fourier series for exponentials Watch

smoothman
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#1
Report Thread starter 11 years ago
#1
hi, there are the fourier series i obtained for e^x, sinhx and coshx, please can you verify if they are correct.

(QUESTION A)

fourier series for e^x:
\displaystyle e^{x}=\frac{e^{\pi}-e^{-\pi}}{2\pi}+\sum_{n=1}^{\infty}\  left[\frac{(-1)^{n}(e^{\pi}-e^{-\pi})}{\pi(1+n^{2})}cos(nx)+\fra  c{(-1)^{n}n(e^{-\pi}-e^{\pi})}{\pi(1+n^{2})}sin(nx)\r  ight]

fourier series for sinh(x):
\displaystyle sinh(x) = \frac{2 \sinh(x)}{\pi}\sum_{n=1}^{\infty  }\left[\frac{(-1)^{n+1} (sin(nx))}{(1+n^2)}\right]

fourier series for cosh(x) = 0..

are these correct?

(QUESTION B)
i need some help working out the An and Bn of the fourier series for e^x:

for example for e^x, how can you show:

\displaystyle a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}e^{x}cos(nx)dx=\frac{(-1)^{n}(e^{\pi}-e^{-\pi})}{\pi(1+n^{2})}

my working doesnt seem to arrive . here is my working:

to integrate \frac{1}{\pi}\int_{-\pi}^{\pi}e^{x}cos(nx)dx

u= e^x, du = e^x ................ dv= cos(nx), v = (sin(nx)/n)
so \displaystyle uv - \int v \hspace{5} du = \frac{e^x(sin(nx))}{n} - \int \frac{e^x(sin(nx))}{n}

but the 2nd part (i.e \displaystyle \int vdu, is a never ending process since the e^x will never vanish. as a result i am not able to go on further with this problem. so could you please explain how we arrive at the final answer being: a_{n}=\frac{(-1)^{n}(e^{\pi}-e^{-\pi})}{\pi(1+n^{2})}
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SsEe
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 \displaystyle \sinh{x} = \frac{2 \sinh(\pi)}{\pi}\sum_{n=1}^{\inf  ty}\left[ \frac{n(-1)^{n+1} \sin(nx)}{(1+n^2)}\right]

 \displaystyle \cosh{x} = \frac{2 \sinh(\pi)}{\pi} \left( \frac{1}{2} + \sum_{n=1}^{\infty}\left[\frac{(-1)^{n} \cos(nx)}{(1+n^2)}\right] \right)

For the next bit, you're on the right lines. By repeated integration by parts (here I'm integrating e^x and differentiating cos(nx)) :



\displaystyle \int_{-\pi}^{\pi}e^{x}\cos(nx)dx\\

=\ [e^{x}\cos(nx)]_{-\pi}^{\pi} + n\int_{-\pi}^{\pi}e^{x} \sin(nx)dx \\

=\ (-1)^n(e^{\pi}-e^{-\pi}) + n \left( [e^{x}\sin(nx)]_{-\pi}^{\pi} - n\int_{-\pi}^{\pi}e^{x} \cos(nx)dx \right)\\

=\ (-1)^n(e^{\pi}-e^{-\pi}) - n^2 \int_{-\pi}^{\pi}e^{x} \cos(nx)dx \right)

Solving for \displaystyle \int_{-\pi}^{\pi}e^{x}\cos(nx)dx and dividing by \pi gives:

\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi}e^{x}\cos(nx)dx = \frac{(-1)^n(e^{\pi}-e^{-\pi})}{\pi(1+n^2)} = \frac{2(-1)^n \sinh{\pi}}{\pi(1+n^2)}
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DFranklin
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Funny how in the last few hours \sum \frac{1}{1+n^2} has come up twice in two unrelated posts!
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smoothman
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#4
Report Thread starter 11 years ago
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thanks a lot
i've worked it out properly now and i get the same answer as you except for one part:

\displaystyle \sinh{x} = \frac{2 \sinh(\pi)}{\pi}\sum_{n=1}^{\inf  ty}\left[ \frac{n(-1)^{n+1} \sin(nx)}{(1+n^2)}\right]

is there a typo in this series: i.e. look at the sum formula...
you have done

\displaystyle \sum_{n=1}^{\infty}\left[ \frac{n(-1)^{n+1} \sin(nx)}{(1+n^2)}\right]

where did the n(-1)^n+1 come from... is it not meant to be just (-1)^n+1 without the extra "n" at the front..

and most important question:
Can the fourier series for sinhx, coshx, and e^x be shown in a more simpler form? or is this the simplest form?
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SsEe
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\displaystyle

\int_{-\pi}^{\pi} \sinh(x)\sin(nx) dx\\

=\ [\cosh(x)\sin(nx)]_{-\pi}^{\pi} - n\int_{-\pi}^{\pi} \cosh(x)\cos(nx) dx\\

=\ -n \left( [\sinh(x)\cos(nx)]_{-\pi}^{\pi} + n\int_{-\pi}^{\pi} \sinh(x)\sin(nx) dx \right) \\

=\ -n(-1)^{n}(\sinh(\pi) - \sinh(-\pi)) - n^2 \int_{-\pi}^{\pi} \sinh(x)\sin(nx) dx

Solving for \displaystyle \int_{-\pi}^{pi} \sinh(x)\sin(nx) dx and dividing by \pi gives:

\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi} \sinh(x)\sin(nx) dx = \frac{-n(-1)^{n}(\sinh(\pi) - \sinh(-\pi))}{\pi(1+n^2)} = \frac{n(-1)^{n+1}2\sinh(\pi)}{\pi(1+n^2)}
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