Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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conceptual doubt in finding $\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}$ While calculating the aforementioned limits I did the following.
$$\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}\\ =
\lim \limits_{x \... | Make life easier using $x=y+\frac{\pi}{4}$
$$\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos (x) + \sin (x) \right )^5}{1-\sin ({2x})}$$ Combine
$$\cos (x) + \sin (x)=\sqrt{2} \cos (y)$$ So, in terms of $y$, the expression reduces to
$$\frac{4\sqrt 2 -4 \sqrt{2} \cos ^5(y) )}{2 \sin ^2(y)}=2\sqrt 2\,\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim\limits_{n\to +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$ How to evaluate:
$$\lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$$
Personally, i always have trouble to evaluate sum of trignometric series, if you know a paper to... | It is possible to prove, by studying the function $ f: x\mapsto 1-x^{2}\cot^{2}{x}-\frac{2 x^{2}}{3} $ on $ \mathcal{D}=\left]-\frac{\pi}{2},\frac{\pi}{2}\right[\setminus\left\lbrace 0\right\rbrace $, that for any $ x\in\mathcal{D} $, we have : $$ \left|1-x^{2}\cot^{2}{x}\right|\leq\frac{2x^{2}}{3} $$
Let $ n \in\mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find minimum of $2x^3+x+\frac3{x^2}$ for $x>0$ Using AM-GM inequality we have:
$$2x^3+x+\frac3{x^2}=2x^3+x+\frac3{2x^2}+\frac3{2x^2}\geq 4\sqrt[4]{2x^3x\frac3{2x^2}\frac3{2x^2}}=4\sqrt[4]{\frac92}$$
As it turns out it is not the minimum, can you spot the mistake?
| There is no flaw in your work - it is true that $$2x^3 + x + \frac{3}{x^2} \ge 4 \sqrt[4]{\frac{9}{2}}$$ for $x > 0$. However, AM-GM inequality won't be enough to find the minimum, as it only attains equality for $$2x^3 = x = \frac{3}{2x^2} = \frac{3}{2x^2}$$
which never happens.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Area of triangle with incircle tangent to semicircle whose diameter is on one triangle side Triangle $\triangle ABC$ with incenter $I$ and inradius $r$ has the following property: the incircle is tangent to the semicircle with diameter $AB$ and is within the semicircle. Find the area of $\triangle ABC$ as a function of... |
The area of the triangle ABC is
\begin{align}
Area & = \frac12 r(AB + BC +CA)
= \frac12 rAB \frac{\sin A+\sin B +\sin C}{\sin C} \\
&= \frac12 rAB \frac{4\cos \frac A2 \cos \frac B2\cos\frac C2 }{2\sin \frac C2\cos \frac C2}
= rAB \frac{\cos \frac A2 \cos \frac B2 }{\cos \frac{A+B}2}
= \frac{rAB }{1- \tan\frac A2\t... | {
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Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've fou... | HINT
Since $x > 0$ and $y > 0$, the proposed inequality is equivalent to
\begin{align*}
x^{3} + y^{3} \geq \sqrt{2}xy & \Longleftrightarrow (x^{3} + y^{3})^{2} \geq 2x^{2}y^{2}\\\\
& \Longleftrightarrow (x+y)^{2}(1 - xy)^{2} \geq 2x^{2}y^{2}\\\\
& \Longleftrightarrow (1 + 2xy)(1-xy)^{2} \geq 2x^{2}y^{2}\\\\
\end{align*... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Maclaurin expansion for sech$(x)$ I am a bit unsure where I have gone wrong in working this out.
Sech$(x)=2/(e^x+e^{-x}).$
Maclaurin expansions:
$e^x = 1+ x + x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 - x^4/24;$
so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$ (I think this is where I have gone wrong.)
The actual... | Since in a neighbourhood of the origin $\cosh(x)=1+\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)$ we have $\operatorname{sech}(x)=1-\frac{x^2}{2}+Ax^4+O(x^6)$
and since
$$ \left(1+\frac{x^2}{2}+\frac{x^4}{24}\right)\left(1-\frac{x^2}{2}+Ax^4\right)=1+\left(A-\frac{5}{24}\right)x^4+O(x^6) $$
we must have $A=\frac{5}{24}$. This un... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify, $\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}$.
Simplify, $\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}$.
What I have tried: Put $x = \frac{1}{9}$. We have, $\sqrt[3]3\big(4x^\frac{1}{3} - 2x^\frac{1}{3} + x^\frac{1... | : $$\sqrt[3]{4/9}-\sqrt[3]{2/9}+\sqrt[3]{1/9}=\frac{(2^{1/3}+1)(2^{2/3}-2^{1/3}+1)}{(2^{1/3}+1)(3^{2/3})}=\frac{3}{3^{2/3}(2^{1/3}+1)}$$ so:$$\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}=\sqrt[3]{3}\cdot \frac{(2^{1/3}+1)(3^{2/3})}{3}=\sqrt[3]{2}+1$$
| {
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What is Riemann sum of this integral and using it to prove this This is Question 11.19 (e) part of Tom Apostol's Mathematical Analysis.
Let $f$ be a function of period $2\pi$ whose values on $[-\pi ,\pi]$ are $f(x)=1$ if $0<x < \pi$, $f(x)=0$ if $x =\pi$, or $0$ and $f(x)=-1$ if $-\pi <x<0$ and $f(x)= 4/\pi \sum_{n=... | The partial sum of the Fourier series for $f$ evaluated at $x = \frac{\pi}{4n}$ is
$$s_n\left(\frac{\pi}{2n}\right) = \frac{4}{\pi}\sum_{k=1}^n\frac{\sin\frac{(2k-1)\pi}{2n} }{2k-1}$$
Consider a partition $P_n: 0 < \frac{\pi}{2n}< \frac{3\pi}{2n} < \frac{5\pi}{2n} < \ldots < \frac{(2n-1)\pi}{2n}< \pi$ of the interval $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Permutation $\alpha^2$ Show that there is no permutation $\alpha \in S_4$ so that $\alpha^2=\begin{pmatrix}
1 &3 &4 &2
\end{pmatrix}$.
$S_4=\begin{pmatrix}
1 &2 &3 &4
\end{pmatrix}$
If $ \alpha=\begin{pmatrix}
2 &4 &3
\end{pmatrix}$, we get:
$$\alpha^2=\alpha\cdot\alpha=\begin{pmatrix}
1 &2 &3 &4 \\
1&4 &2 ... | $\beta = (1 \ 3 \ 4 \ 2)$ is on odd permutation: a cycle of length equal to $4$. As the square of any permutation is even, $\beta$ can't be written as a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3974048",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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$\int \sqrt{x^4+1}\ dx$ I have tried to find $\int \sqrt{x^4+1} dx$ but can not find a way to get the answer.
| By the reduction and transformation formulas mentioned in my Elliptic Integrals and Functions monograph
$$\int_0^y\sqrt{x^4+1}\,dx=\int_0^y\frac{x^4+1}{\sqrt{x^4+1}}\,dx=\frac{y\sqrt{y^4+1}}3+\frac23\int_0^y\frac1{\sqrt{x^4+1}}\,dx$$
$$=\frac{y\sqrt{y^4+1}}3+\frac23F\left(\sin^{-1}\sqrt{y^2+1-\sqrt{y^4+1}},m=\frac12\ri... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to find the series $a_{n}$ which has this generating function: $f(x)=\frac{1}{1-x^{2}}$? I'm given a generating function $f(x)$:
$$
f(x)=\frac{1}{1-x^2}
$$
Naturally, I tried to transform it into the form $f(x)=\sum_0^\infty a_n \cdot x^n$.
So far I figured out what follows:
$$
f(x)=\frac{1}{(1-x)(1+x)}=\frac{\frac... | To repair the approach you were taking:
$$f(x) = \frac{1}{1-x^2} = \frac{1/2}{1-x} + \frac{1/2}{1+x} = \frac{1}{2} \sum_{n=0}^\infty x^n + \frac{1}{2} \sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty \frac{1+(-1)^n}{2} x^n,
$$
so
$$
a_n = \frac{1+(-1)^n}{2} = \begin{cases} 1 &\text{if $2\mid n$} \\0 &\text{otherwise}\\ \en... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that$ \frac{a+b}{\sqrt{ab(1-ab)}} + \frac{b+c}{\sqrt{bc(1-bc)}} + \frac{c+a}{\sqrt{ca(1-ca)}} \le \frac{\sqrt{2}}{abc}$ Prove the following inequality where positive reals $a$, $b$, $c$ satisfies $ab+bc+ca=1$.
$$
\frac{a+b}{\sqrt{ab(1-ab)}} + \frac{b+c}{\sqrt{bc(1-bc)}} + \frac{c+a}{\sqrt{ca(1-ca)}} \le \... | let $x=ab,y=bc,z=ca$ and $x+y+z=1$
after your work we have to show
$$\sum_{cyc} \sqrt{2x(y+z)}\le 2$$ which is true by AM-GM $$\sum_{cyc} \sqrt{2x(y+z)}\le \sum_{cyc}\frac{2x+y+z}{2}=2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Compute the $n$-th power of the difference of the following matrices. Given the matrices $$A=\begin{pmatrix} a & c & b \\ c & b & a \\ b & a & c \end{pmatrix}$$ and $$B=\begin{pmatrix} b & a & c \\ a & c & b \\ c & b & a \end{pmatrix}$$
for $a,b,c \in \mathbb{R}$
find $$(A-B)^n$$ for $n \in \mathbb{N}$.
I can see that ... | Note that $M = A - B$ can be written in the form
$$
\pmatrix{
p&r&q\\
r&q&p\\
q&p&r},
$$
where $p+q+r = 0$. Verify that
$$
M \pmatrix{1\\1\\1} = 0,
$$
which means that $\det(A - B)$ must be zero. With that, we find the characteristic polynomial to be
$$
\det(M - kI) = -k^3 - (pq + pr + qr)k.
$$
Let $\alpha = pq + pr + ... | {
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"timestamp": "2023-03-29T00:00:00",
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Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$
I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.
As an instance, I will justify an example inequality just to show you the intended strategy. Consider th... | mrsamy's method is quite elegant, though it says nothing about how large $x\gg 1$ should be.
If you need a thighter inequality, the best way is to develop
$$f(x)=(5x^2+3x+1)x-10(x^3-x^2-1)$$
Graph it and notice it is negative for some $x$ close to (and below) $3$
Then just calculate $f(u+3)=-5u^3-32u^2-56u-5<0\quad$ fo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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In how many ways can $6$ different candies be divided between three children? I am the freshman in Computer Science at university, and currently struggling with some tasks from the Combinatorics part of the Discrete Math class. Any help would be appreciated.
In how many ways can $6$ different candies be divided between... | Your first solution is correct. There are three possible recipients for each of the six candies, so there are $3^6$ possible distributions of the candies.
If you wish to consider how many candies each child receives, you need to consider cases based on partitions of $6$ into at most three parts.
\begin{align*}
6 & = 6\... | {
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Simplex Algorithm Cycles How can I show that the linear program
\begin{align*}&\min -2x-3y+z+12w\\
&s.t.
\\&-2x-9y+z+9w+s=0\\&1/3x+2y-1/3z-2w+t=0\\&x,y,z,w,s,t\geq 0\end{align*}
can induce cycling in the Simplex Algorithm if we use the original pivot rule of Dantzig? I know that I have to show that the algorithm repea... | To simplify things, let’s denote each basis as $B_0,\ldots,B_n$ for the number of bases we visit until we notice a cycle. Thus, with the given model in the question, the first basis will be:
\begin{array} {|c|c|}
\hline \text{Current Basis} & BV & k & x & y & z & w & s_1 & s_2 & RHS & RT \\ \hline & k & 1 & 2 & 3 & -1... | {
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Proving continuity of $f(x)=\sqrt{1−x^3}$ on $[0,1]$ I have to prove the continuity of the function with $\epsilon, \delta$. For this I have to prove that the function is continuous at every $x \in [0, 1]$ by proving:
$$\forall 0 < \epsilon, \exists 0 < \delta, \forall a \in [0, 1]: |x - a| < \delta \implies \left|\sqr... | If $a\not=1$, then you can continue with
$${3\delta\over\sqrt{1-x^3}+\sqrt{1-a^3}}\lt{3\delta\over\sqrt{1-a^3}}$$
so it suffices to let $\delta=\epsilon\sqrt{1-a^3}/3$ to make the epsilon-delta proof work for $a\in[0,1)$. But for $a=1$ you need a separate argument to show that there is a $\delta$ for which $|x-1|\lt\de... | {
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Find the radius of convergence of the power series representation of $\frac{tan(\frac{z}{2})}{(z+2)(z+3)}$ We are given the following function:
$$\frac{tan(\frac{z}{2})}{(z+2)(z+3)}$$
How do we find the radius of convergence for the power series, representing that function around $z = 0$?
The power series representatio... | The closest singularities of the function are located at $z=-2,-3,\pm\pi$. The closest one to $0$ is $-2$, thus the radius of convergence of the series centered at $0$ would be $2$.
| {
"language": "en",
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Substituting $x = \sec(u)$ to evaluate $\int\sqrt{x^2-1}\ dx$: but why is $\tan(u) = \sqrt{x^2 - 1}$ rather than $\tan(u) = -\sqrt{x^2 - 1}\ $? I tried evaluating $\int\sqrt{x^2-1}\ dx$ using the substitution $x = \sec(u).$
I know my method gets to the same answer as wolframalpha, namely:
$\int\sqrt{x^2-1}\ dx = \frac{... | The integration-by-substitution theorem applies here when the domain of the substitution function $\sec(u)$ is restricted† to $(0,\frac\pi2).$
(This substitution works even if the original domain of integration is, say, $[-2,-1]$.)
This is why only the positive output of $\tan(u)$ is taken.
† due to the discontinuit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3994067",
"timestamp": "2023-03-29T00:00:00",
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How to maximize a function when you cannot solve gradient = 0 I have to maximize this function : $$ f(x,y) = a\sqrt{x} + b\sqrt{y}$$
$$ a, b \in{R+*} $$
Knowing that $$ 0 ≤ x ≤ 2 − 2y $$ with $$ 0 ≤ y ≤ 1 $$
I said that f is a linear combination of 2 concave functions so it has a maximum (for 0<=x<=2-2y and 0<=y<=1). B... | With the help of some slack variables $s_k$ we transform the inequalities into equations and then with $f(x,y) = a\sqrt x+b\sqrt y$ forming the lagrangian
$$
L = f(x,y)+\lambda_1(x-s_1^2)+\lambda_2(y-s_2^2)+\lambda_3(x-2+2y+s_3^2)+\lambda_4(y-1+s_4^2)
$$
The lagrangian stationary points are the solutions for
$$
\nabla ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is the matrix $A^4+A^3-3A^2-3A$ invertible?
Let $A$ be a real $5 \times 5$ matrix satisfying $A^3-4A^2+5A-2I=O$. Is the matrix $A^4+A^3-3A^2-3A$ invertible?
Consider the polynomial $t^3-4t^2+5t-2=(t-1)^2(t-2)$. The minimal polynomial of $A$ divides $(t-1)^2(t-2)$. Let $p(t)$ be the minimal polynomial of $A$.
If $p(x)... | Since $A^3-4A^2+5A-2=0$, $A$ must be invertible ($0$ is not an eigenvalue, else we get $-2v=0$.) Since $$A^4+A^3-3A^2-3A=A(A^3+A^2-3A-3),$$ it is enough to show that $A^3+A^2-3A-3$ is invertible. This follows from \begin{align}
A^3+A^2-3A-3&=(A+1)(A+\sqrt3)(A-\sqrt3)\end{align} Since $-1$, $\pm\sqrt3$ are not eigenvalu... | {
"language": "en",
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$\sqrt a+\sqrt b$ is a root of polynomial, then $\sqrt a -\sqrt b$ is so over finite field Let $a,b$ is not a square element in finite field.
If polynomial over the finite field has a root $\sqrt a+\sqrt b$, then the polynomial has $\sqrt a-\sqrt b$ as a root too?
Thank you for your help.
This question arises from conc... | Let's look at the concrete example, to see what is going on, and assume we are not in characteristic $2$.
Then, using the normal quadratic formula if $$x^4-10x^2+1=0$$ then $$x^2=\frac {10\pm \sqrt{100-4}}{2}=5\pm 2\sqrt 6$$ and if $6$ has a square root in your finite field, then the original equation is reducible.
The... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Showing that a 2-variable function is decreasing with respect to a given variable on an interval For all integers $x\geq 7$ and all real numbers $y$ such that $0.249 \leq y \leq 0.25$ define
$$F(x,y)=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1... | $F(x,y)=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( \frac{x+3}{x}\right)+2\sqrt{\frac{x+3}{x}}+\log\left( \frac{x+1}{x}\right)-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{x}\rig... | {
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"timestamp": "2023-03-29T00:00:00",
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Show $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$. For $n\ge 0$, show $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$.
Over a well ordered set of non-negative integers, induction can be used.
Base case: $n=0$
$\binom{x}{0} \binom{y}{0} = 1.\frac{(y)!}{(y)! \,\,(0)!}=1$.
Taking the rhs,... | $$S=\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k}$$
$$S=\sum_{k=0}^{n}[t^k]~ (1+t)^x~ [t^{n-k}]~ (1+t)^y$$
$$S=[t^n] (1+t)^{x+y}={x+y \choose n}.$$
Here $[t^j]f(t)$ means co-efficient of $t^j$ in $f(t).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Resolve $x+\frac{1}{x}$ While reading an old mathematics question paper of the Unified International Mathematics Olympiad, I hit a question with two very interesting patterns. Here I first mention them before presenting my question.
Pattern 1
When $0 < x < 1$, then the value of $x+\frac{1}{x}$ is greater than $2$.
He... | $x+\frac{1}{x}\geq2$ holds for all $x>0$, with equality at $x=1$ only. As a hint, expand
$$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\geq0.$$
For the second claim, observe that for $1<x<2$, we have
$$(x-2)(x-1)<0\implies x^2+2<3x\implies x+\frac{1}{x}<\frac{x}{2}+\frac{3}{2}<\frac{5}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Arrangement without any monotone section
Here's the question:
"In how many ways can the sequence 45678 be arranged such that there aren't any three consecutive terms increasing or decreasing"
The first thing I thought of is to do complementary counting and then make cases for each number of consecutive terms that the... | My simple Python script enumerates them as
(4, 6, 5, 8, 7)
(4, 7, 5, 8, 6)
(4, 7, 6, 8, 5)
(4, 8, 5, 7, 6)
(4, 8, 6, 7, 5)
(5, 4, 7, 6, 8)
(5, 4, 8, 6, 7)
(5, 6, 4, 8, 7)
(5, 7, 4, 8, 6)
(5, 7, 6, 8, 4)
(5, 8, 4, 7, 6)
(5, 8, 6, 7, 4)
(6, 4, 7, 5, 8)
(6, 4, 8, 5, 7)
(6, 5, 7, 4, 8)
(6, 5, 8, 4, 7)
(6, 7, 4, 8, 5)
(6, 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How do I bypass the limit for the tan function on the calculator? So basically, my partner and I are creating problems for a project. She created one where we aren't sure what the correct answer is.
Problem:
$Find \ cos\frac{\theta}{2} \ if \ tan \ \theta = \frac{3}{4}; \pi < \theta<\frac{3\pi}{2}$
I'll show you what w... | The "limit" you're trying to bypass is the fact that $\tan^{-1}$ only produces output values between $-\frac\pi2$ and $\frac\pi2.$
It will never directly give you an angle in the third quadrant.
The way you "get around" this is you remember that
$$ \tan\theta = \tan(\theta + \pi) = \tan(\theta + 2\pi) = \tan(\theta - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$
If $a,b,c,d$ are real numbers and $a+b+c+d=0$. prove:
$$a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$$
Using Euler's identity: if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$. in this problem substitute $x=a+b,y=c,z=d:$
$$(a+b)^3+c^3+d^3=3cd(a+b)$$
$$a^3+b^3+c^3+d^3=3(a+b)(cd-ab)=3(acd-a^2b+bcd... | $$(a+b)^3=a^3+b^3+3ab(a+b)=a^3+b^3-3ab(c+d)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does this pattern work: $1 \cdot{1} = 1, 11 \cdot{11} = 121, 111 \cdot{111} = 12321\ldots$ I have recently learned about pattern that goes like this:
$$\begin{align}
1^2 &= 1\\
11^2 &= 121\\
111^2 &= 12,321\\
1,111^2 &= 1,234,321\\
11,111^2 &= 123,454,321.
\end{align}$$
It is a very cool pattern, but after a bit it... | Other answers have already provided some intuition behind the pattern. Here is some terminology:
*
*The numbers being squared, e.g. $111111$, are called repunits.
*The resulting products are called Demlo numbers: $\,1, 121, 12321, 1234321,\, \dots$
Repunits can be written as sums of powers of $10$, and from that p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Factoring $a^4+b^4+(a-b)^4$ I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.
I rewrite it as:
$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not nega... | $$(a-b)^4=(a^2+b^2-2ab)^2=(a^2+b^2)^2+4a^2b^2-4ab(a^2+b^2)$$
$$a^4+b^4=(a^2+b^2)^2-2a^2b^2$$
On addition, $$2((a^2+b^2)^2+a^2b^2-2ab(a^2+b^2))=2(a^2+b^2-ab)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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If $\sin x=\frac{2 t}{1+t^{2}}$ and $\cot y=\frac{1-t^{2}}{2 t}$, then the value of $\frac{d^{2} x}{d y^{2}}$
If $\sin x=\frac{2 t}{1+t^{2}}$ and $\cot y=\frac{1-t^{2}}{2 t}$, then the value of $\frac{d^{2} x}{d y^{2}}$
If we take $t=\tan \theta$ then $x=2\theta$ and for $y$ ,I am getting two values as follows ,
if ... | I'm not sure where you're getting $y=\theta+\pi/4$ but it is definitely incorrect as $$\cot(\theta+\pi/4)=\frac{1-\tan\theta\tan\pi/4}{\tan\theta+\tan\pi/4}=\frac{1-\tan\theta}{1+\tan\theta}$$ is not the same as $$\frac{1-t^2}{2t}=\frac{1-\tan^2\theta}{2\tan\theta}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Using $\sec(x)$ for integral.
Find the undefined integral $\int \frac{\sqrt{x^2+4x}}{x^2}\mathrm{dx}$
$$\displaystyle\int \dfrac{\displaystyle\sqrt{x^2+4x}}{x^2}\mathrm{dx}$$
I tried to rearrange the square root and I got:
$$\sqrt{(x+2)^2-4}$$ and I substitute the $x+2$ with $2\sec(u)$ so indeed I got these two:
$$x+... | The difference between the two answers is:
\begin{align}
&\quad\frac {2x\color{red}{+}8}{\sqrt{x^2+4x}}-\frac {8}{x+\sqrt{x^2+4x}}
\\&=\frac {(2x+8)(x+\sqrt{x^2+4x}) - 8\sqrt{x^2+4x}}{(x+\sqrt{x^2+4x})\sqrt{x^2+4x}}
\\&=\frac {2x^2+8x+2x\sqrt{x^2+4x}}
{(x+\sqrt{x^2+4x})\sqrt{x^2+4x}}
\\&=\frac {2(x^2+4x+x\sqrt{x^2+4x})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show $\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx= \int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$ I accidentally found out that the two integrals below
$$I_1=\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx,\>\>\>\>\>\>\>I_2=\int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$$
are equal in value. In fact, they can be evaluat... | Let
$$J=\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx.$$
Using
$$ \arctan x+\arctan\frac1x=\frac{\pi}2$$
and under $x\to \frac1x$, one has
\begin{eqnarray}
J&=&\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx\\
&=&\int_0^\infty \frac{\tan^{-1}x^{-2}-\tan^{-1}x^{-1/2} }{1+x^2}dx\\
&=&\int_0^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Proving $\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$
Prove that (where $a,b,c>0$) $$\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$$
I have found a proof for this problem but it is very lengthy and is not nice.(Its not using computer though).I shall post it later.
Background:
Here however ... | Suppose $c = \min\{a,b,c\},$ then
$$\sum \frac{a^3+abc}{b+c}-(a^2+b^2+c^2) = \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-c)(b-a)}{c+a}+\frac{c(c-a)(c-b)}{a+b}$$
$$\geqslant \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-c)(b-a)}{c+a} = \frac{(a-b)^2(a^2+ab+b^2-c^2)}{(b+c)(c+a)} \geqslant 0.$$
Therefore
$$\sum \frac{a^3+abc}{b+c} \geqslant a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to find $F=\left(\frac{B}{2}\right)^{2H}-1$ when $H$ is maximum of $W_{1}$ and $B$ is minimum of $W_{2}$? The problem is as follows:
A pair of wavelenghts are $W_{1}$ and $W_{2}$:
$W_{1}=5\sin\alpha-5+12\cos\alpha$
$W_{2}=3\sin\beta+\sqrt{3}\cos\beta$
Assuming $H$ and $B$ represent the maximum and the minimum valu... | Let's try to make $a\sin x+ b\cos x$ of the form $\boxed{r\sin(x+\theta)}$ and $r>0$
$\begin{align}&5\sin\alpha+12 \cos\alpha = r(\cos\theta\sin\alpha+\sin\theta\cos\alpha) = r\sin(\alpha+\theta) \\&\Rightarrow r^2 = 5^2+12^2 = 13^2 \\&\Rightarrow r = 13 \text{ and } \theta =\arctan\frac{12}{5} \end{align} $
Now,
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4026064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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The variation of a Ukrainian Olympiad problem: 10982
Given a recursion $a_{n+ 1}= \dfrac{a_{n}}{n}+ \dfrac{n}{a_{n}}$ with $a_{1}= 1.$ Prove that
$$\lim a_{n}^{2}- n= \frac{1}{2}$$
Source: StachMath/@RiverLi _ The limit and asymptotic analysis of $a_n^2 - n$ from $a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}$
The original... | Note that, for $n\ge 1$, $$ n^6+10n^5+38n^4+68n^3+57n^2+18n>n^6+10n^5+37n^4+62n^3+46n^2+12n+1$$ and, factorising,
$$n(n+1)^2(n+2)(n+3)^2>(n^3+5n^2+6n+1)^2.$$
Also note that the function $F_n$ such that $$F_n(x)=\frac{n}{x}+\frac{x}{n}$$ is monotonic decreasing for $0<x\le n$.
Now suppose that $$\frac {n+2}{\sqrt {n+1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Use epsilon-delta definition of limits to show that $\lim_{(x,y) \to (1,2)}(3x+2y-1)=6$ and $\lim_{(x,y) \to (0,0)}(x^2+y^2)=0$ Use epsilon-delta definition of limits to show that
$$\lim_{(x,y) \to (1,2)}(3x+2y-1)=6$$
$$\lim_{(x,y) \to (0,0)}(x^2+y^2)=0$$
We need to show that $$\forall \epsilon >0( \exists \delta >0( \... | The latter one is correct, but the former is not. A big hint is that the choice $\delta \le (\varepsilon - 6) / 5$, which is negative(!) when $\varepsilon < 6$. Your choice of $\delta$ should always be positive, so that at least one $(x, y)$ satisfies $0 < \|(x, y) - (x_0, y_0)\| < \delta$.
Instead, consider that
$$|(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4028984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the chance to get two pairs from a hand of five cards if I have extra card (♠️ king, ♠️ queen and ♦️ queen) added to the standard cards What is the chance to get two pairs if I have extra card (♠️ king, ♠️ queen and ♦️ queen) added to the standard cards, thus I have 55 cards? Can anybody help me !!
Is my soluti... | It seems wise to count hands with pairs of kings or queens separately:
Hands with a pair of kings and not a pair of queens:
$$\binom{5}{2}\binom{11}{1}\binom{4}{2}\binom{46}{1}$$
because we have to pick $2$ kings out of the $5$, then one other type of pair not king or queen (so $11$ ranks to choose from), then $2$ card... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Let $a,b,c$ be positive integers such that $a^3+b^3=2^c.$ Show that $a=b$.
Let $a,b,c$ be positive integers such that $$a^3+b^3=2^c.$$ Show that $a=b$.
I have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2^c =2^x\cdot2^y$$ now it can only be that $a$ and $b$ are both odd or even since they sum to an even number. Thus if $a$ and ... | If $a$ and $b$ are both even, say $a=2^mA$ and $b=2^nB$ with $A$ and $B$ odd, then without loss of generality $n\geq m$ and so
$$2^c=a^3+b^3=2^{3m}\big(A^3+2^{3(n-m)}B^3\big),$$
from which it follows that
$$2^{c-3m}=A^3+2^{3(n-m)}B^3.$$
Because $A$ and $B$ are positive we see that $2^{c-3m}>1$, and so the left hand sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Proving that $f$ verifies $f(1−x) + f(x) = 1$ I have a function which is defined for $x \in (0,1)$ and for $p>1$ with the expression
\begin{align*}
f(x) = \sum_{k=0}^{p-1} \frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right)
\end{align*}
I would like to show that $f(1-x) + f(x) = 1$.
Wh... | Here we have a partial fraction decomposition in disguise.
We obtain
\begin{align*}
\color{blue}{f(x)}&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}
\prod_{\substack{j=k-p+1\\j\neq 0}}^{k+p}(x+k-j)\\
&=\prod_{j=-p+1}^{p}(x-j)\sum_{k=0}^{p-1}
\frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\,\frac{1}{x+k}\tag{1}\\
&\color{blue}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is the approach to find the expression of the partial sum of a series? This question is about computing the limit of a series as $n$ goes to $\infty$.
The series $(x_n)$ is defined as:
$$\sum_{i=0}^n \frac{1}{n^2+i}$$
So the $n$th partial sum will be given by: $\frac{1}{n^2+1} + \frac{1}{n^2+2} + \frac{1}{n^2+3} +... | For the limit itself, you aleady received a good answer.
We can approximate the partial sums using harmonic numbers since
$$S_n=\sum_{i=0}^n \frac{1}{n^2+i}=H_{n^2+n}-H_{n^2-1}$$ Now, using the asymptotics
$$H_p=\log (p)+\gamma +\frac{1}{2 p}-\frac{1}{12 p^2}+\frac{1}{120
p^4}+O\left(\frac{1}{p^6}\right)$$ apply it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Parameterizing both branches of a hyperbola Recently I have been studying parametric equations of surfaces and curves, specifically hyperbolic functions. Given by the equations $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \quad\frac{(x-\alpha)^2}{a^2}-\frac{(x-\beta)^2}{b^2}=1$$
denoting hyperbolic functions centred at $(0,0)$... | Here's a geometric way to derive a parametrization. Consider the family of all lines passing through the point $(a,0)$. Such a line is of the form $y = t(x-a)$ for some slope $t$. Since the hyperbola has degree $2$, each line will intersect it in exactly two points: $(a,0)$ and one other point. (This is an example of B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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convert $(x-x')^2 + (y-y')^2 + (z-z')^2$ to cylindrical coordinates Context
This is an interim problem related to a Green's function solution for a boundary-value problem in the cylindrical coordinate system.
Question
How do I convert $(x-x')^2 + (y-y')^2 + (z-z')^2$ to cylindrical coordinate system?
My worked solution... |
$$(x-x')^2 + (y-y')^2 + (z-z')^2
=r^2 + {r'}^2 -2\,r\,{r}' \,\cos\left(\theta - \theta'\right) + (z-z')^2$$
My attempt
I know that to tranform from Cartesian to cylindrical requires the following three equations
\begin{align}
r &= \sqrt{x^2 + y^2} \\
\theta &= \arctan{\left(\frac{y}{x}\right)} \\
z &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $\tan(\alpha/2)=(1-\cos\alpha)/\sin \alpha, \,?$ Is it true that
$$\bbox[5px,border:2px solid #138D75]{\tan\left(\frac {\alpha}2\right)=\frac{1-\cos\alpha}{\sin \alpha}, \quad ?} \tag1$$
My solution:
First, we note that the expression $\tan \left ( \frac{\alpha}{2} \right )$ makes sense when $\alpha \neq \pi + 2k\pi... | Replace $\alpha$ by $2\beta$ so that the right side is
$${{1-\cos 2\beta}\over {\sin 2\beta}}=
{{1-(\cos^2\beta-\sin^2\beta)}\over {2\sin\beta\,\cos\beta}}$$
$$={{2\sin^2\beta}\over{2\sin\beta\cos\beta}}=
{{\sin\beta}\over {\cos\beta}}=\tan\beta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity $a(x+b)^3+c\equiv4x^3-24x^2+48x+d$
Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity
$$a(x+b)^3+c\equiv4x^3-24x^2+48x+d$$
I have already found $a=4$ and $b=-2$ but I'm struggling to find $c$ and $d$.
... | Expanding we get $ax^3 + 3abx^2 + 3ab^2x +ab^3 + c = 4x^2-24x^2 +48x + d$
Technically what you have going on is 4 equations.
*
*$a = 4$
*$3ab=-24$
*$3ab^2 = 48$.
*$ab^3 + c = d$.
They aren't linear as the involve $b$ to powers $2$ and $3$ but $1,2$ are independent linear so $a,b$ can be solved $a= 4, b=-2$. And ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The triangle formed by the $x$-axis, the tangent to $y=1/x$ at a 1st-quadrant point, and the line from the origin to that point is isosceles Here is Prob. 45, Sec. 3.2, in the book Calculus With Analytic Geometry by George F. Simmons, 2nd edition:
Let $P$ be a point on the first-quadrant part of the curve $y = 1/x$. S... | It looks good to me. A geometric perspective to consider:
When we find point $A =(2a, 0)$, notice that $P$ is directly above the midpoint $OA$. (This midpoint is $C = (a, 0)$.)
The triangles $OCP$ and $ACP$ are congruent, right-angled triangles that are mirror images of each other ($CP$ as a common edge).
Hence $OPA$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the particular solution of a differential equation using at least three different methods. Find the particular solution $(x^2+6y^2)dx-4xydy=0$; when $x=1$, $y=1$ using at least three different methods.
I have done the first two. Can somebody help me with the third method.
Method 1: Homogenous Equation
Let $y=vx... | Compare
$$(x^2+6y^2)dx-4xydy=0~~~~(1)$$
with $$Mdx+Ndy=0 \implies \frac{\partial M}{\partial y}=12y,~~ \frac{\partial N}{dx}=-4y $$ It's integrating factor is
$$I=\exp \left(\int \frac{12y +4y}{-4xy}\right)=x^{-4}$$
Multiplying (1) with $x^{-4}$,(1) becomes an exact ODE:
$$(x^{-2}+6x^{-4} y^2) dx- 4x^{-3} ydy==0$$
It'... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Regularized Incomplete Beta Function I try to solve this property of Regularized Incomplete Beta Function. How can you solve a statement :
$I_x (a,a) = 1 - \frac{1}{2} I_{4x(1-x)}(a,\frac{1}{2}) $ where $1/2 \leq x \leq 1$.
Some definitions :
$I_x (a,b) = \frac{B(x;a,b)}{B(a,b)}$ with B(x;a,b) is incomplete Beta func... | Using the definitions you wrote
$$\text{lhs}=\frac{B_x(a,a)}{B(a,a)}$$
$$\text{rhs}=1-\frac{B_{4 (1-x) x}\left(a,\frac{1}{2}\right)}{2 B\left(a,\frac{1}{2}\right)}$$ Now, for $ \frac 12 \leq x \leq 1$
$$\frac d {dx}\text{[lhs-rhs]}=\frac{(1-x)^{a-1} x^{a-1}}{B(a,a)}-\frac{2^{2 a-3} (4 (1-x)-4 x) ((1-x) x)^{a-1}}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4045119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Least squares solution problem
Let $A = \begin{pmatrix}
2 & 0 & 2 \\
1 & 1 & 2 \\
1 & 1 & 2 \\
2 & 0 & 2
\end{pmatrix}$. Determine the least squares solution for $Ax = e_4.$
The least squares solution can be found from $A^TAx = A^Te_4$. This is problematic since $A^TAx = \begin{pmatrix}10&2&12\\ 2&2&4\\ 12&4&16\end{... | As the matrix
$$B=A^TA=\begin{pmatrix}10&2&12\\ 2&2&4\\ 12&4&16\end{pmatrix}$$ is not invertible, the solution of the least square problem is a linear subspace of dimension greater or equal than $1$. Indeed the kernel of $B$ is a linear line.
You can solve the system
$$Bx = \begin{pmatrix}2\\ 0\\ 2\end{pmatrix}$$ to fi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $2\sin A + \sin B = 2\sin C$ in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$ Given $2\sin A + \sin B = 2\sin C $ where $A,B,C$ are angles in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$
so $\sin C - \sin A = \frac{ \sin B}{2}$ so $2\cos \frac{C+A}{2} \sin \frac{C-A}{2} = \f... | \begin{align*}
\sin B &= 2 (\sin C - \sin A) \\
&=4 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{C-A}{2}\right)
\end{align*}
Moreover
\begin{align*}
\sin B &= \sin(\pi-(A+C)) \\
&=2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A+C}{2}\right)
\end{align*}
Therefore
\begin{align*}
\sin \left(\frac{A+C}{2}\right) ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24]=(z+x-7)[y(z+x)+24]$, find $x^2+y^2+z^2$
Let x, y, z be pairwise distinct real numbers, if $$(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24] $$ $$=(z+x-7)[y(z+x)+24]$$, find $x^2+y^2+z^2$
I've tried many ways but couldn't find a working way to solve it. I tried letting $(x+y-7)[z(x+y)+2... | let $f_z=(x+y-7)(z(x+y)+24)$,$f_y,f_x$ defined similarly $$f_x=f_y=f_z=k$$ $$f_z-f_y=0\to -(y - z) (x^2 - 7 x - y z - 24)=0$$ so $$x^2-7x-yz=24$$and similarly for others yield $$p_x=p_y=p_z=24$$ where $p_x=x^2-7x-yz$ and others defined similarly $$p_x-p_y=0\to (x-y)(x+y+z-7)=0$$ so $$x+y+z=7\to z=7-x-y$$ Plug this in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4048713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$ I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to so... | You can cut some work by just looking at the remainder mod $3$.
Let $n=3q+r$. Easy calculation gives
$$\frac15(3q+r)^5+\frac13(3q+r)^3+\frac{7}{15}(3q+r) = \text{some integer} + \frac{243q^5+7q}{5} + \frac{r^5}5+\frac{r^3}3+\frac{7r}{15}.$$
By Little Fermat we have $q^5 \equiv q \pmod{5}$ so
$$243q^5+7q \equiv 250q \eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the sum of all irrational roots of $4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$
Calculate the sum of all irrational roots of
$$4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$$
I'm not even sure how to begin here, I tried raising it to the power of three, tried writing $8x^{3}+ 3$ with $x^{3}+ y^{3}= \left ( x+ y \right )\left ( x^{2}-... |
Problem. Half-solve
$$4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$$
Substitute $y^{3}= 8x- 3$ in the original equation_
$$\left ( 4y- 8x^{3}- 3 \right )+ \left ( y^{3}- 8x+ 3 \right )= \left ( y- 2x \right )\left ( 4x^{2}+ 2xy+ y^{2}+ 4 \right )= 0$$
With $y= 2x,$ we have the equivalent one
$$8x^{3}- x+ 3=\!\left ( 2x- 1 \right )... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the system of equations: $32y+32x^3=6x+17$, $16z+32y^3=6y+9$, $8x+32z^3=6z+5$ where $x,y,z\in \mathbb{R}$ Solve the system of equations: $$\begin{cases} 32y+32x^3=6x+17 \\16z+32y^3=6y+9 \\8x+32z^3=6z+5 \end{cases}$$ where $x,y,z\in \mathbb{R}$ (Bulgaria 1960)
I attempted to solve this question as follows:
$32y+32... | Proof by Contradiction:
Let, $32x^3-6x>1$ then:
$$\begin{align} 32x^3-6x>1 &\Longrightarrow x>\frac 12\\ 32x^3-6x=17-32y>1 &\Longrightarrow y<\frac 12 \\32y^3-6y=9-16z <1 &\Longrightarrow z>\frac 13\\ 2z^3-6z=5-8x >1 &\Longrightarrow x<\frac 12\end{align}$$ which gives a contradiction.
The same method also works for ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the slope of the tangent line to the graph of the given function at the given value of x. Find the slope of the tangent line to the graph of the given function at the given value of x. Find the equation of the tangent line. : $y=x^4-5x^3+2; x=2$
I understand that the slope of the line follows the equation of
$$\li... | If you use the h-method you start with
$$\lim_{h \rightarrow 0} \frac{y(a+h)-y(a)}{h}=\lim_{h \rightarrow 0} \frac{(a+h)^4-5\cdot (a+h)^3+2-(a^4-5\cdot a^3+2)}{h}$$
To resolve $(a+h)^4$ you can use the binomial theorem: $(x+y)^n=\sum\limits_{k=0}^n\binom{n}x\cdot x^k\cdot y^{n-k}$
$$=\lim_{h \rightarrow 0}\frac{a^4+4a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Point lies inside a triangle ABC with $\measuredangle BAC=45^\circ$ and $\measuredangle ABC=30^\circ$ A triangle $ABC$ is given with $\measuredangle BAC=45^\circ$ and $\measuredangle ABC=30^\circ.$ Point $M$ lies inside the triangle and $\measuredangle MAB=\measuredangle MBA=15^\circ.$ Find $\measuredangle BMC$.
I am ... | Please apply Trigonometric form of Ceva's theorem.
If $\angle BCM = x$,
$\sin \angle BAM \ \sin \angle ACM \sin CBM = \sin \angle MAC \ \sin \angle MCB \ \sin \angle MBA$
$\sin 15^0 \sin (105^0-x) \sin 15^0 = \sin 30^0 \sin x \sin 15^0$
$\cos (15^0-x) \sin 15^0 = 2 \sin 15^0 \cos 15^0 \sin x$
$\sin 30^0 \cos (15^0-x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving a Polynomial Equation by Factoring. $n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.
| You were going in the correct direction, as you said:
$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
If we apply the same here:
$^3+12^2+48+64 = ^3+ 4^3 + 12^2+48 = (n+4)(n^2 - 4n + 16) + 12n(n+4)$
If we take (n+4) common, we get:
$(n+4)(n^2 - 4n + 16 + 12n) = (n+4)(n^2 + 8n + 16) = (n+4)(n+4)^2 = (n+4)^3$Cheers
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 6
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Locus of a point of intersecting lines The perpendicular from the centre of the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ to the tangent at any point $P$ meets the line joining point $P$ to the focus $(ae,0)$ at $Q$. Prove that the locus of $Q$ is $$x^2+y^2-2aex=b^2$$
My attempt
Tangent to an ellipse at $P(\theta)$
$$bx\cos\th... | $$x(a\sin t)-y(b\cos t)=0\ \ \ \ (1)\implies\tan t=\dfrac{by}{ax}\ \ \ \ (3)$$
$$x(b\sin t)+ya(e-\cos t)-abe\sin t=0\ \ \ \ (2)$$
$$\dfrac x{ab^2e\sin t\cos t}=\dfrac y{a^2be\sin^2t}=\dfrac1{a^2e\sin t(1-e\cos t)}$$
$$\implies\dfrac{1-e\cos t}{\cos t}=\dfrac{b^2}{ax}\implies\sec t=\dfrac{b^2}{ax}+e\ \ \ \ (4)$$
Use $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4061022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Decomposition of a sum of cubes I have to decompose the following polynomial:
$$(x+2y)^3+(x-2y)^3$$
Since I have the sum of two cubes, I have thought:
$$(x+2y)^3+(x-2y)^3=(x+2y+x-2y)[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]=2x\color{red}{[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]}$$
Now in my book the result is $2x(x^2+12y^2)$, but I can... | Compare $S = (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
with $T = (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3.$
Adding: $S + T = 2(a^3 + 3ab^2).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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The coefficient of $x^2$ in the expansion of $\left(x^3+2x^2+x+4\right)^{15}$ To find the coefficient of $x^2$ in the expansion of $\left(x^3+2x^2+x+4\right)^{15}$.
I saw a solution to this problem in which the above expression was differentiated 2 times and they put x=0. I really can't understand the significance behi... | We know that, whatever the expansion of $\left(x^3 + 2x^2 + x +4 \right)^{15}$ is, it will look something like
$$
\left(x^3 + 2x^2 + x +4 \right)^{15} = c_0 + c_1 x + c_2 x^2 +c_3x^3+ ... + c_{45} x^{45}
$$
where $c_i$'s are some constants possibly equal to $0$. Now, in the above equation what you want to know is the v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Expected Value of Binomial Distribution What is the expected value of the absolute value of the difference between the number of incoming tails and the number of incoming heads when a coin is tossed 5 times?
Here is what I think : Let $X$ is random variable of number of incoming heads then
$p=\frac{1}{2}$ and $n=5$... | Your random variable $Z=|X-Y|$ can take only the values
$$\{1,3,5\}$$
with probabilities
$$\left\{\frac{20}{32};\frac{10}{32};\frac{2}{32}\right\}$$
This because
$Z=5$ when you get 5 consecutive tails (or heads)
$Z=3$ when you get 4 tails (or heads)
$Z=1$ when you get 2 tails (or heads)
No other situations are possible... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of possible solutions to $2 \le |x-1||y+3| \le 5$, where $x$ and $y$ are negative integers
If $$2 \le |x-1||y+3| \le 5$$ and both $x$ and $y$ are negative integers, find the number of possible combinations of $x$ and $y$ .
Below is my solution approach :-
As $x$ is a negative integer, hence $|x-1|$ in the $2 \... | The possible values of the product are $2,3,4,5$.
Hence the possible factorizations,
$$1\cdot2,2\cdot1,1\cdot3,3\cdot1,1\cdot4,2\cdot2,4\cdot1,1\cdot5,5\cdot1.$$
But $|x-1|\ge2$, which leaves
$$2\cdot1,3\cdot1,2\cdot2,4\cdot1,5\cdot1,$$
while $|y+3|$ can achieve $1$ and $2$ in two ways.
$$10.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding maxima and minima of $f(x,y) = x^{4} + y^{4} - x^{3}$ Finding maxima and minima of $$f(x,y) = x^{4} + y^{4} - x^{3}$$
I tried solving this question but need help.
In this question, I calculated the partial derivatives of first and second order.
I found
$$f_{x} = 4x^{3} - 3x^{2}\\ f_{y}= 4y^{3} \\ f_{xx} = 12x... | For the minimum, if $x\ge 1$ or $x\le 0$ we have $x^4-x^3+y^4=x^3(x-1)+y^4\ge 0$. If $0<x<1$ we have by AM-GM
$$x^3(1-x)=27(x/3)^3(1-x)\le 27\left(\frac{3(x/3)+(1-x)}{4}\right)^4=27/256$$
so $x^4-x^3\ge -27/256$ and so $x^4-x^3+y^4\ge x^4-x^3\ge -27/256$. The minimum value is reached at $x=3/4$, $y=0$.
For the maximu... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Domain of $\frac{x^2-3x-10}{x+2}\div\frac{x^2-25}{x}$ I saw a question asked to simplify
$\dfrac{x^2-3x-10}{x+2}\div\dfrac{x^2-25}{x}$
It is not so hard to do : $$\dfrac{(x-5)(x+2)}{x+2}\times\dfrac{x}{(x-5)(x+5)}=\dfrac x{x+5}$$
But my question is: should we have necessary all the conditions $x\ne-2$ , $x\ne5$ , $x\... | Yes, you have to state all of the conditions $x\neq2$, $x\neq5$, $x\neq-5$ and $x\neq0$. This is because when we write
$$
\frac{x^2-3x-10}{x+2} \div \frac{x^2-25}{x}=\frac{x}{x+5} \, ,
$$
we are asserting that for every value of $x$ for which the LHS and RHS are defined, the LHS represents the same number as the RHS. W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\sqrt{\frac{y}{x}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{x}{z}}$ Suppose $x,y,z$ are positive real numbers that satisfy
\begin{align*}
\frac{x}{y} + \frac{y}{x} + \frac{x}{z} + \frac{z}{x} + \frac{y}{z} + \frac{z}{y} &= 2018 \\
\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{z}} + \sqrt{\frac{z}{x}} &= 17.
\end{align*}
Find ... | Squaring the second equation we get
$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + 2(\sqrt{\frac{x}{z}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{y}{x}}) = 289.$
Subtracting from the first equation
$\frac{x}{z} + \frac{z}{y} + \frac{y}{x} - 2(\sqrt{\frac{x}{z}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{y}{x}}) = 2018-289 = 1729.$
Put $a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Representing all rational numbers between $\dfrac{1}{2}$ and $1$ How do I show that
$$\dfrac{2^{\left\lfloor\frac12 a_1\right\rfloor} + 2^{\left\lfloor\frac12 a_2\right\rfloor} + \ldots + 2^{\left\lfloor\frac12 a_n \right\rfloor}}{2^{\left\lceil\frac12 a_1\right\rceil} + 2^{\left\lceil\frac12 a_2\right\rceil} + \ldots... | Let's group the parameters $a_i$ into two groups, the even and odd numbers (lets call them $e_i$ and $o_i$ resp).
Notice that $2^{\lceil e_i/2 \rceil} = 2^{\lfloor e_i/2 \rfloor} $ while $2^{\lceil o_i/2 \rceil} = 2 \times 2^{\lfloor o_i/2 \rfloor} $
Now, let $\frac12 <Z<1$ with $Z=\frac{X}{Y}$, where $ X, Y \in \mat... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Let $G$ be a group such that for all $x,y\in G,(xy)^3=x^3y^3$ and $(xy)^8=x^8y^8$. Prove that $G$ is Abelian or not? I tried to prove it like $(xy)^3=xyxyxy=x^3y^3 $ which implies
$$ x^{-1}/x^3y^3=xyxyxy$$
$$x^2y^3=yxyxy/y^{-1}$$
$$ x^2y^2=yxyx $$
and I get $x^2y^2=(yx)^2$.
But I dont know how to prove that $(xy)^3=x^3... | I prove commutativity. (The comments
reveal that a more general result was proved in 2017.) The proof is for cancellative semigroups.
$$\ (xy)^3=x^3y^3\tag{A}$$
$$\ (xy)^8=x^8y^8\tag{B}$$
Using cancellation on (A), $(yx)^2=x^2y^2 \tag 1$
Therefore, $x^2y^3x=(yx)^3=y^3x^3$ so that $x^2y^3=y^3x^2\tag 2$
Using cancellati... | {
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Show $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6+\sqrt{14})$ Sifting through my old Galois Theory notes, I found a proof that $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6 + \sqrt{14})$, but my proof was chicken-scratch so I can no longer decipher it.
Can somebody come up with a way of showing this?
| Alternative approach:
If $a,b > 0$ and $a^2 = b^2$, then $a$ must $= b$.
Let $a = \frac{1}{2} \left(\sqrt{6} + \sqrt{14}\right),
~b = \sqrt{5 + \sqrt{21}} \implies 0 < a,b.$
Then $a^2 = \frac{1}{4} \left(6 + 14 + 2\sqrt{84}\right)
= 5 + \sqrt{21} = b^2.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Has the equation $x^2-21 = 17y$ integer solutions? Has the equation $x^2-21 = 17y$ integer solutions?
Attempt:
I saw this: The equation $x ^ 2 + py + a = 0$ can be solved as an integer precisely, if $-a$ is a quadratic remainder modulo p.
I get: $x^2-17y-21=0$
Now i have to show $21$ is quadratic remainder modulo $-17$... | $x=2, y=-1$ is an integer solution.
| {
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"url": "https://math.stackexchange.com/questions/4083984",
"timestamp": "2023-03-29T00:00:00",
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Differentiate $\arcsin(\frac{1}{\sqrt{1 + x^2}})$ This is from Calculus by Michael Spivak, 3rd Edition, Chapter 15, problem 1 (iv):
Differentiate $f(x) = \arcsin\left(\frac{1}{\sqrt{1 + x^2}}\right).$
Here's my attempt:
\begin{align}f'(x) &= \left[\arcsin\left(\frac{1}{\sqrt{1 + x^2}}\right)\right]' \\
&= \arcsin'\le... | Observe that
$$f(\sinh(x))=\arcsin(\frac{1}{\sqrt{1+\sinh^2(x)}})=\arcsin(\frac{1}{\cosh(x)})$$
The differentiation gives
$$f'(\sinh(x))\cosh(x)=$$
$$\frac{1}{\sqrt{1-\frac{1}{\cosh^2(x)}}}\frac{-\sinh(x)}{\cosh^2(x)}=$$
$$\frac{-1}{\cosh(x)}\frac{\sinh(x)}{|\sinh(x)|}$$
$$\implies f'(\sinh(x))=\frac{\pm 1}{\cosh^2(x)}... | {
"language": "en",
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"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Simplify $bx(x-b) + ax(a-x) + ab(b-a)$ to $-(a-b)(a-x)(b-x)$ I was doing some matrices problems when I got to this factoring
$$bx(x-b) + ax(a-x) + ab(b-a)$$
I found this was the answer, but it was
$$-(a-b)(a-x)(b-x)$$
Doing some researches, I found this remarkable product
$$(a–b)(a–c)(b–c) = ab(a–c) + bc(b–c) + ca(c–a)... | Since the target expression has factor $(a-b)$ showing up at the original expression as $(b-a)$, we have a clue how to get this started $\to$ \begin{align}bx(x-b) + ax(a-x) + ab\color{green}{(b - a)}\end{align}
\begin{align}=bx^2-b^2x + a^2x-ax^2 + ab\color{green}{(b - a)} \end{align}
\begin{align}=x^2\color{green}{(b ... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Advanced complex numbers/roots of unity Let $a, b, c, d$ be real numbers, none of which are equal to $-1$, and let $\omega$ be a complex number such that $\omega^3 = 1$ and $\omega \neq 1.$ Given that,
$$
\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega} + \frac{1}{d + \omega} = \frac{2}{\omega},
$$
ho... | The keys is Vieta's formulas.
Let $u$ be a root of $x^3 = 1 $. Then $u+a$ is a root of $(x-a)^3 = 1$ and $\frac{1}{u+a}$ is a root of $(\frac{1}{x}-a)^3=1 $. Multiplying by $x^3$ and expanding the coefficients we have
$$(1+a^3) x^3 -3a^2x^2 + 3ax - 1 = 0$$
Using Vieta's formulas we get that $\frac{3a^2}{1+a^3}$ is eq... | {
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"question_score": "7",
"answer_count": 3,
"answer_id": 2
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How to integrate definite integral of form $\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx$ During WKB approximation maths I typically end up deducing integrals that need to be calculated of the form:
$$\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx$$
where a is a constant.
I know the following standard integral
$$\int_0^{1} \sqrt... | Let $x=a^2y$. Then we get that:
\begin{equation}
\begin{split}
\int_0^{a^2}\sqrt{a-\sqrt{x}}dx & =\int_0^{1}a^2\sqrt{a-\sqrt{a^2y}}dy\\
&=\int_0^{1}a^2\sqrt{a-a\sqrt{y}}dy\\
&=\int_0^1a^2\sqrt{a(1-\sqrt{y})}dy\\
&=\int_0^1a^2a^{\frac{1}{2}}\sqrt{1-\sqrt{y}}dy\\
&=a^{\frac{5}{2}}\int_0^1\sqrt{1-\sqrt{y}}dy\\
&=\frac{8a^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove $\lim\limits_{(x,y) \to (0,0)} \frac{xy(y^2-x^2)}{x^2+y^2}=0$ Prove $\lim\limits_{(x,y) \to (0,0)} \frac{xy(y^2-x^2)}{x^2+y^2}=0$
My attempt: For all $\epsilon > 0$ there is a $\delta > 0$ such that $\left | \frac{xy(y^2-x^2)}{x^2+y^2} \right| < \epsilon \rightarrow \sqrt{x^2+y^2} < \delta$.
Note important facts:... | You should try with polar coordinates, it gets way easier. Impose:
$$\begin{cases} x=\rho\cos\theta\\y=\rho\sin\theta
\end{cases}
$$
and $(x,y)\longrightarrow(0,0)$ will turn into $\rho\longrightarrow0$. As the numerator goes as $\rho^{4}$ and the denominator goes as $\rho^{2}$ your limit is proven. If $\rho$ simplifie... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$a_1Give that $a_1<a_2<a_3<a_4$ are positive integers such that $$\sum_{i=1}^4\frac{1}{a_i}=\frac{11}{6}$$, find the value of $a_4-a_2$.
My try:
Since $11$ is prime, atleast one of $a_1,a_2,a_3,a_4$ should be divisible by $6$.
Now we should express the fraction $\frac{11}{6}$ as $\frac{A+B+C+D}{E}$ such that $E$ is div... | The only choice of $a_1,a_2$ is $a_1=1,a_2=2$ with the reasoning given by Hagen von Eitzen.
So we have $$\frac{1}{a_3}+\frac{1}{a_4}=\frac{11}{6}-\frac{3}{2}=\frac{1}{3}$$
$\implies$
$$(3-a_3)(3-a_4)=9$$
The choice of $a_3=3$ is now ruled out.
So $a_4>a_3>3$ and hence number $9$ can be factored out only in one way viz... | {
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"answer_count": 3,
"answer_id": 2
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Find the matricial representation Let $M_{2}\mathbb{R}$ be the matrix space of $2 \times 2$ with real entries , and $P_{2}[x]$ the space of polinomies with real coeficientes with grade at most 2. Considerate the linear transformation $T: M_{2}(\mathbb{R}) \rightarrow P_{2}[x]$ defined like
$$T \begin{pmatrix} a & b \\ ... | I think it is right:
$$\begin{pmatrix} 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 2\\
0 & 1 &0 & 0
\end{pmatrix}\begin{pmatrix} a \\ b \\ c \\ d \\
\end{pmatrix}\,=\,\begin{pmatrix} a+b \\ 2d \\ b \\
\end{pmatrix}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all $ a,b,c \in \mathbb{R} $ that satisfies the equations, $ \bullet $ $ a+b+c=63$ and $ \bullet $ $ab+bc+ac=2021$? Find all $ a,b,c \in \mathbb{R} $ that satisfies the equations,
$ \bullet $ $ a+b+c=63$
$ \bullet $ $ab+bc+ac=2021$ ?
I try to solve this problem but going to the result that this problem has no solu... | By Vieta's formula we have
$$f(x) = x^3 - 63x^2 + 2021x -abc = 0$$
This function has only one real root since $f'(x) > 0$ for all $x$.
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What does $\sum_{r=1}^{n} r^{k}$ equal in general I know that $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$, $\sum_{r=1}^{n} r^{2} = \frac{n(n+1)(2n+1)}{6}$, $ \sum_{r=1}^{n} r^{3} = \frac{n^2(n+1)^2}{4}$
but what does $\sum_{r=1}^{n} r^{k}$ equal in general ?
Can we express it in some simple predetermined form
| We may also express this in terms of Stirling numbers of the second
kind and falling factorials. We start with
$$\sum_{r=0}^n r^k
= k! [z^k] \sum_{r=0}^n \exp(rz)
= k! [z^k] \frac{\exp((n+1)z)-1}{\exp(z)-1}
\\ = k! [z^k] \frac{1}{\exp(z)-1}
\sum_{q=1}^{n+1} {n+1\choose q} (\exp(z)-1)^q
\\ = k! [z^k]
\sum_{q=1}^{n+1} {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $a$,$b$ and $y$ are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$ If a b and y are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$
This is what I have done so far, but apparently it is incorrect. I want to know why.
$(b+1/y)(y+1/a)(a+1/b)$
$(by+1/y)(ay+1/a)(ab+1/b)$
$(aby^2+1/ay)(ab+1/b)$
which... | Hint:
$$(b+\frac{1}{y})(y+\frac{1}{a})(a+\frac{1}{b})=aby + \frac{1}{aby} + (a + b + y) + (\frac{1}{a} + \frac{1}{b} + \frac{1}{y})$$
(Here is the WolframAlpha verification for this computation.)
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Complex roots of quartic equation during solution of $z^5=1$ While finding the 5th roots of unity $z^5=1$, I arrived at the following
$$(z-1)(z^4+z^3+z^2+z+1)=0$$
Now, I am well aware that I can arrive at the roots by using the fact that each root is separated by a value of $\alpha=2k\pi$
$$z^5=1 +0i \implies arg(z^5)=... | Factorize the equation as follows
\begin{align}
z^5-1= &(z-1)(z^4+z^3+z^2+z+1)\\
=&(z-1)z^2(z^2+z+1+\frac1z+\frac1{z^2})\\
=&(z-1)z^2\left( (z+\frac1z)^2+(z+\frac1z)-1\right)\\
=&(z-1)z^2\left( z+\frac1z -\frac{1+\sqrt5}2\right)\left(z+\frac1z -\frac{1-\sqrt5}2\right)\\
=&(z-1)\left( z^2-\frac{1+\sqrt5}2z+1\right)\lef... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is $\sum_{n=1}^n(-1)^m\frac{m^{n-m+1}}{(n-m)!}$? So consider the sum
$$S=\sum_{m=1}^n(-1)^m\frac{m^{n-m-1}}{(n-m)!},$$
where $n$ is some fixed, positive integer. For specific values of $n$, this gives $-1$, $-\frac{1}{2}$, $\frac{1}{6}$, $\frac{1}{12}$, $-\frac{3}{40}$, $-\frac{1}{120}$, $\frac{31}{1008}$, $-\frac... | The generating function for
$$S_n = \sum_{m=1}^n (-1)^m \frac{m^{n-m-1}}{(n-m)!}$$
may be proved as follows:
$$\sum_{m=0}^{n-1} (-1)^{n-m} \frac{(n-m)^{m-1}}{m!}
= \frac{(-1)^n}{n}
+ \sum_{m=1}^{n-1} (-1)^{n-m} \frac{(n-m)^{m-1}}{m!}
\\ = \frac{(-1)^n}{n}
+ \sum_{m=1}^{n-1} (-1)^{n-m} \frac{1}{m} [z^{m-1}]
\exp((n-m)z)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Prove that $H_n - H_m > \frac{n-m}{n}$ for $n, m \in \mathbb{N}$ and $n > m \geq 1$, by doing induction on $n$ and base case $n = m$ Problem:
Let $H_n = \sum_{i=1}^{n} \frac{1}{i} $ for all natural numbers $n \geq 1$.
Prove that $H_n - H_m \geq \frac{n-m}{n}$ for $n, m \in \mathbb{N}$ and $n > m \geq 1$, by doing induc... | The problem occurs when you are subtracting the sums
$H_{n+1} - H_m = 1 + \frac{1}{2} + ... + \frac{1}{n+1} - \Big(1 + \frac{1}{2} + ... + \frac{1}{m}) = $
$ = 1 + \frac{1}{2} + ... + \frac{1}{m} + \frac{1}{m+1} + ... + \frac{1}{n+1} - \Big(1 + \frac{1}{2} + ... + \frac{1}{m}) $
$ = \frac{1}{m+1} + ... + \frac{1}{n+1} ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $T$ is surjective
Let $T:S_2 \to \mathbb{R}_3[x]$ such that $$T\left(\begin{bmatrix} a &
b\\ b & c \end{bmatrix}\right)=(a+b+c) + (-a+2c)x+(2a+3b+6c)x^2.$$
Prove that $T$ is surjective.
My attempt:
Let $w \in \mathbb{R}_3[x]$ such that $w=(a+b+c) + (-a+2c)x+(2a+3b+6c)x^2$.
So $w=a(1-x+2x^2)+b(1+3x^2)+c(1... | A function $f:X\rightarrow Y$ is said to be surjective if:
For every $y\in Y$ there exists and $x\in X$ such that $f(x)=y$.
We need to show that for every $a+bx+cx^2 \in \mathbb{R}_3[x]$ for arbitrary $a,b,c \in \mathbb{R}$ that there exists some symmetric matrix $S\in S_2$ such that $T(S)=a+bx+cx^2$.
The short answer ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does $\int_1^{\infty}\frac{x^2\tan^{-1}(ax)}{x^4+x^2+1}dx$ have closed form? I have been trying to find the closed form for integral below $$\int_1^{\infty}\frac{x^2\tan^{-1}(ax)}{x^4+x^2+1}dx ,\; \; a>0 $$
My progress to this integral $$\cong\frac{\pi^2}{8\sqrt 3}+\frac{\pi}{8}\log(3)-\frac{\pi}{6a\sqrt 2}+\frac{1}{2... | You could simplify the problem using
$$\frac{x^2}{x^4+x^2+1}=\frac {x^2}{(x^2-r)(x^2-s)}=\frac 1{r-s} \left(\frac{r}{x^2-r}-\frac{s}{x^2-s}\right)$$ where
$$r=-\frac{1+i \sqrt{3}}{2} \qquad \text{and} \qquad s=-\frac{1-i \sqrt{3}}{2}$$ So, the problem boils down to the computation of
$$I(t)=\int_1^\infty \frac{\tan ^{-... | {
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"source": "stackexchange",
"question_score": "3",
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If $(x+1)(x+2)(x+3)(x+k) + 1$ is a perfect square then the value of $k$ is? If $(x+1)(x+2)(x+3)(x+k) + 1$ is a perfect square then the value of $k$ is ?
My approach:
The given expression can be written as $(y-1)(y)(y+1)(y+k-2) + 1$ where $y = x + 2$
This gives me $(y)(y^2 - 1)(y+k-2) = n^2 - 1$ something seems similar ... | $$(x+1)(x+2)(x+3)(x+k) + 1 = f(x)^2$$
for all $x$ then $f(x)=x^2+ax+b$ for all $x$ and some $a,b$.
Since $f(0)^2 = 6k+1$ and $f(-1)=f(-2)=f(-3)=\pm1$ so two of them must be equal and we have next $6$ options:
*
*If $f(-1)=f(-2)=1$ then $f(x) = (x+1)(x+2)+1 = x^2+3x+3$ so $f(-3) = 3$. Impossible.
*If $f(-1)=f(-2)=-1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Improper integral $-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$ If I have to solve $-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$ it is right to write:
$$-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\geq \int_{3}^{+\infty}\frac{-1}{x^{\frac{1}{3}}}$$
and so the integral is diverg... | $$\int_{2n\pi}^{2n\pi+\pi}\frac{\sin^4x}{x^{1/3}}dx\geq\frac{1}{\sqrt[3]{2n\pi+\pi}}\int_{2n\pi}^{2n\pi+\pi}\sin^4xdx\\
=\frac{1}{\sqrt[3]{2n\pi+\pi}}\int_{0}^{\pi}\sin^4xdx\\
=\frac{2}{\sqrt[3]{2n\pi+\pi}}\int_{0}^{\frac\pi2}\sin^4xdx
=\frac{1}{\sqrt[3]{2n\pi+\pi}}\cdot\frac{3\pi}{8}.$$
So
$$\int_{3}^{+\infty}\frac{\s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving for the derivative of absolute value, gone wrong The absolute value $|x|$ can be represented as $\sqrt{x^{2}}$, as per this question.
Let $f(x) = |x| = \sqrt{x^{2}}$. Solving for $f'(x)$, let $u = x^{2}$. Then,
\begin{align*}\frac{df}{dx} &= \frac{df}{du}\cdot\frac{du}{dx} \\ &= \frac{d}{du}(\sqrt{u})\cdot\frac... | Both are fine. What happens is that the second approach only makes sense when $x>0$. And then you do indeed have $(\sqrt{x^2})'=1\left(=\frac x{|x|}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove $9\mid (x^2 + y^2) \iff 3\mid x \text{ and } 3\mid y$ Prove: Let $x, y$ be two non-negative integers. Then $9|(x^2 + y^2)$ if and only if $3|x$ and $3|y$.
General format of proving iff statement.
Proof.
Part 1, proving $p \implies q$
Part 1, proving $q \implies p$
Therefore, $p \iff q. \qed$
| Here's a (very) slightly different approach, presented more to help you with proof-writing than with the math (which I think you understand).
First, assume $3 \mid x$ and $3 \mid y$. Then $9 \mid x^2$ and $9 \mid y^2$, so $9 \mid (x^2+y^2)$. That proves $\Leftarrow$.
Conversely, assume $3 \nmid x$. Then $x^2 \equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Let $02$
Let $0<a<b<1$ and $E=\log_{a}{\frac{2ab}{a+b}} + \log_{b}{\frac{2ab}{a+b}}$
Prove that $E>2$
I tried to change both log bases to $\frac{2ab}{a+b}$ and I got $E=\frac{1}{\log_{\frac{2ab}{a+b}}{a}} + \frac{1}{\log_{\frac{2ab}{a+b}}{b}}$.
Also I noticed that $\frac{2ab}{a+b} < 1$ because $\frac{2}{\frac{1}{a}... | Bringing to $\ln$, we have
$$\begin{align}
E
&=\log_{a}{\frac{2ab}{a+b}} + \log_{b}{\frac{2ab}{a+b}} \\
&= \frac{\ln\frac{2ab}{a+b}}{\ln a} + \frac{\ln\frac{2ab}{a+b}}{\ln b} && \left(1\right) \\
&= \ln\frac{2ab}{a+b}\left(\frac{1}{\ln a} + \frac{1}{\ln b}\right) \\
&> \ln\sqrt{ab}\left(\frac{1}{\ln a} + \frac{1}{\ln b... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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$a_n=(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n}) $ converge Hi guys I have a problem that i need to prove that the sequence:
$$a_n=(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n}) $$ converges
I need to show its a monotone sequence with an upper and lower bound.
For lower bound, 0 seems to fit nicely.
I ... | Note
\begin{eqnarray}
a_n&=&(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n})\\
&=&\frac{(1-\frac12)(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n})}{1-\frac12}\\
&=&2(1-\frac{1}{2^2})(1+\frac{1}{2^2})\dots (1+\frac{1}{2^n})\\
&=&\cdots\\
&=&2(1-\frac{1}{2^{2n}})
\end{eqnarray}
and hence $\{a_n\}$ converges.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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A nonnegative function $f(a,b)\geq 0$ Let $c$ be a fixed positive real umber such that $c\geq 1.$ If $a\geq c^m, b\geq c, $ where $m$ is any positive integer, then is
$$f(a,b)=c(ab+1)(a-c^m)(b-c)+c(ab+1)(a-c^m)(b+1)+c^m(b-c)(ab+1)(a+1)-(ab-c^{m+1})(a+1)(b+1)\geq 0?$$
We can check the validity for the two cases $a=c^m... | Yes, it's true if all my computation is right.
First, the last term can be divided into two part.
$$(ab-c^{m+1})(a+1)(b+1)=(ab-c^{m+1})(ab+1)+(ab-c^{m+1})(a+b) $$
Use the first term to deal the first part and see what remains
$$ c(ab+1)(a-c^m)(b-c)-(ab-c^{m+1})(ab+1)\geq (ab+1)(a-c^m)(b-c)-(ab-c^{m+1})(ab+1)\\
=-(ab+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Possible pattern involving $x$ in the continued fraction expansion of $\frac{1}{\sqrt[3]{x^3+1}-x}$ Consider the expression $$\frac{1}{\sqrt[3]{x^3+1}-x}$$
Plugging in $10$ for $x$ and using $W|A$, we find that the continued fraction expansion is $[300; \mathbf{10}, 450, 8, ...]$.
For $x=11$, it's $[363; \mathbf{11}, 5... | Partial answer: the provided numerical examples strongly suggest that the continued fraction expansion would be of the form $[3x^2,x,\ldots]$. So we only need to show that for $x>0$ we have
$$
3x^2+\frac{1}{x+1} < \frac{1}{\sqrt[3]{x^3+1}-x} < 3x^2+\frac{1}{x}
$$
A few steps of algebraic manipulations on the right-side... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does parity of $f(a)$ and $f(a+1)$ are same whenever $a$ is even? Question related to sum of digits Let $D(a, b)$ denotes sum of digits of $a$ in base $b$.
Example: $D(5,2)=2,D(1227,10)=1+2+2+7=12$
Define $f(a)=\sum_{i=2}^aD(a, i)$ where $a\ge2$.
Example $f(5)=D(5,2)+D(5,3)+D(5,4)+D(5,5)=2+3+2+1=8$
Can it be shown that... | Assume $a$ is even. That means we already know the parity with $i=2j+1$ odd: $D(a, 2j+1)=2k_j$ (even) and $D(a+1,2j+1)=2k'_j+1$ (odd) for all $j=1,...,\frac{a}{2} - 1$.
Now take a look for $i=2j$ even: We don't know the concrete value of $D(a,2j)$, but we can draw a connection between $D(a,2j)$ and $D(a+1,2j)$. As $a$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4135633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Redefine the P matrix so that it’s made of eigenvectors with integer components.
I’m trying to redefine the P matrix so that it’s made of eigenvectors with integer components (which will give the same diagonalizing matrix). But I’m stuck because I’m not sure how to redefine all the columns. I think the first column wi... | $$A = \begin{pmatrix}
\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\
-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
\end{pmatrix}$$
$$A = P D P^{-1} =\begin{pmatrix}
-1 & 1 & -1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How many terms are there containing the term $xyk^2$ in the expansion of $(2x-y+t+3z+4k)^8$
How many terms are there containing the term $xyk^2$ in the expansion of $(2x-y+t+3z+4k)^8$ such as $xyk^2t^2z^2$ or $xyk^2t^4z^0$ or $xyk^2z^3t$ etc.
I made up this question and calculated it , but i do not know whether my so... | Both solutions are correct. The reasoning to obtain $5$ in the first solution is somewhat more detailed. The essence is in both solutions the same, namely to keep the focus on the number of different terms in the expansion of $(t+3z)^4$.
Here is a crosscheck (kinda overkill). We use the coefficient of operator $[x^k]$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\|x^*\|=1$ and $\|x\|=\|y\|=1$ then $\|x-y\|<\epsilon$ Suppose $X$ is a space with inner product. Let $x^{\ast} \in X^{\ast}, \| x^{\ast} \| = 1$, and $0 < \varepsilon < 1$. If
$x, y \in X, \| x \| = \| y \| = 1, x^{\ast} (x) > 1 -
\frac{\varepsilon^2}{8}$, and $x^{\ast} (y) > 1 - \frac{\varepsilon^2}{8}$,
then $\|... | By the parallelogram law we have
\begin{equation}
\| x + y \|^2 + \| x - y \|^2 = 2 (\| x \|^2 + \| y \|^2) \implies \| x +
y \|^2 = 4 - \| x - y \|^2 \tag{1}
\end{equation}
Also by hypothesis
\begin{align}
x^{\ast} (x) &> 1 - \frac{\varepsilon^2}{8}\\
x^{\ast} (y) &> 1 - \frac{\varepsilon^2}{8}\\
\\ \h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How does one find values of $m$ for which the roots of $2x^2-mx-8=0$ differ by $m-1$
Find values of $m$ for which the roots of $2x^2-mx-8$ differ by $m-1$.
When I attempted to solve this I tried to simplify it into something like this: $((m-1)+2)((m-1)-4)$ but when I expanded I got $m^2 -4m -5$ which is nowhere near ... | We may also directly construct the polynomials we seek . The parabola corresponding to $ \ y \ = \ 2x^2 - mx - 8 \ = \ 2·\left(x - \frac{m}{4} \right)^2 \ - \ \left(8 + \frac{m^2}{8} \right) \ $ has its vertex at $ \ x = \frac{m}{4} \ \ , $ which is the midpoint of its $ \ x-$ intercepts. (These always exist, since t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Problem with integration by substitution.
Using the substitution $u=1-x$, compute the integral of
$\int{x(1-x)^2}dx.$
My Work: Let $u=1-x.$ Then $\mathrm dx=-\mathrm du$ and $x=1-u,$ so
$$\int{x(1-x)^2}\,\mathrm dx\\
=-\int{(1-u)u^2}\,\mathrm du\\
=\frac{u^4}{4}-\frac{u^3}{3}+C\\
=\frac{(1-x)^4}{4}-\frac{(1-x)^3}{3}+... | Let the domain of $f$ be an interval.
Then every pair of the infinitely many antiderivatives of $f$ differ by a constant. Since the indefinite integral $$\int f(x)\,\mathrm dx$$ gives the general specification of these antiderivatives, it has the form $F(x)+C,$ where the constant of integration $C$ is arbitrary.
Both y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}$.
Evaluate $$\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}.$$
My work so far and background to the problem.
This question was inspired by the second page of this paper. The author of the paper doesn't mention how he managed to prove the ... | Hint: We find in Evaluations of Binomial Series by J. M. Borwein and R. Girgensohn the formula (47)
\begin{align*}
\sum_{n\geq 1}\frac{1}{n^3\binom{2n}{n}}=\frac{2}{3}\pi\Im\left(L_2\left(e^{i\pi/3}\right)\right)
-\frac{4}{3}\zeta(3)
\end{align*}
with $L_p(z)=\sum_{n>0}\frac{z^n}{n^p}$ the polylogarithms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
On the value of $a+b$ such that $a^2+b^2+3ab=2000$ where $a,b \in \mathbb{Z}^+$
Find the value of $a+b$ such that $a,b \in \mathbb{Z}^+$ such that $a^2+b^2+3ab=2000.$
On the outside, this seems like a very easy problem, but I'm not sure how to do this in a non-messy way. What I have so far is that $$a^2+b^2+3ab=(a+b)... | One less messy way of doing this is to notice that
*
*if $a^2 + 3ab + b^2$ is even, then both $a, b$ must be even.
*if $a^2 + 3ab + b^2$ is a multiple of $25$, then both $a, b$ are multiples of $5$.
Thus $a^2 + 3ab + b^2 = 2000$ leads to $a = 20x, b = 20y$ with $x^2 + 3xy + y^2 = 5$.
Proof of 1:
We have $0\equiv a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.