they match, wow, i think my whole issue was treating the function as an odd function, my integration was indeed correct and the series match up. the function must be treated as an even function, and then its all good
thank you so much for your time guys, really appreciate the help
its ok, i have since done a0 again and am confident i got 0 and can understand that bit.
i think that this question may be just too long to post all the working up, especially when there's a half page of working to type up.
i've got 3 a values being a1 = 24/35pi, a2=0, a3=8/9pi, a4=0
again...
i have redone my a0, using a restricted domain of 0 to pi/6, then multiplying it to get the total area under the curve for the actual range.
This came out to be 8/pi.
correct?
yes I've tried a coupld of times now and i keep ending up at this horrendous thing :
(2/pi){cos(pi(n+6))[-1/(n(n+6)) - 1/(n(n-6)) ] }ummm, yeh, I'm lost
yes, that is what i started with, i wasn't too sure how to proceed with it so i used integration by parts (u=sin(6t), dv/dt = cos(nt) ) and that's how it happened.
well a0 is just a straightforward integration of sin(6t) between -pi and pi, which comes out as -(1/6)cos(6t), which comes out as (-1/6) - (-1/6), which = 0.for an, i started with the equation (1/L)int[ f(t)*cos(nt)].dt, for n>=1. (L is given as pi, and the region is -pi < t < pi.
i am...
Homework Statement
f(t)=sin(|6t|), −pi<t<pi
with f(t) = f(t+2pi)
Homework Equations
Show that the Fourier series for f(t) can be written as (24/pi) time the sum, from n=0 to infinity, of ( 1/( 36 - (2k+1)^2 ) )cos(2k+1)t.
The Attempt at a Solution
I have an answer of a0 being 0...