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lim (1-Πcos(nx))/x^2 的计算思路

计算极限

lim⁡x→01−cos⁡xcos⁡2xcos⁡3x⋯cos⁡nxx2\begin{aligned} \lim_{x\to0}\frac{1-\cos x\cos2x\cos3x\cdots \cos nx}{x^2} \end{aligned}

Method 1: Term-by-term Expansion​

Notice that cos⁡x=1−x22+O(x3)\cos x=1-\frac{x^2}{2}+O(x^3), then

lim⁡x→01−cos⁡xcos⁡2xcos⁡3x⋯cos⁡nxx2=lim⁡x→01−(1−x22+O(x3))(1−(2x)22+O(x3))⋯(1−(nx)22+O(x3))x2\begin{aligned} \lim_{x\to0}\frac{1-\cos x\cos2x\cos3x\cdots \cos nx}{x^2}&=\lim_{x\to0}\frac{1-\left( 1-\frac{x^2}{2}+O(x^3) \right)\left( 1-\frac{(2x)^2}{2}+O(x^3) \right)\cdots\left( 1-\frac{(nx)^2}{2}+O(x^3) \right)}{x^2} \end{aligned}

Comparing coefficients, the coefficient of x0x^0 and x2x^2 is

[x0]=1⋅1⋅⋯⋅1=1[x2]=−122−222−⋯−n22=−112n(n+1)(2n+1)\begin{aligned} [x^0]&=1\cdot1\cdot\dots\cdot1=1\\ [x^2]&=-\frac{1^2}{2}-\frac{2^2}{2}-\dots-\frac{n^2}{2}=-\frac{1}{12} n (n+1) (2 n+1) \end{aligned}

Then the limitation

lim⁡x→01−(1−x22+O(x3))(1−(2x)22+O(x3))⋯(1−(nx)22+O(x3))x2=lim⁡x→01−(1−112n(n+1)(2n+1)x2+O(x3))x2=112n(n+1)(2n+1)+lim⁡x→0O(x)=112n(n+1)(2n+1)\begin{aligned} \lim_{x\to0}\frac{1-\left( 1-\frac{x^2}{2}+O(x^3) \right)\left( 1-\frac{(2x)^2}{2}+O(x^3) \right)\cdots\left( 1-\frac{(nx)^2}{2}+O(x^3) \right)}{x^2}&=\lim_{x\to0}\frac{1-\left( 1-\frac{1}{12} n (n+1) (2 n+1)x^2+O(x^3) \right)}{x^2}\\ &=\frac{1}{12} n (n+1) (2 n+1)+\lim_{x\to0}O(x)\\ &=\frac{1}{12} n (n+1) (2 n+1) \end{aligned}

Method 2: Global Expansion​

donote cos⁡xcos⁡2x⋯cos⁡nx\cos x\cos2x\cdots\cos nx as ∏k=1ncos⁡kx\prod_{k=1}^n\cos kx,

f(x)=∏k=1ncos⁡kxf(0)=1→log⁡f(x)=∑k=1nlog⁡cos⁡kx→f′(x)f(x)=∑k=1n−ksin⁡kxcos⁡kx=∑k=1n(−ktan⁡kx)→f′(x)=f(x)∑k=1n(−ktan⁡kx)f′(0)=f(0)⋅0=0→f′′(x)=f′(x)∑k=1n(−ktan⁡kx)+f(x)∑k=1n(−k2sec⁡2kx)f′′(0)=f(0)∑k=1n(−k2)=−16n(n+1)(2n+1)\begin{aligned} f(x)&=\prod_{k=1}^n\cos kx\\ f(0)&=1\\ \rightarrow\quad\log f(x)&=\sum_{k=1}^n\log \cos kx\\ \rightarrow\quad\frac{f'(x)}{f(x)} &=\sum_{k=1}^n\frac{-k\sin kx}{\cos kx}\\ &=\sum_{k=1}^n(-k\tan kx)\\ \rightarrow\quad f'(x)&=f(x)\sum_{k=1}^n(-k\tan kx)\\ f'(0)&=f(0)\cdot0=0\\ \rightarrow\quad f''(x)&=f'(x)\sum_{k=1}^n(-k\tan kx)+f(x)\sum_{k=1}^n(-k^2\sec^2 kx)\\ f''(0)&=f(0)\sum_{k=1}^n(-k^2)=-\frac{1}{6} n (n+1) (2 n+1)\\ \end{aligned}

Then

f(x)=1+−16n(n+1)(2n+1)2!(x−0)2+O(x3)=1−112n(n+1)(2n+1)x2+O(x3)\begin{aligned} f(x)&=1+\frac{-\frac{1}{6}n(n+1)(2n+1)}{2!}(x-0)^2+O(x^3)\\ &=1-\frac{1}{12}n(n+1)(2n+1)x^2+O(x^3) \end{aligned} lim⁡x→01−f(x)x2=lim⁡x→01−(1−112n(n+1)(2n+1)x2+O(x3))x2=112n(n+1)(2n+1)+lim⁡x→0O(x)=112n(n+1)(2n+1)\begin{aligned} \lim_{x\to0}\frac{1-f(x)}{x^2}&=\lim_{x\to0}\frac{1-(1-\frac{1}{12}n(n+1)(2n+1)x^2+O(x^3))}{x^2}\\ &=\frac{1}{12}n(n+1)(2n+1)+\lim_{x\to0}O(x)\\ &=\frac{1}{12}n(n+1)(2n+1) \end{aligned}