Still revising for exams! Looking at the following, any help would be appreciated!
Consider the function f_i=PC(2PI), i=1,2
f_1(x)=x f_2(x)=x^2 for 0<=x<2PI
Show that f_1(x) ~ PI- 2 SUM [(sin(nx))/n] (n=1 to oo)
Given that the function is on (-PI,PI] and is 2PI periodic
Calculate the Fourier Series
xcosx~-(sinx)/2 + 2 SUM [((-1)^k)(k/((k^2)-1)]sin kx (k=2 to oo)
For the first I think I should be working from the fact that
x~2 SUM [(-1)^(n-1).(Sin nx)/n for -PI<x<PI
but I could be wrong?
It gives almost the same but I'm not sure how or what the differences are.
For the first part I can calculate x~2 SUM [(-1)^(n-1).(Sin nx)/n for -PI<x<PI by the following
compute the fourier coefficients for the function gives a_n=0
Then calculate . The is works out to be
whcih gives from the formula
gives x~2 SUM [(-1)^(n-1).(Sin nx)/n
I'm not sure how this works when you are calculating it with funtions?