Computing an infinite trigonometric sum $\sum \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)$

Question

Let $G(x,y) = \sum_{n=1}^\infty \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)$

I'm trying to compute this sum by understanding it as an integral kernel. This question comes from Dym and Mckean Fourier Series and Integrals Ex 1.7.14:

Consider the cosine basis on $L^2[0,1]$ defined by $f_n(x) = \sqrt 2 \cos(n \pi x)$ for $n \ge 1$ and $f_0 = 1$. Define the operator $F$ by $F f_n = \frac{1}{n^2\pi^2} f_n$ acting only on the subspace $n \ge 1$. It turns out that $F f(x) = \int_0^1 G(x, y) f(y) dy$ for $f \in L^2[0,1]$.

Now let $u = f_n''$ for some unspecified $n \ge 1$ . $$F (f_n'') = -f_n $$ $$(F(f_n''))'' = -f_n''$$ $$(F u)'' = -u$$ $$(Fu)'(x) =-\int_0^x u(y)dy + K_1$$

$K_1 = 0$ because $(F u)'(0) = C \sin(0) = 0$ $$(Fu)(x) = -\int_0^x \int_0^y u(z)dz\,dy + K_2$$

$K_2 = \int_0^1\int_0^x \int_0^y u(z)dz\,dy\,dx$ because $\int_0^1 (Fu)(x) dx = \int_0^1 C \cos (2 \pi n x) dx = 0$

Now changing order of integration

$$ (Fu)(x) = - \int_0^x (x - z)u(z) dy + K_2$$ $$ = - \int_0^x (x - z)u(z) dy + \int_0^1\int_0^x (x - z)u(z) dz\,dx$$ $$ =... + \int_0^1\int_z^1(x -z)u(z)dx\,dz$$ $$ =... + \int_0^1u(z) [(1 - z^2)/2 - z(1-z)] dz $$ $$ = ... + \int_0^1 u(z) (z - 1)^2/2 dz$$ $$ = - \int_0^x (x - z)u(z) dz + \int_0^1 u(z) (z - 1)^2/2 dz$$ $$ = - \int_0^x (x - z)u(z) dz + \int_0^x u(z) (z - 1)^2/2 dz+ \int_x^1 u(z) (z - 1)^2/2 dz$$ $$ = \int_0^1 T(x,z) u(z) dz$$

where $T(x,y) = (y^2 + 1)/2 - x $ for $y < x$ and $(y-1)^2/2$ for $y > x$ and by necessity $T(x,y) = G(x,y)$

Unfortunately, I don't think this is the right function and it doesn't even look symmetric in $x,y$. I am looking for help for the correct derivation which should lead to $G(x,y) = \frac{x^2 - x + y^2 -y - |x-y|}{2} + 1/3$

Edit: Even though $T(x,y)$ gives rise to the same operator $F$, it may not equal to $G(x,y)$ since they can differ by any function $\Delta(x)$ so that $T(x,y) + \Delta(x) = G(x,y)$, since $\int_0^1\Delta(x) f_n(y) dy = 0$. Since $\int G(x,y) dy = 0$, we should require $\Delta(x) = -\int_0^1T(x,y)dy = -1/6 + x^2/2$ so that $G(x,y) = (y^2 + 1)/2 - x - x^2/2 + 1/6 $ for $y < x$ and $(y-1)^2/2 + x^2/2 + 1/6$ for $y > x$ and this is the book's answer.

Answer 1

Note: I am finally correcting my mistake pointed out by Mark.

For $G(x,y) = \sum_{n=1}^\infty \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y) $, since (here's where my mistake was: I had cos-cos instead of the correct cos+cos) $\cos(a)\cos(b) =\frac12(\cos(a-b)+\cos(a+b)) $,

$\begin{array}\\ G(x,y) &= \frac12\sum_{n=1}^\infty \frac{2}{n^2 \pi^2} (\cos(n \pi (x-y))+\cos(n \pi (x+y)))\\ &= \frac1{\pi^2}\left(\sum_{n=1}^\infty \frac{1}{n^2} \cos(n \pi (x-y)) + \sum_{n=1}^\infty \frac{1}{n^2} \cos(n \pi (x+y))\right)\\ &= g(x-y)+g(x+y)\\ \end{array} $

where $g(z) =\frac1{\pi^2}\sum_{n=1}^\infty \frac{1}{n^2} \cos(n \pi z) $.

$g'(z) =-\frac1{\pi}\sum_{n=1}^\infty \frac{1}{n} \sin(n \pi z) $ and, getting into areas of dubious convergence,

$\begin{array}\\ g''(z) &=-\sum_{n=1}^\infty \cos(n \pi z)\\ &=-\Re\sum_{n=1}^\infty \exp(in \pi z)\\ &=-\Re\frac{\exp(i \pi z)}{1-\exp(i \pi z)}\\ &=-\Re\frac{\cos( \pi z)+i\sin( \pi z)}{1-\cos( \pi z)-i\sin( \pi z)}\\ &=-\Re\frac{\cos( \pi z)+i\sin( \pi z)}{1-\cos( \pi z)-i\sin( \pi z)} \frac{1-\cos( \pi z)+i\sin( \pi z)}{1-\cos( \pi z)+i\sin( \pi z)}\\ &=-\Re\frac{\cos( \pi z)+i\sin( \pi z)}{1-\cos( \pi z)-i\sin( \pi z)} \frac{1-\cos( \pi z)+i\sin( \pi z)}{1-\cos( \pi z)+i\sin( \pi z)}\\ &=-\Re\frac{(\cos( \pi z)+i\sin( \pi z))(1-\cos( \pi z)+i\sin( \pi z))}{(1-\cos( \pi z))^2+\sin^2( \pi z)}\\ &=-\Re\frac{\cos( \pi z)(1-\cos( \pi z)-\sin^2( \pi z))+i(....)}{2-2\cos( \pi z)}\\ &=-\frac{\cos( \pi z)-\cos^2( \pi z)-\sin^2( \pi z)}{2(1-\cos( \pi z))}\\ &=-\frac{\cos( \pi z)-1}{2(1-\cos( \pi z))}\\ &=\frac12\\ \end{array} $

This seems to imply that $g(z) =\frac14 z^2+az+b $ for some $a$ and $b$, which I find quite surprising.

I will continue under the assumption that this is correct.

If $z=0$, $b =g(0) =\frac1{\pi^2}\sum_{n=1}^\infty \frac1{n^2} =\frac1{\pi^2}\zeta(2) =\frac{1}{6} $.

If $z=1$,

$\begin{array}\\ g(1) &=\frac14+a+b\\ &=\frac14+a+\frac1{6}\\ &=\frac{5}{12}+a\\ &=\frac1{\pi^2}\sum_{n=1}^\infty \frac1{n^2}\cos(n\pi)\\ &=\frac1{\pi^2}\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\\ &=-\frac1{\pi^2}\frac12\zeta(2)\\ &=-\frac1{12}\\ \end{array} $

so $a =-\frac12 $.

As a check, if $z=2$,

$\begin{array}\\ g(2) &=1+2a+b\\ &=1+2a+\frac1{6}\\ &=\frac{7}{6}+a\\ &=\frac1{\pi^2}\sum_{n=1}^\infty \frac1{n^2}\cos(2n\pi)\\ &=\frac1{\pi^2}\sum_{n=1}^\infty \frac{1}{n^2}\\ &=\frac1{\pi^2}\zeta(2)\\ &=\frac1{6}\\ \end{array} $

so $2a+\frac76 =\frac16 $ or $a= -\frac12 $.

Therefore

$\begin{array}\\ G(x, y) &=g(x-y)+g(x+y)\\ &=\frac14((x-y)^2+(x+y)^2))-\frac12((x-y)+(x+y))+2\frac1{6}\\ &=\frac12(x^2+y^2)-x+\frac13\\ \end{array} $.

Answer 2

Marty Cohen is correct in his approach to this $G(x,y)$ but there is the sign error in the trig expressions.

Defined on the interval $0 < t < P$, it is fairly easy to demonstrate that the function: \begin{equation} h(t)=\pi^2 \left[ \left( \frac{t}{P} \right)^2 -\left( \frac{t}{P} \right) + \frac{1}{6} \right] \end{equation} has a Fourier series expansion of: \begin{equation} h(t) = \sum\limits_{n = 1}^{\infty} \frac{ \cos\left( 2 \pi \frac{n}{P} t \right) }{n^2} \end{equation} The polynomial, $h(t)$, is Bernoulli Polynomial, $B_2(\frac{t}{P})$ which has been scaled appropriately.

Using $h(t)$ above and setting $P=2$ the closed form of $g(z)$ above is given by: \begin{equation} g(z) = \left[ \left(\frac{z}{2}\right)^2 - \left(\frac{z}{2}\right) + \frac{1}{6} \right] \end{equation} This is exactly the same value of $g(z)$ found above, so his exposition on the approch is correct.

Proceeding with the reasoning of Marty Cohen but with the sign correction for $cos(a)cos(b)$: \begin{align} G(x,y) &= g(x-y) + g(x+y) \\ &= \left[ \left( \frac{x-y}{2} \right)^2 - \left( \frac{x-y}{2} \right) + \frac{1}{6} \right] \\ &+ \left[ \left( \frac{x+y}{2} \right)^2 - \left( \frac{x+y}{2} \right) + \frac{1}{6} \right] \\ &= \left[ \frac{x^2}{4} - xy + \frac{y^2}{4} - \frac{x-y}{2} + \frac{1}{6} \right] \\ &+ \left[ \frac{x^2}{4} + xy + \frac{y^2}{4} - \frac{x+y}{2} + \frac{1}{6} \right] \\ &= \frac{x^2 + y^2}{2} -x +\frac{1}{3} \end{align} where the domain for $x$ and $y$ are:

• $0 \le (x + y) \le 2$
• $0 \le (x - y) \le 2$

This is different way to answer to the question: "What is G(x,y) in closed form?" It is not in keeping with the the essence of the question posed by the OP so it is not a proper answer.

None the less, this confirms Marty Cohen's approach (less the sign error) is correct.