# Thread: Real Analysis stuff

1. ## Real Analysis stuff

Hi guys,

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.

2. Originally Posted by musicmental85
Hi guys,

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.

3. 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

b_n= [2(-1)^(n-1)]/n

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?

### fourier series of xcosx from -pie to pie

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