible by itself and not 2.</p>
<p>$\frac{x}{2} != k$</p>
<p>$x$ can not be even.</p>
<hr>
... | Henrik supports the community | 193,386 | <p>What?</p>
<p>Whatever you're trying to do, it gets completely lost because you're not explaining what you're trying to accomplish, but just stating some things.</p>
<p>And for the mathematical content, I gave up after:</p>
<blockquote>
<p>If $x>n$ then $\frac{x}{n}$ does not belong to $\mathbb N$.</p>
</bloc... |
172,366 | <blockquote>
<p>What will be the value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$
provided that $P(1)=10$, $P(2)=20$, $P(3)=30$?</p>
</blockquote>
<p>I put these values and got three simultaneous equations in $a, b, c, d$. What is the smarter way to approach these problems?</p>
| Cocopuffs | 32,943 | <p>I'm not sure this is the smartest way, but here's one way.</p>
<p>Define $Q(x) = P(x) - x^4$ which satisfies the conditions $$Q(1) = 9, Q(2) = 4, Q(3) = -51.$$</p>
<p>So $a,b,c,d$ satisfy the matrix identity</p>
<p>$$\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \\ 1 & 3 & 9 &... |
67,460 | <p>Denote the system in $GF(2)$ as $Ax=b$, where:
$$
\begin{align}
A=&(A_{ij})_{m\times m}\\
A_{ij}=&
\begin{cases}
(1)_{n\times n}&\text{if }i=j\quad\text{(a matrix where entries are all 1's)}\\
I_n&\text{if }i\ne j\quad\text{(the identity matrix)}
\end{cases}
\end{align}
$$
that is, $A$ i... | Robert Israel | 8,508 | <p>An empirical formula: it looks to me like $\det(A) = (-1)^{(m-1)(n-1)} (n+m-1)(n-1)^{m-1}(m-1)^{n-1}$. In particular, $\det(A) \equiv 0 \mod 2$ when $m$ or $n$ is odd (except in the case $m=n=1$) so, as Henning found, $A$ will not be invertible over GF(2) in those cases.
However, it will be invertible when $m$ and ... |
487,171 | <p>Now I tried tackling this question from different sizes and perspectives (and already asked a couple of questions here and there), but perhaps only now can I formulate it well and ask you (since I have no good ideas).</p>
<p>Let there be $k, n \in\mathbb{Z_+}$. These are fixed.</p>
<p>Consider a set of $k$ integer... | Marc van Leeuwen | 18,880 | <p>The number of possible sequences is $k^n$. Then number of favourable outcomes, namely weakly increasing sequences, is $\binom{n+k-1}{k-1}=\binom{n+k-1}n$. The probability of finding the random sequence to be increasing is
$$
\frac{\binom{n+k-1}{k-1}}{k^n} =
\frac{\binom{n+k-1}n}{k^n} = \prod_{i=1}^n\frac{k+i-1}{... |
3,243,406 | <p>I know that the function <span class="math-container">$f(x)=|x(x-1)^3|$</span> is not derivable in <span class="math-container">$x=0$</span>, but why is it derivable in <span class="math-container">$x=1$</span>?</p>
| Sri-Amirthan Theivendran | 302,692 | <p>Note that <span class="math-container">$f(x)=|x|(x-1)^2|x-1|$</span> whence</p>
<p><span class="math-container">$$
f'(1)=\lim_{x\to1}\frac{f(x)-f(1)}{x-1}=\lim_{x\to 1} \frac{f(x)}{x-1}=\lim_{x\to1} |x|(x-1)|x-1|=0.
$$</span>
since <span class="math-container">$|x|(x-1)|x-1|$</span> is a product of continuous funct... |
295,597 | <p>I'm trying to solve this simple integral:</p>
<p>$$\frac12 \int \frac{x^2}{\sqrt{x + 1}} dx$$</p>
<p>Here's what I have done so far:</p>
<ol>
<li><p>$\displaystyle t = \sqrt{x + 1} \Leftrightarrow x = t^2 - 1 \Rightarrow dx = 2t dt$</p></li>
<li><p>$\displaystyle \frac12 \int \frac{x^2}{\sqrt{x + 1}} dx = \int \f... | Thomas Andrews | 7,933 | <p>$$\begin{align}\frac{x^2}{\sqrt{x+1}} &= \frac{x^2-1}{\sqrt{x+1}} + \frac{1}{\sqrt{x+1}}\\&=(x-1)\sqrt{x+1} + \frac{1}{\sqrt{x+1}}\\&=(x+1)^{3/2} - 2\sqrt{x+1} + \frac{1}{\sqrt{x+1}}\end{align}$$</p>
<p>I think you can integrate each of these terms.</p>
|
2,921,927 | <p>We have the following function : <span class="math-container">$$f(z)=\frac{z^2}{1-\cos z}$$</span> where <span class="math-container">$z_0=0$</span> is a removable singularity since the limit as <span class="math-container">$z$</span> goes to <span class="math-container">$0$</span> is <span class="math-container">$2... | Mark | 470,733 | <p>The residue of a removable singularity is always $0$. It simply follows from the fact that if the singularity is removable then the Laurent series is actually Taylor series. </p>
|
3,998,098 | <p>I was asked to determine the locus of the equation
<span class="math-container">$$b^2-2x^2=2xy+y^2$$</span></p>
<p>This is my work:</p>
<blockquote>
<p>Add <span class="math-container">$x^2$</span> to both sides:
<span class="math-container">$$\begin{align}
b^2-x^2 &=2xy+y^2+x^2\\
b^2-x^2 &=\left(x+y\right... | Vajra | 759,757 | <p>The matrix associated to the conic is
<span class="math-container">$$\begin{pmatrix}-2 & -1 & 0\\-1 & -1 & 0\\0 & 0 & b^2 \end{pmatrix}\implies\Bigg|\begin{pmatrix}-2 & -1\\-1 & -1 \end{pmatrix}\Bigg|=2-1=1>0$$</span>
and this shows that the conic is an ellipse.</p>
|
3,998,098 | <p>I was asked to determine the locus of the equation
<span class="math-container">$$b^2-2x^2=2xy+y^2$$</span></p>
<p>This is my work:</p>
<blockquote>
<p>Add <span class="math-container">$x^2$</span> to both sides:
<span class="math-container">$$\begin{align}
b^2-x^2 &=2xy+y^2+x^2\\
b^2-x^2 &=\left(x+y\right... | Amanuel Getachew | 669,545 | <p>The equation
<span class="math-container">$$Ax^2 + Bxy+Cy^2 +Dx + Ey + F = 0$$</span>
is:</p>
<ul>
<li><p>an ellipse if <span class="math-container">$B^2 - 4AC < 0$</span> (could be circle or a point if <span class="math-container">$A = C$</span>, <span class="math-container">$F > 0$</span> and <span class="ma... |
4,386,087 | <p>There exists an elevator which starts off containing <span class="math-container">$p$</span> passengers.</p>
<p>There are <span class="math-container">$F$</span> floors.</p>
<p><em><span class="math-container">$\forall i: P_i = $</span>P(i. passenger exists on any of the floors)</em> = <span class="math-container">$... | true blue anil | 22,388 | <p>I prefer the approach you adopted, as you do not need recourse to a Stirling table, and in a way the formula you use is more basic.</p>
<p>You could, however, learn to condense your expression to</p>
<p><span class="math-container">$$\left(\dfrac1{F^P}\right)\sum_{i=0}^{F}(-1)^i\dbinom{F}{i}(F-i)^P$$</span></p>
<p>a... |
257,821 | <p>The Kullback-Liebler divergence between two distributions with pdfs $f(x)$ and $g(x)$ is defined
by
$$\mathrm{KL}(F;G) = \int_{-\infty}^{\infty} \ln \left(\frac{f(x)}{g(x)}\right)f(x)\,dx$$</p>
<p>Compute the Kullback-Lieber divergence when $F$ is the standard normal distribution and $G$
is the normal distribution ... | VCZ | 49,092 | <p>The pdf of the standard normal distribution is $
\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}$. Similarly, $g(x) = \frac{1}{\sqrt{2\pi}}e^{(x-\mu)^2/2}$.Therefore, $D_{KL}=\int_{-\infty}^{\infty} \frac{{2\mu-\mu}^2}{2\sqrt{2\pi}}e^{-x^{2}/2} dx$. This is equivalent to $\frac{{\mu}^2}{2}$.</p>
<p>This can be easily generalized... |
257,821 | <p>The Kullback-Liebler divergence between two distributions with pdfs $f(x)$ and $g(x)$ is defined
by
$$\mathrm{KL}(F;G) = \int_{-\infty}^{\infty} \ln \left(\frac{f(x)}{g(x)}\right)f(x)\,dx$$</p>
<p>Compute the Kullback-Lieber divergence when $F$ is the standard normal distribution and $G$
is the normal distribution ... | R S | 48,725 | <p>I cannot comment (not enough reputation). </p>
<p>Vincent: You have the wrong pdf for $g(x)$, you have a normal distribution with mean 1 and variance 1, not mean $\mu$. </p>
<p>Hint: You don't need to solve any integrals. You should be able to write this as pdf's and their expected values, so you never need to int... |
1,284,388 | <p>Solving for variable $d$:</p>
<p>$v = \frac{1}{2}hd^2 + 9.9$</p>
<p>$-2(v - 9.9) = hd^2$</p>
<p>$-2v + 19.8 = hd^2$</p>
<p>$d = \sqrt{\frac{-2v + 19.8}{h}}$</p>
<p>The correct answer is:</p>
<p>$d = \pm\sqrt{\frac{2v - 19.8}{h}}$</p>
| Olivier Oloa | 118,798 | <p>From $$V = \frac{1}{2}hd^2 + 9.9$$ you may deduce, with appropriate conditions:
$$
2V = hd^2 + 2\times9.9
$$ $$
2V-19.8 = hd^2
$$$$
\frac{2V-19.8}h = d^2
$$$$
\pm\sqrt{\frac{2V-19.8}h }= d.
$$</p>
|
419,625 | <p>Please help me, I have two functions:</p>
<blockquote>
<pre><code>a := x^3+x^2-x;
b := 20*sin(x^2)-5;
</code></pre>
</blockquote>
<p>and I would like to change a background color and fill the areas between two curves. I filled areas but I dont know how can I change the background, any idea?</p>
<blockquote>
<pre>... | Mikasa | 8,581 | <p>The way in which @acer colored the area between two curves is really elegant, so I don't have any additional points about it for you. Just to say that: when you made the area between two curves colored (following the acer's way); I think the following line could make the background colored as well:</p>
<pre><code>&... |
419,625 | <p>Please help me, I have two functions:</p>
<blockquote>
<pre><code>a := x^3+x^2-x;
b := 20*sin(x^2)-5;
</code></pre>
</blockquote>
<p>and I would like to change a background color and fill the areas between two curves. I filled areas but I dont know how can I change the background, any idea?</p>
<blockquote>
<pre>... | Software | 62,172 | <p><em>Thanks to @Babak S. and @acer</em> $\Huge\color{green}{✔}^\color{red}{+}$</p>
<p>I Just wanted to say that:</p>
<blockquote>
<p>Right click on the background color -> select "<strong>color</strong>" -> You can also use "the default colors".</p>
</blockquote>
<p><img src="https://i.stack.imgur.com/8IcID.jpg"... |
4,280,424 | <p>The PDE:
<span class="math-container">$$\frac1D C_t-Q=\frac2rC_r+C_{rr}$$</span></p>
<p>on the domain <span class="math-container">$r \in [0,\bar{R}]$</span> and <span class="math-container">$t \in [0,+\infty]$</span> and where <span class="math-container">$D$</span> and <span class="math-container">$Q$</span> are R... | Mark | 470,733 | <p>The closure has to contain <span class="math-container">$A$</span>, so there are only two options here: <span class="math-container">$\mathbb{R}$</span> or <span class="math-container">$\mathbb{R}\setminus\{p\}$</span>. Since <span class="math-container">$\mathbb{R}\setminus\{p\}$</span> is not a closed set (its com... |
2,677,134 | <p>Given the definition of Big-O, prove that $f(n) = n^2 - n$ is $O(n^2)$.</p>
<p>When I use the given definition, I get $n^2 - n \leq n^2 - n^2$ which means that $n^2 -n \leq 0$, which is not true. Is there some step I'm missing?</p>
| an4s | 533,556 | <p>The definition that you have stated is incorrect. $f(n)$ is $\mathcal{O}(g(n))$ $\textit{iff}$ there exists a constant $c > 0$ and there exists a value for n, $n_0 > 0$, such that $|f(n)|\leq c\cdot|g(n)|~\forall n \geq n_0$.</p>
<p>In your case, $f(n) = n^2 - n$ and $g(n) = n^2$. Let's use the definition abo... |
2,389,782 | <p>Other than <a href="http://rads.stackoverflow.com/amzn/click/3885380064" rel="nofollow noreferrer">Engelking General Topology</a>, I also come across other graduate general topology text such as <a href="http://rads.stackoverflow.com/amzn/click/0697068897" rel="nofollow noreferrer">Dugundji</a> and <a href="http://w... | Peter Elias | 392,689 | <p>When looking for a result, Engelking is my first choice. But, as <a href="https://math.stackexchange.com/questions/2265141/is-mathbbr-mathord-sim-hausdorff-if-x-y-x-sim-y-is-a-closed-su/2392314#2392314">my recent experience</a> shows, one should also not miss Bourbaki.</p>
|
2,389,782 | <p>Other than <a href="http://rads.stackoverflow.com/amzn/click/3885380064" rel="nofollow noreferrer">Engelking General Topology</a>, I also come across other graduate general topology text such as <a href="http://rads.stackoverflow.com/amzn/click/0697068897" rel="nofollow noreferrer">Dugundji</a> and <a href="http://w... | Henno Brandsma | 4,280 | <p>I'm a fan of "Handbook of Set-theoretic Topology", which I often use, and Engelking as well. It contains many things that are more recent than Engelking and is more research oriented.</p>
<p>Nagata's "Modern General Topology" is also nice. Also in the Japanese school "Topics in General Topology", by Morita and Naga... |
31,158 | <p>To generate 3D mesh <a href="http://reference.wolfram.com/mathematica/TetGenLink/tutorial/UsingTetGenLink.html#167310445" rel="nofollow noreferrer">TetGen</a> can be easily used. Are there similar functions (or a way to use TetGen) to generate 2d mesh? I know that such functionality can be <a href="https://mathemati... | Mark | 9,085 | <p>You can use TetGen and just make a z axis which is set to 0 (or any other constant). The issue with DelaunayTriangulation (if you want to generate a mesh from a list of points) is that it returns an adjacency list of the edges, which is very hard to turn into the polygons. <a href="https://mathematica.stackexchang... |
2,146,457 | <p>Simplify the ring $\mathbb Z[\sqrt{-13}]/(2)$. I have so far:</p>
<p>$$\mathbb Z[\sqrt{-13}]/(2) \cong \mathbb Z[x]/(2, x^2 + 13) \cong \mathbb Z_2[x] / (x^2 + 1)$$</p>
<p>Now how do I simplify it further? I know that $x^2 + 1 = (x + 1)^2$ in $\mathbb Z_2[x]$, is this useful?</p>
| Mustafa | 400,050 | <p>$x^2+1=0 \Rightarrow x^2=-1 \equiv 1(mod 2) \Rightarrow x^3=x$</p>
<p>$f(x)=a_0+a_1x+a_2x^2+..+a_nx^n=a_0+a_1x+a_2(-1)+a_3(x)+...=a_0+a_1x(mod 2)$</p>
<p>$\Rightarrow \mathbb Z_2[x]/(x^2+1) =\{I,1+I, x+I,(1+x)+I \}; I=(x^2+1)$</p>
|
1,130,487 | <p>Jessica is playing a game where there are 4 blue markers and 6 red markers in a box. She is going to pick 3 markers without replacement.
If she picks all 3 red markers, she will win a total of 500 dollars. If the first marker she picks is red but not all 3 markers are red, she will win a total of 100 dollars. Under ... | agha | 118,032 | <p>Remember that:</p>
<p>$$\int \frac{2}{2x+2}dx \neq \ln(2x+2)$$</p>
<p>But:</p>
<p>$$\int \frac{2}{2x+2}dx = \ln(2x+2)+C$$</p>
<p>where $C$ is constant, so:</p>
<p>$$\int \frac{2}{2x+2}dx = \ln(2x+2)+C=\ln(2(x+1))+C=\ln(x+1)+\ln 2+C=\ln(x+1)+(\ln 2+C)=\ln(x+1)+C_1$$</p>
<p>So both calculations are almost correc... |
386,172 | <p>The expression was simplified in the answer to <a href="https://math.stackexchange.com/questions/384592/finding-markov-chain-transition-matrix-using-mathematical-induction">this question</a>. I'm trying to simplify it but I got stuck. Multiplying all the factors and regrouping didn't work, but maybe I'm doing the wr... | Maazul | 74,820 | <p>Just take $\frac{1}{2}(2p-1)^n$ common and simplify.</p>
<p>$(p)(\frac{1}{2}(2p-1)^n) + (1-p)(-\frac{1}{2}(2p-1)^n)$</p>
<p>Taking $\frac{1}{2}(2p-1)^n$ common, we have</p>
<p>$\frac{1}{2}(2p-1)^n\left(p-(1-p)\right)$</p>
<p>$\frac{1}{2}(2p-1)^n\left(2p-1\right)$</p>
<p>$\frac{1}{2}(2p-1)^{n+1}$</p>
|
3,652,102 | <p>Let, <span class="math-container">$(P,\le)$</span> be the poset.<br>
I have begun to solve this in the following way-
Note that, <span class="math-container">$rs-r\le rs-s\iff r\ge s$</span><br>
So, without loss of generality assume that <span class="math-container">$r\ge s$</span>, then <span class="math-container"... | Clive Newstead | 19,542 | <p>Assume that <span class="math-container">$P$</span> has neither an antichain of size <span class="math-container">$r$</span> nor a chain of size <span class="math-container">$s$</span>.</p>
<p>Let <span class="math-container">$\{ a_1, a_2, \dots, a_k \}$</span> be a maximal antichain in <span class="math-container"... |
3,982,937 | <p>To avoid typos, please see my screen captures below, and the red underline. The question says <span class="math-container">$h \rightarrow 0$</span>, thus why <span class="math-container">$|h|$</span> in the solution? Mustn't that <span class="math-container">$|h|$</span> be <span class="math-container">$h$</span>?</... | Vishu | 751,311 | <p><span class="math-container">$h$</span> might tend to zero from either side, so the statement <span class="math-container">$$-\delta \lt h \lt \delta \iff 0\lt |h| \lt \delta$$</span> is required.</p>
|
802,848 | <p>I am reading this book, <em>Gödel's Proof</em>, by James R. Newman, at location 117 (Kindle), it says,</p>
<blockquote>
<p>For <strong>various reasons</strong>, this axiom, (through a point outside a given line only one parallel to the line can be drawn), did not appear "self-evident" to the ancients.</p>
</block... | Mauro ALLEGRANZA | 108,274 | <p>I think that it is not correct to say that "ancient hate the parallel postulate".</p>
<p>For sure, it is not so "self-evident" as others [but please, think at <a href="http://aleph0.clarku.edu/%7Edjoyce/java/elements/bookI/cn.html" rel="nofollow noreferrer">Common notion</a> n°5 : "The whole... |
876,763 | <p>Let $R$ be a commutative ring with identity. Assume that for any two principal ideals $Ra$ and $Rb$ we have either $Ra\subseteq Rb$ or $Rb\subseteq Ra$. Show that for any two ideals $I$ and $J$ in $R$, we have either $I\subseteq J$ or $J\subseteq I$.</p>
<p>Initially i thought that if i could show that any ideal in... | Community | -1 | <p>Suppose $I\not\subseteq J$ and $J\not\subseteq I$ you have $x\in I\setminus J$ and $y\in J\setminus I$</p>
<p>Now consider Principal ideals $Rx$ and $Ry$</p>
|
2,134,167 | <p>So we finished studying chapter 5 of Rudin on differentiation (Mean value theorem, Taylor's theorem etc) and this was given as a homework problem:</p>
<p>Let $ f(x) $ be continuously differentiable on $ [0, \infty) $ such that $ f $ satisfies $ f'(x) = \cos(x^2)f(x) $ for all $ x \geq 0 $, with $ f(0) = 1 $. Prove ... | ztefelina | 326,591 | <p>$cos(x^2)\in [-1,1]$, so $f'(x)\geq-f(x)$ and $f'(x) \leq f(x)$. Multiplying the first relation with $e^x$ and the second with $e^{-x}$ you get $(f(x)e^x)' \geq 0$ and ($f(x)e^{-x})'\leq 0 $, so $f(x)e^x$ is increasing and $f(x)e^{-x}$ is decreasing. <br>
So $ f(x)e^x \geq f(0)e^0=1$, hence $f(x) \geq e^{-x}$.<br>
A... |
2,134,167 | <p>So we finished studying chapter 5 of Rudin on differentiation (Mean value theorem, Taylor's theorem etc) and this was given as a homework problem:</p>
<p>Let $ f(x) $ be continuously differentiable on $ [0, \infty) $ such that $ f $ satisfies $ f'(x) = \cos(x^2)f(x) $ for all $ x \geq 0 $, with $ f(0) = 1 $. Prove ... | victoria | 412,473 | <p>For all real x, $ -1 \leq $cos$ (x) \leq 1$</p>
<p>so $0 \leq $cos$^2 (x) \leq 1 \ \forall x \in R$</p>
<p>Given $f'(x) = $cos$^2 x \ f(x) \ \ \forall x \geq 0$, then</p>
<p>$|f'(x)| \leq |f(x)| \ \ \forall x\geq 0$ </p>
<p>We know if $f'(x) = f(x) $on$ \ R$ with $f(0) = 1$, then $f(x) = e^x$</p>
<p>Y... |
2,697,729 | <p>Suppose that the probability that you will drop a penny on the ground is 1/5, and the probability that you will find a penny on the ground today is 1/4. If the two events are independent, what is the probability that at least one of the two events will occur?</p>
<p>First I tried simply $1/5+1/4$ which was incorrec... | N. F. Taussig | 173,070 | <blockquote>
<p>First, I tried simply $1/5 + 1/4$, which was incorrect. </p>
</blockquote>
<p>Let's see why. </p>
<p>Suppose we want to find $\Pr(A \cup B)$. </p>
<p><a href="https://i.stack.imgur.com/kNcXq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kNcXq.jpg" alt="Venn_diagram_for_two... |
131,322 | <p>A knot in S^3 is small if its complement does not contain a closed incompressible surface. Is it a generic property for knots, meaning that among all knots with less than $n$ crossings, the proportion of small knots goes to 1 when $n$ goes to infinity?</p>
| Robert Bruner | 6,872 | <p>Pardon me for not editing my previous answer, but this is really a different answer, not an improvement on my previous answer, which addresses different aspects of the (neighborhood of) the question.</p>
<p>The fact that $Z[x]/(fg) \longrightarrow Z[x]/(f) \times Z[x]/(g)$ is not iso here makes the attempt to repla... |
2,957,315 | <p><span class="math-container">$j_{1,1}$</span> denotes the first zero of the first Bessel function of the first kind. (That's a lot of firsts!) It's approximately equal to <span class="math-container">$3.83$</span>. My question is, is there any closed form expression for its value? Even a infinite series or infin... | Geremia | 128,568 | <p>The series expansion is:</p>
<p><span class="math-container">$$J_{1}\left(x\right)=\sum_{n=0}^\infty\frac{x^{2n+1}}{2^{2n+1} n! (n+1)!}$$</span>
<span class="math-container">$$=\frac{x}{2} - \frac{x^{3}}{16} + \frac{x^{5}}{384} - \frac{x^{7}}{18432} + …$$</span></p>
<p>The denominators are Sloan Sequence <a href="... |
259,734 | <p>Basically, what the title says. </p>
<p>Presumably, one could use the fact that monoidal categories (resp. strict monoidal categories) are one-object bicategories (resp. 2-categories) and use the Lack model structure on those, but I am unsure if this would work or not.</p>
| john | 8,751 | <p>There is a model structure on the category of monoidal categories and strict monoidal functors. A strict monoidal functor is a fibration / weak equivalence just when its underlying functor is a fibration / weak equivalence in Cat.</p>
<p>More generally for T a 2-monad with rank on Cat you can lift the model struct... |
1,476,946 | <p>So, I'm just starting to peruse "Categories for the Working Mathematician", and there's one thing I'm uncertain on. Lets say I have three objects, $X,Y,Z$ and two arrows $f,g$ such that $X\overset {f} {\to}Y\overset {g} {\to}Z$. Does this necessitate the composition arrow exist so the diagram commutes, i.e mus... | BrianO | 277,043 | <p>By the category axioms, every category is closed under composition of arrows where it's defined. So if $X\overset {f} {\to}Y\overset {g} {\to}Z$ exist, then the category also contains $X\overset {g \dot f}{\to}Z$.</p>
|
3,430,066 | <p><strong>Question:</strong></p>
<p>Calculate the integral </p>
<p><span class="math-container">$$\int_0^1 \frac{dx}{e^x-e^{-2x}+2}$$</span></p>
<p><strong>Attempted solution:</strong></p>
<p>I initially had two approaches. First was recognizing that the denominator looks like a quadratic equation. Perhaps we can ... | zwim | 399,263 | <p>The change in <span class="math-container">$u=e^x$</span> </p>
<p>leads to a denominator of degree <span class="math-container">$3$</span>:</p>
<p><span class="math-container">$\displaystyle\int\dfrac{u}{u^3+2u^2-1}\mathop{du}=-\frac 12\ln\big|u^2+u-1\big|-\frac{3\sqrt{5}}{5}\tanh^{-1}\left(\frac{\sqrt{5}}5(2u+1)\... |
3,757,864 | <p>Given a diagram like this,
<a href="https://i.stack.imgur.com/Xwum0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xwum0.png" alt="enter image description here" /></a></p>
<p>Where <span class="math-container">$O$</span> is the center and <span class="math-container">$OA = \sqrt{50}$</span>, <spa... | Jaap Scherphuis | 362,967 | <p>Assuming <span class="math-container">$\angle ABC=90^o$</span> is given.</p>
<p>You can get there slightly quicker:<br />
By Pythagoras, <span class="math-container">$|AC|=\sqrt{40}$</span>.<br />
<span class="math-container">$OAC$</span> is isosceles, with <span class="math-container">$|OA|=|OC|=\sqrt{50}$</span>.<... |
83,167 | <p>Let $X$ be a compact complex surface of general type which a ball quotient. Is it true that $\pi_{1}(X)$ can not contain ${\mathbb{Z}}^{2}$ as a subgroup? What kind of infinite abelian groups can occur as a subgroup of $\pi_{1}(X)$?</p>
| Community | -1 | <p>I don't know what a ball quotient is, but the decisive value is the genus of $X$.
If the genus is zero, $X$ is the projective space, if the genus is two, $X$ is a torus and the fundamental group is ${\mathbb Z}^2$ if the genus is $\ge 2$, then $X$ is a quotient of the upper half plane and the fundamental group $\pi$... |
3,464,247 | <p>I'm wondering if my reasoning is justified when determining if a vector is in the span of of a set of vectors.</p>
<p><span class="math-container">$$T = \{(1, 1, 0), (-1, 3, 1)\}$$</span></p>
<p>For which <span class="math-container">$a$</span> is <span class="math-container">$(a^2, a+2, 2) \in span(T)$</span></p>... | Yuki.F | 336,662 | <p>Uh yes, the answer should be <span class="math-container">$DNE$</span> if the <span class="math-container">$a$</span> must be real.</p>
<p>You mentioned that <span class="math-container">$T = \{(1, 1, 0), (-1, 3, 1)\}$</span> and the required vectors is <span class="math-container">$(a^2, a+2, 2)$</span>. Assume th... |
3,290,514 | <p>I need to make the navigation and guidance of a vehicle (a quadcopter) in a platform. This platform can be seen like this:</p>
<p><a href="https://i.stack.imgur.com/jeJ34.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jeJ34.png" alt="enter image description here"></a></p>
<p>where the blue dots... | Claude Leibovici | 82,404 | <p>If I properly uunderstand, you have <span class="math-container">$n$</span> data points <span class="math-container">$(x_i,y_i)$</span> <span class="math-container">$(i=1,2,\cdots,n)$</span> and you want to find the coordinates <span class="math-container">$(a,b)$</span> and the radius <span class="math-container">$... |
1,239,806 | <p>I begin by contradiction. Assume that $n$ can be expressed as the sum of three squares. That is $n = a^2 + b^2 + c^2$. Now since $n \equiv 7 \pmod 8$ then $8 \mid n - 7$ so $8 \mid a^2 + b^2 + c^2 - 7$. But then I don't know how to proceed from here. Any ideas</p>
| Elaqqad | 204,937 | <p>Use the equation modulo $8$ for every integer $x$ we have $x^2\equiv 0,1,4 \mod 8$ $ a^2+b^2+c^2\equiv 7 \mod 8$</p>
<p>Can three numbers from $0,1,4$ sum to $7$? </p>
|
1,239,806 | <p>I begin by contradiction. Assume that $n$ can be expressed as the sum of three squares. That is $n = a^2 + b^2 + c^2$. Now since $n \equiv 7 \pmod 8$ then $8 \mid n - 7$ so $8 \mid a^2 + b^2 + c^2 - 7$. But then I don't know how to proceed from here. Any ideas</p>
| Adhvaitha | 228,265 | <p>This means $n$ is odd. Hence, if $n=a^2+b^2+c^2$, we have two possibilities.</p>
<ol>
<li><p>All of $a,b,c$ are odd. This means that $a^2\equiv1\pmod8$, $b^2\equiv1\pmod8$ and $c^2\equiv1\pmod8$. Hence,
$$n = a^2+b^2+c^2 \equiv 3\pmod8 \not\equiv 7\pmod8$$</p></li>
<li><p>Two of $a,b,c$ are even and other one is od... |
4,112,942 | <p>Found this exercise in Serge Lang's <em>Introduction to Linear Algebra</em>:</p>
<blockquote>
<p>Find the rank of the matrix <span class="math-container">$$\begin{pmatrix} 1 &1 &0 &1 \\ 1 &2 &2 &1 \\ 3 &4 &2 &3 \end{pmatrix}$$</span></p>
</blockquote>
<hr />
<p>So my process to so... | Haf | 911,000 | <p>Thanks to the comments I found that I just have to reduce it to row echelon form, I guess I misunderstood ranks, the answer is</p>
<p><span class="math-container">$$\begin{aligned}
\begin{pmatrix}
1 &1 &0 &1 \\
1 &2 &2 &1 \\
3 &4 &2 &3 \end{pmatrix} &\overset{(2... |
4,112,942 | <p>Found this exercise in Serge Lang's <em>Introduction to Linear Algebra</em>:</p>
<blockquote>
<p>Find the rank of the matrix <span class="math-container">$$\begin{pmatrix} 1 &1 &0 &1 \\ 1 &2 &2 &1 \\ 3 &4 &2 &3 \end{pmatrix}$$</span></p>
</blockquote>
<hr />
<p>So my process to so... | Community | -1 | <p>The first two rows are independent (by inspection: they're not multiples of each other).</p>
<p>Then <span class="math-container">$r_3=2r_1+r_2$</span>, as you can see.</p>
<p>So the row rank is <span class="math-container">$2$</span>. So the rank is <span class="math-container">$2$</span>.</p>
|
4,515 | <p>I've been using the sentence:</p>
<blockquote>
<p>If a series converges then the limit of the sequence is zero</p>
</blockquote>
<p>as a criterion to prove that a series diverges (when $\lim \neq 0$) and I can understand the rationale behind it, but I can't find a <strong>formal proof</strong>.</p>
<p>Can you h... | Gadi A | 1,818 | <p>If we know that the sequence converges and merely wish to show it converges to zero, then a proof by contradiction gives a little more intuition here (although the direct proofs are simple and beautiful). Assume $a_n\to a$ with $a>0$, then for all $n>N$ for some large enough $N$ we have $a_n > a/2$ (take $\varepsilo... |
463,239 | <p>Integrate $$\int{x^2(8x^3+27)^{2/3}}dx$$</p>
<p>I'm just wondering, what should I make $u$ equal to?</p>
<p>I tried to make $u=8x^3$, but it's not working. </p>
<p>Can I see a detailed answer?</p>
| GeoffDS | 8,671 | <p>Now, $u = 8x^3 + 27$ is a choice that makes less work for you. But, one thing that students should understand is even if you don't make the best choice, it still might work. You tried $u = 8x^3$. In that case $du = 24x^2 \,dx$ and your integral becomes
$$\int{x^2(8x^3+27)^{2/3}}dx = \frac{1}{24} \int (u+27)^{2/3}... |
877,646 | <p>Friends,I have a set of matrices of dimension $3\times3$ called $A_i$. ,</p>
<p>Following are the given conditions</p>
<p>a) each $A_i$ is non invertible <strong>except $A_0$</strong> because their determinant is zero.</p>
<p>b) $\sum_{n=0}^\infty A_i$ is invertible and determinant is not zero</p>
<p>c) </p>
<... | Urgje | 95,681 | <p>Write the sin-function as the difference of two exponentials and obtain the corresponding integrals by contour integration. This gives the result:
\begin{eqnarray*}
I &=&\int_{-\infty }^{+\infty }dx\frac{1}{x-i}\sin x=\int_{-\infty
}^{+\infty }dx\frac{1}{x-i}\frac{1}{2i}\left\{ e^{ix}-e^{-ix}\right\} \\
\in... |
1,380,697 | <p>I am currently an undergraduate and thinking about applying to graduate school for math. The problem is that I don't know what field I want to go. Taking graduate classes even more confuse me because the more I learn the less I know what specifically I want to do. My question is to where to find an information about... | Sinister Cutlass | 235,860 | <p>If you are in the U.S., you could try attending AMS Sectional Meetings. These sorts of conferences happen often, and they feature talks in a huge number of different topics in current mathematics. Otherwise, I'm sure that in your locale, there are probably analogous conferences.</p>
<p>Also, you could sign up for... |
1,634,520 | <p>I <a href="https://math.stackexchange.com/questions/1632455/mathematical-meaning-of-certain-integrals-in-physics/1633320#1633320">have been told</a> that the Helmholtz decomposition theorem says that</p>
<blockquote>
<p>every <em>smooth</em> vector field $\boldsymbol{F}$ [where I am not sure what precise assumpti... | Self-teaching worker | 111,138 | <p>Let $\boldsymbol{F}:\mathbb{R}^3\to \mathbb{R}^3$, $\boldsymbol{F}\in C^2( \mathbb{R}^3)$ compactly supported and let $V\subset\mathbb{R}^3$ be a region satisfying the hypothesis of Gauss's divergence theorem, with $\boldsymbol{x}\in \mathring{V}$ contained in its interior.</p>
<p>By using <a href="https://math.st... |
1,910,109 | <p>$$\int \frac{1}{\sqrt{x} (1 - 3\sqrt{x})}$$</p>
<p>I tried with the substitution $u = 1-3\sqrt{x}$</p>
<p>I am confused with how to finish this problem I know I am supposed to substitute $u$ and $\text{d}u$ in but I am not sure how to finish it.</p>
| windircurse | 331,921 | <p><strong>Hint:</strong> Try $u=\sqrt x$, $du=\frac{1}{2\sqrt x}$</p>
|
1,057,819 | <p>The number $128$ can be written as $2^n$ with integer $n$, and so can its every individual digit. Is this the only number with this property, apart from the one-digit numbers $1$, $2$, $4$ and $8$? </p>
<p>I have checked a lot, but I don't know how to prove or disprove it. </p>
| Michal Paszkiewicz | 177,915 | <p>No value less than 2^30,000 bigger than 128 follows this pattern, as I have found through this tool:</p>
<p><a href="http://www.michalpaszkiewicz.co.uk/maths/series/powers-of-two.html" rel="nofollow">http://www.michalpaszkiewicz.co.uk/maths/series/powers-of-two.html</a></p>
<p>You can use this tool to try and find... |
1,057,819 | <p>The number $128$ can be written as $2^n$ with integer $n$, and so can its every individual digit. Is this the only number with this property, apart from the one-digit numbers $1$, $2$, $4$ and $8$? </p>
<p>I have checked a lot, but I don't know how to prove or disprove it. </p>
| Fractalic | 197,946 | <p>Here's some empirical evidence (not a proof!).</p>
<p>Here are the first powers which increase the length of 1-2-4-8 runs in the least significant digits (last digits):</p>
<pre>
Power 0: 1 digits. ...0:1
Power 7: 3 digits. ...0:128
Power 18: 4 digits. ...6:2144
Power 19: 5 digits. ...5:24288
Power 90: 6 digits. .... |
345,065 | <p>Let f and g be functions of one real variable and define $F(x,y)=f[x+g(y)]$. Find formulas for all the partial derivatives of F of first and second order.</p>
<p>For the first order, I think we have:</p>
<p>$\frac{\partial F}{\partial x}=\frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}$</p>
<p>$\frac{... | user27182 | 22,020 | <p>you need to differentiate $f$ by its argument, then differentiate the argument by $x$ or $y$</p>
<p>Setting $\xi = x + g(y)$,</p>
<p>$\frac{d F}{d x}=\frac{df}{d \xi} \frac{d \xi}{dx} = \frac{df}{d \xi}$</p>
<p>and </p>
<p>$\frac{d F}{d y}=\frac{df}{d \xi} \frac{d \xi}{dy} = \frac{df}{d \xi} g'(y)$</p>
<p>You s... |
4,622,956 | <p>I think <span class="math-container">$\,9\!\cdot\!10^n+4\,$</span> can be a perfect square, since it is <span class="math-container">$0 \pmod 4$</span> (a quadratic residue modulo <span class="math-container">$4$</span>), and <span class="math-container">$1 \pmod 3$</span> (also a quadratic residue modulo <span cla... | sirous | 346,566 | <p>Comment:</p>
<p>Lets try construction such a number. Suppose we have:</p>
<p><span class="math-container">$k^2-9\times 10^n=4$</span></p>
<p>We use following known Pell's equation:</p>
<p><span class="math-container">$x^2-Dy^1=1$</span></p>
<p>For <span class="math-container">$D=10$</span> we have <span class="math-... |
831,763 | <p>The following equation $$e^{i+z}e^{iz}=1$$ is to be solved for $z$. I have tried
$$
\begin{eqnarray}
e^{i+z+iz} = 1\\
i+z+iz=0\\
z= -{i \over 1+i} = -{i(1+i)\over 2} = \frac12-i\frac12
\end{eqnarray}
$$
However I am absolutely unsure, that's correct. Somehow I suspect trigonometry should creep in the answer.</p>
| Rene Schipperus | 149,912 | <p>You have $$e^{i+(1+i)z}=1$$ this means that
$$i+(1+i)z=2\pi i n$$</p>
<p>So $$z=\frac{(2\pi n -1)i}{1+i}=\left(\pi n -\frac{1}{2}\right) (1+i)$$</p>
|
2,904,603 | <p>I'm working on the following question:</p>
<blockquote>
<p>Show that $G$ is a group if and only if, for every $a, b \in G$,
the equations $xa = b$ and $ay = b$ have solutions $x, y \in G$.</p>
</blockquote>
<p>I'm having trouble getting started because I'm not understanding what it means for "the equations $xa... | Robert Lewis | 67,071 | <p>I think our colleagues stressed out and ACTOH have made clear what we're looking for here. </p>
<p>I would phrase it like this: </p>
<p>Let $G \ne \emptyset$ be a set with an associative binary operation; then $G$ is a group if and only if for any $a, b \in G$ there exist $x, y \in G$ such that $xa = b$ and $ay ... |
3,085,591 | <p>I'm trying to prepare for an exam and came across the following question:</p>
<blockquote>
<p>Given <span class="math-container">$n \times n$</span> matrix <em>A</em>, let <em>U</em> represent row space and <em>W</em>
represent column space. </p>
<p>A) Prove: <span class="math-container">$W \subseteq$</sp... | P Vanchinathan | 28,915 | <p>Let us regard a matrix as a linear transformation. </p>
<p><span class="math-container">$A^2=0$</span> means Range of <span class="math-container">$A$</span> is contained in kernel of <span class="math-container">$A$</span>.
(<span class="math-container">$A^2v=0$</span> means <span class="math-container">$A(Av)=0$... |
2,533,960 | <p>So, the given expression is
$$\binom{2n}{2} = 2\binom{n}{2}+n^2$$</p>
<p>The task is to give a combinatorical proof for it.</p>
<p>Left side of the identity is obviously equal to the number of options for choosing 2 elements out of the set with cardinality $2n$.</p>
<p>What issues me is that I can't think of any ... | Especially Lime | 341,019 | <p>Hint: split the set into two halves and consider whether the selected elements are in the same half.</p>
|
66,068 | <p>I have a list like this. </p>
<pre><code>cdatalist = {{1., 0.898785, Failed, Failed, 50., 25., "serial"}, {1., 1.31175,1., Failed, 50., 25., "serial"}, {1., 18.8025, Failed, 0.490235, 50., 25., "serial"}, {1., 19.6628, 0.990079, Failed, 50., 25., "serial"}, {1., 39.547, Failed, Failed, 50., 25., "serial"}, {1., 39.... | alancalvitti | 801 | <pre><code>cdatalist // Dataset // Query[Select[#[[3]] != "Failed" &], 1 ;; 3] // Normal
</code></pre>
|
2,252,090 | <p>Someone posed this question to me on a forum, and I have yet to figure it out. If $a,b,c,d$ are the zeroes of:</p>
<p>$$x^4-7x^3+2x^2+5x-1=0$$
Then what is the value of $$ \frac1a +\frac1b +\frac1c +\frac1d $$</p>
<p>I can figure out the zeroes, but they are wildly complex. I'm sure there must be an easier way. </... | Semiclassical | 137,524 | <p>Hint: If $f(x)$ has roots $a,b,c,d$ then $f(x)=(x-a)(x-b)(x-c)(x-d)$. Expand this out and compare coefficients with the given quartic. (In particular, the coefficients of $x^3$ and $x^0$.)</p>
|
998,769 | <p>A random variable $X$ distributed over the interval $[0, 2\pi]$</p>
<p>a) the pdf of $X$</p>
<p>b) the cdf of $X$</p>
<p>c) $P(\frac{\pi}{6} \leq X \leq \frac{\pi}{2})$</p>
<p>d) $P(-\frac{\pi}{6} \leq X \leq \frac{\pi}{2})$</p>
<p>my answers:</p>
<p>a) pdf of $X$ is $f(x) = \begin{cases}\frac{1}{2\pi},& 0... | Rey | 73,712 | <p>You want to go from the definition for that one too. For <a href="http://en.wikipedia.org/wiki/Surjective_function" rel="nofollow">surjective</a> you need to show:
$ \forall y\in R , \; \exists x\in R, f(x)=y$ </p>
<p>For example for $h$, we can show that for each $y$, we can set $x = sign(y) \sqrt{y}$, and then w... |
1,579,616 | <p>So I know it's true for $n = 5$ and assumed true for some $n = k$ where $k$ is an interger greater than or equal to $5$.</p>
<p>for $n = k + 1$ I get into a bit of a kerfuffle.</p>
<p>I get down to $(k+1)^2 + 1 < 2^k + 2^k$ or equivalently:</p>
<p>$(k + 1)^2 + 1 < 2^k * 2$.</p>
<p>A bit stuck at how to pro... | barak manos | 131,263 | <p><strong>First, show that this is true for $n=5$:</strong></p>
<p>$5^2+1<2^5$</p>
<p><strong>Second, assume that this is true for $n$:</strong></p>
<p>$n^2+1<2^n$</p>
<p><strong>Third, prove that this is true for $n+1$:</strong></p>
<p>$(n+1)^2+1=$</p>
<p>$n^2+\color\green{2}\cdot{n}+1+1<$</p>
<p>$n^2... |
4,059,489 | <blockquote>
<p>Let <span class="math-container">$ A, B \in M_n (\mathbb{C})$</span> such that <span class="math-container">$(A-B)^2 = A -B$</span>. Then <span class="math-container">$\mathrm{rank}(A^2 - B^2) \geq \mathrm{rank}( AB -BA)$</span>.</p>
</blockquote>
<p>I tried to apply the basic inequalities without resul... | Aderinsola Joshua | 395,530 | <p>The easiest but absurd way to do this is by squaring and squaring and squaring and..... To write out the equation of polynomial in <span class="math-container">$x$</span></p>
<p><span class="math-container">$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}-\sqrt{\frac{x-1388}{8}}-\sqrt{\frac{... |
160,161 | <p>Is there an unlabeled locally-finite graph which is a Cayley graph of an infinitely many non-isomorphic groups with respect to suitably chosen generating sets?</p>
| YCor | 14,094 | <p>Yes. You can even have uncountably many. One recipe is as follows: consider a group $G$ with finite generating subset $S$ and an extension $1\to F\to G'\stackrel{\pi}\to G\to 1$, with $F$ finite. Endow $G'$ with the generating subset $S'=\pi^{-1}(S)$. Then the Cayley graph of $(G',S')$ only depends on the Cayley gra... |
160,161 | <p>Is there an unlabeled locally-finite graph which is a Cayley graph of an infinitely many non-isomorphic groups with respect to suitably chosen generating sets?</p>
| AGenevois | 122,026 | <p>Here is a related construction, but which I find more elementary. The idea is to start from the lamplighter group
<span class="math-container">$$L_2 := \langle a,t \mid a^2=1, [t^nat^{-n},a]=1 \ (n \in \mathbb{N}) \rangle,$$</span>
to fix an arbitrary subset <span class="math-container">$I \subset \mathbb{N}$</span>... |
3,166,999 | <p>I'm reading Kechris' book "Classical Descriptive Set Theory" and the author gives the following definition (pp. <span class="math-container">$49$</span>, row <span class="math-container">$3$</span>):</p>
<blockquote>
<p>A <strong>weak basis</strong> of a topological space <span class="math-container">$X$</span> ... | William Elliot | 426,203 | <p>A base covers the space.<br>
A weak base may not cover the space.<br>
The set of all not empty, open subsets of R that exclude 0 is a weak base. </p>
|
2,150,085 | <p>I've always been puzzled by the value $0^0$. I remembered that one of my professor claimed that it was truely equal to $1$. However I think that most people would say, from an analytic point of view, that it is indeterminate which I agree with. Translating $0^0$ into $e^{0 \ln(0)}$ makes the problem appear explicitl... | M. Winter | 415,941 | <p>I have good and bad news for you. The good news: in the most applied scenarios, $z^z$ indeed converges to $1$ as $z$ converges $0$. The bad news: it is not so easy!</p>
<p>In your question you demonstrated, that approaching the zero on straight lines gives you $1$. But you should not forget the context in which you... |
2,820,464 | <p>Find the limit of the sequence {$a_{n}$}, given by$$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=\dfrac {1}{3}(1+a_{n}+a^{3}_{n-1}), \ for \ n \ > \ 1$$</p>
<p>My try:</p>
<p>$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54$ that is the sequence is incresing and each term is positive. Let the limit of the s... | Tsemo Aristide | 280,301 | <p>Because $f(x)={1\over 3}(1+x+x^3)$ is a strictly increasing function, compute its derivative, show recursively that that $0<a_n$, remark that if $a_{n-1},a_{n-2}$ are strictly inferior to $1$, $a_n\geq f(max(a_{n-1},a_{n-2}))$ and $f(max(a_{n-1},a_{n-2}))<f(1)=1$. _{n-1})$.</p>
|
2,714,450 | <p>Suppose $A$ and $B$ are two square matrices so that $e^{At}=e^{Bt}$ for infinite (countable or uncountable) values of $t$ where $t$ is positive.</p>
<p>Do you think that $A$ <strong>has to be equal to</strong> $B$?</p>
<p>Thanks,
Trung Dung.</p>
<hr>
<p>Maybe I do not state clearly or correctly.</p>
<p>I mean t... | Jens Schwaiger | 532,419 | <p>Note that $t\mapsto e^{tA}-e^{tB}$ is a matrix of power series $c_{ij}(t)=\sum_{l=0}^\infty c_{ij}^l t^{(l)}$ with $\infty$ as its radius of convergence.
By <a href="https://math.stackexchange.com/questions/1577032/identity-theorem-for-power-series">Identity theorem for power series</a> these series are identical ze... |
2,057,813 | <p>How can I prove that for $s \in \mathbb{C}$, with real part of $s$ being equal to 1,
\begin{equation}
\sum_{n=1}^{\infty}\frac{1}{n^{s}}
\end{equation}
diverges?</p>
<p>Thanks a lot!</p>
| reuns | 276,986 | <p>$$\left|n^{-s}-\int_n^{n+1} x^{-s}dx\right| = \left|\int_n^{n+1} \int_n^x s t^{-s-1}dtdx\right| <\int_n^{n+1} \int_n^x |s \, n^{-s-1}|dtdx = \left|\frac{s}{2}n^{-s-1}\right|$$</p>
<p>therefore</p>
<p>$$\left|\sum_{n=1}^{N-1} n^{-s}-\frac{1-N^{1-s}}{s-1}\right| = \left|\sum_{n=1}^{N-1} n^{-s}-\int_n^{n+1} x^{-s... |
2,057,813 | <p>How can I prove that for $s \in \mathbb{C}$, with real part of $s$ being equal to 1,
\begin{equation}
\sum_{n=1}^{\infty}\frac{1}{n^{s}}
\end{equation}
diverges?</p>
<p>Thanks a lot!</p>
| robjohn | 13,854 | <p>Using formula $(10)$ from <a href="https://math.stackexchange.com/a/2027588">this answer</a>,
$$
\zeta(s)=\lim_{n\to\infty}\left[\sum_{k=1}^n\frac1{k^s}-\frac1{1-s}n^{1-s}+\frac12n^{-s}\right]
$$
converges for $\mathrm{Re}(s)\gt-1$.</p>
<p>For $s=1+it$, where $t\in\mathbb{R}$,
$$
\sum_{k=1}^n\frac1{k^s}=\zeta(s)-\f... |
927,188 | <p>This question has been on my mind for a very long time, and I thought I'd finally ask it here. </p>
<p>When I was 6, my dad pulled me out of school. The classes were too easy; the professors, too dull. My father had been man of philosophy his entire life (almost got a PhD in it) and regretted not having a more q... | Luke Willis | 87,726 | <p>Computer Science with Math is a wonderful combination that opens up a lot of opportunities. You could very easily do something for either path and use a fair bit from both along the way.</p>
<p>I had a similar experience with math growing up, though not quite to the degree that you expressed. I don't think I am one... |
2,207,848 | <p>I'm not very familiar with contraposition and so I am having some difficulties proving the statement. </p>
<blockquote>
<p>If $n$ is a positive integer such that $n \equiv 2 \pmod{4}$ or $n \equiv 3 \pmod{4}$, then $n$ is not a perfect square.</p>
</blockquote>
<p>What would be a good way to prove this?<br>
Need... | Louis | 393,349 | <p>To prove $A\implies B$ by contrapositive you need to prove $\bar B \implies \bar A$. Let's prove that if $n$ is a perfect square then $n \equiv 2 [ 4]$ or $n \equiv 3 [4]$ is not true.</p>
<p>Suppose $n$ is a perfect square, say $n=k^2, k\in \mathbb Z$ and suppose $k \equiv 3[4]$ then $n=k^2 \equiv 3 [4]$ which mea... |
3,383,206 | <p><strong>Question</strong>: Can <span class="math-container">$\int_0^\infty \frac{\sqrt{x}}{(1+x)^2} dx$</span> be computed with residue calculus?</p>
<p>The integral comes from computing <span class="math-container">$\mathbb{E}(\sqrt{X})$</span> where <span class="math-container">$X=U/(1-U)$</span> and <span class=... | José Carlos Santos | 446,262 | <p>If <span class="math-container">$x\in\mathbb R\setminus\mathbb N$</span>, let <span class="math-container">$r$</span> be the distance from <span class="math-container">$x$</span> to the closest natural number. Then <span class="math-container">$(x-r,x+r)\cap\mathbb N=\emptyset$</span> and <span class="math-containe... |
1,216,392 | <blockquote>
<p>$P,Q$ are polynomials with real coefficients and for every real $x$ satisfy $P(P(P(x)))=Q(Q(Q(x)))$. Prove that $P=Q$.</p>
</blockquote>
<p>I see only that these polynomials are same degree</p>
| zhw. | 228,045 | <p>Slade, I had the same approach but I'm not sure how you got the constant terms equal so quickly. But if the degree is $1$ that part is easy and you're done. Otherwise, if $p\ne q,$ then $|p_2(x)-q_2(x)|$ blasts off to $\infty$ as $x\to \infty.$ Because the leading coefficients of $p$ and $q$ agree, this would imply ... |
2,088,487 | <p>If $α, β$ are the roots of the equations $x^2 +px+1=0$, and $γ, δ$ are the roots of the equations $x^2+qx+1=0$, then
$( α-γ)(β+ δ)( δ+ α)( β- γ) = $ ?</p>
| Community | -1 | <p>We have $$P= (α - γ)(β + δ) (β - γ)(α + δ)
= (αβ + αδ - βγ - γδ) (αβ + βδ - αγ - γδ)$$</p>
<p>We know that $αβ = γδ = 1$. Using these, we'll have</p>
<p>$$P = (1 + αδ - βγ - 1) (1 + βδ - αγ - 1)$$
$$ = (αδ - βγ) (βδ - αγ)$$
$$= αβδ^2 - γδα^2 - γδβ^2 + αβγ^2$$</p>
<p>Again, using $αβ = γδ = 1$, we get,</p>
<p>$$P... |
2,088,487 | <p>If $α, β$ are the roots of the equations $x^2 +px+1=0$, and $γ, δ$ are the roots of the equations $x^2+qx+1=0$, then
$( α-γ)(β+ δ)( δ+ α)( β- γ) = $ ?</p>
| Hari Shankar | 351,559 | <p>Write the product as $(\gamma - \alpha)(\gamma - \beta)(-\delta - \alpha)(-\delta - \beta) = P(\gamma) P(-\delta)$</p>
<p>$P(\gamma) = x^2+p \gamma+1 = p\gamma - q\gamma = \gamma (p-q)$ since $x^2+1 + q \gamma = 0$</p>
<p>Similarly $P(-\delta) = (q+p) \delta$</p>
<p>Hence $P(\gamma) P(\delta) = (q^2-p^2)\gamma \d... |
2,227,280 | <p>For every positive number there exists a corresponding negative number. Would that imply that the number of positive numbers is "equal" to the number of negative numbers? (Are they incomparable because they both approach infinity?)</p>
| badjohn | 332,763 | <p>The word "infinity" is used in many places in maths and the definitions are not necessarily the same. Different symbols are used but there are still more definitions than symbols. </p>
<p>The most familiar symbol for infinity, $\infty$, is commonly used in calculus but it is more of a suggestive shorthand than an... |
2,894,606 | <p>Please, could you help me with the question below.
Demonstrate that O(log n^k) = O(log n):</p>
| Claude Leibovici | 82,404 | <p>Except for very few specific cases, you cannot get explicit solutions and you need numerical methods.</p>
<p>Consider that you look for the zero(s) of function
$$f(B)=B^S-\frac{B-1}{R}-1$$
$$f'(B)=S B^{S-1}-\frac{1}{R}$$
$$f''(B)=(S-1) S B^{S-2}$$
The first derivative cancels at a point
$$B_*=\left(\frac{1}{R S}\ri... |
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