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Solve for $\sqrt{x}+\sqrt[4]{y}=\sqrt{p}$ with $x$, $y$ are positive integers and $p$ is a prime number For that given problem, I have constructed a solution to it:
Squaring both sides, we get $x+\sqrt{y}+2\sqrt[4]{x^2y}=p$. Suppose $\sqrt{y}$ and $\sqrt[4]{x^2y}$ are integers, we can let $y=y'^2$, $y'$ being a positive integer. Subbing to the equation, we get $x+y'+2\sqrt{xy'}=p$, so $(\sqrt{x}+\sqrt{y'})^2=p$. We can see that $\sqrt{x}+\sqrt{y'}$ is a positive integer, and $x$ and $y'$ are positive, so $\sqrt{x}+\sqrt{y'}>1$, so $(\sqrt{x}+\sqrt{y'})^2$ has at least 3 positive divisors, contradiction.
The only problem here is that I can't show that $\sqrt{y}$ and $\sqrt[4]{x^2y}$ are integers, because $\sqrt[4]{x^2y}=\sqrt{x\sqrt{y}}$, and I cannot guarantee that $x\sqrt{y}$ is an integer, so I can't show that they are both integers.
I would really appreciate it if I can get a hint on this problem, thank you a lot in advance.
| There are not solution. In fact we have
$$y=(\sqrt p-\sqrt x)^4=p^2+6px+x^2-4(p+x)\sqrt{px}$$ One has three possible ways:
►If $x=p$ then $y=8p^2-8p^2=0$
►If $p+x=0$ then $y=2p^2-6p^2\lt0$
►If $x=pz^2$ then $y=p^2(z-1)^4$ which implies the given equation becomes
$(z+z-1)\sqrt p=\sqrt p\Rightarrow z=1\Rightarrow x=p$ which is the first possibility.
| {
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"url": "https://math.stackexchange.com/questions/4245179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trigonometric Substitution Absolute value issue Evaluate $ \, \displaystyle \int _{0}^{4} \frac{1}{(2x+8)\, \sqrt{x(x+8)}}\, dx. $
$My\ work:-$
by completing the square and substitution i.e. $\displaystyle \left(\begin{array}{rl}x+4 & = 4\sec (\theta )\\ dx & = 4\sec \theta \tan \theta \, d\theta \end{array}\right) \qquad$ $\Rightarrow \displaystyle \int \frac{4\sec \theta \tan \theta \, d\theta }{2(4\sec (\theta ))( |4\tan (\theta )|) }$ $\Rightarrow \displaystyle \frac{1}{8} \int \frac{\tan \theta \, d\theta }{|\tan (\theta )|}$ now because my limits are positive so $sec\ \theta \geq 0\ $ and $sec\ \theta\ $ is positive in $\ Ist\ $ and $\ IVth\ $ Quadrant. At this stage i have 2 options either i consider $\ Ist\ $ Quadrant and take postive $|tan\ \theta|\ =\ tan\ \theta\ $ or i consider $\ IVth\ $ Quadrant where $\ |tan\ \theta|\ = -tan\ \theta$ So when in 1st quadrant i.e. $\ |tan\ \theta|\ = tan\ \theta ,$ $\ 0\ \geq\ \theta\ \geq\ \pi/2\ $ i get
$\Rightarrow \displaystyle \frac{1}{8} \theta +C$ $\Rightarrow \displaystyle \frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)+C$ $\Rightarrow \displaystyle \left. \frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)\, \right|_{x=0}^{x=4}$ $\Rightarrow \displaystyle \frac{1}{8} \left(\mathrm{arcsec}(2)-\mathrm{arcsec}(1)\right)$
Now if i consider 4th quadrant i.e. $\ |tan\ \theta|\ = -tan\ \theta ,$ $\ 3 \pi /2 \ \geq\ \theta\ \geq\ 2\pi\ $ i get $\Rightarrow \displaystyle -\frac{1}{8} \theta +C$ $\Rightarrow \displaystyle -\frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)+C$ $\Rightarrow \displaystyle \left. -\frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)\, \right|_{x=0}^{x=4}$ $\Rightarrow \displaystyle -\frac{1}{8} \left(\mathrm{arcsec}(2)-\mathrm{arcsec}(1)\right)$ So i am getting positive value in 1st case whereas in 4th quadrant my answer is Negative. Why is so ? am i making some mistake when considering 4th quadrant ? or both are acceptable answer ? also in MIT lecture they said whatever acceptable quadrant you choose you'll get the same answer. so why i m getting Negative answer ?
| If $\theta$ lies in 4th quadrant, you need to consider the values of $\text{arcsec}$ in fourth quadrant(as the inverse trig functions are multivalued).
$\text{arcsec }(2) = \theta_1 \Rightarrow \theta_1 = \dfrac{5\pi}3$
$\text{arcsec} (1) = \theta_2 \Rightarrow \theta_2 = 2\pi$
So, $-\dfrac{1}{8}(\theta_1-\theta_2) = -\dfrac{1}{8}\left(\dfrac{-\pi}{3}\right) = \dfrac{\pi}{24}$
This is same as the result from 1st quadrant. $\dfrac18(\text{arcsec }(2)-\text{arcsec }(1)) = \dfrac18\left(\dfrac\pi3-0\right) = \dfrac{\pi}{24}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Q: Epsilon-delta proof for $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$ In proving $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$, I have done this working:
Let $ϵ>0$.
$0<|x-1|< \Rightarrow|\frac{x}{(x^2+1)}-\frac{1}{2}|<$
$ |\frac{x}{(x^2+1)}-\frac{1}{2}|< = |\frac{2x-x^2+1}{2(x^2+1)}| = |\frac{-2(x-1)^2}{x(x-1)^2+4x}| = |\frac{-(x-1)^2}{(x-1)^2+2(x-1)+2}| = \frac{(x-1)^2}{|(x-1)^2+2(x-1)+2|}|$
after which I get stuck at trying to implement triangle inequality to push in the modulo in the denominator to get $|x-1|$ terms to substitute .
| As $ x $ goes to $ 1$, we can always assume that
$$|x-1|<2$$
and we look for $ \delta$ such that
$$|x-1|<2 \text{ and } |x-1|<\delta \implies$$
$$ |\frac{-(x-1)^2}{2x^2+2}|<\epsilon$$
but
$$2x^2+2\ge 2$$
and
$$|x-1|<2\implies $$
$$|\frac{(x-1)(x-1)}{2x^2+2}|\le |x-1|$$
So, we just need a $ \delta<2$ such that
$$|x-1|<\delta\implies |x-1|<\epsilon$$
from here, we see that we can take
$$\delta=\min(2,\epsilon)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$ I tried like this:
Let $y=a^{2x}-2\Rightarrow a^{2x}=y+2\Rightarrow 2x\ln a=\ln\left(y+2\right)\Rightarrow x=\dfrac{\ln\left(y+2\right)}{2\ln a}$
Also if $x\longrightarrow0,$ then $y\longrightarrow a^{2(0)}-2=-1.$
But we I put each and every this assumption in the given expression, then I get hanged due to $x^x.$ How to use algebra or any other easy procedure to show that $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$.
| Given
\begin{equation}
\lim_{x\rightarrow 0}\dfrac{a^{2x}-2}{x^x}=-1
\implies
\lim_{x\rightarrow 0}\dfrac{a^{2x}-2}{x^x} +1 =0
\end{equation}
You say,"then I get hanged due to $^$.." but $0^0=1$ and we can even eliminate the denominator if that isn't enough.
\begin{align*}
(\large{x}^x)
\small{\bigg(\dfrac{a^{2x}-2}{x^x}} + 1\bigg)=0
\implies
\space&a^{2 x} + x^x - 2=0\\ \\
a^{2 x} + x^x - 2=0
\implies& a^{2 x} + x^x = 2\\
\end{align*}
We can see by inspection that the derived limit below is valid.
\begin{align*}
&\lim_{x \to 0}a^{2 x} + x^x\space =1+1=2\\
\text{ therefore}&\\
&\lim_{x\to 0}\dfrac{a^{2x}\space -2}{x^x}=-1
\end{align*}
Another approach is using simple substitution , permitted here because there is no possible division by zero.
$$\lim_{x\to 0}\dfrac{a^{2x}\space -2}{x^x}
=\frac{2^0-2}{0^0}=\frac{-1}{1}=-1$$
| {
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"url": "https://math.stackexchange.com/questions/4247266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Caculate $\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$, with D:$\frac{x^2}{2}+y^2\leq1$ I found some difficulty with this exercise:
Calculate
$$\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$$
with $D := \left\{(x,y)\in\mathbb{R}^{2}\mid \dfrac{x^2}{2}+y^2\leq1\right\}$
I use change of Variables in Polar Coordinates, but the integral become so hard to calculate.
I think maybe we change variables $u = x^2 + y^2$, the integral will be easier, but I can't find $v(x,y$) to have the Jacobi easy to calculate.
| Take $$I=\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$$
Consider the change of variables $(x,y)\to(r\cos\theta,r\sin\theta)$.
Let $\partial D=\{(x,y)\in\mathbb{R}^2:\frac{x^2}2+y^2=1\}$. Then for all $(x,y)=(r\cos\theta,r\sin\theta)\in\partial D$, $r^2(1-\frac12\cos^2\theta)=1$. Hence,
$$I=\int_0^{2\pi}\left(\int_0^{R(\theta)}\frac{r^2}{\sqrt{4-r^4}}rdr\right)d\theta$$
where $R(\theta)=\frac1{\sqrt{1-\frac12\cos^2\theta}}$. Now,
$$I=\int_0^{2\pi}\frac12\left(2-\sqrt{4-[R(\theta)]^2}\right)d\theta$$
$$\implies I=2\pi-\frac12\int_0^{2\pi}\sqrt{4-\left(1-\frac12\cos^2\theta\right)^{-2}}d\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I get to this formula for this recursive sum and is my reasoning in proving this correct? I have the following sequence:
$n=1$ $$1$$
$n=2$ $$2+3+4$$
$n=3$ $$5+6+7+8+9$$
$n=4$ $$10+11+12+13+14+15+16$$
$$...$$
So abstracting these sums, we can write:
$n_1+(n_1+1)+...+(n_2-1)+n_2$
, where $n_1$ is the first term an $n_2$ is the last term of the sum.
And the sum of every $n^{th}$ term is:
$\frac{1}{2}(n_2-n_1+1)(n_1+n_2)$
Now I want to show that to calculate that the sum up to a certain $n$ value can be calculated as:
$$\frac{1}{2}n^2(n^2+1)$$
But I get stuck in finding this expression from the prior knowledge I described above.
I see it is closely related to the formula for the sum of nonnegative integers, but I can't figure out the line of reasoning that brings me to this formula for the sum of the above sequence for an $n^{th}$ value.
I could also try to proof that this formula is correct and I immediately thought of induction.
So we can first write
$$\frac{1}{2}(n_2-n_1+1)(n_1+n_2)$$
in terms of $n+1$, like:
$$\frac{1}{2}(n^2+1-(n+1)^2)(2n+1)$$
Then for the base case, $n=1$, we have
$$\frac{1}{2}1^2(1^2+1) = 1$$
Now for the inductive step, I started with:
$$f(n+1) = f(n) + \frac{1}{2}(n^2+1-(n+1)^2)(2n+1)$$
then proceded with the RHS as:
$$\frac{1}{2}n^2(n^2+1)+ \frac{1}{2}(n^2+1-(n+1)^2)(2n+1)$$
Now, to get the inductive step to work, we need to end up with:
$$f(n+1) = \frac{1}{2}(n+1)^2((n+1)^2+1)$$
But I don't see how we can proceed from where I've finished. I can do some algebraic manipulations, but it doesn't brings me to the above equation.
So my questions are:
*
*For the first part: how can I show that the sum of the sequence for a given $n$ is equal to $$\frac{1}{2}n^2(n^2+1)$$
*For the second part/the proof by induction part:
*
*Is my reasoning and are my calculations up to this point correct (in order to prove the thing that I want to prove)?
*If yes, how can I continue to get my inductive step to work. And if no, where are my calculations incorrect or when the reasoning is incorrect, how can I prove, perhaps via an alternative route, that the sum up to a certain $n$-value is indeed calculated as:
$$\frac{1}{2}n^2(n^2+1)$$
| The sum is equal to the sum of integers up to $n^2$ which is trivially given by $\frac{1}{2}n^2(n^2+1)$, then we could prove at first the general result by induction and then use it for this particular case.
Refer also to the related
*
*Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$
Edit
Following your way, for the inductive step we have
$$\frac{1}{2}n^2(n^2+1)+\overbrace{(n^2+1)+(n^2+2)+\ldots+(n+1)^2}^{2n+1\: \text{terms}}=$$
$$=\frac{1}{2}n^2(n^2+1)+n^2(2n+1)+\frac12(2n+1)(2n+2)=$$
$$=\frac{1}{2}n^2(n^2+1)+\frac{1}{2}n^2(2n+1)+\frac{1}{2}n^2(2n+1)+\frac12(2n+1)(2n+2)=$$
$$=\frac{1}{2}n^2(n^2+1+2n+1)+\frac{1}{2}(2n+1)(n^2+2n+2)=$$
$$=\frac12 (n^2+2n+1)(n^2+2n+2=\frac12 (n+1)^2+((n+1)^2+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How would you evaluate the following expression? It seems very difficult to simplify the trig.
I have tried to many different ways and I also end up with a tan inside an arctan which I do not know how to simplify. Please suggest how I can solve this. Am I missing something simple or is it quite tricky and need some sort of manipulation that I am missing.
| By symmetry we can reduce the interval to $[0,\frac{\pi}4]$, the coefficient before the integral sign becomes $8$.
$$I=\frac 8{\pi}\int_0^{\frac{\pi}4}\dfrac{\cos(4x)^2+1}{a\cos(2x)^2+1}dx$$
Then by substitution $u=2x$:
$$I=\frac 4{\pi}\int_0^{\frac{\pi}2}\dfrac{\cos(2u)^2+1}{a\cos(u)^2+1}du$$
The reason behind reducing the interval is to be able to have a bijective change $t=\tan(u)$:
We expand $\cos(2u)$ and replace $\cos(u)^2$ by $\frac 1{1+t^2}$:
$$I=\frac 8{\pi}\int_0^{\infty}\dfrac{t^4+1}{(t^2+a+1)(t^2+1)^2}du$$
The next part is partial fraction decomposition and integration in arctan, it is not difficult but tedious, so I skip it and jump to the result:
$$\dfrac {\tfrac{a^2+2a+2}{a^2}}{t^2+a+1}-\dfrac {\frac{2a+2}{a^2}}{t^2+1}+\dfrac {\frac 2a}{(t^2+1)^2}$$
$$I=4\times\frac{(a^2+2a+2)-\sqrt{1+a}(a+2)}{a^2\sqrt{1+a}}$$
We now have to solve $I=a$ :
$4(a^2+2a+2)-4\sqrt{1+a}(a+2)=a^3\sqrt{1+a}\iff 4(a^2+2a+2)=\sqrt{1+a}(a^3+4a+8)$
We square both sides and simplify to:
$$a^7+a^6+8a^5+8a^4-32a^3-48a^2=0$$
And since $a\neq 0$ we are glad to arrive to the desired expression:
$$a^5+a^4+8a^3+8a^2-32a=48$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4249380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Complex Matrix and its Conjugate terms If $z = \left| {\begin{array}{*{20}{c}}
{3 + 2i}&1&i\\
2&{3 - 2i}&{1 + i}\\
{1 - i}&{ - i}&3
\end{array}} \right|\& \left| {z + \overline z } \right| = k\left| z \right|$, find the value of k
My approach is as follow
$ \Rightarrow z = - \left| {\begin{array}{*{20}{c}}
1&{3 + 2i}&i\\
{3 - 2i}&2&{1 + i}\\
{ - i}&{1 - i}&3
\end{array}} \right|$ where $\left( {{C_1} \leftrightarrow {C_2}} \right)$
$ \Rightarrow \overline z = - \left| {\begin{array}{*{20}{c}}
1&{3 - 2i}&{ - i}\\
{3 + 2i}&2&{1 - i}\\
i&{1 + i}&3
\end{array}} \right|$
How do we proceed from here.
| You already exchanged the first two columns, which is a good start. The corresponding matrix
$$
A = \begin{pmatrix}
1&{3 + 2i}&i\\
{3 - 2i}&2&{1 + i}\\
{ - i}&{1 - i}&3\end{pmatrix}
$$
has the property that $A^T$ is the elementwise conjugate of $A$, so that
$$
\det(A) = \det(A^T) = \overline{\det(A)} \, .
$$
It follows that $z = -\det(A)$ is a real number, and therefore is $|z + \bar z| =2|z|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4249550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compund Inequality with Reciprocal $0<(2x-4)^{-1}<\frac{1}{2}$ I need help with this exercise: $0<(2x-4)^{-1}<\frac{1}{2}$.
$$0<(2x-4)^{-1}<\frac{1}{2}$$
$$0<\frac{1}{2x-4}<\frac{1}{2}$$
I separate this into two inequalities:
*
*$0<\frac{1}{2x-4}$
*$\frac{1}{2x-4}<\frac{1}{2}$
The first one I had no problem solving it.
$$0<\frac{1}{2x-4}$$
By the Reciprocal Property of inequalities
$$0<2x-4$$
$$4<2x$$
$$\frac{4}{2}<\frac{2x}{2}$$
$$2<x$$
$$x>2$$
Is the second that I have a problem with. Is there a way to solve it without using a sign chart/table? I do know how to do it with a sign chart/table. I just want to know if there is another way to do it that is not the sign table.
| \begin{align*}
0<(2x-4)^{-1}<\frac{1}{2}\\ \\
\implies
0<\frac{1}{2x-4}<\frac{1}{2}\\ \\
\implies 0< 2< 2x-4\\ \\
\implies 4< (2)+4< (2x-4) + 4 \\ \\
\implies 4< 6< 2x \\
\implies 2< 3< x\\
\implies 3< x\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\tan(\frac{1}{2}\sin^{-1}\frac{3}{4})$ Evaluate $\tan(\frac{1}{2}\sin^{-1}\frac{3}{4})$
My attempt:
Method:- 1
Let $\sin^{-1}\frac{3}{4} = \theta \implies \sin\theta = \frac{3}{4}$ and $\theta \in [0,\frac{π}{2}]$
Therefore $$\cos\theta = \sqrt{1-\sin^2\theta} = \frac{\sqrt 7}{4}$$
Now $\tan(\frac{1}{2}\sin^{-1}\frac{3}{4}) = \tan\frac{\theta}{2} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \sqrt{ \frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \frac{4-\sqrt7}{3}$
Method:-2
Let $\sin^{-1}\frac{3}{4} = \theta \implies \sin\theta = \frac{3}{4} \implies 2\sin\frac{\theta}{2} \cos\frac{\theta}{2} = \frac{3}{4} \implies
\sin\frac{\theta}{2} \cos\frac{\theta}{2} = \frac{3}{8}$
Now $\tan\frac{\theta}{2} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}$
Please help me in 2nd method. Thanks in advance.
| So you have $\sin\frac{\theta}{2} \cos\frac{\theta}{2} = \sqrt{1-\cos^2\frac{\theta}{2}}\cos\frac{\theta}{2}=\frac{3}{8}$ Solving $(1-x^2)x^2=\frac{9}{64} \implies x^4-x^2+\frac{9}{64}=0 \implies x^2=\frac{1+\sqrt{1-\frac{9}{16}}}{2}=\frac{4+\sqrt{7}}{8} $$\tan\frac{\theta}{2}=\frac{\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{\cos^2\frac{\theta}{2}}=\frac{3\cdot 8}{8(4+\sqrt{7})}=\frac{3}{4+\sqrt{7}}=\frac{4-\sqrt{7}}{3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is rolling a 60 face die (D60) the same as rolling 10 6 faced ones (D6)? I understand that if you roll $10$ $6$-faced dice (D6) the different outcomes should be $60,466,176$ right? (meaning $1/60,466,176$ probability)
but if you are playing a board-game, you don't care about that right? what matters is the sum of the numbers in the end, so it should anyway be $1/60$ instead of whatever right? because its a sum not $10$ individual rolls.
so a $60$ face die (D60) should be the same as rolling all the dice right? if not can someone please explain? I've been looking all over the net for a compelling answer but there is none. I've seen a D60 can be used as $2$ D6 but also there is no explanation for this.
if not, what would be the number of faces needed for an equivalent single die that can replace $10$ D6. I know so far the D120 is the last possible die that can be made, but it doesn't matter, I just want to know too...
EDIT: Sorry I forgot to mention in some tabletop games Dice can have $0$, so you'd treat all $6$ sided dice as going from $0$ to $5$, making minimum value of $0$ possible in all $10$ rolls. But still that changes the maximum in the same way minimum used to be, now the max you get in 6 sided dice is $50$, while on the 60 sided die it would be $59$ if we apply a $-1$ to all results... it gets messy, and this detail doesn't really change much
| No, the probability distribution of the sum of the faces of rolling $10$ $6$-sided dice is different than that of rolling a single $60$-sided die.
For obvious reasons, this is because you can roll any number between $1$ and $9$ with a $60$-sided die, but cannot attain such a sum with $10$ $6$-sided dice.
The probability distribution of a single $6$-sided die can be modeled with the generating function
$$\frac{1}{6}(x+x^2+x^3+x^4+x^5+x^6)$$
Where the coefficient of $x^k$ is the probability of rolling a $k$. Moreover, the probability distribution of the sum of rolling $10$ $6$-sided dice can be modeled as
$$\frac{1}{6^{10}}(x+x^2+x^3+x^4+x^5+x^6)^{10}$$
Where the coefficient of $x^k$ is the probability of rolling a sum of $k$. It can be clearly seen that this expansion is completely different than the generating function that models the probability distribution of rolling a $60$-sided die i.e.
$$\frac{1}{60}(x+x^2+x^3+\ldots+x^{59}+x^{60})$$
However, there are some interesting ideas that you can get from using these generating functions. For example, we have that
$$\frac{1}{6}(x+x^2+x^3+x^4+x^5+x^6)$$
$$=\frac{1}{6}x(1+x+x^2+x^3+x^4+x^5)$$
$$=\frac{1}{6}x(1+x^3)(1+x+x^2)$$
$$=\frac{1}{6}x(1+x)(1-x+x^2)(1+x+x^2)$$
We can take two dice, and say that their generating function for their probability distribution is
$$=\frac{1}{6^2}x^2(1+x)^2(1-x+x^2)^2(1+x+x^2)^2$$
$$=\frac{1}{6^2}[(1+x)(1+x+x^2)x^2][(1+x)(1+x+x^2)(1-x+x^2)^2]$$
$$=\frac{1}{6^2}[x^2+2x^3+2x^4+x^5][1+x^2+x^3+x^4+x^5+x^7]$$
This means we can replace two $6$-sided dice with, for example, a die with faces of $2,3,3,4,4,5$ and a die with faces of $0,2,3,4,5,7$, and the sum from rolling these two dice will have the same probability distribution as rolling two standard $6$-sided dice. We can extend this process to the generating functions for more $6$-sided dice and create some wacky dice that still have the same probability distribution.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Radical expression for $\tan 14.4^\circ$ Given that
$\displaystyle\cos 20^\circ=\frac{\sqrt[3]{\frac{1-i\sqrt{3}}{2}}+\sqrt[3]{\frac{1+i\sqrt{3}}{2}}}{2}$
and
$\displaystyle\sin 20^\circ=\frac{i\left(\sqrt[3]{\frac{1-i\sqrt{3}}{2}}-\sqrt[3]{\frac{1+i\sqrt{3}}{2}}\right)}{2}$
It turns out that there is a nice radical expression for $\tan 20^\circ$:
$\displaystyle\tan 20^\circ=\sqrt{11-\left(1-i\sqrt{3}\right)\sqrt[3]{148+4i\sqrt{3}}-\left(1+i\sqrt{3}\right)\sqrt[3]{148-4i\sqrt{3}}}$.
My question is, how did one get from the first two expressions to the third? It's pretty straightforward to find the minimal polynomial for $\cos 20^\circ$ from the triple-angle formulas, which is $8x^3-6x-1=0$. In other words, where did the three "new" numbers ($11$ and $37\pm i\sqrt{3}$) in the expression come from?
Now a similar situation arises for $\tan 14.4^\circ$:
$\displaystyle\cos 14.4^\circ=\frac{\sqrt[5]{\frac{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}{4}}+\sqrt[5]{\frac{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}{4}}}{2}$
and likewise for $\sin 14.4^\circ$.
My gut instinct is that $\tan 14.4^\circ$ can be written as
$\tan 14.4^\circ=\sqrt{a-\left(-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}\right)\sqrt[5]{b+ci}-\left(-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}\right)\sqrt[5]{b-ci}}$
for some algebraic $a, b,$ and $c.$
My question is, what are the values of a, b, and c? And I'm betting that my "gut instinct" on $\tan 14.4^\circ$ is wrong, so if anyone could help me derive the radical expression for $\tan 14.4^\circ$, that would be great.
| Try using:$$\tan(x)=\pm\frac{\sqrt{1-\cos^2(x)}}{\cos(x)}$$ because you already know $\cos(x),x=14.4^\circ$.
$$\cos 14.4^\circ=\frac{\sqrt[5]{\frac{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}{4}}+\sqrt[5]{\frac{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}{4}}}{2}$$
Therefore
$$\tan(14.4)=\tan\frac{2\pi}{25}= \frac{\sqrt{1-\bigg(\frac{\sqrt[5]{\frac{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}{4}}+\sqrt[5]{\frac{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}{4}}}{2}}\bigg)^2}{\frac{\sqrt[5]{\frac{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}{4}}+\sqrt[5]{\frac{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}{4}}}{2}} = \frac{\sqrt{1-\bigg(\frac{\sqrt[5]{\frac{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}{4}}+\sqrt[5]{\frac{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}{4}}}{2}}\bigg)^2}{\sqrt[5]{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}+\sqrt[5]{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}} $$
Good luck simplifying this. Here is proof of the answer. Please correct me and give me feedback!
| {
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Find value of $x^{5} + y^{5} + z^{5}$ given the values of $x + y + z$, $x^{2} + y^{2} + z^{2}$ and $x^{3} + y^{3} + z^{3}$ If$$x+y+z=1$$
$$x^2+y^2+z^2=2$$
$$x^3+y^3+z^3=3$$
Then find the value of $$x^5+y^5+z^5$$
Is there any simple way to solve this problem ? I have tried all my tricks tried to multiply two equations , substitute $z=1-x-y$ , but things got messy nothing seems to work out .
| HINT
Multiply the third relation by the second so that you obtain
\begin{align*}
(x^{2} + y^{2} + z^{2})(x^{3} + y^{3} + z^{3}) & = x^{5} + y^{5} + z^{5} + x^{2}(y^{3} + z^{3}) + y^{2}(x^{3} + z^{3}) + z^{2}(x^{3} + y^{3})
\end{align*}
Now we can rearrange the last expression as
\begin{align*}
x^{2}y^{2}(x + y) + x^{2}z^{2}(x + z) + y^{2}z^{2}(y + z) & = x^{2}y^{2} + x^{2}z^{2} + y^{2}z^{2} - x^{2}y^{2}z - x^{2}yz^{2} - xy^{2}z^{2}\\\\
& = x^{2}y^{2} + x^{2}z^{2} + y^{2}z^{2} - xyz(xy + xz + yz)
\end{align*}
Hence the problem has been reduced to study the last relation.
Before keep going, notice that
\begin{align*}
x^{2}y^{2} + x^{2}z^{2} + y^{2}z^{2} & = (xy + xz + yz)^{2} - 2xyz(x + y + z)\\\\
& = (xy + xz + yz)^{2} -2xyz
\end{align*}
In order to find the remaining expression, let us observe that
\begin{align*}
(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + xz + yz) \Rightarrow xy + xz + yz = -\frac{1}{2}
\end{align*}
At last but not least, we have to find the value of $xyz$.
Can you take it from here?
| {
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How to find recurrence relation for $n$-digits number where $1$ and $3$ occur an odd number of times
Let $h(n)$ be the number of $n$-digits number with each digit odd, where the digits $1$ and $3$ occur an odd number of times. Find a recurrence relation for $h(n)$.
I have found the recurrence relation:
\begin{equation}
\begin{aligned}h(n+2) &= 6h(n+1) - 5h(n) + 2.3^n \\
h(0) &= h(1) = 0
\end{aligned} \tag{1}
\end{equation}
But the solution I tried was not quite satisfying. I got $(1)$ by using a system of recurrence.
Let $a(n),b(n),g(n)$ respectively be the number of $n$-digits number with even $3$s and odd $1$s, odd $3$s and even $1$s, and both even. Thus,
$a(n) :$ even $3$s odd $1$s
$b(n) :$ odd $3$s even $1$s
$h(n) :$ both odd
$g(n) :$ both even
I got some recurrences by considering the last digit of the number, whether it is $1,3$ or other than them and making connection between the $n$-digits and $(n-1)$digits number.
The system is
\begin{align*}
h(n) &= a(n-1) + b(n-1) + 3h(n-1)\\
b(n) &= g(n-1) + h(n-1) + 3b(n-1)\\
a(n) &= g(n-1) + h(n-1) + 3a(n-1)\\
g(n) &= a(n-1) + b(n-1) + 3g(n-1)\\
\end{align*}
with $a(0) = b(0) =h(0) = 0, g(0) = 1$. Note that $a(n)$ is, ofcourse, equal $b(n)$ since it is only swapping $1$ and $3$. From this system, I got the recurrence $(1)$. It is quite tedious.
Question: How can we proceed directly to get $(1)$ using some combinatorial argument?
The term $-5h(n)$ in $(1)$ suggests that we have to use the inclusion-exclusion principle but I just don't see how. The term $2.3^n$ also seems mysterious.
| The following does not derive $(1)$ from your recurrence but instead finds an explicit formula for $h_n$, which is perhaps the motivation for $(1)$. First replace $b$ with $a$:
\begin{align}
h_n &= 2a_{n-1} + 3h_{n-1}\\
a_n &= g_{n-1} + h_{n-1} + 3a_{n-1}\\
g_n &= 2a_{n-1} + 3g_{n-1}\\
\end{align}
With generating functions $H(z)=\sum_{n \ge 0} h_n z^n$, $A(z)=\sum_{n \ge 0} a_n z^n$, and $G(z)=\sum_{n \ge 0} g_n z^n$, we obtain
\begin{align}
H(z) - 0 &= 2z A(z) + 3z H(z) \\
A(z) - 0 &= z G(z) + z H(z) + 3z A(z) \\
G(z) - 1 &= 2 z A(z) + 3z G(z) \\
\end{align}
Solving this system yields
\begin{align}
H(z) &= \frac{2 z^2}{1-9z+23z^2-15 z^3} \tag2\\
A(z) &= \frac{z}{1-6z+5 z^2} \\
G(z) &= \frac{1-6z+7 z^2}{1-9z+23z^2-15 z^3} \\
\end{align}
Now $(2)$ implies that
\begin{align}
H(z) &= \frac{2 z^2}{(1-z)(1-3z)(1-5z)} \\
&= \frac{1/4}{1-z}-\frac{1/2}{1-3z}+\frac{1/4}{1-5z} \\
&= \frac{1}{4}\sum_{n\ge 0}z^n-\frac{1}{2}\sum_{n\ge 0}(3z)^n+\frac{1}{4}\sum_{n\ge 0}(5z)^n
\end{align}
which immediately yields explicit formula
$$h_n = \frac{1}{4}-\frac{1}{2}\cdot 3^n+\frac{1}{4}\cdot 5^n = \frac{1-2\cdot 3^n+5^n}{4}.$$
If a recurrence relation is required, note that the denominator of $(2)$ implies that $$h_n = 9h_{n-1}-23h_{n-2}+15h_{n-3}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove a Limit Using the $\varepsilon - N$ definition I'm having trouble proving the following limit using the $\varepsilon-N$ definition:
$$\lim_{n \to \infty} \frac{3n}{\sqrt{9n^2 + 11n + 4}} = 1.$$
Here is my work so far.
Take arbitrary $\varepsilon$. We need to prove the existence of a natural number, $N$, such that for all $n \ge N$, $$ | x_n - 1| < \varepsilon.$$
$x_n = \frac{3n}{\sqrt{9n^2 + 11n + 4}}$ and since $x_n < \frac{3n}{\sqrt{9n^2}} = 1$, the equation becomes: $$1 - \frac{3n}{\sqrt{9n^2 + 11n + 4}} < \varepsilon.$$
Rewriting gives: $$\frac{3n}{\sqrt{9n^2 + 11n + 4}} > 1 - \varepsilon.$$
Clearly, when $\varepsilon \ge 1$, the statement is true since the RHS of the inequality becomes non-positive and obviously, the RHS must be positive.
However, I'm having trouble proving the statement when $0 < \varepsilon < 1$.
My idea so far has been to take the square:
$$\frac{9n^2}{9n^2 + 11n + 4} > 1 - 2\varepsilon + \varepsilon^2.$$Then, by taking the reciprocal of both sides, we obtain:
$$\frac{9n^2 + 11n + 4}{9n^2} < \frac{1}{1 - 2\varepsilon + \varepsilon^2}.$$
From here, I thought it might be smart to use the idea that for any real number, $x$, there exists a natural number, $m$, such that $\frac{1}{m} < x$.
Specifially, this would give us $m$ such that: $$\frac{1}{m} < \frac{1}{1 - 2\varepsilon + \varepsilon^2}$$ in which we case we could try to solve for $n$ in terms of $m$.
However, this solution is very complicated and long and bashy. There has to be a more elegant way to prove this limit right?
A pattern I've seen in a lot of limit problems involving square roots is the usage of $$a -b = \frac{a^2 - b^2}{a+b}$$ but I can't see how to apply that here either.
Any ideas or tips for a better solution would be greatly appreciated!
| Let us assume that $n\geq n_{\varepsilon}$. Then it results that
\begin{align*}
\left|\frac{3n}{\sqrt{9n^{2} + 11n + 4}} - 1\right| & = \left|\frac{3n - \sqrt{9n^{2} + 11n + 4}}{\sqrt{9n^{2} + 11n + 4}}\right|\\\\
& \leq\left|\frac{3n - \sqrt{9n^{2} + 11n + 4}}{3n}\right|\\\\
& = \left|\frac{11n + 4}{3n[3n + \sqrt{9n^{2} + 11n + 4}]}\right|\\\\
& \leq\left|\frac{11n + 4}{3n(3n + 3n)}\right|\\\\
& = \frac{11n + 4}{18n^{2}}\\\\
& = \frac{11}{18n} + \frac{2}{9n^{2}}\\\\
& \leq \frac{11}{18n_{\varepsilon}} + \frac{2}{9n^{2}_{\varepsilon}} := \varepsilon
\end{align*}
Then you can take the ceiling function of the corresponding solution.
| {
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What did I do wrong in my proof of $4\nmid (n^2+3)\implies 2\nmid (n^4-3)$? A homework problem is asking me to prove the following statement:
For any $n\in\mathbb{Z}$, $2\mid (n^4-3)$ if and only if $4\mid (n^2+3)$
Quick note: we're not allowed to use the theory of congruences in our solution.
I figured the best way to approach this was to prove that $4\mid (n^2+3)$ implies $2\mid (n^4-3)$ and $4\nmid (n^2+3)$ implies $2\nmid (n^4-3)$.
The proof of "$4\mid (n^2+3)$ implies $2\mid (n^4-3)$" was not an issue. It's proving "$4\nmid (n^2+3)$ implies $2\nmid (n^4-3)$" where I had some trouble. Here's my argument so far:
If $4\nmid (n^2+3)$, then $n^2+3=4k+r$ for some integer $k$, where $r$ is either $1$, $2$, or $3$. This is equivalent to
$$n^2=4k+r-3$$
so
\begin{align}
n^4 &= (4k+r-3)^2\\
&= 16k^2+8k(r-3)+(r-3)^2\\
\end{align}
which implies that
$$n^4-3=16k^2+8k(r-3)+(r-3)^2-3$$
Since $r$ is either $1$, $2$, or $3$, we have that
$$n^4-3=16k^2+8k(1-3)+(1-3)^2-3=16k^2-16k+1\text{,}$$
$$n^4-3=16k^2+8k(2-3)+(2-3)^2-3=\color{red}{16k^2-8k-2}\text{,}$$
$$\text{or}$$
$$\vdots$$
I stopped here because the part in red is problematic. Factoring out a $2$, it can be seen that $n^4-3$ is even in this case. I can't have this because I'm trying to show that $2\nmid (n^4-3)$, so what did I do wrong? I checked my algebra over and over again, and I can't seem to find a mistake anywhere. I also graphed $(4k+r-3)^2-3$ with $16k^2+8k(r-3)+(r-3)^2-3$ on Desmos for $r=1,2,3$ and the graphs seem identical.
In light of this, it seems like either (1) I'm blind, (2) I made a subtle mistake in my assumptions, or (3) there's something deeper going on that I'm failing to recognize. Am I seeing everything? Are my assumptions for the proof correct? If the answer to both of these is yes, does this mean that dividing $n^2+3$ by $4$ will never yield a remainder $r=2$? If not, what's going on here? Any help is appreciated.
| $n^2+3=(2k)^2+3=4k^2+3$ for even $n$, so remainder is $3$.
$n^2+3=(2k+1)^2+3=4k^2+4k+4$ for odd $n$ and is divisible by $4$, so it doesn't qualify.
In other words assumption $4\nmid n^2+3$ implies $n$ is even and $n^4-3$ is odd.
| {
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Evaluating $\lim _{x\to \infty }\tan\left(x^2\sin\frac{\pi }{4x^2}\right)$ without L'hopital's rule How do I solve this without using L'hopital's rule?
$$\lim _{x\to \infty }\left(\tan\left(x^2\sin\left(\frac{\pi }{4x^2}\right)\right)\right)$$
Problem
EDIT:
My progress so far:
$$\lim _{x\to \infty }\left(\tan\left(x^2\sin\left(\frac{\pi }{4x^2}\right)\right)\right)$$
$$=\lim _{x\to \infty }\left(\tan\left(\frac{\sin\left(\frac{\pi }{4x^2}\right)}{\frac{1}{x^2}}\right)\right)$$
$$=\lim _{x\to \infty }\left(\tan\left(\frac{\pi}{4}*\frac{\sin\left(\frac{\pi }{4x^2}\right)}{\frac{\pi}{4x^2}}\right)\right)$$
Now I know $\frac{\sin\left(x\right)}{x} = 1$ but I don't know how to take the above expression to that form.
| You've already taken care of one important bit, namely rewriting
$$\tan\left(x^2\sin\frac{\pi}{4x^2}\right)$$
as
$$\tan\left(\frac{\pi}{4}\cdot\frac{\sin\frac{\pi}{4x^2}}{\frac{\pi}{4x^2}}\right)$$
As you noted, $\lim_{x\to 0}\frac{\sin(x)}{x}=1$, so we should make use of this somehow. We can achieve this goal by making use of another important observation, namely that
$$\lim_{x\to\infty}\frac{\pi}{4x^2}=\frac{\pi}{4}\lim_{x\to\infty}\frac{1}{x^2}=\frac{\pi}{4}\cdot 0=0$$
Knowing this, I hope it makes sense that we should have
$$\lim_{x\to\infty}\frac{\sin\left(\frac{\pi}{4x^2}\right)}{\frac{\pi}{4x^2}}=1$$
After all, as $x$ gets bigger and bigger, $\frac{\pi}{4x^2}$ gets closer and closer to $0$, and $\sin(y)/y$ gets closer and closer to $1$ as its argument gets smaller and smaller, namely $y=\frac{\pi}{4x^2}$.
Alright, so we know that
$$\lim_{x\to\infty}\frac{\sin\left(\frac{\pi}{4x^2}\right)}{\frac{\pi}{4x^2}}=1$$
and thus, by the constant multiple rule for limits
$$\lim_{x\to\infty}\left(\frac{\pi}{4}\cdot\frac{\sin\left(\frac{\pi}{4x^2}\right)}{\frac{\pi}{4x^2}}\right)=\frac{\pi}{4}$$
But the tangent function $\tan$ is continuous near $\pi/4$, so
$$\lim_{x\to\infty}\tan\left(\frac{\pi}{4}\cdot\frac{\sin\left(\frac{\pi}{4x^2}\right)}{\frac{\pi}{4x^2}}\right)=\tan\left(\lim_{x\to\infty}\left(\frac{\pi}{4}\cdot\frac{\sin\left(\frac{\pi}{4x^2}\right)}{\frac{\pi}{4x^2}}\right)\right)=\tan\left(\frac{\pi}{4}\right)=1$$
We conclude that the original limit is
$$\lim_{x\to\infty}\tan\left(x^2\sin\frac{\pi}{4x^2}\right)=1$$
| {
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Nice problem: Prove that: $ab+bc+ca \ge \sum{\sqrt{a^2+b^2+3}}$ Problem: Let $a,b,c>0:a+b+c=abc.$ Prove that: $$ab+bc+ca\ge \sqrt{a^2+b^2+3}+\sqrt{b^2+c^2+3}+\sqrt{c^2+a^2+3}$$
Please help me give a hint to get a nice proof!
My attempts after squaring both side, it is desired to show: $a^2b^2+b^2c^2+c^2a^2+4(ab+bc+ca)\ge9+2\sum{\sqrt{(a^2+b^2+3)(b^2+c^2+3)}}$
It seems I am lost!
| We need to prove that:
$$ab+ac+bc\geq\sum_{cyc}\sqrt{\frac{abc}{a+b+c}\cdot\left(a^2+b^2+\frac{3abc}{a+b+c}\right)}$$ or
$$(ab+ac+bc)(a+b+c)\geq\sum_{cyc}\sqrt{abc((a^2+b^2)(a+b+c)+3abc)}.$$
Now, by AM-GM and C-S
$$\sum_{cyc}\sqrt{(a^2+b^2)(a+b+c)+3abc}=\sum_{cyc}\sqrt{(a^2+b^2)(a+b)+(a^2+b^2+3ab)c}\leq$$
$$\leq\sum_{cyc}\sqrt{(a^2+b^2)(a+b)+\frac{5(a+b)^2c}{4}}=\frac{1}{2}\sum_{cyc}\sqrt{(a+b)(4a^2+4b^2+5(a+b)c)}\leq$$
$$\leq\frac{1}{2}\sqrt{\sum_{cyc}(a+b)\sum_{cyc}(4a^2+4b^2+5(a+b)c)}=\sqrt{(a+b+c)\sum_{cyc}(4a^2+5ab)}$$ and it's enough to prove that:
$$(a+b+c)(ab+ac+bc)^2\geq abc\sum_{cyc}(4a^2+5ab)$$ or
$$\sum_{cyc}c^3(a-b)^2\geq0.$$
| {
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Given two pythagorean triples, generate another I don't know if this has been asked before, but I could not find any existing answer.
I noticed that for any pair of primitive pythagorean triples (not necessarily distinct), let's say:
a² + b² = c²
d² + e² = f²
Then there is at least another primitive triple where:
g² + h² = (cf)²
And there are 2 if the starting triples are distinct.
So, for example:
(3,4,5) and (5,12,13) -> (16, 63, 65) and (33, 56, 65)
(5,12,13) and (8,15,17) -> (21, 220, 221) and (140, 171, 221)
(3,4,5) (5,12,13) (8,15,17) -> (817,744,1105) (943,576,1105) (1073,264,1105) (1104,47,1105)
(3,4,5) and (3,4,5) -> (7,24,25)
I think there is an explanation for that, a property of pythagorean triples, or in general of diophantine equations.
Is it true in every case? Is there a way to calculate the two legs of the resulting triple(s)?
| You can always find new triplets this way, although they are not necessarily primitive. This is just the Brahmagupta–Fibonacci identity in action.
For example, for $(5,12,13)$ and $(8,15,17)$, the identity states that:
$$(5\cdot8 - 12\cdot15)^2 + (5 \cdot 15 + 12 \cdot 8)^2 = (-140)^2 + (171)^2$$
$$= (5\cdot8 + 12\cdot15)^2 + (5 \cdot 15 - 12 \cdot 8)^2 = (220)^2 + (-21)^2$$
$$= (5^2 + 12^2) (8^2 + 15^2) = 13^2 \cdot 17^2 = 221^2$$
When the starting triplets are the same, then $a=d, b=e$. Using the given identity, $(ad - be)^2 + (ae + bd)^2 = (a^2-b^2)^2 + (ab + ba)^2 = a^4+b^4-2a^2b^2+4a^2b^2$ $ = a^4 + b^4 + 2a^2b^2 = (a^2+b^2)^2 = (a^2+b^2)(d^2+e^2)$.
The other possibility is $(ad + be)^2 + (ae - bd)^2 = (a^2 + b^2)^2 + (ab - ba)^2 = (a^2 + b^2)^2$, where one of the numbers of the new triplet is $0$. This explains why there is only one non-trivial triplet in this case.
Here is a counterexample that they are not always primitive. With $(13,84,85)$ and $(16,63,65)$, $ad+be, ae-bd$ gives $(5500,525,5525)$ which are all divisible by $25$. A further conjecture could be made on if $c,f$ are both divisible by $k$, then whether the lowest common divisor would be $k^2$.
| {
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Show that $1998 < \sum_1^{10^6}\frac{1}{\sqrt{n}}<1999$ Show that $1998 < \sum_1^{10^6}\frac{1}{\sqrt{n}}<1999$ by first recalling the inequality:
$$2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1})$$ when $n=1, 2, 3...$
What I have tried to show using the inequality:
By the first step of induction to see if the inequality holds :
$2(\sqrt{1+1}-\sqrt{1})<\frac{1}{\sqrt{1}}<2(\sqrt{1}-\sqrt{1-1})$ is true
Then by induction we have:
$2(\sqrt{n+1+1}-\sqrt{n+1})<\frac{1}{\sqrt{n+1}}<2(\sqrt{n+1}-\sqrt{n+1-1})$
Then by using the fact that if $a>b$ and $c>0$ then $ac>bc$
The following should be true if the induction is right.
$\frac{2(\sqrt{n+1}-\sqrt{n})}{\sqrt{n+1}}>2(\sqrt{n+1+1}-\sqrt{n+1})$ by taking the square of both sides we have after expanding:
$$\frac{8n-8\sqrt{n^2+n}+4}{n+1}>8n+12-8\sqrt{n^2+3n+2}$$
Then placing everything onto one side:
$$\frac{-8n^2+8n\sqrt{n^2+3n+2}-12n+8\sqrt{n^2+3n+2}-8\sqrt{n^2+n}-8}{n+1}>0$$
Thouigh I'm not sure whether I'm going down the right direction. I had assumed that the LHS would cancel to provide something more exact. Any detailed hint son how to proceed?
| More directly, multiply and divide by the conjugate:
$2 ( \sqrt{n+1} - \sqrt{n} ) = 2 \frac{ ( \sqrt{n+1} - \sqrt{n} ) ( \sqrt{n+1} + \sqrt{n} ) }{ ( \sqrt{n+1} + \sqrt{n} )} = \frac{2}{ ( \sqrt{n+1} + \sqrt{n} )} < \frac{2}{2\sqrt{n}} = \frac{1}{ \sqrt{n}}. $
Same for the RHS, can you complete it?
| {
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Dependent equations, with Trigonometry. My questions is: do the first two equations imply the third ?
$$\left(x+\frac{1}{x}\right)\sin\alpha=\frac{y}{z}+\frac{z}{y}+\cos^2\alpha$$
$$\left(y+\frac{1}{y}\right)\sin\alpha=\frac{z}{x}+\frac{x}{z}+\cos^2\alpha$$
$$\left(z+\frac{1}{z}\right)\sin\alpha=
\frac{x}{y}+\frac{y}{x}+\cos^2\alpha$$
This is an old trigonometry problem stated as follows:
Prove that the equations are not independent, and that
they are equivalent to
$$x+y+z=\frac{1}{x}+
\frac{1}{y}+\frac{1}{z}=-\sin\alpha$$
The problem can be solved by subtracing the first two to get
$$\sin\alpha=
\frac{z^2-xy}{z(xy-1)}$$
And by subtracting the last two,
$$\sin\alpha=
\frac{x^2-zy}{x(zy-1)}$$
Note that we are making use of all three equations.
Setting the expressions equal we derive
$$x+y+z=\frac{1}{x}+
\frac{1}{y}+\frac{1}{z}$$
To get the last part
Let
$$S=x+y+z$$ and $$R=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
our condition is $$R=S$$
then if we add all three equations,
$$(R+S)\sin\alpha=RS-3+3\cos^2\alpha$$
and setting $R=S$, this factors as
$$(\sin\alpha+R)(3\sin\alpha-R)$$
For some reason we reject one solution.
Substitution will verify the equations.
Note that this solution makes use of all three equations.
They are dependent in the sense that they imply a relation on the coefficients.
Can the problem be solved using only two of the equations ?
Theoretically one can take the original expression above for $\sin\alpha$ and substitute in the first, this should give a relation. But I cannot make the algebra work, it is horrible.
Also how does some one come up with such a question ? Do they reverse engineer the answer ? I dont have any insight into the process of making such a problem.
|
do the first two equations imply the third ?
No, take $x=\frac 12,y=4,z=2$ and $\alpha=\frac{\pi}{2}$ for which the first two equations hold and the third doesn't.
Subtracting
$$\left(y+\frac{1}{y}\right)\sin\alpha=\frac{z}{x}+\frac{x}{z}+1-\sin^2\alpha\tag1$$
from
$$\left(x+\frac{1}{x}\right)\sin\alpha=\frac{y}{z}+\frac{z}{y}+1-\sin^2\alpha\tag2$$
gives
$$\frac{(x-y)(xy-1)}{xy}\sin\alpha=\frac{(x-y)(z^2-xy)}{xyz}$$
Assuming that $x\not=y$, one gets
$$\sin\alpha=\frac{z^2-xy}{z(xy-1)}\tag3$$
under the condition that
$$xy\not=1\tag4$$
Plugging $(3)$ into $(1)$ gives
$$\frac{(x y - z) (x y + z) (x^2 y z + x y^2 z + x y z^2 - x y-yz - zx)}{x y z^2 (x y - 1)^2}=0\tag5$$
Also,
$$\left(z+\frac{1}{z}\right)\sin\alpha-\bigg(\frac{x}{y}+\frac{y}{x}+\cos^2\alpha\bigg)$$
$$=\frac{(x^2yz +xy^2z -x y z^2 +xy -yz- zx ) (x^2 y z + x y^2 z + x y z^2 - x y-yz - zx)}{x y z^2 (x y - 1)^2}\tag6$$
So, if one can get $(x,y,z,\alpha)$ such that
$$\begin{cases}x\not=y
\\\\xy\not=1
\\\\xy=\pm z
\\\\x^2 y z + x y^2 z + x y z^2 - x y-yz - zx\not=0
\\\\x^2yz +xy^2z -x y z^2 +xy -yz- zx\not=0
\\\\\sin\alpha=\dfrac{z^2-xy}{z(xy-1)}
\end{cases}$$
then one can say that the answer is no to your question.
Here, I take $\alpha=\dfrac{\pi}{2}$ to have
$$z^2-xy=zxy-z\iff (z-xy)(z+1)=0$$
Also, taking $z+1\not=0$, it is enough to find $(x,y,z)$ such that
$$\begin{cases}xy=z
\\\\x\not=y
\\\\xy\not=1
\\\\z(xz + zy + z^2 - 1-y -x)\not=0
\\\\z(xz +zy -z^2 +1 -y- x)\not=0
\\\\z\not=-1
\end{cases}$$
Taking $z=2$, one gets
$$\begin{cases}xy=2
\\\\x\not=y
\\\\x + y + 3\not=0
\\\\x +y -3\not=0
\end{cases}$$
So, $(x,y)=(\frac 12,4)$ works.
Added : I realized that one can write $(5)$ as
$$\frac{(x y - z) (x y + z) }{z(x y - 1)^2}\bigg(x+y+z-\frac 1x-\frac 1y-\frac 1z\bigg)=0$$
and $(6)$ as
$$\left(z+\frac{1}{z}\right)\sin\alpha-\bigg(\frac{x}{y}+\frac{y}{x}+\cos^2\alpha\bigg)$$
$$=\frac{xy}{(x y - 1)^2}\bigg(x +y -z -\frac 1x- \frac 1y+\frac 1z\bigg)\bigg(x+y+z-\frac 1x-\frac 1y-\frac 1z\bigg)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $2\tan^{-1}\left(\sqrt{\frac{a}{b}}\tan\frac{x}{2}\right)=\sin^{-1}\frac{2\sqrt{ab}\sin x}{\left(b+a\right)+\left(b-a\right)\cos x}$ $$
2\tan ^{-1}\left(\sqrt{\frac{a}{b}}\tan \frac{x}{2}\right)=\sin ^{-1}\frac{2\sqrt{ab}\sin x}{\left(b+a\right)+\left(b-a\right)\cos x}
$$
I know within inverse of trigonometric function we have the value.
How Do I solve this.
my approach for this solution was:
1st:
$$
\sin ^{-1}\dfrac{2\sqrt{\dfrac{a}{b}}\tan \dfrac{x}{2}}{1+\left( \sqrt{\dfrac{a}{b}}\tan \dfrac{x}{2}\right) ^{2}}
$$
$$
\sin ^{-1}\dfrac{2\dfrac{\sqrt{a}}{\sqrt{b}}\tan \dfrac{x}{2}}{1+\dfrac{a}{b}\tan ^{2}\dfrac{n}{2}}
$$
$$
\sin \dfrac{2\sqrt{a}\tan \dfrac{x}{2}}{\dfrac{\sqrt{b}\left( b+atan ^{2}\dfrac{x}{2}\right) }{b}}
$$
$$
\sin ^{-1}\dfrac{2\sqrt{a}\tan \dfrac{x}{2}\sqrt{b}}{b+atan ^{2}\dfrac{x}{2}}
$$
$$
\sin ^{-1}\dfrac{2\sqrt{ab}\tan \dfrac{x}{2}}{b+atan ^{2}\dfrac{x}{2}}
$$
$$
\sin ^{-1}\dfrac{2\sqrt{ab}\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}}{b+a\dfrac{\sin ^{2}\dfrac{x}{2}}{\cos ^{2}\dfrac{x}{2}}}
$$
$$
=\sin ^{-1}\dfrac{2\sqrt{ab}\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{\cos ^{2}\dfrac{x}{2}b+asin ^{2}\dfrac{x}{2}}
$$
After this I have tried to approach using
$$
\sin ^{2}\theta +\cos ^{2}\theta =1
$$
formula but could not reach the Right Hand Side.
| Let $$\tan A=\sqrt{\frac{a}{b}}\tan\frac{x}{2}$$ and
$$\sin B=\frac{2\sqrt{ab}\sin x}{(b+a)+(b-a)\cos x}$$
You wish to show $2A=B$. Calculate
$$\sin 2A=2\sin A\cos A=2\frac{\tan A}{\sec^2 A}=2\frac{\tan A}{1+\tan^2 A}$$
so substtitute your value for $\tan A$ into this expression and simplify using $$\tan^2 \frac{x}{2}=\frac{1-\cos x}{1+\cos x}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is $f(x)$ irreducible over $\mathbb{Z}_3$? I am looking for irreducible polynomials to construct finite fields. In this case I need a degree $4$ polynomial over $\mathbb{Z}_3$. Is $f(x)=x^4+x+2$ irreducible over $\mathbb{Z}_3$? I don't have much practice proving the irreducibility of a polynomial. So I am not sure what I have done. It's okay?
First, I have that $f(x)$ has no roots in $\mathbb{Z}_3$. If $f(x)$ was reducible, noting that there are no $x^3$ term in $f(x)$. So the factorization could be $(x^2+2x+c)(x^2+x+d)$, where $c,d \in \mathbb{Z}_3$ ($2 \equiv -1 \pmod 3$). So, $$x^4+x+2=x^4+(c+d+2)x^2+(c+2d)x+cd$$
Then $c+d+2=0$, $c+2d=1$ and $cd=2$. But of the first two equalities I have to $d=0$, wich is impossible since $cd\neq 0$. Thus $f(x)$ is irreducible over $\mathbb{Z}_3$.
| Assume that we have a factorization
$$x^4 + x + e= (x^2 + a x + b)(x^2 + c x + d)$$
Looking at the coefficient of $x^3$ we get $c=-a$. Now the coefficient of $x^2$ on RHS is $-a^2 + b + d$ so $d = a^2- b$. Now we get
$$(x^2 + a x + b)(x^2 - a x + (a^2 - b))= x^4 + (a^3 - 2 a b)x + a^2 b - b^2 $$
Now we get the system
$$a^3 - 2 a b = 1 \\
a^2 b - b^2 = e $$
Express $b$ from the last equation (assume $2$ invertible) and substitute in the second one. We get
$$a^6 - 4 e a^2 -1 = 0$$
Now in our case $e=2$, so the equation is $a^6 - 8 a^2 - 1 = 0$. Moreover, over $\mathbb{F}_3$ we have $x^3 = x$ for all $x$, so $a^6 = a^2$. We get the equation $a^2 = -1$, not possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282589",
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"question_score": "4",
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For positive real numbers $a, b, c$, maximise $(a-3)(b-2)(c-1)$ given $a+b+c=18$. For positive real numbers $a, b, c$, if $a + b + c = 18$, then max value of $(a-3)(b-2)(c-1) =$?
My approach:-
$a-3=x$
$b-2=y$
$c-1=z$
I get $x+y+z=12$ , and I need to find $\max(xyz)$.
So I applied AM-GM inequality here,
$\dfrac{x+y+z}{3} \geq \sqrt[3]{xyz}$,
which gives me the answer as $64$.
But the answer given in the book is $102$, and also it states that AM-GM cannot be applied here. I don't understand why can't we apply it here, the only condition AM-GM requires is that we have positive real numbers, right, then what is the issue ?
| The maximum value of $102$ can be proven if we relax the constraints slightly, such that $a,b,c$ are all non-negative reals, instead of positive reals.
Case $1$: All $3$ terms are greater than or equal to zero.
By AM-GM,
\begin{align}
\sqrt[3]{(a-3)(b-2)(c-1)} & \leq \dfrac{a-3+b-2+c-1}{3} \\
& =4.
\end{align}
This gives us $(a-3)(b-2)(c-1) \leq 64.$
Case $2$:
Exactly $1$ term is less than or equal to zero. Clearly, $(a-3)(b-2)(c-1) \leq 0.$
Case $3$:
All $3$ terms are less than or equal to zero, which is impossible since this would imply $a+b+c \leq 6$.
Case $4$:
Exactly $2$ terms are less than or equal to zero. There are $3$ further sub-cases to consider:
Sub-case $1$:
$a-3 \leq 0, \ c-1 \leq 0.$
Then,
\begin{align}
(a-3)(b-2)(c-1) & =(3-a)(b-2)(1-c) \\
\end{align}
It is obvious that $3-a \leq 3, \ b-2 \leq 16, \ 1-c \leq 1 \Rightarrow (3-a)(b-2)(1-c) \leq 48. $
Sub-case $2$:
$b-2 \leq 0, \ c-1 \leq 0.$
Then,
\begin{align}
(a-3)(b-2)(c-1) & =(a-3)(2-b)(1-c) \\
\end{align}
We have: $a-3 \leq 15, \ 2-b \leq 2, \ 1-c \leq 1 \Rightarrow (a-3)(2-b)(1-c) \leq 30. $
Sub-case $3$:
$a-3 \leq 0, \ b-2 \leq 0.$
Then,
\begin{align}
(a-3)(b-2)(c-1) & =(3-a)(2-b)(c-1) \\
\end{align}
We have: $3-a \leq 3, \ 2-b \leq 2, \ c-1 \leq 17 \Rightarrow (3-a)(2-b)(c-1) \leq 102. $ In particular, this maximum value is achieved when $a=0,\ b=0,\ c=18.$
| {
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Is there a natural number $n$ for which $\sqrt[n]{22-10\sqrt7}=1-\sqrt7$ Is there a natural number $n$ for which $$\sqrt[n]{22-10\sqrt7}=1-\sqrt7$$
My idea was to try to express $22-10\sqrt7$ as something to the power of $2$, but it didn't work $$22-10\sqrt7=22-2\times5\times\sqrt7$$ Since $5^2=25, \sqrt7^2=7$ and $25+7\ne22$. What else can we try?
| Alternate method : Binomial theorem
We have,
$$\begin{align}&\sqrt[n]{22-10\sqrt7}=1-\sqrt7\\
\implies &\left(1-\sqrt 7\right)^n=22-10\sqrt 7\end{align}$$
$22-10\sqrt 7<0$ tells us that, $n$ must be an odd integer. This implies $n≥3$ and $n=2k+1,\thinspace k\in\mathbb Z^{+}$.
Let, $N(n)$ be the integer part of $\left(1-\sqrt 7\right)^{n}$, such that if $\left(1-\sqrt 7\right)^n=a-\sqrt b$, then $N(n)=a$.
Since $\sqrt 7$ is irrational and $\forall k\in\mathbb Z^{+}$ we have,
$$\begin{align}N(2k+1)&≥(2k+1)(\sqrt7)^{2k}+1\\
&=7^k(2k+1)+1\\
&≥22.\end{align}$$
Thus, we conclude that if $N(2k+1)=22$, then $k=1$.
This means, if equality $\left(1-\sqrt 7\right)^n=22-10\sqrt 7 $ is possible, then $n=3$.
Finally, we see that $n=3$ gives us the required equality:
$$\left(1-\sqrt 7\right)^3=22-10\sqrt 7.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the value of $\sqrt[3]{9+2\sqrt5+3\sqrt{15}}$?
$$\sqrt[3]{9+2\sqrt5+3\sqrt{15}}=?$$
Here is my work:
I considered the value is equal to $a+b$. Hence
$$9+2\sqrt5+3\sqrt{15}=(a+b)^3=a^3+b^3+3a^2b+3ab^2$$
$$2\sqrt5+3(3+\sqrt{15})=a^3+b^3+3(a^2b+ab^2)$$
We have
$$ \begin{cases} {a^2b+ab^2=3+\sqrt{15}} \\ {a^3+b^3=2\sqrt5} \end{cases} $$
But I cannot find such $a$ and $b$.
| Remark: Write $$9+2\sqrt{5}+3\sqrt{15}=(a+b\sqrt{3}+c\sqrt{5})^3.$$
Then the equations in rational $a,b,c$ yield immediately a contradiction.
Of course, there might be more complicated expressions. But for such questions, usually the answer is simple.
Edit: The equations are
\begin{align*}
0 & =6abc - 3\\
0 & = 3a^2c + 9b^2c + 5c^3 - 2\\
0 & =3a^2b + 3b^3 + 15bc^2\\
0 & =a^3 + 9ab^2 + 15ac^2 - 9.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many negative real roots does the equation $x^3-x^2-3x-9=0$ have? How many negative real roots does the equation $x^3-x^2-3x-9=0$ have ?
My approach :-
f(x)= $x^3-x^2-3x-9$
Using rules of signs, there is 1 sign change , so there can be at most 1 positive real root
f(-x)= $-x^3-x^2+3x-9$
2 sign changes here, indicating at most 2 negative real roots
I end up with following 2 possibilities:-
1)1 positive, 2 negative real roots
2)1 positive, 2 imaginary roots
how to progress further ?
| $x^3 - x^2 - 3x - 9=0$
$3x = x^3 - x^2 - 9$
When $x<0$ every term on the right hand side is less than 0. If we are gong to find equality, $3x$ must be very negative.
When $-3< x < 0, |3x| > |-9|$
When $x< -3, |3x| < |x^2|$ and $|3x| < |x^3|$
There is no way that $3x$ can be sufficiently negative.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a(n)=\sum_{r=1}^{n}(-1)^{r-1}\frac{1}{r}$. Prove that $a(2n) \neq 1$ for any value $n$. Let $$a(n)=\sum_{r=1}^{n}(-1)^{r-1}\frac{1}{r}.$$
I experienced this function while doing a problem, I could do the problem, but I got stuck at a point where I had to prove that $a(2n)<1$ for all $n$. I proved that $a(2n) \le 1$ for all $n$ so in order to prove what the question demands, I need to prove that $a(2n) \neq 1$ for any value $n$. Nothing striked my mind as of now how to prove it.
So can someone help me proving that $a(2n) \neq 1$ for any value $n$?
| Here's a solution that leverages the identity
$$\sum_{r=1}^{2n}\frac{(-1)^{r-1}}{r}=\sum_{r=1}^{2n}\frac{1}{r}-\sum_{r=1}^{n}\frac{1}{r}$$
I'd suggest you try and prove this yourself. If that's not an option, I've provided a proof below:
Proof:\begin{align} \sum_{r=1}^{2n}\frac{1}{r}-\sum_{r=1}^{n}\frac{1}{r} &= \left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n-1}+\frac{1}{2n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}+\frac{1}{n}\right)\\&= 1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{1}{2}\right)+\frac{1}{5}+\left(\frac{1}{6}-\frac{1}{3}\right)+\cdots+\frac{1}{2n-1}+\left(\frac{1}{2n}-\frac{1}{n}\right)\\&= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}\\&=\sum_{r=1}^{2n}\frac{(-1)^{r-1}}{r}\end{align}
Armed with this identity, we can write
\begin{align}
\sum_{r=1}^{2n}\frac{(-1)^{r-1}}{r} &= \sum_{r=1}^{2n}\frac{1}{r}-\sum_{r=1}^{n}\frac{1}{r}\\
&= \sum_{r=n+1}^{2n}\frac{1}{r}\\
&\leq \sum_{r=n+1}^{2n}\frac{1}{n+1}\\
&= \frac{2n-(n+1)+1}{n+1}\\
&= \frac{n}{n+1}
\end{align}
so $\sum_{r=1}^{2n}\frac{(-1)^{r-1}}{r}\leq\frac{n}{n+1}$. Since $\frac{n}{n+1}<\frac{n+1}{n+1}=1$ for every $n\geq 1$, this immediately gives the desired result.
$$a(2n)=\sum_{r=1}^{2n}\frac{(-1)^{r-1}}{r}<1$$
| {
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"url": "https://math.stackexchange.com/questions/4297176",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Taylor series of $f(x) = \cos x$ centered at $\frac{\pi}{4}$ I am asked to find the Taylor Series that represent the function $f(x) = \cos x$ centered at $\frac{\pi}{4}$.
My process
Finding the first few derivatives and establishing a pattern. Given that sine and cosine functions go back and forth, there needs to be two formulas (sums) to produce all the terms. The even and odd ones are:
$$f^{(2k)}(x) = (-1)^k \cdot \cos(x)$$
$$f^{(2k+1)}(x) = (-1)^{k+1} \cdot \sin(x)$$
Therefore,
$$f^{(2k)} \left( \frac{\pi}{4} \right) = (-1)^k \cdot \frac{1}{\sqrt{2}}$$
$$f^{(2k+1)} \left( \frac{\pi}{4} \right) = (-1)^{k+1} \cdot \frac{1}{\sqrt{2}}$$
The sums will be
$$\sum_{k = 0}^{\infty} (-1)^k \cdot \frac{1}{\sqrt{2}} \frac{\left( x - \frac{\pi}{4}\right)^k}{k!}$$
$$\sum_{k = 0}^{\infty} (-1)^{k+1} \cdot \frac{1}{\sqrt{2}} \frac{\left( x - \frac{\pi}{4}\right)^k}{k!}$$
Now, my question is the following: how do I combine these? Do I need to look at both and try to establish a pattern, which would lead me to the sum shown on the mark scheme? Or is there some algebraic manipulation that I can do in order to get the answer?
Markscheme's answer
$$\frac{\sqrt{2}}{2} \sum_{k = 0}^{\infty} \left( -1 \right)^{\frac{k(k+1)}{2}} \cdot \left( x - \frac{\pi}{4}\right)^k \cdot \frac{1}{k!}$$
Thank you.
Edit
I believe I've found a solution. Interesting to see that, when you combine two sums, the terms show up two by two.
| You basically want the periodic sequence
$$
a_0=1,a_1=1,a_2=-1,a_3=-1,a_4=1,a_5=1,a_6=-1,\dotsc
$$
which obeys the recursion $a_{k+4}=a_k$, with the given initial terms. The characteristic equation is $t^4-1=0$, so the roots are $1,i,-1,-i$ and the general solution is therefore
$$
a_k=\alpha+\beta(-1)^k+\gamma i^k+\delta(-i)^k
$$
and we need
$$
\begin{cases}
\alpha+\beta+\gamma+\delta=1 \\
\alpha-\beta+i\gamma-i\delta=1 \\
\alpha+\beta-\gamma-\delta=-1 \\
\alpha-\beta-i\gamma+i\delta=-1
\end{cases}
$$
from which $\alpha+\beta=0$ and $\alpha-\beta=0$. You can compute $\gamma$ and $\delta$ and using Euler's and De Moivre's formulas, we end up with the essentially obvious
$$
a_k=\cos\dfrac{k\pi}{2}+\sin\dfrac{k\pi}{2}=2\cos\frac{\pi}{4}\cos\Bigl(\frac{k\pi}{2}-\frac{\pi}{4}\Bigr)=\sqrt{2}\cos\frac{2k\pi-\pi}{4}
$$
the last steps using the sum-to-product formulas. Thus you can rewrite your series in the compact form
$$
\cos x=\sum_{k=0}^\infty \Bigl(\cos\frac{2k\pi-\pi}{4}\Bigr)\frac{1}{k!}\Bigl(x-\frac{\pi}{4}\Bigr)^k
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Looking for a way to compute the discrete version of a continuous random distribution I just learn the way to compute the cumulative probability distribution (CDF) of a distribution in terms of some known results. For example, a random variable $Y$ is written in terms of a uniform random variable $U(0,1)$ as
$$
Y \sim \frac{5}{U + 1}
$$
the CDF of $Y$ could be found in the following way
$$
\begin{align*}
\text{CDF}(y) & = F(y) = \mathbb{P}(Y\leq y) = \mathbb{P}\left(\frac{5}{U+1}\leq y\right)\\
& = \mathbb{P}\left(U\geq \frac{5}{y}-1\right) = 1- \mathbb{P}(U<\frac{5}{y}-1)\\
& = 2 - \frac{5}{y}
\end{align*}
$$
so the probability distribution function (PDF) for $Y$ will be
$$
P(y) = \dfrac{dF(y)}{dy} = \frac{5}{y^2}
$$
Compute the expectation value for $y$
$$
\text{E}[y] = \int y\frac{5}{y^2}dy = 3.46574
$$
In the text, it is said that the uniform distribution is continuous so I apply the integral to compute the expectation. I read something online about discrete distribution. If I change the random variable as
$$
Y \sim \left\lfloor \frac{5}{U+1} \right\rfloor
$$
how do we derive the PDF and CDF for the discrete version of $Y$? I am trying a simulation to run 1M random numbers for u as follows:
s = 0
for i=1 to 1000000
s += floor(5/(rand(0, 1) + 1)
end
print s/1000000
this gives me an average of 2.9172 instead of 3.46574 I think the difference is originated from a great number of states that are suppressed by the floor operation. I wonder if there is any analytical way instead of a simulation to compute the result. I try to replace y in PDF with the floor(5/(u+1)) and replace the integral with a summation
$$
\sum \left\lfloor\frac{5}{y+1}\right\rfloor \times \frac{5}{\left\lfloor\frac{5}{y+1}\right\rfloor^2} = \sum \frac{5}{\left\lfloor\frac{5}{y+1}\right\rfloor}
$$
but this gives me 1.8418, very far from 2.9172.
| Since you calculated the original expected value correctly, you already know this, but just to be thorough for other readers, the CDF in the question is not properly specified, as $P(U < \frac{5}{y} - 1)$ is only $\frac{5}{y} - 1$ if this quantity is between 0 and 1. Otherwise, the probability is either 0 or 1. The appropriate range for $y$ where your logic holds is
$$ 0 \le \frac{5}{y} - 1 \le 1 \Longrightarrow y \le 5 \le 2y \Longrightarrow \frac{5}{2} \le y \le 5;$$
if $y < \frac{5}{2}$, we have $P(U < \frac{5}{y} - 1) = 1$ and if $y > 5$, we have $P(U < \frac{5}{y} - 1) = 0$.
Thus, we have
$$
\begin{align*}
CDF(y) &= P(Y \le y) \\
&= P(\frac{5}{U + 1} \le y) \\
&= P(U \ge \frac{5}{y} - 1) \\
&= 1 - P(U < \frac{5}{y} - 1) \\
&= \begin{cases} 0, & y < \frac{5}{2}, \\ 2 - \frac{5}{y}, & \frac{5}{2} \le y \le 5, \\ 1, & y > 5. \end{cases}
\end{align*}
$$
The PDF follows as
$$PDF(y) = \frac{d (CDF(y))}{dy} = \begin{cases} \frac{5}{y^2}, & \frac{5}{2} \le y \le 5, \\ 0, & o/w. \end{cases}$$
And the expected value follows as
$$E[Y] = \int_{-\infty}^{\infty} y \cdot PDF(y) dy = \int_{\frac{5}{2}}^{5} y \cdot \frac{5}{y^2} dy = 5\ln(5) - 5 \ln(5/2) \approx 3.466,$$
as you calculated.
For the discrete case, we can just do casework on where U falls:
*
*If $U = 0$, we have $Y = 5$.
*If $0 < U \le \frac{1}{4}$, we have $Y = 4$.
*If $\frac{1}{4} < U \le \frac{2}{3}$, we have $Y = 3$.
*If $\frac{2}{3} < U \le 1$, we have $Y = 2$.
The probability mass function for the discrete Y is then
$$P(Y = y) = \begin{cases} \frac{1}{3}, & y = 2, \\ \frac{5}{12}, & y = 3, \\ \frac{1}{4}, & y = 4. \end{cases} $$
This yields an expected value of
$$E[Y] = \sum y \cdot P(Y = y) = 2 \cdot \frac{1}{3} + 3 \cdot \frac{5}{12} + 4 \cdot \frac{1}{4} = \boxed{\frac{35}{12}},$$
which is around 2.917.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square. Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square.
I have solved this problem and will post the solution as soon as possible. Hope everyone can check my solution! Thanks very much !
| $(xy+x+2)(xy+1) = a^2$
$\Rightarrow (2xy+x+3)^2 = (2a)^2 +(x+1)^2 $
Because $2 | 2a$ , there exist $3$ positive integers $m , n,d$ satisfying :
$(m,n) = 1$ and : $2a = 2dmn ; x+1 = d(m^2-n^2) ; 2xy+x+3 = d(m^2+n^2)$
$\Rightarrow xy+1 = dn^2 $
$\Rightarrow y = \frac{dn^2-1}{d(m^2-n^2)-1} $
$\Rightarrow \frac {y-1}{d} = \frac {2n^2-m^2}{d(m^2-n^2)-1} $
It is easy to see that $d|(y-1)$ , so $\frac {2n^2-m^2}{d(m^2-n^2)-1} \in \mathbb N^+ $
Let $\frac {2n^2-m^2}{d(m^2-n^2)-1} = k $
Let $ m = \frac{e+f}{2} , n =\frac{e-f}{2}$.
$\Rightarrow k = \frac{e^2+f^2-6ef}{4def-4}$
$\Rightarrow e^2 -ef(6+4kd)+f^2+4k = 0 $
Let $e_1$ is the smallest solution of the above equation .
We have $e_1 + e_2 = (6+4kd)f$ and $e_1 e_2 = f^2+4k >0 $
So $e_2$ is also a solution of the above equation, so $e_1 \le e_2$
$\Rightarrow f^2 \le (e_1-1)(e_2-1) $
$\Rightarrow f(6+4kd) \le 4k+1 $ ( a contradiction )
So there is no $(x,y)$ satisfying the problem.
I just finished it, so no one has checked it yet! Hope you guys can check if I'm wrong!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Extracting coefficients I stuck at the following problem:
Let
\begin{equation}
f(z) = \frac{1 - \sqrt{1 - 4z}}{2}.
\end{equation}
Find $[z^n]$ of $f(z)^r$ for $r \in \mathbb{N}$. Where $[z^n]f(z)$ is the $n$-th coefficient of the power series $f(z) = \sum_{n \geq 0}{a_n z^n}$ therefore $[z^n]f(z) = a_n$.
So far I got
\begin{equation}
\sqrt{1 - 4z} = \sum_{n \geq 0}\binom{1/2}{n}(-4 x)^n.
\end{equation}
and therefore
\begin{align*}
f(z) = \frac{1 - \sqrt{1 - 4z}}{2} &= \frac{\sum_{n \geq 1}\binom{1/2}{n}(-4)^n z^n}{2} \\
&= \sum_{n \geq 1}\binom{1/2}{n}(-1)^n 2^{n} z^n \\
&= \sum_{n \geq 1}\binom{1/2}{n}(-2)^n z^n \\
&= \sum_{n \geq 0}a_n z^n
\end{align*}
with coefficients $a_0 = 0$ and $a_n = \binom{1/2}{n}(-2)^n$.
I wanted to use cauchys integral formula for
$g(z) = f(z)^r$ to extract the $n$ coefficient
but then I get
\begin{align}
\left( \frac{1 - \sqrt{1 - 4z}}{2}\right)^r &= \frac{1}{2^r} \left(1 - \sqrt{ 1 - 4z} \right)^r \\
&= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m \sqrt{1 - 4z}^m \\
&= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m (1 - 4z)^{m/2} \\
&= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m \sum_{k \geq 0}\binom{m/2}{k}(-1)^k 4^k z^k \\
&= \frac{1}{2^r}\sum^{r}_{m=0} \sum_{k \geq 0}\binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k.
\end{align}
Therefore I should have
\begin{align*}
[z^n] (f(z))^r &= [z^n] \sum^{r}_{m=0} \sum_{k \geq 0}\frac{1}{2^r} \binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k \\
&= \sum^{r}_{m=0}[z^n] \sum_{k \geq 0}{\frac{1}{2^r} \binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k} \\
&= \sum^{r}_{m=0}\frac{1}{2^r}\binom{r}{m} \binom{m/2}{n} (-1)^{m + n} 4^n \\
&= \sum^{r}_{m=0}\frac{1}{2^r}\binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n \\
&= \frac{1}{2^r} \sum^{r}_{m=0} \binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n \\
&= \frac{1}{2^r} \sum^{r}_{m=2 n} \binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n
\end{align*}
For $r \geq 2n$ otherwise the coefficient vanishes.
I would be thankful if anyone can give me a hint.
| At page 200 of the the renowned book
"Concrete Mathematics: a foundation for computer science" R. L. Graham - D.E. Knuth - O. Patshnik
the authors analyze the properties of what they call Generalized Binomial Series
(but it seems not to be a widely accepted denomination) defined as
$$
\eqalign{
& {\cal B}_{\,t} (z) = \sum\limits_{0\, \le \,k}
{{{\left( {t\,k} \right)^{\,\underline {\,k - 1\,} } } \over {k!}}z^{\,k} }
= \sum\limits_{0\, \le \,k} {{1 \over {\left( {tk - k + 1} \right)}}
\left( \matrix{ t\,k \cr k \cr} \right)z^{\,k} } = \cr
& = 1 + z\sum\limits_{0\, \le \,k} {{1 \over {k + 1}}
\left( \matrix{ t\,k + t \cr k \cr} \right)z^{\,k} } \cr}
$$
These satisfy the interesting functional identity
$$
{\cal B}_{\,t} (z)^{\,1 - t} - {\cal B}_{\,t} (z)^{\, - t} = z\quad
\Leftrightarrow \quad {\cal B}_{\,t} (z) - z{\cal B}_{\,t} (z)^{\,t} = 1
$$
and their power have a simple expression
$$
\eqalign{
& {\cal B}_{\,t} (z)^{\,r} = \sum\limits_{0\, \le \,k} {{r \over {t\,k + r}}
\left( \matrix{ t\,k + r \cr k \cr} \right)z^{\,k} }
= \sum\limits_{0\, \le \,k} {{{r\left( {t\,k + r - 1} \right)^{\,\underline {\,k - 1\,} } }
\over {k!}}z^{\,k} } = \cr
& = 1 + z\sum\limits_{0\, \le \,k} {{r \over {k + 1}}\left( \matrix{ \,t\,k + t + r - 1 \cr
k \cr} \right)z^{\,k} } = 1 + r\,z\,\sum\limits_{0\, \le \,k}
{{{\left( {\,t\,\left( {k + 1} \right) + r - 1} \right)^{\;\underline {\,k\,} } }
\over {\left( {k + 1} \right)!}}z^{\,k} } \cr}
$$
valid for all real $r$ and $t$ ( with some care in interpreting the case $tk+r=0$).
Putting $t=2$
$$
{\cal B}_{\,2} (z) - z{\cal B}_{\,2} (z)^{\,2} = 1\quad
\Leftrightarrow \quad {\cal B}_{\,2} (z) = {{1 - \sqrt {1 - 4z} } \over {2z}}
$$
where we have taken only the solution that is limited for $z \to 0$.
Then the coefficients you are looking for are
$$
\eqalign{
& \left( {{{1 - \sqrt {1 - 4z} } \over 2}} \right)^{\,r} =
z^{\,r} {\cal B}_{\,2} (z)^{\,r} = \cr
& = z^{\,r} \sum\limits_{0\, \le \,k} {{r \over {2\,k + r}}
\left( \matrix{ 2\,k + r \cr k \cr} \right)z^{\,k} } \cr}
$$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the maximum of $\sin(A)+\sin(B)+\sin(C)$ for $\triangle ABC$. (Without Jensen Inequality)
Find the maximum of $\sin(A)+\sin(B)+\sin(C)$ for $\triangle ABC$. (Without Jensen Inequality)
Proof of Jensen Inequality:
\begin{align}
&\text{let } f(x)=\sin x. \\
\ \\
\Rightarrow & f(A)+f(B)+f(C) \\
& =3 \bigg( \frac 1 3 f(A) + \frac 1 3 f(B)+ \frac 1 3 f(C) \bigg) \\
& \leq 3 \Bigg( f \bigg( \frac 1 3 A + \frac 1 3 B + \frac 1 3 C \bigg) \Bigg) \\
& = 3\Bigg(f\bigg( \frac {A+B+C} 3 \bigg) \Bigg) \\
& = 3\big(f(60)\big) & (\because A+B+C=180) \\
&=3\sin 60 = 3 \cdot \frac {\sqrt{3}} 2 = \frac {3\sqrt{3}}{2}.
\end{align}
I just wondered if there is another precalculus solution to this. Is there another solution to this?
| Note that
$$\sin A+ \sin B=2\sin \frac{A+B}2\cos\frac{{A-B}}2$$
Thus for any fixed $C$ the first 2 terms obtains a maximum for $A=B$. Similarly if we fix $A$ we get that there is a maximum when $B=C$. It follows that the maximum is obtained when:
$$
A=B=C=\frac\pi3
$$
| {
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
Polynomial inequality: show $p(x) = \sum_{n=0}^{2k} x^n - \frac{1}{2k+1} \sum_{n=0}^{2k} (-x)^n\ge 0$ Consider the following polynomial in $x$:
$$p(x) = \sum_{n=0}^{2k} x^n - \frac{1}{2k+1} \sum_{n=0}^{2k} (-x)^n.$$
I want to show that $p(x)\geq 0$ for real $x$. It is trivial to show that $p(-1) = 0$, and by graphing $p(x)$ for some $k$ we can see that $x=-1$ is in fact the global minimum of this polynomial, presumably for any $k\in \mathbb{N}^+$.
How can I show that this is in fact true?
Note that $$p(x)=f(x) - \frac{1}{2k+1}f(-x),$$ with $$f(x)=1 + x + \dots + x^{2k}=\frac{x^{2k+1} - 1}{x-1}.$$
| $$p(x) = \frac{2}{ 2k+ 1} ( k + (k+1) x + k x^2 + (k+1)x^3 + \ldots + (k+1) x^{2k-1} + k x^{2k} ) $$
By considering signs (IE Replace $x$ with $-x$), we WTS that for $ x \geq 0$,
$$ k( 1 + x^2 + \ldots + x^{2k}) \geq (k+1)( x + x^3 + \ldots + x^{2k-1}) .$$
Since the number of terms on each side are equal to $k(k+1)$, and the sum of the degrees are the same, it suggests that we can proceed via smoothing out these terms.
This can by done, say by summing up the following inequalities (overall relevant values of $i$):
*
*$ \frac{k}{2} x^{2i} + \frac{k}{2} x^{2i+2} \geq k x^{2i+1} $
*
*This gives us $ \frac{k}{2}1 + kx^2 + \ldots + kx^{2k-2} + \frac{k}{2} x^{2k} \geq k( x + x^3 + \ldots + x^{2k-1}) .$
*$ \frac{1}{2} x^0 + \frac{1}{2} x^{2k} \geq \frac{1}{2}x^{2i-1 } + \frac{1}{2}x^ { 2k-2i+1} $
*
*This gives us $ \frac{k}{2} 1 + \frac{k}{2} x^{2k} \geq ( x + x^3 + \ldots + x^{2k-1})$.
Notes:
*
*This can be taught of as essentially Muirhead, since it is clear that $ [ k, 0, k, 0, \ldots, 0, k ] $ majorities $ [ 0, k+1, 0, k+2, \ldots, k+1, 0 ] $.
*Of course, there are other ways that one can do the AM-GM / Jensens.
*Equality holds iff $ x = 1$ (, or $ k = 0$).
| {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 2
} |
proving $\binom{n-1}{k} - \binom{n-1}{k-2} = \binom{n}{k} - \binom{n}{k-1} $ The identity is
$$\binom{n-1}{k} - \binom{n-1}{k-2} = \binom{n}{k} - \binom{n}{k-1}. $$
I wrote a matlab code and numerically checked it. It is right.
It is also not difficult to prove it by brute force. But can anyone come up with a simple proof using combinatorics? Many such identities are proven in this elegant way.
ps. It is an identity in the 'Group theory' book by Wigner.
| Hint: You might also find appealing that algebraically this binomial identity is encoded by the relation
\begin{align*}
1-z^2=(1+z)(1-z)
\end{align*}
Denoting with $[z^k]$ the coefficient of $z^k$ of a series we can write
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\tag{1}
\end{align*}
Using (1) and noting that $[z^p]z^qA(z)=[z^{p-q}]A(z)$ we obtain
\begin{align*}
\color{blue}{\binom{n-1}{k}}\color{blue}{ - \binom{n-1}{k-2}}
&=[z^k](1+z)^{n-1}-[z^{k-2}](1+z)^{n-1}\\
&=[z^k]\left(1-z^2\right)(1+z)^{n-1}\\
&=[z^k](1-z)(1+z)^n\\
&=[z^k](1+z)^n-[z^{k-1}](1+z)^{n}\\
&\,\,\color{blue}{=\binom{n}{k}-\binom{n}{k-1}}
\end{align*}
| {
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How do 3 fractions having respectively $(a-x)^2$ , $(a+x)^2$ and $(a-x)$ as denominators add up to a fraction having $(a^2 - x^2)^2$ as denominator? Source : Olry Terquem, Exercices de mathématiques élémentaires , $1842$
( At Archive.org https://archive.org/details/exercicesdemath00terqgoog/page/n85/mode/1up)
I'm working on problem $35$ below, where appears also the solution.
My question is simply : is this solution correct, as far as the denominator is concerned?
When I do $(a-x)^2(a+x)^2(a-x)$, I get
$ (a-x)(a-x)(a+x)(a+x)(a-x) $
$= (a-x)(a+x)(a-x)(a+x)(a-x)$
$= [(a-x)(a+x)] [(a-x)(a+x)](a-x)$
$= (a^2- x^2) ^2 (a-x)$.
Or, am I wrong?
| $\require{cancel}$
We do not need the product of all denominators if one is a factor of another. For instance
$$\quad \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{8}\\
=\dfrac{8}{24}+\dfrac{6}{24}+\dfrac{3}{24}$$
where the denominator is
$\space LCM(3,4,8)=24$ and not $3\times4\times8=96$
$$LCM \big((a-x)^2 , (a+x)^2, (a-x)\big)\\
= (a-x)^2(a+x)^2\cancel{(a-x)}\\
=(a^2-x^2)^2$$
| {
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"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculate the sum of the series $\sum_{n=1}^{\infty} \frac{n+12}{n^3+5n^2+6n}$ They tell me to find the sum of the series
$$\sum a_n :=\sum_{n=1}^{\infty}\frac{n+12}{n^3+5n^2+6n}$$
Since $\sum a_n$ is absolutely convergent, hence we can manipulate it the same way we would do with finite sums. I've tried splitting the general term and I get
$$\frac{n+12}{n(n+2)(n+3)}=\frac{A}{n}+\frac{B}{n+2}+\frac{C}{n+3}=\frac{2}{n}+\frac{-5}{n+2}+\frac{3}{n+3}$$
and so
$$\sum a_n =\sum \frac{2}{n}-\frac{5}{n+2}+\frac{3}{n+3}$$
Now, if I was to split the series in the sum of three different series I would get three different divergent series and so, obviously $\sum a_n$ wouldn't converge.
I also suspect about being a telescopic series althought the numerators of each fraction makes it difficult to find the cancellation terms. I also know that I can rearrange the terms in my series, althought I cannot see how would this solve the problem.
If anyone could give me a hint I would really appreciate it.
| As you said, for every $N \in \mathbb{N}^*$, one has
\begin{align*}\sum_{n=1}^N a_n &=\sum_{n=1}^N \frac{2}{n}-\frac{5}{n+2}+\frac{3}{n+3}\\
&=2 \sum_{n=1}^N \frac{1}{n} - 5 \sum_{n=3}^{N+2} \frac{1}{n} + 3 \sum_{n=4}^{N+3} \frac{1}{n}\\
&=2 \left(1 + \frac{1}{2} + \frac{1}{3} + \sum_{n=4}^N \frac{1}{n} \right) - 5 \left( \frac{1}{3} + \sum_{n=4}^N \frac{1}{n} + \frac{1}{N+1} + \frac{1}{N+2} \right)\\
&\quad + 3 \left( \sum_{n=4}^N \frac{1}{n} + \frac{1}{N+1} + \frac{1}{N+2} + \frac{1}{N+3} \right)\\
&= 2 - 5 \left( \frac{1}{N+1} + \frac{1}{N+2} \right) + 3 \left( \frac{1}{N+1} + \frac{1}{N+2} + \frac{1}{N+3} \right)
\end{align*}
Now just let $N$ tend to $+\infty$ to see that $$\boxed{\sum_{n=1}^{\infty} a_n = 2}$$
| {
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Let $x+y+z = 5$ where $x,y,z \in \mathbb{R}$, prove that $x^2+y^2+z^2\ge \frac{5}{3}$ My thinking:
Since $x+y+z = 5$, we can say that $x+y+z \ge 5$.
By basic fact: $x^2,y^2,z^2\ge 0$
If $x+y+z \ge 5$, then $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$
If $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$ and $x^2,y^2,z^2\ge 0$, hence $\frac{x^2+y^2+z^2}{3}\ge \frac{5}{3}$...???
I'm not sure about this, can someone help?
| I would preface by saying that we can prove a stronger inequality, namely $x^2+y^2+z^2 \geq \frac{25}3$.
You can use the Cauchy-Schwarz inequality. For three variables, it states that $(a_1b_1+a_2b_2+a_3b_3)^2 \leq (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)$. Using $a_1=a_2=a_3=1$ and $b_1=x,b_2=y,b_3=z$ gives $25\leq 3(x^2+y^2+z^2)$.
You can also use Lagrange multipliers. We get $(2x,2y,2z)=\lambda(1,1,1)$, so that an extreme value of $x^2+y^2+z^2$ is achieved when $x=y=z$. If $x+y+z=5$, this means $x=y=z=\frac53$, giving $x^2+y^2+z^2 = \frac{25}3$. To get the inequality in the right direction, we notice that if $x=y=0$ and $z=5$ for example, then $x^2+y^2+z^2=25\geq \frac{25}3$.
| {
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"url": "https://math.stackexchange.com/questions/4327738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the Volume of a Solid Revolution around the y axis Having trouble with this question from my OpenStax Calculus Volume 1 Homework, It is question 89 of Chapter 6 about Solid Revolution. I put my math below:
y=4-x, y=x, x=0 Find the volume when the region is rotated around the y-axis.
$$y= 4-x \Rightarrow x = 4 - y \implies R(y) = 4 - y$$
This is rotated around the y-axis should thus follow the general formula (c and d are position on the y axis):
$$V=\pi\int_{c}^{d} R^2(y) \, dy $$
Using this I inserted the relevant values and variables:
$$
V= \pi\int_{0}^{4} (4-y-y)^2 \,dy
= \pi\int_{0}^{4} (4-2y)^2 \,dy
= \pi\int_{0}^{4} (16 - 16y +4y^2) \,dy
= \pi\Big[16y - \frac{16y^2}{2} + \frac{4y^3}{3}\Big] \Bigg|_{0}^{4}
= \pi \ \Big[16(4) - \frac{16(4^2)}{2}+ \frac{4(4^3)}{3} \Big] -\pi\big[0 \big]
= \pi \Big[64 - 128 - \frac{256}{3} \Big]
= \pi \Big[\frac{64}{3} \Big]
= \frac{64\pi}{3}
$$
The Official Answer given in the textbook is $\frac{16\pi}{3} $ but I am not sure how to get there. Some help would be greatly appreciated.
|
It is symmetrical about $y=2$.
For $0 \le y \le 2$, we have $R(y)=y$.
\begin{align}
V &= 2\pi \int_0^2 y^2 \, dy\\
&= 2\pi \left.\frac{y^3}{3}\right|_0^2\\
&= \frac{16\pi}{3}
\end{align}
Note that $R(y)$ should be the radisu when $y$ takes a certain value.
| {
"language": "en",
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Compute explicitly $\sum_{x=1}^\infty\sum_{y=1}^\infty \frac{x^2+y^2}{x+y}(\frac{1}{2})^{x+y}$ It is possible to compute explicitly the following series?
$$\sum_{x=1}^\infty\sum_{y=1}^\infty \frac{x^2+y^2}{x+y}\Big(\frac{1}{2}\Big)^{x+y}$$
@Edit I tried to sum and subtract $2xy$ in the numerator.
In this way I get the following
\begin{align}
\sum_{x=1}^\infty\sum_{y=1}^\infty (x+y)\Big(\frac{1}{2}\Big)^{x+y}-2\sum_x x\Big(\frac{1}{2}\Big)^x\sum_y\frac{y}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
The first series should be easy to be computed. It remain the second one, in particular the following
\begin{align}
\sum_y\frac{y}{x+y}\Big(\frac{1}{2}\Big)^y=\sum_y\Big(\frac{1}{2}\Big)^y-x\sum_y\frac{1}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
Therefore the problem is reduced to computing:
\begin{align}
\sum_{y=1}^\infty \frac{1}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
But I don't know well how to do it.
| Little easier way is to do a substitution (since $n+m$ arrives many times). What I mean is, we can rewrite (where $x \in (0,1)$ let's say to not bother about changing the order of summation (but any $|x| < 1$ will be good due to absolute convergence of double series) $$ \sum_{m =1}^\infty \sum_{n=1}^\infty \frac{n^2 + m^2}{n+m} x^{n+m} = \sum_{m=1}^\infty \sum_{n=1}^\infty (n+m)x^{n+m} - 2\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{nm}{n+m}x^{n+m} $$
Now, let $k=n+m$. If $n,m$ can be any numbers between $1$ and $+\infty$, then $k$ is any number between $2$ and $+\infty$. If $k \ge 2$ is fixed, $n$ can be any number between $1$ and $k-1$, hence $$ \sum_{m=1}^\infty \sum_{n=1}^\infty (n+m)x^{n+m} = \sum_{k=2}^\infty \sum_{n=1}^{k-1} kx^k = \sum_{k=2}^\infty k(k-1)x^k $$
Similarly $$ \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{nm}{n+m}x^{n+m} = \sum_{k=2}^\infty \sum_{n=1}^{k-1} \frac{n(k-n)}{k}x^k = \sum_{k=2}^\infty \frac{x^k}{k} \sum_{n=1}^{k-1} n(k-n) = \frac{1}{6}\sum_{k=2}^\infty (k-1)(k+1)x^k $$
Both sums can be easily calculated via differentiating under sum sign, I will leave it to you
| {
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"url": "https://math.stackexchange.com/questions/4333629",
"timestamp": "2023-03-29T00:00:00",
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Solving a nonlinear autonomous system using polar coordinates I have the nonlinear system:
\begin{equation}
\begin{array}
fx'=-y-x\sqrt{x^2+y^2}\\
y'=x-y\sqrt{x^2+y^2}
\end{array}
\end{equation}
which I solve by polar coordinates transformation. $x=r\cos\phi \ y=r\sin\phi$ and $x'=dr\cos\phi-r\sin\phi$ and $y'=dr\sin\phi+r\cos\phi$.
\begin{equation}
\begin{array}
fdr\cos\phi-r\sin\phi=-r\sin\phi-r\cos\phi\sqrt{r^2\cos^2\phi+r^2\sin^2\phi}\\
dr\sin\phi+r\cos\phi=r\cos\phi-r\sin\phi\sqrt{r^2\cos^2\phi+r^2\sin^2\phi}\
\end{array}
\end{equation}
Clearly, many terms cancel out and we obtain:
\begin{equation}
\begin{array}
fdr=-r^2\\
dr=-r^2\
\end{array}
\end{equation}
Then, integrating both sides with respect to $r$ we obtain:
\begin{equation}
\begin{array}
fr=-\frac{1}{3}r^3+c\\
r=-\frac{1}{3}r^3+c
\end{array}
\end{equation}
Since $r=\frac{x}{\cos\phi} and r=\frac{y}{\sin\phi}$ we get
\begin{equation}
\begin{array}
fx=-\frac{x^3}{\cos^2\phi}+c\cos\phi\\
y=-\frac{y^3}{\sin^3\phi}+c\sin\phi
\end{array}
\end{equation}
The summing up the two equations, and isolating for y, I get a very scary result:
y≈0.26457 sin^4(ϕ) cos(ϕ) (sqrt((27 c csc^9(ϕ) sec^2(ϕ) + 27 c csc^8(ϕ) sec^3(ϕ) - 27 x^3 csc^9(ϕ) sec^5(ϕ) - 27 x csc^9(ϕ) sec^3(ϕ))^2 + 108 csc^15(ϕ) sec^6(ϕ)) + 27 c csc^9(ϕ) sec^2(ϕ) + 27 c csc^8(ϕ) sec^3(ϕ) - 27 x^3 csc^9(ϕ) sec^5(ϕ) - 27 x csc^9(ϕ) sec^3(ϕ))^(1/3) - (1.2599 csc(ϕ) sec(ϕ))/(sqrt((27 c csc^9(ϕ) sec^2(ϕ) + 27 c csc^8(ϕ) sec^3(ϕ) - 27 x^3 csc^9(ϕ) sec^5(ϕ) - 27 x csc^9(ϕ) sec^3(ϕ))^2 + 108 csc^15(ϕ) sec^6(ϕ)) + 27 c csc^9(ϕ) sec^2(ϕ) + 27 c csc^8(ϕ) sec^3(ϕ) - 27 x^3 csc^9(ϕ) sec^5(ϕ) - 27 x csc^9(ϕ) sec^3(ϕ))^(1/3) and csc(ϕ) sec(ϕ)!=0
What went wrong here?
Thanks
UPDATE, with Jean Maries correction we get:
\begin{equation}
\begin{array}
fx=\frac{\cos\phi}{\phi+c}\\
y=\frac{\sin\phi}{\phi+c}
\end{array}
\end{equation}
Summing up and solving for y:
\begin{equation}
y=\frac{\cos\phi}{\phi+c}+\frac{\sin\phi}{\phi+c}-x
\end{equation}
which plotted , with an arbitrary value of $c$ is:
| $dr=-r^2$ cannot be integrated as you do. You need to write it under the form
$$- \dfrac{dr}{r^2}=1$$
giving :
$$\dfrac{1}{r}=\varphi+c$$
($c$ arbitrary constant), otherwise said:
$$r=\frac{1}{\varphi+c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $(2x^2-3x+1)(2x^2+5x+1)=9x^2$ Solve the equation $$(2x^2-3x+1)(2x^2+5x+1)=9x^2$$
The given equation is equivalent to $$4x^4+4x^3-11x^2+2x+1=9x^2\\4x^4+4x^3-20x^2+2x+1=0$$ which, unfortunately, has no rational roots. What else can I try?
| We have, $$((2x^2+x+1)-4x)((2x^2+x+1)+4x)=9x^2$$
$$(2x^2+x+1)^2-16x^2=9x^2\;\Rightarrow\; (2x^2+x+1)^2=(5x)^2$$
Hence, $\;2x^2+x+1=\pm5x$. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many ways to deal with the integral $\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}$? I tackle the integral by rationalization on the integrand first.
$$
\frac{1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{2 x}
$$
Then splitting into two simpler integrals yields $$
\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\frac{1}{2}\left [\underbrace{\int\frac{\sqrt{1+x}}{x}}_{J} d x+\underbrace{\int\frac{\sqrt{1-x}}{x} d x}_{K}\right]
$$
To deal with $J$, we use rationalization instead of substitution. $$
\begin{aligned}
J &=\int \frac{\sqrt{1+x}}{x} d x \\
&=\int \frac{1+x}{x \sqrt{1+x}} d x \\
&=2 \int\left(\frac{1}{x}+1\right) d(\sqrt{1+x}) \\
&=2 \int \frac{d(\sqrt{1+x})}{x}+2 \sqrt{1+x} \\
&=2 \int \frac{d(\sqrt{1+x})}{(\sqrt{1+x})^{2}-1}+2 \sqrt{1+x} \\
&=\ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1} \right| +2 \sqrt{1+x}+C_{1}
\end{aligned}
$$
$\text {Replacing } x \text { by } -x \text { yields }$
$$
\begin{array}{l} \\
\displaystyle K=\int \frac{\sqrt{1-x}}{-x}(-d x)=\ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|+2 \sqrt{1-x}+C_{2}
\end{array}
$$
Now we can conclude that $$
I=\sqrt{1+x}+\sqrt{1-x}+\frac{1}{2}\left(\ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+\ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|\right)+C
$$
My question is whether there are any simpler methods such as integration by parts , trigonometric substitution, etc…
Please help if you have. Thank you for your attention.
| Let $x=\cos 2 \theta$, for $0\leq 2\theta < \pi$, then $d x=-2 \sin 2 \theta d \theta$ and
$$
\begin{aligned}
I &=\int \frac{-2 \sin 2 \theta d \theta}{\sqrt{1+\cos 2 \theta-\sqrt{1-\cos 2 \theta}}} \\
&=-2 \int \frac{\sin 2 \theta d \theta}{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}} \\
&=-\sqrt{2} \int \frac{\sin 2 \theta}{\cos \theta-\sin \theta} \cdot d \theta \\
&=-\sqrt{2} \int \frac{1-(\cos \theta-\sin \theta)^{2}}{\cos \theta-\sin \theta} d \theta \\
&=\sqrt{2}\left[\underbrace{\int\frac{d \theta}{\sin \theta-\cos \theta}}_{J}+\underbrace{\int(\cos \theta-\sin \theta) d \theta}_{K}\right]
\end{aligned}
$$
$$
\begin{array}{l}
\displaystyle J=\frac{1}{\sqrt{2}} \ln \left|\frac{\sin \theta-\cos \theta}{\sqrt{2}+\cos \theta+\sin \theta}\right|+C_{1} \\
\displaystyle I=\ln \left|\frac{\sin \theta-\cos \theta}{\sqrt{2}+\cos \theta+\sin \theta}\right|+\sqrt{2}(\sin \theta+\cos \theta)+C
\end{array}
$$
Since $0\leq \theta < \dfrac{\pi}{2}$, putting $ \displaystyle \sin \theta=\sqrt{\frac{1-\cos 2 \theta}{2}}=\sqrt{\frac{1-x}{2}}$ $\textrm{ and }$
$\displaystyle \cos \theta=\sqrt{\frac{1+x}{2}}$ yields
$$
I=\sqrt{1-x}+\sqrt{1+x}+\ln \left|\frac{\sqrt{1-x}-\sqrt{1+x}}{2+\sqrt{1+x}+\sqrt{1-x}}\right|+C
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculate $\lim_{x\to \infty}(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}})^x$ Calculate $\lim_{x\to \infty}(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}})^x$.
First off, it's easy to see that $\lim_{x\to \infty}\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}}$ = 1. Therefore, I tried the following:
$$\lim_{x\to \infty}(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}})^x=
\lim_{x\to \infty}(1 +(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}} -1))^x = e^{\lim_{x\to \infty}x(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}} -1)}.$$
Now I find myself stuck at finding $\lim_{x\to \infty}x(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}} -1)$. Keep in mind that I am not allowed to use l'Hopital. Any hint would be appreciated. Thanks.
| Let the limit exist and equal $L$. Then $\ln L$ equals:
$$\lim_{x \to \infty} x\left(\ln\left(2^{1/x}+3^{1/x}\right)-\ln\left(4^{1/x}+5^{1/x}\right)\right)$$
and making the substitution $u \mapsto 1/x$ gives:
$$\lim_{u \to 0^+} \frac{1}{u} \left(\ln\left(2^u+3^u\right)-\ln\left(4^u+5^u\right)\right)$$
Considering the first-order approximation of $f(u) = \ln\left(2^u+3^u\right)$ using Taylor series, $f(0) = \ln 2$. $f'(u) = \frac{1}{2^u + 3^u} \cdot (2^u \ln 2 + 2^u \ln 3)$, hence $f'(0) = \frac{\ln 6}{2}$. Doing this for the other composite function gives:
$$\ln L = \lim_{u \to 0^+} \frac{1}{u} \left(\ln 2 + \frac{\ln 6}{2}u - \ln 2 - \frac{\ln 20}{2} u \right) = \frac{1}{2} \ln \frac{3}{10}$$
hence $L = e^{\ln(3/10) \cdot 1/2} = \sqrt{3/10} \approx 0.548$.
| {
"language": "en",
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Is $\lfloor{\frac{a+b+c+d}{4}}\rfloor=\lfloor\frac{\lfloor{\frac{a+b}{2}}\rfloor+\lfloor{\frac{c+d}{2}}\rfloor}{2}\rfloor$ for $a,b,c,d\in\mathbb R$?
Does the following hold $\forall a,b,c,d\in\mathbb R$?
$$\quad\left\lfloor{\frac{a+b+c+d}{4}}\right\rfloor=\left\lfloor\frac{\left\lfloor{\frac{a+b}{2}}\right\rfloor+\left\lfloor{\frac{c+d}{2}}\right\rfloor}{2}\right\rfloor?$$
It seems like it's enough to restrict $a,b,c,d$ to $[0,4)$, and then we can do a case analysis for all possible values of $\left\lfloor{\frac{a+b}{2}}\right\rfloor,\left\lfloor{\frac{c+d}{2}}\right\rfloor$.
Is there a simpler way?
As suggested in a comment, an equivalent questions is whether the following holds for all $x,y\in\mathbb R$:
$$\left\lfloor\frac{x+y}{2}\right\rfloor = \left\lfloor\frac{\left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor}{2}\right\rfloor.$$
As suggested in the answer (and one of the comments), this fails, e.g., for $a=2.5, c=1.5, b,d=0$.
A followup question:
Does the equality hold for all $a,b,c,d\in\mathbb N$?
| The result is false. As pointed out by J.G., we can apply change of variable $x=\frac{a+b}{2}$ and $y=\frac{c+d}{2}$ to get:
\begin{equation*}
\lfloor \frac{x+y}{2}\rfloor = \lfloor \frac{ \lfloor x \rfloor+\lfloor y\rfloor}{2}\rfloor
\end{equation*}
But this cannot hold for every real $x$ and every real $y$: just put $x=2.5$ and $y=1.5$ for example.
Edit:
If $a$, $b$, $c$, and $d$ are all positive integers, we can just make sure $a+b=5$ and $c+d=3$ to get $x=2.5$ and $y=1.5$ so that the result fails even if we assume they are all positive integers.
| {
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"source": "stackexchange",
"question_score": "1",
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Simplify $A(t)=\frac{1-t}{1-\sqrt[3]{t}}+\frac{1+t}{1+\sqrt[3]{t}}$ Simplify $$A(t)=\dfrac{1-t}{1-\sqrt[3]{t}}+\dfrac{1+t}{1+\sqrt[3]{t}}$$ and calculate $A(3\sqrt3).$ For $t\ne\pm1$ we have, $$A=\dfrac{(1-t)(1+\sqrt[3]{t})+(1+t)(1-\sqrt[3]{t})}{1-\sqrt[3]{t^2}}=\\=\dfrac{2-2t\sqrt[3]{t}}{1-\sqrt[3]{t^2}}$$ What to do next? I can't come up with anything else...
| By using the substitution $\sqrt[3]t=a$, we have, $$\dfrac{1-t}{1-\sqrt[3]{t}}+\dfrac{1+t}{1+\sqrt[3]{t}}=\dfrac{a^3-1}{a-1}+\dfrac{a^3+1}{a+1}=(a^2+a+1)+(a^2-a+1)=2a^2+2$$
Which is equal to $2+2\sqrt[3]{t^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $4\lfloor x \rfloor=x+\{x\}$. What is wrong in my solution? Can anyone tell? $\lfloor x \rfloor$ = Floor Function, and $\{x\}$ denotes fractional part function
Solve for $x$
$$4\lfloor x \rfloor= x + \{x\}$$
$ x = \lfloor x \rfloor + \{x\}$
$\implies x - \{x\} = \lfloor x \rfloor$
$\implies 4x - 4\{x\} = x + \{x\}$
$\implies 3x = 5\{x\}$
Note that ,
$\implies 0 \leq \{x\}<1$
$\implies 0 \leq 5\{x\} < 5 \iff 0 \leq 3x <5 \iff\boxed{ 0 \leq x < \dfrac{5}{3}}$
But the answer is $ x = 0$
| We have :
$$4 \lfloor x \rfloor = x + \{x\} = \lfloor x \rfloor + 2 \{x\}$$
then :
$$2 \{x\} = 3 \lfloor x \rfloor \in \mathbb{Z}$$
We deduce that :
$$\{x\} = 0 \text{ or } \{x\} = \dfrac{1}{2}$$
*
* If $\{x\} = 0$ then $3 \lfloor x \rfloor = 2 \{x\} = 0$. It means that $\lfloor x \rfloor = 0$ then $x = \lfloor x \rfloor + \{x\} = 0$.
* If $\{x\} = \dfrac{1}{2}$ then $3 \lfloor x \rfloor = 2 \{x\} = 1$. It means that $\lfloor x \rfloor = \dfrac{1}{3}$ wich is impossible because $\lfloor x \rfloor \in \mathbb{Z}$.
The unique solution is $x = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4353922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Get derivative with square root in the denominator Given:
$$\begin{align}
f(x) &= \frac{3}{\sqrt{2x^7}} \\
\\
\frac{df}{dx} &=
\end{align}$$
The expected answer is:
$$\begin{align}
f(x) &= \frac{3}{\sqrt{2x^7}} \\
\\
\frac{df}{dx} &= -\frac{21\sqrt{2}}{4x\sqrt{x^7}}
\end{align}$$
But what steps would I need to take to that answer?
| This would use the chain rule. Rewrite the function as:
$$f(x) = 3(2x^7)^{-\frac{1}{2}}$$
Using the chain rule:
$$\frac{df}{dx} = -\frac{1}{2} *3(2x^7)^{-\frac{3}{2}} * 14x^6$$
$$\frac{df}{dx} = -\frac{21x^6}{2^\frac{3}{2} x^\frac{21}{2}}$$
$$\frac{df}{dx} = -\frac{21\sqrt{2}}{\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{2}x^\frac{9}{2}}$$
$$\frac{df}{dx} = -\frac{21\sqrt{2}}{4x^\frac{9}{2}}$$
$$\frac{df}{dx} = -\frac{21\sqrt{2}}{4x \sqrt{x^7}}$$
This is really really weird simplification, like why write $x^\frac{9}{2}$ as $x\sqrt{x^7}$ instead of $\sqrt{x^9}$? Idk but that's how it was done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Easy question on probability I know this is a trivial question but I want to make sure I'm not missing anything:
We have a biased 6-sided die, which brings any of the 6 numbers with equal probability in the first roll, but in the second and all subsequent rolls, brings the previous result with probability $\frac {1}{2}$ and all others with probability $\frac {1}{10}$.
The question is: Suppose we get a 4 in the first roll; what is the probability we also get a 4 in the 3rd roll? In the 4th? And so on.
If we get a 4 in the 1st roll, then for the 2nd roll we have $\frac {1}{2}$ probability to get a 4 and $\frac {1}{10}$ for all other numbers in the 2nd roll.
So in the 3rd roll, we already have the results of the previous roll of getting a 4 with probability $\frac {1}{2}$, so now the probability is $\frac {1}{4}$?
| There are actually two cases:
(1) Get a 4 in the 2nd roll and the 3rd roll
(2) Get a 4 in the 3rd roll but not in the 2nd roll
The probability of (1) is $\frac{1}{4}$ as you have correctly calculated.
In the case of (2), it is $\frac{1}{2} \times \frac{1}{10} = \frac{1}{20}$ because the probability of not getting a 4 in the 2nd roll is 1/2 and getting a 4 in the 3rd roll is 1/10.
Therefore, the probability of getting a 4 in the 3rd roll is $\frac{1}{4} + \frac{1}{20} = \frac{3}{10}$.
We can generalize this by finding the "recurrence formula" for the probability of getting a 4 in the $n$th roll, which we will define $p_n$ to be. That is:
$p_n = \frac{1}{2} p_{n-1} + \frac{1}{10} (1 - p_{n-1}) = \frac{2}{5} p_{n-1} + \frac{1}{10}$
Also, $p_1 = 1$ since we supposed that we got a 4 in the first roll.
Solving this yields $p_n = \frac{25}{12}(\frac{2}{5})^n + \frac{1}{6}$.
So, for example, $p_4 = \frac{25}{12}(\frac{2}{5})^4 + \frac{1}{6} = \frac{11}{50}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proof that $a^n-b^n$ is non prime if n is non prime Hello guys I want to know if my proof is valid, and whether it is coherent and where I can improve it.
Prove that for all natural numbers, $a,b, n$, where $a>b$, and $n>1$ is a not a prime, $a^n-b^n$ is not a prime number.
My attempt:
Since $n$ is a non-prime greater than $1$, we have $n=xy$ where $1<x,y<n. $We know that $$a^n-b^n=a^{xy}-b^{xy}=(a^x-b^x)(a^{x(y-1)}+b^xa^{x(y-2)}+...+b^{x(y-2)}a^x+b^{x(y-1)})$$ Hence $(a^x-b^x)$ will always divide $a^n-b^n$. We merely need to prove that $a^n-b^n>a^x-b^x>1$. That $a^n-b^n>a^x-b^x$ is obvious since $a^{x(y-1)}+b^xa^{x(y-2)}+...+b^{x(y-2)}a^x+b^{x(y-1)}$ is a positive integer greater than $1$ ($y>1$). Furthermore, we can write $a^x-b^x$ as
$(a-b)(a^{x-1}+ba^{x-2}+...+ab^{x-2}+b^{x-1})$. Since $x>1$, $a^{x-1}+ba^{x-2}+...+ab^{x-2}+b^{x-1}$ is a positive integer greater than $1$. Therefore, we have $a^x-b^x=(a-b)(a^{x-1}+ba^{x-2}+...+ab^{x-2}+b^{x-1})>a-b \geq1$. Hence $a^x-b^x>1$. This concludes the proof.
| I don't see anything wrong with your proof. Perhaps I could mention some shortcuts or alternatives that you might find useful.
Let $d$ be a non trivial divisor of $n$, then we may write $n = dk$ for some integer $k > 1$. From $a^d \equiv b^d \pmod{a^d - b^d}$, we obtain $a^n \equiv b^n \pmod{a^d - b^d}$ by raising each side to the $k$-th power. The latter congruence relation is of course equivalent to saying that $a^d - b^d$ is a divisor of $a^n - b^n$.
$x \mapsto x^n - x^d = x^d(x^{n-d} - 1): [1, \infty) \to \mathbb{R}$ is a strictly increasing function whenever $n > d > 0$, since it is the product of strictly increasing functions mapping only to non-negative values. Here $n$ and $d$ need not be integers. Hence both $a^n - b^n > a^d - b^d$ and $a^d - b^d > a - b$ follow from rewriting them as $a^n - a^d > b^n - b^d$ and $a^d - a > b^d - b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Upper bound for a series $\sum_{n=0}^{\infty}2^n r^{2^n}$ for $r\in (0,1)$ It is clear that for $r\in (0,1)$ fixed, the series $$f(r)=\sum_{n=0}^{\infty}2^n r^{2^n}$$ is convergent, and also it is not to hard show that $f(r)$ is not uniformly convergent in $(0,1)$. Since $\sum_{n=0}^{\infty}r^n=\frac{1}{1-r}$, I think this sum function $f(r)$ may be similar.
So can I find some upper bound such as $\frac{1}{1-r^\alpha}$ to dominate $f(r)$? Sincerely thanks!
| This is a different approach with an arbitrary good bound for $x\in(0,1)$.
$$f(x)=\sum_{n=0}^N 2^n x^{2^n} + 2\sum_{n=N+1}^\infty 2^{n-1} x^{2^n} = \sum_{n=0}^N 2^n x^{2^n} + 2\sum_{n=N+1}^\infty \underbrace{x^{2^n} + ... + x^{2^n}}_{2^{n-1} \text{ terms}} \\
\leq \sum_{n=0}^N 2^n x^{2^n} + 2\sum_{n=N+1}^\infty x^{2^{n-1}+1}+x^{2^{n-1}+2}+...+\underbrace{x^{2^{n-1}+2^{n-1}}}_{x^{2^n}} \\
=\sum_{n=0}^N 2^n x^{2^n} + 2\sum_{n=2^N+1}^\infty x^n = \sum_{n=0}^N 2^n x^{2^n} + \frac{2\,x^{2^N+1}}{1-x} \, .$$
For example, for $N=1$ you have $$f(x)\leq \frac{x(x+1)}{1-x} \, .$$
Aside from that, you may want to check these posts
*
*What's the limit of the series $\log_2(1-x)+x+x^2+x^4+x^8+\cdots$.
*Evaluate $\lim_{x\to1^-}\left(\sum_{n=0}^{\infty}\left(x^{(2^n)}\right)-\log_2\frac{1}{1-x}\right)$
which make statements about the behaviour of your function in the vicinity of $x=1$.
Since $$f(x)=\sum_{n=0}^\infty 2^n x^{2^n} = x \frac{{\rm d}}{{\rm d}x} \underbrace{\sum_{n=0}^\infty x^{2^n}}_{g(x)}$$
you may want to study $g(x)$ instead. $g(x)$ (and $f(x)$ alike) is obviously strictly increasing on $(0,1)$. Thus you may seek the behaviour as $x\rightarrow 1$. The first link shows that the function $$G(x)=g(x)+\log_2(1-x)$$ has a mean limit of $$\overline{G(1)}=\overline{\lim_{x\rightarrow 1^-} G(x)} = \frac{1}{2} - \frac{\gamma}{\log 2} \, .$$
However, $G(x)$ does not converge as $x\rightarrow 1$. If you write down the series expression for $G(1)$, you'll find it is still depending on $x$ and in the variable $t$, where $x=e^{-2^{-t}}$, it has a period of $1$. In the second link I calculate the fourier-expansion of $G$ in the variable $t$ with the result
$$G(1)=\frac{1}{2}-\frac{\gamma}{\log 2} + \sum_{n=1}^\infty A_n \cos\left(2\pi nt- \varphi_n\right)$$
where
\begin{align}
A_n&=\frac{2\left|\Gamma\left(\frac{2\pi i \,n}{\log 2}\right)\right|}{\log 2}=\sqrt{\frac{2}{n\sinh\left(\frac{2\pi^2n}{\log 2}\right)\log 2}} \\
-\varphi_n&=\arg \left( \Gamma\left(\frac{2\pi i \,n}{\log 2}\right) \right) \, .
\end{align}
Notice that the first harmonic is already a very good approximation, $A_2/A_1<10^{-6}$.
Thus in the vicinity of $x=1^-$ you can write
$$g(x)=G(1)-\log_2(1-x) \\
f(x)=x \frac{{\rm d}}{{\rm d}x} g(x) = \frac{1}{\log 2} \frac{x}{1-x} - \frac{2\pi}{\log 2} \frac{-1}{\log x} \sum_{n=1}^\infty n A_n \sin\left(2\pi nt - \varphi_n \right) \\
\leq \frac{1}{\log 2} \frac{1}{1-x} \left( x + 2\pi \sum_{n=1}^\infty n A_n \right) < \frac{x+10^{-5}}{\log 2} \frac{1}{1-x}$$
with $\frac{-1}{\log x} \leq \frac{1}{1-x}$ for $0<x<1$.
| {
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"source": "stackexchange",
"question_score": "4",
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How to find the formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}$, where $n\in N$? By the generalization in my post,we are going to evaluate the integral $$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}},$$
where $n\in N.$
First of all, let us define the integral $$I_n(a)=\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} \textrm{ for any positive real number }a.$$
Again, we start with $$I_1(a)=\int_{0}^{\infty} \frac{d x}{x^{2}+a}= \left[\frac{1}{\sqrt{a}} \tan ^{-1}\left(\frac{x}{\sqrt{a}}\right)\right]_{0}^{\infty} = \frac{\pi}{2 }a^{-\frac{1}{2} } $$
Then differentiating $I_1(a)$ w.r.t. $a$ by $n-1$ times yields
$$
\int_{0}^{\infty} \frac{(-1)^{n-1}(n-1) !}{\left(x^{2}+a\right)^{n}} d x=\frac{\pi}{2} \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2 n-3}{2}\right) a^{-\frac{2 n-1}{2}}
$$
Rearranging and simplifying gives $$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} =\frac{\pi a^{-\frac{2 n-1}{2}}}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)}
$$
Putting $a=1$ gives the formula of our integral
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}} =\frac{\pi}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)= \frac{\pi}{2^{2 n-1}} \left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}$$
For verification, let’s try $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{10}} &= \frac{\pi}{2^{19}}\left(\begin{array}{c}
18 \\
9
\end{array}\right) =\frac{12155 \pi}{131072} ,
\end{aligned}
$$
which is checked by Wolframalpha .
Are there any other methods to find the formula? Alternate methods are warmly welcome.
Join me if you are interested in creating more formula for those integrals in the form $$
\int_{c}^{d} \frac{f(x)}{\left(x^{m}+1\right)^{n}} d x.
$$
where $m$ and $n$ are natural numbers.
| Thanks to Mr Nejimban, I am now going to evaluate the integral by converting it into a Wallis integral by the substitution $x=\tan \theta$, which yields
$$I_{n}(1)= \int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}} =\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \theta d \theta}{\sec ^{2 n} \theta} =\int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d \theta
$$
By the Wallis Formula for cosine, $$
\int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d\theta =\frac{\pi}{2} \prod_{k=1}^{n-1} \frac{2 k-1}{2 k},
$$
Simplifying yields $$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta &=\frac{\pi}{2} \prod_{k=1}^{n-1}\left(\frac{2 k-1}{2 k} \cdot \frac{2 k}{2 k}\right) \\
&=\frac{\pi}{2}\cdot \frac{(2 n-2) !}{2^{2 n-2}[(n-1) !]^{2}} \\
&=\frac{\pi}{2^{2 n-1}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)
\end{aligned}
$$
$$
\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} \stackrel{x \rightarrow \frac{x}{\sqrt{a}}}{=} \frac{1}{a^{n-\frac{1}{2}}} I_{n}(1)=\frac{\pi}{a^{\frac{2n-1}{2} } \cdot 2^{2 n-1}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)
$$
In particular, $$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}=\frac{\pi}{2^{2 n-1}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$S_n = \frac{12}{(4-3)(4^2-3^2)} + \frac{12^2}{(4^2-3^2)(4^3-3^3)} + ... + \frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})}$
So I know that in order to find the series we need to change the form because we can cancel the terms $$\frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})} = \frac{A}{4^n-3^n} + \frac{B}{4^{n+1}-3^{n+1}}$$
But my question is how do we get $A = \frac{4^n+3^n}{2}$ and $B = \frac{4^{n+1}+3^{n+1}}{2}$
When I find the common denominator and add them together I got $12^n = A(4^{n+1}-3^{n+1}) + B(4^n-3^n)$. I tried simplifying but it can't seem to work.
Can anyone show me how to find A and B? Thank you in advance.
| $$\frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})}$$
$4^n=a$, $3^n=b$, $12^n=ab$
$$\frac{ab}{(a-b)(4a-3b)}$$
Partitional fractioning in two variables is not unique. Make polynomial partitional fractioning considering $a$ as independent variable
$$\frac{ab}{(a-b)(4a-3b)}=\frac{A}{a-b}+\frac{B}{4a-3b}$$
$$ab=4aA-3bA+Ba-Bb=(4A+B)a-(3A+B)b$$
$$4A+B=b, 3A+B=0 \Rightarrow A=b, B=-3b$$
$$\frac{ab}{(a-b)(4a-3b)}=\frac{b}{a-b}-\frac{3b}{4a-3b}$$
$$\frac{ab}{(a-b)(4a-3b)}=\frac{b}{a-b}+\frac{1}{2}-\left(\frac{3b}{4a-3b}+\frac{1}{2}\right)$$
Combining $\frac{b}{a-b}+\frac{1}{2}$ and $\frac{3b}{4a-3b}+\frac{1}{2}$
$$\frac{ab}{(a-b)(4a-3b)}=\frac{a+b}{2(a-b)}-\frac{4a+3b}{2(4a-3b)}$$
| {
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"url": "https://math.stackexchange.com/questions/4366536",
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Why is this secant substitution allowed? On Paul's Math Notes covering Trig Substitutions for Integrals we start with an integral:
$$\int{{\frac{{\sqrt {25{x^2} - 4} }}{x}\,dx}}$$
Right away he says to substitute $x=\frac{2}{5}\sec(θ)$. Why is that allowed?
Looking further down onto how he approaches the problem, it seems like it's allowed because it's compensated for with a dx:
$$dx = \frac{2}{5}\sec \theta \tan \theta \,d\theta$$
Is that what's going on here? It's fair to say you can substitute x with whatever you want so long as you update dx? Seems like it wouldn't work for constant functions of x, like $x = 5$.. since that'd get you $dx=0$ and clearly be wrong. So what rules are in play here for substitution?
| To add some information to Greg Martin's answer, this is because in some cases, integrals with $\sqrt{\pm(ax)^2 \pm b^2}$ can be solved using trigonometric substitutions. Recall the identities $$\sin^2\theta + \cos^2\theta = 1, \\ 1 + \cot^2\theta = \csc^2\theta, \\ \tan^2\theta + 1 =\sec^2\theta.$$
In the case $\sqrt{(ax)^2 - b^2}$, we can factor out $b^2$ from the square root, giving us $b\sqrt{(\frac{a}{b}x)^2 -1}$. By comparing the term inside the square root to the trig identities, it looks the same to $\csc^2\theta - 1$ and $\sec^2\theta - 1$. Hence, we can use $\frac{a}{b}x = \csc\theta \Leftrightarrow x = \frac{b}{a}\csc\theta$ or $\frac{a}{b}x = \sec\theta \Leftrightarrow x = \frac{b}{a}\sec\theta$.
Let's try to solve $$\int{{\frac{{\sqrt {25{x^2} - 4} }}{x}\,dx}}$$ using the substitution $x = \frac{2}{5}\csc\theta$. Then, $dx = -\frac{2}{5}\csc\theta\cot\theta$. Also, $\sqrt{25x^2 - 4}$ becomes $2\cot\theta$ assuming that $\cot\theta > 0$. This gives us $$\int \frac{2\cot\theta}{\frac{2}{5}\csc\theta}\left(-\frac{2}{5}\csc\theta\cot\theta\right) \\ -2\int\cot^2\theta\,d\theta \\ -2\int(\csc^2\theta - 1)\,d\theta \\ -2(-\cot\theta - \theta) + C \\ 2(\cot\theta + \theta) + C \\ 2\left(\frac{\sqrt{25x^2 - 4}}{2} + \sin^{-1}\left(\frac{2}{5x}\right)\right) + C \\ \sqrt{25x^2 - 4} + 2\sin^{-1}\left(\frac{2}{5x}\right) + C.$$
Comparing the answer obtained when $x = \frac{2}{5}\sec\theta$: $$\sqrt {25{x^2} - 4} - 2{\cos ^{ - 1}}\left( {\frac{2}{{5x}}} \right) + C$$ to the substitution $x = \frac{2}{5}\csc\theta$, we can see that for all $x \in \mathbb{R}$, $$\sin^{-1}\left(\frac{2}{5x}\right) = -\cos^{-1}\left(\frac{2}{5x}\right) + C$$ will hold if and only if $C = \frac{\pi}{2}$. By substituting values of $x$, we can see that it is indeed the case.
| {
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"source": "stackexchange",
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First proof writing practice: writing and style feedback, and is it correct? I'm not even sure if this is correct because there was no answer in the book and I'm new to proofs by induction. I'm mostly interested in feedback about the style and any proof-writing faux-pas'sss I may have made. It felt right to put the lemma up front to improve the flow of the main theorem and it possibly doesn't need to be proven, but I did so because I was advised you should write everything you need to to convince yourself in the early stages. Thanks
So I have to prove that $n(n^2 +5)$ is divisible by $6$ for all $n >=1$, but I start with a lemma I use later.
Lemma 1: $n^2 + n$ is even for all $n \in Z$.
Proof: Assume $n$ is even, then $n = 2m$ for some $m \in Z$, and:
$$n^2 + n = (2m)^2 + 2m = $$
$$ 4m^2 + 2m = $$
$$ 2(2m^2 + m) =$$
$$ n^2 + n = 2k$$
for some integer $k$. If $n$ is even, then $(n^2 + n)$ is also even.
Assume $n$ is odd and let $n = 2m -1$ for some $m \in Z$.
Thus,
$$n^2 + n = (2m -1)^2 + (2m -1)=$$
$$4m^2 - 4m +1 + 2m - 1 =$$
$$4m^2 - 2m =2(2m^2 -1) = 2k$$ for some integer $k$ and again we find:
$$n^2 + n = 2k$$
Thus, $n^2 + n$ is even when $n$ is odd or even, proving $n^2 + n$ is even for all $n \in Z$.
Theorem: $n(n^2 + 5)$ is divisible by 6 for all integers $n>=1 $.
Proof:
By induction. Let $ A(n) = n(n^2 +5)$ and so:
$$A(1) = 1(1^2 + 5) = 6 = 6 \cdot 1$$
By assumption, $A(n) = 6k$ for some $k \in Z$:
$$ n(n^2 + 5) = (n^3 + 5n) = 6k $$
It suffices to show that $A(n+1) = 6r$ for some $r \in Z$:
$$A(n+1) = (n+1)[(n+1)^2 + 5]= $$
$$ (n+1)^3 + 5(n+1) =$$
$$ n^3 + 3n^2 + 8n + 6=$$
$$ n^3 + 3n^2 + (5n + 3n) + 6=$$
$$ (n^3 + 5n) + 3n^2 + 3n + 6=$$
By assumption, $A(n) = (n^3 +5n) = 6k$ for some $k \in Z$ we have:
$$ 6k + 3n^2 +3n + 6 =$$
$$ 6k + 3(n^2 + n) + 6 =$$
By Lemma 1, we let $(n^2 +n) = 2j$ for some positive integer $j$:
$$ 6k + 3(2j) + 6 =$$
$$ 6k + 6j + 6 = $$
$$ 6(k + j + 1)$$
Let $ k + j + 1 = r$, then we have:
$$A(n+1) = 6r$$
Thus proving $A(1)$ is divisible by 6 and that if $A(n)$ is divisible by 6, then $A(n+1)$ is also divisible by 6.
| Instead of editing I decided to make an answer, so you could compare better. In companion to the previous answers and comments, here is a shortened version of the results. Of course, a lot is personal taste how to write things down and you will learn a lot along the way.
Lemma 1. $n^2 + n$ is even for all $n \in \mathbb{Z}.$
Proof: As either $n$ or $n+1$ is even, it follows that $n^2+n = n(n+1)$ is even.
Theorem. $n(n^2 + 5)$ is divisible by 6 for all integers $n\geq 1 $.
Proof: Let $ A(n) = n(n^2 +5).$ We proceed by induction. Notice that $A(1)=6,$ this shows the induction base. Now assume that $A(n)$ is divisible by $6$ and consider
\begin{align*}
A(n+1) &= n^3 + 2n^2+6n+n^2+6 \\ &= (n^3+5n) + 3(n^2+n) +6\\ &= A(n) + 3(n^2+n) +6.
\end{align*}
The first and the second term are divisible by $6$ by the assumption and the lemma respectively, hence $A(n+1)$ is divisible by $6.$
| {
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How do I evaluate $\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}$ How do I evaluate the following integral when where $C$ is the square with vertices at $\pm2, \pm2+4i$,
$$\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}$$
Using Cauchy integral:
$\frac{z^{2}}{z^{2}+4}=\frac{z^2}{(z+2i)(z-2i)}=\frac12\frac{z^2}{z+2i}+\frac12\frac{z^2}{z-2i}$ then $\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4} \implies \frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z+2i}+\frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z-2i}$
$\frac{z^2}{z+2i}$ is analytic on and inside $C$, hence we can apply Cauchy theorem and for the second term we use Cauchy integral formula,
$$\frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z+2i}+\frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z-2i}=0+\frac{1}{4\pi i}2\pi i \times f(2i)=-2$$
Using Residue Theorem:
$2i$ is the only isolated singularity in $C$.
$$\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}=\frac{1}{2 \pi i} 2\pi i \times \text{Res}(f,2i)=i$$
I get correct answer for Residue Theorem but couldn't understand where I do wrong when using Cauchy integral.
It will be great help if someone clear me when to use which method to find the integral.
| Here you don't have to use partial fractions but use $f(z) = \frac{z^2}{z+2i}$ for $f(a)$ in the Cauchy Integral Formula, since $f$ is holomorphic in the smallest open disk that fits your $C$.
Then you get $f(2i) = \frac{(2i)^2}{4i} = i$
| {
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divisibility problem and gcd Let a and b be a positive integers. Proof that if number $ 100ab -1 $ divides number $ (100a^2 -1)^2 $ then also divides number $ (100b^2 -1)^2 $.
My attempt:
Let's notice that \begin{split} b^2(100a^2-1)^2 -a^2(100b^2-1)^2 & =(100a^2b-b)^2-(100ab^2-a)^2\\
& =(100a^2b-b-100ab^2+a)(100a^2b-b+100ab^2-a)\\
& =(100ab(a-b)+(a-b))(100ab(a+b)-(a+b))\\
& =(a-b)(100ab+1)(a+b)(100ab-1).\end{split}
This means that $ 100ab-1 |a^2(100b^2-1)^2$, so if we proof that $\gcd(100ab-1,a^2)=1$ the proof is completed. Now I know that it should be trivial to show that these numbers are relatively prime but somehow i have no idea how to do it.
Also I am intrested if there is a way to solve this problem by using modular arithmetic?
| For a modular arithmetic argument:
Let $N=100ab-1$ and work modulo $N$.
Starting with $(100a^2 -1)^2\equiv 0$, multiply by $10^4b^4$:
$$(10^4a^2b^2-100b^2)^2\equiv 0.$$
Since $10^2ab\equiv 1$, we have $(1-100b^2)^2\equiv 0.$
| {
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When we classify groups of order $n$, can I skip to check if the associativity law holds after I complete Cayley Table? For example, classify groups of order $4$.
Let $G=\{a,b,c,d\}$ be a group whose order is $4$.
$G$ must have an identity element $e$.
Without loss of generality, we can assume $d$ is the identity element of $G$.
So, $G=\{e,a,b,c\}$.
(1) Suppose that there is an element of $G$ whose order is $4$.
Without loss of generality, we can assume $o(a)=4$.
Then, $G=\{e,a,a^2,a^3\}$.
Without loss of generality, we can assume $b=a^2$ and $c=a^3$.
Since $e$ is the identity element of $G$, the Cayley Table for $G$ must be the following:
\begin{array}{c|cccc}
\cdot & e & a & b & c\\
\hline
e & e & a & b & c\\
a & a & \color{red}{?} & \color{red}{?} & \color{red}{?}\\
b & b & \color{red}{?} & \color{red}{?} & \color{red}{?}\\
c & c & \color{red}{?} & \color{red}{?} & \color{red}{?}
\end{array}
$a\cdot a=a^2=b$.
$a\cdot b=a\cdot a^2=a^3=c$.
$a\cdot c=a\cdot a^3=a^4=e$.
$b\cdot a=a^2\cdot a=a^3=c$.
$b\cdot b=a^2\cdot a^2=a^4=e$.
$b\cdot c=a^2\cdot a^3=a^5=a$.
$c\cdot a=a^3\cdot a=a^4=e$.
$c\cdot b=a^3\cdot a^2=a^5=a$.
$c\cdot c=a^3\cdot a^3=a^6=a^2=b$.
So, the Cayley Table for $G$ must be the following:
\begin{array}{c|cccc}
\cdot & e & a & b & c\\
\hline
e & e & a & b & c\\
a & a & b & c & e\\
b & b & c & e & a\\
c & c & e & a & b
\end{array}
Conversely, if the Cayley Table for $(G,\cdot)$ is the above table, then we see that $(G,\cdot)$ is closed under $\cdot$ and $(G,\cdot)$ has the identity element and each element of $(G,\cdot)$ has its inverse.
The Cayley Table for $Z/4Z$ is the following:
\begin{array}{c|cccc}
+ & \overline{0} & \overline{1} & \overline{2} & \overline{3}\\
\hline
\overline{0} & \overline{0} & \overline{1} & \overline{2} & \overline{3}\\
\overline{1} & \overline{1} & \overline{2} & \overline{3} & \overline{0}\\
\overline{2} & \overline{2} & \overline{3} & \overline{0} & \overline{1}\\
\overline{3} & \overline{3} & \overline{0} & \overline{1} & \overline{2}
\end{array}
If we replace $+$ with $\cdot$ and if we replace $\overline{0}$ with $e$ and if we replace $\overline{1}$ with $a$ and if we replace $\overline{2}$ with $b$ and if we replace $\overline{3}$ with $c$, then we get the Cayley Table for $(G,\cdot)$ from the Cayley Table for $Z/4Z$.
And of course, the associativity law holds in $Z/4Z$.
So, the associativity law also holds in $(G,\cdot)$.
So, $(G,\cdot)$ is a group in this case.
Fortunately, we didn't need to check if the associativity law holds in $(G,\cdot)$.
(2) Suppose that there is no element of $G$ whose order is $4$.
Then, $o(a)=o(b)=o(c)=2$ must hold.
Since $e$ is the identity element of $G$, the Cayley Table for $G$ must be the following:
\begin{array}{c|cccc}
\cdot & e & a & b & c\\
\hline
e & e & a & b & c\\
a & a & \color{red}{?} & \color{red}{?} & \color{red}{?}\\
b & b & \color{red}{?} & \color{red}{?} & \color{red}{?}\\
c & c & \color{red}{?} & \color{red}{?} & \color{red}{?}
\end{array}
$a\cdot a=e$.
$b\cdot b=e$.
$c\cdot c=e$.
If $a\cdot b=e$, then $a\cdot a=a\cdot b$.
Since $G$ is a group, the left cancellation law holds in $G$.
So, $a=b$ must hold.
This is a contradiction.
If $a\cdot b=a$, then $a\cdot b=a=a\cdot e$.
Since $G$ is a group, the left cancellation law holds in $G$.
So, $b=e$ must hold.
This is a contradiction.
If $a\cdot b=b$, then $a\cdot b=b=e\cdot b$.
Since $G$ is a group, the right cancellation law holds in $G$.
So, $a=e$ must hold.
This is a contradiction.
So, if $G$ is a group, then $a\cdot b=c$ must hold.
Similarly, $b\cdot a=c$ must hold.
Similarly, $b\cdot c=a$ must hold.
Similarly, $c\cdot b=a$ must hold.
Similarly, $c\cdot a=b$ must hold.
Similarly, $a\cdot c=b$ must hold.
So, the Cayley Table for $G$ must be the following:
\begin{array}{c|cccc}
\cdot & e & a & b & c\\
\hline
e & e & a & b & c\\
a & a & e & c & b\\
b & b & c & e & a\\
c & c & b & a & e
\end{array}
Conversely, if the Cayley Table for $(G,\cdot)$ is the above table, then we see that $(G,\cdot)$ is closed under $\cdot$ and $(G,\cdot)$ has the identity element and each element of $(G,\cdot)$ has its inverse.
Let $H:=\{(),(1\text{ }2)(3\text{ }4),(1\text{ }3)(2\text{ }4),(1\text{ }4)(2\text{ }3)\}$ be a subset of $S_4$.
Then, it is easy to check $H$ is a subgroup of $S_4$ and its Cayley Table is the following:
\begin{array}{c|cccc}
\cdot & () & (1\text{ }2)(3\text{ }4) & (1\text{ }3)(2\text{ }4) & (1\text{ }4)(2\text{ }3)\\
\hline
() & () & (1\text{ }2)(3\text{ }4) & (1\text{ }3)(2\text{ }4) & (1\text{ }4)(2\text{ }3)\\
(1\text{ }2)(3\text{ }4) & (1\text{ }2)(3\text{ }4) & () & (1\text{ }4)(2\text{ }3) & (1\text{ }3)(2\text{ }4)\\
(1\text{ }3)(2\text{ }4) & (1\text{ }3)(2\text{ }4) & (1\text{ }4)(2\text{ }3) & () & (1\text{ }2)(3\text{ }4)\\
(1\text{ }4)(2\text{ }3) & (1\text{ }4)(2\text{ }3) & (1\text{ }3)(2\text{ }4) & (1\text{ }2)(3\text{ }4) & ()
\end{array}
If we replace $()$ with $e$ and if we replace $(1\text{ }2)(3\text{ }4)$ with $a$ and if we replace $(1\text{ }3)(2\text{ }4)$ with $b$ and if we replace $(1\text{ }4)(2\text{ }3)$ with $c$, then we get the Cayley Table for $(G,\cdot)$ from the Cayley Table for $H$.
And of course, the associativity law holds in $H$ since the associativity law holds in $S_4$.
So, the associativity law also holds in $(G,\cdot)$.
So, $(G,\cdot)$ is a group in this case.
Fortunately, we didn't need to check if the associativity law holds in $(G,\cdot)$.
My question is the following:
When we classify groups of order $n$, can we always skip to check if the associativity law holds in $(G,\cdot)$ like the above case in which $n$ is $4$?
| In general one has to check associativity as pointed out in the comment by @ancientmathematician. However, as you have managed to find, there are often quick ways to do this by spotting that the group is simply some well known group.
Moreover, no one is going to ask you to classify groups of order $n$ by completing a Cayley table for anything other than very small $n$. It's just not practicable. So you should always be able to sidestep the issue!
As an example consider the table referenced by @ancientmathematician. This table has order $5$ but the only group of order $5$ is the cyclic group which does not have an element $a$ of order $2$. So there's no need to check associativity - we know it cannot be associative
| {
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"url": "https://math.stackexchange.com/questions/4375477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Show that $\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})$
Question:
Show that, $$\pi =3\arccos(\frac{5}{\sqrt{28}}) +
3\arctan(\frac{\sqrt{3}}{2}) ~~~~~~ (*)$$
My proof method for this question has received mixed responses. Some people say it's fine, others say that it is a verification, instead of a proof.
Proof: $$\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})\iff \frac{\pi}{3} = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2})
$$$$\iff \frac{\pi}{3} = \arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2})$$
As $\arccos(\frac{5}{\sqrt{28}})=\arctan(\frac{\sqrt{3}}{5})$
The plan now is to apply the tangent function to both sides, and show that LHS=RHS using the tangent addition formula to expand it out.
I.e. $$\tan(\frac{\pi}{3}) = \tan\bigg(\arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2}\bigg)$$
$$\iff \sqrt{3} = \frac{\frac{\sqrt{3}}{5}+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{5} \frac{\sqrt{3}}{2}}$$
and the RHS will reduce down to $\sqrt{3}$. Hence LHS=RHS.
Some things that I've noticed about this method of proof:
*
*It could be used to (incorrectly) prove that $$\frac{\pi}{3}+\pi = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2})$$
So because this method of proof can be used to prove things true, that are obviously false, that means it can't be used?
*
*Instead of proving (*), wouldn't this method of proof actually prove that? $$\arccos(\frac{5}{\sqrt{28}})+\arctan(\frac{\sqrt{3}}{2})=\frac{\pi}{3} + \pi k$$
for some $k\in \mathbb{Z}$ which we must find. In this case being when $k=0$.
| To show:
$$\text{arctan}\left(\frac{\sqrt{3}}{5}\right) ~+~ \text{arctan}\left(\frac{\sqrt{3}}{2}\right) ~~<~~ \frac{\pi}{2}.$$
In fact this conclusion is immediate by the following analysis.
Let $~\displaystyle \theta ~~=~~ \text{arctan}\left(\frac{\sqrt{3}}{5}\right) ~+~ \text{arctan}\left(\frac{\sqrt{3}}{2}\right).$
Then,
$$\tan(\theta) =
\frac{\frac{\sqrt{3}}{5} + \frac{\sqrt{3}}{2}}{1 - \left[\frac{\sqrt{3}}{5} \times \frac{\sqrt{3}}{2}\right]}. \tag1 $$
In (1) above, the numerator is clearly positive.
Further, since $~\displaystyle \left[\frac{\sqrt{3}}{5} \times \frac{\sqrt{3}}{2}\right] = \frac{9}{10} < 1$,
the denominator in (1) above is also clearly positive.
Therefore, since $\tan(\theta) > 0$, either $\theta$ is in the 1st quadrant, or $\theta$ is in the 3rd quadrant.
However, by the definition of the arctan function, $\theta$ is the sum of two angles, each of which are between $(0)$ and $(\pi/2)$. Therefore, $\theta$ can not be in the 3rd quadrant. Therefore, $\theta$ must be in the first quadrant.
Edit
Thanks to ryang for pointing out a couple of analytical gaps that need to be filled:
*
*By definition, $\text{arccos}\left(\frac{5}{\sqrt{28}}\right)$ is in the 1st quadrant. Similarly $\text{arctan}\left(\frac{3}{\sqrt{2}}\right)$ is also in the 1st quadrant. Therefore, $\theta$ is in fact the sum of $2$ angles, each of which are in the first quadrant.
*In general $\tan(\theta) = \tan(\pi/3)$ does not imply that $\theta = (\pi/3)$. However, if you add the constraint that $\theta$ is in the 1st quadrant, then the implication holds. Since I deduced that $\theta$ is in the 1st quadrant, the analysis holds.
| {
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"url": "https://math.stackexchange.com/questions/4375994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Limit of a series with interchanging signs Let $S_x=\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}(1-\frac{1}{x^n+1})$. The problem is to find the following limit: $\displaystyle\lim_{x\to1^{-}}S_x$.
I've tried to tackle the problem in several methods so far. Proving that this series is convergent is fairly easy. For evaluating the limit, I've tried bounding the sum and found out that answer is between $0,\frac12$. Here is a sketch of the method:
$$\forall x<1 : S_x=\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}(1-\frac{1}{x^n+1})=\displaystyle\sum_{n=1}^{\infty}\bigg(\frac{1}{x^{2n}+1}-\frac{1}{x^{2n-1}+1}\bigg)$$$$=\displaystyle\sum_{n=1}^{\infty}\frac{x^{2n-1}-x^{2n}}{(x^{2n-1}+1)(x^{2n}+1)}=(1-x)\displaystyle\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(x^{2n-1}+1)(x^{2n}+1)}$$
Hence, $S(x)$ is clearly positive. For the upper bound:
$$\frac{x^{2n-1}}{(x^{2n-1}+1)(x^{2n}+1)}\le x^{2n-1}$$
$$\implies \displaystyle\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(x^{2n-1}+1)(x^{2n}+1)}\le\displaystyle\sum_{n=1}^{\infty}x^{2n-1}=1+x\cdot\displaystyle\sum_{n=1}^{\infty}x^{2n}$$$$=1+x\cdot(\frac{1}{1-x^2}-1)=(1-x)+\frac{x}{1-x^2}$$
$$\implies S_x\le(1-x)\displaystyle\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(x^{2n-1}+1)(x^{2n}+1)}$$$$\le(1-x)\cdot((1-x)+\frac{x}{1-x^2})=(1-x)^2+\frac{x}{1+x}$$
Which approaches $\frac12$ as $x$ gets closer to $1$.
But clearly, this is not enough to evaluate the desired limit. I'm guessing (by the numerical tests) that the answer is zero. Either I shall take another approach, or the bound shall be refined. I couldn't take the problem any further from here.
Any help would be appreciated!
| Write
$$ S_x = \frac{x}{1+x} \sum_{n=1}^{\infty} \underbrace{\frac{x^{2n-2} - x^{2n}}{(1+x^{2n-1})(1 + x^{2n})}}_{=T_{n,x}}. $$
Then we find that
$$ T_{n,x} \geq \frac{x^{2n-2} - x^{2n}}{(1+x^{2n-2})(1 + x^{2n})} = \frac{1}{1+x^{2n}} - \frac{1}{1+x^{2n-2}} $$
and
$$ T_{n,x} \leq \frac{x^{2n-2} - x^{2n}}{(1+x^{2n-1})(1 + x^{2n+1})} = \frac{1}{x}\left( \frac{1}{1+x^{2n+1}} - \frac{1}{1+x^{2n-1}} \right). $$
Summing these for $n = 1, 2, \ldots$, both bounds telescope and hence we obtain
$$ \frac{x}{1+x} \left( 1 - \frac{1}{2} \right) \leq S_x \leq \frac{1}{1+x} \left(1 - \frac{1}{1+x} \right).$$
Therefore, by the squeezing lemma,
$$ \lim_{x \to 1^-} S_x = \frac{1}{4}. $$
More generally, for any $C^1$-function $f$ on $[0, 1]$, we get
$$ \lim_{x \to 1^-} \sum_{n=1}^{\infty} (-1)^{n-1} f(x^n) = \frac{f(1) - f(0)}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4376404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find all pairs (p,q) of real numbers such that whenever $\alpha$ is a root of $x^2 + px+q=0$ $\alpha^2-2$ is also root of the equation.
Find all pairs (p,q) of real numbers such that whenever $\alpha$ is a root of $x^2 + px+q=0$ $\alpha^2-2$ is also root of the equation.
My Approach:
I could not find any elegant method that is why I tried applying quadratic formula: the roots will be $\frac{-p±\sqrt{p^2-4q}}{2}$
Now we have 2 cases:
*
*$$\left(\frac{-p+\sqrt{p^2-4q}}{2}\right) =\left (\frac{-p-\sqrt{p^2-4q}}{2}\right)^2 -2$$
*$$\left(\frac{-p-\sqrt{p^2-4q}}{2}\right)=\left(\frac{-p+\sqrt{p^2-4q}}{2}\right)^2-2$$
It's very tedious to solve these. I searched on Wolfram alpha and got these I II
Is there an elegant way to approach this problem?
| Let $\alpha, \beta$ be roots of this polynomial. If $\alpha=\beta$, then $\alpha^2-2=\alpha$, $\alpha=-1, 2$, and then $(p,q)=(2,1)$ or $(-4, 4)$, both choices work.
Suppose $\alpha\ne \beta$.
If $\alpha^2-2=\beta$, $\beta^2-2=\alpha$, then $(\alpha^2-2)^2-2=\alpha$, $\alpha^4-4\alpha^2-\alpha+2=0$, same for $\beta$. The polynomial $x^4-4x^2-x+2$ is $(x - 2)(x + 1)(x^2+x-1)$, $\alpha,\beta\in\{2, -1\}$ do not work. So $\{\alpha,\beta\}=\{(-1+\sqrt{5})/2, (-1-\sqrt{5})/2\}$. It works, $(p,q)=(1,-1)$.
Finally it could be that $\alpha^2-2=\alpha$ which means $\alpha\in\{-1,2\}$, and $\beta^2-2=\alpha$, so either $\alpha=-1$, $\beta^2-2=-1$, $\beta^2=1$, $\beta=1$ since $\beta\ne \alpha$, $(p,q)=(0,-1)$ or $\alpha=2$, $\beta^2-2=2$, $\beta=-2$, $(p,q)=(0,-4)$
Hence the answer: $(p,q)\in\{(-4,4), (2,1), (1,-1), (0,-1), (0,-4)\}$.
| {
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Evaluate $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2}\arctan{\sqrt{x^2+y^2+z^2}}dxdydz$ I am trying to evaluate this integral:
$$\displaystyle\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2}\arctan{\sqrt{x^2+y^2+z^2}}dxdydz$$
I coverted the region by using polar coordinate system for xy-plane and it became:
$$\displaystyle\int_{0}^{1}\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{1}{\cos\theta}}r\sqrt{r^2+z^2}\arctan{\sqrt{r^2+z^2}}drd\theta dz+\displaystyle\int_{0}^{1}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sin\theta}}r\sqrt{r^2+z^2}\arctan{\sqrt{r^2+z^2}}drd\theta dz$$
After IBP
$$\int r\sqrt{r^2+z^2}\arctan{\sqrt{r^2+z^2}}dr=\frac{1}{3}(r^2+z^2)^{\frac{3}{2}}\arctan{\sqrt{r^2+z^2}}-\frac{1}{6}(r^2+z^2)+\frac{1}{6}\ln(r^2+z^2+1)+C $$ and replace upper and lower limits, it is scary-looking.
My experience in solving this tyle of integral is still weak, I think there is a way to get rid of one variable from the beginning integral, but i can't figure it out. Hope everyone can help me, thanks.
| Very similar steps from (Almost) Impossible Integrals, Sums, and Series, pages 261-261, can be applied where there we have that
$$\int_0^1\left( \int_0^1 \sqrt{x^2+y^2}\arctan\left(\frac{1}{\sqrt{x^2+y^2}}\right)\textrm{d}x\right)\textrm{d}y$$ $$=\int_0^1 \left(\int_0^1 \left(\int_0^1 \frac{x^2+y^2}{x^2+y^2+z^2}\textrm{d}z\right)\textrm{d}x\right)\textrm{d}y.$$
Similarly, we have that
$$\int_0^1 \left( \int_0^1 \left(\int_0^1 \sqrt{x^2+y^2+z^2}\arctan\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)\textrm{d}x\right)\textrm{d}y\right)\textrm{dz}$$
$$=\int_0^1 \left(\int_0^1 \left(\int_0^1\left(\int_0^1 \frac{x^2+y^2+z^2}{x^2+y^2+z^2+w^2}\textrm{d}w\right)\textrm{d}z\right)\textrm{d}x\right)\textrm{d}y,$$
where since the last quadruple integral is immediately done by symmetry, the rest of the work to do is straightforward in various ways.
| {
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$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}$
Evaluate $$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}$$
Wolfram alpha gives $\dfrac{-7}{20}$.
Here is my work
For $x \to 0 $
$\tan x \sim x$
$\sin(x+\tan x) \sim \sin2x$
$\sin2x \sim 2x$ (I wasn't sure about this but I evaluated the limit and got $1$.)
$\sin(x+\tan x) \sim 2x$
$\sin(x+\tan x)-2x\cos x \sim 2x(1-\cos x)$
$(1-\cos x) \sim x$
So I got that
$\dfrac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2} \sim \dfrac{3x}{x(\sin x^2)^2}$
which means I got $\infty$.
Did make a mistake somewhere?
And if so can this problem be solved by this approach or I need to try something else (like l'hopitals or something)?
| You have used the asymptotic relations like $\sin x\sim x$ in an incorrect fashion. The meaning of such relation is $$\lim_{x\to 0} \frac{\sin x} {x} =1$$ It does not mean that you can replace $\sin x$ with $x$ in any limit evaluation as $x\to 0$. I have explained this in detail in this answer of mine.
The simplest approach here is to use Taylor series expansions, but I usually avoid multiplication, division and composition of Taylor series. The approach below uses a little algebraic manipulation before applying Taylor series expansions.
Using L'Hospital's Rule or Taylor series one can establish the following limits
\begin{align}
\lim_{x\to 0}\frac{\sin x-x} {x^3}&=-\frac{1}{6}\tag{1}\\
\lim_{x\to 0}\frac {\tan x-x} {x^3}&=\frac{1}{3}\tag{2}
\end{align}
Replacing $x$ with $\tan x$ in $(1)$ we get $$\lim_{x\to 0} \frac{\sin\tan x - \tan x} {\tan^3x}=-\frac {1}{6}$$ Multiplying this with the cube of $$\lim_{x\to 0}\frac{\tan x} {x} =1$$ we get $$\lim_{x\to 0}\frac{\sin\tan x-\tan x} {x^3}=-\frac{1}{6}$$ Adding the above to equation $(2)$ we get $$\lim_{x\to 0} \frac{\sin\tan x-x}{x^3}=\frac{1}{6}\tag{3}$$ Let $$u=\sin\tan x-x, v=\tan x-x\tag{4}$$ and then we can rewrite the equations $(2),(3)$ as $$u/x^3\to 1/6,v/x^3\to 1/3\tag{5}$$ as $x\to 0$.
Using the identity $$\sin(\tan x+x)+\sin(\tan x-x) =2\sin\tan x\cos x$$ the numerator of the fraction under limit in question can be written as $$2u\cos x-\sin v$$ and the denominator can be replaced by $x^5$ (justify this replacement using the limit $\lim_{x\to 0}(\sin x) /x=1$). Thus we need to evaluate the limit of expression $$\frac{2u\cos x-\sin v} {x^5}$$ The above can be written as $$\frac {2u(\cos x-1)}{x^5}+\frac{2u-v}{x^5}+\frac{v-\sin v} {x^5}\tag {6}$$ The first fraction above can be written as $$\frac{2u}{x^3}\cdot \frac{\cos x-1}{x^2}$$ so that it tends to $$2(1/6)(-1/2)=-1/6$$ via $(5)$ and the last fraction in $(6)$ can be rewritten as $$\frac{v-\sin v} {v^3}\cdot\left(\frac {v} {x^3}\right)^3\cdot x^4$$ which tends to $$(1/6)(1/3)^3\cdot 0^4=0$$ The middle fraction in equation $(6)$ tends to a limit, say $A$, as shown below. And therefore the desired limit is $A-(1/6)$.
The middle fraction in equation $(6)$ is $$\frac{2\sin\tan x-x-\tan x} {x^5}$$ Using the substitution $t=\tan x$ and the limit $$\lim_{t\to 0}\frac{\arctan t} {t} =1$$ the above is reduced to $$\frac{2\sin t-t-\arctan t} {t^5}$$ and using Taylor expansions we see that it tends to $A=-11/60$. The desired limit is then $A-(1/6)=-7/20$.
| {
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In how many ways we can place 9 different balls in 3 different boxes such that in every box at least 2 balls are placed?
In how many ways we can place $9$ different balls in $3$ different boxes such that in every box at least $2$ balls are placed?
Approach 1:
Let $x_1$, $x_2$, $x_3$ denote the number of balls in boxes $1$,$2$, and $3$ respectively. Now,solving $x_1+x_2+x_3 =9$ for $2≤x_1≤x_2≤x_3≤5$ ($5= 9-$(min of $x_1$) $-$ (min of $x_2$)), we get $(x_1,x_2,x_3) ≡ (2,2,5),(2,3,4),(3,3,3)$
Now, we have to divide $9$ objects in these groups and then distribute into boxes. So the required answer is $$\frac{9!}{2!2!5!}×\frac{1}{2!}×3! +\frac{9!}{2!3!4!}×3! + \frac{9!}{3!3!3!}×\frac{1}{3!}×3!$$ which simplifies to be $\boxed{11508}$. This is the correct answer.
Approach 2 (PIE):
At least $2$ $=$ all $-$ at most $1$. Now, considering $1$ box contains no ($0$) elements, the number of cases corresponding to that is ${3 \choose 1}×2^9$ and the number of cases when $2$ boxes contain no elements will be ${3 \choose 2}×1^9$. Now, number of cases when $1$ box contains only $1$ element is ${3 \choose 1}×{9 \choose 1}×2^8$ (I.e., number of ways to choose 1 box × number of ways to choose 1 ball × remaining distribution) and the number of cases when 2 boxes contain only 1 element is ${3 \choose 2}×{9 \choose 2}×1^7$ . So, using PIE, the required answer should be $$3^9 -({3 \choose 1}2^9 - {3 \choose 2}1^9) - ({3 \choose 1}{9 \choose 1}2^8 - {3 \choose 2}{9 \choose 2}1^7)$$ that is $11346$. This answer is wrong but the "closeness" of the answer to the correct one indicates that I have deducted some cases. Please point out my mistake in approach 2.
| $\mathbf{\text{For Alternative Approach:}}$
Distinct balls into distinct boxes with restriction , then it is good time to use exponential generating functions , if each boxes will have at least two objects , then the e.g.f. of each boxes is equal to $$\bigg(\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}\bigg)$$
We restricted it with $x^5$ , because if each boxes will have at least $2$ ball , then a box can have at most $5$ balls , then find the coefficient of $x^9$ in the expansion of $$\bigg(\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}\bigg)^3$$ and multiply it by $9!$.
Then , $$9! \times \frac{137}{4320}= 11,508$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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show this equation $3x^4-y^2=3$ has no integer solution show this diophantine equation
$$3x^4-y^2=3$$ has no integer $(y\neq 0)$ solution?
My try: WLOG Assmue $(x,y)$ is postive integer solution,then $3|y$,let $y=3y'$,then we have
$$x^4-1=3y'^2\tag{1}$$
and following I want $\pmod5$,since $x^4\equiv 0,1\pmod 5$
(1):if $x^4\equiv 0\pmod 5$or $5|x$,then $(1)$ it is clear no solution
(2)if $(x,5)=1$,then $x^4-1\equiv 0\pmod 5$,there also exsit $y'$ such $5|3y'^2$,so How to solve my problem? Thanks
| $(1)$ WLOG we study the impossibility of $A^4-3B^2=1$ with $B\ne0$. Clearly $(A,B)=1$ and by modulo $5$ and modulo $16$ we have $A$ odd and $10|B$.
$(2)$ We need only consider non-negative solutions. The fundamental unit of the ring $\mathbb Z[\sqrt3]$ is $2+\sqrt3$ so all (positive) solution $(x_n,y_n)$ of the equation $X^2-3Y^2=1$ is given by $x_n+y_n\sqrt3=(2+\sqrt3)^n$.
$(3)$ Looking modulo $10$ at the $(x_n,y_n)$ we find that there is a periodicity of order 6. In fact we have
$$\begin{cases}x_1+y_1\sqrt3=2+\sqrt3\\x_2+y_2\sqrt3=7+4\sqrt3\\x_3+y_3\sqrt3=26+15\sqrt3\\x_4+y_4\sqrt3=97+56\sqrt3\\x_5+y_5\sqrt3=362+209\sqrt3\\x_6+y_6\sqrt3=1351+780\sqrt3\end{cases}\Rightarrow\begin{cases}(x_1,y_1)\equiv(2,1)\\(x_2,y_2)\equiv(7,4)\\(x_3,y_3)\equiv(6,5)\\(x_4,y_4)\equiv(7,6)\\(x_5,y_5)\equiv(2,9)\\(x_6,y_6)\equiv(1,0)\end{cases}$$ from which we have $$(x_{6k+h},y_{6k+h})\equiv(x_h,y_h)\pmod{10}$$
$(4)$ Consequently since $A^4-3B^2=1$ and $B\equiv0\pmod{10}$ we must have $$A^2+B\sqrt3=(2+\sqrt3)^{6n}$$ Consider $a_{2n}+b_{2n}\sqrt3=(2+\sqrt3)^{2n}$ so we have
$$A^2+B\sqrt3=(a_{2n}+b_{2n}\sqrt3)^3\Rightarrow A^2=a_{2n}^3+9a_{2n}b_{2n}^2=a_{2n}(a_{2n}^2+(3b_{2n})^2)$$.
$(5)$ Since $a_{2n}^2-3b_{2n}^2=1$ one has $(a_{2n},b_{2n})=1$ and because $3$ can not divide $a_{2n}$ we deduce the factors $a_{2n}$ and $a_{2n}^2+(3b_{2n})^2$ of $A^2$ are coprime therefore
$a_{2n}=\alpha_{2n}^2$ for some positive integer $\alpha$ (less than $a_{2n}$ which is distinct of $1$ because the smallest possible positive value of $a_{2n}$ is $2$) whereby we get $$a_{2n}^2-3b_{2n}^2=1\Rightarrow \alpha_{2n}^4-3b_{2n}^2=1$$
Given this last equation, it is clearly seen that we can apply Fermat's method of infinite descent. We are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof for combinatorial Identity How to prove the following combinatorial identity?
$$\sum_{k=2}^n {k+1\choose3}{2n-2-k\choose n-2}2^k= \frac{1}{3}\cdot\frac{(2n)!}{(n-2)!n!}$$
My attempt:
The LHS is equal to the coefficient of $x^{n-2}$ in $\frac{1}{(1-x)^4\cdot(1-2x)^{n-1}}$. But this doesn't help. I don't see any other approaches right now to tackle this problem.
This is an identity I need to prove to solve another problem, which goes as follows:
Select $n$ intervals uniformly at random from the range $\left[0,1\right]$. Show that the probability that at least one interval intersects every other interval, is equal to $\frac{2}{3}$. The jump from this problem to this identity is non-trivial, but this is basically the background of the problem.
| We seek to prove that
$$\sum_{k=2}^n {k+1\choose 3} {2n-2-k\choose n-2} 2^k
= \frac{1}{3} \frac{(2n)!}{(n-2)! \times n!}.$$
The LHS is
$$4\sum_{k=0}^{n-2} {k+3\choose 3} {2n-4-k\choose n-2} 2^k
= 4\sum_{k=0}^{n-2}
{k+3\choose 3} {2n-4-k\choose n-2-k} 2^k.$$
Writing
$$4 \sum_{k=0}^{n-2} 2^k [w^k] \frac{1}{(1-w)^4} [w^{n-2-k}]
\frac{1}{(1-w)^{n-1}} = 4 [w^{n-2}] \frac{1}{(1-2w)^4}
\frac{1}{(1-w)^{n-1}}$$
we find
$$4 \;\underset{w}{\mathrm{res}} \;
\frac{1}{w^{n-1}} \frac{1}{(1-w)^{n-1}}
\frac{1}{(1-2w)^4}.$$
Next we put $w=(1-\sqrt{1-4v})/2$ so that $w(1-w)=v$ and $dw =
1/\sqrt{1-4v} \; dv$ to get
$$4 \;\underset{v}{\mathrm{res}} \;
\frac{1}{v^{n-1}}
\frac{1}{\sqrt{1-4v}^4} \frac{1}{\sqrt{1-4v}}
= 4 \;\underset{v}{\mathrm{res}} \;
\frac{1}{v^{n-1}}
\frac{1}{(1-4v)^{5/2}}.$$
Extracting the coefficient we find
$$4 [v^{n-2}] (1-4v)^{-5/2} = 4^{n-1} (-1)^n {-5/2\choose n-2}
\\ = 4^{n-1} (-1)^n \frac{n(n-1)}{(-1/2)\times(-3/2)}
{-1/2\choose n}
\\ = 4^n (-1)^n \frac{1}{3} n(n-1)
[z^n] \frac{1}{\sqrt{1+z}}
= \frac{1}{3} n(n-1) [z^n] \frac{1}{\sqrt{1-4z}}
\\ = \frac{1}{3} n(n-1) {2n\choose n}
= \frac{1}{3} \frac{(2n)!}{(n-2)! \times n!}.$$
This is the claim.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Which solution to $\int \frac{x^3}{(x^2+1)^2}dx$ is correct? I tried solving the following integral using integral by parts :
$$\int \frac{x^3}{(x^2+1)^2}dx$$
but I got a different answer from Wolfram Calculator This is the answer that I got :
$$\int \frac{x^3}{(x^2+1)^2}dx=\frac{1}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C$$
I am wondering if I am wrong or the calculator if I am where did I do something wrong.
Here is my solution :
$$\int \frac{x^3}{(x^2+1)^2}dx$$
$$\int x^2\frac{x}{(x^2+1)^2}dx$$
$\implies u =x^2 \qquad u'=2x$
$\displaystyle\implies v'=\frac{x}{(x^2+1)^2}\qquad v=\int\frac{x}{(x^2+1)^2}dx$
$$\int\frac{x}{(x^2+1)^2}dx$$
$\implies \zeta=x^2+1$
$\implies\displaystyle \frac{d\zeta}{2}=x$
$$\boxed{v=\frac{1}{2}\int\frac{d\zeta}{\zeta^2}=-0.5\zeta^{-1}=\frac{-1}{2(x^2+1)}}$$
$$\int x^2\frac{x}{(x^2+1)^2}dx=x^2\frac{-1}{2(x^2+1)}-\underbrace{(\int \frac{-2x}{2(x^2+1)}dx)}_{-\frac{1}{2}\ln(x^2+1)}$$
$$\boxed{\int \frac{x^3}{(x^2+1)^2}dx=\frac{-x^2}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C}$$
| Both are correct. $\frac{-x^2}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C=\frac 1{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C'$ where $C'=C-\frac 1 2$
[Just transfer $\frac 1{2(x^2+1)}$ to the left side and combine it with $\frac{-x^2}{2(x^2+1)}$].
| {
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Proving segments equal $(O)$ is internally tangent to a circle $(O')$ at $F$ i.e $(O')$ is inside $(O)$. Let $CE$ be a chord of $(O)$ which touches $(O')$ at $D$. Let $CO$ meet $(O)$ again at $A$ and $CO'$ meet $AE$ at $B$. It is given that $OO' \perp AC$. Prove that $AB=CD$.
From working backwards and using power of point arguments and a bit of calculation,our goal is to prove that $AB^2=AC\times OO'$ which after some trigonometry implies $\sin^2\alpha=\sin 2\theta$ where $\alpha=\angle ABC$ and $\theta=\angle ACB$,but I can't proceed forward from here.
Also a synthetic or projective solution will be much appreciated.
| Here's a "horrible" proof using algebraic manipulations.
Set things in coordinate plane as in the picture. That is, $O$ is the unit circle centered at origin and center of $O'$ lies on the $y$-axis. Let the circle $O'$ have radius $r$. Denote the x-coordinates of points $D, E, B$ as $x_1, x_2, x_3$, respectively and let the line through $C$ and $D$ have slope $k$.
Let's find useful equations for the quantities. Using the fact that $D$ is a tangent point, we have $k = \frac{x_1}{\sqrt{r^2-x_1^2}}$ and $\frac{x_1}{\sqrt{r^2-x_1^2}}(x_1+1) = 1-r-\sqrt{r^2-x_1^2}$. These can be manipulated to
$$\begin{align*}(x_1+r^2)^2 &= (1-r)^2(r^2-x_1^2) \tag{1} \\
k &= \frac{x_1(1-r)}{x_1+r^2} \end{align*}
$$
Then solve $x_2$ from ($E$ on circle $O$ and the $y$-coord comes from the line equation)
$$x_2^2+k^2(x_2+1)^2=1$$
by factoring $k^2(x_2+1)^2=1-x_2^2 = (1-x_2)(1+x_2)$ and cancelling to get
$$x_2 = \frac{1-k^2}{1+k^2}.$$
Then for $x_3$, we need to solve an intersection of two lines. There's some nice cancelling and we get
$$x_3 = \frac{1-k(1-r)}{1+k(1-r)}.$$
Now we know the coordinates of the points $D$ and $B$ in terms of $r$ and $x_1$. And we have equation (1) to help us prove the equality of the lengths in question. Let's write out their squares
$$d(C,D)^2 = (1+x_1)^2 + k^2(x_1+1)^2 = (1+x_1)^2(1+k^2)$$
$$d(A,B)^2 = (1-x_3)^2 + ((1-r)(x_3+1))^2 = \left(\frac{2(1-r)}{1+k(1-r)}\right)^2(1+k^2).$$
So, it suffices to prove that
$$\frac{2(1-r)}{1+k(1-r)} = 1+x_1.$$
Plugging in $k=\frac{x_1(1-r)}{x_1+r^2}$ and using $(1-r)^2 = \frac{x_1+r^2}{r^2-x_1^2}$ we turn the LHS to $\frac{2(1-r)(r^2-x_1^2)}{r^2(1+x_1)}$.
Finally, multiplying by the denominator, it suffices to prove
$$2(1-r)(r^2-x_1^2) - r^2(1+x)^2 = 0,$$
which when multiplied out, can be seen to be equivalent to equation (1).
| {
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"timestamp": "2023-03-29T00:00:00",
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Union of sets having $\binom{n - 1}{2}$ elements with each two of them having $n - 2$ common elements has cardinality of at least $\binom{n}{3}$.
Let $S_1, S_2, \dots, S_n$ sets that have each of them $n - 1 \choose 2$ elements, with $n - 2$ common elements for each two of them. Prove that their union has at least $n \choose 3$ elements. Find an example for equality case.
Let us examine the simplest case $n = 3$. Each of them has $2 \choose 2$ $= 1$ elements and each two of them have $1$ element in common, so, in fact, the three sets are identical, so their union has exactly $3 \choose 3$ $= 1$. (Equality case confirmed)
Examining $n = 4$, each set should have $3 \choose 2$ $= 3$ elements and $2$ in common every $2$. We may construct $\{a, b, c\}$, $\{a, b, d\}$, and the last set may be $\{a, b, e\}$ or $\{a, c, d\}$, both of them satsifying the conditions.
However, I am unable to generalize the problem
| Here is an alternative approach based on @BillyJoe's formulation. Let $x_j$ be the number of elements that appear in exactly $j$ sets. Then the two conditions in the problem imply the following two linear equality constraints:
\begin{align}
\sum_{j=1}^n j x_j &= n\binom{n-1}{2} = 3\binom{n}{3} \tag1 \\
\sum_{j=1}^n \binom{j}{2} x_j &= (n-2)\binom{n}{2} = 3\binom{n}{3} \tag2
\end{align}
The total number of distinct elements is $\sum_{j=1}^n x_j$, which we want to show is at least $\binom{n}{3}$.
First note that $(j-2)(j-3) \ge 0$, which is true for all integer $j$, is equivalent to
$$1\ge \frac{2}{3}\cdot j-\frac{1}{3}\binom{j}{2}.\tag3$$
Now multiply $(1)$ by $2/3$ and $(2)$ by $-1/3$ and add them up to obtain
$$\sum_{j=1}^n x_j \stackrel{(3)}{\ge}
\frac{2}{3}\sum_{j=1}^n j x_j
-\frac{1}{3}\sum_{j=1}^n \binom{j}{2} x_j
\stackrel{(1),(2)}{=} \frac{2}{3}\cdot3\binom{n}{3}
-\frac{1}{3}\cdot3\binom{n}{3}
= \binom{n}{3}.
$$
This lower bound is attained by taking $x_3=\binom{n}{3}$ and all other $x_j=0$.
| {
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How do I finish proving this inequality?
Given $a, b,c >0$ satisfying $abc=1$. Prove that: $\dfrac{1}{{{a^2} + 2{b^2} + 3}} + \dfrac{1}{{{b^2} + 2{c^2} + 3}} + \dfrac{1}{{{c^2} + 2{a^2} + 3}} \le \dfrac{1}{2}$?
This is my try:
Using the AM–GM inequality, I get: $a^2+b^2 \ge 2ab$ and $b^2+1 \ge 2b$
Therefore, $a^2+2b^2+1 \ge 2ab+2b$ which implies that: $a^2+2b^2+3 \ge 2(ab+b+1)$
Hence, $\dfrac{1}{a^2+2b^2+3} \le \dfrac{1}{2(ab+b+1)}$
Doing the same, I get $\dfrac{1}{b^2+2c+3} \le \dfrac{1}{2(bc+c+1)}$ and $\dfrac{1}{c^2+2a+3} \le \dfrac{1}{2(ca+a+1)}$
Thus,
$\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2(ab+b+1)}+\dfrac{1}{2(bc+c+1)}+\dfrac{1}{2(ca+a+1)}$
I know this is really close to the answer but something messed with my brain and I cannot think any further. How do I continue and is there any better solution than this?
| You've done all the hard work! Now you only need to prove $$\frac{1}2\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\leq\frac{1}2$$
HINT: utilize $abc=1$.
$$\frac{1}{ab+b+1}=\frac{c}{abc+bc+c}=\frac{c}{1+bc+c}$$
$$\frac{1}{ca+a+1}=\frac{bc}{abc^2+abc+bc}=\frac{bc}{c+1+bc}$$
Therefore, $$\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why this Taylor expansion of $ \dfrac{1}{1+\cos x}$ is wrong? I want to expand the function $$ \dfrac{1}{1+\cos x} $$ at $x = 0$
The way I did this is first regard $\cos x $ as a whole and denote it by $y$
The expansion will become $$ \dfrac{1}{1+y}=1-y+y^2+o(y^2)$$
and then return $y=\cos x$ back and I want to expand each $\cos x$ at $ x=0$
Since $$ \cos x = 1-\dfrac{x^2}{2} + o(x^2) $$
$ o(y^2) = o(\big(1-\dfrac{x^2}{2} + o(x^2)\big)^2)=o(x^2)$
and the result is $$ \dfrac{1}{1+\cos x}=1-1+\dfrac{x^2}{2}+1-x^2+o(x^2)=1-\dfrac{x^2}{2}+o(x^2)$$
but this is wrong
Thus, my question is why this is wrong.
Also by the same way, I can correctly get the expansion of $ \tan(\tan x)$ and $\sin(\sin x)$
For example, $ \tan(\tan x)$. I want to expand this at $ x=0$
I regard the inner $\tan x$ as $y$, so the expansion will be $$ \tan(y)=y+\dfrac{1}{3}y^3+o(y^3)$$
Then by some calculations, I can get $ o(y^3)= o(x^3),\ \ y=\tan x=x+\dfrac{1}{3}x^3+o(x^3), \ \ y^3=\tan^3 x=(x+o(x))^3 =x^3+o(x^3)$
Then the result is $$ x+\dfrac{2}{3}x^3+o(x^3)$$ which is right
| This is a good method, and it will work if applied correctly! As Jean-Claude is pointing out in the comments, what you need to do is expand $\frac{1}{1 + y}$ centered at $y = 1$, because $\cos 0 = 1$.
In other words, you need to use the Taylor series in terms $a_0 + a_1 (y - 1) + a_2 (y-1)^2 + \cdots$. Currently you are using the Taylor series at $y = 0$, which is not valid at $y = 1$. It converges with radius $1$, so it's valid only in the interval $(-1, 1)$.
This may be easier to think about if you let $z = 1 - y = 1 - \cos x$. This way, all of the Taylor series stay centered at zero because when $x = 0$, $z = 0$ also. Then we need to expand as follows:
\begin{align*}
\frac{1}{1 + \cos x}
&= \frac{1}{2 - (1 - \cos x)} \\
&= \frac{1}{2 - z} \\
&= \frac{\tfrac12}{{1 - \tfrac12z}} \\
&= \frac12 + \frac14 z + \frac18 z^2 + \frac{1}{16} z^3 + \cdots
\end{align*}
Now, as you note, $\cos x = 1 - \frac{x^2}{2} + o(x^2)$, so $z = 1 - \cos x = \frac12 x^2 + o(x^2)$, and the above becomes
$$
\frac12 + \frac14 \left( \frac12 x^2 + o(x^2) \right) + o(x^4) = \boxed{\frac12 + \frac18 x^2 + o(x^2).}
$$
| {
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Prove : $\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$ Let $a,b,c>0$ satisfy $abc=1$, prove that:
$$\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$$
My attempt:
Let $a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}$, we have $xyz=1$ and using $abc=1$, the inequality can be written as:
$$\dfrac{x}{\sqrt{x+y}}+\dfrac{y}{\sqrt{y+z}}+\dfrac{z}{\sqrt{z+x}}\le \dfrac{xy+yz+zx}{\sqrt{2}}$$
I'm trying to use Cauchy-Schwarz:
$$LHS\le\sqrt{(x+y+z)(\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x})}$$ but now I have to prove $$\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}\le\dfrac{3}{2}$$ because $ab+bc+ca\ge\sqrt{3(a+b+c)}$, but I can't prove it. Can anyone give me a hint? Not necessarily a complete solution.
By the way, I also relized a problem that seems quite similar to the above problem $\sqrt{\frac{2 x}{x+y}}+\sqrt{\frac{2 y}{y+z}}+\sqrt{\frac{2 z}{z+x}} \leq 3$ if $x,y,z>0$ (Vasile Cirtoaje) (and then we can use $3\le xy+yz+zx$ ?Hope it helps)
| As an aside, reversing (or repeating) the change of variables further simplifies the work.
Starting off similar to OP's / River Li's work, we WTS
$$\sqrt{ ( x+y+z) (\frac{x}{x+y} + \frac{y}{y+z} + \frac{z}{z+x}) } \leq \frac{ xy+yz+zx} { \sqrt{2} }. $$
This is equivalent to
$$ \sum x + \sum \frac{xz}{x+y} \leq \frac{ (xy+yz+zx)^2 } { 2}.$$
We revert the change of variables, letting $ x = \frac{1}{a}$ and using $abc = 1$. We WTS
$$ \sum bc + \sum \frac{ ab^2}{a+b} \leq \frac{ (a+b+c)^2}{2} \Leftrightarrow \sum \frac{ab^2}{a+b} \leq \frac{ a^2+b^2+c^2}{2}. $$
This is true because
$$ \frac{ ab^2}{a+b} \leq \frac{ b^2+ab}{4} \Rightarrow \sum \frac{ab^2}{a+b} \leq \sum \frac{b^2+ab}{4} \leq \sum \frac{ a^2}{2}.$$
Notes
*
*And of course, for those who don't want to substitute twice, you can work in just $a, b, c$. However, the steps seem "less obvious".
| {
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How do you evaluate: $\int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$ I want to find the value of
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$
At first, I solved this elementary integral:
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}} \ \mathrm dx$
Using the same method, I couldn't find my asked integral. Are there any ways to connect them? Any help would be appreciated.
| $$ I = \displaystyle \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx = \frac{1}{2}\int_{0}^{\infty} \frac{\ln(x)}{\cosh x +\frac{1}{2}} dx $$
Consider the "discrete" Laplace transform
$$ \sum_{k=1}^{\infty}e^{-kt}\sin(kz) = \frac{1}{2}\frac{\sin z}{\cosh t -\cos z} \quad t>0 $$
If we put $z = \frac{2\pi}{3}$
$$ \sum_{k=1}^{\infty}e^{-kt}\sin\left(\frac{2\pi k}{3}\right) = \frac{1}{2}\frac{\frac{\sqrt{3}}{2}}{\cosh t + \frac{1}{2}} $$
Therefore
\begin{align*} I =& \frac{2}{\sqrt{3}}\int_{0}^{\infty} \frac{1}{2}\frac{\frac{\sqrt{3}}{2}}{\cosh x +\frac{1}{2}} \ln(x) dx\\
=& \frac{2}{\sqrt{3}}\int_{0}^{\infty} \sum_{k=1}^{\infty}e^{-kx} \sin\left(\frac{2\pi k}{3}\right)\ln(x) dx\\
=& \frac{2}{\sqrt{3}}\sum_{k=1}^{\infty}\sin\left(\frac{2\pi k}{3}\right)\int_{0}^{\infty} e^{-kx} \ln(x) dx
\end{align*}
This last integral is the Laplace transform of the $\ln(x)$ function
$$ \mathcal{L}\left\{\ln(x) \right\} = \int_{0}^{\infty} e^{-st} \ln(t)dt = -\frac{\ln(s)+\gamma}{s} \quad \Re(s)> 0 $$
where $\gamma$ is the Euler constant.
Hence
\begin{align*}
I =& \frac{2}{\sqrt{3}}\sum_{k=1}^{\infty}\sin\left(\frac{2\pi k}{3}\right)\int_{0}^{\infty} e^{-kx} \ln(x) dx \\
=& -\frac{2}{\sqrt{3}}\sum_{k=1}^{\infty}\frac{\sin\left(\frac{2\pi k}{3}\right)\left(\ln(k)+\gamma\right)}{k}\\
=& -\frac{2}{\sqrt{3}}\underbrace{\sum_{k=1}^{\infty}\frac{\sin\left(\frac{2\pi k}{3}\right)\ln(k)}{k}}_{S_{1}} - \frac{2\gamma}{\sqrt{3}}\underbrace{\sum_{k=1}^{\infty}\frac{\sin\left(\frac{2\pi k}{3}\right)}{k}}_{S_{2}}
\end{align*}
Recall the Kummer's Fourier series for the log-gamma function:
$$ \ln \Gamma(t) = \frac{1}{2}\ln \left(\frac{\pi}{\sin(\pi t)} \right)+ \left(\gamma + \ln(2\pi) \right)\left(\frac{1}{2}-t\right) + \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{\sin(2\pi k t) \ln(k) }{k} $$
If we put $\displaystyle t=\frac{1}{3}$
$$ S_{1}= \sum_{k=1}^{\infty} \frac{\sin(\frac{2\pi k}{3}) \ln(k) }{k} = \pi\ln\Gamma\left(\frac{1}{3}\right) -\frac{\pi}{2}\ln\left(\frac{2\pi}{\sqrt{3}}\right)-\frac{\pi\ln(2\pi)}{6} -\frac{\gamma\pi}{6} $$
For $S_{2}$ take the principal branch of the $\ln(z)$ function
$$ S_{2} = \sum_{k=1}^{\infty}\frac{\sin\left(\frac{2\pi k}{3}\right)}{k} = \Im\left(\sum_{k=1}^{\infty} \frac{e^{\frac{2\pi ik}{3}}}{k} \right) = \Im\left(-\ln(1-e^{\frac{2\pi i}{3}})\right) = \Im \left(-\frac{\ln(3)}{2}+\frac{i\pi}{6}\right) = \frac{\pi}{6}$$
Therefore
$$I = -\frac{2}{\sqrt{3}}S_{1}-\frac{2\gamma}{\sqrt{3}}S_{2} =-\frac{2\pi}{\sqrt{3}}\ln\Gamma\left(\frac{1}{3}\right)+\frac{\pi}{\sqrt{3}}\ln\left(\frac{2\pi}{\sqrt{3}}\right)+\frac{\pi}{3\sqrt{3}}\ln(2\pi) $$
Therefore, we can conclude
$$\boxed{ \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx = \frac{\pi}{\sqrt{3}}\ln\left(\frac{2\pi}{\sqrt{3}}\right)+\frac{\pi}{3\sqrt{3}}\ln(2\pi)-\frac{2\pi}{\sqrt{3}}\ln\Gamma\left(\frac{1}{3}\right)} $$
| {
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Does FOSD + log concavity of $f(x)$ and $g(x)$ imply MLRP? I am looking for a result on the ordering of distribution functions.
The probability density functions $f(x)$ and $g(x)$ bear the Monotone Likelihood Ratio Property (MLRP) if
$$ \frac{f(x)}{g(x)} $$
is increasing in $x$.
By First Order Stochastic Dominance (FOSD) of the probability distribution function $F(x)$ over $G(x)$, it follows that
$$ F(x) \leq G(x) $$
for all $x$, with strict inequality for some $x$.
It is well known that the MLRP implies First Order Stochastic Dominance (FOSD) but not vice versa.
I am interested in sufficient conditions on $f(x)$ and $g(x)$ that, together with $F(x)$ FOSD $G(x)$ imply the MLRP. It seems to me that assuming that both $f(x)$ and $g(x)$ are log-concave should be sufficient. It holds in the following example:
Consider $g(x) = f(x + a)$, where $a>0$. I.e., $g$ is a version of $f$ shifted to the left. FOSD obviously holds. In this case, MLRP also holds because
$$ \frac{\partial\frac{f(x)}{g(x)}}{\partial x} > 0 \Leftrightarrow \frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)} > 0\Leftrightarrow \frac{f'(x)}{f(x)} - \frac{f'(x+a)}{f(x+a)} >0.$$
The last condition holds by log-concavity, as $\frac{f'(x)}{f(x)}$ is decreasing in $x$.
I'm not so sure how to generalize from this to other cases where $f$ and $g$ are both log-concave but where $g$ might not be a simple left shift of $f$.
Has anyone got ideas how to show this or approach this problem?
| A counterexample:
Let
\begin{align*}
f(x) &= \frac{2x^2 + 2x + 4}{5\sqrt \pi}\mathrm{e}^{-x^2}, \\
g(x) &= \frac{2x^2 + x + 4}{5\sqrt \pi}\mathrm{e}^{-x^2}.
\end{align*}
We have $f, g > 0$ for all $x\in (-\infty, \infty)$. We have, for all $x \in (-\infty, \infty)$,
\begin{align*}
(\ln f)'' &= - \frac{2x^4 + 4x^3 + 12x^2 + 10x + 5}{x^2+x+2} < 0,\\
(\ln g)'' &= - \frac{8x^4 + 8x^3 + 42x^2 + 20x + 17}{2x^2+x+4} < 0.
\end{align*}
Let
\begin{align*}
F(x) &= \int_{-\infty}^x f(t) \mathrm{d} t = \frac12 \mathrm{erf}(x) + \frac12 - \frac{2x + 2}{10\sqrt \pi}\mathrm{e}^{-x^2}, \\
G(x) &= \int_{-\infty}^x g(t) \mathrm{d} t = \frac12 \mathrm{erf}(x) + \frac12 - \frac{2x + 1}{10\sqrt \pi}\mathrm{e}^{-x^2}.
\end{align*}
Clearly, $F(x) < G(x)$ for all $x \in (-\infty, \infty)$.
Also, $\lim_{x\to \infty} F(x) = 1$, $\lim_{x\to \infty} G(x) = 1$, $\lim_{x\to -\infty} F(x) = 0$, and $\lim_{x\to -\infty} G(x) = 0$.
However, we have
$$\frac{f(x)}{g(x)} = \frac{2x^2 + 2x + 4}{2x^2 + x + 4}$$
and
$$\left(\frac{f(x)}{g(x)}\right)'
= \frac{4 - 2x^2}{(2x^2 + x + 4)^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the slope of line intersecting the parabola A line $y=mx+c$ intersects the parabola $y=x^2$ at points $A$ and $B$. The line $AB$ intersects the $y$-axis at point $P$. If $AP−BP=1$, then find $m^2$. where $m > 0$.
so far I know $x^2−mx−c=0,$ and $P=(0,c)$.
$x = \frac{m \pm \sqrt{m^2 + 4c}}{2}$
$A_x = \frac{m + \sqrt{m^2 + 4c}}{2}$, $B_x = \frac{m - \sqrt{m^2 + 4c}}{2} $
$A_y = \frac{m^2 + m\sqrt{m^2 + 4c}}{2} + c$, $B_y = \frac{m^2 - m\sqrt{m^2 + 4c}}{2} + c$
using distance formula(not showing all steps)
$AP = \frac{m + \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$BP = \frac{m - \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$AP - BP = 1$
$(\sqrt{m^2 + 4c})(\sqrt{m^2 + 1}) = 1$
$m^4 + m^2(4c + 1) + 4c - 1 = 0$
well I could manipulate this into quadratic but that doesn't really help me with coefficient with c.
| There is a much simpler solution. At the intersection of $y = mx + c$ and $y = x^2$,
$$x^2 - mx - c = 0$$
We are given $|AP - PB| = 1$ and I am taking the case where $P$ is interior to segment $AB$. Just for completeness sakes, if $P$ is exterior to segment $AB$, then we still have $|AP - PB| = 1$ if we consider $AP$ and $PB$ as signed distances or we can say $|AP| + |PB| = 1$. Coming back to the case where $P$ is interior to $AB$, if the roots are $x = \alpha, \beta$,
$ |\alpha + \beta| = |m|$.
But as $\alpha$ and $\beta$ have opposite signs and they are x-coordinates of points $A$ and $B$, $~|\alpha + \beta|~$ is the absolute difference of the horizontal projections of $AP$ and $PB$, which can also be written as $AP |\cos \theta|$ and $PB |\cos \theta|$, where $\tan \theta = m$ is the slope of the line.
$$ |\cos\theta| = \frac{1}{\sqrt{1+m^2}}$$
$$|AP - PB| \cdot |\cos \theta| = |\alpha + \beta| = |m|$$
As $|AP - PB| = 1$,
$$m \sqrt{m^2 + 1} = 1 \implies (m^2 + \frac 12)^2 = \frac 54$$
We get two real solutions for $m$,
$$m = \pm \sqrt{\frac{\sqrt 5 - 1}{2}}$$
You can draw a few lines with equation $$y = \pm \sqrt{\frac{\sqrt 5 - 1}{2}} ~x + c$$ for different values of $c$ with $c \gt 0$ to confirm $|AP - PB| = 1$
As a side note, for values of $c \lt 0$ as long as the line intersects the parabola at two points, $|AP| + |PB| = 1$
| {
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We have three players A, B and C. They roll an honest die, first A, then B, and lastly C until someone is the first to get an even number...
We have three players A, B and C. They roll an honest die, first A, then B, and lastly C until someone is the first to get an even number. This player will be the winner of the game. If after all 3 players have played no one gets the desired result, they start rolling the dice again in the same order until they get it. What is the probability that A wins the game?
Probability of getting even is 0.5 = p of getting odd is 0.5 = 1-p
A can win in the first round so p=0.5
A can win in the second round so we have $1-p \cdot 1-p\cdot 1-p \cdot p = (1-p)^3 \cdot p$
A can win in the third round so $1-p \cdot 1-p\cdot 1-p \cdot 1-p \cdot 1-p \cdot 1-p \cdot p = (1-p)^6 \cdot p$
A can win in the nth round $(1-p)^{3(n-1)} \cdot p$
$0.5+ (1-p)^3 \cdot p + (1-p)^6 \cdot p + .... +(1-p)^{3(n-1)} \cdot p = 0.57$
| Your answer is correct to the number of decimal places to which you rounded. The probability that player $A$ wins is
\begin{align*}
\sum_{k = 1}^{\infty} \frac{1}{2}\left(1 - \frac{1}{2}\right)^{3(k - 1)} & = \frac{1}{2}\sum_{k = 1}^{\infty} \left(\frac{1}{2}\right)^{3(k - 1)}\\
& = \frac{1}{2}\sum_{k = 1}^{\infty} \left(\frac{1}{8}\right)^{k - 1}\\
& = \frac{1}{2} \cdot \frac{1}{1 - \frac{1}{8}}\\
& = \frac{1}{2} \cdot \frac{1}{\frac{7}{8}}\\
& = \frac{1}{2} \cdot \frac{8}{7}\\
& = \frac{4}{7}
\end{align*}
Here is another approach. Let $p$ be the probability that player $A$ wins. Either player $A$ wins on the first roll with probability $1/2$ or all three players must fail in the first round, in which case we are back in the original situation. Hence,
\begin{align*}
p & = \frac{1}{2} + \left(1 - \frac{1}{2}\right)^3p\\
p & = \frac{1}{2} + \left(\frac{1}{2}\right)^3p\\
p & = \frac{1}{2} + \frac{1}{8}p\\
\frac{7}{8}p & = \frac{1}{2}\\
p & = \frac{4}{7}
\end{align*}
| {
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Generating function for $a_n = \sum_{k=0}^{n-2} a_ka_{n-k-2}$ Let $a_n$ be a sequence following the recurrence relation $$a_n = \sum_{k=0}^{n-2} a_ka_{n-k-2}$$ with initial conditions $a_0 = a_1 = 1$.
We have to find the generating function for $a_n$ that does not contain an infinite series.
Let $f(x) = \sum_{k=0}^{n} a_k$. We know that $$\sum_{k=0}^{n-2} a_ka_{n-k-2}x^{n-2} = \left(\sum a_k x^k \right) \left(\sum a_{n-k-2} x^{n-k-2}\right)$$
After that, I am not able to proceed.
Can someone help?
Note that $a_2 = 1$ and $a_3 = 2$. Thus simplifying from the answer below:
$$x^2f^2(x) -f(x) + (1+x) = 0.$$
And solving for the quadratic function we have $$f(x) = \frac{1-\sqrt{1-4x^2(x+1)}}{2x^2}$$
| Naturally, you would consider the square of the generating function:
$$f^2(x) = \sum_{i=0}^\infty a_ix^i \sum_{j=0}^\infty a_jx^j = \sum_{n=0}^\infty \sum_{k=0}^na_ka_{n-k}x^n = a_0^2 + 2a_0a_1x + \sum_{n=2}^\infty a_{n+2}x^n =$$
$$= 1+2x + x^{-2} \sum_{n=2}^\infty a_{n+2}x^{n+2} = 1+2x+x^{-2}(f(x) - a_0-a_1x-a_2x^2-a_3x^3).$$
You can finish now?
| {
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Show that if $Re{\{z}\}>0$ then $|z+\sqrt{z^{2}-1}|\geq 1$ In Complex Analysis - Bruce P . Palka - Ex: 4.12 I have not been able to prove that $|z+\sqrt{z^2-1}|^2\geq 1+ Re{\{\overline{z}\sqrt{z^2-1}}\},$ what I get is:
$|z+\sqrt{z^2-1}|^2=(z+\sqrt{z^2-1})\overline{(z+\sqrt{z^2-1})}= (z+\sqrt{z^2-1})(\overline{z}+\overline{\sqrt{z^2-1}})=|z|^2+2 Re{\{\overline{z}\sqrt{z^2-1}}\}+|\sqrt{z^2-1}|^2 $
On the other hand, since $z+\sqrt{z^2-1}$ is a root of the polynomial $t^2-2zt+1$ then:
$1=|t(2z-t)|=|(z+\sqrt{z^2-1})(2z-z-\sqrt{z^2-1})|\leq |z|^2+2 |Im{\{\overline{z}\sqrt{z^2-1}}\}|+|\sqrt{z^2-1}|^2.$
But I can not get past here, can someone help me please.
| Suppose $w = re^{i\theta}$ is a complex number. Then $|\sqrt{w}|^2 = |r^{1/2}e^{i\theta/2}|^2 = r^{1/2}e^{i\theta/2}r^{1/2}e^{-i\theta/2} = r = |w|$. Now \begin{align*}
|z + \sqrt{z^2 - 1}|^2 &= |z|^2 + 2\mathfrak{R}\left\{\overline{z}\sqrt{z^2 - 1}\right\} + \left|\sqrt{z^2 - 1}\right|^2\\ &= |z|^2 + 2\mathfrak{R}\left\{\overline{z}\sqrt{z^2 - 1}\right\} + |z^2 -1|\\
&= |z^2| + 2\mathfrak{R}\left\{\overline{z}\sqrt{z^2 - 1}\right\} + |z^2 -1|\\
&\geq|z^2 - (z^2 - 1)| + 2\mathfrak{R}\left\{\overline{z}\sqrt{z^2 - 1}\right\}\\
&= 1 + 2\mathfrak{R}\left\{\overline{z}\sqrt{z^2 - 1}\right\}.
\end{align*}
Then it just remains to prove $\mathfrak{R}\left\{\overline{z}\sqrt{z^2 - 1}\right\} \geq 0$.
| {
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Finding the intervals of increase and decrease of $\frac{x^4 - x^3 -8}{x^2 - x - 6}$ How can I find the intervals of increase and decrease of $\frac{x^4 - x^3 -8}{x^2 - x - 6}$?
I tried to find the derivative by the quotient rule to obtain the critical points but the formula was getting complicated, I know that $D_f = \mathbb{R} \setminus \{-2,3\}$ but then what?
Could anyone help me please?
| One can simplify calculation of derivative by partitioning fraction:
$$\frac{x^4 - x^3 -8}{x^2 - x - 6}=Ax^2+Bx+C+\frac{D}{x-3}+\frac{E}{x+2}$$
$$\frac{x^4 - x^3 -8}{x^2 - x - 6}=x^2+6+\frac{46}{5(x-3)}-\frac{16}{5(x+2)}$$
$$\left(\frac{x^4 - x^3 -8}{x^2 - x - 6}\right)'=2x-\frac{46}{5(x-3)^2}+\frac{16}{5(x+2)^2}$$
Unfortunately, for this problem one needs to have common denominator, which makes almost no sense in proposed way of calculation. Common denominator is $5(x-3)^2(x+2)^2$ and numerator is $$2x(x-3)^2(x+2)^2-46(x+2)^2+16(x-3)^2=2(x^5-2x^4-11x^3+9x^2+8x-4)$$ Then getting answer requires solving fifth degree algebraic equation, which I suppose cannot be done in closed-form.
If there is typo error in some one of signs before $x^3$ or $x$ in problem, then this way of solution leads to cubic equation.
| {
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Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer. Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer.
Attempt:
We have
\begin{equation*}
\frac{3n^2+4n+5}{2n+1} = \frac{4n^2+4n+1 - (n^2-4)}{2n+1} = 2n+1 - \frac{n^2-4}{2n+1}.
\end{equation*}
So, we must have $(2n+1) \mid (n^2-4)$, so $n^2-4 = k(2n+1)$, for some $k \in \Bbb Z$. But, I did not be able to find $n$ from here.
Any ideas? Thanks in advanced.
| Another approach:
By direct division we find:
$3n^2+4n+5=(\frac 32 n+\frac 54)(2n+1)+\frac{15}4$
multiplying both sides by $4$ and simplify we get:
$12n^2+16n+5=(6n+5)(2n+1)$
Or:
$\frac{4(3n^2+4n+5)}{2n+1}=6n+5+\frac {15}{2n+1}$
Therefore $2n+1$ must divide $15$ that is it must be equal to one of its divisors which are $\pm 1$, $\pm 3$, $\pm 5$ and $\pm 15$
1): $2n+1=1\rightarrow n=0$, $2n+1=-1\rightarrow n=-1$
2): $2n+1=3\rightarrow n=1$, $2n+1=-3 \rightarrow n=-2$
3): $2n+1=5\rightarrow n=2$, $2n+1=-5 \rightarrow n=-3$
4): $2n+1=15 \rightarrow n=7$, $2n+1=-15 \rightarrow n=-8$ so solutions are:
$0, \pm 1, \pm 2, -3, 7, -8$
| {
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Quadratic equation: understanding how the absolute values in the derivation correspond to the $\pm$ symbol in the classic quadratic formula expression I'd like if someone could help me understand the typical form of the quadratic formula, which, for the equation $ax^2+bx+c=0$, reads as $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $x \in \mathbb R$ and $a \neq 0$. Throughout this derivation, I will use the definition: $\sqrt{x^2}=|x|$.
Here is my derivation, and I have placed a $\color{red}{\dagger}$ next to the part that I would like some clarificaiton about:
$ax^2+bx+c =0 \iff x^2+\frac{bx}{a}+\frac{c}{a}=0 \iff (x+\frac{b}{2a})^2+(\frac{c}{a}-\frac{b^2}{4a^2})=0 \iff (x+\frac{b}{2a})^2+(\frac{4ac-b^2}{4a^2})=0 $
Bringing the right summand over to the right side of the equation:
$(x+\frac{b}{2a})^2=(\frac{b^2-4ac}{4a^2}) \iff \left| x+\frac{b}{2a}\right|=\sqrt{b^2-4ac}\cdot\sqrt{\frac{1}{4a^2}} \iff \left| x+\frac{b}{2a}\right|=\left|\frac{1}{2a} \right| \cdot \sqrt{b^2-4ac} \quad \quad \color{red}{\dagger}$
My confusion stems from how the final expression above is equivalent to the syntax "$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$" .
For $\color{red}{\dagger}$, we have 4 total cases:
*
*$ a \lt 0$ and $x+\frac{b}{a} \lt 0$
*$ a \lt 0$ and $x+\frac{b}{a} \geq 0$
*$ a \gt 0$ and $x+\frac{b}{a} \lt 0$
*$ a \gt 0$ and $x+\frac{b}{a} \geq 0$
Case 1: the conditions imply that $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$
Case 2: the conditions imply that $x=\frac{-b-\sqrt{b^2-4ac}}{2a}$
Case 3: the conditions imply that $x=\frac{-b-\sqrt{b^2-4ac}}{2a}$
Case 4: the conditions imply that $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$
From the four scenarios, is the way we get to "$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$" simply by noting that for a fixed $a$, we have $x=\frac{-b+\sqrt{b^2-4ac}}{2a} \text{ or } x=\frac{-b-\sqrt{b^2-4ac}}{2a}$? ...where the '$\text{ or }$' here is denoting the logical or.
At this point, we define "$x=\pm \alpha$" as meaning $x = \alpha \text { or } x=-\alpha$...therefore meaning that $x=\beta\pm \alpha$ is equivalent to $x=\beta+\alpha \text{ or } x=\beta - \alpha$.
Is that the proper understanding?
| From your working step $\left(x+\frac{b}{2a}\right)^2=\left(\frac{b^2-4ac}{4a^2}\right),$
take square root (permissible because non-negative);
then apply your definition $\sqrt{x^2}=|x|,$ which implies that $\sqrt{x^2}=\pm x$ (what does $\pm x$ mean);
then simplify;
finally, verify (by substitution) that your two solutions indeed satisfy $ax^2+bx+c=0.$
This considers two instead of four separate cases.
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Evaluating $\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$ I want to find the closed form of:
$\displaystyle \tag*{}\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$
I tried to use the taylor expansion of $\frac{1}{\sqrt{1-x}}$ and $\frac{1}{\sqrt{1-4x}}$ but both of them had $\binom{2n}{n}$ in the numerator and no square in the denominator. I am unsuccesful even in using $\arcsin^2x$ expansion.
However, we know a well known result:
$\displaystyle \tag*{} \sum \limits_{k=1}^{\infty} \frac{\binom{2n}{n}}{k^24^k} = \frac{\pi^2}{6} - 2\ln^2(2)$
So this somehows tells (?) maybe we can decompose our sum into two parts with one being Basel sum. Any help would be appreciated, thanks.
| If you accept that $$f(x)=\sum_{k=1}^\infty \frac{x^{k}}{k\,16^k} \binom{4k}{2k} = f_+(x) + f_-(x)$$
where
$$f_{\pm}(x) = 2\log(2) - 2 \log\left(1+\sqrt{1\pm\sqrt{x}}\right) \, ,$$
which shouldn't be too difficult to obtain (see RobPratt), then
$$I_- =\int_0^1 \frac{f_-(x)}{x} \, {\rm d}x \stackrel{u=\sqrt{1-\sqrt{x}}}{=} 8 \int_0^1 \frac{\log(2)-\log(1+u)}{1-u^2} \, u \, {\rm d}u \\
\stackrel{t=\frac{1-u}{1+u}}{=} \int_0^1 \left( \frac{4\log(1+t)}{t} - \frac{8\log(1+t)}{1+t} \right) {\rm d}t = \frac{\pi^2}{3} - 4\log^2(2)$$
and
$$I_+ = \int_0^1 \frac{f_+(x)}{x} \, {\rm d}x \stackrel{u=\sqrt{1+\sqrt{x}}}{=} 8 \int_1^\sqrt{2} \frac{\log(2)-\log(1+u)}{u^2-1} \, u \, {\rm d}u \\
\stackrel{t=\frac{u-1}{u+1}}{=} \int_0^{\frac{\sqrt{2}-1}{\sqrt{2}+1}} \left( \frac{4\log(1-t)}{t} + \frac{8\log(1-t)}{1-t} \right) {\rm d}t = -4\,{\rm Li}_2\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right) - 4 \log^2\left(\frac{2}{\sqrt{2}+1}\right) \, .$$
| {
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"url": "https://math.stackexchange.com/questions/4427827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration of a piecewise defined discontinuous function In a proof i posted recently on this site (link) i made the mistake of thinking that a function $f$, which is bounded on a closed interval $[a,b]$, would assume its infimum at a certain $x$ in the domain of $f$. I was presented with a counter example which, for my ability, seemed rather complicated. So i tried to find an easier counter example. I found the following function:
$$ f(x) = \begin{cases}
x & x > 0 \\
1 & x = 0
\end{cases} $$
Here, $ inf \{ f(x) : 0 \le x \le 1 \} = 0 $ but $f(x) \neq 0$ for all $x$.
I guess this is suitable as a counter example. Then i came up with the idea of trying to integrate this function. So came to the following proposition:
Proposition: Let $f$ be a function defined on $[a,b]$ as follows:
$$ f(x) = \begin{cases}
x & x > 0 \\
1 & x = 0
\end{cases} $$
This function is integrable with:
$$ \int_0^b = \frac{b^2}{2}$$
Proof: I use a partition $ P = \{t_0, ... , t_n \} $ of $[a,b]$ with
$$ t_i - t_{i-1} = \frac{b}{n} $$
$$ t_i = \frac{ib}{n} $$
$$ t_{i-1} = \frac{(i-1)b}{n} $$
$$ m_i = inf \{ f(x) : t_{i-1} \le x \le t_i , i \neq 1 \} = \frac{(i-1)b}{n} $$
$$ m_1 = inf \{ f(x) : t_0 \le x \le t_1 \} = 0 $$
$$ M_i = sup \{ f(x) : t_{i-1} \le x \le t_i , i \neq 1 \} = \frac{ib}{n} $$
$$ M_1 = sup \{ f(x) : t_0 \le x \le t_1 \} = 1 $$
Then we have
$$ L(f, P) = \sum_{i=2}^n m_i \cdot \frac{b}{n} + m_1 \cdot \frac{b}{n} = \sum_{i=2}^n \frac{(i-1)b}{n} \cdot \frac{b}{n} + 0 \cdot \frac{b}{n} = \frac{b^2}{n^2} \cdot \sum_{j=1}^{n-1} j = \frac{b^2}{2} \cdot \frac{n-1}{n}$$
and
$$ U(f, P) = \sum_{i=2}^n M_i \cdot \frac{b}{n} + M_1 \cdot \frac{b}{n} = \sum_{i=2}^n \frac{ib}{n} \cdot \frac{b}{n} + 1 \cdot \frac{b}{n} = \Biggl[ \frac{b^2}{n^2} \cdot \sum_{j=1}^{n-1} j+1 \Biggl] + \frac{b}{n} = \frac{b^2}{2} \cdot \frac{n+1}{n} + \frac{b}{n}$$
For the difference of the upper and lower sums this results in
$$ U(f, P) - L(f, P) = \frac{b^2}{2} \cdot \frac{n+1}{n} + \frac{b}{n} - \frac{b^2}{2} \cdot \frac{n-1}{n} = \frac{b^2}{2} \cdot \frac{n+1}{n} - \frac{b^2}{2} \cdot \frac{n-1}{n} + \frac{b}{n} = \frac{b^2}{2} \cdot \frac{2}{n} + \frac{b}{n} = \frac{b^2+b}{n}$$
So in order get $ U(f, P) - L(f, P) < \epsilon$ we can choose $ n > \frac{b^2+b}{\epsilon} $. Thus $f$ is integrable and since
$$ \frac{b^2}{2} \cdot \frac{n-1}{n} \le \frac{b^2}{2} \le \frac{b^2}{2} \cdot \frac{n+1}{n} + \frac{b}{n}$$
and the integral is unique, if it exists, we have
$$ \int_0^b = \frac{b^2}{2}$$
as required. $ \blacksquare $
All these equations and manipulations are quite complex regarding my ability. So might anyone tell me if this is correct or point me towards my mistakes? Thanks in advance.
| Yes, the proof seems good and rock solid to me. If I could add something would be this: are you familiar with taking limits of successions?
One equivalent definition of (Riemann) Integrable function would be to check that $\lim_{n\to \infty}L(f,P_n)=\lim_{n\to \infty}U(f,P_n)$ , and using this definition of integrable function could speed up and also make a bit easier the calculations in the last part of your reasoning.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing that $(a^2 - b^2)^2$ $ \ge $ $4ab(a-$ $b)^2$ An inequality problem from Beckenbach and Bellman:
Show that $(a^2 - b^2)^2 \ge 4ab(a-b)^2$
The given answer is simply
Equivalent to $(a - b)^4 \ge 0$
I have tried two approaches, one which agrees with the given answer, and the other which does not.
Approach one. (Agrees with answer)
\begin{align}
(a^2 - b^2)^2 & \ge 4ab(a-b)^2\\
(a^2 - b^2)^2 - 4ab(a-b)^2 & \ge 0\\
((a+b)(a-b))^2 - 4ab(a-b)^2 & \ge 0\\
(a+b)^2(a-b)^2 - 4ab(a-b)^2 & \ge 0\\
(a-b)^2((a+b)^2 - 4ab) & \ge 0 \\
(a-b)^2 (a^2 -2ab + b^2) &\ge 0 \\
(a-b)^2 (a-b)^2 & \ge 0\\
(a - b)^4 & \ge 0
\end{align}
Approach Two
\begin{align}
(a^2 - b^2)^2 & \ge 4ab(a-b)^2\\
((a+b)(a-b))^2 & \ge 4ab(a-b)^2\\
(a+b)^2(a-b)^2 & \ge 4ab(a-b)^2\\
(a+b)^2 & \ge 4ab\\
(a^2 -2ab + b^2) &\ge 0 \\
(a-b)^2 & \ge 0
\end{align}
Could someone point out where the second approach is going wrong?
| Let's examine your second attempt in more detail and check to see what's going wrong with it.
Observation 1
Whenever we are dividing using algebraic terms, it is vital that we make sure that we are not dividing by zero. Otherwise, the proof can end up going very wrong. There are many examples of this going wrong online including an incorrect proof that $0=1$ which is only possible because of the fact that the proof involves a division by a term equal to zero.
Therefore, when you divide by $(a-b)^2$ we want to consider the case where $a=b$ separately for a complete proof.
Observation 2
The inequality $(a-b)^2\ge 0$ which find at the end of your second proof is equivalent to the result you are looking for.
Note: since $(a-b)^2$ is non-negative we can square both sides of the inequality without reversing the "$\ge$" symbol. This gives us the desired inequality $(a-b)^4 \ge 0$.
Conclusion
Therefore, there is no problem with your second proof for the case where $a \neq b$.
However, when $a=b$, we must start again from the first inequality and replace every instance of $b$ with $a$ (or vice versa) to give the inequality:
$(a^2 - a^2)^2 \ge 4a^2(a-a)^2$ $\iff$ $0 \ge 0$ which is clearly true.
As we have shown that the inequalities are equivalent for $a=b$ and $a\neq b$, this completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding rational solutions to $0\neq a^3c + b^3c^2\in\mathbb{Z}$ but $abc\notin\mathbb{Z}$ I'm trying to find $a,b,c\in\mathbb{Q}$ such that $0\neq a^3c + b^3c^2\in\mathbb{Z}$ but that $abc\notin\mathbb{Z}$.
I tried to pick specific values for $a, b$
$$
b^3x^2 + a^3x - n= 0
$$
where I tried different integers $n\neq 0$ and then evaluate the roots of this polynomial. However, this proved unsuccessful and now I'm stuck.
| I don't know offhand of any way to use your approach with $b^3x^2 + a^3x - n= 0$. Instead, since you're looking for just one set of values, consider that $a = \frac{3}{2}$, $b = \frac{5}{2}$ and $c = 1$ gives $a^3c + b^3c^2 = \frac{3^3}{2^3} + \frac{5^3}{2^3} = \frac{27 + 125}{8} = 19 \in \mathbb{Z}$, but $abc = \frac{15}{4} \not\in \mathbb{Z}$.
In general, with $c = \pm 1$, then for any integers $d$, $e$ and $f$, where $\lvert f \rvert \gt 1$ and $\gcd(d,f) = \gcd(e,f) = 1$, plus there's a non-zero integer $n$ such that
$$cd^3 + e^3 = nf^3 \tag{1}\label{eq1A}$$
then $a = \frac{d}{f}$ and $b = \frac{e}{f}$ works since $abc = \frac{dec}{f^2} \not\in \mathbb{Z}$. For example, with $c = 1$, then $e = f^3 - d$ (or, more generally, $e = kf^3 - d$ for any non-zero integer $k$) works in \eqref{eq1A} since $d^3 + (f^3 - d)^3 = d^3 + f^9 - 3f^6d + 3f^3d^2 - d^3 = (f^6 - 3f^3d + 3d^2)f^3$, giving $n = f^6 - 3f^3d + 3d^2$. With $c = -1$, then $e = kf^3 + d$ (with $k \neq 0$) works instead.
There are undoubtedly other techniques to determine different groups of $(a,b,c)$ which also work, with mine likely among the simplest of these techniques.
| {
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"timestamp": "2023-03-29T00:00:00",
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A problem involving a tetrahedron Let $ABCD$ a tetrahedron. We know the angle
$$\angle{ACB}=45^\circ$$the sum
$$\overline{AD}+\overline{BC}+\frac{\overline{AC}}{\sqrt2}=90$$and that the volume is $4500.$ We also know (but I don't know if this can be useful for the solution) that $\overline{CD}^2$ is integer: how much is $\overline{CD}^2?$
My problem is that I don't know how to relate the data I have in my possession. In particular how can I use the given sum of those three edges?
| This is an expanded version of the comment of Zerox. Let $h$ be the (length of the) height from $D$, the distance between $D$ and the plane of $\Delta ABC$. Then:
$$
\begin{aligned}
30^3 &= 27000
=6\operatorname{Volume}[ABCD]
=h\cdot 2\operatorname{Area}[ABC]
=h\cdot AC\cdot BC\cdot\sin 45^\circ
\\
&=h\cdot \frac{AC}{\sqrt 2}\cdot BC\ .
\\
\text{Then:}
\\
30 &=\left(h\cdot \frac{AC}{\sqrt 2}\cdot BC\right)^{1/3}
\\
&\le\left(AD\cdot \frac{AC}{\sqrt 2}\cdot BC\right)^{1/3}
\\
&\qquad\qquad\qquad\text{with equality iff $h=AD$, i.e. $DA\perp (ABC)$}
\\
&\le\frac 13\left(AD+ \frac{AC}{\sqrt 2}+ BC\right)
\\
&\qquad\qquad\qquad\text{with equality iff $AD=\frac{AC}{\sqrt 2}= BC$}
\\
&=\frac 13\cdot 90=30\ .
\end{aligned}
$$
Since the beginning and the end show both the value $30$, we have equalities in between, so we know $DA=30$ is the height from $D$, the segment $DA$ being thus perpendicular on the plane $(ABC)$, and in particular also on $AC$, and $AC/\sqrt 2=30$ and $BC=30$. The triangle $\Delta DAC$ has thus a right angle in $A$, giving
$$
DC^2 = DA^2+AC^2=30^2+(30\sqrt 2)^2=30^2(1+2)=2700\ .
$$
$\square$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Graph of a function $(x^2+2)x(x-2)$ Consider $f(x)=(x^2+2)x(x-2)$. It has two real roots: $0$ and $2$.
If $x<0$, then $f(x)$ is positive, and if $x>2$, then also $f(x)$ is positive.
If $x$ is in $(0,2)$, then $f(x)$ is negative.
I am trying to understand why $f(x)$ has unique local minima in the interval $(0,2)$, so that the graph of $f(x)$ in interval $(0,2)$ is U-shaped. Can one explain, why this happens?
Even if we modify $f(x)$ by
$$f(x)=(x^2+2)x(x-2)(x-4)\cdots (x-2n)$$
still, in each interval of the form $(2k, 2k+2)$, the graph of $f(x)$ is U-shaped or reverse-U shaped.
Why it can't be (smooth) w-shaped? Why this is happening?
| For the general case.
Let $f(x) = (x^2+2)g(x)$ with $g(x) = x(x-2)(x-3)(x-4)...(x-2n)$.
Then $f'(x) = 0$ and $f(x)\neq 0 \iff (x^2+2)g(x) + 2xg'(x) = 0$ and $x\neq 0$ and $g(x)\neq 0 \iff \dfrac{g'(x)}{g(x)} + \dfrac{x^2+2}{2x} = 0$
But $\dfrac{g'(x)}{g(x)} = \dfrac 1 x + \dfrac 1 {x-2} + \ldots + \dfrac 1{x-2n}$.
So the equation $f'(x) = 0$ with $f(x)\neq 0$ is equivalent to $h(x) = x + \dfrac 2x + \dfrac 1 {x-2} + \ldots + \dfrac 1{x-2n} = 0$
This $h(x)$ is a sum of increasing functions, each of which is increasing over each interval it is defined on. So $h(x)$ is increasing over each of the intervals $(-\infty,0)$, $(0,2)$, $(2,4)$, and so on.
Hence it can be equal to zero just once per interval. So the same holds for $f'$.
So $f$ can have only one extremum per interval. QED.
| {
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"timestamp": "2023-03-29T00:00:00",
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Tough integral $\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3) $ How to prove
$$\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3), $$
and does there even exist a closed form of $$\int_0^{ \pi }\frac{x^3(\pi-x)^3}{\sin^3 x} dx \ ? $$
(Note that the easier one $$\int_0^{ \pi }\frac{x (\pi-x) }{\sin x} dx = 7~\zeta(3) ,\text { equivalently }\ \int_{0}^{1}\frac{x - x^{2} }{ \sin(\pi x)}dx = 7\frac{\zeta (3)}{\pi^{3}},$$ has been solved here.)
| Letting $x\mapsto \frac{\pi}{2}-x$ changes
$$
\begin{aligned}
I=& \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right)^{2}\left(\frac{\pi}{2}+x\right)^{2}}{\cos ^{2} x} d x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{\pi^{2}}{4}-x^{2}\right)^{2} d(\tan x)
\end{aligned}
$$
Integration by parts gives
$$
\begin{aligned}
I &=4 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x\left(\frac{\pi^{2}}{4}-x^{2}\right) \tan x d x=-8 \int_{0}^{\frac{\pi}{2}} x\left(\frac{\pi^{2}}{4}-x^{2}\right) d(\ln (\cos x))
\end{aligned}
$$
Integration by parts again yields
$$
I=2 \pi^{2} \int_{0}^{\frac{\pi}{2}} \ln (\cos x) d x-24 \int_{0}^{\frac{\pi}{2}} x^{2} \ln (\cos x) d x
$$
Using my post 1 and post 2, we can conclude that
$$
I=-\pi^{3} \ln 2-24\left(-\frac{\pi^{3} \ln 2}{24}-\frac{\pi}{4} \zeta (3)\right)=\frac{\pi}{6} \zeta(3)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A tower stands vertically in the interior of a field which has the shape of an equilateral triangle of side $a$. If the angles of elevation...
A tower stands vertically in the interior of a field which has the shape of an equilateral triangle of side $a$. If the angles of elevation of the top of the tower are $\alpha$,$\beta$,$\gamma$ from the corners of the field find the height of the tower.
Answer: $$a\sqrt{\frac{p^2+q^2+r^2-\sqrt{2p^2q^2+2p^2r^2+2q^2r^2 -p^4 -q^4 - r^4}}{2(p^4 + q^4 + r^4 -p^2q^2 -p^2r^2 - q^2r^2)}}$$ where $p= \cot \alpha$, $q = \cot \beta$ and $r = \cot \gamma$
My Attempt
$[.]$ denotes area
$$[APB]+[BPC]+[CPA] = [ABC]$$
$$[ABC] = \frac{\sqrt{3}}{4}a^2$$
I tried finding the area of $\triangle APB$,$\triangle BPC$,$\triangle CPA$ by Heron's formula but the result was too complicated because I had to add square roots : something of the form $\sqrt{x} + \sqrt{y} + \sqrt{z} = w$. Moreover, the $x,y,z$ in this expression are also complicated.
I admit that the answer is complicated too but I am searching for an elegant approach towards the solution (something like substitution or anything else). Is there any elegant approach?
| As the figure shows the distances between the base of the tower $P$ and each of the three vertices $A,B,C$ of equilateral $\triangle ABC$ is
$ PA = h \cot \alpha = h p $
$ PB = h \cot \beta = h q $
$ PC = h \cot \gamma = h r $
Now using Barycentric coordinates, and taking $A$ to be the origin, we can express point $P$ in terms of $B $ and $C$ as follows
$ P = c_1 B + c_2 C $
so that
$ AP = P = c_1 B + c_2 C , BP = (c_1 - 1) B + c_2 C , CP = c_1 B + (c_2 - 1) C $
I'll assume that the side length $a = 1$. Taking the magnitude of these vectors, and noting that $|B| = |C| = 1$ and that $B \cdot C = \dfrac{1}{2} $, then we can write the following three equations
$c_1^2 + c_2^2 + c_1 c_2 = h^2 p^2 \hspace{10pt} (*) $
$(c_1 - 1)^2 + c_2^2 + (c_1 - 1) c_2 = h^2 q^2 $
$ c_1^2 + (c_2 - 1)^2 + c_1 (c_2 - 1) = h^2 r^2 $
We want to eliminate $c_1, c_2$, so subtract each pair of equations, i.e. (1)-(2) , and (1) - (3), we get
$ 2 c_1 + c_2 = h^2 (p^2 - q^2) + 1 $
$ c_1 + 2 c_2 = h^2 (p^2 - r^2) +1$
Solving this $2 \times 2$ system, gives us
$ c_1 = \dfrac{1}{3} ( h^2 (p^2 - 2 q^2 + r^2 ) + 1 )$
$ c_2 = \dfrac{1}{3} ( h^2 (p^2 - 2 r^2 + q^2 ) + 1 )$
substituting $c_1, c_2$ into Eq. $(*)$ and expanding, gives us
$ h^4 K + h^2 L + M = 0 $
where
$K = p^4 + q^4 + r^4 - p^2 q^2 - p^2 r^2 - q^2 r^2 $
$ L = - ( p^2 + q^2 + r^2 ) $
$ M = 1 $
Combining the $h^2 $ coefficients, we get the quadratic equation (in $h^2$)
From this, using the quadratic formula, the solution is
$ h^2 = \dfrac{ -L \pm \sqrt{ L^2 - 4 K } }{ 2 K } $
we have
$ - L = p^2 + q^2 + r^2 $
and
$L^2 - 4 K = p^4 + q^4 + r^4 + 2 p^2 q^2 + 2 p^2 r^2 + 2 q^2 r^2 - 4 (p^4 + q^4 + r^4 - p^2 q^2 - p^2 r^2 - q^2 r^2) $
And this reduces to
$ L^2 - 4 K = 3 ( 2 p^2 q^2 + 2 p^2 r^2 + 2 q^2 r^2 - p^4 - q^4 - r^4 ) $
Hence,
$ h^2 = \dfrac{ p^2 + q^2 + r^2 \pm \sqrt{3} \sqrt{2 p^2 q^2 + 2 p^2 r^2 + 2 q^2 r^2 - p^4 - q^4 - r^4} }{ 2 (p^4 + q^4 + r^4 - p^2 q^2 - p^2 r^2 - q^2 r^2)}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$ Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$
Attempt:
I did figure out that $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots = \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$
I have checked on google for $ \sum_{n=1}^\infty \frac{1}{n^2}$ and i do not suppose to know that is equal to $\frac{\pi^2}{6}$.
and I checked how The sum can be given explicitly of $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$
and the result was show that needs the Fourier series and we haven't learned that yet.
So, I have no idea how to prove that equal to $\frac{3}{4}S$ without know the exactly the sum
Thanks.
| $$S_k = \frac1{1^2} + \frac1{2^2} + \frac1{3^2} + \frac1{4^2} + \dots + \frac{1}{(2k+1)^2}$$
$$T_k = \frac1{1^2} + \frac1{3^2} + \frac1{5^2} + \frac1{7^2} + \dots + \frac{1}{(2k+1)^2}$$
Look at $\lim_{k \to \infty} {S_k - T_k}$.
| {
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Calculate the value of expression $Q = \frac{x + 1}{y}$ when $xy > 1$ and expression $P = x + 2y + \frac{5x + 5y}{xy - 1}$ reaches its maximum value.
Consider two positives $x$ and $y$ where $xy > 1$. The maximum value of the expression $P = x + 2y + \dfrac{5x + 5y}{xy - 1}$ is achieved when $x = x_0$ and $y = y_0$. Calculate the value of expression $Q = \dfrac{x_0 + 1}{y_0}$.
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
Isn't being demotivated one of the worst feelings in the world?
Let $x = \cot\alpha \left(n\pi < \alpha < n\pi + \dfrac{\pi}{2}, n \in \mathbb Z\right)$ and $y = \cot\beta \left(n\pi < \beta < n\pi + \dfrac{\pi}{2}, n \in \mathbb Z\right)$. It could be implied that $P = \cot\alpha + 2\cot\beta + \dfrac{5}{\cot(\alpha + \beta)}$.
Hmmm~ never-mind, this leads to nowhere.
Perhaps creating new variables, specifically $a = x + 1 \ (a > 2)$ and $b = \dfrac{1}{y} \ (b > 0)$ might actually be a clue. Then, we have that $a - 1 > 2b$ and $$\begin{aligned} P = a + 5b - 1 + \dfrac{2}{b} + \dfrac{5(b^2 + 1)}{a - b - 1} &= 6b + \dfrac{2}{b} + (a - b - 1) + \dfrac{5(b^2 + 1)}{a - b - 1}\\ &\ge 2\left[3b + \sqrt{5(b^2 + 1)} + \dfrac{1}{b}\right] \end{aligned}$$
Function $f(b) = 3b + \sqrt{5(b^2 + 1)} + \dfrac{1}{b}$ contains its first derivative as $f'(b) = 3 + \dfrac{b\sqrt{5}}{\sqrt{b^2 + 1}} - \dfrac{1}{b^2}$ and the equation $f'(b) = 0$ has roots $b = \dfrac{1}{2}$ and $b = -\sqrt{\dfrac{\sqrt 5 - 1}{2}}$
We can then draw the table of variations for function $f(b)$.
As $f(b) \ge 6, \forall b > 0$, it can be inferred that the minimum value of $P$ is $2 \times 6 = 12$, achieved when $$\left\{ \begin{aligned} a > 0, b > 0&, a - 1 > 2b\\ (a - b - 1)^2 &= 5(b^2 + 1)\\ b &= \dfrac{1}{2} \end{aligned} \right. \iff \left\{ \begin{aligned} a &= 4\\ b &= \dfrac{1}{2} \end{aligned} \right. \implies Q = ab = 4 \times \dfrac{1}{2} = 2$$
Wait... minimum? That's not what the question asks for, although that's one of the given options.
Anyhow, I'm still figuring out how to tackle this problem. Should I give in to the asymmetry or otherwise? As always, thanks for reading, (and even more so if you could help), have a wonderful tomorrow~
By the way, the options are $\sqrt 2, \sqrt 3, 2$ and $1$, so I could have been fooled there.
| With $P = x+2y+5\frac{x+y}{xy-1}$ and $Q=\frac{x+1}{y}$ and forming the lagrangian
$$
L(x,y,\lambda,s) = P+\lambda(x y-1-s^2)
$$
the stationary points are the solutions for
$$
0 = \nabla L = \cases{
\lambda y-\frac{y (5 x+5 y)}{(x y-1)^2}+\frac{5}{x y-1}+1 \\
\lambda x-\frac{x (5 x+5 y)}{(x y-1)^2}+\frac{5}{x y-1}+2 \\
x y-1-s^2 \\
-2 \lambda s }
$$
The solutions give us
$$
\left(
\begin{array}{cccccc}
P & Q & x & y & \lambda & s^2\\
1 & -12 & -3 & -2 & 0 & 5 \\
2 & 12 & 3 & 2 & 0 & 5 \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4445106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Fractional part and greatest integer function Here are a few questions on the fractional part and the greatest integer function.
*
*Find out $[\sqrt[3]{2022^2}-12\sqrt[3]{2022}]$
*If $\{x\}=x-[x],$ find out $[255\cdot x\{x\}]$ for $x=\sqrt[3]{15015}.$
For question 1, using a calculator, I know that the answer is $8.$ But I really do not know how to proceed in a mathematical way.
For question 2, $$255\cdot x\{x\}=255\cdot x( x-[x])=255\cdot \sqrt[3]{15015}(\sqrt[3]{15015}-24)$$
As $25^3>15015>24^3.$ So we have to find
$$[255\cdot \sqrt[3]{15015}(\sqrt[3]{15015}-24)].$$
We can bound it from below as $[xy]\ge [x]\cdot [y].$
So $$[255\cdot \sqrt[3]{15015}(\sqrt[3]{15015}-24)]\ge 255[\sqrt[3]{15015^2}-\sqrt[3]{15015}\cdot 24].$$ This again looks like question 1 but I do not know how to solve it.
We can upper bound it too using $[x-y]\le [x]-[y].$
Now, $$[255\cdot \sqrt[3]{15015}(\sqrt[3]{15015}-24)]=[255\cdot\sqrt[3]{15015^2}-255\cdot\sqrt[3]{15015}\cdot 24]=[255\cdot {15015^2}]-[255\cdot\sqrt[3]{15015}\cdot 24].$$
Any solutions?
| Solution for part 1. Let $a:=\sqrt[3]{2002}$. Then $0<a-12<1$ and
$$0<(a-12)^3 = a^3-36(a^2-12a)-12^3<1.$$
Rearranging the inequality we get
$$
8<\frac{2022-12^3-1}{36}<a^2-12a<\frac{2022-12^3}{36}<9
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4446271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
Computing limits using Taylor expansions and $o$ notation on both sides of a fraction Let's define $o(g(x))$ as usually:
$$
\forall x \ne a.g(x) \ne 0 \\
f(x) = o(g(x)) \space \text{when} \space x \to a \implies \lim_{x \to a} \frac{f(x)}{g(x)}=0
$$
In theorem $7.8$, Tom Apostol in his Calculus Vol. $1$ gave and proved the following basic rules for algebra of o-symbols
*
*$o(g(x)) \pm o(g(x)) = o(g(x)) $
*$o(cg(x)) = o(g(x)) $, if $c \ne 0$
*$f(x) \cdot o(g(x)) = o(f(x)g(x)) $
*$o(o(g(x))) = o(g(x)) $
*$\frac{1}{1 + g(x)} = 1 - g(x) + o(g(x))$
I will use a concrete example to demonstrate the confusion I have, but the question is probably more generally applicable. By using Taylor expansions, compute $\lim_{x \to 0} \frac{1 - \cos x^2}{x^2 \sin x^2}$.
\begin{equation}
\cos x^2 = 1 - \frac{x^4}{2} + o(x^5) \\
\sin x^2 = x^2 + o(x^4)
\end{equation}
$$
\lim_{x \to 0} \frac{1 - \cos x^2}{x^2 \sin x^2} = \lim_{x \to 0} \frac{1 - (1 - \frac{x^4}{2} + o(x^5))}{x^2 (x^2 + o(x^4))} = \lim_{x \to 0} \frac{\frac{x^4}{2} + o(x^5)}{x^4 + o(x^6)}
$$
What would be an easy way to solve that, without flattening the fraction by the application of case $5$ of the theorem $7.8$ above, while using the provided definition for $o$ notation (and without using a more advanced methods, like L'Hopital's rule)?
I saw somewhere $\lim_{x \to 0} \frac{\frac{1}{2} + \frac{o(x^5)}{x^4}}{1 + \frac{o(x^6)}{x^4}} = \frac{1}{2}$, without explanation, suggesting it should be a trivial matter, but I'm not sure why that would be the case.
Appendix in how I solved that, relying only on the definition and the $T7.8$.
One way to proceed could be to use case $3$ from Theorem $7.8$, in reverse (i.e. $o(x^6) = x^4 o(x^2)$)
$$
\lim_{x \to 0} \frac{\frac{x^4}{2} + o(x^5)}{x^4 + o(x^6)} = \lim_{x \to 0} \frac{\frac{1}{2} + o(x)}{1 + o(x^2)}
$$
Then I could apply case $5$ from the above theorem and applying case $4$ of the theorem (i.e. $o(o(x^2)) = o(x^2)$), to get $\frac{1}{1+o(x^2)} = 1 - o(x^2)$.
After applying case $2$ with $c = -1$, we get:
$$
\lim_{x \to 0} \frac{\frac{1}{2} + o(x)}{1 + o(x^2)} = \lim_{x \to 0} (\frac{1}{2} + o(x)) (1 + o(x^2)) = \frac{1}{2}
$$
| Observe that
$$
\frac{o(x^5)}{x^4} = \frac{o(x^5)}{x^5}x \to 0 \quad \text{as } x \to 0$$
since, by definition of $o$, we have $o(x^5)/x^5 \to 0$ as $x \to 0$.
Similarly,
$$
\frac{o(x^6)}{x^4} = \frac{o(x^6)}{x^6}x^2 \to 0 \quad \text{as } x \to 0.
$$
This explains why
$$
\lim\limits_{x \to 0} \frac{\frac{1}{2}+\frac{o(x^5)}{x^4}}{1+\frac{o(x^6)}{x^4}} = 0.
$$
Moreover, the above computations show also that $o(x^5) = o(x^4)$ and $o(x^6)= o(x^4)$ as $x\to 0$.
So you could also write
$$
\begin{align}
\lim_{x \to 0} \frac{\frac{x^4}{2} + o(x^5)}{x^4 + o(x^6)} &= \lim_{x \to 0} \frac{\frac{x^4}{2} + o(x^4)}{x^4 + o(x^4)} \\
& = \lim_{x \to 0} \frac{\frac{1}{2}+o(1)}{1+o(1)} = \frac{1}{2}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4446443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find $\int{\frac{\cos2x}{\sqrt{\sin^3 x}}}dx$? I was trying to solve this indefinite integral;
$$\displaystyle\int{\frac{\cos2x}{\sqrt{\sin^3 x}}}dx.$$
I tried using the online site; Integral Calculator. But it gave a very weird answer:
$$\int{\frac{\cos2x}{\sqrt{\sin^3 x}}}dx=2\operatorname{\Pi}\left(2\,;\dfrac{2x-{\pi}}{4}\,\middle|\,2\right)-4\operatorname{E}\left(\dfrac{2x-{\pi}}{4}\,\middle|\,2\right)+C$$
My attempt:
\begin{align*}
\int{\frac{\cos2x}{\sqrt{\sin^3 x}}}dx
&=\int{\frac{1-2\sin^2x}{\sqrt{\sin^3 x}}}dx\\
&=\int{\frac{1-2\sin^2x}{\sin^{\frac{3}{2}}x}}dx\\
&=\int \left(\frac{1}{\sin^{\frac{3}{2}}x}-\frac{2\sin^2x}{\sin^{\frac{3}{2}}x}\right)dx\\
&=\int(\sin^{\frac{-3}{2}}x-2\sin^{\frac{1}{2}}x)dx\\
&=\int \sin^{\frac{-3}{2}}xdx-2\int \sin^{\frac{1}{2}}xdx.
\end{align*}
What do I do now?
| First, note that $d(\sin x\cos x)/dx = d(\sin(2x)/2)/dx = \cos(2x)$. So we can integrate by parts using $u = \cos^{3/2}x$ and $v = (\sin x \cos x)^{-1/2}$. This gives
\begin{multline}
\int\frac{\cos(2x)}{\sin^{3/2}x}dx = \int\cos^{3/2}x\frac{\cos(2x)dx}{(\sin x\cos x)^{3/2}}dx = -\frac{2\cos x}{\sin^{1/2}x}+\int\frac{2d\left[\cos^{3/2} x\right]}{\sin^{1/2}x\cos^{1/2}x}\\
= -\frac{2\cos x}{\sin^{1/2}x}+\int\frac{-3\sin x \cos^{1/2}x}{\sin^{1/2}x\cos^{1/2}x}dx = -\frac{2\cos x}{\sin^{1/2}x} - 3\int \sin^{1/2}xdx.
\end{multline}
This last integral can be transformed into an incomplete elliptic integral:
$$
\int\sin^{1/2}xdx = \int \cos^{1/2}\left(\frac{\pi}{2}-x\right)d = \int\sqrt{1 - 2\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}dx = -2 E\left(\frac{\pi}{4}-\frac{x}{2}\; \bigg|\; 2\right).
$$
So in total we have
$$
\int\frac{\cos(2x)}{\sin^{3/2}x}dx = 6E\left(\frac{\pi}{4}-\frac{x}{2}\; \bigg|\; 2\right) - \frac{2\cos x}{\sin^{1/2}x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is there a mistake in an old question on this site: "Where does the sum of sin(n) formula come from?" https://math.stackexchange.com/a/1119137/29156
Shouldn't Abel's answer be:
we will use the fact that $$2 \sin 1 \sin k = \cos(k-1) -\cos(k + 1)$$let $S = \sin 1 + \sin 2 \cdots + \sin n,$ then
$\begin{align}
2S \sin 1 &= 2\sin 1 \sin 1+ 2 \sin 1 \sin 2 + 2 \sin 1 \sin 3\cdots +2 \sin 1 \sin n \\
&=(1 - \cos 2) +(\color{red}{\cos 1} - \cos3) +(\color{red}{\cos 2} - \cos 4)+\cdots +(\cos (n- 1) - \cos ( n + 1) )\\
&=1 + \color{red}{\cos 1} - \color{red}{\cos (n)} - \color{red}{\cos(n+1)}
\end{align}$
?
Obviously this is still bounded
| hint
Yes, Abel's result seems to be false.
$$2S\sin(1)=\sum_{k=1}^n\Big((\cos(k-1)-\cos(k))+(\cos(k)-\cos(k+1))\Bigr)=$$
$$1-\cos(n)+\cos(1)-\cos(n+1)=$$
$$2\sin^2(\frac{n+1}{2})+\cos(1)-\cos(n)=$$
$$2\sin(\frac{n+1}{2})\Bigl(\sin(\frac{n+1}{2})+\sin(\frac{n-1}{2})\Bigr)=$$
$$4\sin(\frac{n+1}{2})\sin(\frac n2)\cos(\frac 12)$$
Thus
$$\boxed{S=\frac{\sin(\frac{n+1}{2})\sin(\frac n2)}{\sin(\frac 12)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $z=x+iy$ is the answer of the equation $\cos(z)=-3$ What is the value of $x$?
If $z=x+iy$ is the answer of the equation $\cos(z)=-3$. What is the value
of $x$?
$1)x=2k\pi-\frac{\pi}2 , k\in \mathbb{Z}$
$2)x=2k\pi+\frac{\pi}2 , k\in \mathbb{Z}$
$3)x=k\pi+\frac{\pi}2 , k\in \mathbb{Z}$
$4)x=k\pi-\frac{\pi}2 , k\in \mathbb{Z}$
I saw a similar question here, and I tried to solve it but I got stock at a point. Here is my try:
$$\cos z=\frac{e^{iz}+e^{-iz}}{2}=-3\Rightarrow e^{iz}+e^{-iz}+6=0$$
After using the substitution $e^{iz}=t$ we get $t^2+6t+1=0$. Hence $e^{i(z+2k\pi)}= -3\pm\sqrt8$ and
$$i(z+2k\pi)=\ln(-3\pm2\sqrt2)$$
From here I don't know how to continue.
| \begin{align}
z &= 2\pi k \pm \cos^{-1}(-3), \; k \in \mathbb{Z}\\
&= 2\pi k \pm \frac{\pi}{2} \pm \sin^{-1}(3)\\
& = 2\pi k \pm \frac{\pi}{2} \pm i\ln\left(\sqrt{1 - 3^2} - 3i\right)\\
& = 2\pi k \pm \frac{\pi}{2} \pm i\ln\left(i\left(2\sqrt{2} - 3\right)\right)\\
& = 2\pi k \pm \frac{\pi}{2} \pm i\ln(i) \pm i\ln\left(2\sqrt{2} - 3\right)\\
& = 2\pi k \pm \frac{\pi}{2} \mp \frac{\pi}{2} \pm i\ln\left(2\sqrt{2} - 3\right)\\
& = 2\pi k \pm i\ln\left(2\sqrt{2} - 3\right)\\
& = 2\pi k \mp \pi \pm i\ln\left(3 - 2\sqrt{2}\right)
\end{align}
$$\therefore \boxed{\Re(z) = (2n + 1)\pi, \; n \in \mathbb{Z}}$$
Reference: Inverse trigonometric functions: Logarithmic forms
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Locus of middle points of the chords of conicoid which are parallel to $xy$ plane and touch the given sphere I have the following conicoid before me:
$ax^2+by^2+cz^2=1$
I have to find the locus of the middle points of the chords which are parallel to the plane $z=0$ and touch the sphere $x^2+y^2+z^2=a^2$.
This is what I have tried:
Let $(A,B,C)$ denote the mid point of a chord. Let $<l,m,n>$ be the direction cosines of the chord.
Then equation of chord is given by:
$\dfrac{x-A}{l}=\dfrac{y-B}{m}=\dfrac{z-C}{n}= r$ (say)
Then
$x=lr+A$;
$y=mr+B$;
$z=nr+C$
As the chords are parallel to $z=0$ plane hence $n=0$.
This means that $z=C$.
Putting values of $x$,$y$ and $z$ in the equation of conicoid, we get:
$a(lr+A)^2+b(mr+B)^2+cC^2=1$ which gives:
$r^2(al^2+bm^2)+2r(alA+bmB)+aA^2+bB^2+cC^2-1=0$
For getting equal and opposite values of $r$, sum of roots should be $0$ i.e.
$alA+bmB=0$
Lastly, I tried to use the sphere condition.
I again put the values of $x$,$y$, $z$ in the equation of the sphere:
$(lr+A)^2+(mr+B)^2+C^2=a^2$
which means
$r^2(l^2+m^2)+2r(lA+mB)+A^2+B^2+C^2-a^2=0$
As the chord only touches the sphere, roots must be equal which happens when:
$4(lA+mB)^2-4(l^2+m^2)(A^2+B^2+C^2-a^2)=0$
which gives
$2lmAB-l^2(B^2+C^2-a^2)-m^2(A^2+C^2-a^2)=0$
After this I do not know how to proceed. Please guide me as to how to proceed further. Any suggestions will be highly appreciated.
This question is not getting any answers. At the same time, I do not know what should I do to draw the interest of the community towards this question. Please help me out with this question. I really need to know if I am doing this problem correctly or not. And if yes, what is the way forward from the point I am stuck at.
| I have figured out the answer on my own.
Here it goes:
Dividing final equation by $lm$,we get:
$\dfrac{l}{m}.2AB -(B^2+C^2-a^2)(\dfrac{l}{m})^2-(A^2+C^2-a^2)=0$
From equation $alA+bmB=0$, we get:
$\dfrac{l}{m}=-\dfrac{bB}{aA}$
Putting this value of $\dfrac{l}{m}$ in first equation, we get:
$\dfrac{2B^2b}{a}+(B^2+C^2-a^2)\dfrac{b^2B^2}{a^2A^2}+(A^2+C^2-a^2)=0$
Finally replacing $A$,$B$ and $C$ with $x$, $y$ and $z$ respectively, we get the required locus.
| {
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"url": "https://math.stackexchange.com/questions/4452279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is My Solution Valid? Question from the 1999 Bulgarian Math Olympiad:
Find all pairs $(x,y)\in\mathbb{Z}$ satisfying $$x^3=y^3+2y^2+1$$
My first approach was to take the cube root of both sides: $$x=\sqrt[3]{y^3+2y^2+1}$$
For the sake of comfort, I will switch $x$ and $y$, as I can switch back in the end: $$y=\sqrt[3]{x^3+2x^2+1}$$
This can be seen as a function, which will asymptotically approach $y=x$, which has infinitely many whole whole number $x,y$ pairs for any $x\in\mathbb{Z}$. If we can find the range of $x$ where the distance between that function and $y=x$ is less than 1, then no possible whole number pairs of $x$ and $y$ can exist there. So, we solve the equation
\begin{align*}
1&=\sqrt[3]{x^3+2x^2+1}-x\\
1+x^3+3x^2+3x&=x^3+2x^2+1\\
x^3+3x&=0\\
&\Rightarrow-3\leq x\leq0
\end{align*}
Since this only leaves 4 possible values of $x$, we can test these out manually to find
\begin{align*}
x=-3&\Rightarrow y=-2\\
x=-2&\Rightarrow y=1\\
x=-1&\Rightarrow y=\sqrt[3]{2}\\
x=0&\Rightarrow y=1\
\end{align*}
Answering the question (now with the old variables): The only possible $x,y$ pairs are $(1,0),(1,-2),(-2,-3)$
My question is: Is my reasoning that I wrote in bold correct? Or did I get "lucky" in answering this question?
| With some minor adjustments, your approach seems sensible (you do not have to take cube roots so explicitly and you seem to have swapped $x$ and $y$)
*
*on the reals, cubing is an increasing bijective function
*$2y^2+1 >0$ so $x^3=y^3+2y^2+1 \implies x^3> y^3\implies x>y$
*so any integer solution has $y+1 \le x$ and $(y+1)^3\le x^3$
*$(y+1)^3 \le y^3+2y^2+1 \implies y^2+3y \le 0 \implies -3 \le y \le 0 $ $\implies y \in \{-3,-2,-1,0\}$
*
*$y = -3 \implies y^3+2y^2+1 = -8=(-2)^3$
*$y = -2 \implies y^3+2y^2+1 = 1=1^3$
*$y = -1 \implies y^3+2y^2+1 = 2=(\sqrt[3]2)^3$, and $\sqrt[3]2$ is not an integer
*$y = 0 \implies y^3+2y^2+1 = 1=1^3$
so integer solutions $(x,y)$ are $(-2,-3) ,(1,-2), (1,0)$, as you found
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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General formula for logarithmic cosine integral I am trying to find a general expression for
$$ \int_0^{\pi/2} (\ln(\cos(x)))^n dx $$ for integer $n$, but have not been able to find it online or derive it. I have a good idea that the general form will involve terms with $\zeta(n)$, but little else.
For reference, here are the first few values for $n$ = 1,2,3,4:
$$\int_0^{\pi/2} (\ln(\cos(x))) dx =-\pi\ln(2) $$
$$\int_0^{\pi/2} (\ln(\cos(x)))^2 dx = \frac{\pi}{2} \ln(2)^2 + \frac{\pi^3}{24} $$
$$ = \frac{\pi}{2} \ln(2)^2 + \frac{\pi}{4} \zeta(2) $$
$$\int_0^{\pi/2} (\ln(\cos(x)))^3 dx = - \frac{\pi}{2} \ln(2)^3 -\frac{\pi^3}{8} \ln(2)- \frac{3 \pi}{4} \zeta(3) $$
$$ = - \frac{\pi}{2} \ln(2)^3 -\frac{3\pi}{4}\zeta(2) \ln(2)- \frac{3 \pi}{4} \zeta(3) $$
$$\int_0^{\pi/2} (\ln(\cos(x)))^4 dx = \frac{\pi}{2} \ln(2)^4 + \frac{\pi^3}{4} \ln(2)^2 + {3\pi}\zeta(3)\ln(2) +\frac{19 \pi^5}{480}$$
$$= \frac{\pi}{2}\ln(2)^4 + \frac{3\pi}{2}\zeta(2) \ln(2)^2 + {3\pi}\zeta(3)\ln(2) + \frac{57\pi}{16}\zeta(4)$$
I would enormously appreciate some help. Thank you!
| Recall that $$\int_{0}^{\frac{\pi}{2}} \sin^p(x) \cos^q(x) \: dx = \frac{ \Gamma(\frac{p+1}{2}) \Gamma(\frac{q+1}{2}) }{2 \Gamma(\frac{p+q}{2}+1)} \\ \frac{d^n}{dq^n} \cos^q(x) = \cos^q(x)\ln^n(\cos(x)) $$
so that $$ \int_{0}^{\frac{\pi}{2}} \cos^q(x)\ln^n(\cos(x)) \: dx = \frac{d^n}{dq^n} \frac{ \Gamma(\frac{1}{2})\Gamma(\frac{q+1}{2}) }{2 \Gamma(\frac{q}{2}+1)} \\
\int_{0}^{\frac{\pi}{2}} \ln^n(\cos(x)) \: dx = \lim_{q \to 0}\frac{d^n}{dq^n} \frac{ \Gamma(\frac{1}{2})\Gamma(\frac{q+1}{2}) }{2 \Gamma(\frac{q}{2}+1)}$$
The zeta functions come from the fact that $\psi^{(s)}(1) = \zeta(s+1)(-1)^{s+1} s! $ for $s >0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Sharp Exponential Inequality of $\frac{e^{-a\sqrt{x^2+n^2}}}{\sqrt{x^2+n^2}}$ I am trying to find a sharp inequality of $\frac{e^{-a\sqrt{x^2+n^2}}}{\sqrt{x^2+n^2}}$ s.t. we can write: $$\frac{e^{-a\sqrt{x^2+n^2}}}{\sqrt{x^2+n^2}} \leq \frac{e^{-ax}}{x}f(an) .$$ Here, $a,x>0, n \in \mathbb{N}. $ Any particular technique to answer these kinds of questions will be very helpful. Thanks.
| The RHS approaches infinity as x approaches 0. Also, if $f(an)$ has the value 1, then the RHS is always greater than LHS. The derivative of $\frac{e^{−ax}}{x}$ is negative for all $a,x>0$, and that of the LHS is also negative.
Dividing the LHS by $\frac{e^{−ax}}{x}$ and taking the limit as $x\rightarrow \infty$ would give $f(an)$. We get
\begin{align}\frac{1}{\sqrt{1+\frac{n^2}{x^2}}}\exp{(ax(1-\sqrt{1+\frac{n^2}{x^2}}))}\le f(an)\end{align}
Binomial expansion of the square roots and dismissal of terms of order higher than $\frac{n^2}{x^2}$:
\begin{align}\frac{e^{-\frac{an^2}{2x}}}{1+\frac{n^2}{2x^2}}\le f(an)\end{align}
Thus, \begin{align}1\le f(an)\end{align}
EDIT
A suggestion in the comments states an easy way to do this
\begin{align}\frac{1}{\sqrt{1+\frac{n^2}{x^2}}}\exp{(-\frac{an^2}{\sqrt{n^2+x^2}+x})}\le f(an)\end{align}
which, upon taking the limit as $x\rightarrow \infty$, gives the same result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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