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连续函数$f:\mathbb R\to\mathbb R$满足$f(x+2)=f(x+1)+f(x),f(0)=0,f(1)=1.$
$f(x)=\frac{\left(\sqrt{5}+1\over2\right)^x-\left(\frac{2}{\sqrt{5}+1}\right)^x \cos (\pi x)}{\sqrt{5}}$
\begin{align*}f(x+2)-f(x+1)&=\frac{\left(1+\sqrt{5}\over2\right)^{x+2} -\left(\frac{2}{1+\sqrt{5}}\right)^{x+2}\cos (\pi (x+2))}{\sqrt{5}}-\frac{\left(1+\sqrt{5}\over2\right)^{x+1}-\left(2\over1+\sqrt{5}\right)^{x+1}\cos (\pi (x+1))}{\sqrt{5}}\\
&=\frac{\left(1+\sqrt{5}\over2\right)^{x+2}-\left(1+\sqrt{5}\over2\right)^{x+1}}{\sqrt{5}}-\frac{\left(\frac{2}{1+\sqrt{5}}\right)^{x+2}\cos (\pi x)+\left(2\over1+\sqrt{5}\right)^{x+1}\cos (\pi x)}{\sqrt{5}}
\\&=\frac{\left(1+\sqrt{5}\over2\right)^x-\left(\frac{2}{1+\sqrt{5}}\right)^x \cos (\pi x)}{\sqrt{5}}\\&=f(x)\end{align*} |
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