Periodic Function Fourier Series: Proving with Trigonometric Equations

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Homework Statement



A periodic function of period [tex]2\pi[/tex] is defined by:
[tex]f(t)=\frac{t}{2} , 0<t<2\pi[/tex]

Show that the trigonometric Fourier series of f(t) is given by:
[tex]f(t)=\frac{\pi}{2} - \sum_{n=1}^{\infty} \frac{1}{n}sin(nt)[/tex]

Homework Equations


The Attempt at a Solution



I've gotten [tex]\frac{\pi}{4}[/tex] for constant C instead of [tex]\frac{\pi}{2}[/tex]
 
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Since function is odd, An = 0.

Bn = [tex]- \frac{1}{2\pi} \frac{2\pi}{n} = -\frac{1}{n}[/tex]

Thats for now...
 
My advice is to write:

[tex] f(t)=a_{0}+\sum_{n=-\infty}^{\infty}a_{n}\sin nt+b_{n}\cos nt[/tex]
 
[tex] <br /> f(t)=\sum_{n=-\infty}^{\infty} \frac{(1)^{n+1}}{n}\sin nt<br /> [/tex]
 
So you compute te A_n's and b_n is the usual way, what you have done with this?
 
Handwritten derivation on lined paper: a0 = (1/2π) integral from 0 to 2π of (t/2) dt, then (1/4π)[t²/2] from 0 to 2π, then (1/8π)(4π²), then π/2.


Handwritten Fourier series derivation on two sheets of lined paper: integrals for a_n with cos(nt) and b_n with sin(nt) over 0 to 2π, integration by parts, and the final series f(t) = π/2 − Σ (1/n) sin(nt).
 
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  • Handwritten derivation on lined paper: a0 = (1/2π) integral from 0 to 2π of (t/2) dt, then = (1/4π)[t²/2] from 0 to 2π, then = (1/8π)(4π²), then = π/2.
    2011-03-24 18.10.00.jpg
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  • Two photos of handwritten lined-paper calculations of Fourier coefficients: an a_n integral of t cos nt over 0 to 2π evaluated by integration by parts with the result 0, and a b_n integral of t sin nt giving -2π/n, ending with the series f(t) = π - sum of (2/n) sin(nt).
    2011-03-24 18.10.37.jpg
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