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Consider the formal power series: $$f\left(z,w\right)=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}c_{m,n}z^{m}w^{n}$$

for some complex coefficients $c_{m,n}$. Defining:

$$\alpha_{m}\left(w\right)\overset{\textrm{def}}{=}\sum_{n=0}^{\infty}c_{m,n}w^{n}$$ $$\beta_{n}\left(z\right)\overset{\textrm{def}}{=}\sum_{m=0}^{\infty}c_{m,n}z^{m}$$

we can write: $$f\left(z,w\right)=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}c_{m,n}z^{m}w^{n}=\sum_{m=0}^{\infty}\alpha_{m}\left(w\right)z^{m}=\sum_{n=0}^{\infty}\beta_{n}\left(z\right)w^{n}$$

As such:

(1) If all the $\alpha_{m}$s and $\beta_{n}$s are rational functions of $w$ and $z$, respectively, is $f$ itself then necessarily a rational function of $z$ and $w$?

(2) (Generalization to power series in $d$ indeterminates) Working with a formal power series over $\mathbb{C}$ in $d$ indeterminates $z_{1},\ldots,z_{d}$:

$$f\left(z_{1},\ldots,z_{d}\right)=\sum_{n_{1}=0}^{\infty}\cdots\sum_{n_{d}=0}^{\infty}c_{n_{1},\ldots,n_{d}}z_{1}^{n_{1}}\cdots z_{d}^{n_{d}}$$

suppose that for every $j\in\left\{ 1,\ldots,d\right\}$, the series:

$$\sum_{n_{1}=0}^{\infty}\cdots\sum_{n_{j-1}=0}^{\infty}\sum_{n_{j+1}=0}^{\infty}\cdots\sum_{n_{d}=0}^{\infty}c_{n_{1},\ldots,n_{j-1},n,n_{j+1}\ldots n_{d}}z_{1}^{n_{1}}\cdots z_{j-1}^{n_{j-1}}z_{j+1}^{n_{j+1}}\cdots z_{d}^{n_{d}}$$

is a rational function of $ z_{1},\ldots,z_{j-1},z_{j+1}\ldots,z_{d}$ for all $n\geq0$. Is that sufficient to force $f$ to be a rational function of $z_{1},\ldots,z_{d}$?

MCS
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1 Answers1

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Look at $e^{zw}=\sum_{n=0}^{\infty}\frac{(zw)^n}{n!}$. Each coefficient, as a $z$ series and as a $w$ series, is rational, but $e^{zw}$ isn't.

  • Damn. I should have seen that one coming. That being said, the case I'm really interested in is what happens when the coefficients take on only finitely many distinct values. Any thoughts on that? – MCS Oct 12 '19 at 22:49
  • @MCS What exactly does it mean 'the coefficients take on only finitely many values'? – conditionalMethod Oct 13 '19 at 01:15
  • It means that there is a finite, non-empty set $S\subseteq\mathbb{C}$ so that for every $ \left(n_{1},\ldots,n_{d}\right)\in\mathbb{N}^{d}$, the coefficient $c_{n_{1},\ldots,n_{d}}$ is an element of $S$. The particular case I'm studying is when $S=\left{ 0,1\right} $ – MCS Oct 13 '19 at 19:35
  • @MCS Yes, if you put many gaps in between the non-zero coefficients the function will be transcendental. Like $\sum_n z^{n!}w^{n!}$. – conditionalMethod Oct 13 '19 at 19:39