$$f'(x)=P^{m}(f(x))$$ ,where $P(x)$ is a polynomial, m is a real number.
$x\in\mathbb{C}$ and $f:\mathbb{C}\to\mathbb{C}$
I would like to find a criteria to define if $f(x)$ is a periodic function or not. I also want to find the period value that depends on polynomial constants and $m$ if the function ($f(x)$) is periodic.
My attempt for it:
Example 1:
$P(x)=x$, and $m=1$: Thus our problem is to find period for: $$f'(x)=f(x)$$
If we derivative both side , we can get that
$$f''(x)=f'(x)=f(x)$$
Generally; $$f^{(n)}(x)=f(x)$$
a periodic function must satisfy $f(x+T)=f(x)$ , where $T$ is minimal period .
Taylor expansion of $f(x+T)$:
$$ f(x+T) =f(x)+\frac{Tf'(x)}{1!}+\frac{T^2f''(x)}{2!}+.....=\sum_{n=0}^{\infty} \frac{T^nf^{(n)}(x)}{n!} $$
$$ f(x+T) =f(x)+\frac{Tf(x)}{1!}+\frac{T^2f(x)}{2!}+.....=\sum_{n=0}^{\infty} \frac{T^nf(x)}{n!} =f(x) \sum_{n=0}^{\infty} \frac{T^n}{n!}=f(x) $$
Thus, $$\sum_{n=0}^{\infty} \frac{T^n}{n!}=1$$
We know that $$\sum_{n=0}^{\infty} \frac{T^n}{n!}=e^{T}=1$$
Minimal period : $T=2\pi i$ ,
Example 2:
$P(x)=a^2-x^2$, and $m=\frac{1}{2}$: Thus our problem is to find period for $$f'(x)=\sqrt{a^2-f^2(x)}$$
If we derivative both side , we can get that
$$f''(x)=-\frac{f'(x)f(x)}{\sqrt{a^2-f^2(x)}}=-f(x)$$
$$f'''(x)=-f'(x)$$
$$f^{(4)}(x)=-f''(x)=f(x)$$
$$f^{(5)}(x)=f'(x)$$ . . .
a periodic function must satisfy $f(x+T)=f(x)$ , where $T$ is minimal period .
Taylor expansion of $f(x+T)$:
$$ f(x+T) =f(x)+\frac{Tf'(x)}{1!}+\frac{T^2f''(x)}{2!}+.....=\sum_{n=0}^{\infty} \frac{T^nf^{(n)}(x)}{n!} $$
$$ f(x+T) =f(x)+\frac{Tf'(x)}{1!}-\frac{T^2f(x)}{2!}-\frac{T^3f'(x)}{3!}+.....=f(x)\sum_{n=0}^{\infty} \frac{(-1)^{n}T^{2n}}{(2n)!}+f'(x)\sum_{n=0}^{\infty} \frac{(-1)^{n}T^{2n+1}}{(2n+1)!}=f(x) $$
Thus, $$f(x)(-1+\sum_{n=0}^{\infty} \frac{(-1)^{n}T^{2n}}{(2n)!})+f'(x)\sum_{n=0}^{\infty} \frac{(-1)^{n}T^{2n+1}}{(2n+1)!}=0$$
We get $$\sum_{n=0}^{\infty} \frac{(-1)^{n}T^{2n}}{(2n)!}=1$$ $$\cos{T}=1$$ and $$\sum_{n=0}^{\infty} \frac{(-1)^{n}T^{2n+1}}{(2n+1)!}=0$$ $$\sin{T}=0$$
The common solution of 2 equations above is Minimal period : $T=2\pi$ ,
The solution attempt for General Problem:
$$P(x)=a_0+a_1x+\frac{a_2}{2!}x^2+\frac{a_3}{3!}x^3+....+\frac{a_n}{n!}x^n$$
$$f'(x)=P^{m}(f(x))$$
If we follow the same strategy; Apply derivative both side , so we can get that
$$f''(x)=mP^{m-1}(f(x))P'(f(x))f'(x)$$
$$f''(x)=mP^{2m-1}(f(x))P'(f(x))=m (f'(x))^{(2m-1)/m}P'(f(x))$$
We can see that if $m=1/2$ then it simplifies to
$$f''(x)=\frac{P'(f(x))}{2}$$
$$f'''(x)=f'(x)\frac{P''(f(x))}{2}$$
$$f^{(4)}(x)=\frac{P'(f(x))P''(f(x))}{4}+f'(x)^2\frac{P'''(f(x))}{2}$$
$$f^{(4)}(x)=\frac{P'(f(x))P''(f(x))}{4}+(P^{1/2}(f(x))^2\frac{P'''(f(x))}{2}$$
$$f^{(4)}(x)=\frac{P'(f(x))P''(f(x))}{4}+\frac{P(f(x))P'''(f(x))}{2}$$
But this approach does not help so much like my examples.
Could you please help me what we can do if the function is periodic or not?
Thanks a lot for advice and answers