MHB Is the half interval Fourier series for f(x)=x over (0,L) correct?

[ognik]

Please help me find my mistake - "find the Sine F/series of f(x)=x over the half-interval (0,L)"

I get $ b_n=\frac 2L \int_{0}^{L}x Sin \frac{2n\pi x}{L} \,dx $

$ = \frac 2L \left[ x(-Cos \frac{2n\pi x}{L}. \frac{L}{2n\pi x}\right] + \frac {1}{n\pi} \int_{0}^{L} Cos \frac{2n\pi x}{L} \,dx$ ... and the 2nd term evals to 0

$ = -\frac{L}{n\pi}Cos 2n \pi $ but the book gives something very different.

 

[I like Serena]

Please help me find my mistake - "find the Sine F/series of f(x)=x over the half-interval (0,L)"

I get $ b_n=\frac 2L \int_{0}^{L}x Sin \frac{2n\pi x}{L} \,dx $

$ = \frac 2L \left[ x(-Cos \frac{2n\pi x}{L}. \frac{L}{2n\pi x}\right] + \frac {1}{n\pi} \int_{0}^{L} Cos \frac{2n\pi x}{L} \,dx$ ... and the 2nd term evals to 0

$ = -\frac{L}{n\pi}Cos 2n \pi $ but the book gives something very different.

Hi ognik,

The first term should be:
$$\frac 2L \left[ x(-\cos \frac{2n\pi x}{L} \cdot \frac{L}{2n\pi})\right]_0^L
=-\frac 1{n\pi} \left[ x \cos \frac{2n\pi x}{L}\right]_0^L
=-\frac 1{n\pi} [L\cdot 1 - 0\cdot 1]
= -\frac L{n\pi}
$$

 

[ognik]

Thanks ILS (I have a typo in my last line, left out the 1/L)

The book gives $x=\frac 4\pi \sum_{n-0}^{\infty} \frac{1}{2n+1} Sin \frac{(2n+1)\pi x}{L} $ (I assume they mean f(x)=)

 

[I like Serena]

Thanks ILS (I have a typo in my last line, left out the 1/L)

The book gives $x=\frac 4\pi \sum_{n-0}^{\infty} \frac{1}{2n+1} Sin \frac{(2n+1)\pi x}{L} $ (I assume they mean f(x)=)

Looks like there is a mix-up.
That series belongs to f(x)=1.