Expand f(x)=e^x in a Fourier series on (-l,l)
Expand $f(x)=e^x$ in a Fourier series on $(-l,l)$.
Explanation of Method
$e^x$ is neither even nor odd, so both cosine and sine terms appear. The standard formulas $\int e^{ax}\cos(bx)\,dx$ and $\int e^{ax}\sin(bx)\,dx$ give both coefficients directly in terms of $\sinh l$, since $e^l-e^{-l}=2\sinh l$ appears naturally from evaluating at the endpoints $\pm l$.
- Compute $a_0$ directly.
- Compute $a_n$ using the standard exponential-times-cosine antiderivative.
- Compute $b_n$ using the standard exponential-times-sine antiderivative.
Step 1 · Computing $a_0$
$$ a_0 = \frac1l\int_{-l}^l e^x\,dx = \frac1l\bigl(e^l-e^{-l}\bigr) = \frac{2\sinh l}l. $$
Step 2 · Computing $a_n$
Using $\int e^{x}\cos\frac{n\pi x}l\,dx=\dfrac{e^x}{1+(n\pi/l)^2}\left(\cos\frac{n\pi x}l+\frac{n\pi}l\sin\frac{n\pi x}l\right)$ and evaluating at $x=\pm l$ (where $\sin(n\pi)=0$, $\cos(n\pi)=(-1)^n$): $$ a_n = \frac1l\int_{-l}^le^x\cos\frac{n\pi x}l\,dx = \frac{2(-1)^n l\sinh l}{l^2+n^2\pi^2}. $$
Step 3 · Computing $b_n$
Similarly, using $\int e^x\sin\frac{n\pi x}l\,dx=\dfrac{e^x l^2}{l^2+n^2\pi^2}\left(\sin\frac{n\pi x}l-\frac{n\pi}l\cos\frac{n\pi x}l\right)$: $$ b_n = \frac1l\int_{-l}^le^x\sin\frac{n\pi x}l\,dx = \frac{2n\pi(-1)^{n+1}\sinh l}{l^2+n^2\pi^2}. $$ (Verified numerically at $l=\pi$: $a_0\approx7.3522=2\sinh\pi/\pi$, $a_1\approx-3.6761$, $b_1\approx3.6761$, matching the closed forms.)
Step 4 · Fourier series
$$ \boxed{e^x = \frac{\sinh l}l + \frac{2\sinh l}\pi\sum_{n=1}^\infty\frac{(-1)^n}{l^2+n^2\pi^2}\left[l\cos\frac{n\pi x}l - n\pi\sin\frac{n\pi x}l\right]} $$
Summary Table
| Property | Value |
|---|---|
| $a_0$ | $2\sinh l/l$ |
| $a_n$ | $\dfrac{2(-1)^nl\sinh l}{l^2+n^2\pi^2}$ |
| $b_n$ | $\dfrac{2n\pi(-1)^{n+1}\sinh l}{l^2+n^2\pi^2}$ |
Additional Notes
- Both coefficients decay like $1/n^2$ for large $n$, since $e^x$ is smooth (infinitely differentiable) on $(-l,l)$ — this rapid decay contrasts with the merely $1/n$ decay of Problem 327's series for the (only piecewise-smooth) sawtooth wave.
- Setting $l=\pi$ recovers the classical hyperbolic-sine expansion of $e^x$ found in many textbooks, with $\sinh\pi\approx11.549$ controlling the overall scale of every coefficient.
- Splitting $e^x=\cosh x+\sinh x$ and comparing with Problem 334's $\cosh(ax)$ expansion (at $a=1$) isolates the purely even ($a_n$-only) part of this series; the odd part corresponds to a companion $\sinh(x)$ expansion.