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 |
|---|---|---|---|---|
39,973 | <p>Assume that we have two residually finite groups $G$ and $H$. Which properties of $G$ and $H$ could be used to show that their pro-finite (or pro-p) completions are different?</p>
<p>I asked a while ago in the group-pub mailing list whether finite presentability is such a property but Lubotzky pointed out that it i... | Community | -1 | <p>For polycyclic groups, a result of Grunewald, Pickel and Segal (Polycyclic groups with isomorphic finite quotients. Ann. of Math. (2) 111 (1980), no. 1, 155--195.) says that the class of f.g. polycyclic groups with the same profinite completion (and even with the same sets of finite homomorphic images) is finite. Th... |
565,135 | <p>I have to prove the following theorem :</p>
<blockquote>
<p>Let $p$ be a prime number and let $n \ge 1$,be any integer, then there exists a field of order $p^n$.</p>
</blockquote>
<p><strong>My attempt</strong></p>
<p>I started off by considering the polynomial $f(x)$=$x^{p^n}-x \in \Bbb Z_p[x]$.</p>
<p>I took... | Ayman Hourieh | 4,583 | <p>In $\mathbb Z_p$ and hence its extension $F$ we have
$$
p a = \underbrace{a + \cdots + a}_{p \text{ times}} = (\underbrace{1 + \cdots + 1}_{p \text{ times}})a = 0 \cdot a = 0.
$$</p>
<p>By the binomial theorem
$$
(a + b)^{p^n} = \sum_{k=0}^{p^n}\binom{p^n}{k} a^k b^{p^n - k}.
$$</p>
<p>All terms in the latter expa... |
2,942,681 | <p>In an ordinary function like the temperature—one of the properties of the minimum is that if we go away from the minimum in the first order, the deviation of the function from its minimum value is only second order.</p>
<p>At any place else on the curve, if we move a small distance the value of the function changes... | dxiv | 291,201 | <p>Assuming a sufficiently smooth function, the <a href="https://en.wikipedia.org/wiki/Taylor_series" rel="nofollow noreferrer">Taylor expansion</a> around a point <span class="math-container">$\,a\,$</span> can be written as:</p>
<p><span class="math-container">$$
f(x) = f(a) + {\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)... |
2,760,994 | <p><strong>I have 3 trees.</strong></p>
<ul>
<li>These particular trees are dioecious (male or female).</li>
<li>I don't know the gender of any of the trees. </li>
<li>The chance of a tree being male or female is 50/50. </li>
<li>I need at least one male and one female for successful pollination to occur.</li>
</ul>
... | Przemysław Scherwentke | 72,361 | <p>HINT: Let $M$ denotes male tree, $F$ — female tree. The probability is
$$
1-P(MMM)-P(FFF).
$$
But, from independency, $P(MMM)=P(M)P(M)P(M)=(1/2)^3$.</p>
<p>BTW: What are female trees? Have they a hollow? ;-)</p>
|
2,760,994 | <p><strong>I have 3 trees.</strong></p>
<ul>
<li>These particular trees are dioecious (male or female).</li>
<li>I don't know the gender of any of the trees. </li>
<li>The chance of a tree being male or female is 50/50. </li>
<li>I need at least one male and one female for successful pollination to occur.</li>
</ul>
... | farruhota | 425,072 | <p>This is a binomial distribution with $n=3, p=\frac12$. Let $X$ shows the number of male trees, then:
$$P(X=1)+P(X=2)={3 \choose 1}\left(\frac12\right)^1\left(\frac12\right)^{3-1}+{3 \choose 2}\left(\frac12\right)^2\left(\frac12\right)^{3-2}=\\
3\cdot \frac12\cdot \frac14+3\cdot \frac14\cdot \frac12=3\cdot \frac14=\f... |
878,020 | <p>It's easy to prove that if $I$, $J$ are two-sided ideals and $R/I\cong R/J$ as modules over $R$, then $I=J$. What about left ideals? Is there a simple counterexample?</p>
<p>I believe I've found an answer, but since answering own questions is encouraged, I thought I might post it here. Other examples are obviously ... | Marcin Łoś | 117,513 | <p>Yes. </p>
<p>Let $R$ be ring of endomorphisms of $\mathbb{R}^2$, $I$ and $J$ be annihilators of subspaces spanned by $(1,0)$ and $(0,1)$, respectively. Let $\theta\in R$ be given by $\theta(x,y)=(y,x)$, and $\phi(f)=f\circ \theta$ for $f\in R$. Then $\phi$ is an automorphism of $R$ as a module over itself, and a bi... |
1,658,279 | <p>I'm new here so apologies if I am not clear enough. I am trying to find the zero divisors of the form $ax + b$ in $\mathbb Z_{10}$. Specifically, I need to find the values of $b$. I know that $2,4,5,6$ and $8$ are zero divisors in $\mathbb Z_{10}$ but I am not sure how to translate these into linear divisors.</p>
<... | Hetebrij | 252,750 | <p>The problem is that the original problem does not specify whether Ashmit's and Amisha's problem solving capabilities are independent of each other.<br>
In the answers you cite, it is assumed that their problem solving capabilities are independent.<br>
In your approach, their problem solving capabilities are not inde... |
1,781,225 | <p>I've got an exercise to do and I don't really know what to do.</p>
<p>Exercise : We've got function $f$, where $f(a) = 0$ and $f'(a)$ exists. Also we got function $g$ which is continuous. Does exist $(f-g)'(a)$? Explain it. </p>
<p>My opinion is that exists, but I've got no idea how should I explain it. Some help?... | Joss | 338,303 | <p>Not true, example: take $f=0$, $g=|x|$ and $a=0$.</p>
|
2,024,468 | <p>I need to find the points of intersection of a circle with radius $2$ and centre at $(0,0)$ and a rectangular hyperbola with equation $xy=1$. As per the topic statement is there any way to solve this without the graphical method. I have tried setting the $y$ values equal but I cant solve the resulting equation for $... | Lutz Lehmann | 115,115 | <p>You can also directly combine the equations into complete binomial formulas
$$
(x+y)^2=x^2+y^2+2xy=4+2=6,\\
(x-y)^2=x^2+y^2-2xy=4-2=2
$$
and solve the trivial linear system for each of the 4 sign combinations of the roots.</p>
|
2,950,193 | <p>How one can show that this polynomial.</p>
<p><span class="math-container">$$-2+2x^{2k+1}-x^{3k+2}+x^{k-1}$$</span></p>
<p>is negative for all integer <span class="math-container">$k>1$</span> and real <span class="math-container">$x>2$</span>.</p>
<p>I have no idea to start.</p>
| HackerBoss | 600,971 | <p>Note that
<span class="math-container">$$-2+2x^{2k+1}-x^{3k+2}+x^{k-1}=-2+\left(2x^{k+2}-x^{2k+3}+1\right)x^{k-1}$$</span>
and
<span class="math-container">$$2x^{k+2}-x^{2k+3}+1=1+\left(2-x^{k+1}\right)x^{k+2}$$</span>
If <span class="math-container">$x > 5$</span>, and <span class="math-container">$k>1$</span... |
4,032,422 | <p>This is actually in reference to the question posed here
<a href="https://stackoverflow.com/posts/66285948/edit">https://stackoverflow.com/posts/66285948/edit</a>
but is more appropriate as a question to be posed on a non-coding site.</p>
<p>I provide a partial answer, but not the full answer, relaying the full ques... | NinjaDarth | 181,639 | <p>The Wikipedia page on Cartesian closed categories has a subsection on bi-Cartesian closed categories which asserts that (1) the equational algebra cannot be axiomatized by a finite set of equational identities and (2) the word problem is still open. It cites the following reference, which should be worth a look</p>
... |
3,449,865 | <p>In an informal sense, what does a slice of the Mapping Cylinder look like? There seem to be two feasible choices:
i) Just <span class="math-container">$X$</span>, the slices are exactly the space <span class="math-container">$X \times \{t\}$</span> for some <span class="math-container">$t \in [0,1]$</span>.</p>
<p>... | John Hughes | 114,036 | <p>You might want to consider the map <span class="math-container">$f: S^1 \to S^1 : \theta \mapsto -\theta$</span>. The mapping cylinder of <span class="math-container">$f$</span> is actually diffeomorphic to a standard cylinder, but the map <span class="math-container">$f$</span> and the identity are not homotopic, s... |
4,270,105 | <p>I'd like to prove that this matrix is idempotent using a more algebraic proof for matrices with a similar definition to A, rather than deriving its eigenvalues or calculating <span class="math-container">$A^2$</span> .</p>
<p><span class="math-container">$A=$</span> <span class="math-container">$\begin{bmatrix}0.5&a... | Oscar Lanzi | 248,217 | <p>One way to show an object is idempotent is to double it, subtract the identity and check whether the result is (multiplicatively) self-inverse. For instance in <span class="math-container">$\mathbb Z/10\mathbb Z$</span> we have <span class="math-container">$2×5-1\equiv-1$</span> which is self-inverse, so <span class... |
3,623,250 | <p>My <a href="https://drive.google.com/file/d/1uK4ITDfSHXk-JfBcKZsywkGA1fJ_yGUn/view?usp=drive_open" rel="nofollow noreferrer">textbook</a>(2nd page, at very bottom) states the definition of symmetric relation as follows:</p>
<blockquote>
<p>A relation <span class="math-container">$R$</span> in a set <span class="math... | Qurultay | 338,156 | <p>We have <span class="math-container">$$3x^2-4x+5=3(x-\frac23)^2+\frac{11}{3}$$</span>
which takes the minimum <span class="math-container">$\dfrac{11}{3}$</span> at <span class="math-container">$x=\dfrac{2}{3}$</span>. Since <span class="math-container">$y=\ln x$</span> is increasing,
therefore the range of <span cl... |
2,332,078 | <blockquote>
<p>Why $kerT=Span\{\underline 0\}$ if linear map $T: \mathbb{F}_3[x]\to \mathbb{F}_3[x]$ is defined as $T(p(x))=p(-x)$?</p>
</blockquote>
<p>$\mathbb{F}_3[x]$ is a space of polynomials of form $ax^2+bx+c$.</p>
<p>The solution I saw finds $\text{ker}T$ as follows:</p>
<p>first $p(x)=ax^2+bx+c \Rightarr... | Elad | 433,446 | <p>You don't get to choose $x$. The vector space is a space of functions. Choosing $x$ is changing the function to a constant function. But your function is not constant. The zero in this space is the constant 0 function a function of degree minus infinity(this degree definition is artificial in a way). </p>
|
14,499 | <p><strong>Definition</strong>: n is said to be a highly composite number if and only if $d(n)>d(m)$ for all $m<n$, where $d(n)$ denotes number of divisors of n.</p>
<p><strong>Questions</strong>:<br>
1) Are there any theorems about highest prime that divides n, assuming that n is highly composite?. It seems tha... | Derek Jennings | 1,301 | <p>In this paper <a href="http://www.renyi.hu/~p_erdos/1944-04.pdf" rel="nofollow">On Highly Composite Numbers</a> P. Erdős uses Bertrand's postulate and the Prime Number Theorem to show that if $p$ is the highest prime that divides $n,$ a highly composite number, then</p>
<p>$$ c_1 \log n < p < c_2 \log n $$</p... |
1,662,418 | <p>A local system is a bundle with locally constant sheaf of sections. I have seen several equivalent characterizations (bundles acted upon nicely by the fundamental group of the base, bundles admitting flat connections, etc), and from this I can construct examples of vector bundles that are provably not local systems ... | Eric Wofsey | 86,856 | <blockquote>
<p>On the other hand, as far as I can tell, a local trivialization of a vector bundle over a contractible neighborhood should force the restriction of the sheaf of sections to be constant over the same neighborhood.</p>
</blockquote>
<p>This isn't true at all. When you talk about a vector bundle having... |
2,693,723 | <p>I have started to learn in university, and the way they teach us is very simpled... not the way they tought as geometry. The question is simple, but I want an concrete simple mathematic proof(verbal would be enough) something that would help me grasp the concept:</p>
<p>I have this set $$A = \left\{ \frac 1{2^n} \m... | gt6989b | 16,192 | <p><strong>HINT</strong></p>
<p>Assume $A$ is finite. Certainly $1/2 \in A$. Then $A$ has a minimal element, say $a$. Then, $a^2 \in A$. Since $a$ is minimal, $a \le 1/2$. </p>
<p>Find a contradiction.</p>
|
2,693,723 | <p>I have started to learn in university, and the way they teach us is very simpled... not the way they tought as geometry. The question is simple, but I want an concrete simple mathematic proof(verbal would be enough) something that would help me grasp the concept:</p>
<p>I have this set $$A = \left\{ \frac 1{2^n} \m... | Bernard | 202,857 | <p>Note $\mathbf N$ is <em>not</em> a group (natural numbers have no additive inverse).
Your group $A$ is in bijection with $\mathbf N$, by the map $n\longmapsto\frac1{2^n}$. </p>
<p>Indeed this map is injective, because if $n\ne n'$, say $n<n'$, then $$\frac1{2^{n'}}=\frac1{2^n}\cdot\frac1{2^{\mkern1mu\smash[b]{\... |
565,842 | <p>The common definition of $ \omega $-logic (a.k.a $\mathcal{L}_{\omega_1,\omega}$ logic) is the usual first order logic allowing infinite conjunctions and infinite proof.</p>
<p>Chang and Keisler, in section 2.2 of their book, define $\omega$-logic differently*.</p>
<p>They restrict to the language $\{S,+,\cdot,1\}... | Carl Mummert | 630 | <p>Perhaps there is an inconsistency in the terminology of the literature, but the following are simply different logics:</p>
<ul>
<li><p>$\mathcal{L}_{\omega_1, \omega}$ has infinitary formulas. I would call this <a href="http://plato.stanford.edu/entries/logic-infinitary/#1" rel="nofollow">infinitary logic</a>.</p><... |
2,892,797 | <p>Here's my attempt:</p>
<p>Suppose there is some $L \in \mathbb{R}$ such that $\lim_{ x\to 0 } \sin \frac{2\pi }{x} = L$. Then, if we let $\varepsilon = 1$, there would exist $\delta > 0$ such that $\left| f(x) - L\right| < 1$ for all $0< |x| < \delta $. Now, by the Archimedean property, there is $n \in ... | Arthur | 15,500 | <p>The only mistake I can find is that $x_1 , x_2 < \delta$ ought to be $0<x_1 , x_2 < \delta$. Also, depending on who's grading, "there cannot be such a number" might need some proof or justification. Or it might not. Apart from that, well done!</p>
|
3,652,730 | <blockquote>
<p>Without using L'Hôpital's rule, find:
<span class="math-container">$$\lim_{x\to 0}\dfrac{\cos(\frac{\pi}{2}\cos x)}{\sin(\sin x)}$$</span>
I know that the answer is <span class="math-container">$0$</span>.</p>
</blockquote>
<p>My attempt:</p>
<p>I tried by using the half-angle formula, <span cl... | CHAMSI | 758,100 | <p><span class="math-container">\begin{aligned}\lim_{x\to 0}{\frac{\cos{\left(\frac{\pi}{2}\cos{x}\right)}}{\sin{\left(\sin{x}\right)}}}&=\lim_{x\to 0}{\frac{\sin{\left(\frac{\pi}{2}-\frac{\pi}{2}\cos{x}\right)}}{\sin{\left(\sin{x}\right)}}}\\ &=\lim_{x\to 0}{\left(\frac{\pi x}{2}\times\frac{1-\cos{x}}{x^{2}}\t... |
4,541,198 | <p>I am reading a book and it says any closed set <span class="math-container">$F$</span> in <span class="math-container">$\mathbb{R}^n$</span> can be written as a countable union of compact sets as follows: <span class="math-container">$F=\cup_{k=1}^\infty F\cap B_k$</span> where <span class="math-container">$B_k$</sp... | Paul Frost | 349,785 | <p>No. Consider the circle <span class="math-container">$X = S^1$</span>. No constant map on <span class="math-container">$S^1$</span> is homotopic to the identity.</p>
|
1,480,381 | <pre><code>unsigned long long int H(unsigned long long int n){
unsigned long long int res = 1;
for (unsigned int i = 1; i < n; i++){
res += n / i;
}
return res;
}
</code></pre>
<p>I'm trying to convert this simple loop into a mathematical equation, I did a couple of attempts based on the exa... | BrianO | 277,043 | <p>Suppose $(M, <)$ is a model of your theory "R is irreflexive", "R is transitive", and $\forall x \exists y \: xRy$. </p>
<p>Because $M \models \forall x \exists y xRy$, for every $x \in M$ there's a $y \in M$ for which $x < y$. Using the principle of countable Dependent Choice, we can define a sequence $(a_n)... |
13,460 | <p>I hope that this question is on-topic, though it is not quite technical.</p>
<p>I am curious to hear from people how they approach reading a mathematical paper.</p>
<p>I am not asking specific questions on purpose, though at first I had a few. But I want to keep it rather open-ended.</p>
| Sean Tilson | 627 | <p>I have a couple of pieces of advice:</p>
<ol>
<li><p>Do not be intimidated by the introduction. This is where the whole paper is laid out,so it typically won't be easy to follow. The author will expand sentences into whole pages later on.</p></li>
<li><p>Take notes. I usually get a legal pad out and start copying d... |
889,690 | <p>Is there a formula (with mathematical reasoning) for separating a complex-valued function $f(z)=f(x+iy)$ into the form $
f(z)=u(x,y) + iv(x,y)$?</p>
<p>Thank You,</p>
<p>C.A</p>
| Sidharth Ghoshal | 58,294 | <p>I don't believe there exists one that's quite that clean. But let us assume that over some local area you can model the function with a complex exponential series (the upper bound k need not be finite)</p>
<p>$$ \sum_{j = 0 }^{k} \left[ e^{w_j x} \right] $$</p>
<p>where each $w_j$ is a complex number. We can decom... |
344,269 | <p>Let <span class="math-container">$M$</span> be an <span class="math-container">$n$</span>-dimensional smooth manifold and <span class="math-container">$\Theta$</span> some tensor field on <span class="math-container">$M$</span>, so a smooth section of <span class="math-container">$TM^{\otimes r} \otimes T^*M^{\otime... | Tsemo Aristide | 80,891 | <p>This article gives many examples of manifolds endowed with a structure whose group of automorphisms is finite dimensional.</p>
<p><a href="https://www.ams.org/journals/tran/1964-113-01/S0002-9947-1964-0164299-4/S0002-9947-1964-0164299-4.pdf" rel="nofollow noreferrer">https://www.ams.org/journals/tran/1964-113-01/S0... |
726,650 | <p>a) How does $x^2y''-3xy'+3y=0$ can be solved? I know how to solve for constant coefficients, but in this case they are functions...</p>
<p>b) In which maximum interval there is a solution that confirms $y'(1)=6, y(1)=4$?</p>
| mohammed | 145,939 | <p>$$am^2+(b-a)m+c=0\\
m^2-4m+3=0\\
(m-3)(m-1)=0\\
m_1=3, \, m_2=1\\
m_1 \neq m_2 \text{ so}\\
Y=C_1X^3+C_2X^1 \text{ is general solution}$$</p>
|
1,648,460 | <p>Suppose that $A$ is an m by n matrix and is right invertible, such that there exists and an n by m matrix $B$ such that $AB = I_m.$ Prove that $m\leq n.$ </p>
<p>I'm not really sure how go about this problem; any help would be appreciated.</p>
| Vim | 191,404 | <p>In this answer I'm presenting an opinion that's different from the other answers. </p>
<p>The other two answers seem to be trying to <em>mathematically</em> explain why one is supposed to integrate $\mathrm ds$ instead of $\mathrm dx$, but in my opinion, mathematics itself is unable to account for this, because the... |
4,324,207 | <p>I want to prove <span class="math-container">$\lim\limits_{x \to 5} \sqrt{2x+6} = 4$</span>
using the epsilon-delta definition.</p>
<p>My intuition to proving this is to use the root rule and square both ends of the equation to end up with:</p>
<p><span class="math-container">$\lim\limits_{x \to 5} 2x+6 = 16$</span>... | Michael Rozenberg | 190,319 | <p>For <span class="math-container">$z\rightarrow0^+$</span> and <span class="math-container">$x=y=1$</span> we obtain: <span class="math-container">$t\leq\frac{25}{4}$</span>.</p>
<p>We'll prove that <span class="math-container">$\frac{25}{4}$</span> is valid.</p>
<p>Indeed, we need to prove that <span class="math-con... |
3,966,951 | <p>I am reading about stationary harmonic maps and I came across the following calculation.</p>
<p>Let <span class="math-container">$\mathcal{Q}_t(x) = x+ t\zeta(x),$</span> where <span class="math-container">$\zeta\in C^{\infty}_c(B_{\rho}(x_0), \mathbb{R}^n)$</span> where <span class="math-container">$x_0\in \mathbb{... | Student | 255,452 | <ul>
<li>For the first identity, by the inverse function theorem we have that, <span class="math-container">$DQ_t^{-1}(Q_t(x)) = (DQ_t)^{-1}(x)=I -tD\zeta(x) +O(t^2).$</span></li>
<li>For the second identity, we can use the fact, <span class="math-container">$\det(I+tB)=I+t\operatorname{Tr}(B)+O(t^2)$</span> and <span ... |
1,653,299 | <p>If we have $R/M$ is a field and $M,I$ are ideals of $R$ such that $M\subseteq I \subseteq R$.</p>
<p>If we take $i\in I, i\not\in M$ we have $i+M \ne 0+M$. Since $R/M$ is a field, we have that $i+M$ is invertible, so $(i+M)(l+M)=1+M$</p>
<p>So $il+M=1+M$. If $il=1$ we get that $i$ is invertible, so $I=R$, but what... | D_S | 28,556 | <p>Another way: let $A$ be a ring, and $I$ an ideal of $A$. If $\mathfrak a$ is an ideal of $A/I$, then there exists an ideal $J$ of $A$ such that $I \subseteq J$ and $$\mathfrak a = J/I := \{ x + I : x \in J\}$$ You can obtain $J$ easily: it consists of all $x \in A$ such that the coset $x+I$ lies in $\mathfrak a$. ... |
283,527 | <p>I have a two-server queue with Poisson arrival rate and $\lambda$ exponential services with $\mu$ ( first server service rate) and 2$\mu$ ( 2nd server service rate). Capacity is infinite.</p>
<p>Then why is the number of customers in the queue at time $t$ not a Markov Process?</p>
<p>Can you please help me out?</p... | Math1000 | 38,584 | <p>When there is one customer in the system, we need to know which server is serving that customer. So let the state space be <span class="math-container">$S=\{0,1_A,1_B\}\cup\{2,3,4,\ldots\}$</span>, then <span class="math-container">$\{X(t):t\geqslant 0\}$</span> is a continuous-time Markov chain modelling the queuei... |
1,535,057 | <p>I am confused with the inductive step of this very basic induction example in the book <a href="http://rads.stackoverflow.com/amzn/click/0072899050" rel="nofollow">Discrete Mathematics and Its Applications</a>:</p>
<p>$$1 + 2+· · ·+k = k(k + 1) / 2$$</p>
<p>When we apply $k+1$, the equation becomes:</p>
<p>$$1 + ... | Peter | 82,961 | <p>Assume</p>
<p>$$1+...+k=\frac{k(k+1)}{2}$$</p>
<p>to calculate</p>
<p>$$1+...+k+(k+1)=\frac{k(k+1)}{2}+k+1=\frac{k^2+k+2k+2}{2}=\frac{k^2+3k+2}{2}=\frac{(k+1)(k+2)}{2}$$</p>
<p>completing the proof.</p>
|
1,354,491 | <p>If I wanted to have a die that rolled, for example:</p>
<pre><code>| Roll | Prob (in %) |
|------|-------------|
| 1 | 60 |
| 2 | 25 |
| 3 | 12 |
| 4 | 4 |
| 5 | 1 |
| 6 | 0.2 |
| 7 | 0.04 |
| ... | ... |
</code></pre>
<p>(... | MichaelChirico | 205,203 | <p>The standard approach here is to use the CDF of your distribution, which is defined on $\mathbb{R}$ (regardless of the support of your distribution).</p>
<p>For any finite distribution, you generally use <code>cumsum</code> to calculate the CDF; for an infinite distribution, you'll need an explicit formula for $F:\... |
19,876 | <p>Very important in integrating things like $\int \cos^{2}(\theta) d\theta$ but it is hard for me to remember them. So how do you deduce this type of formulae? If I can remember right, there was some $e^{\theta i}=\cos(\theta)+i \sin(\theta)$ trick where you took $e^{2 i \theta}$ and $e^{-2 i \theta}$. While I am draf... | lhf | 589 | <p>By far, the best approach to trigonometric formulae is to use <a href="http://en.wikipedia.org/wiki/Euler%27s_identity" rel="nofollow">Euler's identity</a> and its corollary, <a href="http://en.wikipedia.org/wiki/De_Moivre%27s_formula" rel="nofollow">de Moivre's formula</a>, recalling that $\sin^2 + \cos^2 = 1$.</p>... |
19,876 | <p>Very important in integrating things like $\int \cos^{2}(\theta) d\theta$ but it is hard for me to remember them. So how do you deduce this type of formulae? If I can remember right, there was some $e^{\theta i}=\cos(\theta)+i \sin(\theta)$ trick where you took $e^{2 i \theta}$ and $e^{-2 i \theta}$. While I am draf... | John Alexiou | 3,301 | <p>All of trig identities can be derived from $\exp{(\hat{i}\theta)}=\cos(\theta)+\hat{i}\sin\theta$. How?</p>
<p><strong>Example</strong></p>
<p>$$\cos(2\,x) = \mathrm{Re}\left(\exp{(2\hat{i}\,x)}\right)$$
$$ = \mathrm{Re}\left(\exp{(\hat{i}x)}\,\exp{(\hat{i}x)}\right) $$
$$ = \mathrm{Re}\left((\cos x+\hat{i}\sin x... |
427,157 | <p>The following problem seems easy at a first glance but I can't see the way to prove it. Actually I don't even know if it's true but it is assumed implicitly in a research paper.</p>
<p>Help highly appreciated.</p>
<p>Given two spd matrices <span class="math-container">$A$</span>, <span class="math-container">$B$</sp... | Carlo Beenakker | 11,260 | <p>I doubt there is an exact closed-form expression, but you could make progress numerically. Note that</p>
<p><span class="math-container">$$\mathbb{E}\left[\mathrm{Tr}\left((a Y_1+bY_2)^{-1}(cY_1+dY_2)\right)\right]=(c/a)\bigl( n+(d/c-b/a)f(b/a)\bigr),$$</span>
<span class="math-container">$$f(x)=\mathbb{E}\left[\mat... |
246,592 | <p>I am having trouble understanding when a function might have a removable singularity over a pole. </p>
<p>For example:
$$f(z)=\frac{\sin^2 z}{z}$$</p>
<p>I believe the pole is at $z=0$. However, if we take the taylor expansion of $f(z)$ apparently the pole vanishes. I do not understand how and where does the pole ... | jim | 44,551 | <p>$$\lim_{z \to 0}(z-0)\cdot f(z) =0$$
hence $0$ is removable singularity</p>
<p>this link may provide more clarity <a href="http://en.wikipedia.org/wiki/Removable_singularity" rel="nofollow">http://en.wikipedia.org/wiki/Removable_singularity</a></p>
|
266,526 | <p>I have a set of quadratic forms.</p>
<p><span class="math-container">$L_{1}=u_1^TJ_{1}u_1$</span></p>
<p><span class="math-container">$L_{2}=u_2^TJ_{2}u_2$</span></p>
<p><span class="math-container">$L_{3}=u_3^TJ_{3}u_3$</span></p>
<p>where <span class="math-container">$u_{i=1,2,3}$</span> - 3<span class="math-conta... | Daniel Huber | 46,318 | <p>First define the ui and ji using helper functions:</p>
<pre><code>u[i_] := {Subscript[u, i, 1], Subscript[u, i, 2], Subscript[u, i, 3]}
{u1, u2, u3} = u /@ {1, 2, 3};
j[i_] := Table[Subscript[j, i, i1, i2], {i1, 3}, {i2, 3}];
{j1, j2, j3} = j /@ {1, 2, 3};
</code></pre>
<p>Now it is easy to create the final matrix:<... |
3,315,227 | <p>I want to solve this differential equation <span class="math-container">$$ u^{(4)}+a u^{(2)} +bu=0$$</span></p>
<p>I put <span class="math-container">$v=u^{(2)}.$</span> I obtain the new equation <span class="math-container">$v^{(2)}+a v+bu=0.$</span> </p>
<p>What to do with <span class="math-container">$u?$</spa... | Witold | 680,456 | <p>Сharacteristic equation <span class="math-container">$t^4+at^2+b=0$</span>. Then <span class="math-container">$t_{1,2,3,4}=\pm \sqrt{\frac{-a \pm \sqrt{a^2-4b}}{2}}$</span>. Then <span class="math-container">$u=C_1e^{t_1x}+C_2e^{t_2x}+C_3e^{t_3x}+C_4e^{t_4x}$</span>.</p>
|
3,315,227 | <p>I want to solve this differential equation <span class="math-container">$$ u^{(4)}+a u^{(2)} +bu=0$$</span></p>
<p>I put <span class="math-container">$v=u^{(2)}.$</span> I obtain the new equation <span class="math-container">$v^{(2)}+a v+bu=0.$</span> </p>
<p>What to do with <span class="math-container">$u?$</spa... | Adrian Keister | 30,813 | <p>I wouldn't do the substitution then, because it's unclear how to "not do" an operation like differentiation. I would go from the original DE, and substitute in the usual ansatz: <span class="math-container">$u=e^{\lambda x}$</span> (assuming <span class="math-container">$u=u(x).$</span>) Then we obtain the quartic e... |
1,221,487 | <p>Problem: Let V be the subspace of all 2x2 matrices over R, and W the subspace spanned by:</p>
<p>\begin{bmatrix}
1 & -5 \\
-4 & 2 \\
\end{bmatrix}
\begin{bmatrix}
1 & 1 \\
-1 & 5 \\
\end{bmatrix}
\begin{bmatrix}
2 & -4 \\
-5 & 7 \\
\end{bmatrix}
\begin{bmatrix}
1 & -7 \\
-5 & 1 \\
\e... | rogerl | 27,542 | <p>(Edited to respond to OP comment):</p>
<p>Take a 9 in the original number and put it aside. Add up the remaining digits and iterate, getting a digital root between 1 and 9. Now show that adding the 9 to that root gives a number with the same digital root. More formally:</p>
<p>First, note that $dr(x+9) = dr(x)$. I... |
1,221,487 | <p>Problem: Let V be the subspace of all 2x2 matrices over R, and W the subspace spanned by:</p>
<p>\begin{bmatrix}
1 & -5 \\
-4 & 2 \\
\end{bmatrix}
\begin{bmatrix}
1 & 1 \\
-1 & 5 \\
\end{bmatrix}
\begin{bmatrix}
2 & -4 \\
-5 & 7 \\
\end{bmatrix}
\begin{bmatrix}
1 & -7 \\
-5 & 1 \\
\e... | Christian Blatter | 1,303 | <p>The essential point is that the operation $n\mapsto S(n)$ does not change the remainder modulo $9$. Write this as
$${\rm rem\,}_9\bigl(S(n)\bigr)={\rm rem\,}_9(n)\ .$$ Since $S'(n)$ is defined as $S(n)$, but ignoring the nines in the decimal representation of $n$, the two numbers $S(n)$ and $S'(n)$ differ by a multi... |
1,221,487 | <p>Problem: Let V be the subspace of all 2x2 matrices over R, and W the subspace spanned by:</p>
<p>\begin{bmatrix}
1 & -5 \\
-4 & 2 \\
\end{bmatrix}
\begin{bmatrix}
1 & 1 \\
-1 & 5 \\
\end{bmatrix}
\begin{bmatrix}
2 & -4 \\
-5 & 7 \\
\end{bmatrix}
\begin{bmatrix}
1 & -7 \\
-5 & 1 \\
\e... | Stephen Howe | 231,455 | <p>You want a simple answer that anyone can understand immediately I presume.
Ok, take a digital sum and add 9 to it. You've added 1 in the tens column and subtracted a 1 from the units so the sum has gone up one and back down. Now point out that the 9 could be anywhere - you can just as easily add 1000 and subtract 10... |
18,844 | <p>To elaborate a bit, I should say that the question of the existence of a complete metric is only of interest in the case of manifolds of infinite topological type; if a manifold is compact, any metric is complete, and if a noncompact manifold has finite topological type(ie is diffeomorphic to the interior of a compa... | Igor Belegradek | 1,573 | <p>By Whitney embedding theorem any smooth manifold embeds into some Euclidean space as a closed subset. The induced metric is complete.</p>
<p>In fact, a good exercise is to show that any Riemannian metric is conformal to a complete metric.</p>
|
309,851 | <p>I was considering the integral $\int_{-1}^1 \frac{1}{x^3}\, \mathrm{d} x$. At first, I suspected that it diverged due to the singularity present at $x = 0$, and WolframAlpha verified my hypothesis. However, I attempted to prove this more rigorously, but was unable. This was my reasoning:</p>
<p>$$\int_{-1}^1 \frac{... | Angel | 109,318 | <p>Let <span class="math-container">$f : [a, b] \rightarrow \mathbb R, a \leqq b$</span>. <span class="math-container">$f$</span> is called Riemann integrable if and only if the following condition is satisfied: for every <span class="math-container">$\epsilon \gt 0$</span>, there exists a tagged partition <span class=... |
1,369,485 | <p>$\log_2(x) = \log_x(2) $ </p>
<p>Using the change of base theorem: $\dfrac{\log(x)}{\log(2)} = \dfrac{\log(2)}{\log(x)}$</p>
<p>Multiplied the denominators on both sides: $\log(x)\log(x) = \log(2)\log(2)$</p>
<p>I kind of get stuck here. I know that you can't take the square root of both sides of the equation, bu... | JimmyK4542 | 155,509 | <p>You manipulated the given equation and got the following equation: $[\log(x)]^2 = [\log(2)]^2$. </p>
<p>Move everything to the left side: $[\log(x)]^2 - [\log(2)]^2 = 0$. </p>
<p>Factor the left side: $[\log(x)-\log(2)][\log(x)+\log(2)] = 0$. </p>
<p>So, for the original equation to be true, you need either $\log... |
912,370 | <p>To prove that a functions has partial derivatives every partial has to exist, and every partial exist only if the limit of definition of partial exist. Is this right?</p>
<p>Then if partials exist ,and the limit of derivability involving the Jacobian is equal to zero, (<a href="http://en.wikipedia.org/wiki/Differe... | gt6989b | 16,192 | <p>You don't need 2 dimensions to understand this. Consider fir your first example
$f(x) = |x|$ in 1D - it is clearly continuous and not differentiable at zero, as your first example was. So such a thing is possible.</p>
<p>Similarly, consider $f(x) = x$ for $x \neq 0$ and $f(0) = 1$. Clearly $f$ is not continuous (an... |
3,439,084 | <p>How to use mean value theorem to show that <span class="math-container">$\sqrt{x}(1+x)\log(\frac{1+x}{x})-\sqrt{x}<1$</span> when <span class="math-container">$x$</span> is positive.</p>
| Community | -1 | <p>The statement in question is showing that <span class="math-container">$gng^{-1}\in N$</span>, where <span class="math-container">$g$</span> and <span class="math-container">$n$</span> are general elements.</p>
<p>There is no problem with using other specific elements to obtain this proof.</p>
<p>In this part of t... |
347,175 | <p>We all know that a mgf of a random variable <span class="math-container">$m_X(t)$</span> is positive and <span class="math-container">$m(0)=1$</span>. My question is: if a positive real function <span class="math-container">$f(t)$</span> satisfies <span class="math-container">$f(0)=1$</span> and the function is smoo... | NWMT | 38,698 | <p>There is a more topological way. If you assume that <span class="math-container">$X$</span> admits a universal covering <span class="math-container">$\tilde X$</span> (so <span class="math-container">$X$</span> is path connected and semilocally simply connected, I believe) then the <span class="math-container">$G=\p... |
110,377 | <p>How can I extract a single dimension from an <code>InterpolatingFunction</code>? As an example:</p>
<pre><code>ClearAll[x];
s = NDSolve[
Evaluate[Derivative[1][x][t] == -x[t]] && x[0] == {10, -10, 4},
x, {t, 0, 5}]
x = x /. First@s
Plot[x[t], {t, 0, 5}]
</code></pre>
<p><a href="https://i.stack.imgur.... | J. M.'s persistent exhaustion | 50 | <p>Using some <em>underdocumented</em> functionality:</p>
<pre><code>x = NDSolveValue[{x'[t] == -x[t], x[0] == {10, -10, 4}}, x, {t, 0, 5}];
pts = Transpose[Append[x["Coordinates"],
Total[Drop[x["ValuesOnGrid"], None, {2}], {2}]]];
xsum = Interpolation[pts, InterpolationOrder -> x["Interpol... |
626,034 | <p>When you first differentiate the above, you get $-8/25$, right? Then you derive the gradient for a normal and proceed so on and so forth.
The textbook I'm using says when you differentiate, you get $-16/25$. I believe that's wrong...</p>
| egreg | 62,967 | <p>Write $f(x)=8(4+x^2)^{-1}$; then
$$
f'(x)=8\cdot(-1)\cdot(4+x^2)^{-2}\cdot 2x
$$
by the chain rule.</p>
<p>Then $f'(4)=-64\cdot20^{-2}=-\dfrac{16}{25}$, so the slope of the normal is $\dfrac{25}{16}$. Since $f(4)=\dfrac{2}{5}$, the normal is
$$
y-\frac{2}{5}=\frac{25}{16}(x-4)
$$</p>
|
1,863,493 | <p>I've been trying to solve this problem but I have an issue with the fact that there is a sum under each absolute value. I'm trying to convert this minimization problem (with respect to $x, y_1, \dots,y_n$)</p>
<p>$$
\sum_{i=1}^{m} \left| x - \sum_{j=1}^{n}\left| y_j - a_{ij} \right| \right|
$$</p>
<p>to a linear ... | Johan Löfberg | 37,404 | <p>You cannot convert this to a linear program. The objective function is not convex (consider the special case $m = n = x = a_{11} = 1$, draw the function and you'll see). You cannot use epigraph models when the expression you outer bound enters in a nonconvex fashion (which the inner absolute value does).</p>
<p>EDI... |
136,413 | <p>Can a single point be a graph? Or is it just a single edge and two vertices? How do you apply this to an induction-proof in graph-theory?</p>
<p>thanks</p>
| user15464 | 15,464 | <p>Define $g(z) = f(z)/z^n$. Then $g$ is analytic on $\mathbb C \setminus \{0\}$, and $g(z) \rightarrow a_n$, the leading coefficient of $f$, as $z\rightarrow \infty$. </p>
<p>Suppose to the contrary that $M(r_1)/r_1 ^n < M(r_2)/r_2^n$ for some $r_1 < r_2$. Equivalently,</p>
<p>$$\sup\{|g(z)| : |z| = r_1 \} < \... |
136,413 | <p>Can a single point be a graph? Or is it just a single edge and two vertices? How do you apply this to an induction-proof in graph-theory?</p>
<p>thanks</p>
| WimC | 25,313 | <p>If $f$ is entire, then by the maximum modulus principle $\max_{|z|=r}|f(z)|$ is non-decreasing in $r$. Take $f(z) = z^n P(\tfrac{1}{z})$. This is an entire function (a polynomial). So $\max_{|z|=r} r^n|P(\tfrac{1}{z})|$ is non-decreasing in $r$. In other words, $\max_{|z|=r}|P(z)|/r^n$ is non-increasing in $r$.<... |
89,675 | <p>Given a geometric series, how would you do this?</p>
<p>For example, how would this be done if the geometric series in question as is as follows?:</p>
<p>$$ \frac{1}{(1 - (-x^2))}$$</p>
| David Mitra | 18,986 | <p>$$
\sum_{n=0}^\infty\, r^n \ = \ \sum_{n=0}^{m-1}\, r^n + \sum_{n=m }^\infty\, r^n
\ = \ \underbrace{ 1-r^m \over 1-r}_{\text{first }m\text{ terms}} + \underbrace{ {r^{m }\over 1-r }}_{\text {error}}.
$$</p>
|
4,438,577 | <p>This question was on a math competition.</p>
<blockquote>
<p>Is there a triangle, which is not equilateral, whose sides form a geometric sequence and whose angles form an arithmetic sequence?</p>
<p>If such a triangle exists, find its sides and angles.</p>
</blockquote>
<p>My attempt:</p>
<p>Assume a triangle with s... | Pierre | 1,051,279 | <p>If the angles are in arithmetic progression, let them be <span class="math-container">$\alpha,\beta,\gamma$</span>.
We then have, for some reason <span class="math-container">$r$</span>, that</p>
<p><span class="math-container">$\beta=\alpha+r$</span> and <span class="math-container">$\gamma=\alpha+ 2r$</span>.</p>
... |
17,429 | <p>I want to define my own little 'Inner Product' function satisfying properties of linearity and commutativity, and I'd like to use the "$\langle$" and "$\rangle$" symbols to output my results. For this I am using <code>AngleBracket</code> which has no built-in meaning.</p>
<p>I was able to use <code>SetAttributes[A... | Jens | 245 | <p>From your question it can be deduced that you're interested only in the Euclidean scalar product for real vector spaces, so I'll make that assumption.</p>
<p>In version 9, I think the cleanest way to do the symbolic manipulations you're after is to use the new capabilities of <code>TensorReduce</code>. The special ... |
2,222,616 | <p>Let $V = P_{2}$ be the vector space of polynomials of degree less than or equal to $2$ with real coefficients, and let $W$ be the subset of polynomials $p(x)$ in $P_{2}$ such that:</p>
<p>$$\int_{-2}^{0}p(x)\,dx = 4\int_{0}^{2}p(x)\,dx.$$</p>
<p>$b)$ Find a basis for $W$, and compute $\dim(W)$.</p>
<p>For $b)$, I... | jorgeegroj | 433,222 | <p>Here's the thing... You're going to have a restricted domain because this is a $\sqrt{*}$ based function.</p>
<p>You need $x-a>0$ and $b-x>0$ so $x>a \cap x<b$. So first of all you HAVE to guarantee $a<b$ </p>
<p>Assuming $a<b$, then you're going to have a function looking like $\cap$, with a max... |
1,859,741 | <p>How do I prove that</p>
<p>$$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$$</p>
<p>without using the calculator?</p>
| Christian Blatter | 1,303 | <p>As in @mweiss 's answer we use repeatedly the approximation
$$\sqrt{a^2+b}\approx a+{b\over 2a}\qquad(|b|\ll a^2)\ .$$In this way we obtain on the one hand
$$\eqalign{
\sqrt{20}&=\sqrt{25-5}\approx 5-{1\over2},\quad
20+\sqrt{20}\approx25-{1\over2},\cr
\sqrt{20+\sqrt{20}}&\approx5-{1\over20},\quad 20+\sqrt{20... |
1,709,013 | <p>Let $T$ be a mobius transformation with exactly one fixed point on $\mathbb{C} \cup \{\infty\}$. What form does $T$ take? Find a fomrula for $T^n(z)$. What happens to $T^n(z)$ as $n \to \infty$.</p>
<h1>Attempt</h1>
<p>(we will assume $w \neq \infty$ since this case I covered already). If we examine a generic Mob... | carmichael561 | 314,708 | <p>After conjugating by an appropriate transformation $S$, you can assume that the fixed point is $\infty$. Then what form does $T$ take?</p>
|
527,568 | <p>I need to integrate, $\int\limits_{|z| = R} \frac{|dz|}{|z-a|^2}$ where $a$ is a complex number such that $|a|\ne R$. </p>
<p>So first I tried polar coordinates, which gives something I cannot continue.</p>
<p>Then I tried to write $|dz| = rd\theta = dz/ie^{i\theta}$ and I have
$$\int\limits_{|z| = R} \frac{dz}{i... | Christian Blatter | 1,303 | <p>You may assume $a=|a|>0$. On $\gamma:=\partial D_R$ one has $\bar z={R^2\over z}$; furthermore the parametrization
$$\gamma:\quad \phi\mapsto z=R e^{i\phi}\qquad(0\leq\phi\leq2\pi)$$
gives $dz=iR e^{i\phi}\ d\phi$ and therefore
$$|dz|=Rd\phi=-i{R\over z}\ dz\ .$$
(Complexifying the real $|dz|$ in this way is a tr... |
107,355 | <p>I am looking for a solver that allows me to solve an optimization problem of the form</p>
<p><span class="math-container">$$\begin{array}{ll} \text{minimize} & x_1 x_2 \cdots x_n\\ \text{subject to} & \color{gray}{\text{(some linear constraints)}}\end{array}$$</span></p>
<p>I've used Gurobi before. However, ... | Yusuf Mustopa | 5,496 | <p>A minor correction: the blowup of $\mathbb{P}^{2}$ at 9 points cannot be del Pezzo, since its anticanonical class has self-intersection equal to 0.</p>
<p>Much of the recent interest in the blowup $X_{k}$ of $\mathbb{P}^{2}$ at $k \geq 9$ points in general position centers around the ample cone of $X_{k},$ rather ... |
729,352 | <p>I am trying to prove that $a_1$, $a_2$, $a_3$ are linearly independent.</p>
<p>I am asked to use vector product and prove that if $c_{1}a_{1} + c_{2}a_{2} + c_{3}a_{3} = 0$ then $c_1 = c_2 = c_3 = 0$</p>
<p>I am completely stuck on where to go with this problem. I would think that linearly independent then the nul... | Community | -1 | <p>Let $(v_1,v_2,v_3)$ be a set of orthogonal <strong>non zero</strong> vectors. Let $\alpha_1,\alpha_2,\alpha_3\in\Bbb R$ such that
$$\sum_{k=1}^3 \alpha_k v_k=0$$
then
$$\require{cancel}\alpha_j \cancelto{\ne0}{||v_j||^2}=\bigg\langle \sum_{k=1}^3 \alpha_k v_k,v_j\bigg\rangle=0,\;\;\forall j=1,2,3$$</p>
<p>Conclude.... |
3,550,293 | <blockquote>
<p>The following are given:
<p><span class="math-container">$X$</span> is a discrete random variable
<p> The probability mass function given is <span class="math-container">$P(X=k)=Clnk$</span>
<p><span class="math-container">$k=e$</span>,<span class="math-container">$e^2$</span>,<span class="math-... | gamma555 | 647,532 | <p>There are several ways one can play with this type of problems.</p>
<p>Here is one:</p>
<p><span class="math-container">$\tan\left(\frac{\pi}{2}-\theta \right) =\frac{\sin\left(\frac{\pi}{2}-\theta \right) }{ \cos\left(\frac{\pi}{2}-\theta \right) }$</span></p>
<p>But </p>
<p><span class="math-container">$\sin... |
3,378,104 | <blockquote>
<p>Let <span class="math-container">$\mathcal{F} \subseteq \mathcal{G}$</span>.</p>
<p>Show that <span class="math-container">$$E((E(X\mid \mathcal{G})-E(X\mid \mathcal{F}))^2=E(E(X\mid \mathcal{G}))^{2}-E(E(X\mid \mathcal{F}))^{2}$$</span></p>
</blockquote>
<p>My idea:</p>
<p><span class="math-container">... | user | 505,767 | <p>We have that</p>
<p><span class="math-container">$$xy\log(x^2+y^2)=\frac{xy}{x^2+y^2}\cdot (x^2+y^2)\log(x^2+y^2) \to 0$$</span></p>
<p>indeed we can se for example by polar coordinates that</p>
<ul>
<li><span class="math-container">$\frac{xy}{x^2+y^2}=\sin \theta\cos \theta$</span> </li>
</ul>
<p>is bounded and... |
1,888,217 | <p>I have an application where I need to resize the users picture to max. 50KB</p>
<p>To resize I use a Method that sizes the picture to a given Width/Height.</p>
<p>The picture is a Bitmap with max. 8BPP.
(Width * Height * 8) = Size in Bits</p>
<p>What I need to know now is: How can I calculate the amount of pixels... | Zubzub | 349,735 | <p>Assuming you want to keep the aspect ratio. Define $r = w_{original}/h_{original}$.</p>
<p>You want to find $w$ and $h$ such that :
$$
w\cdot h \cdot 8 \leq 50\cdot 1024 \cdot 8\\
w/h = r
$$
Inserting the second into the first :
$$
rh^2 \leq 51200 \implies h=\sqrt{51200/r} \\
w = r\cdot h
$$</p>
|
727,664 | <p>I need to evaluate $I = \int^\pi_{-\pi} \cos^3(x) \cos(ax)~dx$ where $a$ is some integer. </p>
<p>I get:
$\dfrac{2a(a^2-7)\sin(\pi a)}{a^4 - 10a^2 + 9}$. However $\sin(\pi a)$ is $0$ for all $a$ so $I=0$. But as noted by an answer, there are answers for $a = 1,3,-1,-3$. </p>
| orion | 137,195 | <p>It tells you everything. After you factorize, it tells you either</p>
<p>$$I=0$$</p>
<p>or</p>
<p>$$1+\frac{9}{a^3}-\frac{3}{a^2}-\frac{7}{a}=0$$</p>
<p>Which is true for $a=1$. For this case, you can perform the integration by hand.</p>
<p>$a=3$ is also nonzero, but your expression seems not to show that. Chec... |
2,001,243 | <p>Suppose I have $6$ fair coins, all facing heads ups. For each turn I will flip each of the $6$ coins. If a coin lands tails up, I set it aside. The next turn I take the remaining coins facing heads up and repeat the process. Now I want to know what is the expected number of turns $T$ until I have exactly $6$ tails. ... | Molossus Spondee | 239,001 | <p>So I originally typed up most of this as my own question but I think maybe you might find this a little helpful.</p>
<p>If you use a tool like <a href="https://github.com/ott-lang/ott" rel="nofollow noreferrer">Ott</a> it extracts grammar and judgments to inductive definitions. This can be useful but this isn't actu... |
766,359 | <p>$12$ women and $10$ men are on the faculty. How many ways are there to pick a
committee of $7$ if </p>
<p>(a) Claire and Bob will not serve together, </p>
<p>(b) at least one woman must be chosen</p>
<p>I'm not sure how to start a. Essentially I need to remove a woman and remove a man?</p>
<p>For b, $\binom{22}{... | ml0105 | 135,298 | <p>(a) The total number of committees are $\binom{22}{7}$. If Claire and Bob serve together, we have two people fixed, and so we have $\binom{20}{5}$ such committees. And so if Claire and Bob will not serve together, we have $\binom{22}{7} - \binom{20}{5}$.</p>
<p>For (b), we use inclusion-exclusion. There are $\binom... |
4,788 | <p>Despite its name, its often claimed that the fundamental theorem of algebra (which shows that the Complex numbers are algebraically closed - this is not to be confused with the claim that a polynomial of degree n has <strong>at most</strong> n roots) is not considered fundamental by algebraists as it's not needed fo... | Agustí Roig | 664 | <p>Well, there is something called "Algebraic Geometry". You may check the first pages of most of the books on that subject and you'll see something like: "Let $k$ be an algebraically closed field..." [for instance, $\mathbb{C}$]. Indeed, this is the beginning of <a href="http://rads.stackoverflow.com/amzn/click/038790... |
4,788 | <p>Despite its name, its often claimed that the fundamental theorem of algebra (which shows that the Complex numbers are algebraically closed - this is not to be confused with the claim that a polynomial of degree n has <strong>at most</strong> n roots) is not considered fundamental by algebraists as it's not needed fo... | Bill Dubuque | 242 | <p>One cannot over-emphasize the point that one of the primary benefits of working over an algebraically closed field such as $\mathbb C$ is the immense simplification gained by <strong>linearizing</strong> what would otherwise be much more complicated <strong>nonlinear</strong> phenomena. The ability to factor polynom... |
119,756 | <p>I hope this question is not completely trivial:</p>
<p>Suppose $V$ is an irreducible projective variety and $U\subset V$ is a Zariski open subset isomorphic to an affine variety. Is it true that $V\setminus U$ is a Cartier divisor in $V$? If not, what conditions should we impose on $V$? (I guess if $V$ is smooth, ... | LMN | 25,854 | <p>aglearner, In his article on abelian varieties Bryden Cais proves the result you mention. You can find the statement/proof on the top of page 4. (math.arizona.edu/~cais/Papers/Expos/AbVar.pdf)</p>
<p>Specifically, he proves that if $X$ is separated, normal, noetherian and $U \subset X$ is a nonempty affine open sub... |
3,593,700 | <p>I have run into a thinking error while trying to solve the following two exercises, here they are:</p>
<p>1) There are 30 fish in the lake. 5 of them have been taken out and marked and then put back into the lake. Later 7 fish have been picked from the lake. What is the probability that 2 out of 7 picked fish were ... | David K | 139,123 | <p>The short answer is <em>there is no reason</em> why you have to compute one probability according to the order of the objects for one problem and without regard to order for the other. You can do it either way for either problem if you count the orderings correctly.</p>
<hr>
<p>Here are some ways you can compute p... |
1,748,211 | <p>I don't understand the determinant condition on SU(3) group, broadly.</p>
<p>I know that the determinant of such matrices should be equal to 1.
But what is the real intention of that 1?</p>
<p>Is it the real number 1 i.e.
$1+i 0$ ?</p>
<p>Or the determinant of such matrices is a complex number per se and the abs... | carmichael561 | 314,708 | <p>The determinant of any unitary matrix is a complex number with absolute value $1$. The special unitary matrices are the subgroup having determinant equal to $1$, i.e. $1+0\cdot i$.</p>
|
1,748,211 | <p>I don't understand the determinant condition on SU(3) group, broadly.</p>
<p>I know that the determinant of such matrices should be equal to 1.
But what is the real intention of that 1?</p>
<p>Is it the real number 1 i.e.
$1+i 0$ ?</p>
<p>Or the determinant of such matrices is a complex number per se and the abs... | user296113 | 296,113 | <p>In the general case, let $A\in \operatorname{U}(n)$ i.e. $A$ is a matrix in $\mathcal M_n(\Bbb C)$ satisfying $A^*A=I_n$ where $A^*=\overline{A^T}$. So we have
$$1=\det(I_n)=\det(A^*)\det(A)=\vert \det(A)\vert$$
hence $\det(A)$ is a complexe number with absolute value $1$. Now it's easy to prove that the map $\varp... |
3,279,492 | <p>It is known that for a surface <span class="math-container">$S \subset \mathbb{R^3}$</span> it can be found the first and the second fundamental form. </p>
<p>I would like to find out if this "first and second fundamental form" can be extended for a surface <span class="math-container">$S \subset \mathbb{R^{n}}$</s... | Saswat Padhi | 531,080 | <p>Suppose <span class="math-container">$N = n^2 + k$</span> such that <span class="math-container">$0 \leq k < 2n+1$</span> so that we have <span class="math-container">$n^2 \leq N < (n+1)^2$</span>. Then <span class="math-container">$\lfloor \sqrt{N} \rfloor = n$</span>.</p>
<p>Now, we can simplify as below:</... |
3,116,768 | <p>The title is preliminary and should be changed if anyone has a better idea how to express this.</p>
<p>This is the series in question:
<span class="math-container">$$\sum_{n=1}^{\infty} \frac{n+4}{n^2-3n+1} := \sum a_n$$</span></p>
<p>What feels natural is a comparison to the harmonic series, since for any <span c... | Bernard | 202,857 | <p>The fast way uses <em>asymptotic equivalents</em>, which is valid since the general term of the series is ultimately positive (to be precise, for <span class="math-container">$n >2$</span>): </p>
<p><span class="math-container">$\;n+4\sim_\infty n$</span>, <span class="math-container">$ \;n^2-3n+1\sim_\infty n^2... |
3,798,729 | <p>Consider some nonzero vector <span class="math-container">$x^1=(x_1,x_2) \in \mathbb Z^2$</span> with integer components. Then by defining
<span class="math-container">$$x^2 = (x_2, -x_1)$$</span>
we can find a vector with integer components orthogonal to <span class="math-container">$x^1$</span>. I am wondering whe... | tomasz | 30,222 | <p>For the first question: <span class="math-container">$M$</span> is an abelian group of exponent <span class="math-container">$4$</span>, and hence so is <span class="math-container">$M^*$</span>. It follows that <span class="math-container">$2M^*$</span> is an abelian group of exponent <span class="math-container">$... |
2,522,951 | <p>So we studied the triangle congruence criteria and we proved the ASA criterium: <em>If two triangles have respectively congruent two angles and the side included in them, then the two triangles are congruent</em>.</p>
<p>The proof is a proof of contradiction, and starts with: <em>Assume the triangles ABC and A'B'C'... | Identicon | 558,191 | <p>If p>3 then, $2^{p-2}+3^{p-2}+6^{p-2}\equiv 1\mod p$. Now, multiply both the sides by 6 to get: $$3 · 2^{p−1} + 2 · 3^{p−1} + 6^{p−1} ≡ 6 \mod p$$ </p>
<p>this is clearly a consequence of Fermat’s little theorem. Therefore p divides $a_{p−2}$. Also, 2|$a_1$ and 3|$a_2$. So, there is no number other than 1
that is ... |
2,159,970 | <p>In Simmons's Calculus with Analytic Geometric, 2nd, pg. 114, there is the following problem:</p>
<blockquote>
<p>$\text{47) Sketch the curve}$ $f(x)=\frac{2}{1+x^2}$ $\text{and find the points on it at which the normal passes through the origin.}$</p>
</blockquote>
<p>This is how I approached it: </p>
<p>let $P... | amd | 265,466 | <p>The error you made stemmed from working with the slopes of the normals. This omitted vertical lines. Using a different form of equation for a line would cover those cases as well.</p>
<p>One form that’s convenient for this problem is $\mathbf n\cdot\mathbf x=\mathbf n\cdot\mathbf p$, where $\mathbf p$ is a known po... |
3,102,210 | <p>I'm trying to solve the following problem:</p>
<blockquote>
<p>Let <span class="math-container">$v_0$</span> be a vertex in a graph <span class="math-container">$G$</span>, and <span class="math-container">$D_0 := \{v_0\}$</span>.</p>
<ol>
<li><p>For <span class="math-container">$n = 1, 2 \dots$</span> ind... | cangrejo | 86,383 | <p>It depends on how you define the neighbourhood, but it seems that here it is defined so that <span class="math-container">$N(V')$</span> comprises the vertices adjacent to some vertex in <span class="math-container">$V'$</span>, excluding vertices in <span class="math-container">$V'$</span>.</p>
<p>So in that case,... |
1,004,534 | <p>I'd like to have a translation (in English) of a paper of Klaus Doerk published
in Journal of Algebra: <strong><a href="http://www.sciencedirect.com/science/article/pii/S0021869384711999" rel="nofollow">http://www.sciencedirect.com/science/article/pii/S0021869384711999</a></strong></p>
<p>It is 4,5 pages long with ... | Hagen von Eitzen | 39,174 | <p>Here's what the preview allows me to see:</p>
<h2>On finite solvable groups that behave like nilpotent groups with respect to the Frattini subgroup.</h2>
<p>Let $\Phi(G)$ denote the Frattini subgroup of a group $G$.
If $G$ is a finite nilpotent group and $U\le G$ and $N\lhd G$, then it is known that $\Phi(U)\le \P... |
2,403,273 | <p>We had the following version of Gronwall's Lemma in our lecture:</p>
<p>Let $a<b, c \in \mathbb{R}$, $v: [a,b] \mapsto \mathbb{R}$ measurable and bounded and $u:[a,b] \mapsto [0, \infty)$ continuous st $v(t) \leq c + \int_{a}^t u(s)v(s)ds$ for all $t \in [a,b]$. Then we have $v(t) \leq c \cdot \exp( \int_{a}^t u... | H. H. Rugh | 355,946 | <ol>
<li><p>The first inequality: If it were false for some $h>0$, then you could replace $t_0$ by $t_0+h$, contradicting the definition of $t_0$.</p></li>
<li><p>The second comes from the fact that for $t\leq t_0$: $v(t)\leq 0$ and using the first inequality.</p></li>
<li><p>Measurability and boundedness imply that... |
2,403,273 | <p>We had the following version of Gronwall's Lemma in our lecture:</p>
<p>Let $a<b, c \in \mathbb{R}$, $v: [a,b] \mapsto \mathbb{R}$ measurable and bounded and $u:[a,b] \mapsto [0, \infty)$ continuous st $v(t) \leq c + \int_{a}^t u(s)v(s)ds$ for all $t \in [a,b]$. Then we have $v(t) \leq c \cdot \exp( \int_{a}^t u... | supinf | 168,859 | <p>For the first inequality:</p>
<p>If there is a small $h$ such that the inequality is not true, then we could choose a bigger $t_0$
(which is defined via supremum).</p>
<p>For the second equality:
You just "forget" the values $t<t_0$ when taking the supremum.
Those dont matter, because $v(t)\leq 0$ for these val... |
638,396 | <p>I'm working through a book to learn trig on my own and I got stuck with the following. This is the image given and the text in the book reads:</p>
<blockquote>
<p>Suppose you are standing an unknown distance away from a cliff of height <span class="math-container">$h$</span>. You need to know the height <span c... | lab bhattacharjee | 33,337 | <p>We have $$\tan A=\frac{t+h}d\iff d=\frac{t+h}{\tan A}$$</p>
<p>Similarly, $$\tan B=\frac hd\iff d=\frac h{\tan B}$$</p>
<p>Compare the two values of $d$</p>
|
1,488,348 | <p>Denote the complex plane by C</p>
<p>Denote the unit sphere centred at the origin by P (edited: i.e. the Riemann sphere)</p>
<p>Lemma: Let A$\subset$C and B$\subset$P with B corresponding to the points in A. Then A is open relative to C if and only if B is open relative to P.</p>
<p>Let {$Q_i$} be an open cover o... | 5xum | 112,884 | <blockquote>
<p>Let $\{Q_i\}$ be an open cover of C. Then there exists $\{K_i\}$, an open cover of P, with each $Q_j$ corresponding to a different $K_j$</p>
</blockquote>
<p>Mistake. $Q_i$ are all subsets of $\mathbb C$, therefore none of them contains $\infty$. So $K_j$ is not an open cover of $P$.</p>
|
1,488,348 | <p>Denote the complex plane by C</p>
<p>Denote the unit sphere centred at the origin by P (edited: i.e. the Riemann sphere)</p>
<p>Lemma: Let A$\subset$C and B$\subset$P with B corresponding to the points in A. Then A is open relative to C if and only if B is open relative to P.</p>
<p>Let {$Q_i$} be an open cover o... | Matematleta | 138,929 | <p>Here's a sketch: </p>
<p>$1):\ $Fix the symbol $\infty $ and consider $C_{\infty}=\mathbb C\cup \left \{ \infty \right \}$.</p>
<p>$2):\ $Define a toplogy $\mathcal T$ on $C_{\infty}$ by declaring the open sets to be either $U$ whenever $U$ is open in $\mathbb C$ or $\mathbb C\cup \left \{ \infty \right \}\setminu... |
154,215 | <p>Prove that</p>
<ol>
<li>Each field of characteristic zero contains a copy of the rational number field.</li>
<li>For an $n$ by $n$ matrix $A,$ if it is not invertible, then there exists an $n$ by $n$ matrix $B$ such that $AB=0$ but $B\ne0.$</li>
</ol>
<p>For (1), I think I have to use the fact that each subfield o... | M.B. | 2,900 | <p>Hint on 2):</p>
<p>You know that since $A$ is not invertible there exists $x\neq 0$ such that $Ax = 0$. Can you use this to construct $B$?</p>
|
44,746 | <p>Okay, I'm not much of a mathematician (I'm an 8th grader in Algebra I), but I have a question about something that's been bugging me.</p>
<p>I know that $0.999 \cdots$ (repeating) = $1$. So wouldn't $1 - \frac{1}{\infty} = 1$ as well? Because $\frac{1}{\infty} $ would be infinitely close to $0$, perhaps as $1^{-\in... | Mikhail Katz | 72,694 | <p>There is one issue that has not been raised in the fine answers given earlier. The issue is implicit in the OP's phrasing and it is worth making it explicit. Namely, the OP is assuming that, just as $0.9$ or $0.99$ or $0.999$ denote <em>terminating</em> decimals with a finite number of 9s, so also $0.999\ldots$ de... |
3,214,662 | <p><strong>Q1</strong> Prove that every simple subgroup of <span class="math-container">$S_4$</span> is abelian.</p>
<p><strong>Q2</strong> Using the above result, show that if <span class="math-container">$G$</span> is a nonabelian simple group then every proper subgroup of <span class="math-container">$G$</span> has... | Narasimham | 95,860 | <p>Vector dot product decides obtuse/acute condition. Tail of arrow of vectors is at <span class="math-container">$A:$</span></p>
<p>If sign <span class="math-container">$(\vec{CA}\cdot\vec{BA})>0,$</span> then <span class="math-container">$\angle A$</span> is acute</p>
<p>If sign <span class="math-container">... |
98,565 | <pre><code>as = <|1 -> 2.2, 2 -> 3.4, 3 -> 8.1|>;
</code></pre>
<p>I don't understand the following:</p>
<pre><code>as[[1]]
</code></pre>
<blockquote>
<p>2.2</p>
</blockquote>
<p>But I expected to see</p>
<blockquote>
<p><|1 -> 2.2|></p>
</blockquote>
<p>On the other hand:</p>
<pre><code>as... | kale | 1,560 | <p>Probably should be handled by <a href="http://reference.wolfram.com/language/ref/KeyTake.html" rel="nofollow"><code>KeyTake</code></a>:</p>
<pre><code>KeyTake[as, 1]
</code></pre>
<blockquote>
<p><code><|1->2.2|></code></p>
</blockquote>
<p>Also works with multiple "selections":</p>
<pre><code>KeyTake... |
306,754 | <p>Looked around a bit and all I see are proofs using the limit definition of a derivative. This is not for an assignment, I could just use the limit definition if I wanted to, but I was wondering how you could go about proving this using the epsilon-delta definition of a derivative ($\forall \epsilon >0$, $\exists ... | DonAntonio | 31,254 | <p>It may depend on what limits you know, but</p>
<p>$$\frac{\sin(x+h)-\sin x}{h}=\frac{\sin x\cos h+\sin h\cos x-\sin x}{h}=$$</p>
<p>$$=\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\xrightarrow[h\to 0]{}\sin x\cdot 0+\cos x\cdot 1 =\cos x$$</p>
<p>You can also try other trigonometric identities with an equivalen... |
67,171 | <p>I am sure <a href="http://en.wikipedia.org/wiki/Modular_multiplicative_inverse">all those symbols</a> are really easy for you guys to understand, but I would appreciate it if someone could bring it down to earth for me.</p>
<p>How could I do this on a basic calculator? or with a few lines of programmer's code which... | ADF | 11,990 | <p>Consider the element $a \in \mathbb{Z}/m\mathbb{Z}$. It is not hard to show that $a^{-1}$ exists in $\mathbb{Z}/m\mathbb{Z}$ if and only if the gcd of $a$ and $m$ is 1, that is, $a$ and $m$ are coprime. Using Euler's theorem (also sometimes called Fermat's theorem),
$a^{\varphi(m)} \equiv 1 \pmod{m}$ for all $a$ cop... |
67,171 | <p>I am sure <a href="http://en.wikipedia.org/wiki/Modular_multiplicative_inverse">all those symbols</a> are really easy for you guys to understand, but I would appreciate it if someone could bring it down to earth for me.</p>
<p>How could I do this on a basic calculator? or with a few lines of programmer's code which... | CopyPasteIt | 432,081 | <p>There is an algorithm besides the extended Euclidean algorithm that can be used to solve</p>
<p><span class="math-container">$\tag 1 ax \equiv b \pmod{p} \quad \text{where } p \text{ is prime} \land p \nmid a \land p \nmid b$</span></p>
<p>An outline for the algorithm's logic was given <a href="https://math.stacke... |
2,983,553 | <p>I've been stuck trying to solve this problem for the whole day. Also, I'm trying to translate the problem as good as I can, as my English skills aren't the greatest; sorry for that.</p>
<p>Problem is as follows: <strong>Points OBDE form a quadrilateral. Points B and D are on the line x=1. Find the value of x that m... | Toby Mak | 285,313 | <p>Use the formula of a trapezoid: <span class="math-container">$A=\frac{1}{2}(a+b)(h)$</span>.</p>
<p>We know that <span class="math-container">$E$</span> is at <span class="math-container">$(\cos x, \sin x)$</span>, so <span class="math-container">$a = 1 - \cos x$</span>, <span class="math-container">$b = 1$</span>,... |
3,828,590 | <p>I am really confused behind the mathematical meaning of a <strong>100th percentile.</strong></p>
<p>What does that mean mathematically? Does it mean that a data point in a sample space is greater in some metric and that is also greater than itself?</p>
<p><strong>That makes no sense. AFAIK, there can be no such thin... | TheSilverDoe | 594,484 | <p><span class="math-container">$$\lim_{|x| \rightarrow +\infty} f(x) = -\infty$$</span></p>
<p>so <span class="math-container">$$\liminf_{|x| \rightarrow +\infty} f(x) = \limsup_{|x| \rightarrow +\infty} f(x) =-\infty$$</span></p>
|
2,929,182 | <p>Let <span class="math-container">$p,q>1$</span>, <span class="math-container">$\frac{1}{p}+\frac{1}{q}=1$</span> and <span class="math-container">$x\in[0,1]$</span>. I would like to show that <span class="math-container">$x^{\tfrac{1}{p}}\leq \frac{1}{p}x+\frac{1}{q}$</span>.</p>
<p>I tried to rewrite both sides... | Dr. Sonnhard Graubner | 175,066 | <p>Hint: Let <span class="math-container">$$f(x)=\frac{1}{p}x+\frac{1}{q}-x^{1/p}$$</span> then <span class="math-container">$$f'(x)=\frac{1}{p}-\frac{1}{p}x^{\frac{1}{p}-1}$$</span></p>
|
2,828,060 | <blockquote>
<p>Let $f(x)=x^3 +4x^2 +6x$ and $g(x)$ be its inverse. Then the value of $g'(-4)$ is?</p>
</blockquote>
<p>My attempt at the solution:
$$
g(x) = f^{-1}(x) \implies f(g(x))=x,
$$
<em>differentiating both sides,</em>
$$
f'(g(x)) \cdot g'(x)=1 \implies g'(x)=\frac1{f'(g(x))}.
$$
I am stuck here as we do no... | g.kov | 122,782 | <p>The three points $P_0,P_1,P_2$ are the control points
of the quadratic Bezier segment.
On the images these points are connected
with straight lines. $P_0$ and $P_2$
are the endpoints of the curve,
$P_1$ (marked with $\times$)
usually is not on the curve.
The formula
\begin{align}
B(t)&=(1-t)^2P_0+2(1-t)tP_1+... |
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