# Thread: Fourier Series, partial differential equations, Gibbs Phenomenon?

1. ## Fourier Series, partial differential equations, Gibbs Phenomenon?

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.

Thankyou x

2. Originally Posted by dazed
I think that
$\displaystyle \pi/N\Bigl(\frac{sin(\pi/(2N))}{\pi/(2N)} + \frac{sin(3\pi/(2N))}{3\pi/(2N)} + \frac{sin(5\pi/(2N))}{5\pi/(2N)} + \ldots\Bigr)$ converges to $\displaystyle \int_0^\pi\frac{\sin u}udu$ when N is large, could someone explain this please?
The series on the left is a Riemann sum approximation for the integral on the right.

Originally Posted by dazed
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.
$\displaystyle \frac2\pi\int_0^\pi\Bigl(1 - \frac{u^2}{3!} + \frac{u^4}{5!} - \ldots\Bigr)du = \frac2\pi\Bigl[u - \frac{u^3}{18} + \frac{u^5}{600} - \ldots\Bigr]_0^\pi$

Originally Posted by dazed
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.
That is exactly what the Gibbs phenomenon says: where a function has a jump discontinuity, the Fourier series will overshoot as it approaches the discontinuity. As the number of terms in the Fourier series increases, the amount of overshoot will converge to a constant percentage (approximately 17.9) of the amount of the jump.