qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
2,265,782 | <p>Number of twenty one digit numbers such that Product of the digits is divisible by $21$</p>
<p>Since product is divisible by $21$ the number should contain the digits $3,6,7,9$ But i am unable to decide how to proceed...can i have any hint</p>
| amWhy | 9,003 | <p>Yes, your initial work is correct:</p>
<p>$$(p\land q)\to p \equiv \lnot (p\land q)\lor p\tag{implication}$$</p>
<p>$$\equiv (\lnot p \lor \lnot q) \lor p\tag{DeMorgan's rule}$$</p>
<hr>
<p>Now using the properties of associativity and commutativity of the $\lor$-operator, we have:</p>
<p>$$\equiv \lnot p \lor ... |
2,691,232 | <p>Let $E$ be a complex Hilbert space.</p>
<blockquote>
<p>I look for an example of $A,B\in \mathcal{L}(E)$ such that $A\neq 0$ and $B\neq 0$ but $AB=0$.</p>
</blockquote>
| José Carlos Santos | 446,262 | <p>In $\ell^2(\mathbb{C})$, define$$A(x_1,x_2,x_3,\ldots)=(x_1,0,x_3,0,x_5,0,\ldots)\text{ and }B(x_1,x_2,x_3,\ldots)=(0,x_2,0,x_4,0,x_6,\ldots).$$</p>
|
1,959,949 | <blockquote>
<p>We introduce new variables as
$\begin{cases}\xi:=x+ct\\\eta:=x-ct\end{cases}
$
which implies that
$
\begin{cases} \partial_ x=\partial_\xi+\partial_\eta\\\partial_t=c\partial_\xi+c\partial_\eta
\end{cases}
$</p>
</blockquote>
<p>This is from page 34 of <em>Partial Differential Equation --- an ... | avs | 353,141 | <p>Use your equations to solve for $x$ and for $t$ in terms of $\xi, \eta$.</p>
|
1,237,077 | <p>For a periodic function we have: $$\int_{b}^{b+a}f(t)dt = \int_{b}^{na}f(t)dt+\int_{na}^{b+a}f(t)dt = \int_{b+a}^{(n+1)a}f(t)dt+\int_{an}^{b+a}f(t)dt = \int_{na}^{(n+1)a}f(t)dt = \int_{0}^{a}f(t)dt.$$ , but I don't understand how we obtain $\int _{b+a}^{\left(n+1\right)a}\:f\left(t\right)\:dt=\int _b^{na}\:f\left(t\... | mathifold.org | 231,554 | <p>If $a$ is the length of one period, then $f(t)=f(t-a)$ and so on. Then</p>
<p>$\int_{b+a}^{na+a}f(t)dt=\int_{b+a}^{na+a}f(t-a)dt=\text{(take $u=t-a$)}=\int_{b}^{na}f(u)du$</p>
|
8,997 | <p>I have a set of data points in two columns in a spreadsheet (OpenOffice Calc):</p>
<p><img src="https://i.stack.imgur.com/IPNz9.png" alt="enter image description here"></p>
<p>I would like to get these into <em>Mathematica</em> in this format:</p>
<pre><code>data = {{1, 3.3}, {2, 5.6}, {3, 7.1}, {4, 11.4}, {5, 14... | Sjoerd C. de Vries | 57 | <p>This imports a whole sheet:</p>
<pre><code>Grid[#, Dividers -> All] &@
Import["http://joliclic.free.fr/html/object-tag/en/data/test.sxc", {"Data", 1}]
</code></pre>
<p><img src="https://i.stack.imgur.com/CE4S2.png" alt="Mathematica graphics"></p>
<p>And this imports three contiguous rows and two non-conti... |
786,086 | <p>For my research I am working with approximations to functions which I then integrate or differentiate and I am wondering how this affects the order of approximation.</p>
<p>Consider as a minimal example the case of $e^x$ for which integration and differentiation doesn't change anything. Now if I would approximate t... | Urgje | 95,681 | <p>Suppose that $f(x)$ has a Taylor series expansion about $x=0$ with a radius
of convergence $r>0$. For convenience we set $f(0)=1$.</p>
<p>We write
$$
f(x)=1+xf^{(1)}(0)+\frac{x^{2}}{2}f^{(2)}(0)+\mathcal{O}%
(x^{3})=1+xf^{(1)}(0)+\frac{x^{2}}{2}f^{(2)}(0)+g(x),
$$
where, in a neighbourhood of $0$,
$$
|x^{-3}g(x)... |
1,109,552 | <p>So the Norm for an element $\alpha = a + b\sqrt{-5}$ in $\mathbb{Z}[\sqrt{-5}]$ is defined as $N(\alpha) = a^2 + 5b^2$ and so i argue by contradiction assume there exists $\alpha$ such that $N(\alpha) = 2$ and so $a^2+5b^2 = 2$ , however, since $b^2$ and $a^2$ are both positive integers then $b=0$ and $a=\sqrt{2}$ h... | azimut | 61,691 | <p>Yes, you are on the right track. All your reasoning makes sense to me.</p>
<p><strong>On your question about the associates</strong></p>
<p>By the properties of the norm, associates have the same norm. So the only possible associates in your list are $1 + \sqrt{-5}$ and $1 - \sqrt{-5}$.</p>
<p>Now determine all u... |
3,058,139 | <p>Let us consider the statement <span class="math-container">$\exists x P(x)$</span> - translated into English, "there exists an <span class="math-container">$x$</span> in our universe of discourse such that <span class="math-container">$P(x)$</span> is true." In writing the negation of this, we are taught to switch q... | hmakholm left over Monica | 14,366 | <p>Which form to consider simpler is basically a matter of taste and convention.</p>
<p>There are some accounts of predicate logic that consider <span class="math-container">$\exists$</span> the only primitive quantifier and treat <span class="math-container">$\forall x\,\varphi$</span> as an abbreviation for <span cl... |
2,117,420 | <p>We all know that particular solution of $A_{n} = A_{(n-1)} + f(n)$</p>
<p>where $f(n)=n^c$ , c is a random positive integer.</p>
<p>Can be set to $(n^c+n^{(c-1)}+.....+1)$</p>
<p>But what about when $c\lt0$?</p>
<p>How do we find a particular solution of the form:</p>
<blockquote>
<p>$A_{n} = A_{(n-1)} + f(n)... | Mark Fischler | 150,362 | <p><strong>HINT</strong></p>
<p>You know how to diagonalize a matrix, that is, to find a diagonal matrix $D$ and an orthogonal matrix $P$ (which will be the matrix whose columns are the eigenvectors of $M$) such that
$$
M = PDP^{-1}
$$
($D$ will have the eigenvalues of $M$ on the diagonal elements.)</p>
<p>Then for ... |
136,021 | <p>There is an equivalence relation between inclusion of finite groups coming from the world of <a href="http://en.wikipedia.org/wiki/Subfactor" rel="noreferrer">subfactors</a>:</p>
<p><strong>Definition</strong>: <span class="math-container">$(H_{1} \subset G_{1}) \sim(H_{2} \subset G_{2})$</span> if <span class="ma... | Sebastien Palcoux | 34,538 | <blockquote>
<p>We give here an easy and purely group-theoretic sufficient condition (but not necessary) : </p>
</blockquote>
<p><strong>Definition</strong> : Let $\sim_2$ be the equivalence relation on inclusions of finite groups, defined by :<br>
$(A \subset B) \sim_2 (C \subset D)$ if $(A/A_B \subset B/A_B) \... |
1,537,881 | <p>Find the values of $a$ and $b$ if $$ \lim_{x\to0} \dfrac{x(1+a \cos(x))-b \sin(x)}{x^3} = 1 $$
I think i should use L'Hôpital's rule but it did not work.</p>
| parsiad | 64,601 | <p>The easiest way (in my opinion) is to plug in the power series expansion of $x(1+a\cos x)-b\sin x$ around zero. Then, the limit becomes</p>
<p>$$\lim_{x\rightarrow 0} \frac{x(a-b+1)+x^3(b-3a)/6+O(x^5)}{x^3}=1.$$</p>
<p>Now you have two equations involving $a$ and $b$ to satisfy (can you figure out what those equat... |
2,409,183 | <p>Good evening all! I'm trying to find the eigenvalues and eigenvectors of the following problem</p>
<p>$$
\begin{bmatrix}
-10 & 8\\
-18 & 14\\
\end{bmatrix}*\begin{bmatrix}
x_{1}\\
x_{2}\\
\end{bmatrix}
$$</p>
<p>I've found that $λ_{1},_{2}=2$ where $... | yeahyeah | 233,771 | <p>Let me show you the universal trick in computing any integral of the form
<span class="math-container">$$
I_{k_1...k_n}= \int d \Omega \, \hat r_{k_1}...\hat r_{k_n}.
$$</span></p>
<p>Just notice that (without being too formal) <span class="math-container">$I_{k_1...k_n}$</span> can be obtained by taking derivativ... |
48,989 | <p>How to prove $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))$?</p>
| xenon | 12,426 | <p>I used a way to prove this, which I thought may not be the most concise way but it feels very intuitive to me.
The matrix $AB$ is actually a matrix that consist the linear combination of $A$ with $B$ the multipliers. So it looks like...
$$\boldsymbol{AB}=\begin{bmatrix}
& & & \\
a_1 & a_2 &... |
772,391 | <p>The formula for the Chi-Square test statistic is the following:</p>
<p>$\chi^2 = \sum_{i=1}^{n} \frac{(O_i - E_i)^2}{E_i}$</p>
<p>where O - is observed data, and E - is expected.</p>
<p>I'm curious why it depends on the absolute values? For example, if we change the units we're measuring we'll get a different sta... | Beojan | 531,438 | <p>The definition of $\chi^2$ you're using is for comparing <em>frequencies</em>, not measurements with units. In the later case, you divide by the square of the error, not the value itself.</p>
|
2,541,044 | <p>I read this argument on the internet about how the solution to the sleeping beauty problem is $\frac{1}{3}$:</p>
<p>All these events are equally likely in the experiment : </p>
<ol>
<li><p>Coin landed Heads, it's Monday and Beauty is awake</p></li>
<li><p>Coin landed Heads, it's Tuesday and Beauty is asleep</p></l... | Qiaochu Yuan | 232 | <p>What's happening in the sleeping beauty problem is much worse than what's happening in the Monty Hall problem. I claim that the solution to the sleeping beauty problem is that in a world where things like the sleeping beauty problem happen to you, there is no such thing as probability. </p>
<p>One way to cash out w... |
735,470 | <p>I am having trouble with integrating the following:</p>
<p>$$\int \frac{\cos2x}{1-\cos4x}\mathrm{d}x$$</p>
<p>I have simplified it using the double angle: </p>
<p>$$\int \frac{1-2\sin^2x}{1-\cos4x}\mathrm{d}x$$</p>
<p>But i am stuck as I am not sure on how to continue on from here. Should i use the double angle ... | Alijah Ahmed | 124,032 | <p>Use the double angle on the $\cos 4x$ term, so you obtain $\cos 4x=1-2\sin^22x$, rather than using the double angle identity on $\cos 2x$. </p>
<p>So your integral will simplify to
$$\int\frac{\cos 2x}{1-\cos 4x}dx=\int\frac{\cos 2x}{1-(1-2\sin^22x)}dx=\frac{1}{2}\int\frac{\cos 2x}{\sin^22x}dx$$</p>
<p>Then use t... |
3,444,556 | <blockquote>
<p>Let <span class="math-container">$f:[-1,1] \to \mathbb{R}$</span> be continuous on <span class="math-container">$[-1,1]$</span>.</p>
<p>Assume <span class="math-container">$\displaystyle \int_{-1}^{1}f(x)x^ndx = 0$</span> for <span class="math-container">$n = 0,1,2,...$</span> </p>
<p>Then s... | ling | 670,949 | <p>Since
<span class="math-container">$$\limsup_{n\to\infty}|f(x)-p_n(x)|=0,$$</span>
we know for any <span class="math-container">$\epsilon>0$</span>, there is <span class="math-container">$N\in\mathbb{N}$</span> such that for <span class="math-container">$\forall n> N$</span>
<span class="math-container">$$|f... |
339,880 | <p>I'm interested in examples where the sum of a set with itself is a substantially bigger set with nice structure. Here are two examples:</p>
<ul>
<li><strong>Cantor set</strong>: Let <span class="math-container">$C$</span> denote the ternary Cantor set on the interval <span class="math-container">$[0,1]$</span>. The... | Nik Weaver | 23,141 | <p>I proved this fact not too long ago: if <span class="math-container">$G$</span> is a finite group of cardinality <span class="math-container">$n$</span>, then there exists a subset <span class="math-container">$S$</span> of <span class="math-container">$G$</span> of cardinality no more than <span class="math-contain... |
339,880 | <p>I'm interested in examples where the sum of a set with itself is a substantially bigger set with nice structure. Here are two examples:</p>
<ul>
<li><strong>Cantor set</strong>: Let <span class="math-container">$C$</span> denote the ternary Cantor set on the interval <span class="math-container">$[0,1]$</span>. The... | Seva | 9,924 | <p>The set <span class="math-container">$Q$</span> of all squares in <span class="math-container">$\mathbb F_p$</span> is definitely thick and very nice. Can it be represented as a difference set <span class="math-container">$A-A$</span>? An open conjecture due to Sárközy is that this is impossible. (It has been <a hre... |
3,370,750 | <p>Show that the moment if inertia of an elliptic area of mass
M
and semi-axis
a
and
b
about a semi-diameter
of length
r
is <span class="math-container">$$\frac{Ma^2b^2}{4r^2}$$</span>.
My attempt.</p>
<ol>
<li>I know that MI about ox is <span class="math-container">${Mb^2 \over 4}$</span>.</li>
<li>MI about oy a... | Connor Harris | 102,456 | <p>Partial answer, based on the fact that the image of any ellipse under affine transformation is another ellipse.</p>
<p>Let the equation of the ellipse be <span class="math-container">$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$</span> or <span class="math-container">$$b^2 x^2 + a^2 y^2 = 1$$</span> and let the desired... |
45,441 | <p>There is a method of constructing representations of classical Lie algebras via Gelfand-Tsetlin bases. It has also been applied to Symmetric groups by Vershik and Okounkov. Does anybody know of any application of the method to complex representations of $GL_n(\mathbb F_q)$? Or, at least, any results in this directio... | Jim Humphreys | 4,231 | <p>My earlier comment was not at all well-focused. After more thought, I'm inclined to be pessimistic about using a Gelfand-Tsetlin approach here (even if it has some success for symmetric groups). Though of course it would be interesting to be proven wrong. </p>
<p>As Matt Davis reminds me, my offhand referenc... |
3,954,410 | <p>I am solving exercises from Loring Tu.</p>
<p>Show that if <span class="math-container">$L : V \rightarrow V$</span> is a linear operator on a vector space V of dimension n, then the pullback <span class="math-container">$L^{\wedge} : A_n(V) \rightarrow A_n(V)$</span> is multiplication by determinant of L.</p>
<p>At... | Robert Shore | 640,080 | <p>Note that <span class="math-container">$7 \vert 7n^3$</span> so the problem reduces to proving <span class="math-container">$n^7 \equiv n \pmod 7$</span>. This is obviously true for <span class="math-container">$n=1$</span>.</p>
<p>Assume <span class="math-container">$k^7 \equiv k \pmod 7$</span>. Then <span clas... |
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 transformati... | Dr. Sonnhard Graubner | 175,066 | <p>$$n+2\left|n^2+5\right. \implies k(n+2) = n^2+5 \implies k = \frac{n^2+5}{n+2}$$
Use that $$k = \frac{n^2+5}{n+2}=n-2+\frac{9}{n+2}$$</p>
|
3,160,563 | <p>My classmates and I were calculating the first homology group of the klein bottle, and we saw that <span class="math-container">$Ker \, \delta_1 \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$</span> and <span class="math-container">$Im \, \delta_2 \cong \mathbb{Z} \oplus \mathbb{Z}$</span>.</p>
<p>However, <... | Joshua Mundinger | 106,317 | <p>Suppose that I have a chain complex <span class="math-container">$(C,\delta)$</span>, i.e. a sequence of groups <span class="math-container">$\{C_n\}_{n\in \mathbb{Z}}$</span> and differentials <span class="math-container">$\delta_n: C_n \to C_{n-1}$</span> such that <span class="math-container">$\delta_{n-1}\delta_... |
3,160,563 | <p>My classmates and I were calculating the first homology group of the klein bottle, and we saw that <span class="math-container">$Ker \, \delta_1 \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$</span> and <span class="math-container">$Im \, \delta_2 \cong \mathbb{Z} \oplus \mathbb{Z}$</span>.</p>
<p>However, <... | yoyostein | 28,012 | <p>I think your question may be related to this more general question:
<a href="https://math.stackexchange.com/questions/40881/isomorphic-quotients-by-isomorphic-normal-subgroups">Isomorphic quotients by isomorphic normal subgroups</a></p>
<p>In general, even if <span class="math-container">$H\cong K$</span>, it is no... |
4,242,116 | <p>I found this question online</p>
<blockquote>
<p>Find the limit:<span class="math-container">$$\lim_{n \rightarrow \infty} n^{3/2} \int _0^1 \frac{x^2}{(x^2+1)^n}dx $$</span></p>
</blockquote>
<p>I've been told that I need to use the gamma function by converting <span class="math-container">$n^{3/2}=n\sqrt{n}$</span... | ReinhardtΩ | 884,092 | <p><span class="math-container">$$ \int_0^{n^{1/2}}\frac{u^2}{(1+n^{-1}u^2)^{n}}\,\mathrm{d}u \overset{n\to\infty}{⟶} \int_0^\infty u^2e^{-u^2}\mathrm{d}u$$</span>
and
<span class="math-container">$$\int_0^\infty u^2e^{-u^2}\mathrm{d}u\overset{t=u^2}{=}\frac{1}{2}\int_0^\infty t^{1/2} e^{-t}\mathrm{d}t= \frac{\Gamma(3/... |
4,242,116 | <p>I found this question online</p>
<blockquote>
<p>Find the limit:<span class="math-container">$$\lim_{n \rightarrow \infty} n^{3/2} \int _0^1 \frac{x^2}{(x^2+1)^n}dx $$</span></p>
</blockquote>
<p>I've been told that I need to use the gamma function by converting <span class="math-container">$n^{3/2}=n\sqrt{n}$</span... | Mark Viola | 218,419 | <p>Using the inequalities <span class="math-container">$x^2-\frac12 x^4\le \log(1+x^2)\le x^2$</span>, we find that</p>
<p><span class="math-container">$$\int_0^1 x^2e^{-nx^2}e^{-nx^4/2}\,dx\le \int_0^1 \frac{x^2}{(x^2+1)^n}\,dx\le \int_0^1 x^2e^{-nx^2}\,dx\tag1$$</span></p>
<p>Then, enforcing the substitution <span cl... |
2,247,798 | <p><strong>Question:</strong> If $\alpha$ is an angle in a triangle and $\tan{\alpha}=-2$, then one of the following is true:</p>
<p>a) $0<\alpha < \frac{\pi}{2}$</p>
<p>b) $\frac{\pi}{2}<\alpha < \pi$</p>
<p>c) Can't be decided.</p>
<p>d) There exist no such angle $\alpha$.</p>
<p>My reasoning was tha... | bjcolby15 | 122,251 | <ul>
<li>Since $\tan \alpha$ is given a value (in this case, $-2$) it would contradict there is no angle $\alpha$ (d) and that we cannot determine the information (c).</li>
<li>If $\tan \alpha$ is negative, it cannot be in the first or third quadrant (all values are positive for $\tan \alpha$), $\therefore \alpha$ is n... |
3,893,908 | <p>I want to compute <span class="math-container">$$\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... | ratatuy | 812,151 | <p>I used @Franklin Pezutti Dyer's hint:</p>
<p><span class="math-container">$-\int_{-\infty}^{+\infty}(1-\cos{x})d\left(\frac{1}{x}\right)=-\int_{-\infty}^{\infty}\frac{\sin{x}}{x}dx=-\pi$</span></p>
|
3,637,526 | <p>I'm trying to understand the proof of Theorem 16.10, Probability and Measure, Patrick Billingsley, I put part of it here exactly as presented in the book</p>
<p>Theorem: If <span class="math-container">$f,g$</span> are nonnegative and <span class="math-container">$\int_Afd\mu=\int_Agd\mu$</span> for all <span class... | Physical Mathematics | 592,278 | <p>I'm not quite familiar with your notation, but I think this is another way writing <span class="math-container">$B_n$</span>:
<span class="math-container">$$B_n:= \{ x\in X \mid 0 \leq g(x) < f(x) \text{ and } g \leq n\}$$</span>
If that's correct, then note that:
<span class="math-container">$$B_n = \{x \in X \m... |
1,638,051 | <p>$$\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... | Community | -1 | <p><strong>Without substitution</strong>:</p>
<p>$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\frac1{36}\int\frac{(x^2-(x^2-36))\,dx}{(x^{2}-36)^{3/2}}=\frac1{36}\int\frac{x^2\,dx}{(x^{2}-36)^{3/2}}-\frac1{36}\int\frac{dx}{(x^{2}-36)^{1/2}}.$$</p>
<p>Then by parts on the first term,</p>
<p>$$\int\frac{x\cdot x\,dx}{(x^{2}-36)^... |
23,994 | <p>For which values of m does the equation:
$$3 \ln x+m x^3 = 17$$
have $1$ solution? $2$ solutions? $0$ solution?</p>
<p>Thanks.</p>
| Jonas Meyer | 1,424 | <p>If $m\geq 0$, note that $f(x)=3\ln x +mx^3$ is always increasing, goes to positive infinity as $x$ does, and goes to minus infinity as $x$ decreases to zero. Also, it is continuous. This is enough information to answer the question in this case.</p>
<p>If $m\lt 0$, then $f'(x)=3(\frac{1}{x}+mx^2)$ has a unique... |
1,261,067 | <p>$$\int \left(\frac15 x^3 - 2x + \frac3x + e^x \right ) \mathrm dx$$</p>
<p>I came up with
$$F=x^4-x^2+\frac{3x}{\frac12 x^2}+e^x$$
but that was wrong.</p>
| atapaka | 79,695 | <p>The first term is $\frac{x^4}{20}$, next $x^2$, next $3 \ln(x)$ the last $\exp(x)$ and a constant</p>
|
1,261,067 | <p>$$\int \left(\frac15 x^3 - 2x + \frac3x + e^x \right ) \mathrm dx$$</p>
<p>I came up with
$$F=x^4-x^2+\frac{3x}{\frac12 x^2}+e^x$$
but that was wrong.</p>
| wythagoras | 236,048 | <p>\begin{align}
\int \frac{1}{5}(x^3)-2x+\frac{3}{x}+e^x dx & = \int \frac{1}{5}x^3 dx- \int 2x dx+ \int \frac{3}{x} dx+ \int e^x dx \\ &= \frac{1}{20}x^4 - x^2 + 3 \ln(x) + e^x + C
\end{align}</p>
<p>Important rules:</p>
<p>•Derivative of $x^n$ is $\frac{1}{n+1}x^{n+1}$ for $n \neq -1$</p>
<p>•Derivative... |
3,408,458 | <p>i have 4 vectors:</p>
<ul>
<li><span class="math-container">$|\vec{AC}|=|\vec{AD}|$</span></li>
<li><span class="math-container">$|\vec{BC}|=|\vec{BE}|$</span></li>
</ul>
<p><span class="math-container">$\angle (\vec{AC}, \vec{AD}) $</span> =<span class="math-container">$\angle (\vec{BC}, \vec{BE}) $</span></p>
<p><... | Kitter Catter | 166,001 | <p>Define <span class="math-container">$X$</span> as the position of <span class="math-container">$C$</span> and <span class="math-container">$\theta$</span> as the angle used</p>
<p>Isn't this defined by the parametric equations:
<span class="math-container">$$x(t) = t + (X-t) \cos\theta$$</span>
<span class="math-co... |
697,402 | <p>I have this limit:</p>
<p>$$ \lim_{x\to\infty}\frac{x^3+\cos x+e^{-2x}}{x^2\sqrt{x^2+1}} $$ I tried to solve it by this:</p>
<p>$$ \lim_{x\to\infty}\frac{x^3+\cos x+e^{-2x}}{x^2\sqrt{x^2+1}} = \lim_{x\to\infty}\frac{\frac{x^3}{x^3}+\frac{\cos x}{x^3}+\frac{e^{-2x}}{x^3}}{\frac{x^2\sqrt{x^2+1}}{x^3}} = \frac{0+0+0}... | Raka | 120,299 | <p>I think just divide with higher x,
like this</p>
<p>(x^4/x^4)*(1/sqrt(1+(x^-2))</p>
<p>so we get ,
(x^4)/sqrt(x^2+1)</p>
<p>and
lim x to->oo
we just divide it again with higher x
and we get,
1/sqrt(1+0)</p>
<p>and the result 1</p>
<p>sorry ,if i have mistake, i'm just trying my logic</p>
|
66,000 | <p>In 2008 I wrote a group theory package. I've recently started using it again, and I found that one (at least) of my functions is broken in Mathematica 10. The problem is complicated to describe, but the essence of it occurs in this line:</p>
<pre><code>l = Split[l, Union[#1] == Union[#2] &]
</code></pre>
<p>He... | MikeLimaOscar | 5,414 | <p>As @Szabolcs points out <code>Dispatch</code> does not interact well with <code>SameQ</code>, etc in Mathematica 10.</p>
<pre><code>Dispatch[1 -> 2] === Dispatch[1 -> 2]
</code></pre>
<blockquote>
<p>False</p>
</blockquote>
<pre><code>Dispatch[1 -> 2] == Dispatch[1 -> 2]
</code></pre>
<blockquote>
... |
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 ... | Shaun | 104,041 | <p>No.</p>
<p>Consider the Klein four group. It has four elements and yet no element of order four.</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 i... | Serge Ballesta | 210,087 | <p>Order is easy and non ambiguous in $\mathbb{R}$, because it is unidimensional. $\mathbb{C}$ on the other hand is generally seen as a plane. So you will easily define pre-order on it, that means transitive and reflexive relations, that do have sense such as the examples of Ross Millikan's answer.</p>
<p>But except f... |
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>
| user1058602 | 1,058,602 | <p>If you evalute the limit of the inverse by using L'Hopital, and you will find that the limit of the inverse goes to zero as h goes to zero. So the limit should goes to infinity.</p>
|
26,823 | <p>Trying to solve for the area enclosed by $x^4+y^4=1$. A friend posed this question to me today, but I have no clue what to do to solve this. Keep in mind, we don't even know if there is a straightforward solution. I think he just likes thinking up problems out of thin air. </p>
<p>Anyway, the question becomes m... | Thomas Kragh | 4,500 | <p>I think this question smells of homework, but another answer, which to me totally obscures the geometric nature of the question has been posted, and I feel that this justifies the following answer (even if the question is closed):</p>
<p>The $l^p$ norms $\lvert(x,y)\rvert_p = (\lvert x\rvert^p+\lvert y \rvert^p)^{1... |
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'... | Bill Dubuque | 242 | <p><strong>Hint</strong> $\ $ Your question amounts to verifying the following polynomial equality</p>
<p>$$\rm (k\!+\!1)\, f(k\!+\!1) - k\, f(k) =\, 2\,(3k\!+\!1)^2\quad for\quad f(k) =\, 6k^2\! - 3k - 1$$</p>
<p>Since LHS and RHS are polynomials of degree $\color{#C00} 2,$ to prove that they are equal it suffices ... |
4,417,901 | <p>In the first chapter of "Differential Equations, Dynamical Systems and an Introduction to Chaos" by Hirch, Smale and Devaney, the authors mention the first-order equation <span class="math-container">$x'(t)=ax(t)$</span> and assert that the only general solution to it is <span class="math-container">$x(t)=... | Eugene | 726,796 | <p>Let us differentiate a function
<span class="math-container">$$
\begin{aligned}
I_n(\lambda) = \int_0^{+\infty}x^ne^{-\lambda x}dx
\end{aligned}
$$</span></p>
<p>with respect to <span class="math-container">$\lambda$</span>:</p>
<p><span class="math-container">$$
\begin{aligned}
\frac{d}{d\lambda}I_n(\lambda) &=... |
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 fa... | Did | 6,179 | <blockquote>
<p>conceptually I understand what a derivative represents. </p>
</blockquote>
<p>So let us start from that. Consider a function $\theta\mapsto F(\theta)$ of a parameter $\theta$, defined at least on an interval $(\theta_*-\varepsilon,\theta_*+\varepsilon)$ around the point $\theta_*$. Then the derivativ... |
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 fa... | Hendy | 39,222 | <p>The answer above is a good one, but I thought I'd add in some more "layman's" terms that helped me better understand concepts of partial derivatives. The answers I've seen here and in the Coursera forums leave out talking about the chain rule, which is important to know if you're going to get what this is doing...</... |
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>... | Artes | 184 | <h2><code>Collect</code></h2>
<p>Since it hasn't been mentioned (and one can interpret the question in another way) I'd recommend to use also <code>Collect</code> (it can be applied not only to polynomials) :</p>
<pre><code>Collect[(x + y)^2, x]
</code></pre>
<blockquote>
<pre><code>x^2 + 2 x y + y^2
</code></pre>
</bl... |
626,958 | <p>I know that $E[X|Y]=E[X]$ if $X$ is independent of $Y$. I recently was made aware that it is true if only $\text{Cov}(X,Y)=0$. Would someone kindly either give a hint if it's easy, show me a reference or even a full proof if it's short? Either will work I think :) </p>
<p>Thanks.</p>
<p>Edit: Thanks for the great... | Did | 6,179 | <p>Assume that $X$ is symmetric Bernoulli, that is, such that $P[X=+1]=P[X=-1]=\frac12$, that $Z$ is symmetric Bernoulli independent of $X$, and that $Y=0$ on $[X=-1]$ while $Y=Z$ on $[X=+1]$. </p>
<p>In other words, the distribution of $(X,Y)$ is $\frac12\delta_{(-1,0)}+\frac14\delta_{(+1,+1)}+\frac14\delta_{(+1,-1)}... |
179,223 | <p>I have posted the same question on the community (<a href="http://community.wolfram.com/groups/-/m/t/1394441?p_p_auth=YV2a4wzw" rel="nofollow noreferrer">http://community.wolfram.com/groups/-/m/t/1394441?p_p_auth=YV2a4wzw</a>).</p>
<p>I tried to register the movie posted below (compressed version here) using Mathem... | kjosborne | 49,349 | <p><strong>Why is <code>ImageAlign</code> breaking?</strong></p>
<p>The message gives an at least partly helpful hint about why <code>ImageAlign</code> is failing here. The source of the failure is that <code>ImageAlign</code> is first using <code>ImageCorrespondingPoints</code> to find a list of points, and then doin... |
2,929,804 | <p>My attempt</p>
<p>First I wanted to show <span class="math-container">$<3,x^2+1>$</span> is maximal
So, I supposed another maximal <span class="math-container">$A$</span> which contain <span class="math-container">$<3,x^2+1>$</span> properly, and choose element <span class="math-container">$a$</span> i... | Stahl | 62,500 | <p>In this case, it is possible to simply compute the quotient ring. Below are some hints to guide you, but first let me make a remark.</p>
<p>In <span class="math-container">$\Bbb Z[x]/(3,x^2 + 1),$</span> the class of <span class="math-container">$3$</span> (i.e., <span class="math-container">$3 + (3, x^2 + 1)\in\Bb... |
125,317 | <p>Consider a (locally trivial) fiber bundle $F\to E\overset{\pi}{\to} B$, where $F$ is the fiber, $E$ the total space and $B$ the base space. If $F$ and $B$ are compact, must $E$ be compact? </p>
<p>This certainly holds if the bundle is trivial (i.e. $E\cong B\times F$), as a consequence of Tychonoff's theorem. It al... | savick01 | 18,493 | <p>I don't get where the problem is. Am I missing something?</p>
<p>Each point $b\in B$ has a neighbourhood $N_b$ such that the bundle over $N_b$ is trivial. Choose a smaller closed (thus compact) neighbourhood $C_b$ (we need some weak assumption here like Hausdorffness of $B$). The bundle over $C_b$ is homeomorphic t... |
2,483,611 | <p>I believe the answer is 13 * $13\choose4$ * $48\choose9$.</p>
<p>There are $13\choose4$ to draw 4 of the same cards, and multiply by 13 for each possible rank (A, 2, 3, ..., K). Then there are $48\choose9$ to choose the remaining cards.</p>
<p>One thing I am not certain of, is whether this accounts for the possibi... | Macavity | 58,320 | <p>For a calculus less approach, we can complete your analysis. As you noted, the roots of $Q(y)$ are real if
$$3m^2+16m + 16 \geqslant 0 \implies m \in (-\infty, -4] \cup [-\tfrac43, \infty)$$</p>
<p>Then for them to be non-negative, we just need $m+2 \geqslant 0 \iff m \geqslant -2$ as then
$$2(m+2) \geqslant \sqr... |
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>
| Dietrich Burde | 83,966 | <p>Yes, the positive rationals are the free abelian group whose basis consists of the primes:
$$
(\mathbb{Q}_{>0},\cdot) \cong \bigoplus_{p\in \mathbb{P}} \mathbb Z.
$$
This follows more or less directly from the Fundamental Theorem of Arithmetic.
Clearly the primes $\mathbb{P}$ are countable (and the whole group i... |
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>
| Shri | 442,962 | <p>I think it is a free abelian group which is generated by prime numbers (basis elements).</p>
|
1,903,717 | <p>This is actually from an Analysis text but i feel its a set theory question.</p>
<p>Proposition for ever rational number $\epsilon > 0$ there exists a non-negative number x s.t $x^2 < 2 < (x+ \epsilon )^2 $</p>
<p>It provides a proof that im having trouble understanding.</p>
<p>Proof: let $ \epsilon >... | Kaligule | 182,303 | <p>We assumed that if $x^2<2$ then $(x+ϵ)^2<2$ (*).</p>
<p>Lets start with $x=0$: $x^2 = 0^2 = 0 < 2$, so we conclude that $(x+ϵ)^2 = ϵ^2<2$.</p>
<p>Now we have learned that $ϵ^2<2$, so we can use (*) again, with $x=ϵ$. We conclude that $(x+ϵ)^2=(ϵ+ϵ)^2=(2ϵ)^2<2$</p>
<p>Now we have learned that $(... |
2,581,135 | <blockquote>
<p>Find: $\displaystyle\lim_{x\to\infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$</p>
</blockquote>
<p>Question from a book on preparation for math contests. All the tricks I know to solve this limit are not working. Wolfram Alpha struggled to find $1$ as the solution, but the solution process pre... | omegadot | 128,913 | <p>If you factor out a $\sqrt{x}$ term from the denominator one has
\begin{align*}
\lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}}} &= \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x} \sqrt{1 + \frac{1}{x} \sqrt{x + \sqrt{x}}}}\\
&= \lim_{x \to \infty} \frac{1}{\sqrt{1 + \sqrt{\frac{1}{x} + \frac... |
3,842,653 | <p>I'm working on a problem for a class and I'm a bit confused on what exactly the question is asking, the question is as follows,</p>
<p>Suppose <span class="math-container">$(x_n)$</span> is a sequence in <span class="math-container">$\Bbb R$</span>. Prove that <span class="math-container">$\bigl\{a \in \Bbb R : \te... | Oliver Díaz | 121,671 | <p>Here is a sketch of a solution. I leave some details (why?) for the OP.</p>
<ul>
<li><p>If <span class="math-container">$a\in \bigcap^{\infty}_{n=1} \overline{\{ x_n,x_{n+1},x_{n+2},...\}}$</span>, there is <span class="math-container">$n_1\geq1$</span> such that <span class="math-container">$|x_{n_1}-a|<\frac12... |
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$</span> which is problematic, but when I checked the graph of the function I have di... | Robert Israel | 8,508 | <p>It goes up to <span class="math-container">$n$</span>, not <span class="math-container">$k+n$</span>.
In this case the product has only one factor namely <span class="math-container">$k+1 = 3$</span>.</p>
|
636,730 | <p>Let $G$ be a group of infinite order . Does there exist an element $x$ belonging to $G$ such that $x$ is not equal to $e$ and the order of $x$ is finite?</p>
| preferred_anon | 27,150 | <p>I can do better: consider the following set under the operation of multiplication:
$$\lbrace x=e^{i\pi t},t \in \mathbb{Q} \rbrace$$
The set is infinite, but every element has finite order (namely, if $t=a/b$ in lowest terms, the order of $x$ is $2b$).</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 <... | Zhanxiong | 192,408 | <p>Regarding to this particular set, you can define $<$ as $a < b$ if $b - a \in \mathbb{N}$ and $b - a \neq 0$.</p>
|
316,699 | <p>If $A,B,C$ are sets, then we all know that $A\setminus (B\cap C)= (A\setminus B)\cup (A\setminus C)$. So by induction
$$A\setminus\bigcap_{i=1}^nB_i=\bigcup_{i=1}^n (A\setminus B_i)$$
for all $n\in\mathbb N$.</p>
<p>Now if $I$ is an uncountable set and $\{B_i\}_{i\in I}$ is a family of sets, is it true that:
$$A\s... | Trevor Wilson | 39,378 | <p>Yes. If $x$ is in the left hand side then it's in $A$ but not in $\bigcap_{i\in I} B_i$, so it's missing from one of the $B_i$'s. Therefore it's in $A \setminus B_i$ for some $B_i$, and it's in the right hand side. The other direction is similar.</p>
|
4,474,806 | <p>I use the following method to calculate <span class="math-container">$b$</span>, which is <span class="math-container">$a$</span> <strong>increased</strong> by <span class="math-container">$x$</span> percent:</p>
<p><span class="math-container">$\begin{align}
a = 200
\end{align}$</span></p>
<p><span class="math-cont... | Ezra | 1,068,060 | <p>The difference between them is with respect to what variable the percentage is being taken. In your first calculation, when you divide, you are saying "<span class="math-container">$a$</span> is <span class="math-container">$x\%$</span> more than <span class="math-container">$c$</span>", while on the latte... |
3,964,237 | <p>I am trying to understand the meaning of this exercise question in a logic textbook.</p>
<blockquote>
<p>For each of the following statement forms,find a statement form that is logically equivalent to its negation and in which negation signs apply only to statement letters.<br/>
i. <span class="math-container">$A \r... | Brian M. Scott | 12,042 | <p>In each problem you have a statement form <span class="math-container">$\varphi$</span>; you are to start with its negation <span class="math-container">$\neg\varphi$</span> and use logical equivalences to transform that into a statement form in which no compound expression is negated: the only expressions that are ... |
3,289,658 | <p>I was solving A-level Further Mathematics paper and I didn't quite understand how to solve the question.</p>
<blockquote>
<p>Question:<a href="https://i.stack.imgur.com/C3UUX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C3UUX.png" alt="Question"></a></p>
</blockquote>
<p>I know the formula ... | Bartek | 671,751 | <p>Just sum the given identity for all <span class="math-container">$n=1,2,...,N$</span>. LHS will telescope and on the RHS you will obtain six times your desired sum plus five times the sum of cubes (which is known) and <span class="math-container">$\frac{3}{8}$</span> of the sum of the first powers which is also know... |
3,289,658 | <p>I was solving A-level Further Mathematics paper and I didn't quite understand how to solve the question.</p>
<blockquote>
<p>Question:<a href="https://i.stack.imgur.com/C3UUX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C3UUX.png" alt="Question"></a></p>
</blockquote>
<p>I know the formula ... | Cornman | 439,383 | <p>With the given identity </p>
<p><span class="math-container">$(n+\frac12)^6-(n-\frac12)^6=6n^5+5n^3+\frac38n\Leftrightarrow n^5=\frac16\left((n-\frac12)^6-(n+\frac12)^6+5n^3+\frac38n\right)$</span></p>
<p>We get:</p>
<p><span class="math-container">$\sum_{n=1}^N n^5=\sum_{n=1}^N \frac16\left((n-\frac12)^6-(n+\fra... |
2,473,951 | <p>The definition I have of a convex function $f: \mathbb{R} \rightarrow \mathbb{R}$ is that for every $x, y \in \mathbb{R}$ and every $\lambda \in [0, 1]$,
$$
f(\lambda x + (1-\lambda )y) \leq \lambda f(x) + (1- \lambda )f(y).$$</p>
<p>By proving that slopes increase I mean that for $x \leq y \leq z$, we get $$\frac... | zhw. | 228,045 | <p>I like to think of this geometrically. The definition of convexity implies $(y,f(y))$ lies on or below the line $L$ through $(x,f(x))$ and $(z,f(z)).$ Suppose it's below. Then the line through through $(x,f(x))$ and $(y,f(y))$ plainly has slope less than the slope of $L.$ And then to move back up to $(z,f(z)),$ the ... |
1,047,544 | <p>I'm doing some research and I'm trying to compute a closed form for $ \mathbb{E}[ X \mid X > Y] $ where $X$, $Y$ are independent normal (but not identical) random variables. Is this known?</p>
| user193702 | 193,702 | <p>Hint: you can directly do the double integrals in the district $x>y$ for $xf(x)$, not so hard, I remember.</p>
|
1,047,544 | <p>I'm doing some research and I'm trying to compute a closed form for $ \mathbb{E}[ X \mid X > Y] $ where $X$, $Y$ are independent normal (but not identical) random variables. Is this known?</p>
| Did | 6,179 | <p>One can reduce the problem to the computation of $E(U\mid U\gt aV-b)$ where $(U,V)$ are i.i.d. standard normal. Let $\varphi$ denote the standard normal PDF, then $u\varphi(u)=-\varphi'(u)$ hence $$E(U;U\gt aV-b)=\int_\mathbb R\varphi(v)\mathrm dv\int_{av-b}^\infty u\varphi(u)\mathrm du=\int_\mathbb R\varphi(v)\varp... |
4,269,414 | <p>In <a href="https://math.stackexchange.com/questions/550764/quotient-space-of-s1-is-homeomorphic-to-s1">this post</a> someone suggested:</p>
<p>"<span class="math-container">$z\mapsto z^2$</span>"</p>
<p>where both <span class="math-container">$z$</span> and <span class="math-container">$z^2$</span> are in... | ryang | 21,813 | <p><span class="math-container">$$z\in S^1 \implies z\in\mathbb C\implies z^2\in\mathbb C;$$</span><span class="math-container">$$\\z\in\mathbb C\kern.6em\not\kern-.6em\implies z^2\in\mathbb C^2.$$</span></p>
|
3,468,336 | <p>Consider the multiplication operator <span class="math-container">$A \colon D(A) \to L^2(\mathbb R)$</span> defined by
<span class="math-container">$$\forall f \in D(A): \quad(Af)(x) = (1+\lvert x \rvert^2)f(x),$$</span>
where <span class="math-container">$$D(A) := \left \{f\in L^2(\mathbb R): (1+\lvert x \rvert^2)... | Henry | 6,460 | <p>The exercise is correct. As Clarinetest says in a comment, your error seems to be with <span class="math-container">$\frac{1}{n^2}\mathbb{E}_\theta\left[\left(\sum_{i=1}^n X_i\right)^2\right]$</span> i.e. with <span class="math-container">$\mathbb E[\bar{X}^2]$</span>. </p>
<p>You should have <span class="math-co... |
11,353 | <p>Thinking about the counterintuitive <em>Monty Hall Problem</em> (stick or switch?),
revisited in <a href="https://matheducators.stackexchange.com/a/11346/511">this ME question</a>,
I thought I would issue a challenge:</p>
<blockquote>
<p>Give in one (perhaps long) sentence a convincing explanation of why <em>swit... | Benoît Kloeckner | 187 | <p>The sticking strategy does not use the additional information revealed by the presenter, and thus cannot have more chance of winning than if the presenter would open no door, which is 1/3.</p>
|
3,270,725 | <p>Hello everyone I read on my notes this proposition: </p>
<p>Given a field <span class="math-container">$K$</span> and <span class="math-container">$R=K[T]$</span>, let <span class="math-container">$M$</span> be a (left) finitely generated <span class="math-container">$R$</span>-module; then <span class="math-contai... | TeM | 247,735 | <p>Given the function <span class="math-container">$f : D_f \to \mathbb{R}$</span> of law:
<span class="math-container">$$f(x) := \frac{1}{\sqrt{|\tan x| - \tan x}}\,,$$</span>
its natural domain is thus determinable:
<span class="math-container">$$
|\tan x| > \tan x
\; \; \; \; \; \; \Leftrightarrow \; \; \; \; \;... |
2,009,557 | <p>I am pretty sure this question has something to do with the Least Common Multiple. </p>
<ul>
<li>I was thinking that the proof was that every number either is or isn't a multiple of $3, 5$, and $8\left(3 + 5\right)$.</li>
<li>If it isn't a multiple of $3,5$, or $8$, great. You have nothing to prove.</li>
<li>But if... | gav | 858,954 | <p>proof by induction</p>
<p>p(8): 3.(1)+5.(1)=8 holds</p>
<p>Assume p(k) holds => 3.a + 5.b = k</p>
<p>we must prove that p(k+1) holds.</p>
<p>p(k+1): 3x+5y = k+1=>
3x+5y = 3a+5b+1=>
3(x-a) + 5(y-a)</p>
<p>thus suffices to prove that 3z+5ω =1 has an integer solution.
That is true because gcd(3,5)=1 | 1 (1 dev... |
4,613,471 | <p>I would like to figure out the power series expansion of <span class="math-container">$f(z)=\frac{1}{(z+1)^2}$</span> around <span class="math-container">$z_0=1$</span>. Somehow expanding this into a geometric series would be the way to go I suppose, however, I fail to see how this can be rearranged in terms of (z-1... | Quanto | 686,284 | <p>Substitute <span class="math-container">$u+\frac12=\frac{\sqrt3}2\tan t$</span>
<span class="math-container">\begin{aligned}
&\int{\frac{1}{u\sqrt{u^{2}+u+1}}}du\\
=& -\int{\frac{1}{\cos(t+\frac\pi3)}}dt
= -\int\frac{d\left[\sin(t+\frac\pi3) \right]}{1-\sin^2(t+\frac\pi3) }\\
=&\ \frac12\ln\frac{1-\sin(t... |
298,481 | <p>Find the general solution of </p>
<p>$y'' + \dfrac{7}{x} y' + \dfrac{8}{x^2} y = 1, x > 0$</p>
<p>I don't even know how to solve the homogeneous version because it involves variables...</p>
<p>Does anyone know how to solve it?</p>
| Mhenni Benghorbal | 35,472 | <p>It is of Euler differential equation type. Here is a <a href="https://math.stackexchange.com/questions/285274/solving-this-second-order-ode">related problem</a>. You should have the following solution </p>
<p>$$ y(x) ={\frac {{\it c_2}}{{x}^{4}}}+{\frac {{\it c_1}}{{x}^{2}}}+\frac{{x}^{2}}{24}.$$</p>
|
3,334,816 | <p>The question first requires me to prove the identity <span class="math-container">$$\sqrt{\frac{1- \sin x}{1+ \sin x}}=\sec x- \tan x, -90^\circ < x < 90^\circ$$</span> I am able to prove this. The second part says “Explain why <span class="math-container">$x$</span> must be acute for the identity to be true”.... | azif00 | 680,927 | <p>The first quadrant is <span class="math-container">$0^\circ \leq x \leq 90^\circ$</span> and the fourth quadrant is <span class="math-container">$270^\circ \leq x \leq 360^\circ$</span> or this is the same as <span class="math-container">$-90^\circ \leq x \leq 0^\circ$</span>. That is
<span class="math-container">$$... |
3,334,816 | <p>The question first requires me to prove the identity <span class="math-container">$$\sqrt{\frac{1- \sin x}{1+ \sin x}}=\sec x- \tan x, -90^\circ < x < 90^\circ$$</span> I am able to prove this. The second part says “Explain why <span class="math-container">$x$</span> must be acute for the identity to be true”.... | Mick | 42,351 | <p>The important point is that “identity” is true only when <span class="math-container">$-90^\circ < x < 90^\circ$</span>.</p>
<p>In another word, it is not true when the terminal arm falls in QII and QIII.</p>
<p>Note that the angle lies in <span class="math-container">$-90^0 < x < 0^0$</span> is also a... |
1,514,388 | <p>$$ \lim_{x\to \infty} \left(\frac{1}{(x^2+x)\left(\ln\frac{x+1}{x}\right)^2}\right) $$</p>
<p>I know the answer is 1, but why does it tend to 1?
Can you manipulate the function and the "$\ln$" to make it obvious? </p>
<p>Much appreciated. </p>
| SchrodingersCat | 278,967 | <p>$$ \lim_{x\to \infty} \left(\frac{1}{(x^2+x)\left(\ln\frac{x+1}{x}\right)^2}\right) $$</p>
<p>Use $u=\frac{1}{x}$</p>
<p>So we have $$\lim_{u\to 0} \left(\frac{u^2}{(1+u)\left(\ln\left[1+u\right]\right)^2}\right)$$
$$=\frac{1}{\lim_{u\to 0}(1+u)}\cdot \frac{1}{\lim_{u\to 0}\frac{\left(\ln\left[1+u\right]\right)^2... |
2,625,763 | <p>I am having trouble with factoring $2x^3 + 21x^2 +27x$. The answer is $x(x+9)(2x+3)$ but not sure how that was done. Obviously I factored out the $x$ to get $x(2x^2+21x+27)$ then from there I am lost. I tried the AC method and grouping. Can someone show the steps? Thanks! </p>
| random | 513,275 | <p>Find the roots $x_1$ and $x_2$ of $2x^2+21x+27$ with the standard abc method.</p>
<p>Then $2x^2+21x+27=c(x-x_1)(x-x_2)$ with an easily computable $c$</p>
|
2,625,763 | <p>I am having trouble with factoring $2x^3 + 21x^2 +27x$. The answer is $x(x+9)(2x+3)$ but not sure how that was done. Obviously I factored out the $x$ to get $x(2x^2+21x+27)$ then from there I am lost. I tried the AC method and grouping. Can someone show the steps? Thanks! </p>
| fleablood | 280,126 | <p>1) </p>
<p>$2x^3 + 21x^2 +27x = x(2x^2 + 21x + 27) = x(2x(x + 9) + 3x + 27) =x(2x(x+9) + 3(x + 9)) = x(2x + 3)(x+9)$.</p>
<p>2) </p>
<p>$2x^3 + 21 x^2 + 27x = x(2x^2 + 21x + 27) = x*2*(x - a)(x+b)$ where $a,b$ are solutions to $2x^2 + 21x + 27=0$. i.e. $x = \frac {-21 \pm {21^2 - 8*27}}{4} = \frac {-21 \pm \sqrt... |
3,473,944 | <p>So i have an object that moves in a straight line with initial velocity <span class="math-container">$v_0$</span> and starting position <span class="math-container">$x_0$</span>. I can give it constant acceleration <span class="math-container">$a$</span> over a fixed time interval <span class="math-container">$t$</s... | NiveaNutella | 695,512 | <p>As one of the comments suggested, it is not possible for all time intervals <span class="math-container">$t$</span>. I ignored the fixing of the time interval and solved the problem without that constraint. Hope it helps:</p>
<p>EDIT: If you want to still have a fixed <span class="math-container">$t_1$</span>, you ... |
3,488,226 | <p>We have <span class="math-container">$$\ln(v-1) - \ln(v+3) = \ln(x) + C$$</span></p>
<p>Multiplying through by e gives:</p>
<p><span class="math-container">$$(v-1)/(v+3) = x + e^C$$</span></p>
<p>But the answer in the textbook is:</p>
<p><span class="math-container">$$(v-1)/(v+3) = Dx$$</span></p>
<p>Where <spa... | Severin Schraven | 331,816 | <p>Because
<span class="math-container">$$ e^{\ln(x)+C} = e^{\ln(x)} e^C = x e^C. $$</span></p>
|
3,488,226 | <p>We have <span class="math-container">$$\ln(v-1) - \ln(v+3) = \ln(x) + C$$</span></p>
<p>Multiplying through by e gives:</p>
<p><span class="math-container">$$(v-1)/(v+3) = x + e^C$$</span></p>
<p>But the answer in the textbook is:</p>
<p><span class="math-container">$$(v-1)/(v+3) = Dx$$</span></p>
<p>Where <spa... | Robert Shore | 640,080 | <p>You're not multiplying through by <span class="math-container">$e$</span>. You're exponentiating both sides of the equation. So your second equation should be:</p>
<p><span class="math-container">$$\frac{v-1}{v+3} = e^Cx.$$</span></p>
|
302,790 | <p>The $j$-ivariant has the following Fourier expansion
$$j(\tau)=\frac 1q +\sum_{n=0}^{\infty}a_nq^n=\frac{1}{q}+744+196884q+21493760q^2+\cdots.$$
Here is $q=e^{2\pi i \tau}$. </p>
<p>Is there some simple <strong>effective</strong> bound on the coefficients $a_n$?</p>
<p><strong>Backround.</strong></p>
<p>This ques... | Noam D. Elkies | 14,830 | <p>Once you know that the coefficients are all positive (see postscript),
it's easy to get an effective upper bound that grows as $\exp(4\pi \sqrt{n})$,
which is within a factor $O(\sqrt n)$ of the correct order of growth.
Start from the inequality
$$
a_n = q^{-n} (a_n q^n) < q^{-n} \sum_{k=-1}^\infty a_k q^k = q^{... |
302,790 | <p>The $j$-ivariant has the following Fourier expansion
$$j(\tau)=\frac 1q +\sum_{n=0}^{\infty}a_nq^n=\frac{1}{q}+744+196884q+21493760q^2+\cdots.$$
Here is $q=e^{2\pi i \tau}$. </p>
<p>Is there some simple <strong>effective</strong> bound on the coefficients $a_n$?</p>
<p><strong>Backround.</strong></p>
<p>This ques... | GH from MO | 11,919 | <p>By a variation of Elkies's answer we can even get $a_n<e^{4\pi\sqrt{n}}$ without using $j(i)=1728$. </p>
<p>For $n=1$ the claim is clear. Now let $0<t<1$ and use the identity $j(it)=j(i/t)$. After expanding and rearranging, we get
$$\sum_{n=1}^\infty a_n(e^{-2\pi nt}-e^{-2\pi n/t})=e^{2\pi/t}-e^{2\pi t}.$$... |
4,424,668 | <p>Suppose <span class="math-container">$V$</span> is a complex vector space with <span class="math-container">$n=\dim V=10$</span> and <span class="math-container">$N∈L(V)$</span> is nilpotent. What are possible values for <span class="math-container">$\dim\ker(N^3)-\dim\ker(N)$</span>? The only two things I know that... | Samuel Adrian Antz | 1,045,826 | <p>Take the matrix <span class="math-container">$N$</span> with only entries of <span class="math-container">$0$</span>, but <span class="math-container">$N_{1i}=1$</span> for <span class="math-container">$i>j$</span> with <span class="math-container">$0<j\leq 10$</span>. We have <span class="math-container">$\op... |
3,009,543 | <p>I am having great problems in solving this:</p>
<p><span class="math-container">$$\lim\limits_{n\to\infty}\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}$$</span></p>
<p>I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I h... | Robert Z | 299,698 | <p>Alternative approach where we do not use of the identity <span class="math-container">$$a^3-b^3=(a-b)(a^2+ab+b^2).$$</span></p>
<p>We have that
<span class="math-container">$$0< \sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}=\frac{\sqrt[3]{n}}{\sqrt{n}}\cdot\frac{
\sqrt[3]{1+\frac{1}{\sqrt{n}}}-1}{\frac{1}{\sqrt{n}}}\leq \fra... |
555,955 | <p>Suppose I have two doors. One of them has a probability of $1/9$ to contain X, the other has a probability of $2/3$ to contain X. Then, supposing I pick randomly one of the two doors, what is the probability that it contains X?</p>
<p>(If one contains X, the other can also contain X. They are independent but not mu... | Peter Halburt | 85,427 | <p>I bet you that the solution is just the average of the probabilities. If they are independent but not mutually exclusive, then if one contains X, the other can too... so you can't just add the probabilities. The next simplest thing to do is average them, but as one knows, in all things probability this must be a wei... |
2,451,092 | <p>I want to solve a Lagrange multiplier problem,</p>
<p>$$f(x,y) = x^2+y^2+2x+1$$
$$g(x,y)=x^2+y^2-16 $$</p>
<p>Where function $g$ is my constraint.
$$f_x=2x+2, \ \ \ f_y=2y, \ \ \ g_x=2x\lambda, \ \ \ g_y=2y\lambda$$</p>
<p>$$
\begin{cases}
2x+2=2x\lambda \\
2y=2y\lambda \\
x^2+y^2-16=0
\end{cases}
$$</p>
<p>See... | Satish Ramanathan | 99,745 | <p>$2x(1-\lambda) = -2\tag 1$</p>
<p>$2y(1-\lambda) = 0\tag 2$</p>
<p>From (2) Either $y = 0$ or $(1-\lambda) = 0$</p>
<p>$(1-\lambda) \ne 0$ because if it were (1) would not be true </p>
<p>Thus $y = 0$</p>
<p>Plug in the value of y in g(x,y) and find x.</p>
<p>and $x = +/- 4$</p>
<p>The points are $(4,0)$ and ... |
376,484 | <p>My questions are motivated by the following exercise:</p>
<blockquote>
<p>Consider the eigenvalue problem
$$
\int_{-\infty}^{+\infty}e^{-|x|-|y|}u(y)dy=\lambda u(x), x\in{\Bbb R}.\tag{*}
$$
Show that the spectrum consists purely of eigenvalues. </p>
</blockquote>
<p>Let $A:L^2({\Bbb R})\to L^2({\Bbb R})$ be ... | Entosider | 972,378 | <p>What you need to prove is true. The problem with the refutations is that they are using functions where <span class="math-container">$\lim_{p^m\to\infty}f(p^m)\neq0$</span> because for some <span class="math-container">$\epsilon>0$</span>, for any <span class="math-container">$N$</span>, there will exist some pri... |
860,247 | <p>Simplify $$\frac{3x}{x+2} - \frac{4x}{2-x} - \frac{2x-1}{x^2-4}$$</p>
<ol>
<li><p>First I expanded $x²-4$ into $(x+2)(x-2)$. There are 3 denominators. </p></li>
<li><p>So I multiplied the numerators into: $$\frac{3x(x+2)(2-x)}{(x+2)(x-2)(2-x)} - \frac{4x(x+2)(x-2)}{(x+2)(x-2)(2-x)} - \frac{2x-1(2-x)}{(x+2)(x-2)(2-x... | lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>As $\displaystyle0\le x\le1,\sqrt{x(x+1)}=\sqrt x\sqrt{x+1}$</p>
<p>So, $$\int_0^1x^2\sqrt{x(x+1)}=\int_0^1x^{\dfrac52}\sqrt{x+1}\ dx$$</p>
<p>Set $x=\tan^2\theta$</p>
<p>Or integrating by parts, $$\int x^{\dfrac52}\sqrt{x+1}\ dx=x^{\dfrac52}\int\sqrt{x+1}\ dx-\int\left(\frac{d x^{\dfrac52}}{dx}\cd... |
1,378,536 | <p>Here is a question that naturally arose in the study of some specific integrals. I'm curious if for such integrals are known <em>nice real analysis tools</em> for calculating them (<em>including here all possible sources<br>
in literature that are publicaly available</em>). At some point I'll add my <em>real analy... | Ron Gordon | 53,268 | <p>Sub $x=\tanh{u}$, $dx = \operatorname{sech^2}{u} \, du$. Then the integral is</p>
<p>$$\int_{-\infty}^{\infty} du \, \frac{\operatorname{sech^2}{u}}{\pi^2+4 u^2} $$</p>
<p>Now, use Parseval. The Fourier transforms of the pieces of the integrand are</p>
<p>$$\int_{-\infty}^{\infty} du \, \frac{e^{i u k}}{\pi^2+4... |
1,436,215 | <p>I'm using the following algorithm (in C) to find if a point lays within a given polygon</p>
<pre><code>typedef struct {
int h,v;
} Point;
int InsidePolygon(Point *polygon,int n,Point p)
{
int i;
double angle=0;
Point p1,p2;
for (i=0;i<n;i++) {
p1.h = polygon[i].h - p.h;
p1.v = polygo... | Community | -1 | <p>Here is a counter-proposal for efficient computation, using only additions and multiplications.</p>
<p>The idea is to count the intersections of the edges of the polygon with the horizontal line through the test point that lie on its right (like in "Solution 1"). An even number means outside.</p>
<p><a href="https... |
2,827,970 | <p>I'm aware this result (and the standard/obvious proof) is considered basic and while I've accepted and used it numerous times in the past, I'm starting to question its validity, or rather that said proof doesn't subtly require a form of the AC. (Disclaimer: it's been some time since I've looked at set theory.)</p>
... | Asaf Karagila | 622 | <p>The definition of $\omega$ is "the least infinite ordinal". So there are no hidden assumptions.</p>
<p>But even ignoring that, the proof I gave in the answer you linked doesn't use any "presupposed assumption". It only uses the fact that if $\alpha$ is an infinite ordinal, then for any finite ordinal $n$, $n<\al... |
2,827,970 | <p>I'm aware this result (and the standard/obvious proof) is considered basic and while I've accepted and used it numerous times in the past, I'm starting to question its validity, or rather that said proof doesn't subtly require a form of the AC. (Disclaimer: it's been some time since I've looked at set theory.)</p>
... | DanielWainfleet | 254,665 | <p>(1). In the absence of Infinity, Choice, and Foundation (Regularity):</p>
<p>Let $On$ be the class of ordinals. Define the class $FOn$ of finite ordinals as $\{x\in On: \forall y\in x\cup \{x\}\;(y=0\lor y\ne \cup y)\}.$ </p>
<p>Prove that $\forall x\in FOn \;\forall y\in x\;(y\in FOn).$ </p>
<p>(1-i). That is, ... |
1,176,435 | <p>Consider $G(4,p)$ - the random graph on 4 vertices. What is the probability that vertex 1 and 2 lie in the same connected component?</p>
<p>So far, I have considered the event where 1 and 2 do not lie in the same component. Then vertex 1 must lie in a component of order 1, 2 or 3 that doesn't contain vertex 2. Howe... | Laars Helenius | 112,790 | <p>For $p=1/2$, since there are a total of six edges in $K_4$, that means there are only $2^6=64$ possible labeled graphs, each of which is equally likely. You could write out the sample space and count.</p>
<p>When $p\ne 1/2$, you could determine a binomial probability for each possible graph and sum the probabilitie... |
362,895 | <p>I have been having a lot of trouble teaching myself rings, so much so that even "simple" proofs are really difficult for me. I think I am finally starting to get it, but just to be sure could some one please check this proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field. Thank you.</p>
<p>Proof: Notice that ... | Martin Brandenburg | 1,650 | <p>Your proof only shows that there are at most two elements. So you also have to check that these two elements differ, i.e. that $1-i$ is not a unit. But instead, you can also do it directly, without any elements at all:</p>
<p>$\mathbb{Z}[i]/(i-1)=\mathbb{Z}[x]/(x^2+1)/(x-1)=\mathbb{Z}/(1^2+1)=\mathbb{F}_2$.</p>
|
1,791,673 | <p>I was wondering about this, just now, because I was trying to write something like:<br>
$880$ is not greater than $950$. <br>
I am wondering this because there is a 'not equal to': $\not=$ <br>
Not equal to is an accepted mathematical symbol - so would this be acceptable: $\not>$? <br>
I was searching around but ... | Community | -1 | <p>Equality is special in that there are two ways that two real numbers $a$ and $b$ can be not equal:</p>
<p>$$a>b,b>a$$</p>
<p>So, instead of saying $a>b \;\textrm {or}\;b>a$, we write $b\neq a$.</p>
<p>For the others, each negation has an existing symbol, so:</p>
<p>$$a\not>b \iff a\leq b,\;\,a\nle... |
177,515 | <p>From <a href="http://mitpress.mit.edu/algorithms/" rel="nofollow">Cormen et all</a>:</p>
<blockquote>
<p>The elements of a matrix or vectors are numbers from a number system, such as the real numbers , the complex numbers , or integers modulo a prime .</p>
</blockquote>
<p>What do they mean by <strong>integers m... | Calvin McPhail-Snyder | 10,104 | <p>In the elementary case, matrices are defined to only contain complex numbers, of which real numbers are treated as a special case. In this sense, matrices only contain complex numbers (since every real number is complex).</p>
<p>However, there are more general notions of number than just complex numbers, so it mak... |
874,946 | <p>What is the remainder when the below number is divided by $100$?
$$
1^{1} + 111^{111}+11111^{11111}+1111111^{1111111}+111111111^{111111111}\\+5^{1}+555^{111}+55555^{11111}+5555555^{1111111}+55555555^{111111111}
$$
How to approach this type of question? I tried to brute force using Python, but it took very long time.... | Darth Geek | 163,930 | <p><strong>Hint</strong></p>
<p>If $a \equiv b \pmod{n}$ then $a^k \equiv b^k \pmod{n}$
So for instance $111 \equiv 11 \pmod{100}$ so $111^{111} \equiv 11^{111} \pmod{100}$</p>
<p>Also note that $11^2 = 121 \equiv 21$ so $11^{111} = 11^{2·65 + 1} \equiv 11·21^{65}$. But $21^2 = 441 \equiv 41$ and so forth.</p>
<p>Co... |
902,522 | <p>How would I simplify a fraction that has a radical in it? For example:</p>
<p>$$\frac{\sqrt{2a^7b^2}}{{\sqrt{32b^3}}}$$</p>
| Kyle Gannon | 152,976 | <p>This paper, called <a href="http://arxiv.org/abs/0903.0340">Physics, Topology, Logic and Computation: A Rosetta Stone</a> does just that in section 3.2. If you have time and interest, I would suggest reading the entire paper (since the whole thing is pretty cool). </p>
|
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