Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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The base of an isosceles triangle; given leg and radius of circumcircle
An isosceles triangle $\triangle ABC$ is given with leg $AC=5$ and $R=\dfrac{25}{6}$ of the circumcircle. Find the base of the triangle.
This was my first sketch. The triangles $AHC$ and $OMC$ are similar, so $\dfrac{CH}{CM}=\dfrac{AC}{CO} \Leftrightarrow \dfrac{CO+OH}{CM}=\dfrac{AC}{CO}.$ When I plug with the values, I get a negative number for $OH$. I realised that this is possible if the triangle is obtuse. How should I notice this from the beginning? I have not studied trigonometry.
| Since in the standard notation $$\frac{abc}{4S}=R,$$ we obtain:
$$\frac{5\cdot5\cdot c}{4\cdot\frac{c\sqrt{25-\frac{c^2}{4}}}{2}}=\frac{25}{6}$$ or
$$\sqrt{25-\frac{c^2}{4}}=3,$$
which gives $$c=8.$$
We'll prove that $$\frac{abc}{4S}=R.$$
Indeed, let $\Phi$ be our circle and $CO\cap\Phi=\{C,D\}$.
Thus, $$\Delta ACH\sim\Delta DCB,$$ which gives
$$\frac{CH}{BC}=\frac{AC}{CD}$$ or
$$\frac{CH}{a}=\frac{b}{2R}$$ or
$$CH=\frac{ab}{2R}.$$
Id est, $$\frac{abc}{4S}=\frac{abc}{4\cdot\frac{c\cdot\frac{ab}{2R}}{2}}=R.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a^2 + b^2 + c^2$ is divisible by $16$, then show that$ a^3 + b^3 + c^3$ is divisible by $64$. Where $a, b, c \in \mathbb{Z}$. If $a^2 + b^2 + c^2$ is divisible by $16$, then show that $a^3 + b^3 + c^3$ is divisible by $64$; where $a, b, c \in \mathbb{Z}$.
I began by proving that if $(a^3+b^3+c^3) -(a^2+b^2+c^2)$ is divisible $16$, then $a^3+b^3+c^3$ will be divisible by $64$. That is equal to $a^3-a^2+b^3-b^2+c^3-c^2$ $= a(a^2-a)+b(b^2-b)+c(c^2-c)$.
But this doesn't give a clear path for the question.
Any help?
| We may write:
$$16\big|a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)$$
⇒ $$4\big| a+b+c$$
Also:
$$8\big| ab+bc+ac$$
Since the phrases are symmetric if one is a multiple of 4 others will also be, Hence a, b and c are multiple of 4 so the sum of their cubes are a multiple $4^3=64$.
| {
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Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$ From Mr. Michael Rozenberg solution:
For $a,b,c>0$$,$ prove that$:$
$$(a+b+c)^2\prod_{cyc}(a+b)\geq4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab,$$
I found two SOS proof:
1) $$\text{LHS-RHS}={\frac { \left( a-b \right) ^{2}\cdot \text{M}+ab \left( {a}^{2
}-2\,ab+ca+{b}^{2}+bc-2\,{c}^{2} \right) ^{2}}{a+b}}$$
Where $$\text{M}=\left( 2\,ab-ca-bc+{c}^{2} \right) ^
{2}+c \left( -c+a+b \right) ^{2} \left( a+b \right)$$
2) $$\text{LHS-RHS}=c \left( a-b \right) ^{2} \left( a+b-c \right) ^{2}+a \left( b-c \right) ^{2} \left( b+c-a \right) ^{2}+b \left( c-a \right) ^{2} \left( c+a-b \right) ^{2}\geqq 0$$
| We write the inequality as
$$\frac{(a+b+c)^2}{ab+bc+ca} \geqslant \frac{\displaystyle 4 \sum (a^2b+ab^2)}{(a+b)(b+c)(c+a)},$$
equivalent to
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+2 \geqslant \frac{4[(a+b)(b+c)(c+a)-2abc]}{(a+b)(b+c)(c+a)},$$
or
$$\frac{a^2+b^2+c^2}{ab+bc+ca} + \frac{8abc}{(a+b)(b+c)(c+a)} \geqslant 2.$$
Which is know.
| {
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Integer solution of $a+b+c=15$ with restrictions on $a, b$ and $c$ I'm checking this problem:
Find the integer solutions of $a+b+c=15$ if $a$ is multiple of 3, $b$ is less than 10 and $c$ is multiple of 2. With $a, b, c ≥ 0$
We can make a series of polynomials with combinatorics. The polynomials would be:
$$P_a(x)=1+x^3+x^6+x^9+\dots$$
$$P_b(x) = 1 +x+x^2+x^3+\dots + x^9 = \frac{1-x^{10}}{1-x}$$
$$P_c(x) = 1+x^2 + x^4 + x^6 + \dots$$
I want to know if my $P_a(x)$ and $P_c(x)$ are stated correctly and how can they be simplified so that I get to $x^{15}$ in an easier way.
Big thanks for your help.
| Um.... why do work when simply counting them is magnitudes of effort easier.
$a = 3k; k=0... 5$. $c$ is even so $b$ is odd if and only if $a$ and $k$ are even. So $b$ can be any even/odd number $\le \min (15-a, 10)$ and $c = 15-a -b$.
Just count them. If $a=0$ then $b=1,3,...,9$ and $c= 15-b$ and there are $5$ such solutions.
If $a=3$ the $b=0,...,8$ and $c = 12-b$ and there are $5$ such solutions.
If $a = 6$ then $b=1,....9$ and $c = 9-b$ and there are $5$ such solutions.
If $a = 9$ then $b=0,2,4,6$ and $c =6-b$ and there are $4$ such solutions.
If $a= 12$ then $b=1,3$ and $c=3-b$ and there are $2$ such solutions.
If $a = 15$ then $b = 0$ and $c = 0$ and there is $1$ such solution.
So there are $5+5+5+4+2 +1=22$ solutions.
| {
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$ x\cos y + y \cos x = \pi$ , how do I find $y''(0)$ in easiest way possible? I considered brute force differentiating but that is very hard. I also tried expanding the cosines and '$y$' as a Taylor series but I don't think that helps much either
| $x\cos y + y\cos x = \pi$
What is $y(0)$?
$0\cos (y(0)) + y(0)\cos 0 = \pi\
y(0) = \pi\
What is $y'$?
$\cos y -y\sin x + (-x\sin y + \cos x)y' = 0\\
y'= \frac {y\sin x - \cos y}{\cos x - x\sin y} = \frac {u}{v}\\
$
And $y'(0)$?
$u(0) = \pi \sin 0 - \cos \pi = 1\\
v(0) = \cos 0 - 0\sin \pi = 1\\
y'(0) = 1$
$y'' = \frac {u'v - uv'}{v^2}\\
u' = y\cos x + (\sin x + \sin y) y'\\
u'(0) = \pi\\
v' = -sin x - sin y + (-x\cos y) y'\\
v'(0) = 0\\
y''(0) = \pi$
| {
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Double integral over $x^2+y^2 \le 1$ I am trying to calculate the double integral
$\displaystyle \iint_{x^2+y^2\leq 1} (\sin x+y+3)\,dA$
Here are my attempts so far:
1) I used polar coordinates
$ x= r \sin(\theta)$
$y= r \cos (\theta)$
where $\theta \in [0,2 \pi]$ and $r \in [0,1]$ which gives
$\displaystyle \int_0^{2\pi} \int_0^1 \bigg(\sin\big(r \cos(\theta)\big)+ r \sin(\theta)+3 \bigg)r\, dr\, d\theta$
and stuck with fiding antiderivative of the function with respect to r
$ r \sin\big(r \cos(\theta)\big)$
2) I tried to divide the region into parts that
$A \cup B = \{ (x,y) : x^2+y^2 \leq 1\}$ where $A=\{ (x,y) : x^2+y^2 \leq 1$ and $x\geq 0 \}$ and $B=\{ (x,y) : x^2+y^2 \leq 1$ and $x <0 \}$ which gives me
$ \quad\displaystyle \int_{-1}^1 \int _0^{\sqrt{1-x^2}} \big( \sin x +y+3 \big)\,dy \,dx + \int_{-1}^1 \int _0^{\sqrt{1-x^2}} \big( \sin x +y+3 \big)\,dy\,dx$
$\displaystyle =\int_{-1}^1 \big( \sqrt{1-x^2} \sin x + \frac{1-x^2}{2}+3 \sqrt{1-x^2} \big)\,dx + \dotsm $
and stuck with finding antiderivative of the function
$\sqrt{1-x^2} \sin x$.
I will be glad to hear any tips. Thanks in advance.
| There is no reason not to split this into three separate integrals:
$$\iint_{x^2+y^2\leq 1} \sin x\;dA+\iint_{x^2+y^2\leq 1}y\;dA+\iint_{x^2+y^2\leq 1}3\;dA$$
You can do the third, I think. For the first, note that $\sin x$ is antisymmetric about the $y$-axis; and for the second, note that $y$ is antisymmetric about the $x$-axis. So there is nothing to evaluate!
| {
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Solve equation: $\log_2 \left(1+ \frac{1}{a}\right) + \log_2 \left(1 +\frac{1}{b}\right)+ \log_2 \left(1 + \frac{1}{c}\right) = 2$ $$
\log_2 \left(1 + \frac{1}{a}\right) + \log_2 \left(1 + \frac{1}{b}\right)+ \log_2 \left(1 + \frac{1}{c}\right) = 2 \quad \text{where $a$, $b$, $c \in N$.}
$$
Apparently, the answer is $a= 1$, $b =2$, and $c\space = 3$.
When I asked my math teacher I was told that the solution involved a bit of number theory, but didn't recieve a complete explanation.
Could someone clear that up for me?
Edit: I had made a mistake in typing the question. I had left it as:
$
\log_2 \left(a + \frac{1}{a}\right) + \log_2 \left(b + \frac{1}{b}\right)+ \log_2 \left(c + \frac{1}{c}\right) = 2 \quad \text{where $a$, $b$, $c \in N$.}
$
My apologies for causing confusion.
| The problem, as you have written it, has no solution.
Simplifying the LHS, we get $$(a^2+1)(b^2+1)(c^2+1) = 4abc$$
But, by the AM-GM inequality, we get $x^2+1\ge2x$, which gives
$$(a^2+1)(b^2+1)(c^2+1) \ge 8abc$$
| {
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Ahmed integral revisited $\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx$ How to prove
$$\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} dx=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{4}+\frac{\pi }{4}\arctan \left(\frac{1}{\sqrt{5}}\right)\;?$$
I came across this Ahmed integral on the site "Art of problem solving", and have found no proof so far. (These two problems seems to be related though). Any help will be appreciated!
| To evaluate that integral we can use Feynman's trick:
$$I=\int _0^1\frac{\arctan \left(\sqrt{x^2+4}\right)}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx$$
$$I\left(a\right)=\int _0^1\frac{\arctan \left(a\sqrt{x^2+4}\right)}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx$$
$$I'\left(a\right)=\int _0^1\frac{1}{\left(x^2+2\right)\left(a^2x^2+4a^2+1\right)}\:dx=\frac{1}{2a^2+1}\int _0^1\frac{1}{x^2+2}-\frac{a^2}{a^2x^2+4a^2+1}\:dx$$
$$=\frac{1}{2a^2+1}\left(\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}}-\frac{a\arctan \left(\frac{a}{\sqrt{4a^2+1}}\right)}{\sqrt{4a^2+1}}\right)$$
Now lets integrate again:
$$\int _1^{\infty }I'\left(a\right)\:da=\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}}\int _1^{\infty }\frac{1}{2a^2+1}\:da-\underbrace{\int _1^{\infty }\frac{a\arctan \left(\frac{a}{\sqrt{4a^2+1}}\right)}{\sqrt{4a^2+1}\left(2a^2+1\right)}\:da}_{a=\frac{1}{x}}$$
$$\frac{\pi }{2}\int _0^1\frac{1}{\left(x^2+2\right)\sqrt{x^2+4}}dx-I\:=\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)}{2\sqrt{2}}\left(\frac{\pi \sqrt{2}}{2}-\sqrt{2}\arctan \left(\sqrt{2}\right)\right)-\int _0^1\frac{\arctan \left(\frac{1}{\sqrt{x^2+4}}\right)}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx$$
$$=\frac{\pi \arctan \left(\frac{1}{\sqrt{2}}\right)}{4}-\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{2}-\frac{\pi }{2}\int _0^1\frac{1}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx+\underbrace{\int _0^1\frac{\arctan \left(\sqrt{x^2+4}\right)}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx}_{I}$$
$$-2I\:=\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{4}-\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{2}-\pi \underbrace{\int _0^1\frac{1}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx}_{t=\arctan \left(\frac{x}{\sqrt{x^2+4}}\right)}$$
$$I\:=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{4}+\frac{\pi }{4}\int _0^{\arctan \left(\frac{1}{\sqrt{5}}\right)}\:dt$$
$$\boxed{I=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{4}+\frac{\pi }{4}\arctan \left(\frac{1}{\sqrt{5}}\right)}$$
This numerically agrees with Wolfram Alpha.
| {
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${\lim_{x\rightarrow \infty}}(\sqrt{x^2+2x+3} - \sqrt{x^2+3})^{x}$ $${\lim_{x\rightarrow \infty}}(\sqrt{x^2+2x+3} - \sqrt{x^2+3})^{x}$$
I tried taking log both sides (on paper).
After taking log how do I proceed? You get $\infty$ * $(\infty-\infty$). But $\infty-\infty$ could be any number. How can I take this as 0? Even if I take it as zero, the question becomes nasty.
Please help me solve this limit. It must be solved without using series expansion as this isn't taught. Only L'Hopital and basic algebra.
| $$A=\left(\sqrt{x^2+2x+3} - \sqrt{x^2+3}\right)^{x}$$ For what is inside the parenthesis, use Taylor
$$\sqrt{x^2+2x+3}=x+1+\frac{1}{x}-\frac{1}{x^2}+O\left(\frac{1}{x^3}\right)$$
$$\sqrt{x^2+3}=x+\frac{3}{2 x}+O\left(\frac{1}{x^3}\right)$$
$$\sqrt{x^2+2x+3} - \sqrt{x^2+3}=1-\frac{1}{2 x}-\frac{1}{x^2}+O\left(\frac{1}{x^3}\right)$$
$$\log(A)=x \log\left(1-\frac{1}{2 x}-\frac{1}{x^2}+O\left(\frac{1}{x^3}\right) \right)=-\frac{1}{2}-\frac{9}{8 x}+O\left(\frac{1}{x^2}\right)$$
$$A=e^{\log(A)}=\frac{1}{\sqrt{e}}-\frac{9}{8 \sqrt{e} x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.
| {
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First trigonometric differential equation
Show that:
$$\tan(x) \frac {dy}{dx}-y=\sin^2(x)+2\sec(x)$$ where $y=\sin^2(x)-2\cos(x)$
I get:
$\frac {dy}{dx}=2sin(x)cos(x)+2sin(x)$
=$tan(x)(2sin(x)cos(x)+2sin(x))-sin^2(x)-2cos(x)$
From here I go into many directions but not towards the RHS. Guidance is much appreciated
| You have a little sign mistake here:
$$E=\tan(x)(2\sin(x)\cos(x)+2\sin(x))-\sin^2(x)\color{red}{+2\cos(x)}$$
$$E=2\sin^2(x)+2\dfrac {\sin^2(x)}{\cos x}-\sin^2(x)+2\cos(x)$$
$$E=\sin^2(x)+2\dfrac {\sin^2(x)}{\cos x}+2\cos(x)$$
$$E=\sin^2(x)+2\dfrac {(\sin^2(x)+\cos^2(x))}{\cos x}$$
$$E=\sin^2(x)+\dfrac 2{\cos x}=\sin^2(x)+ 2{\sec x}$$
You can also solve the DE:
$$\tan(x) \frac {dy}{dx}-y=\sin^2(x)+2\sec(x)$$
Multiply by $\cos x$:
$$\sin(x) \frac {dy}{dx}-y\cos x=\sin^2(x)\cos x+2$$
Divide by $\sin^2 x$:
$$( \frac {y}{\sin x})'=\cos x+2\csc^2(x)$$
Integrate:
$$ \frac {y}{\sin x}=\sin x+2\int \csc^2(x)dx+C_1$$
$$ y(x)=\sin^2 x-2\sin x\cot(x)+C_1\sin x$$
$$ y(x)=\sin^2 x-2\cos(x)+C_1\sin x$$
Take $C_1=0$
| {
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All prime divisors of $\frac{x^m+1}{x+1}$ are of the form $2km+1$. Let $m$ be an odd prime and $x$ be the product of all primes of the form $2km+1$. Then all prime divisors of $\frac{x^m+1}{x+1}$ are of the form $2km+1$.
What I know is that $\frac{x^m+1}{x+1}$ is an integer.
Here is the link to the answer which prompted this question.
Can anyone help me how to prove this.
Any help would be appreciated. Thanks in advance.
| We have $2 \nmid \frac{x^m+1}{x+1}$. Let odd prime $p$ divide $\frac{x^m+1}{x+1}$ :
Case $1$ : $m \mid (p-1)$
We clearly have $p=mq+1$ for some $q \in \mathbb{N}$. As $p$ is an odd prime, $mq+1$ is odd, and thus, $mq$ is even. Moreover, $m$ is an odd prime, thus, $q=2k$ for some $k \in \mathbb{N}$. Substituting:
$$p=2km+1$$
which proves that our prime divisor is of the required form.
Case $2$ : $m \nmid (p-1)$
We have:
$$p \mid (x^m+1) \implies p \mid(x^{2m}-1) \implies p \mid(x^{\gcd(2m,p-1)}-1)$$
by Fermat's Little Theorem.
Since $m$ is an odd prime not dividing $p-1$, it follows:
$$\gcd(2m,p-1)=\gcd(2,p-1)=2$$
This shows us that $p \mid (x^2-1)$.
We thus either have $p \mid (x-1)$ or $p \mid (x+1)$.
Subcase $1$ : $p \mid (x-1)$
We have:
$$p \mid (x-1) \implies p \mid (x^m-1)$$
Since $p \mid (x^m+1)$, it follows that $(x^m+1)-(x^m-1)=2$ is also divisible by $p$ which is a contradiction as $p$ is an odd prime.
Subcase $2$ : $p \mid (x+1)$
This is the same as $x \equiv -1 \pmod{p}$. But then:
$$\frac{x^m+1}{x+1} \equiv x^{m-1}-x^{m-2}+\cdots+1 \equiv 1-(-1)+1-(-1)+\cdots+1 \equiv m \pmod{p}$$
As $p \mid \frac{x^m+1}{x+1}$, it follows that $p \mid m$. Since $p$ and $m$ are both odd primes, we must thus have $p=m$.
However:
$$p \mid (x^m+1) \implies m \mid (x^m+1)$$
Note that as all the prime factors of $x$ are $1 \pmod{m}$, we have $x \equiv 1 \pmod{m}$. Then:
$$0 \equiv x^m+1 \equiv 1+1 \equiv 2 \pmod{m} \implies m \mid 2$$
and this is once again a contradiction since $m$ is an odd prime.
Thus, we have proved that all prime divisors of $\frac{x^m+1}{x+1}$ are of the form $2km+1$.
| {
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Is this series convergent? $1 + 1/2 + 1/2 + 1/4 + 1/4 + 1/4 + 1/4 + ...$ Is this series convergent? How to prove?
$$1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{8} \ (8 \times 1/8) + \frac{1}{16} + ...$$
It's equal to $\{1 + 1 + 1 + 1 + ...\}$, when I take the $(n+1)^{th}$ number and $n^{th}$ number, the ratio of them can be $\frac{1}{2}$ or $1$. I don't think this is a convergent series, but what is it?
Thanks!
| The most direct way to see divergence is by noticing that
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\leq 1+\frac{1}{2}+\frac{1}{2}+\cdots$$
where the series on the right hand side is what you have. Since the LHS diverges as the harmonic series, so must the RHS.
| {
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Limit Using squeeze theorem Find the limit of the following function as $x \rightarrow 0$
$$
\frac{|x|}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}} \sin \left(\frac{1}{3 \sqrt{x}}\right)
$$
$\lim _{x \rightarrow 0} \frac{|x|}{\sqrt{\left(x^{4}+4 x^{2}+77\right)}} \sin \left(\frac{1}{3 \sqrt{x}}\right)$
My approach,
applying squeeze theorem,
$
-1 \leq \sin \left(\frac{1}{3 \sqrt{x}}\right) \leq 1
$
$-\frac{|x|}{\sqrt{x^{2}+4 x^{2}+7}} \leq \frac{|x|}{\sqrt{x^{4}+4 x^{2}+7}} \sin \left(\frac{1}{\sqrt{3x}}\right) \leq \frac{|x|}{\sqrt{x^{4}+4 x^{2}+7}}$
$\operatorname{Now,}_{\operatorname{limit}_{x \rightarrow 0} \frac{-x}{\sqrt{x^{4}+4 x^{2}+7}}}=\ _{x \rightarrow 0} \frac{x}{\sqrt{x^{4}+4 x^{2}+7}}=0$
Hence answer is zero.
Am I correct?
| Other approach:
Let $f(x)=\frac{|x|}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}} \sin \left(\frac{1}{3 \sqrt{x}}\right)$
We note that $f(x)=-f(-x)$, hence $f$ is odd, we can study it for $x > 0$.
Remembering that
$$
\lim_{x \to 0} \frac{\sin x}{x} = 1
$$
we get
$$
\begin{split}
\lim_{x \to 0^+} \frac{x}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}} \sin \left(\frac{1}{3 \sqrt{x}}\right) &= \lim_{x \to 0^+} \frac{x}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}} \frac{\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\frac{1}{3 \sqrt{x}}} \frac{1}{3 \sqrt{x}} \\
&=\lim_{x \to 0^+} \frac{x^{1-\frac{1}{2}}}{3} \cdot \lim_{x \to 0^+} \frac{\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\frac{1}{3 \sqrt{x}}}\cdot \lim_{x \to 0^+} \frac{1}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}} \\
&=\lim_{x \to 0^+} \frac{\sqrt x}{3} \cdot 1 \cdot \frac{1}{\sqrt 7} = 0 \cdot 1 \cdot \frac{1}{\sqrt 7}= 0
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3701891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the percentage of the area of each circle that overlaps.
The diagram shows two overlapping circles with centres $X$ and $Y$. Both circles have radius $r$ cm and the distance between the centres, $XY$, is $1.5r$ cm.
I got the answer $18.2$% (to $3$s.f), but the answer key states otherwise ($14.4$%).
My Steps are:
*
*Find the area of the circle ($\pi r^2$)
*Find the area of $AXBY$ ($r^2$)
*Subtract sector $XAB$ to get the area of $ABY$ ($ABY=(r^2) - (\pi r^2)/4$)
*Area of $AXBY$ - $2$(Area of $ABY$)= Area of overlap = $(r^2)((\pi/2)-1)$
*Percentage= $((\pi-2)/2\pi)x100$% = $18.2$% ($3$s.f.)
| Let's assume that $\text{r}=1$ and we will work through the solution and then the OP can generalize it.
Well, we know that the equation of a circle is given by:
$$\left(x-\text{a}\right)^2+\left(\text{y}-\text{b}\right)^2=\text{r}^2\tag1$$
Where $\left(\text{a},\text{b}\right)$ are the center coordinates of the circle and $\text{r}$ is the radius of the circle.
So, in your case, we have two circles so we write $1=\text{r}_1=\text{r}_2$ and we have $\left(\text{a}_1,\text{b}_1\right)$ and $\left(\text{a}_2,\text{b}_2\right)$. Using Mathematica I used the following code:
ContourPlot[{(x + (-3/4))^2 + (y + 0)^2 ==
1^2, (x + (3/4))^2 + (y + 0)^2 == 1^2}, {x, -2, 2}, {y, -2, 2},
GridLines -> {{0, 3/4, -3/4}, {0}}]
And it gave me:
So, we have $\left(\text{a}_1,\text{b}_1\right)=\left(-\frac{3}{4},0\right)$ and $\left(\text{a}_2,\text{b}_2\right)=\left(\frac{3}{4},0\right)$.
The surface area of the part where the circle's overlap can be found using:
$$\mathcal{A}_1:=4\int_0^\frac{1}{4}\frac{\sqrt{\left(1-4x\right)\left(7+4x\right)}}{4}\space\text{d}x=4\text{arccot}\left(\sqrt{7}\right)-\frac{3\sqrt{7}}{8}\tag2$$
And the total area of both the circle's is given by:
$$\mathcal{A}_2:=2\int_{-\frac{3}{4}-1}^{\frac{3}{4}+1}\left(\text{K}_1+\text{K}_2\right)\space\text{d}x=\frac{3 \sqrt{7}}{8}+4 \arctan\left(\sqrt{7}\right)\tag3$$
Where $\text{K}_1=\theta\left(x\right)\cdot\frac{\sqrt{\left(7-4x\right)\left(1+4x\right)}}{4}$ and $\text{K}_2=\theta\left(-x\right)\cdot\frac{\sqrt{\left(1-4x\right)\left(7+4x\right)}}{4}$.
So, for the percentage we get:
$$\eta=\frac{\mathcal{A}_1}{\mathcal{A}_2}=\frac{4\text{arccot}\left(\sqrt{7}\right)-\frac{3\sqrt{7}}{8}}{\frac{3 \sqrt{7}}{8}+4 \arctan\left(\sqrt{7}\right)}\approx0.0777567\space\rightarrow\space\eta\approx7.77567\text{%}\tag4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Let $a, b, c>0$. Prove that $\sum \limits_{cyc}{\frac{a}{b+c}\left(\frac{b}{c+a}+\frac{c}{a+b}\right)}\le \frac{(a+b+c)^2}{2(ab+bc+ca)}$ Reducing this whole expression i finally came to this
$$\sum \limits_{cyc}\left(ab^4+a^4b+a^2b^2c\right)\geq \sum \limits_{cyc}\left(a^3b^2+a^2b^3+a^3bc\right)$$
Here I am stuck. I can't prove this.
So I thought maybe I should try in another way.
Let $3u=a+b+c$, $3v^2=ab+bc+ca$ and $w^3=abc$
Hence the whole expression comes to this inequality
\begin{align*} (a+b+c)^2&(a+b)(b+c)(c+a) \\
& \geq 4(ab+bc+ca)(ab^2+a^2b+bc^2+b^2c+ca^2+c^2a)\\
\implies (a+b+c)^2&\left((a+b+c)(ab+bc+ca)-abc\right)\\
& \geq 4(ab+bc+ca)\left((a+b+c)(ab+bc+ca)-3abc\right)\\
\implies (3u)^2\left(3u\times3v^2-w^3\right)&\geq 4\times3v^2\left(3u\times3v^2-3w^3\right)\\
\implies 9u^2(9uv^2-w^3)&\geq 12v^2(9uv^2-3w^3)\\
\implies 9u^3v^2-u^2w^3&\geq 12uv^4-4v^2w^3\end{align*}
Here again, I am stuck. How can I prove this inequality?
| We need to prove that:
$$(a+b+c)^2\prod_{cyc}(a+b)\geq4(ab+ac+bc)\sum_{cyc}(a^2b+a^2c)$$ or
$$\sum_{sym}(a^4b-a^3b^2-a^3bc+a^2b^2c)\geq0.$$
Now, let $a\geq b\geq c$.
Thus, $$\sum_{sym}(a^4b-a^3b^2-a^3bc+a^2b^2c)=\sum_{cyc}(a^4b-a^3b^2-a^2b^3+ab^4-abc(a^2-2ab+b^2))=$$
$$=\sum_{cyc}(a-b)^2(ab(a+b)-abc)=\sum_{cyc}(a-b)^2ab(a+b-c)\geq$$
$$\geq(a-c)^2ac(a+c-b)+(b-c)^2bc(b+c-a)\geq$$
$$\geq(b-c)^2ac(a-b)+(b-c)^2bc(b-a)=(a-b)^2(b-c)^2c\geq0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\int \frac{x^2\,dx}{(a-bx^2)^2}$ How do I integrate $\int \frac{x^2\,dx}{(a-bx^2)^2}$
I've tried substitution and partial fraction decomposition, but I'm not getting anywhere.
| This is almost similar to the other integral you posted. The trig substitution $x=\frac{\sqrt a}{\sqrt b} \sin t\implies dx = \frac{\sqrt a}{\sqrt b} \cos t dt $ works here too, after which you get $$ \frac{1}{b\sqrt{ab}}\int\frac{\sin^2t}{\cos^3 t} dt =\frac{}1{b\sqrt{ab}}\int\tan^2t\ \sec t \ dt $$
Now substitute $u=\sec t\implies du = \sec t\tan tdt$ $$\frac{1}{b\sqrt{ab}}\int\sqrt{u^2-1}\ du \\ =\frac{1}{b\sqrt{ab}} \left[ \frac u2\sqrt{u^2-1} -\frac 12\log(u+\sqrt{u^2-1})\right]+C \\ =\frac{1}{b\sqrt{ab}} \left[ \frac{\sec t}{2}\tan t-\frac 12\log (\sec t +\tan t) \right]+C $$
Now, $\sin ^2t =\frac ba x^2 =1-\frac{1}{\sec^2t} \implies \sec t=\sqrt{\frac{a}{a-bx^2}} $
and $\tan t =\sqrt{\frac{a}{a-bx^2}-1} =\sqrt{\frac{bx^2}{a-bx^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Integrate $\int_0^{2\pi}\frac{\ln(a + b\cos x)}{c + d\cos x} dx$, Residue theorem I have recently been given a challenge problem in my Complex Analysis class. Suppose $a > b > 0$ and $c > d > 0$. Evaluate
$$\int_0^{2\pi} \frac{\ln(a + b\cos x)}{c + d\cos x} dx$$
using the Residue Theorem. Unfortunately, I don't even know where to begin with this one. I have managed to solve the integral where the integrand is
$$\frac{a + b\cos x}{c + d\cos x}$$
where the contour I used was the usual a square but I'm not sure whether that can be applied here. If anyone could provide any assistance, that would be greatly appreciated!
| Choose $r, s \in (0, 1)$ so that
$$\frac{b}{a} = \frac{2r}{1+r^2}, \qquad \frac{d}{c} = \frac{2s}{1+s^2}.$$
Then
$$ \left| (1 - re^{i\theta})(1 - re^{-i\theta}) \right| = 1 + r^2 - 2r\cos\theta. $$
From this, we get
\begin{align*}
\log(1 + r^2 - 2r\cos\theta)
&= \log\left|1 - re^{i\theta}\right| + \log\left|1 - re^{-i\theta}\right| \\
&= \operatorname{Re}\left[ \log\left(1 - re^{i\theta}\right) + \log\left(1 - re^{-i\theta}\right) \right] \\
&= 2 \operatorname{Re}\left[ \log\left(1 - re^{i\theta}\right) \right].
\end{align*}
This, together with $z^2 - 2(c/d)z + 1 = (z - s)(z - s^{-1})$, shows that
\begin{align*}
\int_{0}^{2\pi} \frac{\log(a + b\cos\theta)}{c+d\cos\theta} \, \mathrm{d}\theta
&= \int_{0}^{2\pi} \frac{\log(a - b\cos\theta)}{c-d\cos\theta} \, \mathrm{d}\theta \\
&= \int_{0}^{2\pi} \frac{\log\bigl(\frac{a}{1+r^2}\bigr) + \log(1 + r^2 - 2r\cos\theta)}{c-d\cos\theta} \, \mathrm{d}\theta \\
&= \operatorname{Re}\left[ \int_{0}^{2\pi} \frac{\log\bigl(\frac{a}{1+r^2}\bigr) + 2\log(1 - re^{i\theta})}{c-d\cos\theta} \, \mathrm{d}\theta \right] \\
&= \operatorname{Re}\left[ \frac{2i}{d} \int_{|z| = 1} \frac{\log\bigl(\frac{a}{1+r^2}\bigr) + 2\log(1 - rz)}{z^2-2(c/d)z+1} \, \mathrm{d}z \right] \tag{$z=e^{i\theta}$} \\
&= \operatorname{Re}\left[ -\frac{4\pi}{d} \, \underset{z=s}{\mathrm{Res}} \, \frac{\log\bigl(\frac{a}{1+r^2}\bigr) - 2\log(1 - rz)}{z^2-2(c/d)z+1} \right].
\end{align*}
Computing the residue and simplifying, we get
\begin{align*}
&\int_{0}^{2\pi} \frac{\log(a + b\cos\theta)}{c+d\cos\theta} \, \mathrm{d}\theta \\
&= \frac{2\pi}{\sqrt{c^2-d^2}} \left[ \log\left(\frac{a+\sqrt{a^2-b^2}}{2}\right) + 2 \log\left(1 - \frac{a-\sqrt{a^2-b^2}}{b} \cdot \frac{c-\sqrt{c^2-d^2}}{d}\right)\right].
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Find Transition matrix given two basis Consider the ordered bases $B=(\begin{bmatrix}2 & -1\\0 & −1\end{bmatrix},\begin{bmatrix}3 & -3\\0& −1\end{bmatrix}, \begin{bmatrix}-2 & -3\\0 & 2\end{bmatrix})$ and $C=(\begin{bmatrix}-2 & 2\\0& 2\end{bmatrix},\begin{bmatrix}2 & -3\\0& −3\end{bmatrix},\begin{bmatrix}0 & -2\\0& 0\end{bmatrix})$ for the vector space $V$ of upper triangular $2×2$ matrices.
Find the transition matrix from $C$ to $B$.
I have tried solving for linear combinations of $C$ that would create each matrix in $B$. However, the result was not right. Please help with how to proceed.
| I believe the following approach may be what you are looking for. Consider adding the missing basis matrix to each set, unrolling into $4 \times 4$ matrices, then taking the appropriate products.
The procedure I am suggesting is adding $\begin{bmatrix}
0 & 0 \\
1 & 0 \\
\end{bmatrix}$ to each set:
$$
B=
\begin{bmatrix}
2 & -1 \\
0 & -1 \\
\end{bmatrix},
\begin{bmatrix}
3 & -3 \\
0 & -1 \\
\end{bmatrix},
\begin{bmatrix}
-2 & -3 \\
0 & 2 \\
\end{bmatrix},
\begin{bmatrix}
0 & 0 \\
1 & 0 \\
\end{bmatrix}
$$
$$
C=
\begin{bmatrix}
-2 & 2 \\
0 & 2 \\
\end{bmatrix},
\begin{bmatrix}
2 & -3 \\
0 & -3 \\
\end{bmatrix},
\begin{bmatrix}
0 & -2 \\
0 & 0 \\
\end{bmatrix},
\begin{bmatrix}
0 & 0 \\
1 & 0 \\
\end{bmatrix}
$$.
Now, unpacking each basis to form a larger matrix by extracting each entry in a clockwise order from top left to form each column:
$$
B_{4}=
\begin{bmatrix}
2 & 3 & -2 &0 \\
-1 & -3 & -3 & 0\\
-1 & -1& 2 & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
$$
$$
C_{4}=
\begin{bmatrix}
-2 & 2 & 0 &0 \\
2 & -3 & -2 & 0\\
2 & -3& 0 & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
$$
We want to change the basis from $B \rightarrow C$, which requires $C_{4}^{-1}$:
$$
C_{4}^{-1}=
\begin{bmatrix}
-\frac{3}{2} & 0 & -1 &0 \\
-1 & 0 & -1 & 0\\
0 & -\frac{1}{2}& \frac{1}{2} & 0\\
0 &0 & 0 & 1\\
\end{bmatrix}
$$
Lastly, form the product $M=
C_{4}^{-1}B_{4}$:
$$
M=
C_{4}^{-1}B_{4}=
\begin{bmatrix}
-2 & -\frac{7}{2} & 1 &0 \\
-1 & -2 & 0 & 0\\
0 & 1& \frac{5}{2} & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
$$.
I believe this final matrix $M$ is the transition matrix from $B\rightarrow C$.
To test this, consider what the first basis in $B$ would be in $C$. In otherwords, we want to compute:
$$
M\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
\end{bmatrix}=
\begin{bmatrix}
-2 & -\frac{7}{2} & 1 &0 \\
-1 & -2 & 0 & 0\\
0 & 1& \frac{5}{2} & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
\end{bmatrix}
=
\begin{bmatrix}
-2 \\
-1 \\
0 \\
0 \\
\end{bmatrix}
$$.
This means, we take $-2$ of the first basis in $C$ and add it to $-1$ of the second basis in $C$:
$$
\begin{bmatrix}
2 & -1 \\
0 & -1 \\
\end{bmatrix}
=
-2
\begin{bmatrix}
-2 & 2 \\
0 & 2 \\
\end{bmatrix}
-
\begin{bmatrix}
2 & -3 \\
0 & -3 \\
\end{bmatrix}
?
$$
$$
\begin{bmatrix}
2 & -1 \\
0 & -1 \\
\end{bmatrix}
=
\begin{bmatrix}
4 & -4 \\
0 & -4 \\
\end{bmatrix}
+
\begin{bmatrix}
-2 & 3 \\
0 & 3 \\
\end{bmatrix}
$$
This seems to check out.
I hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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solving simultaneous linear congruences and simplifying the mod I wanted to solve the following:
$x\equiv2 (\text{mod 3})$
$x\equiv5 (\text{mod 6})$
$x\equiv7 (\text{mod 8})$
I rewrote the first congruence as $x=2+3a$ and in (mod 6), this can be written $2+3a\equiv5(\text{mod 6})$, which has solution $a\equiv1 (\text{mod 6})$, which implies $a=1+6k$.
Substituting this in for our first expression for $x$ gives $x=5+18k$ which can be written as $x\equiv5 (\text{mod 18})$.
I repeated this process with new expression for $x$ and the expression mod 8 and found that $x\equiv23 (\text{mod 144})$, however apparently the answer is $x\equiv23 (\text{mod 24})$. I am confused about how they simplified the mod?
| Chinese remainder thereom says we can solve a system uniquely for the product of the moduli if the moduli are relatively prime. In this case the are not relatively prime. But, if the equations are consistent, we can solve them uniquely for the least common multiple of the moduli... if the equations are consistent.
We can solve this for $\pmod {\operatorname{lcm}(3*6*8)= 24}$. but breaking $24$ into relatively prime components $3*8$.
The thing to keep in mind is that if $a\equiv k \pmod {mn}$ then $a\equiv k \pmod n$ (because $mn|a-k$ so $n|a-k).
So $x \equiv 2 \pmod 3$
And $x \equiv 5\pmod 6$ so $x\equiv 5\equiv 2\pmod 3$. (Two things to note: 1) this is why the equations must be consistent. if we had $x \equiv 2\pmod 3$ and $x \equiv 4\pmod 6$ we'd have $x \equiv 4\equiv 1 \pmod 3$ and that would not be consistent. $x\equiv 2 \pmod 3$ and $x\equiv 4 \pmod 6$ is not possible. 2) We also have $x\equiv 5\equiv 1\pmod 2$. But we don't need that.)
So $x \equiv 2\pmod 3$.
And $x \equiv 7\pmod 8$.
so we can solve those to and get a unique solution $x \equiv 23\pmod {24}$.
.... oh, I get what they were getting at....
Tangential aside:
$x \equiv 2\equiv -1\pmod 3$ and $x \equiv 5\equiv -1 \pmod 6$ and $x\equiv 7\equiv -1 \pmod 8$
So we have $x \equiv -1 \pmod{ 3,6,8}$.
And if $x\equiv k \pmod m$ and $x\equiv k \pmod n$ and we want to know a (not necessarily unique solution) to $x \equiv ???? \pmod {W}$ then $x\equiv k\pmod W$ is always one solution. (Although and $k + am + bn\pmod{W}$ will also be a solution.
So $x \equiv -1\pmod 24$ will be a solution. And as $\gcd(3,8)=1$ that solution will be unique $\mod 24$.
=====
Your error was in solving
$2+3a = x\equiv 5\pmod 6$ so $2+3a \equiv 5\pmod 6$ so $3a \equiv 3\pmod 6$ does NOT mean $a \equiv 1 \pmod 6$. Modular arithmetic does not hold for divisiong if the moduli are not relatively prime.
$3*1 \equiv 3 \pmod 6$ but $3*3=9\equiv 3 \pmod 6$ and $3*5\equiv 15 \equiv 3\pmod 6$. so $a \equiv 1,3,5 \pmod 6$.
Note if $ma \equiv mb \equiv W$ then $W|ma-mb=m(a-b)$ but that does not mean $W|a-b$. It's possible that $W$ and $m$ have a factor in common.
But if $\gcd(m,W)=1$ then $W$ and $m$ don't have a factor in common and you can divide But not if they do have a factor in common.
So instead of getting $x = 5+18k$ we should have gotten $x = 5,11,17 + 18k$. Three possibible results, not just one.
ANd this way of doing it would eventually have ended with $x \equiv 23,47,71,95,119,143 \pmod{144}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number Theory And Vieta Jumping $\textbf{Question:}$Find all positive integers $a, b$ such that the expression $$\frac{a^2+b^2+1}{ab-1}$$ is an integer.
$$$$As the expression is symmetric in $a, b$, so let $a \geq b$. It is easy to check the cases when $a=b+k$ where $0 \leq k \leq 2$, so let $a \geq b+3$. Suppose $a, b$ satisfies the given condition with $a+b$ minimized. Let us write $$\frac{a^2+b^2+1}{ab-1}=k$$ then our equation becomes $$a^2-kab+b^2+k+1=0$$. Now this equation has a root $a$ and from the Viet's formula we get the second root $$a'=kb-a$$ and $$a'=\frac{b^2+k+1}{a}$$. Now the first equation shows that $a'$ is an integer and the second equation shows that $a'$ is a positive integer. Now to show that $a' < a$ we have to prove that $$a'=\frac{b^2+k+1}{a} < a$$. Now consider $$a^2-(b^2+k+1)=a^2-b^2-1-k=a^2-b^2-1-(\frac{a^2+b^2+1}{ab-1})=\frac{a^3b-ab^3-ab-a^2+b^2+1-(a^2+b^2+1)}{ab-1}=\frac{a(a^2b-b^3-b-2a)}{ab-1}$$. Now as $a \geq b+3$ so we have $$a^2b-2a=a(ab-2) \geq a((b+3)(b)-2)=a(b^2+3b-2) \geq a(b^2+1) > b(b^2+1) > b^3+b$$ and hence we get $\frac{a(a^2b-b^3-b-2a)}{ab-1}>0$ and hence we have $a' < a$ and hence $a'+b < a+b$ contradicting the minimality of $a+b$. So there are no solutions.
$$$$Is My Proof Correct??
| There are infinitely many solutions with $k=3$ and $k=6.$ For 3 the generating point is $(2,2)$ For $k=6$ the points are $(2,1)$ and $(1,2)$
More solutions can be found in either case by Vieta jumping.
Ruling out larger $k$ is by inequalities.
See how, as soon as $k \geq 7,$ the arc of the hyperbola passes through the interior of the square $1 < x < 2, $ $ 1 < y < 2.$ The intersections with the bounding lines $y = \frac{k}{2} x$ and $x = \frac{k}{2} y$ have one coordinate strictly below $1.$ As a result, there are no integer points on the arc between the bounding lines, therefore no integer points in the first quadrant at all. The method is described in HURWITZ 1907
| {
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"timestamp": "2023-03-29T00:00:00",
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Find inverse element of $1+2\alpha$ in $\mathbb{F}_9$
Let $$\mathbb{F}_9 = \frac{\mathbb{F}_3[x]}{(x^2+1)}$$ and consider $\alpha = \bar{x}$. Compute $(1+2 \alpha)^{-1}$
I think I should use the extended Euclidean algorithm: so I divide $x^2 +1 $ by $(1+2x)$:
$$x^2 + 1 = (1+2x)(2x+2)+2$$
$$(2x+2)(1+2x) + 2(x^2+1) = 1$$
Therefore, considering $\text{mod}(x^2+1) $, I have $$(2x+2)(1+2x) = 1\text{mod}(x^2+1)$$
and so $2x+2 = (1+2x)^{-1}$
Is it okay, or did I misunderstood something?
| $(2x+2)(2x+1)=2(x+1)(2x+1)=2(2x^2+3x+1).$
Since we are in $\mathbb{F}_3,$ then $3x=0$ and $4x^2=x^2$ so
$(2x+2)(2x+1)=2(2x^2+1)=4x^2+2=x^2+2=(x^2+1)+1,$
reducing modulo $(x^2+1)$ we get $1,$ so yeah, you are correct.
| {
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$A^2 + B^2 + C^2 = D^2 + E^2 + F^2$ and $A^2 + F^2 = B^2 + E^2 = C^2 + D^2$, distinct positive integers I am looking for parametric formula to this system of equations:
$A^2 + B^2 + C^2 = D^2 + E^2 + F^2$
$A^2 + F^2 = B^2 + E^2 = C^2 + D^2$
for distinct positive integers $A,B,C,D,E,F$. According to my computations, solutions for this system are rare. It looks like this is the smallest one:
$421^2 + 541^2 + 49^2 = 559^2 + 149^2 + 371^2$
$421^2 + 371^2 = 541^2 + 149^2 = 49^2 + 559^2$
I would like to know if there is any known algebraic formula (or method) to find these solutions faster (better than bruteforce).
| Since the equations are all homogeneous of degree $2$, we can multiply a solution by a constant to produce another solution.
We may assume we have a primitive solution, i.e. $\gcd(A,B,C,D,E,F)=1$.
Let's start with $A^2 + F^2 = B^2 + E^2 = C^2 + D^2$. Call this common value $x$. Note that if $x \equiv 0 \mod 4$, all of $A,B,C,D,E,F$ are even, so in a primitive solution $x \equiv 1$ or $2 \mod 4$. Moreover, in a primitive solution $x$ can't be divisible by any prime $\equiv 3 \mod 4$. You need $x$ to be divisible by at least $3$ (not necessarily distinct) primes $\equiv 1 \mod 4$ to allow $x$ to be written as the sum of two squares in $3$ different ways with all distinct squares.
So one approach would be to look at products of three (or more) primes $\equiv 1 \mod 4$ and (optionally) $2$. For each such $x$,
consider all ways to write $x$ as the sum of two squares. Take choices of three of those which involve $6$ distinct squares, and
see if you can get $A^2 + B^2 + C^2 = D^2 + E^2 + F^2$.
Taking three such primes $< 300$ and optionally $2$, I get your solution as well as one other:
$$(A,B,C,D,E,F) = (931, 1541, 1691, 1099, 1301, 1789)$$
Using four primes $< 200$ and $2$, I get another solution:
$$(A,B,C,D,E,F) = (137, 1123, 1523, 283, 1067, 1543)$$
EDIT: And here's another one:
$$(A,B,C,D,E,F) = (15953, 23393, 24553, 18263, 19727, 26113)$$
| {
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Trigonometric inequalities in a triangle What is the proof of
$~~\cos^2 A+\cos^2 B+\cos^2 C \leq 1 ~~$
in an acute triangle ?
This will be of help in finding the answer (if such exists) to finding the minimal T in any ∆ ABC when
$$T \geq \sin^k A+ \sin^k B+ \sin^k C~ ,~~~~ k \geq 3$$
| Let's first assume that $\cos^2A +\cos^2B+ \cos^2C=1$
Let's proceed to solve this
$\cos^2A +\cos^2B+ \cos^2C=1$
$\Rightarrow \cos^2A + \cos^2B - (1−\cos^2C)=0$
$\Rightarrow \cos^2A + cos^2B − sin^2C=0$
$\Rightarrow \cos^2A+(cos(B+C)cos(B−C))=0$
$\Rightarrow \cos^2A+(cos(π−A)cos(B−C))=0$ as $A+B+C = π$
$\Rightarrow \cos^2A−cosAcos(B−C)=0$
$\Rightarrow \cos A(cosA−cos(B−C))=0$
$\Rightarrow cosA(cos(π−(B+C))−cos(B−C))=0 $
$\Rightarrow (−cosA(B+C)−cos(B−C))=0 $
$\Rightarrow cosA(cosA(B+C)+cos(B−C))=0 $
$\Rightarrow −cosA(2cosBcosC)=0 $
$\Rightarrow 2cosAcosBcosC=0 $
$\Rightarrow cosA=0 or cosB=0 or cosC=0 $
$\Rightarrow A=\frac{π}{2} or B=\frac{π}{2} or C=\frac{π}{2} $
With this we can conclude that the triangle ABC will be a right-angled triangle if $\cos^2A +\cos^2B+ \cos^2C=1$.
Now when we say that $\cos^2A +\cos^2B+ \cos^2C \leq 1$ this means that all angles are lesser than $\frac{π}{2}$ which means that the given triangle is Acute angled.
I hope this helped you.
| {
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For how many natural numbers(<=100) is $1111^n +2222^n+3333^n+4444^n$ divisible by 10? For how many natural numbers (0 not included) $n \leq 100$ is $1111^n +2222^n+3333^n+4444^n$ divisible by 10?
I factored out $1111^n$ and got $1111^n(1+2^n+3^n+4^n)$. So $1+2^n+3^n+4^n$ must be divisible by 10. I figured out that this is divisible by 10 for all odd n, but I don't know how to find the other solutions, if any.
| If you divide $1^n$, $2^n$, $3^n$, and $4^n$ by 10, each goes through a cycle of remainders:
$1: 1, 1, 1, 1$
$2: 2, 4, 8, 6$
$3: 3, 9, 7, 1$
$4: 4, 6, 4, 6$
So $1^n+2^n +3^n + 4^n$ goes through the cycle of remainders $0, 0, 0, 4$, and thus will be divisible by $10$ whenever $n$ is not divisible by $4$.
| {
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Evaluate $\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$ Evaluate $$\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$$
I tried substitutions like $u=\frac{\pi}{4}-x$, and trig identities like $\cos^2{x}=1-\sin^2{x}$ after getting a common denominator.
$$4\int_0^{\frac{\pi}{4}} \frac{\sin^2{(5x)\cos^2{x}-\cos^2{(5x)}\sin^2{x}}}{\sin^2{(2x)}} \mathop{dx}$$
Where should I go from here? Any help is appreciated!
| Rewriting the integral where you left off:
$$4\int_0^{\frac{\pi}{4}} \frac{\left(\sin{(5x)}\cos{x}-\cos{(5x)}\sin{x}\right) \left(\sin{(5x)}\cos{x}+\cos{(5x)}\sin{x}\right)}{\sin^2{(2x)}} \;dx$$
$$=4\int_0^{\frac{\pi}{4}} \frac{\sin{(4x)}\sin{(6x)}}{\sin^2{(2x)}} \; dx$$
$$=8\int_0^{\frac{\pi}{4}} \frac{\cos{(2x)}\sin{(6x)}}{\sin{(2x)}} \; dx$$
There are several ways to continue from here. I will let $u=2x$:
$$=4\int_0^{\frac{\pi}{2}} \frac{\cos{u}\sin{(3u)}}{\sin{u}} \; du$$
Again, there are multiple ways to proceed from here. I will use the identity for $\sin{(3x)}$:
$$\sin^3{x}={\left(\frac{e^{ix}-e^{-ix}}{2i}\right)}^3=-\frac{1}{4} \cdot \frac{e^{3ix}-e^{-3ix}-3e^{ix}+3e^{-ix}}{2i}=-\frac{1}{4} \left(\sin{(3x)}-3\sin{x}\right)$$
$$\sin{(3x)}=3\sin{x}-4\sin^3{x}$$
$$=4\int_0^{\frac{\pi}{2}} \frac{\cos{u} \left(3\sin{u}-4\sin^3{u}\right)}{\sin{u}} \; du$$
Let $t=\sin{u}$:
$$=4\left(3t-\frac{4t^3}{3}\right) \bigg \rvert_0^1$$
And so,
$$\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}=\boxed{\frac{20}{3}}$$
| {
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Find $\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$ (I need a review of my resolution please :) ) Find the limit:
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$$
This is what I have, i'm not sure about my answer (I'm just learning limits).
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \lim_{x \to 1} \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1} $$
And I know that cos is a bounded limit, also its $L=-1$.
The other limit:
$$ \lim_{x \to 1} \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1} $$
$$ \lim_{x \to 1} \sqrt{\frac{(x-1) \cdot (x-1)}{(x-1)\cdot (x+1)}} \cdot \frac{1-x}{x^2+x-1} $$
Simplifying:
$$ \lim_{x \to 1} \sqrt{\frac{(x-1) \cdot (x-1)}{(x-1)\cdot (x+1)}} \cdot \frac{1-x}{x^2+x-1} $$
$$ \lim_{x \to 1} \sqrt{\frac{(x-1)}{(x+1)}} \cdot \frac{1-x}{x^2+x-1} $$
Evaluating:
$$ \lim_{x \to 1} \sqrt{\frac{(1-1)}{(1+1)}} \cdot \frac{1-1}{1^2+1-1} $$
$$ = \sqrt{0} \cdot \frac{0}{1} $$
$$\fbox {= 0}$$
So since I have a bounded limit and the other limit function is zero, the whole limit is $0$.
So;
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1} = 0$$
| Your answer is correct.
The limit can be split into parts if each of them exists as follows
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$$
$$=\lim_{x \to 1} \cos(\pi \cdot x) \cdot \lim_{x \to 1}\sqrt{\frac{x-1}{x+1}} \cdot \lim_{x \to 1}\frac{1-x}{x^2+x-1}$$
$$=(-1)\cdot 0\cdot 0=\color{blue}{0}$$
| {
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Calculating $ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$ I need to find the limit of:
$$ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$$
I've tried throught "rationalization" and completing squares... This is my work so far (i'm learning by myself limits since my teachers doesn't respond any email and they are not making lectures, just pdf's... I'm trying to do my best, help pls). Also is there any good book or suggestion to learn limits .
\begin{align*}&\lim_{b \to a} \dfrac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2} \cdot \dfrac{(a-\sqrt{a^2-b^2})}{(a-\sqrt{a^2-b^2})} \\=&
\lim_{b \to a} \dfrac{a[a^2-(a^2-b^2)]-b^2(a-\sqrt{a^2-b^2})}{a[(a^2-\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
= &\lim_{b \to a} \dfrac{-ab^2-ab^2+b^2(\sqrt{a^2-b^2})}{a[(a^2-(\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
=&\lim_{b \to a} \dfrac{-2ab^2+b^2(\sqrt{a^2-b^2})}{a[(a^2-(\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
\end{align*}
| Disclaimer : This method is probably overkill, you can make do with simpler arguments on limits.
What you want to do is find estimates of the numerator and denominator.
The main tool here is the following estimate for $y \rightarrow 0$ : $\sqrt{1+y} = 1 + O(y) $, where $O(y^r)$ means : something that is comparable to or smaller than $y^r$ as $y$ tends to $0$.
For simplicity, we will write $x= b-a$. Note that $x<0$.
So here, the numerator can be estimated as : $a^2 + a\sqrt{a^2-(a+x)^2}-(a+x)^2 = a^2 + a^2\sqrt{1-(1+\frac{x}a)^2}-(a+x)^2 $
$$\begin{aligned}&= a^2 + a^2\sqrt{2\frac{-x}{a} + O(x^2)} - a^2 + O(x)\\&= \sqrt{2}a^{\frac32} \sqrt{-x + O(x^2)} + O(x)\\&= \sqrt{2}a^{\frac32} \sqrt{-x} \sqrt{1 + O(x)} + O(x)\\&= \sqrt{2}a^{\frac32} \sqrt{-x} (1+ O(x)) + O(x)\\&= \sqrt{2}a^{\frac{3}32} \sqrt{-x} + O(x\sqrt{-x}) + O(x)\\&= \sqrt{2}a^{\frac32} \sqrt{-x} + O(x)\end{aligned} $$
Similar computations tells us that the denominator is equivalent to $ -\sqrt{2}a^{\frac32} \sqrt{-x} $.
So the limit should be $-1$.
| {
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Prove that $\det ((A + B + C) (A^3 + B^3 + C^3-3ABC))\geq 0 $
Suppose that A, B and C are 2x2 matrices that switch between each
other. Prove that
$$\det ((A + B + C) (A^3 + B^3 + C^3-3ABC))\geq 0. $$
I did
$$A^3+B^3+C^3-3ABC=\frac12(A+B+C)((A-B)^2+(A-C)^2+(B-C)^2)$$
So, this determinant is equivalent to
$$\frac14[\det(A+B+C)]^2\det((A-B)^2+(A-C)^2+(B-C)^2)$$
But how can I prove that
$$\det((A-B)^2+(A-C)^2+(B-C)^2)\geq0?$$
Can someone help me? Thanks for attention.
| Based on Aryaman Maithani's idea:
Presumably the matrices are real. Let $X=A-B,\,Y=B-C$ and $Z=C-A$. Then $X,Y,Z$ commute and $X+Y+Z=0$. Let $\omega$ be a primitive cube root of unity. Then
\begin{aligned}
\det(X^2+Y^2+Z^2)
&=\det(X^2+Y^2+(-X-Y)^2)\\
&=4\det(X^2+Y^2+XY)\\
&=4\det\left[(X-\omega Y)(X-\bar{\omega}Y)\right]\\
&=4|\det(X-\omega Y)|^2\\
&\ge0.
\end{aligned}
| {
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If $a+b+c=k$ and $a^2+b^2+c^2 =2k$ what is the maximum value of $k$? $a,b,c$ are real numbers and they satisfy the following equations.
$a+b+c=k$
$a^2+b^2+c^2=2k$
Find the maximum value of $k$.
I tried substituting for k in the second equation from the first and got
$a^2+b^2+c^2=2(a+b+c)$
Rearranging the terms I got
$a^2-2a+b^2-2b+c^2-2c=0$
Adding 3 to both sides we get
$a^2-2a+1+b^2-2b+1+c^2-2c+1=3$
This can be simplified to the following
$(a-1)^2+(b-1)^2+(c-1)^2=3$
Therefore,
$0\leq(a-1)^2,(b-1)^2,(c-1)^2\leq3$
From here we can deduce the range of values that a,b,c can take as
$1-\sqrt{3}\leq a,b,c\leq1+\sqrt{3}$
I don't know know if this helps to answer the question.
| Since $c=k-(a+b)$ both inequalities are fulfilled iff the set of points $(a;b)$ such that
$$ a^2 + b^2 + (k-(a+b))^2 = 2k $$
is non-empty. This equation can be written as
$$ 2a^2+2ab+2b^2-2ka-2kb+(k^2-2k) = 0 $$
or, by letting $a=A+\frac{k}{3},b=B+\frac{k}{3}$, as
$$ 2A^2+2AB+2B^2 = 2k-\frac{k^2}{3}.$$
The matrix $\left(\begin{smallmatrix}2 & 1 \\ 1 & 2\end{smallmatrix}\right)$ is positive definite, hence this is the equation of an ellipse provided that $2k-\frac{k^2}{3}>0$.
It follows that the maximum value of $k$ is $\color{red}{6}$.
| {
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Evaluating $2x^3-9x^2-10x+13$ for $x=3+\sqrt{5}$. Is there an efficient way to tell that this reduces to $1$? I was helping my sibling with a math problem from a past year paper for a competitive exam. It requires us to evaluate this cubic expression for a given value of $x$ which has an $a+b$ form where $b$ is a square root, as shown:
I essentially plugged in $x=a+b$ in the expression and expanded each term and finally got the answer as $1$. But this does not seem like the fastest way to do this, especially because there is only 1 minute given to solve each multiple choice question.
Is there a better way to reduce the original expression that gives the answer as $1$ or suggests that the square root term is going to evaluate to $0$?
Thanks!
| if $5 = (x-3)^2$ then $-10x = -2x(x-3)^2 = -2x^3+12x^2-18x$
Substituting this into $2x^3 - 9x^2 - 10x + 13$
we have
$$\begin{align}
2x^3 - 9x^2 +(-2x^3+12x^2-18x) + 13 &= 3x^2-18x+13\\
&= 3x^2-18x+27-14\\
&= 3(x^2-6x+9)-14\\
&= 3(x-3)^2-14\\
&= 3\cdot 5-14\\
&= 15-14\\
&= 1\\
\end{align}$$
| {
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$ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?
| Well you got
$\lim_{x\to 0}\frac{\sec^2x-\cos x}{3x^2}$
$\frac{1}{3}\lim_{x\to 0}\frac{\frac{1}{\cos^2x}-\cos x}{x^2}$
$\frac{1}{3}\lim_{x\to 0}\frac{1-\cos^3x}{x^2\cos^2x}$
$\frac{1}{3}\lim_{x\to 0}\frac{(1-\cos x)(1+\cos x+\cos^2x)}{x^2\cos^2x}$
$\frac{1}{3}\lim_{x\to 0}\frac{(1+\cos x+\cos^2x)}{\cos^2x}\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
$\frac{1}{3}\frac{(1+1+1)}{1}\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
$\frac{1}{3}3\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
$\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
Multiplying the numerator and denominator of the limit by $(1+\cos x)$ we get
$\lim_{x\to 0}\frac{(1-\cos x)(1+\cos x)}{x^2}\frac{1}{(1+\cos x)}$
$\lim_{x\to 0}\frac{(1-\cos^2x)}{x^2}\lim_{x\to 0}\frac{1}{(1+\cos x)}$
We know $\sin^2x+\cos^2x=1$
From this we get $(1-\cos^2x)=\sin^2x$
$\lim_{x\to 0}\frac{\sin^2x}{x^2}\frac{1}{2}$
$\frac{1}{2}\lim_{x\to 0}\frac{\sin^2x}{x^2}$
$\frac{1}{2}[\lim_{x\to 0}\frac{\sin x}{x}]^2$
Using L'Hopital's rule to evaluate the limit we get,
$\frac{1}{2}[\lim_{x\to 0} \frac{\cos x}{1}]^2$
$\frac{1}{2}(1)^2$
$\frac{1}{2}$
| {
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Bound for $\sum_{k=0}^{n}(-1)^k{3n\choose k}{n\choose k}$. In the book Complex analysis by Bak J. & Newman J., chapter 11, talks about Sums Involving Binomial coefficients and find a bound $\frac{16}{9}\sqrt{3}$ for $|(z-1)^2(z+1)|$ in the "Example 3" on the unit circle, my way was using lagrange multipliers in this form:
We want find $\max{|(z-1)^2(z+1)|}$ ahnd have that $|z-1|^2+|z+1|^2=4$. Let be $a=|z-1|$ and $b=|z+1|$ and then the exercise is: "Maximize $f(a,b)=a^2b$ subject to $a^2+b^2=4$" then $\nabla f=\lambda\nabla g$ so $\begin{cases}2ab=\lambda(2a)\\a^2=\lambda(2b)\end{cases}$, i.e., $ab=\lambda a$ if $a(b-\lambda)=0$ therefore
i)$a=0$ or $b=\lambda$, $b^2=4$ then $b=2$ then $|z+1|=2$ or $|z-1|=0$ then $z=1$ and $a^2b=0$.
ii)$b=\lambda$, $a^2=2b^2$ then $4=a^2+b^2=3b^2$ then $b^2=4/3$ then $b=\frac{2\sqrt{3}}{3}$ then $a^2=\frac{8}{3}$ then $a=\frac{2\sqrt{2}}{\sqrt{3}}$
So $a^2b=\frac{8}{3}\frac{2\sqrt{3}}{3}=\frac{16}{9}\sqrt{3}$.
After i want to use this idea for exercise 17.b with $\sum_{k=0}^{n}(-1)^k{3n\choose k}{n\choose k}$ but, i don´t see what should be $a$ and $b$.
Edit: this exercise says that $|\sum_{k=0}^{n}(-1)^k{3n\choose k}{n\choose k}|\leq4^n$.
| The first term in the asymptotic expansion follows from the analysis given by 'Asymptotics of a Family of Binomial Sums' by R. Noble. Putting this in a form for comparison one gets
$$ \sum_{k=0}^n (-1)^k\binom{3n}{k} \binom{n}{k} \sim (-4)^n \frac{2^{1/4}}{\sqrt{\pi n}} \cos(n \tan^{-1}(10\sqrt{2}/23) + \tan^{-1}((1-\sqrt{2})/(1+\sqrt{2})) $$
This is not a bound, of course. However, when $n$ is large enough,
$$ \Big|\sum_{k=0}^n (-1)^k\binom{3n}{k} \binom{n}{k} \Big| < C\frac{4^n}{\sqrt{n}}.$$
Is $C=2^{1/4}/\sqrt{\pi} \approx 0.671$ ? Checking up through $n=1000$ says it is. To be rigorous the next term in the asymptotic expansion needs to be generated.
For comparison of the asymptotic formula and the exact: n=200, true=-1.2130 x $10^{119},$ appx=-1.2145 x $10^{119},$ and absolute % error between them is 0.12%.
| {
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Solution verification: $ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $ This limit is not too difficult but I was just wondering if my work/solution looked good?
Thanks so much for your input!!
$$ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $$
$$ 2 x - 6 = 2 x \left( 1 - \frac 6 { 2 x } \right) $$
$$ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } =
\lim _ { x \to 3 } \frac { 2 x \left( 1 - \frac 6 { 2 x } \right) } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \frac { x \left( 1 - \frac 6 { 2 x } \right) } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \frac { x - 3 } { \sqrt x - \sqrt 3 } $$
By rationalizing the denominator:
$$ \frac { x - 3 } { \sqrt x - \sqrt 3 } = \sqrt x + \sqrt 3 $$
$$ 2 \cdot \lim _ { x \to 3 } \frac { x - 3 } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \left( \sqrt x + \sqrt 3 \right) $$
By plugging in $ x = 3 $:
$$ 2 \cdot \lim _ { x \to 3 } \left( \sqrt x + \sqrt 3 \right) =
2 \left( \sqrt 3 + \sqrt 3 \right) = 4 \sqrt 3 $$
| Your solution is ok, but a bit verbose.
To remedy to that, I personally suggest working around zero by setting $x=3+u$ with $u\to 0$, I find it triggers natural reflexes more. Also for presentation purposes, I prefer working on the expression and then make use of $\to$ to specify the limit rather than carrying the $\ \lim\limits_{x\to 3}\ $operator everywhere, and the fact the the limit is now in zero helps a lot (it makes the context obvious).
Compare how much shorter this is:
$\require{cancel}f(x)=\dfrac{2x-6}{\sqrt{x}-\sqrt{3}}=\dfrac{2u}{\sqrt{3+u}-\sqrt{3}}\overset{(*)}{=}\dfrac{2\cancel u}{\cancel u}(\overbrace{\sqrt{3+u}}^{\to\ \sqrt{3}}+\sqrt{3})\to4\sqrt{3}$
$(*)$ multiply by conjugated quantity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Modified central binomial coefficients generating function Given $n \in \mathbb{N}$, I would like to find the ordinary generating function of the sequence $a_k = \binom{2n-2k}{n-k}$.
If
\begin{align}
A(x) = \sum_{k = 0}^\infty a_kx^k,
\end{align}
then I find that
\begin{align}
A(x) &= \sum_{k = 0}^n \binom{2n-2k}{n-k}x^k \\
&= \sum_{k = 0}^n \binom{2k}{k}x^{n-k} \\
&= x^n\sum_{k = 0}^n \binom{2k}{k} \left(\frac{1}{x}\right)^k
\end{align}
but I am stuck here. My understanding is that you cannot extend the last finite sum into an infinite series, so I cannot use the generating function of $\binom{2k}{k}$.
I have also tried rewriting $A(x)$ as
\begin{align*}
A(x) &= [y^n]\left(1 + xy + (xy)^2 + \cdots\right)\left(\sum_{i \ge 0} \binom{2i}{i}y^i\right)\\
&= [y^n] \frac{1}{1-xy}\frac{1}{\sqrt{1-4y}}
\end{align*}
but I have no idea how to proceed from here.
Any idea is greatly appreciated.
| We consider $a_{n,k}=\binom{2n-2k}{n-k}$ with $n,k\geq 0$ non-negative integers.
*
*Horizontal GF: First of all we note that
\begin{align*}
A_n(x)=\sum_{k=0}^na_{n,k}x^k=\sum_{k=0}^n\binom{2n-2k}{n-k}x^k\qquad\qquad n\geq 0
\end{align*}
is a polynomial in $x$ and as such a perfect ordinary generating function, a so-called horizontal generating function. Since it is a polynomial having a finite number of terms $a_{n,k}x^k$ not equal to zero, we do not expect a representation via $\frac{1}{\sqrt{1-4x}}$ which is an infinite series.
*Vertical GF: On the other hand we can consider the vertical generating function for fixed $k\geq 0$:
\begin{align*}
B_k(y)&=\sum_{n=k}^\infty a_{n,k}y^n=\sum_{n=k}^\infty\binom{2n-2k}{n-k}y^n\\
&=\sum_{n=0}^\infty\binom{2n}{n}y^{n+k}\\
&=\frac{y^k}{\sqrt{1-4y}}
\end{align*}
*Bivariate GF: We have the bivariate generating function $G(x,y)$ with $A_n(x)$ and $B_k(y)$ as horizontal resp. vertical section:
\begin{align*}
\color{blue}{G(x,y)}&=\sum_{k=0}^\infty\sum_{n=k}^\infty a_{n,k}x^ky^n\\
&=\sum_{k=0}^\infty\sum_{n=k}^\infty\binom{2n-2k}{n-k}x^ky^n\\
&=\sum_{k=0}^\infty\sum_{n=0}^\infty\binom{2n}{n}x^ky^{n+k}\\
&=\sum_{k=0}^\infty(xy)^k\sum_{n=0}^\infty \binom{2n}{n}y^n\\
&\,\,\color{blue}{=\frac{1}{1-xy}\,\frac{1}{\sqrt{1-4y}}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Closed form sought for $a_1 = a_2 = 1, a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i $ where $n>2$ I've been working through a problem that I've got as far as getting a recursive answer to. I was hoping to turn this into more of a "closed form" answer, but haven't really gotten anywhere. I'm hoping that someone can help with this, though anything would be greatly appreciated.
The recursive answer I have is a sequence of real numbers given by $$\begin{gather}
a_1 = a_2 = 1 \\
a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i \qquad (n > 2)
\end{gather}$$
The first few non-trivial members of this sequence are
*
*$a_3 = \frac{5}{3}$
*$a_4 = 2$
*$a_5 = \frac{37}{15}$
*$a_6 = \frac{26}{9}$
*$a_7 = \frac{349}{105}$
I've tried to express these in terms of $a_1$ and $a_2$ and and constants and arrived at
*
*$a_3 = 1 + \frac{2}{3} a_1$
*$a_4 = 1 + \frac{2}{4} a_1 + \frac{2}{4} a_2$
*$a_5 = (1 + \frac{2}{5}) + (\frac{2}{5} + \frac{2^2}{3\cdot5} ) a_1 + \frac{2}{5} a_2$
*$a_6 = (1 + \frac{2}{6} + \frac{2}{6}) + (\frac{2}{6} + \frac{2^2}{3 \cdot 6} + \frac{2^2}{4 \cdot 6}) a_1 + (\frac{2}{6} + \frac{2^2}{4 \cdot 6}) a_2$
*$a_7 = (1 + \frac{2}{7} + \frac{2}{7} + \frac{2}{7} + \frac{2^2}{5 \cdot 7}) + (\frac{2}{7} + \frac{2^2}{3\cdot7} + \frac{2^2}{4 \cdot 7} + \frac{2^2}{5 \cdot 7} + \frac{2^3}{3 \cdot 5 \cdot 7}) a_1 + (\frac{2}{7} + \frac{2^2}{4 \cdot 7} + \frac{2^2}{5 \cdot 7}) a_2$
I am not seeing a pattern developing here.
I also rearranged the above noting that $a_1 = a_2 = 1$ and got
*
*$a_3 = 1 + \frac{2}{3} $
*$a_4 = 1 + 2 (\frac{2}{4})$
*$a_5 = 1 + 3 (\frac{2}{5}) + \frac{2^2}{3\cdot5}$
*$a_6 = 1 + 4 (\frac{2}{6}) + \frac{2^2}{3 \cdot 6} + 2 (\frac{2^2}{4 \cdot 6})$
*$a_7 = 1 + 5 (\frac{2}{7}) + \frac{2^2}{3 \cdot 7} + 2 (\frac{2^2}{4 \cdot 7}) + 3 (\frac{2^2}{5 \cdot 7}) + \frac{2^3}{3 \cdot 5 \cdot 7}$
Here I do notice a couple of things
*
*The expression for $a_n$ begins with "$1 + (n-2) \frac{2}{n}$".
*The remaining terms of the expression look like "$k \dfrac{2^{i+1}}{b_1 \cdots b_{i} \cdot n}$" where each $b_j$ is between $3$ and $n-2$ and consecutive numbers cannot appear among them. The $k$ seems to be determined by the smallest number among the $b_j$, but this is more of a guess than anything right now.
These observations are not really helping me much at all.
| This might help by converting it into an order 2 recurrence:
$$
a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i
$$
$$
a_{n-1} = 1 + \frac{2}{n-1} \sum_{i=1}^{n-3} a_i
$$
therefore
$$
\sum_{i=1}^{n-3} a_i = \frac{(a_{n-1}-1)(n-1)}{2}
$$
and
$$
\sum_{i=1}^{n-2} a_i = a_{n-2} + \sum_{i=1}^{n-3} a_i
$$
$$
a_n = 1 + \frac{2}{n} a_{n-2} + \frac{n-1}{n}(a_{n-1}-1)
$$
I'll see if I can get any further...
Some ideas/notes here:
Formally, the generating function:
$$
G(x) = \sum_{n=1}^\infty a_n x^n = \sum_{n=1}^\infty (1 + \frac{2}{n} a_{n-2} + \frac{n-1}{n}(a_{n-1}-1)) x^n
$$
$$
G(x) = \sum_{n=1}^\infty a_n x^n = \sum_{n=1}^\infty x^n + \sum_{n=1}^\infty (\frac{2}{n} a_{n-2} + \frac{n-1}{n}(a_{n-1}-1)) x^n
$$
$$
G(x) = \sum_{n=1}^\infty a_n x^n = \frac{x}{1-x} + \sum_{n=1}^\infty (\frac{2}{n} a_{n-2} + \frac{n-1}{n}(a_{n-1}-1)) x^n
$$
$$
G(x) = \sum_{n=1}^\infty a_n x^n = \frac{x}{1-x} - \log (1-x)-\frac{x}{x-1} + \sum_{n=1}^\infty (\frac{2}{n} a_{n-2} + \frac{n-1}{n}a_{n-1}) x^n
$$
$$
G(x) = \sum_{n=1}^\infty a_n x^n = \frac{2x}{1-x} - \log (1-x) + \sum_{n=1}^\infty (\frac{2}{n} a_{n-2} + \frac{n-1}{n}a_{n-1}) x^n
$$
Now there is a quite interesting interpretation of a term like
$$
H(x) = \sum_{n=1}^\infty \frac{n-1}{n} a_{n-1} x^n
$$
but it requires some kind of 'parallel' or umbral universe.
Normally we differentiate a generating function to get
$$
G'(x) = \frac{d}{dx}\sum_{n=1}^\infty a_n x^n = \sum_{n=1}^\infty n a_{n}x^{n-1}
$$
if we consider a transform where functions are mapped to new functions whose series are ratios of the previous coefficients
$$
\mathcal{T}[G(x)](t) = \sum_{n=1}^\infty \frac{a_n}{a_{n-1}}t^n
$$
where the differential operator ends up turning into a 'shift' operator and the effect on the coefficients looks similar to that in $H(x)$. Seeing as we are apparently already in that domain, it might be worth stepping back out to the domain consistent with differentiation and consider the generating function that takes iterated products of terms
$$
F(x) = \sum_{k=1}^\infty \left(\prod_{l=1}^k a_l\right) x^k
$$
Mathematica has managed to solve the recurrence and I believe for the above reasons, the answer is very ugly. There may be a strategy for simplifying it down.
$$
a(n)\to \frac{-\frac{2 (8 E_{-n-3}(-2) \Gamma (n+2)+\Gamma (n+4,-2))}{e^2
\Gamma (n+2)}+n (n+5)+\frac{(-2)^{n+3}}{\Gamma (n+2)}+2}{4
(n+2)}+\frac{1}{3} (n+3) \sum _{K[1]=0}^{n-1} -\frac{3\ 2^{-K[1]-5} e^{-2-i
\pi K[1]} \left(e^2 (-1)^{K[1]} 2^{K[1]+6}+K[1]^2 (-\Gamma (K[1]+5,-2))-8
K[1] \Gamma (K[1]+5,-2)+K[1] \Gamma (K[1]+6,-2)-15 \Gamma (K[1]+5,-2)+7
\Gamma (K[1]+6,-2)\right)}{(K[1]+2) (K[1]+3) (K[1]+5)}$$
Edit: Based on @Gary's development the coefficients appear to be
$$
a_n = \frac{e^2 \left((-1)^n 2^{n+1}+\Gamma (n+2)\right)-(n+3) \Gamma (n+1,-2)}{2
e^2 n!}
$$
this is obtained by the inverse Z-transform of $G(\frac{1}{x})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Is the inequality true for all $n\geq 2$? Let $x,y,z>0$. I am wondering if the following inequality is true?
$$\sum_{cyc}\frac{x^n}{y^2+yz+z^2}\geq\frac{x^{2n-2}+y^{2n-2}+z^{2n-2}}{x^n+y^n+z^n},\qquad n\geq 2$$
If not, is it known for which $n$ it is true?
$\displaystyle(1)\qquad\sum_{cyc}\dfrac{x^2}{y^2+yz+z^2}\overset{AGM}{\geq} \frac{2}{3}\sum_{cyc}\frac{x^2}{y^2+z^2}\overset{Nesbitt}{\geq}1=\frac{x^2+y^2+z^2}{x^2+y^2+z^2}$
$\displaystyle(2)\qquad\sum_{cyc}\dfrac{x^3}{y^2+yz+z^2}=\sum_{cyc}\dfrac{3x^4}{3xy^2+3xyz+3xz^2}\overset{AGM}{\geq}\sum_{cyc}\frac{3x^4}{3x^3+3y^3+3z^3}=\frac{x^4+y^4+z^4}{x^3+y^3+z^3}$
$\displaystyle(3)\qquad\sum_{cyc}\dfrac{x^4}{y^2+yz+z^2}\overset{AGM}{\geq}\frac{2}{3}\sum_{cyc}\dfrac{x^4}{y^2+z^2}\overset{(*)}{\geq}\frac{x^6+y^6+z^6}{x^4+y^4+z^4}$
$(*)\iff\forall$ $a,b,c>0$, $\,\,\displaystyle \frac{2}{3}\sum_{cyc}\dfrac{a^2}{b+c}\geq\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$
$\iff \displaystyle \frac{2}{3}\left(\sum_{cyc}\dfrac{a^2}{b+c}-\frac{a+b+c}{2}\right)\geq\frac{a^3+b^3+c^3}{a^2+b^2+c^2}-\frac{a+b+c}{3}$
$\iff\displaystyle\frac{(a+b+c)}{3(a+b)(b+c)(c+a)}\sum_{cyc}(a+b)(a-b)^2\geq\frac{1}{3(a^2+b^2+c^2)}\sum_{cyc}(a+b)(a-b)^2$
$\iff\displaystyle\left((a+b+c)(a^2+b^2+c^2)-(a+b)(b+c)(c+a)\right)\sum_{cyc}(a+b)(a-b)^2\geq0$
where
$$(a+b+c)(a^2+b^2+c^2)-(a+b)(b+c)(c+a)$$
$$\ge(a+b+c)(ab+bc+ca)-(a+b)(b+c)(c+a)=abc>0$$
Done!
| It's wrong for any $n\geq5$.
For $n=5$ try $x=1.1$ and $y=z=1$.
For $n=4$ and $n=3$ it's true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Perturbation Integral How to show that for $f(x,m)$ = $\frac{mx^2}{2} + \frac{x^4}{4}$,
the integral $Z(\lambda)$ = $\frac{1}{\sqrt \lambda}$$\int_{-\infty}^{\infty} dz e^{-f(z)/\lambda}$ is $\sqrt{\frac{m}{2\lambda}}$$e^{\frac{m^2}{8\lambda}}$
$K_{1/4}(\frac{m^2}{8\lambda})$ for $m>0$?
$m$ here is a mass term, $\lambda$ is a perturbation parameter, the integral is to be asymptotically evaluated near $\lambda = 0$.
I found this reading a paper on the "Power of perturbation series" https://arxiv.org/abs/1702.04148 .
I tried using steepest descent, but not getting this answer.
| Start with
\begin{eqnarray}
A &=& \frac{1}{\sqrt{\lambda}} \exp\left[-\frac{1}{\lambda}\left(\frac{m}{2} x^2 +\frac{1}{4}x^4\right)\right]
\\
&=& \frac{1}{\sqrt{\lambda}} \exp\left[\frac{m^2}{8 \lambda}\right]\exp\left\{-\frac{m^2}{8 \lambda}\left[2 \left(\frac{x^2}{m}+1 \right)^2 -1 \right]\right\} \, .
\end{eqnarray}
Then, writing $u=\frac{x}{\sqrt{m}}$ we get
\begin{eqnarray}
\int_{-\infty}^\infty \mathrm{d}x \, A &=& \sqrt{\frac{m}{\lambda}} \exp\left[\frac{m^2}{8 \lambda}\right] \int_{-\infty}^\infty \mathrm{d}u \,\exp\left\{-\frac{m^2}{8 \lambda}\left[2 \left(u^2+1 \right)^2 -1 \right]\right\}
\\
&=& \sqrt{\frac{m}{2 \lambda}} \exp\left[\frac{m^2}{8 \lambda}\right] \int_{0}^\infty \mathrm{d}u \, 2\sqrt{2} \exp\left\{-\frac{m^2}{8 \lambda}\left[2 \left(u^2+1 \right)^2 -1 \right]\right\} \, .
\end{eqnarray}
Finish with $\cosh t = 2 \left(u^2+1 \right)^2 -1$, which corresponds to $u = \sqrt{\cosh\left[\frac{t}{2}\right]-1}$. Hence $2 \sqrt{2}\mathrm{d}u = \frac{1}{\sqrt{2}} \frac{\sinh\left[\frac{t}{2}\right]}{\sqrt{\cosh\left[\frac{t}{2}\right]-1}} \mathrm{d}t = \cosh\left[\frac{1}{4}t\right] \mathrm{d}t$. Therefore
\begin{eqnarray}
\int_{-\infty}^\infty \mathrm{d}x \, A &=& \sqrt{\frac{m}{2 \lambda}} \exp\left[\frac{m^2}{8 \lambda}\right] \int_{0}^\infty \mathrm{d}t \, \cosh\left[\frac{1}{4}t\right] \exp\left[-\frac{m^2}{8 \lambda} \cosh t\right]
\\
&=& \sqrt{\frac{m}{2 \lambda}} \exp\left[\frac{m^2}{8 \lambda}\right] K_{\frac{1}{4}} \left[\frac{m^2}{8 \lambda}\right] \,,
\end{eqnarray}
where we used the integral representation in link.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Evaluating Sum at bounds I have to find an expression in terms of n using standard results for $$\sum_{r=n+1}^{2n} r(r+1)$$
And have found the general equation
$$\sum_{r=n+1}^{2n} r(r+1) = \frac{2n^3+6n^2+4n}{6}$$
However evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n+1)^3+6(n+1)^2+4(n+1)}{6}$$
does not yield the correct answer, yet evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n)^3+6(n)^2+4(n)}{6}$$
gives the correct answer
Im at a loss here, why am I not getting the correct answer by finding the difference of the sum between the two bounds?
| Another approach: it's clear that the result is a polynomial of $n$ of degree $3$, let $$\sum\limits_{r=n+1}^{2n}(r^2+r)=An^3+Bn^2+Cn+D=P(n)$$
thus
\begin{align*}P(n)-P(n-1)&=\sum\limits_{r=n+1}^{2n}(r^2+r)-\sum\limits_{r=n}^{2n-2}(r^2+r)\\
&=-(n^2+n)+((2n-1)^2+(2n-1))+((2n)^2+2n)\\
&=7n^2-n\\
&\equiv A(3n^2-3n+1)+B(2n-1)+C\\
&=3An^2+(-3A+2B)n+(A-B+C)
\end{align*}
$$
\begin{cases}
3A=7\\
-3A+2B=-1\\
A-B+C=0\\
A+B+C+D=P(1)=2^2+2=6
\end{cases}$$
$$
\begin{cases}
A=\frac{7}{3}\\
B=3\\
C=\frac{2}{3}\\
D=0
\end{cases}$$
Thus $$\sum\limits_{r=n+1}^{2n}(r^2+r)=\frac{7}{3}n^3+3n^2+\frac{2}{3}n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754358",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to decouple this system of 2nd order partial differential equations? Solving a problem I found this system of equations:
$$ (\partial _t ^2 + \partial _x ^2 + \partial_y ^2 + \partial_z ^2)a_1(\vec{x},t) - 4gB(x\partial_y - y\partial_x)a_{\color{Red}2}(\vec{x},t) - 4g^2B^2(x^2 + y^2)a_1 - \lambda a_1 = 0 $$
$$ (\partial _t ^2 + \partial _x ^2 + \partial_y ^2 + \partial_z ^2)a_2(\vec{x},t) + 4gB(x\partial_y - y\partial_x)a_{\color{Red}1}(\vec{x},t) - 4g^2B^2(x^2 + y^2)a_2 - \lambda a_2 = 0 $$
taking into account that $g,B,\lambda=$ constants,
Any ideas to separate these equations in terms of $a_1$ and $a_2$?
Thanks a lot!
| Here's a linear algebra approach. I'll rearrange this slightly to make the next steps a bit clearer. I'm going to use $\Delta$ to represent $\partial_t^2 + \partial_x^2 + \partial_y^2 + \partial_z^2$. If it were $-\Delta_t^2$, I'd use the standard $\square$ d'Alembert notation instead. I'll also write $x\partial_y - y\partial_x$ as $iL_z$, taking cues from quantum mechanics, where $L$ is the angular momentum operator. (I suspect this is a relativistic quantum particle in an external magnetic field anyway..)
\begin{align}
(\Delta - 4g^2 B^2(x^2+y^2) - \lambda) a_1 - 4igB L_z a_2 &= 0 \\
(\Delta - 4g^2 B^2(x^2+y^2) - \lambda) a_2 + 4igB L_z a_1 &= 0
\end{align}
Writing this as a matrix equation, this would be
$$ \begin{pmatrix} \Delta - 4g^2 B^2 (x^2+y^2) & -4igB L_z \\ 4igB L_z & \Delta - 4g^2 B^2(x^2+y^2) \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$
This matrix is of the form
$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix}. $$
A quick check shows that this matrix is diagonalized by
$$ P = \frac{1}{\sqrt{2}} \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix}, \qquad P^{-1} = \frac{1}{\sqrt{2}} \begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix} $$
so we can turn the above matrix equation into..
$$ \frac{1}{\sqrt{2}} \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix} \begin{pmatrix} \Delta - 4g^2 B^2 (x^2+y^2) & -4igB L_z \\ 4igB L_z & \Delta - 4g^2 B^2(x^2+y^2) \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$
simply by multiplying on the left by $P$. Multiplying by $P^{-1}P$ on the left of $\begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$, we get
$$ \begin{pmatrix} \Delta - 4g^2 B^2 (x^2+y^2) + 4gB L_z & 0 \\ 0 & \Delta - 4g^2 B^2 (x^2+y^2) - 4gB L_z\end{pmatrix} \begin{pmatrix} ia_1 + a_2 \\ -ia_1 + a_2\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$
This matches with what you have. The strength of this approach is it gives you a prescription to go by. Decoupling a system of linear ODEs/PDEs is tantamount to diagonalizing the matrix-of-operators system.
| {
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Problem with proving inequalities Question:
Prove that if $x,y,z$ are positive real numbers such that $x+y+z=a$ then $(a-x)(a-y)(a-z)>\frac8{27}a^3$ is not true.
My Approach:
$$\frac{a-x}{2}=\frac{y+z}2$$
$$\frac{a-y}{2}=\frac{x+z}2$$
$$\frac{a-z}{2}=\frac{x+y}2$$
Using $AM>GM$ we get $$\frac{x+y+z}{3}>\root 3 \of {xyz}$$ Cubing both sides and multiplying by $8$, $$\frac{8a^3}{27}>8xyz$$
Also, by $AM>GM$, $$(\frac{y+z}2)(\frac{x+z}2)(\frac{x+y}2)>8xyz$$
Now, how do I find the relation between $(\frac{y+z}2)(\frac{x+z}2)(\frac{x+y}2)$ and $\frac{8a^3}{27}$?
| Using AM-GM identity,
$\implies$ $[(a-x)(a-y)(a-z)]^{1/3} \le \frac{(a-x)+(a-y)+(a-z)}{3}$$
Then we end up with our desired result
$$\implies (a-x)(a-y)(a-z) \le \frac{8a^3}{27}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers? $\textbf{Question:}$Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers?
that is pairwise relatively prime $a,b,c \in \mathbb N $($0$ not included) such that $a+b \mid c^2,b+c \mid a^2,c+a \mid b^2$
$\textbf{Stuffs I have tried:}$
I have tried a bunch of constructions like :$(p,\frac{p+1}{2},\frac{p-1}{2})$,one of them are the sum of other two,tried small values.But was unable to find any valid triple.So,I began trying to prove such triple can't exist.I tried to bound the numbers,tried to use modular arithmatic but failed again.
Any kind of hint or solution will both be appreciated.
| Observe that
*
*Clearly $ a > 1, b > 1 , c > 1$.
*WLOG $ a < b < c$
*$ 0 \equiv c^2 \equiv c(c+a+b) \equiv (a+b+c)(a+b+c) \pmod{a+b}$ (and similar expressions)
*$\gcd(a+b, b+c ) \mid \gcd( c^2, a^2) = 1$ so $\gcd(a+b, b+c) = 1$. (and similar expressions)
Hence
$$ (a+b)(b+c)(c+a) \mid (a+b+c)^2 \Rightarrow (a+b)(b+c)(c+a) \leq (a+b+c)^2. $$
Thus
$$ 4 (a+b)( b+c)^2 < 4 (a+b)(b+c)(c+a) \leq \left[2(a+b+c)\right]^2 < [3(b+c)]^2$$
which implies that $ (a+b) < \frac{ 9}{4} $, which is a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\int_0^{\pi/2} \frac{\sin x}{\sin^{2n+1}x +\cos^{2n+1}x} dx$? I have an exercise to evalute the following integral for all $n\geq 1 $
$$I(n)=\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin^{2n+1} x+\cos^{2n+1} x}dx$$
I attempted to find the closed form for the integral above in the following manner, where I used the integral identity $\int_a^bf(x)=\int_a^b f(a+b-x)dx$. $$I(\bar{n})=\int_0^{\frac{\pi}{2}}\frac{\cos x}{\cos^{2n-1} x+\sin ^{2n-1} x}dx$$ adding $I(n)$ and $I(\bar{n})$ its reduces to $$\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\cos x +\sin x}{\cos^{2n+1}x +\sin^{2n+1}x}dx$$ using the algebraic identity $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2+\cdots +b^{n-1})$ for odd integers $n$, I get $$\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos^{2n}x-\cos^{2n-1}\sin x+\cdots +\sin^{2n}x}dx $$
I'm now stuck here. How can I continue now?. Thanks in advance.
| This integral appear in the Jozsef Wildt International mathematical competition proposed by Ovidui Furdui and Alina Sintamarian which I solved in the following way , couples of months back.
For all $n\geq 2$ we shall show that
$$I(n)=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin^{2n-1}x+\cos^{2n-1}x}dx=\frac{\pi}{2n-1}\sum_{k=0}^{n-2}{n-2\choose k}\operatorname{csc}\left(\frac{(2\pi(n-k-1)}{2n-1}\right)$$.
Before we prove the above closed form we shall be using classical result.
Lemma: For all $n>1, m\in\mathbb {N}$following holds
$$ \int_0^{\infty}\frac{x^{m-1}}{1+x^n}dx=\frac{\pi}{n}\operatorname{csc}\left(\frac{m}{n}\pi\right)$$
Proof: We make subbing of $\frac{1}{1+x^n} = y$ and the integral takes the form of beta function, i.e; $$ \frac{1}{n}\int_0^{1}y^{1-\frac{m-n}{n}}(1-y)^{\frac{m}{n}-1}dy=\frac{1}{n}\Gamma\left(1-\frac{m}{n}\right)\Gamma\left(\frac{m}{n}\right)=\frac{\pi}{n}\operatorname{csc}\left(\frac{m}{n}\pi\right) $$
We evaluate the main integral $I(n)$ as follows $$\int_{0}^{\frac{\pi}{2}}\frac{\sin x\sec^{2n-1}x}{1+\tan^{2n-1} x}dx=\int_0^{\frac{\pi}{2}}\frac{\tan x (\sec^{2}x)^{n-2}\sec^2x}{1+\tan^{2n-1}x}dx$$ substitute $\tan x =u\implies \sec^{2}x dx=du$ and hence $$\int_0^{\infty}\frac{u(u^2+1)^{n-2} du}{1+u^{2n-1}} =\int_0^{\infty}\frac{u^2}{1+u^{2n-1}}\sum_{k=0}^{n-2}{n-2\choose k} u^{2(n-2-k)}du =\sum_{k=0}^{n-2}{n-2\choose k}\left(\int_0^{\infty}\frac{u^{2n-2k-4+1}}{1+u^{2n-1}} du\right) \underbrace{=}_{Lemma}\frac{\pi}{2n-1}\sum_{k=0}^{n-2}{n-2\choose k}\operatorname {csc}\left(\frac{2(n-k-1)\pi}{2n-1}\right)$$ we are done.
For $n=3$ we have a beautiful closed form for above integral
$$\int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin ^{5} x+\cos^5 x}dx = \frac{2}{5} \sqrt{1+\frac{2}{\sqrt 5}}\pi\approx 1.729 $$
| {
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Using partial information to factor $x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ I wish to find exact expressions for all roots of $p(x)=x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ By observing that for the roots $x_0 \pm iy_0, x_0 \approx -0.15883609808599033632, y_0 \approx 0.27511219196092896700,$ we have that $x_0$ is the unique real root of $r(x) = x^3+12x^2+8x+1,$ I was able to prove that all roots of the original sextic can be expressed in radicals. The process is as follows:
*
*Divide $p(x+iy)$ by $r(x)$ to get $\frac{1}{8}x^3 + \frac{3}{16}x^2 + x\left(\frac{7}{32}-\frac{15y^2}{8}\right) + \left(\frac{95}{32}-\frac{15y^2}{16}\right) + \frac{R(x,y)}{p(x)}$ where $R(x,y) = A(y)x^2 + B(y)x + C(y)$ and $A(y) = 15y^4 - \frac{15y^2}{4} - \frac{201}{16}, B(y) = 15y^8 - 30y^6 + 12y^4 + \frac{75y^2}{8} - \frac{767}{32}, C(y) = -y^6+5y^4-\frac{193y^2}{16}-\frac{63}{32}.$
*The equation $R(x_0, y_0) = 0$ is a quartic in $y_0^2,$ which we can solve exactly to obtain $y_0^2$ and hence $y_0.$
*Polynomial division reduces $p(x)$ to a quartic, and now we apply the quartic formula again to find the other $4$ roots.
However, I don't want to perform the rest of the computations. Is there a cleaner way to use the observation that $r(x_0) = 0,$ perhaps in the realm of abstract algebra?
| Just for the fun, starting from @Michael Rozenberg's answer, we need to solve the two cubic equations
$$x^3+\frac{3}{2} \left(1-i \sqrt{3}\right) x^2-2 \left(1+i \sqrt{3}\right) x-1=0$$
$$x^3+\frac{3}{2} \left(1+i \sqrt{3}\right) x^2-2 \left(1-i \sqrt{3}\right) x-1=0$$
Prepare yourself for nasty solutions.
| {
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Convergence of $\sum \limits_{n=1}^{\infty}\sqrt{n^3+1}-\sqrt{n^3-1}$ Hello I am a high school student from germany and I am starting to study math this october. I am trying to prepare myself for the analysis class which I will attend so I got some analysis problems from my older cousin who also studied maths. But I am stuck on this problem.
Check the following series for convergence/divergence $$\sum \limits_{n=1}^{\infty}\sqrt{n^3+1}-\sqrt{n^3-1}$$ I tried to prove the convergence by comparison test $$\sqrt{n^3+1}-\sqrt{n^3-1}= \frac{2}{\sqrt{n^3+1}+\sqrt{n^3-1}}=\frac{1}{n^2} \cdot \frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}$$ and then compare it with $$\sum \limits_{n=1}^{\infty}\frac{1}{n^2}$$ But in order to do that, I need to prove that $$\frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} \leq 1$$ But I am having problems to prove that. Does anyone have tip how to solve this problem?
| You cannot prove
$$\frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} \leq 1
$$
because the left-hand side tends to $+\infty$ for $n \to \infty$.
You should compare your series with $\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ instead:
$$
\frac{2}{\sqrt{n^3+1}+\sqrt{n^3-1}}=\frac{2}{n^{3/2}} \cdot \frac{1}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} <\frac{2}{n^{3/2}} \, .
$$
| {
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Solve $x^2+3y = u^2$ and $y^2+3x=v^2$ in positive integers. The question is from the pg - 59 from ' An Introduction to Diophantine Equations ' by Titu Andreescu , Dorin Andrica , Ion Cucurezeanu.
Example 1 : Solve in positive system of equations in positive integers
$$\begin{cases} x^2+3y = u^2 \\ y^2 + 3x = v^2 \end{cases}$$ $\;\;\;\;\;\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\text{(Titu Andreescu)}$
Solution. The inequality $x^2 + 3y ≥ (x + 2)^2 , y^2 + 3x ≥ (y + 2)^2$
cannot both be true, because adding them would yield a contradiction. So at least one of the inequalities $x^2 + 3y < (x + 2)^2$ and
$y^2 + 3x < (y + 2)^2$ is true. Without loss of generality, assume
that $x^2 + 3y < (x + 2)^2$. Then $$x^2 < x^2 + 3y < (x + 2)^2 \implies
x^2 + 3y = (x+1)^2$$ or, $3y = 2x+ 1$ . We obtain $x = 3k + 1, y = 2k + 1$
for some nonnegative integer $k$ and $y^2 + 3x = 4k^2 + 13k + 4$. For
$k > 5, (2k+ 3)^2 < 4k^2 + 13k+ 4 < (2k+ 4)^2$ ; hence $y^2 + 3x$ cannot be
a perfect square. Thus we need only consider $k ∈ {\{0, 1, 2, 3, 4\}}$ . Only
$k = 0$ makes $y^2 + 3x$ a perfect square; hence the unique solution is
$$x = y = 1,\;\;\;\;\;\; u = v = 2.$$
But if we take , $$4k^2+13k + 4 = v^2$$ $$\implies k = \dfrac{-13 \pm\sqrt{105+16v^2}}{8}$$
Since $105+16v^2 = a^2 \implies 105 = (a-4v)(a+4v)$ which gives $a \in \{\pm11 , \pm13 , \pm19 ,\pm53\}$ .
Out of these , only $a \in \{ \pm13 , \pm53\}$ works which gives $k=0,5$ , And so the the answer should be $$(x,y,u,v) = (1,1,2,2)\;\;\;,(16,11,17,13)\;\;\;\;,(11,16,13,17)$$
Who is correct here?
| $\begin{cases} x^2+3y = u^2 \\ y^2 + 3x = v^2 \end{cases}$ -----(1)
"OP" gave numerical solution to equation (1) as:
$(v,u,x,y)=(2,2,1,1)=(13,17,16,11)$
There is another numerical solution & is given below:
$(v,u,x,y)=[(10),(25/4),(13/4),(19/2)]$
| {
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On Resolving a Fuss on Squares and Fractions over a few Inequalities Firstly, only AM-GM and C-S are to be sought.
$1.$Let $a, b, c, d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Show that
$${a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge 4$$
In my textbook, this problem is credited to Michael Rozenberg, but I couldn't find a solution to it by myself or on this site, so decided to ask.
I tried my best with fallacy-
$$4(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2 \Rightarrow 4 \ge a+b+c+d$$ with the constraint given. Proceeding-
$$(a+b+c+d)\left({a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a}\right) \ge (a+b+c+d)^2 $$
$$\Rightarrow {a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge a+b+c+d$$
That means, I need $a+b+c+d\ge 4$ to complete the proof but instead got $4 \ge a+b+c+d$ !
$2.$Let $a, b, c$ be positive real numbers such that $abc = 1$.
Show that
$${1\over b(a+b)}+{1\over c(b+c)}+{1\over a(c+a)}\ge \frac{3}{2}.$$
$3.$If a, b, c and d are positive real numbers such that $a + b + c + d = 4$. Prove that
$$ {a \over 1+b^2c}+{b \over 1+c^2d}+{c \over 1+d^2a}+{d \over 1+a^2b} \ge 2. $$
The #2 is credited to the Zhautykov Olympiad 2008 and there is no need for revealing my attempts as I've no idea what to do, Lastly,
This is a doubt, not an question that definitely has an answer, but if the doubt comes out to have an answer, nevertheless, this is a problem.
$4.$ Prove without Induction:
$$\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2} + ...+\sqrt{a_n^2+b_n^2} \ge \sqrt{(a_1+a_2+...+a_n)^2+(b_1+b_2+...+b_n)^2}$$
| The fourth problem it's just Minkowski (triangle inequality), but also, after squaring of the both sides, we can use C-S:
$$\sqrt{(a_i^2+b_i^2)(a_j^2+b_j^2)}\geq a_ia_j+b_ib_j$$ and we obtain an identity.
| {
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What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
Initially, this seemed like one could work it out with $AM-GM$, but it doesn't seem so.
From $AM-GM$ one gets that $$\frac{1^2+2^2+ \dots+n^2}{n} \geqslant \sqrt[\leftroot{-1}\uproot{2}n]{1^2\cdot2^2\dots\cdot n^2}
$$
is this of any help here?
Remark. Thanks to Favst, the source of the problem is Problem 1 of 1994 British Mathematical Olympiad Round 2
| The mean of the squares $1^2, \ldots, n^2$ is
$$ f(n) = \frac{1}{n} \sum_{i=1}^n i^2 = \frac{2n^2+3n+1}{6}$$
It is an integer if and only if $n \equiv 1$ or $5 \mod 6$.
The first $n > 1$ for which it is a square is $337$, where $f(337) = 38025 = 195^2$. There are infinitely many. See OEIS sequence A084231.
| {
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2010 USAMO #5:Prove that if $\frac{1}{p}-2S_q = \frac{m}{n}$ for integers $m$ and $n$, then $m - n$ is divisible by $p$. Let $q = \frac{3p-5}{2}$ where $p$ is an odd prime, and let $S_q = \frac{1}{2\cdot 3 \cdot 4} + \frac{1}{5\cdot 6 \cdot 7} + \cdots + \frac{1}{q(q+1)(q+2)}
$
Prove that if $\frac{1}{p}-2S_q = \frac{m}{n}$ for coprime integers $m$ and $n$, then $m - n$ is divisible by $p$.
My Progress till now: $$2S_q = 2\sum_{x=1}^{\frac{q+1}{3}} \frac{1}{(3x-1)(3x)(3x+1)} = \sum_{x=1}^{\frac{p-1}{2}} \left[\frac{1}{3x(3x-1)}-\frac{1}{3x(3x+1)}\right]\\
=\sum_{x=1}^{\frac{p-1}{2}} \left[ \frac{1}{3x-1} - \frac{2}{3x} +\frac{1}{3x+1}\right]\\
=\sum_{x=1}^{\frac{p-1}{2}}\left[ \frac{1}{3x-1} + \frac{1}{3x} +\frac{1}{3x+1}\right] - \sum_{x=1}^{\frac{p-1}{2}} \frac{1}{x} $$
With the help of @user10354138 , I have got $\frac{1}{p} - 2S_q = \frac{1}{p} + \frac{1}{1} - \sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k} = \frac{m}{n}$
But then I am stuck.
Please give me some hints rather than a solution.
Thanks in advance.
PS: I didn't post it in AOPS, because there we don't get any guidance.
| With the help of @user10354138 's hints, I think I got the solution.I will be grateful if someone proof reads it.
Note that $$2S_q = 2\sum_{x=1}^{\frac{q+1}{3}} \frac{1}{(3x-1)(3x)(3x+1)} = \sum_{x=1}^{\frac{p-1}{2}} \left[\frac{1}{3x(3x-1)}-\frac{1}{3x(3x+1)}\right]\\
=\sum_{x=1}^{\frac{p-1}{2}} \left[ \frac{1}{3x-1} - \frac{2}{3x} +\frac{1}{3x+1}\right]\\
=\sum_{x=1}^{\frac{p-1}{2}}\left[ \frac{1}{3x-1} + \frac{1}{3x} +\frac{1}{3x+1}\right] - \sum_{x=1}^{\frac{p-1}{2}} \frac{1}{x}$$ .
Proceeding further we get that,$$\frac{1}{p} - 2S_q = \frac{1}{p} + \frac{1}{1} - \sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k} = \frac{m}{n}$$
or we get that $$- 2S_q = \frac{1}{1} - \sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k} = \frac{m}{n}-\frac{1}{p}$$
Now, note that $$\sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k}\equiv \sum_{k=1}^{p-1}\frac1k \equiv \sum_{k=1}^{p-1}k \equiv 0$$ mod $p$
So we get that $$\frac{m}{n}\equiv 1$$ mod $p$ .
Hence we have $$1-\frac{m}{n}\equiv 0$$ mod p $$\implies m-n \equiv 0 $$ mod $p$.
And we are done!
| {
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Probability of exactly $2$ sixes in $3$ dice rolls where $2$ dice have $6$ on $2$ faces? Three dice are rolled. One is fair and the other two have 6 on two faces.
Find the probability of rolling exactly 2 sixes.
My textbook gives an answer of $\frac{20}{147}$ but I get an answer of: $$\frac{1}{6}\frac{2}{6}\frac{4}{6}+\frac{1}{6}\frac{4}{6}\frac{2}{6}+\frac{5}{6}\frac{2}{6}\frac{2}{6}=\frac{8}{216}+\frac{8}{216}+\frac{20}{216}=\frac{36}{216}=\frac{1}{6}$$
I just want to know where I am going wrong or could the textbook be mistaken ?
| Your solution seems correct, indeed also by the naif definition of probability we obtain
$$p=\frac{\text{#favorauble cases}}{\text{#total cases}}=\frac{8+8+20}{6^3}=\frac16$$
| {
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$6$-digit permutation problem How many $6$-digit different numbers greater than $400000$ can be formed from the digits $2$, $2$, $5$, $6$, $7$, $7$, $8$?
Note: $22$ and $77$ are allowed in the number. For instance, $622775$ is allowed.
My approach: $5$ ways to choose the first digit, $6$ ways to choose the second, $5$ ways to choose the third, $4$ ways to choose the fourth, $3$ ways to choose the fifth and $2$ ways to choose the sixth. So
the total arrangement is $5\times 6\times 5\times 4\times 3\times 2 = 3600$. But with the $2,2, 7, 7$ coming up, I think there is more do be done with it. I tried dividing $3600$ by $(2! \times 2!)$. I tried this with a small number of digits and it gave the wrong answer. I am stuck. Can anyone help me with the answer or hint. Thanks
| We apply generating function to the problem. First, we find how many 6-digit number can be formed from 2, 2, 5, 6, 7, 7, 8 starting with any digit. Notice that 2, 2, 7, 7, can appeal at most 2, which corresponds to $(1 + x + \frac{x^{2}}{2!})^{2}$. Also, 5, 6, 8 can appeal at most 1, which corresponds to $(1 + x)^{3}$. Now find the coefficient of $\frac{x^{6}}{6!}$ in the product $(1 + x + \frac{x^{2}}{2!})^{2}(1 + x)^{3} = \frac{x^{7}}{4} + \frac{7x^{6}}{4} + \frac{23x^{5}}{4} + \frac{45x^{4}}{4} + 14x^{3} + 11x^{2} + 5x + 1$, which is $\frac{6! \times 7}{4} = 1260$.
Next, find the 6-digit numbers that begin with 2. In that case, 5 slots to fill using 2, 5, 6, 7, 7, 8. Again, 7, 7, can appeal at most 2 giving $(1 + x + \frac{x^{2}}{2!})$ and 2, 5, 6, 8 can appeal at most 1 giving (1 + x)^{4}. Since the have 5 choices, we find the coefficient of $\frac{x^{5}}{5!}$ in the product $(1 + x + \frac{x^{2}}{2!})(1 + x)^{4} = \frac{x^{6}}{2} + 3x^{5} + 8x^{4} + 12x^{3} + \frac{21x^{2}}{2} + 5x + 1$, which is $5! \times 3 = 360$.
Finally, the 6-digit number can't start with 2, and hence total number of arrangement is $1260 -360 = 900$ ways.
| {
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Prove $\sum_{b=0}^{n-1}b\binom{n+1-b}{2} = \binom{n+2}{4}$
An equilateral triangle of side $n$ is divided into $n^2$ equilateral
triangles of side $1$ such that each side of the congruent triangles
is parallel to the original triangle. Find the number of
parallelograms that can be formed by the segments.
The answer to this problem is $3 \binom{n+2}{4}$. I have a different solution than the intended one and arrived at the following sum:
$$3\sum_{b=0}^{n-1}b\binom{n+1-b}{2}$$
Wolfram Alpha verifies that it is equal to $3\binom{n+2}{4}$, however I can't prove it. I tried pairing the terms, applying some identities, but failed.
| Marcus M has given a very nice combinatorial proof, which in general I prefer. Here’s a computational proof, in case you’re more comfortable with that approach.
$$\begin{align*}
\sum_{b=0}^{n-1}b\binom{n-1-b}2&\overset{(0)}=\sum_{k=0}^{n-1}(n-1-k)\binom{k}2\\
&=n\sum_{k=0}^{n-1}\binom{k}2-\sum_{k=0}^{n-1}(k+1)\binom{k}2\\
&\overset{(1)}=n\binom{n}3-3\sum_{k=0}^{n-1}\binom{k+1}3\\
&=n\binom{n}3-3\sum_{k=1}^n\binom{k}3\\
&\overset{(1)}=n\binom{n}3-3\binom{n+1}4\\
&=\frac{n^2(n-1)(n-2)}6-\frac{(n+1)n(n-1)(n-2)}8\\
&=\frac{n(n-1)(n-2)}2\left(\frac{n}3-\frac{n+1}4\right)\\
&=\frac{n(n-1)(n-2)}2\cdot\frac{n-3}{12}\\
&=\frac{n(n-1)(n-2)(n-3)}{24}\\
&=\binom{n}4\;.
\end{align*}$$
At $(0)$ I substituted $k=n-1-b$, and at $(1)$ I used the hockey stick identity.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Solving : $5^{x^2+6x+8}$ = 1
Solve for $x$:
$$5^{x^2+6x+8} = 1$$
So, I took the natural logarithm on both sides,
$$(x^2+6x+8)\ln(5) = \ln(1)$$
then I divide both sides by $\ln(5)$ to set the polynomial to zero because we know $\ln(1) = 0$.
I will be left with:
$$x^2+6x+8 = 0$$
Factoring this will give:
$$(x+2)(x+4) = 0 \implies x = -2, -4 $$
Then I checked my $x$ values I got $1$. So my question is did I do it correctly?
| Another way to reach the same equation consists in noticing that the exponential function $a^{x}$ is injective:
\begin{align*}
5^{x^{2} + 6x + 8} = 1 = 5^{0} \Longleftrightarrow x^{2} + 6x + 8 = 0 \Longleftrightarrow \ldots
\end{align*}
| {
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"question_score": "5",
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What is $\lim_{N\to\infty}\frac{-2}{\pi}\sum_{n=1}^N \frac{(-1)^n}{n} \sin(n\frac{N\pi}{N+1})$? Here is what I have so far:
$$\lim_{N\to\infty} f_N \left(\frac{N\pi}{N+1}\right)$$
$$f_N (x) = \frac{-2}{\pi}\sum_{n=1}^N \frac{(-1)^n}{n} \sin(nx)$$
$$\implies \lim_{N\to\infty} f_N \left( \frac{N\pi}{N+1}\right)=\lim_{N\to\infty}\frac{-2}{\pi}\sum_{n=1}^N \frac{(-1)^n}{n} \sin\left(n\frac{N\pi}{N+1}\right)$$
$$=\lim_{N\to\infty}\frac{-2}{\pi}\left(-\sin\left(\frac{N\pi}{N+1}\right)+\frac{1}{2}\sin\left(\frac{2N\pi}{N+1}\right)-\frac{1}{3}\sin\left(\frac{3N\pi}{N+1}\right)+\cdots \pm \frac{1}{N}\sin\left(\frac{N^2 \pi}{N+1}\right) \right)$$
$$=\frac{2}{\pi}\lim_{N\to\infty}\frac{N\pi}{N+1}\left(\frac{\sin( \frac{N\pi}{N+1})}{\frac{N\pi}{N+1}} -\frac{\sin( \frac{2N\pi}{N+1})}{\frac{2N\pi}{N+1}}+\frac{\sin( \frac{3N\pi}{N+1})}{\frac{3N\pi}{N+1}} ... \pm \frac{\sin( \frac{N^2 \pi}{N+1})}{\frac{N^2 \pi}{N+1}}\right)$$
And in the last step I suppose I use the Riemann sum using midpoints to find the corresponding integral and evaluate it however I am a bit confused as to what integral I get.
If there is a better way to evaluate this limit, I am open to suggestions.
The expected answer is approximately 1.18.
| Since
$$
\sin \left( {n\frac{{N\pi }}{{N + 1}}} \right) = \sin \left( {\pi n - \frac{{\pi n}}{{N + 1}}} \right) = ( - 1)^{n+1} \sin \left( {\frac{{\pi n}}{{N + 1}}} \right),
$$
we have
$$
\frac{2}{{\pi (N + 1)}}\sum\limits_{n = 1}^N {\frac{{N + 1}}{n}\sin \left( {\frac{{\pi n}}{{N + 1}}} \right)} \to \frac{2}{\pi }\int_0^1 {\frac{\sin (\pi x)}{x}dx} = 1.1789797\ldots .
$$
| {
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how to find the geometric centroid of a trapezoid? I'm doing a project involving a trapezoidal prism
I need to find the center of mass. To do this, I need to first find the geometric centroid of the trapezoid.
I have found the formulas online. According to Wolfram Mathworld,
$$\begin{align}
\bar{x} &= \frac{b}{2} + \frac{(2a+b)(c^2-d^2)}{6(b^2-a^2)}\\
\bar{y} &= \frac{b+2a}{3(a+b)}h
\end{align}$$
https://mathworld.wolfram.com/Trapezoid.html
The project I'm doing is too serious for me to just take these formulas as given off the internet. I want to derive them from scratch.
However, I cannot find a derivation.
Can someone derive the geometric centroid of a trapezoid? If it's a long proof, then just sketch the proof for me and I can derive it myself.
| If you choose a coordinate system where $b$ is on the positive $x$ axis, between origin $(0, 0)$ and $(b, 0)$, the four vertices of the trapezoid are
$$(0, 0), \quad (b, 0), \quad (f+a, h), \quad (f, h)$$
Since trapezoids are simple polygons, we can use the shoelace formula for the area of a 2D polygon,
$$A = \displaystyle \frac{1}{2} \sum_{i=0}^{n-1} x_i y_{i+1} - x_{i+1} y_i \tag{1}\label{None1}$$
and the centroid of a simple polygon:
$$\left\lbrace ~ \begin{aligned}
\overline{x} &= \displaystyle \frac{1}{6 A} \sum_{i=0}^{n-1} (x_i + x_{i+1})(x_i y_{i+1} - x_{i+1} y_i) \\
\overline{y} &= \displaystyle \frac{1}{6 A} \sum_{i=0}^{n-1} (y_i + y_{i+1})(x_i y_{i+1} - x_{i+1} y_i) \\
\end{aligned} \right . \tag{2}\label{None2}$$
where $x_0 = 0$, $x_1 = b$, $x_2 = f + a$, $x_3 = f$, and $x_4 = x_0 = 0$; and $y_0 = 0$, $y_1 = 0$, $y_2 = h$, $y_3 = 0$, and $y_4 = y_0 = 0$.
These yield
$$\begin{aligned}
\overline{x} &= \displaystyle \frac{f (b + 2 a) + (a + b)^2 - a b}{3 (a + b)} \\
\overline{y} &= \displaystyle h \frac{b + 2 a}{3 (a + b)} \\
\end{aligned} \tag{3}\label{None3}$$
You can solve $f$ and $h$ by splitting the trapezoid into two right triangles and a rectangle in between, and solving the system of three equations and three unknowns (the third being $g$, such that $f + a + g = b$). It yields
$$\begin{aligned}
f &= \displaystyle \frac{b - a}{2} + \frac{c^2 - d^2}{2 (b - a)} \\
g &= \displaystyle \frac{b - a}{2} - \frac{c^2 - d^2}{2 (b - a)} \\
h &= \displaystyle \frac{\sqrt{ 2 (d^2 + c^2)(b - a)^2 - (d^2 - c^2)^2 - (b - a)^4}}{2 ( b - a ) } \\
\end{aligned} \tag{4}\label{None4}$$
Substituting $f$ in the centroid $x$ coordinate indeed yields, after simplification, $\overline{x} = (b/2) + (2 a + b)(c^2 - d^2) / (6 (b^2 - a^2))$.
As to references, the above links to Wikipedia have references you can use.
Or, you can start with the integral definition.
Given the characteristic function $g(x, y)$ of the shape, $g(x,y)$ being $1$ inside the shape and $0$ outside, the centroid is
$$\begin{aligned}
\overline{x} &= \displaystyle \frac{ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x g(x, y) ~ d x ~ d y }{ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} g(x, y) ~ d x ~ d y } \\
\overline{y} &= \displaystyle \frac{ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} y g(x, y) ~ d x ~ d y }{ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} g(x, y) ~ d x ~ d y } \\
\end{aligned} \tag{5}\label{None5}$$
The characteristic function is the weight function, so one might consider this the definition of the 2D centroid. (For references, look for "centroid integral".)
For the $y$ axis, the integral simplifies to
$$\begin{aligned}
\overline{y} &= \frac{ \int_{0}^{h} y \left(a + \frac{b - a}{h} y\right) ~ dy }{ \int_{0}^{h} a + \frac{b - a}{h} y ~dy } \\
~ &= \frac{ ~ \frac{ 2 b + a }{ 6 } h^2 ~ }{ ~ \frac{ b + a }{2} h ~ } \\
~ &= \frac{ 2 b + a }{ 3 ( b + a ) } h \\
\end{aligned} \tag{6a} \label{None6a}$$
For the $x$ axis, we need to split the integral into three parts. From above, we already know the divisor integral evaluates to $h (b + a) / 2$:
$$\begin{aligned}
\overline{x} &= \frac{2}{h (b + a)} \biggr( \int_{0}^{f} x \left( \frac{h}{f} x \right) ~ d x ~ + \int_{f}^{f+a} x h ~ d x ~ + \int_{f+a}^{b} x \left(\frac{b - x}{b - f - a} h \right) ~ d x \biggr) \\
~ &= \frac{2}{h (b + a)} \biggr( \frac{f^2 h}{3} ~ + ~ \frac{2 a f h + a^2 h}{2} ~ + ~ \frac{ h }{6} \left( f ( b - 4 a - 2 f ) + b^2 + a b - 2 a^2 \right) \biggr) \\
~&= \frac{(2 a + b) f + (a + b)^2 - a b}{3 (a + b)} \\
\end{aligned}$$
which is exactly the same as we got in the polygon form.
| {
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Better methods to approximate $2^{2\over 3}$ Recently while solving a problem on thermodynamics I ended up with $2^{2\over 3}$ .
Now the problem was on a test where no calculators were allowed and answer was required upto $2$ decimal digits.
I then resorted to binomial theorem for help (for $x\lt 1$) $$\left. \begin{array} { l } { ( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) x ^ { 2 } } { 2 ! } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } + \ldots \ldots + \frac { n ( n - 1 ) \ldots \ldots ( n - r + 1 ) } { r ! } x ^ { r } \ldots } \\ \end{array} \right.\text{upto}\, \, \infty$$
So the original problem can be written as :
$$2^{2\over 3}=4^{1\over 3}=(8-4)^{1\over 3}=2\left(1-\frac{1}{2}\right)^{1\over 3}$$
Now after evaluation first $3$ terms I ended up with $\left(2-\frac{1}{3}-\frac{1}{18}\right) \approx 1.61$ but the correct answer was $1.59$.
Also the average time you got per question was around $2$ minutes and I had already used more than half of it so I did not think of calculating more terms.
Now I am looking for a method which can help me evaluate $2^{2\over 3}$ faster and more precisely.
Can anyone please help me with this ?
| Perhaps more elaboration on how to use Newton's method, since I personally find it very easy to use to find the first few digits of $n$th roots using only basic calculations.
From the binomial expansion, we know that
$$(x+\Delta x)^3=x^3+3x^2\Delta x+\mathcal O((\Delta x)^2)$$
Our goal is essentially finding the $\Delta x$ that fixes $x^3$ so that $(x+\Delta x)^3$ is closer to $4$. In this way, we can find $\sqrt[3]4$.
Setting it equal and solving, we end up with
$$x^3+3x^2\Delta x=4\implies\Delta x=\frac{4-x^3}{3x^2}=\dfrac{\dfrac4{x^2}-x}3$$
Adding $x$ to both sides to find our new $x$ gives us
$$x+\Delta x=\frac{4+2x^3}{3x^2}=\dfrac{\dfrac4{x^2}+2x}3$$
Starting with values of $x$ closer to $\sqrt[3]4$ such as the $1.61$ you found will lead to faster correction, but for example purposes we start far away at $x=1$, which makes the next estimate become
$$x+\Delta x=\frac{4+2}3=2$$
Substituting $x=2$ in now gives us
$$x+\Delta x=\frac{4+16}{12}=\frac53\simeq1.66$$
The next estimate is a tad messier but still manageable:
$$x+\Delta x=\frac{4+2\times\frac{125}{27}}{3\times\frac{25}9}=\frac{4\times27+2\times125}{3\times3\times25}=\frac{108+250}{225}=\frac{358}{225}\simeq1.591$$
In general you will find that
$$x+\Delta x=\frac{a+(n-1)x^n}{nx^{n-1}}$$
for computing $\sqrt[n]a$, which works out to give the first few digits fairly fast and easily provided a good initial estimate (as you had done) and that $n$ is not too large.
| {
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"timestamp": "2023-03-29T00:00:00",
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Need help with $\arccos$ equation I have the equation
$$ \cos(2x + \frac{\pi}{9}) = 0.5$$
I know that in order to solve for $x\in \Bbb R$, I need to use
$$\arccos(0.5) = 2x + \frac{\pi}{9} $$
This yields
$$ 2x + \frac{\pi}{9} =
\begin{cases}
\frac{\pi}{3} + 2k\pi, & \text{Positive angle} \\
2 \pi - \frac{\pi}{3}+ 2k\pi, & \text{Negative angle}
\end{cases} $$
I would then subtract $\frac{\pi}{9}$ from both sides and get:
$$ 2x =
\begin{cases}
\frac{2\pi}{9} + 2k\pi, & \text{Positive angle} \\
\frac{14\pi}{9}+ 2k\pi, & \text{Negative angle}
\end{cases} $$
However according to the handout the correct solution is:
$$ 2x =
\begin{cases}
\frac{4\pi}{9} + 2k\pi, & \text{Positive angle} \\
\frac{16\pi}{9}+ 2k\pi, & \text{Negative angle}
\end{cases} $$
Can anyone help me?
| Remember that $\;\cos x=\alpha\implies x=\pm\arccos x\;$. Besides this, we only need basic trigonometry to solve that equation:
$$\cos t=\frac12\iff t=\pm\frac\pi3+2k\pi\;,\;\;k\in\Bbb Z\implies$$
puting $\;t=2x+\frac\pi9\;$ we get
$$2x+\frac\pi9=\pm\frac\pi3+2k\pi\implies\begin{cases}2x=\cfrac{2\pi}9+2k\pi,&\text{(positive solution)}\\{}\\
2x=-\cfrac{4\pi}9+2k\pi=\cfrac{14\pi}9+2k\pi,&\text{(negative solution)}\end{cases}\;,\;k\in\Bbb Z$$
Thus you're right, the handout's solution is wrong.
| {
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Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + 2$ = $(x - 1)(x - 2)$
and I can see $(x - 1)^2 \equiv 1$ $($mod $x - 2)$ . We also have :-
$$\frac{(x - 1)^{100}}{(x - 1)(x - 2)} = \frac{(x - 1)^{99}}{(x - 2)}.$$
We have :- $(x - 1)^{98} \equiv 1$ $($mod $x - 2).$ $\rightarrow (x - 1)^{99} \equiv (x - 1)$ $($mod $x - 2)$. Now for the case of $(x - 2)^{200}$ we have :-
$$\frac{(x - 2)^{200}}{(x - 1)(x - 2)} = \frac{(x - 2)^{199}}{(x - 1)}.$$
We have :- $(x - 2) \equiv (-1)$ $($mod $x - 1)$ $\rightarrow (x - 2)^{199} \equiv (-1)$ $($mod $x - 1)$.
Adding all these up we have :- $(x - 1)^{100} + (x - 2)^{200} \equiv (x - 2)$ $($mod $x² - 3x + 2)$ .
On checking my answer with wolfram alpha , I found the remainder to be $1$, so I messed up in some step .
Can anyone help me?
| You are right that $(x-1)^{98}\equiv1\pmod{(x-2)}$. But that implies
$$(x-1)^{100}\equiv(x-1)^2=x(x-2)+1\equiv1\pmod{x-2}.$$
More naively, as
$$x-1\equiv1\pmod{x-2}$$
then
$$(x-1)^{100}\equiv1^{100}=1\pmod{x-2}.$$
Similarly,
$$x-2\equiv-1\pmod{x-1}$$
and
$$(x-2)^{200}\equiv(-1)^{200}=1\pmod{x-1}.$$
So $(x-1)^{100}+(x-2)^{200}$ is congruent to $1$ modulo both $x-1$ and $x-2$,
and so also modulo $(x-1)(x-2)$.
| {
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Given $\sinh x$, find the exact value of $\cosh x, \cos 2x$ and $\tan 2x$ Given $\sinh x = 8/14$, find the exact value of $\cosh x, \cos 2x$ and $\tan 2x$.
I have been getting two answers which has made me confused. I keep getting $\sqrt{65/7}$ or $\sqrt{4/7}.$
That's what I got: $$\cosh^2 x - \sinh^2 x = 1$$ $$\cosh^2 x = 1 + \sinh^2 x$$ $$\cosh^2 x = 1 +\left(\frac{8}{14}\right)^2$$ $$\cosh^2 x=\frac{65}{49}$$ $$\cosh (x) = \sqrt{\frac{65}{49}}$$ $$\cosh x = \frac{\sqrt{65}}{7}$$
| $$\sinh x =\frac{e^x-e^{-x}}{2}= \frac{4}{7}$$
We want
$$\cosh x = \frac{e^x+e^{-x}}{2} = q$$
Adding these:
$$e^x = \frac{4}{7}+q$$
Subtracting the first equation from the second:
$$e^{-x} = q - \frac{4}{7}$$
So $$\frac{4}{7}+q = \frac{1}{q-\frac{4}{7}}$$
$$q^2 - \left(\frac{4}{7}\right)^2 =1$$
$$q^2 =1+\frac{16}{49}$$
$$\cosh x = q=\frac{\sqrt{65}}{7}$$(q must be positive because each exponential is positive.)
| {
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Elegant way of finding the least perimeter of triangle A triangle $ABC$ has positive integer sides, $\angle A = 2\angle B$ and $\angle C > \pi/2$ , then the minimum length of the perimeter of $ABC$ is?
We have $\angle A = 2\angle B$
$\Rightarrow \sin A=\sin 2B=2 \sin B \cos B $
$\sin C=\sin(\pi-3B)=\sin(3B)=3\sin B-4\sin^3B$
Using $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
$a=2b\cos B $
$c=b(3-4\sin^2 B)$
From above two equations $a^2=b(c+b) $
NOTE: similar questions have been asked earlier but I want to do this question analysing with $a^2=b(c+b)$ and no fancy inequalities
| First of all, it is consider very bad style to just throw random equations around. Write down in English what exactly you are doing --- is it an assumption you make? A given condition? Some logical deduction from earlier? And every sentence should start with an English word not an equation, unless you absolutely have to.
Note that it suffices to consider $a,b,c$ coprime (otherwise you get a smaller triangle by scaling). Now $a^2=b^2+bc$ is equivalent to $c^2+4a^2=(2b+c)^2$. If $c$ is even we have $a^2+(\frac12c)^2=(b+\frac12c)^2$, and we must have $a$ odd (otherwise $a,c$ even gives $b$ also even, and so not primitive). So applying the classification of primitive Pythagorean triples, we have:
\begin{align*}
(c,2a,2b+c)&=(m^2-n^2,2mn,m^2+n^2)\text{ or }\\
(\tfrac12c,a,b+\tfrac12c)&=(2mn,m^2-n^2,m^2+n^2)\\
\end{align*}
for some $m,n$ coprime, $m>n$ opposite parity. Now analyse each case separately:
Case 1: $(c,2a,2b+c)=(m^2-n^2,2mn,m^2+n^2)$, so $(a,b,c)=(mn,n^2,m^2-n^2)$. We want $c^2>a^2+b^2$, so $m^2>3n^2$. Also triangle inequality $c<a+b$ gives $(m+n)(m-2n)<0$. So seek $\sqrt3<\frac{m}n<2$ (it is probably not a good idea to bring in $\sqrt3$ but you know what I mean) and we will have perimeter $m(m+n)$. $\frac{m}n=\frac74$ is obviously the best candidate here with least $n$ and least $m$. So we have $(a,b,c)=(28,16,33)$ and perimeter $77$.
Case 2: $(\tfrac12c,a,b+\tfrac12c)=(2mn,m^2-n^2,m^2+n^2)$ so $(a,b,c)=(m^2-n^2,(m-n)^2,4mn)$. Then $c^2>a^2+b^2$ gives $16m^2n^2>(m+n)^2(m-n)^2+(m-n)^4$, i.e., $(m+n)^2 (m^2-4mn+n^2)<0$ and $c<a+b$ gives $m>3n$, so $3<\frac{m}{n}<2+\sqrt3$ and the perimeter is $2m(m+n)$. The best candidate here is the choice $\frac{m}n=\frac72$, which gives $(a,b,c)=(45,25,56)$ and perimeter $126$.
So the least perimeter is 77.
| {
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Homogenous Equation with Complex Vector Solution: Converting to Real Functions Solving the system
$$
\begin{array}{l}\frac{d x}{d t}=6 x-y \\ \frac{d y}{d t}=5 x+4 y\end{array}
$$
we get $\lambda_{1}=5+2 i, \lambda_{2}=5-2 i$ eigenvalues. So eigenvectors and corresponding solutions is:
$$
\mathbf{K}_{1}=\left(\begin{array}{c}1 \\ 1-2 i\end{array}\right), \quad \mathbf{X}_{1}=\left(\begin{array}{c}1 \\ 1-2 i\end{array}\right) e^{(5+2 i) t}
$$
$$
\mathbf{K}_{2}=\left(\begin{array}{c}1 \\ 1+2 i\end{array}\right), \quad \mathbf{X}_{2}=\left(\begin{array}{c}1 \\ 1+2 i\end{array}\right) e^{(5-2 i) t}
$$
Thus general solution:
$$
\mathbf{X}=c_{1}\left(\begin{array}{c}1 \\ 1-2 i\end{array}\right) e^{(5+2 i) t}+c_{2}\left(\begin{array}{c}1 \\ 1+2 i\end{array}\right) e^{(5-2 i) t}
$$
So converting the above solution to a real function via Euler's:
$$
\begin{array}{l}e^{(5+2 i) t}=e^{5 t} e^{2 t i}=e^{5 t}(\cos 2 t+i \sin 2 t) \\ e^{(5-2 i) t}=e^{5 t} e^{-2 t i}=e^{5 t}(\cos 2 t-i \sin 2 t)\end{array}
$$
My text states to collect terms and replace $ c_1 + c_2 $ by $ C_1 $ and $ (c_1 - c_2)i $ by $ C_2 $ the solution becoming $\mathbf{X}=C_{1} \mathbf{X}_{1}+C_{2} \mathbf{X}_{2}$ where
$$
\mathbf{X}_{1}=\left[\left(\begin{array}{l}1 \\ 1\end{array}\right) \cos 2 t-\left(\begin{array}{r}0 \\ -2\end{array}\right) \sin 2 t\right] e^{5 t}
$$
$$
\mathbf{X}_{2}=\left[\left(\begin{array}{r}0 \\ -2\end{array}\right) \cos 2 t+\left(\begin{array}{l}1 \\ 1\end{array}\right) \sin 2 t\right] e^{5 t}
$$
I seem to be making a mistake somewhere and don't get to the same simplification. I'm looking for some of the steps to the books specification. I've tried a few times by multiplying the general solution across rows, fully expanding outside of its matrix form.
| We have
$$\lambda_1 = 5 + 2 i, v_1 = \begin{pmatrix} 1 \\ 1 - 2 i \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} +i \begin{pmatrix} 0 \\ - 2 \end{pmatrix}$$
We form $x_1(t) = e^{\lambda_1 t}v_1$, while expanding using Euler's Formula and this approach
$$x_1(t) = e^{5t}(\cos 2 t + i \sin 2t) \left(\begin{pmatrix} 1 \\ 1 \end{pmatrix} +i \begin{pmatrix} 0 \\ - 2 \end{pmatrix} \right) = e^{5t}\left(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\cos 2t +i^2 \begin{pmatrix} 0 \\ - 2 \end{pmatrix}\sin 2t \right)$$
This is enough to solve the system, but you can try the second eigenvector.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\lim_{n\to \infty} \sum_{k=1}^n\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$ exist? Let $x \in \mathbb{R}.$ Is is true that the following limit exists : $$\lim_{n \to \infty} \sum_{k=1}^n\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$$ What is the value of this limit?
I tried the Integral test for convergence, but nothing came out.
Any suggestions?
| We can write
$$
\sum\limits_{k = 1}^n {\log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)} = - \frac{{x^2 }}{{2n}}\sum\limits_{k = 1}^n {\sin ^2 k} + \sum\limits_{k = 1}^n {\left[ {\frac{{x^2 \sin ^2 k}}{{2n}} + \log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)} \right]} .
$$
Here
$$
\sum\limits_{k = 1}^n {\sin ^2 k} = \frac{n}{2} + \mathcal{O}(1).
$$
Suppose that $n$ is so large that $x^2 \le n$. Then
$$
\left| {\frac{{x^2 \sin ^2 k}}{{2n}} + \log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)} \right| \le \frac{{x^4 \sin ^4 k}}{{4n^2 }}\le \frac{{x^4}}{{4n^2 }}.
$$
Hence,
$$
\left| {\sum\limits_{k = 1}^n {\left[ {\frac{{x^2 \sin ^2 k}}{{2n}} + \log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)} \right]} } \right| \le \frac{{x^4 }}{{4n }}.
$$
From these estimates, we can see that
$$
\sum\limits_{k = 1}^n {\log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)} \to - \frac{{x^2 }}{4}
$$
uniformly on compact subsets of $\mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Systems of polynomial equations involving sums of equal powers Given the following system of polynomial equations:
$$
\left\{\begin{array}{lclclcr}
x & + & y & + & z & = & 1
\\
x^{2} & + & y^{2} & + & z^{2} & = & 14
\\
x^{3} & + & y^{3} & + & z^{3} & = & 36
\end{array}\right.
$$
What is $x^{5} + y^{5} + z^{5}\ {\large ?}$ .
How should I approach this? Is there a general formula for this kind of system?
| Let $p=x+y+z$, $q=xy+yz+zx$, $r=xyz$. So we have
$$\begin{cases}
p=1\\p^2-2q=14\\p^3-3pq+3r=36
\end{cases}$$
$$\begin{cases}
p = 1\\q = -\frac{13}{2}\\ r = \frac{31}{6}
\end{cases}$$
By consecutive eliminating the highest powers terms we get
$$x^5+y^5+z^5-(x+y+z)^5+5(xy+yz+xz)(x+y+z)^3-5(xy+yz+xz)^2(x+y+z)-5(x+y+z)^2xyz+5(xy+yz+xz)xyz=0$$
In other words, $x^5+y^5+z^5=p^5-5qp^3+5q^2p+5p^2r-5qr$ $=\frac{877}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786631",
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"source": "stackexchange",
"question_score": "2",
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The number of three digit numbers $abc$, which satisfy $a≤b>c$ is My approach in this question is as follows,
⇒When $a=1$
$-> 1-1-(0) / 1-2-(0,1) / 1-3-(0,1,2) ... / 1-9-(0,1...,8)$ ⇒Total numbers = $1+2+3+...+9$
⇒When $a=2$
$-> 2-2-(0,1) / 2-3-(0,1,2) ... / 2-9-(0,1...,8)$ ⇒Total numbers = $2+3+...+9$
⇒When $a=3$
$-> 3-3-(0,1,2) ... / 3-9-(0,1...,8)$ ⇒Total numbers = $3+...+9$
So on counting, we get total numbers as,
$$(1+2+...9)+(2+3+...9)+...(8+9)+(9)$$
$$⇒1^2+2^2+3^2+...9^2$$
$$⇒\frac{(9)(9+1)(2*9+1)}{6}$$
$$⇒285$$
But the actual solution says the answer is,
$$\bigl(2*({9 \choose 1}+{9 \choose 2}+{9 \choose 3})\bigr)$$
$$⇒258$$
I think I have recounted cases, can anyone explain which cases have I recounted??
| ⇒When $a=1$
$-> 1-1-(0) / 1-2-(0,1) / 1-3-(0,1,2) ... / 1-9-(0,1...,8)$ ⇒Total numbers = $1+2+3+...+9$
⇒When $a=2$
$-> 2-2-(0,1) / 2-3-(0,1,2) ... / 2-9-(0,1...,8)$ ⇒Total numbers = $2+3+...+9$
⇒When $a=3$
$-> 3-3-(0,1,2) ... / 3-9-(0,1...,8)$ ⇒Total numbers = $3+...+9$
So on counting, we get total numbers as,
$$(1+2+...9)+(2+3+...9)+...(8+9)+(9)$$
$$⇒1^2+2^2+3^2+...9^2$$
$$⇒\frac{(9)(9+1)(2*9+1)}{6}$$
$$⇒285$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Minimizing a function by finding its critical points Let $f_n(x)$ equal:
$$(2^n+2) \left(2x+1-\sqrt{2x^2+2x} \right)^n-x^n-(x+1)^n-\left(3x+1-2\sqrt{2x^2+2x}\right)^n-\left(3x+2-2\sqrt{2x^2+2x}\right)^n$$
Mathematica suggests that this function has two critical points on $(0,\infty)$, namely $x_1=1$ and $x_2=1/\sqrt{2}-1/2$. This seems to be independent of $n$ (well, I checked for all $4\leq n\leq 7$; also $f_1(x),f_2(x),f_3(x)$ identically equal to zero, a fact that I don't quite understand yet).
Problem: Prove that the only critical points of $f_n(x)$, $n\geq 4$, are $x_1=1$ and $x_2=1/\sqrt{2}-1/2$.
My goal is to show that $f_n(x)\geq 0$ for all $x\geq 0$. If I can show the only critical points are $1$ and $1/\sqrt{2}-1/2$, then by noting that $f_n(1)=0$ and $f_n(1\sqrt{2}-1/2)>0$ we have proved the claim. Of course, if you can prove $f_n(x)\geq 0$ in any way, please do share your proof.
| Some thoughts
It suffices to prove that $f_n(x) \ge 0$ for $x > 0$.
With the substitution $x = \frac{2}{u^2+2u-1}$ for $u > \sqrt{2} - 1$ (correspondingly, $u = \sqrt{2 + \frac{2}{x}} - 1$), we have
$$f_n(x) = \frac{1}{(u^2+2u-1)^n}
[(2^n+2)(u^2+1)^n - 2^n - (u+1)^{2n} - (u-1)^{2n} - 2^n u^{2n}].$$
It suffices to prove that, for $u > 0$,
$$(2^n+2)(u^2+1)^n - 2^n - (u+1)^{2n} - (u-1)^{2n} - 2^n u^{2n} \ge 0.$$
Denote the LHS by $g_n(u)$. Note that $u^{2n} g_n(\frac{1}{u}) = g_n(u)$ for $u > 0$.
Thus, it suffices to prove that, for $u \ge 1$,
$$(2^n+2)(u^2+1)^n - 2^n - (u+1)^{2n} - (u-1)^{2n} - 2^n u^{2n} \ge 0.$$
Let $u = 1 + v$. It suffices to prove that, for $v \ge 0$,
$$(2^n+2)((1+v)^2+1)^n - 2^n - (v+2)^{2n} - v^{2n} - 2^n (1+v)^{2n} \ge 0.$$
Denote the LHS by $h_n(v)$. It suffices to prove that
the polynomial $h_n(v)$ has non-negative coefficients.
I did some numerical experiments which support this claim.
| {
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"url": "https://math.stackexchange.com/questions/3788045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Nested functions Assuming that we have some function $L(x)$ such that $L(x) = x - \frac{x^2}{4}.$ Now, define $a_n$ as $$L \Bigl( L \Bigl( L \Bigl( \cdots L \Bigl( \frac{17}{n} \Bigr) \cdots \Bigr) \Bigr) \Bigr),$$ where we have $n$ iterations of $L.$ My question here is, what value does $n \cdot a_n$ approach as $n$ approaches infinity?
I tried to find some sort of pattern but it got nasty fast. I than tried to find a few small values and test them out, but they didn't quite work. How should I approach this problem? Thanks.
| I'd like to post the solution I came up with.
Note that $0 < L(x) < x$ for $0 < x < 2.$ Assuming $n$ is sufficiently large, i.e. $n \ge 9,$ we have that $0 < a_n < \frac{17}{n} < 2.$
From $L(x) = x - \frac{x^2}{2},$ we can write
$$\frac{1}{L(x)} = \frac{1}{x - \frac{x^2}{2}} = \frac{2}{2x - x^2} = \frac{2}{x(2 - x)} = \frac{x + (2 - x)}{x(2 - x)} = \frac{1}{x} + \frac{1}{2 - x},$$ so
$$\frac{1}{L(x)} - \frac{1}{x} = \frac{1}{2 - x} \quad (*).$$ For a nonnegative integer $k,$ let $L^{(k)}(x)$ denote the $k$th iterate of $L(x).$ Then $0 < L^{(k)}(x) < x,$ so
$$0 < L^{(k)} \left( \frac{17}{n} \right) \le \frac{17}{n}.$$ Hence,
$$\frac{1}{2} < \frac{1}{2 - L^{(k)} (\frac{17}{n})} \le \frac{1}{2 - \frac{17}{n}} = \frac{n}{2n - 17}.$$ By equation $(*),$
$$\frac{1}{L^{(k + 1)} (\frac{17}{n})} - \frac{1}{L^{(k)} (\frac{17}{n})} = \frac{1}{2 - L^{(k)} (\frac{17}{n})},$$ so
$$\frac{1}{2} < \frac{1}{L^{(k + 1)} (\frac{17}{n})} - \frac{1}{L^{(k)} (\frac{17}{n})} \le \frac{n}{2n - 17}.$$ Summing over $0 \le k \le n - 1,$ we get
$$\frac{n}{2} < \frac{1}{L^{(n)} (\frac{17}{n})} - \frac{1}{\frac{17}{n}} \le \frac{n^2}{2n - 17}.$$ Since $a_n = L^{(n)} \left( \frac{17}{n} \right),$ this becomes
$$\frac{n}{2} < \frac{1}{a_n} - \frac{n}{17} \le \frac{n^2}{2n - 17}.$$ Dividing by $n,$ we get
$$\frac{1}{2} < \frac{1}{na_n} - \frac{1}{17} \le \frac{n}{2n - 17}.$$ As $n$ approaches infinity, $\frac{n}{2n - 17}$ approaches $\frac{1}{2},$ so if $L$ is the limit of $na_n,$ then
$$\frac{1}{L} - \frac{1}{17} = \frac{1}{2}.$$ Solving, we find $L = \boxed{\frac{34}{19}}.$
Thanks to the people who helped me with this problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Proof that if $x,y>0$ and $x+y=1$, then $(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$ For positive reals $x$ and $y$ such that $x+y=1$, prove that $$(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$$
I have tried using Jensen’s inequality but it won’t cover all the possible choices for $x$ and $y$ since the concavity varies. I am trying to find a neat solution so that a generalization could also be made. Thank you.
| $$(2x)^{\frac 1 x}=\frac{1}{\left( \frac{1}{2x}\right) ^{\frac{1}{x}}}\leq \frac{1}{1+\frac{1}{x}\left( \frac{1}{2x}-1\right)}=\frac{2x^2}{2x^2-2x+1}$$ by Bernoulli’s inequality. The same holds for $y$ and one immediately computes that if $x+y=1$, $$\frac{2x^2}{2x^2-2x+1}+\frac{2y^2}{2y^2-2y+1}=2$$ and the result follows.(just observe that the denominator are the same $=x^2+y^2$)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Knowing that $ \sin(2x) \equiv 2 \sin(x)\cos(x) $ show that $\cos(2x) \equiv \cos^2x - \sin^2x $ Apparently one option is to differentiate the identity $ \sin(2x) \equiv 2 \sin(x)\cos(x) $ to get the identity $\cos(2x) \equiv \cos^2x - \sin^2x $. Which is surprising as I didn't realize that differentiating an identity produces another identity.
However, I'd like to know how it can be done without involving calculus.
I have the list of all trigonometric properties and operations at hand, but can't find the right way to relate them to go from one to the other.
Could I get a hint?
| $$\cos^2 (2x) = 1 - \sin^2 (2x) = 1 - 4 \sin^2 x \cos^2 x = 1 - 4 \sin^2 x (1 - \sin^2 x)$$
$$ = 4 \sin^4 x - 4 \sin^2 x +1$$
Now if we let $u = \sin x$, we have $4u^4 - 4u^2 + 1 = (2u^2-1)^2$. Thus:
$$\cos^2 (2x) = (2 \sin^2 x - 1)^2$$
$$\cos^2 (2x) = (2 \sin^2 x - (\sin^2 x +\cos^2 x) )^2$$
and now you are very close to the desired result.
One last note: from $a^2 = b^2$ we cannot directly conclude that $a = b$. However, we can disprove that $\cos(2x) = (\sin^2 x - \cos^2 x)$ by substituting in $x = 0$ for example: this gives $1 = -1$ which is false. Since what we have done is true no matter which $x$ we choose, the other possibility $\cos(2x) = -(\sin^2 x - \cos^2 x) = \cos^2 x - \sin^2 x$ must be true.
| {
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"source": "stackexchange",
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Show that $x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded, monotone, and find its limit Prove that $x_1 = 0, x_2 = 0, x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded and monotonic. Then find its limit.
My attempt at boundedness:
(Using induction) For the base case we have $0 \leq x_1 = 0 \leq 2$. Assume that the sequence is bounded for $n = k$. Then,
\begin{align*}
0 \leq x_k &\leq 2 \\
\vdots \\
\text{lower bound } \leq x_{k + 1} &\leq \text{upper bound}
\end{align*}
I am thrown off by the term $x_{n + 2}$ in the recursive formula and I can't see the algebra to produce the above steps without getting $x_{n + 2}$ in the expression of the upper / lower bound.
Thank you.
Update:
I have added this to the prove:
We have $0 \leq x_1 = 0 \leq 2$ and $0 \leq x_2 = 0 \leq 2$. Assume that the sequence is bounded for $k+1$,
\begin{align*}
0 &\leq x_{k + 1} \leq 2 \\
0 &\leq x_k + x_{k+1} \leq 4 \\
0 &\leq x_k + \frac{1}{3} x_{k+1} \leq 4 \\
0 &\leq \frac{1}{6} x_{k} + \frac{1}{3} x_{k+1} \leq 4 \\
0 &\leq x_{k+2} \leq 4
\end{align*}
Therefore, by the principle of mathematical induction, the sequence is bounded.
Is this valid?
| For boundedness we use Strong Induction, it is trivial that the sequence is positive.
We want to show that for all $n \in \mathbb{N}$ we have $x_{n} < 2$
*
*For k = 1 we have: $x_{1} = 0 < 2$
*Let $n \in \mathbb{N}$ and suppose that for all $k \leq n$ we have: $x_{k} < 2$
*We have: $x_{n-1} < 2$ and $x_{n} < 2$
Then: $\frac{1}{3}x_{n} + \frac{1}{6}x_{n-1} + 1 < \frac{2}{3} + \frac{2}{6} + 1$
Hence: $x_{n+1} < 2$
For monotony, Let use again induction to prove that for all $n \in \mathbb{N}$, $x_{n+1} \geq x_{n}$
*
*For n = 1, it is clearly that $x_{2} = 0 \geq x_{1}$ since $x_{1} = 0$
*Let $n \geq 2$ and suppose that for all $k \leq n$ we have: $x_{k+1} \geq x_{k}$
We have: $x_{n} \geq x_{n-1}$ and $x_{n+1} \geq x_{n}$
Hence: $\frac{1}{3}x_{n+1} + \frac{1}{6}x_{n} + 1 \geq \frac{1}{3}x_{n} + \frac{1}{6}x_{n-1} + 1$
Thus: $x_{n+2} \geq x_{n+1}$
We conclude that the sequence is increasing and thus it is monotone, And since it is bounded then the sequence converge.
Let $L$ be the limit of the sequence, then $L$ is solution to the equation $x = \frac{1}{3}x + \frac{1}{6}x + 1$, which gives that $L = 2$
| {
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How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$ Here is a new challenging problem:
Show that
$$I=\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx=2\ln(2)G-\frac{\pi}{8}\ln^2(2)-\frac{5\pi^3}{32}+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
My attempt:
With Weierstrass substitution we have
$$I=2\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1-x^2}{1+x^2}\right)dx\overset{x\to \frac{1-x}{1+x}}{=}4\int_0^1\frac{\frac{\pi}{4}-\arctan x}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx$$
$$=\pi\underbrace{\int_0^1\frac{1}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx}_{I_1}-4\underbrace{\int_0^1\frac{\arctan x}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx}_{I_2}$$
By setting $x\to \frac{1-x}{1+x}$ in the first integral we have
$$I_1=\frac12\int_0^1\frac{1}{x}\ln\left(\frac{1-x^2}{1+x^2}\right)dx$$
$$=\frac14\int_0^1\frac{1}{x}\ln\left(\frac{1-x}{1+x}\right)dx=\frac14\left[-\text{Li}_2(x)+\text{Li}_2(-x)\right]_0^1=-\frac38\zeta(2)$$
For the second integral, write $\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$
$$I_2=\frac12\int_0^1\frac{\arctan x}{1-x}\ln\left(\frac{2x}{1+x^2}\right)dx+\frac12\int_0^1\frac{\arctan x}{1+x}\ln\left(\frac{2x}{1+x^2}\right)dx$$
The first integral is very similar to this one
$$\int_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{192}\,\pi^{3} +
2\Im\left\{%
\text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}$$
So we are left with only $\int_0^1\frac{\arctan x\ln(1+x^2)}{1+x}dx$ as $\int_0^1\frac{\arctan x\ln x}{1+x}dx$ is already nicely calculated by FDP here. Any idea?
I noticed that if we use $x\to\frac{1-x}{1+x}$ in $\int_0^1\frac{\arctan x\ln(1+x^2)}{1+x}dx$ we will have a nice symmerty but still some annoying integrals appear.
In $I$, I also tried the Fourier series of $\ln(\cos x)$ but I stopped at $\int_0^{\pi/2} \frac{x\cos(2nx)}{\sin x}dx$. I would like to see different approaches if possible.
Thank you.
| Many ways to go are possible!
A simple way would be to exploit the known result,
$$\int_0^1 \frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)=\frac{\pi^3}{16},\tag 1$$
since with the Weierstrass subs the main integral reduces to
$$\mathcal{I}=2\int_0^1\frac{\arctan(x)}{x}\log\left(\frac{1-x^2}{1+x^2}\right)\textrm{d}x$$
$$=-2 \int_0^1 \frac{ \arctan(x)}{x}\log \left(\frac{1+x^2}{(1-x)^2}\right) \textrm{d}x-2 \int_0^1 \frac{\arctan(x)\log (1-x)}{x} \textrm{d}x$$
$$+2 \int_0^1 \frac{\arctan(x)\log (1+x) }{x} \textrm{d}x$$
$$=2\log(2)G-\frac{\pi}{8}\log^2(2)-\frac{5}{32}\pi^3+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\},$$
where the last two integrals are calculated by Ali Shather in this answer https://math.stackexchange.com/q/3261446.
End of story
Credit for this approach goes to Cornel.
A first note: Interestingly, different ways make the problem very difficult. It would be nice to have in place more ways to go.
A second note: The generalization of the key integral in $(1)$ may be found in the book, (Almost) Impossible Integrals, Sums, and Series, page $17$,
$$ \int_0^x \frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t-2 \int_0^1 \frac{\arctan(xt)\log (1-t)}{t}\textrm{d}t$$
$$=2\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{2n-1}}{(2n-1)^3}, \ |x|\le1.$$
| {
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Simplify $\sum_{k = 0}^n \left[ \binom{m + n + k}{k} 2^{n + 1 - k} - \binom{m + n + k + 1}{k} 2^{n - k} \right]$. This is Exercise 6 from page 44 of Analysis I by Amann and Escher.
Exercise:
Simplify the sum
\begin{align*}
S(m, n) := \sum_{k = 0}^n \left[ \binom{m + n + k}{k} 2^{n + 1 - k} - \binom{m + n + k + 1}{k} 2^{n - k} \right]
\end{align*}
for $m, n \in \mathbb N$.
Hint: for $1 \leq j < \ell$ we have $\binom{\ell}{j} - \binom{\ell}{j - 1} = \binom{\ell + 1}{j} - 2\binom{\ell}{j - 1}$.
My attempt:
Unfortunately I don't understand how to use the hint. I don't see how it corresponds with the expression in the sum.
\begin{align*}
\sum_{k = 0}^n \Bigg[ \binom{m + n + k}{k} 2^{n + 1 - k} - \binom{m + n + k + 1}{k} 2^{n - k} \Bigg] &= \sum_{k = 0}^n \Bigg[ 2^{n - k} \Big[ \binom{m + n + k}{k} 2 - \binom{m + n + k + 1}{k} \Big] \Bigg]\\
&= \sum_{k = 0}^n \Bigg[ 2^{n - k} \Big[ \binom{m + n + k}{k} + \binom{m + n + k}{k} - \binom{m + n + k + 1}{k} \Big] \Bigg]\\
&= \sum_{k = 0}^n \Bigg[ 2^{n - k} \Big[ \binom{m + n + k}{k} - \binom{m + n + k}{k - 1} \Big] \Bigg] \text{ (Pascal)}.
\end{align*}
At this point I'm stuck. I'm not sure if this is a dead end, especially since I didn't use the hint. I appreciate any help.
| Starting with
$$
\sum_{k = 0}^n \Bigg[ 2^{n - k} \Big[ \binom{m + n + k}{k} - \binom{m + n + k}{k - 1} \Big] \Bigg],
$$
and using the hint with $\ell=m+n+k$ and $j=k$, we get
$$
\sum_{k = 0}^n \Bigg[ 2^{n - k} \Big[ \binom{m+n+k+1}{k} - 2\binom{m+n+k}{k - 1} \Big] \Bigg]=\sum_{k = 0}^n\left(2^{n-k}\binom{m+n+k+1}{k}-2^{n-k+1}\binom{m+n+k}{k - 1}\right).
$$
This is a telescoping sum, so it can easily evaluated. Namely, letting
$$
a_k=2^{n-k}\binom{m+n+k+1}{k},
$$
then the sum in question is equal to
$$
\sum_{k=0}^n (a_k-a_{k-1}),
$$
which telescopes to $a_n-a_{-1}$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Ramanujan's $\sqrt{\frac{\pi e}{2}}$ formula The following identity is due to Ramanujan:
$$\DeclareMathOperator{\k}{\vphantom{\sum}\vcenter{\LARGE K}} \sqrt{\frac{\pi e}{2}}=\frac{1}{1+\k_{n=1}^\infty \frac{n}{1}}+\sum_{n=0}^\infty\frac{1}{(2n+1)!!}$$
or
$$\sqrt{\frac{\pi e}{2}}=\cfrac{1}{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{\ddots}}}}+1+\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\cdots $$
I'm interested in the proof of this identity, but I couldn't manage to find any reference except for the linked page.
| As pointed out by the linked page, it suffices to prove
$$ 1+\dfrac{1}{1+\dfrac{2}{1+\dfrac{3}{1+\ddots}}} = \sqrt{\frac{2}{\pi e}} \frac{1}{\operatorname{erfc}(1/\sqrt{2})}. \tag{1} $$
To this end, we will resort to the standard theory of continued fraction. Define $(p_n)$ and $(q_n)$ by the following relation:
$$
\begin{pmatrix} p_n \\ q_n \end{pmatrix} = A_1A_2\dots A_n \begin{pmatrix} 1 \\ 0 \end{pmatrix}
\quad\text{where}\quad
A_n = \begin{pmatrix} 1 & n \\ 1 & 0 \end{pmatrix}.
$$
Then it is routine to check that
\begin{align*}
p_0 &= 1, & p_1 &= 1, & p_{n+2} &= p_{n+1} + (n+1) p_n, \\
q_0 &= 0, & q_1 &= 1, & q_{n+2} &= q_{n+1} + (n+1) q_n.
\end{align*}
Moreover, if $f_A(z) = \frac{a_{11}z+a_{12}}{a_{21}z+a_{22}}$ denotes the linear fractional transformation induced by the $2\times2$ matrix $A=[a_{ij}]_{1\leq i,j\leq 2}$, then we have:
$$ \frac{p_n}{q_n} = f_{A_1\dots A_n}(\infty) = (f_{A_1}\circ\dots\circ f_{A_n})(\infty) = 1+\dfrac{1}{1+\dfrac{2}{\ddots+\dfrac{\ddots}{1+\dfrac{n-1}{1}}}} $$
The standard theory also affirms that this converges as $n\to\infty$. So it suffices to compute the limit as $n\to\infty$. To this end, note that both $p_n$ and $q_n$ are increasing and diverges to $\infty$. Moreover, if we introduce the exponential generating functions of $(p_n)$ and $(q_n)$ by
$$ y_p (x) = \sum_{n=0}^{\infty} \frac{p_n}{n!}x^n \quad\text{and}\quad y_q (x) = \sum_{n=0}^{\infty} \frac{q_n}{n!}x^n, $$
then they satisfy
$$ y_p' = (1+x)y_p \quad\text{and}\quad y_q' = 1 + (1+x)y_q. $$
These equations, together with the initial conditions $y_p(0) = p_0 = 1$ and $y_q(0) = q_0 = 0$, can be solved by the integrating factor method, and we obtain
$$ y_p(x) = e^{x+\frac{x^2}{2}} \quad \text{and} \quad y_q(x) = e^{x+\frac{x^2}{2}}\sqrt{\frac{\pi e}{2}} \left( \operatorname{erf}\left(\frac{1+x}{\sqrt{2}}\right) - \operatorname{erf}\left(\frac{1}{\sqrt{2}}\right) \right). $$
Now by invoking the standard argument for abelian theorem,
$$ \lim_{n\to\infty} \frac{p_n}{q_n} = \lim_{x\to\infty} \frac{y_p(x)}{y_q(x)} = \sqrt{\frac{2}{\pi e}} \frac{1}{\operatorname{erfc}\left(1/\sqrt{2}\right)} $$
as required.
| {
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"source": "stackexchange",
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$ f $ is differentiable in $ (0,0). $ Definition: Let $V\subseteq{\mathbb{R}^{m}}$ an open set, $a\in V$ y $f\colon V\to\mathbb{R}^{n}$ a function. We will say that $f$ is differentiable in $a,$ if exists a linear transformation $f'(a)\colon\mathbb{R}^{m}\to\mathbb{R}^{n}$
such that
\begin{equation}
f(a+h)=f(a)+f'(a)(h)+r(h),\qquad\lim_{h\rightarrow 0}{\dfrac{r(h)}{\lVert h\rVert}}=0.
\end{equation}
Let $ a \in \mathbb {R}$ be. Define the function $ f \colon \mathbb {R}^ {2} \to \mathbb {R} $ given by
\begin{equation}
f(x,y)=\left\{\begin{matrix}
\dfrac{x\sin^{2}(x)+axy^{2}}{x^{2}+2y^{2}+3y^{4}} & (x,y)\neq(0,0)\\
0 & (x,y)=(0,0)
\end{matrix}\right.
\end{equation}
Find the value of $ a $ so that $ f $ is differentiable by $ (0,0). $
My attempt:
We observed that
\begin{equation}
\dfrac{\partial f}{\partial x}(0,0)=0=\dfrac{\partial f}{\partial y}(0,0).
\end{equation}
If $(x,y)\in\mathbb{R}^{2}\setminus\{(0,0)\},$ then
\begin{equation}
\dfrac{\partial f}{\partial x}(x,y)=\dfrac{\sin^{2}(x)(2y^{2}+3y^{4}-x^{2})+x\sin(2x)(x^{2}+2y^{2}+3y^{4})+ay^{2}(2y^{2}+3y^{4}-x^{2})}{(x^{2}+2y^{2}+3y^{4})^{2}}
\end{equation}
\begin{equation}
\dfrac{\partial f}{\partial y}(x,y)=\dfrac{2axy(x^{2}-3y^{4})-4xy\sin^{2}(x)(1+3y^{2})}{(x^{2}+2y^{2}+3y^{4})^{2}}
\end{equation}
If $\dfrac{\partial f}{\partial y}(x,y)=0,$ then
\begin{align}
2axy(x^{2}-3y^{4})-4xy\sin^{2}(x)(1+3y^{2})=0&\quad\Longleftrightarrow\quad a(x^{2}-3y^{4})=2\sin^{2}(x)(1+3y^{2})\\
&\quad\Longleftrightarrow\quad a=\dfrac{2\sin^{2}(x)(1+3y^{2})}{x^{2}-3y^{4}}
\end{align}
\begin{equation}
f(x,y)=\left\{\begin{matrix}
x\sin^{2}(x) & (x,y)\neq(0,0)\\
0 & (x,y)=(0,0)
\end{matrix}\right.
\end{equation}
\begin{equation}
\dfrac{\partial f}{\partial x}(0,0)=0=\dfrac{\partial f}{\partial y}(0,0)
\end{equation}
From this it follows that $\dfrac{\partial f}{\partial x}(x,y)$ and $\dfrac{\partial f}{\partial y}(x,y)$ are continuous by $(0,0)$ y $f$ is differentiable by $(0,0).$
Are my arguments correct?
Any suggestion is welcome.
| A somewhat different approach:
In order to be differentiable, a function must be continuous and have a continuous derivative (or have a derivative with an essential singularity). Continuity requires that the limit as you approach the point be the same, regardless of direction of your approach.
Suppose we approach along the line $x=y=\epsilon$. Then we have (using the fact that $\frac{d}{da}\sin^2(a)=\sin(2a)$:
$$g(\epsilon)=f(\epsilon,\epsilon) = \frac{\epsilon\sin^2(\epsilon)+a\epsilon^3}{\epsilon^2+2\epsilon^2+3\epsilon^4} = \frac{\sin^2(\epsilon)+a\epsilon^2}{3\epsilon+3\epsilon^3}=\frac{1}{3}\frac{\sin^2(\epsilon)+a\epsilon^2}{\epsilon+\epsilon^3}$$
$$g'(\epsilon)=\frac{1}{3}\frac{(\epsilon+\epsilon^3)(\sin(2\epsilon)+2a\epsilon)-(\sin^2(\epsilon)+a\epsilon^2)(1+3\epsilon^2)}{\epsilon^2+2\epsilon^4+\epsilon^6} = \frac{1}{3}\frac{\epsilon\sin(2\epsilon)+2a\epsilon^2+\epsilon^3\sin(2\epsilon)+2a\epsilon^4-\sin^2(\epsilon)-a\epsilon^2-3\epsilon^2\sin^2(\epsilon)-3a\epsilon^5}{\epsilon^2+2\epsilon^4+\epsilon^6}$$
$$\lim_{\epsilon\rightarrow0}g'(\epsilon)=\frac{1}{3}\lim_{\epsilon\rightarrow0}\frac{\epsilon\sin(2\epsilon)+2a\epsilon^2+\epsilon^3\sin(2\epsilon)+2a\epsilon^4-\sin^2(\epsilon)-a\epsilon^2-3\epsilon^2\sin^2(\epsilon)-3a\epsilon^5}{\epsilon^2+2\epsilon^4+\epsilon^6} = \frac{1}{3} \lim_{\epsilon\rightarrow0} \frac{\sin(2\epsilon)+2\epsilon\cos(2\epsilon)+4a\epsilon+3\epsilon^2\sin(2\epsilon)+2\epsilon^3\cos(2\epsilon)+8a\epsilon^3-\sin(2\epsilon)-2a\epsilon-6\epsilon\sin^2(\epsilon)-3\epsilon^2\sin(2\epsilon)-15a\epsilon^4}{2\epsilon+8\epsilon^3+6\epsilon^5} = \frac{1}{3} \lim_{\epsilon\rightarrow0} \frac{2\epsilon\cos(2\epsilon)+2a\epsilon+2\epsilon^3\cos(2\epsilon)+8a\epsilon^3-6\epsilon\sin^2(\epsilon)-15a\epsilon^4}{2\epsilon+8\epsilon^3+6\epsilon^5} = \frac{1}{3} \lim_{\epsilon\rightarrow0} \frac{2\cos(2\epsilon)+2a+2\epsilon^2\cos(2\epsilon)+8a\epsilon^2-6\sin^2(\epsilon)-15a\epsilon^3}{2+8\epsilon^2+6\epsilon^4} = \frac{1}{3} \frac{2+2a}{2} = \frac{1+a}{3}$$
Suppose we approach along the line $-x=y=\epsilon$. Then we have:
$$h(\epsilon)=f(-\epsilon,\epsilon) = \frac{-\epsilon\sin^2(-\epsilon)-a\epsilon^3}{\epsilon^2+2\epsilon^2+3\epsilon^4} = \frac{-\epsilon\sin^2(\epsilon)-a\epsilon^3}{\epsilon^2+2\epsilon^2+3\epsilon^4}= -g(\epsilon)$$
$$h'(\epsilon)=-g'(\epsilon)$$
$$\lim_{\epsilon\rightarrow0}h'(\epsilon)=-\lim_{\epsilon\rightarrow0}g'(\epsilon)=-\frac{1+a}{3}$$
In both directions, the limit of the derivative existed, and therefore because the direction of approach doesn't matter, we require that the limits be the same.
$\frac{1+a}{3}=-\frac{1+a}{3}$, which means that $a=-1$.
| {
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"source": "stackexchange",
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Change this integral $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$. It is given that $f(a+b+1 -x) = f(x)$ where $a$ and $b$ are positive real numbers then $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$ is equal to
*
*$\int_{a-1}^{b-1} f(x) dx$
*$\int_{a+1}^{b+1} f(x+1) dx$
*$\int_{a-1}^{b-1} f(x+1) dx$
*$\int_{a+1}^{b+1} f(x) dx$
I tried using integration by parts along with this property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ and the one given in the question but I'm unable to get anything. It's that extra $x$ and the fraction $\frac{1}{a+b}$ which is making everything quite uneasy.
| $$I=\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx=I_1+I_2$$
$$I_1=\frac{1}{a+b}\int_{a}^{b}xf(x) dx$$
$$I_2=\frac{1}{a+b} \int_{a}^{b} x f(x+1) dx = \frac{1}{a+b} \int_{a+1}^{b+1} (u-1)f(u) du$$
N0w apply the property:
$$\int_{p}^{q} g(x) dx=\int_{p}^{q} g(p+q-x) dx$$
Then $$I_2=\frac{1}{a+b} \int_{a+1}^{b+1} (a+1+b+1-u-1)f(a+b+2-u) du$$
Use $f(a+b+2-u)=f(a+b+1-(u-1))= f(u-1)$. then
$$I_2=\frac{1}{a+b} \int_{a+1}^{b+1} (a+b+1-u) f(u-1) du.$$
Let $u-1=w$, then $$I_2=\frac{1}{a+b} \int_{a}^{b} (a+b-w)f(w) dw$$
Finally $$I=I_1+I_2=\int_{a}^{b} f(w) dw =\int_{a-1}^{b-1} f(x+1)dx$$
| {
"language": "en",
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"source": "stackexchange",
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Is this proof of $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ incomplete? So, for any angle $\alpha$ :
$$\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = \dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha} = \dfrac{\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}}{\dfrac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}= \dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}$$
Now, $\cos\alpha = \cos\Big(2\cdot\dfrac{\alpha}{2}\Big) = \dfrac{1-\tan^2\dfrac{\alpha}{2}}{1+\tan^2\dfrac{\alpha}{2}}$
Now, using the componendo and dividendo rule, we get :
$$\dfrac{\cos\alpha+1}{\cos\alpha-1} = \dfrac{2}{-2\tan^2\dfrac{\alpha}{2}} = \dfrac{-1}{\tan^2\dfrac{\alpha}{2}} \implies \tan^2\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{1+\cos\alpha}$$
$$\implies \tan^2\dfrac{\alpha}{2} = \dfrac{(1-\cos\alpha)(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)} = \Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)^2$$
$$\implies \Bigg|\tan\Big(\dfrac{\alpha}{2}\Big)\Bigg| = \Bigg|\dfrac{1-\cos\alpha}{\sin\alpha}\Bigg|$$
Now, only if $\mathrm{sign}\Big(\tan\dfrac{\alpha}{2}\Big) = \mathrm{sign}\Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)$ is true, we can say that $\tan\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{\sin\alpha}$
So, I think that without proving that, the proof will be incomplete but my Math textbook doesn't prove it.
So, is it necessary to prove it? If not, why not?
Thanks!
| Here's a simple way of proving the identity:
$$\frac{1-\cos x}{\sin x}=\frac{1-(1-2\sin^2\frac{x}{2})}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2} $$
as required. I hope that was helpful:)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Factor $x^8 + x^2 + 13$ into irreducible polynomials in $\Bbb F_{23} [x]$. I want to factor $x^8+ x^2 + 13$ into irreducible polynomials in $\Bbb F_{23} [x]$. I am trying using the method given in this link but not able to find its factors. Any help would be appreciated.
Thanks.
Factor $X^4 + 3$ into irreducible factors in $F_7[X]$
| I'd start by viewing this as $y^4+y+13$, where $y=x^2$. Does that factor? Search for linear factors by testing numbers as roots. It takes a little patience (or a calculator) but $y=8$ is a root. So you have $$(y-8)\left(y^3+8y^2-5y+7\right)$$
Does this factor further? Only if the cubic has a root. Checking for roots, it does not. So you have:
$$(x^2-8)\left(x^6+8x^4-5x^2+7\right)$$
Does $x^2-8$ factor? Again checking for linear factors, we find a pair.
$$(x-10)(x-13)\left(x^6+8x^4-5x^2+7\right)$$
Could that sixth degree polynomial factor further?
*
*There are no linear factors. (If it had a root, then the cubic from before would have had a root.)
*It could factor into a quadratic and an irreducible quartic. But for the original polynomial $f$, $f(x)=f(-x)$. So the map $x\to -x$ must permute the factors (while negating any odd-degree factor). In this case, the trivial permutation. So that quadratic would have to have no linear term, and the quartic would have no linear term nor a cubic term. This would make our cubic in $y$ from earlier factorable, which we know it is not.
*It could factor into three quadratics. Again, the map $x\to-x$ might preserve all three quadratics, but that means none of them have linear terms and our cubic in $y$ would have been factorable. The other possibility is that it only preserves one quadratic, and we have $$\begin{align}x^6+8x^4-5x^2+7&=(x^2+a)(x^2+bx+c)(x^2-bx+c)\\&=x^6+(2c+a)x^4+(c^2+2ac)x^2+ac^2\end{align}$$ from which we deduce $a=8-2c$, so $$\begin{align}x^6+8x^4-5x^2+7&=x^6+8x^4+(c^2+2(8-2c)c)x^2+(8-2c)c^2\\&=x^6+8x^4+(16-3c^2)x^2+(8-2c)c^2\end{align}$$ Can $16-3c^2$ equal $-5$? Then $3c^2=21$, and $c^2=7$. But $7$ is not a square mod $23$, which can be checked by inspection.
*It could factor into two cubics. Again , $x\to -x$ permutes the factors (and since both are odd-degree, negates them at the same time). It is impossible for the map $x\to-x$ to preserve the cubics while negating them, or each factor is of the form $x^3+ax$, and $0$ would be a root. It follows that they are negated and nontrivially permuted. So $$\begin{align}x^6+8x^4-5x^2+7&=(x^3+ax^2+bx+c)(x^3-ax^2+bx-c)\\&=x^6+(2b-a^2)x^4+(b^2-2ac)x^2-c^2\end{align}$$ Since $-7$ is a square, we can deduce wlog that $c=4$. So $$\begin{align}x^6+8x^4-5x^2+7&=x^6+(2b-a^2)x^4+(b^2-8a)x^2+7\end{align}$$ Now can we solve $2b-a^2\equiv8$ with $b^2-8a\equiv-5$? You can try each value of $a$ in the first relation to get a value of $b$, and try that in the second. You find $a\equiv1$, $b\equiv16$ works. So we have a factorization: $$\begin{align}x^6+8x^4-5x^2+7&=(x^3+x^2+16x+4)(x^3-x^2+16x-4)\end{align}$$
And then we conclude with: $$(x-10)(x-13)(x^3+x^2+16x+4)(x^3-x^2+16x-4)$$
| {
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"source": "stackexchange",
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What is the lowest value of $m$ if $m>2$ and $m^3-3m^2+2m$ is divisible by $79$ and $83$?
$m^3-3m^2+2m$ is divisible by $79$ and $83$ where $m>2$. Find the
lowest value of $m$
$m^3-3m^2+2m$ is the product of three consecutive integers. Both $79$ and $83$ are prime numbers. The product of three consecutive positive integers is divisible by $6$. So, $m^3-3m^2+2m$ is a multiple of $lcm(6,79,83)=39342$. But I can't go any further.
What would be the correct approach to solve problems like this?
| The brute force method: as $m^3−3m^2+2m=m(m−1)(m−2)$, and $79,83$ are prime, you can just solve the following nine congruences: $m\equiv\alpha\pmod{79}$, $m\equiv\beta\pmod{83}$, where $\alpha,\beta\in\{0,1,2\}$. This is possible as per Chinese Remainder Theorem, and the smallest of the nine $m$'s you will get (greater than $2$) is the solution.
It is easy to solve all those congruences simultaneously: per Wikipedia, we first express $1$ as $1=79u+83v$, where $u,v$ can be found using Euclidean algorithm. In this case, as $4=83-79$ and $1=20\cdot 4-79$, we have $1=20\cdot 83-21\cdot 79$.
Now, $m\equiv\alpha\pmod{79}$ and $m\equiv\beta\pmod{83}$ resolves as $m\equiv 20\cdot 83\alpha-21\cdot 79\beta\pmod{79\cdot 83}$, i.e. $m\equiv 1660\alpha-1659\beta\pmod{6557}$. This gives us the following table:
$$\begin{array}{r|r|r|r}\alpha&\beta&m\pmod{6557}&\text{smallest }m\gt 2\\\hline0&0&0&6557\\0&1&4898&4898\\0&2&3239&3239\\1&0&1660&1660\\1&1&1&6558\\1&2&4899&4899\\2&0&3320&3320\\2&1&1661&1661\\2&2&2&6559\end{array}$$
so the smallest solution seems to be $m=1660$.
| {
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How to evaluate $\lim_{n\to\infty} a_n$, where $a_{n+1} = \sqrt{1+\frac12 a_n}$? Note: A similar question (same recursive function) has been asked here, but none of the answers is relevant to my question.
I am trying to evaluate $\lim_{n\to\infty} a_n$. The sequence $a_n$ is given by the recursive function $$a_{n+1} = \sqrt{1+\frac12 a_n}$$ with $$a_1 =0$$
*
*I proved using induction that the monotonicity is: $a_n \nearrow$
*and that the upper bound is $a_n> \sqrt{2}$
Theorem 1
If $a_n$ is monotonous and bounded then it converges, thus its limit exists
Theorem 2
If $\lim_{n\to\infty} a_n = M$ then every subsequence has the same limit.
Therefore applying limit to $(1)$:
$$ \lim_{n\to\infty} a_{n+1} = \sqrt{1+\frac12 a_n} \iff M = \sqrt{1+\frac12 \lim_{n\to\infty} a_n} \iff M^2 = 1 +\frac{M}{2} \iff $$
$$ \boxed{\lim_{n\to\infty} a_{n} = M=\frac{1+\sqrt{17}}{4}} $$
This value of the limit is smaller than the upper bound $\sqrt{2}$ and this concerns me.
Does this fact mean that the upper bound I've found is some upper bound but not the supremum or did I make a mistake in calculating the limit?
Edit: Upper bound proof
We will prove by induction that $a_n < \sqrt{2}$.
*
*For $n=1$: $a_2 = \sqrt{1 +\frac12 1} < \sqrt{2}$
*For $n=k$: Let $a_k < \sqrt{2}$
*For $n=k+1:$ $a_{k+1} = \sqrt{1 + \frac12 a_k} < \sqrt{1 + \frac{\sqrt{2}}{2}} = \sqrt{1 + \frac1{\sqrt{2}}}< \sqrt{2}$
Hence indeed $a_n < \sqrt{2}$
| Let $L = \frac{1+\sqrt{17}}{4}$. Then $L^2 = 1 + \frac{1}{2}L$ and $\sqrt{1 + \frac{1}{2}L} = L$.
Let us use the mathematical induction to prove: $0 < a_n < a_{n+1} < L$ for all $n \ge 2$.
For $n = 2$, it is easy to verify it.
Assume that the inequality is true for $n = k$ ($k\ge 2$), i.e., $0 < a_k < a_{k+1} < L$.
Let us prove that the inequality is also true for $n = k + 1$, i.e., $0 < a_{k+1} < a_{k+2} < L$.
First, since $a_{k+1} < L$, we have $a_{k+2} = \sqrt{1 + \frac{1}{2}a_{k+1}} < \sqrt{1 + \frac{1}{2}L} = L$.
Second, since $0 < a_{k+1} < L$ and $L > \frac{1}{2}$, we have $(a_{k+1} - L)(a_{k+1} + L - \frac{1}{2}) < 0$
that is $a_{k+1}^2 - \frac{1}{2}a_{k+1} - 1 < 0$ (using $L^2 = 1 + \frac{1}{2}L$), which results in
$a_{k+2} = \sqrt{1 + \frac{1}{2}a_{k+1}} > a_{k+1}$. This completes the proof.
$\phantom{2}$
Thus, $\lim_{n\to \infty} a_n$ exists (finite). Let $\lim_{n\to \infty} a_n = M > 0$.
Taking limit on both sides of $a_{n+1} = \sqrt{1 + \frac{1}{2}a_n}$, we have
$M = \sqrt{1 + \frac{1}{2}M}$ which results in $M = \frac{1+\sqrt{17}}{4}$. We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$ (without L'Hopital)? I am trying to evaluate the following limit:
$$ \lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$$
Approach #1
$ \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \\ \sqrt{x+2} + \frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}}\cdot\frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{\sqrt{x^2-2x}-\sqrt{2x-4}} = \\ \sqrt{x+2} + \frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{x-2} $
But I still end up at the indefinite form $\frac12 + \frac{0}{\infty}$
My Approach #2
$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot\frac{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}= \frac1{\sqrt{x-2}}\frac{x^2-x+2\sqrt{2x}-2}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}$
Which also seems to be a dead end.
Any ideas on how to evaluate this?
| For
$x\ne 2$,$$\sqrt{x^2-4}=\sqrt{x+2}\sqrt{x-2}$$
and
$$\sqrt{x-2}=\sqrt{\sqrt x+\sqrt2}\sqrt{\sqrt x-\sqrt2}$$
so that by simplification
$$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}
=\frac{\sqrt{x+2}\sqrt{\sqrt x+\sqrt2}+\sqrt{\sqrt x-\sqrt2}}{\sqrt{\sqrt x+\sqrt2}}.$$
The limit is $\sqrt{2+2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Counting solutions to $x_1 + x_2 + x_3 + x_4 = 12$ with at least one $x_i\ge 5$.
Count the non-negative integral solutions of $x_1 + x_2 + x_3 + x_4 = 12$ with at least one $x_i\ge 5$.
I got really confused about that problem and really would like to know about methods to solve this.
I have tried to substitute the variables and the result for $7$ and then adding the $5$ later. I also thought about subtracting and using the pigeonhole principle.
| We can first get $4$ numbers adding to $7 = ^{(7+4-1)}C_{(4-1)} = 120$. As $5$ can be added to any of the $4$ numbers, multiply the answer by $4$.
Then we need to subtract duplicate arrangements -
$\{7,0,0,0\}$ arrangements that make arrangements of $\{7,5,0,0\}$ by adding $5$ are already covered in $\{2,5,0,0\}$ arrangements. So for each place of $7$, the only valid placement of $5$ is with $7$. Other $3$ are duplicates.
$S1 = 4 \times 3 = 12$
$\{6,1,0,0\}$ arrangements that make arrangements of $\{6,1,5,0\}$ by adding $5$ are already covered in $\{5,1,1,0\}$ arrangements.
$S2 = 2 \times \dfrac{4!}{2!} = 24$
$\{6,1,0,0\}$ arrangements that make arrangements of $\{6,6,0,0\}$ by adding $5$ are counted twice.
$S3 = \dfrac{1}{2} \times \dfrac{4!}{2!} = 6$
$\{5,2,0,0\}$ arrangements that make arrangements of $\{5,2,5,0\}$ by adding $5$ are counted twice.
$S4 = \dfrac{4!}{2!} = 12$
$\{5,1,1,0\}$ arrangements that make arrangements of $\{5,1,1,5\}$ by adding $5$ are counted twice.
$S5 = \dfrac{1}{2} \times \dfrac{4!}{2!} = 6$
Total valid arrangements $= 480 - (S1+S2+S3+S4+S5) = 420$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $x_{n+1}=x_n(2-ax_n)$ converges and find the limit
Let $a>0$ and $x_0\in I=\left (\frac{1}{2a}, \frac{3}{2a}\right )$. Show that the sequence $(x_n)$, $n\geq 0$, $$x_{n+1}=x_n(2-ax_n), \quad n \geq 0$$ converges. Which is the limit? Hint: Consider $\phi (x)=x(2-ax)$ and show that $\phi (I)\subset \left [\frac{3}{4a}, \frac{1}{a}\right ]$.
So that I understand that correctly, with the hint we want to show that $\phi(x)$ is bounded and monotonic, which means that the sequence converges?
We have the following: \begin{align*}\frac{1}{2a}<x< \frac{3}{2a} &\Rightarrow a\cdot \frac{1}{2a}<ax<a\cdot \frac{3}{2a}\Rightarrow \frac{1}{2}<ax< \frac{3}{2} \Rightarrow -\frac{3}{2}<-ax< -\frac{1}{2}\\ & \Rightarrow 2-\frac{3}{2}<2-ax< 2-\frac{1}{2}\Rightarrow \frac{1}{2}<2-ax< \frac{3}{2} \\ & \Rightarrow x\cdot \frac{1}{2}<x(2-ax)< x\cdot \frac{3}{2}\\ & \Rightarrow \frac{1}{2a}\cdot \frac{1}{2}<x\cdot \frac{1}{2}<x(2-ax)< x\cdot \frac{3}{2}< \frac{3}{2a}\cdot \frac{3}{2}\\ & \Rightarrow \frac{1}{4a}<x(2-ax)< \frac{9}{4a}\end{align*}
That is not the interval that we want to get. Do we have to do that maybe using derivatives?
|
P1. $\phi(x)$ has a maximum at $x_m=\frac{1}{a}$ and $\phi(x_m)=x_m$. Thus $\phi(x)\leq \frac{1}{a}$.
P2. $\phi(x)$ is ascending on $\left(-\infty,\frac{1}{a}\right]$ and descending on $\left(\frac{1}{a},\infty\right)$.
Since $\phi'(x)=2-2ax$
P3. For $x\in\left[\frac{1}{2a},\frac{3}{2a}\right] \Rightarrow
\phi(x)\in\left[\frac{3}{4a},\frac{1}{a}\right]$
Since $\frac{1}{2a}<\frac{1}{a}<\frac{3}{2a}$ and considering Pr2
*
*if $x\in\left[\frac{1}{2a},\frac{1}{a}\right] \Rightarrow \phi(x)\geq\phi\left(\frac{1}{2a}\right)=\frac{3}{4a}$
*if $x\in\left[\frac{1}{a},\frac{3}{2a}\right] \Rightarrow \phi(x)\geq\phi\left(\frac{3}{2a}\right)=\frac{3}{4a}$
And from Pr1, $\phi(x)\in\left[\frac{3}{4a},\frac{1}{a}\right]$.
Pr4. $\left[\frac{3}{4a},\frac{1}{a}\right]\subset\left[\frac{1}{2a},\frac{3}{2a}\right]$
Since
$\frac{1}{2a}<\frac{3}{4a}<\frac{1}{a}<\frac{3}{2a}$.
Pr5. $\phi(x)$ is a contraction map on $\left[\frac{3}{4a},\frac{1}{a}\right]$.
Since $$\forall x\in\left[\frac{3}{4a},\frac{1}{a}\right] \overset{Pr4}{\Rightarrow}
x\in\left[\frac{1}{2a},\frac{3}{2a}\right] \overset{Pr3}{\Rightarrow}
\phi(x)\in\left[\frac{3}{4a},\frac{1}{a}\right]$$
and $\exists \epsilon$ in between $x,y$ such that
$$|\phi(y)-\phi(x)|=|\phi'(\epsilon)|\cdot|y-x|=
|2-2a\epsilon|\cdot|y-x|\leq \color{red}{\frac{1}{2}}\cdot|y-x|$$
Summary. Although $x_0$ may not fall inside $\left[\frac{3}{4a},\frac{1}{a}\right]$ (as seen in Pr4), $x_1=\phi(x_0)$ does (as seen in Pr3). And so does any $x_n$ for $\forall n\geq 1$ (as seen in Pr5). Applying Banach fixed-point theorem, the limit exists and it's a solution of
$$L=L(2-aL)$$
| {
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"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$ If $a+b+c=1$ and $a,b,c>0$ then prove that $$\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$$
My try : We have to prove: $$ab+bc+ca+36abc(ab+bc+ca)\ge 21abc$$ or after homogenising we get :
$$\sum a^4b+3\sum a^3b^2+6\sum a^2b^2c\ge 14\sum a^3bc$$
I don't know what to do next. I think SOS may be helpful but find it difficult to factorize. Any other methods are also welcome but, if possible, can someone help me continue from here?
| Now, we need to prove that:
$$\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2)+4\sum_{cyc}(a^3b^2+a^3c^2-2a^3bc)-6\sum_{cyc}(a^3bc-a^2b^2c)\geq0$$ or
$$\sum_{cyc}(a^4b-a^3b^2-a^2b^3+ab^4)+$$
$$+4\sum_{cyc}(c^3a^2+c^3b^2-2c^3bc)-3abc\sum_{cyc}(a^2+b^2-2ab)\geq0$$ or
$$\sum_{cyc}(a-b)^2(ab(a+b)+4c^3-3abc)\geq0,$$ which is true by AM-GM:
$$ab(a+b)+4c^3\geq2\sqrt{a^3b^3}+c^3\geq3\sqrt[3]{a^3b^3c^3}=3abc$$
The following inequality a bit of stronger.
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=1$. Prove that:
$$\frac{1}{abc}+48\geq\frac{25}{ab+ac+bc}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Finding $\cos ( 2 \sin^{-1}( \frac{5}{ 13} )) $ The following problem is from the $8$th edition of the book Calculus, by James Stewart. It is problem number $9$ in section $6.6$.
Problem:
Find an exact value for the expression:
$$ \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } $$
Answer:
\begin{align*}
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - \sin^2{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
%
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - 2 \sin^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
%
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - 2 \left( \frac{25}{13^2} \right) \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
%
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - \left( \frac{50}{13^2} \right) \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
\end{align*}
\begin{align*}
\cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
1 - \sin^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } \\
\cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &= 1 - \frac{25}{169} = \frac{169 - 25}{169} \\
\cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \frac{144}{169} \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - \left( \frac{50}{13^2} \right) \left( \frac{144}{169} \right) } \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ \frac{13^4 - 50(144)}{13^4} } \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ \frac{21361}{13^4} } \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \frac{ \sqrt{ 21361 } } { 169 }
\end{align*}
The book's answer is $\frac{119}{169}$ and SciLab matches the book. Where did I go wrong?
| Alternative solution:
Let $sin^{-1}(\frac{5}{13})=x$, then $\sin x=\frac{5}{13},\space \cos x=\frac{12}{13}$
$\cos 2x=1-2\sin^2x=1-2\times(\frac{5}{13})^2=\frac{119}{169}$
| {
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"answer_id": 2
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Proving $Q=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3}$ For $a,b,c\geqslant 0.$ Prove: $$\text{Q}=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3},$$
where $k={\frac {27}{8}}+\frac{9\sqrt{3}}{4}$ is a Root Of $64{k}^{2}-432k-243=0.$
By computer (Maple) I found this inequality is equivalent to$:$ $$\sum \Big[ a+b+ ( 1-\frac{\sqrt {3}}{2} ) c\Big] \Big[ 2(a+b)- ( 1+\sqrt {3} ) c \Big] ^{2} ( \,a-b \,) ^{2} \geqslant 0$$
But I hope for alternative proof$?$
| Another way
Let $p = a + b + c = 3, q = ab+bc+ca, r = abc$.
Since $(a-b)^2(b-c)^2(c-a)^2 \ge 0$ that is $-4p^3r+p^2q^2+18pqr-4q^3-27r^2 \ge 0$,
we have
\begin{align}
r &\ge -\frac{2}{27}p^3 + \frac{1}{3}pq - \frac{2}{27}\sqrt{(p^2 - 3q)^3}\\
&= q - 2 - 2\sqrt{(1-q/3)^3}.\tag{1}
\end{align}
We have to prove that
$$\frac{p^3 - 3pq + 3r}{pq - r} - \frac{3}{8} \ge k\left(\frac{1}{3} - \frac{q}{p^2}\right).$$
From (1), it suffices to prove that
$$\frac{p^3 - 3pq + 3(q - 2 - 2\sqrt{(1-q/3)^3})}{pq - (q - 2 - 2\sqrt{(1-q/3)^3})} - \frac{3}{8} \ge k\left(\frac{1}{3} - \frac{q}{p^2}\right)$$
or
$$\frac{27(1-q/3)(3 - \sqrt{1-q/3})}{8q + 8 + 8\sqrt{(1-q/3)^3}}\ge \frac{k}{3}(1 - q/3).$$
It suffices to prove that
$$\frac{81(3 - \sqrt{1-q/3})}{8q + 8 + 8\sqrt{(1-q/3)^3}}\ge k.$$
With the substitution $q = 3 - \frac{t^2}{3}$ for $0 \le t \le 3$, it suffices to prove that
$$\frac{729(9-t)}{8(t+3)(6-t)^2} \ge k.$$
It is easy to prove that
$$\frac{729(9-t)}{8(t+3)(6-t)^2} - \frac{27}{8}
= \frac{27t^2(9 - t) + 27^2(5 - t)}{8(t+3)(6-t)^2}
> 0$$
and
$$\left(\frac{729(9-t)}{8(t+3)(6-t)^2} - \frac{27}{8}\right)^2 - \frac{243}{16}
= \frac{243(99-6t-t^2)(t^2 - 12t + 9)^2}{64(t+3)^2(6-t)^4} \ge 0$$
which results in
$$\frac{729(9-t)}{8(t+3)(6-t)^2} \ge \frac{27}{8} + \frac{9\sqrt{3}}{4}.$$
(Alternatively, we may use derivatives to find the minimum of $\frac{729(9-t)}{8(t+3)(6-t)^2}$ on $[0, 3]$.)
We are done.
| {
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"url": "https://math.stackexchange.com/questions/3809195",
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"question_score": "2",
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Sequence and polynomial power relationship I recently encountered this relationship between polynomial powers and a certain associated sequence and I am seeking any help or idea that might answer why the relationship is true.
Let $P(x)$ be a polynomial, say for instance $P(x)=1+3x+2x^2$. Consider the consecutive powers of $P(x)$ and arrange the numerical coefficients in order of appearance. For the given $P(x)$ we have for up to fifth power:
$P(x)^1=1+3x+2x^2$
$P(x)^2=1 + 6 x + 13 x^2 + 12 x^3 + 4 x^4$
$P(x)^3=1 + 9 x + 33 x^2 + 63 x^3 + 66 x^4 + 36 x^5 + 8 x^6$
$P(x)^4=1 + 12 x + 62 x^2 + 180 x^3 + 321 x^4 + 360 x^5 + 248 x^6 + 96 x^7 +
16 x^8$
$P(x)^5=1 + 15 x + 100 x^2 + 390 x^3 + 985 x^4 + 1683 x^5 + 1970 x^6 +
1560 x^7 + 800 x^8 + 240 x^9 + 32 x^{10}$.
Also, consider the sequence defined by
$a_{m,n}=a_{(m-1),n}+3a_{(m-1),(n-1)}+2a_{(m-1),(n-2)}$ with $a_{1,1}=1,a_{1,2}=3$ and $a_{1,3}=2$.
Observe that the sequence above completely determines the entries in the expansion of the polynomial power. For instance the number $63$ in the third power of $P(x)$ is equal to $63=12+3(13)+2(6)$.
I am wondering why is this TRUE. Thanks for your help and suggestion.
| The sequence you encountered is used to construct a particular extension of the $a-b$-based triangle. In particular, your sequence generates the $1-3-2$ triangle. In general, for $a_0a_1\ldots a_{r-1}$-based triangle, the entry in the $m^{th}$ row and $n^{th}$ column which is exactly the term $a_{mn}$ in your recurrence is given by the numerical coefficient of $x^n$ in the expansion of $(a_0+a_1x+\ldots+a_{r-1}x^{r-1})^m$.
Hence, $a_{mn}$ is the coefficient of $x^n$ in the expansion of $[P(x)]^m$.
| {
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Find the power series of $\frac{3x+4}{x+1}$ around $x=1$. I'm trying to find the power series of
$$
\frac{3x+4}{x+1}
$$
around $x=1$.
My idea was to use the equation
$$
\left(\sum_{n\ge0}a_n (x-x_0)^n\right)\left(\sum_{k\ge0}b_k (x-x_0)^k\right) = \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-x_0)^n \tag{1}
$$
taken from here, to solve this problem.
My attempt
I can write the numerator of the expression as
$$
3x +4 = 3(x-1)+7 = \sum_{n\ge0}a_n (x-1)^n
$$
where
$$
a_n=\begin{cases}
7 & n=0 \\
3 & n=1 \\
0 & n>1
\end{cases}
$$
On the other hand, given $|x-1| <2$ we get
\begin{align*}
\frac{1}{x+1} = \frac{1}{2} \cdot \frac{1}{1-\left(\frac{1-x}{2}\right)} = \frac{1}{2} \sum_{k\ge0} \left(\frac{1-x}{2}\right)^n = \sum_{k\ge0}\underbrace{\frac{-1}{(-2)^{k+1}}}_{\color{blue}{b_k}} (x-1)^k
\end{align*}
And then, using equation $(1)$ we get
\begin{align*}
\frac{3x+4}{x+1} &= \left(\sum_{n\ge0}a_n (x-1)^n\right)\left(\sum_{k\ge0}b_k (x-1)^k\right)\\
&= \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-1)^n\\
&= \sum_{n\ge0}\left(a_0 b_n + a_1 b_{n-1} \right)(x-1)^n\\
&= \sum_{n\ge0}\left(7 \frac{-1}{(-2)^{n+1}} + 3 \frac{-1}{(-2)^{n}} \right)(x-1)^n\\
&= \sum_{n\ge0}\left( \frac{-7 +3(-1)(-2)}{(-2)^{n+1}} \right)(x-1)^n\\
&= \sum_{n\ge0} \frac{-1}{(-2)^{n+1}} (x-1)^n\\
\end{align*}
And this seems to imply that $\frac{3x+4}{x+1} = \frac{1}{x+1}$, at least for $|x-1| <2$, which is clearly not true.
I've gone over the steps, but I don't see where my mistake is. Could anyone tell me where my solution went wrong? Thank you!
| Correction to your answer
Numerator:
$$
3x+4=7+3(x-1)\tag1
$$
Denominator:
$$
\begin{align}
\frac1{x+1}
&=\frac1{2+(x-1)}\tag2\\[6pt]
&=\frac12\frac1{1+\frac{x-1}2}\tag3\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{2^{k+1}}(x-1)^k\tag4
\end{align}
$$
Multiplying the series:
$$
\begin{align}
&(7+3(x-1))\left(\sum_{k=0}^\infty\frac{(-1)^k}{2^{k+1}}(x-1)^k\right)\\
&=\color{#C00}{\sum_{k=0}^\infty7\frac{(-1)^k}{2^{k+1}}(x-1)^k}+\color{#090}{\sum_{k=0}^\infty3\frac{(-1)^k}{2^{k+1}}(x-1)^{k+1}}\tag5\\
&=\color{#C00}{\frac72+\sum_{k=1}^\infty7\frac{(-1)^k}{2^{k+1}}(x-1)^k}+\color{#090}{\sum_{k=1}^\infty3\frac{(-1)^{k-1}}{2^k}(x-1)^k}\tag6\\
&=\frac72+\sum_{k=1}^\infty7\frac{(-1)^k}{2^{k+1}}(x-1)^k-\sum_{k=1}^\infty6\frac{(-1)^k}{2^{k+1}}(x-1)^k\tag7\\
&=\frac72+\sum_{k=1}^\infty\frac{(-1)^k}{2^{k+1}}(x-1)^k\tag8
\end{align}
$$
Explanation:
$(5)$: the red sum is from the $7$, the green from the $3(x-1)$
$(6)$: pull out the constant term from the red sum
$\phantom{\text{(6):}}$ substitute $k\mapsto k-1$ in the green sum
$(7)$: make the right sum look like the left sum
$(8)$: combine the sums
How I would do it
$$
\begin{align}
\frac{3x+4}{x+1}
&=3+\frac1{x+1}\tag9\\[9pt]
&=3+\frac1{(x-1)+2}\tag{10}\\[9pt]
&=3+\frac12\frac1{1+\frac{x-1}2}\tag{11}\\
&=\frac72+\sum_{k=1}^\infty\frac{(-1)^k}{2^{k+1}}(x-1)^k\tag{12}
\end{align}
$$
Explanation:
$\phantom{1}(9)$: divide the polynomials
$(10)$: put in terms of $x-1$
$(11)$: put the fraction in the form $\frac1{1+u}$
$(12)$: use the series $\frac1{1+u}=\sum\limits_{k=0}^\infty(-1)^ku^k$
$\phantom{\text{(12):}}$ and combine the $k=0$ term with $3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how can i find $ \frac{1}{2\pi}\left ( \frac{\pi^{3}}{1!3}-\frac{\pi^{5}}{3!5}+\frac{\pi^{7}}{5!7}-... \right ) $ Consider a sequence $ s_{n} = \frac{1}{2\pi}\left ( \frac{\pi^{3}}{1!3}-\frac{\pi^{5}}{3!5}+\frac{\pi^{7}}{5!7}-...+\frac{\left ( -1 \right )^{n-1}\pi^{2n+1}}{\left ( 2n-1 \right ) ! \left ( 2n+1 \right )} \right ) $
How can I attack this to find $ \lim_{n \to \infty}s_{n} \ $ ?
As the series $ \lim_{n \to \infty}s_{n} \ $ converges absolutely, so by simple manipulation I arrived to this :
$ \lim_{n \to \infty}s_{n} = \frac{1}{\pi}\left \{ \left ( \pi \right ) + \left ( \pi - \frac{\pi^{3}}{3!} \right ) + \left ( \pi - \frac{\pi^{3}}{3!} + \frac {\pi^{5}}{5!} \right )+...\right \} $
What to do next? Anyone please?
| Consider
$$f(x)=\frac{x^3}{1!3}-\frac{x^5}{3!5}+\frac{x^7}{5!7}-\cdots$$
(an infinite series). We want to find $f(\pi)/(2\pi)$. Then
$$f'(x)=\frac{x^2}{1!}-\frac{x^4}{3!}+\frac{x^6}{5!}-\cdots
=x\left(\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)$$
which is a series you should recognise. Now integrate to find $f(x)$ etc.
| {
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"source": "stackexchange",
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Showing that $\sqrt{2} \in \mathbb{Q}[\varepsilon^{2}]$ If we work in the field of complex numbers and let $\varepsilon$ be a primitive 16th root of unity and set $b=\frac{\varepsilon}{\sqrt{2}}$ and $A=\mathbb{Q}[\varepsilon]$ where $\mathbb{Q}$ represents the rationals, and also set $f(X)=X^{8}+16 \in \mathbb{Q}[X]$ and note that $b$ is a root of $f(X)$, then how can we prove that $\sqrt{2} \in \mathbb{Q}[\varepsilon^{2}]$?
| $\epsilon$ is a root of $p(x)=x^8+1$ as it is a primitive root of $x^{16}-1 = (x^8-1)(x^8+1)$.
Therefore $\epsilon^2$ is a root of
$$q(x) = x^4+1= x^2\left(x^2+\frac{1}{x^2}\right)= x^2\left(x+\frac{1}{x} - \sqrt 2\right)\left(x+\frac{1}{x} + \sqrt 2\right)$$
As $\epsilon^4 \neq 0$, $\epsilon^2$ is a zero of
$$r(x) = \left(x+\frac{1}{x} - \sqrt 2\right)\left(x+\frac{1}{x} + \sqrt 2\right)$$
This implies $\epsilon^2 + \frac{1}{\epsilon^2} \in \lbrace \sqrt 2, - \sqrt 2 \rbrace$ and the desired result $\sqrt 2 \in \mathbb Q(\epsilon^2)$
| {
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"timestamp": "2023-03-29T00:00:00",
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$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$; find my mistake solve :$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$
My try:
$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}\\\frac{y'}{y^{2}}+\frac{2}{y(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\\left(t\frac{1-x}{1+x}\right)'=\left(x-1\right)\frac{1-x}{1+x}\\\left(t\frac{1-x}{1+x}\right)=\frac{1}{2}(7+6x-x^{2}-8log(1+x))+c\\t=\frac{(1+x)(c+1/2(7+6x-x^{2}-8log(1+x))))}{1-x}\\y=-\frac{1-x}{(1+x)(c+1/2(7+6x-x^{2}-8log(1+x))))}\\\\\\\\\\$
| Your 5th and 4th lines are not equivalent. You're missing a -ve sign in the second term on LHS.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving infinite nested square roots of 2 converging to finite nested radical Can anyone explain to solve the identity posted by my friend $$2\cos12°= \sqrt{2+{\sqrt{2+\sqrt{2-\sqrt{2-...}}} }}$$ which is an infinite nested square roots of 2. (Pattern $++--$ repeating infinitely)
Converging to finite nested radical of $2\cos12° = \frac{1}{2}\times\sqrt{9+\sqrt5+\sqrt{(30-6\sqrt5)}}$
The finite nested radical, I was able to derive $\cos12° = \cos(30-18)°$ as follows
$$\cos30°\cdot\cos18° + \sin30°\cdot\sin18°$$
$$= \frac{√3}{2}\cdot\frac{\sqrt{2+2\cos36°}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2-2\cos36°}}{2}$$
Where $\cos18° = \frac{\sqrt{2+2\cos36°}}{2}$ (by Half angle cosine formula) and
$\sin18° = \frac{\sqrt{2-2\cos36°}}{2}$ (solving again by half angle cosine formula)
$2\cos36° =\frac{ \sqrt5 +1}{2}$ which is golden ratio
$\frac{\sqrt3}{2}\cdot\frac{\sqrt{10+2\sqrt5}}{4}+ \frac{1}{2}\cdot\frac{\sqrt{5}-1}{4} = \frac{\sqrt{30+6\sqrt5}}{8}+ \frac{\sqrt5-1}{8}$
Further steps finally lead to the finite nested radical
Method actually I tried to solve infinite nested square roots of 2 is as follows.
$2\cos\theta = \sqrt{2+2\cos2\theta}$ and
$2\sin\theta = \sqrt{2-2\cos2\theta}$
Now simplifying infinite nested square roots of 2, we will get the following as simplified nested radical
$$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2\cos12°}}}}$$
Simplifying step by step as follows
$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\sin6°}}}$ then
$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\cos84°}}}$ (by $\sin\theta = \cos(90-\theta)$
$2\cos12° = \sqrt{2+\sqrt{2+2\sin42°}}$
$2\cos12° = \sqrt{2+\sqrt{2+2\cos48°}}$
$2\cos12° = \sqrt{2+2\cos24°}$
$2\cos12° = 2\cos12°$
We are back to $\sqrt1$
Actually this is how I got stuck!
But for infinite nested square roots of 2(as depicted), if I run program in python I am able to get good approximation ( Perhaps if we run large number of nested square roots in python we get more number of digits matching the finite nested radical), because I'm not able get anywhere solving such a kind of infinite cyclic nested square roots of 2.
Dear friends, is there anyway to find the solution by any other means like solving infinite nested square roots
Thanks in advance.
| If the value of the radical is $x$, then we have $$x=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-x}}}}\tag1$$ Repeated squaring gives
$$\left(\left(\left(x^2-2\right)^2-2\right)^2-2\right)^2=2-x\tag2$$
Now, $(2)$ has $8$ solutions, and notice for all choices of the first three signs in $(1)$, repeated squaring gives $(2)$. Thus, the solutions of $(2)$ are the eight solutions to $$x=\sqrt{2\pm\sqrt{2\pm\sqrt{2\pm\sqrt{2-x}}}}$$
So, we must first show that $2\cos12^\circ$ satisfies $(2)$, and then to show that it is the root given by the choice of signs in the question.
To verify that $2\cos12^\circ$, we use the formula $$(2\cos\theta)^2-2 = 2(2\cos^2\theta-1)=2\cos2\theta\tag3$$
Then setting $x=2\cos12^\circ$, $(3)$ gives
$$\begin{align}
x^2-2&=2\cos24^\circ\\
(x^2-2)^2-2&=2\cos48^\circ\\
((x^2-2)^2-2)^2-2&=2\cos96^\circ\\
(((x^2-2)^2-2)^2-2)-2&=2\cos192^\circ=-2\cos12^\circ=-x\\
\end{align}$$
as required.
ADDENDUM
Since $0\leq x\leq 2$, there is a value $0\leq\theta\leq\frac\pi2$ such that $x=2\cos\theta$. The argument above gives $2\cos16\theta=-2\cos\theta$ so either $$16\theta=(2n+1)\pi+\theta$$or$$16\theta=(2n+1)\pi-\theta$$ The condition $0\leq\theta\leq\frac\pi2$ gives $8$ possibilities for $\theta$: either$$\theta=\frac{(2n+1)\pi}{15},\ n=0,1,2,3$$ or
$$\theta=\frac{(2n+1)\pi}{17},\ n=0,1,2,3$$
so if you can sort the values of the nested radicals for the $8$ choices of sign in decreasing order, you not only evaluate the given infinite nested radical, but $7$ more. For example, it seem clear that choosing the choosing the $3$ plus signs would give the larges value, which would mean that the value of that radical would be $2\cos\frac\pi{17}$. If the second largest value comes from choosing the first $2$ signs as $+$ and the third as $-$, that would complete the proof for the original question.
Actually, there's still something missing. For this argument to work, we also have to show that the infinite nested radical converges for all $8$ sign choices. I'm having trouble coming up with an economical way to do that.
I carried out the numerical work, and found that
$$\begin{align}
2\cos\frac{7\pi}{15}&=
\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}\\
2\cos\frac{7\pi}{17}&=
\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\cdots}}}}\\
2\cos\frac{5\pi}{15}&=
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}\\
2\cos\frac{5\pi}{17}&=
\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\cdots}}}}\\
2\cos\frac{3\pi}{15}&=
\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\cdots}}}}\\
2\cos\frac{3\pi}{17}&=
\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}\\
2\cos\frac{\pi}{15}&=
\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\cdots}}}}\\
2\cos\frac{\pi}{17}&=
\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}\\
\end{align}$$
Still haven't found a nice argument for convergence, though it's clear numerically that all sequences converge rapidly.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^{-1}$ is an algebraic integer Let $b = (4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^{-1}$, then $1 = b(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})$, and $1 - 4b = b(3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})$.
Therefore $1 - 12b + 48b^2 - 64b^3 = (1 - 4b)^3 = b^3(3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^3$
$= b^3(81 + 18 \cdot 3^{2/3} \cdot 3^{2/3} + 12 \cdot 3^{1/3} \cdot 3^{4/3} + 72) = b^3(153 + 18 \cdot 3 \cdot 3^{1/3} + 18 \cdot 2 \cdot 3^{2/3})$
$= 18b^3(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3}) + 81b^3 = 18b^3(b^{-1}) + 81b^3= 18b^2 + 81b^3$. Therefore $b$ is a root of $1 - 12b + 48b^2 - 64b^3 - 18b^2 - 81b^3 = - 145b^3 + 30b^2 - 12b + 1$
My problem is $145b^3 - 30b^2 + 12b - 1$ is not monic and is also not reducible, since if it were reducible being of degree $3$ we could factor out a root, which we can't by the rational roots theorem,
Any advice on what I am doing wrong?
| Your mistake occurs at this step
$$(3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^3=(81 + 18 \cdot 3^{2/3} \cdot 3^{2/3} + 12 \cdot 3^{1/3} \cdot 3^{4/3} + 72)$$
Here you have forgotten the binomial coefficients $\binom{3}{1}$ and $\binom{3}{2}$.
| {
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Proving $\sum_{i=1}^n (1-\frac{1}{(i+1)^2}) = \frac{n+2}{2n+2}$ using induction. My textbook has the following question:
Prove the follwing statement using induction for all natural numbers $n$
$$(1- \frac{1}{4})+(1- \frac{1}{9})+.......+(1- \frac{1}{(n+1)^2})=\frac{n+2}{2n+2}$$
So, I check both the sides for $n=1$. In that case LHS=RHS=$\frac{3}{4}$.
Now I assume the statement to be true for $n=k$ which gives
$$
\sum_{i=1}^k (1-\frac{1}{(i+1)^2}) = \frac{k+2}{2k+2}
$$
Now I evaluate the original statement for $n=k+1$ which leaves us with the
$$ LHS= \frac{3k^3+16k^2+26k+14}{2(k+1)(k+2)^2} $$
And we have to prove this LHS to be equal to RHS which is
$$
RHS= \frac{k+3}{2k+4}
$$
But these (new) LHS and RHS don't seem to be equal. And hence I am not able to complete the proof.
How should I proceed?
A solution without induction is also welcome.
Book: Comprehensive Algebra VOL-1
Author: Vinay Kumar
Publisher: McGraw Hill Education.
| The formula you are trying to prove is wrong. The actual correct formula is
$$ \prod_{i=1}^n \left(1-\frac{1}{(i+1)^2}\right)=\frac{n+2}{2n+2}=\frac{n+2}{2(n+1)} $$
For $n=1$, this is easy to see.
$$\left(1-\frac{1}{(1+1)^2}\right)=\frac{3}{4}=\frac{1+2}{2\cdot 1+2}$$
For the induction step:
$$\prod_{i=1}^{n+1} \left(1-\frac{1}{(i+1)^2}\right)=\left(1-\frac{1}{(n+2)^2}\right)\prod_{i=1}^{n} \left(1-\frac{1}{(i+1)^2}\right)=\left(1-\frac{1}{(n+2)^2}\right)\cdot \frac{n+2}{2n+2}$$
Further computing leads to
$$\left(1-\frac{1}{(n+2)^2}\right)\cdot \frac{n+2}{2n+2}=\frac{n+2}{2(n+1)}-\frac{1}{2(n+1)(n+2)}=\frac{n^2+4n+4-1}{2(n+1)(n+2)}=\frac{n^2+4n+3}{2(n+1)(n+1+1)}$$
Finally by dividing by $n+1$ follows
$$\frac{n^2+4n+3}{2(n+1)(n+1+1)}=\frac{n+1 +2}{2(n+1+1)}$$
which is what we wanted to show.
| {
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"timestamp": "2023-03-29T00:00:00",
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Rate of convergence for a sequence (Preferably without Taylor series) I am trying to solve the following problem:
Knowing that the sequence $(a_{n})$ with:
$$a_{n+1}=\frac{1}{2}(a_{n}+\frac{3}{a_{n}})$$
converges to $\sqrt{3}$, find it's rate of convergence.
After doing some searching, I found this formula from wikipedia:
$$\lim\limits_{n \to \infty} \frac{|a_{n+1}-L|}{|a_{n}-L|} = μ$$
And I think that our L is $\sqrt{3}$.
Do I need to find the value of $a_{n}$ to find the rate of convergence (μ)? And how do I find $a_{n}$ ?
UPDATE: I can simply use the formula above but I need to make my limit approach to $\sqrt{3}$ because we have $a_{n} \to \sqrt{3}$:
$$\lim\limits_{x \to \sqrt{3}} \frac{|\frac{1}{2}(x+\frac{3}{x})-\sqrt{3}|}{|x-\sqrt{3}|}$$
But my problem is that this limit is resulting in a indeterminate form because of $\frac{0}{0}$
How can I solve this limit without expanding series?
UPDATE 2 - ANSWER:
Using @user's approach we can write our limit as:
$$\lim\limits_{x \to \sqrt{3}} \frac{\frac{1}{2}(x+\frac{3}{x})-\sqrt{3}}{x-\sqrt{3}}=\frac{x^2-2\sqrt 3x+3}{2x(x-\sqrt{3})}=\frac{(x-\sqrt{3})^2}{2x(x-\sqrt{3})}=\frac{x-\sqrt{3}}{2x}\to 0$$
and then the sequence converges Q-superlinearly to $\sqrt 3$. Look at here.
| We have that
$$\frac{a_{n+1}-\sqrt 3}{a_{n+1}+\sqrt 3}=\frac{a_n^2-2\sqrt 3a_n+3}{a_n^2+2\sqrt 3a_n+3}=\left(\frac{a_{n}-\sqrt 3}{a_{n}+\sqrt 3}\right)^2$$
therefore by induction with $a_0=a>0$ we have that
$$\frac{a_{n}-\sqrt 3}{a_{n}+\sqrt 3}=\left(\frac{a-\sqrt 3}{a+\sqrt 3}\right)^{2^{n}}$$
and therefore
$$a_n=\frac{\sqrt 3\left(1+\left(\frac{a-\sqrt 3}{a+\sqrt 3}\right)^{2^{n}}\right)}{1-\left(\frac{a-\sqrt 3}{a+\sqrt 3}\right)^{2^{n}}}$$
Refer to the related
*
*Finding $\lim_{n \to \infty} x_n$
By a limit approach we have
$$\frac{\frac{1}{2}(x+\frac{3}{x})-\sqrt{3}}{x-\sqrt{3}}=\frac{x^2-2\sqrt 3x+3}{2x(x-\sqrt{3})}=\frac{(x-\sqrt{3})^2}{2x(x-\sqrt{3})}=\frac{x-\sqrt{3}}{2x}\to 0$$
and then the sequance converges Q-superlinearly to $\sqrt 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Solve $\frac{\sin (10^\circ) \sin (30^\circ)}{\sin 40^\circ \sin (80^\circ-x^\circ)} = \frac{\sin 20^\circ}{\sin x}$ The context to this is trivial I think, I was solving a geometry problem using the trigonometric version of Ceva, I got here and I was stuck, I tried using the sum-difference, product to sums, sums to products identities but my attempts failed and was rather tedious.
| You can use
\begin{equation}
\sin a\sin b\sin c=\frac{1}{4}(\sin(a+b-c)+\sin(a-b+c)+\sin(-a+b+c)-\sin(a+b+c))
\end{equation}
With this you get
\begin{equation}
\sin(40-x)+\sin(-20+x)+\sin(20+x)-\sin(40+x) \\
=
\sin(-20+x)+\sin(60-x)+\sin(100-x)-\sin(140-x)
\end{equation}
Then you can use $\sin(140 - x) = \sin(180 - 140 + x) = \sin(40+x)$ to reduce this to
\begin{equation}
\sin(40-x)+\sin(20+x) =\sin(60-x)+\sin(100-x) \\
\sin(70-(30+x))+\sin(-10+(30+x)) - \sin(90-(30+x)) - \sin(130-(30+x)) = 0 \\
\end{equation}
Now I use formula $\sin(a\pm(30+x))=\sin a\cos (30+x)\pm \cos a\sin (30+x)$.
\begin{equation}
\cos(30+x)(\sin 70 - \sin 10 - 1 -\sin130)+\sin(30+x)(-\cos 70 +\cos 10 + \cos 130) = 0 \\
-\cos(30+x)(\sin 250 + \sin 10 + 1 + \sin130)+\sin(30+x)(\cos 250 +\cos 10 + \cos 130) = 0
\end{equation}
Now we can notice the following
$(\cos 250 +\cos 10 + \cos 130) = \Re e^{i 10^\circ}(e^{i 0^\circ} + e^{i 120^\circ} + e^{i 240^\circ})$ and $(\sin 250 +\sin 10 + \sin 130) = \Im e^{i 10^\circ}(e^{i 0^\circ} + e^{i 120^\circ} + e^{i 240^\circ})$. But $e^{i 0^\circ} + e^{i 120^\circ} + e^{i 240^\circ} = 0$, which yields
\begin{equation}
-\cos(30+x) = 0
\end{equation}
which has two solutions $x=60$ and $x=-120$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $x^5-x^3+x-2=0$, find $\lfloor x^6\rfloor$.
If $\alpha$ is a real root of the equation $x^5-x^3+x-2=0$, find the value of $\lfloor\alpha^6\rfloor$, where $\lfloor x\rfloor$ is the least positive integer not exceeding $x$.
My approach is to bound the value of $\alpha^6=\alpha^4-\alpha^2+2\alpha$.
First I proved the function $f(x)=x^5-x^3+x-2$ is monotone increasing by derivative. Then I argue that $1<\alpha<\frac32$ as $f(1)<0$ and $f(\frac32)>0$.
Then I tried to create an upper and lower bound for $\alpha$, as such
$$\alpha^6=\alpha^4-\alpha^2+2\alpha<\frac94\alpha^2-\alpha^2+2\alpha=\frac54\alpha^2+2\alpha<\frac{45}{16}+3=\frac{93}{16}$$
and
$$\alpha^6=\alpha^4-\alpha^2+2\alpha>\alpha^3-\alpha^2+2\alpha=\alpha^2(\alpha-1)+2\alpha>2$$
Now we know that $\lfloor\alpha\rfloor\in\{2,3,4,5\}$. But I cannot proceed any further.
Any idea, help, hint, or answer is appreciated. Thank you.
| Since $f'(x)=5x^4-3x^2+1$, which is always greater than $0$, $f$ is strictly increasing. Since, furthermore, $\lim_{x\to\pm\infty}f(x)=\pm\infty$, $f$ has one and only one real root. On the other hand, $f(1)=-1<0<24=f(2)$. So, the only real root $\alpha$ belongs to $(1,2)$ and $\lfloor\alpha\rfloor=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $\lim_{\Delta\rightarrow 0}\int_{-\Delta}^{\Delta}\frac{\mathrm{d}\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\frac{\pi}{r^2}$ While I'm working on the proof of Bertrand's Theorem, I stuck at a limits calculation. I want to prove:
$$
\lim_{\Delta\rightarrow 0}\int_{-\Delta}^{\Delta}\frac{\mathrm{d}\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\frac{\pi}{r^2}
$$
With the help of the software Mathematica, I got:
$$
\int_{-\Delta}^{\Delta}\dfrac{\mathrm{d}\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\dfrac{\pi}{(1-\frac{\Delta^2}{r^2})^{\frac{3}{2}} \ r^2}
$$
but I have no clue how to formally calculate the limit or the integral. Thanks in advance if you can offer some help.
| Let
$\omega=\Delta \sin(x)$
$$I=\int\dfrac{d\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\int\dfrac{dx}{(r+\Delta \sin (x))^2}$$ Now, tanent half-angle substitution
$$I=\int \frac{2 \left(t^2+1\right)}{\left(r t^2+2 \Delta t+r\right)^2}\,dt$$
Write
$$r t^2+2 \Delta t+r=r (t-a)(t-b)$$ to make
$$I=\frac 2 {r^2} \int \frac{\left(t^2+1\right)}{(t-a)^2(t-b)^2}\,dt$$ Partial fracion decomposition
$$\frac{\left(t^2+1\right)}{(t-a)^2(t-b)^2}=\frac{a^2+1}{(a-b)^2 (t-a)^2}+\frac{b^2+1}{(a-b)^2 (t-b)^2}-\frac{2 (a
b+1)}{(a-b)^3 (t-a)}+\frac{2 (a b+1)}{(a-b)^3 (t-b)}$$ Now, it is very simple.
When done, replace
$$a=\frac{\sqrt{\Delta ^2-r^2}-\Delta }{r} \qquad \text{and} \qquad b=\frac{-\sqrt{\Delta ^2-r^2}-\Delta }{r}$$
Edit
Since
$$I=\int_{-\Delta}^\Delta\dfrac{d\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\frac{\pi r^2 \sqrt{1-\frac{\Delta ^2}{r^2}}}{\left(r^2-\Delta ^2\right)^2}$$ the expansion around $\Delta=0$ is
$$\frac{\pi}{r^2} \left(1+\frac{3 \Delta ^2}{2 r^2}+\frac{15 \Delta ^4}{8 r^4}+\frac{35 \Delta ^6}{16
r^6}+O\left(\Delta ^{8}\right) \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's the rotation matrix needed to align a rhombic dodecahedron such that six of its contact points line up to form a hexagon?
I want to know what the rotation matrix is I need to make this mesh have a hexagonal shape from a top-side orthographic viewpoint. The point where it has at least something resembling this is when viewing through an axis moving through points [.5, .5, .5] and the origin, shown here.
Obviously the fastest solution would be to rotate .5,.5,.5 to 1, but I also want to make sure it's rotated at an angle where one of the other points aligns with an X or Y axis.
I also probably forgot how to rotate source points to destinations.
This should be a simple question but I can't really find the answer here.
| If the eight threefold vertices are $$(\pm 0.5, \pm 0.5, \pm 0.5),$$ and the six fourfold vertices are $$(\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm 1),$$ then a rotation that maps $v_1 = (0.5, 0.5, 0.5)$ to $v_3 = (0, 0, \frac{\sqrt{3}}{2})$ will work. A rotation about the $z$-axis by $\pi/4$ radians transforms $v_1 = (0.5, 0.5, 0.5)$ to $v_2 = (0, \frac{\sqrt{2}}{2}, 0.5)$. Then we require a rotation about the $x$-axis by an angle $\varphi$ that satisfies $$\cos \varphi = \frac{v_2 \cdot v_3}{|v_2||v_3|} = \frac{1}{\sqrt{3}}.$$ So the complete rotation matrix is $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \varphi & -\sin \varphi \\ 0 & \sin \varphi & \cos \varphi \end{bmatrix} \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} & 0 \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & - \sqrt{\frac{2}{3}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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probability that first $2$ drawn balls are red
A bag contain $9$ red and $12$ blue balls. If $4$ balls are selected randomly without replacement, then find the probability that the first $2$ balls are red.
What I tried: Let $A$ be the event the first drawn ball is red and $B$ be the event that the second drawn ball is red. Then $P(A)=8/21$ and $P(B)=7/20$.
If first $2$ drawn balls are red. Then other $2$ drawn balls are both red or blue or one red one blue.
But I don't understand how this helps.
Please help me! Thanks!
| What happens after the first two balls are drawn has no effect on whether the first two balls are red, so we only need to consider what happens on the first two draws.
The probability that the first ball is red is $9/21$. If the first ball is red, the probability that the second ball is also red is $8/20$. Therefore, the probability that the first two balls are red is
$$\frac{9}{21} \cdot \frac{8}{20} = \frac{6}{35}$$
It looks like you were thrown by the fact that four balls are drawn. To convince yourself that the above answer is correct, consider what happens in the third and fourth draws. There are four possible sequences: RRRR, RRRB, RRBR,
RRBB. Adding the probabilities yields
\begin{align*}
& \frac{9}{21} \cdot \frac{8}{20} \cdot \frac{7}{19} \cdot \frac{6}{18} + \frac{9}{21} \cdot \frac{8}{20} \cdot \frac{7}{19} \cdot \frac{12}{18} + \frac{9}{21} \cdot \frac{8}{20} \cdot \frac{12}{19} \cdot \frac{7}{18} + \frac{9}{21} \cdot \frac{8}{20} \cdot \frac{12}{19} \cdot \frac{11}{18}\\
& \quad = \frac{9}{21} \cdot \frac{8}{20}\left(\frac{7}{19} \cdot \frac{6}{18} + \frac{7}{19} \cdot \frac{12}{18} + \frac{12}{19} \cdot \frac{7}{18} + \frac{12}{19} \cdot \frac{11}{18}\right)\\
& \quad = \frac{9}{21} \cdot \frac{8}{20} \cdot 1\\
& \quad = \frac{9}{21} \cdot \frac{8}{20}\\
& \quad = \frac{6}{35}
\end{align*}
since the probability that the third and fourth balls are red or blue is $1$.
| {
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"url": "https://math.stackexchange.com/questions/3835703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimum value of $f(x,y,z) = x^z + y^z - (xy)^{\frac{z}{4}}, x > 0, y > 0, z > 0$ $$f(x,y,z) = x^{z}+y^{z}-(xy)^{\frac{z}{4}}$$
for all real positive numbers x, y, z
Does anyone have a clue to find the minimum value of $f(x,y,z)$?
I honestly don't know where to start the solution, I just come up with $AM \geq GM$
$\frac{x^z + y^z}{2} \geq \sqrt{{x}^{z}{y}^{z}} \\ x^z + y^z \geq 2{x}^{\frac{z}{2}}{y}^{\frac{z}{2}}$
With the equality holds if and only if $x^z = y^z$
$x^{z}+y^{z}-(xy)^{\frac{z}{4}} \\ \geq 2{x}^{\frac{z}{2}}{y}^{\frac{z}{2}} - (xy) ^{\frac{z}{4}} \\ = (xy)^{\frac{z}{4}}(2(xy)^{\frac{z}{4}} - 1)$
Set x^z = y^z for minimum value
$(x^{\frac{z}{2}})(2x^{\frac{z}{2}} - 1)$
From here, I set the function $\leq$ 0.
Since x > 0, It's obvious that $x^{\frac{z}{2}}$ can't be $\leq$ 0
$2x^{\frac{z}{2}} - 1 \leq 0 \\ (\sqrt{2} \cdot {x}^{\frac{z}{4}} + 1)(\sqrt{2} \cdot {x}^{\frac{z}{4}} - 1) \leq 0 \\ -\frac{1}{\sqrt{2}} \leq x^{\frac{z}{4}} \leq \frac{1}{\sqrt{2}}$
Since x > 0
$0 < x^{\frac{z}{4}} \leq \frac{1}{\sqrt{2}}$
I don't know what to do after this, I probably did a wrong method to solve the problem. Does anyone have a hint to solve it?
| An alternative approach: set $a = x^{z/4}$ and $b = y^{z/4}$. Then we want to minimize $a^4 + b^4 - ab$. Due to symmetry, this quantity achieves the minimum value when $a = b$ so we have $2a^4 - a^2$. The derivative is equal to $8a^3 - 2a$ and its equal to $0$ when $4a^2 - 1 = 0$ (since $a > 0$). And we find $a = 1/2$ so the min is $2(1/16) - 1/4 = -1/8$.
| {
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"url": "https://math.stackexchange.com/questions/3837048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Computation of a complex series I got stuck in a computation for hours and don't know where I did wrong. Let $N$ be a positive integer, I want to show the identity,
\begin{equation*}
\sum_{n=1}^{N}e^{inx} = e^{ix}\frac{e^{iNx}-1}{e^{ix}-1} = e^{i\frac{1}{2}(N+1)x}\frac{\sin (\frac{1}{2}Nx)}{\sin (\frac{1}{2}x)}
\end{equation*}
Here's what I did, using $\sin{x} = \frac{e^{ix}-e^{-ix}}{2i}$, we have
\begin{align*}
\frac{\sin [\frac{1}{2}Nx]}{\sin [\frac{1}{2}x]} &= \frac{e^{\frac{1}{2}Nix} - e^{-\frac{1}{2}Nix}}{e^{\frac{1}{2}ix}- e^{-\frac{1}{2}ix}}\\
& = \frac{e^{\frac{1}{2}(N+1)ix} - e^{-\frac{1}{2}(N-1)ix}}{e^{ix}- 1}\\
& = e^{-i\frac{1}{2}(N-1)x} \frac{ e^{iNx}-1}{e^{ix}- 1}
\end{align*}
But
\begin{equation*}
e^{ix} e^{-i\frac{1}{2}(N-1)x} = e^{i(\frac{3}{2}-\frac{N}{2})x}
\end{equation*}
where did this go wrong?
| You didn't do anything wrong, just forgot to multiply by $e^{i\frac{1}{2}(N+1)x}$.
$$\begin{align*}
\frac{\sin [\frac{1}{2}Nx]}{\sin [\frac{1}{2}x]} &= \frac{e^{\frac{1}{2}Nix} - e^{-\frac{1}{2}Nix}}{e^{\frac{1}{2}ix}- e^{-\frac{1}{2}ix}}\\
& = \frac{e^{\frac{1}{2}(N+1)ix} - e^{-\frac{1}{2}(N-1)ix}}{e^{ix}- 1}\\
& = e^{-i\frac{1}{2}(N-1)x} \frac{ e^{iNx}-1}{e^{ix}- 1}
\end{align*}$$
is correct and then to obtain the result you multiply by $e^{i\frac{1}{2}(N+1)x}$
to obtain $$e^{i\frac{1}{2}(N+1)x}\frac{\sin[\frac{1}{2}Nx]}{\sin[\frac{1}{2}x]}$$
$$=e^{i\frac{1}{2}(N+1-N+1)x}\frac{ e^{iNx}-1}{e^{ix}- 1}$$
$$=e^{ix}\frac{ e^{iNx}-1}{e^{ix}- 1}= \sum_{n=1}^{N}e^{inx}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3839345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ z,w\in\mathbb{C},|z|=|w|=R\gt0$. Show that $\left(\frac{z+w}{R^2+zw}\right)^2+\left(\vcenter{\frac{z-w}{R^2-zw}}\right)^2\ge\frac1{R^2}$ Let $ z, w \in \mathbb{C} $ be such that $ |z| = |w| = R > 0 $. Show that
$ \left(\frac{z + w}{R^2 + zw}\right)^2 + \left(\frac{z - w}{R^2 - zw}\right)^2 \geq \frac{1}{R^2} $
Well, i could only proof that
$ u=\left(\frac{z + w}{R^2 + zw}\right) $ , $ v = \left(\frac{z - w}{R^2 - zw}\right) \in \mathbb{R} $
By showing that
$ u = \overline{u}$ and $ v = \overline{v}$
Which leads to
$ u^2 + v^2 = |u|^2 + |v|^2 $
But i can't find a way to compute $|u|$ or $|v|$
|
From the picture, if $\arg{(w)}=a$ and $\arg{(z)}=b$, then
$$
\begin{aligned}
|z+w|&=2R\left|\cos{\left(\frac{a-b}{2}\right)}\right|\\
|z-w|&=2R\left|\sin{\left(\frac{a-b}{2}\right)}\right|\\
\\
|R^{2}+zw|&=2R^{2}\left|\cos{\left(\frac{a+b}{2}\right)}\right|\\
|R^{2}-zw|&=2R^{2}\left|\sin{\left(\frac{a+b}{2}\right)}\right|
\\
\\
\left(\frac{z+w}{R^{2}+zw}\right)^{2}+\left(\frac{z-w}{R^{2}-zw}\right)^{2}&=\frac{1}{R^{2}}\frac{\cos^{2}{\left(\frac{a-b}{2}\right)}}{\cos^{2}{\left(\frac{a+b}{2}\right)}}+\frac{1}{R^{2}}\frac{\sin^{2}{\left(\frac{a-b}{2}\right)}}{\sin^{2}{\left(\frac{a+b}{2}\right)}}
\end{aligned}
$$
Which is always greater than $\frac{1}{R^{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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