Expand the 2 pi -periodic function f(x)=sin dfrac 5x6, x
Expand the $2\pi$-periodic function $f(x)=\sin\dfrac{5x}6$, $x\in(-\pi,\pi)$, in a Fourier series.
Explanation of Method
This is a special case of Problem 332's general formula for $\sin(ax)$ with non-integer $a=5/6$, simplified using the fact that $\sin\frac{5\pi}6=\sin\frac\pi6=\frac12$ has a clean closed form. Only sine terms appear (oddness), and the coefficients decay like $1/n$.
- Note oddness, so only sine terms appear.
- Apply Problem 332's general formula with $a=5/6$, simplifying $\sin(5\pi/6)$.
- Verify the resulting closed form numerically.
Step 1 · Symmetry
$f(x)=\sin\frac{5x}6$ is odd, so $a_0=a_n=0$; only sine terms appear.
Step 2 · Computing $b_n$
By Problem 332's general result with $a=\frac56$: $$ b_n = \frac{2n(-1)^n\sin\frac{5\pi}6}{\pi\left(\frac{25}{36}-n^2\right)}. $$ Since $\sin\frac{5\pi}6=\sin\left(\pi-\frac\pi6\right)=\sin\frac\pi6=\frac12$: $$ b_n = \frac{n(-1)^n}{\pi\left(\frac{25}{36}-n^2\right)} = \frac{36n(-1)^n}{\pi(25-36n^2)} = \frac{36n(-1)^{n+1}}{\pi(36n^2-25)}. $$ (Verified: this matches an independent symbolic computation of $\frac1\pi\int_{-\pi}^\pi\sin\frac{5x}6\sin(nx)\,dx$ exactly.)
Step 3 · Fourier series
$$ \boxed{\sin\frac{5x}6 = \frac{36}\pi\sum_{n=1}^\infty\frac{n(-1)^{n+1}}{36n^2-25}\sin(nx) = \frac{36}\pi\left[\frac{\sin x}{11}-\frac{2\sin2x}{119}+\frac{3\sin3x}{299}-\cdots\right]} $$
Summary Table
| Property | Value |
|---|---|
| Coefficient | $b_n=\dfrac{36n(-1)^{n+1}}{\pi(36n^2-25)}$ |
| Decay rate | $\sim1/n$ (conditional convergence) |
Additional Notes
- This is a direct application of Problem 332's general non-integer-frequency formula, simplified only by the clean value $\sin(5\pi/6)=\frac12$ — a good illustration of how a general result specializes.
- Because $a=5/6$ is close to the integer $1$, the coefficient has a near-resonance: the denominator $36n^2-25$ is small relative to $n^2$ specifically near $n=1$ (where $36-25=11$), making $b_1$ the dominant term by far.
- Since the coefficients decay only like $1/n$, this series — like Problem 324's Leibniz-type series and Problem 327's sawtooth series — converges conditionally rather than absolutely.