Expand f(x)=sin(ax) in a Fourier series on (- pi,pi)
Expand $f(x)=\sin(ax)$ in a Fourier series on $(-\pi,\pi)$, where $a$ is not an integer.
Explanation of Method
Because $\sin(ax)$ is odd, only sine terms appear. The product-to-sum identity turns $\int\sin(ax)\sin(nx)\,dx$ into a combination of $\sin((a\pm n)\pi)$, which (since $a$ is not an integer) does not vanish, giving a genuinely infinite, non-terminating expansion — unlike the integer-frequency case of Problem 331.
- Note oddness, so $a_0=a_n=0$.
- Compute $b_n$ via the product-to-sum identity and the expansion $\sin((a\pm n)\pi)=\pm\sin(a\pi)\cos(n\pi)$.
- Combine into a single closed-form coefficient with a pole structure at $a=\pm n$.
Step 1 · Symmetry
$f(x)=\sin(ax)$ is odd, so $a_0=a_n=0$; only sine terms appear.
Step 2 · Computing $b_n$
$$ b_n = \frac2\pi\int_0^\pi\sin(ax)\sin(nx)\,dx = \frac1\pi\int_0^\pi\bigl[\cos((a-n)x)-\cos((a+n)x)\bigr]dx = \frac1\pi\left[\frac{\sin((a-n)\pi)}{a-n}-\frac{\sin((a+n)\pi)}{a+n}\right]. $$ Since $\sin(n\pi)=0,\cos(n\pi)=(-1)^n$: $\sin((a\mp n)\pi)=\sin(a\pi)\cos(n\pi)=(-1)^n\sin(a\pi)$ (both signs give the same value). So: $$ b_n = \frac{(-1)^n\sin(a\pi)}\pi\left[\frac1{a-n}-\frac1{a+n}\right] = \frac{(-1)^n\sin(a\pi)}\pi\cdot\frac{2n}{a^2-n^2}. $$
Step 3 · Fourier series
$$ \boxed{\sin(ax) = \frac{2\sin(a\pi)}\pi\sum_{n=1}^\infty\frac{n(-1)^n}{a^2-n^2}\sin(nx)} $$ (Verified numerically at $a=1.7$: computed $b_1\approx0.27251$, $b_2\approx0.92799$, $b_3\approx-0.25288$, matching the closed form to 6+ significant figures.)
Summary Table
| Property | Value |
|---|---|
| Coefficient | $b_n=\dfrac{2n(-1)^n\sin(a\pi)}{\pi(a^2-n^2)}$ |
| Fourier series | $\sin(ax)=\dfrac{2\sin(a\pi)}\pi\sum\dfrac{n(-1)^n}{a^2-n^2}\sin(nx)$ |
Additional Notes
- As $a\to m$ (an integer), $\sin(a\pi)\to0$ while the pole at $n=m$ makes that single term finite in the limit, and the series collapses to $\sin(mx)$ itself — recovering the trivial case of Problem 331 as a limiting case, even though the formula here assumes $a\notin\mathbb Z$.
- This is the sine-series counterpart of the cosine expansion for $\cosh(ax)$ in Problem 334: both involve a coefficient of the form $\frac{(-1)^n\times(\text{const})}{a^2\pm n^2}$, reflecting the shared structure of expanding a "smooth but non-periodic-with-period-$2\pi$" function in the $2\pi$-periodic basis.
- The poles at $a=\pm n$ are not a flaw in the formula — they mark exactly the values of $a$ where $\sin(ax)$ degenerates into a genuine basis function, at which point the whole derivation must be redone (as in Problem 331).