Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then what values of $(a,b)$ would work? Let $a$ and $b$ be distinct integers. If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then which of the following statements are true?
1) $c*(a+c)=-b$
2) $a+b=-1$
3) $a+b+c=0$
4) $c=0$
Update
I just tried to su... | Factor Theorem implies $x-c$ is a factor of $x^2+ax+b-(x^2+bx+a)$, and this is equal to $(a-b)(x-1)$, so $c=1$, let $r$ be the another root of $x^2+ax+b$ then $(x-1)(x-r)=x^2-(1+r)x+r=x^2+ax+b$, so $r=b$ and $a=-(1+r)=-1-b$. Therefore $a+b=-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the limit: limit x tends to zero [arcsin(x)-arctan(x)]/(x^3) I'm having difficulty in finding the following limit.
$$\lim_{x\to 0}\frac{\arcsin(x)-\arctan(x)}{x^3}$$
I tried manipulating the given limit in standard limit(s) but I got nowhere. I tried L'Hôpital's rule and then realized it would be very lengthy and ... | It's worth mentioning that l'Hopital does work here. Your limit becomes
$$\lim_{x \rightarrow 0} {{1 \over \sqrt{1 - x^2}} - {1 \over 1 + x^2} \over 3x^2}$$
$$= \lim_{x \rightarrow 0}{(1 + x^2) - \sqrt{1 - x^2} \over 3x^2(1 + x^2)\sqrt{1 - x^2}}$$
Multiply numerator and denominator by $(1 + x^2) + \sqrt{1 - x^2}$ and y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$n$ is some natural number. Let $x$ be the integer part of $\sqrt n$ and $y$ be the decimal part. If $x^2 - y^2 = 1+4y$ what is $y^x$? $n$ is some natural number. Let $x$ be the integer part of $\sqrt n$ and $y$ be the decimal part. If $x^2 - y^2 = 1+4y$ what is $y^x$?
This is some high school problem but I can't solve... | Since $y$ is the decimal part, we know that $0\leq{y}<1$, then from $x^2=1+4y+y^2$ we get $$1\leq{x^2}<6$$. $x$ is an integer, then $x$ can be $1$ or $2$.
For $x=1$, $y=0$, then $y^x=0$;
for $x=2$, $y=\sqrt{7}-2$, and we have $n=7$ which satisfies the requirement. Then $y^x=11-4\sqrt{7}$.
| {
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Let $n$ be a positive integer. Show that if $2^n -1$ is a prime number, then $n$ is a prime number. Let $n$ be a positive integer. Show that if $2^n -1$ is a prime number, then $n$ is a prime number.
This is how I started to tackle this question:
Assume that instead of $n$ being a prime number, it is a composite number... | It's probably easier to prove the contrapositive: if $n$ is composite then $2^n - 1$ is composite. So suppose $n$ is composite, i.e. $n=ab$ with $a$ and $b$ integers greater than $1$. Using the factorization
$$
y^a - 1 = (y - 1)(y^{a-1} + y^{a-2} + \cdots + y + 1)
$$
and writing $2^n - 1 = 2^{ab} - 1 = (2^b)^a - 1$, we... | {
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Find the value of $(a+b+c)$ when $\cos\theta+\cos^2\theta+\cos^3\theta=1$ and $\sin^6\theta=a+b\sin^2\theta+c\sin^4\theta$ Given: $\cos\theta+\cos^2\theta+\cos^3\theta=1$
and $\sin^6\theta=a+b\sin^2\theta+c\sin^4\theta$
Then find the value of $(a+b+c)$
| Clearly we need to eliminate $\cos\theta$
We have $\displaystyle\cos\theta(1+\cos^2\theta)=1-\cos^2\theta\iff\cos\theta(1+1-\sin^2\theta)=\sin^2\theta$
Squaring we get $$\cos^2\theta(2-\sin^2\theta)^2=(\sin^2\theta)^2$$
$$\iff(1-\sin^2\theta)(2-\sin^2\theta)^2=\sin^4\theta$$
Rearrange to form a six-degree equation i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving Non-homogenous System (repeated case) I have the following system. $\vec{x^{'}}(t)=\begin{pmatrix}4&-2\\8&-4 \end{pmatrix} \vec{x}+ \begin{pmatrix}t^{-3}\\-t^{-2}\end{pmatrix}$
I get $\lambda=0$ and the eigenvector of $$4x_1=2x_2 \implies \begin{pmatrix} 1\\2 \end{pmatrix}$$
Next I solved $$\begin{pmatrix}4&-2\... | Given:
$$\vec{x^{'}}(t)=\begin{pmatrix}4&-2\\8&-4 \end{pmatrix} \vec{x}+ \begin{pmatrix}t^{-3}\\-t^{-2}\end{pmatrix}$$
We have as eigenvalues and eigenvectors:
$$\lambda_{1,2} = 0, v_1 = (1,2), v_2 = \left(\dfrac{1}{4},0\right)$$
Update 1: Note that eigenvectors are not unique and your book chose $v_2 = \left( 0, -\dfr... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Calculation of $\int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$ Calculation of $\displaystyle \int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$
$\bf{My\; Try::}$ Let $x=\tan \psi\;,$ Then $\displaystyle dx = \sec^2 \psi$
So Integral convert into $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi-1+\... | Let $x=\sqrt{y^2-1}$ to reduce the integral
\begin{align}
&\int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx
=\int_1^{\sqrt2}\frac y{1+y}dy=\sqrt2-1+\ln[2(\sqrt2-1)]
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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If $x\equiv2\pmod{3}$ prove that $3|4x^2+2x+1$ I've tried many different things to get a factor of $k-2$ but keep failing.
If $x\equiv2\pmod{3}$ prove that $3 \mid 4x^2+2x+1$
| Since $x \equiv 2 \pmod 3$,
$4x^2+2x+1 \equiv (16) + (4) + (1) \equiv 21 \equiv 0 \pmod 3$
Hence, $3\mid4x^2+2x+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\int_0^{2 \pi} \sqrt{1-(a+b \sin\phi)^2} d\phi $ have a closed form in terms of elliptic integrals? Consider the following integral for real $a, b$ such that the square root is real:
\begin{equation}
I=\int_0^{2 \pi} \sqrt{1-(a+b \sin\phi)^2} d\phi
\end{equation}
For $a = 0$, the integral is easily expressed in t... | Result:
$$\boxed{\displaystyle \mathcal{I}=4\sqrt{\frac{b}{k}}\,\biggl[\mathbf{E}\left(k^2\right)-\left(1-k^2\right)\mathbf{K}\left(k^2\right)+\left(1-k^2\right)\Pi\left(c^{-2}|k^2\right)\biggr]} \tag{$\heartsuit$}$$
where I follow Mathematica conventions for arguments of $\mathsf{EllipticE}$, $\mathsf{EllipticK}$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve this equation or system of equations? I want to solve the equation
$$(5 x-4) \cdot\sqrt{2 x-3}-(4 x-5)\cdot \sqrt{3 x-2}=2.$$
I tried. Put $a = \sqrt{2 x-3}\geqslant 0$ and $b =\sqrt{3 x-2}\geqslant 0 $.
Suppose
$$5x-4=m(2x-3)+n(3x-2)$$
then $m=\dfrac{2}{5}$ and $n=\dfrac{7}{5}$.
Therefore
$$5x-4=\dfrac{2}... | Hint: try integer values...usually in this kind of problem first try to see when you get integer result, so both $\sqrt{3x-2} \text{ and } \sqrt{2x-3}$ should be integers...solution: $x=6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
inverse of an infinite matrix How to find inverse of an infinite lower triangular matrix all of whose diagonal entries are 1 and the entries of each column are given by coefficients of some power series rings?
| Just use gaussian elimination. You have an infinite matrix $(a_{ij})_{i,j \in \mathbb{N}}$ where $a_{ii} = 1$ for all $i$, and $a_{ij} = 0$ if $i < j$. We're looking for a (lower triangular) matrix $(b_{ij})$ such that $$
\begin{pmatrix}
1 & 0 & \ldots \\
a_{21} & 1 & 0 & \ldots \\
a_{31} & a_{32} & 1 & \ddots \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/772074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Generating function of $\binom{3n}{n}$ Wolfram alpha tells me the ordinary generating function of the sequence $\{\binom{3n}{n}\}$ is given by
$$\sum_{n} \binom{3n}{n} x^n = \frac{2\cos[\frac{1}{3}\sin^{-1}(\frac{3\sqrt{3}\sqrt{x}}{2})]}{\sqrt{4-27x}}$$
How do I prove this?
| As was already mentioned in the comment section the Lagrange Inversion Formula is a proper method to prove this identity. In the following I use the notation from R. Sprugnolis (etal) paper Lagrange Inversion: when and how.
Let us suppose that a formal power series $w=w(t)$ is implicitely defined by a relation $w=t\Ph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/774434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Integral $\int_0^{\pi/4} \frac{\ln \tan x}{\cos 2x} dx=-\frac{\pi^2}{8}.$ $$I:=\int_0^{\pi/4} \frac{\ln \tan x}{\cos 2x} dx=-\frac{\pi^2}{8}.$$
I am trying to see nice solutions to this integral. I tried the following
$$
I=\int_0^{\pi/4}\frac{\ln \sin x}{\cos 2x} dx-\int_0^{\pi/4} \frac{\ln \cos x }{\cos 2x}dx
$$
but... | $$
\begin{aligned}
\int_0^{\frac{\pi}{4}} \frac{\ln (\tan x)}{\cos 2 x} d x=&\int_0^{\frac{\pi}{4}} \frac{\ln \left(\tan \left(\frac{\pi}{4}-x\right)\right)}{\cos 2\left(\frac{\pi}{4}-x\right)} d x \\
=& \int_0^{\frac{\pi}{4}} \frac{\ln \left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)}{\sin 2 x} d x\\=&\frac{1}{2} \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/775219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 5
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Exploring 3-cycle points for quadratic iterations How do you factorise $z^8 +4cz^ 6 + (6c^2 +2c)z^4 + (4c^3 +4c^2 )z^2 - z+ (c^4 +2c^3 +c^2 +c)$?
I want to find the 3-cycle points for the quadratic iteration $z \rightarrow z^2 + c$. In order to find it I have to solve the above polynomial. I have factorised into two fa... | The Galois group of such a polynomial of degree $6$ will not be $S_6$.
As soon as you know one of the roots $z$, you know two of the others, by applying $z \mapsto z^2+c$ twice.
The splitting field of $P$ should then generally be of degree $6 \times 3 = 18$.
Let $\zeta$ be a primitive thrid root of unity.
If you call t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $f(t)$ be defined by $ f(t) = \left\{ \begin{array}{c} t, &0 \le t < 1 \\ b - t^2, &1 \le t \le 2 \end{array} \right. $ Let $f(t)$ be defined by
$$
f(t) =
\left\{
\begin{array}{c}
t, &0 \le t < 1 \\
b - t^2, &1 \le t \le 2
\end{array}
\right.
$$
and let $F(x)$ be defined by $F(x) = \int_0^xf(t)dt,\,\,\,\,\, 0 \le... | This seems correct to me. Now you must check for which $b$ the right and left derivatives at $x = 1$ are equal. The left derivative is $1$ and the right is $-1+b$ so $1 = -1+b$ iff $b =2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\cot(x) +\cot(\frac{\pi}{3}+x) + \cot(\frac{2\pi}{3}+x) = \frac{3-9\tan^2x}{3\tan x-\tan^3x}$ As the title says. I tried simplifying the LHS but got:
$$\frac{\tan^2x+7\tan x- \sqrt{3}}{\tan^2x-\sqrt{3}}$$
What should I do next?
| From this, $\displaystyle\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}$
$\displaystyle\implies\cot3x=\frac{1-3\tan^2x}{3\tan x-\tan^3x}\ \ \ \ (1)$
Multiplying the numerator & the denominator by $\cot^3x,$
$\displaystyle\cot3x=\frac{\cot^3x-3\cot x}{3\cot^2x-1}\ \ \ \ (2)$
If $\displaystyle\cot3x=\cot3A\iff\tan3x=\tan3A\i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To find relatively prime ordered pairs of positive integers $(a,b)$ such that $ \dfrac ab +\dfrac {14b}{9a}$ is an integer How many ordered pairs $(a,b)$ of positive integers are there such that g.c.d.$(a,b)=1$ , and
$ \dfrac ab +\dfrac {14b}{9a}$ is an integer ?
| If $(a,b)$ is an ordered pair of positive integers such that $\gcd(a,b)=1$ and
$$\frac{a}{b}+\frac{14b}{9a}=\frac{9a^2+14b^2}{9ab},$$
is an integer, then $9ab$ divides $9a^2+14b^2$. In particular $a$ and $b$ both divide $9a^2+14b^2$, and so $a$ divides $14b^2$ and $b$ divides $9a^2$. Because $\gcd(a,b)=1$ it follows th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/779541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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$\int_{0}^{\infty}\frac{x}{x^3+1}dx$ =? So guys I have this improper integral $\int_{0}^{\infty}\frac{x}{x^3+1}dx$. I checked that it converges by $ \int_{0}^{1}\frac{x}{x^3+1}dx + \int_{1}^{\infty}\frac{x}{x^3+1}dx $ and using the $\frac{c}{(x-a)^\lambda} $ and $\frac{1}{x^\lambda} $ criteria. But for finding the val... | If $I=\int_{0}^{\infty}\frac{x}{x^3+1}dx$ then $2I=\int_{0}^{\infty}\frac{2x-x^2+x^2}{x^3+1}dx$=$\int_{0}^{\infty}\frac{2x-x^2}{x^3+1}dx$+$\int_{0}^{\infty}\frac{x^2}{x^3+1}dx.$
The secod itegral is easy and can be solved by substituting $u=x^3+1$. For the first write $$\int_{0}^{\infty}\frac{2x-x^2}{x^3+1}dx=\int_{0... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $y =\sqrt{5+\sqrt{5-\sqrt{5+ \cdots}}}$, what is the value of $y^2-y$? If $$y =\sqrt{5+\sqrt{5-\sqrt{5 + \cdots}}},$$
what is the value of $y^2-y$ ?
I am unable to get the clue due to these alternative signs of plus and minus please help on this thanks...
| Consider $z_{-+} = -y = -\sqrt {5 + \sqrt{5 - \ldots}}$.
$z_{-+}$ is clearly a solution to the equation $(z^2 -5)^2-5 = z$,
and the other three solutions of this degree $4$ polynomial equation, should be
$z_{+-} = +\sqrt {5 - \sqrt{5 + \ldots}}, z_{++} = +\sqrt {5 + \sqrt{5 + \ldots}}$, and $z_{--} = -\sqrt {5 - \sqrt{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the determinant of the following;
Find the determinant of the following matrix, and for which value of $x$ is it invertible;
$$\begin{pmatrix}
x & 1 & 0 & 0 & 0 & \ldots & 0 & 0 \\
0 & x & 1 & 0 & 0 & \ldots & 0 & 0 \\
0 & 0 & x & 1 & 0 & \ldots & 0 & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \l... | Develop the determinant by the first column...
$$\begin{vmatrix}\color{red}x&1&0&0&\ldots&0\\
\color{red}0&x&1&0&\ldots&0\\
\color{red}\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\
\color{red}1&0&0&0&\ldots&x\end{vmatrix}=$$$${}$$
$$x\begin{vmatrix}x&1&0&0&\ldots&0\\
0&x&1&0&\ldots&0\\
\ldots&\ldots&\ldots&\ldots&\ldots&... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Sum of Digits Question If A is the sum of the digits of $5^{10000}$, B is the sum of the digits of A, and C is the sum of the digits of B, what is C? I know it has something to do with mod 9, but I'm not sure how do use it to solve the problem. I found this question in my Math Challenge II Number Theory Packet, and I g... | Let $S(n)$ denote the sum of digits of $n$. Then, note that $S(n) \le 9(\lfloor\log_{10}(n)\rfloor + 1)$.
Thus,
$(S(5^{10000}) \le 9(\lfloor\log_{10}(5^{10000})\rfloor + 1) = 9(\lfloor10000\log_{10}(5)\rfloor + 1) = 9(\lfloor1000\log_{10}(5^{10})\rfloor + 1) = 9(\lfloor1000\log_{10}(9765625)\rfloor + 1) \le 9(\lfloor1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/785733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Points on $(x^2 + y^2)^2 = 2x^2 - 2y^2$ with slope of $1$ Let the curve in the plane defined by the equation: $(x^2 + y^2)^2 = 2x^2 - 2y^2$
How can i graph the curve in the plane and determine the points of the curve where $\frac{dy}{dx} = 1$.
My work:
First i found the roots of this equation with a change of variable ... | For determining the graph is better plot it as Lemur says in polar coordinates and see what happen at different angles with $r$.
$\hskip2in$
For the points with derivative equal to one:
$$(x^2 + y^2)^2 = 2x^2 - 2y^2$$
$$x^4+y^4+2x^2y^2=2x^2-2y^2$$
$$4x^3+4y^3\frac{dy}{dx}+4xy^2+4x^2y\frac{dy}{dx}=4x-4y\frac{dy}{dx}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/785900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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How to solve for polynomial fractions? I'm self-studying. I have this problem I can't wrap my head around.
$\frac{x}{x+1} + \frac{4}{1-x} + \frac{x^2-5x-8}{x^2-1}$
The answer is $\frac{2\left(x-6\right)}{x-1}$
How do I get that answer? I keep getting $2(x^2-5x-2)$...
This is how I solved it:
$\frac{x}{x+1}-\frac{4}{x+1... | Find the common denominator, and simplify. (You'll be able to utilize the fact that $x^2-1$ is a difference of squares: $$x^2 - 1= (x+1)(x-1)$$
$$\begin{align} \frac{x}{x+1} + \frac{4}{1-x} + \frac{x^2-5x-8}{x^2-1}& = \dfrac {x}{x+1} -\frac{4}{x-1} + \frac{x^2 - 5x-8}{x^2-1}\\ \\& =\dfrac{x(x-1)-4(x+1) +x^2 - 5x -8}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/789226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The closed form of $\int_0^{\infty} \frac{\log(\cosh(x))}{x} e^{-x} \ dx$ An integral I discussed last days in a chat, and it looks like a hard nut since
after some manipulations of the initial form we reach an integral where the integrand is expressed in
terms of
digamma function that can be further reduced to som... | The integral to evaluate is
\begin{align}
I(1) = \int_{0}^{\infty} e^{-x} \, \ln(\cosh(x)) \, \frac{dx}{x}.
\end{align}
The evaluation is as follows. Consider
\begin{align}
\int_{1}^{\infty} e^{-\alpha x} d\alpha = \left[ - \frac{1}{x} e^{-\alpha x} \right]_{1}^{\infty} = \frac{e^{-x}}{x}.
\end{align}
Let
\begin{align... | {
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"url": "https://math.stackexchange.com/questions/789321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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prove that if $\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma=2$ then the triangle has a right angle prove that if $\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma=2$ then the triangle has a right angle.
$\alpha,\beta,\gamma$ are the angles of the triangle.
I tried to use all kinds of trigonometric identities but it didn't w... | Let $\alpha$, $\beta$ and $\gamma$ be the angles of a triangle; then it holds
$$
\boxed{\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2+2\cos\alpha\cos\beta\cos\gamma}
$$
If this equals $2$, we conclude
$$
\cos\alpha\cos\beta\cos\gamma=0
$$
so one of the angles is a right angle.
Proof of the claim
Let's use that $\gamma=\pi-\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/793591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding invertible 3x3 matrix A such that Stab(M)A=AStab(N) for given matrices M,N I'm reading that it is possible to find an invertible $3 \times 3$ matrix $A$ such that for the matrices
$M = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ and
$N = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 2 & 2 \\ 1 & 1 &... | Your ideas would be good if this was a general property of $M$ and $N$, but I don't think it is. You need to use the specific $M$ and $N$.
The names of these two ideas are “orbit-stabilizer” and “Gauss–Jordan elimination.”
GJ:
Notice that $\begin{bmatrix} 1& 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. N
=\begin{bmat... | {
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"source": "stackexchange",
"question_score": "2",
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How do I calculate the new matrix from the new basis? $\bf(1)$ Let $\mathbb{R}^{2\times2}$ denote the vector space of all $2\times2$ matrices with real number entries. Set $A=\pmatrix{1&2\\-2&-4}$. Define a linear transformation $T:\mathbb{R}^{2\times2}\longrightarrow\mathbb{R}^{2\times2}$ by $T(B)=AB$ for any $B\in\ma... | $$T\begin{pmatrix}0&0\\1&1\end{pmatrix}:=\begin{pmatrix}\;\;2&\;\;2\\-4&-4\end{pmatrix}=\color{red}{-5}\begin{pmatrix}0&0\\1&1\end{pmatrix}+\color{red}2\begin{pmatrix}1&0\\0&0\end{pmatrix}+\color{red}{(-1)}\begin{pmatrix}\;\;0&0\\-1&1\end{pmatrix}+\color{red}2\begin{pmatrix}0&1\\0&1\end{pmatrix}$$
and that's how you ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/796224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Compute $ \int\sin(x^2)\, dx + \int \sqrt{\arcsin t}\, dt$ Compute the sum of two integrals
$$
\int_{\large\sqrt{\frac{\pi}{6}}}^{\large\sqrt{\frac{\pi}{3}}}\sin(x^2)\ \ dx +
\int_{\large\frac{1}{2}}^{\large\frac{\sqrt{3}}{2}} \sqrt{\arcsin t}\ \ dt.
$$
| Let $x=\sqrt{\arcsin t}$, then
$$
\int\limits_{\sqrt{\frac{\pi}{6}}}^{\sqrt{\frac{\pi}{3}}}\sin(x^2)\, dx
= \int\limits_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} t\,d\sqrt{\arcsin t}
= t\sqrt{\arcsin t}\vert_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}
- \int\limits_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \sqrt{\arcsin t}\, dt.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/801094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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To solve $2^x+4^x+2^{\lfloor x \rfloor}+4^{\lfloor x \rfloor}+2^{x- \lfloor x \rfloor}-4^{x-\lfloor x \rfloor}=50+\sqrt{50}$ How to solve for positive real $x$: $$2^x+4^x+2^{\lfloor x \rfloor}+4^{\lfloor x \rfloor}+2^{x- \lfloor x \rfloor}-4^{x-\lfloor x \rfloor}=50+\sqrt{50}$$ ?
| Notice the function on the left hand side is increasing (for $x>0$). [And for $x<0$ the left hand side is $\leq6$ so we don't care about it]. The only term that pushes down is $4^{x-[x]}$ but this is $\leq4$.
For $x=2$ we get $4+16+4+16+1-1=48<50+\sqrt{50}$.
For $x=3$ we get $8+64+8+64+1-1>50+\sqrt{50}$.
Therefore the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/802803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solve by using modulo. Or something else An infinite sequence of positive integers $a_1, a_2,\ldots$ has the properties that for $k\geq2$, the $k^\text{th}$ element is equal to $k$ plus the product of the first $k-1$ elements of the sequence. Suppose $a_1 =1$. what is the smallest prime number that does not divide $a_{... | Modular arithmetic is congruent under multiplication, so we can define sequences $a_k^{(2)}$, $a_k^{(3)}$, $a_k^{(5)}$, etc. to represent the sequences modulo 2, 3, and 5 respectively.
Thanks to congruency we still have the recurrence:
$$a_k^{(p)} \equiv \prod_{i=1}^{k-1}a_i^{(p)} + k \mod p$$
Now it remains to look at... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/806091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How prove $\left(\frac{b+c}{a}+2\right)^2+\left(\frac{c}{b}+2\right)^2+\left(\frac{c}{a+b}-1\right)^2\ge 5$ Let $a,b,c\in R$ and $ab\neq 0,a+b\neq 0$. Find the minimum of:
$$\left(\dfrac{b+c}{a}+2\right)^2+\left(\dfrac{c}{b}+2\right)^2+\left(\dfrac{c}{a+b}-1\right)^2\ge 5$$
if and only if $$a=b=1,c=-2$$
My idea: Since ... | Remark
$$\displaystyle\sum f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$$
means cyclic sum.
$$\displaystyle\prod f(a,b,c)=f(a,b,c)\cdot f(b,c,a) \cdot f(c,a,b)$$
means cyclic product.
$$
\left( \frac{b+c}{a}+2 \right) ^2+\left( \frac{c}{b}+2 \right) ^2+\left( \frac{c}{a+b}-1 \right) ^2
$$
$$
=\left( \frac{a+b+c}{a}+1 \right) ^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/807404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Evaluation of $\int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx$ The value of $\displaystyle \int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx = $
$\bf{My\; Try::}$ Let $\displaystyle I = \int 4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx$
Using Int... | This appeared in my test paper today i.e JEE-Advanced 2014, I am curious as to how you got your hands on this one. :P
As for the solution, apply IBP once to get (I am dropping the factor of 4 for the moment):
$$I=\left(x^3 \frac{d}{dx}(1-x^2)^5 \right|_0^1-\int_0^1 3x^2\frac{d}{dx}(1-x^2)^5\,dx$$
Notice that the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve this weird inequality? $\frac{x-1}{x+1} < x$
Thanks!
I did the following.
$\frac{x-1}{x+1} - x< 0 /-x$
$\frac{x-1 - x(x+1)}{x+1} < 0$
$\frac{-x^2-1}{x+1} < 0$
What to do next?
| Start with: $x\le\frac{1}{2}(x^2+1)\Rightarrow x-1\le\frac{1}{2}(x^2-1)$
Thus $\frac{x-1}{x+1}\le\frac{1}{2}(x-1)\le x-1\lt x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Simplifying $(1/x-1/5 )/( 1/x^2-1/25)$ How do I get $\frac{5x}{x+5}$ from simplifying the following?
$$\frac{(\frac{1}{x}-\frac{1}{5} )}{( \frac{1}{x^2}-\frac{1}{25})}$$
My work:
I multiplied the top and bottom by the LCD: $25x^2$ (to get the same denominators).
Then I got: $\frac{25x-5x}{25-x^2}$
Then I got this for... | You should have gotten to
$$\frac{25x-5x^2}{25-x^2} = \frac{5x(5 - x)}{(5+x)(5-x)}$$
Now cancel the common factor to get the desired result, assuming $x\neq 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$ I need to prove that $\frac{1+\tan^2\theta}{1+\cot^2\theta}= \tan^2\theta.$
I know that $1+\tan^2\theta=\sec^2\theta$ and that $1+\cot^2\theta=\csc^2\theta$, making it now $$\frac{\sec^2\theta}{\csc^2\theta,}$$ but I don't know how to get it do... | $$\frac{1+\tan^2\theta}{1+\cot^2\theta}=\frac{\sec^2\theta}{\csc^2\theta}=\frac{\frac{1}{\cos^2\theta}}{\frac{1}{\sin^2\theta}}=\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$$
As for your second question that $\tan\theta+\cot\theta=\sec\theta\csc\theta,$ just use the fact that $\tan\theta=\frac{\sin\theta}{\cos\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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$4 \sin 72^\circ \sin 36^\circ = \sqrt 5$ How do I establish this and similar values of trigonometric functions?
$$
4 \sin 72^\circ \sin 36^\circ = \sqrt 5
$$
| Using Werner Formula, $$2\sin36^\circ\sin72^\circ=\cos36^\circ-\cos108^\circ$$
$$\cos108^\circ=\cos(180^\circ-72^\circ)=-\cos72^\circ$$
Now, $\displaystyle\cos36^\circ\cos72^\circ=\frac{2\sin36^\circ\cos36^\circ}{2\sin36^\circ}\cos72^\circ=\frac{2\sin72^\circ\cos72^\circ}{4\sin36^\circ}=\frac{\sin144^\circ}{4\sin(180^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/816195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Please help to prove $\sum\limits_{n=1}^\infty {1\over a_n-1}<{2\over 7}$,in which $a_n=4n(n+1)$ Please help to prove this.
Suppose $a_n=4n(n+1)$,then
$$\sum_{n=1}^{+\infty}{1\over a_n-1}<{2\over7}.$$
Thanks.
| $$4n(n+1)-1=(2n+1)^2-2>(2n+1)^2-4=(2n+3)(2n-1)$$So, given sum is less than $$\frac{1}{7}+\sum_{n=2}^{\infty}\frac{1}{(2n+3)(2n-1)}=\frac{1}{7}+\frac{1}{4}\sum_{n\ge 2}\left(\frac{1}{2n-1}-\frac{1}{2n+3}\right)=\frac{1}{7}+\frac{1}{4}\left(\frac{1}{3}+\frac{1}{5}\right)\\=\frac{1}{7}+\frac{2}{15}<\frac{2}{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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With how many ways can we choose $9$ balls of a box? With how many ways can we choose $9$ balls of a box that contains $12$ balls, of which $3$ are green, $3$ are white, $3$ are blue and $3$ are red?
$$$$
I have done the following:
$x_1=\# \text{ green balls that we choose }$
$x_2=\# \text{ white balls that we choose }... | Assuming we do not care about the order in which the balls are selected, the number of ways of choosing $9$ balls from the box is the same as the number of ways of leaving $3$ balls in the box, which is much easier to calculate.
If we number the balls $1$ to $12$, then there are $12\choose 3$ ways of doing this. But r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Math Olympiad Algebraic Question Comprising Square Roots
If $m$ and $n$ are positive real numbers satisfying the equation $$m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n=3$$
find the value of $$\frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}$$
I came across this question in a Math Olympiad Competition and had no idea h... | Given: $m$ and $n$ are positive real numbers satisfying the equation
$$m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n=3$$
Just to get a better feeling, substitute
$\sqrt{m}=x$
$\sqrt{n}=y$
Now your equation becomes
$x^2+4xy-2x-4y+4y^2=3$
Combining first, second and last term of L.H.S. , we get,
$(x+2y)^2-2(x+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Singular Values of negative eigenvalue? Consider a $4 \times 4$ matrix with eigenvalues $0,1,-2,4$. What are the singular values of this matrix?
I am aware that singular values are just square root of the eigenvalues, but what will be the answer for $-2$ eigenvalue? Is it undefined?
| @Alexander Vigodner provides the punchline
$$
\sigma \left( \mathbf{A} \right) = \sqrt{\lambda \left( \mathbf{A} \right)}
$$
only when $\mathbf{A} = \mathbf{A}^{*}$.
We can provide a bit more insight with a smaller matrix. What are the $2\times 2$ matrices with the eigenvalue spectrum $\left( 1, 0 \right)$?
The charac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/819686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How to integrate $\int \frac{1}{\sin^4x + \cos^4 x} \,dx$?
How to integrate
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx$$
I tried the following approach:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{\sin^4x + (1-\sin^2x)^2} \,dx = \int \frac{1}{\sin^4x + 1- 2\sin^2x + \sin^4x} \,dx \\
= \frac{1}{2}\int \... | Another approach:
We have
$$
\frac{1}{\sin^4x+\cos^4x},\tag1
$$
Multiply $(1)$ by $\dfrac{\tan^4x}{\tan^4x}$ we obtain
$$
\frac{\tan^4x}{\sin^4x(1+\tan^4x)}=\frac{\sec^4x}{1+\tan^4x}=\frac{(1+\tan^2x)\sec^2x}{1+\tan^4x}.\tag2
$$
Letting $t=\tan x$, the integral turns out to be
$$\eqalign
{
\int\frac{1+t^2}{1+t^4}\ dt&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/820830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 0
} |
number of non-negative integer solutions of $ax+by+cz=k$ How to find the number of non-negative integer solutions of $2x+3y+z=21$ ?
Please Help , thanks in advance
| We can look at the number of $(x,y)$ pairs such that $2x+3y\le 21$.
We have:
$$(0,0),(0,1),\dots,(0,7)$$
$$(1,0),\dots,(1,6)$$
$$(2,0),\dots,(2,5)$$
$$\vdots$$
$$(10,0)$$
The maximum $y$ takes the value $\lfloor \frac{21-2x}{3} \rfloor$, i.e.
$$\begin{array} {c|c}
x&\lfloor \frac{21-2x}{3} \rfloor\\
\hline
0&7\\
1&6\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
| Note that
$$\sqrt{x^2+1}-x=\frac{1}{\sqrt{x^2+1}+x}\rightarrow 0$$ as $x\rightarrow \infty$
Your expression is equal to
$$\sqrt{(x+1)^2+1}-(x+1)+2 \rightarrow 2$$ as $x\rightarrow \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/822017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Divisibility of sum of exponents Consider the sequence
$$r, \ ra, \ ra^2, \ ra^3, ... \ , ra^n \mod M $$
such that:
$$ ra^{n+1} \equiv r \mod M$$
and $a \ne 1$ and $a,r$ are both coprime to $M$
Is it always true then that:
$$ r + ra + \ ra^2 + \ ra^3 ... + \ ra^n \equiv 0 \mod M$$
Consider the case of $M = 7$ jus... | Your example of $M = 7$ does not make sense unless you mean $ra^{n+1} \equiv r$. Then, since $\gcd(M,r) = 1$, we have $a^{n+1} \equiv 1 \pmod M$. You also probably mean $a$ is not $\equiv 1$.
Since $a$ is not $\equiv 1$, the sum is equal to $r \cdot \frac{a^{n+1}-1}{a-1} \equiv 0 \pmod M$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate the integral of $\operatorname{sgn}(\sin\pi/x)$ in the interval $(0,1)$? How can I calculate the integral of $\operatorname{sgn}(\sin\pi/x)$ in the interval $(0,1)$?
I need to calculate this integral, thanks
| Outline: In the interval $(1/2,1)$ our function is $-1$.
In the interval $(1/3,1/2)$, our function is $1$.
In the interval $(1/4,1/3)$, our function is $-1$.
In the interval $(1/5,1/4)$, our function is $1$.
And so on. The intervals have length $\frac{1}{1\cdot 2}$, $\frac{1}{2\cdot 3}$, $\frac{1}{3\cdot 4}$, and so ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$ If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$
My 1st approach :
$\tan(\alpha +2\alpha +4\alpha) = \f... | (1) Note first that $$\tan x\tan(2x)=\frac{\sin x\sin(2x)}{\cos x\cos(2x)}=\frac{2\sin^2x}{\cos(2x)}=\frac{1}{\cos(2x)}-1
$$
(2) It follows that
$$\eqalign{
S~&\buildrel{\rm def}\over{=}~\tan\alpha\tan{2\alpha}+\tan2\alpha\tan{4\alpha}+\tan4\alpha\tan{\alpha}\cr
&=-3+\frac{1}{\cos\alpha}+\frac{1}{\cos2\alpha}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Prove the sum $\sum_{n=1}^\infty \frac{\arctan{n}}{n}$ diverges. I must prove, that sum diverges, but...
$$\sum_{n=1}^\infty \frac{\arctan{n}}{n}$$
$$\lim_{n \to \infty} \frac{\arctan{n}}{n} = \frac{\pi/2}{\infty} = 0$$
$$\lim_{n \to \infty} \frac{ \sqrt[n]{\arctan{n}} }{ \sqrt[n]{n} } = \frac{1}{1} = 1$$
Cauchy's co... | For large enough $n$, $\text{arctan}(n)\geqslant \pi/4$ and $\sum \frac{\pi/4}{n} $ diverges.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?
Prove or disprove that if $t$ is a positive integer, $$f(x,y)=\dfrac{x^2+y^2}{xy-t},$$ then $f(x,y)$ has only finitely many distinct positive integer values with $x,y$ positive integer... | October 14, 2015. This is with $$ \frac{x^2 + y^2}{xy - t} = q > 0, $$
which I believe to be the intent of the question.
THEOREM: $$ \color{red}{ q \leq (t+1)^2 + 1 } $$
I got some help from Gerry Myerson on MO to finish the thing.
https://mathoverflow.net/questions/220834/optimal-bound-in-diophantine-representatio... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Drawing Marbles Probability My daughter had a math problem in her textbook that we are having a hard time understanding. You have $3$ white and $4$ black marbles in a bag and $3$ people will draw a marble and not replace it in the bag. The first person to draw the white marble wins. What is the probability that the fir... | I'll expand a bit on what David said. Firstly, it is logical that the first person has the highest probability of winning, as that person can stop the contest before anyone else has a chance. All the other contestants' chances are dependent on all prior contestants NOT winning prior to their turn.
Round 1:
There are fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/831621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluate the limit or show that is does not exist $$\lim_{x \to 0}\frac{1-\cos(3x)}{2x^2}$$
So far I have noted
$$\frac{1-\cos(3x)}{2x^2} = \frac{1-\cos(3x)}{2x^2} \cdot \frac{1+\cos(3x)}{1+\cos(3x)}$$
and then used the identity $$\sin^2x = 1-\cos^2x$$ to reduce this to
$$\frac{\sin^2(9x)}{2x^2+\cos(9x)}.$$
Any tips on... | You had the right idea, but your algebra is suspect. For example, $\cos 3x \cdot \cos 3x \ne \cos^2 9x$
$$\lim_{x\to 0} \frac{1-\cos 3x}{2x} = \frac{3}{2}\cdot\lim_{x\to 0} \frac{1-\cos 3x}{3x}$$
$$= \frac{3}{2}\cdot\lim_{3x\to 0} \frac{1-\cos 3x}{3x}$$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{1-\cos \theta}{\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Interesting line integral using green's theorem Find $$\int_C \frac{2x^3+2xy^2-2y}{ x^2+y^2} \, dx+\frac{2y^3+2x^2y+2x}{ x^2+y^2} \, dy$$
Where $C$ is any simple closed loop which contains the origin.
What I figured out
I cannot use the direct version of Green's theorem.
I know that there is another version of greens ... | You can write your integral as
$$2\int_{C} \left(-\frac{y}{ x^2+y^2}dx+\frac{x}{ x^2+y^2}dy \right) + \int_{C} \left( 2x\ dx + 2y\ dy\right)\ ,$$
the second form is exact, so its contribution is 0, while for the first for you can do the following. Note that $$\left(-\frac{y}{ x^2+y^2}dx+\frac{x}{ x^2+y^2}dy \right) =d... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Proof by induction that $(1^3 + 2^3 + 3^3+\cdots+n^3) = (1 + 2 + \cdots + n)^2$ I'm sitting with the proof in front of me, but I do not understand it.
$$A = \{n \in Z^{++} \mid (1^3 + 2^3 + 3^3+\cdots+n^3) = (1 + 2 + \cdots + n)^2\}$$
The first step of proof by induction is simple enough,to prove that $1 \in A$
$1^3 = ... | For the inductive step:
$$(\underbrace{1+2\cdots+n}_{=A=\frac{n(n+1)}{2}}+(n+1))^2=\underbrace{A^2}_{=1^3+2^3+\cdots+n^3}+\underbrace{(n+1)^2+n(n+1)^2}_{=(n+1)^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/832756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find a generating function for $\sum_{k=0}^{n} k^2$
Find a generating function for $\sum_{k=0}^{n} k^2$
I know my solution is wrong, but why?
My solution:
If $F(x)$ generates $\sum_{k=0}^{n} k^2$ then $F(x)(1-x)$ generates $k^2$.
$\frac{x}{(1-x)^4}: \left\{ 0,1,4,9,16,25... \right\}$
$\frac{x}{(1-x)^3}: \left\{ ... | According to the example, how to get the generating function for squares you can do the following:
Use $$
\sum_{n=0}^{\infty}\binom{n+k}k x^n= \frac{1}{(1-x)^{k+1}} $$
and set
$$
a\binom{n+3}{3}+ b\binom{n+2}{2}+c\binom{n+1}{1}+d=\frac{n^3}{3}+\frac{n^2}{2}+\frac n2=\sum_{k=1}^n k^2
$$
You'll find $a=2,b=-3,c=1$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Partial fraction (doubt) I have this partial fraction
$$\displaystyle\frac{1}{(2+x)^2(4+x)^2}$$
I tried to resolve using this method:
$$\displaystyle\frac{A}{2+x}+\displaystyle\frac{B}{(2+x)^2}+\displaystyle\frac{C}{4+x}+\displaystyle\frac{D}{(4+x)^2}$$
$$1=A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2$$
When x=-2
$$1=... | take LCM of second expression
you get and sum it
{A(2+x)(4+x)^2 +B(4+x)^2+C(2+x)^2(4+x)+D(2+x)^2}/{(x+2)^2(x+4)^2}= your 1st expression
so you have {A(2+x)(4+x)^2 +B(4+x)^2+C(2+x)^2(4+x)+D(2+x)^2}=1
compare coefficients of x^3 ,x^2,x^1 and x^0 which on right side of aboveequation are equal to 0,0,0 and 1 respective... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/833170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Real solution of the equation $(x^2-2x+2)^2-2(x^2-2x+2)+2 = x$
Calculate all real solutions $x\in\mathbb{R}$ of the equation
$$
\tag1(x^2-2x+2)^2-2(x^2-2x+2)+2 = x
$$
My Attempt:
I used the concept of a composite function. Let $f(x) = x^2-2x+2$. Then equation $(1)$ converts into $f(f(x)) = x$. Both $f(x) = x$ and $... | Setting $T=x-1$ you have $(T^2+1)^2-2(T^2+1)+1-T=0$ or $T^4-T=0$ that is
$T(T-1)(T^2+T+1)=0$ so for real solutions you have $T=0$ or $T=1$, that is $x=1$ or $x=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/834867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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How to solve this quadratic form equation?
Let $Q(x,y,z)=7x^2+7y^2-2z^2-10xy+8xz+8yz$ be a quadratic form and
$A = \begin{bmatrix}
7 & -5 & 4 \\
-5 & 7 & 4 \\
4 & 4 & -2
\end{bmatrix}$ its matrix. Such that $Q=X^TAX$ for $X=\begin{bmatrix} x \\y \\ z \end{bmatrix}$.
Find an $X$ suc... | I found the follow technic for this kind of problem, that is very similar to the Mark Bennet's answer.
If $Q(x,y,z)=7x^2+7y^2-2z^2-10xy+8xz+8yz$ then $A$ is the associated matrix.
Let $X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} \neq0 \text{ and } \lambda \in \mathbb{R}$. If $AX=\lambda X$, then $\lambda$ is an eigenva... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/835821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$ F(x) = \int_0^2 \sin(x+l)^2\ dl$ Consider the function : $ F(x) = \int_0^2 \sin(x+l)^2\ dl$, calculate $ \frac{dF(x)}{dx}|_{x=0}$ the derivative of $F(x)$ with respect to $x$ in zero.
Let $g(x) = \sin (x)$ and $h(x) = (x+l)^2$ then $F(x) = \int_0^2 g(h(x))\ dl$, so $F´(x) = g(h(x)) h´(x)$ then I can evaluate this i... | If you can justify permuting the order of limit and integral then
\begin{align}
\frac{dF(x)}{dx}
&= \lim_{h \to 0} \frac{1}{h}\left( \int_0^2 \sin(x + h+ \ell)^2 d\ell -
\int_0^2 \sin(x + \ell)^2 d\ell \right) \\
&= \lim_{h \to 0} \frac{1}{h} \int_0^2 (\sin(x + h+ \ell)^2 - \sin(x + \ell)^2) d\ell \\
&= \int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/838103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proof of Cauchy's functional equation for rational arguments We have thesis that for every $c\in\mathbb{Q}$ every additive function has form of $f(x)=cx$. In the proof we're showing that $f(nx)=nf(x)$. Then we're supposed to replace $nx$ by $\frac{1}{n}x$. Why can we do this?
| It follows from $x = \underbrace{\frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \cdots}_{n\text{ times}}$
$$f(x) = f(\underbrace{\frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \cdots}_{n\text{ times}}) = \underbrace{f(\frac{1}{n}x) + f(\frac{1}{n}x) + f(\frac{1}{n}x) + \cdots}_{n\text{ times}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/840802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$\forall \ n \in \mathbb{N}: 6\mid 5n^3+n$ Show $\forall \ n \in \mathbb{N}: 6\mid 5n^3+n$
Indeed,
we've to show $5n^3+n=0\pmod 6 $
note that for $n\in \{0,1,2,3,4,5\}$ we've: $n^{3}=n\pmod 6$ then
$5n^{3}=5n\pmod 6 \implies 5n^{3}+n=6n\pmod 6=0\pmod 6$
AM i right ?
| Use induction
$$
5\Big(n+1\Big)^3 + \Big(n+1\Big) = 5 n^3 + n + 6 \Big( \tfrac{5}{2} n(n+1) +1 \Big)
$$
and as $n(n+1)$ is even we see that
$$
5\Big(n+1\Big)^3 + \Big(n+1\Big) = 5 n^3 + n + 6 k
$$
As it is true for $n=1$, it is true for all $n \ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/841102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a^{12} \equiv 1 \pmod{35}$,knowing that $(a,35)=1$ Prove that $\forall a \text{ with } (a,35)=1:$
$$a^{12} \equiv 1 \pmod{35}$$
$$35 \mid a^{12}-1 \Leftrightarrow 5 \cdot 7 \mid a^{12}-1 \overset{(5,7=1)}{ \Leftrightarrow} 5 \mid a^{12}-1 \text{ and } 7 \mid a^{12}-1$$
Therefore, $\displaystyle{ a^{12} \equiv 1 \pmod{... | Hint $\ $ If $\,a\,$ is coprime to primes $\,p_1\!\ne p_2\,$ and $\,\color{#0a0}{p_i\!-1\mid n}\,$ then by $\rm\color{#c00}{little\ Fermat}$
$\qquad\qquad\qquad {\rm mod}\ p_i\!:\,\ a^n\equiv (\color{#c00}{a^{\,\large p_i-1}})^{\Large\color{#0a0}{\frac{n}{p_i-1}}\!}\equiv \color{#c00}1^{\Large\color{#0a0}{\frac{n}{p_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Trigonometric equation with complex numbers Let $x$, $y$, and $z$ be real numbers such that $\cos x+\cos y+\cos z=\sin x+\sin y+\sin z=0$.
Prove that $\cos 2x+\cos 2y+\cos 2z=\sin 2x+\sin 2y+\sin 2z=0$.
Starting with the given equation, I got that $i\sin x+i\sin y+i\sin z=0$.
Adding this to the other part of the given... | In the same way you have: $(\cos x-i \sin x)+(\cos y-i\sin y)+(\cos z-i \sin z)=0$, it's equal: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$, so $ab+ac+bc=0$ and $0=(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/843022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Optimization of parallelepiped inside an ellipsoid Let $K \in R^3$ the ellipsoid given by the equation $ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $ with $a,b,c > 0$ , let $(x,y,z) \in K$ on the first octant, consider the parallelepiped of vertices $(\pm x,\pm y,\pm z)$ inscribed on $K$ with volume $V =... | Because
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$
We have
$$\left(\frac{x^2}{a^2}\frac{y^2}{b^2}\frac{z^2}{c^2}\right)^{1/3} \le \frac{1}{3}\left(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}\right) = \frac{1}{3}$$
Thus
$$V=8|xyz| \le \frac{8abc}{\sqrt{27}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/844193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integral: $\int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx$ Please help me in proving the following result:
$$\displaystyle \int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx=\frac{\pi}{b}e^{-b\sqrt{a^2+c^2}}\sinh (ab)$$
I found this integral from here: http://integralsandser... | In order to prove the final result I will need to state a lemma that will be used later.
Lemma$\require{autoload-all}$ $1$:
$$\int_0^\infty \! \frac{\cos(bx)}{x^2+\alpha} \mathrm{d}x = \frac{\pi e^{-b\sqrt{\alpha}}}{2b\sqrt{\alpha}}\tag{1}$$
Proof here.
Consider
$$I = \int_0^\infty\!\! \tan^{-1}\left(\frac{2ax}{x^2+c... | {
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When is $(a^2+b)(b^2+a)$ a power of $2$? When is $(a^2+b)(b^2+a)$ a power of $2$?
($a$, $b$ positive integers)
I tried some values on the computer and it seems the only solutions are $a=1$, $b=1$. But I am not sure how to prove it?
Thanks for any help.
| The product $(a^2+b)(a+b^2)$ is a power of two if and only if both factors are:
$$\begin{array}{rcl}
a^2+b&=&2^r\\
a+b^2&=&2^s
\end{array}$$
If $a=b$, then the equation $a^2+a=a(a+1)=2^r$ has only one solution, namely, $a=1$. So we can WLOG assume that $a>b$. Thus, $r>s$.
Since $2^s=a^2+b\geq2^2+1$, we have that $s>1$.... | {
"language": "en",
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"source": "stackexchange",
"question_score": "10",
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Sketch the set $\{ z \in \mathbb{C} | \left|\frac{z-i}{z+i}\right|<1 \}$ My question is to sketch the set $\{ z \in \mathbb{C} | \left|\frac{z-i}{z+i}\right|<1\}$ in the complex plane.
I substituted $z$ for $a+bi$, but did not get anywhere:
$\left|\frac{a+(b-1)i}{a+(b+1)i}\right|<1\\
\left|\frac{(a+(b-1)i)(a-(b+1)i)}{... | $|z-i|<|z+i| \implies |x+(y-1)i|<|x+(y+1)i| \implies x^2+y^2-2y+1<x^2+y^2+2y+1\implies y>0$
is just an algebraic way to show what Hans Lundmark said
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the PDF of Y given Y=X(2-X) and X's PDF Suppose that the continuous random variable $X$ has probability density function
$f_X(x)=\begin{cases}\frac{1}{2}x & \text{if } 0<x<2\\0&\text{otherwise}\end{cases}$
Let $Y=X(2-X)$. Calculate $P(Y>y)$ and hence find the probability density function of $Y$.
So far I have trie... | We complete the square. Note that $X(2-X)=2X-X^2=1-(X-1)^2$. Suppose that $0\le y\le 1$. Then
$$\Pr(X(2-X)\gt y)=\Pr(1-(X-1)^2\gt y)=\Pr((X-1)^2\lt 1-y)=\Pr\left(1-\sqrt{1-y}\le X\le 1+\sqrt{1-y}\right).$$
But between $0$ and $2$, the random variable $X$ has cdf $\frac{1}{4}x^2$. So
$$\Pr(X(2-X)\gt y)=\frac{1}{4}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/847453",
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"source": "stackexchange",
"question_score": "1",
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To find the right most non zero digit When expanded 30! ends in 7 zeroes. Find the first non zero digit from right?
| We desire the last digit of $n=\dfrac{30!}{10^7}$.
Note that $n\equiv0\pmod2$. Now we just need to find $n\pmod5$. By Wilson's theorem, or direct computation, we know $1(2)(3)(4)\equiv-1\pmod5$. This gives us
$$\dfrac{30!}{5^7}\equiv (-1)\frac{5}{5}(-1)\frac{10}{5}\cdots\frac{25}{5^2}(-1)\frac{30}{5}\equiv(-1)^61(2)(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/848501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Geometric intuition for $\pi /4 = 1 - 1/3 + 1/5 - \cdots$? Following reading this great post : Interesting and unexpected applications of $\pi$, Vadim's answer reminded me of something an analysis professor had told me when I was an undergrad - that no one had ever given him a satisfactory intuitive explaination for wh... | By elementary means:
In the figure below, we have drawn a quarter circle inside a unit square and divided the top side in $n$ equal parts. Then we have chosen the height of the blue rectangles in such a way that the area of a rectangle equals the area of the corresponding sector (the aperture angles go decreasing).
If... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/849348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
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Express a complex number in modulus amplitude form Express a complex number in modulus amplitude form
$\displaystyle 1+\sin \alpha +i\cos \alpha $
My Attempt:
$\displaystyle r\cos \theta= 1+\sin \alpha $
$\displaystyle r\sin \theta= \cos \alpha $
Squaring and adding.. $\displaystyle r^2= (1+\sin \alpha)^2+ \cos^2 \alp... | It is better to solve it this way:
$$\begin{aligned}
1+\sin\alpha+i\cos \alpha &=1+\cos\left(\frac{\pi}{2}-\alpha\right)+i\sin\left(\frac{\pi}{2}-\alpha\right)\\
& \stackrel{*}{=}2\cos^2\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)+i2\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/850249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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I need help finding the derivative of this natural logarithm function. Okay so $$f(x)=\ln[x\ln(x+2)]$$
so $$\ln(x)+\ln(\ln(x+2))$$so $$1.a\frac{dy}{dx}\ln(x)=\frac{1}{x}$$and I thought by chain rule that $$1.b\frac{dy}{dx}\ln(\ln(x+2))=\frac{1}{\ln(x+2)}\cdot\frac{1}{x}$$ so $$2.af'(x)=\frac{1}{x}+\frac{1}{\ln(x+2)}*\f... | \begin{align*}
f'(x) &= \frac{d}{dx} f(x) \\
&= \frac{d}{dx} ln[xln(x+2)] \\
&= \frac{d}{dx}(ln(x) + ln(ln(x+2))) \\
&= \frac{d}{dx}ln(x)+\frac{d}{dx}ln(ln(x+2)) \\
&= \frac{\frac{d}{dx}x}{x}+\frac{\frac{d}{dx}ln(x+2)}{ln(x+2)} \;\; \text{(Chain Rule)} \\
&= \frac{1}{x}+\frac{\frac{1}{x+2}}{ln(x+2)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the minimum of $\displaystyle \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$ Is it possible to find the minimum value of $E$ where
$$E = \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$$for any $\triangle ABC$.
I've got the feeling that $\... | HINT: Use Lagrange multipliers to find the minimum of the function $f(A,B,C) = \frac{1}{\sin^2(A)} + \frac{1}{\sin^2(B)} +\frac{1}{\sin^2(C)}$ with the constraint $A+B+C = \pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Problem calculating the argument of a complex variable In Signals & Systems 2nd Ed. written by A. V. Oppenheim,
there is a result of Fourier transformation:
$ \begin{align} H (j \omega) = \frac{1 + (j \omega / \omega_{0})^2 - 2 j \zeta (\omega / \omega_0)}{1 + (j \omega / \omega_{0})^2 + 2 j \zeta (\omega / \omega_0)} ... | You have probably a mistake in your last calculation:
$$\alpha = -2 \tan^{-1} \left( \frac{y}{1 + \sqrt{1+y^2}} \right) , \text{with } y = \frac{4 \zeta x(1-x^2)}{(1-x^2)^2 - 4 (\zeta x)^2}$$
I got
$$
\sqrt{1+y^2} = \frac{(1-x^2)^2 + 4(\zeta x)^2}{(1-x^2)^2-4(\zeta x)^2}
$$
and therefore
$$
1+\sqrt{1+y^2} = \frac{2(1-... | {
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"timestamp": "2023-03-29T00:00:00",
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} |
Limit of $x_n = 0.5(x_{n-1} + x_{n-2})$ - help finishing proof... EDIT: Fixed geometric proof off-by-one error.
I am looking for the limit of $x_n = 0.5(x_{n-1} + x_{n-2})$ with $x_2 > x_1$ arbitrary. I can show $\forall n: |x_{n}-x_{n+1}| = \frac{c}{2^{n-1}}$ (where $c = x_2 - x_1$) and thus existence of the limit (ca... | $$\sum_{k=1}^{n/2} \frac1{2^{2k-1}}= \sum_{k=1}^{n/2} \frac{1}{2}\frac{1}{2^{2k-2}}= \sum_{k=1}^{n/2} \frac{1}{2}\frac{1}{4^{k-1}}= \sum_{k=0}^{n/2-1} \frac{1}{2}\frac{1}{4^k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/856398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt$ I'd like to ask something about the following integral:
$$
\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt
$$
I rewrote and took another variable.
$$
-i\int_0^{2\pi}\frac{e^{is}-e^{-is}+8i}{e^{is}+e^{-is} + \frac{10}3} dt
\quad = \quad
-i\int_{C(0,... | You can do this way. Clearly
$$
\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt=\int_0^{2\pi}\frac{\sin t}{\cos t + \frac{5}3} dt+4\int_0^{2\pi}\frac{1}{\cos t + \frac{5}3} dt
$$
and
$$
\int_0^{2\pi}\frac{\sin t}{\cos t + \frac{5}3} dt=-\int_0^{2\pi}\frac{1}{\cos t + \frac{5}3} d(\cos t + \frac{5}3)=0.
$$
You on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/860003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $x > 0$ for which $\int_{0}^{x} [t]^2 \ dt = 2 (x-1)$. What are all possible $x > 0$ for which the following equation is satisfied?
$$\int_{0}^{x} [t]^2 \ dt = 2 (x-1),$$ where $[.]$ denotes the bracket (or floor) function.
I guess we will have to consider different cases depending on whether $\sqrt{n-1} \leq x ... | Differentiate both sides, we get $[x]^2 = 2$. Le us denote by $\{x\}$ the fractional part of the number $x$, so that $\{x\}=x-[x]$. It is clear that $0\le \{x\} <1$ and $[x]=x-\{x\}$. Thus, the equation becomes $$[x]^2=(x-\{x\})^2=2$$ that is $$x^2-2x\{x\}+\{x\}^2=2.$$ Setting $p=\{x\}$, and write $$x^2-2px+p^2-2=0.$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/862951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Imaginary part of $\int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x$ and $\int_{0}^{\pi/2} \frac{\log \cos x}{x^2}\:\mathrm{d}x$ I have found the following new result connecting two rational log-cosine integrals.
Proposition. \begin{align}
\displaystyle & {\Im} \int_{0}^{\pi/2} \frac{x^2}{x^2+\log ... | Numeric confirmation
From Mathematica. Left hand side:
Right hand side:
$$
f(z) = \frac{\ln \cos x}{x^2}
$$
$$
\text{Im }f(z) = \frac{z^2}{z^2+\ln^{2} \left(-2 \cos z \right)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/863496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 1,
"answer_id": 0
} |
What is the closed form for $\sum_{n=1}^\infty \frac1n - \frac1{n+1/p}$? A while ago, I started to look at expressions of the following form:
$$
S_p:=\sum_{n=1}^\infty \frac1n - \frac1{n+1/p},
$$
where $p$ is prime, because otherwise things get too complicated for me at the moment.
What I found so far is the followin... | Write your expression as the limit at $x \to 1$ of
$$\sum_{n = 1}^{\infty} \frac{p x^{pn}}{pn} - \frac{p x^{pn+1}}{pn+1}.$$
If we differentiate this power series, we get
$$\sum_{n = 1}^{\infty} p x^{pn - 1} - p x^{pn}
=
p x^{p-1} \frac{1-x}{1 - x^p}.
$$
Integrating this (say via a partial fraction decomposition)
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/864609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
A closed form for $\int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:\mathrm{d}x$ We already know that
\begin{align}
\displaystyle & \int_{0}^{\pi/2} x \ln(2 \cos x)\:\mathrm{d}x = -\frac{7}{16} \zeta(3),
\\\\ & \int_{0}^{\pi/2} x^2 \ln^2(2 \cos x)\:\mathrm{d}x = \frac{11 \pi}{16} \zeta(4). \end{align}
Does the following... | We can try the harmonic analysis path. Since:
$$\log(2\cos x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}\cos(2nx),$$
$$\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n},\tag{1}$$
we have, as an example:
$$\int_{0}^{\pi/2}\log^3(2\sin x)dx=-\frac{3\pi}{4}\zeta(3)$$
since
$$\int_{0}^{\pi/2}\cos(2n_1 x)\cos(2n_2 x)\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 4,
"answer_id": 3
} |
Simplify expression $(x\sqrt{y}- y\sqrt{x})/(x\sqrt{y} + y\sqrt{x})$ I'm stuck at the expression: $\displaystyle \frac{x\sqrt{y} -y\sqrt{x}}{x\sqrt{y} + y\sqrt{x}}$.
I need to simplify the expression (by making the denominators rational) and this is what I did:
$$(x\sqrt{y} - y\sqrt{x}) \times (x\sqrt{y} - y\sqrt{x}) ... | The answer can be simplified even further (using the difference of two squares):
$$\frac{x-2\sqrt{xy} + y}{x-y} = \frac{(\sqrt{x} - \sqrt{y})^2}{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})} = \frac{1}{\sqrt{x} - \sqrt{y}} \text{ or } \frac{\sqrt{x} - \sqrt{y}}{x+y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/869037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$ $$\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$
My approaches so far has been using substitution with $\tan x = t$ and $\tan \frac x2 = t$ but the calculations has been harder than I think they should.
I've also tried using ordinary ... | $$\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$
$$\int \frac{\tan x(\tan^2(x)+1)}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$
$$\int \frac{\tan x(\sec^2(x))}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$
Frome here, let $u=\tan(x)$, $\dfrac{du}{dx} = \sec^2(x)$., thus,
$$\int \frac{u}{u^3 + 3u^2 + 2u + 6}$$
From there, use... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to solve this: $\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left(\frac{2n+3k}{n} \right)^2$ How to solve this:
$$\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left(\frac{2n+3k}{n} \right)^2$$
The answer is supposed to be 39.
My attempt:
$$\frac{3}{n}\sum\left(4+\frac{12}{n}k+\frac{9}{n^{2}}k^{2}\r... | $$\lim_{n\to\infty}\frac3n\sum_{k=1}^n\left(2+3\frac kn\right)^2$$
$$=3\int_0^1(2+3x)^2\ dx$$
$$\text{as }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/871586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculating the area For the two graphs $ \frac{x^3+2x^2-8x+6}{x+4} $ and $ \frac{x^3+x^2-10x+9}{x+4} $, calculate the area which is confined by them;
Attempt to solve: Limits of the integral are $1$ and $-3$, so I took the definite integral of the diffrence between $ \frac{x^3+2x^2-8x+6}{x+4} $ and $ \frac{x^3+x^2-10x... | @Mary knows that because $$\frac{x^3+2x^2-8x+6}{x+4} - \frac{x^3+x^2-10x+9}{x+4}=\frac{x^2+2x-3}{x+4}=\frac{(x-1)(x+3)}{x+4}$$ and since on $-3<x<1$ so the last fraction in negative. A simple figure could clear the point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/871646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
find a polynomial whose roots are inverse of squares of roots of $x^3+px+q$ Question is :
Given a polynomial $f(x)=x^3+px+q\in \mathbb{Q}[x]$ find a polynomial whose roots are inverse of sqares of roots of $f(x)$
Supposing $a,b,c$ as roots of $f(x)$ we have :
*
*$a+b+c=0$
*$ab+bc+ca=p$
*$abc=-q$
Now i need to kn... | Note that
$$
0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) = a^2 + b^2 + c^2 + 2p
$$
Thus
$$
\dfrac{1}{a^2b^2} + \dfrac{1}{a^2b^2} + \dfrac{1}{a^2b^2} = \dfrac{a^2 + b^2 + c^2}{a^2b^2c^2} = -\dfrac{2p}{q^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Is $2^{3^n}+1$ always divisible by $3^{n+1}$? Because it certainly seems to be the case for all positiv $n$ but how can I prove it?
| Suppose
$$2^{3^n} + 1 \equiv 0 \pmod{3^{n+1}}$$
Then
$$2^{3^n} \equiv -1 + 3^{n+1} a \pmod{3^{n+2}}$$
for some value of $a$. And thus
$$\begin{align}2^{3^{n+1}} &\equiv \left(-1 + 3^{n+1} a\right)^3
\\&\equiv (-1)^3 + 3 \cdot (-1)^2 \left(3^{n+1} a \right) + \ldots
\\&\equiv -1 + 3^{n+2}(\ldots) + \ldots
\\&\equiv -1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
A Binet-like integral $\int_{0}^{1} \left(\frac{1}{\ln x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{x^s }{1-x}\mathrm{d}x$ I met this integral
$$
\int_{0}^{1}
\left(\frac{1}{\ln x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{ \mathrm{d}x}{1-x} \qquad (*)
$$
while evaluating this log-cosine integral. I made several at... | Recall Binet's formula
$$
\log \Gamma(z)= \left( z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) +
\int_0^{\infty} \!
\left(\frac{1}{2} - \frac{1}{x} + \frac{1}{e^{x}-1} \right)\frac{e^{-zx}}{x} \mathrm{d}x,\quad \Re z >0
$$
which, upon making $x = - \log v$, can be written as
$$
\log \Gamma(z)= \! \left( z-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/874029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
If $a+b+c+d=4$, then $\sum\frac{1}{a+3}\le \frac{1}{abcd}$ This is somehow related to this problem but I don't have any idea about it.
Let $a$, $b$, $c$ and $d$ be positive reals such that $a+b+c+d=4$. Prove that:
$$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}+\frac{1}{d+3}\le \frac{1}{abcd}$$
Now I also tried to prov... | Let $a+b+c+d=4u$ and $ab+ac+ad+bc+bd+cd=6v^2$.
Hence, by AM-GM $abcd\leq v^4$ and by C-S
$$\frac{1}{abcd}-\sum_{cyc}\frac{1}{a+3}=\frac{1}{abcd}-\frac{4}{3}-\sum_{cyc}\left(\frac{1}{a+3}-\frac{1}{3}\right)=$$
$$=\frac{1}{abcd}-\frac{4}{3}+\sum_{cyc}\frac{a}{3(a+3)}\geq\frac{1}{abcd}-\frac{4}{3}+\frac{(a+b+c+d)^2}{3\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx$
Compute the indefinite integral
$$
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx
$$
My Attempt:
$$
\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx &= \int\frac{\cos 2x}{\sin^2 x\sqrt{\cos 2x}}\sin xdx\\
&= \int\frac{2\cos^2 x-1}{(1-\cos^2 x)\sqrt{2\cos^2 x-1} }\sin ... | Put $$\cos 2x =\frac{1 -\tan^2x}{1+\tan^2x}$$
$$\int\frac{\sqrt{1-\tan^2 x}}{\tan x}$$
$$1-\tan^2 x =t^2$$
$$\implies -2\tan x \sec^2 x dx=2tdt$$
$$\int\frac{t}{\tan x}.\frac{-tdt}{\tan x\sec^2 x}=-\int\frac{t^2}{(1-t^2)(2-t^2)}$$
$$=-\int\left[\frac{1}{(1-t^2)} -\frac{2}{(2-t^2)}\right]dt$$
$$=-\frac{1}{2}\log\left|{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 2
} |
How to factorize $x^4+2x^2+4$ to a product of polynomials with real coefficients? How do you factor
$$x^4+2x^2+4 $$
so it can be written as
$$ (x^2+2x+2)(x^2-2x+2) $$
| Certainly the solution using completing the square (already provided in two answers) is in this case the easiest one.
I will just point out that in general if you have a real polynomial which is simple enough so that we are able to find all complex roots, this might also help you find factorization over $\mathbb R$. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
In a triangle ABC, prove that cot(A/2)+cot(B/2)+cot(C/2) =cot(A/2)cot(B/2)cot(C/2) In a triangle ABC, prove that $\cot \left ( \frac{A}{2} \right )+\cot \left ( \frac{B}{2} \right )+\cot \left ( \frac{C}{2} \right )=\cot \left ( \frac{A}{2} \right )\times \cot \left ( \frac{B}{2} \right )\times \cot \left ( \frac{C}{2}... | Here’s a somewhat weird proof using inverse trigonometric functions
If $$A+B+C=π$$ then $$\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{π}{2}$$
Let $$\frac{A}{2}=tan^{-1} x, \>\frac{B}{2}=tan^{-1} y ,\> \frac{C}{2}=tan^{-1} z$$
Now,
$$tan^{-1} x+ tan^{-1} y+ tan^{-1} z= π/2$$
$$tan^{-1} x+ tan^{-1} y= cot^{-1} z$$
B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
How to do integral $\int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$ and $\int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$? I met these two integrals but don't know how to do them:
$$I_1 = \int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t$$
$$I_2 = \int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}... | Hint :
Try the substitution $$t=\cfrac{T}{u^2+1}$$
The first integral has the shape of the gaussian.
The second one leads you to $$I(\beta) = \alpha \int_{\mathbb{R}^+} \cfrac{1}{u^2+1} \exp(-\beta(u^2+1) ) \,du$$
Considering $$\begin{cases} I'(\beta)=\alpha \int_{\mathbb{R}^+} \exp(-\beta(u^2+1) ) \,du=\alpha\exp(-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Scalar triple product in terms of scalar products only I am trying to express the scalar triple product $\bf a \cdot (b \times c)$ only in terms of scalar products $\bf a \cdot b$, $\bf b \cdot c$, $\bf c \cdot a$ and the lengths of vectors $a$, $b$ and $c$. I start by writing $$\mathbf{a} = B\, \mathbf{b} + C\, \math... | A "similar" expression is the determinant representation
$$\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=\operatorname{det} S,\qquad S=\left(
\begin{array}{ccc}
a_x & a_y & a_z \\
b_x & b_y & b_z \\
c_x & c_y & c_z
\end{array}\right).$$
The expression that you have obtained can be easily deduced from this one:
$$\Bigl(\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Series for logarithms This is more of a challenge than a question, but I thought I'd share anyway. Prove the following identities, and prove that the pattern continues.
\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)=\ln2
\end{equation*}\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{3n+... | So far, there is a proof involving differentiation/integration, and a proof involving nothing more than the knowledge that $\gamma$amma exists. Here is my proof, requiring the Taylor series expansion of the logarithm.
The well known Taylor series for $\ln(x)$ is as follows:
$$-\ln(1-x)=\frac{x}{1}+\frac{x^2}{2}+\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 2
} |
How can I find $\lim_{x \to 0}\frac{\tan(3x)}{\sin(8x)}$ without L'Hospital's Rule Is there a way to solve $\lim_{x \to 0}\frac{\tan(3x)}{\sin(8x)}$ without using the trig identity $\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-\tan^2(x)}$. I want to know because I had to look up this trig identity in order to solve this limit,... | This problem is simpler and does not require the use of $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$. We have $$\begin{aligned}L &= \lim_{x \to 0}\frac{\tan 3x}{\sin 8x}\\
&= \lim_{x \to 0}\frac{\tan x(3 - \tan^{2}x)}{1 - 3\tan^{2}x}\cdot\frac{1}{2\sin 4x\cos 4x}\\
&= \lim_{x \to 0}\frac{3 - \tan^{2}x}{1 - 3\tan^{2}x}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/884852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How do I use Euler's result to find the sum of a series? So I am given:
$$ \zeta(4) = \sum_{n=1}^\infty {1\over n^4}={\pi^4 \over 90} $$
I need to use it to find the sum of the following series using the above information.
$$ \sum_{k=1}^\infty {1\over{(k+2)^4}} $$
So, this is what I have so far:
$$ \sum_{k=1}^\infty {1... | Compare $$\sum_{k=3}^{\infty}\frac{1}{k^4}=\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\cdots$$
with$$\sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{1}{1^4}+\frac{1}{2^4}+\color{red}{\frac{1}{3^4}+\frac{1}{4^4}+\cdots}.$$
Hence we have $$\sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{1}{1^4}+\frac{1}{2^4}+\sum_{k=3}^{\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$a,b,c \geq 0$ and $a+b+c=3$ prove that $\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$ $a,b,c \geq 0$ and $a+b+c=3$ prove that $\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$
can anyone help me solve this problem,i've tried to use C-S and also AM-GM for couple in cycle... | $$\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$$
$$\iff \frac{a^2+bc}{3b+3ac} + \frac{b^2+ac}{3c+3ab} + \frac{c^2+ab}{3a+3bc} \geq 1$$
$$\iff \frac{a^2+bc}{(a+b+c)b+3ac} + \frac{b^2+ac}{(a+b+c)c+3ab} + \frac{c^2+ab}{(a+b+c)a+3bc} \geq 1$$
$$\Leftarrow \frac{a^2+bc}{(a+b+c)b+(a^2+ac+c^2)} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
If $~a^3 + b^3 = c^3~$ has nonzero integer solutions and $~c-b~$ is a cubic number and $~c-b \neq 1$ If $~a^3 + b^3 = c^3~$ has nonzero integer solutions, because:
$c^3 - b ^ 3 = (c - b)((c - b) ^ 2 + 3cb) = a ^ 3,\quad (1)$
if $~c-b~$ is a cubic number and $~c-b \neq 1~$, divide both side of $~(1)~$ by $~c - b~$ get... | Let $~(c−b)^2=y^3,~$ from $~(2)~$ get:
$x^3−y^3=(x−y)((x−y)^2+3xy)=3cb,\quad (3) $
if $~x-y \neq 3,c,b~$ suppose $~(x-y) \mid c,~$ divide both side of $(3)\ $ by $~3$:
$(x−y)^2+3xy=3c_1b,\quad (4) $
form $~(4)~$ we see $~3 \mid (x−y)^2,~$ because $~(x-y) \neq 3,~$then $~3 \mid c_1,~$
form $~(4)~$ get $~3^2k^2+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Arithmetic Progression. Q. The ratio between the sum of $n$ terms of two A.P's is $3n+8:7n+15$. Find the ratio between their $12$th term.
My method:
Given:
$\frac{S_n}{s_n}=\frac{3n+8}{7n+15}$
$\frac{S_n}{3n+8}=\frac{s_n}{7n+15}=k$
$\frac{T_n}{t_n}=\frac{S_n-S_{n-1}}{s_n-s_{n-1}}=\frac{k\left(\left(3n+8\right)-\left(3\... | The trick is to notice that we can safely cancel constant common factors (if any), and stick in the $n$ in both numerator and denominator:
$\dfrac{T_n}{t_n} = \dfrac{S_n - S_{n - 1}}{s_n - s_{n - 1}} = \dfrac{n(3n + 8) - (n - 1)(3(n - 1) + 8)}{n(7n + 15) - (n - 1)(7(n - 1) + 15)} = \dfrac{6n + 5}{14n + 8}$
Plug in $n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/894079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Question on Factoring I have very basic Question about factoring, we know that,
$$x^2+2xy+y^2 = (x+y)^2$$
$$x^2-2xy+y^2 = (x-y)^2$$
But what will
$$x^2-2xy-y^2 = ??$$
$$x^2+2xy-y^2 = ??$$
| You can use the quadratic formula:
$$x^2 + (-2y)x + (-y^2) = 0\to \\ x= \frac{2y \pm \sqrt{4y^2 - 4(1)(-y^2)}}{2} \\ =y \pm\sqrt{2}y = (1 \pm \sqrt{2})y.$$
So $x^2 - 2xy - y^2 = [x - (1+\sqrt{2})y][x - (1-\sqrt{2})y].$
For the other case,
$$x^2 + (2y)x + (-y^2) = 0\to \\ x= \frac{-2y \pm \sqrt{4y^2 - 4(1)(-y^2)}}{2} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/894392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
If a fair die is thrown three times, what is the probability that the sum of the faces is 9?
If a fair die is thrown thrice, what is the probability that the sum of the faces is 9?
I did like this.
The total number of cases is $6^3=216$
Now,the number of solutions of the equation $x + y + z = 9$ with each of $x,y,z$ ... | I get that there are only $25$ ways of writing $9$ as a sum of three integers in $[1,6]$, since:
$$[x^9](x+x^2+x^3+x^4+x^5+x^6)^3 = 25.$$
Hence the probability is $\frac{5^2}{6^3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/895803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to prove that a diffrensiation of a formula equals to another formula. QUESTION 1) if $y =\dfrac{ \sin x-x\cos x}{x\sin x+\cos x}$ show that $\dfrac{dy}{dx}= \dfrac{x^2}{(x\sin x+\cos x)^2}$
QUESTION 2) if $y = \dfrac{\tan x+1}{\tan x-1}$ show that $\dfrac{dy}{dx}= \dfrac{-2}{1-\sin 2x}$
| for the second, we get
$$\begin{align}
y&=\frac{\tan x+1}{\tan x-1}\\
\frac{dy}{dx}&=\frac{d}{dx}\left(\frac{\tan x+1}{\tan x-1}\right)\\&
=\frac{\frac{d}{dx}(\tan x+1)(\tan x-1)-(\tan x+1)\frac{d}{dx}(\tan x-1)}{(\tan x-1)^2}\\&
=\frac{\sec^2 x(\tan x-1)-(\tan x+1)\sec^2 x}{(\tan x-1)^2}\\&
=\frac{\sec^2x\tan x-\sec^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"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.