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 |
|---|---|---|---|---|
2,426,659 | <p>I have a very basic question on wether or not we use the axiom of choice when we prove the very simple fact that the union of open sets of $\mathbb{R}$ (defined as unions of open intervals) is an open set of $\mathbb{R}$. </p>
<p>Say that $(U_i)_{i\in I}$ is a family of open sets of $\mathbb{R}$. So each $U_i$ is o... | Hagen von Eitzen | 39,174 | <p>Your doubt arises because we are only given for $U_i$ that there <em>exists</em> an index set $J$ and intervals $I_j$ such that $U_i=\bigcup_{j\in J}I_j$. But we are not <em>given</em> this information.</p>
<p>However, we can describe the union without invoking choice as
$$\bigcup_{i\in I}U_i = \bigcup_{\langle a,b... |
754,392 | <p>Consider the following integrals in variables $x,y$ over the whole $\mathbb{R}$, where $a,b\in\mathbb{R}/0$ are constants:</p>
<p>$$\int dx \int dy ~\delta(x-a)\delta(y-b\,x)=\int dy ~\delta(y-b\,a)=1$$</p>
<p>In the following we will evaluate the above integrals in a slightly different way and obtain a completely... | Community | -1 | <p>We have</p>
<p>$$F'(x)=e^{-x^2}=\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{n!}$$
with radius of convergence $+\infty$ hence by integrating term by term we find:</p>
<p>$$\int_0^xF'(t)dt=F(x)-F(0)=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)n!}$$</p>
|
3,508,730 | <p>I was trying to prove that <span class="math-container">$(N+\sqrt{N^2-1})^k$</span>, where k is a positive integer, differs from the integer nearest to it by less than <span class="math-container">$(2N-\frac{1}{2})^{-k}$</span>. Note: N is an integer greater than 1. </p>
<p>So, I tried to look for the answer of the... | Somos | 438,089 | <p>Given that <span class="math-container">$\,N>1\,$</span> is an integer,
define the real numbers
<span class="math-container">$\, u := (N+\sqrt{N^2-1}),\,$</span>
<span class="math-container">$\, v := (N-\sqrt{N^2-1}),\,$</span>
and thus <span class="math-container">$\,u+v=2N, u\,v=1.\,$</span>
Define the sequen... |
127,643 | <p>Hello everybody,</p>
<p>I'm a math student who has just got his first degree, and I am studying algebraic geometry since a few months. Something I have noticed is the (to my eyes) huge amount of commutative algebra one needs to push himself some deeper than the elementary subjects. This can be seen just counting th... | user15425 | 15,425 | <p>I would recommend the new book "Computational Commutative Algebra and Algebraic Geometry: Course and exercises with detailed solutions" <a href="https://rads.stackoverflow.com/amzn/click/com/1096374447" rel="nofollow noreferrer" rel="nofollow noreferrer">https://www.amazon.com/dp/1096374447?ref_=pe_3052080_397514860... |
2,498,424 | <p>Consider double sequences $a_{n,m}\in\mathbb R$ where $n,m\in\mathbb Z,$ satisfying</p>
<ol>
<li>$a_{n,m}=a_{n-1,m}+a_{n,m-1}$ for all $n,m\in\mathbb Z,$ and</li>
<li>$\sup_\limits{m\in\mathbb Z}|a_{n,m}|<\infty$ for all $n\in\mathbb Z.$</li>
</ol>
<p>An example solution is $a_{n,m}=(-1)^m2^{-n}.$ A more genera... | mathworker21 | 366,088 | <p>I wanted to post an answer just containing what is needed to answer the OP's question. This answer is due to Josse van Dobben de Bruyn and not me. </p>
<p>Yes, there is a solution not of the form (x). Define the measure <span class="math-container">$\mu$</span> on <span class="math-container">$\mathbb{T} := \{z \in... |
422,196 | <p>After a long reflection, I've decided I won't go to graduate school and do a thesis, among other things. I personally can't cope with the pressure and uncertainty of an academic job.</p>
<p>I will therefore move towards a master's degree in engineering and probably work in industry. However, I'm still passionate abo... | Timothy Chow | 3,106 | <p>While I agree with others that it is possible to pursue mathematical research as an amateur, and I don't think you'll be "ostracized," I do think that there are some potential sociological obstacles that you should be aware of.</p>
<p>If you want to publish a paper, then you have to write your paper in a w... |
422,196 | <p>After a long reflection, I've decided I won't go to graduate school and do a thesis, among other things. I personally can't cope with the pressure and uncertainty of an academic job.</p>
<p>I will therefore move towards a master's degree in engineering and probably work in industry. However, I'm still passionate abo... | Arnab | 23,911 | <p>Anything is possible, but what you propose is very very very hard. You can certainly learn new stuff for fun, but producing research may be challenging. Amongst many, a big factor could be the lack of collaboration. When you are inside a circle, you have access to experts. Without quick access to an expert, even a s... |
422,196 | <p>After a long reflection, I've decided I won't go to graduate school and do a thesis, among other things. I personally can't cope with the pressure and uncertainty of an academic job.</p>
<p>I will therefore move towards a master's degree in engineering and probably work in industry. However, I'm still passionate abo... | Count Iblis | 52,954 | <p>After I left academia, I kept up with doing academic work. What works well for me is to work on a wide range of subjects. So, while my background is in mathematical physics, I use the freedom I have by virtue of not being in academia, to work on whatever I want to work on in my free time. I'm not subject to any pres... |
2,148,484 | <p>Find the value of the series $$\sum_{n=0}^\infty \frac{n^{2}}{2^{n}}.$$ I tried the problem but not getting the sum. Please help.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>for the infinity sum we get $$\sum_{i=0}^n\frac{i^2}{2^i}=2^{-n}(-6+3\cdot2^{n+1}-4n-n^2)$$
can you finish this?</p>
|
1,377,595 | <p>Compute this integral:
$$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$$</p>
<p>my solution is I used derivative of Cauchy integral formula, which is
$$f^{(n)}(z_0) = \frac{n!}{2\pi i}\int \frac{f(z_0)}{(z-z_0)^{n+1}}$$
Then, I got
$$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$$
$$ = \int_{|z|=1} \frac{(z-2)^3... | Harish Chandra Rajpoot | 210,295 | <p>Notice,
$$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz=\frac{1}{8}\oint\frac{(z-2)^3}{\left(z-\frac{1}{2}\right)^3}dz$$ We see that $z=2$ & $z=\frac{1}{2}$ are two points out of which point $z=\frac{1}{2}$ is a pole of third order as it is lying inside the unit circle $|z|=1$. Hence, $f(z)=(z-2)^3$</p>
<p>henc... |
494,932 | <p>Why can trigonometry as a geometrically defined concept be used to algebraic operations between complex numbers? What connects the two things together and how ?</p>
| Ricardo Manrubia | 94,163 | <p>$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}...$</p>
<p>$e^{i\theta}=1+{i\theta}+\frac{{(i\theta)}^2}{2!}+\frac{{(i\theta)}^3}{3!}+\frac{{(i\theta)}^4}{4!}...=1+{i\theta}-\frac{{\theta}^2}{2!}-i\frac{{\theta}^3}{3!}+\frac{{\theta}^4}{4!}+i\frac{\theta^5}{5!}...=$
$=[1-\frac{{\theta}^2}{2!}+\frac{{\theta}^4... |
4,058,947 | <p>Have got into a pretty heated debate with a friend, and looking online there's lacking proof
<a href="https://i.stack.imgur.com/S30uM.png" rel="nofollow noreferrer">Line contained in a plane</a></p>
<p>Is a line that is contained within a plane, considered parallel to it? By my understanding it is parallel , if at a... | Ethan Mark | 722,965 | <p>I assume you are talking about planes and lines in <span class="math-container">$\mathbb{R^3}$</span>. If so, then all planes in are defined by a normal vector and a point, while all lines are defined by a direction vector and a point. It follows that some line will be parallel to some plane if that line's direction... |
4,058,947 | <p>Have got into a pretty heated debate with a friend, and looking online there's lacking proof
<a href="https://i.stack.imgur.com/S30uM.png" rel="nofollow noreferrer">Line contained in a plane</a></p>
<p>Is a line that is contained within a plane, considered parallel to it? By my understanding it is parallel , if at a... | CyclotomicField | 464,974 | <p>In Euclidean space for two lines to be parallel they must never intersect. Since a line in a plane intersects the plane it cannot be parallel to it. Note that the lack of intersection is necessary but not sufficient in higher dimensions.</p>
|
2,512,212 | <p>When we work on a metric space, the concepts of open set is defined as</p>
<blockquote>
<p>A subset $A$ of the metric space (X,d) is called open iff $Int A = A$</p>
</blockquote>
<p>And in topology books (see Munkress), a open set is defined as </p>
<blockquote>
<p>We say that a subset $U$ of $X$ is an open s... | drhab | 75,923 | <p>A metric space $(X,d)$ induces a topological space $(X,\tau_d)$.</p>
<p>The collection $\tau_d\subseteq\wp(X)$ is actually defined as the the collection of subsets of $X$ that are open according to the definition of open on metric space $(X,d)$.</p>
<p>That guarantees that "open in metric space $(X,d)$" will be th... |
3,614,178 | <p>I was going through this article on geometric series on Wikipedia and found this diagrammatic representation of an infinite geometric series with a said common factor of (1/2) but shouldn't the common factor be (1/4) based on the diagram?
link to the diagram:
<a href="https://commons.wikimedia.org/wiki/File:Geometr... | SagarM | 142,677 | <p><span class="math-container">$$\log^a_b = \frac{\ln^a}{\ln^b}$$</span>
In the following, if <span class="math-container">$a,b,c,d$</span> are all positive, then
<span class="math-container">$$\frac{a}{b} < \frac{c}{d} \rightarrow ad< bc$$</span>
<a href="https://brilliant.org/wiki/does-cross-multiple-always-w... |
3,677,967 | <p>I've got a <span class="math-container">$\prod$</span> (product operator) function that I'm trying to make explicit. I've managed to convert everything else to explicit form, which we can call <span class="math-container">$g(x)$</span>, except for this one part, so overall I've got:</p>
<p><span class="math-contain... | M1183 | 531,544 | <p>For <span class="math-container">$x\in\mathbb{N}\cup \{0\} $</span>, we have <span class="math-container">$h(x)=(x!)^k$</span> which
for general <span class="math-container">$x\geq 0$</span> (and even more general, for <span class="math-container">$x\ \in \{y\ | \ y\in\mathbb{C} \wedge \Re(y)>-1$</span> } ) beco... |
1,919,584 | <p>Given two non-colinear real unit vectors $v,w$, I believe the rank of $M=vv^\top + ww^\top$ is 2 and I'd like to prove it. $vv^\top$ and $ww^\top$ are obviously of rank one, $v$ is not in the kernel of $M$ because $vv^\top v=v$ and $\|ww^\top v\|<1$ (because $v$ and $w$ are not colinear), same with $w$. So the ra... | heptagon | 357,340 | <p>Choose a basis in which $e_1=v$ and $e_2=w$. Then $vv^\top+ww^\top=E_{11}+E_{22}$, and it has rank two.</p>
|
1,919,584 | <p>Given two non-colinear real unit vectors $v,w$, I believe the rank of $M=vv^\top + ww^\top$ is 2 and I'd like to prove it. $vv^\top$ and $ww^\top$ are obviously of rank one, $v$ is not in the kernel of $M$ because $vv^\top v=v$ and $\|ww^\top v\|<1$ (because $v$ and $w$ are not colinear), same with $w$. So the ra... | Christopher A. Wong | 22,059 | <p>Hint: Take a vector $x$ and write down an expression for $Mx$. Then, regardless of $x$, $Mx$ is a linear combination of...</p>
|
2,490,654 | <p>$$\int_0^1\sqrt\frac x{1-x}\,dx$$
I saw in my book that the solution is $x=\cos^2u$ and $dx=-2\cos u\sin u\ du$.<br>
I would like to see different approaches, can you provide them?</p>
| Bernard | 202,857 | <p>The <em>standard way</em> for rational functions <span class="math-container">$f\biggl(x,\sqrt{\dfrac{ax+b}{cx+d}}\biggr)$</span> consists in setting <span class="math-container">$t=\sqrt{\dfrac{ax+b}{cx+d}}$</span>. So we'll set
<span class="math-container">$$t=\sqrt{\frac{x}{1-x}}\iff x=\frac{t^2}{1+t^2},\qquad \m... |
2,490,654 | <p>$$\int_0^1\sqrt\frac x{1-x}\,dx$$
I saw in my book that the solution is $x=\cos^2u$ and $dx=-2\cos u\sin u\ du$.<br>
I would like to see different approaches, can you provide them?</p>
| omegadot | 128,913 | <p>Here is another way that involves <em>rationalising the numerator</em> first.</p>
<p>For $0 \leqslant x < 1$ we can write
\begin{align*}
\int_0^1 \sqrt{\frac{x}{1 - x}} \, dx &= \int_0^1 \sqrt{\frac{x}{1 - x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} \, dx\\
&= \int^1_0 \frac{x}{\sqrt{x - x^2}} \, dx
\end{align*}... |
1,946,144 | <p>I have a rough idea of how to solve this nonautonomous equation.</p>
<p>$x'=3x+sin(2t)$ </p>
<p>$\int 1\, dx=\int (3x+sin(2t))\, dt$</p>
<p>$x = 3xt - \frac{cos(2t)}{2} + constant$</p>
<p>$(1-3t)x = -\frac{1}{2}cos(2t) +constant$</p>
<p>$x = -\frac{cos(2t)}{2(1-3t)}+\frac{constant}{1-3t}$</p>
<p>Does this look... | Robert Z | 299,698 | <p>No. It is not correct. This is a <a href="https://en.wikipedia.org/wiki/Linear_differential_equation" rel="nofollow">linear differential equation</a>. Note that
$$x'(t)-3x(t)=\sin(2t)\implies
\frac{d}{dt}\left(e^{-3t}x(t)\right)=e^{-3t}(x'(t)-3x(t))=e^{-3t}\sin(2t).$$</p>
<p>Can you take it from here?</p>
|
1,946,144 | <p>I have a rough idea of how to solve this nonautonomous equation.</p>
<p>$x'=3x+sin(2t)$ </p>
<p>$\int 1\, dx=\int (3x+sin(2t))\, dt$</p>
<p>$x = 3xt - \frac{cos(2t)}{2} + constant$</p>
<p>$(1-3t)x = -\frac{1}{2}cos(2t) +constant$</p>
<p>$x = -\frac{cos(2t)}{2(1-3t)}+\frac{constant}{1-3t}$</p>
<p>Does this look... | Community | -1 | <p>By educated guess, the solution might be a sum of complex exponentials of the form</p>
<p>$$x=e^{zt}$$ where $z$ is a constant.</p>
<p>Plugging in the equation,</p>
<p>$$(z-3)e^{zt}=\frac{e^{i2t}-e^{-i2t}}{2i}=\frac{e^{i2t}}{2i}-\frac{e^{-i2t}}{2i}+0.$$</p>
<p>By identification, the solution will have terms with... |
3,829,280 | <p>Let <span class="math-container">$H$</span> be a semisimple algebraic subgroup of <span class="math-container">$GL(V)$</span> without compact factors (I am not sure if this part is relevant) where <span class="math-container">$V$</span> is a finite dimensional vector space. From a paper I have read, it follows that ... | Mehta | 21,974 | <p>Let <span class="math-container">$H$</span> be a semisimple algebraic subgroup of <span class="math-container">$GL_n.$</span> Since <span class="math-container">$H$</span> is semisimple, <span class="math-container">$H$</span> is its own derived group, so it is in the derived group of <span class="math-container">$G... |
2,752,377 | <p>Let $p$ be a prime number. I could already show, that $\mathbb{Z}[\sqrt{-2}]$ is an euclidean domain and that $$p\text{ reducible }\Leftrightarrow\text{ }p=a^2+2b^2\text{ for some $a,b\in\mathbb{Z}$.}$$</p>
<p>Now I wonder how to prove</p>
<p>$$p\equiv 1,2,3\text{ mod }8 \Rightarrow p=a^2+2b^2\text{ for some $a,b\... | Dave huff | 230,217 | <p>1.</p>
<p>I'll start by dealing with odd primes p.</p>
<p>$(p)$ is a prime in $\mathcal{O}=\mathbb{Z}[\sqrt{-2}]$ (doesn't split) if and only if $\mathcal{O}/(p)$ is a field. If $\mathcal{O}/(p)$ has any zero divisors, then it implies $(p)=\mathfrak{a}\cdot\bar{\mathfrak{a}}$; it is a product of prime ideals. Her... |
1,492,477 | <p>The sequence $\frac{1}{n}$ is convergent under euclidean metric. But not convergent with discrete metric.</p>
<p>Is there a non-constant convergent sequence with discrete metric ?</p>
| amd | 265,466 | <p>If you expand the sums for the first few values of $p$, a clear pattern emerges: $$\begin{align}
\sum_{i-1}^2(y_i-\bar y)^2 &= \frac12(y_1^2+y_2^2-2y_1y_2) \\
\sum_{i=1}^3(y_i-\bar y)^2 &= \frac13(2y_1^2+2y_2^2+2y_3^2-2y_1y_2-2y_1y_3-2y_2y_3) \\
\sum_{i=1}^4(y_i-\bar y)^2&=\frac14(3y_1^2+3y_2^2+3y_3^2+3y... |
731,624 | <p>For a set $\mathbb{X}$ given order relation extended from that of $\mathbb{R}$, if $\mathbb{X} \supseteq \mathbb{R}$ then $\mathbb{X} = \mathbb{R}$ ?</p>
<p>Motivation for this question rose from an intuitive question that when I draw a virtual continuous straight line between $1$ and $2$ and virtually pick a point... | André Nicolas | 6,312 | <p>One can extend the order of $\mathbb{R}$ to proper supersets of $\mathbb{R}$. </p>
<p>For example, consider the set of <em>rational functions</em> with coefficient in $\mathbb{R}$. Any such function $f(x)$ can be written as $\frac{P(x)}{Q(x)}$ where the lead coefficient of $Q(x)$ is positive. Define an order on th... |
2,632,392 | <blockquote>
<p>What is the remainder of $23! +29!$ divided by $13! +19!$?</p>
</blockquote>
<p><strong>Attempt</strong></p>
<p>$$23!+29!=23![1+(29×28×27×26×25×24)]$$ and </p>
<p>$$13!+19!=13![1+(19×18×17×16×15×14)]$$</p>
<p>Now how do I proceed..I am stuck at this initial stage.</p>
<p>Please help.</p>
| Joffan | 206,402 | <p>Well... as a simple check, we have the common factor of $13!$ which I will ignore initially, then </p>
<p>$$ \begin{align}
14\cdot 15\cdot 16\cdot 17\cdot 18\cdot 19 + 1 &= 19535041 \\
(23!/13!) \equiv 19535040\cdot 20 \cdot 21 \cdot 22 \cdot 23 &\equiv 19322521 \bmod 19535041 \\
(29!/13!) \equiv 19322521\... |
3,504,777 | <p>Here's what I initially started with:</p>
<blockquote>
<blockquote>
<p>Find a 2x2 non zero matrix <span class="math-container">$A$</span>, satisfying <span class="math-container">$A^2=A$</span>, and <span class="math-container">$A\neq I$</span>.</p>
</blockquote>
</blockquote>
<p>I understand that this is ... | egreg | 62,967 | <p>Let <span class="math-container">$\{v_1,v_2,\dots,v_n\}$</span> any basis of <span class="math-container">$\mathbb{R}^n$</span> (or the field of your choice). Fix <span class="math-container">$k$</span> with <span class="math-container">$1\le k<n$</span> (to avoid trivial cases) and define the linear map <span cl... |
3,591,372 | <p>I have the following question: </p>
<blockquote>
<p>Let D be the region bounded below by the cone <span class="math-container">$z=\sqrt{x^2+y^2}$</span> and above the paraboloid <span class="math-container">$z=2-x^2-y^2$</span>. Set-up the volume in polar coordinates to get the the volume of D. </p>
</blockquote... | Christian Blatter | 1,303 | <p>Introducing <span class="math-container">$S_1'$</span> and <span class="math-container">$S_2'$</span> was a good move. But the second figure should look more like the following:</p>
<p><a href="https://i.stack.imgur.com/EYq0W.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EYq0W.jpg" alt="enter i... |
3,591,372 | <p>I have the following question: </p>
<blockquote>
<p>Let D be the region bounded below by the cone <span class="math-container">$z=\sqrt{x^2+y^2}$</span> and above the paraboloid <span class="math-container">$z=2-x^2-y^2$</span>. Set-up the volume in polar coordinates to get the the volume of D. </p>
</blockquote... | Rezha Adrian Tanuharja | 751,970 | <p>Elementary solution will be connecting the points to their circle center to obtain 4 isosceles triangle. Checking the angle, opposite angle sums up to <span class="math-container">$180$</span> so they are concyclic.<img src="https://i.stack.imgur.com/kSYXc.jpg" alt="enter image description here"></p>
|
3,775,647 | <p>The maths book I'm using shows:<br/>
<span class="math-container">$$ a^2b \div \frac13a^2b^3 $$</span>
Which would be something like:<br/>
<span class="math-container">$$ a^2 \cdot b ÷ \frac13 \cdot a^2 \cdot b^3 $$</span>
My understanding of order of operations it would equal:<br/>
<span class="math-container">$$ 3... | VIVID | 752,069 | <p>All books I have seen state:</p>
<ul>
<li><span class="math-container">$x:yz = \cfrac{x}{y}z$</span></li>
<li><span class="math-container">$x:(yz) = \cfrac{x}{yz}$</span></li>
</ul>
<p>And I know only these as a rule.</p>
|
3,775,647 | <p>The maths book I'm using shows:<br/>
<span class="math-container">$$ a^2b \div \frac13a^2b^3 $$</span>
Which would be something like:<br/>
<span class="math-container">$$ a^2 \cdot b ÷ \frac13 \cdot a^2 \cdot b^3 $$</span>
My understanding of order of operations it would equal:<br/>
<span class="math-container">$$ 3... | user | 505,767 | <p>In this form</p>
<p><span class="math-container">$$a^2b \div \frac13a^2b^3$$</span></p>
<p>my interpretation is to consider the two terms as a whole, that is
<span class="math-container">$$\left(a^2b\right) \div \left(\frac13a^2b^3\right)=\frac{3a^2b}{a^2b^3}=\frac 3 {b^2}$$</span></p>
<p>Otherwise, in this form</p>... |
2,978,793 | <p>I want to show that</p>
<p><span class="math-container">$$\sum_{i=1}^n i^{2-\alpha} \sim c n^{3-\alpha}$$</span></p>
<p>for some constant <span class="math-container">$c$</span>. Here, <span class="math-container">$\alpha$</span> is assumed <span class="math-container">$0 < \alpha < 1$</span>.</p>
<p>I know... | kludg | 42,926 | <p>By the law of iterated expectations
<span class="math-container">$$\mathbb{E}[\mathbb{E}[Y|Z,X,W]|X] = \mathbb{E}[Y|X]$$</span></p>
<p><span class="math-container">$$\mathbb{E}[\mathbb{E}[\mathbb{E}[Y|Z,X,W]|Z,X]|X]=\mathbb{E}[\mathbb{E}[Y|Z,X]|X]=\mathbb{E}[Y|X]$$</span></p>
|
2,800,416 | <p>this is my first question and I don't quite understand how do I confront this equation:</p>
<p>$z^2+i\sqrt{32}z-6i=0$</p>
<p>I tried using the quadratic formula but it doesn't seem to give me a correct answer, any help will be much obliged.</p>
<p>Thank you! :)</p>
| Henry Lee | 541,220 | <p>if z is a complex number let $z=a+bj$,</p>
<p>$$(a+bj)^2+\sqrt{32}(a+bj)j-6j=0$$
$$\therefore a^2-b^2+2abj+\sqrt{32}aj-\sqrt{32}b-6j=0$$
$$\therefore a^2-b^2-\sqrt{32}b=0, 2abj+\sqrt{32}aj-6j=0$$
These can then be solved simultaneously to find the roots</p>
|
1,991,822 | <p>Suppose $f:[0,1]\rightarrow\mathbb{R}$. defined by $f(x)=(-1)^n n $ when $x\in(1/(n+1),1/n]$ and $f(0)=0$. Show that the improper Riemann integral $$\int_{0}^{1} f(x) dx $$ is real number. </p>
<p>First note that $f$ has infinite points when $f$ is discontinuous. I have trouble to compute the integral. I have s... | Yarden Sharabi | 568,884 | <p>Let $\epsilon>0$ and let $n_0(\epsilon)\in\mathbb{N}$such that $\frac{1}{n_0+1}<\epsilon<\frac{1}{n_0}$:Notice $n_0(\epsilon)\underset{\epsilon\rightarrow0}{\rightarrow}\infty$ and $\frac{n_0(\epsilon)}{\epsilon}\underset{\epsilon\rightarrow0}{\rightarrow}1$ </p>
<p>Now, $$\underset{\epsilon\rightarrow0}{l... |
1,399,921 | <p>Suppose that $A$ and $B$ are invertible, $p \times p$ matrices. If $A^n = B$ and I know all of the entries in $B$, can I find an $A$ for some or all integers $n \ge 0$? How many solutions for $A$ exist? If I'm thinking correctly, then $A = B * (A^{-1})^{n-1},$ but this is sort of self referential. Thanks!</p>
| Martin Argerami | 22,857 | <p>You cannot do it even when $p=1$, $n=2$: take $B=1$. If $A^2=B$, then maybe $A=1$, or maybe $A=-1$. </p>
|
27,233 | <p>A finitely presented, countable discrete group $G$ is amenable if there is a finitely additive measure $m$ on the subsets of $G \backslash${$e$} with total mass 1 and satisfying $m(gX)=mX$ for all $X\subseteq G \backslash${$e$} and all $g \in G.$ </p>
<p>A countable discrete group $G$ is inner amenable if there is ... | Denis Osin | 10,251 | <p>Here is a construction of a countable i.c.c. nonamenable simple group that is inner amenable. First consider the following condition:</p>
<p>(*) For every finite subset $S\subseteq G$, there exists $g\in G\setminus \{ 1\} $ such that $[g,s]=1$ for every $s\in S$.</p>
<p>Using paradoxical decomposition for non-inne... |
98,348 | <p><a href="https://math.stackexchange.com/a/47261">A</a> <a href="https://math.stackexchange.com/a/15210">few</a> <a href="https://math.stackexchange.com/a/69164">answers</a> here on math.SE have used as an intermediate step the following inequality that is due to Walter Gautschi:</p>
<p>$$x^{1-s} < \frac{\Gamma(x... | Jack D'Aurizio | 44,121 | <p>Given $a,b\geq 0$, let us consider the function $f(x)=x^{a}(1-x)^{b}$ on the interval $[0,1]$.<br>
Its maximum value is given by $\frac{a^a b^b}{(a+b)^{a+b}}$, since $f'$ only vanishes at $x=\frac{a}{a+b}$.<br>
For any $p>0$ we have
$$ \| f\|_p^p = \int_{0}^{1}x^{pa}(1-x)^{pb}\,dx=\frac{\Gamma(ap+1)\,\Gamma(bp+1)... |
450,370 | <blockquote>
<p>Use Lagrange multipliers to determine the shortest distance from a point $\,x \in R^n\,$ to a plane $\{y\mid b^Ty = c\}.$</p>
</blockquote>
<p>I don't even know where to start!</p>
| pitfall | 36,624 | <p>Assume that you know something about optimization, your described problem then can be rewritten as
$${\rm{minimize}} \rho(x,y)=\|x_{n\times 1}-y_{n\times 1}\|\\ {\rm{such\,that}}\,\, b^Ty=c$$
where $\|.\|$ is your preferred distance measure, e.g. euclidean distance. </p>
<p>To solve this optimization problem, then... |
246,492 | <p>I am running an iMac Pro with Intel silicon under the latest version of Big Sur 11.3.1. I upgraded my Mathematica from 12.2 to 12.3 yesterday, but now I have suddenly lost the ability to use the QuickLook feature with Mathematica notebooks. I click on a notebook's icon in Finder to highlight it and then touch the sp... | Alan | 19,530 | <pre><code>test = <|"key1" -> <|"key11" -> 11, "key12" -> 12, "key13" -> 13|>,
"key2" -> <|0 -> <|"X" -> 5, "Y" -> 0|>, 1 -> <|"X" -> 6, "Y" -> 0|>,
2 -> <|&... |
159,851 | <p>If Im not wrong when we use the command <code>N</code> we are using float point numbers in machine precision, right? And the machine precision depends on the bits of the CPU where Mathematica is installed. Then, by example, irrational numbers are truncated at some place and there is some error associated to the magn... | TimRias | 10,587 | <p>This should give you what you need:</p>
<pre><code>Accumulate@Tan[Range[10^8]]-SetPrecision[Accumulate@Tan[N@Range[10^8]],30]
</code></pre>
|
3,926,603 | <p>So im learning about limits from tutorials and at the same time im solving the examples myself. Im getting solutions that sometimes differ from the explainer's.</p>
<p>Lets take this limit and the way it is solved on the video:</p>
<p><span class="math-container">$$\lim_{x \to -2} \dfrac { (x^3 - x^2 - 6x)}{(x^2+2x)... | Sivakumar Krishnamoorthi | 686,991 | <p>Substituting <span class="math-container">$x$</span> with <span class="math-container">$2\cos\theta$</span> will suffice to get the result to your question <span class="math-container">$$\sqrt{2+\sqrt{2-\sqrt{2+x}}} = x$$</span> will be <span class="math-container">$\sqrt{2+\sqrt{2-\sqrt{2+2\cos\theta}}} = 2\cos\the... |
3,640,587 | <blockquote>
<p>How do I find the <span class="math-container">$\angle BCA$</span>?</p>
<p><a href="https://i.stack.imgur.com/zOux6m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zOux6m.png" alt="enter image description here" /></a></p>
</blockquote>
<p>I tried to find the formula of the parabola, ... | Community | -1 | <p>Hint :</p>
<p>By maximality of the interior we can say that if there is an open set that is include in the set then the open set is in the interior</p>
|
294,725 | <p>I've been out of school for a very long time, and can't wrap my head around calculating 2 whole numbers from a fraction. For instance, if I have 2 ratios -- </p>
<pre><code>4:3 1.33333...
16:9 1.77777...
?:? (automatically generated)
</code></pre>
<p>I want to be able to look up the whole number represent... | Ross Millikan | 1,827 | <p>You can compute it. Multiply the decimal by $10^n$, where $n$ is the number of digits in the repeat of the decimal and subtract. Example: given $ x=5.181818\ldots \\100x=518.1818\ldots,\\99x=513,\\x=\frac{513}{99}$</p>
|
2,424,649 | <blockquote>
<p>Show that $\sqrt{2} + \sqrt[3]{5}$ is algebraic of degree $6$ over
$\mathbb{Q}$</p>
</blockquote>
<p>What is the degree of a root? Is it the smallest polynomial that gives this thing as root?</p>
<p>What I tried:</p>
<p>$x = \sqrt{2} + \sqrt[3]{5} \implies x^2 = 2 + 2\sqrt{2}\sqrt[3]{5} + \sqrt[3... | C. Falcon | 285,416 | <p>It is easily seen that a sum of two algebraic numbers $\alpha$ and $\beta$ is algebraic since one has:
$$\mathbb{Q}(\alpha+\beta)\subseteq\mathbb{Q}(\alpha,\beta).$$
Hence, $\mathbb{Q}(\alpha+\beta)$ is finite-dimensional since $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha)(\beta)$ is, indeed recall that:
$$[\mathbb{Q... |
2,134,121 | <p>I can't figure out an algebraic proof for the following identity, (and I don't know if I can use the binomial theorem for this one):
<span class="math-container">$$\sum_{k=0}^m(-1)^{m-k}{{n \choose {k}}}= {{n-1}\choose m}$$</span></p>
<p>Thank you, for your help in advance.</p>
| Brevan Ellefsen | 269,764 | <p>We desire to prove, by induction, that
$$\sum_{k=0}^m (-1)^{k} {n \choose k} = (-1)^m{n-1 \choose m}$$
We start with the case $m=0$
$$\sum_{k=0}^0 (-1)^{k} {n \choose k} = {n \choose 0}= {n-1 \choose 0}$$
We now apply induction. We first assume that
$$\sum_{k=0}^{m-1} (-1)^{k} {n \choose k} = (-1)^{m-1}{n-1 \choose ... |
2,134,121 | <p>I can't figure out an algebraic proof for the following identity, (and I don't know if I can use the binomial theorem for this one):
<span class="math-container">$$\sum_{k=0}^m(-1)^{m-k}{{n \choose {k}}}= {{n-1}\choose m}$$</span></p>
<p>Thank you, for your help in advance.</p>
| David Holden | 79,543 | <p>$$
(1+x)^{n-1} = \sum_{m=0}^n \binom{n-1}m x^m
$$
but also
$$
\begin{align}
(1+x)^{n-1} =(1+x)^n \frac1{1+x} & = \sum_{k=0}^n \binom{n}{k}x^k\sum_{l=0}^{\infty}(-1)^lx^l \\
& =\sum_{m=0}^{\infty}x^m\sum_{k=0}^m (-1)^{m-k}\binom{n}k \tag{2}
\end{align}
$$
now compare the coefficients of $m$ in the two expansi... |
2,134,121 | <p>I can't figure out an algebraic proof for the following identity, (and I don't know if I can use the binomial theorem for this one):
<span class="math-container">$$\sum_{k=0}^m(-1)^{m-k}{{n \choose {k}}}= {{n-1}\choose m}$$</span></p>
<p>Thank you, for your help in advance.</p>
| Hypergeometricx | 168,053 | <p>$$\begin{align}\require{cancel}
\sum_{k=0}^m(-1)^{m-k}\binom nk
&=(-1)^m\sum_{k=0}^m (-1)^k\left[\binom {n-1}{k-1}+\binom {n-1}k\right]
&&\text{as $(-1)^{-k}=(-1)^k$}\\
&=(-1)^m\bigg\lbrace \cancel1-\left[\cancel{\binom {n-1}0}+\bcancel{\binom {n-1}1}\right]\\
&\qquad\quad\quad\quad+\left[\bcance... |
3,945,610 | <p>If <span class="math-container">$a, b, c, d>0$</span>, such that <span class="math-container">$a+b+c=1$</span>, prove that: <span class="math-container">$$a^3+b^3+c^3+abcd\ge \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).$$</span></p>
<p>I tried solving it as follows:
<span class="math-container">$$a^3+b... | TheSimpliFire | 471,884 | <p>This is an easy application of Lagrange as the first equality of <span class="math-container">$$3a^2+bcd=3b^2+acd=3c^2+abd=\lambda$$</span> gives <span class="math-container">$3(a+b)(a-b)=(a-b)cd$</span> so either <span class="math-container">$a=b$</span> or <span class="math-container">$3(a+b)=cd$</span>. Thus ther... |
3,945,610 | <p>If <span class="math-container">$a, b, c, d>0$</span>, such that <span class="math-container">$a+b+c=1$</span>, prove that: <span class="math-container">$$a^3+b^3+c^3+abcd\ge \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).$$</span></p>
<p>I tried solving it as follows:
<span class="math-container">$$a^3+b... | River Li | 584,414 | <p>WLOG, assume <span class="math-container">$c = \min(a, b, c)$</span>.</p>
<p>Let <span class="math-container">$P(x) = a^3 + b^3 + c^3 + abc x$</span> and <span class="math-container">$Q(x) = 2(\frac{a+b}{2})^3 + c^3 + (\frac{a+b}{2})^2 c x$</span> where <span class="math-container">$x \ge 0$</span>.
We have
<span cl... |
1,533,762 | <blockquote>
<p>Prove that $ 16^{20}+29^{21}+42^{22}$ is divisible by $13$.</p>
</blockquote>
<p>This is not a homework question. I would like to know how to solve this type of problems, I solved similar problem with n in exponent, but that could be proved by induction. Here I guess, Euler's theorem could be useful,... | Prabath Hewasundara | 243,089 | <p>$(16^{20}+29^{21}+42^{22})\text{ mod }13=(3^{20}+3^{21}+3^{22})\text{ mod }13$</p>
<p>$3^{20}(3^{0}+3^{1}+3^{2}) \text{ mod }13=3^{20}13\text{ mod }13$</p>
<p>$16^{20}+29^{21}+42^{22}=0\text{ mod }13$</p>
|
4,325,440 | <p>I need to compute <span class="math-container">$$I:=\int_{-\infty}^{\infty} \frac{\sin(ax)}{x(\pi^2-a^2x^2)}dx$$</span>
(<span class="math-container">$a>0$</span>)</p>
<p>Probably there is a way to compute it with residue theorem.</p>
<p>My thoughts:</p>
<ul>
<li>The singularity at <span class="math-container">$x... | Célestin | 365,426 | <p>Assume without loss of generality that <span class="math-container">$a$</span> is positive. Using the change of variable <span class="math-container">$ax=\pi t$</span> we get rid of <span class="math-container">$a$</span>
<span class="math-container">$$ I = \frac{1}{\pi^2} \int_{-\infty}^\infty \frac{\sin \pi t}{t(1... |
31,562 | <p>How to evaluate the number of ordered partitions of the positive integer <span class="math-container">$ 5 $</span>?</p>
<p>Thanks!</p>
| Mitch | 1,919 | <p>So $4+1$ is one example. $2+2+1$ is another</p>
<ul>
<li><p>What kinds of things add up to 5? (only numbers greater than or equal to 1 are used).</p></li>
<li><p>What's the least number of numbers you can use? What's the greatest number?</p></li>
<li><p>What if you rearrange the order of something you already have?... |
3,080,402 | <p>I need to show that the series <span class="math-container">$\sum_{n=1}^{\infty}\frac{c^{-n}}{n!}$</span> is convergent.</p>
<p>I invoked the limit comparison with the series <span class="math-container">$\sum_{n=1}^{\infty}\frac{c^n}{n!}$</span> which is absolutely convergent (and hence convergent).</p>
<p>I got ... | Michael Rozenberg | 190,319 | <p>If <span class="math-container">$a=-b$</span> so <span class="math-container">$c=-d$</span> and we are done.</p>
<p>Let <span class="math-container">$a+b\neq0.$</span></p>
<p>Thus, <span class="math-container">$$(a+b)^3-3ab(a+b)=(c+d)^3-3cd(c+d)$$</span> or
<span class="math-container">$$ab(a+b)=cd(c+d)$$</span> o... |
3,080,402 | <p>I need to show that the series <span class="math-container">$\sum_{n=1}^{\infty}\frac{c^{-n}}{n!}$</span> is convergent.</p>
<p>I invoked the limit comparison with the series <span class="math-container">$\sum_{n=1}^{\infty}\frac{c^n}{n!}$</span> which is absolutely convergent (and hence convergent).</p>
<p>I got ... | nonuser | 463,553 | <p>Prove: for all <span class="math-container">$n$</span>: <span class="math-container">$a^{2n+1}+b^{2n+1}=c^{2n+1}+d^{2n+1}$</span></p>
<p>If <span class="math-container">$a+b\ne 0$</span> then we get <span class="math-container">$$a^2-ab+b^2 =c^2-cd+d^2\implies (a+b)^2-3ab = (c+d)^2-3cd$$</span></p>
<p>so <span cla... |
3,520,313 | <p>Let <span class="math-container">$I\subset\mathbb R$</span> be an interval and <span class="math-container">$\left(f_n\right)_{n\in\mathbb N}:I\to\mathbb R$</span> be a sequence that is uniformly convergent on <span class="math-container">$I$</span>. Is the sequence <span class="math-container">$g_n=\left\{\frac{f_n... | José Carlos Santos | 446,262 | <p>If <span class="math-container">$x\in I$</span>, and if <span class="math-container">$f$</span> is the uniform limit of the sequence <span class="math-container">$(f_n)_{n\in\mathbb N}$</span>, then<span class="math-container">\begin{align}\left\lvert \frac f{1+f^2}-\frac{f_n}{1+{f_n}^2}\right\rvert&=\left\lvert... |
1,452,903 | <p>Has the aforementioned constant ever been used in any major proofs? Can it be expressed in terms of $e$, $\pi$, or both? Does it appear in any sort of geometric sense like $\pi$ does? Or is it just a kind of pretty number somebody thought would be nice to experiment with?</p>
| Joseph Weissman | 4,300 | <p>For what it's worth I'm a software engineer and have ended up using Champernowne's constant (expressed in base 2) as a cheap stand-in for a random binary sequence. </p>
<p>For instance, to simulate grass growing on the floor of a dungeon, we can initialize the floor cells with Champernowne and then apply game of li... |
3,220,543 | <p>A natural number <span class="math-container">$n$</span> is said to be a good number if and only if <span class="math-container">$4$</span> times the sum of its digits equals the original number. We have to find out the sum of all such good numbers.</p>
<p>I'm 99% sure the only such good natural numbers are <span ... | farruhota | 425,072 | <p>Assuming <span class="math-container">$n>0$</span>, </p>
<p>1-digit: <span class="math-container">$n=a_1=4a_1 \iff a_1=0 \not >0$</span>.</p>
<p>2-digit: <span class="math-container">$n=\overline{a_2a_1}=10a_2+a_1=4a_2+4a_1 \iff a_1=2a_2 \iff 12, 24, 36, 48$</span>.</p>
<p>3-digit: <span class="math-contain... |
110,599 | <blockquote>
<p>Why isn't $M = \mathbb{C}[x,y,z]/(xz-y)$ a flat $R = \mathbb{C}[x,y]$-module?</p>
</blockquote>
<p>The reason given on the book is "the surface defined by $y-xz$ doesn't lie flat on the $(x,y)$-plane". But I don't understand why this can be a reason.</p>
<p>Since</p>
<blockquote>
<p>An $R$-module... | Zhen Lin | 5,191 | <p>The question seems to be about gaining geometric intuition for flat morphisms of (affine) schemes. Let $A = \mathbb{C}[x, y, z]/(x z - y)$, $X = \operatorname{Spec} A$ and $\mathbb{A}^2_\mathbb{C} = \operatorname{Spec} \mathbb{C}[x, y]$. The claim is that the projection $\pi : X \to \mathbb{A}^2_\mathbb{C}$ is not a... |
1,157,306 | <p>Is the following system has any positive integer solution $(x,y,u,v)$?
$$\begin{cases} x^2+y^2=u^2\\ x^2-y^2=v^2 \end{cases}$$
I can prove that any pair of these integers can be relatively prime, but I couldn't find any solution. Any hint?</p>
<p>Thanks in advance!</p>
| ali | 206,957 | <p>From solutions of the Pythagorean equation we have $x = 2ab, y = a^2 - b^2$, so we must have $$(a^2 - b^2 - 2ab)(a^2 - b^2 + 2ab) = v^2$$ we have $(a^2 - b^2 - 2ab, a^2 - b^2 + 2ab) = 1$, so $$a^2 - b^2 - 2ab = (a - b)^2 - 2b^2 = x^2$$ $$a^2 - b^2 + 2ab = (a + b)^2 - 2b^2 = y^2$$ your equation has solution if and on... |
1,157,306 | <p>Is the following system has any positive integer solution $(x,y,u,v)$?
$$\begin{cases} x^2+y^2=u^2\\ x^2-y^2=v^2 \end{cases}$$
I can prove that any pair of these integers can be relatively prime, but I couldn't find any solution. Any hint?</p>
<p>Thanks in advance!</p>
| poetasis | 546,655 | <p>A rearrangement of variables in the "given" shows two Pythagorean triples where the odd leg of one triple matches the hypotenuse of another.
<span class="math-container">\begin{equation}
\qquad\qquad\qquad\qquad x^2+y^2=u^2\\
x^2-y^2=v^2\implies x^2=y^2+v^2
\end{equation}</span>
We can find matching x an... |
1,350,147 | <p>I was wondering whether, for a given finite-dimensional manifold, the connection $\nabla$ exists and is uniquely defined?</p>
<p>Afais for Riemannian manifolds, there exists always exactly one Levi-Civita connection, but the calculation is rather cumbersome.</p>
<p>Now, if we consider manifolds without a metric, i... | Travis Willse | 155,629 | <p>No, on a general manifold $M$:</p>
<ul>
<li>there is no preferred linear connection,</li>
<li>there is always at least one connection $\nabla$, and</li>
<li>the connections on $M$ are precisely the maps $$\bar{\nabla}_X Y := \nabla_X Y + A(X, Y),$$ where $A \in \Gamma(TM \otimes T^*M \otimes T^*M)$ is a $(2, 1)$-te... |
56,337 | <p>Suppose I have the following list:</p>
<pre><code>list = {a, b, c, d}
</code></pre>
<p>I want to generate this result:</p>
<pre><code>{{f[a, a], f[a, b], f[a, c], f[a, d]}, {f[b, b], f[b, c],
f[b, d]}, {f[c, c], f[c, d]}, {f[d, d]}}
</code></pre>
<p>What could be the shortest way?</p>
<p>The list elements can... | halirutan | 187 | <p>How about a simple table?</p>
<pre><code>Table[f @@ list[[{i, j}]], {i, 4}, {j, i, 4}]
</code></pre>
<p>If you want to use this for a general list, you should use <code>Length[list]</code> in the table iterators or maybe:</p>
<pre><code>With[{n = Length[list]},
Table[f @@ list[[{i, j}]], {i, n}, {j, i, n}]
]
</c... |
56,337 | <p>Suppose I have the following list:</p>
<pre><code>list = {a, b, c, d}
</code></pre>
<p>I want to generate this result:</p>
<pre><code>{{f[a, a], f[a, b], f[a, c], f[a, d]}, {f[b, b], f[b, c],
f[b, d]}, {f[c, c], f[c, d]}, {f[d, d]}}
</code></pre>
<p>What could be the shortest way?</p>
<p>The list elements can... | Karsten 7. | 18,476 | <p>Building up an <code>Association</code></p>
<pre><code>len=Length@list;
asso=<||>;
(asso[#]=list[[-#]])&/@Range@len;
</code></pre>
<p>and then</p>
<pre><code>Array[Function[x,Array[f[x,#]&,x,{x,1}]],len,{len,1}]/.asso
</code></pre>
|
56,337 | <p>Suppose I have the following list:</p>
<pre><code>list = {a, b, c, d}
</code></pre>
<p>I want to generate this result:</p>
<pre><code>{{f[a, a], f[a, b], f[a, c], f[a, d]}, {f[b, b], f[b, c],
f[b, d]}, {f[c, c], f[c, d]}, {f[d, d]}}
</code></pre>
<p>What could be the shortest way?</p>
<p>The list elements can... | Michael E2 | 4,999 | <p>A few obfuscations via <code>Listable</code>:</p>
<pre><code>Block[{f, op},
SetAttributes[f, Listable];
op[x_] := {f[x, x]};
op[x_, y__] := Sequence[f[x, {x, y}], op[y]];
{op @@ list}
]
Block[{f},
SetAttributes[f, Listable];
f @@@ Table[{list[[i]], list[[i ;;]]}, {i, 4}]
]
Module[{op1, op2},
op1 = Func... |
56,337 | <p>Suppose I have the following list:</p>
<pre><code>list = {a, b, c, d}
</code></pre>
<p>I want to generate this result:</p>
<pre><code>{{f[a, a], f[a, b], f[a, c], f[a, d]}, {f[b, b], f[b, c],
f[b, d]}, {f[c, c], f[c, d]}, {f[d, d]}}
</code></pre>
<p>What could be the shortest way?</p>
<p>The list elements can... | Simon Woods | 862 | <p>A few more just for fun:</p>
<pre><code>ReplaceList[list, {___, a__} :> Thread @ f[#& @ a, {a}]]
Thread @* f ~MapThread~ {list, NestList[Rest, list, 3]}
Pick[Outer[f, list, list], # <= #2 & ~Array~ {4, 4}]
</code></pre>
|
2,310,413 | <p>A coin is flipped twice. $A$ is the result of the first coin flip and $B$ the result of the second one. I know that $A$ and $B$ are independent. Following $$P(A\cap B)=P(A)P(B),$$ when it comes to write down the event of the intersection of $A$ and $B$ I am stuck because the event $A$ excludes the event $B$, and yet... | Especially Lime | 341,019 | <p>It doesn't make sense to talk about the intersection of $A$ and $B$ as you have defined them, because the result of a coin flip is a <em>variable</em>, not an <em>event</em>.</p>
<p>If you define them as events, say $A$ is the event that the first flip is a head, and $B$ is the event that the second flip is a head,... |
126,251 | <p>Suppose one has a finite number of distances $d_1,\ldots,d_k$ on the Euclidean plane all of which metricize the usual Euclidean topology.</p>
<p>Define for each pair of points $x$ and $y$ in the plane
$$d(x,y) = \inf\left\lbrace d_{i_1}(x_0,x_1) + \cdots d_{i_l}(x_{l-1},x_l) \right\rbrace$$
where the infimum is tak... | Anton Petrunin | 1,441 | <p>It can be zero.</p>
<p>Take the standard metric on $\mathbb R^3$ and the one given in <a href="https://mathoverflow.net/questions/125283/on-lipschitz-embeddability-of-certain-compact-metric-spaces-into-mathbbrn/125295#125295">this example</a> by S. Ivanov.
Below I give simplification of his example which works in y... |
119,981 | <p>Let $C/\mathbb Q$ be a smooth projective curve of genus $g\geq 2$ or a smooth affine curve of genus $g \geq 1$. The exact sequence</p>
<p>$1 \to \pi_1^{et}(C \otimes_\mathbb Q \bar{\mathbb Q}) \to \pi_1^{et}(C) \to \operatorname{Gal}(\bar{\mathbb Q}|\mathbb Q) \to 1$</p>
<p>gives a homomorphism from $\operatorname... | Steven Landsburg | 10,503 | <p>Turning my last comment into an answer:</p>
<p>The simplest model of money demand is $M=M(P,Y,i)$ where $P$ is the price level (if all prices rise, you'll probably want more money in your pocket), $Y$ is real income (if you're richer, you might want more money in your pocket) and $i$ is the nominal interest rate (... |
119,981 | <p>Let $C/\mathbb Q$ be a smooth projective curve of genus $g\geq 2$ or a smooth affine curve of genus $g \geq 1$. The exact sequence</p>
<p>$1 \to \pi_1^{et}(C \otimes_\mathbb Q \bar{\mathbb Q}) \to \pi_1^{et}(C) \to \operatorname{Gal}(\bar{\mathbb Q}|\mathbb Q) \to 1$</p>
<p>gives a homomorphism from $\operatorname... | none | 37,063 | <p>I feel like I'm missing something dumb but it seems to me that fractional-reserve banking is not referenced or explained in the question or any of the answers. People here DO understand it, right? It is where money actually comes from. Basically if you borrow \$1000 from a bank, they can issue the loan by typing ... |
3,872,174 | <p>Let <span class="math-container">$a\in \mathbb{R}^n$</span> column vector and <span class="math-container">$A\in \mathbb{R}^{n\times n}$</span> a matrix. What are some bounds for the number</p>
<p><span class="math-container">$$ |a^T A a| $$</span></p>
<p>In terms of norms of <span class="math-container">$a$</spa... | sven svenson | 803,521 | <p>Essentially, what you're asking is whether it is possible to have a non-monotonic sequence of positive numbers that add up to <span class="math-container">$1$</span>. We can take the sequence <span class="math-container">$\left\{2^{-k}\right\}^{\infty}_{k=1}$</span> and reverse the order of each pair of terms, i.e.,... |
1,649,076 | <p>Could anyone guide along with this question?</p>
<p>I was trying $(A−I)(A−I)^{−1} = I$ and was figuring if there's a way out to expand $(A−I)^{−1}$. I also tried $(A−I)x=0$ but to no avail.</p>
| N74 | 288,459 | <p>For excercises like this it is always a good idea to start with the classical expansions. In this case: $$
A^3-I=(A-I)(A^2+A+I)
$$
We can then write: $$
(A-I)(A^2+A+I)=A^3-I=2I-I=I
$$
By comparison with $(A-I)(A-I)^{-1}=I$ we can find that $$
(A-I)^{-1}=(A^2+A+I)
$$</p>
|
2,224,980 | <p>Apologies if this kind of question is not allowed here - if so please delete it.</p>
<p>I was just wondering if anyone could recommend a book on mathematical analysis that is interesting enough to sit down and read for enjoyment alone? Something not written in the style of a textbook?</p>
<p>All the best.</p>
| J. Deff | 261,331 | <p>There are two books according to me</p>
<ol>
<li>How to Think About Analysis 1st Edition
by Lara Alcock </li>
</ol>
<p><a href="http://rads.stackoverflow.com/amzn/click/0198723539" rel="nofollow noreferrer">https://www.amazon.com/Think-About-Analysis-Lara-Alcock/dp/0198723539</a></p>
<p>2.
The Real Analysis Life... |
2,224,980 | <p>Apologies if this kind of question is not allowed here - if so please delete it.</p>
<p>I was just wondering if anyone could recommend a book on mathematical analysis that is interesting enough to sit down and read for enjoyment alone? Something not written in the style of a textbook?</p>
<p>All the best.</p>
| CoffeeCCD | 433,261 | <p>For an easy and fun book read Understanding Analysis by Abbott. For a tougher read that is thorough and intuitive read Real Mathematical Analysis by C.C.Pugh.I would recommend this over Rudin anyday.</p>
|
2,224,980 | <p>Apologies if this kind of question is not allowed here - if so please delete it.</p>
<p>I was just wondering if anyone could recommend a book on mathematical analysis that is interesting enough to sit down and read for enjoyment alone? Something not written in the style of a textbook?</p>
<p>All the best.</p>
| epi163sqrt | 132,007 | <blockquote>
<p>I recommend</p>
<ul>
<li><em><a href="https://rads.stackoverflow.com/amzn/click/com/0387770313" rel="nofollow noreferrer" rel="nofollow noreferrer">Analysis by its History</a></em> by E. Hairer and G. Wanner</li>
</ul>
<p>and</p>
<ul>
<li><em><a href="https://rads.stackoverflow.com/amzn/click/com/088385... |
2,344,259 | <blockquote>
<p>$$\int_0^\infty \frac{x^2}{x^4+1} \; dx $$</p>
</blockquote>
<p>All I know this integral must be solved with beta function, but how do I come to the form $$\beta (x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\;dt \text{ ?}$$</p>
| Dr. Sonnhard Graubner | 175,066 | <p>note that $$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$$</p>
|
2,847,359 | <p>So say you had $5^x=25$ where $x$ is obviously $2$, how would you work $x$ out if the question wasn't obvious?</p>
<p><strong>edit:</strong> What if the question was something like $a^x=-1$ (where $a$ is any number).</p>
<p>PS to all the maths elitists out there: Feel free to down vote, I just want to know how to ... | Deepak | 151,732 | <p>Your primary question has been answered.</p>
<p>To answer the additional question in your edited version, on how to solve $a^x = -1$, the short version is that you can't, at least in real numbers. Such an $x$ does not exist amongst the real numbers.</p>
<p>The long answer is that you can have $x$ taking on imagina... |
2,110,681 | <p>I recently started a Discrete Mathematics course in college and I am having some difficulties with one of the homework questions. I need to learn this, so please guide me through at least two steps to get the ball rolling. </p>
<p>The question reads: Show that if $A$ and $B$ are sets, then: $(A \cap B) \cup (A \cap... | Chris | 164,598 | <p>Consider an element of $A$ - either it is in $B$, or it isn't, and thus is in the complement of $B$. Thus $A \subset (A \cap B)$ $\cup$ $(A \cap B')$. Now, try to argue on your own that the reverse "inclusion" holds: that we have $A \supset (A \cap B)$ $\cup$ $(A \cap B')$. </p>
|
1,517,189 | <p>My first question here..sorry if I'm not very specific but I try to be.</p>
<p>A T-tetromino has three connected blocks in a line and another one above the middle block. How many ways can one be painted on the grid if orientation matters? What about if it doesn't?</p>
| Community | -1 | <p>This means that for any given constant $\epsilon $ , you can always find constants, $n_0 >0$, $c > 0$ such that :</p>
<p>$lg(n) < c n^\epsilon $ for all $n\geq n_0$.</p>
|
2,192,400 | <p>In Durrett's Probability Theory and Examples theorem 5.1.8 says:</p>
<blockquote>
<p>Suppose that $\mathbb{E} X^2 <\infty$. Then $\mathbb{E}(X| \mathcal{F})$ is the variable $Y \in \mathcal{F}$ that minimizes the "mean square error" $\mathbb{E}(X-Y)^2$. </p>
</blockquote>
<p>To explain notation, it is assumed... | Nate Eldredge | 822 | <p>See Durrett's Theorem 5.1.4 which, taking $p=2$, says that $E[E[X \mid \mathcal{F}]^2] \le E[X^2]$. So $E[X \mid \mathcal{F}]$ is in $L^2$. In particular, since the random variable $E[X \mid \mathcal{F}]$ has finite expectation, it must be finite almost surely.</p>
|
657,162 | <p>Is there any good approximation for $\prod_{i=3}^k (n-i)$? $(n \gg k)$</p>
<p>I know that $\prod_{i=3}^k (n-i) < \prod_{i=3}^k n = n^{k-2}$</p>
<p>Also a tighter upper bound is appreciated.</p>
| Claude Leibovici | 82,404 | <p>Using Peter's answer and Stirling approxiamtion, you end with $$e^{2-k} (n-3)^{n-\frac{5}{2}} (n-k-1)^{k-n+\frac{1}{2}}$$</p>
|
3,283,101 | <p>I was working on a problem:</p>
<blockquote>
<p>"Find the number of those <span class="math-container">$5$</span>-digit numbers whose digits are non-decreasing." </p>
</blockquote>
<p>I was able to calculate the number of decreasing <span class="math-container">$5$</span>-digit numbers, which I found to be <span... | Parcly Taxel | 357,390 | <p><span class="math-container">$89748$</span> numbers is way too large. Here is the correct derivation.</p>
<p>Zero cannot be in any of the five-digit numbers in question. If there was one, there would have to be zeros all the way left, making it not five-digit.</p>
<p>Any admissible such number thus corresponds to ... |
1,177,871 | <p>I need to find a bijective map from $A=[0,1)$ to $B=(0,1).$ Is there a standard method for coming up with such a function, or does one just try different functions until one fits the requirements?</p>
<p>I've considered some "variation" of the floor function, but not sure if $\left \lfloor{x}\right \rfloor$ is bije... | BigMathTimes | 220,890 | <p>Consider the function
$\psi: A \rightarrow B$ given by
$$
\psi(x) = \begin{cases} x & \text{ if}\; \; \; x \neq 0 \; \; \text{and}\; \; x \neq \frac{1}{n} \in \mathbb{N} \\
\frac{1}{2} & \text{if} \;\; x = 0 \\
\frac{1}{n+1} & \text{if } \; \; x= \frac{1}{n}, \; n \in \mathbb{N}
\end{cases}
$$
It is not... |
1,227,759 | <p>$$x \mod 1000 \mod 5$$</p>
<p>I would have thought that it was $x \mod 5000$ except that it doesn't hold true for $x = 5005$ since you'll get zero, but $5005 \mod 5000 = 5$.</p>
| Adhvaitha | 228,265 | <p>Note that $x \pmod {1000}$ is same as $1000k+x$, where $k \in \mathbb{Z}$. Hence, $x \pmod {1000} \pmod{5}$ is same as $5m+x\pmod{1000} = 5m+1000k+x = 5(m+200k)+x$ is same as $x \pmod5$.</p>
|
119,375 | <p>Modular Arithmetic (MA) has the same axioms as first order Peano Arithmetic (PA) except $\forall x (Sx \ne 0)$ is replaced with $\exists x(Sx = 0)$.
(<a href="http://en.wikipedia.org/wiki/Peano_axioms#First-order_theory_of_arithmetic" rel="nofollow noreferrer"><a href="http://en.wikipedia.org/wiki/Peano_axioms#Firs... | Emil Jeřábek | 12,705 | <p>The answer is no. It is enough to find a model of MA which is an integral domain of characteristic $0$ (whence O1 is true and E1 false) such that $2$ is not invertible (whence E2 is true and O2 false).</p>
<p>One example of such a model is the ring of $2$-adic integers $\mathbb Z_2$. This is clearly a domain, and $... |
2,639,304 | <p>I'm looking for a monotonic increasing function $f(x)$ defined on $[0,\infty)$ so that $f(0)=0$ and with $f'(x)$ approaching $0$ as $x$ approaches $0$, and which approaches 1 as $x \rightarrow \infty$. </p>
<p>The function might look roughly like the figure below, but that particular function is just a shifted and... | Alex Jones | 350,433 | <p>Let $g:[0,\infty)\rightarrow[0,\infty)$ be a continuous function with $g(0)=0$ and $\int_0^\infty g(s)ds = M < \infty$. Then, simply define $f(x) := \frac{1}{M}\int_0^x g(s)ds$. </p>
<p>Since $g \geq 0$, $f$ is increasing (choose $g(x)> 0$ for $x> 0$ to be strictly increasing). Since $g(x) \rightarrow 0$ a... |
2,216,161 | <p>How to prove with identities:
(A - B) ∩ (C - D) = (A ∩ C) - (B ∪ D) </p>
| DMcMor | 155,622 | <p>This is not true. Consider $X = \lbrace a,b, c\rbrace$ and $Y = \lbrace y \rbrace$ both with the discreet topology. Let $f\colon X\to Y$ be defined by $f(x) =y$ for all $x\in X.$ Then $f$ is clearly continuous, but $f^{-1}(U) = X,$ for all nonempty $U\subseteq Y$, and there is no way to write, for example, $\lbra... |
66,485 | <p>How do I evaluate the mean end-to-end squared distance of a FENE ideal chain at fixed inverse temperature $\beta$ in the canonical ensemble?</p>
<p>This quantity is defined as the mean value
$$\left<\left(\sum_{i=1}^{N-1} \vec r_i\right)\cdot\left(\sum_{j=1}^{N-1} \vec r_j\right)\right>$$</p>
<p>Under the c... | Ferdinando Randisi | 22,378 | <p>First, let's rewrite the mean value as the sum of the mean values of the scalar products of the single particle relative positions:</p>
<p>\begin{align*}r^2_{e2e} & = \left<\left(\sum_{i=1}^{N-1} \vec r_i\right)\cdot\left(\sum_{j=1}^{N-1} \vec r_j\right)\right>\\
&=\sum_{i=1}^{N-1} \sum_{j=1}^{N-1}... |
55,509 | <p>I have a measured experimental dataset which is well approximated by the sum of several basis functions in linear combinations. Linear least squares of course gives me the optimal weight for each basis function. These basis functions are all unrelated and may or may not be correlated (or even repeated). That still ... | zyx | 14,120 | <blockquote>
<p>These basis functions are all unrelated and may or may not be correlated (or even repeated). That still doesn't cause any problem when fitting.. least squares is well defined and I always get optimal weights.</p>
</blockquote>
<p>The optimal least-squares approximation of the target is uniquely defin... |
7,787 | <p>I'm trying to expand the following polynomial </p>
<pre><code> Expand[ (A1 a1 + A2 a2 + A3 a3 + A4 a4 + A5 a5 + A6 a6 + A7 a7 + A8 a8)
(D1 a1 + D2 a2 + D3 a3 + D4 a4 + D5 a5 + D6 a6 + D7 a7 + D8 a8)
+ (H1 a1 + H2 a2 + H3 a3 + H4 a4 + H5 a5 + H6 a6 + H7 a7 + H8 a8)
(E1 a1 + E2 a2 + ... | Jens | 245 | <p><strong>Original attempt, experimentation</strong></p>
<p>OK, I have an answer based on an alpha channel but it required more morphological manipulations first. </p>
<p>The good thing is that it should work for other shapes, and requires <em>only</em> the final product <code>newimage</code>. However, to make it wo... |
192,788 | <p>In case you are not familiar with a <em>fullhouse</em>, it is when you have <code>XXXYY</code> (three of a kind + a pair).</p>
<p>I am not that good calculating odds, but after time I manage to write all the possible combinations:</p>
<p><img src="https://i.stack.imgur.com/yOuJG.png" alt="enter image description h... | André Nicolas | 6,312 | <p>There are $\binom{52}{5}$ ways to choose $5$ cards from $52$. All these ways are <strong>equally likely</strong>. Now we will count the number of "full house" hands.</p>
<p>For a full house, there are $\binom{13}{1}$ ways to choose the kind we have three of. For <strong>each</strong> of these ways, the actual car... |
192,788 | <p>In case you are not familiar with a <em>fullhouse</em>, it is when you have <code>XXXYY</code> (three of a kind + a pair).</p>
<p>I am not that good calculating odds, but after time I manage to write all the possible combinations:</p>
<p><img src="https://i.stack.imgur.com/yOuJG.png" alt="enter image description h... | mjqxxxx | 5,546 | <p>The tree you have drawn has each state where a random choice occurs as an internal node, each edge weighted by the probability of descending from one state to another, and each final outcome as a leaf. To find the probability of a particular final outcome, you would take the product of the probabilities along the p... |
6,208 | <p>Let $f$ be an analytic function on the open unit disk domain $D$. Suppose also that $f$ is bounded.</p>
<p>Since $f$ is bounded I believe that $f$ can be continuously extended to the closed unit disk.
I know that the zeros of $f$ in the open disk $D$ are isolated. Are the zeros of $f$ in the closed unit disk als... | Pete L. Clark | 299 | <p>Jonas Meyer's answer goes much deeper, but let me say the following as well.</p>
<p>Even if your (nonzero -- you didn't say that, but of course you meant it!) function does extend continuously to the closed unit disk it need not have isolated zeros. Indeed, suppose that $f$ has infinitely many zeros in the open un... |
391,669 | <p>Let <span class="math-container">$X_1,\dots,X_n$</span> be non commutative variables such that <span class="math-container">$\operatorname{tr} f(X_1,\dots,X_n) = 0$</span> whenever the <span class="math-container">$X_i$</span> are specialized to square matrices in <span class="math-container">$M_r(k)$</span> for any... | Will Sawin | 18,060 | <p>The reformulation suggested by Christian Remling and Benjamin Steinberg is true (at least over a field <span class="math-container">$k$</span> of characteristic zero):</p>
<p>If <span class="math-container">$\operatorname{tr} f(X_1,\dots, X_n)=0$</span> for all <span class="math-container">$X_1,\dots, X_n$</span> i... |
1,355,502 | <p>We inductively define $a^1=a, a^{n+1}=a^n a$. I want to show that $a^{n+m}=a^n a^m$. </p>
<p>By definition, this is true if $m=1$. Now for $m=2$, we have
$$
\begin{align}
a^{n+2} =& a^{(n+1)+1}\\
=& a^{n+1}a \\
=& \left(a^{n}a\right)a \\
=& a^{n}\left(aa\right) \\
=& a^{n}a^2
\end{align}
$$
Ho... | pancini | 252,495 | <p>Step one: show that $a^n\cdot a=a^{n+1}$ (given).</p>
<p>Step two: assume that $a^n\cdot a^k=a^{n+k}$.</p>
<p>Then $a^n\cdot a^{k+1}=a^n\cdot a^k\cdot a=a^{n+k}\cdot a=a^{n+k+1}$</p>
<p>then add a bunch of useless words and you're done!</p>
|
1,355,502 | <p>We inductively define $a^1=a, a^{n+1}=a^n a$. I want to show that $a^{n+m}=a^n a^m$. </p>
<p>By definition, this is true if $m=1$. Now for $m=2$, we have
$$
\begin{align}
a^{n+2} =& a^{(n+1)+1}\\
=& a^{n+1}a \\
=& \left(a^{n}a\right)a \\
=& a^{n}\left(aa\right) \\
=& a^{n}a^2
\end{align}
$$
Ho... | Community | -1 | <p>Fix any $a\in\mathbb R$ and any $m\in\mathbb N$ and let $P(n)$ be the statement $a^ma^n=a^{m+n}$ and show that for all $n\in\mathbb N$, $P(n)$ is true. </p>
<p>For $n=1$, $P(n)$ is true since by definition, $a^ma^1=a^ma=a^{m+1}$. </p>
<p>Now suppose that for some $n=k$, $P(n)$ is true, i.e., $a^ma^k=a^{m+k}$. The... |
2,895,648 | <blockquote>
<p>Suppose $f: \mathbb{R} \to \mathbb{R}$ is continuous such that for any
real $x$, </p>
<p>$|f(x) - f(f(x))| \leq \frac{1}{2} |f(x) -x|$.</p>
<p>Must $f$ have a fixed point?</p>
</blockquote>
<p>The question seems to invite an eventual application of the standard contraction mapping theorem... | enedil | 126,823 | <p>To further improve on Marcos answer:</p>
<p>the sequence is Cauchy, since for all $n<m$, we have
$$
|x_m - x_n| \leq \Sigma_{i=n}^{m-1}|x_{i+1}-x_n| \leq 2^{-n+1} |x_1-x_0|
$$</p>
|
1,650,333 | <p>Can someone help me figure out this ODE, its driving me crazy. I dont need a full solution beacuse that would take hours but maybe just the final answer?</p>
<p>Find the general solution of the ODE
$xy′′ − y′ + 4x^3y = 0$
assuming $x > 0$ and given that $y_1(x) = \sin(x^2)$ is a solution.</p>
| Nitin Uniyal | 246,221 | <p>Hint: The second solution is $y_2(x)=v(x)·y_1(x)$ where
$$
v(x)=\int \frac{e^{-\int P(x)dx}}{y_1(x)^2}\;dx
$$ where $P(x)=\frac{-1}{x}$ in your case.</p>
|
160,355 | <p>If i import a .xyz file in Mathematica that will show me the moelcule in the default way. But i want to visualise the .xyz file in the following manner.
<a href="https://i.stack.imgur.com/7SwPp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7SwPp.jpg" alt="enter image description here"></a></p>
| J. M.'s persistent exhaustion | 50 | <p>Here's a starting point, whose finessing/generalization I will leave to somebody else. First, import the elements and coordinates separately:</p>
<pre><code>elem = Import["https://pastebin.com/raw/QqrhhvMj", {"XYZ", "VertexTypes"}];
coords = Import["https://pastebin.com/raw/QqrhhvMj", {"XYZ", "VertexCoordinates"}];... |
1,869,167 | <p>The least value of $a \in R$ for which $4ax^2 + \frac{1}{x} \ge 1 $for all $x \gt 0 $, is </p>
<p>Using AM-GM inequality
$$\frac{4ax^2 + \frac{1}{2x} + \frac{1}{2x}}{3} \ge \sqrt[3]{a}$$
$$4ax^2 + \frac{1}{x} \ge 3\sqrt[3]{a}$$</p>
<p>Now my question start from here .</p>
<p>Can I do that for least value of $a$... | Michael Rozenberg | 190,319 | <p>No, I think. Because does not exist $a>0$, for which $3\sqrt[3]a>3\sqrt[3]a$. </p>
<p>For the original problem we see that since for $4ax^2=\frac{1}{2x}$ we get an equality and we obtain $a\geq\frac{1}{27}$, so the answer is $\frac{1}{27}$.</p>
|
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