# Fourier Series Representation $e^{ax}$

a) Compute the full Fourier series representation of $f(x) = e^{ax}, −π ≤ x < π.$

b) By using the result of a) or otherwise determine the full Fourier series expansion for the function $g(x)=\sinh(x), −π ≤ x < π.$

For part a) this is what I did and I'm pretty sure it's all correct. (Please let me know if it isn't!)

\begin{align} a_0\ &=\frac{1}{π}\int_{-π}^{π} f(x)\cos(0x)\;dx\\ &=\frac{1}{π}\int_{-π}^{π} e^{ax}\;dx=\frac{1}{π}\left[\frac{e^{ax}}{a}\right]^π_{-π}=\frac{e^{aπ}-e^{-aπ}}{aπ}\\ a_m\ &=\frac{1}{π}\int_{-π}^{π} e^{ax}\cos(mx)\;dx\\ &=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{ae^{ax}\sin(mx)}{m} dx\\ &=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{ae^{ax}\cos(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{a^2e^{ax}\cos(mx)}{m^2} dx\\ &=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}+\frac{ae^{ax}\cos(mx)}{m^2}\right]^π_{-π}-\frac{a^2}{m^2}a_m\\ a_m\left(\frac{m^2+a^2}{m^2}\right)&=\frac{ae^{aπ}(-1)^m-e^{-aπ}(-1)^m}{πm^2}\\ a_m&=\frac{(-1)^m\left(ae^{aπ}-e^{-aπ}\right)}{π\left(m^2+a^2\right)} \end{align}

Then \begin{align} b_m&=\frac{1}{π}\int_{-π}^{π} e^{ax}sin(mx)\;dx\\ &=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}\right]^π_{-π}+\frac{1}{π}\int_{-π}^{π} \frac{ae^{ax}\cos(mx)}{m}\;dx\\ &=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{ae^{ax}\sin(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{a^2e^{ax}\sin(mx)}{m^2}\;dx\\ &=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}+\frac{ae^{ax}\sin(mx)}{m^2}\right]^π_{-π}-\frac{a^2}{m^2}b_m\\ b_m\left(\frac{m^2+a^2}{m^2}\right)&=\frac{(-1)^{m+1}\left(e^{aπ}-e^{-aπ}\right)}{πm}\\ b_m&=\frac{m(-1)^{m+1}\left(e^{aπ}-e^{-aπ}\right)}{π\left(m^2+a^2\right)} \end{align}

so \begin{align} f(x)&=\frac{1}{2}a_0+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx)\\ e^{ax}&=\frac{e^{aπ}-e^{-aπ}}{π}\left[\frac{1}{2a}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+a^2}(\cos(nx)-n\sin(nx))\right]\\ e^{ax}&=\frac{2}{π}\sinh(aπ)\left[\frac{1}{2a}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+a^2}(\cos(nx)-n\sin(nx))\right] \end{align}

but for part b) I wasn't sure of how to get the answer from part a) so instead I did this. I am also unsure if this is right. How do you do part b) using part a) ?

\begin{align} g(x)&=\sinh(x)\\ a_0\ &=\frac{1}{π}\int_{-π}^{π}\sinh(x)\cos(0x)\;dx\\ &=\frac{1}{π}\left[\cosh(x)\right]^π_{-π}=0\\ a_m\ &=\frac{1}{π}\int_{-π}^{π}\sinh(x)\cos(mx)\;dx\\ &=\frac{1}{π}\left[\frac{\sinh(x)\sin(mx)}{m}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π}\frac{\cosh(x)\sin(mx)}{m}\;dx\\ &=\frac{1}{π}\left[\frac{sinh(x)\sin(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{\cosh(x)\cos(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{\sinh(x)\cos(mx)}{m^2}\\ a_m\left(\frac{m^2+1}{m^2}\right)&=\frac{1}{π}\left[\frac{\sinh(x)sin(mx)}{m}+\frac{\cosh(x)\cos(mx)}{m^2}\right]^π_{-π}=0 \end{align}

Then \begin{align} b_m&=\frac{1}{π}\int_{-π}^{π}\sinh(x)\sin(mx)\;dx\\ b_m\left(\frac{m^2+a^2}{m^2}\right)&=\left[\frac{-\sinh(x)(-1)^{m}}{πm}\right]^π_{-π}\\ b_m&=\frac{-2m\sinh(π)(-1)^m}{π(m^2+1)} \end{align}

so \begin{align} g(x)&=\frac{1}{2}a_0+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx)\\ \sinh(x)&=\frac{-2\sinh(π)}{π}\sum_{n=1}^{\infty}\frac{(-1)^nn\sin(nx)}{n^2+1} \end{align}

• Congrats! You posted a lot of questions and never accepted an answer! – Jack D'Aurizio Sep 28 '16 at 15:46

We have: $$\int_{-\pi}^{\pi}e^x \sin(nx)\,dx = \frac{2n}{1+n^2}(-1)^{n+1}\sinh \pi,$$ $$\int_{-\pi}^{\pi}e^x \cos(nx)\,dx = \frac{2}{1+n^2}(-1)^{n}\sinh \pi,$$ hence:
$$e^{x} = \frac{\sinh \pi}{\pi}+\frac{2\sinh \pi}{\pi}\sum_{n\geq 1}(-1)^n\frac{\cos(nx)-n\sin(nx)}{1+n^2}$$ and since $\sinh x$ is just the odd part of $e^x$ it follows that: $$\sinh x = \frac{2\sinh \pi}{\pi}\sum_{n\geq 1}(-1)^{n+1}\frac{n\sin(nx)}{1+n^2}.$$ For $e^{ax}$ we have: $$e^{ax} = \frac{\sinh(a\pi)}{a\pi}+\frac{2\sinh(a \pi)}{\pi}\sum_{n\geq 1}(-1)^n\frac{a\,\cos(nx)-n\sin(nx)}{a^2+n^2}.$$
• For which values of $a$ does this work? – clathratus Jan 1 at 8:20
$\sinh x=\frac{e^x-e^{-x}}{2}$, so if you know the representations for $e^x$ and $e^{-x}$, you can find the one for $\sinh$