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Integrating the function $f(x,y)=x$ over the area inside the disc $x^2+y^2\le 4$ and outside the disc $(x-1)^2+y^2=1.$
Integrate the function $f(x,y)=x$ over the area inside the disc $x^2+y^2\le 4$ and outside the disc $(x-1)^2+y^2=1.$
My attempt:
I chose shifted polar coordinates $$\begin{cases}x=2+r\cos\varphi\\ y=r\sin\varphi.\end{cases}$$ Then $$\begin{aligned}x^2+y^2&=4\\\iff (2+r\cos\varphi)^2+(r\sin\varphi)^2&=4\\\iff r(2\cos\varphi+r)&=0\end{aligned}$$ and $$\begin{aligned}(x-1)^2+y^2&=1\\\iff (1+r\cos\varphi)^2+(r\sin\varphi)^2&=1\\\iff r(4\cos\varphi)&=0\end{aligned}$$
So, I'm integrating over the set $$\left\{(\varphi,r),\frac\pi2\le\varphi\le\frac{3\pi}2, -2\cos\varphi\le r\le -4\cos\varphi\right\}.$$
The integral becomes $$\begin{aligned}\int_{\pi/2}^{3\pi/2}\int_{-2\cos\varphi}^{-4\cos\varphi}r(2+r\cos\varphi)drd\varphi&=\int_{\pi/2}^{3\pi/2}\left(r^2+\frac{r^3}3\cos\varphi\Big|_{-2\cos\varphi}^{-4\cos\varphi}\right)d\varphi\\&=\int_{\pi/2}^{3\pi/2}12\cos^2\varphi-\frac{56}3\cos^4\varphi d\varphi\\&=\int_{\pi/2}^{3\pi/2}\frac{56}3\cos^2\varphi(1-\cos^2\varphi)d\varphi-\int_{\pi/2}^{3\pi/2}\frac{20}3\cos^2\varphi d\varphi\\&=\frac{14}3\int_{\pi/2}^{3\pi/2}(2\sin\varphi\cos\varphi)^2-\frac{10}3\int_{\pi/2}^{3\pi/2}(1+\cos(2\varphi))d\varphi\\&=\frac{14}3\int_{\pi/2}^{3\pi/2}\frac{1-\cos(4\varphi)}2d\varphi-\frac{10}3\left(\varphi+\frac{\sin(2\varphi)}2\Big|_{\pi/2}^{3\pi/2}\right)\\&=\frac{14}6\left(\varphi-\frac{\sin(4\varphi)}4\Big|_{\pi/2}^{3\pi/2}\right)-\frac{10}3\pi\\&=\frac{14}6\pi-\frac{10}3\pi\\&=\frac{-6}6\pi\\&=-\pi\end{aligned}$$
UPDATE:
I've just found my mistake, instead of dividing by $3,$ I divided by $2$ after the first integration in the $r$ variable.
| You can just exploit symmetries without having to do any explicit integral (or have to deal with polar coordinates) to find the answer. Here's an explaination.
Let $D:=\big\{(x,y) \in \mathbb{R}^2 \mid x^2+y^2\le4\big\}$ and $E:=\big\{(x,y) \in \mathbb{R}^2 \mid (x-1)^2+y^2\le1\big\}$. Since $D = \big\{(x,y) \in \mathbb{R}^2 \mid (-x)^2+y^2\le1\big\}$ and $\forall x,y\in \mathbb{R}^2, f(-x,y)=-f(x,y)$, it follows that
\begin{equation*}
\int_Df(x,y)\mathrm{d}x\mathrm{d}y=\int_Df(-x,y)\mathrm{d}x\mathrm{d}y = -\int_Df(x,y)\mathrm{d}x\mathrm{d}y
\end{equation*}
which implies $\int_Df(x,y)\mathrm{d}x\mathrm{d}y=0$.
On the other hand $E\subset D$. It follows that
\begin{equation*}
\int_{D\backslash E}f(x,y)\mathrm{d}x\mathrm{d}y=\int_Df(x,y)\mathrm{d}x\mathrm{d}y-\int_Ef(x,y)\mathrm{d}x\mathrm{d}y = -\int_Ef(x,y)\mathrm{d}x\mathrm{d}y.
\end{equation*}
Now, define $g \colon \mathbb{R}^2\to \mathbb{R}, (x,y) \mapsto x-1 $ and $F:= \big\{(x,y) \in \mathbb{R}^2 \mid x^2+y^2\le1\big\}$. Since $F = \big\{(x,y) \in \mathbb{R}^2 \mid (-x)^2+y^2\le1\big\}$, it follows that:
\begin{equation*}
\int_E g(x,y)\mathrm{d}x\mathrm{d}y= \int_F f(x,y)\mathrm{d}x\mathrm{d}y = \int_F f(-x,y)\mathrm{d}x\mathrm{d}y =-\int_F f(x,y)\mathrm{d}x\mathrm{d}y
\end{equation*}
which implies $\int_E g(x,y)\mathrm{d}x\mathrm{d}y=0$.
It follows that
\begin{equation*}
\int_Ef(x,y)\mathrm{d}x\mathrm{d}y = \int_E g(x,y)\mathrm{d}x\mathrm{d}y + \int_E 1\mathrm{d}x\mathrm{d}y = 0+ |E|=\pi.
\end{equation*}
In conclusion
\begin{equation*}
\int_{D\backslash E}f(x,y)\mathrm{d}x\mathrm{d}y= -\int_Ef(x,y)\mathrm{d}x\mathrm{d}y = -\pi
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Upper bound of $\sum_{n=1}^N |1-z^n|$ where $|z| \leq 1$ How to derive an upper bound of
$$\sum_{n=1}^N |1-z^n|$$ where $z\in\mathbb{C}$ and $|z| \leq 1$?
A trivial upper bound would be $2N$ since each $|1-z^n| \leq 2$. But I am hoping for tighter bounds. I ran some numerical experiments and believe the bound should be $\frac{3}{2}N$ but don't know how to prove it.
| Some thoughts:
As already noted, the maximum of
$\sum_{n=1}^N |1 - z^n|$ is attained on $|z| = 1$.
Let $z = \mathrm{e}^{\mathrm{i}\theta}$
with $\theta \in [0, 2\pi]$.
We have
$$\sum_{n=1}^N |1 - z^n| = \sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|.$$
Conjecture 1: The maximum of $\sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|$ on $[0, 2\pi]$ is attained at some $x_0 \in [0, 2\pi/N]$.
(Note: Numerical experiment supports the claim. )
If Conjecture 1 is true, when
$\theta \in [0, 2\pi/N]$,
we have
$$
\sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|
= \sum_{n=1}^N 2\sin \frac{n\theta}{2}
= \frac{2\sin^2 \frac{N\theta }{4} \cos \frac{\theta}{4}}{\sin \frac{\theta}{4}} + \sin \frac{N\theta}{2}
= \frac{2\sin^2 y \cos \frac{y}{N}}{\sin \frac{y}{N}} + \sin 2y
$$
where $y = N\theta/4 \in [0, \pi/2]$.
Then we have
$$\frac{2\sin^2 y \cos \frac{y}{N}}{\sin \frac{y}{N}} + \sin 2y
\le \frac{2\sin^2 y}{\frac{2y}{\pi}\sin \frac{\pi}{2N}} + 1 = \frac{\pi}{\sin \frac{\pi}{2N}}\frac{\sin^2 y}{y} + 1 \le \frac{\pi}{\sin \frac{\pi}{2N}}\, \frac{29}{40} + 1$$
where we have used
$\sin \frac{y}{N} \ge \frac{2y}{\pi}\sin \frac{\pi}{2N}
$ and $\frac{\sin^2 y}{y} \le \frac{29}{40}$ for all $y\in [0, \pi/2]$.
It is not difficult to prove that, for all $N > 21$,
$$\frac{\pi}{\sin \frac{\pi}{2N}}\, \frac{29}{40} + 1 < \frac{3}{2}N.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Error in computing a limit I was given the following limit to compute,
$$\lim_{k\to\infty}\left[k-\sqrt{k^{2}+1}\right]$$
My approach:
$$= \lim_{k\to\infty}\left[k-\sqrt{k^{2}(1+k^{-2})}\right]$$
$$= \lim_{k\to\infty}[k-k]=0$$
So the following evaluates to $0$. But my book gives that answer is $\frac{-1}{2}$. Where did my process go wrong?
| $\require{\cancel}$
Your book is incorrect. Since $(a-b)(a+b) = a^2-b^2$ we get for $k>0$
\begin{align}
k - \sqrt{k^2 +1}& = k - \sqrt{k^2 +1} \left(\frac{k + \sqrt{k^2 +1}}{k + \sqrt{k^2 +1}} \right) \\
&= \frac{\cancel{k^2} - (\cancel{k^2} +1)}{k + \sqrt{k^2 +1}} \\
&= - \frac{1}{k + k\sqrt{ 1+\frac{1}{k^2}}} \\
& = -\frac{1}{k}\cdot\frac{1}{1+\sqrt{ 1+\frac{1}{k^2}}}
\end{align}
And since both $\lim_{k \to \infty}-\frac{1}{k} =0$ and $\lim_{k \to \infty}\frac{1}{1+\sqrt{ 1+\frac{1}{k^2}}} = \frac{1}{2}$ exist, your original limit is the product of these limits. Hence
$$
\lim_{k \to \infty}k - \sqrt{k^2 +1} =0\cdot \frac12 =0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Solve the integral $\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx$ Question
Solve the integral,$$\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx$$
Attempt
I converted the equation in terms of $\sin(x)$ and $\cos(x)$ using the definition of $\tan(x)$ and $\cot(x)$, and then applied the substitution $t=\sin(x)$, however, this has proved itself to be quite a difficult integral to resolve in and of itself.
I would be great full for any suggestions of a more compact method. Any hints would be also be greatly appreciated.
| Clear
$$ \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx=2\int_{0}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x} dx. $$
Let
$$ I=\int_{0}^{\frac{\pi}2} \frac{\sin^3x}{\tan^3x+\cot^3x}dx. \tag1$$
Then
$$ I=\int_{0}^{\frac{\pi}2} \frac{\cos^3x}{\tan^3x+\cot^3x}dx. \tag2$$
Adding (1) to (2) gives
\begin{eqnarray}
2I&=&\int_{0}^{\frac{\pi}2} \frac{\sin^3x+\cos^3x}{\tan^3x+\cot^3x} dx\\
&=&\int_{0}^{\frac{\pi}2} \frac{(\sin x+\cos x)(1-\sin x\cos x)}{(\tan x+\cot x)(\tan^2x+\cot^2x-1)} dx\\
&=&\int_{0}^{\frac{\pi}2} \frac{\sin^3 x\cos^3 x(\sin x+\cos x)(1-\sin x\cos x)}{\sin^4x+\cos^4x-\sin^2x\cos^2x} dx\\
&=&\int_{0}^{\frac{\pi}2} \frac{\sin^3 x\cos^3 x(\sin x+\cos x)(1-\sin x\cos x)}{1-3\sin^2x\cos^2x} dx\\
&=&\int_{0}^{\frac{\pi}2} \frac{\sin^4 x\cos^3 x}{1-3\sin^2x\cos^2x} dx+\int_{0}^{\frac{\pi}2} \frac{\sin^3 x\cos^4 x}{1-3\sin^2x\cos^2x} dx-\int_{0}^{\frac{\pi}2} \frac{\sin^5 x\cos^4 x}{1-3\sin^2x\cos^2x} dx-\int_{0}^{\frac{\pi}2} \frac{\sin^4 x\cos^5 x(1-\sin x\cos x)}{1-3\sin^2x\cos^2x} dx\\
&=&2\int_{0}^{\frac{\pi}2} \frac{\sin^4 x\cos^3 x}{1-3\sin^2x\cos^2x} dx-2\int_{0}^{\frac{\pi}2} \frac{\sin^5 x\cos^4 x}{1-3\sin^2x\cos^2x} dx\\
&=:&2J_1-2J_2.
\end{eqnarray}
Note that, under $t=\sin x$,
\begin{eqnarray}
J_1&=&\int_{0}^{\frac{\pi}2} \frac{\sin^4 x\cos^3 x}{1-3\sin^2x\cos^2x} dx\\
&=&\int_{0}^{\frac{\pi}2} \frac{\sin^4 x\cos^2 x}{1-3\sin^2x\cos^2x} d\sin x\\
&=&\int_{0}^{1} \frac{t^4(1-t^2)}{1-3t^2(1-t^2)} dt
\end{eqnarray}
which is not hard to handle by partial fractions. You can do the same thing for $J_2$. I omit the details.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Prove that $\binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m$ for $m\in \mathbb{N}\backslash \{0,1\}$ Prove the following inequality for every $m\in \mathbb{N}\backslash \{0,1\}$:
$$ \binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m.$$
By some computational arguments, the inequality seems to be true and in particular the difference between the first term $\binom{13+m}{m}$ and the second one $(m+1)\binom{6+m}{m}$ is very large, so obviously is more that $m$. Hence the use of the induction technique, in order to find a way to give a formal proof, leads to a disaster consequence.
In fact, for $m=2$ the inequality is trivial. Fix $m> 2$ and suppose the the inequality is true for $m$. We want to prove that
$$ \binom{14+m}{m+1}-(m+2)\binom{7+m}{m+1}\geq m+1.$$
Start from the first member and apply Stifiel's formula
$$ \binom{14+m}{m+1}-(m+2)\binom{7+m}{m+1}=\binom{13+m}{m+1}+\binom{13+m}{m}-(m+1+1)\Big[\binom{6+m}{m+1}+\binom{6+m}{m}\Big]=
\binom{13+m}{m}-(m+1)\binom{6+m}{m}+\binom{13+m}{m+1}-(m+1)\binom{6+m}{m+1}-\binom{6+m}{m+1}-\binom{6+m}{m}\geq...
$$
Applying the inductive hypothesis we have:
$$...\geq m+\binom{13+m}{m+1}-(m+1)\binom{6+m}{m+1}-\binom{6+m}{m+1}-\binom{6+m}{m},$$
but this is less than $m+1$. Therefore we don't get the desired conclusion.
Which can be an useful stategy to prove this inequality?
| I agree that the problem has been reduced to proving that for all $m \in \Bbb{Z_{\geq 2}}$,
$$\binom{13+m}{m+1}-(m+1)\binom{6+m}{m+1}-\binom{6+m}{m+1}-\binom{6+m}{m} \geq 1. \tag1 $$
In (1) above, on the LHS, the two rightmost terms may be collapsed into
$$\binom{13+m}{m+1}-(m+1)\binom{6+m}{m+1}-\binom{7+m}{m+1} \geq 1. \tag2 $$
Empirically, (2) above holds for $m = 2$. Induction may be used to show that (2) above holds for all $m \in \Bbb{Z_{\geq 2}}.$
This is done by showing that for all $m \in \Bbb{Z_{\geq 2}}$, you have that
$$\binom{14+m}{m+2} - \binom{13+m}{m+1} \tag{A-1}$$
$$\geq \left[(m+2)\binom{7+m}{m+2} - (m+1)\binom{6+m}{m+1} \right] \tag{A-2}$$
$$+ \binom{8+m}{m+2} - \binom{7+m}{m+1}. \tag{A-3}$$
(A-1) above collapses into $~\displaystyle \binom{m+13}{m+2}.$
(A-2) above collapses into $~\displaystyle \left[(m+1)\binom{m+6}{m+2}\right] + \binom{m+7}{m+2}.$
(A-3) above collapses into $~\displaystyle \binom{m+7}{m+2}.$
So, the inequality in (A-1),(A-2) and (A-3) above is equivalent to the assertion that for all $m \in \Bbb{Z_{\geq 2}},$
$$\binom{m+13}{m+2} \geq \left[(m+1)\binom{m+6}{m+2}\right] + \binom{m+7}{m+2} + \binom{m+7}{m+2} $$
which is equivalent to the assertion that
$$\binom{m+13}{m+2} \geq \left[(m+1)\binom{m+6}{m+2}\right] + 2\binom{m+7}{m+2}. \tag3 $$
The exploration of (3) above may be simplified by dividing each term by $~\displaystyle \binom{m+6}{m+2}.$
This yields the equivalent assertion of
$$\frac{[(m+13)!]~~(4!)}{[(m+6)!]~~ (11!)} \geq (m+1) + (2)\frac{m+7}{5} = \frac{7m + 19}{5}. \tag4 $$
Empirically, (4) above is true for $m=2$.
Assume that (4) above is true for $m$.
As $m \to (m+1)$, the LHS of (4) above is multiplied by
$~\displaystyle \frac{m+14}{m+7}$,
while the RHS of (4) above is multiplied by
$~\displaystyle \frac{7m+26}{7m+19}.$
Note that $~\displaystyle (m+14)(7m+19) = 7m^2 + 117m + 266$
while $~\displaystyle (m+7)(7m+26) = 7m^2 + 75m + 182.$
Therefore, in (4) above, as $m \to (m+1)$, the LHS is multiplied by a larger factor than the RHS.
Therefore, (4) above is true for any $m \in \Bbb{Z_{\geq 2}}$, as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate the sum of $\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}$ I hava an infinite sum
$$\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}$$
I factored the denominator
$$\sum_{n=0}^{\infty}\frac{1}{\left(3n+1\right)\left(n+1\right)}$$
Then I separated the fraction
$$\frac{1}{2}\sum_{n=0}^{\infty}\frac{3}{\left(3n+1\right)}-\frac{1}{\left(n+1\right)}$$
Then I set 1 (the numerator) to be equal to x to some power which I don't know if I can do
$$\frac{3}{2}\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1}-\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$$
Then I set the integral which would satisfy the previous terms
$$\frac{3}{2}\sum_{n=0}^{\infty}\int_{0}^{1}x^{3n}dx$$
and
$$-\frac{1}{2}\sum_{n=0}^{\infty}\int_{0}^{1}x^{n}dx$$
Then I changed the order of summation and integration and I got
$$\frac{3}{2}\int_{0}^{1}\frac{1}{1-x^{3}}dx$$ and $$-\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}dx$$
The first integral can be factored to
$$\frac{3}{2}\int_{0}^{1}\frac{1}{\left(1-x\right)\left(1+x+x^{2}\right)}dx$$
Then separated
$$\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}+\frac{x+2}{1+x+x^{2}}dx$$
The first one will cancel out with
$$-\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}dx$$
And I'm left with
$$\frac{1}{2}\int_{0}^{1}\frac{x+2}{1+x+x^{2}}dx$$
which is
$$\frac{\sqrt{3}\pi}{12}+\frac{\ln\left(3\right)}{4}$$
But the correct answer is
$$\frac{\sqrt{3}\pi}{12}+\frac{3\ln\left(3\right)}{4}$$
So I would like to ask if this approach is invalid or if I'm just missing something.
| Hi welcome to MSE:As a hint for telescopic series$$\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}
\\=\sum_{n=0}^{\infty}\frac{1}{(3n+1)(n+1)}
\\=\sum_{n=0}^{\infty}\frac{3}{(3n+1)(3n+3)}
\\=3\sum_{n=0}^{\infty}\frac{(3n+2)}{(3n+1)(3n+2)(3n+3)}
\\=3\sum_{n=0}^{\infty}\frac{(3n+3)-1}{(3n+1)(3n+2)(3n+3)}
\\=3\sum_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)}-3\sum_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Quadratic With Periodic Coefficents I've come across a problem that results in the equation:
$t^2 -2t\sin t -2\cos t -2 = 0$
I've tried to do this analytically but I can't figure it out. At this point, I just want to know if it's even possible for something like this to be solved exactly. So can an equation $ax^2+bx+c=0$, where $a, b, \text{or } c$ are a trig function, be solved?
| Using graphics or inspection, the zero of function
$$f(t)=t^2 -2t\sin( t) -2\cos (t) -2 $$is just above to $\frac {2\pi}3$
$$f\left(\frac{2 \pi }{3}\right)=-1-\frac{2 \pi }{\sqrt{3}}+\frac{4 \pi ^2}{9}=-0.241108\cdots$$
Expanded as series
$$f(t)=\left(-1-\frac{2 \pi }{\sqrt{3}}+\frac{4 \pi ^2}{9}\right)+2\pi \left(t-\frac{2 \pi }{3}\right)+\left(\frac{3}{2}+\frac{\pi }{\sqrt{3}}\right)\left(t-\frac{2 \pi }{3}\right)^2 +$$
$$\frac 13 \sum_{n=2}^\infty \frac { \left(2 \pi -3 \sqrt{3} (n-1)\right) \sin \left(\frac{\pi n}{2}\right)-\left(3 (n-1)+2 \sqrt{3} \pi \right) \cos \left(\frac{\pi n}{2}\right)} { n!} \left(t-\frac{2 \pi }{3}\right)^n$$
Truncate to some low order and use series reversion to obtain, as an approximation,
$$t=\frac{2 \pi }{3}+x-\left(\frac{1}{2 \sqrt{3}}+\frac{3}{4 \pi }\right)
x^2+\left(\frac{2}{9}+\frac{9}{8 \pi ^2}+\frac{1}{\sqrt{3} \pi }\right)
x^3-$$ $$\frac{\left(3645+1350 \sqrt{3} \pi +1152 \pi ^2+176 \sqrt{3} \pi ^3\right)
}{1728 \pi ^3}x^4+O\left(x^{5}\right)$$ with $x=\frac{1}{\sqrt{3}}+\frac{1}{2 \pi }-\frac{2 \pi }{9}$.
This truncated series gives an explicit expression : its decimal representation is
$t=\color{red}{2.132020}087\cdots$ while, as given by Newton method, the solution is $t=\color{red}{2.132020146}\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do you find this limit with a relationship to $e$ using Taylor series? The limit in question is
$$
\lim_{x \to 0}\left(\frac{\sin(x)}{x}\right)^{1/x^2}
$$
When I replace $\sin(x)$ with its Taylor series about $0$ and cancel out the $x$, I get
$$
\lim_{x \to 0}\left(1-\frac{x^2}{6} + \frac{x^4}{5!} \mp \cdots\right)^{1/x^2}
$$
The answer in the book is $e^\frac{-1}{6}$. If I only look at the $x^2$ term, I can see where $-1/6$ comes from. I'm just not sure how I can definitively say the answer is $e^\frac{-1}{6}$. Why can I discount the other powers of $x$?
| To handle the higher power terms, you can use the big O notation.
Note that
$$
\frac{\sin x}{x}=\frac1x\left(x-\frac{x^3}{6}+O(x^5)\right)=1-\frac{x^2}{6}+O(x^4)
$$
and
$$
\ln(1-y)=-y-\frac{y^2}{2}-\frac{y^3}{3}+O(y^5)
$$
So
$$
\begin{align}
\left(\frac{\sin x}{x}\right)^{1/x^2}
=&\exp\left(\frac{1}{x^2}\ln(\frac{\sin x}{x})\right)\\
=&\exp\left(\frac{1}{x^2}\ln(1-\frac{x^2}{6}+O(x^4))\right)\\
=&\exp\left(\frac{1}{x^2}(-\frac{x^2}{6}+O(x^4))\right)\\
=&\exp\left(-\frac{1}{6}+O(x^2)\right)\to e^{-1/6}
\end{align}
$$
as $x\to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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If $ab+1 = r^2$ for $a,b,r \in \mathbb{N},$ how to show that $\gcd(2a(r+b)+1,2b(r+a)+1) = 1?$ Let $a<b$ be positive integers such that $ab+1 = r^2$ for some $r \in \mathbb{N}.$ If $m_1 = 2a(r+b)+1$ and $m_2 = 2b(r+a)+1.$ I want to find the possible values of $\gcd(m_1,m_2).$ I had taken some examples in the range $a,b\in\{1,2,\dots 1000\},$ and found the corresponding values as $\gcd(m_1,m_2) = 1.$ Is $\gcd(m_1,m_2) = 1$? If yes, how to prove it? If no, provide a counter example.Only thing that I could see is $\gcd(a,r)=\gcd(b,r) = 1.$
| We are given that $ab+1=r^{2}$.
Note
$$2a(r+b)+1 = 2ar + 2ab+1 = 2ar+2r^2-1$$
$$2b(r+a)+1= 2br+2r^2-1$$
We claim that the gcd of the above two quantities is $1$.
$\textbf{For proof by contradiction}$ suppose that there is some prime $p$ that divides both $2ar+2r^2-1$ and $2br+2r^2-1$. Then note that $ p \not \div 2r, p \not \div a, p \not \div b$.
If $p \div 2r^2-1$, then $p|a$ and $p|b$, but $(ab, 2r^2-1) = (r^2-1,2(r^2-1)+1) = 1$ $\Rightarrow$ Thus $p \not \div 2r^2-1$
From $2ar+2r^2-1 \equiv 0 \mod p$ we get $a \equiv (1-2r^2)(2r)^{-1}\mod p $
From $ab = r^2 -1$ we get $b \equiv (r^2-1)(a)^{-1} \equiv (2r^3-2r)(1-2r^2)^{-1} \mod p$
Note that we also have $a \equiv b \mod p$ as $2ar+2r^2-1 \equiv 2br+2r^2-1 \mod p$. Thus
$$(1-2r^2)(2r)^{-1} \equiv (2r^3-2r)(1-2r^2)^{-1} \mod p$$
$$ \Rightarrow (1-2r^2)^2 \equiv 4r^4-4r^2 \mod p$$
$$\Rightarrow 1 \equiv 0 \mod p$$
Which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to show that $\frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{n^2+1} \le \frac{3\pi}{4}$ How to show that $\frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{n^2+1} \le \frac{3\pi}{4}$ ?
My Attempt : I was using Integral Test of a Series. I got $\int_0^\infty \frac{1}{1+x^2} \le \sum_{n=0}^\infty \frac{1}{1+n^2} \le \frac{1}{1+0^2} + \int_0^\infty \frac{1}{1+x^2}$ which gives $ \frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{1+n^2} \le 1+ \frac{\pi}{2}$.
Can anyone please help me by giving any hint ?
| We shall give a stronger bound. Note that
$$ 2=1+\sum^\infty_{n=1}\frac{1}{n^2+n}<\sum^\infty_{n=0}\frac{1}{n^2+1}<1+\frac{1}{2}+\sum^\infty_{n=2}\frac{1}{n^2}=\frac{1}{2}+\frac{\pi^2}{6}$$
which is a stronger bound.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find all values of $z$ such that $z^3=−8$ The question is
Solve the equation
$z^3=-8$
My attempt
I attempt to write it out in polar co-ordinates
since $z = r(\sin(x) + i\sin(x)) \\ z^3 = r^3(\cos(3x) + i\sin(3x))$
then
$r^3\sin(3x) = -8$ and $r^3\cos(3x) = 0$
but from here I'm not really sure where to go , I've searched up the solution to this before and people have written $r^3 = 8$ so $\cos(3x) = -1$ and $\sin(3x) = 0$ .
| We could also apply the "triple-angle" identities to write $ \ \sin(3 \theta) \ = \ \sin \theta · (3 - 4 \sin^2 \theta) \ = \ 0 \ \ $ and $ \ \cos(3 \theta) \ = \ 4 \cos^3 \theta - 3 \cos \theta \ = \ -1 \ \ . $
From the first of these, we have either $ \ \sin \theta \ = \ 0 \ \Rightarrow \ \theta \ = \ 0 \ , \ \pi \ \ $ or $ \ \sin^2 \theta \ = \ \frac34 \ \Rightarrow \ \cos^2 \theta \ = \ \frac14 \ \ . $ Inserting these results into the second equation, we find that $ \ \theta \ = \ 0 \ $ is not a solution, but $ \ \theta \ = \ \pi \ $ is $ \ ( \ \cos(3 \pi) \ = \ 4 · [-1]^3 - 3 ·[-1] \ = \ -1 \ ) \ . $ For $ \ \cos \theta \ = \ \pm \frac12 \ \ , $ we see that $ \ 4 · \left[-\frac12 \right]^3 - 3 ·\left[-\frac12 \right] \ = \ +1 \ \ , $ but $ \ 4 · \left[+\frac12 \right]^3 - 3 ·\left[+\frac12 \right] \ = \ -1 \ \ . $
Hence the three angle solutions are $ \ \theta \ = \ \pi \ , \ \pm \ \frac{\pi}{3} \ \ , $ which gives the three solutions to the original equation (the three complex cube-roots of $ \ -8 \ $ as
$$ 2·(cis \ \pi) \ \ = \ \ -2 \ \ \ , \ \ \ 2·\left(cis \ \pm \frac{\pi}{3} \right) \ \ = \ \ 1 \ \pm \ i\sqrt3 \ \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$\int \frac{5x^3+2}{\sqrt{x^3+1}}dx$ How do I integrate $\int \frac{5x^3+2}{\sqrt{x^3+1}}dx$ ?
I know that the result is $2x\sqrt{x^3+1}$, but I cannot think of a way to get to it.
| This is a subtle problem with integration by parts.
We begin by noting that integrals with $\sqrt{x^3+1}$ ordinarily require elliptic functions. Our objective is to try to eliminate the integral.
Begin by decomposing the integrand thusly, separating a term that has a higher power ($+1/2$ instead of $-1/2$) of the radicand:
$\dfrac{5x^3+2}{\sqrt{x^3+1}}=\dfrac{5x^3+5}{\sqrt{x^3+1}}-\dfrac{3}{\sqrt{x^3+1}}$
$=5\sqrt{x^3+1}-\dfrac{3}{\sqrt{x^3+1}}.$
So
$\int\dfrac{(5x^3+2)dx}{\sqrt{x^3+1}}=5\int\sqrt{x^3+1}dx-3\int\dfrac{dx}{\sqrt{x^3+1}}.$
We now integrate the first term on the right by parts:
$\int\dfrac{(5x^3+2)dx}{\sqrt{x^3+1}}=5x\sqrt{x^3+1}-5\int x[d\sqrt{x^3+1}]-3\int\dfrac{dx}{\sqrt{x^3+1}}+C$
$\text{[Properly, an indefinite integration by parts should include the constant.}$
$\text{Doing so avoids a fallacy in certain cases.]}$
$=5x\sqrt{x^3+1}-(5/2)\int\dfrac{3x^3 dx}{\sqrt{x^3+1}}-3\int\dfrac{dx}{\sqrt{x^3+1}}+C$
$=5x\sqrt{x^3+1}-\int\dfrac{[(15/2)x^3+3] dx}{\sqrt{x^3+1}}+C$
And then
$\color{blue}{\int\dfrac{(5x^3+2)dx}{\sqrt{x^3+1}}}=5x\sqrt{x^3+1}-\dfrac32\color{blue}{\int\dfrac{(5x^3+2)dx}{\sqrt{x^3+1}}}+C,$
in which the integrals in blue are now identical. We may therefore combine them on the left side and solve algebraically. Note that since $C$ is arbitrary, we need not multiply it by $2/5$ in the final result, which matches that given in the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given a triangle ABC inscribed in the unit circle , the 3 vertices could be described via 3 complex number, namely, $a$, $b$, and $c$.
Now $AD$ is an altitude, $D$ is the foot of $AD$ on $BC$.
Prove: $$D = \frac{a+b+c}2 - \frac{bc}{2a}$$
So far my progress is --
*
*the circumcentre of triangle $ABC$ is just $O = 0$.
*its centroid $G = \frac{a+b+c}3$.
*by Euler line, the orthocentre $H = a+b+c$.
*also, we can see that the centre of the nine point circle $N =
\frac{a+b+c}2$.
But then I'm a bit stuck. $D$ is on the line $AH$ and $BC$, but I couldn't reach the conclusion to be proved.
| line $AH$:
$$\frac{z-a}{b+c} = \overline{(\frac{{z-a}}{b+c})}$$
line $BC$:
$$\frac{z-b}{c-b} = \overline{(\frac{{z-b}}{c-b})}$$
so in the first equation:
$$\bar z = \overline{b+c} \cdot \frac{z-a}{b+c} + \bar a$$
in the second equation:
$$\bar z = \overline{c-b} \cdot \frac{z-b}{c-b} + \bar b$$
now $$a \bar a = b \bar b = c \bar c = 1$$
so:
$$\overline{c-b} \cdot \frac{z-b}{c-b} + \bar b = \overline{b+c} \cdot \frac{z-a}{b+c} + \bar a$$
$$(\frac1c-\frac1b) \cdot \frac{z-b}{c-b} + \frac1b = (\frac1b+\frac1c)\cdot \frac{z-a}{b+c} + \frac1 a$$
$$\frac{b-z}{bc} + \frac1b = \frac{z-a}{bc} + \frac1 a$$
$$\frac1c+ \frac1b -\frac1a= \frac{2z-a}{bc} $$
$$z = \frac{a+b+c}2 - \frac{bc}{2a}$$
our answers are a bit different. Did I miss something?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Evaluate the line integral along a parabola
Evaluate the line integral of $f(x,y)=-y+x$ along part of the parabola $y=2(x+1)^2$ from the point $(0,2)$ to the point $(-1,0)$
I need help trying to find a good parameterization for this because what I've done just lands me in a mess.
My work so far:
Let $x=t, y=2(t+1)^2$
$$
\begin{split}
r(t) &= \left<t,2(t+1)^2\right>, \quad -1\leq t \leq 0 \\
r'(t) &= \left<1,4(t+1)\right>, \quad -1\leq t \leq 0 \\
\|r'(t)\| &= \sqrt{1+16(t+1)^2} \\
-y+x &= -2(t+1)^2+t\\
&=-2t^2-3t-2\\
\end{split}
$$
So we have
$$
\int_0^{-1}\left(-2t^2-3t-2\right)\sqrt{1+16(t+1)^2}dt
$$
This integral is really ugly. so then I tried a different method:
$$
\begin{split}
y &= 2(x+1)^2 \\
\frac{dy}{dx} &= 4x+4 \\
dS &= \sqrt{1+(4x+4)^2} dx\\
&=\sqrt{16x^2+32x+17} dx
\end{split}
$$
So we get
$$\int_0^{-1} \left(-2(x+1)^2+x\right)\sqrt{16x^2+32x+17}dx$$
Again, very ugly. Can someone please help me solve this?
| Here we apply a substitution to get rid of the square root. Nevertheless the evaluation of the integral is somewhat cumbersome. We have a function
\begin{align*}
&f:\mathbb{R^2}\to\mathbb{R}\\
&f(x,y)=x-y
\end{align*}
and a curve
\begin{align*}
C=\{(x,y):y=2(x+1)^2,-1\leq x\leq 0\}
\end{align*}
where we want to evaluate the line integral clockwise along $C$ from the point $(0,2)$ to $(-1,0)$. The parametrisation in this case is
\begin{align*}
x&=x(t)\\
y&=y(t)=2(x(t)+1)^2\\
r^{\prime}(t)&=\left(x^{\prime}(t),4\left(x(t)+1\right)x^{\prime}(t)\right)\\
\|r^{\prime}(t)\|&=\left|x^{\prime}(t)\right|\,\sqrt{1+16\left(x(t)+1\right)^2}\tag{1}
\end{align*}
Looking at (1) indicates the substitution
\begin{align*}
\color{blue}{x}&=\color{blue}{\frac{1}{4}\sinh(t)-1}\\
y&=\frac{1}{8}\sinh^2(t)\\
\|r^{\prime}(t)\|&=\frac{1}{4}\left|\cosh(t)\right|\,\sqrt{1+\sinh^2(t)}\\
&=\frac{1}{4}\cosh^2(t)
\end{align*}
The variable change transforms the interval of integration into
\begin{align*}
-1\leq &x\leq 0\\
0\leq &t\leq \sinh^{-1}(4)
\end{align*}
We so obtain with some help of Wolfram Alpha
\begin{align*}
&\color{blue}{\int_{\sinh^{-1}(4)}^0\left(-\frac{1}{8}\sinh^2(t)+\frac{1}{4}\sinh(t)-1\right)\frac{1}{4}\cosh^2(t)dt}\\
&\qquad=\frac{1}{32}\int_{0}^{\sinh^{-1}(4)}\left(\sinh^4(t)-2\sinh^3(t)+9\sinh^2(t)-2\sinh(t)+8\right)dt\\
&\qquad\,\,=\frac{1}{768}\left(16+508\sqrt{17}+93\sinh^{-1}(4)\right)\color{blue}{\simeq 3.001\,8}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Find $\tan\frac{\alpha}{2},$ if $\sin\alpha+\cos\alpha=\frac{\sqrt7}{2}$ and $\alpha\in(0^\circ;90^\circ)$. Find $\tan\dfrac{\alpha}{2},$ if $\sin\alpha+\cos\alpha=\dfrac{\sqrt7}{2}$ and $\alpha\in(0^\circ;90^\circ)$.
So we have $$\sin\alpha+\cos\alpha=\dfrac{\sqrt7}{2}$$ Then $$\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha=\dfrac74\\2\sin\alpha\cos\alpha=\dfrac74-1=\dfrac34\\\sin2\alpha=\dfrac34\\\dfrac{2\tan\alpha}{1+\tan^2\alpha}=\dfrac34\\\tan\alpha=\dfrac{4\pm\sqrt{7}}{3},$$ Is there a more direct approach?
| Hint:
$\sin(x)= \sin(2\frac{x}{2})=\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})},$
and $\cos(x)= \cos(2\frac{x}{2})=\frac{1-\tan^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}.$
Try to do this in $\sin(\alpha)+\cos(\alpha)=\frac{\sqrt{7}}{2}$,and after some algebraic manipulation you can put $ \tan(\frac{x}{2})=t $ and then solve the equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Difficult integral from general relativity $\int_{a}^{0} \frac{1+mu}{\sqrt{u-a} \sqrt{b-u}} \,du$ When solving for the angular deflection of a light ray passing close to a star in general relativity, one needs to compute an elliptic integral. By some clever factoring and application of a relevant approximation, one reduces the problem to
$$\Delta \phi = -2\int_{a}^{0} \frac{1+mu}{\sqrt{u-a} \sqrt{b-u}} \,du$$
I know (and have verified with Mathematica) that this has exact solution
$$\Delta \phi = 2m\sqrt{-ab}+[4+2m(a+b)]\arctan\left(-\sqrt\frac{a}{b}\right),$$
but do not see how to replicate this solution. I've tried reverse engineering this in the case integration by parts was used but can't quite figure it out. Also, is there a complex contour integration approach - perhaps on the dogbone contour? Any hint or point to relevant literature would be appreciated. Thanks.
| Hmm. We have
\begin{align*}
-2\int_{a}^{0} \frac{1+mu}{\sqrt{u-a} \sqrt{b-u}} \,du
&=2\int_{0}^{a} \frac{1+mu}{\sqrt{(u-a)(b-u)}} \,du \\
&=2\int_{0}^{a} \frac{1+mu}{\sqrt{-(u^2-u(b+a)+ab)}} \,du \\
&=2\int_{0}^{a} \frac{1+mu}{\sqrt{-\left(\left(u-\frac{b+a}{2}\right)^{\!2}-\frac{(b-a)^2}{4}\right)}} \,du \\
&=2\int_{0}^{a} \frac{1+mu}{\sqrt{\frac{(b-a)^2}{4}-\left(u-\frac{b+a}{2}\right)^{\!2}}} \,du.
\end{align*}
From here, it seems advisable to substitute
\begin{align*}
v&=u-\frac{b+a}{2}\\
dv&=du
\end{align*}
to obtain
\begin{align*}
\Delta\phi
&=2\int_{-(b+a)/2}^{(a-b)/2} \frac{1+m(v+(b+a)/2)}{\sqrt{\frac{(b-a)^2}{4}-v^2}} \,dv \\
&=2\int_{-(b+a)/2}^{(a-b)/2} \frac{(1+m(a+b)/2)+mv}{\sqrt{\frac{(b-a)^2}{4}-v^2}} \,dv \\
&=(2+m(a+b))\int_{-(b+a)/2}^{(a-b)/2} \frac{1}{\sqrt{\frac{(b-a)^2}{4}-v^2}} \,dv+2m\int_{-(b+a)/2}^{(a-b)/2} \frac{v}{\sqrt{\frac{(b-a)^2}{4}-v^2}} \,dv.
\end{align*}
The antiderivative of the first gives you the $\arctan$ function via trigonometric substitution, and the second is the square root function via substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$ given $x = \frac{1}{2-\sqrt{3}}$? It is given that $x = \frac{1}{2-\sqrt{3}}$. Find the value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$.
Well I tried rationalising and I came to know that $x = 2 + \sqrt{3}$. And I know that directly putting the values wont help either but I am not able to factorize the polynomial or manipulate it to help me.
I would be grateful if anybody could help me.
| I calculated it longhand:
$$ \begin{align} x & = 2 + \sqrt 3 \\ x^2 & = 2^2 + 3 + 2 \cdot 2 \sqrt 3 \\ & = 7 + 4 \sqrt 3 \\ x^3 & = 2 \cdot 7 + 4 \cdot 3 + 2 \cdot 4 \sqrt 3 + 7 \sqrt 3 \\ & = 26 + 15 \sqrt 3 \\ x^4 & = 97 + 56 \sqrt 3 \\ x^5 & = 362 + 209 \sqrt 3 \\ x^6 & = 1351 + 780 \sqrt 3 \end{align} $$
We also have
$$ 2 \sqrt 3 x^5 = 209 \cdot 6 + 362 \cdot 2 \sqrt 3 = 1254 + 724 \sqrt 3 $$
Then
$$ \begin{align} & x^6 - 2 \sqrt 3 x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt 3 \\ & = 1351 - 1254 - 97 + 26 - 28 + 4 + (780 - 724 - 56 + 15 - 16 + 2 - 1) \sqrt 3 \\ & = 2 \end{align} $$
It was only after doing so that I noticed @JoséCarlosSantos's short cut.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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In how many ways can a number be chosen from 1 to 20 such that it is a multiple of 2 or 3?
In how many ways can a number be chosen from $1$ to $20$ such that it is a multiple of $2$ or $3$?
Numbers that are multiple of $2=\{4,6,8,10,12,14,16,18,20\}$ $=>9$ choices
Numbers that are multiple of $3=\{6,9,12,15,18\}$ $=>5$ $=>5$ choices
Hence there are $14$ choices. But this answer is scored as the wrong answer. How to solve this problem correctly?
| There are $10$ numbers,$[\frac{20}{2}]=10$, can divisible by $2$
There are $6$ numbers,$[\frac{20}{3}]=6$ can divisible by $3$
There are $3$ numbers,$[\frac{20}{6}]=3$, can divisible by $6=2\times 3$
by inclusion-exclusion princible
$n(A\cup B) = n(A)+n(B)-n(A \cap B)= 10+6-3=13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Compute $f^{(2020)}(0)$ Problem :
Let
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$
Then find $f^{(2020)}(0)$.
My Attempt :
From partial fraction decomposition,
$$f(x)=\frac{1}{4(1-x)}+\frac{1}{4(x+1)}-\frac{1}{2(x+1)^2}$$
and,
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n,\quad \frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n, \quad\frac{1}{(1+x)^2}=\sum_{n=1}^\infty (-1)^{n+1}nx^{n-1}=\sum_{n=0}^\infty(-1)^n(n+1)x^n$$
From these,
$$f(x)=\sum_{n=0}^\infty \frac{x^n}{4}+\sum_{n=0}^\infty \frac{(-1)^nx^n}{4}+\sum_{n=0}^\infty\frac{(-1)^{n+1}(n+1)x^n}{2} \\ =\sum_{n=0}^\infty\frac{x^n+(-1)^nx^n+(-1)^{n+1}(n+1)x^n}{4} \\ =\sum_{n=0}^\infty\frac{(1+(-1)^n+(-1)^{n+1}(n+1))x^n}{4}$$
$f^{(2020)}(0)=(2020)!\times a_{2020}$ where $f(x)=\sum a_nx^n$
In this case, $a_n = 1+(-1)^n+(-1)^{n+1}(n+1), a_{2020}=-2019$ but there is no same answer.
Where did I make a mistake?
| You have$$\sum_{n=0}^\infty\left(\frac{x^n}{4}+\frac{(-1)^nx^n}{4}+\frac{(-1)^{n+1}(n+1)x^n}2\right)=\sum_{n=0}^\infty\frac{1+(-1)^n+\color{red}2(-1)^{n+1}(n+1)}4x^n$$So, the coefficient of $x^{2020}$ is $-1010$. In other words, $f^{(2020)}(0)=-1010\times2020!$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\
&= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)
\end{align}$$
now,
$$\begin{align}
a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\
&= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\
&= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\
&= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\
&= 2017
\end{align}$$
so
$$\begin{align}
&\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\
&= 7^6 - (6\cdot2\cdot(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\
&= 90585
\end{align}$$
correct?
| Why work so hard? From the quadratic formula we have:
$\{ a,b \} = \left\{ \frac{7-\sqrt{41}}{2}, \frac{7+\sqrt{41}}{2} \right\}$ so $a^6 + b^6 = 90585$.
My goodness!!! A downvote on a simple, clear, correct answer.
Well, if the approach demands that the solution be done "by hand"—which was NOT part of the problem statement!— then here:
$$a^6 + b^6 = \frac{1}{2^6} \left[ ((7 - \sqrt{41})^2 )^3 + ((7 + \sqrt{41})^2 )^3\right]$$
$$= \frac{1}{64} \left[ (90 - 14 \sqrt{41})^3 + (90 + 14 \sqrt{41})^3\right]$$
Now cancel the terms with odd powers of $(14 \sqrt{41})$ (because they have opposite signs in the pairs), leaving
$$\frac{1}{64} \left[ 2 \cdot 90^3 + 6 \cdot 90 \cdot 14^2 \cdot 41\right] = 90585$$
Satisfied?
| {
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"timestamp": "2023-03-29T00:00:00",
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What purely real analytic techniques are there to evaluate $\int_{-\pi/2}^{\pi/2}\frac{1}{1+\sin^4(x)}\,\mathrm{d}x$? $\newcommand{\d}{\,\mathrm{d}}$Last night, I evaluated the following integral: $$\begin{align}I:&=\int_{-\pi/2}^{\pi/2}\frac{1}{1+\sin^4(x)}\d x\\&=\int_{-1}^1\frac{1}{(1+x^4)\sqrt{1-x^2}}\d x\\&=\frac{\pi}{2^{3/4}}
(\sin(\pi/8)+\cos(\pi/8))\\&=\frac{\pi}{2}\sqrt{1+\sqrt{2}}\end{align}$$
Using a "double keyhole" (as I phrase it) contour method involving a management of branch cuts and residues at infinity, here. Although I was happy to have succeeded in this, I wondered afterwards if I would have had any hope of evaluating $I$ with real analytic technique only.
The challenge:
Evaluate $I$ without use of complex analysis or even of complex arithmetic (e.g. for partial fraction decompositions involving $i$)
I posed this to some friends and they came up with the following method which I wanted to share with MSE:
$$\begin{align}I&\overset{x\mapsto\tan x}{=}\int_{-\infty}^\infty\frac{1+x^2}{(1+x^2)^2+x^4}\d x\\&\overset{x\mapsto1/x}{=}2\int_0^\infty\frac{1+x^2}{(1+x^2)^2+1}\d x\\&=2\int_0^\infty\int_0^\infty e^{-t(1+x^2)}\cos(t)\d t\d x\quad\text{Repr. with IBP}\\&=\sqrt{\pi}\int_0^\infty\frac{e^{-t}\cos(t)}{\sqrt{t}}\d t\end{align}$$
$$\begin{align}J:&=\int_0^\infty\frac{e^{-t}\cos(t)}{\sqrt{t}}\d t\\J^2&=\int_0^\infty\int_0^\infty\frac{e^{-(t+x)}\cos(t)\cos(x)}{\sqrt{tx}}\d t\d x\\&\overset{x\mapsto tx}{=}\int_0^\infty\int_0^\infty\frac{e^{-t(1+x)}\cos(t)\cos(tx)}{\sqrt{x}}\d x\d t\\&=\frac{1}{2}\int_0^\infty\frac{1}{\sqrt{x}}\cdot\frac{1+x+x^2}{(1+x)(1+x^2)}\d x\\&\overset{x\mapsto x^2}{=}\frac{1}{2}\int_0^\infty\left(\frac{1+x^2}{1+x^4}+\frac{1}{1+x^2}\right)\d x\\&=\frac{1}{2}\left[\frac{\pi}{4}\csc\left(\frac{\pi}{4}\right)+\frac{\pi}{4}\csc\left(\frac{3\pi}{4}\right)+\frac{\pi}{2}\right]\\&=\frac{\pi}{4}(1+\sqrt{2})\end{align}$$
Referencing this answer by Sangchul.
We conclude:
$$\begin{align}I&=\sqrt{\pi}\cdot\sqrt{J^2}\\&=\sqrt{\pi}\cdot\sqrt{\frac{\pi}{4}(1+\sqrt{2})}\\&=\frac{\pi}{2}\sqrt{1+\sqrt{2}}\end{align}$$
Among those who helped me, who use MSE, I credit @TheSimpliFire and @KStarGamer who are much better at real integration than I am!
My question is less of a question and more of a request for a list - a list of other, purely real, methods to attack this integral. I hope the outcome of this will be an interesting selection of advanced integration techniques that I and others can learn from.
Note 1: I am aware of this posting by Quanto but it uses complex numbers.
Note 2: You must expand the cosine product as a sum of cosines and use the same integral representation (which is classically gotten from complex arithmetic but can be done with integration by parts): $$\int_0^\infty e^{-tx}\cos(t)\d t=\frac{x}{x^2+1},\,x\gt0$$
| Working the antiderivative
Using your first step
$$\int \frac{u^2+1}{\left(u^2+1\right)^2+u^4} \,du$$
$$\left(u^2+1\right)^2+u^4=2 \left(u^2-u\sqrt{\sqrt{2}-1} +\frac{1}{\sqrt{2}}\right)
\left(u^2+u\sqrt{\sqrt{2}-1} +\frac{1}{\sqrt{2}}\right)$$ Partial fraction decomposition
$$ \frac{u^2+1}{\left(u^2+1\right)^2+u^4}=$$
$$\frac 1 {2 \sqrt{2 \left(\sqrt{2}-1\right)} }\Bigg[\frac {\left(\sqrt{2}-2\right) u+2
\sqrt{\sqrt{2}-1} } {2 u^2-2 \sqrt{\sqrt{2}-1} u+\sqrt{2} }+\frac {\left(2-\sqrt{2}\right) u+2
\sqrt{\sqrt{2}-1}} {2 u^2+2 \sqrt{\sqrt{2}-1} u+\sqrt{2} }\Bigg]$$
$$4\int \frac {\left(\sqrt{2}-2\right) u+2
\sqrt{\sqrt{2}-1} } {2 u^2-2 \sqrt{\sqrt{2}-1} u+\sqrt{2} }\,du=$$ $$\left(\sqrt{2}-2\right) \log \left(2 u^2-2 \sqrt{\sqrt{2}-1}
u+\sqrt{2}\right)+2 \sqrt{3-2 \sqrt{2}} \left(2+\sqrt{2}\right) \tan
^{-1}\left(\frac{2 u-\sqrt{\sqrt{2}-1}}{\sqrt{1+\sqrt{2}}}\right)$$
$$4\int\frac {\left(2-\sqrt{2}\right) u+2
\sqrt{\sqrt{2}-1}} {2 u^2+2 \sqrt{\sqrt{2}-1} u+\sqrt{2} }\,du$$
$$2 \sqrt{3-2 \sqrt{2}} \left(2+\sqrt{2}\right) \tan ^{-1}\left(\frac{2
u+\sqrt{\sqrt{2}-1}}{\sqrt{1+\sqrt{2}}}\right)-\left(\sqrt{2}-2\right) \log
\left(2 u^2+2 \sqrt{\sqrt{2}-1} u+\sqrt{2}\right)$$
$$\color{red}{8 \sqrt{2 \left(\sqrt{2}-1\right)} \int \frac{u^2+1}{\left(u^2+1\right)^2+u^4} \,du=}$$
$$\color{red}{-2 \sqrt{2} \tan ^{-1}\left(\frac{2 \sqrt{\sqrt{2}-1} u}{2 \left(\sqrt{2}-1\right)
u^2+\sqrt{2}-2}\right)-\left(2-\sqrt{2}\right) \log \left(\frac{2 u
\left(u-\sqrt{\sqrt{2}-1}\right)+\sqrt{2}}{2 u
\left(u+\sqrt{\sqrt{2}-1}\right)+\sqrt{2}}\right)}$$ gives the desired result.
| {
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Given $0\leq a\leq b\leq c$ and $b > 0$, prove that: $2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}$ This is an exercise that got me stuck. It's from the book "Functions of Several Real Variables 1.1 - exercise 10" by Martin Moskowitz and Fotios Paliogiannis.
Exercise:
Given $0\leq a\leq b\leq c$ and $b > 0$, prove that:
$2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}$
What I did so far:
$2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}$
$4\frac{a}{c}\leq \frac{b^2}{c^2} +2\frac{a}{c} + \frac{a^2}{b^2}\leq 1 + 2\frac{a}{c}+\frac{a^2}{c^2}$ [power of 2]
$2\frac{a}{c}\leq \frac{b^2}{c^2} + \frac{a^2}{b^2}\leq 1 +\frac{a^2}{c^2}$ [minus $2\frac{a}{c}$]
$2acb^2 \leq b^4+a^2c^2 \leq c^2b^2+a^2b^2$ [multiply by $c^2b^2$]
$2acb^2 -2a^2c^2\leq b^4\leq c^2b^2+a^2b^2 - a^2c^2$
And now I have to prove the inequalities for $b^4$. However, I can not advance (actually I have no way of knowing if this is the right path since I have very little experience writing proofs).
| It is easier to prove the two inequalities separately.
*
*$2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\quad$ This is just AM-GM $\,\sqrt{uv} \le \frac{u+v}{2}\,$ with $\,u=\frac{b}{c}\,$, $\,v=\frac{a}{b}\,$.
*$\frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}\quad$ This is $\,1 + \frac{a}{c} -\frac{b}{c} - \frac{a}{b} \ge 0 \iff \left(\frac{b}{c} - \frac{a}{c}\right)\left(\frac{c}{b} - 1\right) \ge 0\,$.
[ EDIT ] $\;$ This works the same with the inequalities on the last-but-one line in OP's post.
$2acb^2 \leq b^4+a^2c^2 \leq c^2b^2+a^2b^2$
*
*$2acb^2 \leq b^4+a^2c^2\iff (b^2 - ac)^2 \ge 0\,$ which always holds.
*$b^4+a^2c^2 \leq c^2b^2+a^2b^2$ $\iff c^2b^2+a^2b^2-b^4-a^2c^2 \ge 0$ $\iff (b^2 - a^2) (c^2 - b^2) \ge 0\,$ which holds because $\,c \ge b \ge a \ge 0\,$.
| {
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Finding the sum of the series $\sum_{n=2}^{\infty} ((n^2+1)^{1/2} - (n^3+1)^{1/3})$ I need help in finding the sum of the series $\sum_{n=2}^{\infty} ((n^2+1)^{1/2} - (n^3+1)^{1/3})$ if it converges.
I can't even prove convergence. I tried comparison test. I tried telescoping or even sandwich the partial sums. But I could not get anything.
I also tried using the identity
$a-b = \frac{a^3-b^3}{a^2+ab+b^2}$
and even the identity for $a^6-b^6$
but it won't yield.
I found this one while going through some really old question papers of my college.
Please help.
| Without series, we have
$$\sqrt{n^2+1}-\sqrt[3]{n^3+1} = \sqrt{n^2+1}- n + n - \sqrt[3]{n^3+1}$$
which allows us to use those conjugates without appealing to sixth order
$$= \frac{1}{n+\sqrt{n^2+1}} - \frac{1}{n^2+n\sqrt[3]{n^3+1}+(n^3+1)^{\frac{2}{3}}}$$
The divergence comes from the square root term.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the indefinite integral $\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}$ without trigonometric substitution. In order to find
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}
$$
we set $t=\arctan x$. Then $x=\tan t$ and $dt=\frac{dx}{x^2+1}$, so
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}=\int\frac{dt}{\sqrt{\frac{1}{\cos^2t}}}=\int\cos tdt\\
=\sin t+C=\sin(\arctan x)+C
$$
Now, since
$$
\sin(\arctan x)=\sqrt{\frac{\tan^2(\arctan x)}{\tan^2(\arctan x)+1}}
$$
the answer is $\frac{x}{\sqrt{x^2+1}}+C$.
My Question: Is there another way to find this integral without using trigonometry?
| Let $u=x^2+1$.
$$I=\frac{1}{2}\int{u^{-\frac{3}{2}}(u-1)^{-\frac{1}{2}}}du=\frac{1}{2}\int{u^{-2}\left(1-\frac{1}{u}\right)^{-\frac{1}{2}}}du$$
Using the general formula $\int{(f(x))^nf'(x)}dx=\dfrac{(f(x))^{n+1}}{n+1}+c$ (which can be verified by differentiating the right side),
$$I=\left(1-\frac{1}{u}\right)^{\frac{1}{2}}+c=\frac{x}{\sqrt{x^2+1}}+c$$
| {
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Let $n=4\cdot p^t+1.$ Is $r_k=4$ for some $1\le kLet $n=4\cdot p^t+1$, $t\ge1$ and $p\ge5$ prime.
Define sequence $R$ as follows;
$r_0=4,$ $r_{k+1}\equiv r_k^2-2\pmod n$, $0\le r_k<n$ for $k\ge1$.
It appears that whenever $n$ is prime, $r_k=4$ for some positive integer $k<n$. How does one go about proving this result for all primes $p\ge5$ and positive integers $t$? Are there any counter examples? (I have done a search for all $p<10^6$ with $t=1,2$. No counter examples.)
| I can't seem to find a solution using only elementary number theory. From what I did find, it's obvious that we need
$$r_{k-1}^2-2\equiv 4 \text{ mod } n$$
$$r_{k-1}^2\equiv 6 \text{ mod } n$$
By the Law of Quadratic Reciprocity
$$\left( \frac{2}{n} \right) \equiv (-1)^\frac{n^2-1}{8}\equiv (-1)^{\frac{(4p^k)(4p^k+2)}{8}}\equiv (-1)^{(p^k)(2*p^k+1)}=-1 $$
Additionally,
$$\left( \frac{3}{n} \right) =-1 $$ since by the Law of Quadratic Reciprocity,
$$\left( \frac{3}{n} \right)\left( \frac{n}{3} \right) =(-1)^{\frac{n-1}{2}\frac{3-1}{2}} = (-1)^{2*p^k}=1$$ and
$p \text{ mod } 3 \equiv 1,2$
$p^k \text{ mod } 3 \equiv 1,2$
$n \equiv 4p^k+1 \text{ mod } 3 \equiv 2,0$
Since $2$ is a quadratic nonresidue mod $3$, and n is defined as a prime greater than or equal to 5, then n cannot be a multiple of 3. Therefore,
$$\left( \frac{n}{3} \right)=-1$$
Since the Legendre Function is multiplicative, we get
$$\left( \frac{6}{n} \right)=\left( \frac{3}{n} \right)\left( \frac{2}{n} \right)=1$$, and so 6 is a quadratic residue mod n. This implies that it is possible for the number 4 to be reached again in a sequence like that. From there however, I can't seem to find any literature on how to continue, other than references to the Pisano Period (which has a derivation referring to algebraic number theory) and a closed-form for the recursion $r_k=r_{k-1}^2-2$ in the page for "Proof of correctness" in the Wikipedia page for the Lucas Lehmer primality test, which introduces the same recursion.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate triple integrals $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt[3]{\log{(xyz)}}dxdydz$ I am trying to evaluate this integral:
$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt[3]{\log{(xyz)}}dxdydz$$
Honestly, I have no ideas to deal with this.
I hope I can be helped by everyone.
I just need a hint to process; thank you.
| My second solution:
Using the identity ($q > 0, \alpha \in (0, 1)$)
$$q^\alpha = \frac{\alpha}{\Gamma(1 - \alpha)}\int_0^\infty \frac{1 - \mathrm{e}^{-qs}}{s^{1 + \alpha}}\,\mathrm{d} s,$$
we have
$$\sqrt[3]{\ln \frac{1}{xyz}}
= \frac{1}{3\Gamma(2/3)}\int_0^\infty \frac{1 - (xyz)^s}{s^{4/3}}\,\mathrm{d} s.$$
Denote the integral by $I$.
We have
\begin{align*}
I &= - \int_0^1 \int_0^1 \int_0^1 \frac{1}{3\Gamma(2/3)}\int_0^\infty \frac{1 - (xyz)^s}{s^{4/3}}\,\mathrm{d} s \,\mathrm{d}x \,\mathrm{d} y\, \mathrm{d} z \\
&= - \frac{1}{3\Gamma(2/3)}\int_0^\infty
\frac{1}{s^{4/3}}\left(\int_0^1 \int_0^1 \int_0^1 (1 -x^s y^s z^s) \,\mathrm{d}x \,\mathrm{d} y\, \mathrm{d} z\right)\,\mathrm{d} s\\
&= - \frac{1}{3\Gamma(2/3)}\int_0^\infty
\frac{1}{s^{4/3}}\left(1 - \frac{1}{(s + 1)^3}\right)\,\mathrm{d} s \\
&= -\frac{1}{3\Gamma(2/3)}\int_0^\infty
\frac{s^2 + 3s + 3}{s^{1/3}(s + 1)^3}\,\mathrm{d} s\\
&= -\frac{1}{3\Gamma(2/3)}\int_0^\infty
\frac{3u(u^6 + 3u^3 + 3)}{(u^3 + 1)^3}\,\mathrm{d} u \qquad\qquad (s = u^3)\\
&= - \frac{1}{3\Gamma(2/3)}\cdot \frac{28}{27}\pi\sqrt 3
\end{align*}
where we have used Fubini's theorem to interchange the order of integration; and the integral $\int_0^\infty
\frac{3u(u^6 + 3u^3 + 3)}{(u^3 + 1)^3}\,\mathrm{d} u$ is easily evaluated using partial fraction decomposition.
(Note: $\frac{1}{3\Gamma(2/3)}\cdot \frac{28}{27}\pi\sqrt 3 = \frac12 \Gamma(10/3)$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Using complex number in integration I know,
$$I_1 = \int \dfrac{dx}{\sqrt{x^2 - 1}} = \ln|x + \sqrt{x^2-1}| + c$$
But if I factor out $i$ from the denominator, I get:
$$I_2 = -i \int \dfrac{dx}{\sqrt{1 - x^2}} = -i \sin^{-1}x + ic$$
Are these 2 expressions equivalent?
| Let $\sin^{-1}x = θ$
$$ \sin θ = x\\
\frac{e^{iθ} - e^{-iθ}}{2i} = x \\ $$
Solving the quadratic,
$$ e^{iθ} = ix \mp \sqrt{1 - x^2} = i(x \mp \sqrt{x^2 - 1})\\ $$
Taking log on both sides and then multiplying the numerator and denominator by the congugate,
$$\begin{align}
iθ &= \ln i + \ln |x \mp \sqrt{x^2 - 1}| \\
i \sin^{-1}x &= \ln e^{\frac{iπ}2} + \ln |\frac 1{x \pm \sqrt{x^2 - 1}}|\\
&= \frac{iπ}2 - \ln |x \pm \sqrt{x^2 - 1}|\\
\end{align}$$
So $ - i \sin^{-1}x$ is equivalent to $\ln |x ± \sqrt{x^2 - 1}|$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda} \leq \frac{3}{\lambda +1}$
If $a+b+c=3$ ,$ a,b,c>0$ and $\lambda \geq 1$,
prove that : $$\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda} \leq \frac{3}{\lambda +1}.$$
my attempt:
using CBC twice and AM_AG inequality
$\frac{3}{3}\Big(\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda}\Big)^2\leq 3 \Big(\frac{a}{a^2+\lambda}\Big)^2+\Big(\frac{b}{b^2+\lambda}\Big)^2+\Big(\frac{c}{c^2+\lambda}\Big)^2\leq 3\Bigg( \bigg(\frac{a}{\frac{(a+\sqrt{\lambda})^2}{2}}\bigg)^2+ \bigg(\frac{b}{\frac{(b+\sqrt{\lambda})^2}{2}}\bigg)^2+\bigg(\frac{c}{\frac{(c+\sqrt{\lambda})^2}{2}}\bigg)^2\Bigg) =12\bigg(\frac{a^2}{(a+\sqrt{\lambda })^4}+\frac{b^2}{(b+\sqrt{\lambda})^4}+\frac{c^2}{(c+\sqrt{\lambda })^4}\bigg)\leq12 \bigg(\frac{a^2}{\sqrt{\lambda}^4}+\frac{b^2}{\sqrt{\lambda}^4}+\frac{c^2}{\sqrt{\lambda}^4}\bigg)=\frac{48(a^2+b^2+c^2)}{4\lambda ^2 }\leq \frac{48\cdot9}{4\lambda^2} \leq \frac{9}{(\lambda +1)^2}$
beacuse $f(\lambda )=\frac{9}{(\lambda+1)^2}-\frac{48\cdot9}{4\lambda^2} \geq 0 $ for $\lambda \geq 1$
so finally
$\frac{3}{3}\Big(\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda}\Big)^2\leq\frac{9}{(\lambda +1)^2}$$\Leftrightarrow$$\Big(\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda}\Big)\leq\frac{3}{(\lambda +1)}$
I have just one question:
-does my attempt is true?
| Tangent Line (TL) method:
It suffices to prove that
$$\frac{a}{a^2 + \lambda}
\le \frac{1}{1 + \lambda} + \frac{\lambda - 1}{(1 + \lambda)^2}(a - 1). \tag{1}$$
(The desired result follows by summing cyclically.)
(1) is true since
$$\frac{a}{a^2 + \lambda}
- \frac{1}{1 + \lambda} - \frac{\lambda - 1}{(1 + \lambda)^2}(a - 1) = - \frac{(a - 1)^2(a\lambda - a + 2\lambda)}{(a^2 + \lambda)(1 + \lambda)^2} \le 0.$$
We are done.
| {
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Prove $\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} dx = \frac{\pi}{2} \sec (\frac{\pi}{2n})$ without complex techniques Prove
$$\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} \,dx = \frac{\pi}{2} \sec \left(\frac{\pi}{2n}\right)$$
for all natural numbers $n \ge 2$.
There are several answers (A1 A2) to this integral but they all involve the gamma function or the beta function or contour integration etc. Can one solve this using only 'real' 'elementary' techniques? For $n = 2$ and $n = 3$ it can be solved using only elementary substitutions and partial fractions.
| \begin{aligned}
\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} d x&=\int_{0}^{\frac{\pi}{2}} \sin ^{\frac{1}{n}} x \cos ^{-\frac{1}{n}} x d x\\
&= \int_{0}^{\frac{\pi}{2}} \sin ^{2\left(\frac{1}{2 n}+\frac{1}{2}\right)-1} x \cos^ {2\left(-\frac{1}{2n}+\frac{1}{2}\right)-1 }x d x\\
&=\frac{1}{2} B\left(\frac{1}{2 n}+\frac{1}{2},-\frac{1}{2 n}+\frac{1}{2}\right)\\
&=\frac{1}{2} \pi \csc \left[\pi\left(\frac{1}{2 n}+\frac{1}{2}\right)\right]\\
&=\frac{\pi}{2} \sec \frac{\pi}{2 n}
\end{aligned}
where the second last line using the property of Beta function:
$$B(x, 1-x)=\pi \csc (\pi x) \quad x \notin \mathbb{Z}.$$
| {
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Trying to solve $|2x-15| = -x^2 - 5x -8$ My first instinct was to take the positive and negative of the right hand side, resulting in
$2x-15 = -x^2 - 5x - 8$, and $2x-15 = x^2 + 5x + 8$, which results in the first giving me two real answers using the quadratic equation, and the second being two imaginary solutions. The problem is that, when graphed, these 2 graphs do not intersect at all, so there should be no real solutions. As for the given complex solutions, neither were considered by wolfram alpha.
Knowing that there are no real solutions, I'm confused as to how I might get complex solutions through means not already attempted and explained above.
| The "is $2x - 15$ positive or negative" dichotomy doesn't work in $\Bbb{C}$. What we can say is that, if $-x^2 - 5x - 8 = |2x - 15|$, then $-x^2 - 5x - 8$ must be a non-negative real number, let's call it $k$. Let's start by solving
$$-x^2 - 5x - 8 = k$$
for $x \in \Bbb{C}$. Using the quadratic formula,
$$x = \frac{-5 \pm \sqrt{5^2 - 4(8 + k)}}{2} = \frac{-5 \pm i\sqrt{7 + 4k}}{2}.$$
Next, we need $|2x - 15|$ to be equal to this $k$. In particular,
$$k = \left|2\frac{-5 \pm i\sqrt{7 + 4k}}{2} - 15\right| = \left|-20 \pm i\sqrt{7 + 4k}\right| = \sqrt{407 + 4k}.$$
Thus, we get a real equation:
$$k^2 - 4k - 407 = 0, \quad k \ge 0.$$
This has a unique solution (bearing in mind $k \ge 0$):
$$k = 2 + \sqrt{411}.$$
Thus, our solutions for $x \in \Bbb{C}$ must be:
$$x = \frac{-5 \pm i\sqrt{7 + 4k}}{2} = \frac{-5 \pm i\sqrt{15 + 4\sqrt{411}}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Integral $\int_0^1 \frac{(1+t^2)}{1-t^2+t^4}\ln(t)dt$ $$I=\int_0^1 \frac{(1+t^2)}{1-t^2+t^4}\ln(t)dt=-\frac{4}{3}G $$ G is Catalan's constant.
$$I=\int_0^1 \frac{(1+t^2)^2}{(1+t^2)(1-t^2+t^4)}\ln(t)dt=\int_0^1 \frac{(1+t^2)^2}{(1+t^6)}\ln(t)dt$$
Do series expansion:
$$I=\sum_{n=0}^\infty (-1)^n \int_0^1(1+2t^2+t^4)t^{6n}\ln(t)dt$$
Integrate term by term:
$$I=-\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}+\frac{2}{(6n+3)^2}+\frac{1}{(6n+5)^2} \right)$$
Re-organize them:
$$I=-\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}-\frac{1}{(6n+3)^2}+\frac{1}{(6n+5)^2}\right)-\sum_{n=0}^\infty (-1)^n \frac{3}{(6n+3)^2} $$
The first part is Catalan's constant:
$$I=-G-\sum_{n=0}^\infty (-1)^n \frac{3}{(6n+3)^2}=-G-\frac{3}{9}G=-\frac{4}{3}G$$
Done.
| You can arrive to the desired result starting at
$$I=-\sum_{n=0}^\infty (-1)^n\left( \frac{1}{(6n+1)^2}+\frac{2}{(6n+3)^2}+\frac{1}{(6n+5)^2} \right)$$
Using the Hurwitz zeta function
$$S_a=\sum_{n=0}^\infty \frac{ (-1)^n}{(6n+a)^2}=\frac{1}{144} \left(\zeta \left(2,\frac{a}{12}\right)-\zeta \left(2,\frac{a+6}{12}\right)\right)$$
$$S_1+2S_3+S_5=S_1+\frac{2}{9}C+S_5=$$ $$\frac{2}{9}C+\frac{1}{144}\left(\zeta \left(2,\frac{1}{12}\right)+\zeta \left(2,\frac{5}{12}\right)-\zeta
\left(2,\frac{7}{12}\right)-\zeta \left(2,\frac{11}{12}\right)\right)$$ and use
$$\zeta \left(2,\frac{1}{12}\right)+\zeta \left(2,\frac{5}{12}\right)-\zeta
\left(2,\frac{7}{12}\right)-\zeta \left(2,\frac{11}{12}\right)=160 C$$ You could do the same using the first derivative of the digamma function.
Edit
We could do the same without series expansion writing
$$\frac{(1+t^2)}{1-t^2+t^4}=\frac 1{a-b}\left(\frac{a+1}{t^2-a}-\frac{b+1}{t^2-b} \right)$$ with $a=\frac{1+i \sqrt{3}}{2}$ and $b=\frac{1-i \sqrt{3}}{2}$ and use, for a complex $c$,
$$J(c)=\int_0^1 \frac{\log(t)}{t^2-c}\,dt=\frac{1}{4 \sqrt{c}}\left(4
\text{Li}_2\left(\frac{1}{\sqrt{c}}\right)-\text{Li}_2\left(\frac{1}{c}\right)\right)$$
| {
"language": "en",
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"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int_0^\infty dx \, x^2 e^{-\frac{a^2}{2} x^2} \sqrt{x^2 + b^2}$ Does anyone know how to evaluate $\int_0^\infty dx \, x^2 e^{-\frac{a^2}{2} x^2} \sqrt{x^2 + b^2}$ by hand?
I've been relying on Mathematica for these kinds of integrals, which in this case evaluates to $$\frac{b^2}{2 a^2} e^{\frac{a^2 b^2}{4}} K_1\left( \frac{a^2 b^2}{4} \right) \, ,$$ where $K_\nu(z)$ is the modified Bessel function of the second kind. I haven't found integral tables (e.g. Gradshteyn and Ryzhik) to be particularly helpful here either, but I may just need to look harder.
A plausible approach I thought might work would be to start by Fourier transforming the root, since $$\sqrt{x^2 + b^2} = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} -b \sqrt{\frac{2}{\pi}} \frac{K_1(b |q|)}{|q|} e^{-i x q} \, dq \, ,$$ or perhaps more neatly with a Fourier cosine transformation $$\sqrt{x^2 + b^2} = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} -b \sqrt{\frac{2}{\pi}} \frac{K_1(b q)}{q} \cos(x q) \, dq \, ,$$
and then re-ordering integrals, but this doesn't seem to quite work out how I want (or at least I'm not sure how to progress from here). Any help would be much appreciated.
| $$
\int_0^\infty dx \, x^2 e^{-\frac{a^2}{2} x^2} \sqrt{x^2 + b^2}
$$
is equivalent to
$$
b^3\int_0^\infty d\left(\frac{x}{b}\right) \, \left(\frac{x}{b}\right)^2 e^{-\frac{a^2b^2}{2} \left(\frac{x}{b}\right)^2} \sqrt{\left(\frac{x}{b}\right)^2 + 1} = b^3\int_0^\infty dz \, z^2 e^{-k z^2} \sqrt{z^2 + 1}
$$
where $k = \frac{a^2b^2}{2}$ we can move $z^2$ under the sqrt
$$
\sqrt{z^4 +z^2}
$$
this leads to
$$
b^3\int_0^\infty dz \, z e^{-k z^2} \sqrt{z^4 + z^2}
$$
if we set $z^2 = v$ we have
$$
\frac{b^3}{2}\int_0^\infty dv e^{-k v} \sqrt{v^2 + v}
$$
we can use completing the square
$$
v^2 + v = \left(v + \frac{1}{2}\right)^2 - \frac{1}{4} = \left(\frac{v + \frac{1}{2}}{1/2}\right)^2 - 1
$$
setting $t = \frac{v + \frac{1}{2}}{1/2}$
$$
\frac{b^3}{2}\int_1^\infty 2dt e^{-k \left(\frac{t}{2} - \frac{1}{2}\right)} \sqrt{t^2 - 1}
$$
which becomes
$$
b^3\mathrm{e}^{\frac{k}{2}}\int_1^\infty dt e^{-\frac{k}{2}t} \sqrt{t^2 - 1}
$$
now we can use the identity here
$$
K_n(z) = \dfrac{\sqrt{\pi}}{\left(n-\frac{1}{2}\right)!}\left(\frac{1}{2}z\right)^{n}\int_1^\infty dx\mathrm{e}^{-zx}\left(x^2-1\right)^{n-1/2}
$$
setting $n=1, z =\frac{k}{2}$ we find
$$
K_1(\frac{k}{2}) = \dfrac{\sqrt{\pi}}{\left(-\frac{1}{2}\right)!}\frac{k}{4}\int_1^\infty dx\mathrm{e}^{-\frac{k}{2}x}\sqrt{x^2-1}
$$
with $(-1/2)! = \Gamma(1/2) = \sqrt{\pi}$
thus
$$
\frac{K_1(\frac{k}{2})}{k/4} = \int_1^\infty dx\mathrm{e}^{-2kx}\sqrt{x^2-1}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the $(X - \bar{X})^2$ of first 5 data of dataset given mean and population variance Mean and population variance of the dataset $x_{1},x_{2}..x_{10}$ are $19$ and $49$ respectively. If the value $\sum_{i=6}^{10}x_{i}^{2} = 1900$, what is the value of $\sum_{i=1}^{5}x_{i}^{2} = ?$.
I've solved it as following and it is wrong:
Population variance:
$$S^2 = \frac{\sum (x_i - \bar{x})^2}{n}\\
49 = \frac{\sum_{i=1}^{5}x_{i}^{2}}{10} + \frac{1900}{10}\\
49 = \frac{\sum_{i=1}^{5}x_{i}^{2}}{10} + 190\\
-190+49 = \frac{\sum_{i=1}^{5}x_{i}^{2}}{10}\\
\sum_{i=1}^{5}x_{i}^{2} = -141*10 = -1410$$
And this solution is wrong. How to solve this problem?
| You don't use the whole definition of the population variance which is
$$s^2 = \frac{\sum\limits_{i=1}^{10} (x_i - \bar{x})^2}{10}=\frac1{10}\cdot \left( \sum\limits_{i=1}^{10} x_i^2 -2 x_i\bar{x} +\overline x^2 \right)$$
$$\frac1{10}\cdot \left( \sum\limits_{i=1}^{10}x_i^2 -\sum\limits_{i=1}^{10}2 x_i\bar{x} +\sum\limits_{i=1}^{10}\overline x^2 \right)$$
$2, \overline x$ and $\overline x^2$ are constants. They can be put in front the sigma signs.
$$\frac1{10}\cdot \left( \sum\limits_{i=1}^{10}x_i^2 -2 \bar{x}\sum\limits_{i=1}^{10} x_i +\overline x^2 \sum\limits_{i=1}^{10} 1 \right)$$
We have $\frac1{10} \sum\limits_{i=1}^{10} x_i=\overline x\Rightarrow \sum\limits_{i=1}^{10} x_i=10\cdot \overline x$
$$\frac1{10}\cdot \left( \sum\limits_{i=1}^{10}x_i^2 -2 \cdot \bar{x}\cdot 10\cdot \overline x +\overline x^2 \cdot 10 \right)=\frac1{10}\cdot \left( \sum\limits_{i=1}^{10}x_i^2 -10\cdot \overline x^2 \right)$$
$$\frac1{10}\cdot \sum\limits_{i=1}^{10}x_i^2 -\overline x^2 =\frac1{10}\sum\limits_{i=1}^{10} x_i^2-\left(\frac1{10}\sum\limits_{i=1}^{10} x_i\right)^2$$
$=\frac1{10}\left(\sum\limits_{i=1}^{5} x_i^2+\sum\limits_{i=6}^{10} x_i^2\right)-\left(\underbrace{\frac1{10}\sum\limits_{i=1}^{10} x_i}_{\textrm{population mean}}\right)^2=s^2$
Now you can insert the values: $\frac1{10}\sum\limits_{i=1}^{10} x_i=19, s^2=49$ and $\sum\limits_{i=6}^{10} x_i^2=1900$. Finally solve for $\sum\limits_{i=1}^{5} x_i^2$.
| {
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"answer_count": 1,
"answer_id": 0
} |
Region area with double integral I need to find the area $\{(x, y) \in \mathbb{R}^2 : 0 < 2x < y < 3-x^2\}$ as the double integral $\int(\int dx) dy$.
I came up with:
$$\int_0^2\int_0^{\frac{1}{2}y}dxdy + \int_2^3\int_0^\sqrt{3-y}= \int_0^2\frac{1}{2}ydy + \int_2^3\sqrt{3-y}dy = \frac{1}{2}\frac{y^2}{2}\biggr\rvert^2_0 + \frac{2}{3}(3-y)^\frac{3}{2}\biggr\rvert_2^3 = \frac{1}{3}$$
But the solution gives $\frac{5}{3}$. I'm around this exercise a couple hours now and I can't seem to find my mistake.
| Your answer is almost correct.
You made a single error when you took the antiderivative of the second integreal.
It should be $- \frac{2}{3}(3-y)^\frac{3}{2}\biggr\rvert_2^3 $ not $+ \frac{2}{3}(3-y)^\frac{3}{2}\biggr\rvert_2^3 $
That will give you the correct result of $\frac{5}{3}$.
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
Number of Triangles in this figure How many triangles are in this figure? I suspect there are $45$, but I would like someone to confirm or correct me. I arrived at $45$ after manually counting them. If anyone has a formula by which this number can be obtained, please share.
| I get only $43$ triangles. Here's how I did the counting: consider the collection of sets of points on each segment
$$S=\{ AKDHC, CEB, BFA, ALGJE, CIGMF, BMLK, BGD, BJIH \}$$
(here for simplicity I'm denoting the points of a segment as a concatenated string).
Now make a graph $g$ where vertices are the points $A,\dots, M$ and put an edge if the points are on a same segment (loop over $S$ and for each pair on a segment put an edge).
Number of closed $3$-walks on this graph gives the number of "triangles" but some of these are degenerated, i.e. all the points lie an a line. But these are easily counted as
$$\sum_{L\in S} { |L| \choose 3} = 41$$
So the answer is $1\over 6$(trace of third power of the adjacency matrix of $g$) minus $41$
$$
\frac{1}{6} tr\left(
\displaystyle \left(\begin{array}{rrrrrrrrrrrrr}
0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 0 \\
1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 \\
1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\
1 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 0 & 0 & 1 & 0 \\
1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 \\
1 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 1 \\
0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 1 & 0
\end{array}\right)^3
\right)
- \sum_{L\in S} { |L| \choose 3} = 43
$$
Here are the triangles that I get by iterating all closed $3$-walks, accepting only non-degenerated and considering permutations the same:
ABC
ABD
ABE
ABG
ABH
ABJ
ABK
ABL
ACE
ACF
ACG
ADG
AFG
AHJ
AKL
BCD
BCF
BCG
BCH
BCI
BCK
BCM
BDH
BDK
BEG
BEJ
BEL
BFG
BFI
BFM
BGI
BGJ
BGL
BGM
BHK
BIM
BJL
CDG
CEG
CHI
CKM
GIJ
GLM
So I get
*
*$15$ triangles that include the point $A$
*$22$ that include $B$ but not $A$
*$4$ that include $C$ but not $A$ or $B$
*$2$ that don't include any of $A,B,C$
Can you spot the $2$ that I've missed or you have over-counted?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Two hard integrals: $\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx$ and $\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx$ I found two integrals that seem hard to evaluate:
$$I_1=\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx$$
$$I_2=\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx$$
I am just a beginner in logarithmic integral. So, I searched to find substitution like $x=\frac{1}{1+x}$, $x=\frac{1}{1-x}$, or $x=\frac{1-x}{1+x}$, but they didn't work.
Can I ask some ideas from every one? Thank you.
EDIT: After using Mathematica, with MZIntegrate paclet gives closed-form:
$$\begin{align}I_1&=\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx\\&=G^2+\frac{5 \text{Li}_4\left(\frac{1}{2}\right)}{4}+\frac{35}{32} \zeta (3) \log (2)-\frac{119 \pi ^4}{5760}+\frac{5 \log ^4(2)}{96}-\frac{5}{96} \pi ^2 \log ^2(2)\\I_2&=\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx\\&=2 G^2+\frac{35}{16} \zeta (3) \log (2)-\frac{199 \pi ^4}{5760}\end{align}$$
where $G$ is Catalan's constant.
| The routine integrals $\int_0^1 \frac{\ln^2t}{1-t}dt = 2\zeta(3)$,
$\int_0^1 \frac{\ln^2t}{1+t}dt = \frac32\zeta(3)$,
$\int_0^1 \frac{\ln^3t}{1-t}dt = -\frac{\pi^4}{15}$,
$\int_0^1 \frac{\ln^2t}{1+t^2}dt =\frac{\pi^3}{16}$,
$\int_0^1 \frac{\ln t}{1+t^2}dt =-G$, and
$\int_0^1 \frac{\ln t}{1+t}dt =-\frac{\pi^2}{12}$ are used below without elaboration.
\begin{align}
I_2=&\int_0^1\frac{\ln^2x\ln(1+x^2)}{1-x}dx\\
=& \int_0^1\ln(1+x^2)\ d\left( \int_0^x \frac{\ln^2t}{1-t}dt\right)\\
=& \ \ln2 \int_0^1 \frac{\ln^2t}{1-t}dt
-\int_0^1\frac{2x}{1+x^2}\int_0^x\frac{\ln^2t}{1-t} \overset{t=xy}{dt}\\
=&\ 2\ln2\zeta(3) +2\int_0^1\int_0^1\frac{\ln^2xy}{1+y^2}\left(\frac{1+xy}{1+x^2}-\frac1{1-xy}\right)dxdy\tag1
\end{align}
Note that, with $\ln^2xy =\ln^2x +2\ln x\ln y +\ln^2 y$
\begin{align}
&\int_0^1\int_0^1\frac{(1+xy)\ln^2xy}{(1+x^2)(1+y^2)}dxdy\\
=& \int_0^1 \int_0^1\frac{\ln^2x +2\ln x\ln y +\ln^2 y}{(1+x^2)(1+y^2)}+\frac1{16}\frac{\ln^2x +2\ln x\ln y +\ln^2 y}{(1+x)(1+y)}\ dxdy\\
=&\ 2G^2 +\frac3{16}\ln2\zeta(3)+\frac{37\pi^4}{1152}\\
\\ & \int_0^1\int_0^1\frac{\ln^2xy}{(1+y^2)(1-xy)}\overset{x=t/y}{dx}dy\\
=& \ \frac12\int_0^1 d\left(\ln\frac{y^2}{1+y^2}\right)\int_0^y \frac{\ln^2t}{1-t}dt
\overset{ibp}=-\ln2\zeta(3)+\frac{\pi^4}{15}+\frac12I_2
\end{align}
Plug into (1) to obtain
\begin{align}
I_2 = &\ 2\ln2\zeta(3)+2 \left(2G^2 +\frac3{16}\ln2\zeta(3)+\frac{37\pi^4}{1152} \right)+2\ln2\zeta(3)-\frac{2\pi^4}{15}-I_2\\
=&\ 2G^2 +\frac{35}{16}\ln2\zeta(3)-\frac{199\pi^4}{5760}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4495798",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to calculate the determinant of $4 \times 4$ matrix with a variable on the diagonal? I need to calculate the determinant of the following $4 \times 4$ matrix:
\begin{bmatrix}x&1&1&1\\1&x&1&1\\1&1&x&1\\1&1&1&x\end{bmatrix}
I heard there is a way by separating the matrix into blocks, but I couldn't succeed in doing that.
| In the special case $$
\left( {\begin{array}{*{20}c}
A & B \\
B & A \\
\end{array}} \right)
$$
where $A,B$ are $2×2$ blocks you can use the formula
$$
\det \left( {\begin{array}{*{20}c}
A & B \\
B & A \\
\end{array}} \right) = \det \left( {A - B} \right)\det \left( {A + B} \right)
$$
In our case you have
$$
\det \left( {A - B} \right) = \det \left( {\begin{array}{*{20}c}
{x - 1} & 0 \\
0 & {x - 1} \\
\end{array}} \right) = (x - 1)^2
$$
while
$$
\det \left( {A + B} \right) = \det \left( {\begin{array}{*{20}c}
{x + 1} & 2 \\
2 & {x + 1} \\
\end{array}} \right) = x^2 + 2x - 3 = \left( {x - 1} \right)\left( {x + 3} \right)
$$
whence
$$
\det \left( {\begin{array}{*{20}c}
A & B \\
B & A \\
\end{array}} \right) = \left( {x - 1} \right)^3 \left( {x + 3} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Quadratic formula $x = \frac{- (b +\sqrt{b^2- 4ac})}{ \pm2a}$ In the proof of the quadratic formula
$$x = \frac{- b +\sqrt{b^2- 4ac}}{2a}$$
shouldn't there be $\pm 2a$ instead of $2a$, since both can be the square root of $4a^2$?
| $$ax^2+bx+c=0,~~~a\neq0$$
$$\begin{align}
&\Rightarrow x^2+\frac{b}{a}x+\frac{c}a=0\\
\\
&\Rightarrow\left(x+\frac{b}{2a} \right)^2+\frac{c}{a}-\frac{b^2}{4a^2}=0\\
\\
&\Rightarrow\left(x+\frac{b}{2a} \right)^2=\frac{b^2-4ac}{4a^2}\\
\\
&\Rightarrow\left(x+\frac{b}{2a} \right)=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\\
\\
&\Rightarrow x+\frac{b}{2a} =\pm\frac{\sqrt{b^2-4ac}}{2|a|}\\
\end{align}$$
Case.(1) $a>0\Rightarrow |a|=a$
$$x+\frac{b}{2a} =\pm\frac{\sqrt{b^2-4ac}}{2a} \Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Case.(2) $a<0\Rightarrow |a|=-a$
$$x+\frac{b}{2a} =\pm\frac{\sqrt{b^2-4ac}}{-2a} \Rightarrow x=\frac{-b\mp\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Therefore, no matter what cases, you get the same formula.
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Issues with integrating $\int x^2 \sqrt{x^3 +1}$ via integration by parts I want to integrate
$$\int x^2 \sqrt{x^3 +1}~dx$$
I tried it with integration by parts (because we have a product here), but an online calculator did it with integration by substition.
Would this still be correct?
$$\frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int \frac{1}{3} x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} \cdot 3~dx \\
= \frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} ~dx\\
= \frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \frac{1}{4} x^4\cdot 2(x^3 +1)^{-\frac{1}{2}} \\
= \frac{1}{3}x^3 \sqrt{x^3+1} - \frac{1}{2} x^4 \frac{1}{\sqrt{x^3+1}}$$
I think this is wrong because when $x=1$ I get a different result than when I insert $x=1$ into
Can someone tell me where I went wrong and why we rather use integration by substition instead of integration by parts here?
| $$\frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int \frac{1}{3} x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} \cdot 3 \tag{1}$$
In your first line, you miss $x^2$, which is from the chain rule. It should be
$$\frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int \frac{1}{3} x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} \cdot 3x^2$$
Here is a trick, note that $dx^3=d(x^3+1)$, so we have
$$\begin{align}
\int x^2 \sqrt{x^3 +1}~dx&=\frac{1}{3}\int \sqrt{x^3 +1}~d(x^3+1)\tag{2}\\
\\
&=\frac{1}{3}\cdot\frac{2}3\cdot (x^3+1)^{3/2}+C
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$ $( \frac{1}{a} +\frac{1}{b} +\frac{1}{c}) +a +b +c \geq 4\sqrt{3}$
If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$
Show that:
$$\frac{1}{a} +\frac{1}{b} +\frac{1}{c}+a +b +c \geq 4\sqrt{3}.$$
My attempt:
First , I used Holder's : $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{27}{3(a+b+c)}$$
similarly:
$$ a+b+c \geq \frac{(a+b+c)^3}{3(a^2+b^2+c^2)} = \frac{(a+b+c)^3}{3}$$
Which gives us:
$$LHS \geq \frac{27}{3(a+b+c)}+ \frac{(a+b+c)^3}{3}$$
By AM-GM:
$$LHS \geq 2\sqrt{3} \times (a+b+c)$$
So all we need to do is prove that $$2\sqrt{3} \times (a+b+c)\geq 4\sqrt{3}$$
And this means we need to prove $$a+b+c \geq 2 $$
Which I can't.
I am not looking for a solution , only hints that would guide me through solving it .
Thanks in advance for any communicated help !
| Alternative approach using the point of incidence technique.
We will first employ the AM-GM inequality, followed by QM-AM:
\begin{align*}
a+b+c+\frac1a+\frac1b+\frac1c&=a+b+c+\frac13\left(\frac1a+\frac1b+\frac1c\right) + \frac23\left(\frac1a+\frac1b+\frac1c\right)\\
&\geqslant \frac{6}{\sqrt{3}}\sqrt[6]{abc\cdot\frac1{abc}}+\frac{2}{\sqrt[3]{abc}}\\
&\geqslant 2\sqrt{3}+\frac2{\sqrt{\frac{a^2+b^2+c^2}3}}\\
&= 2\sqrt{3}+\frac2{\sqrt{\frac13}}=4\sqrt{3}
\end{align*}
Equality holds iff $a=b=c=\frac1{\sqrt{3}}$.
Observation. This problem is a stronger version of an inequality that appeared in Macedonia's National Olympiad back in $1999$. Notwithstanding, a similar idea to that of Macedonia yields a solution for this problem.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating the limit of $\lim_{n \to \infty} \frac{4n^2 +9}{2n^2+2n+3}$ with definition I have troubles with "estimating".
We want to evaluate the limit with the definition. For all $\epsilon >0$ we have to find an $N$, such that $|a_n-a|<\epsilon$ holds for $n>N$.
The limit in question is:
$$\lim_{n \to \infty} \frac{4n^2 +9}{2n^2+2n+3}$$
We know the limit is $2$.
My approach:
Let $\epsilon > 0$.
$$\begin{align*}
|a_n-2|
&= \left|\frac{4n^2+9}{2n^2+2n+3} -2\right| \\
&= \left| \frac{4n^2+9 - 2(2n^2+2n+3)}{2n^2+2n+3} \right| \\
&= \left| \frac{4n^2+9 - 4n^2 - 4n -6}{2n^2+2n+3} \right| \\
&= \left| \frac{-4n+3}{2n^2+2n+3} \right| \\
&\leq \left| \frac{-4n}{2n^2+2n+3}\right| \\
&\leq \frac{4n}{2n^2} \\
&= \frac{2}{n} < \epsilon
\end{align*}$$
If we choose $N = \frac{2}{\epsilon}$ it holds that $\forall \epsilon > 0: |a_n -a| < \epsilon$, if $n > N$.
My professors approach:
$$\begin{align*}
\left| \frac{-4n+3}{2n^2+2n+3} \right|
&\leq \left|\frac{3-4n}{2n^2} \right| \\
&\leq \left| \frac{3}{2n^2} \right| + \left|\frac{4n}{2n^2}\right| \\
&= \frac{3}{2} \cdot \frac{1}{n^2} + 2 \cdot \frac{1}{n}
\end{align*}$$
We note that:
$$\begin{align*}
&\frac{3}{2} \cdot \frac{1}{n^2} < \frac{\epsilon}{2} \Leftrightarrow \frac{3}{n^2} < \epsilon \iff n > \sqrt{\frac{1}{\epsilon}} \\
&2 \cdot \frac{1}{n} < \frac{\epsilon}{2} \Leftrightarrow 4 \cdot \frac{1}{n} < 2 \iff \frac{4}{\epsilon} < n
\end{align*}$$
Then:
$$\begin{align*}
&\text{Let } N > \max\bigg\{\sqrt{\frac{1}{\epsilon}}, \frac{4}{\epsilon} \bigg\} \\
&\implies\forall n > N \text{ it holds that } |a_n-a| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
\end{align*}$$
I have the following questions:
$1.$ What is wrong with my estimation?
$2.$ How does my prof get to $\frac{3}{2} \cdot \frac{1}{n^2} < \frac{\epsilon}{2}$
$3.$ What exactly does $\max\bigg\{\sqrt{\frac{1}{\epsilon}}, \frac{4}{\epsilon} \bigg\}$ tell us?
| Your working has no errors as far as I can tell. Sometimes proofs like these have multiple ways to approach it; this is one such instance.
To get to the line about $\newcommand{\ve}{\varepsilon}
\ve/2$, consider momentarily: we have two terms at this point in the working,
$$\frac 3 2 \cdot \frac{1}{n^2} \text{ and } 2 \cdot \frac 1 n$$
If $n$ is large enough, such that
$$\frac 3 2 \cdot \frac{1}{n^2} < \frac \ve 2 \text{ and } 2 \cdot \frac 1 n < \frac \ve 2$$
then clearly
$$\frac 3 2 \cdot \frac{1}{n^2} + 2 \cdot \frac 1 n < \frac \ve 2 + \frac \ve 2 = \ve$$
which is enough for what we want.
"Okay, but why bother?" It's just somewhat of a common trick in analysis to make each term in a $\ve-N$ or $\ve-\delta$ proof work individually, and then bring it all together in the end to make them all work together. It's a good trick to keep in mind, especially at points where the triangle inequality applies, which is something your professor used in getting to this point.
The goal, then, is to see what values of $n$ ensure this in each case. Well, solving for $n$, these respectively give
$$\begin{align*}
\frac 3 2 \cdot \frac{1}{n^2} < \frac \ve 2
&\implies \frac{1}{n^2} < \frac \ve 2 \cdot \frac 2 3 = \frac \ve 3 \\
&\implies n^2 >\frac 3 \ve \\
&\implies n > \sqrt{ \frac 3 \ve } > \sqrt{ \frac 1 \ve } = \frac{1}{\sqrt \ve} \tag{$\ast$} \\
2 \cdot \frac 1 n < \frac \ve 2
&\implies \frac 1 n < \frac \ve 4 \\
&\implies n > \frac 4 \ve \\
\end{align*}$$
Note: The choice to go to $1/\sqrt \ve$ as in $(\ast)$ is purely aesthetic, and serves no real purpose. (I certainly can't think of a meaningful reason why to do it, and I don't see why your instructor didn't do likewise for the latter equality and have $n > 1/\ve$ there.)
Anyhow. We know that if we choose $n>1/\sqrt \ve$, the first inequality is given, and if $n>4/\ve$, the second one is too. But we want both to be satisfied together. So we want $n$ greater than both of them.
Note: This ties precisely into "what does this tell us?" It tells us we broke down a necessary inequality up term by term, optimized each individually, and that each member of the $\max \{\cdots\}$ function makes (at least) one of those terms work.
The $\max$ function is ideal for this purpose, and ensures both are satisfied. It's as simple as realizing that
$$\begin{align*}
&\max \{x,y\} \ge x \\
&\max \{x,y\} \ge y
\end{align*}$$
All in all, this - optimizing individual terms, and using $\max \{\cdots\}$ to let it all work - might seem clunky at first, but it's a bread-and-butter trick for analysis. It's usually a lot easier to handle individual terms, rather than complicated fractions.
That doesn't invalidate your method, of course, but it would be a very good idea to be aware of this trick for future use.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
find $\int_0^\infty \frac{|\cos (\pi x)|}{4x^2 - 1} dx$
Find (with proof) $\displaystyle\int_0^\infty \frac{|\cos (\pi x)|}{4x^2 - 1}dx$
It's actually not even clear that the integral converges. If there were only sines/cosines in the integral, a standard technique would be to use the trigonometric identity $\sin \theta = \frac{2t}{1+t^2}$, $\cos \theta = \frac{1-t^2}{1+t^2}$, $t = \tan(\theta/2)$. I know that $\frac{2}{4x^2-1} = \frac{1}{2x-1} -\frac{1}{2x+1}$, so maybe one could plug this into the given integral to obtain an integral that's easier to evaluate?
Edit: from a comment, I think it would be useful to note that $\cos (\pi x) \leq 0$ iff $(2k+1/2)\pi \leq \pi x \leq (2k + 3/2)\pi\iff(2k + 1/2)\leq x\leq 2k+3/2$ (for some integer $k$) and $\cos(\pi x) \ge 0$ iff $(2k - 1/2) \leq x \leq (2k+1/2)$ (for some integer $k$). So we can split the integral according to these ranges. Then it might be useful to apply integration by parts. Let $a\in \mathbb{R}$. Then
$$\begin{align}\int \frac{\cos(ax)}{4x^2-1}dx &= \frac{1}2\left(\int \frac{\cos(a x)}{2x-1}dx - \int \frac{\cos(a x)}{2x+1}dx\right)\\
\\
&= \frac{1}2\left(\frac{1}2 \ln(2x-1)\cos (ax) +\frac{1}2a \int \sin(ax)\ln(2x-1)dx\right) \\
\\&-\frac{1}2\left(\frac{1}2 \ln(2x+1)\cos (ax) +\frac{1}2a \int \sin(ax)\ln(2x+1)dx\right)\end{align}$$
but I'm not sure how to simplify the result.
For the bounty, I'm looking for formal proofs. In particular, I'd like to see justifications for interchanging an integral and a sum. One can freely interchange an integral and a sum if the terms are nonnegative (as $\sum \int f_n = \int \sum f_n$ for nonnegative Lebesgue measurable functions $f_n$ and the Lebesgue integral equals the Riemann integral for Riemann integrable functions).
Also I'd like to see justifications for why $$\int_0^\infty \frac{\cos^2(\pi (2m+1) x)}{4x^2 - 1}dx = 0 = \int_0^\infty \frac{\sin^2(\pi (2m+1)x)}{4x^2-1} dx,$$ where $m$ is any nonnegative integer.
| $$I=\int_0^{\frac{1}2} \frac{\cos(\pi x)}{4x^2-1}dx-\int_{\frac{1}2}^{\frac{3}2} \frac{\cos(\pi x)}{4x^2-1}dx+\int_{\frac{3}2}^{\frac{5}2} \frac{\cos(\pi x)}{4x^2-1}dx-\int_{\frac{5}2}^{\frac{7}2} \frac{\cos(\pi x)}{4x^2-1}dx+\cdots$$
Look at the first term:
$$\int_0^{\frac{1}2} \frac{\cos(\pi x)}{4x^2-1}dx=\frac{1}{2}\int_0^{\frac{1}2} \frac{\cos(\pi x)}{2x-1}-\frac{\cos(\pi x)}{2x+1}dx=\frac{1}2\int_{-\frac{1}2}^{\frac{1}2} \frac{\cos(\pi x)}{2x-1}dx$$
Similarly, the second term:
$$-\int_{\frac{1}2}^{\frac{3}2} \frac{\cos(\pi x)}{4x^2-1}dx=-\frac{1}2\int_{-\frac{3}2}^{-\frac{1}2} \frac{\cos(\pi x)}{2x-1}dx-\frac{1}2\int_{\frac{1}2}^{\frac{3}2} \frac{\cos(\pi x)}{2x-1}dx$$
So we can write the integral as:
$$I=\frac{1}2\sum_{k=-\infty}^\infty(-1)^k\int_{k-\frac{1}2}^{k+\frac{1}2}\frac{\cos(\pi x)}{2x-1}dx$$
Let $z=\pi x-k\pi+\frac{\pi}2$
$$I=\frac{1}4\int_{0}^{\pi}\sum_{k=-\infty}^\infty\frac{\sin(z)}{z+(k-1)\pi}dz=\frac{1}4\int_{0}^{\pi}\sin(z)\cot(z)dz=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 2
} |
Evaluating $\int_0^\pi x\frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx$
How am I supposed to solve the following definite integral?
$$
\mathcal{I} = \int_0^\pi x \cdot \frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx
$$
This definite integral is solved if the minus sign is replaced by a plus sign, and it yields $\pi^2$.
$$
\mathcal{I} = \int_0^\pi x \cdot \frac{\sin{\frac{x}{2}} + \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx \text{
— (I)} \\
\implies \mathcal{I} = \int_0^\pi (\pi - x) \cdot \frac{\cos{\frac{x}{2} + \sin{\frac{x}{2}}}}{\sqrt{\sin{x}}} dx \text{
— (II)}
$$
On (I) + (II), we have,
$$
\mathcal{I} = \frac{\pi}{2}\int_0^\pi \frac{\sin{\frac{x}{2}} + \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx = \frac{\pi}{2} \int_0^\pi \frac{\sin{\frac{x}{2}}+\cos{\frac{x}{2}}}{\sqrt{1 - (\sin{\frac{x}{2}-\cos{\frac{x}{2}}})^2}} dx
$$
On substitution,
$$
\sin{\frac{x}{2}} - \cos{\frac{x}{2}} = u \implies \left(\sin{\frac{x}{2}} + \cos{\frac{x} {2}}\right) dx = 2 \cdot du
$$
The upper and lower limits changes to 1 and -1. Now, we have
$$
\mathcal{I} = \frac{\pi}{2} \int_{-1}^1 \frac{2 \cdot du}{\sqrt{1 - u^2}} du = \pi \cdot \left[\arcsin{u}\right]_{-1}^1 = \pi^2
$$
But...
*
*The sign was not supposed to be changed. We get $(2x - \pi)$ instead of $\pi$ in the nominator when adding both integrals. It complicates the problem.
*Using integral-calculator.com or a scientific calculator is helpless.
*The answer to the original problem should be $2\pi \cdot \ln{2}$ (approx 4.35.)
| $$I=\int_0^\pi\frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}}x dx=2\sqrt2\int_0^\frac{\pi}{2}\Big(\sqrt{\tan x}-\sqrt{\cot x}\Big)xdx=2\sqrt2(I_1-I_2)$$
Both terms converge, so we can evaluate them separately.
Making the substitution $x=\arctan t$ and $\arctan t=\int_0^1\frac{t}{1+t^2x^2}dx$
$$I_1=\int_0^\infty\frac{\sqrt t}{1+t^2}\arctan tdt=\int_0^1\frac{dx}{x^2}\int_0^\infty\frac{t\sqrt t}{(1+t^2)\big(t^2+\frac{1}{x^2}\big)}dt$$
To evaluate the first integral we go in the complex plane and integrate along a keyhole contour. We have four simple poles ($\,\pm i; \,\,\pm\frac{i}{x}\,$).
The straightforward evaluation gives
$$\int_0^\infty\frac{t\sqrt t}{(1+t^2)\big(t^2+\frac{1}{x^2}\big)}dt=\frac{2\pi i}{2}\underset{z=\pm i; z=\pm\frac{i}{x}}{\operatorname{Res}}\frac{z\sqrt z}{(1+z^2)\big(z^2+\frac{1}{x^2}\big)}=\frac{\pi}{\sqrt2}\frac{1-\sqrt x}{\sqrt x(1-x^2)}x^2$$
Therefore,
$$I_1=\frac{\pi}{\sqrt2}\int_0^1\frac{dx}{\sqrt x(1+\sqrt x)(1+x)}=\sqrt2\pi\int_0^1\frac{dt}{(1+t)(1+t^2)}$$
$$=\frac{\pi}{\sqrt2}\int_0^1\Big(\frac{1}{1+t}+\frac{1-t}{1+t^2}\Big)dt$$
Integration is straightforward and gives
$$I_1=\frac{\pi}{2\sqrt2}\Big(\frac{\pi}{2}+\ln2\Big)$$
In the similar way we evaluate $I_2$ (using $\sqrt{\cot x}=\frac{1}{\sqrt{\tan x}}$ and making the same substitution $x=\arctan t\,$)
$$I_2=\int_0^1\frac{dx}{x^2}\int_0^\infty\frac{\sqrt t}{(1+t^2)\big(t^2+\frac{1}{x^2}\big)}dt=\frac{\pi}{\sqrt2}\int_0^1\frac{dx}{(1+\sqrt x)(1+x)}=\frac{\pi}{2\sqrt2}\Big(\frac{\pi}{2}-\ln2\Big)$$
$$\boxed{\,\,I=2\sqrt2(I_1-I_2)=2\pi\ln2\,\,}$$
And, as a bonus,
$$J=2\sqrt2(I_1+I_2)=\pi^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 0
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Eliminate $\alpha$ by combining two integrals I have a function $f(t,x)$:
$$
f(t, x) = | \sin(2 \cdot \pi \cdot (t - x)) - \sin(2 \cdot \pi \cdot t) |
$$
and two values for $x$:
$$
l = \cos(\alpha)\cdot b \\ m = \sin(\alpha)\cdot b
$$
Can I "combine" $f(t, l)$ and $f(t, m)$ via some function $g(u, v)$ such that
$$
r = g(\int_0^{\frac12} f(t,l) \, dt, \int_0^{\frac12} f(t,m) \, dt)
$$
is independent of the value of $\alpha$? (I still need it to depend on both $u$ and $v$ in order to be able to say something about $b$ in the end; so "trivial" solutions such as $g(u,v) = 0$ or $g(u,v) = b$ are not what I'm looking for.)
I know that $cos^2(\alpha) + sin^2(\alpha) = 1$, but don't know if I can make use of this somehow. I've plotted both integrals for different values of $\alpha$ and somehow feel that there should be a way ...
For reference, here's how the two integrals look plotted with varying $\alpha$ values (on horizontal axis, in degree):
Background are calculations about the difference in amplitude between multiple (spacial, 2D) points with respect to a wave. The spacial difference is transferred into a "time difference" (because speed of wave is known) $x$. Depending on the direction of impact of the wave ($\alpha$) that $x$ is different; of interest are differences between a point $p_l = (-b, 0)$ and the origin as well as between a point $p_m = (0, b)$ and the origin. I assume that I can get rid of $\alpha$ somehow (I don't care really about the direction atm).
A bit more background: I basically get 2 values ($u$ and $v$, the results of the two integrals) and (atm) don't know $\alpha$. But I need a (constant, independent of $\alpha$) value $r = g(u,v)$.
| The first step would be to solve the integral before trying to find such a function $g$.
Since $f(t,x+1) = f(t,x)$, we can assume that $x \in (-\frac{1}{2},\frac{1}{2}]$. Then for $t \in (0, \frac{1}{2}]$ we have $f(t,x) = 0$ if either $ x \in \mathbb{Z}$ or $t = t_* = \frac{1}{4} + \frac{x}{2}$. In particular, you cannot differentiate between $b$ and $b+1$ with any function $g$.
If $x = 0$, then the desired integral is $0$, so let assume for now that $x \neq 0$. Then the $f(\cdot,x)$ has precisely one zero in $(0,\frac{1}{2})]$, hence we can split the integral in two parts:
\begin{align*}
\int_0^{\frac{1}{2}} f(t,x) \text{d}t &= \int_0^{t_*} f(t,x) \text{d}t + \int_{t_*}^{\frac{1}{2}} f(t,x) \text{d}t \\
&= \pm \int_0^{t_*} \sin( 2 \pi t - 2 \pi x) \text{d}t \\
&\quad \mp \int_0^{t_*} \sin( 2 \pi t ) \text{d}t \\
& \quad \mp \int_{t_*}^{\frac{1}{2}} \sin( 2 \pi t - 2 \pi x) \text{d}t \\
& \quad \pm \int_{t_*}^{\frac{1}{2}} \sin( 2 \pi t) \text{d}t.
\end{align*}
Where the sign is determined by $x$. Solving the $4$ integrals we get, up to a constant $\frac{- 1}{2 \pi}$ for simplicity,
\begin{align*}
\int_0^{\frac{1}{2}} f(t,x) \text{d}t &\propto \pm \cos( 2 \pi t_* - 2 \pi x) \mp \cos( - 2 \pi x) \\
&\quad \mp \cos( 2 \pi t_* ) \pm \cos(0) \\
& \quad \mp \cos( \pi - 2 \pi x) \pm \cos( 2 \pi t_* - 2 \pi x) \\
& \quad \pm \cos( \pi ) \mp \cos(2 \pi t_*) \\
&= \pm \cos(\frac{\pi}{2} - \pi x) \mp \cos( - 2 \pi x) \\
&\quad \mp \cos( \frac{\pi}{2} + \pi x) \pm 1 \\
& \quad \mp \cos( \pi - 2 \pi x) \pm \cos( \frac{\pi}{2} - \pi x) \\
& \quad \pm (-1) \mp \cos( \frac{\pi}{2} + \pi x).
\end{align*}
If we group corresponding terms, we obtain
\begin{align*}
\int_0^{\frac{1}{2}} f(t,x) \text{d}t &\propto \pm 2 \cos(\frac{\pi}{2} - \pi x) \mp 2 \cos( \frac{\pi}{2} + \pi x) \\
&\quad \mp \cos( - 2 \pi x) \mp \cos( \pi - 2 \pi x).
\end{align*}
Since $\cos(\alpha) = - \cos(\pi + \alpha)$ and $\cos(\frac{\pi}{2} - \alpha) = -\cos(\frac{\pi}{2} + \alpha) = \sin(\alpha)$ we find
\begin{align*}
\int_0^{\frac{1}{2}} f(t,x) \text{d}t &= \pm \frac{2}{\pi} \sin(\pi x).
\end{align*}
In particular, if $x = 0$, then $\frac{2}{\pi} \sin(\pi x) = 0$ which is precisely the value of $\int_0^{\frac{1}{2}} f(t,0) \text{d}t$, so for all $x \in (\frac{1}{2},\frac{1}{2}]$ we find
\begin{align}
\int_0^{\frac{1}{2}} f(t,x) \text{d}t &= \pm \frac{2}{\pi} \sin(\pi x).
\end{align}
You can now try to find a function $g$ which satisfies your desired properties.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing $\int_{0}^{\pi} \ln (\sin x+2) d x$ and $\int_{0}^{\pi} \ln (2-\sin x) d x$ I first encountered this integral
$$
I=\int_{0}^{\pi} \ln (\sin x+2) d x
$$
several months ago without any idea and had tried many methods such as integration by parts, substitution and Fourier series etc. but all are in vain. Today, I tried the tangent substitution and succeeded. Now I want to share with you and seek any other alternatives.
For simplicity, I convert, by symmetry, the integral into
$$
I=2 \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x+2) d x
$$
and substitute $t=\tan x$, and get
$$
\begin{aligned}
I &=2 \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x+2) d x \\
&=2 \int_{0}^{\infty} \ln \left(\frac{2 t}{1+t^{2}}+2\right) \frac{d t}{1+t^{2}} \\
&=2 \int_{0}^{\infty} \frac{\ln 2+\ln \left(t^{2}+t+1\right)-\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=\pi \ln 2+2 \int_{0}^{\infty} \frac{\ln \left(t^{2}+t+1\right)}{1+t^{2}}-2 \int_{0}^{\infty} \frac{\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=\pi \ln 2+2 \underbrace{ \left[\frac{\pi}{3} \ln (2+\sqrt{3})+\frac{4}{3} G\right]}_{(*)} -2 \underbrace{\pi \ln 2}_{(**)} \\
&=-\pi \ln 2+\frac{2 \pi}{3} \ln (2+\sqrt{3})+\frac{8}{3} G
\end{aligned}
$$
Note:
(*): post 1, (**):post 2
For the second integral, we use the similar technique and arrive at
\begin{aligned}
J&=2 \int_{0}^{\infty} \frac{\ln 2+\ln \left(1-t+t^{2}\right)-\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=2 \int_{0}^{\infty} \frac{\ln 2}{1+t^{2}} d t+2 \int_{0}^{\infty} \frac{\ln \left(1-t+t^{2}\right)}{1+t^{2}} d t-2 \int \frac{\ln \left(1+t^{2}\right)}{1+t^{2}} d t \\
&=\pi \ln 2+2 \underbrace{\left(\frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G\right)}_{(***)}-2 \pi \ln 2 \\
&=-\pi \ln 2+\frac{4 \pi}{3} \ln (2+\sqrt{3})-\frac{8}{3} G
\end{aligned}
Note:(***) post 3
Eager to know whether there are any alternatives!
Comments and alternative solutions are highly appreciated.
| $$I(a)=\int_{0}^{\pi} \log (a\sin (x)+2)\, dx$$
$$I'(a)=\int_{0}^{\pi} \frac{\sin (x)}{a \sin (x)+2}\, dx=\frac \pi a-\frac{2 \pi }{a \sqrt{4-a^2}}+\frac{4 \tan ^{-1}\left(\frac{a}{\sqrt{4-a^2}}\right)}{a \sqrt{4-a^2}}$$
$$\int_0^1 \Bigg[\frac \pi a-\frac{2 \pi }{a \sqrt{4-a^2}}\Bigg]\,da=\pi \log \left(\frac{2+\sqrt{3}}{4} \right)$$
$$\frac{4 \tan ^{-1}\left(\frac{a}{\sqrt{4-a^2}}\right)}{a \sqrt{4-a^2}}=\sum_{n=0}^\infty \frac{(n!)^2}{(2 n+1)!} a^{2n}$$
$$\int_0^1 \frac{4 \tan ^{-1}\left(\frac{a}{\sqrt{4-a^2}}\right)}{a \sqrt{4-a^2}}\, da=\sum_{n=0}^\infty \frac{(n!)^2}{(2 n+1) (2 n+1)!}=\frac{1}{3} \left(8 C-\pi \log \left(2+\sqrt{3}\right)\right)$$
Edit
Trying to generalize
$$I(a,b)=\int_{0}^{\pi} \log (a\sin (x)+b)\, dx$$
$$I'(a,b)=\int_{0}^{\pi} \frac{\sin (x)}{a \sin (x)+b}\, dx=\frac{\pi }{a}-\frac{\pi b}{a \sqrt{b^2-a^2}}+\frac{2 b}{a \sqrt{b^2-a^2}}\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)$$
$$\int_0^1 \Bigg[\frac{\pi }{a}-\frac{\pi b}{a \sqrt{b^2-a^2}}\Bigg]\,da=\pi \log \left(\frac{b+\sqrt{b^2-1}}{2 b}\right)$$
Using series expansion again
$$\int_0^1\frac{2 b}{a \sqrt{b^2-a^2}}\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\,da=\sum_{n=0}^\infty \frac {\alpha_n}{\beta_n}\, b^{-(2n+1)}$$ the $\alpha_n$ and $\beta_n$ corresponding respectively to sequences $A101926$ and $A321234$ in $OEIS$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
$f: \Bbb{R} \to \Bbb{R}, f(-y^2)+f(x^2+y)-y=(x+y)f(x-y).$
$f: \Bbb{R} \to \Bbb{R}, f(-y^2)+f(x^2+y)-y=(x+y)f(x-y).$
From some Olympiad problem book.
My attempt:
\begin{align}
P(0, 0): \; & f(0)+f(0)=0, f(0)=0. \\
\ \\
P(x, x): \; & f(-x^2)+f(x^2+x)-x=0. \\
P(x, -x): \; & f(-x^2)+f(x^2-x)+x=0. \\
\therefore \; & f(x^2+x)-f(x^2-x)=2x. \\
\ \\
P(x, 0): \; & f(x^2)=xf(x). \\
\ \\
P(x, -x^2): \; & f(-x^4)+x^2=(x-x^2)f(x+x^2). \\
& =(x-x^2)(f(x^2-x)+2x) \\
& =-f(x^4-2x^3+x^2)+2x^2-2x^3.
\end{align}
I can't proceed more. Can anyone give some help or answer this question?
| $$\begin{cases}P(x,y):\quad f(-y^2)+f(x^2+y)-y=(x+y)f(x-y)\\
P(-x,y):\quad f(-y^2)+f(x^2+y)-y=(-x+y)f(-x-y)
\end{cases}$$
Then $P(x,y)-P(-x,y)$ is $$(x+y)f(x-y)+(x-y)f(-x-y)=0.$$
Take $(a,b)\in\mathbb{R}^2$, and let $(x,y)$ be the solution of $$\begin{cases}x-y=a\\ x+y=-b \end{cases}$$
so that $P(x,y)-P(-x,y)$ becomes
$$-bf(a)+af(b)=0 \iff af(b)=bf(a).$$
You gave a formula for $P(x,-x)$, if you put $x=1$ in that formula you find $f(2)=2$.
Substituting $a=2$ in the formula above, $$2f(b)=bf(2)=2b,$$
meaning that $f$ is the identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute derivative of $\sin(x^3)$ by definition? I am trying to proof that derivative of $\sin(x^3) = 3x^2\cos(x^3)$ by definition.
But I don't know an identity for $\sin(x^3)$ for getting $\cos(x^3)$.
Even I try to find a quantity similar to $\frac{\sin(x)}{x}$.
| Computation using the differentiability of $\sin$ on $\mathbb R$
Let $f(x)=\sin(x^3)$. Then, for every $a\in\mathbb R$ and $x\neq a$, we have
$$
\frac{f(x)-f(a)}{x-a}
= \frac{\sin(x^3)-\sin(a^3)}{x-a}
= \frac{\sin(x^3)-\sin(a^3)}{x^3-a^3} \frac{x^3-a^3}{x-a}
$$
Let $X=x^3$ and $A=a^3$. Then $\frac{\sin(x^3)-\sin(a^3)}{x^3-a^3}=\frac{\sin(X)-\sin(A)}{X-A}$ and since $\sin$ is differentiable at $A$ and $X\to A$ when $x\to a$, we get
$$\lim_{x\to a}\frac{\sin(x^3)-\sin(a^3)}{x^3-a^3} = \lim_{X\to A}\frac{\sin(X)-\sin(A)}{X-A} = \cos(A) = \cos(a^3)$$
Moreover, $\frac{x^3-a^3}{x-a}=x^2+ax+a^2 \to 3a^2$ as $x\to a$. By Limit Laws, we have
$$
\lim_{x\to a} \frac{f(x)-f(a)}{x-a}=3a^2\cos(a^3)
$$
This means that $f$ is differentiable at $a$ and $f'(a)=3a^2\cos(a^3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove $\int_{0}^{\infty}{e}^{a\cos\left(bx\right)}\sin\left(a\sin\left(bx\right)\right)\frac{\mathrm{d}x}{x}=\frac{\pi}{2}\left({e}^{a}-1\right)$ Show that if $a > 0$ and $b > 0$ that
\begin{align}
\int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} = \frac{\pi}{2} \left({e}^{a} - 1\right) \\
\end{align}
Attempt:
I attempted to simplify the integral by putting it into a form of complex exponentials.
\begin{align}
& \int_{0}^{\infty} {e}^{a {e}^{i b x}} \frac{\mathrm{d}x}{x} \\
= & \int_{0}^{\infty} {e}^{a \cos \left(b x\right) + i a \sin \left(b x\right)} \frac{\mathrm{d}x}{x} \\
= & \int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \cos \left(\sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} + i \int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} \\
\end{align}
We then have the result by linearity that
\begin{align}
& \int_{- \infty}^{\infty} {e}^{a {e}^{i b x}} - {e}^{a {e}^{- i b x}} \frac{\mathrm{d}x}{x} \\
& = \int_{- \infty}^{\infty} {e}^{a \cos \left(b x\right)} \left({e}^{i a \sin \left(b x\right)} - {e}^{i a \sin \left(b x\right)}\right) \frac{\mathrm{d}x}{x} \\
& = 2 i \int_{- \infty}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} \\
& = 4 i I \\
\end{align}
where
\begin{align}
I = \int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} \\
\end{align}
I am confused because when I take the residue at $0$, I find
\begin{align}
& {\text{Res}}_{z = 0} \frac{{e}^{a {e}^{i b z}} - {e}^{a {e}^{- i b z}}}{z} \\
& = {e}^{a} - {e}^{a} = 0 \\
\end{align}
| Let $bx\to x$ to simplify the integral to\begin{align}
& \int_{0}^{\infty} {e}^{a \cos x}\ \sin (a \sin x)\ \frac{dx}{x} \\
= &\ \Im \int_{0}^{\infty} {e}^{a e^{ix}}\ \frac{dx}{x}
= \Im \int_{0}^{\infty} \sum_{k=0}^\infty\frac{(a e^{ix})^k}{k!}\frac{dx}x\\
= &\sum_{k=1}^\infty\frac{a^k}{k!} \int_{0}^{\infty} \frac{\sin(kx)}x\ dx
= \sum_{k=1}^\infty\frac{a^k}{k!}\cdot \frac\pi2 = \frac{\pi}{2} \left({e}^{a} - 1\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Exponential equation in $\mathbb{Z}$ Solve the equation:
$$5^{18x}+12=13^{18y+1},$$ where x, y are positive integer.
I notice that $x=y=0$ is the only solution. I tried to reduce modulo primes but I didn't succeed.
| Suppose $x \neq 0$. Modulo $5$ we find that
$$2 \equiv 3^{18y+1} \equiv 3 \cdot 4^y\hspace{10pt} \mbox{ mod } 5$$
since by Euler-Fermat's theorem the exponent can be reduced modulo $\varphi(5)=4$. So looking at the powers of $4$ mod $5$ ($4^2 \equiv 1$) we conclude that $y$ must be odd.
Now, modulo 13 we have
$$5^{18x} \equiv 1 \hspace{10pt} \mbox{ mod } 13$$
and therefore if we examine the powers of $5$ modulo 13 ($5^2 \equiv -1$, $5^3 \equiv -5$, $5^4 \equiv 1$) we obtain
$$18x \equiv 0 \hspace{10pt} \mbox{ mod } 4$$
and so $x$ must be even.
Finally if we look at the equation modulo $37$ substituting $x=2k$ and $y=2h+1$ we get
$$5^{36k}+12 \equiv 13^{36h+19} \hspace{10pt} \mbox{ mod } 37$$
and hence, since $\varphi(37)=36$ and $13$ has an inverse mod $37$
$$1+12=13 \equiv 13^{19} \Rightarrow 13^{18} \equiv 1 \hspace{10pt} \mbox{ mod } 37$$
But $$13^{18} \equiv 21^9 \equiv 11^3 \equiv 110 \equiv -1 \hspace{10pt} \mbox{ mod } 37$$
so this is absurd. Therefore we cannot have $x \neq 0$, and if $x=0$ then immediately $y=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving $\cos(2\theta)=\sin\left(\frac{\theta}{2}\right)$ Consider an acute angle $\theta$, which has the property that
$$\cos(2\theta)=\sin\left(\frac{\theta}{2}\right).$$
I am trying to find the value of $\theta$. Using the identity
$$\cos(2\theta)=\sin\left(\frac{\pi}{2}-2\theta\right),$$
I can write the first equation as
$$\sin\left(\frac{\pi}{2}-2\theta\right)=\sin\left(\frac{\theta}{2}\right).$$
Equating the argument of the $\sin$ functions, we see that
\begin{align}
\frac{\pi}{2}-2\theta&=\frac{\theta}{2} \\
\frac{5}{2}\theta&=\frac{\pi}{2} \\
\theta&=\frac{\pi}{5}.
\end{align}
While this yields the correct answer, I do not know how we can justify equating the arguments of the $\sin$ functions.
| In general we have
$$\cos x = \sin y \iff y = \frac \pi 2 \pm x + 2k\pi$$
that is in this case
$$\frac \theta 2 = \frac \pi 2 \pm 2\theta + 2k\pi$$
which leads to the general solutions
*
*$\theta_1 = \frac \pi 5+\frac 4 5 k \pi$
*$\theta_2 = -\frac \pi 3+\frac 4 3 k \pi$
and the solution for an acute angle is indeed $\theta = \frac \pi 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in which series Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in
(1)AP
(2) GP
(3) HP
(4) none of these
My approach is as follow
$2{b^2} = {a^2} + {c^2} \Rightarrow 2{b^2} + {b^2} = {a^2} + {c^2} + {b^2} \Rightarrow 3{b^2} = {a^2} + {c^2} + {b^2}$
${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)$
${\left( {a + b + c} \right)^2} = 3{b^2} + 2 \Rightarrow {\left( {a + b + c} \right)^2} - {b^2} = 2{b^2} + 2$
$\Rightarrow \left( {a + b + c - b} \right)\left( {a + b + c + b} \right) = 2{b^2} + 2$
$\Rightarrow \left( {a + c} \right)\left( {a + c + 2b} \right) = 2{b^2} + 2 \Rightarrow {\left( {a + c} \right)^2} + 2b\left( {a + c} \right) = 2{b^2} + 2$
$ \Rightarrow {\left( {a + c} \right)^2} + 2\left( {ab + bc} \right) = 2{b^2} + 2 \Rightarrow {\left( {a + c} \right)^2} + 2\left( {1 - ac} \right) = 2{b^2} + 2$
$\Rightarrow {\left( {a + c} \right)^2} - 2ac = 2{b^2} \Rightarrow {\left( {a + c} \right)^2} = 2\left( {{b^2} + ac} \right)$
$ \Rightarrow \frac{{a + c}}{{{b^2} + ac}} = \frac{2}{{\left( {a + c} \right)}}$
Not able to proceed further.
| $$b^2-a^2=c^2-b^2=k(\ne0)\text{say}\implies b^2=a^2+k, c^2-a^2=2k$$
$$(c-b)(c+b)=k\iff b+c=\dfrac k{c-b}, \text{similarly } a+b=\dfrac k{b-a}, c-a=\dfrac{2k}{c+a}$$
$$\dfrac1{b+c}+\dfrac1{a+b}=\dfrac{c-b+b-a}k=\dfrac{2k}{k(c+a)}=? $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Advanced Functions, How to simplify $\tan \frac{5\pi}{12}$? I was asked to find the exact value of $\,\tan \left(\frac{5\pi}{12}\right)$, so I did up until this point where I got completely stuck.
I split the ratio into two, so $\;\dfrac{5\pi}{12} = \dfrac{\pi}4 + \dfrac{\pi}6\,.$
using the formula, $\tan(x+y) = \dfrac{\tan x+\tan y}{1-\tan x\tan y}$:
$$\tan(\pi/4 + \pi/6) = \frac{1+1/\sqrt 3 }{1-(1)(1/\sqrt 3)
}$$
...
$$\tan(\pi/4 + \pi/6) = \frac{3+\sqrt3}{ 3-\sqrt3 }$$
I'm stuck here, I looked up the answer online, and apparently, you have to times numerator and denominator both by $\,3+\sqrt3\,,\,$ but it's cheating, I don't understand why you have to do that.
| So, you've remembered (or re-derived) that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$, and plugging this into your sum-of-angles formula gives:
$$\tan(\frac{5\pi}{12}) = \frac{\tan(\frac{\pi}{4}) + \tan(\frac{\pi}{6})}{1 - \tan(\frac{\pi}{4})\tan(\frac{\pi}{6})} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - (1)(\frac{1}{\sqrt{3}})}$$
But fractions within fractions are just ugly, so let's multiply numerator and denominator by $\sqrt{3}$.
$$\tan(\frac{5\pi}{12}) = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$
And this is a perfectly valid way of expressing the value (which numerically is approximately 3.732051). But it's often considered good style to “rationalize the denominator”, i.e., have √ signs only on top and never on bottom.
Recall the FOIL method for multiplying two binomials, and that $$(a - b)(a + b) = a^2 + ab - ab - b^2 = a^2 - b^2$$
So, if you multiply a binomial by the same thing but with the opposite sign, then only squared terms occur in the product, so any √ signs go away. So let's multiply the denominator by $\sqrt{3} + 1$. Of course, we need to multiply the numerator by $\sqrt{3} + 1$ too so that the value of the expression doesn't change
$$\frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$$
$$= \frac{3 + 2\sqrt{3} + 1}{3 - 1}$$
$$= \frac{4 + 2\sqrt{3}}{2}$$
$$= 2 + \sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Is there an analytic solution to this set of cubic equations? I have three parameters $(a,b,c)$ that define equations I am looking for an analytical solution.
$$q_1=b/c$$
$$q_2=\frac{(a+b)^3}{3c^2b}$$
$$q_3=\frac{(a+b)^3}{3b^3}$$
I want to solve these equations for $a$, $b$, and $c$ in terms of $q_1$, $q_2$, and $q_3$. The first equation can easily be solved for $b=q_1c$. Inserting this into either of the two remaining is third degree of each parameter.
$$q_2=\frac{a^3}{3c^3q_1}+\frac{a^2}{c^2}+\frac{aq_1}{c}+\frac{q_1^2}{3}$$
$$q_3=\frac{a^3}{27c^3q_1^3}+\frac{a^2}{9c^2q_1^2}+\frac{a}{cq_1}+\frac{1}{3}$$
This doesn't seem like a problem where I could apply the cubic formula, unless there is no error in defining a new dummy variable $x=a/c$ first and solving for its constituents later.
If there isn't an "easy" analytical solution to find I'll linearize the last two equations, but I'm looking for something more exact.
| Slightly rewriting the equations to avoid fractions:
$$q_1 c = b \tag{1}$$
$$3 q_2 bc^2 = (a + b)^3 \tag{2}$$
$$3 q_3 b^3 = (a + b)^3 \tag{3}$$
From (2) and (3), we get $3 q_2 bc^2 = 3 q_3 b^3$, or $q_2 c^2 = q_3 b^2$. But from (1), $b = q_1 c$, so plugging that in gives $q_2c^2 = q_3 q_1^2 c^2$, or $q_2 = q_1^2 q_3$. So we don't have three independent $q$'s. I'll deal with this redundancy later.
Taking the cube root of (3) gives:
$$b\sqrt[3]{3q_3} = a + b$$
Or equivalently, $a = (\sqrt[3]{3q_3} - 1)b$.
From (1), we get $c = \frac{b}{q_1}$. And with explicit formulas for $a$ and $c$ in terms of $b$, we can now write (2) in terms of $b$ alone.
$$3 q_2 b(\frac{b}{q_1})^2 = 3q_3 b$$
Cancelling $b$ from each side gives an easily-solved quadratic, with $b = q_1 \frac{\sqrt{q_3}}{\sqrt{q_2}}$.
Thus, the solution is:
$$a = (\sqrt[3]{3q_3} - 1)q_1\frac{\sqrt{q_3}}{\sqrt{q_2}}$$
$$b = q_1 \frac{\sqrt{q_3}}{\sqrt{q_2}}$$
$$c = \frac{\sqrt{q_3}}{\sqrt{q_2}}$$
However, as noted earlier, the three $q$'s aren't independent, but related by $q_2 = q_1^2 q_3$. So we only need two of these to express our solution.
If we substitute $q_1 = \frac{\sqrt{q_2}}{\sqrt{q_3}}$, then:
$$a = (\sqrt[3]{3q_3} - 1)$$
$$b = 1$$
$$c = \frac{\sqrt{q_3}}{\sqrt{q_2}}$$
If we substitute $q_2 = q_1^2 q_3$, then:
$$a = (\sqrt[3]{3q_3} - 1)$$
$$b = 1$$
$$c = \frac{1}{q_1}$$
Finally, if we substitute $q_3 = \frac{q_2}{q_1^2}$,
$$a = (\sqrt[3]{3\frac{q_2}{q_1^2}} - 1)$$
$$b = 1$$
$$c = \frac{1}{q_1}$$
So $b$ isn't really a variable at all!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculate the integral of $\frac{1}{x^2 +x + \sqrt x}$ How to correctly calculate the integral:
$$\int_0^\infty \frac{1}{x^2 +x + \sqrt x}dx$$
Edit: I tried to figure out if the limit exists:
Step 1: break the integral to two parts: from 0 to 1, from 1 to infinity.
Step 2: use limit comparison test for both of the integral: the first integral compared at 1 to 1/sqrt(x) and the second is compared at infinity to 1/x^2.
Step 3: conclude that both converge, hence the original integral also converges.
Step 4: (this is the one im trying to figure out, how to actually calculate it, because the limit exists).
| We have
\begin{align*}
I &= \int_0^\infty \frac{1}{x^2 + x + \sqrt{x}} \ \mathrm{d}x\\
&= \int_0^\infty \frac{2}{y^3 + y + 1} \ \mathrm{d}y && \text{using $y = \sqrt{x}$}\\
\end{align*}
Clearly, this only has one real root, denoted $r$, where $r < 0$. Since $r^3 + r + 1 = 0$, we have $r^{-1} = -r^2 - 1$, so $y^3 + y + 1 = (y - r)\left (y^2 + ry + (r^2 + 1) \right)$. Then we have
\begin{align*}
\frac{2}{y^3 + y + 1} &= \frac{A}{y - r} + \frac{By + C}{y^2 + ry + (r^2 + 1)}\\
2 &= A\left (y^2 + ry + (r^2 + 1) \right ) + \left (By + C \right )(y - r)\\
y=r \implies A &= -\frac{2r}{2r + 3}\\
B &= \frac{2r}{2r + 3}\\
C &= \frac{4r^2}{2r+3}\\
\implies \frac{2}{y^3 + y + 1} &= \frac{2r}{2r + 3}\left (\frac{y + 2r}{y^2 + ry + (r^2 + 1)} - \frac{1}{y - r} \right )
\end{align*}
Thus we have
\begin{align*}
I &= \frac{2r}{2r + 3} \int_0^\infty \frac{y + 2r}{y^2 + ry + (r^2 + 1)} - \frac{1}{y - r} \ \mathrm{d}y\\
&= \frac{2r}{2r + 3} \int_0^\infty \frac{y + \frac{r}{2}}{y^2 + ry + (r^2 + 1)} + \frac{\frac{3r}{2}}{\left ( y + \frac{r}{2}\right )^2 + \left (\frac{3r^2}{4} + 1\right )} - \frac{1}{y - r} \ \mathrm{d}y\\
&= \frac{6r}{2r + 3} \left ( \frac{1}{2} \ln(-r) - \frac{(-r)^{\frac{3}{2}}}{\sqrt{3 - r}} \left (\frac{\pi}{2} + \tan^{-1}\left ( \frac{(-r)^{\frac{3}{2}}}{\sqrt{3 - r}}\right )\right )\right )
\end{align*}
Taking $r = -\frac{2}{\sqrt{3}}\sinh\left ( \frac{1}{3} \sinh^{-1} \left (\frac{3\sqrt{3}}{2}\right ) \right )$ gives a numerical evaluation for the above term of about $1.8435267\dots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Cartesian equation for a transcendental / trigonometric curve
Hello! Please see figure above. I am searching for the cartesian equation for the curve in green, similar to how the equation for a semicircle is $f(x) = √(1 - x^2)$. I'm not sure if this is even possible, so any feedback is appreciated. My several attempts have failed.
The construction is set up like so (upon a circle of radius=1):
$$\frac{m}{k} = \frac{k}{t}$$
or
$$\frac{sin(α)} {sin(β)} = \frac{sin(β)} {sin(α+β)}$$
As you can see, $x$ is in blue and $f(x)$ is in green, and as α grows, then the point H traces out the curve in green, which I'm calling a gourd curve. Is it possible to obtain a cartesian equation for the gourd curve and not something which contains trigonometric identities? Trigonometric identities are also welcome, but I'd prefer to analyze the equation without them.
| Relabelling your figure a bit, I'm taking center $O$, and points $A$ and $B$ on the circle with $M$ the foot of the perpendicular from $B$ to $\overline{OA}$. If the circle meets the $x$-axis at $R$, then define $\alpha :=\angle ROA$ and $\beta :=\angle AOB$.
If the radius of the circle is $r$, then we can write the coordinates of $M$ as
$$x = r \cos\beta \cos\alpha \qquad y = r\cos\beta \sin\alpha \tag1$$
The imposted relation on $\alpha$ and $\beta$ is
$$\frac{\sin\alpha}{\sin\beta}=\frac{\sin\beta}{\sin(\alpha+\beta)} \tag2$$
Expanding $(2)$, and using the parameterizations
$$\cos\alpha=\dfrac{1-a^2}{1+a^2} \quad \sin\alpha=\dfrac{2a}{1+a^2} \quad
\cos\beta=\dfrac{1-b^2}{1+b^2} \quad \sin\beta=\dfrac{2b}{1+b^2} \tag3$$ we can use, say, the method of resultants to eliminate $a$ and $b$ to obtain this cartesian equation for the gourd curve:
$$\begin{align}
0 &= (x^2 + y^2)^8 \\
&\quad- 2 r^2\, (x^2 + y^2)^5\, (2 x^4 + 3 x^2 y^2 + 3 y^4) \\
&\quad+\phantom{2}r^4\, (x^2 + y^2)^4\, (6 x^4 + 6 x^2 y^2 + 11 y^4) \\
&\quad- 2 r^6\, (x^2 + y^2)^2\, (2 x^6 + 3 x^4 y^2 + 5 x^2 y^4 + 3 y^6) \\
&\quad+\phantom{2}r^8\, ((x^2+y^2)^2 - x^2 y^2)^2
\end{align} \tag{$\star$}$$
(This form was used to generate the figure above. Because $x$ and $y$ appear exclusively to even powers, the solution set includes a horizontally-mirrored copy of the gourd.)
In polar form (using $\rho$ for the radial variable since I'm using $r$ for the radius of the circle):
$$\begin{align}
0 &= \rho^8 \\
&\quad-2\rho^6 r^2\, (3 - 3 \cos^2\theta + 2 \cos^4\theta) \\
&\quad+\phantom{2}\rho^4 r^4\, (11 - 16 \cos^2\theta + 11 \cos^4\theta) \\
&\quad-2 \rho^2 r^6\, (3 - 4 \cos^2\theta + 2 \cos^4\theta + \cos^6\theta) \\
&\quad+\phantom{2\rho^2} r^8\, (1 - \cos^2\theta + \cos^4\theta)^2
\end{align} \tag{$\star\star$}$$
(Because $\rho$ appears exclusively to even powers, real roots occur in $\pm$ pairs; taking just one of these should give a single "gourd".)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Proving a function is continuous using rigorous definition of a limit Im trying to prove that the function $$\begin{cases}f(x,y)=\dfrac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}, & (x,y)≠0\\ 0, & (x,y)=0\end{cases}$$ is continuous at point (0,0) using the rigorous defintion of a limit.
Attempting to find the upper limit of the function:
$$|f(x)-f(x_0)|= \left|\frac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}-0\right|$$
I see the denominator is always positive so this is equal to
$\dfrac{|2x^2y^4-3xy^5+x^6|}{(x^2+y^2)^2}$.
Using the triangle inequality i know that this is equal or less than
$\dfrac{|(2x^2y^4)-(3xy^5)|+|x^6|}{(x^2+y^2)^2}$.
From here I would like to continue finding expressions which are equal or greater than this, which allow me to cancel some terms against $((x^2+y^2)^2)$.
Im thinking i can write
$$x^6 = (x^2)^3 ≤ (x^2+y^2)^3 $$
for instance, but i am unsure of how to "handle" $|(2x^2y^4)-(3xy^5)$|.
Could someone give me any pointers?
| Using the triangle inequality like this:
$$|f(x,y)|\le \frac{|2x^2y^4|+|-3xy^5|+|x^6|}{(x^2+y^2)^2}$$
Notice that
$$\frac{|x^6|}{(x^2+y^2)^2}\le \frac{|x|^6}{x^4}=\frac{x^6}{x^4}=x^2$$
$$\frac{|2x^2y^4|}{(x^2+y^2)^2} \le \frac{2x^2y^4}{y^4}=2x^2$$
$$\frac{|-3xy^5|}{(x^2+y^2)^2} \le \frac{3|x||y|y^4}{y^4}=3|x||y|$$
It is $|f(x,y)| \le 3(x^2+|x||y|)$.
So, noticing that $\sqrt{x^2+y^2}<\delta$ implies $|x|<\delta$ and $|y|<\delta$, you shold be able to conclude with an appropriate choice of $\delta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Trouble with the Binomial sum I guess the following sum is equal to $1$ since you can compute the baby cases for $t=1,2$ straightforward to get the desired result. So, in general, we should have the following sum to be $1$, I think. But, I am stuck in the last equality. Any comments or advice would be appreciated.
\begin{align*}
&\frac{t}{n}+t\left(\frac{\binom{n-t}{1}}{2\binom{n}{2}}+\frac{\binom{n-t}{2}}{3\binom{n}{3}}+\frac{\binom{n-t}{3}}{4\binom{n}{4}}+\cdots+\frac{\binom{n-t}{n-t}}{(n-t+1)\binom{n}{n-t+1}}\right)\\\\
=~~&\frac{t}{n}+t\left(\frac{\binom{n-t}{1}}{n\binom{n-1}{1}}+\frac{\binom{n-t}{2}}{n\binom{n-1}{2}}+\frac{\binom{n-t}{3}}{n\binom{n-1}{3}}+\cdots+\frac{\binom{n-t}{n-t}}{n\binom{n-1}{n-t}}\right)\\\\
=~~&\frac{t}{n}+\frac{t}{n}\left(\frac{\binom{n-t}{1}}{\binom{n-1}{1}}+\frac{\binom{n-t}{2}}{\binom{n-1}{2}}+\frac{\binom{n-t}{3}}{\binom{n-1}{3}}+\cdots+\frac{\binom{n-t}{n-t}}{\binom{n-1}{n-t}}\right)
\end{align*}
where the first equality is used $k\binom{n}{k}=n\binom{n-1}{k-1}$ and $t\in\{1,2,...,n\}.$
| $$\frac tn\sum_{i=0}^{n-t}\frac{\binom{n-t}i}{\binom{n-1}i}=\frac1{\binom nt}\sum_{i=0}^{n-t}\binom{n-1-i}{t-1}=\frac1{\binom nt}\sum_{j=0}^{n-t}\binom{t-1+j}{t-1}=1.$$
The first equality is obtain by using factorials, the second one by letting $j=n-t-i,$ and the last one is the hockey-stick identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A precalculus solution for $x^6 - 3 x^4 + 2 x^3 + 3 x^2 - 3 x + 1 =0$
Using algebra (precalculus) and suggest the solution method for the polynomial $$x^6 - 3 x^4 + 2 x^3 + 3 x^2 - 3 x + 1 =0$$
I'm solving problems on polynomials. I'm stuck here.
My attempts.
First, I tried the Rational root theorem, then I failed.
Then I tried factorise the polynomial e.g. $(x^2+ax+b)(x^4+cx^3+dx^2+ex+f)$, but I failed again.
At the end I tried
$$P(x)/x^3=x^3-3x+2+\frac 3x-3\frac {1}{x^2}+\frac {1}{x^3}=x^3+\frac {1}{x^3}-3\bigg(x-\frac 1x\bigg)-\frac {3}{x^2}+2=0$$
I failed again.
| We can rewrite the given polynomial as
$$p(x) = x^6 - 3 x^4 + 2 x^3 + 3 x^2 - 3 x + 1 = \\
(x^6 + 2x^3 + 1) - 3x (x^3 - x + 1) = \\
(x^3 + 1)^2 - 2x (x^3 + 1) + 2x^2 - x (x^3 - x + 1) = \\
(x^3 - x + 1)^2 - x (x^3 - x + 1) + x^2.$$
Therefore, if $x$ is a root of $p$, then either
$$x^3 - x + 1 = -\omega x$$
or
$$x^3 - x + 1 = -\omega^2 x$$
where $\omega = e^{2\pi i / 3} = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$ and $\omega^2 = \bar\omega = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$.
Since each of these is a cubic whose coefficients are expressible as radicals, the solutions to these individual cubics are also expressible as radicals.
(Honestly, though, the above sequence of steps was somewhat reverse-engineered from the eventual solution that I found based on numerical experimentation. The route we came to this in the comments was: through using a CAS to calculate the Galois group of the polynomial, use that to conjecture what solutions in terms of radicals might look like, and then do numerical experimentation based on that form to try to find elements of the form.)
(Note also that we can see a relation of this solution to the CAS-computed Galois group of the polynomial. The solution above gives the splitting field of $p$ as a compositum of two extensions of $\mathbb{Q}(\sqrt{-3})$, each of degree at most 6. Assuming the CAS is correct that the Galois group has order 72, that implies that the two extensions have minimal intersection $\mathbb{Q}(\sqrt{-3})$ and each have degree 6.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
If $x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$, show that $\frac{201}{403}
If $$x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$$
show that
$$\frac{201}{403}<x<\frac{2014}{2015}$$
So, I manage to do the RH inequality using that $$x<\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2014\cdot2015}=1-\frac{1}{2015}=\frac{2014}{2015}$$ Unfortunately I can't find a way to do the LH inequality. I saw that
$$\frac{1}{2015}+\frac{1}{2015}+\cdots+\frac{1}{2015}$$ for 1005 times is equal to $\frac{201}{403}$ but I don't see why
$$x> \frac{1}{2015}+\frac{1}{2015}+\cdots+\frac{1}{2015}$$ thx!
| We have the bound
$$\sum_{n=2}^{2015}\frac{1}{n^2}>\int_2^{2016}\frac{1}{x^2}dx=\frac{1}{2}-\frac{1}{2016}>\frac{1}{2}-\frac{1}{806}=\frac{402}{806}=\frac{201}{403}$$
by thinking about the graph of the function $y=1/x^2$ and noticing that if you draw rectangles of base length $1$ with four vertices $(n,0),(n+1,0),(n,1/n^2),(n+1,1/n^2)$ for $n=2,\dots,2015$ then the union of the resulting rectangles have an area bounded below by the integral given above.
By the way, this also works to establish the upper bound $2014/2015$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
An alternating sum I ran into an alternating sum in my research and would like to know if the following identity is true:
$$
\sum_{i = 0}^{\left\lfloor \left(n + 1\right)/2\right\rfloor} \frac{\left(n + 1 - 2i\right)^{n + 1}}{2^{n}\left(n + 1\right)!}\binom{n + 1}{i}\left(-1\right)^{i} = 1\quad
\forall\ \mbox{positive integers}\ n\geq 3
$$
Any help would be appreciated!.
Edit.
We might try to use an Iverson bracket $[[2q\le n]]$ in attempting to
evaluate
$$S_n = \sum_{q=0}^{\lfloor n/2\rfloor}
(n-2q)^n {n\choose q} (-1)^q.$$
We obtain
$$[v^n] \frac{1}{1-v}
\sum_{q\ge 0} v^{2q} (n-2q)^n {n\choose q} (-1)^q$$
Using a coefficient extractor,
$$n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{n+1}} \frac{1}{1-v}
\sum_{q\ge 0} v^{2q} \exp(-2qz) {n\choose q} (-1)^q
\\ = n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{n+1}} \frac{1}{1-v} (1-v^2\exp(-2z))^n.$$
Now residues sum to zero and the residue at one yields
$$- n! [z^n] \exp(nz) (1-\exp(-2z))^n.$$
We have that since $(1-\exp(-2z))^n = (2z-2z^2\pm\cdots)^n = 2^n z^n +\cdots$ this evaluates to $-2^n n!.$ We find for the residue at infinity
$$- n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^2}
v^{n+1} \frac{1}{1-1/v} (1-\exp(-2z)/v^2)^n
\\ = n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^n} \frac{1}{1-v} (v^2-\exp(-2z))^n
\\ = n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^n} \frac{1}{1-v}
\sum_{q=0}^n {n\choose q} (-1)^{n-q} \exp(-2(n-q)z) v^{2q}
\\ = \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^n} \frac{1}{1-v}
\sum_{q=0}^n {n\choose q} (-1)^{n-q} (2q-n)^n v^{2q}
\\ = \sum_{q=0}^n {n\choose q} (-1)^q (n-2q)^n [[2q\le n-1]].$$
Now when $n$ is odd this gives the upper limit $\lfloor n/2\rfloor$ and
when $n$ is even $\lfloor n/2\rfloor -1$ however in the latter case we
may raise to $\lfloor n/2\rfloor$ because the added term is zero in the
sum per $(n-2q)^n = 0$. We have obtained
$$\sum_{q=0}^{\lfloor n/2\rfloor} {n\choose q} (-1)^q (n-2q)^n
= S_n.$$
Collecting everything we have shown that $S_n - 2^n n! + S_n = 0$ or
$S_n = 2^{n-1} n!.$ The question now becomes, is there a simpler proof?
| We seek to show that
$$\sum_{q=0}^{\lfloor n/2\rfloor}
(n-2q)^n {n\choose q} (-1)^q
= 2^{n-1} n!.$$
Observe that
$$(n-2q)^n {n\choose q} (-1)^q
= \frac{1}{2}
\left[ (n-2q)^n {n\choose q} (-1)^q
+ (2q-n)^n {n\choose n-q} (-1)^{n-q} \right].$$
Hence we get for our sum
$$\frac{1}{2}
\sum_{q=0}^{\lfloor n/2\rfloor}
(n-2q)^n {n\choose q} (-1)^q
+ \frac{1}{2}
\sum_{q=0}^{\lfloor n/2\rfloor}
(2q-n)^n {n\choose n-q} (-1)^{n-q}
\\ = \frac{1}{2}
\sum_{q=0}^{\lfloor n/2\rfloor}
(n-2q)^n {n\choose q} (-1)^q
+ \frac{1}{2}
\sum_{q=n-\lfloor n/2\rfloor}^n
(n-2q)^n {n\choose q} (-1)^q
\\ = \frac{1}{2}
\sum_{q=0}^n
(n-2q)^n {n\choose q} (-1)^q.$$
Introducing a coefficient extractor,
$$\frac{1}{2} \sum_{q=0}^n {n\choose q} (-1)^q
n! [z^n] \exp((n-2q) z)
\\ = \frac{1}{2} n! [z^n] \exp(nz)
\sum_{q=0}^n {n\choose q} (-1)^q
\exp((-2q) z)
\\ = \frac{1}{2} n! [z^n] \exp(nz)
(1-\exp(-2z))^n.$$
Note however that $(1-\exp(-2z))^n = (2z-2z^2\pm\cdots)^n$ so the only
contribution to the coefficient extractor $[z^n]$ is from the first
term of the series so that $[z^n] \exp(nz) (2z-2z^2\pm\cdots)^n = 2^n$ and we
finally have
$$2^{n-1} n!$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
What is the pattern in the powers of $\sqrt{2}-\sqrt{1}$? What is the pattern in this?
$$\begin{align}
\left(\sqrt{2}-\sqrt{1}\right)^1 &= \sqrt{2}-\sqrt{1}\\
\left(\sqrt{2}-\sqrt{1}\right)^2 &= \sqrt{9}-\sqrt{8}\\
\left(\sqrt{2}-\sqrt{1}\right)^3 &= \sqrt{50}-\sqrt{49}\\
\left(\sqrt{2}-\sqrt{1}\right)^4 &= \sqrt{289}-\sqrt{288}\\
\end{align}$$
I thought of applying the binomial theorem
| We have $$(\sqrt{1}-\sqrt{2})^k = \sqrt{a_k^2} - \sqrt{2b_k^2}$$
where $a_k,b_k$ obey the following recurrence relationships:
$$a_{k+1}=a_{k}+2b_{k}\\
b_{k+1}=a_{k}+b_{k}$$
with $a_{1}=1$ and $b_1 = 1$. This can be proved inductively. To understand the recurrence relationship we can recast it as a matrix equation:
$${\begin{pmatrix}
a_{k+1}\\
b_{k+1}
\end{pmatrix}} = \begin{pmatrix}
1 & 2 \\
1 & 1
\end{pmatrix} \begin{pmatrix}
a_k\\
b_k
\end{pmatrix} $$
The eigenvectors are $\begin{pmatrix}
\sqrt{2} \\
1
\end{pmatrix}$, $\begin{pmatrix}
\sqrt{2} \\
-1
\end{pmatrix}$ with coresponding eigenvalues: $1+\sqrt{2}$ and $1-\sqrt{2}$.
This allows us to solve:
$${\begin{pmatrix}
a_{n}\\
b_{n}
\end{pmatrix}} = {\begin{pmatrix}
1 & 2 \\
1 & 1
\end{pmatrix}}^{n-1} \begin{pmatrix}
1\\
1
\end{pmatrix} $$ with: $$a_{n} = \frac{(1+\sqrt{2})^n + (1-\sqrt{2})^n}{2}\\
b_{n} = \frac{(1+\sqrt{2})^n - (1-\sqrt{2})^n}{2\sqrt{2}}$$
A recurrence relationship can also be written for each term separately as:
$$a_{n} = 2a_{n-1} + a_{n-2}\\
b_{n} = 2 b_{n-1} + b_{n-2}$$
with starting terms $a_1 = 1$, $a_2 = 3$, $b_1 = 1$, $b_2 = 2$.
We can write it as a recurrence relationship or a matrix equation if we wish to use only integers, or we can write the terms individually, but that causes the reappearance of $\sqrt{2}$ in the formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to find n for $n = 2^{\frac {n^2}{16}}$ I have this equation $n = 2^{\frac{n^2}{16}}$
I need to find n.
I tried using the Lambert W function, but I don't know how to get one side into the form $W(xe^x)=x$ without having an $n$ on the other side.
I have done:
$n = e^{\frac{n^2ln2}{16}}$
$1 = \frac{1}{n}e^{\frac{n^2ln2}{16}}$
I'm stuck because I am multiplying $\frac{1}{n}$ by $\frac{n^3}{16}ln2$ but that leaves n on the other side.
| $n = 2^{(\frac {n}{4})^2}$
We need to get the right hand side into the form ${f(n)}e^{f(n)} = c$
Then $f(n) = W(c)$
$1 = n^{-1}e^{(\frac {n}{4})^2\ln 2}$
$1^{-1} = ne^{-(\frac {n}{4})^2\ln 2}$
$\frac 14 = \frac {n}{4}e^{-(\frac {n}{4})^2\ln 2}$
$\frac 1{16} = (\frac {n}{4})^2e^{-2(\frac {n}{4})^2\ln 2}$
$\frac {\ln 2}{8} = (2(\frac {n}{4})^2\ln 2)e^{-2(\frac {n}{4})^2\ln 2}$
$-\frac {\ln 2}{8} = (-2(\frac {n}{4})^2\ln 2)e^{-2(\frac {n}{4})^2\ln 2}$
Now we have the right-hand side in the correct form.
$-2(\frac {n}{4})^2\ln 2 = W(-\frac {\ln 2}{8})$
And unwind to find $n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Finding all $x \in \mathbb{R}$ with $|x+1| + |x-1| < 4$ We want to find all $x \in \mathbb{R}$ with
$$|x+1| + |x-1| < 4$$
I tried the following, but I'm uncertain if this is correct:
Critical points: $x= -3$ and $x = 1$.
Case $1$:
$$x \in (-\infty, -3) \\
|x+1| = -(x+1) = -x-1 \\
|x-1| = -(x-1) = -x+1 \\
|x+1| + |x-1| < 4\\
\iff -x-1-x+1 < 4\\
\iff -2x < 4 \\
\iff x < -2 \\ \Rightarrow \text{ holds } \forall x \in (-\infty,-2[$$
Case $2$:
$$x \in [-3, 1) \\
|x+1| + |x-1| < 4\\
\iff -x-1-x+1 < 4\\
\iff -2x < 4 \\
\iff x < -2 \\ \Rightarrow \text{ holds } \forall x \in [-3,-2[$$
Case $3$:
$$x \in (1, \infty) \\
|x+1| = x+1 \\
|x-1| = x-1 \\
x+1+x-1 < 4\\
\iff 2x < 4 \\
\iff x < 2 \\ \Rightarrow \text{ holds } \forall x \in [1,2)$$
The solution would then be $x \in (-\infty,-2) \cup [1,2)$.
Is this correct/wrong? Is this the common approach for questions like these?
| Sol 1)
This image shows the graph of $y=|x+1|+|x-1|$ and $y=4$.
From this graph, we can easily get the answer: $x \in (-2, 2).$
Sol 2)
Case 1. $x < -1.$
$|x-1|+|x+1|=1-x-x-1=-2x < 4.$
$\therefore \;-2 < x < -1.$
Case 2. $-1 \leq x \leq 1.$
$|x-1|+|x+1|=1-x+x+1<4.$
$\therefore \; -1 \leq x \leq 1. $
Case 3. $1 <x. $
$|x-1|+|x+1| = 2x < 4, x < 2.$
$\therefore 1<x<2.$
$\therefore x \in (-2, 2).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Sum of Fibonacci Number and Product of Fibonacci Number are equal? Let $a,b,c,d$ be nonnegative integers. Consider the equation
$$F_a+F_b=F_cF_d$$
WLOG, I can assume that $a \geq b$ and $c \geq d$. I try to solve for all subscripts $a,b,c,d$ in the above equation.
It is definitely solvable. For example, we have $F_d=1$ for $d=1,2$, the equation turns into
$$F_a+F_b=F_c$$
By definition of Fibonacci number, $$(a,b,c)=(n-1,n-2,n)$$ is one of the solutions and other solutions can be found in lemma 2.3 in this link. https://www.researchgate.net/publication/337918351_On_the_sum_of_three_arbitrary_Fibonacci_and_Lucas_numbers.
For $d=3$, the equation becomes
$$F_a+F_b=2F_c$$
Clearly, if $a=b=c=n$, we have another set of the solutions.
However, my hunch tells me there are other solutions for $d \geq 3$, I even try writing program on MATHEMATICA to look out other forms of solution.
Anyone has a nice idea to attack this problem? Actually, I try to do for Lucas numbers, too.
$$L_a+L_b=L_cL_d$$
$$L_a+L_b=F_cF_d$$
$$F_a+F_b=L_cL_d$$
| All solutions to $F_a+F_b=F_cF_d$ with $a \geq b$, $c \geq d$ are given by ($n \geq 1$):
\begin{align}
(a,b,c,d) \in \{(n+5,n+1,n+3,4),(n+3,n,n+2,3),(n+1,n,n+2,2),(n+1,n,n+2,1),(n+2,n+2,n+2,3),(1,1,3,1),(1,1,3,2),(2,2,3,1),(2,2,3,2),(3,1,4,1),(3,1,4,2),(4,2,3,3),(5,1,4,3),(6,1,4,4)\}.
\end{align}
To prove this we use Zeckendorf's theorem: every positive integer can be represented uniquely(!) as the sum of one or more distinct non-consecutive Fibonacci numbers. Together with known formula for Zeckendorf's representation of a product $F_cF_d$, this gives the complete solution.
Proof.
The left hand side $F_a+F_b$ is already in Zeckendorf's form unless $a=b$ or $a=b+1$. We omit these cases for now and assume $a \geq b+2$. The right hand side $F_cF_d$ has well-known Zeckendorf representation for $c \geq d \geq 2$ (see for example here):
$$
F_cF_d=\sum_{r=1}^{\lfloor d/2 \rfloor } F_{c+d+2-4r}
$$
when $d$ is even. When $d$ is odd, we have to add $F_{c-d+1}$ if $c>d$ and $F_2$ if $c=d$. Going case by case, we can infer that by Zeckendorf's theorem the sum must contain exactly two terms and they need to be $F_a$ and $F_b$.
If $d$ is even, then the sum has exactly $d/2=2$ terms, so we must have $d=4$. The terms are then $F_{c+4+2-4}=F_{c+2}$ and $F_{c+4+2-8}=F_{c-2}$ and we conclude $a=c+2$, $b=c-2$ (recall $a \geq b$). This corresponds to
$$
F_{c+2}+F_{c-2}=F_cF_4=3F_c.
$$
Here we must be careful and notice that the identity we used works for $d\geq 2$ only, but since $F_1=F_2$, we get additional solution from above when $c=4$ and replacing $F_2$ with $F_1$. Namely $F_6+F_1=F_4F_4$.
If $d$ is odd and $c>d$, then we have $\lfloor d/2 \rfloor+1=2$ terms, hence $d=1$ or $d=3$. However by assumption $d \geq 2$, hence we consider only $d=3$. The two terms are then $F_{c+d+2-4}=F_{c+1}$ and $F_{c-d+1}=F_{c-2}$. This corresponds to
$$
F_{c+1}+F_{c-2}=F_{c}F_3=2F_c.
$$
As before, for $c=4$ and replacing $F_2$ with $F_1$ we get additional solution $F_5+F_1=F_4F_3$.
If $d$ is odd and $c=d$, then again we have $d=3$. The two terms are now $F_{c+d+2-4}=F_{4}$ and $F_2$. This corresponds to
$$
F_4+F_2=F_3F_3.
$$
This also gives $F_4+F_1=F_3F_3$, but this is already covered by another parametrization so we do not mention this case explicitly in the summary.
If $d=1$, then you have already found all solutions to $F_a+F_b=F_cF_d=F_c$, so we are done there.
So the only remaining cases are then $a=b$ and $a=b+1$.
The case $a=b+1$ is simple as then $F_a+F_b=F_{b+2}$ is its own Zeckendorf representation, so there must be exactly one term in representation $F_cF_d$ and so $d$ is even, $d/2=1$, thus $d=2$. The only term is $F_{c+d-2}=F_{c}=F_{b+2}$ and so $c=b+2$. This retrieves
$$
F_{b+1}+F_b=F_{b+2}F_{2}=F_{b+2}.
$$
So the recurrent relation is the only solution in this case.
If $a=b$, then we want to find all solutions of $2F_a=F_cF_d$. We can view the left side as $F_3F_a$. First for $a > 3$, we have
$$
F_aF_3=F_{a+3+2-4}+F_{a-3+1}=F_{a+1}+F_{a-2}.
$$
So similar as before, looking at representations of $F_cF_d$ with $c \geq d \geq 2$ and have exactly two terms, we find either $d=4$ and $F_cF_d=F_{c+2}+F_{c-2}$ (incompatible $F_{a+1}+F_{a-2}$), $d=3$ and $F_{c+1}+F_{c-2}$ (which corresponds to $a=c$ trivially) or $d=3$ and $F_4+F_2$ (incompatible with $F_{a+1}+F_{a-2}$).
For $a \leq 3$, we can consider just case by case $a=b=1$, $a=b=2$ and $a=b=3$ and we find few cases $(a,b,c,d) \in \{(1,1,3,1),(1,1,3,2),(2,2,3,1),(3,3,3,3)\}$.
$\square$
| {
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"question_score": "2",
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Classify, up to similarity, real matrices of dimension $6\times6$ with minimal polynomial $(t-1)^2(t+1)(t-2)$ Classify, up to similarity, real matrices of dimension $6\times6$ with minimal polynomial $(t-1)^2(t+1)(t-2)$
My attempt:
If the minimal polynomial is $(t-1)^2(t+1)(t-2)$ then we only have the following invariant factors:
$(t-1)^2(t+1)(t-2)=t^4-3t^3+t^2+3t-2, (t-1)^2=t^2-2t+1$
So, every matrix of dimension $6\times6$ with minimal polynomial $(t-1)^2(t+1)(t-2)$ is similar to
$$A=\begin{bmatrix}
0 & 0 & 0 & 2 & 0 & 0 \\
1 & 0 & 0 & -3 & 0 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 \\
0 & 0 & 1 & 3 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 \\
0 & 0 & 0 & 0 & 1 & 2 \\
\end{bmatrix}$$.
Is this reasoning correct?
Thanks in advance.
| It will be simpler to consider $3 \times 3$ block matrices.
In the spirit of the answer of Masacroso here, here is "prototypical case" helping (so I think) to understand what is going on is as follows:
$$M=\begin{bmatrix}
B & 0 & 0\\
0 & B & 0 \\
0 & 0 & D
\end{bmatrix} \ \text{where} \ B:=\begin{bmatrix}
\color{red}{1} & 1\\
0 & \color{red}{1}
\end{bmatrix}, \ D:=\begin{bmatrix}
\color{red}{-1} & 0\\
0 & \color{red}{2}
\end{bmatrix}$$
(eigenvalues have been written in red in order to spot their placement ; please note in particular the typical Jordan block $B$).
Let us now consider decomposition :
$$M=\underbrace{\begin{bmatrix}
B & 0 & 0\\
0 & B & 0 \\
0 & 0 & 0
\end{bmatrix}}_P+\underbrace{\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0 \\
0 & 0 & D
\end{bmatrix}}_Q$$
We have, separately, $(P-I_6)^2=0$ (but not $(P-I_6)^1=0$) and $(Q+I_6)*(Q-2*I_6)=0$.
Therefore:
$$(M-I_6)^2*(M+I_6)*(M-2*I_6)=0$$
giving the expression of the minimal polynomial.
| {
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find the smallest positive integer $n$ so that there exist $n\times n$ matrices $A,B$ with $(AB)^k = 0$ and $(BA)^k\neq 0$
For any positive integer k, find the smallest positive integer $n$ so that there exist $n\times n$ matrices $A,B$ with $(AB)^k = 0$ and $(BA)^k\neq 0$.
I know that by the rank nullity theorem, for any linear map $T : V\to W$ between finite-dimensional vector spaces $V$ and $W, \mathrm{rank}(T) + \mathrm{nullity}(T) = \dim V.$ Also, for any matrices $A,B, \det(AB) = \det(A)\det(B)$ and $\mathrm{rank}(AB) \leq \min\{\mathrm{rank}(A),\mathrm{rank}(B)\}, \mathrm{rank}(A+B) \leq \mathrm{rank}(A) + \mathrm{rank}(B).$ Also, if $M_n(\mathbb{R})$ denotes the ring of matrices of size $n\times n$ with real entries, then $M_n(\mathbb{R})$ is commutative for $n=1$ and noncommutative for $n > 1$ (for instance we have for $n = 2$ that $\begin{pmatrix}1 & 0\\
0 & 0\end{pmatrix}\cdot \begin{pmatrix}0 & 1\\
0 & 0\end{pmatrix} = \begin{pmatrix}0 & 1\\
0 & 0\end{pmatrix}\neq \begin{pmatrix}0 & 0\\
0 & 0\end{pmatrix} = \begin{pmatrix}0 & 1\\
0 & 0\end{pmatrix} \cdot \begin{pmatrix}1 & 0\\
0 & 0\end{pmatrix} $. For the general case of $n > 2,$ we can consider the following two matrices $A = \begin{pmatrix}1 & O_{1\times (n-1)}\\
O_{(n-1)\times 1} & O_{(n-1)\times (n-1)}\end{pmatrix}, B = \begin{pmatrix} O_{1\times (n-1)} & 1\\
O_{(n-1)\times 1} & O_{(n-1)\times (n-1)}\end{pmatrix}$ where $O_{m\times n}$ denotes the $m\times n$ zero matrix ($AB$ has a $1$ in row 1, column n while $BA$ is the zero matrix). There is also the Cayley Hamilton theorem, which states that the characteristic polynomial of any matrix evaluated at that matrix equals the zero matrix. I also know that if $A$ is a matrix such that $A^2$ has rank k = rank A, then $A^n$ has rank k for all $n\ge 1$.
$n=1$ will never work for any $k$ since $M_1(\mathbb{R})$ is commutative. Also neither $A$ nor $B$ can be in the center of $M_n(\mathbb{R})$ for otherwise $AB=BA\Rightarrow (AB)^k = (BA)^k.$ For $k=2,$ we need to find some $n$ so that $(AB)^2 = 0$ while $(BA)^2\neq 0.$ Supposing $n=2$ works we have $A = \begin{pmatrix}a& b\\
c&d \end{pmatrix}, B = \begin{pmatrix}a_2& b_2\\
c_2&d_2 \end{pmatrix}$ for some real numbers $a,b,c,d,a_2,b_2,c_2,d_2.$ It seems fairly tedious to solve the resulting system of equations to determine the values of $A$ and $B$ directly so there's likely a shorter approach. We need $\det(AB) = 0$ for any $k$.
| If $(AB)^k = 0$ then $B(AB)^k A = (BA)^{k+1} = 0$. So $BA$ is nilpotent with nilpotence degree exactly $k+1$. An $n \times n$ matrix is nilpotent iff its characteristic polynomial is $t^n$, meaning its nilpotence degree is at most $n$; this gives that $n \ge k+1$.
Here is a construction showing that we can take $n = k+1$. Take $B$ to be diagonal with entries $\underbrace{1, 1,}_{k \text{ times}} \dots 0$, and take $A$ to be a nilpotent Jordan block of size $k+1$. Then $BA = A$ so $(BA)^{k+1} = 0$ while $(BA)^k \neq 0$, but $AB$ is the nilpotent Jordan block of size $k$ (plus some zeroes) so $(AB)^k = 0$.
| {
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evaluate $\lim\limits_{N\to\infty} \dfrac{\ln^2 N}{N}\sum_{k=2}^{N-2} \dfrac{1}{\ln k \cdot \ln (N-k)}$
Evaluate $\lim\limits_{N\to\infty} \dfrac{\ln^2 N}{N}\sum_{k=2}^{N-2} \dfrac{1}{\ln k \cdot \ln (N-k)}$ where $\ln$ denotes the natural logarithm.
Let $A_N = \dfrac{\ln^2 N}{N}\sum_{k=2}^{N-2} \dfrac{1}{\ln k \cdot \ln (N-k)}.$ Clearly $A_N \ge \dfrac{\ln^2 N}N \cdot \dfrac{N-3}{\ln^2 N} = 1-3/N$ for all $N$ so the main question is whether $A_N$ converges to 1 as $N\to\infty.$ Fix $2\leq M < N/2.$ We want to find an upper bound for $A_N$ in terms of $N$ and $M$ that converges to 1 as $N\to\infty$ when $M$ is chosen carefully enough. Note that by differentiating $f(x)=\dfrac{1}{\ln x\cdot \ln (N-x)},$ one can conclude that it is decreasing on $(1,N/2]$ and increasing on $[N/2,N-2]$ (one can equivalently analyze the behaviour of $\dfrac{1}{f(x)}$ by differentiating to simplify the calculations). Using this observation, we have $\sum_{k=2}^{N-2} \dfrac{1}{\ln k\cdot \ln(N-k)} = (\sum_{k=2}^{M} + \sum_{k=M+1}^{N-M-1} + \sum_{k=N-M}^{N-2} )\dfrac{1}{\ln k\cdot \ln(N-k)}\leq \dfrac{2(M-1)}{\ln 2 \ln (N-2)} +\sum_{k=M+1}^{N-M-1} \dfrac{1}{\ln k\cdot \ln(N-k)}.$
But I'm not sure how to simplify the above sum.
| Assume that $N > 4$. Let
$$A_N = \frac{\ln^2 N}{N}\sum_{k=2}^{N-2} \frac{1}{\ln k \cdot \ln(N - k)}.$$
Let $f(x) := \frac{1}{\ln x \ln (N - x)}$. Note that $f(x)$ is strictly decreasing on $[2, N/2]$,
and strictly increasing on $[N/2, N-2]$. We have
\begin{align*}
\sum_{k=2}^{N-2} \frac{1}{\ln k \cdot \ln(N - k)}
&\le
2f(2) + 2\int_2^{N/2} f(x)\,\mathrm{d} x\\
&\le 2f(2) + 2\int_2^{N/2} \frac{1}{\ln x \ln (N - N/2)}\,\mathrm{d} x\\
&= 2f(2) + \frac{2}{\ln(N/2)}\int_2^{N/2} \frac{1}{\ln x}\, \mathrm{d} x.
\end{align*}
Also, we have
$$\sum_{k=2}^{N-2} \frac{1}{\ln k \cdot \ln(N - k)}
\ge \sum_{k=2}^{N-2} \frac{1}{\ln N \cdot \ln N} = \frac{N-3}{\ln^2 N}.$$
Thus, we have
$$1 - \frac3N \le A_N
\le 2f(2)\frac{\ln^2 N}{N} + \frac{2\ln^2 N}{N\ln(N/2)}\int_2^{N/2} \frac{1}{\ln x}\, \mathrm{d} x.$$
We have
$$\lim_{N\to \infty} 2f(2)\frac{\ln^2 N}{N} = 0$$
and (using L'Hopital rule)
\begin{align*}
\lim_{N\to \infty} \frac{2\ln^2 N}{N\ln(N/2)}\int_2^{N/2} \frac{1}{\ln x}\, \mathrm{d} x
= \lim_{N \to \infty}
\frac{2\int_2^{N/2} \frac{1}{\ln x}\, \mathrm{d} x}{\frac{N}{\ln N}} = 1.
\end{align*}
We are done.
| {
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Solve the equation $(x+1)^4=2(1+x^4)$ Solve the equation $$(x+1)^4=2(1+x^4)$$ The most intuitive approach for me was to use the formula $$(a+1)^4=a^4+4a^3+6a^2+4a+1,$$ so our equation is $$x^4+4x^3+6x^2+4x+1-2-2x^4=0\\-x^4+4x^3+6x^2+4x-1=0$$ $\pm1$ aren't solutions, so this equation does not have whole roots.
Another thing I tried is to factor $$1+x^4\ne(x+1)(x^3-x^2+x-1)=x^4-1,$$ but then I remembered it is for odd $n$. I don't know what else to try.
| Based on the coefficients and the polynomial $ x^4 - 4 \, x^3 - 6 \, x^2 - 4 \, x + 1 = 0$ try a product of the type
$$( x^2 - 2 \, a \, x + 1)(x^2 - 2 \, b \, x + 1) = 0$$
which gives
$$ a + b = 2 \hspace{5mm} \text{and} \hspace{5mm} a \, b = -2.$$
The roots of the original polynomial take the form
$$ x \in \left\{ a + \sqrt{a^2 - 1}, \, a - \sqrt{a^2 - 1}, \, b + \sqrt{b^2 - 1}, \, b - \sqrt{b^2 - 1}\;\right\}.$$
What is left is to solve for $a$ and $b$.
| {
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Finding $x-\frac{1}{x}$, given $x^3 - \frac{1}{x^3} = 108+76\sqrt{2}$
If $x^3 - \dfrac{1}{x^3} = 108+76\sqrt{2}$, find the value of $x-\dfrac{1}{x}$.
Here's what I've tried so far.
$$\begin{align}
\left(x-\dfrac{1}{x}\right)^3&=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right) \\
\rightarrow \quad \left(x-\dfrac{1}{x}\right)^3&=108+76\sqrt{2}-3\left(x-\dfrac{1}{x}\right) \\u:=x-\dfrac{1}{x} \quad\rightarrow \quad u^3+3u-108-76\sqrt{2}&=0
\end{align}$$
Got stuck here since I didn't know how to solve this cubic equation.
I also tried factorizing $x^3-\dfrac{1}{x^3}$.
$$\begin{align}x^3-\dfrac{1}{x^3}&=\left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+1\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+2-1^2\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(\left(x+\dfrac{1}{x}\right)^2-1^2\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}+1\right)\left(x+\dfrac{1}{x}-1\right)
\end{align}$$
Again, I didn't know what I could do with this.
|
$u^3+3u=108+76\sqrt{2}$
A few posts here assumed $u = a + b\sqrt{2}$, then search for rational $(a, b)$
Let "slope", $\displaystyle \;m_1 = b/a$
$\displaystyle u^2 = (a^2+2b^2) \;+\; (2ab)\,\sqrt{2}\qquad\; →m_2 = \frac{2\,m_1}{1+2\,m_1^2}$
Note that $m_2$ has same sign as $\displaystyle m_1 \quad →m_∞ = \frac{sgn(m1)}{\sqrt{2}}$
Slope of RHS = $\displaystyle \frac{76}{108} ≈ 0.7037
\;<\; \frac{1}{\sqrt{2}} ≈ 0.7071$
$u^3 ≈ RHS,\;m_1$ likely convergent of RHS slope.
C:\>spigot -C 76/108
0/1
1/1
2/3
5/7
7/10
19/27
Convergent 2/3 work! Upon checking, $(b,a) = (2,3)$, not just the ratio.
$u^3 + 3u = (99+70\sqrt{2}) + 3×(3+2\sqrt{2}) = 108+76\sqrt{2}$
| {
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Finding anti-derivative of $f(x)= \sin^3 x \cos^2 x $
Finding anti-derivative of $f(x)= \sin^3 x \cos^2 x $
so, integrate $\int \sin^3 x \cos^2 x dx = \int \sin x (1- \cos^2 x) \cos^2 (x) dx $
let $u = \cos x$
$\int -u^2 (1-u^2) du = \int -u^2 dx + \int u^4 dx = \frac{-u^3}{3} + \frac{u^5}{5} + C $
Therefore $ \frac{\cos^3 x}{3}+\frac{\cos^5 x}{5} + C$
Why am I wrong? My teacher said the answer is $ \frac{-\cos x}{8} - \frac{\cos (3x)}{48} + \frac{\cos (5x)}{80} $
| First of all, Both answers $I=-\frac{\cos ^3 x}{3}+\frac{\cos ^5 x}{5}+C= \frac{1}{80} \cos 5 x-\frac{1}{48} \cos 3 x-\frac{1}{8} \cos x+C$ are correct, which can be confirmed by the following proof:
Let $z=\cos x+i\sin x$, then $\sin x= \frac{1}{2i}(z-\frac{1}{z})$ and $\cos x= \frac{1}{2}(z+\frac{1}{z})$.
Expressing $\sin^3x\cos^2x$ in terms of $z$ and then sine of multiples of $x$ gives
$$
\begin{aligned}
\sin ^3 x \cos ^2 x =&-\frac{1}{32 i}\left(z-\frac{1}{z}\right)^3\left(z+\frac{1}{z}\right)^2 \\
=&-\frac{1}{32 i}\left(z^2-\frac{1}{z^2}\right)^2\left(z-\frac{1}{z}\right) \\
=&-\frac{1}{32 i}\left[\left(z^5-\frac{1}{z^5}\right)-\left(z^3-\frac{1}{z^3}\right)-2\left(z-\frac{1}{z}\right)\right] \\
=&-\frac{1}{32 i}(2 i \sin 5 x-2 i \sin 3 x-4 i \sin x) \\
=&-\frac{1}{16} \sin 5 x+\frac{1}{16} \sin 3 x+\frac{1}{8}\sin x
\end{aligned}
$$
Integrating back yields
$$
\boxed{\int \sin ^3 x \cos ^2 x d x=\frac{1}{80} \cos 5 x-\frac{1}{48} \cos 3 x-\frac{1}{8} \cos x+C}
$$
Wish it helps!
| {
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In how many ways can we select $6$ cards from a pack of $52$ cards such that we get all $4$ suits? Number of ways to select $1$ card from each suit $= 13^4=28561$
Total cards remaining $=52-4=48$
Cards remaining in each suit $=13-1=12$
Now I have to choose $2$ more cards and this can happen in two ways. First, either both cards are selected from the same suit or both cards are selected from two different suits.
Number of ways to choose $2$ cards from the same suit $=^{12}C_2+^{12}C_2+^{12}C_2+^{12}C_2 = 66 \times 4 = 264$
Number of ways to choose $2$ cards from two different suit = $=(^{12}C_1 \times ^{12}C_1\times^{12}C_0\times^{12}C_0 ) \times ^{4}C_2 = 864$
So the total number of ways should be $=28561 \times (264 + 864) = 32,216,808$
But it seems that this answer is wrong. Please help me to understand what I am doing wrong.
Thanks in advance !!!
| JMoravitz has provided you with two nice solutions to the problem. To understand where you made your mistakes, we will consider how your method compares with solving by cases.
Solution:
There are two cases:
*
*three cards are drawn from one suit and one card apiece is drawn from each of the other three suits
*two cards each are drawn from two of the suits and and one card apiece is drawn from each of the other two suits
Three cards are drawn from one suit and one card apiece is drawn from each of the other three suits: There are four ways to select the suit from which three cards are drawn and $\binom{13}{3}$ ways to draw three cards from that suit. For each of the other three suits, there are $13$ ways to select a card from that suit. Hence, the number of such cases is
$$\binom{4}{1}\binom{13}{3}\binom{13}{1}^3$$
Two cards each are drawn from two of the suits and and one card apiece is drawn from each of the other two suits: There are $\binom{4}{2}$ ways to select the suits from which two cards are drawn. For each of those two suits, there are $\binom{13}{2}$ ways to select two cards from that suit. For each of the other two suits, there are $13$ ways to select a card from that suit. Hence, the number of such cases is
$$\binom{4}{2}\binom{13}{2}^2\binom{13}{1}^2$$
Total: Since these cases are mutually exclusive and exhaustive, the number of six-card hands which contain at least one card from each suit is
$$\binom{4}{1}\binom{13}{3}\binom{13}{1}^3 + \binom{4}{2}\binom{13}{2}^2\binom{13}{1}^2 = 8,682,544$$
What errors did you make?
As lulu indicated in the comments, you counted hands multiple times.
By designating a particular card as a representative of each suit and then selecting additional cards, you counted each case in which three cards are drawn from one suit and one card each is drawn from each of the other three suits three times, once for each way you could have designated one of the three cards from which three cards are drawn as the representative of that suit.
Let's illustrate this with the following example: Suppose the hand you selected is $5\clubsuit, 8\clubsuit, J\clubsuit, \color{red}{5\diamondsuit}, \color{red}{7\heartsuit}, K\spadesuit$. Your method counts this hand in three ways.
\begin{array}{c c c c c}
\text{clubs} & \text{diamonds} & \text{hearts} & \text{spades} & \text{additional cards}\\ \hline
5\clubsuit & \color{red}{5\diamondsuit} & \color{red}{7\heartsuit} & K\spadesuit & 8\clubsuit, J\clubsuit\\
8\clubsuit & \color{red}{5\diamondsuit} & \color{red}{7\heartsuit} & K\spadesuit & 5\clubsuit, J\clubsuit\\
J\clubsuit & \color{red}{5\diamondsuit} & \color{red}{7\heartsuit} & K\spadesuit & 5\clubsuit, 8\clubsuit\\
\end{array}
By designating a particular card as a representative of each suit and then selecting additional cards, you count each hand in which two cards each are drawn from two suits and one card each is drawn from the other two suits four times, two times each for each way you could designate one of the two cards from the suits from which two cards are drawn as the card of that suit.
Let's illustrate this with the following example: Suppose the hand you selected is $4\clubsuit, \color{red}{5\diamondsuit}, \color{red}{9\diamondsuit}, \color{red}{Q\heartsuit}, 8\spadesuit, 10\spadesuit$. Your method counts this hand in four ways:
\begin{array}{c c c c c}
\text{clubs} & \text{diamonds} & \text{hearts} & \text{spades} & \text{additional cards}\\ \hline
4\clubsuit & \color{red}{5\diamondsuit} & \color{red}{Q\heartsuit} & 8\spadesuit & \color{red}{9\diamondsuit}, 10\spadesuit\\
4\clubsuit & \color{red}{5\diamondsuit} & \color{red}{Q\heartsuit} & 10\spadesuit & \color{red}{9\diamondsuit}, 8\spadesuit\\
4\clubsuit & \color{red}{9\diamondsuit} & \color{red}{Q\heartsuit} & 8\spadesuit & \color{red}{5\diamondsuit}, 10\spadesuit\\
4\clubsuit & \color{red}{9\diamondsuit} & \color{red}{Q\heartsuit} & 10\spadesuit & \color{red}{5\diamondsuit}, 8\spadesuit\\
\end{array}
Note that
$$\color{red}{3}\binom{4}{1}\binom{13}{3}\binom{13}{1}^3 + \color{red}{4}\binom{4}{2}\binom{13}{2}^2\binom{13}{1}^2 = \color{red}{32,216,808}$$
| {
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"url": "https://math.stackexchange.com/questions/4557246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying $\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)}$ using Lagrange’s polynomial I have a question of symplifying this expression
$$
A=\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)} \tag{1}
$$
where $a$, $b$ and $c$ are distinct nonzero real numbers. There is a quite similar problem, which is simplifying
$$
B=\frac{a^2}{(a-b)(a-c)} + \frac{b^2}{(b-a)(b-c)} + \frac{c^2}{(c-a)(c-b)}. \tag{2}
$$
In this problem, they use Lagrange’s interpolation polynomial for $P(x) =x^2$ at nodes $a$, $b$ and $c$. Then, they compare the coefficient of $x^2$ to obtain that $B=1$.
My effort is to use the similar method with $P(x)=x^4$, but I am struggling with the fact that there are only three nodes here. So the polynomial I need to choose must be a second degree polynomial but it is hard for me to find such one.
Can someone have some idea about using different polynomial to use the Lagrange’s interpolation polynomial? Thank you very much!
| You want to write down a quadratic polynomial $p$ for which $p(a)=a^4$, $p(b)=b^4$, and $p(c)=c^4$.
We see that $p(x)=x^4$ whenever $x\in\{a,b,c\}$. This means that
$$(x-a)(x-b)(x-c)\mid (x^4-p(x)).$$
In particular, we can reduce $x^4$ modulo $(x-a)(x-b)(x-c)$, and we'll have the polynomial we need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find slope of the tangent line of $4\sqrt x + 2e^\frac {3x-12}{x+2}$ at $ x_0$ Find the slope and the equation of the tangent line to the
graph $y = f(x)$ at $x_0=4$, $$4\sqrt x + 2e^\frac {3x-12}{x+2} $$
$$\lim_{h\to 0}\tfrac{4\sqrt {4+h} + 2e^\frac {12+3h-12}{4+h+2} - 10}{h} = \lim_{h\to 0}\frac{4\sqrt {4+h} + 2e^\frac {3h}{6+h} - 10}{h} $$ I am stuck at this part
| $$\lim_{h\to 0}\frac{4\sqrt{4+h}-8}{h}+\lim_{h\to 0}\frac{2e^{\tfrac{3h}{6+h}}-2}{h}$$$$=4\lim_{h\to 0}\frac{\sqrt{4+h}-\sqrt4}{h}+2\lim_{h\to0}\frac{e^{\tfrac{3h}{6+h}}-1}{h}$$Rationalising the first limit and adjusting the denominator of the second limit, $$ =4\lim_{h\to 0}\frac{h}{h\cdot(\sqrt{4+h}+\sqrt4)}+2\times 3\lim_{h\to0}\frac{e^{\tfrac{3h}{6+h}}-1}{\frac{3h}{6+h}\cdot (6+h)}$$ Using direct substitution of $h=0$ on the first limit and the formula $\lim \limits_{U\to 0}\dfrac{e^U-1}{U}=1$ on the second limit,$$=4\cdot\frac{1}{2+2}+\frac{2\times3}{6\times 1}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve for $n: \cos\left(\frac\pi4(n^2+2n)\right)=\sin\left(\frac\pi4(n^2+n+1)\right), n\in \mathbb Z$
Solve for $n: \cos\left(\frac\pi4(n^2+2n)\right)=\sin\left(\frac\pi4(n^2+n+1)\right), n\in \mathbb Z$
My Attempt:
$$\cos\left(\frac\pi4(n^2+2n)\right)=\cos\left(\frac\pi2-\frac\pi4(n^2+n+1)\right)\\\implies\frac\pi4(n^2+2n)=2p\pi\pm\left(\frac\pi2-\frac\pi4(n^2+n+1)\right), p\in \mathbb Z$$
With minus sign,
$$\frac\pi4(n^2+2n)=2p\pi-\left(\frac\pi2-\frac\pi4(n^2+n+1)\right)\\\implies n^2+2n=8p-(2-n^2-n-1)\\\implies n=8p-1$$
With plus sign,
$$\frac\pi4(n^2+2n)=2p\pi+\left(\frac\pi2-\frac\pi4(n^2+n+1)\right)\\\implies n^2+2n=8p+(2-n^2-n-1)\\\implies 2n^2+3n=8p+1$$
How to conclude from this?
The answer given is $8p-1$ or $8p-3$.
| It follows from $2n^2+3n=8p+1$ that $8p=2n(n+1)+(n-1)$. Since $2\mid n(n+1)$, we have $4\mid 2n(n+1)$, hence $4\mid(n-1)=8p-2n(n+1)$. Let's assume that $n=4k+1$ for some integer $k$, then we have
$$8p=4(2k+1)(4k+1)+4k\\\implies 2p=(2k+1)(4k+1)+k.$$
Since $2p$ is even and $(2k+1)(4k+1)$ is odd, we know that $k=2p-(2k+1)(4k+1)$ is odd, hence $k=2m-1$ for some integer $m$, and thus $n=4k+1=8m-3$.
On the other hand, it is a direct calculation to check that all integers of the form $n=8m-3$ for some integer $m$ satisfies
$$8\mid(2n^2+3n-1).$$
| {
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Use the substitution $u=\frac{1}{\sqrt{3}}\tan x$ to find $\int\frac{dx}{3-2\sin^2x}$ Question:
Use the substitution $u=\frac{1}{\sqrt3}\tan x$ to find $$\int\frac{dx}{3-2\sin^2x}$$
My Working:
Let $u=\frac{1}{\sqrt3}\tan x$, then
\begin{align}
\frac{du}{dx}&=\frac{1}{\sqrt3}\cdot\sec^2x+0=\frac{\sec^2x}{\sqrt3}\\
du&=\frac{\sec^2x}{\sqrt3}\cdot dx
\end{align}
Unfortunately, after that, I do not know how to proceed. Could anyone please help? Thank you!
| Hint: $\sec^{2}x=1+\tan^{2}x$ and
$$3-2 \sin^{2}x=3-2 (1-\cos^{2}x)=3-2 \left(1-\frac{1}{\sec^{2}x}\right)=3-2 \left(1-\frac{1}{1+\tan^{2}x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569523",
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"source": "stackexchange",
"question_score": "1",
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Is there any better way of finding the required value If $$z=\cos\theta+i\sin\theta$$ find the value of $$\frac{1+z}{1-z}$$
The solution that I have is this
$$z=\cos\theta+i\sin\theta \implies$$
$$\frac{1+z}{1-z}=\frac{1+(\cos\theta+i\sin\theta)}{1-(\cos\theta+i\sin\theta)}=\frac{(1+\cos\theta)+i\sin\theta}{(1-\cos\theta)+i\sin\theta}$$
$$=\frac{2\cos^2\frac{\theta}{2}+i\:\:2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}-i\:\:2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}\cdot \frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}$$
$$=\cot\frac{\theta}{2}\cdot\frac{\color{red}{i}}{\color{red}{i}}×\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}$$
$$=\color{red}{i}\cot\frac{\theta}{2}\cdot\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\color{red}{i}\sin\frac{\theta}{2}-\color{red}{i}i\cos\frac{\theta}{2}}$$
$$=\color{red}{i}\cot\frac{\theta}{2}\cdot\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}+\color{red}{i}\sin\frac{\theta}{2}}$$
$$=\color{red}{i}\cot\frac{\theta}{2}$$
Now how on earth one will imagine the steps written in red. I am looking for an easy and a logical answer to this question. The solution that I have is impractical as you all can see. I tried it doing by $e^{i\theta}$ but no good happen.
Any help is greatly appreciated.
| Rather than have to guess to use a special trick like multiplying by $\frac{i}{i}$, just multiply top and bottom by the conjugate of the denominator. This can be done in trig form, or exponential form, or just like this:
$$\frac{1+z}{1-z}\times\frac{1-\bar{z}}{1-\bar{z}}$$
$$=\frac{1+z-\bar{z}-z\bar{z}}{1-z-\bar{z}+z\bar{z}}$$
$$=\frac{1+2i\sin\theta-1}{1-2\cos\theta+1}$$
Now use the half-angles:
$$=\frac{4i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{4\sin^2\frac{\theta}{2}}$$
$$=i\cot\frac{\theta}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof that $\frac{V_n(k_1,...,k_n)}{V_n(1,...,n)}$ is a integer. Let $k_1 < k_2 < ... < k_n$ will be integers. Prove that the quotient
$\frac{V_n(k_1,...,k_n)}{V_n(1,...,n)}$ is a integer.
where $V_n(x_1,...,x_n)$ is vandermonde determinant
My idea: Show that the numerator is divisible by the denominator
Can anyone give a hint or help solve it? Thank u.
| Without loss of generality, we can assume that $k_1\geq n-1$ since $V_n(k_1,\ldots,k_n) = V_n(k_1+p,\ldots,k_n+p)$ for any $p\in\mathbb{Z}$.
Since
$$V_n(1,\ldots,n) = 1!2!\cdots (n-1)!.$$
We have that
$$\frac{V_n(k_1,\ldots,k_n)}{V_n(1,\ldots,n)}
= \det\begin{pmatrix}
1 & \cdots & 1 \\
\frac{k_1}{1!} & \cdots & \frac{k_n}{1!} \\
\vdots & & \vdots \\
\frac{k_1^{n-1}}{(n-1)!} & \cdots & \frac{k_n^{n-1}}{(n-1)!}
\end{pmatrix}.$$
Note that $\binom{k_i}{n-1} = \frac{k_i(k_i-1)\cdots (k_i-n+2)}{(n-1)!}$
can be regarded as a polynomial of $k_i$ with the highest order term $\frac{k_i^{n-1}}{(n-1)!}$.
So we can add and subtract the proper multiples of the first $n-1$ rows
to the last row to transform it into
$$\binom{k_1}{n-1}\cdots \binom{k_n}{n-1}.$$
Similarly, we can transform the $(n-1)$-th row into
$$
\binom{k_1}{n-2}\cdots\binom{k_n}{n-2},
$$
and $(n-2)$-th row ... Finally, the quotient equals to
$$
\det\begin{pmatrix}
\binom{k_1}{1} & \cdots & \binom{k_n}{1} \\
\binom{k_1}{2} & \cdots & \binom{k_n}{2} \\
\vdots & & \vdots \\
\binom{k_1}{n-1} & \cdots & \binom{k_n}{n-1} \\
\end{pmatrix},
$$
which is an integer since each entry in the matrix is integer.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I show $\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} = \frac{1}{4}$ with the definition of $\lim\limits_{x \to \infty}f(x)$? How do I show $\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} = \frac{1}{4}$ with the definition of $\lim\limits_{x \to \infty}f(x)$?
What I have done so far:
$\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} =\lim\limits_{x\to\infty}{\frac{x^{3}(1+\tfrac{1}{x^2})}{x^3(2\tfrac{1}{x^3}+4)}}=\frac{1+0}{0+4}=\frac{1}{4}$
At this point I wanted to use the definition of $\lim\limits_{x \to \infty}f(x)$
Given: $f : D \subset \mathbb{R} \rightarrow \mathbb{R}$
$\exists r \in \mathbb{R}$ with $(r,\infty) \subset D$.
for $\eta \in \mathbb{R}$ define:
$\lim\limits_{x\to\infty}{f(x)}=\eta : \Leftrightarrow \forall \epsilon >0\ \exists \delta > r \forall x > \delta : f(x) \in B_{\epsilon}(\eta) $\
to show: $\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} = \frac{1}{4}$
let $\epsilon >0$
$\exists \delta>6 \ \forall x > \delta: \left | \frac{x^{3}+x}{2+4x^3}-\frac{1}{4} \right |<\epsilon$
The plan was to go on and find $\delta$ depending on $\epsilon$ but I didn't get it.
| First note that $$\frac{x^3+x}{2+4x^3}-\frac{1}{4} = \frac{4(x^3+x)-(2+4x^3)}{4(2+4x^3)} = \frac{4x-2}{16x^3+8} = \frac{\frac{1}{4x^2}-\frac{1}{8x^3}}{x^3+\frac{1}{2}}$$
and so $$\left|\frac{x^3+x}{2+4x^3}-\frac{1}{4}\right| \leq \frac{\left|\frac{1}{4x^2}\right|+\left|\frac{1}{8x^3}\right|}{\left|x^3+\frac{1}{2}\right|} = \frac{\frac{1}{4x^2}+\frac{1}{8x^3}}{x^3+\frac{1}{2}}.\tag{1}\label{eq1}$$
Now let $\epsilon>0$. Using that $$\lim_{x\to\infty}\frac{1}{x^n} = 0$$ for any integer $n$, we know that there exists $r_1$ such that $x> r_1$ implies that $$\frac{1}{4x^2}<\frac{1}{x^2}<\epsilon.$$
Similarily you can find an $r_2$ such that $x>r_2$ implies that $$\frac{1}{8x^3}<\frac{1}{x^3}<\epsilon.$$
Note that we can pick $r_1$ and $r_2$ so that both are greater than or equal to one. This is handy, because then e.g. if $x>r_1$ one has that the denominator $x^3+1/2$ is also greater than $1$.
Now let $r=\max\{r_1,r_2\}$. What can you conclude about the right hand side of \eqref{eq1} if $x>r$?
| {
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"timestamp": "2023-03-29T00:00:00",
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Verify that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})\le\frac{1}{8}$ in a general triangle $\triangle ABC$ So, this problem is inspired by a contest preparation problem I saw back in Japan, and it is as follows:
In a general triangle $\triangle ABC$, show that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})\le\frac{1}{8}$
Now, while I still haven't figured out a geometric interpretation of this inequality, here is my attempt to prove this:
Recall that:
$$\sin^2(\frac{A}{2})=\frac{(1-\cos(A))}{2}$$
$$\sin^2(\frac{A}{2})=\frac{1}{2}(1-\frac{b^2+c^2-a^2}{2bc})$$
$$\sin^2(\frac{A}{2})=\frac{1}{2}(\frac{a^2-b^2-c^2+2bc}{2bc})$$
$$\sin^2(\frac{A}{2})=\frac{a^2-(b-c)^2}{4bc}$$
Now, obviously $\frac{a^2-(b-c)^2}{4bc} \le \frac{a^2}{4bc}$, therefore:
$$\sin^2(\frac{A}{2}) \le \frac{a^2}{4bc}$$
$$\sin(\frac{A}{2}) \le \frac{a}{2\sqrt{bc}}$$
This can be done for $\sin(\frac{A}{2}), \sin(\frac{B}{2})$ and $\sin(\frac{C}{2})$.
$$\sin(\frac{A}{2}) \le \frac{a}{2\sqrt{bc}}$$
$$\sin(\frac{B}{2}) \le \frac{b}{2\sqrt{ac}}$$
$$\sin(\frac{C}{2}) \le \frac{c}{2\sqrt{ab}}$$
Therefore:
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le (\frac{a}{2\sqrt{bc}})(\frac{b}{2\sqrt{ac}})(\frac{c}{2\sqrt{ab}})$$
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le \frac{abc}{8abc}$$
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le \frac{1}{8}$$
However I'm not sure if this is correct or if it is, I don't think this brute force approach is good. Are there any better options to prove this inequality? Please share your answers!
| To add some more insight, there is indeed a geometric intepretation of this inequality.
As shown in the picture, $AI$ is an angle bisector. $$\sin{A\over 2}={BH\over AB}\leq {BI\over AB} = {CI\over AC} = {BC\over AB+AC}$$
Therefore, with similar argument for $B$ and $C$ angles,
$$\sin{A\over 2}\sin{B\over 2}\sin{C\over 2}\leq{AB\cdot BC\cdot CA \over (AB+AC)(BA+BC)(CA+CB)}\leq {AB\cdot BC\cdot CA \over (2\sqrt{AB\cdot AC})(2\sqrt{BA\cdot BC})(2\sqrt{CA\cdot CB})}={1\over 8}$$
| {
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Calculating $\sum_{k = 1}^{\infty} k^{-8}$ using Parseval´s identity I calculated the Fourier series of $x^2$ getting
$$x^2=\frac{\pi^2}{3}+4\sum_{k=1}^\infty \frac{(-1)^k}{k^2} \cos(kx)$$
Then, integrating this equation two times i got
$$\frac{x^4-2\pi^2x^2}{12}=4\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^4} \cos(kx)$$
Using the Parseval's identity $\sum c_k^2=\int f^2$ i obtained that
$$\sum_{k=1}^\infty \frac{1}{k^8}=\frac{107}{362880}\pi^8 $$
which, according to Wolfram Alpha should be $\sum_{k=1}^\infty \frac{1} {k^8}=\frac{\pi^8} {9450}$
I can't find where my error is. Is my proceeding wrong or am I just missing something?
| Starting with
$$ \frac{\theta^2}{4} - \frac{\pi^2}{12} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \, \cos(n \theta) $$
then, by integration, the series
$$ \frac{\theta^3}{12} - \frac{\zeta(2) \, \theta}{2} + c_{1} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} \, \sin(n \theta) $$
is obtained. The constant of integration, $c_{1}$ is found to be zero by evaluating the series at $\theta = 0$ or $\theta = \pi$. Now integrating once again the resulting series is
$$ \frac{\theta^4}{48} - \frac{\zeta(2) \, \theta^2}{4} + c_{2} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^4} \, \cos(n \theta). $$
Evaluating this series at $\theta = 0$, or $\theta = \pi$, the constant $c_{2}$ can be found and leads to the series
$$ \frac{\theta^4}{48} - \frac{\zeta(2) \, \theta^2}{4} + \frac{7 \, \zeta(4)}{8} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^4} \, \cos(n \theta). $$
Now, by Pasreval's theorem for Fourier Series, it can be seen that
\begin{align}
\sum_{n=1}^{\infty} \left( \frac{(-1)^{n-1}}{n^4} \right)^2 &= \frac{1}{\pi} \, \int_{-\pi}^{\pi} \left( \frac{\theta^4}{48} - \frac{\zeta(2) \, \theta^2}{4} + \frac{7 \, \zeta(4)}{8} \right)^2 \, d\theta \\
\sum_{n=1}^{\infty} \frac{1}{n^8} &= \frac{2}{\pi} \, \int_{0}^{\pi} \left( \frac{\theta^4}{48} - \frac{\zeta(2) \, \theta^2}{4} + \frac{7 \, \zeta(4)}{8} \right)^2 \, d\theta \\
\zeta(8) &= \frac{2}{\pi} \, \left[ \frac{\theta^9}{20736} - \frac{\zeta(2) \, \theta^7}{672} + \frac{37 \, \zeta^{2}(2) \, \theta^5}{2400} - \frac{7 \, \zeta^{3}(2) \, \theta^3}{120} + \frac{49 \, \zeta^{4}(2) \, \theta}{400} \right]_{0}^{\pi} \\
&= \frac{\pi^8}{9450}.
\end{align}
Comparing this result to that from MathWorld Zeta Function then it is determined that
$$ \zeta(8) = \frac{\pi^8}{9450} $$
is the correct result.
| {
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How to solve $\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n$ Question:
$$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=?$$
My work:
$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=\lim_{n\to\infty}\left(\dfrac{n^2(1+5/n+3/n^2}{n^2(1+1/n+2/n^2)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)^n$
$\log L=n\log\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)$
Is this equal to 0? Then the answer would be $e^0=1$.
The answer was given as $e^4$ and I have no idea how to get to that.
| By using $$ \lim_{n \to \infty} \, \left(1 + \frac{x}{n}\right)^n = e^{x}, $$ $n^2 + 5 n + 3 = (n+a_{1})(n+a_{2}),$ and $n^2 + n + 2 = (n+b_{1})(n+b_{2})$ then
\begin{align}
L &= \lim_{n \to \infty} \, \left( \frac{n^2 + 5 n + 3}{n^2 + n + 2} \right)^n \\
&= \lim_{n \to \infty} \, \left( \frac{(n+a_{1})(n+a_{2})}{(n+b_{1})(n+b_{2})} \right)^n \\
&= \lim_{n \to \infty} \, \left( \frac{\left(1+\frac{a_{1}}{n}\right)\left(1+\frac{a_{2}}{n}\right)}{\left(1+\frac{b_{1}}{n}\right)\left(1+\frac{b_{2}}{n}\right)} \right)^n \\
&= \frac{ e^{a_{1}} \, e^{a_{2}} }{ e^{b_{1}} \, e^{b_{2}} } = e^{a_{1}+a_{2}-b_{1}-b_{2}}.
\end{align}
Now,
$$ a_{1} = \frac{5+\sqrt{13}}{2}, \, a_{2} = \frac{5 -\sqrt{13}}{2}, \, b_{1} = \frac{1+i \, \sqrt{7}}{2}, \, b_{2} = \frac{1 - i \, \sqrt{7}}{2} $$
which gives $a_{1} + a_{2} = 5$, $b_{1} + b_{2} = 1$, and $L = e^{4}$, or
$$ \lim_{n \to \infty} \left(\frac{n^2 + 5 n + 3}{n^2 + n + 2}\right)^n = e^{4}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Find the volume of the part of the cylinder $\frac{x^2}{a^2}+\frac{z^2}{c^2}=1$ which lies between the plane $y=0 ,y=mx$ Find the volume of the part of the cylinder $\frac{x^2}{a^2}+\frac{z^2}{c^2}=1$ which lies between the plane $y=0, y=mx$
The limits of integration which I set went as follows:
$${\int \int}_{D} \int_{y=0}^{mx}\mathrm dy\mathrm dx\mathrm dz$$ where $$D=\big \lbrace \: (x,z) \mid \frac{x^2}{a^2}+\frac{z^2}{c^2}=1 \: \big \rbrace$$
Now let $x=au,z=bv$, then $D$ transforms into $\big \lbrace \: (u,v) \mid u^2 + v^2=1 \: \big \rbrace$
Then the integral becomes $$\int_{u=-1}^1 \int_{v=-\sqrt{1-u^2}}^{v=\sqrt{1-u^2}} ma^2cu\mathrm du\mathrm dv$$
| From $ \dfrac{x^2}{a^2} + \dfrac{z^2}{c^2} = 1 $, take
$x = a u \cos t$
$z = c u \sin t$
$y = y$
where $ 0 \le t \le 2 \pi $ and $ 0 \le u \le 1 $
The Jacobian of the above transformation is
$ J = \dfrac{\partial(x, y, z) }{\partial(u, t, y)} = \big| \begin{vmatrix} a \cos t && - a u \sin t && 0 \\ 0 && 0 && 1 \\ c \sin t && c u \cos t && 0 \end{vmatrix} \big| = a c u$
Now the volume is given by the following integral
$\begin{equation} \begin{split}
V &= \displaystyle \int_{0}^{2\pi} \int_0^1 \int_{y=0}^{a m u |\cos t|} (a c u) \ dy \ du \ dt =\int_{0}^{2\pi} \int_0^1 (a^2 m c u^2) |\cos t | \ du \ dt \\ &= \dfrac{1}{3} a^2 c m \int_{0}^{2 \pi} | \cos t | dt = \dfrac{4}{3} a^2 c m \end{split} \end{equation}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Fair Value of a Dice Game 1 die. Up to 3 rolls. Your winnings are equal to the value of a single roll. You can stop after roll 1 or roll 2. If you proceed to the next roll (i.e the 2nd or 3rd), you forfeit the previous value (no memory). Therefore, if you roll 3 times, the value of the 3rd roll is what you win.
What is the fair price of this game?
Strategy
Since the expected value of a single roll is 3.5, I will stop on any roll if I get a 4, 5, or 6.
If Fair Price = Expected Value
And Expected Value = P(i_th roll) x E(i_th roll) i=1 to 3
Then Fair Value = P(1 roll) E(1st roll) + P(2 rolls) E(2nd roll) + P(3 rolls) * E(3rd roll)
E(i_th roll) = 3.5 for any roll
P(only 1 roll) = 1/2 :: must roll 4|5|6
P(only 2 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 4|5|6
P(3 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 1|2|3
$\frac{1}{2} * 3.5 + \frac{1}{4} * 3.5 + \frac{1}{4} * 3.5 = $
$\frac{1}{2}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} = $
$\frac{7}{4} + \frac{7}{8} + \frac{7}{8} = $
$\frac{14}{8} + \frac{7}{8} + \frac{7}{8} = $
$\frac{28}{8} = 3.50$
Whn trying to solve this before, I think I made a mistake and end up with something like $4\frac{5}{8} = 4.625$
Should 3.5 be the answer, or should the 1st and 2nd rolls be "wieghted" by what values you must roll to move on (1,2 or 3) instead of the general 3.5 expected value for a single roll?
| If you only have one throw left, then the expected amount you will have after the throw, if you do use it, is $\ \frac{7}{2}\ $. Therefore, if your previous throw was $3$ or less, you should use your last throw, but if it was $4$ or more you should not use your last throw, but accept the value of the previous one.
Thus, when you have two throws left, your expected winnings, if you decide to take the next throw, will be
$$
\frac{1}{2}\times\frac{7}{2}+\frac{4+5+6}{6}=\frac{17}{4}\ .
$$
Thus, if your first throw was four or less you should throw again, but if it was $5$ or $6$ you should not throw again, but accept the value your first throw. Your expected winnings, and the fair price of the game, is therefore
$$
\frac{2}{3}\times\frac{17}{4}+\frac{5+6}{6}=\frac{14}{3}\ .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Will the perpendicular bisector between the line connecting two cubic roots of the same arc never intersect its turning point? The quadratic graph: $$ f(x) = (x+2)(x+1)$$
would have a midpoint between its roots at $x = -1.5$. This line would intersect its turning point.
However the cubic graph: $$ f(x) = (x+1)(x-2)(x+3)$$
would have a midpoint between the roots -1 and -3 at $x = -2$ , which does not intersect the turning point as shown below.
Is there any particular proof or intuition for this?
| We can take as a requirement that the perpendicular bisector of two adjacent zeroes of a polynomial pass through the "turning point" [relative extremum] between them and see what polynomials may satisfy it. For a quadratic polynomial $ \ x^2 + a_1 x + a_0 \ $ (it suffices to use a monic polynomial, since a "leading coefficient" $ \ a_2 \ $ will only affect the "vertical scale" of the function curve), we will place the zeroes at $ \ x \ = \ c \ \ , \ $ $ d \ = \ c + 2 \Delta \ \ , \ $ with the turning point at $ \ x \ = \ c + \Delta \ = \ d - \Delta \ \ . \ $ The value of the quadratic function at the turning point is then
$$ (c + \Delta)^2 \ + \ a_1·(c + \Delta) \ + \ a_0 \ \ = \ \ (d - \Delta)^2 \ + \ a_1·(d - \Delta) \ + \ a_0 $$
$$ \Rightarrow \ \ c^2 \ + \ 2c \Delta \ + \ \Delta^2 \ + \ a_1 c \ + \ a_1 \Delta \ + \ a_0 \ \ = \ \ d^2 \ - \ 2d \Delta \ + \ \Delta^2 \ + \ a_1 d \ - \ a_1 \Delta \ + \ a_0 $$
$$ \Rightarrow \ \ c^2 \ + \ 2c \Delta \ + \ a_1 c \ + \ a_1\Delta \ + \ a_0 \ \ = \ \ d^2 \ - \ 2d \Delta \ + \ a_1d \ - \ a_1\Delta \ + \ a_0 \ \ , $$
or, since $ \ c^2 + a_1c + a_0 \ = \ d^2 + a_1d + a_0 \ = \ 0 \ \ , \ $
$$ 2c \Delta \ + \ a_1\Delta \ \ = \ \ - 2d \Delta \ - \ a_1\Delta \ \ . $$
As we are taking $ \ \Delta \ \neq \ 0 \ \ , \ $ we obtain $ \ a_1 \ = \ -(c + d) \ \ , \ $ with no dependence on the separation between the adjacent zeroes. The location of the turning point is given by setting the first derivative equal to zero, thus, $ \ 2 x + a_1 \ = \ 0 \ \Rightarrow \ x \ = \ -\frac12·a_1 \ = \ \frac{c \ + \ d}{2} \ \ , \ $ which is just where we set it to be by the bisector requirement. (Unsurprisingly, a little manipulation would show this to be the $ \ x-$coordinate of the "vertex" of the quadratic parabola.)
If we make a similar calculation for a general monic cubic polynomial $ \ x^3 + a_2 x^2 + a_1 x + a_0 \ \ , \ $ the bisector requirement produces
$$ (c + \Delta)^3 \ + \ a_2·(c + \Delta)^2 \ + \ a_1·(c + \Delta) \ + \ a_0 $$ $$ = \ \ (d - \Delta)^3 \ + \ a_2·(d - \Delta)^2 \ + \ a_1·(d - \Delta) \ + \ a_0 $$
$$ \Rightarrow \ \ 2 a_2 (d + c) \ + \ 2 a_1 \ + \ 3 (c^2 + d^2) \ - \ 3 (d - c) \Delta \ + \ 2 \Delta^2 \ \ = \ \ 0 \ \ , \ \ \Delta \ \neq \ 0 \ \ , $$
indicating that there is not only a dependence upon the coefficients of the polynomial, but also upon the separation of adjacent zeroes. The turning point(s) of the curve for the cubic polynomial (when they exist) are found from
$$ 3 x^2 \ + \ 2 a_2 x \ + \ a_1 \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ \ = \ \ \frac13 · \left(-a_2 \ \pm \ \sqrt{a_2^2 \ - \ 3 a_1} \right) \ \ , \ $$
which do not have this relation to the separation of its zeroes. So it would be only for very contrived conditions at best that the perpendicular bisector between adjacent zeroes could pass through a turning point of the cubic polynomial curve.
This dependence on separation of zeroes will persist for all higher-degree polynomials as well, except in particular circumstances. For instance, the quartic polynomial $ \ x^4 \ - \ (c^2 + d^2) x^2 \ + \ c^2 d^2 \ \ \ $ $ = \ (x^2 - c^2)·(x^2 - d^2) \ \ , \ \ c^2 \ < \ d^2 \ \ $ has its zeroes at $ \ x \ = \ \pm c \ \ , \ \ \pm d \ \ $ and its turning points at
$$ 4x^3 \ - \ 2·(c^2 + d^2) x \ \ = \ \ 2x · [ \ 2x^2 \ - \ (c^2 + d^2) \ ] \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ \ = \ \ 0 \ \ , \ \ \pm \sqrt{\frac{c^2 \ + \ d^2}{2}} \ \ . \ $$
The turning point at $ \ x \ = \ 0 \ $ is certainly half-way between $ \ x \ = \ -c \ $ and $ \ x \ = \ +c \ \ , \ \ $ but the bisectors between adjacent zeroes, $ \ \pm \frac{c \ + \ d}{2} \ , \ $ are closer to the $ \ y-$axis than the turning points $ \ \pm \sqrt{\frac{c^2 \ + \ d^2}{2}} \ \ $ by the "root-mean-square-arithmetic-mean inequality" (for which equality only holds when $ \ c \ = \ d \ \ ) \ . \ $ So it is not even sufficient for the polynomial to be an even function, much less of even degree, to have bisectors between successive zeroes pass through its turning points; only quadratic polynomials have a simple enough symmetry.
| {
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"url": "https://math.stackexchange.com/questions/4590347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln(1-x)$? For $|x|<1,$ we have
$$
\begin{aligned}
& \frac{1}{1-x}=\sum_{k=0}^{\infty} x^k \quad \Rightarrow \quad \ln (1-x)=-\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1}
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x)}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k dx \\
& =-\sum_{k=0}^{\infty} \frac{1}{(k+1)^2} \\
& =- \zeta(2) \\
& =-\frac{\pi^2}{6}
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln xdx \\
& =\sum_{k=0}^{\infty} \frac{1}{k+1}\cdot\frac{1}{(k+1)^2} \\
& =\zeta(3) \\
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^2 x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^2 xdx \\
& =-\sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{2}{(k+1)^3} \\
& =-2 \zeta(4) \\
& =-\frac{\pi^4}{45}
\end{aligned}
$$
In a similar way, I dare guess that
$$\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x =(-1)^{n+1}\Gamma(n)\zeta(n+2),$$
where $n$ is a non-negative real number.
Proof:
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^n xdx \\
\end{aligned}
$$
Letting $y=-(k+1)\ln x $ transforms the last integral into a Gamma function as
$$
\begin{aligned}
\int_0^1 x^k \ln ^n x d x & =\int_{\infty}^0 e^{-\frac{k}{k+1}}\left(-\frac{y}{k+1}\right)^n\left(-\frac{1}{k+1} e^{-\frac{y}{k+1}} d y\right) \\
& =\frac{(-1)^n}{(k+1)^{n+1}} \int_0^{\infty} e^{-y} y^n d y \\
& =\frac{(-1)^n \Gamma(n+1)}{(k+1)^{n+1}}
\end{aligned}
$$
Now we can conclude that
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =(-1)^{n+1} \Gamma(n+1) \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+2}} \\
& =(-1)^{n+1} \Gamma(n+1)\zeta(n+2)
\end{aligned}
$$
Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln (1-x)$?
Your comments and alternative methods are highly appreciated?
| Starting with the integral
$$ \int_{0}^{1} (1-x)^m \, x^{-t} \, dx = B(m+1, 1-t), $$
a case of the Beta function, then
\begin{align}
\partial_{m} \int_{0}^{1} (1-x)^m \, x^{-t} \, dx &= \partial_{m} \, B(m+1, 1-t) \\
\int_{0}^{1} \ln(1-x) \, (1-x)^m \, x^{-t} \, dx &= B(m+1, 1-t) \, \left(\psi(m+1) - \psi(m-t+2) \right),
\end{align}
where $\psi(x)$ is the digamma function. Setting $m=0$ leads to
$$ \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx = \frac{\psi(1) - \psi(1-t)}{1-t}. $$
Now using
$$ \psi(x+1) = - \gamma + \sum_{k=1}^{\infty} (-1)^{k+1} \, \zeta(k+1) \, x^{k} $$
then
$$ \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx = \sum_{k=1}^{\infty} (-1)^k \, \zeta(k+1) \, (1-t)^{k-1}. $$
Using the operator $(-1)^n \, \partial_{t}^{n}$ on both sides of this last expression the following is obtained.
\begin{align}
I_{n} &= \int_{0}^{1} \ln(1-x) \, \ln^{n}(x) \, x^{-t} \, dx \\
&= \sum_{k=1}^{\infty} (-1)^{k+n} \, \zeta(k+1) \, \partial_{t}^{n} \, (1-t)^{k-1} \\
&= \sum_{k=1}^{\infty} (-1)^{k} \, \zeta(k+1) \, \frac{(k-1)!}{(k-n-1)!} \, (1-t)^{k-n-1} \\
&= \sum_{k=n}^{\infty} (-1)^{k+1} \, \zeta(k+2) \, \frac{k!}{(k-n)!} \, (1-t)^{k-n} \\
&= (-1)^{n+1} \, \zeta(n+2) \, n! + \sum_{k=n+1}^{\infty} (-1)^{k+1} \, \zeta(k+2) \, \frac{k!}{(k-n)!} \, (1-t)^{k-n}.
\end{align}
Setting $t = 1$ gives the desired result
$$ \int_{0}^{1} \frac{\ln(1-x) \, \ln^{n}(x)}{x} \, dx = (-1)^{n+1} \, n! \, \zeta(n+2). $$
or
\begin{align}
(-1)^{n+1} \, n! \, \zeta(n+2) &= \int_{0}^{1} \frac{\ln(1-x) \, \ln^{n}(x)}{x} \, dx \\
&= (-1)^n \, \partial_{t}^{n} \, \left. \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx \right|_{t=1} \\
&= (-1)^n \, \partial_{t}^{n} \, \partial_{m} \, \left. \int_{0}^{1} (1-x)^{m} \, x^{-t} \, dx \right|_{t=1}^{m=0} \\
&= (-1)^n \, \partial_{t}^{n} \, \partial_{m} \, \left. B(m+1, 1-t) \right|_{t=1}^{m=0}.
\end{align}
| {
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Does $(x^2+xy+y^2)$ divide $(x+y)^n-x^n-y^n$ if and only if n has no prime factors less than 5 I noticed that for $n$ with prime factors greater than 5, that $xy(x+y)(x^2+xy+y^2)$ always seems to divide $(x+y)^n-x^n-y^n$. The $xy(x+y)$ factors seem fairly obvious, but I can't figure out where the $x^2+xy+y^2$ term comes from. Strangely, if $n$ is congruent to 6 mod 1, then $(x^2+xy+y^2)^2$ seems to divide $(x+y)^n-x^n-y^n$.
Is this true for all $n$ as specified? Is the converse true, i.e. if $x^2+xy+y^2$ divides $(x+y)^n-x^n-y^n$ then $n$ has no prime factors less than 5, and if $(x^2+xy+y^2)^2$ divides $(x+y)^n-x^n-y^n$, then n is congruent to 1 mod 6?
| By homogeneity, it’s enough to show that $x^2+x+1$ divides $(x+1)^n-x^n-1$. Because these polynomials are rational and $x^2+x+1$ is irreducible over $\mathbb{Q}$, it’s enough to show that $(j+1)^n-j^n-1=0$, where $j=e^{2i\pi/3}$ is a primitive third root of unity.
Now, $(j+1)^n-j^n-1=(-j^2)^n-j^n-1=-(1+j^n+j^{2n})$. Now, $3$ doesn’t divide $n$ so $j^n \in \{j,j^2\}$ and both of these elements are roots of $X^2+X+1$.
By the same reasoning, $(x^2+xy+y^2)^2|(x+y)^n-x^n-y^n$ iff $x^2+x+1$ divides the derivative $D$ of $(x+1)^n-x^n-1$, iff $D(j)=0$.
But $D(X)=n((X+1)^{n-1}-X^{n-1})$, so $D(j)=0$ iff $(j+1)^{n-1}=j^{n-1}$ iff $(-j^2)^{n-1}=j^{n-1}$ iff ($n$ is odd) $j^{2n-2}=j^{n-1}$ iff $3|n-1$, so iff $6|n-1$.
| {
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Sove the recurrence relation $a_n = 5a_{n-1}-8a_{n-2}+4a_{n-3}$ and $a_0 = 0$, $a_1 = 0$, $a_2 = 1$
Consider the recurrence relation $a_n = 5a_{n-1}-8a_{n-2}+4a_{n-3}$ and $a_0 = 0$, $a_1 = 0$, $a_2 = 1$. Find a closed form expression for $a_n$.
I have computed the OGF $A(z)$ of $(a_n)_n$ as
$$A(z) = \frac{z^2}{1-5z+8z^2-4z^3}.$$
According to Wolfram Alpha this has the following partial fraction decomposition:
$$= \frac{-3}{2(1-2z)} + \frac{1}{2(1-2z)^2} + \frac{1}{1-z}$$
and using the Generalised Binomial Theorem yields:
$$= -\frac{3}{2} \sum_{n = 0}^\infty 2^nz^n + \frac{1}{2} \sum_{n=0}^\infty \binom{-2}{n}2^nz^n + \sum_{n=0}^\infty z^n.$$
Thus we should have
$$a_n = [z^n]A(z) = -\frac{3}{2}2^n + \frac{1}{2} \binom{-2}{n}2^n +1 = -3 \cdot 2^{n-1} + \binom{-2}{n}2^{n-1} +1 $$
However, this does not seem to be true. Could you please tell me what I am doing wrong?
| Characteristic equation is $r^3-5r^2+8r-4=(r-1)(r-2)^2$
So $a_n=a\,1^n+(bn+c)\,2^n$
Applying initial conditions give $\begin{cases}a_0=a+c=0\\a_1=a+2(b+c)=0\\a_2=a+4(2b+c)=1\end{cases}\iff a=1,\ c=-1,\ b=\frac 12$
$$a_n=2^n(\tfrac n2-1)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594786",
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Solving a system of equations with logarithms in exponents I am interested in what approach could be applied to solve such a system:
$$ 2^{x-\log_2{y}} - 8^{x+\log_2{y}} = 5y - 2^{x-\log_2{y}} $$
$$ \frac{2^{x-\log_2{y}}}{8^{x+\log_2{y}}} = \frac{5y}{2^{x-\log_2{y}}} $$
| Notice that the first equation is
$$\frac{2^x}y-8^x\cdot y^3=5\cdot y-\frac{2^x}y$$
Multiply this equation for $y$ and get
$$8^x\cdot y^4+5\cdot y^2-2^{x+1}=0$$
so you have a way to express $y$ purely in function of $x$. Now, the second equation is
$$\frac{2^x}{8^x\cdot y^4}=\frac{5\cdot y^2}{2^x}$$
so you can rewrite it as $1=5\cdot 2^x\cdot y^6$. Now you know that $y^6=5^{-1}\cdot 2^{-x}$ so you have the system
$$\left\{\begin{array}{l} (2^x)^3\cdot y^4+5\cdot y^2-2^{x+1}=0;\\
y^6=5^{-1}\cdot 2^{-x}
\end{array}\right.$$
that is easier than the original one. For example, you could call $z:=2^x$ and get $z=1/(5y^6)$ from the second equation, and plug it into the first to get
$$\frac{y^4}{5^3y^{18}}+5y^2-\frac2{5y^6}=0$$
from which you get $y^4+5^4y^{20}-50y^{12}=0$ so $5^4y^{16}-50y^8+1=0$ that is $(25y^8-1)^2=0$ from which you get $y=\pm1/\sqrt[4]5$. Now, $z=\sqrt5$ and so $x=\log_2(5)/2$.
| {
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How to prove identity $\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$? Looking at Jolley, Summation of Series, formula 445:
$\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$
How can one prove this?
Considering $\left( \sum^{n-1}_{j=0}Z_j\right)^2 = \sum^{n-1}_{j=0} Z_j^2 +\sum^{n-1}_{j=0}\sum^{n-1}_{i\neq j} Z_jZ_i$ from here
I thought one could take the log differential of the well known sine product:
$\begin{align}
\prod_{k=0}^{n-1}\sin\left(\theta + \frac{k\pi}{n}\right)
&= 2^{1-n} \sin(n\theta)
\end{align}$
to get
$n \cot(n\theta) = \sum_{k=0}^{n-1}\cot\left(\theta+\frac{k\pi}{n}\right)$
or
$-n \cot(\frac{n\pi}{2}+n\theta) = \sum_{k=0}^{n-1}\tan\left(\theta+\frac{k\pi}{n}\right)$
then square to get
$n^2 \cot(\frac{n\pi}{2}+n\theta)^2 = (\sum_{k=0}^{n-1}\tan\left(\theta+\frac{k\pi}{n})\right)^2$
but I cannot see how to show the
$\sum _{j=0}^{n-1} \sum _{k=0}^{n-1} \tan \left(\theta +\frac{\pi j}{n}\right) \tan \left(\theta +\frac{\pi k}{n}\right) [j\neq k]$
necessary for the $n(n-1)$ part of the identity.
Perhaps somebody could show a more successful method.
| If $\frac{2n\theta}{\pi}$ is an odd integer, one $\tan^2$ value in the sum is undefined (or "infinity"), so assume that's not the case.
$$ z = \tan \left(\theta + \frac{k\pi}{n}\right); 0 \leq k \leq n-1 $$
satisfy $\arg(z+i) = \frac{\pi}{2} + \theta + \frac{k\pi}{n}$. Defining $\alpha = \theta + \frac{\pi}{2}$ just for brevity, this is $\arg(z+i) = \alpha + \frac{k\pi}{n}$. These $n$ values of $z$ also satisfy
$$ \operatorname{Im}\left(\left((z+i)e^{-i\alpha}\right)^n\right) = 0$$
$$\left(((z+i)e^{-i\alpha}\right)^n - \left((z-i)e^{i\alpha}\right)^n = 0$$
Since this is a polynomial in $z$ of degree $n$ (call it $p$), there are no other solutions, and
$$ \begin{align*}
p(z) &= (z+i)^n e^{-in\alpha} - (z-i)^n e^{in\alpha} \tag{1}\label{1} \\
p(z) &= (e^{-in\alpha} - e^{in\alpha})\prod_{k=0}^{n-1} \left(z-\tan\left(\theta+\frac{k\pi}{n}\right)\right) \\
p(z) &= -2i \sin(n\alpha) \prod_{k=0}^{n-1} \left(z-\tan\left(\theta+\frac{k\pi}{n}\right)\right) \tag{2}\label{2}
\end{align*} $$
Now augment the polynomial's set of roots by the negatives of its $n$ roots, to get polynomial $q(z) = (-1)^n p(z)p(-z)$. From $\ref{1}$:
$$ \begin{align*} (-1)^n q(z) &= \left((z+i)^n e^{-in\alpha} - (z-i)^n e^{in\alpha}\right)\left((-z+i)^n e^{-in\alpha} - (-z-i)^n e^{in\alpha}\right) \\
q(z) &= (z^2+1)^n e^{-2in\alpha} - (z-i)^{2n} - (z+i)^{2n} + (z^2+1)^n e^{2in\alpha} \\
q(z) &= 2(z^2+1)^n \cos(2n\alpha) - (z+i)^{2n} - (z-i)^{2n} \tag{3}\label{3} \end{align*} $$
From $\ref{2}$:
$$ \begin{align*}
(-1)^n q(z) &= -4 \sin^2(n\alpha) \prod_{k=0}^{n-1} \left(z-\tan\left(\theta+\frac{k\pi}{n}\right)\right) \left(- z - \tan\left(\theta + \frac{k\pi}{n}\right)\right) \\
q(z) &= -4 \sin^2(n\alpha) \prod_{k=0}^{n-1} \left(z^2 - \tan^2\left(\theta + \frac{k\pi}{n}\right) \right) \tag{4}\label{4}
\end{align*} $$
Now to get the sum of the $\tan^2$ values, equate the coefficients of $z^{2n-2}$ in $\ref{4}$ and $\ref{3}$:
$$ \begin{align*}
a_{2n-2} &= 4 \sin^2(n\alpha) \sum_{k=0}^{n-1} \tan^2\left(\theta + \frac{k\pi}{n}\right) \\
a_{2n-2} &= 2{n \choose 1} \cos(2n\alpha) - {2n \choose 2}i^2 - {2n \choose 2}(-i)^2 \\
&= 2n(1-2\sin^2(n\alpha)) + 2n(2n-1) \\
&= -4n \sin^2(n\alpha) + 4n^2
\end{align*} $$
$$ \begin{align*}
\sum_{k=0}^{n-1} \tan^2\left(\theta + \frac{k\pi}{n}\right) &= n^2 \csc^2(n\alpha) - n \\
&= n^2 (1+\cot^2(n\alpha)) - n \\
&= n^2 \cot^2\left(\frac{n\pi}{2} + n\theta\right) + n(n-1)
\end{align*} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of Sequence For the second time I've had trouble calculating this limit
$$\lim\limits_{n\rightarrow\infty}\frac{(n＋2)(n＋5)\cdot\ldots\cdot(n+3r-1)\cdot\ldots\cdot(4n－1)}{(n＋1)(n＋4)\cdot\ldots\cdot(n+3r-2)\cdot\ldots\cdot(4n－2)}$$
This limit can be calculated by limitation method but it is not a smart method.
When using the inventory method, you get a result:
$$\lim\limits_{n\rightarrow\infty}\frac{(n＋2)(n＋5)\cdot\ldots\cdot(n+3r-1)\cdot\ldots\cdot(4n－1)}{(n＋1)(n＋4)\cdot\ldots\cdot(n+3r-2)\cdot\ldots\cdot(4n－2)}\quad＝\sqrt[3]{4}$$
I'm looking for a clever way to calculate the limit in a straightforward way
It is the second time that I try to ask a question on this site. I apologize if my question is not accurate.
| Using
$$ \Gamma(x) \approx \sqrt{2 \pi} \, x^{x - 1/2} \, e^{-x} \, \left(1 + \frac{1}{12 \, x} + \mathcal{O}\left(\frac{1}{x^2}\right) \right) $$
then it can be shown that
$$ \frac{\Gamma\left(\frac{n+1}{3}\right)}{\Gamma\left(\frac{n+2}{3}\right)} \approx \left(\frac{n}{3}\right)^{-1/3}. $$
Now,
\begin{align}
L &= \lim_{n \to \infty} \, \prod_{j=1}^{n} \left\{ \frac{n + 3 j -1}{n + 3 j -2} \right\} \\
&= \lim_{n \to \infty} \frac{\left( \frac{n+2}{3} \right)_{n} }{ \left(\frac{n+1}{3}\right)_{n} } = \lim_{n \to \infty} \frac{\Gamma\left( \frac{n+1}{3} \right) \, \Gamma\left( \frac{4n+2}{3} \right)}{\Gamma\left( \frac{n+2}{3} \right) \, \Gamma\left( \frac{4n+1}{3} \right)} \\
&= \lim_{n \to \infty} \left( \left(\frac{n}{3}\right)^{-1/3} \, \left(\frac{4 n}{3}\right)^{1/3} \right) \\
&= \lim_{n \to \infty} \sqrt[3]{4}
\end{align}
which gives
$$ \lim_{n \to \infty} \, \prod_{j=1}^{n} \left\{ \frac{n + 3 j -1}{n + 3 j -2} \right\} = \sqrt[3]{4}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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In any triangle $\triangle ABC$, show that $4R\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})=r$ This is a problem problem I found in a JEE examination prep textbook, it was a "starred" question which I believe implies that it is more challenging than usual. It goes as follows:
In any triangle $\triangle ABC$, show that $$4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=r$$
Hint: $$2R^2\sin\left(A\right)\sin\left(B\right)\sin\left(C\right)=\Delta$$
Here is my attempt at it. I want to know if this is correct and if there any better alternative approaches to achieve the same result, please do share them!
We know that:
$$\Delta=rs$$
Using the given hint:
$$2R^2\sin\left(A\right)\sin\left(B\right)\sin\left(C\right)=rs$$
$$16R^2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=r\left(\frac{a+b+c}{2}\right)$$
$$16R\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=r\left(\sin\left(A\right)+\sin\left(B\right)+\sin\left(C\right)\right)$$
Now, focusing on the equation of the right hand side for a bit, we know that:
$$A+B+C=\pi$$
$$\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$$
$$\sin\left(A\right)+\sin\left(B\right)+\sin\left(C\right)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)+2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)$$
$$2\cos\left(\frac{C}{2}\right)\left(\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{A+B}{2}\right)\right)$$
$$4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$$
Now substituting this back into the original problem:
$$16R\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=\left(4r\right)\left[\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)\right]$$
And that gives us:
$$4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=r$$
| Here is a rather simple geometrical approach that does not use the hint
Let the incenter of the triangle be $I$. Draw a perpenduclar line from $I$ to $BC$ and call the foot of the perpendicular $D$. Now, $IB$ is the angle bisector of $\angle CBA$ and $\triangle DBI$ is a right-angled triangle, thus,
$$ \cot\frac{B}{2} = \frac{BD}{ID} = \frac{BD}{r} \implies BD = r \cot\frac{B}{2} $$
Similarly
$$ \cot\frac{C}{2} = \frac{CD}{ID} = \frac{CD}{r} \implies CD = r \cot\frac{C}{2}$$
Now
$$\begin{align}
BC &= BD + CD \\
\implies a &= r \left(\cot\frac{B}{2} + \cot\frac{C}{2}\right) \\
&= r \left( \frac{\cos\frac{B}{2}}{\sin\frac{B}{2}} + \frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}\right) \\
&= r \left( \frac{\sin\frac{B+C}{2}}{\sin\frac{B}{2} \sin\frac{C}{2}} \right) \\
\implies 2R\sin A \left(\sin\frac{B}{2} \sin\frac{C}{2} \right) &= r \sin\frac{\pi - A}{2} \\
\implies 2R \left( 2\sin\frac{A}{2} \cos\frac{A}{2} \right) \sin\frac{B}{2} \sin\frac{C}{2} &= r \cos\frac{A}{2} \\
\therefore 4R \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} &= r
\end{align}$$
Note that $\cos\dfrac{A}{2} \ne 0$ because $A \ne \pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof of a tighter inequality than Cauchy-Schwarz inequality A few days ago, I found this post which discuss about some tighter versions of Cauchy-Schwarz (or some might prefer the name AM-GM) inequality. User @Michael Rozenberg proposed a very interesting inequality at the comment section
Let $a$, $b$ and $c$ be three positive numbers, then the following inequality holds
$$
3(a^2+b^2+c^2) - (a+b+c)^2 \ge \dfrac{25(a-b)^2(b-c)^2(c-a)^2}{a^4+b^4+c^4}.
$$
He also mentioned that if we replace number $25$ at the R.H.S with $26$ then we'll get a wrong inequality, so perhaps $25$ is the "best" number, although I don't know what happen if we replace with $25 + \varepsilon$ for $\varepsilon > 0$ sufficiently small.
What I'm curious about is how to fully prove the above inequality. What I tried is to use $pqr$ method: Let $p = a + b + c$, $q = ab + bc + ca$, $r = abc$ and without loss of generality, we can assume that $p = 1$: Since if $p = M$ for some real number $M$, using the change of variables $a' = \frac{a}{M}$, $b' = \frac{b}{M}$ and $c' = \frac{c}{M}$, we get the same inequality as before but this time $a' + b' + c' = 1$. With the $pqr$ substitution and the assumption $p = 1$, we'll have
\begin{align*}
a^2 + b^2 + c^2 &= 1 - 2q \\
a^4 + b^4 + c^4 &= 2q^2 - 4q + 4r + 1 \\
(a-b)^2(b-c)^2(c-a)^2 &= -4q^3 + q^2 + 18qr - 27r^2 - 4r
\end{align*}
The inequality we need to prove now is
$$
104q^3 - 13q^2 - 458qr + 675r^2 - 10q + 108r + 2 \ge 0.
$$
My idea to continue from here is to use Schur inequality of order 1: $r \ge \max\{0, \frac{1}{9}(4q-1)\}$, but this idea is not working since the L.H.S contains the term $-458qr$ makes "$\ge$" sign become "$\leq$" sign. I still don't know how to continue from here, and I'm not sure if $pqr$ method is a good idea.
There are three questions that I really want the answer:
(1) How can we prove the above inequality?
(2) How do we know that $25$ is the "best" constant for the R.H.S?
(3) Is there any other function $f(a,b,c) \ge 0$ such that $$3(a^2 + b^2 + c^2) - (a+b+c)^2 \ge f(a,b,c), \quad \forall a,b,c > 0?$$ If there exists, how do we actually come up with such function?
| *
*Let $c=\min\{a,b,c\}$,$a=c+u$,$b=c+v$ and $u^2+v^2=2tuv$.
Thus, $$(a^4+b^4+c^4)(3(a^2+b^2+c^2)-(a+b+c)^2)-25\prod_{cyc}(a-b)^2=$$
$$=((c+u)^4+(c+v)^4+c^4)\sum_{cyc}(a-b)^2-25\prod_{cyc}(a-b)^2\geq$$
$$=(u^4+v^4)(u^2+v^2+(u-v)^2)-25(u-v)^2u^2v^2=$$
$$=u^3v^3((4t^2-2)(4t-2)-25(2t-2))=u^3v^3(16t^3-8t^2-58t+54)=$$
$$=u^3v^3(t(4t-5)^2+32t^2-83t+54)\geq0$$
because $$83^2-4\cdot32\cdot54=-23<0.$$
*From the above reasoning we obtain that the maximal value of $k$, for which the inequality $$3(a^2+b^2+c^2) - (a+b+c)^2 \ge \dfrac{k(a-b)^2(b-c)^2(c-a)^2}{a^4+b^4+c^4}$$ is true for any positives $a$, $b$ and $c$ it's
$$k_{max}=\min_{t>1}\frac{2(2t^2-1)(2t-1)}{t-1}=25.367...$$
*I got the following inequality in this type.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$3(a^2+b^2+c^2) - (a+b+c)^2 \ge \dfrac{48(a-b)^2(b-c)^2(c-a)^2}{a^4+b^4+c^4+5(a^2b^2+a^2c^2+b^2c^2)}.$$
The number $48$ is a best constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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What is wrong with the calculation $ \lim\limits_{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $? We are given
$$ \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $$
We can write
$$ L = \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} = \lim _{x \to 0} \frac {x (cos x - \sin x/x)} {x^2 \sin x} $$
It can be written as
$$ L = \lim _{x \to 0} \frac { \cos x - \sin x/x} {x^2 \sin x/x} $$
We know
$$ \lim _{x \to 0} \frac {\sin x} {x} = 1 $$
The original limit can be written as
$$ L = \lim _{x \to 0} \frac { \cos x - 1} {x^2} $$
Now using $ ( \cos x ) $ expansion formula, the answer will be $ L = -\frac12$ , but the answer is $ L = -\frac13$ , which is not matching.
I want to know what I did wrong in this method. I know the answer using other methods, but I want to know what went wrong with this one.
| To check where it went wrong , let us use Series Expansion :
$ L = \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $
$ L = \lim _{x \to 0} \frac { x ( 1-x^2/2!+x^4/4!-x^6/6! \cdots ) - ( x-x^3/3!+x^5/5!-x^7/7! \cdots ) } {x^2 ( x-x^3/3!+x^5/5!-x^7/7! \cdots ) } $
$ L = \lim _{x \to 0} \frac { ( x-x^3/2!+x^5/4!-x^7/6! \cdots ) - ( x-x^3/3!+x^5/5!-x^7/7! \cdots ) } { ( x^3-x^5/3!+x^7/5!-x^9/7! \cdots ) } $
$ L = \lim _{x \to 0} \frac { -x^3/3 + (x^5/4!-x^7/6! \cdots ) - (x^5/5!-x^7/7! \cdots ) } { ( x^3 + (-x^5/3!+x^7/5!-x^9/7! \cdots) ) } $
$ L = \lim _{x \to 0} \frac { -1/3 + (x^2/4!-x^4/6! \cdots ) - (x^2/5!-x^4/7! \cdots ) } { ( 1 + (-x^2/3!+x^4/5!-x^6/7! \cdots ) ) } $ [[ Cancelling $x^3$ through out ]]
$ L = \lim _{x \to 0} \frac { -1/3 + 0 - 0 } { ( 1 + 0 ) } $ [[ Setting $x=0$ through out ]]
$ L = -1/3 $ & there is no Doubt about this Answer.
Now we get a clue or hint about what went wrong.
The given limit is $ ( \infty - \infty ) $ where we can not take the Individual limits.
Numerator is in the format $ (x+A_1x^3+A_2x^5 \cdots ) - (x+B_1x^3+B_2x^5 \cdots ) $
Denominator is in the format $ (x^3+C_1x^5+C_2x^7 \cdots ) $
Numerator will be cancelling the $ (x) - (x) $ terms.
Denominator & Numerator will then have Common $ (x^3) $ which will be cancelling throughout.
We want both $ A_1 = -1/2 $ & $ B_1 = -1/6 $ to get the Exact limit $ L = A_1 - B_1 = (-1/2) - (-1/6) = (-1/3) $ which is Correct.
Whereas by taking Individual limit earlier , we are keeping $ A_1 = -1/2 $ while losing $ B_1 = -1/6 $ , hence getting the limit $ L = A_1 = -1/2 $ which is wrong.
Developing that Core Point with Extreme Detail :
We have $ L = \frac{ (x+A_1x^3+A_2x^5 \cdots ) - (x+B_1x^3+B_2x^5 \cdots ) }{ (x^3+C_1x^5+C_2x^7 \cdots ) } $
In the Denominator , we take $x^3=x \cdot x^2$ & leave the $x^2$ in the Denominator , to use the $x$ on the Numerator $A$ & $B$ terms to get :
$ L = \frac{ (x+A_1x^3+A_2x^5 \cdots )/x - (x+B_1x^3+B_2x^5 \cdots )/x }{ (x^2+C_1x^4+C_2x^6 \cdots ) } $
We have $ L = \frac{ \cos x - (1+B_1x^2+B_2x^4 \cdots ) }{ (x^2+C_1x^4+C_2x^6 \cdots ) } $
It is Correct till here.
The trouble starts now.
We are now using $ ( \sin x ) / ( x ) = 1 $ , thereby losing $B_1$ , yet keeping $A_1$ in the $ ( \cos x ) $ term. That is WRONG ! We have to Consistently keep both terms having the Same Power !
When we take over-all limit while keeping both $A$ term & $B$ term having the $x^3$ Power , we get $L=-1/3$ , which is Correct.
| {
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"url": "https://math.stackexchange.com/questions/4602507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area of Quadrilateral $ABCD$. A puzzle for 10th graders. As the title suggests, the problem in this post was meant to be a puzzle for 10th graders, so claims the person who posted this on a language exchange platform: The problem is as follows:
Given a Quadrilateral $ABCD$ with internal point $P$, where $AP=1$, $BP=2$ and $CP=3$, and unknown sides $k$ and $2k$, compute the area of this Quadrilateral.
I first tried to inscribe this quadrilateral into a square but that approach did not turn out successful, I was thinking if there are any other ways to solve it, perhaps via setting up a coordinate system, or via a trigonometric method. I will share my own successful approach below as an answer!
| I'm pretty bad at geometry so I just spammed a bunch of algebra.
First, note that the area is clearly $3k^2.$ Then, draw a line through $P$ perpendicular to $AD$ and $BC,$ which intersects them at $X$ and $Y,$ respectively. Let $PX=n,$ so $PY=2k-n.$ Similarly, let $AX=BY=m$ so $CY=2k-m.$ Then, note that $m^2+n^2=1.$ Also, we have $m=1-n^2=4-(2k-n)^2,$ using the second equality, we have $4k^2-4kn=3$ so $n=\frac{4k^2-3}{4k}.$ Similarly, we have $n=4-m^2=9-(2k-m)^2$ and similar to above, this yields $m=\frac{4k^2-5}{4k}.$ Therefore, we have $$\left(\frac{4k^2-3}{4k}\right)^2+\left(\frac{4k^2-5}{4k}\right)^2=1.$$ This yields $$16k^4-24k^2+9+16k^4-40k^2+25=16k^2$$ $$32k^4-80k^2+34=0$$ $$16k^4-40k^2+17=0.$$ This is a quadratic in $k^2$ and solving for $k^2$ gives $$k^2=\frac{40\pm\sqrt{1600-1088}}{32}=\frac{5\pm 2\sqrt{2}}{4}.$$ However, note that $k>1,$ so $k^2>1$
and therefore, we only take the $+$ and we have that the area is $$3k^2=\frac{15+6\sqrt{2}}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proofs about a recursive sequence. Let $x_1=a>2$, and $$x_{n+1}=\frac{x_n^2}{2(x_{n}-1)}, n=1,2,...$$
Prove
(i) If $a\leq 3$, we have $x_n\leq 2+\frac1{2^{n-1}},(n=1,2...)$
(ii) If $a> 3$, then when $n\geq\frac{lg(\frac a3)}{lg \frac 43 }$, we have $x_{n+1}<3$.
My attempts:
I can use induction to show that $x_n>2$, and get $\frac{x_{n+1}}{x_n}=\frac{x_{n}}{2(x_n-1)}< 1$, where I checked the graph of the function $f(x)=\frac{x}{2(x-1)}$. For (i), I used $x_n>2$ and tried the induction to obtained that, under the assumption $x_n\leq 2+\frac1{2^{n-1}},(n=1,2...)$, we have $$x_{n+1}=\frac{x_n^2}{2(x_n-1)}\leq \frac{x_n^2}{2(2-1)}=\frac{x_n^2}{2}\leq \frac{(2+\frac1{2^{n-1}})^2}{2}.$$ My target next is to show $x_{n+1}\leq 2+\frac1{2^{n}}$ but I have no idea to go further. For(ii), I have no idea so far and any suggestions or hints would be very appreciated!
| The claimed inequality $$x_n \le 2 + \frac{1}{2^{n-1}}$$ suggests defining an auxiliary sequence $$y_n = x_n - 2,$$ hence the original recursion may be written
$$y_{n+1} = x_{n+1} - 2 = \frac{(x_n-2 + 2)^2}{2(x_n - 2 + 1)} - 2 = \frac{(y_n + 2)^2 - 4(y_n + 1)}{2(y_n + 1)} = \frac{y_n^2}{2(y_n + 1)}. \tag{1}$$ In turn, this suggests that when $y_n > 0$, $$y_{n+1} = \frac{y_n^2}{2(y_n+1)} \le \frac{y_n^2}{2y_n} = \frac{y_n}{2}. \tag{2}$$ Consequently, $$x_n - 2 = y_n \le \frac{y_{n-1}}{2} \le \frac{y_{n-2}}{2^2} \le \cdots \le \frac{y_1}{2^{n-1}} = \frac{x_1 - 2}{2^{n-1}} = \frac{a-2}{2^{n-1}}.\tag{3}$$
Therefore, if $2 < a \le 3$, then $0 < a-2 \le 1$ and we have proven (i).
After illustrating the solution for the first part, I encourage you to reconsider how you might approach (ii).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608819",
"timestamp": "2023-03-29T00:00:00",
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Find Laurent series of $\frac{x^2}{(1+x^2)^2}$ I would like to find the Laurent series of the function
$$
f(x)=\frac{x^2}{(1+x^2)^2}.
$$
I already know that there are two poles of order $2$, namely $x=\pm i$. And I also know that, by partial fraction decomposition,
$$
\frac{x^2}{(1+x^2)^2}=\frac{1}{x^2+1}-\frac{1}{(1+x^2)^2}.
$$
| Here are two variations to calculate the residue of $f$ at $x=i$. We already know the pole of $f$ at $x=i$ is of order $2$. A common way is to use the
Limit formula for higher order poles:
We obtain
\begin{align*}
\color{blue}{\mathrm{res}_{x=i}f(x)}&=\mathrm{res}_{x=i}\frac{x^2}{\left(1+x^2\right)}\\
&=\lim_{x\to i}\frac{d}{dx}\left((x-i)^2f(x)\right)\\
&=\lim_{x\to i}\frac{d}{dx}\left(\frac{x^2}{(x+i)^2}\right)\\
&=\lim_{x\to i}\frac{2ix}{(x+i)^3}\\
&=\frac{2i^2}{(2i)^3}\\
&\,\,\color{blue}{=-\frac{1}{4}i}\tag{1}
\end{align*}
It is somewhat more cumbersome, but also not so difficult, to calculate the remainder of $f$ at $x=i$ by consequent expansion at $x=i$.
Manual expansion in terms of $(x-i)$:
We start with
\begin{align*}
f(x)&=\frac{x^2}{\left(1+x^2\right)^2}=\frac{x^2}{(x+i)^2(x-i)^2}\\
&=\frac{(x-i+i)^2}{(x-i+2i)^2(x-i)^2}\\
&=\frac{(x-i)^2+2i(x-i)-1}{(2i)^2\left(1+\frac{x-i}{2i}\right)^2(x-i)^2}\\
&=-\frac{1}{4}\frac{1}{\left(1+\frac{x-i}{2i}\right)^2}
+\frac{1}{2i(x-i)}\,\frac{1}{\left(1+\frac{x-i}{2i}\right)^2}\\
&\qquad+\frac{1}{4(x-i)^2}\,\frac{1}{\left(1+\frac{x-i}{2i}\right)^2}\tag{2}
\end{align*}
The expansion of the square of the geometric series in (2) gives
\begin{align*}
\frac{1}{\left(1+\frac{x-i}{2i}\right)^2}
&=\left(1-\frac{x-i}{2i}+\cdots\right)\left(1-\frac{x-i}{2i}+\cdots\right)\\
&=1-\frac{x-i}{i}+\mathrm{higher\ order\ terms}\tag{3}
\end{align*}
We derive from (2) and (3)
\begin{align*}
\color{blue}{\mathrm{res}_{x=i}f(x)}=0+\frac{1}{2i}+\frac{1}{4}\left(-\frac{1}{i}\right)\,\,\color{blue}{=-\frac{1}{4}i}
\end{align*}
in accordance with (1).
| {
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"url": "https://math.stackexchange.com/questions/4611521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Finding $\cot(\beta)$, knowing $\sin(\alpha+\beta)=4/5$ and $\sin(\alpha-\beta)=3/5$ Let's assume that for $0<\beta<\alpha<\frac{\pi}{2}$, $\sin(\alpha+\beta) = \frac{4}{5}$, and $\sin(\alpha-\beta) = \frac{3}{5}$. Then, how could we find $\cot(\beta)$?
$$\sin(\alpha+\beta)+\sin(\alpha-\beta) = 2\sin(\alpha)\cos(\beta) = \frac{7}{5}$$
$$\sin(\alpha+\beta)-\sin(\alpha-\beta) = 2\sin(\beta)\cos(\alpha) = \frac{1}{5}$$
$$\tan(\alpha)\cot(\beta) = 7$$
But I am not sure where this would lead us.
| $\displaystyle sin( a+b) =4/5\ \ \ sin( a-b) =3/5.$
$\displaystyle So\ cos( a+b) \ =\ 3/5$ but that means $\displaystyle sin\left(\frac{\pi }{2} -( a+b)\right) =sin( a-b) =3/5$
So $ $$\displaystyle \frac{\pi }{2} =a+b+a-b$
Or $\displaystyle a\ =\ \frac{\pi }{4}$
So $cot(a) = 1 $
And since you have already figured out $tan(a)cot(b)=7$ so $cot(b)=7$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the value of $\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n\frac{1}{n+k^{\alpha}}\right)$
Find the value of the expression $$\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n\frac{1}{n+k^{\alpha}}\right)$$
where $\alpha$ is a positive number.
It is my first time seeing a question like this. Normally, we just put $n=\infty$ in the summand which is same as calculating the series till infinite terms.
But here it is different. The variable in the bound is also in the expression, meaning $n=\infty$ in the summand is not applicable in this case.
I expanded the series as $$\frac{1}{n+1}+\frac{1}{n+2^{\alpha}}+\cdots$$
But had no luck.
Any help is greatly appreciated.
| Actually the person who posted the problem has also given its solution, that I would like to share.
The limit equals $1$ when $0<\alpha<1$, $\ln2$ when $\alpha=1$ and $0$ when $\alpha>1$
$$\frac{1}{n+n^{\alpha}}<\frac{1}{n+1^{\alpha}}+\frac{1}{n+2^{\alpha}}+\cdots+\frac{1}{n+n^{\alpha}}<\frac{n}{n+1}$$ and hence when $\alpha<1$ the limit equals $1$.
When $\alpha=1$ we have
$$\lim_{n\rightarrow\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right)=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}=\int_0^1\frac{\operatorname{dx}}{1+x}$$ $$=\ln2$$
When $\alpha>1$ we have since $n+k^{\alpha}\ge2\sqrt{n}k^{\frac{\alpha}{2}}$ that
$$0<\frac{1}{n+1^{\alpha}}+\frac{1}{n+2^{\alpha}}+\cdots+\frac{1}{n+n^{\alpha}}<\frac{1}{2\sqrt{n}}\left(\frac{1}{1^{\frac{\alpha}{2}}}+\frac{1}{2^{\frac{\alpha}{2}}}+\cdots+\frac{1}{n^{\frac{\alpha}{2}}}\right)$$
An application of Stolź-Cesaro Lemma (the $\frac{\infty}{\infty}$ case) shows that
$$\lim_{n\rightarrow\infty}\frac{\frac{1}{1^{\frac{\alpha}{2}}}+\frac{1}{2^{\frac{\alpha}{2}}}+\cdots+\frac{1}{n^{\frac{\alpha}{2}}}}{\sqrt{n}}=\lim_{n\rightarrow\infty}\frac{\frac{1}{(n+1)^{\frac{\alpha}{2}}}}{\sqrt{n+1}-\sqrt{n}}=\lim_{n\rightarrow\infty}\frac{\sqrt{n+1}+\sqrt{n}} {(n+1)^{\frac{\alpha}{2}}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Prove that $(\sum x^2)^3\ge9\sum x^4yz$
Prove that $\displaystyle\left(x^2+y^2+z^2\right)^3\ge9\left(x^4yz+y^4xz+z^4xy\right)$, for $x$, $y$, $z\in\Bbb R_+$.
The $pqr$ method doesn't seem possible because the power is too high.
$$\iff\left(p^2-2q\right)^2\ge9r\left(p^3-3pq+3r\right).$$
Then expand the expression to get
$$\sum x^6+3\sum\left(x^4y^2+x^2y^4\right)+6x^2y^2z^2\ge9\sum x^4yz.$$
I wanted to use SOS but cannot find the weight of three squares, my progress:
$$3\sum x^4(y-z)^2=3\sum\left(x^4y^2+x^2y^4\right)-6\sum x^4yz.$$
Whats left is $\displaystyle\sum x^6+6x^2y^2z^2-3\sum x^4yz$. I have trouble dealing with it.
| Hint :
Due to homogeneization we have :
$$\left(x^{2}+1^{2}+c^{2}\right)^{3}-9xc\left(x^{3}+1^{3}+c^{3}\right)$$
We need to show for $x\ge 1$ and WLOG $c\geq 1$
$$f(x)=\left(x^{2}+1^{2}+c^{2}\right)^{3}-9xc\left(x^{3}+1^{3}+c^{3}\right)\geq 0$$
The function is convex on $(0,\infty)$ because we have :
$$f''(x)=6(c^{6}+6c^{3}x^{2}+2c^{3}-18cx^{2}+5x^{4}+6x^{2}+1)>0$$
So we use the tangent line method we have :
$$f(x)\geq f(1)(x-1)+f(1)=(6 (c^3 + 2)^2 - 9 c (c^3 + 2) - 27 c)(x-1)+(c^3 + 2)^3 - 9 c (c^3 + 2)\geq 0$$
Because :
$$(6(c^{3}+2)^{2}-9c(c^{3}+2)-27c)\geq 0,(c^3 + 2)^3 - 9 c (c^3 + 2)=(c^3 + 2) (c - 1) (c^5 + c^4 + c^3 + 5 c^2 + 5 c - 4)\geq 0$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
How to show $k \cdot a + c \cdot b \ge \frac{1}{2} \cdot (k+c) \cdot (a+b)$? Assume k, c, a, b $\gt$ 0 and that $k \ge c$ and $a \ge b$. Prove $k \cdot a$ + $c \cdot b$ $\ge$ $\frac{1}{2}\cdot (k+c) \cdot (a+b)$. This is what I have so far:
$2\cdot k\cdot a +2\cdot c \cdot b$ = $(k\cdot a + k\cdot a)$ + $(c\cdot b + c\cdot b)=(k\cdot a + c\cdot b)$ + $(k\cdot a + c\cdot b)\ge(k\cdot a + c\cdot b)$ + $(k\cdot b + c\cdot b)$
| $$
\begin{aligned}
2(k a+c b)-(k+c)(a+b)
= & 2 k a+2 c b-k a-c a-k b-b c \\
= & -k a+c b-c a-k b \\
= & (k-c) a+(c-k) b \\
= & (k-c)(a-b)\\ \geqslant & 0 \\
\end{aligned}
$$
Therefore $$k a + cb \ge \frac{1}{2} (k+c)(a+b)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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