Fourier series of |t| on (-π,π) with period 2π

[ediggity]

Homework Statement

f(t) defined by f(t) = |t| for (-pi,pi) and f(t+2pi)=f(t)

the graph is just ^^^

where w=2pi/T = 1

Homework Equations

Periodic function using Trigonometric from

Even Function f(t) = (1/2)anot + (the sum from n=1 to inf) (an)*COS(nwt), where an = 4/T Integrated from 0 to T/2 f(t)*COS(nwt)dt, where T= 2pi

The Attempt at a Solution

My answer: I used integration by parts and calculated pi/2 +(the sum from n=1 to inf) (4/(pi)n^2)*COS(nt),

where anot/2 = 1/T integrated from -T/2 to T/2 f(t)dt, anot=pi

The book answer has pi/2- [4/pi *the sum from n=1 to inf (1/(2n-1)^2*COS(2n-1)t

Can anyone tell me if I basically have the same thing?

Thanks.

 

[HallsofIvy]

No, they are not at all the same thing. Your sum has cos(nt) for all n. The second sum only odd integers times t.