问题及解答

证明:三角多项式的 Fourier 展开就是自身.

设 $P(x)=a_0+\sum_{k=1}^{n}(a_k\cos kx+b_k\sin kx)$. 设其 Fourier 展开式为 $\frac{A_0}{2}+\sum_{k=1}^{\infty}(A_k\cos kx+B_k\sin kx)$. 则

\[
\begin{split}
A_0&=\frac{1}{\pi}\int_{-\pi}^{\pi}P(x)dx=\frac{1}{\pi}\int_{-\pi}^{\pi}\Bigl[a_0+\sum_{k=1}^{n}(a_k\cos kx+b_k\sin kx)\Bigr]dx\\
&=\frac{1}{\pi}a_0\cdot 2\pi+\frac{1}{\pi}\sum_{k=1}^{n}\Bigl[\int_{-\pi}^{\pi}a_k\cos kx dx+\int_{-\pi}^{\pi}b_k\sin kxdx\Bigr]\\
&=2a_0+\frac{1}{\pi}\sum_{k=1}^{n}\Bigl[a_k\cdot\frac{1}{k}(\sin kx)\bigr|_{-\pi}^{\pi}+b_k(-\cos kx)\bigr|_{-\pi}^{\pi}\Bigr]\\
&=2a_0.
\end{split}
\]

\[
\begin{split}
A_k&=\frac{1}{\pi}\int_{-\pi}^{\pi}P(x)\cos kxdx\\
&=\frac{1}{\pi}\int_{-\pi}^{\pi}\Bigl[a_0+\sum_{j=1}^{n}(a_j\cos jx+b_j\sin jx)\Bigr]\cos kxdx\\
&=\frac{1}{\pi}\biggl[\int_{-\pi}^{\pi}a_0\cos kxdx+\sum_{j=1}^{n}\bigl(a_j\int_{-\pi}^{\pi}\cos jx\cos kxdx+b_j\int_{-\pi}^{\pi}\sin jx\cos kxdx\bigr)\biggr]\\
&=\frac{1}{\pi}\sum_{j=1}^{n}a_j\int_{-\pi}^{\pi}\cos jx\cos kxdx\\
&=\delta_{jk}\frac{a_j}{\pi}\int_{-\pi}^{\pi}\cos jx\cos kxdx
\end{split}
\]

因此, 当 $k > n$ 时, $A_k=0$. 当 $1\leqslant k\leqslant n$ 时, 

\[
A_k=\frac{a_k}{\pi}\int_{-\pi}^{\pi}\cos^2 kxdx=\frac{a_k}{\pi}\cdot 2\int_{0}^{\pi}\cos^2 kxdx=\frac{a_k}{\pi}\cdot \int_{0}^{\pi}(1+\cos 2kx)dx=a_k.
\]

类似可证明, 当 $k > n$ 时, $B_k=0$. 当 $1\leqslant k\leqslant n$ 时, 

\[
B_k=\frac{1}{\pi}\int_{-\pi}^{\pi}P(x)\sin kx dx=b_k.
\]