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又一类微分函数方程

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青青子衿 posted 2013-11-22 20:27 |Read mode
$f'(x)=f^{-1}(x)$
又一类微分函数方程(系数与黄金分割比有关)

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original poster 青青子衿 posted 2013-11-22 20:52
$f'(x)=f^{-1}(x)$
$f(f'(x))=x$
$[f(f'(x))]'=1$
$f''(x)f'(f'(x))=1$
$f'(x)=\beta x^\alpha$
$f''(x)=\alpha\beta x^{\alpha-1}$
$f'(f'(x))=\beta^{\alpha+1}x^{\alpha^2}$
$f''(x)f'(f'(x))=1$
$\begin{cases}
\alpha\beta^{\alpha+2}=1 \\
\alpha^2+\alpha-1=0  \end{cases}$
$\Longrightarrow$
$\begin{cases}
\alpha=\frac{-1+\sqrt 5}{2} \\
\beta=(\frac{1+\sqrt 5}{2})^{\frac{3-\sqrt 5}{2}}  \end{cases}$
$f(x)=\int f'(x)dx=\frac{\beta}{\alpha+1}x^{\alpha+1}=$
$(\frac{1+\sqrt 5}{2})^{\frac{1-\sqrt 5}{2}}x^{\frac{1+\sqrt 5}{2}}$

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icesheep posted 2013-11-23 21:41

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