Expand f(x)=sin dfrac x2 in a sine series on (0,pi)
Expand $f(x)=\sin\dfrac x2$ in a sine series on $(0,\pi)$.
Explanation of Method
Since the half-frequency $\sin\frac x2$ is not itself one of the basis functions $\sin(nx)$, a genuine infinite sine expansion is required — analogous to Problem 332's non-integer-frequency case, but here restricted to a sine (not full) series on $(0,\pi)$. The product-to-sum identity reduces $b_n$ to two elementary sine integrals that combine via $\sin(n\pi\pm\pi/2)=\pm\cos(n\pi)$.
- Apply product-to-sum to $\sin\frac x2\sin(nx)$.
- Integrate and simplify using $\sin\left(\frac{1\mp2n}2\pi\right)=\cos(n\pi)$.
- Combine into a single closed-form coefficient.
Step 1 · Computing $b_n$
$$ b_n = \frac2\pi\int_0^\pi\sin\frac x2\sin(nx)\,dx = \frac1\pi\int_0^\pi\left[\cos\left(\frac{1-2n}2x\right)-\cos\left(\frac{1+2n}2x\right)\right]dx. $$ Using $\sin\left(\frac{1-2n}2\pi\right)=\sin\left(\frac\pi2-n\pi\right)=\cos(n\pi)=(-1)^n$ and similarly $\sin\left(\frac{1+2n}2\pi\right)=(-1)^n$: $$ b_n = \frac{2(-1)^n}\pi\left[\frac1{1-2n}-\frac1{1+2n}\right] = \frac{2(-1)^n}\pi\cdot\frac{4n}{1-4n^2} = \frac{8n(-1)^{n+1}}{\pi(4n^2-1)}. $$
Step 2 · Fourier series
$$ \boxed{\sin\frac x2 = \frac8\pi\sum_{n=1}^\infty\frac{n(-1)^{n+1}}{(2n-1)(2n+1)}\sin(nx) = \frac8\pi\left(\frac{\sin x}3-\frac{2\sin2x}{15}+\frac{3\sin3x}{35}-\cdots\right)} $$ (Verified numerically at $x=0.3,1.0,2.0,2.8$: partial sums to $20{,}000$ terms match $\sin\frac x2$ to 4+ decimal places.)
Summary Table
| Property | Value |
|---|---|
| Coefficient | $b_n=\dfrac{8n(-1)^{n+1}}{\pi(4n^2-1)}$ |
| Decay rate | $\sim1/n$ (conditional convergence) |
Additional Notes
- This is structurally the sine-series analogue of Problem 337's full-range expansion of $\sin\frac{5x}6$: both involve non-integer frequencies expanded in the integer harmonic basis, with coefficients that decay only like $1/n$.
- Since $\sin\frac x2$ is smooth on $(0,\pi)$ but its odd $2\pi$-periodic sine-series extension has a corner at $x=0$ (jumping from the extension's behavior on the left to $\sin(x/2)$'s positive slope on the right), the coefficients cannot decay faster than $1/n$.
- All harmonics are present with no vanishing terms (unlike Problem 351's $\cos(2x)$ sine series, where only odd harmonics survive) — a consequence of $\sin\frac x2$ having no special symmetry relative to the integer-frequency basis.