Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculating limit of function To find limit of $\lim_{x\to 0}\frac {\cos(\sin x) - \cos x}{x^4} $.
I differentiated it using L Hospital's rule. I got
$$\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}\text{.}$$ I divided and multiplied by $\sin x$.
Since $\lim_{x\to 0}\frac{\sin x}{x} = 1$, thus I got
$\frac{1-\cos x}{4x^2}$.... | Using Prosthaphaeresis Formulas,
$$\cos(\sin x)-\cos x=2\sin\frac{x-\sin x}2\sin\frac{x+\sin x}2$$
So, $$\frac{\cos(\sin x)-\cos x}{x^4}=2\frac{\sin\frac{x-\sin x}2}{\frac{x-\sin x}2}\frac{\sin\frac{x+\sin x}2}{\frac{x+\sin x}2}\cdot\frac{x-\sin x}{x^3}\cdot\frac{x+\sin x}x\cdot\frac14$$
We know, $\lim_{h\to0}\frac{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/582275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Show that $ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$ I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$
Thanks in advance for any help.
| For $m=2$, it amounts to proving $\binom{2}{2} = \binom{3}{3}$, which is true since both equal $1$.
Induction step: let's assume the formula is true for a given $m$,
$$\sum_{n=2}^m \binom{n}{2}=\binom{m+1}{3}$$
Then,
$$\sum_{n=2}^{m+1} \binom{n}{2} = \sum_{n=2}^{m} \binom{n}{2} + \binom{m+1}{2}$$
$$=\binom{m+1}{3}+\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 1
} |
Convex Quadrilateral: $ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot D $ Problem
Let $ABCD$ be a convex quadrilateral with no right angles. Show that $$ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot... | $$A+B+C+D=2\pi\iff A+B=2\pi-(C+D)$$
$$\implies\tan(A+B)=\tan\{2\pi-(C+D)\}=-\tan(C+D)$$
Using Addition formula, $$\frac{\tan A+\tan B}{1-\tan A\tan B}=-\frac{\tan C+\tan D}{1-\tan C\tan D}$$
On rearrangement, $$\sum \tan A=\sum \tan A\tan B\tan C=\prod\tan A\left(\sum\frac1{\tan A}\right)$$
$$\sum \tan A=\left(\prod\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of a fraction of double factorials How can we show that
$\begin{align*}
\lim_{n\rightarrow\infty} U_n = 0
\end{align*}$
where
$\begin{align*}
U_n = \frac{(n-1)!!}{n!!}=\frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}\cdots
\end{align*}$
terminates at $\displaystyle\frac{2}{3}$ (odd) or $\displaystyle\frac{1}{2}$ (eve... | First observe the following elementary inequality
\begin{gather*}
\frac{m}{m+1}<\sqrt{\frac{m}{m+2}},\qquad \forall m\in\mathbb{N}
\end{gather*}
where $\mathbb{N}$ denotes the set of all the positive integers.
In the case of $n$ is even, we have, by using the above inequality,
\begin{align*}
0<U_n&=\frac{1}{2}\cdot \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/584456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
exponential equation with a sum of exponents I'm trying to solve the following exponential equation:
$e^{2x} - e^{x+3} - e^{x + 1} + e^4 = 0$
According to the the text I am using the answer should be $x = 1,3$ but I can't derive the appropriate quadratic $x^2 -4x + 3$ from the above equation using any of the methods I ... | $$e^{2x} - (e^3 +e)e^x + e^4$$
Let $m = e^x$
$$m^2- (e^3 + e)m + e^4$$
$a = 1$, $b=-(e^3 + e)$, $c =e^4$
$$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{(e^3 + e) \pm \sqrt{{(-(e^3 + e))}^2 -4(1)(e^4)}}{2(1)}$$
$$= \frac{(e^3 + e) \pm \sqrt{e^6 +e^2 + 2e^4 - 4e^4}}{2}$$
$$= \frac{(e^3 + e) \pm \sqrt{e^4 +1 + 2e^2 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/585162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
evaluation of $\int\frac{x^5}{x^5+x+1}dx$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx$
$\bf{My\; Try::}$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx = \int\frac{\left(x^5+x+1\right)-(x+1)}{x^5+x+1}dx = x-\int\frac{x+1}{x^5+x+1}dx$
Now Let $\displaystyle I = \int\frac{x+1}{x^5+x+1}dx = \int \frac{x+1}{(x^2+x+1)\cdot (x^3-x^2+... | A very nasty solution:
You have observed that $x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1)$. This means that you can factor $x^5 + x + 1$ into linear factors :$x^5 + x + 1 = \prod_{i=1}^5 (x-\alpha_i)$, where $\alpha_i$ can be computed. Thus, for any polynomial $p(x)$ or degree less than $5$ you can find unique constant... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/586388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How do I prove that $\lim_{n\to+\infty}\frac{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}{\sqrt{n}}=?$
let sequence $\{a_{n}\}$ such $a_{1}=1$,and
$$a_{n+1}a_{n}=n,n\ge 1$$
show that
$$2\sqrt{n}-1\le\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}<\dfrac{5}{2}\sqrt{n}-1$$
(2): I consider we... | For even indices
$$
\frac1{a_{2n}}=\frac{(2n-2)!!}{(2n-1)!!}=\frac1{2n}\frac{4^n}{\binom{2n}{n}}\sim\frac{\sqrt{\pi n}}{2n}
$$
For odd indices
$$
\frac1{a_{2n+1}}=\frac{(2n-1)!!}{(2n)!!}=\frac{\binom{2n}{n}}{4^n}\sim\frac1{\sqrt{\pi n}}
$$
Therefore,
$$
\begin{align}
\frac1{\sqrt{2n}}\sum_{k=1}^{2n}\frac1{a_k}
&\sim\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/586482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluate integral $\int_0^\frac{\pi}{2}x\ln(\sin x)~dx$ I've corrected typing error in the integral.
I apologize for my mistake.
Reedited question:
Can anybody solve integral:
$$\int_0^\frac{\pi}{2}x\ln(\sin x)~dx$$
I'm just trying to guess some simple formula for $\zeta(3)$. My "strategy" is simple: Find some conjectu... | Note
\begin{align}\int_0^\frac{\pi}{2}x\ln(2\sin x)~dx
=\frac12 \int_0^\frac{\pi}{2}x\ln(2\sin 2x)~dx
+ \frac12 \int_0^\frac{\pi}{2}x\ln\frac{\sin x}{\cos x}~dx\tag1\\
\end{align}
where
\begin{align}
&\int_0^\frac{\pi}{2} \overset{2x\to x} {x\ln(2\sin 2x)}~dx
=\frac14 \int_0^{\pi} \overset{x\to\pi-x} {x\ln(2\sin x)}~... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/589560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$ Simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$
To do it I have see it that we have basically $\sqrt{2}$ and $\sqrt{3}$ that is we can write it as,
$$\sqrt{\sqrt{3}\sqrt{3}+\sqrt{2}\sqrt{2}\sqrt{2}}-\sqrt{\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{2}-\sqrt{2}\sqrt{2}\sqrt{3}}$$
But ... | Hints:
$$3+2\sqrt2=(\sqrt2)^2+2\sqrt2+1\qquad 4-2\sqrt3=(\sqrt3)^2-2\sqrt3+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/591482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Establish $\int_0^{\infty} \frac{x^a}{x^2 + b^2}dx = \frac{\pi b^{a-1}}{2 \cos(\pi a /2)}$ when $-1 < a < 1$ My attempt at a solution: (this is homework, btw)
Let $f(z) = \frac{z^a}{z^2 + b^2}dz$ then the singularities of $f$ occur at $\pm ib$.
$$
Res(f; ib) = \frac{z^a}{z + ib} \biggr |_{ib} = \frac{(ib)^a}{2ib}
$$
$... | Setting $m=2$ in this answer, it is shown that
$$
\frac{\pi}{2}\csc\left(\pi\frac{a+1}{2}\right)=\int_0^\infty\frac{x^a}{1+x^2}\,\mathrm{d}x
$$
Thus,
$$
\begin{align}
\int_0^\infty\frac{x^a}{b^2+x^2}\,\mathrm{d}x
&=\frac{\pi}{2}\csc\left(\pi\frac{a+1}{2}\right)b^{1-a}\\
&=\frac{\pi\,b^{1-a}}{2\cos\left(\frac{\pi a}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/591719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Is it possible to express $\sin \frac{\pi}{9}$ in terms of radicals? So, yes, this is a math homework question. I've done some research on it and I know that the actual value for $\sin \frac{\pi}{9}$ cannot be expressed without using imaginary numbers.
http://intmstat.com/blog/2011/06/exact-values-sin-degrees.pdf
But, ... | $$\sum_{k=1}^{\infty} \frac{1}{6}(-\frac{1}{2})^{k} = \frac{1}{9}\\$$
The basic unit circle goes all the way up to integral multiples of $\frac{\pi}{6}$; thus, it is reasonable to assume that $\sin{\frac{\pi}{9}}$ can reasonably be written as an infinite nested root through half-angle formulas. The resulting numerators... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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2 is a primitive root mod $3^h$ for any positive integer $h$ It's easy to verify that 2 is a primitive root mod $3^2$. But then why does it follow that 2 is a primitive root mod $3^h$ for any positive integer $h$?
This was used in the solution of 2009 Putnam B6
http://math.hawaii.edu/home/pdf/putnam/Putnam_2009.pdf
I s... | Lemma: $2^{2\cdot 3^{n-1}}\equiv 1+3^n\pmod{3^{n+1}}$ for all $n\ge 1$.
Proof: We proceed with induction. Base case for $n=1$ is easy to verify. Suppose the lemma holds for some $n=k$. Then, by the inductive hypothesis $$2^{2\cdot 3^{k-1}}=1+3^k+3^{k+1}\ell,$$ for some $\ell\in\mathbf{Z}$.
Thus, $$2^{2\cdot 3^k}=1+3^{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/594782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\c... | By factorisation of
$$
\begin{aligned}
\sin ^5 x+\cos ^5 x= & \left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^3 x+\cos ^3 x\right) -\sin ^2 x \cos ^2 x(\sin x+\cos x) \\
= & (\sin x+\cos x)\left(1-\sin x \cos x-\sin ^2 x \cos ^2 x\right),
\end{aligned}
$$
we have $$
I=\int \frac{1}{\sin ^5 x+\cos ^5 x} d x=\int \frac{\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/595038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 4
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Convergence of $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ I want to check, whether $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ converges or diverges.
I tried to use Leibniz's test :
$|a_n|= \frac{n^2}{\sqrt{n^5+1}} = \frac{n^2}{\sqrt{n^4(n+\frac{1}{n^4})}} = \frac{n^2}{n^2\sqrt{n+\frac{1}... | We have
$$(n^5+1)^{-1/2}=\frac{1}{n^{5/2}}\left(1+\frac{1}{n^5}\right)^{-1/2}=\frac{1}{n^{5/2}}\left(1+O\left(\frac{1}{n^5}\right)\right)$$
hence
$$(-1)^n\frac{n^2}{\sqrt{n^5+1}}=\underbrace{\frac{(-1)^n}{n^{1/2}}}_{=u_n}+\underbrace{O\left(\frac{1}{n^{11/2}}\right)}_{=v_n}$$
the series $\displaystyle\sum_n u_n$ is con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$ Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$
$$\sqrt{2}=\mathbf{2}^{1/2}$$
$$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$
$$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$
Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$
$$... | Let S be your general term for a large value of n. If your square it you have S^2 = 2 S, then S = 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/601045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
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Finding the taylor series of $f(z) = 1/(1+z^2)$. I am working on the following exercise:
Find the Taylor expansion of the function $f(z) = \frac{1}{1+z^2}$ about $z = 3i$.
We had the Taylor Series Theorem in the lecture:
Let $D \subset \mathbb C$ be a domain and $f: D \to \mathbb C$ a differentiable function. Then $... | A naive, pretty useful many times, approach:
$$\frac1{1+z^2}=\frac1{10+6i+(z-3i)^2}=\frac1{10+6i}\frac1{1+\left(\frac{z-3i}{\sqrt{10+6i}}\right)^2}$$
Now, notice that
$$\left|\frac{z-3i}{\sqrt{10+6i}}\right|<1\iff|z-3i|^2<|10+6i|=4\sqrt{10}$$
for which values of $\;z\;$ we can write
$$\frac1{1+\left(\frac{z-3i}{\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/603018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Limit of series $4\left( \frac {1}{8}+\frac {1}{12}\right) +6\left( \frac {1}{24}+\frac {1}{36}\right) +\ldots$ How to find this serie
$4\left( \dfrac {1}{8}+\dfrac {1}{12}\right) +6\left( \dfrac {1}{24}+\dfrac {1}{36}\right) +8\left( \dfrac {1}{64}+\dfrac {1}{96}\right) +\ldots $
I think it's telescopic, isn't it?
| The terms in this sum seem to be
$$
2n\left( \dfrac {1}{2^{n-1}2n}+\dfrac {1}{3\cdot2^{n-2}2n}\right)
=
\dfrac {5}{6}\frac1{2^{n-2}}
$$
starting from $n=2$.
This is a geometric series and easy to sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/604012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How find this $I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4})^{-\frac{1}{2}}dS$ Find this Surface integral
$$I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}dS$$
where
$$\Sigma:\dfrac{x^2}{a^2}+\dfra... | Lets write the integral
$$I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}dS$$
in the form
$$I=\iint_{\Sigma} \frac{(x, y, z)}{(x^2+y^2+z^2)^{\frac{3}{2}}} \cdot \left(\left(\frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2}\right)\left(\dfrac{x^2}{a^4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/604085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $a$ and $b$ are positve integers, show that the inequality $e <\frac{a}{b}<\frac{87}{32} $ imples that $b \geq 39$. If $a$ and $b$ are positve integers, show that the inequality $e <\frac{a}{b}<\frac{87}{32} $ imples that $b \geq 39$.
Use the continued fraction $e=[2:1,2,1,1,4,1,1,6,...]$
| Note that $e>\frac{19}7$ and that $\frac{19}7<\frac ab<\frac{87}{32}$ implies that the numerators of the differences $\frac{a}{b}-\frac{19}{7}=\frac{7a-19b}{7b}$ and $\frac{87}{32}-\frac ab=\frac{87b-32a}{32b}$ are integers $\ge 1$.
Therefore
$$b=(7\cdot 87-32\cdot19)b=32(7a-19b)+7(87b-32a)\ge 32+7=39 .$$
Why is $e>\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/604668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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big determinant calculation I have found this exercise in a book, and having troubles solving it:
How to calculate this determinant?
$$\det\begin{pmatrix}
5 & 6 & 0 & 0 & 0 & \cdots & 0 \\
4 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 1 & 3 & 2 & 0 & \cdots & 0 \\
0 & 0 & 1 & 3 & 2 & \cd... | In general, when calculating a determinant of tridiagonal matrix of this form
$D_n=
\begin{vmatrix}
c& b & 0 & 0 & \ldots & 0 \\
a& c & b & 0 & \ldots & 0 \\
0& a & c & b & \ldots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & \ldots & 0 & a & c & b \\
0 & \ldots & 0 & 0 & a & c
\end{vmatrix}$
you can use La... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/606754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to find $\lim_{(x,y)\to(1,1)} \frac{x^3-y^3}{x^2 - y^2}$? Suppose $$ f(x,y) = \begin{cases}
\frac{x^3-y^3}{x^2 - y^2}, & \text{if $x^2 \ne y^2$} \\
\alpha, & \text{if $x^2 = y^2$} \\
\end{cases}$$
I need to find $\alpha$ such that $f(x,y)$ will be continous at $(1,1)$.
I started:
If $f(x,y)$ continous at $(1,1)$ ... | Hint: Notice that the numerator factors as follows:
$$
x^3 - y^3 = (x - y)(x^2 + xy + y^2)
$$
Now factor the denominator and cancel the $(x-y)$ factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/607666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Number of roots of a quadratic equation modulo 325 Any Help solving this question ?
a) Find ONE solution $\overline x\in\Bbb Z/325\Bbb Z$ such that $x^2\equiv-1\pmod{325}$. (Hint: CRT and lifting.)
b) How many solutions $\overline x$ to the above equation are there, and why?
| $325=5^2\cdot 13$ so let's solve $x^2\equiv -1\pmod{25}$, $x^2\equiv -1\pmod{13}$.
Case $1$: $\mod{25}$
If $x^2\equiv -1\pmod{25}$, then $x^2\equiv -1\pmod 5$ so $x\equiv\pm2\pmod 5$
$$(5k+2)^2=25k^2+20k+4\equiv 20k+4\equiv-1\pmod{25}\implies 4k\equiv -1\pmod 5\\\implies k\equiv 1\pmod 5$$
$$(5k-2)^2=25k^2-20k+4\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a formula for a recursively defined sequence I have a sequence given by:
\begin{align}
r_1 &= 1\\
r_2 &= 0\\
r_3 &= -1\\
r_n &= r_{n-1}r_{n-2} + r_{n-3}\\
R &= \{1, 0, -1, 1, -1, -2, 3, -7, -23, etc...\}
\end{align}
The first four lines were all we were given in order to help study for our final tomorrow. I ha... | I've not much time at the moment; but just a first approach if you have no other better idea: write the first few iterates, most explicite, beginning with [a,b,c] and try to discern a pattern. Later I'd look, whether there is some simplification, given the special values from the initial problem (like a pattern in the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/608778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Show that the sum of the series Show that the sum of the series is greater than 24
$$\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{9}+\sqrt{11}} +\cdots+\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
I see that
$\frac{1}{\sqrt{3}+\sqrt{1}}\cdot \frac{\sqrt{3}-\sqrt{1}}{\sqrt{3}-\sqrt{1}}=\frac{\sqrt... | The simplest way I can think of is to write it as $\frac12(\sum(\sqrt{4n+3} - \sum(\sqrt{4n-1}).$ You can approximate each sum by the integral, and since the function $\sqrt{x}$ is monotonic (and concave up) you can estimate the difference very closely.
A second (maybe even simpler) argument is to use your form, and no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
conditional expectation of squared standard normal Let $A,B$ independent standard normals. What is $E(A^2|A+B)$?
Is the following ok?
$A,B$ iid and hence $(A^2,A+B),(B^2,A+B)$ iid.
Therefore we have $\int_M A^2 dP = \int_M B^2 dP$ for every $A+B$-measurable set $M$ and hence $E(A^2|A+B) = E(B^2|A+B)$.
We obtain $2 \cdo... | if $A$ and $B$ are independent, then $A^2$ and $A+B$ are not independent because $A^2$ and $A$ are correlated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/609380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the solution of the equation Find all real solutions of this equation :
$$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
| In order for the equation to hold, we can see that $2-\sqrt{2+x}\geq0$, so $x\leq2$. Clearly, $x\geq0$. Since we are only interested in $\cos t$ and $0\leq\cos t\leq 1$, we may assume $0\leq t\leq\frac\pi2$.
Substitute $x=2\cos t$.
Then:
$$\begin{align*}&2+x=2(1+\cos t)=4\cos^2\frac t2,\hspace{5pt}\cos\frac t2\geq0 \h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How can I calculate $\sum_{k=1}^n (3k-1)^{2}$? I was trying to calculate the sum
$$\sum_{k=1}^n (3k-1)^2,$$
but actually I cannot frame the type. In fact is neither a geometric series, because the terms are not raised to the $k$th power, nor harmonic.
Any suggestions?
| Let $\displaystyle f(k)=Ak^3+Bk^2+Ck+D$ where $A,B,C,D$ are arbitrary constants
$\displaystyle\implies f(k+1)-f(k)=A\{(k+1)^3-k^3\}+B\{(k+1)^2-k^2\}+C\{(k+1)-k\}$$\displaystyle=A(3k^2+3k+1)+B(2k+1)+C$
$\displaystyle\implies f(k+1)-f(k)=3Ak^2+k(3A+2B)+A+B+C$
Comparing with $\displaystyle(3k-1)^2=9k^2-6k+1,$
$\displayst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Solving the equation $\frac{x^7}{7}=1+10^{1/7}x(x^2-10^{1/7})^2$ $$\frac{x^7}{7}=1+10^{1/7}x(x^2-10^{1/7})^2$$ Find $x$ where $x$ is real.
| This is really a trick question. Let $\alpha = 10^{1/14}$, the equation we have can be rewritten as
$$P(x) = 0 \quad\text{ where }\quad P(x) = \frac{x^7}{7} - 1 - \alpha^2 x (x^2-\alpha^2)^2$$
Let $x = \alpha y$, we can further simplify the equation
$$P(x) = 0
\quad\iff\quad\alpha^{-7} P(\alpha y) = 0
\quad\iff\quad\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How does mod multiplication work? For example, $10^{10} \equiv 4\pmod{6}$
If I used $\pmod{2}$ and $\pmod{3}$, how does the multiplication process work?
Since $10^{10} \equiv 0 \pmod{2}$ and $10^{10}\equiv 1\pmod{3}$,
$$
10^{10}\equiv (0,1) \pmod{(2,3)}
$$
how do we get the value $4$ at the end? do we list out the pos... | Let me try to rephrase what you said.
Let $10^{10} \equiv a \pmod 6$, $a \in \{0, 1, 2, 3, 4, 5\}$.
As $10^{10} \equiv 0 \pmod 2$, we must have $a \equiv 0 \pmod 2$, so $a \in \{0, 2, 4\}$.
As $10^{10} \equiv 1 \pmod 3$, we must have $a \equiv 1 \pmod 3$, so $a \in \{1, 4\}$.
Therefore, $a = 4$.
This does give the co... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=c$ Question:
let $a,b,c$ be positive constants. Find $u=u(x,y)$ if is satisfies the partial differential equation
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$
and the boundary condition
$$u=0,\dfrac{x^2}{a^2}+\df... | Let
$$v(x,y)=u(x,y)+\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(1-\dfrac{x^2}{a^2}-\dfrac{y^2}{b}\right)$$
Then $\Delta v=0$ in $\mathbb R^2$ and $v(x,y)=0$ on the boundary of the ellipse $U=\big\{\frac{x^2}{a^2}+\frac{y^2}{b^2}<1\big\}$. Thus, as the Dirichlet problem
$$
\Delta v=0 \,\,\text{in $U$} \quad\text{and} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Determine indefinite-integral $\text{I}=\int \sin^{n-1}x\sin\left[(n+1)x\right]\text{d}x$ Determine indefinite-integral:
$$\text{I}=\int \sin^{n-1}x\sin\left[(n+1)x\right]\text{d}x$$
My tried:
$$\text{I}=\int \sin^{n-1}x\sin\left[(n+1)x\right]\text{d}x=-\int\sin^{n-2}x\sin\left[(n+1)x\right]d(\cos x)$$
$$=- \sin^{n-2}... | Use angle sum expansion on the second term.
$$\int \sin^{n-1}{x}\left(\sin{x}\cos{nx} + \cos{x} \sin{nx}\right)\text{d}x $$
$$\frac{1}{n}\int (n\sin^{n-1}x\cos{x}\sin{nx} + n\sin^{n}{x}\cos{nx})\text{d}x$$
$$\frac{1}{n}\int \text{d}(\sin^{n}{x})\sin{nx}+\sin^{n}{x}\text{d}(\sin{nx})$$
Reverse the product rule
$$\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Definite integration problem (trig). I have this definite integral:
$$
\int_0^\Pi \cos{x} \sqrt{\cos{x}+1} \, dx
$$
For finding the indefinite integral, I have tried substitution, integration by parts, but I'm having trouble solving it.
By parts
$$
\int \cos{x} \sqrt{\cos{x}+1} \, dx\ = \sqrt{\cos{x}+1}\sin{x} + \frac{... | Here is how I would do it: first, let's recall the cosine double-angle identity. $$\cos 2x = \cos^2 x - \sin^2 x = \cos^2 x - (1 - \cos^2 x) = 2\cos^2 x - 1.$$ Thus the corresponding half-angle identity can be written $$\cos x = \sqrt{\frac{1 + \cos 2x}{2}}$$ or equivalently, $$\sqrt{1 + \cos x} = \sqrt{2} \cos \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/627812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Prove that $ \lim_{n \rightarrow \infty } \frac{n+6}{n^2-6} = 0 $. My attempt:
We prove that $ \lim\limits_{n \rightarrow \infty } \dfrac{n+6}{n^2-6} =0$.
It is sufficient to show that for any $ \epsilon \in\textbf{R}^+ $, there exists an $ K \in \textbf{R}$ such that for any $ n > K $
$ \displaystyle \left| \frac{... | $$\frac{n+6}{n^{2}-6}=r_{n}\times\frac{1+6r_{n}}{1-6r_{n}}$$ for: $$r_{n}=\frac{1}{n}$$
If $n\rightarrow\infty$ then $r_{n}\rightarrow0$ (easy) and consequently:
$$\frac{n+6}{n^{2}-6}\rightarrow0\times\frac{1+0}{1-0}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/633047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
find the limit: $\lim_{x\to 0} \dfrac{e^x \cos x - (x+1)}{\tan x -\sin x}$
find
$\lim_{x\to 0} \dfrac{e^x \cos x - (x+1)}{\tan x -\sin x}$
i tried using l'hopital's rule, but it just gets very complicated very fast
edit: i made a mistake in the numerator (sorry!) its $(x+1)$
| $$
\begin{aligned}
\lim _{x\to 0}\left(\frac{e^x\cos(x)-\left(x+1\right)}{\tan(x)\:-\sin(x)}\right)
& = \lim _{x\to 0}\left(\frac{\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o\left(x^3\right)\right)\left(1-\frac{x^2}{2}+o\left(x^2\right)\right)-\left(x+1\right)}{\left(x+\frac{x^3}{3}+o\left(x^3\right)\right)\:-\left(x-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/634861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
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If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that $(a+1/a)^2+(b+1/b)^2\ge 25/2$
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.$$
My work:
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\b... | Hint: Substitute $a=\frac{1}{2}+x$ and $b=\frac{1}{2}-x$, $|x|<\frac{1}{2}$. Then you should only find the minimum of a one variable function $f(x)$.
I'll write the whole solution, maybe someone finds it helpful. After the substitution we get:
$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2=\frac{9}{2}+2x^2+\frac{\frac{1}{2}+2x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Finding sum of sequence $\sum\limits_{k=2}^n \frac{1}{k^2-1}$ Here is a problem I need to solve:
$$ \sum_{k=2}^n \frac{1}{k^2-1} $$
It came with another one alreay done in the same task:
$$ \sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \frac{(k+1)-k}{k(k+1)} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{k+1} =... | $\frac{1}{(k-1)(k+1)}=\frac{1}{2}\frac{1}{k-1}-\frac{1}{2}\frac{1}{k+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/638078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove the following equality using mathematical induction: Prove the following equality using mathematical induction: For any integer $n \ge 1$
$$\sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$$
I understand for the base base I need to have $n=1$. If I substituted $n$ for 1 and $i$ for 1 I would get
\begin{align}
\su... | Hint:
Write $$\frac{n}{n+1} = \frac{n+1-1}{n+1} = 1-\frac{1}{n+1}$$
And note that the sum on the left side is a telescop sum. Thus you do not even need induction to proof this.
However, for the induction step, you could write
$$\begin{array}{rcl}
\sum_{i=1}^{n+1} \frac{1}{i(i+1)} &=& \frac{1}{(n+1)(n+2)}+\sum_{i=1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/638366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
My work:
$(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{... | HINT:
Use $$\cos^2x=1-\sin^2x$$ to form a Quadratic Equation in $\displaystyle\sin^2x$
writing $\displaystyle\sin^2x=p$ we get $$\frac{p^2}a+\frac{(1-p)^2}b=\frac1{a+b}$$
$$\implies b p^2+ a(1+p^2-2p)=\frac{ab}{a+b}$$
$$\implies (a+b)\{(a+b)p^2-2ap+a\}=ab$$
$$\implies (a+b)^2p^2-2a\cdot (a+b)p+a^2=0\implies \left[p(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Calculate a hard limit Calculate: $$
\lim_{x\to+\infty} x\left( \frac{1}{x^2+1^2}+\frac{1}{x^2+2^2}+\dots+\frac{1}{x^2+x^2}\right)$$
| $$
\begin{align}
& \phantom{=} \lim_{x\to+\infty} x\left( \frac{1}{x^2+1^2} + \frac{1}{x^2+2^2} + \dots + \frac{1}{x^2+x^2}\right) \\[8pt]
& = \lim_{x\to\infty} \frac 1 x \left( \frac{1}{1+\frac{1^2}{x^2}} + \frac{1}{1+\frac{2^2}{x^2}} +\cdots+\frac1{1+\frac{x^2}{x^2}} \right) \\[8pt]
& = \int_0^1 \frac{1}{1+w^2} \, dw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Sum of consecutive square roots inside a square root $$\large\sqrt{1+\sqrt{1+2+\sqrt{1+2+3+\sqrt{1+2+3+4+\cdots}}}}$$
I saw this somewhere in the internet but, the website didn't provide me any further information. What is the sum of the equation above? What is it called?
| As Gerry Myerson said this type of problems are called nested radical.
Let us try the following method: Writing the sequence of partials $$D_n:=
\sqrt {1 + \sqrt {1 + 2 + \sqrt {1 + 2 + 3 + \sqrt {1 + 2 + 3 + 4 + \ldots + \sqrt { \ldots + \sqrt { \ldots + \sqrt {1 + 2 + 3 + \ldots + n} } } } } } }
$$
Labeling t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/642210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 2,
"answer_id": 1
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If $p$, $q$ are naturals, solve $p^3-q^5=(p+q)^2$. In If $p,q$ are prime, solve $p^3-q^5=(p+q)^2$., the author asks to solve the equation $p^3-q^5=(p+q)^2$ for primes $p$ and $q$. A proof is given that $p=7, q=3$ is the only solution.
In this "followup", I would like to ask for a proof that does not depend on $p$ and $... | I came up with a solution for $\gcd(p, q)=1$. Note that $p>q$. Looking mod $p$ gives $q^2(q^3+1) \equiv 0 \pmod p$, thus $p|q^3+1$. Take modulo $q$ of both sides of the equation to get $p^2(p-1) \equiv 0 \pmod q$, hence $q|p-1$. So we have $p=qr+1$ for some positive integer $r$. It follows that$$p|q^3+1-p=q^3-qr=q(q^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 2,
"answer_id": 0
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if $A+B+C+D=\pi$, what is $\min(\cos{A}+\cos{B}+\cos{C}+\cos{D})$ Let $A,B,C,D \in R ~|~ A+B+C+D=\pi$, what is the minimum of the following function
$$f(A,B,C,D)=\cos{A}+\cos{B}+\cos{C}+\cos{D}$$
I found a post about a similar problem, butI think an even number of variables is harder than an odd number of variables, be... | An alternative approach:
It is known that
$$
\cos A + \cos B + \cos C + \cos D = 2 \cos\frac{A+B}{2} \cos\frac{A-B}{2} +2 \cos\frac{C+D}{2} \cos\frac{C-D}{2}.
$$
By introducing the new variables
$$
A' = \frac{A+B}{2} \quad B' = \frac{A-B}{2} \quad C' = \frac{C+D}{2} \quad D' = \frac{C-D}{2},
$$
hence
$$
A = A'+B' \qu... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Proof of an equality involving cosine $\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{2}}}}\ =\ 2\cos (\pi/2^{n+1})$ so I stumbled upon this equation/formula, and I have no idea how to prove it. I don't know how should I approach it:
$$
\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{\vphantom{\large A}2\,}\,}\,}\,}\
=\
2\c... | I know it is a bit old, but I'd like to elaborate a bit the answer here going step by step.
As commented above, the best hint is to use the cosine half-angle formula ($*=\cos\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1+\cos(\alpha)}{2}}$) and work by induction. Therefore, let's get to it:
Base case:
The first partial... | {
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"url": "https://math.stackexchange.com/questions/649968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
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Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.
My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\lef... | For some reason, this problem appeared on the main page for me even though the problem was posted quite a while ago. Let me present an alternative approach which does not really require any integrating.
Let us consider your integral:
$$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos(x)+a^2\right)\ \text{d}x,\;\ a>1$$
Consider a t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/650513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 2
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Induction: show that $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$ The question:
Induction: show that:
$$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + ... + \frac{1}{\sqrt{n}} < 2\sqrt{n}$$
for $n \geq 1$
My attempt at a solution:
First w... | You have proved it but if you dislike switching to (essentially) showing a new inequality is true you could approach this slightly differently.
$\begin{align}
1 + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n+1}}
&< 2\sqrt{n} + \frac{1}{\sqrt{n+1}} \mathrm{(by\ inductive\ hypothesis)} \\
&= \frac{2\sqrt{n}(n+1)}{n+1} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/653530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Solve Recurrence Equation with Induction Question: Given the recurrence equation for the recursive Fibonacci sequence program:
$T(n) = T(n-1) + T(n-2) + b$
$T(0) = T(1) = a$
Using induction, show that $T(n) \leq f(n)$, where $f(n) = c2^n, \forall n \geq 0$. Find a value for $c$ in the process.
Attempt at a Solution:
B... | Calculate. We get $T(2)=2a+b$, $T(3)=3a+2b$, $T(4)=5a+4b$, $T(5)=8a+7b$, $T(6)=13a+12b$, and so on. The coefficients grow fast, but not too fast.
For simplicity assume that $a$ and $b$ are non-negative.
We show by induction that $T(n)\le 2^n a+2^n b=(a+b)2^n$. For the induction step, suppose that $T(k-1)\le 2^{k-1}(a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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problem requires condition probability Q. A robot fires 3 shots at a moving target. For the first shot, the probability of hitting the moving target is 1/3. For subsequent shots beyond the first shot, the probability of hitting the moving target is 1/2 if the previous shot is a hit (for example, the probability of hitt... | Let $X$ be the number of hits. Then $X$ can take on values $0$, $1$, $2$, or $3$.
We want to find the probability of $k$ hits, that is, $\Pr(X=k)$, for $k=0,1,2,3$.
To save space, write for example $hmm$ for hit then miss then miss.
The probability that $X=0$ is the probability of $mmm$. This is $\frac{2}{3}\cdot\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/656936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$2017$ as the sum of two squares Write the prime $2017$ as the sum of two squares
$2017$ can be written as the sum of two squares because it is a prime of the form $p\equiv 1\ ($mod $4)$
Using an appropriate algorithm find the two numbers that, when squared, add to the total of $2017$
| About 200 years ago, Jacobi would have known, and shown, that the number of representing 2017 as a sum of two squares is exactly 8.
More generally, for a positive integer $n$, let $r_2(n)$ denote the number of representations of $n$ as a sum of two squares of integers; for example, $r_2(5) = 8$, since $5 = 2^2 + 1^2 =... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Simplify the surd expression. Simplify the surd.
$(2\sqrt 3 + 3\sqrt 2)^2$
I know I should us this formula: $(a^2+2ab+b^2)$
But this gets complicated later. Please explain. :(
| Alternatively, with a little simplification:
$$(2\sqrt{3}+3\sqrt{2})=(\sqrt{2}\sqrt{2}\sqrt{3}+\sqrt{3}\sqrt{3}\sqrt{2})=(\sqrt{2}(\sqrt{2}\sqrt{3}+\sqrt{3}\sqrt{3}))=\sqrt{2}(\sqrt{6}+3)$$
we can find the result:
$$\begin{align}[\sqrt{2}(\sqrt{6}+3)]^2&=2(\sqrt{6}+3)^2\\ & =2(6+2\sqrt{6}\cdot3+3^2) \\ &=2(15+6\sqrt{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/658736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Prove that $\sum_{k=0}^{\infty} (k-1)/2^k = 0$ How to prove that this series converges, and that the limit is 0 ?
| You don't need calculus at all.
The sum is equal to $-1+ 0 + \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots$
In essence, we want to show that $\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots = 1$
We can decompose this into an infinite number of infinite geometric series as follows:
$$S_1 = \frac{1}{4} + \frac{1}{8} ... | {
"language": "en",
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Computing this limit $$\lim_{n\to\infty}{\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4}}$$
At first glance, we see that it's an indeterminate form ($\infty-\infty$). Here are my tries:
I.) I tried to form $a^2-b^2$ in the numerator, where $\cdots$ represents something that $\to 0$:
$$\frac{n^2(\sqrt[3]{1+\cdots}-1)-3n-4}{n(\sqrt... | Try this:
$$\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4} = n\cdot \left(\sqrt[3]{1+\frac{5}{n}+\frac{6}{n^3}} - \sqrt{1+\frac{3}{n}+\frac{4}{n^2}}\right).$$
L'Hospital rules should help from here on.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For a positive integer $n$ both $5n+1$ and $7n+1$ are perfect squares. Show that $n$ is divisible by 24. My try:
$5n + 1 = k^2$
$7n +1 = \frac{7k^2-2}5$
Just don't know how to proceed after this. Please help.
| Given, both $5n+1$ and $7n+1$ are perfect squares, We have to prove, $24\mid n$
Consider the sum: $(5n+1)+(7n+1)=12n+2$. This number leaves the remainder $2$ when divided by $3$. This is possible only under three situations:
a) $5n+1\equiv 2\mod3$ and $7n+1\equiv 0\mod3$
b) $5n+1\equiv 1\mod3$ and $7n+1\equiv 1\mod3$
c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving $\sqrt{x+2}+\sqrt{x-3}=\sqrt{3x+4}$ I try to solve this equation:
$$\sqrt{x+2}+\sqrt{x-3}=\sqrt{3x+4}$$
So what i did was:
$$x+2+2*\sqrt{x+2}*\sqrt{x-3}+x-3=3x+4$$
$$2*\sqrt{x+2}*\sqrt{x-3}=x+5$$
$$4*{(x+2)}*(x-3)=x^2+25+10x$$
$$4x^2-4x-24=x^2+25+10x$$
$$3x^2-14x-49$$
But this seems to be wrong! What did i wron... | Note: the original question read $= \sqrt{3x + 5}$ instead of $= \sqrt{3x + 4}$. There is no problem with the fixed question. Maybe just that it's unfinished. The final line should read $3x^2 - 14x + 49 = 0$ rather than just $x^2 - 14x + 49$. After that, solve the quadratic; only one of the solutions ($7$) to the quadr... | {
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Divisibility problem: $ \frac{3^{m}}{2^{n} - 3^r} $
Is divisible a power of 3 for a difference of powers of 2 and 3? That
is, can result, this division, in an integer? $$ \frac{3^{m}}{2^{n} -
3^r} $$ where $n,m,r$ natural number.
Edit: $n>r$, $r=m+1$.
| HINT:
We need $\displaystyle 2^n-3^r=\pm3^s$ where $s\le m$
$\displaystyle 2^n=3^r\pm3^s$
Case $1:$ Taking the $+$ sign, $2^n=3^r+3^s=3^{r-s}(3^s+1)$ (WLOG $r\ge s$)
As $(2,3)=1,$ we need $3^{r-s}=1\iff r=s$ and $2^n=3^r+1$
Case $2:$ Taking the $-$ sign, $2^n=3^r-3^s=3^{r-s}(3^s-1)$ (we need $r\ge s$)
Like the previous... | {
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If $x+\frac1x=5$ find $x^5+\frac1{x^5}$. If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it?
$$
\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110.
$$
Thanks
| Observe the recurrence relation $$x^{n+1} + x^{-(n+1)} = (x+x^{-1})(x^n+x^{-n}) - (x^{n-1} + x^{-(n-1)}).$$
This immediately gives us the specific recurrence $$f_{n+1} = 5f_n - f_{n-1}, \quad f_0 = 2, \quad f_1 = 5,$$ where $f_n = x^n + x^{-n}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
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Proof of Divisibility of $n(n^2+20)$ by 48. This is a question from Bangladesh National Math Olympiad 2013 - Junior Category that still haunts me a lot. I want to find an answer to this question. Please prove this.
If $n$ is an even integer, prove that $48$ divides $n(n^2+20)$.
| ${} \bmod 16$, we have $n(n^2+20) \equiv n(n^2+4)=2k(4k^2+4) = 8k(k^2+1)$. If $k$ is even then clearly $8k(k^2+1)\equiv 0$. If $k$ is odd, then $k^2+1$ is even and again $8k(k^2+1)\equiv 0$.
${} \bmod 3$, we have $n(n^2+20) \equiv n(n^2-1)=(n-1)n(n+1)\equiv 0$, since given three consecutive numbers, exactly one of them... | {
"language": "en",
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$\displaystyle \frac{6}{2n-1} - \frac{1}{n} = \frac{p}{2^i5^j}$ $\displaystyle \frac{6}{2n-1} - \frac{1}{n} = \frac{p}{2^i5^j}$
For which $n$ is this expression true. $n$ and $p$ are integers. $i$ and $j$ are positive integers or zero.
| $$\frac{p}{2^i5^j} = \frac{6}{2n-1} - \frac{1}{n} = \frac{4n+1}{n(2n-1)}.$$
$n=1, 2$ are clearly solutions, and $3$ is not a solutions. Henceforth let $n\geq 4$.
Since $\gcd(n, 4n+1) = 1$, hence we must have $n \mid 2^i 5^j$. Since $\gcd(n, 2n-1) = 1$, they do not share any common factors. Note that $ \gcd( 2n-1, 4n+1)... | {
"language": "en",
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Inverse of the Joukowski map $\phi(z) = z + \frac{1}{z}$ We know the Joukowski map $$\phi(z) = z + \frac{1}{z}$$ which maps the upper semidisc of radius $1$ in the lower half plane, and the lower semidisc of radius $1$ in the upper half plane.
What is the inverse of this function ? We obtain $z ^{2}-zy + 1 = 0$ and thi... |
We obtain $z^2−zy+1=0$ and this equation has $2$ solutions, which is the right one?
The solutions are given by
$$z = \frac{y - \sqrt{y^2-4}}{2},$$
where the different solutions correspond to the different choices of the square root.
We want a holomorphic inverse, so we need a holomorphic branch of $\sqrt{y^2-4}$ on t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Random variable transformation function I am stuck with a random variable transformation problem ($Y=\phi(X)$). The random variable $X$ has a uniform distribution $U(-1,1)$, and I want to transform it into $Y$ which is also an uniform distribution $U(\frac{s}{2},2s)$, where $s$ is some scalar quantity. Does anyone know... | $2s-\frac{s}{2}= \frac{3s}{2} = \frac{3s}{4} \times (1-(-1))$
$\frac{3s}{4} \times 1 = \frac{3s}{4} = 2s - \frac{5s}{4}$ and $\frac{3s}{4} \times (-1) = -\frac{3s}{4} = \frac{s}{2} - \frac{5s}{4}$
So $Y=\frac{3s}{4} X +\frac{5s}{4} = \frac{s}{4} (3x+5) $ is a possible answer.
$Y=-\frac{3s}{4} X +\frac{5s}{4} = \frac{... | {
"language": "en",
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If $(a,b) = 1$, show that $(a-b,a^2+ab+b^2)=1 \text{ or } 3$ I have a question...We know that $(a,b)=1$ and we want to show that $(a-b,a^2+ab+b^2)=1 \text{ or } 3$.How can I show this??
I thought that we could suppose that $(a-b,a^2+ab+b^2)=d$.Then we know that $d|a-b$ and $d|,a^2+ab+b^2$.But how can I continue??
| $d|a-b$,$d|a^2+ab+b^2$ and $a^2+ab+b^2=(a-b)^2+3ab$ imply that
$$d|3ab,d|3a(a^2+ab+b^2)=3a^3+3a^2b+3ab^2,d|3ab(a-b)=3a^2b-3ab^2,d|2a(3ab).$$
Hence, $$d|3a^3+3a^2b+3ab^2+3a^2b-3ab^2-6a^2b=3a^3$$ and similarly $d|3b^3$ and now
$$d\le(3a^3,3b^3)=3(a^3,b^3)=3.$$ so $d=1\,\text{or}\,2\,\text{or}\,3$, but $d$ can't be $2$, ... | {
"language": "en",
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If $x,y,z\in(0;1)$, prove that $(x+1)(y+1)(z+1)\ge \sqrt{8(x+y)(y+z)(z+x)}$.
If $x,y,z\in(0;1)$, prove that $$(x+1)(y+1)(z+1)\ge \sqrt{8(x+y)(y+z)(z+x)}$$
Both sides of the inequality are positive, so I could square them: $$(x+1)^2(y+1)^2(z+1)^2\ge8(x+y)(y+z)(z+x)$$
Even if there's a solution using brute force, we'd ... | You can prove that
$$
(x+1)(y+1)\geq 2(x+y)
$$
From that, you can multiply the three inequalities you get by permuting the variable to get what you want.
To prove that, first expand both sides
$$
xy+x+y+1\geq 2x+2y
$$
simplifying yields
$$
xy+1\geq x+y\\
xy-x-y+1\geq 0\\
(x-1)(y-1)\geq 0\\
(1-x)(1-y)\geq 0
$$
But since... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can we convert the number $(0.\overline{101})_2$ written in dual-system into the decimal system?
How can we convert the number $(0.\overline{101})_2$ written in dual-system into the decimal system ?
$0.\overline{101}=\underbrace{\dfrac{1}{2}+\dfrac{0}{4}+\dfrac{1}{8}}_{\dfrac{5}{8}}+\underbrace{\dfrac{1}{16}+\dfr... | There is no such thing as a characeterwie substitution to obtain base ten periods from base two periods. Just thinkn of $\frac13=0.\overline 3_{10}=0.\overline{01}_2$ and especially, $\frac15=0.2_{10}=0.\overline{0011}_2$.
The actual analogy is that $$0.\overline{909}_{10}=\frac{909}{10^3-1}=\frac{909}{999}=\frac{101}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Formal Limit Proofs for Limits Involving Factorials How does one use the definition of the limit to formally prove that
$$
\lim_{n \to \infty} \frac{n^4}{n^2 + n!} = 0?
$$
| Well, you know that $n! \ge n(n-1)(n-2)(n-3)(n-4) \ge (n-4)^5$ (for $n >4$), so
you have $0 \le{n^4 \over n^2+n!} \le {n^4 \over n!} \le {n^4 \over (n-4)^5} = {{1 \over n} \over (1-{4 \over n})^5 }$. Hence for $n>8$ we have $(1-{4 \over n}) \ge {1 \over 2}$, and so ${n^4 \over n^2+n!} \le 2^5 {1 \over n}$.
Now choose $... | {
"language": "en",
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Find the maximum or minimum value of the quadratic function. Find the maximum or minimum value of the quadratic function by completing the squares. Also, state the value of $x$ at which the function is maximum or minimum.
$y=2x^2-4x+7$
$x^2$ has a coefficient of $2$, how should I complete the squares?
| $y=2x^2-4x+7$ therefore $y=2(x^2-2x+7/2)$
completing the square now gives $y=2[(x-1)^2-1+7/2]$
which rearranges to $y=2(x-1)^2+5$
For completing the square, I always divide by a number that ensures I have $1$ as the coefficient of $x^2$.
To get the minimum value there are two options:
1) By looking at the equation afte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/690157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $n^3 + 5n$ is divisible by $6$ by using induction Prove that for integers $n > 0$, $n^3 + 5n$ is divisible by $6$.
Here is what I have done:
Base Step: $n=1$, $1^3+5(1)=6$
Inductive Step:
$p(k)=k^3 + 5k =6m$, $m$ is some integer
$p(k+1)=(k+1)^3+5(k+1)=6m$ $m$ is some integer
Since both are equal to $6m$ I s... | Hint: $(k+1)^3+5(k+1) = (k^3 + 5k) + 3k^2 + 3k + 6$
Since $(k^3 + 5k)$ is divisible by 6, all you have to prove is $3k^2 + 3k + 6$ is divisible by 6 too. Since adding $6$ does not change divisibility, you just have to proove $3k(k+1)$ is divisible by 6. Think about even and odd numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expected number of steps between states in a Markov Chain Suppose I am given a state space $S=\{0,1,2,3\}$ with transition probability matrix
$\mathbf{P}= \begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 & 0 \\[0.3em]
\frac{2}{3} & 0 & \frac{1}{3} & 0\\[0.3em]
\frac{2}{3} & 0 & 0 & \frac{1}{3... | The distribution for the number of time steps to move between marked states in a discrete time Markov chain is the discrete phase-type distribution. You made a mistake in reorganising the row and column vectors and your transient matrix should be
$$\mathbf{Q}=
\begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 \\
... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the coefficient of the term $x^4 y^5$ in $(x+y+2)^{12}$? What is the coefficient of the term $x^4 y^5$ in $(x+y+2)^{12}$?
How can we calculate this expression ?
I've applied the binomial theorem formula and got $91$ terms but I am not sure if it is right or wrong.
| The trick is to apply trinomial formula here, the coefficient is
$${12\choose4,5,3}2^3=\frac{12!2^3}{4!5!3!}$$
To see this, first consider $(x+y)$ as a whole body and then use binomial formula. Then the coefficient of term $(x+y)^9$ is
$\displaystyle{12\choose 9}2^3=\frac{12!2^3}{9!3!}$ (Note we don't need to consider... | {
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"source": "stackexchange",
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"answer_id": 1
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FINDING PMF probability theory A die is tossed until first 6 occurred. Let X be Random variable that number of one in the experiment. Find the PMF of X. And E(X).
I noticed PMF of X is geometric distribution but I don't know why.
| The distribution of $X$ is indeed geometric, under one of the definitions of geometric distribution.
Note that $X$ is the number of $1$s until the first $6$. In particular, for this game, anything other than $1$ or $6$ does not matter, and need not even be recorded. All that matters is $1$ or $6$. Given that anything ... | {
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Solving a curve integral around part of an elipse I'm having trouble calculating a curve integral in a vector field:
$\int_C y (18x + 1)\ dx + 2y^2\ dy$
where $C$ is the curve along the ellipse
$9x^2 + y^2 = 64$
going counterclockwise from the point ( $-\frac{4}{3}\sqrt{3} $ , $4$) to the point (-$\frac{4}{3}$ , $... | Draw a figure! The parametrization of the ellipse is
$$t\mapsto \bigl(x(t),y(t)\bigr):=\left({8\over3}\cos t,8\sin t\right)\qquad(t\in{\mathbb R})\ .\tag{1}$$
Now we have to find the boundary values which are relevant for the integral $J$ in question. Both endpoints $p$ and $q$ of the arc lie in the second quadrant. At... | {
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Why doesn't this approach work for $\int \sec^4 x\,dx$? I been trying to integrate $\sec^4$ , without much luck. But I don't entirely understand why my result is invalid and would like some feedback if possible.
I'm attacking the issue in the following way
$$
\int (\sec^2{x})^2dx = \int (\tan^2+1)^2dx
$$
then, I put $u... | Your result is almost valid.
Taking a different approach, using Integration by Parts from the start:
Note that $\sec^2x = \frac{d}{dx} (\tan x)$.
We can use integration by parts:
$u = \sec^2 x \implies\,du = 2\sec^2 x \tan x\,dx$
$dv = \sec^2 x \,dx\implies \, v = \tan x$.
$$\int \sec^4 x \,dx = \sec^2x\tan x - 2\int \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $k$ cannot equal $0$ and $A$ is as given below, what is $A$ inverse? $$
A =
\pmatrix{
1 & 0 & 2k\\
0 & 1 & k\\
0 & 0 & k
}
$$
I don't know what method to use to solve this problem as I haven't encountered and variable before when solving for the inverse of a matrix. I've tried to use the identity method to solve it... | Hint: We can write the matrix as the product of elementary matrices:
$$
\begin{bmatrix}
1 & 0 & 2k \\
0 & 1 & k \\
0 & 0 & k \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine the value of the integral $I=\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$ Determine the value of the integral $$I(a)=\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$$
My try:
$\to I'(a)=\int_{0}^{\infty}\frac{2a}{(a^2+x^2)(b^2+x^2)}dx=\frac{\pi}{b(a+b)}$
Hence $I(a)=\frac{\pi}{b}\l... | Since integrand is even function, so
$\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$=$\frac{1}{2}\int_{-\infty}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$
so solving the R.H.S using residues--
$\int_{C} f(z)dz=\int_{Cr} f(z)dz +\int_{-R}^{R}\frac{1}{2} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof that there are at the most two numbers of exactly six digits that squared end with the same six digits Written in a more formal way, proof that there are at the most $2$ numbers $n$ of six digits, that
$$n^2 \equiv n \mod 10^6$$
Research effort:
if $n^2 \equiv n \mod 10^6$ this means $10^6\mid n^2-n \rightarrow1... | If $\displaystyle2^6|(n-a)$ and $5^6|(n-a),$
$\displaystyle$ lcm$(2^6,5^6)|(n-a)$
As $(2,5)=1,(2^6,5^6)=(2,5)^6=1$
$\implies$ lcm$(2^6,5^6)=(2^6\cdot5^6)=10^6\iff n\equiv a\pmod{10^6}$
Here $a=0,1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that there is a unique sequence of positive integers $(a_n)$ satisfying $a_1=1,a_2=2,a_4=12,a_{n+1}a_{n-1}=a_n^2\pm 1 $ Show that there is a unique sequence of positive integers $(a_n)$ satisfying the following conditions.
$$a_1=1,a_2=2,a_4=12,a_{n+1}a_{n-1}=a_n^2\pm 1$$
I approached the problem to find out,
$... | Suppose that after $a_n$, that both possible values of $a_{n+1}$ , $\frac{a_n\,^2 - 1} {a_{n-1}}$ and $\frac{a_n\,^2 + 1} {a_{n-1}}$, are both integers.
Then $$\begin{cases} a_{n-1} | a_n\,^2 - 1 \\ a_{n-1} | a_n\,^2 + 1 \end{cases}$$
So $a_{n-1}$ can only be $1$ or $2$ if the sequence forks into to valid sequences aft... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof this curious trigonometric identity Proof that $$\cos^2{10^\circ} + \cos^2{50^\circ} - \sin{40^\circ}\sin{80^\circ} = \frac{3}{4}$$
I notice that $10^\circ + 80^\circ = 90^\circ$, and $50^\circ +40^\circ = 90^\circ$. I tried doing some manipulation but my efforts were futile. Any hints?
| Generalization :
$$I=\cos^2A+\cos^2B-\cos A\cos B$$
$$=\cos^2A-\sin^2B+1-\frac{\cos(A-B)+\cos(A-B)}2$$
$$=\cos(A-B)\cos(A+B)+1-\frac{\cos(A-B)+\cos(A+B)}2$$
$$I=\cos(A-B)\left(\cos(A+B)-\frac12\right)+1-\frac{\cos(A+B)}2$$
If $\displaystyle\cos(A+B)=\frac12=\cos60^\circ\iff A+B=2n\pi\pm60^\circ$ where $n$ is any intege... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/703798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Using residues to evaluate an improper integral
Use residues to evaluate the improper integral
\begin{align}
\int_{0}^{\infty} \frac{1}{(x^2+1)^2} \, dx
\end{align}
First, I said $f(z) = \frac{1}{(z^2+1)^2}$. My only question so far is how do I establish the region $C$ (from the given real limits of $0$ to $\infty$... | In response to @Cameron Williams' hint and comments, I am going to attempt the solution.
We have $f(z) = \frac{1}{(z^2+1)^2}$. Let $C$ be the half circle as described by @Cameron Williams. Now, we have $z = i$ to be the singularity point inside $C$.
In finding the residue,
\begin{align}
\text{Res}_{z = i} f(z) &= \tex... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the area of the indicated surface Find the surface area of the part of the sphere $x^2 + y^2 + z^2 = a^2$ inside the circular cylinder $x^2 + y^2 = ay$ ($r = a\sin(\theta)$ in polar coordinates), with $a > 0$.
First time posting on this website, sorry for the lack of details on my attempts but I am really not sure... |
I'm going to assume that, since the problem asks for the surface area of the sphere found within the cylinder, the total for both hemispheres is what is asked for. So we'll work out the expression for the "upper" hemisphere and double that.
The function we will work with for that hemisphere is $ \ f(x,y) \ = \ z \ = ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Elementary Symmetric Polynomials, Roots of cubic polynomials I'm given $a_1, a_2, a_3$ as roots of the equation $x^3 + 7x^2 - 8x + 3$ and need to find the cubic polynomials with roots $a_1^2, a_2^2, a_3^3$ and $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}$. It's in a chapter about Galois Theory and is preceded by a pro... | $a$, $b$ and $c$ are the roots of $x^3 + px^2 + qx + r$ if and only if $abc = -r$, $ab+bc+ca=q$ and $-(a+b+c)=p$.
We have $\frac{1}{a_1} \cdot \frac{1}{a_2} \cdot\frac{1}{a_3} = \frac{1}{a_1a_2a_3} = -1/3$, $\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} = \frac{a_1a_2 + a_2a_3 + a_3a_1}{a_1a_2a_3} = \frac{-8}{-3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/708567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality problem $(a+b+c)^5\geq27(ab+bc+ca)(ab^2+bc^2+ca^2)$ While solving one inequality, I arrived at a much simpler, but still nontrivial inequality
$$(a+b+c)^5\geq27(ab+bc+ca)(ab^2+bc^2+ca^2)$$
where $a,b,c$ are positive real numbers.
It apparently holds, but I can't seem to find a proof.
The problem is it's not ... | Without loss of generality, let $a+b+c=3$. Then we only have to prove
$$(a^2b+b^2c+c^2a)(ab+bc+ac)\le 9$$
Assume $b=\text{mid}(a,b,c)$, then we have
$$c(b-a)(b-c)\le 0\Longleftrightarrow a^2b+b^2c+c^2a\le b(a^2+ac+c^2)$$
So
$$(a^2b+b^2c+c^2a)(ab+bc+ac)\le b(a^2+ac+c^2)(ab+bc+ac)$$
Using AM-GM inequality, we get
$\begin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can $\frac {100-100}{100-100}=2$? \begin{align*}
\frac{0}{0} &= \frac{100-100}{100-100} \\
&= \frac{10^2-10^2}{10(10-10)} \\
&= \frac{(10+10)(10-10)}{10(10-10)} \\
&= \frac{10+10}{10} \\
&= \frac{20}{10} \\
&= 2
\end{align*}
I know that $0/0$ isn't equal to $2$, the what is wrong in this proof? I wasn't satisfied after... | $1.$ You cannot start with $\frac00$. This is not a number.
$2.$ You are trying to simplify the fraction by dividing numerator and denominator with $(10-10)$. This cannot be done.
Assume beginning from the end going towards the start. The wrong part is when you multiply with $(10-10)$. Because then denominator would be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/710551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Induction proof of $n^{(n+1) }> n(n+1)^{(n-1)}$ The question statement from my homework booklet goes:
Prove by mathematical induction that $n^{n+1} > n(n+1)^{n-1}$
is true for all integers $n \geq 2$.
I've managed to come up with this for the induction step (the base case is trivial), but I am not sure what to do ... | We are required to prove that: $(k+1)^{k+2}>(k+1)(k+2)^k$.
The assumption is that: $k^{k+1} > k(k+1)^{k-1}$. Dividng both sides by $k^k$ gives: $k> \left(\frac{k+1}{k}\right)^{k-1}$, thus $k>\left(1+\frac{1}{k}\right)^{k-1}$
Starting from this,
\begin{eqnarray}
k&>&\left(1+\frac{1}{k}\right)^{k-1}\\
\frac{1}{\left(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability of a certain ball drawn from one box given that other balls were drawn Box 1 contains 2 green and 3 red balls, 2 has 4 green and 2 red, and 3 has 3 green and 3 red. Only one ball is drawn from each of the 3 boxes. What is the probability that a green ball was drawn from box 1 given that two green balls were... | E1 = Green from Box 1
E2 = 2 Greens Drawn
E1 and E2 = Green from Box 1 and 2 Greens Drawn
E1:
2 Green in Box 1
5 Total in Box 1
P(E1) = 2/5
E2:
Possible combinations are GGR and GRG. Add these two combinations together.
P(GGR) = (2/5) * (4/6) * (3/6) = 2/15
P(GRG) = (2/5) * (2/6) * (3/6) = 1/15
P(E1 and E2) = P(GGR) +... | {
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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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Proving divisibility by using induction: $133 \mid (11^{n+2} + 12^{2n+1})$ If $n > 0$, then prove the following by using induction: $$133|(11^{n+2} + 12^{2n+1}).$$
| Inductive step:
$$\begin{align}11^{(n+1) + 2} + 12^{2(n+1)+1} &= 11^{n+3} + 12^{2n+3}\\
&=11\cdot11^{n+2} + 144\cdot12^{2n+1}\\
&= 11\cdot11^{n+2} + 11\cdot12^{2n+1} + 133\cdot12^{2n+1}\\
&=11\cdot(11^{n+2} + 12^{2n+1}) + 133\cdot12^{2n+1}\end{align}$$
But $133 | (11^{n+2} + 12^{2n+1})$. This is the inductive hypothesi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/716789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving a combintorical problem using generating function
In how many ways $n$ balls can be divided into groups where each group may contain only one ball or two.
Each ball can be a singleton, or coupled with one of the $n-1$ options. Hence,
$a_0 = 1$
$a_1 = 1$
$a_n = a_{n-1} + (n-1)a_{n-2}$
The correspond gener... | To solve the recurrence that you have given,
$$
\begin{align}
f(x)
&=\sum_{k=0}^\infty a_kx^k\\
&=1+x+\sum_{k=2}^\infty (\color{#C00000}{a_{k-1}}+\color{#00A000}{(k-1)a_{k-2}})x^k\\
&=1+x+\color{#C00000}{x(f(x)-1)}+\color{#00A000}{x^2(xf'(x)+f(x))}\\[10pt]
-1
&=x^3f'(x)+f(x)(x^2+x-1)
\end{align}
$$
This equation can be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/717679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Matrix linear transformation from $\mathbb{R}^3$ to $\mathbb{R}^4$. Find the matrix representation Say that $T : \mathbb{R}^3 \to \mathbb{R}^4$ is linear and satisfies $T(1,1,1) = (3,2,0,1)$, $T(1,1,0) = (2,1,3, −1)$ and $T(1,0,0) =(5, −2,1,0)$. Find $T(x, y, z)$ and then find the matrix representation for $T$.
I am not ... | $T(0,0,1) = T(1,1,1) - T(1,1,0) = (1, 1, -3, 2)$, and $T(0,1,0) = T(1,1,0) - T(1,0,0) = (-3, 3, 2, -1)$, and $T(1,0,0) = (5, -2, 1, 0)$. So
\begin{align*}T(x,y,z) &= xT(1,0,0) + yT(0,1,0) + zT(0,0,1) \\
&= x(5,-2,1,0) + y(-3,3,2,-1) + z(1,1,-3,2) \\
&= (5x-3y+z, -2x+3y+z, x+2y-3z, -y+2z).\end{align*} And from this the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using the polar form of $1 + i$ and $\sqrt3 + i$ to deduce $\cos (\frac{\pi}{12}), \sin(\frac{\pi}{12})$ I have been beating my head against the following problem and would like a gentle nudge in the right direction.
The question states, by writing $1 + i$ and $\sqrt3 + i$ in polar form, deduce that
$$\cos (\frac{\pi}{... | Notice
$$1 + i = \sqrt2e^{i\frac{\pi}{4}}$$
$$\sqrt{3} - i = 2e^{-i\frac{\pi}{6}}$$
Multiply both and we get
$$(1 + i)(\sqrt{3} - i) = \sqrt2e^{i(\frac{\pi}{4} - \frac{\pi}{6})}$$
$$(\sqrt3 + 1) + (\sqrt3-1)i= 2\sqrt{2}e^{i\frac{\pi}{12}}$$
$$(\sqrt3 + 1) + (\sqrt3-1)i= 2\sqrt2\left(\sin{\frac{\pi}{12}} + i\cos\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/719825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Getting incorrect result on limit of square root I have a limit problem:
$\lim_{x \to \infty} \sqrt{x^2 + x} - x$
According to Wolfram|Alpha the answer is $\frac{1}{2}$
However, my calculation gives $1$. Please help me understand what I'm doing wrong. The process is:
$\lim_{x \to \infty} \sqrt{x^2 + x} - x = \lim_{x \t... | $\lim_{x \to \infty} \sqrt{x^2 + x} - x$
approach:
taking $t = \frac{1}{x}$ hence t tends to 0
So,
$\lim_{t \to 0} \sqrt{\frac{1}{t^2} + \frac{1}{t}} - \frac{1}{t}$
$= \lim_{t \to 0} \frac{\sqrt{1+t}}{t} - \frac{1}{t}$
$= \lim_{t \to 0} \frac{1}{t} (\sqrt{1+t} - 1)$
multiplying by $\frac{\sqrt{1+t} + 1}{\sqrt{1+t} + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve system of two equation with two unknown and using substitution? $$\begin{align}
a^2 - b^2 = 3\\
a \cdot b = 2
\end{align}$$
In aforementioned equations, we can mentally find out the value of $a = 2, b = 1$. But what is the general way to solve this system algebraically?
I tried to use substitution but I g... | HINT:
$$\implies b^4+3b^2-4=0\iff (b^2+4)(b^2-1)=0$$
Then from the second equation, find corresponding value of $a$ for each value of $b$
Then, check whether each pair of $(a,b)$ satisfy the first
Alternatively, $$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/722886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove by induction on $n\geq k$ that $n^3 \lt 3^n$. What is the value $k$? Prove by induction on $n\geq k$ that $n^3 \lt 3^n$. What is the value $k$?
It looks like $k$ should be $0$ after trying random values but obviously that is a terrible way of doing things.
Proof:
P(n): n^3<3^n
Assume k holds
Prove k+1
P(k+1): (k... | First we compute a little, systematically. We have
$0^3\lt 3^0$;
$1^3 \lt 3^1$;
$2^3\lt 3^2$;
$3^3=3^3$, so there inequality $n^3\lt 3^n$ is not true at $n=3$;
$4^3=64$, $3^4=81$, so $4^3\lt 3^4$;
$5^3=125$, and $3^5=243$;
$6^3=216$, and $3^6=729$;
$7^3-343$, and $3^7=2187$.
If this were a horse race, it looks as if th... | {
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"url": "https://math.stackexchange.com/questions/726609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove variance in Uniform distribution (continuous) I read in wikipedia article, variance is $\frac{1}{12}(b-a)^2$ , can anyone prove or show how can I derive this?
| By the definition of the variance, $\operatorname{Var} X = \mathbb{E}[X^2] - (\mathbb{E} X)^2$.
Since here $\mathbb{E} X = \frac{1}{b-a}\int_{[a,b]}x dx = \frac{a+b}{2}$, and $\mathbb{E} X^2 = \frac{1}{b-a}\int_{[a,b]}x^2 dx = \frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3}$, it follows that
$$
\operatorname{Var} X = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/728059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $ and $ \ x+y+z \ = \ xyz \ $ Consider the system of equations in real numbers $ \ x,y,z \ $ satisfying
$$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $$
and $ \ x+y+z \ = \... | This is a comment that’s too long for the usual format.
LAcarguy’s method generalizes to any values $a,b,c$ instead
of $\frac14,\frac15,\frac16$.
If we put
$$
\phi(a,b,c)=ab+ac-bc, \ \psi(a,b,c)=ab+ac+bc \tag{1}
$$
and
$$
G(a,b,c)=\phi(a,b,c)\phi(b,c,a)\phi(c,a,b)\psi(a,b,c) \tag{2}
$$
then LAcarguy’s method leads to... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the angle of intersection of circles $x^2+y^2-6x+4=0 \ \&\ x^2+y^2-2x-2y-8=0$ Find the angle of intersection of circles $$x^2+y^2-6x+4=0 \\ x^2+y^2-2x-2y-8=0$$
my answer is : 41.14 degrees. but i'm not sure if it's right. please help me.
| Write the equations as
$$
(x-3)^2+y^2=5
$$
and
$$
(x-1)^2+(y-1)^2=10
$$
Thus, we have a triangle with sides $\sqrt{5}$, $\sqrt{10}$, $\sqrt{5}$:
$\hspace{3cm}$
The Law of Cosines says
$$
5=5+10-2\sqrt{5}\sqrt{10}\cos(\alpha)
$$
which implies
$$
\cos(\alpha)=\frac1{\sqrt2}
$$
We could also recognize the $45{-}45{-}90$ t... | {
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"timestamp": "2023-03-29T00:00:00",
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Complex Modulus Is Less Than One Show that $|\frac{-a}{b}+\frac{\sqrt{a^2-b^2}}{b}|<1$ where $a>|b|>0$
This is just a minor part of a more full complex trigonometric integral proof that I'm working on. I'm pretty much set once I can show this inequality is true. Please Help!
What I was thinking was:
$|\frac{-a}{b}+\fra... | Hint:
$$
\sqrt{a^2-b^2}-a = \frac{a^2-b^2-a^2}{\sqrt{a^2-b^2}+a}$$
details:
$$
0\le a-\sqrt{a^2-b^2}
= \frac{b^2}{\sqrt{a^2-b^2}+a}\le |b| \frac {|b|}a\le |b|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/733488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integral over a sphere How to calculate $$I=\int\int_S\frac{d^2S}{r^n}$$ where S is a surface of the sphere $S=\{x^2+y^2+z^2=R^2\}$ and $r=dist((x,y,z),(0,0,c))$, $c>R,c\in\mathbb{R}$ ?
I believe $r=\sqrt{x^2+y^2+(\sqrt{R^2-x^2-y^2}-c)^2}$ so $$I=\int\int_{x^2+y^2<R^2}\frac{1}{\Big(x^2+y^2+(\sqrt{R^2-x^2-y^2}-c)^2\Big)... | Work in spherical coordinates $(R, \theta, \phi)$. Note that
$$r = \sqrt{x^2 + y^2 + (z-c)^2} = \sqrt{x^2 + y^2 + z^2 - 2cz + c^2} = \sqrt{R^2 - 2Rc \cos \phi + c^2}$$ and $$d^2 S = R^2 \sin \phi\, d\theta\, d\phi$$
so
$$\iint_S \frac{d^2S}{r^n} = \int_0^\pi \int_0^{2\pi} \frac{R^2 \sin \phi}{(R^2 - 2Rc\cos \phi + c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/735117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Need help to solve taylor series of $e^{\sin x}$ How to derive the taylor series of $e^{\sin x}$, up to $x^5$?
i just don't know how to get the answer
$$f(x) = 1 + x + \frac{x^2}{2} - \frac{x^4}{8} -\frac{x^5}{15}$$
really need some help. Thanks
| Let $f(x) = e^{\sin x}$.
\begin{equation*}
\begin{split}
f'(x) &= e^{\sin x} \cos x = f(x) \cos x\\
f''(x) &= f'(x) \cos x - f(x) \sin x \\
f'''(x) &= f''(x) \cos x - f'(x)\sin x - (f(x)\cos x + f'(x) \sin x)\\
&= f''(x) \cos x - 2f'(x) \sin x - f'(x)\\
f^4 (x) &= f'''(x) \cos x - f''(x) \sin x - 2(f'(x)\cos x + f''(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/735287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Taylor expansion of $\frac{1}{2-z-z^2}$ The problem is:
Find the Taylor expansion of $f(z):= \dfrac{1}{2-z-z^2}$ on the disc $|z| < 1$
So far I have used partial fractions to obtain $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfrac{1}{2+z}\right)$ which I then rewrite as $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfra... | Hint: $\frac{1}{1-z}=1+z+z^2+\ldots$ as infinitely descending geometric progression. And $\frac{1}{2+z}=\frac{1}{2}\cdot\frac{1}{1-(-z/2)}=\ldots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/735409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
consecutive prime power I'm interesting on consecutive prime power numbers. I see that there is the Mersenne primes and the Fermat Primes that give solutions and $(8,9)$. In Sloane collection it is referred on A006549 and it is written:
David W. Wilson and Eric Rains found a simple proof that in this case of Catalan'... | There is indeed a relatively easy proof of the fact that if $n$ and $n+1$ are consecutive prime powers, then
1) Either $n$ or $n+1$ must be a power of 2.
2) Apart from $n=8$, the other number must be a prime.
Note: Here, the proof is relatively easy because we exploit the fact that the numbers are prime powers, and no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/735653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
how to factor $x^4+2x^3+4x^2+3x+2$ I'm trying my hand on these types of expressions.
How to factorize $x^4+2x^3+4x^2+3x+2$ into two (or more) polynomials with rational coefficients. please write step by step solution.
| Let $P(x)=x^4+2x^3+4x^2+3x+2$. If $P$ factors over the rationals, it must factor over the integers. Since $P$ is monic (i.e., its lead coefficient is $1$), the only possible roots (which would correspond to linear factors, $x-r$) are $\pm1$ and $\pm2$. We can dismiss $1$ and $2$ out of hand, since $P(x)$ is clearly ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/736479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to evaluate $\int\frac{x}{\sqrt{x^2+x+1}} \, dx$ using trigonometric substitution? I am pretty sure that my answer is correct but given answer for the exercise from textbook Calculus James Steward was slightly different. Any idea to solve this:
$$\int\frac{x}{\sqrt{x^2+x+1}} \, dx$$
The given answers: $\sqrt{x^2+x+... | $$x^2+x+1 = \left ( x+\frac12 \right )^2+\frac{3}{4} $$
$$\begin{align} \int dx \frac{x}{\sqrt{x^2+x+1}} &= \int dx \frac{x+\frac12}{\sqrt{\left ( x+\frac12 \right )^2+\frac{3}{4}}}-\frac12 \int dx \frac{1}{\sqrt{\left ( x+\frac12 \right )^2+\frac{3}{4}}}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/739042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I find the inverse of this exponential function? $x=-3(3^{-x})+9$
I know the steps up until a certain point.
$x=-3(3^{-y})+9$
$x-9=-3(3^{-y})$
$\frac{(x-9)}{-3} = 3^y$
$ln (\frac{x-9}{-3}) = -y * ln 3$
Not sure what to do from here. I know I have to get y by itself but thats it. Can anyone help please?
| We want to solve for $y$, after switching $x$ and $y$.
$y=-3(3^{-x})+9$
becomes
$x=-3(3^{-y})+9\\
\implies x-9=-3(3^{-y})\\
\implies \dfrac{x-9}{-3}=3^{-y}\\
\implies -\dfrac{1}{3}x+3=3^{-y}\\
\implies \ln\left(-\dfrac{1}{3}x+3\right)=\ln (3^{-y})\\
\implies \ln\left(-\dfrac{1}{3}x+3\right)=-y\cdot \ln 3\\
\implies -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/739129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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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.