Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$ and $\lim_{n\to\infty} \frac{1}{\ln(n+1)}=0$
Use the definition of the limit of a sequence to prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$.
We have
$$\begin{align}
\left|\frac{n^2-1}{2n^2+3}-\frac{1}{2}\right| & =\left|\frac{2... | If you are using the definition of a limit at infinity, you should include a few more references to the definition in the proof:
Prove: $\lim_{n \to \infty} \left(\frac{n^2-1}{2n^2+3}\right)=\frac{1}{2}$
Proof: Let $\epsilon>0$. Show that there is a positive integer $n_0$ such that if $n>n_0$ then $\left|\frac{n^2-1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Evaluating $\frac{2013^3-2\cdot 2013^2\cdot 2014+3\cdot 2013\cdot 2014^2-2014^3+1}{2013\cdot 2014}$ What is the value of $$\frac{2013^3-2\cdot 2013^2\cdot 2014+3\cdot 2013\cdot 2014^2-2014^3+1}{2013\cdot 2014}?$$
What I have tried:
$$\implies\frac{2013^2(2013-2\cdot2014)+2014^2(3\cdot 2013-2014)+1}{2013\cdot 2014}$$
$... | HINT:
Let $a=2013$ and $b=2014$. You have:
$$\frac{a^3 - 2a^2b + 3ab^2 - b^3 + 1}{ab}$$
Recall that $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ and use that to simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate the integral $\int_2^\infty \frac{1}{\sqrt x}log(\frac{x+1}{x-1})\,dx$
I actually need to check whether the following series is convergent
$\sum_{n=2}^\infty \frac{1}{\sqrt n}log(\frac{n+1}{n-1})$.
The question specifically asks to use Cauchy's integral method, but I am not able to evaluate the integral no... | $$\int_2^\infty \log\left(\frac{x+1}{x-1}\right)\frac{dx}{\sqrt{x}}=2\sqrt{x}\log\left(\frac{x+1}{x-1} \right)|_2^\infty-2\int_2^\infty\sqrt{x}\left(\frac{1}{x+1}-\frac{1}{x-1}\right)dx\\=-\log(9)\sqrt{2}-2\int_2^\infty \frac{\sqrt{x}}{x+1}-\frac{\sqrt{x}}{x-1}dx$$ Let $u^2=x$, $2udu=dx$.
$$-2\int_2^\infty \frac{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103447",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Compute $\sum_{k=1}^{25} (\frac{1}{k}-\frac{1}{k+4})$
Compute $\sum_{k=1}^{25} (\frac{1}{k}-\frac{1}{k+4})$
I know that some of the terms will cancel each other. Have it been $k+1$ instead of $k+4$, I could have easily see the pattern in which the terms cancel each other. I don't know what would I do if it was say $k... | You may write
$$\frac{1}{k} - \frac{1}{k+4} = \left( \frac{1}{k} - \frac{1}{k+1} \right) + \left( \frac{1}{k+1} - \frac{1}{k+2} \right)+ \left( \frac{1}{k+2} - \frac{1}{k+3} \right) + \left( \frac{1}{k+3} - \frac{1}{k+4} \right)$$
Now, apply telescoping and you get for the sum
$$\left( \frac{1}{1} - \frac{1}{26} \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103771",
"timestamp": "2023-03-29T00:00:00",
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Finding $\lim_{n\to\infty}\frac{\log(1^1 +2^2 +\cdots+ n^n)}{\sqrt{n^4 + 2n^3\log(n)}-\sqrt{n^4-n^3}}$ I have the limit and I tried to find the limit but I am stuck after few steps:
$\lim_{n\to\infty}\dfrac{\log(1^1 +2^2 + \cdots + n^n)}{\sqrt{n^4 + 2n^3\log(n)}-\sqrt{n^4-n^3}} $
$= \lim_{n\to\infty}\dfrac{\log(1^1 +2^... | Try to use the inequality
$$
n^{n} < 1^{1} + 2^{2} + \cdots + n^{n} < n^{n} + n^{n} + \cdots + n^{n} = n^{n+1}
$$
and apply the squeeze theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem... | Note that
$$x^4-2x^3+x-2 = x^3(x-2)+1(x-2)=(x-2)(x^3+1)=\color{red}{(x-2)(x+1)}(x^2-x+1).$$
Regarding the taking a factor out part, we use
$$ab+ac = a(b+c)$$
to factor $a$ out of the terms. In your case, it is $ab+ac+ad = a(b+c+d)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Test $\sum_{n=1}^\infty [n^4 \sin^2 (\frac{2n}{3n^3 - 2n^2 + 5})]^n$ for convergence Any hints regarding this question would be appreciated - is the ratio test the best place to start?
| Hint:
$$a_n^{1/n} = n^4 \sin^2 \bigg( \frac{2n}{3n^3-2n^2+5} \bigg) \sim n^4 \bigg(\frac{2n}{3n^3-2n^2+5} \bigg)^2 = \frac{4n^6}{(3n^3-2n^2+5)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding all real $x$ such that $1+\sum_{j=1}^n\sin{\frac{j\pi x}{n+1}} = 0$, where $n=18$.
The task is to find all $ x \in \mathbb R $ such that
$$ 1 + \sum_{j=1}^n \sin{\frac{j\pi x}{n + 1}} = 0, \qquad n = 18 $$
What I have tried
Using the following formulas:
$$1. \sin{x} + \sin{y} = 2\sin{\frac{x + y}{2}}\cos{\fra... | You want to find $x$ s.t. $-(n+1) = \sum_{j=0}^{n} \sin (j\pi x)$.
If $x\not = 1$, then the RHS is the imaginary part of
$$
X = \sum_{j=0}^n (\cos(j\pi x) + i\sin(j\pi x))= \sum_{j=0}^n \left( \cos(\pi x) + i\sin(\pi x) \right)^j = \frac{1 - \left( \cos(\pi x) + i\sin(\pi x) \right)^{n+1}}{1 - \left( \cos(\pi x) + i\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the particular solution of the differential equation $\sqrt{x} + \sqrt{y}y' =0$, with $y(1) = 9$.
I get:
$\sqrt{y}dy = -\sqrt{x}dx \Rightarrow \frac{2}{3}y^{\frac{3}{2}} = -\frac{2}{3}x^{\frac{3}{2}}+C$.
What should I do from here?
| Solve it by separation of variables because it's a separable differential equation:
$$
\sqrt{x} + \sqrt{y}y' =0\\
\sqrt{y}\frac{dy}{dx}=-\sqrt{x}\\
\int\sqrt{y}\frac{dy}{dx}\,dx=-\int\sqrt{x}\,dx\\
\int\sqrt{y}\,dy=-\int\sqrt{x}\,dx\\
\frac{2\sqrt{y^3}}{3}+C_1=-\frac{2\sqrt{x^3}}{3}+C_2\\
\sqrt{y^3}=-\sqrt{x^3}+C\\
y=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Let $x=1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{...}}}}$; then the value of $(2x-1)^2$ equals... Let $$x=1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{...}}}};$$ then the value of $(2x-1)^2$ equals...
I don't how to start this question. Please help.
| If you only want to know the value that the continued fraction converges to, you use a simple technique:
$$
x=1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{...}}}}=1+\frac{1}{2+\frac{1}{x}}
$$
With some manipulation you could come up with the value of $x$, but you want $(2x-1)^2:$
$$
x\left(2+\frac{1}{x} \right)=\left(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the value of expression $\frac{1}{\cos 290^{\circ}} + \frac{1}{\sqrt3 \sin250^{\circ}}$.
Find the value of expression $\frac{1}{\cos 290^{\circ}} + \frac{1}{\sqrt3 \sin250^{\circ}}$.
Though this question seems easy, I just seem to get stuck somewhere. This is what I have tried so far:
$$\frac{1}{\cos70^{\circ}}-... | $$\frac{\sqrt{3}\sin 70^\circ-\cos 70^\circ }{\sqrt{3}\sin 70^\circ \cos 70^\circ}=\frac{\frac{\sqrt{3}}{2}\sin 70^\circ-\frac{1}{2}\cos 70^\circ }{\frac{\sqrt{3}}{2}\sin 70^\circ \cos 70^\circ}=$$
$$=\frac{\sin 70^\circ \cos 30^\circ -\sin 30^\circ\cos 70^\circ }{\frac{\sqrt{3}}{4}(2\sin 70^\circ \cos 70^\circ)}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For three positive numbers verifying $xyz=2+x+y+z$, is there an upper bound on $x+y+z$? And, if so, which is that? For three positive numbers verifying $xyz=2+x+y+z$, is there an upper bound on $x+y+z$? And, if so, which is that?
I've managed to find only a lower bound, from $ xyz=2+x+y+z \le \frac{(x+y+z)^3}{27} $, f... | No, there is no upper bound. We solve for $x$ and get
$$ x = \frac{2+y+z}{yz - 1} $$
Pick $z=\frac{2}{y}$, then we get
$$ x = 2 + y + \frac{2}{y} $$
and hence for this choice
$$ x+y+z = 2 + 2y + \frac{4}{y} $$
Choosing $y$ large will get you a large number for $x+y+z$.
Added: To make it more fun. Let us consider the sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$ = a - b? I'm given the complex rational expression
$$\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$$
and asked to simplify. The solution provided is $a - b$; however I get $\frac{a^2 - b^2}{a+b}$.
My working:
Numerator first:
$$\frac{a}{b} - \frac{b}{a}$$
Le... | Not that $$\frac{a}{b}-\frac{b}{a}=\frac{(a-b)(a+b)}{ab}$$ so we get
$$\frac{(a-b)(a+b)ab}{(a+b)ab}=a-b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Ramanujan's radical and how we define an infinite nested radical I know it is true that we have
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$
The argument is to break the nested radical into something like
$$3 = \sqrt{9}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cd... | $4=\sqrt{16}=\sqrt{1+3\sqrt{25}}=\sqrt{1+3\sqrt{1+4\sqrt{36}}}=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{49}}}}=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{64}}}}}=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{81}}}}}}$
$=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+8\sqrt{1+\cdots}}}}}}}$
$5=\sqrt{25}=\sqrt{1+4\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
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Find envelope of $x \sin \theta - y \cos \theta + z = a \theta$, where $\theta$ is a parameter. I have tried to find envelope for
$$x \sin \theta - y \cos \theta + z = a \theta$$
First I find derivative w.r.t. $\theta$
$$F(\theta)=x \sin \theta - y \cos \theta + z - a \theta = 0$$
$$\frac{\partial F(\theta)}{\partial \... | Hint:
You get rid of the annoying trigonometric functions by
$$(z-a\theta)^2+a^2=x^2+y^2.$$
From this you draw $\theta$, which you plug in one of the initial equations. Not a really nice expression.
Alternatively, in cylindrical coordinates $(\phi,\rho)$ we rewrite
$$\begin{cases}\rho\sin(\phi-\theta)=z-a\theta,\\\rho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119885",
"timestamp": "2023-03-29T00:00:00",
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Proving the identity $\tan x+2\tan2x+4\tan4x+8\cot8x=\cot x$ by considering a more general form I came across this exercise:
Prove that
$$\tan x+2\tan2x+4\tan4x+8\cot8x=\cot x$$
Proving this seems tedious but doable, I think, by exploiting double angle identities several times, and presumably several terms on the l... | Let $\displaystyle S =\sum^{2}_{k=0}2^k\tan(2^k x)+8\cot (8x).$
Then $\displaystyle \int Sdx=-\sum^{2}_{k=0}\ln\bigg(\cos(2^k x\bigg)+\ln(\sin 8x)$
Using $\displaystyle \prod^{n-1}_{r=0}\cos(2^r x)=\frac{1}{2^n}\frac{\sin(2^nx)}{\sin x}.$
$\displaystyle \int Sdx=-\ln\bigg(\frac{1}{2^3}\frac{\sin 8x}{\sin x}\bigg)+\ln(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$?
The number of pairs of $(a,b)$ of positive real numbers satisfying
$a^4+b^4<1$ and $a^2+b^2>1$ is -
$(i)$0
$(ii)$1
$(iii)$2
$(iv)$ more than 2
Solution:We have $a,b>0$,
According to the given situation,$0<a^4+b^... | Geometrically, the solution set is the set of points outside of the circle $x^2+y^2=1$ but inside the superellipse $x^4+y^4=1$. There are an infinite number of such points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Integral $\int^{1}_{1/2}\frac{\ln(x)}{1-x}dx$
Evaluate $$\int^{1}_{1/2}\frac{\ln(x)}{1-x}dx$$
Try: Let $$ I =\int^{1}_{1/2}\frac{\ln x}{1-x}dx=\int^{1}_{1/2}\sum^{\infty}_{k=0}x^k\ln(x)dx$$
$$I =\sum^{\infty}_{k=0}\int^{1}_{1/2}x^k\ln(x)dx$$
Using integration by parts, gives: $$I =\sum^{\infty}_{k=0}\frac{\ln(x)x^{k+... | Here we will address the integral:
\begin{equation}
I = \int_{\frac{1}{2}}^1 \frac{\ln(x)}{1 - x}\:dx
\end{equation}
We first observe that:
\begin{equation}
\frac{\partial}{\partial a} x^a = x^a \ln(x) \Longrightarrow \ln(x) = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a
\end{equation}
Thus,
\begin{equati... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the $n^{th}$ term derivative of $f(x) = (x^2-x-1)(\ln(1-x))$? I have the first three terms but am struggling with finding the $n^{th}$ term derivative of the function. Here is my work:
$$\\$$
$$f(x) = (x^2-x-1)(\ln(1-x)) $$
$$f'(x) = (2x-1)(\ln(1-x))-\left(\dfrac{x^2-x-1}{1-x}\right)$$
$$f''(x) = \dfrac{3x^2-5x... | As you noticed,
$$
f^{(3)}(x)=\frac{2x^{2}-5x+1}{\left(x-1\right)^{3}}.
$$
Wolfram claims that
$$
f^{(3+n)}(x)=n!\frac{n^{2}+n\left(x+2\right)-2x^{2}+5x-1}{\left(x-1\right)^{3+n}}.
$$
To convince yourself that this is indeed the case, fix $n\geq0$ and differentiate the expression for $f^{(3+n)}$ to get the expression f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Taylor series, Convergence I need to find the radius of convergence (R) of the Taylor series expansion of the following function $f(x) =\frac{x^3 -2x+1}{x+7}$, $x_0=5$
The Taylor series of $f(x) = \frac{29}{3}+\frac{\frac{95}{18}}{1!}\left(x-5\right)+\frac{\frac{175}{108}}{2!}\left(x-5\right)^2+\frac{\frac{41}{432}}{3!... | For simplicity, I let $x=y+5$ to make
$$\frac{x^3 -2x+1}{x+7}=\frac{y^3+15 y^2+73 y+116}{y+12}$$ and using the long division
$$\frac{y^3+15 y^2+73 y+116}{y+12}=\frac{29}{3}+\frac{95 y}{18}+\frac{175 y^2}{216}+\frac{41 y^3}{216 (y+12)}$$
Focusing on the last term and using the classical expansion, we then have
$$\frac{ ... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the value of $2x+3y$ if $x+y=7$ and $x^2-y^2=21$? If $x+y=7$ and $x^2-y^2=21$, then what is $2x+3y$?
I solved it like this:
\begin{align}
y& =7-x \\
x^2-(7-x)^2-21&=0 \\
x^2-49+14x-x^2-21&=0 \\
14x&=70 \\
x&=5
\end{align}
Then I solved for $y$ and I got $2$. I plugged in the values of $x$ and $y$ to $2x+3y$, a... | Quicker way is to note that
$$
21=(x-y)(x+y)=7(x-y)
$$
so
$$
x-y=3
$$
Combined with $x+y=7$, we get that $2x=10$ i.e. $x=5$ and $y=2$.
So your answer is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What's maximum value of $x (1-x^2)$ for $0 < x <1$? Since we are being taught about AM-GM inequality, I decided to use the method however I am getting two different answers by slightly different methods.
Method 1
\begin{equation}
v=x (1-x^2)$
\implies v^2=x^2 (1-x^2)^2
\end{equation}
Using the AM-GM-inequality we obtai... | Hint:
$$\dfrac{ax+b(1+x)+c(1-x)}3\ge ?$$
for $a,b,c>0$
Set $a+b-c=0\ \ \ \ (1)$
The equality will occur if $ax=b(x+1)=c(x-1)=k$(say)
Put the values of $a,b,c$ in$(1)$
| {
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"url": "https://math.stackexchange.com/questions/3131672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Evaluate $\iint_{\{(x,y)\mid (x-1)^2+y^2\leq1\}}\sqrt{x^2+y^2} dxdy$ [Background]: I'm trying to find the volume of the region bounded by the $xy$-plane, the cone $z^2=x^2+y^2$ and the cylinder $(x-1)^2+y^2=1$.
[Attempt]: I tried to use the polar coordinate:
\begin{align*}
\iint_{\{(x,y)\mid (x-1)^2+y^2\leq1\}}\sqrt{x^... | By using the usual polar coordinates $x=r\cos(\theta)$ and $y=r\sin(\theta)$ then the domain is
$$1\geq (x-1)^2+y^2=(r\cos(\theta)-1)^2+r^2\sin^2(\theta)\Leftrightarrow r\leq 2\cos(\theta),$$
and the integral becomes
$$\iint_{\{(x,y)\mid (x-1)^2+y^2\leq 1\}}\sqrt{x^2+y^2} dxdy=\int_{\theta=-\pi/2}^{\pi/2} \int_{r=0}^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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maximum value of $ab$ If $a,b\in R$ and $\displaystyle a^2+b^2=1+\frac{2ab}{a-b}$ and $\sqrt{a-b}=a^2+5b$, what is the maximum of $ab$?
Here is what I tried:
$a=r\cos\alpha$ and $b=r\sin \alpha$
$\displaystyle r^2=1+\frac{r\sin 2\alpha}{\cos \alpha-\sin \alpha}$
and $\sqrt{r(\cos \alpha-\sin \alpha)}=r^2\cos^2\alpha+5... | The domain gives $a-b>0$ and from the first equation we obtain:
$$(a-b)^2+2ab=1+\frac{2ab}{a-b}$$ or
$$(a-b)^2-1+2ab\left(1-\frac{1}{a-b}\right)=0$$ or
$$(a-b-1)\left(a-b+1+\frac{2ab}{a-b}\right)=0$$ or
$$a=b+1.$$
Can you end it now?
I got the following answer: $42$.
| {
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"url": "https://math.stackexchange.com/questions/3135830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing that $\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$ Since I don't have the answer to this one, I want to make sure I've done this correctly.
Show that
$$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$
Since
$$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$
we have
$$\ta... | $\tan \arcsin \frac{x}{\sqrt{x^2+1}} = \frac{\sin}{\cos} \arcsin \frac{x}{\sqrt{1+x^2}} = \frac{\frac{x}{\sqrt{x^2+1}}}{\cos \arcsin \frac{x}{\sqrt{x^2+1}}}$ and we know that $\cos^2 x+\sin^2 x=1 $ so $\cos x= \sqrt{1-\sin^2 x}$
so $ \frac{\frac{x}{\sqrt{x^2+1}}}{\cos \arcsin \frac{x}{\sqrt{x^2+1}}} = \frac{\frac{x}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
*
*If $n$ is od... | Just to be different:
$(a-1)(a^{n-1} + a^{n-2} + ..... + a + 1)=(a^n - 1)$.
$(a+1)(a^{n-1} -a^{n-2} + ........ \mp a \pm 1) = (a^n \pm 1)$.
with $n$ even yielding $(a+1)(a^{n-1} -a^{n-2} + ........ + a - 1)=a^n -1$
and $n$ odd yielding $(a+1)(a^{n-1} -a^{n-2} + ...... -a +1) = a^n+1$
So we $a = 7$ we get $7+1=8|a^n+1$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrate $\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ via trig substitution $x = a\sec\theta, dx = \sec\theta \tan\theta$
$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{(\tan^2\theta)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{\tan^{\frac{7}{2}}\theta}$ =
$\int \frac{\sec\theta \tan\theta}{\tan\theta ^{\frac{7}{2}}}$... | If you do $x=\sec\theta\,\mathrm d\theta$ and $\mathrm dx=\sec(\theta)\tan(\theta)\,\mathrm d\theta$, what you get is$$\int\frac{\sec(\theta)\tan(\theta)}{\tan^3\theta}\,\mathrm d\theta=\int\frac{\cos^2\theta}{\sin^3\theta}\,\mathrm d\theta=\int\frac{\cos^2(\theta)\sin(\theta)}{\bigl(1-\cos^2(\theta)\bigr)^2}\,\mathrm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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minimum value of $2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$ Let $\alpha,\beta$ be real numbers ; find the minimum value of
$2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$
What I tried :
$\bigg|4\cos \beta+(2\cos \alpha+3\sin \alpha)\sin \beta\bigg|^2 \leq 4^2+(2\cos \alpha+3\sin \alpha)^2$... | In
$$(2\cos\alpha+3\sin\alpha)\sin\beta+4\cos\beta$$ the parenthesised factor takes values in $[-\sqrt{2^2+3^2},\sqrt{2^2+3^2}]$. Using the largest value, the minimum of $$\sqrt13\sin\beta+4\cos\beta$$ is
$$-\sqrt{13+4^2}.$$
Justification:
$$a\cos t+b\sin t$$ is the scalar product of $(a,b)$ with the unit rotating ve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Computing $m-n$
When $P(x+4)$ is divided by $P(x)$, the remainder is $3x+m$. When
$P(x)$ is divided by $P(x+4)$, the remainder is $nx-6$. Compute $m-n$
Here I wrote down the equations as follows
$$P(x+4) = P(x)Q_1(x)+3x+m$$
$$P(x) = P(x+4)Q_2(x) + nx-6$$
However, these will not give me anything useful. Could you as... | firstly, we write :
$m=P(x+4)-P(x)Q_1(x)-3x,$
$n=[P(x)-P(x+4)Q_2(x)+6]/x=(-3)[P(x)-P(x+4)Q_2(x)+6]/[m-P(x+4)+P(x)Q_1(x)]$
pick $n=\pm{3}$ by irreducible and improve above formula to :
$P(x)(Q_1(x)-1)+P(x+4)(Q_2(x)-1)=6-m,$$P(x)(Q_1(x)+1)-P(x+4)(Q_2(x)+1)=-6-m$
also note that:
$P(x)Q_1(x)-P(x)=P(x+4)-3x-m-P(x+4)Q_2(x)-n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate integral $\int \sin^4(t)\cos^3(t)dt$ $$\int \sin^4(t)\cos^3(t)dt = \int \sin^4(t)(1-\sin^2(t))\cos(t) dt $$
$$u = \sin(t) \\ du = \cos(t)dt$$
$$ \int \sin^4(t)\cos^3(t)dt = \int u^4(1-u^2) du \\
= u^4 - u^6 = \frac{1}{5}u^5 - \frac{1}{7}u^7 + C \\
= \frac{1}{5}\sin^5(t) - \frac{1}{7}\sin^7(t) + C $$
This s... | I think your answer is all right since $$\cos 2 \theta =1-2 \sin^2 \theta$$
You see : $$\dfrac{1}{5} \sin^5 (x )+ \dfrac{1}{7} \sin^7 (x) =\dfrac{1}{70} \sin^5( x) (14+10 \sin^2 (x)) =\dfrac{1}{70} (9+5 \cos (2x))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An interesting indefinite integral Here's an interesting integral that I'm struggling with.
$$ \int \frac{x\ dx}{\sqrt{x^4 + 4x^3 - 6x^2 + 4x + 1}} $$
MY EFFORTS:
I'm very very close in reaching a closed form.
\begin{align}
&= \frac12\int \frac{2\ dx}{\sqrt{\left(x^2+\dfrac{1}{x^2}\right) + 4\left(x+\dfrac{1}{x}\right)... | I don't think that your effort leads to a simpler integral that the original one. They are two radicals in your last integral. This makes it more difficult to integrate. So, you are not close to solve it.
HINT :
$$x^4+4x^3-6x^2+x^4+1= (x-r_1)(x-r_2)(x-r_3)(x-r_4)$$
$r_1=\frac12\left(1+\sqrt{2}+\sqrt{7-4\sqrt{2}} \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Let $A=[1 2 0 1 ]$. Find all $2×2$ matrices B, $B≠O_2$ and $B≠I_2$ such that $AB=BA$.
Let $A=[1 2 0 1 ]$. Find all $2×2$ matrices B, $B≠O_2$ and $B≠I_2$ such that $AB=BA$
Explain your answer.
I know two ways to find B
A) As det of A is 1, it is invertible. So A = B (inverse)
B) Or by assigning a,b,c,d as four elemen... | So $$A = \begin{pmatrix}1&2 \\ 0&1 \end{pmatrix}$$
If $$B = \begin{pmatrix}a&b \\ c&d \end{pmatrix}$$
From $$\begin{pmatrix}1&2 \\ 0&1 \end{pmatrix} \begin{pmatrix}a&b \\ c&d \end{pmatrix} = \begin{pmatrix}a&b \\ c&d \end{pmatrix} \begin{pmatrix}1&2 \\ 0&1 \end{pmatrix}$$
we have $$ a+2c = a\implies c=0$$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the number of roots of the equation, $x^3 + x^2 +2x +\sin x = 0$ in $[-2\pi , 2\pi]$.
Find the number of roots of the equation,
$$x^3 + x^2 +2x +\sin x = 0$$
in $[-2\pi , 2\pi]$.
What I have tried:
$$x^3 + x^2 +2x = -\sin x$$
$$x^2 +x +2 = \frac{-\sin x }{x}$$
$$(x + \frac{1}{2})^2 + \frac{7}{4} = \frac{-\sin ... | From the comments, I knew the solution to this question. So, I am just writing it out as an answer.
For the case where $x \ne 0$,
$$(x+\frac{1}{2})^2 + \frac{7}{4} = - \frac{\sin x}{x}$$
Since, LHS is at least $\frac{7}{4}$
From the comment of Eric Yau, $0<\frac{\sin x}{x}<1$ and hence, $-1 < \frac{\sin x}{x} < 0$
Hen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Solve for $x$ in $x\sqrt{1-4x^2}+2x\sqrt{1-x^2}=\frac{\sqrt 3}2$ Find out the value of $x$ in $x\sqrt{1-4x^2}+2x\sqrt{1-x^2}=\dfrac{\sqrt 3}2$.
I tried squaring both the sides but it only makes it more complicated. Is there any other way?
| Put $2x = \cos(\alpha), x = \cos(\beta) \implies \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha) = \dfrac{\sqrt{3}}{2}\implies \sin(\alpha+\beta) = \dfrac{\sqrt{3}}{2}\implies \alpha+\beta=\dfrac{\pi}{3}, \dfrac{2\pi}{3}\implies \beta = \dfrac{\pi}{3} - \alpha\implies x = \cos(\dfrac{\pi}{3} -\alpha)= \dfrac{1}{2}(2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$? How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$?
Let's suppose that we have a function defined as follo... | $$
\begin{align}
\sum_{q=3}^p\frac{q^2-3}{q}
&=\int_{3^-}^{p^+}\frac{x^2-3}{x}\,\mathrm{d}\pi(x)\\
&=\left[\frac{p^2-3}{p}\pi(p)-2\right]-\int_{3^-}^{p^+}\pi(x)\left(1+\frac3{x^2}\right)\mathrm{d}x\\[3pt]
&=\frac{p^2}{2\log(p)}+\frac{p^2}{4\log(p)^2}+\frac{p^2}{4\log(p)^3}+O\!\left(\frac{p^2}{\log(p)^4}\right)
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Prove that $\left\lfloor \frac {n+3}{2} \right\rfloor=\left\lceil \frac {n+2}{2}\right\rceil$ Prove that $$\left\lfloor \frac {n+3}{2} \right\rfloor=\left\lceil \frac {n+2}{2}\right\rceil$$
I have tried to solve this on my own, and I want to check my solution.
My steps:
Set $x=\left\lfloor \frac {n+3}{2}\right\rfloor $... | If $n$ is an integer, then $\epsilon$ is either $0$ or $\frac12$ and hence
$$ \bigg\lceil\epsilon-\frac12\bigg\rceil=0. $$
Otherwise, the identity doesn't hold. For example, $n=4.3$ and then
$$ \bigg\lfloor\frac{4.3+3}{2}\bigg\rfloor=\lfloor3.65\rfloor=3, \bigg\lceil\frac{4.3+2}{2}\bigg\rceil=\lceil3.25\rceil=4. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $x^n(x-y)(x-z)+y^n(y-x)(y-z)+z^n(z-x)(z-y)>0$ Let $x,y,z>0$ and $n\in\mathbb{N}$, prove that:
$$x^n(x-y)(x-z)+y^n(y-x)(y-z)+z^n(z-x)(z-y)>0$$
Some idea? I have tried to develop the multiplication and certain assumptions buy I can't see it clearly. Note that this is not true if $x=y=z$ or $x=y=z=0$.
| Well. It's NOT true. If $x = y = z > 0$ then $x^n(x-y)(x-z) + y^n(y-x)(y-z) + z^n(z-x)(z-y) = 0$.
So you can not prove it.
But maybe we can prove it is we stipulate that at least two of the terms are distinct.
Note that we can label the three variables whatever we want and by symmetry we might as well, and we can ass... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of PA+PB+AB If $P(2,1) $ and $A $ and $B $ lie on $x$ axis and $y=x $ respectively,
then find the minimum value of $PA+PB+AB $ .
If $A$ was given , I could have worked geometrically, by using that image of $P$ and point $A$ and $B$ should be collinear.
But here vary two things , $A$ and $B$....
| Another way.
By Minkowski we obtain:
$$PA+PB+AB=\sqrt{(x-2)^2+1}+\sqrt{(y-x)^2+y^2}+\sqrt{(1-y)^2+(2-y)^2}\geq$$
$$\geq\sqrt{(x-2+y-x+1-y)^2+(1+y+2-y)^2}=\sqrt{10}.$$
The equality occurs for $A\left(\frac{5}{3},0\right)$ and $B\left(\frac{5}{4},\frac{5}{4}\right),$ which says that $\sqrt{10}$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3162753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Trouble calculating probability Machine generates one random integer in range ${[0;40)}$ on every spin.
You should choose 5 numbers in that range.
Then the machine will spit out 5 numbers (numbers are independent of each other).
what is the probability that you will get exactly two numbers correct?
My logic:
You ... | If the machine is choosing with replacement, but the person is choosing five distinct numbers, and if success means two distinct numbers the person chose are picked by the machine, then the problem is much more difficult:
Probability machine chooses five distinct numbers:
$$\dfrac{40!}{35!40^5}$$
Probability machine ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Finding intercalates within a (reduced) latin square I need confirmation on if my intuition on finding intercalates is correct, suppose we have the following reduced latin square
\begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 4 & 1 & 3 \\
3 & 1 & 4 & 2\\
4 & 3 & 2 & 1
\end{bmatrix}
we know fr... | None of these are intercalates. As defined in this paper (for example), we're looking for two rows $i,j$ and two columns $x,y$ such that $L_{i,x} = L_{j,y}$ and $L_{j,x} = L_{i,y}$. There are four of these in your example:
*
*$\{i,j\} =\{1,4\}$ and $\{x,y\} = \{1,4\}$: $$\begin{bmatrix}1 & 4 \\ 4 & 1\end{bmatrix}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A probability question: Poor Alex
Alex remembers all but the last digit of his friend's telephone number. He decides to choose the last digit at random in an attempt to reach him. Given that, Alex has only enough money to make two phone calls, the probability that he dials the right number before running out of money ... | Alex tries two numbers. Then the probability that the correct number belongs to this pair is simply
$$\frac{2}{10} = 20 \%$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding the value of $x-y$. Given two equations \begin{align} & x^4+y^4=\dfrac {-7}{9} \\ & x^3-y^3=3. \end{align} From these equations find the value of $(x-y).$
I have just factorize $x^3-y^3=(x-y)(x^2+xy+y^2)=3$. But don't know how to proceed from here. One way to find the values of $x$ and $y$ and then compute poss... | Here is a hint to get you started:
$$\frac{-7}9= (x^2+y^2)^2-2(xy)^2$$$$3=(x-y)(x^2+y^2+xy)=(x-y)^3+3(xy)(x-y)$$ If we substitute $a=x^2+y^2$ and $b=xy$ and $c=x-y$, we get $$\frac{-7}9=a^2-2b^2$$$$3=c\cdot(a+b)=c^3+3bc$$
Now, you have three equations and three variables. Can you figure it out?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Can $a(n) = \frac{n}{n+1}$ be written recursively? Take the sequence $$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \dots$$
Algebraically it can be written as $$a(n) = \frac{n}{n + 1}$$
Can you write this as a recursive function as well?
A pattern I have noticed:
*
*Take $A_{n-1}$ an... | \begin{align*}
a_{n+1} &= \frac{n+1}{n+2} \\
&= \frac{n+2-1}{n+2} \\
&= 1 - \frac{1}{n+2} \text{, so } \\
1 - a_{n+1} &= \frac{1}{n+2} \text{, } \\
\frac{1}{1 - a_{n+1}} &= n+2 &[\text{and so } \frac{1}{1 - a_n} = n+1]\\
&= n+1+1 \\
&= \frac{1}{1- a_n} +1 \\
&= \frac{1}{1- a_n} + \frac{1-a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation... | Construct the Groebner basis of your system, with the variables ordered $x$, $y$:
$$ \mathrm{GB}(\{3x+2y-36, 5x+4y-64\}) = \{y-6, x-8\} $$
and read out the solution. (If we reverse the variable order, we get the same basis, but in reversed order.) Under the hood, this is performing Gaussian elimination for this pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 10
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If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$. Find $\lim_{n \to \infty} nt_n$ If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$
Find $\lim_{n \to \infty} nt_n$
First attempt: $t_n$ is positive(grouping two... | Another approach
$$t_n=\sum_{k=1}^n\left(\frac{1}{2n+2k-1}-\frac1{2n+2k}\right)=\sum_{k=1}^n\frac1{(2n+2k-1)(2n+2k)}$$
and then
$$nt_n=\frac1{n}\sum_{k=1}^n\frac1{(2+(2k-1)/n)(2+2k/n)}$$
which is a Riemann sum for
$$\int_0^1\frac{dx}{(2+2x)^2}=\frac18.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the v... | You can use
$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)$,
$(ab+ac+bc)^2=a^2+b^2+c^2+2(ab+ac+bc)$,
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$,
$(a+b+c)^5-a^5-b^5-c^5=(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
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Distance Formula Problem If two vertices of an equilateral triangle are $(1, -1)$ and $(-\sqrt{3}, - \sqrt{3})$, find the coordinates of the third vertex.
Step by Step procedure to get the answer.
Take $A=(1, -1)$, $B=(-\sqrt{3}, - \sqrt{3})$.
Let the third vertex be $C=(x, y)$.
The distance d between the two points $A... | Distance between two given points will be the side of the triangle $a=\sqrt{(1+\sqrt{3})^2+(\sqrt{3}-1)^2}=2\sqrt{2}$. Now you just need to draw two circles of radius $a$ and with centers at the known vertices. The equations of the circles are:
$(x-1)^2+(y+1)^2=8$
$(x+\sqrt{3})^2+(y+\sqrt{3})^2=8$
The circles will int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find the minimum of $\space\frac{1}{x}+\frac{1}{y}+c\cdot xy\space$ subject to $\space x+y-c=0$ Let $f(x,y):\mathbb{D}\rightarrow\mathbb{R}$ be the function:
$$f(x,y)=\frac{1}{x}+\frac{1}{y}+c\cdot xy\space\space|\space\space c\in(0,\sqrt[4]8)\text{ $\space$constant}$$
$$\mathbb{D}=\{(x,y)\space|\space x>0,\space y>0\}... | $\begin{array}\\
f(x,y)
&=\dfrac{1}{x}+\dfrac{1}{y}+c xy\\
&=\dfrac{c}{x(c-x)}+c x(c-x)\\
&=\dfrac{c+cx^2(c-x)^2}{x(c-x)}\\
&=c\dfrac{1+x^2(c-x)^2}{x(c-x)}\\
f_x(x,y)
&= \dfrac{c^3 x^2 - 4 c^2 x^3 + 5 c x^4 - c - 2 x^5 + 2 x}{x^2 (c - x)^2}
\quad\text{according to Wolfy}\\
&= \dfrac{(c - 2 x) (c x - x^2 - 1) (c x - x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
My try:Let $y=x+1$. Then: $$f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}=\frac{1}{(2y-1)(4-\frac{1}{y^2})^2}=-\frac{1}{2}\cdot\frac{1}{... | $\begin{array}\\
f(x)
&=\dfrac{(1+x)^4}{(1+2x)^3(1-2x)^2}\\
&=\dfrac{(1+x)^4(1-2x)}{(1+2x)^3(1-2x)^3}\\
&=\dfrac{(1+x)^4(1-2x)}{(1-4x^2)^3}\\
&=(1+4x+6x^2+4x^3+x^4)(1-2x)(1-4x^2)^{-3}\\
&=(1+2x-2x^2+...)\sum_{k=0}^{\infty} \binom{k+2}{2})(4x^2)^k
\qquad\text{generalized binomial theorem}\\
&=(1+2x-2x^2+...)(1+3(4x^2)+6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving ${\lim\limits_ {n\to\infty}}\frac{6n^3+5n-1}{2n^3+2n+8} = 3$ I'm trying to show that $\exists \,\varepsilon >0\mid\forall n>N\in\mathbb{N}$ such that:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| < \varepsilon$$
Let's take $\varepsilon = 1/2$:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| = \left|\frac{6n^3+5n-1... | Note that for $n\geq 1$ we have $n+25\leq n+25n=26n$ and $2n^3+2n+8\geq 2n^3$ so you have that $$\left|\frac{n+25}{2n^3+2n+8}\right|\leq \frac{26n}{2n^3}=\frac{13}{n^2}$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
System of quadratic equations with three variables (generic form) Try solve a system of equation like this one.
\begin{cases}
(O_x -A_x)^2+(O_y-A_y)^2+(O_z-Az)^2=x^2 \\
(O_x -B_x)^2+(O_y-B_y)^2+(O_z-Bz)^2=y^2 \\
(O_x -C_x)^2+(O_y-C_y)^2+(O_z-Cz)^2=z^2
\end{cases}
Is there any way to express $O_x, O_y, O_z$ in terms of... | OK lets check this out I'll replace $x \ y \ z$ with $a\ b\ c$ to prevent name collision with function independent variables.
\begin{cases}
(O_x -A_x)^2+(O_y-A_y)^2+(O_z-A_z)^2=a^2 \ ①\\
(O_x -B_x)^2+(O_y-B_y)^2+(O_z-B_z)^2=b^2 \ ②\\
(O_x -C_x)^2+(O_y-C_y)^2+(O_z-C_z)^2=c^2 \ ③
\end{cases}
$④=①-②$ and $⑤=②-③$ we get
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How to show that $ab+bc+ca\le \frac34$ Let $a,b$ and $c$ be positive real numbers such that $(a+b)(b+c)(c+a) = 1$ , hen show that $$ab+bc+ca\le \frac34$$
I believe I need to use AM-GM inequality and use the fact $(a+b)(b+c)(c+a) = 1$ Using AM-GM in $a+b,b+c$ & $c+a$, I get $a+b+c\ge \frac32$. Any hint will be thankful.... | $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ it's
$$\sum_{cyc}(a^2b+a^2c-2abc)\geq0,$$ which is true by AM-GM:
$$\sum_{cyc}(a^2b+a^2c-2abc)=\sum_{cyc}(a^2c+b^2c-2abc)\geq\sum_{cyc}(2\sqrt{a^2c\cdot b^2c}-2abc)=0.$$
Thus, it's enough to prove that
$$(a+b+c)^2\geq3(ab+ac+bc),$$ which is true by AM-GM again:
$$(a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluating $\int_{0}^{\frac{\pi}{4}}\frac{x\sin(x)}{\cos^{3}(x)}\,dx$ $$\int_{0}^{\frac{\pi}{4}}\frac{x\sin(x)}{\cos^{3}(x)}\,dx$$
I started like this:
$$\int_{0}^{\frac{\pi}{4}}\dfrac{x\sin(x)}{\cos^{3}(x)}\,dx=\int_{0}^{\frac{\pi}{4}}\dfrac{1}{\cos^{2}(x)}\cdot \dfrac{\sin(x)}{\cos(x)}\cdot (x)\,dx=\int_{0}^{\frac{\... | An alternative solution.
Observe that:
\begin{align}\dfrac{\partial}{\partial x}\left(\frac{1}{2\cos^2 x}\right)=\frac{\sin x}{\cos^3 x}\end{align}
Therefore,
\begin{align}\int_{0}^{\frac{\pi}{4}}\frac{x\sin(x)}{\cos^{3}(x)}\,dx&=\left[\frac{x}{2\cos^2 x}\right]_{0}^{\frac{\pi}{4}}-\frac{1}{2}\int_{0}^{\frac{\pi}{4}} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Cant find an eigenvector for an eigenvalue For the matrix
$$
\begin{pmatrix}
1 & 2 & 3 \\
0 & 3 & 4 \\
0 & 0 & 5 \\
\end{pmatrix}
$$
I know that $5, 2+\sqrt3, 2-\sqrt3$ are eigenvalues. I am trying to find an eigenvector for $2+\sqrt3$ using $(A-\lambda I)V=0$. But this gives me:
$$
\begin{pmatr... | Indeed, as Jose has rightly pointed out, the eigenvalues are the diagonal entries of a triangular matrix:
$$ \det(A - \lambda I) = 0$$
$$ \begin{vmatrix} 1-\lambda & 2 & 3 \\ 0 & 3-\lambda & 4 \\ 0 & 0 & 5- \lambda \end{vmatrix} = 0$$
$$ (1-\lambda) \begin{vmatrix} 3-\lambda & 4 \\ 0 & 5-\lambda \end{vmatrix} - 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\sum_{i=10}^n (\sum_{r=10}^i \binom{30}{r}\binom{20}{r-10})$=? Evaluate:
$$\sum_{i=10}^n \left(\sum_{r=10}^i \binom{30}{r}\binom{20}{r-10}\right)$$
I tried this and my result come:
$$n\binom{30}{10}\binom{20}{1}+ (n-1)\binom{30}{11}\binom{20}{2} + (n-2)\binom{30}{12}\binom{20}{3} + ...... +(n-(n-1))\binom{30}{n-1... | Note
$$
\sum_{i=10}^n \left(\sum_{r=10}^i \binom{30}{r}\binom{20}{r-10}\right) = \\
\sum_{i=10}^n \left(\sum_{k=0}^{i-10 } \binom{30}{k+10}\binom{20}{k}\right)= \\
\sum_{j=0}^{n-10} \left(\sum_{k=0}^{j} \binom{30}{k+10}\binom{20}{k}\right)= \\
\sum_{j=0}^{20} \left(\sum_{k=0}^{j} \binom{30}{k+10}\binom{20}{k}\right) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $x = 2$ is a root of $\det\left[\begin{smallmatrix}x&-6&-1\\2&-3x&x-3\\-3&2x&x+2\end{smallmatrix}\right]=0$, find other two roots
If $x = 2$ is a root of equation
$$ \begin{vmatrix}
x & -6 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} = 0 $$
Then find the other two roots.
I solved it and got a cubic eq... | $$
\begin{align*}
\begin{vmatrix}
x & -6 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} & =
\begin{vmatrix}
x-2 & 3x-6 & 2-x \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix}
\\
& = (x-2)
\begin{vmatrix}
1 & 3 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix}
\\
& = (x-2)
\begin{vmatrix}
1 & 0 & 0 \\
2 & -3x-6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving limit without L'Hôpital's rule: $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}$ How can I solve this limit without L'Hôpital's rule?
$$\begin{align}\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}&=\lim\limits_{x \to 0} \frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x... | Hint:
$\frac{\tan x- \sin x}{x^3}=\frac{1}{ \cos x} \cdot \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
$x^3-8x^2+30x-20=0$ has roots $a$, $b$, $c$. Find the equation with roots $a+2$, $b+2$, $c+2$ I have the equation $$x^3-8x^2+30x-20=0$$ let's call the roots $a,b,c$
It's easy to find the equation with roots $(a+2)(b+2)(c+2)$
these are the steps in my books
so the equation is $$ aX^3+ bX^2 + cX + D = 0 $$
$$ x^3 -14x^2... | Alternatively: for the roots $a,b,c$ of $x^3-8x^2+30x-20=0$, by Vieta's we have:
$$a+b+c=8\\
ab+bc+ca=30\\
abc=20$$
Now we want the roots $a+2,b+2,c+2$ of $X^3+kX^2+lX+m=0$. Hence:
$$a+2+b+2+c+2=a+b+c+6=14=-k\\
(a+2)(b+2)+(b+2)(c+2)+(c+2)(a+2)=\\
(ab+bc+ca)+4(a+b+c)+12=74=l\\
(a+2)(b+2)(c+2)=\\
abc+2(ab+bc+ca)+4(a+b+c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find $1 \le a < b \le n$ such that $a\cdot b + a + b = n\cdot (n+1)/2$
Find $1 \le a < b \le n$ such that $$a\cdot b + a + b = \frac{n\cdot (n+1)}2$$
Is there a more efficient way than picking $a$ or $b$ and trying all values between $1$ and $n$ ?
| $(a+1)(b+1)=\dfrac{n^2+n+2}{2}$.
If $a$ and $b$ are restricted to integers, consider the factors of $\dfrac{n^2+n+2}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Having the identity: $\int_0^1 \frac{x^{2n}}{1+x^2} dx= \sum_{k=0}^{\infty} \frac{(-1)^k}{2n+2k+1}$, how do I square the fraction inside the sum? I have the integral:
\begin{align}
\int_0^1 \frac{x^{2n}}{1+x^2} dx &= \int_0^1 x^{2n} \sum_{k=0}^{\infty}(-1)^kx^{2k} dx\\
&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2n+2k+1}
\en... | HINT: Manipulate this identity involving the Digamma Function $$\sum_{k=0}^\infty \frac{(-1)^k}{zk+1}= \frac{1}{2z}\left( \psi_0(\frac{1}{2}+\frac{1}{2z})-\psi_0(\frac{1}{2})\right)$$
to get $$\frac{1}{4}\left(\psi_0(\frac{n}{2}+\frac{3}{4})-\psi_0(\frac{n}{2}+\frac{1}{4})\right)$$ for the value of the integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof that $ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $ for $p>5$ Proof that $$ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $$
for $p>5$ and $p$ is prime.
$\newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)}$
My try
Let show that
$$p^2 - 1 \equiv 0 \Mod{30} \vee p^2 - 1\equiv 18 \Mod{30}$$
Let check $$p^2 -1 = (p... | $\ 5< p$ prime $\,\Rightarrow\, \begin{align} p&\equiv\pm1\ \ \ \ \ \pmod{\!6}\\ p&\equiv \pm1,\pm2\!\!\!\pmod{\!5}\end{align}$
$\,\Rightarrow\,\begin{align}\color{#0a0}{p^2}&\: \color{#0a0}{\equiv\, 1 \pmod{\!6}}\\ p^2&\equiv \color{#c00}{\pm1}\!\!\!\!\pmod{\!5}\end{align}\!$
$\overset{\rm CRT}\iff p^2\equiv 1,19\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Computing to volume of ball intersection with cone and another ball I am trying to calculate the integral $\ \int\int\int_V z dV $ where $\ V $ is the volume inside the ball $\ x^2 + y^2 + (z-2)^2 = 4 $ and also inside the cone $\ z^2 = x^2 + y^2 $ and outside the ball $\ x^2 + y^2 + z^2 = 4 $ .
So my attempt it to set... | $x = rsin\theta cos\phi$
$y = rsin\theta sin\phi$
$z = rcos\theta$
$x^2+y^2+z^2 = r^2$
$dxdydz = r^2sin\theta \ dr \ d\theta \ d\phi$
Case1:
$x^2+y^2+z^2 = r^2 = 4$
$$r = 2$$
Case2:
$x^2+y^2+(z-2)^2 = x^2+y^2+z^2 - 4z +4 = 4$
$r^2 - 4rcos\theta = 0$
$$r = 4cos\theta$$
(refer the graph for $r\not=0$)
$\theta$ varies fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving differential equation with power series $y'' - xy' - y = 0$ I need a work check
$$y'' -xy' - y = 0$$ given $y(0) = 1$ and $y'(0) = 0$
so we have:
$$
\begin{split}
y &= \sum_{n=0}^\infty C_nx^n\\
y' &= \sum_{n=1}^\infty nC_nx^{n-1}\\
y' &= \sum_{n=2}^\infty n(n-1)C_nx^{n-2}
\end{split}
$$
so making a substitu... | You're given initial conditions so you must define $c_0$ and $c_1$ to satisfy these initial conditions.
Notice that you can write your series as
$$y(x) =c_0+ \sum_{n=1}^\infty \frac{c_0}{2^n \cdot n!}x^{2n} + \sum_{n=0}^\infty \frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n+1}$$
where I have simply taken the first t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show the following problem of determinant If the matrices $A,B\in M_{3}(\Bbb{Z})$ are singular and $AB=BA$, show that the number $$\det(A^3+B^3)+\det(A^3-B^3)$$ is the double of a perfect cube.
I have considered the polynomial $$\det(A+xB)=\det A+mx+nx^2+x^3\det B$$
From this how I can show
| Following Anurag A's suggestion in the comments, let $\omega$ be a primitive third root of unity (so $\omega^2+\omega + 1=0$).
Then observe that since $A$ and $B$ commute, we have $$A^3+B^3 = (A+B)(A+\omega B)(A+\omega^2B)$$ and $$A^3-B^3 = (A-B)(A-\omega B)(A-\omega^2 B).$$
Then using that $\det(A)=\det(B)=0$, we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
show this inequality with $xy+yz+zx=3$ let $x,y,z>0$ and such $xy+yz+zx=3$,show that
$$\dfrac{x}{x^3+y^2+1}+\dfrac{y}{y^3+z^2+1}+\dfrac{z}{z^3+x^2+1}\le 1$$
To prove this inequality,I want use following Cauchy-Schwarz inequality
$$(x^3+y^2+1)(\frac{1}{x}+1+z^2)\ge (x+y+z)^2$$
$$\dfrac{x}{x^3+y^2+1}\le \dfrac{1+x+xz^2}{... | By Muirhead
$$\sum_{cyc}\frac{x}{x^3+y^2+1}\leq\sum_{cyc}\frac{x}{x^2+y^2+x}.$$
Thus, it's enough to prove that
$$\sum_{cyc}\frac{x}{x^2+y^2+x}\leq1$$ or
$$\sum_{cyc}\left(x^4y^2+x^4z^2+\frac{2}{3}x^2y^2z^2-x^3y-x^2yz-\frac{2}{3}xyz\right)\geq0$$ or
$$\sum_{cyc}(3x^4y^2+3x^4z^2+2x^2y^2z^2-x^4y^2-x^4yz-x^3y^2z-x^3y^2z-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the sum: $\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) $ $$S_n=\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) = \frac{1}{r-s}\left( \frac{1}{s+1}+\frac{1}{s+2}\cdots+\frac{1}{r}\right)$$
I fail to see how this:
$$S_n = \frac{1}{r-s}\left( \frac... | Since $r > s$,
$$\begin{array}{rl} & \frac{1}{n+s+1} & - \frac{1}{n+r+1}\\
= & \frac{1}{n+s+1} &+ \color{red}{\frac{1}{n+s+2}} &+ \cdots &+ \color{blue}{\frac{1}{n+r}}\\
& & - \color{red}{\frac{1}{n+s+2}} &- \cdots &- \color{blue}{\frac{1}{n+r}} &- \frac{1}{n+r+1}\end{array}$$
we have
$$\begin{align}\frac{1}{n+s+1} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Can $\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha}$ be simplified? I am trying to simplify the following but I cannot.
$$
\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha}
$$
Can it be simplified?
Edit
My last result is
$$
- \frac{\cos \alpha \left( 1 + \sin \alpha \cos... | First, let us state of the fundamentals.
$$\csc \alpha = \frac{1}{\sin \alpha};\ \tan \alpha = \frac{\sin \alpha}{\cos \alpha};\ \cot \alpha = \frac{\cos\alpha}{\sin \alpha};\ \sec \alpha = \frac{1}{\cos \alpha};\ \sin^2\alpha + \cos^2\alpha = 1;$$
So,
$$\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $\frac{2 f'}{(f-1)^2}=1$ with initial condition $f(0)=c$ I'm trying to solve this by writing it as the derivative of a log of a polynomial but I can't make it work. Any hints?
| With
$g = f - 1, \; g(0) = f(0) - 1 = c - 1, \tag 1$
we have
$g' = f'; \tag 2$'
thus,
$\dfrac{2f'}{(f - 1)^2} = \dfrac{2g'}{g^2} = 1, \tag 3$
whence
$g^{-2}g' = \dfrac{1}{2}, \tag 4$
or
$(-g^{-1})' = \dfrac{1}{2}; \tag 5$
we integrate 'twixt $0$ and $x$:
$-g^{-1}(x) - (-g^{-1}(0)) = \displaystyle \int_0^x (-g^{-1}(s))\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate the derivative of $\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ at $x=0$? Let $F(x):=\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ ,where $[\frac{1}{t}]$ is the largest integer no more than $\frac{1}{t}$.Prove $F'(0)=\frac{1}{2}$.
I have tried in this way:
\begin{equation}
\begin{aligned}
\l... | With the help of reuns,I can complete the proof.
\begin{equation}
\begin{aligned}
\lim_{n\to\infty}nF\left(\frac{1}{n}\right)&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-\left[\frac{1}{t}\right]\right)dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
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Anders, Bodil, Cecilia, and David shall receive 4 oranges. In how many ways is this possible if Anders should have at least one? Anders, Bodil, Cecilia, and David shall receive 4 oranges. In how many ways is this possible if Anders should have atleast one?
Correct answer: 29
My solution:
How many solutions are there t... | Your solution is correct.
If we ignore the restriction that Anders must receive at least one orange, the number of ways we could distribute the oranges is the number of solutions of the equation
$$a + b + c + d = 4$$
in the nonnegative integers. A particular solution of this equation corresponds to the placement of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Does dividing a common factor out from numerator and denominator of a rational function create a new function with different domain? Suppose a function is defined as $ f(x) =\frac{x^2 - 9}{x-3}. $
If we divide the common factor $ x-3 $ from both the numerator and denominator :-
$$ \frac{x^2 - 9}{x - 3} \\ = \frac{(x+3... | Two functions are equal if and only if they have the exact same domain, and the exact value at each element of the domain. (If you like your functions to also have a specified codomain, then you also require the functions to have the exact same codomain).
“Cancelling” a common factor includes the implicit assertion tha... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why do we take the positive square root only from $\sqrt{a^2-x^2}$ when integrating using trig substitutions? $$\int\frac{\mathrm dx}{x^2 \sqrt{16-x^2}}$$
when substituting $\,x=4\sin\theta ,\;\mathrm dx=4\cos\theta\, \mathrm d\theta,\,$ it becomes
$$\int\frac{4\cos\theta\, \mathrm d\theta}{4^2\sin^2\theta\sqrt{16-4^2\... | You wrote that “$\sqrt{16}$ is $\pm4$”. This is wrong. Yes, the number $16$ has two square roots: $4$ and $-4$. But $\sqrt{16}$ stands for the non-negative root of $16$, which is $4$. For the same reason, $\sqrt{16-4^2\sin^2\theta}=4\sqrt{1-\sin^2\theta}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing the matrix powers of a non-diagonalizable matrix Define
\begin{equation}
A = \begin{pmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{pmatrix}.
\end{equation}
Note that the sum of the dimensions of the eigenspaces of $A$ is only two. $A$ is thus not diag... | Here is a different way using a rather classical trick, converting the issue into a binomial expansion. Indeed, we can write :
$$A=\frac12(I+B) \ \text{where} \ B:=\begin{pmatrix}0&1&0 \\0&1/2&1/2\\0&1/2&1/2\end{pmatrix}$$
where matrix $B$ has the following particularity
$$B^n=C \ \text{for all} \ n>1 \ \text{where} \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $x + y + z = 2$, then show $\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$, added a second question(Problem 2). Problem number 1:
The problem is that $x, y, z$ are proper fractions, and each one of them is greater than zero.
Given $x + y + z = 2$, prove
$$\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$$
I have tried to solve this... | $\dfrac ab-\dfrac cd=\dfrac{ad-bc}{be}$ will be $\ge0$
only if $ad-bc>0\iff \dfrac ab>\dfrac cd$
Alternatively
Let $a=c+p,b=d+q,p,q>0$
$\dfrac ab-\dfrac cd=\dfrac{c+p}{d+q}-\dfrac cd=\dfrac{d(c+p)-c(d+q)}{...}$
$=\dfrac{dp-cq}{...}$
will be $>0$
only if $dp-cq>0\iff \dfrac dc>\dfrac qp$
as $c,d,p,q>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim\limits_{n \to \infty} \int_0^1 \frac{\ln^n (1+x)}{1+x^2}dx$ Can you please verify my proof below?
My solution
Notice that $f(x):=\ln^n(x)$ and $g(x):=\dfrac{1}{1+x^2}$ is continuous over $[0,1]$, and $g(x)>0$. As per the first integral mean value theorem, on can have
$$0\leq \int_0^1 \frac{\ln^n (1+x)}... | It's correct, but you can get the same bound easier, without using MVT:
For $0\le x \le 1$ we have $0 \le \ln(1+x) \le \ln 2$, so $0\le \frac{\ln^n(1+x)}{1+x^2} \le \frac{\ln^n 2}{1+x^2}$ and
$$ 0\le \int_0^1 \frac{\ln^n(1+x)}{1+x^2} dx \le \int_0^1 \frac{\ln^n 2}{1+x^2} dx=\frac{\pi}{4} \ln^n 2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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For every integer $n$, the quantity $n^2 + 2n \equiv 0\pmod 4$ or $n^2 + 2n \equiv 3\pmod 4$ I'm trying to prove this question using induction
So far I have
Base Case
Let $n = 1$, $(1^2 + 2)\equiv 3 \pmod 4 $
Claim holds for base case
Induction
Assume $n = k$ holds, that is $k^2 + 2k \equiv 0\pmod 4$ or $k^2 + 2k \equ... | You only need one equation.
$\begin{array}\\
(n+2)^2+2(n+2)
-(n^2+2n)
&=n^2+4n+4+2n+4
-(n^2+2n)\\
&=4n+8\\
\end{array}
$
which is divisible by $4$,
so whatever remainder $n$ has
mod 4, $n+2$ also has.
Since $n=0 \implies n^2+2n = 0$,
all even numbers have a
remainder of 0 mod 4.
Since $n=1 \implies n^2+2n = 3$,
all odd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Is $x = \frac{a}{b} = \frac{b}{a/3}$ a continued fraction? How to solve for $x$? I believe following is a continued fraction. I'm stumped on how to solve for x
$x = \frac{a}{b} = \frac{b}{a/3}$
I know it can be re-written as
$x = \frac{a}{b} = \frac{3b}{a}$
$x = a^2 = 3b^2$
I'm unsure where to go from here.
Below are ... | The part where you wrote $x=a^2=3b^2$ is not correct.
From $$ \frac{a}{b} = \frac{b}{a/3} \implies a^2/3 = b^2 \implies a^2=3b^2$$
So $a = \sqrt{3}b$ and thus $$x = \frac{a}{b} = \sqrt{3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove for all positive a,b,c that $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$ Prove for all positive a,b,c $$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$$
My Try
I tried taking common denominator of the expression,
$\frac{a^2b+ab^2+b^2c+c^2b+ac^2+a^2c}{abc}$
How to proceed? Is there a way to write them as per... | Using Cauchy Schwarz, we can write $\displaystyle (a + b + c) \left(\frac{1}{a} +\frac{1}{b} + \frac{1}{c} \right) \geq 9$. Thus, we have,
$$ \frac{a+b+c}{a} +\frac{a+b+c}{b} + \frac{a+b+c}{c} \geq 9 \implies 1+ \frac{b+c}{a} + 1 + \frac{a+c}{b} + 1 + \frac{a+b}{c} \geq 9$$
Hence, $$ \frac{a+b}{c} + \frac{b+c}{a} + ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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$\frac{x}{4} = \frac{y}{x} = \frac{7}{y}$ Could someone please explain to me what's wrong here?
$\cfrac{x}{4} = \cfrac{y}{x} = \cfrac{7}{y}$
So $$\cfrac{x^2}{4} = y \ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \ y = \sqrt{7x}$$
Hence $$\cfrac{x^2}{4} =\sqrt{7x}$$
With a solution of $x = 2.57$
But $$\cfrac{2.57^2}{4} \neq \sqr... | $x\neq 2.57$. You want to solve for $x^2/4 = \sqrt{7x}$, which corresponds to solving the polynomial equation $x^4/16 - 7x = 0$, which has a real and nontrivial zero at $\sqrt[3]{112}$. Hence, $x = \sqrt[3]{112}$ and $y = (112)^{2/3} / 4$. You can check that this in fact, is a solution to your system.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the minimum value of $\sum_\mathrm{cyc}\frac{a^2}{b + c}$ where $a, b, c > 0$ and $\sum_\mathrm{cyc}\sqrt{a^2 + b^2} = 1$.
$a$, $b$ and $c$ are positives such that $\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} = 1$. Calculate the minimum value of $$\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}... | SOS helps!
Let $a=b=c=\frac{1}{3\sqrt2}.$
Thus, we get a value $\frac{1}{2\sqrt2}.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{a^2}{b+c}\geq\frac{1}{2\sqrt2}\sum_{cyc}\sqrt{a^2+b^2}$$ or
$$\sum_{cyc}\left(\frac{a^2}{b+c}-\frac{a}{2}\right)\geq\frac{1}{4}\sum_{cyc}\left(\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Calculation of a series using series definition I have to calculate the series $$\sum_{k=2}^\infty \left(\frac 1k-\frac 1{k+2}\right)$$
Using the definition: $$L = \lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=0}^na_k$$
Obviously $\lim_{n\to\infty} (\frac 1n-\frac 1{n+2})=0$, but I don't think that this is the right w... | As suggested from the comment by JMoravitz, you should write down the series explicitly and see if you can see something interesting:
$$ \sum_{2}^\infty \big( \frac{1}{k} - \frac{1}{k+2} \big) = \bigg(\frac{1}{2} - \frac{1}{4} \bigg) + \bigg(\frac{1}{3} - \frac{1}{5} \bigg) + \bigg(\frac{1}{4} - \frac{1}{6} \bigg) + \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\lim_{x\to 1}\frac{x^{x}-x^{x^2}}{(1-x)^2}$ $$L=\lim_{x\to 1}\frac{x^{x}-x^{x^2}}{(1-x)^2}$$
Any hints on how to approach this problem in the first place? The answer should be $-1$.
I tried adding and subtracting $1$, then evaluating
$$L_1=\lim_{x\to 1}\frac{x^{x}-1}{(1-x)^2}$$ $$and$$ $$L_2=\lim_{x\to 1}\fra... | We have \begin{align} \frac{x^x - x^{x^2}}{(x-1)^2} &= x^x \cdot \frac{1 - x^{x(x-1)}}{(x-1)^2} = \\
&= x^x \cdot \frac{1-e^{x(x-1)\ln x}}{(x-1)^2} = \\
&= x^x \cdot\frac{1-e^{x(x-1)\ln x}}{x(x-1)\ln x}\cdot\frac{x\ln x}{x-1} \end{align}
We calculate the limits separately: $$ \lim_{x \rightarrow 1} x^x = 1^1 = 1$$
$$ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve integer equation $2^m.m^2=9n^2-12n+19$
Problem: Find $m,n\in \mathbb{N}^*$ satisfies: $2^m.m^2=9n^2-12n+19$.
This is my attempt:
We have $9n^2-12n+19\equiv 1 \pmod 3$, so: $2^{m}m^2\equiv 1 \pmod 3\tag{1}$
In addition, we have: $$m^2\equiv 0\text{ or }1 \pmod 3 $$
So $(1)\implies m\equiv \pm 1\pmod 3$.
Suppose ... | First of all, let's complete the square on the RHS and write it as $(3n-2)^2+15$. Next, suppose $m$ is odd. Then $2^mm^2$ will be $\equiv 2\pmod 4$ (can you see why?) but the RHS — as a sum of a square and a number $\equiv 3\pmod 4$ — will be congruent to either $3$ or $0$, so they can't be equivalent.
Now, suppose $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Double Integration Problem $\int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$
Compute
$$I = \int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$$
Here are my steps:
$$\begin{split}
I &=\int_{0}^{1} \left(\int_0^1 \frac{dy}{1+y(x^2-x)}\right)dx\\
&=\int_{0}^{1} \left[\frac{\ln(1+y(x^2-x))}{x^2-x}\right]_0^1dx\\
&=\int_... | $$I=\int_0^1 \frac{\ln(1-x+x^2)}{x(x-1)}dx=-\int_0^1 \frac{\ln(1-x(1-x))}{x}dx-\int_0^1\frac{\ln(1-x(x-1))}{1-x}dx$$
The substitution $1-x\to x $ in the second integral reveals that:
$$\int_0^1\frac{\ln(1-x(1-x))}{1-x}dx=\int_0^1\frac{\ln(1-x(1-x))}{x}dx$$
$$\Rightarrow I=-2\int_0^1\frac{\ln(1-x(1-x))}{x}dx=2\int_0^1\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Evaluation of $\int_{0}^{1}{\int_{0}^{1}{\frac{\ln \left( 1-x \right)-\ln \left( 1-y \right)}{x-y-1}dx}dy}$ I trying to verify the following:
$$
\int_{0}^{1}{\int_{0}^{1}{\frac{\ln \left( 1-x \right)-\ln \left( 1-y \right)}{x-y-1}dx}dy}=\frac{{{\pi }^{2}}}{12}+{{\ln }^{2}}\left( 2 \right)
$$
I thought the solution may ... | With
$$\int\frac{\ln(x+z)}{x-a}dx = \text{Li}_2\left(\frac{x+z}{a+z}\right) - \ln(x+z)\,\text{Li}_1\left(\frac{x+z}{a+z}\right) + C $$
we get
$$\int\limits_0^1\frac{\ln(1-x)}{x-y-1}dx = \int\limits_0^{-1}\frac{\ln(x+1)}{x+y+1}dx = -\text{Li}_2\left(-\frac{1}{y}\right)$$
and the other part is:
$$\int\limits_0^1\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Simplifying $\sum_{cyc}\tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)$. I get $0$, but the answer is $\pi$. So the question is
$$ \tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)+\tan^{-1}\left(\sqrt{\frac{y(x+y+z)}{xz}}\right)+\tan^{-1}\left(\sqrt{\frac{z(x+y+z)}{yx}}\right) =\ ? $$
So my take on the question is to... | Use https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values,
$-\dfrac\pi2\le \tan^{-1}a\le\dfrac\pi2$
Now $\sqrt b\ge0$ for real $\sqrt b$
$\implies0\le\tan^{-1}\sqrt b\le\dfrac\pi2$
So, here the sum will lie in $\in[0,3\pi/2]$
Now the sum will be $=0$ only if each term under is individually $=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Solve $(y^2 + xy)(x^2 - x + 1) = 3x - 1$ over the integers.
Solve $$(y^2 + xy)(x^2 - x + 1) = 3x - 1$$ over the integers.
There are many solutions to this problem, and perhaps I chose the worst one possible. I hope that someone could come up with a better answer.
This problem is adapted from a recent competition (whi... | After getting $(x^2-x+1)\mid(3x-1)$, and bounding $-2\leq x\leq 3$ as a result, you should first go back and check the condition $(x^2-x+1)\mid(3x-1)$:
*
*$x=-2$: $x^2-x+1=7$, $3x-1=-7\quad\checkmark$
*$x=-1$: $x^2-x+1=3$ so doesn't divide $3x-1$.
*$x=0,1$: $x^2-x+1=1\quad\checkmark$
*$x=2$: $x^2-x+1=3$, so doesn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Calculate the maximum value of $\sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}$ where $x + 2y + 3z = 2$
$x$, $y$ and $z$ are positives such that $x + 2y + 3z = 2$. Calculate the maximum value of $$ \sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}... | We have that
$$\sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}$$
$$ = \sqrt{\frac{3yz}{3yz + (2 - 2y - 3z)}} + \sqrt{\frac{3zx}{3zx + 2(2 - 3z - x)}} + \sqrt{\frac{xy}{xy + (2 - x - 2y)}}$$
$$ = \sqrt{\frac{3yz}{(1 - y)(2 - 3z)}} + \sqrt{\frac{3zx}{(2 - 3z)(2 - x)}} + \sqrt{\frac{xy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Showing that two curves do not intersect I want to show that
$$x+1 \neq (x^3(x+2))^{1/4} + \sqrt{x+1-\sqrt{x^2+2x}}$$
for any real $x>0$.
There are two approaches I've taken: showing they are equal and arriving at a contradiction (but this hasn't worked) and computing the derivative of the RHS and showing it is stri... | Suppose that there exists an $x\gt 0$ such that
$$x+1 - \sqrt{x+1-\sqrt{x^2+2x}}= (x^3(x+2))^{1/4} $$
Squaring the both sides gives
$$(x+1)^2-2(x+1)\sqrt{x+1-\sqrt{x^2+2x}}+x+1-\sqrt{x^2+2x}=x\sqrt{x^2+2x},$$
i.e.
$$(x+1)^2+x+1-(x+1)\sqrt{x^2+2x}=2(x+1)\sqrt{x+1-\sqrt{x^2+2x}}$$
Let $x+1=y\ (\gt 1)$. Then, we have
$$y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all possible integers $n$ such that $\sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.
Find all possible integers $n$ such that $m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.
Guess what? This problem is adapted from a recent competition. There have been a solution below for you to check out. I ... | A shorter approach:
Generally for numbers $m=\sqrt{n+a}+\sqrt{n+\sqrt{n+a}}$ we must have:
$n+a=b^2$
$b^2+b-a=c^2$
If this system of equations, with given a, has integer solutions for n then m is integer.For example $a=3$; $m=\sqrt{n+3}+\sqrt{n+\sqrt{n+3}}$ we have:
$n+3=b^2$
$b^2+b-3=c^2$
It can be seen that with $c=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Investigate convergence of $\int_0^{\pi} \frac{\sin x}{x^{\frac{3}{2}}(\pi-x)^{\alpha}} dx$ for $\alpha \in \mathbb R$
Investigate convergence of $\int_0^{\pi} \frac{\sin x}{x^{\frac{3}{2}}(\pi-x)^{\alpha}} dx$ for $\alpha \in \mathbb R$
My try:
I want to use Direct comparison test so: $$0\le \frac{\sin x}{x^{\frac{3... | Divide the integral $$ \int_0^\pi \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx = \int_0^{\pi/2} \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx + \int_{\pi/2}^\pi \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx $$
For $x\in[0,\pi/2]$ we have $$ \frac{2}{\pi} \le \frac{\sin x}{x} \le 1$$
$$ 0 < M_1(\alpha) = \min\{\frac{1}{\pi^\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int\limits_0^\infty \frac{\ln^2(1+x)}{1+x^2}\ dx$ This problem was already solved here (in different closed form).
But how can you prove $\ \displaystyle\int\limits_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=2\Im\left(\operatorname{Li}_3(1+i)\right)\ $
Where $\displaystyle \operatorname{Li}_3(x)=\sum_{n=1}^\inft... | The following solution is by Cornel Valean:
\begin{gather*}
\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\textrm{d}x\overset{x= \frac1y}{=}\int_0^\infty\frac{\ln^2\left(\frac{y}{1+y}\right)}{1+y^2}\textrm{d}y\\
\overset{\frac{y}{1+y}=x}{=}\int_0^1\frac{\ln^2(x)}{x^2+(1-x)^2}\textrm{d}x\\
\left\{\text{write $\frac{1}{x^2+(1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Constraints on the roots to a quadratic equation If both roots of the quadratic equation
$$2x^2 +kx -(k+1)=0$$
are greater than $1$, then $k$ lies in what interval?
I tried to solve this using using different graphical and algebraic method but i seem to miss a crucial insight which is stopping me from getting the answ... | Let $x_1,x_2$ be the two roots of your equation, so $x_1 \geq 1$, $x_2 \geq 1$.
In general, given a quadratic equation $ax^2+bx+c=0$ you have that $\frac{c}{a}$ is the product of the two roots and that $-\frac{b}{a}$ is the sum of the two roots.
In our case $x_1+x_2=-\frac{k}{2}$ and $x_1 x_2=-\frac{k+1}{2}$ and $x_1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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How to convert the following Quadratic optimization problem to a linear one? I have an optimization problem
$$Min : \ 3x_{11} + 5x_{12} + 4x_{21} + 3 x_{22} - (10x_{11}x_{22} + 2x_{12}x_{21})$$
subject to the following constraints :
$$ x_{11} + x_{12} = 1$$
$$ x_{21} + x_{22} = 1$$
where $x_{11}. x_{12}, x_{21}, x_{2... | The general linearization of a product of binary variables is overkill here. Instead, first substitute $x_{12} = 1 - x_{11}$ and $x_{22} = 1 - x_{21}$ to obtain objective function
\begin{align}
&3x_{11} + 5x_{12} + 4x_{21} + 3 x_{22} - 10x_{11}x_{22} - 2x_{12}x_{21}\\
&=3x_{11} + 5(1 - x_{11}) + 4x_{21} + 3 (1 - x_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3265561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Reasoning about $x(x+2)$ relatively prime to $\dfrac{p\#}{w}$ Let:
*
*$p_k$ be the $k$th prime so that $p_3 = 5$
*$p\#$ be the primorial of $p$.
*$\text{gcd}(a,b)$ be the greatest common divisor of integer $a$ and integer $b$.
*$w \ge 1$ be an integer that divides $p\#$
Does it follow that for all integers $k \ge ... | We can solve for any integer $A$, not just for primorial divided by something, and proof is clearly easier.
Here, we should state that, for any positive number $m$, there is a number $x$ in $[m, m+A)$ which $gcd(x(x+2), A) = 1$.
The necessary/sufficient condition for $gcd(x(x+2), A) = 1$ is, $gcd(x, A)=1$ and $gcd(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Given three positive numbers $a,\,b,\,c$ . Prove that $(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$ . Given three positive numbers $a,\,b,\,c$ . Prove that $$(\!abc+ a+ b+ c\!)^{3}\geqslant 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!).$$
My own problem is given a solution, and I'm looking forward to seeing... | Hint: By AM-GM,
$$(abc+a) + (b+c) \geqslant 2a\sqrt{bc}+2\sqrt{bc} = 2\sqrt{bc}(1+a)$$
Now multiply three such inequalities (after cyclical shift).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Given $a,b,c,d\in\mathbb{Z}_{+}$ with $a\geqq b>c>d\geqq 0$ and $ac+bd=(\!b+d+a-c\!)(\!b+d-a+c\!)$. Prove $ab+cd$ is not prime. Given positve integers $a, b, c, d$. For $a\geqq\!b>\!c>\!d\geqq\!0$ and $ac+\!bd= (\!b+ d+ a- c\!)(\!b+ d- a+ c\!)$
Prove $ab+ cd$ is not a prime number.
I have a solution, and I'm looking fo... | We prove that $a^nb^m+c^md^n$ is composite for odd $n$ and arbitrary $m.$
$ a + b - c + d | ac + bd \rightarrow a + b - c + d |(a + b)(a + d)$
$ \therefore \exists k \in N$ ; $ (a + b)(a + d) = k(a + b - c + d)$
Assume that $ gcd(a + d,a + b - c + d) = 1 \implies a + b = l(a + b - c + d)$, where $l$ is a divisor of $ k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Integral by the residue theorem
Check that the value of
$$I(\alpha)=28\int_{-\infty}^\infty \frac{1-x^2}{2+3x^2+2\alpha^2x^4}\text{d}x$$
is
$$I(\alpha)=14\pi\left(1-\frac{1}{\alpha}\right)\sqrt{\frac{2}{4\alpha+3}}$$
for some $\alpha>0$.
I've tried to apply the Residue Theorem to this integral with the contou... | You are on the right track, and only needs a bit of extra input to get to the answer. Assume for a moment that $\alpha > 3/4$. Then the zeros of $2+3t+3\alpha^2 t^2 = 0$ are
$$ t = \frac{-3 \pm i\sqrt{16\alpha^2-9}}{4a^2} = \left( \frac{\sqrt{4\alpha-3} \pm i\sqrt{4\alpha+3}}{\sqrt{8}\alpha} \right)^2. $$
Now, among th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3272678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Stuck on this - In ABC right triangle AC= $2+\sqrt{3}$ and BC = $3+2\sqrt{3}$. circle touches point C and D, Find the Area of $AMD$ In ABC right triangle $AC= 2+\sqrt{3}$ and $BC = 3+2\sqrt{3}$. circle touches point C and D, Find the Area of $AMD$
Here's my strategy of solving this, I'm not sure if it's correct, if y... | In a video by Mathematics professor Michael Penn, this is solved first by finding out the hypotenuse of the triangle and writing it in the form $a+b\sqrt{3}$, where $a, b \in \mathbb{R}$. Then, he finds the area of the triangle ($AOD$) and the sector ($EOD$) as shown in Figure 1 and subtracts the area of the sector fro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3272856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Finding general solution of $3{\times}3$ system I am given the following:
$$
x'=
\begin{bmatrix}
2 &0 &0 \\
-7&9 &7 \\
0&0 &2
\end{bmatrix}
x
$$
Solving $\det(A-\lambda I)$, I get $\lambda = 2,2,9$. Solving $\det(A-2\lambda)$, I get
\begin{bmatrix}
0&0 &0 \\
-7&7 &7 \\
0&0 &0
\end{bmatrix}
So we have g... | The system $$x'=
\begin{bmatrix}
2 &0 &0 \\
-7&9 &7 \\
0&0 &2
\end{bmatrix}$$
is simply
$$ x_1'=2x_1$$
$$ x_2'=-7x_1+9x_2+7x_3$$
$$ x_3'=2x_3$$
Therefore $$x_1=c_1e^{2t}$$ and $$x_3=c_2 e^{2t}$$
Solving for $$x_2'=9x_2+7(c_2-c_1)e^{2t}$$
we get $$x_2=c_3e^{9t}+(c_1-c_2)e^{2t}$$
Thus we have $$x(t) =
c_1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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