Forgot password?
 Register account
View 1426|Reply 2

又一类微分函数方程

[Copy link]

459

Threads

952

Posts

9843

Credits

Credits
9843

Show all posts

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)

459

Threads

952

Posts

9843

Credits

Credits
9843

Show all posts

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

7

Threads

52

Posts

393

Credits

Credits
393

Show all posts

icesheep Posted 2013-11-23 21:41

Mobile version|Discuz Math Forum

2025-6-5 01:04 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit