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 |
|---|---|---|---|---|
1,849,588 | <p>The exercise asks if exists a definition of $Nat(x)$ such that $Nat(x) \Rightarrow Nat(S(x))$, and $\exists x $such that $ Nat(x)$ is false without the use of axiom of infinity. Here $Nat(x) \Leftrightarrow x$ is a natural number, and $S(x)$ is the successor of $x$. </p>
<p>I tried to define $$S(a)=\{a\}.$$ Then I ... | edgar alonso | 329,621 | <p>No, you cannot define the set of natural numbers without infinity; but you can define the property of being a natual number:</p>
<p>Define
\begin{equation}
Or(x)\equiv\forall y\in x(y\subset x) \wedge(\forall u,v\in x)(u\in v \vee v\in u \vee v=u);
\end{equation}
"$x$ is and ordinal". Then define,
\begin{equation}
... |
206,780 | <p>Let $f:X\to Y$ is a measurable function. Banach indicatrix
$$
N(y,f) = \#\{x\in X \mid f(x) = y\}
$$
is the number of the pre-images of $y$ under $f$. If there are infinitely many pre-images then $N(y,f) = \infty$. </p>
<p>Let $X\subset\mathbb R^n$, $Y\subset\mathbb R^m$ with Lebesgue measure.</p>
<p><em>I am inte... | Gerald Edgar | 454 | <p>A counterexample.</p>
<p>Let $F : \mathbb R \to \mathbb R$ be the Cantor singular function. $F$ is continuous. Let $C \subseteq \mathbb R$ be the middle-thirds Cantor set. $C$ has Lebesgue measure zero. $F$ maps $C$ onto $Y = [0,1]$. Let $M \subseteq [0,1]$ be a non-measurable set. For simplicity, remove from... |
1,477,325 | <p>It is known that an integrable function is a.e. finite. Is an a.e. finite function integrable? What if the measure is finite?</p>
| Joe | 234,473 | <p>No. $\frac{1}{x}$ is an example.</p>
|
3,368,402 | <p>I am utilizing set identities to prove (A-C)-(B-C).</p>
<p><span class="math-container">$\begin{array}{|l}(A−B)− C = \{ x | x \in ((x\in (A \cap \bar{B})) \cap \bar{C}\} \quad \text{Def. of Set Minus}
\\
\quad \quad \quad \quad \quad =\{ x | ((x\in A) \wedge (x\in\bar{B})) \wedge (x\in\bar{C})\} \quad \text{Def. o... | richard1941 | 133,895 | <p>All of the above looks like hard math requiring actual thought. I did it by means of an excel spreadsheet. Easy, as there are only 8 possibilities. The two highlighted columns are identical.</p>
<p><a href="https://i.stack.imgur.com/dvI3R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dvI3R.... |
3,492,376 | <p>Can anyone explain to me why the below expression:</p>
<p><span class="math-container">$$\int\frac{2\cos x}{{(4-4\sin^2x})^{3/2}}\:dx$$</span></p>
<p>is equal to this:</p>
<p><span class="math-container">$$\frac{2}{8}\int\frac{\cos x}{{(1-\sin^2x})^{3/2}}\:dx$$</span></p>
<p>a) Why the constant <span class="math... | A. Goodier | 466,850 | <p>Use that <span class="math-container">$4^{3/2}=8$</span> as pointed out in the comments and also
<span class="math-container">$$(1-\sin^2x)^{3/2}=(\cos^2x)^{3/2}=\cos^3x$$</span></p>
|
3,492,376 | <p>Can anyone explain to me why the below expression:</p>
<p><span class="math-container">$$\int\frac{2\cos x}{{(4-4\sin^2x})^{3/2}}\:dx$$</span></p>
<p>is equal to this:</p>
<p><span class="math-container">$$\frac{2}{8}\int\frac{\cos x}{{(1-\sin^2x})^{3/2}}\:dx$$</span></p>
<p>a) Why the constant <span class="math... | J. W. Tanner | 615,567 | <p><span class="math-container">$$\int\frac{2\cos x}{{(4-4\sin^2x})^{(3/2)}}\:dx$$</span></p>
<p><span class="math-container">$$ \int\frac{2\cos x}{{4^{3/2}(1-\sin^2x})^{(3/2)}}\:dx$$</span></p>
<p><span class="math-container">$$=\frac{2}{8}\int\frac{\cos x}{{(1-\sin^2x})^{(3/2)}}\:dx$$</span></p>
<p><span class="ma... |
65,886 | <p>It is clear that Sylow theorems are an essential tool for the classification of finite groups.
I recently read an article by Marcel Wild, <em>The Groups of Order Sixteen Made Easy</em>, where he gives a complete classification of the groups of order $16$ that is based on
elementary facts, in particular, he does not ... | Community | -1 | <p>Here is a proof that any simple group $G$ of order $360$ is isomorphic to a specific subgroup of $A_{10}$ (and hence there can be only one [insert Highlander pun]).</p>
<p>Let $G$ be a simple group of order $360$, and let's ask how many Sylow 3-subgroups there can be. A quick check shows the possibilities are : $1... |
3,299,296 | <p>The question is </p>
<blockquote>
<p>When <span class="math-container">$~2x^3 + x^2 - 2kx + f~$</span> is divided by <span class="math-container">$~x - 1~$</span>, the remainder is
<span class="math-container">$~-4~$</span>, and when it is divided by <span class="math-container">$~x+2~$</span>, the remainder i... | evaristegd | 447,617 | <p>After you replace the value of <span class="math-container">$x$</span> with <span class="math-container">$1$</span> and then with <span class="math-container">$-2$</span>, you will get two equations. The first equation is equal to <span class="math-container">$-4$</span> , by Bézout's theorem. Analogously, the secon... |
898,495 | <p>A standard pack of 52 cards with 4 suits (each having 13 denominations) is well shuffled and dealt out to 4 players (N, S, E and W).</p>
<p>They each receive 13 cards.</p>
<p>If N and S have exactly 10 cards of a specified suit between them. </p>
<p>What is the probability that the 3 remaining cards of the suit a... | drhab | 75,923 | <p>It can be thought of as drawing $3$ distinct numbers out of the set $\{1,\dots,26\}$ and to find the probability that the numbers all three belong to set $\{1,\dots,13\}$ or to set $\{14,\dots,26\}$. </p>
<p>This results in $2\times\frac{13}{26}\times\frac{12}{25}\times\frac{11}{24}=\frac{11}{50}$</p>
<p>Actually ... |
1,868,440 | <p>In a game , there are <code>N</code> numbers and <code>2</code> player(<code>A</code> and <code>B</code>) . If <code>A</code> and <code>B</code> pick a number and replace it with one of it's divisors other than itself alternatively, how would I conclude who would make the last move? (Notice that eventually when the ... | Francesco Alem. | 175,276 | <p>Conclusion comes when the player is faced with a list of 1's and a prime number somewhere, the player is forced to replace the prime number with 1. Ending the game (since a prime number can only be divided by 1 and itself)</p>
<p>EDIT.</p>
<p>unless the game requires some criterion that the player must respect for... |
1,218,238 | <p>Describe explicitly a subgroup $H$ of order 8 of the permutation group $S_5$.</p>
<p>How could I find such a subgroup? I don't know how to start with. Should I start with some transition $(i,j)$ and use them to generate a subgroup?</p>
| Pedro | 37,702 | <p>The statement is true for $n \geq 1$, otherwise the statement is false.</p>
<p>We are in a subspace when if we add two vectors, we still are in the subspace and if we multiply the vector by any scalar multiple (including $0$), we are still in the subspace. Because we can multiply always by $0$, the null vector alwa... |
1,209,934 | <p>So I am given two points $A=(-.5,2.3,-7.3)$ and $B=(-2,17.1,-0.3)$ and then using $AB = OB - OA$ to give me $(1.5,-14.8,-7)$. The plane is $$x+23y+13z=500$$ From there I got $r.n$ where $r=(1.5,-14.8,-7)$ and $n=(1,28,13)$. From here I do not know how to check if the vector is perpendicular to the plane.</p>
| Harish Chandra Rajpoot | 210,295 | <p>In general two vectors: <span class="math-container">$(a_1, a_2, a_3)$</span> and <span class="math-container">$(b_1, b_2, b_3)$</span> are said to be parallel if
<span class="math-container">$$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}$$</span></p>
<p>Now, the vector <span class="math-container">$\vec{OB}=(-1.... |
2,725,455 | <p>Probably this is pretty simple (or even trivial), but I'm stucked.</p>
<p>If $H\leq G$ is a subgroup, does it follow that $hH=Hh$, if $h\in H$ ? I can't prove or find a counter-example. If anyone could help me, I'd be grateful!</p>
| Joaquin Liniado | 237,607 | <p>If $H\leq G$ is a subgroup, then for any $h,h' \in H$ you have that $hh'\in H$. </p>
<p>Therefore $hH=H=Hh$. To see this, let's see the double inclusion:</p>
<p>An element of $hH$ is of the form $hg$ with $g\in H$. Since $H$ is a subgroup, then $hg \in H$. </p>
<p>The other way around, take $h' \in H$ you want to... |
1,196,317 | <p><a href="https://math.stackexchange.com/questions/1196261/let-g-be-a-group-where-ab3-a3b3-and-ab5-a5b5-prove-that-g-is/1196295#1196295">Let $G$ be a group, where $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$. How to prove that $G$ is an abelian group?</a></p>
<p>P.S Why cannot not we just cancel ab out of the middle of these... | grand_chat | 215,011 | <p>The middle cancellation property does not hold for every group. See <a href="https://math.stackexchange.com/a/1196399/215011">this example</a>.</p>
<p>In fact a group possesses the middle cancellation property if and only if it is abelian:</p>
<p>$(\Rightarrow)$ If the middle cancellation property holds, then for ... |
39,597 | <p>There was a recent question on intuitions about sheaf cohomology, and I answered in part by suggesting the "genetic" approach (how did cohomology in general arise?). For historical material specific to sheaf cohomology, what Houzel writes in the Kashiwara-Schapira book <em>Sheaves on Manifolds</em> for sheaf theory ... | roy smith | 9,449 | <p>As explained to us by Alan Mayer, sheaf cohomology is a generalization of cech cohmology. I found this very helpful.</p>
<p>As to the question of how ordinary cohomology arose, Hermann Weyl implies in the revised version of his book Concept of a Riemann surface, that it is a generalization of the Weierstrass, Hens... |
646,032 | <p>I'm wondering where the notation for the quotient of a ring by an ideal comes from. I.e., why do we write $R/I$ to denote a ring structure on the set $\{r+I: r\in I\}$, wouldn't $R+I$ be more natural?</p>
| Sempliner | 122,727 | <p>We do so because in general what we are doing is arranging the object $R$ into equivalence classes (in such a way that the set of equivalence classes has a structure analogous to that of $R$), in a manner very similar to what happens when one takes one integer modulo another (in fact this can be reconceptualized as ... |
2,569,267 | <p><a href="https://gowers.wordpress.com/2011/10/16/permutations/" rel="nofollow noreferrer">This</a> article claims:</p>
<blockquote>
<p>we simply replace the number 1 by 2, the number 2 by 4, and the number 4 by 1</p>
<p>....I start with the numbers arranged as follows: 1 2 3 4 5 6. After doing the permutation (124) ... | hmakholm left over Monica | 14,366 | <p>Perhaps a better example would be:</p>
<p>If you apply the permutation $(1\,2\,4)$ to (each element of) the sequence "6, 5, 3, 1, 2, 4" you get "6, 5, 3, 2, 4, 1".</p>
<p>The 6, 5, and 3 are unchanged by the permutation, 1 becomes 2, 2 becomes 4, and 4 becomes 1.</p>
|
499,652 | <p>I saw this a lot in physics textbook but today I am curious about it and want to know if anyone can show me a formal mathematical proof of this statement? Thanks!</p>
| Glen O | 67,842 | <p>The formal proof would involve demonstrating that, for any value $\epsilon<1$, one can find a value $\omega>0$ such that</p>
<p>$$
\forall \alpha\in(-\omega,\omega), |\tan(\alpha)-\alpha|<\epsilon |\alpha|
$$</p>
<p>That is, for all values of $\alpha$ between $-\omega$ and $\omega$, the difference between... |
298,913 | <p>Suppose you have $n$ triangles whose corners are random points on a sphere $S$
in $\mathbb{R}^3$.
Viewing the triangles as built from rigid bars as edges,
two triangles are <em>linked</em> if they cannot be separated without two
edges passing through one another.
A triangle that is not topologically linked with any... | Gerhard Paseman | 3,402 | <p>Here is an idea which suggests why the probability is near zero as the number of triangles gets large.</p>
<p>Among the many random triangles, there is one which has the smallest area as measured when embedded in the plane. Let us call this triangle T, and consider how likely that another triangle links with it.</p... |
2,943,973 | <p>I'm trying to prove there is some <span class="math-container">$N$</span> such that for all <span class="math-container">$n > N$</span>, it is the case that <span class="math-container">$$2n^{3/4} + 2(n-\sqrt{n})^{3/2} + n - 2n^{3/2} \leq 0$$</span></p>
<p>I know that this is true, since I graphed this function ... | mathlove | 78,967 | <p>Let <span class="math-container">$m:=n^{1/4}$</span>.</p>
<p>The inequality is equivalent to
<span class="math-container">$$2m^3(m^2-1)^{3/2} \leq 2m^6-m^4-2m^3$$</span></p>
<p>Dividing the both sides by <span class="math-container">$m^3$</span> gives
<span class="math-container">$$2(m^2-1)^{3/2} \leq m(2m^2-1)-... |
2,943,973 | <p>I'm trying to prove there is some <span class="math-container">$N$</span> such that for all <span class="math-container">$n > N$</span>, it is the case that <span class="math-container">$$2n^{3/4} + 2(n-\sqrt{n})^{3/2} + n - 2n^{3/2} \leq 0$$</span></p>
<p>I know that this is true, since I graphed this function ... | Claude Leibovici | 82,404 | <p>Consider the Taylor series of the expression for large values of <span class="math-container">$n$</span>. You should get
<span class="math-container">$$2n^{3/4} + 2(n-\sqrt{n})^{3/2} + n - 2n^{3/2}=-2 n+2 n^{3/4}+\frac{3 }{4}n^{1/2}+\frac{1}{8}+O\left(\frac{1}{n^{1/2}}\right)$$</span> that is to say
<span class="mat... |
1,759,836 | <p>It is well-known that on a smooth manifold $M$, the Lie derivative commutes with the exterior derivative, i.e.
$${\cal L}_Xd\alpha=d{\cal L}_X\alpha$$
for any vector field $X$ and differential form $\alpha$.</p>
<p>If $M$ is a complex manifold, is there a similar result for the partial derivative
$${\cal L}_X\parti... | Michael Albanese | 39,599 | <p>This is not true as stated.</p>
<p>Suppose $\alpha$ is a $d$-closed $(p, q)$-form, then $\partial\alpha = 0$, so $\mathcal{L}_X\partial\alpha = 0$. On the other hand, </p>
<p>$$\partial\mathcal{L}_X\alpha = \partial(di_X + i_Xd)\alpha = \partial di_X\alpha = \partial(\partial + \bar{\partial})i_X\alpha = \partial\... |
1,759,836 | <p>It is well-known that on a smooth manifold $M$, the Lie derivative commutes with the exterior derivative, i.e.
$${\cal L}_Xd\alpha=d{\cal L}_X\alpha$$
for any vector field $X$ and differential form $\alpha$.</p>
<p>If $M$ is a complex manifold, is there a similar result for the partial derivative
$${\cal L}_X\parti... | jws | 71,874 | <p>It's true if the complex structure is invariant under the vector field X. In that case: $\mathcal{L}_X \partial a = \mathcal{L}_X (1 - \imath J) da = (1 - \imath J)\mathcal{L}_X da = \partial \mathcal{L}_X a$. (Note: This is written with $a$ being a scalar. For a $(p,q)$ form you just need to pick a suitable project... |
1,423,449 | <p>Find all extrema for the function $f(x)=-\frac{x^{3}}{3}+x^{2}-x+4$ on the domain $x \in [-3.3]$.</p>
<p><strong>Solution:</strong> $f'(x)=-x^{2}+2x-1 = 0 \implies (x-1)^{2}=0 \implies x^{*}=1$. </p>
<p>Is that it? </p>
| Emilio Novati | 187,568 | <p>Hint:</p>
<p>note that $f'(x)=-(x-1)^2 \le 0 \quad \forall x \in \mathbb{R}$, so the function is motononic decreasing and $x=1$ can not be an extremum.</p>
<p>You have to find the values of the function in $x=-3$ and $x=3$ to find the extrema in the given interval.</p>
|
2,842,217 | <p>im looking to understand the tangent taylor series, but im struggling to understand how to use long division to divide one series (sine) into the other (cosine). I also can't find examples of the Tangent series much beyond X^5 (wikipedia and youtube videos both stop at the second or third term), which is not enough ... | Travis Willse | 155,629 | <p>You might find it conceptually easier to set up the identity of power series and compare the first few coefficients, and solve. This is algebraically equivalent to long division, though the order of some of the arithmetic operations is somewhat rearranged.</p>
<p>Write the desired Taylor series at $x = 0$ as
$$\tan... |
1,392,340 | <p>Suppose there is a function $f:\mathbb R^n \to \mathbb R$. One way to find a stationary value is to solve the ODE $\dot x = - \nabla f(x)$, and look at $\lim_{t\to\infty} x(t)$.</p>
<p>However I want to consider a variation of this method where we solve
$$ dx = - \nabla f(x) dt + C(t,x) \cdot dW_t ,$$
where $C(x,t... | Sergio Almada | 172,877 | <p>Freidlin & Wentzell and its community is interested in a set of topics a little bit different from yours (metastability and exit problems). Your kind of cases have been studied, see for example <a href="http://repository.ias.ac.in/1132/1/323.pdf" rel="nofollow">http://repository.ias.ac.in/1132/1/323.pdf</a> and ... |
3,115,347 | <p>Let <span class="math-container">$f:(0,\infty) \to \mathbb R$</span> be a differentiable function and <span class="math-container">$F$</span> on of its primitives. Prove that if <span class="math-container">$f$</span> is bounded and <span class="math-container">$\lim_{x \to \infty}F(x)=0$</span>, then <span class="m... | Jonas De Schouwer | 581,053 | <p><span class="math-container">$a)$</span>
The probability that one dice rolls a number equal to <span class="math-container">$3$</span> or lower is <span class="math-container">$\frac{3}{6}=\frac{1}{2}$</span>.
Hence, the probability that all of them roll <span class="math-container">$3$</span> or lower is <span clas... |
1,064,115 | <p><strong>UPDATE:</strong> Thanks to those who replied saying I have to calculate the probabilities explicitly. Could someone clarify if this is the form I should end up with:</p>
<p>$G_X$($x$) = P(X=0) + P(X=1)($x$) + P(X=2) ($x^2$) + P(X=3)($x^3$)</p>
<p>Then I find the first and second derivative in order to calc... | heropup | 118,193 | <p>You don't need to use the formula for a hypergeometric distribution. Simply observe that the most number of balls you can draw before obtaining the first red ball is $3$, so the support of $X$ is $X \in \{0, 1, 2, 3\}$. This is small enough to very easily compute explicitly $\Pr[X = k]$ for $k = 0, 1, 2, 3$.</p>
|
684,076 | <blockquote>
<p>In the past, practical applications have motivated the development of mathematical theories, which then became the subject of study in pure mathematics, where mathematics is developed primarily for its own sake. Thus, the activity of <a href="http://en.wikipedia.org/wiki/Applied_mathematics" rel="nofo... | baffld | 128,321 | <p>I think it is important to note that NP-complete problems have solutions to them. An algorithm exists to solve any NP-complete problem. All NP-complete problems are NP. If one were to implement an algorithm solving an NP problem (that is not in P), then the implementation will have at least exponential running time ... |
3,354,566 | <p>I see integrals defined as anti-derivatives but for some reason I haven't come across the reverse. Both seem equally implied by the fundamental theorem of calculus.</p>
<p>This emerged as a sticking point in <a href="https://math.stackexchange.com/questions/3354502/are-integrals-thought-of-as-antiderivatives-to-avo... | Stella Biderman | 123,230 | <p>From the point of view of analysis (as hinted at in Henning Makholm's answer) the issue is that the mapping <span class="math-container">$I:f'\to f$</span> is <strong>extremely</strong> not one-to-one. When you try to invert it, you find that a great deal of functions are possible "anti-integrals" of a given functio... |
553,845 | <p>Could we assert that if $H$ is a subgroup of $G$, then the factor group $N_G(H)/C_G(H)$ is isomorphic to a subgroup of ${\rm Inn}(H)$ instead of ${\rm Aut}(H)$?</p>
| anon | 11,763 | <p>No, it is not necessary for (the image of) $N_G(H)/C_G(H)$ to be inner in ${\rm Aut}(H)$. </p>
<p>This means that "external" conjugation can defy imitation by "internal" conjugation.</p>
<p>In fact, every element of ${\rm Aut}(H)$ can be realized as conjugation by an element in some overgroup containing $H$. In pa... |
2,935,743 | <p>Given two <strong>independent</strong> random variables X, Y, the expectation of their product XY is:</p>
<p><span class="math-container">$\mathrm{E}[XY] = \mathrm{E}[X]\cdot\mathrm{E}[Y]$</span></p>
<p>Similarly, the variance of the product of these variables is:</p>
<p><span class="math-container">$\mathrm{Var}... | StubbornAtom | 321,264 | <p>The first formula for independent random variables <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>, sometimes called the product law of expectation, is not hard to find. It can be found in most undergrad probability and statistics textbooks. For instance, you can find it <a href="... |
2,935,743 | <p>Given two <strong>independent</strong> random variables X, Y, the expectation of their product XY is:</p>
<p><span class="math-container">$\mathrm{E}[XY] = \mathrm{E}[X]\cdot\mathrm{E}[Y]$</span></p>
<p>Similarly, the variance of the product of these variables is:</p>
<p><span class="math-container">$\mathrm{Var}... | Teemu Sarapisto | 1,108,398 | <p><a href="https://www.tandfonline.com/doi/abs/10.1080/01621459.1960.10483369" rel="nofollow noreferrer">On the Exact Variance of Products</a> by Leo A. Goodman, published in
Journal of the American Statistical Association, Volume 55, 1960 - Issue 292</p>
<p>is a peer-reviewed article focusing on this problem with the... |
1,708,900 | <p>Does a closed form exist for </p>
<blockquote>
<p>$$\sum \limits_{n=0}^{\infty} \frac{1}{(kn)!}$$</p>
</blockquote>
<p>in terms of $k$ and other functions? The best that I have been able to do is solve the case where $k=1$, since the sum is just the infinite series for $e$. I would guess that any closed form mus... | David C. Ullrich | 248,223 | <p>If $(c_n)$ is any sequence with period $k$ (that is, $c_{n+k}=c_n$) then it's possible to evaluate $\sum c_n/n!$ using tricks involving $k$-th roots of unity.</p>
<p>Let $\omega=e^{2\pi i/k}$. Consider the $k$ sequences</p>
<p>$s_0:1,1,1\dots$</p>
<p>$s_1: 1, \omega,\omega^2,\omega^3,\dots$</p>
<p>$s_2: 1, \omeg... |
2,799,439 | <blockquote>
<p>Prove that if $p$ is a prime in $\Bbb Z$ that can be written in the form $a^2+b^2$ then $a+bi$ is irreducible in $\Bbb Z[i]$ .</p>
</blockquote>
<p>Let $a+bi=(c+di)(e+fi)\implies a-bi=(c-di)(e-fi)\implies a^2+b^2=(c^2+d^2)(e^2+f^2)\implies p|(c^2+d^2)(e^2+f^2)\implies p|c^2+d^2 $ or $p|e^2+f^2$
since... | Tsemo Aristide | 280,301 | <p>Hint: Use the norm $N(a+ib)=a^2+b^2$, $N(zz')=N(z)N(z')$, so if $zz'=a+ib$, $N(z)N(z')=p$ implies that $N(z)=1$ or $N(z')=1$, you deduce that $z$ is a unit or $z'$ is a unit.</p>
|
38,731 | <p>The <a href="http://en.wikipedia.org/wiki/Ramanujan_summation">Ramanujan Summation</a> of some infinite sums is consistent with a couple of sets of values of the Riemann zeta function. We have, for instance, $$\zeta(-2n)=\sum_{n=1}^{\infty} n^{2k} = 0 (\mathfrak{R}) $$ (for non-negative integer $k$) and $$\zeta(-(2n... | Sumit Kumar Jha | 37,260 | <p>Ramanujan summation arises out of <a href="http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula">Euler-Maclaurin summation formula</a>. Ramanujan summation is just (C, 1) summation. (See <a href="http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation">Cesàro summation</a>)</p>
<p>You can find out easily from Eu... |
3,485,441 | <p>I don't quite understand why Burnside's lemma
<span class="math-container">$$
|X/G|=\frac1{|G|}\sum_{g\in G} |X_g|
$$</span>
should be called a "lemma". By "lemma", we should mean there is something coming after it, presumably a theorem. However, I could not find a theorem which requires Burnside as a lemma. In ever... | Community | -1 | <p>An closely related fact, by looking at the <em>class equation</em>, is that a nontrivial <span class="math-container">$p$</span>-group has nontrivial center. </p>
|
200,777 | <p>I have a question regarding sums in arrays.</p>
<p>So I have the following array:</p>
<pre><code>list=RandomReal[{0,1},{5,2}]
(*{{0.693551,0.447185},{0.274842,0.637526},{0.745271,0.0288363},{0.894933,0.937219},{0.605447,0.0337067}}*)
</code></pre>
<p>And from that I want to have the splitting for each pair like ... | Armani42 | 66,191 | <p>Thanks, but now I have another problem, now I have</p>
<pre><code>ener[r2_] = 4* (1/r2^12 - 1/r2^6);
</code></pre>
<p>And I want to make now</p>
<pre><code>dist = DistanceMatrix[list, DistanceFunction->(#1-#2 &)];
</code></pre>
<p>Which produces me 2x2 vectors in a matrix.
But now I want to do that</p>
<... |
982,021 | <p>$$ f(x) = (1+x)^\frac35, a= (1.2)^\frac35 $$</p>
<p>I got the linear approximation equation of $$1+ \frac35 x$$
What do I do with the value of a? </p>
| David | 119,775 | <p><strong>Hint</strong>: $f(x)=a$ when $x=\cdots\,$.</p>
<p>Find $f(x)$ approximately by using this $x$ value in the linear approximation.</p>
|
982,021 | <p>$$ f(x) = (1+x)^\frac35, a= (1.2)^\frac35 $$</p>
<p>I got the linear approximation equation of $$1+ \frac35 x$$
What do I do with the value of a? </p>
| Paul | 17,980 | <p>Since $$ f(x) = (1+x)^\frac35 \approx 1+\frac35x$$, then $f(a)=f(1.2)=f(1+0.2)\approx1+\frac35\times 0.2=1.12$</p>
|
1,987,507 | <p>I find this question, which comes from section 2.2 of Dummit and Foote's algebra text, to be somewhat confusing:</p>
<blockquote>
<p>Let $G = S_n$, fix $i \in \{1,...,n\}$ and let $G_i = \{\sigma \in G ~|~ \sigma(i) = i\}$ (the stabilizer of $i$ in $G$). Use group actions to prove that $G_i$ is a subgroup of $G$.... | msm | 350,875 | <p>Your mistake (in your own calculation) is that you assume all tickets are sold and the winner is the last person who buys the last ticket. This is not what always happens. <em>Any</em> of the tickets can win (including the first one!) and it is likely that some of them are not sold at all.</p>
|
114,895 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/21282/show-that-every-n-can-be-written-uniquely-in-the-form-n-ab-with-a-squa">Show that every $n$ can be written uniquely in the form $n = ab$, with $a$ square-free and $b$ a perfect square</a> </p>
</blockq... | Bill Dubuque | 242 | <p><strong>Hint</strong> $\ $ The problem is <em>multiplicative</em>, thus it suffices to show that it is true for a prime power $\rm\ P^N\:.\ $ But that's trivial: $\rm\ P^{2N} =\ (P^N)^2,\ \ P^{\:2N+1} =\ P\ (P^N)^2\:,\ $ uniquely. $\ $ <strong>QED</strong></p>
<p>Alternatively, examining the power of each prime in ... |
4,581,539 | <p>Consider the task of proving that <span class="math-container">$|z+w|\leq |z|+|w|$</span>, where <span class="math-container">$z$</span> and <span class="math-container">$w$</span> are complex numbers.</p>
<p>We can consider three cases:</p>
<ol>
<li><span class="math-container">$|z|$</span> or <span class="math-con... | Ryszard Szwarc | 715,896 | <p>The function can be calculated explicitly. Namely let <span class="math-container">$$f(t)=\begin{cases} \phantom{-} 1 & \phantom{-} 0<t < \pi \\
-1 & -\pi <t<0
\end{cases}$$</span>
Then the Fourier coefficient are <span class="math-container">$a_0=0$</span> and
<span class="math-container">$$a_n... |
201,122 | <p>A little bit of <em>motivation</em> (the question starts below the line): I am studying a proper, generically finite map of varieties $X \to Y$, with $X$ and $Y$ smooth. Since the map is proper, we can use the Stein factorization $X \to \hat{X} \to Y$. Since the composition is generically finite, $X \to \hat{X}$ is ... | Karl Schwede | 3,521 | <p>For what it's worth, one can say the following sort of thing.</p>
<p>Since $Y$ is log terminal so is $(\hat{X}, -\mathrm{Ram})$. This doesn't mean much since in the pair, the boundary has a negative coefficients (ie, the singularities of $\hat{X}$ can be arbitrarily bad). But it does say things like:</p>
<p><em>... |
2,424,508 | <p>One textbook exercise asks to prove $$|a|+|b|+|c|-|a+b|-|a+c|-|b+c|+|a+b+c| \geq 0.$$</p>
<p>The textbook's solution is:</p>
<blockquote>
<p>If $a$, $b$ or $c$ is zero, the equality follows. Then, we can assume
$|a| \geq |b| \geq |c| > 0$. </p>
<p>Dividing by $|a|$, the inequality is equivalent
to</p... | Marios Gretsas | 359,315 | <p>To answer the first question the set $\{0,2\}$ is compact in the discrete topology and thus from Tychonov's theorem the set $X=\{0,2\}^{\mathbb{N}}$ is compact with respect to the product topology.</p>
<p>So $X$ cannot be a discrete topological space because a discrete topological space $(X,\mathcal{T})$ is comp... |
2,544,864 | <p>I have been trying to prove the continuity of the function:
$f:\mathbb{R}\to \mathbb{R}, f(x) =x \sin(x) $ using the $\epsilon -\delta$ method. </p>
<p>The particular objective of posting this question is to understand <strong>the dependence of $\delta$ on $\epsilon$ and $x$</strong>. I know that $f(x) =x \sin(x) $... | user284331 | 284,331 | <p>So you want to further show that it is not uniform? Assume it were, then there exists some $\delta>0$ such that for every $x,y$ with $|x-y|<\delta$, we have $|x\sin x-y\sin y|<1$. Now take $x_{n}=2n\pi+\eta$, $y_{n}=2n\pi$, $n=1,2,...$ where $\eta>0$ is so small that $\eta<\min\{\delta,\pi/2\}$, then ... |
2,544,864 | <p>I have been trying to prove the continuity of the function:
$f:\mathbb{R}\to \mathbb{R}, f(x) =x \sin(x) $ using the $\epsilon -\delta$ method. </p>
<p>The particular objective of posting this question is to understand <strong>the dependence of $\delta$ on $\epsilon$ and $x$</strong>. I know that $f(x) =x \sin(x) $... | Community | -1 | <p>Hint: $|x\sin x-y\sin y|=|x\sin x-x \sin y+x\sin y-y\sin y|\le 2|x||\cos\frac {x+y}{2}||\sin\frac {x-y}{2}|+|x-y||\sin y|\le |x||x-y|+|x-y|=(|x|+1)|x-y|$ so use $\delta=\frac {\varepsilon}{|x|+1} $. I have used the facts that, for every $x $, $|\sin x|, |\cos x|\le 1$, $|\sin x|\le|x|$.</p>
|
52,657 | <p>I have a pair of points at my disposal. One of these points represents the parabola's maximum y-value, which always occurs at x=0. I also have a point which represents the parabola's x-intercept(s). Given this information, is there a way to rapidly derive the formula for this parabolic curve? My issue is that I ... | Eric Naslund | 6,075 | <p>I'll assume you meant you know the $x$-intercepts and maximum height.</p>
<p>If you have any parabola with $x$-intercepts $a,b$, $a\neq b$, and maximum height $c$, then you can write it as $$y=k(x-a)(x-b)$$ where $$k=-c\left(\frac{4}{(a-b)^2}\right).$$</p>
<p>(Notice that the value $c$ must be positive)</p>
<p>If... |
4,444,669 | <p>I'm unsure about the problem below</p>
<hr>
Under which conditions is the following linear equation system solvable ?
<span class="math-container">$$x_1 + 2x_2 - 3x_3 = a$$</span>
<span class="math-container">$$3x_1 - x_2 + 2x_3 = b$$</span>
<span class="math-container">$$x_1 - 5x_2 + 8x_3 = c$$</span>
<hr>
<p>We se... | Greg Nisbet | 128,599 | <p>If you're allowed to use the ceiling, then you can use the following.</p>
<p>Let <span class="math-container">$x$</span> be a positive real.</p>
<p>Suppose <span class="math-container">$x > 1$</span>, choose <span class="math-container">$n = 1$</span>.</p>
<p>Suppose <span class="math-container">$x = 1$</span>, c... |
172,058 | <p>I'm wondering whether there is certain relationship between the largest eigenvalue of a positive matrix(every element is positive, not neccesarily positive definite) $A$, $\rho(A)$ and that of $A∘A^T$, $\rho( A∘A^T)$, where $∘$ denotes hadamard product.</p>
<p>Here's a result I find for many numerical cases. I crea... | Suvrit | 8,430 | <p>The following information may be useful, though probably you already know it.</p>
<ol>
<li>$\rho(A\circ A^T) \le \rho(A)\rho(A^T)=\rho^2(A)$</li>
<li>The other direction of course fails easily; though it is interesting to note that $\rho(A) \le \rho( (A+A^T)/2)$</li>
<li>Section 5.7 of <em>Topics in matrix analysis... |
8,052 | <p>I wonder how you teachers walk the line between justifying mathematics because of
its many—and sometimes surprising—applications, and justifying it as the study
of one of the great intellectual and creative achievements of humankind?</p>
<p>I have quoted to my students G.H. Hardy's famous line,</p>
<bl... | Benoît Kloeckner | 187 | <p>My answer is mostly about higher education, and mostly suited for undergrads (which make the largest part of my students).</p>
<h3>The main purpose of their studies is to become more intelligent.</h3>
<p>This is what I say my undergrad student a lot. It applies to math as well as to any other fields; as was mentio... |
8,052 | <p>I wonder how you teachers walk the line between justifying mathematics because of
its many—and sometimes surprising—applications, and justifying it as the study
of one of the great intellectual and creative achievements of humankind?</p>
<p>I have quoted to my students G.H. Hardy's famous line,</p>
<bl... | jhocking | 5,160 | <p>I wish I could find an excerpt to link you to, but the book <a href="http://rads.stackoverflow.com/amzn/click/1400064287" rel="nofollow">Made to Stick</a> had a great reason as one of its examples. You learn math in order to make you better at thinking. It's like athletes lifting weights; they don't lift barbells be... |
8,052 | <p>I wonder how you teachers walk the line between justifying mathematics because of
its many—and sometimes surprising—applications, and justifying it as the study
of one of the great intellectual and creative achievements of humankind?</p>
<p>I have quoted to my students G.H. Hardy's famous line,</p>
<bl... | dtldarek | 42 | <p>Some reasons (random order):</p>
<ul>
<li><p><em>Math is fun and beatiful.</em> Not all marvel at the beauty of mathematics and not all enjoy working with math. In fact, daily mathematical work can be tedious, boring, uninspiring even for those in love with it. But every now and then you discover something great, an... |
716,122 | <p>Using the following deductive system D1:</p>
<pre><code>(A1) A → (B → A)
(A2) (A → B) → ((A → (B → C)) → (A → C))
(A3) (A → B) → ((A → ¬B) → ¬A)
(A4) ¬¬A → A
(MP) A A → B
---------
B
</code></pre>
<p><strong>Premise: p → q . Conclusion: ¬q → ¬p</strong></p>
<p><strong>My attempt:</strong> </p>
<ol>
... | Doug Spoonwood | 11,300 | <p>I use <a href="http://en.wikipedia.org/wiki/Polish_notation" rel="nofollow">Polish notation</a>.</p>
<p>Your axioms are</p>
<p>A1 CaCba</p>
<p>A2 CCabCCaCbcCac</p>
<p>A3 CCabCCaNbNa</p>
<p>A4 CNNaa</p>
<p>Note that {A1, A2} give you a deduction metatheorem. Thus, if we make a hypothesis h and we can then deri... |
716,122 | <p>Using the following deductive system D1:</p>
<pre><code>(A1) A → (B → A)
(A2) (A → B) → ((A → (B → C)) → (A → C))
(A3) (A → B) → ((A → ¬B) → ¬A)
(A4) ¬¬A → A
(MP) A A → B
---------
B
</code></pre>
<p><strong>Premise: p → q . Conclusion: ¬q → ¬p</strong></p>
<p><strong>My attempt:</strong> </p>
<ol>
... | Mauro ALLEGRANZA | 108,274 | <p>Your <em>axiom system</em> is that of Elliott Mendelson, <em>Introduction to Mathematical Logic</em> (4th ed - 1997). page 35.</p>
<p>With <em>axioms</em> (A1) and (A2) - as said by Doug - you may prove <em>Deduction Theorem</em> [see Mendelson, page 37 for a proof].</p>
<p>We need an "intermediate result" (we cal... |
2,412,959 | <p>In <a href="https://math.stackexchange.com/questions/170362/pointwise-convergence-implies-lp-convergence">this</a> question a user asks if pointwise convergence implies convergence in $L^p$. I would have thought that the answer is yes. I am not experienced with measure theory, which is how that question is framed. T... | Angina Seng | 436,618 | <p>For $\Omega=\Bbb R$ and Lebesgue measure I like this example.
$$f_n(x)=e^{-(x-n)^2}.$$
Then $f_n\to0$ pointwise, but $$\|f_n\|_p=\|f_1\|_p>0.$$</p>
|
1,014,987 | <p>I need to solve the bound for $n$ from this inequality: </p>
<p>$$c \leq 1.618^{n+1} -(-0.618)^{n+1},$$</p>
<p>where $c$ is some known constant value. How can I solve this? At first I was going to take the logarithm, but the difference of the two exponentials trouble me...</p>
<p>Any hints? :) Thnx for any help !... | LinAlgMan | 49,785 | <p><strong>Hints:</strong></p>
<p>First note that $$\phi = \frac{1 + \sqrt{5}}{2} \approx 1.618$$ and $$-\frac1\phi = \frac{1 - \sqrt{5}}{2} \approx -0.618$$ both are the roots of $x^2 - x - 1 = 0$.</p>
<p>Note that
$$ \lim_{n \to \infty}(-0.618)^{n+1} = 0 $$
as $|-0.618| < 1$.</p>
<p>Now note the $n$ element in ... |
4,537,489 | <p>Assume that <span class="math-container">$a>0$</span>, Suppose we have :<br />
<span class="math-container">$$X = \{x\in \mathbb{R} \ : \ x^2 < a \}$$</span><br />
We should prove that this set has a supremum, and that's <strong><span class="math-container">$\sqrt{a}$</span></strong> .<br />
I saw <a href="htt... | José Carlos Santos | 446,262 | <p><a href="https://math.stackexchange.com/questions/4080221/a-closed-subset-of-a-lindel%c3%b6f-space-is-lindel%c3%b6f">Every closed subspace of a Lindelöf space is also a Lindelöf space</a>. But an uncountable discrete space is not Lindlöf.</p>
|
3,712,094 | <p><a href="https://i.stack.imgur.com/S3n1g.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S3n1g.jpg" alt="enter image description here"></a></p>
<p>For part (a) these are clearly two parallel lines so no points of intersection.<br>
For part (b) this has one point of intersection because these two ... | hdighfan | 796,243 | <p><span class="math-container">$z=0$</span> and <span class="math-container">$x=2y+1$</span> gives the equation of a line; <span class="math-container">$x=2y+1$</span> is clearly a line in the <span class="math-container">$xy$</span>-plane, and <span class="math-container">$z=0$</span> forces us to stay in this plane.... |
639,449 | <p>I've seen on Wikipedia that for a complex matrix $X$, $\det(e^X)=e^{\operatorname{tr}(X)}$.</p>
<p>It is clearly true for a diagonal matrix. What about other matrices ?</p>
<p>The series-based definition of exp is useless here.</p>
| not all wrong | 37,268 | <p>A alternative to doing this by normal forms which perhaps assumes more but is much more natural to me is (as suggested in the comment on <a href="https://math.stackexchange.com/questions/299528/det-exp-x-e-mathrmtr-x-for-2-dimensional-matrices?rq=1">$\det(\exp X)=e^{\mathrm{Tr}\, X}$ for 2 dimensional matrices</a>) ... |
258,205 | <p>I want to know if $\displaystyle{\int_{0}^{+\infty}\frac{e^{-x} - e^{-2x}}{x}dx}$ is finite, or in the other words, if the function $\displaystyle{\frac{e^{-x} - e^{-2x}}{x}}$ is integrable in the neighborhood of zero.</p>
| copper.hat | 27,978 | <p>Let $f(x) = \frac{e^{-x} - e^{-2x}}{x}$.</p>
<p>L'Hopital gives $\lim_{x \to 0} f(x)= 1$. Hence in some neighborhood $B(0,\epsilon)$ , $|f(x)| <2$. For $x\geq \epsilon$, we have $\frac{1}{x} \leq \frac{1}{\epsilon}$, and the function $x \mapsto e^{-x} - e^{-2x}$ is clearly integrable.</p>
<p>Hence $\int_0^\inft... |
258,205 | <p>I want to know if $\displaystyle{\int_{0}^{+\infty}\frac{e^{-x} - e^{-2x}}{x}dx}$ is finite, or in the other words, if the function $\displaystyle{\frac{e^{-x} - e^{-2x}}{x}}$ is integrable in the neighborhood of zero.</p>
| Douglas B. Staple | 65,886 | <p>Claim: $$\int_0^{\infty} \frac{e^{-u}-e^{-2u}}{u} du = \ln(2).$$</p>
<p>Proof: Let
\begin{align}
C &\equiv \int_0^{\infty} \frac{e^{-u}-e^{-2u}}{u} du\\\ \\
&=\lim_{x=0}\left[ \operatorname{Ei}(1,x) - \operatorname{Ei}(1,2x)\right],
\end{align}
where
$$
\operatorname{Ei}(1,x) \equiv \int_x^\infty \frac{e^{-... |
2,937,990 | <p>I need to prove or disprove that in any Boolean algebra: if <span class="math-container">$a+ab=b$</span> then <span class="math-container">$a=b=1$</span> or <span class="math-container">$a=b=0$</span>.</p>
<p>I build the following truth table:
<span class="math-container">$$
\begin{array}{|c|c|c|}
\hline
a & b ... | Carl Schildkraut | 253,966 | <p>Your <em>reasoning</em> is correct, but the claim that <span class="math-container">$d_n$</span> is decreasing and <span class="math-container">$\lim_{n\to\infty} d_n = 0$</span> needs to be proven, not just stated. (In fact, it is not actually true, although it may <em>look</em> true.)</p>
|
2,937,990 | <p>I need to prove or disprove that in any Boolean algebra: if <span class="math-container">$a+ab=b$</span> then <span class="math-container">$a=b=1$</span> or <span class="math-container">$a=b=0$</span>.</p>
<p>I build the following truth table:
<span class="math-container">$$
\begin{array}{|c|c|c|}
\hline
a & b ... | Jack D'Aurizio | 44,121 | <p>If <span class="math-container">$a_n=\frac{(-1)^n}{\sqrt{n+1}}$</span> then <span class="math-container">$f(x)=\sum_{n\geq 0}a_n x^n$</span> has a singularity of the <span class="math-container">$\frac{1}{\sqrt{x+1}}$</span> kind at <span class="math-container">$x=-1$</span> and <span class="math-container">$g(x)=\s... |
221,428 | <p>Is there any pair of random variables (X,Y) such that Expected value of X goes to infinity, Expected value of Y goes to minus infinity but expected value of X+Y goes again to infinity?</p>
| user642796 | 8,348 | <p>Your approach won't work, since your $x$ might not belong to $G_i$, but then it would be a limit point of $E_i$, and so $x \in \overline{E_i} \setminus G_i$. </p>
<hr>
<p>In order to prove this, note the following facts.</p>
<ul>
<li>If $G$ is a dense open set, and $U$ is any nonempty open set, then $G \cap U$ is... |
141,655 | <blockquote>
<p>What is the chance that at least two people were born on the same day
of the week if there are 3 people in the room?</p>
</blockquote>
<p>I'm wondering if my solution is accurate, as my answer was different than the solution I found:</p>
<p>Probability that there are at least 2 people in the room ... | Dennis Gulko | 6,948 | <p>Another possible approach: name the guys $A,B,C$. Look at the ordered triple $d_1,d_2,d_3$ of days on which $A,B,C$ were born respectively. You have $7^3$ such triples. Now, count the "good" triples. i.e triples in which at least two guys have birthday on the same day:<br>
1) exactly two people have birthday on the ... |
4,019,119 | <p>im struggeling to find <span class="math-container">$$\lim _{x\to 0}\left(2-e^{\arcsin^{2}\left(\sqrt{x}\right)}\right)^{\frac{3}{x}}$$</span></p>
<p>Ive tried the following:
<span class="math-container">$$\lim_{x \to x_0} ax^{bx} = \lim_{x \to x_0} e^{ax^{bx}} = \lim_{x \to x_0} e^{bx \ln(ax)} = e^{\lim_{x \to x_... | RavenclawPrefect | 214,490 | <p>Shortly after reading Edward H's answer, I realized that there is a self-contained compactness argument, which I missed the first time around; I thought I would present it here.</p>
<p>Let <span class="math-container">$T$</span> be an infinite set of polyominoes. Say that an infinite "polyomino" <span clas... |
3,087,570 | <p>The "school identities with derivatives", like
<span class="math-container">$$
(x^2)'=2x
$$</span>
are not identities in the normal sense, since they do not admint substitutions. For example if we insert <span class="math-container">$1$</span> instead of <span class="math-container">$x$</span> into the identity abov... | H Huang | 604,218 | <p>One way to do it is to say <span class="math-container">$f’(x)$</span>=2x, and define <span class="math-container">$f(x)=x^2$</span>. This way, it should be relatively clear that x is not some placeholder variable, but rather the independent variable in an equation. Additionally, if you introduce the Chain Rule, it ... |
3,087,570 | <p>The "school identities with derivatives", like
<span class="math-container">$$
(x^2)'=2x
$$</span>
are not identities in the normal sense, since they do not admint substitutions. For example if we insert <span class="math-container">$1$</span> instead of <span class="math-container">$x$</span> into the identity abov... | Somos | 438,089 | <p>There is no one right answer to this question. It depends on what the students are willing to accept. One approach mentioned comes from the Wikipedia article on <a href="https://en.wikipedia.org/wiki/Dual_number" rel="nofollow noreferrer">Dual numbers</a>. The key idea is that
<span class="math-container">$\, f(x+\e... |
39,551 | <p>How can I use <em>Mathematica</em> to equate coefficients in a non-power-series equation?</p>
<p>For example, I would like to take an equation like the following:
$$af_x+\frac{b}{2}f_xf_y+chf_x=f_x+e^af_x+3f_xf_y+2bhf_x$$
and produce the following system:
$$a=1+e^a$$
$$\frac{b}{2}=3$$
$$c=2b$$
<strong>EDIT:</strong... | Daniel Lichtblau | 51 | <p>Also there is <code>MonomialList</code>.</p>
<pre><code>coefficientRelations[expr_, params_] := Module[
{vars},
vars = DeleteCases[Variables[expr],
vv_ /; Internal`DependsOnQ[vv, params]];
MonomialList[expr, vars] /. Thread[vars -> 1]
]
</code></pre>
<p>Your example is then as follows.</p>
<pre><c... |
61,798 | <p>Are there any generalisations of the identity $\sum\limits_{k=1}^n {k^3} = \bigg(\sum\limits_{k=1}^n k\bigg)^2$ ?</p>
<p>For example can $\sum {k^m} = \left(\sum k\right)^n$ be valid for anything other than $m=3 , n=2$ ?</p>
<p>If not, is there a deeper reason for this identity to be true only for the case $m=3 , ... | Gerben | 6,004 | <p>Not a completely rigorous answer, but you should be able to turn it into one.</p>
<p>By comparing the sums to their corresponding integrals $\int_0^n \mathrm{d}x x^m$, you can see that $$\sum k^m = \frac{1}{m+1} n^{m+1} + \mathcal{O}(n^m).$$ Also, $$(\sum k)^q = \frac{1}{2^{q}} n^{2q} + \mathcal{O}(n^{2q-1}).$$ By ... |
2,603,799 | <p>Good morning, i need help with this exercise.</p>
<blockquote>
<p>Prove all tangent plane to the cone $x^2+y^2=z^2$ goes through the origin</p>
</blockquote>
<p><strong>My work:</strong></p>
<p>Let $f:\mathbb{R}^3\rightarrow\mathbb{R}$ defined by $f(x,y,z)=x^2+y^2-z^2$</p>
<p>Then,</p>
<p>$\nabla f(x,y,z)=(2x... | user284331 | 284,331 | <p>The equation for the plane should be $2a(x-a)+2b(y-b)-2c(z-c)=2ax-2a^{2}+2by-2b^{2}-2cz+2c^{2}=0$. Now the point $(a,b,c)$ lies on the cone, so $a^{2}+b^{2}-c^{2}=0$, so simplifying the equation for the plane, we have then $2ax+2by-2cz=0$ and this equation goes through the origin since $2a\cdot 0+2b\cdot 0-2c\cdot 0... |
2,590,068 | <p>$$\epsilon^\epsilon=?$$
Where $\epsilon^2=0$, $\epsilon\notin\mathbb R$.
There is a formula for exponentiation of dual numbers, namely:
$$(a+b\epsilon)^{c+d\epsilon}=a^c+\epsilon(bca^{c-1}+da^c\ln a)$$
However, this formula breaks down in multiple places for $\epsilon^\epsilon$, yielding many undefined expressions l... | PC1 | 960,197 | <p>The problem is that you can't isolate <span class="math-container">$\epsilon^\epsilon$</span> as a standalone expression like what you're doing.</p>
<p>You need to consider the whole expression so it keeps its sense. So if we go back to your expression:
<span class="math-container">\begin{align}
(a+b\epsilon)^{c+d\e... |
55,482 | <p>I write a code that creates a compiled function, and then call that function over and over to generate a list. I run this code on a remote server via a batch job, and will run several instances of it. Sometimes when I make changes to the code, I make a mistake, and inside the compiled function is an undefined vari... | Karsten 7. | 18,476 | <p>You can add</p>
<pre><code>RuntimeOptions -> {"EvaluateSymbolically" -> False}
</code></pre>
<p>to your <code>Compile</code> function.<br>
Consult <a href="http://reference.wolfram.com/language/ref/RuntimeOptions.html" rel="nofollow">RuntimeOptions</a> for more details.</p>
|
3,853,351 | <p>Given an n-dimensional ellipsoid in <span class="math-container">$\mathbb{R}^n$</span>, is any orthogonal projection of it to a subspace also an ellipsoid? Here, an ellipsoid is defined as</p>
<p><span class="math-container">$$\Delta_{A, c}=\{x\in \Bbb R^n\,:\, x^TAx\le c\}$$</span></p>
<p>where <span class="math-co... | Arnaud | 122,865 | <p>Yes they do. You can prove it by induction on the codimension of the subspace you project to. For <span class="math-container">$x\in Vect(e_1,\ldots e_{n-1})$</span> there exists <span class="math-container">$t \in \mathbb{R}$</span> such that <span class="math-container">$x+te_n$</span> belongs to <span class="math... |
2,476,194 | <p>I am trying to prove that $I =(x^2+1,y-1)$ is a maximal ideal in $\mathbb{Q}[x,y]$, but I am having a hard time understanding what this ideal even looks like. I know that I can prove it's a maximal ideal by proving $\mathbb{Q}[x,y]/I$ is a field, but I'm also having a hard time understanding what this quotient looks... | Zach Teitler | 343,280 | <p>@luthien Your idea is correct. The maps you are thinking of are indeed isomorphisms. One way to show they are isomorphisms is to start with the map $f : \mathbb{Q}[x,y] \to \mathbb{Q}[i]$ given by $x \mapsto i$, $y \mapsto 1$. Then $x^2+1$ and $y-1$ are in the kernel of $f$.</p>
<p><em>Claim:</em> The kernel of $f$... |
318,983 | <p>$$\int_{-\infty}^{\infty} \frac{x^2}{x^6+9}dx$$ I'm a bit puzzled as how to go about solving this integral. I can see that it isn't undefined on -infinity to infinity. But I just need maybe a hint on how to go about solving the problem.</p>
| amWhy | 9,003 | <p>Hint: put $\;u = x^3$, so $\,du \;=\; 3x^2\, dx \;\implies\; x^2\, dx \;= \;\frac 13\, du$</p>
<p>This gives you $$\frac 13 \int_{-\infty}^\infty \frac {du}{u^2 + (3)^2} $$</p>
<p>Look familiar?:</p>
<p>Using one more substitution, let $\quad u =3\tan(\theta),\quad du =3\sec^2(\theta)\,d\theta, \quad \theta=\arct... |
2,929,203 | <p>Suppose we define the relation <span class="math-container">$∼$</span> by <span class="math-container">$v∼w$</span> (where <span class="math-container">$v$</span> and <span class="math-container">$w$</span> are arbitrary elements in <span class="math-container">$R^n$</span>) if there exists a matrix <span class="mat... | Yanko | 426,577 | <p>The relation works like that: The zero vector is only in relation with itself. All the other vectors are in relation with each other</p>
<p>Proof: if <span class="math-container">$v=0$</span> then <span class="math-container">$Av=0$</span> for every matrix and therefore <span class="math-container">$0\sim w$</span>... |
4,577,266 | <blockquote>
<p>Let <span class="math-container">$X_n$</span> be an infinite arithmetic sequence with positive integers term. The first term is divisible by the common difference of successive members. Suppose, the term <span class="math-container">$x_i$</span> has exactly <span class="math-container">$m>1$</span> ... | kabenyuk | 528,593 | <p>You have drawn the picture incorrectly. The vertex <span class="math-container">$w$</span> lies on the path <span class="math-container">$P$</span> and does not coincide with <span class="math-container">$x_1$</span>, so moving along <span class="math-container">$P$</span> until you meet <span class="math-container"... |
1,419,209 | <p>How do I evaluate this (find the sum)? It's been a while since I did this kind of calculus.</p>
<p>$$\sum_{i=0}^\infty \frac{i}{4^i}$$</p>
| Ángel Mario Gallegos | 67,622 | <p>For $-1<x<1$, the series $\sum_{i=0}^{\infty}x^i$ converges absolutely to $\frac{1}{1-x}$
$$\sum_{i=0}^{\infty}x^i=\frac{1}{1-x}$$
Then
\begin{align*}
\sum_{i=0}^{\infty}ix^{i-1} &= \frac{d}{dx}\left(\frac{1}{1-x}\right)\\
&=\frac{1}{(1-x)^2}\\
\sum_{i=0}^{\infty}ix^i&=\frac{x}{(1-x)^2}
\end{align*... |
3,536,061 | <p>Find the number of ways you can invite <span class="math-container">$3$</span> of your friends on <span class="math-container">$5$</span> consecutive days, exactly one friend a day, such that no friend is invited on more than two days. </p>
<p>My approach: Let <span class="math-container">$d_A,d_B$</span> and <span... | David G. Stork | 210,401 | <p>A hint:</p>
<p>The only configurations that obey your constraint are: </p>
<p>person A: 2 days</p>
<p>person B: 2 days</p>
<p>person C: 1 day</p>
<p>(We'll assign names to these different people below.)</p>
<p>Suppose you start with the "1 day" person. Then there are just two legal sequences:</p>
<p>CABA... |
3,726,772 | <p>For finite-dimensional vector space <span class="math-container">$V$</span>, there exist linear operators <span class="math-container">$A$</span> and <span class="math-container">$B$</span> on <span class="math-container">$V$</span> such that <span class="math-container">$AB=BA$</span> commutative relation holds.</p... | user1551 | 1,551 | <p><span class="math-container">$Om(nom)^3$</span>'s elegant answer really shows the key: <span class="math-container">$\deg(A)$</span> is the dimension of the subspace of all polynomials in <span class="math-container">$A$</span>. I cannot do better. However, if you want to use binomial expansion to solve the problem,... |
251,182 | <p>Is 13 a quadratic residue of 257? Note that 257 is prime.</p>
<p>I have tried doing it. My study guide says it is true. But I keep getting false. </p>
| André Nicolas | 6,312 | <p>We use somewhat heavy machinery, Quadratic Reciprocity. For typing convenience, we use the notation $(a/p)$ for the Legendre symbol. By Reciprocity,
$$(13/257)=(257/13)=(10/13)=(2/13)(5/13).$$
This is because at least one of $13$ and $257$ (indeed both) is of the shape $4k+1$. </p>
<p>Note that $(2/13)=-1$ because... |
3,514,547 | <p>The problem is as follows:</p>
<p>The figure from below shows vectors <span class="math-container">$\vec{A}$</span> and <span class="math-container">$\vec{B}$</span>. It is known that <span class="math-container">$A=B=3$</span>. Find <span class="math-container">$\vec{E}=(\vec{A}+\vec{B})\times(\vec{A}-\vec{B})$</s... | Robert Z | 299,698 | <p>Hint. By expanding the cross product we find
<span class="math-container">$$(\vec{A}+\vec{B})\times(\vec{A}-\vec{B})=\vec{A}\times\vec{A}+\vec{B}\times\vec{A}-\vec{A}\times\vec{B}-\vec{B}\times\vec{B}.$$</span>
Are you able to find each of the 4 cross-products on the right-hand side?</p>
<p>Recall the <a href="http... |
1,006,562 | <p>So I am trying to figure out the limit</p>
<p>$$\lim_{x\to 0} \tan x \csc (2x)$$</p>
<p>I am not sure what action needs to be done to solve this and would appreciate any help to solving this. </p>
| Aaron Maroja | 143,413 | <p>$$\lim_{x \to 0} \tan x \csc (2x) = \lim_{x \to 0} \frac{\sin x}{\cos x} \frac{1}{\sin 2x} = \lim_{x \to 0} \frac{\sin x}{\cos x} \frac{1}{2\sin x\cos x}$$ </p>
<p>Can you take from here?</p>
|
489,562 | <p>I am teaching a "proof techniques" class for sophomore math majors. We start out defining sets and what you can do with them (intersection, union, cartesian product, etc.). We then move on to predicate logic and simple proofs using the rules of first order logic. After that we prove simple math statements via dir... | dfeuer | 17,596 | <p>Sometimes it makes sense to teach a little bit backwards. Rather than always teaching the foundations first and then building on top of them, it sometimes pays to build a little higher-level context first, and then build foundations underneath. One way is to use a partially historical approach. That is, start by tea... |
1,711,653 | <p>Let's define:</p>
<p>$f(t) = A_1 \cos(\omega_1t) + A_2 \cos(\omega_2t) $</p>
<p>I am interested in finding an expression for the peak of this function. It is not true in general that this peak will have the value:</p>
<p>$max{f(t)} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2}$</p>
<p>To find the value of max(f), I did the f... | Community | -1 | <p>If $A_1, A_2>0$, the maximum $A_1+A_2$ occurs at $t=0$ !</p>
|
422,233 | <p>I was asked to find a minimal polynomial of $$\alpha = \frac{3\sqrt{5} - 2\sqrt{7} + \sqrt{35}}{1 - \sqrt{5} + \sqrt{7}}$$ over <strong>Q</strong>.</p>
<p>I'm not able to find it without the help of WolframAlpha, which says that the minimal polynomial of $\alpha$ is $$19x^4 - 156x^3 - 280x^2 + 2312x + 3596.$$ (True... | DanielWainfleet | 254,665 | <p>(Part answer).Start by rationalizing the denominator.Let $x=1-\sqrt 5+\sqrt 7.$ Then $$1/x=(1-\sqrt 5-\sqrt 7)/(x(1-\sqrt 5-\sqrt 7))=(1-\sqrt 5-\sqrt 7)/(19-2\sqrt 5).$$ Let $y=19-2\sqrt 5.$ Then $$1/y=(19+2\sqrt 5)/(y(19+2\sqrt 5))=(19+2\sqrt 5)/341.$$ So $$1/x=(1-\sqrt 5-\sqrt 7)(19+2\sqrt 5)/341.$$ Inserting th... |
666,217 | <p>If $a^2+b^2 \le 2$ then show that $a+b \le2$</p>
<p>I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\... |
1,748,751 | <p>By K values, I mean the values described here:</p>
<p><a href="https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods#Explicit_Runge.E2.80.93Kutta_methods" rel="nofollow">https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods#Explicit_Runge.E2.80.93Kutta_methods</a></p>
<p>I know how the K values in the Rung... | Jan Peter Schäfermeyer | 399,820 | <p>This is how Runge himself taught the method to the students of Columbia University, where he was guest professor in 1909/10: <a href="https://archive.org/stream/graphicalmethod01runggoog#page/n135/mode/2up" rel="nofollow noreferrer">https://archive.org/stream/graphicalmethod01runggoog#page/n135/mode/2up</a></p>
|
1,827,080 | <p>Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=0$ and $|f'(x)|\leq1 \forall x\in\mathbb R$. Then there exists $C$ in $\mathbb R $ such that </p>
<ol>
<li>$|f(x)|\leq C \sqrt |x|$ for all $ x$ with $|x|\geq 1$</li>
<li>$|f(x)|\leq C |x|^2$ for all $ x$ with $|x|\geq 1$</li>
<li>$f(x)=x+... | Lærne | 252,762 | <p>Compute the derivative of the right hand side of the inequality. Then if the derivative is such that you can set $C$ so that this derivative it is always superior to 1, then, since the right hand side of the of inequality always grow larger than the left hand side, and since both begins at $(0,0)$, the inequality m... |
2,713,873 | <p>We know that if a real valued function $f$ is continuous over an interval $[a,b]$ then the following integral $$\int_a^bf(x)dx$$ represents the area between horizontally the line $y=0$ and the curve of $f$, vertically between the lines $x=a$ and $x=b$. So what represent the following $$\int_{[a,b]\times [c,d]}g(x... | Redsbefall | 97,835 | <p>karimath, I may contribute.</p>
<hr>
<p>1D integral can be used to calculate area. For example, if we want to calculate an area bounded by $y=f(x)$ and $y=0$ in $x \in [a, b]$, then we can do approximate this by :</p>
<p>$$ A \approx \sum_{i=0}^{N} |f(x_{i})-0| \triangle x, \:\:\: \triangle x = \frac{b-a}{N} $$<... |
325,186 | <p>If <span class="math-container">$p$</span> is a prime then the zeta function for an algebraic curve <span class="math-container">$V$</span> over <span class="math-container">$\mathbb{F}_p$</span> is defined to be
<span class="math-container">$$\zeta_{V,p}(s) := \exp\left(\sum_{m\geq 1} \frac{N_m}{m}(p^{-s})^m\right)... | Vivek Shende | 4,707 | <p>In <a href="https://arxiv.org/abs/math/0001005" rel="nofollow noreferrer">this article of Kapranov</a>, you will find the motiv-ation of the zeta function of an algebraic curve.</p>
|
4,506 | <p>I keep producing bits of code like the following:</p>
<pre><code>stuff = Module[
{curTarget = #},
getRowsForUserAndTarget[u, #, curTarget] & /@ validUsers
] & /@ allTargets;
</code></pre>
<p>Basically, I'm iterating through all the targets and all the users. Using a For loop it would look somethin... | Rojo | 109 | <p>Is this what you want?</p>
<pre><code>l1 = {a, b, c};
l2 = {aa, bb, cc};
sth[#1, #2] & @@@ Tuples[{l1, l2}]
</code></pre>
<blockquote>
<pre><code>{sth[a, aa], sth[a, bb], sth[a, cc], sth[b, aa], sth[b, bb],
sth[b, cc], sth[c, aa], sth[c, bb], sth[c, cc]}
</code></pre>
</blockquote>
|
4,506 | <p>I keep producing bits of code like the following:</p>
<pre><code>stuff = Module[
{curTarget = #},
getRowsForUserAndTarget[u, #, curTarget] & /@ validUsers
] & /@ allTargets;
</code></pre>
<p>Basically, I'm iterating through all the targets and all the users. Using a For loop it would look somethin... | Mr.Wizard | 121 | <p>A couple of other options:</p>
<pre><code>allTargets = {"a", "b", "c", "d"};
validUsers = {1, 2, 3};
Table[getRowsForUserAndTarget[u, j, i], {i, allTargets}, {j, validUsers}]
Outer[getRowsForUserAndTarget[u, #2, #] &, allTargets, validUsers, 1]
</code></pre>
<p>Both of these methods produce nested lists sepa... |
1,517,456 | <blockquote>
<p>Rudin Chp. 5 q. 13:</p>
<p>Suppose <span class="math-container">$a$</span> and <span class="math-container">$c$</span> are real numbers, <span class="math-container">$c > 0$</span>, and <span class="math-container">$f$</span> is defined on <span class="math-container">$[-1, 1]$</span> by</p>
<p><span... | Bernard | 202,857 | <p>$\mathbf{Q}(\sqrt[4] 5)$ is a $\mathbf{Q}$-vector space with dimension $4$, while $\mathbf{Q}(\sqrt[3] 5)$ is a $\mathbf{Q}$-vector space with dimension $3$. As they're both field extensions, if the latter were contained in the former, its dimension should divide $4$.</p>
|
3,673,613 | <p>I have to find out if <span class="math-container">$\displaystyle\sum_{n=2}^{\infty}$$\dfrac{\cos(\frac{\pi n}{2}) }{\sqrt n \log(n) }$</span> is absolute convergent, conditional convergent or divergent. I think it's divergent while the value for <span class="math-container">$\cos\left(\dfrac{\pi n}{2}\right)$</span... | Ben Grossmann | 81,360 | <p>In fact, the dimensions are always equal. Because we generally have <span class="math-container">$\dim \ker(AB) = \dim\ker B + \dim(\operatorname{im}(B) \cap \ker A)$</span>, it suffices to note that the image of <span class="math-container">$(A - \psi I)$</span> contains the kernel of <span class="math-container">... |
2,523,112 | <p>Let $f\left(x\right)$ be differentiable on interval $\left(a,b\right)$ and $f'\left(x\right)>0$ on that interval. If $\underset{x\rightarrow a+}{\lim}f\left(x\right)=0$, $f\left(x\right)>0$ on that interval?</p>
<p>I think this proposition is true by my intuitive, but I wonder whether intuitive is mathematica... | Arkya | 276,417 | <p>By Taylor's theorem (upto first order), for any $x\in(a+\epsilon,b)$, $\exists \zeta\in(a+\epsilon,b)$
$$f(x)=f(a+\epsilon)+(x-a-\epsilon)f'(\zeta) $$</p>
<p>Since $f'(\zeta)>0$, and $x>a+\epsilon$, this shows that $f(x)>f(a+\epsilon)$. </p>
<p>Now take the limit $\epsilon\rightarrow0^+$ to get that $f(x)... |
4,612 | <p>I would like to make a slope field. Here is the code</p>
<pre><code>slopefield =
VectorPlot[{1, .005 * p*(10 - p) }, {t, -1.5, 20}, {p, -10, 16},
Ticks -> None, AxesLabel -> {t, p}, Axes -> True,
VectorScale -> {Tiny, Automatic, None}, VectorPoints -> 15]
</code></pre>
<p>I solved the diffe... | Dr. belisarius | 193 | <p>For example:</p>
<pre><code>VectorPlot[{1, .005*p*(10 - p)}, {t, -1.5, 20}, {p, -10, 16},
FrameTicks -> {
Join[{{0, "ZERO", {0, .1}, Red}}, Table[{i, ""}, {i, -1.5, 20, 3}]],
Table[{i, ""}, {i, -10, 16, 3}]},
AxesLabel -> {t,... |
1,954,411 | <p>Let $N>0$ be a large integer, and $n<N$, then how to simply the following sum
$$\sum\limits_{k=1}^n\frac{N-n+k}{(N-k+1)(N-k+1)(N-k)}.$$
Thank you very much, guys.</p>
<p>Actually for another similar sum $\sum\limits_{k=1}^n\frac{1}{(N-k+1)(N-k)}=\sum\limits_{k=1}^n\frac{1}{N-k}-\frac{1}{N-k+1}=\frac{1}{N-n}-\... | Felix Marin | 85,343 | <p>$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}... |
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