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 |
|---|---|---|---|---|
33,430 | <p>Using <kbd>ctrl</kbd><kbd>/</kbd> you can make a fraction. If you have selected something it will appear in the numerator.</p>
<p>Does there exist a shortcut to make the selected text appear in the denominator instead?
If not, is it possible to create a shortcut that does this?</p>
| Anton Antonov | 34,008 | <p>The data in the question presents a good case for <a href="https://mathematicaforprediction.wordpress.com/2016/06/03/making-chernoff-faces-for-data-visualization/" rel="nofollow noreferrer">visualization with Chernoff faces</a>. For that data, actually, the Chernoff faces work "out of the box" pretty well!</p>
<h2>... |
1,216,419 | <p>We know that if the given equation were $\quad y = x^3 + ax^2 + bx$, $\quad$ then the derivative would be $3x^2 + 2ax + b$.</p>
<p>Since the given equation is different so the derivative will be: $$2(x^3 + ax^2 + bx)(3x^2 + 2ax + b) \implies 2y(3x^2 + 2ax + b)$$</p>
<p>What does: $$\frac{3x^2 + 2ax + b}{2y}$$
mean... | pjs36 | 120,540 | <p>$L$ will be the $3 \times 3$ array</p>
<p>$$L = \begin{pmatrix}L_{11}&L_{12}&L_{13}\\L_{21}&L_{22}&L_{23}\\L_{31}&L_{32}&L_{33}\end{pmatrix}$$ such that</p>
<p>$$L = \begin{pmatrix}L_{11}&L_{12}&L_{13}\\L_{21}&L_{22}&L_{23}\\L_{31}&L_{32}&L_{33}\end{pmatrix}\begin{pm... |
272,193 | <p>It is well known that the Petersen Graph is not Hamiltonian. I can show it by case distinction, which is not too long - but it is not very elegant either.</p>
<p>Is there a simple (short) argument that the Petersen Graph does not contain a Hamiltonian cycle? </p>
| Michael Tarsi | 184,603 | <p>Assume there exists a Ham. cycle. Color the vertices along the cycle alternately $5$ Red and $5$ Blue. Each vertex now has at least two neighbors of the opposite color.</p>
<p>But starting with two adjacent vertices of the same color (must exist on a non-bipartite graph), there is only one way to complete a $2$-co... |
717,084 | <p>I'm slightly confused as to how </p>
<p>$$\{\emptyset,\{\emptyset,\emptyset\}\} = \{\{\emptyset\},\emptyset,\{\emptyset\}\}$$</p>
<p>are equivalent. I thought two sets were equivalent if and only if "$A$" and "$B$" have exactly the same elements. In this case, we have one element which is in both sets but then two... | user85798 | 85,798 | <p>A set is a collection of distinct objects. Every element of a set must be unique; no two members may be identical.</p>
|
897,815 | <p>how do I go forward with sketching the graphs of the following two functions?</p>
<p>i) $y(t)=|2+t^3|$</p>
<p>ii) $f(x)=4x+|4x-1|$</p>
<p>thanks in advance!</p>
| Edward ffitch | 26,243 | <p>The double coset arises in representation theory of finite groups, in particular <a href="http://en.wikipedia.org/wiki/Character_theory#Mackey_decomposition" rel="nofollow">Mackey's Decomposition Theorem</a> (a very important and useful result!). </p>
|
669,405 | <p>Let $P \to M$ be a principal $G$-bundle, equipped with a principal connection $D$.</p>
<p>Let $Q \subset P$ be a principal subbundle with fiber $H$, where $H \leq G$ is a (let's say closed and connected) Lie subgroup.</p>
<blockquote>
<p><strong>Question:</strong> Suppose that $D$ restricts to a principal connec... | Paul Siegel | 1,509 | <p>Let $p$ denote the basepoint for the holonomy group. The Ambrose-Singer theorem says that the Lie algebra of the holonomy group of $D$ is the subspace of the Lie algebra of $G$ spanned by elements of the form $\Omega_q(X,Y)$ where $q$ is a point in the holonomy bundle at $p$, $X$ and $Y$ are horizontal tangent vect... |
669,405 | <p>Let $P \to M$ be a principal $G$-bundle, equipped with a principal connection $D$.</p>
<p>Let $Q \subset P$ be a principal subbundle with fiber $H$, where $H \leq G$ is a (let's say closed and connected) Lie subgroup.</p>
<blockquote>
<p><strong>Question:</strong> Suppose that $D$ restricts to a principal connec... | Gil Bor | 118,580 | <p>The answer is essentially "yes". More precisely, $\text{Hol}(D)$ is only well-defined up to conjugation in $G$ and the statement is that if $D$ restricts to some $H$ subbundle $Q\subset P$ then $\text{Hol}(D)$ is the conjugacy class of a subgroup of $H$. </p>
<p>This is an immediate consequence of the definitions... |
3,475,674 | <p>If the objective function is <span class="math-container">$\min\limits_{x} \sum\limits_{i=1}^n e^{-a_ix_i}$</span>, can I transform the objective into <span class="math-container">$\max\limits_{x} \sum\limits_{i=1}^n a_ix_i$</span>?</p>
| cruijf | 707,527 | <p>Yes. What you're doing is basically making use of two general facts: </p>
<ol>
<li><p>Any monotonic transformation preserves the max/min (note that <span class="math-container">$\ln$</span> is a monotonic transformation). This should be pretty intuitive to see. </p></li>
<li><p>Maximizing <span class="math-containe... |
818,512 | <p>I have tried to find posts that are related to the question but they end up with the terms like ‘find a distance’. What I want is not to find the distance: I already have the distance, I want something else.</p>
<p>Assume <span class="math-container">$(x_1,y_1)$</span> and <span class="math-container">$(x_2,y_2)$</s... | user2566092 | 87,313 | <p>The vector that traverses parallel to the line is $(1,a)$. Normalize this vector by dividing by $\sqrt{1 + a^2}$ to get vector $v$ in the same direction that has length $1$. Then your desired point is $w \pm d\cdot v$ where $w$ is the original point and $d$ is the desired distance.</p>
|
1,786,306 | <p>Let $\sum\limits_{n=1}^\infty$ $a_n$ be a convergent series of positive terms. What can be said about the convergence of the following series: </p>
<p>$\sum\limits_{n=1}^\infty$ $\frac{a_1 + a_2 + ... + a_n}{n}$</p>
<p>The series above diverges. We know that $\sum\limits_{n=1}^\infty$ $\frac{a_1 + a_2 + ... + a_n}... | David C. Ullrich | 248,223 | <p>Hint: What's the sum of the coefficients of $a_1$ in the second sum?</p>
|
1,074,740 | <p>We know that torsion-free plus finitely generated <span class="math-container">$\rightarrow$</span> free and that <span class="math-container">$\mathbf{Q}$</span> is torsion-free is easy. </p>
<blockquote>
<p>But how to show <span class="math-container">$\mathbf{Q}$</span> is not finitely generated and not free?<... | egreg | 62,967 | <p>If an abelian group $G\ne\{0\}$ is free, then it is isomorphic to $\mathbb{Z}^{(X)}$ (direct sum of copies of $\mathbb{Z}$), for some set non empty set $X$.</p>
<p>Then $\mathbb{Z}$ is an epimorphic image of $G$. Since $\mathbb{Z}$ is not divisible, $G$ can't be divisible. But $\mathbb{Q}$ is divisible.</p>
<hr>
... |
2,869,753 | <p>I get that $∅$ is subset of every set thus $∅ ⊆ \{\{∅\}\}$.
However, I'm not sure if $∅ ⊂ \{\{∅\}\}$.
From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
What I'm confused is, does $\{\{∅\}\}$ have an element that $∅$ doesn't have?... | AlvinL | 229,673 | <p>The set $\{\{\emptyset\}\}$ contains the element $\{\emptyset\}$. The empty set contains no elements, thus the containment is proper i.e $\emptyset\subsetneq \{\{\emptyset\}\}$.</p>
|
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | Benjamin Steinberg | 15,934 | <p>Every finite simple group can be generated by at most <span class="math-container">$2$</span> elements. This is another famous consequence of the classification.</p>
|
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | kabenyuk | 173,068 | <p>It seems to me appropriate to name the following totally unexpected result, which is too good to be true, yet is true.</p>
<blockquote>
<p>Are there only finitely many finite groups with <span class="math-container">$m$</span> generators of
exponent <span class="math-container">$n$</span>, up to isomorphism (Restric... |
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | KhashF | 128,556 | <p>For a group with finitely many elements of finite order, the set of elements of finite order is a subgroup.</p>
|
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | Shahab | 165,074 | <p>If <span class="math-container">$k$</span> is an algebraic number field then for every positive integer <span class="math-container">$n$</span> there exist infinitely many field
extensions of <span class="math-container">$k$</span> of degree <span class="math-container">$n$</span> having no proper subfields over <sp... |
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | Dave Benson | 460,592 | <p>The projective dimension of <span class="math-container">${\mathbb C}(x,y,z)$</span> as a module over <span class="math-container">${\mathbb C}[x,y,z]$</span> is two if the continuum hypothesis holds, and three otherwise.</p>
|
3,380,849 | <p>Is the equation I wrote in the title true for positive integers <span class="math-container">$x,y$</span>? I checked some cases and it seems to hold, but how do I prove it? I am trying to solve another problem and it turns out that if this equation holds, the problem is solved.</p>
| saulspatz | 235,128 | <p>Yes, it's true. Recall that <span class="math-container">$$\text{lcm}(x,y)={xy\over\gcd(x,y)}$$</span> Since <span class="math-container">$\gcd(x,y)=\gcd(x,x+y)$</span>, a simple substitution proves your formula.</p>
|
3,380,849 | <p>Is the equation I wrote in the title true for positive integers <span class="math-container">$x,y$</span>? I checked some cases and it seems to hold, but how do I prove it? I am trying to solve another problem and it turns out that if this equation holds, the problem is solved.</p>
| Bill Dubuque | 242 | <p>It can be proved directly by lcm laws (without using gcds) as follows, with <span class="math-container">$\,[a,b] := {\rm lcm}(a,b)$</span></p>
<p><span class="math-container">$$\begin{align}
\color{#c00}{\dfrac{n}{yz}}\:\ +\:\ \dfrac{n}{xz}\:\ =\:\ \dfrac{n}{xy},\,\ \ z = x\!+\!y \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\... |
2,626,597 | <p>In both my textbook (Hungerford's Algebra), and in class, it is claimed that Monoid Homomorphisms are not required to preserve the identity. Interestingly enough, the Wikipedia page for Monoids requires Monoid Homomorphisms to preserve the identity element: <a href="https://en.wikipedia.org/wiki/Monoid#Monoid_homomo... | StefanH | 33,817 | <p>Given an algebraic structure $(A, f_1, f_2, \ldots)$ in the sense of <a href="https://en.wikipedia.org/wiki/Universal_algebra" rel="noreferrer">universal algebra</a> morphism are always required to respect <em>all</em> operations (this also includes the constants, as these are modelled as $0$-ary operations). Simila... |
319,059 | <p>I am helping out a friend who can't seem to get these proofs; unfortunately, I can't find them either. Can someone tell me how to solve this or point me in the right direction with resources? </p>
<p>Question 1: </p>
<blockquote>
<p>Prove that for all real numbers x, y, and z, if x + y + z greater than or equal... | Zhiyong Wang | 34,197 | <p>Proof 1. </p>
<p>By contradiction, x<1 and y<1 and z<1 implies x+y+z <3, so this assumption is false. Thus we have x>=1 or y>=1 or z>=1.</p>
<p>Proof 2 can be done in a similar way.</p>
|
319,059 | <p>I am helping out a friend who can't seem to get these proofs; unfortunately, I can't find them either. Can someone tell me how to solve this or point me in the right direction with resources? </p>
<p>Question 1: </p>
<blockquote>
<p>Prove that for all real numbers x, y, and z, if x + y + z greater than or equal... | Berci | 41,488 | <p>$$(A\Rightarrow B) \equiv (\lnot B\Rightarrow\lnot A)$$</p>
<ol>
<li>The statement is equivalent to its <em>contraposition</em>: if $x< 1$ <em>and</em> $y< 1$ <em>and</em> $z< 1$ then $x+y+z< 3$.</li>
<li>Similarly, use contraposition.</li>
</ol>
|
1,168,402 | <p>Solve DE: $$y'' + 2y' + 5y = x + 4$$</p>
<p>I have the correct general solution
$$y(x) = c_1\,e^{-x}\cos(2x) + c_2\,e^{-x}\sin(2x)\ .$$</p>
<p>So I take my 'guess' , take a couple of derivatives, and plug in for the equation $$2A +5(Ax + B) = x + 4$$</p>
<p>But at this point I'm stuck on solving for the constant... | BaronVT | 39,526 | <p>You need your equation to hold for <strong>all</strong> $x$, right? Then the parts that depend on $x$ must be equal ($5Ax = x$) and the parts that do not depend on $x$ must be equal; thus $A$ and $B$ must satisfy</p>
<p>$$
5A = 1 \\
2A + 5B = 4.
$$</p>
<p>Can you solve for $A$ and $B$ from here?</p>
|
23,224 | <p>How to integrate $x^{1/2}e^{-x}$ using integration by parts?</p>
<p>Answer should be $\left(-\sqrt{x} e^{-x}+(1/2)\sqrt{\pi} \mbox{erf}(\sqrt{x})\right)+c$</p>
| kcrisman | 24,113 | <p>In the comment:</p>
<blockquote>
<p>How you get erf(\sqrt(x)) from there without substitutions</p>
</blockquote>
<p>If you leave the $1/2$ in the integral and do the most obvious substitution (recall your first step?), you will get something that should look like the definition of the function in @t.b.'s link.</... |
28,321 | <p>Let $K$ be a nonempty compact convex subset of $\mathbb R^n$ and let $f$ be the function that maps $x \in \mathbb R^n$ to the unique closest point $y \in K$ with respect to the $\ell_2$ norm. I want to prove that $f$ is continuous, but I can't seem to figure out how.</p>
<p>My thoughts: Suppose $x_n \to x$ in $\mat... | Ben Derrett | 2,934 | <p>Another way, for the record:</p>
<p>Since $f(a)$ is optimal, $$(f(b) - f(a))\cdot (a - f(a)) \le 0.$$ Similarly, since $f(b)$ is optimal, $$(f(b)-f(a))\cdot(f(b) - b)\le 0.$$ </p>
<p>When we sum these two inequalities and rearrange, we get
$$\begin{align}\|f(a)-f(b)\|^2&\le (f(b)-f(a))\cdot(b-a) \\
... |
993,132 | <p>I have been looking into this question :
we have two surfaces :</p>
<p>$$\big\{(x,y,z)\in \mathbb{R}^3 \mid\;\; S_1\colon\;\; x+z=1 ,\;\; S_2\colon\;\; x^2+y^2=1 \big\}$$</p>
<p>we need to draw or describe the "shape" that we get .
I tried to solve it by drawing the two surfaces and imagining the intersection whic... | Lolman | 160,018 | <p>The approach is wrong. An equation in $\mathbb{R}^3$ is a surface, but an intersection of two surfaces usually is something with lower dimension.
$$x^2+y^2=1$$ $$x+z=1$$ are your equations, the right way to go would be to get the parametric equations for the plane $x+z=1$ and put them inside the other. $$(a,b,1-a)$$... |
2,604,377 | <p>Given an orientation-preserving homeomorphism of the circle $$f : S^1 \rightarrow S^1,$$ I want to define a homeomorphism of the cylinder $$F : S^1 \times \mathbb{I} \rightarrow S^1 \times \mathbb{I}$$ such that for all $x \in S^1$, we have: $F(x,0) = (f(x),0)$ and $F(x,1) = (x,1)$.</p>
<p>The same goes with 'homeo... | Eric Wofsey | 86,856 | <p>This is possible. Let $e:\mathbb{R}\to S^1$ be the covering map $e(s)=\exp(2\pi i s)$. We can then lift $f$ to a map $g:\mathbb{R}\to \mathbb{R}$ such that $eg=fe$. The assumption that $f$ is an orientation-preserving homeomorphism implies that $g$ is strictly increasing with $g(s+1)=g(s)+1$ for all $s$. Now we o... |
88,034 | <p>V10.1:</p>
<pre><code><|a -> 1|>[Key @ a]
(* Missing["KeyAbsent", Key[a]]*)
</code></pre>
<p>V10.0:</p>
<pre><code><|a -> 1|>[Key @ a]
(* 1 *)
</code></pre>
<p>Bug or Design Change?</p>
<p><strong>Update:</strong>
This bug/design change remains in 10.2.</p>
<p>Even though my conclusions in a ... | m_goldberg | 3,066 | <p>In V10.1</p>
<pre><code>a = 42;
<|a -> 1|>[a]
</code></pre>
<p>gives</p>
<blockquote>
<pre><code>1
</code></pre>
</blockquote>
<p>I look at this as a fix for a bug or design flaw that made the use of symbol keys awkward in V10.0. The loss of the <code><|a -> 1|>[Key @ a]</code> is a small price... |
88,034 | <p>V10.1:</p>
<pre><code><|a -> 1|>[Key @ a]
(* Missing["KeyAbsent", Key[a]]*)
</code></pre>
<p>V10.0:</p>
<pre><code><|a -> 1|>[Key @ a]
(* 1 *)
</code></pre>
<p>Bug or Design Change?</p>
<p><strong>Update:</strong>
This bug/design change remains in 10.2.</p>
<p>Even though my conclusions in a ... | Ronald Monson | 2,249 | <p><strong>Summary: Yes it is a "bug" and it also suggests changing <code>[]</code>, <code>#&</code> and <code>KeyTake</code>'s key extracting semantics:</strong></p>
<p><strong>Background: A Disambiguating Imperative Emerges:</strong></p>
<p>While related to language design, subtlities such as these do impact de... |
3,209,740 | <p>In an article the author consider the square matrix <span class="math-container">$(A^{ij}(x))_{i,j=1}^d$</span> defined by
<span class="math-container">$$
A^{ij}(x):= \frac{x_ix_j}{|x|^2} \qquad (x\in R^d).
$$</span>
He writes that the matrix is elliptic.
I don't know to prove that there exists <span class="math-co... | Στέλιος | 403,502 | <p>Τhe inequality, that you wrote down, is not true in general. First of all, it is equivalent to</p>
<p><span class="math-container">$$\langle x,\xi\rangle^2=\left(\sum_{i=1}^dx_i{\xi}_i\right)^2=\sum_{i,j=1}^d(x_i{\xi}_i)(x_j{\xi}_j)\geq c|x|^2|\xi|^2,\ \ \forall x,\xi\in \mathbb{R}^d.$$</span></p>
<p>If this is tr... |
1,511,477 | <p>So I tried and found that $$7^{100} \equiv 1 \pmod{100}$$ but I got stuck with $8^{100}$. Help me out please. </p>
| Quang Hoang | 91,708 | <p>First, note that $7^2\equiv8^2\pmod{5}$. So $$7^{100}-8^{100}=(7^2-8^2)\left((7^2)^{49}+(7^2)^{48}(8^2)+\cdots+(8^2)^{49}\right)\equiv 0\pmod{25},$$
since the second term consists of $50$ summands of the same modulo $5$. </p>
<p>Second $7^{100}-8^{100}\equiv 7^{100}\pmod{4}\equiv 1\pmod{4}$.</p>
<p>Chinese remaind... |
2,664,483 | <p>Why is $\mathcal{C}=\{\{ X_T \in A \} \text{ or } \{ X_T \in A\} \cup \{T=\infty \},A \in \mathcal{B}(R) \}$ closed under complementation where X is a random measurable process and T a random time? </p>
<p>This is (Problem 1.17) in Karatzas and Shreve. </p>
<p><strong>My problem</strong>:
If I choose $D=\{ X_T ... | José Carlos Santos | 446,262 | <p>Indeed, your answer is wrong, since it maps $3$ into $48$ instead of $0$.</p>
<p>The correct answer is $-2x^3-25x^2+48x+135$.</p>
<p>Note that your polynomial must be a multiple of $x-3$ and its constant term must be $135$. Therefore, it must be of the form $(x-3)(ax^2+bx-45)$. Now, plugging in $x=1$ and $x=2$, it... |
2,664,483 | <p>Why is $\mathcal{C}=\{\{ X_T \in A \} \text{ or } \{ X_T \in A\} \cup \{T=\infty \},A \in \mathcal{B}(R) \}$ closed under complementation where X is a random measurable process and T a random time? </p>
<p>This is (Problem 1.17) in Karatzas and Shreve. </p>
<p><strong>My problem</strong>:
If I choose $D=\{ X_T ... | amd | 265,466 | <p>If you plug the coordinates of each of the given points into the generic cubic equation $Ax^2+Bx^2+Cx+D=y$, you will end up with a system of linear equations in the coefficients of this cubic. I presume that you know how to solve such a system of equations. </p>
<p>However, you can instead generate the cubic equat... |
107,882 | <p>Can someone recommend a good basic book on Geometry? Let me be more specific on what I am looking for. I'd like a book that starts with Euclid's definitions and postulates and goes on from there to prove thereoms about triangles, circles and other plane shapes. I'm not interested (at this time) in a book that tie... | lhf | 589 | <ul>
<li><em>Introduction to Geometry</em> by Coxeter.</li>
<li><em>Elementary Geometry From An Advanced Viewpoint</em> by Moise.</li>
<li><em>Geometry: Euclid and Beyond</em> by Hartshorne.</li>
</ul>
|
107,882 | <p>Can someone recommend a good basic book on Geometry? Let me be more specific on what I am looking for. I'd like a book that starts with Euclid's definitions and postulates and goes on from there to prove thereoms about triangles, circles and other plane shapes. I'm not interested (at this time) in a book that tie... | BlakeDavis | 21,723 | <p>You might want to look at "The Harpur Euclid". It's a free Google book which covers basic Euclidean Geometry based on Euclid's five postulates, has a large number of exercises, and some enrichment material on what was called "Modern Geometry" in 1894, topics like Simson's line, the radical axis, poles and polars, i... |
3,090,052 | <p>My question concerns the answer to exercise 1.3:</p>
<blockquote>
<p>Given a partition <span class="math-container">$P$</span> on a set <span class="math-container">$S$</span>, show how to define a relation <span class="math-container">$\sim$</span> on <span class="math-container">$S$</span> such that <span class... | Yiorgos S. Smyrlis | 57,021 | <p>Observe that
<span class="math-container">$$
\frac{n}{1\cdot 3\cdot 5\cdots (2n+1)}=\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5\cdots (2n-1)}-\frac{1}{1\cdot3\cdot 5\cdots (2n+1)}\right)
$$</span>
Hence
<span class="math-container">$$\sum_{n=1}^\infty\frac{n}{1\cdot 3\cdot 5\cdots (2n+1)}=\sum_{n=1}^\infty\frac{1}{2}\... |
552,307 | <blockquote>
<p>If A $\cap$ B $\cap$ C = $\emptyset$, then the sum principle applies so |A $\cup$ B $\cup$ C| = |A|+|B|+|C|.</p>
</blockquote>
<p>I think it would be true since there is nothing in common among A,B and C, but just wondering if there is any exceptions to this problem so it would be false?</p>
| ncmathsadist | 4,154 | <p>Use the inclusion-exclusion principle<br>
$$|A\cup B \cup C| = |A| + |B| + |C| - |A\cap B| - |A\cap C| - |B\cap C|
+ |A\cap B \cap C|.$$</p>
|
4,238,492 | <p>A practical example lead me to believe that a geographical projection, such as the <a href="https://en.wikipedia.org/wiki/Mercator_projection" rel="nofollow noreferrer">Mercator projection</a>, is an <a href="https://en.wikipedia.org/wiki/Affine_transformation" rel="nofollow noreferrer">affine transformation</a>.</p... | José Carlos Santos | 446,262 | <p>You have <span class="math-container">$\sup A=1$</span> because <span class="math-container">$1\in A$</span> and because <span class="math-container">$a\in A\implies a\leqslant1$</span> (so, in fact, <span class="math-container">$1=\max A$</span>).</p>
<p>On the other hand, if <span class="math-container">$a\in A$</... |
3,209,497 | <p>The problem is:</p>
<p>Suppose <span class="math-container">$H$</span> is Hilbert and <span class="math-container">$\{e_n\}_{n = 1}^\infty$</span> is its orthonormal basis. Prove <span class="math-container">$x_n \rightharpoonup x_0$</span> if and only if:</p>
<ol>
<li><span class="math-container">$||x_n||$</span>... | PrincessEev | 597,568 | <p>Usually whenever <span class="math-container">$+\infty$</span> is written in lieu of merely <span class="math-container">$\infty$</span>, it seems mostly meant to just clarify that "this is the positive infinity". Typically when left signless, people tend to assume <span class="math-container">$\infty$</span> refers... |
1,245,499 | <p>In the proof we say $\left\{\left(\frac1n,1\right):n\geq 1\right\}$ is an infinite cover with no finite subcover.</p>
<p>But, $(0,1)$ set also belongs to cover mentioned above.
We can say $\{(0,1)\}$ is a subcover of mentioned above cover.</p>
<p>I am not able to understand what I am doing wrong.</p>
| Chappers | 221,811 | <p>$(0,1)$ is not in your open cover (for which integer $n$ is $0$ of the form $1/n$?). Therefore it can't be in a subcover.</p>
<p>To show that it has no finite subcover, take such a finite subcover. Being finite, it would have a maximum integer $n$ included, say $N$. But then the subcover is a subset of
$$ \left( \f... |
487,872 | <p>assume that $\frac xy \in \mathbb N$ , is it correct that $\frac xy mod (n)=\frac{x \ mod \ n}{y\ mod \ n} mod \ n $ ? if not then how to compute it ?</p>
<p>explination : </p>
<p>I am dealing with large numbers and I want to compute $\frac xy mod \ n$ , assume that x is very big integer ( can reach $100^{100}$) w... | Caleb Stanford | 68,107 | <p>I'm assuming that by "$a \mod b$" you mean "the remainder when $a$ is divided by $b$" rather than the (more mathematical) equivalence class definition. To avoid confusion, I'll instead write "% n" for the remainder modulo $n$, and $\mod n$ for mathematical equivalence modulo $n$.</p>
<p>Since you're assuming $\fra... |
386,899 | <p>Show that $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^{n+2}=\frac{2^{n}n(n+2)!}{6}.$$</p>
| robjohn | 13,854 | <p>Using these three identities for $n^{\text{th}}$ differences:
$$
\begin{align}
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^n&=n!\\
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^{n+1}&=n!\binom{n+1}{2}\\
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^{n+2}&=n!\left(3\binom{n+2}{4}+\binom{n+2}{3}\right)
\end{align}
$$
and the fact tha... |
1,280,882 | <p>This may be a silly question, but one that I am confused about nonetheless.</p>
<p>With regards to the compound trig identities such as $\cos(A+B)=\cos A\cos B - \sin A\sin B$ etc., I'd like to know why they are used. What's the purpose? Surely, one would ask themselves that if we can just add $A$ and $B$ together ... | wythagoras | 236,048 | <p>They aren't really used in geometry itself, but more in solving geometric equations and when doing trig substitutions in calculus. Furthermore, it is just interesting to know these properties, even if we rarely use them. </p>
|
1,470,879 | <p>I want to solve this system of advective-diffusive-reactive equations analytically:</p>
<p>$$\left(\alpha - k_0c_B\right)c_A+v\frac{dc_A}{dx}-D\frac{d^2c_A}{dx^2} = f_A $$
$$\left(\alpha - k_0c_A\right)c_B+v\frac{dc_B}{dx}-D\frac{d^2c_B}{dx^2} = f_B $$
$$k_0c_Ac_B+\alpha c_C+v\frac{dc_C}{dx}-D\frac{d^2c_C}{dx^2} = ... | Chinny84 | 92,628 | <p>$$
\frac{1}{\sqrt{n} +\sqrt{n+1}} = \frac{\sqrt{n} -\sqrt{n+1}}{(\sqrt{n} +\sqrt{n+1})(\sqrt{n} -\sqrt{n+1})}
$$
The denominator becomes
$$
(\sqrt{n} +\sqrt{n+1})(\sqrt{n} -\sqrt{n+1}) = n -n -1 = -1
$$
Thus the sum becomes
$$
-\sum_{n=0}^k \sqrt{n}-\sqrt{n+1}
$$
Which is a telescope sum.</p>
|
2,233,369 | <p>If there exists a homeomorphism $f$ between the closed unit interval and some cartesian product $A\times B$, either $A$ is a singleton or $B$ is a singleton.</p>
<p>The proof I have argues as follows:</p>
<p>Since $[0,1]$ is connected, $A\times B$ is, so $A$ and $B$ are as well. If $a_1,a_2 \in A$ and $b_1,b_2\in ... | shrinklemma | 67,953 | <p>Since $f^{-1}:A\times B\to [0,1]$ is a homeomorphism and $\{a_1\}\times B$ is closed in $A\times B$ and connected, $f^{-1}(\{a_1\}\times B)$ is also closed in $[0,1]$ and connected. But a closed and connected subset of $[0,1]$ is a closed interval. </p>
<p>In response to the comments, here is an argument to show th... |
2,233,369 | <p>If there exists a homeomorphism $f$ between the closed unit interval and some cartesian product $A\times B$, either $A$ is a singleton or $B$ is a singleton.</p>
<p>The proof I have argues as follows:</p>
<p>Since $[0,1]$ is connected, $A\times B$ is, so $A$ and $B$ are as well. If $a_1,a_2 \in A$ and $b_1,b_2\in ... | Kernel_Dirichlet | 368,019 | <p>We have by hypothesis that $f$ is a homeomorphism. In particular, we have $f$ continuous. This means by continuity, the pre-image of an open (closed) set is open (closed). Also, if we assume the standard topology, we have that $[0,1]$ is compact, and the continuous image of a compact set is also always compact. Name... |
4,421,862 | <p>I am trying to prove that <span class="math-container">$\frac{d}{dx}x^e=ex^{e-1}$</span>.</p>
<p><span class="math-container">$\displaystyle \frac{d}{dx}x^e=\lim_{h\to 0}\dfrac{(x+h)^e-x^e}{h}=\lim_{h\to 0}x^e \cdot \dfrac {e^{e \ln \left({1 + \frac h x} \right)} - 1} {e \ln {\left(1 + \dfrac h x\right)} } \cdot \... | gt6989b | 16,192 | <p><span class="math-container">$(x+h)^e$</span> can be expanded using the Binomial theorem to get
<span class="math-container">$$
(x+h)^e = x^e + \binom{e}{1}x^{e-1} h + h^2 (\cdot)
$$</span>
and now <span class="math-container">$\binom{e}{1} = e$</span>, the term <span class="math-container">$x^e$</span> is subtracte... |
1,740,981 | <blockquote>
<p>Let $B$ be a group of all the continuous functions in the interval $[a,b]$ such that $f(a)=0$. Prove that $A$ is close group in the metric space $C[a,b]$</p>
</blockquote>
<p><strong>My attempt:</strong></p>
<p>Metric space $C[a,b]$ defined by $d(x(t),y(t))=\max\limits_{a\leqslant t\leqslant b} \mid... | Alex M. | 164,025 | <p>The fact that $B$ is a group shouldn't be difficult: if $f,g \in B$, then $f(a) + g(a) = 0 + 0 = 0$ and $f+g$ is obviously continuous. Next, if $f \in B$ then $-f \in B$ because it is continuous and $(-f) (a) = - (f(a)) = -0 = 0$, and $0$ is obviously the neutral element. That addition is associative follows from th... |
1,467,946 | <p><strong>10 fair coins are tossed. How many outcomes have 3 Heads?</strong>
I'm supposed to solve it with combination C(10, 3). But...</p>
<p>How do you know it's a combination that will solve it? I'm <strong>not</strong> interested in what makes it a combination, instead of a permutation. I know the answer is (some... | WW1 | 88,679 | <p>think about naming the order of the tosses ... toss#1, toss#2 etc.</p>
<p>e.g. The number of ways of getting 3 heads when tossing 5 coins is the same as the number of ways of deciding which 3 of the 5 tosses came up heads</p>
<p>e.g. the choice ${2,4,5}$ corresponds to the sequence <code>THTHH</code></p>
<p>the ... |
1,556,609 | <p>Let $C$ be the curve of intersection of the two surfaces $x+y=2 , x^2+y^2+z^2=2(x+y)$ . The curve is to be traversed in clockwise direction as viewed from the origin . The what is the value of $\int_Cydx+zdy+xdz$ ? I am not even able to parametrize the curve of intersection . Please help . Thanks in advance </p>
| Siddharth Joshi | 288,487 | <p>I think a curve along the intersection of the two surfaces can be thought of as consisting of two curves $C_1$ and $C_2$ with parameterizations $r_1: \left [ \ 0, 2 \right ] \to \mathbb{R^3}$ where $$r_1(x) := \left ( x, 2 - x, \sqrt{4x - 2x^2} \right)$$ and $r_2: \left [ \ 0, 2 \right ] \to \mathbb{R^3}$ where $r_2... |
1,480,010 | <p>Express $A \cup B$ in set buildier notation when $A=\{2n+1 \mid n \in \mathbb{Z}\}$ and $B=\{3n+2 |n \in \mathbb{Z}\}$</p>
<p>We know that the rooster notation version of the union set is $A \cup B = \{1,2,3,5,7,8,9,11,13,14,15,17,19,20,21,\ldots\}$</p>
<p>It is a combination of every odd integer and every multipl... | Epsilon | 269,233 | <p>Consider $f_n(t)=t^n, t \in [0,1]$</p>
<p>Here, $||t^n|| \le1$. Also $||f_n||=1,\ \forall n \in \mathbb{N}$.</p>
<p>But, $||T(f_n)||=||T(t^n)||=||nt^{n-1}||=n$ which is not bounded.</p>
<p>Now, for $T: (C^1[0,1],|||.|||)\to\ (C[0,1],||.||_u)$ you get</p>
<p>$||T(f)||_u = ||f'||_u \le ||f||_u+||f'||_u= |||f||| \... |
2,242,634 | <ol>
<li>Let $\mathcal{P}$ be the plane containing the points $(-3,4,-2)$, $(1,4,0)$, and $(3,2,-1)$.
Find the point in this plane that is closest to $(0,3,-1)$.</li>
</ol>
<hr>
<ol start="2">
<li>Let $\mathcal{P}$ be the plane containing the points $(-3,4,-2)$, $(1,4,0)$, and $(3,2,-1)$.
Let $\ell$ be the line conta... | The Dead Legend | 433,379 | <p>Hint: Let $(p,q,r)$ be any arbitrary point on the plane.<br>
Let plane be $Ax+By+Cz+D=0$ (You can easily solve the plane, I hope?</p>
<p>Now, A closest point to the $(0,3,-1)$ will be actually nothing but lying on perpendicular to the plane. Thus simply draw a line with points $(0,3,-1)$ and $(p,q,r)$ and intersect... |
2,870,910 | <blockquote>
<p>I wish to show that if $z$ is real, then
$$\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$$</p>
</blockquote>
<p>I have shown this result, although my inequality is the wrong way around.</p>
<p>I considered
\begin{align}
|z^2+1|&\leq |z^2|+|1| \ \ \ \ \ \ \ \text{(triangle inequality)... | user | 505,767 | <p>We have</p>
<p>$$\left|\frac{e^{iz}}{z^2+1}\right|=\frac{\left|e^{iz}\right|}{\left|z^2+1\right|}= \frac{1}{\left|z^2+1\right|}$$</p>
<p>and </p>
<p>$$0\le\left|{z^2+1}\right|= |z|^2+1$$</p>
<p>therefore the result follows.</p>
|
1,722,628 | <p>I assume that I need to use the theorem that states that the sum of the degrees of the vertices is equal to twice the number of edges. Then, because $k$ must be greater than or equal to 3, the there must be 1.5 edges, but that is impossible. How else can I continue this proof, using the stated theorem?</p>
| bof | 111,012 | <p>I address the part of your question which asks "how would you show that any consistent extension of ZFC doesn't have a prime model?" In fact (assuming of course that ZF is consistent), some consistent extensions of ZFC do have prime models.</p>
<p>Let $M$ be a model of ZFC in which each element is definable. (Such ... |
11,070 | <p>I am a game developer currently making a game based on basic arithmetic problems for kids and teens. </p>
<p>Are there any research papers on the average time to solve basic arithmetic problems or on how to evaluate success and improvement rate. </p>
<p>That would help us a lot in creation of this game! </p>
<p>E... | Stefan Schmiedl | 1,479 | <p>I'd expect that such data would be quite useless for your game. "Reliable" numbers would originate from large-scale tests which are usually quite far away from the situation of a kid competing in a game.</p>
<p>To appeal to the broadest population, you should keep individual timings for each player and only use the... |
11,070 | <p>I am a game developer currently making a game based on basic arithmetic problems for kids and teens. </p>
<p>Are there any research papers on the average time to solve basic arithmetic problems or on how to evaluate success and improvement rate. </p>
<p>That would help us a lot in creation of this game! </p>
<p>E... | Daniel R. Collins | 5,563 | <p>This question is almost totally unanswerable unless you clarify what is meant by "basic arithmetic problems for kids and teens". </p>
<p>But let's consider a lower-bound case where you're talking about basic one-digit addition and multiplying facts. In instances like this, the knowledge should be <em>automatic</em>... |
3,865,636 | <p>I was wondering if one can explicitly solve the following equation
<span class="math-container">$$
x^2 y'' + x y' + k^2 x^2 (x^\beta+1) y = a^2 y
$$</span>
for real constants <span class="math-container">$k,a,\beta$</span>?
If <span class="math-container">$\beta = 0$</span> then the solution is a linear combination ... | markvs | 454,915 | <p>Yes. Take any integer non-identity <span class="math-container">$2\times 2$</span> matrix <span class="math-container">$A$</span> with <span class="math-container">$A^3=I$</span>.</p>
<p>For example <span class="math-container">$\begin{pmatrix} -1 & 1 \\ -1 & 0\end{pmatrix}$</span></p>
|
259,129 | <p>In dimension 3 we have that for $T=\int_{[0,\infty)}1_{B_{t}\in B(0,1)}dt$ has the Laplace transform
$$E[e^{-\lambda T}]=sech(\sqrt{2\lambda}).$$</p>
<p>And in dimension 1 we have the same for $\tau=\min\{t: |B(t)|=1\}$:</p>
<p>$$E[e^{-\lambda \tau}]=sech(\sqrt{2\lambda}).$$</p>
<p>Inverting this in Mathematica d... | Russ Lyons | 108,178 | <p>See p. 342 of Feller, vol. 2: Here, B starts at x and is absorbed at 0 and a.</p>
<p><a href="https://i.stack.imgur.com/Dol0V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dol0V.png" alt="half page from Feller"></a></p>
<p>What you wrote about T is not correct: They are not equal in dimensions... |
1,547,097 | <p>So I have</p>
<p><strong>1.</strong> $$\frac{r}{3\tan \theta} = \sin \theta$$</p>
<p><strong>2.</strong> $$r=3\cos \theta$$</p>
<p>What would be the Cartesian equation???</p>
| Justpassingby | 293,332 | <p>You could substitute for $r$ and $\theta$ their expressions in terms of $x$ and $y$. But remember that trigonometric functions of $\theta$ have easier expressions than $\theta$ itself.</p>
|
64,716 | <p>We know that by using Stirling approximation:
$\log n! \approx n \log n$</p>
<p>So how to approximate $\log {m \choose n}$?</p>
| Thomas Ahle | 7,072 | <p>If you expand stirling's approximation for $n\choose\alpha n$, you may also notice that it can be written as
$$\log{n\choose\alpha n} = n H(a) - \tfrac12\log(2\pi n\,\alpha(1-\alpha)) + O\left(\tfrac1n\right)$$,
where
$$H(a) = \alpha \log\tfrac1\alpha + (1 - \alpha) \log\tfrac1{1-\alpha}$$
is the binary entropy fun... |
93,068 | <p>Suppose that for $n \geq 4$ we have $F(x_1, \cdots, x_n) \in \mathbb{Z}[x_1, \cdots, x_n]$ is a homogeneous polynomial. Consider a large prime $p$, and suppose that we consider points of the variety $F(x_1, \cdots, x_n) \equiv 0 \pmod{p}$. If we consider non-singular points, then it is easy to see that the number of... | Denis Chaperon de Lauzières | 21,428 | <p>If we identify $\mathbf{F}_p$ with $X_p=\{0,1,\ldots, (p-1)\}\subset [0,p]$, one can probably show that for many varieties $Y/\mathbf{Z}$ in $\mathbf{A}^n$ (e.g, many hypersurfaces), the intersection $Y\cap [0,B]^n$ has not much more than the expected number $B^{n-1}$ of points, for $B$ slightly larger than $\sqrt{p... |
93,068 | <p>Suppose that for $n \geq 4$ we have $F(x_1, \cdots, x_n) \in \mathbb{Z}[x_1, \cdots, x_n]$ is a homogeneous polynomial. Consider a large prime $p$, and suppose that we consider points of the variety $F(x_1, \cdots, x_n) \equiv 0 \pmod{p}$. If we consider non-singular points, then it is easy to see that the number of... | Igor Rivin | 11,142 | <p>Just to add to @Denis' references: results of this sort with very mild hypotheses on the variety are proved in <a href="http://dl.dropbox.com/u/5188175/fouvry.pdf" rel="nofollow">Fouvry's paper</a> "Consequences of a result of N. Katz and G. Laumon..." $B$ has to be a fair bit bigger than $\sqrt{p},$ but there are v... |
3,413,166 | <blockquote>
<p>We are interested in estimating <span class="math-container">$J=\int_0^1 g(x)\ \mathsf dx$</span>, where <span class="math-container">$0\leqslant g(x)\leqslant 1$</span> for all <span class="math-container">$x$</span>. Let <span class="math-container">$X$</span> and <span class="math-container">$Y$</s... | Simon Fraser | 717,270 | <p>The CDF of <span class="math-container">$f(x)$</span> is simply the probability that <span class="math-container">$f(x)$</span> is less than or equal to a given value of <span class="math-container">$x.$</span> Expressed more formally, this is</p>
<p><span class="math-container">$g(x) =\displaystyle\int_{-\infty}^x... |
235,332 | <p>My assignment is to translate mathematical statements into formula of predicate logic. But before I can write formulas about these statements, I'm really confused about their meaning. </p>
<p>Given that the mathematical notation of a quadratic polynomial with leading coefficient 1 is $P(x) = a_0 + a_1x + x^2$</p>... | Rick Decker | 36,993 | <p>For (1), you know what a quadratic polynomial with leading coefficient 1 is: it's $P(x)=x^2+a_1x+a_0$, so each such polynomial is completely described by giving values for $a_1\text{ and }a_0$. (A polynomial with leading coefficient $1$ is called a "monic" polynomial, by the way.) You also know what it means for a n... |
174,938 | <p>Adam and Bob are running on a circular track of circumference 1500 m. They start simultaneously from point A in the same direction. Ratio of their speeds is 5:3 respectively. If they keep running , then at how many different points can they meet?</p>
<p>a)Two b)One c)Three d)Data Insufficient</p>
| Aang | 33,989 | <p>Here, $|\alpha|^2+|z|^2-|\alpha z|^2=|\alpha|^2+|z|^2-|\alpha|^2|z|^2-1+1=1-(1-|\alpha|^2)(1-|z|^2)$. Now, Since $|\alpha|\lt 1$ and $|z|\lt 1\implies (1-|\alpha|^2)(1-|z|^2)\gt 0\implies 1-(1-|\alpha|^2)(1-|z|^2)\lt 1$ $$\implies |\alpha|^2+|z|^2-|\alpha z|^2\lt 1$$ $$\implies |\alpha|^2+|z|^2\lt|\alpha z|^2 + 1$$ ... |
847,093 | <p>Given that $-2\pi≤\theta≤0$ and $\theta$ has a reference angle of $\cfrac{\pi}{6}$ , find $\theta$ if it is in the</p>
<p>a) 1st quadrant</p>
<p>b) 2nd quadrant</p>
<p>c) 3rd quadrant</p>
<p>d) 4th quadrant</p>
<p>I need help on this problem which i'm unfamiliar with negative in radian..</p>
| 1233dfv | 102,540 | <p>Since $G$ is bipartite we can split the vertex set of $G$ into two sets, say $x$ and $y$. We also know that each edge of $G$ has one end in $x$ and one end in $y$. So since $G$ is $k$-regular we have $k|x|=k|y|$ edges in $G$ which implies that $|x|=|y|$.</p>
|
4,473,584 | <p>As per the title, im asked to Prove that <span class="math-container">$p \implies (q \lor (\neg r \implies p)) \equiv q \lor r$</span>.</p>
<p>However, im stuck on this logical equivalence question, my working so far is as follows;</p>
<p><span class="math-container">$$p \implies (q \lor (\neg r \implies p)) \equiv ... | Hamdiken | 728,796 | <p><span class="math-container">\begin{align}
p \implies (q \lor (\neg r \implies p)) &\equiv \neg p \lor (q \lor r\lor p) \hspace{0.4cm} \text{Using }(p\implies q)\iff (\neg p\lor q)\\
&\equiv (\neg p \lor p) \lor (q\lor r) \hspace{0.4cm} \text{Associativity and commutativity of }\lor\\
\end{align}</span></p>
... |
4,473,584 | <p>As per the title, im asked to Prove that <span class="math-container">$p \implies (q \lor (\neg r \implies p)) \equiv q \lor r$</span>.</p>
<p>However, im stuck on this logical equivalence question, my working so far is as follows;</p>
<p><span class="math-container">$$p \implies (q \lor (\neg r \implies p)) \equiv ... | ryang | 21,813 | <blockquote>
<blockquote>
<p><span class="math-container">$p \implies (q \lor (\neg r \implies p)) \equiv q \lor r$</span></p>
</blockquote>
<p>im stuck on this logical equivalence question</p>
</blockquote>
<p>Neither <span class="math-container">$$P → \bigg((Q ∨ (¬R → P)) ↔ Q ∨ R\bigg)\tag1$$</span> nor <span class="... |
2,330,438 | <blockquote>
<p>$$x\frac{dy}{dx}=x^2 +y$$ </p>
</blockquote>
<p>given that $\\ y\left( 1 \right) =0$</p>
<p>When i got partial derivatives of both sides, found it's not an exact equation..please can anybody can give a clue to solve this..</p>
| haqnatural | 247,767 | <p>$$x\frac { dy }{ dx } =x^{ 2 }+y\\ xdy-ydx={ x }^{ 2 }dx\\ \frac { xdy-ydx }{ { x }^{ 2 } } =dx\\ \\ d\left( \frac { y }{ x } \right) =dx\\ \frac { y }{ x } =x+C\\ y=x\left( x+C \right) \\ $$ and since $y(1)=0$ we get from that $$y\left( x \right) =x\left( x+C \right) \\ y\left( 1 \right) =0\\ 1+C=0\\ C=-1$$
finall... |
2,330,438 | <blockquote>
<p>$$x\frac{dy}{dx}=x^2 +y$$ </p>
</blockquote>
<p>given that $\\ y\left( 1 \right) =0$</p>
<p>When i got partial derivatives of both sides, found it's not an exact equation..please can anybody can give a clue to solve this..</p>
| John Wayland Bales | 246,513 | <p>Two good methods of solution have been given. Here is a third using an integration factor.</p>
<p>Since the initial equation is not exact we can check to see whether either</p>
<ol>
<li><p>$\dfrac{\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}}{N}$ is a function of $x$ alone or whether </p></li>
<li... |
2,282,063 | <p>This is Exercise 2.6.11 of Howie's <em>"Fundamentals of Semigroup Theory"</em>.</p>
<h2>The Details:</h2>
<p>Let <span class="math-container">$S$</span> be a semigroup.</p>
<blockquote>
<p><strong>Definition 1:</strong> We say <span class="math-container">$S$</span> is <em>regular</em> if for all <span cla... | Asinomás | 33,907 | <p>Let $e$ be the unique idempotent.</p>
<p>For every $c$ there is an $x$ so that $cxc=c$, this means $cx=cxcx$, we conclude $cx=e$, and so $c=cxc=ec$.</p>
<p>Hence $e$ is a left identity, you can show analogously that $e$ is right identity, so $e$ is identity.</p>
<p>We also showed that $cx=e$ and analogously $xc=e... |
3,816,808 | <p>I was solving some functions problems and those exercises asked for stating the domain and range of the functions. In this process, I had my doubts about the function notation. I would like something to relate the Domain and Range. Considering the function <span class="math-container">$f$</span> I've seen notations ... | morrowmh | 732,532 | <p>Let <span class="math-container">$A$</span> and <span class="math-container">$B$</span> be sets. The notation <span class="math-container">$f:A\to B$</span> says precisely that <span class="math-container">$f$</span> is a function whose domain is <span class="math-container">$A$</span> and whose codomain is B. This,... |
3,506,899 | <p>I need to find the solutions of the following equations:</p>
<p><span class="math-container">$$\ddot{x}=-g+\alpha\dot{x}$$</span></p>
<p>and
<span class="math-container">$$\ddot{x}=-g+\alpha\dot{x}^2$$</span></p>
<p>Considering <span class="math-container">$g$</span> and <span class="math-container">$\alpha$</sp... | Izaak van Dongen | 473,276 | <p>Well, you can always just try some functions to see if it works. What happens if you let <span class="math-container">$f(x) = x$</span> and <span class="math-container">$g(x) = x$</span>?</p>
<p>In general, it is true that
<span class="math-container">\begin{equation*}
\int uv' \,\mathrm dx = uv - \int u'v\,\mathr... |
1,266,507 | <p>Let $K_{a,b}$ be the complete bipartite graph. Show that
$K_{a,b}$ is a tree if and only if $a = 1$ or $b = 1$.</p>
<p>The way my professor showed us for a complete graph is as below. I just don't know how to start for a complete bipartite graph. </p>
<blockquote>
<p>$K_a$ is a tree if and only if $a=2$ or $a=1$... | advocateofnone | 77,146 | <p>Let the first set have nodes $v_1,v_2,...v_a$ and the second set have nodes $w_1,w_2,....w_b$. </p>
<ol>
<li>Now if both $a$ and $b$ are greater than one then you always find a cycle in the graph , take the minimal case $a=b=2$, then you would have a cycle $(v_1,w_1),(w_1,v_2),(v_2,w_2),(w_2,v_1)$ , here $(i,j)$ me... |
3,766,918 | <p>In this section, Strang converts the constant-coefficient differential equation into linear algebra in order to solve them. I was in trouble reading the difference equation in this section which demands to provide an alternative solution to Example 3 which is a second differential equation of the motion around a cir... | ssuraj | 1,006,332 | <p>I believe this is how they derived the relation between matrices <span class="math-container">$U_{n+1}$</span> and <span class="math-container">$U_n$</span> <br />
First Consider Finite difference stencil for the first derivative <span class="math-container">$$y^{'}_{n} = \frac{y_{n+1} - y_{n}}{\Delta t}$$</span>
<s... |
3,420,677 | <p>I have been studying on cubic equations for a while and see that the cubic equation needs to be in the form of <span class="math-container">$mx^3+px+q=0$</span> so that we can find the roots easily. In order to obtain such an equation having no quadratic term for a cubic equation in the form of <span class="math-con... | Arnaud Mortier | 480,423 | <p>There is not one relation, but two.</p>
<p>You noticed the first one: <span class="math-container">$y''$</span> is equal to zero exactly once, and this happens at the centre of symmetry of the curve, which is also its only inflection point. If you look at the abscissa (<span class="math-container">$x$</span>-value)... |
239,127 | <p>Can the series $\dfrac1{i-0} + \dfrac1{i-1} + \dfrac1{i-2} + \cdots$ be reduced to $\log(i)$?</p>
<p>It looked similar to the harmonic series, so I checked wikipedia for Harmonic series, and found the following info. </p>
<p><a href="https://i.stack.imgur.com/L9Gty.png" rel="nofollow noreferrer"><img src="https://... | Berci | 41,488 | <p>It's just a partial answer, I think the answer is <em>no</em>, it should be considered in som other way..</p>
<p>We have $\displaystyle\frac1{1-x}=-\sum_nx^n\ $ (if $|x|<1$), hence $\log(1-x)=\displaystyle\sum_{n\ge 1}\frac{x^n}n$. </p>
<p>Plugging in $x=1$, we arrive to $\displaystyle\sum_{n\ge 1}\frac1n=\log ... |
239,127 | <p>Can the series $\dfrac1{i-0} + \dfrac1{i-1} + \dfrac1{i-2} + \cdots$ be reduced to $\log(i)$?</p>
<p>It looked similar to the harmonic series, so I checked wikipedia for Harmonic series, and found the following info. </p>
<p><a href="https://i.stack.imgur.com/L9Gty.png" rel="nofollow noreferrer"><img src="https://... | xavierm02 | 10,385 | <p>$\cfrac{1}{i-n} \sim \cfrac{-1}{n}$</p>
<p>So $\sum \cfrac{1}{i-n}$ converges $\Leftrightarrow$ $\sum \cfrac{-1}{n}$ converges</p>
<p>But as you know, $\sum\limits_{k=1}^n \cfrac{-1}{n}\xrightarrow{n \rightarrow +\infty} - \infty$</p>
<p>$\sum\limits_{k=1}^n \cfrac{1}{i-n}$ doesn't converge.</p>
<hr>
<p>Another... |
239,127 | <p>Can the series $\dfrac1{i-0} + \dfrac1{i-1} + \dfrac1{i-2} + \cdots$ be reduced to $\log(i)$?</p>
<p>It looked similar to the harmonic series, so I checked wikipedia for Harmonic series, and found the following info. </p>
<p><a href="https://i.stack.imgur.com/L9Gty.png" rel="nofollow noreferrer"><img src="https://... | GregRos | 47,775 | <p>No. A finite series of partial sums reduces to the special <a href="http://en.wikipedia.org/wiki/Digamma_function" rel="nofollow">digamma function</a>. It is not possible to express this function in terms of elementary functions</p>
<p>$$\sum_{x=0}^n{\frac{1}{a-x}}=-\sum_{x=0}^n{\frac{1}{x-a}}=~ψ(-a) - ψ(n-a+1)$$</... |
2,167,265 | <p>Zeno, a follower of Parmenides, reasoned that any unit of space or time is infinitely divisible or not. If they be infinitely divisible, then how does an infinite plurality of parts combine into a finite whole? And if these units are not infinitely divisible, then calculus wouldn't work because $n$ couldn't tend t... | dgnuff | 464,005 | <p>Consider the following thought experiment, which I will frame in terms that Zeno would most likely have understood.</p>
<p>Take an urn, and fill it with water. It will contain a finite amount of water.</p>
<p>Now, transfer half the water in the urn to a second urn.</p>
<p>Then transfer half the water in the seco... |
53,001 | <p>Consider one of the standard methods used for defining the <a href="http://en.wikipedia.org/wiki/Riemann_integral" rel="noreferrer">Riemann integrals</a>: </p>
<blockquote>
<p>Suppose $\sigma$ denotes any subdivision $a=x_0<x_1<x_2\cdots<x_{n-1}<x_n=b$, and let $x_{i-1}\leq \xi_i\leq x_i$. Then if
... | minimalrho | 13,352 | <p>One way of thinking about it is that you have a function defined on the set of partitions of $[a,b]$ into the real numbers called the Riemann sum. You put an order on partitions by defining the notion of mesh ($|\sigma|$ in your notation) and defining an order on the set of partitions by $\sigma\succeq\tau$, if and ... |
53,001 | <p>Consider one of the standard methods used for defining the <a href="http://en.wikipedia.org/wiki/Riemann_integral" rel="noreferrer">Riemann integrals</a>: </p>
<blockquote>
<p>Suppose $\sigma$ denotes any subdivision $a=x_0<x_1<x_2\cdots<x_{n-1}<x_n=b$, and let $x_{i-1}\leq \xi_i\leq x_i$. Then if
... | Toby Bartels | 63,003 | <p>Besides taking the limit of a function, you can take the limit of any relation, thought of as a multi-valued function. Recall that <span class="math-container">$ \lim _ { x \to x _ 0 } y = L $</span>, where <span class="math-container">$ y = f ( x ) $</span> for some function <span class="math-container">$ f $</spa... |
1,025,499 | <p>The unit circle is defined to be <span class="math-container">$x^2 + y^2 = 1$</span>. Makes sense. Its an equation of a circle. Now from here if we think about cosine as the <span class="math-container">$x$</span> value and sine as the <span class="math-container">$y$</span> value then we get a trig identity most of... | Travis Willse | 155,629 | <p>Another way to say this is that the map $\theta \mapsto (\cos \theta, \sin \theta)$ <em>parameterizes</em> the unit circle. (The fact that the image of this map is contained in the unit circle is exactly a consequence of the identity you mention.)</p>
<p>Geometrically, the point corresponding to $\theta$ is the int... |
1,025,499 | <p>The unit circle is defined to be <span class="math-container">$x^2 + y^2 = 1$</span>. Makes sense. Its an equation of a circle. Now from here if we think about cosine as the <span class="math-container">$x$</span> value and sine as the <span class="math-container">$y$</span> value then we get a trig identity most of... | AJY | 192,914 | <p>If we see $( \cos \theta, \sin \theta)$ as a point in $\mathbb{R}^{2}$, then we know that the distance from $(0, 0)$ to $( \sin \theta, \cos \theta)$ is (according to the Pythagorean Theorem)</p>
<p>\begin{align*}
\sqrt{(\cos \theta - 0)^{2} + (\sin \theta - 0)^{2}} & = \sqrt{(\cos \theta)^{2} + (\sin \theta)^{... |
1,329,112 | <p>Given a measurable $E\subset \Bbb R^d $ and a measurable function $f:E\rightarrow \Bbb R^d $, prove that :</p>
<p>$$
\int (\left\lvert f \right\rvert)^r d\mu = r\int_{0}^\infty t^{r-1} \mu(\{x \in E \mid \left\lvert f(x) \right\rvert>t\})\,dt
$$<br>
where $r \ge 1$</p>
<p>The $r=1$ case is simple; howe... | David C. Ullrich | 248,223 | <p>You can derive the general case from the case $r=1$. Because $|f|>t$ if and only if $|f|^r>t^r$.</p>
|
4,371 | <p>I would like to learn Graph Theory from the beginning. It seems to me that one does not need to be familiar with many abstract type subjects to be able to understand the more basic concepts of graphs.</p>
<ol>
<li><p>Which subjects should one know prior to learn Graph Theory at the introductory level?</p></li>
<li>... | Community | -1 | <p>For easy read:</p>
<blockquote>
<ol>
<li><p>Combinatorics and Graph Theory by Harris</p></li>
<li><p>Introduction to Graph Theory by Wilson</p></li>
</ol>
</blockquote>
|
1,541,859 | <p>Trying to find the sum of the following infinite series:</p>
<p>$$ \displaystyle\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{(2n-1)3^{n-1}}$$</p>
<p>Any ideas on how to find this sum?</p>
| Michael Hardy | 11,667 | <p>$$
\frac{x^{n-1}}{2n-1} = \frac{y^{2n-2}}{2n-1} \quad \text{where }y=\sqrt x
$$
and
$$
\frac d {dy}\, \frac{y^{2n-1}}{2n-1} = y^{2(n-1)} = x^{n-1}.
$$
$$
\sum_{n=1}^\infty \frac{x^{n-1}}{2n-1} = \sum_{n=1}^\infty \frac{y^{2n-1}}{2n-1}.
$$
The derivative of this with respect to $y$ is
$$
\sum_{n=1}^\infty y^{2n-1} = ... |
287,203 | <blockquote>
<p>Prove the statement:</p>
<p><span class="math-container">$\forall n \in \mathbb{N}$</span>,<span class="math-container">$\forall m \in \{2, 3,...,floor(\sqrt{n})\}$</span>, <span class="math-container">$m$</span> does not divide <span class="math-container">$n \implies n$</span> is prime</p>
</blockquot... | user55514 | 55,514 | <p>If n is not prime, then we only need to prove n is a product of 2 factors, one of them prime-because trial dividing by primes is faster than trial dividing by all the natural numbers- but the other factor can be composite or prime. </p>
|
4,533,858 | <h2>Problem :</h2>
<p>Show that :</p>
<p><span class="math-container">$$\frac{1}{2\ln2}\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right)>\sqrt{2}-1$$</span></p>
<hr />
<hr />
<p>Using some approximation using itself algoritm found here (<a href="https://en.wikipedia.org/wiki/Methods_of_computing_square_roots" rel="nofollo... | Enrico M. | 266,764 | <p>A nice trick for your last question is this: take the exponential of both terms, that is</p>
<p><span class="math-container">$$e^{\ln(2)} < e^{1/\sqrt{2}}$$</span> that is</p>
<p><span class="math-container">$$2 < e^{1/\sqrt{2}}$$</span></p>
<p>Now if <span class="math-container">$x$</span> is small enough, yo... |
419,807 | <p>Given a homomorphism of rings <span class="math-container">$S \rightarrow R$</span>, for a pair of <span class="math-container">$R$</span>-modules <span class="math-container">$M, N$</span> the machinery of relative homological algebra defines relative <span class="math-container">$Ext$</span>-groups</p>
<p><span cl... | Mark Grant | 8,103 | <p>This is only a partial answer to question 1. In case <span class="math-container">$S=\mathbb{Z}H$</span> and <span class="math-container">$R=\mathbb{Z}G$</span> are group rings, and <span class="math-container">$S\to R$</span> is induced by an inclusion <span class="math-container">$H\leq G$</span> of groups, there ... |
4,443,941 | <p>By the Lagrange Inversion Theorem, one can derive the series expansion for the principal branch of <span class="math-container">$W_0(x)$</span>:
<span class="math-container">$$W_0(x)= \sum_{n=1}^{\infty} \frac{(-n)^{n-1}x^n}{n!}, \, |x| \leq \frac1e$$</span></p>
<p>For <span class="math-container">$x \in \mathbb{R}$... | GEdgar | 442 | <p>Let <span class="math-container">$y=xe^x$</span>; so <span class="math-container">$x=W_{-1}(y)$</span>, where <span class="math-container">$y\to 0^-$</span> as <span class="math-container">$x\to-\infty$</span>. Note
that <span class="math-container">$y=0$</span> is a logarithmic branch point for <span class="math-c... |
4,443,941 | <p>By the Lagrange Inversion Theorem, one can derive the series expansion for the principal branch of <span class="math-container">$W_0(x)$</span>:
<span class="math-container">$$W_0(x)= \sum_{n=1}^{\infty} \frac{(-n)^{n-1}x^n}{n!}, \, |x| \leq \frac1e$$</span></p>
<p>For <span class="math-container">$x \in \mathbb{R}$... | Tyma Gaidash | 905,886 | <p>Here are <span class="math-container">$2$</span> series expansions:</p>
<p><span class="math-container">$$\text W_{-1}(z)=\ln(z)-2i\pi-\ln(\ln(z)-2i\pi)-\sum_{n=1}^\infty\sum_{m=1}^n\frac{\text S_n^{(m-n+1)}\ln^m(\ln(z)-2i\pi)}{(2\pi i-\ln(z))^nm!} $$</span></p>
<p>from:</p>
<p><a href="https://functions.wolfram.com... |
10,730 | <p>While the question is stated with reference to the iPhone, my actual question is about phones in general. Just as there was much talk about the use of Computers in the classroom over the past fifty years (or so), how much talk is there about the use of cellphones in the classroom?</p>
<p>What benefits are there to ... | Mikhail Katz | 1,385 | <p>I have had very good results with banning both iphones and computers in the classroom. Usually I announce that university regulations prohibit the use of iphones, computers, guns, pistols, and invite the students to close them all. Both the attention span and the grades go up.</p>
|
298,179 | <p>Are these sets conformally equivalent, That is, is there a conformal bijection between them. $S_1=\{z\in\mathbb C\mid0<|z|<1\}$ and $S_2=\{z\in \mathbb C\mid1<|z|<2\}$?<br/><br/>The only thing that I could find is a theorem of (F.H. Schottky, 1877) see <a href="https://www.google.ca/search?q=F.H.+Schottk... | Qiao | 21,943 | <p>I like the suggestion by Meyer. Suppose $f: S_1 \rightarrow S_2$ is biholomorphic. Then it is bounded and has a removable singularity at $0$. Therefore, $f$ can be extended to a analytic map on the unit disk centered at the origin. This is a contradiction as no point near $S_2$ can be the image of $0$.</p>
|
2,172,949 | <p>I have the integral:
$$\int _{\partial D(a, r)} \frac{e^z}{z^3 + 2z^2 + z} dz$$</p>
<p>which I have to find for different cases:</p>
<p>1 - $a = 0$ and $r =1/2$</p>
<p>2 - $a = -i - 1$ and $r = 1/2$</p>
<p>3 - $a = -1$ and $r = 1/2$</p>
<p>4 - $a = 0$ and $r = 2$</p>
<p>My attempt is this:</p>
<p>$$ \frac{e^z... | xunitc | 412,134 | <p>if $a=0$ and $r=2$, look the $f(z) = \frac{e^z}{(z+1)^2}$, it has singularity $z=-1$ in $D(a,r)$, so can not use Cauchy formula.</p>
|
3,455,482 | <p>I'm stuck trying to solve an inequality: Let <span class="math-container">$X \in \mathcal{L}^{2}$</span>, then </p>
<p><span class="math-container">$$\mathbb{E}[\vert X-\mathbb{E}[X\mid \mathcal{G}]\vert^{2}] \leq \mathbb{E}[\vert X - \mathbb{E}[X]\vert^{2}].$$</span></p>
<p>My attempts:</p>
<p>First I (mistakenl... | Kavi Rama Murthy | 142,385 | <p>Let <span class="math-container">$Y=E(X|\mathcal G)$</span>. Then <span class="math-container">$E(X-EX)^{2}=E((X-Y)+(Y-EX))^{2}$</span>. Expand this as <span class="math-container">$E(X-Y)^{2}+E(Y-EX)^{2}+2E(X-Y)(Y-EX) \geq E(X-Y)^{2}+2E(X-Y)(Y-EX)$</span> If we show that the last term is <span class="math-container... |
2,863,265 | <p>Let a and b be elements of a group, with $a^2 = e, b^6 = e $ and $ ab =b^4 a $ . Find the order of ab, and
express its inverse in each of the forms $a^mb^n$ and $b^ma^n?$</p>
<p>Though it seems very simple I'm unable to find a power such that $(ab)^x =e$. Please help.</p>
<p>Im all confused with this problem.</p>... | Claudius | 218,931 | <p>From $b^6=e$ and $ab = b^4a$ we can conclude $b^3 = e$: we have
$$
b^3 = a^{-1}(aba^{-1})^3a = a^{-1}(b^4)^3a = a^{-1}b^{12}a = a^{-1}ea = e.
$$
Hence, $ab = b^4a = ba$, i.e. $a$ and $b$ commute. Now, it is easy to see that $(ab)^6 = e$. From this, we can only conclude $\lvert ab\rvert \in \{1,2,3,6\}$.</p>
<p>For ... |
1,242,541 | <p>I am trying to prove the following, and here is what I have done:
Can somebody help to complete this?</p>
<p>$2^n \ge n^2$ for $n\ge 4$</p>
<p>$n=4$, LHS: $2^4 = 16$, RHS: $4^2=16$, $16=16$ Therefore TRUE</p>
<p>Assume True for $n=k$, for $k\ge 4$</p>
<p>$2^k \ge k^2$</p>
<p>Should be true for $n=k+1$ for $k\ge... | Daniel W. Farlow | 191,378 | <p>mathlove's answer addresses the core part of the induction step. It seems like you're still struggling though; thus, I'll try to flesh things out a bit more. Let me know if a step(s) does not make sense.</p>
<hr>
<p>For $n\geq 4$, let $S(n)$ denote the statement
$$
S(n) : 2^n\geq n^2.
$$
<strong>Base step ($n=4$):... |
597,680 | <p>Calculate $|A|$ with $A=\{ F \subset \mathcal P(\mathbb N): F \text{ is a partition of } \mathbb N\}$. A partition of $\mathbb N$ is a family of subsets $F=\{P_i\}_{i \in I} \subset \mathcal P(\mathbb N)$, non-empty such that $\bigcup_{i \in I} P_i= \mathbb N$ and $P_i \cap P_j= \emptyset$</p>
<p>My attempt at a so... | Clive Newstead | 19,542 | <p>You've done the hard part.</p>
<p>$A$ has as elements countable sequences from $\mathcal{P}(\mathbb{N})$, i.e. it's a subset of $\mathcal{P}(\mathbb{N})^{\mathbb{N}}$, so all you need to do is exhibit bijections
$$\left(\{0,1\}^{\mathbb{N}}\right)^{\mathbb{N}} \cong \{ 0, 1 \}^{\mathbb{N} \times \mathbb{N}} \cong \... |
19,069 | <p>In this weird pandemic school year, I'm doubly interested in technology integration to help my virtual (high school) students as much as my in-person students. I've been particularly eager to get that working with compass-and-straightedge constructions. Obviously, students need a little familiarity with holding a ... | Brad Franklin | 15,192 | <p>I think <a href="https://www.robocompass.com/" rel="nofollow noreferrer">https://www.robocompass.com/</a> fits the bill... if you are ok with something code-driven, which for me is just an amazing bonus for my students and for exposition.</p>
<p>So, the way in which robocompass deviates from rusty compass is directl... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.