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 |
|---|---|---|---|---|
1,857,630 | <p>Let $\mathcal{A}$ be a non (necessarily) unital commutative Banach algebra, and let
$$ M_{\mathcal{A}} = \{ \phi:\mathcal{A} \to \mathbb{C} : \phi \mbox{ is multiplicative and not trivial}\} $$
and
$$ \mathrm{Max}(\mathcal{A})=\{ I \lhd \mathcal{A} : I \mbox{ maximal} \}.$$
If $\mathcal{A}$ is unital, it is well kn... | Tomasz Kania | 17,929 | <p>Linear-multiplicative functionals (<em>aka</em> characters) on complex Banach algebras are automatically continuous, so their kernels are closed. (You will find a slick proof of this fact on p. 181 of Allan's and Dales' <em>Introduction to Banach Spaces and Algebras</em>.) However, in the non-unital case it may well... |
1,444,216 | <p>An LP is given in the form :</p>
<p>$max\langle c,x\rangle$ subject to $Ax=b,x\geq 0$.</p>
<p>I'm trying to show that any such problem can be reduced to a problem of the form </p>
<p>$max\langle c,x\rangle$ subject to $Ax=b,x\geq 0,\displaystyle \sum {x_j}=1$.</p>
<p>I think I'll somehow have to modify the probl... | Ittay Weiss | 30,953 | <p>The intrinsic metric structure is what the ant is using. Generally, given a metric space $X$ and a susbet $S$, the ambient metric in $X$ means $S$ is a metric space, using the exact same metric. Thinking of $S\subseteq \mathbb R^2$ with the Euclidean metric on $\mathbb R^2$ and where $S$ is the unit circle (only the... |
1,444,216 | <p>An LP is given in the form :</p>
<p>$max\langle c,x\rangle$ subject to $Ax=b,x\geq 0$.</p>
<p>I'm trying to show that any such problem can be reduced to a problem of the form </p>
<p>$max\langle c,x\rangle$ subject to $Ax=b,x\geq 0,\displaystyle \sum {x_j}=1$.</p>
<p>I think I'll somehow have to modify the probl... | JMoravitz | 179,297 | <p>As mentioned already, the metric to use is called the <em>intrinsic metric</em>, but without having worked with it before, it might be hard to use.</p>
<p>This is a good example to start using the intrinsic metric on though.</p>
<p>Let us label the four walls, ceiling, and floor as $N,E,S,W,C,F$ respectively.</p>
... |
2,903,429 | <p>Show that the equation $$\frac{4}{3}{x^2} + \frac{4}{3}{y^2} + \frac{4}{3}{z^2} + \frac{4}{3}{xy} + \frac{4}{3}{xz} + \frac{4}{3}{yz} = 1$$ represents an ellipsoid. Find the position and lengths of its principal half-axes.</p>
<p>Workings
$$(x+y)^2+(y+z)^2+(z+x)^2=\frac32\,.$$
Common factor $\left(\dfrac32\right)^{... | Deepesh Meena | 470,829 | <p>$$\binom{n}{2}=\frac{n!}{2!(n-2)!}=\frac{n(n-1)(n-2)!}{2!(n-2)!}=\frac{n(n-1)}{2}$$</p>
<p>$$\sum_{k=1}^{j}1=j$$
$$\sum_{k=1}^{n}1=n$$
$$\sum_{k=1}^{n}1=\sum_{k=1}^{j}1+\sum_{k=j+1}^{n}1$$
$$\sum_{k=j+1}^{n}1=\sum_{k=1}^{n}1-\sum_{k=1}^{j}1=n-j$$</p>
<p>$$\sum_{j=1}^n \sum_{k=j+1}^{n} \ 1 =\sum_{j=1}^n\left(n-j\r... |
754,644 | <p>This question is related to a solved problem in Gilbert Strang's 'Introduction to Linear Algebra'(Chapter 3,Question 3.6A, Page 190).</p>
<blockquote>
<p>Q) Find bases and dimensions for all four fundamental subspaces of A if you know that</p>
<p><span class="math-container">$A = \begin{bmatrix}1 & 0 & 0\\2 ... | yago | 141,261 | <p>If you think of the product $E^{-1}R$ as the composition of linear applications, then $E^{-1}$ acts "last" and its columns determine somehow the image of $E^{-1}R$ (of course, it also depends on $R$).</p>
<p>More generally, for any functions from any sets that you can compose (assuming $Im(g) \subset D(f)$ where $D... |
2,006,870 | <p>Is the following inequality true?</p>
<p>$\left( \sum \limits_{i=1}^\infty \sum \limits_{j=1}^\infty \sum \limits_{k=1}^\infty \sum \limits_{l=1}^\infty a_{ij}\,a_{ik}\,a_{jl}\,a_{kl} \right) \leq \left( \sum \limits_{i=1}^\infty \sum \limits_{j=1}^\infty a_{ij}^2 \right)^{1/2}\left( \sum \limits_{i=1}^\infty \sum ... | B. Goddard | 362,009 | <p>Pick two primes $p$ and $q$ larger than $999,999,999$. Solve the system $t \equiv X \pmod{p}$, $t\equiv Y \pmod{q}$ by the Chinese Remainder Theorem. Then $t$ will be unique for each pair $(X,Y)$ modulo $pq>10^{18}.$</p>
|
126,270 | <p>How many seven-digit numbers divisible by 11 have the sum of their digits equal to 59?</p>
<p>I am able to get the seven-digit numbers divisible by 11 </p>
<p>and </p>
<p>I am also able to get the seven-digit numbers whose sum of their digits equal to 59.</p>
<p>But i am not able to get how i can get the count o... | Brian M. Scott | 12,042 | <p>Since I promised vikiiii that I’d answer, here’s my version.</p>
<p>A number is divisible by $11$ if and only if the alternating sum of its digits is divisible by $11$. Say that the seven-digit number is $abcdefg$, where the letters represent the individual digits; then it’s divisible by $11$ if and only if $(a+c+e... |
767,282 | <blockquote>
<p><strong>Problem</strong>: Show that $f: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ with $f(x,y)=\|x-y\|_2^2$ is differentiable and compute its differential at every point in the domain of $f$<br><strong>Note</strong>: $\| \cdot \|_2$ denotes the euclidian norm $d(x,y)_2 =\sqrt{\sum\limits_{i=1}^... | Pedro | 23,350 | <p>Note this is the map $(x,y)\mapsto (x-y)\cdot (x-y)$, that is $(x,y)\mapsto x\cdot x-2x\cdot y+y\cdot y$. It suffices we show each of the summands is differentiable. But two of the summands are $\lVert \cdot \rVert ^2$; which we know how to differentiate, and the central summand is bilinear, so it is not too hard to... |
184,142 | <p>I am currently an undergraduate math student. (In fact, freshmen.)</p>
<p>I know that usually abstract algebra is taught somehow late in the undergraduate course, and curious how studies of abstract algebra at graduate level differ from studies at undergraduate level.</p>
<p>So, things like what gets new treatment... | M Turgeon | 19,379 | <p>There are several ways in which it is/can be different.</p>
<ul>
<li>There might be greater emphasis on proofs and rigor, depending on the level of the undergraduate abstract algebra.</li>
<li>The material is covered more quickly, and as such, more material can be covered.</li>
<li>As an extension of the previous p... |
279,238 | <p>I want to create a regular polygon from the initial two points <span class="math-container">$A$</span>, <span class="math-container">$B$</span> and number of vertices <span class="math-container">$n$</span>,<br />
<a href="https://i.stack.imgur.com/dKr9A.png" rel="noreferrer"><img src="https://i.stack.imgur.com/dKr9... | cvgmt | 72,111 | <ul>
<li><code>AnglePath</code>.</li>
</ul>
<p>For two point <code>{x1,y1}</code> and <code>{x2,y2}</code>, at first we locate at <code>{x2,y2}</code> and toward to the direction <code>{x2,y2}-{x1,y2}</code>, then turn to a new direction by rotate a angle<code>2 Pi/n</code>, after <code>n</code> rotation we go back to... |
4,112,011 | <p>Let <span class="math-container">$f:X\rightarrow Y$</span> be a map and let <span class="math-container">$T'$</span> be a topology on <span class="math-container">$Y$</span>.</p>
<p>Show that <span class="math-container">$$T=\{U\subseteq X \mid \exists V\in T' \text{ with }U=f^{-1}(V)\} $$</span> is a topology on <s... | Community | -1 | <p>So, we made two corrections, and you are well on your way!</p>
<p>Left is just to prove the intersection and union properties. But they follow from the general facts:</p>
<ul>
<li><span class="math-container">$f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$</span></li>
<li><span class="math-container">$f^{-1}(A\cap B)=f^{... |
301,038 | <p>If you want to show that a sequence $(a_{n})$ in $\mathbb{R}$ is convergent, when is it sufficient to show that there is a number $b\in\mathbb{R}$ such that
$$ \liminf a_{n} \geq b \geq \limsup a_{n}$$</p>
<p>In particular, I have a situation where my sequence is bounded and I wanted to use this approach, but I'm n... | Amr | 29,267 | <p>$$\forall n\in\mathbb{Z}^+[\inf\{a_m|m\geq n\}\leq\sup\{a_m|m\geq n\}]$$
Thus:
$$\lim\inf a_n\leq\lim \sup a_n$$</p>
<p>Now if you can show that $\lim\sup a_n\leq\lim\inf a_n$, then this inequality and the previous one show that $\lim\inf a_n=\lim\sup a_n$</p>
<p>The last equation can be used to show that $\lim a_... |
2,096,296 | <p>Recently I discovered that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{π^2}{6}$, and we know that sum of reciprocals of naturals to the first power diverges to infinity. So I was wondering just out of curiosity, whether there was a number between $1$ and $2$ where this sum of reciprocals raised to that power started dive... | Simply Beautiful Art | 272,831 | <p>We have the integral test, which has</p>
<p>$$\int_1^\infty\frac1{x^s}\ dx\sum_{n=1}^\infty\frac1{n^s}<1+\int_1^\infty\frac1{x^s}\ dx$$</p>
<p>Which quickly shows it converges only if $s>1$ and diverges for $s\le1$.</p>
|
109,307 | <p><strong>Problem.</strong> Let $d=(a,b)$, $b=\beta d$ and $n>1$. If $\beta$ is odd number, prove that $(n^a+1,n^b-1)\le 2$.</p>
<p><em>Solution (from the book).</em> Each common divisor of numbers $n^a+1$ and $n^b-1$ has to be divisor of their sum $n^a+n^b=n^b(n^{a-b}+1)$, that is, it has to be a common divisor o... | André Nicolas | 6,312 | <p>The proof you were given is unnecessarily convoluted. In particular, the first part is totally unnecessary! We give an alternate proof in great detail. </p>
<p>We use congruence notation, because it is convenient. In a remark at the end, we show how you can avoid congruence notation, if you really want to, which ... |
3,020,825 | <p>let A be a subset of <span class="math-container">$\mathbb{R}^P$</span> and <span class="math-container">$ x \in \mathbb{R}^P$</span> denotes <span class="math-container">$d(x,A) = \inf \{ d(x,y) : y \in A\}$</span>.There exists a point <span class="math-container">$y_0 \in A $</span> with <span class="math-c... | Noble Mushtak | 307,483 | <p>Primitive recursion allows us to define <span class="math-container">$F(x+1)$</span> in terms of both <span class="math-container">$x$</span> and <span class="math-container">$F(x)$</span> like this:</p>
<p><span class="math-container">$$F(0)=0$$</span>
<span class="math-container">$$F(S(x))=g(x,F(x))$$</span></p>
... |
1,017,320 | <p>Assume ZFC. Let $B\subseteq\mathbb R$ be a set that is not Borel-measurable. Clearly, $B$ must be uncountable, since countable sets are always Borel being a countable union of measurable singletons.</p>
<p><strong>Question:</strong> can one conclude that $B$ necessarily has the cardinality of the continuum <em>with... | Francis Adams | 29,633 | <p>You can still have an idea of the cardinality of sets that are definable in a more general sense than being Borel. Say a set is analytic $(\bf{\Sigma}^1_1)$ if it is the continuous image of a Borel set, coanalytic $(\bf{\Pi^1_1})$ if its complement is analytic, and $\bf{\Sigma}^1_2$ if it is the continuous image of ... |
1,037,632 | <p>How to find the last 2 digits of $2014^{2001}$? What about the last 2 digits of $9^{(9^{16})}$?</p>
| Swapnil Tripathi | 117,387 | <p>Finding the last two digits it equivalent to finding</p>
<p>$2014^{2001}\pmod{100}$.</p>
<p>Note $2014\equiv 14\pmod{100}$</p>
<p>Finding this requires using $100=25\times 4$ (coprime factors)</p>
<p>Is there a common solution to the system below?</p>
<p>$x\equiv 2014^{2001}\pmod{25}\equiv 14^{2001}\pmod{25}$</... |
15,702 | <p>One way to prove that a field $K$ has no ideals except the entire field and the trivial ideal is to note the fact that every element $x$ has an inverse. By the definition of an ideal, if $x$ is in the ideal then $x^{-1}x$ is because $x^{-1} \in K$. But now we have that 1 is in the ideal, and so again by the definiti... | Arturo Magidin | 742 | <p>Yes, it's true that any ideal in a ring which contains a unit must be the whole ring. (Also: every left ideal that contains a left-invertible element must be the whole ring).</p>
<p>Now, there are two items here. One is the notion of generalizing "multiples of $n$", and the second is the notion of making quotients... |
1,761,273 | <p>Good morning. I have a problem with this:</p>
<p>Find the maximum and minimum distances from the origin to the curve* <span class="math-container">$$g\left(x,y\right)=5x^{2}+6xy+5y^{2}$$</span></p>
<p>I have done this:</p>
<p>Function to optimize:<span class="math-container">$f\left(x,y\right)=x^{2}+y^{2}$</span></p... | Reynan Henry | 628,467 | <p>divide the first equation with the second one to get rid of the <span class="math-container">$\lambda$</span> then you can solve for <span class="math-container">$x$</span> and <span class="math-container">$y$</span> with the third equation</p>
|
1,430,447 | <p>sketch the graph of the integrand function and use it to help evaluate the integral.</p>
<p>integration from(1 , -1) |x|-1 </p>
<p><a href="https://i.stack.imgur.com/NY3Ei.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NY3Ei.png" alt="enter image description here"></a></p>
<p>I think I can ev... | nathan.j.mcdougall | 181,447 | <p>Another method would be to use the graph alone. <a href="https://i.stack.imgur.com/sh6Du.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sh6Du.png" alt="Created using Desmos"></a>
Created using <a href="http://www.desmos.com" rel="nofollow noreferrer">Desmos</a></p>
<p>From the graph, it is clear... |
3,270,504 | <blockquote>
<p>Prove that <span class="math-container">$\log|e^z-z|\leq |z|+1$</span> where <span class="math-container">$z\in\mathbb{C}$</span> with <span class="math-container">$|z|\geq e$</span>.</p>
</blockquote>
<p><strong>Background:</strong></p>
<p>This is from a proof that <span class="math-container">$e^z... | Philip Hoskins | 41,421 | <p>Use the definition: <span class="math-container">$e^z = \sum_{n=0}^\infty z^n/n!$</span></p>
<p><span class="math-container">$$
\begin{align}
\vert e^z - z \vert &= \vert 1+z^2/2 + z^3/6 + \cdots \vert \\
&\leq 1+\vert z \vert ^2/2 + \vert z^3 \vert/6 + \cdots\\
&= e^{\vert z \vert} - \vert z \vert \\
... |
3,281,965 | <p>It is known that a way to check whether a number <span class="math-container">$n$</span> is prime, is to check for divisors of <span class="math-container">$n$</span> from <span class="math-container">$2$</span> to <span class="math-container">$\lfloor\sqrt{n}\rfloor$</span>. If we find any divisor, then <span class... | Peter | 82,961 | <p>For very large numbers we can estimate the probability as follows :</p>
<p>Let <span class="math-container">$N$</span> be the given number , then <span class="math-container">$M:=\frac{\ln(N)}{\ln(2)}$</span> is the limit of trial division.</p>
<p>Denote <span class="math-container">$$P:=\prod{\frac{p}{p-1}}$$</sp... |
275,526 | <p>What would be an easy example of a sequence of functions defined on a compact interval so that $f_n$ goes to $f$ pointwise but $\sup f_n$ does not go to $sup f$.</p>
<p>I thought of the usual example we take to show that the limits in integration can't be interchanged when we only have pointwise convergence. Is th... | David Mitra | 18,986 | <p>Consider the functions $f_n$ defined on $[0,1]$, where $f_n$ is the function whose graph consists of the following straight line segments: from $(0,0)$ to $(1/n,1)$, from $(1/n,1)$ to $(2/n,0)$, and from $(2/n,0)$ to $(1,0)$.</p>
<p>Note that $(f_n)$ converges pointwise to the zero function on $[0,1]$.</p>
|
2,111,841 | <p>A geodesic is a line representing the shortest route between two points on a sphere, for example on the Earth treated here as a perfect sphere.
Two points on Earth having the same latitude can be also connected with the line being a part of a circle for selected constant latitude.
Differences between these two lines... | Andrew D. Hwang | 86,418 | <p><strong>Summary</strong>: If $A$ and $B$ lie on the latitude line at angle $0 < \alpha < \pi/2$ north of the equator on a sphere of unit radius, and at an angular separation $0 < \theta = \beta_{2} - \beta_{1} < \pi$, then the "digon" bounded by the latitude and the great circle arc $AB$ (in bl... |
1,694,991 | <p>I tried doing $u$-substitution and got $-20e$ as my final answer, but I think the correct answer is just $20$. I'm not sure what I did wrong, but probably had to do with plugging in infinity... could someone explain the process of solving this integral?</p>
| Michael Hardy | 11,667 | <p>\begin{align}
\int_0^\infty \frac x{20}e^{-x/20} \, dx & = 20 \int_0^\infty \left(\frac x{20}\right) e^{-x/20} \, \left(\frac{dx}{20}\right) \\[6pt]
& = 2\underbrace{0 \int_0^\infty u e^{-u}\,du}_\text{substitution} = \underbrace{20 \int u\,dv = 20\left( uv - \int v\,du \right)}_\text{integration by parts} \... |
3,299,863 | <p>Let <span class="math-container">$f_{1}(x) = e^{x^5}$</span> and <span class="math-container">$f_{2}(x) = e^{x^3}$</span>. Let <span class="math-container">$g(x) = f_{1}f_{2}$</span>. Find <span class="math-container">$g^{(18)}(0)$</span>.</p>
<p>By series expansion at <span class="math-container">$x = 0$</span>:</p... | Con | 682,304 | <p>Not sure whether this is the kind of answer that you wanted or expected.</p>
<p>You only need a topology to talk about convergence in general. To get a nice and convenient topology on non-standard sets <span class="math-container">${^*}Y$</span>, one usually takes superstructures (actually in a way such that <span ... |
1,282,592 | <p>I am often asked to prove properties of regular bipartite graphs, and beyond the two parts having equal size nothing seems obvious. Are these graphs more intuitive than they first seem?</p>
<p>In particular, right now I can't work out why an r-regular bipartite graph is r-edge-colourable.</p>
<p>Thanks</p>
| Exodd | 161,426 | <p>You can reduce this problem (through induction on $r$) to looking for a perfect matching that covers all the nodes. </p>
<p>Thanks to the <a href="http://en.wikipedia.org/wiki/Hall%27s_marriage_theorem" rel="nofollow">Hall's marriage theorem</a>, you can prove there's one by verifying the condition that for every s... |
299,140 | <p>Is there a closed form sum of </p>
<p>$\sum_{k=0}^{\infty} \frac{x^k}{(k!)^2}$</p>
<p>It is trivial to show that it is less than $e^x$ but is there a tighter bound?</p>
<p>Thanks</p>
| Ryan O'Donnell | 658 | <p>The first way I try to solve questions like this is to "ask Maple" (or Mathematica). If you have access to, say, Maple, then you can type</p>
<p>"sum(x^k/k!^2, k=0..infinity)"</p>
<p>and it will report BesselI(0,2*sqrt(x)). [It's impossible to tell from the font, but that is "Bessel" followed by capital-i.] If y... |
15,093 | <p>For example, I am confident that very few students majoring in pure mathematics can write a complete proof to the <a href="https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem" rel="noreferrer">Abel–Ruffini theorem</a> (there is no algebraic solution to general polynomial equations of degree five or higher wit... | Nicola Ciccoli | 465 | <p>It just depend on what do you mean by "knowing very well".</p>
<p>To me "knowing very well" means: if you give me 1 hour I can tell you the basic ingredients that go in the proof (and the reason why the result is relevant) and if you give me 1 full day I can sketch a reasonably detailed proof.</p>
<p>This means:</... |
199,842 | <p>I understand the reasoning behind $\pi r^2$ for a circle area however I'd like to know what is wrong with the reasoning below:</p>
<p>The area of a square is like a line, the height (one dimension, length) placed several times next to each other up to the square all the way until the square length thus we have heig... | copper.hat | 27,978 | <p>Here is a similar approach. Split the radius into $n$ equal parts, and form concentric circles of radius $0, \frac{r}{n}, \frac{2r}{n},...,r$. Think of the cross section of an onion. Then estimate the area by unrolling each circle, approximating its area by a rectangular strip of length given by the outer radius and... |
199,842 | <p>I understand the reasoning behind $\pi r^2$ for a circle area however I'd like to know what is wrong with the reasoning below:</p>
<p>The area of a square is like a line, the height (one dimension, length) placed several times next to each other up to the square all the way until the square length thus we have heig... | JavaMan | 6,491 | <p>Actually, you can use your method, but you have to be a bit careful. Start with a circle of radius $r$ centered at the origin. We can think of the total area of the circle as adding the "lengths" of all the circumferences of all the circles centered at the origin which have a radius $x \leq r$:</p>
<p>$$\text{Are... |
2,217,295 | <p>I thought about defining a function $ g(x) = \dfrac{1}{\epsilon} x^{\epsilon} - \ln(x) $ and show that it's larger than $0$. I saw that if I substitute
$1$, I get $\dfrac{1}{\epsilon}$ which is positive and so on. I also tried to derive the function and I got $g'(x) = x^{\epsilon -1} - \dfrac{1}{x} $</p>
<p>I'... | Zubzub | 349,735 | <p>Let $c > 0$ and $\epsilon \in ]0,1[$. We start with the trivial inequality :
$$
\begin{align}
c &< c + \frac{1}{\epsilon} &\Longleftrightarrow \\
c & < \frac{1}{\epsilon}(1 + \epsilon c) & \overset{(*)}{\implies} \\
c & < \frac{1}{\epsilon} e^{\epsilon c} & \overset{(**)}{\implies... |
2,217,295 | <p>I thought about defining a function $ g(x) = \dfrac{1}{\epsilon} x^{\epsilon} - \ln(x) $ and show that it's larger than $0$. I saw that if I substitute
$1$, I get $\dfrac{1}{\epsilon}$ which is positive and so on. I also tried to derive the function and I got $g'(x) = x^{\epsilon -1} - \dfrac{1}{x} $</p>
<p>I'... | Nathanael Skrepek | 423,961 | <p>Alternative you could use L'Hôpital's rule to show
$$\lim_{x\to\infty} \frac{\ln(x)}{x^{\epsilon}} = \lim_{x\to\infty} \frac{x^{-1}}{\epsilon x^{\epsilon-1}} = \lim_{x\to\infty} \frac{1}{\epsilon} x^{-\epsilon} = 0.$$
From this follows that from some point $x^\epsilon > \ln(x)$ otherwise the limit can't be $0$.</... |
813,301 | <p>$f$ is strictly increasing and $g$ is decreasing. How to find whether $f\circ g$ and $g\circ f$ are increasing, decreasing, strictly increasing or strictly decreasing?</p>
<p>This is what I did,</p>
<p>$f \circ g=f(g(x))$</p>
<p>If we take $x_1 < x_2$,</p>
<p>$f(x_1) < f(x_2) $ and $\ g(x_1) \ge g(x_2) $<... | Empy2 | 81,790 | <p>You have shown that $f\circ g(x_1)<f\circ g(x_2)\implies x_2<x_1$ - that is, $f\circ g$ is decreasing. </p>
<p>Yes, the same method shows that $g\circ f$ is also decreasing.</p>
|
2,265,368 | <p>Let $R$ be a commutative ring with unit and $N$ its nilradical. If $R$ is connected then $R/N$ is also connected; the quotient map induces a homeomorphism between the spectra. I'd like to see a more hands-on proof, but I'm unable to make it work. That is to say, I'd like to show that if $e^2-e\in N$ for some $e\in R... | E.R | 325,912 | <p>It is well known that every idempotent modulo $N$ can be lifted to an idempotent of $R$. Thus, if $R$ has no non trivial idempotent, then $R/N$ also has no non trivial idempotent. (See Anderson's "Rings and Categories of modules")</p>
|
4,318,842 | <p>Thanks for your time.</p>
<p>I am studying Euclidean Geometry through a text book and they have a corollary that afirms that given <span class="math-container">$r \parallel s$</span>, if t intersects r then it must intersect s, although the proof they give is not much convinceable. It says: "If one line can int... | Community | -1 | <p>I must assume that two lines "intersect" when they have a point in common and they don't coincide, otherwise the claim is falsified by the case <span class="math-container">$t=r$</span>.</p>
<p>As per your work, using the fact that parallelism is a transitive relation defeats the purpose of disproving the ... |
3,836,662 | <p>I understand basic group theory. I would say that I've seen most of the standard stuff up to, say, the quotient group.</p>
<p>I feel like I've seen in more than one place the suggestion that group theory is the study of symmetries, or actions that leave something (approximately) unchanged. Unfortunately I can only f... | Community | -1 | <p>The equivalence relation <span class="math-container">$(h,k)\sim (h',k')\stackrel{(def.)}{\iff} hk=h'k'$</span> induces a partition of <span class="math-container">$H\times K$</span> into equivalence classes each of cardinality <span class="math-container">$|H\cap K|$</span>, and the quotient set <span class="math-c... |
4,563,135 | <p>Denote the linear space of linear operators on the linear space <span class="math-container">$V$</span> with field <span class="math-container">$\mathbb{F}$</span> by <span class="math-container">$L(V)$</span> and the linear space of <span class="math-container">$n \times n$</span> matrices with entries in <span cla... | user8675309 | 735,806 | <p>The idea is that you can get the result by using two standard operations you are already familiar with: (i) transposition <span class="math-container">$^T$</span> (which you know doesn't change rank) and (ii) conjugate transposition <span class="math-container">$^*$</span>. These are both involutions.</p>
<p>Now co... |
4,024,871 | <p>I'm looking to build a function <span class="math-container">$f:S^2 \to \mathbb R^2$</span> such that <span class="math-container">$f(x)\neq f(−x)$</span> for all <span class="math-container">$x\in S^2$</span>.</p>
<p>By Borsuk-Ulam Theorem, this function must be discontinuous. I was trying to build a not too compli... | Aryaman Maithani | 427,810 | <p>Here's an overkill using the Axiom of Choice:</p>
<p>Let <span class="math-container">$\mathcal A = \{\{\mathbf x, -\mathbf x\} \mid \mathbf x \in S^2\}$</span> be the collection of all antipodal pairs. From each set in <span class="math-container">$\mathcal A$</span>, pick exactly one element. Map that to <span cla... |
20,470 | <p>I couldn't find a more descriptive title, but I guess an example will explain my problem.</p>
<p>I set up some customized <code>Grid</code> function including some additional functionalities which I control with custom Options. Additionally, I would like to change some of the standard <code>Grid</code> Options, e.... | Mr.Wizard | 121 | <p><a href="http://reference.wolfram.com/mathematica/ref/OptionsPattern.html" rel="noreferrer">OptionsPattern</a>:</p>
<p><img src="https://i.stack.imgur.com/8Dpvd.png" alt="Mathematica graphics"></p>
<p>Therefore declare <code>Options</code> for both <code>myGrid</code> and <code>Grid</code> as valid:</p>
<pre><cod... |
366,589 | <p>how do I solve this one? </p>
<p>$$\lim_{q \to 0}\int_0^1{1\over{qx^3+1}} \, \operatorname{d}\!x$$</p>
<p>I tried substituting $t=qx^3+1$ which didn't work, and re-writing it as $1-{qx^3\over{qx^3+1}}$ and then substituting, but I didn't manage to get on. </p>
<p>Thanks in advance! </p>
| Abel | 71,157 | <p>For any $q>0$, $$\frac{1}{q+1} = \int_0^1\frac{1}{q\cdot 1+1}\,\mathrm{d}x\leq\int_0^1\frac{1}{qx^3+1}\,\mathrm{d}x \leq \int_0^1\frac{1}{q\cdot 0+1}\,\mathrm{d}x = 1.$$
For any $q\in(-1,0)$,
$$1 = \int_0^1\frac{1}{q\cdot 0+1}\,\mathrm{d}x\leq \int_0^1\frac{1}{qx^3+1}\,\mathrm{d}x\leq \int_0^1\frac{1}{q\cdot 1+1}... |
2,940,306 | <p>I understand that it would be n! permutations for the given amount of elements, but I am not sure calculate it with these parameters.</p>
| Saketh Malyala | 250,220 | <p>Choose the first character. </p>
<p>Choose the last character.</p>
<p>How many ways can you order the remaining <span class="math-container">$7$</span>? </p>
<p><span class="math-container">$3 x 3 x 7!$</span></p>
|
1,091,549 | <p>How to compute $\int_{0}^{+\infty} \frac{dx}{e^{x+1} + e^{3-x}}$?</p>
<p>My partial solution:
$$
\int_{0}^{+\infty} \frac{dx}{e^{x+1} + e^{3-x}} = \int_{0}^{+\infty} \frac{dx}{e^{3-x}(1 + e^{2x-2})} \\
= \int_{0}^{+\infty} \frac{e^{x-3}dx}{1 + e^{2x-2}}.
$$
Thank you very much.</p>
| André Nicolas | 6,312 | <p>It is natural to let $y=e^x$. Then $dy=e^x\,dx$, so $dx=\frac{1}{y}\,dy$. Thus we want
$$\int_1^\infty \frac{1}{y}\cdot \frac{1}{e\cdot y+e^3 \cdot \frac{1}{y}}\,dy,$$
which simplifies to
$$\int_1^\infty \frac{1}{e}\cdot\frac{1}{y^2+e^2}\,dy.$$
Let $y=et$, and we are nearly finished.</p>
|
939,911 | <p>Question Re-phrased:</p>
<hr>
<p>I'm having a lot of trouble wrapping my head around this problem. While I've looked through similar posts, It's difficult understanding the maths because I currently have another approach stuck in my head. Is this valid?</p>
<p>Question:
Prove that an at most countable union of at... | Asaf Karagila | 622 | <p>This is not a proof. And had I been grading it, I'd have given it no points. </p>
<ol>
<li><p>At most countable includes the infinite case. All you did in your proof is to handle the case all the sets are finite.</p></li>
<li><p>Even if there is a maximal index in $F$, what difference does it make? You're just poin... |
3,443,672 | <p>The equation is
<span class="math-container">$$\tan\frac{5\pi}{6} \cos x=1-\sin x$$</span>
<span class="math-container">$$\sin\frac{5\pi}{6} \cos x=\cos\frac{5\pi}{6}-\cos\frac{5\pi}{6} \sin x$$</span>
<span class="math-container">$$\sin\left(\frac{5\pi}{6}+x\right)=\cos \frac{5\pi}{6}$$</span> which looks weird to... | Calum Gilhooley | 213,690 | <p>One set of solutions is given by <span class="math-container">$\cos x = 0$</span> and <span class="math-container">$\sin x = 1$</span>, i.e.:
<span class="math-container">$$
x = \left(2n + \tfrac12\right)\pi.
$$</span></p>
<p>If <span class="math-container">$\cos x \ne 0$</span>, then <span class="math-container">$... |
2,013,650 | <p>I am trying to prove that
$$n^{n/2}<n!,\text{ for } n\ge2.$$
I can't really figure it out. </p>
| Chill2Macht | 327,486 | <p>$\newcommand{\Vec}{\mathsf{Vec}}$$\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\id}{\operatorname{id}}$This is for my future reference so I never have to write out these proofs again.</p>
<blockquote>
<p><strong>Claim:</strong> $\Hom(-,-):\Vec^{op}\times \Vec \to \Vec$ is a functor.</p>
</blockquote>
<p>(a... |
471,710 | <p>Why do small angle approximations only hold in radians? All the books I have say this is so but don't explain why.</p>
| Meow | 39,568 | <p>It's because this relationship
$$\lim_{x\rightarrow0}\frac{\sin(x)}{x}=1$$ (i.e. $\sin(x) \approx x$) only holds if $x$ is in radians, as, using L'Hoptial's rule,
$$\lim_{x\rightarrow0}\frac{\sin(x)}{x}=\lim_{x\rightarrow0}\frac{\frac{d}{dx}\sin(x)}{\frac{d}{dx}x}=\lim_{x\rightarrow0}\frac{d}{dx}\sin(x)=\begin{case... |
2,698,555 | <p><strong>Question</strong></p>
<p>Find all the rational values of $x$ at which $y=\sqrt{x^2+x+3}$</p>
<p><strong>My attempt</strong></p>
<p>Since we only have to find the rational values of $x$ and $y$, we can assume that
$$ x \in Q$$
$$ y \in Q$$
$$ y-x \in Q $$
Let$$ d = y-x$$
$$d=\sqrt{x^2+x+3}-x$$
$$d+x=\sqrt... | King Tut | 520,782 | <p>There are two red parts and two blue parts. Let one blue part be called $B$ and red be $R$. Radius of small circle be $r$. Now write area of big quarter circle with radius $2r$ in terms of these variables using <em>inclusion exclusion principle</em>:</p>
<p>$$2 \cdot \frac{\pi r^2}{2} -2B+2R = \frac{\pi (2r)^2}{4}\... |
2,698,555 | <p><strong>Question</strong></p>
<p>Find all the rational values of $x$ at which $y=\sqrt{x^2+x+3}$</p>
<p><strong>My attempt</strong></p>
<p>Since we only have to find the rational values of $x$ and $y$, we can assume that
$$ x \in Q$$
$$ y \in Q$$
$$ y-x \in Q $$
Let$$ d = y-x$$
$$d=\sqrt{x^2+x+3}-x$$
$$d+x=\sqrt... | CiaPan | 152,299 | <p>See the image with blue parts shifted:</p>
<p><a href="https://i.stack.imgur.com/LazQz.png" rel="noreferrer"><img src="https://i.stack.imgur.com/LazQz.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/f00Mb.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/f00Mb.gif" al... |
2,698,555 | <p><strong>Question</strong></p>
<p>Find all the rational values of $x$ at which $y=\sqrt{x^2+x+3}$</p>
<p><strong>My attempt</strong></p>
<p>Since we only have to find the rational values of $x$ and $y$, we can assume that
$$ x \in Q$$
$$ y \in Q$$
$$ y-x \in Q $$
Let$$ d = y-x$$
$$d=\sqrt{x^2+x+3}-x$$
$$d+x=\sqrt... | Thomas Pornin | 38,580 | <p>If you forgot formulas, but remember that areas are quadratic (i.e. "double the length, quadruple the area"), then you know that a circle of radius $2r$ uses four times the area of circle of radius $r$. Here, you have one quarter of a circle of radius $2r$, and one circle of radius $r$ (in two halves), so they shoul... |
2,698,555 | <p><strong>Question</strong></p>
<p>Find all the rational values of $x$ at which $y=\sqrt{x^2+x+3}$</p>
<p><strong>My attempt</strong></p>
<p>Since we only have to find the rational values of $x$ and $y$, we can assume that
$$ x \in Q$$
$$ y \in Q$$
$$ y-x \in Q $$
Let$$ d = y-x$$
$$d=\sqrt{x^2+x+3}-x$$
$$d+x=\sqrt... | Community | -1 | <p>I assume the sides of the square are $2R$, and that the blue area is the area of the intersection of two half circles of radius $R$</p>
<p>$RedArea = Sa_{Big_{quartercircle}} - 2 * Sa_{Small_{halfcircles}} + intersection $</p>
<p>But </p>
<p>$Sa_{Big_{quartercircle}} = \pi R^2$</p>
<p>and:</p>
<p>$Sa_{Small_{ha... |
1,639,521 | <p>Let <span class="math-container">$f:[0,\infty)\to\mathbb{R}$</span> differentiable and suppose that <span class="math-container">$$\lim_{x\to\infty}f'(x)=L.$$</span> How can I prove that <span class="math-container">$$\lim_{x\to\infty}\frac{f(x)}{x} = L\;?$$</span></p>
<p>I have solved some similar problems using t... | Robert Israel | 8,508 | <p>Hint: If for $x > N > 0$, $ f'(x) < L+\epsilon$, then for such $x$,
$f(x) < f(N) + (x-N) (L + \epsilon)$ and
$$\dfrac{f(x)}{x} < L + \epsilon + \dfrac{f(N)- N(L+\epsilon)}{x} $$
What's the limit of the right side as $x \to \infty$?</p>
<p>Similarly in the other direction...</p>
|
1,639,521 | <p>Let <span class="math-container">$f:[0,\infty)\to\mathbb{R}$</span> differentiable and suppose that <span class="math-container">$$\lim_{x\to\infty}f'(x)=L.$$</span> How can I prove that <span class="math-container">$$\lim_{x\to\infty}\frac{f(x)}{x} = L\;?$$</span></p>
<p>I have solved some similar problems using t... | Eugene Zhang | 215,082 | <p>Since <span class="math-container">$\lim_{x\to\infty}f'(x)=L$</span>, for any <span class="math-container">$\epsilon>0$</span>, there is <span class="math-container">$M>0$</span> such that for any <span class="math-container">$x>M$</span>, there is
<span class="math-container">$$
|f'(x)-L|<\epsilon
$$</s... |
35,375 | <p>Good morning,
today I have read that "number theory is nothing but the study of $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$", here <a href="http://www.math.uconn.edu/~alozano/elliptic/finding%20points.pdf" rel="nofollow">http://www.math.uconn.edu/~alozano/elliptic/finding%20points.pdf</a>
can anyone give a very naiv... | Bill Dubuque | 242 | <p>Perhaps the most accessible introduction for a "very naive layperson" is Ash and Gross's <a href="http://press.princeton.edu/titles/8141.html" rel="noreferrer">Fearless Symmetry.</a> I recall reading many glowing reviews from non-experts, so this may be precisely the exposition that you seek. See also <a ... |
3,480,857 | <p>For <span class="math-container">$x \in \mathbb{R}^n$</span> we define <span class="math-container">$\Vert x \Vert _\infty := \sup_{k = 1,..,n} |x_k|$</span> (meaning that <span class="math-container">$\Vert x \Vert _\infty $</span> is the biggest component of <span class="math-container">$x$</span> according to amo... | Prabath Hewasundara | 243,089 | <p>Let <span class="math-container">$\Vert x \Vert_\infty=l=\sup_\limits{i\in{1,2,...n}}\vert x_i \vert$</span></p>
<p>Note that <span class="math-container">$0 \leq l, \Vert x \Vert _2$</span>.</p>
<p>Then <span class="math-container">$l^2 \leq \sum \limits_{i=1}^n \vert x_i \vert^2 \implies \Vert x \Vert_\infty=l\l... |
1,955,505 | <blockquote>
<p>$\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$ Given hint: $(n^2+3n+2) = (n+2)(n+1)$</p>
</blockquote>
<p><strong>I've tried</strong> converting the series to a geometric one but failed with that approach and don't know other methods for normal series that help determine the actual conver... | Dr. Sonnhard Graubner | 175,066 | <p>prove with induction that $$\sum_{n=0}^k \frac{n^2+3n+2}{4^n}=\frac{1}{27} 4^{-k} \left(-9 k^2-51 k+2^{2 k+7}-74\right)$$</p>
|
1,955,505 | <blockquote>
<p>$\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$ Given hint: $(n^2+3n+2) = (n+2)(n+1)$</p>
</blockquote>
<p><strong>I've tried</strong> converting the series to a geometric one but failed with that approach and don't know other methods for normal series that help determine the actual conver... | Olivier Oloa | 118,798 | <p>One may start with the standard <strong>finite</strong> evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by differentiating once more one gets
$$
... |
1,955,505 | <blockquote>
<p>$\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$ Given hint: $(n^2+3n+2) = (n+2)(n+1)$</p>
</blockquote>
<p><strong>I've tried</strong> converting the series to a geometric one but failed with that approach and don't know other methods for normal series that help determine the actual conver... | E.H.E | 187,799 | <p>Hint:
$${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\quad {\text{ for }}|x|<1\!$$
differentiate it and multiply by $x$</p>
<p>$${\frac {x}{(1-x)^{2}}}=\sum _{n=0}^{\infty }nx^{n}$$
differentiate it again and multiply by $x$
$${\frac {x(x+1)}{(1-x)^{3}}}=\sum _{n=0}^{\infty }n^2x^{n}$$</p>
|
499,496 | <p>We have the diff. eq:</p>
<p>$$ y' = \dfrac{2y}{t}$$</p>
<p>I tried to do the following:</p>
<p>$$ y' = \dfrac{1}{t} \cdot 3y$$</p>
<p>$$3y = \dfrac{1}{t} \cdot y'$$</p>
<p>$$ \dfrac{3}{2} y^2 = \ln(t)$$</p>
<p>Here I stopped because I noticed the answer had to be $c \cdot t^3$. I don't understand how to get t... | the_candyman | 51,370 | <p>You have to use the separability of the variables.
$$y' = \frac{dy}{dt} = \frac{2y}{t}$$
We can use $dy$ and $dt$ as "normal" number, and we must move all "$y$"s on a side, and all "$t$"s on the other.
$$\frac{dy}{2y} = \frac{dt}{t}$$
Then, we integrate!
$$\int_{y(t_0)}^{y(t)}\frac{1}{2y} dy = \int_{t_0}^{t}\frac{1}... |
499,496 | <p>We have the diff. eq:</p>
<p>$$ y' = \dfrac{2y}{t}$$</p>
<p>I tried to do the following:</p>
<p>$$ y' = \dfrac{1}{t} \cdot 3y$$</p>
<p>$$3y = \dfrac{1}{t} \cdot y'$$</p>
<p>$$ \dfrac{3}{2} y^2 = \ln(t)$$</p>
<p>Here I stopped because I noticed the answer had to be $c \cdot t^3$. I don't understand how to get t... | Avitus | 80,800 | <p>You can write the original O.D.E (I assume $y=y(t)$ is a $C^1$ function) as follows</p>
<p>$$\frac{y'}{y}=\frac{2}{t};$$</p>
<p>integrating both sides you arrive at</p>
<p>$$\ln y=2\ln(t)+C, $$</p>
<p>with $C$ constant. Then,</p>
<p>$$y(t)=C't^2,$$</p>
<p>as $2\ln t=\ln t^2$ and $C'=e^C$.</p>
|
593,746 | <p>In general, if a random process is ergodic, does it imply that it is also stationary in any sense?</p>
| user127022 | 127,022 | <p>In the theory of Dynamical Systems, one usually only defines ergodicity for invariant measures, so the short answer could be 'yes'.</p>
<p>But, consider the following theorem (taken from Norris' Markov Chains):</p>
<p><strong>Theorem 1.10.2</strong>: let $P=(p_{ij}:\,i,j\in I)$ be an irreducible, positive recurren... |
3,161,874 | <p>Find the vector equation of the plane which contains points <span class="math-container">$A(0,1,1)$</span>, <span class="math-container">$B(-1,2,1)$</span> and <span class="math-container">$C(2,0,2)$</span>. </p>
<p>I did this by first finding <span class="math-container">$AB$</span> and <span class="math-container... | inavda | 134,481 | <p>Arthur's answer is very good, but here is another (equally valid) way of visualizing the problem:</p>
<p>Let us graph the positions of the wall, fly, and car over time. The wall doesn't move, so it is represented by a horizontal line. The car starts at some distance away from the wall but moves towards it at a cons... |
273,580 | <p>I notice that the built-in "Coordinates Tool" has a very efficient zoom tooltip displaying an enlarged portion of the image along with coordinates and row/column indices:</p>
<p><a href="https://i.stack.imgur.com/KteX5.png" rel="noreferrer"><img src="https://i.stack.imgur.com/KteX5.png" alt="screenshot" />... | Alexey Popkov | 280 | <p>Answering the first part of my question (thanks to <a href="https://mathematica.stackexchange.com/users/36508/lukas-lang">Lukas Lang</a> for pointing me in the right direction!).</p>
<p>The source code for the <a href="http://reference.wolfram.com/language/workflow/GetCoordinatesFromAnImage.html" rel="nofollow noref... |
507,867 | <p>I saw an inequality for $n\times n$ matrices. I was wondering if the inequality is true or not?</p>
<p>Does $\det(A)>0$ imply $\det(I+A)>0$?</p>
| Asaf Karagila | 622 | <p>Consider $-I$ in $M_n(\Bbb R)$ for an even $n$.</p>
|
2,713,555 | <p>I have a question stating</p>
<blockquote>
<p>Let $A= \{ x_1 , x_2 , x_3 , \ldots , x_n \}$ be a set consisting of $n$ distinct elements. How many subsets does it have? How many proper subsets?</p>
</blockquote>
<p>My thought is that there would be subsets with $1$ element, $2$ elements, $3$ element and so on, u... | fleablood | 280,126 | <p>Why do you think there will be $n-(k-1)$ subsets of size $k$?</p>
<p>Let's say you have a set of $25$ elements, $\{1,2,3,....., 25\}$ and you want is see who many subset of size $2$ there are.</p>
<p>You can have $\{1,2\}, \{1,3\}......, \{2, 25\}$. That's $24 = n- 1$. But those are all with the element $1$. Wh... |
2,713,555 | <p>I have a question stating</p>
<blockquote>
<p>Let $A= \{ x_1 , x_2 , x_3 , \ldots , x_n \}$ be a set consisting of $n$ distinct elements. How many subsets does it have? How many proper subsets?</p>
</blockquote>
<p>My thought is that there would be subsets with $1$ element, $2$ elements, $3$ element and so on, u... | Arian | 172,588 | <p>There are
$${n\choose 0},{n\choose 1},...,{n\choose n-1},{n\choose n}$$
subsets from $\{x_1,...,x_n\}$ with no element, one element,..., $(n-1)$ elements, and $n$ elements respectively. Therefore the total number of subsets is
$${n\choose 0}+{n\choose 1}+...+{n\choose n-1}+{n\choose n}=(1+1)^n=2^n$$</p>
|
1,031,304 | <p>Let $\varphi$ be a bounded, differentiable function on $\mathbb{R}$ such that $\varphi'$ is bounded and uniformly continuous on $\mathbb{R}$.</p>
<p>We want to prove that $\displaystyle\frac{\varphi(x+h)-\varphi(x)}{h}\to\varphi'(x)$ uniformly as $h\to 0$</p>
<p>I can prove $\varphi$ is uniformly continuous, but I... | mookid | 131,738 | <p><strong>Hint:</strong> use the mean value theorem.
$$
\phi(x+h) - \phi(x) - h\phi'(x) = h (\phi'(x + \theta_x h) - \phi'(x))
$$
for some $\theta_x\in(0,1)$.</p>
|
203,212 | <p>Q1: When a question asking me to show a set consists of countably many point, do it mean infinite countably many or both infinite or finite case. </p>
<p>Q2:Also, to show that a set consist of countably many point, is it enough to show that the set is countable or i have to show somethings more?</p>
| Belgi | 21,335 | <p><strong>For 1:</strong> the common use of the word countable includes both finite and
countably infinite (the definition is that $A$ is countable if $|A|\leq\aleph_{0}$)</p>
<p><strong>For 2:</strong> yes, it is sufficient as you show this by definition </p>
|
11,882 | <p>Prove that $\lim\limits_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ using an epsilon delta proof.</p>
<p>So I have most of the work done. I choose $\delta = min{\frac{1}{2}, y}$,<br>
$f(x)$ factors out to $\frac{|x-3||x-2|}{|x+1|}$
But $|x-3| \lt \frac{3}{2}$ for $\delta = \frac{1}{2}$ and also $|x+1| > 5/2$ (I'll spar... | Zarrax | 3,035 | <p>I'm going to go out on a limb and guess that you're trying to show the limit is 3 and that $f(x) = {x^2 - 2x + 9 \over x + 1} - 3$. I suggest trying to translate what you've done into the fact that $|{x^2 - 2x + 9 \over x + 1} - 3| < {3 \over 5}|x - 2|$ whenever $|x - 2| < {1 \over 2}$. </p>
<p>This means th... |
4,065,868 | <p>Can anyone here recommend a low-dimensional topology textbook that contains knot theory and 3,4-manifolds?Or should I look for these subjects in separate textbooks?</p>
| Harambe | 357,206 | <p>My book preferences are different from the other answer! My preferences are the following:</p>
<p>Knot theory: Lickorish "An introduction to knot theory". You don't need much background apart from singular homology/fundamental groups.</p>
<p>3-manifolds: Saveliev "Lectures on the topology of 3-manifol... |
4,065,868 | <p>Can anyone here recommend a low-dimensional topology textbook that contains knot theory and 3,4-manifolds?Or should I look for these subjects in separate textbooks?</p>
| Strongly Negative Amphicheiral | 881,176 | <p>In addition to what others have mentioned, I would like to add Hatcher's "Notes on Basic 3-Manifold Topology" which don't cover too much material but are in my opinion the best introduction to 3-Manifolds around.</p>
|
1,691,685 | <p>The answer for it is $$3 + \sum_{k=1}^n (3+k(k-1)2^{k-2})\frac{(-1)^k}{k!} x^k + o(x^n)$$
Well, I've tried to change every $e^x$ to $1 + x + \frac{x}{2!} + ... + o(x^n)$ and got nothing useful. I know how to get Maclaurin series for polynomials but I don't really know what should I do with fractions and how get that... | OKPALA MMADUABUCHI | 98,218 | <p>First Write the function as $3e^{-x}+x^2e^x$. Then add the Maclaurin expansions.</p>
|
4,431,764 | <p>I was solving one problem in which I had to find the value of <span class="math-container">$x$</span> and at the last step my result came out to be <span class="math-container">$x^{\log_b{a}}=2$</span> but I was not able to get to the correct answer after this. Here's what I did:</p>
<p><span class="math-container">... | Koro | 266,435 | <p>Take <span class="math-container">$\rm\log$</span> w.r.t. <span class="math-container">$\rm \, base\,2$</span> so that <span class="math-container">$$\rm\log_ba \log_2x=1\implies \log_2x=\frac 1{\log_ba}=\log_ab\implies x=2^{\log_ba}$$</span></p>
<p>The first implication assumes <span class="math-container">$\rm a\n... |
38,597 | <p>Or equivalently, if $G$ is a group, do the projective and injective dimension of $Z$ (viewed as a $ZG$-module) agree?</p>
<p>Thanks! </p>
| David White | 11,540 | <p>So, Simon Wadsley’s comment clearly answers this question but a hypothetical future user will not have his/her eye drawn to that answer. That’s why I’m posting this, which is all Simon’s idea, in the hope that the OP will come back on at some point and accept this answer (thus preventing this problem from being bump... |
292,159 | <blockquote>
<p>Find the relation between <span class="math-container">$a, b, c, d$</span> if the roots of <span class="math-container">$ax^3+bx^2+cx+d=0$</span> are in geometric progression.</p>
<p>By considering <span class="math-container">$(\alpha+\beta)(\beta+\gamma)(\alpha+\gamma)$</span> show that the above cubi... | lab bhattacharjee | 33,337 | <p>If $\alpha, \beta, \gamma$ are the roots of $ax^3+bx^2+cx+d=0$</p>
<p>Using <a href="http://mathworld.wolfram.com/VietasFormulas.html" rel="nofollow">Vieta's Formulas</a>, $$\alpha+\beta+\gamma=-\frac ba, \alpha\beta+\beta\gamma+\gamma\alpha=\frac ca,\alpha\beta\gamma=-\frac da$$</p>
<p>If $\alpha,\beta,\gamma$ a... |
3,672,948 | <p>Consider the series
<span class="math-container">$$
S = \sum_{n=1}^\infty e^{-n^2x}
$$</span>
then I have to argue for that <span class="math-container">$S$</span> is convergent if and only if <span class="math-container">$x>0$</span>.</p>
<p>As this is if and only if I think I have to assume first that S is con... | mathemagician99 | 784,628 | <p>If S is convergent then <span class="math-container">$e^{-n^2x}\to 0$</span>, which it only does if <span class="math-container">$x>0$</span>.
If <span class="math-container">$x\le 0$</span> then <span class="math-container">$-n^2x\ge 0$</span> and hence <span class="math-container">$\sum_{n\in\mathbb{N}} e^{-n^2... |
997,116 | <p>If one root of the equation $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ is common and $a \ne b$ then:</p>
<p>The options are as follows:
$$\begin{array}{ll}
(A)\quad& a + b = 0\\
(B)& a + b = -1\\
(C)& a - b = 1\\
(D)& a + b = 1
\end{array}$$</p>
<p>Idk how to solve this, please help me.</p>
| bankrip | 96,097 | <p>$x^2+ax+b=x^2+bx+a$ => $(a-b)(x-1)=0$. Since $a \neq b$, $x=1$. Thus $1^2+a+b=0$. Hence, $a+b=-1$. </p>
<p>Thus, C is the answer. </p>
|
276,725 | <p>Is there a positive number $n$ of distinct odd integers $z_1,z_2, \ldots, z_n \geq 3$ such that $\frac{1}{z_1} + \frac{1}{z_2} + \cdots + \frac{1}{z_n} = 1$?</p>
| Calvin Lin | 54,563 | <p>In 1954, it was shown by <a href="http://en.wikipedia.org/wiki/Egyptian_fraction#Open_problems">Stewart and Breusch</a> (independently) that if $\frac {p}{q} >0$ and $q$ is odd, then it can be written as the sum of finitely many reciprocals of odd numbers.</p>
<p>As a specific example,</p>
<p>$$1=\frac {1}{3} +... |
18,691 | <p>Let's say I have a folowing set of data:</p>
<ul>
<li>k = 1 : list of values </li>
<li>k = 3 : list of values </li>
<li>k = 10 : list of values</li>
</ul>
<p>I know that to make a <code>BoxWhiskerChart</code> I have to give it a list of listen of these values as data and ks as labels. </p>
<p>How do I force the o... | kglr | 125 | <p>Using a custom <code>ChartElementFunction</code> that translates the graphics primitives based on specified <code>x</code> values passed as metadata:</p>
<pre><code>ClearAll[ceF]
ceF[func_: "GlassBoxWhisker"][delta_] :=
ChartElementData[func][{#3[[1]] + {-delta, delta}, #[[2]]}, ##2] &
salaries = ExampleData... |
1,166,127 | <p>I want to select an even rows of matrix. How can I show it using mathematical notations.</p>
<p>Something like this:</p>
<p>$X_{i:}$ where <code>i</code> is even.</p>
| philipp | 136,005 | <p>even indices:</p>
<p>$X_{2i}$</p>
<p>uneven indices can be expressed as</p>
<p>$X_{2i+1}$</p>
<p>Of course this changes the range for $i$.</p>
|
42,854 | <p>How much money would you have if the amount of money you started with was 5 and it increased by 5 a day for 365 days.
So January 1st you receive 5, Jan 2nd you receive 10, the third 15.. etc.
I'm wondering what the formula is</p>
| Eric Naslund | 6,075 | <p><strong>Hint:</strong>
On the 365th day, you will receive $365*5=1825$ dollars. Then the total amount of money is </p>
<p>$$5+10+15+\cdots+1820+1825.$$ Consider double this amount. That is consider $$\begin{array}{ccccccc}
5 & +10 & +15 & +\cdots & +1815 & +1820 & +1825\\
1825 & +18... |
1,970,103 | <p>I'm trying to show that for any countable ordinal $\alpha$, there is a subset of $(\mathbb{R},<)$ that has order type $\alpha$. In this post I'm not asking for a solution, but instead for a proof-check. If it's wrong I'll go back to the drawing board.</p>
<p>I tried using induction: assume that every $\beta<\... | DanielWainfleet | 254,665 | <p>The proof is incorrect. I think you meant to say that $\cup_kS_k$ has order-type $\alpha$ where $S_k$ is the image of $\beta_k$ under $f_k.$ But it's still wrong. For example if $\alpha=\omega + \omega=\omega \cdot 2$ then for some $k',k'',k'''$ we have $\beta_k'=\omega,\; \beta_k''=\omega +1, \beta_k'''=\omega + ... |
1,970,103 | <p>I'm trying to show that for any countable ordinal $\alpha$, there is a subset of $(\mathbb{R},<)$ that has order type $\alpha$. In this post I'm not asking for a solution, but instead for a proof-check. If it's wrong I'll go back to the drawing board.</p>
<p>I tried using induction: assume that every $\beta<\... | Mirko | 188,367 | <p>Here is an alternative way of showing that every countable ordinal is order-isomorphic to a subset of the real line (or a subset of <span class="math-container">$[0,1]$</span>, below).</p>
<p>For finite ordinals this is clear (just take a finite subset of <span class="math-container">$[0,1]$</span> with the same num... |
300,154 | <p>I am considering an optimization problem of the form:
\begin{equation}
\begin{split}
f(s) &= \min_{X} \mathrm{tr}(C(s)X) \\
&\;\;\;\;\;\;\;\;\;\;\; X \ge 0, \\
&\;\;\;\;\;\;\;\;\;\;\; \mathrm{tr}(A_iX) = a_i, \;\; 1 \le i \le M,
\end{split}
\end{equation}
where the minimization is over $n\times n$ Her... | Mark L. Stone | 75,420 | <p>Yes, this is convex because the objective function and all constraints are convex.</p>
<p>The objective function is affine (linear), which is convex. The semidefinjite constraint on X is convex. The trace equality constraint on X is affine (linear), and therefore is convex.</p>
|
2,936,994 | <p>There are 7 white balls in a row and a fair die (a cube with numbers 1, 2,..., 6 in its six faces). We roll the die 7 times and paint the ith balls into black if we get either 5 or 6 in the ith roll.</p>
<p>(a) What is the expected number of black balls?</p>
<p>(b) What is the chance that there exist 6 or more con... | lulu | 252,071 | <p>Here is a recursive method:</p>
<p>Let <span class="math-container">$P(n)$</span> denote the answer for a string of length <span class="math-container">$n$</span> . We remark that <span class="math-container">$$n≤3\implies P(n)=1 \quad \&\quad P(4)=1-\left(\frac 23\right)^4$$</span></p>
<p>For <span class="mat... |
839,516 | <p>What is the number of binary sequences of length $n$, with no two consecutive zeros, and if starts with $0$ has to end with $1$.</p>
<p>Would appreciate suggestions and help.</p>
<p>I tried counting the total sequences and than substractung the ones containing 2 consecutive zeros, and then substracting the ones st... | André Nicolas | 6,312 | <p>Call a sequence of length $n$ <em>good</em> if it has no two consecutive $0$ (no other restriction).</p>
<p>Let $g_n$ be the number of good sequences of length $n$.</p>
<p>Now look at our more restricted collection, with the added condition that if we start with $0$ we must end with $1$. There are two types of suc... |
1,973,724 | <p>Now this was the explanation on how to solve a prove in my book : </p>
<p>We will prove that the two sets complement (A∩B) and complement(A) ∪ complement(B) are equal by showing that each is a subset of the other. </p>
<p>First,we will show that complement (A∩B) ⊆ complement(A) ∪ complement(B). We do this by show... | user2825632 | 250,232 | <p>If you are asking for an intuitive explanation of why $(A \cap B)^c = (A^c \cup B^c)$, consider a set $X$ containing $A$ and $B$, so that $A^c = X - A$ and $B^c = X - B$. Then we see that any element in $X$ lies in exactly one of the following sets (they are all mutually exclusive):</p>
<p>$$A \cap B$$</p>
<p>$$A^... |
98,317 | <p>This comes from Artin Second Edition, page 219. Artin defined $G = \langle x,y\mid x^3, y^3, yxyxy\rangle$, and uses the Todd-Coxeter Algorithm to show that the subgroup $H = \langle y\rangle$ has index 1, and therefore $G = H$ is the cyclic group of order 3.</p>
<p>That being the case, $x$ cannot be either $y$ or ... | Andrés E. Caicedo | 462 | <p>Let's see. We have $yxyxy=1$, so (multiplying by $y$ on the left) $y=y^2xyxy$, so (cancelling $y$ on the right) $y^2xy=x^{-1}$.</p>
<p>Also, $yxyxy=1$, so $yxyxy^2=y$, or $xyxy^2=1$. So $yxy^2=x^{-1}$. </p>
<p>It follows that $y^2xy=yxy^2$, or $yx=xy$. (So the group is Abelian.)</p>
<p>But then $1=yxyxy=x^2y^3=x^... |
573,484 | <p>Prove that if $A \setminus B = \emptyset$, then $A \subseteq B$.</p>
<p>The Venn Diagram helped me to visualize what I'm trying to show (thanks @GA316), but the book asks for a written proof (step by step) by contradiction. Sorry if I wasn't more specific at first, is just that I've had many troubles in the past w... | MarnixKlooster ReinstateMonica | 11,994 | <p>Here is a complete proof in a calculational style: we start at the most complex side, and then just expand the definitions and simplify, and see where that leads us.
\begin{align}
& A \setminus B = \emptyset \\
\equiv & \qquad \text{"basic property of $\;\emptyset\;$"} \\
& \langle \forall x :: x \not\in... |
573,484 | <p>Prove that if $A \setminus B = \emptyset$, then $A \subseteq B$.</p>
<p>The Venn Diagram helped me to visualize what I'm trying to show (thanks @GA316), but the book asks for a written proof (step by step) by contradiction. Sorry if I wasn't more specific at first, is just that I've had many troubles in the past w... | Joshua P. Swanson | 86,777 | <p>It seems to me you could benefit from writing a proof in multiple stages. Here's an example of my own thought processes on this problem.</p>
<ol>
<li><p>Translate the math symbols into English (or Spanish, as the case may be). I literally read "If $A \setminus B = \emptyset$, then $A \subseteq B$" as "If A with the... |
573,484 | <p>Prove that if $A \setminus B = \emptyset$, then $A \subseteq B$.</p>
<p>The Venn Diagram helped me to visualize what I'm trying to show (thanks @GA316), but the book asks for a written proof (step by step) by contradiction. Sorry if I wasn't more specific at first, is just that I've had many troubles in the past w... | Hicham Amarir | 900,374 | <p>We assume <span class="math-container">$A \subset E$</span>, <span class="math-container">$B \subset E$</span> and <span class="math-container">$A \setminus B = \varnothing$</span></p>
<p><span class="math-container">\begin{align*}% Utiliser & pour aligner en colonne. Enlever * pour numéroter chaque ligne et \no... |
414,400 | <p>Let $ \mathbb{Z}[i]$ denote the ring of the Gaussian intergers. For which of the following value of n is the quotient ring $ \mathbb{Z}[i]/n\mathbb{Z}[i]$ an integral domain?</p>
<p>$ a. 2$</p>
<p>$ b. 13$</p>
<p>$ c. 19$</p>
<p>$ d. 7$</p>
<p>I'm doubtful with the following attempt I made.</p>
<ul>
<li><p><st... | davidlowryduda | 9,754 | <p>You claim that $n$ prime $\implies (n)$ is a prime ideal in $\mathbb{Z}[i]$. Let's consider $(5)$.</p>
<p>Clearly, $5$ is prime. But $(5) = (2+i)(2-i)$, and thus $(5)$ is not a Gaussian prime.</p>
<p>I encourage you to trace your proof over on this example and see where it goes wrong.</p>
|
256,785 | <p>Let $T_0$ be the set theory axiomatized by $ZFC^-$ (that is $ZFC$ without powerset) + every set is countable + $\mathbb{V}=\mathbb{L}$. </p>
<p><strong>Question 1:</strong> Suppose $\phi$ is a sentence of set theory. Must there be a large cardinal axiom $A$ such that $\phi$ is decided by $T_0$ + "there are transiti... | Danielle Ulrich | 26,705 | <p>Overnight the following occurred to me...</p>
<p>The answer to Question 2 is negative (with an asterisk), and so the same is true of Question 1. Namely, let $T$ be a set of $\Pi_2$ sentence with $ZFC^- \cup T$ consistent; I claim that $ZFC^- \cup T$ does not resolve whether the class of ordinals $\alpha$ such that ... |
2,947,218 | <p>I know that if a curve <span class="math-container">$C$</span> is parametrize as <span class="math-container">$$s(t)=(x(t),y(t)),\quad t\in [a,b]$$</span>
then <span class="math-container">$$\ell(C)=\int ds=\int_a^b \sqrt{\dot x^2(t)+\dot y^2(t)}dt.\tag{*}$$</span></p>
<p>In the <a href="https://fr.wikipedia.org/w... | Ethan Bolker | 72,858 | <p>On your curve the expression
<span class="math-container">$$
dx = \dot{x}(t)dt
$$</span>
says (informally) that if you change time by the "infinitesimal" amount <span class="math-container">$dt$</span> the <span class="math-container">$x$</span> value change <span class="math-container">$dx$</span> in <span class="m... |
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just get... | Brian M. Scott | 12,042 | <p>Here’s one to get things started.</p>
<p>Let $U$ be a non-empty open subset of $\Bbb R$. For $x,y\in U$ define $x\sim y$ iff $\big[\min\{x,y\},\max\{x,y\}\big]\subseteq U$. It’s easily checked that $\sim$ is an equivalence relation on $U$ whose equivalence classes are pairwise disjoint open intervals in $\Bbb R$. (... |
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just get... | Yoni Rozenshein | 36,650 | <p>A variant of the usual proof with the equivalence relation, which trades in the ease of constructing the intervals with the ease of proving countability (not that either is hard...):</p>
<ol>
<li>Define the same equivalence relation, but only on $\mathbb Q \cap U$:
$q_1 \sim q_2$ iff $(q_1, q_2) \subset U$ (or $(q_... |
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just get... | P.. | 39,722 | <p>Let <span class="math-container">$U\subseteq\mathbb R$</span> open. It is enough to write <span class="math-container">$U$</span> as a disjoint union of open intervals.<br>
For each <span class="math-container">$x\in U$</span>, we define <span class="math-container">$\alpha_x=\inf\{\alpha\in\mathbb R:(\alpha,x+\epsi... |
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just get... | Community | -1 | <p>The proof that every open set is a disjoint union of countably many open intervals relies on three facts:</p>
<ul>
<li>$\Bbb R$ is locally-connected</li>
<li>$\Bbb R$ is ccc</li>
<li>The open connected sets in $\Bbb R$ are open intervals</li>
</ul>
<p>Let $U\subseteq \Bbb R$ be open. Then there is a collection of ... |
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just get... | Theorist Reflectionist | 211,474 | <p>This proof is an extended version of the nice proof proposed by
<strong>Stromael</strong> and it serves best for beginners who want to understand every detail(that one that for any established mathematician logically seems trivial) of the proof.</p>
<p><span class="math-container">$ \textbf{Proof:} $</span></p>
<p>L... |
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just get... | Nathan A.S. | 451,262 | <p>Essentially nothing differs here from the two previous responses which rely principally on the fact that $\mathbb{R} $ is locally connected. However I present here a proof that hopefully will feel accessible to readers with a slightly lower level of topological literary at the cost of appearing cumbersome to an expe... |
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