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 |
|---|---|---|---|---|
2,239,956 | <p>As we can always hear that topology is related to graph theory. But I can only read some conceptual relations between them.
Technically, is it possible to write down a topology $\mathscr{T}_V$ (collection of open sets) of any given graph $V$ to consider the graph as a topological space?</p>
| D1811994 | 173,609 | <p>The response is affirmative. One way to assign a topology to a graph is to study the graph as a CW-complex (or even a $\Delta$-complex). You can look at definitions of these spaces in Hatcher's book, freely available online (<a href="https://www.math.cornell.edu/~hatcher/AT/ATpage.html" rel="nofollow noreferrer">her... |
2,379,932 | <p>Since the the powerset is defined as $\mathcal {P}(A)=\{x\mid x\subseteq A\} $, it seems obvious that $A \subseteq \mathcal P(A)$. But I don't know how to show it formally.<br>
Is it okay to say that $x\subseteq A$ (definition of a powerset), thus $A\subseteq \mathcal P(A)$?</p>
<p>It seems trivial, yet I don't kno... | ajotatxe | 132,456 | <p>Let $A=\{1\}$. Then $\mathcal P(A)$ has two elements: $\emptyset$ and $\{1\}$. But $A$ has one element: $1$. This element is not an element of $\mathcal P(A)$, so $A\not\subseteq\mathcal P(A)$. Note that $\{1\}$ and $1$ are not equal.</p>
|
1,410,498 | <p>How would I prove this using mathematical induction:
$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$.</p>
<p>Here is what I have tried:</p>
<p>$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$</p>
<p>Base case: $\dfrac{6^6}{3^6} < 6!$. Yes, this is true.</p>
<p>Assume for $k$ it's true so $\dfrac{k^k}{3^k} < k!$... | lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>If $m!>\dfrac{m^m}{3^m}$</p>
<p>$\implies(m+1)!>m\cdot\dfrac{m^m}{3^m}$</p>
<p>It is sufficient to show $m\cdot\dfrac{m^m}{3^m}>\dfrac{(m+1)^{m+1}}{3^{m+1}}$</p>
<p>$\iff3>\left(1+\dfrac1m\right)^{m+1}$ </p>
<p>Now $\left(1+\dfrac1m\right)^{m+1}>\left(1+\dfrac1{m+1}\right)^{m+1}$</p>... |
1,410,498 | <p>How would I prove this using mathematical induction:
$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$.</p>
<p>Here is what I have tried:</p>
<p>$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$</p>
<p>Base case: $\dfrac{6^6}{3^6} < 6!$. Yes, this is true.</p>
<p>Assume for $k$ it's true so $\dfrac{k^k}{3^k} < k!$... | user84413 | 84,413 | <p>We will use induction to show that $\;\;\color{red}{3^n n!>n^n}$ for all $\color{blue}{n\ge1}$:</p>
<p>1) This is true for $n=1$, since $3>1$.</p>
<p>2) Let $n\in\mathbb{N}$ with $3^n n!>n^n$. Then $3^{n+1}(n+1)!=3(n+1)\big(3^n n!\big)>3(n+1)\big(n^n\big)$,</p>
<p>$\;\;\;$and $3(n+1)(n^n)>(n+1)^{n... |
2,328,775 | <p>let $f(x,y)$ be a homogeneous function of order 2. What is the order of
$-x\cdot (f(x,y))^{2}$?</p>
<p>More generally: </p>
<blockquote>
<p><strong>Q.</strong> How do you determine the order of the product of two homogeneous functions, and what happens when you multiply a homogeneous function by a variable such... | Community | -1 | <p>The associated polynomial is:</p>
<p>$$r^2 - 2r + 2 = (r-1)^2 +1$$</p>
<p>So, for the roots we get:</p>
<p>$$(r-1)^2 = -1 \iff r-1 = \pm i \iff r = 1 \pm i$$</p>
<p>So, you should change your homogeneous equation and particular suggestion accordingly. If you have any problems, let me know and I will elaborate.<... |
1,339,390 | <p>I know that Szemerédi's theorem states that any set of integers with positive natural density contains arbitrary long arithmetic progressions. However, does this imply that such a set contains an <em>infinite</em> arithmetic progression? It seems to me that they are not exactly equivalent because the longer progre... | Brian M. Scott | 12,042 | <p>For $n\in\Bbb Z^+$ let $A_n=\{n!-k:0\le k<n\}$, and let $S=\Bbb Z^+\setminus\bigcup_{n\in\Bbb Z^+}A_n$. Since $S$ contains arbitrarily long gaps, $S$ cannot contain an infinite arithmetic sequence. However,</p>
<p>$$\frac{|S\cap[n!]|}{n!}=1-\frac{n(n+1)}{2n!}=1-\frac{n+1}{2(n-1)!}$$</p>
<p>for $n>1$, so $S$ ... |
3,429,075 | <blockquote>
<p>Evaluate <span class="math-container">$\lim\limits_{x\to 0} \cot ^2 (x)-\dfrac{1}{x^2}.$</span> </p>
</blockquote>
<p>I was thinking of using L'Hôpital's Rule, but things got very, very ugly so I wasn't able to solve it. I know from experimentation that the limit is <span class="math-container">$-\df... | Quanto | 686,284 | <p>The L'Hôpital's rule can still be applied as shown below,</p>
<p><span class="math-container">$$\lim\limits_{x\to 0} \left(\>\cot ^2 x-\dfrac{1}{x^2}\right)=\lim\limits_{x\to 0} \left(\frac1{\tan^2 x}-\dfrac{1}{x^2} \right)$$</span>
<span class="math-container">$$=\lim\limits_{x\to 0}\frac{x-\tan x}{x^2\tan x}\c... |
4,515,063 | <p>I am working my way through lecture notes from an econometrics course taught at Ohio State University on time series analysis. The current set of notes I am on (<a href="https://www.asc.ohio-state.edu/de-jong.8/note2.pdf" rel="nofollow noreferrer">https://www.asc.ohio-state.edu/de-jong.8/note2.pdf</a>) discusses ARM... | Quertiopler | 1,056,750 | <p>There is already an answer to this question, but since I also had trouble understanding AR, MA, and ARMA models when I first learned about them, I want to share what helped me understand the models better:</p>
<p>Consider the equation <span class="math-container">$$\Psi(L)x_t = \Xi(L)\epsilon_t,$$</span> where <span... |
3,317,887 | <p>domain and support are not the same thing.</p>
<p>Per <a href="http://mathworld.wolfram.com/Domain.html" rel="nofollow noreferrer">mathworld</a></p>
<blockquote>
<p>The term <code>domain</code> is most commonly used to describe the set of values
for which a function (map, transformation, etc.) is defined. For ... | Joe | 693,577 | <p>Technically, I believe the support is the smallest closed subset of the domain containing all points that are not mapped to zero.</p>
<p>But yes, both the domain and the support for the normal distribution are all real numbers.</p>
|
2,394,582 | <blockquote>
<p>Show that $$f(x) = \ln \left(x+\sqrt{x^2+1}\right)$$ is an odd function.</p>
</blockquote>
<p>My attempt:</p>
<p>$$f(-x)=\ln\left(-x+\sqrt{(-x)^2+1}\right)=\ln\left(-x+\sqrt{x^2+1}\right).$$</p>
<p>How should I proceed? I know that if $f(-x)=-f(x)$, the function is odd. </p>
| MAN-MADE | 422,850 | <p>Note that $\sqrt{1+x^2}$ is even function.</p>
<p>Now, $$\int_{-x}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\int_{0}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\ln(x+\sqrt{1+x^2})$$</p>
<p>Again, $$\int_{-x}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\int_{-x}^{0}\frac{1}{\sqrt{1+t^2}}dt=-2\ln(-x+\sqrt{1+x^2})$$</p>
<p>Hence $\ln(x+\sqrt{1+x^2})=\ln(-x... |
2,394,582 | <blockquote>
<p>Show that $$f(x) = \ln \left(x+\sqrt{x^2+1}\right)$$ is an odd function.</p>
</blockquote>
<p>My attempt:</p>
<p>$$f(-x)=\ln\left(-x+\sqrt{(-x)^2+1}\right)=\ln\left(-x+\sqrt{x^2+1}\right).$$</p>
<p>How should I proceed? I know that if $f(-x)=-f(x)$, the function is odd. </p>
| Peter Szilas | 408,605 | <p>One more:</p>
<p>$f(x) = \ln (x + \sqrt{x^2+1})$, $x \in \mathbb{R}.$</p>
<p>$\star)$ $f(-x) = \ln(-x + \sqrt{x^2 +1}) =$</p>
<p>$\ln (\frac{1}{x+\sqrt{x^2+1}} )=$</p>
<p>$- \ln (x + \sqrt{x^2+1})$. </p>
<p>Combining:</p>
<p>$f(-x) = - f(x)$.</p>
<p>Used : </p>
<p>$ \ln [( -x +\sqrt{x^2+1}) \frac{x + \sqrt{... |
12,394 | <p>There have been a couple questions recently regarding metric spaces, which got me thinking a bit about representation theorems for finite metric spaces.</p>
<p>Suppose $X$ is a set equipped with a metric $d$. I had initially assumed there must be an $n$ such that $X$ embeds isometrically into $\mathbb{R}^n$, but t... | Andrew Stacey | 45 | <p>Since the paper referred to by Hagen Knaf is published by Springer, it may not be available to one and all. The (publicly viewable) MathSciNet reference is: <a href="http://www.ams.org/mathscinet-getitem?mr=355836">MR355836</a>.</p>
<p>It's a very short paper (7 pages) and the main theorem is:</p>
<p><strong>Theo... |
12,394 | <p>There have been a couple questions recently regarding metric spaces, which got me thinking a bit about representation theorems for finite metric spaces.</p>
<p>Suppose $X$ is a set equipped with a metric $d$. I had initially assumed there must be an $n$ such that $X$ embeds isometrically into $\mathbb{R}^n$, but t... | David Eppstein | 440 | <p>It's probably worth pointing out that (unlike for $L_2$) any finite metric space can be embedded into a finite-dimensional $L_\infty$ space with no distortion. Simply map each of the $n$ points of the metric space to the $n$-dimensional vector of distances from all points. (For a closely related construction, see th... |
12,394 | <p>There have been a couple questions recently regarding metric spaces, which got me thinking a bit about representation theorems for finite metric spaces.</p>
<p>Suppose $X$ is a set equipped with a metric $d$. I had initially assumed there must be an $n$ such that $X$ embeds isometrically into $\mathbb{R}^n$, but t... | zouzias | 3,756 | <p>For Question 0 :</p>
<p>For isometric embedding a standard reference is </p>
<p>Deza, M.; Laurent, M. (1997), Geometry of cuts and metrics, Algorithms and Combinatorics, 15, Springer, MR1460488, ISBN 354061611X </p>
<p>For approximate embeddings a beautiful exposition from Matouseks' book:</p>
<p>Lectures on Dis... |
2,724,100 | <p>Let $x, y \in M$ be two different points in a metric space $M$, show that there are two open disjoint sets $U, V$ with $x \in U$ and $y \in V$. Moreover, show that $U$ and $V$ can be chosen such that $cl(U)$ and $cl(V)$ are disjoint.</p>
<p>If $r=d(x,y)$ then the open balls $B_{r/2}(x)$ and $B_{r/2}(y)$ are disjoin... | William Elliot | 426,203 | <p>First show B[r,a] = { x : d(x,a) <= r } is closed.<br>
Then use B(r',x) and B(r',y) with r' < d(x,y)/2.</p>
|
1,840,997 | <p>I'm looking at a question out of Lee's Smooth Manifolds:</p>
<blockquote>
<p>Show that a disjoint union of uncountably many copies of $\Bbb{R}$ is locally Euclidian and Hausdorff but not second countable.</p>
</blockquote>
<p>My question is not about the problem itself, but rather what a common notation would be... | Eric Wofsey | 86,856 | <p>I don't think there's any notation that is more standard than $\coprod_{i\in I}\mathbb{R}$ that makes it clear you are talking about a disjoint union of copies of $\mathbb{R}$. But if you want a nice way to be able to refer to the individual copies, you could just write it as $I\times\mathbb{R}$, with the $i$th cop... |
2,068,545 | <p>The fixed point B has position vector <strong>$b$</strong> relative to a fixed point $O$. A variable point $M$ has position vector <strong>$m$</strong> relative to $O$. Find the locus of M if <strong>m $\cdot$ (m – b)=0</strong>.</p>
<p>I am told the answer is derived as such:
<a href="https://i.stack.imgur.com/kIa... | Michael Seifert | 248,639 | <p>If $O$ and $B$ are two antipodal points on a circle in the plane, and $M$ is another point on the circle, then $\angle OMB = 90^\circ$. In 3-D space, the locus of all such points will be obtained by rotating this circle around the line $OB$, obtaining a sphere.</p>
<p>As an aside, note that you can also solve this... |
2,068,545 | <p>The fixed point B has position vector <strong>$b$</strong> relative to a fixed point $O$. A variable point $M$ has position vector <strong>$m$</strong> relative to $O$. Find the locus of M if <strong>m $\cdot$ (m – b)=0</strong>.</p>
<p>I am told the answer is derived as such:
<a href="https://i.stack.imgur.com/kIa... | hamam_Abdallah | 369,188 | <p>If $\vec{OM}=(x,y,z)$ and $\vec{OB}=(a,b,c)$ then</p>
<p>$$\vec{OM}.\vec{BM}=0\implies$$</p>
<p>$$x(x-a)+y(y-b)+z(z-c)=0 \implies$$</p>
<p>$$(x-\frac{a}{2})^2+(y-\frac{b}{2})^2+(z-\frac{c}{2})^2=\frac{a^2+b^2+c^2}{4}$$</p>
<p>the point M is in the sphere of center $(\frac{a}{2},\frac{b}{2},\frac{c}{2})$ and radi... |
2,582,863 | <p>Bounty ending tomorrow:</p>
<p>Consider the following function defined by two infinite integrals:
$$
F(\epsilon) = \int_0^\infty
\frac{q^3 (1+q)e^{-2q}}{q^4+\epsilon^4} \, \mathrm{d} q
+ \int_0^\infty \frac{\, G_q \, e^{-q}}{q^2(q^4+\epsilon^4)} \, \mathrm{d} q
$$
where
$$
G_q := G \left( \left[ \left[\frac{1}... | H. H. Rugh | 355,946 | <p>You may simultaneously convert it into linear recurrence with a "pure" polynomial RHS by substituting: $b_n^2=4^n a_n$</p>
|
1,153,054 | <p>So I have </p>
<ol>
<li><p>$i=e^{i\pi/2}=e^{5×i\pi/2}$</p></li>
<li><p>If I raise it to the power of $i$,
$i^i=[e^{i\pi/2}]^i=[e^{5×i\pi/2}]^i$</p></li>
</ol>
<p>Hence I have $e^{-\pi/2}=e^{-5\pi/2}$?</p>
<p>Can someone please tell me what I did wrong in my steps?</p>
| xpaul | 66,420 | <p>Hint: divide both the numerator and the denominator by $x^5$.</p>
|
1,153,054 | <p>So I have </p>
<ol>
<li><p>$i=e^{i\pi/2}=e^{5×i\pi/2}$</p></li>
<li><p>If I raise it to the power of $i$,
$i^i=[e^{i\pi/2}]^i=[e^{5×i\pi/2}]^i$</p></li>
</ol>
<p>Hence I have $e^{-\pi/2}=e^{-5\pi/2}$?</p>
<p>Can someone please tell me what I did wrong in my steps?</p>
| Alex Silva | 172,564 | <p>From Taylor's expansions at $x=0$, $\log(1+x^5) \approx x^5$, $e^{x^3}-1 \approx x^3$ and $\sin x^2 \approx x^2$, as $x \rightarrow 0$. Thus,</p>
<p>$$\lim \limits_{x \rightarrow 0} \frac{\log(1+x^5)}{(e^{x^3}-1)(\sin x^2 )} = \lim \limits_{x \rightarrow 0} \frac{x^5}{x^{3} x^{2}} =1.$$</p>
|
2,890,965 | <p>Let's begin with the definition of the inner product. It's as follows:</p>
<p>$$
A . B = ||A|| \ ||B||\cos(\theta) .
$$ </p>
<p>My understanding is that the $\theta$ is the angle between the two vectors. As such this can be used to find the axis of rotation about a line, given that you have the <strong>line</s... | Mauricio Cele Lopez Belon | 485,657 | <p>Derivation of the result:</p>
<p>A rotation matrix $A$ in $R^3$ rotates 3d vectors around an axis of rotation $l$. So that the axis $l$ is not affected by the rotation $A$. In other words, the vector $l$ is an eigenvector of $A$ with eigenvalue equals to $1$ since $A l = l$. </p>
<p>Conversely, vectors that lie i... |
2,940,288 | <p>Suppose that <span class="math-container">$f:[0,1]\to \mathbb{R}$</span> is a continuous function and that <span class="math-container">$f(c)=\min\{f(x):0 \le x \le 1\}$</span> for some <span class="math-container">$c\in(0,1)$</span> . Prove that <span class="math-container">$f(x)=f(y)$</span> for some <span class=... | coffeemath | 483,139 | <p>If <span class="math-container">$f(c)=f(0)$</span> or <span class="math-container">$f(c)=f(1)$</span> done. Otherwise <span class="math-container">$f(c)<f(0)$</span> and <span class="math-container">$f(c)<f(1).$</span> So <span class="math-container">$f$</span> takes on each real value between <span class="mat... |
4,551,007 | <blockquote>
<p>Let <span class="math-container">$A$</span> be a <span class="math-container">$2 \times 2$</span> matrix with <span class="math-container">$-3$</span> and <span class="math-container">$-1$</span> as eigenvalues. The eigenvectors are <span class="math-container">$v_1=[-1,1]$</span> and <span class="math-... | Lutz Lehmann | 115,115 | <p>You misunderstood something from the start. The matrix <span class="math-container">$A$</span> is not explicitly given. From what is known of the general solution, you get
<span class="math-container">$$
\pmatrix{x\\y}=c_1v_1e^{-3t}+c_2v_2e^{-t}.
$$</span>
There is no further differentiation or integration required.... |
4,523,587 | <p>If anyone can help me will be eternally grateful. I'm studying indices and I need a clarification about this problem solved by my teacher
<span class="math-container">$$\frac{10y^{18}}{8y^6}$$</span></p>
<p>and said , "well, <span class="math-container">$10$</span> and <span class="math-container">$8$</span> a... | Nuclear Hoagie | 240,172 | <p>You <em>can</em> divide both the top and the bottom of a fraction by "any old number", so long as that number isn't zero. You are simply multiplying the fraction by <span class="math-container">$(1/X)/(1/X)$</span>, which is <span class="math-container">$1$</span> - multiplying any number by <span class="m... |
132,134 | <p>The Temperley-Lieb algebra has the same generators as the $S_n$ group algebra, and the same commuting relations, but its other relations are different. A nice diagrammatic interpretation can be seen in <a href="http://en.wikipedia.org/wiki/Temperley-Lieb_algebra" rel="nofollow">the Wikipedia article</a>.</p>
<block... | F. C. | 10,881 | <p>There are some ad-hoc definitions for some types. Type B can be defined using diagrams that have a left-right symmetry. Tammo tom Dieck has proposed a definition for type D here: (<a href="http://www.uni-math.gwdg.de/tammo/preprints/1dk.pdf" rel="nofollow">http://www.uni-math.gwdg.de/tammo/preprints/1dk.pdf</a>).</p... |
1,172,576 | <p>I've seen that the <a href="http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation">Cesaro sum</a> is $1/2$ but haven't been able to find the steps for figuring that out.</p>
| robjohn | 13,854 | <p>The Cesàro Sum of a series $\sum\limits_{k=1}^\infty a_k$ is
$$
\begin{align}
\lim_{n\to\infty}\frac1n\sum_{m=1}^n\sum_{k=1}^ma_k
&=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\sum_{m=k}^na_k\\
&=\lim_{n\to\infty}\frac1n\sum_{k=1}^n(n-k+1)a_k
\end{align}
$$
Since $a_k=(-1)^{k-1}$, we get
$$
\begin{align}
&\lim_{... |
1,412,428 | <blockquote>
<p>Is it possible to embed $\mathbb Z^n$ inside $ \mathbb Z^m$ as a $ \mathbb Z$-module for $m < n$ ?</p>
</blockquote>
<p>I think it's not possible. It might be a easy problem for some of you, but I really don't have any clean way to show this. Any ideas?</p>
| Caleb Stanford | 68,107 | <p>Let $m < n$. If $\mathbb{Z}^n$ embeds in $\mathbb{Z}^m$,
then the $n$ linearly independent basis elements of $\mathbb{Z}^n$ map to $n$ linearly independent elements of $\mathbb{Z}^m$.
The fact that they are linearly independent over $\mathbb{Z}$ implies they are linearly independent over $\mathbb{Q}$, in $\mathbb... |
848,565 | <p>someone could explain to me this: </p>
<p>$$\int { \arctan { \left( \frac { 1 }{ { u }^{ 2 } } \right) } } \,du=\int { \frac { \pi }{ 2 } } -\arctan { \left( { u }^{ 2 } \right) } \, du$$</p>
| M. Vinay | 152,030 | <p>$$\boxed{\tan^{-1} \dfrac 1 x = \cot^{-1}x = \dfrac {\pi} 2 - \tan^{-1} x}$$</p>
<p>Proof of first equality: $y = \tan^{-1} \dfrac 1 x \Rightarrow \tan y = \dfrac 1 x \Rightarrow \cot y = x \Rightarrow y = \cot^{-1} x$.</p>
<p>Proof of second equality: $\cot \left(\dfrac {\pi} 2 - \theta\right) = \tan \theta \Righ... |
260,276 | <p>I just started learning about parametric curves and I find it confusing that we have a 3rd variable but this 3rd variable "t" is some imaginary variable....I dont get what the difference is between this parameter "t" and the good ole 3 variable functions...This gets confusing for me especially when I do area in a bo... | mboratko | 17,349 | <p>In your example, you could actually do what you suggest - namely find $x$ bounds for the curve in question - but in order to do that you would have to rewrite the parametric curve in terms of $y$ and $x$ only:</p>
<p>$$x^2=t^2-2+t^{-2}\\
y^2=t^2+2+t^{-2}\\
\Rightarrow y^2=x^2+4.$$</p>
<p>From this point you could ... |
3,485,208 | <p>the reverse of the number <span class="math-container">$129$</span> is <span class="math-container">$921$</span> and these add to <span class="math-container">$1050$</span> which is divisible by <span class="math-container">$30$</span>. How many three-digit numbers have the property that, when added to their reverse... | Alessio K | 702,692 | <p>Any <span class="math-container">$3$</span> digit number can be expressed in the form <span class="math-container">$x=100+10+$</span>, so reversing we number we obtain <span class="math-container">$y=100+10+$</span>.</p>
<p>Now lets add these numbers together to obtain <span class="math-container">$$x+y=100(a+c)+20... |
622,616 | <p>I have attempted a proof of this, but upon looking at my notes, I feel it might be incorrect: it is noticeably simpler than the one in my notes.</p>
<p><strong>Proposition:</strong> any permutation in $S_n$ can be written as a product of disjoint cycles.</p>
<p><em>Proof (attempt):</em> $s\in S_n$. Let $\langle s\... | Ben Grossmann | 81,360 | <p>If, in proving this theorem, you have available the previously explained notion of group-actions and orbits, and if you have defined "cycles" in this manner, then yes your proof is valid. However, in the progression of a typical algebra course, this theorem is covered before those "tools" are available, hence the ne... |
2,323,605 | <p>I am very confused on this question.</p>
<p>How does $(4^{-3} \cdot 2^{-3})^0 = (8^{-3})^0$? Should it not be $8^{-6}$ since $-3+-3 = -6$?</p>
<p>I just don't get where the other power of $-3$ goes off to?</p>
<p><a href="https://i.stack.imgur.com/xC5pY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.img... | lioness99a | 401,264 | <p>The rule you need to use is:</p>
<p>$$a^xb^x=(ab)^x$$</p>
<hr>
<p>Consider $6^4\times 7^4$. </p>
<p>We could write this as \begin{align}6^4\times 7^4&=(6\times 6\times 6\times 6)\times (7\times 7\times 7\times 7)\\
&=(6\times7)\times(6\times7)\times(6\times7)\times (6\times7)\\
&=(6\times7)^4\\
&... |
2,138,631 | <p>If you reverse the binary digits of a multiple of $3$, the result is always a multiple of $3$. The same is not true for $7$, but it does appear to happen more often than $\frac{1}{7}$ of the time. For fun, I'll call a number $n$ such that the reverse of the binary digits of $7n$ represents a multiple of $7$ a "seven... | marty cohen | 13,079 | <p>Here is my reasoning on why
the number of times
the remainders mod $7$
match more than
randomly.</p>
<p>The key things,
to me,
are that
mod $7$,
$2^{3m} \equiv 1,
2^{3m+1} \equiv 2,
2^{3m+2} \equiv 4
$.</p>
<p>Therefore,
if
$m$ is one less than
the number of bits in $n$,
write
$n = \sum_{k=0}^m b_k 2^k
$.</p>
<p... |
4,332,703 | <p>I am reviewing a homework problem that is supposed to be really easy but I have trouble wrapping my head around it.
For <span class="math-container">$j=0, \ldots, n \quad t_{j} \neq t_{i}$</span> if <span class="math-container">$ i \neq j $</span>
we define the <span class="math-container">$n$</span> Lagrange Polyno... | PierreCarre | 639,238 | <p>The interpolating polynomial of <span class="math-container">$f$</span> on the distinct points <span class="math-container">$x_0, \cdots, x_n$</span> is given by
<span class="math-container">$$
p_n(t) = \sum_{i=0}^n f(t_i) L_i(t)
$$</span></p>
<p>In particular, if you take the constant function <span class="math-con... |
4,191,936 | <p>I'm not a mathematician so please forgive me if I have a poor understanding of some math concept.</p>
<p>I see a lot of tutorial on how to generate fractals and Mandelbrot images from a pixel-wise perspective but I would like to know if it is possible to find some kind of parametric equation to see contours of image... | Adam | 44,654 | <p><a href="https://i.stack.imgur.com/nQfVp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nQfVp.png" alt="enter image description here" /></a>
These curves are called <a href="https://commons.wikimedia.org/wiki/File:Lemniscates5.png" rel="nofollow noreferrer">boundaries of level sets, equipotentia... |
1,324,045 | <p>Here $(x_i) \in \ell^2$. All I have on paper right now is
$$\rVert Tx\rVert = \sqrt{\sum_{i=1}^{\infty} \frac{x_i ^2}{i^2}}$$</p>
<p>and I'm not sure about what's next. Please help. </p>
| marty cohen | 13,079 | <p>This (and more)
is all explained
in excruciating detail
in Landau's
"Foundations of Analysis".</p>
<p>You can buy it here:</p>
<p><a href="http://rads.stackoverflow.com/amzn/click/082182693X" rel="nofollow">http://www.amazon.com/Foundations-Analysis-AMS-Chelsea-Publishing/dp/082182693X</a></p>
|
72,220 | <p>I am a graduate student thinking about some stuff.</p>
<p>1) If I have an idea for a result but it is not yet complete, is it advisable to mention it to others (especially more experienced people)? Is it possible that he/she will take it, develop it quicker than I do and publish it? How should one behave in these s... | Hailong Dao | 2,083 | <p>My answer is fairly simple, perhaps due to my own ignorance and inexperience, but since you asked:</p>
<p>There is a non-zero risk. But the rewards for talking and sharing with people about math <em>far</em> outweight this risk. </p>
<p>A more practical advice if you are really worried: I would try to cultivate an... |
236,711 | <p>It is "well known" that a matrix system $Ax=b$ where $A\in \Bbb Z^{m\times n}$, $x\in \Bbb Z^n,b\in\Bbb Z^m$ for some $m,n \in \Bbb N$, can be solved in polynomial time, using Smith/Hermite Normal Form computation. By "solved" I mean either "decide, for a given $A,b$, whether a solution exists" or "decide whether a ... | Dima Pasechnik | 11,100 | <p>An improvement of Storjohann's result is due to
Micciancio and Warinschi,
<a href="http://dl.acm.org/citation.cfm?doid=384101.384133" rel="nofollow">A linear space algorithm for computing the Hermite normal form</a>. </p>
<p>As to recovery of transformation matrices, in the classical algorithm one merely has to tr... |
1,722,670 | <p>In a 52-card deck, Jack is worth $1$ point, Queen $2$ points, King $3$ points, Ace $4$, and everything else is worth $0$.</p>
<p>How to calculate the probability you have a hand with sum $C\geq13$ points? where one hand is $13$ cards.</p>
| Tryss | 216,059 | <p>Another possible way to show the divergence :</p>
<p>For $n > 2$, you have that $n-1\geq \frac{n}{2}$ and that imply that $\frac{n-1}{n^2} \geq \frac{1}{2}\frac{1}{n}$, but $\sum \frac{1}{n}$ diverge, so $\sum_{n-1}{n^2}$ will diverge too</p>
|
4,474,923 | <p>I am confused as to the relation between Markov chains and random walks. Some sources I look at claim all random walks are markov chains and then other sources say the opposite. What makes a markov chain a random walk?</p>
| Vons | 274,987 | <p>Epistemologically speaking, a random walk does not possess the essential properties of a Markov chain: transition probabilities. An instantiation of a random walk is, to say the least, an utterly random path in mathematical space. From this path, we cannot go back and deduce the transition probabilities. However, we... |
763,956 | <p>While playing around with cyclotomic fields, I started to wonder about taking the roots of unity in higher dimensional analogues of the complex plane. Are the roots of unity well defined in the quaternions, octonions, and other hypercomplex algebras? Are there higher dimensional cyclotomic fields, and if so do they ... | Anixx | 2,513 | <p>In tessarines they are well defined. In 4-dimensional tessarines a polynomial of the order <span class="math-container">$n$</span> has <span class="math-container">$n^2$</span> roots (accounting for root multiplicity). This is called fundamental theorem of tessarine algebra.</p>
|
4,114,479 | <p>I would like to make and <strong>visualize</strong> graph library of <span class="math-container">$n$</span> vertices that each vertex can only have maximum of four edges. The reason behind this is that the results will be used for designing chemical libraries where each vertex is a carbon and a carbon can only form... | Igor Rivin | 109,865 | <p>This looks very nice and fast:</p>
<p><a href="https://github.com/takemaru/graphillion#graphillion---fast-lightweight-library-for-a-huge-number-of-graphs" rel="nofollow noreferrer">https://github.com/takemaru/graphillion#graphillion---fast-lightweight-library-for-a-huge-number-of-graphs</a></p>
|
948,306 | <p>Assuming a standard English 26-letter alphabet, and $k<26$, what's the number of $k$-letter words where the letters don't repeat? So for example 'house' has no repeat letters, while 'rolls' has a repeat. </p>
<p>The way I solved this was to first select the $k$ letters which will make up the word, and then count... | David | 175,745 | <p>Another answer is you could take the formula for getting $5$+ consecutive out of $10$ random draws and then just add in the case $4$s ($139$ of those).</p>
<p>The formula for $5$+ consecutive out of 10 random draws is $p^5(6-5p)$ which for p=$0.5$ gives us $7/64$ which is $112/1024$. </p>
<p>Another answer is you... |
3,299,532 | <p>Could you please give me a hint on how to prove the inequality below for <span class="math-container">$k \geq 3$</span>?</p>
<blockquote>
<p><span class="math-container">$$\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2} $$</span> </p>
</blockquote>
<p>Thank you in advance.</p>
| MaximusFastidiousIrreverence | 419,507 | <p>If <span class="math-container">$k \geq 3, $</span> then...</p>
<p><span class="math-container">$$\begin{align} \frac{\sqrt{k(k+1)} }{k-1} &=\frac{\sqrt{k^2+k}}{k-1} \\
&\lt \frac{\sqrt{k^2+k+\frac 1 4}}{k-1}= \frac{k+ \frac 1 2}{k-1}=1+\frac{3}{2k-2} \\
\end{align}$$</span></p>
<p>Have to compare <span cl... |
3,299,532 | <p>Could you please give me a hint on how to prove the inequality below for <span class="math-container">$k \geq 3$</span>?</p>
<blockquote>
<p><span class="math-container">$$\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2} $$</span> </p>
</blockquote>
<p>Thank you in advance.</p>
| Matteo | 686,644 | <p>You have to solve this inequality system:</p>
<p><span class="math-container">$$\left\{\begin{matrix}
& \\k \neq 1 \: \land \: k\neq0
& \\k(k+1)\geqslant0
& \\1+\frac{2}{k}+\frac{3}{4k^2}\geq 0
& \\ \frac{\sqrt{k(k+1)}}{k-1} \geq 0
& \\\left (\frac{\sqrt{k(k+1)}}{k-1}\right )^2\leq \left ... |
3,299,532 | <p>Could you please give me a hint on how to prove the inequality below for <span class="math-container">$k \geq 3$</span>?</p>
<blockquote>
<p><span class="math-container">$$\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2} $$</span> </p>
</blockquote>
<p>Thank you in advance.</p>
| Piquito | 219,998 | <p><span class="math-container">$$\dfrac{\sqrt{k(k+1)} }{k-1} \leq1 + \dfrac{2}{k} + \dfrac{3}{4k^2}=\dfrac{(2k+1)(2k+3)}{4k^2}$$</span>
<span class="math-container">$$\sqrt{k(k+1)}\le\dfrac{(2k+1)(2k+3)(k-1)}{4k^2}=\frac{2k+1}{2}\times\dfrac{(2k+3)(k-1)}{2k^2}$$</span> We know that <span class="math-container">$\sqrt{... |
1,048,379 | <p>The equation is $f(x) = 2^{x-3}$</p>
<p>I know you switch the $x$ and $y$ and then solve for $y$, giving $x = 2^{y-3}$, but I'm not sure how to solve the equation...?</p>
<p>Thanks in advance!</p>
<p>And yes, I'm familiar with logarithms and I know that you would take the log of each side, giving $\ln x = \ln 2^{... | N. F. Taussig | 173,070 | <p>To find the inverse of $f(x) = 2^{x - 3}$, first interchange $x$ and $y$ to obtain </p>
<p>$$x = 2^{y - 3}$$</p>
<p>If we take the logarithm base $2$ of each side, we obtain</p>
<p>\begin{align*}
\log_2 x & = y - 3\\
\log_2 x + 3 & = y
\end{align*}</p>
<p>The inverse is $f^{-1}(x) = \log_2 x + 3$. </p>
... |
3,962,608 | <p>A travel agency offers the chance to its customers to win vouchers for 2 free excursions during the Covid crisis. Out of the x total customers, y will be randomly chosen to participate in the first excursion and z in the second. In how many of all possible draws we will have a total of w customers who will participa... | true blue anil | 22,388 | <p>There are <span class="math-container">$3$</span> distinct groups: <span class="math-container">$(y-w)$</span> for only excursion <span class="math-container">$1$</span>, <span class="math-container">$w$</span> for both excursions, and <span class="math-container">$(z-w)$</span> for only excursion <span class="math-... |
301,259 | <p>Problem : How many ways are there to make $n$ by adding $k$ non-negative integers, where order matters. Suppose $n=4$ and $k=3$. There are 15 solutions using $0, 1, 2, 3, 4$:</p>
<p>$(0,0,4), (0,1,3), (0,2,2), (0, 3, 1), (0, 4, 0), (1, 0, 3), (1, 1, 2), (1, 2, 1), (1, 3, 0), (2, 0, 2), (2, 1, 1), (2, 2, 0), (3, 0, ... | Ross Millikan | 1,827 | <p>You could see <a href="http://en.wikipedia.org/wiki/Composition_%28number_theory%29" rel="nofollow">Wikipedia on compositiions</a> for a simple proof. Note that if you allow zeros, you can covert the problem to one that doesn't allow zeros by adding one to each piece. So your problem of breaking $4$ into $3$ piece... |
301,259 | <p>Problem : How many ways are there to make $n$ by adding $k$ non-negative integers, where order matters. Suppose $n=4$ and $k=3$. There are 15 solutions using $0, 1, 2, 3, 4$:</p>
<p>$(0,0,4), (0,1,3), (0,2,2), (0, 3, 1), (0, 4, 0), (1, 0, 3), (1, 1, 2), (1, 2, 1), (1, 3, 0), (2, 0, 2), (2, 1, 1), (2, 2, 0), (3, 0, ... | ncmathsadist | 4,154 | <p>Do this to see all of the solutions. Imagine you have $n$ characters that are letter o and $k-1$ characters that are letter x. Every solution to your problem is can be represented uniquely by an anagram of the word composed of $n$ os and $k - 1$ xs. </p>
<p>How. Let me show and example If $n=5$ and $k = 3$, t... |
2,282,738 | <p>I was pondering the PMI and there's something bothering me about proofs that I've seen that the well-ordering principle (WOP) and the PMI imply each other. It seems to me that you could use the PMI on a set $S \subseteq \Bbb{Z}$ that has a maximum element $a$. Let $p:S\to\{T,F\}$ be a propositional function on S. Th... | Janet | 966,786 | <p>What if you had an uncountable set that was not well ordered, and you could place the elements in concentric rings. Could you do an induction proof on the indexes of the rings and show that if something were true for elements in ring with index <span class="math-container">$n$</span>, and was also true for elements ... |
3,834,355 | <p><a href="https://i.stack.imgur.com/KqPj7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KqPj7.png" alt="enter image description here" /></a></p>
<p>As shown above, given <span class="math-container">$AB=BC=AC$</span> and <span class="math-container">$BE=CF=AD$</span>, to prove <span class="math-c... | Disintegrating By Parts | 112,478 | <p>Suppose <span class="math-container">$A^mx \ne 0$</span>, but <span class="math-container">$A^{m+1}x=0$</span> for some <span class="math-container">$x\in\mathbb{R}^m$</span> (or <span class="math-container">$x\in \mathbb{C}^m$</span> if you are working with a complex space.) Then <span class="math-container">$\{ x,... |
3,977,672 | <p>I have seen in many examples that using the fundamental definition of derivative only gives the correct results while checking for differentiability at a point. But why do we get wrong results while we directly apply the rules of differentiation and taking the limit at the point. Here is an example</p>
<p><span clas... | Joe | 623,665 | <p>When considering limits, you always have to look at what happens as <span class="math-container">$h$</span> approaches <span class="math-container">$0$</span> both from the left hand side and right hand side. The limit
<span class="math-container">$$
\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
$$</span>
exists if and only i... |
18,771 | <p>Show that if a manifold $M$ is contractible, then every vector bundle over $M$ is equivalent to the trivial bundle.</p>
<p><em>Got this as homework but I'm kind of lost in the hole vector bundle subject, so if anyone could give some hints to this one as well as to understanding the subject, it would be highly appre... | Community | -1 | <p>This may be a repetition of what Mariano and Sean were suggesting: you will most likely
ultimately need to either: produce a bundle morphism between your bundle and
the trivial bundle, or produce a global trivialization of your top space.</p>
|
1,446,439 | <p>Find all cluster points for the sequence $x_{n}$ = The $n$-th rational number</p>
<p>Note: In this problem a labeling of rational numbers by positive integers is used. Such labellings do exist because $\mathbb{Q}$ is countable and we choose any of them. The answer does not depend on the choice.</p>
<p>Def: $c$ is ... | themo123 | 273,091 | <p>As you said Q is dense in R meaning that for any real number there exists a sequence of rationals converging toward that real number, here the sequence converging toward any real number is a sub-sequence of Xn.Now use the definition of the limit of a sequence to conclude </p>
|
1,267,956 | <p>I have the $$\int_{16}^{500} \frac{1}{x^{0.25} - 2} dx,$$ and am trying to find whether it converges or diverges. </p>
<p>I have sketched the graph and noticed that their is an asymptote at $x=16$ (hence why the integral is improper for these boundaries).</p>
<p>I am now trying to evaluate the limits to see if it ... | Hasan Saad | 62,977 | <p>Let $x=t^4$</p>
<p>Then, $dx=4t^3dt$</p>
<p>Now, we have:</p>
<p>$I=\int{ \frac{1}{x^{0.25}-2}dx}=\int{\frac{4t^3}{t-2}dt}$</p>
<p>We use long division to get,</p>
<p>$I=\int{4t^2+8t+\frac{32}{t-2}+16dt}=\frac{4}{3}t^3+4t^2+32\ln(t-2)+16t$</p>
<p>We can ignore the constant of integration since we're going to e... |
2,899,798 | <p>My work so far:</p>
<p>Letting $z = x+iy$, where $x,y$ are real,</p>
<p>then $\Im(z) = y > 0$</p>
<p>$y+1 > 0$</p>
<p>$1/(y+1) > 0$</p>
<p>But I am not sure how to proceed. I also attempted to multiply $-1/(z+1)$ by its conjugate, but that led to $(-x-1+iy)/((x+1)^2 + y^2)$, which I feel is the wrong w... | Angina Seng | 436,618 | <p>Let $w=z+1$. The imaginary part of $w$ is also positive; it lies above the
$x$-axis. Write $w=re^{it}$. We can take $0<t<\pi$. Then $-1/w=(1/r)(-e^{-it})$.
Can you prove $-e^{-it}$ has positive imaginary part.</p>
|
2,594,453 | <p>I'm at the very beginning in algebra and also I don't speak English much. Maybe this question is very simple but I can't understand how to determine parameters in polynomials in order to find out real, rational or integer roots. Here's the polynomial:</p>
<p>$$x^{3} - 33x^{2} + 315x -k$$</p>
| Mostafa Ayaz | 518,023 | <p>If a 3rd-order polynomial has 2 real roots the 3rd one must be so either. Since we generally want two distinct roots there must be one duplex such root. Then we can decompose $f(x)$ as follows:
$$f(x)=(x-a)^2(x-b)$$
By differentiating we have:
$$f^{'}(x)=3x^2-66x+315=3x^2-(a+5b)x+b(a+2b)$$
therefore we get:
$$a+5b=6... |
2,594,453 | <p>I'm at the very beginning in algebra and also I don't speak English much. Maybe this question is very simple but I can't understand how to determine parameters in polynomials in order to find out real, rational or integer roots. Here's the polynomial:</p>
<p>$$x^{3} - 33x^{2} + 315x -k$$</p>
| Community | -1 | <p>If you write the equation as</p>
<p>$$x^3-33x^2+315x=k,$$</p>
<p>you need to find the possible ordinates of an horizontal line that intersects the curve in the LHS exactly twice.</p>
<p>If we look for the extrema,</p>
<p>$$3x^2-66x+315=0$$ yields the solutions $x=7,k=931$ and $x=15,k=675$ (an horizontal by an ex... |
33,821 | <p>The following differentiated implicitly with respect to $\theta$:</p>
<p>$3x = \tan \, \theta $</p>
<p>The book says </p>
<p>$3 dx = \sec^2 \theta \, d \theta$</p>
<p>One could start the calculation like this (I think):</p>
<p>$ \frac{d}{d \theta}3x = \frac{d}{d \theta} \tan \, \theta $</p>
<p>From there I'm n... | Juan S | 2,219 | <p>It is simply that $$\frac{d}{d \theta} \left( 3x \right) = \frac{d}{d \theta} \left( \tan \theta \right)$$</p>
<p>Then we know that $$\frac{d}{d \theta } \left( \tan \theta \right) = \mbox{sec}^2 \theta $$</p>
<p>and</p>
<p>$$\frac{d}{d \theta} \left( 3x \right) = 3 \frac{dx}{d \theta}$$</p>
<p>(Can you see why?... |
33,821 | <p>The following differentiated implicitly with respect to $\theta$:</p>
<p>$3x = \tan \, \theta $</p>
<p>The book says </p>
<p>$3 dx = \sec^2 \theta \, d \theta$</p>
<p>One could start the calculation like this (I think):</p>
<p>$ \frac{d}{d \theta}3x = \frac{d}{d \theta} \tan \, \theta $</p>
<p>From there I'm n... | quanta | 4,290 | <p>I don't know what you are asking but I will explain how I view this sort of thing.</p>
<p>What is meant by $$3x = \tan(\theta)$$ is that both sides are the same function of $x$ or $\theta$ and $x$ and $\theta$ are related somehow. Writing out the relation explicitly we have $$3x = \tan(\theta(x)).$$ Now define $f(x... |
2,363,381 | <p>I tried to prove two directions. </p>
<p>The ($\Leftarrow$) direction is not so hard, it's just in $\mathbb R$, since $\mathbb R$ is complete, Heine-Borel Theorem holds. However, I have a hard time proving ($\Rightarrow$) direction. Which is using Heine-Borel Theorem we can prove the Completeness Axiom. </p>
<p>Th... | Xiao | 131,137 | <p>We show if Heine-Borel theorem holds on $\mathbb{R}$, then $\mathbb{R}$ has the least upper bound property. </p>
<p>Let $F$ be a bounded subset of $\mathbb{R}$, we want to show that $F$ has a least upper bound.</p>
<p>Note that the clousure $\overline F$ is closed and bounded. </p>
<p>If $\overline F$ has a maxim... |
627,394 | <p>Let $R$ be a commutative ring and consider the ideal $I = (r_{1}, r_{2})$ where $r_{1}, r_{2}$ are 2 linear independent generators of $R$.</p>
<p>Now consider the $R-$modules. Here $I$ is a $R-$submodule of the $R-$module $R$. Can I then conclude that $I$ is free? </p>
<hr>
<p><strong>Argument</strong></p>
<p>I ... | Keenan Kidwell | 628 | <p>Any two elements in a commutative ring $R$ are linearly <em>dependent</em>. If $r,s\in R$ are non-zero then $0=s\cdot r-r\cdot s$ is a non-trivial linear dependence relation. Free modules of any rank over $R$ exist, but are not submodules of $R$ (except in the case of rank zero or one). Namely, for any set $X$, the ... |
1,083,708 | <p>If k is an odd and positive integer, please prove that any group whose order is 2k has a subgroup whose order is k.</p>
<p>I have been struggling for this, I tried to use mathematical induction to prove it, but maybe it doesn't work. Can someone help me solve this question? thank you for your help.</p>
| robjohn | 13,854 | <p>Given that $\sum\limits_{j=1}^nx_j^2=n$ and $\sum\limits_{j=1}^nx_j=k$
we want to minimize $\sum\limits_{j=1}^n\dfrac1{x_j}$.</p>
<p>That is, we need to find $x_j$ so that for all $\delta x_j$ where $\sum\limits_{j=1}^nx_j\,\delta x_j=0$ and $\sum\limits_{j=1}^n\delta x_j=0$, we also have $\sum\limits_{j=1}^n\dfrac... |
3,979,598 | <p>Quick question. I was wondering whether the following limit is simply equal to zero or it is rather indeterminate:</p>
<p><span class="math-container">$$\lim_{x\to0}\frac0x$$</span></p>
<p>For more detail: I was studying the differentiability of the function <span class="math-container">$\lfloor x\rfloor\sin(\pi x)^... | José Carlos Santos | 446,262 | <p>It is <span class="math-container">$0$</span>, since, for each <span class="math-container">$x\ne0$</span>, <span class="math-container">$\frac0x=0$</span>. So, <span class="math-container">$\lim_{x\to0}\frac0x=\lim_{x\to0}0=0$</span>.</p>
|
1,275,779 | <p>I am taking Abstract Algebra now and it seems most of its examples come from elsewhere. For example, I know what permutations are in counting, but the kind Algebra studies seem to be somewhat different. And to me these concepts Algebra comes bundled with is the most difficult part of studying it. It's like studying ... | user222031 | 222,031 | <p>Try this book. It seems like a good book to bridge the gap and it's free.
<a href="http://abstract.ups.edu/" rel="nofollow">http://abstract.ups.edu/</a>
Also have you looked on the site, there are a quite few abstract algebra book recommendations. There may be book suggestions which meet your needs.</p>
|
3,248,162 | <p>I have an exercise in my Exam that says: Prove that the set V=R (doesn't specify real positive numbers, only Real) is a vector space in the Real Field given the operations:</p>
<p><span class="math-container">$\ x+y = \sqrt[3]{x^3+y^3}$</span></p>
<p><span class="math-container">$ c * x = \sqrt[3]{c*x}$</span></p>... | AndrewM | 679,025 | <p>Distribution : </p>
<p><span class="math-container">$ c \otimes x = \sqrt[3]{c * x} = \\ c \otimes (x \oplus y) =\\ (c \otimes x) \oplus (c \otimes y) = \\ \sqrt[3]{c*x} \oplus \sqrt[3]{c*y} = \\ \sqrt[3]{(\sqrt[3]{c*x})^3 + (\sqrt[3]{c*y})^3} =
\\ \sqrt[3]{c*x + c*y} $</span> </p>
<hr>
<h2><span class="math... |
1,907,997 | <p>Prove $3^n \geq 2n^2 +1$ for $n = 1,2,\ldots$ using induction</p>
<p>This is what I have so far</p>
<p>Base case - $n=1,3^1 \geq 2^1 + 1 = 3$ true</p>
<p>Induction step - Assume true for some n, then,</p>
<p>$ 3*3^n \geq 3*(2n^2 +1)=6n^2 +3 $ </p>
<p>I have to somehow manipulate and show its $\geq 2*(n+1)^2 +1$... | Clement C. | 75,808 | <p>You're almost there.
$$2(n+1)^2+1 = 2(n^2+2n+1)+1 = 2n^2+4n +3 \underbrace{\leq}_{(\ast)} 2n^2+4n^2 +3 = 6n^2+3,$$
where $(\ast)$ holds since $n\geq 1$.</p>
|
1,787,867 | <p>Prove that $\dfrac{2\cos x}{\cos 2x + 1 }= \sec x$. </p>
<p>So far I have:
$\dfrac{2\cos x}{\cos 2x + 1 }= \dfrac 1 {\cos x}$</p>
<p>Where do I go from here?</p>
| Tesla | 338,187 | <p>Multiply both sides with $\cos(x)$ and make use of the identity Olivier Oloa gave you and you are done.
$\dfrac{2\cos x}{\cos 2x + 1 }= \sec x \to \dfrac{2\cos^2x}{\cos 2x + 1 }= 1 \to \dfrac{1+\cos 2x}{\cos 2x + 1 } = 1$ </p>
<p>Well or just use the identity the other way around right at the beginning.</p>
|
996,868 | <p>I am currently learning about vector spaces and have a slight confusion.</p>
<p>So I know that a vector space is a set of objects that are defined by addition and multiplication by scalar, and also a list of axioms.</p>
<p>I know that a subspace is created from the subset of a vector space and also defined by 3 pr... | brick | 187,522 | <p>Yes, but be careful. If you take one random set and show that elements are closed under substraction, and scalar multiplication (the conditions for subspace) it doesn't mean that it's a vector space. You can say this only if you know that this set is contained in a vector space.</p>
|
3,272,824 | <p>If <span class="math-container">$V$</span> is an inner product space, <span class="math-container">$T:V\to V$</span> linear transformation,
show that</p>
<p><span class="math-container">$(\ker(T))^{\bot}=\operatorname{Im}(T^*)$</span> </p>
<p>when <span class="math-container">$T^*=\overline{(T)^T}. $</span></p>
... | Clara B. | 572,257 | <p>First let me say that <span class="math-container">$\ker(T)^{\bot}=\operatorname{Im}(T^*)$</span> is not true in general for infinite dimensional <span class="math-container">$V$</span>. For example the operator <span class="math-container">$T:(x_n)\mapsto(\frac {x_n}{2^n} )$</span> on <span class="math-container"... |
1,127,117 | <p>Determine the integers $a$ such that the congruence $ax^4 \equiv b \pmod{13}$
has a solution for $b = 2, 5, 6$ </p>
<hr>
<p>I think the problem wants $a$'s that work for all $b=2,5,6$. Can I please have a hint on how to approach this problem ? </p>
| Bernard | 202,857 | <p>The nonzero $4$-th powers in $\mathbf 13$ are $\{1, 3, -4\}$ as @Joffan noticed. Let $c$ be one of these values. We have to solve the equation with unknown $a$: $\enspace a=c^{-1}b$.</p>
<p>Now, an inverse of a $4$-th power is again a $4$-th power and the set of $c^{-1}$ is just $\{1,3,-4\}$. When we compute all ... |
702,481 | <p>I am trying to solve the following problem: Let $a$ and $b$ be positive integers and let $p$ be a prime number. Prove that if $a^p \equiv$ $b^p$ (mod $p$), then $a \equiv b$ (mod $p$).</p>
<p>My attempt to solve this starts off by using the idea that $b^p-a^p$ is a multiple of $p$, i.e. $pm=b^p-a^p$, for some $m \i... | Stella Biderman | 123,230 | <p>This follows more or less immediately from FLT. Think about what FLT tells you about $a^p$ (mod p) and $b^p$ (mod p). If $a^p\cong b^p$ (mod p) then those two expressions are equivalent by transitivity.</p>
<p>Alternatively, you don't need FLT to prove this at all. Try to factor $a^p-b^p$ and see what you get. $p$... |
4,044,564 | <p>calculation<span class="math-container">$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{1}{k}\sin\frac{k\pi}{n+1}.$$</span>
I try to use <span class="math-container">$|\sin x-x|\leqslant\frac{x^2}{2}$</span>, but I don't feel like it's going to work. Is there any other way?</p>
| Unit | 196,668 | <p>This is essentially a Riemann sum of the function <span class="math-container">$$f(x) = \frac{\sin \pi x}{x}$$</span> over the unit interval <span class="math-container">$[0,1]$</span>.
The partition seems to be <span class="math-container">$x_k = k/(n+1)$</span> for <span class="math-container">$0 \le k \le n+1$</s... |
7,367 | <p>This question and the described solution are copied from a test-paper :</p>
<p>For the equation $x^2$ + |x| - 6 = 0 analyze the four statements below for correctness.</p>
<ol>
<li>there is only one root</li>
<li>sum of the roots is + 1</li>
<li>sum of the roots is zero</li>
<li>the product of the roots is +4</li>
... | Bill Dubuque | 242 | <p>$\rm\ 0\ =\ x^2 + |x| -6\ =\ (|x| - 2)\:(|x| + 3)\ \Rightarrow\ |x| = 2\ \Rightarrow\ x = \pm2 $</p>
|
521,412 | <p>So I tried this out and got stuck with this:</p>
<p>$$0 = 3x^{(7/6)} + 2x - 10$$</p>
<p>I didn't think I could use a quadratic for this since its to the power of $7/6$</p>
<p>Here is the working I did:</p>
<p>We know its a critical point when f'(a) = 0</p>
<p>So I found the derivative of f(x) which is $$2*(5-x)... | amWhy | 9,003 | <p>If $$f(x) = x^{2/3}(5-x) =5x^{2/3}-x^{5/3},$$</p>
<p>then $$f'(x) = \frac{10}{3}x^{-1/3} - \frac{5}{3}x^{2/3}$$
Of course, to find critical points, we need to solve for
$$f'(x) = 0 \iff \frac{10}{3}x^{-1/3} = \frac{5}{3}x^{2/3}$$</p>
<p>Assuming if $x\neq 0$, multiply both sides of the equation by $\;\frac 35 x^{1... |
85,024 | <p>There are two results famously associated with Cantor's celebrated diagonal argument. The first is the proof that the reals are uncountable.</p>
<p><img src="https://i.stack.imgur.com/v9G8r.png" alt="Cantor's Diagonal Argument"></p>
<p>This clearly illustrates the namesake of the diagonal argument in this case... | Michael Hardy | 11,667 | <p>The proof of the first result you give mentions</p>
<p>the 1st digit of the 1st number, and<br>
the 2nd digit of the 2nd number, and<br>
the 3rd digit of the 3rd number, and<br>
the 4th ditit of the 4th number, and<br></p>
<p>........ etc. You see the "dual occurrence" of something here. You also see that the fi... |
1,639,381 | <p>I have a data set</p>
<pre><code> x y
--------------------
505.000000 77.000000
507.000000 88.000000
509.000000 99.000000
509.000000 110.000000
511.000000 121.000000
511.000000 132.000000
511.000000 143.000000
513.000000 154.000000
513.000000 165.000000
513.000000 176.000000
515.000000 187.000000
515.0... | mm8511 | 180,904 | <p>You can just remove the outliers from your data set manually, and then run the regression on the remaining data. <em>I don't have enough rep to comment</em> Otherwise, you're pretty much stuck with the result you first obtained.</p>
|
2,390,862 | <p>Let $f :A \to B, \ g : C \to D$ be bijections then is there always a bijection $h: A\cup C \to B \cup D$ or $h : A \cap C \to B \cap D$ ?</p>
<p>Clearly, if $A, C$ are disjoint and $B, D$ are disjoint, then union works. I can't see how to split up the sets though since if we split up $A \cup B = A \setminus B \upl... | Brethlosze | 386,077 | <p>If the spheres centers are forming a square, the side of the square will be $2r$. </p>
<p>Referring a coordinate system by putting the square vertex at $(\pm r,\pm r)$, then the cylinder will contact each sphere at angles $45°$, $135°$, $225°$ and $315°$. </p>
<p>So the contact will happen at distance $R=r\sqrt{2}... |
1,607,670 | <p>I'm getting stuck at solving the following ODE:</p>
<p>$\ddot{f}(x)-\dfrac{\dot{f}(x)^2}{f(x)}+C\left(\dfrac{\dot{f}(x)}{f(x)}\right)^2=0$, where $C\neq 0$ constant, and let $f(0)=P$, $\dot{f}(0)=Q$.</p>
<p>Is it possible to find a closed-form solution to this problem? Thanks for help!</p>
| Claude Leibovici | 82,404 | <p>I suppose that the trick is to work with $x(y)$ and to use the fact that $$\frac{d^2x}{dy^2}=-\frac{\frac{d^2 y}{dx^2}}{\left({\frac{dy}{dx}}\right)^3}$$ So, replacing $\frac{d^2 y}{dx^2}$ by its expression as functions of $\frac{dx}{dy}$ and $\frac{d^2x}{dy^2}$ leads to somethink looking like $$-\frac {x''}{x'}-\f... |
255,807 | <p>With <code>x</code>, <code>y</code>, <code>z</code> all positive, how can one simplify the following expression:</p>
<pre><code> f[x_,y_,z_]=(Sqrt[1 + x - y -
Sqrt[(-1 + x + y)^2 + 4 z^2]] (-1 + x + y -
Sqrt[(-1 + x + y)^2 + 4 z^2]))/(2 Sqrt[2] z Sqrt[
2 - (2 (-1 + x + y))/(-1 + x + y + Sqrt[(-1 + x + y)^... | Michael E2 | 4,999 | <p>You might add rationalization to the transformation functions:</p>
<pre><code>xfs = {
# /. a_ - Sqrt[b_] :> Simplify[(a^2 - b)/(a + Sqrt[b])] &,
# /. a_ + Sqrt[b_] :> Simplify[(a^2 - b)/(a - Sqrt[b])] &
};
ClearSystemCache[]; (* this is for testing, since simplification
... |
1,302,459 | <p>I think I understand the proof for square matrices, such that $(AB)^{-1}=B^{-1}A^{-1}$, but I'm not sure if I can just say the same for the bounded linear operators A and B.</p>
<p>Does the existence of $(AB)^{-1}:X\rightarrow X$ imply the existence of $B^{-1}$ and $A^{-1}$?</p>
| Prahlad Vaidyanathan | 89,789 | <p>For the question in the title, the answer is No. Take $B : \ell^2 \to \ell^2$ be the right-shift operator
$$
(x_n) \mapsto (0,x_1,x_2,\ldots)
$$
and $A$ to be the left shift
$$
(x_n) \mapsto (x_2,x_3,\ldots)
$$
Then $AB$ is the identity operator, and so invertible, but neither $A$ nor $B$ are invertible.</p>
<p>But... |
4,226,115 | <p>I was going through some practice problems for an upcoming calculus exam and I came across the question: "Prove the following statement: if <span class="math-container">$\lim _{n\to \infty }a_n = L $</span> and <span class="math-container">$\lim _{n\to \infty }a_n = M$</span>, then <span class="math-container">... | Theo Bendit | 248,286 | <p>The proof is subtly circular, but there is something that can be salvaged from it. As it turns out, yes, you can actually apply algebra of limits to show that <span class="math-container">$0 = a_n - a_n \to L - M = k$</span>. The proof of the algebra of limits theorem doesn't rely on uniqueness at all; that is if <s... |
4,332,748 | <p>This is slightly different question. First I need to mention that I am neither a mathematician nor a researcher. As an ordinary student the separation " with and without Axiom of choice " trouble me a lot. But I am really interested to know further.</p>
<ol>
<li><p>Every ideal in a ring contained in some ... | Couchy | 87,768 | <p>As HallaSurvivor pointed out, each (most?) of these statements are equivalent to the axiom of choice. However I want to point out that just because <span class="math-container">$\phi$</span> is provable from <span class="math-container">$ZF + \neg AC$</span> (or <em>consistent</em> with <span class="math-container">... |
2,658,073 | <p>I've got a problem where I have to find the degree of the Maclaurin polynomial (meaning it's centered at $c = 0$, correct?) of $\ln(1.25)$ such that the error is less than 0.001.</p>
<p>I think I have a fairly good understanding of how you find the degree required to match that criteria for other functions, but I'm... | copper.hat | 27,978 | <p>For $|x| <1$ we have $\log(1+x) = \sum_{k=1}^\infty (-1)^{k+1} {x^k \over k}$.</p>
<p>Since the series is alternating, we see that
$|\log(1+x) - \sum_{k=1}^n (-1)^{k+1} {x^k \over k} | \le {|x|^{n+1} \over n+1}$.</p>
<p>Choose $n$ such that ${ {1 \over 4^{n+1}} \over n+1} < {1 \over 1000}$. I believe $n=3 $ ... |
12,922 | <p>European universities are running the so called "Bologna plan" which includes the recommendation of keeping some percentage of the final grades to be filled with various activities during the lessons period. Like everything in life, it has pros and cons.</p>
<p>I will tell you the cons I have seen (in the particula... | guest | 8,804 | <p>In addition to 138's points, it can serve as a motivation to push some work getting done by the student. Not everyone is a perfectly rational time utilizer balancing work and study. </p>
<p>This doesn't mean I agree that all the pros overcome the cons. I do think there can be some busy work on being in a class v... |
16,521 | <blockquote>
<p><strong>Is there an easy way to tell how many roots $f(x)$ has in $\Bbb{Q}[x]/(f)$ given the coefficients of the polynomial $f$ in $\Bbb{Q}[x]$?</strong></p>
<p><strong>Is there an easy way to find the roots as rational expressions in $x$?</strong></p>
</blockquote>
<p>The easiest example is a p... | Qiaochu Yuan | 232 | <p>If $f$ has abelian Galois group and you can find an explicit embedding of its splitting field into $\mathbb{Q}(\zeta_n)$, you get a quotient map $(\mathbb{Z}/n\mathbb{Z})^{\ast} \to \text{Gal}(f)$ which makes the action of the Galois group quite explicit. Applying elements of $(\mathbb{Z}/n\mathbb{Z})^{\ast}$ to a ... |
1,980,012 | <p>I am facing issue to evaluate this particular type of definite integral,</p>
<p>$$I=\int_{0}^{\infty}\exp(-\alpha x^2)\cos(\beta x)dx$$</p>
<p>Please suggest a way to this.</p>
<p>Thanks</p>
| robjohn | 13,854 | <p>$$
\begin{align}
\int_0^\infty e^{-\alpha x^2}e^{i\beta x}\,\mathrm{d}x
&=\int_0^\infty e^{-\alpha\left(x-i\frac\beta{2\alpha}\right)^2-\frac{\beta^2}{4\alpha}}\mathrm{d}x\tag{1}\\
&=e^{-\frac{\beta^2}{4\alpha}}\int_{-i\frac\beta{2\alpha}}^{\infty-i\frac\beta{2\alpha}}e^{-\alpha x^2}\mathrm{d}x\tag{2}\\
&... |
934,343 | <p>How can I show the matrix
$$A = \left(
\begin{array}{c c}
-2 & 0 \\
0 & -1 \\
\end{array}
\right)$$</p>
<p>can't be written as $A = exp(a)$?</p>
<p>I've tried to write A like</p>
<p>$$A = \left(
\begin{array}{c c}
2 & 0 \\
0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{c c}
-1 & 0 ... | Robin Goodfellow | 176,079 | <p>Perhaps you would be interested in <a href="http://en.wikipedia.org/wiki/Logarithm_of_a_matrix#Calculating_the_logarithm_of_a_diagonalizable_matrix" rel="nofollow">this link</a>? By this method, we know that the matrix would have to be complex to cope with the logarithms of negative numbers.</p>
|
813,971 | <p>I have the following exercise:</p>
<p>We want to integrate the function $w=x+y^2$ and we have a path that begins from $A(0,0)$ and reaches at $B(1,1)$.</p>
<p>$$$$</p>
<p>Could you give me some hint what I am supposed to do?</p>
<p>Do I have to use the following formula?
$$\int_C w ds= \int_a^b w(x(t), y(t), z(t... | Klaas van Aarsen | 134,550 | <p>Yes!
That's the formula to use, which integrates a function along a curve.</p>
<p>You're supposed to think up a formula for a (presumably straight) path $(x(t), y(t), z(t))$ that starts at A and ends at B.
When you have it, you're supposed to substitute it in the formula and evaluate.</p>
|
4,123,074 | <p>Let <span class="math-container">$\phi: X \rightarrow Y$</span> be a measurable function between measurable spaces. Let <span class="math-container">$\mu$</span> be a <span class="math-container">$\sigma$</span>- finite measure on <span class="math-container">$X$</span>. In general I have seen that the push forward ... | Lorenzo Panebianco | 351,960 | <p>It's been a while since a worked with measure theory, but try like this (it's a bit naive).</p>
<p>Consider subsets <span class="math-container">$X_n$</span> of <span class="math-container">$X$</span> with finite measure and such that <span class="math-container">$X = \bigcup_n X_n$</span>. Suppose <span class="math... |
147,739 | <p>How would one show that if $\displaystyle \lim_{z \to 0} z \exp(f(z))$ exists, where $f: \mathbb{C}^\ast \longrightarrow \mathbb{C}$ is holomorphic, then it must be zero? Is this even true?</p>
| Qiaochu Yuan | 232 | <p>No. As it turns out, $MN$ and $NM$ must have the same nonzero eigenvalues (although the multiplicities need not agree), so if $(MN)^n = 0$ then $MN$ has only zero eigenvalues, so the same must be true of $NM$, which must therefore have characteristic polynomial $t^n$.</p>
<p>One elegant proof uses the following lem... |
1,387,170 | <p>Let $V$ be a vector space with dimension $n$ such that $\{v_1,\cdots,v_n\}$ is its basis. Take $A\equiv\{a_1,\cdots,a_p\}\subset V$ with $p>n$.</p>
<p>How do can I show that $A$ is linearly dependent <strong>without having to do all those summations</strong>. Is there an easier more direct proof than that?</p>
... | gt6989b | 16,192 | <p><strong>Hint</strong>: If <span class="math-container">$A$</span> is linearly independent, then <span class="math-container">$\dim(\text{span}(A)) = p$</span>, but <span class="math-container">$\text{span}(A) \subseteq V$</span>. Can you finish the argument?</p>
|
3,557,252 | <p>By " same base, same exponent " rule I mean : </p>
<p><span class="math-container">$$b^x = b^n \iff x=n$$</span></p>
<p>Can the equation:
<span class="math-container">$$ 2^x+4^x = 6$$</span>
be solved by saying
<span class="math-container">$$ 2^x+4^x = 6 = 2+4= 2^1 + 4^1$$</span>
and consequently, that
<span c... | Allawonder | 145,126 | <p>Well, this is one way to obtain a solution, especially if you don't know any other way to proceed -- we usually say the solution has been obtained <em>by observation.</em></p>
<p>However, the drawback with this method is that it may not work with a sufficiently complicated equation. Also, it doesn't guarantee that ... |
3,557,252 | <p>By " same base, same exponent " rule I mean : </p>
<p><span class="math-container">$$b^x = b^n \iff x=n$$</span></p>
<p>Can the equation:
<span class="math-container">$$ 2^x+4^x = 6$$</span>
be solved by saying
<span class="math-container">$$ 2^x+4^x = 6 = 2+4= 2^1 + 4^1$$</span>
and consequently, that
<span c... | Acccumulation | 476,070 | <p>The term "solve" generally means a systematic method of finding solutions, and showing that they are exhaustive (that is, no further solutions exist). This method fails on both counts. </p>
<p>First, while for the particular example you give, it's not too difficult to see that 1 is a solution, I don't see any way t... |
5,751 | <p>I've just been asked for a good example of a situation in maths where using infinity can greatly shorten an argument. The person who wants the example wants it as part of a presentation to the general public. The best I could think of was Goodstein sequences: if you take a particular instance of Goodstein's theorem,... | Kristal Cantwell | 1,098 | <p>If you want a situation where there is only an infinite proof then there is the strengthened finite Ramsey theorem which is not provable in PA but can be deduced from the infinite Ramsey theorem. see:</p>
<p><a href="http://en.wikipedia.org/wiki/Paris%E2%80%93Harrington_theorem">http://en.wikipedia.org/wiki/Paris%E... |
5,751 | <p>I've just been asked for a good example of a situation in maths where using infinity can greatly shorten an argument. The person who wants the example wants it as part of a presentation to the general public. The best I could think of was Goodstein sequences: if you take a particular instance of Goodstein's theorem,... | Scott Carter | 36,108 | <p>One of my colleagues (Silver) gave a general purpose talk last week. You can't untie a knot $A$ by tying another knot $B$ into the rope. </p>
<p>Proof. First show that the connect-sum of long knots is abelian by sliding $A$ through $B$. Then form the decreasing-in-size connect-sum $(AB)(AB)(AB) \dots$. Regroup: $A(... |
5,751 | <p>I've just been asked for a good example of a situation in maths where using infinity can greatly shorten an argument. The person who wants the example wants it as part of a presentation to the general public. The best I could think of was Goodstein sequences: if you take a particular instance of Goodstein's theorem,... | Chris Pressey | 29,482 | <p>Short version of "there exist functions not computable by Turing machine" using infinities:</p>
<p>There are an uncountable number of predicates, but only a countable number of Turing machines, so there are some predicates that are not computable by a Turing machine.</p>
<p>(That's weaker than the usual proof of t... |
157,187 | <p>I got thrown off by an interview question recently, </p>
<blockquote>
<p>What structure do you use to find all the ways up a ladder with N
steps in 1 or 2 steps moves?</p>
</blockquote>
<p>It occured to me today that a graph expansion could be used but it's not optimal because when N = 30, it evalates almost 7... | Brian M. Scott | 12,042 | <p>If $N=0$ there is one way: $\langle\rangle$. (No steps at all.) If $N=1$, there is one way, $\langle 1\rangle$. For $N>1$, take each way up a ladder of $N-2$ steps and append a $2$, and each way up a ladder of $N-1$ steps and append a $1$. Thus, for $N=2$ you get $\langle 2\rangle$ and $\langle 1,1\rangle$. The f... |
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