qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,845,448 | <p>I know that $(AB)^T$ = $B^TA^T$ and that $(A^T)^{-1}= (A^{-1})^T$ but couldn't reach any convincing answer. Can someone demonstrate the expression.</p>
| Community | -1 | <p>you have the next property too
$$(AB)^{-1}=B^{-1}A^{-1}$$
then the inverse of $(AB)^{T}$ is $\left[(AB)^{T}\right]^{-1}$
$$\left[(AB)^{T}\right]^{-1}=\left[B^{T}A^{T}\right]^{-1}=\left[A^{T}\right]^{-1}\left[B^{T}\right]^{-1}=\left[A^{-1}\right]^{T}\left[B^{-1}\right]^{T}$$</p>
|
4,429,607 | <blockquote>
<p>Minimum value of <span class="math-container">$M$</span> such that <span class="math-container">$\exists a, b, c \in \mathbb{R}$</span> and
<span class="math-container">$$
\left|4 x^{3}+a x^{2}+b x+c\right| \leq M ,\quad \forall|x| \leq 1
$$</span></p>
</blockquote>
<p>What i considered was that puttin... | dshin | 16,006 | <p>Let <span class="math-container">$P[x]$</span> be the set of all cubic polynomials with leading coefficient <span class="math-container">$4$</span>, and let <span class="math-container">$P_M[x]$</span> be the subset of <span class="math-container">$P[x]$</span> consisting of polynomials <span class="math-container">... |
4,429,607 | <blockquote>
<p>Minimum value of <span class="math-container">$M$</span> such that <span class="math-container">$\exists a, b, c \in \mathbb{R}$</span> and
<span class="math-container">$$
\left|4 x^{3}+a x^{2}+b x+c\right| \leq M ,\quad \forall|x| \leq 1
$$</span></p>
</blockquote>
<p>What i considered was that puttin... | River Li | 584,414 | <p><strong>Remarks</strong>: It is related to the Chebyshev polynomials of the first kind. I used it before e.g. in <a href="https://math.stackexchange.com/questions/3940136/minimizing-the-maximum-of-x2-x-k/3949211">Minimizing the maximum of $|x^2 - x - k|$</a></p>
<hr>
<p>The condition is written as
<span class="math-... |
3,843,352 | <p>It’s called a Diophantine Equation, and it’s sometimes known as the “summing of three cubes”.</p>
<blockquote>
<p>A Diophantine equation is a polynomial equation, usually in two or
more unknowns, such that only the integer solutions are sought or
studied (an integer solution is such that all the unknowns take
intege... | poetasis | 546,655 | <p>This is the low-tech approach I took for this question. Note the optional output file in lines 140 and 220. The user may copy/paste or import the data into a spreadsheet for sorting as I did for the other answer.</p>
<pre><code> 100 print "Enter limit";
110 input l1
120 print time<span class="math-contai... |
3,455,967 | <p>Solve: (Hint: use <span class="math-container">$x\ln y=t$</span>)<span class="math-container">$$(xy+2xy\ln^2y+y\ln y)\text{d}x+(2x^2\ln y+x)\text{d}y=0$$</span></p>
<p>My Work:</p>
<p><span class="math-container">$$x\ln y=t, \text{ d}t=\ln y \text{ d}x+\frac{x}{y} \text{ d}y$$</span></p>
<p><span class="math-cont... | Maverick | 171,392 | <p>I am using the product rule of differentiation in reverse</p>
<p>Divide throughout by <span class="math-container">$y$</span></p>
<p><span class="math-container">$$\left(x+2x(lny)^2+lny\right)dx+\left(2x^2\frac{lny}{y}+\frac{x}{y}\right)dy=0$$</span></p>
<p><span class="math-container">$$xdx+(lny)^2d(x^2)+x^2d((l... |
3,455,967 | <p>Solve: (Hint: use <span class="math-container">$x\ln y=t$</span>)<span class="math-container">$$(xy+2xy\ln^2y+y\ln y)\text{d}x+(2x^2\ln y+x)\text{d}y=0$$</span></p>
<p>My Work:</p>
<p><span class="math-container">$$x\ln y=t, \text{ d}t=\ln y \text{ d}x+\frac{x}{y} \text{ d}y$$</span></p>
<p><span class="math-cont... | user577215664 | 475,762 | <p><span class="math-container">$$(xy+2xy\ln^2 y+y\ln y)dx+(2x^2\ln y+x)dy=0$$</span>
<span class="math-container">$$(x+2x\ln^2 y+\ln y )+(2x^2\ln y+x)(\ln y)'=0$$</span>
2. How we could recognize that we should use xlny=t. If question didn't Hint? </p>
<p>Only <span class="math-container">$\ln y $</span> terms in th... |
4,093,406 | <p>Which of the equations have at least two real roots?
<span class="math-container">\begin{aligned}
x^4-5x^2-36 & = 0 & (1) \\
x^4-13x^2+36 & = 0 & (2) \\
4x^4-10x^2+25 & = 0 & (3)
\end{aligned}</span>
I wasn't able to notice something clever, so I solved each of the equations. The first one h... | Gregory | 197,701 | <p>Descartes' rule of signs says that:</p>
<ul>
<li>1 pos and 1 neg real root for the first one.</li>
<li>2,0 pos and 2,0 neg for the second and third one.</li>
</ul>
<p>Thus the first one you know already that it has two real roots. For the second and third, you can use the discrimant to see that the third has no real... |
4,093,406 | <p>Which of the equations have at least two real roots?
<span class="math-container">\begin{aligned}
x^4-5x^2-36 & = 0 & (1) \\
x^4-13x^2+36 & = 0 & (2) \\
4x^4-10x^2+25 & = 0 & (3)
\end{aligned}</span>
I wasn't able to notice something clever, so I solved each of the equations. The first one h... | Aditya Singh | 1,012,095 | <p>All you need to do is</p>
<p>let <span class="math-container">$x^2=a$</span></p>
<p>The above equations will become so</p>
<p><span class="math-container">$a^4−5a^2−36=0$</span></p>
<p><span class="math-container">$a^4−13a^2+36=0$</span></p>
<p><span class="math-container">$4a^4−10a^2+25=0$</span></p>
<p>And now cal... |
2,707,299 | <p>Suppose that the conditional distribution of $Y$ given $X=x$ is Poisson with mean $E(Y|x)=x$, $Y|x\sim POI(x)$, and that $X\sim EXP(1)$</p>
<p>Find $V[Y]$</p>
<p>We know that</p>
<p>$$V[Y]=E_x[Var(Y|X)] + Var_x[E(Y|X)]$$</p>
<p>we also know that </p>
<p>$$V(Y|x)=E(Y^2|x)-[E(Y|x)]^2$$</p>
<p>I know we can find ... | zoli | 203,663 | <p>If the conditional distribution of $Y$ given $X$ is Poisson with parameter $X$ then the conditional expectation is also $X$. That is, if $X$ is of exponential distribution with parameter one then</p>
<p>$$E[Y]=\int_0^{\infty}E[Y\mid X=x\,]e^{-x}\ dx=\int_0^{\infty}xe^{-x}\ dx=E[X]=1.$$</p>
<p>Also,</p>
<p>$$E[Y^2... |
2,078,264 | <p>I've been see the following question on group theory:</p>
<p>Let $p$ be a prime, and let $G = SL2(p)$ be the group of $2 \times 2$ matrices of determinant $1$ with entries in the field $F(p)$ of integers $\mod p$.</p>
<p>(i) Define the action of $G$ on $X = F(p) \cup \{ \infty \}$ by Mobius transformations. [You nee... | Andreas Caranti | 58,401 | <p>$\newcommand{\Set}[1]{\left\{ #1 \right\}}$
You have $$f(z) = \frac{a z + b}{c z + d},$$ and
$$f(\infty) = \frac{a \infty + b}{c \infty + d} = \frac{a}{c}.$$</p>
<p>This is best understood by letting the group act on the projective line
$$
\mathcal{P} = \Set{[x, y] : x, y \in F(p), (x, y) \ne (0, 0)},
$$
by
$$
f([x... |
3,007,664 | <p>I can't find the right approach to tackle the question whether
<span class="math-container">$$\lim_{N\to\infty} \sum_{n=1}^N \exp\Bigg(-N \sin^2\left(\frac{n\pi}{2N}\right)\Bigg)$$</span>
and
<span class="math-container">$$\lim_{N\to\infty} \sum_{n=1}^N \exp\Biggl(-\sin^2\left(\frac{n\pi}{2N}\right)\Biggr)$$</span>
... | Myunghyun Song | 609,441 | <p>This is only a partial result (EDIT: full result is given below), but I'll post it anyway. Let us write <span class="math-container">$(x,y)$</span> if <span class="math-container">$y = f(x)$</span> holds. Since <span class="math-container">$f(f(x)) = x$</span>, <span class="math-container">$(x,y)$</span> is equivale... |
3,007,664 | <p>I can't find the right approach to tackle the question whether
<span class="math-container">$$\lim_{N\to\infty} \sum_{n=1}^N \exp\Bigg(-N \sin^2\left(\frac{n\pi}{2N}\right)\Bigg)$$</span>
and
<span class="math-container">$$\lim_{N\to\infty} \sum_{n=1}^N \exp\Biggl(-\sin^2\left(\frac{n\pi}{2N}\right)\Biggr)$$</span>
... | Tianlalu | 394,456 | <p>There is a proof given by Professor Wu Wei-chao in 2001, you may download the paper <a href="http://xueshu.baidu.com/usercenter/paper/show?paperid=c09c9061b96b50fcbf62d3e3f4963de6&site=xueshu_se" rel="nofollow noreferrer">here</a>. </p>
<p>Here is the overview of the proof:</p>
<ul>
<li><p><span class="math-co... |
1,761,189 | <p>Suppose that voters arrive to a voting booth according to a Poisson process with rate $\lambda = 100$ voters per hour. The voters will vote for two candidates, candidate $1$ and candidate $2$ with equal probability $\left(\frac{1}{2}\right)$. I am assuming that we start the voting at time $t=0$ and it continues inde... | Henry | 6,460 | <p>Each voter who turns up in ${10}$ hours has a probability of $\frac{4}{10}$ of being in the first $4$ hours and independently a probability of voting for the first candidate of $\frac12$, so a probability of $\frac15$ of both. </p>
<p>So, given $1000$ voters in total, you have a binomial distribution $A_4 \sim Bin(... |
2,814,793 | <p>I don't know what I did wrong. Can anyone point out my mistake? The problem is:</p>
<blockquote>
<p>For $\lim\limits_{x\to1}(2-1/x)=1$, finding $\delta$, such that if $0<|x-1|<\delta$, then $|f(x)-1|<0.1$</p>
</blockquote>
<p>Here is what I did:</p>
<p>Since $|f(x)-1|=|2-1/x-1|=|1-1/x|=|x-1|/x<0.1$,... | fleablood | 280,126 | <p>$\delta$ must be a constant determined by $.1$ and not dependent on the value of $x$.</p>
<p>(How could you chose an $x$ so that $|x -1| <\frac x{10}$ is we don't know what $x$ is in the first place?)</p>
<p>So you need to find a <em>constant</em> $\delta$ so that $|x - 1| < \delta \implies |2-\frac 1x-1|=|1... |
2,230,417 | <p>I need to formally prove that $f(x)=x^2$ is a contraction on each interval on $[0,a],0<a<0.5$. From intuition, we know that its derivative is in the range $(-1,1)$ implies that the distance between $f(x)$ and $f(y)$ is less then the distance between $x$ and $y$. </p>
<p>But now I need an explicit $\lambda$ su... | Hamza | 185,441 | <p>If $x\in[0,a]$ we get
$$
\sup_{t\in[0,a]}|f'(t)|=\sup_{t\in[0,a]}|2t|=2a
$$
So
$$
d(f(x),f(y))\leq 2a d(x,y) \qquad \forall x,y\in [0,a]
$$
and we can't find a constant less than $2a$, in fact for $x=a$ and $y=a-\epsilon$ for $a>\epsilon>0$ we get :
$$
\frac{d(f(x),f(y))}{d(x,y)}=\frac{d(a^2,(a-\epsilon)^2)... |
1,920,776 | <p>As I know there is a theorem, which says the following: </p>
<blockquote>
<p>The prime ideals of $B[y]$, where $B$ is a PID, are $(0), (f)$, for irreducible $f \in B[y]$, and all maximal ideals. Moreover, each maximal ideal is of the form $m = (p,q)$, where $p$ is an irreducible element in $B$ and $q$ is an irred... | egreg | 62,967 | <p>Let's look at a decimal example. You want to do $735-78$.</p>
<p>Borrow 1000 from the Number Bank; the loan is subject to no interest, but you <em>must</em> give back what you got as soon as you have used it.</p>
<p>Now consider that
$$
735-78=735+(1000-78)-1000
$$
The subtraction $1000-78$ is very easy to do: jus... |
1,920,776 | <p>As I know there is a theorem, which says the following: </p>
<blockquote>
<p>The prime ideals of $B[y]$, where $B$ is a PID, are $(0), (f)$, for irreducible $f \in B[y]$, and all maximal ideals. Moreover, each maximal ideal is of the form $m = (p,q)$, where $p$ is an irreducible element in $B$ and $q$ is an irred... | fleablood | 280,126 | <p>Basic assumption is that we are doing modulo arithmetic over some $2^{n+1}$. i.e. $x \equiv x + 2^{n+1}$ for all $x$. (I suppose I should write $x \equiv x + 2^{n+1} \pmod{2^{n+1}}$ but I'm not going to write the $(\operatorname{mod}{2^{n+1}})$ part.) Especially $0 \equiv 2^{n+1}$.</p>
<p>So what we want to show i... |
1,624,888 | <h2>Question</h2>
<p>In the following expression can $\epsilon$ be a matrix?</p>
<p>$$ (H + \epsilon H_1) ( |m\rangle +\epsilon|m_1\rangle + \epsilon^2 |m_2\rangle + \dots) = (E |m\rangle + \epsilon E|m_1\rangle + \epsilon^2 E_2 |m_2\rangle + \dots) ( |m\rangle +\epsilon|m_1\rangle + \epsilon^2 |m_2\rangle + \dots) $... | Dac0 | 291,786 | <p>I would say there's nothing preventing you from using a matrix as perturbation of a matrix equation as long as you let the limit of the norm converge to $0$.</p>
|
3,520,057 | <p>how would you draw <span class="math-container">$A\cup(B\cap C\cap D)$</span>
I tried with paint doesn't look good and I don't know if there are tools to make this easier would appreciate any help or tips</p>
<p><img src="https://i.stack.imgur.com/fc6Pq.png" alt="enter image description here"></p>
| Alex Kruckman | 7,062 | <blockquote>
<p>How can we say that the direct limit is the tensor product of the given family? </p>
</blockquote>
<p>Well, Atiyah and Macdonald are <em>defining</em> the tensor product of an infinite family of <span class="math-container">$A$</span>-algebras. We're free to make whatever definitions we want...</p>
... |
273,463 | <p>I want to solve this differential equation</p>
<p><span class="math-container">$$
\frac{1}{\Phi(\phi)} \frac{d^2 \Phi}{d \phi^2}=-m^2
$$</span></p>
<p>with b.c.</p>
<p><span class="math-container">$$
\Phi(\phi+2 \pi)=\Phi(\phi)
$$</span></p>
<p>it is easy to solve by wolfram</p>
<pre><code>eq1 = D[Phi[phi], {phi, 2}... | user293787 | 85,954 | <p>There is <a href="https://reference.wolfram.com/language/ref/PeriodicBoundaryCondition.html" rel="nofollow noreferrer">PeriodicBoundaryCondition</a>. The syntax is a bit complicated:</p>
<pre><code>PBC=PeriodicBoundaryCondition[Phi[phi],phi==2*Pi,phi|->phi-2*Pi];
</code></pre>
<p>This can be used with some solver... |
273,463 | <p>I want to solve this differential equation</p>
<p><span class="math-container">$$
\frac{1}{\Phi(\phi)} \frac{d^2 \Phi}{d \phi^2}=-m^2
$$</span></p>
<p>with b.c.</p>
<p><span class="math-container">$$
\Phi(\phi+2 \pi)=\Phi(\phi)
$$</span></p>
<p>it is easy to solve by wolfram</p>
<pre><code>eq1 = D[Phi[phi], {phi, 2}... | xzczd | 1,871 | <p>First of all, the deduction in <em>Quantum chemistry</em> by <em>McQuarrie</em> is not rigorous (if we don't call it incorrect). As shown in Nasser's answer, the general solution (when no b.c. is imposed) is</p>
<pre><code>generalsol[phi_] =
With[{ϕ = Phi[phi]},
Collect[DSolveValue[{D[ϕ, {phi, 2}]/ϕ == -m^2}, ϕ... |
3,299,072 | <p>I want to prove that
<span class="math-container">$$\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2}\geq 0$$</span>
for positive numbers <span class="math-container">$x,y,z$</span>.
I don't know how to even begin. I must say I'm not 100% certain the inequality always holds.</p>
<p>I tried the so... | Michael Rozenberg | 190,319 | <p>Let <span class="math-container">$x\geq y\geq z$</span>.</p>
<p>Thus, <span class="math-container">$$\sum_{cyc}\frac{(x-y)(x-z)}{x^2}\geq\frac{(x-z)(y-z)}{z^2}-\frac{(x-y)(y-z)}{y^2}=$$</span>
<span class="math-container">$$=(y-z)\left(\frac{x-z}{z^2}-\frac{x-y}{y^2}\right)\geq0$$</span>
because <span class="math-c... |
3,299,072 | <p>I want to prove that
<span class="math-container">$$\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2}\geq 0$$</span>
for positive numbers <span class="math-container">$x,y,z$</span>.
I don't know how to even begin. I must say I'm not 100% certain the inequality always holds.</p>
<p>I tried the so... | Gibbs | 498,844 | <p>There is no loss of generality in assuming <span class="math-container">$0 < x \leq y \leq z$</span>. Rewrite the inequality as
<span class="math-container">$$\frac{(x-y)^2+(y-z)(x-y)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-y)^2+(y-x)(z-y)}{z^2} \geq 0$$</span>
and rearrange the terms as follows:
<span class="math-... |
1,781,467 | <p>First, I know what the right answer is, and I know how to solve it. What I'm trying to figure out is why I can't get the following process to work.</p>
<p>The probability that we get 2 consecutive heads with one flip is 0. The probability that we get 2 consecutive heads with 2 flips = 1/4. The probability of gettin... | Jonathan | 338,908 | <p>The definition of expected value in this case is:</p>
<p>$$E[x]=\sum_{x=2}^{\infty}xP[x]$$</p>
<p>and</p>
<p>$$P[x]=\frac{x-1}{2^x}$$</p>
<p>Combining these yields:</p>
<p>$$E[x]=\sum_{x=2}^{\infty}\frac{x^2-x}{2^x}$$</p>
<p>Therefore $E[x]=4$.</p>
|
3,422,861 | <p>The problem is as follows:</p>
<blockquote>
<p>The acceleration of an oscillating sphere is defined by the equation
<span class="math-container">$a=-ks$</span>. Find the value of <span class="math-container">$k$</span> such as <span class="math-container">$v=10\,\frac{cm}{s}$</span> when <span class="math-conta... | IamWill | 389,232 | <p>The term oscillating is very generic, so this could be a pendulum for example. I'm assuming <span class="math-container">$s$</span> stands for the position of the sphere. Then, notice that <span class="math-container">$$a =\frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt} = \frac{dv}{ds}v$$</span>
Therefore, we have <span ... |
264,450 | <p>We know that the class of open intervals $(a,b)$, where $a,b$ are rational numbers is a countable base for $\mathbb R$. </p>
<p>But, $[a,b]$ where $a,b$ are rational numbers does not produce a base for $\mathbb R$.</p>
<p>Can we say that any $(a,b)$ or $[a,b]$ where <strong>$a$ is rational number and $b$ is an ir... | Daron | 53,993 | <p>I am not sure if this is related to your question at all, but we could say that the family $\{ [a,b]\ \colon a,b \in \mathbb R, a <c< b\}$ form a <strong>neighborhood basis</strong> at the point $c$ with respect to the Euclidean topology. This is saying that $c$ is in the interior of $U$ iff there is some c... |
2,094,698 | <p>I am trying to figure out how to create a formula to solve this problem I have.</p>
<p>OK, the problem</p>
<p>I need to work how much money to give people bassed on a percentage of a total value, but the problem is that the percentage varies from person to person.</p>
<p>Ie,</p>
<p>Total amount = $1,000,000</p>
... | KyloRen | 406,318 | <p>So after all this time I figured out how to solve the issue. And I am also going to say that there is an issue with the wording of my question which will throw confusion as to what I really wanted in the first place. </p>
<p>So as the question states,</p>
<blockquote>
<p>Total amount = $1,000,000</p>
<p>Tot... |
1,526,014 | <p>I have no idea how to compute this infinite sum. It seems to pass the convergence test. It even seems to be equal to $\frac{p}{(1-p)^2}$, but I cannot prove it. Any insightful piece of advice will be appreciated.</p>
| Mark Viola | 218,419 | <p>Assume that $|p|<1$.</p>
<p><strong>METHOD $1$:</strong></p>
<p>$$p\left(\frac{d}{dp}\sum_{k=1}^\infty p^k\right)=\sum_{k=1}^{\infty}kp^k$$</p>
<hr>
<p><strong>METHOD $2$:</strong></p>
<p>$$\begin{align}
\sum_{k=1}^\infty kp^k&=\sum_{k=1}^{\infty}p^k\sum_{j=1}^k (1)\\\\
&=\sum_{j=1}^{\infty}\sum_{k=j... |
3,070,127 | <p>Let <span class="math-container">$B := \{(x, y) \in \mathbb{R}^2
: x^2 + y^2 \le 1\}$</span>be the closed ball in <span class="math-container">$\mathbb{R^2}$</span> with center at the origin.
Let I denote the unit interval <span class="math-container">$[0, 1].$</span> Which of the following statements are true?</p>
... | Caleb Stanford | 68,107 | <p>You can also arrive at jmerry's closed form answer directly from your answer. You got:
<span class="math-container">$$\sum_{j=0}^{n-1}\binom{2n}{j}$$</span></p>
<p>Now, notice that
<span class="math-container">$$\sum_{j=0}^{n-1}\binom{2n}{j} = \sum_{j=0}^{n-1} \binom{2n}{2n-j} = \sum_{j=n+1}^{2n} \binom{2n}{j}.$$</... |
3,072,274 | <p><a href="https://i.stack.imgur.com/WaOF8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WaOF8.png" alt="enter image description here"></a></p>
<blockquote>
<p>Let <span class="math-container">$n\in\mathbb N$</span> be fixed. For <span class="math-container">$0\le k \le n$</span>, et <span clas... | Robert Lewis | 67,071 | <p>Yes, <span class="math-container">$H$</span> is normal in <span class="math-container">$G$</span>; for</p>
<p><span class="math-container">$h \in H, \; g \in G \tag 1$</span></p>
<p>we have</p>
<p><span class="math-container">$$\begin{align}
\det (g^{-1}hg) &= \det(g^{-1}) \det(h) \det(g)\\
& = \det(h) \d... |
307,287 | <p>According to <a href="https://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem" rel="nofollow noreferrer">Brouwer's fixed point theorem</a>, for compact convex $K\subset\mathbb{R}^n$, every continuous map $K\rightarrow K$ has a fixed point.</p>
<p>However, these fixed points cannot be chosen continuously, even for... | Martin Kell | 123,897 | <p>There can be no such function even in the category of smooth functions. </p>
<p>Here an example for functions $f:[0,1]\to [0,1]$ with $f(0)=\epsilon$ and $f(1)=1-\epsilon$: </p>
<p>(1) Let $\operatorname{id}:x\mapsto x$ and choose a smooth function $f_0$ with $f_0=\operatorname{id}$ on $[\frac{1}{3},\frac{2}{3}]$... |
3,497,328 | <p>I'm stuck. Can I get a hint? I heard the answer is zero.</p>
<p>I'm guessing we use the SSA congruent triangle theorem. </p>
<p>If <span class="math-container">$m∠A = 50°$</span>, side <span class="math-container">$a = 6$</span> units, and side <span class="math-container">$b = 10$</span> units, what is the maximu... | N. F. Taussig | 173,070 | <p>The <a href="https://en.wikipedia.org/wiki/Law_of_sines" rel="nofollow noreferrer">Law of Sines</a> states that
<span class="math-container">$$\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}$$</span>
Therefore,
<span class="math-container">$$\sin\beta = \frac{b\sin\alpha}{a} = \frac{10\sin(50^\cir... |
286,852 | <p>I'm currently trying to understand time continuous Fourier transformations for my Signal Analysis class (Electrical Engineering major). In my Textbook we have a set of "useful FT-pairs", one of which being
$x(t)<=FT=>X(\omega)$</p>
<p>$\sin(\omega_0*t)<=>i\pi[\delta(\omega+\omega_0)-\delta(\omega-\omega... | Dirk | 3,148 | <p>Your observations are right. The Fourier transform of $x\mapsto\sin(ax)$ is not defined by the integral but rather in a <em>weak sense</em> or in the sense of distributions (http://en.wikipedia.org/wiki/Distribution_(mathematics)#Tempered_distributions_and_Fourier_transform). This is also explained on the Wikipedia ... |
1,926,593 | <p>I am currently taking a course in Discrete Math. The first part of our lesson this week is regarding sequences. I am stuck on formulas like the ones shown in the images I attached... I was hoping someone might be able to help me learn how to solve them. :)</p>
<p>Ps: What does it mean when <span class="math-containe... | JMP | 210,189 | <p>For the inductive step, assume:</p>
<p>$$\sum_{i=0}^n \dbinom{i}{m} = \dbinom{n+1}{m+1}$$</p>
<p>Add $\dbinom{n+1}{m}$ to both sides to give:</p>
<p>$$\sum_{i=0}^{n+1} \dbinom{i}{m} = \dbinom{n+1}{m}+\dbinom{n+1}{m+1}=\dbinom{n+2}{m+1}$$</p>
|
2,811,908 | <p>I have this monster as the part of a longer calculation. My goal would be to somehow make it... nicer.</p>
<p>Intuitively, I would try to somehow utilize the derivate and the integral against each other, but I have no idea, exactly how.</p>
<p>I suspect, this might be a relative common problem, i.e. if we want to ... | GEdgar | 442 | <p>This is probably in your calculus textbook ... under reasonable conditions,</p>
<p>$$
\frac{d}{db}\int_0^1e^{bx}f(x)dx = \int_0^1 \frac{\partial}{\partial b}e^{bx}f(x)dx = \int_0^1 x e^{bx}f(x)dx
$$</p>
|
2,806,164 | <p>I recently came across a question that asked for the derivative of $e^x$ with respect to $y$. I answered $\frac{d}{dy}e^x$ but the answer was $e^x\frac{dx}{dy}$. How is that the answer? I am confused.</p>
| peterh | 111,704 | <p>Informally, the question understood it on the way, that <em>how the $y$ coordinate changes of the curve $(x;e^x)$ for the infinitesimally small changes of the $x$ coordinate</em>. This is the common interpretation of the derivate, but it is not the only one. You have probably got this question in differential calcul... |
2,806,164 | <p>I recently came across a question that asked for the derivative of $e^x$ with respect to $y$. I answered $\frac{d}{dy}e^x$ but the answer was $e^x\frac{dx}{dy}$. How is that the answer? I am confused.</p>
| mr_e_man | 472,818 | <p>The question you have been asked is a bad question, if you weren't given more information.</p>
<p>If $e^x$ is considered a function of two (independent) variables $x$ and $y$, then "derivative" probably means "partial derivative", and $\frac{\partial}{\partial y}e^x=0$.</p>
<p>If there is some relation between $x$... |
4,480,276 | <p>[This question is rather a very easy one which I found to be a little bit tough for me to grasp. If there is any other question that has been asked earlier which addresses the same topic then kindly link this here as I am unable to find such questions by far.]</p>
<p>let <span class="math-container">$f(x)=x.$</span... | Átila Correia | 953,679 | <p>I think it is worth working with the definition of limit.</p>
<p>Let <span class="math-container">$f:X\to Y$</span> be a real-valued function with real domain s.t. <span class="math-container">$a\in\mathbb{R}$</span> is an accumulation point of <span class="math-container">$X$</span>.</p>
<p>We say the limit of <spa... |
1,648,354 | <p>We have been taught that linear functions, usually expressed in the form $y=mx+b$, when given a input of 0,1,2,3, etc..., you can get from one output to the next by adding some constant (in this case, 1).
$$
\begin{array}{c|l}
\text{Input} & \text{Output} \\
\hline
0 & 1\\
1 & 2\\
2 & 3
\end{array}
... | Brady Gilg | 188,927 | <p>Each of your tables is found by taking 2 to the power of the previous table (possibly with a shift by 1). As the formula for your second table is $2^x$, the formula for your third table is $2^{2^{x-1}}$ (for $x \geq 1$).</p>
|
2,253,676 | <p>$f(x)$ continuous in $[0,\infty)$. Both $\int^\infty_0f^4(x)dx, \int^\infty_0|f(x)|dx$ converge. </p>
<p>I need to prove that $\int^\infty_0f^2(x)dx$ converges.</p>
<p>Knowing that $\int^\infty_0|f(x)|dx$ converges, I can tell that $\int^\infty_0f(x)dx$ converges.</p>
<p>Other than that, I don't know anything. I ... | Angina Seng | 436,618 | <p>By AM/GM
$$3f(x)^2\le f(x)^4+|f(x)|+|f(x)|.$$</p>
|
2,253,676 | <p>$f(x)$ continuous in $[0,\infty)$. Both $\int^\infty_0f^4(x)dx, \int^\infty_0|f(x)|dx$ converge. </p>
<p>I need to prove that $\int^\infty_0f^2(x)dx$ converges.</p>
<p>Knowing that $\int^\infty_0|f(x)|dx$ converges, I can tell that $\int^\infty_0f(x)dx$ converges.</p>
<p>Other than that, I don't know anything. I ... | zhw. | 228,045 | <p>Let $A = \{|f|\le 1\}, B = \{|f|>1 \}.$ Then $\int_A f^2 \le \int_A |f|<\infty$ and $\int_B f^2 \le \int_B f^4<\infty.$ The result follows.</p>
|
2,013,382 | <blockquote>
<p>If $AX=\lambda X$ for every eigenvector $X\in \mathbb{R}^n$ then show that $A=\lambda I_n$.</p>
</blockquote>
<p>Given, $AX=\lambda X$ for every $X\in \mathbb{R}^n$.</p>
<p>Then $(A-\lambda I_n)X=0$ is true for every $X\in \mathbb{R}^n$.</p>
<p>But how to show $A=\lambda I_n$? Please help.</p>
| Emilio Novati | 187,568 | <p>Hint:</p>
<p>If $\vec e_i$ is the vector of the standard basis with components $\delta_{i,j}$, than
$A\vec e_i$ is the column $i$ of the matrix $A$.</p>
<p>And we have $A\vec e_i=\lambda \vec e_i=\lambda \delta_{i,j}$, so....</p>
<hr>
<p>For $\vec e_1=[1,0,0]^T$ we have
$$
A=\begin{bmatrix}
a_{11}&a_{12}&a... |
3,447,872 | <p>Subtracting both equations
<span class="math-container">$$x(a-b)+b-a=0$$</span>
<span class="math-container">$$(x-1)(a-b)=0$$</span>
Since <span class="math-container">$a\not = b$</span>
<span class="math-container">$$x=1$$</span>
Substitution of x gives
<span class="math-container">$$1+a+b=0$$</span> which is con... | Elke Hennig | 728,267 | <p>You assume that both equations have the same solution. This is not stated in the problem. Only the difference of the two solutions to each equation is identical.</p>
|
3,447,872 | <p>Subtracting both equations
<span class="math-container">$$x(a-b)+b-a=0$$</span>
<span class="math-container">$$(x-1)(a-b)=0$$</span>
Since <span class="math-container">$a\not = b$</span>
<span class="math-container">$$x=1$$</span>
Substitution of x gives
<span class="math-container">$$1+a+b=0$$</span> which is con... | lab bhattacharjee | 33,337 | <p>Hint:</p>
<p>You have wrongly assumed a common root, then only the subtraction makes sense.</p>
<p>If <span class="math-container">$p,q;(p\ge q)$</span> are the roots of <span class="math-container">$$ct^2+dt+e=0$$</span></p>
<p><span class="math-container">$p-q=\dfrac{-d+\sqrt{d^2-4ce}-(-d-\sqrt{d^2-4ce)}}2=?$</... |
3,301,387 | <p>I don't understand what it is about complex numbers that allows this to be true.</p>
<p>I mean, if I root both sides I end up with Z=Z+1. Why is this a bad first step when dealing with complex numbers?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>if so then it would be
<span class="math-container">$$z^6+6z^5+15z^4+20z^3+15z^2+6z+1=z^6$$</span>
or factorized
<span class="math-container">$$ \left( 2\,z+1 \right) \left( {z}^{2}+z+1 \right) \left( 3\,{z}^{2}+
3\,z+1 \right)
=0$$</span></p>
|
286,082 | <p><a href="https://en.wikipedia.org/wiki/Budan's_theorem" rel="nofollow noreferrer">Budan's theorem</a> gives an upper bound for the number of real roots of a real polynomial in a given interval $(a,b)$. This bound is not sharp (see the example in Wikipedia).</p>
<p>My question is the following: let us suppose th... | Harry Richman | 81,295 | <p>You could take
$$ f(x) = x^4 + 1$$
as a counterexample. Budan's theorem would tell you there's at most 4 roots in the interval $(-\epsilon, \epsilon)$ for any positive $\epsilon$. But $f(x)$ doesn't have any roots with real part 0.</p>
<hr>
<p>Also $x^3 + 1$ works for the same reason. But degree 3 is minimal -- so... |
1,377,927 | <p>Prove using mathematical induction that
$(x^{2n} - y^{2n})$ is divisible by $(x+y)$.</p>
<p><strong>Step 1:</strong> Proving that the equation is true for $n=1 $</p>
<p>$(x^{2\cdot 1} - y^{2\cdot 1})$ is divisible by $(x+y)$ </p>
<p><strong>Step 2:</strong> Taking $n=k$</p>
<p>$(x^{2k} - y^{2k})$ is divisible ... | Bernard | 202,857 | <p><strong>Hint:</strong> Rewrite:
$$x^{2k+2}-y^{2k+2}=(x^{2k+2}-y^{2k}x^2)+(y^{2k}x^2-y^{2k+2})=x^2(x^{2k}-y^{2k})+y^{2k}(x^2-y^2).$$</p>
<p><strong>Added:</strong> Note it can be proved without induction:
$$x^{2n}-y^{2n}=(x^2)^n-(y^2)^n=(x^2-y^2)(x^{2(n-1)}+x^{2(n-2)}y^2+\dots+x^2y^{2(n-2)}+y^{2(n-1)}),$$
and the fi... |
1,377,927 | <p>Prove using mathematical induction that
$(x^{2n} - y^{2n})$ is divisible by $(x+y)$.</p>
<p><strong>Step 1:</strong> Proving that the equation is true for $n=1 $</p>
<p>$(x^{2\cdot 1} - y^{2\cdot 1})$ is divisible by $(x+y)$ </p>
<p><strong>Step 2:</strong> Taking $n=k$</p>
<p>$(x^{2k} - y^{2k})$ is divisible ... | Harish Chandra Rajpoot | 210,295 | <p>Step 1: putting $n=1$, we get $$x^{2n}-y^{2n}=x^2-y^2=(x-y)(x+y)$$ above number is divisible by $(x+y)$. Hence the statement is true $n=1$ </p>
<p>step 2: assuming that for $n=m$, $(x^{2n}-y^{2n})$ is divisible by $(x+y)$ then we have $$(x^{2m}-y^{2m})=k(x+y) \tag 1$$
Where, $k$ is an integer </p>
<p>step 3: put... |
91,766 | <p>Hello,
I am looking for a proof for the Chern-Gauss-Bonnet theorem. All I have found so far that I find satisfactory is a proof that the euler class defined via Chern-Weil theory is equal to the pullback of the Thom class by the zero section, but I would like a proof of the fact that this class gives the Euler char... | Renato G. Bettiol | 15,743 | <p>One reference that seems fairly good and that I just found by googling those key words is <a href="https://web.archive.org/web/20100524152105/http://www.math.upenn.edu/~alina/GaussBonnetFormula.pdf" rel="nofollow noreferrer">https://web.archive.org/web/20100524152105/http://www.math.upenn.edu/~alina/GaussBonnetFormu... |
4,219,193 | <p>The <code>p implies q</code> statement is often described in various ways including:<br />
(1) <code>if p then q</code> (i.e. whenever p is true, q is true)<br />
(2) <code>p only if q</code> (i.e. whenever q is false, p is false)</p>
<p>I see the truth table for (1) as</p>
<pre><code>p | q | if p then q
-----------... | Al Brown | 955,529 | <p>Truth table last</p>
<p>First, on necessary and sufficient:</p>
<p>If A, then B. This means that anytime you have A, you will have B. It does not say that anytime you have B, you <strong>will</strong> have A. There could be times with B but not A. If anytime you have A you have B, then B is <strong>necessary</strong... |
533,628 | <p>I'm learning how to take indefinite integrals with U-substitutions on <a href="https://www.khanacademy.org/math/calculus/integral-calculus/u_substitution/v/u-substitution" rel="nofollow">khanacademy.org</a>, and in one of the videos he says that: $$\int e^{x^3+x^2}(3x^2+2x) \, dx = e^{x^3+x^2} + \text{constant}$$
I ... | Elchanan Solomon | 647 | <p>Here is the intuition behind the proof. As you have done, we split the sum into two pieces. Each piece is controlled differently.</p>
<p>To control the first piece of the sum, we note that our sequence is bounded by some constant $M$, so that this sum is at most</p>
<p>$$M\frac{m}{n}$$</p>
<p>That is, $m$ terms o... |
2,629,115 | <blockquote>
<p>We will define the convolution (slightly unconventionally to match Rudin's proof) of $f$ and $g$ as follows:
$$(f\star g)(x)=\int_{-1}^1f(x+t)g(t)\,dt\qquad(0\le x\le1)$$</p>
<ol>
<li>Let $\delta_n(x)$ be defined as $\frac n2$ for $-\frac1n<x<\frac1n$ and 0 for all other $x$. Let $f(x)$... | Quoka | 410,062 | <p>By definition of convolution and $\delta_{10}$, we have;
\begin{align*}
\left(f*\delta_{10}\right)(x) = \int_{-1}^1 f(x+t)\delta_{10}(t)\, dt
=5\int_{-1/10}^{1/10} f(x+t)\, dt
=5\int_{x-1/10}^{x+1/10} f(t)\, dt
\end{align*}
Now, $f(t) = 0$ outside of the interval $(0.4, 0.7)$. Thus, the above integral evaluates to $... |
3,766,146 | <p>Show that:</p>
<p><span class="math-container">$$\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\int\limits_1^2 \frac{dx}{x}=\ln(2)$$</span></p>
<hr />
<p><strong>My attempt:</strong></p>
<p>We build a Riemann sum with:</p>
<p><span class="math-container">$1=x_0<x_1<...<x_{N-1}<x_N=2$</... | Claude Leibovici | 82,404 | <p><em>Without Rieman sums.</em>
<span class="math-container">$$S_N=\sum\limits_{n=1}^N\frac{1}{N+n}=H_{2 N}-H_N$$</span> Using the asymptotics of harmonic numbers
<span class="math-container">$$S_N=\log (2)-\frac{1}{4 N}+\frac{1}{16 N^2}+O\left(\frac{1}{N^4}\right)$$</span></p>
|
3,766,146 | <p>Show that:</p>
<p><span class="math-container">$$\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\int\limits_1^2 \frac{dx}{x}=\ln(2)$$</span></p>
<hr />
<p><strong>My attempt:</strong></p>
<p>We build a Riemann sum with:</p>
<p><span class="math-container">$1=x_0<x_1<...<x_{N-1}<x_N=2$</... | Community | -1 | <p>Your reasoning is correct, but you make it more complicated than needed.</p>
<p><span class="math-container">$$\frac1N\sum_{i=1}^N\frac1{1+\dfrac nN}\to\int_0^1\frac{dx}{1+x}=\left.\ln(1+x)\right|_0^1.$$</span></p>
|
31,790 | <p>See following three snippet code:</p>
<pre><code>(*can compile*)
range = Range[-2, 2, 0.005];
Compile[{},
With[{r = range}, Table[ArcTan[x, y], {x, r}, {y, r}]]][];
(*can compile*)
Compile[{},
With[{r = Range[-2, 2, 0.005 - 10^-8]}, Table[ArcTan[x, y], {x, r}, {y, r}]]][];
(*can't compile*)
Compile[{},
... | Jens | 245 | <p>In addition to Mr. Wizard's analysis, one can also avoid the indeterminacy by replacing <code>ArcTan</code> as follows:</p>
<pre><code>Compile[{},
With[{r = Range[-2, 2, 0.005]},
Table[Arg[Complex[x, y]], {x, r}, {y, r}]]][];
</code></pre>
<p>The fact that <code>ArcTan[0,0]</code> is undefined is a real n... |
4,396,765 | <p>What would be an efficient algorithm to determine if <span class="math-container">$n \in \mathbb{N}$</span> can be written as <span class="math-container">$n = a^b$</span> for some <span class="math-container">$a,b \in \mathbb{N}, b>1$</span>?</p>
<p>So far, I've tried:</p>
<pre><code>def ispower(n):
if n<... | J.-E. Pin | 89,374 | <p>Since <span class="math-container">$n = n^1$</span>, every number can be written in the required form. Thus an efficient algorithm is to always answer yes.</p>
|
2,411,081 | <p>How can I show (with one of the tests for convergence , <strong>not by solving</strong>) that the integral $$\int _{1}^\infty\frac{\ln^5(x)}{x^2}dx$$ converges?</p>
| Enrico M. | 266,764 | <p>Abel's theorem: let $g = fh$ be defined on $[a, b)$ with $b \in (a, \infty]$. If it holds that:</p>
<ul>
<li><p>$h$ is continuous, positive, decreasing and $\lim_{x \to b} h(x) = 0$</p></li>
<li><p>$f$ is locally integrable on $[a, b)$ (i.e. integrable in the neighbourhood of any point) and $F$ is bounded on $[a, b... |
2,411,081 | <p>How can I show (with one of the tests for convergence , <strong>not by solving</strong>) that the integral $$\int _{1}^\infty\frac{\ln^5(x)}{x^2}dx$$ converges?</p>
| marty cohen | 13,079 | <p>For any $a, b >0$,
$(\log x)^a/x^b \to 0$
as $x \to \infty$.</p>
<p>Therefore
$\int_1^{\infty}\dfrac{(\log x)^adx}{x^{1+b}}
$
converges for
$a,b >0$
(and diverges for
$b=0$).</p>
|
2,411,081 | <p>How can I show (with one of the tests for convergence , <strong>not by solving</strong>) that the integral $$\int _{1}^\infty\frac{\ln^5(x)}{x^2}dx$$ converges?</p>
| Stu | 460,772 | <p>$\displaystyle \mathcal{I}=\int _{1}^\infty\frac{\ln^5(x)}{x^2}dx$</p>
<p>As $\displaystyle \lim_{x\to +\infty}\frac{\ln^5(x)}{\sqrt{x}}=0\implies \exists a>0, \text{such that } \forall x>a \quad \frac{\ln^5(x)}{\sqrt{x}}<1\iff\frac{\ln^5(x)}{x^2}<\frac{1}{x^{\frac{3}{2}}}$</p>
<p>$\displaystyle\implie... |
1,397,646 | <blockquote>
<p>Given $f_n:= e^{-n(nx-1)^2} $, any suggested approaches for showing that $\displaystyle \lim_{n \to \infty}\int^1_0 f_n(x)dx = 0$? </p>
</blockquote>
<p>I have already shown that $f_n \to 0$ pointwise (and not uniformly) on $[0,1]$ and have been trying to show that $$ \lim_{n \to \infty}\int^1_0 f_n(... | Zhanxiong | 192,408 | <p>Break the interval to $[0, 1/n]$ and $[1/n, 1]$ is a good idea, but may be still too stringent, so why don't try $[0, 2/n]$ and $[2/n, 1]$?</p>
<p>On the interval $[0, 2/n]$, the integrand is bounded by $1$.</p>
<p>On the interval $[2/n, 1]$, the integrand is bounded by
$$\exp(-n(2 - 1)^2) = e^{-n}.$$
Therefore, ... |
2,409,377 | <p>I'm trying to prove that if $F \simeq h_C(X)$ or "$X$ represents the functor $F$", then $X$ is unique up to unique isomorphism. I already know that if $h_C(X) \simeq F \simeq h_C(Y)$ that $s: X \simeq Y$ since Yoneda says that $h_C(X)$ is fully faithful, so reflects isomorphisms (in either direction). If $h_C(X) ... | D_S | 28,556 | <p>The Yoneda lemma gives you a canonical bijection $$d: \textrm{Hom}_{C}(X,Y) \rightarrow \textrm{Hom}_{\textrm{nat}}(\textrm{Hom}_C(-,X),\textrm{Hom}_C(-,Y))$$</p>
<p>Specifically, to each morphism $f: X \rightarrow Y$, associate the natural transformation $d(f): \textrm{Hom}_C(-,X) \rightarrow {Hom}_{C}(-,Y)$ which... |
2,662,792 | <p>I'm looking for a Galois extension $F$ of $\mathbb{Q}$ whose associated Galois group $\mbox{Gal}(F, \mathbb{Q})$ is isomorphic to $\mathbb{Z}_3 \oplus \mathbb{Z}_3$. I'm wondering if I should just consider some $u \notin Q$ whose minimum polynomial in $Q[x]$ has degree 9. In that case we'd have $[\mathbb{Q}(u) : \ma... | David Lui | 445,002 | <p><a href="http://www.math.ku.edu/~dlk/abgal.pdf" rel="nofollow noreferrer">http://www.math.ku.edu/~dlk/abgal.pdf</a></p>
<p>This paper might help. It solves the inverse Galois problem for all abelian groups, and your group is abelian.</p>
|
2,662,792 | <p>I'm looking for a Galois extension $F$ of $\mathbb{Q}$ whose associated Galois group $\mbox{Gal}(F, \mathbb{Q})$ is isomorphic to $\mathbb{Z}_3 \oplus \mathbb{Z}_3$. I'm wondering if I should just consider some $u \notin Q$ whose minimum polynomial in $Q[x]$ has degree 9. In that case we'd have $[\mathbb{Q}(u) : \ma... | Adrián Barquero | 900 | <p>This is something that nowadays can be easily done by using the <a href="http://www.lmfdb.org/" rel="nofollow noreferrer">LMFDB Database</a>.</p>
<p>For example, if you go to the above link, on the left hand side you'll find a column with the words Number Fields and the options <a href="http://www.lmfdb.org/NumberF... |
2,662,792 | <p>I'm looking for a Galois extension $F$ of $\mathbb{Q}$ whose associated Galois group $\mbox{Gal}(F, \mathbb{Q})$ is isomorphic to $\mathbb{Z}_3 \oplus \mathbb{Z}_3$. I'm wondering if I should just consider some $u \notin Q$ whose minimum polynomial in $Q[x]$ has degree 9. In that case we'd have $[\mathbb{Q}(u) : \ma... | Angina Seng | 436,618 | <p>$F=\Bbb Q(\cos(2\pi/7),\cos(2\pi/9))$ works.</p>
<p>It's the compositum of two linear disjoint Galois cubic extensions.</p>
|
9,540 | <p>I'm following <a href="http://reference.wolfram.com/mathematica/ref/FinancialData.html">http://reference.wolfram.com/mathematica/ref/FinancialData.html</a></p>
<p>I get the following:</p>
<pre><code>In[6]:= DateListLogPlot[FinancialData["^DJI", All]]
</code></pre>
<blockquote>
<p>During evaluation of In[6]:= Da... | Vitaliy Kaurov | 13 | <p>If one looks for curated data accessible from Mathematica, the <code>WolframAlpha</code> function should always be considered as an option, because it links to curated data bases with frequent continuous updating:</p>
<pre><code>data = WolframAlpha["^DJI price history",
{{"HistoryDaily:Close:FinancialData... |
1,897,212 | <p>Does there exist any surjective group homomorphism from <span class="math-container">$(\mathbb R^* , .)$</span> (the multiplicative group of non-zero real numbers) onto <span class="math-container">$(\mathbb Q^* , .)$</span> (the multiplicative group of non-zero rational numbers)? </p>
| quid | 85,306 | <p>Let me add a more abstract view to the concrete arguments already given. </p>
<p>Recall that a group $(G,\cdot)$ is called divisible if for each $a\in G$, positive integer $n$, the equation $X^n = a$ has a solution. </p>
<p>(The terminology makes more sense in additive notation where it means $nX=a$ has a solution... |
2,439,278 | <p>I wanted to solve the following problem.</p>
<blockquote>
<p>In $\triangle ABC$ we have $$\sin^2 A + \sin^2 B = \sin^2 C + \sin A \sin B \sin C.$$ Compute $\sin C$.</p>
</blockquote>
<p>Since it's an equation for a triangle, I assumed that $\pi = A + B + C$ would be important to consider.</p>
<p>I've tried solv... | Michael Rozenberg | 190,319 | <p>By the law of sines the triangle $ABC$ is a similar to the triangle with sides-lengths $\sin{A}$, $\sin{B}$ and $\sin{C}$. </p>
<p>Thus, by the law of cosines
$$\sin^2C=\sin^2A+\sin^2B-2\sin{A}\sin{B}\cos{C}$$ and by the given
$$\sin^2C=\sin^2A+\sin^2B-\sin{A}\sin{B}\sin{C}.$$
Id est, $$2\cos{C}=\sin{C}$$ or
$$4(1... |
2,439,278 | <p>I wanted to solve the following problem.</p>
<blockquote>
<p>In $\triangle ABC$ we have $$\sin^2 A + \sin^2 B = \sin^2 C + \sin A \sin B \sin C.$$ Compute $\sin C$.</p>
</blockquote>
<p>Since it's an equation for a triangle, I assumed that $\pi = A + B + C$ would be important to consider.</p>
<p>I've tried solv... | Dr. Sonnhard Graubner | 175,066 | <p>using the Theorem of sines we get
$$\sin^2(C)\left(\frac{a^2+b^2}{c^2}\right)=\sin^2(C)+\frac{ab}{c^2}\sin^3(C)$$
since $$\sin(C)\neq 0$$ we obtain
$$\sin(C)=\frac{a^2+b^2-c^2}{ab}$$ using $$2\cos(C)=\frac{a^2+b^2-c^2}{ab}$$ we get $$\tan(C)=2$$</p>
|
1,063,774 | <blockquote>
<p>Prove that the <span class="math-container">$\lim_{n\to \infty} r^n = 0$</span> for <span class="math-container">$|r|\lt 1$</span>.</p>
</blockquote>
<p>I can't think of a sequence to compare this to that'll work. L'Hopital's rule doesn't apply. I know there's some simple way of doing this, but it ... | marty cohen | 13,079 | <p>For fun,
I'll try to do
a proof by contradiction.</p>
<p>Suppose
$\lim_{n \to \infty} r^n
\ne 0
$.
Then,
since $S =(r^n)_{n=1}^{\infty}$
is a decreasing sequence
(since $0 < r < 1$),
$s = \lim S$
exists
and $s > 0$.
(Only $\lim\inf$
is needed
for what follows.)</p>
<p>Then,
for any $c > 0$,
there is an... |
47,188 | <p>I am an amateur mathematician, and I had an idea which I worked out a bit and sent to an expert. He urged me to write it up for publication. So I did, and put it on arXiv. There were a couple of rounds of constructive criticism, after which he tells me he thinks it ought to go a "top" journal (his phrase, and he wen... | sleepless in beantown | 8,676 | <p>What do you have to lose by submitting an article for publication? You'll have an even better record/credentialing/verification of the work you've put into it by being published in a Journal of good reputation. In the worst case, you will get a rejection letter, perhaps with a good explanation of why they are reje... |
1,659,224 | <p>Find a sequence $a_n$ such that $\lim_{n \to \infty}|a_{n+1} - a_n | = 0$ while the sequence does not converge.</p>
<p>I am thrown for a loop. For a sequence not to converge it means that for all $\epsilon \geq 0$ and for all $N$, when $n \geq N$, $|a_n - L| \geq \epsilon$ as $n \longrightarrow \infty$</p>
<p>or I... | Arkady | 23,522 | <p>Hint: Take your favourite example of a <strong>series</strong> that does not converge but whose terms tend to zero. Think partial sums as elements of a new sequence..</p>
|
1,659,224 | <p>Find a sequence $a_n$ such that $\lim_{n \to \infty}|a_{n+1} - a_n | = 0$ while the sequence does not converge.</p>
<p>I am thrown for a loop. For a sequence not to converge it means that for all $\epsilon \geq 0$ and for all $N$, when $n \geq N$, $|a_n - L| \geq \epsilon$ as $n \longrightarrow \infty$</p>
<p>or I... | Omran Kouba | 140,450 | <p>I would take $a_n=\sqrt{n}$. This would do the job.</p>
|
1,108,787 | <p><img src="https://i.stack.imgur.com/YfmKh.png" alt="enter image description here"></p>
<p>I know we have to prove these 10 properties to prove a set is a vector space. However, I don't understand how to prove numbers 4 and 5 on the list.</p>
| MathMajor | 113,330 | <p>I think it will help to see examples of sets that fulfill $(4)$ and $(5)$, along with some that don't.</p>
<p>To prove $(4)$, you just need to identify element called the zero element. For example, if $V = \mathbb{R}^3$, then $\vec 0 = (0, 0, 0)$ would be the zero element, because for any point $(a, b, c) \in \math... |
1,535,914 | <p>When I plot the following function, the graph behaves strangely:</p>
<p><span class="math-container">$$f(x) = \left(1+\frac{1}{x^{16}}\right)^{x^{16}}$$</span></p>
<p>While <span class="math-container">$\lim_{x\to +\infty} f(x) = e$</span> the graph starts to fade at <span class="math-container">$x \approx 6$</span>... | Gottfried Helms | 1,714 | <p>Here is a plot using the (arbitrary precision) calculator Pari/GP. I use 200 dec digits precision as default in my computations and got this plot without oscillation up to x=20: </p>
<p><a href="https://i.stack.imgur.com/nDTtY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nDTtY.png" ... |
3,217,130 | <p>How can I prove this by induction? I am stuck when there is a <span class="math-container">$\Sigma$</span> and two variables, how would I do it? I understand the first step but have problems when i get to the inductive step.</p>
<p><span class="math-container">$$\sum_{j=1}^n(4j-1)=n(2n+1)$$</span></p>
| Maxime Ramzi | 408,637 | <p><span class="math-container">$xyx = xyx$</span> therefore with <span class="math-container">$a=x, b=yx, c=xy$</span> we get <span class="math-container">$ab = ca$</span>, so <span class="math-container">$b=c$</span>, that is <span class="math-container">$xy=yx$</span></p>
|
3,059,020 | <p>The integral is<span class="math-container">$$\int_0^{2\pi}\frac{\mathrm dθ}{2-\cosθ}.$$</span>Just to skip time, the answer of the indefinite integral is <span class="math-container">$\dfrac2{\sqrt{3}}\tan^{-1}\left(\sqrt3\tan\left(\dfracθ2\right)\right)$</span>.</p>
<p>Evaluating it from <span class="math-contai... | Mark Viola | 218,419 | <p>Note that that tangent function, <span class="math-container">$\tan(x)$</span>, is discontinuous when <span class="math-container">$x=\pi/2+n\pi$</span>. So, the antiderivative <span class="math-container">$\frac2{\sqrt{3}} \arctan\left(\sqrt 3 \tan(\theta/2)\right)$</span> is not valid over the interval <span clas... |
603,291 | <p>Suppose $f:(a,b) \to \mathbb{R} $ satisfy $|f(x) - f(y) | \le M |x-y|^\alpha$ for some $\alpha >1$
and all $x,y \in (a,b) $. Prove that $f$ is constant on $(a,b)$. </p>
<p>I'm not sure which theorem should I look to prove this question. Can you guys give me a bit of hint? First of all how to prove some function... | André Nicolas | 6,312 | <p><strong>Hint:</strong> Show that $f'(y)$ exists and is equal to $0$ for all $y$. Then as usual by the Mean Value Theorem our function is constant. </p>
|
3,127,795 | <p>So I’m struggling to understand how to find the angle in this circle, we’ve recently learnt about trigonometry and like finding the area of a circle and all that but I can’t seem to remember which formula I have to use to find this angle. Can anyone lend a helping hand?
<img src="https://i.stack.imgur.com/V2Sm7.jp... | Dr. Sonnhard Graubner | 175,066 | <p>You can use the theorem of cosines:
<span class="math-container">$$10^2=7^2+7^2-2\cdot 7^2\cos(\angle{AOB})$$</span></p>
|
597,986 | <p>It's a classical theorem of Lie group theory that any compact connected abelian Lie group must be a torus. So it's natural to ask what if we delete the connectedness, i.e. the problem of classification of the compact abelian Lie groups.</p>
| Mariano Suárez-Álvarez | 274 | <p>The quotient of your Lie group $G$ by the connected component of the identity $G_0$ is a finite discrete abelian group $A$, because the Lie group is compact. It follows that we have an extension $$0\to G_0\to G\to A\to 0$$ Such a thing is classified (forgetting topologies) by an element of $H^2(A,G_0)$. This group i... |
3,752,167 | <p>Freshman question, really, but the more I think about it, the more I doubt.</p>
<p>Suppose that two sets belong to the same equivalence class. Are they in effect interchangeable? (I understand that there is no axiom of `interchangeability' in the definition of an equivalence relation.)</p>
<p>For example, consider t... | Ethan Bolker | 72,858 | <p>The answer to your question is in the question, here:</p>
<blockquote>
<p>They are interchangeable as long as what matter is their age.</p>
</blockquote>
<p>If all that matters in any particular context is the condition that specifies the equivalence relation then any representative will do.</p>
|
2,009,134 | <p>Suppose $$\frac{{{{\sin }^4}(\alpha )}}{a} + \frac{{{{\cos }^4}(\alpha )}}{b} = \frac{1}{{a + b}}$$ for some $a,b\ne 0$. </p>
<p>Why does $$\frac{{{{\sin }^8}(\alpha )}}{{{a^3}}} + \frac{{{{\cos }^8}(\alpha )}}{{{b^3}}} = \frac{1}{{{{(a + b)}^3}}}$$</p>
| Piquito | 219,998 | <p>Solving the system
$$\begin{cases}X^2+Y^2=1\\\dfrac{X^4}{a}+\dfrac{Y^4}{b}=\dfrac{1}{a+b}\end{cases}$$ where obviously $X$ and $Y$ are the sinus and cosinus respectively, one has $$X=-\sqrt{\dfrac{a}{a+b}}\\Y=\pm\sqrt{\dfrac{b}{a+b}}$$</p>
<p>This gives directly the equalities $\dfrac{X^8}{a^3}=\dfrac{a^4}{a^3(a+b... |
694,090 | <p>Let $V$ and $W$ be vector spaces over $\Bbb{F}$ and $T:V \to W$ a linear map. If $U \subset V$ is a subspaec we can consider the map $T$ for elements of $U$ and call this the restriction of $T$ to $U$, $T|_{U}: U \to W$ which is a map from $U$ to $W$. Show that</p>
<p>$$\ker T|_{U} = \ker T\cap U.$$</p>
<p>I know ... | Community | -1 | <p>Notice that by definition</p>
<p>$$\forall u\in U,\quad T_{|U}(u)=T(u)$$</p>
<p>Now if $u\in U$ such that $u\in \ker T_{|U}$ then we have
$$T_{|U}(u)=T(u)=0$$
and this means that $u\in \ker T$ hence</p>
<p>$$\ker T_{|U}\subset U\cap \ker T$$
Conversely if $u\in U\cap \ker T$ then
$$0=T(u)= T_{|U}(u)$$
hence $u\... |
268,482 | <p>One has written a paper, the main contribution of which is a few conjectures. Several known theorems turned out to be special cases of the conjectures, however no new case of the conjectures was proven in the paper. In fact, no new theorem was proven in the paper. </p>
<p>The work was reported on a few seminars, an... | John Pardon | 35,353 | <p>The main contribution of the paper "The McKay conjecture and Galois automorphisms" by Gabriel Navarro is to propose various conjectures. It is published in the Annals of Mathematics.</p>
<p><a href="http://www.ams.org/mathscinet-getitem?mr=2144975" rel="noreferrer">http://www.ams.org/mathscinet-getitem?mr=2144975<... |
19,848 | <p><strong>{Xn}</strong> is a sequence of independent random variables each with the same
<strong>Sample Space {0,1}</strong> and <strong>Probability {1-1/$n^2$ ,1/$n^2$}</strong>
<br> <em>Does this sequence converge with probability one (Almost Sure) to the <strong>constant 0</strong>?</em>
<br><br>Essentially th... | Shai Covo | 2,810 | <p>Let's elaborate on the answers already given, by considering the more general case where $X_n = 1$ with probability $a_n$ and $X_n = 0$ with probability $1-a_n$, where $a_n \downarrow 0$ (the typical cases being $a_n = 1/n$ and $a_n = 1/n^2$) and the $X_n$ are not necessarily independent.</p>
<p>The most natural ap... |
87,091 | <p>Find $C$ such that</p>
<p>$$Ce^{(-4x^2-xy-4y^2)/2}$$</p>
<p>is a joint probability density of a $2$-variable Gaussian.</p>
<p>If someone could give me a jumping off point, or a process as to how to go about this, I'd really appreciate it.</p>
| Steven Gamer | 47,540 | <p>It is easy to show that the commutator subgroup is a characteristic subgroup , hence it is a normal subgroup. Proving G/C abelian is straightforward recalling that C is the subgroup of G generated by all commutators.</p>
|
87,091 | <p>Find $C$ such that</p>
<p>$$Ce^{(-4x^2-xy-4y^2)/2}$$</p>
<p>is a joint probability density of a $2$-variable Gaussian.</p>
<p>If someone could give me a jumping off point, or a process as to how to go about this, I'd really appreciate it.</p>
| Community | -1 | <p>Recall that $gh = hg[g,h]$ and that $[G,G]$ is the subgroup of $G$ generated by all commutators. $[G,G] \unlhd G$ because $$[h,k]g = g[h,k][[h,k],g].$$</p>
<p>The map $$G \longrightarrow \frac{G}{[G,G]}$$ is called abelianization (precisely because of the theorem we are about to prove).</p>
<p>Every element of $G/... |
545,728 | <p>So I have $X \sim \text{Geom}(p)$ and the probability mass function is:</p>
<p>$$p(1-p)^{x-1}$$</p>
<p>From the definition that:</p>
<p>$$\sum_{n=1}^\infty ns^{n-1} = \frac {1}{(1-s)^2}$$</p>
<p>How would I show that the $E(X)=\frac 1p$</p>
| hejseb | 70,393 | <p>What is $s$ in your case? Consider:</p>
<p>$$
E(X)=\sum_{x=1}^\infty x p(1-p)^{x-1}=p\sum_{x=1}^\infty x(1-p)^{x-1}=\cdots=?
$$</p>
<p>Can you see how to take it from here?</p>
|
2,161,463 | <p>I am in the final stages of a proof and need help. I have simplified my starting expression expression down to $\dfrac{v\Gamma(-1+\frac{v}{2})}{2\Gamma(\frac{v}{2})}$</p>
<p>I know the above expression is to equal $\frac{v}{v-2}$</p>
<p>I am having ahard time getting there.</p>
<p>I know $\Gamma(n) = (n-1)!$</p>
... | Jack D'Aurizio | 44,121 | <p>$\Gamma\left(\frac{v}{2}\right) = \left(\frac{v}{2}-1\right)\,\Gamma\left(\frac{v}{2}-1\right)$ hence
$$ \frac{v\,\Gamma\left(\frac{v}{2}-1\right)}{2\,\Gamma\left(\frac{v}{2}\right)}= \frac{v\,\Gamma\left(\frac{v}{2}-1\right)}{2\,\left(\frac{v}{2}-1\right)\,\Gamma\left(\frac{v}{2}-1\right)}=\frac{v}{2\left(\frac{v}{... |
3,752,948 | <p>Find <span class="math-container">$\displaystyle \lim_{x \to\frac{\pi} {2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{\cos^2 x} $</span><br />
I tried to do like this :
Let <span class="math-container">$A=\displaystyle \lim_{x \to\frac{\pi}{2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{... | marty cohen | 13,079 | <p>Since
<span class="math-container">$\sec(x) = 1/\cos(x)$</span>
and
<span class="math-container">$\cos(\pi/2+y)
=\cos(\pi/2)\cos(y)-\sin(\pi/2)\sin(y)
=-\sin(y)
$</span>,</p>
<p><span class="math-container">$\begin{array}\\
a(n)
&=\lim_{x \to \pi/2}\left(\sum_{k=1}^n k^{\sec^2(x)}\right)^{\cos^2(x)}\\
&=\lim... |
478,523 | <p>I'm trying to reason through whether $\int_{x=2}^\infty \frac{1}{xe^x} dx$ converges.</p>
<p>Intuitively, It would seem that since $\int_{x=2}^\infty \frac{1}{x} dx$ diverges, then multiplying the denominator by something to make it smaller would make it converge.</p>
<p>If I apply a Taylor expansion to the $\fra... | user1337 | 62,839 | <p>Using the Taylor expansion was fine, the problem is in the "leading order" approach.</p>
<p>You chose to express $\frac{1}{e^x}$ as an <em>infinite</em> sum - you cannot truncate it for no reason. </p>
|
2,620,700 | <p>I want to prove or disprove that ($\mathbb{R},\mathcal{E}_1$) with equivalence relation $x\mathcal{R}y \Leftrightarrow a \neq 0$ y = ax is homeomorphic to $(\{0,1\}, \tau)$ where $\tau$ is the discrete topology. </p>
<p>To me it seems that there are two classes of equivalence: every point is equivalent to every poi... | Alex Kruckman | 7,062 | <p>You've observed that the quotient space $(\mathbb{R},\mathcal{E}_1)/\mathcal{R}$ consists of two points, the equivalence classes $C_0 = \{0\}$ and $C_1 = \mathbb{R}\backslash \{0\}$. Now what's the topology on the quotient space?</p>
<p>$(\mathbb{R},\mathcal{E}_1)/\mathcal{R}$ is homeomorphic to the two-point space... |
1,130,029 | <p>I am not sure if this question is off topic or not but a question like this has been asked on this site before - <a href="https://math.stackexchange.com/questions/409408/insertion-sort-proof">Insertion sort proof</a></p>
<p>Here is an example of insertion sort running a on a set of data
<a href="https://courses.cs.w... | subash rajaa | 254,060 | <p>The intuition relies on the fact that for every position in the array on an average half the elements are greater than the number in that position, and hence they will contribute to a comparison.</p>
|
373,313 | <p>Given a manifold <span class="math-container">$M$</span>, we can always embed it in some Euclidian space (general position theorem). Hence we can define the minimal embedding space of <span class="math-container">$M$</span> to be the smallest euclidean space that we can embed <span class="math-container">$M$</span> ... | skupers | 798 | <p>Yes, it depends on the category of manifolds you are considering.</p>
<p>For example, by Corollary 1.4 of <a href="https://www.sciencedirect.com/science/article/pii/0040938365900418" rel="noreferrer">Hsiang-Levine-Szczarba</a>, the 16-sphere with non-standard smooth structure does not admit a smooth embedding into <... |
2,390,215 | <p>There exist a bijection from $\mathbb N$ to $\mathbb Q$, so $\mathbb Q$ is countable. And by well ordering principle $\mathbb N$ has a least member say $n_1$ which is mapped to something in $\mathbb Q$, In this way $\mathbb Q$ can be well arranged? Is my argument good? </p>
| rschwieb | 29,335 | <p>If you assume the axiom of choice, <em>any</em> set can be well ordered using this strategy. See <a href="https://math.stackexchange.com/questions/2283566/what-is-the-largest-well-ordered-set">this post</a></p>
<p>But if the original question was really "is $\mathbb Q$ well-ordered <em>by its natural ordering?</em>... |
2,390,215 | <p>There exist a bijection from $\mathbb N$ to $\mathbb Q$, so $\mathbb Q$ is countable. And by well ordering principle $\mathbb N$ has a least member say $n_1$ which is mapped to something in $\mathbb Q$, In this way $\mathbb Q$ can be well arranged? Is my argument good? </p>
| Alec Rhea | 232,121 | <p>The answer is yes, but how we prove it depends on our assumptions -- are you assuming the axiom of choice (AC)? Your method of finding a bijection between $\mathbb{Q}$ and $\mathbb{N}$ is a good idea, however it is easiest to 'pass through' $\mathbb{Z}$ along the way in my opinion. I will sketch a possible proof w... |
815,418 | <p>Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the </p>
<pre><code>nth x = √(1st x √ 2nd x ... √nth x);
</code></pre>
<p>Then</p>
<p>$$\text{the "infinith" } x = x$$</p>
<p>Example: </p>
<p>$$\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4$$<br>
Try it yourself, type calc... | Nicky Hekster | 9,605 | <p>Suppose $y = \sqrt{x\sqrt{x\sqrt{x\cdots}}}$. Then obviously $y^2=xy$, whence $y=x$ or $y=0$. But $y \gt 0$, so $y=x$.</p>
|
261,156 | <p>Well, I am trying to write a code that makes the number:</p>
<p><span class="math-container">$$123456\dots n\tag1$$</span></p>
<p>So, when <span class="math-container">$n=10$</span> we get:</p>
<p><span class="math-container">$$12345678910$$</span></p>
<p>And when <span class="math-container">$n=15$</span> we get:</... | kglr | 125 | <pre><code>f1 = FromDigits @ StringRiffle[Range[#], ""] &;
f1 /@ {4, 10, 15}
</code></pre>
<blockquote>
<pre><code>{1234, 12345678910, 123456789101112131415}
</code></pre>
</blockquote>
<p>And</p>
<pre><code>f2 = FromDigits @* StringJoin @* IntegerString @* Range;
f2 /@ {4, 10, 15}
</code></pre>
<blockq... |
261,156 | <p>Well, I am trying to write a code that makes the number:</p>
<p><span class="math-container">$$123456\dots n\tag1$$</span></p>
<p>So, when <span class="math-container">$n=10$</span> we get:</p>
<p><span class="math-container">$$12345678910$$</span></p>
<p>And when <span class="math-container">$n=15$</span> we get:</... | AccidentalFourierTransform | 34,893 | <p>For fun, here's some more options, which are quite distinct from the already existing ones.</p>
<p>First, a recursive definition:</p>
<pre><code>f[1] = 1;
f[n_] := f[n] = f[n - 1]*10^Floor[Log[10, 10*n]] + n
</code></pre>
<p>And, second, one using <a href="https://mathematica.stackexchange.com/users/3056/kirma">kirm... |
188,139 | <p>Let $f$ be entire and non-constant. Assuming $f$ satisfies the functional equation $f(1-z)=1-f(z)$, can one show that the image of $f$ is $\mathbb{C}$?</p>
<p>The values $f$ takes on the unit disc seems to determine $f$...</p>
<p>Any ideas?</p>
| Robert Israel | 8,508 | <p>If $f({\mathbb C})$ misses $w$, then it also misses $1-w$. The only case where $w = 1-w$ is $1/2$, which is $f(1/2)$. Now use Picard's "little" theorem.</p>
|
1,110,652 | <p>Would anyone happen to know any introductory video lectures / courses on partial differential equations? I have tried to find it without success (I found, however, on ODEs). </p>
<p>It does not have to be free material, but something not to expensive would be nice.</p>
| Sophie Clad | 190,787 | <p>Here is answer to your Question .Complete course of PDE are available there</p>
<p><a href="http://nptel.ac.in/courses.php" rel="nofollow">http://nptel.ac.in/courses.php</a></p>
<p>These lectures are there in youtube channel 'nptel" but contents and syllabus can be seen from link above</p>
|
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