Real Analysis stuff
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.
Are you asking to find the Fourier series? If that is the case post what you have done. These problems that a long time to do.
Originally Posted by musicmental85
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 http://www.mathhelpforum.com/math-he...39887410-1.gif. 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?