Let \(f:\mathbb R\to\mathbb R\) be the \(2\pi\)-periodic function satisfying \(f(x)=x^3\) for \(-\pi\le x<\pi\).
Prove that the Fourier series of \(f\) has the form \[ \sum_{n=1}^{\infty}b_n\sin(nx), \] and give an integral formula for \(b_n\) without evaluating it.
Prove that the Fourier series converges for every \(x\).
Prove that \[ \sum_{n=1}^{\infty}b_n^2=\frac{2\pi^6}{7}. \]
Solution
Use the Fourier-coefficient convention \[ a_0=\frac1\pi\int_{-\pi}^{\pi}f(x)\,dx, \] \[ a_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx, \] and \[ b_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx. \]
1. The periodic function \(f\) is odd almost everywhere.
For \(-\pi<x<\pi\), \[ f(-x)=(-x)^3=-x^3=-f(x). \] The endpoint values form a set of measure zero and do not affect the Fourier coefficients.
2. One has\[ a_0=0. \]
The function \(f\) is odd almost everywhere on \([-\pi,\pi]\), so its integral over that symmetric interval is zero.
3. For every \(n\geq1\),\[ a_n=0. \]
The function \(f\) is odd almost everywhere, while \(\cos(nx)\) is even. Therefore \(f(x)\cos(nx)\) is odd almost everywhere, and its integral over \([-\pi,\pi]\) is zero.
4. For every \(n\geq1\),\[ \boxed{ b_n = \frac2\pi \int_0^\pi x^3\sin(nx)\,dx. } \]
The functions \(x^3\) and \(\sin(nx)\) are both odd, so their product is even. Thus \[ \begin{aligned} b_n &= \frac1\pi \int_{-\pi}^{\pi}x^3\sin(nx)\,dx\\ &= \frac2\pi \int_0^\pi x^3\sin(nx)\,dx. \end{aligned} \]
5. The Fourier series of \(f\) has the form\[ \boxed{ \sum_{n=1}^{\infty}b_n\sin(nx). } \]
The general real Fourier series is \[ \frac{a_0}{2} + \sum_{n=1}^{\infty} \bigl( a_n\cos(nx)+b_n\sin(nx) \bigr). \] Steps <1>2–<1>3 show that the constant and cosine coefficients vanish.
6. The function \(f\) is piecewise \(C^1\) on every period and has finiteone-sided limits at every point.
On each interval \[ (-\pi+2k\pi,\pi+2k\pi), \qquad k\in\ZZ, \] the function is a translate of the polynomial \(x^3\), hence is smooth. At the endpoints the one-sided limits exist and are finite.
7. At every point where \(f\) is continuous, its Fourier series convergesto \(f(x)\).
By step 6, the hypotheses of the Dirichlet convergence theorem hold. At a continuity point, the two one-sided limits agree with \(f(x)\), so their average is \(f(x)\).
8. At every discontinuity point\[ x=(2k+1)\pi, \qquad k\in\ZZ, \]the Fourier series converges to \(0\).
At each such point the one-sided limits are \(\pi^3\) and \(-\pi^3\), whose average is \(0\). The Dirichlet convergence theorem therefore gives convergence to \(0\).
9. The Fourier series converges for every real \(x\).
Every real point is either a continuity point, handled by step 7, or a jump point, handled by step 8.
10. Parseval’s identity reduces here to\[ \sum_{n=1}^{\infty}b_n^2 = \frac1\pi \int_{-\pi}^{\pi}\abs{f(x)}^2\,dx. \]
The function \(f\) is bounded on one period and therefore belongs to \(L^2([-\pi,\pi])\). Parseval’s identity states \[ \frac1\pi \int_{-\pi}^{\pi}\abs{f(x)}^2\,dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} \left( a_n^2+b_n^2 \right). \] Steps <1>2–<1>3 make all \(a\)-terms vanish.
11. One has\[ \frac1\pi \int_{-\pi}^{\pi}\abs{f(x)}^2\,dx = \frac{2\pi^6}{7}. \]
On \([-\pi,\pi)\), \[ \abs{f(x)}^2=x^6. \] Therefore \[ \begin{aligned} \frac1\pi \int_{-\pi}^{\pi}\abs{f(x)}^2\,dx &= \frac1\pi \int_{-\pi}^{\pi}x^6\,dx\\ &= \frac2\pi \int_0^\pi x^6\,dx\\ &= \frac2\pi \frac{\pi^7}{7}\\ &= \frac{2\pi^6}{7}. \end{aligned} \]
12. Therefore\[ \boxed{ \sum_{n=1}^{\infty}b_n^2 = \frac{2\pi^6}{7}. } \]
Combine steps <1>10–<1>11.
13. Q.E.D.
Steps <1>4–<1>5 prove part (1), step 9 proves part (2), and step 12 proves part (3).