Hey! I have spilt the question up into parts and I have tried to show my sad attempt, please help.

a) Show the partial sum S = 4/pi Sum from n=1 to N ((sin((2n-1)t)/2n-1)) which may also be written as 2/pi integral from 0 to x (sin(2Nt)/sin(t))dt has extrema at x= m*pi/2N where m is any positive integer except m=2kN, k also integer.

Solution: The derivative of S = 2/pi sin(2Nx)/sin(x) = 0 where sin(2Nx)=0, sin(x) cannot equal zero. sin(x) = 0 where x is a multiple of pi. Therefore, sin(2Nx)=0 where x=m*pi/2N; however sin(x) cannot equal zero, sin(m*pi/2N) cannot equal zero so m is any positive integer except m=2kN, k also integer. Is this complete?!

b) Consider the first extrema to the right of the discontinuity, located at x=pi/2N. By considering a suitable small angle formula show that the value of the sum at this point S(pi/2N) is approximately 2/pi integral from 0 to pi (sin(u)/u)du.

Solution: I'm not sure which small angle formula i'm meant to considering?! Taylor series of sin? or how to consider it?

I see that

S(pi/2N) = 4/pi (sin(pi/2N)+sin((3*pi)2N)/3+sin((5*pi)/2N)/5+...

=2/pi(pi/N((sin(pi/2N))/(pi/2N)+sin((3*pi)2N)/((3*pi)/2N)+sin((5*pi)/2N)/((5*pi)/2N)+...

I think that

(pi/N((sin(pi/2N))/(pi/2N)+sin((3*pi)2N)/((3*pi)/2N)+sin((5*pi)/2N)/((5*pi)/2N) converges to integral from 0 to pi (sin(u)/u)du when N is large, could someone explain this please?

Also, could you explain how 2/pi integral from 0 to pi (sin(u)/u)du = 2 - pi^2/9 + pi^4/300- ...

I know that the taylor of sin(x) = x - x^3/3! + x^5/5! - ...

I get the wrong coefficients.

c) Evaluating this numerically S(pi/2N) = approx. 1.1790

independently of N. I am meant to comment on the accuracy of Fourier series at discontinuities and say whether the Fourier Convergence theorem been violated? So there is an error of approx 18% ... Pretty lost here except for a reference of Gibbs' phenomenon.

Any ideas please?

Thankyou x