Expand f(x)=(pi²-x²)², x in[- pi,pi], period 2 pi
Expand $f(x)=(\pi^2-x^2)^2$, $x\in[-\pi,\pi]$, period $2\pi$, in a Fourier series, and calculate $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^4}$.
Explanation of Method
Expanding $f(x)=\pi^4-2\pi^2x^2+x^4$, $f$ is even, so only cosine terms appear. Integrating each monomial term against $\cos(nx)$ (using the standard $\int x^2\cos(nx)\,dx$ and $\int x^4\cos(nx)\,dx$ antiderivatives) and combining, the $\pi^2$-dependent pieces cancel exactly, leaving a clean $1/n^4$ coefficient. Evaluating at $x=0$ then extracts the target sum.
- Expand $f(x)$ as a polynomial and compute $a_0$.
- Compute $a_n$ by integrating each monomial term against $\cos(nx)$ and combining.
- Evaluate the series at $x=0$ to extract $\sum(-1)^{n-1}/n^4$.
Step 1 · Expansion and $a_0$
$f(x)=(\pi^2-x^2)^2=\pi^4-2\pi^2x^2+x^4$ is even, so $b_n=0$. $$ a_0 = \frac2\pi\int_0^\pi(\pi^4-2\pi^2x^2+x^4)\,dx = 2\pi^4\left[1-\frac23+\frac15\right] = 2\pi^4\cdot\frac8{15} = \frac{16\pi^4}{15}. $$
Step 2 · Computing $a_n$
Using $\int_0^\pi\cos(nx)\,dx=0$, $\int_0^\pi x^2\cos(nx)\,dx=\dfrac{2\pi(-1)^n}{n^2}$, and $\int_0^\pi x^4\cos(nx)\,dx=\dfrac{4\pi^3(-1)^n}{n^2}-\dfrac{24\pi(-1)^n}{n^4}$: $$ a_n = \frac2\pi\left[-2\pi^2\cdot\frac{2\pi(-1)^n}{n^2}+\frac{4\pi^3(-1)^n}{n^2}-\frac{24\pi(-1)^n}{n^4}\right] = \frac2\pi\left[-\frac{24\pi(-1)^n}{n^4}\right] = \frac{48(-1)^{n+1}}{n^4}. $$ (The $\pi^2/n^2$ terms cancel exactly, leaving pure $1/n^4$ decay.) $$ \boxed{(\pi^2-x^2)^2 = \frac{8\pi^4}{15} + 48\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^4}\cos(nx)} $$ (Verified numerically: $a_0\approx103.903=16\pi^4/15$, $a_1=48$, $a_2=-3$, $a_3\approx0.59259=48/81$, matching exactly.)
Step 3 · The sum
At $x=0$: $f(0)=\pi^4$, so $$ \pi^4 = \frac{8\pi^4}{15}+48\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^4} \implies \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^4} = \frac{\pi^4-\frac{8\pi^4}{15}}{48} = \frac{7\pi^4}{720}. $$ $$ \boxed{\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^4} = \frac{7\pi^4}{720}} $$ (Verified numerically: partial sums to $200{,}000$ terms give $\approx0.947033$, matching $7\pi^4/720\approx0.947033$.)
Summary Table
| Property | Value |
|---|---|
| Fourier series | $(\pi^2-x^2)^2=\frac{8\pi^4}{15}+48\sum\frac{(-1)^{n+1}}{n^4}\cos(nx)$ |
| Sum | $\sum\frac{(-1)^{n-1}}{n^4}=\frac{7\pi^4}{720}$ |
Additional Notes
- This is the quartic analogue of Problem 326's $x^2$ Basel-problem evaluation: each extra factor of smoothness in $f$ buys an extra power of $n$ decay in the Fourier coefficients, here reaching $1/n^4$.
- The value $\frac{7\pi^4}{720}$ is the Dirichlet eta function $\eta(4)$, related to the Basel-type sum $\sum1/n^4=\pi^4/90=\zeta(4)$ by $\eta(4)=(1-2^{1-4})\zeta(4)=\frac78\zeta(4)$.
- Because $f(x)=(\pi^2-x^2)^2$ vanishes (along with its first derivative) at $x=\pm\pi$, the periodic extension is $C^1$ with no jump in slope, which is exactly why the coefficients decay two full powers faster than Problem 326's merely-continuous-with-a-corner extension of $x^2$.