Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Does midpoint-convexity at a point imply midpoint-convexity at other points? This question is a follow-up of this one.
Let $f:\mathbb R \to \mathbb [0,\infty)$ be a $C^{\infty}$ function satisfying $f(0)=0$.
Suppose that $f$ is strictly decreasing on $(-\infty,0]$ and strictly increasing on $[0,\infty)$, and that $f$ i... | Problem: Let $r < s < 0$ be given. Find a $C^\infty$ function $f : \mathbb{R} \to [0, \infty)$ with $f(0)=0$ such that:
i) $f$ is strictly decreasing on $(-\infty, 0)$ and strictly increasing on $(0, \infty)$;
ii) $f(x) + f(2r - x) \ge 2f(r)$ for all $x\in \mathbb{R}$;
iii) $f(y) + f(2s - y) < 2f(s)$ for some $y \in \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $a,b,c$ are sides of a triangle, then find range of $\frac{ab+bc+ac}{a^2+b^2+c^2}$ $$\frac{ab+bc+ac}{a^2+b^2+c^2}$$
$$=\frac{\frac 12 ((a+b+c)^2-(a^2+b^2+c^2))}{a^2+b^2+c^2}$$
$$=\frac 12 \left(\frac{(a+b+c)^2}{a^2+b^2+c^2}-1\right)$$
For max value, $a=b=c$
Max =$1$
How do I find the minimum value
| Since we have triangle sides, we better use $$a=u+v,\quad b=v+w,\quad c=w+u$$ with $u,v,w>0.$ Arithmetically, $u=(a+c-b)/2, v=(a+b-c)/2, w=(b+c-a)/2,$ so taking $u,v,w>0,$ the triangle inequality is satisfied, automagically. Geometrically speaking, $u,v,w$ are the segments the sides are divided into by the touch-down p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Can someone explain the limit $\lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)$? $$\begin{aligned}
&\text { Find the following limit: } \lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\rig... | You cannot apply Cauchy's first theorem because
$$\sum_{k=1}^n\frac{1}{\sqrt{n^2+k}}\ne \frac{\sum_{k=1}^nu_k}{n}$$
since
$$\sum_{k=1}^nu_k=\sum_{k=1}^n\frac{k}{\sqrt{k^2+k}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute Sum by Rows and Columns (Double series) I am trying to solve this problem.
If
$$a_{m,n} = \frac{m-n}{2^{m+n}}\frac{(m+n-1)!}{m!n!}, (m, n > 0)$$
$$a_{m,0}=2^{-m}, a_{0, n} = -2^{-n}, a_{0, 0} =0,$$
Show that $\sum_{m=0}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right) = -1, \sum_{n=0}^{\infty}\left(\sum_{m=0}^{\... | Let $b_{m,n}=\frac{(m+n-1)}{2^{m+n}m!n!}$ so $a_{m,n}=(m-n)b_{m,n}$. Yhe sums $(n=1,\infty),(m=1,\infty)$ and switched, over $mb_{m,n}$ and $nb_{m,n}$ are divergent. However the sums over $(n-m)b_{n,m}$ are both zero.
The final pieces are $\sum_{m=1}^\infty \frac{1}{2^m}=1$ and $\sum_{n=1}^\infty -\frac{1}{2^n}=-1$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ .
And then there are some question related to this.
In ... | Consider the limit
$$\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}$$
$$\lim_{x \to\infty}\frac{f(x)}{x^4+\frac{1}{x^4}}$$
Since limit is finite and is equal to 3,we observe that $f(\infty) \rightarrow \infty $
Also $f^2(\infty)\rightarrow \infty$ , $f^3(\infty)\rightarrow \infty$
But after diffrentiating four times (L` Hosp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Parametric solution of a Diophantine equation of three variables I came across this Diophantine equation $$4x^2+y^4=z^2$$
Primitive solutions of this equation can be found by
\begin{align}
\begin{split}
x&=2ab(a^2+b^2)\\
y&=a^2-b^2\\
z&=a^4+6a^2b^2+b^4\\
\end{split}
\end{align}
where $a$, $b$ are relatively prime and $... | As JCAA's question comment suggests, rewriting the equation as $(2x)^2 + (y^2)^2 = z^2$ shows $2x$, $y^2$ and $z$ form a Pythagorean triple. Primitive solutions are obtained from
$$2x = 2mn \implies x = mn \tag{1}\label{eq1A}$$
$$y^2 = m^2 - n^2 \implies y^2 + n^2 = m^2 \tag{2}\label{eq2A}$$
$$z = m^2 + n^2 \tag{3}\lab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $\cos(a) +\cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s = \tan(a/2)$ and $t = \tan(b/2)$ Given $\cos(a) + \cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s = \tan(a/2)$ and $t = \tan(b/2)$.
I have tried using the identity $\cos(a) = \frac{1-t^2}{1+t^2}$. but manipulating this s... | $\frac{1-t^2}{1+t^2} +\frac{1-s^2}{1+s^2}=1$
$1+s^2-t^2-s^2t^2+1+t^2-s^2-s^2t^2=1+s^2+t^2+s^2t^2$
$\Rightarrow 1-s^2-t^2-3s^2t^2=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$ Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$
My attempt :
\begin{align*}
f(x)&=\dfrac{5}{9\cos^2x-6\sin x\cos x+\sin^2x-6\cos^2x}\\
&= \dfrac{5}{(3\cos x+\sin x)^2-6\cos^2x}
\end{align*}
The problem is if I'm going to use
$$-1\leqslant\sin x... | The denominator can be written
$$\sin^2x-6\sin x\cos x+3\cos^2x
=\frac{1-\cos 2x}2-3\sin 2x+3\frac{\cos 2x+1}2
\\=2+\cos2x-3\sin2x,$$
which varies continuously in $[2-\sqrt{10},2+\sqrt{10}]$.
Hence as the interval straddles $0$, the range of the function is
$$\left(-\infty,\frac5{2-\sqrt{10}}\right]\cup\left[\frac5{2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to use Picard-Lindelöf Fixpoint Iteration on this differential equation system I have this system of differential equations
$$ x'=yz $$
$$y'=-xz $$
$$z'=2 $$
with $ x(0)=1, y(0)=1, z(0)=0 $
I do know how to do the iteration with one equation.
My question is, how do I do it with the system of the 3 equations?
Any he... | We have the system
$$\begin{align} x'&=yz, \\ y'&=-xz \\ z'&=2\end{align} $$
The IC is given as $x(0)=1, y(0)=1, z(0)=0$.
We can solve this by finding $z$ and then substituting into $y'$, solving for $y$ and then for $x$ as
$$\begin{align} x(t) &= \sin \left(t^2\right)+\cos \left(t^2\right)\\ y(t) &= \cos \left(t^2\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find x intercepts of a higher degree polynomial $2x^4+6x^2-8$ I am to factor and then find the x intercepts (roots?) of $2x^4+6x^2-8$
The solutions are provided as 1 and -1 and I am struggling to get to this.
My working:
$2x^4+6x^2-8$ =
$2(x^4+3x^2-4)$
Focus on just the right term $(x^4+3x^2-4)$:
Let $u$ = $x^2$, then:... | $$2x^4+6x^2-8=2(x^2+4)(x^2-1)=2(x^2+4)(x-1)(x+1)=0$$
is true when either $$x^2+4=0$$ or $$x+1=0$$ or $$x-1=0.$$
The first condition is not possible in the reals as $x^2+4\ge4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Binomial Expansion Of $\frac{24}{(x-4)(x+3)}$ Can somebody help me expand $\frac{24}{(x-4)(x+3)}$ by splitting it in partial fractions first and then using the general binomial theorem?
This is what I've done so far:
$$\frac{24}{(x-4)(x+3)}$$
$$=\frac{24}{7(x-4)}-\frac{24}{7(x+3)}$$
Now I know I have to find the binomi... | $\dfrac{24}{7(x-4)}=\dfrac{-6}{7\left(1-\frac x4\right)}=-\dfrac67\left(1+\frac x4+(\frac x4)^2+(\frac x4)^3+\cdots\right)$
$\dfrac{24}{7(x-(-3))}=\dfrac{8}{7\left(1-\left(-\frac x3\right)\right)}=\dfrac87\left(1-\frac x3+(\frac x3)^2-(\frac x3)^3+\cdots\right)$
$\therefore\dfrac{24}{(x-4)(x+3)}=-2+\dfrac16x-\dfrac{13}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Purely geometric proof of inverse trigonometric functions derivatives Can you compute the derivatives of $\sin^{-1}(x),\cos^{-1}(x),$ and $\tan^{-1}(x)$ using only geometry?
I know how to use geometry to find the derivatives of $\sin x$ and $\cos x$ like this:
We can use the fact that we know the tangent of the circle... | It is similar to the derivation of cos(x) or sin(x) geometrically.
Let $\theta_1 = \arcsin(x)$
$$\sin(\arcsin(x)) = x = P$$
$$\cos(\arcsin(x)) = \sqrt{1 - \sin^2(\arcsin(x))}= \sqrt{1-x^2}= B $$
$$H = 1$$
A very important result used in all the cases,
$$\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$
And assume
$$\lim_{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
AM/GM inequalities I need some help to prove this inequality... I guess one can use Jensen's then AM/GM inequalities.
Let $x_1, x_2, x_3, x_4$ be non- negative real numbers such that
$x_1 x_2 x_3 x_4 =1$.
We want to show that
$$x_1^3 + x_2^3 + x_3^3 + x_4^3 \ge x_1+x_2+x_3+x_4,$$
and also
$$x_1^3 + x_2^3 + x_3^3 + x_4^... | The first inequality.
We need to prove that:
$$\sum_{cyc}x_1^3\geq\sum_{cyc}x_1\sqrt{\prod_{cyc}x_1}$$ or
$$\sum_{cyc}x_1^3\geq\sum_{cyc}\sqrt{x_1^3x_2x_3x_4},$$which is true by Muirhead because
$$(3,0,0,0)\succ(1.5,0.5,0.5,0.5).$$
We can use also AM-GM:
$$\sum_{cyc}x_1^3=\frac{1}{6}\sum_{cyc}(3x_1^3+x_2^3+x_3^3+x_4^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Prove that $\tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac 12 \cos^{-1}x^2$ Let the above expression be equal to $\phi$
$$\frac{\tan \phi +1}{\tan \phi-1}=\sqrt{\frac{1+x^2}{1-x^2}}$$
$$\frac{1+\tan^2\phi +2\tan \phi}{1+\tan^2 \phi-2\tan \phi}=\frac{1+x^2}{1-x^2}$$
$$\frac{1+\t... | Because for $x\neq0$ and $-1\leq x\leq1$ easy to see that:
$$0<\frac{\pi}{4}+\frac 12 \cos^{-1}x^2<\frac{\pi}{2}$$ and we obtain: $$\tan\left(\frac{\pi}{4}+\frac 12 \cos^{-1}x^2\right)=\frac{1+\tan\frac{1}{2}\arccos{x^2}}{1-\tan\frac{1}{2}\arccos{x^2}}=$$
$$=\frac{\cos\frac{1}{2}\arccos{x^2}+\sin\frac{1}{2}\arccos{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
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How to evaluate $\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx$ without complex analysis This particular integral evaluates to,
$$\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx=\frac{\pi }{8}\ln \left(2\right)-\frac{3\pi }{8}+\frac{\pi }{3}\ln \left(2+\sqrt{3}\right)... | With subbing $t=\frac{1-x}{1+x}$ we have
$$\int_0^1\frac{\ln(1+t^3)}{1+t^2}dt=\int_0^1\frac{\ln\left(\frac{2(1+3x^2)}{(1+x)^3}\right)}{1+x^2}dx$$
$$=\ln2\int_0^1\frac{dx}{1+x^2}+\int_0^1\frac{\ln(1+3x^2)}{1+x^2}dx-3\int_0^1\frac{\ln(1+x)}{1+x^2}dx$$
where the first integral
$$\int_0^1\frac{dx}{1+x^2}=\arctan(1)=\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}}\ dx$? I am trying to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}} dx$
This is apparently a binomial integral of the form $\int x^m (a+bx^k)^ndx$. Therefore, we can use Euler's substitutions in order to evaluate it. Since $\dfrac{m+1}{k} = \dfrac{5+1}{2} = 3 \in \mathbb{Z}$ w... | Do the substitution $u^{3/5} = 1+ x^2$. Then $x = \sqrt{u^{3/5}-1}$, so $\mathrm{d}x = \frac{3\,u^{-2/5}}{5\,\sqrt{u^{3/5}-1}}\,\mathrm d u$ and
$$
∫ x^5 \,(1+x^2)^{2/3}\,\mathrm d x = \frac{3}{5}∫ (u^{3/5}-1)^2 \,\mathrm d x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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How to evaluate $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx$? I am trying to evaluate
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx \quad (1)$$
The typical way to confront this kind of integrals are the conjugates i.e:
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx = $$... | Multiplying by the conjugate is not a dead end. I'm not sure why you multiplied by the conjugate of the numerator and denominator, you can easily evaluate this integral just by multiplying by the conjugate of the denominator:
$$I=\int \left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)\left(\frac{\sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Other way to evaluate $\int \frac{1}{\cos 2x+3}\ dx$? I am evaluating
$$\int \frac{1}{\cos 2x+3} dx \quad (1)$$
Using Weierstrass substitution:
$$ (1)=\int \frac{1}{\frac{1-v^2}{1+v^2}+3}\cdot \frac{2}{1+v^2}dv =\int \frac{1}{v^2+2}dv \quad (2) $$
And then $\:v=\sqrt{2}w$
$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)... | Long way but doable
$$I=\int \frac{dx}{\cos (2x)+3}$$
Let
$$\cos(2x)=t \implies x=\frac{1}{2} \cos ^{-1}(t)\implies dx=-\frac{1}{2 \sqrt{1-t^2}}$$
$$I=-\frac{1}{2}\int \frac{dx}{(t+3) \sqrt{1-t^2}}=-\frac{1}{2}\int \frac{\sqrt{1-t^2}}{(t+3) (1-t^2)}\,dt$$
$$\frac{1}{(t+3) (1-t^2)}=-\frac{1}{8 (t+1)}+\frac{1}{16 (t+3)}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Is it possible to show that the fifth roots of 1 add up to 0 simply by using trigonometric identities? You can't use geometric sums, minimal polynomials, pentagon, and exact values with radicals.
All the five, fifth-roots of unity are :$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\righ... | Good question. Shall we try?
So we have the five roots: $$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\right),\left(\cos \left(\frac{4 \pi}{5}\right)+i \sin \left(\frac{4 \pi}{5}\right)\right),\left(\cos \left(\frac{6 \pi}{5}\right)+i \sin \left(\frac{6 \pi}{5}\right)\right), \left(\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Show that $\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\int\limits_1^2 \frac{dx}{x}=\ln(2)$ Show that:
$$\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\int\limits_1^2 \frac{dx}{x}=\ln(2)$$
My attempt:
We build a Riemann sum with:
$1=x_0<x_1<...<x_{N-1}<x_N=2$
$x_n:=\frac{n}{N}+1... | Your reasoning is correct, but you make it more complicated than needed.
$$\frac1N\sum_{i=1}^N\frac1{1+\dfrac nN}\to\int_0^1\frac{dx}{1+x}=\left.\ln(1+x)\right|_0^1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Lagrange not returning all critical points? I'm trying to identify the critical points of $f(x,y)=x^3+xy^2-4xy$ on the constraint $(x+1)^2+(y-2)^2\leq1$.
Setting up $\nabla f = \lambda\nabla g$ we have
x: $3x^2+y^2-4y=\lambda(2x+2) \implies \lambda=\tfrac{3x^2+y^2-4y}{2x+2}$
y: $2xy-4x=\lambda(2y-4) \implies \lambda=\t... | To make our lives easier we can rewrite $f$ as
$$f(x,y) = x(x^2+y^2-4y) = x(x^2+(y-2)^2-4)$$
which gives us the system of equations
$$\begin{cases}3x^2+(y-2)^2-4 = 2\lambda(x+1) \\ x(y-2) = \lambda(y-2) \\ (x+1)^2+(y-2)^2=1\end{cases}$$
Now if you assume $y\neq 2$ then $x=\lambda$, giving the points which are the inter... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int _0^1\frac{\ln \left(x^3+1\right)}{x+1}\:dx$ What methods would work best to find $\displaystyle \int _0^1\frac{\ln \left(x^3+1\right)}{x+1}\:dx$
As usual with this kind of integral i tried to differentiate with the respect of a parameter
$$\int _0^1\frac{\ln \left(ax^3+1\right)}{x+1}\:dx$$
$$\int _0^1\... | For once, Feynman's trick makes the problem more difficult.
Making the problem more general, I would write
$$I=\int _0^1\frac{\log \left(x^n+1\right)}{x+1}\,dx=\sum_{i=1}^n\int _0^1\frac{\log \left(x-r_i\right)}{x+1}\,dx$$ where the $r_i$ are the roots of unity. Now using
$$\int \frac{\log \left(x-r_i\right)}{x+1}\,dx=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Ray-Triangle intersection algorithm Havel-Herout I currently implement the HH algorithm to detect the intersection of a limited ray and a triangle.
My problem is that this implemention does not seems to work.
My understanding of this algorithm:
Triangles are defined with three planes:
$ \overrightarrow{n_{0}} = \overri... | I think you are making two mistakes. First, if you compute by hand,
you'll see that the intersection occur at the barycentric coordinate
$(1,0)$ which is exactly at one of the corners of the triangle. Fast
ray-triangle intersection algorithms have numerical issues and may
give incorrect results in such (literal) edge c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is the convergence point of $ \sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} $ negative? I am trying to evaluate $\frac1{2^1} - \frac1{2^2} + \frac1{2^3} - \frac1{2^4} + \cdots$
I re-wrote the sum using sigma notation as:
$$ \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) \quad (1) $$
Hen... | The series you are trying to evaluate is
$$\sum_{n=1}^\infty(-1)^{n-1}\left(\frac12\right)^n=\frac12\frac1{1+\dfrac12}=\frac13.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrix representation of square matrix function with respect to a matrix basis The mapping $f$ from $V$ to $V$ of the vector space on $\mathbb{C}$ formed by the complex square matrices,
\begin{align}
f(x) = \begin{pmatrix}
3& 4&\\
-2& -3
\end{pmatrix}\begin{pmatrix}X\end{pmatrix}\begin{pmatrix}
1&2\\
-1&-1
\end{pmatri... | To elaborate a bit on the comment: by calculating $f(e_1),\dots,f(e_4)$, you have done the "hard part". From there, we use each of these outputs to build the columns of our matrix.
For example, we find that plugging in the second basis element yields
$$
f(e_2) = \pmatrix{-3&-3\\2&2} = -3e_1 + -3e_2 + 2e_3 + 2e_4.
$$
T... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a^2 + b^2 + c^2 = 1$, what is the the minimum value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$?
Suppose that $a^2 + b^2 + c^2 = 1$ for real positive numbers $a$, $b$, $c$. Find the minimum possible value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$.
So far I've got a minimum of $\sqrt {3}$. Can anyo... | Trivially, we have $(x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$, so we get
$$(x+y+z)^2 \geq 3(xy+yz+xz)$$
by adding to both sides of the equation. Thus by plugging in $x = \frac{ab}{c}$, $y = \frac{bc}{a}$, $z = \frac{ca}{b}$, we get
$$\left(\frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b}\right)^2 \geq 3(b^2 + c^2 + a^2) = 3$$
and ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Problems regarding the ratio of two definite integrals These are two interesting problems:
(1) Find the ratio of $\int_0^1 (1-t^4)^{-0.5} \,dt$ and $\int_0^1 (1+t^4)^{-0.5} \,dt$
(2) Find the ratio of $\int_0^x e^{xt-t^2} \,dt$ and $\int_0^x e^{\frac{-t^2}4} \,dt$
Both are not supposed to be solved via solving individu... | *
*Use the change of variable formula, respectively $t=\sin u$ and $t=\tan u$:
\begin{align}
&\int_0^1{\dfrac{1}{\sqrt{1-t^4}}\text{d}t}\\
=&\int_0^{\pi /2}{\dfrac{\cos u}{\sqrt{1-\sin ^4u}}\text{d}u}\\
=&\int_0^{\pi /2}{\dfrac{1}{\sqrt{1+\sin ^2u}}\text{d}u},
\end{align}
\begin{align}
&\int_0^1{\dfrac{1}{\sqrt{1+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculation of $\int\frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}dx=$
$$\int\frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}\,dx=\text{?}$$
My work :
$$\int\frac{e^{-\sin^2(x)} \tan^3(x)}{\cos(x)} \, dx = \left\{\left(\int e^{-\sin^2(x)}\tan^3(x)\,dx\right)\frac{1}{\cos(x)}\right\}-\int\frac{\sin(x)}{\cos^2(x)}\left(\int e^{-\sin^2... | Let $u=\sec(x)$ to make
$$\int\frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}\,dx=\frac 1e\int e^{\frac{1}{u^2}} \left(u^2-1\right) \,du$$
$$\int e^{\frac{1}{u^2}}\,du=u\mathrm{e}^\frac{1}{u^2}-{\displaystyle\int}-\dfrac{2\mathrm{e}^\frac{1}{u^2}}{u^2}\,\mathrm{d}u$$
Let $v=\frac 1u$
$${\displaystyle\int}-\dfrac{2\mathrm{e}^\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of $\sum_{k=1}^n \left(\frac{1}{k^2}\right)\le \:\:2-\frac{1}{n}$ Proof of $\displaystyle\sum_{k=1}^n\left(\frac{1}{k^2}\right)\le \:\:2-\frac{1}{n}$
The following proof is from a book, however, there is something that I don't quite understand
for $k\geq 2$ we have:
(1): $\displaystyle\frac{1}{k^2}\le \frac{1}{k... | Note that $k(k-1)<k^2$ for $k>0$ but when $k=1$, $\dfrac{1}{n(n-1)}$ is not defined.
$\begin{aligned}\displaystyle \sum_{k=2}^{n} \frac{1}{k^2} &<\sum_{k=2}^{n} \frac{1}{k(k-1)} \\
&= \sum_{k=2}^{n} \frac{1}{k-1}-\frac{1}{k} \\
&=1-\frac{1}{n} \\ &\Rightarrow \sum_{k=1}^{n} \frac{1}{k^2}
=1+\sum_{k=2}^{n} \frac{1}{k^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Square equal to sum of three squares For which integers $n$ there exists integers $0\le a,b,c < n$ such that $n^2=a^2+b^2+c^2$?
I made the following observations:
*
*For $n=1$ and $n=0$ those integers doesn't exist.
*If $n$ is a power of 2 those integers doesn't exist. Let $n=2^m$ with $m>0$ the smallest power of 2... | You are correct: If $p > 2$ is prime, then $p^2$ can always be written as the sum of three squares at least two of which are non-zero.
Let $s(n)$ denote the number of ways of writing $n = a^2 + b^2 + c^2$, where $a$, $b$, and $c$ are integers (positive or negative) and not accounting for symmetries. One has $s(1) = 6$.... | {
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How to find $a$, $b$, $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$ I'm trying to find $a$, $b$ and $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$.
After expanding the binomial in $P(x-1)$, I end up getting
$3ax^2-3ax+2bx+a-b=x^2$. What next? Using $3a = 1$... | Sure it does work. As you let $P(x)=ax^3+bx^2+cx$, we have $P(x)-P(x-1)=3ax^2-(3a-2b)x+a-b+c$
Now just compare the coefficients. That is, $3a=1$, $3a-2b=0$, $a-b+c=0$ which after solving gives $a=\frac13 , b=\frac12 , c=\frac16$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove: $\int_0^2 \frac{dx}{\sqrt{1+x^3}}=\frac{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)}{6\Gamma\left(\frac{1}{2}\right)}$ Prove:
$$
\int_{0}^{2}\frac{\mathrm{d}x}{\,\sqrt{\,{1 + x^{3}}\,}\,} =
\frac{\Gamma\left(\,{1/6}\,\right)
\Gamma\left(\,{1/3}\,\right)}{6\,\Gamma\left(\,{1/2}\,\right)}
$$
First... | A hypergeometric solution: Modulo Beta function $I_0=\int_0^{\infty } \frac{1}{\sqrt{x^3+1}} \, dx=\frac{2 \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{7}{6}\right)}{\sqrt{\pi }}$ one may evaluate $I_1=\int_2^{\infty } \frac{1}{\sqrt{x^3+1}} \, dx$ instead. Substitute $x\to\frac 1x$ and binomial expansion gives
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Proving $\sum_{n=0}^{\infty}\frac{(\phi-1)^n}{(2n+1)^2}=\frac{\pi^2}{12}-\frac{3\ln^2(\phi)}{4}$
How do we prove this?
$$\sum_{n=0}^{\infty}\frac{(\phi-1)^n}{(2n+1)^2}=\frac{\pi^2}{12}-\frac{3\ln^2(\phi)}{4}$$
where $\phi:=\frac12(1+\sqrt{5})$ is the Golden Ratio.
My attempt:
\begin{align*}
\displaystyle\sum_{n=0}^{\... | Your formula has a typo, it should be :
$$\sum_{n=0}^\infty\frac{(\phi-1)^{\color{red}{2n+1}}}{(2n+1)^2}=\frac{\pi^2}{12}-\frac{3\ln^2(\phi)}{4}$$
It´s an instance of Legendre's Chi-Function evaluated at $x=\phi-1$
Here is the proof:
$$
\begin{aligned}
\mathrm{Li}_{2}(x)&=\sum_{n=1}^\infty\frac{x^n}{n^2}\\
&=\frac14\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\frac{1}{d_1} + \dots + \frac{1}{d_k} = 1,$ and $\gcd(d_i,d_j)>1 \, \forall i,j$ implies $\gcd(d_1, \dots, d_k) > 1$ for distinct $d_i.$
Conjecture: If $d_i \in \mathbb{N}$ are distinct, $\frac{1}{d_1} + \dots + \frac{1}{d_k} = 1,$ and $\gcd(d_i,d_j)>1 \, \forall i,j,$ then $\gcd(d_1, \dots, d_k) > 1.$
Motive: In th... | The conjecture is false. It is well-known that any positive rational can be obtained as a sum of a finite subsequence of the harmonic series. Let $$\frac{1}{c_1} + \frac{1}{c_2} + \dots + \frac{1}{c_k} = 30(1 - 1/6 - 1/10 - 1/15) = 20$$ where the $c_i$ are distinct. Now we can let $d_1 = 6, d_2 = 10, d_3 = 15, d_{3+i} ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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How to compute $\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$?
How to compute the following integral?
$$\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$$
So what I did is to change $\sin(x)$ to $\cos(x)$ with cofunction identity, which is $\sin(\frac{\pi}{2} -x) ... | Your first method, setting $u=\tan x$, is the substitution suggested by Bioche's rules Your problem is that there's an error in you computation. You should obtain the integral
$$\int_0^{\infty}\frac{\color{red}u\,\mathrm du}{1+u^3},$$
which is convergent since $\:\frac u{1+u^3}\sim_\infty\frac1{u^2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\tan\frac{\pi}{7}\tan\frac{2\pi}{7}\tan\frac{3\pi}{7}=\sqrt 7$ I'm trying to show that
$$\tan\frac{\pi}{7}\tan\frac{2\pi}{7}\tan\frac{3\pi}{7}=\sqrt 7$$
My attempt:
Since $\tan x=\frac{\sin x}{\cos x} $ so immediately the denominator is recognized as $$\prod_{k=1}^{3}\cos\frac{k\pi}{7}=\frac{1}{2^3}=\frac{1... | Here by @ Darth Geek, in general we have $$S(n)=\prod_{k=1}^{n}\sin\left(\frac{k\pi}{2n+1} \right)=\frac{\sqrt{2n+1}}{2^n}$$ set $n=3$ we have $S(3)=\frac{\sqrt{7}}{8}$ and we are done. Also $$C(n)=\prod_{k=1}^{n}\cos\left(\frac{k \pi}{2n+1} \right)=\frac{1}{2^n}$$.
Alternative approach:
We exploit reflection fo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$? Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c... | Observe that by AM-GM inequality for $3$ positive reals: $\dfrac{a+b+c}{3} \ge \sqrt[3]{abc}$. Thus you need to show: $\dfrac{a^{n+1}+b^{n+1} + c^{n+1}}{a^n+b^n+c^n} \ge \dfrac{a+b+c}{3}$, which is the same as: $2(a^{n+1}+b^{n+1}+c^{n+1}) \ge a^n(b+c)+b^n(c+a)+c^n(a+b)$. But this is quite clear because you can use the ... | {
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find the complex integral: $\int_0^\infty \frac{z^6}{(z^4+1)^2}dz$. Problem with integral formula.... Question I am trying to find the complex integral: $\displaystyle\int_0^\infty \frac{z^6}{(z^4+1)^2}dz$.
My Attempt (and eventual question): $\int_0^\infty \frac{z^6}{(z^4+1)^2}dz=\frac{1}{2}\int_\infty^\infty\frac{z^6... | @Vercassivelaunos has the right idea; it's not even that computationally demanding. A root $a$ of $z^4+1$ has residue$$\begin{align}\lim_{z\to a}\tfrac{d}{dz}\tfrac{(z-a)^2z^6}{(z^4+1)^2}&=\lim_{z\to a}\tfrac{2(z-a)z^5(az^4+4z-3a)}{(z^4+1)^3}\\&=2a^5\lim_{\epsilon\to0}\tfrac{\epsilon(a(a+\epsilon)^4+4(a+\epsilon)-3a)}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Expansion of generating function $\frac{1}{ \sqrt{1-12x+4x^2 } }$ I came across this generating function
$$\frac{1}{ \sqrt{1-12x+4x^2 } }$$
How exactly does one expand this series? I have read through some notes, it seems like we need to factorize the denominator, but it doesn't look like this one can be factorized?
| Make the long division
$$\frac{1}{ 1-12x+4x^2 }=1+12 x+140 x^2+1632 x^3+19024 x^4+221760 x^5+2585024 x^6+O\left(x^7\right)$$
Now (being patient), the binomial expansion
$$\frac{1}{ \sqrt{1-12x+4x^2 } }=1+6 x+52 x^2+504 x^3+5136 x^4+53856 x^5+575296 x^6+O\left(x^7\right)$$
| {
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"url": "https://math.stackexchange.com/questions/3789124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the sum of all possible values of $a$ such that the following equation $(x - a)^2 + (x^2 - 3x + 2)^2 = 0$ has real root in $x$ :-
Find the sum of all possible values of $a$ such that the following equation $(x - a)^2 + (x^2 - 3x + 2)^2 = 0$ has real root in $x$ :-
What I Tried :- I know $(x^2 - 3x + 2) = (x - 1)... | Just note that $$ (x - a)^2 + (x - 3x + 2)^2 = 0 $$ $$ \implies (x -a) = (x^2 - 3x + 2) = (x - 2)(x - 1) = 0 \quad [\text{As }x, a \in \mathbb{R}] $$
Then $a = 2, a = 1$. Then $$ {\text{Sum of all possible valus of } a} = 2 + 1 = 3 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
I tried going along the path of computing $(x+y+z)... | Hint: $$(xy)(yz)(xz)=(xyz)^2,$$
take square root, then divide by $xy$ or $yz$ or $xz$ to find $z,\,x,\,y$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How precise is an inequality Problem: Let $a, b, c$ be positive real numbers such that
$a + b + c = \frac{1}{a^2} +\frac{1}{b^2} +\frac{1}{c^2}$.
Prove that
$$\begin{align} 2(a + b + c) \geq \sqrt[3]{7a^2b+1}+\sqrt[3]{7b^2c+1}+\sqrt[3]{7c^2a+1} \end{align}$$
Source: MEMO 2013
Original proof:
Using AM-GM:
$$\begin{align... | Because we always need to save the case of the equality occurring.
In your first proof it happens.
In your second proof it does not happen, which says that there is a mistake in your second proof.
Another way to save the equality case:
By AM-GM twice we obtain:
$$\sum_{cyc}\sqrt[3]{7a^2b+1}=\frac{1}{4}\sum_{cyc}\sqrt[3... | {
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"timestamp": "2023-03-29T00:00:00",
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How to convert the recursive function $f(n)=n\cdot f(n-1)+1$ to an explicit function $f(0)=1$
$f(n)=n\cdot f(n-1)+1$
How can an explicit function be derived from this?
| The general strategy here is to make a "guess" and prove it:
$$ f(0)=1$$
$$ f(1)=1\cdot f(0)+1=2$$
$$ f(2)=2\cdot f(1)+1=5$$
$$ f(3)=3\cdot f(2)+1=16$$
$$ f(4)=4\cdot f(3)+1=65$$
$$ f(5)=5\cdot f(4)+1=326$$
We can write out the computations to get an idea of what's going on:
$$ f(2)=2\cdot 2+1=2\cdot 2!+1$$
$$ f(3)=3\c... | {
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"url": "https://math.stackexchange.com/questions/3792233",
"timestamp": "2023-03-29T00:00:00",
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Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}... | First, observe
$$3x+2x^3-2(x^2+1)^{3/2}=\frac{3x^2+4}{-3x-2x^3-2\sqrt{x^2+1}-2x^2\sqrt{x^2+1}}.$$
The top is a quadratic, while the bottom grows on the order of $x^3$, hence the limit as $x\to \infty$ is zero.
| {
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"url": "https://math.stackexchange.com/questions/3794325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
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Finding $\lim _{n\to\infty}(1+\frac1{a_{1}})(1+\frac1{a_{2}})\cdots(1+\frac1{a_{n}})$, where $a_1=1$ and $a_n=n(a_{n-1}+1)$
Let $a_{1}=1$ and $a_{n}=n\left(a_{n-1}+1\right)$ for $n=2, 3, \ldots$. Define
$$
P_{n}=\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right) \cdots\left(1+\frac{1}{a_{n}}\right)
$$
for $n... | Use the fact that:
$$
1+\frac{1}{a_{k}} =
\frac{a_{k} + 1}{a_{k}} =
\frac{a_{k} + 1}{k(a_{k-1} + 1)}
$$
This will allow us to express $P_n$ as $\frac{a_{n+1}}{(n+1)!}$. Now, expand it using the recurrence relation of the sequence $\left(a_n\right)$:
\begin{align}
\frac{a_{n+1}}{(n+1)!} &=
\frac{a_{n}}{n!} + \frac{1}{... | {
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Where I made a mistake during factoring $x^6+x^5+x^4+x^3+x^2+x+1$? In order to factor the expression, due to symmetry of coefficients if $r_1,r_2,r_3$ are zeros of $x^6+x^5+x^4+x^3+x^2+x+1$ then $\frac{1}{r_1}, \frac{1}{r_2} , \frac{1}{r_3}$ are also zeros. So we can rewrite:
$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2-(r_1+\frac{... | Noe that
$$
(x^2+1)(x+1)^2(x^2-x+1)=
x^6 + x^5 + x^4 + 2x^3 + x^2 + x + 1,
$$
so you forgot a factor $2$, as remarked above. The polynomial
$$
\Phi_7(x)=1+x+x^2+x^3+x^4+x^5+x^6
$$
is irreducible over the integers and over the rational numbers by the Gauss lemma and Eisenstein's criterion. Therefore such a factorization... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\cdots+\frac{1}{3001}<\frac{4}{3}$ Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\cdots+\frac{1}{3001}<\frac{4}{3}$
Using AM- HM inequality,
$\left(\sum_{k=1001}^{3001} k\right)\left(\sum_{k=1001}^{3001} \frac{1}{k} \right) \geq(2001)^{2}$
But $\sum_{k=1001}^{3001} k=(2001)^{2... | A way to do it without calculus and mostly reasoning:
The sum is $\frac{1}{1001}+\frac{1}{1002}+\dots+\frac{1}{3001}$.
Split the sum into eight parts: $$\sum_{k=1001}^{1250} \frac{1}{k}, \sum_{k=1251}^{1500} \frac{1}{k}, \dots, \sum_{k=2750}^{3000} \frac{1}{k}$$
In the first summation, the smallest value is $\frac{1}{1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Challenging integral: $\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\sin x}dx$ How to tackle
$$I=\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\sin x}dx\ ?$$
This integral popped up in my solution ( see the integral $\mathcal{I_3}\ $ at the end of the solution.)
My attempt: By Weierstrass substitution we have
$$I=2\int_0^1\frac{\arctan^... | We have
*
*$\int \frac{\log ^3(1+i x)}{x} \, dx=6 \text{Li}_4(i x+1)+3 \text{Li}_2(i x+1) \log ^2(1+i x)-6 \text{Li}_3(i x+1) \log (1+i x)+\log (-i x) \log ^3(1+i x)$
So
*
*$\Re\left(\int_0^1 \frac{\log ^3(1+i x)}{x} \, dx\right)=\int_0^1 \frac{\frac{1}{8} \log ^3\left(x^2+1\right)-\frac{3}{2} \log \left(x^2+1\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluating $\int_0^\infty \left| \frac{\sin t}{t} \right|^n \, \mathrm{d}t$ for $n = 3, 5, 7, \dots$ I would like to determine the general term of the following sequence defined by an infinite integral:
$$
I_n = \int_0^\infty \left| \frac{\sin t}{t} \right|^n \, \mathrm{d}t \, ,
$$
wherein $n =3, 5, 7, \dots$ is an odd... | Here is a partial answer: If $n \geq 2$ is an integer, then
$$ I_n = \frac{1}{(n-1)!2^{n-1}} \sum_{l=0}^{\lfloor n/2 \rfloor} (-1)^l \binom{n}{l} (n-2l)^{n-1} J_{n-2l}, \tag{1} $$
where $J_p$ is defined by
$$ J_{p} = \begin{cases}
\displaystyle \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{\log(2\pi j)}{4j^2-p^2}, & \text{i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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System of Three Equations - Prove that at least two numbers' absolute value is equal. Question :
Prove that at least two of the numbers $a,b,c$ must have the absolute values equal in order that the system of following equations in $x,y$ may be consistent
$$ a^2x+b^2y+c^2=0$$
$$a^4x+b^4y+c^4 =0$$
$$x+y+1=0$$
I found $... | As $$\begin{pmatrix}1&1&1\\a^2&b^2&c^2\\a^4&b^4&c^4\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix},$$
the matrix must be singular, i.e., a non-trivial linear combinations of the rows is zero, i.e., there are numbers $\alpha,\beta,\gamma$, not all $=0$, such that the polynomial $$
\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF
I have done this problem by inspection as $$\frac{x^2+6}{6}=[x] \implies x>0.$$
Let [x]=0, then $x$ is non real. Let $[x]=1$, then $x=0$ which contradicts. Let $[x]=2$, it gives $x=\sqrt{6}$, in agrrement. Simi... | You have got all real solutions. Because in
$$\frac{x^2+6}{6}=[x]$$
LHS increases quadratically (parabolic) and RHS variies roughly as a line, so after $x=5$ the parabola leaves out the line.
Method I: You can use $$x-1 \le [x] \le x \implies x^2-6x+6 \le 0 ~~~~(1),~~~ x^2-6x+12 >0~~~(2)$$
(1) gives $3-\sqrt{3}< x \le ... | {
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"timestamp": "2023-03-29T00:00:00",
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Algebraic Manipulation with Cube Roots Let $a,$ $b,$ $c$ be the real roots of $x^3 - 4x^2 - 32x + 17 = 0.$ Solve for $x$ in $$\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c} = 0.$$
We probably have to manipulate the $\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c}$ into something more convenient, so we can actu... | Elevating to third power we get
$$
3 \left[x+\left(\sqrt[3]{x-a}+\sqrt[3]{x-b}\right) \left(\sqrt[3]{x-a}+\sqrt[3]{x-c}\right) \left(\sqrt[3]{x-b}+\sqrt[3]{x-c}\right)\right]-(a+b+c)=0
$$
using the original equation gives
$$
3 \left[x-\sqrt[3]{x-a}\sqrt[3]{x-b}\sqrt[3]{x-c}\right]-(a+b+c)=0\\
3 \left[x-\sqrt[3]{(x-a)(x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$|_{x=1}$ notation? I had the equation:
$$\frac{1}{(1+x)(1-x)^2}=\frac{A}{(1-x)^2}+\frac{B}{1-x}+\frac{C}{1+x}$$
$$A=\left.\frac{1}{1+x}\right|_{x=1} = \frac{1}{2} \tag2$$
What does $(2)$ mean? Particularly, the notation $|_{x=1}$
| It means you substitute $x=1$ into the expression. Thus $$A=\left.\frac{1}{1+x}\right|_{x=1} =\frac{1}{1+1} = \frac{1}{2}. $$
This notation can also be used as follows:
$$\int_{0}^{1}\frac{1}{x+1}dx=\ln(x+1)\bigg|_{x=0}^{x=1}=\ln(1+1)-\ln(0+1)=\ln(2).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of squares of numbers equals product of numbers
Find number of tuples such that the sum of the squares of the numbers equals the product of the numbers.
I tried it and found some tuples like $(3,3,3)$ satisfying $3^2+3^2+3^2=3\times 3\times 3$ but I don't know the real approach to find all such numbers. Can anyo... | A partial answer, just to set the ball rolling on this question. As written in the comments the question is too open for a complete answer. This answer is limited to pairs and triples of numbers.
*
*There are no numbers such that $a^2+b^2=ab$ (except the trivial solution $0,0$). A quick way to see this is $ab=a^2+b^2... | {
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Prove $\lim\limits_{x^2 + y^2 \to +\infty} x^2 -2xy + 2y^2 = +\infty$
Prove that $$\lim_{x^2 + y^2 \to +\infty} x^2 -2xy + 2y^2 = +\infty$$
My attempt:
$$x^2 + 2y^2 = x^2+y^2 + y^2 \implies \lim_{x^2 + y^2 \to +\infty}x^2 +2y^2 = +\infty$$
Then, from Cauchy-Schwarz:
$$x^2 + 2y^2 \geq 2\sqrt2xy \geq 2xy $$
Thus,
$$x... | We have that
$$x^2 -2xy + 2y^2 =(x-y)^2+y^2\ge y^2$$
$$x^2 -2xy + 2y^2 =\left(\frac{x}{\sqrt 2}-\sqrt 2y\right)^2+\frac{x^2}2\ge \frac{x^2}2 $$
therefore
$$x^2 -2xy + 2y^2 \ge \frac{x^2+2y^2}4 \ge \frac{x^2+y^2}4\to \infty$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Computing the genus of $y^2=x(x^2-1)$ using 1-forms I'm trying to compute the genus of the projective curve $C:=V(Y^2Z-X(X^2-Z^2))\subset\Bbb{P}^2_\Bbb{C}$ explicitly using differential forms.
I know beforehand that this is an elliptic curve, so the expected answer is $g=1$.
So I must find a globally defined differenti... | I think you've made a small mistake in your calculation. I get
\begin{align*}
\omega := \frac{dx}{y}=v\cdot d\left(\frac{u}{v}\right)=v \frac{v du - u dv}{v^2} = du - \frac{u}{v} dv \, .
\end{align*}
Differentiating the affine equation $v = u^3 - uv^2$ you found in a neighborhood of $(0:1:0)$, we have
\begin{align*}
dv... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find a matrix $A$ such that $X$ generates the subspace $W$ (the solution space of the system $AX=0$.) Consider $W$ a subspace of $\mathbb{R}^{5}$ generated by:
\begin{align*}
X=\left \{(1,-1,0,5,1), (1,0,1,0,-2),(-2,0,-1,0,-1)\right \}
\end{align*}
Find a system of linear equations $AX=0$ such that W be the solution sp... | Since $W$ is $3$-dimensional, you need the $2$-dimensional orthogonal space, putting the generating vectors as rows of $A$.
$$0=\begin{pmatrix}1&-1&0&5&1\\1&0&1&0&-2\\-2&0&-1&0&-1\end{pmatrix}\mathbf{x}=\begin{pmatrix}1&0&0&0&3\\0&1&0&-5&2\\0&0&1&0&-5\end{pmatrix}\mathbf{x}$$ so $\mathbf{x}=(-3t,5s-2t,5t,s,t)=t(-3,-2,5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3825772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\cos(4x)=8\cos^4(x)-8\cos^2(x)+1$ without using LHS and RHS
Prove that
$$\cos(4x)=8\cos^4(x)-8\cos^2(x)+1$$
My Attempt
$$\Rightarrow \cos^2(2x)-\sin^2(2x) =8\cos^4(x)-8\cos^2(x)+1$$
Add 1 to both sides
$$\cos^2(2x)-\sin^2(2x) +1=8\cos^4(x)-8\cos^2(x)+2$$$$\Rightarrow \cos^2(2x)=4\cos^4(x)-4\cos^2(x)+1$$
$$\R... | Your proof is not wrong, it's just backwards. You need to start with $\cos^2(2x) = \cos^2(2x)$, then do the same manipulations you did in reverse order to end up with $ \cos(4x)=8\cos^4(x)-8\cos^2(x)+1$.
But your question is can a backwards proof ever be wrong? Yes it can. Consider this very simple "proof" that $1=2$:
... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that binomial coefficient equals to sum of arithmetic progression I have to prove that
$\sum^{n}_{r=1}r^2 + \sum^{n}_{r=1}r = 2$${n+2}\choose{3}$
So far the only thing I can come up with is
${n(n+1)(2n+1)}\over{6}$ $+ {n(n+1)\over{2}} =$ $2{{n+2}\choose{3}}$
${n(n+1)(2n+1)}\over{3}$ $+ {n(n+1)} =$ ${{n+2}\choose{... | Going from the first line of what you came up with to your second, you multiplied the left side by $2$ but divided the right side by $2$. Multiplying that side by $2$ instead gives that you're trying to prove
$$\frac{n(n + 1)(2n + 1)}{3} + n(n+1) = 4\binom{n+2}{3} \tag{1}\label{eq1A}$$
You have the right idea of factor... | {
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The polynomial $ x^7 + x^2 +1$ is divisible by (A) $ x^5 - x^4 + x^2 -x +1 \quad$ (B) $ x^5 + x^4 +1 \quad$ (C) $ x^5 + x^4 + x^2 +x +1\quad$
(D) $ x^5 - x^4 + x^2 +x +1$
My effort: Looking at the polynomial, I know that it will have only one real root, which is negative. All other 6 roots should be imaginary. And that... | $$x^7+x^2+1=x^7-x+x^2+x+1=$$
$$=(x^2+x+1)(x(x-1)(x^3+1)+1)=$$
$$=(x^2+x+1)(x^5-x^3+x^2-x+1).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the rank of a matrix $A$ Find the rank of the $n \times n$ matrix $A = [i + j]_{i,j \le n}$ (over $C$).
C here should be the complex space; although i am having trouble interpreting what A exactly is, I do not understand the middle bracket notation too well, is it ixn+jxn matrix? Any hint on this problem would ... | The first row is
$$R_1=\begin{bmatrix}2 & 3 & \cdots & n+1 \end{bmatrix}$$
the second one
$$R_2=\begin{bmatrix}3 & 4 & \cdots & n+2 \end{bmatrix}$$
the difference is
$$R_2-R_1=\begin{bmatrix}1 & 1 & \cdots & 1 \end{bmatrix}$$
and so on by row operation any other row is a combination of this first two
$$ \begin{bmatrix}... | {
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"timestamp": "2023-03-29T00:00:00",
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A tricky inequality: $n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>2n,\ n\geq3.$ Let $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$ be the $n-$th harmonic number,
it is not difficult to prove that:
(1) $n(n+1)^{\frac{1}{n}}<n+H_n,$ for $n>1$; (use AM-GM inequality)
(2) $(n-2)n^{\frac{1}{n-2}}>n-H_n$, for $n>2$. (Hint: $n=3$ is ... | Using that for $n>1$
$$1+n\geq3>e>\left(1+\frac1n\right)^n\implies (n+1)^{\frac{1}{n}}>1+\frac1n.$$
We have
$$ n(n+1)^{\frac{1}{n}}>n+1,\ n>1.$$
And then for $n>3$,
$$(n-2)n^{\frac{1}{n-2}}>(n-2)(n-1)^{\frac{1}{n-2}}>(n-2)+1=n-1,$$
when $n=3$, we also have
$$(n-2)n^{\frac{1}{n-2}}=3>2=n-1$$
therefore
$$n(n+1)^{\frac{1}... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding $x$ when $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$ Find $x$ if $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1.$
I was thinking of trying to substitute some number $y$ written in terms of $x$ than solving for $y$ to solve for $x.$ However, I'm not sure what $y$ to input, so can someone give me a hint?
| By $x=\frac {y^3} 3+\frac 53$
$$\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1 \iff \sqrt {y^3+1}-y=1 $$
$$\iff y^3+1=(1+y)^2 \iff y^3-y^2-2y=0 \iff y\in \{-1,0,2\}$$
that is $x\in \{\frac43,\frac53,\frac{13}3\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Completing the Square : $2+0.8x-0.04x^2$
Write $2+0.8x-0.04x^2$ in the form $A-B(x+C)^2$, where A, B and C are constants to be determined.
Here's how I have tried it out:
$2+0.8x-0.04x^2$
$-0.04x^2+0.8x+2$
$-0.04[(x-10)^2-10]+2$
$-0.04(x-10)^2+o.8+2$
$2.8-0.04(x-10)^2$
So the answer should be, $A=2.8$ $B=0.04$ $C=-1... | Let
$$2+0.8x-0.04x^2=-\frac4{100}\left(x^2-20x-50\right)=-\frac4{100}\left(x^2-20x+100-150\right)$$
$$-\frac4{100}\left((x-10)^2-150\right)=-0.04(x^2-10)^2+6$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of sequence differ from 2 by 0.02 For what value of $n≥0$, does $(2n+3)/(n+4)$ differ from 2 by 0.02?
I tried to solve this by applying the definition of limit,
$|(2n+3)/(n+4)-2|<0.02$
But didn't get the answer. I think I am misunderstanding the question. Please tell me how to solve this?
Edit:
Answer is $n=196$
... | If $n\ge 0$ then $n+4>0$ so $|n+4|=n+4,$ and $$\left|\frac {2n+3}{n+4}-2\right|=\left|\frac {2n+3}{n+4}-\frac {2(n+4)}{n+4}\right|=$$ $$=\left|\frac {2n+3-2(n+4)}{n+4}\right|=$$ $$=\left|\frac {2n+3-2n-8}{n+4}\right|=$$ $$=\left|\frac {-5}{n+4}\right|=\frac {|-5|}{|n+4|}=\frac {5}{n+4}.$$ Therefore if $n\ge 0$ then $$\... | {
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"url": "https://math.stackexchange.com/questions/3831573",
"timestamp": "2023-03-29T00:00:00",
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Finite part distribution Let $\varphi$ be a test function such that $Supp(\varphi)\subseteq [-M, M]$, $a\in \mathbb R$. We define the distribution principal value of $\frac{1}{x-a}$ :
$\left\langle\operatorname{P.\!v.}\left(\frac{1}{x - a}\right),\varphi\right\rangle: =\lim_{\varepsilon\to 0^+}\int_{|x - a|\geq \vareps... | If you know that $\left( \operatorname{pv}\frac{1}{x} \right)' = -\operatorname{fp}\frac{1}{x^2}$:
$$
\left< \frac{1}{2a} \left( \operatorname{pv}\frac{1}{x-a} - \operatorname{pv}\frac{1}{x+a} \right), \varphi(x) \right>
= \frac{1}{2a} \left( \left< \operatorname{pv}\frac{1}{x-a}, \varphi(x) \right> - \left< \operatorn... | {
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"url": "https://math.stackexchange.com/questions/3840468",
"timestamp": "2023-03-29T00:00:00",
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Find the length of $x$. Let $ABC$ be a right angled triangle right-angled at $B$. The angle bisectors of angles $A$ and $C$ are drawn, intersecting the opposite sides at $D$ and $E$ respectively. Given that $CD=12$ and $AE=8$ find the length of $AC$ (which was given as $x$ in the question). Furthermore, let the points ... | Let $BC=a$, $AB=c$ and $a=tc$.
Thus, since $\frac{cx}{x+a}=8$ and $\frac{ax}{x+c}=12,$ we obtain:
$$\frac{a(x+a)}{c(x+c)}=\frac{3}{2}$$ or
$$\sqrt{a^2+c^2}(2a-3c)=3c^2-2a^2$$ or
$$\sqrt{t^2+1}(2t-3)=3-2t^2,$$ which gives $$\sqrt{1.5}<t<1.5$$
Now, after squaring we obtain:
$$t(4t-3)(3t-4)=0,$$ which gives $$t=\frac{4}{3... | {
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"timestamp": "2023-03-29T00:00:00",
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By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x... | You can use $a^3+b^3=a^2-ab+b^2$ and proceed
$$\begin{equation}\begin{aligned}
&\sin^6 x+\cos^6 x\\[3ex]
&=(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)\\[2ex]
&=\color{green}{\sin^4x}-\sin^2x\cos^2x+\color{green}{\cos^4x}\\[2ex]
&=\color{green}{(\sin^2x+\cos^2x)^2}-3\sin^2x\cos^2x\\[2ex]
&=1-\frac{3}{\color{red}4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 0
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Calculate ${S_n} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + kn + {k^2}}}} ;{T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{n}{{{n^2} + kn + {k^2}}}} $ Let ${S_n} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + kn + {k^2}}}} ;{T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{n}{{{n^2} + kn + {k^2}}}} $, for n=1,2,3,.....Then
(A) ${S_n} ... | This is a beautiful question asked in JEE Advanced which tests your basic knowledge about definite integrals- Riemann Sum.
We have:
$${S_n} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + kn + {k^2}}}} ;{T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{n}{{{n^2} + kn + {k^2}}}} $$
Note that both these functions are decreasing.
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find all values for $x$ such that $|x^2|>|3x-2|$ I have absolutely no idea how to even start this inequality - I have seen some methods involving squaring both sides and rearranging to get $x^4 - 9x^2 + 12x - 4$, but have no idea where to go from there.
Any help would be greatly appreciated!
| Clearly $x^2\geq0$ for all real $x$ (and $x=0$ doesn't satisfy the inequality).
So you only need to consider the inequalities $x^2>3x-2$ or $x^2>2-3x$.
From the first you have $x^2-3x+2=(x-2)(x-1)>0$, so $x<1$ or $x>2.$
The second gives $x^2+3x-2>0,$ so $x<\frac{1}{2}(-3-\sqrt{17})$ or $x>\frac{1}{2}(-3+\sqrt{17}).$ T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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if $x^5=1$ with $x\neq 1$ then find value of $\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$
if $x^5=1$ with $x\neq 1$ then find value of $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$$
So my first observation was x is a non real fifth root of unity. Also $$x^5-1=(x-1)(1... | $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3} =$$
$$=\frac{x}{1+x^2}\cdot\frac{x^4}{x^4}+\frac{x^2}{1+x^4}\cdot\frac{x^3}{x^3}+\frac{x^3}{1+x}\cdot \frac{x^2}{x^2}+\frac{x^4}{1+x^3}\cdot\frac{x}{x}= $$
(remember that $x^5=1$, so $x^6=x$ and $x^7=x^2$)
$$=\frac{1}{x^4+x}+\frac{1}{x^3+x^2}+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$
I came across... | Multiplying the eq. to prove by $\frac{(1-x^2)(1-y^2)(1-z^2)}{2}$ you obtain :
$$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2} \iff$$
$$x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)= 4xyz$$
but expanding the left side under the hypothesis $x+y+z=xyz\ $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 1
} |
$A^{n}$ matrix problem without using $A=PDP^{-1}$ Eigendecomposition i have the following problem
Find $A^n$ for the following matrix
\begin{equation}
A=\begin{pmatrix}
0 & 0 & 1\\
0 & 1 & 1 \\
1 & 1 & 1
\end{pmatrix}
\end{equation}
I have tried the following, calculating for $n=1,2,3,4,5,6$
\begin{equation}
A^1=\begin... | With $\{e_1, e_2, e_3\}$ as your base, check what happens when you operate on them
$$Ae_1 = e_3 \implies A^ne_1 = A^{n-1}e_3$$
$$Ae_2 = e_2+e_3 \implies A^ne_2 = A^{n-1}(e_2+e_3) = A^{n-1}e_3 + A^{n-1}e_2$$
$$Ae_3 = e_1+e_2+e_3\implies A^ne_3 = A^{n-1}(e_1+e_2+e_3) = $$
$$ = A^{n-1}e_1+A^{n-1}e_2+A^{n-1}e_3 = 2A^{n-2}e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$.
I tried to substitute some basic values like $-1,0,1$ and try to find the roots of the function but couldn't.
Then I graphed the function on desmos and this is the graph.
So from this, we can say that $x^{... | My SOS is ugly.
$$x^{12}-x^9+x^4-x+1$$
$$={\frac { ( 8x^6-4x^3-1)^2}{64}}+{\frac { \left( 80x^
2-5x-72 \right) ^{2}}{6400}}+{\frac {2299}{1280} \left( x-{\frac {
712}{2299}} \right) ^{2}}+{\frac {7741}{3678400}}.$$
Remark. From Mr. Mike solution we can get
$$x^{12}-x^9+x^4-x+1={\frac {{x}^{13} \left( {x}^{2}+x+1 \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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use mathematical induction to show that $n^3 + 5n$ is divisible by $3$ for all $n\ge1$ What I have so far
Base: $n^3 + 5n$
Let $n=1$
$$
1^3 + 5(1) = 6
$$
$6$ is divisible by $3$
Induction step: $(k+1)^3 + 5(k+1)$
$(k^3 + 3k^2 + 8k + 6)$ is divisible by $3$
I kind of get lost after this point. For starters, how do I pro... | HINT
$$
\begin{split}
k^3 + 3k^2+8k+6
&= 3(k^2+3k+2) + k^3-k \\
&= 3(k^2+3k+2) + k(k^2-1) \\
&= 3(k^2+3k+2) + k(k-1)(k+1)
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Limit of $f(x)=x^4$ as $x\to a$ by definition
$\displaystyle\lim_{x\to a} x^4=a^4$
Note that $|x^4-a^4|=|x-a||x+a||x^2+a^2|$
If $|x-a|<1,$ then $|x+a|<1+2a.$
Then, $|x-a|<1$ implies $|x^4-a^4|<(1+2a)(|x^2+a^2|)$
Also, if $|x^2+a^2|<\frac{\varepsilon}{1+2|a|}$, then $|x^3-a^4|<\varepsilon$. Consequently, taking $\delt... | Assume $a>0$.Taking $|x-a|<\delta$ gives $0<a-\delta < x< a+\delta$, when $\delta < a$. So $0<x+a=|x+a|<2a+\delta<3a$.
Analogically $0< x^2+a^2 =|x^2+a^2|<(a+\delta)^2+a^2<5a^2$. Now you need $|x^4-a^4|<\delta 15a^3<\varepsilon$ . Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3848414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Three fair dice are rolled. What is the probability that the sum of the three outcomes is 10 given that the three dice show different outcomes? I know how to solve this question using conditional probability. I tried using another method of solving it which gives me another answer(which i know is wrong), help me find t... | You should divide the number of favourable cases, $3$ (that is $(2,3,5),(1,4,5),(1,3,6)$), by the number of possible distinct outcomes for three dice, $\binom{6}{3}$:
$$\frac{3}{\binom{6}{3}}=\frac{3}{20}.$$
or, equivalently, divide $3\cdot 3!$ by $6\cdot 5\cdot 4$:
$$\frac{3\cdot 3!}{6\cdot 5\cdot 4}=\frac{3}{20}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3848525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Please check my proof for $x^2+|x-2|>1$ Let $f(x)=x^2+|x-2|-1$.
If $x <2,$ then
$f(x)=x^2-x+1 \implies f'(x)=2x-1 \implies f_{min}=f(1/2)=\frac{3}{4} >1.$
If $x>2$, then $f(x)=x^2+x-3 \implies f'(x)=2x+1>0$. So the function is increasing for $x>2$
The function $f(x)$ has just one min, so $f(x)>f(1/3)=3/4 >0$, hence fo... | A standard way to prove such an inequality without derivatives would be to complete the square.
For $x \ge 2$ the case is simpler:
$$x^2 + |x-2| \ge 4 + 0 > 1$$
For $x < 2$ we have $|x - 2| = 2-x$. Hence:
$$\begin{align}x^2 + |x-2| &= x^2-x+2 \\&= x^2-x+\frac14 + \frac74 \\&= \left(x-\frac12\right)^2+\frac 74 \\&\ge \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3848989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Show that $a_{n+1} = \frac{1}{2}(a_n+b_n), $ $b_{n+1} = \frac{1}{2}(b_n+c_n) $and $c_{n+1} = \frac{1}{2}(c_n+a_n)$ are convergent,
Question: Let $a_0,b_0,c_0$ be real numbers.
Define the sequences $(a_n)_n,(b_n)_n, (c_n)_n$ recursively by
$$a_{n+1} = \frac{1}{2}(a_n+b_n), \quad b_{n+1} = \frac{1}{2}(b_n+c_n) \quad \te... | To expand on my comment. C_M's comment, together with your calculation, says we are looking at $XE_{11}X^{-1}$. The effect of $E_{11}$ is to multiply the first column of $X$ by the first row of $X^{-1}$.
Both of those are the eigenvector connected to $\lambda=1$, so are multiples of $(1,1,1)$. Suppose the first colum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at...
The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at...
What I tried...
The equation $x^2=y-6$ is of a parabola. To find the slope of the tangent to the parabola at the point $(1,7)$,
$$\... |
As you've simplified the question, we need to find
a point, where the line
\begin{align}
y&=2x+5
\tag{1}\label{1}
\end{align}
touches the circle of unknown radius
centered at $O=(-8,-6)$.
A convenient point on the tangent line below the center $O$ is $A(-8,-11)$, $|OA|=5$.
Let $\phi=\angle T_cOA$,
\begin{align}
\phi&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Show convergence of the following series. I honestly am not sure how to start. I think we will use convergence of geometric series. This section has to do with rearrangement of series.
Prove that if $$0 \leq x \lt 1$$ then $$\sum_{n=0}^\infty (n+1)x^n= (\frac{1}{1-x})^2$$
| Hints-We have:
$$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$
and the convergence is absolute. Hence
$$\sum_{n=0}^\infty nx^{n-1}=\sum_{n=0}^\infty \frac{d}{dx}x^n=\frac{d}{dx}\sum_{n=0}^\infty x^n=\frac{d}{dx}\frac{1}{1-x}=\frac{1}{\left(1-x\right)^2}.$$
Added: $$F(x) = \sum_{n=0}^\infty (n+1)x^n = 1+x + 2x^2 + 3x^3 + ...$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Simplify $\frac{x^3+1}{x+\sqrt{x-1}}$
Simplify $$A=\dfrac{x^3+1}{x+\sqrt{x-1}}.$$
Firstly, $x-1\ge0$ and $x+\sqrt{x-1}\ne0:\begin{cases}x-1\ge0\\x+\sqrt{x-1}\ne0\end{cases}.$ The first inequality is equivalent to $x\ge1$. Can we use that in the second inequality? I mean can we say that $x+\sqrt{x-1}>0$ because $x>0$ ... | The inequality $x>=1$ covers it all. You can simplify further by noting that
$\frac{1}{x+\sqrt{x-1}}=\frac{x-\sqrt{x-1}}{x^2-x+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Using integrating factors to solve difficult differential equations Consider the differential equatioN:
$$(2x^2 y -2y^4)dx+ ( 2x^3 + 3xy^3) dy = 0$$
This equation is of the form:
$$ Q dx + P dy=0$$
Now, it's easy to see that this differential is not exact by using the commutativity of second order partial condition. ... | $(2x^2y-2y^4)~dx+(2x^3+3xy^3)~dy=0$
$(2x^3+3xy^3)~dy=(2y^4-2x^2y)~dx$
$\dfrac{dy}{dx}=\dfrac{2y^4-2x^2y}{2x^3+3xy^3}$
Let $r=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=2x\dfrac{dy}{dr}$
$\therefore2x\dfrac{dy}{dr}=\dfrac{2y^4-2x^2y}{2x^3+3xy^3}$
$\dfrac{dy}{dr}=\dfrac{y^4-x^2y}{x^4+3x^2y^3}$
$\dfrac{dy}{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this exp... | Rationalization is a standard way to manipulate such kind of limits when they lead to an indeterminate form.
The aim for this kind of manipulation is to eliminate the term which leads to the indetermination, indeed by $(A-B)(A+B)=A^2-B^2 \implies A-B= \frac{A^2-B^2}{A+B}$ we have that
$$\sqrt{6-x}-2=\frac{2-x}{\sqrt{6-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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"answer_id": 0
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$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $ Calculate below limit
$$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $$
Using L'Hôpital's rule might be too tedious. I wonder if there is a trick given the ... | $$L=\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} }$$
With L'Hospital's rule:
$$L=\lim_{x \rightarrow 1} \frac{ 2x + 3/2x^2(x^3+1)^{-1/2} - 2x^3(x^4+1)^{-1/2} }{ 1 + 1/2(x+1)^{-1/2} -x (x^2+1)^{-1/2} }$$
$$L=\frac{ 2 + 3/2(2)^{-1/2} - 2(2)^{-1/2} }{ 1 + 1/2(2)^{-1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3862143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Range of $f(z)=|1+z|+|1-z+z^2|$ when $ |z|=1$
Find the Range of $f(z)=|1+z|+|1-z+z^2|$ when $z$ is a complex $ |z|=1$
I am able to get some weaker bounds using triangle inequality.
$f(z)< 1+|z|+1+|z|+|z^2|=5$
also $f(z)>|1+z+1-z+z^2|=|z^2+2|>||z^2|-2|=1$,but
these are too weak!.as i have checked with WA
Is there an ... | Let $z=x+yi,$ where $x$ and $y$ are reals.
Thus, $x^2+y^2=1$ and
$$|z+1|+|1-z+z^2|=\sqrt{(x+1)^2+y^2}+\sqrt{(1-x+x^2-y^2)^2+(-y+2xy)^2}=$$
$$=\sqrt{2+2x}+\sqrt{(2x^2-x)^2+(1-x^2)(2x-1)^2}=\sqrt{2+2x}+|2x-1|$$ and we got a function of one variable $-1\leq x\leq 1.$
Can you end it now?
I got that the maximal value it's $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove $(x^2 + xy + y^2)^2$ divides $(x + y)^{6n + 1} - x^{6n + 1} - y^{6n + 1}$. As stated in the title, I want to show the (bivariate) polynomial $g(x, y) = (x^2 + xy + y^2)^2$ divides the polynomial $f(x, y) = (x + y)^{6n + 1} - x^{6n + 1} - y^{6n + 1}$, where $n \geq 0$.
Naturally, I resorted induction. For $n = 1$,... | The key insight is that suffices to take $y=1$ because the polynomials are homogeneous. The precise details are below.
Write $z=x/y$.
Then $x^2 + xy + y^2 = y^2(z^2+z+1)$ and $(x + y)^{6n + 1} - x^{6n + 1} - y^{6n + 1}=y^{6n + 1}((z + 1)^{6n + 1} - z^{6n + 1} - 1)$.
Thus, it suffices to prove that $(z^2+z+1)^2$ divides... | {
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"timestamp": "2023-03-29T00:00:00",
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We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$ We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$
I just did the question above in the following way:
$t=\frac{\sqrt{6}+\sqrt{3}+1}{2}$
From Heron w... | $$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{3+1-6}{2\cdot\sqrt 3\cdot 1}=-\frac1{\sqrt 3}$$
Negative sign means $A>90^\circ$. If an angle is greater than $90^\circ$, the other two must be less.
$$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{6+1-3}{2\cdot{\sqrt 6}\cdot 1}=\frac{2}{\sqrt 6}$$
Then $$\sin B=\sqrt{1-\cos^2 B}=\sqrt{1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865515",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Show that the equation $\sqrt{x+5}+\sqrt{x}=2$ has no real roots
Show that the equation $$\sqrt{x+5}+\sqrt{x}=2$$ has no real roots.
What is the fastest approach to solve the problem? We have $$\sqrt{x+5}+\sqrt{x}=2 \iff \sqrt{x+5}=2-\sqrt{x}$$$$D_x:\begin{cases}x+5\ge0\\x\ge0\\2-\sqrt{x}\ge0\end{cases}\iff x\in[0;4]... | $$\sqrt{x+5}+\sqrt{x}=2\implies$$
$$\sqrt{x+5}=2-\sqrt{x}\implies$$
$$(\sqrt{x+5})^2=(2-\sqrt{x})^2\implies$$
$$x+5=4+x-4\sqrt{x}\implies$$
$$1=-4\sqrt{x}$$
which is impossible since $1>0$ and $-4\sqrt{x}\le0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the inequality $(1-a)^b>(1-b)^a$ if $1>b>a>0$ by using binomial expansion For this question Proving or disproving: If $0<a<b<1$, then $(1-a)^b>(1-b)^a$ it is already proved. But now I want to proof it using binomial expansion, please help to verify if the proof valid.
Given $1>b>a>0$ and let $x$ bigger or equal t... | You're on the right track. You have two base cases and now you need to prove your claim that it holds for the $n^{th}$ term using induction.
| {
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"url": "https://math.stackexchange.com/questions/3869986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Claculate limit $\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$ I have a problem to calculte this limit:
$$\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$$
I used Taylor expansion for $\sin(x), \cos(x)$ and considered also $1-\cos(\alpha)=2\sin^2(\frac{\alpha}{2})$ and $\alpha=2-2\sqrt{\f... | Since there is $x^4$ on denominator we have to go at least the same order on numerator.
$\sin(x)=x-\frac 16 x^3+\frac 1{120}x^5+o(x^5)$
$\dfrac{\sin(x)}x=1-\frac 16 x^2+\frac 1{120}x^4+o(x^4)$
$S=\left(\frac{\sin(x)}x\right)^\frac 12=1+\frac 12\left(-\frac 16 x^2+\frac 1{120}x^4+o(x^4)\right)-\frac 18\left(-\frac 16 x^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $\mathbb{E}\left[e^{\int_t^TB_\tau dB_\tau}\right]$ I just learned basic stochastic calculus and found it extremely hard to calculate $\mathbb{E}\left[e^{\int_t^TB_\tau dB_\tau}\right]$, where $B_\tau$ is a standard Brownian motion.
I can do the calculation that $\int_t^TB_\tau dB_\tau=\frac{1}{2}(B_T^2-B_t^2... | Writing $$B_T = (B_T-B_t)+B_t$$ we find that
$$\int_t^T B_s \, dB_s = \frac{1}{2} (B_T-B_t)^2+ (B_T-B_t)B_t - \frac{1}{2} (T-t).$$
By the independence of the increments of Brownian motion, we know that $B_t$ and $B_T-B_t$ are independent, and it follows that
\begin{align*} \mathbb{E}\exp\left( \int_t^T B_s \, dB_s \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870811",
"timestamp": "2023-03-29T00:00:00",
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Use mathematical induction to prove that (n+2)(n+3)(n+7) is divisible by 6. Use mathematical induction to prove that $q(n)=(n+2)(n+3)(n+7)$ is divisible by $6$.
I have already proved the base case at n=1. I need help on the second part to prove $n=k+1$.
What I did: $(n+2)(n+3)(n+7)=6P$
\begin{align*}
((k+2)+1)&((k+1)+3... | Hint:
$k^2+9k+18=(k+3)(k+6)$
Now if k is odd then $k+3$ is even.
If k is even then $k+6$is even. So $k^2+9k+18$is always even.
| {
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"url": "https://math.stackexchange.com/questions/3871936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $ f(f(x)) \geq 0$ for all real x Let $f(x)= a x^2 + x +1 , x \in \mathbb{R} $. Find all values of parameter $a \in \mathbb{R} $ such that $f(f(x)) \geq 0 $ holds for all real $x$.
$f(x)> 0 $ iff $a> 0 $ and $ 1- 4a \leq 0$ which gives $a \geq \frac{1}{4} $ . But we have:
$f(f(x))= a ( a x^2 + x +1)^2 + ... | First we require $a \ge 0$ because the quartic $f(f(x))$ has the term $a^3x^4$. If a is negative, this term will dominate at high x-values and make the quartic negative.
We look at the range of
$f(x)=ax^2+x+1$ by completing the square
$f(x) = a(x+\frac{1}{2a})^2 + 1 - \frac{1}{4a}$
So $f(x) \ge 1-\frac{1}{4a}$
Now we w... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I do not know how to solve for Induction Conclusion A sequence has $x_1=8$, $x_2=32$, $x_{n}= 2x_{n-1}+3x_{n-2}$ for $n \ge 3.$ Prove, for all $i$ of Naturals, $X_i = 2 (-1)^i + 10 \cdot 3^{i-1}$
I got bases covered, and I got the inductive step as $X_{i} = 2 * (-1)^k + 10 * 3^{k-1}$. I do not know how to follow up wit... | You were close.
The base cases of $X_1 = 8$ and $X_2 = 32$ check out.
The formula to be verified is
$$X_k = [2 \times (-1)^{k}] + [10 \times 3^{(k-1)}].$$
The algorithm for expressing $X_{(k+1)}$ in terms of $X_k$ and $X_{(k-1)}$ is
$$X_{(k+1)} = [2 \times X_k] + [3 \times X_{(k-1)}].$$
Inductively assume that the form... | {
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"timestamp": "2023-03-29T00:00:00",
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Partial fraction decomposition $\frac{1}{(x-y)^2}\frac{1}{x^2}$ I want to integrate $$\frac{1}{(x-y)^2}\frac{1}{x^2}$$ with respect to $x$.
I know that I have to apply partial fraction decomposition but which ansatz do I have to make to arrive at
$$\frac{1}{(x-y)^2}\frac{1}{x^2}=-\frac{2}{y^3(x-y)}+\frac{1}{y^2(x-y)^2}... | Treat $y$ as a constant. So in the denominator we have $2$ terms $x$ and $(x-y)$, both raised to the power $2$.
So, let $$\frac{1}{(x-y)^2x^2} = \left(\frac{a}{x} + \frac{b}{x^2}\right) + \left(\frac{c}{x-y}+\frac{d}{(x-y)^2}\right)$$
By method of inspection, we can easily get $b$ and $d$.
$x= 0 \Rightarrow b = \frac... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Integral $\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx$ vanishes Come across
$$I=\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx$$
and break the integrand as
$$\frac{1-x(2-\sqrt x)}{1-x^3}=\frac{1-x}{1-x^3}- \frac{x(1-\sqrt x)}{1-x^3}
$$
The first term simplifies and the second term transforms with $\sqrt x\to x$. Then, t... | Note
\begin{align}
&\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx=\int_0^\infty \frac{1+x^{\frac32}-2x}{1-x^3}dx\\= &\int_0^\infty \frac{1+x^{\frac32}}{1-x^3}dx
-\int_0^\infty \underset{x\to x^{\frac12} }{\frac{2x}{1-x^3}}dx= \int_0^\infty \frac{dx}{1-x^{\frac32}}
-\int_0^\infty \frac{dx}{1-x^{\frac32}}=0
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3879991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find a useful variable change for this integral I would like to find the area of the following region
$$
D=\left \{(x,y): -\sqrt{1+y^2}\leq x\leq \sqrt{1+y^2}; -1\leq y\leq (x+1)/2\right \}.
$$
I try to calculate the double integral brute force, but following this path
I came across some very unpleasant integral... | split the area into 5 separate areas:
$$\int_{\pi}^{\pi+\tan^{-1}(1/\sqrt{2})}\frac{1}{2}r^2d\theta+
\int_{\pi+\tan^{-1}(1/\sqrt{2}))}^{-\tan^{-1}(1/\sqrt{2}))}\frac{1}{2}r^2d\theta+
\int_{-\tan^{-1}(1/\sqrt{2}))}^{0}\frac{1}{2}r^2d\theta+
\int_{0}^{\tan^{-1}(4/5)}\frac{1}{2}r^2d\theta+
\int_{\tan^{-1}(4/5)}^{\pi}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3882212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the nature of $\sum_{n = 2}^\infty (\sqrt{n + 1} - \sqrt{n})^\alpha \ln \frac{n + 1}{n - 1}$ I need to find whether the following series converges or diverges:
$$\sum_{n = 2}^\infty (\sqrt{n + 1} - \sqrt{n})^\alpha \ln \frac{n + 1}{n - 1}$$
By graphing the sum, it seems that it converges if and only if $\alpha \gt... | You have $$(\sqrt{n + 1} - \sqrt{n})^\alpha \ln \left(\frac{n + 1}{n - 1} \right)= (\sqrt{n})^{\alpha}\left(\sqrt{1 + \frac{1}{n}} - 1\right)^\alpha \ln \left(1 +\frac{2}{n - 1} \right)$$
$$ \sim (\sqrt{n})^{\alpha}\left(\frac{1}{2n}\right)^\alpha\frac{2}{n-1} = \frac{2}{(2\sqrt{n})^{\alpha}(n+1)} \sim \frac{2^{1-\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3883312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
(AIME 1994) $ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $
$($AIME $1994)$ Find the positive integer $n$ for which
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $$ where $\lfloor x \rflo... | We'll find a maximal $m$ for which $$\sum_{k=1}^mk2^k\leq1994.$$
Indeed, $$\sum_{k=1}^mk2^k=2\sum_{k=1}^mk2^{k-1}=2\left(\sum_{k=1}^mx^{k}\right)'_{x=2}=2\left(\frac{x(x^m-1)}{x-1}\right)'_{x=2}=$$
$$=2\cdot\frac{(m+1)2^m-1-2^{m+1}+2)}{(1-1)^2}=(m+1)2^{m+1}-2^{m+2}+2.$$
Id est, $$(m+1)2^{m+1}-2^{m+2}+2\leq1994,$$ which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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