Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

I've been having little problems getting Fourier series of [itex]e^x[/itex].

I have given

[tex]

f(x) = e^{x}, x \in [-\pi, \pi)

[/tex]

Then

[tex]

a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} e^{x}\ dx = \frac{2\sinh \pi}{\pi}

[/tex]

[tex]

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

\frac{1}{\pi}\left\{\left[ \frac{1}{n}\ e^{x} \sin (nx)\right]_{-\pi}^{\pi} -

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

\frac{(-1)^{n+1}\left(e^{-\pi} - e^{\pi}\right)}{\pi(1+n^2)}

[/tex]

[tex]

b_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}e^{x}\sin (nx)\ dx = \frac{1}{\pi}\left\{

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

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

\frac{n(-1)^{n}\left(e^{-\pi} - e^{\pi}\right)}{\pi(1+n^2)}

[/tex]

So the Fourier series looks like this:

[tex]

S(f,x) = \frac{\sinh \pi}{\pi}\ +\ \frac{2\sinh \pi}{\pi}\sum_{n=1}^{\infty}

\frac{(-1)^{n+1}}{1+n^2}\left(\cos(nx) - n\sin(nx))

[/tex]

Anyway, our professor gave us another right (I hope so) result:

[tex]

S(f,x) = \frac{\sinh \pi}{\pi}\ +\ \frac{2\sinh \pi}{\pi}\sum_{n=1}^{\infty}

\frac{(-1)^{n}}{1+n^2}\left(\cos(nx) - n\sin(nx))

[/tex]

Obviously my series is just of opposite sign than it should be, but I can't find

the mistake, could you help me please?

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# Homework Help: Fourier series of e^x

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