I would like to get some information about the following functional equation:
$$f(z)=\frac{1}{24} - \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ f \left (\frac{-4\pi^2}{z}\right )\right )$$
This functional relationship must hold only for $\mathfrak{R}(z)>0$. I would like to know wether it is possible to know what types of functions hold it.
To be clear, I do not know where to start from. I have barely worked with functional equations in my life.
Any help or bibliography will be welcomed.
Edit: Answering the comments, we are supposed to know that $f(z)$ as $z \to 0^+$ is $ O\left ( \frac{1}{z^2} \right )$. Moreover, we know that $f(-2\pi)=f(2\pi)=\frac{1}{24} - \frac{1}{8\pi}$. Thank you for the help received.
Edit 2: $f(z)$ is even: $f(z)=f(-z)$. From here, we could get the follwing functional equation for $\mathfrak{R}(z)<0$:
$$f(z)=\frac{1}{24} + \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ f \left (\frac{4\pi^2}{z}\right )\right )$$
So that we can remove the $-$ sign from the first equation to get for $\mathfrak{R}(z)>0$:
$$f(z)=\frac{1}{24} - \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ f \left (\frac{4\pi^2}{z}\right )\right )$$
That formula puts no restrictions at all on your function on the half-plane $\Re(z) > 0$.
If you know $f(z_1)$ for some $z_1$ with $\Re(z_1) > 0$, then it tells you the value of $f(z_2)$ where $z_2 = \frac{-4\pi^2}{z_1}$. But $\Re(z_2) < 0$. And if you try to use the formula again, the new value is for $\frac{-4\pi^2}{z_2} = z_1$ (where it gives the same value for $f(z_1)$ that you started with).
So, you can choose any function $g$ on $\Re(z) > 0$, then define $$f(z) = \begin{cases} g(z) & \Re(z) > 0\\\frac{1}{24} - \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ g \left(\frac{-4\pi^2}{z}\right )\right) & \Re(z) < 0\end{cases}$$ $f$ will be defined everywhere except the imaginary line and will satisfy your equation.