傅里叶级数¶
1.傅里叶级数¶
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三角函数系：\(\{1,\cos,\sin x, \cos 2x, \sin 2x\}\) 在定义为 \(\int_{-\pi}^\pi f(x)g(x)\,dx\) 的内积下是正交的
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傅里叶级数：\(f(n) \sim \frac{a_0}{2} \sum_{n=1}(a_n\cos nx + b_n \sin nx)\)
Warning
右侧函数不一定收敛于 \(f\)，即不能直接画等号
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若 \(\frac{|a_0|}{2} + \sum_{n=1}(|a_n| + |b_n|)\)，收敛，则傅里叶级数绝对收敛且一致收敛
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系数公式：
\[a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x)\,dx\]
\[a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x)\cos nx\,dx\]
\[b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x)\sin nx\,dx\]
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收敛定理：若 \(f(x)\) 按段光滑（导函数连续），则傅里叶级数收敛于 \(\frac{f(x+0)+f(x-0)}{2}\)
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一致收敛定理：若在上述基础上 \(f(x)\) 连续，则其傅里叶级数绝对且一致收敛于 \(f(x)\)
2. 以 \(2l\) 为周期的函数的展开式¶
- 系数公式：
\[a_0 = \frac{1}{l} \int_{-l}^l f(x)\,dx\]
\[a_n = \frac{1}{l} \int_{-l}^l f(x)\cos \frac{nx\pi}{l}\,dx\]
\[b_n = \frac{1}{l} \int_{-l}^l f(x)\sin \frac{nx\pi}{l}\,dx\]
- 奇偶延拓：
- 奇函数：仅含正弦项
- 偶函数：仅含余弦项
3. 收敛定理的证明¶
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Parseval 等式： \(\frac{a_0^2}{2} + \sum_{n=1}(a_n^2 + b_n^2) = \frac{1}{\pi} \int_{-\pi}^\pi f(x)^2\,dx\)
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黎曼-勒贝格定理 \(a_n\to 0, b_n\to 0\)
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傅里叶级数部分和的积分表示：\(S_n(x)=\frac{1}{\pi}\int_{-\pi}^\pi f(x+t)\frac{\sin(n+\frac{2}{2})t}{2\sin\frac{t}{2}}\,dt\)
Note
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Parseval 等式是比书上 Bessel 不等式更强的一个结论
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常用函数的傅里叶级数：
\[\sum_{n=1}^{\infty}\frac{\sin nx}{n}=\frac{\pi-x}{2},0<x<2\pi\]
\[\sum_{n=1}^{\infty}\frac{\cos nx}{n^{2}}=\frac{1}{4}x^{2}-\frac{\pi}{2}x+\frac{\pi^{2}}{6},0\leqslant x\leqslant2\pi \]
\[ \sum_{n=1}^{\infty} \frac{\sin nx}{n^3}=\frac{1}{12}x^3-\frac{\pi}{4}x^2+\frac{\pi^2}{6}x,0\leqslant x\leqslant2\pi \]
\[ \sum_{n=1}^\infty\frac{\cos nx}{n^4}=-\frac{1}{48}x^4+\frac{\pi}{12}x^3-\frac{\pi^2}{12}x^2+\frac{\pi^4}{90},0\leqslant x\leqslant2\pi \]
\[ \sum_{n=1}^{\infty}\frac{\sin nx}{n^5} =-\frac{1}{240}x^5+\frac{\pi}{48}x^4-\frac{\pi^2}{36}x^3+\frac{\pi^4}{90}x,0\leqslant x\leqslant2\pi \]
\[ \sum_{n=1}^\infty\frac{\cos nx}{n^6}=\frac{1}{1440}x^{6}-\frac{\pi}{240}x^{5}+\frac{\pi^{2}}{144}x^{4}-\frac{\pi^{4}}{180}x^{2}+\frac{\pi^{6}}{945},0\leqslant x\leqslant2\pi\]
\[\sum_{n=1}^\infty\frac{\left(-1\right)^{n-1}}{n^3}\sin nx=\frac{1}{12}x\left(\pi^2-x^2\right),\quad-\pi\leqslant x\leqslant\pi\]
\[\sum_{n=1}^\infty\frac{\sin nx}{n^3}=\frac{1}{12}x\left(x-\pi\right)\left(x-2\pi\right),\quad0\leqslant x\leqslant2\pi\]
\[\frac{1}{1-a^2}+\frac{2}{1-a^2}\sum_{n=1}^{\infty}a^n\cos nx=\frac{1}{1-2a\cos x+a^2}, |a|<1\]
\[\sum_{n=1}^{\infty}a^n\sin nx=\frac{a\sin x}{1-2a\cos x+a^2}, |a|<1\]
\[-\sum_{n=1}^\infty\frac{\cos nx}{n}=\ln\left(2\sin\frac{x}{2}\right),\quad0<x<2\pi\]
\[\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\cos nx=\ln\left(2\cos\frac{x}{2}\right),\quad-\pi<x<\pi\]
\[2\sum_{n=1}^\infty\frac{\cos(2n-1)x}{2n-1}=\ln\left(\cot\frac{|x|}{2}\right),\quad-\pi<x<\pi,x\neq0\]
\[2\sum_{n=1}^\infty\frac{\left(-1\right)^{n-1}}{2n-1}\mathrm{sin}(2n-1)x=\ln\left(\cot\left|\frac{x}{2}-\frac{\pi}{4}\right|\right),\quad-\frac{\pi}{2}<x<\frac{3\pi}{2}\]
\[\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}a^n\sin nx=\arctan\frac{a\sin x}{1+a\cos x},|a|<1\]