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令\(\,x=n,\;y=1,\;a_n=f(n)\,\)得\(\;4 a_n a_1=a_{n+1}+a_{n-1},\)
\(a_{n+1}-a_n+a_{n-1}=0,\; \)特征方程\(\;x^2-x+1=0,\)
\(a_n = \alpha e^{i\frac{n\pi}{3}}+\beta e^{-i\frac{n\pi}{3}}.\;\;\alpha=\beta=\frac{1}{4}\,(a_1=\frac{1}{4}).\)
\(\therefore\;\;a_n=\frac{1}{2}\cos\frac{n\pi}{3}.\) 容易验证\(\,f(x)=\frac{1}{2}\cos\small\dfrac{\pi x}{3}\)
是原函数方程的解. 进而\(\,f(2010)=\large\frac{1}{2}.\)
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