Fourier series expansion of Sin(x)

[therimalaya]

May be simple, but I'm getting problem with doing Fourier series expansion of Sin(x) for -pi\leqx\leqpi

 

[CompuChip]

What does the formula for the Fourier expansion of a general function f(x) look like?

 

[therimalaya]

What does the formula for the Fourier expansion of a general function f(x) look like?

f(x)=a_o+\sum(a_ncos(nx)+b_nsin(nx))

where,
a_o=\frac{1}{\pi}\intf(x) dx
a_n=\frac{1}{2\pi}\intf(x)cos(nx) dx
b_n=\frac{1}{2\pi}\intf(x)sin(nx) dx

 

[Defennder]

Isn't it supposed to be 1/pi for both a_n and b_n and 1/2pi for a_0? And in the formulae for a_n, b_n, these are supposed to be definite integrals not indefinite ones.

 

[CompuChip]

Anyway, for the sine you can either calculate all these coefficients, but you can also just read off the values from the general expression.

HallsOfIvy, you may be right, didn't notice that. Asked the mods to look into it and maybe split the thread.[/size]

 

[therimalaya]

Isn't it supposed to be 1/pi for both a_n and b_n and 1/2pi for a_0? And in the formulae for a_n, b_n, these are supposed to be definite integrals not indefinite ones.

Yes there is definite intergral from -infinity to +infinity. I missed that.
Sorry

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[therimalaya]

Anyway, for the sine you can either calculate all these
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HallsOfIvy, you may be right, didn't notice that. Asked the mods to look into it and maybe split the thread.[/size]

I've tried that, May be I might have done some mistake, The integral term while calculating a_o, a_n and b_n vanishes and results zero. That is why I kept this in this forum. Any way I'll try that.
Thank You

 

[CompuChip]

You mean they all vanished? Check again the ones for n = -1 and n = 1.
Also, check the integration boundaries. They are not (-\infty, \infty).