Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$. Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$.
My solutions: the homogeneous portion is $a_n=c3^n$, and the inhomogeneous portion is $a^*_n=-1/2n^2-3/4n+9/8$.
This results in a final recurrence relation of
$$a_n=-\frac{1}{8}3^n-\frac{... | Given the homogenous solution: $a_n^{(h)} = c_03^n$ and the particular solution: $a_n^{(p)} = c_1n^2+c_2n+c_3$, and we have: $a_n = a_n^{(h)}+a_n^{(p)}$. First we find the coefficients to the particular solution:
$$
\begin{align}
&a_n^{(p)} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 3a_{n-1}^{(p)} &&= n^2+0n-3\\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How can I prove the last two digits of $1+2^{2^{n}}+3^{2^n}+4^{2^n}$ always are $54$ How can I prove the last two digits of $$1+2^{2^{n}}+3^{2^n}+4^{2^n}$$ are $54$ when $n$ is a positive integer number if $n>1$
| These things go in cycles with periods of length $4$.
The last two digits of $4^{2^n}$ are $16$, $56$, $36$ or $96$.
For $n > 4$, the last two digits of $3^{2^n}$ are $41$, $81$, $61$ or $21$.
For $n > 4$, the last two digits of $2^{2^n}$ are $96$, $16$, $56$ or $36$.
Then notice that $16 + 41 + 96 + 1 = 154$, and ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Taylor Series of $2xe^x$ I have to find the Taylor Series for $2xe^x$ centred at $x=1$. I came up with the following.
$$e^x = e^{x-1} \times e = e \bigg( \sum_{n=0}^\infty \frac{(x-1)^n}{n!}\bigg)$$
Then consider $2xe^x$.
$$2xe^x = 2xe \bigg( \sum_{n=0}^\infty \frac{(x-1)^n}{n!}\bigg) = \bigg( \sum_{n=0}^\infty \frac{2... | Write $2xe^x=2(x-1)e^x+2e^x$ and $e^x=e\cdot e^{x-1}$ to obtain $2xe^x=2e(x-1)e^{x-1}+2e\cdot e^{x-1}$. Since
$$
e^{x-1}=\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k,
$$
we have
\begin{align*}
2xe^x&=2e(x-1)e^{x-1}+2e\cdot e^{x-1} \\ &= 2e(x-1)\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k+2e
\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k \\ &=
2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Using an Integral to Solve for a Variable a I am struggling to use the following equation:
$$
\int_0^a \sqrt{a^2-x^2}\,\,\text{sgn}(|x|-1)\, dx = 0
$$
where $a > 1$, to deduce that $a = \text{cosec}(\frac{\pi}{4} - \frac{\alpha}{2})$, where $\alpha$ satisfies $\alpha = \cos(\alpha)$.
I integrate the integrand, via
$$
\... | First, $a>1$ and $x\in[0,a]$, then $|x|=x\ge0$
$$
\renewcommand\sgn{\operatorname{sgn}}
\renewcommand\arcsec{\operatorname{arcsec}}
\begin{array}{ll}
0\!\!\!&=\int_0^a\sqrt{a^2-x^2}\sgn(x-1)\,\mathrm dx\\
&=\int_1^a\sqrt{a^2-x^2}\,\mathrm dx-\int_0^1\sqrt{a^2-x^2}\,\mathrm dx
\end{array}
$$
Then
$$\int_1^a\sqrt{a^2-x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Express this linear optimization problem subject to a circular disk as a semidefinite program I have to express the following problem as a semidefinite program
$$\begin{array}{ll} \text{minimize} & F(x,y) := x + y +1\\ \text{subject to} & (x-1)^2+y^2 \leq 1 \tag{1}\end{array}$$
Only affine equality conditions should be... | First of all, let's simplify a bit:
$$(x-1)^2+y^2\leq 1 \quad\Longleftrightarrow\quad x^2-2x+1+y^2\leq 1
\quad\Longleftrightarrow\quad x^2+y^2\leq 2x$$
Note that this implies that $x\geq 0$, which makes sense since the original inequality describes a disc centered at $(1,0)$ with radius $1$.
Let's go a bit further: add... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\frac{1}{\sqrt{x^2-x}}dx$ For a differential equation I have to solve the integral $\frac{dx}{\sqrt{x^2-x}}$. I eventually have to write the solution in the form $ x = ...$ It doesn't matter if I solve the integral myself or if I use a table to find the integral. However, the only helpful integral in an in... | $$
\displaylines{
\sqrt {x^2 - x} = t + x \cr
\Leftrightarrow x^2 - x = t^2 + 2xt + x^2 \cr
\Leftrightarrow - x = t^2 + 2xt \cr
- x = \frac{{t^2 }}{{1 + 2t}} \Rightarrow - dx = \frac{{2\left( {t^2 + 1} \right)}}{{\left( {1 + 2t} \right)^2 }}dt \cr
\frac{{dx}}{{\sqrt {x^2 - x} }} = - \frac{{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 1
} |
Prove that $\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$
Prove that $$\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$$
I tried to to prove the above statement using the AM-HM inequality:
$$\begin{align}\frac{1}{2^n - 2^{n-1}}\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} &\ge \f... | Assuming $a,b\gt0$, we get
$$
\begin{align}
\sum_{i=2^{n-1}+1}^{2^n}\left(\frac1{a+ib}-\frac{2^{-n+1}}{a+2b}\right)
&\ge\sum_{i=2^{n-1}+1}^{2^n}\left(\frac1{a+2^nb}-\frac{2^{-n+1}}{a+2b}\right)\\
&=\sum_{i=2^{n-1}+1}^{2^n}\frac{a(1-2^{-n+1})}{(a+2^nb)(a+2b)}\\
&=\frac{a(2^{n-1}-1)}{(a+2^nb)(a+2b)}\\[12pt]
&\ge0
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integration involving $\sin x/x$ Let $f$ be a differentiable function satisfying:
$$\int_0^{f(x)}f^{-1}(t)dt-\int_0^x(\cos t-f(t))dt=0$$
and $f(\pi)=0$, Considering $g(x)=f(x)\forall x\in\mathbb R_0=\mathbb R+{0}$.
If $$\int_0^{\infty}(g(x))^3dx=A$$ and $$\int_0^{\infty}\frac{1-g(x)}{x^2}dx=\frac{kA}{k+1}$$ then k is?
... |
But how can someone calculate $\quad\quad\quad\displaystyle\int_0^{\infty}\frac{\sin^3x}{x^3}dx$
Assuming that you know $$\int_0^{\infty}\frac{\sin x}{x}=\frac{\pi}{2}$$
Use$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\displaystyle\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$
And then Integrate b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &... | Notice that $\tan x\times \cot\ x=1$, and
according to the inequality of means we get $$\tan x+\cot x \geqslant 2\sqrt{\tan x\times \cot x}=2$$
which is what we wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
Integral of constant divided by polynomial and another constant $$\int_{-\infty}^{-1}\frac{4}{\sqrt{x^6+2}}\,dx$$
What are the steps to integrate?
| $\int_{-\infty}^{-1}\dfrac{4}{\sqrt{x^6+2}}dx$
$=\int_\infty^1\dfrac{4}{\sqrt{(-x)^6+2}}d(-x)$
$=\int_1^\infty\dfrac{4}{\sqrt{x^6+2}}dx$
$=\int_1^\sqrt[6]2\dfrac{4}{\sqrt{x^6+2}}dx+\int_\sqrt[6]2^\infty\dfrac{4}{\sqrt{x^6+2}}dx$
$=\int_1^\sqrt[6]2\dfrac{2\sqrt2}{\sqrt{1+\dfrac{x^6}{2}}}dx+\int_\sqrt[6]2^\infty\dfrac{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Compute $\int \frac{\mathrm{d}x}{49x^2+1}$ So I tried solving this by taking a substitute for the integrand, $t=49x$, so its derivative is $dx = \frac {dt} {49}$. Then you insert it into the integrand and get $$\int \frac{\mathrm dt}{49(t^2 +1)} = \frac{1}{49}\int\frac{1}{t^2 +1}dt = \frac{1}{49}\arctan t + c = \frac... | \begin{align}
& \int\frac{dx}{49x^2+1} = \int\frac{dx}{(7x)^2+1} = \frac 1 7\int \frac{7\,dx}{(7x)^2+1} = \frac 1 7 \int\frac {dt}{t^2+1} \\[10pt]
= {} & \frac 1 7 \arctan t + C = \frac 1 7 \arctan(7x)+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to prove $\lim \limits_{n \rightarrow \infty} \frac{2^n n!}{n^{n+1}} = 0$ I want to prove the following and wanted to ask, if my proof is correct.
$$\lim \limits_{n \rightarrow \infty} \frac{2^n n!}{n^{n+1}} = 0$$
Remark (i): For $n \in \mathbb{N}$ and $n > 1$ holds: $2 < (1+\frac{1}{n})^n$
Define $b_n := \left\{... | I present an elementary solution.
Write $$\ell = \lim_{n \to \infty} \frac {2^n n!} {n^{n + 1}} = \lim_{n \to \infty} \frac {1} {n} \prod_{k = 1}^{n} \frac {2k} {n}.$$ For $k < n/2$, write $2k/n = 1 - a_k$, where $0 < a_k < 1$. Then, obviously, for $k > n/2$ we'll have $2k/n = 1 + a_k$. Hence, $$\ell = \lim_{n \to \inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Non-linear equation system Resolution I want to prove (I am not sure it is true though) that
$$\left\{\begin{array}{c} x-y-x\left( x^2+5y^2\right)=0\\ x+y-y\left( x^2+y^2\right) =0\end{array}\right.$$
Admits no other solution in the plane than the origin.
================================================================... | As you have tried $x^2+y^2= \frac{x+y}{y}$, and then I tried, $x^2+5y^2=\frac{x-y}{x}$.. Now, from that I 've found $y^2=-\frac{x^2+y^2}{4xy}$. But $x^2+y^2>0$[assuming $\left( x,y\right)\in \mathbb{R}^2-\{ \left( 0,0\right)\}$]. So, $xy>0$.
So, $x>0 \leftrightarrow y>0$ and $x<0 \leftrightarrow y<0$
Now, remark that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use Laplace Transform to solve the following IVP: I know that this is a somewhat simple problem but I have been having trouble coming up with the little "tricks" that help with Laplace.
The problem is:
$y''+2y' +5y = e^{-t}\sin(2t)$ where $y(0) = 2, y'(0) = -1$
Attempt at Solution
$(s^2+2s+5)Y = \frac{2}{(s+1)^2+4} + ... | You need the inverse Laplace transform of
$$ \dfrac{2}{(s^2 + 2 s + 5)^2}$$
This may help:
$$ {\mathcal L} \{t e^{at} \cos(bt)\} = \dfrac{(s-a)^2 - b^2}{((s-a)^2+b^2)^2}
= \dfrac{1}{(s-a)^2 + b^2} - \dfrac{2b^2}{((s-a)^2+b^2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving $\sin\theta -1=\cos\theta $
Solve$$\sin\theta -1=\cos\theta $$
Steps I took to solve this:
$$\sin^{ 2 }\theta -2\sin\theta +1=1-\sin^2\theta $$
$$2\sin^{ 2 }\theta -2\sin\theta =0$$
$$(2\sin\theta )(\sin\theta -1)=0$$
$$2\sin\theta =0, \sin\theta -1=0$$
$$\quad \sin\theta =0, \sin\theta =1$$
$$\theta =0+\pi ... | use that $$\sin(\theta)=2\,{\frac {\tan \left( \theta/2 \right) }{1+ \left( \tan \left( \theta
/2 \right) \right) ^{2}}}
$$
and $$\cos(\theta)={\frac {1- \left( \tan \left( \theta/2 \right) \right) ^{2}}{1+
\left( \tan \left( \theta/2 \right) \right) ^{2}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Mixed Conditioning - Two Normal Distributions
Let $Z \sim \mathcal{N}(0,1)$ and $Y|Z \sim \mathcal{N}(Z, 1)$.
Show that $f_{Z|Y}(z|y)$ is a normal density, and find the parameters of this density.
What I have so far:
\begin{align*}
f_Z(z) &= \phi(z)\\
f_{Y|Z}(y|z) &= \frac{1}{\sigma} \phi\left( \frac{y - \mu}{\sigma}... | $$
f_{Z|Y}(z|y) =
\frac{f_Z(z)f_{Y|Z}(y|z)}{\int_{-\infty}^{+\infty}{f_Z(a)f_{Y|Z}(y|a)da}} =
\frac{\frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-z)^2}{2}}}
{\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}} e^{-\frac{a^2}{2}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-a)^2}{2}}da}} =
\frac{e^{-z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Convergence of a complex function I need to proof if the following function is bounded and convergent.
$f(n)=\left(\frac{10+in}{n^{2}+2in}\right)^{n}$
Status:
This should be correct. Can anybody confirm this?
I tried it with Bernoulli now:
$\lim_{n\rightarrow\infty}q^{n}=0$ for $\left|q\right|<1,\quad q\in\mathbb{C}$
... | $$f(n)=\left(\frac{10+in}{n^{2}+2in}\right)^{n}$$
Gives:
$$f(n)=\left(\frac{\sqrt{n^2+100}e^{(arg(10+in))i}}{\sqrt{n^4+4n^2}e^{(arg(n^2+2in))i}}\right)^{n}=$$
$$f(n)=\left(\frac{\sqrt{n^2+100}}{\sqrt{n^4+4n^2}}e^{((arg(10+in))-(arg(n^2+2in)))i}\right)^{n}=$$
$$f(n)=\sqrt{\frac{{n^2+100}}{{n^4+4n^2}}}^ne^{((arg(10+in))-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x→0}\left(\frac{1+\tan x}{1+\sin x}\right)^{1/x^2} $ I have the following limit to evaluate:
$$ \displaystyle \lim_{x→0}\left(\frac{1+\tan x}{1+\sin x}\right)^{1/x^2} $$
What's the trick here?
| Using Bernoulli's Inequality,
$$
\begin{align}
\left(\frac{1+\tan(x)}{1+\sin(x)}\right)^{\large1/x^2}
&=\left(1+\frac{\tan(x)-\sin(x)}{1+\sin(x)}\right)^{\large1/x^2}\\
&=\left(1+\frac{1-\cos(x)}{x^2}\frac{\tan(x)}{x}\frac1{1+\sin(x)}x^3\right)^{\large1/x^2}\\
&\ge1+\underbrace{\vphantom{\frac1{\sin(x)}}\frac{1-\cos(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Finding an interval containing the values of $ abc(a + b + c) $, given a quadratic constraint Suppose $ a, b, c $ are real numbers such that $$ (ab)^2 + (bc)^2 + (ca)^2 = k $$
then all possible values of $ abc(a + b + c) $ lies between which interval?
| Note $$x^2+y^2+z^2\ge xy+yz+zx$$
then we have
$$k=(ab)^2+(bc)^2+(ac)^2\ge abc(a+b+c)$$
other hand
$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)\ge 0$$
so
$$xy+yz+xz\ge -\dfrac{1}{2}(x^2+y^2+z^2)$$
so
$$abc(a+b+c)\ge -\dfrac{1}{2}[(ab)^2+(bc)^2+(ac)^2]=-\dfrac{k}{2}$$
so
$$abc(a+b+c)\in [-\dfrac{k}{2},k]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1062995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano
Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$
Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.
Also, can w... | Substitute $x=\frac1{t-1}$ to translate $x^3-9 x^2-15x-6 =0$ into
$$6t^3-3t^2-3t-1=0$$
and then rewrite it as $7t^3=(t+1)^3$,
which yields $t=\frac1{\sqrt[3]7-1}$ and in turn the solution
$$x= \frac1{t-1} =\frac{\sqrt[3]7-1}{2-\sqrt[3]7}=\sqrt[3]{49}+\sqrt[3]7+3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
A problem with proving using definition that $\lim_{n\to\infty}\frac {n^2-1}{n^2+1}=1$
Prove using the definition that: $$\displaystyle\lim_{n\to\infty}\frac {n^2-1}{n^2+1}=1 $$
What I did:
Let $\epsilon >0$, finding $N$: $\mid\frac {n^2-1}{n^2+1}-1\mid=\mid\frac {-2}{n^2+1}\mid\le\mid\frac {-2}{n^2}\mid=\frac {2}{n^... | Hint:-
$\dfrac{n^2-1}{n^2+1}=1-\dfrac{2}{n^2+1}<1-\dfrac{1}{n^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
prove that $\sqrt{2} \sin10^\circ+ \sqrt{3} \cos35^\circ= \sin55^\circ+ 2\cos65^\circ$ Question:
Prove that: $\sqrt{2} \sin10^\circ + \sqrt{3} \cos35^\circ = \sin55^\circ + 2\cos65^\circ$
My Efforts:
$$2[\frac{1}{\sqrt{2}}\sin10] + 2[\frac{\sqrt{3}}{2}\cos35]$$
$$= 2[\cos45 \sin10] + 2[\sin60 \cos35]$$
| $$\sqrt{2}\sin10+\sqrt{3}\cos35=2(\frac{\sqrt{2}}{2}\sin10+\frac{\sqrt{3}}{2}\cos35)=2(\sin45\sin10+\cos30\cos35)$$
using the product rule for sine and cosine yields
$$2(\sin45\sin10+\cos30\cos35)=2(\frac{\cos(45-10)-\cos(45+10)}{2}+\frac{\cos(30+35)+\cos(35-30)}{2})$$$$=\cos35-\cos55+\cos65+\cos5$$
$$=\cos(90-55)+\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find maximum without calculus Let $f:(0,1]\rightarrow\mathbb{R}$ with $f(x)=2x(1+\sqrt{1-x^2})$. Is it possible to find the maximum of this function without calculus? Possibility through some series of inequalities?
| Putting in some more details to make the answer better:
\begin{align*}
f(x) &= 2x + 2x\sqrt{1-x^2}\\
&= 2\sin(\theta) + 2\sin(\theta)\sqrt{\smash[b]{1-\sin(\theta)^2}} &\text{via \( x = \sin(\theta)\), $\theta \in (0,\frac{\pi}{2}]$}\\
&= 2\sin(\theta)+2\sin(\theta)\cos(\theta)\\
&= 2\sin(\theta) + \sin(2\theta)\\
&= 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Solve a second order nonlinear equation I have a second order nonlinear equation:
$$-u''+ \frac{1}{4}(u')^2+au=x^2.$$
I am only interested in the solutions in $[0, \frac{x^2}{a}+\frac{1}{a^2}]$. One paper claims without proof that the unique solution should be
$$u(x)=(\sqrt{a^2+4}-a)\big(\frac{x^2}{2}+\frac{1}{a}\big).... | we can trivially see that a solution could be of the form
$$
u(x) = \alpha x^2 + \beta x + \gamma \tag{1}
$$
trialing wih Eq. (1)
we find
$$
-2\alpha + \frac{1}{4}\left(2\alpha x + \beta\right)^2 + a\left(\alpha x^2 + \beta x + \gamma\right) = x^2
$$
comparing coefficents we find
$$
-2\alpha + \frac{\beta^2}{4} + a\ga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\sin 4x +\sqrt{3} \sin 3 x + \sin 2 x=0$ This question is from a 2012 VMK entrance exam
I was trying to solve it first by expanding $\sin 4 x = 2 \sin 2 x \cos 2x$, then by noticing that if divided by 2, one can get, e.g. $ \frac{\sqrt{3}}{2} \cos 3 x = \sin \frac{\pi}{6} \cos 3 x$ and then trying to find some way to... | If you don't see any trick, you can always try
\begin{align*}
&\sin(4x)=\text{Im}(\cos x+i\sin x)^4=4\cos^3\!x\sin x-4\cos x\sin^3\!x\\
&\sin(2x)=\text{Im}(\cos x+i\sin x)^3=3\cos^2\!x\sin x-\sin^3\!x\\
&\sin(3x)=\text{Im}(\cos x+i\sin x)^2=2\cos x\sin x.
\end{align*}
Plugging it in yields
$$(4\cos^3\!x-4\cos x\sin^2\!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
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Calculating a limit of integral Computing the limit: $$\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n$$
I made the substiution $t=nx$ then, we have: $$I=\frac{1}{n^2}\int_{n\pi}^{2n\pi}\frac{t}{\arctan t}dt$$ where $I$ is the inside integral. How would you continue?
| Since $\arctan$ is increasing we get
$$\frac{1}{ n^2}\int_{n\pi}^{2n\pi }\frac{t}{\arctan 2\pi n}dt \leq I \leq \frac{1}{n^2}\int_{n\pi}^{2n\pi}\frac{t}{\arctan n \pi}dt$$
Now we can calculate both sides and we get
$$ \frac{4\pi ^2 - \pi^2}{2\arctan 2\pi n} \leq I \leq \frac{4\pi ^2 - \pi^2}{2\arctan \pi n} $$
So
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
Prove $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $ for all $n\ge 1$ and $b,c$ are integers.
Is it possible to prove this without induction?
| Let $x_1$ and $x_2$ be the roots of the quadratic equation $$x^2 - bx + c = 0$$
Then $$x_1 = \frac{b + \sqrt{b^2 - 4c}}{2}$$ $$x_2 = \frac{b - \sqrt{b^2 - 4c}}{2} $$
The above statement is equivalent to $$2^n \,\big| \, 2^n\big({x_1}^n + {x_2}^n \big) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Integrating $\int_0^\pi \frac{x\cos x}{1+\sin^2 x}dx$ I am working on $\displaystyle\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\,dx$
First: I use integrating by part then get
$$ x\arctan(\sin x)\Big|_0^\pi-\int_0^\pi \arctan(\sin x)\,dx $$
then I have $\displaystyle -\int_0^\pi \arctan(\sin x)\,dx$ because $x\arctan(\sin x)... | Split the integral at $\dfrac{\pi}{2}$, we get
$\displaystyle \begin{align} \int_0^\pi \frac{x\cos x}{1+\sin^2 x}\,dx &= \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\,dx + \int_{\pi/2}^{\pi} \frac{x\cos x}{1+\sin^2 x}\,dx \\ & = \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\,dx - \int_0^{\pi/2} \frac{(x+\pi/2)\sin x}{1+\cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 1
} |
Maximum $\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$ where $y\in \left[0,1\right]$? How find maximum this integral $$\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$$ where $y\in \left[0,1\right]$?
| HINT:
By taking derivative with respect to $y$ and setting it to be equal to $0$ we have :
$$\int_{0}^{y}\frac{(y-y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}}dx+\sqrt{y^4+(y-y^2)^2}=0$$
Hence:
$$\int_{0}^{y}\frac{dx}{\sqrt{x^4+(y-y^2)^2}}dx=\frac{\sqrt{y^4+(y-y^2)^2}}{(y-y^2)(2y-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Inequality between AM-GM Prove that for $x>y>0$ $$\sqrt {xy} <\frac {x-y}{\ln x-\ln y}<\frac {x+y}2$$ Using $x=y+k$, we can turn the inequality into $$\sqrt{y^2+ky}<\frac k{\ln(1+\frac ky)}<y+\frac k2$$ Now what do i do to the $\ln\left(1+\frac ky\right) $ term? We can't approximate it using the expansion of $\ln(1+x)$... | For $ 0<a<b$ note $ A(a,b)=\frac{a+b}{2}, G(a,b)=\sqrt{ab}, L(a,b)=\frac{b-a}{\ln b-\ln a}$
Proof 1.:
With $\frac{b}{a} = x > 1$ we have:
$\sqrt{x}< \frac{x - 1}{\ln x} < \frac{1}{2}( x + 1)$
Let functions $f, g : [ 1, \infty) \rightarrow \textbf{R}$:
$f( x ) = \ln x -\sqrt{x}+ \frac{1}{\sqrt{x}}, g( x ) = \ln x - \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$ I have these problems :
How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$
For some reason this is incorrect I'll be glad to understand why, This is what I done :
I used this formula : $(\alpha,\beta)*(\gamm... | *
*$$\left(1+i\right)\left(\frac{1}{2}+\frac{1}{2}i\right)\ne\left(\frac{1}{2}\underbrace{\color{red}{\bf -}}_{\small\text{wrong}}\frac{i^2}{2}+\frac{1}{2}i+\frac{1}{2}i\right)$$
*$$((1+\sqrt{3}i)(1-\sqrt{3}i)^{-1})^{10}=
(1+\sqrt{3}i)\left(\frac{1}{\color{red}4}+\frac{\sqrt{3}i}{\color{red}4}\right)^{10}$$
Actually... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of $\sum_{n=1}^{\infty }\frac{1}{\pi n }\sin ^k\left(\frac{2\pi n}{k}\right)$ We have: $$S_k=\sum_{n=1}^{\infty }\frac{1}{\pi n }\sin ^k\left(\frac{2\pi n}{k}\right)$$
where $k$ is an odd number greater than $1$.
I was able to find the sum of the series when $k=3,5$ as follows:
$$S_3={\sum_{n=1}^{\infty }\frac{1}{n... | We can treat this problem with Fourier series. For instance, since:
$$ \sin^3 x = \frac{1}{4}\left(3 \sin x - \sin(3x)\right) $$
we have:
$$\begin{eqnarray*}S_3 &=& \sum_{n\geq 1}\frac{\sin^3(2\pi n/3)}{\pi n} = \frac{1}{4}\left(3\sum_{n\geq 1}\frac{\sin(2\pi n/3)}{\pi n}-\sum_{n\geq 1}\frac{\sin(2\pi n)}{\pi n}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$ Good evening everyone,
how can I prove that
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}\;?$$
Well, I know that $\displaystyle\frac{1}{x^4+x^2+1} $ is an even function and the interval $(-\infty,+\infty)$ is sy... | $$x^4 + x^2 + 1 = 0 \implies x = \frac{1 \pm i\sqrt{3}}{2}, \frac{-1 \pm i\sqrt{3}}{2}$$
Only consider the two positive roots,
$$a, b = \frac{1 + i\sqrt{3}}{2}, \frac{-1 + i\sqrt{3}}{2}$$
Consider a semi circle contour $C$, with radius $R$ and upper semi-circle $\Gamma$
In other words, $C = \text{line} + \Gamma$
The in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Sum $S=\sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^3}?$ We know that
$$\sum _{k=1}^{\infty } \frac{H_k}{k^3} = \frac{\pi^4}{72}.$$
Is there a closed form for the sum
$$S=\sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^3}?$$
Mathematica doesn't give anything resembling a closed form and I have no idea if one exists.
| Since:
$$ \sum_{k\geq 1} H_k x^k = \frac{-\log(1-x)}{1-x} $$
we have:
$$ \sum_{k\geq 1} \frac{H_k}{k+1} x^{k+1} = \frac{1}{2}\log^2(1-x), $$
$$ \sum_{k\geq 1} \frac{(-1)^{k+1}H_k}{k+1} x^{k} = \frac{1}{2x}\log^2(1+x), $$
and since $\int_{0}^{1} x^k\log(x) = -\frac{1}{(k+1)^2}$, it follows that:
$$ S=\sum_{k\geq 1} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Sum of infinite series $1+\frac22+\frac3{2^2}+\frac4{2^3}+\cdots$ How do I find the sum of $\displaystyle 1+{2\over2} + {3\over2^2} + {4\over2^3} +\cdots$
I know the sum is $\sum_{n=0}^\infty (\frac{n+1}{2^n})$ and the common ratio is $(n+2)\over2(n+1)$ but i dont know how to continue from here
| Once you've determined that the sum converges (you can do this by the ratio test or by integration as other users have pointed out), you can find the value quite nicely. Let
$$S=1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\cdots$$
Then
$$2S=2+\frac{2}{1}+\frac{3}{2}+\frac{4}{4}+\frac{5}{8}+\frac{6}{16}+\cdots$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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For how many natural numbers $X(X+1)(X+2)(X+3)$ has exactly three different prime factors?
For how many natural numbers $X(X+1)(X+2)(X+3)$ has exactly three different prime factors?
My attempt:
I have used a hit and trial approach. I found out that only for x=2 and x=3 this is happening. But how can I be sure of it? ... | It's obvious that $x$ can only be the form $2^a,3^b,p^c,2^a3^b,3^bp^c$ or $2^ap^c$. I solved the case when $x=2^a3^b$ (and hope the others are similar).
First, if $a\ge 1$ and $b\ge 1$, then $2^a3^b+1$ must be the power of a prime $p$ different from $2$ and $3$. Also $2^a3^b+2=2(2^{a-1}3^b+1)$, and $2^{a-1}3^b+1$ can n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1094874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simplifying fractions with exponents I'm revising for an exam, which I have the solutions to. One of the questions asks me to prove that a sequence is a Cauchy sequence, sequence is written as:
$ a_n = \frac{2^{n+2}+1}{2^n},$
and then in the solutions, it has:
$\left | a_n - a_m \right | = \left | \frac{2^{n+2}+1}{2^n}... | Notice that
$$
\frac{2^{n+2}+1}{2^n} = \frac{2^{n+2}}{2^n} + \frac{1}{2^n} = 2^2 + \frac{1}{2^n} = 4 + \frac{1}{2^n}.
$$
Doing the same with the second term (the one involving $m$), we find that
$$
\left| \frac{2^{n+2}+1}{2^n} - \frac{2^{m+2}+1}{2^m} \right|
= \left| 4 + \frac{1}{2^n} - \left( 4 + \frac{1}{2^m} \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1096711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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integration of $\large\int \frac{u^2}{(1-u^2)^2}$ $ du$ Is there a way to integrate $$\large\int \frac{u^2}{(1-u^2)^2} du$$ without using partial fraction decomposition?
| For the diversity of the solution set, I present an exotic method that works almost exclusively for this integral:
Decompose the integrand as
\begin{align*}
\frac{u^{2}}{(u^{2}-1)^{2}}
&= \frac{1}{u^{2} + u^{-2} - 2} \\
&= \frac{1}{2}\frac{1+u^{-2}}{(u-u^{-1})^{2}} + \frac{1}{2}\frac{1-u^{-2}}{(u+u^{-1})^{2} - 4} \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Pythagorean triple problem I am doing research on perfect cuboids, and I'm looking for values $a,b,c$ such that the following is integer, and I'm not sure how to continue this. Any suggestions are appreciated!
$PED$ is a very large constant=$899231100768000$
$$
\begin{align}
&\exp\left(\sigma_1+\sigma_2+\frac{\ln(a^2+b... | If $a=b=1$, $c=4$, then $$(a^2+b^2)(a^2+c^2)(b^2+c^2)(a^2+b^2+c^2)=2\times17\times17\times18=102^2,$$ so $$abc\sqrt{a^2+b^2}\sqrt{a^2+c^2}\sqrt{b^2+c^2}\sqrt{a^2+b^2+c^2}=408$$ is an integer. Multiply each of $a,b,c$ by $899231100768000$, and you have an example.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Calculated $f(2004)$ where the function $f:N \rightarrow Q^*$ It is considered that the function $f:N \rightarrow Q^*$ which has properties:
$a) f(7) = 4;$
$b) f(8013) = 8015;$
$c) f(n+2).f(n) = 1+ f(n+1).$
Calculated $f(2004)$.
All my attempts to find the required number had no result. Can anyone help me? Thanks so mu... | Using (c) to express a simple recurrence $f(n+2)=\frac{f(n+1)+1}{f(n)}$, it's easy to demonstrate the sequence operates a cycle of length 5 (provided zeros are avoided):
$$ f(1)=x $$
$$f(2)=y$$
$$ f(3) = \frac{y+1}{x}$$
$$ f(4) = \frac{x+y+1}{xy}$$
$$f(5) = \frac{x(xy+x+y+1)}{xy(y+1)} = \frac{(x+1)(y+1)}{y(y+1)} = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Surface area of a sphere above the plane $z=1$ The question is: Find the surface area of the part of the sphere $x^2+y^2+z^2=4$ that lies above the plane $z=1$. I got $4\pi(\sqrt3-1)$ but the answer key says $4\pi(\sqrt2-1)$. Am I doing something wrong or is the answer key wrong?
Thanks in advance.
Edit: This is my wor... | For those who want to know why the answer is $4\pi$, let's do it!
First, plot the figure:
Since the goal is to find that area, the integral is just:
$$\iint_S dS,$$
with the appropriate limits. It will be easier if we use spherical coordinates:
$$x = 2 \sin \phi \cos \theta, \quad y = 2 \sin \phi \sin \theta, \quad z ... | {
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"url": "https://math.stackexchange.com/questions/1100364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{1}{90} \sum_{n=1}^{90} 2n \cdot \sin((2n)^\circ) = \cot (1^{\circ})$ Show that $$\frac{(2\sin(2^\circ)) + (4\sin(4^\circ))+ (6\sin(6^\circ)) + \ldots +(180\sin(180^\circ))}{90} = \cot(1^\circ).$$
I used a lot of steps, and typing it all down on here would take me an hour, but here are my last few step... | $$S=\sum_{n=1}^{90}2(91-n)\sin(2n^\circ) = \sum_{n=1}^{90}2\sum_{k=1}^{n}\sin(2k^\circ).$$
Since:
$$2(\sin 1^{\circ})\sum_{k=1}^{n}\sin(2k^\circ)=\sum_{k=1}^{n}\left(\cos((2k-1)^{\circ})-\cos((2k+1)^{\circ})\right)=\cos 1^\circ-\cos((2n+1)^\circ)$$
it follows that:
$$ S = \frac{1}{\sin 1^\circ}\left(90\cos 1^\circ-\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To show sequence $a_{n+1}= \frac{a_n^2+1}{2 (a_n+1)}$ is convergent Let $a_1=0$
and $$a_{n+1}= \dfrac{a_n^2+1}{2 (a_n+1)}$$ $\forall n> 1.$
Show that sequence $a_n$ convergent.
I tried to prove $a_n$ is less than 1 by looking at few terms. But i failed to prove my claim.
| since
$$a_{n+1}-(\sqrt{2}-1)=\dfrac{a^2_{n}+1-2(\sqrt{2}-1)(a_{n}+1)}{2a_{n}+2}=\dfrac{[a_{n}-(\sqrt{2}-1)]^2}{2a_{n}+2}\tag{1}$$
simaler
$$a_{n+1}+\sqrt{2}+1=\dfrac{(a_{n}+\sqrt{2}+1)^2}{2a_{n}+2}\tag{2}$$
$\dfrac{(1)}{(2)}$ we have
$$\dfrac{a_{n+1}-(\sqrt{2}-1)}{a_{n+1}+\sqrt{2}+1}=\left(\dfrac{a_{n}-(\sqrt{2}-1)}{a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Solve the pde $(x^2-y^2-yz)p+ (x^2-y^2-zx)q=z(x-y)$ I'm solving this by Lagrange's method.
Lagrange's auxiliary equation is:
$\frac{dx}{x^2-y^2-yz}=\frac {dy}{x^2-y^2-zx}= \frac{dz}{z(x-y)}$
From the first two ratio and the last ratio:
$\frac{dx-dy}{z(x-y)}=\frac {dz}{z(x-y)}$
Solving which I get $z-x+y=c_1$
I'm unable... | since
$$\dfrac{dx}{x^2-y^2-yz}=\dfrac{dy}{x^2-y^2-xz}=\dfrac{dz}{z(x-y)}$$
$$\Longrightarrow \dfrac{xdx}{x^3-y^3-xyz}=\dfrac{ydy}{x^2y-y^3-xyz}=\dfrac{\dfrac{dz}{z}}{x-y}$$
$$\Longrightarrow\dfrac{xdx-ydy}{(x^3+y^3)-xy(x+y)}=\dfrac{\dfrac{dz}{z}}{x-y}$$
$$\Longrightarrow \dfrac{xdx-ydy}{(x+y)(x-y)^2}=\dfrac{\dfrac{dz}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $|z+w|^2$ + $|z-w|^2$ = $2|z|^2 + 2|w|^2$
Show that $|z+w|^2$ + $|z-w|^2$ = $2|z|^2 + 2|w|^2$.
Every time I work this problem out I have $2|z|^2$ but I can not get the other piece.
Here is what I got so far. Let $z = a+ bi$ and $w = c + di$, then
\begin{align*}
|z+w|^2 + |z-w|^2 &= a^2 + b^2 + c^2 + d^2 + a... | You seem to be using $|z + w|^2 = a^2 + b^2 + c^2 + d^2$ and $|z - w|^2 = a^2 + b^2 - c^2 - d^2$. These are both false. Instead we have
$$|z + w|^2 = |(a+bi) + (c+di)|^2 = |(a + c) + (b + d)i|^2 = (a + c)^2 + (b + d)^2$$
and a similar calculation gives $|z - w|^2 = (a-c)^2 + (b - d)^2$. If you use these, you will find ... | {
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"url": "https://math.stackexchange.com/questions/1105580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
extracting the middle term of $ (z \cos \theta + w\sin \theta )^m(- z\sin \theta + w\cos \theta )^m $ Is there a systematic way to extract the middle term of the following expression?
$$ (z \cos \theta + w\sin \theta )^m(- z\sin \theta + w\cos \theta )^m $$
This is homogeneous polynomial of degree $2m$, so I am looki... | Continuing mickep's answer, these are, indeed, the Legendre Polynomials
$$
\sum_{k=0}^m (-1)^k{m\choose k}^2 \bigl[\tfrac{1}{2}(1+\cos2\theta)\bigr]^{m-k}\bigl[\tfrac{1}{2}(1-\cos2\theta)\bigr]^k
= P_m(\cos(2\theta))
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Hypergeometric 2F1 with negative c I've got this hypergeometric series
$_2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
where $a,n>0$ and $a,n\in \mathbb{N}$
The problem is that $-a-n+1$ is negative in this case. So when I try to use Gauss's identity
$_2F_1 \left[ \begin{array}{ll}
a & b \\
... | Here I post the full solution
The problem: we are looking for the closed form of this sum:
$\sum\limits_{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$
The first term of the sum for $i=0$ is $\binom{a+n-1}{n}$
The ratio two consecutive terms:
$\dfrac{t_{i+1}}{t_i} = \dfrac{P(i)}{Q(i)} = \dfrac{\binom{a+i}{i+1} \binom{a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Combination questions Q: A team of $11$ is to be chosen out of $15$ cricketers of whom $5$ are bowlers and $2$ others are wicket keepers. In how many ways can this be done so that the team contains at least $4$ bowlers and at least $1$ wicket keeper?
| The number of ways to choose exactly $4$ bowlers and exactly $1$ keeper:
$$\binom{5}{4}\cdot\binom{2}{1}\cdot\binom{15-5-2}{11-4-1}=280$$
The number of ways to choose exactly $5$ bowlers and exactly $1$ keeper:
$$\binom{5}{5}\cdot\binom{2}{1}\cdot\binom{15-5-2}{11-5-1}=112$$
The number of ways to choose exactly $4$ bow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How find the number of zeros at the end of the sum $4^{5^6}+6^{5^4}$? The problem is to find the number of zeros at the end of the sum $4^{5^6}+6^{5^4}$.
I tried $2^{2 \cdot 5^6}+3^{5^4} \cdot 2^{5^4}= 2^{5^4} \cdot ( 2^{2 \cdot 5^6 -5^4}+ 3^{5^4} )$.
| Use the following lemma, which can be proven by induction 'on $b$':
For any odd prime $p$ and $a,b \in \mathbb{N}_{>0}$ such that $p^a \mid\mid x-y$ and $p^b \mid\mid k$:
$p^{a+b} \mid\mid x^k-y^k$.
For this question we don't need the "strictly divides" in the conclusion, but in general this lemma is very useful.
$4^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to solve this linear equations using gauss jordan method? How to solve the linear equations in Gauss elimination method
$$2x+3y+z=1\\
x+y+z=3\\
3x+4y+2z=4$$
| \begin{align*}
\left(\begin{array}{ccc|c}
1&1&1&3\\
2&3&1&1\\
3&4&2&4
\end{array}\right)\\\\
\left(\begin{array}{ccc|c}
1&1&1&3\\
0&1&-1&-5\\
0&1&-1&-5
\end{array}\right)\\\\
\begin{array}{ccc}
x&y&z
\end{array}\qquad\qquad\\
\left(\begin{array}{ccc|c}
1&1&1&3\\
0&1&-1&-5\\
0&0&0&0
\end{array}\right)
\end{align*}
Z is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$?
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$
This thing doesn't make sense how should I use first identity to find the second one.
| HINT: Use $a^3+b^3=(a+b)(a^2+b^2-ab)$ and also the fact that if $x^2+x+1=0$, then $x+\frac1x$=$-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to prove $\sum_{n=0}^{\infty} \frac{1}{1+n^2} = \frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}$ How can we prove the following
$$\sum_{n=0}^{\infty} \dfrac{1}{1+n^2} = \dfrac{\pi+1}{2}+\dfrac{\pi}{e^{2\pi}-1}$$
I tried using partial fraction and the famous result $$\sum_{n=0}^{\infty} \dfrac{1}{n^2}=\frac{\pi^2}{6}$$ ... | First consider the following expansion of $\pi \cot(\pi z)$:
$$\pi \cot(\pi z) = \frac{1}{z} + \sum_{n = 1}^\infty \frac{2z}{z^2 - n^2} \quad (z \neq 0, \pm 1, \pm 2,\ldots)$$
Replacing $z$ by $iz$, we have
$$-i\pi \coth(\pi z) = \frac{1}{iz} - \sum_{n = 1}^\infty \frac{2iz}{z^2 + n^2} = -i\left(\frac{1}{z} + 2\sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 2,
"answer_id": 0
} |
Limit $\lim_{x\to\infty}(x(\log(1+\sqrt{1+x^2}-\log(x)))$ can someone give me a hint/solution for:
$$\lim_{x\to\infty}\left(x(\log(1+\sqrt{1+x^2}-\log(x))\right)$$
Shall I do a derivative ?
But there's no L'Hospital to use here..
Shall I change its form ?
$x(\log(1+(1+x^2)^{1/2})-\log(x))$ (minimal change)
Or what shal... | L'Hospital's rule:
$$\lim_{x \to \infty} x \{\ln(1+\sqrt{x^2 + 1}) - \ln x \} =\lim_{x \to \infty} \frac{(\ln(1+\sqrt{x^2 + 1}) - \ln x)' }{(\frac{1}{x})'} =$$$$= \lim_{x \to \infty} \frac{\frac{1}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}} - \frac{1} {x} }{-\frac{1}{x^2}} = \lim_{x \to \infty}(x-\frac{x^2}{1+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to Solve Trigonometric Equations? How are you supposed to go about solving equations such as:
$$-\sqrt{3} = \frac{\sin{4\theta}}{\sin{7\theta}}.$$
I know that $\theta = 30^{\circ}$ is one such solution, but how do I find all solutions using algebra?
Thanks
Edit: I figured out one possible method of reasoning. For $... | Use the identities
$$\sin4\theta = 4\cos\theta\left (\sin\theta - 2\sin^3\theta\right)$$
$$ \sin7\theta = -64\sin^7\theta + 112\sin^5\theta -56\sin^3\theta+7\sin\theta,$$ obtained from De Moivre's formula and by taking the substitution $\cos^2\theta
=1-\sin^2\theta$.
Hence,
$$ \frac{4\cos\theta\left (\sin\theta - 2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$ Let
$$\left({x+\sqrt{x^2+1}}\right)\left({y+\sqrt{y^2+1}}\right)=1$$
Prove that $x+y=0$.
This is my solution:
Let
$$a=x+\sqrt{x^2+1}$$
and
$$b=y+\sqrt{y^2+1}$$
Then $x=\dfrac{a^2-1}{2a}$ and $y=\dfrac{b^2-1}{2b}$. Now $ab=1\implies b=\dfrac1a$. Then ... | $x+\sqrt{x^2+1}$ is a strictly increasing, continuous function so that a single $x$ corresponds to a single $y$, and $y=-x$ is the one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
Convergence of $\sum_{n=1}^{\infty} \log~ ( n ~\sin \frac {1 }{ n })$ Convergence of $$\sum_{n=1}^{\infty} \log~ ( n ~\sin \dfrac {1 }{ n })$$
Attempt:
Initial Check : $\lim_{n \rightarrow \infty } \log~ ( n ~\sin \dfrac {1 }{ n }) = 0$
$\log~ ( n ~\sin \dfrac {1 }{ n }) < n ~\sin \dfrac {1 }{ n }$
$\implies \... | Since $$\log\left(n\sin \frac{1}{n}\right) = \log\left(n\left(\frac{1}{n} + O\left(\frac{1}{n^3}\right)\right)\right) = \log\left(1 + O\left(\frac{1}{n^2}\right)\right) = O\left(\frac{1}{n^2}\right)$$
and $\sum_{n = 1}^\infty \frac{1}{n^2}$ converges, so does $\sum_{n = 1}^\infty \log(n\sin\frac{1}{n})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Finding an order of a coset in $A/B$ where $A$ is a free abelian group and $B$ is a subgroup. Let $A$ be a free abelian group with basis $x_1,x_2,x_3$ and let $B$ be a subgroup of A generated by $x_1+x_2+4x_3, 2x_1-x_1+2x_3$. In the group $A/B$ find the order of the coset $(x_1+2x_3)+B$.
How can I find this order? Jus... | $Proof$: since no one has yet answered, I shall make an attempt and would appreciate your correction: Let us denote the basis $\{x_1+x_2+4x_3=a,\ 3 \cdot (x_2+2x_3)=b\}$ then $3a-b=3x_2+6x_6$. Then obviously, since $3x_2+6x_6\in B$ then the order of $(x_1+2x_3)+B$ is $3$, as $(x_1+2x_3)+B\ne B$, $2(x_1+2x_3)+B\ne B$ bu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove, inequality ,positive numbers $$\frac{a}{e+a+b}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{c+d+e}+\frac{e}{d+e+a}<2$$
Prove that for positive numbers $a,b,c,d,e$ there is such inequality
| For convenience of notation I'll use $a_i$'s ($1 \le i \le 5$) instead of $a,b,c,d,e$.
Wlog assume $a_1+a_2+a_3 = \min\limits_{1 \le i \le 5}\{a_{i-1}+a_i+a_{i+1}\}$
(where, the indexing is cyclic $a_{0} = a_{5}$ and $a_6 = a_1$ and so on)
Then, $\displaystyle \sum\limits_{i=1}^{3} \frac{a_i}{a_{i-1}+a_i+a_{i+1}} \le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(1+c\sin x) dx$, where $0I am trying to evaluate the following integral:
$$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(1+c\sin x) dx,$$
where $0<c<1$.
I can't really think of a way to find it so please give me a hint.
| THIS IS ONLY A PARTIAL ANSWER
First we apply DUIS $$I'(a)=\int_{-\pi/2}^{\pi/2} \frac{\sin(x)}{a\sin(x)+1}$$
This can be integrated with a Weierstrass substitution, which I'll leave for you to verify:
$$\large\left.I'(a)=\frac{x-\frac{2\tan^{-1}\left(\frac{\tan(x/2)+y}{\sqrt{1-y^2}}\right) }{\sqrt{1-y^2}}}{y}\right|_{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Is there an obvious reason why $4^n+n^4$ cannot be prime for $n\ge 2$? I searched a prime of the form $4^n+n^4$ with $n\ge 2$ and did not find one with
$n\le 12\ 000$.
*
*If $n$ is even, then $4^n+n^4$ is even, so it cannot be prime.
*If $n$ is odd and not divisible by $5$ , then $4^n+n^4\equiv (-1)+1\equiv 0 \pm... | If $n$ is even, so will be $4^n+n^4$ and the later will definitely be $>2$ and hence composite
Else
$$4^n+n^4=(2^n)^2+(n^2)^2=(n^2+2^n)^2-2\cdot2^n\cdot n^2$$
$$=(n^2+2^n)^2-(n2^{\frac{n+1}2})^2$$
As $n$ is odd $\iff n+1$ is even $\implies\dfrac{n+1}2$ is an integer
$$4^n+n^4=(n^2+2^n+n2^{\frac{n+1}2})(n^2+2^n-n2^{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Probability of Seven (Distinct) car accidents occurred on the same day Seven (Distinct) car accidents occurred in a week. What is probability that they all occurred on the same day?
My Solution:
All 7 accident occurs in 1 day in $\binom{7}{1}$ ways
All 7 accident occurs in 2 days in $\binom{7}{2}$ ways
All 7 accident ... | There are seven days and seven accidents (which are distinguishabla from one another), so the number of possible outcomes is $7^7$. However, only seven of them fulfil our criteria: all accidents on Monday, all on Tuesday, ..., all on Saturday. So the answer is $1/7^6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality $(a+b)^2 + (a+b+4c)^2\ge \frac{kabc}{a+b+c}$ for $a,b,c \in\mathbb{R}$ Find biggest constans k such that $(a+b)^2 + (a+b+4c)^2\ge \frac{kabc}{a+b+c}$ is true for any $a,b,c \in\mathbb{R}$
Could you check up my solution? I'm not sure it's ok -
$(a+b)^2 + (a+b+4c)^2 \ge 0$
and
$0\ge \frac{kabc}{a+b+c}$
so
$0... | Using $\text{arithmetic mean} \geq \text{geometric mean}$, we get
$$
\begin{eqnarray}
(a+b)^2+(a+b+4c)^2&=&(a+b)^2+(a+2c+b+2c)^2 \\
&\geq& \big(2\sqrt{ab}\big)^2+\big(2\sqrt{2ac}+2\sqrt{2b}\big)^2 \\
&=& 4ab+8ac+8bc+16c\sqrt{ab}
\end{eqnarray}
$$
Therefore
$$
\begin{eqnarray}
\frac{(a+b)^2+(a+b+4c)^2}{abc}(a+b+c)&\geq&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $ Let $a,b,c\in\mathbb{R^+}$ such that $ a^2 + b^2 + c^2 = 3 $. Prove that
$$
(a+b+c)(a/b + b/c + c/a) \geq 9.
$$
My Attempt
I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.
| From Cauchy-Schwarz Inequality: $$\sum\limits_{cyc} \frac{a}{b}\ge \frac{(a+b+c)^2}{ab+bc+ca}$$
Hence, it suffices to prove: $\displaystyle \frac{(a+b+c)^3}{ab+bc+ca} \ge 9$
Squaring both sides and using $a^2+b^2+c^2 = 3$,
$$\begin{align} & (a+b+c)^6 \ge 81(ab+bc+ca)^2 = 27(a^2+b^2+c^2)(ab+bc+ca)^2 \end{align}$$
which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Why does solving the spherical Bessel equation using Frobenius series produce two quadratic equations for the exponents at the singularity? The spherical Bessel equation is:
$$x^2y'' + 2xy' + (x^2 - \frac{5}{16})y = 0$$
If I seek a Frobenius series solution, I will have:
\begin{align*}
&\quad y = \sum_{... | You set the value of $r$ found in the first equation into the second equation. It then defines $a_n$ in terms of $a_{n-2}$ for all $n \geq 2$. So, once we know $a_0$, $a_1$ we get $a_2$ from the second equation and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Defining the equation of an ellipse in the complex plane
Usually the equation for an ellipse in the complex plane is defined as $\lvert z-a\rvert + \lvert z-b\rvert = c$ where $c>\lvert a-b\rvert$. If we start with a real ellipse, can we define it in the manner below?
For $x,y,h,k,a,b\in\mathbb{R}$ such that $a,b\ne... | Setting $z_0=h+ik$, you can't do much better than
$$\frac{\left(\Re(z-z_0)\right)^2}{a^2}+\frac{\left(\Im(z-z_0)\right)^2}{b^2}=1.$$
The function in the LHS isn't analytic, so taking the real and imaginary parts isn't worse nor better than using the modulus.
You could also try starting from the parametric form $(h+a\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Elementary number theory , when is $12n^2 + 1$ a square Prove that if
$$k = 2 + 2\sqrt{12n^2 + 1}$$
is an integer then it is a square.
Can anyone help me with this? All I know is that k is an integer if and only if ${12n^2 + 1}$ is a square. What do I do next?
| let
$$\sqrt{12n^2+1}=m\Longrightarrow 12n^2+1=m^2\Longrightarrow \dfrac{m-1}{2}\cdot\dfrac{m+1}{2}=3n^2$$
because $m$ is odd numbers,so $\dfrac{m-1}{2},\dfrac{m+1}{2}\in N^{+}$
since
$$\gcd\left(\dfrac{m+1}{2},\dfrac{m-1}{2}\right)=1$$
case 1:
$$\dfrac{m-1}{2}=3u^2,\dfrac{m+1}{2}=v^2,uv=n$$
$$\Longrightarrow 2+2\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
System of exponential equations If $x,y,z \in \mathbb{R}$ and
$$
\begin{cases}
2^x+3^y=5^z \\
2^y+3^z=5^x \\
2^z+3^x=5^y
\end{cases}
$$
does it imply that $x=y=z=1$?
| I think so.
Let $f(x) = 5^x - 2^x - 3^x$, you have that
$$x < 1 \Leftrightarrow f(x) < 0$$
$$x = 1 \Leftrightarrow f(x) = 0$$
$$x > 1 \Leftrightarrow f(x) > 0$$
Suppose that $x \geq y \geq z$:
If $x > 1$, then $5^x > 2^x + 3^x \geq 2^y + 3^z$, therefore $5^x \neq 2^x + 3^y$.
This implies that $1 \geq x \geq y \geq z$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Expected first return time of Markov Chain Given the following Markov Chain:
$$M = \left(
\begin{array}{cccccc}
\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 \\
\frac{1}{4} & \frac{3}{4} & 0 & 0 & 0 & 0 \\
\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & 0 & 0 \\
\frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{4} & 0 &... | I think Zoe's approach is the easiest but that she has the equations wrong.
Let
$$T_{11} = E(T_1\mid X_0=1) \\
T_{21} = E(T_1\mid X_0=2).$$
We want to find $T_{11}$. Considering the possible transitions between states $1$ and $2$ and their probabilities, we get equations:
$$T_{11} = 1+\dfrac{1}{2}T_{21}\qquad\qquad\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$x^2+y^2+9=3(x+y)+xy$ Find all pairs of real $x,y$ that meet this equation $\frac{(x-y)^2}{(y-3)(3-x)} = 1$
That was my attempt, I can't think of anything else here. I'd prefer a hint
| Just did this yesterday: What's so special about the form $ax^2+2bxy+cy^2$?
Given some $g$ that is going to be set equal to $0,$ with
$$ g = a x^2 + b x y + c y^2 + d x + e y + f $$
and taking the usual
$$ \Delta = b^2 - 4 a c, $$
we get, assuming both $a, \Delta \neq 0$ for this amount of work to be necessary,
$$ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
The pedantic function $\frac{y \cdot \sin(x^5y^3+x^3)}{(x^4y^8+x^6+3y^2)\cos(x^2y)^2}$ I was shown the following "pedantic" function.
$$
f(x,y) := \frac{y \cdot \sin(x^5y^3+x^3)}{(x^4y^8+x^6+3y^2)\cos(x^2y)^2}
$$
The question is what happens as the function approaches origo. So in order for the limit to exists we ... | First, notice that
$$f(x,kx^n) = \frac{kx^n \cdot \sin(x^5(kx^n)^3+x^3)}{(x^4(kx^n)^8+x^6+3(kx^n)^2)\cos(x^2(kx^n))^2} = \frac{kx^n\sin(x^3+k^3x^{n+5})}{(x^6+3k^2x^{2n}+k^8x^{8n+4})[\cos(kx^{n+2})]^2}$$
As $x\to 0$, lower powers of $x$ converge to $0$ faster than higher powers. So only the lowest order powers of $x$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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exercise: ordinary differential equations I am struggling with an exercise. Can you please give me a hint?
Exercise:
Show that the solution curves of the differential equation:
$\frac{dy}{dx}=-\frac{y(2x^3-y^3)}{x(2y^3-x^3)}$, are of the form $x^3+y^3=3Cxy$.
I tried the substitution $u=y/x \rightarrow y=xu, \frac{dy}{d... | here is a cheap way of doing this knowing that $x^3 + y^3 = Cxy$ is a solution. we will make the change of variables $$u = x^3+y^3, t = xy$$
the deifferential equation satisfied by $u$ and $t$ are:
$$\frac{du}{dx} = 3x^2 + 3y^2 \frac{dy}{dx} = \frac{3(y^6-x^6)}{x(2y^3 - x^3}, \frac{dt}{dx} = x\frac{dy}{dx} + y = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Proof: $n^p < \frac{(n+1)^{p+1}-n^{p+1}}{p+1} < (n+1)^p$ I've edited the post in order to add at the end what, I think, is the complete proof of these inequalities. I want to apologize by not having given a reply as early as those given by the users who gave me hints about this exercise. It took me a little of time but... | $$b^{p+1} - a^{p+1} = (b-a)(b^{p}+ b^{p-1}a+b^{p-2}a^{2}+ \cdots+a^{p})$$ Substitute $$b=n+1,\,\,\,\,\,\,\,a=n$$ Then $$(n+1)^{p+1}-n^{p+1}=(n+1)^p+n(n+1)^{p-1}+n^2(n+1)^{p-2}+\cdots+n^p$$ Since $$n<n+1$$ we have $$(p+1)n^p<(n+1)^p+n(n+1)^{p-1}+n^2(n+1)^{p-2}+\cdots+n^p<(p+1)(n+1)^p.$$ Hence we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Problem for system of equations Find the solution of the following set of equations
\begin{equation}
\begin{cases}
yz-2z+x-1=0 \nonumber \\
zx+y-z-2=0 \nonumber \\
xy-2x-y+z+2=0
\end{cases}
\end{equation}
| By summing (2) and (3) you end up with the equation:
$$
x(z+y-2)=0.
$$
Which has two solutions, either $x=0$ or $z+y-2=0$.
For the first case you end up with the following system:
\begin{align}
z(y-2)&=1\\
y-z&=2,
\end{align}
which is equivalent to $(y-2)^2=1$ and whose solutions are $(0,1,-1)$ and $(0,3,1)$.
For the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve the following systems of equations ( Matrices) Solve the following systems of equations:
\begin{Bmatrix}
x_1 & -x_2 & -x_3 & +0x_4 & = 2 \\
-x_1 & +2x_2 & +0x_3 & +3x_4 &= 1 \\
x_1 & +0x_2 & +x_3 & +0x_4 & =5\\
\end{Bmatrix}
I try to make a zero triangle but always reach a step whe... | when i row reduce your matrix $\pmatrix{1&-1&-1&0&2\\-1&2&0&3&1\\1&0&1&0&5}$
i get $\pmatrix{1&0&0&1&5\\0&1&0&2&3\\0&0&1&-1&0}.$
let us call the variables $x_1, x_2, x_3$ and $x_4$ corresponding to the four columns. the fourth variable $x_4$ is called the free variable. to find a particular solution we set it to zero... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determine matrix A from linear transformations Let $T:R^3\rightarrow R^3$ be the linear transformation such that
$$
T
\left[
\begin{array}{c}
1\\
1\\
1\\
\end{array}
\right]
=\left[
\begin{array}{c}
-2\\
5\\
-2\\
\end{array}
\right],\,\,
T
\left[
\begin{array}{c}
1\\
1\\
0\\
\end{array}
\right]
=\left[
\begin{array}{c}... | You should be able to see that the set of vectors
$$ u_1=\left[
\begin{array}{c}
1\\
1\\
1\\
\end{array}
\right], u_2=\,\left[
\begin{array}{c}
1\\
1\\
0\\
\end{array}
\right], u_3 =\, \left[
\begin{array}{c}
1\\
0\\
0\\
\end{array}
\right] $$
spans $\mathbb{R^3}$. Let $X\in \mathbb{R^3}$ so we can write it as
$$ X = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $a$, $r$, and $s$ are odd and $b$ is even Question: Suppose that $a$, $b$, $r$, $s$ are pairwise relatively prime and that $a^2 + b^2 = r^2$ and $a^2 - b^2 = s^2$. Prove that $a$, $r$, and $s$ are odd and $b$ is even.
My incomplete answer: Suppose $r$ is even then $a$ and $b$ are both even or both odd; then ... | Use the fact that if $x$ is odd, then $x^2 \equiv 1 \pmod{4}$. This is because $(2t + 1)^2 = 4(t^2 + t) + 1$.
Now first look at $a^2 + b^2 = r^2$. If $r$ is even, then $r^2 \equiv 0 \pmod{4}$. Then either (i) $a$ and $b$ are both even or (ii) $a$ and $b$ are both odd. In Case (i), we violate the gcd condition. In Case ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Is there $a,b,c,d\in \mathbb N$ so that $a^2+b^2=c^2$, $b^2+c^2=d^2$? Question:
Are there $a,b,c,d \in \mathbb N$ such that $$a^2 + b^2 = c^2 \ \ \text{and} \ \ b^2 + c^2 = d^2$$
I'm a bit lost here.
| Suppose there exist such $a,b,c,d$ with $a\le b\le c\le d,$ then
$$c^2-a^2=d^2-c^2=b^2$$
$$2c^2=a^2+d^2$$
Suppose $$a=x-y,\,\,\,\,d=x+y.$$ Then $$c^2=x^2+y^2.$$
Therefore there exist $A,B$ such that $$c=A^2+B^2,\,\,\,x=A^2-B^2,\,\,\,y=2AB.$$
Now we have $$a=A^2-2AB-B^2\le(A-B)^2$$
$$d=A^2+2AB-B^2\le(A+B)^2$$
Also $$c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How to factorize $n^5+n+1$? How to factorize $n^5+n+1$ ?
I think I should break $a^5$ and use a factorization formula. but how is it done?
| For completeness sake, the brute force approach:
Suppose there is a linear factor. Then
$$n^5+n+1=(n+a)(n^4+bn^3+cn^2+dn+e),$$
for some integers $a$, $b$, $c$, $d$ and $e$. In particular $ae=1$ so $a=\pm1$. Then plugging in $n=-a=\mp1$ yields
$$(-a)^5+(-a)+1=((-a)+a)((-a)^4+b(-a)^3+c(-a)^2+d(-a)+e)=0.$$
But this does n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
When is $(12x+5)/(12y+2)$ not in lowest terms? I am struggling to solve this problem and would appreciate any help:
When is $\frac{12x+5}{12y+2}$ NOT in lowest terms? ($x$,$y$ are nonnegative integers)
I have found that it is not in lowest terms for $x=6$ and $y=9$ because numerator and denominator are divisible by $... | When is $\dfrac{12x+5}{12y+2} $ irreducible?
This is really just a partial answer.
So, when does $12x + 5 = A$ and $12y + 2 = B$ where $\gcd(A, B)=1$
Note that
\begin{align}
12 &\mid 2A - 5B \\
5B &\equiv 2A \pmod{12} \\
B &\equiv 10A \pmod{12} \\
B &= 12n + 10A
\end{align}
So $\gcd(A,B)=1$ becomes $\gcd(12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}... | The sum $\sum_{k=2}^n\frac{1}{k^2}$ is a lower Riemann sum for the function $1/x^2$ in the interval $[1,n]$, so
$$
\sum_{k=2}^n\frac{1}{k^2}\le\int_{1}^n\frac{1}{x^2}\,dx=
\left[-\frac{1}{x}\right]_1^n=1-\frac{1}{n}
$$
Therefore
$$
\sum_{k=1}^n\frac{1}{k^2}\le 1+1-\frac{1}{n}<2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
For odd $m\ge3$, does it follow: $\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$ Unless I am making a mistake, I am calculating that:
$$\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$$
Here's my reasoning:
*
*$\dfrac{x^m + y^m}{x+y} = x^{m-1} - x^{m-2}y ... | Your argument is fine, possibly a bit more complicated than necessary. And the theorem is true for $m$ even, too, so a better proof wouldn't use the closed form for $\frac{x^m+y^m}{x+y}$, which is only true for $m$ odd.
A more advanced way to prove this uses linear recurrences. Show that $a_n=x^n+y^n$ satisfies:
$$a_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Argument of $3+4i$ for $z = \sqrt{3 + 4i}$, I am trying to put this in Standard form, where z is complex. I let $w = 3+4i$ and find that the modulus, $|w|=r$, is 5. I am having trouble solving for arg(w).
I find that $\tan^{-1}{\theta} = \frac{4}{3}$. However, this is not an angle well known. How do I find it?
| Suppose $\sqrt{3+4i}$ were in standard form, say $x+yi$. Then we would have $$\begin{align}
x+yi & = \sqrt{3+4i}\\
(x+yi)^2 & = 3+4i\\
(x^2-y^2) + 2xyi & = 3+4i
\end{align}
$$
so
$$\begin{align}
x^2 -y^2 &= 3 \\
2xy &= 4 \\
\end{align}
$$
From the second equation we have $y = \frac2x$. Putting this into the first equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Math Induction $N$ greater than or equal to $2$ $(1-\frac{1}{2})$$(1-\frac{2}{3})$$(1-\frac{1}{4})$...$(1-\frac{1}{n})$=$(\frac{1}{n})$ So I'm trying to make it equal.
so $n$ is equal or greater than 2.
When i substitute $2$ to n, $(1-\frac{1}{2})$ = $(\frac{1}{2})$
then
$(1-\frac{1}{2})$$(1-\frac{2}{3})$$(1-\frac{1}{... | As yashg mentions, you are adding and not multiplying. This might help:
$$\left(\prod_{k=2}^n\left(1-\frac1k\right)\right)\left(1-\frac1{n+1}\right)=\prod_{k=2}^{n+1}\left(1-\frac1k\right)$$
And
$$\begin{align}\left(\frac1n\right)\left(1-\frac1{n+1}\right)&=\frac1n-\frac1{n(n+1)}\\
&=\frac{n+1}{n(n+1)}-\frac1{n(n+1)}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$
A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\g... | $\sum\limits_{cyc}\frac{a}{b+c-a}-3=\sum\limits_{cyc}\left(\frac{a}{b+c-a}-1\right)=\sum\limits_{cyc}\frac{a-b-(c-a)}{b+c-a}=$
$=\sum\limits_{cyc}(a-b)\left(\frac{1}{b+c-a}-\frac{1}{a+c-b}\right)=\sum\limits_{cyc}\frac{2(a-b)^2}{(b+c-a)(a+c-b)}\geq0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
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Prove that $m+\frac{4}{m^2}\ge3$ How to prove that $m+\frac{4}{m^2}\geq3$ for every $m>0$? I multiplied both sides by $m^2$ and finally got $m^3+4-3m^2\geq0$, yet don't know how to move further. Hint?
| Using $A.M\geq G.M$
$$\frac{\frac{m}{4}+\frac{m}{4}+\frac{m}{4}+\frac{m}{4}+\frac{m}{4}+\frac{2}{m^2}+\frac{2}{m^2}}{6}\geq\frac{1}{2}$$
$$m+\frac{4}{m^2}\geq3$$
Equality occurs if
$$\frac{m}{4}=\frac{m}{4}=\frac{m}{4}=\frac{m}{4}=\frac{m}{4}=\frac{2}{m^2}=\frac{2}{m^2}$$
$i.e.$ for $m=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1157949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Evaluate $\int{\frac{x^3}{x^3-3x^2+9x-27}}dx$ So for the equation below I used long division first, and got $\int{(1+\frac{-3x^2+9x-27}{x^3-3x^2+9x-27}})dx$ (So my signs should be the other way around, thank you to Alijah Ahmed for pointing it out, see answer below). Now I'm not sure how to proceed, completing the squa... | As the comments have mentioned, use partial fractions.
Also, note that the denominator $x^3-3x^2+9x-27$ factorises to $(x-3)(x^2+9)$ which helps, so you want to find $A$, $B$ and $C$ such that
$$\frac{A}{x-3}+\frac{Bx+c}{x^2+9}=\frac{\color{red}{+}3x^2\color{red}{-}9x\color{red}{+}27}{(x-3)(x^2+9)}$$
(you had the sig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to solve the summation of series $a^{i}(x+i)$ where $i$ is from $1$ to $N$ I have the following series and I am unable to figure out which series it belongs to and how to solve it
$a(x+1)+a^{2}(x+2)+…+a^{N}(x+N)$
Above series is a generalization of my actual series
$\dfrac{1}{2}(x+1)+\dfrac{1}{4}(x+2) +\cdots +\dfr... | Read all sums as sums from $0$ to $n$
Using the identity $\sum_k a^k=\frac{1-a^{N+1}}{1-a}$ we can calculate $\sum_k ka^k$
$\sum_k (k+1)a^k=1+a+a^2+...+a^N+a+a^2+....+a^N+a^2+a^3+...+a^N+...+a^N=\frac{1-a^{N+1}}{1-a}+a\frac{1-a^{N}}{1-a}+a^2\frac{1-a^{N-1}}{1-a}+....a^N\frac{1-a}{1-a}=\frac{1+a+a^2+...+a^N - Na^{N+1}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$ Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$
$$\begin{align}\frac{1}{n^4+n^2+1}&
=\frac{1}{n^4+2n^2+1-n^2}\\
&=\frac{1}{(n^2+1)^2-n^2}\\
&=\frac{1}{(n^2+n+1)(n^2-n+1)}\\
&=\frac{1-n}{2(n^2-n+1)}+\frac{1+n}{2(n^2+n+1)}\\
\end{align}$$
For $n={1,2,3}$ it is not givi... | An observation:
We can reduce this series to another one. Note that $$\frac{1}{n(n\pm 1)+1}=\frac{|n\pm 1 -n|}{n(n\pm 1)+1}=|\tan(\arctan (n\pm 1)-\arctan n)|$$ Thus given summation is $$S=\sum_{n=1}^{50}\frac{1}{2n}\left(\tan(\arctan(n)-\arctan(n-1))-\tan(\arctan(n+1)-\arctan n)\right)\\=\sum_{n=1}^{50}\frac{1}{2n}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integration by De Moivre Theorem For example , integrate $\sin ^4(x) dx$.
I solved this question by reduction formula which is fairly easy. But my senior said that it would be easier if you expand the $\sin ^4(x) $by using De Moivre Theorem then integrate it. I've no idea of how De Moivre's Theorem would apply. Can an... | In order to evaluate the integral we have by De Moivre
$$(\cos x+i\sin x)^4=\cos4x+i\sin4x$$
Expanding the left hand side results in
$$\cos^4x+4i\cos^3x\sin x-6\cos^2x\sin^2x-4i\cos x\sin^3x+\sin 4x=\cos4x+i\sin4x$$
Equating the real component on both sides of the equation, we have
$$\cos^4x-6\cos^2x\sin^2x+\sin^4x=\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Simplification of $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$ I was trying to simplify $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$. Numerical evaluation suggested that the answer is $\sqrt{2}$ and it checked out when I substituted $\sqrt{2}$ in the equation $x= \sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$.
But I still cannot simplify the ini... | \begin{align}
\Big(\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}\Big)^2&=14+16-4\sqrt{7}-2\sqrt{14}\sqrt{16 - 4 \sqrt{7}}\\
&=30-4\sqrt{7}-4\sqrt{14}\sqrt{4 - \sqrt{7}}\\
&=30-4\sqrt{7}-4\sqrt{56 - 14\sqrt{7}}\\
&=30-4\sqrt{7}-4\sqrt{49 - 2\times 7 \sqrt{7}+(\sqrt{7})^2}\\
&=30-4\sqrt{7}-4\sqrt{(7- \sqrt{7})^2}\\
&=30-4\sqrt{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue? If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue ?
I notice quadratic residues of odd prime $13$ :
$$1^2\equiv 12^2\equiv 1\\2^2\equiv 11^2\equiv ... | Maps like $\,x\mapsto x^{-1}$ that are multiplicative $\,h(ab) = h(a)h(b)$ preserve squares $\,h(a^2)= (h(a))^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What am i doing wrong when solving this differential equation $$
f(x) = \frac{\frac{d^2}{dx^2}[e^y]}{\frac{d}{dx}[e^y]}
$$
Given that $f(x) = cx$
$$
\frac{c}{2}x^2 + k_1 = \ln(e^y y')
$$
$$
k_2\int e^{\frac{c}{2}x^2} dx = e^y
$$
$$
y = \ln(k_2\int e^{\frac{c}{2}x^2} dx)
$$
Therefore
$$
y'^2 - cxy' + y'' = 0
$$
$$
y' = ... | $(e^{y(x)})'=k_2e^{\frac12Cx^2}$ implies $e^{y(x)}=k_2\int_a^xe^{\frac12Ct^2}dt+k_3$, where $k_3=e^{y(a)}$ and which for $C>0$ does not involve the error function.
Then,
$$y(x)=\ln\left(k_2\int_a^xe^{\frac12Ct^2}dt+k_3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$ How to calculate the following limit?
$$\lim\limits_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$$
| Recall that, as $u \to 0$, by the Taylor series expansion, we readily have
$$
\begin{align}
e^u& =1+u+\mathcal{O}(u^2)\\
\ln (1+u)&=u+\mathcal{O}(u^2)
\end{align}
$$ giving, as $x \to \infty$,
$$
x^2\ln \left(1+\frac {2}{x^2-1}\right)=x^2 \left(\frac {2}{x^2-1}+\mathcal{O}\left(\frac {1}{x^4}\right)\right)=2+\mathcal{O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Proving that $7^n(3n+1)-1$ is divisible by 9 I'm trying to prove the above result for all $n\geq1$ but after substituting in the inductive hypothesis, I end up with a result that is not quite obviously divisible by 9.
Usually with these divisibility induction problems, it falls apart nicely and we can easily factorise... | Here's the inductive step. Assume the claim is true for $n=k$. Then we know that
$$9\mid 7^k(3k+1)-1.$$
Consider the case where $n=k+1$. In this case, the expression is
$$7^{k+1}(3(k+1)+1)-1.$$
Now, let's simplify this expression to
$$7[7^k(3k+1)+7^k\cdot 3]-1.$$
We observe that $7^k(3k+1)$ appears in this expressio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
Number of distinct terms in the expansion of $\big(x+\frac{1}{x}+x^2+\frac{1}{x^2}\big)^{15}$
Number of distinct terms in the expansion of $\bigg(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\bigg)^{15}$ is equal to ?
We can write the above as,
$$ \bigg(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\bigg)^{15} = \dfrac{1}{x^{30}}(1+x+x^3+x^4... | A general term (without the coefficient) of the polynomial $(1+x+x^3+x^4)^{15}$ will look like
$$x^{a+3b+4c} \qquad \text{ where } 0 \leq a,b,c, \leq 15.$$
So your question is now to show that $a+3b+4c$ can take on all integer values between $0$ and $60$.
For numbers in $0 \to 15$ let $a$ vary and have $b=c=0$.
Now tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Remainder of division by $23$ Find
$$(2014^{16}+2016^{32})^9 \ \ \ (\text{mod }23)$$
I know that
$$2014=2\times 19 \times 53$$
$$2016=2^5\times 3^2\times 7$$
This means that $2014 \equiv 2 \times (-4)\times (-3)$. I've tried solving this in multiple ways but I can't get the same answer as Wolframalpha, which is $(2014... | This can be done simply and quickly with purely mental arithmetic mod $\,23.\,$
We transform powers to powers of $\,2\,$ and use $\,\color{#c00}{a^{22}\equiv 1}\,$ if $\,23\nmid a\,$ by little Fermat.
$2016 = 20(100)+16 \equiv -3(8)+16 \equiv -8 \ \Rightarrow\ 2014\equiv 2016-2\equiv -10$
$2016^{32}\equiv (-8)^{32} \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
When is $20q^4-40q^3+30q^2-10q$ a square for positive integer $q$?
For what $q$ is the following polynomial a square?
$$
\begin{align}
&20q^4-40q^3+30q^2-10q\\
=\:&10q(q - 1)(2q^2 - 2q + 1)
&q\in\mathbb N
\end{align}
$$
I know of two single cases, $q=1$ gives $0$ and $q = 2$ gives $100$. I tested $3\le q\le12$ and ... | A similar Pell approach.
We want solutions to
$$10q(q-1)(2q^2-2q+1)=n^2$$
So $q\in{0,1}$ with $n=0$ are trivial examples. We might get solutions as linear combinations of these.
Let $2p=q(q-1)$.
Then $$5(4p)(4p+1)=n^2$$
We know $n$ must be a multiple of 5, so let n=5k.
$$5(4p)(4p+1)=25k^2$$
$$(4p)^2+4p=5k^2$$
Let x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find all integral solutions to $a+b+c=abc$. Find all integral solutions of the equation $a+b+c=abc$.
Is $\{a,b,c\}=\{1,2,3\}$ the only solution? I've tried by taking $a,b,c=1,2,3$.
| Of course, if one of $a,b,c$ are zero, then you can only have $\{a,b,c\}=\{0,k,-k\}$ for some integer $k$. So suppose none of $a,b,c$ are zero. Since $a,b,c$ satisfies $a+b+c=abc$ if and only if $-a,-b,-c$ satisfies $a+b+c=abc$, we can assume that either exactly one of $a,b,c$ is negative or none are negative.
Now divi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 2
} |
Proving $(|a|+|b|)^2(|c|+|d|)^2 \leq 4(a^2+b^2)(c^2+d^2)$. Given $a,b,c,d \in \Bbb R$, I want to prove that: $$(|a|+|b|)^2(|c|+|d|)^2 \leq 4(a^2+b^2)(c^2+d^2).$$
But the bound I'm getting is just too weak. I know that: $$|a| = \sqrt{a^2} \leq \sqrt{a^2+b^2} \implies |a|+|b| \leq 2\sqrt{a^2+b^2}\implies (|a|+|b|)^2\leq ... | Note that
$$
(|a|+|b|)^2=a^2+b^2+2|a||b|\leq a^2+b^2+a^2+b^2=2(a^2+b^2)
$$
and similarly for $(|c|+|d|)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show $x^2+4x+18\equiv 0\pmod{49}$ has no solution My method was to just complete the square:
$x^2+4x+18\equiv 0\pmod{49}$
$(x+2)^2\equiv -14\pmod{49}$
$x+2\equiv \sqrt{35}\pmod{49}$
So $x\equiv\sqrt{35}-2\pmod{49}$, which has no real solutions.
I feel that this may be too elementary, is this the correct way to solve t... | This isn't elegant, but it doesn't take that long. Suppose, for contradiction, that there is a solution modulo $49$. In order for there to be a solution modulo $49$, there must be a solution modulo $7$.
$$x^2+4x+4\equiv \begin{cases}
4 \pmod{7} &\mbox{if } x \equiv 0 \pmod{7}\\
2 \pmod{7} &\mbox{if } x \equiv 1 \pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
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.