Question about the lower limit of sums in Fourier series

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July 27th 2010, 02:23 PM
arbolis

Question about the lower limit of sums in Fourier series

According to many websites including wikipedia (Fourier series - Wikipedia, the free encyclopedia), the fourier series of f is defined as $\frac{a_0}{2} + \sum_{n=1}^\infty \, [a_n \cos(nx) + b_n \sin(nx)]$ .
According to my class notes and wikipedia (Fourier series - Wikipedia, the free encyclopedia), $f(x)=\sum _{n= -\infty} ^{+\infty} \hat f(n)e^{inx}$ .
I don't understand why there's a difference of the lower limit of the sums. I'd love to get an explanation.
July 28th 2010, 02:17 AM
Ackbeet

The cosine function is even, and the sine function is odd. Therefore, negative n's are either superfluous (in the cosine case) or can be captured in the b_n's (in the sine case). The exponential function is neither even nor odd. Therefore, to capture all the required information, you must sum over all the integers.
July 28th 2010, 11:04 AM
arbolis

Ok so if I understand well, both definitions are equivalent, right?
July 28th 2010, 11:40 AM
Opalg

Quote:

Originally Posted by arbolis

Ok so if I understand well, both definitions are equivalent, right?

Yes, they are equivalent.
You can convert the real Fourier seies into the complex one by writing $\cos nx = \frac12(e^{inx} + e^{-inx})$ and $\sin nx = \frac1{2i}(e^{inx} - e^{-inx})$ . Notice that when you do that, you get –n as well as n in the complex exponentials. That is where the negative terms in the complex summation come from.

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