求 $ f(x)=\frac{1-r^2}{1-2r\cos x+r^2}\ (|r|\lt 1)$ 的 Fourier 级数.
求 $\displaystyle f(x)=\frac{1-r^2}{1-2r\cos x+r^2}\ (|r|\lt 1)$ 的 Fourier 级数.[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
$$\begin{aligned} f(x)=&\frac{2-2r\cos x-(1-2r\cos x+r^2)}{1-2r\cos x+r^2} =\frac{2-2r\cos x}{1-2r\cos x+r^2}-1\\ =&\frac{2-r(\mathrm{e}^{\mathrm{ i} x}+\mathrm{e}^{-\mathrm{ i} x})}{1-r(\mathrm{e}^{\mathrm{ i} x}+\mathrm{e}^{-\mathrm{ i} x})+r^2}-1 =\frac{(1-r\mathrm{e}^{\mathrm{ i} x})+(1-r\mathrm{e}^{-\mathrm{ i} x})}{(1-r\mathrm{e}^{\mathrm{ i} x})\cdot(1-r\mathrm{e}^{-\mathrm{ i} x})}-1\\ =&\frac{1}{1-r\mathrm{e}^{-\mathrm{ i} x}}+\frac{1}{1-r\mathrm{e}^{\mathrm{ i} x}}-1 =\sum_{k=0}^\infty r^k \mathrm{e}^{-\mathrm{ i} kx} +\sum_{k=0}^\infty r^k \mathrm{e}^{\mathrm{ i} kx}-1\\ =&1+2\sum_{k=1}^\infty r^k \cos kx. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
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