TITLE: Motivation behind this proof of the Basel problem?
QUESTION [3 upvotes]: On a past paper from our examination committee there seems to be a proof to the Basel problem that is plucked out of thin air. It doesn't match any of the proofs on Wikipedia (at least by first glance) and I can't find it anywhere on Google through searching. It follows like so:
Consider $ \displaystyle  A_n=\int_0^\frac{\pi}{2}\cos^{2n}{x}\ dx$ and $\displaystyle B_n=\int_0^\frac{\pi}{2}x^2\cos^{2n}x\ dx$
Through integration by parts performed twice on $A_n$, one with $u=\cos^{2n-1}x,\  v'=\cos{x}$ to show that $A_n=\frac{2n-1}{2n}A_{n-1}$ and the other with $u=\cos^{2n}x,\  v'=1$ to show that $\frac{1}{n^2}=\frac{1}{A_n}\left(\frac{2n-1}{n}B_{n-1}-2B_n\right)$ we show that $\displaystyle \sum_{k=1}^n\frac{1}{k^2}=2\frac{B_0}{A_0}-2\frac{B_n}{A_n}$
which is $\displaystyle \sum_{k=1}^n\frac{1}{k^2}=\frac{\pi^2}{6}-2\frac{B_n}{A_n}$.
Then, using $\sin{x}≥\frac{2}{\pi}x$ for $0≤x≤\frac{\pi}{2}$, we know that $\displaystyle B_n≤\int_0^{\frac{\pi}{2}}x^2\left(1-\frac{4x^2}{\pi^2}\right)^n\ dx$
Through integration by parts where $u=x, \ v'=x\left(1-\frac{4x^2}{\pi^2}\right)$ it is then shown that $\displaystyle B_n≤\frac{\pi^2}{8(n+1)}\int_0^{\frac{\pi}{2}}\left(1-\frac{4x^2}{\pi^2}\right)^{n+1}\ dx$
Then through substituting $x=\frac{\pi}{2}\sin t$ it follows that $B_n≤\frac{\pi^3}{16(n+1)}A_n$.
$\therefore \displaystyle \frac{\pi^2}{6}-\frac{\pi^3}{8(n+1)}≤\sum_{k=1}^n\frac{1}{k^2}<\frac{\pi^2}{6}$, which then arrives at the solution $\displaystyle \sum_{k=1}^\infty\frac{1}{k^2}=\frac{\pi^2}{6}$ by the squeeze theorem.
So I just want to know what motivates this proof, which unlike most of the other proofs on Wikipedia doesn't use multivariable calculus or complex analysis? What hints at considering the specific integrals of $A_n$ and $B_n$ and is this proof original or a simplified version of a more complex/rigorous proof that is already widely available?

REPLY [4 votes]: The error term in partial sums of $\zeta(2)$ must be $O\left(\frac1n\right)$ (this admits an obvious $1$-variable calculus motivation). Such terms are frequently expressed as pointwise ratios of two sequences of definite integrals. A result of the form$$\zeta(2)-f(n)<\sum_{k=1}^nk^{-2}<\zeta(2)$$would imply an upper bound on this ratio, hopefully computable by upper-bounding the numerator integral with some inequality involving its integrand. My guess is $\sin x\le x$ was tried at first, but powers of $\sin x$ didn't pay off, so even powers of $\cos x$ were attempted instead to exploit an upper bound on$$\cos^{2n}x=\left(1-\sin^2x\right)^n.$$This requires a lower bound on $\sin x$ to work, but this has one advantage: $\sin x\ge\frac{2x}{\pi}$ fixes the integration range we'll try as $\left[0,\,\frac{\pi}{2}\right]$. So at some point you'll consider$$A_n:=\int_0^{\pi/2}\cos^{2n}xdx\le\int_0^{\pi/2}x^2\left(1-\frac{4x^2}{\pi^2}\right)^ndx.$$That upper bound looks susceptible to IBP, which eventually gives us the extra $x^2$ factor (from which we work backwards to the more trigonometric $B_n$), and fortunately gets the $O\left(\frac1n\right)$ ratio we wanted.


TITLE: Show that $X+Y$ and $X-Y$ are independent from joint density.
QUESTION [0 upvotes]: I have a random vector $X,Y$ (with respect to the Lebesgue measure):
$$f_{X,Y}=1/2 \mathbb I _{|x|+|y|\leq1}$$
And I want to show that $X+Y$ and $X-Y$ are independent. One way to show it, is to show that:
$$f_{X+Y}f_{X-Y}=f_{X+Y,X-Y}$$
I understand that I can calculate the densities for $X+Y$, $X-Y$ with the usual convolution formula. Although maybe there is the better way... in any case how do I get the joint density?

REPLY [4 votes]: Notice that $|x|+|y| \leq 1$ if and only if $-1 \leq x+y \leq 1$ and $-1 \leq x-y \leq 1$, and so,
$$ \mathbf{1}_{\{|x|+|y|\leq 1\}} = \mathbf{1}_{\{x+y \in [-1,1]\}} \mathbf{1}_{\{x-y \in [-1,1]\}}. $$
This already tells that the joint density of $X+Y$ and $X-Y$ must split. Indeed, the change of variables $(u, v) = (x+y, x-y)$ shows
\begin{align*}
f_{X,Y}(x,y)\, dxdy
&= \frac{1}{4} \mathbf{1}_{\{u\in [-1,1]\}} \mathbf{1}_{\{v\in [-1,1]\}} \, dudv.
\end{align*}
Therefore the joint density of $X+Y$ and $X-Y$ splits and they are independent.