Expand f(x)=cosh(ax) in a Fourier series on (- pi,pi)
Expand $f(x)=\cosh(ax)$ in a Fourier series on $(-\pi,\pi)$.
Explanation of Method
$\cosh(ax)$ is even, so only cosine terms appear. Writing $\cosh(ax)=\frac12(e^{ax}+e^{-ax})$ and applying the standard exponential-cosine antiderivative to each piece combines cleanly into a single coefficient involving $\sinh(a\pi)$.
- Note evenness, so $b_n=0$.
- Compute $a_0$ directly.
- Compute $a_n$ by splitting into exponentials and combining.
Step 1 · Symmetry and $a_0$
$\cosh(ax)$ is even, so $b_n=0$ for all $n$. $$ a_0 = \frac2\pi\int_0^\pi\cosh(ax)\,dx = \frac2{\pi a}\sinh(a\pi). $$
Step 2 · Computing $a_n$
Writing $\cosh(ax)=\frac12(e^{ax}+e^{-ax})$ and using $\int e^{ax}\cos(nx)\,dx=\frac{e^{ax}}{a^2+n^2}[a\cos(nx)+n\sin(nx)]$ on $[0,\pi]$: $$ \int_0^\pi e^{ax}\cos(nx)\,dx = \frac a{a^2+n^2}\bigl[e^{a\pi}(-1)^n-1\bigr], \qquad \int_0^\pi e^{-ax}\cos(nx)\,dx = \frac a{a^2+n^2}\bigl[1-e^{-a\pi}(-1)^n\bigr]. $$ Adding and multiplying by $\frac1\pi$: $$ a_n = \frac a{\pi(a^2+n^2)}(-1)^n\bigl(e^{a\pi}-e^{-a\pi}\bigr) = \frac{2a(-1)^n\sinh(a\pi)}{\pi(a^2+n^2)}. $$
Step 3 · Fourier series
$$ \boxed{\cosh(ax) = \frac{\sinh(a\pi)}\pi\left[\frac1a + 2a\sum_{n=1}^\infty\frac{(-1)^n\cos(nx)}{a^2+n^2}\right]} $$
Summary Table
| Property | Value |
|---|---|
| $a_0$ | $2\sinh(a\pi)/(\pi a)$ |
| $a_n$ | $\dfrac{2a(-1)^n\sinh(a\pi)}{\pi(a^2+n^2)}$ |
| $b_n$ | $0$ |
Additional Notes
- The coefficients here decay like $1/n^2$, exactly matching the decay rate of $a_n$ in Problem 333's $e^x$ expansion — unsurprising, since $\cosh(ax)$ is just the even half of $e^{ax}$.
- Setting $a=1$ recovers precisely the even (cosine) part of Problem 333's expansion with $l=\pi$, isolating $a_n=\dfrac{2(-1)^n\sinh\pi}{\pi(1+n^2)}$ from that more general result.
- Because $\cosh(ax)$ has no jump discontinuities anywhere (unlike Problems 328–329's piecewise sawtooth/parabolic waves), its Fourier coefficients decay by a full extra power of $n$ compared to those merely-continuous, corner-having functions.