Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Is there any similar solutions including $\pi$ like $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\cdots=\frac{\pi}{3\sqrt{3}}$? First equation is very popular - there are only odd numbers. Other words, numbers, which are coprime with $2$.
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{7}+\cdots=\frac{\pi}{4}$$... | I was going to write more or less the same answer of Lord Shark, so let us steer in a more elementary direction. The fact that Gregory series equals $\frac{\pi}{4}$ can be seen as a consequence of
$$ \sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\int_{0}^{1}\sum_{n\geq 0}(-1)^n x^{2n}\,dx = \int_{0}^{1}\frac{dx}{1+x^2}=\arctan(1)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
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Can either of these two simultaneous equations be solved? I have been struggling with the elimination of the variables.
$\begin{array}{c|c}
\left(1-2^k\right) a+2^k b=-2^k+3^k & \left(-1+2^k\right) a-2^k b=2^k-3^k\\
b+\left(-1+2^k\right) c=-1+3^k & a+2^k c=3^k\\
a+2^k c=3^k & b+\left(-1+2^k\right) c=-1+3^k \\
\end{a... | Note that using Cramer’s rule may be much better. But, let us go with a brute-force method.
From the last equation, we have: $c = \frac{3^k-a}{2^k}$. Substituting this in the second equation, gives us: $$b + (2^k-1)[\frac{3^k-a}{2^k}]=3^k -1$$ $$\implies b -a+\frac{a}{2^k}=\frac{3^k}{2^k} -1 \implies (1-2^k)a+2^kb= 3^k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate the determinant? I need help with this determinant ($n \times n)$:
$D_n = \begin{vmatrix}
a & x & x & \dots & x & x \\
y & a & x & \dots & x & x \\
y & y & a & \dots & x & x \\
\vdots \\
y & y & y & \dots & a & x \\
y & y & y & \dots & y & a \\
\end{vmatrix}$
I tried to simplify it and derive a reсcu... | Subtract the 2nd column from the first, the 3rd from the second, etc.
Here is what you get in the $4\times4$ case:
$$D_4 = \begin{vmatrix}
a-x & 0 & 0 & x \\
y-a & a-x & 0 & x \\
0 & y-a & a-x & x \\
0 & 0 & y-a & a
\end{vmatrix}$$
Laplace expansion on the first column will yield:
$$D_4=(a-x)D_3-(y-a) \begin{vmatrix}
... | {
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"timestamp": "2023-03-29T00:00:00",
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$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$ and other sums of quadratic reciprocals What is the exact value for $$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$$ and can other infinite sums of quadratic reciprocals have specific values.
| According to Maple,
$$ \frac{\pi}{\sqrt{3}} \tanh \left(\frac{\pi \sqrt{3}}{2}\right) - 1 $$
It looks like all the sums $\sum_{n=1}^\infty \frac{1}{n^2+tn+1}$ for integers $t\ne -2 $ have closed forms.
EDIT: Here are a few examples:
$$ \eqalign{\sum_{n=1}^{\infty } \frac{1}{ {n}^{2}+n+1} & =\frac{\pi \tanh(\pi \sqrt{3}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\sqrt{a+ib}$ How do to find the square root of a complex number?
$z=a+ib$
$z=\sqrt{3+i4}=a+ib$
remove the square by squaring both side
$a^2+2iab+b^2=3+i4$
$a^2+b^2=3$
$2ab=4$
So I just solve the simultaneous equation. But my friend is telling me this way I am doing is wrong. Can anyone give a hand, thank in ad... | The method is correct, you identify the real and imaginary parts:
$$\begin{cases}x^2-y^2=a,\\2xy=b.\end{cases}$$
Then multiplying the first equation by $4x^2$,
$$4x^4-4ax^2-4x^2y^2=4x^4-4ax^2-b^2=(2x^2-a)^2-a^2-b^2=0,$$
giving
$$x^2=\frac{\pm\sqrt{a^2+b^2}+a}2.$$
As we need $a$ to be real, only the $+$ sign is useful a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the largest value of the expression $P=\frac{3x+2y+1}{x+y+6}$ For real numbers $x, y$ satisfy the condition $x>0, y>0$ and
$$\log _{\sqrt{3}}\left(\frac{x+y}{x^2+y^2+xy+2}\right)=x\left(x-3\right)+y\left(y-3\right)+xy$$
Find the largest value of the expression $$P=\frac{3x+2y+1}{x+y+6}$$
| I think this is a very nice question.
We have $$P(x,y)=\frac{3x+2y+1}{x+y+6}\tag{1}$$ so by the Quotient Rule $$P'(x,y)=\frac{(3+2y')(x+y+6)-(3x+2y+1)(1+y')}{(x+y+6)^2}=\frac{y+(11-x)y'+17}{(x+y+6)^2}=0$$ for stationary points. Thus $$(11-x)y'+y+17=0\implies y'+\frac1{11-x}y=-\frac{17}{11-x}$$ is a first-order linear O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581555",
"timestamp": "2023-03-29T00:00:00",
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X ways to choose R objects in generating functions I'm currently doing generating functions, and the purpose of those functions is to find how many ways there are to choose an x amount of objects.
I just don't understand how they get to the final conclusion.
Take for example this exercise:
You have to go to the baker ... | The generating function approach is as follows. We would like to know the number of integer solutions to
$$x_1 + x_2 + x_3 = r$$
subject to $0 \le x_1 \le 3$, $0 \le x_2 \le 2$, and $0 \le x_3 \le4$. To solve the problem with a generating function, let $a_r$ be the number of solutions for $r = 0, 1, 2, 3, \dots$ and ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the limit of the sequence $a_n\cdot a_{n+1}=n,\,n=1,2,3,\cdots.$ Let $(a_n)_{n>=1}$ be a sequence of real numbers defined by the below recurrence relation:
$$a_n\cdot a_{n+1}=n,\quad n=1,2,3,\cdots.$$
Prove that $\lim_{n\to \infty}a_n=+\infty.$
Edit: $a_1>0$
| Since $a_1 > 0$, it follows that $a_n > 0$, for all $n$.
Since $a_n a_{n+1} = n$, at least one of $a_n,a_{n+1}$ is greater or equal to $\sqrt{n}$, hence the sequence $(a_n)$ is unbounded above.
For all positive integers $n > 2$, we have
\begin{align*}
a_n &= \frac{n-1}{a_{n-1}}\\[4pt]
a_{n-2} &= \frac{n-2}{a_{n-1}}\\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to prove $2\le\frac{g(x)}{g(y)}+\frac{g(y)}{g(x)}$ using AM-GM? In this question Prove that, if $g(x)$ is concave, for $S = {x : g(x) > 0}$, $f(x) = 1/g(x)$ is convex over $S$. , in the proof of Math536,
how to prove $2\le\frac{g(x)}{g(y)}+\frac{g(y)}{g(x)}?$
This is the proof of Math536:
$$\begin{align}
1 &= (a... | There is no need for the general AM-GM here. Suppose you have two positive real numbers $a, b > 0$ (justified since $g(x), g(y) > 0$). Then,
$$2 \leq \frac{a}{b} + \frac{b}{a} = \frac{a^2+b^2}{ab} \iff$$
$$2ab \leq a^2+b^2 \iff a^2-2ab+b^2 \geq 0 \iff (a-b)^2 \geq 0$$
with equality iff $a=b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{n\rightarrow\infty}$ $\left[\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}\right]^{\frac{1}{n}}$
Question: Evaluate lim$_{n\rightarrow\infty}$$\left[\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}\right]^{\frac{1}{n}}$
My Approach Let $a_{n}=\frac{\left(n+1\rig... | Let the limit be $L$ then:
$$\begin{align}
\ln(L) &= \lim_{n\to \infty}\frac{1}{n} \sum_{i=1}^{n} \ln\left( 1 + \frac{i}{n} \right) \\
&= \int_{0}^{1}\ln( 1+x ) dx \\
&= (x+1)\ln(1+x) -x |_{0}^{1} \\
&= 2\ln(2) - 1
\end{align}$$
So the required limit is $\exp({2\ln (2) -1}) = \color{blue}{\dfrac{4}{e}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $3\cos x - 4 \sin x = 2$, find $3 \sin x + 4 \cos x$ without first solving for $x$ If $$3\cos{x}-4\sin{x}=2$$
find $$3\sin{x} +4\cos{x} $$
I have solved the equation for $x$, then calculated the required value, but I think there is a direct solution without solving the equation.
| Think of it as a rotation equation. This is equivalent to the matrix formulation $$\begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x\end{bmatrix} \begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 \\ y \end{bmatrix}$$ where y is the unknown. Since the magnitude of a vector is preserved in rotation, $2^2 + y^2 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Partial Fraction Decomposition of $\prod\limits_{j=0}^{n}\frac 1{1+Xa^{j-k}}$
Question: Decompose the fraction$$\frac 1{\left\{1+X\right\}\left\{1+Xa\right\}\left\{1+Xa^2\right\}\cdots\left\{1+Xa^n\right\}}$$where $X=x^2$.
My work: Obviously, the right-hand side has to take the form$$\prod\limits_{j=0}^n\frac 1{1+Xa^... | Consider
$$\dfrac{1}{(1+X)(1+Xa)(1+Xa^2)}
=\dfrac{c_0}{1+X} + \dfrac{c_1}{1+aX} + \dfrac{c_2}{1+a^2}$$
Multiplying both sides by $(1+X)(1+Xa)(1+Xa^2)$, we get
$$c_0(1+Xa)(1+Xa^2) + c_0(1+X)(1+Xa^2) + c_0(1+X)(1+Xa) = 1$$
Let $X=-1$, and we get
$$c_0(1-a)(1-a^2) = 1$$
$$c_0 = \dfrac{1}{(1-a)(1-a^2)}$$
Let $X=-\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Justify that the vectors are linearly dependent Let $V$ be a real vector space and $a,b,c,d,e\in V$.
We have the vectors \begin{align*}&v_1=a+b+c \\ &v_2=2a+2b+2c-d \\ &v_3=a-b-e \\ &v_4=5a+6b-c+d+e \\ &v_5=a-c+3e \\ &v_6=a+b+d+e\end{align*}
I want to justify that these vectors are linearly dependent.
$$$$
These si... | If you select the set $\{v_k \}_{k=1}^{\color{red}{5}}$ you can write
$$
\left(\begin{array}{c}v_1\\ v_2 \\ v_3 \\v_4 \\ v_5 \end{array}\right) = \left(\begin{array}{ccccc}
1 & 1 & 1 & 0 & 0 \\
2 & 2 & 2 & -1 & 0 \\
1 & -1 & 0 & 0 & -1 \\
5 & 6 & -1 & 1 & 1 \\
1 & 0 & -1 & 0 & 3
\end{array}\right) \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Sum of the infinite series The series is $$\frac{5}{1\cdot2}\cdot\frac{1}{3}+\frac{7}{2\cdot3}\cdot\frac{1}{3^2}+\frac{9}{3\cdot4}\cdot\frac{1}{3^3}+\frac{11}{4\cdot5}\cdot\frac{1}{3^4}+\cdots$$
This is my attempt:
$$T_n=\frac{2n+3}{n(n+1)}\cdot\frac{1}{3^n}$$
Assuming $$\frac{2n+3}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$$
... | Just try to write the expression for $T_{n+1}$ and note that the series telescopes.
$$T_n=\frac{1}{n}\cdot\frac{1}{3^{n-1}}-\color{purple}{\frac{1}{n+1}\cdot\frac{1}{3^{n}}} \\
T_{n+1}=\color{purple}{\frac{1}{n+1}\cdot\frac{1}{3^{n}}}-\frac{1}{n+2}\cdot\frac{1}{3^{n+1}}$$
We see the terms are cancelling, so
$$\begin{... | {
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"timestamp": "2023-03-29T00:00:00",
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Find ED and (I am so so sorry for my bad English in advance, it is my math homework problem, I tried to solve it but I've got no idea whether the solution's right or wrong:))) It is given that $∠B=120^{o}$, $AB=6 cm$, $BC=4cm$, $CE=BE$, $AC=CD$. Find the lenght of ED and angle $∠CED$. Look
Solution:
$AC=\sqrt{6^2+4^2-... | $$ED^2=4+CD^2-4CD\cos(180^{\circ}-\gamma)$$
$$CD^2=AC^2=36+16-48\cos(120^{\circ})$$
$$\frac{\sin(\gamma)}{\sin(120^{\circ})}=\frac{6}{AC}$$
$$AC^2=4+ED^2-4ED\cos(\delta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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using $x^9 - 1$ as a product of linear and quadratic factors with real coefficients to solve trig So here's where I'm at I know i can find the ninth roots of unity by using the nth roots of unity formula $$ \zeta_n = e^{\frac{i2\pi k}{n}} ,$$ when $k = 1,2,3,...,n-1$
so the roots of unity where $e^{i\frac{2\pi}{9}},e... | Note sure where this problem came from or what kind of solution is expected, but this runs up against the Casus Irreducibilus. Let $\zeta=\zeta_9$ be an irreducible $9$th root of unity, then you want the polynomial
$$(x-\zeta)(x-\zeta^{-1})=x^2-(\zeta+\zeta^{-1})x+1$$
So to find the coefficient we have
$$(\zeta+\zeta^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability-Tossing of die What's the probability of tossing a die 4 times, and getting a higher value than the previous value in all the 4 outcomes ? How can we I the method for such a case in the case of 3 tosses?
| There is no method. you have to list all the possibilities.
The first thought is that the first result cannot higher that $3$. So, there are three groups of favorable outcomes:
$$\begin{matrix}
\color{red}1&2&3&4\\
\color{red}1&2&3&5\\
\color{red}1&2&3&6\\
\color{red}1&2&4&5\\
\color{red}1&2&4&6\\
\color{red}1&2&5&6\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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lim$_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+...+\frac{1}{\sqrt{3n}+\sqrt{3n+3}}\right)$
Question
Edit:
My Approach:
$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+...+\frac{1}{\sqrt{3n}+\sqrt{3n+3}... | We have: $$\lim_{n \to \infty} \frac1{\sqrt n} \left( \frac1{\sqrt 6 + \sqrt 3} + \frac1{\sqrt 9 + \sqrt 6} +\ldots \right) = \lim_{n \to \infty} \frac1{\sqrt n} \left( \frac{\sqrt 6- \sqrt 3}3 + \frac{\sqrt 9 - \sqrt 6}3 + \ldots \right) $$
Can you see it is a telescoping sum?
| {
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"timestamp": "2023-03-29T00:00:00",
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In how many ways can $6$ red and $6$ blue balls be distributed among $10$ persons such that every person gets at least one ball? In how many ways can $6$ red and $6$ blue balls be distributed among $10$ persons such that every person gets at least one ball? (Balls of the same colour are identical).
My Attempt
There app... | Here is a solution by way of the Principle of Inclusion / Exclusion (PIE).
Suppose we ignore, for now, the restriction that each person must get at least one ball. Then by a bars-and-stars argument, there are $\binom{10+6-1}{6}$ ways to distribute the six red balls among the ten people, and similarly for the six blue ... | {
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"timestamp": "2023-03-29T00:00:00",
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Behaviour of $\sum\limits_{n=1}^\infty \frac1{n^2}\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n x^n$ for $|x|=2$ I want to determine the radius of convergence for the power series
$$\sum_{n=1}^\infty \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2} x^n$$
and determine what happens at the boundaries. I determined the ratio of c... | $$
\begin{align}
&\sum_{n=1}^\infty \frac{\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n}{n^2}x^n\\
&=\sum_{n=1}^\infty \frac{n^n\left(\sqrt{1+\frac1n} - \sqrt{1+\frac1{n^2}}\right)^n}{n^2}x^n\\
&=\sum_{n=1}^\infty \frac{n^n\left(\left(1+\frac1{2n}-\frac1{8n^2}+O\!\left(\frac1{n^3}\right)\right) - \left(1+\frac1{2n^2}+O\!\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim\limits_{n\to+\infty}\sum\limits_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}$
Compute
$$\lim_{n\to+\infty}\sum_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}.$$
My Approach
Since $k^{3}+6k^{2}+11k+5= \left(k+1\right)\left(k+2\right)\left(k+3\right)-1$
$$\lim_{n\rightarrow\infty}\sum_{k=1}... | Good start!
$$ \begin{align}
\lim_{n \to \infty}\sum_{k=1}^n\left(\frac{1}{k!} - \frac{1}{(k+3)!}\right) &=
\lim_{n \to \infty}\left(\sum_{k=1}^n\frac{1}{k!} - \sum_{k=1}^n\frac{1}{(k+3)!} \right) \\ &= \lim_{n \to \infty}\left(\sum_{k=1}^n\frac{1}{k!} - \sum_{k=4}^{n+3}\frac{1}{k!} \right) \\ &= \lim_{n\to\infty}\lef... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Beginner troubleshooting an eigenvector calculation I am having some difficulty identifying the error in my eigenvector calculation. I am trying to calculate the final eigenvector for $\lambda_3 = 1$ and am expecting the result $ X_3 = \left(\begin{smallmatrix}-2\\17\\7\end{smallmatrix}\right)$
To begin with, I set up ... | You have solved the problem correctly up to $$x_1 + \frac{2}{7}x_3 = 0 \rightarrow x_1 = -\frac{2}{7}x_3$$ and $$x_1 + \frac{2}{17}x_2 = 0 \rightarrow x_1 = -\frac{2}{17}x_2$$ Now when you let $x_1=1$ you get $x_3=-\frac {7}{2}$ and $x_2=-\frac {17}{2}$ which give you the correct eigenvector.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that this triangle is equilateral? Question: If $\cos A +\cos B +\cos C=\frac{3}{2}$, prove that the triangle is equilateral.
My attempt: According to cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and $\cos C=\frac{a^2+b^2-c^2}{2ab}$.
$\cos A +\cos B +\cos C=\frac{3}{2}$
$... | To prove that the triangle is equilateral, i have done the following: I have used the cosine rule to simplify the given equation, algebraic manipulation to bring it into the required form, triangle inequality to deduce that each term in the manipulated equation is non-negative and the fact that when the sum of non-nega... | {
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} |
Prove that if three real numbers $a, b, c$ satisfy $a^2+b^2+c^2=ab+bc+ac$, then $a=b=c$. I don't have much to start the problem, as I have trouble approaching it. What method of solving should I use here?
| $$(a^2+b^2+c^2)-(ab+ac+bc) = \frac{1}{2}\left[(a-b)^2+(a-c)^2+(b-c)^2\right]\geq 0 $$
and the middle term equals zero iff $a=b=c$.
Alternative approach: by the Cauchy-Schwarz inequality
$$\left|ab+ac+bc\right|\leq \sqrt{a^2+b^2+c^2}\sqrt{b^2+c^2+a^2} = a^2+b^2+c^2$$
and the equality holds iff the vectors $(a,b,c)$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find non-zero real numbers $a,b,c,d$ such that $a^2+c^2=b^2+d^2$ and $ab+cd=0$. There are the obvious cases where $a=c=0$ and $b=d$. Are there any cases for this where all four numbers are non-zero?
Note, this problem is equivalent to saying the matrix $A^{-1}JA$ is a rotation invariant linear complex structure where ... | $$d=-\frac{ab}{c}$$ and we obtain:
$$a^2+c^2=b^2+\frac{a^2b^2}{c^2}$$ or
$$(a^2+c^2)(b^2-c^2)=0.$$
If $b=c$ then we get $d=-a$.
if $b=-c$ then we get $d=a$.
Id est, we got the answer:
$$\{(a,b,b,-a),(a,b,-b,a)|a\neq0,b\neq0\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2608031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Value of $a_2+a_6+a_{10}+\cdots+a_{42}$ If $(1+x+x^2+\cdots+x^9)^4(x+x^2+x^3+\cdots+x^9)=\sum_{r=1}^{45} a_rx^r ,$ then what is the value of $a_2+a_6+a_{10}+\cdots+a_{42}$
| Calling your expression $f(x)$, we have
$$ a_2 + a_6 + \ldots + a_{42} = \dfrac{f(1) + f(-1) - f(i) - f(-i)}{4}
$$
Now $f(1) = 9 \cdot 10^4$, $f(-1) = 0$, $f(i) = -4 i$ and $f(-i) = 4i$, so
this is $(9 \cdot 10^4)/4 = 22500$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2608564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int_0^\pi \frac{dx}{1+a^2\cos^2(x)}$
I am trying to compute the following integral
$$\int_0^\pi \frac{dx}{1+a^2\cos^2(x)}$$
But I got stuck on my way.
Indeed, enforcing the change of variables $t =\cos^2x$ leads to
$$\int_0^\pi \frac{d x}{1+a^2\cos^2(x)}= \int_{-1}^1 \frac{d x}{(1+a^2t^2)(1-t^2)^{1/2... | Here is another way. Consider the identity $$\cos^2x=\frac{1+\cos 2x}{2}$$ and the integral is then transformed into $$2\int_{0}^{\pi}\frac{dx}{2+a^2+a^2\cos 2x}$$ and using substitution $$2x=t$$ we get $$\int_{0}^{2\pi}\frac{dt}{2+a^2+a^2\cos t} $$ Since $\cos(2\pi - t) =\cos t$ we can see that the above integral is e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2609216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove that $a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$ One of my friend had just given me an inequality to solve which is stated below.
Consider the three positive reals $a, b, c$ then prove that
$$a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$$
I have solved this inequality very e... | This is the same as
$$a^4+b^4+c^4\ge abc(a+b+c).$$
By AM/GM,
$$2a^4+b^4+c^4\ge 4a^2bc.$$
Now add the cyclic permutations of this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the value of a third order circulant type determinant To calculate the value of the determinant\begin{pmatrix}
(b+c)^2 &c^2 &b^2 \\
c^2 &(c+a)^2 &a^2 \\
b^2 &a^2&(a+b)^2
\end{pmatrix} .
I multiplied R1,R2,R3 by a²,b²,c² respectively and then used the operations R1=R1-R2,R2=R2-R3,R3=R3-R1 which got me a zero v... | It's $$\prod_{cyc}(a+b)^2+2a^2b^2c^2-\sum_{cyc}a^4(b+c)^2=$$
$$=2\sum_{cyc}(a^3b^3+3a^3b^2c+3a^3c^2b+2a^2b^2c^2)=2(ab+ac+bc)^3.$$
I used the following.
$$\prod_{cyc}(a+b)=\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right),$$
$$\left(\sum_{cyc}(a^2b+a^2c)\right)^2=\sum_{cyc}(a^4b^2+a^4c^2+2a^3b^3+2a^4bc+2a^3b^2c+2a^3c^2b+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things... | $$x^3+y^3+z^3-3xyz=(x+y+z)\{(x+y+z)^2-3(xy+yz+zx)\}$$
Now $$2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 2
} |
Formulas involving pairs of coefficients of expansion of $(1+x+x^2)^n$ If $a_{0},a_{1},a_{2},\cdots\cdots$be coefficient in the expansion of $(1+x+x^2)^n$ in ascending power of $x$.Then prove that
$(1)\;a_{0}\cdot a_{1}-a_{1}\cdot a_{2}+a_{2}\cdot a_{3}-\cdots\cdots -a_{2n-1}\cdot a_{2n}=0$
$(2)\;a_{0}\cdot a_{2}-a_{1}... | Substituting $x$ by $\frac 1x$ into
$$(1+x+x^2)^n=\sum_{j=0}^{2n}a_{j}x^j\tag3$$
we get
$$\left(1+\frac 1x+\frac{1}{x^2}\right)^n=\sum_{j=0}^{2n}a_{j}\left(\frac 1x\right)^j\tag4$$
Multiplying the both sides by $x^{2n}$ gives
$$(x^2+x+1)^n=\sum_{j=0}^{2n}a_{j}x^{2n-j}\tag5$$
Substituing $x$ by $-x$ into $(5)$ gives
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find major & minor axes of the ellipse $10x^2+14xy+10y^2-7=0$?
What are major & minor axes of the ellipse: $10x^2+14xy+10y^2-7=0$ ?
My trial:
from given equation: $10x^2+14xy+10y^2-7=0$
$$10x^2+14xy+10y^2=7$$
$$\frac{x^2}{7/10}+\frac{xy}{7/14}+\frac{y^2}{7/10}=1$$
I know the standard form of ellipse: $\frac{... | Rewriting the equation in polar coordinates, with $x=r\cos\theta$ and $y=r\sin\theta$, we get
$$r^2(10\cos^2\theta+14\cos\theta\sin\theta+10\sin^2\theta)-7=0$$
Using the trig identities $\sin^2\theta+\cos^2\theta=1$ and $2\sin\theta\cos\theta=\sin2\theta$, we find this simplifies to
$$r^2(10+7\sin2\theta)-7=0$$
or
$$r=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Calculate the triple integral: $\iiint_W(x^2+y^2)z\ dx\,dy\,dz$ I am trying to solve the following triple integral:
$$
\iiint_W(x^2+y^2)z\,dx\,dy\,dz \\
W=\{(x,y,z) \in \mathbb{R}: x^2+y^2+z^2 \le 9; x^2+y^2\le 1; x \ge 0; y \ge 0; z \ge 0\}
$$
From which I know that there are two surfaces:
$$
x^2+y^2+z^3 = 9 \rightarr... | Hint:
from $\rho^2\le 1$ you have the limits for $\rho$, that is : $0\le \rho \le 1$
so you can find the limits for $z$ that is: $0\le z \le \sqrt{9-\rho^2}$
Finally, from the condition that $x,y$ are positive, you have the limits $0\le \theta \le \frac{\pi}{2}$
Can you do from this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is this a correct derivation of completing the square? $x^2 + bx$
$=x^2 + bx + c - c$
$=(x + k)^2 - c$
$=x^2 + 2kx + (k^2 - c) = x^2 + bx + 0$
This implies:
$2k = b$, so $k = b/2$, and:
$k^2 - c = 0$, or $k^2 = c$, or $(b/2)^2 = c$
So to complete the square we are making the transformation:
$x^2 + bx \implies (x + b/2)... | It's true, but you can get it much more easier:
$$x^2+bx=x^2+bx+\frac{b^2}{4}-\frac{b^2}{4}=\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that the intersection of a cylinder and a plane forms an ellipse I've seen the nice proof of this using spheres, but I'm looking for a way to prove it parametrically if possible. Using a cylinder $x^2+y^2=r^2$ and a plane $ax+by+cz+d=0$ I got:
$x=r\cos(\theta), y=r\sin(\theta), z=\dfrac{-ar\cos(\theta)+br\sin(\th... | Without loss of generality let's assume the cylinder has radius $1$, so that it has equation $x^2 + y^2 = 1$. Assuming that the plane is not parallel to the cylinder, we can always rearrange the coordinate system so that the plane goes through the origin, or even better, make the plane go through the $x$-axis after a s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Sum of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $ What is the limit of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $?
The $n$th summand is $\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2... | The $n$-th term is
$$\frac12\left(\frac1{2n}-\frac{2}{2n+1}+\frac1{2n+2}\right).$$
The whole series does not telescope but is
$$\frac12\left(\frac12-\frac23+\frac24-\frac25+\frac26-\frac27+\cdots
\right).$$
This is very similar (not identical) to a well-known series...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find lengths of tangents drawn from $(3,-5)$ to the Ellipse Find lengths of tangents drawn from $A(3,-5)$ to the Ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$
My try: I assumed the point of tangency of ellipse as $P(5\cos a, 4 \sin a)$
Now Equation of tangent at $P$ is given by
$$\frac{x \cos a}{5}+\frac{ y \sin a}{4}=1$$... | The slope of tangent at a point (x,y) on the ellipse $$\frac{x^2}{25}+\frac{y^2}{16}=1$$ is $$\frac {dy}{dx} = \frac {-16x}{25y}$$
On the other hand slope of the line passing through $(x,y)$ and $(3,-5)$ is $m = \frac {y+5}{x-3} $
$$ \frac {-16x}{25y}= \frac {y+5}{x-3} \implies 48x-125y=400 $$
Solve the system $$ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find all solutions in positive integers of the Diophantine equation $x^2+2y^2 =z^2$ I can't solve the following exercise. It's already on this site, but the solution method is not the same as in my solutions manual.
Find all solutions in positive integers of the Diophantine equation $x^2+2y^2 =z^2$.
The solution is g... | *
*$2y^2=z^2-x^2$, the left-hand side is an even number so $z,x$ must either both be even (even minus even is even) or they must both be odd (odd minus odd is even). They did, however, mistype. They typed $x,y$ have the same parity (false for some solutions) but meant $x,z$ gave the same parity (proven true here).
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why is $\int_{-\infty}^{\infty}\left\lvert\log\frac{a^2+x^2}{b^2+x^2}\right\rvert dx=2\pi(b-a)$?
If $a,b\in\mathbb R_{>0}$ and $ b>a$, then why is $\int_{-\infty}^{\infty}\left\lvert\log\frac{a^2+x^2}{b^2+x^2}\right\rvert dx=2\pi(b-a)$ ?
Usual change of variables doesnt bring anything, I think. Is there a special fun... | \begin{align*}
\int_{-M}^{M}\left|\log\dfrac{a^{2}+x^{2}}{b^{2}+x^{2}}\right|dx&=\int_{-M}^{M}\log\dfrac{b^{2}+x^{2}}{a^{2}+x^{2}}dx\\
&=\int_{-M}^{M}\log(b^{2}+x^{2})dx-\int_{-M}^{M}\log(a^{2}+x^{2})dx,
\end{align*}
where
\begin{align*}
\int_{-M}^{M}\log(b^{2}+x^{2})dx&=2M\log(b^{2}+M^{2})-\int_{-M}^{M}\dfrac{2x}{b^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
how to get the limit of $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)$ How to get the limit $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) = \frac{1}{2}$ ?
$\begin{align}
\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) &=
\lim_{n \to \infty}\left( \frac{(\sqrt{n+\sqrt{n}}-\sq... | In a simpler way:
Let $n=m^2$.
$$\lim_{m\to\infty}\sqrt{m^2+m}-m=\lim_{m\to\infty}\frac m{\sqrt{m^2+m}+m}=\lim_{m\to\infty}\frac 1{\sqrt{1+\dfrac1m}+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Compute the 100th power of a given matrix So, I was asked this question in an interview.
Given a matrix
$$M = \begin{bmatrix} 0 & 1& 1 \\ 1& 0& 1 \\ 1&1& 0\end{bmatrix},$$
find $M^{100}$.
How does one approach this question using pen and paper only?
| Trying to avoid orthogonal diagonalization and square roots. The matrix of all ones has eigenvalues $(3,0,0),$ with eigenvectors (NOT normalized) as the columns of
$$
W =
\left(
\begin{array}{ccc}
1 & -1 & -1 \\
1 & 1 & -1 \\
1 & 0 & 2
\end{array}
\right)
$$
Oh, we need the inverse,
$$
W^{-1} = \frac{1}{6}
\left(
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to prove this geometry question on quadrilateral and circles? In quadrilateral $ABCD$, $AC$ and $BD$ crossed at point $E$. Circle $O$ passes through $A, D$, and $E$ and its center is $O$. $P, Q, R$ are the midpoint of $AB$, $BC$ and $CD$, respectively. Circle $O_2$ passes through $P, Q$, and $R$, and crosses $BC$ a... | It suffices to prove that$$
BO^2 - CO^2 = BF^2 - CF^2. \tag{1}
$$
Denote the radius of circle $ADE$ by $r$. The power of point $B$ with respect to circle $ADE$ is$$
BO^2 - r^2 = BE \cdot BD,
$$
and the power of point $C$ with respect to circle $ADE$ is$$
CO^2 - r^2 = CE \cdot CA,
$$
also $BD = 2 QR$, $CA = 2 QP$, there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Is it always a positive value? The values of $x$ can be from $0$ to $\infty$. So for these values of $x$ is the following function always positive $$f(x)=x^2+x+\exp(\frac{1}{x})(2x+1)Ei(-\frac{1}{x})$$ where $Ei(x)$ is the exponential integral. My checking in the WA shows that it is a positive value for arbitrarily lar... | \begin{align}
\mathrm{Ei}(-1/x) &= - \int_{1/x}^\infty \frac{e^{-t}}{t} \, dt \\
f(x) &> 0 \\
x (x + 1) &> - (2x+1) \exp(1/x) \, \mathrm{Ei}(−1/x) \\
x (x + 1) &> (2x+1) \exp(1/x) \, \int_{1/x}^\infty \frac{e^{-t}}{t} \, dt \\
\end{align}
Since by requirement $x$ is positive, we can bring everything except the integral... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solving $\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)$ Solve
$$
\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)
$$
My Attempt:
From the domain consideration,
$$
\boxed{0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}}
$$
$$
\cos\big(\tan^{-1}x\big)=\cos\big(\frac{\pi}{2}-\cot^{-1}\frac... | $$\cos(\arctan x)=\cos\left(\arctan\dfrac34\right)$$
$$\implies\arctan x=2m\pi\pm\arctan\dfrac34=2m\pi+\arctan\left(\pm\dfrac34\right)$$ where $m$ is any integer as $\arctan(-a)=-\arctan(a)$
As $-\dfrac\pi2<\arctan y\le\dfrac\pi2,$
$$\arctan x= \arctan\left(\pm\dfrac34\right)$$
Apply tan on both sides
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Eigenvalues of symmetric $\mathbb{R}^{p\times p}$ matrix I want to prove that
$$A^{(p)}
=
\begin{pmatrix}
a & 1... | This is an answer not by induction, in response to a comment/answer by OP.
$$
\newcommand{\bu}{ {\mathbf u}}
\newcommand{\bv}{ {\mathbf v}}
\newcommand{\bM}{ {\mathbf M}}
\newcommand{\bI}{ {\mathbf I}}
\newcommand{\bU}{ {\mathbf U}}
\newcommand{\bQ}{ {\mathbf Q}}
$$
Your matrix $\bM$ is
$$
\bM = \bU + (a-1)\bI
$$
whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to tell if this integral converges? $$\int_{0}^{\infty} \frac{\sqrt{1-x+x^2}}{1-x^2+x^4} dx$$
What's the method for determining if this integral converges or diverges? The integral seems to converge if I put it into Wolfram Alpha.
But do we assume it's similar to $$\int_{0}^{\infty} \frac{1}{x^4} dx$$
Because if so... | To prove convergence of $\int_0^{\infty}f(x)dx$ for a continuous $f(x)$ it often helps to find a crude but effective upper bound for $|f(x)|$ for large $x$.
When $x>1$ we have $\sqrt {1-x+x^2}\; <\sqrt {1+2x+x^2}=x+1<2x .$
And when $x>2$ we have $x^4-x^2+1>x^4-x^2>x^4/2.$
So when $x>2$ we have $$0<\frac {\sqrt {1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Reduction formula for $\int\frac{dx}{(x^2+a^2)^n}$ How can I use integration by parts to write $$\int\frac{dx}{(x^2+a^2)^n} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}\int\frac{dx}{(x^2+a^2)^{n-1}}?$$
I would try $$u=\frac{1}{(x^2+a^2)^n},du=\frac{-2nx}{(x^2+a^2)^{n+1}}dx; dv=dx,v=x.$$ Integration by pa... | Hint:
Use integration by parts for the computation of $\;\displaystyle\int\frac{\mathrm dx}{(x^2+a^2)^{\color{red}{n-1}}} $, setting
$$ \begin{cases}u=\dfrac{1}{(x^2+a^2)^{n-1}},\\[1ex]\mathrm dv=\mathrm dx,\end{cases}
\;\text{ whence}\quad\begin{cases}\mathrm du=\dfrac{-2(n-1)x\,\mathrm dx}{(x^2+a^2)^{n}},\\[1ex] v=x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the nth term of the given series The series is $$1^3 + (2^3 +3^3)+(4^3+5^3+6^3)+...$$
i.e., $1^3$, $(2^3 +3^3)$ , $(4^3+5^3+6^3)$ , $(7^3+8^3+9^3+10^3)$ , and so on are the $1^{st}$, $2^{nd}$, $3^{rd}$ and so on terms respectively. ($(4^3+5^3+6^3)$ is the third term). So we need to find the $n^{th}$ term.
So, the ... | Let me see if I understand you correctly for I am a little bit confused about your notation. I guess that by the first term of the sequence you mean $1^3$ and $(2^3 + 3^3)$ is the second term, $(4^3 + 5^3 + 6^3)$ is the third term and so on. If this is so, it seems correct that the firs summand of the $n$-th term is gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that $x^2+y^2+z^2+2xyz=1$
If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that:
$$x^2+y^2+z^2+2xyz=1$$
My Attempt:
$$\sin^{-1} (x) + \sin^{-1} (y) + \sin^{-1} (z)=\dfrac {\pi}{2}$$
$$\sin^{-1} (x\sqrt {1-y^2}+y\sqrt {1... | Let: $\sin^{-1}x=\alpha$, $\sin^{-1}y=\beta$, $\sin^{-1}z=\gamma$. We know that $\alpha+\beta+\gamma={\pi\over2}$, that is: $\gamma={\pi\over2}-\alpha-\beta$ and $$\sin\gamma=\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$
We have then:
$$
\begin{align}
&x^2+y^2+z^2+2xyz=\\
&\sin^2\alpha+\sin^2\beta+\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Limit using Taylor series expansion
Can someone help me with this limit. I know I have to expand in Taylor series
$\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+o(x^3)$
$a^x=e^{x\ln a}=1+x\ln a+\frac{1}{2!}x^2(\ln a)^2+\frac{1}{3!}x^3(\ln a)^3+o(x^3)$.
But how to proceed next?
| $$\lim_{ x \to 0} \frac{a^{\sqrt{1+x}}-a^{1+x/2-x^2/8}}{x^3}=\lim_{ x \to 0} \;\underbrace{a^{1+x/2-x^2/8}}_{\to a} \cdot\left(\frac{a^{\sqrt{1+x}-1-x/2+x^2/8}-1}{x^3}\right)$$
Using Taylor series $\displaystyle \sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+\mathcal{O}(x^3)$
;
$$\lim_{ x \to 0} \;\underbra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$ Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$.
My try :
$$\begin{align}f(-x)&=|-x+a|-|-x+2|+b|-x+5| \\ &=-f(x) \\&= -\big(|x+a|-|x+2|+b|x+5| \big) \end{align}$$
$$|-x+a|-|-x+2|+b|-x+5|+ \big(|x+a|-|x+2|+b|x+5|) =0... | You can equate the slopes for small and large $x$ and
$$-1+1-b=1-1+b$$ so that $b=0$.
Then
$$f(x)=|x+a|-|x+2|$$ odd requires $|a|=2$ to ensure that $f(0)=0$.
By direct check, $f(x)=|x-2|-|x+2|$ or $f(x)=|x+2|-|x+2|=0$ are indeed odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
I need to find the general nth degree polynomial expansion of $x(x+1)(x+2)\dots(x+n)$ in order to solve an integral. Expanding $x(x+1)(x+2)\dots(x+n)$ has been challenging for me.
Say $n=5$, you will notice many patterns like $x^n+(n-1)x^{n-1}+\dotsb$ et cetera, but does there exist a general polynomial expansion for a... | $f(x) = \frac{1}{(x-a)(x-b)\cdots(x-k)} = \frac {A}{x-a} + \frac {B}{x-b} \cdots \frac {K}{(x-k)}$
$A = \lim_\limits {x\to a} (x-a)f(x) = \frac {1}{(a-b)(a-c)\cdots(a-k)}$
$f(x) = {1}{(x)(x+1)\cdots(x+n)} = \frac 1{n!}\frac {1}{x} - \frac {1}{(n-1)!}\frac {1}{x+1} + \frac {1}{2(n-2)!}\frac {1}{x+2} \cdots\\
f(x) = \sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove that $\sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy} \le 3$ Prove that
$$\sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy} \le 3$$
for $x \ge 0$, $y \ge 0$, $z \ge 0$ and $x+y+z \le 2$.
My work:
\begin{align*}
&\mathrel{\phantom{=}} \sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy}\\
&\le\sqrt3 \sqrt{x^2+yz+y^2+xz+z^2+xy}\\
&\le\sq... | By C-S we have:
$$\left(\sum_{cyc}\sqrt{x^2+yz}\right)^2\leq\sum_{cyc}\frac{x^2+yz}{2x^2+y^2+z^2+yz}\sum_{cyc}(2x^2+y^2+z^2+yz).$$
Id est, it remains to prove that
$$\sum_{cyc}\frac{x^2+yz}{2x^2+y^2+z^2+yz}\sum_{cyc}(4x^2+yz)\leq\frac{9}{4}(x+y+z)^2,$$ which is
$$\sum_{sym}(x^8+2x^7y+9x^6y^2-23x^5y^3+10x^4y^4)+$$
$$+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve the equation $ { \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $ Prove that one root of the Equation
$${ \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $$
is
$$x=2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right) $$
My progress:
After simplyfying the given Equation I got
$$ x^8-16x^6+88x^4-192x^2+x+140=0$$
Then I t... | $\pm 2,\pm 3,\pm 5$ are precisely twice the cubic residues $\!\pmod{19}$. Let $\omega=\exp\left(\frac{2\pi i}{19}\right)$. $\omega$ is al algebraic number over $\mathbb{Q}$ with degree $18$, hence by Galois theory both
$$ \sum_{k\in\{\pm 1,\pm 7,\pm 8\}}\omega^k,\qquad \sum_{k\in\{\pm 2,\pm 3,\pm 5\}}\omega^k $$
(relat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solving logarithmic equation with different bases $\log_2\left(x-5\right)=\log_5\left(2x+7\right)$ I have to solve this but my answer vary from x = 7, x = -9, x = 5/3 and I don't know why this happens.
Here is my approach:
$\frac{\log_5\left(x-5\right)}{\log_5\left(2\right)}=\log_5\left(2x+7\right)$
$\frac{\log_5\left(... | $\log_2(x-5) = \log_5(2x+7)=k$
$2^k = x-5$ so $x = 2^k + 5$. $5^k =2x - 7$ so $x = \frac {5^k -7}2$
Or $2*2^k +10 = 5^k - 7$
$2^{k+1} = 5^k-17$
[cute that $2^{k+1} -8 = 5^k -25$... but, I'll assume I wasn't clever enough to see that.]
At this point we can really only guess but clearly if we increase $k$ by small incre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$ Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$
My try
I found that $0 \lt x,y,z \lt 6$
Multiplying the equations we got $x(6-y)y(6-z)z(6-x)=9^3$
$x(6-x)y(6-y)z(6-x)=9^3$
And here is the problem,... | Repeated substitution gives
\begin{eqnarray*}
x \left( 6- \frac{9}{6- \frac{9}{6-x}} \right) =9
\end{eqnarray*}
and this simplifies to $(x-3)^2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
quadrilateral inside a parallelogram
the diagonals of the parallelogram creates four equal trianges, so △ ABD and △ BOC together becomes 3/4 s. But stuck here, how do I find area of △DEM and △MFC, so as to solve to find out the area of MEOF??
| Let $DM=xDC$.
Thus, $$S_{\Delta ADM}=xS_{\Delta ADC}=\frac{1}{2}xS$$ and
$$\frac{S_{\Delta ADE}}{\frac{1}{2}xS}=\frac{AE}{AM}=\frac{AE}{AE+EM}=\frac{1}{1+\frac{EM}{AE}}=\frac{1}{1+\frac{DM}{AB}}=\frac{1}{1+x}.$$
Hence, $$S_{\Delta ADE}=\frac{\frac{1}{2}xS}{1+x}.$$
In another hand,
$$S_{\Delta BCM}=(1-x)S_{\Delta BDC}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$(3x- 3)\prod_{n = 1}^∞\frac{\exp(2^{-n})+1}{x^{2^{-n}}+1}=(x^3-1)\prod_{n = 1}^∞\frac{\exp(2^{-n})+1}{x^{3 \cdot 2^{-n}}+1}$ Show that
$$ (3x - 3) \prod_{n = 1}^{\infty} \frac{ \exp(2^{-n}) +1} {x^{2^{-n}} + 1}
= (x^3 - 1) \prod_{n = 1}^{\infty} \frac{ \exp(2^{-n}) +1} {x^{3\cdot2^{-n}} + 1} $$
| Hint: $\;\;
\left(x^{2^{-n}}-1\right) \cdot \left(x^{2^{-n}}+1\right)\left(x^{2^{-(n-1)}}+1\right)\ldots\left(x^{2^{-1}}+1\right) = x - 1
\,$ by telescoping.
[ EDIT ] It then follows that the $\,n^{th}\,$ partial product on the LHS is:
$$\require{cancel}
3\cancel{(x - 1)} \cdot \frac{\left(x^{2^{-n}} - 1\right)}{\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is? If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is?
My attempt:
Solving the above quadratic equa... | I agree with your answer, indeed note that
$$2\tan^2x - 5\sec x = 1\iff2(1-\cos^2x)-5\cos x=\cos^2x\\\iff3\cos^2x+5\cos x-2=0$$
and
$$3t^2+5t-2=0\implies t=\frac{-5\pm\sqrt{25+24}}{6}\implies t=\frac{-5+\sqrt{47}}{6}= \frac13$$
thus we have 2 solution on the interval $[0,2\pi]$ and notably one in the interval $[0,\pi/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Area of a triangle using a matrix Find the area of a triangle whose vertices are (1,0), (2,2), and (4,3)
$$\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}1&0&1\\2&2&1\\4&3&1\end{vmatrix}$$
$$=1(-1)^2\begin{vmatrix}2&1\\3&1\end{vmatrix}+0(-1)^3\begin{vmatrix}2&1\\4&1\end{vmatrix}+1(-1)^4\begi... | It is from Laplace expansion for determinant.
As an alternative to calculate the area or to check the result you can also use
$$A=\frac12 |\det(u,v)|$$
with, for example, $u=(2,2)-(1,0)=(1,2)$ and $v=(4,3)-(1,0)=(3,3)$ that is
$$A=\frac12\begin{vmatrix}1&2\\3&3\end{vmatrix}=\frac32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using DeMoivre's Theorem to prove some identities regarding trigonometric functions The following question is from a previous post. The reason that I am posting this again is because it had two questions which were not really related and hence, one of the questions was not answered. The question is from a book, "Mathem... | Hints
For part 2, since you already showed the existence of the polynomial s.t. $$\sin \left( 2m + 1 \right)\theta = \left(\sin^{2m + 1}\theta \right)P_m(\cot^2\theta)$$
replace $\theta$ with $\frac{k\pi}{2m+1}$ and
$$\left(\sin^{2m + 1}\frac{k\pi}{2m+1} \right)\color{red}{P_m\left(\cot^2\frac{k\pi}{2m+1}\right)}=\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
For $x, y \in \mathbb{R}$, prove that $\max(x, y) = \frac{x + y + |x - y|}{2},$ and $\min(x, y) = \frac{x + y - |x - y|}{2}$. Prove that for all real numbers $x$ and $y$,
$$\max(x, y) = \dfrac{x + y + |x - y|}{2},$$
and
$$\min(x, y) = \dfrac{x + y - |x - y|}{2}.$$
For any real number $x$, the absolute value of $x$, d... |
If $\displaystyle x>y$ then $x-y>0$:
$|x-y|=x-y$
$\max(x,y) = (x+y+x−y)/2 = x$
$\min(x,y) = (x+y-(x−y))/2 = y$
If $\displaystyle x<y$ then $x-y<0$:
$|x-y|=y-x$
$\max(x,y) = (x+y+y-x)/2 = y$
$\min(x,y) = (x+y-(y-x))/2 = x$
If $x=y$ then:
$|x-y|=0$
$\max(x,y) = (x+y)/2 = x=y$
$\min(x,y) = (x+y)/2 =x=y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Computing $\int_{0}^{+\infty}\arctan\left(\frac{1}{x^n}\right)\,\mathrm{d}x$ Let $n \in (1,+\infty)$, I've shown that the function
$$
I(n)=\int_{0}^{+\infty}\arctan\left(\frac{1}{x^n}\right)\,\mathrm{d}x
$$
is defined on $(1,+\infty)$ and that it can be expressed by
$$
I(n)=\frac{1}{n}\int_{0}^{+\infty}\frac{\arctan\le... | Here is yet another approach that uses Feynman's trick of differentiating under the integral sign.
We begin by enforcing a substitution of $x \mapsto x^{1/n}$. This gives
$$\int_0^\infty \tan^{-1} \left (\frac{1}{x^n} \right ) \, dx = \frac{1}{n} \int_0^\infty \frac{\tan^{-1} x}{x^{1 + \frac{1}{n}}} \, dx, \quad n > 1.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $n^2 - 2n + 7$ is even then $n + 1$ is even I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even.
Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$.
What I have done so far:
\begin{align}
& n + 1 = (2k+1)^2 - 2(2... | As pointed out bt @Bram28 and others you use contradiction by making the appropriate changes.
However, I suggest that you try to use contrapositive instead of contradiction, whenever possible as it makes fewer assumptions.
The easiest (and cleanest) way to solve this is proving the contrapositive instead i.e. if $n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Variable inequality: Show that $ {(\sum\limits_{i = 1}^{2n + 1} {{a_i}} )^2} \geqslant 4n\sum\limits_{i = 1}^{n + 1} {{a_i}{a_{i + n}}}.$ For any positive integer $n$, and real numbers (not necessarily positive) $a_1\geqslant a_2 \geqslant …\geqslant a_{2n+1}$, show that $$
{(\sum\limits_{i = 1}^{2n + 1} {{a_i}} )^2} ... | I'm close to a solution,
but I can't go all the way,
so I'll show what I've got
in the hope that
someone else
can complete the proof.
Let
$a_i = a-b_i$,
where
$a = a_1$ and
$b_1 = 0$
so $b_i \ge 0$
and
$b_i \le b_{i+1}$.
The inequality becomes
${(\sum\limits_{i = 1}^{2n + 1} {(a-b_i)} )^2}
\ge 4n\sum\limits_{i = 1}^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
A committee of 4 is to be formed out of 6 superheroes, 7 supervillains, and 4 citizens. How many ways to form a committee? I am, once again, working on a problem which I have a disagreement with the given solution and I was wondering which solution is correct.
A committee of four people is to be formed from a set of 6 ... | Your solution looks good.
You can cross-check against the less-elegant method of assembling valid committees and then excluding the $\binom{15}{2}$ of the Fantastic/Doom constraint.
$\binom{6}{1}\binom{7}{1}\binom{4}{2} + \binom{6}{2}\binom{7}{1}\binom{4}{1} + \binom{6}{1}\binom{7}{2}\binom{4}{1} + \binom{6}{3}\binom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Binomial Harmonic Numbers Prove this equation for $0 \leq m \leq n$:
$$
\frac{1}{\binom{n}{m}}\sum_{k=1}^m \binom{n-k}{n-m} \frac{1}{k} = H_n - H_{n-m}
$$
where $H_k$ denotes the k-th harmonic number $\left(~H_k := \sum_{n=1}^k \frac{1}{n}~\right)$.
Tried to use Abels partial summation $\big(\sum_{k=1}^m a_k b_k = a_... | Prove by Induction:
Base Case: n=0
0 = 0 $\checkmark$
Induction Step:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\begin{eqnarray*}\frac{1}{\binom{n+1}{m}} \sum_{k=1}^m \binom{n+1-k}{n+1-m} \frac{1}{k} &=& \frac{1}{\binom{n+1}{m}} \sum_{k=1}^m \frac{n+1-k}{n+1-m} \binom{n-k}{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Third Moment of Geometric Series? $\sum_{x=1}^{\infty}x(1-p)^{x-1} = \frac{1}{p^2}$
$\sum_{x=1}^{\infty}x^2(1-p)^{x-1} = \frac{2-p}{p^3}$
$\sum_{x=1}^{\infty}x^3(1-p)^{x-1} =$ ... ?
The textbook gives the first 2 moments, but not the third one. I could not find the forumla on Google as well. I think it'll be useful in ... | Using summation of series.
$$S=\sum_{k=1}^\infty x^3(1-p)^{k-1}\\
-(1-p)S=-\sum_{k=1}^\infty (x-1)^3(1-p)^{k-1}$$
We add these two and get
$$pS=\sum_{k=1}^\infty (x^3-(x-1)^3)(1-p)^{1-k} \\ =\sum_{k=1}^\infty (x^2+x(x-1)+(x-1)^2)(1-p)^{1-k} \\ =\sum_{k=1}^\infty x^2(1-p)^k+\sum_{k=1}^\infty (x-1)^2(1-p)^{1-k}+\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove that $x^4 + y^4 - 3xy = 2$ is compact The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.
In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \i... | If you prove that $\lim_{x^2+y^2\to\infty}(x^4+y^4-2xy)=\infty$, then there exists $K>0$ such that $x^2+y^2-2xy>2$ for $x^2+y^2>K$. Hence the solutions of $x^2+y^2-2xy-2=0$ are contained in the circle $x^2+y^2\le K$ and so they form a bounded set.
For the limit, you can use polar coordinates; given $x=r\cos\varphi$ and... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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>Finding range of $f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$
Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$
Try: put $\sin x=t$ and $-1\leq t\leq 1$
So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$
$$2yt^2+8yt+8y=t^2+4t+5$$
$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$
For real roots $D\geq 0$
So $$16(2y-1)... | Let $t=sin(x)$, $$y=\frac{t^2+4t+5}{2t^2+8t+8}=\frac {1}{2}[1+ \frac {1}{(t+2)^2}]$$
Note that $$-1\le t \le 1$$ Thus $$ 1\le y \le5/9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a pattern and prove it by mathematical induction: $1 = 1$
$3 + 5 = 8$
$7 + 9 + 11 = 27$
$13 + 15 + 17 + 19 = 64$
Etc...
I am having trouble seeing a pattern with this, I know it is relatable with Fibonacci Numbers but I am having trouble grasping this topic
| We can continue the above pattern as follows:
$$ 21 + 23 + 25 + 27 + 29 = 125 = 5^3.$$
$$ 31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3. $$
We note that the first odd natural number is $1$, which we can write as $$ 1 = 2(1) - 1;$$
the second odd natural number is $3$, which can be written as
$$ 3 = 2(2) - 1; $$
the third od... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
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Equilateral triangle $ABC$ with $O$ is the circumcenter and M is a point on $(O)$, is $MA^x+MB^x+MC^x$ a constant?
Given an inscribed equilateral triangle $ABC$ with circumcenter $O$. M can be a point on any position of the circle $(O)$, except $A,B,C$. Prove or disprove that $MA^x+MB^x+MC^x$ is a constant for all int... | A necessary condition for the equality to hold in general is that the sums when $\,M =A\,$ and when $\,M=A'\,$ be equal, where $\,A'\,$ is the point diametrically opposed to $\,A\,$:
$$
\begin{align}
0^x + \left(R \sqrt{3}\right)^x + \left(R \sqrt{3}\right)^x = (2R)^x + R^x + R^x \;&\iff\; \left(2^x - 2 \left(\sqrt{3}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How would one solve a linear equation in two integer variables? For example, how would I find integers $a$ and $b$ that satisfy the following equation?
$$5a - 12b = 13$$
I always resorted to trial and error when doing something like this and more often than not I would finally reach my answer. But for this one I just k... | The method here is to repeatedly putting new variables:
$5a-12b=13 \Leftrightarrow 5a=12b+13 \Leftrightarrow a=\frac{12b+13}{5}=2b+2+\frac{2b+3}{5}$
Put $\frac{2b+3}{5}=c$ with $c$ is an integer (because $a$ or $2b+2+c$ is an integer)
$\Leftrightarrow 2b+3=5c \Leftrightarrow 2b-5c=-3$ or $b=\frac{5c-3}{2}$.
$b$ is an i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
How to calculate the summation of $w_i$, where $w_i = {n(n-1)\over (n-i)(i+1)}+{i-1\over i+1}w_{i-1}$ I am having trouble to calculate $$\sum_{i=1}^{n-1} w_i$$
where n is constant and $$w_i = {n(n-1)\over (n-i)(i+1)}+{i-1\over i+1}w_{i-1}$$
I am given the hint that
$$ \sum_{i=1}^{n-1} w_i = n(n-1)\sum_{i=1}^{n-1} {1... | Let have a look at the first terms.
$w_1=\dfrac{n(n-1)}{2(n-1)}+0=\dfrac{n(n-1)}{2}\left[\dfrac 1{n-1}\right]$
$w_1+w_2=w_1+\dfrac{n(n-1)}{3(n-2)}+\dfrac{1}{3}w_1=\dfrac{n(n-1)}{3(n-2)}+\dfrac 43\times\dfrac{n(n-1)}{2(n-1)}\\\phantom{w_1}=\dfrac{n(n-1)}{3}\left[\dfrac 2{n-1}+\dfrac 1{n-2}\right]$
I let you do the calc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Counting Principle - Dependent Events There are 7 red and 5 yellow fish in an aquarium. Three fish are randomly caught in a net. Find the probability that the fish were:
a) All red
b) Not all of the same color.
I have solved this question using the method for dependent events as follows:
a) $\frac{7}{12} \cdot \frac... | a) $$\frac{\binom73\binom50}{\binom{12}3}$$
b) $$1-\frac{\binom73\binom50}{\binom{12}3}-\frac{\binom70\binom53}{\binom{12}3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2678292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$ How can we find the value of $$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$$
My Approach:
By Euler's Theorem: $$\huge \cos \frac{2 \pi k}{11}+i \sin \frac{2 \pi k}{11} =e^{
\frac{i2 \pi k}{11}}$$
$$\therefore \sin \frac... | $$-i e^{
\frac{i2 \pi }{11}} \frac{e^{
\frac{i20 \pi }{11}} -1}{e^{
\frac{i2 \pi }{11}}-1}=
-i \frac{e^{
\frac{i2 \pi }{11}} e^{
\frac{i20 \pi }{11}} -e^{
\frac{i2 \pi }{11}} }{e^{
\frac{i2 \pi }{11}}-1}=-i \frac{1-e^{
\frac{i2 \pi }{11}} }{e^{
\frac{i2 \pi }{11}}-1}=i
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Area of largest inscribed rectangle in an ellipse. Can I take the square of the area to simplify calculations? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can... | Yes, it is valid. You want to determine the maximum of a non-negative function $f$. But, since $f$ is non-negative, asserting the $\max f=M$ is equivalent to asserting that $\max f^2=M^2$. Besides, $f(x)=M\iff f^2(x)=M^2$. So, the points at which the functions $f$ and $f^2$ attain their maximal value are the same.
Note... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Inverse of a modular matrix I have the matrix$$
\begin{pmatrix}
1 & 5\\
3 & 4
\end{pmatrix} \pmod{26}
$$
and I need to find its inverse. I do it according to this website.
I find the modular multiplicative inverse (of the matrix determinant, which is $1×4-3×5=-11$) with the extended Euclid algorithm (it is $-7 \equiv ... | The determinant is $-11 \pmod{26}$.
$$(-7)(-11)=77=26\cdot 3-1\equiv -1 \pmod{26}$$
Hence the inverse is suppose to be $7$ rather than $-7$.
$$7 \begin{bmatrix} 4 & -5 \\ -3 & 4\end{bmatrix} \equiv \begin{bmatrix} 2 & 17 \\ 5 & 3\end{bmatrix} \pmod{26}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2686150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_\limits{(x,y)\to (0,0)}\frac{x^\alpha y^4}{x^2+y^4}$ where $\alpha > 0$ How can we find the following limit? $$\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4}\qquad \alpha>0$$
By using the polar coordinate, we get
$$\lim_{r\to 0}r^{\alpha+2}\frac{\cos^\alpha\theta \sin^4\theta}{\cos^2\theta+r^2\sin^4{\theta... | Just notice that since $\alpha > 0$ and $x^2 \geq 0$ you have that
$$ \frac{x^\alpha y^4}{x^2+y^4} \leq \frac{x^\alpha y^4}{y^4} = x^\alpha \overset{(x,y) \to 0}{\longrightarrow} 0.$$
Thus it is clear that one has $\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4} = 0$. You rarely need polar coordinates for these kinds ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Help with $\int\frac{1+\sin x}{(1+2\cos x)\sin x}\;dx$ $$\int\frac{1+\sin x}{(1+2\cos x)\sin x}\;dx$$
I tried to solve this question of integral many times but I don't understand. How do I solve it?
| This calls for the hyperbolic Kepler angle! We make the substitution
$$\begin{align}\sinh\theta=\frac{\sqrt{e^2-1}\sin x}{1+e\cos x},&&\cosh\theta=\frac{\cos x+e}{1+e\cos x},&&d\theta=\frac{\sqrt{e^2-1}}{1+e\cos x}dx\end{align}$$
With $e=2$ for us. We will also need the inverse relation:
$$\sin x=\frac{\sqrt{e^2-1}\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdo... | $$2079000 = 2^3*3^3*5^3*7*11$$
You want to have at least one of each numbers $2$,$3$,and $5$
Thus you should have
$$3*3*3*(1+1)*(1+1)=108$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
Basis over GF(2^5) with sum of its elements = 1 I need to work over $GF(2^5)$ starting from $GF(2)$ and I would need a basis $\{a_0, a_1, \cdots , a_4\}$ over $GF(2^5)$ such that $a_0+a_1+\cdots +a_4=1$. Do you know how could i proceed?
Thank you!
| The sums of the roots of the primitive polynomials
$$x^5+x^4+x^3+x^2+1, \quad x^5+x^4+x^3+x+1, \quad x^5+x^4+x^2+x+1\tag{1}$$
in $\mathbb F_2[x]$ all equal $1$ (the coefficient of $x^4$ is $1$ in each case) and the five roots are a normal basis of $\mathbb F_{32}$ over $\mathbb F_2$. If you are using $x^5+x^2+1$ for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2692681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of $\tan^2\alpha+\tan^2\beta+\tan^2\gamma$
If $\alpha,\beta,\gamma$ be a variable and $k$ be a constant such that $a\tan\alpha+b\tan\beta+c\tan\gamma=k$.Then find minimum value of $\tan^2\alpha+\tan^2\beta+\tan^2\gamma$ is
Try: Using Cauchy Schwarz Inequality:
$(a^2+b^2+c^2)(\tan^2\alpha+\tan^2\beta+\ta... | Let
\begin{align}
f(x,y,z)&=\tan^2 x + \tan^2 y+\tan^2 z\\
g(x,y,z)&=a\tan x +b\tan y+c\tan z=k
\end{align}
We want to find the minimum of $f$ constrained to $g$, and we can apply Lagrange Multiplier. Consider the equation $\nabla f=\lambda \nabla g$. Since
\begin{align}
\nabla f(x,y,z)&=(2\tan x\sec^2 x, 2\tan y\sec^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric identity. How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$? How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$ ?
My failed take on this matter is:
$$
\sin A =
\sin\frac{A}2 + \sin\frac{A}2 = 2\sin\Big(\frac{\frac{A}2+\frac{A}2}2\Big)\cos\Big(\frac{\frac{A}2-\frac{A}2}2\Big)=
2\sin\Big... | In your work, you used the sum-to-product formula, $$\sin\frac{A}2 + \sin\frac{A}2 = 2\sin\Big(\frac{\frac{A}2+\frac{A}2}2\Big)\cos\Big(\frac{\frac{A}2-\frac{A}2}2\Big),$$
which is correct. The mistake is that $$\sin A\ne\sin\frac{A}2 + \sin\frac{A}2$$
You can still use the sum-to-product formula.
$$\sin A + \sin 0 = 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2700397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Evaluate $\int_0^1 \frac 1 {\lfloor 1-\log_2(x) \rfloor}\,\mathrm{d}x$ Evaluate $$\int_{0}^{1} \frac{1}{\lfloor 1-\log_2(x) \rfloor}\,\mathrm{d}x$$ where $\lfloor \cdot \rfloor$ is the Greatest Integer Function (GIF)
I learnt that the solution is probably $\displaystyle\sum_{r=1}^\infty \frac 1 {r \cdot 2^r}.$
But how... | Let $$f(x)=\frac {1}{\lfloor 1-\log_2 x\rfloor}$$
For $x\in \left(1/2,1\right)$, $f(x)=1$
For $x\in \left(1/4,1/2\right)$, $f(x)=1/2$
For $x\in \left(1/8,1/4\right)$, $f(x)=1/3$
So the integral turns out to be $$\frac 11\cdot\frac 12+\frac 12\cdot\frac 14+\frac 13\cdot\frac 18\cdots= \frac {1}{1\cdot 2^1}+\frac {1}{2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2701236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Formula for the large derivative Is there any formula for the large number derivative?
I need to find $y^{(100)}$ at $x=0$, if $y=(x+1)2^{x+1}$
I tried to find a pattern, but 2nd and 3rd derivatives are already too hairy. I see no pattern, how it evolves.
1st derivative $2^{x+1}+\ln \left(2\right)\cdot \:2^{x+1}\left(x... | Use Taylor series:
$$y=2(x+1)\left(1+x\ln 2+\frac{x^2}{2!}\ln^2 2+\cdots \right)=\\
[2x+\cdots +\frac{2x^{100}}{99!}\ln^{99} 2+O(x^{101})]+\\
[2+\cdots+\frac{2x^{100}}{100!}\ln^{100} 2+O(x^{101})].$$
$$y^{(100)}(0)=200\ln^{99} 2+2\ln^{100} 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2701950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Double integral of maximum function. Let $D:= \lbrace (x,y) \in [0,\infty)^2: 1 \le x^2+y^2\le9 \rbrace$.
Determine the integral :
$$\int\int_D \max(3x^2,y^2)\;dx\,dy.$$
I have a little problem, because I'm not sure where the maximum is $3x^2$ or $y^2$.
| Let $x=r\cos\theta$ and $y=r\sin\theta$, then
$$
\iint_D \max\{3x^2,y^2)\ \mathsf dx\ \mathsf dy = \iint_D \det J\cdot\max\{3r^2\cos^2\theta, r^2\sin^2\theta \}\ \mathsf d\theta\ \mathsf d r,
$$
where
$$
J = \begin{bmatrix}\frac{\partial x}{\partial r}&\frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2702610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What is the limit of $\frac{\left ( 2n \right )!}{(n!)^{2}4^{n}}$? What is the limit of the following sequence?.
$$\frac{\left ( 2n \right )!}{(n!)^{2}4^{n}}$$
I guess the limit is zero but I don't know exactly how to prove it.
Any help please.
| Because I think there are people interested in an elementary solution:
The ratio between terms $a_n=\binom{2n}{n}\frac{1}{4^n}$ is given by
$$\begin{array}{ll} \displaystyle \frac{a_{n+1}}{a_n} & =\frac{\displaystyle\frac{(2n+2)!}{(n+1)!(n+1)!}\frac{1}{4^{n+1}}}{\displaystyle\frac{(2n)!}{n!n!}\frac{1}{4^n}} \\[6pt] & \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to approximate a function involving arc-tangent for a sufficiently large argument? I have a function, $f(x) = {\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x)$. Let's assume the argument $x$ is sufficiently high. If so, how can I approximate $f(x)$ in terms of $x$? I have computed $f(x)$ with MATLAB. It looks like that $f(x... | As Martin Argerami already answered, you are totally write.
If you want more accuracy, you could use Taylor series
$$\sqrt{1+x^2}=x+\frac{1}{2 x}-\frac{1}{8 x^3}+\frac{1}{16
x^5}+O\left(\frac{1}{x^7}\right)$$
$$\sqrt{1+x^2}-x=\frac{1}{2 x}-\frac{1}{8 x^3}+\frac{1}{16
x^5}+O\left(\frac{1}{x^7}\right)$$ Now, using
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $\frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1$, find $\frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}$.
If $\displaystyle \frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1$, find $\displaystyle \frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}$.
I tr... | \begin{cases}
\dfrac{\cos\alpha}{\cos\beta}+\dfrac{\sin\alpha}{\sin\beta}+1=0\\[4pt]
\sin2\alpha + \sin2\beta = 2\sin(\alpha+\beta)\cos(\alpha - \beta),
\end{cases}
\begin{cases}
2\sin(\alpha+\beta) = - \sin2\beta\\
2\sin(\alpha+\beta)(1+\cos(\alpha - \beta)) = \sin2\alpha.
\end{cases}
\begin{align}
\dfrac{\cos^3\beta}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $ $$\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $$
What method should I use to evaluate it. I can't use the ${a^3}$-${b^3}$ formula because it is positive. I also tried to separate limits and tried multiplying with $\frac {\sqrt[3]{(1-x^3)^2}}{\sqrt[3]{(1-x^3... | How about without calculus? Does $x+\sqrt[3]{1-x^3}$ ever cross the x axis?
$$\begin{align}
x+\sqrt[3]{1-x^3}&=0 \\
x&=-\sqrt[3]{1-x^3} \\
x&=\sqrt[3]{x^3-1} \\
x^3&=x^3-1 \\
0&=-1
\end{align}$$
It doesn't. It's also defined for all $x$. And one point on the curve $f(x)=x+\sqrt[3]{1-x^3}$ is $f(1)=1\ge0$. Therefor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimum of $\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}+\frac{z}{\sqrt{1-z^2}}$ Let $0\leq x,y,z<1$ and $x^2+y^2+z^2=1$. What is the minimum value of
$$\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}+\frac{z}{\sqrt{1-z^2}}?$$
From the condition, the point $(x,y,z)$ lies on a sphere with radius $1$. At the equality poin... | By AM-GM $$\sum_{cyc}\frac{x}{\sqrt{1-x^2}}=\sum_{cyc}\frac{2x^2}{2\sqrt{x^2(1-x^2)}}\geq\sum_{cyc}\frac{2x^2}{x^2+1-x^2}=2.$$
The equality occurs for $z=0$ and $x=y=\frac{1}{\sqrt2},$ which says that $2$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$ Give $a, b, c$ be positive real numbers such that $a+ b+ c= 3$. Prove that:
$$\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$$
My try
We have:
$$\left ( a+ b \right )^{2}\geq 4ab$$
$$\l... | Set $$f=27(a+b)^2(b+c)^2(c+a)^2-64abc(a+b+c)^3$$
Assume $a=\max \{a,b,c\}$ thus $$f=27\,(a-b)^2(b-c)^2(c-a)^2+4abc\left[11(a^3+b^3+c^3-3abc)+6\sum_{cyc} c(a-b)^2\right]+108\left\{a^2\Big[ab+c(a-b)\Big](b-c)^2+b^2c^2(a-c)(a-b)\right\} \ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
My Attempt
$$
\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\
\implies \frac{dy}{dx}\bigg[a-\fra... | Both $x,y\in[-1,1]$. So, $x=\sin\theta$ and $y=\sin\phi$ for some $\displaystyle \theta,\phi\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$. Note that $\cos\theta, \cos\phi\ge0$.
We have $$\cos\theta+\cos\phi=a(\sin\theta-\sin\phi)$$
So, $$2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}=2a\cos\frac{\theta+\phi}{2}\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Interesting 3 Variable Inequality for Real Numbers I recently saw an olympiad style inequality that seemed very difficult. I tried to use elementary inequalities such as AM-GM or Cauchy-Schwarz, but neither have helped in making significant progress. Could anyone provide a rigorous proof, preferably using more elementa... | This follows almost immediately by a direct application of Hölder's inequality (generalised Cauchy-Schwarz):
$$\begin{aligned} \left ( \sum \limits_{\text{cyc}} \frac{1}{y(x+y)} \right ) (y + z + x) ((x+y) + (y+z) + (z+x)) &\geq (1^{\frac{1}{3}} + 1^{\frac{1}{3}} + 1^{\frac{1}{3}})^{3}\\
\iff \sum \limits_{\text{cyc}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. Where $a,b,c$ are the sides of triangle and $R$ is the circum radius
My Attempt
From sine law,
$$\dfrac {a}{\sin A}=\dfrac {b}{\sin B}=\dfrac {c}{\sin C}=2R$$
So,
$$a=2R \sin A$$
$$... | HINT:
Use the equality valid for any $A$, $B$, $C$ with sum $\pi$
$$1-(\cos^2 A + \cos^2 B + \cos^2 C)- 2 \cos A \cos B \cos C = 0$$
ADDED: The equality follows from the following formula valid for any angles $\alpha$, $\beta$, $\gamma$ with sum $2s$
$$1-(\cos^2\alpha + \cos^2 \beta + \cos^2 \gamma)- 2 \cos \alpha \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Where’s the error in my factorial proof? On one of our tests, the extra credit was to find which number you would take out from the set $\{1!,2!,3!,...(N-1)!,N!\}$ such that the product of the set is a perfect square, for even $N$ My answer was as follows:
Assume $N$ is even. First note that $(n!)=(n-1)!\cdot n$.
Appl... | You are only showing the case $N/2$ is even, but missing when it is odd.
Alternative to your method, the shortcut is:
$$1!\cdot 2!\cdot 3!\cdot 4!\cdots (N-1)!\cdot N!=(1!)^2\cdot 2\cdot (3!)^2\cdot 4\cdots ((N-1)!)^2\cdot N=(1!3!\cdots(N-1)!)^2\cdot 2^{N/2}\cdot \left(\frac{N}{2}\right)!$$
When $\frac{N}{2}$ is even, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Two disjoint random events You roll twice with four-sided die in which the numbers one and two occur with probability $\frac{1}{3}$, and the numbers three and four each with probability $\frac{1}{6}$. Let X be the number of singles and Y the number of fours that occurred after two throws.
How do I create a table of pr... | You cannot say $P(X,Y)=P(X)\cdot P(Y)$ as these events are not independent.
Further, you need to account for different orderings. For example,
$$P(X=1, Y=1)=2\cdot\left(\frac{1}{3}\cdot\frac{1}{6}\right)=\frac{1}{9}$$
since we can get a one and then a four or a four and then a one.
Similarly
$$P(X=0, Y=1)=2\cdot\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Complete sequence of sequences Who can continue (complete) the following sequence:
$$1,n-1,\frac{(n-2)(n-1)}{2},\frac{(n-4)(n-3)(n+1)}{6},\frac{(n-7)(n-4)(n-2)(n+3)}{24},\dots$$
This was emerging in the course of this question as crucial coefficients in the transformation of Fibonacci polynomials.
I am pretty sure I ha... | As you reference in another question, $$F_n(x) = \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} x^k$$
If we consider $F_n(x+1)$ directly we get $$\begin{eqnarray}
F_n(x+1) & = & \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} (x+1)^k \\
& = & \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} \sum_{i=0}^k \binom{k}{i} x^i \\
& = & \sum_{i=0}^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solve the Goursat problem $xy^3u_{xx} - x^3yu_{yy} -y^3u_x + x^3u_y=0$
Solve the following Goursat problem
$xy^3u_{xx} - x^3yu_{yy} -y^3u_x + x^3u_y=0,$
$u(x,y)=f(x) \; \; on \; \; y^2+x^2=16 \; \; for \; \; 0 \leq x \leq 4
> $-- eq 1
$u(x,y)=g(y) \; \; on \; \; x=0 \; \; for \; \; 0 \leq y \leq4$ -- eq
2
$f(0) = ... | $$xy^3u_{xx} - x^3yu_{yy} -y^3u_x + x^3u_y=0$$
HINT :
We observe that the change of $x$ into $-x$ doesn't change the equation, as well as the change of $y$ into $-y$. This draw us to change of variables :
$\begin{cases}X=x^2\\Y=y^2\end{cases} \quad;\quad u_x=2xu_X\quad;\quad u_y=2yu_Y$
$u_{xx}=2u_X+4x^2u_{XX}\quad;\qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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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.