1^()circ Expand f(x)=cases pi x-x², x in(0,pi), x²-3 pi
$1^\circ$ Expand $f(x)=\begin{cases}\pi x-x^2, & x\in(0,\pi)\\x^2-3\pi x+2\pi^2, & x\in(\pi,2\pi)\end{cases}$ in a Fourier series. $2^\circ$ Show $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi^3}{32}$.
Explanation of Method
Both pieces are congruent parabolic arcs, and the function is continuous and odd (as a $2\pi$-periodic extension) about $x=0$, so only sine terms appear. Two integrations by parts give coefficients decaying like $1/n^3$, all with the same sign. Evaluating at $x=\pi/2$ then produces the cubed-reciprocal alternating series.
- Confirm odd symmetry, so $a_n=0$.
- Compute $b_n$ via double integration by parts on each piece.
- Evaluate the series at $x=\pi/2$ to extract the target sum.
Step 1 · Symmetry and continuity
At $x=\pi$: both pieces give $f(\pi^-)=f(\pi^+)=0$, and at $x=0,2\pi$: $f(0^+)=0=f(2\pi^-)$, so $f$ is continuous on $(0,2\pi)$. As a $2\pi$-periodic function it is odd about $0$, so $a_0=a_n=0$ for all $n$; only sine terms appear.
Step 2 · Computing $b_n$
$$ b_n = \frac1\pi\left[\int_0^\pi(\pi x-x^2)\sin(nx)\,dx+\int_\pi^{2\pi}(x^2-3\pi x+2\pi^2)\sin(nx)\,dx\right]. $$ After integration by parts (twice, for the quadratic pieces), the terms combine to give, for odd $n=2k-1$: $$ b_{2k-1} = \frac8{\pi(2k-1)^3}, \qquad b_{2k}=0. $$ (Verified numerically: direct numerical integration gives $b_1\approx2.5465=\frac8\pi$, $b_3\approx0.09431=\frac8{27\pi}$, $b_5\approx0.02037=\frac8{125\pi}$ — confirming the essential factor of $1/\pi$ that is easy to drop in the hand computation.)
Step 3 · Fourier series
$$ \boxed{f(x) = \frac8\pi\sum_{k=1}^\infty\frac{\sin((2k-1)x)}{(2k-1)^3}} $$ (Verified numerically at several points, e.g. $x=1.5$: series gives $\approx2.46239$, matching $f(1.5)=\pi(1.5)-1.5^2\approx2.46239$ exactly.)
Part 2° · The sum $\sum\frac{(-1)^{n-1}}{(2n-1)^3}$
At $x=\dfrac\pi2$: $f\!\left(\dfrac\pi2\right)=\pi\cdot\dfrac\pi2-\dfrac{\pi^2}4=\dfrac{\pi^2}4$, and $\sin\dfrac{(2k-1)\pi}2=(-1)^{k-1}$, so $$ \frac{\pi^2}4 = \frac8\pi\sum_{k=1}^\infty\frac{(-1)^{k-1}}{(2k-1)^3} \implies \boxed{\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi^3}{32}}. $$ (Verified numerically: partial sums to $200{,}000$ terms give $\approx0.968946$, matching $\pi^3/32\approx0.968946$.)
Summary Table
| Property | Value |
|---|---|
| Fourier series | $f(x)=\frac8\pi\sum\frac{\sin((2k-1)x)}{(2k-1)^3}$ |
| Sum | $\sum\frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi^3}{32}$ |
Additional Notes
- The cubic decay $1/(2k-1)^3$ makes this series converge considerably faster than the linear-decay series of Problem 327 or the quadratic-decay series of Problem 325 — a direct consequence of $f$ being continuous with a continuous derivative except at the corners.
- This sum is the odd-cubed analogue of Problem 324's Leibniz series ($1/(2n-1)$, from $x$) and Problem 325's ($1/(2n-1)^2$, from $|x|$): each extra degree of smoothness in the underlying function buys one more power of decay in its Fourier coefficients.
- Like Problem 328, the underlying wave here is built from congruent arcs (parabolic rather than linear) meeting smoothly at their endpoints — the shared structure is why both series involve only odd harmonics.