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 |
|---|---|---|---|---|
1,869,167 | <p>The least value of $a \in R$ for which $4ax^2 + \frac{1}{x} \ge 1 $for all $x \gt 0 $, is </p>
<p>Using AM-GM inequality
$$\frac{4ax^2 + \frac{1}{2x} + \frac{1}{2x}}{3} \ge \sqrt[3]{a}$$
$$4ax^2 + \frac{1}{x} \ge 3\sqrt[3]{a}$$</p>
<p>Now my question start from here .</p>
<p>Can I do that for least value of $a$... | mathlove | 78,967 | <blockquote>
<p>Can I do that for least value of $a$, the value of $3\sqrt[3]{a} $ must greater than minimum value of $4ax^2 + \frac{1}{x}$</p>
</blockquote>
<p>I find this strange because "minimum value of $4ax^2 + \frac{1}{x}$" is $3\sqrt[3]{a}$. So I guess that you meant that "the value of $3\sqrt[3]{a} $ must... |
1,822,362 | <p><a href="https://i.stack.imgur.com/jOOyV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jOOyV.jpg" alt="enter image description here"></a></p>
<p>I'm not sure if I understand this question and was wondering if anyone could provide any insight to an answer. The only thing I can think of adding is... | Kuifje | 273,220 | <p>To have a Linear Mixed Integer Programming, replace the second constraint by
$$
4x_1+4(y_1+3y_2+4y_3+10y_4)\ge 10\\
y_1+y_2+y_3+y_4=1\\
y_1,y_2,y_3,y_4 \in \{0,1\}
$$</p>
<p>With $y_1+y_2+y_3+y_4=1$, you force exactly one term of the sum $y_1+3y_2+4y_3+10y_4$ to be active. The active term is precisely $x_2$.</p>
|
1,907,182 | <p>I have got short question: How to draw Arc in 2 demension know only start and end point in OpenGL?</p>
<p>Actually, something already wrote, however, this algorithm does not accomplish its purpose. Here we go:</p>
<pre><code> 1. Get startXY
2. Get endXY
3. Const num_segments = numeric below than 0
4. Calc midPo... | Jean Marie | 305,862 | <p>Let $(x_s,y_s)$ and $(x_e,y_e)$ denote the coord. of start and end points. Let $(x_m,y_m)$ be those of their midpoint.</p>
<p>Choose $\varepsilon=+1$ (resp. $\varepsilon=-1$) if you want a direct (resp. inverse) half circle.</p>
<p>This half-circle can be plotted with the simple formulas:</p>
<blockquote>
<p>$$... |
1,907,182 | <p>I have got short question: How to draw Arc in 2 demension know only start and end point in OpenGL?</p>
<p>Actually, something already wrote, however, this algorithm does not accomplish its purpose. Here we go:</p>
<pre><code> 1. Get startXY
2. Get endXY
3. Const num_segments = numeric below than 0
4. Calc midPo... | Narasimham | 95,860 | <p>By arc do you mean a semi-circle? </p>
<p>If so,the given points are at either end of the diameter of a circle.</p>
<p>$$ (x-h)^2 + (y-k)^2 = R^2 $$</p>
<p>where for full circle we have: </p>
<p>$$ h = ( x_{start} + x_{end})/2,\,k = (y_{start} + y_{end})/2 ,\, 4 R^2 = ( x_{start} - x_{end}) ^2 + ( y_{start} - y_... |
192,058 | <p>A covering design $(v,k,t)$ is a family of subsets of $[v]$ each having $k$ elements such that given any subset of $[v]$ of $t$ elements it is a subset of one of the sets of the family. A problem is to find the minimum number of subsets such a family can have.</p>
<p>I am interested in the case $(v,k,2)$. It seems ... | Thomas Kalinowski | 12,674 | <p>Fort and Hedlund determined the minimum size of a $(v,3,2)$-covering design: <a href="http://projecteuclid.org/euclid.pjm/1103039697">Minimal coverings of pairs by triples, Pacific J. Math. 8(4), 709-719, 1958.</a> </p>
<p>The case $(v,4,2)$ was solved by Mills: On the covering of pairs by quadruples <a href="http:... |
3,535,725 | <p>I have started working on this <span class="math-container">$\int_0^1 \frac{x^2}{\sqrt{1+x^5}} \, dx \leq \frac{1}{3}$</span> by using the fact that <span class="math-container">$\int_0^1 \frac{x^2}{\sqrt{1+x^5}} \, dx < \int_0^1 \frac{x^2}{\sqrt{x^5}} \, dx$</span>
But that didn't work.</p>
| aud098 | 744,646 | <p>Use the fact that <span class="math-container">$\frac{x^2}{\sqrt {1+x^5}}\leq x^2$</span> on <span class="math-container">$[0,1]$</span>. </p>
<p>now just integrate both side. </p>
|
2,068,643 | <p>I'm solving a problem about a <a href="https://en.wikipedia.org/wiki/Chemical_reactor#PFR_.28Plug_Flow_Reactor.29" rel="nofollow noreferrer">plug flow reactor</a> and I have this limit to compute. Just to control my result I asked <a href="http://www.wolframalpha.com/input/?i=lim_%7BR%20%5Cto%20%2B%5Cinfty%7D%20(1-e... | Gono | 384,471 | <p>You cannot consider different expressions depending on the same variable one by one!</p>
<p>With your same reason the well known sequence $\left(1 + \frac{1}{n}\right)^n$ would converge to $1$ because the expression in the brackets tends to 1 and $1^n = 1$. But this is wrong and $$\lim_{n\to \infty} \left(1 + \frac... |
349,680 | <p>The particular solution $Y_p(t)$ of this problem is actually in the form of $Ae^tt$, but isn't it supposed to be $Ae^t$ ? Since there is no homogenous root = 0, why do we need to multiply $t$.</p>
| Amzoti | 38,839 | <p>The homogeneous solution is dictated by:</p>
<p>$m^2 - 4m +3 = 0$.</p>
<p>That gives one root that equals $1$ and gives one solution $= e^t$.</p>
<p>Clear?</p>
|
174,853 | <p>In the <a href="http://en.wikipedia.org/wiki/Free_variable" rel="nofollow noreferrer">Free Variable</a> article on Wikipedia, it lists these:</p>
<p><img src="https://i.stack.imgur.com/wAfKv.png" alt="variable-binding operators"></p>
<p>as variable-binding operators. I have seen all of them during my math studies,... | c.w.chambers | 16,959 | <p>My first impulse in seeing ‘$\varphi x$’ is to read it as a variation on the logical schema (‘$\varphi(x)$’, ‘$F x$’, etc.), which, if that was the intented meaning, should, I believe, preclude it from a list of variable-binding operators. I would vote up for removal from the article if I could.</p>
|
265,728 | <p>Consider the polynomial:</p>
<p>$$p(x) = \sum_{k=0}^{r}(-1)^{r-k} {r \choose k} x^{k(k-1) / 2}$$</p>
<p>I want to show that</p>
<p>$$p(x) = (x - 1)^{\lceil r/2 \rceil} \, q(x)$$</p>
<p>That is, $(x - 1)^{\lceil r/2 \rceil}$ is a factor of $p(x)$. Even better, find a formula for the quotient polynomial $q(x)$.</p... | Qayum Khan | 16,862 | <p>The <em>countability of the number of conjugacy classes of closed subgroups of any compact Lie group</em> is, in fact, a simple corollary of ideas that predate the end of World War II. Unfortunately, the earliest quotable reference that I could find was given already in Igor Belegradek's comment, as follows.</p>
<p... |
2,060,793 | <p>I need to know wich answer is right</p>
<p><a href="https://i.stack.imgur.com/QcxdH.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/QcxdH.jpg</a></p>
<p>I tried to solve it using recursivity but I didn't get any one of them</p>
<ul>
<li>$y_1=\sqrt{x}$ </li>
<li>$y'=\frac{1}{2\sqrt{x}}=\frac{1}{2y_1}$</li... | Community | -1 | <p>As @WW1 mentioned in his comments, $$y=\sqrt {x+{\sqrt {x+\cdots}}} $$ $$\Rightarrow y=\sqrt {x+y} $$ $$\Rightarrow y^2=x+y $$ Now using the techniques of implicit differentiation, we have $$2y\frac{dy}{dx} = 1+\frac{dy}{dx} $$ Rearranging, we get, $$\Rightarrow \frac{dy}{dx} =\frac {1}{2y-1} $$ Hope it helps. </p>
|
3,479,780 | <p>I don't understand those explanations that <span class="math-container">$F(a)$</span> is a subset of <span class="math-container">$F(b)$</span>, so <span class="math-container">$F(a)$</span> is <span class="math-container">$\leq F(b)$</span>. And A event is contained in the event B, so <span class="math-container">$... | lab bhattacharjee | 33,337 | <p>If <span class="math-container">$$f(a,b)=\dfrac1{x^a(x+c)^b},$$</span></p>
<p><span class="math-container">$$f(a,b)=\dfrac1c\cdot\dfrac{x+c-x}{\cdots}=\dfrac{f(a,b-1)}c-\dfrac{f(a-1,b)}c$$</span></p>
<p>We can use this reduction formula repeatedly until at least one of <span class="math-container">$a,b$</span> bec... |
1,924,259 | <p>That is possible?, can you show me some theorem and who worked on these.</p>
<p>If we have the sum of n cubes, can we express that like the sum of m squares?</p>
<p>Thanks!</p>
| Dietrich Burde | 83,966 | <p>Note that
$$
74 = −284650292555885^3 + 66229832190556^3 + 283450105697727^3
$$
is also a sum of three squares:
$$
74=8^2+3^2+1^2.
$$</p>
<p>In general, it is known which numbers can be written as a sum of $m$ squares. In fact, for $m\ge 4$, all positive integers. The case $m=2$ is due to Fermat, and $m=3$ to Gauss.... |
459,553 | <p>How do I compute this integral?
$$\int_{\mathbb {R}}\sin \left(\frac {\pi x}{x^2+1}\right)\frac{1}{x^2+1} dx$$</p>
| Simar | 86,694 | <p>$$\left|\;\frac3{x^3-8}\;\right|=\left|\;\frac1{x-2}\;\right|\iff3|x-2|=|x-2||x^2+2x+4|$$</p>
<p>Since $x-2\ne0$ we can cancel both sides and get a quadratic $|x^2+2x+4|$ which has no solution in the real axis.</p>
|
4,548,865 | <p>I wish to determine whether the limit <span class="math-container">$L = \lim_{z \rightarrow i} \frac{z^3 + i}{|z| - 1}$</span> exists. Noticing it to be of the form <span class="math-container">$0/0$</span>, I separate the expression into its real and imaginary parts:
<span class="math-container">$$L = \lim_{(r, \th... | Anne Bauval | 386,889 | <p>I'd rather translate to complex variable <span class="math-container">$h:=z-i\to0$</span>:</p>
<p><span class="math-container">$$\frac{z^3 + i}{|z|-1}=\frac{(i+h)^3+i}{|i+h|-1}=\frac{-3h+o(h)}{\sqrt{a^2+(1+b)^2}-1}$$</span>
hence there is no limit as <span class="math-container">$(a,b)\to(0,0)$</span>. This is becau... |
1,998,810 | <p>There exists $x \in \mathbb{R}$ such that the number $f(x)=x^2 +5x +4$ is prime.</p>
<p>I can't understand where to start. </p>
<p>This is what I have so far: </p>
<p>Let P(x) be the statement "$x^2 + 5x +4$ is prime".
Then we have $\exists x \in \mathbb{R}, P(x)$.</p>
<p>I built a table and I suspect that this ... | 5xum | 112,884 | <p>$f$ is continuous, and since $f(-1)=0$ and $$\lim_{x\to\infty} f(x)=\infty,$$ you can show that for every $p>0$, there exists some $x\in\mathbb R$ such that $f(x)=p$. You don't even have to restrict yourself to primes, the function reaches <em>every positive number</em>!</p>
|
24,321 | <p>I'm trying to figure out the probability of a 3rd failure occurring on the 5th attempt of doing something. Let's just call the probability of success of failure P(S) or P(F), I won't put numbers as I want to actually learn.</p>
| curious | 223 | <p>(2) Your guess is correct, except I think it is an $(n-1)\times (n+1)$ matrix. Think of your line as the intersection of $n-1$ hyperplanes in non-degenerate position.</p>
<p>(3) Assuming you meant the <em>union</em> of two lines $L_1, L_2$, then the defining variety is given by $I_1I_2$.</p>
<p>As for the Hilbert ... |
1,198,627 | <p>I have the following approximation: </p>
<p>$$f(x) \simeq f(a) + f^{'}(a)(x - a)$$</p>
<p>Letting $a = \mu_{x}$, the mean of $X$, a Taylor seties expansion of $y=f(x)$ about $\mu_{x}$ gives the approximation:
$$y=f(x) \simeq f(\mu_{x}) +f^{'}(\mu_{x})(x - \mu_{x})$$
Taking the variance of both sides yields:</p>
... | John Hughes | 114,036 | <p>$$
Var(c + Y) = var(Y)
$$
for a constant c (like $c = f(\mu_x)$). </p>
<p>$$
Var(cY) = c^2 Var(Y)
$$
from the definition. Apply this to the second term to pull out the $(f'(\mu))^2$.</p>
<p>Then once more observe that $Var(X - c) = Var(X)$ to get rid of the $\mu_x$ in the remaining parenthesized term.</p>
<p>By t... |
638,048 | <p>I am looking at an exercise,where given that $a_{n}=\sqrt{2+\sqrt{2+...+\sqrt{2}}}$,I have to show that $\lim_{n \to \infty}a_{n}=2$.
We find that $a_{0}=0,a_{1}=\sqrt{2},a_{2}=\sqrt{2+a_{1}} \text{ and so on and we conclude that }a_{n+1}=\sqrt{2+a_{n}}$.Then we take the function $\varphi(x)=\sqrt{2+x}$,in order to ... | Community | -1 | <p>The interval is taken because it works; there isn't really a deeper reason. One of the nice things about real analysis -- especially when you take limits -- is that it doesn't matter too much what you choose for bounds on things. You usually have a great deal of freedom to make simple choices, or choices that simpli... |
344,119 | <p>$$\int_{0}^{\pi/2}\int_{x}^{\pi/2}\frac{\cos y}{y} \, \operatorname{d}\!y\, \operatorname{d}\!x$$</p>
<p>$\iint_R2xy^2 \, \operatorname{d}\!A$ where R is the right half of the unit circle</p>
| M. Strochyk | 40,362 | <p>Сhange the order of integration in the first integral:
$$\int_{0}^{\frac{\pi}{2}}\int_{x}^{\frac{\pi}{2}}\frac{\cos y}{y} \, \operatorname{d}\!y\, \operatorname{d}\!x=\int_{0}^{\frac{\pi}{2}}\int_{0}^{y}\frac{\cos y}{y} \, \operatorname{d}\!x\, \operatorname{d}\!y=\int_{0}^{\frac{\pi}{2}}\frac{\cos y}{y} \int_{0}^... |
344,119 | <p>$$\int_{0}^{\pi/2}\int_{x}^{\pi/2}\frac{\cos y}{y} \, \operatorname{d}\!y\, \operatorname{d}\!x$$</p>
<p>$\iint_R2xy^2 \, \operatorname{d}\!A$ where R is the right half of the unit circle</p>
| Mhenni Benghorbal | 35,472 | <p>$$\int_{0}^{\pi/2}\int_{x}^{\pi/2}\frac{\cos y}{y} \, \operatorname{d}\!y\, \operatorname{d}\!x = \int_{0}^{\pi/2}\frac{\cos y}{y}\int_{0}^{y} \, \operatorname{d}\!x\, \operatorname{d}\!y=\ldots. $$</p>
|
2,598,289 | <p><a href="https://i.stack.imgur.com/eCeNh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eCeNh.jpg" alt="enter image description here"></a></p>
<p>This is the definition of the real projective space in John Lee's book. However what I know is that the real projective space is defined by the quotie... | BallBoy | 512,865 | <p>The quotient $\pi: \mathbb{R}^{n+1} \setminus \{0\} \to \mathbb{RP}^n$ can be broken up into two quotients $\mathbb{R}^{n+1} \setminus \{0\} \to S^{n} \to \mathbb{RP}^n$. The first sends $x$ to $\frac{x}{|x|}$ (considering $S^{n}$ as the unit circle embedded in $\mathbb{R}^{n+1}$); the second identifies the antipode... |
3,778,195 | <p>This is Exercise 4 from Section 2.2 of <em>Topology and Groupoids</em> by Brown.</p>
<p><strong>Exercise:</strong></p>
<blockquote>
<p>Let <span class="math-container">$X$</span> be a topological space and let <span class="math-container">$A \subseteq B \subseteq X$</span>. We
say that <span class="math-container">$... | Novice | 543,177 | <p>Joshua P. Swanson pointed out the key error in my proof, which was the conflation of the intersection with <span class="math-container">$A$</span> and <span class="math-container">$U$</span> with the intersection <em>with the intersection</em> <span class="math-container">$A \cap U$</span>. However, his answer used ... |
534,670 | <p>I am a bit unclear about underflowing in terms of binary representation.</p>
<p>Let's say that an unsigned 8-bit variable gets overflown from the addition of $150+150$. </p>
<p>A signed 8-bit variable gets underflown after the subtraction of $-120-60$.</p>
<p>Now my point is let's think of 8-bit variable, we are ... | davcha | 174,996 | <p>Say you have $8$-bits signed integers. The range of representable integers start at $-128$ and ends at $127$.</p>
<p>If you perform $127+1$, you obtain $-128$ : $0111 1111+0000 0001 = 1000 0000$ and the overflow flag is turned on.</p>
<p>If you perform $-128-1$, you obtain $127$ : $1000 0000-0000 0001 = 0111 1111$... |
1,236,572 | <p>I've been stuck on this question for a long time now and was wondering if anyone could show me how it's done. So far I have done the following:
Since $\lim_{x\to \infty}f'(x)=M$ then $\forall \epsilon ,\exists A >0$ s.t. if $x>A$ then $|f'(x)-M|<\epsilon$. So we have that $M-\epsilon<f'(x)<M+\epsilon... | Adhvaitha | 228,265 | <p>Here is an easier way out. By Mean Value theorem, given any $x$, there exists $y \in [x,x+1]$ such that
$$f(x+1) - f(x) = \dfrac{f(x+1)-f(x)}{(x+1)-x} = f'(y)$$
Now as $x \to \infty$, since $y \in [x,x+1]$, we have $y \to \infty$. Hence, we obtain that
$$\lim_{x \to \infty} \left(f(x+1) - f(x)\right) = \lim_{y \to \... |
2,693,430 | <p>Problem:</p>
<p>$$
\begin{align}
5y+w=1 \\
2x+5y-4z+w=1
\end{align}
$$</p>
<p>What I've done:
-1 times first equation in order to get rid of $5y$ and $w$.
Then, when I add first and second equation, the equation becomes $2x-4z=0$
And I'm stuck right here.</p>
<p>Thanks.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Yes you system is consistent and you have infinitely many solutions.</p>
<p>You may pick $x$ and $z$ such that $x=2z$ and solve for $y$ and $w$ such that $5y+w=1.$</p>
<p>For example $$(x,y,z,w)=(2,1,1,-4)$$ is such a solution.</p>
|
2,693,430 | <p>Problem:</p>
<p>$$
\begin{align}
5y+w=1 \\
2x+5y-4z+w=1
\end{align}
$$</p>
<p>What I've done:
-1 times first equation in order to get rid of $5y$ and $w$.
Then, when I add first and second equation, the equation becomes $2x-4z=0$
And I'm stuck right here.</p>
<p>Thanks.</p>
| Bernard | 202,857 | <p>The system is consistent because the matrix of the system and the augmented matrix:
$$\begin{bmatrix}0&5&0&1\\2&5&-4&1\end{bmatrix}\enspace\text{ and }\enspace\begin{bmatrix}0&5&0&1&1\\2&5&-4&1&1\end{bmatrix}\enspace \text{ resp. }$$
have the same rank, which i... |
2,340,231 | <p>I have the following ODE for complex $z$:</p>
<p>$$\dot{z}=i|z|^2z$$</p>
<p>and I would like to find the most general exact solution possible. It is easy to see that $z=0$ and $z=e^{it}$ are two solutions, but I am hoping for a way to see if these are the only ones or if more may be found.</p>
<p>Thanks</p>
| Weaam | 1,746 | <p>Let $z = x + iy$, where $x, y$ are real functions in $t$. Then the ODE</p>
<p>$$\dot{x} + i\dot{y} = i(x^2 + y^2)(x+iy)$$</p>
<p>In particular, $$\dot{x} = -y(x^2 + y^2), \dot{y} = x(x^2 + y^2)$$
Then, $$\frac{dy}{dx} = -\frac{x}{y} \implies \int ydy = -\int x dx$$
Solving, we find that any solution must have cons... |
142,481 | <ol>
<li><p>Which term is used for model categories whose homotopy categories are triangulated? Stable proper model categories?</p></li>
<li><p>I want $Ho(Pro-M)$ to be triangulated ($Pro-M$ is the category of pro-objects of M) and the functor $Ho(M)\to Ho(Pro-M)$ to be an exact full embedding. Which restrictions on M ... | Andrew D. King | 4,580 | <p>This notion is complementary to the notion of <em>vertex multiplication</em>, in which every vertex is replaced with a <em>homogeneous clique</em>. This goes back to Lovasz' proof of the Weak Perfect Graph Theorem, and probably even earlier.</p>
|
200,723 | <p>Trying to get the modulus of the five numbers immediately before a prime, added together in there factorial form; I'll call this operation $S(p)$. For example,</p>
<p>$$S(p) = ((p-1)! + (p-2)! + (p-3)! + (p-4)! + (p-5)!) \bmod p$$</p>
<p>$$S(5) = (4!+3!+2!+1!+0!) \bmod 5$$</p>
<p>$$S(5) = 4$$</p>
<p>However, I h... | André Nicolas | 6,312 | <p>We produce a concrete and easy to calculate "formula" that is useful if we need to compute the answer for a number of values of $p$. </p>
<p>Let $x=(p-1)!+(p-2)!+(p-3)!+(p-4)!+(p-5)!$. Then, as shown by Michael Albanese, we have
$$x\equiv 9(p-5)!\pmod{p}.$$
But $p-3\equiv -3\pmod{p}$, and therefore
$$x\equiv (-... |
368,129 | <p>30 years ago, Yves Colin de Verdière introduced the algebraic graph invariant <span class="math-container">$\mu(G)$</span> for any undirected graph <span class="math-container">$G$</span>, see [1]. It was motivated by the study of the second eigenvalue of certain Schrödinger operators [2,3]. It is defined in purely ... | gwynneth-m.sc. | 163,173 | <p>Perhaps I can contribute to the history part the question, since I was quite close to the Institut Fourier at that time and was very interested in their work (I am a physicist). Grenoble now has several different research groups doing graph theory (like G-SCOP, Institut Fourier, GIPSA-lab, LIG) but I think L'Institu... |
368,129 | <p>30 years ago, Yves Colin de Verdière introduced the algebraic graph invariant <span class="math-container">$\mu(G)$</span> for any undirected graph <span class="math-container">$G$</span>, see [1]. It was motivated by the study of the second eigenvalue of certain Schrödinger operators [2,3]. It is defined in purely ... | M. Winter | 108,884 | <p>Embeddability in any surface but the sphere (or plane) can probably not be characterized via the Colin de Verdière number.</p>
<p>Suppose that <span class="math-container">$K_n$</span> is the largest complete graph that embedds into a surface <span class="math-container">$S$</span>.
This shows that the best we can h... |
42,326 | <p>Does there exist a nowhere monotonic continuous function from some open subset of $\mathbb{R}$ to $\mathbb{R}$?
Some nowhere differentiable function sort of object?</p>
| Andrés E. Caicedo | 462 | <p>The <a href="http://andrescaicedo.wordpress.com/2013/11/07/weierstrass-function/" rel="noreferrer">Weierstrass function</a>, mentioned in other answers, is indeed an example of a nowhere monotone function, meaning that $f$, even though continuous and bounded, is increasing at no point, decreasing at no point (and di... |
3,695,173 | <p><span class="math-container">$M$</span> is a <span class="math-container">$2\times2$</span> matrix. <span class="math-container">$M$</span> is diagonalizable over <span class="math-container">$\mathbb{R}$</span>. <span class="math-container">$M$</span> has the values <span class="math-container">$1$</span> and <span... | Didier | 788,724 | <p>For a <span class="math-container">$2\times2$</span> matrix, its characteristic polynomial is <span class="math-container">$\chi_M = X^2 - \mathrm{trace}(M)X + \det (M)$</span>. Suppose <span class="math-container">$M$</span> is diagonalisable, so it has two real eigenvalues. Here, they are distinct. Otherwise, <spa... |
1,984,849 | <p>Can you please clarify whether, for the following question, I need to use the definition of linear transformation, or something else?</p>
<blockquote>
<p>Compute the inverse of the function
$f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$, where $f(x_1,x_2,x_3) := (x_2+x_3, x_1+x_3, x_1+x_2)$.</p>
</blockquote>
| Khosrotash | 104,171 | <p>$$f(x_1,x_2,x_3)=(0x_1+x_2+x_3,x_1+0x_2+x_3,x_1+x_2+0x_3)=\\
\begin{bmatrix}0 & 1 &1\\1 &0 & 1 \\1 &1 &0\end{bmatrix}\times
\begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix}=AX$$now
$$f^{-1}=A^{-1}X$$</p>
|
142,220 | <p>Fermat proved that <span class="math-container">$x^3-y^2=2$</span> has only one solution <span class="math-container">$(x,y)=(3,5)$</span>.</p>
<p>After some search, I only found proofs using factorization over the ring <span class="math-container">$Z[\sqrt{-2}]$</span>.</p>
<p>My question is:</p>
<p>Is this Fermat'... | Franz Lemmermeyer | 3,503 | <p>Fermat did not prove this result; he claimed that the only solution is the obvious one and conjectured (in words that seem to suggest he knew how to prove it, but without explicitly saying so) that this can be proved by descent. I am sure that Fermat, if he really believed
to have a proof (in my opinion he did not)... |
4,578,252 | <p>On the MIT <span class="math-container">$2021$</span> <em>Integration Bee Qualifying Exam</em>, it asked to approximate
<span class="math-container">$$
\int_{0}^{\pi}{\rm e}^{{\rm e}^{x}}\,{\rm d}x
$$</span>
I got <span class="math-container">$\displaystyle{\rm e} + {\rm e}^{\rm e} + {\rm e}^{{\rm e}^{2}} + {\rm e}^... | openspace | 243,510 | <p>I am not sure if the answer supposed to be this, but:</p>
<p><span class="math-container">$$
\int_0^\pi e^{e^x} dx = \int_0^\pi \sum_{k \ge 0} \dfrac{e^{kx}}{k!} dx = \sum_{k \ge 1} \dfrac{e^{k\pi} - 1}{k \cdot k!} = S_1 - S_2
$$</span></p>
<p>For <span class="math-container">$S_1$</span> and <span class="math-conta... |
2,650,913 | <p>I was looking at the solid of revolution generated by revolving $\cos(x)$ about the $x$-axis on the interval $[0, 2\pi]$, and I noticed that when the volume of the solid was approximated with $3$ or more cylinders via the disk method the approximation would equal the true volume. To prove this, I deduced that it suf... | cansomeonehelpmeout | 413,677 | <p>Not a very elegant solution, but it becomes easy if you use: $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$</p>
<p>Then $\cos(x)^2=\frac{e^{2ix}+2+e^{-2ix}}{4}=\frac{e^{2ix}}{2}+\frac{1}{2}+\frac{e^{-2ix}}{2}$.</p>
<p>Your sum becomes $$\sum_{n=1}^{k}\frac{e^{2ix}}{2}+\sum_{n=1}^{k}\frac{1}{2}+\sum_{n=1}^{k}\frac{e^{-2ix}}{... |
3,167,336 | <p><img src="https://i.stack.imgur.com/FLO6E.jpg" alt="enter image description here">
It took me much time to reach the solution where I find the value of X as 2 but still not sure whether this is correct or not.
please help me with the solution</p>
| mjw | 655,367 | <p>May not be the answer you are looking for, and probably not the forum for it, but you can plot the two curves using software, for example <em>Mathematica</em> or MATLAB or Python or Java or ...</p>
<p>Here is a quick way to do it in <em>Mathematica</em>, for example:</p>
<pre><code> a[x_, y_] := (x^2 + y^2 - 1);
... |
1,814,855 | <blockquote>
<p>How many ways are there to hand out 24 cookies to 3 children so that they each get an even number, and they each get at least 2 and no more than 10? Use generating functions. </p>
</blockquote>
<p>So the first couple steps are easy.</p>
<p>The coefficient is $x^{24}$</p>
<p>$g(x) = x^6(1+x^2+x^4+x^... | Semiclassical | 137,524 | <p>When computing $1+x^2+(x^2)^2+(x^2)^3+(x^2)^4$, the series is in powers of $x^2$ not $x$. So the proper expression is $$1+x^2+(x^2)^2+(x^2)^3+(x^2)^4=\frac{1-(x^2)^5}{1-(x^2)}=\frac{1-x^{10}}{1-x^2}.$$ By contrast, $$\frac{1-x^{9}}{1-x}=1+x^1+x^2+\cdots +x^8$$ which differs from the above series in it has both odd a... |
1,895,248 | <p><a href="http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/exams/prac1a.pdf" rel="nofollow noreferrer">Problem 5. a)</a> Find the area of the space triangle with vertices $P_0, P_1, P_2$:</p>
<p>$$ P_0 = (2,1,0),\ \ P_1=(1,0,1),\ \ P_2=(2,-1,1) $$</p>
<p>My current solution is to use $\... | Christopher Carl Heckman | 261,187 | <p>You shouldn't be assigning propositional variables to the pieces of the sentences; they should be predicates, so that you can substitute the day of the week.</p>
<blockquote>
<ul>
<li>"If I take the day off, it either rains or snows."</li>
</ul>
</blockquote>
<p>Let's call this one $$(\forall d)(T (d) \to (R... |
100,551 | <p>I need to get a matrix $\{a(x_i-x_j)\}$, where $x_i$ form a partition of an interval, $a(x)$ is a given function. I use </p>
<pre><code>In[67]:= a[x_?NumericQ] := N[Exp[-Abs[x]]];
x = Table[-10 + 0.02 (j - 1), {j, 1, 1001}];
A = Outer[a[#1 - #2] &, x, x]; // AbsoluteTiming
Out[69]= {2.99032, ... | Szabolcs | 12 | <p>Vectorization will help a lot:</p>
<pre><code>a[x_?NumericQ] := N[Exp[-Abs[x]]];
x = Table[-10 + 0.02 (j - 1), {j, 1, 1001}];
A = Outer[a[#1 - #2] &, x, x]; // AbsoluteTiming
(* {2.11988, Null} *)
B = Exp[-Abs[x - #]] & /@ x; // AbsoluteTiming
(* {0.016182, Null} *)
A == B
(* True *)
</code></pre>
<p>Not... |
3,419,850 | <p>Does <span class="math-container">$\displaystyle\sum_{n\geq 0} \dfrac{n!k^n}{(n+1)^n}$</span> converge or diverge for <span class="math-container">$k=\dfrac{19}{7}$</span>?</p>
<p>I'm not sure what convergence test I should use for this one. <span class="math-container">$k=\dfrac{19}{7}$</span> also seems randomly ... | user | 505,767 | <p>By ratio test we obtain</p>
<p><span class="math-container">$$\left|\dfrac{(n+1)!k^{n+1}}{(n+2)^{n+1}}\dfrac{(n+1)^n}{n!k^n}\right|=k\frac{n+1}{n+2}\left(1-\frac1{n+2}\right)^n\to \frac ke$$</span></p>
<p>indeed recall that</p>
<p><span class="math-container">$$\left(1-\frac1{n+2}\right)^n=\left[\left(1-\frac1{n+... |
3,419,850 | <p>Does <span class="math-container">$\displaystyle\sum_{n\geq 0} \dfrac{n!k^n}{(n+1)^n}$</span> converge or diverge for <span class="math-container">$k=\dfrac{19}{7}$</span>?</p>
<p>I'm not sure what convergence test I should use for this one. <span class="math-container">$k=\dfrac{19}{7}$</span> also seems randomly ... | Mohammad Riazi-Kermani | 514,496 | <p>Apply the ratio test <span class="math-container">$$ \left|\dfrac{(n+1)!k^{n+1}}{(n+2)^{n+1}}\dfrac{(n+1)^n}{n!k^n}\right|=$$</span></p>
<p><span class="math-container">$$k\frac{n+1}{n+2}\left(1-\frac1{n+2}\right)^n\to \frac ke =0.9985...<1$$</span></p>
<p>Thus the series converges. </p>
|
1,366,462 | <p>The line $x + y − 1 = 0$ intersects the circle $x^2 + y^2 = 13$ at $A(\alpha_1, \alpha_2)$ and $B(\beta_1, \beta_2)$.
Without finding the coordinates of A and B, find the length of the chord AB. </p>
<p><strong>Hint:</strong> Form a quadratic equation in $x$ and evaluate $|\alpha_1 − \beta_1 |$, and similarly find ... | mathlove | 78,967 | <p>Note that
$$|\alpha_1-\beta_1|^2=(\alpha_1+\beta_1)^2-4\alpha_1\beta_1.$$
For $x^2-x-6=0$, by <a href="https://en.wikipedia.org/wiki/Vieta%27s_formulas" rel="nofollow">Vieta's formulas</a>, you can have
$$\alpha_1+\beta_1=-\frac{-1}{1}=1,\ \ \ \alpha_1\beta_1=\frac{-6}{1}=-6.$$</p>
<hr>
<p>Another way is to use
$$... |
1,751,955 | <p>$ S = \{(1/2)^n : n \in \mathbb{N} \} \cup \{ 0 \} $ is obviously bounded and infinite. It also looks totally disconnected to me (it is not and does not contain as its subset an interval with more than one element). But we know that a compact, totally disconnected set must be finite. Hence we know that $ S $ is not ... | Exit path | 161,569 | <p>A subset $A \subseteq \mathbb{R}$ under the standard topology is compact if and only if it is closed and bounded. Since the set $S$ is equal to the points of the sequence $(x_n)=(1/2^n)$ together with its limit $0$, it is closed. It is also clearly bounded and therefore compact. </p>
|
3,177,596 | <p>I'd be grateful for some help with this problem I am trying to solve.</p>
<p>Let's say that I have an object travelling at a velocity v. I want that object to come to a halt in time t AND travel exactly distance d within that time.</p>
<p>So if we are at t0 when we are at velocity v and apply the brakes, the dista... | cphys | 661,323 | <p>To solve this type of problem we need to do two steps. First we find the time it takes for the object to come to a stop using the equation</p>
<p><span class="math-container">$$v=v_0+at$$</span></p>
<p>where <span class="math-container">$v_0$</span> is the initial velocity. Solving this equation for <span class="m... |
139,125 | <p>This is a variation on an earlier question resolved by <em>user35353</em>: <a href="https://mathoverflow.net/questions/139105/can-a-tangle-of-arcs-interlock">Can a tangle of arcs interlock?</a> In that question, the arcs were restricted to circular arcs, and <em>user35353</em>'s proof that one arc can be removed wit... | Cristi Stoica | 10,095 | <p>Four tangled ellipses that cannot be unlocked.
<img src="https://i.stack.imgur.com/xp8IG.gif" alt="enter image description here"></p>
|
2,176,656 | <p>C.T. is comparison test</p>
<p>TYPE II is when a improper integral is improper but not at $\infty$. </p>
<p>a)</p>
<p>$$\int_{1}^{\infty} \frac{\sin\left(\frac{\pi}{x}\right)}{x^2}dx$$</p>
<p>Let g(x) = $\frac{1}{x^2}$ because $|sin(\frac{\pi}{x})| \leq 1$. Since the numerator has been maximized and denominator ... | GEdgar | 442 | <p>Let $y=\sqrt{x}$. Then
\begin{align}
f(y) &= \sum_{n=0}^\infty \frac{y^{2n}}{(2n)!}
\\
f'(y) &= \sum_{n=0}^\infty \frac{2n y^{2n-1}}{(2n)!}
= \sum_{n=1}^\infty \frac{ y^{2n-1}}{(2n-1)!}
\\
f''(y) &= \sum_{n=1}^\infty \frac{(2n-1) y^{2n-2}}{(2n-1)!}
= \sum_{n=1}^\infty \frac{ y^{2n-2}}{(2n-2)!}
=\sum_{k=... |
2,734,442 | <p>Prove in natural deduction (Negation of existential quantifier):</p>
<ul>
<li>∀x ¬P(x) ⊢ ¬∃x P(x)</li>
</ul>
<p>Inference rules:</p>
<ul>
<li>(∀−) If $Σ ⊢ ∀xA(x)$, then $Σ ⊢ A(t)$ where $t$ is any term.</li>
<li>(∀+) If $Σ ⊢ A(u)$ and $u$ does not occur in $Σ$, then $Σ ⊢ ∀xA(x)$.</li>
<li>(∃−) If $Σ, A(u) ⊢ B$ an... | Taroccoesbrocco | 288,417 | <p>The following is a proof in natural deduction of $\forall x \lnot P(x) \vdash \lnot \exists x P(x)$:</p>
<p>$$\dfrac{\dfrac{[\exists x P(x)]^* \qquad \dfrac{[P(x)]^{**} \qquad \dfrac{\forall x \lnot P(x)}{\lnot P(x)}\forall_\text{elim}}{\bot}\lnot_\text{elim}}{\bot}\exists_\text{elim}^{**}}{\lnot \exists x P(x)}\ln... |
1,000,349 | <p>$\def\Li{\operatorname{Li}}$
I wonder how to prove:
$$
\int_{0}^{-1} \frac{\Li_2(x)}{(1-x)^2} dx=\frac{\pi^2}{24}-\frac{\ln^2(2)}{2}
$$
I'm not used to polylogarithm, so I don't know how to tackle it. So any help is highly appreciated.</p>
| Anastasiya-Romanova 秀 | 133,248 | <p><span class="math-container">$\def\Li{{\rm{Li}}_2}$</span>Set <span class="math-container">$x\mapsto -x$</span> followed by integration by parts, we have
<span class="math-container">\begin{align}
\int_0^{-1}\frac{\Li(x)}{(1-x)^2}\,dx&=-\int_0^{1}\frac{\Li(-x)}{(1+x)^2}\,dx\qquad\Rightarrow\qquad u=\Li(-x)\,\,\m... |
1,557,097 | <p>Show that a finite domain $F$ is a field.</p>
<p>Let $I$ a proper ideal of $F$ and let $a\in I$. In particular, $a$ is not invertible, otherwise $I$ wouldn't be proper. </p>
<p>I would like to show that $I=(a)=(0)$, but without success. </p>
| quid | 85,306 | <p>Let continue your idea. Suppose $I$ is an ideal and it contains $a \neq 0$. Then $aF \subset I$. However as $F$ is a domain $af \neq af'$ for $f \neq f'$. Thus $|aF| = |F|$ and $|F| \le |I| $. Since $|I| \le |F|$ is trivial, we have $|I|=|F|$. Now since $F$ is finite this means $I=F$. </p>
|
143,601 | <p>I am looking for a way to either get or set a vertex property for all vertices of the graph at once. The two functions will be:</p>
<pre><code>getVertexProp[g_?GraphQ, prop_] := ...
setVertexProp[g_?GraphQ, prop_, values_] := ...
</code></pre>
<p><code>getVertexProp</code> must return a list of values. Each valu... | WReach | 142 | <p>I haven't run this through all of the use cases, but just to get the creative juices flowing:</p>
<pre><code>setVertexPropSemiImperative[g_?GraphQ, prop_, values_List] :=
Module[{vertices = VertexList[g] // Developer`FromPackedArray, i = 0}
, If[Length[values] =!= Length[vertices]
, $Failed
, SetPropert... |
1,386,677 | <p>Proving that $$\sin x > \frac{(\pi^{2}-x^{2})x}{\pi^{2}+x^{2}}, \qquad\forall x>\pi$$</p>
| Jack D'Aurizio | 44,121 | <p>We just need to prove the inequality:
$$\frac{\sin x}{x}\geq \frac{\pi^2-x^2}{\pi^2+x^2}\tag{1}$$
over the $[\pi,2\pi]$ interval, since the RHS is a decreasing function and its value at $x=2\pi$ is $-\frac{3}{5}<-\frac{1}{2\pi}$. Translating and rearranging, we just need to prove that:
$$ \forall x\in[0,\pi],\qqu... |
2,326,259 | <p>I tried the following $$I = \langle X^2,X+1\rangle =\langle X^2,X+1,X^2+2(X+1)\rangle =\langle X^2,X+1,(X+1)^2+1 \rangle$$</p>
<p>Yet no matter how I arrange it, I cannot obtain $1$. Can someone help me out?</p>
| Community | -1 | <p>I am using $\langle f,g \rangle = \langle r,g \rangle$ when $f=h\,g+r$ for $h,r \in K[X]$. </p>
<p>So lets do the polynomial division:</p>
<pre><code>X^2 : X + 1 = X - 1
X^2 + X
-------
- X
- X - 1
-------
1
</code></pre>
<p>So we get $\langle 1,g \rangle$ for both $\mathbb{Z}[X]$... |
180,672 | <p>Let $f: C \to C$ be a smooth function and $C$ be a compact set, subset of $\mathbb{R}^n$.
We assume that all the fixed points are hyperbolic. Is it true that the number of fixed points is finite or countable?</p>
| Jaap Eldering | 3,928 | <p>Building upon what others have already said, the number of hyperbolic fixed points is indeed countable, but need not be finite.</p>
<p>First, it follows from the definition of a hyperbolic fixed point $x$ that they are isolated: its total derivative $Df(x)$ does not have any eigenvalues on the unit circle, which pr... |
353,327 | <p>Can someone please help me out with this question? I have been at it for hours and I can't wrap my head around this one.</p>
<blockquote>
<p>Karen and Kurt's backyard has a width of $20$ meters and a length of $30$ meters. They want to put a rectangular flower garden in the middle of the backyard,leaving a strip... | Brian M. Scott | 12,042 | <p>If $x$ is the width of the strip, your initial setup of $(20-2x)(30-2x)=336$ is fine. </p>
<p><strong>Correction:</strong> Like you, I didn’t read closely enough. It’s the <strong>grass</strong>, not the garden, that is to have an area of $336$ square metres. If you make a sketch, you’ll see that the area of the st... |
2,090,885 | <p>I need to calculate the value of the integral:
$$\int_T\frac 1 {\sqrt{x^2+y^2}} \, dx \, dy$$ where $T=\{(x,y) : x\in[-2,2], x^2<y<4\}$.</p>
<p>Specifically, I need to know how to set integration extremes.</p>
| kryomaxim | 212,743 | <p>You have to integrate over $y$ first with the bounds between $x^2$ and $4$. After that integrate over $x$ with the bounds between $-2$ and $2$.</p>
|
954,147 | <p>I want to solve the congruence for $k$ such that $k^2\equiv 5k\pmod {15}, 2\leq k\leq 30$.</p>
<p>For this, if $\gcd(15,k)=1$, then $k\equiv 5\pmod{15}$. Is my approach correct?
How can I get the values of $k$.</p>
| lhf | 589 | <p>$k^2\equiv 5k\bmod{15}$</p>
<p>$\iff$
$k^2\equiv 0\bmod{5}$ and $k^2\equiv 2k\bmod{3}$</p>
<p>$\iff$
$k\equiv 0\bmod{5}$ and $k\equiv 0,2\bmod{3}$</p>
<p>$k\equiv 0\bmod{5}$ and $k\equiv 0\bmod{3}$ $\iff$ $k\equiv 0\bmod{15}$,
which gives us $k=15$ and $k=30$.</p>
<p>$k\equiv 0\bmod{5}$ and $k\equiv 2\bmod{3}$ $... |
265,494 | <p>I have two lists given by:</p>
<pre><code>t1 = {{1, 2}, {3, 4}, {5, 6}};
t2 = {a, b, c};
</code></pre>
<p>and want to replace the second parts of <code>t1</code> with <code>t2</code> to get</p>
<pre><code>{{1,a},(3,b},{5,c}}
</code></pre>
<p>I tried</p>
<pre><code>t1 /. {u_, v_} -> {u, #} & /@ t2
</code></pre... | bmf | 85,558 | <p>The answer by @lericr is optimal with regards to minimality and I like it a lot. Another way to go about it is the following:</p>
<pre><code>t1 = {{1, 2}, {3, 4}, {5, 6}};
t2 = {a, b, c};
ArrayReshape[Riffle[First /@ t1, t2], {3, 2}]
</code></pre>
<blockquote>
<p><code>{{1, a}, {3, b}, {5, c}}</code></p>
</blockquo... |
2,743,266 | <p>I am able to find the sixth derivative of $\cos(x^2)$ by simply replacing the $x$ in the Taylor series for $\cos(x)$ with $x^2$ but beyond simple substitutions, I am struggling... </p>
<p>Thanks for any help!</p>
| Bernard | 202,857 | <p><strong>Hint</strong>:
Linearise first:
$$\cos^2x=\frac12(1+\cos 2x),$$
and differentiate $\cos 2x$ six times, remembering that
$$(\cos u)'=\cos\Bigl(u+\frac\pi2\Bigr)u'=\cos\Bigl(2x+\frac\pi2\Bigr)\cdot 2.$$</p>
|
386 | <p>The Reshetikhin-Turaev construction take as input a Modular Tensor Category (MTC) and spits out a 3D TQFT. I've been told that the other main construction of 3D TQFTs, the Turaev-Viro State sum construction, factors through the RT construction in the sense that for each such TQFT Z there exists a MTC M such that the... | Manuel Bärenz | 13,767 | <p>If your TQFT does not extend to the circle, (EDIT: <em>and you're willing to generalise to lax TQFTs</em>,) then the answer is <em>no</em>. A reference is: <a href="https://arxiv.org/abs/1408.0668" rel="nofollow noreferrer">https://arxiv.org/abs/1408.0668</a></p>
<p>The proof goes by an explicit generators and rela... |
2,776,912 | <p>Consider the sequences $\{a_n\}_{\forall n \in \mathbb{N}}<0$, $\{b_n\}_{\forall n \in \mathbb{N}}>0$, $\{c_n\}_{\forall n \in \mathbb{N}}>0$ and suppose
$$
\begin{cases}
\lim_{n\rightarrow \infty} (a_n+b_n)=0\\
\lim_{n\rightarrow \infty} c_n=L<\infty
\end{cases}
$$</p>
<p>Could you help me to show that... | Tommaso Seneci | 320,104 | <p>You can just compute
$$ |e^{a_n}c_n - L e^{-b_n}| = e^{a_n}|c_n - L e^{-(a_n+b_n)}| \leq |c_n - L e^{-(a_n+b_n)}|. $$
Now $c_n\to L$ and $a_n+b_n\to 0$. So, because $e^x$ and $|x|$ are continuous functions, you can "pass to the limit inside the functions" and get
$$ |e^{a_n}c_n - L e^{-b_n}| \leq |c_n - L e^{-(a_n+b... |
1,134,612 | <p>A speaks truth in $75\%$ cases and B in $80\%$ cases. In what percentage of cases are they likely to contradict each other in stating the same fact?</p>
<p>(a) $70\%$ (b)$35\%$</p>
<p>(c) $25\%$ (d)$20\%$</p>
<p><strong>what i have tried:</strong><br/>
I would like to show about what i think, If QA is in Yes/No b... | Hagen von Eitzen | 39,174 | <p>Since it is not stated that they decide independantly whether or not to tell the truth, anything between $5\%$ and $45\%$ is possible.</p>
|
1,134,612 | <p>A speaks truth in $75\%$ cases and B in $80\%$ cases. In what percentage of cases are they likely to contradict each other in stating the same fact?</p>
<p>(a) $70\%$ (b)$35\%$</p>
<p>(c) $25\%$ (d)$20\%$</p>
<p><strong>what i have tried:</strong><br/>
I would like to show about what i think, If QA is in Yes/No b... | Curious | 141,191 | <p>The probability that they contradict each other is given by (TF, FT) as you mentioned in the table. Assuming both the speakers choose to speak truth or false independently, the probability can be calculated as follows-</p>
<p>Suppose the probability A and B speak truth is given by $P_A$, $P_B$ respectively. Their p... |
2,891,062 | <p>My textbooks lists the set of values b may take as $b>-4 $ and $b≠4$</p>
<p>One of the roots is given as x=2 in the first part of the question, which asks for the value of a such that, for all values of b, one root of the equation $2x^3+ax+4=b(x-2)$ I calculated this as a=-10</p>
<p>If the RHS of the equation w... | Stefan4024 | 67,746 | <p>Cancel out the $x-2$ factors on the both sides to end up with:</p>
<p>$$2x^2 + 4x -2 = b$$</p>
<p>This is a quadratic equation and will have two distinct solutions if the discriminant is positive. The discriminant in our case is $D = 16 + 8(b+2)$. And thus we obtain one condition which says that $D>0 \iff b>... |
2,544,705 | <p>I got introduced to the idea of fractals, and the idea that fractals can have dimensions that are non-integers. </p>
<p>This got me thinking, the space of real-valued functions has a dimensionality as well, and it seems likely to me that the space of continuous functions has a dimensionality strictly lower than tha... | Nosrati | 108,128 | <p>\begin{align}
\int_\mu^\infty z^{k-1}e^{-z+\mu}dz
&= \int_0^\infty (\mu+u)^{k-1}e^{-u}du \\
&= \int_0^\infty \sum_{x=0}^{k-1}{k-1\choose x}u^{k-1-x}\mu^xe^{-u}du \\
&= \sum_{x=0}^{k-1}{k-1\choose x}\mu^x\int_0^\infty u^{k-1-x}e^{-u}du \\
&= \sum_{x=0}^{k-1}\dfrac{\Gamma(k)}{\Gamma(x+1)\Gamma(k-x)}\m... |
1,558,530 | <p>Let $\alpha, \beta$ be random variables, $P(\alpha = i) = P(\beta = i) = \frac{1}{N}$, $i \in \{1, \ldots, N\}$.</p>
<p>What is the probability that $\alpha^3 + \beta^3 = 3 t, t \in \mathbb{N}$? </p>
| triton21 | 279,514 | <p>like person above me said use the fact $a^3 = a mod 3$ (i dont know how to write equivalent so i used = instead).</p>
<p>this mean $a^3$ = $a$ mod 3 and $b^3$= $b$ mod 3 </p>
<p>then just divide the case</p>
<p>case 1 : if a divisible by 3 then b should divisible by 3 </p>
<p>in this case just search how man... |
1,558,530 | <p>Let $\alpha, \beta$ be random variables, $P(\alpha = i) = P(\beta = i) = \frac{1}{N}$, $i \in \{1, \ldots, N\}$.</p>
<p>What is the probability that $\alpha^3 + \beta^3 = 3 t, t \in \mathbb{N}$? </p>
| Zubin Mukerjee | 111,946 | <p>In words, your question could be written like this:</p>
<blockquote>
<p>What is the probability that two integers independently and uniformly chosen at random from the first $N$ positive integers have a sum of cubes that is divisible by $3$?</p>
</blockquote>
<p>As @djechlin pointed out, the cube of an integer a... |
1,533 | <p>What is up with this site: <a href="https://mathoverflow.net/">https://mathoverflow.net/</a> ? Is it a clone or something? I wasn't paying attention and went to login, and it says that my name is unknown... What's up? I revoked access from my Google account just to be sure.</p>
| Jeff Atwood | 153 | <p>That site runs on the Stack Exchange 1.0 platform.</p>
<p>These sites run on the Stack Exchange 2.0 platform. </p>
<p>For more detail and context see:<br>
<a href="http://blog.stackoverflow.com/2010/04/changes-to-stack-exchange/" rel="nofollow">http://blog.stackoverflow.com/2010/04/changes-to-stack-exchange/</a></... |
745,095 | <p>For a second order ODE </p>
<p>y''+10y'+ 21y=0</p>
<p>which was reduced to this quadratic expression
x^2+10x+21=0</p>
<ul>
<li>is there any way to tell whether the expression is bounded that is y(x) is either periodic or has a limit 0 as x tends to infinity?</li>
</ul>
<p>*Does periodic means having only comple... | Yiyuan Lee | 104,919 | <p>Suppose that you want to solve the quadratic equation, $ax^2 + bx + c = 0$. The quadratic formula gives the two roots as</p>
<p>$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$</p>
<p>Now, take a look at the surd, $\sqrt{b^2 - 4ac}$. If $b^2 - 4ac$ was negative, then the roots would be complex, because the square root o... |
745,095 | <p>For a second order ODE </p>
<p>y''+10y'+ 21y=0</p>
<p>which was reduced to this quadratic expression
x^2+10x+21=0</p>
<ul>
<li>is there any way to tell whether the expression is bounded that is y(x) is either periodic or has a limit 0 as x tends to infinity?</li>
</ul>
<p>*Does periodic means having only comple... | Lutz Lehmann | 115,115 | <p>To answer the second part, in the differential equation
$$
ay''(x)+by'(x)+cy(x)=0
$$
you would need $b=0$ and $ac>0$ to get periodic solutions. With complex, but not purely imaginary eigenvalues you get oscillating solutions where the amplitude changes with an exponential function.</p>
|
2,488,165 | <p>Find the line integral of $x^2+y^2$ over the polar curve $r=e^{\theta}$</p>
<p>Not really sure on how to find the curve on terms of a parameter in order to evaluate the integral</p>
| mechanodroid | 144,766 | <p>You can parameterize your curve <span class="math-container">$c$</span> by the polar angle theta as <span class="math-container">$$c(\theta) = (x(\theta),y(\theta)) = (r(\theta)\cos\theta, r(\theta)\sin\theta) = (e^\theta\cos\theta, e^\theta\sin\theta)$$</span></p>
<p>so <span class="math-container">$\dot{c}(\theta... |
1,715,945 | <p>I need help solving/understanding this question:</p>
<p>L (x,y) : "x loves y".
Translate "there are exactly two people whom Lynn loves".
Its answer includes a variable "z". I do not get that part with the variable "z". How did it come here when it was not introduced in the question? Detailed solution is appreciated... | Tsemo Aristide | 280,301 | <p>Hint: $F_n$ is a free object in the category of groups i.e $Hom_{Group}(F_n,G)=Hom_{Set}(\{1,...,n\},G)$.</p>
<p>This is equivalent to saying that the forgetful functor from the category of sets to the category of groups has a left adjoint which associates to a set $S$ the free group $F_S$.</p>
|
3,628,919 | <blockquote>
<p>Give an example of a <span class="math-container">$T\in\mathcal L\left(\mathbb R^2\right)$</span> s. t. <span class="math-container">$Ker(T) = Im(T)$</span>.</p>
</blockquote>
<p><strong>MY APPROACH</strong></p>
<p>According to the rank-nullity theorem, <span class="math-container">$\dim Ker(T) = \o... | xpaul | 66,420 | <p>Let
<span class="math-container">$$ T: \mathbb{R}^2\to \mathbb{R}^2$$</span>
by <span class="math-container">$x\to Ax$</span>. Here
<span class="math-container">$$ A=\left[\begin{matrix}1&-1\\1&-1\end{matrix}\right]. $$</span>
It is easy to check
<span class="math-container">$$ N(T)=R(T)=\text{span}\{e\}, e=... |
3,366,071 | <p>For which conditions of <span class="math-container">$\alpha_1 $</span> and <span class="math-container">$\alpha_2 $</span> the following set is convex.</p>
<p><span class="math-container">$ \{x_1,x_2 \in \mathbb{R} : (\alpha_1 +2)x_1^2 +\alpha_1 x_2^2 +2 \alpha_2 x_1 x_2 \le 1 \}$</span></p>
<p>I have started fro... | Mick | 42,351 | <p>It seems the answer is quite trivial. (Correct me if my logic is wrong.)</p>
<p><a href="https://i.stack.imgur.com/A70JP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A70JP.png" alt="enter image description here"></a></p>
<p>We have a symmetric drawing with MNK as the axis of symmetry when we ... |
3,366,071 | <p>For which conditions of <span class="math-container">$\alpha_1 $</span> and <span class="math-container">$\alpha_2 $</span> the following set is convex.</p>
<p><span class="math-container">$ \{x_1,x_2 \in \mathbb{R} : (\alpha_1 +2)x_1^2 +\alpha_1 x_2^2 +2 \alpha_2 x_1 x_2 \le 1 \}$</span></p>
<p>I have started fro... | bajun65537 | 709,210 | <p>Alternatively:</p>
<p><a href="https://i.stack.imgur.com/JLf5M.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JLf5M.png" alt="enter image description here"></a></p>
<p>We observe that quadrilaterals <span class="math-container">$DHMI \sim AIMF$</span> (are similar) as the corresponding angles a... |
267,045 | <p>The Fibonacci recurrence $F_n=F_{n-1}+F_{n-2}$ allows values for all indices $n\in\mathbb{Z}$. There is an almost endless list of properties of these numbers in all sorts of ways. The below question might even be known. Yet, if true, I like to ask for alternative proofs.</p>
<blockquote>
<p><strong>Question.</str... | Ira Gessel | 10,744 | <p>More generally,
$$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$
which is equivalent to Will Sawin's identity.</p>
<p>Similarly,
$$e^x\sum_{k=0}^\infty F_{n+k}\frac{x^k}{k!}= \sum_{k=0}^\infty F_{n+2k}\frac{x^k}{k!},\tag{2}$$
and more generally, from the formula
$$(-... |
1,725,828 | <p>I am having trouble understanding the following proof on sized biased picking. We have the following situation:</p>
<p>Let $ X_1, \cdots , X_n $ be i.i.d. and positive, and $S_i = X_1 + \cdots + X_i$ for $ 1 \leq i \leq n $. The values of $S_i/S_n$ are used to partition the interval $[0,1]$, each sub-interval has s... | Tsemo Aristide | 280,301 | <p>The oriented surface with infinite genus does not have a finite atlas.</p>
|
1,725,828 | <p>I am having trouble understanding the following proof on sized biased picking. We have the following situation:</p>
<p>Let $ X_1, \cdots , X_n $ be i.i.d. and positive, and $S_i = X_1 + \cdots + X_i$ for $ 1 \leq i \leq n $. The values of $S_i/S_n$ are used to partition the interval $[0,1]$, each sub-interval has s... | littleO | 40,119 | <p>There's a relevant discussion on pp. 12-13 of Introduction to Smooth Manifolds by Lee:</p>
<p>"Our plan is to define a "smooth structure" on $M$ by giving a smooth atlas, and to define a function $f:M \to \mathbb R$ to be smooth if and only if $f \circ \phi^{-1}$ is smooth in the sense of ordinary calculus for each... |
2,397,249 | <p>Well, to quote from Wolfram MathWorld directly,</p>
<blockquote>
<p>Given an affine variety $V$ in the $n$-dimensional affine space $K^n$, where $K$ is an algebraically closed field, the coordinate ring of $V$ is the quotient ring $K[V] = K[x_1 , \dots , x_n] / I(V)$.</p>
</blockquote>
<p>My question is simply t... | Eric Wofsey | 86,856 | <p>The $i$th "coordinate function" on $K^n$ is the function $K^n\to K$ taking $(x_1,\dots,x_n)$ to $x_i$, its $i$th coordinate. Thinking of elements of $K[x_1,\dots,x_n]$ as functions on $K^n$ in the usual way, $x_i$ is the $i$th coordinate function. The ring $K[V]$ is then the $K$-algebra of functions on $V$ generat... |
2,117,481 | <p>what must be added to $x^3-6x^2+11x-8$ to make a polynomial having factor $x-3$?</p>
<p>If the required expression to be added be $K$ then $x^3-6x^2+11x-8+K$ is exactly divisible by $x-3$ but how do I find $K$??</p>
| L Parker | 410,830 | <p>Let $f(x)=x^3-6x^2+11x-8+K$</p>
<p>By factor theorem, $f(3)=0$</p>
<p>$27-54+33-8+K=0$</p>
<p>$K=2$</p>
|
1,184,963 | <p>Toss two fair dice. There are $36$ outcomes in the sample space $\Omega$, each with probability $\frac{1}{36}$. Let:</p>
<ul>
<li>$A$ be the event '$4$ on first die'.</li>
<li>$B$ be the event 'sum of numbers is $7$'.</li>
<li>$C$ be the event 'sum of numbers is $8$'.</li>
</ul>
<p>It says here $A$ and $B$ are ind... | bof | 111,012 | <p>How do you think '$4$ on the first die' affects the chances of getting a total of $7$? Do you think it increases it, or decreases it? Let's do the math.</p>
<p>$6$ on the first die: we need $1$ on the second die. One chance in $6$.<br>
$5$ on the first die: we need $2$ on the second die. One chance in $6$.<br>
$4$ ... |
1,279,829 | <p>I know that other groups like $\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_5$ are isomorphic to $\mathbb{Z}_{60}$. </p>
| Alex Fok | 223,498 | <p>There is only one order 2 non-identity $\overline{30}$ in $\mathbb{Z}_{60}$ whereas there are more than one (more precisely, 3) order 2 non-identity elements in $\mathbb{Z}_2\times\mathbb{Z}_{30}$, e.g. $(\overline{1}, 0)$ and $(0, \overline{15})$.</p>
|
1,279,829 | <p>I know that other groups like $\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_5$ are isomorphic to $\mathbb{Z}_{60}$. </p>
| Kaj Hansen | 138,538 | <p>In general, $\mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{nm} \iff \operatorname{gcd}(n, m) = 1$. </p>
<p>Even more generally, $\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \cdots \times \mathbb{Z}_{n_k} \cong \mathbb{Z}_{n_1n_2...n_k} \iff \operatorname{gcd}(n_k, n_j) = 1$ whenever $k \neq j$.</p>
<hr>
... |
1,279,829 | <p>I know that other groups like $\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_5$ are isomorphic to $\mathbb{Z}_{60}$. </p>
| Alex W | 230,729 | <p>$\mathbb{Z}_2\times\mathbb{Z}_{30}\not\cong\mathbb{Z}_{60}$ because, for example, maximum of element's orders in $\mathbb{Z}_2\times\mathbb{Z}_{30}$ not greater than 30, whereas maximum of elements orders in $\mathbb{Z}_{60}$ is 60. Indeed, let $(a,b)\in\mathbb{Z}_2\times\mathbb{Z}_{30}$. Since $a\in\mathbb{Z}_2$, t... |
1,279,829 | <p>I know that other groups like $\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_5$ are isomorphic to $\mathbb{Z}_{60}$. </p>
| Marc van Leeuwen | 18,880 | <p>It suffices to look at the subgroups of elements whose order is a power of$~2$. For $\Bbb Z_2 \times \Bbb Z_{30}$ this subgroup is $\Bbb Z_2 \times \Bbb Z_2$ while for $\Bbb Z_{60}$ is it $\Bbb Z_4$, and these subgroups $\Bbb Z_2 \times \Bbb Z_2$ and $\Bbb Z_4$ are not isomorphic.</p>
<p>In general you can decompos... |
1,125,862 | <p>Proof: There exists $a = 0$ (For every $b$, an element of the set of positive numbers, such that: $b > a$)</p>
<p>$$a + b > 0 \implies b > 0 \implies a < b.$$</p>
<p>Thus, we have shown that $0 < b$ for every $b$ that is an element of the set of positive numbers. Now, we want to show that $ab = 0 \i... | vuur | 85,875 | <p>The associated bilinear form reads
$$\begin{align}
B(x,y) &=\tfrac12 (q(x+y)-q(x)-q(y)) \\
&=(a_x-b_x)(a_y-b_y)+(b_x-c_x)(b_y-c_y)+(c_x-d_x)(c_y-d_y)
\end{align}$$
Thus a vector $v$ of the form $\begin{pmatrix}\alpha & \alpha \\ \alpha & \alpha\end{pmatrix}$ is orthogonal to every other vector... |
2,176,080 | <p>'$\Leftrightarrow$' Is very much important in this question . Actually, it seems very obvious to me.</p>
<p>We say a function is differentiable at $x=a$ iff </p>
<p>$\lim_{ h\rightarrow 0 }{ \frac { f(a+h)-f(a) }{ h } } = lim_{ h\rightarrow 0 }{ \frac { f(a-h)-f(a) }{ -h } }$</p>
<p>Now, let</p>
<p>$f'(x)=g(x)... | mathcounterexamples.net | 187,663 | <p>$f^\prime(x)$ may not even exist for $x \neq 0$.
Consider the function defined by
$$f(x)=x^2 \cdot 1_{\mathbb Q}(x) = \begin{cases}
0 & \text{ for } x \in \mathbb{Q}\\
x & \text{ for } x \notin \mathbb{Q}
\end{cases}$$</p>
<p>$f$ is differentiable at $0$ (with $f^\prime(0)=0$) but $f^\prime$ doesn't exist f... |
450,686 | <p>I am aware that to check if a linear transformation is injective, then we must simply check if the kernel of that linear transformation is the zero subspace or not. If the kernel is the zero subspace, then the linear transformation is indeed injective.</p>
<p>Is there a similar way to check for surjectivity?</p>
| A.E | 82,485 | <p>Yes, a linear transformation $T \in L(V,W)$ is surjective iff $\text{range} (T) = W$. For finite dimensional spaces, another necessary and sufficient condition is that the dimension of range$(T)=\dim (W)$.</p>
|
497,344 | <p>I would like to ask for a little help or a hint about a set theory exercise i am stuck in.</p>
<blockquote>
<p>Let $f: \mathbb{N}\rightarrow \mathbb{P}(\mathbb{N})$, $\mathbb{P}(\mathbb{N})$ is the power set of the natural numbers, be a map. Consider the subset $A\subset \mathbb{N}$ defined by $A=\{ m\in \mathbb{... | Brian M. Scott | 12,042 | <p>Yes, your ideas are correct. The only mistake is a small technical one: when you write $r\in A=\{r\in\Bbb N:r\notin f(r)\}$, the first $r$ is the specific one such that $f(r)=A$, but the rest are dummy variables. You shouldn’t use the same symbol with two different meanings in the same expression; you could change i... |
1,596,297 | <p>I must prove that $7^n-1$ $(n \in \mathbb{N})$ is divisible by $6$.</p>
<p>My "inductive step" is as follows:</p>
<p>$7^{n+1}-1 = 7\times 7^n-1 = (6+1)\times 7^n-1 = 6\times 7^n+7^n-1$</p>
<p>So now, $6\times7^n$ is divisible by 6, that's obvious. But what about the other part, the $7^n-1$ ? How do we know that i... | Sean English | 220,739 | <p>The first step of induction is a base case. This is what allows us to make further assumptions. Before we can even begin to use induction, we first have to show the statement for a specific value. In this case since we are trying to show a statement for all $n\in\mathbb{N}$, let's start with $1$. It is clear that $7... |
14,761 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://mathematica.stackexchange.com/questions/2157/customize-front-end-to-add-notifications-when-evaluation-finishes">Customize front end to add notifications when evaluation finishes?</a> </p>
</blockquote>
<p>How to setup sound alert notifica... | Andy Ross | 43 | <p>I'd do something like this.</p>
<pre><code>Pause[5];
Speak["Done Pausing for 5 Seconds"]
</code></pre>
|
13,635 | <p>I want to define</p>
<pre><code>isGood[___] = False;
isGood[#] = True & /@ list
</code></pre>
<p>where <code>list</code> is a list of several million integers. What's the fastest way of doing this?</p>
| István Zachar | 89 | <p>Are you sure you want to use <code>UpValues</code>? You can use <code>Dispatch</code> which is pretty fast when generating the lookup table and is equally fast when accessing values:</p>
<pre><code>n = 6;
list = RandomInteger[{0, 10^(n + 1)}, {10^n}];
AbsoluteTiming[disp = Dispatch@Thread[list -> True];]
</code... |
13,635 | <p>I want to define</p>
<pre><code>isGood[___] = False;
isGood[#] = True & /@ list
</code></pre>
<p>where <code>list</code> is a list of several million integers. What's the fastest way of doing this?</p>
| Rojo | 109 | <p>This seems to be faster to define downvalues</p>
<pre><code>list = RandomInteger[{-100000000, 100000000}, 1000000];
DownValues[isGood] =
HoldPattern[isGood[#]] :> True & /@ list; // AbsoluteTiming
isGood[___] = False;
</code></pre>
|
1,408,893 | <p>How can I find the derivative of $\sqrt{x}$ using first principle. Specifically I'm having difficulty expanding $\sqrt{x + h}$ or rather $(x + h)^.5$. </p>
<p>Is there any generalized formula for the expansion of non integer exponents less than 1? </p>
| Harish Chandra Rajpoot | 210,295 | <p>Notice, $$\frac{d\sqrt x}{dx}=\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$$
$$=\lim_{h\to 0}\frac{\sqrt x\left(1+\frac{h}{x}\right)^{1/2}-\sqrt{x}}{h}$$
Since, $h\to 0 \iff \frac{h}{x}\to 0$ hence using binomial expansion & neglecting higher power terms, we get
$$=\lim_{h\to 0}\frac{\sqrt x\left(1+\frac{1}{2}\fr... |
971,648 | <p>How does this $||x-x'||$ expand to the equation below? </p>
<p>$\|x-x'\|^2 = (x^T)x + (x')^T x' - 2x^T x'$</p>
| Community | -1 | <p>Using the fact that the inner product is a <strong>bilinear symmetric form</strong> we get</p>
<p>$$||x-x'||^2=\langle x-x',x-x'\rangle=\langle x,x\rangle-\langle x,x'\rangle-\langle x',x\rangle+\langle x',x'\rangle\\=\langle x,x\rangle-2\langle x,x'\rangle+\langle x',x'\rangle$$</p>
|
20,773 | <p><strong>Background</strong></p>
<p>The students in my country are supposed to be able to work with and answer questions about functions at the age of around 15. This is asserted in the standard mathematics curriculum for middle school students.</p>
<p>Personally, I think the definition of a function is extremely abs... | Daniel R. Collins | 5,563 | <p>In the U.S. Common Core standards, functions are supposed to be introduced in the 8th grade, i.e., around age 13-14. So arguably age 15 is a year or two behind where they ought to be.</p>
<p>The standard for the 8th grade says:</p>
<blockquote>
<p>Understand that a function is a rule that assigns to each input
exact... |
20,773 | <p><strong>Background</strong></p>
<p>The students in my country are supposed to be able to work with and answer questions about functions at the age of around 15. This is asserted in the standard mathematics curriculum for middle school students.</p>
<p>Personally, I think the definition of a function is extremely abs... | Beefster | 15,559 | <p>If 10-11 year olds can learn programming (which some have done even before Scratch was a thing), then it's hardly a leap at all to suggest that 15 year olds can learn functions since they are a common element of programming languages.</p>
<p>Every student is going to be different and some are going to be better at m... |
3,197,362 | <p>this is a proof by contradiction
let y and z be least upper bounds of a set A, such that y != z
so, according to a theorem, L - ε < x,for all x in A. where L is the least upper bound and ε is a positive real number.
so my proof goes like this
according to that theorem, we have</p>
<p>1) y - ε < x, for all ... | Community | -1 | <p><strong>The proof is easier.</strong></p>
<p>Let <span class="math-container">$y$</span> and <span class="math-container">$z$</span> be two distinct least upper bounds. Assume <span class="math-container">$y<z$</span> (otherwise swap them). But as <span class="math-container">$z$</span> is least, for any upper b... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.