I noticed an interesting property of holomorphic functions and I'm wondering if it forms the basis of divergent sum renormalisation.
Let $f,g:\mathbb C \rightarrow \mathbb C$ be holomorphic functions.
Let the power series of $f(z)$ and $g(z)$ respectively be: $$\sum_{n=0}^\infty a_n z^n \quad , \quad \sum_{n=0}^\infty b_n z^n$$
Using only the fact that $f$ and $g$ are uniquely defined by their ($n^{th}$) derivatives at a point, it can be shown that if these power series coincide term by term when evaluated at $z_1$ and $z_2$ respectively, then the values $f(z_1)$ and $g(z_2)$ are equal: $$a_n z_1^n = b_n z_2^n \quad \forall n \in \mathbb N_0 \quad \Rightarrow\quad f(z_1) = g(z_2)$$
Now let $(d_n)$ be the terms of some divergent series.
In theory, we should be able to find a holomorphic function $h$ whose sequence of power series terms at some point $c$ is equal (on a term by term basis) to $(d_n)$.
For instance, let $(d_n) = (-1)^n$. This is the sequence $(1,-1,1,-1,1,\dots)$. We could let $h(z) = \frac 1 {1-z}$ and then the power series of $h$ would be $1+z+z^2+\dots$ Evaluating the power series of $h$ at $z=-1$ gives the series: $$1-1+1-1+\dots$$ which is the same as $(d_n)$ on a term-by-term basis.
$h$ of course has a well defined value of $\frac 1 2$ at $z=-1$ and so we might assign this divergent series $(d_n)$ the value $\frac 1 2$.
The beauty of this is that no matter which holomorphic function $h$ we choose, as long as its power series at a point $c$ is equal to $1-1+1-1+\dots$, then $h(c) = \frac 1 2$. So the value we assign to a divergent series in this way is unique.
- Is this the basis of divergent sum renormalisation?
- If we remove all zero terms from $(d_n)$ and from the power series of $h$, then does this uniqueness survive? i.e. will a function $g$ with a power series $(1 + 0 + 0 - 1 + 1 + 0 - 1...)$ admit the same value as $h$ above?