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$f_n(n)=f(n+1)$

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青青子衿 Posted 2013-11-22 19:12 |Read mode
已知$f_n(x)$表示函数$f(x)$的$n$次迭代,则求满足$f_n(n)=f(n+1)$的函数$f(n)$

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icesheep Posted 2013-11-22 21:58
n=0 得出 f(0)=f(1)=0
n=1 得出 f(1)=f(2)=0
n=2 得出 f(2)=f(3)=0
...

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 Author| 青青子衿 Posted 2013-11-23 08:45
回复 2# icesheep
n=0 得出 f(0)=f(1)=0
n=1 得出 f(1)=f(2)=0
n=2 得出 f(2)=f(3)=0
...
icesheep 发表于 2013-11-22 21:58
如果是定义在正整数集上的呢?

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icesheep Posted 2013-11-23 21:53
解不唯一,懒得想了。

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