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I've noticed in my free time when the functional mapping $f(z+c)=-1/(f(z)+1)$ is iterated twice, it yields the original function $f(z)$ (i.e. $f(z+3c)=f(z)$). So I thought to study it as a periodic function...but I don't know enough about it to evaluate its Fourier series. I went back to its original equation, and saw that if it had a zero (I'll call it $\alpha$), it would also have a singularity at $\alpha-c$; I realized that likewise if $f(z)$ doesn't vanish it can't have any singularities (but I'm not sure that's possible since its equation cannot be satisfied by an exponential function).

I would like to know the most general kind of function $f(z)$ can be. It looks to me like a trigonometric function could describe it...but nothing appears to disqualify it having a second period. I am assuming $f(z)$ is meromorphic and continuous in the interval $(0,c)$, and I'm allowing $c$ to be complex (though not forcing it).

Antonio DJC
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  • How are you getting that f is periodic? – Tim Raczkowski Jul 29 '16 at 02:10
  • yes with $h(z) = \frac{-1}{z+1}$ we have $h(h(h(z))) = z$, so $f(x+3c) = h(h(h(f(x)))) = f(x)$ @TimRaczkowski – reuns Jul 29 '16 at 02:10
  • Ah I see how you got that. – Tim Raczkowski Jul 29 '16 at 02:14
  • If $x$ is real, assign any values to $f$ on $[0,c)$, and set the rest such that it fulfills your equation, I don't see what you can say more. Now if $x$ is a complex number $\in U$ and $f$ is meromorphic on $U$, it could be different – reuns Jul 29 '16 at 02:15
  • I am considering $f$ to be a meromorphic function, and $x\in\mathbb{C}$, so I can determine the most general form for $f(x)$. If I, say, set $f(0)=0$, then I could say $f(x)$ includes $\sin(2\pi x/(3\alpha))$ and, due to my equation, $1/\sin(2\pi x/(3\alpha)-2\pi/3)$. – Antonio DJC Jul 29 '16 at 02:24
  • You should add the meromorphic hypothesis to the question itself. Right now you don't even ask for $f$ to be continuous, so one can simply set arbitrary values for every $x$ with $0\le\Re(x)<c$. (Is $c$ positive? real?) For that matter, it's more suggestive to use $z$ instead of $x$, if you mean the input to be complex. – Greg Martin Jul 29 '16 at 07:43

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There was a mistake in my calculations and it now doesn't seem dolvable anymore for me, but I don't want to completely delete the andere.

First of all, let $x_0$ be a solution of $x^2 + x + 1 = 0$. We have \begin{align*} x^2+x+1=0 \iff x(x+1)=-1 \iff x = - \frac{1}{x+1}. \end{align*} Thus, $f(z) = x_0$ is a solution for your problem. Now \begin{align*} f'(z) = f'(z-c+c) = -(\frac{1}{f(z-c)+1})' = \frac{f'(z - c)}{(f(z-c)+1)^2}. \end{align*}

  • I see how you got $x^2+x+1=0$, but how did you get $f'(z)$ in the numerator of $-(-1/(f(z-c)+1))'$? Shouldn't it be instead $f'(z-c)$? – Antonio DJC Aug 10 '16 at 02:27
  • You are right, I made a mistake there. Then I have no idea how to solve it :/ –  Aug 10 '16 at 06:53
  • I'll look at how $f(z)$ can have zeros and poles, since its functional equation implies they'll have the same order. – Antonio DJC Aug 10 '16 at 21:34