Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to find the sum of the series $\sum_{k=2}^\infty \frac{1}{k^2-1}$? I have this problem :
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}$$
My solution
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\f... | your choice to use the "telescoping sum" technique is fine. what is wrong is just an arithmetic prob.
since the series is absolutely convergent try first rewriting as two sums:
$$
\sum_{k=1}^\infty \frac{1}{(2k)^2-1} + \sum_{k=1}^\infty \frac{1}{(2k+1)^2-1}
$$
and evaluate these separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to prove the inequality $2\sqrt{n + 1} − 2 \le 1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n} \le 2\sqrt n − 1$? Prove that for any positive integer $n$,
$$2\sqrt{n + 1} − 2 \le 1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n} \le 2\sqrt n − 1$$
Progress
I think Riemann sum should ... | There's a tricky solution using integral estimates:
$$2\sqrt{n+1}-2=\int_1^{n+1}\frac1{\sqrt n}\le\sum_{k=1}^n\frac1{\sqrt n}=1+\sum_{k=2}^n\frac1{\sqrt n}\le 1+\int_1^n\frac1{\sqrt {n}}=1+2\sqrt n-2=2\sqrt n-1$$
But you can easily proceed by induction, the base case is trivial and
$$(2\sqrt{n+1}-1)-(2\sqrt n-1)=2(\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/995110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove statement about determinants. $A$ is a $3\times 3$ matrix over $\mathbb{R}$, I want to show that if $$\det(A + I_3)=\det(A+2I_3),$$ then $$2\det(A+I_3) + \det(A-I_3) + 6 = 3\det A.$$
Can you help me?
| If $a,b,c$ are the eigenvalues of $A$, then $\det(A)=abc$, and $\det(A+dI)=(a+d)(b+d)(c+d)$.
So we are given that
$$
(a+1)(b+1)(c+1)=\det (A+I)=\det (A+2I)=(a+2)(b+2)(c+2),
$$
and hence
$$
(ab+bc+ca)+(a+b+c)+1=2(ab+bc+ca)+4(a+b+c)+8,
$$
or simpler
$$
(ab+bc+ca)+3(a+b+c)+7=0.\tag{1}
$$
We need to show that
$$
2\det (A+I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/995191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Remainders on division of polynomials I am told that the remainder on division of a polynomial $p(z)$ by $z^3+z^2+z+1$ is $z^2-z+1$. I am also given that $p(1)=2$, and then asked to determine the remainder when $p(z)$ is divided by $z^4-1$.
I have expressed $p(z)$ as $p(z) = (z^3+z^2+z+1)(q(z)) + z^2-z+1$, for some pol... | We write the Euclidean division of $q(z)$ by $z-1$
$$q(z)=(z-1)u(z)+v$$
and since $q(1)=\frac14$ so $v=\frac14$ hence
\begin{align}p(z) &= (z^3+z^2+z+1)q(z) + z^2-z+1 \\&= (z^3+z^2+z+1)((z-1)u(z)+\frac14) + z^2-z+1\\&=(z^4-1)u(z)+\underbrace{z^3+\frac54z^2-\frac34z+\frac54}_{\text{the remainder}}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/995765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Implicit Diff. check I have the expression ${\dfrac{x^2+y^2}{x+y}}=3$ and I wanna find $dy/dx$. Here's my approach:
$x^2+y^2=3x+3y$
$\implies (x^2-3x)+(y^2-3y)=0$
$\implies (2x-3)+\dfrac{dy}{dx}(2y-3)=0$
${\implies \dfrac{dy}{dx}=\dfrac{3-2x}{2y-3}}$
Wolfram Alpha gives me a very different answer. This seems to work (i... | implicit differentiating of $\frac{x^2+y^2}{x+y}=3$ gives
$\frac{2x+2yy')(x+y)-(x^2+y^2)(1+y')}{(x^2+y^2)^2}=0$
plugging $x^2+y^2=3(x+y)$ in this term we get
$(2x+2yy')(x+y)-3(x+y)(1+y')=0$
with $x+y\ne 0$ we get $2x+2yy'-3-y'=0$ and this is your result
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/995916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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linear equations in a matrix form Considering
$$x_1 − x_2 + x_3 − x_4 = 2$$
$$x_1 − x_2 + x_3 + x_4 = 0$$
$$4x_1 − 4x_2 + 4x_3 = 4$$
$$−2x_1 + 2x_2 − 2x_3 + x_4 = −3$$
We have the following matrix
$$
\begin{pmatrix}
1 & -1 & 1 & -1 & 2 \\
1 & -1 & 1 & 1 & 0 \\
4 & -4 & 4 & 0 & 4\\
... | Applying Gauss-Jordan to reach an upper triangular matrix:
$$\left(\begin{array}{ccccc} \boxed{1}&-1&1&-1&2\\0&0&0&\boxed2&-2\\0&0&0&0&0\\0&0&0&0&0 \end{array}\right).$$
I have boxed two positions in the matrix. These are called pivots. Recall that when you want set all the positions under a pivot to $0$, you use that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/996544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Line tangent to the circle Find the equation of the lines which are tangent to the circle with equation $x^2+y^2=9$ and parallel to the line of equation $x-y+1=0$
| Lines parallel to $x-y+1$ have equation $y=x+a$ for $a \in \mathbb{R}$, so common point of circle and $y=x+a$ are solutions of equation:
$$x^2+y^2=x^2+(x+a)^2=9$$
If $y=x+a$ is tangent line, then has only one common point with circle, so equation should has only one solution, so you must have $\Delta=0$.So:
$$x^2+(x+a)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find common ratio of $\sum_{k=1}^\infty \frac{1}{k(k+1)}$ I have this problem, I need to find the sum.
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{k(k+1)}$$
The problem is that the ratio is not conclusive, Any idea how to find the ratio?
Thanks!
| Since you speak of the ratio test, maybe you're only concerned with whether the series converges rather than with what the sum is. That is the most that the ratio test can give you. If that is what you're concerned with, then you can say
$$
\sum_{k=1}^\infty \frac 1 {k(k+1)} \le \sum_{k=1}^\infty \frac 1 {k^2}
$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.
Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ im... | $$x^2 \geq 0 \implies 1+x^2 > 1 \implies \frac{1}{1+x^2} < 1$$
Using the above inequality,
$$
\begin{align}
|f(x)-f(a)| &\leq |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}\\
&\leq |x - a||x+a|\\
&\leq |x - a|(|x|+|a|)
\end{align}
$$
Choose $$|x-a| \leq 1 \implies |x|\leq |a|+1$$
Now,
$$
\begin{align}
|f(x)-f(a)| &\leq |x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 5
} |
derivative of $y=\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$ $y=\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$
The answer is $\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5} \left(\dfrac{2}{x}+\dfrac{1}{2(x+1)}-\dfrac{1}{x+2}-\dfrac{5}{x-3}\right)$
I know that the quotient rule is used but I don't know how to do this problem. Would you multipl... | $$y'=\dfrac{(x^2\sqrt{x+1})'(x+2)(x-3)^5-x^2\sqrt{x+1}((x+2)(x-3)^5)'}{(x+2)^2(x-3)^{10}}=
\dfrac{((2x\sqrt{x+1}+\frac{x^2}{2\sqrt{x+1}})(x+2)(x-3)^5-x^2\sqrt{x+1}((x-3)^5)+5(x+2)(x-3)^4)}{(x+2)^2(x-3)^{10}}=\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}(\dfrac{2}{x}+\dfrac{1}{\sqrt{x+1}\sqrt{x+1}}-(\dfrac{1}{x+2}+\dfrac{5}{x-3})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Proving/disproving that $1=\frac {1}{2}(x)^{-\frac {1}{2}} \Longleftrightarrow 2\sqrt{x}=0$ I was looking at the solution to this problem:
Find $\frac {dy}{dx}$ for $x=10+\sqrt {x^2+y^2}$.
The solution given was as follows:
$$
\frac {d}{dx}(x)=\frac {d}{dx}(10)+\frac {d}{dx}(\sqrt {x^2+y^2})
$$
$$
1=\frac {1}{2}(x^2+y^... | Rewriting from step 2 might help:$$1=\frac {1}{2}(x^2+y^2)^{-\frac {1}{2}}(2x+2y\frac {dy}{dx})$$$$1=\frac {1}{2\sqrt{x^2+y^2}}(2x+2y\frac {dy}{dx})$$$$1=\frac{(2x+2y\frac {dy}{dx})}{2\sqrt{x^2+y^2}}$$Now just multiply both sides by $2\sqrt{x^2+y^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I want help to Prove $\pi \tanh(\frac{\pi }{2})=\sum_{n=0}^{\infty }\frac{1}{n^2+n+0.5}$ It is difficult, so I need how to start to prove it
$$\pi \tanh(\frac{\pi }{2})=\sum_{n=0}^{\infty }\frac{1}{n^2+n+0.5}$$
| I am just mimicking the answer of @robjohn for this question, all credits go to him
let $x^2 + x + 0.5 = (x-a)(x-b)$, then we have $a+b = -1$, i.e. $b = -1-a$
\begin{align}
&\sum_{n=0}^{+\infty}\dfrac{1}{n^2 + n + 0.5} \\
= &\sum_{n=0}^{+\infty}\dfrac{1}{(n-a)(n-b)}\\
= & \dfrac{1}{a-b}\sum_{n=0}^{+\infty}\left(\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving equations with mod So, I'm trying to solve the following equation using regular algebra, and I don't think I'm doing it right: $3x+5 = 1\pmod {11}$
I know the result is $x = 6$, but when I do regular algebra like the following, I do not get 6:
$3x=1 - 5\pmod{11}$
$x = \dfrac{(1 \pmod{11} - 5)} 3$
So, I figured... | When we are working with congruence, $\text{mod } n$, for some $n$ (here, 11), we are interested in integral relations, i.e., relationships between integer values, and instead of division, we use the fact, in this case, that $3x +5= 11k + 1$, where $k$ is any integer (the quotient), and $1$ is the remainder when $3x +5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Inequality proof by induction, what to do next in the step I have to prove that for $n = 1, 2...$ it holds: $2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$
Base: For $n = 1$ holds, because $2\sqrt{2}-2 < 1$
Step: assume holds for $n_0$.
$2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt2} + \f... | \begin{align}
\frac1{\sqrt{n+1}}+2\sqrt{n+1}-2&=\frac{1+2(n+1)}{\sqrt{n+1}}-2\\
&=\frac{\sqrt{4n^2+12n+9}}{\sqrt{n+1}}-2\\
&>\frac{\sqrt{4n^2+12n+8}}{\sqrt{n+1}}-2\\
&=\frac{2\sqrt{n^2+3n+2}}{\sqrt{n+1}}-2\\
&=\frac{2\sqrt{(n+2)(n+1)}}{\sqrt{n+1}}-2\\
&=2\sqrt{n+2}-2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}......
$$
and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$
Now,find the positive integer numbers $(a_{0}, a_{... | $$\begin{align}\frac{\pi^2}{9}&=1+\frac{1}{3}\sum_{k=0}^\infty\frac{1}{(k+1)^2(k+2)^2}\\&=1+\frac{1}{3(1·2)^2}+\frac{1}{3(2·3)^2}+\frac{1}{3(3·4)^2}+\frac{1}{3(4·5)^2}+\ldots\\&=1+\frac{1}{12}+\frac{1}{108}+\frac{1}{432}+\frac{1}{1200}\ldots...\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
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Trigonometric equation, missing some solutions I'm missing part of the answer, and I'm not quite sure why. The given answer doesn't even seem to hold...
Solve for x: $$\tan 2x = 3 \tan x $$
First some simplifications:
$$\tan 2x = 3 \tan x $$
$$\tan 2x - 3 \tan x = 0$$
$$\frac{\sin 2x}{\cos 2x} - \frac{3 \sin x}{\cos ... | Hint:
You should use $$\tan 2x=\frac{2\tan x}{1-\tan^2x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Volume of rotated region (integration) Let $T$ be a right-angled triangular region with vertices $(0,−b)$,$(1,0)$ and $(0,a)$ where $a$ and $b$ are positive numbers. When $T$ is rotated about the line $x=2$, it generates a solid with volume $V=\dfrac{410\pi}{27}$
Find $a$ and $b$.
Really having trouble with this one. I... | Your rotated volume is made of two parts ($y>0$ and $y<0$): both are made of a cylinder minus a truncated cone. Thus the volume is easy to compute, if you know $a$ and $b$: this gives you an equation between $a$ and $b$.
For the second equation, you know your triangle is right angled: the right angle must be at vertex ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Indefinite integral of $\frac{\arctan x}{x^2+1}$ EDIT: I was studying from a site that uses really ambiguous notation so I misread $\arctan\ (x)^2$ as $\arctan\ (x^2)$. Now I can see why the integral is actually $\frac{1}{2} \arctan^2\ x + c $. Thanks to everyone who answered and corrected me!
Why is $$\int\frac{\arcta... | instead of $\int u\ du$ where $u=\arctan\ x$ and $du=\frac{1}{x^2+1} dx$, thus $$\int\frac{\arctan\ x}{x^2+1} dx=\int u du= \frac12u^2+C=\frac12(\arctan x)^2+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How find this maximum of this $(1-x)(1-y)(10-8x)(10-8y)$
let $x,y\in (0,1)$, and such
$$(1+x)(1+y)=81(1-x)(1-y)$$
Prove
$$(1-x)(1-y)(10-8x)(10-8y)\le\dfrac{9}{16}$$
I ask $\dfrac{9}{16}$ is best constant?
PS:I don't like Lagrange Multipliers,becasue this is Hight students problem.
My idea:
$$(1-x)(1-y)(10-8x)... | Hint. Let $X=\frac{1+x}{1-x},\ Y=\frac{1+y}{1-y}$. We have $XY=81$ and $1\le X,\ 1\le Y$. Since $$(1-x)(10-8x)=\left(1-\frac{X-1}{X+1}\right)\left(10-8\frac{X-1}{X+1}\right)=\frac{4(X+9)}{(X+1)^2},$$ what we need to show is
$$\frac{(X+9)}{(X+1)^2}\cdot\frac{(Y+9)}{(Y+1)^2}\le \frac9{256}$$
and we would like to know ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Alternative ways to evaluate $\int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$ In the following link here I found the integral & the evaluation of
$$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$
I'll also include a simpler version together with the question: is it possible to find some easy
ways of computing both i... | Here is a slightly different approach to evaluating
$$\int^1_0 \frac{\text{Li}^2_2 (x)}{x} \, dx,$$
which, like the answer given by @Zaid Alyafeai, makes use of the result
$$\sum^\infty_{n = 1} \frac{H_n}{n^4} = 3 \zeta (5) - \zeta (2) \zeta (3).$$
We start by writing
$$I = \int^1_0 \frac{\text{Li}^2_2 (x)}{x} \, dx = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
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Solve this triple integral I am trying to solve this triple integral:
$$\int_0^{ 2\pi} \int_0^4 \int_3^\sqrt{25-r^2} r \: dz \: dr \: d \theta$$.
I get stuck at
$$\int_0^{ 2\pi} \int_0^4 \ r(\sqrt{25-r^2}-3) \: dr \: d \theta$$
Should I use u-sub to find the rest? And if so, how?
| Firstly, note that
$$\int_0^{2 \pi} \int_0^4 r(\sqrt{25-r^2} -3)dr d \theta = 2 \pi \int_0^4 r(\sqrt{25-r^2} -3)dr$$
Now, call $5 \sin u = r$. So
$$\int_0^4 r(\sqrt{25-r^2} -3)dr = \int_0^{\arcsin \frac{4}{5}} 5 \sin u(5 \sqrt{1- \sin^2 u} -3) 5 \cos u \ du =$$
$$= 25 \int_0^{\arcsin \frac{4}{5}} \sin u \cos u (5 \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
On sums and identities I am given the following problem set:
(a) the Riemann $\zeta$-function for $s > 1$ is defined through the convergent sum: $$\zeta(s) := \sum_{n = 1}^{\infty} \frac{1}{n^s}$$
show the identity $$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{3}{4}\zeta(2)$$
(b) show that $$\sum_{n=1}^{\infty} \... | In the first identity you are interested in the infinite sum of just the odd integers, where $s = 2$. One way you can get this is to observe that $$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \sum_{n=1}^{\infty} \frac{1}{n^2}-\sum_{n=1}^{\infty} \frac{1}{(2n)^2} \\ =\zeta(2)-\sum_{n=1}^{\infty} \frac{1}{4n^2} \\ = \zeta(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Tetrahedron- Inscribed Sphere In tetrahedron $ABCD, AB=BC=CA$ and $DA=DB=DC$. Given that the altitude of $ABCD$ from point $D$ is $24$ and that the radius of the inscribed sphere of $ABCD$ is $11$, determine $AB$.
| If we call $A1$, $A2$, $A3$, and $A4$ the areas of the faces of a tetrahedron, the radius $R$ of the inscribed sphere can be calculated using the formula
$$V = \frac{1}{3} R (A1 + A2 + A3 + A4)$$
where V is the volume of the tetrahedron.
In this case, calling $s$ the length of $AB=BC=CA$, we have that:
*
*the area ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How prove this inequality $\left[\frac{n}{\sqrt{3}}\right]+1>\frac{n^2}{\sqrt{3n^2-5}}$
let $k$ is postive integer,and for any postive integer $n\ge 2$,
show that:
$$\left[\dfrac{n}{\sqrt{3}}\right]+1>\dfrac{n^2}{\sqrt{3n^2-5}}>\dfrac{n}{\sqrt{3}}$$
where $[x]$ is the largest integer not greater than $x$
| Let $q_n = \left\lfloor \frac{n}{\sqrt{3}}\right\rfloor + 1$. When $n \ge 11\sqrt{3}$, we have
$$\frac{n}{\sqrt{3}q_n} \ge \frac{\frac{n}{\sqrt{3}}}{\frac{n}{\sqrt{3}}+1} = 1 - \frac{\sqrt{3}}{n+\sqrt{3}} \ge \frac{11}{12}
\quad\implies\quad \frac{n^2}{3q_n^2} \ge \left(\frac{11}{12}\right)^2 > \frac{5}{6}
$$
Be defini... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the surface integral from the paraboloid
Evaluate the surface integral $$\iint\limits_S xy \sqrt{x^2+y^2+1}\,\mathrm d\sigma,$$ where $S$ is the surface cut from the paraboloid $2z=x^2+y^2$ by the plane $z=1$.
Is it possible for the answer to be $0$ ? I am not so sure. Would anyone mind telling me the answer... | Notice that your bounds are arranged as such:
\begin{equation}
2z=x^2+y^2\bigg|_{z=1}\implies y=\pm\sqrt{2-x^2},
\end{equation}
therefore the bounds on your integrals are
\begin{equation}
\int_{-\sqrt{2}}^{\sqrt{2}}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}xy\sqrt{x^2+y^2+1}\:\:dy\:dx,
\end{equation}
because it appears as tho... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How does $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)}$ simplify? $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + ...$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... = 1$
My professor wrote this the other day. But I'm wondering...how does the series become $= 1 - \frac{1}{2} + \frac{1... | Write
$$\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1},$$
split the sum into two sums, shift the index of one, recombine, simplify, and you are done.
N.B. You need to do this for partial sums and only at the end take the limit to be rigorous.
| {
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Find the volume of the region bounded by the planes $ z=8-y^2, y = 8-x^2, x=0, y=0, z=0$ I figured out the bounds for z:
$z=0$ to $z=8-y^2$
The bounds for y:
$y=0$ to $y=8-x^2$
The bounds for x:
$x=0$ to $x=\sqrt{8}$ (Since $8-x^2 = 0$)
So, the volume by using triple integral:
$\int_{0}^{2\sqrt{2}}\int_{0}^{8-x^2}\i... | For $x : 0 < x < \sqrt{8-2\sqrt{2}}$, the upper boundary in $y$ is $2\sqrt{2}$ not $8-x^2$ (which is bigger). The $z = 8 - y^2$ cylinder has "chopped off" part of the $y = 8 - x^2$ cylinder.
So the integral should be:
$I = \displaystyle\int\limits_{0}^{\sqrt{8-2\sqrt{2}}}\int\limits_{0}^{2\sqrt{2}}\int\limits_{0}^{8-y^... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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How to evaluate $\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$ Im tempted to say that the limit of this sequence is 1 because infinite root of infinite number is close to 1 but maybe Im mising here something? What will be inside the root?
This is the sequence: ... | You may write
$$
\begin{align}
\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n} &=\frac{1\cdot 2\cdot 3\cdot 4 \cdot 5\cdot6\cdots(2n-1)\cdot 2n}{(2\cdot 4\cdot 6\cdots (2n))^2}\\
&=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2}\\
& =\frac{(2n)!}{2^{2n} (n!)^2 }\\
& =\frac{1}{\sqrt{\pi n}}+\... | {
"language": "en",
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"source": "stackexchange",
"question_score": "13",
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How can I evaluate $\lim_{n \to\infty}\left(1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)\right)/\left(1^2+2^2+3^2+\dots+n^2\right)^2$? How can I evaluate this limit? Give me a hint, please.
$$\lim_{n \to\infty}\frac{1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)}{\left(1^2+2^2+3^2+\dots+n^2\right)^2}$$
| $$ \lim_{n\to \infty} \frac{\sum\limits_{k=1}^n k(k+1)(k+2)}{\left[\sum\limits_{k=1}^n k^2\right]^2}= \lim_{n\to \infty} \frac{\sum\limits_{k=1}^n k^3+3\sum\limits_{k=1}^n k^2+2\sum\limits_{k=1}^n k}{\left[\sum\limits_{k=1}^n k^2\right]^2} = \lim_{n\to \infty} \frac{\frac{n^2(n+1)^2}{4}+ \frac{n(n+1)(2n+1)}{2}+n(n+1)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$ So, a,b,c are non-negative real numbers for which holds that $a+b+c=3$.
Prove the following inequality: $$4\ge a^2b+b^2c+c^2a+abc$$
For now I have only tried to write the inequality as $$4\left(\frac{a+b+c}3\right)^3\ge a^2b+b^2c+c^2a+abc$$ but I don't know what to do af... | So we need to show
$$4(a+b+c)^3 \ge 27(a^2b+b^2c+c^2a+abc)$$
One way is to use the cyclic symmetry and WLOG assume $a$ to be the min of $a, b, c$. Then we can write $b = a+x, c = a+y$, where $x, y \ge 0$. Now the inequality reduces to
$$9a(x^2-xy+y^2)+(x-2y)^2(4x+y) \ge 0$$
which is obvious. Also from the above, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Proving binomial coefficients identity: $\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}$
Let $n$ and $r$ be positive integers with $n \ge r$. Prove that:
$$\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}.$$
Tried proving it by induction but got stuck. Any help with p... | Proving by induction
inductive step
$$\begin{pmatrix}r\\r\end{pmatrix} + \begin{pmatrix}r+1\\r\end{pmatrix} + \dots + \begin{pmatrix}n\\r\end{pmatrix} =\begin{pmatrix}n\\r+1\end{pmatrix} + \begin{pmatrix}n\\r\end{pmatrix}$$ [as the identity holds for natural numbers less than n.]
We Know that, $$\begin{pmatrix}n\\r+1\e... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how many positive integer solutions to the following equation? $a^2 + b^2 + 25 = ab + 5a + 5b$
I have tried looking for a factorisation that could solve this question but couldn't find anything useful - found $(a+b+5)^2$ - don't know if this is useful
The equation does look similar to an equation of a circle - can you ... | Let $x = a+b \to x^2 - 2ab + 25 = ab + 5x \to x^2 -5x + 25 = 3ab \leq 3\cdot \dfrac{(a+b)^2}{4} = \dfrac{3x^2}{4} \to 4x^2 - 20x + 100 \leq 3x^2 \to x^2 - 20x + 100 \leq 0 \to (x-10)^2 \leq 0 \to x = 10, a = b = \dfrac{10}{2} = 5$.
| {
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Use induction to prove $\sum_{k=1}^n\frac{k}{2^k}=2-\frac{n+2}{2^n}$
Use induction on $n\in\Bbb N$ to prove that $$\sum_{k=1}^n\frac{k}{2^k}=2-\frac{n+2}{2^n}\;.$$
I have got as far as to the induction step where I have:
$$S(n+1)= 2-\frac{n+3}{2^{n+1}}$$ and this should be equal to
$$S(n) +\frac{n+1}{2^{n+1}} = 2-\fr... | When making a common denominator, don't forget to distribute the negative sign!
$$
\frac{-(n + 2)}{2^n} + \frac{n+1}{2^{n+1}}
= \frac{-2(n + 2) + (n + 1)}{2^{n+1}}
= \frac{(-2n - 4) + (n + 1)}{2^{n+1}}
= \frac{-n - 3}{2^{n+1}}
= -\frac{n + 3}{2^{n+1}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate these indefinite integrals with $\sqrt{1+x^4}$? These integrals are supposed to have an elementary closed form, but Mathematica only returns something in terms of elliptic integrals. I got them from the book Treatise on Integral Calculus by Edwards. How can we evaluate them?
$$
I = \int{\frac{\sqrt{1+x... | The problem is on p. 319 in the 1921 edition of Volume I. Write
$$
I = \int \frac{1+x^4}{(1-x^4)} \frac{dx}{\sqrt{1+x^4}}
,\qquad
J = \int \frac{x^2}{(1-x^4)} \frac{dx}{\sqrt{1+x^4}}
,
$$
and note that $I=\frac{1}{2}(A+B)$ and $J=\frac{1}{4}(A-B)$, where
$$
A = \int\frac{1+x^2}{1-x^2} \frac{dx}{\sqrt{1+x^4}}
,\qquad
B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041531",
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"source": "stackexchange",
"question_score": "10",
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Help me, a doubt $f(x)=\cot^{-1} \frac{1-x}{1+x}$ I have a doubt
$$f(x)=\cot^{-1} \frac{1-x}{1+x}$$
$$f´(x)=\frac{1}{(\frac{1-x}{1+x})^2}\cdot\frac{(-1)(1+x)-(1-x)}{1+\frac{(1-x)^2}{(1+x)^2}}$$
mm this could to be really easy but I do not understand in the first denominator gives one, someone who can explain,
| The derivative is incorrect.
The usual way is to use the chain rule on $\cot^{-1}$ then the quotient rule on the fraction. You would then get
$$f'(x)=(\cot^{-1})'\left( \frac{1-x}{1+x} \right) \cdot \left( \frac{1-x}{1+x} \right)'$$
$$=-\frac{1}{1+\left(\frac{1-x}{1+x} \right)^2} \cdot \frac{(1-x)'(1+x)-(1-x)(1+x)'}{... | {
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How to get $\sqrt {k} + \frac{1}{\sqrt{k+1}}$ in the form $\frac{\sqrt{k^2} + 1}{\sqrt{k+1}}$? I was wondering if it is possible to get $\sqrt {k} + \dfrac{1}{\sqrt{k+1}}$ in the form $\dfrac{\sqrt{k^2} + 1}{\sqrt{k+1}}$, and if so, how?
I ask this, because I'm following this answer, and I get lost at how they arrive a... | That answer says
$$\sqrt {k} + \frac{1}{\sqrt{k+1}}= \frac{(\sqrt {k})(\sqrt {k+1}) +1}{\sqrt{k+1}}=\frac{\sqrt {k(k+1}) +1}{\sqrt{k+1}}=\frac{\sqrt {k^2+\color{red}{k}} +1}{\sqrt{k+1}}\color{blue}{\ge}\frac{\sqrt {k^2} +1}{\sqrt{k+1}}$$
It's just like saying $\sqrt{a+b}\ge\sqrt{a}$ for $b\ge0$
Answer to Edit:
Autho... | {
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Let f(x) be a non negative continuous function on R such that $f(x) +f(x+\frac{1}{3})=5$ then calculate ........ Problem :
Let f(x) be a non negative continuous function on R such that $f(x) +f(x+\frac{1}{3})=5$
then calculate the value of the integral $\int^{1200}_0 f(x) dx$
My approach :
Given that : $f(x) +f(x+\f... | Check this
$\int_0 ^{1200}f=\int_0 ^{\frac{2}{3}}f+\int_{\frac{2}{3}}^{\frac{4}{3}}f+...$
$\int _{\frac{2}{3}}^{\frac{4}{3}}f(x)=\int_0 ^{\frac{2}{3}}f(x)$
$\int _0^ {1200} f=800\times \int _0^{\frac{2}{3}}f$
$\int _0^{\frac{2}{3}}f=\int _0^{\frac{1}{3}}f+\int _{\frac{2}{3}}^{\frac{4}{3}}f$
Let $y=x-{\frac{1}{3}} \impl... | {
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"url": "https://math.stackexchange.com/questions/1046573",
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"source": "stackexchange",
"question_score": "5",
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Factor $55 - 88 \sqrt{-2}$ as a product of primes in $\mathbb{Z}[\sqrt{-2}]$ To solve this problem, I let $K = \mathbb{Q}(\sqrt{-2})$, and I thought to take the norm $$N(55 - 88 \sqrt{-2}) = 55^2 + 2 \cdot 88^2 = 18513 = 3^2\cdot11^2 \cdot 17$$ If $a \in \mathbb{Z}[\sqrt{-2}]$ is irreducible, then $N(a) = p^f$, where $... | $55 - 88\sqrt{-2} = 11(5 - 8\sqrt{-2})$ so $11=\alpha_2\sigma(\alpha_2)$ is a factor, rather than $\alpha_2^2$ or $\sigma(\alpha_2)^2$, and we only need to factor $5 - 8\sqrt{-2}$.
I think the best thing to do now is to try dividing by the possible factors and check if the result is in $\Bbb{Z}[\sqrt{-2}].$
$$\frac{5 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047337",
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What am I doing wrong? (Trigonometric Identity) $$\frac { \cos\theta }{ 1-\sin\theta } =\frac { \sin\theta -\csc\theta }{ \cos\theta -\cot\theta } $$
Steps I took:
$$\frac { \sin\theta -\frac { 1 }{ \sin\theta } }{ \cos\theta -\frac { \cos\theta }{ \sin\theta } } $$
$$\frac { \frac { \sin^{ 2 }\theta -1 }{ \si... | You've already solved the problem:
$$\begin{align}
\dfrac { (\sin^{ 2 }\theta -1) }{ (\cos\theta )(\sin\theta -1) } &= \dfrac{-\cos^2\theta}{(\cos\theta)(\sin\theta - 1)} \\
&= \dfrac{-\cos\theta}{\sin\theta-1} \\
&= \dfrac{\cos \theta}{-(\sin\theta-1)} \\
&= L.H.S.
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Did I correctly verify the convergence of this series? I want to find if the following series is convergent.
$$\sum_{n=1}^\infty \frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} $$
I use the asymptotic criterion for series convergence.
$$ a_n=\frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} $$
I take such $b_n$ that $a_n$ and $b_n... | Your conclusion is correct. You could get there using the comparison test too if you are interested. Namely, that $$1 < \left(1+\frac{1}{n} \right)^n \\ \frac{1}{n^3+3n^3+n^3}\leq \frac{1}{n^3+3n^2+1}$$ for all $n \geq 1$. Hence, $$\sum_{n=1}^\infty \frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} \geq \sum_{n=1}^\infty \f... | {
"language": "en",
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Proving $\lim_{x \to -2}$ $\frac{(x-1)}{(x+1)}$ = 3 by definition Here we go. $\lim_{x \to -2}$ $\frac{(x-1)}{(x+1)}$ = 3
Proof: $\vert \frac{x-1}{x+1} -3 \vert$ =...= $\frac{\vert-2\vert \vert x-(-2) \vert}{(x+1)}$$\lt \frac{1}{2} \vert x -(-2) \vert$...now let $\epsilon$ $\gt$ 0 arbitrary and $\delta$ = min{1,2$\eps... | If your $\delta$ is $\min\{1, 2\epsilon\}$ then it is not always true that $$\frac{2|x+2|}{|x+1|} < \frac{1}{2}|x+2|.$$
When $\epsilon$ is large, then $0 < |x + 2| < \delta = 1$ implies only that $-1 < x + 2 < 1$, or $-2 < x + 1 < 0$, so that $2 > |x+2| > 0$ and $1/2 < 1/|x+2|$ is not bounded above. The problem is $|x ... | {
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Multinomial Expansion Question: when variables appear more than once Find the coefficient of $x^{12}y^{24}$ in $(x^3 + 2xy^2 +y + 3)^{18}$.
I have been working on this problem for a while now and I cannot figure out how to use the multinomial theorem to solve it. I tried using auxiliary variables but then I did not kno... | Break it into cases based on how the $x^{12}$ was formed from some combination of $x^3$ and $2xy^2$. In each case, you must also check that it is possible that $y$ could be raised to the 24th power using the remaining available options.
Case 1: $(x^3)^4$: You might have had an $x^{12}y^{24}$ term because of $(x^3)^4\c... | {
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Solving $2\cos^2 x-2\sin^2 x-2\cos x=0$ $$f(x) = 2\cos^2 x-2\sin^2 x-2\cos x$$
Need values of x that which make $f(x) = 0$
Tried $a^2-b^2 = (a+b)(a-b)$ with no luck
Really just need a hint that could bring me in the right direction
Thanks
EDIT: Solution thanks to everyones help! :D
$$f(x) = 2\cos^2 x-2\sin^2 x-2\cos x... | We have $$\cos x=\cos^2x-\sin^2x=\cos2x$$
$$\implies2x=2m\pi\pm x$$ where $m$ is any integer
Taking the '-' sign, $3x=2m\pi\iff x=\dfrac{2m\pi}3\ \ \ \ (1)$
Taking the '+' sign, $x=2m\pi $ which is a proper subset of $(1)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Laurent series of $f(z)=\frac{1}{z(z-1)}$ given four different conditions Expand $f(z)=\frac{1}{z(z-1)}$ in a Laurent series valid for the follwing annular domains.
$a)0\lt \vert z \rvert \lt 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)1\le\lvert z \rvert\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c)0\le \lvert z-1 \rvert \lt1\,\... | For (c)
$$\frac{1}{z(z-1)}= \frac1{z-1}\frac{1}{1+(z-1)}= \frac1{z-1}\sum_{k=0}^{\infty}(-1)^k(z-1)^{k}$$
For (d)
$$\frac{1}{z(z-1)}= \frac{\frac1{(z-1)^2}}{1+\frac1{z-1}}=\frac1{(z-1)^2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(z-1)^k}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of the sum $$\arctan\left(\dfrac{1}{2}\right)+\arctan\left(\dfrac{1}{3}\right)$$
We were also given a hint of using the trigonometric identity of $\tan(x + y)$
Hint
$$\tan\left(x+y\right)\:=\:\dfrac{\tan x\:+\tan y}{1-\left(\tan x\right)\left(\tan y\right)}$$
| Yep, sum of tangents identity is the way to go. Write $x=arctan\frac{1}{2}$ and $y=arctan\frac{1}{3}$.
Then $tan(arctan\frac{1}{2}+arctan\frac{1}{3})=\frac{tan(arctan\frac{1}{2})+tan(arctan\frac{1}{3})}{1-(tan(arctan\frac{1}{2}))(tan(arctan\frac{1}{3}))}$
This is equal to $\frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2}... | {
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Find the minimum possible value of $x(1-z)+y(1-x)+z(1-y)$ It is given that $$xyz=(1-x)(1-y)(1-z)$$ and $$x, y, z \in (0,1)$$
Find the minimum possible value of the expression: $$x(1-z)+y(1-x)+z(1-y)$$
Using the AM-GM inequality concepts, I can write that the value is minimum when
$$x(1-z)=y(1-x)=z(1-y)$$
What else can... | From $xyz=(1-x)(1-y)(1-z)$ we get
$$1-2xyz=\sum_{cyc}x(1-y)\geq 3\sqrt[3]{xyz(1-x)(1-y)(1-z)}=3\sqrt[3]{x^2y^2z^2},$$
Which, on setting $xyz=\dfrac {t^3}{8},$ gives us $4\ge t^3+3t^2$. This factors into
$$0\ge (t-1)(t+2)^2,$$
Hence $t\le 1$ and $xyz\le \dfrac 18$. So,
$$\sum_{cyc}x(1-y)=1-2xyz\ge\frac34.$$
Equality occ... | {
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asymptotics of sum I wanna find asymptotic of sum below
$$\sum\limits_{k=1}^{[\sqrt{n}]}\frac{1}{k}(1 - \frac{1}{n})^k$$
assume I know asymptotic of this sum (I can be wrong):
$$\sum\limits_{k=1}^{n}\frac{1}{k}(1 - \frac{1}{n^2})^k \sim c\ln{n}$$
So I use Stolz–Cesàro theorem and wanna show that
$$\sum\limits_{k=1}^{[\... | One error that I see
is your computation of
$x_n-x_{n-1}$.
You have
$x_n = \sum\limits_{k=1}^{[\sqrt{n}]}\frac{1}{k}(1 - \frac{1}{n})^k$.
Note that the individual terms
depend on both $k$ and $n$.
Therefore,
$\begin{array}\\
x_n-x_{n-1}
&=\sum\limits_{k=1}^{[\sqrt{n}]}\frac{1}{k}(1 - \frac{1}{n})^k
-\sum\limits_{k=1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding exact value of trigonometric functions I was wondering, how do I get the exact fraction (the value) of this trigonometric function:
$$\cos\left(\sin^{-1}(12/13)+\sin^{-1}(4/5)\right)$$
Usually, I would evaluate the inverse sin in degree mode and multiply (by hand) by $\pi/180$.
But in this case, I don't get ex... | We have:
$$\arcsin\frac{4}{5} = \arg(3+4i),\qquad \arcsin\frac{12}{13}=\arg(5+12i), $$
hence:
$$\arcsin\frac{4}{5}+\arcsin\frac{12}{13}=\arg((3+4i)(5+12i))=\arg(-33+56i)$$
and:
$$ \cos\left(\arcsin\frac{4}{5}+\arcsin\frac{12}{13}\right)=\frac{-33}{\sqrt{33^2+56^2}}=-\frac{33}{65}.$$
Avoiding complex numbers.
$$\arcsin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Question on the sum $\sum_{n=1}^{\infty}\frac{x^n}{n} = -\ln(1-x)$ $f(x) = \displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + ... = -\ln(1-x)$ for $|x| < 1$.
$f'(x) = \displaystyle\sum_{n=1}^{\infty}x^{n-1} = 1 + x + x^2 + x^3 +... = \frac{1}{1-x}$ for $|x| < 1$
If $f'(x) = \displaystyl... | You're right: $f'(x)=\frac1{1-x}$, so $f(x)$ must be $-\ln(1-x)+C$ for some $C$. Now, to find what $C$ is, try letting $x$ be $0$ and solving for $C$.
Or, if you want to do a definite integral, start with:
$$1+t+t^2+\dotsb=\frac1{1-t}$$
Take the definite integral from $0$ to $x$:
\begin{align}
\int_0^x(1+t+t^2+\dotsb)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How find $a,b$ if $\int_{0}^{1}\frac{x^{n-1}}{1+x}dx=\frac{a}{n}+\frac{b}{n^2}+o(\frac{1}{n^2}),n\to \infty$ let
$$\int_{0}^{1}\dfrac{x^{n-1}}{1+x}dx=\dfrac{a}{n}+\dfrac{b}{n^2}+o(\dfrac{1}{n^2}),n\to \infty$$
Find the $a,b$
$$\dfrac{x^{n-1}}{1+x}=x^{n-1}(1-x+x^2-x^3+\cdots)=x^{n-1}-x^n+\cdots$$
so
$$\int_{0}^{1}\dfrac... | Thank you,I have use two parts integral following
$$\int_{0}^{1}x^nf(x)dx=\dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o(1/n^2)$$
where $f\in C^{2}[0,1]$
$$I=\int_{0}^{1}f(x)d\dfrac{x^{n+1}}{n+1}=\dfrac{f(1)}{n+1}-\dfrac{1}{n+1}\int_{0}^{1}f'(x)d\dfrac{x^{n+2}}{n+2}$$
so
$$I=\dfrac{f(1)}{n+1}-\dfrac{f'(1)}{(n+1)(n+2)}+\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such th... | Hint note that: if $k$ is a even number then also the next number $k+2$ is even
$$2^{3(k+2)-1}+5\cdot3^{k+2}=2^{3k-1+6}+5\cdot3^{k+2}=64\cdot2^{3k-1}+9\cdot5\cdot3^{k}$$$$=55\cdot2^{3k-1}+9\cdot2^{3k-1}+9\cdot5\cdot3^{k}=55\cdot2^{3k-1}+9\cdot(2^{3k-1}+5\cdot3^{k})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 2
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ordered pairs $(A,B)$ of subsets of $X$ >such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
Let $X$ be a set of $5$ elements. Then the number of ordered pairs $(A,B)$ of subsets of $X$
such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
$\bf{My\; Try::}$ Let $X = \left\{1,2,3,4,5\right\}\;,$
th... | Phicar’s answer gives you a nice, short calculation if you know about Stirling numbers of the second kind. If not, you can still organize your argument a bit more efficiently.
Suppose that the set $A\cup B$ has $n$ elements; clearly $n$ must be $2,3,4$, or $5$. For each of these four possible values of $n$ we can argue... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
I am working on proving or disproving $\cos^5(x)-\sin^5(x)=\cos(5x)$ True or false? $$\cos^5(x)-\sin^5(x)=\cos(5x)$$ for all real x.
I have no idea how to prove or disprove this. I tried to expand $\cos(5x)$ using double angle formula but I wasn't sure how to go from that to $$\cos^5(x)-\sin^5(x)$$
| It is false. Quick counterexample: Note that, at $x=\frac{\pi}{5}$, we have $ \cos (5x) = -1 $, but clearly${}^\dagger$, $ \cos^5 \left( \frac {\pi}{5} \right) - \sin^5 \left( \frac {\pi}{5} \right) \ne 1 $. $\Box$
${}^\dagger$ It is clear that, since $ \frac {\pi}{5} $ is acute, $ 0 < \cos^5 \left( \frac {\pi}{5} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$\lim\limits_{n\to\infty}a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}$ converges for $a$ in what range?
$\displaystyle\lim_{n\to\infty}a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}$
converges for $a$ in what range?
I tried $\displaystyle\lim_{n\to\infty}\ln \left[a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}\right]=\lim_{n\to\infty... | Let $c_n=a^n\left(1+\frac{1}{n}\right)^{n^2}$ and let $t=\frac{1}{a}$, so $\ln c_n=n\ln a+n^2\ln(1+\frac{1}{n})=n^2\ln(1+\frac{1}{n})-n\ln t$
where $\frac{1}{n}-\frac{1}{2n^2}<\ln(1+\frac{1}{n})<\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}\implies n-\frac{1}{2}<n^2\ln(1+\frac{1}{n})<n-\frac{1}{2}+\frac{1}{3n}$
$\implies (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Why this happen only with $4,8,12,.......$ We will take some examples to illustrate my question:
If we take a set of numbers, for example $1,2$, and $3$
$$1+2^4+3^4=98$$
$$1+2^8+3^4=338$$
$$1+2^8+3^8=6818$$
$$1+2^8+3^{12}=531698$$
We note that all numbers resulting end 8.This will not change for any selected power fro... | Because for any integer $x$ whatsoever, and any $r,s\geq 1$,
$$x^{4r}\equiv x^{4s}\bmod 10$$
As an example, with $r=3$, and $s=5$ this says that
$$1^{12}\equiv 1^{20}\bmod 10,\qquad 2^{12}\equiv 2^{20}\bmod 10,\qquad 3^{12}\equiv 3^{20}\bmod 10$$
and therefore
$$\begin{align*}
(\text{last digit of $1^{12}+2^{12}+3^{12}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Question about converting a polar equation to a rectangular equation $$\sec\theta =2$$
So I went through all the steps and got:
$$\cos\theta =\frac { 1 }{ 2 } $$
$$\sin\theta =\pm \sqrt { 1-\frac { 1 }{ 4 } } $$
$$\sin\theta =\pm \frac { \sqrt { 3 } }{ 2 } $$
$$y=\pm {\sqrt { 3 } }$$
Now why is it that the correct a... | $$ \sec \theta = 2 $$
$$ 1 = 2 \cos \theta $$
$$ r = 2 r\cos \theta $$
$$ \sqrt{x^2 + y^2} = 2x $$
$$ x^2 + y^2 = 4x^2 $$
$$ 3x^2 = y^2 $$
$$ y = \pm \sqrt{3} \,x$$
Another method:
$$ \cos \theta = \frac{1}{2} \Rightarrow \tan \theta = \pm \sqrt{3} = \frac{y}{x} \Rightarrow y = \pm \sqrt{3} \,x $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_{0}^{\pi/2}\frac{x\sin x\cos x\;dx}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}$ How to evaluate the following integral
$$\int_{0}^{\pi/2}\frac{x\sin x\cos x}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}dx$$
For integrating I took $\cos^{2}x$ outside and applied integration by parts.
Given answer is $\dfrac{\pi}{4ab^... | Integrating by parts,
$$\int\frac{x\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
$$=x\int\frac{\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx-\int\left[\frac{dx}{dx}\int\frac{\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx\right]dx$$
Now $a^2\cos^2x+b^2\sin^2x=u\implies2(b^2-a^2)\sin x\cos x\ dx=du$
For $\displaystyle\int\frac{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
Find an $\epsilon$ such that the $\epsilon$ neighborhood of $\frac{1}{3}$ contains $\frac{1}{4}$ and $\frac{1}{2}$ but not $\frac{17}{30}$ I am self studying analysis and wrote a proof that is not confirmed by the text I am using to guide my study. I am hoping someone might help me comfirm/fix/improve this.
The prob... | Compute the distances from $1/3$,
$$\Big|\frac14-\frac13\Big|=\frac1{12}=\frac5{60},$$
$$\Big|\frac12-\frac13\Big|=\frac1{6}=\frac{10}{60},$$
$$\Big|\frac{17}{30}-\frac13\Big|=\frac{7}{30}=\frac{14}{60}.$$
So any $$\frac{10}{60}<\epsilon\le\frac{14}{60}$$will do, say $1/5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$ Today I discussed the following integral in the chat room
$$\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$$
where $0\leq a, b\leq \pi$ and $k>0$.
... | First note that by substituting $x\mapsto kx$, we get
$$
\int_0^\infty\log\left(\frac{x^2+2kx\cos(a)+k^2}{x^2+2kx\cos(b)+k^2}\right)\frac{\mathrm{d}x}{x}
=\int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x\cos(b)+1}\right)\frac{\mathrm{d}x}{x}
$$
Let $u=\frac{x+\cos(a)}{\sin(a)}$. Then
$$
\begin{align}
\frac{\mathrm{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 1
} |
Limits using Maclaurins expansion for $\lim_{x\rightarrow 0}\frac{e^{x^2}-\ln(1+x^2)-1}{\cos2x+2x\sin x-1}$ $$\lim_{x\rightarrow 0}\frac{e^{x^2}-\ln(1+x^2)-1}{\cos2x+2x\sin x-1}$$
Using Maclaurin's expansion for the numerator gives:
$$\left(1+x^2\cdots\right)-\left(x^2-\frac{x^4}{2}\cdots\right)-1$$
And the denominator... | Doing almost the same as mixedmath, I have very slightly different results since
$$e^{x^2} = 1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\frac{x^8}{24}+O\left(x^9\right) $$
$$\ln(1 + x^2) = x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+O\left(x^9\right)$$
$$\cos(2x)=1-2 x^2+\frac{2 x^4}{3}-\frac{4 x^6}{45}+\frac{2 x^8}{315}+O\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Volume of solid of revolution by shell method consider the region bounded by $ \displaystyle y=4{{x}^{2}}$ and $ \displaystyle 2x+y=6$. What is the volume of solid of revolution about $\displaystyle x$-axis.
What is thought about setting the integral:
I split the region into two parts
$\displaystyle V=4\pi \int\limits_... | I don't see why you need to split the region. Notice that $y = 4x^2$ and $y = 6 - 2x$ both intercept at: $$4x^2 = 6-2x \implies2x^2 + x - 3 =0 \implies x = \frac{-1\pm\sqrt{1+24}}{4} \implies x = \frac{-1\pm 5}{4},$$ so $x = -3/2$ and $x = 1$. Since the area of revolution of $y = f(x)$ around the $x$ axis is $$A = \pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
In triangle ABC, Find $\tan(A)$.
In triangle ABC, if $(b+c)^2=a^2+16\triangle$, then find $\tan(A)$ . Where $\triangle$ is the area and a, b , c are the sides of the triangle.
$\implies b^2+c^2-a^2=16\triangle-2bc$
In triangle ABC, $\sin(A)=\frac{2\triangle}{bc}$, and $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$,
$\implies \ta... | You have: $\dfrac{\sin A}{\cos A} =\tan A = \dfrac{4\triangle}{16\triangle - 2bc} = \dfrac{2bc\sin A}{8bc\sin A-2bc} = \dfrac{\sin A}{4\sin A-1} \Rightarrow \cos A = 4\sin A -1 \Rightarrow \cos^2A = \left(1-4\sin A\right)^2 = 1 - 8\sin A + 16\sin^2A \Rightarrow 1 - \sin^2A = 1 - 8\sin A + 16\sin^2A \Rightarrow \sin A\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How does $\log(x^2 + 1)$ become $\log(2x^2)$? My textbook attempts to take the big O of $\log(x^2 +1)$. It proceeds by saying $x^2 + 1 \le 2x^2$ when $x \ge 1$. But I don't know how it came up with this idea.
Question:
Why set $x^2+1$ to a random value to be $2x^2$? Why $2$ of all numbers? Why not $x^2$ or $x^3$?
| We’re not setting $x^2+1$ to anything. We want to find an upper bound on $\log(x^2+1)$ that can be expressed as a fairly simple function of $x$. Provided that $x\ge 1$, we know that $x^2+1\le 2x^2$, so $\log(x^2+1)\le\log(2x^2)$, and $\log(2x^2)$ can be expressed as a simple function of $\log x$:
$$\log(2x^2)=\log 2+2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the fraction where the decimal expansion is infinite? Find the fraction with integers for the numerator and denominator, where the decimal expansion is $0.11235.....$
The numerator and denominator must be less than $100$.
Find the fraction.
I believe I can use generating functions here to get $1+x+2x^2+3x^3+5x^... | Like you hint, we can observe that the value
$$0.11235955056\ldots = \sum_{k = 1}^{\infty} 10^{-k} F_k,$$
where $F_k$ is the $k$th Fibonacci number (with the convention that $F_0 = 0$, $F_1 = 1$),
is simply the value of the series
$$F(x) := \sum_{k = 1}^{\infty} F_k x^k = x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Transforming linear combination of the cosine and sine function In the proof of Transforming $a\cos\left(\, x\,\right)+b\sin\left(\, x\right)$ to $r\cos\left(\,\phi - x\,\right)$
\begin{align}
a\cos\left(\, x\,\right) + b\sin\left(\, x\,\right)
&=\,\sqrt{\,a^{2} + b^{2}\,}\,
\left[\,\frac{a}{\,\sqrt{\, a^{2} + b^{2}\,}... | Start at "the other end" of the problem. If we want
$$a\cos x+b\sin x=r\cos(\phi-x)$$
and we expand the right hand side, then what we are looking for is
$$a\cos x+b\sin x=r\cos\phi\cos x+r\sin\phi\sin x\ .$$
If this is to be true for all values of $x$ then we need
$$a=r\cos\phi\quad\hbox{and}\quad b=r\sin\phi\ .$$
Squ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove $4^k - 1$ is divisible by $3$ for $k = 1, 2, 3, \dots$ For example:
$$\begin{align}
4^{1} - 1 \mod 3 &=
\\
4 -1 \mod 3 &=
\\
3 \mod 3 &=
\\3*1 \mod 3 &=0
\\
\\
4^{2} - 1 \mod 3 &=
\\
16 -1 \mod 3 &=
\\
15 \mod 3 &=
\\3*5 \mod 3 &= 0
\\
\\
4^{3} - 1 \mod 3 &=
\\
64 -1 \mod 3 &=
\\
21 \mod 3 &=
\\3*7 \mod 3 &=
... | $4^k-1=(4-1)(4^{k-1}+\cdots+4+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 3
} |
$b_{n}$ is increasing I think there is misunderstanding in my last post because its contain three questions so i will post question by question step by step
An inequality for the product $\prod_{k=2}^{n}\cos\frac{\pi }{2^{k}}$
Let $n\geq 2\quad a_{n}=\prod\limits_{k=2}^{n}\cos\left(\dfrac{\pi }{2^{k}}\right)$ and $b_... | A variant: $$\begin{align*} b_n <b_{n+1} &\iff \cos\Bigl(\frac{\pi}{2^n}\Bigr) < \cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr)\\ &\iff 2\cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr)-1 < \cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr)\\ &\iff \cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr) <1 .\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Number of $ 6 $ Digit Numbers with Alphabet $ \left\{ 1, 2, 3, 4 \right\} $ with Each Digit of the Alphabet Appearing at Least Once Find the number of 6 digit numbers that can be made with the digits 1,2,3,4 if all the digits are to appear in the number at least once.
This is what I did -
I fixed four of the digits to ... | Really easy using generating functions. The generating function is:
$$G(x)=\left(x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\right)^4$$
We need the coefficient of x^6 in that and we make use of the fact that $x^4 = x^2 \cdot x^2$
$\left(x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\right) \left(x+\frac{x^2}{2}+\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$mn | m^2+n^2+m \implies$ $(n-1)$ is a square Let $m;n \in \mathbb{Z^+}$ such that $mn | m^2+n^2+m$
Prove that $(n-1)$ is a square number.
P/s : I don't have any ideas about this problems :(
Thanks :)
| Consider the equation $kmn = m^2+n^2+m$ for fixed $k$. Suppose this equation has solutions in $(m,n)$ in positive integers, and take the solution with $m+n$ minimized. We distinguish two cases.
*
*Suppose $m \geq n$. Note that $m$ satisfies the quadratic equation
$$ X^2 - (kn-1)X+n^2 = 0$$
which also has $X=\frac{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Equation with three variables I am confused as how to solve an equation with three squared variables to get its integer solutions? As:
$$x^2+y^2+z^2=200$$
Thanks!
| I would do it this way: (it takes just few minutes)
Let's assume $x<=y<=z$, without loss of generality.
$z$ must be larger than $8$, since $3*8^2 = 192 < 200$.
$z$ must be smaller than $15$, since $15^2 = 225 > 200$.
Now, there are six possibilities:
$$x^2+y^2=200-9^2=119$$
$$x^2+y^2=200-10^2=100$$
$$x^2+y^2=200-11^2=7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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The locus of points $z$ which satisfy $|z - k^2c| = k|z - c|$, for $k \neq 1$, is a circle Use algebra to prove that the locus of points z which satisfy $|z - k^2c| = k|z - c|$, for $k \neq 1$ and $c = a + bi$ any fixed complex number, is a circle centre $O$.
Give the radius of the circle in terms of $k$ and $|c|$.
I s... | instead of squaring let me see if conjugates help.
$\begin{eqnarray}
0 = |z - k^2c|^2 -k^2 |z -c|^2 &=& (z - k^2 c)(\bar z - k^2 \bar c) -k^2(z-c)(\bar z - \bar c) \\
&=& z \bar z - k^2 c \bar z - k^2\bar c z + k^4 c \bar c -k^2(z \bar z - c \bar z -z \bar c + c \bar c) )\\
&=&(1-k^2)z \bar z - k^2(1-k^2)c\bar c \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Two questions about divisible I have two questions.
1) Why the relation $a^n\equiv a^{n+4k}\pmod{10}$ is true?
2) Let $(a,90)=1$. Which number can be $ x$ in $ x\mid a^4-1$? Answer is $240$. Why?
| \begin{align}
3^0 & \equiv 1 \pmod{10} \\
3^1 & \equiv 3 \pmod{10} \\
3^2 & \equiv 9 \pmod{10} \\
3^3 & = 27 \equiv 7 \pmod{10} \\
3^4 & = 81 \equiv 1 \pmod{10}
\end{align}
We've returned to $1$ when the exponent is $4$.
Try this with other numbers is $\{1,3,7,9\}$ in place of $3$. In each case we get the $4$th power ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Show that f is periodic if $f(x+a)+f(x+b)=\frac{f(2x)}{2}$? Suppose $a$ and $b$ are distinct real numbers and $f$ is a continuous real function such that $\frac{f(x)}{x^2}$ goes to 0 when $x$ goes to infinity or minus infinity. Suppose that$ f(x+a)+f(x+b)=\frac{f(2x)}{2}$. How show that $f$ is periodic?
| There are not so many possibilities for equation transformations, so we go straight forward. For the determinancy assume that $a<b$.
Let $x$ be an arbitrary real number. Then
$$f(2x+a)+f(2x+b)=\frac{f(4x)}{2}.$$
But
$$f(2x+a)=2\left(f\left(x+a+\frac a2\right)+ f\left(x+b+\frac a2\right)\right),$$
$$f(2x+b)=2\left(f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving that a number is an integer. Prove that the following number is an integer:
$$\left( \dfrac{76}{\dfrac{1}{\sqrt[\large{3}]{77}-\sqrt[\large{3}]{75}}-\sqrt[\large{3}]{5775}}+\dfrac{1}{\dfrac{76}{\sqrt[\large{3}]{77}+\sqrt[\large{3}]{75}}+\sqrt[\large{3}]{5775}}\right)^{\large{3}}$$
How can I prove it?
| Following the simplifications suggested by MathBot, we have:
$$\left( \dfrac{76}{\dfrac{1}{b-a}-ab}+\dfrac{1}{\dfrac{76}{b+a}+ab}\right)^{\large{3}}$$
Let's just take the part inside the parentheses, and put it over a common denominator.
$$\dfrac{\dfrac{76^2}{b+a} + 76 ab + \dfrac{1}{b-a} - ab}{\left(\dfrac{1}{b-a}-ab\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
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Units in a ring of fractions
Let $R$ be a UFD and $D \subseteq R$ multiplicative set. What are the units in $D^{-1}R$?
I assume the answer should be $D^{-1}R^{\times}$, but I get stuck:
If $a/b$ is a unit, then there exists $c/d$ so that
$$\frac{a}{b} \cdot \frac{c}{d} = \frac{1}{1} \Longleftrightarrow ac = bd,$$
bu... | It is clear that $\dfrac{1}{1} = \dfrac{d}{d}, d \in D$ is the identity element of $D^{-1}R.$ We want to find the units. Let $u \in R$ is a unit. So there exists $v \in R$ such that $uv = vu = 1.$ Then obviously, $\dfrac{u}{1} \cdot \dfrac{v}{1} = \dfrac{v}{1} \cdot \dfrac{u}{1} = \dfrac{1}{1},$ showing that $\dfrac{u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ \sum_{n=1}^\infty \ln\big(n\sin \frac{1}{n}\big)$ converges.
Prove that $\displaystyle \sum_{n=1}^\infty \;\ln\left(n\sin\frac{1}{n}\right)$ converges.
My Work:
$$\left|\ln \left(n \sin \frac{1}{n}\right)\right| \leq\left|\ln \left(n \sin \frac{1}{n^{2}}\right)\right| \leq\left|\ln \left(\sin \frac{1}{n^... | In the same spirit as mixedmath's answer, let us massively use Taylor expansions for large values of $n$. Then $$\sin(\frac 1n)=\frac{1}{n}-\frac{1}{6 n^3}+\frac{1}{120 n^5}-\frac{1}{5040
n^7}+O\left(\left(\frac{1}{n}\right)^8\right)$$ $$n\sin(\frac 1n)=1-\frac{1}{6 n^2}+\frac{1}{120 n^4}-\frac{1}{5040
n^6}+O\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$ Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$
I seem to struggle with this simple proof. All I managed to find is that ac+bd=-4 (which might not even be correct).
| Here's another cool geometric approach if $a,b,c,d$ are positive. Consider the following diagram where $BD=1$ and is the diameter of the circle.
By Pythagoras, we have $a^2+b^2=1$ and $c^2+d^2=1$. However, by Ptolemy's Theorem, we also have $ac+bd=BD\cdot AC=AC$. Since $AC$ is any chord in a circle with diameter $1$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
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How do I integrate: $\int\sqrt{\frac{x-3}{2-x}} dx$? I need to solve:
$$\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$
What I did is:
Substitute: $x=2\cos^2 \theta + 3\sin^2 \theta$. Now:
$$\begin{align}
x &= 2 - 2\sin^2 \theta + 3 \sin^2 \theta \\
x &= 2+ \sin^2 \theta \\
\sin \theta &= \sqrt{x-2} \\
\theta &=\sin^{-1}\sqrt{... | $2(\cos x)^2=1+\cos(2x)$ but $2(\cos x)^2\neq1-\sin(2x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find the angle if the area of the two triangles are equal?
Let $I$ be the incenter of $\triangle ABC$, and $D$, $E$ be the midpoints of $AB$, $AC$ respectively. If $DI$ meets $AC$ at $H$ and $EI$ meets $AB$ at $G$, then find $\measuredangle A$ if the areas of $\triangle ABC$ and $\triangle AGH$ are equal.
I played a... | Here's my trigonometric endeavors:
Take $\alpha=\measuredangle AID$. Using the fact that $ID$ is the median of $AIB$, we have that $\cot \frac{A}2+\cot \alpha= \cot \frac{B}2+\cot (180^{\circ}-\alpha)$, or thus: $$\cot \alpha=\frac{\cot \frac{B}2 - \cot \frac{A}2}2=\frac{(s-b)-(s-a)}{2r}=\frac{a-b}{2r}=\frac{s(a-b)}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit involving square roots, more than two "rooted" terms The limit is
$$\lim_{x\to\infty} \left(\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\right)$$
which has a value of $\dfrac{27}{4}$.
Normally, I would know how to approach a limit of the form
$$\lim_{x\to\infty}\left( \sqrt{a_1x^2+b_1x+c_1}\pm\sqrt{a_2x^2+b_2... | $\lim_{x\to\infty} \left(\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\right) = 27/4 $
i will use the binomial theorem in the form
$(BIG + small)^{1/2} = BIG^{1/2} + {1 \over 2}BIG^{-1/2}small+\cdots $ to each.
$$\sqrt{x^2+5x-2} = x + {5x - 2 \over 2x} + \cdots \\
-\sqrt{4x^2-3x+7} = -\left( 2x + {-3x + 7 \over 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Factorise a matrix using the factor theorem Can someone check this please?
$$
\begin{vmatrix}
x&y&z\\
x^2&y^2&z^2\\
x^3&y^3&z^3\\
\end{vmatrix}$$
$$C_2=C_2-C_1\implies\quad
\begin{vmatrix}
x&y-x&z\\
x^2&y^2-x^2&z^2\\
x^3&y^3-x^3&z^3\\
\end{vmatrix}$$
$$(y-x)
\begin{vmatrix}
x&1&z\\
x^2&y+x&z^2\\
x^3&y^2+xy+x^2&z^3\\
\... | you are not factoring the matrix. you are using the properties of the determinants to simplify. for example, you could write your first step as
$$AE = \pmatrix{x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3}
\pmatrix{1 & -1 & 0\\0&1&0\\0&0&1} =
\pmatrix{x & y-x & z \\ x^2 & y^2-x^2 & z^2 \\ x^3 & y^3 -x^3& z^3} = B
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given $a+b+c$, Can I calculate $a^2+b^2+c^2$? I want to calculate $a^2 + b^2 + c^2$ when I am given $a+b+c$.
It is known that a,b,c are positive integers.
Is there any way to find that.
| No. Suppose that $a+b+c = 6$.
If $a = b = c = 2$, then $a^2 + b^2 + c^2 = 12$.
If $a = 4$ and $b = c = 1$, then $a^2 + b^2 + c^2 = 18$
The sum of the squares varies for a given sum of $a$, $b$, and $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
a way to integrate: $\int (\sqrt{x} +3)/(2+ x^ \frac{1}{3}) dx$ Im looking for a way to integrate:
$$
\int \frac{ \sqrt{x} +3}{2+ x^ \frac{1}{3}} dx
$$
that would make it efficient and not too difficult...
Any suggestions?
| The substitution $x = t^6$ works. The integral becomes
$$ \int \frac{t^3 + 3}{t^2 + 2} \,6t^5\,dt = 6 \int \frac{t^8 + 3t^5}{t^2 + 2} dt = 6\int \left( t^6 - 2t^4 + 3t^3 + 4t^2 - 6t - 8 + \frac{12t+16}{t^2+2} \right) \, dt $$
You can take it from here
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Sum: $\sum_{n=1}^\infty\prod_{k=1}^n\frac{k}{k+a}=\frac{1}{a-1}$ For the past week, I've been mulling over this Math.SE question. The question was just to prove convergence of
$$\sum\limits_{n=1}^\infty\frac{n!}{\left(1+\sqrt{2}\right)\left(2+\sqrt{2}\right)\cdots\left(n+\sqrt{2}\right)}$$
but amazingly Mathematica to... | Here is a completely elementary proof, which only needs introductory calculus concepts:
$$a_{n+1} = \frac{(n+1)!}{(a+1)(a+2)\dots(a+n+1)} = \frac{1}{a-1}\left(\frac{(n+1)!}{(a+1)(a+2) \dots (a+n)} -\frac{(n+2)!}{(a+1)(a+2) \dots (a+n+1)}\right) = b_{n+1} - b_{n+2}$$
where $a_n$ is the term of our series and $$b_n = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Area of the overlap between a triangle and a square $ABC$ is an equilateral triangle, each side has length 4.
$M$ is the midpoint of $\overline{BC}$, and $\overline{AM}$ is a diagonal of square $ALMN$. Find the area of the region common to both $ABC$ and $ALMN$.
I'm not sure how to solve this problem, or even where t... | Use this diagram.
The equilateral triangle $ABC$ has sides of length $4$, so $CM=2$ and $AM=2\sqrt 3$.
The area you are trying to find is shaded and is clearly twice the area of triangle $AEM$. We know the base, $AM$, so we want the height $EF$. If we let $x=EF$ then $MF=x$ due to $45-45-90$ triangle $EFM$. We also ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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$\lim_{n \rightarrow \infty} \frac{1-(1-1/n)^4}{1-(1-1/n)^3}$ Find
$$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}$$
I can't figure out why the limit is equal to $\dfrac{4}{3}$ because I take the limit of a quotient to be the quotient of their limits.
I'm taking th... | Hint: $\dfrac{1-x^4}{1-x^3}= \dfrac{(1-x)(1 + x + x^2 + x^3))}{(1-x)(1+x+x^2)} = \dfrac{1 + x + x^2 + x^3}{1 + x+ x^2}$ (provided $x \neq 1$.)
Now put $x = 1 - \frac{1}{n}$ and take $n \to \infty.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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How many divisors of N ended by 5 I must know how to find how many divisor of N ended by 5 ? In my exercise, I have $\ N=63'000 = 2^3*3^2*5^3*7 $ and I can found the number of divisors of N using $\ (3+1)*(2+1)*(3+1)*(1+1)=96$
Among these 96 divisors, how many ended by 5 ? How can calculate this ?
Thank you so much
| *
*Take out the $2$s, because when multiplied by $5$, the result will end with $0$.
*Find the number of divisors of $3^2\cdot7^1$, which is $(2+1)\cdot(1+1)=6$:
*
*$3^0\cdot7^0$
*$3^1\cdot7^0$
*$3^2\cdot7^0$
*$3^0\cdot7^1$
*$3^1\cdot7^1$
*$3^2\cdot7^1$
*Multiply each divisor by each one of the following $3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Linear Combinations of Fibonacci Numbers (integer coefficients) While working on problem #2 on Project Euler, I came across the need to express $F_n$ as a linear combination of $F_{n-3}$ and $F_{n-6}$. This is relatively simple to do:
$$\begin{align} F_n &= F_{n-1}+F_{n-2}\\ &= F_{n-1}+F_{n-3}+F_{n-4}\\ &= F_{n-1}+F_{n... | Note that we can't always do this with integer coefficients. For example,
$$
F_{n}=\frac52F_{n-2}+\frac12F_{n-5}\tag{1}
$$
and
$$
F_{n}=\frac{13}3F_{n-3}-\frac23F_{n-7}\tag{2}
$$
We can use the fact that
$$
\left(\frac{1\pm\sqrt5}2\right)^n=\frac1{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}5^k\pm\frac{\sqrt5}{2^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Basic exercise with exponents and radicals I'm trying to solve a simple high school algebra problem, I would like to know if my result is correct.
Convert the radicals into exponents, solve and then express the result as a radical
$2\sqrt[3]{2}:\sqrt{(\dfrac{1}{2}\sqrt[5]{4})^{\frac{-1}{3}})}$
If I call $A$ to the num... | You could simplify the result to $2\sqrt[30]{2^7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$ As the title,
By considering $\bigtriangleup$ABC, Prove
$$\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$$
Thanks
| Assume that $A$ is the largest angle (the modifications when this is not the case should be minor). Drop a perpendicular from $A$ to the side $BC$. Let $A'$ and $A''$ be the angles formed by dividing $A$ with this perpendicular, where $A'$ is adjacent to $B$ and $A''$ is adjacent to $C$. The trigonometry of right trian... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Considering $ (1+i)^n - (1 - i)^n $, Complex Analysis I have been working on problems from Complex Analysis by Ahlfors, and I got stuck in the following problem:
Evaluate:
$$
(1 + i)^n - (1-i)^n
$$
I have just "reduced" to:
$$
(1 + i)^n - (1-i)^n = \sum_{k=0} ^n i^k(1 - (-1)^k)
$$
by using expansion of each term.
Th... | ok. we can use the binomial theorem. we will check out for small values of $n.$
case $n = 1, \ z_1 = 1+ i -(1-i) = 2i$
case $n = 2, \ z_2 = (1+i)^2 - (1-i)^2 =(1 + 2i + i^2) -(1 - 2i + i^2) = 2(2i) = 4i $
case $n = 3, \ z_2 = (1+i)^3 - (1-i)^3 =(1 + 3i + 3i^2 + i^3) -(1 - 3i + 3i^2 - i^3) = 2(3i +i^3) = 4i$
case $n = 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Solving Diophantine equation $1/x^2+1/y^2=1/z^2$ How can we find positive integers solutions $(x,y,z)$, where $\gcd(x,y,z)=1$ for the equation:
$$1/x^2+1/y^2=1/z^2$$
Can we conclude that $x$ and $y$ are not coprimes for it to have solution?
| Multiply both sides by $x^2y^2z^2$.
Then you get
$$y^2z^2+x^2z^2=x^2y^2$$
Now use that each of $x^2, y^2,z^2$ divide two of the terms hence the third.
Added Here is the rest of the solution. Let $a=gcd(x,y), b=gcd(x,z), c=gcd(y,z)$.
Then, $gcd(a,b)=1$ and hence $ab|x$. We claim $ab=x$.
Indeed write $x=abd$. Assume by c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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prove the given inequality (for series ) For any given $n \in \Bbb N,$ prove that, $$1+{1\over 2^3}+\cdots+{1\over n^3} <{3\over 2}.$$
| For $n\ge 3$ the sum is less than
$$1+\frac{1}{8}+\frac{1}{3}\left(\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\cdots+\frac{1}{(n-1)(n)}\right).$$
Note that $\frac{1}{(2)(3)}=\frac{1}{2}-\frac{1}{3}$ and $\frac{1}{(3)(4)}=\frac{1}{3}-\frac{1}{4}$ and $\frac{1}{(4)(5)}=\frac{1}{4}-\frac{1}{5}$ and so on. It follo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
} |
False proof of 0=1 using Laurent series I found the following proof that 0 = 1:
\begin{align*}
\sum_{n=-\infty}^{\infty} 0\cdot z^n = 0 = \frac{1}{z-1} + \frac{1}{1-z} = \frac{1}{z}\frac{1}{1-\frac{1}{z}} + \frac{1}{1-z} \\
= \frac{1}{z} \sum_{n=0}^{\infty}\frac{1}{z^n} + \sum_{n=0}^{\infty}z^n = \sum_{n=-\infty}^{\inf... | The equality $\dfrac{1}{z}\dfrac{1}{1-\frac{1}{z}} + \dfrac{1}{1-z}
= \dfrac{1}{z} \sum \limits_{n=0}^{\infty}\dfrac{1}{z^n} + \sum \limits_{n=0}^{\infty}z^n$ only holds if $z\not \in \{0,1\}$, $\left|\frac 1 z\right|<1$ and $|z|<1$, i.e., never.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Matrix gauss continuation Suppose we have the matrix $4\times4$, $R_1 =(0,0,1,1)$, $R_2=(0,1,1,0)$, $R_3=(1,1,0,0)$, $R_4(1,0,0,0)$ and suppose we want to see the rank.The difficult part is that you cannot order them somehow to use gauss.Any ideas what should be done in such situations?
| You can subtract any row from any other without changing the rank:
$$\begin{pmatrix}0&0&1&1\\0&1&1&0\\1&1&0&0\\1&0&0&0\end{pmatrix}\sim\begin{pmatrix}0&0&1&1\\0&1&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}\sim\begin{pmatrix}0&0&1&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}\sim\begin{pmatrix}0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is this divisibility problem correct? Let $n$ be a natural number and let $1 \le a_1<a_2<...<(a_k=n)$ be all of its divisors.
Find all $n$ such that $a_2^3+a_3^2-15=n$ .
It seems impossible to find all such numbers.
| We must have $a_2 = p$, where $p$ is the smallest prime dividing $n$. Then $a_3$ must be either the next smallest prime $q$ dividing $n$, or $p^2$. But if we had $a_3 = p^2$, then from $p^2 | n$ we would obtain $p^2 | 15$, which is absurd. Therefore $a_3 = q$, and moreover $p < q$.
Thus we obtain $p^3 + q^2 - 15 = n = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Fredholm integral equation How can I solve the following fredholm integral equation
$$ψ(x)=x+λ\int_{0}^{2π}|x-t|ψ(t)dt$$
The kernel contains absolute value
| First we express it in the form
\begin{align}
(I-\lambda K)\psi = x
\end{align}
where
\begin{align}
K\psi(x):= \int_{0}^{2\pi} |x-t|\psi(t)\, \mathrm{d}t
\end{align}
Existence and uniqueness is guaranteed if $|\lambda| < \frac{1}{\left\|K\right\|}$, so assuming $K\colon C[0,2\pi] \to C[0,2\pi]$ we can estimate
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit of logarithms exponential $$
\lim_{x\to\infty}\biggl(\frac{\ln(x-2)}{\ln(x-1)}\biggr)^{x\ln x}.
$$
L'Hopital seems like a very hardcore solutions given the situation.Are the any other options?
| Let $$f(x) = \biggl(\frac{\ln(x-2)}{\ln(x-1)}\biggr)^{x\ln x}.$$
Then
$$
\begin{align*}
\ln f(x) &= x\ln x[\ln \ln (x-2) - \ln \ln (x-1)] \\
&= x \ln x\left[\ln \left(1 + \frac{\ln(1 - 2/x)}{\ln x}\right) - \ln \left(1 + \frac{\ln(1 - 1/x)}{\ln x}\right) \right]\\
&= x \ln x\left[\ln \left(1 - \frac{2}{x \ln x} + o \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How prove $\sin \left( \alpha+\frac{\pi }{n} \right) \cdots \sin \left( \alpha+\frac{n\pi }{n} \right) =-\frac{\sin n\alpha}{2^{n-1}}$? How prove
$$\prod_{k=1}^{n}\sin \left( \alpha+\frac{\pi k }{n}\right) =-\frac{\sin n\alpha}{2^{n-1}}$$
for $n \in N$?
| Since, $\sin \theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}$,
Hence, $$\begin{align}\prod_{j=1}^{n-1}\sin \left(\alpha+\frac{j\pi}{n}\right)&=\prod_{j=1}^{n-1}\left(\frac{e^{i(\alpha+j\pi/n)}-e^{-i(\alpha+j\pi/n)}}{2i}\right) \\&= \left(\frac{-1}{2i}\right)^{n-1}e^{-((n-1)\alpha + i\frac{(n-1)\pi}{2})}\prod_{j=1}^{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Markov chain doesn't sum up to 1
Let $\{X_n\}$ be a Markov chain on $S=\{1,2,3,4,5,6\}$ with the matrix
suppose we define a new sequence $\{Y_n\}$ by $$Y_n=\cases{1\quad X_n=1\vee X_n=2\\2\quad X_n=3\vee X_n=4\\3\quad X_n=5\vee X_n=6}$$does for $a=1$ this is a Markov chain?
I thought summing up and computing for exa... | Let's write out the transition matrix of $X$ when $a = 1$: $$\mathcal P = \frac{1}{10} \begin{bmatrix} 1 & 3 & 0 & 1 & 5 & 0 \\ 2 & 2 & 1 & 0 & 1 & 4 \\ 6 & 0 & 1 & 1 & 2 & 0 \\ 6 & 0 & 1 & 1 & 0 & 2 \\ 3 & 0 & 4 & 1 & 0 & 2 \\ 3 & 0 & 1 & 4 & 2 & 0 \end{bmatrix}$$ Now split it up into $2 \times 2$ blocks, and observ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.