$g$ is a meromorphic function on $\mathbb{C}$, $g(\frac1n)=\frac{1+n^4}{1-2n^4}$, please find out the analytical expression of $g$.

Question

Problem: If $g$ is a meromorphic function on $\mathbb{C}$, and for every $n\in\mathbb{N+}$, we have $$g \left (\frac1n \right)=\frac{1+n^4}{1-2n^4}$$, please find out the analytical expression of $g$.

I think maybe I can use Laurent series to solve the problem, but I just get stuck here.

The answer is easy to guess, but I cannot prove it strictly:( — LonnerT, Jul 11 '20 at 03:19
The only question seems to be "is this the only expression for $g$?" Here you can use the identity theorem — Brian Moehring, Jul 11 '20 at 03:24
I think I've got the point... but I still have a problem here, the theorem holds when $g$ is holomorphic function, but $g$ is just meromorphic, does the theorem still hold? — LonnerT, Jul 11 '20 at 03:32
@LonnerT: https://math.stackexchange.com/questions/905376/identity-theorem-for-meromorphic-functions. — Martin R, Jul 11 '20 at 03:35

Rivers McForge · Accepted Answer · 2020-07-11T05:17:43.587

Set $z = 1/n$, then $n = 1/z$ and

$$g(z) = \frac{1 + \frac{1}{z^4}}{1 - \frac{2}{z^4}} = \frac{z^4 + 1}{z^4 - 2}.$$

Note that $g$ is holomorphic at $0$: in fact, $g(0) = -1/2$.

If $h(z)$ is any other function holomorphic at $0$ which agrees with $g(z)$ on the sequence $z_n = 1/n$, then $h(z_n) - g(z_n) = 0$ for all $n$, and since $0$ is an accumulation point of $z_n$, the identity theorem shows $h(z) = g(z)$ for all $z$ for which $h(z)$ is defined.

$g$ is a meromorphic function on $\mathbb{C}$, $g(\frac1n)=\frac{1+n^4}{1-2n^4}$, please find out the analytical expression of $g$.

1 Answers1