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What is the difference (if any) between using:

$\cos \left( \frac{n\pi x}{L} \right)$ and $ \sin \left( \frac{n\pi x}{L} \right)$

or using:

$\cos \left( \frac{2n\pi x}{L} \right)$ and $ \sin \left( \frac{2n\pi x}{L} \right)$

In a fourier series?

My guess would be that the first approaches faster to the actual $f(x)$ (in computing for example), but in the infinity sum they are both equvalent.

Thanks.

Daniel Duque
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  • You don't have a choice, the terms must have a period multiple of $L$, hence $2\pi$. (And by the way "faster approach" is surrealistic.) –  Sep 28 '17 at 10:40

1 Answers1

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Let $f(t)$ be a function with period $T$ then if we want to express $f(t)$ as a Fourier series

$$f(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left[a_n\cos(nt)+b\sin(nt) \right]$$

we can calculate the coefficients by

$$a_n = \frac{2}{T}\int_{0}^{T}f(t)\cos(nt)dt$$

$$b_n = \frac{2}{T}\int_{0}^{T}f(t)\sin(nt)dt.$$

If you take different expressions for cosine and sine you will need to adapt the formulas for the coefficients. Hence, I would always try to stick to one formula.

MrYouMath
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