MHB Integrating Fouries series problem

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The discussion centers on the Fourier sine series expansion of the Dirac delta function and its integration to derive the cosine expansion of a square wave function. The initial equation provided for the delta function is correct, but a discrepancy arises in the sign of the result after integration. The correct form of the cosine expansion should include a positive sign in front of the first term, which indicates a potential typo in the original equation. Participants confirm the majority of the calculations are accurate, emphasizing the importance of careful attention to signs in mathematical expressions. The conversation highlights the need for precision in mathematical derivations.
ognik
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As the 2nd part of a question, we start with the Fourier sin series expansion of dirac delta function $\delta(x-a)$ in the half-interval (0,L), (0 < a < L):

$ \delta(x-a) = \frac{2}{L} \sum_{n=1}^{\infty} sin \frac{n \pi a}{L} sin \frac{n \pi x}{L} $

The questions goes on "By integrating both sides of this eqtn from 0 to x, show that the cos expansion of the square wave $ f(x) =\begin{cases}0,\; 0\le x \lt a \\ 1,\; a \lt 0 \lt L\end{cases}$

is $ f(x) = \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) - \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) sin(\frac{n\pi x}{L}) $

But when integrating both sides I get:

$ \int_{0}^{x}\delta(x-a).1 \,dx = \frac{2}{L} \sum_{n=1}^{\infty} sin \frac{n \pi a}{L} \int_{0}^{x} sin \frac{n \pi x}{L} \,dx$

$ = \frac{2}{L} \sum_{n=1}^{\infty} sin \frac{n \pi a}{L} (\frac{L}{n\pi}) [-cos \frac{n \pi x}{L}]^x_0 $

$ = -\frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) - \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) cos(\frac{n\pi x}{L}) $

It looks to me that I am correct, could someone please check this?
 
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ognik said:
But when integrating both sides I get:

$ \int_{0}^{x}\delta(x-a).1 \,dx = \frac{2}{L} \sum_{n=1}^{\infty} sin \frac{n \pi a}{L} \int_{0}^{x} sin \frac{n \pi x}{L} \,dx$

$ = \frac{2}{L} \sum_{n=1}^{\infty} sin \frac{n \pi a}{L} (\frac{L}{n\pi}) [-cos \frac{n \pi x}{L}]^x_0 $

$=$ -$\frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) - \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) cos(\frac{n\pi x}{L}) $

It looks to me that I am correct, could someone please check this?

The first sign of you result is wrong. It should be $$\frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} \sin(\frac{n\pi a}{L}) - \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} \sin(\frac{n\pi a}{L}) \cos(\frac{n\pi x}{L}) $$
The rest is correct! (Smile)
ognik said:
The questions goes on "By integrating both sides of this eqtn from 0 to x, show that the cos expansion of the square wave $ f(x) =\begin{cases}0,\; 0\le x \lt a \\ 1,\; a \lt 0 \lt L\end{cases}$

is $ f(x) = \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) - \frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n} sin(\frac{n\pi a}{L}) sin(\frac{n\pi x}{L}) $
So there must be a typo...
 

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