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 |
|---|---|---|---|---|
1,885,492 | <p>Question.</p>
<blockquote>
<p>prove that if ${ a }_{ 1 },{ a }_{ 2 },...{ a }_{ n }>0$ then $$ \frac { { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n } }{ n } \ge \frac { n }{ \frac { 1 }{ { a }_{ 1 } } +\frac { 1 }{ { a }_{ 2 } } +...+\frac { 1 }{ { a }_{ n } } } $$ </p>
</blockquote>
<p><strong>Proof</strong>
$$... | Tito Piezas III | 4,781 | <p>The first has no solutions with positive integer $m$. However, the second case,</p>
<p>$$\begin{aligned}
m&=\dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)\\
&=\dfrac{1}{18}\left(x-1\right)\\
&\equiv {3\pmod{4}}
\end{aligned}$$</p>
<p>has infinitely many given by,</p>
<p>$$x= \frac{ (7+\sqrt{48})^{6k+3}+(7... |
4,187,498 | <p>I am studying the proof of the Prime Number Theorem and I want to show that the function <span class="math-container">$\frac{\zeta'(s)}{\zeta(s)}$</span> has a simple pole at <span class="math-container">$s=1$</span>.</p>
<p>I think that if I can find the Laurent series expansion of <span class="math-container">$\ze... | robjohn | 13,854 | <p>This answer was posted to <a href="https://math.stackexchange.com/questions/123531/how-to-show-that-the-laurent-series-of-the-riemann-zeta-function-has-gamma-as">another question</a> which dealt only with the constant term of the Laurent expansion, but it actually answers this question, so I will repost it here.</p>... |
970,409 | <p>Let matrices $A$, $B\in{M_2}(\mathbb{R})$, such that $A^2=B^2=I$, where $I$ is identity matrix. </p>
<p>Why can be numbers $3+2\sqrt2$ and $3-2\sqrt2$ eigenvalues for the Matrix $AB$? </p>
<p>Can be numbers $2,1/2$ the eigenvalues of matrix $AB$? </p>
| Ben Grossmann | 81,360 | <p>Note that
$$
\pmatrix{1&0\\0&-1}
\pmatrix{a& b\\-(a^2-1)/b & -a} =
\pmatrix{
a & b\\
(a^2 - 1)/b & a
}
$$
Every such product is similar to a matrix of this form or a triangular matrix with 1s on the diagonal. The associated characteristic polynomial is
$$
x^2 - 2ax + 1
$$
Check the possible r... |
3,955,162 | <p>I was trying to write this linear transformation as a matrix,
the exercise is the following:
let "f: <span class="math-container">$R^3 --> R^4$</span>" be a linear transformation and Kerf = (x, y , z) in <span class="math-container">$R^3$</span>: <span class="math-container">$x-2y+z = 0$</span> is its ... | Peter Morfe | 711,689 | <p>If <span class="math-container">$f([0,1])$</span> is an immersed closed curve (suitably defined --- you need <span class="math-container">$f'(1) = f'(0)$</span> --- or, in other words, <span class="math-container">$f$</span> should descend to a <span class="math-container">$C^{1}$</span> map of <span class="math-co... |
3,955,162 | <p>I was trying to write this linear transformation as a matrix,
the exercise is the following:
let "f: <span class="math-container">$R^3 --> R^4$</span>" be a linear transformation and Kerf = (x, y , z) in <span class="math-container">$R^3$</span>: <span class="math-container">$x-2y+z = 0$</span> is its ... | Krishnarjun | 722,463 | <p>This answer is motivated by what Peter wrote in his answer. I realize now that when I was thinking about this question, I was thinking of something along the lines of intermediate value theorem, but didn't know how to apply it for a curve in <span class="math-container">$\mathbb{R}^2$</span>.</p>
<p>Suppose that <sp... |
1,242,317 | <p>If I have a unitary square matrix $U$ ie. $U^{\dagger}U=I$ ( $^\dagger$ stands for complex conjugate and transpose ), then for what cases is
$U^{T}$ also unitary. One simple case I can think of is $U=U^{T}$ ( all entries of $U$ are real, where $^T$ stands for transpose ). Are there any other cases ?</p>
| TZakrevskiy | 77,314 | <p>$$(U^T)^\dagger = \bar U = (U^\dagger)^T, $$
where $\bar U$ is the complex conjugate of $U$.</p>
<p>Moroever $$(U^T)^\dagger U^T = (U^\dagger)^T U^T = UU^\dagger = I.$$</p>
<p>Therefore, your proposition is always true.</p>
|
815,770 | <p>Compute $F_{1000} \bmod F_{11}$, where $F_n$ denote the Fibonacci numbers.</p>
<p>Progress:</p>
<p>$F_{11}=89$ . I believe you should find the period of $F_n \bmod 89$ and use that to solve it. But I'm not not getting anywhere from that.</p>
<p>Thanks!</p>
| Matt Gallagher | 148,015 | <p>Assuming we're starting with $F_0=0$ and $F_1 = 1$, an easy induction shows that for $n \geq 11$ $$F_n = F_{n-10}\cdot F_{11} + F_{n-11}\cdot F_{10}$$</p>
<p>So $F_n \equiv F_{n-11} \cdot F_{10} \bmod F_{11}$ for $n \geq 11$.</p>
<p>Edit: I made a mistake from here on. See heropup's answer for the rest.</p>
|
3,403,272 | <p><p> I'm currently taking abstract algebra and I'm very lost.</p>
<blockquote>
<p>Let <span class="math-container">$G = (\Bbb Z/18\Bbb Z, +)$</span> be a cyclic group of order <span class="math-container">$18$</span>.</p>
<p>(1) Find a subgroup <span class="math-container">$H$</span> of <span class="math-cont... | Bernard | 202,857 | <p>For the last question, you can use the <em>third isomorphism theorem</em>:</p>
<p><span class="math-container">$g=\mathbf Z/18\mathbf Z$</span>, <span class="math-container">$H==6\mathbf Z/18 \mathbf Z$</span>, so
<span class="math-container">$$G/H=\mathbf Z/18\mathbf Z\Big/6\mathbf Z/18\mathbf Z\simeq\mathbf Z/6\... |
3,050,295 | <p>I'm having a problem with following equation:</p>
<p>I'm applying <span class="math-container">$(a+b)^2$</span> and <span class="math-container">$(a-b)^2$</span>, but am unable to get the correct answer.</p>
<blockquote>
<p><span class="math-container">$$(\sqrt{3}+i)^{2017} + (\sqrt{3} - i)^{2017}$$</span></p>
<... | cansomeonehelpmeout | 413,677 | <p><strong>Hint</strong>:</p>
<p>It's probably easier to write <span class="math-container">$\sqrt 3+i=re^{i\theta}$</span> and <span class="math-container">$\sqrt 3-i=re^{-i\theta}$</span>, by finding <span class="math-container">$r,\theta$</span>, and then use <span class="math-container">$$r\cos(\theta)=r\frac{e^{i... |
3,050,295 | <p>I'm having a problem with following equation:</p>
<p>I'm applying <span class="math-container">$(a+b)^2$</span> and <span class="math-container">$(a-b)^2$</span>, but am unable to get the correct answer.</p>
<blockquote>
<p><span class="math-container">$$(\sqrt{3}+i)^{2017} + (\sqrt{3} - i)^{2017}$$</span></p>
<... | Mohammad Riazi-Kermani | 514,496 | <p>Let <span class="math-container">$z=(\sqrt 3 +i)^{2017}$</span>, then <span class="math-container">$$z=2^{2017}e^{2017 i\pi /6}$$</span></p>
<p><span class="math-container">$$Re(z)=2^{2017}\cos {2017\pi /6} =2^{2017} \cos {\pi/6}$$</span></p>
<p><span class="math-container">$z+\bar z =2Re(z)=2^{2017}\sqrt 3$</span... |
1,103,239 | <p>For example, if I multiply the value of a base squared by four, I also get twice the base if it's squared. Look:$$6^2\cdot4=12^2$$ because $$36\cdot4=144$$and $36$ is the square of $6$ and $144$ is the square of $12$. Why does this always happen?</p>
| agha | 118,032 | <p>Suppose that you have $a^2$, then:</p>
<p>$$4 \cdot a^2 = 2^2 \cdot a^2= (2 \cdot a)^2=(2a)^2$$</p>
|
2,264,791 | <p>I have a problem that I'm having trouble figuring out the distribution with given condition.</p>
<p>It is given that 1/(<span class="math-container">$X$</span>+1), where <span class="math-container">$X$</span> is an exponentially distributed random variable with parameter 1.</p>
<blockquote>
<p><strong>Original Prob... | callculus42 | 144,421 | <p>Keep it short and simple. Let <span class="math-container">$Y=\frac{1}{1+X}$</span> where <span class="math-container">$0<Y<1$</span>. Then</p>
<p><span class="math-container">$P(Y<y)=P\left(\frac{1}{1+X}<y\right)=P\left(\frac{1}{1+X}<y\right)=P(1<y+Xy)$</span></p>
<p><span class="math-container">$... |
267,707 | <p>The weight 2, level 1 Eisenstein series $E_2(z)$ is a non-holomorphic automorphic form. It is defined as the analytic continuation to $s = 0$ of the series
$$ E_2(z, s) = \sum_{\substack{m, n \in \mathbf{Z} \\ (m, n) \ne (0,0)}} \frac{\operatorname{Im}(z)^s}{(mz + n)^2 |mz + n|^{2s}} $$
which is convergent for $\ope... | GH from MO | 11,919 | <p>Excellent question indeed. The quick answer is that $E_2(z)$ is an <em>almost holomorphic modular form</em> of weight $2$ and level $1$, so the automorphic representation generated by it is not irreducible. For more details (and my thought process), read below.</p>
<p>Consider the Maass raising operator in weight $... |
2,661,443 | <p>For the equation $2^x = 7$</p>
<p>The textbook says to use log base ten to solve it like this $\log 2^x = \log 7$. </p>
<p>I then re-arrange it so that it reads $x \log 2 = \log 7$ then divide the RHS by $\log 2$ to isolate the $x$. I understand this part.</p>
<p>I can alternatively solve it in an easier way by s... | Tony Ma | 526,038 | <p>Question 1) As you see, both method arrive at the same answer, however $\log$ base 10 is a natural choice, as stated in the comment by Matti P above.</p>
<p>Question 2) It is because for $a>0,a\neq 1,b>0, c>0, c\neq 1$,$$\log_a b=\frac{\log_c b}{\log_c a}$$In particular in your example, $c=10$, so both m... |
2,661,443 | <p>For the equation $2^x = 7$</p>
<p>The textbook says to use log base ten to solve it like this $\log 2^x = \log 7$. </p>
<p>I then re-arrange it so that it reads $x \log 2 = \log 7$ then divide the RHS by $\log 2$ to isolate the $x$. I understand this part.</p>
<p>I can alternatively solve it in an easier way by s... | Martin Argerami | 22,857 | <p>Other answers mention the change of base $$ \log_a x = \frac {\log_b x}{\log_b a}. $$ An immediate consequence is that a quotient of logarithms does not depend on the base: $$\frac {\log_ax}{\log_ay}=\frac {\frac {\log_bx}{\log_ba}}{\frac {\log_by}{\log_ba}}
=\frac {\log_bx}{\log_by}. $$</p>
|
2,661,443 | <p>For the equation $2^x = 7$</p>
<p>The textbook says to use log base ten to solve it like this $\log 2^x = \log 7$. </p>
<p>I then re-arrange it so that it reads $x \log 2 = \log 7$ then divide the RHS by $\log 2$ to isolate the $x$. I understand this part.</p>
<p>I can alternatively solve it in an easier way by s... | Mark Miller | 534,643 | <p>The only reason we use base ten for anything (including decimal numbers) is that biological evolution endowed humans with ten fingers. The Mayans are supposed to have <a href="https://en.wikipedia.org/wiki/Maya_numerals" rel="nofollow noreferrer">used base twenty</a>, apparently because they were barefoot and though... |
77,379 | <p>It is to show for an $a\in \mathbb{C}^{\ast}$ that $aB_{1}(1)= B_{|a|}(a)$ </p>
<p>where B denotes a disc </p>
<p>Okay, maybe this is correct: </p>
<p>$aB_{1}(1) = a(e^{i\phi}) = ae^{i\phi} = |a|e^{i\phi} = B_{|a|}(a)$</p>
<p>But this seems very wrong! </p>
<p>V</p>
| Adam Smith | 15,426 | <p>Since this looks like homework, I'll just give you a hint. You are on the right track, but it is not true that $a e^{i\phi} = |a| e^{i \phi}$. Indeed, remember that $a = |a| e^{i \theta}$ for some $\theta$ (the modulus of $a$). Geometrically, what is going on is that multiplication by $a = |a| e^{i \theta}$ scale... |
218,933 | <p>The space $L^2 (\mathbb{R}^2; \mathbb{C})$ can be decomposed as
$$
L^2 (\mathbb{R}^2; \mathbb{C}) = \bigoplus_{k \in \mathbb{Z}} L^2_k (\mathbb{R}^2; \mathbb{C}),
$$
where
$$
L^2_k (\mathbb{R}^2; \mathbb{C}) = \bigl\{ f \in L^2 (\mathbb{R}^2;\mathbb{C}) : \text{for almost every \(z \in \mathbb{R}^2 \simeq \mathb... | Yemon Choi | 763 | <p>Thinking about this a bit more, here is a <strong>guess</strong> — I haven't got time right now to check the details, so comments and corrections are welcome from others.</p>
<p>The Euclidean motion group $E(2)$ is the group of all isometries of ${\bf R}^2$ with its usual Euclidean metric. Since $E(2)$ acts o... |
2,579,140 | <p>N. Elkies' page <a href="http://www.math.harvard.edu/~elkies/trinomial.html" rel="nofollow noreferrer">http://www.math.harvard.edu/~elkies/trinomial.html</a> ends with an information about octic trinomials "whose Galois group is contained in $G_{1344}$".</p>
<p>One of reported trinomials, $x^8+324x+567$, "has a sma... | ahulpke | 159,739 | <p><code>TransitiveGroup</code> will give you <em>one</em> representative from a conjugacy class of subgroups. For two subgroups $A$, $B$ in the library it is not guaranteed (in fact not guaranteeable) that if a conjugate of $A$ is contained in $B$, then automatically $A$ is contained in $B$.</p>
<p>To test for such i... |
4,188,020 | <p>I know that there exists a connection on a principal bundle and via parallel transport it is possible to define a a covariant derivative on the associated bundle.</p>
<p>However, can we also define a covariant derivative on the principal bundle. I.e. something that can differentiate a section along a vector field? O... | nicrot000 | 932,402 | <p>So from my point of view what a covariant derivative should be or should do I'd say that indeed you need a linear structure.</p>
<p>So in principle, when passing from simply vector space-valued functions <span class="math-container">$M\to W$</span> (or sections in the trivial bundle <span class="math-container">$M\t... |
359,742 | <p>I have a mathematical problem that leads me to a particular necessity. I need to calculate the convolution of a function for itself for a certain amount of times. </p>
<p>So consider a generic function $f : \mathbb{R} \mapsto \mathbb{R}$ and consider these hypothesis:</p>
<ul>
<li>$f$ is continuos in $\mathbb{R}$.... | user268922 | 268,922 | <p>You can use the cumulants of the original distribution and make the inverse Fourier Transform. Being m_1 the mean and k_n the cumulants of f, after N self convolutions, the resulting mean is N<em>m_1 and the cumulants N</em>k_n.
If f have all cumulants well defined, the result tends to a gaussian with mean N<em>m_1 ... |
230,971 | <p>At the moment I use <code>Length[ DeleteDuplicates[ array ] ] == 1</code> to check whether an array is constant, but I'm not sure whether this is optimal.</p>
<p>What would be the quickest way to test whether an array consists of equal elements?</p>
<p>What if the elements would be integers?</p>
<p>What if they are ... | Henrik Schumacher | 38,178 | <p><code>Equal@@MinMax[array]</code> might be quite fast if <code>array</code> is a packed list of integers. But it cannot short-circuit like <code>Statistics`Library`ConstantVectorQ</code> does. And it is also not very robust with regard to (machine) floating point numbers: <code>Equal</code> and <code>SameQ</code> bo... |
3,988,180 | <p>The task:
<span class="math-container">$$
\text{Calculate } \int_{C}{}f(z)dz\text{, where } f(z)=\frac{\bar{z}}{z+i}\text{, and } C \text{ is a circle } |z+i|=3\text{.}
$$</span>
Finding the circle's center and radius:
<span class="math-container">$$
|z+i|=|x+yi+i|=|x+(y+1)i|=3
\\
x^2+(y+1)^2=3^2
$$</span>
Parametri... | Milo Moses | 630,231 | <p>You cannot directly use the Residue Theorem in this case since <span class="math-container">$\frac{\overline{z}}{z+i}$</span> is not holomorphic, so your approach is entirely reasonable. Your calculation of the integral is correct since</p>
<p><span class="math-container">$$\int_{0}^{1}e^{2k\pi ti}dt=\left.\frac{1}{... |
158,916 | <p>Let $M$ be a closed riemannian 3-manifold. I think that the following fact should be true and should have a relatively simple proof, but I cannot figure it out.</p>
<blockquote>
<p>For every $\varepsilon>0$ there is a $\delta>0$ such that every smooth 2-sphere in $M$ of area smaller than $\delta$ bounds a b... | Misha | 21,684 | <p>This is a direct corollary of Federer -Flemming deformation Lemma saying that vary small area sphere can be homotoped to 1-skeleton of the fixed triangulation of the manifold. The dimension assumption is irrelevant. Proof of this Lemma can be found somewhere in Federer's book (I will chase down the precise reference... |
2,080,716 | <p>I have the quadratic form
$$Q(x)=x_1^2+2x_1x_4+x_2^2 +2x_2x_3+2x_3^2+2x_3x_4+2x_4^2$$</p>
<p>I want to diagonalize the matrix of Q. I know I need to find the matrix of the associated bilinear form but I am unsure on how to do this.</p>
| Vidyanshu Mishra | 363,566 | <p>First posted as a hint, but now it is the compilation to @Atul Mishra's answer.</p>
<p>Let us see what is happening in diagrammatic way:</p>
<p><img src="https://i.stack.imgur.com/CHw7J.png" alt="Image Describing unions and intersections" /></p>
<p>Now, look at the image carefully.The coloured regions represent foll... |
2,804,129 | <p>To Evaluate the Limit $$L=\lim_{n \to \infty}\left(1+\sum_{k=1}^{n} \frac{1}{\binom{n}{k}}\right)^n \tag{1}$$</p>
<p>My try:</p>
<p>I tried to use $$\frac{1}{\binom{n}{k}}+\frac{1}{\binom{n}{k+1}}=\frac{n+1}{n} \frac{1}{\binom{n-1}{k}} $$</p>
<p>taking summation both sides from $k=1$ to $k=n$ we get</p>
<p>$$\su... | Fred | 380,717 | <p>Your martrix $M$ is o.k.</p>
<p>The equation in the question has a solution $ \iff$ the linear system</p>
<p>$(*)$ $M(x,y,z)^T=(1,2,3)^T$ </p>
<p>has a solution. Since $M$ is invertible, the equation $(*)$ has exactly one solution. Hence the equation</p>
<p>$y+e^{-2t}\frac{dy}{dt}=1+2e^{2t}+3e^{4t}$ has xactly o... |
2,804,129 | <p>To Evaluate the Limit $$L=\lim_{n \to \infty}\left(1+\sum_{k=1}^{n} \frac{1}{\binom{n}{k}}\right)^n \tag{1}$$</p>
<p>My try:</p>
<p>I tried to use $$\frac{1}{\binom{n}{k}}+\frac{1}{\binom{n}{k+1}}=\frac{n+1}{n} \frac{1}{\binom{n-1}{k}} $$</p>
<p>taking summation both sides from $k=1$ to $k=n$ we get</p>
<p>$$\su... | Atif Farooq | 451,530 | <p><em>Solution</em>. The matrix representation $M$ implies that the dimension of the column space and consequently that of $\operatorname{range}T = 3$ thus by the rank-nullity theorem $$\dim\operatorname{null}T = \dim\operatorname{span}\{1,e^{2t},e^{4t}\}- \dim\operatorname{span}\{T(1),T(e^{2t}),T(e^{4t})\}
= 0$$ imp... |
21,243 | <p>I know that this question is some sort of bridge between Informatics and Mathematics, not knowing the best place where to post this question, I opted for this place because of the type of answer I want (which concerns Math more than anything).</p>
<p>Consider to have a graph represented by a collection of nodes and... | Sam | 6,582 | <p>There are a <strong>lot</strong> of properties on eigenvalues of adjacency matrix.</p>
<p>If the diameter of $G$ is d, then $A$ has at least $d+1$ distinct eigenvalues.</p>
<p>For instance, if a graph is $d$-regular, then its largest eigenvalue is bound by d. He is connected if and only the multiplicity of d is 1.... |
1,515,478 | <p>Given a quadratic equation with one and only one root (for example $6-\sqrt{2}$ ). Does there exist integers $a,b$ and $c$ where $ax^2 + bx + c = 0$ for that root?</p>
| marty cohen | 13,079 | <p>Not even if the coefficients are rational.
If a quadratic equation
has $r$ and $s$ as roots,
it can be written as
$(x-r)(x-s)
=x^2-(r+s)x + rs
$.</p>
<p>If the coefficients are rational,
then
$r+s$ is rational,
so $r$ and $s$
are both rational
or both irrational.</p>
|
211,290 | <p>Is it possible to import graphs generated by <code>geng</code> (a tool from <a href="http://pallini.di.uniroma1.it/" rel="noreferrer">the nauty suite</a>) one by one, rather than all at once. If one could also specify not only the order but also the number of edges that would be great, but the main thing is to be a... | Mariusz Iwaniuk | 26,828 | <p>Try this:</p>
<pre><code>a = 7.06; b = 0; c = 0.107763;(*initial values to obtain the root*)(*defining the equations*)
Eq1 = ((a - b)*((628*d)^(1 - c))*Cos[c*Pi/2]);
Eq2 = 1 + 2*((628*d)^(1 - c))*Sin[c*Pi/2] + ((628*d)^(2*(1 - c)));
(*define the equation which the root must be calculated*)
Eqt = Eq1 - 0.08*Eq2 ==... |
4,041,140 | <p>this is a problem which is for homework in my math course. The problem states that you must find two distinct, non-zero matrices, (Size 2x2) such that A * B + A + B = 0.</p>
<p>I'm not really looking for an answer, but rather the methodology I should be using to come to this answer. It seems like the easiest way to ... | Kenta S | 404,616 | <p><span class="math-container">$AB+A+B=0$</span> is equivalent to <span class="math-container">$(A+I)(B+I)=AB+A+B+I=I$</span> for the identity matrix <span class="math-container">$I=\begin{bmatrix}1 &0\\0&1\end{bmatrix}$</span>.</p>
<p>Thus, picking any invertible matrix <span class="math-container">$M$</span>... |
2,301,198 | <p>Solve the initial value problem for the sequence $\left \{ u_{n}| n \in \mathbb{N} \right \}$ satisfying the recurrence relation:
$u_n − 5u_{n-1} + 6u_{n−2} = 0 $ with $u_0 = 1$ and $u_1 = 1$.</p>
<p>Ive gotten the general solution to be $u_n = A(2)^n + B(3)^n$. </p>
<p>Once I sub the initial values: </p>
<p>$u_0... | Cye Waldman | 424,641 | <p>These problems of generalized Fibonacci sequences seem to show up here weekly. Here, then, is a generalized solution for solving <strong>all</strong> such problems with a recursion formula of the type $f_n=af_{n-1}+bf_{n-2}$ and arbitrary initial conditions.</p>
<p>There have been many extensions of the sequence wi... |
1,545,929 | <p>Assume $0<\alpha\leq 1$ and $x>0$. Does the following inequality hold?
$$(1-e^{-x})^{\alpha}\leq (1-\alpha e^{-x})$$
I know that the reverse inequality holds if $\alpha\ge 1$.</p>
| layman | 131,740 | <p>First of all, all norms on $\Bbb R^{n}$ are equivalent (do you know what that means?). As a consequence, if a function is continuous with respect to one norm, it is continuous with respect to every norm, so you can just pick the norm that makes the proof of continuity easiest based on the given function.</p>
<p>Al... |
2,511,095 | <p>Let $p$ be an odd prime. We know that the polynomial $x^{p-1}-1$ splits into linear factors modulo $p$. If $p$ is of the form $4k+1$ then we can write
$$x^{p-1}-1=x^{4k}-1=(x^{2k}+1)(x^{2k}-1).$$
The theorem of Lagrange tells us that any polynomial congruence of degree $n$ mod $p$ has at most $n$ solutions. Hence we... | John Alexiou | 3,301 | <p>You can try eliminating the $a^3$ term, by multiplying the first equation by $b$ and the second equation by $\frac{a}{3}$ and subtracting the two.</p>
<p>$$ \begin{cases}
b \left(a^3 + 39 ab^2 - 18 = 0 \right) \\
\frac{a}{3} \left( 3a^2 b + 13 b^3 - 5 = 0 \right)
\end{cases} \Rightarrow $$</p>
<p>$$ \begin{c... |
2,292,324 | <p>I know what the answer to this question is, but I am not sure how the answer was reached and I would really like to understand it! I am omitting the base case because it is not relevant for my question.</p>
<p>Inductive hypothesis:</p>
<p>$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac... | fleablood | 280,126 | <p>"What I am confused about is where the 1/(k+1)*(k+2) comes from in the first line of the inductive step."</p>
<p>It comes because you are trying to evaluate for $n = k+1$</p>
<p>You want to prove the statement $\frac 1{1*2} + .....+ \frac 1{(n-1)(n)} + \frac 1{n*(n+1)} = \frac n{n+1}$.</p>
<p>For $n = k$ if you... |
878,785 | <p>I know that the common approach in order to find an angle is to calculate the dot product between 2 vectors and then calculate arcus cos of it. But in this solution I can get an angle only in the range(0, 180) degrees. What would be the proper way to get an angle in range of (0, 360)?</p>
| AgvaniaRekuva | 530,850 | <p><strong>Before reading this answer</strong> - Imagine your angle in a 3D space - you can look at it from the "front" and from the "back" (front and back are defined by you). The angle from the front will be the opposite of the angle that you see from the back. So there is no real sense in a value in a range larger t... |
878,785 | <p>I know that the common approach in order to find an angle is to calculate the dot product between 2 vectors and then calculate arcus cos of it. But in this solution I can get an angle only in the range(0, 180) degrees. What would be the proper way to get an angle in range of (0, 360)?</p>
| theodore panagos | 615,306 | <p>I write the formula like I wrote in excel . (xa,ya,xb,yb put in the cells a2,b2,c2,d2).</p>
<p>angle(vector.a,vector.b)</p>
<p>=(180/pi())* abs(pi()/2*((1+sign(a2))* (1-sign(b2^2))-(1+sign(c2))* (1-sign(d2^2)))</p>
<p>+pi()/4*((2+sign(a2))*sign(b2)-(2+sign(c2))*sign(d2))</p>
<p>+sign(a2*b2)*atan((abs(a2)-abs(b2))/(a... |
436,225 | <p><a href="http://en.wikipedia.org/wiki/Incidence_matrix">The incidence matrix</a> of a graph is a way to represent the graph. Why go through the trouble of creating this representation of a graph? In other words what are the applications of the incidence matrix or some interesting properties it reveals about its grap... | dtldarek | 26,306 | <p>There are many advantages, especially if the total number of edges is $|E| = \Omega(|V|^2)$. First of all, worst-case constant time for adding, deleting edges, also testing if edge exists (adjacency lists/sets might have some additional $\log n$ factors). Second, simplicity: no advanced structures needed, easy to wo... |
2,241,100 | <p>Please someone help me solve the following equation in terms of $y$:</p>
<blockquote>
<p><strong>$\frac{y^2}{2}+y = \frac{x^3}{3}+\frac{x^2}{2}+c_1$</strong></p>
</blockquote>
<p>The calculator gives me:</p>
<blockquote>
<p>$y = \frac{1}{3}(\sqrt{3}\sqrt{c_1+2x^3+3x^2+3}-3), -\frac{1}{3}(\sqrt{3}\sqrt{c_1+2x^... | Ángel Mario Gallegos | 67,622 | <p>Since $$\frac{y^2}{2}+y = \frac{x^3}{3}+\frac{x^2}{2}+c_1\qquad\iff\qquad \frac12(y+1)^2-\frac12= \frac{x^3}{3}+\frac{x^2}{2}+c_1$$
It follows
$$(y+1)^2=2\left(\frac{x^3}{3}+\frac{x^2}{2}+c_1\right)+1$$
Then
$$y=-1\pm\sqrt{2\left(\frac{x^3}{3}+\frac{x^2}{2}+c_1\right)+1}$$</p>
|
2,106,003 | <p>I was just reading about the <a href="https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox">Banach–Tarski paradox</a>, and after trying to wrap my head around it for a while, it occurred to me that it is basically saying that for any set A of infinite size, it is possible to divide it into two sets B and C su... | Paul Sinclair | 258,282 | <p>The fullest version of the paradox (that is known to me) says that if you have two sets $A, B \subseteq \Bbb R^3$, both of which have non-empty interior, then you can divide $A$ up into some finite number $n$ of disjoint subsets, then move those subsets around isometrically so that they remain disjoint and their uni... |
413,165 | <p>I am a graduate student and I've been thinking about this fun but frustrating problem for some time. Let <span class="math-container">$d = \frac{d}{dx}$</span>, and let <span class="math-container">$f \in C^{\infty}(\mathbb{R})$</span> be such that for every real <span class="math-container">$x$</span>, <span class=... | Terry Tao | 766 | <p>I was able to adapt <a href="https://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial/34067#34067">the accepted answer</a> to <a href="https://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial">this MathOverflow ... |
2,506,279 | <blockquote>
<p>If $\lim_{x\to \infty}xf(x^2+1)=2$ then find
$$\lim_{x\to 0}\dfrac{2f'(1/x)}{x\sqrt{x}}=?$$</p>
</blockquote>
<p>My Try :
$$g(x):=xf(x^2+1)\\g'(x)=f(x^2+1)+2xf'(x^2+1)$$
Now what?</p>
| Lutz Lehmann | 115,115 | <p>Using Python to replicate your calculations, and enumerating the bit string from left-to-right,</p>
<p><code>sum(int(b)*2**(1-k) for k,b in enumerate("110010010000111111011011"))
</code> </p>
<p>correctly returns <code>3.1415927410125732</code>. However, enumerating the bit string in the wrong direction (<em>from ... |
2,613,484 | <p>Give an example of a vector space which has 125 elements.
I don't know how proceed!!!
Is there any technique about the field?? </p>
| GhD | 191,008 | <p>Just consider the field $\mathbb Z_5$ and mutiply it $3$ times with itself. Then new set over the field $\mathbb Z_5$ is a vector space with $125$ elements.</p>
|
1,401,516 | <p>Given is the unit circle in the plane. Choose randomly point in it, such that $P(\left(x,y\right)\in A)$ is proportional to area of $A$, where $A$ is measurable set in plane. Find density function of random variable $X$ which represents the $x$ coordinate of this point.</p>
<p>My idea was to find $P(X\leq x)$ and t... | Community | -1 | <p>It is easier to find the marginal density directly. For each $-1<x<1$,
we have
$$f_X(x)=\int_{-\infty}^\infty f(x,y)\,dy=
\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}{1\over \pi}\,dy=
{2\sqrt{1-x^2}\over \pi},$$
that is, $f_X(x)$ is the length of the slice at $x$
divided by the area of the circle. </p>
|
1,724,812 | <p>I'm struggling with the following limit:</p>
<p>$$\lim_{x \to 0} \frac{1-(\cos x)^{\sin x}}{x^2}$$</p>
<p>Don't know where to start with this. Hints/solutions very appreciated.</p>
| jim | 289,829 | <p>For a series approach, just look at $y(x) = \cos(x)^{\sin(x)}$ and consider the logarithm $\ln y = \sin(x) \ln \cos(x)$. For $x \to 0$ then $\ln y \approx (x + x^3/3) \ln (1 - x^2/2)$ and the expansion for the $\ln$ term is well known.</p>
|
3,830,204 | <p>Working through <em>Spivak's Calculus</em> and using old assignments from the course offered at my school I'm working on the following problem, asking me to find the integral <span class="math-container">$$\int \frac{1}{x^{2}+x+1} dx$$</span></p>
<p>Looking through Spivak and previous exercises I worked on, I though... | Karagum | 428,319 | <p>It's actually very simple to identify this integral as this:</p>
<p><span class="math-container">$$\int \frac{1}{x^2+x+1} dx = \int \frac{1}{\left(x+\frac{1}{2} \right)^2+ \frac{3}{4}} dx.$$</span></p>
<p>Now you can see that you can use the following rule for integration: <span class="math-container">$$\int \frac{1... |
3,830,204 | <p>Working through <em>Spivak's Calculus</em> and using old assignments from the course offered at my school I'm working on the following problem, asking me to find the integral <span class="math-container">$$\int \frac{1}{x^{2}+x+1} dx$$</span></p>
<p>Looking through Spivak and previous exercises I worked on, I though... | Community | -1 | <p>Though the efficient way has been given several times, decomposition in simple fractions remains your good friend.</p>
<p>The polynomial <span class="math-container">$x^2+x+1$</span> has complex roots, let <span class="math-container">$\omega$</span> and <span class="math-container">$\omega^*$</span>, which, inciden... |
3,552,915 | <p>Determine the point on the plane <span class="math-container">$4x-2y+z=1$</span> that is closest to the point <span class="math-container">$(-2, -1, 5)$</span>. This question is from Pauls's Online Math Notes. He starts by defining a distance function: </p>
<p><span class="math-container">$z = 1 - 4x + 2y$</span></... | MilesB | 744,186 | <p>In general, if <span class="math-container">$$d(x,y)=\sqrt {f(x,y)}$$</span> for some positive function <span class="math-container">$f(x,y)$</span> then the minima of d will correspond to the minima of f. So, if <span class="math-container">$f(x,y)$</span> is differentiable, it makes sense to search for solutions ... |
1,284,039 | <p>What function satisfies $f(x)+f(−x)=f(x^2)$?</p>
<p>$f(x)=0$ is obviously a solution to the above functional equation.</p>
<p>We can assume f is continuous or differentiable or similar (if needed).</p>
| Lukas Geyer | 43,179 | <p>Define $f(x)$ any way you want for $x > 0$, then define $f(-x) = f(x^2) - f(x)$ also for $x>0$. If you want continuity, make sure that $\lim_{x\to 0} f(x) = 0$.</p>
|
2,438,362 | <p>We have, $\rho(A) \leq \|A\|$
where $\rho(A)$ denotes the spectral radius of $A$.</p>
<p>Now there is a corollary
that $\rho(A) < 1$ iff $\|A\|<1$
it is clear that when $\|A\|<1$ then $\rho(A)<1$</p>
<p>but how to show that if $\rho(A)<1$ then $\|A\|<1$,
perhaps it is because of this one</p>
<p>... | GAVD | 255,061 | <p>You can use <a href="https://en.wikipedia.org/wiki/Spectral_radius#Gelfand.27s_formula" rel="nofollow noreferrer">Gelfand's formula</a>: $$\rho(A) = \lim_{k\to \infty} \|A^k\|^{\frac{1}{k}}.$$</p>
<p>Otherwise, follow <a href="https://mathoverflow.net/questions/179105/operator-norm-vs-spectral-radius-for-positive-m... |
277,594 | <p><a href="https://i.stack.imgur.com/yX9my.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/yX9my.gif" alt="enter image description here" /></a></p>
<pre><code>Manipulate[
ParametricPlot[{Sec[t], Tan[t]}, {t, 0, u}, PlotStyle -> Dashed,
PerformanceGoal -> "Quality", Exclusions -> All,
... | Syed | 81,355 | <p>As a possibility, let:</p>
<pre><code>expr = a[1, 1] x[1]^2 + a[1, 2] x[1] x[2] + a[2, 3] x[2] x[3]
rule = {Times[y__, x_[a_Integer] x_[b_Integer]] :> y C[a, b]
, Times[y__, x_[a_Integer] x_[a_Integer]] :> y G[a, a]
};
</code></pre>
<p>Test:</p>
<pre><code>expr /. rule
</code></pre>
<blockquote>
<pre><... |
277,594 | <p><a href="https://i.stack.imgur.com/yX9my.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/yX9my.gif" alt="enter image description here" /></a></p>
<pre><code>Manipulate[
ParametricPlot[{Sec[t], Tan[t]}, {t, 0, u}, PlotStyle -> Dashed,
PerformanceGoal -> "Quality", Exclusions -> All,
... | Trev | 52,996 | <p>I'm not sure whether this would work for you here, but this reminds me of similar cases where <code>Plus</code> and <code>Times</code> would give me headaches. An easy fix was to just replace as <code>expr /. Times -> times</code> (and/or <code>Plus</code> to <code>plus</code>), do whatever, and replace back when... |
3,371,339 | <p>How to show <span class="math-container">$Pr(X>2E(X))\le 1/2$</span> given that <span class="math-container">$X$</span> is a continuous random variable and <span class="math-container">$P(X\le 0)=0$</span>? <span class="math-container">$E(X)$</span> here is the mean of <span class="math-container">$X$</span>.</p>... | Reveillark | 122,262 | <p>Here's another way of doing it. Define</p>
<p><span class="math-container">$$
g:=\liminf_{n\to\infty} f_n
$$</span></p>
<p><span class="math-container">$$
h:=\limsup_{n\to\infty} f_n
$$</span></p>
<p>The functions <span class="math-container">$g$</span> and <span class="math-container">$h$</span> are both measura... |
1,710,304 | <p>I have a boolean algebra equation that i'm not able to simplify fully.</p>
<p>\begin{align}
&(c+ab)(d+b(a+c))\\
&(c+ab)(d+ba+bc)\\
&cd+ abc + bc^2+abd+a^2 b^2 + ab^2 c\\
&\text{using boolean laws $x^2=x$ and $x+x=x$}\\
&cd + bc + abd + ab + (abc + abc)\\
&cd + bc + abd + ab + abc
\end{align... | CAGT | 119,244 | <p>You have backwards the AND's and the OR's entered in Mathematica, you should use:</p>
<p>BooleanMinimize[(C || (A && B)) && (D || (B && (A || C)))</p>
<p>|| = AND (*)</p>
<p>&& = OR (+)</p>
|
2,172,975 | <p>I am reading <a href="http://www.deeplearningbook.org/contents/linear_algebra.html" rel="nofollow noreferrer">http://www.deeplearningbook.org/contents/linear_algebra.html</a> Chapter $2$, page $44$ ($3$rd paragraph) of this book and got confused. Can any body help me to understand this paragraph? Thanks in advance.<... | Ethan Bolker | 72,858 | <p>I'll start by assuming that you understand that if an $n \times n$ symmetric matrix has $n$ distinct eigenvalues then the eigenspaces are all one dimensional and any basis of eigenvectors is essentially unique: it contains one nonzero vector from each eigenspace.</p>
<p>One way to understand the nonuniqueness when ... |
1,307,159 | <p>Let $B$ denote the unit ball of $L^\infty$.
Question: is $B$ sequentially compact for the topology of convergence in measure ? I am not necessarily assuming that the measure is finite (but $\sigma$ finite is fine).</p>
<p>(I have looked a bit, and a counterexample to a lot of similar questions was the sequence of ... | Martin Argerami | 22,857 | <p>With $\sigma$-finite there is a counterexample, which shows that the ball is not sequentially compact: consider $L^\infty(\mathbb R)$ with Lebesgue measure, and let $I_n=(n,n+1)$, $n\in\mathbb Z$, and $f_n=\chi^{\phantom{I_n}}_{I_n}$. Then
$$
\mu(\{x:\ |f_m-f_n|\geq\varepsilon\})=2
$$
for any $\varepsilon\leq1$. </... |
1,307,159 | <p>Let $B$ denote the unit ball of $L^\infty$.
Question: is $B$ sequentially compact for the topology of convergence in measure ? I am not necessarily assuming that the measure is finite (but $\sigma$ finite is fine).</p>
<p>(I have looked a bit, and a counterexample to a lot of similar questions was the sequence of ... | Albert | 19,331 | <p>Even in the finite measure case this seems false. </p>
<p>Take $f_n(x)=\mathrm{sign}(2^n x)$ in $L^\infty(0,2\pi)$.
Then for all $n \neq m$, $\mathrm{Leb}\{ x , |f_n(x)-f_m(x)|>1\}=\pi$.</p>
|
4,338,285 | <p>I have been thinking about the problem of finding the sum of the first squares for a long time and now I have an idea how to do it.
However, the second step of this technique looks suspicious.</p>
<ol>
<li><p><span class="math-container">$$\sum_{i=1}^n i = \frac{n^2+n}{2}$$</span></p>
</li>
<li><p><span class="math-... | PavelDev | 709,857 | <p>After 4 months, I suspect that the proof of this assumption is very simple. Currently, this answer is a draft, but it will demonstrate all the necessary ideas to generalize to an arbitrary case. We will start with the next special case:</p>
<ol>
<li><p><span class="math-container">$$\sum_{i=1}^{n}i=\frac{n^{2}+n}{2}... |
2,384,538 | <p>I am studying Linear Algebra Done Right, chapter 2 problem 6 states:</p>
<blockquote>
<p>Prove that the real vector space consisting of all continuous real valued functions on the interval $[0,1]$ is infinite dimensional.</p>
</blockquote>
<p><strong>My solution:</strong></p>
<p>Consider the sequence of functio... | Kamil Maciorowski | 331,040 | <p>Two identical functions have identical derivatives. Differentiate your equality $n$ times and you will see the two sides cannot be identical.</p>
|
1,970,305 | <p>I have just begun reading through Section 3.2 of Hatcher's Algebraic Topology. While I reasonably understood the computations relating to the cup product, I was unsure of the purpose of the cup product. From what I knew, it does not help us to compute cohomology groups, given that we need the cohomology groups to co... | Len West | 377,227 | <p>Evaluating equation mod 5 we get
x $\equiv$ 3 (mod 5)
Evaluating equation mod 3 we get
z $\equiv$ 2 (mod 3)
Setting x=5m+3, z=3n+2 in equation we get
y=1-2m-2n
Integer values for m & n will generate
Integer solutions for x,y,z</p>
|
97,788 | <p>What exactly is the connection between knots and operator algebra? I heard that Jones established such a connection while discovering the celebrated Jones Polynomial. </p>
<p>Now Jones Polynomial is probably understood out of that context on its own, but what was it with Operator algebras in this space ? Can someon... | MTS | 703 | <p>I would recommend you to look at Jones' survey paper from 1986, entitled <em>A New Knot Polynomial and Von Neumann Algebras.</em> It is very readable. Let me try to make a brief summary, though.</p>
<p>The basic object you start with is a $II_1$ factor. This is a von Neumann algebra $M$ with trivial center $Z(M)... |
4,459,722 | <p>I have the vector field
<span class="math-container">\begin{align*}
X:\mathbb R^d&\to\mathbb R^d\\
x&\mapsto\frac{x}{\|x\|}
\end{align*}</span>
which is a differentiable vector field outside of the origin, and I am interested in its divergence. After some easy computation we get
<span class="math-container">... | peek-a-boo | 568,204 | <p>The divergence you wrote is correct distributionally as well. To see this, let <span class="math-container">$\phi$</span> be any test function, then using the definition of distributional divergence, dominated convergence, and the divergence theorem, we get
<span class="math-container">\begin{align}
\langle\text{div... |
779,696 | <p>So my problem is:</p>
<p>$$\arcsin (x) = \arccos (5/13)$$ </p>
<p><strong>^ Solve for $x$.</strong></p>
<p>How would I begin this problem? Do I draw a triangle and find the $\sin(x)$ or is there a more algebraic way of doing this? Thanks in advance for any help.</p>
| drhab | 75,923 | <p><strong>Hint</strong>:</p>
<p>It is sufficient to prove that $P\left[B_{1}\cap\cdots\cap B_{n-1}\right]=P\left(B_{1}\right)\times\cdots\times P\left(B_{n-1}\right)$
whenever $B_{i}\in\left\{ A_{i},A_{i}^{c}\right\} $ for $i=1,\dots,n-1$. </p>
<p>You can probably do that with induction to $n$.</p>
|
112,021 | <p>Let $n$ be a positive integer.
The $n$ by $n$ Fourier matrix may be defined as follows:</p>
<p>$$
F^{*} = (1/\sqrt{n}) (w^{(i-1)(j-1)})
$$</p>
<p>where </p>
<p>$$
w = e^{2 i \pi /n}
$$</p>
<p>is the complex $n$-th root of unity with smaller positive argument
and $*$ means transpose -conjugate.</p>
<p>It is we... | Balu Santhanam | 350,201 | <p>There are several known eigenvector basis for the DFT using
matrices that commute with the DFT matrix:</p>
<ol>
<li>Alberto Grunbaum (Tridiagonal Commuting matrix), Journal of Mathematical Analysis and Applications, 1982.</li>
<li>B. Santhanam and T. S. Santhanam (quantum mechanics in finite dimensions)(Symmetric ma... |
3,295,021 | <p>The Hopf fibration is a mapping <span class="math-container">$h:\mathbb{S^3} \mapsto\mathbb{S}^2$</span> defined by <span class="math-container">$r\mapsto ri\bar{r}$</span> where <span class="math-container">$r$</span> is a unit quaternion in the form <span class="math-container">$r=a+bi+cj+dk $</span> where <span c... | Thomas Andrews | 7,933 | <p><strong>Part ii:</strong></p>
<p>In general: <span class="math-container">$$\lfloor z\rfloor \leq z< \lfloor z\rfloor+1.$$</span> </p>
<p>For ii, set <span class="math-container">$z=x/2$</span> then we get:
<span class="math-container">$$\left\lfloor\frac{x}{2}\right\rfloor\leq \frac{x}{2}<\left\lfloor\frac{... |
3,295,021 | <p>The Hopf fibration is a mapping <span class="math-container">$h:\mathbb{S^3} \mapsto\mathbb{S}^2$</span> defined by <span class="math-container">$r\mapsto ri\bar{r}$</span> where <span class="math-container">$r$</span> is a unit quaternion in the form <span class="math-container">$r=a+bi+cj+dk $</span> where <span c... | G Cab | 317,234 | <p>Let's denote with <span class="math-container">$\{ x \}$</span> the fractional part:
<span class="math-container">$$
x = \left\lfloor x \right\rfloor + \left\{ x \right\}
$$</span></p>
<p>Then I would suggest that you first master the addition
<span class="math-container">$$
\eqalign{
& \left\lfloor {x + y}... |
1,081,447 | <p>I'm talking about a Roulette wheel with $38$ equally probable outcomes. Someone mentioned that he guessed the correct number five times in a row, and said that this was surprising because the probability of this happening was $$\left(\frac{1}{38}\right)^5$$</p>
<p>This is true if you only play the game $5$ times. H... | awkward | 76,172 | <p>Feller, "An Introduction to Probability Theory and Its Applications", Third Edition, gives a useful approximation on p. 325, equation 7.11.</p>
<p>Suppose we toss a possibly biased coin $n$ times, where the probability of a head is $p$ and $q = 1-p$. Let $q_n$ be the probability there is no run of $r$ successive h... |
832,710 | <p>Does there exist an algebraic structure $(\mathbb{K},+)$ such that equations of the form $x+a=x+b$, $a\neq b$ have solutions for all $a,b\in \mathbb{K}$?</p>
| Community | -1 | <p>This is a version of the <a href="http://mathworld.wolfram.com/MovingLadderProblem.html" rel="nofollow">moving ladder problem</a>, though more difficult because the road and lane have different widths. Let $\alpha$ be the angle between the pole and the direction of the road. In the process of moving, $\alpha$ takes ... |
3,152,021 | <p>I'm wondering if there are well known sorting techniques for the following problem.</p>
<p><strong>Problem</strong>:</p>
<p>Suppose you would like to sort a list of integer numbers <span class="math-container">$0, 1, 2, \ldots, d$</span>.
If one is only allowed to use swaps of adjacent positions the major part
of ... | angryavian | 43,949 | <p>If <span class="math-container">$D$</span> is a diagonal matrix with diagonal entries <span class="math-container">$\lambda_1, \ldots, \lambda_n$</span>, then the eigenvectors are the standard basis vectors <span class="math-container">$e_1 = (1,0,\ldots, 0)$</span>, <span class="math-container">$e_2 = (0,1,0, \ldot... |
2,939,163 | <p>I want to find a certain <span class="math-container">$x$</span> that belongs to <span class="math-container">$\mathbb R$</span> so that </p>
<p><span class="math-container">$$\left|\begin{array}{r}1&x&1\\x&1&0\\0&1&x\end{array}\right|=1$$</span></p>
<p>This should be easy enough. I apply t... | Siong Thye Goh | 306,553 | <p>Hint:</p>
<p>Notice that <span class="math-container">$x=1$</span> is a solution, hence you can reduce the problem to a quadratic equation.</p>
|
4,062,667 | <p>Let <span class="math-container">$A$</span> be a <span class="math-container">$n^{th}$</span> order square and skew-symmetric matrix, if <span class="math-container">$(E-A)$</span> is an invertible matrix show that
<span class="math-container">$(E+A)(E-A)^{-1}$</span> is an invertible matrix (where <span class="math... | Priya | 741,323 | <p>I am writing an answer to my own question please let me know if I am wrong</p>
<p>1.
<span class="math-container">\begin{align*}
||(E-A)u|| &= \sqrt{((E-A)u)^T(E-A)u}\\
&= \sqrt{u^T (E-A)^T (E-A)u}\\
&= \sqrt{u^T(E^T- A^T)(E-A)u}\\
&= \sqrt{u^T (E^TE - E^TA -A^TE + A^T... |
736,036 | <p><strong>Problem</strong> - The least number which leaves remainders 2, 3, 4, 5 and 6 on dividing by 3, 4, 5, 6 and 7 is?</p>
<p><strong>Solution</strong> - Here 3-2 = 1, 4-3 = 1, 5-4 = 1 and so on.</p>
<p>So required number is (LCM of 3, 4, 5, 6, 7) - 1 = 419</p>
<p><strong>My confusion</strong> - </p>
<p>I didn... | Bill Dubuque | 242 | <p>$\, x\!+\!1\equiv 0\pmod{m_i}\iff m_i\mid x\!+\!1\iff {\rm lcm}\{m_i\}\mid x\!+\!1\iff x\equiv -1\pmod{{\rm lcm}\{m_i\}}$</p>
<p>Or, equivalently: $\,\ x\equiv -1\pmod{m_i}\iff x\equiv -1\pmod{{\rm lcm}\{m_i\}},\ $ whihc may be viewed as the special <a href="https://math.stackexchange.com/a/73541/242"><em>constant... |
736,036 | <p><strong>Problem</strong> - The least number which leaves remainders 2, 3, 4, 5 and 6 on dividing by 3, 4, 5, 6 and 7 is?</p>
<p><strong>Solution</strong> - Here 3-2 = 1, 4-3 = 1, 5-4 = 1 and so on.</p>
<p>So required number is (LCM of 3, 4, 5, 6, 7) - 1 = 419</p>
<p><strong>My confusion</strong> - </p>
<p>I didn... | Man_From_India | 134,937 | <p>I just found this <a href="https://math.stackexchange.com/a/731302/134937">answer</a> in another similar <a href="https://math.stackexchange.com/questions/731299/find-the-least-value-of-x-which-when-divided-by-3-leaves-remainder-1">question</a></p>
<p>And this explain it.</p>
<p>Thanks.</p>
|
1,765,530 | <p>How many $5$-digit numbers (including leading $0$'s) are there with no digit appearing exactly $2$ times? The solution is supposed to be derived using Inclusion-Exclusion.</p>
<p>Here is my attempt at a solution:</p>
<p>Let $A_0$= sequences where there are two $0$'s that appear in the sequence.</p>
<p>...</p>
<p... | Christian Blatter | 1,303 | <p>The partitions of $5$ showing the admissible multiplicities of occurring digits are $(1,1,1,1,1)$, $(3,1,1)$, $(4,1)$, and $(5)$.</p>
<p>There are $10\cdot9\cdot 8\cdot 7\cdot 6=30\,240$ numbers with five different digits.</p>
<p>There are $10\cdot{9\choose 2}=360$ ways to choose three digits whereby the first ch... |
1,649,320 | <p>I am having a hard time proving this simple and natural identity of sets. what I do is go round and round in circles:</p>
<p>$$A\cup( A\cap B) = (A\cup A) \cap (A\cup B)$$
$$= A \cap(A\cup B)$$</p>
<p>Now what? I apply the distributive property again and reach the first expression. How can I show this using set pr... | J.ALI | 296,390 | <p>$A∪(A∩B)=(A∪A)∩(A∪B)$=$A∩(A∪B)$=$A$, the first equality using the distribution law and the last equality since $A\subset A∪B$.</p>
|
2,994,296 | <p>I'm trying to figure out how to prove, that <span class="math-container">$$\lim_{n\to \infty} \frac{n^{4n}}{(4n)!} = 0$$</span>
The problem is, that <span class="math-container">$$\lim_{n\to \infty} \frac{n^{n}}{n!} = \infty$$</span>
and I have no idea how to prove the first limit equals <span class="math-container"... | Samvel Safaryan | 534,896 | <p>Define
<span class="math-container">$$
\\a_n=\frac{n^{4n}}{(4n)!}=>
\\\frac{a_{n+1}}{a_n}=\frac{(n+1)^{4(n+1)}}{n^{4n}}\cdot\frac{(4n)!}{(4n+4)!}=
\\(n+1)^4\big(1+\frac{1}{n}\big)^{4n}\cdot\frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}=b_n
$$</span>
<span class="math-container">$$
\\\dfrac{(4n+1)(4n+2)(4n+3)(4n+4)}{(n+1)^4}... |
79,084 | <p>Let $X$ be a topological space (say a manifold). A result of R. Thom states that the pushforwards of fundamental classes of closed, smooth manifolds generate the rational homology of $X$. This work of Thom predates the development of bordism. Is there now a more elementary proof of this result that does not rely ... | Tom Goodwillie | 6,666 | <p>I suspect that the "obvious" proof used an Atiyah-Hirzebruch spectral sequence, so it's not obvious unless you are happy with spectral sequences.</p>
<p>Here is an argument with no spectral sequence in it.</p>
<p>There is a homology theory $\Pi_\ast$ called stable homotopy theory. It has a natural map to ordinary ... |
2,458,863 | <p>I tried to find the critical points of the function</p>
<p>$$f(x,y) = x^2y-2xy + \arctan y $$</p>
<p>And I found that is $P(1,0)$, the problem is that the Hessian is null, and I don't know how to procede to determine the nature of that point.
Can you help me ?</p>
<p><strong>Update:</strong> Thanks you all, and I... | Arthur | 15,500 | <p>More generally: the product of all elements in <em>any</em> finite abelian group has order $2$ or $1$, and the order is $2$ iff there is exactly one element of order $2$ in the group.</p>
<p>The first claim is easy to show: For each element that does not have order $2$, its inverse is also in the product, so they p... |
2,458,863 | <p>I tried to find the critical points of the function</p>
<p>$$f(x,y) = x^2y-2xy + \arctan y $$</p>
<p>And I found that is $P(1,0)$, the problem is that the Hessian is null, and I don't know how to procede to determine the nature of that point.
Can you help me ?</p>
<p><strong>Update:</strong> Thanks you all, and I... | lhf | 589 | <p>The product of all units mod $m$ can be written as $P=\prod_{x^2\ne1} x \prod_{y^2=1} y$.</p>
<p>The first product is $1$ because each $x$ is paired with $x^{-1}$ and they are different.</p>
<p>Therefore, $P=y_1 \cdots y_n$ with $y_i^2=1$.</p>
<p>If $n=1$, then $P=1$ because $y=1$ is certainly there.</p>
<p>If $... |
1,012,344 | <p>My knowledge of $C^*$-algebras is very little.</p>
<p>We call an element positive if $a=b^*b$ for some $b$ and make a relation on all positive elements by saying
$$
b \geqslant a \iff b-a \text{ is positive}.
$$
I can't figure out why this gives us a poset.</p>
| Martin Argerami | 22,857 | <p>Below is one avenue. </p>
<p>Since $a=b^*b$, we have $a^*=a$. Then $C^*(a)$, the C$^*$-subalgebra generated by $a$, is commutative. In a commutative Banach algebra, we have
$$
\sigma(a)=\{\tau(a):\ \tau \text{ is a character }\}.
$$
So, when $a=b^*b$, $\tau(a)=\tau(b^*)\tau(b)=\overline{\tau(b)}\tau(b)=|\tau(b)|^2\... |
38,439 | <p>I've mentioned before that I'm using this forum to expand my knowledge on things I know very little about. I've learnt integrals like everyone else: there is the Riemann integral, then the Lebesgue integral, and then we switch framework to manifolds, and we have that trick of using partitions of unity to define inte... | Paul Siegel | 4,362 | <p>As the others have mentioned, integration over a connected oriented smooth manifold $M$ can be characterized (modulo some technicalities) according to the fact that it fits into an exact sequence:</p>
<p>$\Omega_c^{n-1}(M) \to \Omega_c^n(M) \to \mathbb{R} \to 0$ </p>
<p>where $\Omega_c^*$ refers to the De Rham com... |
2,713,201 | <p>How would you work something like this out? </p>
<p>Are there similar problems to
$$\frac{d\left( (\cos(x))^{\cos(x)}\right)}{dx}$$
which could be worked out the same way?</p>
| kayush | 532,597 | <p><strong>Hint</strong>: Given proper domain for the function so that $\cos(x) >0$ we can write:<br>
$$(\cos x)^{\cos x} = e^{(\cos x) \ln(\cos x)}$$</p>
|
2,658,691 | <p>I have one last question regarding permutation. I understand the problem and the rule of product but this problem seems to be in a different format compared with the two questions I asked before.</p>
<p>A committee of eight is to form a round table discussion group. In how many ways
may they be seated if the 2 memb... | gerw | 58,577 | <p>You can easily prove the reverse of the sentence after your question: If an $S$ as in the question exists, then $T$ is weak*-weak* continuous.</p>
|
3,443,094 | <blockquote>
<p>If <span class="math-container">$$\lim_{x\to 0}\frac{ae^x-b}{x}=2$$</span> the find <span class="math-container">$a,b$</span></p>
</blockquote>
<p><span class="math-container">$$
\lim_{x\to 0}\frac{ae^x-b}{x}=\lim_{x\to 0}\frac{a(e^x-1)+a-b}{x}=\lim_{x\to 0}\frac{a(e^x-1)}{x}+\lim_{x\to 0}\frac{a-b}{... | Andrei | 331,661 | <p>You are almost done. <span class="math-container">$a+0=2$</span>.</p>
|
1,390,423 | <p>An acquaintance of mine proposed a scenario. Imagine parents who ground their child. Initially, the grounding is for 5 days, but for every day the child misbehaves while they're grounded, the parents will tack on an extra 2 days. The child is very predictable and has a 30% chance of misbehaving on any given day, ... | mjqxxxx | 5,546 | <p>Let $f(n)$ be the expected value of the remaining sentence given that the current sentence is $n$ days. At the end of the day, the new sentence is either $n-1$ days if the kid behaved (with probability $0.7$), or $n+1$ days if the kid didn't behave (with probability $0.3$). So
$$
f(n)=1+\frac{7}{10}f(n-1)+\frac{3}... |
1,246,356 | <p>Let $A,B \in {M_n}$ . suppose $A$ is normal matrix and has distinct eigenvalue, and $AB=0$. why $B$ is normal matrix?</p>
| JJacquelin | 108,514 | <p>On have to take care to the boundaries. One little picture says more than a long speech!</p>
<p><img src="https://i.stack.imgur.com/0jWkj.jpg" alt="enter image description here">
$$
\sum_{k=5}^m\frac{\log(\log(k))}{k\log(k)}
$$
$$\int_5^{m+1}\frac{\ln(\ln (x))}{x\ln(x)}dx<\sum_{k=5}^m\frac{\log(\log(k))}{k\log(k... |
601,951 | <p><em><strong>2</strong> + <strong>5</strong> + <strong>8</strong> + . . . + <strong>(6n-1)</strong> = <strong>n(6n+1</strong>)</em></p>
<p>This is what I have so far. </p>
<p>The <strong>sum</strong> of <strong>(3j-1)</strong> from <strong>j=1</strong> to <em>something I`m not sure of</em>.</p>
| Marko Riedel | 44,883 | <p>This can also be done using a basic complex variables technique, which
gives a close variation on what was posted by @GrigoryM.</p>
<p>Suppose we seek to evaluate
<span class="math-container">$$\sum_{k=0}^n {2n+1\choose 2k+1} {m+k\choose 2n}.$$</span></p>
<p>Introduce the integral representation
<span class="math-co... |
324,594 | <p>You have three buckets, two big buckets holding <code>8 litres</code> of water each and one small empty bucket that can hold <code>3 litres</code> of water. How will you split the <code>16 litres</code> of water to <code>four people</code> evenly? Each person has a container but once water is distributed to someone ... | Ravi Gupta | 528,046 | <p>Above can be completed in 13 steps:
<code>
8,8,0[0,0,0,0]
8,5,3[0,0,0,0]
8,5,0[3,0,0,0]
5,5,3[3,0,0,0]
5,5,0[3,3,0,0]
5,2,3[3,3,0,0]
5,2,0[3,3,3,0]
5,0,2[3,3,3,0]
4,0,3[3,3,3,0]
0,0,3[3,3,3,4]
0,0,2[4,3,3,4]
0,0,1[4,4,3,4]
0,0,0[4,4,4,4]</code></p>
|
1,034,698 | <p>I have an assignment with the following question:</p>
<pre><code>Does an Orthogonal Matrix exist such that its first row consists of the
following values:
</code></pre>
<p>($1$/$\sqrt{3}$, $1$/$\sqrt{3}$, $1$/$\sqrt{3}$)</p>
<pre><code>If there is, find one.
</code></pre>
<p>I know I can solve this question wi... | JohnD | 52,893 | <p>To show $W$ is a subspace, use the Subspace Theorem: $0\in W$, $W$ is closed under addition and scalar multiplication.</p>
<p>To compute the dimension of $W$, just note that $2x+3y-z=0$ implies $x=-3y+z$. So you have two free variables: $y$ and $z$. Thus, $W$ is two dimensional. </p>
<p>Since $\mathbb{R}^3$ is thr... |
619,890 | <p>I have a question.There is a group of 5 men and a group of 7 women.With how many ways can each of the 5 men get married with one of the 7 women?</p>
| Emanuele Paolini | 59,304 | <p>I would define the projection on a affine subspace of an euclidean space as the map which sends any point to the closest one in the affine subspace.</p>
<p>In your case the subspace is defined as the smallest affine subspace containing a given set of points $v_1,\dots, v_k$. Let $w_j = v_{j}-v_k$ for $j=1,\dots, k-... |
3,583,879 | <blockquote>
<p>a) $P_5=11$$</p>
<p>b) <span class="math-container">$P_1+P_2+P_3+P_4+P_5 =26$</span></p>
</blockquote>
<p>For the first part
<span class="math-container">$$\alpha^5+\beta ^5$$</span>
<span class="math-container">$$=(\alpha^3+\beta ^3)^2-2(\alpha \beta )^3$$</span></p>
<p>I found the value of <... | heropup | 118,193 | <p>First, prove the identity <span class="math-container">$$P_{k+1} = P_1 P_k - \alpha\beta P_{k-1}.$$</span> Then note that if <span class="math-container">$\alpha, \beta$</span> are roots of <span class="math-container">$x^2 - x - 1$</span>, then they admit the factorization <span class="math-container">$$x^2 - x - ... |
916,963 | <p>I am reading an introductory book on mathematical proofs and I don't seem to understand the mechanics of proof by contradiction. Consider the following example. </p>
<p>$\textbf{Theorem:}$ If $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$.</p>
<p>$\textbf{Proof:}$ (by contradiction)
Ass... | paw88789 | 147,810 | <p>In proving $A\rightarrow B$ by contradiction, you assume $\neg(A\rightarrow B)$. The negation of $A\rightarrow B$ is $A\wedge \neg B$ (the one false case of an implication).</p>
<p>In your case this means you get to assume $P\rightarrow Q$, $R\rightarrow \neg Q$, and $\neg(P\rightarrow \neg R)$. However $\neg(P\r... |
183,749 | <p>Considering two functions <span class="math-container">$\psi_{1}(u,v)$</span> and <span class="math-container">$\psi_{4}(u,v)$</span>. we have these two parial differential equation for them</p>
<p><span class="math-container">$(-2 i Sech[\frac{u}{\alpha}] \frac{\partial\psi_{4}}{\partial u}+2 i Sech[\frac{v}{\alp... | Alex Trounev | 58,388 | <p>An example of a numerical solution</p>
<pre><code>eq = {((-2 I Sech[u/α] D[ψ4[u, v], u] + 2 I*Sech[v/α] D[ψ4[u, v], v])
+ (-m ψ1[u, v] + 2 I Sech[u/α] D[ψ1[u, v], u]
+ 2 I Sech[v/α] D[ψ1[u, v], v])) == 0, ((2 I*Sech[u/α] D[ψ1[u, v], u] -
2 I Sech[v/α] D[ψ1[u, v], v]) + (-m ψ4[u, v] - 2 I ... |
2,805,312 | <p>Suppose $E$ is a vector bundle over $M, d^E$ a covariant derivative, $\sigma\in\Omega^p(E)$ and $\mu$ a q-form.</p>
<p>I have seen the following pair of formulae for wedge products:</p>
<p>$d^E(\mu\wedge \sigma)=d\mu\wedge\sigma+(-1)^q\mu\wedge d^E\sigma$</p>
<p>$d^E(\sigma\wedge\mu)=d^E\sigma\wedge\mu+(-1)^p\si... | Jackozee Hakkiuz | 497,717 | <p>Just to add to the discussion, I'd like to write the derivation I went through using abstract index notation. It also allows one to work without writing $\sigma = \omega\otimes s$.</p>
<p>I introduce a bit of notation</p>
<blockquote>
<p>$$\begin{align}
\sigma &\equiv {\sigma^{c}}_{a_1\dots a_p} & \sigma... |
131,283 | <p>I came across a question which required us to find $\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^5-5n^3+4n}$. I simplified it to $\displaystyle\sum_{n=3}^{\infty}\frac{1}{(n-2)(n-1)n(n+1)(n+2)}$ which simplifies to $\displaystyle\sum_{n=3}^{\infty}\frac{(n-3)!}{(n+2)!}$. I thought it might have something to do with pa... | Did | 6,179 | <p><strong>Hint:</strong> There exists some $c_k$ independent of $n$ such that
$$
\frac1{(n-2)(n-1)n(n+1)(n+2)}=\sum_{k=-2}^2\frac{c_k}{n+k}.
$$
To find $c_k$, multiply both sides by $n+k$ and evaluate the result at $n=-k$. For example,
$$
c_{-2}=\left.\frac1{(n-1)n(n+1)(n+2)}\right|_{n=2}=\frac1{24}.
$$
<em>Sanity che... |
2,713,038 | <p>I've seen some solutions to this problem but I'm wondering what is incorrect about an argument like this:</p>
<p>$S = \{x \in \mathbb{R}^d: |x| = 1\}$, then $\delta S = \{x \in \mathbb{R}^d: |\delta x| = 1\}$, and so</p>
<p>\begin{align*}
\{ x \in \mathbb{R}^d: |\delta ||x| = 1\} & = \{ x \in \mathbb{R}^d: |x|... | David C. Ullrich | 248,223 | <p>The argument you give is wrong, as already noted. One <em>can</em> show that <span class="math-container">$m(S)=0$</span> by considering the sets <span class="math-container">$\delta S$</span>.</p>
<p>For example, say <span class="math-container">$(\delta_j)\subset(1,2)$</span> and <span class="math-container">$\del... |
462,397 | <p>So, I read the John Baez essay "Lectures on n-categories and cohomology" and I understand the notion of a (-1)-category" and a (-2)-category" and how to derive them. However, I'm not totally clear on what a (-1)-morphism is.</p>
<p>At nLab at <a href="http://ncatlab.org/nlab/show/k-morphism" rel="nofollow noreferr... | Zhen Lin | 5,191 | <ol>
<li>An <strong>object</strong> is always a 0-morphism, so the identity morphism of an object is a 1-morphism.</li>
<li>Yes, the empty category has a $(-1)$-morphism. The convention is that the free $\infty$-category generated by a $(-1)$-morphism is the empty $\infty$-category, and so every $\infty$-category has a... |
4,281,654 | <p>My professor gave me an exercise where I had to show that the special linear group <span class="math-container">$SL(2,\mathbb{R})$</span> is a lie subgroup of <span class="math-container">$GL(2,\mathbb{R})$</span>. I was able to do this part. However, I was then asked to do the following:</p>
<p>All real <span class... | Theone | 463,966 | <p>You can actually do this with Lagrange multipliers. In the following, I always identify the matrix <span class="math-container">$\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}$</span> with the corresponding point <span class="math-container">$(a, b, c, d) \in \mathbb{R}^4$</span>.</p>
<p>First, you have to reco... |
86,536 | <p>Considering we have a an association:</p>
<pre><code>assc = <|"A" -> <|"a" -> 1, "aa" -> 2|>, "B" -> 0, "C" -> 5,"D" -> <|"d" -> 2, "dd" -> 12|>|>
</code></pre>
<p>Let's also consider we have 2 known lists and one list for nested keys
(**
- <em>how to create this list?</... | SquareOne | 19,960 | <p>I've found a simple approach using the function <code>Part</code> (its shorthand is <code>[[ ]]</code>):</p>
<h2>1.</h2>
<p>If you write your list <code>lst1</code> rather that way:</p>
<pre><code>newlst1 = {{"A", "a"}, {"D", "dd"}, {"B"}}
</code></pre>
<p>you can assign any value (for example here <code>999</co... |
812,345 | <p><a href="https://math.stackexchange.com/a/87705/53259">This</a> proves: Similar matrices have the same characteristic polynomial. (Lay P277 Theorem 4)</p>
<p>I prefer <a href="https://math.stackexchange.com/a/8407/53259">https://math.stackexchange.com/a/8407/53259</a>, but this proves that they have the same eigenv... | Frank Wan | 70,425 | <p>"What about in general, even for matrices which are NOT similar?"
You can get matrices A and B, jordan matrices, with the same characteristic polynomial and same eigenvalues and the matrices ARE NOT EQUIVALENT Jordan matrices.</p>
|
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