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 |
|---|---|---|---|---|
742,577 | <p>It's been a while since I've had to do math and I've been stuck around a problem for two good hours. I hate asking questions but I can't figure it out. </p>
<p>I have the following problem: </p>
<p>Find the zero/domain of </p>
<p>$$ f(x) = \frac{9x^3-4x}{(x-3)(x^2-2x+1)} $$</p>
<p>So far, I've been able to find ... | TonyK | 1,508 | <p>Surface area of a sphere = $4\pi r^2$. Surface area of a die = $24r^2$. Therefore $\pi = 6$. </p>
<p>Or: Volume of a sphere = $\frac43 \pi r^3$. Volume of a die = $8r^3$. Therefore $\pi=6$. </p>
<p>Looks watertight to me.</p>
|
742,577 | <p>It's been a while since I've had to do math and I've been stuck around a problem for two good hours. I hate asking questions but I can't figure it out. </p>
<p>I have the following problem: </p>
<p>Find the zero/domain of </p>
<p>$$ f(x) = \frac{9x^3-4x}{(x-3)(x^2-2x+1)} $$</p>
<p>So far, I've been able to find ... | user114438 | 114,438 | <p>You can do it using simple geometry like someone said. Take a square of sides R and cut it into smaller equalsized squares, preferably the size of the dice. So it's advisable that R is a multiple of the length of a side of the dice. Then stick the squares randomly in a circle on radius R, without any overlapping. Wh... |
3,173,122 | <p>I'm trying to get the set of solutions of the following equation, whose unknowns are the rational functions <span class="math-container">$f$</span> and <span class="math-container">$g$</span> :</p>
<p><span class="math-container">$\forall x\in\mathbb{R}$</span> such that the LHS and RHS are both defined, <span class... | Ivan Neretin | 269,518 | <p>A fairly broad (sub?)set of solutions is given by the following obvious formula (<strong>updated</strong>): <span class="math-container">$$f(x)={R(x+1)\over R(x)},\\[4ex]
g(x)={R(x)\cdot R(x+1)}
$$</span>
where <span class="math-container">$R(x)$</span> is an arbitrary rational function.</p>
<blockquote class="spoi... |
3,173,122 | <p>I'm trying to get the set of solutions of the following equation, whose unknowns are the rational functions <span class="math-container">$f$</span> and <span class="math-container">$g$</span> :</p>
<p><span class="math-container">$\forall x\in\mathbb{R}$</span> such that the LHS and RHS are both defined, <span class... | Helmut | 540,853 | <p>Edit: The answer has been modified following a comment of Harmonic Sun and to include a second, simpler proof.</p>
<p>Any solution is essentially of the form in Ivan Neretin's answer:
<span class="math-container">$$ f(x)=\pm\frac{R(x+1)}{R(x)},g(x)=\pm R(x+1)R(x), $$</span>
where <span class="math-container">$R(x)... |
1,068,068 | <p>Can someone please explain to me how I am supposed to approach this question:</p>
<p>If $f: [0,1] \to \mathbb{ R}$ is continuous, and has only rational values, then $f$ must be a constant.</p>
| Alex Ravsky | 71,850 | <p>Hint: a continuous image of a connected space should be connected.</p>
|
283,077 | <p>For a $4$-digit string, where each digit can be from $\{0,1,..,9\}$, what is the probability that the number contains exactly two different digits (e.g. $3555$)?</p>
<p>The answer is $\binom{10}{2}\cdot (2^4-2)/10^4$. I couldn't get anything close to that answer. I'm guessing I'm under-counting, but I don't see wha... | TMM | 11,176 | <p>To create a four-digit string with two different digits, you first need to pick two digits, and then decide how to choose each of your $4$ digits. The number of ways to choose two digits is exactly
$${10 \choose 2}.$$
Now, to choose which digits to put on each of the four positions, you basically have two options ... |
3,007,962 | <blockquote>
<p>Prove
<span class="math-container">$$
\lim_{n \to \infty}\frac{q^n}{n} = 0
$$</span> for <span class="math-container">$|q| < 1$</span> using <span class="math-container">$\epsilon$</span> definition.</p>
</blockquote>
<p>Using the definition of a limit:</p>
<p><span class="math-container">$$
\... | José Carlos Santos | 446,262 | <p>Your proof is almost correct, but there is a small problem concerning the inequality <span class="math-container">$\dfrac1{n^2t}<\dfrac1n$</span>. This is equivalent to <span class="math-container">$nt>1$</span>. Why would that be true? All you know about <span class="math-container">$t$</span> is that <span c... |
786,239 | <p>Can you find two functions $f$ and $g$ defined on a closed interval $[a, b]$, with real values, such that $\exists (x_n) $ an infinite sequence of distinct points in $[a, b] $ such that $$\forall n, f(x_n) =g(x_n) $$ but $$f\neq g$$</p>
<p><strong>EDIT</strong>: the question is not interesting as it is stated. I ... | nadia-liza | 113,971 | <p>$$f(x,y)=f(x,y+0)=f(0+x,0)+yx=f(0,0)+x\cdot0+yx=K+xy$$</p>
|
2,652,562 | <p><a href="https://i.stack.imgur.com/Okrbw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Okrbw.jpg" alt="enter image description here"></a></p>
<p>When is $\ln x = x^{\frac13}$?</p>
<p>Is there any way to find this solution that isn't just an estimate? </p>
<p>--</p>
<p>Wolfram says it has to ... | robjohn | 13,854 | <p>Similar to Argon's answer, though the path is a bit different</p>
<p>If $\log(x)=x^{1/3}$, then
$$
x^{-1/3}\log\left(x^{-1/3}\right)=-\frac13
$$
Thus, if we let $u=\log\left(x^{-1/3}\right)$, then
$$
u\,e^u=-\frac13
$$
Therefore,
$$
x=e^{-3u}=-27u^3=-27\operatorname{W}\!\left(-\frac13\right)^3
$$
There are two real... |
766,272 | <p>I am having a problem with this assignment. So the task says:</p>
<pre><code>Show that every planar graph with at least 4 vertices has at least 4
vertices of degree less than or equal to 5.
</code></pre>
<p>I know about the theorem stating that every planar graph has to have at least one vertex of degree 5 or les... | Perry Elliott-Iverson | 125,899 | <p>Consider a maximally planar graph $G$ - that is, $G+e$ is not planar for any $e \not \in E(G)$. Note that proving that a maximally planar graph has at least 4 vertices of degree 5 or less is sufficient to prove that any planar graph has at least 4 vertices of degree 5 or less, since we could successively add edges t... |
11,178 | <p>I have the trigonometric equation
\begin{equation*}
\sin^8 x + 2\cos^8 x -\dfrac{1}{2}\cos^2 2x + 4\sin^2 x= 0.
\end{equation*}
By putting $t = \cos 2x$, I have
\begin{equation*}
\dfrac{3}{16} t^4+ \dfrac{1}{4}t^3 + \dfrac{5}{8}t^2 -\dfrac{7}{4}t + \dfrac{35}{16} = 0.
\end{equation*}
How do I tell Mathematica to ... | Vitaliy Kaurov | 13 | <p>One way to do this is:</p>
<pre><code>Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0 /.
Solve[t == Cos[2 x], x] //FullSimplify // Expand // Union // Column // TraditionalForm
</code></pre>
<p><img src="https://i.stack.imgur.com/pKMpF.png" alt="enter image description here"></p>
<p>It gives exactly yo... |
1,648,119 | <p>Let $(X_n)$ be a sequence of positive valued rvs on a probability space $(\Omega,\mathcal{F},P),$ such that $(X_n)$ converges ae to a rv $X.$ If $E(X_n)\leq c<+\infty$ for all $n$, then $X$ is integrable and $E(X)\leq c.$</p>
<p>Some thoughts: the dominated (bounded) convergence theorem can not be applied (at le... | Community | -1 | <p>(From my comment above.)</p>
<p>This follows from <a href="https://en.wikipedia.org/wiki/Fatou%27s_lemma" rel="nofollow">Fatou's lemma</a>. </p>
|
2,063,524 | <p>I am trying to prove:</p>
<p>$$\lim_{x\to c}Ax^k=Ac^k.$$</p>
<p>What I have:</p>
<p>For $k > 0$</p>
<p>For all $\varepsilon >0$ there exists $\delta>0$ such that $0<|x−c|<\delta$ implies $|Ax^k−Ac^k|<\varepsilon$</p>
<p>$|Ax^k−Ac^k|=|A||x^k−c^k|<|A||(\delta-c)^k−c^k|=k.root((((|A||\varepsil... | Mark Viola | 218,419 | <p>Note that we can write</p>
<p>$$\begin{align}
\frac{1}{1-e^t}&=\frac{1}{-t\left(1+\frac12t+\frac16t^2+O(t^3)\right)}\\\\
&=\frac{1-\frac12t+\frac{1}{12}t^2+O(t^3)}{-t}\\\\
&=-\frac1t+\frac12-\frac1{12}t+O(t^2)
\end{align}$$</p>
<p>Hence, we have</p>
<p>$$\frac{\frac{1}{1-e^2}+\frac1t-\frac12+\frac1{12... |
2,362,804 | <p>I'd like to know if there is a formula for finding integer solutions to equations of the form
$$x^{2}+1=ay$$
where $a\in\mathbb{Z}$, and $x$ and $y$ are unknowns.</p>
| Robert Israel | 8,508 | <p>Of course you need $a \ne 0$. You need $x$ to be a square root of $-1 \mod a$, and then $y = (x^2+1)/a$. For $-1$ to have a square root $\mod a$, you need $a$ to be either odd or $2$ times an odd number,
and to have no prime factors $\equiv 3 \mod 4$. Then if $a$ has $k$ distinct odd prime factors, there are $2^k... |
3,657,887 | <p>I have been doing some practice questions for an upcoming Maths Challenge. There's one question I can't seem to grasp. I'm not sure entirely sure where to start. I don't know how to approach this one. Any help would be appreciated</p>
<p><span class="math-container">$n$</span> in the form <span class="math-containe... | Angina Seng | 436,618 | <p>We know that
<span class="math-container">$$\pi=4\int_0^1\frac{dx}{1+x^2}.$$</span>
A lower Riemann sum of this integral is
<span class="math-container">$$L_n=\frac4n\sum_{k=1}^{n}\frac1{1+k^2/n^2}.$$</span>
An upper Riemann sum of this integral is
<span class="math-container">$$U_n=\frac4n\sum_{k=0}^{n-1}\frac1{1+k... |
25,163 | <p>You know the error, when you're watching a student work through an algebraic calculation to solve for a variable trapped in the argument of a function, usually <span class="math-container">$\log$</span> or a trig function, and you watch them write this:
<span class="math-container">$$
\log x = 42 \qquad\text{so}\qqu... | guest troll | 19,795 | <p>Function concept and various types of symbols and terms (e.g. log) are new to the students. It is not uncommon for weaker ones to struggle. The solution is not some secret aha revelation, not some change to convention, log(x) vice logx. But rather drill and repetition and corrective feedback for mistakes and prai... |
25,163 | <p>You know the error, when you're watching a student work through an algebraic calculation to solve for a variable trapped in the argument of a function, usually <span class="math-container">$\log$</span> or a trig function, and you watch them write this:
<span class="math-container">$$
\log x = 42 \qquad\text{so}\qqu... | Ralf Kleberhoff | 15,862 | <p>Writing from a software engineer's point of view, it's a fact that mathematics uses a notation that's highly ambiguous.</p>
<p>If you don't know that <span class="math-container">$log$</span> is used to denote some logarithm function, then seeing a term like <span class="math-container">$log x$</span> might as well ... |
348,567 | <p>This may be a dumb question, but what are the surjective maps
$$f_n:\operatorname{Sym}^n(V)\to \operatorname{Sym}^{n-2}(V),$$
where Sym$^n$ denotes the $n$-th symmetric power of $V$? Wouldn't it just be a projection? And what is the kernel of such a map (if it's not trivial)?</p>
| Qiaochu Yuan | 232 | <p>Suppose $V$ is a real vector space equipped with an inner product. Then, thinking of $\text{Sym}^n(V)$ as homogeneous polynomials of degree $n$ in $\dim V$ variables, an example of such a map which is $\text{O}(V)$-invariant is given by the <a href="http://en.wikipedia.org/wiki/Laplace_operator" rel="nofollow">Lapla... |
2,854,189 | <blockquote>
<p>If $\lfloor\log _71\rfloor+\lfloor\log _72\rfloor+\lfloor\log_73\rfloor+\dots+\lfloor\log_7N\rfloor=N$ then find $N$.</p>
</blockquote>
<p>Honestly speaking,this time i do not have any clue to move forward.
I was thinking to apply only one logarithmic rule -:</p>
<blockquote>
<p>$\lfloor\log _71\r... | Parcly Taxel | 357,390 | <p>Because the logarithms are enclosed in floors, the multiplication rule is not valid, but the question also reduces to one of whole numbers.</p>
<p>For $1\le k\le 6$, $x=\lfloor\log_7k\rfloor=0$. For $7\le k\le48$, $x=1$ and the sum of floored logarithms to this point is 42, so we continue with the numbers for which... |
145,255 | <p>I am currently working on a proof involving finding bounds for the f-vector of a simplicial $3$-complex given an $n$-element point set in $\mathbb{E}^3$, and (for a reason I won't explain) am needing to find the answer to the following embarrassingly easy (I think) question.</p>
<blockquote>
<p>What is the area o... | Isaac | 72 | <p>From <a href="http://mathworld.wolfram.com/SphericalTriangle.html" rel="noreferrer">Mathworld</a> (edited slightly):</p>
<blockquote>
<p>Let a spherical triangle have angles $A$, $B$, and $C$ (measured in radians at the vertices along the surface of the sphere) and let the sphere on which the spherical triangle s... |
253,306 | <p>Let $\beta: H^n(X, \mathbb{Z}_2)\to H^{n+1}(X, \mathbb{Z})$ be the Bockstein homomorphism. Is it possible to define a cohomology operation $f: H^{n+1}(X, \mathbb{Z})\to H^{n+k+1}(X, \mathbb{Z})$ such that
\begin{eqnarray*}\beta \text{Sq}^k=f\beta\end{eqnarray*}
where $\text{Sq}^k: H^n(X, \mathbb{Z}_2)\to H^{n+k}(X,... | Tyrone | 54,788 | <p>From the Adem relations we get $Sq^1Sq^{k-1}=(k-2)Sq^k$. It follows that for $k$ odd we have $Sq^k=Sq^1Sq^{k-1}$. Note that we have $Sq^1=\rho\beta$, where $\rho$ is mod 2 reduction, so if $k$ is odd $\beta Sq^k=\beta Sq^1Sq^{k-1}=\beta\rho\beta Sq^{k-1}=0$ since $\beta\rho\simeq \ast$. Therefore we are forced to ta... |
3,850,537 | <p>Suppose <span class="math-container">$f(x) = \textbf{x}^TA\textbf{x}$</span>. I want to verify the product rule using this. So I decided to try a set up like <span class="math-container">$\mathcal{F}(\textbf{x}) = f(\textbf{x})g(\textbf{x})$</span> where <span class="math-container">$f(\textbf{x}) = \textbf{x}^T$</s... | vb628 | 828,410 | <p>I'm going to use the Einstein summation notation to suppress sums and make things look nicer. The idea is that a product with matching upper and lower indices will be summed over with one sum per matching index. For example we could write that <span class="math-container">$$C_k^m = A_{ijk}B^{ijm} = \sum_{i=1}^n\sum_... |
3,850,537 | <p>Suppose <span class="math-container">$f(x) = \textbf{x}^TA\textbf{x}$</span>. I want to verify the product rule using this. So I decided to try a set up like <span class="math-container">$\mathcal{F}(\textbf{x}) = f(\textbf{x})g(\textbf{x})$</span> where <span class="math-container">$f(\textbf{x}) = \textbf{x}^T$</s... | copper.hat | 27,978 | <p>Note that <span class="math-container">$f(x)=x^T$</span> and <span class="math-container">$g(x) = Ax$</span> are both linear hence they are their own derivatives. Then the chain rule gives
<span class="math-container">$D(f \circ g)(x)h = Df(x)h f(x) + f(x)Dg(x)h = f(h)g(x)+f(x)g(h) = x^T(A+A^T) h$</span>.</p>
<p>I h... |
3,597,611 | <p>I know that <span class="math-container">$\{a_i\}=R(pq)$</span>, and the title is step <span class="math-container">$(b)$</span>, here is step <span class="math-container">$(a)$</span> (maybe a hint?): </p>
<p>Let <span class="math-container">$p$</span> and <span class="math-container">$q$</span> be two distinct... | Calvin Lin | 54,563 | <p><strong>Claim:</strong> <span class="math-container">$\{ a_i \} \pmod{p}$</span> (as a multi-set) consists of <span class="math-container">$ (q-1)$</span> 1's, <span class="math-container">$ (q-1)$</span> 2's, <span class="math-container">$ (q-1)$</span> 3's, <span class="math-container">$\ldots$</span>, <span class... |
2,207,862 | <p>Reading a paper, I came across the statement that there are no transversely orientable codimension-one foliations on even dimensional spheres $\mathbb{S}^{2n}$. Could anyone explain why?</p>
| Doug M | 317,162 | <p>A series is said to be "absolutely convergent" if $\sum_\limits{n=1}^{\infty} |a_n|$ converges. In order for this to happen $a_n$ must approach $0$ fast enough that the tail is effectively $0.$ The classic example would be the infinite geometric series. e.g. $\sum_\limits{n=1}^{\infty} \frac 1{2^n} = 1$</p>
<p>I... |
2,052,645 | <blockquote>
<p>Why is $\log\left(\frac{2a+2b+3}{2a+2b+1}\right)>\frac{1}{a+b+1}$ ? with $a,b\ge 0$</p>
</blockquote>
<p>If this is true (I don't know whether it is true, just inserted some values in wolfram alpha) then I can show that $\frac{1}{a+b+1}<\int_{a}^{a+1}\int_{b}^{b+1}\frac{1}{x+y}dxdy$ </p>
<p>Is... | Jason | 164,082 | <p>I was also typing this up for myself - sorry if some of the notations don't align perfectly<br>
Let <span class="math-container">$S_w,S_b \in \mathcal{S}_+^m$</span>, where <span class="math-container">$S_w$</span> is invertible, and <span class="math-container">$\text{rank}S_b=K-1 \leq m$</span>, so <span class="ma... |
2,382,224 | <blockquote>
<p>Maximize $f(x, y) = x + 2y$ with constraint $x^2 + y^2 \le 4$.</p>
</blockquote>
<p>$f(x, y)$ has no CP's so thats something gone.</p>
<p>I considered the boundary/edge of $g(x, y) = x^2 + y^2$. And I got the point $(0, 2), (2, 0)$, which gave $f(0, 2) = f(2, 0) = 4$. However, the point $(2/\sqrt(5)... | uniquesolution | 265,735 | <p>$f(x,y)=x+2y$ is extremal on the circle $x^2+y^2=4$ when the gradient of $f$ is proportional to the vector $(x,y)$, that is: $(1,2)=\lambda(x,y)$ for some $\lambda$, from which it follows that $\lambda=\sqrt{5}/2$, so $f$ is maximized at the point $(2,4)/\sqrt{5}$.</p>
<p>Another way to look at it is to use lineari... |
2,382,224 | <blockquote>
<p>Maximize $f(x, y) = x + 2y$ with constraint $x^2 + y^2 \le 4$.</p>
</blockquote>
<p>$f(x, y)$ has no CP's so thats something gone.</p>
<p>I considered the boundary/edge of $g(x, y) = x^2 + y^2$. And I got the point $(0, 2), (2, 0)$, which gave $f(0, 2) = f(2, 0) = 4$. However, the point $(2/\sqrt(5)... | Mundron Schmidt | 448,151 | <p>So you have a problem to compute the critical points of $f\mid_{S}$ where
$S=g^{-1}(4)$ with $g(x,y)=x^2+y^2$.<p>
You need to solve the equation $\nabla f(x,y)=\lambda \nabla g(x,y)$ for some $\lambda\in\mathbb R$. You get
$$
\nabla f(x,y)=\lambda\nabla g(x,y)\Leftrightarrow \begin{pmatrix}1\\2\end{pmatrix}=\lambda\... |
2,382,224 | <blockquote>
<p>Maximize $f(x, y) = x + 2y$ with constraint $x^2 + y^2 \le 4$.</p>
</blockquote>
<p>$f(x, y)$ has no CP's so thats something gone.</p>
<p>I considered the boundary/edge of $g(x, y) = x^2 + y^2$. And I got the point $(0, 2), (2, 0)$, which gave $f(0, 2) = f(2, 0) = 4$. However, the point $(2/\sqrt(5)... | Rodrigo de Azevedo | 339,790 | <p>$$\begin{array}{ll} \text{maximize} & x + 2y\\ \text{subject to} & x^2 + y^2 \leq 4\end{array}$$</p>
<p>Since the objective function is <em>linear</em> and <em>nonzero</em>, its gradient never vanishes. Thus, the maximum is attained at the boundary of the feasible region, i.e., on the circle of radius $2$ c... |
2,382,224 | <blockquote>
<p>Maximize $f(x, y) = x + 2y$ with constraint $x^2 + y^2 \le 4$.</p>
</blockquote>
<p>$f(x, y)$ has no CP's so thats something gone.</p>
<p>I considered the boundary/edge of $g(x, y) = x^2 + y^2$. And I got the point $(0, 2), (2, 0)$, which gave $f(0, 2) = f(2, 0) = 4$. However, the point $(2/\sqrt(5)... | farruhota | 425,072 | <blockquote>
<p>Max $z=x+2y$ subject to $x^2+y^2\le4$.</p>
</blockquote>
<p>The contour lines: $y=-\frac{1}{2}x+\frac{1}{2}z$, the largest value of $z$ occurs when $y$ is tangent to the constraint circle $x^2+y^2=4$ or $y=\sqrt{4-x^2}$.</p>
<p>Thus the slope of tangent line must be equal to the slope of $y=-\frac{1... |
1,036,176 | <p>I need to solve Problem 3.5 - 11 p. 164 of the book <em>Partial Differential Equations</em> by Lawrence C. Evans (2nd ed., AMS, 2010):</p>
<blockquote>
<ol start="11">
<li>Show that
<span class="math-container">$$
u(x,t) = \begin{cases}
-\dfrac{2}{3}\left(t+\sqrt{3x+t^2}\right); & \text{if } 4x ... | EditPiAf | 418,542 | <p>As suggested by @Saad, setting <span class="math-container">$t=0$</span> in <span class="math-container">$u(x,t)$</span> gives the initial data
<span class="math-container">$$
u(x,0) = g(x) =
\left\lbrace
\begin{aligned}
&{-{2}}\sqrt{{x}/{3}} & &\text{if}\quad x>0, \\
&0 & &\text{if}\quad ... |
3,343,730 | <p>There is a explicit form that admit Bernoulli numbers rationals but there is another definition where the Bernoulli numbers are <span class="math-container">$B_n$</span>, such that <span class="math-container">$\displaystyle \frac{x}{e^x-1}= \sum_{n=0}^\infty B_n \frac{x^n}{n!}$</span>. </p>
<p>How can I prove that... | J. W. Tanner | 615,567 | <p>From the definition you gave for Bernoulli numbers, <span class="math-container">$$x=\left(\sum\limits_{n=0}^{\infty} B_n\dfrac{x^n}{n!}\right)(e^x-1)=\left(\sum\limits_{n=0}^{\infty} B_n\dfrac{x^n}{n!}\right)\left(\left(\sum_{n=0}^{\infty} \dfrac{x^n}{n!}\right)-1\right)$$</span></p>
<p><span class="math-container... |
2,566,633 | <p>We say that the sequense $\{T_\alpha\}_{\alpha<\lambda}\subseteq[\omega]^\omega$ (where $\lambda$ is a ordinal) is a tower iff</p>
<ul>
<li>$\alpha<\beta<\lambda\rightarrow T_\beta\subseteq^*T_\alpha$.</li>
<li>$\neg\exists K\in[\omega]^\omega\forall\alpha<\lambda(K\subseteq^*T_\alpha)$.</li>
</ul>
<p>... | Asaf Karagila | 622 | <p>You could weaken this even more. Why require that $\lambda$ is an ordinal? Why not ask about just a chain without a lower bound, and not necessarily a well-ordered chain? The answer is that all of this doesn't matter.</p>
<p>This follows from the following observations:</p>
<ol>
<li>Every linear order has a cofina... |
3,093,712 | <p><span class="math-container">$$k^{\log_{2}5}=16$$</span> then find
<span class="math-container">$$k^{{(\log_{2}5})^2}$$</span></p>
<p>Note: the exponent(entire log) is squared unlike the value
Inside the log squared.</p>
| lab bhattacharjee | 33,337 | <p>Let <span class="math-container">$\log_25=a,\implies2^a=5$</span></p>
<p>and <span class="math-container">$k^a=16$</span></p>
<p><span class="math-container">$k^{a^2}=(k^a)^a=16^a=(2^a)^4=?$</span></p>
|
1,277,567 | <p>Q: Let R be a commutative ring with unity. Prove that if A is an ideal of R and A contains a unit, then A=R.</p>
<p>This is my attempt at an answer:
It suffices to show that all the elements in R are in A.</p>
<p>Let a be and element of A, and r be an element of R.
Since A is an ideal of R then all elements ar and... | quid | 85,306 | <p>The idea is correct. But, the details are lacking. </p>
<p>You know that $A$ contains a unit (not <em>the</em> unit element $1_R$). </p>
<p>Recall that a unit is an element $u$ that has a multiplicative inverse, that is there exists a $v$ such that $uv =1_R$. </p>
<p>So, you have such an element $u$. It might be ... |
3,626,214 | <p>Is there any simple way to know how to reverse a percentage?</p>
<p>For example if I have 100 and it goes down by 10% I end up with 90. If I then add to it by 10% I end up with 99, not the 100 that you would think of. Is there a simple trick to quickly work out the reverse of a percentage change (even if it only wo... | DonAntonio | 31,254 | <p>Because it is always true that <span class="math-container">$\;T\subset\overline T\;$</span> for any subspace <span class="math-container">$\;T\;$</span> of a topological space, since by definition</p>
<p><span class="math-container">$$\overline T:=\bigcap_{T\subset C}\left\{C\;\text{ is closed}\right\}$$</span></p... |
1,767,773 | <p>I want to find whether the expression $D = \sqrt{5t^2 - 40t+125}$ is increasing or decreasing when $t=5$.</p>
<p>My logic is I want to find whether is $f'(5)>0$ or $f'(5) < 0$.</p>
<p>I need to use the chain rule $h'(x) = g'(f(x))f'(x)$</p>
<p>$g'(f(x)) = \frac{1}{2}(5t^2 - 40t+125)^\frac{-1}{2}$</p>
<p>$f... | Roman83 | 309,360 | <p>$D = \sqrt{5t^2 - 40t+125}=\sqrt{5(t^2-8t+25)}$</p>
<p>Let $f(t)=t^2-8t+25=(t-4)^2+9$</p>
<p>If $t\ge4$ and $t_1<t_2$ then $f(t_1)<f(t_2)$</p>
|
2,034,050 | <p>I am new to Stackexchange so hope this problem is already known and a simple proof to the answer exists.</p>
<p>Flatland is a plane extending infinitely in all directions. It has an infinite number of airfields no two of which are exactly the same distance apart. </p>
<p>A training execise involves a single auropl... | Gerry Myerson | 8,269 | <p>Suppose the planes from B, C, D, E, F, and G all land at A. Suppose the distance from A to B is 1. Then C must lie on the A-side of the perpendicular bisector of AB (in order to land at A instead of at B), and it must be outside the circle of radius 1 centered at B (so B lands at A, and not at C). Draw a picture and... |
1,868,495 | <p>Consider $\lim_{ x \to -\infty} \sqrt{x^2-x+1}+x$ </p>
<p>Rationalising, one will get, $\lim_{x \to -\infty} \frac{1-x}{\sqrt{x^2-x+1}-x}$, which after taking x common and cancelling out gives $-\infty$.</p>
<p>Now, replace $x$ by $-x$, so the limit becomes, $\lim_{x \to \infty} \frac{1+x}{\sqrt{x^2+x+1}+x}$, whic... | DonAntonio | 31,254 | <p>$$\frac{1-x}{\sqrt{x^2-x+1}-x}\cdot\frac{-x}{-x}=\frac{-\frac1x+1}{\sqrt{1-\frac1x+\frac1{x^2}}+1}\xrightarrow[x\to-\infty]{}\frac1{2}$$</p>
<p>If you don't show your work, who knows what "cancelling out" gives you $\;-\infty\;$ in the limit and where your mistake is.</p>
|
1,868,495 | <p>Consider $\lim_{ x \to -\infty} \sqrt{x^2-x+1}+x$ </p>
<p>Rationalising, one will get, $\lim_{x \to -\infty} \frac{1-x}{\sqrt{x^2-x+1}-x}$, which after taking x common and cancelling out gives $-\infty$.</p>
<p>Now, replace $x$ by $-x$, so the limit becomes, $\lim_{x \to \infty} \frac{1+x}{\sqrt{x^2+x+1}+x}$, whic... | Community | -1 | <p>Your mistake in the first method is that there is no canceling. The square root is a positive number, and so is $-x$. If by "taking $x$ common" you mean pulling an $x$ factor everywhere ($x^2$ under the square root), you need to insert a minus before the square root, as $\sqrt{x^2}=-x$.</p>
<p>The change of variabl... |
5,249 | <p>Everybody knows that there are $D_n=n! \left( 1-\frac1{2!}+\frac1{3!}-\cdots+(-1)^{n}\frac1{n!} \right)$ <a href="http://en.wikipedia.org/wiki/Derangement" rel="nofollow">derangements</a> of $\{1,2,\dots,n\}$ and that there are $D_n(q)=(n)_q! \left( 1-\frac{1}{(1)_q!}+\frac1{(2)_q!}-\frac1{(3)_q!}+\cdots+(-1)^{n}\fr... | Greg Stevenson | 310 | <p>I think the main reason is the flexibility of working in the category of $R$-modules rather than just with the ring $R$. For instance suppose we stick to rings - we have some ways of building new rings like localization and taking factor rings and limited ways of "building new things" - basically linear combinations... |
5,249 | <p>Everybody knows that there are $D_n=n! \left( 1-\frac1{2!}+\frac1{3!}-\cdots+(-1)^{n}\frac1{n!} \right)$ <a href="http://en.wikipedia.org/wiki/Derangement" rel="nofollow">derangements</a> of $\{1,2,\dots,n\}$ and that there are $D_n(q)=(n)_q! \left( 1-\frac{1}{(1)_q!}+\frac1{(2)_q!}-\frac1{(3)_q!}+\cdots+(-1)^{n}\fr... | klj | 34,851 | <p>A simple example: If you have a representation of a group presented by generators and relations, you have an explicit algorithm to solve the word problem for this group. Another example: the proof of <a href="http://en.wikipedia.org/wiki/Burnside_theorem" rel="nofollow">Burnside's theorem</a>.</p>
<p>A representati... |
19,766 | <p>I know that for a series of non-negative, continuous functions $u_{n}(x)$, a sufficient condition for uniform convergence of $\sum u_{n}(x)$ to $u(x)$ is for $u(x)$ to be continuous in $I\subset \mathbb{R}$. </p>
<p>But I can't think of an example where $\sum f_{n}(x)\to f(x)$ uniformly in $I\subset \mathbb{R}$ but... | milcak | 5,105 | <p><strong>EDIT:</strong> If you want <em>discontinuous</em> functions whose sum converges uniformly to a <em>discontinuous</em> function, consider the characteristic function of the rationals: </p>
<p>$$
\lambda(x) = \left\{
\begin{array}{lr}
1 & : x \in \mathbb{Q}\\
0 & : x \notin \... |
19,766 | <p>I know that for a series of non-negative, continuous functions $u_{n}(x)$, a sufficient condition for uniform convergence of $\sum u_{n}(x)$ to $u(x)$ is for $u(x)$ to be continuous in $I\subset \mathbb{R}$. </p>
<p>But I can't think of an example where $\sum f_{n}(x)\to f(x)$ uniformly in $I\subset \mathbb{R}$ but... | Jesse Madnick | 640 | <p>Suppose your interval $I = [-1,1]$.</p>
<p>Let $u_n(x) = 2^{-n}$ for $x \in [-1,0)$ and $u_n(x) = 2^{-n} + 1$ for $x \in [0,1]$. Then $\sum u_n(x) \to u(x)$ uniformly, where $u(x) = 1$ on $[-1,0)$ and $u(x) = 2$ on $[0,1]$.</p>
<p>For general intervals $I$... well, just change the bounds.</p>
|
155,365 | <p>It is well-known that for any presheaf $F \colon \mathcal{C}^{\mathrm{op}} \to \mathrm{Set}$, the category of elements (obtained by the so-called Grothendieck construction) of $F$ is a comma category $y/\ulcorner F \urcorner$ in the category $\mathrm{CAT}$ of </p>
<p>$$\mathcal{C} \xrightarrow{y} \widehat{\mathcal{... | Emily | 130,058 | <p>Another way of phrasing the isomorphism of categories <span class="math-container">$\int^{\mathcal{C}}\mathcal{F}\congよ\downarrow[\mathcal{F}]$</span> is by saying that <span class="math-container">$\int^{\mathcal{C}}\mathcal{F}$</span> is the full subcategory of <span class="math-container">$\mathsf{PSh}(\mathcal{C... |
283,310 | <p>For $n\in\mathbb{N}^{+}$, let $c_{n}$ denote the number of simple non-isomorphic cycle matroids of graphs on $n$ vertices. That is, let</p>
<p>$$A(n)=\{M(G)\;;\;G\text{ is a graph on }n\text{ vertices}\},$$</p>
<p>and let $B(n)$ be a largest subset of $A(n)$ such that no two elements of $B(n)$ are isomorphic (as m... | Ilya Bogdanov | 17,581 | <p>It seems now that the question really boins down to the number of non-isomorphic graphs on $n$ vertices. Denote this number by $f(n)$. Clearly,
$$
2^{n\choose 2}\geq f(n)\geq \frac{2^{n\choose 2}}{n!}=\frac{2^{n\choose 2}}{n^{O(n)}},
$$
these two numbers being sufficiently close on the logarithic scale. (Moreover,... |
708,694 | <p>Find values for $a$ and $b$ so that $z=a+bi$ satisfies $\displaystyle \frac{z+i}{z+2}=i$. Below are my workings:</p>
<p>so far i simplify $\displaystyle \frac{z+i}{z+2}=i$ to $z=zi+i$ </p>
<p>which $a=i$, $b=z$ </p>
| David Mitra | 18,986 | <p>Let $A$ be a dense $G_\delta$. Then $A =\bigcap\limits_{i=1}^\infty O_i$ where each $O_i$ is open. So, $A^C=\bigcup\limits_{i=1}^\infty O_i^C $. </p>
<p>Each $O_i^C$ is closed.
Show, using the denseness of $A$, that each $O_i^C$ is nowhere dense. </p>
<p>Once you do this, it follows by definition that $A^C$ is o... |
3,481,384 | <blockquote>
<p>Find all integers <span class="math-container">$m,\ n$</span> such that both <span class="math-container">$m^2+4n$</span> and <span class="math-container">$n^2+4m$</span> are perfect squares.</p>
</blockquote>
<p>I cannot solve this, except the cases when <span class="math-container">$m=n$</span>.</p... | ViHdzP | 718,671 | <p>We will prove that <a href="https://math.stackexchange.com/a/3481451/718671">@David G.</a>’s set of solutions is complete.</p>
<p>We can suppose without loss of generality that <span class="math-container">$|m|\geq|n|$</span>, as the system of equations is symmetric in both variables. For every solution <span class=... |
17,836 | <p>Contrasting 2D and 3D in my field (Discrete & Computational Geometry<sup>1</sup>) is essential. For example, every 2D polygon can be triangulated (with vertex-to-vertex diagonals), but not every 3D polyhedron can be similarly tetrahedralized.</p>
<p>I experience significant push-back when I move to 3D
(with und... | guest | 13,427 | <p>(long comment) I think you just have to give them time. </p>
<ol>
<li><p>Solidify your algorithms and the rest of it within the 2-D world. Don't mention 3-D at all then. Lay a solid foundation on the tools themselves.</p></li>
<li><p>Then move to 3-D and start by doing some refresher on solid geometry itself (fo... |
191,738 | <p>I have the following limit:</p>
<p>$$\lim_{n\rightarrow\infty}e^{-\alpha\sqrt{n}}\sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose k}\sum_{m=0}^{n-1-k}\frac{(\alpha\sqrt{n})^m}{m!}$$</p>
<p>where $\alpha>0$.</p>
<p>Evaluating this in Mathematica suggests that this converges, but I don't know how to evaluate it. Any he... | marty cohen | 13,079 | <p>I would start even more simple-mindedly by replacing the inner sum
with its infinite $n$ value of $e^{\alpha \sqrt{n}}$.
This cancels out the outer expression, so we are left with</p>
<p>$\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose k}$.</p>
<p>Doing some manipulation,
$\sum_{k=0}^{n-1}2^{-n-k... |
4,245,883 | <p>Here's a problem from my probability textbook:</p>
<blockquote>
<p>Of three independent events the chance that the first <em>only</em> should happen is <em>a</em>; the chance of the second <em>only</em> is <span class="math-container">$b$</span>; the chance of the third <em>only</em> is <span class="math-container">... | Sandipan Dey | 370,330 | <p>This is an interesting problem and all the above solutions I think can be obtained using the law of the total expectation, with an assumption that the expected number of flips gets doubled when we have an additional tail to start with.</p>
<p>Let <span class="math-container">$X$</span> be the r.v. denoting the numbe... |
654,130 | <p>prove that if a sequence $\{a_n\}$ is decreasing and there exists a subsequence $\{a_{n_k}\}$ so that $\lim_{k\to\infty} \{a_{n_k}\} =L$ then $\lim_{n\to\infty} \{a_n\}=L$.</p>
<p>I don´t know how to do this can someone help me please.</p>
| Benjamin | 118,815 | <p>Element chasing, element chasing, element chasing.
You are right in realizing that the truth table will do little for you.
Since sets are uniquely determined by their elements, if you can show that an arbitrary element, say $x$ of one set is also in another you have shown set inclusion in one direction, show it in t... |
868,582 | <p>Suppose that $u$ and $w$ are defined as follows: </p>
<p>$u(x) = x^2 + 9$</p>
<p>$w(x) = \sqrt{x + 8}$</p>
<p>What is: </p>
<p>$(u \circ w)(8) = $</p>
<p>$(w \circ u)(8) = $</p>
<p>I missed this in math class. Any help?</p>
| Jam | 161,490 | <p>$$w\circ u(8)=\sqrt{\left(8^2+9\right)+8}=9$$
$$u\circ w(8)=\left(\sqrt{8+8} \right)^2+9=25$$
Note that as Ivo says, they aren't equal.</p>
|
868,582 | <p>Suppose that $u$ and $w$ are defined as follows: </p>
<p>$u(x) = x^2 + 9$</p>
<p>$w(x) = \sqrt{x + 8}$</p>
<p>What is: </p>
<p>$(u \circ w)(8) = $</p>
<p>$(w \circ u)(8) = $</p>
<p>I missed this in math class. Any help?</p>
| Graham Kemp | 135,106 | <p>$(u\circ w)$ is the <strong>composition</strong> of function $u$ over $w$. </p>
<p>It means: apply function $u$ to the result applying function $w$. That is all.</p>
<p>$$(u\circ w)(x) = u(w(x))$$</p>
<p>And conversely: $(w\circ u)(x) = w(u(x))$</p>
|
20,025 | <p>Let $D$ be a PID, $E$ a domain containing $D$ as a subring. Is it true that if $d$ is a gcd of $a$ and $b$ in $D$, then $d$ is also a gcd of $a$ and $b$ in $E$?</p>
| Pete L. Clark | 299 | <p>In any domain $D$, for $a,b \in D \setminus \{0\}$, if the ideal $\langle a,b \rangle_D = \{xa + yb \ | \ x,y \in D\}$ of $D$ is principal, then any generator $d$ of the ideal is a gcd of $a$ and $b$. (Note that in general gcd's are unique precisely up to units, i.e., the corresponding principal ideal is unique.)</... |
1,754,820 | <p>$\frac{n^4}{\binom{4n}{4}}$</p>
<p>$= \frac{n^4 4! (4n-4)!}{(4n)!}$</p>
<p>$= \frac{24n^4}{(4n-1)(4n-2)(4n-3)}$</p>
<p>$\rightarrow \infty$ as $n \rightarrow \infty$</p>
<p>However, the answer key says that</p>
<p>$\frac{n^4}{\binom{4n}{4}}$</p>
<p>$= \frac{6n^3}{(4n-1)(4n-2)(4n-3)}$ this is the part I don't u... | heropup | 118,193 | <p>It is worth generalizing the given limit: for positive integers $m$,</p>
<p>$$\begin{align*} a_n(m) &= \frac{n^m}{\binom{mn}{m}} \\ &= \frac{n^m m! \, (m(n-1))!}{(mn)!} \\ &= \frac{n^m m!}{(mn)(mn-1)\cdots(m(n-1)+1)} \\ &= \prod_{k=1}^m \frac{nk}{m(n-1) + k} \\ &= \prod_{k=1}^m \frac{k}{m + (k-... |
3,329,085 | <p>The only subspace of <span class="math-container">$\mathbb{R}$</span> that is homeomorphic to <span class="math-container">$\mathbb{R}$</span> and complete (with the restricted metric) is <span class="math-container">$\mathbb{R}$</span>.</p>
<p>My work-</p>
<p>Let <span class="math-container">$A$</span> be a subsp... | egreg | 62,967 | <p>If <span class="math-container">$A$</span>, with the induced metric, is a complete subspace of <span class="math-container">$\mathbb{R}$</span>, then it is closed. Since <span class="math-container">$A$</span> is by assumption is homeomorphic to <span class="math-container">$\mathbb{R}$</span>, it is an interval. A ... |
1,450,737 | <p>I think that this expression very easy, but i don't know how resolve it. Please, help me, guys. So, there is:
$$
\dfrac{\cos 3\alpha - \sin 3\alpha}{\cos \alpha + \sin \alpha}, \;\;\; \mbox{if} \;\;\; \sin \left(\dfrac{\pi}{4} - \alpha\right) = 0,1.
$$
I make following transformation:
\begin{gather}
\dfrac{\cos 3\al... | Sonner | 175,269 | <p>Notice that
$$1-2\sin 2\alpha= 1-2\cos\left(2\left(\frac \pi 4 -\alpha\right)\right)=1-2\sqrt{1-\sin^2\left(2\left(\frac \pi 4 -\alpha\right)\right)}$$</p>
|
3,274,974 | <p>I want to read Carlitz paper on arithmetic functions and he defines a 'complement' of a function <span class="math-container">$f$</span> by <span class="math-container">$f'$</span>. He defines it this way: </p>
<p>If <span class="math-container">$f$</span> is a function on the set <span class="math-container">$N$<... | cmk | 671,645 | <p>Here's a sketch:</p>
<p>If <span class="math-container">$Df(x,y)=0$</span> for all <span class="math-container">$(x,y)\in\mathbb{R}^2,$</span> then we're done. So, suppose instead that there exist <span class="math-container">$(a,b)$</span> so that <span class="math-container">$Df(a,b)\neq 0,$</span> and suppose, W... |
1,843 | <p>I don't think the <a href="https://math.stackexchange.com/questions/tagged/proof" class="post-tag" title="show questions tagged 'proof'" rel="tag">proof</a> tag conveys any useful information. Sometimes, it excuses the asker from adding a more descriptive tag, like in <a href="https://math.stackexchange.com/... | Community | -1 | <p>The <a href="https://math.stackexchange.com/questions/tagged/proof" class="post-tag" title="show questions tagged 'proof'" rel="tag">proof</a> tag <a href="http://meta.math.stackexchange.com/a/2126/856">has been removed and blacklisted</a>, so this question can be closed.</p>
|
4,389,911 | <p>everyone; I have a question to ask you about this.
I have to solve this <span class="math-container">$$\int\limits_0^1\int_\limits0^{(1-x)}(y-2x)^2\sqrt{(x+y)}\ \mathrm dy\mathrm dx$$</span> with using change of variables to UV plane. we have <span class="math-container">$u=x+y$</span> and <span class="math-containe... | Math Lover | 801,574 | <p>The original bounds are <span class="math-container">$0 \leq x + y \leq 1, x, y \geq 0$</span></p>
<p>As <span class="math-container">$u = x + y$</span>, first find the bounds of <span class="math-container">$u$</span> which is straightforward.</p>
<p>Now, <span class="math-container">$v = y - 2x = (x + y) - 3x$</sp... |
2,883,848 | <p>Find all functions $f:\Bbb {R} \rightarrow \Bbb {R} $ such that:
$$f(f(xy-x))+f(x+y)=yf(x)+f(y)$$</p>
<p>Source: <a href="https://artofproblemsolving.com/community/c6h1532563" rel="nofollow noreferrer">2018 Hong Kong TST 2 problem 3</a></p>
<hr>
<p>I recently proved that $f(x+1)=f(x)+f(1).$</p>
| nonuser | 463,553 | <p>If we put <span class="math-container">$x=0$</span> then we get <span class="math-container">$$f(f(0)) = yf(0)\quad \forall y \implies f(0)=0$$</span></p>
<p>If we put <span class="math-container">$y= 0$</span> we get <span class="math-container">$$ f(f(-x)) =-f(x)\implies \boxed{f(f(x))= -f(-x)}$$</span></p>
<p>If ... |
1,118,458 | <p>A fairly pretty technique of showing that<br>
$$\int_{-\infty}^{+\infty}e^{-x^2}dx = \sqrt{\pi}$$ is to square the integral, writing that square as the product of two integrals with integration variables $x$, and $y$, treating that as an integral over the whole plane $\Bbb{R}^2$, and then changing to polar coordina... | Predrag Punosevac | 32,077 | <p>IIRC the textbook we used for the Analysis II course in Belgrade 25 years ago (roughly corresponding to honour Calc III and non-existing Calc IV in U.S.) refereed to that integral and technique you described as Poisson integral. Siméon Denis Poisson and Carl Friedrich Gauss are contemporaries as you probably already... |
737,250 | <p>Need help! I'm having some problem with understanding this equation! We have a similiar example in the book, but I dont really get what they mean.</p>
<p>So here is the question. Given 3u = 1 (mod 5), find u?</p>
<p>I mean, as far as I understand, 3 = 3 (mod 5), right? So how about "u"?</p>
<p>Could you help me w... | Mark Bennet | 2,906 | <p>$1$ mod $5$ means $\dots -9, -4, 1, 6, 11, 16, 21 \dots$</p>
<p>To get divisibility by $3$ you choose the representatives $\dots -9, 6, 21 \dots$</p>
<p>Divide by $3$ to get $\dots -3, 2, 7 \dots$</p>
<p>Notice that these are all equivalent to $2$ mod $5$</p>
<p>Can you use the pattern to prove this?</p>
|
2,754,870 | <p>I've been studying naive set theory and I have been told that Russell's paradox causes problems in Cantor's set theory when sets get "too big". </p>
<p>I don't understand why this causes a problem. I know how the paradox effected Frege's work in terms of logic but not Cantor. </p>
<p>Any help will be appreciated, ... | Bram28 | 256,001 | <p>Maybe they meant that due to Russell's paradox there cannot be a 'universal set': a set of 'everything'.</p>
<p>Here's why: Assuming there would be a set of everything $U$, then we can consider the set $D$ defined as $\{ x \in U | x \not \in x \}$. We could thus say that $D$ contains all 'normal' sets, where a 'no... |
3,663,019 | <blockquote>
<p>If <span class="math-container">$a, b$</span> and <span class="math-container">$c$</span> (all distinct) are the sides of a triangle ABC opposite to the angles <span class="math-container">$A, B$</span> and <span class="math-container">$C$</span>, respectively, then <span class="math-container">$\frac... | user | 293,846 | <p>One can easily show that
<span class="math-container">$$
\frac{c\sin(A-B)}{a^2-b^2}=\frac1{2R},\tag1
$$</span>
where <span class="math-container">$R$</span> is the radius of the circumscribed circle. </p>
<p>Indeed substituting in LHS of (1)
<span class="math-container">$$a=2R\sin A,\quad b=2R\sin B,\quad c=2R\si... |
3,663,019 | <blockquote>
<p>If <span class="math-container">$a, b$</span> and <span class="math-container">$c$</span> (all distinct) are the sides of a triangle ABC opposite to the angles <span class="math-container">$A, B$</span> and <span class="math-container">$C$</span>, respectively, then <span class="math-container">$\frac... | Yuta | 590,171 | <p>Assume that <span class="math-container">$a>b$</span>.</p>
<p>There exists a point <span class="math-container">$D$</span> on line segment <span class="math-container">$AB$</span> such that <span class="math-container">$CD=b$</span>, Produce <span class="math-container">$CD$</span> to meet the circumscribed circ... |
3,663,019 | <blockquote>
<p>If <span class="math-container">$a, b$</span> and <span class="math-container">$c$</span> (all distinct) are the sides of a triangle ABC opposite to the angles <span class="math-container">$A, B$</span> and <span class="math-container">$C$</span>, respectively, then <span class="math-container">$\frac... | mathlove | 78,967 | <p>On OP's request, I am converting my comment into an answer.</p>
<p>If you want to put values of angles and sides, you can use
<span class="math-container">$$a=\sqrt 3,\quad b=1,\quad c=2,\quad A=60^\circ,\quad B=30^\circ,\quad C=90^\circ$$</span>
for which we have
<span class="math-container">$$\frac{c\sin(A-B)}{a... |
3,663,019 | <blockquote>
<p>If <span class="math-container">$a, b$</span> and <span class="math-container">$c$</span> (all distinct) are the sides of a triangle ABC opposite to the angles <span class="math-container">$A, B$</span> and <span class="math-container">$C$</span>, respectively, then <span class="math-container">$\frac... | Quanto | 686,284 | <p>Use sine and cosine rules to evaluate </p>
<p><span class="math-container">\begin{align}
& \frac{\sin(A-B)}{\sin(A-C)}
= \frac{\sin A\cos B - \cos A \sin B}{\sin A\cos C - \cos A \sin C}\\
=& \frac{a\frac{a^2+c^2-b^2}{2ac} - b\frac{b^2+c^2-a^2}{2bc} }
{a\frac{a^2+b^2-c^2}{2ab} - c\frac{b^2+c^2-a^2}{2bc} }
... |
126,914 | <p>It is probably a stupid question, but I can't figure out how to generate a mesh from picture like this</p>
<pre><code>Graphics[{{EdgeForm[Thin], FaceForm[LightGray],
Rectangle[{-3, -2}, {3, 2}]}, {Circle[{0, 0}, 1.]}, {EdgeForm[
Thin], FaceForm[White], Disk[{0, 0}, .5]}}]
</code></pre>
<p><a href="https://... | Simon Woods | 862 | <p>I think this is a good question. Here's an example showing what I believe is the desired result, but using a laborious method specifying the boundary coordinates and line elements explicitly.</p>
<pre><code>Needs["NDSolve`FEM`"]
rectanglepoints = {{-3, -2}, {-3, 2}, {3, 2}, {3, -2}};
circlepoints = CirclePoints[1,... |
126,914 | <p>It is probably a stupid question, but I can't figure out how to generate a mesh from picture like this</p>
<pre><code>Graphics[{{EdgeForm[Thin], FaceForm[LightGray],
Rectangle[{-3, -2}, {3, 2}]}, {Circle[{0, 0}, 1.]}, {EdgeForm[
Thin], FaceForm[White], Disk[{0, 0}, .5]}}]
</code></pre>
<p><a href="https://... | Michael E2 | 4,999 | <p>I once helped someone with something similar from a micrograph.</p>
<pre><code>img = Import["http://i.stack.imgur.com/B8B1D.jpg"];
cleaned = Thinning@ColorNegate@Binarize[img, 0.7];
imgComp = MorphologicalComponents[cleaned];
imgComp // Colorize
</code></pre>
<p><img src="https://i.stack.imgur.com/fJmt0.png" alt=... |
1,798,687 | <blockquote>
<p>Suppose $X$ is a continuous random variable with PDF: $$\begin{cases}
e^{-(x-c)}\ \ \text{when }x > c \\ 0\ \quad \quad\text{when}\ x \leq c
\end{cases}$$</p>
<p>a. Find $\mathbb{E}(X)$<br>
b. Find $\mathbb{E}(X-c)$<br>
c. Find $\mathbb{E}(X/c)$</p>
</blockquote>
<p>I know how to calcul... | Graham Kemp | 135,106 | <p>In general, the Linearity of Expectation says that for any constants $a,b$ and random variable $X$, then: $$\mathsf E(aX+b) ~=~ a~\mathsf E(X)+b$$</p>
<p>If you can find $\mathsf E(X)$, then you can find $\mathsf E(X-c)$ and $\mathsf E(\tfrac 1 c X)$</p>
<blockquote class="spoiler">
<p> $$\begin{align} \mathsf E... |
2,723,763 | <p>The integral is </p>
<p>$$\int^3_1\sqrt{16-x^2}dx$$</p>
<p>I've used the trig substitution method, replacing $x$ with $4\sin\theta$:</p>
<p>$$x=4\sin\theta, \quad \theta=\arcsin\left(\frac x4\right), \quad dx=4\cos\theta \ d\theta$$
(I've excluded the intervals for the definite integral for now)
\begin{align}
I&a... | user284331 | 284,331 | <p>$\sin(2\sin^{-1}(3/4))=2\sin(\sin^{-1}(3/4))\cos(\sin^{-1}(3/4))=2\cdot(3/4)\cdot(\sqrt{7}/4)$ because of the identity that $\cos(\sin^{-1}(3/4))=\sqrt{1-\sin^{2}(\sin^{-1}(3/4))}=\sqrt{1-9/16}=\sqrt{7}/4$.</p>
|
2,723,763 | <p>The integral is </p>
<p>$$\int^3_1\sqrt{16-x^2}dx$$</p>
<p>I've used the trig substitution method, replacing $x$ with $4\sin\theta$:</p>
<p>$$x=4\sin\theta, \quad \theta=\arcsin\left(\frac x4\right), \quad dx=4\cos\theta \ d\theta$$
(I've excluded the intervals for the definite integral for now)
\begin{align}
I&a... | Mehrdad Zandigohar | 521,055 | <p>Hint: </p>
<p>$$\sin(2\alpha)=2sin\alpha cos\alpha$$ where here $\alpha=\arcsin(1/4),\arcsin(1/4)$</p>
<p>and </p>
<p>$$\cos(\arcsin\beta)=\sqrt{1-\beta^2},\sin(\arcsin\beta)=\beta$$</p>
<p>where $\beta=1/4,3/4$</p>
|
1,397,779 | <p>To solve $-3x^2 +2x +1=0$, I'd normally break the middle term and then factorise. But I was wondering if there was a way to skip the factorising step? The factors I'd use in place of the middle term would be $3$ and $-1$. If I were to flip the sign of each, I would get $-3$ and $1$. And then, were I to divide by the... | DeepSea | 101,504 | <p>Observe $a+b+c=-3+2+1=0 \Rightarrow 1$ is a root, thus $x-1$ is a factor of $p(x)=-3x^2+2x+1$, and since its a quadratic, you can guess the other factor: $-3x-1$.</p>
|
1,397,779 | <p>To solve $-3x^2 +2x +1=0$, I'd normally break the middle term and then factorise. But I was wondering if there was a way to skip the factorising step? The factors I'd use in place of the middle term would be $3$ and $-1$. If I were to flip the sign of each, I would get $-3$ and $1$. And then, were I to divide by the... | Mike | 17,976 | <p>As Paul Sinclair brings up in his answer, by the Rational Root Theorem, it is only necessary to check $\pm1$ and $\pm\frac13$. You can use the theorem whenever all coefficients are integers, though of course if all coefficients are rational, you can always just multiply the entire equation by the lowest common deno... |
2,141,153 | <p>Let A=$\left[ \begin {array}{cc} 0&-1\\ 1&0\end {array}
\right] $</p>
<p>$L:Mat_2\mathbb{R}\rightarrow Mat_2\mathbb{R}$</p>
<p>$B\rightarrow AB-BA$</p>
<p>(1) Show that it is a linear transformation.</p>
<p>I really have no idea how to this, expect I know that:</p>
<p>$L(u+w)=L(u)+L(w)$</p>
<p>$\alpha... | Giulio | 401,660 | <p>Let $B,C \in Mat_2\Bbb R$, $\alpha,\beta \in\Bbb R$.</p>
<p>$\begin{align}L(\alpha B + \beta C)& = A(\alpha B + \beta C)-(\alpha B + \beta C)A\\
& = \alpha AB + \beta AC - \alpha BA - \beta CA \\
& = (\alpha AB-\alpha BA) + (\beta AC - \beta CA) \\
& = \alpha(AB-BA) + \beta (AC-CA) \\
& = \alpha... |
2,589,143 | <p>Let $(a_n)_n$ be a Cauchy sequence in $\mathbb C$. I want to show</p>
<blockquote>
<p>$(a_n)_n$ is Cauchy $\iff$ the real part $(\Re(a_n))_n$ and the imaginary part $(\Im(a_n))_n$ are both Cauchy</p>
</blockquote>
| amd | 265,466 | <p>You’ve substituted for $a$ and $b$ in your formula for the image of a point, but not for $c$. Then, for some reason you didn’t multiply this extraneous $c$ (which should be $1$) by $b$ or $a$ in the expressions that you derived for $bx_2$ and $ay_2$. </p>
<p>You should have $x_2+k+1=y_2+h+1=0$, from which $bx_2+ay... |
4,309,551 | <p>Circle formula that touches <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> axes and passes <span class="math-container">$P = (2, -1)$</span>.</p>
<p>Firstly I tried to solve the problem using perpendicular bisector conjunctions. P.B of <span class="math-container">$XP$</span> and... | Shooting Stars | 964,104 | <p>Alternatively, remember that the <span class="math-container">$x$</span> and <span class="math-container">$y$</span> intercepts can be given by:</p>
<p><span class="math-container">$y = \pm\sqrt{r^2-h^2} + k$</span></p>
<p><span class="math-container">$x = \pm\sqrt{r^2-k^2} + h$</span></p>
<p>Because the constraint ... |
2,249,320 | <p>We have to find </p>
<blockquote>
<p>$$g(x)=\cos{x}+\cos{3x}+\cos{5x}+\cdots+\cos{(2n-1)x}$$</p>
</blockquote>
<p>I could not get any good idea .</p>
<p>Intialy I thought of using </p>
<p>$$\cos a+\cos b=2\cos(a+b)/2\cos (a-b)/2$$</p>
| Frenzy Li | 32,803 | <p>Following @TheDeadLegend's answer I found this telescoping technique. Turns out that you need a similar identity to make it work:</p>
<p>$$ \sin(\alpha + \beta) - \sin(\alpha - \beta) =2\cos \alpha \sin \beta$$</p>
<hr>
<p>\begin{align}
g(x) &= \sum_{k=1}^n \cos(2k-1)x \\
&= \frac{1}{2\sin x}\sum_{k=1}^n ... |
3,897,460 | <p>If <span class="math-container">$F$</span> has characteristic <span class="math-container">$P>0$</span> prove that <span class="math-container">$$A=\begin{pmatrix}1&\alpha\\0&1\end{pmatrix}$$</span> satisfies <span class="math-container">$A^p=I$</span>.</p>
<p>I am trying to solve this problem but I could... | Antonio Ficarra | 589,812 | <p>You can check that
<span class="math-container">$$
A^2=\left(\begin{matrix}
1&2\alpha\\ 0&1
\end{matrix}\right).
$$</span>
By induction
<span class="math-container">$$
A^p=\left(\begin{matrix}
1&p\alpha\\ 0&1
\end{matrix}\right),
$$</span>
but <span class="math-container">$p\alpha=0$</span> for <span... |
3,328,397 | <p>I want to know how many combinations of three dice given as a product an even number. If these dice are different from each other the answer would be <span class="math-container">$6^3- 3^3$</span>. What about if these dice are equal to one another? Like having <span class="math-container">$2, 4, 6$</span> is the sam... | saulspatz | 235,128 | <p>First you need to know how many combinations you can get from three dice if the order doesn't matter. This is what Steven Stadnicki meant in his comment.</p>
<p>There are <span class="math-container">$6$</span> ways if all three numbers are the same.
If two of the numbers are the same, and the other is different, ... |
1,553,754 | <p>There is a highly believable theorem:</p>
<p>Let $A, B$ be disjoint sets of generators and let $F(A), F(B)$ be the corresponding free groups. Let $R_1 \subset F(A)$, $R_2 \subset F(B)$ be sets of relations and consider the quotient groups $\langle A | R_1 \rangle$ and $\langle B | R_2 \rangle$.</p>
<p>Then $\lang... | jgon | 90,543 | <p>This an old question that recently got bumped, but there's another nice proof that uses the fact that colimits commute with one another and left adjoints, so I thought I'd add it for potential future visitors.</p>
<p>Let <span class="math-container">$G_i =\langle X_i\mid R_i\rangle$</span> for <span class="math-cont... |
2,910,243 | <p>How to prove that $\sin(\sqrt{x})$ is not periodic?
THe definition of a periodic function is $f(x+P)=f(x)$. </p>
<p>So I assume that $\sin(\sqrt{x+P})=\sin(\sqrt{x})$. This is equivalent to $\sin(\sqrt{x+P})-\sin(\sqrt{x})=0$. This implies $2cos(\frac{\sqrt{x+P}+\sqrt{x}}{2})\sin(\frac{\sqrt{x+P}-\sqrt{x}}{2})$. Wh... | Mohammad Riazi-Kermani | 514,496 | <p>$$\sin(\sqrt{x+P})=\sin(\sqrt{x})$$</p>
<p>$$\implies \sqrt{x+P}=\sqrt{x}+2k\pi \text { or }\sqrt {x+P}=2k\pi+ \pi - \sqrt{x} $$</p>
<p>Upon squaring we get $$ x+P = x+4 k^2\pi ^2 +4k\pi \sqrt x $$</p>
<p>or $$x+P =((2k+1) \pi) ^2 + x-2(2k+1)\pi \sqrt x$$</p>
<p>Note that neither of the above holds for a const... |
229,286 | <p>Which of the following groups are cyclic? For each cyclic group, list all the generators of the group.
$$G_1 = \langle \mathbb{Z},+\rangle\;\;G_2 = \langle\mathbb{Q}, +\rangle\;\;G_3=\langle\mathbb{Q}^+, \cdot\rangle\;\;G_4 = \langle 6\mathbb{Z}, +\rangle$$
$$G_5 = \{6^n \mid n\in\mathbb{Z}\} \text{ under multiplica... | Brian M. Scott | 12,042 | <p>HINT: For $G_2$, show that if $0\ne q\in\Bbb Q$, the group generated by $q$ does not contain $q/2$. </p>
<p>For $G_3$ you can use the same idea: show that if $0<q\in\Bbb Q$, there is some $r\in\Bbb Q^+$ that is not in $\langle q\rangle$, the group generated by $q$. First work out just what <strong>is</strong> in... |
1,359,098 | <p>Let $x_1=c$ and $x_{n+1}=\frac{x_n^2 + \frac{a}{b}}{2x_n}$. How prove $\lim x_n = \sqrt{\frac{a}{b}}$ if $b \neq 0$ ?</p>
| Chiranjeev_Kumar | 171,345 | <p>Let $v=\sqrt{\frac{a}{b}},b\gt0,x_1=c\gt 0$ Now we first show that $x_n^2\ge v$ for $n\ge2$,</p>
<p>since $x_n$ satisfied the quadratic equation $$x_n^2-2x_nx_{n+1}+v=0$$ this equation has a real root. Hence the discriminant $$4x_{n+1}^2-4v$$ must be non-negative, that is ,$x_{n+1}^2\ge v$ for $n\ge 1$</p>
... |
2,618,524 | <p>I managed to prove this using a direct proof but my prof suggested I try proving it using the contrapositive. Here's what I have so far:</p>
<p>Contrapositive: $(x \le 0) \lor (x \ge 1) \Rightarrow x^4 + 2x^2 - 2x \ge 0$</p>
<p>Splitting this into two, ($P_1 \Rightarrow Q)\land(P_2 \Rightarrow Q)$:</p>
<p>$$x \ge... | Kcronix | 322,722 | <p>I'd like to offer an alternative explanation as to why the divergence is equal to the sum of the partial derivatives of <span class="math-container">$M(x,y)$</span> and <span class="math-container">$N(x,y)$</span> in the <span class="math-container">$x$</span> and <span class="math-container">$y$</span> directions r... |
1,987,685 | <p>How do I find the following sum?
$\sum_{k=0}^n(-1)^k{2n\choose2k}$</p>
<p>Tried to simplify it somehow but got nothing less complicated.</p>
| Tom | 103,715 | <p><b>Hint</b>:</p>
<p>\begin{align}
\sum\limits_{k=0}^{n} {{2n}\choose{2k}} (-1)^k &= \text{ even terms of } \sum\limits_{k=0}^{2n} {{2n}\choose{k}} i^k \\
&= \text{ real part of } \sum\limits_{k=0}^{2n} {{2n}\choose{k}} i^k \\
&= \text{ real part of } (1+i)^{2n}
\end{align}</p>
|
1,632,006 | <p>The book <em>Discrete Mathematics</em> by Kenneth A. Ross says: "Let's verify the logic implication $(p\wedge q) \implies p$. For that, we need to consider only just the case when $p\wedge q$ is true; <em>i.e.</em>, both, $p$ and $q$, are true. This gives us the truncated table:"</p>
<p><img src="https://i.stack.im... | OKPALA MMADUABUCHI | 98,218 | <p>The relation $p\rightarrow r$ is logically equivalent to $\neg p\vee r$. So for the case where $p$ is false, $\neg p$ is true and the outcome is then true. So the statement would be false only if $p$ is true and $r$ is false. Now replace this $p$ with $p\wedge q$ and $r$ by $p$ and you have a tautology.</p>
|
1,632,006 | <p>The book <em>Discrete Mathematics</em> by Kenneth A. Ross says: "Let's verify the logic implication $(p\wedge q) \implies p$. For that, we need to consider only just the case when $p\wedge q$ is true; <em>i.e.</em>, both, $p$ and $q$, are true. This gives us the truncated table:"</p>
<p><img src="https://i.stack.im... | hardmath | 3,111 | <p>Recall that an implication $R\implies S$ is "automatically true" in cases where the hypothesis $R$ is <em>false</em>. So the only interesting cases to check are those for which the hypothesis is <em>true</em>.</p>
<p>In this case the hypothesis is $p \wedge q$, so you can proceed just with analyzing what would be ... |
1,224,268 | <p>I was given this problem (I have to prove) and not sure if I use fermat's theorem
$24^{31} ≡ 23^{32} (mod 19)$</p>
<p>If I do use fermat's is this right:</p>
<p>I would do the LHS first:</p>
<p>$24^{18}·24^{13} ≡ 1·24·24^{12}$</p>
<p>RHS:</p>
<p>$23^{18}·23^{14} ≡ 1·23^{14}$</p>
<p>I am not sure where to from... | Elaqqad | 204,937 | <p>As $126$ is equal to $18\cdot 7$ and $19-1=18$ we have:
$$\color{#0a0}{23\equiv 2^2}\,\Rightarrow\,\color{#0a0}{23}^{31+32}\equiv \color{#0a0}{23}^{63}\equiv (\color{#0a0}{2^2})^{63}\equiv 2^{2\cdot 9\cdot 7}\equiv (\color{#c00}{2^{18}})^7\equiv \color{#c00}1^7\equiv 1\mod 19 $$</p>
<p>Now we multiply both side by ... |
2,454,663 | <blockquote>
<p>Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$</p>
</blockquote>
<p>What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because of parity of the integrand (nothing else worked for me): $$2\int_0^\infty \frac{x^4 +... | Carl Schildkraut | 253,966 | <p>Hint: The difference of two squares $a^2-b^2$ can be factored as $(a-b)(a+b)$. A sum of two squares $a^2+b^2$ can be thought of as $a^2-(ib)^2$...</p>
|
3,324,817 | <p>Angles <span class="math-container">$(\psi -\theta) $</span> relation is <em>unique</em> for Conics in a 3D situation. </p>
<p>// It is to seek <span class="math-container">$r- \psi$</span> of planar sections aka Conics on Gauss curvature <span class="math-container">$K=0$</span> cones for uniqueness and generality... | Intelligenti pauca | 255,730 | <p>WARNING: this solution refers only to the case when the plane is perpendicular to the base of the cone. See Blue's answer for the general case.</p>
<p>Let <span class="math-container">$G'$</span> be the projection of <span class="math-container">$G$</span> (vertex of the hyperbola) on the axis of the cone, <span cl... |
532,820 | <p>I've been stuck on this problem for a few hours now and haven't been able to make progress. The original problem is below with my work and progress beneath it. </p>
<blockquote>
<p>$$\frac{dy}{dx}=-\frac{y(x^3-y^3)}{x(2y^3-x^3)}$$</p>
</blockquote>
<p>First:<br>
Let $y=ux$ then $\frac{dy}{dx}=\frac{du}{dx}x + u... | Amzoti | 38,839 | <p>You were doing fine, but somehow something went South in the simplification.</p>
<p>We have:</p>
<p>$$\dfrac{dy}{dx} = -\dfrac{y}{x} \dfrac{x^3-y^3}{2y^3-x^3}$$</p>
<p>We make the substitution:</p>
<p>$$y = vx \rightarrow \dfrac{dy}{dx} = v + xv'$$</p>
<p>When we substitute, we get:</p>
<p>$$ xv' = -\dfrac{v^4... |
42,682 | <blockquote>
<p>$\tan(\theta) = -\frac{15}{8}$ given that $\theta$ is in quadrant II</p>
</blockquote>
<p>I know that $x= -8$ and $y= 15$ since it is in quadrant II $x$ has to be the negative. Where do I go from here? I tried $\tan^2\theta - \sec^2\theta = 1$ got some nonsensical answers.</p>
<p>Not sure how the $-... | ncmathsadist | 4,154 | <p>Draw a picture! You can then get all six triggies easily.</p>
|
1,313,504 | <p>I have come to know that Pythagorean triples' behaviour is unique in this way after verification.</p>
| Mark Viola | 218,419 | <p><strong>REVISED BY REQUEST</strong></p>
<p>We begin with the integral $\int_0^1 x(1-x)^ndx$. Integrating by parts gives</p>
<p>$$\begin{align}
\int_0^1 x(1-x)^ndx&=\left.-\frac{x(1-x)^{n+1}}{n+1}\right|^{^1}_{_0}+\frac{1}{n+1}\int_0^1 (1-x)^{n+1}dx\\\\
&=\frac{1}{n+1}\int_0^1 (1-x)^{n+1}dx
\end{align}$$</... |
1,313,504 | <p>I have come to know that Pythagorean triples' behaviour is unique in this way after verification.</p>
| DeepSea | 101,504 | <p>It can be done by a simple move: Let $u = 1-x$, then the integral becomes a much simpler one: $I = \displaystyle \int_{0}^1 (u^n-u^{n+1})du$. And you can take it from here...</p>
|
19,958 | <p>I read a paper saying the Mobius transformation from $\mathbb{R}^n\cup \infty \to \mathbb{R}^n\cup \infty$is generated by translations, scalar multiplications and inversions $x\to \frac{x}{|x|^2}$. So how do we get rotations from above mentioned generators?</p>
| kakaz | 3,811 | <p>Moebius Transformation is kind of group action obtained by homomorphism between complex matrices 2x2 and linear fractions of the shape</p>
<p>$
f(z) = \frac{(az +b)}{(cz+d)}
$</p>
<p>It works in case of complex numbers but it is different from general rotation group. It is only kind of <em>representation</em> for ... |
4,606,642 | <p>We have the given problem</p>
<p><span class="math-container">$$xu_x-yu_y=u,\ x>0, \ y>0 \ (1)$$</span>
<span class="math-container">$$u(x,x)=x^2, \ x>0, \ (2)$$</span></p>
<p>They ask to check if the problem is well-posed and solve it next.</p>
<p>I know that a problem is well-posed if:</p>
<ol>
<li>It ha... | Athanasios Paraskevopoulos | 1,004,010 | <p>To check for the existence of a unique solution to the above problem, we will need the tangent vector at each point <span class="math-container">$(x,y)$</span> of the characteristic projections which is
<span class="math-container">$$\tau=(\frac{dx}{dt},\frac{dy}{dt})=(x,-y)$$</span></p>
<p>The data curve is</p>
<p... |
2,720,460 | <p>I have a question regarding a improper double integral which will diverge but I cannot seem to understand how to reach that conclusion. The integral is the following: $$\iint_{\mathbf{ℝ^2}}\frac{x}{1+x^2+y^2}dxdy$$ </p>
<p>I can see that $f(x,y) \geq 0 \space \forall x\geq0$ and $f(x,y)\leq 0 \space \forall x\leq 0... | Robert Z | 299,698 | <p>Hint. Since the integrand is non negative in $\Omega_1$, we have that, for $R>0$,
$$\iint_{\Omega_1}\frac{x}{1+x^2+y^2}dxdy\geq \iint_{\Omega_1\cap B(0,R)}\frac{x}{1+x^2+y^2}dxdy=
\int_{-\pi/2}^{\pi/2}\int_{\rho=0}^R\frac{\rho^2\cos(\theta)}{1+\rho^2}d\theta d\rho\\=2\int_{\rho=0}^R\frac{\rho^2}{1+\rho^2} d\rho=2... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.