**Alexander Frolkin**

**3 May 2001**

Our aim here is to derive value of
for even .
From Fourier theory, we have:

where is periodic in with period , and

and

We start by finding the Fourier series for .

We need to find

We can see that this vanishes, since

Next, we need to find

We will first deduce a formula for

for . Integrating by parts,

We can easily find :

This gives

and

We can now see the pattern emerging and after some educated guesswork, we can say:

We are now in a position to formally prove the stated identity, by induction. We first assume that the result holds true for . We then have, from the recurrence relation:

Looking at the sine coefficient,

Similarly, for the cosine coefficient,

Hence, we can conclude that our guess is correct, by induction, .

To find , we note that for
, and
. Hence, the part vanishes
and we are left with

We now have

Now put and to obtain

We now look at the summand:

Substituting back,

and so we have

and therefore,

Similarly, we can deduce the value of by putting .

Now substitute into the Fourier series,

Rearranging,

Hence,

To find , put .

Plugging back into the Fourier series,

Rearrange,

Therefore,

Given enough patience and time, we can proceed as far as we like in the same way to find , having already found , .