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Does $\int_0^\infty\frac{\ln(1+x)}{x(1+x^n)}dx$ have a general form?
Does $$I_n=\int_0^\infty\frac{\ln(1+x)}{x(1+x^n)}dx$$ have a general form?
I tried to evaluate some small $n$s.
For $n=1$, $I_1$ is obviously $\frac16\pi^2$.
For $n=2$, see here.
$I_2=\frac5{48}\pi^2$.
For $n=3$, I put it in Mathematica and get $$\small{\frac{1}{108} \left(9 \left(4 \left(\text{Li}_2\left(\frac{\sqrt[6]{-1}}{\sqrt{3}}\right)+\text{Li}_2\left(-\frac{(-1)^{5/6}}{\sqrt{3}}\right)\right)+\log ^2(3)\right)+5 \pi ^2\right)}$$ Use the result $$\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)=\frac1{12}\pi^2-\frac12\arctan^2t-\frac18\ln^2\frac{1+t^2}4,$$ I'm able to show $I_3=\frac5{54}\pi^2$.
For $n=4$, I numerically found $I_4=\frac{17}{192}\pi^2$.
I'm not able to find the general form with $n\in \mathbb{Z}^+$.
| $$\begin{aligned}
I_n &= \int_0^1 \frac{\ln(1+x)}{x(1+x^n)}dx + \int_1^\infty \frac{\ln(1+x)}{x(1+x^n)}dx \\
&= \int_0^1 \frac{\ln(1+x)}{x(1+x^n)}dx + \int_0^1 \frac{x^n \ln(1+x)}{x(1+x^n)}dx - \int_0^1 \frac{x^{n-1} \ln x}{1+x^n}dx \\
&= \int_0^1 \frac{\ln(1+x)}{x}dx + \frac{1}{n}\int_0^1 \frac{\ln(1+x^n)}{x}dx \\
&= \int_0^1 \frac{\ln(1+x)}{x}dx+\frac{1}{n^2}\int_0^1 \frac{\ln(1+u)}{u}du = \color{blue}{(1+\frac{1}{n^2})\frac{\pi^2}{12}}
\end{aligned}$$
*
*Second line: $x\mapsto 1/x$
*Third line: Integration by parts
*Fourth line: $x=u^{1/n}$
Note that $n$ need not be an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit of the sequence: $\lim_{n_\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$ Evaluate the limit of the sequence:
$$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$$
My try:
Stolz-cesaro: The limit of the sequence is $\frac{\infty}{\infty}$
$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
For our sequence:
$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}=\lim_{n\to\infty}\frac{\sqrt{n!}-\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})\cdot(1+\sqrt{n+1})-(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}=\lim_{n\to\infty}\frac{\sqrt{(n-1)!}\cdot(\sqrt{n-1})}{\left((1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})\right)\cdot(\sqrt{n}+1)}$
Which got me nowhere.
| Note that
$$
\frac{\sqrt{(n-1)!}}{\prod_{k=1}^n\big(1+\sqrt{k}\big)}=\frac{1}{1+\sqrt{n}}
\prod_{k=1}^{n-1}\frac{\sqrt{k}}{1+\sqrt{k}}<\frac{1}{1+\sqrt{n}}\to 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit $\lim_{(x, y) \to (\infty, \infty)} \frac{x+\sqrt{y}}{x^2+y}$ Show whether the limit exists and find it, or prove that it does not.
$$\lim_{(x, y) \to(\infty,\infty)}\frac{x+\sqrt{y}}{x^2+y}$$
WolframAlpha shows that limit does not exist, however, I do fail to conclude so.
$$\lim_{(x, y) \to(\infty,\infty)}\frac{x+\sqrt{y}}{x^2+y} = [x=r\cos\theta, y = r\sin\theta] = \lim_{r\to\infty}\frac{r\cos\theta+\sqrt{r\sin\theta}}{r^2\cos^2\theta+r\sin\theta} = \lim_{r\to\infty}\frac{\cos\theta\frac{\sqrt{\sin\theta}}{\sqrt{r}}}{r\cos^2\theta+\sin\theta} = 0.$$
Having gotten the exact results for whatever the substitution is made (such as $y = x, y = x^2, [x = t^2, y = t])$, my conclusion is that limit does exist and equals $0.$
Did I miss something?
| Since $(x,y)\rightarrow (\infty,\infty),$ assume $x,y>0.$ $$0<\frac{x+\sqrt{y}}{x^2+y}=\frac{x\sqrt {y}\left(\frac{1}{\sqrt y}+\frac 1x\right)}{x\sqrt {y}\left(\frac{x}{\sqrt y}+\frac {\sqrt y}{x}\right)}\leq
\frac{\frac{1}{\sqrt y}+\frac 1x}{2} \rightarrow 0$$
as $\frac ab + \frac ba \geq 2$ for $a,b>0.$ Thus by the squeeze theorem is our limit $0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3040482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Calculate the limit $\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}$. We could use L'Hospital here, because both numerator as well as denominator tend towards 0, I guess. The derivative of the numerator is $$x^2\cdot \left(-\sin\left(\frac{1}{x}\right)\right) \cdot \left( -\frac{1}{x^2}\right) + 2x \cos\left(\frac{1}{x}\right)=\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right) $$ The derivative of the denominator is $\cos(x)$. So, $$\lim\limits_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)} = \lim\limits_{x\rightarrow 0}\displaystyle\frac{\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right)}{\cos(x)}$$
Is that right so far?
Thanks for the help in advance.
Best Regards,
Ahmed Hossam
| With the Taylor power series, $\sin x= x+o(x)$
$$\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}=\lim_{x\to0}{x \cos\left(\frac{1}{x}\right)}=0$$
Because $x\to0$ and $\cos(1/x)$ is bounded from $-1$ to $1$ as $x\to0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integral$\int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} (x)}{\sin^{1/3} (x)+\cos^{1/3} (x)}$ $$\int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} (x)}{\sin^{1/3} (x)+\cos^{1/3} (x)}$$
I tried using the substitution $t^3=\tan x$. Which gives me
$$\int\frac{3t^3}{(t+1)(t^6+1)}$$
How should I proceed?
| I will put you on the path:
\begin{align}
\frac{3x^3}{\left(x + 1\right)\left(x^6 + 1\right)} &= \frac{3x^3}{\left(x + 1\right)\left(x^2 + 1\right)\left(x^2 + \sqrt{3}x + 1\right)\left(x^2 - \sqrt{3}x + 1\right)} \\
&= -\frac{3}{2}\frac{1}{x + 1} - \frac{1}{2}\left[\frac{1}{x^2 + 1} +\frac{x}{x^2 + 1}\right] + \frac{1}{2\left(2 + \sqrt{3}\right)}\frac{x + 1}{x^2 - \sqrt{3}x + 1} \\
&\qquad + \frac{2 +\sqrt{3}}{2}\frac{x + 1}{x^2 + \sqrt{3}x + 1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
How to find the other asymptote of $y=\sqrt{x^2+x}$ The task is to find the asymptotes of $y=\sqrt{x^2+x}$.
I first calculated the limits to infinity and found that $\lim_{x \to \pm}y= \infty$.
Next, to find $m_{1,2}$: $$m_1=\lim_{x \to +\infty}\frac{y}{x}=\frac{\sqrt{x^2+x}}{x}=\sqrt{1+ \frac{1}{x}}=1=m_1$$
and $$m_2=\lim_{x \to -\infty}\frac{y}{x}=\frac{\sqrt{x^2+x}}{x}=\sqrt{1+ \frac{1}{x}}=1=m_2$$
Now to find $c_{1,2}$: $$c_1=\lim_{x \to +\infty}y-mx={\sqrt{x^2+x} - x}=\frac{(\sqrt{x^2+x} - x)({\sqrt{x^2+x} +x})}{{\sqrt{x^2+x} +x}}=\frac{x}{{\sqrt{x^2+x} +x}}= \frac{x}{x+{|x|\sqrt{1+\frac{1}{x}}}} \to \frac{x}{x+{|x|}}
\approx \frac{1}{1+\frac{|x|}{x}}=\frac{1}{2} $$
$$c_2=\frac{1}{0}?$$
From here, it is clear that as $x$ tends to positive infinity, we have $c_1=1/2$, therefore, $y=\frac{1}{2}(2x+1)$ is an asymptote. However, when I try to calculate the limit to negative infinity, I have an indeterminate $\frac{1}{0}$. However, according to the answer, the other asymptote should be $y=-\frac{1}{2}(2x+1)$.
How can I find it and how did I miss it in my calculations? Do I need to try and calculate the limit another way?
| You may try also in this way. By definition $y=mx+q$ is an asymptote of $f$ as $x\to \pm\infty$ if $f(x)=mx+q+o(1)$. Now, as $|x|\to +\infty$,
$$\sqrt{x^2+x}=|x|\sqrt{1+1/x}=|x|\left(1+\frac{1}{2x}+o(1/x)\right)\\
=|x|+\frac{|x|}{2x}+o(1).$$
Hence if $x>0$ then
$$\sqrt{x^2+x}=x+\frac{1}{2}+o(1)$$
and for $x<0$
$$\sqrt{x^2+x}=-x-\frac{1}{2}+o(1).$$
What may we conclude?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Divisor of $x^2+x+1$ can be square number? $$1^2+1+1=3$$
$$2^2+2+1=7$$
$$8^2+8+1=73$$
$$10^2+10+1=111=3\cdot37$$
There is no divisor which is square number.
Is it just coincidence? Or can be proved?
*I'm not english user, so my grammer might be wrong
| What about $x=653$, where
$$x^2+x+1=427063=7\cdot (13\cdot 19)^2\,?$$
How did I find this $x$? I first suppose that $x^2+x+1=ay^2$ for some positive integers $a$ and $y$. This means
$$(2x+1)^2-a(2y)^2=-3\,.$$
Let $u:=2x+1$ and $v:=2y$. Then, we are to solve the Pell-type equation $$u^2-av^2=-3\,,$$ where $u,v\in\mathbb{Z}_{>0}$. In particular, the case $x=2$ yields $a=7$ and $y=1$. Therefore, I attempt to solve
$$u^2-7v^2=-3\,,\tag{*}$$
where a minimal solution is $(u,v)=(5,2)$. Since $u^2-7v^2=1$ has the minimal solution $(u,v)=(8,3)$, we obtain an infinite family of solutions $(u,v)$ of (*):
$$u+v\sqrt{7}=(5+2\sqrt{7})\,(8+3\sqrt{7})^k\,,\tag{#}$$
where $k$ is a positive integer. We want $u$ to be odd, so $k=1$ does not work. With $k=2$, we get $(u,v)=(1307,494)$, so $$x=\frac{1307-1}{2}=653\text{ and }y=\frac{494}{2}=13\cdot 19$$
form a counterexample. (Using $k=-2$, we get lhf's counterexample $x=18$. Indeed, with even values of $k$ in $(\#)$, we obtain infinitely many counterexamples.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How to solve equations of this type My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:
Q1. If: $x - \frac{1}{x} = 3$ then what is $x^2 + \frac{1}{x^2}$ equal to?
The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.
Q2. If: $\frac{x}{x + y} = 5$ then what is $\frac{y}{x + y}$ equal to?
The answer for this is -5, I also don't know how.
Q3. If: $x^4 + y^4 = 6 x^2 y^2 \land x\neq y$ then what is $\frac{x^2 + y^2}{x^2 - y^2}$ equal to?
I guess the answer for this was $\sqrt{2}$ but I'm not sure.
Any body can explain how to solve these questions, and questions of the same pattern?
| For the first one :
$$\left(x-\frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2} \Leftrightarrow x^2 + \frac{1}{x^2} = \left(x-\frac{1}{x}\right)^2 + 2 \implies x^2+ \frac{1}{x^2} = 11$$
For the second one, observe that :
$$\frac{x}{x+y} + \frac{y}{x+y} = 1 \Rightarrow 5 + \frac{y}{x+y} = 1 \Leftrightarrow \frac{y}{x+y} = -4 $$
For the third one, a small hint :
$$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$
Alternativelly, observe that it also is :
$$x^4 + y^4 = 6 x^2 y^2 \Leftrightarrow \frac{x^2}{y^2} + \frac{y^2}{x^2} = 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Radius of convergence of $\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n$ I want to determine the convergence of the following series in dependency of $x$:
$\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n=\frac{3}{4}x+\frac{2}{5}x^2+\frac{1}{4}x^3+\frac{3}{17}x^4+ ... $
How can I solve this?
EDIT:
@Winther said, I should try the ratio test:
$q := \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n}\right| $
So we get
$q = \lim_{n \to \infty} \left| \frac{\frac{n+3}{2(n+1)^2+2}}{\frac{n+2}{2n^2+2}}\right| = \lim_{n \to \infty} \left| \frac{n+3}{2(n+1)^2+2}\frac{2n^2+2}{n+2} \right| = \lim_{n \to \infty} \left| \frac{2n^3+2n+6n^2+6}{2(n+1)^2n+4(n+1)^2+2n+4}\right| = \lim_{n \to \infty} \left| \frac{n^3+3n^2+n+3}{n^3+4n^2+6n+4}\right| $
With the tip from @Alex Vong to divide by $n^3$ the ratio test becomes:
$ q = \lim_{n \to \infty} \left| \frac{1+3/n+1/n^2+3/n^3}{1+4/n+6/n^2+4/n^3}\right|= 1$
So now there is no clear statement if the series is convergent ($q<1$ convergent, $q>1 $ divergent).
Should I try another test (e. g. the root test)?
EDIT 2: Corrected the first coefficients.
| If the ratio test works, there's no need to check with other tests; in your case you want to compute, for $x\ne0$,
$$
\lim_{n\to\infty}
\left|\,\frac{\dfrac{(n+1)+2}{2(n+1)^2+2}x^{n+1}}{\dfrac{n+2}{2n^2+2}x^n}\,\right|
=\lim_{n\to\infty}\frac{n+3}{n+2}\frac{2(n+1)^2+2}{2n^2+2}|x|=|x|
$$
This limit is $<1$ if and only if $|x|<1$.
Thus the radius of convergence is $1$.
There are cases where the limit for the ratio test doesn't exist; other tests should be used. The “universal” test is Hadamard's:
$$
\frac{1}{R}=\limsup_{n\to\infty}\sqrt[n]{a_n}
$$
but this can be difficult to compute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve the system $x^2(y+z)=1$ ,$y^2(z+x)=8$ and $z^2(x+y)=13$ Solve the system of equations in real numbers
\begin{cases} x^2(y+z)=1 \\ y^2(z+x)=8 \\z^2(x+y)=13 \end{cases}
My try:
Equations can be written as:
\begin{cases}\frac{1}{x}=xyz\left(\frac{1}{y}+\frac{1}{z}\right)\\
\frac{8}{y}=xyz\left(\frac{1}{x}+\frac{1}{z}\right)\\
\frac{13}{z}=xyz\left(\frac{1}{y}+\frac{1}{x}\right)\end{cases}
Let $p=xyz.$ Then we have:
\begin{cases}\frac{1}{x}-\frac{p}{y}-\frac{p}{z}=0\\
\frac{p}{x}-\frac{8}{y}+\frac{p}{z}=0\\
\frac{p}{x}+\frac{p}{y}-\frac{13}{z}=0\end{cases}
Then we get
$$\frac{p+1}{x}=\frac{p+8}{y}=\frac{p+13}{z}$$
Any clue here?
| I would substitute $$y+z=a,z+x=b,x+y=c$$ then $$z=\frac{a+b-c}{2}$$ and so on.
Eliminating the variables $$y,z$$ we get for $x$:
$$25 x^9-3828 x^6+109 x^3=-42$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $a_n \in [0,2)$
Let $(a_n)_{n \in \mathbb{N}}$ be a sequence, with $a_0=0$, $a_{n+1}=\frac{6+a_n}{6-a_n}$.
Prove that $a_n \in [0,2)$ $\forall n \in \mathbb{N}$
Here's what I did:
I tried to prove this by induction:
Base case:
$0 \leq a_0 (=0) < 2$.
Inductive step:
Suppose that $0 \leq a_n < 2$
So $$\begin {split} 0 \leq a_n < 2 &\iff 0 \geq -a_n > -2 \\ &\iff 6 \geq 6-a_n > 6-2 \\ &\iff \frac{1}{6} \geq \frac{1}{6-a_n} > \frac{1}{4} \\ &\iff \frac{a_n}{6}+1 \geq \frac{6+a_n}{6-a_n} > \frac{3}{2} + \frac{a_n}{4} \end{split}$$
To be fair I have no idea if this is going somewhere.
| First, let's rewrite the recurrence formula:
$$a_{n+1}=\frac{6+a_n}{6-a_n}=\frac{a_n-6+12}{6-a_n}=-1+\frac{12}{6-a_n}$$
Now, we have:
$$0\leq a_n<2$$
$$6\geq 6-a_n > 4$$
$$2\leq \frac{12}{6-a_n} < 3$$
$$1\leq -1+\frac{12}{6-a_n}=a_{n+1} < 2$$
Thus, $a_{n+1} \in [1, 2)$ and since $[1, 2) \subset [0, 2)$, $a_{n+1} \in [0, 2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $B^*$, the dual basis Find a basis $B$ for
$$V = \left\{ \left[
\begin{array}{cc}
x\\
y\\
z
\end{array}
\right] \in \mathbb{R}^3 \vert x+y+z = 0\right\}$$ and then find $B^*$, the dual basis for $B$.
The way we learned it was that given a basis, we build a matrix A whose columns is the vectors, find $A^{-1}$, and those build linear functionals that are the rows of the inverse matrix.
For example, if $V=\mathbb{R}^2$ and $ B = $$\left\{ \left[
\begin{array}{cc}
3\\
4
\end{array}
\right], \left[
\begin{array}{c}
5\\
7
\end{array}
\right]\right\} $, then $ A = $$ \left[
\begin{array}{cc}
3&5\\
4&7
\end{array}
\right]$, then $A^{-1} = $$ \left[
\begin{array}{cc}
7&-5\\
-4&3
\end{array}
\right]$, and the dual basis $B^* = (l_1, l_2)$ when :
$ l_1= \left( \left[
\begin{array}{cc}
x\\
y\\
\end{array}
\right]\right) = 7x_1 - 5x_2 $
$ l_2= \left(\left[
\begin{array}{cc}
x\\
y\\
\end{array}
\right]\right) = -4x_1 + 3x_2 $
However, when finding a basis for $V$, I get to the following solution vector
$$\left[\begin{array}{cc}
-y-z\\
y\\
z
\end{array}
\right] = y \left[\begin{array}{cc}
-1\\
1\\
0
\end{array}
\right] + z \left[\begin{array}{cc}
-1\\
0\\
1
\end{array}
\right]$$
If I build a matrix out of the basis$ \left\{ \left[\begin{array}{cc}
-1\\
1\\
0
\end{array}
\right] , \left[\begin{array}{cc}
-1\\
0\\
1
\end{array}
\right] \right\}$ I will have a matrix that is $3\times2$ and I cannot inverse this.
How to proceed from here?
| By definition, the dual basis functionals $f_1, f_2$ are given on your basis $\{v_1, v_2\}$ as $$f_1(\alpha_1v_1 + \alpha_2v_2) = \alpha_1, \quad f_2(\alpha_1v_1 + \alpha_2v_2) = \alpha_2$$
Now $$f_1\left(\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right) = f_1\left(\begin{bmatrix} -y-z \\ y \\ z\end{bmatrix}\right) = f_1\left(y\begin{bmatrix} -1 \\ 1 \\ 0\end{bmatrix} + z\begin{bmatrix} -1 \\ 0 \\ -1\end{bmatrix}\right) = f_1(yv_1 + zv_2) = y$$
$$f_2\left(\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right) = f_2\left(\begin{bmatrix} -y-z \\ y \\ z\end{bmatrix}\right) = f_2\left(y\begin{bmatrix} -1 \\ 1 \\ 0\end{bmatrix} + z\begin{bmatrix} -1 \\ 0 \\ -1\end{bmatrix}\right) = f_2(yv_1 + zv_2) = z$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limlit of $\frac{x-x\ln(1+x)-\ln(1+x)}{x^3+x^2}$ at $0$ without l'Hôpital or Taylor series I have $f : x \mapsto \frac{\ln(1+x)}{x}$ which derivative is $$\frac{1}{x(1+x)}-\frac{\ln(1+x)}{x^2}$$
I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$\frac{x - x\ln(1+x) - \ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)
| First, you can decompose the first term in your derivative as:
$\frac{1}{x(1+x)} = \frac{1}{x}-\frac{1}{1+x}$.
Second, you can expand the second term using Taylor series:
$\frac{ln(1+x)}{x^2}$
$ = \frac{1}{x^2} \left( x - \frac{x^2}{2} + \frac{x^3}{3} - .. \right)$
$ = \frac{1}{x} - \frac{1}{2} + \frac{x}{3} - $
Putting it together, your derivative is:
$\frac{1}{x}-\frac{1}{1+x} - \frac{1}{x} + \frac{1}{2} - \frac{x}{3} +...$
*
*The $1/x$ terms cancel out.
*$\lim_{x \rightarrow 0} \frac{1}{1+x} = 1$.
*$\lim_{x \rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.
So you are left with $-1 + \frac{1}{2} = -\frac{1}{2}$.
| {
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Prove that $ \int\limits_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\dfrac{3\pi}{8} $ Is this conjecture true?
Conjecture:
$$
\int_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\dfrac{3\pi}{8}
$$
I found it myself based on numerical evidence. Need help in analytical proof. Thanks.
| Well, we can just rewrite the integral as: $$\int_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\int_0^\infty \frac{x^2}{e^{2x}\left(\frac{1-x}{1+x}\right)^2+e^{-2x}}\frac{1}{(1+x)^2}dx$$
$$=\int_0^\infty \frac{1}{e^{4x}\left(\frac{1-x}{1+x}\right)^2+1}\frac{1}{e^{-2x}}\frac{x^2}{(1+x)^2}dx =\int_0^\infty \frac{\color{blue}{e^{2x}}}{\left(\color{red}{e^{2x}\cdot\frac{1-x}{1+x}}\right)^2+1}\color{blue}{\frac{x^2}{(1+x)^2}dx}$$ Now set $\,\displaystyle{\color{red}{e^{2x} \frac{1-x}{1+x}}=t\Rightarrow \color{blue}{e^{2x}\frac{x^2}{(1+x)^2}dx}=-\frac12 dt}\,$ and the result follows.
| {
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"question_score": "1",
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} |
calculate flux through surface I need to calculate the flux of the vector field $\vec{F}$ through the surface $D$, where
$$\vec{F} = \left<z, \, y \sqrt{x^2 + z^2}, \, -x \right> \\
D = \{x^2+6x+z^2\le 0 \,| -1\le y \le 0\}.$$
So there should be a cylinder (height on $y$ axis) shifted by 3 units (center $x=-3$), with cut on $ -1\le y \le 0$.
Cylindrical parametrization:
$$\begin{cases}
x=r\cos\theta -3\\
y=y\\
z=r\sin\theta
\end{cases}
$$
I think I can do this problem in two ways: The first one by calculating the flux for each of the 3 surfaces (1 cylinder, 2 disks), and the second one by using the divergence theorem.
$$\operatorname{div}F = \sqrt{x^2+y^2} \overset{\text{cylindrical}}{\underset{\text{coordinates}}{=}} \sqrt{r^2-9-6r\cos\theta},$$
so with the divergence theorem, $$\int_{0}^{2\pi}\int_{0}^{3\sqrt{3}}\int_{-1}^{0}{r\sqrt{r^2-9-6r\cos\theta} \, dy \, dr \, d\theta}.$$
Is this correct?
| I'm not exactly sure where the $3\sqrt{3}$ comes from in your result, but there is indeed more than one way to evaluate this problem.
(1) Direct method
Here is some technical information about this method from MIT's open notes, and some visualization for what the flux of a vector field through a surface is. Let the flux of a vector field $\vec{\mathbf{V}}$ through a surface $\Sigma$ be denoted $\Phi$ and defined
$$
\Phi := \iint_\Sigma \mathbf{\vec{V}} \cdot \mathbf{\hat{n}} \, d\sigma.
$$
The vector $\mathbf{\hat{n}}$ is the unit outward normal to the surface $\Sigma$. Suppose $\Sigma$ is given by $z = f(x,y).$ Let $\mathbf{\vec{r}}(x,y)$ trace $\Sigma$ such that
$$
\mathbf{\vec{r}}(x,y) = \begin{pmatrix}
x \\ y \\ f(x,y)
\end{pmatrix}.
$$
Then the unit normal $\mathbf{\vec{n}}$ is given by
$$
\mathbf{\vec{n}} = \frac{\mathbf{\vec{r}}_x \times \mathbf{\vec{r}}_y}{|| \mathbf{\vec{r}}_x \times \mathbf{\vec{r}}_y ||} =
\frac{1}{\sqrt{f_x^{\,2} + f_y^{\,2} + 1}}
\begin{pmatrix}
-f_x \\
-f_y \\
1
\end{pmatrix}.
$$
So given that $\mathbf{\vec{V}} = u(x,y,z) \mathbf{\hat{i}} + v(x,y,z) \mathbf{\hat{j}} + w(x,y,z) \mathbf{\hat{k}}$, the corresponding flux of $\mathbf{\vec{V}}$ through $\Sigma$ is
$$
\Phi = \iint_\Sigma \frac{-uf_x - vf_y + w}{\sqrt{f_x^{\,2} + f_y^{\,2} + 1}} \, d\sigma.
$$
For the given field, we have
$$\mathbf{\vec{V}} = z \mathbf{\hat{i}} + y \sqrt{x^2 + z^2} \mathbf{\hat{j}} - x \mathbf{\hat{k}},
$$
and the surface $\Sigma$ is given such that $(x + 3)^2 + z^2 = 9\ \forall y \in (-1,0).$
Thus we choose to trace the surface of the cylinder with
$$
\mathbf{\vec{r}}(x,z) =
\begin{pmatrix}
x \\
(x + 3)^2 + z^2 - 9 \\
z
\end{pmatrix},
$$
where the unit outward normal on the cylinder is
$$\mathbf{\hat{n}} = \frac{1}{\sqrt{4(x+3)^2 + 4z^2 + 1}}
\begin{pmatrix}
-2(x+3) \\
1 \\
-2z
\end{pmatrix}
=
\frac{1}{\sqrt{37}}
\begin{pmatrix}
-2(x+3) \\
1 \\
-2z
\end{pmatrix}
.
$$
The flux is then
\begin{align}
\Phi &= \underbrace{\iint_{\Sigma_1} \mathbf{\vec{V}} \cdot \begin{pmatrix}
0 \\ -1 \\ 0
\end{pmatrix} \, d\sigma}_{y = -1} + \overbrace{\iint_{\Sigma_2} \mathbf{\vec{V}} \cdot \mathbf{\hat{n}} \, d\sigma}^{\mathbf{\vec{V}}_2\text{ cannot contribute}} + \underbrace{\iint_{\Sigma_3} \mathbf{\vec{V}} \cdot \begin{pmatrix}
0 \\ 1 \\ 0
\end{pmatrix} \, d\sigma}_{\text{nothing since }y = 0} \\
&= \iint_{\Sigma_1} \sqrt{x^2 + z^2} \, d\sigma + \frac{1}{\sqrt{37}}\iint_{\Sigma_2} -2z(x+3) + y\sqrt{x^2 + z^2} + 2xz \, d\sigma \\
&= \int_0^{2\pi}\int_0^3 r\sqrt{(r \cos\theta - 3)^2 + (r \sin\theta)^2} \, dr \, d\theta - \frac{6}{\sqrt{37}} \int_0^3 r^2 \, dr \, \underbrace{\int_0^{2\pi} \sin\theta \, d\theta}_{0} \\
&= 96.
\end{align}
(2) Divergence Theorem
Technical information about this method can also be found in MIT's open notes, and no visualization from the site this time, but the divergence theorem here is explained in casual language. Given everything is nice, the flux of the field through the surface is
$$
\iint_\Sigma \mathbf{\vec{V}} \cdot \mathbf{\hat{n}} \, d\sigma = \iiint_M \nabla \cdot \mathbf{\vec{V}} \, dV,
$$
where $M$ is the bounded region contained within $\Sigma$.
Applying it to this problem, the divergence theorem takes us straight to the end result of the direct approach.
\begin{align}
\Phi &= \iint_\Sigma \nabla \cdot \mathbf{\vec{V}} \, dV \\
&= \iiint_M \sqrt{x^2 + z^2} \, dV \\
&= \int_{-1}^0 \int_0^{2\pi} \int_0^3 r\sqrt{(r \cos\theta - 3)^2 + (r \sin\theta)^2} \, dr \, d\theta \, dy \\
&= 96.
\end{align}
From the comments, $18\pi$ falls out as the solution if $x$ is not properly shifted over from the origin. Also, Wikipedia is a fairly good source for this material as well.
| {
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Seeking methods to solve: $\int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt$ Seeking Methods to solve the following two definite integrals:
\begin{equation}
I_(n) = \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt \qquad J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt
\end{equation}
For $n \in \mathbb{R},\:n \gt 1$
The method I took was to take the following integral:
\begin{equation}
\int_0^\infty \frac{t^k}{\left(t^n + 1\right)^m}\:dt = \frac{1}{n}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)
\end{equation}
Where: $0 \leq k \lt n$. Here let $m = 1$.
Differentiate under the curve with respect to $k$ and taking the limit as $k \rightarrow 0^+$ (via the Dominated Convergence Theorem), i.e.
\begin{align}
\lim_{k \rightarrow 0+} \frac{d}{dk} \left[ \int_0^\infty \frac{t^k}{t^n + 1}\:dt \right] &= \lim_{k \rightarrow 0+} \frac{d}{dk} \left[\frac{1}{n}B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n} \right) \right] \\
\lim_{k \rightarrow 0+} \int_0^\infty \frac{t^k \ln(t)}{t^n + 1}\:dt &= \lim_{k \rightarrow 0+} \left[\frac{1}{n^2} B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\left[\psi^{(0)}\left(\frac{k + 1}{n} \right) - \psi^{(0)}\left(1 - \frac{k + 1}{n} \right)\right] \right] \\
\int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt&= \frac{1}{n^2} B\left(1 - \frac{1}{n}, \frac{1}{n}\right)\left[\psi^{(0)}\left(\frac{1}{n} \right) - \psi^{(0)}\left(1 - \frac{1}{n} \right)\right] \\
&=- \frac{\pi^2}{n^2}\operatorname{cosec}\left(\frac{\pi}{n}\right)\cot\left(\frac{\pi}{n} \right)
\end{align}
Which is our expression for $I_n$. Taking the same approach but differentiating twice with respect to $k$ we arrive at our expression for $J_n$:
\begin{equation}
J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt = \frac{\pi^3}{n^3}\operatorname{cosec}\left(\frac{\pi}{n} \right)\left[\operatorname{cosec}^2\left(\frac{\pi}{n}\right) + \cot^2\left(\frac{\pi}{n}\right) \right]
\end{equation}
And in fact we may generalise:
\begin{equation}
\int_0^\infty \frac{\ln^p(t)}{\left(t^n + 1\right)^m}\:dt = \lim_{k\rightarrow 0}\frac{d^p}{dk^p}\left[\frac{1}{n} B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\right]
\end{equation}
Where $p \in \mathbb{N}$
This method however was just an extension of another integral. I'm curious, if I had just started with $I_n, J_n$ what alternative methods could be used?
| You can use the pole expansions
\begin{align}
\csc^2 (z) &= \sum_{m \in \mathbb{Z}} \frac{1}{[m \pi + z]^2} \, , \, z \in \mathbb{C}\setminus\pi\mathbb{Z} \, , \\
\sec^2 (z) &= \sum_{m \in \mathbb{Z}} \frac{1}{\left[\left(m+\frac{1}{2}\right) \pi + z\right]^2} \, , \, z \in \mathbb{C}\setminus\pi \left(\mathbb{Z} + \frac{1}{2}\right) \, ,
\end{align}
and their derivatives to find the integrals. We have
\begin{align}
-I_n &= \int \limits_0^\infty \frac{-\ln(t)}{1+t^n} \, \mathrm{d}t = \int \limits_0^1 [-\ln(t)] \frac{1-t^{n-2}}{1+t^n} \, \mathrm{d} t = \sum \limits_{k=0}^\infty (-1)^k \int\limits_0^1 [-\ln(t)] (1-t^{n-2})t^{n k} \, \mathrm{d} t \\
&= \sum \limits_{k=0}^\infty (-1)^k \left[\frac{1}{(nk+1)^2} - \frac{1}{(n k + n - 2
+ 1)^2}\right] = \sum \limits_{k=0}^\infty (-1)^k \left[\frac{1}{(nk+1)^2} - \frac{1}{(n(k+1)-1)^2}\right] \\
&= \frac{\pi^2}{4 n^2} \sum \limits_{l=0}^\infty \left[\frac{1}{\left(l \pi + \frac{\pi}{2n}\right)^2} - \frac{1}{\left(\left(l+\frac{1}{2}\right) \pi + \frac{\pi}{2n}\right)^2} + \frac{1}{\left((l+1) \pi - \frac{\pi}{2n}\right)^2} - \frac{1}{\left(\left(l+\frac{1}{2}\right) \pi - \frac{\pi}{2n}\right)^2}\right] \\
&= \frac{\pi^2}{4 n^2} \sum_{m \in \mathbb{Z}} \left[\frac{1}{\left(m \pi + \frac{\pi}{2n}\right)^2} - \frac{1}{\left(\left(m+\frac{1}{2}\right) \pi + \frac{\pi}{2n}\right)^2} \right] = \frac{\pi^2}{4n^2} \left[\csc^2 \left(\frac{\pi}{2n}\right) - \sec^2 \left(\frac{\pi}{2n}\right)\right] \\
&= \frac{\pi^2}{n^2} \frac{\cos\left(\frac{\pi}{n}\right)}{\sin^2\left(\frac{\pi}{n}\right)}
\end{align}
and, similarly,
\begin{align}
J_n &= \int \limits_0^1 \ln^2(t) \frac{1+t^{n-2}}{1+t^n} \, \mathrm{d} t = 2 \sum \limits_{k=0}^\infty (-1)^k \left[\frac{1}{(nk+1)^3} + \frac{1}{(n(k+1)-1)^3}\right] \\
&= \frac{\pi^3}{4 n^3} \sum_{m \in \mathbb{Z}} \left[\frac{1}{\left(m \pi + \frac{\pi}{2n}\right)^3} - \frac{1}{\left(\left(m+\frac{1}{2}\right) \pi + \frac{\pi}{2n}\right)^3}\right] \\
&= \frac{\pi^3}{4 n^3} \left[\csc^2 \left(\frac{\pi}{2n}\right) \cot \left(\frac{\pi}{2n}\right) + \sec^2 \left(\frac{\pi}{2n}\right) \tan \left(\frac{\pi}{2n}\right)\right] \\
&= \frac{\pi^3}{n^3} \frac{1+\cos^2 \left(\frac{\pi}{n}\right)}{\sin^3 \left(\frac{\pi}{n}\right)}
\end{align}
for $n > 1$ .
| {
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Pythagorean triples where the sum of the two cubes is also a square Are there any Primitive Pythagorean triple solutions $(a,b,c)$ where the sum of the two cubes is also a square? In other words are there coprime $a,b>0 \in \mathbb{N} \;, (a,b)=1$ where $a^2+b^2=c^2$ and $a^3+b^3=d^2$ for some $c,d \in \mathbb{N}$
| Disclaimer: These are some unfinished thoughts I will leave here to work on later, or for others to continue.
Given that $a$ and $b$ are coprime, it follows that $\gcd(a+b,a^2-ab+b^2)$ divides $3$ because
$$\gcd(a+b,a^2-ab+b^2)=\gcd(a+b,3b^2)=\gcd(a+b,3).$$
Suppose towards a contradiction that the gcd equals $3$: Then the factorization
$$d^2=a^3+b^3=(a+b)(a^2-ab+b^2),$$
shows that there exist $e,f\in\Bbb{Z}$ such that
$$a+b=3e^2\qquad\text{ and }\qquad a^2-ab+b^2=3f^2,$$
from which it quickly follows that
$$9e^4=(a+b)^2=a^2+2ab+b^2=3c^2-6f^2,$$
and reducing mod $8$ yields a contradiction, so the gcd is $1$. Hence there exist $e,f\in\Bbb{Z}$ such that
$$a+b=e^2\qquad\text{ and }\qquad a^2-ab+b^2=f^2,$$
and in the same way as before we find that
$$e^4=(a+b)^2=a^2+2ab+b^2=3c^2-2f^2.$$
Luckily $\Bbb{Z}[\sqrt{6}]$ is a UFD, and we have
$$N((3c-2f)+(c-f)\sqrt{6}):=
\left((3c-2f)+(c-f)\sqrt{6}\right)\left((3c-2f)-(c-f)\sqrt{6}\right)
=3c^2-2f^2=e^4.$$
The gcd of two conjugate factors divides $2(3c-2f)$ and $2(c-f)$, and because $c$ and $f$ are coprime it follows that the gcd divides $2$. Because their product $e^4=(a+b)^2$ is odd, the two conjugate factors are in fact coprime. This means there exists some $x+y\sqrt{6}\in\Bbb{Z}[\sqrt{6}]$ such that
$$(3c-2f)+(c-f)\sqrt{6}=(x+y\sqrt{6})^4.\tag{1}$$
This immediately tells us that
$$a+b=e^2=\sqrt{N((3c-2f)+(c-f)\sqrt{6})}=(x+y\sqrt{6})^2(x-y\sqrt{6})^2=(x^2-6y^2)^2.\tag{2}$$
Furthermore, expanding equation $(1)$ yields the two equations
$$3c-2f=x^4+36x^2y^2+36y^4\qquad\text{ and }\qquad c-f=4x^3y+24xy^3.$$
Because $c-f>0$, without loss of generality $x,y>0$. The above tells us that
\begin{eqnarray*}
c&=&x^4-\ 8x^3y+36x^2y^2-48xy^3+36y^4,\\
f&=&x^4- 12x^3y+36x^2y^2-72xy^3+36y^4,
\end{eqnarray*}
and hence that
$$ab=c^2-f^2=(c-f)(c+f)=8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).\tag{3}$$
This means $a$ and $b$ are the roots of the quadratic polynomial
$$Z^2-(x^2-6y^2)^2Z+8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).$$
This polynomial has integer roots if and only if its discriminant $\Delta$ is a square, where
$$\Delta=(x^2-6y^2)^4-32xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4),$$
which leaves me with the question of when this homogeneous octic polynomial takes on square values.
| {
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Householder transformations to upper triangular form Let $A=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}$.
How to transform this matrix with Householder transformations to an upper triangular matrix?
I started like this:
$\alpha_1=$sgn$(0)(0^2+0^2+1^2+0^2)^{\frac{1}{2}}=1$.
So $v_1=\begin{pmatrix} 0\\0\\1\\0 \end{pmatrix}+\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix}=\begin{pmatrix} 1\\0\\1\\0 \end{pmatrix}$
Then $\frac{2vv^T}{v^Tv}=\begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
$Q_1= I-\frac{2vv^T}{v^Tv}=\begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$
So I get:
$Q_1A=\begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} -1 & 0 & -1 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
Then I used the lower-right $(3 \times 2)$-submatrix and I got:
$\alpha_2=$sgn$(0)(0^2+(-1)^2+0^2)^{\frac{1}{2}}=1$
$v_2=\begin{pmatrix} 0\\-1\\0 \end{pmatrix}+\begin{pmatrix} 1\\0\\0 \end{pmatrix}=\begin{pmatrix} 1\\-1\\0 \end{pmatrix}$
$Q_2=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$
and $Q_2Q_1A=\begin{pmatrix} -1 & 0 & -1 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}$
How to continue from here? I'm not sure which submatrix I have to take.
If I take the lower-right $(2 \times 1)$-submatrix I get $\sqrt2$ for $\alpha_3$ and a matrix $Q_3$ which looks wrong to me.
| You are full on track towards
$\,\left(\begin{smallmatrix}\ddot\smile& *& *\\ 0& \ddot\smile& *\\ 0& 0& \ddot\smile\\ 0& 0& 0\end{smallmatrix}\right)\;$ just go ahead:
$$v_3\:=\: \begin{pmatrix}0\\ 0\\ \sqrt 2+1\\ 1\end{pmatrix}\quad\Longrightarrow\;
-2\frac{v_3\,v_3^T}{v_3^Tv_3}\:=\: -\frac 2{\left(\sqrt 2+1\right)^2+1}\:
\begin{pmatrix}0\\ 0\\ \sqrt 2+1\\ 1\end{pmatrix}
\begin{pmatrix}0&0&\sqrt 2+1&1\end{pmatrix}$$
Thus,
$$Q_3\:=\:\mathbb 1_{4\times 4}\: -\frac{2-\sqrt 2}2\:
\begin{pmatrix}0\\ 0\\ \sqrt2+1\\ 1\end{pmatrix}
\begin{pmatrix}0 &0 &\sqrt 2+1 &1\end{pmatrix}$$
and $$Q_3Q_2Q_1A\,=\,
\begin{pmatrix}-1& 0& -1\\ 0& -1& 0\\ 0& 0& 1\\ 0& 0& 1\end{pmatrix}
\,-\frac{2-\sqrt2}2\:\begin{pmatrix}0\\ 0\\ \sqrt2+1\\ 1\end{pmatrix}
\begin{pmatrix}0&0&2+\sqrt2\end{pmatrix}
=\begin{pmatrix}-1& 0& -1\\ 0& -1& 0\\ 0& 0& -\sqrt2\\ 0& 0& 0\end{pmatrix}$$
"Householder updates never entail the explicit formation of the Householder matrix."$\ $ is a quotation from Golub & Van Loan's book "Matrix computations" (on page 236, 4th edition 2013). So $Q_3$ was intentionally left in the form above, because treating it as a general matrix considerably increases work.
| {
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Find the formula of the linear transformation Consider the linear transformation $f:R^3\rightarrow R^3$ with $f(0,1,2)=(2,1,1)$ and $\ker f=\{(0,2,1),(1,0,1)\}.$ Find:
(i) $f(0,1,0)$ (ii) the general formula of $f$.
Any help, please.
| Now,
$$
(0,1,0)=\frac{1}{3}\big[2(0,2,1)-(0,1,2)\big],
$$
hence
$$f(0,1,0)
=\frac{1}{3}\big[2 \cdot f(0,2,1)-f(0,1,2)\big]
=\frac{1}{3}\big[2 \cdot 0 -(2,1,1)\big]
=\left(-\frac{2}{3},-\frac{1}{3},-\frac{1}{3}\right).
$$
Now,
$$
(0,0,1)
=\frac{1}{2}\big[(0,1,2)-(0,1,0)\big],
$$
hence
$$
f(0,0,1)
=\frac{1}{2}\left[(2,1,1)-(-\frac{2}{3},-\frac{1}{3},-\frac{1}{3})\right]
=\left(\frac{4}{3},\frac{2}{3},\frac{2}{3}\right)
$$
Furthermore,
$$
(1,0,0)
=(1,0,1)-(0,0,1),
$$
hence
$$f(1,0,0)
=f(1,0,1)-f(0,0,1)
=-f(0,0,1)
=\left(-\frac{4}{3},-\frac{2}{3},-\frac{2}{3}\right).
$$
We can get the formula of $f$:
$$
f(x,y,z)
= \left(-\frac{4}{3}x-\frac{2}{3}y+\frac{4}{3}z,-\frac{2}{3}x-\frac{1}{3}y+\frac{2}{3}z,-\frac{2}{3}x-\frac{1}{3}y+\frac{2}{3}z\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i}$ To prove $$\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i},$$
I demonstrated that the equality is true for any $n,$ for $m = 0, 1,$ and for any $r < n + m$ simply by fixing $n$ and $r$ and inserting $0,1$ for $m.$ Then, I proceed to induct on $m$ (and on $m$ only).
I am not perfectly confident in my self, however, for I see two placeholders, $n$ and $m.$ Is this a case where double induction is needed (first on $m$ and then on $n$)?
Whether or not this proof requires double induction, may someone explain when double induction is needed?
Consider any fixed $n, r \geq 0$ and the following two cases (I know that only one case is needed to complete this inductive proof).
CASE 1
\begin{align}
\binom{n + 0}{r} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{0}{r - i} \\ &= \binom{n}{0}\binom{0}{r} + \binom{n}{1}\binom{0}{r-1} + \cdots + \binom{n}{r}\binom{0}{0} \\ &= 0 + 0 + \cdots + \binom{n}{r} \\ &= \binom{n}{r}
\end{align}
CASE 2
\begin{align}
\binom{n + 1}{r} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{1}{r - i} \\ &= \binom{n}{0}\binom{0}{r} + \binom{n}{1}\binom{0}{r-1} + \cdots + \binom{n}{r-1}\binom{1}{r - (r-1)} + \binom{n}{r}\binom{1}{r - r} \\ &= 0 + 0 + \cdots + \binom{n}{r-1} + \binom{n}{r} \\ &= \binom{n}{r-1} + \binom{n}{r}
\end{align}
INDUCTION
Suppose it is true for $m \leq k.$ Now, consider $$\binom{n + (k + 1)}{r}.$$ It follows from Pascal's Identity that
$$\binom{n + (k+1)}{r} = \binom{n + k}{r} + \binom{n + k}{r-1}$$
And,
\begin{align}
\binom{n + k}{r} + \binom{n + k}{r-1} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{k}{r - i} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - 1 - i} \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - i} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - 1 - i} \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\bigg[\binom{k}{r - i} + \binom{k}{r - 1 - i}\bigg] \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k+1}{r-i} \\ &= \sum_{i = 0}^{r} \binom{n}{i}\binom{k+1}{r-i}
\end{align}
Hence, the equality holds for $m = k + 1.$ Given that the equality holds for $m = 0, 1,$ and that if equality holds for $m = k,$ it then holds for $m = k + 1,$ it follows that the equality holds $\forall m \in \mathbb{N}.$
| You can think of a combinatoric proof as follows.
Suppose we have $n$ men and $m$ women, and we wish to form a committee of $r$ people. We can count in two different ways.
Case 1 $\binom{n+m}{r}$ is the number of r-subsets of a set with n+m elements.
Case 2 First we pick $i$ males. That leaves $r-i$ female to choose. So given $0\le i \le r$ We have $\binom{n}{i}\binom{m}{r-i}$ such committees. If we let $i$ range, we add up all these committees, so we get $\displaystyle\sum_{i=0}^r \binom{n}{i}\binom{m}{r-i}$ .
Since both cases count the same number, they must be equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
$f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$. True or false? $f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$
$($ true or false$)?$
Here
$f(0) = -46$ and $f(6) = 2$ since function is continuous so it must have at least one root between $0$ and $6$, but how to check if it has all its roots between $0$ and $6$, without really finding out the roots?
| Consider $F(x):=f(x)-2= (x-2)(x-4)(x-6).$
1)$F(x), f(x)$ polynomials of degree $3.$
2) $x \rightarrow \infty$ $F(x), f(x) \rightarrow \infty.$
3) $x \rightarrow -\infty$ $F(x), f(x) \rightarrow -\infty.$
4) Roots of $F(x)$ at $x=2,4,6$. That's it.
5) $x> 6:$ $F(x) > 0,$ $f(x)= F(x)+2$.
No roots of $f$ for $x>6$.
6) For $x <2$, $F(x) <0$, and
$F(x)= -(2-x)(4-x)(6-x)$ is strictly decreasing.
7)Likewise for $x <2$ $f(x)= F(x)+2$ is strictly decreasing
$F(1)=-(1)(3)(5)=-15$, and $f(1)= -15+2=-13$.
No root of $f$ for $x <1$.
Hence?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Functional equation involving $f(x^4)+f(x^2)+f(x)$
Find all increasing functions $f$ from positive reals to positive reals satisfying $f(x^4) + f(x^2) + f(x) = x^4 + x^2 + x$.
It's easy to show that $f(1)=1$, and I was also able to show that
$$f(x)-x = f(x^{(8^k)}) - x^{(8^k)}$$
for all integers $k$.
But where to go from here? Any hints would be much appreciated.
| First we show that $\lim_{x \to 1} f(x) = f(1) = 1$. Note that the limit of an increasing function from below or from above always exists. Let the limit from below be $a$ and the limit from above be $b$. If we take the limit from below of $f(x) + f(x^2) + f(x^4) = x + x^2 + x^4$, we get $a + a + a = 3$, so $a = 1$ (since as $x$ approaches $1$, $x^2$ and $x^4$ also approach $1$). Similarly from above, $b + b + b = 3$, so $b = 1$. Finally, we already know $f(1) = 1$. So
$$
\lim_{x \to 1} f(1) = 1.
$$
(In fact, we could show $f$ is continuous for all $x$, using a similar argument.)
To finish, consider your equation $f(x) + f(x^{8^k}) = x + x^{8^{k}}$. Take the limit as $k \to -\infty$. Regardless of $x$, $x^{8^k}$ approaches $x^{0} = 1$. So, we get
\begin{align*}
f(x) + 1 = x + 1 \\
\implies & \boxed{f(x) = x},
\end{align*}
which holds for all $x$. So $f(x) = x$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Calculus : Plug in value into a derivative? $$ f(x) = x^3-2x^2-6x $$
after derivative
$$ 3x^2-4x-6 $$
If u plugin x=1,2,3,4,5
as a result:
$$ 3(1)^2-4(1)-6=-7 $$
$$ 3(2)^2-4(2)-6=-2 $$
$$ 3(3)^2-4(3)-6=9 $$
I know it means the slope of the current point on the curve or instantaneous rate of change
Look at the answer 9 what is it actually mean?Is there any relationshiop between x=3 and x=4? an increment?
| For $x=1$ :
$$f(1)=(1)^3-2(1)^2-6(1)=-7$$
$$f'(1)=3(1)^2-4(1)-6=-7$$
This means that at point $(1\:,\:-7)$ the slope of the tangent is $=-7$ .
For $x=2$ :
$$f(2)=(2)^3-2(2)^2-6(2)=-12$$
$$f'(2)=3(2)^2-4(2)-6=-2$$
This means that at point $(2\:,\:-12)$ the slope of the tangent is $=-2$ .
For $x=3$ :
$$f(3)=(3)^3-2(3)^2-6(3)=-9$$
$$f'(3)=3(3)^2-4(3)-6=9$$
This means that at point $(3\:,\:-9)$ the slope of the tangent is $=9$ .
For $x=4$ :
$$f(4)=(4)^3-2(4)^2-6(4)=8$$
$$f'(4)=3(4)^2-4(4)-6=26$$
This means that at point $(4\:,\:8)$ the slope of the tangent is $=26$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding sum of none arithmetic series I have a question to find the sum of the following sum:
$$
S = \small{1*1+2*3+3*5+4*7+...+100*199}
$$
I figured out that for each element in this series the following holds:
$$
a_n = a_{n-1} + 4n - 3
$$
But I don't know where to go from here, I tried subtracting some other series but that did not work very well
| Via generating functions we first find the generating function for each term and then sum them up. I'll build from the ground up. It may look overcomplicated but it also doesn't require remembering too many special identities. Our goal is to find $S_{99}$ where
\begin{align*}
S_m = \sum_{n=0}^{m} a_n
\end{align*}
where
\begin{align*}
a_0 &= 1 \\
a_n &= (n + 1)(2n + 1) \\
&= a_{n-1} + 4n + 1
\end{align*}
Now to find the generating function for our $a_n$ we find
\begin{align*}
A(x) &= \sum_{n=0}^{\infty} a_n x^n \\
A(x) - a(0) &= \sum_{n=1}^{\infty} (a_{n-1} + 4n + 1) x^n \\
A(x) - 1 &= \sum_{n=1}^{\infty} a_{n-1}x^n + \sum_{n=1}^{\infty} (4n +1) x^n \\
A(x) - 1 &= x\sum_{n=0}^{\infty} a_{n}x^n + (\sum_{n=0}^{\infty} (4n +1) x^n) - (4*0 + 1) \\
A(x) &= xA(x) + (4\frac{x}{(x - 1)^2} + \frac{1}{x-1}) \\
(1 - x)A(x) &= \frac{4x + (1 - x)}{(1 - x)^2} \\
A(x) &= \frac{3x + 1}{(1 - x)^3}
\end{align*}
Summing them all up is easy enough, since $S_0 = 0$:
\begin{align*}
S(x) &= \sum_{n=0}^{\infty} S_n x^n \\
S(x) - S(0) &= \sum_{n=1}^{\infty} (S_{n - 1} + [y^n]A(y)) x^n \\
(1 - x)S(x) &= A(x) \\
S(x) &= \frac{3x + 1}{(1 - x)^4}
\end{align*}
Now we want to find the coefficient for $S(x)$
\begin{align*}
[x^n]S(x) &= [x^n]\frac{3x + 1}{(1 - x)^4} \\
&= [x^n]\left(3x + 1\right) \sum_{n=0}^{\infty}\binom{n + 3}{3}x^n \tag{1} \\
&= \left(3[x^{n-1}] + [x^n]\right) \sum_{n=0}^{\infty}\binom{n + 3}{3}x^n \tag{2}\\
&= 3\binom{n+2}{3} + \binom{n+3}{3}\\
&= \frac{1}{2}(n+2)(n+1)n + \frac{1}{6}(n+3)(n+2)(n+1)
\end{align*}
*
*In (1) we use the binomial series representation
*In (2) we use the linearity of the coefficient of operator and $[x^n]x^kS(x)=[x^{n-k}]S(x)$.
Finally, plugging in $n = 99$: $\frac{1}{2}*101*100*99 + \frac{1}{6}*102*101*100 = 671 650$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
$| \sin x| > \frac{\sqrt{2-\sqrt{2}}}{2}$ iff $1/8< \{ \frac{x}{\pi}\} < 7/8$ where $\{x\}$ is the fractional part of $x$ This is a problem that arose while reading the book "Putnam and Beyond":
Why is $| \sin x| > \frac{\sqrt{2-\sqrt{2}}}{2}$ iff $\frac18 < \{ \frac{x}{\pi}\} < \frac78$ where $\{x\}$ is the fractional part of $x?$
This was used in proving the divergence of $\sum \frac{ |\sin n|}{n}$, although the divergence is not in question here.
| Because we need to prove that:
$$|\sin x|>\frac{\sqrt{2-\sqrt2}}{2},$$ which is
$$\frac{\pi}{8}+2\pi k<x<\frac{7\pi}{8}+2\pi k$$ or
$$\frac{9\pi}{8}+2\pi k<x<\frac{15\pi}{8}+2\pi k,$$ which is
$$\frac{\pi}{8}+\pi k<x<\frac{7\pi}{8}+\pi k,$$
where $k$ is an integer numbers, which is true because
$$\frac{1}{8}<\left\{\frac{x}{\pi}\right\}<\frac{7}{8}$$ it's
$$\frac{1}{8}<\frac{x}{\pi}-\left[\frac{x}{\pi}\right]<\frac{7}{8}$$
or
$$\frac{\pi}{8}+\pi\left[\frac{x}{\pi}\right]<x<\frac{7\pi}{8}+\pi\left[\frac{x}{\pi}\right].$$
Now, take $k=\left[\frac{x}{\pi}\right].$
Actually, $$\sin\frac{\pi}{8}=\sin\frac{7\pi}{8}=\sqrt{\frac{1-\cos45^{\circ}}{2}}=\sqrt{\frac{1-\frac{1}{\sqrt2}}{2}}=\frac{\sqrt{2-\sqrt2}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3089031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Problem when convert $\sqrt{A}+\sqrt{B}=\sqrt{C}+\sqrt{D}$ to A+B=C+D. This is what my lecturer taught me.
If you have $\sqrt{A}+\sqrt{B}=\sqrt{C}+\sqrt{D}$
You can easily convert to $A+B=C+D$ or $AB=CD$
And then he gave me an example.
$\sqrt{8x+1}+\sqrt{3x-5}=\sqrt{7x+4}+\sqrt{2x-2}$ Find x.
Which is totally work.
After that, I've tried to make my own problem.
But there are some of the problems doesn't work with this technique.
For example, $\sqrt{5x+25}+\sqrt{x}=\sqrt{4x}+\sqrt{3x-5}$.
If I use this technique I'll get $x=30$.
But if I solve the problem normally by power 2 both side I'll get $x = \frac { 40 } { 11 } + \frac { 10 \sqrt { 115 } } { 11 }$.
Questions:
Does anyone know how this technique works?
And what're the limitations?
| Try $A=25$, $B=1$ and $C=D=9$.
We see that $$A+B\neq C+D$$ and $$AB\neq CD,$$
but $$\sqrt{A}+\sqrt{B}=\sqrt{C}+\sqrt{D}.$$
We can try to understand, when it happens.
Firstly, $A$, $B$, $C$ and $D$ are non-negatives.
If $A=B$ and $C=D$ we obtain $A+B=C+D$ and $AB=CD.$
Now, let $A>B$, $C>D$, but $A+B=C+D$.
Thus, after squaring of the both sides we obtain
$$A+B+2\sqrt{AB}=C+D+2\sqrt{CD},$$ which gives
$$AB=CD.$$
Thus, $$(A+B)^2-4AB=(C+D)^2-4CD$$ or
$$(A-B)^2=(C-D)^2$$ or
$$A-B=C-D,$$ which after summing with $$A+B=C+D$$ gives $$A=C$$ and $$B=D.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Integral $\int\frac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
Integrate $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$.
| After substituting $\cos(2x)=\cos^2x-\sin^2x$ and $\sin(2x)=2\sin x\cos x$ and doing a bit of algebra, we find that
$$2x\cos(2x)+(x^2-1)\sin(2x)=2(x\sin x+\cos x)(x\cos x-\sin x)$$
and thus
$$\begin{align}{2x^2\over 2x\cos(2x)+(x^2-1)\sin(2x)}
&={x^2\over(x\sin x+\cos x)(x\cos x-\sin x)}\\
&={x\cos x\over x\sin x+\cos x}+{x\sin x\over x\cos x-\sin x}\end{align}$$
The substitution $u=x\sin x+\cos x$, $du=x\cos x\,dx$ tells us
$$\int{x\cos x\over x\sin x+\cos x}\,dx=\int{du\over u}=\ln|u|+C=\ln|x\sin x+\cos x|+C$$
while the substitution $u=x\cos x-\sin x$, $du=-x\sin x$ tells us
$$\int{x\sin x\over x\cos x-\sin x}\,dx=-\int{du\over u}=-\ln|u|+C = -\ln|x\cos x-\sin x|+C$$
and thus
$$\int{2x^2\over 2x\cos(2x)+(x^2-1)\sin(2x)}\,dx=\ln\left|x\sin x+\cos x\over x\cos x-\sin x\right|+C$$
Remark: The key here was to get split the integral into two pieces, each of which is easily handled with a simple substitution. The two pieces' substitutions are similar, but not the same. It was not obvious (to me, at least) that the double-angle formulae would lead to a nice factorization, nor that "partial fractions" on the factored denominator would produce such nice results, but the individual steps seemed natural to consider.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find the density of $Y = a/(1 + X^{2})$, where $X$ has the Cauchy distribution. Find the density of $Y = a/(1 + X^{2})$, where $X$ has the Cauchy distribution.
MY SOLUTION
To begin with, let us remember that the Cauchy probability density function is given by
\begin{align*}
f_{X}(x) = \frac{1}{\pi(1+x^{2})}\quad\text{for}\quad x\in\textbf{R}
\end{align*}
Consequently, the sought distribution is described as
\begin{align*}
F_{Y}(Y\leq y) &= \textbf{P}(Y\leq y) = \textbf{P}\left(\frac{a}{1+X^{2}} \leq y\right) = \textbf{P}\left(1+X^{2}\geq \frac{a}{y}\right)\\\\
& = \textbf{P}\left(|X|\geq \sqrt{\frac{a-y}{y}}\right) = \textbf{P}\left(X\geq \sqrt{\frac{a-y}{y}}\right) + \textbf{P}\left(X \leq -\sqrt{\frac{a-y}{y}}\right)\\\\
& = 1 - F_{X}\left(\sqrt{\frac{a-y}{y}}\right) + F_{X}\left(-\sqrt{\frac{a-y}{y}}\right)
\end{align*}
where $y\in(0,a]$. Finally, we have
\begin{align*}
F_{X}(x) = \int_{-\infty}^{x}f_{X}(u)\mathrm{d}u = \int_{-\infty}^{x}\frac{\mathrm{d}x}{\pi(1+x^{2})} = \frac{\arctan(x)}{\pi} + \frac{1}{2}\end{align*}
I have two questions. Firstly, I would like to know if this approach is correct. Secondly, I would like to know if there is another way to solve this problem. Thanks in advance!
| Your approach is fine.
Perhaps you are aware of this result (very useful to generate a Cauchy variable): if $Z \sim U(-\frac{\pi}{2},\frac{\pi}{2})$ then $X = \tan(Z)$ follows a canonical Cauchy distribution.
Now, we have $Y = \frac{a}{1+X^2}= a \cos^2(Z)$ which is slightly easier.
Then, assuming $a>0$ $$\begin{align}
f_Y(y) &= 2 \frac{f_Z(z)}{|2 a \cos(z) \sin(z)|} \\
&=\frac{1}{a \pi} \frac{1}{|\cos^2(z)\tan(z)|}\\
&=\frac{1}{a \pi} \frac{1}{|\frac{y}{a}x|}\\
&=\frac{1}{ \pi} \frac{1}{y\sqrt{a/y-1}} \\
&=\frac{1}{ \pi} \frac{1}{\sqrt{y(a-y)}}
\end{align}
$$
for $0<y<a$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
I integrated $\int_0^1 \sqrt{x-x^2} dx$ after a $u$-substitution and trig substitution, but I'm not sure what to do with the limits I'm a bit lost on the technique in this problem and could use some insight. When do we have to change the limits?
$$\int_0^1 \sqrt{x-x^2} dx$$
completing the square gives me:
$$\int \sqrt{-x^2 +x - \frac{1}{4} + \frac{1}{4} } dx$$
$$\int \sqrt{-(x-\frac{1}{2})^2 + \frac{1}{4}} dx$$
$$ \int \sqrt{-u^2 + \frac{1}{4}} du$$
$$\int \sqrt{\frac{1}{4} - u^2} du$$'
Is trig sub the best way to go?
if $u = \frac{1}{2} \sin \theta$ then $du = \frac{1}{2} \cos \theta d \theta$
$$ = \int \sqrt{\frac{1}{4} - \frac{1}{4} \sin^2\theta} \frac{1}{2} cos \theta d \theta$$
$$ \int \frac{1}{2} cos \theta \frac{1}{2} cos \theta d \theta$$
$$\frac{1}{4} \int cos^2 \theta d \theta$$
using half angle identity I eventually get:
$$ \frac{1}{4} \int \frac{1}{2} ( 1 + cos(2 \theta)) d \theta$$
$$\frac{1}{8} \int (1 + cos(2 \theta) d \theta$$
$$ \frac{1}{8} ( \theta + \frac{1}{2} sin 2 \theta)$$
Where do I go from here? I'm a bit lost as to what to do with the u sub and how to get back to x from here.
| Just so you are aware, this can be handled with the Beta and by extension the Gamma Function.
Here you have:
\begin{align}
I &= \int_0^1 \sqrt{x - x^2}\:dx = \int_0^1 \sqrt{x\left(1 - x\right)}\:dx = \int_0^1 x^{\frac{1}{2}}\left(1 - x\right)^{\frac{1}{2}}\:dx\\& = B\left(\frac{1}{2} + 1, \frac{1}{2} + 1 \right)= B\left(\frac{3}{2} , \frac{3}{2} \right)
\end{align}
Using the relationship between the Beta and the Gamma Function this becomes:
\begin{equation}
I = B\left(\frac{3}{2} , \frac{3}{2} \right) = \frac{\Gamma\left(\frac{3}{2} \right)\cdot \Gamma\left(\frac{3}{2} \right)}{\Gamma\left(\frac{3}{2} + \frac{3}{2} \right)} = \frac{\Gamma\left(\frac{3}{2} \right)^2}{\Gamma\left(3\right)} = \frac{\left(\frac{\sqrt{\pi}}{2} \right)^2}{2} = \frac{\pi}{8}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I rearrange this? Hi please could some one explain the steps taken to rearange the below. I've had a dabble but don't seem to be getting it.
$$\frac{C}{\sqrt{C^2 - V^2}} $$
To
$$\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}$$
Any help would be greatly appreciated.
Thanks
| Assuming $C > 0$ then
$\frac C{\sqrt{C^2 - V^2}}=$
$\frac {C*\frac 1C}{(\sqrt{C^2 - V^2})\frac 1C}=$
$\frac {1}{(\sqrt{C^2 - V^2})\frac 1{\sqrt{C^2}})}=$
$\frac {1}{\sqrt{(C^2 - V^2)\frac 1{C^2}}}=$
$\frac {1}{\sqrt{\frac {C^2}{C^2} - \frac {V^2}{C^2}}}=$
$\frac 1{\sqrt{1 - \frac {V^2}{C^2}}}$
.... or ...
$\frac {C}{\sqrt {C^2 - V^2}}=$
$\frac {C}{\sqrt{C^2 - C^2\frac {V^2}{C^2}}}=$
$\frac {C}{\sqrt{C^2(1 - \frac {V^2}{C^2})}}=$
$\frac {C}{\sqrt{C^2}\sqrt{1-\frac {V^2}{C^2}}}=$
$\frac {C}{C\sqrt{1-\frac {V^2}{C^2}}}=$
$\frac 1 {\sqrt{1-\frac {V^2}{C^2}}}$
=====
Addendum. It should be pointed out that if $C < 0$ then $\sqrt{C^2} \ne C$ but $\sqrt{C^2} = |C| = -C> 0$. And $C = -\sqrt{C^2}$. Thus if $C < 0$ then $\frac C{\sqrt{C^2 - V^2}} = \frac {C*\frac 1C}{\sqrt{C^2 -V^2}\frac 1{-\sqrt{C^2}}} = -\frac 1{\sqrt{1 - \frac{V^2}{C^2}}}$
This also requires that $C \ne 0$ because we can't divide by $0$. However for $\frac C{\sqrt{C^2 - V^2}}$ to be defined then $\sqrt{C^2 - V^2} \ne 0$ which means $C^2 \ne V^2$ and $C^2 - V^2 \ge 0$ which means $C^2 \ge V^2$ so $C^2 > V^2$ and $V^2 \ge 0$ so $C^2 > 0$ so $C \ne 0$.
But we must assume $|C| > |V|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3102159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving Trig Function with both $\sin$ and $\cos$ I am trying to understand the function $h(t) = 65 + 36\sin\left(\frac{3t}{2}\right) - 15\cos\left(\frac{3t}{2}\right)$, where $t$ is the time in seconds and $\frac{3t}{2}$ is expressed in radians.
A question asks that I find the times in the first revolution when the height is exactly $65$m. Using basic algebra I have been able to solve in order to get time which is $t = 0.26319$. But the question asks for times in the first revolution and I am having trouble understanding how to find these values.
Hence, I need to find the time to complete one revolution and the angular speed of the function.
How would you determine the times when the height is $65$m and from that be able to determine the time it takes to complete one revolution and the angular speed of the function?
I assume that this equation is not in the form of $a + b\sin(cx)$?
| You are looking for $t$ such that
$$
\begin{align}
&\quad h(t) = 65\\
\Rightarrow &\quad 36\sin\left(\frac{3t}{2}\right) = 15\cos\left(\frac{3t}{2}\right)\\
\Rightarrow &\quad 12\sin\left(\frac{3t}{2}\right) = 5\cos\left(\frac{3t}{2}\right)\\
\Rightarrow &\quad 12^2\sin^2\left(\frac{3t}{2}\right) = 5^2\cos^2\left(\frac{3t}{2}\right)\\
\Rightarrow &\quad 144\sin^2\left(\frac{3t}{2}\right) = 25\left({1-\sin^2\left(\frac{3t}{2}\right)}\right)\\
\Rightarrow &\quad 169\sin^2\left(\frac{3t}{2}\right) = 25\\
\Rightarrow &\quad 13\sin\left(\frac{3t}{2}\right) = 5\\
\Rightarrow &\quad \sin\left(\frac{3t}{2}\right) = \frac{5}{13}\\
\Rightarrow &\quad \frac{3t}{2} = \sin^{-1}\left(\frac{5}{13}\right)\\
\Rightarrow &\quad t = \frac23\sin^{-1}\left(\frac{5}{13}\right)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability that the sum of three dice is not greater than 9
What is the probability that the sum of 3 indistinguishable dice is less than or equal to 9?
I have tried counting the pairs $X_1, X_2$ such that their sum is less than or equal to 8, but I seem to be overcounting because I get that for each pair of dice I have 46 possibilities for their sum to be less than 8.
| Consider the sum of the first two dice. If the sum is lower than or equal to 3, the third die can have any value. As the sum increases from 3 to 8, fewer possible values can be thrown by the third die. Since the number of ways to throw a sum of $n$ with two dice equals $6 - |n - 7|$, the number of ways to throw a total of 9 or less equals:
$$1 \cdot 6 + \sum_{i = 3}^{8} (6 - |i - 7|) (9 - i) = 1 \cdot 6 + 2 \cdot 6 + 3 \cdot 5 + 4 \cdot 4 + 5 \cdot 3 + 6 \cdot 2 + 5 \cdot 1 = 81$$
Since there are $6^3 = 216$ ways to throw the three dice, the probability of throwing 9 or less equals:
$$\frac{81}{216} = 0.375$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to find the radius of convergence ? $\sum_{n=1}^{\infty} (\frac{x^n}{n} - \frac{x^{n+1}}{n+1})$ How can i find the radius of convergence ? i dont know where to start i cant use $\frac{a_n}{a_{n+1}}$ test here.
$\sum_{n=1}^{\infty} (\frac{x^n}{n} - \frac{x^{n+1}}{n+1})$
the question looks simple but i dont know how to solve it
i got that $R = 0$ but its wrong because when $x=1$ it is convergent
| Let us consider the partial sums.
\begin{align}
s_N &= \sum_{n=1}^N\left(\frac{x^n}n - \frac{x^{n+1}}{n+1}\right)\\
&= \left(x-\frac{x^2}{2}\right) + \left(\frac{x^2}{2} - \frac{x^3}{3}\right) + \left(\frac{x^3}{3} - \frac{x^4}{4}\right) + \text{ ... } + \left(\frac{x^{N}}{N} - \frac{x^{N+1}}{N+1}\right)\\
&= x - \frac{x^{N+1}}{N+1}
\end{align}
To evaluate the sum, we take the limit $\lim\limits_{N\to\infty}s_N$ and find that the limit is equal to $x$ if $|x|\leq1$ and does not exist otherwise.
This means that the radius of convergence is $1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\frac{2a}{a^2-4} - \frac{1}{a-2}-\frac{1}{a^2+2a}$
Evaluate
$$\dfrac{2a}{a^2-4} - \dfrac{1}{a-2}-\dfrac{1}{a^2+2a}$$
We have to see what their common term is. Therefrom, we can evaluate the simplified expression by canceling out.
$$a^2 - 4 = (a)^2 - (2)^2 = (a-2)(a+2) \tag {1}$$
$$a^2 +2a = a(a+2)\tag{2}$$
Rewriting the expression
$$\dfrac{2a}{(a-2)(a+2)} - \dfrac{1}{a-2}-\dfrac{1}{a(a+2)}$$
$$\dfrac{2a}{(a-2)(a+2)} - \dfrac{(a+2)}{(a-2)(a+2)}-\dfrac{1}{a(a+2)}$$
Factoring $\dfrac{1}{a+2}$
$$\dfrac{1}{a+2}\biggr (\dfrac{2a}{a-2} - \dfrac{a+2}{a-2}-\dfrac{1}{a}\biggr )$$
This is where I'm stuck. Could you assist me?
Regards
| Hint: You can write $$\frac{2a^2}{a(a-2)(a+2)}-\frac{a(a+2)}{a(a-2)(a+2)}-\frac{a-2}{a(a+2)(a-2)}$$
| {
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"source": "stackexchange",
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If $a_{n} = ((-1)^{n})/(\sqrt{1+n})$ then is $\sum_{k=0}^{n} (a_{n-k})(a_{k})$ convergent? If $a_{n} = ((-1)^{n})/(\sqrt{1+n})$ then is $C_{n} = \sum_{k=0}^{n} (a_{n-k})(a_{k})$ convergent?
$a_{n}$ is conditionally convergent, but how do I check the convergence of $C_{n}$?
I could transform $C_{n}$ in following way
$C_{n} = \sum_{k=0}^{n} \frac{(-1)^{n}}{(1+k)^{1/2}(1+n-k)^{1/2}}$
How to proceed?
| $C_{n}
= \sum_{k=0}^{n} \frac{(-1)^{n}}{(1+k)^{1/2}(1+n-k)^{1/2}}
$
$\begin{array}\\
(1+k)(1+n-k)
&=1+n+kn-k^2\\
&=1+n-(k^2-kn)\\
&=1+n-(k^2-kn+n^2/4)+n^2/4\\
&=1+n-(k-n/2)^2+n^2/4\\
&= (n+2)^2/4-(k-n/2)^2\\
\text{so}\\
\dfrac1{\sqrt{(1+k)(1+n-k)}}
&= \dfrac1{\sqrt{(n+2)^2/4-(k-n/2)^2}}\\
\end{array}
$
so
$\begin{array}\\
|C_{n}|
&= \sum_{k=0}^{n} \dfrac{1}{\sqrt{(n+2)^2/4-(k-n/2)^2}}\\
&= \sum_{k=0}^{n} \dfrac{1}{n\sqrt{(1+2/n)^2/4-(k/n-1/2)^2}}\\
&\approx \int_0^1 \dfrac{dx}{\sqrt{1/4-(x-1/2)^2}}
\quad\text{Riemann sum}\\
&= \pi
\quad\text{since }\int \dfrac{dx}{\sqrt{1/4-x^2}}=\arcsin(2x)\\
\end{array}
$
Wolfy says that
$(n, |C_n|)
=(10, 2.293...), (100, 2.852...), (1000, 3.049...), (10000, 3.1123...)
$
so the convergence is there
but slow.
| {
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"url": "https://math.stackexchange.com/questions/3105882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that |z|=1. I am trying to prove that
If $z\in \mathbb{C}-\mathbb{R}$ such that $\frac{z^2+z+1}{z^2-z+1}\in \mathbb{R}$. Show that $|z|=1$.
1 method , through which I approached this problem is to assume $z=a+ib$ and to see that $$\frac{z^2+z+1}{z^2-z+1}=1+\frac{2z}{z^2-z+1}$$.
So problem reduces to show that $|z|=1$ whenever $\frac{2z}{z^2-z+1}\in \mathbb{R}$
I put $z=a+ib$ and then rationalise to get the imaginary part of $\frac{2z}{z^2-z+1}$ be $\frac{b-b^3-a^2b}{something}$. I equated this to zero and got my answer.
Is there any better method?
| Let $w=\frac{z^2+z+1}{z^2-z+1}=1+\frac{2z}{z^2-z+1}$
Since $\frac{z^2+z+1}{z^2-z+1}\in \mathbb{R}$ , so $\operatorname{Im}(w)=0$
$$\iff w-\bar{w}=0$$
Now, let's solve $1+\frac{2z}{z^2-z+1}=\overline{1+\frac{2z}{z^2-z+1}}$
$$\implies \frac{z}{z^2-z+1}=\overline{\frac{z}{z^2-z+1}}$$
Since $z^2-z+1=0 \ when\ z= \frac{1 \pm i \sqrt{3}}{2}$
Assume $z \neq \frac{1 \pm i \sqrt{3}}{2}$
$$\implies \frac{z}{z^2-z+1}= \frac{\overline{z}}{\overline{z^2}-\overline{z}+1}$$
$$\implies z(\overline{z^2}-\overline{z}+1)=\overline{z}(z^2-z+1) $$
After some simplifications, we have
$$ (z-\overline{z})(|z|-1)=0$$
$$\implies (z-\overline{z})=0 \ or \ |z|-1=0$$
Since $(z-\overline{z})=0\ $must be true by the above statement, so we just need to solve $|z|-1=0 \implies |z|=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof Explanation - Odd prime division
Prove that if $p$ is an odd prime, then $$p\mid \lfloor(2+\sqrt5)^p\rfloor - 2^{p+1}$$
The solution posted by another user is as follows:
Let $=(2+\sqrt5)^p + (2-\sqrt5)^$.
Note that $N$ is an integer. There are various ways to see this. One can, for example, expand using the binomial theorem, and observe that the terms involving odd powers of $\sqrt5$ cancel.
Because $(2−\sqrt5)^$ is a negative number close to $0$, it follows that $=⌊(2+\sqrt5)^⌋$.
In the two binomial expansions, all the binomial coefficients $\binom{p}{k}$ apart from the first and last are divisible by $p$. The first term in each expansion is $2^p$. We conclude that $N\equiv 2\cdot 2^p\pmod{p}$, and the result follows.
Could someone explain this proof in more detail please? I don't understand the following:
1) How the odd powers of $\sqrt5$ cancel,
2) How $N=\lfloor(2+\sqrt5)^p\rfloor$ (I understand that $(2-\sqrt5)^p$ is negative),
3) How $N\equiv 2\cdot 2^p\pmod{p}$, and the result follows.
How would one go about generalising to a result about $p^n$?
Thanks
| 1) We have
$(2+\sqrt 5)^p=2^p+\binom p12^{p-1}\sqrt 5+\binom p22^{p-2}\sqrt 5^2+\binom p23^{p-3}\sqrt 5^3+...+\binom pp\sqrt 5^p$, and
$(2-\sqrt 5)^p=2^p-\binom p12^{p-1}\sqrt 5+\binom p22^{p-2}\sqrt 5^2-\binom p23^{p-3}\sqrt 5^3+...-\binom pp\sqrt 5^p$.
When you add these together, all the odd terms are equal and opposite, so you get
$N=2\times2^p+2\times\binom p22^{p-2}\sqrt 5^2+...+2\times\binom p{p-1}2\sqrt 5^{p-1}$.
Since this only has even powers of $\sqrt 5$, it is an integer.
2) As well as being an integer, $N$ is slightly less than $(2+\sqrt 5)^p$. In fact it is $-(2-\sqrt 5)^p=(\sqrt 5-2)^p$ less. Since $0<(\sqrt 5-2)^p<1$, $N$ is the integer between $(2+\sqrt 5)^p$ and $(2+\sqrt 5)^p-1$, i.e. it is $\lfloor(2+\sqrt 5)^p\rfloor$.
3) Going back to $N=2\times2^p+2\times\binom p22^{p-2}\sqrt 5^2+...+2\times\binom p{p-1}2\sqrt 5^{p-1}$, each term except the first has a factor of $\binom pr$ for some $0<r<p$. Writing this in factorial form, it is $\frac{p!}{r!(p-r)!}$. Since $p$ is prime, $p$ divides the top but not the bottom. This means every term except the first is a multiple of $p$, so $N\equiv 2\times2^p$ mod $p$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ .
I think that must I use from $\dfrac{a^2}{2}+\dfrac{b^2}{2} \geq ab$ but no result please help me .!
| $$\frac{a^2}{2}+\frac{b^3}{3}+\frac{c^6}{6}=\frac{a^2}{6}+\frac{a^2}{6}+\frac{a^2}{6}+\frac{b^3}{6}+\frac{b^3}{6}+\frac{c^6}{6}.$$
Then use AM-GM with these $6$ variables.
| {
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"url": "https://math.stackexchange.com/questions/3110691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find closed form of composition. Consider $f(z) = \frac{1-a}{1-az}$, where $a \in(0,1)$
. Now we want to find $f(f\dots(f(z)\dots)$. It will begood if there is a closed form of it.
I've consider first step : $f(f(z)) = \frac{1 - a}{1 -a\frac{1-a}{1-az}} = \frac{(1-a)(1-az)}{1-az-a+a^2}$ , but it looks non-simplified.
I guess I miss something , any hints?
| If we identify linear fractional transformation with the matrix (up to multiplicative constant), i.e.
$$
f(z) = \frac{az+b}{cz+d}\ \ \ \Longleftrightarrow \ \ \ \left[\begin{array}{cc}a&b\\c&d\end{array}\right],
$$ then composition corresponds to matrix multiplication. Thus the problem boils down to calculating
$$
A^n =\left[\begin{array}{cc}0&1-a\\-a&1\end{array}\right]^n.
$$ Fortunately, $A$ is diagonalizable with
$$
A= \left[\begin{array}{cc}1&1-a\\1&a\end{array}\right]\left[\begin{array}{cc}1-a&0\\0&a\end{array}\right]\cdot\frac{1}{2a-1}\left[\begin{array}{cc}a&a-1\\-1&1\end{array}\right],
$$ hence
$$
A^n =\left[\begin{array}{cc}1&1-a\\1&a\end{array}\right]\left[\begin{array}{cc}(1-a)^n&0\\0&a^n\end{array}\right]\cdot\frac{1}{2a-1}\left[\begin{array}{cc}a&a-1\\-1&1\end{array}\right].
$$ Up to multiplicative constant, we have
$$
A^n \sim \left[\begin{array}{cc}a(1-a)^n-(1-a)a^n&-(1-a)^{n+1}+(1-a)a^n\\a(1-a)^n-a^{n+1}&-(1-a)^{n+1}+a^{n+1}\end{array}\right],
$$ hence
$$
f^n(z) = \frac{[a(1-a)^n-(1-a)a^n]z-(1-a)^{n+1}+(1-a)a^n}{[a(1-a)^n-a^{n+1}]z-(1-a)^{n+1}+a^{n+1}}.
$$
| {
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"url": "https://math.stackexchange.com/questions/3111799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\cot{142\frac{1}{2}^\circ} = \sqrt2 + \sqrt3 - 2 - \sqrt6$.
Show that $\cot{142\frac{1}{2} ^\circ} = \sqrt2 + \sqrt3 - 2 - \sqrt6$.
What I have tried:
Let $\theta = 142\frac{1}{2}^\circ \text{ and } 2\theta = 285^\circ$.
$$\cos 285^\circ = \cos 75^\circ$$
$$\cos 75^\circ = \frac{\sqrt3 - 1}{2\sqrt2}$$
$$\cot \theta = \sqrt{\frac{1 + \cos 2\theta}{1 - \cos 2\theta}}$$
From here, pls help me proceed further. Thank you :)
| First render $\cot(142\frac{1}{2}°)=-\cot(37\frac{1}{2}°)$. Then for the latter angle put in:
$\cot x= \dfrac{2\cos^2 x}{2\sin x \cos x}=\dfrac{1+\cos 2x}{\sin2x}$
with $x=37\frac{1}{2}°$ so $2x=75°$. If you know the sine and cosine of the latter the rest is algebra.
| {
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"url": "https://math.stackexchange.com/questions/3113366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A formula for tan(2x) Help with solving...
Suppose that $\tan^2x=\tan(x-a)·\tan(x-b)$, show that $$\tan(2x)=\frac{2\sin(a)·\sin(b)}{\sin(a+b)}$$
As far as I know, so far the $\tan2x$ can be converted to $\frac{2\tan x}{1-\tan^2x}$ using the double angle formula and the $\tan 2x$ can be further be substituted to the following given item ($\tan(x-a·)\tan(x-b)$) however the problem now is that there are no ways to simplify this to my knowledge...
| Variation of lab bhattacharjee's answer:
$$\tan^2x=\tan(x-a)·\tan(x-b) \Rightarrow \\
\tan^2x+1=\tan(x-a)·\tan(x-b)+1 \Rightarrow \\
\frac{1}{\cos^2 x}=\tan(x-a)·\tan(x-b)+1 \Rightarrow \\
\cos^2x =\frac1{\tan(x-a)·\tan(x-b)+1} \Rightarrow \\
\frac{1+\cos 2x}{2}=\frac1{\tan(x-a)·\tan(x-b)+1} \Rightarrow \\\\
\cos 2x=\frac2{\tan(x-a)·\tan(x-b)+1}-1 \Rightarrow \\
\cos 2x=\frac{1-\tan (x-a)\cdot \tan (x-b)}{1+\tan (x-a)\cdot \tan (x-b)}=\\
\cos 2x=\frac{\cos (x-a)\cos (x-b)-\sin (x-a)\cos (x-b)}{\cos (x-a)\cos (x-b)+\sin (x-a)\cos (x-b)} \Rightarrow \\
\cos 2x=\frac{\cos (2x-(a+b))}{\cos (a-b)} \Rightarrow \\
\cos 2x=\frac{\cos 2x\cos (a+b)+\sin 2x\sin (a+b)}{\cos (a-b)}\Rightarrow \\
\cos 2x\cos (a-b)=\cos 2x\cos (a+b)+\sin 2x\sin (a+b) \Rightarrow \\
\cos 2x[\cos (a-b)-\cos (a+b)]=\sin 2x\sin (a+b)\Rightarrow \\
\tan 2x=\frac{\cos (a-b)-\cos(a+b)}{\sin (a+b)}=\\
\frac{2\sin a\sin b}{\sin(a+b)}.$$
| {
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"url": "https://math.stackexchange.com/questions/3113894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving an equality using given ones; no need for differentiation Prove that $(\frac ab+\frac bc+\frac ca)(\frac ba+\frac cb+\frac ac)\geqslant9$
The formulas given were
$$\frac{a+b} {2}\geqslant\sqrt {ab}$$
$$a^2+b^2\geqslant2ab$$
$$\frac{a+b+c} {3}\geqslant\root3\of{abc}$$
$$a^3+b^3+c^3\geqslant3abc$$
for all $a\gt0, b\gt0, c\gt0.$
I couldn't think of any way to do this; please help.
| Method 1: AM-GM
Apply the last given inequality, but with the fractions instead.
$$\frac ab+\frac bc+\frac ca\ge 3 \sqrt[3]{\frac ab\,\frac bc\,\frac ca} = 3.$$
Do the same for $\frac ba+\frac cb+\frac ac$
Method 2: Cauchy Schwarz
$$9 = 3^2 = \left(\frac{\sqrt{a}}{\sqrt{b}} \, \frac{\sqrt{b}}{\sqrt{a}} + \frac{\sqrt{b}}{\sqrt{c}} \, \frac{\sqrt{c}}{\sqrt{b}} + \frac{\sqrt{c}}{\sqrt{a}} \, \frac{\sqrt{a}}{\sqrt{c}}\right)^2 \\ \le \left(\frac ab + \frac bc + \frac ca \right) \left(\frac ba + \frac cb + \frac ac \right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{0}^{\pi} x\sin\big(\frac{1}{x}\big)\cos x \,dx$ I wonder if an integral of the form $$\int_{0}^{\pi} x\sin\Bigl(\frac{1}{x}\Bigr)\cos x \,dx$$ which can be further simplified to
$$\int_{0}^{\pi} \frac{\sin (x^{-1})}{(x^{-1})}\,\cos x\, dx=\cos x\biggl(\int_{0}^{\pi} \frac{\sin (x^{-1})}{(x^{-1})}\,dx\biggr)\Bigg|_{0}^{\pi}+\int_{0}^{\pi}\sin x \bigg(\int \frac{\sin (x^{-1})}{(x^{-1})}dx\bigg)dx$$
will have an explicit form. One simplifies the integral
$$\int_{0}^{\pi} \frac{\sin (x^{-1})}{(x^{-1})}dx=\int_{0}^{\pi} \int_{0}^{\infty}e^{-(x^{-1}t)}\sin (x^{-1})\,dx\,dt$$
where we know that $\int_{0}^{\infty}e^{-xt}dt=\frac{1}{x}$.
We also know that $$\int_{0}^{\infty}e^{-(xt)}\sin (x)\,dx=\frac{1}{t^2+1}.$$
Any idea on how to compute
$$\int_{0}^{\infty}e^{-(x^{-1}t)}\sin (x^{-1})\,dx?$$
| NOT THE SOLUTION:
DAMN, I just got to the of the solution as below and realised I had the bounds wrong the whole time - Thought I would leave in case you're interested!
Here we are addressing the integral:
\begin{equation}
I = \int_0^\infty x \sin\left(\frac{1}{x}\right)\cos(x)\:dx
\end{equation}
Here we will employ Feynman's Trick and introduce a new function with two variables:
\begin{equation}
J(a,b) = \int_0^\infty x \sin\left(\frac{a}{x}\right)\cos(bx)\:dx
\end{equation}
We see that $J(1,1) = I$. Now $\frac{\partial}{\partial b}\sin(bx) = x\cos(bx)$ and so,
\begin{equation}
J(a,b) = \int_0^\infty x \sin\left(\frac{a}{x}\right)\cos(bx)\:dx = \int_0^\infty \sin\left(\frac{a}{x}\right)\cdot x\cos(bx)\:dx = \int_0^\infty \sin\left(\frac{a}{x}\right)\cdot\frac{\partial}{\partial b}\sin(bx)\:dx
\end{equation}
By using Leibniz's Integral Rule we find:
\begin{equation}
J(a,b) = \int_0^\infty \sin\left(\frac{a}{x}\right)\cdot\frac{\partial}{\partial b}\sin(bx)\:dx = \frac{\partial}{\partial b} \underbrace{\int_0^\infty \sin\left(\frac{a}{x}\right)\sin(bx)\:dx}_{H(a,b)}
\end{equation}
We now will address $H(a,b)$. To proceed we employ Fubini's Theorem and take the Laplace Transform with respect to $a$:
\begin{align}
\mathscr{L}_{a \rightarrow s} \left[H(a,b)\right] &= \int_0^\infty \mathscr{L}_{a \rightarrow s} \left[\sin\left(\frac{a}{x}\right)\right]\sin(bx)\:dx = \int_0^\infty \frac{\frac{1}{x}}{s^2 + \left(\frac{1}{x}\right)^2}\cdot \sin(bx)\:dx \\
&= \int_0^\infty \frac{x}{s^2x^2 + 1}\cdot \sin(bx)\:dx
\end{align}
We now use the same trick from before: $ \frac{\partial}{\partial b} -\cos(bx) = x\sin(bx) $. Thus,
\begin{align}
\mathscr{L}_{a \rightarrow s} \left[H(a,b)\right] &= -\frac{\partial}{\partial b} \int_0^\infty \frac{\cos(bx)}{s^2x^2 + 1}\:dx
\end{align}
Or
\begin{align}
H(a,b) &= -\frac{\partial}{\partial b}\left( \mathscr{L}_{s \rightarrow a}^{-1} \left[\int_0^\infty \frac{\cos(bx)}{s^2x^2 + 1}\:dx\right]\right)
\end{align}
Or, in terms of $J(a,b)$
\begin{align}
J(a,b) &= \frac{\partial}{\partial b} H(a,b) = -\frac{\partial^2}{\partial b^2}\left( \mathscr{L}_{s \rightarrow a}^{-1} \left[\int_0^\infty \frac{\cos(bx)}{s^2x^2 + 1}\:dx\right]\right)
\end{align}
We now need only address the much simpler integral (below) and then taken it's inverse Laplace Transform:
\begin{equation}
K(a,b) = \int_0^\infty \frac{\cos(bx)}{s^2x^2 + 1}
\end{equation}
Here we again use Fibini's Theorem and take the Laplace Transform with respect to $b$:
\begin{align}
\mathscr{L}_{b \rightarrow w} \left[ K(a,b) \right] &= \int_0^\infty \frac{\mathscr{L}_{b \rightarrow w} \left[ \cos(bx) \right]}{s^2x^2 + 1} = \int_0^\infty \frac{1}{s^2x^2 + 1} \cdot \frac{w}{w^2 + x^2}\:dx \\
&= \frac{w}{s^2w^2 - 1} \int_0^\infty \left[\frac{s^2}{s^2 x^2 + 1} - \frac{1}{w^2 + x^2 } \right]\:dx = \frac{w}{s^2w^2 - 1} \left[s\arctan(sx) - \frac{1}{w}\arctan\left(\frac{x}{w}\right)\right]_0^\infty \\
&= \frac{w}{s^2w^2 - 1}\left[ s\cdot\frac{\pi}{2} - \frac{1}{w} \cdot \frac{\pi}{2} \right] = \frac{sw - 1}{s^2w^2 - 1} \cdot \frac{\pi}{2} = \frac{\pi}{2\left(sw + 1\right)}
\end{align}
We now take the Inverse Laplace Transform with respect to $w$ to resolve $K(s,b)$:
\begin{align}
K(s,b) &= \mathscr{L}_{w \rightarrow b}^{-1} \left[ \frac{\pi}{2\left(sw + 1\right)} \right] = \frac{\pi}{2s}e^{-\frac{b}{s}}
\end{align}
From here we can now address J(a,b)
\begin{align}
J(a,b) &= \frac{\partial}{\partial b} H(a,b) = -\frac{\partial^2}{\partial b^2}\left( \mathscr{L}_{s \rightarrow a}^{-1} \left[ K(s,b) \right] \right) = -\frac{\partial^2}{\partial b^2}\left( \mathscr{L}_{s \rightarrow a}^{-1} \left[\frac{\pi}{2s}e^{-\frac{b}{s}} \right] \right) = -\frac{\partial^2}{\partial b^2} J_{0}\left(2\sqrt{ab} \right) \\
&= -\frac{\pi}{4b}\left[t\left( J_{0}\left(2\sqrt{ab} \right) + J_{2}\left(2\sqrt{ab} \right) \right) - \frac{ J_{1}\left(2\sqrt{ab} \right)}{\sqrt{b}} \right]
\end{align}
Where $J_{\alpha}(x)$ is the Modified Bessel Function. From here we now need only let $a,b = 1$ to solve $I$
\begin{equation}
I = J(1,1) = -\frac{\pi}{4\cdot 1}\left[t\left( J_{0}\left(2\sqrt{1\cdot 1} \right) + J_{2}\left(2\sqrt{1\cdot 1} \right) \right) - \frac{ J_{1}\left(2\sqrt{1\cdot 1} \right)}{\sqrt{b}} \right] = -\frac{\pi}{4}\big[J_0(2) + J_2(2) - J_1(2)\big]
\end{equation}
Thus,
\begin{equation}
\int_0^\infty x \sin\left(\frac{1}{x}\right)\cos(x)\:dx = -\frac{\pi}{4}\big[J_0(2) + J_2(2) - J_1(2)\big]
\end{equation}
In fact, we we can go further using this method and have:
\begin{align}
J_p(a,b) = \int_0^\infty x^{2p + 1} \sin\left(\frac{a}{x}\right)\cos(bx)\:dx = (-1)^{p + 1} \frac{\partial^{2p + 2}}{\partial b^{2p + 2}} J_{0}\left(2\sqrt{ab} \right) \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Applying comparison theorem to limits when inequality is known I somehow confuse myself whenever I apply the comparison lemma. I know there is the comparison lemma that says the following:
Let the sequence $\{a_{n}\}$ converge to the number $a$. Then the sequence $\{b_{n}\}$ converges to the number $b$ if there is a non-negative number $C$ and an index $N$ such that $$\lvert b_{n} - b \rvert \leq C \lvert a_{n}-a\rvert$$
Somehow i accidentally apply this wrong. For example let's say I'm applying it to the sequence $a_n$, to say that it converges to $0$,
$$a_n=\frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}$$
and
$$b_n=\frac{1}{\sqrt{1+\frac{1}{n}}+1}$$
Obviously, since $n\ge1$, then I can say
$$\frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}<\frac{1}{\sqrt{1+\frac{1}{n}}+1}$$
Subtracting $1/2$ from both sides, I can see that
$$\frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}-\frac{1}{2}<\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2}$$
Since I know that $$\frac{1}{\sqrt{1+\frac{1}{n}}+1} \longrightarrow \frac{1}{2}$$ then the comparison lemma seems to be telling me that $a_n$ converges to $1/2$, when in fact I know that it converges to $0$. Where am I going wrong in this logic?
| The problem is that $$\sqrt{1+\frac{1}{n}} + 1 > 2$$ for all $n\in\mathbb N$, and therefore
$$\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2} < 0$$
When you compare the absolute values of both sides of your last inequality, the "$<$" becomes "$>$", and the theorem does not apply.
Namely, because the values inside the absolute value sign are negative, we have that
$$\left| \frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)} - \frac{1}{2} \right| = \frac{1}{2} - \frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}$$
$$\left| \frac{1}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{2} \right| = \frac{1}{2} - \frac{1}{\sqrt{1+\frac{1}{n}}+1}$$
But, starting from the inequality (that does indeed hold):
$$\frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}-\frac{1}{2}<\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2}$$
and multiplying both sides by -1, and remembering that we have to flip the inequality sign, we arrive at
$$\left| \frac{1}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+1\right)}-\frac{1}{2} \right| > \left|\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2} \right|$$
Therefore, the sequences $a_n$ and $b_n$ do not satisfy the conditions of the theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can anybody help with the integral: $\int \frac{\sqrt{1-x^2}}{\sqrt{1+x^2}} dx$ please? I have searched online. I have got a solution from Wolfram but I don't understand it, or how to reach it. If anyone has a method that would be fantastic, thanks!
| If the integral is a definite integral with the interval of integration being from $0$ to $1$, here is an approach that makes use of the following form for the Beta function $\operatorname{B} (x,y)$:
$$\operatorname{B} (x,y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt, \qquad x,y > 0. \tag1$$
Let
$$I = \int_0^1 \sqrt{\frac{1 - x^2}{1 + x^2}} \, dx.$$
Enforcing a substitution of $x \mapsto \sqrt{x}$ yields
$$I = \frac{1}{2} \int_0^1 \sqrt{\frac{1 - x}{1 + x}} \, \frac{dx}{\sqrt{x}}.$$
Rearranging the integrand we have
\begin{align}
I &= \frac{1}{2} \int_0^1 \sqrt{\frac{1 - x}{1 + x}} \cdot \sqrt{\frac{1 - x}{1 - x}} \, \frac{dx}{\sqrt{x}}\\
&= \frac{1}{2} \int_0^1 \frac{1 - x}{\sqrt{1 - x^2}} \, \frac{dx}{\sqrt{x}}\\
&= \frac{1}{2} \int_0^1 \frac{dx}{\sqrt{1 - x^2} \sqrt{x}} - \frac{1}{2} \int_0^1 \frac{\sqrt{x}}{\sqrt{1 - x^2}} \, dx. \qquad \qquad \qquad \qquad (*)
\end{align}
Enforcing a substitution of $x \mapsto \sqrt{x}$ in each of the above two integrals leads to
\begin{align}
I &= \frac{1}{4} \int_0^1 x^{-3/4} (1 - x)^{-1/2} \, dx - \frac{1}{4} \int_0^1 x^{-1/4} (1 - x)^{-1/2} \, dx\\
&= \frac{1}{4} \int_0^1 x^{1/4 - 1} (1 - x)^{1/2 - 1} \, dx - \frac{1}{4} \int_0^1 x^{3/4 - 1} (1 - x)^{1/2 - 1} \, dx\\
&= \frac{1}{4} \operatorname{B} \left (\frac{1}{4}, \frac{1}{2} \right ) - \frac{1}{4} \operatorname{B} \left (\frac{3}{4}, \frac{1}{2} \right ) \tag2\\
&= \frac{1}{4} \left [\frac{\Gamma (\frac{1}{4}) \Gamma (\frac{1}{2})}{\Gamma (\frac{3}{4})} - \frac{\Gamma (\frac{3}{4}) \Gamma (\frac{1}{2})}{\Gamma (\frac{1}{4})} \right ] \tag3\\
&= \frac{\Gamma^2 (\frac{1}{4})}{4 \sqrt{2 \pi}} - \frac{\pi \sqrt{2 \pi}}{\Gamma^2 (\frac{1}{4})}.\tag4
\end{align}
Explanation
(2) Each integral is of the form of the Beta function given in (1).
(3) Using the result $\operatorname{B}(x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$.
(3) Using $\Gamma (\frac{1}{2}) = \sqrt{\pi}$ and $\Gamma (\frac{3}{4}) = \dfrac{\pi \sqrt{2}}{\Gamma (\frac{1}{4})}$, a result that comes immediately from Euler's reflexion formula for the Gamma function.
If it is the indefinite integral you are really after, then as suggested in the comments, your integral can be expressed in terms of elliptic functions of the first and second kinds.
From ($*$), namely
$$I = \frac{1}{2} \int \frac{du}{\sqrt{1 - u^2} \sqrt{u}} - \frac{1}{2} \int \frac{\sqrt{u}}{\sqrt{1 - u^2}} \, du,$$
where $u = x^2$, substituting $u = \sin \theta$ into each of the above integrals one arrives at
\begin{align}
I &= \frac{1}{2} \int \sqrt{\operatorname{cosec} x} \, d\theta - \frac{1}{2} \int \sqrt{\sin \theta} \, d\theta\\
&= - F \left (\frac{\pi - 2 \theta}{4} \Big{|} 2 \right ) + E \left (\frac{\pi - 2 \theta}{4} \Big{|} 2 \right ) + C\\
&= E \left (\frac{\pi - 2 \sin^{-1} (x^2)}{4} \Big{|} 2 \right ) - F \left (\frac{\pi - 2 \sin^{-1} (x^2)}{4} \Big{|} 2 \right ) + C, \qquad -1 \leqslant x \leqslant 1.
\end{align}
Here $F( \phi | k^2)$ is an elliptic integral of the first kind while $E(\phi | k^2)$ is an elliptic integral of the second kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Distributing pairs of coloured socks to people Say you have $100$ socks, where $50$ are black, $30$ are white, $20$ are blue. You want to distribute them to $10$ different people such that none of them end up with an odd number of any colour, but there is no requirement that all $10$ people get socks. We are assuming the socks are identical other than colour and the people are all distinguishable. My solution is:
There are $25$ pairs of black socks, $15$ white, $10$ blue. There are
$\binom{10+10-1}{10}=\binom{19}{10}$ ways to distribute the blue socks,
$\binom{25+10-1}{25}=\binom{34}{25}$ ways to distribute the black socks, and
$\binom{15+10-1}{15}=\binom{24}{15}$ ways to distribute the white socks.
So there are $\binom{19}{10}\binom{34}{25}\binom{24}{15}$ total ways to distribute the $100$ socks such that nobody gets an odd number of any of the socks.
Would anybody be able to check my solution for this?
| I was looking into the Generating Function approach to this question.
The 25 black pairs of socks could be represented by;
$$(1+x+x^2+x^3+x^4+...+x^{25})$$
and the ten boxes by raising this to the power of 10;
$$(1+x+x^2+x^3+x^4+...+x^{25})^{10}$$
So, essentially, the answer could be obtained from expanding the brackets of;
$$(1+x+x^2+...+x^{25})^{10} \times(1+y+y^2+...+y^{15})^{10} \times(1+z+z^2+...+z^{10})^{10}$$
and looking at the coefficient of $x^{25}y^{15}z^{10}$ where $x,y,z$ represent black, white and blue pairs of socks respectively. This looks heavy going, although each part is a geometric progression;
$$\bigr(\frac{(1-x^{25})}{(1-x)}\bigr)^{10} \times \bigr(\frac{(1-y^{15})}{(1-y)}\bigr)^{10} \times\bigr(\frac{(1-z^{10})}{(1-z)}\bigr)^{10}$$
Keeping in mind that we're only after the coefficient of $x^{25}y^{15}z^{10}$ it might be possible to extract it$\dots$
Anyone want to take it on from here ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trisected Triangle Perimeter ICTM State 2017 Division AA Precalculus Individual Question 17 reads:
$C$ and $D$ lie on $BE$ such that $AC$ and $AD$ trisect $\angle$BAE in $△ABE$. $BC = 2$, $CD = 3$, and $DE = 6$. Then the perimeter of $△ABC$ may be expressed as $P = f + k \sqrt{w} + p \sqrt{q}$ in simplified and reduced radical form. Determine the sum $(f+k+w+p+q)$. See figure below
What I tried.
I tried to create a system of equations equating the angles of the three triangles to 180°. I also equated the supplementary angles.
There were not enough equations to solve for every variable.
| Let $AB=x$.
Thus, since $$\frac{AD}{AB}=\frac{CD}{BC},$$ we obtain
$$\frac{AD}{x}=\frac{3}{2},$$ which gives
$$AD=\frac{3}{2}x.$$
Also, let $AC=y$.
Thus, since $$\frac{AE}{AC}=\frac{DE}{DC},$$ we obtain $AE=2y$.
But, $$AC^2=AB\cdot AD-BC\cdot CD$$ and
$$AD^2=AC\cdot AE-CD\cdot DE,$$ which gives
$$y^2=\frac{3}{2}x^2-6$$ and
$$\frac{9}{4}x^2=2y^2-18.$$
After solving of this system we obtain: $x=2\sqrt{10},$ $y=3\sqrt{6}$, $P=11+2\sqrt{10}+6\sqrt{6},$
which gives the answer: $35$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus. Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus. This inequality is equivalent to
$$
2018^{1/2018}>2019^{1/2019}
$$
One of my 'High school' student asked why the inequality is true. The whole class became interested in the problem.The demonstration that such inequality is true, using calculus, can be found here. But my students are not familiar with calculus.
I can also show by induction and Newton's binomial formula that $ n^{(n + 1)}> (n + 1)^n $, to $ n> 3$, but my students are not familiar with mathematical induction.
Another limitation of my students is that they have not yet learned the Newton's binomial formula.
How to prove the inequality $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without calculus? That is how to prove this inequality for High school students without using Newton's binomial formula?
| Compared to my first answer, this answer makes a simpler use of the same familiar identity, which is proved without explicit use of induction, thus:
\begin{align*}
x^m - y^m & = x^m - (x^{m-1}y - x^{m-1}y) - (x^{m-2}y^2 - x^{m-2}y^2) - \cdots - (xy^{m-1} - xy^{m-1}) - y^m \\
& = (x^m - x^{m-1}y) + (x^{m-1}y - x^{m-2}y^2) + (x^{m-2}y^2 - x^{m-3}y^3) + \cdots + (xy^{m-1} - y^m) \\
& = (x - y)x^{m-1} + (x - y)x^{m-2}y + (x - y)x^{m-3}y^2 + \cdots + (x - y)y^{m-1} \\
& = (x - y)(x^{m-1} + x^{m-2}y + x^{m-3}y^2 + \cdots + y^{m-1}).
\end{align*}
The rest of the proof is a straightforward calculation. (It is based on a trick which was inspired loosely by nbarto's answer, but I am to blame for it!)
Suppose $n \geqslant 3$.
In the above identity, take $x = m = n+1$, $y = n$, and group together all but the first two terms in the brackets, obtaining:
$$
(n+1)^{n+1} - n^{n+1} = (n+1)^n + (n+1)^{n-1}n + [(n+1)^{n-2}n^2 + \cdots + n^n].
$$
There are $n-1$ terms in the square brackets, they are in strictly decreasing order, and the largest of them is $(n+1)^{n-2}n^2$. Therefore:
\begin{align*}
(n+1)^{n+1} - n^{n+1} & < (n+1)^n + (n+1)^{n-1}n + (n-1)(n+1)^{n-2}n^2 \\
& = (n+1)^n + (n+1)^{n-2}[n(n+1) + n^2(n-1)] \\
& = (n+1)^n + (n^3+n)(n+1)^{n-2}.
\end{align*}
On the other hand:
\begin{align*}
(n+1)^{n+1} - (n+1)^n & = n(n+1)^n \\
& = (n+1)^n + (n-1)(n+1)^n \\
& = (n+1)^n + (n-1)(n+1)^2(n+1)^{n-2} \\
& = (n+1)^n + (n^2-1)(n+1)(n+1)^{n-2} \\
& = (n+1)^n + (n^3 + n^2 - n - 1)(n+1)^{n-2}.
\end{align*}
But:
$$
(n^3 + n^2 - n - 1) - (n^3 + n) = n^2 - 2n - 1 = (n - 1)^2 - 2 > 0,
$$
therefore:
$$
(n+1)^{n+1} - n^{n+1} < (n+1)^{n+1} - (n+1)^n,
$$
therefore $n^{n+1} > (n+1)^n$. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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Find the number of ways to express 1050 as sum of consecutive integers I have to solve this task:
Find the number of ways to present $1050$ as sum of consecutive
positive integers.
I was thinking if factorization can help there:
$$1050 = 2 \cdot 3 \cdot 5^2 \cdot 7 $$
but I am not sure how to use that information (if there is a sense)
example
I can solve something similar but on smaller scale:
\begin{align} 15 &= 15 \\ &= 7+8 \\ &=4+5+6 \\ &= 1+2+3+4+5 \end{align}
($4$ ways)
| We want to find the number of solutions of
$$n+(n+1)+\ldots + (n+k) = 1050,\ n\in\mathbb Z_{>0},\ k\in\mathbb Z_{\geq 0}.\tag{1}$$
Rewrite the sum as $$n(k+1) + 0 + 1 +\ldots + k = n(k+1) + \frac{k(k+1)}{2}= \frac 12(2n+k)(k+1).$$
Thus, the number of solutions to $(1)$ is the same as the number of solutions of
$$(2n+k)(k+1) = 2100,\ n\in\mathbb Z_{>0},\ k\in\mathbb Z_{\geq 0}.\tag{2}$$
Let $a$ and $b$ be divisors of $2100$ such that
\begin{align}
2n+k &= a,\\
k+1 &= b.\tag{3}
\end{align}
Solving it we get
\begin{align}
n &= \frac{a-b+1}2,\\
k &= b -1.\tag{4}
\end{align}
From here we see that not every choice of integers $a$ and $b$ such that $ab = 2100$ will give us a solution to $(2)$. Since $a-b+1$ must be even, $a$ and $b$ are of opposite parities. Also, $a\geq b > 0$ since $n> 0$ and $k \geq 0$.
First determine the number of ways to factor $2100 = 2^2\cdot 3\cdot 5^2 \cdot 7$ such that one of the factors is odd. For this to be fulfilled, we shouldn't allow $4 = 2^2$ to be factored, so consider $2100 = 4\cdot 3 \cdot 5^2 \cdot 7$ instead. Thus, there are $2\cdot 2\cdot 3\cdot 2 = 24$ positive integral solutions to $2100 = a'b'$ such that one factor is odd. Because of commutativity, it means there are $12$ distinct ways to factor $2100$ into product of two factors, one of which is odd, and for every such factorization there is a unique choice for $a$ and $b$ such that $a\geq b$.
Thus, there are $12$ positive integral solutions to $(2)$.
| {
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"url": "https://math.stackexchange.com/questions/3133472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 2
} |
Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$ where a, b, c and d are positive real numbers
I have to prove the following inequality using the Cauchy-Schwarz inequality:
$$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$$
where a, b, c and d are positive real numbers.
But I am not able to do it, I am hitting dead-ends with every method I try. Please help!
| By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(ab+ac)}=2+\frac{(a+b+c+d)^2-2\sum\limits_{cyc}(ab+ac)}{\sum\limits_{cyc}(ab+ac)}=$$
$$=2+\frac{a^2+c^2+b^2+d^2-2ac-2bd}{\sum\limits_{cyc}(ab+ac)}\geq2+\frac{2\sqrt{a^2c^2}+2\sqrt{b^2d^2}-2ac-2bd}{\sum\limits_{cyc}(ab+ac)}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve these trigonometry equations? I have to work with the following 5 equations:
*
*$(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)=8$
*$(2\cos 2x+1)(2\cos 6x+1)(2\cos 18x+1)=1$
*$\dfrac{\cos 2x}{\sin 3x}+\dfrac{\cos 6x}{\sin 9x}+\dfrac{\cos 18x}{\sin 27x}=0$
*$\dfrac{\cos x}{\sin 3x}+\dfrac{\cos 3x}{\sin 9x}+\dfrac{\cos 9x}{\sin 27x}=0$
*$\dfrac{1}{\cos x\cos 2x}+\dfrac{1}{\cos 2x\cos 3x}+\dfrac{1}{\cos 3x\cos 4x}=0$
These equations have patterns, and I know if we can use the pattern we will solve the equations very easily. I managed to use the pattern on the first equation to find a telescoping series and get this (it is not a full solution but it is the way to solve the first equation):
We have $\frac{2\tan x}{1-\tan^2x}=\tan 2x$ therefore
\begin{align*}
(1)&\Leftrightarrow\dfrac{1}{(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)}=\dfrac18\\
&\Leftrightarrow\left(2\tan x\cdot\dfrac{1}{1-\tan^2x}\right)\cdot\dfrac{1}{1-\tan^22x}\cdot\dfrac{1}{1-\tan^24x}=\dfrac14\tan x\\
&\Leftrightarrow\left(2\tan 2x\cdot\dfrac{1}{1-\tan^22x}\right)\cdot\dfrac{1}{1-\tan^24x}=\dfrac12\tan x\\
\end{align*}
and so on.
However, I can't manage to solve the last four. I can't find the key equalities like $\frac{2\tan x}{1-\tan^2x}=\tan 2x$ in the first equation. So here are my questions:
*
*How to solve equations 2, 3, 4, and 5?
*What is the strategy to find the key equalities like $\frac{2\tan x}{1-\tan^2x}=\tan 2x$ to get a telescoping series for each equation?
Thank you in advance.
| Using that
$$\tan(2x)=2\,{\frac {\tan \left( x \right) }{1- \left( \tan \left( x \right)
\right) ^{2}}}
$$
$$\tan(4x)={\frac {4\,\tan \left( x \right) -4\, \left( \tan \left( x \right)
\right) ^{3}}{1-6\, \left( \tan \left( x \right) \right) ^{2}+
\left( \tan \left( x \right) \right) ^{4}}}
$$
we get for your first equation
$$ \left( 8\, \left( \cos \left( x \right) \right) ^{3}+4\, \left(
\cos \left( x \right) \right) ^{2}-4\,\cos \left( x \right) -1
\right) \left( 8\, \left( \cos \left( x \right) \right) ^{3}-4\,
\left( \cos \left( x \right) \right) ^{2}-4\,\cos \left( x \right) +
1 \right)
=0$$
for #5:
Your equation is equivalent to
$$\cos \left( 3\,x \right) \cos \left( 4\,x \right) +\cos \left( 2\,x
\right) \cos \left( x \right) +\cos \left( 4\,x \right) \cos \left( x
\right)
=0$$
and this can be written in the form
$$\cos \left( x \right) \left( 32\, \left( \cos \left( x \right)
\right) ^{6}-48\, \left( \cos \left( x \right) \right) ^{4}+22\,
\left( \cos \left( x \right) \right) ^{2}-3 \right)
=0$$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Minimum value of $(x^2+y^2)^2$ if $x,y$ are real number such that $x^2+2xy-y^2=6$ Then find minimum value of $(x^2+y^2)^2$
what i try : $x^2+2xy+y^2-2y^2=6$ or $(x+y)^2-\bigg(\sqrt{2} y\bigg)^2=6$
put $\displaystyle (x+y)=\sqrt{6}\cos \alpha$ and $\displaystyle \sqrt{2}y=\sqrt{6}\sin \alpha$
$\displaystyle x=\sqrt{6}\cos \alpha-\sqrt{3}\sin \alpha$ and $\displaystyle y =\sqrt{3}\sin \alpha$
$\displaystyle x^2+y^2=3\bigg[\bigg(\sqrt{2}\cos \alpha-\sin \alpha\bigg)^2+\sin^2\alpha\bigg)\bigg]$
$\displaystyle x^2+y^2=3\bigg[2\cos^2\alpha+\sin^2\alpha-2\sqrt{2}\cos \alpha\sin \alpha+\sin^2\alpha\bigg]$
$\displaystyle x^2+y^2=3\bigg(2-\sqrt{2}\sin 2\alpha\bigg)\geq 3(2-\sqrt{2})$
but answer is $\sqrt{18}$ How do i solve it Help me please
| $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=(6-2xy)^2+4x^2y^2=2(2xy-3)^2+18\ge18$$
The equality occurs if $2xy=3$
| {
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"source": "stackexchange",
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$I_1=\int_0^1 \frac 1 {1+\frac 1 {\sqrt x}} dx$ If we let $I_2= \int_0^1 \frac 1 {1+\frac 1 {1+\sqrt x}} dx$ and so on, we’re tasked with solving $I_n$. We can show that $I_2 = \int \frac {1+\sqrt x} {1-\sqrt x + 1} = \int \frac {1+\sqrt x} {2+\sqrt x }$
And that $I_3=\int \frac {2+\sqrt x}{3+2\sqrt x}$, and that in general, using $F_n$ to denote the nth Fibonacci number,
$$I_n=\int^1_0 \frac{F_n+F_{n-1}\sqrt x}{F_{n+1}+F_n \sqrt x}dx$$
However, at this point I’m uncertain how to proceed; we obviously have some ratio of Fibonacci’s which looks like the golden ratio, but I’m unsure what substitutions to make to have the golden ratio more clearly emerge as a useful quantity.
| We define $$j=j(a,b)=\int_0^1\frac{a+\sqrt x}{b+\sqrt x}dx$$
So we see that
$$j^*(a_1,a_2,a_3,a_4)=\int_0^1\frac{a_1+a_2\sqrt x}{a_3+a_4\sqrt x}dx=\frac{a_2}{a_4}j\left(\frac{a_1}{a_2},\frac{a_3}{a_4}\right)$$
So then we see that
$$\begin{align}
j&=\int_0^1\frac{a-b+b+\sqrt x}{b+\sqrt x}dx\\&=(a-b)\int_0^1\frac{dx}{b+\sqrt x}+\int_0^1\frac{b+\sqrt x}{b+\sqrt x}dx\\&=(a-b)\int_0^1\frac{dx}{b+\sqrt x}+1
\end{align}$$
Then we set $x=u^2\Rightarrow dx=2udu$:
$$\begin{align}
\int_0^1\frac{dx}{b+\sqrt x}&=2\int_0^1\frac{u}{b+u}du\\
&=2\int_0^1\frac{-b+b+u}{b+u}du\\
&=2-2b\int_0^1\frac{du}{b+u}du\\
&=2-2b\ln|u+b|]_0^1\\
&=2+2b\ln\left|\frac{b}{b+1}\right|
\end{align}$$
So $$j(a,b)=1+2(a-b)\left(1+b\ln\left|\frac{b}{b+1}\right|\right)$$
And your integral is given by $$I_n=j^*(F_n,F_{n-1},F_{n+1},F_n)=\frac{F_{n-1}}{F_n}j\left(\frac{F_n}{F_{n-1}},\frac{F_{n+1}}{F_n}\right)$$
| {
"language": "en",
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Find all positive integer $n$ such that there exist $m\in\mathbb{Z}$ with $2^n-1|m^2+9$ I have an problem with elementary number theory:
Find all positive integer $n$ such that there exist $m\in\mathbb{Z}$ with $2^n-1|m^2+9$
It's look like the problem in this link, but there some differences: Find all positive integer $n$ such that there exists $m$ with $2^n-1|m^2+17^2$.
| If $2^n-1$ has a prime divisor $p\equiv3\pmod4$, then $m^2+3^2\equiv0\pmod p$. If $p\neq 3$, let $a$ be an inverse of $3$ modulo $p$. Then $(am)^2\equiv-1\pmod p$, contradicting $p\equiv3\pmod4$.
Thus $3$ must be the only prime divisor of $2^n-1$ that is $3$ mod $4$. First, $n=1$ clearly works. Suppose $n \geq 2$. Then $2^n-1$ has at least one prime divisor congruent to $3$ mod $4$, which must be $3$. Thus $n$ is even. Write $n = 2k$. Then $2^k-1 \mid m^2+9$, so again, $k=1$ or $k$ is even... By induction, we see that $n$ must be a power of $2$. Write $n=2^m$, then
$$2^{2^m}-1 = 3 \cdot \prod_{k=1}^{m-1}\left( 2^{2^k}+1 \right)$$
Because every prime divisor of the second factor is $1$ mod $4$, each $n=2^m$ is a solution.
| {
"language": "en",
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Value of $(p,q)$ in indefinite integration Finding value of $(p,q)$ in $\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx=\frac{px^3+qx^8}{x^{10}-2x^5+1}+c$
what i try
$\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx$
put $x=1/t$ and $dx=-1/t^2dt$
$\displaystyle -\frac{2t^5+3x^{10}}{t^{10}-2t^5+1}dt$
How do i solve it Help me please
| \begin{equation}
\left(\frac{px^3+qx^8}{x^{10}-2x^5+1}\right)^\prime=\frac{(8qx^7+3px^2)(x^5-1)^2-10x^7(qx^5+p)(x^5-1)}{(x^5-1)^4}
\end{equation}
Let $p=1,\quad q=-1$ and we get
\begin{eqnarray}
&&\frac{(-8x^7+3x^2)(x^5-1)^2-10x^7(-x^5+1)(x^5-1)}{(x^5-1)^4}\\
&=&\frac{(-8x^7+3x^2)(x^5-1)^2+10x^7(x^5-1)^2}{(x^5-1)^4}\\
&=&\frac{2x^7+3x^2}{(x^5-1)^2}
\end{eqnarray}
| {
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"answer_id": 1
} |
How to simplify $\sqrt{2+\sqrt{3}}$ $?$
Simplify $\dfrac{2\left(\sqrt2 + \sqrt6\right)}{3\sqrt{2+\sqrt3}}$
The answer to this question is $\frac{4}{3}$ in a workbook.
How would I simplify $\sqrt{2+\sqrt3}$ $?$ If it was something like $\sqrt{3 + 2\sqrt2}$ , I would have simplified it as follows:
$\sqrt{3 + 2\sqrt2}$ $=$ $\sqrt{(\sqrt2)^2 + 2(\sqrt2)(1) + (1)^2}$ $=$
$\sqrt{(\sqrt2 + 1)^2}$ $=$
$\sqrt2 + 1$
But I can't simplify $\sqrt{2+\sqrt3}$ like that as $2+\sqrt3$ is can't be written as squares of two numbers. Is there any other method?
| $$\left(\frac{2\left(\sqrt2+\sqrt6\right)}{3\sqrt{2+\sqrt3}}\right)= \frac{(2√2+2√6)\sqrt{2+\sqrt{3}}}{3(2+√3)}=\frac{8+4√3}{2+√3}=\frac{4}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$ Qestion: $\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$
Answer: $K=(ab+bc+ca)$
My attempt: $$\begin{align}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3-a^3\end{vmatrix}\\&=\begin{vmatrix}b^2-a^2&c^2-a^2\\b^3-a^3&c^3-a^3\end{vmatrix}\\&=(b-a)(c-a)\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix}\\&=(c-b)(a-b)(b+a)(c+a)\begin{vmatrix}1&1\\(b+a)-ba&(c+a)-ca\end{vmatrix}\\\end{align}$$
I don't know what should I do next, maybe I've made a mistake but I didn't notice it
| The mistake is here
$$\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix}=(c-b)(a-b)(b+a)(c+a)\begin{vmatrix}1&1\\(b+a)-ba&(c+a)-ca\end{vmatrix}$$
These aren't equal to each other. If you want to continue row reduction methods, you should continue with
$$\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix} = \begin{vmatrix}b-c& c+a\\(b-c) a + b^2 - c^2 &c^2+ca+a^2\end{vmatrix} = (b-c)\begin{vmatrix}1& c+a\\a+b+c &c^2+ca+a^2\end{vmatrix}$$
And you should be able to take it from there.
| {
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"source": "stackexchange",
"question_score": "1",
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Prove that $(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\dfrac{3}{4}$ where $a+b+c=3$
If $a$, $b$, $c$ are non-negative numbers such that $a + b + c = 3$ then prove
$$(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\frac{3}{4}$$
Here's what I did.
Let $c \ge a \ge b$.
We have that
\begin{align*}
(c - 1)^3 + (a - 1)^3 &= (c + a - 2)(c^2 + a^2 - ca - c - a + 1)\\
&\ge (1 - b)\left[\dfrac{3}{4}(c + a)^2 - b + 4\right]\\
&=(1 - b)\left[\dfrac{3}{4}(3 - b)^2 - b + 4\right]\\
&= \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43)
\end{align*}
That means
\begin{align*}
(a - 1)^3 + (b - 1)^3 + (c - 1)^3 &\ge \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43) + (b - 1)^3\\
&= \dfrac{1}{4}(b - 1)(b^2 + 14b - 39)
\end{align*}
Since $c \ge a \ge b \implies b \le \dfrac{a + b + c}{3} = 1$.
And this is where I am stuck right now.
| Perhaps a different approach than you are using. Using the Lagrange multipliers, we define the auxiliary problem
$$ F = (x-1)^3 + (y-1)^3 + (z-1)^3 - \lambda(x + y + z - 3). $$
Differentiating and setting to $0$ we obtain the conditions
\begin{align}
3(x-1)^2 = \lambda \\
3(y-1)^2 = \lambda \\
3(z-1)^2 = \lambda
\end{align}
It is easy to see that if we set $x,y,z$ to the critical point obtained by these conditions we will reach a contradiction with the constraint $x + y + z = 3$ (check this yourself). Then, it must be that the extremum if it exists must have at least one variable (say $z$) which is $0$. This leads to a value of $\lambda = 3/4$ and finally, one can show that any permutation of the triplet $(3/2,3/2,0)$ leads to the minimum value. You should also check that this is indeed the minimum.
| {
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Prove an inequality with positives $a$, $b$ and $c$.
If $a$, $b$ and $c$ are positives such that $(a + b + c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) = x$ ($x \ge 9$) then prove that $$\large(a^2 + b^2 + c^2)\left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}\right) \le x(\sqrt x - 2)^2$$
The equality sign occurs when $a : b : c = m^2 : m : 1$ with
$$\large m = \dfrac{\pm \sqrt{x - 2\sqrt x - 3} + \sqrt x - 1}{2}$$, which is very irregular.
Here's what I did.
We have that $$(a + b + c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) = (ab + bc + ca)\left(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}\right) = x$$
and
$$(a^2 + b^2 + c^2)\left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} \right) = x^2 - 2\left[(ab + bc + ca)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \right)^2 + (a + b + c)^2\left(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca} \right)\right] + 4x$$
$$\le x^2 - 4x\sqrt x + 4x = x(\sqrt x- 2)^2$$
Wait, did I just solve this problem by accident? I am saying the truth, I never write down anything at all, I usually type as I think.
Well... I can't calculate when does the equality sign occur. Someone help me.
| Let $a+b+c=3u$, $ab+ac+bc=v^2$ and $abc=w^3$.
Thus, $a$, $b$ and $c$ are roots of the equation:
$$t^3-3ut^2+3v^2t-w^3=0.$$
Now, by your work the equality in your inequality occurs for
$$(ab+ac+bc)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=(a+b+c)^2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)$$ or
$$\frac{(ab+ac+bc)^3}{a^2b^2c^2}=\frac{(a+b+c)^3}{abc}$$ or
$$ab+ac+bc=(a+b+c)\sqrt[3]{abc}$$ or $$v^2=uw.$$
Id est, $x=\frac{9u^2}{w^2}$ and $a$, $b$ and $c$ are roots of the following equation.
$$t^3-3ut^2+3uwt-w^3=0$$ or
$$(t-w)(t^2+tw+w^2)-3ut(t-w)=0$$ or
$$(t-w)(t^2+(w-3u)t+w^2)=0.$$
Can you end it now?
| {
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Prove without induction that $2×7^n+3×5^n-5$ is divisible by $24$. I proved this by induction. But I want to show it using modular arithmetic. I tried for sometime as follows
$$2×7^n-2+3×5^n-3\\
2(7^n-1)+3(5^n-1)\\
2×6a+3×4b\\
12(a+b)$$
In this way I just proved that it is divisible by 12 but it is not enough. Am I missing something or it will solved by some other method.
| You may split it up by calculating $\mod 8$ and $\mod 3$:
*
*$\mod 8$:
\begin{eqnarray*} 2×7^n+3×5^n-5
& \equiv_8 & 2\times (-1)^n + 3\times (-3)^n +3 \\
& \equiv_8 & 2\times (-1)^n + 3((-3)^n + 1)\\
& \stackrel{3^2 \equiv_8 1}{\equiv_8}& \begin{cases} 2+3\times (1+1) & n = 2k \\ -2 +3 (-3 + 1) & n= 2k+1\end{cases}\\
& \equiv_8 & 0
\end{eqnarray*}
*$\mod 3$:
\begin{eqnarray*} 2×7^n+3×5^n-5
& \equiv_3 & 2\times 1^n + 3\times (-1)^n +1 \\
& \equiv_3 & 3\times (1 + (-1)^n)\\
& \equiv_3 & 0
\end{eqnarray*}
| {
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Three dice: conditional probability After throwing $3$ dice, we know that on every die there is a different number. What is the probability that there is a $6$ on exactly one dice?
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then
$$P(B)=\dfrac{6∗5∗4}{6^3}\text{ and }
P(A\cap B)=\dfrac{1*5*4}{6^3}$$
| If I understood right from question and also from the discussion before. That every throw it is ensured that you get a different number (so all three different numbers). So you can get 6 only in one throw.
Probability of getting 6 in first throw is $\dfrac{1}{6}$, so it is guaranteed that next two throws there will not be any 6. So the total probability in this case is $\dfrac{1}{6}.1.1 = \dfrac{1}{6}$.
Now for getting 6 in second throw; in the first throw the required probability is $\dfrac{5}{6}$, in the second throw the probability is $\dfrac{1}{5}$ and we do not bother about the third throw (as it is guaranteed that 6 will not be there). So the total probability in this case is $\dfrac{5}{6}.\dfrac{1}{5}.1 = \dfrac{1}{6}$.
For getting 6 in third throw we have, not getting a 6 in first throw is $\dfrac{5}{6}$, not getting a 6 in second throw is $\dfrac{4}{5}$ , getting a 6 in third throw is $\dfrac{1}{4}$, so total probability is $\dfrac{5}{6}.\dfrac{4}{5}.\dfrac{1}{4} = \dfrac{1}{6}$.
So finally total probability is $\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}= \dfrac{1}{2}$
| {
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Create generating function: $(1, 1, -2, -2, 10, 3, -4, -4....)$
Find generating function (without using infinite series):
a) (0, 1, 4, 9, 16, 25, 36...)
b) (1, 1, -2, -2, 10, 3, -4, -4, 5, 5, -6, -6, 7...) (Only irregularity is the 10)
Here's what I got:
a) $(0, 1, 4, 9, 16...) = \frac{x(1+x)}{(1-x)^3}$
(Begin with (1, 1, 1, 1, ...) differentiate and shift by 1, then differentiate and shift by 1 again.)
Any help with b)?
| The function is
$$
7x^4+(1+x)(1-2x^2+3x^4-4x^6+\dots)
$$
Can you find a closed form for $1-2x^2+3x^4-4x^6+\dots$? Perhaps you can find a closed form for $1+2y+3y^2+4y^3+\dots$, then substitute $y\gets -x^2$?
| {
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can not find root of the equation $(0,0)$ with semi-major axis a and semi-minor axis $b$">
I want find out for each point $(x,y)$ in the white region, the nearest point $(a \cos \theta, b \sin \theta )$ on the ellipse curve.I tried with the following approach.
$$
\text {distance} = \sqrt{(x-a \cos \theta)^2+(y-b \sin \theta)^2}
$$
I find out the first derivative of the $$F(\theta)=(x-a \cos \theta)^2+(y-b \sin \theta)^2.$$ The equation after first derivative is:
$$
ax \sec \theta- by \csc \theta=a^2-b^2.
$$
How to calculate theta for this equation?
| You can get it by solving a constrained optimization problem. You want the point $(u,v)$ that minimizes $f(u,v)=(u-x)^2+(v-y)^2$ subject to the constraint $\frac{u^2}{a^2} + \frac{v^2}{b^2} = 1$. The point $(x,y)$ doesn't need to be inside the ellipse.
The Lagrangian is given by
$$
L(u,v,\lambda) = (u-x)^2+(v-y)^2-\lambda\left(\frac{u^2}{a^2} + \frac{v^2}{b^2}-1\right)
$$
Now you just need to compute the critical points of $L$ and choose the ones that yield the smallest distance to $(x,y)$.
You can solve this by hand but, out of laziness, I give you Wolfram's solution... Two critical points
WARNING: Wolfram is not giving the full set of solutions. After computing $u,v$ in terms of $\lambda$ from the first two equations of the system $\nabla L = 0$ and substituting them in the last equation we get a fourth degree polynomial equation in $\lambda$. The solution must be conmputed in an alternative way if this polynomial has four real roots.
$$
\left( -\frac{a \sqrt{b^2-y^2}}{b}, \frac{-a^2 b^4 y+a^2 b^2 y^3+a b^3 x y \sqrt{b^2-y^2}+b^6 y-b^4
y^3}{a^4 b^2-a^4 y^2-2 a^2 b^4-a^2 b^2 x^2+2 a^2 b^2 y^2+b^6-b^4
y^2}\right)
$$
and
$$
\left( \frac{a \sqrt{b^2-y^2}}{b}, \frac{-a^2 b^4 y+a^2 b^2 y^3-a b^3 x y \sqrt{b^2-y^2}+b^6 y-b^4
y^3}{a^4 b^2-a^4 y^2-2 a^2 b^4-a^2 b^2 x^2+2 a^2 b^2 y^2+b^6-b^4
y^2}\right)
$$
One will correspond to the minimum distance and the other will correspond to the maximum distance.
This is not relevant to your question, but the Lagrange multipliers are
$$
\frac{a^2 b^2-a^2 y^2+a b x \sqrt{b^2-y^2}}{b^2-y^2}
$$
and
$$
\frac{a^2 b^2-a^2 y^2-a b x \sqrt{b^2-y^2}}{b^2-y^2},
$$
respectively.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of Infinite series $\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+......$ Prove that the sum of the infinite series $\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+......$ is 23.
My approach
I got the following term
$S_n=\sum_1^\infty\frac{4n^2}{2^n}-\sum_1^\infty\frac{1}{2^n}$.
For $\sum_1^\infty\frac{1}{2^n}$ the answer is 1 as it forms a geometric series but I am bot able to find the solution to $\sum_1^\infty\frac{4n^2}{2^n}$.
| $$\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+…=\sum_{n=1}^{\infty }\frac{(2n-1)(2n+1)}{2^n}=\sum_{n=1}^{\infty }\frac{(4n^2-1)}{2^n}$$
depending on the geometric series
$$\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$
$$(\frac{1}{1-x})'=\sum_{n=1}^{\infty }nx^{n-1}$$
$$x(\frac{1}{1-x})'=\sum_{n=1}^{\infty }nx^{n}$$
$$(x(\frac{1}{1-x})')'=\sum_{n=1}^{\infty }n^2x^{n-1}$$
$$x(x(\frac{1}{1-x})')'=\sum_{n=1}^{\infty }n^2x^{n}$$
$$4x(x(\frac{1}{1-x})')'=\sum_{n=1}^{\infty }4n^2x^{n}$$
$$4x(x(\frac{1}{1-x})')'-\frac{1}{1-x}=\sum_{n=1}^{\infty }4n^2x^{n}-\sum_{n=0}^{\infty }x^n$$
$$4x(x(\frac{1}{1-x})')'-\frac{1}{1-x}+1=\sum_{n=1}^{\infty }4n^2x^{n}-\sum_{n=1}^{\infty }x^n$$
so
$$\sum_{n=1}^{\infty }x^n(4n^2-1)=\frac{4x^2+4x}{(1-x)^3}-\frac{x}{1-x}$$
now let $x=1/2$ to get $23$
| {
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"timestamp": "2023-03-29T00:00:00",
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exist infinitely many positive integers $n$ such $\omega (n)+\omega (n+1)\equiv 0\pmod 3$ For integer $n>1$, $\omega (n)$ denotes number of distinct prime factors of $n$, and $\omega (1)=0$. Prove that:there exist infinitely many positive integers $n$ satisfying
$$\omega (n)+\omega (n+1)\equiv 0\pmod 3$$
Positive integers $n$ satisfying $\omega(n)<\omega(n+1)<\omega(n+2)$
| With $\Omega$ instead of $\omega$ I have a solution.
For any $n$ then $a = \Omega(2n) \bmod 3,b=\Omega(2n+1)\bmod 3,c=\Omega(2n+2)\bmod 3$, $\Omega((2n+1)^2) = 2b\bmod 3, \Omega((2n+1)^2-1) = a+c\bmod 3,\Omega(n)+\Omega(n+1) = a+c-2 \bmod 3$
Assume we chose $n$ such that $a=2\bmod 3$
The claim $\Omega(l)+\Omega(l+1) = 0 \bmod 3$ for some $l \in \{ 2n,2n+1, (2n+1)^2-1,n\}$ would follow from
$2+b=0\bmod 3$ or $b+c=0\bmod 3$ or $2+c+2b=0\bmod 3$ or $2+c-2=0\bmod 3$.
The only fail is when $c=2\bmod 3$.
But if $c=2\bmod 3$ we can redo it replacing $n$ by $n+1$, finding that for the claim to fail we need $\Omega(2(n+m)) = 2\bmod 3$ for every $m\ge 0$, a contradiction.
| {
"language": "en",
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"source": "stackexchange",
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Why do these laurent series approaches conflict? I was working on a problem of finding the Laurent series of $\frac{1}{z-3}$ that converges where $|z-4| > 1$
So I had one approach, let $u=z-4$ then:
$$\frac{1}{z-3} = \frac{1}{1+u} $$
$$ = \frac{1}{u} - \frac{1}{u^2} + \frac{1}{u^3}...$$
$$ = \frac{1}{z-4} - \frac{1}{(z-4)^2} + ...$$
But this apparently incorrect.
The correct answer is found by noting:
$$ \frac{1}{z-3} = \frac{1}{z-4 + 1} = \frac{1}{z-4} \frac{1}{1 - \frac{-1}{z-4}} = -\frac{1}{(z-4)^2} + ... $$
Where did I go wrong?
| ${1\over 1+x} = 1 - x + x^2 - x^3 + \cdots$ when $|x| < 1$.
This is derived from ${1 \over 1-x} = 1 + x + x^2 + \cdots$ with $|x| < 1$.
| {
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Expand $\frac{1}{z^2}$ as a series on $|z - 2| < 2$ I am asked to show that
$$\frac{1}{z^2} = \frac{1}{4} + \frac{1}{4}\sum_{n = 1}^{\infty}(-1)^{n+1}(n+1)\left(\frac{z-2}{2}\right)^n$$
on $|z - 2| < 2$.
My plan is to consider a series expansion of $\frac{-1}{z}$ in $|z - 2| < 2$ and differentiate this series term by term to yield the desired series (This is valid from theorems we have proved in class).
Indeed,
$$\frac{-1}{z} = \frac{-1}{2-2+z} = \frac{-1}{2}\cdot\frac{1}{1+\left(\frac{z-2}{2}\right)}$$
$$ = \frac{-1}{2}\sum_{n = 0}^{\infty}(-1)^n\left(\frac{z-2}{2}\right)^n $$
$$ = \frac{1}{2}\sum_{n = 0}^{\infty}(-1)^{n+1}\left(\frac{z-2}{2}\right)^n $$
Differentiating term by term gives:
$$ \frac{1}{z^2} = \frac{1}{2}\sum_{n = 1}^{\infty}(-1)^{n+1}n\left(\frac{z-2}{2}\right)^{n-1}\cdot\frac{1}{2} $$
$$ = \frac{1}{4}\sum_{n = 1}^{\infty}(-1)^{n+1}n\left(\frac{z-2}{2}\right)^{n-1}$$
Now, if we let $k = n - 1$, we can re-index the series as:
$$ = \frac{1}{4}\sum_{k = 0}^{\infty}(-1)^{k+2}(k+1)\left(\frac{z-2}{2}\right)^{k}$$
$$ = \frac{1}{4} + \frac{1}{4}\sum_{k = 1}^{\infty}(-1)^{k}(k+1)\left(\frac{z-2}{2}\right)^{k}$$
Unfortunately, the power of $(-1)$ present in my series does not match what the prompt reads. I am off by one. Did I make a mistake or is the prompt incorrect?
| Perhaps a simpler approach would be to write
$$
{1 \over z^2} = {1 \over 1 - \left(1 - z^2\right)}.
$$
Letting $a = 1 - z^2$, we see that the above expression has (at least formally) the geometric series expansion:
$$
{1 \over 1 - \left(1 - z^2\right)} = {1 \over 1 - a} = \sum_{n \geq 0} a^{n}.
$$
Now expand each power
$$
a^{n} = \left(1 - z^2\right)^n
$$
using the Binomial Theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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Alternate complex binomial series sum
Calculation of $\displaystyle \sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot \frac{1}{\binom{2n}{r}}$ is
My Try: Using $$\int^{1}_{0}x^m(1-x)^ndx = \frac{1}{(m+n+1)}\cdot \frac{1}{\binom{m+n}{n}}$$
So $\displaystyle \int^{1}x^{2n-r}(1-x)^r=\frac{1}{2n}\cdot \frac{1}{\binom{2n}{r}}$
Sum convert into $\displaystyle 2n\sum^{2n-1}_{r=1}(-1)^{r-1}r\int^{1}_{0}x^{2n-r}(1-x)^rdx$
$\displaystyle \Longrightarrow 2n \int^{1}_{0}x^{2n}\sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r \cdot \bigg(1-\frac{1}{x}\bigg)^rdx$
Could some help me to solve it , Thanks
| Here is a more elementary method to find the sum which gives a more general result. Let $\displaystyle f(n)=\sum^{n-1}_{r=1}(-1)^{r-1}\frac{r}{\binom{n}{r}}$. We have been asked to find $\displaystyle f(2n)$. Note that $\displaystyle \frac{1}{\binom{n+1}{k+1}}+\frac{1}{\binom{n+1}{k}}=\frac{n+2}{n+1}\frac{1}{\binom{n}{k}}$. We will use this identity twice to telescope the sum.
$$f(n)=\sum^{n-1}_{r=0}(-1)^{r-1}\frac{r}{\binom{n}{r}}=\frac{n+1}{n+2}\sum^{n-1}_{r=0}\bigg((-1)^{r-1}\frac{r}{\binom{n+1}{r}}-(-1)^{r}\frac{r}{\binom{n+1}{r+1}}\bigg)$$$$=\frac{n+1}{n+2}\sum^{n-1}_{r=0}\bigg((-1)^{r-1}\frac{r}{\binom{n+1}{r}}-(-1)^{r}\frac{r+1}{\binom{n+1}{r+1}}+\frac{(-1)^r}{\binom{n+1}{r+1}}\bigg)=\frac{n+1}{n+2}\bigg((-1)^n\frac{n}{n+1}+\sum^{n}_{r=1}\frac{(-1)^{r-1}}{\binom{n+1}{r}}\bigg)$$$$=\frac{n+1}{n+2}\Bigg((-1)^n\frac{n}{n+1}+\frac{n+2}{n+3}\sum^{n}_{r=1}\bigg(\frac{(-1)^{r-1}}{\binom{n+2}{r}}-\frac{(-1)^r}{\binom{n+2}{r+1}}\bigg)\Bigg)=\frac{n+1}{n+2}\Bigg((-1)^n\frac{n}{n+1}+\frac{1-(-1)^n}{n+3}\Bigg)$$
$$\implies f(n)=\frac{n}{n+2}(-1)^n+\frac{n+1}{(n+2)(n+3)}\big(1-(-1)^n\big)=\left \{
\begin{aligned}
&\ \ \ \ \ \frac{n}{n+2}, && \text{if}\ n \text{ is even} \\
&-\frac{n-1}{n+3}, && \text{if } n \text{ is odd}
\end{aligned} \right.$$
$$\therefore f(2k+1)=-\frac{k}{k+2}\text{ and }f(2k)=\frac{k}{k+1}\text{, as desired.}$$
$\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Proving $\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c \geq \frac{6abc}{3ab+bc+2ca}$ for positive $a$, $b$, $c$ I'm at the end of an inequality proof that started out complex and I was able to simplify it to:
$$\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c \geq \frac{6abc}{3ab+bc+2ca} \quad\text{where}\quad a, b, c > 0$$
I'm able to plug in very small values close to 0 and the inequality holds, but I'm having trouble finding a way to prove it and how to start off this problem.
| Since $\frac{1}{6}+\frac{1}{3}+\frac{1}{2}=1,$ by AM-GM we obtain:
$$\left(\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c\right)\left(\frac{1}{2}ab+\frac{1}{6}bc+\frac{1}{3}ca\right)\geq$$
$$\geq a^{\frac{1}{6}}b^{\frac{1}{3}}c^{\frac{1}{2}}(ab)^{\frac{1}{2}}(bc)^{\frac{1}{6}}(ca)^{\frac{1}{3}}=abc,$$ which gives your inequality.
Also, by C-S
$$\left(\frac{a}{6}+\frac{b}{3}+\frac{c}{2}\right)\left(\frac{1}{6a}+\frac{1}{3b}+\frac{1}{2c}\right)\geq\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)^2=1,$$ which gives your inequality again.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum: $\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)$ Let $0<a<b$, I would like to compute the sum
$$\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right).$$
But first I am worrying that a test convergence might lead to the divergence of this series
What do I miss here?
$$\begin{split}\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)&= \sum_{n=1}^{\infty} \int^{\frac1{a+n}}_{\frac1{b+n}}\frac{dx}{x+1}\\
&= \sum_{n=1}^{\infty} \int_{a+n}^{b+n}\frac{dt}{t(t+1)}~~~~(t= 1/x)\\
&= \sum_{n=1}^{\infty} \int_{a}^{b}\frac{dt}{(t+n)(t+n+1)}\\
&=\int_{a}^{b}dt \sum_{n=1}^{\infty} \frac{1}{t+n}-\frac{1}{t+n+1}~~~(\text{Monotone convergence})\\
&= \int_{a}^{b}\frac{dt}{t+1} ~~~~(\text{by Telescoping sum})\\
&= \ln\left(\frac{b+1}{a+1}\right) \end{split}$$
However the series seems $\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)$ not to be convergent.
Have I missed something ?
| HINT:
Note that if $a\ne b$, then
$$\begin{align}
\int_{1/(b+n)}^{1/(a+n)}\frac1{x+1}\,dx&=\log\left(\frac{1+1/(a+n)}{1+1/(b+n)}\right)\\\\
&=\log\left(\frac{a+n+1}{a+n}\frac{b+n}{b+n+1}\right)\\\\
&\ne \log\left(\frac{a+n+1}{b+n+1}\right)
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the exact length of the curve $y = \ln(1-x^2)$ Edit: Update with the full question for context
Find the exact length of the curve $y = \ln(1-x^2), 0 \leq x \leq \frac{1}{2}$
The integral below is what I got after finding the derivative $\frac{-2}{1-x^2}$ via the chain rule.
Can someone give me a hint on how to evaluate this integral with a range of $0 \leq x \leq \frac{1}{2}$?
$$ I = \int_{0}^{\frac{1}{2}} \sqrt{1+\bigg(\frac{-2x}{1-x^2}\bigg)^2}\mathrm dx$$
More specifically, how to deal with this fraction? That is what I am struggling the most with.
I tried to simplify it into $\frac{1}{1-2x^2 + x^4} = 1(1-2x^2+x^4)^{-1} = 1 - 2x^{-2} + x^{-4}$ which gives me
$$</s>I = \int_{0}^{\frac{1}{2}} \sqrt{2 - 2x^{-2} + x^{-4}} \mathrm dx$$
I looked at $u$-substitution, where $u = 2-2x^{-2} + x^{-4}$, but the $\mathrm du$ value didn't work out for me.
I am not entirely sure how trig substitution would work for a fraction//polynomial either.
| Partial fraction seems to be a good tool for your quesiton.
\begin{align}I &= \int_{0}^{\frac{1}{2}} \sqrt{1+\bigg(\frac{-2x}{1-x^2}\bigg)^2}\,\mathrm dx\\
&=\int_{0}^{\frac{1}{2}} \frac{\sqrt{(1-x^2)^2+4x^2}}{1-x^2}\mathrm \,dx\\
&=\int_0^\frac12 \frac{\sqrt{1+2x^2+x^4}}{1-x^2}\mathrm\, dx \\
&=\int_0^\frac12 \frac{1+x^2}{1-x^2}\mathrm\, dx \\
&=\int_0^\frac12 \frac{x^2-1+2}{1-x^2}\, dx \\
&= \int_0^\frac12 -1+\frac{2}{1-x^2}\, dx\\
&= \int_0^\frac12 -1 + \frac{1}{1-x} + \frac{1}{1+x} \, dx \\
&= -x - \ln (1-x)+\ln (1+x) |_{0}^{\frac12}\\
&= - \frac12 - \ln \frac12 + \ln \frac32 \\
&= \ln 3 - \frac12
\end{align}
| {
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Describe all integral solutions of the equation $x^2 + y^2 = 2z^2$ such that $x,y,z > 0$, gcd$(x,y,z) = 1$, and $x > y$. As the title states, the question tasks me with finding all the integral solutions of the equation under the specified constraints. I have an idea of where to start due to a somewhat similar problem in my notes, but I'm having trouble adapting it to this new equation, and knowing where I need to change my process. My professor has provided the hint: "Use the circle $^2 + ^2 = 2$ and the lines
passing through the point $(1,1)$." Here's what I have so far:
$x^2 + y^2 = 2z^2$
$\frac{x^2}{z^2} + \frac{y^2}{z^2} = 2$
Let $X = \frac{x}{z}$ and $Y = \frac{y}{z}$
$X^2 + Y^2 = 2$
So we now have a circle with an origin at $(0,0)$ and radius of $\sqrt{2}$. I then drew up the circle and the line passing through $(1,1)$.
So the slope $\lambda$ of this line would be: $\lambda = \frac{Y-1}{X-1}$. This is the part where I get lost, in class we went off on a tangent related to these problems and I'm having trouble knowing exactly how to proceed. I have an idea of what the final answer will look like. The form we found for $x^2 + y^2 = z^2$ was: $(x,y,z) = (a^2 - b^2, 2ab, a^2 + b^2)$ Any help would be greatly appreciated!
| The trick to this one is to create three new variables
$$
a = x+y\\
b = x-y\\
c = 2z
$$
Then $x^2+y^2 = 2z^2 \implies a^2+b^2 = c^2$
Temporarily ignoring the condition that $\gcd(x,y,z) = 1$ we can now list all solutions of
$a^2+b^2 = c^2$ as
$$
a = k(n^2 - m^2)\\
b = 2kmn\\
c = k(n^2 + m^2)
$$
with $m \neq n \pmod 2$ and $\gcd(m,n) = 1$.
To get back to $x,y,z$ we have
$$ x = \frac{b+a}2 = k\frac{(n^2 + 2mn - m^2)}2
\\ y = \frac{b-a}2 = k\frac{(m^2 + 2mn - n^2)}2
\\ z = \frac c2 = k\frac{(n^2 + m^2)}2
$$
However, since $m$ and $n$ are of opposite parity, $(n^2 + m^2)$ is odd, so in order for $z$ to be an integer, $k$ must be even: $k = 2p$.
Then the generic triplet is
$$ x = p(n^2 + 2mn - m^2)
\\ y = p(m^2 + 2mn - n^2)
\\ z = p(n^2 + m^2)
$$
$x$ and $y$ are not coprime unless $p = 1$. So we have
$$ x = n^2 + 2mn - m^2
\\ y =| m^2 + 2mn - n^2|
\\ z = n^2 + m^2
$$
Finally, we can ask when $\gcd(x,y) = 1$ but it is easier to ask when $(x,z) = 1$ which is equivalent for our purposes.
The difference $z-x = m(m-2)$ is a multiple of $m$ and thus is coprime with $z$ because $\gcd(n^2,m) = 1)$ So all such triplets are coprime, and that is our answer:
$$
x = n^2 + 2mn - m^2
\\ y =| m^2 + 2mn - n^2|
\\ z = n^2 + m^2
$$ with $ n,m\in \Bbb N \wedge n > m \wedge m \neq n \pmod 2 \wedge \gcd(m,n) = 1$.
| {
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Show that a triangle is right angled In $\triangle ABC$ (not isosceles) $CH$, $CL$ and $CM$ are respectively height, bisector and median. Show that $\angle ACB = 90^\circ$ if and only if $\angle HCL = \angle MCL$.
I think that I have to show that $\triangle MCB$ (or $\triangle ACM$) is isosceles, but I can't figure it out. I will be very grateful if you give me a hint.
|
Let's assume $\angle C > 90 ^\circ$ as show in the picture. $CL1$ is the bisector of $\angle ACB$ , $CL2$ is the bisector of $\angle HCM$ , now we show $AL2>AL1$
let $l1=AL1,l2=AL2$
$l1=\frac{AC}{AC+BC}*(c1+c2)$,
$AC=\sqrt{h^2+c1^2},BC=\sqrt{h^2+c2^2}$,
$l1=\frac{\sqrt{h^2+c1^2}}{\sqrt{h^2+c1^2}+\sqrt{h^2+c2^2}}*(c1+c2)=\frac{\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}-(h^2+c1^2)}{c2^2-c1^2}*(c1+c2)=\frac{\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}-(h^2+c1^2)}{c2-c1}$
$HM=\frac{c2-c1}{2},m=\sqrt{h^2+(\frac{c2-c1}{2})^2}$
$AL2=c1+HL2=c1+\frac{h}{h+m}*(\frac{c2-c1}{2})=c1+\frac{h}{h+\sqrt{h^2+(\frac{c2-c1}{2})^2}}*(\frac{c2-c1}{2})=c1+\frac{h*(\sqrt{h^2+(\frac{c2-c1}{2})^2}-h)}{(\frac{c2-c1}{2})^2}*(\frac{c2-c1}{2})=c1+\frac{h*(\sqrt{4h^2+(c2-c1)^2}-2h)}{c2-c1}=\frac{c2c1-c1^2+h*(\sqrt{4h^2+(c2-c1)^2}-2h)}{c2-c1}$
$AL2-AL1=\frac{D}{c2-c1}$
$D=c2c1-c1^2+h*(\sqrt{4h^2+(c2-c1)^2}-2h)-(\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}-(h^2+c1^2))=c1c2-h^2+h*\sqrt{4h^2+(c2-c1)^2}-\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}=c1c2-h^2+\frac{3h^2(h^2-c1c2)}{h*\sqrt{4h^2+(c2-c1)^2}+\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}}=(c1c2-h^2)(1-\frac{3h^2}{h*\sqrt{4h^2+(c2-c1)^2}+\sqrt{h^2+c2^2}\sqrt{h^2+c1^2}})$
I left last step , note $h^2<c1c2$, when it is right angle $h^2=c1c2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
How to find the power series of $\sqrt{1+x^4}$? The complete question is to find the integral from $0$ to $1$ of $$\sqrt{1+x^4}$$
I am unsure of how to find the power series of this equation in order to do that. I haven't dealt with square root power series equations yet and any help would be appreciated. Thank you!
| $$\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n\tag{1} $$
for any $x\in[-1,1)$ is a standard result. By applying termwise integration one gets
$$ \sqrt{1-x} = \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\frac{x^n}{1-2n}\tag{2} $$
for any $x\in[-1,1]$. By replacing $x$ with $-x^4$ one gets
$$ \sqrt{1+x^4} = \sum_{n\geq 0}\frac{(-1)^n}{4^n}\binom{2n}{n}\frac{x^{4n}}{1-2n}\tag{3}$$
and finally
$$ \int_{0}^{1}\sqrt{1+x^4}\,dx=\sum_{n\geq 0}\frac{(-1)^n}{4^n}\binom{2n}{n}\frac{1}{(1-2n)(4n+1)}.\tag{4} $$
By partial fraction decomposition, the RHS of $(4)$ only depends on $\sqrt{2}$ and
$$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\frac{(-1)^n}{(4n+1)}=\frac{\Gamma^2\left(\tfrac{1}{4}\right)}{8\sqrt{\pi}}.\tag{5}$$
The last equality can be derived through Euler's Beta function. Summarizing,
$$ \int_{0}^{1}\sqrt{1+x^4}\,dx = \phantom{}_2 F_1\left(-\tfrac{1}{2},\tfrac{1}{4};\tfrac{5}{4};-1\right)= \frac{\sqrt{2}}{3}+\frac{\Gamma^2\left(\frac{1}{4}\right)}{12\sqrt{\pi}}.\tag{6}$$
A numerical computation is extremely simple via $\frac{\Gamma^2\left(\frac{1}{4}\right)}{12\sqrt{\pi}}=\frac{\pi}{3\,\text{AGM}(\sqrt{2},2)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3197148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$11x + 13 \equiv 4$ (mod 37) $11x + 13 \equiv 4$ (mod 37)
My "solution" to the problem
$11x + 13 \equiv 4$ (mod 37) $\rightarrow$
$11x + 13 = 4 + 37y $
$11x - 37y = - 9$
Euclid's algorithm.
$37 = 11*3 + 4$
$11 = 4*2 + 3$
$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$
$3 = 1*3 + 0$
Write as linear equation
$1 = 4 - 1*3$
$3 = 11 - 2*4$
$4 = 37-3*11$
$1 = 4 -1*3 = 4-1(11-2*4) = 3*4-1*11 = 3(37-3*11)-1*11 = 3*37 -10*11$
$1 = 3*37 -10*11$
$11(10) - 37(3) = -1$
$11(90) - 37(27) = -9$
x = 90
That is my answer. But the correct answer should be:
x = 16
| $\bmod 37\!:\,\ x\equiv 90\equiv 16\ $ so both are correct. Gauss's algorithm is simpler:
$\bmod 37\!:\,\ x\equiv\dfrac{-9}{11}\equiv\dfrac{-27}{33}\equiv\dfrac{10}{-4}\equiv\dfrac{5}{{-}2\,}\equiv\dfrac{-32}{-2}\equiv 16$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluating limit of a function with integral involved. We need to evaluate the following limit:
$$\lim\limits_{x\to 0} \frac{1}{x} \int_0^x \sin^2\left(\frac{1}{y}\right)\,\mathrm dy.$$
I am finding hard to solve this integral as I cannot see a clear method to evaluate such integral. I have tried using substitutions, integration by parts, expansion formulas etc. but I don't see them be of any help. Also, I cannot take the limit inside. The best I could do was to reduce it to cosine function as follows:
$$\lim\limits_{x\to 0}\frac{1}{x} {\int}_0^x \sin^2\left(\frac{1}{y}\right)\,\mathrm dy = \lim\limits_{x\to 0}\frac{1}{x} {\int}_0^x \frac{\left(1-\cos\left(\frac{2}{y}\right)\right)}{2}\,\mathrm dy = \frac{1}{2}-\lim\limits_{x\to 0}\frac{1}{x} {\int}_0^x \cos\left(\frac{2}{y}\right)\,\mathrm dy.$$
But again the integrand has no standard antiderivative. Is there some easy method to solve this or I am making some mistakes? Any hint/help will be of great help.
Thanks in advance.
| I am posting an answer on the basis of hints given in comments, hope it helps if someone has the same question. Please tell if this is correct.
\begin{align*}
\lim\limits_{x \to 0}\frac{1}{x} \int_{0}^{x} \sin^2\left( \frac{1}{u}\right) \mathrm du & = \lim\limits_{x \to 0}\frac{1}{x} \int_{0}^{x} \frac{\left(1-\cos\left( \frac{2}{u}\right)\right)}{2}\ \mathrm du\\
& = \lim\limits_{x \to 0}\left(\frac{1}{x}\cdot \frac{1}{2}\int_{0}^{x} du - \frac{1}{x}\cdot \frac{1}{2} \int_{0}^{x} \cos\left(\frac{2}{u} \right)\mathrm du \right)\\
& = \lim\limits_{x \to 0} \left(\frac{1}{2x}\cdot x \right) - \lim\limits_{x \to 0}\left( \frac{1}{2x} \int_{0}^{x} \cos \left( \frac{2}{u}\right)\mathrm du \right)\\
& = \frac{1}{2} - \frac{1}{2}\ \lim\limits_{x \to 0} \frac{1}{x} \int_{0}^{x} \cos \left( \frac{2}{u}\right) \mathrm du.\\
\end{align*}
{Claim} : $$ \lim\limits_{x \to 0} \frac{1}{x} \int_{0}^{x} \cos \left( \frac{2}{u} \right)\mathrm du = 0$$
{Proof} : Consider the function $F(x) = x^2\sin \left( \frac{2}{x} \right).$ Then,
$F(0) = 0$ as $\lim\limits_{x \to 0} x^2\sin \left(\frac{2}{x} \right) = 0\ (\because -1 \le \sin \left( \frac{2}{x} \right)\le 1 \Rightarrow -x^2 \le x^2 \sin \left( \frac{2}{x}\right) \le x^2.) $ Hence, we define function value to be $0$ at $x =0$ making $F(x)$ continuous at $x= 0$.
Also, we can calculate using limit definition that $F'(0) = 0 $ as follows :
\begin{align*}
F'(0) & = \lim\limits_{x \to 0} \frac{F(x) - F(0)}{x-0}\\
& = \lim\limits_{x\to 0} \frac{x^2 \sin \left( \frac{2}{x}\right)}{x}\\
& = \lim\limits_{x \to 0} x \sin \left( \frac{2}{x}\right)\\
& = 0 \ \ \ \ \ \ \ \ \ \text{(using sandwitch theorem as done above).}\\
\end{align*}
We also know that $F'(x) = 2x \sin \left( \frac{2}{x} \right)- 2 \cos \left( \frac{2}{x}\right)$ whenever $x \ne 0$. To prove our claim, assume $$G(x) = \int_{0}^{x} \cos \left( \frac{2}{u} \right) \mathrm du$$
Now, by Fundamental theorem of calculus (FTOC), $$F(x) = \int_{0}^{x} F'(u) \mathrm du = 2 \int_{0}^{x} u\sin \left(\frac{2}{u}\right) \mathrm du - 2 G(x)$$
Next, dividing by $x$ and taking $\lim\limits_{x \to 0}$ we get,
\begin{align*}
\lim\limits_{x \to 0} \frac{F(x)}{x} & = \lim\limits_{x \to 0}\frac{2}{x} \int_{0}^{x} u \sin \left(\frac{2}{u}\right) \mathrm du - \lim\limits_{x \to 0} \frac{2}{x} G(x)\\
\Rightarrow F'(0) & = \lim\limits_{x \to 0}\frac{2}{x} \int_{0}^{x} u \sin \left(\frac{2}{u}\right) \mathrm du - \lim\limits_{x \to 0} \frac{2}{x} G(x)\\
\end{align*}
We see that $$\lim\limits_{x \to 0}\frac{2}{x} \int_{0}^{x} u \sin \left(\frac{2}{u}\right) \mathrm du = 0$$
To see this, we apply similar method as done above. Let $h(x) = x\sin \left(\frac{2}{x}\right).$ Since, $\lim\limits_{x \to 0}x\sin \left(\frac{2}{x} \right) = 0$, we define function value to be $0$ at $x =0$, making it continuous there. Next, let
\begin{align*}
H(x) &= \int_{0}^{x} h(u)\mathrm du\\
\Rightarrow \lim\limits_{x \to 0}\frac{H(x)}{x} &= \lim\limits_{x \to 0} \frac{1}{x}\int_{0}^{x} u\sin\left(\frac{2}{u} \right) \mathrm du\\
\Rightarrow H'(0) &= \frac{d}{dx} \int_{0}^{x} u\sin \left(\frac{2}{u} \right) \mathrm du\\
\Rightarrow H'(0) &= h(0) \ \ \ \ \ \text{(using FTOC)}\\
\Rightarrow H'(0) &= 0\\
\end{align*}
Therefore we have that,
\begin{align*}
F'(0) &= 0- \lim\limits_{x \to 0} \frac{2}{x}G(x)\\
&\Rightarrow \lim\limits_{x \to 0} \frac{2}{x}G(x) = 0\\
&\Rightarrow \lim\limits_{x \to 0} \frac{1}{x} \int_{0}^{x} \cos \left( \frac{2}{u} \right)\mathrm du = 0.\\
\end{align*}
Hence, the claim is proved. Next, using the claim we obtain finally,
$$ \lim\limits_{x \to 0}\frac{1}{x} \int_{0}^{x} \sin^2\left( \frac{1}{u}\right) \mathrm du = \frac{1}{2}\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I prove this combinatorial identity Show that
$${2n \choose n} + 3{2n-1 \choose n} + 3^2{2n-2 \choose n} + \cdots + 3^n{n \choose n} \\ = {2n+1 \choose n+1} + 2{2n+1 \choose n+2} + 2^2{2n+1 \choose n+3} + \cdots + 2^n{2n+1 \choose 2n+1}$$
One way that I did it was to use the idea of generating functions.
For the left hand side expression, I can find 2 functions. Consider;
$$f_1 (x) = \frac{1}{(1-3x)} \\ = 1 + 3^1x + 3^2x^2 + 3^3x^3 + \cdots + 3^nx^n + \cdots \\ f_2(x) = \frac{1}{(1-x)^{n+1}} \\ = {n \choose n} + {n+1 \choose n}x + {n+2 \choose n}x^2 + \cdots + {2n-1 \choose n}x^{n-1} + {2n \choose n}x^n + \cdots + $$
Consider the coefficient of $x^n$ in the expansion of $f_1 (x) . f_2 (x)$. Then the coefficient will be the expression on the left hand side.
Now we further consider 2 functions for the right-hand side expression.
Consider;
$$f_3 (x) = \frac {1}{(1-2x)} \\ = 1 + 2^1x + 2^2x^2 + \cdots + 2^{n-1}x^{n-1} + 2^nx^n + \cdots \\ f_4 (x) = (1+x)^{2n+1} \\= 1 + {2n+1 \choose 1}x + \cdots + {2n+1 \choose n-1}x^{n-1} + {2n+1 \choose n}x^n +\cdots + {2n+1 \choose 0}x^{2n +1} \\ = {2n+1 \choose 2n+1} + {2n+1 \choose 2n}x + {2n+1 \choose 2n-1}x^2 + \cdots + {2n+1 \choose n+2}x^{n-1} + {2n+1 \choose n+1}x^{n} + \\
+ {2n+1 \choose n}x^{n+1} +\cdots + {2n+1 \choose 0}x^{2n +1}$$
Hence the coefficient of $x^n$ is the coefficient of $x^n$ in the expansion of $f_3(x) . f_4(x)$
This is what I managed to do so far. I'm not sure if $f_1(x) .f_2(x) = f_3(x).f_4(x)$. If the two functions are indeed equal, then I can conclude that their coefficient of $x^n$ must be equal, which will immediately answer the question. If they are equal, how do I show that they are?
If the two functions are not equal? How do I proceed to show this question?
Edit: It might not be true that the product of the two functions are equal. I tried substituting $x=0.1, n=1$. Seems like the two values are not equal. How do I proceed with this question?
| Here is a combinatorial proof. Both sides of the equation answer the following question:
How many sequences are there of length $2n+1$, with entries in $\{0,1,2\}$, such that
*
*at least one of the entries is a $2$, and
*there are exactly $n$ zeroes to the left of the leftmost $2$?
LHS:
Suppose the leftmost $2$ occurs in spot $k+1$. Among the $k$ spots before hand, you must choose $n$ of the entries to be zero. The $2n+1-(k+1)=2n-k$ spots afterward can be anything. There are $\binom{k}n3^{2n-k}$ ways to do this. Then sum over $k$.
RHS:
Suppose there are $j$ entries which are equal to $0$ or $2$. Choose those entries which are equal to $0$ or $2$ in $\binom{2n+1}j$ ways. The leftmost $n$ of these entries must be zero, the $(n+1)^{st}$ entry must be two, then the remaining $j-(n+1)$ entries can be chosen freely among $0$ and $2$. There are $\binom{2n+1}{j}2^{j-(n+1)}$ ways to do this, then sum over $j$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3201479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
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Minimizing $\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2$ While solving a problem I came across this task, minimizing
\begin{align}
\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2.
\end{align}
One can easily do it with calculus to show that the minimum value is $12.5$.
I tried to do it using trigonometric identities and fundamental inequalities (like AM-GM, Cauchy-Schwarz, etc.) but failed. Can someone help me to do it using trig identities and inequalities?
| AM-GM says $a^2+b^2\geq 2ab$ with equality when $a=b.$ Thus we cannot obtain a lower value than which gives GM.
With a bit of algebra we get that the equality really occurs, and this is iff $|\sin x|=|\cos x|,$ from where the minimum $$2.5^2+2.5^2=12.5$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Describe the region $\{|z^2 - 1|<1\}$ of the complex plane Since it's complex analysis, I assume there is an easy solution without much algebra. But even in my algebra heavy solution there should be a more streamlined approach... Either approach is appreciated as an answer, or verification that my approach is the standard one, thanks!
My Attempt:
We can equivalently look at the condition $|z^2 - 1|^2 - 1< 0$. Note that $z^2 - 1 = (x^2 - y^2 - 1) +i(2xy)$, so
\begin{align*}
|z^2 - 1|^2 - 1 & = (x^2 - y^2 - 1)^2 + 4x^2 y^2 - 1 \\
& = x^4 + y^4 -2x^2 - 2y^2 + 2x^2 y^2 \\
& = (x^2 + y^2)(x^2 + y^2 - 2) \\
& <0
\end{align*}
Case 1: $x^2 + y^2 < 0$ and $x^2 + y^2 - 2 >0$
This case is not possible
Case 2: $x^2 + y^2 > 0$ and $x^2 + y^2 - 2 <0$
Here, at most one of $x$ and $y$ can be zero, so $z\neq 0$. Further, $x^2 + y^2 <2$ is the open disk radius $2$ centered at $0$
Therefore, The region is the punctured open disk $D(0,2)\setminus \{0\}$
| Credit to @Reinhard Meier in the comments
There is an algebra heavy answer that you can find to give an exact, graphable answer: $$ x^4 - 2x^2 + y^4 + 2y^2 + 2x^2 y^2 < 0 $$
Although we will take another approach. Rewrite the condition to be $$ |(z+1)(z-1)|<1 \implies |z+1|\cdot|z-1|<1 $$
We can see clearly 3 things:
*
*$z\neq 0$
*If $z\in\mathbb{R}$, then $0< z < \sqrt{2}$
*The region is has 2 ``centers'' $z=1$ and $z=-1$, and for any point in the region the product of distances to $1$ and $-1$ is never greater than $1$
From this we can describe the region as the interior of a "figure 8" that is is "centered" at the complex numbers $1$ and $-1$, and is not perfectly circular due to the third bullet point
Here is a picture of the region, obtained by drawing random points in the $[-2,2] \times [-2,2]$ square and painting them according to the desired
inequality.
R code:
n = 1e4
x = runif(n,-2,2)
y = runif(n,-2,2)
i = complex(real=0,imaginary=1)
z = x+i*y
ind = abs(z^2-1)<1
plot(x,y,col=ind+1,xlab='Re(z)',ylab='Im(z)')
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x$ and $y$, The equations are $x \cos^{3} y+3x \cos y \sin^{2} y =14 $ and $ x \sin^{3} y+3x \cos^{2} y \sin y = 13 $ Consider the system of equations
$$x \cos^{3} y+3x \cos y \sin^{2} y =14 $$
$$ x \sin^{3} y+3x \cos^{2} y \sin y = 13 $$
$1)$ the values of $x$ is /are..
Answer is $ \pm\sqrt 5 $
The number of values of $y$ in $(0,6 \pi)$is
Answer is $6$
$Sin^2 y + 2cos^2 y $ is
Answer is $ \frac95 $
I added both the equation and make whole cube
I subtracted both the equation and make whole cube then I divided these equations to find the value of $ \tan y $ = $ \dfrac{1}{2} $
I divided the first equation with $ cos^2 y $ and plugged the value of $\tan y$ which gives $x= +5\sqrt 5 $. But I also need -$ 5\sqrt 5 $ I have no idea where I go wrong.
| When you render $\tan y =1/2$, it follows that
$1+\tan^2 y = \sec^2 y = 1/\cos^2 y = 5/4$
But then you have to allow both the positive and negative square roots for the cosine. One period of the cosine is $2\pi$ which is two periods of the tangent function, so both the positive and negative cosine values occur in different periods of the tangent function. When you allow the negative value, $\cos y = -2/\sqrt{5}$, you pick up the negative value for $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Simplifying fractional surds I have this fractional surd:
$$\frac{5\sqrt{7}+4\sqrt{2}}{3\sqrt{7}+5\sqrt{2}}$$
I can calculate this with a calculator fairly easily obviously but what is the best tactic without one?
Thank you!
| So first remember some key rules:
*
*$\sqrt{a} \times \sqrt{a}=a$
*$\sqrt{a} \times \sqrt{b}=\sqrt{ab}$ (which you can simplify usually)
(There are more but we only really need these!)
So a good technique to use in this question is called rationalising the denominator. Which basically means take the denominator of the surd and change the middle sign so $3\sqrt{7}+5\sqrt{2}$ turns into $3\sqrt{7}-5\sqrt{2}$. So we have:
$$\frac{5\sqrt{7}+4\sqrt{2}}{3\sqrt{7}+5\sqrt{2}} \times \frac{3\sqrt{7}-5\sqrt{2}}{3\sqrt{7}-5\sqrt{2}}$$
Which we can do as the $2^{nd}$ fraction is equal to $1$.
Looking at the top now we have $({5\sqrt{7}+4\sqrt{2}}) \times ({3\sqrt{7}-5\sqrt{2}})= 65-13\sqrt{14}$
Then doing the same thing with the bottom we get it just equal to $13$,
So putting this together we get: $$\frac{65-13\sqrt{14}}{13}= 5-\sqrt{14}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan C $ and $ \tan B $
Settings discriminant greater than equal to zero.
$ { (p-1)}^2 - 4p \ge 0 $
This gives $ p \le 3 - 2\sqrt2 $. Or $ p \ge 3 + 2\sqrt2 $
solving both equation
$ A + B + C = \pi $
$ C + B + \frac{\pi}{4} = \pi $
$ C + B = \frac{3\pi}{4} $
Using this to solve both the equation give $ p \in $ real
I found this on Quora.
https://www.quora.com/Let-A-B-C-be-three-angles-such-that-A-frac-pi-4-and-tan-B-tan-C-p-What-are-all-the-possible-value-of-p-such-that-A-B-C-are-the-angles-of-the-triangle
the right method
$ 0 \lt B , C \lt \frac{3\pi}{4} $
Converting tan into sin and cos gives
$ \dfrac {\sin B \sin C}{\cos B \cos C} = p $
Now using componendo and dividendo
$ \frac{\cos (B-C) }{- \cos(B+C) } = \frac{p+1}{p-1} $
We know $ \cos (B+C) = 1/\sqrt2 $
We know the range of $B$ and $C$ $(0, 3π/4)$
Thus the range of $B - C$. $(0, 3π/4 )$
Thus range of $\cos(B+C)$ is $ \frac{ -1}{\sqrt2} $ to $1$
Thus using this to find range gives
$ P \lt 0 $ or $ p \ge 3+ 2\sqrt2 $
| We should consider discriminants but in the way they arise directly:
Let $tx = \tan(x)$ for shorthand notation
$$t_A= t{(\pi-B-C)}$$
$$1=\frac{-(tA+tB)}{1-tA\, tB} $$
$$tA \,tB=p\quad$$
plug in and simplify to find quadratic equation roots
$$ tB^2+(1-p)tB+p=0 $$
$$ 2\,tA=-(1-p)-\sqrt{1-6p+p^2} $$
$$ 2\,tB=-(1-p)+\sqrt{1-6p+p^2}$$
For this to be real, quantity under radical sign should be positive.
The roots of quadratic equation then supply upper and lower limits automatically.
$$p_{lower}= 3+2\sqrt{2}\quad p_{lower}= 3-2\sqrt{2} $$
when $p$ should not lie between these limits as the quantity under radical sign is:
$$ \sqrt{p^2 -6p +1 } = \sqrt{(p-p_{lower})(p-p_{upper})} $$
| {
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"url": "https://math.stackexchange.com/questions/3223162",
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"source": "stackexchange",
"question_score": "6",
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Prove that $2^{2^{2^{\cdot^{\cdot^{2}}}}} \mod 9 = 7$
Prove that $\underbrace{2^{2^{2^{\cdot^{\cdot^{2}}}}}}_{2016 \mbox{
times}} \mod 9 = 7$
I think that it can be done by induction:
Base:
$2^{2^{2^{2}}} \equiv 2^{16} \equiv 2^8 \cdot 2^8 \equiv 2^4 \cdot 2^4 \cdot2^4 \cdot2^4 \equiv 7^2 \cdot 7^2 \equiv 4 \cdot 4 \equiv 16 \equiv 7$
Assume that it is true for $n$ times.
Let $a_n = \underbrace{2^{2^{2^{\cdot^{\cdot^{2}}}}}}_{n \mbox{
times}}$
$$ a_{n+1} = 2^{a_n} = 2^{9k+7} = 2^7 \cdot 2^{9k} \equiv 128 \cdot 1 \equiv 2 \neq 7 $$
Where did I go wrong?
| For the correct way to do modular exponent reduction see here. Using that method
$\!\!\bmod 9\!:\,\ 2^{\Large\color{#c00} 6}\equiv 1\ \ {\rm thus}\ \ 2^{\large N}\equiv\, 2^{\large N\Large \bmod\, \color{#c00}6}\ $
$\text{Applying that here}\!:\ 2^{\Large {2^{\Large 2K}}}\!\!\equiv\, 2^{\Large \color{#0a0}{2^{\Large 2K}}\!\bmod\color{#c00} 6}\!\equiv 2^{\Large\color{#0a0} 4}\equiv 7$
$\text{since we have that:}$ $\bmod \color{#c00}6\!:\,\ \color{#0a0}{2^{\large 2K}}\!\equiv 4^{\large K}\!\equiv\color{#0a0} 4,\,$ by $\,4^{\large 2}\equiv 4\,$ and induction (or by here).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $a \equiv 9 \pmod {12}$, find all possible values for $\gcd(a^2+21a+72,252)$ We know that
$$a^2+21a+72 \equiv 9^2 + 21 \cdot 9 + 0 \equiv 6 \pmod {12}$$
So we know that that expression, let's say $\alpha$, is such that $12 \mid \alpha - 6$. But then $12 \mid 2(\alpha - 6)=2a+12$, which means that $\alpha$'s prime factorization contains $2$ as a factor (with multiplicity $1$) and 3 as a factor.
We know that $3$ has multiplicity $2$ because $72=2^3 \cdot 3^2$, $21a = 7 \cdot 3^2 \cdot k, a^2= 3^2k^2$. So $18$ is the lowest possible candidate to gcd. If $a^2 \equiv 5 \pmod 7$, we know that $126$ is the other candidate solution. Is there any way to check whether the system
$$a^2 \equiv 9 \pmod {12} \\ a^2 \equiv 5 \pmod 7$$
has no solution? If it does, does this mean that the gcd is $126$ unless $a^2 \not\equiv 5 \pmod 7$?
| Write $a=12b+9$. Then $a^2+21a+72=18 (8 b^2 + 26 b + 19)$ and so $$\gcd(a^2+21a+72,252) = 18 \gcd(8 b^2 + 26 b + 19,14)$$
Since $8 b^2 + 26 b + 19$ is always odd, we have
$$\gcd(8 b^2 + 26 b + 19,14) = \gcd(8 b^2 + 26 b + 19,7) = \gcd(b^2 -2b -2,7) = \gcd((b-1)^2 -3,7) = 1
$$
because $3$ is not a square mod $7$.
Therefore, $\gcd(a^2+21a+72,252) = 18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to factorized this 4th degree polynomial? I need your help to this polynomial's factorization.
Factorize this polynomials which doesn't have roots in Q.
$ \ f(x) = x^4 +2x^3-8x^2-6x-1 $
P.S.) Are there any generalized method finding 4th degree polynomials factor?
| For all real $k$ we have:
$$x^2+2x^3-8x^2-6x-1=(x^2+x+k)^2-x^2-k^2-2kx^2-2kx-8x^2-6x-1=$$
$$=(x^2+x+k)^2-((2k+9)x^2+2(k+3)x+k^2+1).$$
Now, we'll choose a value of $k$ such that
$$(2k+9)x^2+2(k+3)x+k^2+1=(ax+b)^2.$$
Easy to see that $k=0$ is valid.
In the general we need to solve the following equation.
$$(k+3)^2-(2k+9)(k^2+1)=0.$$
Id est, $$x^2+2x^3-8x^2-6x-1=(x^2+x+k)^2-x^2-k^2-2kx^2-2kx-8x^2-6x-1=$$
$$=(x^2+x+k)^2-((2k+9)x^2+2(k+3)x+k^2+1)=$$
$$=(x^2+x)^2-(9x^2+6x+1)=(x^2+x)^2-(3x+1)^2=(x^2-2x-1)(x^2+4x+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to get sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$ using mathematical induction Prehistory: I'm reading book. Because of exercises, reading process is going very slowly. Anyway, I want honestly complete all exercises.
Theme in the book is mathematical induction. There were examples, where were shown how with mathematical induction prove equations like $(1+q)(1+q^2)(1+q^4)\dots(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$.
Now I'm trying to complete exercise where I have to find sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$.
I tried to do it with mathematical induction. Like this:
n=1: $\frac{1}{1+x^2}$
n=2: $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}=\frac{1+1+x^2}{(1+x^2)^2}$
n=3: $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}=\frac{(1+x^2)^2+2+x^2}{(1+x^2)^3}$
...
And so on (I've calculated til n=5). But I don't see any consistent pattern to evaluate sum of progression.
After that I found formulas of geometrical progression:
$q=\frac{b_{n+1}}{b_n}$ and $S_n=\frac{b_1(1-q^n)}{1-q}$, so I've evaluated:
$q=\frac{\frac{1}{(1+x^2)^2}}{\frac{1}{1+x^2}}=\frac{1}{1+x^2}$
and
$S_n=\left(\frac{1}{1+x^2}\left[1-\left(\frac{1}{1+X^2}\right)^n\right]\right):\left(1-\left[\frac{1}{1+x^2}\right]\right)=\left[\frac{1}{1+x^2}-\left(\frac{1}{1+x^2}\right)^{n+1}\right]\frac{1+x^2}{x^2}=\frac{\left[1-\left(\frac{\sqrt[n+1]{1+x^2}}{1+x^2}\right)^{n+1}\right]}{x^2}$
First of all, I'd like to know how to find sum of geometrical progression with mathematical induction.
Secondly, I'd like to know what is wrong with my evaluations.
| For any $n$ in your calculations(which are right but missed the right trick) use the following:
first set
$${{S}_{n}}=q+{{q}^{2}}+\ldots +{{q}^{n}}$$
then
$$q{{S}_{n}}={{q}^{2}}+{{q}^{3}}+\ldots +{{q}^{n+1}}$$
al last
$$\begin{align}
& {{S}_{n}}-q{{S}_{n}}=q-{{q}^{n+1}} \\
& {{S}_{n}}=\frac{q\left( 1-{{q}^{n}} \right)}{\left( 1-q \right)} \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Given three positve numbers $a,b,c$. Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{b(a+b)}}\geqq \sum\limits_{cyc}\frac{a}{\sqrt{b(c+a)}}$ .
Given three positive numbers $a, b, c$, prove that $$\sum\limits_{cyc}\frac{a}{\sqrt{b(a+ b)}}\geqq \sum\limits_{cyc}\frac{a}{\sqrt{b(c+ a)}}.$$
I tried on Holder inequality and https://artofproblemsolving.com/community/c6h194103p1065812 but it's hard! I need to the hints and hope to see the $uvw$ help here! Thanks a lot!
| The inequality is true. The Buffalo Way works.
Here is a hint. Currently, I do not give the Buffalo Way part for people to attempt.
$\phantom{2}$
With the substitutions $(a, b, c) \to (a^2, b^2, c^2)$, the inequality becomes
$$\sum_{\mathrm{cyc}} \frac{a^2}{b\sqrt{a^2+b^2}} \ge \sum_{\mathrm{cyc}} \frac{a^2}{b\sqrt{c^2+a^2}}.$$
WLOG, assume that $c = \min(a,b,c) = 1$.
Let $X = \frac{a^2+b^2}{2}, Y = \frac{b^2+c^2}{2}, Z = \frac{c^2 + a^2}{2}$. We need to prove that
$$\frac{a^2}{b\sqrt{X}} + \frac{b^2}{c\sqrt{Y}} + \frac{c^2}{a\sqrt{Z}}
\ge \frac{a^2}{b\sqrt{Z}} + \frac{b^2}{c\sqrt{X}} + \frac{c^2}{a\sqrt{Y}}$$
or
$$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)\frac{1}{\sqrt{Y}}
\ge \big(\frac{b^2}{c} - \frac{a^2}{b}\big)\frac{1}{\sqrt{X}} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{1}{\sqrt{Z}}$$
Since $\frac{b^2}{c} - \frac{c^2}{a} \ge 0$, it suffices to prove that
$$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)^2\frac{1}{Y}
\ge \Big[\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\frac{1}{\sqrt{X}} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{1}{\sqrt{Z}}\Big]^2$$
or
$$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)^2\frac{1}{Y}
\ge \big(\frac{b^2}{c} - \frac{a^2}{b}\big)^2\frac{1}{X} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)^2\frac{1}{Z}
+ 2\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{1}{\sqrt{ZX}}.$$
There are three possible cases:
Case I $b \ge a^3$: Clearly $\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big) \le 0$.
Since $\sqrt{ZX}\le \frac{Z+X}{2}$, it suffices to prove that
$$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)^2\frac{1}{Y}
\ge \big(\frac{b^2}{c} - \frac{a^2}{b}\big)^2\frac{1}{X} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)^2\frac{1}{Z}
+ 2\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{2}{Z+X}.$$
After clearing the denominators, it suffices to prove that $f(a,b,c)\ge 0$ where $f(a,b,c)$ is a homogeneous polynomial of degree $14$. The Buffalo Way works.
Case II $b \le a^{2/3}$: Clearly $\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big) \le 0$.
Since $\sqrt{ZX}\le \frac{Z+X}{2}$, it suffices to prove that
$$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)^2\frac{1}{Y}
\ge \big(\frac{b^2}{c} - \frac{a^2}{b}\big)^2\frac{1}{X} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)^2\frac{1}{Z}
+ 2\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{2}{Z+X}.$$
After clearing the denominators, it suffices to prove that $g(a,b,c)\ge 0$ where $g(a,b,c)$ is a homogeneous polynomial of degree $14$. The Buffalo Way works.
Case III $a^{2/3}< b < a^3$: Clearly $\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big) \ge 0$.
Since $\sqrt{ZX} \ge \frac{ac + ab}{2}$, it suffices to prove that
$$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)^2\frac{1}{Y}
\ge \big(\frac{b^2}{c} - \frac{a^2}{b}\big)^2\frac{1}{X} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)^2\frac{1}{Z}
+ 2\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{2}{ab+ac}.$$
After clearing the denominators, it suffices to prove that $h(a,b,c)\ge 0$ where $h(a,b,c)$ is a homogeneous polynomial of degree $13$.
The Buffalo Way works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^3$.
| First let's get rid of the square root:
$$\lim_{n \to \infty} (\sqrt[3]{n^3-3n^2}-\sqrt{n^2+2n})=\lim_{n \to \infty} \frac{\sqrt[3]{(n^3+3n^2)^2}-(n^2+2n)}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}\\= -\lim_{n \to \infty} \frac{n^2+2n-\sqrt[3]{(n^3+3n^2)^2}}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}.$$
Now let's get rid of the cube root:
$$-\lim_{n \to \infty} \frac{n^2+2n-\sqrt[3]{(n^3+3n^2)^2}}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}$$
$$=-\lim_{n \to \infty} \frac{(n^2+2n)^3-(n^3+3n^2)^2}{(\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n})((n^2+2n)^2+(n^2+2n)\sqrt[3]{(n^3+3n^2)^2}+\sqrt[3]{(n^3+3n^2)^4})}.$$
The numerator yields $-3n^4$ and a bunch of lower-order terms, and the denominator yields $6n^5$ and a bunch of lower order terms, so dividing numerator and denominator by $n^5$ shows that the answer is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
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Even stronger than Sophomore's dream Sophomore's dream states that:
$$
\int_0^1x^{-x}dx=\sum_{n=1}^\infty n^{-n}
$$
and
$$
\int_0^1x^{x}dx=-\sum_{n=1}^\infty(-n)^{-n}
$$
A friend of mine noticed that numerically:
$$
\int_0^1\int_0^1(xy)^{xy}dxdy\approx 0.7834... \approx \int_0^1x^{x}dx
$$
Are these two integrals really equal?
| Yes they are equal! The trick is to use the binomial theorem to separate the integrals:
\begin{align}
\int_0^1\int_0^1(xy)^{xy}dxdy &= \int_0^1\int_0^1\exp(xy\log{xy})dxdy \\
&= \sum_{n=0}^\infty\frac{1}{n!}\int_0^1\int_0^1(xy\log{xy})^ndxdy \\
&= \sum_{n=0}^\infty\frac{1}{n!}\int_0^1\int_0^1(xy)^n(\log{x}+\log{y})^ndxdy \\
&= \sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^nC_n^k\left(\int_0^1 x^n(\log{x})^kdx\right)\left(\int_0^1 y^n(\log{y})^{(n-k)}dy\right)
\end{align}
Then using this result:
$$
\int_0^1u^n\log{u}^m du=-m!(-\frac{1}{1+n})^{1+m}
$$
we get:
\begin{align}
\int_0^1\int_0^1(xy)^{xy}dxdy &= \sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^nC_n^k\left(k!(-\frac{1}{1+n})^{1+k}\right)\left((n-k)!(-\frac{1}{1+n})^{1+n-k}\right) \\
&= \sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^nn!(-\frac{1}{1+n})^{2+n} \\
&= \sum_{n=0}^\infty (1+n)(-\frac{1}{1+n})^{2+n} \\
&= \sum_{n=0}^\infty (-1)^{2+n}(\frac{1}{1+n})^{1+n} \\
&= \sum_{n=1}^\infty -(\frac{-1}{n})^{n} \\
& = \int_0^1x^xdx
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Can we show a sum of symmetrical cosine values is zero by using roots of unity? Can we show that
$$\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{5\pi}{7}+\cos\frac{6\pi}{7}=0$$
by considering the seventh roots of unity? If so how could we do it?
Also I have observed that
$$\cos\frac{\pi}{5}+\cos\frac{2\pi}{5}+\cos\frac{3\pi}{5}+\cos\frac{4\pi}{5}=0$$
as well, so just out of curiosity, is it true that $$\sum_{k=1}^{n-1} \cos\frac{k\pi}{n} = 0$$
for all $n$ odd?
| Pointing at the link I left in the comments
$$1 + \cos\theta + \cos2\theta +... + \cos n\theta = \frac{1}{2} + \frac{\sin[(n+\frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})}$$
Then for $\forall n\in\mathbb{N}, n>0$
$$\cos\frac{\pi}{n+1}+ \cos\frac{2\pi}{n+1} +... + \cos \frac{n\pi}{n+1} = \frac{\sin\left[(n+\frac{1}{2})\frac{\pi}{n+1}\right]}{2\sin\left(\frac{\pi}{2(n+1)}\right)}-\frac{1}{2}=\\
\frac{\sin\left[\frac{2n+1}{2(n+1)}\pi\right]}{2\sin\left(\frac{\pi}{2(n+1)}\right)}-\frac{1}{2}=
\frac{\sin\left[\pi-\frac{\pi}{2(n+1)}\right]}{2\sin\left(\frac{\pi}{2(n+1)}\right)}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Determine this limit $\lim_{n\to\infty} \frac{\ln\left(\frac{3\pi}{4} + 2n\right)-\ln\left(\frac{\pi}{4}+2n\right)}{\ln(2n+2)-\ln(2n)}.$ how can I determine the following limit? $$\lim_{n\to\infty} \frac{\ln\left(\frac{3\pi}{4} + 2n\right)-\ln\left(\frac{\pi}{4}+2n\right)}{\ln(2n+2)-\ln(2n)}.$$
This question stems from this question. The proof presented there is incorrect, and it would be trivial to show that the mentioned integral diverges if the above limit is $>0$ by using the comparison test (anyone feels free to do this by the way).
| $$a_n= \frac{\ln\left(\frac{3\pi}{4} + 2n\right)-\ln\left(\frac{\pi}{4}+2n\right)}{\ln(2n+2)-\ln(2n)}=\frac{\log \left(1+\frac{\pi }{4 n+\frac{\pi }{2}}\right) } {\log \left(1+\frac 1n \right)}$$ Using Taylor expansions
$$a_n=\frac{\frac{\pi }{4 n}-\frac{\pi ^2}{16 n^2}+O\left(\frac{1}{n^3}\right) } {\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right) }=\frac{\pi }{4}-\frac{(\pi -2) \pi }{16 n}+O\left(\frac{1}{n^2}\right) $$ which shows the limit and how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(1+AB)(A^{3} + B^{3})^{2} = (AB)^{2} + 4AB + 4 $$
$$ A^{6} + B^{6} + BA^{7} + A B^{7} = -2 (AB)^{4} - 2(AB)^{3} + (AB)^{2} + 4AB + 4 $$
I also have tried using$A = (1+x), B = (1-x)$ , and some others, but none solves the problem.
I am now trying $A = (1+x)$ and $(1-x) = -(1+x) + 2 = 2 - A$, so:
$$\sqrt{1 + \sqrt{A(2-A)}}\left(\sqrt{(A)^{3}} + \sqrt{(2-A)^{3}} \right) = 2 + \sqrt{A(2-A)} $$
| Substituting $$a=\sqrt{1+x},b=\sqrt{1-x}$$ and using that $$a^2+b^2=2$$ and $$a^3+b^3=(a+b)(a^2+b^2-ab)$$ we get$$\sqrt{1+ab}(a+b)(2-ab)=2+ab$$
And we get by squaring $$(1+ab)(2+2ab)(2-ab)^2=(2+ab)^2$$
and with $$u=ab$$ we get
$$2(1+u)^3(2-u)^2=(2+u)^2$$
The solutions are $$\left\{\left\{x\to
-\sqrt{-\frac{7}{4}+\frac{3}{\sqrt{2}}-\frac{1}{4}
\sqrt{97-68 \sqrt{2}}}\right\},\left\{x\to
\sqrt{-\frac{7}{4}+\frac{3}{\sqrt{2}}-\frac{1}{4}
\sqrt{97-68 \sqrt{2}}}\right\}\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 2
} |
Proof that $ \frac{3\pi}{8}< \int_{0}^{\pi/2} \cos{\sin{x}} dx < \frac{49\pi}{128}$ Proof that
$$
\frac{3\pi}{8} <
\int_{0}^{\pi/2} \cos\left(\sin\left(x\right)\right)\,\mathrm{d}x <
\frac{49\pi}{128}
$$
Can somebody give me some instruction how to deal with inequality like that? My current idea is:
I see $\frac{3\pi}{8}$ on the left. So I think that I can prove that
$$ \frac{3}{4}<\cos{\sin{x}} $$
And after take integral:
$$ \frac{3x}{4} \rightarrow \frac{3}{4} \cdot \frac{\pi}{2} = \frac{3\pi}{8} $$
But it is not true because
$$ \cos{\sin{x}} \geqslant \cos{1} \approx 0.5403 < 3/4$$
What have I do in such situation?
| Hint for one part, using this inequality and this one
$$\cos{x} \geq 1 - \frac{x^2}{2}$$
we have
$$\int\limits_{0}^{\frac{\pi}{2}}\cos{\sin{x}} dx > \int\limits_{0}^{\frac{\pi}{2}} \left(1-\frac{\sin^2{x}}{2}\right)dx=\frac{3 \pi}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Finding the angle between two tangents of a curve I just got back from my math test. In the test I encountered the following question:
A curve is defined by the equation $$x^3+y^3=3xy$$ Find the angle between the tangents at points (-1,1) and (1,2).
After differentiating implicitly, we get:
$\frac{dy}{dx} = \frac{y-x^2}{y^2-x}$
And the equation of the tangents at the above points are:
$y = \frac{x}{3} + \frac{5}{3}$ and $y = 1$
So, I formed a triangle representing one part of the intersection between the two lines, which looks something like this.
Therefore, the angle between them would be $$\arcsin(4/5)$$ Is my approach right or wrong? If its wrong then how would we do this question?
| Slope formula: $m = \frac{y-x^2}{y^2-x}$
At $P_1 \equiv( -1,1)$, $m_1 = \frac{1-0}{1+1} = 0$
At $P_2\equiv(1,2)$, $m_2 = \frac{2-1}{4-1}=\frac{1}{3}$
Acute angle between the 2 tangents,
$\phi = \tan^{-1}\big\vert\frac{m_1-m_2}{1+m_1m_2}\big\vert = \tan^{-1}\big\vert\frac{0-\frac{1}{3}}{1+0} \big\vert = \tan^{-1}\frac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$ How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$
| Writing $\cos^2\theta=c,\sin^2\theta=s\implies c+s=1$
$\cos^32\theta + 3\cos2\theta$
$=(c-s)^3+3(c-s)$
$=c^3-s^3-3cs(c-s)+3(c-s)$
$=c^3-s^3+3(c-s)(1-cs)$
$=c^3-s^3+3(c-s)((c+s)^2-cs)$
$=c^3-s^3+3(c^3-s^3)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the slope of a normal line to a curve given curve equation and x coordinate What are the steps to solving this problem? (Thank you in advance)
Find the slope of the normal line to the curve (x^2)-(xy)+(y^2) = 7 at the point where x=1. There are two possible answers - accept the SMALLER answer.
Where I started:
y' = (y-2x) / (2y-x)
|
Slope of the normal at a point $P =m= -\frac{dx}{dy}|_P = -\frac{1}{\frac{dy}{dx}}\bigg|_P = -\frac{1}{y'|_P}$
As $x^2 - xy + y^2 =7$, by differentiating w.r.t $x$,
$2x - xy' - y + 2yy' = 0$
$y' = \frac{2x-y}{x-2y}$
and $$-\frac{1}{y'} = -\frac{x-2y}{2x-y}$$
At $x=1$, $1 -y+y^2 = 7 \implies y^2 -y -6 = 0\implies y = 3,-2$
At $(1,3)$ the slope is,
$m_1 = -\frac{1}{y'} = -\frac{1-6}{2-3} = -5$
At $(1,-2)$,
$m_2 = -\frac{1}{y'} = -\frac{1+4}{2+2} = -\frac{5}{4} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the roots of equation based on some geometry hints Plots of the equations $y = 8 - x^2$ and $|y|=\sqrt{8+x}$ are symmetric w.r.t. the line $y=-x$. We have to solve the equation $$8-x^2=\sqrt{8+x}$$
| $y_1=8-x^2$ and $|y_2|=\sqrt{8+x}$ are symmetric w.r.t. the line $y=-x$ implies the tangent lines at the intersection points are reflected over $y=-x$.
Refer to the graph:
$\hspace{3cm}$
The line $y_1=ax+b$ is reflected over $y=-x$ to the line $y_2=\frac1ax+\frac ba$. It implies $a\cdot \frac1a=1$.
Reference: $y_1=8-x^2$ and $|y_2|=\sqrt{8+x}$ (or $y_2=\pm \sqrt{8+x}$).
Finding $x_2$ (blue lines):
$$y_1'(x_2)\cdot y_2'(x_2)=1 \Rightarrow -2x_2\cdot \frac1{2\sqrt{8+x_2}}=1 \Rightarrow x_2=\frac{1-\sqrt{33}}{2}.$$
Finding $x_4$ (purple lines):
$$y_1'(x_4)\cdot y_2'(x_4)=1 \Rightarrow -2x_4\cdot \left(-\frac1{2\sqrt{8+x_4}}\right)=1 \Rightarrow x_4=\frac{1+\sqrt{33}}{2}.$$
Finding $x_1$ and $x_3$ (red and green lines):
$$\begin{cases}y_1'(x_1)\cdot y_2'(x_3)=1\\y_1'(x_3)\cdot y_2'(x_1)=1 \end{cases} \Rightarrow \begin{cases}-2x_1\cdot \frac1{2\sqrt{8+x_3}}=1 \\ -2x_3\cdot \left(-\frac1{2\sqrt{8+x_1}}\right)=1\end{cases} \Rightarrow \begin{cases}x_1^2=8+x_3\\ x_3^2=8+x_1\end{cases} \Rightarrow \\
(x_1-x_3)(x_1+x_3)=x_3-x_1 \Rightarrow x_1=-1-x_3 \Rightarrow x_3^2+x_3-7=0 \Rightarrow \\
x_3=\frac{-1+\sqrt{29}}{2} \Rightarrow x_1=\frac{-1-\sqrt{29}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$ I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$
My Attempt:
I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$
where $I=I(1)$. I then differentiate to get $$I'(a)=\int_0^1 \frac{ax\ln(1+x)}{1-ax}dx=\int_0^1 ax\ln(1+x)\sum_{n=0}^\infty(ax)^ndx=\sum_{n=1}^\infty a^{n+1}\int_0^1 x^{n+1}\ln(1+x)dx$$
Evaluating this integral by integration by parts and geometric series I get
$$\int_0^1 x^{n+1}\ln(1+x)dx=\frac{x^{n+2}}{n+2}\ln(1+x)|_0^1-\frac{1}{n+2}\int_0^1 \frac{x^{n+2}}{1+x}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\int_0^1 x^{n+2}\sum_{k=0}^\infty(-x)^kdx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty(-1)^k\int_0^1 x^{k+n+2}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty\frac{(-1)^k}{k+n+2}=\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)$$ So I arrive at $$I'(a)=\sum_{n=0}^\infty a^{n+1}\left(\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)\right)$$
Re-indexing I get
$$I'(a)=\frac{\ln(2)}{a}\sum_{n=2}^\infty \frac{a^n}{n}+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n}a^{n-1}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n}a^{n-1}$$Integrating both sides from $0$ to $1$ I recover $I(1)$
$$I(1)=\int_0^1 \frac{\ln(2)}{a}\left(-\ln(1-a)-a\right)da+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ Then using the integral equation for the Dilogarithm I arrive at
$$I(1)=\ln(2)\int_0^1 -\frac{\ln(1-a)}{a}da-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$
$$I(1)=\frac{\ln(2)\pi^2}{6}-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$
At this point I could not continue further as I did not know how to simplify the Digamma terms in the sums. I think that by using the Digamma function's relation to the Harmonic Numbers it could be possible to exploit known values of Harmonic sums to arrive at the answer but I could not get the sums in a form where this would work. If anyone could help me continue further or let me know if I am on the right track I would greatly appreciate it. Thank you in advance.
| \begin{align}J&=\int_0^1 \ln(1+x)\ln(1-x)\,dx\\
&=\Big[\left(\left(1+x\right)\ln(1+x)-x-2\ln 2+1\right)\ln(1-x)\Big]_0^1+\int_0^1\frac{\left(1+x\right)\ln(1+x)-x-2\ln 2+1}{1-x}\,dx\\
&=\int_0^1\frac{\left(1+x\right)\ln(1+x)-x-2\ln 2+1}{1-x}\,dx\\
\end{align}
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
\begin{align}J&=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}-\frac{1-x}{1+x}-2\ln 2+1}{x(1+x)}\\
&=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{x(1+x)}\,dx\\
&=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{x}\,dx-\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{1+x}\,dx\\
&=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{x}\,dx+2+2\ln^2 2-4\ln 2\\
&=\int_0^1 \left(\frac{2\ln\left(\frac{2}{1+x}\right)}{x(1+x)}-\frac{2\ln 2}{x}\right)\,dx+2+2\ln^2 2-2\ln 2\\
&=\int_0^1 \left(\frac{2\ln\left(\frac{2}{1+x}\right)}{x}-\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}-\frac{2\ln 2}{x}\right)\,dx+2+2\ln^2 2-2\ln 2\\
&=-2\int_0^1 \frac{\ln(1+x)}{x}\,dx-2\int_0^1 \frac{\ln\left(\frac{2}{1+x}\right)}{1+x}\,dx+2+2\ln^2 2-2\ln 2\\
&=-2\int_0^1 \frac{\ln(1+x)}{x}\,dx+2+\ln^2 2-2\ln 2\\
\end{align}
But,
\begin{align}\int_0^1 \frac{\ln(1+x)}{x}\,dx=\frac{1}{2}\int_0^1 \frac{2x\ln(1-x^2)}{x^2}\,dx-\int_0^1 \frac{\ln(1-x)}{x}\,dx\end{align}
In the second integral perform the change of variable $y=x^2$,
\begin{align}\int_0^1 \frac{\ln(1+x)}{x}\,dx&=-\frac{1}{2}\int_0^1 \frac{\ln(1-x)}{x}\,dx\\
&=-\frac{1}{2}\times -\frac{\pi^2}{6}\\
&=\frac{\pi^2}{12}
\end{align}
Thus,
\begin{align}\boxed{J=2+2\ln^2 2-2\ln 2-\dfrac{\pi^2}{6}}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3246020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Inverse Laplace Transform of $\frac{1}{\lambda\cosh(\sqrt{as}+\sqrt{bs})+(1-\lambda)\cosh(\sqrt{as}-\sqrt{bs})}$. I am trying to find the inverse Laplace transform of some function of the form:
$$
\mathrm{F}\left(s\right) =
\frac{1}{\lambda\cosh\left(\,\sqrt{\, as\,}\, + \,\sqrt{\, bs\, }\right) +
\left(1 - \lambda\right)\cosh\left(\,\sqrt{\, as\,}\, -\,\sqrt{\, bs\,}\right)},
$$
where $\lambda \in \left[0,1\right]$. I've tried to use the residue theorem, however, finding the poles ( roots of the denominator ) seems quite complicated, which I have not figured out, yet.
I've also tried to calculate the integral:
$$
\int_{\sigma - \mathrm{i}\infty}^{\sigma + \mathrm{i}\infty}
\mathrm{F}\left(s\right)\mathrm{e}^{st}\,\mathrm{d}s,
$$
by using some substitution of the form
$u = \mathrm{e}^{\,\sqrt{\, s\,}\,}$, which, however, yields another complicated form from which I can proceed further.
Any ideas, hints, partial solutions, are appreciated!
| Concerning the zeros for
$$
\lambda\cosh(\sqrt{as}+\sqrt{bs})+(1-\lambda)\cosh(\sqrt{as}-\sqrt{bs}) = 0
$$
making $s = x + i y$ and taking the real and the imaginary parts
$$
R=\lambda \cos \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin \left(\frac{1}{2} \arg (x+i
y)\right)\right) \cosh \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \cos \left(\frac{1}{2} \arg
(x+i y)\right)\right)-(\lambda -1) \cos \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin
\left(\frac{1}{2} \arg (x+i y)\right)\right) \cosh \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2}
\cos \left(\frac{1}{2} \arg (x+i y)\right)\right)=0\\
I = \lambda \sin \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin \left(\frac{1}{2} \arg (x+i
y)\right)\right) \sinh \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \cos \left(\frac{1}{2} \arg
(x+i y)\right)\right)-(\lambda -1) \sin \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin
\left(\frac{1}{2} \arg (x+i y)\right)\right) \sinh \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2}
\cos \left(\frac{1}{2} \arg (x+i y)\right)\right)=0
$$
and analyzing the root locus for some values of $a,b,\lambda$ we have good information.
For instance with $a = 3, b = 2, \lambda = 0.9$ no zeroes.
with $a = 3, b = 2, \lambda = 0.1$ the zeroes are at the intersections.
with $a = 3, b = 2, \lambda = 0.5$ no zeroes.
So we can visualize the zeroes position depending on the parameters. This can help with the residues method. Follows the MATHEMATICA script to make the plots
re0 = -(-1 + lambda) Cos[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
Sin[1/2 Arg[x + I y]]] Cosh[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
Cos[1/2 Arg[x + I y]]] + lambda Cos[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
Sin[1/2 Arg[x + I y]]] Cosh[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
Cos[1/2 Arg[x + I y]]]
im0 = -(-1 + lambda) Sin[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
Sin[1/2 Arg[x + I y]]] Sinh[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
Cos[1/2 Arg[x + I y]]] + lambda Sin[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
Sin[1/2 Arg[x + I y]]] Sinh[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
Cos[1/2 Arg[x + I y]]]
a = 3;
b = 2;
lambda = 0.1;
r = 50;
gr1 = ContourPlot[re0 == 0, {x, -r, r}, {y, -r, r}, ContourStyle -> Red, PlotPoints -> 25];
gr2 = ContourPlot[im0 == 0, {x, -r, r}, {y, -r, r}, ContourStyle -> Blue, PlotPoints -> 25];
Show[gr1, gr2]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3248835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Rationalizing $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$ Problem
Rationalize $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$
So, I'm training for the Mexican Math Olimpiad. This is one of the algebra problems from my weekly training.
Before this problem, there was other very similar, after proving it, there's an useful property:
If $a+b+c=0$ then $a^3+b^3+c^3=3abc$
It can be easily proved using the following factorization
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
I tried using this one for the rationalization. I got
$$\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}*\frac{(a^2+b^2+c^2-ab-bc-ac)}{(a^2+b^2+c^2-ab-bc-ac)}=\frac{\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}-\sqrt[3]{ab}-\sqrt[3]{bc}-\sqrt[3]{ac}}{a+b+c-3\sqrt[3]{abc}}$$
But i didn't know how to proceed. I tried looking into it with wolfram alpha and i got there is no racionalization, so i assume $a+b+c=0 $ (not $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$, because the first expression wouldn't have sense) if we want a solution (In fact, if this happens, i've already rationalized it).
So my truly question is not the answer, but how to prove that $a+b+c$ needs to be 0
I would appreciate any help, ideas or suggestions, thanks.
| Maple rationalizes it as
$$ -{\frac { \left( {x}^{2/3}+2\,\sqrt [3]{x}\sqrt [3]{y}-\sqrt [3]{z}
\sqrt [3]{x}+{y}^{2/3}-\sqrt [3]{z}\sqrt [3]{y}+{z}^{2/3} \right)
\left( 3\,{x}^{5/3}\sqrt [3]{y}-6\,{x}^{2/3}{y}^{4/3}+3\,{x}^{2/3}
\sqrt [3]{y}z-6\,{x}^{4/3}{y}^{2/3}+3\,\sqrt [3]{x}{y}^{5/3}+3\,\sqrt
[3]{x}{y}^{2/3}z-{x}^{2}+7\,xy-2\,xz-{y}^{2}-2\,yz-{z}^{2} \right) }{{
x}^{3}+3\,{x}^{2}y+3\,{x}^{2}z+3\,x{y}^{2}-21\,xyz+3\,x{z}^{2}+{y}^{3}
+3\,{y}^{2}z+3\,y{z}^{2}+{z}^{3}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
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