Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding conjugates of all z $\in$ C that satisfy $z^3$ $$ z^3 = \frac{16e^{i\frac{3\pi}4}}{(1-\sqrt3)+\sqrt6e^{i\frac\pi4}} $$
Anyone know of a good way to simplify this expression?
| $\sqrt{6}e^{i\pi/4} = \sqrt{6}\left(\cos(\pi/4) + i\sin(\pi/4)\right) = \sqrt{6}\left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i\right) = \sqrt{3} + \sqrt{3}i \Rightarrow 1 - \sqrt{3} + \sqrt{6}e^{i\pi/4} = 1 +\sqrt{3}i = 2e^{i\pi/3} \Rightarrow z^3 = \dfrac{16e^{i3\pi/4}}{2e^{i\pi/3}} = 8e^{i5\pi/12} = 8e^{i5\pi/12+2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find bases for eigenspaces of A $$A = \begin{pmatrix} 6 & 4 \\ -3 & -1\end{pmatrix}$$
Find the bases for eigenspaces $E_{\lambda_1}$ and $E_{\lambda_2}$ of $A$.
I don't really know where to start on this problem.
| First step: find the eigenvalues, via the characteristic polynomial $$\det(A - \lambda I) = \begin{vmatrix} 6 - \lambda & 4 \\ -3 & -1-\lambda \end{vmatrix} = 0 \implies \lambda^2 - 5\lambda + 6 = 0.$$
One of the eigenvalues is $\lambda_1 = 2$. You find the other one.
Second step: to find a basis for $E_{\lambda_1}$,... | {
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"timestamp": "2023-03-29T00:00:00",
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Equation $a^5+15ab+b^5=1$ What are the integer solutions of $a^5+15ab+b^5=1$?
The equation is symmetric in $a$ and $b$, so let's assume $a\geq b$. When $a=b$, we have $2a^5+15a^2=1$, which has no solution by the Rational Root Theorem. Else, $a>b$. It is degree-five, so we cannot use the quadratic formula.
| Clearly $a \gt 0, b \le 0, a \gt |b|$ because the $15ab$ is negative. We can see that $a=1,b=0$ is a solution. If $a =1-b,$ the left side becomes $a^5+15a(1-a)-(a-1)^5=5a^4-10a^3-5a^2+10a+1$, which is greater than $1$ when $a \gt 2$ so we only need to check $a=2$ We find $a=2,b=-1$ is the only other solution.
| {
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Proving the inequality $a^4+b^4+c^4+2abc(a+b+c)\ge \frac{(a+b+c)^4}9$ If $a,b,c$ are non-negative real numbers prove the following inequality $a^4+b^4+c^4+2abc(a+b+c)\ge \frac{(a+b+c)^4}9$.
| It's obviously true for all $a,b,c$ real numbers.
| {
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Simplify - Help I am supposed to simplify this:
$$(x^2-1)^2 (x^3+1) (3x^2) + (x^3+1)^2 (x^2-1) (2x)$$
The answer is supposed to be this, but I can not seem to get to it:
$$x(x^2-1)(x^3+1)(5x^3-3x+2)$$
Thanks
| $$(x^2-1)^2 (x^3+1) (3x^2) + (x^3+1)^2 (x^2-1) (2x)\\=x(x^2-1)(x^3+1)\left[3x(x^2-1)+2(x^3+1)\right] \color{blue}{\text{(Factor out common factors)}}\\=x(x^2-1)(x^3+1)\left[3x^3-3x+2x^3+2\right] \color{blue}{\text{ (Distribute inside brackets)}}\\=x(x^2-1)(x^3+1)(5x^3-3x+2) \color{blue}{\text{ (Simplify)}}$$
| {
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"url": "https://math.stackexchange.com/questions/1031464",
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Indefinite Integral of Reciprocal of Trigonometric Functions How to evaluate following integral
$$\int \frac{\mathrm dx}{\sin^4x+\cos^4x\:+\sin^2(x) \cos^2(x)}$$
Can you please also give me the steps of solving it?
| Apolozies for any typos:
My solution:
$$
\int \frac{dx}{\sin^4x+\cos^4x\:+\:\sin^2\left(x\right)\cos^2\left(x\right)}\\
=\int\frac{dx}{(\sin^2x+\cos^2x)^2-\sin^2x\cos^2x}\\
=\int\frac{dx}{1-\sin^2x\cos^2x}\\
=\int\frac{(1+\tan^2x)\sec^2xdx}{1+\tan^4x+\tan^2x}\\
=\int\frac{(1+y^2)dx}{1+y^2+y^4}\quad y:=\tan x\\
=\frac12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035014",
"timestamp": "2023-03-29T00:00:00",
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Prove that one of x,y,z is smaller than 3 and one is bigger than 5 if... If $x+y+z=12$ and $x^2+y^2+z^2=54$ then prove that one has to be smaller or equal to 3 and one has to be bigger or equal than 5.
So I got that $xy+yz+zx=45$ and with that I had a function with x,y,z as zeros: $f(a)=a^3-12a^2+45a-k$, where k=xyz.
I... | A sketch that needs fleshing out: We might as well require $x \ge y \ge z$. If $z \gt 3$, let $w=3-z$ and we are thinking of $w$ as small. The largest $x^2+y^2+z^2$ can be is $(6-2w)^2+2(3+w)^2=54-12w+6w^2$, so $z \lt 3$ If $x \lt 5$, Let $x=5-w$, then the largest $x^2+y^2+z^2$ can be is $2(5-w)^2+(2+2w)^2=54-16w+6w... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Fourier series for $f(x)=\begin{cases} 0 & -\pi\leq x<0 \\\sin x & 0\leq x\leq \pi \end{cases}$
Find the Fourier series for $$f(x)=\begin{cases} 0 & -\pi\leq x<0 \\\sin x & 0\leq x\leq \pi \end{cases}$$
I found an answer, I'm not completly sure if it's right.
The solution would be given by $$f(x)=a_0+\sum_n\left(a_n... | To avoid partial integration, a complex approach:
\begin{align}
a_n&=\frac{1}{\pi}\int_0^\pi\sin(x)\cos(nx)dx\\
&=\frac{1}{4\pi i}\int_0^\pi(e^{ix}-e^{-ix})(e^{inx}+e^{-inx})dx\\
&=\frac{1}{4\pi i}\int_0^\pi e^{i(n+1)x}+e^{i(1-n)x}-e^{i(n-1)x}-e^{i(-1-n)x}dx\\
&=-\frac{1}{4\pi}\left[\frac{e^{i(n+1)x}}{n+1}+\frac{e^{i(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036275",
"timestamp": "2023-03-29T00:00:00",
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A second order problem on recurrence relation equals 3^n I had this Recurrence Relation problem:
$a_{n+2} + a_{n+1} - 12a_n = 0$
And I solved in a form like this $a_n = A(r_1)^n + B(r_2)^n$
$r^{n+2} + r^{n+1} - 12r^n = 0$ -> Divided the hole eqation by the lowest $r$
Result of this:
$r^2+r -12 = 0$
And then i found the... | You didn't mentioned the initial conditions of the recurrence relation, but by the solution that you obtained
$$a_n = \frac{2}{7}(3)^n − \frac{7}{2}(−4)^n$$
one can infer the initial conditions
$$a_0 = -\frac{45}{14} \;\;\;\;\;\;\;\; a_1 = \frac{104}{7}$$
That made me think that maybe there is a typo and you intended
$... | {
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Find a value for "c"
For what value of $c$ is $$\lim_{x\to\infty} \left(\frac{x+c}{x-c}\right)^x = e?$$
I am unsure of how to start this question in any sense.
| You also have a solution using Taylor series.
Let $$A=\left(\frac{x+c}{x-c}\right)^x$$ $$\log(A)=x \log \left(\frac{x+c}{x-c}\right)=x \log \left(\frac{1+\frac cx}{1-\frac cx}\right)$$ For small values of $y$, we also have $$\log \left(\frac{1+y}{1-y}\right)=2 \Big(\frac{y}{1}+\frac{y^3}{3}+\cdots\Big)$$ Replace $y$ by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$ I would like to understand why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$
I am particularly concerned with the term, $-4$.
| Here's another way without using fancy formulas
$$ (x-5)^2-4 $$
$$=x^2-5x-5x+25-4 $$
$$=x^2-10x+25-4 $$
$$=x^2-10x+21 $$
$$=(x-3)(x-7) $$
$$=(x-5+2)(x-5-2) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Factorising quadratics - coefficient of $x^2$ is greater than $1$ In factoring quadratics where the coefficient of $x^2$ is greater than $1$, I use the grouping method where we multiply the coefficient and constant together and then factor.
My question is can someone explain the math behind that?
Example:
$5x^2+11x+2,\... | Consider the factorization of $9x^2 - 18x - 16$. To split the linear term, we must find two numbers with product $9 \cdot -16 = -144$ and sum $-18$. They are $-24$ and $6$. Hence,
\begin{align*}
9x^2 - 32x - 16 & = 9x^2 - 24x + 6x - 16 && \text{split the linear term}\\
& = 3x(3x - 8) + 2(3x - 8) && \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to integrate $\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$? How to integrate $$\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$$
I tried both $t=\sqrt{x^2-1}$ and $t=\sin x$ but didn't reach the right result.
| Here's the easiest way:
$$\int \frac{dx}{x^2\sqrt{x^2-1}} = \int \frac{dx}{x^3 \sqrt{1 - \frac{1}{x^2}}}$$
Now substitute $u = 1 - \frac{1}{x^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Taylor series of a definite integral Consider the function of a parameter $\alpha > 0$, given by
$$f(\alpha) = \frac{2}{\sqrt 2\pi} \int_0^\infty e^{\dfrac{-x^2}{2\alpha^2}}\cosh(x)\log\cosh(x) dx.$$
From Wolfram-alpha, it seems that for small values of $\alpha$,
$$f(\alpha) = \frac{\alpha^3}{2}+\frac{\alpha^5}{2} + o... | Considering $$f(\alpha) = \frac{2}{\sqrt {2\pi}} \int_0^\infty e^{\frac{-x^2}{2\alpha^2}}\cosh(x)\log\big(\cosh(x)\big) dx$$ start developing $\cosh(x)\log\big(\cosh(x)\big)$ as an infinite Taylor series built at $x=0$. We then get $$\cosh(x)\log\big(\cosh(x)\big)=\frac{x^2}{2}+\frac{x^4}{6}+\frac{x^6}{720}+\frac{x^8}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Showing that $\lim_{x\to\infty}\left(\sqrt{x^2+c}-x\right)=0$ A limit I had to compute recently boiled down to the following limit:
$$\lim_{x\to\infty}\left(\sqrt{x^2+c}-x\right)=0\quad\mbox{for $c\ge0$}$$
How can I show that this limit is correct?
| Note that \begin{gather*}
\sqrt{x^2+c}-x=(\sqrt{x^2+c}-x)\frac{\sqrt{x^2+c}+x}{\sqrt{x^2+c}+x}\\
=\frac{x^2-x^2+c}{\sqrt{x^2+c}+x}\\
=\frac{c}{\sqrt{x^2+c}+x}
\end{gather*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Showing $\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$ In my textbook they asked me to show that
$$\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$$
but this is not true, I think.
I put down
$$\begin{align}
\frac12(-1+\sqrt{3}i)(-1+\sqrt{3}i)(-1+\sqrt{3}i)&=\frac12\left[(1-2\sqrt3i-3)(-1+\sqrt3i\right)]\\
&=\frac12\left[-1+2\sqrt... |
Now you know your mistake!
But there is an alternative way to show that
Consider this cubic
$$x^3-1=0$$
$$(x-1)(x^2+x+1)=0$$
By applying quadratic formula to
$x^2+x+1=0$
$$x=\frac{-1- i\sqrt{3}}{2} \quad\text{or}\quad x=\frac{-1+ i\sqrt{3}}{2}$$
Since $x=\dfrac{-1+ i\sqrt{3}}{2}$ is root of $x^3-1=0$ we have
$$\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043294",
"timestamp": "2023-03-29T00:00:00",
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Let $z_1,z_2$ be complex numbers such that $Im(z_1z_2)=1$ Find the minimum value of $|z_1|^2+|z_2|^2+Re(z_1z_2)$ Question :
Let $z_1,z_2$ be complex numbers such that $Im(z_1z_2)=1$ Find the minimum value of $|z_1|^2+|z_2|^2+Re(z_1z_2)$
I know that $|z_1+z_1| \leq |z_1|+|z_2|$
Also if I consider two complex number... | Let $z_{1} = a+ib$ and $z_{2} = c+id$. Thus, $|z_{1}|^2 = a^2+b^2, |z_{2}|^2 = c^2+d^2, Re(z_{1}z_{2}) =ac-bd$ and $Im(z_{1}z_{2}) =ad+dc$.
The problem can be posed as
\begin{equation}
\begin{array}{c}
minimize \hspace{1cm} a^2+b^2+ c^2+d^2 + ac-bd \\
s.t. \hspace{3cm} ad+bc = 1. \\
\end{array}
\end{equation}
Let $\m... | {
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"url": "https://math.stackexchange.com/questions/1043772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the value of $a+b+c$?
What is the value of $a+b+c$?
if $$a^4+b^4+c^4=32$$
$$a^5+b^5+c^5=186$$
$$a^6+b^6+c^6=803$$
How to approach this kind of problem. Any help.
UPDATE: Thank you all for answers. Now I realize there is no integer solution. But is there any real number solution? I am curious to know.
| If $x$ is an integer then $x^4$ and $x^6$ have the same parity. Therefore $a^4+b^4+c^4$ and $a^6+b^6+c^6$ are either both odd or both even.
Therefore there is no solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Non linear effect in differential equation Suppose I want to study the non linear ODE
$\frac{d^2}{dx^2}f(x) + a f^2(x) + b f(x) + c=0$
I know the solution for $a=0$ and I know $a$ is a small parameter. How can I study the effect of the non linear term?
| Multiply by $f'$ to yield a integrable equation. After it depends on $f$ if you can do the last integral.
$\textbf{Edit:}$
$$
f'' f' + \left[af^2+bf + c\right]f' = f'\dfrac{d}{dx}f' + \frac{d}{dx}\left[\frac{1}{3}af^3 + \frac{1}{2}bf^2+cf\right] = 0
$$
now we can re-write the last but one equation as
$$
\dfrac{d}{dx}\l... | {
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"timestamp": "2023-03-29T00:00:00",
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Computing High Powers of a Matrix With Polynomial Entries I have a matrix where the terms are from a polynomial with two variables. Specifically it's
$C = \left(\begin{array}{cccc}
1 & 0 & 1 & 0 \\
s & 0 & s & 0 \\
0 & 1 & 0 & 1 \\
0 & st & 0 & st
\end{array}\right)$
Is there a good method for computing (exactly) th... | As noted by @Simon S, the matrix can be diagonalized. By the spectral theorem
$$
%
\mathbf{A} = \mathbf{P}^{-1} \mathbf{\Lambda} \, \mathbf{P} \\
%
$$
Instead of multiplying with the full matrix $\mathbf{A}$, work with the diagonal matrix of eigenvalues. For $k\in\mathbb{N}$:
$$
\mathbf{A}^{k} = \mathbf{P}^{-1} \math... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it w... | Put:
$$f(x) = \sum_{k=0}^{n-1}\exp\left[(3 k +1)x\right]=\exp(x)\frac{\exp(3 n x)-1}{\exp(3 x)-1}$$
If we expand $f(x)$ in a series expansion around $x=0$, then the coefficient of $x$ will give us the desired summation. We can do this as follows, we put:
$$g(x) = \log\left[f(x)\right] = x + \log\left[\exp(3 n x)-1\righ... | {
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"timestamp": "2023-03-29T00:00:00",
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Concise induction step for proving $\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$ I recently got a book on number theory and am working through some of the basic proofs. I was able to prove that $$\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$$ with the help of the identity $$\sum_{i=1}^ni = \frac{n(n+1)}{2}$$ M... | An idea for your fourth line:
$$\frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4}=\frac{(n+1)^2}4\left(n^2+4n+4\right)=\frac{(n+1)^2}4(n+2)^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Writing an alternating series as two non-alternating series? How does one calculate $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$$
if given this information: $$\sum_{n=1}^{\infty} \frac1{n^2} = \frac{\pi^2}{6}.$$
How does one account for the $(-1)^{n+1}$ in the first series?
Note: in this earlier post, user David Hodde... | Notice that
$$\frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \ ... \ = \ \frac{1}{1^2} + \left( \frac{1}{2^2} - 2 \frac{1}{2^2}\right) + \frac{1}{3^2} + \left( \frac{1}{4^2} - 2 \frac{1}{4^2}\right) + \ ...$$
See where this is going?
| {
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Prove that there are infinitely many $m,n$ for which $\frac{m+1}{n}+\frac{n+1}{m}$ is an integer Prove that there are infinitely many pairs of positive integers (m,n) such that
$\frac{m+1}{n}+\frac{n+1}{m}$
is a positive integer.
I tried following:
clearly $(1,1)$ satisfies the condition. we assume that $(a,b)$ satis... | Apply the ideas of Vieta's root jumping to $ \frac{m+1}{n} + \frac{n+1}{m} = k $.
OP found $(1,1)$ as a solution, which give us the equation $ \frac{m+1}{n}+ \frac{n+1}{m} = 4$.
This simplifies to the quadratic $m^2 + (1-4n)m + n^2 + n = 0$.
Divide by $ m \neq 0$ to get: $ m - (4n-1) + \frac{n^2 + n}{m} = 0$.
So, if we... | {
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"timestamp": "2023-03-29T00:00:00",
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The minimum value of $\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$ The problem is to find the minimum of $A$, which I attempted and got a different answer than my book:
$$A=\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$$ where $a$ is a constant
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}... | For the ease of computation I use $$x=a\sec2\theta=a\cdot\frac{1+t^2}{1-t^2}$$ where $t=\tan\theta$ and $0\le\theta\le\dfrac\pi2$
So, $f(t)=\dfrac{4a^3}{t(1-t^2)}$
We need to attain the maximum positive value of $g(t)=t-t^3$ which can be achieved using Second derivative test
| {
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"url": "https://math.stackexchange.com/questions/1053927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$ How do I evaluate
$$\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$$
I've started doing this problem by taking $u=\tan x$.
| $\bf{Another\; Solution::}$Let
$$\displaystyle I = \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx = \int\frac{\cos ^{\frac{5}{2}}x}{(1+\sin x)^{\frac{9}{2}}}dx = \int\frac{\cos^{\frac{3}{2}}x\cdot \cos x}{(1+\sin x)^{\frac{9}{2}}}dx$$
Now Let $\sin x = t\;,$ Then $\cos xdx = dt\;,$ So $$\displaystyle I = \int\fra... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 3
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Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| Using complex numbers, you can factor $x^6-1$ from the roots of unit.
$$(x^6-1)=(x-1)(x-e^{i\pi/3})(x-e^{2i\pi/3})(x+1)(x-e^{4i\pi/3})(x-e^{5i\pi/3}).$$
Dropping the factor $(x-1)$ and grouping for the conjugate roots*,
$$(x^2-x+1)(x+1)(x^2+x+1)=(x^4+x^2+1)(x+1)=x^5+x^4+x^3+x^2+x+1.$$
*Using
$(x-z)(x-z^*)=x^2-(z+z^*)x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 11
} |
Evaluating $ \int \frac{1}{\sin x} dx $ Verify the identity
$$\sin x = \frac {2 \tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}$$
Use this identity and the substitution $t = \tan\frac{x}{2}$ to evaluate the integral of $$ \int \frac{1}{\sin x} dx $$
My answer:
$$ A = \frac{x}{2} \Rightarrow \sin 2A= \frac{2\tan A}{1+\tan^2 A} ... | $$\int\frac{1}{\sin (x)} dx=\int\frac{\sin(x)}{\sin^2(x)}dx=\int\frac{\sin(x)}{1-\cos^2(x)} $$ Use substitution $t=\cos(x) \to dt=\sin(x)dx\to dx=\frac{dt}{\sin(x)}$
$$\int \frac{\sin(x)}{1-t^2}*\frac{dt}{\sin(x)}=\int \frac{dt}{1-t^2}=arth(t)=arth(\cos(x))+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Laurent expansion with different annuli Find the Laurent expansion about $0$ of $$f(z)= \frac{1}{(z-i)(z-2)}$$ on the annuli:
*
*$0 \lt \lvert z \rvert \lt 1 $,
*$ 1 \lt \lvert z \rvert \lt 2$,
*$ 2 \lt \lvert z \rvert \lt \infty $.
So far I have put the function into partial fractions so $ \frac{i+2}{5} \left(\... | First, we need to address your partial fraction decomposition since it is missing a multiplication by $-1$. You should have obtained
$$
\frac{1}{(z-i)(z-2)}=\frac{2+i}{5}\Bigl[\frac{1}{z-2}-\frac{1}{z-i}\Bigr]
$$
Recall that for $\lvert z\rvert < 1$, the geometric series $\sum_{n=0}^{\infty}z^n$ converges to $\frac{1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Column space of a 3x3 matrix Let $A$ be a $3 \times 3$ matrix $A$ with reduced row echelon form $
\left[ {\begin{array}{c}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{array} } \right]$.
Then
$\left[ {\begin{array}{c}
1 \\
0 \\
0
\end{array} } \right]$ and
$\left[ {\begin{array}{c}
0 \\
1 \\
0
\end{array} } \ri... | The column space of the original matrix is spanned by the columns of the original matrix which, when reduced to row echelon form, contain a pivot. So for a counterexample, consider:
$$
A = \begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 1 & 0
\end{bmatrix}
$$
whose row reduced echelon form has the required form. Notice t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x+1}\right)=x$ $\forall x\neq -1$ Given function $y=f(x)$
such that
$$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x+1}\right)=x \quad\forall x\neq -1$$
find $f(x)$ and $f(2007)$.
| $f\left(\dfrac{x-3}{x+1}\right)+f\left(\dfrac{x+3}{x+1}\right)=x$
$f\left(1-\dfrac{4}{x+1}\right)+f\left(1+\dfrac{2}{x+1}\right)=x$
$f\left(1-\dfrac{4}{x}\right)+f\left(1+\dfrac{2}{x}\right)=x-1$
$f(1-2x)+f(1+x)=\dfrac{2}{x}-1$
$f(3-2x)+f(x)=\dfrac{2}{x-1}-1~......(1)$
$f(3-2(3-2x))+f(3-2x)=\dfrac{2}{3-2x-1}-1$
$f(4x-3... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such th... | This might be irrelevant but it is equal to $2^{6n-1}+5 \cdot 9^n=2^{6n-6} \cdot 2^5+5\cdot 9^n$
Now since $2^6=64 \equiv -2 (mod 11)$, and $9^n \equiv -2(mod 11)$, the whole expression is equivalent to $(-2)^{n-1} \cdot 32+5 \cdot (-2)^n (mod 11) \equiv (-2)^n \cdot -16+5 \cdot (-2)^n (mod 11) \equiv 11 \cdot (-2)^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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If $\left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right)$ If $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = $
$\bf{My\; Try::}$ Given $... | HINT:
$$\dfrac54=1+\sin x+\cos x+\sin x\cos x=1+\sqrt2\cos\left(\frac\pi4-x\right)+\dfrac{2\cos^2\left(\dfrac\pi4-x\right)-1}2$$
Write $\cos\left(\frac\pi4-x\right)=u$
$$4u^2+4\sqrt2u-3=0$$
We need $-1\le u\le1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If both $a,b>0$, then $a^ab^b \ge a^bb^a$ Prove that $a^a \ b^b \ge a^b \ b^a$, if both $a$ and $b$ are positive.
| Given that both $a$ and $b$ are positive integers, let us consider the case where $b > a$.
$b$ can be expressed as $a+x$, where $x$ is some positive integer.
to prove $a^a b^b > a^b b^a$,we need to prove that $a^a b^b - a^b b^a > 0$
Rewrite $a^a b^b - a^b b^a$, by substituting $b = (a+x)$
$= a^a (a+x)^{a+x} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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"answer_id": 2
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$24x^7 + 5y^2 = 15$ has no integer solutions, having none $\!\bmod 12$ Prove $5a^2\equiv k \pmod{12}$, where $k\in \{0,5,8,9\}$. Hence show that the equation $24x^7 + 5y^2 = 15$ has no integer solutions.
My lecturer used a table containing $a$, $a^2$, and $5a^2$ from $1$ to $11$ to which he applied $\pmod{12}$
for ex.
... | Your lecturer calculated the remainder when $5a^2$ is divided by $12$, for $a=0$ to $11$. Actually we can stop at $6$, since for any $a$ we have $(12-a)^2\equiv a^2\pmod{12}$. (Indeed we could stop at $a=3$, since $(6-a)^2\equiv a^2\pmod{12}$.)
So calculating from $0$ to $6$, we find that modulo $12$, $5y^2$ can tak... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $\sum_{n=1}^{\infty}\left(\, \frac{1}{n} - \frac{1}{n + 2}\,\right)$ What criteria can I use to prove the convergence of
$$
\sum_{n=1}^{\infty}\left(\,{1 \over n} - {1 \over n + 2}\,\right)\ {\large ?}
$$
My idea was to use ratio test:
$$\displaystyle{1 \over n} - {1 \over n+2} = {2 \over n^{2} + 2n}$$
$... | Here's another way to prove the convergence
$$ \sum \limits_{n=1}^{\infty} \left(\frac1n - \frac{1}{n+2}\right)= \sum \limits_{n=1}^{\infty} \frac{n+2-n}{n(n+2)} $$
$$= \sum \limits_{n=1}^{\infty} \frac{2}{n^2+2n}=2 \sum \limits_{n=1}^{\infty} \frac{1}{n^2+2n} $$
Also note that
$$ \left|\frac{1}{n^2+2n}\right|\leq \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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Evaluate this Trigonometric Expression: $\sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$
Evaluate
$$ \sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$$
I found the following
*
*$\large{\cos \frac{2\pi}{7}+\cos \frac... | My solution:
Let $\alpha= \sqrt [3] {\cos \frac{2\pi}{7}}, \beta=\sqrt [3] {\cos \frac{4\pi}{7}}, \gamma=\sqrt [3] {\cos \frac{8\pi}{7}}$ so $(x-{\alpha}^3)(x-{\beta}^3)(x-{\gamma}^3)$ $=8x^3+4x^2-4x-1$.
hence we get $${\alpha}^3+{\beta}^3+{\gamma}^3=-\frac{1}{2} \tag{1}$$ $${\alpha}^3{\beta}^3+ {\beta}^3{\gamma}^3+ {\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Minimum value of $|z+1|+|z-1|+|z-i|$ How to find the minimum value of $|z+1|+|z-1|+|z-i|$.
I have tried geometrically etc but failed.
| The post has received good and simple answers based on triangles.
However, if you really want to see the calculus based problem, let me start using Dr. Sonnhard Graubner's approach to the solution. Writing $$f(a,b)=\sqrt{(a+1)^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}$$ Computing the partial derivatives lead to the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Rewrite $\sin(\cos^{-1}(x)-\tan^{-1}(y))$ as an algebraic function of $x$ and $y$. Rewrite the expression as an algebraic function of $x$ and $y$:
$$\sin(\cos^{-1}(x)-\tan^{-1}(y)).$$
I am unsure of how to change this into an algebraic function, yet I am able to simplify inso sin and cos.
thank you
| Let $\theta = \arccos x$. Then $\cos\theta = x$, so we can draw a right triangle with adjacent side of length $|x|$ and hypotenuse of length $1$. By the Pythagorean Theorem, the opposite side has length $\sqrt{1 - x^2}$.
Let $\varphi = \arctan y$. Then $\tan\varphi = y$, so we can draw a right triangle with oppos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many distinct values of floor(N/i) exists for i=1 to N. Say we have a function $F(i)=\text{floor}(N/i)$.
Then how many distinct values of $F(i)$ will exist for all $0 \leq i \leq N$
e.g. We have $N=25$ then.
$F(1)=25$
$F(2)=12$
$F(3)=8$
$F(4)=6$
$F(5)=5$
...
...
...
$F(24)=1$
$F(25)=1$
So total distinct values of $... | If $k \leqslant \sqrt{N}$, then $\lfloor N/\lfloor N/k\rfloor\rfloor = k$, since, letting $m = \lfloor N/k\rfloor$, we have $$mk\leqslant N < (m+1)k = mk + k \leqslant mk + m = m(k+1),$$ so that gives you $\lfloor \sqrt{N}\rfloor$ values. And the values of $\lfloor N/k\rfloor$ for $k \leqslant \lfloor \sqrt{N}\rfloor$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using trig substitution to solve for integration? So I used a trig sub for this problem:
$$\int \frac{1}{x^2\sqrt{9-x^2}}dx.$$
${x=3\sin\theta}$
${dx=3\cos\theta\ d\theta}$
${\sqrt{9-x^2}= 3\cos\theta}$
I ended up with
$$\frac19 \int \frac{ d\theta}{\sin^2\theta}$$
From there how do I solve that?
| $${\int{1\over x^2\sqrt{9-x^2}}}dx$$
Your substitutions are fine:
$$x = 3\sin \theta \implies dx = 3\cos \theta \,d\theta$$
But be careful about $$\sqrt{9 - (3\sin \theta)^2}=\sqrt{9\cos^2 \theta} = 3|\cos\theta|$$
So we have $$\int \frac{3\cos \theta d\theta}{9\sin^2 \theta\cdot 3|\cos \theta|} \;= \;\frac{1}9\int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to separate variables in this equation: $\;y\frac{dy}{dx} = (x+7)(y^2+6)\;?$ I need to solve the differential equation $$y\frac{dy}{dx} = (x+7)(y^2+6)$$
I know that the first step is to isolate both term each side and then integrate...
But I can't figure out how to isolate term on this one, I would probably be able... | $$y\dfrac{dy}{dx}=(x+7)(y^2+6)\implies \dfrac{y}{y^2+6}\dfrac{dy}{dx}=(x+7)\implies \dfrac{y}{y^2+6} dy=(x+7) dx\implies \int\frac{y}{y^2+6}dy=\int x+7 \,dx\implies \frac12\log|y^2+6|=\dfrac{x^2}2+7x\implies \log|y^2+6|=x^2+14x+c\implies y^2+6=e^{x^2+14x+c}\implies y=\sqrt{e^{x^2+14x+c}-6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Showing that a function diverges for large x Consider the function $$f(x) = \frac{x^{3}}{x^{2}+1}$$ Show that as $x\to\infty$ we have that $f(x)\to\infty$. ie. I want to prove that $$\lim_{x\to\infty} \frac{x^3}{x^2+1} = \infty.$$
So given $k>0$ I want to find $M>0$ such that $$x>M\implies |f(x)|>k.$$
First note that
... | Write $f(x) = x-{x \over x^2+1}$, and since $|{x \over x^2+1}| \le {1 \over 2}$, you have
$f(x) \ge x-{1 \over 2}$. Then for all $k>0$, if you let $M=k+1$, then you have the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How can I write this power series as a power series representation? How can I write this power series ($1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8....$) as a power series representation (like $\dfrac{1}{1-x}$ or something neat like that)?
| I would go about this by first splitting the series up:
$$1+x+2x^2+2x^3+3x^4+3x^5+...=(1+x)(1+2x^2+3x^4+...)$$
Letting $s=1+2x^2+3x^4$ we can do a few tricks:
$$s-x^2s=\begin{array}{c} 1&+2x^2&+3x^4+... \\ &-x^2&-2x^4-...\end{array}$$
$$=1+x^2+x^4+...$$
Which converges to $\frac{1}{1-x^2}$ for $-1 < x < 1$ (proving thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Series expansion of $\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}$? How would I find the series expansion $\displaystyle\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}$ so that it will turn into an infinite power series again??
| $$\frac{1}{1-x^k}=1+x^k+x^{2k}+x^{3k}\cdots$$
$$\frac{1}{1+x^k}=1-x^k+x^{2k}-x^{3k}\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to find LCM of this equation? Can $(x+1)(2x-1)$ be the LCM of this biquadratic equation
$$\frac{5x-1}{x+7}=\frac{3x+1}{x+5}$$
| The lowest common denominator (LCD) of rational fractions
$$\frac{5x-1}{x+7}\quad {\rm and} \quad \frac{3x+1}{x+5}$$
is the lowest common multiple (LCM) of denominator polynomials
$$ x+7 \quad {\rm and} \quad x+5, $$
which is
$$ (x+7)(x+5).$$
So, formally, the original equality of rational fractions can be re-written... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving the convergence of $\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$ How to prove that the series
$$\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$$
is convergent? What about finding the sum?
My attempt:
$$\ln (1-1/n^2)= \ln(n-1) -2\ln n + \ln(n+1)$$
The term in the log is decreasing so I get the feeling that th... | $$log(1-\frac{1}{n^2})= \\log(\frac{n^2-1}{n^2})=\\log(\frac{n-1}{n})+log(\frac{n+1}{n})\\s_{1}+s_{2}\\s_{1}=log(\frac{2-1}{2})+log(\frac{3-1}{3})+log(\frac{4-1}{4})+...++log(\frac{n-1}{n})=\\+log(\frac{1}{2}\frac{2}{3}\frac{3}{4}...\frac{n-1}{n})\\=log(\frac{1}{n})\\$$ $$
s_{2}=log(\frac{2+1}{2})+log(\frac{3+1}{3})+lo... | {
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"url": "https://math.stackexchange.com/questions/1078043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Cube roots escape $ \sqrt{\sqrt[3]{5}-\sqrt[3]{4}} \times 3 = \sqrt[3]{a} + \sqrt[3]{b} - \sqrt[3]{c}, $
where $ a, b $ and $ c $ are positive integers. What is the value of $ a+b+c $?
This question appeared in one of my exams
| You can check the following expression works by squaring the right hand side
$$
\sqrt{\sqrt[3]{5}-\sqrt[3]{4}} \times 3 = \sqrt[3]{2} + \sqrt[3]{20} - \sqrt[3]{25} \, .
$$
So $ a + b + c = 47 $.
As for finding it: It is initially a good idea to try working in the field extension $ \mathbb{Q}(\sqrt[3]{4}, \sqrt[3]{5}) =... | {
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"answer_id": 0
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Generalisation of $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ After seeing the neat little identity $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ somewhere on MSE, I tried generalising this to higher consecutive powers in the form
$\sum_{k=0}^a\epsilon_k(n+k)^p=C$, where $C$ is a constant and $\epsilon_k=\pm1$. I discovered a relatively simple ... | Let $X = \mathbb{R}^{\mathbb{Z}}$ be the space of real valued sequences defined over $\mathbb{Z}$. Let $R : X \to X$ be the operator on $X$ replacing the terms of a sequence by those on their right. More precisely,
$$X \ni (\ell_n)_{n\in\mathbb{Z}} \quad\mapsto\quad ( (R\ell)_n = \ell_{n+1} )_{n\in\mathbb{Z}} \in X$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 2,
"answer_id": 1
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How can find this limit? $\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$ $$\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$$
plugging in infinity gives infinity - infinity, what kind of manipulation can I do to solve this?
| $$\sqrt{x^2+6x}=\sqrt{x^2+6x+9-9}=\sqrt{(x+3)^2-9}\approx |x|+3,\text{ as }x\to-\infty$$
$$\sqrt{x^2-2x}=\sqrt{x^2-2x+1-1}=\sqrt{(x-1)^2-1}\approx |x|-1,\text{ as }x\to-\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.
| Rewrite the given term as
$$x\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}=\frac{x}{\sqrt{x^2+1}+x}$$
Assuming $x>0$, we have
$$\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}$$
This clearly tends to $\frac{1}{2}$ as $x\rightarrow\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
How to integrate $f(x) = \frac{1}{a + b \cos x + c \sin x }$ over $x \in (0,\pi/2)$
Conjecture 1
$$ \begin{align*} I_{T}=\int_0^{2\pi} \frac{\mathrm{d}x}{a + b \cos x + c
\sin x} & = \frac{2\pi}{\rho} \tag{1} \\ I_{T/4} = \int_0^{\pi/2} \frac{\mathrm{d}x}{a +
b \cos x + c \sin x} & = \frac{2}{\rho} \arctan \lef... | First we need a lemma or two.
Lemma 1:
$ \hspace{3.75cm} \displaystyle
\int \frac{\,\mathrm{d}x}{1+x^2} = \arctan x+\mathcal{C}
$
Proof: Use the substitution $x \mapsto \tan u$. Then we have
$$
\mathrm{d}x
= \left( \frac{\sin u}{\cos u} \right)'\mathrm{d}u
= \frac{(\sin u)' \cdot \cos u - \sin u \cdot (\cos u)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
What is wrong with the sum of these two series? Could anyone help me to find the mistake in the following problem? Based on the formula of the sum of a geometric series:
\begin{equation}
1 + x + x^{2} + \cdots + x^{n} + \cdots = \frac{1}{1 - x}
\end{equation}
\begin{equation}
1 + \frac{1}{x} + \frac{1}{x^{2}} + \cdots ... | The first series only applies when $|x| < 1$ whereas the second series only applies when $\left|\frac{1}{x}\right| < 1$ (i.e. $|x| > 1$). By adding them, you are assuming that they both apply simultaneously, but they don't (for any $x$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 0
} |
Find Min Values Of $P=\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{4}{(1+x)(1+y)(1+z)}$ Given $x,y,z>0$ and $y+z=x(y^2+z^2)$
Find Min Values Of
$P=\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{4}{(1+x)(1+y)(1+z)}$
Could Someone give me an idea ?
| let $y+z=2t\implies y^2+z^2\ge 2t^2 \implies xt\le 1$
$P\ge \dfrac{1}{(1+x)^2}+\dfrac{2}{(1+y)}\dfrac{1}{(1+z)}+\dfrac{4}{(1+x)(1+y)(1+z)}=\dfrac{1}{(1+x)^2}+\dfrac{2}{(1+y)(1+z)}(1+\dfrac{2}{(1+x)})\ge \dfrac{1}{(1+x)^2}+\dfrac{2*4}{(1+y+1+z)^2}(1+\dfrac{2}{(1+x)})={(1+x)^2}+\dfrac{2}{(1+t)^2}(1+\dfrac{2}{(1+x)})\ge {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluate Complex Integral with $\frac{\Gamma(\frac{s}{2})} {\Gamma\big({\beta +1\over 2} - {s\over 2}\big)}$ I am proving this integral:
$$ \int_{c\ -\ j\infty}^{c\ +\ j\infty} \left(\,x^{-1}\sigma\beta^{1 \over 2}\,\right)^{s}\ \Gamma\left(\,s \over 2\,\right) \Gamma\left(\,{\beta +1 \over 2} - {s \over 2}\,\right)\,{... | The equation should actually be
$$\frac{1}{4\pi j} \int_{c - j\infty}^{c + j\infty} (x^{-1}\sigma \beta^{\frac{1}{2}})^s\, \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{\beta + 1}{2} - \frac{s}{2}\right)\, \mathrm{d}s = \Gamma\left(\frac{\beta + 1}{2}\right) \left[1 + \frac{1}{\beta}\left(\frac{x}{\sigma}\right)^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Interesting Surd Problem If $ x $, $ y $ and $ z $ are rational numbers such that
$ \sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}$ then find $ x,y,z $
| According to Srinivasa Ramanujan,
$\sqrt[3]{\sqrt[3]{2}-1}
=\sqrt[3]{\frac{1}{9}}
-\sqrt[3]{\frac{2}{9}}
+\sqrt[3]{\frac{4}{9}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Double integral $ \iint \limits_D \frac{y}{x^2+(y+1)^2}dxdy$, $D$=$\{(x,y): x^2+y^2 \le1 , y\ge0\}$
Solve $$ \iint \limits_D \frac{y}{x^2+(y+1)^2}dxdy \ \ \ \ . . . \ (*)$$
where $D$=$\{$$(x,y): x^2+y^2 \le1 , y\ge0 $$\}$
$$
$$
Here is my attempt.
$$\begin{align}
&(1).\ \ \ (*)=\int_{-1}^1 \int_{0}^{\sqrt{1... | I shall introduce new coordinates (again denoted by $x$, $y$) such that the point $(0,-1)$ becomes the origin, and your vertical axis is my horizontal axis. Your integral then appears as
$$J:=\int_H{x-1\over x^2+y^2}\>{\rm d}(x,y)\ ,$$
where $H$ is the right half of the unit disk with center $(1,0)$. Introducing polar ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
Arithmetic Error in Calculation of the Limit of a Given Function I consider a function $f(x)$ which is equal to $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6}$
While trying to evaluate the $\lim_{x \to 6} f(x)$
It is true that $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6} = \dfrac{\frac{3}{x}}{x-6}-\dfrac{\frac{1}{2}}{x-6}$
It is also ... | $$
\frac 3 x\cdot 2x = 6\text{ since $x$ cancels, and }\frac 1 2\cdot 2x = x\text{ since $2$ cancels.}
$$
Therefore
$$
\frac{\frac{3}{x}-\frac{1}{2}}{x-6}=\frac{\left(\frac{3}{x}-\frac{1}{2}\right)2x}{(x-6)2x} = \frac{6-x}{(x-6)2x} = \frac{-1(x-6)}{(x-6)2x} = \frac{-1}{2x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Computing the exponential of the operator $ H(\textbf{x}) = \alpha \textbf{n} \times\textbf{x}$ Define the vector operator: $$ H(\textbf{x}) \equiv \alpha \textbf{n} \times\textbf{x}$$ For unit vector $\textbf{n}$ and some constant $\alpha$. We define further the operator: $$G \equiv I + H + \frac{H^2}{2!} + \frac{H^3}... | $H x = \alpha \, n \times x$ gives the matrix:
$$
H = \alpha \left(
\begin{matrix}
0 & -n_3 & n_2 \\
n_3 & 0 & -n_1 \\
-n_2 & n_1 & 0
\end{matrix}
\right) = \alpha N
$$
Then for your proof
$$
G = I + \sin(\alpha) N + (1-\cos\alpha) N^2
$$
needs to be shown.
So we calculate $N^2$ and $N^3$:
\begin{align}
N^2
&=
\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$? The inequality
$$
e_n:=\left(1+\frac1n\right)^n\leq3-\frac1n,
$$
where $n\in\mathbb{N}_+$, is certainly true, because we know, how LHS is connected with $e$. The other argument is the standard proof of boundedness of $(e_n)$, which uses the binomial theorem.
A... | You can use induction.
For $n=1$, clearly $2 \leq 2$. Assume it holds for some $n-1 \in \mathbb{N}$. Then,
$$\left(1+ \frac{1}{n} \right)^n = \left( 1 + \frac{1}{n} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 1 + \frac{1}{n-1} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 3- \frac{1}{n-1} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
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How to find sum of this series? $\sum_{n=1}^{+\infty}(-1)^{n+1}\prod_{k=1}^n\frac{3k-2}{5k}$ How to find sum of this series?
$$\frac{1}{5}-\frac{1\cdot 4}{5\cdot 10}+\frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15}-\ldots = \sum_{n=1}^{+\infty}(-1)^{n+1}\prod_{k=1}^n\frac{3k-2}{5k}$$
Any hint is welcome.
| The sum of the series is: $$1-\frac{1}{2}\sqrt[3]{5}.$$
As suggested by Abhishek Bakshi, this comes from considering the Taylor series of
$$ f(x) = \left(1+\frac{3}{5}x\right)^{1/3} $$
around $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1088355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$y=\frac{1}{x}$, show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find $\frac{d^2y}{dt^2}$ If $y=\frac{1}{x}$, where $x$ is a function of $t$ show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find an expression for $\frac{d^2y}{dt^2}$.
For $\frac{d^2y}{dt^2}$ I keep getting
$$\frac{d^2y}{dt^2}=\frac{-... | You are misapplying the product rule. Note that
$$\begin{aligned}
\frac{d}{dt} \left( - \frac{1}{x^2} \frac{dx}{dt} \right) & = \left( \frac{d}{dt} \frac{(-1)}{x^2} \right) \frac{dx}{dt} - \frac{1}{x^2} \left( \frac{d}{dt} \frac{dx}{dt} \right) \\
& = \left( \frac{2}{x^3} \cdot \frac{dx}{dt} \right) \frac{dx}{dt} - \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1088976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Formalize a proof without words of the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$ This website gives the following proof without words for the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$.
I find it interesting but have trouble seeing the proof behind it. Could anyone could give me ... | Notice that the base of the $n^\text{th}$ cube is the diagonal square prism $n^2\times 1$, and that each cube is build out of the diagonal square prism and rectangular prisms above and to the left of the diagonal square prism. Since adding another term to the left hand sum will give us one more diagonal square and set... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
How to solve $\tan x =\sin(x+45^{\circ})$? How do I solve $\tan x = \sin(x +45^{\circ})$?
This is how far I have come: $\sqrt{2}\sin x = \sin x\cdot\cos x + \cos^2 x$
| we will show that the only solution to $\tan x = \sin(x+45^\circ)$ is $x = 45^\circ$
here is another way to solve $\tan t = \sin (t + 45^\circ).$ i have changed the independent variable to $t$ because i want to use $x = \cos t$ and $y = \sin t.$
using the addition formula for $\sin$ we can rewrite the equation as two ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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How does polynomial long division work?
I get normal long division but this doesn't make sense. How can doing it by only dividing by the leading term work? The problem is $$\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3},$$ not $$\dfrac{3x^3 - 2x^2 + 4x - 3} {x^2}.$$
| As said, "it works the same way as euclidean division works for integers" (quoting Bernard). So, if you take $$A=\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}$$ The division of $3x^3$ by $x^2$ (the highest terms) gives $3x$. So $$A=3x+\Big(\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}-3x\Big)=3x -\frac{11 x^2+5x+3}{x^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
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$\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}=\prod_{k=1}^{n}\bigl(1-\frac{1}{2k}\bigr)$
i cant see why we have :
*
*$$\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}$$
*$$\dfrac{(2n-1)!!}{(2n)!!} =\prod_{k=1}^{n}\left(1-\dfrac{1}{2k}\right),$$
Even i see the notion of Double factorial
this question is relat... | Note first that $(2n)!=(2n)!!(2n-1)!!$, because $(2n)!!$ gives you the even factors in $(2n)!$, and $(2n-1)!!$ gives you the odd factors. Now
$$\begin{align*}
(2n)!!&=(2n)(2n-2)(2n-4)\ldots(4)(2)\\
&=\big(2n\big)\big(2(n-1)\big)\big(2(n-2)\big)\ldots\big(2(2)\big)\big(2(1)\big)\\
&=2^nn!\;,
\end{align*}$$
so
$$\frac{(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on I am noticing this pattern:
\begin{align}
1+2&=3\\
4+5+6&=7+8\\
9+10+11+12&=13+14+15 \\
16+17+18+19+20&=21+22+23+24 \\
&\vdots
\end{align}
Is there a general formula that expresses a pattern here, such as for the $n$th term, or the closed form of the entire... | Proof of the Identities
$$
\begin{align}
n^2&=\sum_{k=1}^nn\tag{1}\\
n^2+\sum_{k=1}^n\left(n^2+k\right)&=\sum_{k=1}^n\left(n^2+n+k\right)\tag{2}\\
\sum_{k=0}^n\left(n^2+k\right)&=\sum_{k=n+1}^{2n}\left(n^2+k\right)\tag{3}
\end{align}
$$
Explanation:
$(1)$: multiplication written as a sum
$(2)$: add $\sum\limits_{k=1}^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 1
} |
Trigonometric Integration with Negative Exponents How do you integrate $\csc^4 x/\cot^2 x$? I know that this is the same as $\csc^4 x \cot^{-2} x$ and when you use techniques in trig integrals you end up with
$$\int \csc^2 x \csc^2 x \cot^{-2} x \,dx = \int \left(1 + \cot^2 x\right)\left(\cot^{-2} x\right)\csc^2 x \,dx... | Another approach is to use
$\displaystyle\int\frac{\csc^{4}x}{\cot^{2}x}dx=\int\frac{1}{\sin^{4}x}\cdot\frac{\sin^{2}x}{\cos^{2}x}dx=\int\frac{1}{\sin^{2}x\cos^{2}x}dx=\int\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}dx$
$\displaystyle=\int\left(\frac{1}{\cos^{2}x}+\frac{1}{\sin^{2}x}\right)dx=\int(\sec^{2}x+\csc^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate $ \int_{-2}^{1}\sqrt{\frac{\left ( 1-x \right )\left ( x+2 \right )^2}{x+3}}\,dx$ I want to evaluate the integral:
$$\int_{-2}^{1}\sqrt{\frac{\left ( 1-x \right )\left ( x+2 \right )^2}{x+3}}\,dx$$
but I have no clue of how to start in order to attack it.
I cannot even split that root because it does not make ... | Hint:
Notice that
$$\int_{-2}^{1}\sqrt{\frac{\left ( 1-x \right )\left ( x+2 \right )^2}{x+3}}\,dx = \int_{-2}^{1} (x+2)\sqrt{\frac{1-x}{x+3}}\,dx, \text{because $ x\geq -2$}.$$
Take $t^2 = \frac{1-x}{x+3} \implies x= \frac{1-3t^2}{t^2+1}\implies dx = \frac{-8t}{(t^2+1)^2}\,dt.$
Hence,
$$\int_{-2}^{1}\sqrt{\frac{\left ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1096502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this limits $\lim_{n\to\infty}\frac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\frac{1}{8}$ Interesting Question:
Let denote by $v_{p}(a)$ the exponent of the prime number $p$ in the prime factorization of $a$,
show that
$$\lim_{n\to\infty}\dfrac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\c... | Exploiting:
$$\sum_{k=1}^{+\infty}\left\lfloor\frac{n}{5^k}\right\rfloor = \frac{n}{4}+O(1)$$
there is left very little to do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$ I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$
$$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{... | $$\begin{align}
\text{Your sum} &= \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} \dots \\
\\
{\rm e} &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}\dots
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 2
} |
Numbers that are the sum of the squares of their prime factors A number which is equal to the sum of the squares of its prime factors with multiplicity:
*
*$16=2^2+2^2+2^2+2^2$
*$27=3^2+3^2+3^2$
Are these the only two such numbers to exist?
There has to be an easy proof for this, but it seems to elude me.
Thanks
| Maple confirms that $A$ and $B$ are prime, and that $N$ is a solution to the question. $N$ has 179 digits.
$$N=45AB\\
A=(C^{107}+D^{107})/(C+D)\\
B=(C^{109}+D^{109})/(C+D)\\
C=(\sqrt{47}+\sqrt{43})/2\\
D=(\sqrt{47}-\sqrt{43})/2$$
The general solution to $3^2+3^2+5^2+A^2+B^2=3×3×5×A×B$ is a sequence with recursion
$$a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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How many roots does $(1+z^3)^8 = (1 + z^4)^6$ have? I got stuck on this question:
How many roots does $(1+z^3)^8 = (1 + z^4)^6$ have? (including complex roots and roots with multiplicity)
My attempt at a solution:
First we can write the equation as:
$$(1+z^3)^8 - (1 + z^4)^6 = 0$$
We can then factor $(1+z^3)$
$$(z+1)^... | $(z^3+1)^8 = (z^4+1)^6$
$z^{24} + 8z^{21} + 28z^{18} + 56z^{15} + 70z^{12} + 56z^9 + 28z^6 + 8z^3 + 1 = z^{24} + 6z^{20} + 15z^{16} + 20z^{12} + 15z^8 + 6z^4 + 1$
$z^{24} + 8z^{21} + 28z^{18} + 56z^{15} + 70z^{12} + 56z^9 + 28z^6 + 8z^3 + 1 - z^{24} - 6z^{20} - 15z^{16} - 20z^{12} - 15z^8 - 6z^4 - 1 = 0$
$z^3(8z^{18} -... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Limit $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$ I try to calculate $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$. So, $\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4} = \frac{(x-2)x}{(x+2)(x+\sqrt{(x-2)^2(x+1)}-2)}$ but I don't know what to do next.
| $$\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}} =$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 2x^2-x^2 + 4}-(x -2)}{(x - 2)(x+2)}}=$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{x^2(x-2)-(x-2)(x+2)}}{(x - 2)(x+2)}}-\lim \limits_{x \to 2}{\frac{1}{x+2}}=$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{(x-2)(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a polynomial given the remainders of the division of that polynomial with 3 other polynomials
A polynomial from $ \mathbb{C}[x]$ divided by $ x - 1$, $x + 1$, $ x -2$ has the remainders 2, 6 and 3. Find the remainder of the division of that polynomial by $(x-1)(x+1)(x-2)$
The degree of the expression $(x-1)(x+1)... | Lets call this polynomial $P(x)$ than by the conditions you have that the polynomial $P(x)$ can be written as $$P(x)=(x-1)Q_1(x)+2\\P(x)=(x+1)Q_2(x)+6\\P(x)=(x-2)Q_3(x)+3$$ From this you can see that $P(1)=2,P(-1)=6,P(2)=3$
Now you can also write your polynomial as
$$P(x)=(x-1)(x+1)(x-2)Q_4(x)+ax^2+bx+c$$
Where $ax^2+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determinant equation. let
A = [[x,-1],
[3,(1-x)]]
and
B = [[1,0,-3],
[2,x,-6],
[1,3,(x-5)]]
solve for x;
det(A) = det(B)
on expanding both the determinants and equating them, i obtain;
$-2x^2+3x+3 = 0$
completeing the square
$2[x^2-\frac{3x}{2}] -3 = 0$
$2[(x-\frac{3}{4})^2 - \frac{9}{16}] -3 = 0$
$2(x-\... | You have properly solved this exercise. Actually you can simplify the result:
$$x = \frac{3 \pm \sqrt{33}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1104432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove that $\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1$ From this geometry problem, I can not find geometry solution.
However the answer is $X=\frac{2\pi}{15}$ by geometry method.
Then I get the identity $$\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\ta... | I'd try using that
$$\frac{1}{5} - \frac{1}{2} + \frac{1}{3} = \frac{1}{30},$$
$$\frac{4}{3} - \frac{6}{5} = \frac{2}{15}.$$
And the formula:
$$
\tan(a\pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a\tan b}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Show that x is independent of this trig expression For some reason I cannot get the solution.
Show that $x$ is independent of: $\sin^2(x+y)+\sin^2(x+z)-2\cos(y-z)\sin(x+y)\sin(x+z)$
I have used all the identities but seem to be missing something.
| $\begin{align}
& \sin^2(x+y)+\sin^2(x+z)-2\cos(y-z)\sin(x+y)\sin(x+z) \\
&=\dfrac{1}{2}(1-\cos(2x+2y)) + \dfrac{1}{2}(1-\cos(2x+2z)) - 2\cos(y-z)\sin(x+y)\sin(x+z)\\
&=1-\cos(2x+y+z)\cos(y-z) - 2\cos(y-z)\sin(x+y)\sin(x+z)\\
&=1-\cos(y-z)\left[\cos(2x+y+z) + 2\sin(x+y)\sin(x+z) \right]\\
&=1-\cos(y-z)\left[\cos(2x+y+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate the double integral $\int _0^1\int _0^1\frac{x+i}{(1-ix y) \ln (x y)} \,dx\,dy$ We know that
$$\int _0^1\int _0^1\frac{x-1}{(1+x y) \ln (x y)} \, dx\,dy=\gamma$$
$$\int _0^1\int _0^1\frac{x+1}{(1-x y) \ln (x y)}\,dx\,dy=\ln \frac4\pi$$
I wonder what would be
$$\int _0^1\int _0^1\frac{x+i}{(1-ix y) \ln (x y... | It is quite easy to prove that:
$$\iint_{(0,1)^2}\frac{(xy)^n}{\log(xy)}\,dx\,dy = -\frac{1}{n+1},\tag{1}$$
$$\iint_{(0,1)^2}\frac{x^{n+1}y^n}{\log(xy)}\,dx\,dy = -\log\frac{n+2}{n+1},\tag{2}$$
hence:
$$\begin{eqnarray*} I &=& \iint_{(0,1)^2}\frac{(x+i)}{(1-i xy)\log(xy)}\,dx\,dy = \sum_{n\geq 0}\iint_{(0,1)^2}\frac{i^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Convert the Polar Equation to Cartesian Coordinates $$
r^2=\sec 4\theta
$$
I graphed this equations using Wolfram Alpha and found it to be 2 hyperbolas. I'm having difficulty showing this using the standard equations
$$
x=r\cos\theta \;, \; y=r\sin\theta \;, and \; x^2 +y^2 =r^2
$$
My work so far:
$$
r^2 = \sec4\theta... | Hint
$$\dfrac{y^2}{x^2}=\tan^2{\theta}=\dfrac{\sin^2{\theta}}{1-\sin^2{\theta}}\Longrightarrow \sin^2{\theta}=\dfrac{y^2}{x^2+y^2}$$
and you have
$$x^2+y^2=r^2=\dfrac{1}{1-8\sin^2{\theta}+8\sin^4{\theta}}=\dfrac{1}{1-\dfrac{8y^2}{x^2+y^2}+8\left(\dfrac{y^2}{x^2+y^2}\right)^2}$$
so
$$x^2+y^2=\dfrac{(x^2+y^2)^2}{(x^2+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How to show $\lceil ( \sqrt3 +1)^{2n}\rceil$ where $n \in \mathbb{ N}$ is divisible by $2^{n+1}$
Show that $ \left\lceil( \sqrt3 +1)^{2n}\right\rceil$ where $n \in \mathbb{N}$ is divisible by $2^{n+1}$.
I wrote the binomial expansion of $ ( \sqrt3 +1)^{2n}$ and $( \sqrt3 -1)^{2n}$ and then added them to confirm th... | Let:
$$ A_n = (\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n} = (4+2\sqrt{3})^n+(4-2\sqrt{3})^{n}.\tag{1} $$
Since $0<4-2\sqrt{3}<\frac{2}{3}$, we have that $A_n$ is the integer giving the ceiling of $(\sqrt{3}+1)^{2n}$.
Now we have:
$$ A_0 = 2, \quad A_1 = 8,\qquad A_{n+2} = 8 A_{n+1} - 4 A_{n}\tag{2} $$
and if we take:
$$ \nu_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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How to find the x intercepts
$\frac{4}{3} e^{3x} + 2 e^{2x} - 8 e^x$
I have some confusion especially because of the e
how can I approach the solution?
The solution of the x-intercept is 0.838
Many thanks
| \begin{align}
0&=\frac{4}{3}e^{3x}+2e^{2x}-8e^x\tag{1} \\[1em]
& = \frac{\frac{4}{3}e^{3x}+2e^{2x}-8e^{x}}{2e^x}\tag{2} \\[1em]
& = \frac{2}{3}e^{2x}+e^x-4\tag{3} \\[1em]
\end{align}
Now let $\xi=e^x,\therefore e^{2x}=\left(e^x\right)^2=\xi^2.$ This gives us
\begin{align}
0&=\frac{2}{3}\xi^2 +\xi-4\tag{4} \\[1em]
\ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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the general solution of $y(n+3)-\frac{2}{3}y(n+1)+\frac{1}{3}y(n) = 0$ I have some trouble finding the correct solution for the difference equation
$$y(n+3)-\frac{2}{3}y(n+1)+\frac{1}{3}y(n) = 0$$
I've found that the characteristic equation of the difference equation is $\lambda^3-\frac{2}{3}\lambda+\frac{1}{3}$.
By co... | Wolfram Alpha gives the roots as
$-1$ and
$1/6 (3\pm i \sqrt(3))$.
This last is
$\frac{1\pm i\sqrt{1/3}}{2}$
which is your answer.
My guess is
either you misread WA's answer
or you entered the equation incorrectly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$\sum_1^n 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} $ converge or not? how to check if this converge? $$\sum_{n=1}^\infty a_n$$
$$a_n = 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}$$
what i did is to show that:
$$a_n =2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} > 2\sqrt{n} - \sqrt{n+1} - \sqrt{n+1} = 2\sqrt{n} - 2\sqrt{n+1} = -2({\sqrt{n+1... | Hint
It's a telescoping sum. With $a_n := \sqrt n - \sqrt{n+1}$, $a_n - a_{n-1} = 2\sqrt n - \sqrt{n-1} - \sqrt{n+1}$ so
$$\sum_{i=1}^n 2\sqrt i - \sqrt{i-1} -\sqrt{i+1} = 1 + \sqrt n - \sqrt{n+1}$$
NB that your estimate contains an error in the last step, [fixed now]
$$2 \sqrt n - 2 \sqrt{n+1} = 2 ( \sqrt n - \sqrt{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find smallest $n \in \mathbb{N}$ s.t $A^n=I$ Let $A$ be $2 \times 2$ matrix:
$$
\left(
\begin{matrix}
\sin\frac{\pi}{18} \\ \sin\frac{4\pi}{9}
\end{matrix}
\begin{matrix}
-\sin\frac{4\pi}{9} \\ \sin\frac{\pi}{18}
\end{matrix} \right) $$
Then find smallest $n \in \mathbb{N}$ s.t $A^n=I$.
Looking at this questio... | Computing directly gives that the eigenvalues of $A$ are $\lambda_{\pm} := \exp \left(\pm \frac{4 \pi i}{9}\right)$, so there is a some matrix $P$ such that
$$A := P^{-1} \begin{pmatrix} \lambda_+ & 0 \\ 0 & \lambda_-\end{pmatrix} P.$$
So, for any integer $n$,
$$A^n = (P^{-1} A P)^n = P^{-1} \begin{pmatrix} \lambda_+ &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Modular Arithmetic, Pythagorean triples I'm not sure if this question should be under Modular Arithmetic, but that's where it was in my book.
Show that, if $x$, $y$, and $z$ and integers such that $x^2 + y^2 = z^2$, then at least one of $\{x, y, z\}$ is divisible by $2$, at least one of $\{x, y, z\}$ is divisible by $3... | If $x^2$ and $y^2$ are both odd $z^2$ is even.
If $x^2$ and $y^2$ are not multiples of $3$ they are $1 \bmod 3$ so $z^2$ is $2\bmod 3$, which is not a quadratic residue.
If $x^2$ and $y^2$ are not multiple of $3$ they are either $1$ or $-1 \bmod 5$. So $z^2$ is $0,2$ or $3\bmod 5$. $2$ and $3$ are impossible because th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Convergence of $ \sum_{n=1} ^\infty \frac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$ Convergence of $$ \sum_{n=1} ^\infty \dfrac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$$
Attempt: I believe not a nice attempt: $ n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) \leq n( 1+1+... | Using $GM\ge HM$ $$\left( 1\cdot\frac{1}{2}\cdot \frac{1}{3}\cdots \frac{1}{n}\right)^{\frac{1}{n}}\ge \frac{n}{\frac{1}{1+\frac{1}{2}+ \frac{1}{3}+\cdots \frac{1}{n}}}$$ so that $$\frac{1}{n}\cdot\frac{1}{1+\frac{1}{2}+ \frac{1}{3}+\cdots \frac{1}{n}}\ge (n!)^{\frac{1}{n}}$$ and as $(n!)^{\frac{1}{n}}>1 \forall n\ge2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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If $a^2+b^2+c^2=1$ then prove the following. If $a^2+b^2+c^2=1$, prove that $\frac{-1}{2}\le\ ab+bc+ca\le 1$.
I was able to prove that $ ab+bc+ca\le 1$. But I am unable to gain an equation to prove that
$ \frac{-1}{2}\le\ ab+bc+ca$ .
Thanks in advance !
| $0\leq(a+b+c)^2=(a^2+b^2+c^2)+2(ab+ac+bc)\implies ab+ac+bc\geq-\dfrac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Working Out Easy Equations does anyone know how to do this equation? I know it's easy but I can't work out what the question means.
When I expanded the first equation:
$(y+4)-(y-3)$
$y^2 -3y +4y - 12$
$y^2-1y-12$
Not sure what I should do after.
Can someone explain how would you work it out in easy terms?
Thank you
| $$(y + 4) - (y - 3) = (y + 4) - (y) - (-3) = (y+4) + (3 - y) = (y - y) + (3 + 4) = 7$$
$$(y-2) - (y - 3) = (y - 2) - y - (-3) = (y-2) + (3 - y) = (y - y) + (3 - 2) = 1$$
In general,
$$(y-a) - (y - b) = b - a$$
In the same way that $x + (y + z) = x + y + z$, $x - (y + z) = x - y - z$. The minus sign distributes to all t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 0
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Combinatorics, surjective functions with conditions Question: $A=\left\{ 1,2,3,4,5\right\} $
, $B=\left\{ 1,2,3\right\} $
. How many surjective functions are there such that $ f(1)\neq1$
,$f(2)\neq2$
,$ f(3)\neq2$
.
Solution: Overall we have $3^{5}-{3 \choose 1}2^{5}+{3 \choose 2}1^{5}=150$
functions that is s... | Your calculation for $|A_{11}|$ gives the number of surjective maps $f:\{1,2,3,4,5\}\to\{1,2,3\}$ such that $f(1)=1$ and $f(\{2,3,4,5\})=\{1,2,3\}$. However, because $f(1)=1$, you do not need $f(\{2,3,4,5\})=\{1,2,3\}$, you only need $f(\{2,3,4,5\})\supseteq\{2,3\}$. You need to make similar corrections to various ot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rationality and triangles
Consider a triangle with angles $\alpha, 5\alpha, 180-6\alpha$. What is the minimum perimeter of that triangle, if it has integer sides and $5\alpha<90$?.
Let's call the sides that face each angle $x,y,z$ respectively, and set $\cos \alpha =t$. Using the law of sines , $x=2R\sin\alpha, y=2R\... | Finally solved the problem! for shortness, let's set $2\cos18=x$. From the inequalities $a\le2b$ and $2\cos(18)b\le a$, we see the trivial inequalities:
$$a^5\ge(xb)^5,a^4\ge(xb)^4, -a^3\ge-8b^3,-a^2\ge-4b^2,a\ge xb$$
So
\begin{align*}
P(a,b)&=a^5+a^4b-4a^3b^2-3a^2b^3+3ab^4+2b^5\\
&\ge b^5(x^5+x^4-32-12+3x+2)\\
&\appro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$ Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$.
I am not sure how to approach this problem. Should I divide this problem into multiple cases based on whether $a$,$b$,$c$ and odd/even or i... | Another way:
$$a^2 + b^2 + c^2 = \frac{a^2 + b^2}{2} + \frac{b^2 + c^2}{2} + \frac{c^2 + a^2}{2} \ge ab + bc + ca$$
The last part follows from the fact that $x^2 + y^2 \ge 2xy$ for all $x, y$ (you can get this by observing $x^2 + y^2 - 2xy = (x - y)^2 \ge 0$ for all $x,y$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Trig substitution fails for evaluating $ \int \frac{\cos x \sin x}{\sin^2{x} + \sin x + 1} dx$? Evaluate the integral
\begin{equation}
\int \frac{\cos x \sin x}{\sin^2{x} + \sin x + 1} dx
\end{equation}
Basically I could substitute: $t = \sin x$ and get:
$$\int \frac{t}{t^2 +t + 1} dt$$
But, although it seems a reason... | $$\int \frac{t}{t^2 +t + 1} dt = \int \frac{t}{\left(t+\frac 12\right)^2 + \frac 34}\,dt$$
$$\frac 34 = \left(\frac{\sqrt 3}2\right)^2$$
So , in therms of substitution, $$t+\frac 12 = \frac {\sqrt 3}2 \tan\theta\implies dt = \frac {\sqrt 3}2 \sec^2\theta\,d\theta,\quad \text{and}\;t = \frac{\sqrt 3}2\tan\theta - \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Determine the set of points $z$ that satisfy the condition $|2z|>|1+z^2|$ Determine the set of points $z$ that satisfy the condition $|2z|>|1+z^2|$
I tried to redo this problem and got to this point
$|2z|>|1+z^2|$ $\Rightarrow$ $2|z|>1+|z^2|$ $\Rightarrow$ $2|z|>1+z\overline z$
Let $z=x+iy$ then
$$2\sqrt{x^2+y^2}>1+(x... | Let $z=re^{i\theta}$. Then after squaring both sides we have
$$
4r^2 > r^4 + 2r^2\cos(2\theta) + 1\Rightarrow
0>(r^2-2r\sin(\theta)-1)(r^2+2r\sin(\theta)-1)\tag{1}
$$
where I used the following identity $\cos(2\theta) = 1 -2\sin^2(\theta)$. Let's write $(1)$ in Cartesian coordinates so recall that $r^2 = x^2 + y^2$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x\rightarrow\pi}\frac{\sin x}{x^2-\pi ^2}$ without L'Hopital I need to calculate the following limit (without using L'Hopital - I haven't gotten to derivatives yet):
$$\lim_{x\rightarrow\pi}\frac{\sin x}{x^2-\pi ^2}$$
We have $\sin$ function in the numerator so it looks like we should somehow make thi... | \begin{align}
\frac{\sin(x)}{x^2 - \pi^2} = \frac{x}{x^2(1-\pi^2/x^2)} \prod_{n=1}^{\infty}\left(1 -\frac{x^2}{n^2\pi^2}\right) = \frac{1}{x} \frac{1-x^2/\pi^2}{1-\pi^2/x^2} \prod_{n=2}^{\infty} \left(1 -\frac{x^2}{n^2\pi^2}\right)
\end{align}
Since no L'Hospital available, we consider of the sign of numerator and deno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Calculation of determinant of an arrowhead matrix Is there any easier way to make sure the determinant of the following $n \times n$ matrix is $n$?
$$\begin{vmatrix}
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 1 & 0 & \cdots & 0 \\
1 & 0 & 0 & 1 & \cdots & 0 \\
\vdots &... | $$
\begin{vmatrix}
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 1 & 0 & \cdots & 0 \\
1 & 0 & 0 & 1 & & 0 \\
\vdots & \vdots & \vdots & & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 & 1
\end{vmatrix} =
\begin{vmatrix}
n & 0 & 0 & 0 & \cdots & 0 \\
1 & 1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Transformation Matrix for cube in 2D My task is to transform the cube from the left corner to the big cube in the middle:
What I did was:
First i scale the cube:
$$
\begin{pmatrix}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
Next i rotate it $45^\circ$:
$$
\b... | There is a mistake in your multiplication.
$$
T_{total}=\begin{pmatrix}
4\cos(45^\circ) & -4\sin(45^\circ) & 40\cos(45^\circ) - 28\sin(45^\circ) \\
4\sin(45^\circ) & 4\cos(45^\circ) & \color{red}{28\cos(45^\circ) + 40\sin(45^\circ)} \\
0 & 0 & 1 \\
\end{pmatrix}
$$
But, since $si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$ Find $$\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$$
with $n \in \mathbb{N}$.
My tried:
I think that, I need to find the value of
$$I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}$$
because:
$$\begin{align} I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}} &= \frac{... | Given $$I = \int_{0}^{1}\frac{1}{(1+x^n)^2\sqrt[n]{1+x^n}}dx\;,$$ Now Put $\displaystyle x = (\tan \theta )^{\frac{2}{n}}\;,$ Then $\displaystyle dx = \frac{2}{n}(\tan \theta)^{\frac{2}{n}-1}\cdot \sec^2 \theta d\theta$ and changing limit, We get
$$I = \frac{2}{n}\int_{0}^{\frac{\pi}{4}}\frac{(\tan \theta)^{\frac{2}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
no. of real roots of the equation $ 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$
The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0 $
$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle ... | $f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$ So, as you note,
$f'(x)=1+x+x^2+x^3+..........+x^6$. Now consider $x$ in three ranges
*
*For $x \ge 0$ then clearly $f'(x) >0$
*For $ -1 \le x < 0$, $f'(x)=(1+x) + (x^2+x^3) + (x^4+x^5) + x^6$. Each bracketed term is non-negative, and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
$a^2 + b^2 + c^2 = 1 ,$ then $ab + bc + ca$ gives =? In a recent examination this question has been asked, which says:
$a^2+b^2+c^2 = 1$ , then $ab + bc + ca$ gives = ?
What should be the answer? I have tried the formula for $(a+b+c)^2$, but gets varying answer like $0$ or $0.25$, on assigning different values to vari... | $$
(a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab + bc + ca)
$$
The LHS of the above identity is a perfect square, hence it is always positive or 0.
Thus,
$$
a^2 + b^2 +c^2 + 2(ab + bc + ca) ≥ 0
$$
It is given that $a^2 + b^2 +c^2 =1 $ .
$$
1 + 2(ab + bc + ca) ≥ 0
$$
Therefore,
$$
ab + bc + ca ≥ -1/2
$$
It is possible to determine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
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.