Find the Fourier series for the function
f(x) = 1/cos(x)
on the segment [-pi/4, pi/4].
A Fourier series for a function f(x) with period 2L has the form:
(a0/2) + SUM(n=0 to infinity) [ an*cos((n*pi*x)/L) + bn*sin((n*pi*x)/L) ], where:
an = (1/L) INTEGRAL (-L to L) f(x)cos((n*pi*x)/L) dx for n = 0, 1, 2, ....
bn = (1/L) INTEGRAL (-L to L) f(x)sin((n*pi*x)/L) dx for n = 0, 1, 2, ...
The Attempt at a Solution
By treating the function as (pi/2)-periodic, I used the equation above to get an expression for a0:
a0 = (4/pi) ln [(sqrt(2)+1)/(sqrt(2) - 1)]
Also, because f(x) is an even function, we can deduce without any calculation that bn will be zero for all n. So the only thing is to work out an.
But this is far as I've got. By using the formula above, I have this expression for an:
an = (4/pi) INTEGRAL (-pi/4 to pi/4) (cos(4nx)/cos(x)) dx
I don't know how to solve this integral. I know how to do it if n is some small integer; for instance, I can find a1 by integrating cos(4x)/cos(x), using trigonometric identities, but I can't see how to generalise the method. Can anyone help?
P.S. Sorry I don't know how to use latex, I did try but it came out as a mess.