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 |
|---|---|---|---|---|
948,329 | <p>I have come across this trig identity and I want to understand how it was derived. I have never seen it before, nor have I seen it in any of the online resources including the many trig identity cheat sheets that can be found on the internet.</p>
<p>$A\cdot\sin(\theta) + B\cdot\cos(\theta) = C\cdot\sin(\theta + \Ph... | heropup | 118,193 | <p>The trigonometric angle identity $$\sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta$$ is exactly what you need. Note that $A^2 + B^2 = C^2$, or $$(A/C)^2 + (B/C)^2 = 1.$$ Thus there exists an angle $\phi$ such that $\cos\phi = A/C$ and $\sin\phi = B/C$. Then we immediately obtain from the aforem... |
3,846,717 | <p>Denote <span class="math-container">$\mathbb{F}=\mathbb{C}$</span> or <span class="math-container">$\mathbb{R}$</span>.</p>
<p><strong>Theorem (Cauchy - Schwarz Inequality).</strong> <em>If <span class="math-container">$\langle\cdot,\cdot\rangle$</span> is a semi-inner product on a vector space <span class="math-con... | Math Lover | 801,574 | <p>Take example of <span class="math-container">$x^2-3x+2 = (x-1)(x-2) = 0$</span>. It has two roots and between <span class="math-container">$x = 1$</span> and <span class="math-container">$x = 2$</span>, it is negative, while positive for <span class="math-container">$x \lt 1, x \gt 2$</span>. As the quadratic in the... |
1,017,989 | <p>Consider three disjoint circles not necessarily of same radii. How do you draw the smallest circle enclosing all these three circles? Where is its centre, and what is its radius? </p>
| Axel Kemper | 58,610 | <p><strong>Not a full solution but a sketch:</strong></p>
<p>Let us assume the three disjoint circles are described by:</p>
<p>$$\begin{aligned}
(x - x_1)^2 + (y - y_1)^2 &= r_1^2 \\
(x - x_2)^2 + (y - y_2)^2 &= r_2^2 \\
(x - x_3)^2 + (y - y_3)^2 &= r_3^2
\end{aligned}$$</p>
<p><img src="https://i.stack.... |
2,146,929 | <p>Let $f:S^n \to S^n$ be a homeomorphism. I know the result that a rigid motion in $\mathbb R^{n+1}$ is always <a href="https://math.stackexchange.com/a/866471/185631">linear</a>, but can we get more information from the assumption that $f:S^n \to S^n$ is a homeomorphism?</p>
| Tsemo Aristide | 280,301 | <p>Let $f:S^n\rightarrow S^n$ an homeomorphism. define $F:R^n\rightarrow R^n$ by $F(x)=\|x\|f({x\over{\|x\|}}), x\neq 0$, $F(0)=0$.</p>
|
101,953 | <p>I've been asked (by a person not by a homework) about how to compute the following limit:</p>
<p>$$ \lim_{x \to 10^-} \frac{[x^3] - x^3}{[x] - x}$$</p>
<p>where $[\cdot]$ is used to denote the floor function:</p>
<p>$$ [x] := \begin{cases} x && x \in \mathbb{Z} \\ 
\text{biggest integer smaller ... | Davide Giraudo | 9,849 | <p>We have, putting $x=N+\delta$:
$$\lim_{x\to N^-}\frac{[x^3]-x^3}{[x]-x}=\lim_{\delta\to 0^-}\frac{[(N+\delta)^3]-(N+\delta)^3}{[N+\delta]-(N+\delta)}.$$
We have $[N+\delta]-N=N+[\delta]-N=[\delta]=-1$ when for example $-\frac 12\leq\delta<0$, and
$$[(N+\delta)^3]-(N+\delta)^3=[N^3+3N^2\delta+3N\delta^2+\delta^3]... |
1,785,633 | <p><em>(First, I am very aware of the fact that Brownian motion is actually probably more difficult to understand than at least basic complex analysis, so the pedagogical merits of such an approach would be questionable for anyone besides a probabilist wanting to refresh or reshape their already existing complex analys... | Lee Mosher | 26,501 | <p>Hint: It can be easily be reduced to a simpler problem: find three vectors $a,b,c$ in $\mathbb{R}^2$ such that $\{a,b\}$, $\{a,c\}$ and $\{b,c\}$ are each linearly independent sets. Once you've found those vectors, simply include $\mathbb{R}^2$ as a vector subspace of $\mathbb{R}^3$.</p>
|
3,592,747 | <p>How to solve this equation for <span class="math-container">$x$</span> in reals, without using theory of complex number?
<span class="math-container">$$\frac{a}{\left(x+\frac{1}{x}\right)^{2}}+\frac{b}{\left(x-\frac{1}{x}\right)^{2}}=1$$</span>
Where <span class="math-container">$a$</span> and <span class="math-cont... | Matti P. | 432,405 | <p>I would start by doing this:
<span class="math-container">$$
\frac{a}{\left(x+\frac{1}{x}\right)^{2}}+\frac{b}{\left(x-\frac{1}{x}\right)^{2}}=
\frac{ax^2}{\left(x+\frac{1}{x}\right)^2 x^2}+\frac{bx^2}{\left(x-\frac{1}{x}\right)^2 x^2}=
\frac{ax^2}{\left(x^2+1\right)^2}+\frac{bx^2}{\left(x^2-1\right)^2}=1
$$</span... |
3,012,416 | <p>I know the answer is obvious: In <span class="math-container">$\mathbb{Z}$</span> the only solutions of <span class="math-container">$xy=-1$</span> are <span class="math-container">$x=-y=1$</span> and <span class="math-container">$x=-y=-1$</span>.
My problem is that I want to formally prove it and I don't know how t... | Joel Pereira | 590,578 | <p>If xy = -1, then x = <span class="math-container">$\frac{-1}{y}$</span>. If x is an integer, y must divide -1. Therefore, y = <span class="math-container">$\pm$</span>1, which implies x <span class="math-container">$\mp$</span>1.</p>
|
3,012,416 | <p>I know the answer is obvious: In <span class="math-container">$\mathbb{Z}$</span> the only solutions of <span class="math-container">$xy=-1$</span> are <span class="math-container">$x=-y=1$</span> and <span class="math-container">$x=-y=-1$</span>.
My problem is that I want to formally prove it and I don't know how t... | Ovi | 64,460 | <p>For positive integers (same can be done for negative, or just use absolute value): </p>
<p>If <span class="math-container">$x >1$</span> and <span class="math-container">$y>1$</span>, then <span class="math-container">$xy>1$</span>. So either <span class="math-container">$x =1$</span> or <span class="math-... |
645,340 | <p>I am not sure how to start tackling this question and would love a hint:</p>
<blockquote>
<p>Let $<$ the regular order relation on $\mathbb{R}$, and $<_w$ well ordering on $\mathbb{R}$. We define a coloring function $h:[\mathbb{R}]^2 \rightarrow 2$ like so:
$$
h(\{x,y\}) = \left\{\begin{matrix}
1 &((x... | Clive Newstead | 19,542 | <p><strong>Hint:</strong> Suppose there is such a homogeneous set, and use it to construct a subset of $\mathbb{R}$ whose order-type under $<$ is $\omega_1$. This must be a contradiction because, by selecting rational points between points in this sequence, you get a strictly increasing $\omega_1$-sequence of ration... |
645,340 | <p>I am not sure how to start tackling this question and would love a hint:</p>
<blockquote>
<p>Let $<$ the regular order relation on $\mathbb{R}$, and $<_w$ well ordering on $\mathbb{R}$. We define a coloring function $h:[\mathbb{R}]^2 \rightarrow 2$ like so:
$$
h(\{x,y\}) = \left\{\begin{matrix}
1 &((x... | Asaf Karagila | 622 | <p>The question is not related to large cardinals. Rather it is closer to Ramsey theory, and its various degrees of failure in infinite sets.</p>
<p>Suppose that there was such homogeneous $H$ of size $\aleph_1$.</p>
<p>If $H$ is homogeneous and its color is $1$ then there is an order embedding of an uncountable ordi... |
4,463,105 | <p>Write down the radius of convergence of the power series, obtained by the Taylor expansion of the analytic functions about the stated point, in</p>
<p><span class="math-container">$f(z) = \frac{(z+20)(z+21)}{(z-20i)^{21} (z^2 +z+1)}$</span> about <span class="math-container">$z = 0$</span>.</p>
<p>My attempt: Since ... | Kavi Rama Murthy | 142,385 | <p><span class="math-container">$f$</span> has poles at <span class="math-container">$\frac {-1\pm i\sqrt 3} 2$</span> also and these pole have modulus <span class="math-container">$1$</span>. So the radius of convergence is <span class="math-container">$1$</span>.</p>
|
22 | <p>By matrix-defined, I mean</p>
<p>$$\left<a,b,c\right>\times\left<d,e,f\right> = \left|
\begin{array}{ccc}
i & j & k\\
a & b & c\\
d & e & f
\end{array}
\right|$$</p>
<p>...instead of the definition of the product of the magnitudes multiplied by the sign of their angle... | Noldorin | 56 | <p>The obvious but slightly trite answer is "because that's just how the cross-product works as an operation".</p>
<p>If you're looking for an intuitive reason, the cross-product by definition produces a vector that is orthogonal to the two operand (input) vectors. You know that the base vectors $\mathbf{i}$, $\mathbf... |
500,931 | <p>I have this theorem which I can't prove.Please help.</p>
<p>"Show that every positive rational number $x$ can be expressed in the form $\sum_{k=1}^n \frac{a_k}{k!}$ in one and only one way where each $a_k$ is non-negative integer with $ a_k≤ k − 1$ for $k ≥ 2$ and $a_n>0$."</p>
<p>I think the ONE way is <a href... | Calvin Lin | 54,563 | <p>First, we show existence. For any rational <span class="math-container">$x$</span>, define a sequence of integers as follows:</p>
<p><span class="math-container">$a_1 = \lfloor x \rfloor$</span>, observe that <span class="math-container">$ 0 \leq x - a_1 < 1$</span>. If it is 0, stop.<br>
<span class="math-conta... |
2,640,763 | <p>Let $\{x_i\}_{i=1}^{n}$ and $\{y_i\}_{i=1}^{n}$ two positive sequences, the first one is monotonic, the second one is strictly increasing .</p>
<p>I noticed that in many cases if $\{x_i\}_{i=1}^{n}$ is increasing $$\frac{\frac{1}{n}\sum_{i=1}^{n}{x_iy_i}}{\left(\frac{1}{n}\sum_{i=1}^{n}{x_i}\right)\left(\frac{1}... | user | 505,767 | <p>Suppose that $T$ has rank k, then you can construct a basis made of $n-k$ vectors in Ker(T) and of $k$ vectors orthogonal to them</p>
<p>$$\alpha=\{\alpha_1,...\alpha_k,\alpha_{k+1},...\alpha_n\}$$</p>
<p>then assume</p>
<p>$$\beta=\{\beta_1,...\beta_k,\beta_{k+1},...\beta_n\}$$</p>
<p>with $\beta_i=T(\alpha_i)... |
3,182,587 | <p>I think it is simpler if I just focus on the derivation of <span class="math-container">$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$</span>.</p>
<p>I start from the formulae for the complex exponential, assuming <span class="math-container">$z=x+iy$</span>: <span class="math-container">$e^{iz}=e^{-y}*(\cos(x)+i\sin(x))$</spa... | Peter Foreman | 631,494 | <p>The Taylor series expansions for <span class="math-container">$e^z$</span>, <span class="math-container">$\sin{(z)}$</span> and <span class="math-container">$\cos{(z)}$</span> which are valid for all <span class="math-container">$z\in\mathbb{C}$</span> are supplied below
<span class="math-container">$$e^z=1+\frac{z}... |
354,100 | <p>Does the expression <span class="math-container">$$\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n,$$</span> which gives the volume of an <span class="math-container">$n$</span>-dimensional ball of radius <span class="math-container">$R$</span> when <span class="math-container">$n$</span> is a nonnegative integer... | Anixx | 10,059 | <p>I want to make another answer to this question, although it is is not really an answer, and I do not know whether it is relevant or not, but it may be useful.</p>
<p>Let's take </p>
<p><span class="math-container">$$V(n,R)=\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n$$</span></p>
<p>Also, using the reflecti... |
1,346,039 | <p>Let <span class="math-container">$F$</span> be a field of characteristic prime to <span class="math-container">$n$</span>, and let <span class="math-container">$F^a$</span> be an algebraic closure of <span class="math-container">$F$</span>. Let <span class="math-container">$\zeta$</span> be a primitive <span class="... | user149792 | 149,792 | <ol>
<li>$\zeta^{m_1}$ and $\zeta^{m_2}$ are conjugate if and only if there exists $g \in \text{Gal}(F(\zeta)/F)$ such that $g(\zeta^{m_1}) = \zeta^{m_2}$. This is the case if and only if there exists $g \in \text{Gal}(F(\zeta)/F)$ such that $\chi(g)m_1 = m_2$, which happens if and only if $m_1$ and $m_2$ lie in the sa... |
1,421,740 | <p>Let $90^a=2$ and $90^b=5$, Evaluate </p>
<h1>$45^\frac {1-a-b}{2-2a}$ </h1>
<p>I know that the answer is 3 when I used logarithm, but I need to show to a student how to evaluate this without involving logarithm. Also, no calculators.</p>
| JUMAR VELASCO | 622,772 | <p>Given <span class="math-container">$90^a = 2$</span>, <span class="math-container">$90^b = 5$</span></p>
<p><span class="math-container">$$(90/2)^{(1-a-b)/(2–2a)}$$</span>
<span class="math-container">$$=(90/90^a)^{(1-a-b)/(2–2a)}$$</span>
<span class="math-container">$$=(90^{1-a})^{(1-a-b)/(2(1–a))}$$</span>
<span... |
2,589,019 | <p>I am aware that there is a result saying that if $L_1, L_2$ are two algebraically closed fields of the same characteristic, then either $L_1$ is isomorphic to the subfield of $L_2$, or $L_2$ is isomorphic to a subfield of $L_1$.</p>
<p>I am unsure however of how to prove this result. I am quite surprised that this ... | Stella Biderman | 123,230 | <p>What follows is meant to illuminate the intuition, and is neither a proof nor entirely formally accurate.</p>
<p>Both fields contain the same copy of either the algebraic closure of $\mathbb{Q}$ or of $\mathbb{F}_p$ depending on what the characteristic is. This is a straightforward consequence of the fact that the ... |
1,102,324 | <p>I could use some pointers solving this problem:</p>
<blockquote>
<p>Given a certain s.v. $X$ with cdf $F_x(x)$ and pdf $f_X(x)$. Let s.v. $Y$ be the lower censored of $X$ at $x=b$. Meaning:</p>
<p>$$Y = \begin{cases}0 & \text{if }X<b\\
X & \text{if } X \geq b\end{cases}$$</p>
<p>Find cdf $F_Y... | abel | 9,252 | <p>the parabola is symmetric about $x = 0$ has roots $\pm \dfrac{1}{\sqrt a}$ so the area bounded by the parabola and $y = 0$ is $$\int_{-1/\sqrt a}^{1/\sqrt a}(1-ax^2) dx=2 \int_0^{1/\sqrt a} 1 - ax^2 dx = 2\left[ x - \dfrac{ax^3}{3}\right]_0^{1/\sqrt a}= 2(\dfrac{1}{\sqrt a} - \dfrac{1}{3\sqrt a}) = \dfrac{4}{3 \sqrt... |
1,700,590 | <p>Let $A=\{a_1\ldots,a_m \}$ be a set of linearly independent vectors. Suppose that each $a_j$ $(j=1,\ldots,m)$ can be written as a linear combination of vectors in the set $B=\{b_1,\ldots,b_n\}$.</p>
<p>Then how to show that $m \le n$?</p>
<p>I have tried as follows:</p>
<p>Since $a_j \in A$, we have $a_j \in \o... | Darío G | 27,454 | <p>Suppose for a contradiction that <span class="math-container">$n<m$</span> and write the equations:</p>
<p><span class="math-container">$$\begin{align*}a_1&=\alpha_{11}b_1+\ldots+\alpha_{1n}b_n\\\vdots & \hspace{2cm}\vdots\\ a_m&=\alpha_{m1}b_1+\cdots +\alpha_{mn}b_n\end{align*}$$</span></p>
<p>From h... |
1,781,117 | <h1>The question</h1>
<p>Prove that:
$$\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$$</p>
<hr>
<h2>What I've tried</h2>
<p>Knowing that:
$$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$$
evaluating at $z=i$ gives
$$ \frac{e^π - e^{-π}}{2i} = \sin(πi) = πi \prod_{n=1}^∞ \... | Jack D'Aurizio | 44,121 | <p>$$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right),\qquad \frac{\sinh(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1+\frac{z^2}{n^2}\right)\tag{1}$$
give:
$$ \frac{\sin(\pi z)\sinh(\pi z)}{\pi^2 z^2(1-z^4)}=\prod_{n\geq 2}\left(1-\frac{z^4}{n^4}\right)\tag{2} $$
hence by considering $\lim_{z\to 1}LHS$ ... |
2,077,694 | <p>How to find $A = M^{A}_{B}$ in linear transformation $F = \mathbb{P_{2}} \rightarrow \mathbb{R^{2}} $, where $ F(p(t)) = \begin{pmatrix} p(0) \\ P(1) \end{pmatrix},$
$ A = \{1,t,t^{2}\},$
$B=\left \{ \begin{pmatrix} 1\\ 0\end{pmatrix},\begin{pmatrix} 0\\ 1\end{pmatrix} \right \}$?</p>
| Robert Israel | 8,508 | <p>If $n$ has proper divisors $d_1, \ldots, d_k$ with $d_1 + \ldots + d_k > n$,
then the proper divisors of $mn$ include $md_1, \ldots, m d_k$ with ... </p>
|
3,338,885 | <p>I want to prove this theorem using Binomial theorem and I've got trouble in understanding 3rd step if anyone knows why please explain :) Prove that sum: </p>
<p><span class="math-container">$\sum_{r=0}^{k}\binom{m}{r}\binom{n}{k-r}=\binom{m+n}{k}$</span> </p>
<p>1st step:<br>
<span c... | Kavi Rama Murthy | 142,385 | <p>If <span class="math-container">$c_0+c_1y+\cdots +c_{n+m}y^{n+m}=d_0+d_1y+\cdots +d_{n+m}y^{n+m}$</span> for all <span class="math-container">$y$</span> then <span class="math-container">$c_i=d_i$</span> for all <span class="math-container">$i$</span>. This is a basic fact about polynomials (and comes from the fac... |
3,338,885 | <p>I want to prove this theorem using Binomial theorem and I've got trouble in understanding 3rd step if anyone knows why please explain :) Prove that sum: </p>
<p><span class="math-container">$\sum_{r=0}^{k}\binom{m}{r}\binom{n}{k-r}=\binom{m+n}{k}$</span> </p>
<p>1st step:<br>
<span c... | Nick Shih | 698,771 | <p>By the equation of 2nd step,</p>
<p>Left:
<span class="math-container">$$\binom{m+n}{0}y^{0}+\binom{m+n}{1}y^{1}+\cdots+\binom{m+n}{m+n-1}y^{m+n-1}+\binom{m+n}{m+n}y^{m+n}$$</span></p>
<p>The coefficient of <span class="math-container">$y^{k}$</span> is <span class="math-container">$\binom{m+n}{k}$</span></p>
<p>... |
156,479 | <p>Let $S$ be a compact oriented surface of genus at least $2$ (possibly with boundary). Let $X$ be a connected component of the space of embeddings of $S^1$ into $S$.</p>
<p>Question : what is the fundamental group of $X$? My guess is that the answer is $\mathbb{Z}$ with generator the loop of embeddings obtained by... | Sam Nead | 1,650 | <p>$
\newcommand{\Homeo}{\operatorname{Homeo}}
\newcommand{\SO}{\operatorname{SO}}
$Since you are interested in the topological category, then I think it will suffice to prove the necessary facts about $\Homeo_0(S)$ and about the curve stabilizer. Now, there is a topological proof that $\Homeo_0(S)$ is contractible --... |
2,618,804 | <p>Let $V$ be a vector space of dimension $m\geq 2$ and $ T: V\to V$ be a linear transformation such that $T^{n+1}=0$ and $T^{n}\neq 0$ for some $n\geq1$ .Then choose the correct statement(s):</p>
<p>$(1)$ $rank(T^n)\leq nullity(T^n)$</p>
<p>$(2)$ $rank(T^n)\leq nullity(T^{n+1})$</p>
<p><strong>Try:</strong></p>
<... | Community | -1 | <p>$0=T^{n+1}(V)=T(T^n(V)) \implies rank(T^n)\le nullity (T)$. But of course $nullity T\le nullity T^{n+1}$</p>
<p>So (2) is true.</p>
<p>Oh and (1) is true also...</p>
<p>For (1), see <a href="https://math.stackexchange.com/a/400927"> here</a>... </p>
|
216,532 | <p>How do I find the limit of something like</p>
<p>$$ \lim_{x\to \infty} \frac{2\cdot3^{5x}+5}{3^{5x}+2^{5x}} $$</p>
<p>?</p>
| filmor | 7,247 | <p>Divide both the upper and the lower term by $3^{5x}$, that will do it.</p>
|
3,522,867 | <p>Consider the sequence <span class="math-container">$\{x_n\}_{n\ge1}$</span> defined by <span class="math-container">$$x_n=\sum_{k=1}^n\frac{1}{\sqrt{k+1}+\sqrt{k}}, \forall n\in\mathbb{N}.$$</span> Is <span class="math-container">$\{x_n\}_{n\ge 1}$</span> bounded or unbounded. </p>
<p>I solved the problem as stated... | Kavi Rama Murthy | 142,385 | <p><span class="math-container">$\sqrt {k+1}+\sqrt k \leq \sqrt {2k}+\sqrt k<3\sqrt k$</span>. Hence the given sum is at least <span class="math-container">$\sum\limits_{k=1}^{n} \frac 1 {3\sqrt k}$</span> Now use the fact that the series <span class="math-container">$\sum\limits_{k=1}^{n} \frac 1 {3\sqrt k}$</span>... |
2,512,556 | <p>What would be the solution of $ y''+y=\cos (ax) \ $ if $ \ a \to 1 \ $. </p>
<p><strong>Answer:</strong></p>
<p>I have found the complementary function $ y_c \ $ </p>
<p>$ y_c(x)=A \cos x+B \sin x \ $</p>
<p>But How can I find the particular integral if $ a \to 1 \ $ </p>
| Plutoro | 108,709 | <p>For $a\neq 1$, we would choose $ P\cos(ax)+Q\sin(ax)$ for our particular solution, and solve for $P$ and $Q$. But if $a=1$, then this form will never solve the particular solution because it is also a solution to the homogeneous equation. The standard strategy in this situation is to multiply the standard form for t... |
1,844,374 | <p>Why does the "$\times$" used in arithmetic change to a "$\cdot$" as we progress through education? The symbol seems to only be ambiguous because of the variable $x$; however, we wouldn't have chosen the variable $x$ unless we were already removing $\times$ as the symbol for multiplication. So why do we? I am very cu... | hjhjhj57 | 150,361 | <p>I believe the reason must be mostly pedagogical:</p>
<ul>
<li>As other answers mention, a kid learning about decimals may get confused between a product and a decimal.</li>
<li><p>It may be easier for kids to think about $\times$ than $\cdot$, given that they already know the sum symbol and they are quite similar.<... |
70,976 | <blockquote>
<p>I'm considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it's not a UFD.</p>
</blockquote>
<p>I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able to show that $2$, $\sqrt{-n}$ and $1+\sqrt{-n}$ are all irreducible. Is there some... | Mr. Brooks | 162,538 | <p>It really depends on what $-n$ is, but if you have to guess for a random $n$, it's a safe bet to say it's not a UFD, because only nine of the imaginary quadratic rings are UFDs. The general principle is to find an example of a number with two distinct factorizations, thereby proving the domain is not a unique factor... |
1,925,245 | <p>Find the eigenvalues of
$$
\left(\begin{matrix}
C_1 & C_1 & C_1&\cdots&C_1 \\
C_2 & C_2 & C_2&\cdots &C_2 \\
C_3 & C_3 & C_3&\cdots&C_3 \\
\vdots&\ & \ & \ & \vdots\\
C_n & C_n & C_n&\cdots&C_n \\
\end{matrix}\right)
$$</p>
<p>My approa... | Nicholas Stull | 28,997 | <p>I'll work it out for $n=2$, $n=3$ and you can try to extrapolate (and figure out the pattern) from there:</p>
<p>$n=2$:</p>
<p>$$A = \left(\begin{array}{cc}a&a\\b&b\end{array}\right)$$
$$\det(A-\lambda I) = \left|\begin{array}{cc}a-\lambda&a\\b&b-\lambda\end{array}\right|= (a-\lambda)(b-\lambda)-ab... |
2,718,286 | <p>I have came across this problem and after trying to answer it for some time, I thought my solution was correct but apparently it is not. Can you please explain to me what I have done wrong ?</p>
<p>Problem:</p>
<p>A train traveling from Aytown to Beetown meets with an accident after 1 hour. The train is stopped fo... | saulspatz | 235,128 | <p>The problem is double counting. Let's just consider the example you gave. Since we know that there are $\binom{12}{6}$ strings in all, we just have have to count the strings with three consecutive ones. There are $10$ positions for the first one in the substring and then $\binom{9}{3}$ places to put the other thr... |
2,718,286 | <p>I have came across this problem and after trying to answer it for some time, I thought my solution was correct but apparently it is not. Can you please explain to me what I have done wrong ?</p>
<p>Problem:</p>
<p>A train traveling from Aytown to Beetown meets with an accident after 1 hour. The train is stopped fo... | Remy | 325,426 | <p>If you're just interested in the probability of getting at least three consecutive ones, then this problem is analogous to finding the probability that the length of the longest run of heads in $n$ coin tosses, say $\ell_n$, exceeds a given number $m$. This probability is given by</p>
<p>$$\mathbb{P}(\ell_n \geq m)... |
2,495,440 | <p>If $f'(x) + f(x) = x,\;$ find $f(4)$.</p>
<p>Could someone help me to solve this problem ? </p>
<p>The answer is 3 but I don't know why.
<em>with no use of integration or exponential functions</em> and
<em>the function is polynomial</em></p>
| Guy Fsone | 385,707 | <p>hint: multiplying both side by $e^x$ we have </p>
<p>$$f'(x) + f(x) = x\Longleftrightarrow ( f(x)e^x)'= xe^x
\Longleftrightarrow f(x)e^x= xe^x -e^x +c.
$$
can you continue from here ?</p>
|
2,495,440 | <p>If $f'(x) + f(x) = x,\;$ find $f(4)$.</p>
<p>Could someone help me to solve this problem ? </p>
<p>The answer is 3 but I don't know why.
<em>with no use of integration or exponential functions</em> and
<em>the function is polynomial</em></p>
| ℋolo | 471,959 | <p>if $f'(x) + f(x) = x$ then $f'(x)e^x + f(x)e^x = xe^x$ now remember derivative by parts? $(f(x)g(x))'=f'(x)g(x)+g'(x)f(x)$. if $g(x)=e^x$ then you have $(f(x)e^x)'=f'(x)e^x+\left[e^x\right]'f(x)$ and because $\left[e^x\right]'=e^x$ you have $(f(x)e^x)'=f'(x)e^x+e^xf(x)$ so you left with:$$f'(x) + f(x) = x\implies f'... |
3,173,636 | <p>I have been trying to prove that there is no embedding from a torus to <span class="math-container">$S^2$</span> but to no avail.</p>
<p>I am completely stuck on where to start. The proof is supposed to be based on Homology theory. I know how to prove that <span class="math-container">$S^n$</span> cannot be embedde... | Lee Mosher | 26,501 | <p>Suppose that there exists an embedding <span class="math-container">$f : T^2 \mapsto S^2$</span>. </p>
<p>Each point <span class="math-container">$x \in T^2$</span> has an open neighborhood <span class="math-container">$U$</span> homeomorphic to the open unit disc in <span class="math-container">$S^2$</span>, and i... |
3,950,808 | <p><em>(note: this is very similar to <a href="https://math.stackexchange.com/questions/188252/spivaks-calculus-exercise-4-a-of-2nd-chapter">a related question</a> but as I'm trying to solve it without looking at the answer yet, I hope the gods may humor me anyways)</em></p>
<p>I'm self-learning math, and an <a href="h... | Albus Dumbledore | 769,226 | <p><strong>Hint</strong>:</p>
<p>Think about the sum as the <strong>coefficient of <span class="math-container">$x^l$</span></strong> in the expansion <span class="math-container">${(1+x)}^n{(x+1)}^m$</span></p>
|
3,308,291 | <p>I have an array of numbers (a column in excel). I calculated the half of the set's total and now I need the minimum number of set's values that the sum of them would be greater or equal to the half of the total. </p>
<p>Example:</p>
<pre><code>The set: 5, 5, 3, 3, 2, 1, 1, 1, 1
Half of the total is: 11
The least a... | Ethan Bolker | 72,858 | <p>It's not closed because <span class="math-container">$1+1$</span> isn't in it.</p>
|
2,294,548 | <p><strong>Problem:</strong> Solve $y'=\sqrt{xy}$ with the initial condition $y(0)=1$.</p>
<p><strong>Attempt:</strong> Using $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$, I get that the DE is separable by dividing both sides by $\sqrt{y}:$ $$y'=\sqrt{x}\cdot\sqrt{y}\Leftrightarrow\frac{y'}{\sqrt{y}}=\sqrt{x}$$</p>
<p>which can... | Lutz Lehmann | 115,115 | <p>Yes, you can make that restriction based on the initial point.</p>
<p>You could avoid using two integration constants, using one integration constant $D$ (or $C$) when integrating both sides of an equation is standard.</p>
|
2,663,303 | <blockquote>
<p>Let $G$ be finite. Suppose that $\left\vert \{x\in G\mid x^n =1\}\right\vert \le n$ for all $n\in \mathbb{N}$. Then $G$ is cyclic.</p>
</blockquote>
<p>What I have attempted was the fact that every element is contained in a maximal subgroup following that <a href="https://groupprops.subwiki.org/wiki/... | Community | -1 | <p>Although <a href="https://math.stackexchange.com/questions/1593222/a-finite-abelian-group-a-is-cyclic-iff-for-each-n-in-bbbn-a-in-a">the duplicate</a> supposes $G$ to be abelian, Ihf’s answer doesn’t. It’s a perfect answer and please see an even more complete one <a href="https://math.stackexchange.com/questions/599... |
4,537,685 | <p>Let <span class="math-container">$H$</span> be a Hilbert space, and let
<span class="math-container">$$H_n = \otimes_n H = \Big\{\sum_{i_1,\ldots, i_n} \alpha_{i_1, \ldots, i_n} \big(e_{i_1} \otimes \cdots \otimes e_{i_n}\big) : \sum_{i_1,\ldots, i_n} |\alpha_{i_1, \ldots, i_n}|^2<\infty \Big\}$$</span>
denote th... | Nicholas Todoroff | 1,068,683 | <p>The symmetric tensors are all symmetrized tensor products of vectors:
<span class="math-container">$$
H^s_n = \left\{\sum_{\sigma \in S_n}v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(n)} \;:\; v_1,\dotsc,v_n \in H\right\}.
$$</span>
It follows that each basis element of <span class="math-container">$H^s_n$</span> c... |
17,480 | <p>I have asked a question at <a href="https://academia.stackexchange.com/">academia.stackexchange</a> with three sub-questions recently and I was told that it was not proper there. I just wonder if it is acceptable if one asks multiple (related) questions at math.stackexchange? </p>
<p>To mathematicians, if the answe... | user642796 | 8,348 | <p>I generally dislike it when users ask multiple questions within a single question, and this is backed up by the <a href="https://meta.stackexchange.com/q/39223/214632">MSE faq</a>. I believe the main reasons for this are the following:</p>
<ul>
<li><p>If people provide answers of varying quality to the different pa... |
224,970 | <p>$\newcommand{\Int}{\operatorname{Int}}\newcommand{\Bdy}{\operatorname{Bdy}}$
If $A$ and $B$ are sets in a metric space, show that:
(note that $\Int$ stands for interior of the set)</p>
<ol>
<li>$\Int (A) \cup \Int (B) \subset \Int (A \cup B)$.</li>
<li>$(\overline{ A \cup B}) = (\overline A \cup \overline B )$. (n... | Luke Mathieson | 35,289 | <p>You're off to the right start, it just has to be a bit more elaborate, the rough idea is that you want to add a character to $a$ and $c$ for each one you put into $b$, then the last step is to stick the "middle" $1$ in.</p>
<p>More detail in the spoiler, but try to get it yourself first.</p>
<blockquote class="spo... |
3,744,560 | <p>Suppose I have a function <span class="math-container">$\Lambda(t)$</span> for any <span class="math-container">$t>0$</span>. This function has the following three properties:</p>
<ol>
<li><span class="math-container">$\Lambda(t)$</span> is differentiable.</li>
<li><span class="math-container">$\Lambda(t)$</span>... | Community | -1 | <p>Let <span class="math-container">$f (x)$</span> be such a function.From the second property we get,
<span class="math-container">$$f(x+y)=f(x)+f(y)$$</span>
<strong>Step 1</strong> Differentiate partially w.r.t. <span class="math-container">$x$</span>,
<span class="math-container">$$f'(x+y)=f'(x)$$</span>
<strong>St... |
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| Reid Barton | 126,667 | <p>Another answer is that a groupoid is a space which has no homotopy groups in dimension ≥ 2. (Analogously a set is a space which has no homotopy groups in dimension ≥ 1.) They arise from taking (homotopy) orbits of group actions on sets, as well as from categories (by discarding the noninvertible morphisms an... |
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| Tom Leinster | 586 | <p>As other people have mentioned, a groupoid can be defined as a category in which every map is invertible. A groupoid with only one object is exactly a group, and a groupoid in which there are no maps between distinct objects is simply a family of groups.</p>
<p>But there's another class of examples, orthogonal to ... |
929,532 | <p>Okay so I want some hints (not solutions) on figuring out whether these sets are open, closed or neither.</p>
<p>$A = \{ (x,y,z) \in \mathbb{R}^3\ \ | \ \ |x^2+y^2+z^2|\lt2 \ and \ |z| \lt 1 \} \\ B = \{(x,y) \in \mathbb{R}^2 \ | \ y=2x^2\}$</p>
<p>Okay so since this question is the last part of the question where... | saz | 36,150 | <p><strong>Hint</strong> Consider <span class="math-container">$\Omega = \mathbb{N}$</span>, the power set <span class="math-container">$F=\mathcal{P}(\mathbb{N})$</span> and the mapping <span class="math-container">$\mu: F\to [0,\infty]$</span>, <span class="math-container">$$\mu(A) := \begin{cases} 0, & \text{$A... |
94,501 | <p>The well-known Vandermonde convolution gives us the closed form <span class="math-container">$$\sum_{k=0}^n {r\choose k}{s\choose n-k} = {r+s \choose n}$$</span>
For the case <span class="math-container">$r=s$</span>, it is also known that <span class="math-container">$$\sum_{k=0}^n (-1)^k {r \choose k} {r \choose n... | Phira | 9,325 | <p>You can use the Zeilberger algorithm to find a recurrence for the sum and then use Petkovsek's algorithm to verify that there are no hypergeometric (i.e. essentially products of factorials) closed formulas for the general case.</p>
<p>So, you cannot expect to do better than the expressions proposed in the other ans... |
780,611 | <p>The differential equation in question is a FODE,</p>
<p>$$
\frac{dy}{dt} = -\frac{a^2\sqrt{2g}}{\sqrt{(R+y)(R-y)}}
$$</p>
<p>Upon first inspection, this is separable, but I don't know how to proceed from there.</p>
<p>Thanks.</p>
| Claude Leibovici | 82,404 | <p>Hint</p>
<p>Just integrate $\sqrt{(R-y) (R+y)}$ with respect to $y$ after having set $dt=...dy$. You will have a nice first change of variable such as $y=R x$</p>
|
2,930,292 | <p>I'm currently learning the unit circle definition of trigonometry. I have seen a graphical representation of all the trig functions at <a href="https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/unit-circle-definition-of-trig-functions/a/trig-unit-circle-review" rel="nofollow noreferrer">khan academ... | Doug M | 317,162 | <p>While this figure is elegant in its way, it has a little bit too much going on for the beginning student.
<a href="https://i.stack.imgur.com/1fNWH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1fNWH.png" alt="enter image description here"></a></p>
<p>Lets start with a simpler figure.
<a href="h... |
140,134 | <p>Does anyone know whether there is any geometric applications of the Iwaniec's conjecture on $ l^p $ bound of Beurling Alfhors transform (or the complex Hilbert transform). One application could have been was Behrling's conjecture (solved). Is there is any thing else that might be of geometric significance.
For p=2, ... | Community | -1 | <p>As already pointed out, the Iwaniec conjecture has close links with calculus of variations. It has already been settled for "stretch" functions (with a rotation $e^{i\theta}$ introduced), ie, functions of the form
\begin{equation*}
f(re^{i\theta})=g(r)e^{-i\theta}
\end{equation*}
for positive and smooth $g$ on the ... |
2,125,206 | <ol>
<li><p>Let $W$ be the region bounded by the planes
$x = 0$, $y = 0$, $z = 0$, $x + y = 1$, and $z = x + y$.</p></li>
<li><p>$(x^2 + y^2 + z^2)\, \mathrm dx\, \mathrm dy\, \mathrm dz$; $W$ is the region bounded by $x + y + z = a$ (where $a > 0$), $x = 0$, $y = 0$, and $z = 0$.</p></li>
</ol>
<p>$x,y,z$ being $0... | W.R.P.S | 390,844 | <p>1)</p>
<p><a href="https://i.stack.imgur.com/QwKSM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QwKSM.png" alt="enter image description here"></a></p>
<p>$$\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}\int_{z=0}^{z=x+y}\ dzdydx $$</p>
<p>2)</p>
<p><a href="https://i.stack.imgur.com/OKL9e.png" rel="nof... |
1,644,905 | <p>How to simplify the following equation:</p>
<p>$$\sin(2\arccos(x))$$
I am thinking about:</p>
<p>$$\arccos(x) = t$$</p>
<p>Then we have:</p>
<p>$$\sin(2t) = 2\sin(t)\cos(t)$$</p>
<p>But then how to proceed?</p>
| Harish Chandra Rajpoot | 210,295 | <p>Notice,
let $\cos^{-1}x=\theta\iff \cos\theta=x$
$$\sin(2\theta)=2\sin\theta\cos \theta$$$$\sin(2\theta)=2\cos\theta\sqrt{1-\cos^2\theta}$$
setting the value of $\theta$,
$$\sin(2\cos^{-1}x)=\color{red}{2x\sqrt{1-x^2}}$$
$\forall \ -1\le x\le 1$</p>
|
108,372 | <p>Given a map $\psi: S\rightarrow S,$ for $S$ a closed surface, is there any algorithm to compute its translation distance in the curve complex? I should say that I mostly care about checking that the translation distance is/is not very small. That is, if the algorithm can pick among the possibilities: translation dis... | Lee Mosher | 20,787 | <p>In the case that $\psi$ is pseudo-Anosov, the best one can do in general, as far as I know, is to get upper and lower bounds which are linear in translation length. These come from train track considerations. Assuming you have an invariant train track $T$ for $\psi$ in your hands (obtained by some algorithmic method... |
2,743,288 | <p>I need help with this exponential equation: $5^{x+2}\ 2^{4-x} = 1000 $</p>
<p>We know that $ 1000 = 10^3$, so:</p>
<p>$$\ln(5^{x+2}\cdot2^{4-x}) = \ln10^3 \implies\ln(5^{x+2}) + \ln(2^{4-x}) = \ln10^3$$</p>
<p>In the next step I use that: $\ln(a^x) = x\ln(a)$</p>
<p>$$(x+2)\ln 5 + (4-x)\ln 2 = 3\ln 10$$</p>
<p... | B. Mehta | 418,148 | <p>You can simplify a little more:
$$\begin{align}(x+2)\ln 5 + (4-x)\ln2 &= 3\ln 10 \\
&=3\ln2 + 3 \ln 5 \\
(x-1)\ln5+(1-x)\ln2&=0.\end{align}$$
Can you finish it off from here?</p>
<p>Alternatively, you can do this without logs: $5^{x+2} \times 2^{4-x} = 2^3 5^3$, so $5^x 2^{-x} = \frac{5}{2}$, from wher... |
2,743,288 | <p>I need help with this exponential equation: $5^{x+2}\ 2^{4-x} = 1000 $</p>
<p>We know that $ 1000 = 10^3$, so:</p>
<p>$$\ln(5^{x+2}\cdot2^{4-x}) = \ln10^3 \implies\ln(5^{x+2}) + \ln(2^{4-x}) = \ln10^3$$</p>
<p>In the next step I use that: $\ln(a^x) = x\ln(a)$</p>
<p>$$(x+2)\ln 5 + (4-x)\ln 2 = 3\ln 10$$</p>
<p... | user247327 | 247,327 | <p>I see no reason to use logarithms here.</p>
<p>Yes, $1000= 10^3$ and $10= (2)(5)$ so $1000= (2^3)(5^3)$.</p>
<p>Your equation is $(2^{4- x})(5^{x+ 2})= (2^3)(5^3)$.</p>
<p>2 and 5 are prime numbers and prime factorization is unique so we must have 4- x= 3 and x+ 2= 3. That is two equations in one unknown but fort... |
2,831,270 | <p>I am quite fascinated by the formula for the Mellin transform of the Gaussian Hypergeometric Function, which is given by:</p>
<blockquote>
<p><span class="math-container">$$\mathcal M [_2F_1(\alpha,\beta;\gamma;-x)] = \frac {B(s,\alpha-s)B(s,\beta-s)}{B(s,\gamma-s)}$$</span></p>
</blockquote>
<p><em>Source : <a ... | Maxim | 491,644 | <p>The inverse Mellin transform is given by
$$\mathcal M^{-1}[F] =
\frac 1 {2 \pi i} \int_{\sigma -i \infty}^{\sigma + i \infty} F(s) x^{-s} ds.$$
For $F(s) = \Gamma(s) \Gamma(\alpha - s) \Gamma(\beta - s) / \Gamma(\gamma - s)$, the line $\operatorname{Re} s = \sigma$ should separate the poles of $\Gamma(s)$ from the p... |
2,377,946 | <blockquote>
<p>The integral is:
$$\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$$</p>
</blockquote>
<p>I used an approach that involved substitution of x by $a\tan\theta$. No luck :\ . Help?</p>
| Bernard | 202,857 | <p><strong>Hint</strong></p>
<p>You obtain the integral: $\;\displaystyle\int_0^\tfrac\pi4\!\frac{\tan^4\theta}{a^3(1+\tan^2\theta)^3}\,\mathrm d\mkern1mu\theta=\frac1{a^3}\int_0^\tfrac\pi4\frac{\sin^4\theta}{\cos\theta}\,\mathrm d\mkern1mu\theta$.</p>
<p><em>Bioche's rules</em> say you should set $u=\sin\theta$.</p>... |
1,836,190 | <p>I've been working on a problem and got to a point where I need the closed form of </p>
<blockquote>
<p>$$\sum_{k=1}^nk\binom{m+k}{m+1}.$$</p>
</blockquote>
<p>I wasn't making any headway so I figured I would see what Wolfram Alpha could do. It gave me this: </p>
<p>$$\sum_{k=1}^nk\binom{m+k}{m+1} = \frac{n((m+2... | Marco Cantarini | 171,547 | <p>Using the integral representation of the binomial coefficient $$\dbinom{s}{k}=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{s}}{z^{k+1}}dz$$ we have
$$ \sum_{k=1}^{n}k\dbinom{m+k}{m+1}=\frac{1}{2\pi i}\sum_{k=1}^{n}k\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{m+k}}{z^{m+2}}dz
$$ $$=\frac{1}{2... |
1,836,190 | <p>I've been working on a problem and got to a point where I need the closed form of </p>
<blockquote>
<p>$$\sum_{k=1}^nk\binom{m+k}{m+1}.$$</p>
</blockquote>
<p>I wasn't making any headway so I figured I would see what Wolfram Alpha could do. It gave me this: </p>
<p>$$\sum_{k=1}^nk\binom{m+k}{m+1} = \frac{n((m+2... | Marko Riedel | 44,883 | <p>For those who enjoy integrals here is another approach using the
Egorychev method as presented in many posts by @FelixMarin and also by @MarkusScheuer, where we focus on finding an answer that differs from the approaches that have already been seen.</p>
<p><P>Suppose we seek to compute</p>
<p>$$S(n,m) = \... |
1,836,190 | <p>I've been working on a problem and got to a point where I need the closed form of </p>
<blockquote>
<p>$$\sum_{k=1}^nk\binom{m+k}{m+1}.$$</p>
</blockquote>
<p>I wasn't making any headway so I figured I would see what Wolfram Alpha could do. It gave me this: </p>
<p>$$\sum_{k=1}^nk\binom{m+k}{m+1} = \frac{n((m+2... | epi163sqrt | 132,007 | <p>Here is a slightly different variation of the theme. It is convenient to use the <em>coefficient of</em> operator $[x^k]$ to denote the coefficient of $x^k$ of a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}\tag{1}
\end{align*}</p>
<blockquote>
<p>We obtain
\begin{align*}
\sum_{k=1... |
512,768 | <p>I am trying to intuitively understand why the solution to the following problem is $-2$. $$\lim_{x\to\infty}\sqrt{x^2-4x}-x$$
$$\lim_{x\to\infty}(\sqrt{x^2-4x}-x)\frac{\sqrt{x^2-4x}+x}{\sqrt{x^2-4x}+x}$$
$$\lim_{x\to\infty}\frac{x^2-4x-x^2}{\sqrt{x^2-4x}+x}$$
$$\lim_{x\to\infty}\frac{-4x}{\sqrt{x^2-4x}+x}$$
$$\lim_{... | egreg | 62,967 | <p>Change the variable: set $t=1/x$, so you want to compute
$$
\lim_{t\to0^+}\sqrt{\frac{1}{t^2}-\frac{4}{t}}-\frac{1}{t}
=
\lim_{t\to0^+}\sqrt{\frac{1-4t}{t^2}}-\frac{1}{t}
=
\lim_{t\to0^+}\frac{\sqrt{1-4t}-1}{t}
$$
Now it should be clearer why the limit can't be $0$. The square root can be written
$$
\sqrt{1-4t}=1+\f... |
57,232 | <p>Given a Heegaard splitting of genus $n$, and two distinct orientation preserving homeomorphisms, elements of the mapping class group of the genus $n$ torus, is there a method which shows whether or not these homeomorphisms, when used to identify the boundaries of the pair of handlebodies, will produce the same $3$-m... | Matt Young | 2,627 | <p>Here is an answer to 1. It is known that for any $A > 0$ that $\sum_{m \leq x} \mu(m) e(\alpha m) = O_A(x (\log{x})^{-A})$ uniformly in $\alpha$. For instance, consult Theorem 13.10 of Iwaniec and Kowalski's book, Analytic Number Theory. This uniform bound comes from combining the zero free region of Dirichlet... |
1,419,483 | <p>Can anyone please help me in solving this integration problem $\int \frac{e^x}{1+ x^2}dx \, $?</p>
<p>Actually, I am getting stuck at one point while solving this problem via integration by parts.</p>
| Will R | 254,628 | <p>According to Wolfram|Alpha, no closed form exists. However, that doesn't mean we can't make any progress at all.</p>
<p>We can use the substitution $x=\tan{u}$, expand the resulting integrand as a power series in $\tan{u}$ and then each term can be expressed by way of a reduction formula.</p>
<p>Let $x=\tan{u}$. T... |
443,475 | <p>I am reading some geometric algebra notes. They all started from some axioms. But I am still confused on the motivation to add inner product and wedge product together by defining
$$ ab = a\cdot b + a \wedge b$$ Yes, it can be done like complex numbers, but what will we lose if we deal with inner product and wedge ... | joeA | 79,432 | <p>Here's an excerpt from Lasenby, Lasenby and Doran, 1996, <a href="http://www.mrao.cam.ac.uk/~clifford/publications/ps/dll_millen.pdf" rel="nofollow"><em>A Unified Mathematical Language for Physics and Engineering in the 21st Century</em></a>:</p>
<blockquote>
<p>The next crucial stage of the story occurs in 1878 ... |
443,475 | <p>I am reading some geometric algebra notes. They all started from some axioms. But I am still confused on the motivation to add inner product and wedge product together by defining
$$ ab = a\cdot b + a \wedge b$$ Yes, it can be done like complex numbers, but what will we lose if we deal with inner product and wedge ... | benrg | 234,743 | <p>It isn't so much that the Clifford product is useful as that it's natural.</p>
<p>Instead of defining the Clifford product as <span class="math-container">$v\cdot w + v\wedge w$</span>, you can define the Clifford algebra as the algebra in which S+S and V+V addition, SS and SV multiplication, and V<sup>2</sup> squar... |
635,989 | <p>What counterexample can I use to prove that ($ \mathbb{R}_{[x]}$is any polynomial):</p>
<p>$L :\mathbb{R}_{[x]}\rightarrow\mathbb{R}_{[x]},(L(p))(x)=p(x)p'(x)$ is not linear transformation. I have already proven this using definition but it is hard to think about example. I would be grateful for any help.</p>
| Dan Rust | 29,059 | <p>Let $p$ be given by $p(x)=x$ and let $q=2p$. We note that $L(p)(x)=x$ but $L(q)(x)=2x\cdot 2=4x$ But if $L$ was linear then $L(q)(x)=L(2p)(x)=2L(p)(x)=2x\neq 4x$ and so we reach a contradiction. Hence $L$ is not linear.</p>
|
3,306,089 | <p>I came across this meme today:</p>
<p><a href="https://i.stack.imgur.com/RfJoJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RfJoJ.jpg" alt="enter image description here"></a></p>
<p>The counterproof is very trivial, but I see no one disproves it. Some even say that the meme might be true. Wel... | Clayton | 43,239 | <p>An infinite non-repeating decimal does <strong>not</strong> imply that every possible number combination exists somewhere. Consider the number: <span class="math-container">$0.101001000100001\ldots$</span>, the pattern is easy to spot, but this is an irrational number because...you guessed it...it's an infinite, non... |
2,713,311 | <p>$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $</p>
<p><strong>Answer:</strong></p>
<p>$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \\ \Rightarrow \lim_{x \to \infty} [\frac{x^2+1-ax^2-ax-bx-b}{x+1}]=0 \\ \Rightarrow \lim_{x \to \infty} \frac{2x-2ax-a-b}{1}=0 \\ \Righta... | farruhota | 425,072 | <p>The limit is:
$$L=\lim_\limits{x\to \infty} ((2-2a)x-(a+b))$$
Consider the cases:
$$L=\begin{cases} \infty, \ 2-2a\ne 0 \\
-(1+b), \ 2-2a=0\end{cases}$$
Further note:
$$L=\begin{cases} 0, \ b=-1 \\ -(1+b)\ne 0, \ b \ne -1\end{cases}$$</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Tom Leinster | 586 | <p>77 instructional videos on category theory:</p>
<p><a href="https://www.youtube.com/TheCatsters" rel="nofollow noreferrer">https://www.youtube.com/TheCatsters</a></p>
<p>I know you said "only one video per post", but I'm not posting 77 times...</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Harrison Brown | 382 | <p>You probably won't learn much actual math from it, but <a href="http://www.youtube.com/watch?v=E3tdimtxqic">One Geometry</a> is funnier and catchier than a Snoop Dogg parody about 3-manifolds has any right to be.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Harrison Brown | 382 | <p>I believe this was mentioned elsewhere, but for completeness, here's <a href="https://www.youtube.com/watch?v=ECQyFzzBHlo" rel="nofollow">Serre on writing</a>.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Scott Carter | 36,108 | <p>My good friend Professor Elvis Zap has the "Calculus Rap," the "Quantum Gravity Topological Quantum Field Theory Blues," a vid on constructing "Boy's Surface," "Drawing the hypercube (yes he knows there is a line missing in part 1)," A few things on quandles, and a bunch of precalculus and calculus videos. In orde... |
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | sanokun | 797 | <p>This one is quite old but it was fun when I watched a few years ago.
It's about Fermat's Last theorem.</p>
<p><a href="http://www.archive.org/details/fermats_last_theorem">http://www.archive.org/details/fermats_last_theorem</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Jan Weidner | 2,837 | <p>Lots of Lie Theory talks:
<a href="http://sms.cam.ac.uk/collection/533438?mediaOffset=20&mediaMax=20&mediaOrder=asc&mediaSort=title#Media">http://sms.cam.ac.uk/collection/533438?mediaOffset=20&mediaMax=20&mediaOrder=asc&mediaSort=title#Media</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Kelly Davis | 1,400 | <p>The series of videos from <a href="http://video.ias.edu/sm">IAS School of Mathematics</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Hans-Peter Stricker | 2,672 | <p><a href="http://www.youtube.com/watch?v=OmSbdvzbOzY" rel="nofollow">The Dot and the Line: A Romance in Lower Mathematics</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | graveolensa | 4,672 | <p>I'd like to think that <a href="http://www.youtube.com/diproton" rel="nofollow">my math art</a> is awesome, and start <a href="http://www.youtube.com/diproton#p/u/19/FWG5JblSsps" rel="nofollow">here</a>. </p>
<p>the mapping behind that video is $(x,y,z)\rightarrow(2*cos(z-y),2*sin(x-z),7*cos(y-x))$, and has a singu... |
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | pi2000 | 4,778 | <p>The Abdus Salam International Centre for Theoretical Physics has lots of lectures in mathematics and physics.Some of them are difficult to find in other places(Complex Analysis,Abstract Algebra,Topology,Functional Analysis,Algebraic Geometry..).For the same topic(ex:Complex Analysis)there are lectures by 2 ore more... |
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Vivek Shende | 4,707 | <p><a href="http://www.youtube.com/watch?v=SyD4p8_y8Kw" rel="nofollow noreferrer">Hitler learns topology</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Gerald Edgar | 454 | <p>Searching for a video relating to another question, I found this:
<a href="http://www.youtube.com/watch?v=kklb1J6Ij_U&hd=1" rel="nofollow">My Calculus Project</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Peter Luthy | 5,751 | <p>Richard Feynman gave the 1964 Messenger Lectures at Cornell University --- this is an endowed lecture series to which a number of famous scholars have been invited, including several physicists. His lectures were recorded, and Bill Gates bought the rights to them and has provided them to the public for free.</p>
<... |
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Sharanjit Paddam | 15,071 | <p><a href="http://www.youtube.com/user/njwildberger" rel="nofollow">http://www.youtube.com/user/njwildberger</a></p>
<p>Excellent lectures by Norman Wildberger on topics including: Geometry, Algebraic Topology, Linear Algebra, Foundations of Mathematics, and history of Mathematics</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Aleks Vlasev | 15,493 | <p>I am not sure if this will qualify as math exactly, but it's amazing nonetheless. It is a film with Richard Feynman called "Feynman: Take the wold from another point of view". Here is part 1</p>
<p><a href="http://www.youtube.com/watch?v=PsgBtOVzHKI" rel="nofollow">Feynman: Take the wold from another point ov view ... |
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Carl Najafi | 19,027 | <p><a href="https://www.youtube.com/watch?v=9MV65airaPA" rel="nofollow noreferrer">John Stillwell - ET Math: How different could it be?</a> A nice talk given at the SETI Institute.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Justin Hilburn | 333 | <p>They filmed the <a href="http://nd.edu/~cmnd/conferences/topology/" rel="nofollow">FRG Conference on Topology and Field Theories</a> and put the lectures on <a href="http://www.youtube.com/user/NDdotEDU/videos" rel="nofollow">youtube</a>.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | David Corwin | 1,355 | <p>The first <a href="http://www.math.princeton.edu/events/seminars/minerva-lectures/inaugural-minerva-lectures-i-equidistribution" rel="nofollow">Minerva Lecture</a> by Jean-Pierre Serre at Princeton in Fall 2012 is online. There were two other lectures, and they did videotape them, but I can't find them online.</p>
|
736,624 | <blockquote>
<p>Heaviside's function $H(x)$ is defined as follows:</p>
<p>$H(x) = 1$ if $x > 0$, </p>
<p>$H(x) = \frac 1 2$ if $x= 0$</p>
<p>$H(x) = 0$ if $x < 0$ </p>
</blockquote>
<p>Define $F(x) := \int^x_0 H(t) \ dt, \ x \in \mathbb R$. </p>
<p>$H(x)$ is continuous except at $x=0$ so the in... | Community | -1 | <p>$F(x)$ is not differentiable at $0$. Graph the function and you'll see why. </p>
<p>The FTC states that if $H(x)$ is integrable, then $F(x)=\int_0^x H(t) dt$ is continuous and if $H(x)$ is continuous at $x_0$ then $F(x)$ is differentiable at $x_0$.</p>
<p>This is the case here, no contradiction.</p>
|
51,341 | <p>I have a function that is a summation of several Gaussians. Working with a 1D Gaussian here, there are 3 variables for each Gaussian: <code>A</code>, <code>mx</code>, and <code>sigma</code>:</p>
<p>$A \exp \left ( - \frac{\left ( x - mx \right )^{2}}{2 \times sigma^{2}} \right )$</p>
<pre><code>A*Exp[-((x - mx)^2/... | Bob Hanlon | 9,362 | <p>Since all of the component functions are Listable</p>
<pre><code>f[{a_, m_, s_}, x_] := Total[a*Exp[-(x - m)^2/(2*s^2)]]
n = 5;
amp = Array[a, n];
mean = Array[m, n];
sigma = Array[s, n];
f[{amp, mean, sigma}, x]
</code></pre>
<blockquote>
<p>a[1]/E^((x - m[1])^2/
(2*s[1]^2)) +
a[2]/E^((x ... |
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?<... | thepiercingarrow | 274,737 | <p>I will be using the definition of a set most commonly used in mathematics and the real world.</p>
<p>A set is, simply put, a list. Anything is a set. One way to denote a set is to surround a list of elements separated with commas with brackets. A set of my favorite ice-cream types might be $\{Chocolate, Vanilla, St... |
1,722,948 | <blockquote>
<p>$$\frac{1}{x}-1>0$$</p>
</blockquote>
<p>$$\therefore \frac{1}{x} > 1$$</p>
<p>$$\therefore 1 > x$$</p>
<p>However, as evident from the graph (as well as common sense), the right answer should be $1>x>0$. Typically, I wouldn't multiple the x on both sides as I don't know its sign, bu... | N. F. Taussig | 173,070 | <p>You have
$$\frac{1}{x} - 1 > 0$$
Forming a common denominator yields
$$\frac{1 - x}{x} > 0$$
The inequality is true if the numerator and denominator have the same sign.</p>
|
3,002,325 | <p>Proof <span class="math-container">$t$</span> is irrational <span class="math-container">$ t = a-bs $</span> , Given <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are rational numbers, <span class="math-container">$b \neq 0$</span> and <span class="math-container">$s$</span> i... | Eric kioko | 577,696 | <p>For the first part</p>
<p><span class="math-container">$a/b -s = t/b$</span></p>
<p><span class="math-container">$(a-t)/b=s$</span></p>
<p>Assuming t was rational would mean <span class="math-container">$ s $</span> is rational since rational numbers are closed under addition and subtraction.</p>
<p>This contrad... |
1,130,302 | <p>I am struggling with following challenge in my free time programming project $-$ how is it possible to make reflection vector that reflects along normal with angle that is not larger than some $\alpha$?</p>
<p><img src="https://i.stack.imgur.com/6bmgv.png" alt="Example of normal and limited angle reflection"></p>
... | MooS | 211,913 | <p>Let $G$ act on the cosets of $H$, we obtain a morphism $G \to S_3$ with Kernel $N$. It is well known that we have $N \leq H$, we need to prove equality. If not, then we have $[G:N]= 6$ and $G \to S_3$ is surjective. By the sign we obtain a surjection $G \to S_3 \to \{-1,1\}$, hence a subgroup of index $2$.</p>
|
2,628,149 | <p>I am having trouble finding the general solution of the following second order ODE for $y = y(x)$ without constant coefficients: </p>
<p>$3x^2y'' = 6y$<br>
$x>0$</p>
<p>I realise that it may be possible to simply guess the form of the solution and substitute it back into the the equation but i do not wish to us... | Parcly Taxel | 357,390 | <p>As mentioned in the comments (though I did not see them while typing this):
$$15^{1/4}=16^{1/4}\cdot\left(\frac{15}{16}\right)^{1/4}=2\cdot\left(1-\frac1{16}\right)^{1/4}$$
Thus the binomial series can be applied to $\left(1-\frac1{16}\right)^{1/4}$, the result to be doubled afterwards. $\frac1{16}$ is small, so thi... |
1,335,878 | <p>Let $N\unlhd K$ be a normal subgroup of a given group $K$ and let $$q:K\to K/N$$ be the natural quotient map. Let $A\subseteq K$ be a subset of $K$ and let the <a href="https://en.wikipedia.org/wiki/Conjugate_closure" rel="nofollow">conjugate closure</a> of $A$ in $K$ be denoted by $\langle A^K\rangle$.</p>
<p><str... | Hagen von Eitzen | 39,174 | <p>Assume $f_0\in X$ and $\|f_0\|_\infty=1$. Assume wlog. that $c:=\int_0^1 f(x)\,\mathrm dx>0$.
As $f_0(0)=0$ there exists $a>0$ such that $|f_0(x)|<\frac12$ for $0\le x\le a$. Then $$c\le \int_0^1 |f_0(x)|\,\mathrm dx\le \int_0^a |f_0(x)|\,\mathrm dx+ \int_a^1 |f_0(x)|\,\mathrm dx\le \frac a2+1-a<1$$
Now... |
2,832,374 | <p>I'm given the following definition asked to prove the following theorem:</p>
<p>Definition: Let $X$ be a set and suppose $C$ is a collection of subsets of $X$. Then $\cup \mathbf{C}=\{x \in X : \exists C\in \mathbf{C}(x\in C)\}$</p>
<p>Theorem: Let $\mathbf{C,D}$ be collections of subsets of a set $X$. Prove that ... | mathcounterexamples.net | 187,663 | <p><strong>Let's abstract a bit the question...</strong></p>
<p>We are asked to prove that a closed ball $A$ with radius equal to $1$ in a normed vector space is closed. This is very generic.</p>
<p>A simple proof is to consider the complement $V \setminus A$ and to prove that it is open. This is simple as for $f \in... |
3,157,338 | <p>If we have a linear recurrence relation on a sequence <span class="math-container">$\{x_n\}$</span>, then I know how to find the worst case asymptotic growth. We consider the largest absolute value <span class="math-container">$\alpha$</span> of any root of the characteristic polynomial. Then, independent of the ini... | Sangchul Lee | 9,340 | <p><strong>Method 1.</strong> Under the unitary Fourier transform <span class="math-container">$ \mathcal{F}[f](\xi) = \int_{\mathbb{R}} f(x)e^{-2\pi i \xi x} \, \mathrm{d}x $</span> defined on the Schwarz space <span class="math-container">$\mathcal{S}(\mathbb{R})$</span>,</p>
<p><span class="math-container">$$ \math... |
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