qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,738,579 | <blockquote>
<p>What is the cardinality of set <span class="math-container">$\big\{(x,y,z)\mid x^2+y^2+z^2= 2^{2018}, xyz\in\mathbb{Z} \big\}$</span>?</p>
</blockquote>
<p>Since I have very limited knowledge in number theory, I tried using logarithms and then manipulating the equation so that we get <span class="mat... | DonAntonio | 31,254 | <p><span class="math-container">$$\frac{2}{3}(4^n - 1) + 2^{2n + 1}= \frac23\left(\color{red}{4^n}-1+\overbrace{\color{red}{3\cdot2^{2n}}}^{=3\cdot4^n}\right)=\frac23\left(\color{red}{4\cdot4^n}-1\right) = \frac{2}{3}(4^{n + 1} - 1)$$</span></p>
|
3,738,579 | <blockquote>
<p>What is the cardinality of set <span class="math-container">$\big\{(x,y,z)\mid x^2+y^2+z^2= 2^{2018}, xyz\in\mathbb{Z} \big\}$</span>?</p>
</blockquote>
<p>Since I have very limited knowledge in number theory, I tried using logarithms and then manipulating the equation so that we get <span class="mat... | Community | -1 | <p>You want to prove</p>
<p><span class="math-container">$$\frac23(4^{n+1}-1)-\frac23(4^n-1)=2^{2n+1}.$$</span></p>
<p>Simplifying the <span class="math-container">$-1$</span> and dividing by <span class="math-container">$2\cdot4^n$</span>,
<span class="math-container">$$\frac13\cdot4-\frac13=1.$$</span></p>
<hr />
<p>... |
25,853 | <p>With regard to an undergraduate statistics course, I am developing a standardized list of point deductions with the TAs (doctoral students) so that graders are consistent in what they are taking off intermediate points for. For example, most problems are 10 points total, and my proposed point deductions for interme... | Xander Henderson | 8,571 | <p>Part of the problem is writing the exam questions in the first place. Others have noted that, when designing a grading rubric, you should identify what the key skills in the problem are. This seems backwards to me. <em>First</em>, identify the key skill that you want to test, and <em>then</em> write the exam ques... |
4,309,812 | <p>Recently I knew about <a href="https://en.m.wikipedia.org/wiki/Heron%27s_formula" rel="nofollow noreferrer">Heron's formula</a> for the area of some triangle, and its generalizations to quadrilaterals by Bretschneider's formula. According to Wikipedia there are also generalizations for pentagons and hexagons inscrib... | Ninad Munshi | 698,724 | <p>I'm assuming your bounds were actually</p>
<p><span class="math-container">$$y=x \hspace{15 pt} y=x+2 \hspace{15 pt} y = \frac{1}{x} \hspace{15 pt} y = \frac{2}{x}$$</span></p>
<p>It is completely possible to find <span class="math-container">$x+y$</span>, the trick is noticing the degree of the terms. <span class="... |
4,187,932 | <p>Is there a general <strong>algebraic</strong> form to the integral <span class="math-container">$$\int_{k_1}^{k_2} x^2 e^{-\alpha x^2}dx?$$</span> I know that if this integral is an improper one, then the integral can be calculated quite easily (i.e. is a well known result). However, when these bounds are not impose... | Timur Bakiev | 855,963 | <p><span class="math-container">$dv$</span> is not even a number in math, this is another kind of object, so physicists clearly abuse some natural ways of understanding and visualising the differential. In math, expressions like <span class="math-container">$v^2 + (dv)^2$</span> or <span class="math-container">$v + dv$... |
1,626,362 | <p><code>The following is a short extract from the book I am reading:</code> </p>
<blockquote>
<p>If given a Homogeneous ODE:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+5\frac{\mathrm{d} y}{\mathrm{d}x}+4y=0\tag{1}$$
Letting
$$D=\frac{\mathrm{d}}{\mathrm{d}x}$$ then $(1)$ becomes
$$D^2 y + 5Dy + 4y=(D^2+5D+4)... | Valentin | 31,877 | <p>The equation can be written in the following form:
$$L[y]=0 \tag{1}$$
where $L=\frac{d^2}{dx^2} + 5\frac{d}{dx} + 4$ is a second-order linear operator on the space of twice differentiable functions $C^2$. Let $y_1,y_2\in C^2$ be two solutions, that is $L[y_1] = L[y_2] =0$ (assuming we know they exist). By linearity ... |
1,626,362 | <p><code>The following is a short extract from the book I am reading:</code> </p>
<blockquote>
<p>If given a Homogeneous ODE:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+5\frac{\mathrm{d} y}{\mathrm{d}x}+4y=0\tag{1}$$
Letting
$$D=\frac{\mathrm{d}}{\mathrm{d}x}$$ then $(1)$ becomes
$$D^2 y + 5Dy + 4y=(D^2+5D+4)... | Community | -1 | <p>If you recall from linear algebra, abstract functional spaces can be considered as vector spaces. We define the zero function to serve as the zero vector, and pointwise addition/multiplication as the vector space operations.</p>
<p>For a collection of normal vectors, to show linear independence, we want to show non... |
2,746,153 | <p>Assume $m\ \mathrm{and}\ n\ \mathrm{are\ two\ relative\ prime\ positive\ integers.}$</p>
<p>Given $x \equiv a\ \pmod m$ and $x \equiv a\ \pmod n$.</p>
<p>Prove that $x \equiv a\ \pmod {mn}\ \mathrm{by\ using\ Chinese\ Remainder\ Theorem}.$<br/></p>
<p>And I did the following:
<br>
$$ \mathrm {M_1 = }\ n\ \ and\... | thesmallprint | 438,651 | <p>$x\equiv a\bmod n$ implies there exists a $k\in\Bbb Z$ such that $x=nk+a$. Now, we have $$nk+a\equiv a\bmod m\Rightarrow nk\equiv0\bmod m\Rightarrow k\equiv0\bmod m,$$ so then there exists a $j\in\Bbb Z$ such that $k=jm$. Substituting this in our equation for $x$ gives $$x=njm+a,$$ which means that $x\equiv a\bmod n... |
3,167,261 | <blockquote>
<p>Let <span class="math-container">$\mathcal{O}$</span> be an open subset of the plane <span class="math-container">$\mathbb{R}^{2}$</span> and
let the mapping <span class="math-container">$F : \mathcal{O} \rightarrow \mathbb{R}^{2}$</span> be
represented by <span class="math-container">$F(x, y) = ... | Ernie060 | 592,621 | <p>Yes. For instance,
<span class="math-container">$$
\frac{\partial^2 u}{\partial^2 x} + \frac{\partial^2 u}{\partial^2 y} =
\frac{\partial }{\partial x}\frac{\partial u}{\partial x} + \frac{\partial }{\partial y}\frac{\partial u}{\partial y}
= \frac{\partial }{\partial x}\frac{\partial v}{\partial y} - \frac{\... |
2,669,278 | <p>I noticed a strange thing with my calculator.<br>
When I start with any number like 1,2,3 or 1.2, 1.34 .... or even 0.<br>
And repeatedly take the cosine function of this number.<br>
I get the same following number. I don't thing this is a coincidence since it's happening with any number I try. </p>
<pre>0.9998477... | Travis Willse | 155,629 | <p>First, the number is a (the) fixed point $x_0$ of the map $x \mapsto \cos x^{\circ}$; here, $\cdot^{\circ}$ denotes interpreting $x$ as an angle measure of $x$ degree. Alternatively, we can avoid mention of degrees by saying this number is (the) fixed point of the map $T : x \mapsto \cos \frac{\pi x}{180}$.</p>
<p>... |
168,020 | <p>Let $R$ be an local Artinian ring, with maximal ideal $\mathfrak{m}$.</p>
<p>Let $e$ be the smallest positive integer for which $\mathfrak{m}^e=(0)$.</p>
<p>Let $t$ be the smallest positive integer for which $x^t=0$ for all $x \in \mathfrak{m}$.</p>
<p>We know $t \leq e$, with equality holding whenever $\mathfrak... | Neil Epstein | 19,045 | <p>To complement Mohan's answer, it is worth noting that there are counterexamples when $R$ contains a field $k$ of prime characteristic $p$. Indeed, when $p\geq 3$, let $R=k[\![X,Y]\!]/(X^p, Y^p)$, and denote the images of $X$, $Y$ in $R$ by $x$, $y$ respectively. Then I claim that $t=p$ but $e\geq 2p-2>p$. To s... |
114,122 | <p>I am trying to figure out the maximum possible combinations of a (HEX) string, with the following rules:</p>
<ul>
<li>All characters in uppercase hex (ABCDEF0123456789)</li>
<li>The output string must be exactly 10 characters long</li>
<li>The string must contain at least 1 letter</li>
<li>The string must contain a... | Syed | 321,426 | <p>Four years and no answer? To continue with your original reasoning...<br><p><ol>
<li>Total amount of numbers: $10^{10}$ or 10,000,000,000</li>
<li>Total amount of letters: $6^{10}$ or $60,466,176$</li>
<li>Subtract the above two numbers $10^{10}$ and $6^{10}$ from...(drum roll please)...Your last criteria! ("A numb... |
784,753 | <p>In spherical coordinates, we have</p>
<p>$ x = r \sin \theta \cos \phi $;</p>
<p>$ y = r \sin \theta \sin \phi $; and </p>
<p>$z = r \cos \theta $; so that</p>
<p>$dx = \sin \theta \cos \phi\, dr + r \cos \phi \cos \theta \,d\theta – r \sin \theta \sin \phi \,d\phi$;</p>
<p>$dy = \sin \theta \sin \phi \,dr + r ... | Community | -1 | <p>$dV=dxdydz=|\frac{\partial(x,y,x)}{\partial(r,\theta,\phi)}|drd\theta d\phi$</p>
|
1,238,210 | <p>How we can solve that $\lim _{_{x\rightarrow \infty }}\int _0^x\:e^{t^2}dt$ ?</p>
<p>P.S: This is my method as I thought:
$\int _0^x\:\:e^{t^2}dt>\int _1^x\:e^tdt=e^x-e$ which is divergent, so all your answers, helped me to think otherwise, maybe my method help something else :D</p>
| Teoc | 190,244 | <p>$$e^{t^2}> e^t\text{ from 1 to $\infty$, and the part from 0 to 1 is finite.}$$ and the integral$ \int_1^\infty e^t $ diverges. Therefore, by the comparison test, it diverges too.</p>
|
1,934,033 | <p>I'm new here. I wish to ask a question regarding predicate logic:</p>
<p>I was given three predicates:</p>
<p><strong>parent(p,q): p is the parent of q.</strong></p>
<p><strong>female(p): p is a female.</strong></p>
<p><strong>p = q: p and q are the same person.</strong></p>
<p>Now, I was tasked with translatin... | Parcly Taxel | 357,390 | <p>Intuitively, a predicate in a predicate doesn't make sense; predicates only take terms as arguments. Using the normal convention of abbreviating predicates and terms by letters ($P(p,q)$ for parent, $F(q)$ for female, $a$ for Alice), your example is
$$\exists q\ P(a,F(q))$$
which is interpreted as "Alice is the pare... |
2,616,280 | <p>This question seems obvious, but I'm not secure of my proof.</p>
<blockquote>
<p>If a compact set $V\subset \mathbb{R^n}$ is covered by a finite union of open balls of common radii $C(r):=\bigcup_{i=1}^m B(c_i,r)$, then is it true that there exists $0<s<r$ such that $V\subseteq C(s)$ as well? The centers ar... | Umberto P. | 67,536 | <p>Let $X$ denote the set of centers: $X = \{c_1,\ldots,c_m\}$. </p>
<p>The function $\phi(x) = \mathop{\rm dist} (x,X)$ is continuous on $\mathbb R^n$ and attains a maximum value on $V$ because $V$ is compact. </p>
<p>Note that if $x \in V$, then by definition $\phi(x) < r$. Whatever maximum it attains must be le... |
2,616,280 | <p>This question seems obvious, but I'm not secure of my proof.</p>
<blockquote>
<p>If a compact set $V\subset \mathbb{R^n}$ is covered by a finite union of open balls of common radii $C(r):=\bigcup_{i=1}^m B(c_i,r)$, then is it true that there exists $0<s<r$ such that $V\subseteq C(s)$ as well? The centers ar... | Mikhail Katz | 72,694 | <p>Replace each open ball $B_i$ of radius $r$ in the cover by the union of concentric open balls of radii strictly smaller than $r$. You get an infinite cover of $V$. By compactness there is a finite subcover. By construction the radii are smaller than before. Finally we choose the maximal radius (for all of the fi... |
79,869 | <p>Let <span class="math-container">$(X,\mu,\mathcal{F})$</span> be a probability space. The paper <em><a href="http://projecteuclid.org/euclid.aoms/1177693405" rel="nofollow noreferrer">Equiconvergence of Martingales</a></em> by Edward Boylan introduced a pseudometric on sub-<span class="math-container">$\sigma$</spa... | Yuri Bakhtin | 2,968 | <p>I suspect that the following is a dense set:</p>
<p>For each $n\in\mathbb{N}$ take all sub-algebras of the finite sigma-algebra generated by intervals of the form $[i/2^n,(i+1)/2^n)$, $i=0,\ldots,2^n-1$.</p>
|
39,423 | <ul>
<li><p>case1</p>
<pre><code>Options[f] = {"t" -> "0"};
f[___, OptionsPattern[]] := StringReplace["content", "t" :> OptionValue["t"]]
f[]
(*
con0en0
*)
</code></pre></li>
<li><p>case2</p>
<pre><code>rule = {"t" -> OptionValue["t1"]};
Options[gg] = {"t1" -> "T1", "t2" -> "1"};
gg[___, OptionsPat... | Mr.Wizard | 121 | <p>It would help if you outlined you intended use of this behavior, as without that it is not clear what is and is not helpful.</p>
<h3>Single function case</h3>
<p>You can use the two-argument form of <code>OptionValue</code>:</p>
<pre><code>rule = {"t" :> OptionValue[gg, "t1"]}; (* note RuleDelayed *)
Options... |
28,877 | <p>Since I self-study mathematical analysis without <em>formal</em> teacher, I can only appeal to help from out site most of the time. It's obvious that to grasp the underlying concepts in mathematics, we must roll the sleeves and solve problems.</p>
<p>It's clear that there are actually mistakes and misunderstanding ... | Andres Mejia | 297,998 | <p>I'm not sure about appealing, but if you're looking for an answer, I think that there are some alternatives. I'm mostly just speaking from personal experience, in hopes to add to a nice list of suggestions by Arnaud Mortier.</p>
<ol>
<li><p>Explain succinctly the proof idea. Usually, if it is novel/ different and s... |
108,010 | <p>It is not necessarily true that the closure of an open ball $B_{r}(x)$ is equal
to the closed ball of the same radius $r$ centered at the same point $x$. For a quick example, take $X$ to be any set and define a metric
$$
d(x,y)=
\begin{cases}
0\qquad&\text{if and only if $x=y$}\\
1&\text{otherwise}
\end{case... | JDH | 413 | <p>Here is a characterization that is straight from the definitions, but which it seems may be useful when verifying that a particular space has the property.</p>
<p>For any metric space <span class="math-container">$(X,d)$</span>, the following are equivalent:</p>
<ul>
<li>For any <span class="math-container">$x\in X$... |
2,612,308 | <p>Obviously we can rearrange for <span class="math-container">$x$</span> in a polynomial of degree 2. </p>
<p>Let <span class="math-container">$y=ax^2+bx+c$</span></p>
<p>then </p>
<p><span class="math-container">$x=\frac{-b\pm\sqrt{b^2-4ac+4ay}}{2a}$</span></p>
<p>Similarly, for <span class="math-container">$y=ax... | Ross Millikan | 1,827 | <p>You have to define what equations you care about and desirably but it seems likely the answer is no. You are presumably familiar with the solutions of linear and quadratic equations. Your equation is a cubic, so you can feed it to <a href="https://en.wikipedia.org/wiki/Cubic_function" rel="noreferrer">Cardano's fo... |
1,285,774 | <p>I have looked at similar questions under 'Questions that may already have your answer" and unless I have missed it, I cannot find a similar question.</p>
<p>I am trying to answer the following:</p>
<p>Let $A = \left(\begin{matrix}
a & b \\
b & d \\
\end{matrix}\right)$ be a symmetric 2 x 2 matrix. Prove t... | abel | 9,252 | <p>suppose $\pmatrix{a&b\\b&d} $ is positive definite. we need $$ax^2 + 2bxy + dy^2 > 0 \text{ for all } x, y, x^2 + y^2 \neq 0.$$ taking $x = 1, y = 0$ gives $a > 0.$ </p>
<p>taking $x = -b, y = a$ gives $ab^2 -2ab^2+da^2 > 0 \to ad - b^2 > 0$ </p>
<p>now, for other directions. suppose $a > 0... |
2,342,124 | <p><a href="https://i.stack.imgur.com/QdbFG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QdbFG.png" alt="enter image description here"></a></p>
<p>Well this seems like <span class="math-container">$1-|t|$</span> for <span class="math-container">$|t|<1$</span> and <span class="math-container">... | Michael Hardy | 11,667 | <p>If you have $dx\,dy\,dz = \rho^2 \sin\varphi \, d\rho\, d\theta\,d\varphi$ Then this is
\begin{align}
& \int_0^\pi \int_0^{2\pi} \int_0^1 \rho (\rho^2\sin\varphi \, d\rho \, d\theta \, d\varphi) \\[10pt]
= {} & \int_0^\pi\left( \int_0^{2\pi} \left( \int_0^1 \rho^3 \sin\varphi \, d\rho \right) d\theta \righ... |
3,832,484 | <p>Title's all there is to say.
I'm very new to linear algebra and haven't wrapped my head around determinant rules yet.
Any help would be appreciated.</p>
| fleablood | 280,126 | <p>You should wrestle this to the ground to get a feel for it.</p>
<p>If <span class="math-container">$A = (a_{ij})$</span> then <span class="math-container">$A + A = (a_{ij}+a_{ij}) = (2a_{ij}) = (b_{ij})$</span> where <span class="math-container">$b_{ij} = 2a_{ij}$</span></p>
<p>Now the determinate of an <span class=... |
4,017,964 | <p>Write down in roster notation a set of cardinality 3, of which all elements are sets, and which
satisfies the following property:
∀, ∈ ( ⊆ ⇒ ⊆ ).
No justification is needed. Do not use ellipses (“…”) in your answer.</p>
<p>What I understands from this question is that all the 3 sets have to be the same to fulfi... | Ross Millikan | 1,827 | <p><span class="math-container">$U$</span> is a set of sets. The inner sets have to have three members each but nothing is specified about the size of <span class="math-container">$U$</span>. Your example has two problems. First, the elements of a set have to be distinct. Second, you need a set of outer braces for ... |
3,186,627 | <p>Proposition: Let A be a subset of R which is bounded below. Let B be a subset of R which is bounded above. If <span class="math-container">$\inf(A) < \sup (B) $</span> then there is some <span class="math-container">$a \in A$</span> and <span class="math-container">$b \in B$</span> such that <span class="math-con... | Clayton | 43,239 | <p>Your proof works fine; note that what you've done in case <span class="math-container">$2$</span> is actually sufficient to prove the full statement, though. As you say, assume that there do not exist elements <span class="math-container">$a\in A$</span> and <span class="math-container">$b\in B$</span> such that <sp... |
79,542 | <p>The <em>polarization identity</em> expresses a symmetric bilinear form on a vector space in terms of its associated quadratic form:
$$
\langle v,w\rangle = \frac{1}{2}(Q(v+w) - Q(v) - Q(w)),
$$
where $Q(v) = \langle v,v\rangle$. More generally (over fields of characteristic $0$), for any homogeneous polynomial
... | Scot Adams | 812,550 | <p>If you are looking at a single tangent space V to a Riemannian 2-manifold, then there is a positive definite quadratic form Q on V, and you can use that quadratic form to define a function r that represents lengths of the tangent vectors in V. Specifically, for any point v in V, r(v) will be the square root of Q(v).... |
2,138,916 | <p>My question read: </p>
<p>Show that $S_{10}$ contains elements of orders $10,20,30$. Does it contain an element of order $40$? </p>
<p>I am not too sure what the question is asking. Would I have to explicitly write out all the permutations in $S_{10}$ first and then find the orders for all of them? </p>
<p>Update... | Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\exp... |
3,438,048 | <p>I've recently obtained my University entrance papers from 1967 (yes,52 years ago!) and I found the question below difficult. I presume the answer is a symmetric expression in the differences between alpha,beta and gamma.Am I missing some obvious trick? Any help would be appreciated.</p>
<p>Simplify and evaluate the... | Mark | 470,733 | <p>I'm not sure you really understood the definition of <span class="math-container">$\tau$</span>. A set <span class="math-container">$U$</span> is in <span class="math-container">$\tau$</span> if for each <span class="math-container">$x\in U$</span> there is some <span class="math-container">$\epsilon>0$</span> (w... |
88,363 | <p>It is easy to truncate Series upto some order, say $n$. My question is how do I remove low orders? Let us say my series is a power series in $x$. I want to remove the terms with negative powers because they diverge at $x = 0$. I can simply write</p>
<p>s1-s2, where</p>
<p>s1=Normal[Series[blah, {x, 0, n}]</p>
<p>... | Marius Ladegård Meyer | 22,099 | <p>I'm not sure this approach is applicable to all series, but from a quick test it seems to work for rational exponents:</p>
<p>Looking at the <code>FullForm</code> of</p>
<pre><code>ser = Series[Exp[x]/x^(2/3), {x, 0, 5}]
(* x^(-2/3) + x^(1/3) + x^(4/3)/2 + x^(7/3)/6 + x^(10/3)/24 + x^(13/3)/120 + O[x]^(16/3) *)
</... |
545,003 | <p>I have a proof that I am trying to prove and I am getting stuck at the inductive hypothesis. This is my theorem:</p>
<blockquote>
<p>For all real numbers $n>3$, the following is true: $n + 3 < n!$.</p>
</blockquote>
<p>I have proven true for $n = 4$, and will assume true for some arbitrary value $k$, i.e.,... | JessicaK | 102,435 | <p>Since by induction hypothesis,</p>
<p>$$k+3< k!$$</p>
<p>for $k>4$, multiply both sides by $(k+1)$ to get</p>
<p>$$(k+3)(k+1) < k! (k+1)$$</p>
<p>or</p>
<p>$$(k+3)(k+1) < (k+1)!$$</p>
<p>I'll leave the rest for you to think about, as a hint, remember that's an inequality.</p>
|
545,003 | <p>I have a proof that I am trying to prove and I am getting stuck at the inductive hypothesis. This is my theorem:</p>
<blockquote>
<p>For all real numbers $n>3$, the following is true: $n + 3 < n!$.</p>
</blockquote>
<p>I have proven true for $n = 4$, and will assume true for some arbitrary value $k$, i.e.,... | sundaycat | 102,804 | <p>Suppose $k! \gt k+3$ is true:</p>
<blockquote>
<p>\begin{align*}
\ (k+1)! &=k!\cdot(k+1)
\\ &\gt(k+3)(k+1)
\\ &=k^2+4k+3
\\ &\gt k^2+k+3
\\ &\gt (k+1)+3\ldots(\text{where}\space k\gt 3)
\end{align*}</p>
</blockquote>
|
2,030,739 | <p>Find <span class="math-container">$\frac{d^2y}{dx^2}$</span> of:</p>
<blockquote>
<p><span class="math-container">$$4y^2+2=3x^2$$</span></p>
</blockquote>
<h2>My Attempt</h2>
<p>I attempted the probelm my first solving for the first derivative:</p>
<blockquote>
<p><span class="math-container">$8y*y'=6x$</span><br>
<... | Shraddheya Shendre | 384,307 | <p>Your final answer is wrong and since you only ask for the correct final answer, here you go :
$$\frac{d^2y}{dx^2} = \frac{12y^2-9x^2}{16y^3}$$</p>
|
2,068,906 | <p>Recall, with the birthday problem, with 23 people, the odds of a shared birthday is APPROXIMATELY .5 (correct?)</p>
<p>P(no sharing of dates with 23 people) = $$\frac{365}{365}*\frac{364}{365}*\frac{363}{365}*...*\frac{343}{365} $$</p>
<p>$$= \frac{365!}{342!}*\frac{1}{365^{23}} $$</p>
<p>I want to do this multip... | Beni Bogosel | 7,327 | <p>You can use Pari Gp in order to do this. You need multiple precision arithmetic due to the large powers and factorials. Pari GP is usually the right way to go if you need to do this kind of computations. Just open the program, type <code>1.0* 365!/342!/365^23</code> and you'll get the result
$$ 0.4927027656760145927... |
2,068,906 | <p>Recall, with the birthday problem, with 23 people, the odds of a shared birthday is APPROXIMATELY .5 (correct?)</p>
<p>P(no sharing of dates with 23 people) = $$\frac{365}{365}*\frac{364}{365}*\frac{363}{365}*...*\frac{343}{365} $$</p>
<p>$$= \frac{365!}{342!}*\frac{1}{365^{23}} $$</p>
<p>I want to do this multip... | heropup | 118,193 | <p>You can certainly do this in Excel, and here's how you would do it:</p>
<p>$$\begin{array}{|c|c|c|c|} \hline & \text{A} & \text{B} & \text{C} \\ \hline
1 & 365 & \text{=A1} & \text{=B1/A1} \\
2 & \text{=A1} & \text{=B1-1} & \text{=B2/A2} \\
3 & \text{=A2} & \text{=B2-1} &... |
2,122,389 | <p>The problem goes so : you have a parking lot with 8 parking spaces and 8 cars, of which 4 are red and 4 are white. What is the probability of :</p>
<p>a) 4 white cars being parked next to each other ?</p>
<p>b) 4 white cars and 4 red cars being parked next to each other ?</p>
<p>c) red and white cars being parked... | Brevan Ellefsen | 269,764 | <p><strong>Mathematica Output</strong> </p>
<hr>
<p>Mathematica instantly produces the following:
$$\frac{1}{2} x \left(2 \log \left(m^2+2 m \cos (x)+1\right)-2 \log \left(\frac{m+e^{i
x}}{m}\right)-2 \log \left(1+m e^{i x}\right)+i x\right)+i
\operatorname{Li}_2\left(\frac{-e^{i x}}{m}\right)+i \operatorname{... |
23,566 | <p>I love math, and I used to be very good at it. The correct answers came fast and intuitively. I never studied, and redid the demonstration live for the tests (sometimes inventing new ones). I was the one who answered the tricky questions in class (8 hours of math/week in high school)... You get the idea.</p>
<p>As ... | Derek Jennings | 1,301 | <p>I think the key to your problem is in your first paragraph. You say, "The correct answers came fast and intuitively. I never studied." This is the classic high school con that can lead one to doubt one's own abilities as soon as the going gets more challenging.</p>
<p>No matter what your abilities, to do worthwhile... |
23,566 | <p>I love math, and I used to be very good at it. The correct answers came fast and intuitively. I never studied, and redid the demonstration live for the tests (sometimes inventing new ones). I was the one who answered the tricky questions in class (8 hours of math/week in high school)... You get the idea.</p>
<p>As ... | Community | -1 | <p>there are some great answers / advise here already ... what i would like to add is that it demonstrates, in my opinion the clear link between communication, speech, language patterns and numerical skills, that part of the brain that visualizes these patterns and helps you make sense of the answer ... </p>
|
23,566 | <p>I love math, and I used to be very good at it. The correct answers came fast and intuitively. I never studied, and redid the demonstration live for the tests (sometimes inventing new ones). I was the one who answered the tricky questions in class (8 hours of math/week in high school)... You get the idea.</p>
<p>As ... | Community | -1 | <p>I'm in my late 30s and still remember being crap at maths while at school (this might sound like cold comfort, but run with this a little) because it's like I said in a Yahoo Q&A some time ago now. a little each day will keep your brain sharp and active and you'll still be able to do a little of the advanced sta... |
1,988,191 | <p>Today I coded the multiplication of quaternions and vectors in Java. This is less of a coding question and more of a math question though:</p>
<pre><code>Quaternion a = Quaternion.create(0, 1, 0, Spatium.radians(90));
Vector p = Vector.fromXYZ(1, 0, 0);
System.out.println(a + " * " + p + " = " + Quaternion.product(... | Emilio Novati | 187,568 | <p>I dont well understand your code, but it seems that you have multiplied only one way the quaternion by the vector, and this is wrong.</p>
<p>The rotation of the vector $\vec v = \hat i$ by $\theta=\pi/2$ around the axis $\mathbf{u}=\hat j$ is represented by means of quaternions as ( see <a href="https://math.stacke... |
1,012,236 | <p>A continuous time process it's nule for t < 0. In which conditions is it stationary (WSS)?</p>
<p>I know that E[x(t)] must be a constant and the autocorrelation function must depend only on the time difference t2-t1. Are there any other conditions?</p>
| user2345215 | 131,872 | <p>Both sequences have $e$ as the limit, so it suffices to show that the left sequence is increasing and the right sequence is decreasing.</p>
<p>Use the AM-GM inequality to get
$$\frac{n+2}{n+1}=\frac{\frac{n+1}n+\ldots+\frac{n+1}n+1}{n+1}>\sqrt[n+1\,]{\left(\frac{n+1}n\right)^n}$$
It follows that
$$\left(1+\frac1... |
99,572 | <p>One of the most useful tools in the study of convex polytopes is to move from polytopes (through their fans) to toric varieties and see how properties of the associated toric variety reflects back on the combinatorics of the polytopes. This construction requires that the polytope is rational which is a real restrict... | Fiammetta Battaglia | 32,295 | <p>Dear Gil, in addition to Dan's answer let me mention that the construction of toric varieties à la Cox has been generalized to arbitrary convex polytopes in <a href="http://www.worldscientific.com/doi/abs/10.1142/S0129167X11007562">Geometric spaces from arbitrary convex polytope, Int. J. Math., 23, (2012)</a> (the s... |
3,362,654 | <p>Let's say <span class="math-container">$C$</span> is a category, and <span class="math-container">$\mathscr{C}$</span> is a collection of morphisms in <span class="math-container">$C$</span>. I have come across the following sentence </p>
<p>"<span class="math-container">$C$</span> admits pullbacks along morphisms ... | Henry | 6,460 | <p>You asked about how to find this in R</p>
<p>You could do something like:</p>
<pre><code>twelve <- 12
probwin <- numeric(twelve)
for (n in 1:twelve){
if(n == 1){
probwin[n] <- 1 / twelve
}else{
probwin[n] <- (1 - sum(probwin[1:(n-1)])) * n / twelve
}
}
probwin
</code></pre>
<p>which wou... |
2,798,207 | <p>This problem needs also to be extended to $n*m$ chessboard. I tried to think like this:</p>
<p>First I choose a place for the first king in $64$ ways. Then I have a choice $64-5 = 59$ squares for the second king . But this solution is not right because this is not the case if I place the first king in the sidemost... | Donald Splutterwit | 404,247 | <p>The first King could be on a corner square($4$ ways), leaving $60$ other squares for the next King.</p>
<p>The first King could be on a edge square($24$ ways), leaving $58$ other squares for the next King.</p>
<p>The first King could be on a central square($36$ ways), leaving $55$ other squares for the next King.<... |
1,714,654 | <p>Show that a box (rectangular parallelopiped) of maximum volume V with prescribed surface area is a cube.
Let $$V=xyz$$
$$S=2xy + 2yz + 2zx$$
$S$ is constant.</p>
<p>Using Lagrange method, I am stuck at $V_x$$_x$=$0$=$V_y$$_y$=$V_z$$_z$ at the (only) critical point. How to approach this. </p>
| mrprottolo | 84,266 | <p>You can use polar coordinates here. Set $x=r\cos\theta$, $y=r\sin\theta$, then notice that $x^2-y^2=r^2\cos 2\theta$. Then the limit becomes
$$\lim_{r \to 0} \frac{\sin (r^2\cos 2\theta)}{r^2\cos 2 \theta}.$$</p>
<p>Clearly you have to exclude the case $\theta=\pm \pi/4$ because $f$ is not defined there, even if yo... |
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field prove... | YorkT | 135,410 | <p>some important conjectures in matroid theory, for instance the Rota conjecture on excluded minors (by Geelen, Gerards and Whittle, still unpublished, note claiming proof <a href="http://www.ams.org/notices/201407/rnoti-p736.pdf" rel="noreferrer">here</a>) and the log-concavity conjecture (also due to Rota) for the c... |
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field prove... | Sean Lawton | 12,218 | <p>A Margulis spacetime is the quotient of the Minkowski space by a free proper orientation-preserving isometric action of a free group of rank at least two.</p>
<p>From <a href="https://arxiv.org/pdf/1306.2240.pdf" rel="noreferrer"> Danciger, Kassel, and Guéritaud</a>:</p>
<blockquote>
<p>"Based on a question of M... |
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field prove... | Neal | 20,796 | <p>A "hot spot" on a sufficiently regular domain is an interior extremum of the first nonconstant Neumann eigenfunction of the Laplace operator. The Hot Spots conjecture states that hot spots do not exist on convex planar domains.</p>
<p>Chris Judge and Sugata Mondal have settled the Hot Spots conjecture in the affirm... |
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field prove... | David White | 11,540 | <p>The <a href="https://en.wikipedia.org/wiki/Kervaire_invariant" rel="nofollow noreferrer">Kervaire Invariant One Problem</a> (1969) is a question about which framed manifolds can be converted into spheres via surgery. It's related to the classification of exotic smooth structures on spheres (like Milnor's Fields Meda... |
316,770 | <p>It is known that for a subring $R$ of some (commutative) ring $S$, the nilradical of $R$ $$\text{nil }R=R\cap\text{nil }S.$$ Moreover for Jacobson rings $R\subset S$, this means that the Jacobson radical of $R$ can also be written in this way, i.e., $J(R)=R\cap J(S)$.</p>
<p><strong>Edit.</strong> Are there separat... | Community | -1 | <p>Take $R$ an integral domain, $\mathfrak m$ a maximal ideal, and $S=R_{\mathfrak m}$. Then $J(S)\cap R=\mathfrak m$ and obviously $J(R)$ does not necessarily contain $J(S)\cap R$. </p>
|
316,770 | <p>It is known that for a subring $R$ of some (commutative) ring $S$, the nilradical of $R$ $$\text{nil }R=R\cap\text{nil }S.$$ Moreover for Jacobson rings $R\subset S$, this means that the Jacobson radical of $R$ can also be written in this way, i.e., $J(R)=R\cap J(S)$.</p>
<p><strong>Edit.</strong> Are there separat... | zacarias | 35,464 | <p>Let $\mathbb Z_4=\left\{0, 1, 2, 3\right\}$ be the residual classes modulo $z$. Considere the matrix ring $S=M_2(\mathbb Z_4)$. Since for a ring $A$ the nilradical of the matrix ring $M_n(A)$ is $M_n(\mathcal N(A))$ where $\mathcal N(A)$ is the nilradical of $A$, it follows that $\mathcal N(M_n(\mathbb Z_4))=M_n(I)$... |
122,848 | <p>Is my calculation correct for this rotation around a point?</p>
<p>A point a(-19.94,392.11) is rotated -49.45 degrees, what is the new coordinates of point a?</p>
<p>My solution:</p>
<pre><code>x' = x*cos(0) - y*sin(0)
y' = x*sin(0) + y*cos(0)
x' = (-12.961) - (-298.0036)
y' = (15.15) + (254.92)
x' = 285.04
y' ... | Raymond Manzoni | 21,783 | <p>I don't know why this old point emerged... Anyway let's try this using complex numbers :
$$(-19.94+392.11 i)\cdot e^{2\pi i\dfrac{-49.45}{360}}\approx 284.98 + 270.07i$$</p>
<p>So that the OP's answer looked not so bad!</p>
|
25,100 | <p>Suppose one has a set $S$ of positive real numbers, such that the usual numerical ordering on $S$ is a well-ordering. Is it possible for $S$ to have any countable ordinal as its order type, or are the order types that can be formed in this way more restricted than that?</p>
| gowers | 1,459 | <p>You can get any order type. Let's assume you can get all order types up to but not including alpha, using subsets of (0,1]. If alpha=beta + 1 then squash your representation of beta and add an extra point. If alpha is a limit ordinal, choose a sequence of ordinals that converges to alpha and put the first one into (... |
25,100 | <p>Suppose one has a set $S$ of positive real numbers, such that the usual numerical ordering on $S$ is a well-ordering. Is it possible for $S$ to have any countable ordinal as its order type, or are the order types that can be formed in this way more restricted than that?</p>
| Andrew Marks | 6,151 | <p>Using wellorderings of positive reals is actually the standard way to construct an Aronszajn tree.</p>
|
3,536,822 | <p>A man has three bags filled with balls. One bag contains balls weighing <span class="math-container">$9$</span> grams, the second bag contains balls weighing <span class="math-container">$10$</span> grams and the third bag contains balls weighing <span class="math-container">$11$</span> grams. The man got confused a... | Bram28 | 256,001 | <blockquote>
<p>Since <span class="math-container">$(A\vdash B)$</span> and <span class="math-container">$(A \mkern-2mu\not\mkern2mu\Rightarrow B)$</span>, therefore, <span class="math-container">$\text{If}\ A \vdash B\ \text{then}\ A \mkern-2mu\not\mkern2mu\Rightarrow B$</span> can be concluded.</p>
</blockquote>
<... |
3,479,883 | <p>I know that (I might be wrong):</p>
<ul>
<li>Symbol for empty or null set : {Ø} or {}</li>
<li>Null or empty set is 'subset of all sets' as well as 'empty or null set' set</li>
<li>So, { {} } is same as { Ø }</li>
</ul>
<p>I just want to know { {} } or { Ø } is an empty set or not ? And if yes then we can conclud... | StackTD | 159,845 | <blockquote>
<p>I know that (I might be wrong):</p>
<ul>
<li>Symbol for empty or null set : {Ø} or {}</li>
</ul>
</blockquote>
<p>You write the empty set as "Ø" or "{}" so your first notation, "{Ø}" is already <em>a set containing the empty set</em>!</p>
<p>So don't mix:</p>
<ul>
<li>the ... |
1,015,826 | <p>For $r>1$, prove the sequence $$X_n=\left(1+r^n\right)^{1/n}$$ is decreasing. I understand the limit is decreasing and that the limit of this sequence is $r$. I am just not sure on the algebra. My thought is to show $X_n>X_{n+1}$ by showing $X_n-X_{n+1}>0$ for all $n$. I could also use induction; however, I... | orangeskid | 168,051 | <p>Here is how you show that if $x_1$, $\ldots$, $x_k >0$ and $k \ge 2$ then the function
$$(0 , \infty) \ni s \mapsto (x_1^s + \cdots +x_k^s)^{\frac{1}{s}}$$ is strictly decreasing. </p>
<p>Let $0< s< t$. Want to show </p>
<p>$$ (x_1^{s} + \cdots +x_k^s)^{\frac{1}{s}}> (x_1^{t} + \cdots +x_k^t)^{\frac{1... |
4,593,212 | <p>Question: A coin is tossed where the probability it lands on heads is <span class="math-container">$1/3$</span>. What is the expected number of heads before tails?</p>
<p>My answer: number of heads before tails = <span class="math-container">$\frac{1}{3}^1+\frac{1}{3}^2+\frac{1}{3}^3+...$</span> <span class="math-co... | Daniel S. | 362,911 | <p>Almost there, it should be <span class="math-container">$3/2-1=0.5$</span>. There is a chance you will get no heads before tails! So, you subtract 1 from your solution. There are two versions for the geometric distribution: <a href="https://en.wikipedia.org/wiki/Geometric_distribution" rel="nofollow noreferrer">http... |
4,593,212 | <p>Question: A coin is tossed where the probability it lands on heads is <span class="math-container">$1/3$</span>. What is the expected number of heads before tails?</p>
<p>My answer: number of heads before tails = <span class="math-container">$\frac{1}{3}^1+\frac{1}{3}^2+\frac{1}{3}^3+...$</span> <span class="math-co... | RyRy the Fly Guy | 412,727 | <p>Let <span class="math-container">$X$</span> be a geometric random variable such that <span class="math-container">$X=k$</span> if and only if the first instance of tails is the <span class="math-container">$k$</span>th flip.</p>
<p>With probability <span class="math-container">$p = \frac{2}{3}$</span> of getting tai... |
3,768,198 | <p>Show that <span class="math-container">$\|uv^T-wz^T\|_F^2\le \|u-w\|_2^2+\|v-z\|_2^2$</span>, assuming <span class="math-container">$u,v,w,z$</span> are all unit vectors.</p>
| Daniel Li | 294,291 | <p>Let <span class="math-container">$a=u^Tw,b=v^Tz.$</span></p>
<p><span class="math-container">$\|uv^T-wz^T\|_F^2=tr((uv^T-wz^T)^T(uv^T-wz^T))=2-2ab$</span></p>
<p>And RHS=<span class="math-container">$4-2(a+b).$</span></p>
<p>Check that <span class="math-container">$2-2ab\le4-2(a+b) \iff a+b-ab\le1,$</span> using tha... |
3,768,198 | <p>Show that <span class="math-container">$\|uv^T-wz^T\|_F^2\le \|u-w\|_2^2+\|v-z\|_2^2$</span>, assuming <span class="math-container">$u,v,w,z$</span> are all unit vectors.</p>
| user1551 | 1,551 | <p>Let <span class="math-container">$A=(u-w)v^T$</span> and <span class="math-container">$B=w(v-z)^T$</span>. The inequality in question is then equivalent to
<span class="math-container">$$
\|A+B\|_F^2\le\|A\|_F^2+\|B\|_F^2.
$$</span>
It is true if and only if <span class="math-container">$\langle A,B\rangle_F\le0$</s... |
2,094,123 | <p>A plane curve is printed on a piece of paper with the directions of both axes specified. How can I (roughly) verify if the curve is of the form $y=a e^{bx}+c$ without fitting or doing any quantitative calculation?</p>
<p>For example, for linear curves, I can choose two points on the curve and check if the midpoint ... | Anonymous | 399,787 | <p><strong>Are the doubling points evenly spaced?</strong></p>
<p>Assume $a$ and $b$ are positive (if not, it's easy to see and readjust - $a$ is positive if the curve flattens to the left, $b$ has the same sign as $a$ if the curve is increasing). Mentally reset the $x$ axis at the height to which the curve tends to b... |
1,802,515 | <blockquote>
<p>Say you have a bank account in which your invested money yields 3% every year, continuously compounded. Also, you have estimated that you spend $1000 every month to pay your bills, that are withdrawn from this account.</p>
<p>Create a differential model for that, find its equilibriums and determine its ... | mvw | 86,776 | <p>The money flow consists of two contributions
$$
\dot{S} = \dot{S}_y + \dot{S}_m
$$
with the continous contribution
$$
\dot{S}_y = a S \quad (*)
$$
where $a$ must be adjusted to give the yearly interest rate such that
$$
S_y(1\text{y}) = (1 + p) S_y(0\text{y})
$$
for $p = 3\% = 3/100$ and the monthly part
$$
\dot{S... |
433,403 | <ol>
<li>Let F(x,y) be the statement, “x can fool y,” where the domain consists of all of the people in the world. Translate this statement into symbolic logic.
a. Everyone can be fooled by somebody.</li>
</ol>
<p>Would it be: For every x.y in W, F(x,y) is in W?</p>
<p>I am not getting the gist of this...</p>
| Amr | 29,267 | <p>$$\forall x \exists y F(y,x)$$</p>
|
229,161 | <p>A sequence of positive integer is defined as follows</p>
<blockquote>
<ul>
<li>The first term is $1$.</li>
<li>The next two terms are the next two even numbers $2$, $4$.</li>
<li>The next three terms are the next three odd numbers $5$, $7$, $9$.</li>
<li>The next $n$ terms are the next $n$ even numbers if... | Jean-Sébastien | 31,493 | <p>Not a general term, but an interesting reformulation of your recursion. Given a(1)=1</p>
<p>$$
a(n):=\begin{cases}
a(n-1)+1& \text{if} \,\, a(n-1) \text{ is a square}\\
a(n-1)+2 &\text{otherwise}
\end{cases}
$$</p>
|
138,243 | <p><a href="https://i.stack.imgur.com/kRmeb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kRmeb.png" alt="Area and Perimeter"></a></p>
<p>How can I draw the figure shown above in rectangular coordinates, calculate the area and perimeter of the shaded region as a function of radius <code>r</code> o... | m_goldberg | 3,066 | <p>This isn't really a Mathematica problem. It is a Euclidean geometry problem and can be solve by a little classic geometry reasoning. Like ubpdqn I will work in the 1st quadrant and invoke symmetry. </p>
<p><a href="https://i.stack.imgur.com/0fAM8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0f... |
138,243 | <p><a href="https://i.stack.imgur.com/kRmeb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kRmeb.png" alt="Area and Perimeter"></a></p>
<p>How can I draw the figure shown above in rectangular coordinates, calculate the area and perimeter of the shaded region as a function of radius <code>r</code> o... | Carl Woll | 45,431 | <p>I would use <a href="http://reference.wolfram.com/language/ref/BooleanRegion.html" rel="noreferrer"><code>BooleanRegion</code></a>:</p>
<pre><code>reg = BooleanRegion[
Xor,
{Disk[{-1,0},1], Disk[{0,1},1], Disk[{1,0},1], Disk[{0,-1},1], Disk[{0,0},2]}
];
RegionMeasure @ reg
RegionMeasure @ RegionBoundary @... |
842,271 | <p>Evaluation of $\displaystyle \int \frac{\sqrt[3]{x+\sqrt[4]{x}}}{\sqrt{x}}dx$</p>
<p>$\bf{My\; Try::}$ Let $x=t^4\;,$ Then $dx = 4t^3dt$</p>
<p>So Integral is $\displaystyle \int\frac{\sqrt[3]{t^4+t}}{t^2} \cdot 4t^3dt$</p>
<p>So Integral is $\displaystyle 4\int t^{\frac{7}{3}}\cdot (1+t^{-3})^{\frac{1}{3}}$</p>
... | Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\... |
207,865 | <p>It is known that all $B$, $C$ and $D$ are $3 \times 3$ matrices. And the eigenvalues of $B$ are $1, 2, 3$; $C$ are $4, 5, 6$; and $D$ are $7, 8, 9$. What are the eigenvalues of the $6 \times 6$ matrix
$$\begin{pmatrix}
B & C\\0 & D
\end{pmatrix}$$
where $0$ is the $3 \times 3$ matrix whose entries are all $... | hadi | 743,402 | <p><span class="math-container">$BE=-CD^{-1}$</span>
By your assumptions <span class="math-container">$B$</span> is invertible, hence we have:
<span class="math-container">$-B^{-1}CD^{-1}$</span></p>
|
2,293,600 | <p>How to calculate X $\cap$ $\{X\}$ for finite sets to develop an intuition for intersections?</p>
<p>If $X$ = $\{$1,2,3$\}$, then what is $X$ $\cap$ $\{X\}$? </p>
| Affineline | 448,123 | <p>$\phi$ ... The set $\{1,2,3\} \notin \{1,2,3\}$.</p>
<p>a collection which contains itself is NOT a set...</p>
|
1,081,417 | <p>This is exercise number $57$ in Hugh Gordon's <em>Discrete Probability</em>. </p>
<hr>
<p>For $n \in \mathbb{N}$, show that</p>
<p>$$\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$$</p>
<hr>
<p>My algebraic solution:</p>
<p>$$\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$$</p>
<p>$$\binom{\frac{n(n-1)}{2}}{2}=\frac{3n(n+1... | Marc van Leeuwen | 18,880 | <p>For the right hand side, add a special element $s$ to your $n$-element set; then choose $4$ elements from the extended set, and a partition of those $4$ into $2$ sets of size $2$ (the latter is possible in $3$ ways). If $s$ was not among the selected elements retain the two disjoint pairs; otherwise let the pairs be... |
2,752,511 | <p>Prove that if $X$ is Hausdorff, $\Delta=\{(x, x)\mid x\in X\}$ is closed in $X\times X$ (with the product topology).</p>
<p><strong>My attempt:</strong></p>
<p>Let $x_1, x_2\in X$ s.t. $x_1\ne x_2$.</p>
<p>There exist neighborhoods $U_1$ and $U_2$ of $x_1$ and $x_2$ that are disjoint.</p>
<p>$U_1\times U_2$ is a... | Mike Earnest | 177,399 | <p>Your work shows that
$$
\Delta^c=\bigcup_{\substack{(x_1,x_2)\in X\times X\\x_1\neq x_2}} U_1(x_1)\times U_2(x_2),
$$
where $U_1(x_1)$ and $U_2(x_2)$ are separating sets for $x_1,x_2$. This shows the complement of $\Delta$ is a union of open sets, so the complement of $\Delta$ is open, so $\Delta$ is closed.</p>
|
1,114,258 | <p>I am new to differential geometry and Riemannian geometry. </p>
<p>I have on two separate occasions (separated by 6 months) encountered exercises where I feel like I am not giving a complete answer. </p>
<p>Problem 1: </p>
<p><em>Show that the Gaussian curvature of the surface of a cylinder is zero.</em></p>
<p>... | Steven Stadnicki | 785 | <p>Unfortunately, there is no more elementary argument than going through some form of AC, because the result actually does depend on some amount of choice. As shown by e.g. C.J. Ash (<a href="http://journals.cambridge.org/download.php?file=%2FJAZ%2FJAZ1_19_03%2FS1446788700031505a.pdf&code=680c42651efdb40c1e3a1a9f... |
1,029,485 | <p>I wish to show the following statement:</p>
<p>$
\forall x,y \in \mathbb{R}
$</p>
<p>$$
(x+y)^4 \leq 8(x^4 + y^4)
$$</p>
<p>What is the scope for generalisaion?</p>
<p><strong>Edit:</strong></p>
<p>Apparently the above inequality can be shown using the Cauchy-Schwarz inequality. Could someone please elaborate,... | DeepSea | 101,504 | <p>Apply Cauchy-Schwarz inequality twice: $x^4 + y^4 \geq \dfrac{1}{2}\left(x^2+y^2\right)^2 \geq \dfrac{1}{2}\left(\dfrac{1}{2}\left(x+y\right)^2\right)^2 = \dfrac{1}{8}\left(x+y\right)^4$.</p>
|
1,029,485 | <p>I wish to show the following statement:</p>
<p>$
\forall x,y \in \mathbb{R}
$</p>
<p>$$
(x+y)^4 \leq 8(x^4 + y^4)
$$</p>
<p>What is the scope for generalisaion?</p>
<p><strong>Edit:</strong></p>
<p>Apparently the above inequality can be shown using the Cauchy-Schwarz inequality. Could someone please elaborate,... | Milly | 182,459 | <p>A more general result is ($x,y\geq 0$, $p\geq 1$)
$$(x+y)^p \leq 2^{p-1} (x^p+y^p),$$
which is direct consequence of convexity of $t\mapsto t^p$.</p>
|
187,545 | <p><span class="math-container">$\DeclareMathOperator\GL{GL}\DeclareMathOperator\L{\mathfrak{L}}$</span>The free Lie algebra <span class="math-container">$\L(V)$</span> generated by an <span class="math-container">$r$</span>-dimensional vector space <span class="math-container">$V$</span> is, in the
language of <a href... | F. C. | 10,881 | <p>Here is the result, computed using sage:</p>
<pre><code>sage: def lie(n):
....: p = SymmetricFunctions(QQ).p()
....: return p.sum_of_terms((Partition([d for j in range(ZZ(n / d))]),
....: moebius(d) / n) for d in divisors(n))
sage: s = SymmetricFunctions(QQ).schur()
sage: [s(lie(i... |
1,241,970 | <p>A fair coin is tossed three times. Let $X$ be the number of heads that turn up on the first two tosses and $Y$ the number of heads that turn up on the third toss. Give the distribution of $X$, $Y$, $X + Y$, $X − Y$ and $XY$.</p>
| Brian Tung | 224,454 | <p>The distribution of $X+Y$ is a binomial distribution, as expected:</p>
<p>$$
q_0 = q_3 = 1/8 \\
q_1 = q_2 = 3/8
$$</p>
<p>where $q_i = P(X+Y = i)$. This can be reasoned out as follows: You can obtain $0$ only if $X = Y = 0$, so the probability is $(1/4)(1/2) = 1/8$. You can obtain $3$ only if $X = 2, Y = 1$, so a... |
340,886 | <p>Suppose $x=(x_1,x_2),y = (y_1,y_2) \in \mathbb{R}^2$. I noticed that
\begin{align*}
\|x\|^2 \|y\|^2 - \langle x,y \rangle^2 &=
x_1^2y_1^2 + x_1^2 y_2^2 + x_2^2 y_1^2 + x_2^2 y_2 ^2 - (x_1^2 y_1^2 + 2 x_1 y_1 x_2 y_2 + x_2^2 y_2^2) \\
&=(x_1 y_2)^2 - 2x_1 y_2 x_2 y_1 + (x_2 y_2)^2 \\
&=(x_1 y_2 - x_2 y_1)... | Community | -1 | <p>$\sum x_i^2\sum y_i^2-(\sum x_iy_i)^2$</p>
<p>$=\sum x_i^2y_i^2+\sum_{i\neq j} x_i^2y_j^2-\sum x_i^2y_i^2-\sum_{i\neq j} x_iy_ix_jy_j$</p>
<p>$=\sum_{i<j} (x_iy_j-x_jy_i)^2$</p>
|
2,185,118 | <p>Find the sum of the series. For what values of the variable does the series converge to this sum?</p>
<p>$$1+\frac{x} {2}+\frac{x^2} {4}+\frac{x^3} {8}...$$</p>
<p>Summation notation: $\sum_{n=0}^\infty \frac{x^n} {2^n}$</p>
<p>I know you use the formula $\frac{a} {1-r}$ to find the sum of geometric series but I'... | joeb | 362,915 | <p>$\sum_{n=0}^\infty \frac{x^n}{2^n} = \sum_{n=0}^\infty y^n$, where $y = x/2$. The series converges when $|x|/2 = |y| < 1$.</p>
|
2,185,118 | <p>Find the sum of the series. For what values of the variable does the series converge to this sum?</p>
<p>$$1+\frac{x} {2}+\frac{x^2} {4}+\frac{x^3} {8}...$$</p>
<p>Summation notation: $\sum_{n=0}^\infty \frac{x^n} {2^n}$</p>
<p>I know you use the formula $\frac{a} {1-r}$ to find the sum of geometric series but I'... | Dr. Sonnhard Graubner | 175,066 | <p>Hint: prove by induction that $$\sum_{n=0}^m\frac{x^n}{2^n}=-\frac{2^{-m} \left(2^{m+1}-x^{m+1}\right)}{x-2}$$</p>
|
231,887 | <p>I'm learning to do proofs, and I'm a bit stuck on this one.
The question asks to prove for any positive integer $k \ne 0$, $\gcd(k, k+1) = 1$.</p>
<p>First I tried: $\gcd(k,k+1) = 1 = kx + (k+1)y$ : But I couldn't get anywhere.</p>
<p>Then I tried assuming that $\gcd(k,k+1) \ne 1$ , therefore $k$ and $k+1$ are no... | Bob Dobbs | 221,315 | <p>For <span class="math-container">$k>0$</span>, consider the polynomial <span class="math-container">$p(x)=x^{k+1}-x^k=x^k(x-1)$</span> whose only roots are <span class="math-container">$0$</span> and <span class="math-container">$1$</span>. Let <span class="math-container">$d=\gcd(k,k+1)\geq 1$</span>. Then <span... |
40,474 | <p>there is a binomial formula:</p>
<p>$$(x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$$</p>
<p>When operations are done in $GF(2^m)$ then all positive integers are reduced $\bmod2$, so binomial formula for $n=2^i$ in $GF(2^m)$ is:</p>
<p>$$(x+y)^{2^i}=x^{2^i} + y^{2^i} $$</p>
<p>So now the question. I... | Luboš Motl | 10,599 | <p>It's because multiplication by a coefficient is periodic in the coefficient with the right period $P$:
$$ (a+P)\cdot b = a\cdot b + P\cdot b \equiv a\cdot b \quad {\rm mod} \quad P$$
because the equivalence modulo $P$ is defined so that it allows me to subtract multiples of $P$ such as $P\cdot b$ above - while power... |
40,474 | <p>there is a binomial formula:</p>
<p>$$(x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$$</p>
<p>When operations are done in $GF(2^m)$ then all positive integers are reduced $\bmod2$, so binomial formula for $n=2^i$ in $GF(2^m)$ is:</p>
<p>$$(x+y)^{2^i}=x^{2^i} + y^{2^i} $$</p>
<p>So now the question. I... | Phira | 9,325 | <p>Regard the simplest non-trivial example to get a better understanding.</p>
<p>For example $\{0,1,x,x+1\}$ with $2=0$ and $x^2=x+1$.</p>
|
2,952,014 | <p>So I was doing some self study and came across a proposition in one of my chemical engineering course's prescribed textbooks. I can't quite get the proof out. It's to do with a particle moving through a medium such that when it makes contact with to either of two plates <span class="math-container">$L$</span> units ... | amd | 265,466 | <p>Write your equation as <span class="math-container">$\mathbf\pi(P-I)=0$</span>. The first and last rows of <span class="math-container">$P-I$</span> are zero, so <span class="math-container">$(1,0,\dots,0)$</span> and <span class="math-container">$(0,\dots,0,1)$</span> are obvious independent solutions of the equati... |
1,700 | <p>Is there an algorithm in literature to compute an efficient (pareto optimal) and envy-free cake cutting when there are only $n=2$ players and a mediator?</p>
| TonyK | 767 | <p>Huh? I cut, you choose. Why do we need a Mediator?</p>
|
4,155,718 | <p>I'm given <span class="math-container">$n$</span> points <span class="math-container">$(p_1, p_2, \ldots, p_n)$</span>, lying on the boundary of a polygon and constituting this polygon (not necessarily convex), whereby those points are given me in clockwise order and I want to compute the convex hull of this polygo... | Glärbo | 933,372 | <p>Given an unordered set of points, <a href="https://en.wikipedia.org/wiki/Graham_scan" rel="nofollow noreferrer">Graham scan</a> yields their convex hull in <i>O</i>(<i>N</i> log <i>N</i>) time complexity. The limiting factor is the sort phase; everything else is linear in time, as the Wikipedia article explains.</p... |
2,712,631 | <p>The problem is to use a power series to evaluate the integral to six decimal places. The upper limit of integration is one and the lower limit of integration is zero.</p>
<p>To start the problem I factored $x$ out and focused on $\arctan(3x)$.
I knew that by taking the derivative I could get this equation
in the f... | geometryfan | 546,282 | <p>You method is good. Although, I think at the beginning you lost some exponents that were supposed to be even; an $x^{2n}$ that became $x^n$. As a consequence some of the coefficients and exponents are not right.</p>
<p>I would have probably computed the primitive $$g(x)=\frac{1}{18}((9x^2 + 1)\arctan(3 x) - 3 x)$$<... |
469,947 | <blockquote>
<p>Show that the presentation $G=\langle a,b,c\mid a^2 = b^2 = c^3 = 1, ab = ba, cac^{-1} = b, cbc^{-1} =ab\rangle$ defines a group of order $12$.</p>
</blockquote>
<p>I tried to let $d=ab\Rightarrow G=\langle d,c\mid d^2 =c^3 = 1, c^2d=dcdc\rangle$. But I don't know how to find the order of the new pre... | Community | -1 | <p>Consider $D := \langle a, b\rangle$ and $C = \langle c \rangle$. Note that</p>
<p>$$ab = ba \implies D = \langle a \rangle \langle b \rangle$$</p>
<p>so $|D| \leq 4$. Further, the last two relations tell you that $c \in N_{G}(D)$ (the normalizer in $G$ of $D$), so $C \leq N_{G}(D)$. It follows that $DC \leq G$, bu... |
469,947 | <blockquote>
<p>Show that the presentation $G=\langle a,b,c\mid a^2 = b^2 = c^3 = 1, ab = ba, cac^{-1} = b, cbc^{-1} =ab\rangle$ defines a group of order $12$.</p>
</blockquote>
<p>I tried to let $d=ab\Rightarrow G=\langle d,c\mid d^2 =c^3 = 1, c^2d=dcdc\rangle$. But I don't know how to find the order of the new pre... | Mikasa | 8,581 | <p>Although my approach is not the way you expected, it is a nice way for your group. This way is called <a href="http://en.wikipedia.org/wiki/Coset_enumeration" rel="nofollow noreferrer">Coset enumeration</a> or <a href="http://en.wikipedia.org/wiki/Todd%E2%80%93Coxeter_algorithm" rel="nofollow noreferrer">Todd-Coxete... |
469,947 | <blockquote>
<p>Show that the presentation $G=\langle a,b,c\mid a^2 = b^2 = c^3 = 1, ab = ba, cac^{-1} = b, cbc^{-1} =ab\rangle$ defines a group of order $12$.</p>
</blockquote>
<p>I tried to let $d=ab\Rightarrow G=\langle d,c\mid d^2 =c^3 = 1, c^2d=dcdc\rangle$. But I don't know how to find the order of the new pre... | AG. | 80,733 | <p>We can construct a directed graph called the Cayley diagram of the group $G := \langle a,b,c | a^2=b^2=c^3=1, ab=ba, ca=bc, cb=abc \rangle$ with respect to the generators $a,b,c$. The vertices of this digraph will be the group elements. The set of arcs will be of of the form $\{(g,gs): g \in G, s \in \{a,b,c\} \}$... |
288,974 | <p>Alright this maybe really funny but I want to know why is this wrong. We often come across identities which we prove by multiplying both the sides of the identity by a certain entity but why don't we multiply it by $0$. That way every identity will be proved in one single line. That is so stupid. I mean, by that way... | Ross Millikan | 1,827 | <p>When you go from $ax=b$ to $x=\frac ba$ you have multiplied both sides by $\frac 1a$ and it would be good to remark that this can only be done if $a \ne 0$ (though people often forget this). What you are wanting to do under <strong>Funny Way</strong> is to start with $0=0$, then divide both sides by zero (which is ... |
288,974 | <p>Alright this maybe really funny but I want to know why is this wrong. We often come across identities which we prove by multiplying both the sides of the identity by a certain entity but why don't we multiply it by $0$. That way every identity will be proved in one single line. That is so stupid. I mean, by that way... | achille hui | 59,379 | <p>One can look at the issue from the angle of information.</p>
<p>When we multiply $b$ and $c$ by a non-zero number $a$, no information is loss:</p>
<p>$$b = c \to a b = a c\\
b \ne c \to a b \ne a c$$
Since no information is lost, we can reverse the "logic" and cancel $a$ in both side of equation.
In contrast, wh... |
288,974 | <p>Alright this maybe really funny but I want to know why is this wrong. We often come across identities which we prove by multiplying both the sides of the identity by a certain entity but why don't we multiply it by $0$. That way every identity will be proved in one single line. That is so stupid. I mean, by that way... | Barbara Osofsky | 59,437 | <p>A function actually consists of two things, a domain (in first courses in calculus the domain is assumed to be the natural domain) and a rule for assigning a unique value to any real number in that domain. The domain of $\sin^2(\theta)$ is all of $\mathbb R$ and the domain of $\tan(\theta)$ is all of the reals that ... |
288,974 | <p>Alright this maybe really funny but I want to know why is this wrong. We often come across identities which we prove by multiplying both the sides of the identity by a certain entity but why don't we multiply it by $0$. That way every identity will be proved in one single line. That is so stupid. I mean, by that way... | Jim | 23,117 | <p>Suppose</p>
<p>$a \neq b$</p>
<p>$a\cdot0 \neq b\cdot0$</p>
<p>But, since $a\cdot0=0$ and $b\cdot0=0$</p>
<p>We get, by substitution </p>
<p>$0\neq0$ </p>
<p>Combine this with your discovery, and I expect that this site will self-desctruct in 3,...2,...1,... (Wait, don't do it, it's just a jo........</p>
|
288,974 | <p>Alright this maybe really funny but I want to know why is this wrong. We often come across identities which we prove by multiplying both the sides of the identity by a certain entity but why don't we multiply it by $0$. That way every identity will be proved in one single line. That is so stupid. I mean, by that way... | user217187 | 217,187 | <p>This is an interesting point. I will do this on the fly....</p>
<p>OK. So, false proof.</p>
<p>\begin{align*}
1 & = 2\\
1 \cdot 0 & =2 \cdot 0\\
0 & =0
\end{align*}</p>
<p>Q.E.D.</p>
<p>Um... I can't see anything wrong with this. So $0 = 0$, ok. So $1 \cdot 0 = 2 \cdot 0$... ok. So $1 = 2$... Not ok.... |
1,303,577 | <p>I have started to learn about the properties of the <a href="http://en.wikipedia.org/wiki/Quadratic_residue" rel="nofollow">quadratic residues modulo n (link)</a> and reviewing the list of quadratic residues modulo $n$ $\in [1,n-1]$ I found the following possible property:</p>
<blockquote>
<p>(1) $\forall\ p \gt ... | fretty | 25,381 | <p>The Chinese Remainder theorem lets us conclude that counting squares mod $mn$ is the same as counting pairs of squares mod $m$ and mod $n$ separately...whenever $m,n$ are coprime.</p>
<p>Note that there are exactly two squares mod $2$ and same mod $3$.</p>
<p>It is also known that modulo an odd prime there are $\f... |
1,148,760 | <p>$\displaystyle \int x^7\cos x^4 dx$</p>
<p>I tried first by letting $x^4 = u$ and then using integration by parts by assigning f(x) to $u^\frac74$ and cos(u) to g'(x) and I end up getting after applying parts twice, the same integral on the RHS as what we are looking for. So I bring it in on the LHS and add it over... | Claude Leibovici | 82,404 | <p><strong>Hint</strong></p>
<p>$$I=\int x^7 \cos(x^4)\,dx=\frac 14\int (4x^3)\, x^4 \cos(x^4)\,dx$$ So, let $x^4=u$ and then $$I=\frac 14\int u \cos(u) \, du$$ I am sure that you can take from here.</p>
|
934,660 | <p>Prove that for $ n \geq 2$, n has at least one prime factor.</p>
<p>I'm trying to use induction. For n = 2, 2 = 1 x 2. For n > 2, n = n x 1, where 1 is a prime factor. Is this sufficient to prove the result? I feel like I may be mistaken here.</p>
| Adriano | 76,987 | <p>If $n \ge 2$ is prime, then we're done, since $n$ is our desired prime factor of $n$. Otherwise, if $n \ge 2$ is not prime, then $n = ab$ for some $a,b \in \mathbb N$, where $1 < a \leq b < n$. But then since $a \geq 2$, it follows by the induction hypothesis that $a$ has at least one prime factor, say $p$, so... |
3,604,745 | <p>(I Prefer to open new question because those are my homework and i want to understand my way)</p>
<p>In my homework i need to solve the integral: </p>
<p><span class="math-container">$$
\int \frac{e^x}{2e^x + \sqrt{e^x}}dx
$$</span></p>
<p>I tried the substitution method: </p>
<p><span class="math-container">$$
... | Surb | 154,545 | <p><strong>Hint</strong></p>
<ul>
<li><p>Make the substitution <span class="math-container">$u=\sqrt t$</span> in your last integral.</p></li>
<li><p>Or do at the beginning the substitution <span class="math-container">$t=\sqrt{e^x}$</span> instead of <span class="math-container">$t=e^x$</span>.</p></li>
</ul>
|
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