Identity theorem and computing complex value of entire function Let 
$A = \{ z \in \mathbb{C} : z = \frac{1}{n} \ \textrm{ with } n \in \mathbb{N} \}$
$f : A \to \mathbb{C}$, $f(z) = z^2$
I want the analytic continuation of the function $f$ to a function $h$ entire in the complex plane.  I know by the identity theorem the only solution is $h : \mathbb{C} \to \mathbb{C}$, $h(z) = z^2$
But I want a formula to calculate $h(1+i)$ in function of only the values of $f$ over $A$.  
This formula has to exist (again by the identity theorem), but is it know how to find it?
(I literally want to make a python script that numerically aproximates this result only using the values of $f$ over $A$)
Thanks.
 A: If you are given $h$ on a sequence $z_n \to z_0$, then you can use the sequence to approximate all derivatives of $h$ at $z_0$ to arbitrary accuracy. Then you can expand in Taylor series centered at $z_0$ to approximate other values. The error will be from both approximating the derivatives as well as truncating the Taylor series. To be a bit more explicit, if $z_n=1/n$ and you are given $h(0)$, then you can define the sequence of polynomial approximations
$$h_d(z)=\sum_{n=0}^d h[0,1/d,\dots,1/(d+1-n)] z^n$$
where the notation used there is the divided difference (and by convention the $n=0$ term is just $h[0]$). Divided differences are defined by the recursion:
$$f[x_0]=f(x_0) \\
f[x_0,\dots,x_n]=\frac{f[x_1,\dots,x_n]-f[x_0,\dots,x_{n-1}]}{x_n-x_0}.$$
A variant of Taylor's theorem tells you that $f[x_0,x_1,\dots,x_n]=\frac{f^{(n)}(\xi_n)}{n!}$ for some $\xi_n$ between $\min x_i$ and $\max x_i$. (Or if the $x_i$ were complex, in their convex hull.) Thus this is sort of like the Taylor polynomial of degree $d$ at $z_0$ but it deviates from it because $\xi_n \neq z_0$ in general.
Note that the coefficients for a given $n$ depend also on $d$, so this formula requires you to choose $d$ up front (instead of just continuing to produce coefficients until you have achieved desired accuracy). 
Remarks:


*

*In the above I assumed that you are given $h(z_0)$ as well as $h(z_n)$. If you are not (and strictly speaking you do not need to be), then you need to approximate $h(z_0)$ through some extrapolation procedure (which will introduce a third source of error).

*This is weaker than analytic continuation: for example, one cannot use this to compute $\zeta(0)$ if given $\zeta(2+1/n)$, because the pole at $s=1$ is "in the way". But it will suffice for entire functions.

