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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 $f$ is strictly convex in some neighbourhood of $0$.
Given $c \in \mathbb R$, we say that $f$ is midpoint-convex at the point $c$ if
$$
f((x+y)/2) \le (f(x) + f(y))/2,
$$
whenever $(x+y)/2=c$, $x,y \in \mathbb R$.
Question: Let $r<s<0$, and suppose that $f$ is midpoint-convex at $r$. Is $f$ midpoint-convex at $s$?
In the example given here (essentially $(f(x)=-x^3$) $f$ is concave after its global minimum point, while here I assume that there is a convex neighbourhood.
| 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 $y \in \mathbb{R}$;
iv) $f$ is strictly convex in some neighbourhood of $0$.
Solution: Let $f(x) = x^4 + bx^3 + cx^2$. Clearly $f(0)=0$.
We claim that $f$ satisfies i), ii), iii) and iv) if
\begin{align}
32c - 9b^2 &> 0, \tag{1}\\
6r^2 + 3br + c &> 0, \tag{2}\\
6s^2 + 3bs + c &< 0. \tag{3}
\end{align}
Indeed, first, we have $f'(x) = x(4x^2 + 3bx+2c)$, and if $32c - 9b^2 > 0$,
then $4x^2 + 3bx+2c > 0$ for all $x\in \mathbb{R}$, so i) is satisfied;
second, we have $f(x) + f(2r-x) - 2f(r) = 2(x-r)^2((x-r)^2 + 6r^2 + 3br + c)$,
and if $6r^2 + 3br + c > 0$, then ii) is satisfied;
third, we have $f(y) + f(2s-y) - 2f(s) = 2(y-s)^2((y-s)^2 + 6s^2 + 3bs + c)$,
and if $6s^2 + 3bs + c < 0$, then iii) is satisfied.
fourth, we have $f''(x) = 12x^2 + 6bx + 2c$, and if $c > 0$ (follows from $32c - 9b^2 > 0$),
then $f''(0) = 2c > 0$ and $f$ is strictly convex in some neighbourhood of $0$ (due to continuity of $f''(x)$).
Then, we prove that there exist $b, c$ such that (1), (2) and (3) are satisfied.
We simply choose $b = -3s$ and
$$c = 3s^2 - \frac{1}{2}\min\left(3(2r-s)(r-s), \ \frac{15}{32}s^2\right).$$
Indeed, first, since $c \ge 3s^2 - \frac{1}{2}\cdot \frac{15}{32}s^2$,
we have $32c - 9b^2 = \frac{15}{2}s^2 > 0$;
second,
since $c \ge 3s^2 - \frac{1}{2} \cdot 3(2r-s)(r-s)$, we have
$6r^2 + 3br + c \ge \frac{3}{2}(2r-s)(r-s) > 0$;
third, since $c < 3s^2$,
we have $6s^2 + 3bs + c < 6s^2 + 3(-3s)s + 3s^2 = 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 points of the incircle.
Then, your expression becomes (after some simple algebra)
$$\frac{1+3\,p/s}{2\,(1+p/s)},$$ where $s=u^2+v^2+w^2$ and $p=uv+vw+wu.$ Obviously, $0<p\le s,$ so the range is $\left(\frac12,1\right].$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can someone explain the limit $\lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)$? $$\begin{aligned}
&\text { Find the following limit: } \lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)\\
&\text { **In a solution I got** Ans:- Let } u_n=\frac{n}{\sqrt{n^2+n}}\\
&\therefore \lim _{n \rightarrow \infty} u_n=\lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^{2}+n}}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{n}}}=1\\
&\text { By Cauchy's first theorem:- } \lim _{n \rightarrow \infty} \left(\frac{u_1+\cdots+u_n}{n}\right)=1\\
&\text{So, } \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)=1
\end{aligned}$$
I am unable to understand this. How the general term comes like this! Any such explanation to the whole problem would be greatly appreciated.
| 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
$$\sum_{k=1}^nu_k=\sum_{k=1}^n\frac{k}{\sqrt{k^2+k}}$$
| {
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"url": "https://math.stackexchange.com/questions/3741401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 $\sum_{m=0}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right) = -1, \sum_{n=0}^{\infty}\left(\sum_{m=0}^{\infty}a_{m,n}\right) = 1$.
Here is my attempt for $\sum_{m=0}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right)$.
$$\sum_{m=0}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right)= \sum_{m=1}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n}.$$ Then, we have
\begin{align*}
\sum_{m=1}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right) &=
\sum_{m=1}^{\infty}\left(\sum_{n=1}^{\infty}\left[\frac{m-n}{2^{m+n}}\frac{(m+n-1)!}{m!n!}\right]+\frac{1}{2^m}\right) \\
&= \sum_{m=1}^{\infty}\left(\sum_{n=1}^{\infty}\left[\frac{m}{2^{m+n}}\frac{(m+n-1)!}{m!n!}\right]-\sum_{n=1}^{\infty}\left[\frac{n}{2^{m+n}}\frac{(m+n-1)!}{m!n!}\right]+\frac{1}{2^m}\right) \\
&= \sum_{m=1}^{\infty}\left(1-\frac{1}{2^m}-1+\frac{1}{2^m} \right) \\
&= 0.
\end{align*}
See below about how I get $0$.
\begin{align*}
\sum_{n=1}^{\infty}\frac{m}{2^{m+n}}\frac{(m+n-1)!}{m!n!}
&= \frac{1}{2^m} \sum_{n=1}^{\infty} \frac{(m+n-1)!}{(m-1)!n!}\left(\frac{1}{2}\right)^n \\
&= \frac{1}{2^m} \sum_{n=1}^{\infty}{m+n-1 \choose n}\left(\frac{1}{2}\right)^n \\
&= \frac{1}{2^m} \sum_{n=1}^{\infty}(-1)^n{-m \choose n}\left(\frac{1}{2}\right)^n \\
&=
\frac{1}{2^m}\left[(1-\frac{1}{2})^{-m}-1\right] (\text{binomial expansion})\\
&= \frac{1}{2^m}\left[2^m-1\right] \\
&= 1-\frac{1}{2^m}.
\end{align*}
Similarly,
\begin{align*}
\sum_{n=1}^{\infty}\frac{n}{2^{m+n}}\frac{(m+n-1)!}{m!n!} &=
\frac{1}{2^m}\sum_{n=1}^{\infty}\frac{1}{2^{n}}\frac{(m+n-1)!}{m!(n-1)!} \\
&= \frac{1}{2^m}\sum_{n=1}^{\infty}{m+n-1 \choose n-1}\frac{1}{2^{n}} \\
&= \frac{1}{2^m}\sum_{n=1}^{\infty}(-1)^{n-1}{-(m+1) \choose n-1}\frac{1}{2^{n-1}}\frac{1}{2} \\
&=\frac{1}{2^m}\frac{1}{2}\left(1-\frac{1}{2}\right)^{-(m+1)} \\
&= \frac{1}{2^{m+1}}2^{m+1}\\
&=1.
\end{align*}
Thus, we have
$$\sum_{m=0}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right)= \sum_{m=1}^{\infty}\left(\sum_{n=0}^{\infty}a_{m,n}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n} = 0-1.$$
I just answer my own question.
| 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 $\sum_{n=1}^\infty -\frac{1}{2^n}=-1$, which cancel.
| {
"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 the solution it is given that according to the given information, $f(x)$ is a polynomial of degree 4 with leading coefficient 3.
$f(x)=3(x-2)(x-3)(x+1)(x+6)+(x^2+1)$
Now what i think is that he got the coefficient 3 using the limits and the rest of the terms$[(x-2)(x-3)(x+1)(x+6)]$ from the respective $f(x)$ values from what was given and also the $(x^2+1)$ was added for the remainders, as they satisfy all the values that have been given to us in the question. I just wanted to know if my approach is right? or is there another elegant way to do it?.
| 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` Hospital rule),we see that $f^4(\infty)=72$ which implies that $ f(x)$ must be four degree polynomial
Given that $f(2)=5,f(3)=10,f(-1)=2,f(-6)=37$
Observe that output of function is input squared plus 1, so we must have $x^2+1$ involving in function
Let $f(x)=k(x-2)(x-3)(x+1)(x+6)+x^2+1$, since we don't know leading cofficient
Using, $f^4(\infty)=72$ ,we got $k=3 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 $1 \leq b < a$. One of these two, one is odd, and the other one is even.
I would like to know the intermediate steps that are required to find such a parametrization. I tried to manipulate the identity.
$$4x^2+y^4=z^2 \implies 4x^2=(z+y^2)(z-y^2)$$ and then using parity check to further simplify it.
Another approach was
$$4x^2+y^4=z^2 \implies 4xy^2=(2x+y^2+z)(2x+y^2-z)$$
I could not make any more progress. Some help will be appreciated.
| 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
$$2x = 2mn \implies x = mn \tag{1}\label{eq1A}$$
$$y^2 = m^2 - n^2 \implies y^2 + n^2 = m^2 \tag{2}\label{eq2A}$$
$$z = m^2 + n^2 \tag{3}\label{eq3A}$$
where $m$, $n$ are relatively prime and $1 \leq n < m$.
Note \eqref{eq2A} shows $y$, $n$ and $m$ form another Pythagorean triple. In this case, it's another primitive solution since $\gcd(m,n) = 1$ means $y$ is also relatively prime to $m$ and $n$. Also, since $y$ is odd (due to $2x$ being even in the original equation), you get
$$y = a^2 - b^2 \tag{4}\label{eq4A}$$
$$n = 2ab \tag{5}\label{eq5A}$$
$$m = a^2 + b^2 \tag{6}\label{eq6A}$$
where $a$, $b$ are relatively prime and $1 \leq b < a$. Plugging \eqref{eq5A} and \eqref{eq6A} into \eqref{eq1A} gives
$$x = 2ab(a^2 + b^2) \tag{7}\label{eq7A}$$
while plugging \eqref{eq5A} and \eqref{eq6A} into \eqref{eq3A} gives
$$\begin{equation}\begin{aligned}
z & = (a^2 + b^2)^2 + (2ab)^2 \\
& = a^4 + 2a^2b^2 + b^4 + 4a^2b^2 \\
& = a^4 + 6a^2b^2 + b^4
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 seems to have got me nowhere.
| $\frac{1-t^2}{1+t^2} +\frac{1-s^2}{1+s^2}=1$
$1+s^2-t^2-s^2t^2+1+t^2-s^2-s^2t^2=1+s^2+t^2+s^2t^2$
$\Rightarrow 1-s^2-t^2-3s^2t^2=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754249",
"timestamp": "2023-03-29T00:00:00",
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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
$$-1\leqslant\sin x\leqslant1\;\text{and}-1\leqslant\cos x\leqslant1$$
I think I need to have only one term.
Edit : I have made some more progress
$$-3\leqslant 3\cos x\leqslant 3$$
$$\therefore -4\leqslant 3\cos x+\sin x\leqslant 4$$
$$ 0\leqslant (3\cos x+\sin x)^2\leqslant 16$$
| 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
$$\left(-\infty,\frac5{2-\sqrt{10}}\right]\cup\left[\frac5{2+\sqrt{10}},\infty\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755235",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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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 help, to start is very appreciated because I don't know how to tackle those.
| 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
$$\begin{align} x(t) &= \sin \left(t^2\right)+\cos \left(t^2\right)\\ y(t) &= \cos \left(t^2\right)-\sin \left(t^2\right)\\ z(t) &= 2 t \end{align}$$
Using eighteen terms, we can write this as a series as
$x(t) = (t^2-\frac{t^6}{6}+\frac{t^{10}}{120}-\frac{t^{14}}{5040}+\frac{t^{18}}{362880}+O\left(t^{19}\right)) + (1-\frac{t^4}{2}+\frac{t^8}{24}-\frac{t^{12}}{720}+\frac{t^{16}}{40320}+O\left(t^{19}\right))$
$y(t) = (1-\frac{t^4}{2}+\frac{t^8}{24}-\frac{t^{12}}{720}+\frac{t^{16}}{40320}+O\left(t^{19}\right)) -(t^2-\frac{t^6}{6}+\frac{t^{10}}{120}-\frac{t^{14}}{5040}+\frac{t^{18}}{362880}+O\left(t^{19}\right))$
$z(t) = 2t$
The Picard-Lindelöf iteration is given by
$$
f\left(\begin{pmatrix} x \\ y \\ z\end{pmatrix}\right) = \begin{pmatrix} y z \\ - x z \\ 2\end{pmatrix}, ~~~~\phi_0 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$$
Then
$$
\phi_1(t) = \phi_0 + \int_0^t f(\phi_0(s)) \,ds \\
= \phi_0 + \int_0^t f\left(\begin{pmatrix} 1\\1 \\ 0\end{pmatrix}\right) \,ds \\
= \phi_0 + \int_0^t \begin{pmatrix} 0 \\ 0 \\ 2\end{pmatrix} \,ds \\
= \begin{pmatrix}1 \\ 1\\ 2t\end{pmatrix}
$$
$$
\phi_2(t) = \phi_0 + \int_0^t f(\phi_1(s)) \,ds \\
= \phi_0 + \int_0^t f\left(\begin{pmatrix} 1\\1 \\ 2t\end{pmatrix}\right) \,ds \\
= \phi_0 + \int_0^t \begin{pmatrix} 2s \\ -2s \\ 2\end{pmatrix} \,ds \\
= \begin{pmatrix} 1+t^2 \\ 1-t^2 \\ 2t\end{pmatrix}
$$
You can compare that to the series written above and see that we are converging to that solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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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:
$u^2+3u-4$ =
master term is 1 * -4 = -4. Factors that give minus 4 and sum to 3 are 4 and -1...
$(u^2-u)+(4u-4)$ =
$u(u-1)+4(u-1)$ =
$(u+4)(u-1)$
I don't know where to go from here. If I write $u$ back into it's original $x^2$ I get:
$(x^2+4)(x^2-1)$
Where do I go from here to arrive at x intercepts of 1 and -1?
| $$2x^4+6x^2-8=2(x^2+4)(x^2-1)=2(x^2+4)(x-1)(x+1)=0$$
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",
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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 binomial expansion for this; I just don't know how.
Can anyone help me with this?
| $\dfrac{24}{7(x-4)}=\dfrac{-6}{7\left(1-\frac x4\right)}=-\dfrac67\left(1+\frac x4+(\frac x4)^2+(\frac x4)^3+\cdots\right)$
$\dfrac{24}{7(x-(-3))}=\dfrac{8}{7\left(1-\left(-\frac x3\right)\right)}=\dfrac87\left(1-\frac x3+(\frac x3)^2-(\frac x3)^3+\cdots\right)$
$\therefore\dfrac{24}{(x-4)(x+3)}=-2+\dfrac16x-\dfrac{13}{72}x^2 +\dfrac{25}{864}x^3\cdots$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 to show that $\frac{d}{dx}\cos(x)=\sin(x)$.
Wondering if you can do the same with the inverse functions.
| 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
$$\lim_{x \rightarrow 0} {\sin x} = x$$
You wouldn't even need to use $\sin'(x) = \cos(x)$
Sorry for the bad drawing and not drawing a unit circle:(
If carefully notice
$$\theta_1 + \theta_2 = \arcsin(x+h)$$
As in other differential equations we will take $h \rightarrow 0$ at the end
$$\theta_2 = \arcsin(x+h)-\arcsin(x)$$
$$AO' = \sin(\theta_1 + \theta_2)$$
$$AO' = \sin(\arcsin(x+h)) = x+h$$
$$CB = AO' - AO = x+h -x = h$$
$$AC = OO'$$
$$AC = \cos(\arcsin(x))-\cos(\arcsin(x+h) )$$
$$AC = \sqrt{1-x^2} - \sqrt{1-(x+h)^2}$$
$$AC = h \lim_{h\rightarrow 0}\frac{\sqrt{1-x^2} - \sqrt{1-(x+h)^2}}{h}$$
$$AC =- h \frac{d}{dx}(\sqrt{1-x^2})$$
Use Chain Rule
$$AC = -h\frac{-x}{\sqrt{1-x^2}}$$
$$AC = h\frac{x}{\sqrt{1-x^2}}$$
Using Pythagoras Theorem for ABC
$$AB^2 = AC^2 + BC^2$$
$$AB^2 = h^2\frac{x^2}{ 1-x^2}+ h^2$$
$$AB = h\sqrt{\frac{x^2 +1-x^2}{1-x^2}}$$
$$AB = h\sqrt{\frac{ 1 }{1-x^2}}$$
$$AB = \sin(\theta_2)$$
$$AB = \lim_{h\rightarrow 0}\sin(\arcsin(x+h)-\arcsin(x) )= \lim_{\theta_2 \rightarrow 0}\sin(\theta_2)$$
$$AB = \theta_2$$
$$\theta_2 = h \frac{ 1 }{\sqrt{1-x^2}}$$
$$\theta_2/h = \frac{ 1 }{\sqrt{1-x^2}}$$
$$\lim_{h \rightarrow 0} \frac{\arcsin(x+h)-\arcsin(x)}{h} = \frac{ 1 }{\sqrt{1-x^2}}$$
Now you can do the same for $\arccos (x)$, but since this proof was done by only using $\sin(x)/x = 1$,
Let $\theta_2 = \arccos(x-h) - \arccos(x)$
You can also assume h to be positive but this is a lot easier
$B = x$
$P = \sqrt{1-x^2}$
$H= 1$
$$BC = \sqrt{1-(x-h)^2}- \sqrt{1-x^2}$$
$$BC^2 = h^2 \frac{x^2}{ {1-x^2}}$$
$$AC^2 = h^2$$
AB remains the same
$$AB = \theta_2 = h \frac{ 1 }{\sqrt{1-x^2}}$$
$$\lim_{h \rightarrow 0} \frac{\arccos(x-h)-\arccos(x)}{h} = \frac{ 1 }{\sqrt{1-x^2}}$$
But....
$$\lim_{h \rightarrow 0} \frac{\arccos(x+h)-\arccos(x)}{h}= \lim_{h \rightarrow 0} \frac{\arccos(x-h)-\arccos(x)}{-h}$$
$$\frac{\mathsf d}{\mathsf {dx}}(\arccos(x)) = -\frac{ 1 }{\sqrt{1-x^2}}$$
I found out that arctan also requires a similar approach
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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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
$$x_1^3 + x_2^3 + x_3^3 + x_4^3 \ge \frac1{x_1}+ \frac1{x_2} +\frac1{x_3}+\frac1{x_4}.$$
Since $x↦x^3$ is convex on R+ by Jensen's Inequality we have $x_1^3+x_2^3+x_3^3+x_4^3≥4^{-2}(x_1+x_2+x_3+x_4)^3$ then using AM/GM inequality and since $x_1x_2x_3x_4=1,$ we can show that $(x_1+x_2+x_3+x_4)^3≥4$
Many thanks for your help.
| 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:
$$\sum_{cyc}x_1^3=\frac{1}{6}\sum_{cyc}(3x_1^3+x_2^3+x_3^3+x_4^3)\geq\frac{1}{6}\sum_{cyc}6\sqrt[6]{x_1^9x_2^3x_3^3x_4^3}=\sum_{cyc}x_1.$$
Also, the Tangent Line method works:
$$\sum_{cyc}(x_1^3-x_1)=\sum_{cyc}\left(x_1^3-x_1-2\ln{x_1}\right)\geq0.$$
Indeed, let $f(x)=x^3-x-2\ln{x},$ where $x>0$.
Thus, $$f'(x)=3x^2-1-\frac{2}{x}=\frac{(x-1)(3x^2+3x+2)}{x},$$ which gives $x_{min}=1$, $$f(x)\geq f(1)=0$$ and we are done!
The second inequality we can prove by similar ways.
| {
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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 $\phi$
$$\frac{\tan \phi +1}{\tan \phi-1}=\sqrt{\frac{1+x^2}{1-x^2}}$$
$$\frac{1+\tan^2\phi +2\tan \phi}{1+\tan^2 \phi-2\tan \phi}=\frac{1+x^2}{1-x^2}$$
$$\frac{1+\tan^2\phi}{2\tan \phi }=\frac{1}{x^2}$$
$$\sin 2\phi=x^2$$
$$\phi=\frac{\pi}{4}-\frac 12 \cos^{-1}x^2$$
Where am I going wrong?
| 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: $$\tan\left(\frac{\pi}{4}+\frac 12 \cos^{-1}x^2\right)=\frac{1+\tan\frac{1}{2}\arccos{x^2}}{1-\tan\frac{1}{2}\arccos{x^2}}=$$
$$=\frac{\cos\frac{1}{2}\arccos{x^2}+\sin\frac{1}{2}\arccos{x^2}}{\cos\frac{1}{2}\arccos{x^2}-\sin\frac{1}{2}\arccos{x^2}}=\frac{\sqrt{\frac{1+x^2}{2}}+\sqrt{\frac{1-x^2}{2}}}{\sqrt{\frac{1+x^2}{2}}-\sqrt{\frac{1-x^2}{2}}}=\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}.$$
Your mistake in the last line.
Indeed, since $$\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}>1$$
and from here $$\frac{\pi}{4}<\phi<\frac{\pi}{2},$$ we obtain:
$$2\phi=\pi-\arcsin{x^2}=\frac{\pi}{2}+\arccos{x^2}.$$
| {
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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,
$$\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx=\frac{\pi }{8}\ln \left(2\right)-\frac{3\pi }{8}+\frac{\pi }{3}\ln \left(2+\sqrt{3}\right)-\frac{G}{6}$$
And its been proven here.
But i'd like to know how to evaluate this without complex analysis.
One of the answers uses differentiation under the integral sign directly and partial fraction decomposition on a similar integral, but doing it that way doesnt help me with this case here
I tried to evaluate this way but got stuck,
$$\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx=\int _0^1\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx+\int _1^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx\:\:\:\:\:\: \text{then sub}\:\:x=\frac{1}{t}\:\:\text{for the 2nd integral}$$
$$=\int _0^1\frac{\ln \left(t^3+1\right)}{\left(t^2+1\right)^2}\:dt+\int _0^1\frac{t^2\ln \left(t^3+1\right)}{\left(t^2+1\right)^2}\:dt-3\int _0^1\frac{t^2\ln \left(t\right)}{\left(t^2+1\right)^2}\:dt$$
$$=\int _0^1\frac{\ln \left(t^3+1\right)}{t^2+1}\:dt+3G+3\int _0^1\frac{\ln \left(t\right)}{\left(t^2+1\right)^2}\:dt$$
I managed to evaluate the last integral expanding the denominator but i cant think of a way to evaluate the 1st integral, please help me.
| 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
$$\int_0^1\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4},$$
and the second integral is already calculated in the comments by Dennis or it can be found calculated in details by Zacky in this solution (see the integral $J$);
$$\int_0^1\frac{\ln(1+3x^2)}{1+x^2}dx=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac{\pi}{4}\ln 2-\frac53G.$$
For the third integral, let $x\mapsto \frac{1-x}{1+x}$
$$\int_0^1\frac{\ln(1+x)}{1+x^2}dx=\int_0^1\frac{\ln\left(\frac{2}{1+x}\right)}{1+x^2}dx=\ln2\int_0^1\frac{dx}{1+x^2}-\int_0^1\frac{\ln(1+x)}{1+x^2}dx$$
$$\Longrightarrow \int_0^1\frac{\ln(1+x)}{1+x^2}dx=\frac{\pi}{8}\ln2$$
Combine the three results we have
$$\int_0^1\frac{\ln(1+t^3)}{1+t^2}dt=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac{\pi}{8}\ln 2-\frac53G.$$
All credit goes to Zacky and Dennis for evaluating the second integral as its the key to crack the integral in the question.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $\dfrac{m+1}{k} = \dfrac{5+1}{2} = 3 \in \mathbb{Z}$ we will use the substitution: $$ a+bx^k = u^{\frac{1}{n}}$$
Therefore,
$$ u^3 = 1 + x^2 \iff x = \sqrt{u^3 +1} \text{ (Mistake Here. Check the comments) } $$ $$\iff dx = \frac{3u^2}{2\sqrt{u^3}+1}$$
So the new integral is,
$$ \int x^5 (1+x^2)^{\frac{2}{3}} dx = \frac{3}{2} \int u^4 (u^3+1)^{\frac{7}{6}} du$$
Instead of simplifying the integral, the substitution did nothing by keeping it at the same form, with different values on the variables $m,k,n$.
I tried to substitute once again and it doesn't seem to lead in any known paths, anytime soon.
Any ideas on how this could be evaluated?
| 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
$$
∫ x^5 \,(1+x^2)^{2/3}\,\mathrm d x = \frac{3}{5}∫ (u^{3/5}-1)^2 \,\mathrm d x.
$$
| {
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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:
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx = $$
$$ \int \left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right) dx = $$
$$\int \left(\frac{(\sqrt{1+x})^2-(\sqrt{1-x})^2}{(\sqrt{1+x})^2-(\sqrt{1-x})^2}\right)\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)dx = $$
$$\int 1*\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)dx $$
That's a dead end.
I also tried other conjugate approaches (only the numerator, only the denominator etc) with no better luck.
Any ideas?
| 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:
$$I=\int \left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) dx$$
$$I=\int \frac{{(1+x)}+(1-x)+2\sqrt{1-x^2}}{2x} dx$$
$$I=\ln{|x|}+\int \frac{\sqrt{1-x^2}}{x} \; dx$$
Let $x=\sin{\theta}$ to evaluate the integral:
$$I=\ln{|x|}+\int \frac{\cos^2{\theta}}{\sin{\theta}} \; d\theta $$
$$I=\ln{|x|}+\int \csc{\theta} \; d\theta - \int \sin{\theta} \; d\theta$$
$$I=\ln{|x|}-\ln{\big | \csc{\theta}+\cot{\theta}\big |}+ \cos{\theta}+C$$
Substitute $\theta$ back for $x$:
$$I=2\ln{|x|}-\ln{\big | \sqrt{1-x^2}+1\big |} + \sqrt{1-x^2}+C$$
| {
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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 $\:v=\sqrt{2}w$
$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)^2+2}\sqrt{2} dw$$$$= \frac{1}{2} \int \frac{1}{\sqrt{2}\left(w^2+1\right)}dw$$$$ = \frac{1}{2\sqrt{2}}\arctan \left(w\right) + C$$$$= \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C$$
Therefore,
$$\int \frac{1}{\cos 2x+3} dx = \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C $$
That's a decent solution but I am wondering if there are any other simpler ways to solve this (besides Weierstass). Can you come up with one?
| Long way but doable
$$I=\int \frac{dx}{\cos (2x)+3}$$
Let
$$\cos(2x)=t \implies x=\frac{1}{2} \cos ^{-1}(t)\implies dx=-\frac{1}{2 \sqrt{1-t^2}}$$
$$I=-\frac{1}{2}\int \frac{dx}{(t+3) \sqrt{1-t^2}}=-\frac{1}{2}\int \frac{\sqrt{1-t^2}}{(t+3) (1-t^2)}\,dt$$
$$\frac{1}{(t+3) (1-t^2)}=-\frac{1}{8 (t+1)}+\frac{1}{16 (t+3)}+\frac{1}{16 (t-1)}$$ and we face three integrals
$$J_a=\int \frac {\sqrt{1-t^2} }{t+a}$$
$$J_a=-\sqrt{1-a^2} \log \left(\sqrt{1-a^2} \sqrt{1-t^2}+a t+1\right)+\sqrt{1-a^2} \log (a+t)+a \sin ^{-1}(t)+\sqrt{1-t^2}$$
This leads to
$$I=\frac{i \left(\log (t+3)-\log \left(2 i \sqrt{2-2 t^2}+3 t+1\right)\right)}{4 \sqrt{2}}=-\frac{i}{4 \sqrt{2}}\log \left(1+i\frac{2 \sqrt{2} \sqrt{1-t^2}}{t+3}\right)$$
| {
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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 :$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\right),\left(\cos \left(\frac{4 \pi}{5}\right)+i \sin \left(\frac{4 \pi}{5}\right)\right),\left(\cos \left(\frac{6 \pi}{5}\right)+i \sin \left(\frac{6 \pi}{5}\right)\right), \left(\cos \left(\frac{8 \pi}{3}\right)+i \sin \left(\frac{8 \pi}{5}\right)\right)$
| Good question. Shall we try?
So we have the five roots: $$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\right),\left(\cos \left(\frac{4 \pi}{5}\right)+i \sin \left(\frac{4 \pi}{5}\right)\right),\left(\cos \left(\frac{6 \pi}{5}\right)+i \sin \left(\frac{6 \pi}{5}\right)\right), \left(\cos \left(\frac{8 \pi}{3}\right)+i \sin \left(\frac{8 \pi}{5}\right)\right)$$
I am just going to write $\frac{2\pi}{5}$ as $\alpha$. It's going to save time and make this easier to read. In this notation, the roots are :
$1, \ \cos (\alpha) + i\cdot\sin(\alpha), \ \cos (2\alpha) + i\cdot\sin(2\alpha), \ \cos (3\alpha) + i\cdot\sin(3\alpha)$ and $\cos (4\alpha) + i\cdot\sin(4\alpha)$
OK let's add them all up, we'll start with the imaginary part because that is going to be easy. We have $$i \times \left[\sin \alpha+\sin 2\alpha+\sin 3\alpha+\sin 4\alpha\right]$$ Well we know that $5\alpha = 2\pi$ and so $\sin \alpha = \sin(2\pi-\alpha) = \sin(-4 \alpha) =-\sin (4\alpha$). Also $\sin(2 \alpha)=-\sin(3 \alpha)$ for similar reasons and so all of these terms cancel to zero.
In the real part, we have $$1+\cos(\alpha)+\cos(2\alpha)+\cos(3\alpha)+\cos(4\alpha)$$ $\cos(\alpha)=\cos(2\pi-\alpha)=\cos(4\alpha)$ Also $\cos(2 \alpha)=\cos(3 \alpha)$ for similar reasons. So we just need to show that $$1+2\cos(\alpha)+2\cos(2\alpha)=0$$
How hard can this be? We know, from the trig identities that $\cos(2\alpha)=2 \cos^{2} (\alpha)-1$. So now we need $1+2 \cos(\alpha)+4 \cos^{2}(\alpha)-2=0$ or $$4 \cos^{2}(\alpha)+2 \cos(\alpha)-1=0$$
Trigonometry tells us that $\cos \alpha = \frac{\sqrt 5-1}{4}$. If we pop this in then we get.
$$\frac{6-2 \sqrt 5}{4}+\frac{\sqrt 5-1}{2}-1=0$$
Hooray and Phew.
| {
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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:
$1=x_0<x_1<...<x_{N-1}<x_N=2$
$x_n:=\frac{n}{N}+1,\,\,\,n\in\mathbb{N}_0$
That gives us:
$$\sum\limits_{n=1}^N(x_n-x_{n-1})\frac{1}{x_n}=\sum\limits_{n=1}^N \left(\frac{n}{N}+1-\left(\frac{n-1}{N}+1\right)\right)\frac{1}{\frac{n}{N}+1}=\sum\limits_{n=1}^N \frac{1}{N}\frac{N}{N+n}=\sum\limits_{n=1}^N\frac{1}{N+n}$$
We know from the definition, that:
$$\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N(x_n-x_{n-1})\frac{1}{x_n}=\int\limits_1^2 \frac{dx}{x}$$
Now we show that,
$$\int\limits_1^2 \frac{dx}{x}=\ln(2)$$
First we choose another Rieman sum with:
$1=x_0<x_1<...<x_{N-1}<x_N=2$
$x_n:=2^{\frac{n}{N}},\,\,\,n\in\mathbb{N}_0$
We get:
$$\sum\limits_{n=1}^N(x_n-x_{n-1})\frac{1}{x_n}=\sum\limits_{n=1}^N\left(2^{\frac{n}{N}}-2^{\frac{n-1}{N}}\right)\frac{1}{2^{\frac{n-1}{N}}}=\sum\limits_{n=1}^N 2^{\frac{1}{N}}-1=N\left(2^{\frac{1}{N}}-1\right)$$
Since we know that (with $x \in \mathbb{R})$:
$$\lim\limits_{x\rightarrow0}\frac{2^x-1}{x}=\ln(2)\Longrightarrow \lim\limits_{x\rightarrow \infty}x(2^{\frac{1}{x}}-1)=\ln(2)\Longrightarrow \lim\limits_{N\rightarrow \infty}N(2^{\frac{1}{N}}-1)=\ln(2)$$
We get:
$$\ln(2)=\lim\limits_{N\rightarrow \infty}N(2^{\frac{1}{N}}-1)=\lim\limits_{N\rightarrow \infty}\sum\limits_{n=1}^N\left(2^{\frac{n}{N}}-2^{\frac{n-1}{N}}\right)\frac{1}{2^{\frac{n-1}{N}}}=\int\limits_1^2 \frac{dx}{x}=\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}$$
$\Box$
Hey it would be great, if someone could check my reasoning (if its correct) and give me feedback and tips :)
| Your reasoning is correct, but you make it more complicated than needed.
$$\frac1N\sum_{i=1}^N\frac1{1+\dfrac nN}\to\int_0^1\frac{dx}{1+x}=\left.\ln(1+x)\right|_0^1.$$
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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: $2xy-4x=\lambda(2y-4) \implies \lambda=\tfrac{2xy-4x}{2y-4}=\tfrac{x(2y-4)}{2y-4}=x$
Setting the lambdas equal to each other we get
$\tfrac{3x^2+y^2-4y}{2x+2}=x \implies 3x^2+y^2-4y=2x^2+2x \implies -x^2+2x=y^2-4y$
Now completing the square gives
$-((x^2-2x+1)-1)=(y^2-4y+4)-4 \implies -(x-1)^2+1=(y-2)^2-4$
$\implies (x-1)^2+(y-2)^2=5$
Creating a system of equations with our original constraint gives
$(x+1)^2+(y-2)^2=1$
$(x-1)^2+(y-2)^2=5$
$\implies 1-(x+1)^2=5-(x-1)^2 \implies x=-1$
And plugging back into the constraint gives
$((-1)+1)^2+(y-2)^2=1 \implies y=1,3$
So Lagrange has given the points $(-1,1)$ and $(-1,3)$. However, it's pretty obvious geometrically that critical points exist at $(0,2)$ and $(-2,2)$ (one can also verify that they satisfy $\nabla f=\lambda\nabla g$), and it seems like you can find these points with one of the circles $(x+1)^2+(y-2\pm2)^2=5$, given how it intersects the constraint. But to get a circle like this you would have to alter the inputs to the Lagrange, something I'm pretty sure you can't do, so I'm stuck. Is there some property of Lagranges that allows you to alter their inputs?
| 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 intersections of the circles
$$\begin{cases}(x-1)^2+(y-2)^2=5 \\ (x+1)^2+(y-2)^2 = 1\end{cases} \implies (x,y) = (-1,1),(-1,3)$$
But if $y=2$ that would leave us with the points
$$(x+1)^2 = 1 \implies (x,y) = (0,2),(-2,2)$$
which we have to consider separately because otherwise we are just dividing by zero. And we must consider the extrema of the function as well
$$\begin{cases}3x^2+(y-2)^2 = 4 \\ x(y-2) = 0 \\ \end{cases} \implies (x,y) = (0,0),(0,4),\left(\pm\frac{2}{\sqrt{3}},0\right)$$
but none of these are in the region of consideration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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
$$\int _0^1\frac{\ln \left(ax^3+1\right)}{x+1}\:dx$$
$$\int _0^1\frac{x^3}{\left(x+1\right)\left(ax^3+1\right)}\:dx=\frac{1}{a-1}\int _0^1\left(\frac{ax^2-ax+a}{ax^3+1}-\frac{1}{x+1}\right)\:dx$$
in the end $2$ of these integrals are nice but the other $2$ are not
$\displaystyle \int _0^1\frac{-ax}{ax^3+1}\:dx$ and $\displaystyle \int _0^1\frac{a}{ax^3+1}\:dx$
Is there a better approach to this?
| 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
$$\int \frac{\log \left(x-r_i\right)}{x+1}\,dx=\text{Li}_2\left(\frac{r_i-x}{r_i+1}\right)+\log \left(\frac{x+1}{r_i+1}\right) \log (x-r_i)$$
$$\int_0^1 \frac{\log \left(x-r_i\right)}{x+1}\,dx=\text{Li}_2\left(\frac{r-1}{r+1}\right)-\text{Li}_2\left(\frac{r}{r+1}\right)-$$ $$\log (-r) \log \left(\frac{1}{r+1}\right)+\log (1-r) \log
\left(\frac{2}{r+1}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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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}} = \overrightarrow{AB} \times \overrightarrow{AC},\ d_{0} = -A \cdot{} \overrightarrow{n_{0}} $
$ \overrightarrow{n_{1}} = \frac{\overrightarrow{AC} \times \overrightarrow{n_{0}}}{|\overrightarrow{n_{0}}|^{2}},\ d_{1} = -A \cdot{} \overrightarrow{n_{1}} $
$ \overrightarrow{n_{2}} = \frac{\overrightarrow{n_{0}} \times \overrightarrow{AB}}{|\overrightarrow{n_{0}}|^{2}},\ d_{2} = -A \cdot{} \overrightarrow{n_{2}} $
The ray has this definition:
$ P(t) = \overrightarrow{o} + t \cdot{} \overrightarrow{d} $
The algorithm uses additional variables:
$ det = \overrightarrow{d} \cdot{} \overrightarrow{n_{0}} $
$ t' = d_{0} - (\overrightarrow{o} \cdot{} \overrightarrow{n_{0}}) $
$ P(t)' = det \cdot{} \overrightarrow{o} + t' \cdot{} \overrightarrow{d} $
$ u' = P(t)' \cdot{} \overrightarrow{n_{1}} + det \cdot{} \overrightarrow{d_{1}} $
$ v' = P(t)' \cdot{} \overrightarrow{n_{2}} + det \cdot{} \overrightarrow{d_{2}} $
$ \begin{pmatrix}t \\ u \\ v \end{pmatrix} = \frac{1}{det} \cdot{} \begin{pmatrix}t' \\ u' \\ v' \end{pmatrix} $
An intersection isdetected at these conditions:
$ sign(t') = sign(det \cdot{} t_{max} - t') $
$ sign(u') = sign(det - u') $
$ sign(v') = sign(det - u' - v') $
Which are equivalent with:
$ 0 \le t \le t_{max} $
$ u \ge 0 $
$ v \ge 0 $
$ (u + v) \le 1 $
My problem: I must misunderstand something
Let's calculate the planes for the following triangle:
$ A(2, 1, 1),\ B(1, 2, 2),\ C(3, 2, 1) $
This results in these values for $ n_{0} $ and $ d_{0} $:
$ n_{0} = \begin{pmatrix}-1 \\ 1 \\ -2\end{pmatrix},\ d_{0} = 3 $
And this is my ray that is aiming to hit point B:
$ \overrightarrow{o} = \begin{pmatrix}1 \\ 4 \\ 2\end{pmatrix},\ \overrightarrow{d} = \begin{pmatrix}0 \\ -1 \\ 0\end{pmatrix},\ t_{max} = 10 $
But these values fail the first condition:
$ det = -1,\ t' = 4 $
$ sign(t') = sign(det \cdot{} t_{max} - t') $
$ sign(4) \not= sign(-14) $
If I calculate all values, all other conditions fail as well.
What did I understand wrong?
| 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 cases.
Second, you have switched the sign on the $d_0$ parameter. With the
triangle $(2.5, 1, 1)$, $(1, 2, 2)$, $(3, 2.1, 1)$ and the ray $(1, 1,
1)$, $(3, 2, 2)$ the calculations work out as follows:
$$
n_0 = (v_1 - v0) \times (v_2 - v_0) = (-1.1, 0.5, -2.15)
$$
$$
det = n_0 \cdot d = (-1.1, 0.5, -2.15)\cdot (3, 2, 2) = -6.6
$$
$$
dett = d_0 - o\cdot n_0 = -4.4 - (1,1,1)\cdot(-1.1, 0.5, -2.15) = -1.65
$$
$$
r = det\cdot o + d\cdot dett = -6.6\cdot(1,1,1) + -1.65\cdot(3,2,2) = (-11.55, -9.9, -9.9)
$$
$$
x = r\cdot n_1 + det\cdot d_1 = (-11.55, -9.9, -9.9)\cdot(-0.39, 0.18, 0.24) -6.6\cdot0.56 = -3.3
$$
$$
y = r\cdot n_2 + det\cdot d_2 = -0.00000057
$$
$$
det_0 = det - x - y = -3.3
$$
$x$, $y$, and $det_0$ all have the same sign so the ray intersects the
triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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:
$$ \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) \quad (1) $$
Hence,
$$ (1) = \sum _{n=1}^{\infty } \frac{2^{n+1} - 2^n }{2^n2^{n+1}} = \sum _{n=1}^{\infty } \frac{2^n(2-1) }{2^n2^{n+1}} = - \sum _{n=1}^{\infty } \frac1{2^{n+1}} = - \sum _{n=2}^{\infty } \frac1{2^{n}} = -\left( \sum _{n=0}^{\infty } \left(\frac12\right)^n -1-\frac12\right) = -2+\frac32 =-\frac12 $$
Therefore,
$$ \boxed{\sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} =-\frac12} $$
I can't spot any algebraic mistakes, thus I assume the sum is correct. But I don't understand the result. How can a real valued sum converge to a negative number?
| 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",
"url": "https://math.stackexchange.com/questions/3773308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
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{pmatrix} \qquad\qquad X\in V
\end{align}
Find the matrix representation $A$ of $f$ with respect to the basis $e_1=\begin{pmatrix}1&0\\0&0\end{pmatrix},\,\, e_2=\begin{pmatrix}0&1\\0&0\end{pmatrix},\,\, e_3=\begin{pmatrix}0&0\\1&0\end{pmatrix},\,\, e_4=\begin{pmatrix}0&0\\0&1\end{pmatrix}$ 0f $V$
Note: I tried to find the matrix $A$, but stuck after find the $f(e_1),\,f(e_2),\,f(e_3),\, f(e_4)$. How to construct the matrix $A$ (it cannot find with the traditional way like vectors or polynomials because $f(e_i)$`s are also matrices) please give an idea. Thank you
| 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
$$
f(e_2) = \pmatrix{-3&-3\\2&2} = -3e_1 + -3e_2 + 2e_3 + 2e_4.
$$
Thus, the matrix $A$ of your transformation $f$ relative to the given basis will have the form
$$
A = \pmatrix{
?&-3&?&?\\
?&-3&?&?\\
?&2&?&?\\
?&2&?&?
}.
$$
Ultimately, you should find that
$$
A = \begin{pmatrix}
3& 4\\
-2& -3
\end{pmatrix}
\otimes
\begin{pmatrix}
1&2\\
-1&-1
\end{pmatrix}^T = \left(\begin{array}{cccc} 3 & -3 & 4 & -4\\ 6 & -3 & 8 & -4\\ -2 & 2 & -3 & 3\\ -4 & 2 & -6 & 3 \end{array}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 anyone confirm this? However, I've been having trouble actually proofing that this is the lower bound. Typically, I've solved problems where I need to prove an inequality as true, but this problem is a bit different asking for the minimum of an inequality instead, and I'm not sure how to show that $\sqrt {3}$ is the lower bound of it. Any ideas?
| 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 thus $\frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b} \geq\sqrt{3}$.
We attain equality by setting $a=b=c=\frac{\sqrt{3}}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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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 individual integrals.
I also know that the answer to (2) is $e^{\frac{x^2}4}$, but how it is arrived I am confused.
Furthermore, what's the general way of solving problems like these, the fraction of two definite integrals with same upper and lower limits?
I thank in advance any help I receive!
| *
*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+t^4}}\text{d}t}\\
=&\int_0^{\pi /4}{\dfrac{\sec ^2u}{\sqrt{1+\tan ^4u}}\text{d}u}\\
=&\int_0^{\pi /4}{\dfrac{1}{\sqrt{\sin ^4u+\cos ^4u}}\text{d}u}\\
=&\int_0^{\pi /4}{\dfrac{1}{\sqrt{\sin ^4u+\cos ^4u+2\sin ^2u\cos ^2u-2\sin ^2u\cos ^2u}}\text{d}u}\\
=&\int_0^{\pi /4}{\dfrac{1}{\sqrt{\left( \sin ^2u+\cos ^2u \right) ^2-\frac{1}{2}\sin ^22u}}\text{d}u}\\
=&\int_0^{\pi /4}{\dfrac{1}{\sqrt{\sin ^22u+\cos ^22u-\frac{1}{2}\sin ^22u}}\text{d}u}\\
=&\int_0^{\pi /4}{\dfrac{1}{\sqrt{\frac{1}{2}+\frac{1}{2}\cos ^22u}}\text{d}u}\\
=&\int_0^{\pi /2}{\dfrac{1}{2\sqrt{\frac{1}{2}+\frac{1}{2}\cos ^2u}}\text{d}u}\\
=&\int_0^{\pi /2}{\dfrac{1}{\sqrt{2\left( 1+\sin ^2u \right)}}\text{d}u}.
\end{align}
The ratio is $\sqrt{2}$.
*Use the change of variable formula $t-\frac x2=\frac u2$, thus $\text{d}t=\frac 12\text{d}u$:
\begin{align}
&\int_0^x{\text{e}^{xt-t^2}\text{d}t}\\
=&\int_0^x{\text{e}^{\frac{x^2}{4}}\text{e}^{-\frac{x^2}{4}+xt-t^2}}\text{d}t\\
=&\text{e}^{\frac{x^2}{4}}\int_0^x{\text{e}^{-\left( t-\frac{x}{2} \right) ^2}}\text{d}t\\
=&\text{e}^{\frac{x^2}{4}}\int_{-x}^x{\text{e}^{-\frac{u^2}{4}}\dfrac{1}{2}\text{d}u}\\
=&\text{e}^{\frac{x^2}{4}}\int_0^x{\text{e}^{-\frac{u^2}{4}}\text{d}u}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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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 :
$$\int\frac{e^{-\sin^2(x)} \tan^3(x)}{\cos(x)} \, dx = \left\{\left(\int e^{-\sin^2(x)}\tan^3(x)\,dx\right)\frac{1}{\cos(x)}\right\}-\int\frac{\sin(x)}{\cos^2(x)}\left(\int e^{-\sin^2(x)}\tan^3(x) \, dx\right)$$
where
$$\int e^{-\sin^2(x)} \tan^3(x)\,dx=\frac{e^{-\sin^2(x)}}{2\cos^2(x)}+c$$
Proof
\begin{align*}
\int e^{-\sin^2(x)}\tan^3(x) \,dx &=\int \frac{e^{-\sin^2(x)}(-\sin^2(x))(-2\sin(x)\cos(x)}{2\cos^4(x)}\,dx\\
&=\int \frac{e^{-\sin^2(x)}(-\sin^2(x))(\cos^2(x))^{'}}{2\cos^4(x)}dx\\
&=\int \frac{e^{\cos^2(x)-1}(\cos^2(x)-1)(\cos^2(x))^{'}}{2\cos^4(x)} \, dx\\
&=\frac{1}{2}\int e^{t-1}\frac{t-1}{t^2} \, dt & (t=\cos^2(x))\\
&=\frac{1}{2}\int e^{t-1}(\frac{1}{t}-\frac{1}{t^2}) \, dt\\
&=\frac{1}{2} \left(\int \frac{e^{t-1}}{t}dt+\frac{e^{t-1}}{t}-\int \frac{e^{t-1}}{t} \, dt\right)\\
&=\frac{1}{2}\frac{e^{t-1}}{t}\\
&=\frac{1}{2}\frac{e^{\cos^2(x)-1}}{\cos^2(x)}+c
\end{align*}
So:
$$\int \frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}dx=\frac{1}{2} \frac{e^{\cos^2(x)-1}}{\cos^3(x)}-\frac{1}{2} \int \frac{e^{\cos^2(x)-1}\sin(x)}{\cos^4(x)} \, dx=\frac{1}{2} \frac{e^{\cos^2(x)-1}\sin(x)}{\cos^4(x)}-\frac{1}{2}J$$
where:
$$J=\int\frac{e^{\cos^2(x)-1}\sin(x)}{\cos^4(x)} \, dx$$
But I have not found a way to calculate $J$ after so many attempts
Any help please and thanks in advance
| 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 $v=\frac 1u$
$${\displaystyle\int}-\dfrac{2\mathrm{e}^\frac{1}{u^2}}{u^2}\,\mathrm{d}u=\class{steps-node}{\cssId{steps-node-1}{\sqrt{{\pi}}}}{\displaystyle\int}\dfrac{2\mathrm{e}^{v^2}}{\sqrt{{\pi}}}\,dv=\sqrt{{\pi}}\operatorname{erfi}\left(v\right)=\sqrt{{\pi}}\operatorname{erfi}\left(\dfrac{1}{u}\right)$$
For $$\int u^2\,e^{\frac{1}{u^2}} \,du=-{\displaystyle\int}\dfrac{\mathrm{e}^{v^2}}{v^4}\,dv$$ just continue with integrations by parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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): $\displaystyle\frac{1}{k^2}\le \frac{1}{k\left(k-1\right)}$
so
(2):$\displaystyle \sum _{k=1}^n\left(\frac{1}{k^2}\right)\le \sum _{k=2}^n\left(\frac{1}{k\left(k-1\right)}\right)$
(3):$\displaystyle \sum_{k=1}^n\left(\frac{1}{k^2}\right)\le \:1+\left(1-\frac{1}{n}\right)$
(4):$\displaystyle\sum_{k=1}^n\left(\frac{1}{k^2}\right)\le \:\:2-\frac{1}{n}$
I don't understand why in line (2), on the right, we start the sum at $k=2$ while
on the left, we start on $k=1$. Why start at $k=2$?
| Note that $k(k-1)<k^2$ for $k>0$ but when $k=1$, $\dfrac{1}{n(n-1)}$ is not defined.
$\begin{aligned}\displaystyle \sum_{k=2}^{n} \frac{1}{k^2} &<\sum_{k=2}^{n} \frac{1}{k(k-1)} \\
&= \sum_{k=2}^{n} \frac{1}{k-1}-\frac{1}{k} \\
&=1-\frac{1}{n} \\ &\Rightarrow \sum_{k=1}^{n} \frac{1}{k^2}
=1+\sum_{k=2}^{n} \frac{1}{k^2} \\
&<1+1-\frac{1}{n} \\
&=2-\frac{1}{n} \end{aligned}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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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 for which there exists $a,b,c$ such that $\left (2^m\right )^2=4^m=a^2+b^2+c^2$. Since $4^m$ is divisible by 4, $a^2+b^2+c^2$ has to be divisible by 4 too. This is only possible if $a^2\equiv b^2\equiv c^2\equiv 0\pmod 4$, so we can write $a=2a',b=2b',c=2c'$ with $a',b',c'\in \mathbb{N}$. But then we get $\left (2^{m-1}\right )^2=4^{m-1}=a'^2+b'^2+c'^2$, so $m=1$, otherwise $2^m$ wouldn't be the smallest power of two with this property. It is easy to check that $n=2$ doesn't work, so for $n=2^m$ the statement doesn't hold.
*I suspect (but can't prove) that for all other values the statement holds. It would be enough to prove that for all odd primes $p$ there exists $a,b,c$ such that $p^2=a^2+b^2+c^2$, since for all other values of $n$ there exist some $p,m$ such that $n=pm$. Then we get $n^2=(pm)^2=(ma)^2+(mb)^2+(mc)^2$.
| 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 $s(1) = 6$.
If $p > 2$ is prime, then $p^2$ can be written as a sum of three squares (including degenerate examples) in
$$6\left(p + 1 - \left( \frac{-1}{p} \right)\right)$$
ways. (For a reference, see https://mathoverflow.net/questions/3596/is-there-a-simple-way-to-compute-the-number-of-ways-to-write-a-positive-integer). For example, if $p = 3$, then $(-1/3) = -1$ so we get $30$ ways, and indeed
$$3^2 = (\pm 3)^2 + 0^2 + 0^2 = 0^2 + (\pm 3)^2 + 0^2 = 0^2 + 0^2 + (\pm 3)^2,$$
giving $3 \times 2 = 6$ ways, and
$$3^2 = (\pm 2)^2 + (\pm 2)^2 + (\pm 1)^2 = (\pm 2)^2 + (\pm 1)^2 + (\pm 2)^2 = (\pm 1)^2 + (\pm 2)^2 + (\pm 2)^2$$
giving $3 \times 8 = 24$ ways. The examples you wish to rule out at the ones where either $a$, $b$, or $c$ is $\pm p$, and this gives $6$ solutions. Hence you simply need to observe that $p + 1 - (-1/p) > 1$, which is always true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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 $3a = 1$ doesn't work.
| 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",
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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 obvious sub is $t = 1 + x^{3}$:
$$
\frac{1}{3}\int_{1}^{9}{\left(\,{t - 1}\,\right)}^{-2/3}\, t^{-1/2}\, \mathrm{d}t
$$
From here I tried many things like $\frac{1}{t}$, $t-1$, and more. The trickiest part is the bounds! Reversing it from the answer the integral should be like
$$
\frac{1}{6}\int_{0}^{1}
x^{-2/3}\left(\,{1 - x}\,\right)^{-5/6}\,\mathrm{d}t
$$
I'm not sure where the $1/2$ comes from and the $0$ to $1$ bounds. Any idea or tip please ?.
| 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
$$I_1=\sqrt{2} \, _2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\frac{1}{8}\right)=\frac{2 \sqrt{\frac{\pi }{3}} \Gamma \left(\frac{7}{6}\right)}{\Gamma \left(\frac{2}{3}\right)}$$
Where the last step has invoked the following formula $$\, _2F_1\left(a,a+\frac{1}{3};\frac{4}{3}-a;-\frac{1}{8}\right)=\frac{\left(\frac{2}{3}\right)^{3 a} \Gamma \left(\frac{2}{3}-a\right) \Gamma \left(\frac{4}{3}-a\right)}{\Gamma \left(\frac{2}{3}\right) \Gamma \left(\frac{4}{3}-2 a\right)}$$
Computing $I_0-I_1$ gives the desired result.
Update: Hypergeometric method can also establish @pisco's result (the case of $4$-torsion)
$$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{12 \sqrt{\pi }} \qquad \alpha = \sqrt[3]{2 \left(3 \sqrt{3}-5\right)} \approx 0.732$$
Since by binomial expansion again, it equals $$\left(\sqrt{3}-1\right) {_2F_1}\left(\frac{1}{3},\frac{1}{2},\frac{4}{3},10-6 \sqrt{3}\right)=\frac{\sqrt{\frac{1}{2} \left(6 \sqrt{3}-9\right) \pi } \Gamma \left(\frac{1}{3}\right)}{3\ 3^{3/4} \left(\sqrt{3}-1\right) \Gamma \left(\frac{5}{6}\right)}$$ due to certain transformation of hypergeometric series (see Special values of hypergeometric series by Akihito Ebisu). The rest are trivial.
| {
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"timestamp": "2023-03-29T00:00:00",
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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}^{\infty}\frac{(\phi-1)^n}{(2n+1)^2}&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{\phi-1})^{2n}}{2n+1}\left(1-\frac{2n}{2n+1}\right)\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{\phi-1})^{2n}}{2n+1}-\displaystyle\sum_{n=0}^{\infty}\frac{(2n)(\sqrt{\phi-1})^{2n}}{(2n+1)^2}\\
&=J-I\\
\text{where}\\
J&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{\phi-1})^{2n}}{2n+1}\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n}}{2n+1}|_{x=\sqrt{\phi-1}}\\
\text{and}\\
\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n}}{2n+1}&=\frac{1}{x}\displaystyle\sum_{n=0}^{\infty}\displaystyle\int x^{2n}dx\\
&=\frac{1}{x}\displaystyle\int\frac{1}{1-x^2}dx\\
&=\frac{1}{2x}\ln\left(\frac{1+x}{1-x}\right)\\
\text{So}\\
J&=\frac{1}{2x}\ln\left(\frac{1+x}{1-x}\right)|_{x=\sqrt{\phi-1}}\\
&=\frac{1}{2\sqrt{\phi-1}}\ln\left(\frac{1+\sqrt{\phi-1}}{1-\sqrt{\phi-1}}\right)
\end{align*}
But how do we calculate $I$ to get the result.
| 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:
$$
\begin{aligned}
\mathrm{Li}_{2}(x)&=\sum_{n=1}^\infty\frac{x^n}{n^2}\\
&=\frac14\sum_{n=1}^\infty\frac{x^{2n}}{n^2}+\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)^2}\\
&=\frac14\mathrm{Li}_{2}(x^2)+\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)^2}\\
\end{aligned}
$$
Hence
$$
\begin{aligned}
\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)^2}&=\mathrm{Li}_{2}(x)-\frac14\mathrm{Li}_{2}(x^2)\\
&=\frac12\left(\mathrm{Li}_{2}(x)-\mathrm{Li}_{2}(-x)\right)
\end{aligned}
$$
Letting $x=\phi-1$ and noting that $\phi-1=\frac{1}{\phi}$ we obtain
$$
\begin{aligned}
\sum_{n=0}^\infty\frac{(\phi-1)^{2n+1}}{(2n+1)^2}&=\frac12\left(\mathrm{Li}_{2}\left(\frac{1}{\phi}\right)-\mathrm{Li}_{2}\left(-\frac{1}{\phi}\right)\right)\\
&=\frac12\left(\frac{\pi^2}{10}-\ln^2(\phi)+\frac{\pi^2}{15}-\frac{\ln^2(\phi)}{2}\right)\\
&=\frac{\pi^2}{12}-\frac{3\ln^2(\phi)}{4} \qquad \blacksquare
\end{aligned}
$$
Where we used that
$
\mathrm{Li}_{2}(x)+\mathrm{Li}_{2}(-x)=\frac{1}{2} \mathrm{Li}_{2}\left(x^{2}\right)
$
and
$
\begin{aligned}
\mathrm{Li}_{2}\left(\frac{1}{\phi}\right) &=\frac{\pi^{2}}{10}-\ln ^{2} \phi \\
\mathrm{Li}_{2}\left(-\frac{1}{\phi}\right) &=\frac{\ln ^{2} \phi}{2}-\frac{\pi^{2}}{15} \\
\end{aligned}
$
A proof of these last two relations between dilogs and the golden ratio can be found here in my blog
| {
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"timestamp": "2023-03-29T00:00:00",
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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 the process of solving a question related to covering $\mathbb{N}$ with arithmetic progressions, I found that the following statement would allow me to finish my proof by contradiction.
Clearly, the conjecture is true if any of $d_1, \dots, d_k$ are prime powers. If none of them are prime powers, then they must be among $6, 10, 12, 14, 15, 16, 18, 20, 21, \dots$ Since $S = \frac{1}{6}+ \frac{1}{10}+ \frac{1}{12}+ \frac{1}{14}+ \frac{1}{15}+ \frac{1}{16}+ \frac{1}{18}+ \frac{1}{20}+ \frac{1}{21} + \frac{1}{22}+\frac{1}{24}+\frac{1}{26}+\frac{1}{28}+\frac{1}{30}+\frac{1}{33}+\frac{1}{34}+\frac{1}{35} < 1 < S + \frac{1}{36},$ any counterexample must have $k \ge 18.$ But including a number with $2$ distinct prime divisiors severely restricts the numbers we are allowed to take afterwards, so we could perform further casework and make the bound for a potential counterexample higher and higher.
Warning: You cannot rely solely on analysis because if $=1$ is replaced with $\ge 1,$ the result is false. There are $2$ constructions that demonstrate this:
$\{6, 10, 15\} \cup \{30a : a \le n\}$ for $n$ large enough.
$\{2 \cdot 3 \cdot 5 \cdot 7 = 210\} \cup S'$ where $S'$ is a finite subset of $S = \{2^a 3^b 5^c : \text{at most one of } a, b, c \text{ is zero}\}$ with sum $\ge 209/210.$ This is possible since $\sum\limits_{s \in S} \frac{1}{s} = 1.$
| 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 $d_1 = 6, d_2 = 10, d_3 = 15, d_{3+i} = 30c_i$ for $1 \le i \le k.$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $\sin(\frac{\pi}{2} -x) = \cos(x)$. The integral changes into easier:
$$\int_{0}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^3(x)+ \cos^3(x)} \, \mathrm{d}x$$
And then I divided by $\cos^3(x)$. It will turn everything to
$$\int_{0}^{\frac{\pi}{2}} \frac{\sec^2(x)}{\tan^3(x)+1} \, \mathrm{d}x.$$
And I used $u$-substitution setting $u = \tan(x)$ $\Rightarrow$ $\mathrm{d}u = \sec^2(x) \, \mathrm{d}x$ and the bounds $u = \tan(\frac{\pi}{2}) = \infty$ and $u =\tan(0) = 0$ and the integral changed into integral to
$$ \int_{0}^{\infty} \frac{\mathrm{d}u}{1+u^3} $$
This is where I used partial fraction decomposition and my answer is divergent and my answer is wrong:
\begin{align*}
\int_{0}^{\infty} \frac{1}{3(u+1)} + \frac{-u+2}{3(u^2-u+1)} \, \mathrm{d}u
\end{align*}
by this it looks like it will divergent.
The correct answer is $\frac{2\pi}{3\sqrt{3}}$. So, what do I do, next?
Then I did the another method which gives me divergent again,
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\sin^3(x)+\cos^3(x)} \, \mathrm{d}x
&= \int_{0}^{\frac{\pi}{2}} \frac{\tan(x)\sec^2(x)}{1+\tan^3(x)} \, \mathrm{d}x
\end{align*}
and $u = \tan(x)$ $\Rightarrow$ $\mathrm{d}u = \sec^2(x) \, \mathrm{d}x$, the integral
$$\int_{0}^{\frac{\pi}{2}} \frac{u}{1+u^3} \, \mathrm{d}u$$
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{-1}{3(u+1)} + \frac{u+1}{3(u^2-u+1)} \, \mathrm{d}u
\end{align*}
and did partial fraction and this gives divergent.
I have no idea what to do next for the first method of work, or the second method of work.
| 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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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 $$\prod_{k=1}^{3}\cos\frac{k\pi}{7}=\frac{1}{2^3}=\frac{1}{8}$$ as known since elementary classes. To tackle with sine product I tried as $\sin(x)=\cos\left(\frac{\pi}{2}-x\right)$. But I fail with this ideas.
How do I deal with $\sin x$ product?
| 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 formula of the gamma function ,
$$ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin{\pi}{ z}}$$ giving us $$\sin\frac{\pi}{7}\sin\frac{3\pi}{7}\sin\frac{3\pi}{7}=\frac{\pi^3}{\Gamma\left(\frac{1}{7}\right)\Gamma\left(\frac{6}{7}\right)\Gamma\left(\frac{2}{7}\right)\Gamma\left(\frac{5}{7}\right)\Gamma\left(\frac{3}{7}\right)\Gamma\left(\frac{4}{7}\right)}=\frac{\sqrt 7 \pi^3}{8\pi^3 } $$ We deduce the latter result using multiplication theorem for gamma function.
Now it is just simplification which gives us $\sqrt 7$.
| {
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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 $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c^{n+1}+c^na+c^nb \ge 3c^n\sqrt[3]{abc}$
But from these things, i can't prove the problem.
| 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 inequality:
$(x-y)(x^n-y^n) \ge 0 \implies x^{n+1}+y^{n+1} \ge x^ny+xy^n$ three times for the pairs $(a,b), (b,c), (c,a)$ and add up.
| {
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"timestamp": "2023-03-29T00:00:00",
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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): $\int_0^\infty \frac{z^6}{(z^4+1)^2}dz=\frac{1}{2}\int_\infty^\infty\frac{z^6}{(z^4+1)^2}dz$. Now, the singularities of $z^4+1$ are of the form $z_k=e^{\frac{i(\pi+2\pi k)}{4}}$, where $k=0,1,2,3$. Drawing the contour in the upper half plane, we see tht the only two singularities in our contour are $z_0=e^{\frac{i\pi}{4}}$ and $z_1=e^{\frac{i3\pi}{4}}$.
Let $f(z)=\frac{g(z)}{h(z)}$ where $g(z)=z^6$ and $h(z)=(z^4+1)^2$. Then, $h'(z)=8z^3(z^4+1)$. So, the value of the integral is $2\pi i\frac{1}{2}\sum_{k=0}^1\frac{z_k^6}{8z_k^3(z_k^4+1)}=\frac{\pi i}{8}\Big(\frac{e^{i\frac{3\pi}{2}}}{e^{i\frac{3\pi}{4}}(e^{i\pi}+1)}+\frac{e^{i\frac{\pi}{2}}}{e^{i\frac{\pi}{4}}(e^{i\pi}+1)}\Big)$. But, $e^{i\pi}+1=0$, so I must have done something wrong....? Any insight would be great! Thank you.
| @Vercassivelaunos has the right idea; it's not even that computationally demanding. A root $a$ of $z^4+1$ has residue$$\begin{align}\lim_{z\to a}\tfrac{d}{dz}\tfrac{(z-a)^2z^6}{(z^4+1)^2}&=\lim_{z\to a}\tfrac{2(z-a)z^5(az^4+4z-3a)}{(z^4+1)^3}\\&=2a^5\lim_{\epsilon\to0}\tfrac{\epsilon(a(a+\epsilon)^4+4(a+\epsilon)-3a)}{(4a^3\epsilon)^3}\\&=\tfrac{1}{32a^4}\lim_{\epsilon\to0}\tfrac{a(a+\epsilon)^4+a+4\epsilon}{\epsilon^2}.\end{align}$$Of course, the numerator's $\epsilon^0$ and $\epsilon^1$ terms vanish, and the $\epsilon^2$ coefficient is $a\binom42a^2$, so the residue is $\frac{3}{16a}$. So the integral on $\Bbb R$ is $2\pi i\frac{3}{16}(e^{-\pi i/4}+e^{-3\pi i/4})=\frac{3\pi\sqrt{2}}{8}$, while the original integral is $\frac{3\pi\sqrt{2}}{16}$.
For what it's worth, you could also have solved this with $z=\tan^{1/2}t$ if you know your Beta & Gamma functions; the original integral becomes$$\tfrac12\int_0^{\pi/2}\sin^{5/2}t\cos^{-1/2}tdt=\tfrac14\operatorname{B}(\tfrac74,\,\tfrac14)=\tfrac14\Gamma(\tfrac74)\Gamma(\tfrac14)=\tfrac{3\pi}{16}\csc\tfrac{\pi}{4}=\tfrac{3\pi\sqrt{2}}{16}.$$
| {
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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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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 $(x^2 - 3x + 2) = (x - 1)(x - 2)$ .
So :- $(x - a)^2 = - (x^2 - 3x + 2)^2$
=> $(x - a) = -(x^2 - 3x + 2)$
=> $(x - a) = -(x - 1)(x - 2) = (1 - x)(x - 2)$ .
From here I don't have a good hint or a clue to move forward . Can anyone help ?
| 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 $$ {\text{Sum of all possible valus of } a} = 2 + 1 = 3 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"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 $(x+y+z)^2$, which expands to $(x^2+y^2+z^2) + 2\cdot (xy+yz+xz)$, but I couldn't go anywhere from there (other than substituting xy, yz, xz, but that's not enough information.
| 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",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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:
$$\begin{align} \sqrt[3]{7a^2b+1} = 2 \sqrt[3]{a \cdot a \cdot \left (\ \frac{7b}{8} + \frac{1}{8a^2} \right ) } \end{align} \leq \frac{2}{3} \left ( a+a+\frac{7b}{8} + \frac{1}{8a^2} \right )$$
Summing up alll three inequalities with $\sqrt[3]{7a^2b+1},\sqrt[3]{7b^2c+1},\sqrt[3]{7c^2a+1}$ we get as desired.
The first equation uses a trick where we devide the variables by 8. I tried the same thing with $1^3$ from which we got a value higher than $2(a+b+c)$. Then I tried $3^3$ and that led to an inequality which was even smaller than $2(a+b+c)$. Thus we got $<$ instead of $\leq$. To be more exact summing up the three inequalities led to $\frac{62(a+b+c)}{3\cdot 27} < 2(a+b+c)$. Why is this the case? For $a=b=c=1$ we have equality. Why is this version so strange? What happens when we use different numbers in AM-GM inequalities at tricks like these? How do we know which is truly the lowest value we can get of an expression.
| 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:
$$\sum_{cyc}\sqrt[3]{7a^2b+1}=\frac{1}{4}\sum_{cyc}\sqrt[3]{8a\cdot8b\left(a+\frac{7}{ab}\right)}\leq\frac{1}{12}\sum_{cyc}\left(8a+8b+a+\frac{7}{ab}\right)\leq$$
$$\leq\frac{1}{12}\sum_{cyc}\left(8a+8b+a+\frac{7}{2}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\right)=2(a+b+c).$$
Also, by Holder:
$$\sum_{cyc}\sqrt[3]{7a^2b+1}=\sum_{cyc}\sqrt[3]{a^2\left(7b+\frac{1}{a^2}\right)}\leq$$
$$\leq\sqrt[3]{(a+b+c)^2\sum_{cyc}\left(7b+\frac{1}{a^2}\right)}=2(a+b+c).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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:
$$ f(2)=2\cdot 2+1=2\cdot 2!+1$$
$$ f(3)=3\cdot f(2)+1=3(2\cdot 2+1)+1=2\cdot 3!+3+1$$
$$ f(4)=4(3(2\cdot 2+1)+1)+1=2\cdot 4!+4\cdot 3+4+1=2\cdot 4!+\frac{4!}{2!}+\frac{4!}{3!}+\frac{4!}{4!}.$$
Now, this predicts a general trend that for $n\ge 2$,
$$ f(n)=2n!+\sum_{k=2}^n \frac{n!}{k!}.$$
Indeed, this is true for $f(2)=2\cdot 2!+1=5$. Suppose the formula holds for $n$, then
$$ f(n+1)=(n+1)\bigg(2n!+\sum_{k=2}^n \frac{n!}{k!}\bigg)+1=2(n+1)!+\sum_{k=2}^n\frac{(n+1)!}{k!}+1=2(n+1)!+\sum_{k=2}^{n+1}\frac{(n+1)!}{k!}.$$
This completes the proof. This rule does not apply (in its current form) to the cases $n=0,1$, but we get
$$ f(n)=
\begin{cases}
1&n=0\\
2&n=1\\
2n!+\sum_{k=2}^n \frac{n!}{k!}&n\ge 2.
\end{cases}$$
Edit: Now that I've seen the other answer here, it is clear that mine can be improved. Indeed, we have that $2n!=\frac{n!}{1!}+\frac{n!}{0!}$. So, the final result is that
$$ f(n)=\sum_{k=0}^n \frac{n!}{k!}$$
and this formula holds for all $n$.
| {
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"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
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ output is $\infty - \infty$ which is indeterminate form.
So next I basically but it on the same denominator: $\frac{1}{3}$ $((3x + 2x^3 - 2(x^2+1)^{\frac{3}{2}})$ and turned $2(x^2+1)^{\frac{3}{2}}$ into something easier to work with $2\sqrt{x^2+1}+2x^{2}\sqrt{x^2+1}$
now the limit is $\frac{1}{3} \lim_{x \to \infty} ((3x + 2x^3-2\sqrt{x^2+1} -2x^{2}\sqrt{x^2+1})$ and this is where I am stuck to do next and lost.
| 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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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $n=1, 2, \ldots$. Find $\lim _{n \rightarrow \infty} P_{n}$.
My work:
Using the expression I am getting
$$a_{2} = 4$$
$$a_{3} = 15$$
$$a_{4} = 64$$
What to do next?
Edit 1. After taking log, I am getting the final series as $\log(1+1/n^{2})$. So, the answer should be $0$.
| 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}{n!} \\&=
\frac{a_{n-1}}{(n-1)!} + \frac{1}{(n-1)!} + \frac{1}{n!} \\&=
\frac{a_{n-2}}{(n-2)!} + \frac{1}{(n-2)!} + \frac{1}{(n-1)!} + \frac{1}{n!} \\&=
\cdots \\&=
\frac{a_1}{1!} + \sum^{n}_{k=1} \frac{1}{k!}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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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:
$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2-(r_1+\frac{1}{r_1})x+1)+(x^2-(r_2+\frac{1}{r_2})x+1)(x^2-(r_3+\frac{1}{r_3})x+1)$$
And it can be written as:$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2+ax+1)(x^2+bx+1)(x^2+cx+1)$$
$$=x^6+(a+b+c)x^5+(3+ab+ac+bc)x^4+(abc+a+b+c)x^3+(3+ab+ac+bc)x^2+(a+b+c)x+1$$
In orther to find$a,b,c$ I should solve the system :
${ \begin{cases}{a+b+c=1} \\ {ab+ac+bc=-2} \\ {abc=0}\end{cases} }$
from the equation $abc=0$ We have $a=0$ (Since our derivation was symmetric in $a,b,c$ it doesn't matter to put $a=0$ or $b=0$ or $c=0$).
${ \begin{cases}{b+c=1} \\ {bc=-2}\end{cases} }$
and from that we have $b=2$ , $c=-1$ , $a=0$
So I get :
$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2+1)(x^2+2x+1)(x^2-x+1)=(x^2+1)(x+1)^2(x^2-x+1)$$
But my answer is wrong because if we multiply the equation by $x-1$ We should obtain $x^7-1$. But:
$$(x-1)(x^2+1)(x+1)^2(x^2-x+1)=(x-1)(x+1)(x^2+1)(x+1)(x^2-x+1)=(x^4-1)(x^3+1) \ne x^7-1$$
I checked my answer several times but I couldn't find my mistake.
| 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 cannot hold.
| {
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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 $\sum_{k=1001}^{3001} k=(2001)^{2}$
Hence, $\sum_{k=1001}^{3001} \frac{1}{k}>1$
How to prove the next part?(Is there any inequality I am missing)
| 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 $\frac{1}{1250}$, and there are $250$ total values to sum, all greater than or equal to $\frac{1}{1250}$. So we know the sum is at least $\frac{1}{1250}\cdot 250 = \frac{1}{5}$.
Next, the smallest value is $\frac{1}{1500}$, with $250$ values to sum, so the total sum is at least $\frac{1}{6}$.
Continuing on, you should get $$\frac{1}{5}, \frac{1}{6}, \dots,\frac{1}{12}$$
Now we add all of these split sums to get the total value of the original summation.
$$\frac{1}{5}+\frac{1}{6}+\dots+\frac{1}{12} \approx 1.02$$
And $1.02>1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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
$$I=2\int_0^1\frac{\arctan^2(x)}{x}\ln\left(\frac{2x}{1+x^2}\right)dx$$
$$=2\int_0^1\frac{\ln(2)+\ln x}{x}\arctan^2(x)dx-2\int_0^1\frac{\ln(1+x^2)}{x}\arctan^2(x)dx$$
The first integral simplifies to known harmonic series using the identity
$$\arctan^2(x)=\frac12\sum_{n=1}^\infty\frac{(-1)^n\left(H_n-2H_{2n}\right)}{n}x^{2n}$$
But using this series expansion in the second integral yields very complicated harmonic series. Also integrating by parts, yields the integrand $\frac{\text{Li}_2(-x^2)\arctan(x)}{1+x^2}$ which complicates the problem. Any thought how to approach any of these two integrals?
Thank you.
| 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
*
*$\Re\left(\int_0^1 \frac{\log ^3(1+i x)}{x} \, dx\right)=\int_0^1 \frac{\frac{1}{8} \log ^3\left(x^2+1\right)-\frac{3}{2} \log \left(x^2+1\right) \tan ^{-1}(x)^2}{x} \, dx\\=-\frac{3}{4} \pi C \log (2)+\frac{3}{64} \pi \Im\left(-32 \text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\Re\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right) \log (8)-6 \text{Li}_4\left(\frac{1}{2}-\frac{i}{2}\right)\right)-\frac{5}{64} \left(42 \zeta (3) \log (2)+\log ^4(2)\right)+\frac{1249 \pi ^4}{15360}+\frac{21}{128} \pi ^2 \log ^2(2)$
Also one have
*
*$\int_0^1 \frac{\log ^3\left(x^2+1\right)}{x} \, dx=-3 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{8} \zeta (3) \log (8)+\frac{\pi ^4}{30}-\frac{1}{8} \log ^4(2)+\frac{1}{8} \pi ^2 \log ^2(2)$
So
*
*$\int_0^1 \frac{\log \left(x^2+1\right) \tan ^{-1}(x)^2}{x} \, dx=\frac{1}{2} \pi C \log (2)+\pi \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{8} \zeta (3) \log (2)-\frac{421 \pi ^4}{11520}+\frac{\log ^4(2)}{24}-\frac{7}{96} \pi ^2 \log ^2(2)$
So
*
*$\int_0^{\frac{\pi }{2}} \frac{x^2 \log (\sin (x))}{\sin (x)} \, dx=-4 \pi \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{3 \pi ^4}{32}+\frac{1}{8} \pi ^2 \log ^2(2)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 integer.
It can be checked that the integral is convergent for all values of $n$ in the prescribed range. The case of even $n$ is solved in A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$. Also, $I_1 = \infty$.
I have tried to use the method of multiple integrations by parts but in vein.
I was wondering whether there exists a suitable approach to address this problem more effectively.
| 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
$$ J_{p} = \begin{cases}
\displaystyle \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{\log(2\pi j)}{4j^2-p^2}, & \text{if $p$ is odd}, \\
\displaystyle \frac{\pi}{2}, & \text{if $p$ is even}.
\end{cases} $$
Proof of $\text{(1)}$. The case of even $n$ has already been discussed in other postings, so we focus on odd $n$. We first note that, if $n \geq 1$ is an odd integer, then
\begin{align*}
\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}} \sin^n x
&= \frac{1}{(2i)^n} \sum_{l=0}^{\frac{n-1}{2}} (-1)^l \binom{n}{l} \frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}} (e^{(n-2l)ix} - e^{-(n-2l)ix}) \\
&= \frac{1}{2^{n-1}} \sum_{l=0}^{\frac{n-1}{2}} (-1)^l \binom{n}{l} (n-2l)^{n-1} \sin((n-2l)x).
\end{align*}
So by applying integration by parts $(n-1)$-times, we get
\begin{align*}
I_n
&= \sum_{k=0}^{\infty} \int_{0}^{\pi} \frac{\sin^n x}{(x+k\pi)^n} \, \mathrm{d}x \\
&= \frac{1}{(n-1)!} \sum_{k=0}^{\infty} \int_{0}^{\pi} \biggl( \frac{1}{x+k\pi} - \frac{1}{(k+1)\pi} \biggr) \biggl( \frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}} \sin^n x \biggr) \, \mathrm{d}x \\
&= \frac{1}{(n-1)!2^{n-1}} \sum_{l=0}^{\frac{n-1}{2}} (-1)^l \binom{n}{l} (n-2l)^{n-1} J_{n-2l},
\end{align*}
where $J_{p}$ is defined by
\begin{align*}
J_{p}
&= \sum_{k=0}^{\infty} \int_{0}^{\pi} \biggl( \frac{1}{x+k\pi} - \frac{1}{(k+1)\pi} \biggr) \sin(px) \, \mathrm{d}x.
\end{align*}
If $p$ is odd, then the above definition is recast as
\begin{align*}
J_{p}
&= \sum_{k=0}^{\infty} \biggl( \int_{0}^{\pi} \frac{1}{x+k\pi} \sin(px) \, \mathrm{d}x - \frac{2}{p\pi(k+1)} \biggr) \\
&= \lim_{N \to \infty} \biggl( \int_{0}^{N \pi} \frac{\sin(p(x \text{ mod } \pi))}{x} \, \mathrm{d}x - \frac{2}{p\pi} H_N \biggr),
\end{align*}
where $H_N = 1 + \frac{1}{2} + \dots + \frac{1}{N}$ is the $N$-th harmonic number. Still assuming that $p$ is an odd integer, Fourier series computation shows that
\begin{align*}
\sin(p(x \text{ mod } \pi))
&= \frac{2}{p\pi} - \frac{4p}{\pi} \sum_{n=1}^{\infty} \frac{\cos(2\pi n x)}{4n^2-p^2} \\
&= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1 - \cos(2\pi j x)}{4j^2-p^2},
\end{align*}
and so,
\begin{align*}
\int_{0}^{N \pi} \frac{\sin(p(x \mathrm{ mod } \pi))}{x} \, \mathrm{d}x
&= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} \int_{0}^{N \pi} \frac{1 - \cos(2 j x)}{x} \, \mathrm{d}x \\
&= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} (\gamma + \log(2\pi j N) - \operatorname{Ci}(2\pi j N) ).
\end{align*}
Plugging this back and using the identity $\frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} = \frac{2}{p\pi}$, which itself follows from the Fourier series of $\sin(p(x \text{ mod } \pi))$, we finally obtain
\begin{align*}
J_{p}
&= \frac{4p}{\pi} \lim_{N \to \infty} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} (\gamma + \log(2\pi j N) - \operatorname{Ci}(2\pi j N) - H_N ) \biggr) \\
&= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{\log(2\pi j)}{4j^2-p^2}
\end{align*}
as desired. $\square$
Addendum. Here is a Mathematica code for numerical verification of $\text{(1)}$:
n = 5; (* Choose your favorite odd integer >= 3*)
NIntegrate[Evaluate[Sum[1/(x + k Pi)^n, {k, 0, Infinity}] Sin[x]^n], {x, 0, Pi}, WorkingPrecision -> 20]
TermJ[p_] := (4 p)/Pi NSum[Log[2 Pi j]/(4 j^2 - p^2), {j, 1, Infinity}, WorkingPrecision -> 20];
1/((n - 1)! 2^(n - 1)) Sum[Binomial[n, l] (-1)^l (n - 2 l)^(n - 1) TermJ[n - 2 l], {l, 0, (n - 1)/2}]
Clear[n, TermJ];
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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 $x=\frac{c^2(b^2-c^2)}{a^2(a^2-b^2)}$ and $y=\frac{c^2(c^2-a^2)}{b^2(a^2-b^2)}$ from first two equation.
After putting $x$ and $y$ in $3rd$ equation.
$$\frac{c^2(b^2-c^2)}{a^2(a^2-b^2)}+\frac{c^2(c^2-a^2)}{b^2(a^2-b^2)}+1=0 $$
I did prove it but I am not satisfied with the process as I have to use an identity(never seen and found in the simplification)$$a^2b^2(a^2-b^2)+b^2c^2(b^2-c^2)+c^2a^2(c^2-a^2)=-(a^2-b^2)(b^2-c^2)(c^2-a^2)$$ and the simplification nearly took 30 minutes(which is not good in exam)
Is there any other fast and efficient way to prove or simplify this ?
| 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 $$
\alpha+\beta t+\gamma t^2$$
has at least three roots $a^2,b^2,c^2$. But such a quadratic (or perhaps even linear) polynomial cannot have more than two distinct roots.
Hence two of these squares must be equal, hence two of $a,b,c$ differ by at most a factor of $-1$ (and therefore have the same absolute value).
| {
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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. Similarly assuming $[x]=3,4$; we get correct roots as $\sqrt{12}$ and $\sqrt{18}$. But if er let $[x]=5$, it gives $x=\sqrt{24}$. which contradicts. So I get the sum of roots as $\sqrt{6}(1+\sqrt{2}+\sqrt{3}).$
The question is: Have I found all the real roots. In any case, what is(are) more appropriate method(s) of doing it.
| 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 $3-\sqrt{3}< x \le 3+\sqrt{3}$ and $(2)$ is always true. So your choices of $[x]=2,3,4$ get justified and work well.
Method II: Let $$x=n+q,~~ n \in I^+, 0\le q< 1$$. Putting it in the equation you get
$$n^2-6n+6=-q^2-2nq \le 0 \implies n=2,3,4.$$
For $n=2$ get $$q=\pm \sqrt{6}-2 \implies q=\sqrt{6}-2>0 \implies x=2+\sqrt{6}-2= \sqrt{6}.$$ Similarly, you get other two roots.
Method III: Graphically, the LHS is a parabola and RHS is a starcase function. These two cut each other in the first quadrant at three points. See the Fig. below
| {
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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 actually use it to solve the problem. The first thing I tried was cubing the equation; I quit midway through realizing it was a bad idea (It was really messy.)
Next, I got the stupid idea of trying to actually solve the cubic. I got nowhere. (What I was hoping for were some nice solutions for $x.$)
Now, I'm stuck. Help? Thanks!
| 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
$$
3 \left[x-\sqrt[3]{x-a}\sqrt[3]{x-b}\sqrt[3]{x-c}\right]-(a+b+c)=0\\
3 \left[x-\sqrt[3]{(x-a)(x-b)(x-c)}\right]-(a+b+c)=0\\
$$
but
$$
a+b+c=4\\
(x-a)(x-b)(x-c)=x^3-4x^2-32x+17
$$
so the equation becomes
$$
3 \left[x-\sqrt[3]{x^3-4x^2-32x+17}\right]-4=0\\
$$
and isolating the cube root and elevating again to the third power
$$
x^3-4x^2-32x+17=\left(x-\frac{4}{3}\right)^3\\
$$
simplifying we get
$$
1008x=523\quad\implies\quad x_0=\frac{523}{1008}
$$
Be aware that the cube root is intended to be defined on the whole $\mathbb{R},$ otherwise no real solution would exist, given that one of $x_0-a,x_0-b,x_0-c$ is negative.
| {
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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:
$$\int_{0}^{1}\frac{1}{x+1}dx=\ln(x+1)\bigg|_{x=0}^{x=1}=\ln(1+1)-\ln(0+1)=\ln(2).$$
| {
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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 anyone try it?
| 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 $ab=a^2+b^2\ge2ab\ge ab$ so $a,b=0$.
*If $a^2+b^2+c^2=abc$ then all three numbers must be divisible by $3$. Otherwise if $a=b=c=\pm1\pmod{3}$ then the LHS is divisible by $3$ but not the RHS. More generally, if $m$ divides any pair, say $a,b$, then it must divide the other, $c$. In fact, let $3m$ be the gcd of $(a,b,c)$, with $a=3mA$, $b=3mB$, $c=3mC$, so $$A^2+B^2+C^2=3mABC$$ In this form, no pair can have common factors (if $p|A,B$ then $p^2|C^2$), so all $A,B,C$ consist of distinct primes; also $m$ is not even.
In fact there are an infinite number of solutions with $C=1$ $$A^2+B^2+1=3AB$$ Take any integer solutions of $x^2-5y^2=-4$ (there are an infinite number of these from the theory of quadratic forms). Then $A=\frac{3y\pm x}{2}$, $B=y$, $C=1$, satisfy the above equation, corresponding to the solutions of the original problem $$(3A)^2+(3B)^2+3^2=(3A)(3B)3$$ Thus there are an infinite number of such solutions. The first few combinations for $(A,B,C)$ are $(1,1,1)$, $(2,1,1)$, $(5,2,1)$, $(13,5,1)$, $(34,13,1)$, $(89,34,1)$, $(233,89,1)$, $(610,233,1)$, etc.
Some other observations:
*
*If a prime power $p^k$ divides $A$ say, then it also divides $B^2+C^2=A(3mBC-A)$. The sum of squares theorem implies that $p=2$ or $p=1\pmod{4}$; $p=3\pmod{4}$ cannot divide $A$ only, else $B^2+C^2=0\pmod{p}$.
*If $A$ is even, then $B,C$ are odd; but $B^2+C^2$ is not divisible by $4$, so $A$ is not divisible by $4$. Thus either exactly one of $A,B,C$ is divisible by $2$ or none are.
*For each prime $p=1\pmod{4}$, one can solve the problem $A^2+B^2+p^2=3pAB$ using a similar approach as above. Solve $x^2-(9p^2-4)y^2=-4p^2$; then take $A=\frac{3py\pm x}{2}$, $B=y$, $C=p$.
| {
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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,
$$x^2+2y^2 -2xy \geq 0$$
I think I am on the correct path, but I don't know how to proceed.
| 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
$$x^2 -2xy + 2y^2 \ge \frac{x^2+2y^2}4 \ge \frac{x^2+y^2}4\to \infty$$
| {
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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 differential $1$-form $\omega\in\Omega_C$.
In $U_Z:=\{Z\neq 0\}$, we define $x:=\frac{X}{Z}$ and $y:=\frac{Y}{Z}$, so that $y^2=x(x^2-1)$ $(*)$.
I've read in more than one source that the desired form is $\omega:=\frac{dx}{y}$.
Using $(*)$, we have $\frac{dx}{y}=\frac{2dy}{3x^2-1}$ and since $3x^2-1\neq 0$ at the points $(0:0:1),(1:0:1),(-1:0:1)\in C$, we see that $\omega$ has no poles in $U_Z$.
We still have to check that $\omega$ has no pole at infinity $(0:1:0)$.
So we restrict to $U_Y$ and define $u:=\frac{X}{Y}, v:=\frac{Z}{Y}$, so that $v-u^3+uv^2=0$ $(**)$. This way:
$$\frac{dx}{y}=v\cdot d\left(\frac{u}{v}\right)=du-\frac{dv}{v}$$
I still can't see how to use $(**)$ to rewrite $\frac{dv}{v}$ so that the pole will vanish.
Am I missing something?
| 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 = 3u^2du - \left(v^2 du + 2uv dv \right) &\implies (1 + 2uv)dv = (3u^2 - v^2)du\\
&\implies dv = \frac{3u^2 - v^2}{1 + 2uv} du \, .
\end{align*}
I think it's a little easier to start with the other representation for $\omega$ in $x,y$ coordinates:
\begin{align*}
\omega = \frac{2dy}{3x^2 - 1} = \frac{2 d(1/v)}{3(u/v)^2 - 1} = 2 \frac{-\frac{1}{v^2} dv}{3\frac{u^2}{v^2} - 1} = -2 \frac{dv}{3u^2 - v^2} \, .
\end{align*}
Rewriting $dv$ in terms of $du$ yields
\begin{align*}
\omega = -2 \frac{1}{3u^2 - v^2} \frac{3u^2 - v^2}{1 + 2uv} du = \frac{-2}{1 + 2uv} du \, .
\end{align*}
Since the denominator doesn't vanish at $(u,v) = (0,0)$, we see that $\omega$ doesn't have a pole at $\infty$.
| {
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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 space of the system.
I understand that what I've to do is to find a matrix $A_{5x5}$ such that:
\begin{align*}
A \begin{pmatrix}
1\\
-1\\
0\\
5\\
1
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0\\
0\\
0
\end{pmatrix} \ \text{ , } \ A \begin{pmatrix}
1\\
0\\
1\\
0\\
-2
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0\\
0\\
0
\end{pmatrix} \ \text{ and, } \ A \begin{pmatrix}
-2\\
0\\
-1\\
0\\
-1
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0\\
0\\
0
\end{pmatrix}
\end{align*}
So I think that the solution can be to propose a new system of linear equations but I'm not sure how can I do it. How can I find the matrix $A$?
| 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 $\mathbf{x}=(-3t,5s-2t,5t,s,t)=t(-3,-2,5,0,1)+s(0,5,0,1,0)$.
Thus $$A=\begin{pmatrix}-3&-2&5&0&1\\0&5&0&1&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{pmatrix}$$ is one possible matrix. (Any equivalent matrix will do.)
| {
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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
$$\cos^2(2x)-\sin^2(2x) +1=8\cos^4(x)-8\cos^2(x)+2$$$$\Rightarrow \cos^2(2x)=4\cos^4(x)-4\cos^2(x)+1$$
$$\Rightarrow \cos^2(2x)=(2\cos^2(x)-1)^2$$ $$\Rightarrow \cos^2(2x)=(\cos^2(x)-\sin^2(x))^2=(\cos(2x))^2$$
$$\therefore \cos(4x)=8\cos^4(x)-8\cos^2(x)+1$$
My professor says that this is an invalid proof as it was not proven using LHS and RHS and assumed that they were equivalent to add 1 to each side. My question: are there any proofs that would not work if you assumed that the two side were equivalent? Any examples will be much appreciated!
| 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 $1=2$:
(1): $1=2$
(2): $1*0=2*0$ (multiplying both sides by 0)
(3) $0=0$
$\therefore 1 = 2$.
| {
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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$${n+2}\choose{3}$
So far the only thing I can come up with is
${n(n+1)(2n+1)}\over{6}$ $+ {n(n+1)\over{2}} =$ $2{{n+2}\choose{3}}$
${n(n+1)(2n+1)}\over{3}$ $+ {n(n+1)} =$ ${{n+2}\choose{3}}$
$n(n+1)($${(2n+1)}\over{3}$ $+ 1) =$ ${{n+2}\choose{3}}$
And from here on I am stumped. Can anyone direct me or give me a hint as to what I should do next?
| 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 factoring on the left side. Doing that gives
$$\begin{equation}\begin{aligned}
\frac{n(n + 1)(2n + 1)}{3} + n(n+1) & = n(n + 1)\left(\frac{2n + 1}{3} + 1\right) \\
& = n(n + 1)\left(\frac{2n + 4}{3}\right) \\
& = \frac{2n(n + 1)(n + 2)}{3}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Using the standard definition of the binomial coefficient being $\binom{a}{b} = \frac{a!}{b!(a - b)!} = \frac{a(a-1)\ldots (a - b + 1)}{b!}$ gives the right side of \eqref{eq1A} to be
$$\begin{equation}\begin{aligned}
4\binom{n+2}{3} & = 4\left(\frac{(n+2)(n+1)n}{1(2)(3)}\right) \\
& = \frac{2n(n + 1)(n + 2)}{3}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Thus, the left side and right sides of \eqref{eq1A} match, showing it's always true.
| {
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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's how it is in the four options as well. Without actually dividing each polynomials in the options, which looks a tedious way, is there a method for finding it?
| $$x^7+x^2+1=x^7-x+x^2+x+1=$$
$$=(x^2+x+1)(x(x-1)(x^3+1)+1)=$$
$$=(x^2+x+1)(x^5-x^3+x^2-x+1).$$
| {
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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 be apprecaited. Thanks.
| 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
$$ \begin{bmatrix}2 & 3 & \cdots & n+1 \\ 3 & 4 & \cdots & n+2 \\4 & 5 & \cdots & n+3 \\ \vdots & \vdots & \ddots & \vdots \\ n+1 & n+2 & \cdots & 2n\end{bmatrix} \stackrel{R_2-R_1}\to \begin{bmatrix}2 & 3 & \cdots & n+1 \\ 1 & 1 & \cdots & 1 \\4 & 5 & \cdots & n+3 \\ \vdots & \vdots & \ddots & \vdots \\ n+1 & n+2 & \cdots & 2n\end{bmatrix} \stackrel{R_3-(R_1+2R_2)}\to$$
$$ \to \begin{bmatrix}2 & 3 & \cdots & n+1 \\ 1 & 1 & \cdots & 1 \\0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ n+1 & n+2 & \cdots & 2n\end{bmatrix} \stackrel{R_4-(R_1+3R_2)}\to \ldots \stackrel{R_n-(R_1+(n-1)R_2)}\to \begin{bmatrix}2 & 3 & \cdots & n+1 \\ 1 & 1 & \cdots & 1 \\0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0\end{bmatrix} $$
| {
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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 obvious,
when $n>3$, $n-(n-2)n^{\frac{1}{n-2}}<2<H_n$)
So from (1), we know
$$H_n>n(n+1)^{\frac{1}{n}}-n, n>1;\ (3)$$
from (2), we know
$$H_n>n-(n-2)n^{\frac{1}{n-2}},n>2.\ (4)$$
It seems inequality $(3)$ is better than $(4)$, that is to say,
$$\boxed{n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>2n,\ n\geq3.}$$
My concern is whether there are tricky proofs for above inequality without complicated derivative calculations.
Any helps, hints and comments will welcome.
| 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
$$n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>n+1+n-1=2n,n\geq 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 $x\in \{\frac43,\frac53,\frac{13}3\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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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, $A=2.8$ $B=0.04$ $C=-10$.
But on my book solution sheet, $A = 6$. Could you please help me out what have I done wrong?
| Let
$$2+0.8x-0.04x^2=-\frac4{100}\left(x^2-20x-50\right)=-\frac4{100}\left(x^2-20x+100-150\right)$$
$$-\frac4{100}\left((x-10)^2-150\right)=-0.04(x^2-10)^2+6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830798",
"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 $n=196$
This question is from the book engineering mathematics by K A Stroud
| 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 $$\left|\frac {2n+3}{n+4}-2\right|=0.02 \iff $$ $$\iff \frac {5}{n+4}=0.02 \iff$$ $$\iff \frac {n+4}{5}=(0.02)^{-1}=50 \iff $$ $$\iff n+4=(5) \frac {n+4}{5}=(5)(50)=250 \iff$$ $$\iff n=(n+4)-4=250-4=246.$$
This is not about limits. "$X$ and $Y$ differ by $0.02$" means $|X-Y|$ is $exactly$ $0.02.$
If you want to know which $n\ge 0$ satisfy $|\frac {2n+3}{n+4}-2|<0.02$ then from the first part above, we have $$n\ge 0\implies 0<\frac {5}{n+4}=\left|\frac {2n+3}{n+4}-2\right|.$$ So if $n\ge 0$ then $$\left|\frac {2n+3}{n+4}-2\right|<0.02 \iff $$ $$ \iff 0<\frac {5}{n+4}<0.02 \iff$$ $$ \iff \frac {n+4}{5}>(0.02)^{-1}=50 \iff n>246.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $\frac{1}{x-a}$ :
$\left\langle\operatorname{P.\!v.}\left(\frac{1}{x - a}\right),\varphi\right\rangle: =\lim_{\varepsilon\to 0^+}\int_{|x - a|\geq \varepsilon}\frac{\varphi(x)}{x-a} \, \mathrm{d}x$
How do I show that :
$$\lim_{a\to 0} \frac{1}{2a}\left(\operatorname{P.\!v.}\left(\frac{1}{x - a}\right) - \operatorname{P.\!v.}\left(\frac{1}{x + a}\right)\right) = \operatorname{F.\!p.}\left(\frac{1}{x^2}\right)$$
I proved that :
$$\left\langle\operatorname{P.\!v.}\left(\frac{1}{x - a}\right) - \operatorname{P.\!v.}\left(\frac{1}{x + a}\right), \varphi \right\rangle = \lim_{\varepsilon\to 0^+}\int_{ \varepsilon}^M\frac{2\varphi(x)-\varphi(2a - x) - \varphi(x-2a)}{x-a} \, \mathrm{d}x $$
Is my reasoning correct?Thanks for any help.
| If you know that $\left( \operatorname{pv}\frac{1}{x} \right)' = -\operatorname{fp}\frac{1}{x^2}$:
$$
\left< \frac{1}{2a} \left( \operatorname{pv}\frac{1}{x-a} - \operatorname{pv}\frac{1}{x+a} \right), \varphi(x) \right>
= \frac{1}{2a} \left( \left< \operatorname{pv}\frac{1}{x-a}, \varphi(x) \right> - \left< \operatorname{pv}\frac{1}{x+a}, \varphi(x) \right> \right) \\
= \frac{1}{2a} \left( \left< \operatorname{pv}\frac{1}{x}, \varphi(x+a) \right> - \left< \operatorname{pv}\frac{1}{x}, \varphi(x-a) \right> \right)
= \left< \operatorname{pv}\frac{1}{x}, \frac{\varphi(x+a) - \varphi(x-a)}{2a} \right> \\
\to \left< \operatorname{pv}\frac{1}{x}, \varphi'(x) \right>
= - \left< \left( \operatorname{pv}\frac{1}{x} \right)', \varphi(x) \right>
= \left< \operatorname{fp}\frac{1}{x^2}, \varphi(x) \right>
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 at which the incircle is tangent to the side $AB$ and $BC$ be $F$ and $G$ respectively. (Diagram given at end of question)
I tried two approaches:
Approach 1 : I tried to use trigonometry for this one. Let $\angle BAO= \alpha$. And let $r$ be the magnitude of inradius of the triangle $ABC$. Also let $EF=y$ and $DG=z$. Then in $\triangle AGO$ and $\triangle OEG$, $z=r \tan \alpha$ and $z=\frac{r}{\tan \left(\frac{\pi}{4}-\alpha \right)} - 12 $. Combing both of them I get this:
$$12=r \left( \frac{1}{\tan \left(\frac{\pi}{4}-\alpha \right)}- \tan \alpha \right) $$
Similarly in $\triangle AGO$ and $\triangle OFD$ :
$$8= r \left( \dfrac{1}{\tan \alpha} - \tan \left( \frac{\pi}{4} - \alpha\right) \right)$$
Dividing those two equations and simplifying by letting $t=\tan \alpha$, I get the following expression:
$t^3+4t^2-2t+2=0$ which has an extremely horrifying solution, so I left it.
Edit: I am a complete fool and did the calculations wrong: after dividing it simplifies to $(2t-1)(t+3)=0$ after which it immediately follows that $x=24$.
Approach 2 (incomplete) : So instead of using trigonometry, I tried to use elementary techniques this time. Since $\triangle AFO \sim \triangle OGD$, we get
$$\frac{r}{y}=\frac{z+12}{r} \implies r^2 = y(z+12)$$
Similarly $r^2=z(8+y)$. Combining those two, we get $3y=2z$. The value of $x=20+y+z$. Now only if I could find $y+z$ the question would be finished. Would please anybody give any hint or could provide a better solution?
Diagram of original question:
Labelled diagram:
| 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 $$t=\frac{4}{3},$$
$$a=\frac{4}{5}x$$ and $$c=\frac{3}{5}x,$$ which gives $x=24.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = (-3/2)(2\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = (-3/2)(\sin^22x)$
How can I get it into $ 1 - (3/4)\sin^2(2x)$?
| You can use $a^3+b^3=a^2-ab+b^2$ and proceed
$$\begin{equation}\begin{aligned}
&\sin^6 x+\cos^6 x\\[3ex]
&=(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)\\[2ex]
&=\color{green}{\sin^4x}-\sin^2x\cos^2x+\color{green}{\cos^4x}\\[2ex]
&=\color{green}{(\sin^2x+\cos^2x)^2}-3\sin^2x\cos^2x\\[2ex]
&=1-\frac{3}{\color{red}4}(\color{red}4\sin^2x\cos^2x)&\text{($\because 4\sin^2x\cos^2x=\sin^22x$)}\\[2ex]
&=\fbox{$1-\frac{3}{4}(\sin^22x)$}\\
\end{aligned}\end{equation}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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) ${S_n} < \frac{\pi }{{3\sqrt 3 }}$
(B) ${S_n} > \frac{\pi }{{3\sqrt 3 }}$
(C) ${T_n} < \frac{\pi }{{3\sqrt 3 }}$
(D) ${T_n} > \frac{\pi }{{3\sqrt 3 }}$
The official Answer is A and D
My approach is as follows
$\mathop {\lim }\limits_{n \to \infty } {S_n} = \sum\limits_{k = 1}^n {\frac{1}{{n\left( {1 + \frac{k}{n} + {{\left( {\frac{k}{n}} \right)}^2}} \right)}}} $
${S_n} < \int\limits_0^1 {\frac{{dx}}{{{x^2} + x + 1}}} = \int\limits_0^1 {\frac{{dx}}{{{x^2} + x + \frac{1}{4} + \frac{3}{4}}}} = \int\limits_0^1 {\frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} = \frac{1}{{\frac{{\sqrt 3 }}{2}}}} \left( {\left. {{{\tan }^{ - 1}}\frac{{x + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}}} \right|_0^1} \right) = \frac{2}{{\sqrt 3 }}\left( {{{\tan }^{ - 1}}\sqrt 3 - {{\tan }^{ - 1}}\frac{1}{{\sqrt 3 }}} \right) = \frac{2}{{\sqrt 3 }}\left( {\frac{\pi }{3} - \frac{\pi }{6}} \right) = \frac{\pi }{{3\sqrt 3 }}$
So A is correct
$\mathop {\lim }\limits_{n \to \infty } {T_n} = \sum\limits_{k = 0}^{n - 1} {\frac{1}{{n\left( {1 + \frac{k}{n} + {{\left( {\frac{k}{n}} \right)}^2}} \right)}}} $
$t = k + 1\mathop {\lim }\limits_{n \to \infty } {T_n} = \sum\limits_{t = 1}^n {\frac{1}{{n\left( {1 + \frac{{t - 1}}{n} + {{\left( {\frac{{t - 1}}{n}} \right)}^2}} \right)}}} $
The answer is D but don't know to proceed.
| 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.
Therefore
$\lim_{n\to\infty}\sum_{k = 1}^n f(x)$ represents left Riemann sum
and
$\lim_{n\to\infty}\sum_{k = 0}^{n-1} f(x)$ the right Riemann sum.
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ Left Riemann Sum $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ Right Riemann Sum
Now show that the value of integral is $\boxed{\frac{\pi}{3\sqrt3}}$
Conclude that $T_n$ is greater than this and $S_n$ is lesser than this
| {
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"url": "https://math.stackexchange.com/questions/3842432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $x>\frac{1}{2}(-3+\sqrt{17}).$ This is a quadratic, so find the roots and determine which points $x$ satisfy the inequality (you can draw a sketch).
Thus combining the solutions we have $x<\frac{1}{2}(-3-\sqrt{17}),$ $x>2$ or $\frac{1}{2}(\sqrt{17}-3)<x<1.$
See the graph here.
| {
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"url": "https://math.stackexchange.com/questions/3842882",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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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 $$x^5-1=(x-1)(1+x+x^2+x^3+x^4)=0$$
Thus $$1+x+x^2+x^3+x^4=0$$ I tried using this condition to simplify the above expression but nothing interesting simplified. Please note i am looking for hints rather than complete solutions.
EDIT:I came to know its a duplicate ,but i feel that the answers given below are different from those in the original.
| $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3} =$$
$$=\frac{x}{1+x^2}\cdot\frac{x^4}{x^4}+\frac{x^2}{1+x^4}\cdot\frac{x^3}{x^3}+\frac{x^3}{1+x}\cdot \frac{x^2}{x^2}+\frac{x^4}{1+x^3}\cdot\frac{x}{x}= $$
(remember that $x^5=1$, so $x^6=x$ and $x^7=x^2$)
$$=\frac{1}{x^4+x}+\frac{1}{x^3+x^2}+\frac{1}{x^2+x^3}+\frac{1}{x+x^4}=
2\left(\frac{1}{x+x^4}+\frac{1}{x^2+x^3}\right)=$$
$$= 2\left(\frac{x^2+x^3+x+x^4}{(x+x^4)(x^2+x^3)} \right) = 2\left(\frac{x+x^2+x^3+x^4}{x^3+x^4+x^6+x^7} \right) = 2\left(\frac{x+x^2+x^3+x^4}{x^3+x^4+x+x^2} \right) =2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 this question in an ancient ($19$th century) Trigonometry book, and this is the method they use to prove the result (please note: I understand this method fully):
Let $x=\tan A$ ,$~y=\tan B$ and $z=\tan C$ which is acceptable without any loss of generality.
This means that we are saying that
$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$
Consider $\tan (A+B+C)$:
$$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1- \tan A \tan B- \tan C \tan A - \tan B \tan C}$$
So if $~\tan A+\tan B+\tan C=\tan A\tan B\tan C~~$ then $~~\tan (A+B+C)=0$. Hence, let $~A+B+C=\pi$.
Now consider $\tan (2A+2B+2C)$:
$$\tan (2A+2B+2C)=\frac{\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C}{1- \tan 2A \tan 2B- \tan 2C \tan 2A - \tan 2B \tan 2C}=0$$
$$\implies \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$$
$$\implies \frac{2\tan A}{1-\tan^2 A}+\frac{2\tan B}{1-\tan^2 B}+\frac{2\tan C}{1-\tan^2 C}=\frac{2\tan A}{1-\tan^2 A}\times\frac{2\tan B}{1-\tan^2 B}\times\frac{2\tan C}{1-\tan^2 C}$$
$$\therefore\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$$
as required.
My question is, is there any other way of proving this without this rather heavy use of trigonometry? I also would prefer not to work through heaps of algabraic manipulation and expansion to obtain the required result, although if that's necessary I will put up with it ;)
Thank you for your help.
| 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\ $, you get :
$$x y^2 z^2 - x y^2 - x z^2 + x +x^2 y z^2 - x^2 y - y z^2 + y+ x^2 y^2 z - x^2 z - y^2 z + z =$$
$$x+y+z+xyz(yz+xy+xz)- x y^2 - x z^2- x^2 y - y z^2 - x^2 z - y^2 z=$$
$$(x+y+z)(yz+xy+xz+1)- x y^2 - x z^2- x^2 y - y z^2 - x^2 z - y^2 z=$$
$$xyz+xyz+xyz+x+y+z=3xyz+x+y+z = 4xyz$$
| {
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"url": "https://math.stackexchange.com/questions/3844870",
"timestamp": "2023-03-29T00:00:00",
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$A^{n}$ 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{pmatrix}
0 & 0 & 1\\
0 & 1 & 1 \\
1 & 1 & 1
\end{pmatrix} \qquad \qquad
A^2=\begin{pmatrix}
1 & 1 & 1\\
1 & 2 & 2 \\
1 & 2 & 3
\end{pmatrix} \qquad \qquad
A^3=\begin{pmatrix}
1 & 2 & 3\\
2 & 4 & 5 \\
3 & 5 & 6
\end{pmatrix} \\
A^4=\begin{pmatrix}
3 & 5 & 6\\
5 & 9 & 11 \\
6 & 11 & 14
\end{pmatrix} \qquad \qquad
A^5=\begin{pmatrix}
6 & 11 & 14\\
11 & 20 & 25 \\
14 & 25 & 31
\end{pmatrix} \qquad \qquad
A^6=\begin{pmatrix}
14 & 25 & 31\\
25 & 45 & 56 \\
31 & 56 & 70
\end{pmatrix}
\end{equation}
That gives the following terms
\begin{equation}
A_{11} = 0,1,1,3,6,14,...\\
A_{12} = 0,1,2,5,11,25,...\\
A_{22} = 1,2,4,9,20,45,...
\end{equation}
But i can't figure out the succesion in terms of n.
| With $\{e_1, e_2, e_3\}$ as your base, check what happens when you operate on them
$$Ae_1 = e_3 \implies A^ne_1 = A^{n-1}e_3$$
$$Ae_2 = e_2+e_3 \implies A^ne_2 = A^{n-1}(e_2+e_3) = A^{n-1}e_3 + A^{n-1}e_2$$
$$Ae_3 = e_1+e_2+e_3\implies A^ne_3 = A^{n-1}(e_1+e_2+e_3) = $$
$$ = A^{n-1}e_1+A^{n-1}e_2+A^{n-1}e_3 = 2A^{n-2}e_3+A^{n-2}e_2 +A^{n-1}e_3$$
This implies a recursive expression for each of your columns, maybe it can be tweaked a bit further.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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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 $x^{12}-x^9+x^4-x+1>0$ for all values of $x$.
But I want to know how to find the required values of $x$ without graphing
| 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
$$x^{12}-x^9+x^4-x+1={\frac {{x}^{13} \left( {x}^{2}+x+1 \right) \left( {x}^{8}+1 \right)
+ \left( {x}^{2}+x+1 \right) \left( {x}^{6}+{x}^{3}-x+1 \right) }{
\left( {x}^{2}+x+1 \right) \left( x+1 \right) \left( {x}^{2}+1
\right) \left( {x}^{2}-x+1 \right) \left( {x}^{4}-{x}^{2}+1
\right) }}$$
For $x>0,$ it's clearly true!
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
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 prove that this isn't applicable for any number less than $1$? Also, where do I go after this? Thank you!
| HINT
$$
\begin{split}
k^3 + 3k^2+8k+6
&= 3(k^2+3k+2) + k^3-k \\
&= 3(k^2+3k+2) + k(k^2-1) \\
&= 3(k^2+3k+2) + k(k-1)(k+1)
\end{split}
$$
| {
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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 $\delta =min\left(min\left(1,1+2a\right),\frac {\varepsilon}{1+2|a|}\right)$ would work.
Is this correct? Is my choosing of $\delta$ ok?
| 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?
| {
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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 the fault in it.
The $3$ cases where the sum of $3$ distinct outcomes of the die is $10$ are: $A=(3,5,2)$ $B=(4,5,1)$ $C=(1,3,6)$.
Probability $(A) = 3!\cdot (1/6)^3$ .
Reasoning.
$(1/6)^3$=probability to get $3/5/2$ (they are independent events)
$3!$ = number of ways to arrange $3,5,2$
This is the same for $B,C$ as well. Hence the final answer should be $3\cdot p(A)=1/12$.
The actual answer is $3/20$ (solve using conditional probability). What are the wrong steps I have assumed or taken?
| 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",
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"source": "stackexchange",
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} |
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 for all real values $x^2+|x-2| >1.$.
Is my proof correct. What can be other methods to prove this?
| 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:
$$\begin{align}x^2 + |x-2| &= x^2-x+2 \\&= x^2-x+\frac14 + \frac74 \\&= \left(x-\frac12\right)^2+\frac 74 \\&\ge \frac74>1\end{align}$$
using the fact that squares are nonnegative.
| {
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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
$$a_{n+1} = \frac{1}{2}(a_n+b_n), \quad b_{n+1} = \frac{1}{2}(b_n+c_n) \quad \text{and}\quad c_{n+1} = \frac{1}{2}(c_n+a_n).$$
Prove that the sequences are convergent and find their limits.
My attempt: I try to find the closed-form solution for $a_n,b_n,c_n$.
Note that
$$
\begin{pmatrix}
a_{n+1} \\
b_{n+1} \\
c_{n+1} \\
\end{pmatrix} = \begin{pmatrix}
\frac{1}{2} & \frac{1}{2} & 0 \\
0 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & 0 & \frac{1}{2}
\end{pmatrix} \begin{pmatrix}
a_n \\
b_n \\
c_n
\end{pmatrix} = A\begin{pmatrix}
a_n \\
b_n \\
c_n
\end{pmatrix}.
$$
Clearly
$$
\begin{pmatrix}
a_{n} \\
b_{n} \\
c_{n} \\
\end{pmatrix} = A^{n}\begin{pmatrix}
a_0 \\
b_0 \\
c_0 \\
\end{pmatrix}.
$$
To compute $A^n$, I try to diagonalize $A$ over complex numbers.
From Wolfram alpha, $A$ has eigenvalues $1, \frac{1}{4}(1+i\sqrt3)$ and $\frac{1}{4}(1-i\sqrt3)$.
Therefore,
$$
\begin{pmatrix}
a_{n} \\
b_{n} \\
c_{n} \\
\end{pmatrix} = X \begin{pmatrix}
1 & 0 & 0 \\
0 & \left( \frac14 (1+i\sqrt 3) \right)^n & 0 \\
0 & 0 & \left( \frac14 (1-i\sqrt 3) \right)^n
\end{pmatrix} X^{-1} \begin{pmatrix}
a_0 \\
b_0 \\
c_0 \\
\end{pmatrix}
$$
for some invertible matrix $X$ (in fact, columns of $X$ are eigenvectors of $A$).
However, I stuck at here.
I do not know how to proceed to show that the sequences $(a_n), (b_n), (c_n)$ convergent.
It would be good if someone can solve this problem by continuing my method above.
EDIT: After reading comments below, I attempted the following.
From Wolfram alpha, we have
$$\lim_{n\to\infty} A^n = \frac13 J_3$$
where $J_3$ is the $3\times 3$ matrix with all entries $1$.
Therefore,
$$\lim_{n\to\infty} \begin{pmatrix}
a_{n} \\
b_{n} \\
c_{n} \\
\end{pmatrix} = \frac13 J_3 \begin{pmatrix}
a_{0} \\
b_{0} \\
c_{0} \\
\end{pmatrix} = \frac13 \begin{pmatrix}
a_{0} + b_{0} + c_{0} \\
a_{0} + b_{0} + c_{0} \\
a_{0} + b_{0} + c_{0}
\end{pmatrix}.$$
Hence, the sequences $(a_n),(b_),(c_n)$ are convergent and their limits are $\frac13 (a_0 + b_0 + c_0)$.
| 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 column of $X$ is $(a,a,a)^t$ and the first row of $X^{-1}$ is $(b,b,b)$. From $X^{-1}X=I$, we see $3ab=1$. Then $XE_{12}X^{-1}$ has $ab=1/3$ in every entry.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
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 $(1,7)$,
$$\frac{dy}{dx}\Bigg|_{(1,7)}=2\tag{Slope of the line tangent to the parabola}$$
So the equation of the line is $2x-y+5=0\implies y=2x+5$
Substituting this in the equation of circle to find the point of intersection of the line with the circle, we get,
$$x^2+(2x+5)^2+16x+12(2x+5)+c=0$$
Solving this, I get a complicated equation and then the answer comes out in terms of $c$ but the actual answer does not contain $c$ at all.
I would prefer a more analytical/geometrical approach if possible
|
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&=\arctan2
=\arccos\tfrac{\sqrt5}5
=\arcsin\tfrac{2\sqrt5}5
\tag{2}\label{2}
.
\end{align}
Then the radius of the circle
\begin{align}
r=|OT_c|&=|OA|\cdot\cos\phi
=\sqrt5
\tag{3}\label{3}
,\\
T_c&=O+r\cdot(\sin\phi,\,-\cos\phi)
\\
&=O+(2,-1)=(-6,-7)
.
\end{align}
| {
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"answer_count": 3,
"answer_id": 2
} |
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: $$F(x) = \sum_{n=0}^\infty (n+1)x^n = 1+x + 2x^2 + 3x^3 + ...$$
$$xF(x) = \sum_{n=0}^\infty (n+1) x^{n+1} = x+ x^2 +2x^3 + ...$$
$$F(x) - xF(x) =1+ x + x^2 + x^3 + x^4... = \sum_{n=0}^\infty x^n = \dfrac{1}{1 - x}.$$
$$F(x)=\sum_{n=0}^\infty(n+1) x^n=\dfrac{1}{(1 - x)^2}.$$
| {
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"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 $x>0$ and $\sqrt{x-1}\ge0.$ (I am asking if we can use in the inequalities after $x\ge1$ that $x$ is actually greater than or equal to $1$.)
After that we have $A=\dfrac{(x+1)(x^2-x+1)}{x+\sqrt{x-1}}.$ What from here? Thank you in advance!
| The inequality $x>=1$ covers it all. You can simplify further by noting that
$\frac{1}{x+\sqrt{x-1}}=\frac{x-\sqrt{x-1}}{x^2-x+1}$
| {
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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. Multiply it by an integrating factor $\eta(x,y)$ such that it does:
$$ \eta_x P + P_x \eta = \eta_y Q + Q_y \eta$$
Or,
$$ \eta_x P - \eta_y Q + ( P_x - Q_y) \eta = 0$$
This becomes:(*)
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 6x^2 +3 y^3 - 2x^2 -8y^3) \eta =0$$
Or,
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 4x^2 - 11y^3) \eta =0$$
Now... I'm not quite sure what to do.. I don't think I can write $ \eta$ as function solely of $x$ or $y$.. did I miss something or... ?
As correctly observed by @Aleksas Domarkas. There is a mistake after the place where I put *. The current working is as follows:
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 6x^2 +3 y^3 - 2x^2 +8y^3) \eta =0$$
Or,
$$ \eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 4x^2 + 11y^3) \eta =0$$
| $(2x^2y-2y^4)~dx+(2x^3+3xy^3)~dy=0$
$(2x^3+3xy^3)~dy=(2y^4-2x^2y)~dx$
$\dfrac{dy}{dx}=\dfrac{2y^4-2x^2y}{2x^3+3xy^3}$
Let $r=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=2x\dfrac{dy}{dr}$
$\therefore2x\dfrac{dy}{dr}=\dfrac{2y^4-2x^2y}{2x^3+3xy^3}$
$\dfrac{dy}{dr}=\dfrac{y^4-x^2y}{x^4+3x^2y^3}$
$\dfrac{dy}{dr}=\dfrac{y^4-ry}{r^2+3ry^3}$
Let $s=y^3$ ,
Then $\dfrac{ds}{dr}=3y^2\dfrac{dy}{dr}$
$\therefore\dfrac{1}{3y^2}\dfrac{ds}{dr}=\dfrac{y^4-ry}{r^2+3ry^3}$
$\dfrac{ds}{dr}=\dfrac{3y^6-3ry^3}{r^2+3ry^3}$
$\dfrac{ds}{dr}=\dfrac{3s^2-3rs}{r^2+3rs}$
| {
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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 expression by the conjugate of the denominator.
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it still gives the indeterminate form $\frac{0}{0}$ .
But multiplying the expression by the conjugate of the demoninator and numerator we get
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right) $$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{1} \cdot \dfrac{1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{2-x}{2-x} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
Now we can evaluate the limit:
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-2}+1}{\sqrt{6-2}+2}\right) = \dfrac{1}{2}$$
Taking this example, I would like to understand why rationalization was used. What did it change in the expression so the evaluation was possible? Especially, why multiplying by the numerator's and denominator's conjugate?
I am still new to limits and Calculus, so anything concerning concepts I'm missing is appreciated. I still couldn't understand how a limit supposedly tending to $\frac{0}{0}$ went to be $\frac{1}{2}$, I really want to understand it.
Thanks in advance for you answer.
| 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
$$\sqrt{6-x}-2=\frac{2-x}{\sqrt{6-x}+2}$$
$$\sqrt{3-x}-1=\frac{2-x}{\sqrt{3-x}+1}$$
and by the ratio the problematic $x-2$ term cancel out.
As an alternative we can also use binomial first order approximation (i.e. Taylor's series) at $x=2$ to obtain
$$\sqrt{6-x}=\sqrt{4-(x-2)}=2\sqrt{1-\frac{(x-2)}4}=2\left(1-\frac {x-2}{8}+o(x-2)\right)$$
$$\implies \sqrt{6-x}-2=-\frac {x-2}{4}+o(x-2)$$
$$\sqrt{3-x}=\sqrt{1-(x-2)}=1-\frac {x-2}{2}+o(x-2)$$
$$\implies \sqrt{3-x}-1=-\frac {x-2}{2}+o(x-2)$$
which gives evidence of the same problematic term and the final result.
| {
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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 resemblence of numerator and denominator?
| $$L=\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} }$$
With L'Hospital's rule:
$$L=\lim_{x \rightarrow 1} \frac{ 2x + 3/2x^2(x^3+1)^{-1/2} - 2x^3(x^4+1)^{-1/2} }{ 1 + 1/2(x+1)^{-1/2} -x (x^2+1)^{-1/2} }$$
$$L=\frac{ 2 + 3/2(2)^{-1/2} - 2(2)^{-1/2} }{ 1 + 1/2(2)^{-1/2} - (2)^{-1/2} }$$
$$L=\frac{ 2\sqrt 2 + 3/2 - 2 }{ \sqrt 2 + 1/2 -1 }$$
$$ \implies L=\frac{ 4\sqrt 2 -1 }{ 2\sqrt 2 - 1 }=\frac{ 15+2\sqrt 2 }{ 7 }$$
| {
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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 elgant solution using minimum calculus?. Substituiting $x+iy$ makes things very complex indeed.
Answer:
$[\sqrt{\frac{7}{2}},3\sqrt{\frac{7}{6}}]$
| 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 $\frac{13}{4}$ and occures for $x=\frac{-7}{8}.$
The minimal value it's $\sqrt3$ and occurs for $x=\frac{1}{2}.$
For the minimal value there is the following.
We need to prove that:
$$\sqrt{2x+2}+|2x-1|\geq\sqrt3$$ or
$$|2x-1|\geq\frac{1-2x}{\sqrt3+\sqrt{2+2x}},$$ which is true because
$$|2x-1|\left(\sqrt3+\sqrt{2+2x}\right)\geq|2x-1|\geq-(2x-1)=1-2x.$$
For the maximal value there is the following.
Let $x\geq\frac{1}{2}.$
Thus, $$\sqrt{2x+2}+|2x-1|=\sqrt{2x+2}+2x-1\leq\sqrt{2+2}+2-1=3<\frac{13}{4}.$$
Let $x\leq\frac{1}{2}$ and $\sqrt{2+2x}=y$.
Thus, $$\sqrt{2x+2}+|2x-1|=\sqrt{2x+2}-2x+1=\sqrt{2x+2}-2x-2+3=$$
$$=y-y^2+3=-\left(y-\frac{1}{2}\right)^2+\frac{13}{4}\leq\frac{13}{4}.$$
The equality occurs for $y=\frac{1}{2}$ or $x=-\frac{7}{8}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $n = 1$, $f(x, y) = 7xy(x + y)g(x, y)$. From $n$ to $n + 1$,
I am only able to show that $x^2 + xy + y^2$ is a factor of $f(x, y)$. Is there any trick I am missing?
| 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 $(z + 1)^{6n + 1} - z^{6n + 1} - 1$.
Let $h(z)=z^2+z+1$.
We have $(z+1)^{6n} \equiv 1 \bmod h(z)$.
because $(z+1)^6=(z^4 + 5 z^3 + 9 z^2 + 6 z)h(z)+1$.
We have $z^{6n} \equiv 1 \bmod h(z)$
because $z^3= (z-1)h(z)+1$.
Therefore,
$f(z)=(z + 1)^{6n + 1} - z^{6n + 1} - 1 \equiv (z+1) -z -1 = 0 \bmod h(z)$
Now
$f'(z)=(6n+1)((z + 1)^{6n} - z^{6n}) \equiv (6n+1)(1-1) = 0\bmod h(z)$
Therefore, $h$ divides both $f$ and $f'$, and so $h^2$ divides $f$.
All this can be done exactly the same without introducing $z$ but it is easy only in hindsight.
| {
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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 we have that:
$=\sqrt{\frac{\sqrt{6}+\sqrt{3}+1}{2}*\frac{\sqrt{3}+1-\sqrt{6}}{2}*\frac{\sqrt{6}-\sqrt{3}+1}{2}*\frac{\sqrt{6}+\sqrt{3}-1}{2}}$
$=\frac{1}{4}\sqrt{8}$
I state that AK is a height of the triangle.
We have that it is enough if $sin(90+B)=sin(A)$
However we also have that $sin(90+B)=sin(BAK)$ and $sin(BAK)=\frac{BK}{1}=BK$ (BAK is a right angled triangle)
Also:
$\frac{1}{2}*sin(A)*\sqrt{3}=\frac{1}{4}\sqrt{8}$
$sin(A)=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}$
So we just have to prove that $BK=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}$
We have that:
$\frac{AK*BC}{2}=\frac{1}{4}\sqrt{8}$
$AK=\frac{\sqrt{4}}{2\sqrt{3}}$
Hence from Pythagoras we have that:
$BK=\sqrt{\frac{8}{12}}=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}=sin(A)$
Hence it is proved.
I realize that my method of proving it is complex. Could you please show some simple solutions to this problem?
| $$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{3+1-6}{2\cdot\sqrt 3\cdot 1}=-\frac1{\sqrt 3}$$
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 $$\sin B=\sqrt{1-\cos^2 B}=\sqrt{1-\frac 46}=\frac 1{\sqrt 3}$$
Using the formula for $$\cos(x+y)=\cos x\cos y-\sin x\sin y$$will get you the result
| {
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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 $$\sqrt{x+5}+\sqrt{x}=2 \iff \sqrt{x+5}=2-\sqrt{x}$$$$D_x:\begin{cases}x+5\ge0\\x\ge0\\2-\sqrt{x}\ge0\end{cases}\iff x\in[0;4].$$$$x+5=4-4\sqrt{x}+x \text{ ...}$$ We will get that the equation has no roots.
Can I ask you what steps should I follow when I am solving a radical equation with 2 square roots?
| $$\sqrt{x+5}+\sqrt{x}=2\implies$$
$$\sqrt{x+5}=2-\sqrt{x}\implies$$
$$(\sqrt{x+5})^2=(2-\sqrt{x})^2\implies$$
$$x+5=4+x-4\sqrt{x}\implies$$
$$1=-4\sqrt{x}$$
which is impossible since $1>0$ and $-4\sqrt{x}\le0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 to 1 then
$\frac{x-a}{a}>\frac{x-b}{b}$
According to binomial expansion
$(1-a)^b=1-ab+\frac{b(b-1)}{2!}(-a)^2+\frac{b(b-1)(b-2)}{3!}(-a)^3+...$
and
$(1-b)^a=1-ab+\frac{a(a-1)}{2!}(-b)^2+\frac{a(a-1)(a-2)}{3!}(-b)^3+...$
I compare each nth term and multiply the nth term by $\frac{n!}{(ab)^n}$ to simplify the evaluation
Example for the 2nd term of $(1-a)^b$
$\frac{b(b-1)}{2!}(-a)^2.(\frac{2!}{(ab)^2})=\frac{-1(1-b)}{b}$
and 2nd term of $(1-b)^a$
$\frac{a(a-1)}{2!}(-b)^2.(\frac{2!}{(ab)^2})=\frac{-1(1-a)}{a}$
From above inequality
$\frac{1-a}{a}>\frac{1-b}{b}$ times negative one then $\frac{-1(1-b)}{b}>\frac{-1(1-a)}{a}$ which show the 2nd term of $(1-a)^b$ from binomial expansion bigger than $(1-b)^a$
Repeat the procedure and for 3rd term will get
$\frac{-1(1-b)(2-b)}{b^2}>\frac{-1(1-a)(2-a)}{a^2}$ since from above equality $\frac{1-a}{a}>\frac{1-b}{b}$ and $\frac{2-a}{a}>\frac{2-b}{b}$ it follow that $\frac{1-a}{a}.\frac{2-a}{a}>\frac{1-b}{b}.\frac{2-b}{b}$
This proof can be repeat for subsequent term which conclude the proof $(1-a)^b>(1-b)^a$ since all the nth term in $(1-a)^b>(1-b)^a$
| 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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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 $\alpha=2-2\sqrt{\frac{\sin(x)}{x}}$ (I have no clue, what to do next with it), but with Taylor and ended up with:
$$\lim_{x\to 0}\frac{\sqrt{1-\frac{x^2}{6}+o(x^2)}+o(\sqrt{1-\frac{x^2}{6}+o(x^2)})}{x^4} $$
which tends to infinity
| Since there is $x^4$ on denominator we have to go at least the same order on numerator.
$\sin(x)=x-\frac 16 x^3+\frac 1{120}x^5+o(x^5)$
$\dfrac{\sin(x)}x=1-\frac 16 x^2+\frac 1{120}x^4+o(x^4)$
$S=\left(\frac{\sin(x)}x\right)^\frac 12=1+\frac 12\left(-\frac 16 x^2+\frac 1{120}x^4+o(x^4)\right)-\frac 18\left(-\frac 16 x^2+\frac 1{120}x^4+o(x^4)\right)^2+o(x^4)=1+(-\frac 1{12})x^2+(\frac 1{2\times 120}-\frac 1{8\times 6^2})x^4+o(x^4)$
$\cos(1-S)=\cos(\frac 1{12}x^2-\frac 1{1440}x^4+o(x^4))=1-\frac 12\left(\frac 1{12}x^2-\frac 1{1440}x^4+o(x^4)\right)^2+o(x^4)=1+(\frac {-1}{2\times 12^2})x^4+o(x^4)$
$\dfrac{1-\cos(1-S)}{x^4}=\dfrac{\frac 1{288}x^4+o(x^4)}{x^4}=\frac 1{288}+o(1)\to\frac 1{288}$
Note that many terms disappear quickly because they are negligible compared to $x^4$. In particular only the $-\frac 16x^2$ term of $\sin$ development brings something to the final result, but we need nevertheless to make the calculation up to $o(x^4)$ along the entire chain to ensure coherence.
| {
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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 $\int_t^TB_\tau dB_\tau=\frac{1}{2}(B_T^2-B_t^2)-\frac{1}{2}(T-t)$, but I don't know how to continue. I have tried Ito's lemma, but it seemed to make the calculation harder; I also tried Feynman-Kac Formula, but I cannot find any proper function form to solve the PDE...
Appreciate any help/hint!
| 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 \right) &= \exp \left(- \frac{1}{2}(T-t) \right) \mathbb{E}f(B_t) \tag{1} \end{align*}
where
$$f(x) := \mathbb{E}\exp \left( \frac{1}{2} (B_T-B_t)^2 + (B_T-B_t) x \right).$$
Using the scaling property, $B_T-B_t \sim B_{T-t} \sim \sqrt{T-t} B_1$, we see that
$$f(x) = \mathbb{E}\exp \left( \frac{T-t}{2} B_1^2 + \sqrt{T-t} B_ 1 x \right).$$
To compute this function, we need to do some computations involving the Gaussian density. As $B_1 \sim N(0,1)$, we have
$$f(x) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \exp \left( \frac{T-t}{2} y^2 + \sqrt{T-t} yx \right) \exp \left(- \frac{1}{2} y^2 \right) \, dy.$$ In particular, we see that the integral on the right-hand side is infinite if $T-t \geq 1$, and therefore we will assume from now on that $T-t<1$. If we set $\sigma^2 := 1/(1+t-T)$, then
\begin{align*} f(x) &= \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \exp \left(- \frac{1}{2} \frac{y^2}{\sigma^2} + yx \sqrt{T-t} \right) \, dy \\ &= \sqrt{\sigma^2} \exp \left( \frac{1}{2} x^2 \sigma^2 (T-t) \right) \underbrace{\frac{1}{\sqrt{2\pi \sigma^2}} \int_{\mathbb{R}} \exp \left(- \frac{1}{2 \sigma^2} (y- x \sqrt{T-t} \sigma^2)^2 \right) \, dy}_{=1} \\ &= \frac{1}{\sqrt{1+t-T}} \exp \left( \frac{1}{2} \frac{T-t}{1+t-T} x^2 \right). \end{align*}
Plugging this into $(1)$, we obtain that
\begin{align*} \mathbb{E}\exp \left( \int_t^T B_s \, dB_s \right) &= \exp \left(-\frac{1}{2} (T-t) \right) \frac{1}{\sqrt{1+t-T}} \mathbb{E}\exp \left( \frac{1}{2} \frac{T-t}{1+t-T} B_t^2 \right) \\ &= \exp \left(-\frac{1}{2} (T-t) \right) \frac{1}{\sqrt{1+t-T}} \mathbb{E}\exp \left( \frac{t}{2} \frac{T-t}{1+t-T} B_1^2 \right). \end{align*}
As $\mathbb{E}e^{rB_1^2} = 1/\sqrt{1-2r}$ for any $r<1$ (you can check this using the density of $B_1$), we conclude that
\begin{align*} \mathbb{E}\exp \left( \int_t^T B_s \, dB_s \right)&= \exp \left(-\frac{1}{2} (T-t) \right) \frac{1}{\sqrt{1+t-T}} \frac{1}{\sqrt{1- t \frac{T-t}{1+t-T}}} \\ &= \exp \left(-\frac{1}{2} (T-t) \right) \frac{1}{\sqrt{1+t-T-(T-t) t}}. \end{align*}
| {
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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)((k+1)+7) = (k+3)(k+4)(k+8)\\
= &(k+3)[(k+2)+2][(k+7)+1]\\
= &[(k+3)(k+2)+(2)(k+3)][(k+7)+1]\\
= &(k+2)(k+3)(k+7)+2(k+3)(k+7)+(k+2)(k+3)+2(k+3)\\
= &6P+2k^2+20k+42+k^2+5k+6+2k+6\\
= &6P+3k^2+27k+54\\
= &6p+3(k^2+9k+18)
\end{align*}
I'm not sure what to do, my proof turned out to be divisible by 3 instead of 6. Please let me know how I can move forward with this. Thank you!
| 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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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 + a x^2 + x +1 +1 $ , now the degree is 4 and I am not sure what to do...Can anyone help? Thanks in advance.
| 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 want the range of $f(f(x))$
We are plugging $f(x)$ back into to $f(x)$.
So we can just look at $f(x)$ with the restriction that $x \ge 1-\frac{1}{4a}$
We want to see if $1-\frac{1}{4a}$ is less than $\frac{-1}{2a}$.
Suppose
$1-\frac{1}{4a}<\frac{-1}{2a}$
$4a-1 < -2$
$a<\frac{-1}{4}$
This is impossible since we already know that $a \ge 0$. So we know that
$1-\frac{1}{4a} \ge \frac{-1}{2a}$
So we're on the right side of the vertex. And as x gets greater than $1-\frac{1}{4a}$, $f(x)$ will only get bigger.
So $f(f(x))$ achieves a minimum when $f(x) = 1-\frac{1}{4a}$
We just need to guarantee $f(1-\frac{1}{4a}) \ge 0$
$a(1-\frac{1}{4a})^2+1-\frac{1}{4a}+1 \ge 0$
Simplify this to get
$a - \frac{3}{16a} + 1.5 \ge 0$
$16a^2 + 24a -3 \ge 0$
$a \ge \frac{-3 + 2\sqrt{3}}{4}$
or
$a \le \frac{-3 - 2\sqrt{3}}{4}$
But we know that $a \ge 0$ so the solution is:
$a \ge \frac{-3 + 2\sqrt{3}}{4}$
| {
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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 with k+1, as I get complete gibberish. Any suggestions
Could someone guide me through each step?
My work so far:—
Base Cases include $x = 1, 2.$
$X_{1} = 2(-1)^1 + 10 \cdot 3^0$
$8 = 8$
$X_2 = 2(-1)^2 + 10 \cdot 3^{1}$
$32 = 32$
Inductive Step: $i = k$
$X_k = 2(-1)^k + 10 \cdot 3^{k-1}$
Inductive Conclusion:
$X_{k+1} = X_k + X_{k-1}$
$X_{k+1} = 2(2(-1)^k + 10 \cdot 3^{k-1}) + 3(2(-1)^{k-1} + 10 \cdot 3^{k-2})$
$X_{k+1} = 4(-1)^k + 6(-1)^{k-1} + 20 \cdot 3^{k-1} + 10 \cdot 3^{k-1}$
At this point. I know I should factor, but the 2nd half of $3^{k-1}$ does not create $10\cdot3^{k+1}$ as needed.
| 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 formula holds for $X_k$ and $X_{(k-1)}.$
Then $$X_{(k+1)} = [2 \times X_k] + [3 \times X_{(k-1)}]$$
$$= 2 \times \{[2 \times (-1)^{k}] + [10 \times 3^{(k-1)}]\} +
3 \times \{[2 \times (-1)^{(k-1)}] + [10 \times 3^{(k-2)}]\}$$
$$= [2 \times (-1)^{(k-1)} \times (3-2)] + [10 \times 3^{(k-2)} \times (3 + <2\times 3>)]$$
$$= [2 \times (-1)^{(k+1)}] + [10 \times 3^{(k-2)} \times (9)]$$
$$= [2 \times (-1)^{k+1)}] + [10 \times 3^{(k)}].$$
This completes the inductive step.
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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
$$\frac{1}{(x-y)^2}\frac{1}{x^2}=-\frac{2}{y^3(x-y)}+\frac{1}{y^2(x-y)^2}+\frac{2}{y^3x}+\frac{1}{y^2x^2}$$
| 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 $d$.
$x= 0 \Rightarrow b = \frac{1}{(0-y)^2} = \frac{1}{y^2}$
$x =y \Rightarrow d = \frac{1}{y^2}$
Now you have,
$\frac{1}{(x-y)^2x^2} = \left(\frac{a}{x} + \frac{1}{y^2x^2}\right) + \left(\frac{c}{x-y}+\frac{1}{y^2(x-y)^2}\right)$
$\Rightarrow1 = ax(x-y)^2 + c(x-y)x^2 + \frac{(x-y)^2+x^2}{y^2}$
Let us consider only coefficients of $x^2$ and $x^3$
$ (a+c) = 0 $ and $(-2ay-cy+\frac{2}{y^2}) = 0$
Using $c = -a , -ay + \frac{2}{y^2} = 0 \Rightarrow a= \frac{2}{y^3}$ and $c= -\frac{2}{y^3}$
Thus,
$$\frac{1}{(x-y)^2x^2} = \left(\frac{2}{y^3x} + \frac{1}{y^2x^2}\right) + \left(\frac{-2}{y^3(x-y)}+\frac{1}{y^2(x-y)^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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, the integral becomes
$$I=\int_0^\infty \frac{1}{1+x+x^2}dx
- 2\int_0^\infty \frac{x^3}{1+x+x^2+x^3+x^4+x^5}dx
$$
The first integration is straightforward $\frac{2\pi}{3\sqrt3}$ and the second is carried out via partial fractional decomposition. After some tedious and lengthy evaluation, it surprisingly yields the same value $\frac{2\pi}{3\sqrt3}$.
Given its vanishing value, there may exist a shorter and more direct derivation, without evaluation of any explicit intermediate values.
| 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}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3879991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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 integrals. So I think that maybe a variable
change may be an appropriate approach here.
Can someone suggest me a useful variable change to calculate the area of this region?
| split the area into 5 separate areas:
$$\int_{\pi}^{\pi+\tan^{-1}(1/\sqrt{2})}\frac{1}{2}r^2d\theta+
\int_{\pi+\tan^{-1}(1/\sqrt{2}))}^{-\tan^{-1}(1/\sqrt{2}))}\frac{1}{2}r^2d\theta+
\int_{-\tan^{-1}(1/\sqrt{2}))}^{0}\frac{1}{2}r^2d\theta+
\int_{0}^{\tan^{-1}(4/5)}\frac{1}{2}r^2d\theta+
\int_{\tan^{-1}(4/5)}^{\pi}\frac{1}{2}r^2d\theta
$$
the second and the last integral are triangles, and the first and the third are equal:
$$
\sqrt{2}+
2\int_{-\tan^{-1}(1/\sqrt{2}))}^{0}\frac{1}{2}r^2d\theta+
\int_{0}^{\tan^{-1}(4/5)}\frac{1}{2}r^2d\theta+
\frac{2}{3}
$$
the polar function that defines the hyperbola is:
$$r=\left(\frac{1}{\cos2\theta}\right)^{\frac{1}{2}}$$
so putting it into the integral then evaluating:
$$
\sqrt{2}+
2\int_{-\tan^{-1}(1/\sqrt{2}))}^{0}\frac{1}{2}\left(\frac{1}{\cos2\theta}\right)d\theta+
\int_{0}^{\tan^{-1}(4/5)}\frac{1}{2}\left(\frac{1}{\cos2\theta}\right)d\theta+
\frac{2}{3}
$$
$$
=\sqrt{2}+
\frac{1}{2}\ln \left(3\right)-\frac{1}{2}\ln \left(3-2\sqrt{2}\right)+
\frac{2}{3}
$$
| {
"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 $\alpha \gt 1$. I thought that Dirichlet's Criterion is the most suitable for this series, because $\ln \frac{n + 1}{n - 1}$ is decreasing and $\lim_{n \rightarrow \infty} \ln \frac{n + 1}{n - 1} = 0$. But $\sum_{n = 2}^\infty (\sqrt{n + 1} - \sqrt{n})^\alpha$ seems to be divergent regardless of $\alpha$, so I can't use this idea.
| You have $$(\sqrt{n + 1} - \sqrt{n})^\alpha \ln \left(\frac{n + 1}{n - 1} \right)= (\sqrt{n})^{\alpha}\left(\sqrt{1 + \frac{1}{n}} - 1\right)^\alpha \ln \left(1 +\frac{2}{n - 1} \right)$$
$$ \sim (\sqrt{n})^{\alpha}\left(\frac{1}{2n}\right)^\alpha\frac{2}{n-1} = \frac{2}{(2\sqrt{n})^{\alpha}(n+1)} \sim \frac{2^{1-\alpha}}{n^{1+\frac{\alpha}{2}}}$$
So the series converges iff $1+\frac{\alpha}{2} > 1$, i.e. iff $\alpha > 0$.
| {
"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 $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.
The first few terms of this series shows me that summation $\lfloor \log_2 n \rfloor$ for $n=1$ to $n=10$ give $2^{n +1}$.
| 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 gives $m=7$.
Now, $$\frac{1994-((7+1)2^{7+1}-2^{7+2}+2)}{8}=57$$ and we obtain:
$$n=1+2^1+...+2^7+57=2^8-1+57=312.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
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