Differentiate Fourier Sine Series

1. Oct 6, 2007

Mindscrape

The problem is that if f(x) is continuous function, except for a jump discontinuity at $$x = x_0$$, where $$f(x_0^-) = \alpha$$ and $$f(x_0^+) = \beta$$, and df/dx is piecewise smooth, determine the Fourier cosine series of df/dx in terms of the Fourier sine series coefficients of f(x).

Let me preface this by saying that I split f(x) into f1 from 0 to x0 and f2 from x0 to L. So that would mean, which might not necessarily be true, that f'(x) would f1' from 0 to x0 and f2' from x0 to L. Is this alright so far? I could see the derivative gaining extra discontinuities that I am not accounting for properly, in which case I don't know how to account for them. But if it is okay, then read on.

Start off by saying that f'(x) is a FCS

$$f'(x) = \hat{A}_0 + \sum_1^\infty \hat{A}_n cos \frac{n \pi x}{L}$$

*I put hats on just to clarify these are not regular FCS coefficients. So that would mean that

$$\hat{A}_0 = \frac{1}{L}[ \int_0^{x_0} f_1' dx + \int_{x_0}^L f_2' dx] = \frac{1}{L} [(f_1(x_0) - f(0)) + (f_2(L) - f_2(x_0)]$$

so that would end up being

$$\hat{A}_0 = \frac{1}{L} [(\alpha - f(0)) + (f_2(L) - \beta)]$$

Then do a similar treatment for other coefficient

$$\hat{A}_n = \frac{2}{L} \int_0^L f'(x) cos\frac{n \pi x}{L} dx$$

Here is another part I am uncertain about, but I'm pretty sure it is okay. Integration by parts can only be used on continuous functions with continuous derivatives, but I should be alright using IBP on f1' from 0 to x0 and f2' from x0 to L. So, use integration by parts (u = cosnπx/L dv = f'(x)) on each of the functions to get

$$\hat{A}_n = \frac{2}{L} [ (\alpha cos\frac{n\pi\alpha}{L} - f_1(0)) + (f_2(L) - \beta cos\frac{n\pi\beta}{L}) + \int_0^{x_0} \frac{n\pi}{L} sin\frac{n\pi x}{L} f_1(x) + \int_{x_0}^L \frac{n\pi}{L} sin\frac{n\pi x}{L} f_2(x)]$$

Quick detour to the FSS coefficient of f(x)

$$f(x) = \sum_0^\infty B_n sin\frac{n\pi x}{L}$$

which means that

$$B_n = \frac{L}{2}[\int_0^{x_0} {L} sin\frac{n\pi x}{L} f_1(x) + \int_{x_0}^L sin\frac{n\pi x}{L} f_2(x)]$$

Substitute B_n into A_n and it should be

$$\hat{A}_n = \frac{2}{L} [ (\alpha cos\frac{n\pi\alpha}{L} - f_1(0)) + (f_2(L) - \beta cos\frac{n\pi\beta}{L})] + \frac{n\pi}{L} B_n$$

Would this be correct?

Last edited: Oct 6, 2007