Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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determining $n$ in a given sequence $\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $ Given that: $$\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $$
Determine $n$.
The memorandum says the answer is 2011 but how is that so? Where did I go wrong?
| $1 + 3 + 5 + .. + 2n - 1 = n^2$, and $2 + 4 + ... + 2n = n(n + 1) = n^2 + n$. So:
$\dfrac{n^2}{n^2 + n} = \dfrac{2011}{2012}$. So: $2012n^2 = 2011n^2 + 2011n$, and $n^2 = 2011n$. So $n = 2011$
| {
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"url": "https://math.stackexchange.com/questions/765664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help with integral from Apostol I've been working on this all day and I'm still stumped. To state the problem (ref: Apostol Section 5.11, Question 29):
Show
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{2}{a(2n+3)}(x^n(ax+b)^{3/2} - nb\int x^{n-1}\sqrt{ax+b}\,\,dx) + c$$
My attempts so far have been as follows (NB: Im avoiding... | I would start with integration by parts with the choice $$u = x^n, \quad dv = (ax+b)^{1/2} \, dx,$$ which gives $$du = nx^{n-1}, \quad v = \frac{2}{3a}(ax+b)^{3/2}.$$ Thus if $I$ is the given integral, then $$\begin{align*} I &= \int u \, dv = uv - \int v \, du \\ &= \frac{2}{3a}x^n (ax+b)^{3/2} - \frac{2n}{3a} \int x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/767106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maclaurin series: $\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^{11}}{11!}+\frac{x^{12}}{12!}+...$ The Taylor series of a real or complex-valued function ƒ(x) that is infinitely differentiable at a real or complex number a is the power series
$$f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a... | Note that
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots.\tag{1}$$
A little less familiar is
$$\sinh x=\frac{e^x-e^{-x}}{2}=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots.\tag{2}$$
Subtract (1) from (2) and divide by $2$. We get
$$\frac{1}{2}(\sinh x-\sin x)=\frac{x^3}{3!}+\frac{x^7}{7!}+\frac{... | {
"language": "en",
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Find $b-d$ when $\log_ab={3\over2}$ and $\log_cd={5\over4}$ $a,b,c$ are three natural numbers such that $\log_ab={3\over2}$ and $\log_cd={5\over4}$.
Given: $a-c=9$
Find $b-d$
| Hint: $$\log_ab={3\over2} \Rightarrow b=a^{\frac32}\Rightarrow a \text{ is a perfect square, } a= \alpha^2 \to b=\alpha^3$$
$$\log_cd={5\over4} \Rightarrow d=c^{\frac54} =(a-9)^{\frac54} \Rightarrow a-9 \text{ is a perfect 4-th power} \to \alpha^2-9=\beta^4 \to \alpha^2-\beta^4=(\alpha+\beta^2)(\alpha-\beta^2)=9$$
Sinc... | {
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Using the method of Lagrange multipliers, find the extreme values of a function Using the method of Lagrange multipliers, find the extreme values of the function
$f(x,y)= \frac{2y^3}{3} + 2x^2 +1$ on the ellipse $5x^2 + y^2 = 1/9$
. Identify the (absolute) maximal and minimal values of f
taken on the ellipse.
currentl... | $\dfrac{\partial f}{\partial x} = 4x= \gamma\cdot 10x$, and $\dfrac{\partial f}{\partial y} = 2y^2 = \gamma\cdot 2y$. So: $2x(2 - 5\gamma) = 0$, and $2y(y - \gamma) = 0$. So
case 1: $x = 0$, so $y^2 = \dfrac{1}{9}$. So $y = \dfrac{1}{3}$ or $-\dfrac{1}{3}$. So:
$f(0,\frac{1}{3}) = \dfrac{83}{81}$, and $f(0,-\frac{1}{3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Which conic is represented by $r = a \cos \theta$ The polar equation $r = a \cos \theta$ represents which conic?
| Multiply both sides by $r$ to get $r^2 = a r \cos \theta$. Then we use the substitution $r^2 = x^2 + y^2$ and $r \cos \theta = x$ to get $x^2 + y^2 = ax$. Then we get $\displaystyle \left(x - \frac{a}{2}\right)^2 + y^2 = \frac{a^2}{4}$. Looks like it's a circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771829",
"timestamp": "2023-03-29T00:00:00",
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How prove this $100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$
let $a,b,c\ge 0$, and such $$a+b+c=6$$
show that
$$100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$$
My idea: since
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=36-2(ab+bc+ac)$$
$$a^2b^2+b^2c^2+a^2c^2=(ab+ca+bc)^2-2abc(a+b+c)=(ab+bc+ac)^2-12a... | Let $G(a,b,c)=100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2$. We will
show that $G(a,b,c) \geq G(2,2,2)=0$.
From the identity
$$
4\frac{G(a,b,c)-G(a,\frac{b+c}{2},\frac{b+c}{2})}{(c-b)^2}
+\frac{G(a,b,c)-G(a,b+c,0)}{bc}=\frac{(a^2+2)(b+c)^2}{4}
$$
we see that at least one of the two numbers
$G(a,b,c)-G(a,\frac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solve $\cos x+8\sin x-7=0$
Solve $\cos x+8\sin x-7=0$
My attempt:
\begin{align}
&8\sin x=7-\cos x\\
&\implies 8\cdot \left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)=7-\cos x\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=7-1+2\sin ^2\frac{x}{2}\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=6+2\sin^2 \frac{x}{2}\\
&\i... | $1.) \; \cos x+8\sin x-7=0; \tag{1}$
$2.) \; \cos^2 x = (7- 8\sin x)^2; \tag{2}$
$3.) \; \cos^2 x = 1 - \sin^2 x; \tag{3}$
$4.) \; 1 - \sin^2 x = (7- 8\sin x)^2; \tag{4}$
$5.) \; 1 - \sin^2 x = 49 - 112 \sin x + 64 \sin^2 x; \tag{5}$
$6.) \; 65 \sin^2 x -112\sin x + 48 = 0; \tag{6}$
$7.) \; \sin x = \dfrac{112 \pm \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/776603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Algebra question: Finding inverse function This question is about finding the inverse function of $f(x)=-\sqrt{9-x^2}$
I seem to be making an error with one of the manipulations. Here is my attempt.
$$x=-\sqrt{9-y^2}$$
$$x^2=(-\sqrt{9-y^2})^2$$
$$x^2= -(9-y^2)$$
$$x^2=y^2-9$$
$$x^2+9=y^2$$
$$\sqrt{x^2+9}=\sqrt{y^2}$$
$... | \begin{align}
y=-\sqrt{9-x^2}
\end{align}
Change $x$ to $y$ and vice versa.
\begin{align}
x&=-\sqrt{9-y^2}\\
-x&=\sqrt{9-y^2}\\
(-x)^2&=\left(\sqrt{9-y^2}\right)^2\\
x^2&=9-y^2\\
y^2&=9-x^2\\
y&=\pm\sqrt{9-x^2}
\end{align}
The last step, change $y$ to $f^{-1}(x)$.
\begin{align}
f^{-1}(x)&=\pm\sqrt{9-x^2}
\end{align}
Si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/776680",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Elegant proof of $\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$? Let $a, b > 0$ satisfy $a^2-4b^2 \geq 0$. Then:
$$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$$
One way to calculate this is by computing the residues at the poles in the upper h... | Let $b=\frac{a}{2}\sin2\theta$ ($0<\theta<\frac{\pi}{2}$), then
\begin{eqnarray}
\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + \frac14a^2\sin^22\theta}&=&\int_{-\infty}^{\infty}\frac{dx}{(x^2+a\sin^2\theta)(x^2+a\cos^2\theta)}\\
&=&\frac{1}{a(\cos^2\theta-\sin^2\theta)}\int_{-\infty}^{\infty}\left(\frac{1}{x^2+a\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/776812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Simplifying $\frac{a}{(a-b)(a-c)(x-a)}+\frac{b}{(b-c)(b-a)(x-b)}+\frac{c}{(c-a)(c-b)(x-c)}$ We need to simplify $$\dfrac{a}{(a-b)(a-c)(x-a)}+\dfrac{b}{(b-c)(b-a)(x-b)}+\dfrac{c}{(c-a)(c-b)(x-c)}$$
The biggest problem is that the above expression has four variables.I transformed the expression into $$\dfrac{a}{-(a-b)(c... | This answer is to show that this can be done by hand without any particular clever trick, but by being clearheaded and organised. With the original expression being symmetric, you should expect the appearance of symmetric expressions in the expansion, and often a fair amount of cancellation.
If we put everything over a... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\cdots$ Evaluate
$$
\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\frac{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}{4}\cdots .
$$
First, it is clear that terms tend to $1$.
It seems that the infinity product is not 0. This is related to the post Se... | Define the sequence $x_n$ by $x_0=\dfrac{1}{2}$ and $x_{n+1}=\dfrac{1}{2}+\dfrac{\sqrt{x_n}}{2}$,
and let $y_n=x_0x_1\cdots x_{n}$. The question is to evaluate
$\lim\limits_{n\to\infty}y_n$.
It is easy to see by induction that $0<x_n<1$ for every $n$, so we can define
$$\theta_n=\arccos(\sqrt{x_n})$$
So that
$$\cos^2(\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Evaluate the sum $1! + 2! + 3! + \cdots + n! \le ?? < (2n)!$ How to evaluate the sum below as close as possible?
$$1! + 2! + 3! + \cdots + n! \le ??? $$
Is the next evaluation $ 1! + 2! + 3! + \cdots + n! \le n n! < (2n)! $ correct?
| $1!+2!+3!+\ldots+n!\le 2n!$ follows by induction (using $1+\ldots+n!+(n+1)!\le 2n!+(n+1)n!=(n+2)n!\le2(n+1)n!=2(n+1)!$). Of course, for $n\ge 2$, we have $2n!\le nn!<(n+1)!<(2n)!$.
A somewhat better (because $\frac{n+1}{n-1}\to1$) approximation is given by
$$ 1!+2!+3!+\ldots+n!\le\frac{n+1}{n-1}n!\qquad\text{for }n\ge... | {
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Find all solutions for $\cos(2x)\cos x-\sin(2x)\sin x=\frac{1}{\sqrt{2}}$ if $0\leq x<\pi$ Find all solutions for $\cos(2x)\cos(x)-\sin(2x)\sin(x)=\frac{1}{\sqrt{2}}$ if $0\leq x< \pi$
Can you verify my work? Thanks!
$$\cos(2x)\cos(x)-\sin(2x)\sin(x)=\frac{1}{\sqrt{2}}$$
$$\cos(2x+x)=\frac{1}{\sqrt{2}}$$
$$\cos(3x)=\fr... | The solution $x=\frac{\pi}{12} + \frac{2k\pi}{3}$ gives, in $[0, \pi[$, the angles $\frac{\pi}{12}$ and $\frac{9\pi}{12}=\frac{3\pi}{4}$.
And the solution $x=-\frac{\pi}{12} + \frac{2k\pi}{3}$ gives, in $[0, \pi[$, the angle $\frac{7\pi}{12}$.
And that's all.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of a sequence involving roots How can I find the limit of the following sequence:
$b_n = \sqrt[4]{n^4+n^3} - \sqrt[4]{n^4+1}$?
I succeeded in finding the limits of other sequences involving roots using the binomic formulas but I don't know what to apply to fourth roots.
Thanks for your help!
| This calls for multiplication with the conjugate: $\sqrt a-\sqrt b=\frac{a-b}{\sqrt a+\sqrt b}$.
$$ \begin{align}b_n&=\sqrt[4]{n^4+n^3}-\sqrt[4]{n^4+1}\\
&=\frac{\sqrt{n^4+n^3}-\sqrt{n^4+1}}{\sqrt[4]{n^4+n^3}+\sqrt[4]{n^4+1}}\\
&=\frac{(n^4+n^3)-(n^4+1)}{(\sqrt[4]{n^4+n^3}+\sqrt[4]{n^4+1})(\sqrt{n^4+n^3}+\sqrt{n^4+1})}... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of the integral Find the value of the following integral conatining a term with natural logarithm$$\int_0^1 (1-y) \ln\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)\, dy.$$
| $$I=\int_{0}^{1}(1-y)\log{\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)}\,dy$$
Substitute $x=1-y$.
$$I=\int_{0}^{1}(1-y)\log{\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)}\,dy=\int_{0}^{1}x\,\log{\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}\right)}\,dx.$$
This can be tackled by integration by parts. First we'll need the deriv... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Solving $315 x \equiv 5 \pmod {11}$ I have to solve this: $$315 x \equiv 5 \pmod {11}$$
Isn't it like that?
$$315 \equiv (22+8) \cdot 10+15 \equiv 8 \cdot 3+4 \equiv 5+8+4 \equiv 6$$
Or have I done something wrong?
| If we use Fermat's little theorem and the fact that $315 \equiv 7\text{ (mod 11)}$, then $7^{11 - 1}\equiv 1 \text{ (mod } 11) \implies 7^{11 - 1} \cdot 5 \equiv 5 \text{ (mod } 11) \implies 7 \cdot 7^{11-2} \cdot 5\equiv 5\text{ (mod 11)} \implies 7 \cdot 8 \cdot 5 \equiv 5 \text{ (mod 11)} \implies x \equiv 40 \text{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Maximum and Minimum value of an inverse function
Find the maximum and minimum value of
$\arcsin \left(x\right)^3+\arccos \left(x\right)^3$.
given that $-1\le x\le 1$
I have solved the problem but i am just curious to know if there are any other ways to solve this particular problem other than the method i used belo... | Help:
a^3+b^3=(a+b)((a+b)^2-3ab)
Let p=Pi
ASinx+ACosx=p/2
f(x)=(ASinx)^3+(ACosx)^3
=p/2 (p^2/4-3.ASinx.ACosx)
=p/2 (p^2/4-3.(p/2).ASinx+3.(ASinx)^2)
This shall be maximum for ASinx= -(p/2)
Hence max value of the function will be:
f (x)max=(7/8) (p^2)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integral $\int \sqrt{x^2-3x+2}\ dx$ How to evaluate $$\int_3^{17} \sqrt{x^2-3x+2}\ dx \ ?$$ I tried Euler's substitution, that is $$\sqrt{x^2-3x+2}=x+t \Longleftrightarrow \frac{t^2-2}{-3-2t}+t=\frac{t^2+3t+2}{2t+3}\ ,$$ which I obtained from $$x^2+2tx+t^2=x^2-3x+2\Longleftrightarrow x=\frac{t^2-2}{-3-2t}$$ $$dx=\frac{... | Hint
I think that you made your life very complicated.
If I may suggest, start completing the square $$x^2-3x+2=\Big(x-\frac{3}{2}\Big)^2-\frac {9}{4}+2=\Big(x-\frac{3}{2}\Big)^2-\frac {1}{4}$$ and now change variable such $$x-\frac{3}{2}= \frac{1}{2}\cosh(y)$$ You will arrive to something very simple for the integrand... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Cubic trig equation I'm trying to solve the following trig equation:
$\cos^3(x)-\sin^3(x)=1$
I set up the substitutions $a=\cos(x)$ and $b=\sin(x)$ and, playing with trig identities, got as far as $a^3+a^2b-b-1=0$, but not sure how to continue. Is there a way to factor this so I can use the zero product property to sol... | $$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x\cos x)$$
If we set $\displaystyle\cos x-\sin x=u, u^2=1-2\sin x\cos x\implies (\cos x-\sin x)(1+\sin x\cos x)=u\left(1+\frac{1-u^2}2\right)=\frac{3u-u^3}2$
So, the problem reduces to $\displaystyle u^3-3u+2=0$
Clearly, $1$ is a root and $\displaystyle\displaystyle\frac{u^3-3u... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Double integral for $\int_{0}^{1} \int_{-1}^{0} \frac {xy}{x^2 + y^2 + 1}\ dy\ dx$ I'm trying to evaluate this
$$\int_{0}^{1} \int_{-1}^{0} \frac {xy}{x^2 + y^2 + 1}\ dy\ dx$$
tried substition
$$ u = {(x^2+y^2+1)}^{-1} \ \ du = \ln {(x^2+y^2+1)}$$
but du is not found in the given equation. I have the feeling that I sh... | $\int_{0}^{1}\int_{-1}^{0}\dfrac{xy}{1 + x^2 + y^2}dydx = \int_{0}^{1}\dfrac{x}{2}\int_{-1}^{0}\dfrac{2ydy}{1+x^2+y^2}dx = \int_{0}^{1}\dfrac{x}{2}\int_{-1}^{0}\dfrac{d(1 + x^2 + y^2)}{1 + x^2 + y^2}dx$. Thus,
$\int_{0}^{1}\int_{-1}^{0}\dfrac{xy}{1 + x^2 + y^2}dydx = \int_{0}^{1}\dfrac{x}{2}\ln(1 + x^2 + y^2)\mid_{-1}^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
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Finding generators for an ideal of $\Bbb{Z}[x]$ We know that $\Bbb{Z}$ is Noetherian. Hence, we can conclude that $\Bbb{Z}[x]$ is Noetherian, too. Consider the ideal generated by $\langle 2x^2+2,3x^3+3,5x^5+5,…,px^p+p,…\rangle$ for all prime natural numbers $p$. How can I determine a finite number of elements which gen... |
The arguments of the following proof relies on computations within $\mathbb Z[x]$ without any reference to bigger rings.
Let $I=\langle 2x^2+2,3x^3+3,5x^5+5,\dots,px^p+p,\dots\rangle$. Since $$px^p+p=p(x^p+1)=p(x+1)(x^{p-1}-\cdots+1)$$ for any prime $p\ge 3$, we have $$I\subseteq \langle 2x^2+2,x+1\rangle.$$ For $2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/802806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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System of linear first order DEs Question:
$$3\dot{x} + \dot{y} + 5x - y = 2e^{-t}+4e^{-3t}$$
$$\dot{x} + 4\dot{y} - 2x + 7y = -3e^{-t}+5e^{-3t}$$
Subject to:
$$x(0)=y(0)=0$$
Attempt at a solution:
I have gotten to:
$$\left(\begin{array}{cc}1&-1&3&-3\\ 1&1&1&1\end{array}\right)\left(\begin{array}{cc}\dot{x}\\ \dot{y}\\... | One way to rewrite:
$$
\left[\begin{array}{cc} 3 & 1 \\ 1 & 4\end{array}\right]
\frac{d}{dt}\left[\begin{array}{c} x \\ y\end{array}\right]+
\left[\begin{array}{cc} 5 & -1 \\ -2 & 7\end{array}\right]
\left[\begin{array}{c} x \\ y\end{array}\right]=
\left[\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/802895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}$ I am trying to prove this interesting integral
$$
I:=\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}.
$$
I tried using $y=1+x^3$ but that didn't help.
We c... | Letting $x\mapsto \frac{1}{x}$ simplify the integral as
$$
I=\int_0^{\infty} \ln \left(\frac{1+x^3}{x^3}\right) \frac{x}{1+x^3} d x =\int_0^{\infty} \frac{\ln \left(x^3+1\right)}{x^3+1} d x
$$
Next consider another integral
$$
I(a)=\int_0^{\infty}\left(x^3+1\right)^a d x
$$
and transform $I(a)$, by putting $y=\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Compute $\det(A^n+B^n)$
Let $A, B $ be two real $3\times 3 $ matrices, $AB=BA$, and $ \det(A-B)=\det(A^2+B^2)=1,\det(A+B)=3, \det(B)=0 $, then, what is ?
$$\det(A^n+B^n)$$
here $n$ is a positive integer.
The problem looks very horrible. Any help will be appreciated
| Although I don't know the real tricks behind the problem, the problem is not as terrible as it appears.
As $A$ and $B$ commute, they are simultanesouly triangularisable over $\mathbb C$. Given that $\det(B)=0$, we may assume that $A$ and $B$ are triangular matrices whose diagonal entries are respectively $a_1,a_2,a_3$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
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Prove that if x and y are odd natural numbers, then $x^2+y^2$ is never a perfect square. Prove that if x and y are odd natural numbers, then $x^2+y^2$ is never a perfect square.
Let $x=2m+1$ and $y=2l+1$ where m,l are integers.
$x^2+y^2=(2m+1)^2+(2l+1)^2=4(m^2+m+l^2+l)+2$
Where do I go from here?
| You can now look at all the natural numbers modulo $4$. We know that numbers must either be even or odd, hence they have the form
$$2n\text{ or } 2n+1.$$
In modulo $4$, they are just $2n\text{ or } 2n+1\mod4$. Now, look at the squares of these numbers, we have that
$$(2n)^2\equiv 4n^2\equiv0\mod4.$$
Also, the odds giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/804871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to prove this inequality $\frac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\frac{\pi}{n+1}}$
Let $a_{1},a_{2},\cdots,a_{n},n\ge 2$ be real numbers,show that
$$\dfrac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\dfrac{\pi}{n+1}}$$
I thi... | Let $M_1 = \dfrac{1}{4\cos\dfrac{\pi}{n+1}}$ and $M_{k+1} = \dfrac{1}{4[\cos\dfrac{\pi}{n+1} - M_k]}$. If we can prove $0 <M_k < \cos\dfrac{\pi}{n+1}$ for $k < n -1$, then we have
\begin{align}
a_1a_2 \leq \cos\dfrac{\pi}{n+1}a_1^2 + M_1 a_2^2 \\
a_2 a_3 \leq (\cos\dfrac{\pi}{n+1} - M_1)a_2^2 + M_2 a_3^2 \\
\vdots \\
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/807326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 2
} |
Asymptotic expansions for the roots of $\epsilon^2x^4-\epsilon x^3-2x^2+2=0$ I'm trying to compute the asymptotic expansion for each of the four roots to the following equation, as $\epsilon \rightarrow 0$:
$\epsilon^2x^4-\epsilon x^3-2x^2+2=0$
I'd like my expansions to go up through terms of size $O(\epsilon^2)$.
I´ve... | Your analysis is correct.
Let's look at the original equation
$${\epsilon ^2}{x^4} - \epsilon {x^3} - 2{x^2} + 2 = 0$$
and plot the Newton-Kruskal diagram:
There are only two possible placements of straight lines passing through two or more points with all remaining points "above the line"; those two lines correspond ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$ I need to prove that $\frac{1+\tan^2\theta}{1+\cot^2\theta}= \tan^2\theta.$
I know that $1+\tan^2\theta=\sec^2\theta$ and that $1+\cot^2\theta=\csc^2\theta$, making it now $$\frac{\sec^2\theta}{\csc^2\theta,}$$ but I don't know how to get it do... | $$\large\frac{1+\tan^2\theta}{1+\cot^2\theta}$$
$$\implies\large \frac {\frac{cos^2\theta+sin^2\theta}{cos^2\theta}}{\frac{sin^2\theta+cos^2\theta}{sin^2\theta}}$$
$$\implies \large \frac {\frac{1}{cos^2\theta}}{\frac{1}{sin^2\theta}} \\$$
$$\implies \large \frac {sin^2\theta}{cos^2\theta} $$
$$\implies \large \tan^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 5
} |
proving $\tan^{-1} \left( \frac{x}{1-x^2} \right) = \tan^{-1}x + \tan^{-1}(x^3)$ Two related questions, one easy, one just a bit harder:
1) Prove the identity
$$
\tan^{-1} \left( \frac{x}{1-x^2} \right) =
\tan^{-1}x + \tan^{-1}(x^3)
$$
2) Now try to find a geometric or trigonometric proof of that same geometry, without... | Part I)
Taking the tan of the RHS, we have a form $tan(A+B)$ where $A = tan^{-1}(x)$, $B = tan^{-1}(x^3)$.
$$
tan(A+B) = \frac{tan(A)+tan(B)}{1-tan(A)*tan(B)}
= \frac{x + x^3}{1 - x^4}
= \frac{x(1 + x^2)}{(1-x^2)(1+x^2)}
= \frac{x}{1-x^2}
$$
Now the tan of LHS is also $\frac{x}{1-x^2}$.
Therefore the identity is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/812123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute and find 2009th decimal(2009th digit after the point), without automation, the following sum Compute and find 2009th decimal of (2009th digit after the point), without automation, the following sum
$$\frac{10}{11}+\frac{10^2}{1221}+\frac{10^3}{123321}+ \cdots +\frac{10^9}{123456789987654321}$$
| You can write your series as
$$f(x)=\sum_{1}^{x}\frac{81\times10^k}{(10^k-1)(10^{k+1}-1)}=9\sum_{1}^{x}\frac{1}{10^k-1}-\frac{1}{10^{k+1}-1}$$
Where $x=9$. By telescopy we can show:
$$f(x)=\frac{10^{x+1}-10}{10^{x+1}-1}=1-\frac{9}{10^{x+1}-1}$$
So your sum is $S=1-1111111111^{-1}=0.\overline{9999999990}$. So since $200... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/816589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why does $\frac{(x^2 + x-6)}{x-2} = \frac{(x+3)(x-2)}{x-2}$? I'm not the best at algebra and would be grateful if someone could explain how you can get from,
$$\frac{x^2 + x-6}{x-2}$$
to,
$$\frac{(x+3)(x-2)}{x-2}$$
| \begin{align}
x^2+x-6 = x^2 + \underbrace{3x - 2x} - 6 & = \underbrace{x^2+3x}+\underbrace{{}-2x-6} \\[8pt]
& = x(x+3) + (-2)(x+3) \\[8pt]
& = x(\cdots\cdots) + (-2)(\cdots\cdots) \\[8pt]
& = x(\cdots\cdots) -2(\cdots\cdots) \\[8pt]
& = (x-2)(\cdots\cdots) \\[8pt]
& = (x-2)(x+3).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Cannot solve by hand:$ x + y = 2; 4x^2 + y^2 = 5(2x - y)(xy)^{\frac12}$ Firstly, this is not my homework. I am well past high school (finished graduate school several years ago) but I am mentoring a high schooler and I want to explain how to solve this by hand using just pen and paper.
A more presentable form + solutio... | since $xy>0,x+y=2>0, \implies x>0,y>0$
let $x=a^2,y=b^2,a>0,b>0 \implies 4a^4+b^4=5(2a^2-b^2)ab \iff \\ 4a^4-10a^3b+5ab^3+b^4=0$
by observation, $a=b$ is a solution as $4-10+5+1=0$
so we get $(a-b)(4a^3-6a^2b-6ab^2-b^3)=0$
now to check $4a^3-6a^2b-6ab^2-b^3=0$,a factor $b=-2a$ is existed. $\implies$
$(2a+b)(2a^2-4ab-b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$ Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$
My approach :
I used $\sin A +\sin B = 2\sin(A+B)/2\times\cos(A-B)/2 $
$\Rightarrow \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = 2\sin(3\pi)/7\times\co... | Let $\Phi_7(x)$ the minimal polynomial of $\xi=\exp\left(\frac{2\pi i}{7}\right)$. Its Galois group over $\mathbb{Q}$ is cyclic and generated by $\xi\mapsto\xi^3$, since $3$ is a generator of $\mathbb{Z}/(7\mathbb{Z})^*$. The quadratic residues $\!\!\pmod{7}$ are $1,2,4$ and the quadratic non-residues are $3,5,6$, henc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
Area of triangle having an inscribed circle
The radius of an inscribed circle in a triangle is $2 cm$. A point of tangency divides a side into $3 cm$ and $4 cm$. Find the area of the triangle.
We know that a side is $7 cm$, the others are $4+x$, $3+x$. I tried finding $2$ equations of the area, $S=pr \Rightarrow S=2x... |
$AR=AP=3$, $CR=CQ=4$, $BP=BQ=x$, $OP=OQ=OR=2$.
$AO=\sqrt{3^2+2^2}=\sqrt{13}$;
$CO=\sqrt{4^2+2^2}=\sqrt{20}$.
$$S = \dfrac{1}{2} \cdot AB\cdot AC \cdot \sin \angle BAC = \dfrac{1}{2} \cdot CA\cdot CB \cdot \sin \angle ACB$$
Using formula $\sin 2\alpha = 2 \sin \alpha \cos \alpha$, we get
$$
S=AB\cdot AC\cdot \sin \an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Answer to simple algebraic formula manipulation I have to show that
$2y+(x + 1) = 3 \cdot 2^{x+1}− (x + 1) − 2$
is equal to
$y=3⋅2^x−x−2$
I can get this far:
$2y+(x+1)=3⋅2^{x+1}−(x+1)−2$
$2y+(x+1)=3⋅2^{x+1}−x−1−2$
$2y+(x+1)=3⋅2^{x+1}−x−3$
$2y+x+1=3⋅2^{x+1}−x−3$
$2y=3⋅2^{x+1}−x−3−1−x$
$2y=3⋅2^{x+1}−2x−4$
Now I sh... | If you divide $3 \cdot 2^{x+1}$ by $2$ you obtain $3 \cdot 2^x$, since $2^{x+1}=2 \cdot 2^{x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the number of values of $x$ in $(\sqrt{2})^x+(\sqrt{3})^x = (\sqrt{13})^{\frac{x}{2}}$. Question:
Find the number of values of $x$ with $$(\sqrt{2})^x+(\sqrt{3})^x = (\sqrt{13})^{\frac{x}{2}}.$$
My attempt:
I tried setting $p = \sqrt{2}$, $q = \sqrt{3}$, and resetting the equation to:
$$p^x + q^x = \sqrt{p^4+q^4... | Divide by $(\sqrt{13})^{\large\frac{x}{2}}$, and write $x = 4y$. You obtain
$$\left(\frac{4}{13}\right)^y + \left(\frac{9}{13}\right)^y = 1.$$
For $0 < a < 1$, the sole solution of $a^y + (1-a)^y = 1$ is $y = 1$, since $y\mapsto a^y$ and $y\mapsto (1-a)^y$ are both strictly decreasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/820477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Integrate : $\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}dx$
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$
My approach :
Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then
$$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\t... | HINT :
Rewrite the integrand
$$
\frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}
$$
as
$$
\frac{x\cos x}{x\sin x +\cos x}+\frac{x\sin x}{x\cos x -\sin x}
$$
then
$$
\frac{\color{red}{\sin x}+x\cos x-\color{red}{\sin x}}{x\sin x +\cos x}+\frac{\color{blue}{\cos x}+x\sin x-\color{blue}{\cos x}}{x\cos x -\sin x}.
$$
Now let... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 3
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minimum and maximum of $f(x,y)=\sin(x)+\sin(y)-\sin(x+y)$ we are asked to find the minimum and maximum of the function$f:A \to A$ $f(x,y)=\sin(x)+\sin(y)-\sin(x+y)$
Where $A$ is the triangle bound by $x=0$,$y=0$ and $y=-x+2\pi$
I'd like someone to review my answer.
What I did:
$A$ is a closed and bounded set, $f(x,y)$ ... | You are correct in saying that if $f$ attains a minimum/maximum at $(x,y)$ in the interior of $A$, then $(x,y)$ must satisfy $\cos x = \cos y = \cos(x+y)$.
However, $(0,2\pi)$ and $(2\pi,0)$ are not the only points in $A$ at which this condition can be met.
Suppose $(x,y) \in A$ satisfies $\cos x = \cos y = \cos(x+y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/827058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Intersection of ellipse and hyperbola at a right angle Need to show that two functions intersect at a right angle.
Show that the ellipse
$$ \frac{x^2}{a^2} +\frac{y^2}{b^2} = 1
$$
and the hyperbola
$$
\frac{x^2}{α^2} −\frac{y^2}{β^2} = 1
$$
will intersect at a right angle if
$$α^2 ≤ a^2 \quad \text{and}\quad a^2 − b^2... | Apparently, if $\alpha^2>a^2$, the two curves never intersect in the first place. Now let's just find the derivatives at the point of intersection and show that their product is $-1$.
Ellipse:
$$y=b\sqrt{1-\left({x\over a}\right)^2}\\
y'=b\cdot{-2x/a^2\over2\sqrt{1-\left({x\over a}\right)^2}}=-{b^2x\over a^2y}\tag{1}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
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Prove by induction that $(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}$ Prove by induction that $$(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}.$$
I got up to:
$n=1$ is true, and assuming $n=k$ prove for $n=k+1$.
Prove...
$$\frac{k(2k+1)(7k+1)+6(2(k+1))^2}{6} = \frac{(k+1)(2k+3)(7k+8)}{6}$$
I... | Check this out
#
$(k+1)^2 +(k+2)^2+ ............ (k+k)^2 = \frac{k(2k+1)(7k+1)}{6}$
#
$p(k+1): (k+1)^2+(k+2)^2...........(2k)^2+(2k+1)^2+.(2k+2)^2-(k+1)^2= \frac{(k+1)(2k+3)(7k+8)}{6}$
#
$p(k+1): \frac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2$=RHS
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/831521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How can I find the roots of a quartic equation, knowing one of its roots? I need to decompose (in $\Bbb{C}[x]$) the function
$$
f(x) = x^4 + 4x^3 - 4x^2 + 24x + 15
$$
in its simplest form, knowing that $1 - 2i$ is one of its roots. Any ideas?
| The answer has already been given in the comments, so I'll make a community answer out of it.
So we have the function $$f(x) = x^4 + 4x^3 - 4x^2 + 24x + 15$$
One root has been found at $x = 1-2i$.
Since the coefficients are real, there must be another root at $x = 1+2i$. We can now reduce this equation to a quadratic o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/835108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate the limit $\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$ I need to evaluate the limit without using l'Hopital's rule.
$$\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$$
| $\frac{\sqrt{1+x\sin(x)}-\sqrt{\cos(2x)}}{\tan^{2}(\frac{x}{2})}=\bigg(\frac{1-\cos(2x)}{\tan^{2}(\frac{x}{2})}+\frac{x\sin(x)}{\tan^{2}(\frac{x}{2})}\bigg)\frac{1}{\sqrt{1+x\sin(x)}+\sqrt{\cos(2x)}}=\bigg(\frac{1-\cos(2x)}{(2x)^{2}}\frac{(\frac{x}{2})^{2}}{\sin^{2}(\frac{x}{2})}(16\cos^{2}(\frac{x}{2}))+\frac{(\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/835333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Use induction to show that $a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0) .$
Let $a_0$ and $a_1$ be distinct real numbers. Define
$a_n=\frac{a_{n-1}+a_{n-2}}{2}$ for each positive integer $n\geq 2$. Prove that $$a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0) $$
This is what I have so far:
Let $P(n)$ be ... | You should use the induction hypothesis (with two base cases) immediately after your second equality when computing $a_{k + 1} - a_{k + 1}$, for we then have
\begin{align*}
\frac{a_{k + 1} + a_k}{2} - \frac{a_k + a_{k - 1}}{2} &= \frac 1 2\left(- \frac 1 2\right)^k (a_1 - a_0) - \frac 1 2\left(-\frac 1 2\right)^{k - 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/835487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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GCD of polynomials over $\mathbb{Z}_3$ $f$ and $g$ are polynomials over field $\space \mathbb{Z}_3$. $f=X^4+X^3+X+2, \space g=X^4+2X^3+2X+2$. And I been asked to find the GCD of them.
What I have done is using Euclidean algorithm. After long division I get,
$$X^4+2X^3+2X+2=(1)(X^4+X^3+X+2)+(X^3+X)$$
So GCD$(f,g)=X^3+X... | This is for the completeness, ever step is done by long division.
$$X^4+2X^3+2X+2=(1)(X^4+X^3+X+2)+(X^3+X) \\
X^4+X^3+X+2 = (X+1)(X^3+X)+(X^2+1)(2) \\
X^3+X = (2X)(2X^2+2)+0$$
Therefore, the GCD of $f$ ad $g$ is $X^2+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/836156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Proving $n^2(n^2+16)$ is divisible by 720 Given that $n+1$ and $n-1$ are prime, we need to show that $n^2(n^2+16)$ is divisible by 720 for $n>6$.
My attempt:
We know that neither $n-1$ nor $n+1$ is divisible by $2$ or by $3$, therefore $n$ must be divisible by both $2$ and $3$ which means it must be divisible by $6$.
... | HINT (since this is worth solving yourself): You need to use that $n-1, n+1$ are prime. If either of these are equal to $5$, you have the excluded values $n=6$ or $n=4$.
Otherwise, note that the product of any five successive integers is divisible by $5$. You are given $n+1$ and $n-1$ - can you see which five successiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/836482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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Type of this Conic section I want to determine, to which type the following Conic sections belong to:
$$
\begin{align}
\textrm{(i)}&\quad-8x^2+12xy-6x+8y^2-18y+8=0\\
\textrm{(ii)}&\quad5x^2-8xy+2x+5y^2+2y+1=0
\end{align}
$$
To (i)
Matrix notation:
$$
\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}-8&6\\6&8\end{pmatrix}\... | Well the second equation
$$ 5 x^2 - 8 x y + 5 y^2 + 2 x + 2 y + 1 = 0 $$
can be written as
$$
5 \Big( \big[ x - x_o \big] + x_o \Big)^2
- 8 \Big( \big[ x - x_o \big] + x_o \Big) \Big( \big[ y - y_o \big]
+ y_o \Big) + 5 \Big( \big[ y - y_o \big] + y_o \Big)^2\\
+ 2 \Big( \big[ x - x_o \big] + x_o \Big)
+ 2 \Big( \big[ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to evaluate the following indefinite integral? $\int\frac{1}{x(x^2-1)}dx.$ I need the step by step solution of this integral
please help me!
I can't solve it!
$$\int\frac{1}{x(x^2-1)}dx.$$
| We use partial fraction decomposition:
$$\int\frac{1}{x(x^2-1)}dx = \int \frac 1{x(x-1)(x+1)}\,dx = \int \left(\frac A{x} + \frac{B}{x - 1} + \frac C{x+1}\right)\,dx$$
Solving for $A, B, C$:
$$A(x-1)(x+1) + Bx(x+1) + Cx(x-1) = 1$$
When $x = 1 \implies 2B = 1 \implies B = \frac 12$
$x = -1 \implies 2C = 1 \iff C = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/839510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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Find the last two digits of $9^{{9}^{9}}$ I have to find the last two decimal digits of the number $9^{{9}^{9}}$.
That's what I did:
$$m=100 , \phi(m)=40, a=9$$
$$(100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}$$
$$9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \c... | Let $k$ be the order of $9$ mod $100$. Then
$$1\equiv 9^k=(10-1)^k \equiv 10\binom{k}{1}\cdot 10(-1)^{k-1}+(-1)^k\pmod{100}$$
It implies that $1=\pm (10k-1)\pmod{100}$. The minimal integer which satisfies this condition is $k=10$. It follows that $$9^{9^9} \equiv 9^9 \equiv 9^{-1}\equiv 89\pmod{100}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Solve for x if $z$ is a complex number such that $z^2+z+1=0$ I was given a task to solve this equation for $x$:
$$\frac{x-1}{x+1}=z\frac{1+i}{1-i}$$
for a complex number $z$ such that $z^2+z+1=0$.
Solving this for $x$ is trivial but simplifying solution using the given condition is what's bothering me. Thanks ;)
| You said that solving for $x$ was trivial, so I imagine you were able to get to:
$$x=\frac{1+zi}{1-zi}$$
This is equal to:
$$x=\frac{\left(1+zi\right)^2}{(1-zi)(1+zi)}=\frac{1+2zi-z^2}{1+z^2}$$
Since $1+z+z^2=0$ we can conlude that $1+z^2=-z$ and also that $1=-z-z^2$. We substitute that into the expression to get:
$$x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/843719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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analytical solution for linear 1st order PDE using laplace and seperation of variables I am looking for the solution of the following pde:
$\frac{\partial y(x,t)}{\partial t} = a* \frac{\partial y(z,t)}{\partial x} + b* y(x,t) + c$
and need help with the boundary and initial conditions:
$y(x=0,t)=0 $ bzw $y(x=0,t)=con... | The solution is
$$ y(x,t) = \left( e^{-\frac{b}{a}x}y_l + e^{-\frac{b}{a}x}\frac{c}{b}\left( 1 - e^{b(t+\frac{x}{a})} \right)\right)\sigma\left(t + \frac{x}{a}\right) - \frac{c}{b}\left(1-e^{bt}\right) - \mathcal{L}^{-1}\{ F(x,s) \}, $$
where
$$F(x,s) = \frac{1}{as}e^{\frac{s-b}{a}x} \int\limits_0^x e^{\frac{b-s}{a}\x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/848252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Sum of infinite series $\sum_{n=0}^{\infty}=\frac{n^2}{4n^2-1}t^n$ I have this problem, finding infinite sum of this series:
$$\sum_{n=0}^{\infty}\frac{n^2}{4n^2-1}t^n$$
It should be done using derivatives and integrals, like for example:
$$\sum_{n=1}^{\infty}\frac{t^n}{n}=\sum_{n=0}^{\infty}\frac{t^{n+1}}{n+1}=\sum_{n... | The series
\begin{align}
S = \sum_{n=0}^{\infty} \frac{ n^{2} \ t^{n} }{ 4n^{2}-1}
\end{align}
can be reduced as follows.
\begin{align}
S &= \frac{1}{4} \sum_{n=0}^{\infty} \frac{(4n^{2}-1) + 1}{4n^{2}-1} \ t^{n} \\
&= \frac{1}{4} \sum_{n=0}^{\infty} t^{n} + \frac{1}{8} \sum_{n=0}^{\infty} \left( \frac{1}{2n-1} - \fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that all real roots of the polynomial $P (x) = x^5 − 10x + 35$ are negative. I got this problem out of Andreescu's Putnam and Beyond. I solved it differently from the given solution and could not understand the solution. Can you explain what is happening in the last step of the solution?
Because P (x) has odd degr... | To see what can be done without calculus, we can use a little bit from the "theory of equations" to show that $ \ x^5 \ - \ 10x \ - \ 35 \ $ has either five, three, or one real roots (complex conjugate zeroes) and that either two or none of those are positive and just one is negative (Descartes' Law of Signs).
If we ju... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Simplifying polynomial fraction Working through an old book I got and am at this problem: Simplify:
$$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}.$$
The answer is supposed to be $\frac{3(x - 1)}{2(x - 1)}$. I thought I had all this polynomial stuff figured out well enough, but I'm having trouble seeing how they got to that answ... | $$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4} = \frac{3(x^2 + x -2)}{2(x^2 + 3x + 2)}= \frac{3\require{cancel}\cancel{(x+2)}(x-1)}{2(x+1)\cancel{(x + 2)}}=\frac{3(x-1)}{2(x+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Newtonian potential at (0, 0, – a) I found this problem in the book Advanced Calculus, written by Friedman.
"Newtonian potential at (0, 0, – a) due to a mass with constant densinty $\sigma$ on the hemisphere S: $x^2 + y^2 + z^2 = a^2$, $z \geq 0$, is
$$\int\int_S \frac{\sigma}{[x^2+y^2+(z+a)^2]^{1/2}}dS$$
Compute U."
I... | The surface integral,
$$\Phi(0,0,-a)=\iint_{\Sigma}\frac{\sigma}{\left[x^2+y^2+(z+a)^2\right]^{1/2}}\mathrm{d}S,$$
is most easily done using spherical coordinates: $\langle x,y,z\rangle=\langle a\cos{\varphi}\sin{\theta}, a\sin{\varphi}\sin{\theta}, a\cos{\theta}\rangle$, where $\varphi\in[0,2\pi]$ and $\theta\in[0,\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Exercise about linear independence A) Let $V$ a vector space over $\mathbb{R}$ and $T: V \rightarrow V$ a linear mapping, such that $T^n=0$ for a natural $n \geq 2$. If $x \in V$, then the set $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly independent if and only if $T^{n-1}x \neq 0$.
B) Let $V$ a vector space over $\... | To prove (A), it is clear that if $T^{n-1} x = 0$, then the set $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly dependent.
Conversely, suppose the set is linearly dependent, and let
$$
c_{0} x + c_{1} Tx + \dots + c_{n-1} T^{n-1} x = 0\tag{eq}
$$
be a relation where not all coefficients are zero.
If $c_{0} = c_{1} = \d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculation of Trigonometric Limit with Summation. If $\displaystyle f(x)=\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$ Then value of $f(x)$ is
$\bf{My\; Try::}$ Let $\displaystyle \left(... | Continuing on from your answer,
$$\eqalign{\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}
&=\frac{1}{2}\sum_{r=0}^{n}\Bigl(\tan\Bigl(\frac{x}{2^r}\Bigr)-\tan\Bigl(\frac{x}{2^{r+1}}\Bigr)\Bigr)\cr}$$
which is a telesc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to solve this linear equation? which has an x on each side I have made this equation.
$5x + 8 = 10x + \dfrac{3}{6}$
And I have achieved this result:
$x = 9$
Is my result correct?
I have already posted two other questions related to this topic, I'm a programmer and am learning Math out of my interest, this is not h... | A good tip would be to put all of the variables on the left hand side of the equation and all of the numbers on the right side equation.
$5x + 8 = 10x + \dfrac{3}{6}$
First we would reduce the fraction
$5x + 8 = 10x + \dfrac{1}{2}$
and then we would subtract $10x $ from both sides, so we will have
$ 5x-10x + 8 = \df... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Eliminate $\theta$ from the equations $\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$
Eliminate $\theta$ from the equations
$$\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$$
Ans: $m^2+m\cos\alpha-2=0$.
I tried using the following two ide... | We have, $$\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$$
Then, $$m=\frac{\sin\theta\cos(\alpha-3\theta)+\cos\theta\sin(\alpha-3\theta)}{\sin\theta\cos^3\theta+\cos\theta\sin^3\theta}=\frac{\sin(\alpha-2\theta)}{\sin\theta\cos\theta}$$
Expanding $\sin(\alpha-2\theta)$ and using ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Is this procedure for $5^{300} \bmod 11$ correct? I'm new to modular exponentiation. Is this procecdure correct?
$$5^{300} \bmod 11$$
$$5^{1} \bmod 11 = 5\\
5^{2} \bmod 11 = 3\\
5^{4} \bmod 11 = 3^2 \bmod 11 = 9\\
5^{8} \bmod 11 = 9^2\bmod 11 = 4\\
5^{16} \bmod 11 = 4^2 \bmod 11 = 5\\
5^{32} \bmod 11 = 5^2 \bmod 11 = ... | You could also notice $11$ is prime, so by Fermat's Little Theorem $ 5^{10}\equiv 1 \bmod 11$
from here we get $5^{300}=(5^{10})^{30}\equiv1^{30}\equiv 1 \bmod 11$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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I need help solving these limits I've been struggling to solve the following limits for about an hour. I've tried using conjugates as well as common factors, but it gets me nowhere. Wolfram Alpha does not provide steps for these limits. I could really use some help.
(x greater than 1) (SOLVED)
(SOLVED)
Thanks for ta... | For question 1: You can factorize $x^2+2x-3 = (x+3)(x-1)$ to get $\frac{1}{\sqrt{x-1}}(\frac{1}{2}-\frac{1}{\sqrt{x+3}})$. After that, I'd suggest calculating Taylor expansions at x=1 for $\frac{1}{\sqrt{x-1}}$ and $\frac{1}{2}-\frac{1}{\sqrt{x+3}}$ (you can do this with Wolfram Alpha). Btw, the limit is zero, I think ... | {
"language": "en",
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"source": "stackexchange",
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If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$
Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?
M... | I'm sure this answer is not rigorous, but perhaps someone can make it so.
By manipulating factorials we get
$$\frac{1}{K}=\frac{1}{2^{100}}\binom{100}{50}=P(X\,{=}\,50)\ ,$$
where $X$ is a binomial random variable with $n=100$ and $p=\frac{1}{2}$. Approximating $X$ by a normal random variable $Y$ in the usual way, we ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Commutant Algebra of Matrix Representation I am currently working on Bruce Sagan's The Symmetric Group.
In the following example they show that for a representation that contains 2 different subrepresentations the commutative algebra Com$X$ has the form
\begin{equation} \text{Com}X = \left\{ c_1I_{d_1} \oplus c_2I_{d_2... | Write $V=V_1\oplus V_2$, a direct sum of two irreducible representations of $G$ with dimensions $d_1,d_2$, respectively. We are showing the intertwiners in ${\rm End}_G(V)$ look like $c_1I_{V_1}\oplus c_2 I_{V_2}$.
(1) Every $(n+m)\times(n+m)$ matrix can be written as a block matrix $(\begin{smallmatrix} A & B \\ C & D... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find orthogonal projection of vector on a subspace? Well, I have this subspace: $V = \operatorname{span}\left\{ \begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix},\begin{pmatrix}1 \\3 \\4\end{pmatrix}\right\}$
and the vector $v = \begin{pmatrix}9 \\0 \\0\end{pmatrix}$
How can I find the orthog... | That would be the correct method...if $v_1$ and $v_2$ were orthogonal and unit length. Unfortunately, they're not.
Three alternatives:
*
*Compute $w = v_1 \times v_2$, and the projection of $v$ onto $w$ -- call it $q$. Then compute $v - q$, which will be the desired projection.
*Orthgonalize $v_1$ and $v_2$ usin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Calculate $\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$ I have homework questions to calculate infinity sum, and when I write it into wolfram, it knows to calculate partial sum...
So... How can I calculate this:
$$\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$$
| Lets solve the problem generally;
$$\begin{array}{l}\sum\limits_{k = i}^\infty {\frac{1}{{\left( {k + a} \right)\left( {k + b} \right)}}} = \\\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = i}^n {\frac{1}{{\left( {k + a} \right)\left( {k + b} \right)}}} = \\\mathop {\lim }\limits_{n \to \infty } \sum\limits_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Tridiagonal Symmetric Matrix Could anyone help me to find the determinant of a $N\times N$ tri-diagonal symmetric matrix, named "$A[i,j]$" with $i,j \le N$, that has all the elements in the super-diagonal and sub-diagonal equal, and in the main diagonal all the elements are also equal, except for $A[1,N]$ and $A[N,N]$,... | For a general $N \times N$ tridiagonal matrix $A$, define:
$$d_i = A_{i,i}, \, i=1 \dots N \\
u_i = A_{i,i+1}, \, i=1 \dots N-1 \\
l_i = A_{i+1,i}, \, i=1 \dots N-1 \\
f_0 = 1 \\
f_1 = d_1 \\
f_n = d_n f_{n-1} - u_{n-1} l_{n-1} f_{n-2}, \, n = 2 \dots N$$
Then $\text{det}(A)=f_N$.
In your situation, $u_i \equiv s$, $l_... | {
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"timestamp": "2023-03-29T00:00:00",
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Fastest way to integrate $\int_0^1 x^{2}\sqrt{x+x^2} \hspace{2mm}dx $ This integral looks simple, but it appears that its not so.
All Ideas are welcome, no Idea is bad, it may not work in this problem, but may be useful in some other case some other day ! :)
| Note that
$$x^2\sqrt{x+x^2}=\frac{x^3+x^4}{\sqrt{x(1+x)}}$$
So, let us look for a polynomial $P(x)=a x^3+bx^2+cx+d$ such that the derivative
$(P(x)\sqrt{x(1+x)})^\prime$ is as close as we can to this function. An easy calculation shows that
$$
\left(P(x)\sqrt{x(1+x)}\right)^\prime=\frac{4 a x^4+(\frac{7 a }{2}+3 b)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/860244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The number of distinct multiples of composites greater than $n$ that can be factored into two naturals less than or equal to $n$ Given a list of composites between $n$ and $\lfloor \frac{n^2}{2} \rfloor$:
What would be the most efficient way to count, for each composite, the number of its distinct multiples that can b... | Let the composite be $m$. Factor $m$, then find all factorizations with no factor greater than $n$. For each one, see how much room you have. I'll take the larger example $n=25, m=144=2^43^2$ From the prime factorization we can find $144=6\cdot 24=8 \cdot 18 = 9 \cdot 16 = 12 \cdot 12$ are possible, while $1 \cdot ... | {
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How find this P(x) if $ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $ Let $m \neq 0 $ be an integer. Find all polynomials $P(x) $ with real coefficients such that
$$ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $$
This problem is IMO Shortlist 2013
let $$P(x)=\sum_{i=0}... | Note that, $P(x)=cx$ is a trivial solution, we will prove that it is the only one.
Suppose that degree of this polynomial is $>1$.
From your work, comparing coefficients of $x^{n+1}$ gives , $n=2m$, and so $m>0$.
Now,after doing some manipulations(multiply by $x$ and then regroup) , we get,
$$ (x^3-mx^2+1)F(x)= (x^... | {
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Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c$ are nonnegative Hope someone can help on this inequality using nonanalytical method (i.e. simple elementary method leveraging basic inequalities are prefered).
Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c... | Proof.
(i) if abc=0, W.L.O.G assume c=0, $\Longleftrightarrow a+b \geq 2\sqrt{ab}$, it is clearly true.
(ii) in the following, assume $abc > 0$.
Case 1: $a^2+b^2+c^2 \leq 2(ab+bc+ca)$
according to Cauchy-Schwarz inequality, we have:
$$LHS \leq \sqrt{3(a^2+b^2+c^2+ab+bc+ca)}$$
$$\Longleftrightarrow \sqrt{3(a^2+b^2+c^2+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/864002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the derivative of $y=\cos(x) - 2\sin(x),$ when the gradient is $1$ I need to find the smallest positive value of $x$ for which the gradient of the curve has value 1.
For this equation:
$$
y =\cos(x)-2\sin(x)
$$
The answer is 2.5c grad.
The following is my working out but I’ve stopped at the factor formula, how ca... | $$\sin x + 2\cos x + 1 = 0\tag{1}$$ Now use the Weierstrass substitution $y = \tan\left(\frac x2\right).$ From this, it follows that $$\sin x = \frac {2y}{y^2 + 1},\;\cos x = \frac{1-y^2}{y^2 + 1}\tag{2}$$
Then, substituting the values from $(2)$ into $(1)$ gives us $$ \frac {2y}{y^2 + 1} + \frac{2(1-y^2)}{y^2 + 1} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$ \int \frac{1}{(x-a)(x+b)} dx $ Could you please explain how to integrate this integral:
$$ \int \frac{1}{(x-a)(x+b)} dx $$
| $$\int\frac1{(x-a)(x+b)}dx$$
Use partial fractions
$$\int\frac1{x(a+b)-a(a+b)}+\frac1{-x(a+b)-b(a+b)}dx$$
Integrate the sum by term
$$\int\frac1{x(a+b)-a(a+b)}dx+\int\frac1{-x(a+b)-b(a+b)}dx$$
Substitute $u=-x(a+b)-b(a+b)$ and $du=(-a-b)dx$
$$\int\frac1{x(a+b)-a(a+b)}dx + \frac1{-a-b}\int\frac1u du$$
Since $\int\frac1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of series with triangular numbers Can you please tell me the sum of the seires
$ \frac {1}{10} + \frac {3}{100} + \frac {6}{1000} + \frac {10}{10000} + \frac {15}{100000} + \cdots $
where the numerator is the series of triangular numbers?
Is there a simple way to find the sum?
Thank you.
| I thought I might add another derivation (devised by me). This one is long and involves dissecting the sequence into its simplest terms.
$1/10 + 3/100 + 6/1000 + \ldots$
$= 1/10 + (1+2)/100 + (1+2+3)/1000 + \ldots$ (from the definition of
triangular numbers.)
$= 1/10 + 1/100 + 2/100 + 1/1000 + 2/1000 + 3/1000 + \ldo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Homework - Resolve the recurrence relation What's the closed formula of this recurrence relation?
$$a_n = a_{n-1}+2a_{n-2}+2^n \text{ with } a_0=1, a_1=2 $$
| The best course of action for simple inhomogeneous recurrences is to make use of a smart "change of variables" (read: substitute with another recurrence relation) to turn it into a homogeneous recurrence. A nice observation here is to notice that $2^n$ is itself a recurrence relation, namely:
$$y_n=2y_{n-1},\,\,(y_0=1... | {
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Calculating the area For the two graphs $ \frac{x^3+2x^2-8x+6}{x+4} $ and $ \frac{x^3+x^2-10x+9}{x+4} $, calculate the area which is confined by them;
Attempt to solve: Limits of the integral are $1$ and $-3$, so I took the definite integral of the diffrence between $ \frac{x^3+2x^2-8x+6}{x+4} $ and $ \frac{x^3+x^2-10x... | Note that
$$
\left|\frac{x^2+2x-3}{x+4}\right| = -x + 2 - \frac{5}{x+4},\ \textrm{for}\ x \in [-1,3].
$$
So we obtain
\begin{eqnarray}
\int_{-3}^1 dx \left| \frac{x^2+2x-3}{x+4} \right|
&=& \int_{-3}^1 dx \Big( -x + 2 - \frac{5}{x+4} \Big)\\
&=& \left[ - \frac{1}{2} x^2 + 2 x - 5 \ln(x+4) \right]_{-3}^1\\
&=& \Big( - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Nth value of Function Given x and y we define a function as follow :
f(1)=x
f(2)=y
f(i)=f(i-1) + f(i+1) for i>2
Now given x and y, how to calculate f(n)
Example : If x=2 and y=3 and n=3 then answer is 1
as f(2) = f(1) + f(3), 3 = 2 + f(3), f(3) = 1.
Constraints are : x,y,n all can go upto 10^9.
| The general way to solve $$F(i+1)= F(i) - F(i - 1)$$
Is to first to see that if $F(i + 1)$ can be written as the sum of 2 geometric series, so that $$F(n) = ar^n + bs^n$$
So $$ar^{n+1} + bs^{n+1} = ar^{n} + bs^n - ar^{n-1} - bs^{n-1}$$
$$ar^{n-1}(1 - r + r^2) = -bs^{n-1}(1 - s + s^2)$$
Since $r\ne s$, it follows that ... | {
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Show that this expression is a perfect square? Show that this expression is a perfect square?
$(b^2 + 3a^2 )^2 - 4 ab*(2b^2 - ab - 6a^2)$
| Once you factored it out and got $9a^4+24a^3b+10a^2b^2−8ab^3+b^4$, you'd suspect that this is the square of a sum. What are the parts of the sum?
If we let b = 0, the sum is $9a^4$ so there should be a term $±3a^2$. If we let a = 0, the sum is $b^4$, so there should be a term $±b^2$.
We know that $(x+y)^2=x^2 + 2xy +... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_0^1 \sqrt{2x-1} - \sqrt{x}$ $dx$ I'm trying to calculate the area between the curves $y = \sqrt{x}$ and $y= \sqrt{2x-1}$
Here's the graph:
I've already tried calculating the area with respect to $y$, i.e.
$\int_0^1 (\frac{y^2+1}{2} - y^2)$ $ dx$
[since $y^2=x$ for the first curve and $\frac{y^2+1}{2}... | As noted in the comments the correct way is
$$
A
= \int_0^{1/2} \color{blue}{\sqrt{x}} \, \mathrm{d}x +
\int_{1/2}^1 \color{red}{\sqrt{x} - \sqrt{2x-1}} \,\mathrm{d}x
= \int_0^1 \frac{y^2+1}{2} - y^2 \mathrm{d}y
= 1/3
$$
Since $\sqrt{2x-1}$ is not defined for $x<1/2$. Note that you coul... | {
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Area of a triangle adjacent to two similar triangles This GMAT problem states:
In the figure above $AD = 4$, $AB = 3$ and $CD = 9$. What is the area of triangle $AEC$
The solution states:
to find the base we need to see that triangles $AEB$ and $CDE$ are similar. The ratio $AB: CD$, is therefore equal to the ratio $A... | Apply Componendo Dividendo on fraction from similar triangles
$$ \frac{AE}{ED}= \frac{AB}{CD}=\dfrac{3}{9}= \dfrac{1}{3}$$
$$\dfrac{AE}{AE+ED}= \dfrac{1}{1+3}= \dfrac{1}{4} $$
$$ =\dfrac{AE}{AD}= \dfrac{1}{4} $$
Since $AD=4$
$$AE=1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove that $ \sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\sqrt[3]{\sqrt[3]2-1} $ How check that $ \sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\sqrt[3]{\sqrt[3]2-1} $?
| use three equations:
$$a^3+b^3=(a+b)(a^2-ab+b^2)\quad (1)$$
$$a^3-b^3=(a-b)(a^2+ab+b^2)\quad (2)$$
$$(a+b)^3=a^3+3a^2b+3ab^2+b^3\quad (3)$$
for your problem:
$$left\\=(\sqrt[3]{\frac{1}{3}})^2-\sqrt[3]{\frac{1}{3}}\sqrt[3]{\frac{2}{3}}+(\sqrt[3]{\frac{2}{3}})^2\\=\frac{\frac{1}{3}+\frac{2}{3}}{\sqrt[3]{\frac{1}{3}}+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/873582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Longest path in a grid I recently saw a computer programming question that asked for the longest path that one can build in a $3\times3$ unit grid connecting the vertexes, with the following rules(the same rules of a pattern password):
*
*Each vertex may be used at most once.
*A vertex cannot be skipped in a line i... | I suggest to start with longest distances and then add shorter.
You have the following vertex sequence
$$
\begin{array}{cccc}
1 & \cdots & 6 & \cdots & 9 & \cdots & 3 \\
\vdots & & \vdots & & \vdots & & \vdots \\
10 & \cdots & 13 & \cdots & 16 & \cdots & 8 \\
\vdots & & \vdots & & \vdots & & \vdots \\
4 & \cdots & 15 &... | {
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"timestamp": "2023-03-29T00:00:00",
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How find $x$ in a right triangle $ABC$ (${\measuredangle}A=90^\circ$) where ${\measuredangle}DBC={\measuredangle}DCA=x$,${\measuredangle}BAD=5x$? In a right triangle $ABC$ (${\measuredangle}A=90^\circ$) taken in point $D$ such that $BD=AC$, ${\measuredangle}DBC={\measuredangle}DCA=x$,${\measuredangle}BAD=5x$. How find ... | W.l.o.g. you may assume the following coordinates:
$$A=(0,0)\quad B=(a,0)\quad C=(0,1)\quad D=(b,c)$$
If you define $d:=\cos(x)$ then you can compute the cosines of multiple angles using a Chebyshev polynomial as
$$\cos(5x)=T_5(d)=16d^5 - 20d^3 + 5d\;.$$
The dot product of two vectors is proportional to the cosine, so ... | {
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"source": "stackexchange",
"question_score": "3",
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What is the value of this double integral? Let $C$ be the subset of the plane given by
$$ C \colon= \{ \ (x,y) \in \mathbb{R}^2 \ | \ 0 \leq x^2 + y^2 \leq 1 \}.$$
Then what is the value of the double integral
$$ \int_{C} \int (x^2 + y^2) \ dx \ dy?$$
My work:
In $C$, we have $-1 \leq x \leq 1$ and $-\sqrt{1-x^2}... | Use polar coordinates,
We know that
$$r^2 = x^2 + y^2$$
So our double integral becomes
$$\int_{0}^{2\pi} \int_0^{1}r^2\cdot rdrd\theta$$
Now solve.
EDIT
I see that your computation is correct, I am simply offering another alternative and more easier way to solve this double integral.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{1}^{e}\frac{u}{u^3+2u^2-1}du.$ I'm trying to solve
$$\int_{1}^{e}\frac{u}{u^3+2u^2-1}du.$$
My first approach was to factorise and then do a partial integration. However the factorisation $(u+1)\left(u+\frac{1}{2}-\frac{\sqrt{5}}{2} \right)\left(u+\frac{1}{2}+\frac{\sqrt{5}}{2} \right)$ leads me to heavy... | As $$u^3+2u^2-1=(u+1)(u^2+u-1)$$
Write,
$$\frac u{u^3+2u^2-1}=\frac A{u+1}+B\frac{\dfrac{d(u^2+u-1)}{du}}{u^2+u-1}+\frac C{u^2+u-1}$$
$$u=A(u^2+u-1)+(2Bu+B)(u+1)+C(u+1)$$
Set $\displaystyle u+1=0$ to find $A$
Now comparing the coefficients of $u^2$, $0=A+2B\iff B=?$
Again comparing the constants $0=-A+B+C\implies C=?$
... | {
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"timestamp": "2023-03-29T00:00:00",
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Euler's proof of divergence of sum of reciprocals of primes On Wikipedia at link currently is:
\begin{align}
\ln \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \ln\left( \prod_p \frac{1}{1-p^{-1}}\right)
= -\sum_p \ln \left( 1-\frac{1}{p}\right) \\
& {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p... | $
\ln \left( \sum_{n=1}^\infty \frac{1}{n}\right)$ has no meaning.
It is like you say $$\ln (\infty)=...$$
But $\ln x$ is defined on real numbers and not on "anything we understand".
But everyone understands that Euler "understood" that the method was correct.
This is why the proof is attributed to him.
| {
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Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$
Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rare... | $\sum \limits_{cyc}\dfrac {a}{(b+c)^2}=\sum \limits_{cyc}\dfrac {2a^2}{2a(b+c)^2} \ge \dfrac{2(a+b+c)^2}{2a(b+c)^2+2b(a+c)^2+2c(a+b)^2}\ge \dfrac{2(a+b+c)^2}{3\times \left(\dfrac{2a+2b+2c}{3}\right)^3}=\dfrac {9}{4(a+b+c)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$? I was going through answers on this question and came across this answer and I was wondering how the user arrived at the first line where they state:
$$f(x) \equiv x^3 - \sin^3 x = x^3 + {1 \over 4} \,\sin {3x} - {3 \over 4}\,\sin x$$
How ... | Hint:
$$\sin x =\frac{opposite}{hypotenuse}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A certain “harmonic” sum Is there a simple, elementary proof of the fact that:
$$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$
I have thought of a very simple notation for "harmonic" sums like these: just write down the numerators. So, for... | I think we can "squeeze" something out of this:
$$0=\sum_{n=0}^\infty\left(\frac{1}{6n+6}+\frac{-1}{6n+6}+\frac{-2}{6n+6}+\frac{-1}{6n+6}+\frac{1}{6n+6}+\frac{2}{6n+6}\right)\le\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)\le\sum_{n=0}^\infty\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Integration of $x\cos(x)/(5+2\cos^2 x)$ on the interval from $0$ to $2\pi$
Compute the integral
$$\int_{0}^{2\pi}\frac{x\cos(x)}{5+2\cos^2(x)}dx$$
My Try: I substitute $$\cos(x)=u$$
but it did not help. Please help me to solve this.Thanks
| This is not an answer to the post but a reply to David's comment
The antiderivative does not express in terms of elementary functions. For your curiosity, I write it down, but, as said, it looks like a nightmare.
$$4 \sqrt{14}\int\frac{x\cos(x)}{5+2\cos^2(x)}dx=-2 i \text{Li}_2\left(-\frac{i \left(-7+\sqrt{35}\right) e... | {
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Limit of a definite integral We need to calculate $$\lim_{x \to 0}\int_{\sin x}^{x}\frac{dt}{t^3(1+\sqrt{t})}$$
Integral itself doesn't seem to be the problem here. When making a substitution $\sqrt{t}=u$, we get $$\lim_{x \to 0}2\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5(1+u)}=2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\s... | This is not much different from Sami Ben Romdhane's answer.
Since the function $g(t)=\frac{1}{t^3(1+\sqrt{t})}$ is decreasing over $\mathbb{R}^+$, by the mean value theorem:
$$(x-\sin x) \frac{1}{\sin^3 x\,(1+\sqrt{\sin x})}\leq\int_{\sin x}^{x}\frac{dt}{t^3(1+\sqrt{t})}\leq (x-\sin x)\frac{1}{x^3(1+\sqrt{x})},$$
but w... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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prove by induction $7 \mid 3^{3^n}+8$ Okay so ive been trying to prove this for about 5 hours...
really need salvation from the geniouses around here.
prove by induction
$7\mid 3^{3^n}+8$
i really need some directions on what to do here...
| First note that this is equivalent to $7$ dividing $3^{3^n}+1$. Clearly if $n=1$, then $3^{3^1}+1 = 28$ and $7$ divides this.
Let's suppose it's true for $n$. Then $3^{3^{n+1}} = 3^{3\cdot3^n} = (3^{3^n})^3$. We want then to show that $7$ divides $(3^{3^n})^3+1$. A nice factorization is that $a^3+b^3 = (a+b)(a^2-ab+b^2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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Maximum and minimum of $z=\frac{1+x-y}{\sqrt{1+x^2+y^2}}$ Find the maximum and minimum of the function
$$z=\frac{1+x-y}{\sqrt{1+x^2+y^2}}$$
I have calculated
$\frac{\partial z}{\partial x}=\frac{1+y^2+xy-x}{(1+x^2+y^2)^{\frac{3}{2}}}$
$\frac{\partial z}{\partial y}=\frac{-1-x^2-xy-y}{(1+x^2+y^2)^{\frac{3}{2}}}$
I have... | $z^2 = \dfrac{(1+x+(-y))^2}{1+x^2+(-y)^2} \leq 3$ because $(1-x)^2 + (x+y)^2 +(y+1)^2 \geq 0$.
The second inequality becomes $ = $ when $x = 1, y = -1$. From this we have: $z_\text{min} = -\sqrt{3}$, and $z_\text{max} = \sqrt{3}$.
Note: The CS inequality is: $(a_1b_1 + a_2b_2 +a_3b_3)^2 \leq (a_1^2+a_2^2+a_3^2)(b_1^2+b... | {
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Given $\sum_{n=0}^\infty \frac{1}{2^n}$ and $\sum_{n=0}^\infty \frac{1}{4^n}$, what is the Cauchy product? Given $\sum_{n=0}^\infty \frac{1}{2^n}$ and $\sum_{n=0}^\infty \frac{1}{4^n}$, what is the Cauchy product?
The definition of the Cauchy product is for two given series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\inft... | I think you mean
$c_n=\sum\limits_{j=0}^n\frac{1}{2^j}\frac{1}{4^{n-j}}=\sum\limits_{j=0}^n\frac{1}{2^j}\frac{1}{2^{2n-2j}}=\sum\limits_{j=0}^n\frac{1}{2^{2n-j}}=\sum\limits_{j=0}^n2^{j-2n}=2^{-2n}\sum\limits_{j=0}^n2^j$
The sum on the right hand side is a geometric series, which has the formula
$\sum\limits_{j=0}^nar^... | {
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Prove $OD^2+OG^2=3AB^2$
$AOB$ is cicular sector of $90^{\circ}$.
$C$ is a point on $\stackrel \frown {AB}$.
$ACDE$ and $CBFG$ are squares.
Prove $OD^2+OG^2=3AB^2$
My attempt :
$OA=OC=OB=r$
$O$ is central angle so :
$$\angle O_{1}+\angle O_{2}=\stackrel \frown {BC}=m \hspace{20pt}\angle O_{3}+\angle O_{4}=\stackrel \fr... | Hint: Note that $CG = CB$ and $CD = AC$ so:
$CG^2 + CD^2 = CB^2 + AC^2 = AB^2 +2CB·AC·\cos 135 = r^2 + 2CB·AC·\cos 135$.
All you have to do now is to prove that:
$r·CB\cos x + r·AC\cos y = AC·CB·\cos 135$.
Note also that $x = 90 + \angle OCA$ so, applying the formula of cosine of sum of angles: $\cos x = -\sin \angle O... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find a reduced echelon basis from a reduced echelon matrix. The reduced row matrix was this ---> $\begin{pmatrix}1&2&0&1&0\\0&0&1&3&0\\0&0&0&0&1\\0&0&0&0&0&\end{pmatrix} = 0$
So i computed the basis to be such that
$\begin{pmatrix}x\\y\\z\\t\\u\end{pmatrix} = y\begin{pmatrix}-2\\1\\0\\0\\0\end{pmatrix} + t\begin{pmatr... | I suppose they simply made a rref from this:
$$
\begin{pmatrix}
-1 & 0 &-3 & 1 & 0 \\
-2 & 1 & 0 & 0 & 0 \\
\end{pmatrix}\sim
\begin{pmatrix}
1 & 0 & 3 &-1 & 0 \\
-2 & 1 & 0 & 0 & 0 \\
\end{pmatrix}\sim
\begin{pmatrix}
1 & 0 & 3 &-1 & 0 \\
0 & 1 & 6 &-2 & 0 \\
\end{pmatrix}
$$
i.e.
$$
\begin{pmatrix}
f_2 \\
... | {
"language": "en",
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"question_score": "2",
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Can we expect to find some constant $C$; so that, $\sum_{n\in \mathbb Z} \frac{1}{1+(n-y)^{2}} Fix $y\in \mathbb R;$ and consider the series:
$$\sum_{n\in \mathbb Z}\frac{1}{1+(n-y)^{2}}.$$
My Question is: Can we expect to find some constant $C$; so that,
$$\sum_{n\in \mathbb Z} \frac{1}{1+(n-y)^{2}} <C$$ for all $... | \begin{align}
\sum_{n=-\infty}^{[y]} \frac{1}{1+(n-y)^2} = \sum_{n=-\infty}^{[y]-1} \frac{1}{1+(n-y)^2} + \frac{1}{1+([y]-y)^2} \leq \sum_{n=-\infty}^{1} \frac{1}{1+n^2} + 1
\end{align}
\begin{align}
\sum_{[y]+1}^{n=+\infty} \frac{1}{1+(n-y)^2} = \sum_{[y]+2}^{n=+\infty} \frac{1}{1+(n-y)^2} + \frac{1}{1+([y]+1 -y)^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888809",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Elementary ways to calculate the arc length of the Cantor function (and singular function in general) Cantor's function: http://en.wikipedia.org/wiki/Cantor_function
There is an elementary way to prove that the arc length of the Cantor function is 2?
In this article (http://www.math.helsinki.fi/analysis/seminar/esitelm... | Here is an elementary way to derive the arc length of the Cantor Function:
We know that the following sequence of functions converges uniformly to the Cantor Function. Letting $c_0(x) = x$ and $n\ge 0$, we define
$$c_{n+1}(x)= \begin{cases}
\frac{1}{2}c_n(3x) & 0 \le x \le \frac{1}{3},\\
\frac{1}{2} & \fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Throwing dice twice, with unlike probability of occourence? A loaded dice has the property that when the dice is thrown the probability of showing a given number is proportional to the number. For example $2$ is twice as likely to show up compared to $1$ and $3$ is thrice as likely to show up compared to $1$, And so on... | Since $1+2+\cdots+6=21$, the probabilities of $1,2,3,4,5,6$ in $1$ toss are respectively $\frac{1}{21}$, $\frac{2}{21}$, and so on up to $\frac{6}{21}$.
The probability of a sum of $2$ in $2$ throws is $\frac{1}{21}\cdot\frac{1}{21}$.
The probability of a sum of $3$ is $2\cdot \frac{1}{21}\cdot\frac{2}{21}$.
The pro... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Prove that $\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$ Question:
$$\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$$
Prove that L.H.S.=R.H.S.
My Efforts:
L.H.S.$$=\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}\times\f... | You are almost there.
Hint:
$$\sin^2\theta+\cos^2\theta=1\implies \left(\frac{1}{\sin^2\theta}\right)(\sin^2\theta+\cos^2\theta)=\frac{1}{\sin^2\theta}$$
Also note that in the calculation you have reproduced, there is an error in the final denominator. It should be $(\csc\theta +1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/890870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Is this function decreased with $x$? Given three positive integers $a,b,c\ge 1$, I am wondering if the following $f(x)$ is decreased with $x$ ?
$$f(x)=\frac{c+2x}{(a+x)(b+x)}, \quad x \in Z^+ \cup \{0\}$$
where $1\le c \le ab$.
| since my calculations are error-prone, i will merely show some working which someone might check. if this result is correct then at present i cannot see see how it can be non-trivially related to the given condition $1 \le c \le ab$. might one expect a condition involving $c^2$ rather than $c$? apologies in advance
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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