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because normally the question i'd encountered has the same interval, eg -2

maybe we can do a shift of the function by substituting u=t-2

|dw:1343832213995:dw|
this can be transformed into regular Heaviside function

you mean\[f(t)=5(1-u_2(t))\]?

but that doesn't fix the intervals, which I do think need to be symmetric

why 1/5 ?

sorry ... (t - 2)/2

|dw:1343833073250:dw|

the Fourier expansion of left is given by
http://mathworld.wolfram.com/FourierSeriesSquareWave.html

|dw:1343833287038:dw|

I still don't see why you need the /2 part

that will make the interval of |dw:1343833578897:dw|

however that was not necessary ... i guess

but why does it have to be from -1 to 1?
can't it just be symmetric?

i guess yes i can be ... i thought unit would make easier.

I just did \(x=t-2\)\[A_0=\frac1{2(2)}\int_{-2}^2f(x)dx=\frac14\int_{-2}^05dx=\frac52\]

|dw:1343833717021:dw|

|dw:1343833858118:dw|

\[B_n=\frac52\int_{-2}^0\sin\left(n\pi x\over2\right)dx=\frac5{n\pi}\left(\cos(n\pi)-1\right)\]

doesn't anyone know how to plug this into wolf?

Plot[5/2 +
5*4/(2*Pi)*Sum[Sin[(2 n + 1)*Pi*(x)/2]/(2 n + 1), {n, 0, 10}], {x ,
0, 10}]

|dw:1343836422904:dw|
the example at wolframmathworld was for 0 to 2L