How to determine if $f_n(x) $ is periodic function or not for $n>2$

Question

$f_n(x)$ is defined as

$$\int_0^{f_n(x)}\! \frac{\mathbb{d}t}{1+t^n}=x$$

$$\frac{d}{dx}\int_0^{f_n(x)}\! \frac{\mathbb{d}t}{1+t^n}=\frac{d (x)}{dx}$$

$$f'_n(x) \frac{1}{1+f^n_n(x)}=1$$

$$f'_n(x) =1+f^n_n(x) \tag 1$$

if $n=1$ then $f_1(x) =e^{x}-1$ and its period is $T=2\pi i$

$f_1(x+2\pi i) =f_1(x)$

if $n=2$ then $f_2(x) =\tan (x)$ and its period is $T=\pi$

$f_2(x+\pi )=f_2(x )$

Is $f_n(x )$ also periodic if $n>2$ ?

How can their periods be found if they are periodic functions?

Thanks a lot for answers and hints

Answer 1

Since $t\mapsto \frac{1}{1+t^n}$ is not an analytic function,$\displaystyle\int_0^f \frac{\mathrm{d}t}{1+t^n}$ depends on the path taken from $0$ to $f$. The multiple values of $\displaystyle\int_0^f \frac{\mathrm{d}t}{1+t^n}$ differ exactly by linear combinations over $\mathbb{Z}$ of $2\pi i$ times the residues of $t\mapsto \frac{1}{1+t^n}$ at its poles. The residue of $t\mapsto \frac{1}{1+t^n}$ at $\sqrt[n]{-1}$ is $-\frac{1}{n}\sqrt[n]{-1}$ for any $n$-th root of $-1$.

Thus if $\displaystyle\int_0^f \frac{\mathrm{d}t}{1+t^n} = x$ say, then also $\displaystyle\int_0^f \frac{\mathrm{d}t}{1+t^n} = x-\frac{1}{n}\sqrt[n]{-1}$ for any $n$-th root of $-1$.

This means that $f$ defined by $$\int_0^{f(x)} \frac{\mathrm{d}t}{1+t^n} = x,$$ has periods $-\frac{2\pi i}{n}\sqrt[n]{-1}$, for any $n$-th root of $-1$. This reduces nicely to your result for $n=1$ and $2$.

Answer 2

As mentioned in the comment above: assuming you are looking for solutions of the form $f\colon\mathbb R\to \mathbb R$ (which in particular rules out complex periods and $\tan$)

• if $n$ is even, say $n=2k$, you cannot have a periodic function as a solution. Indeed, from your the differential equation, $$ f_{n}^\prime = 1 + f_{n}^{2k} \geq 1$$ so $f_{n}(x) \geq x+f_{n}(0) \xrightarrow[x\to\infty]{} \infty$; and cannot be periodic. To see why, observe that by contradiction if you had a $T$-periodic solution with $T > 0$, then for any $x\in\mathbb{R}$ $$f_{n}(x) = f_{n}(x+m T)\xrightarrow[m\to\infty]{} \infty$$ so $f_n(x) = \infty$ (which does not make sense).

• if $n$ is odd, say $n=2k+1$, then (this is not finished) one can at least prove that any solution $f_n$ either:

  • is the constant function $f_n\equiv -1$;
  • always is below $-1$: $f_n(x)< -1\ \forall x\in\mathbb R$;
  • or always is above $-1$: $f_n(x)> -1\ \forall x\in\mathbb R$. This is a consequence of the Cauchy-Lipstchitz theorem for the differential equation $$ f_{n}^\prime = 1 + f_{n}^{2k+1}$$ as any solution taking value $-1$ for some $x_0\in\mathbb R$ satisfies the equation with conditions $(f'(x_0),f(x_0)) = (0,1)$; and the constant function $-1$ is a maximum solution of this Cauchy-Lipstchiz problem. I guess from this, one needs a lot more to build up further to get qualitative results about the possibility of a periodic solution, though.