MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Lets define a discrete analytic function such a function that is equal to its Newton series:

$$f(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k f\left (0\right)$$

Is function $g(x)=e^{f(x)}$ also discrete-analytic?

This question arose from the following considerations.

As you know the difference equation

$$\Delta y(x) = F(x)$$

has multiple solutions that differ only by an arbitrary 1-periodic function $C(x)$:


At the same time there can be no more than one (up to a constant term) discrete-analytic solution which we can consider to be the natural solution of the equation.

But when considering multiplicative-difference equation $\frac{y(x+1)}{y(x)}=F(x)$ we come to a similar situation, this equation has multiple solutions which differ by an arbitrary 1-periodic factor:


Of these solutions, similarly, no more than one (up to a constant factor) is discrete-analytic which allows us to define the distinguished solution.

But on the other hand the following rule holds for indefinite product and sum:

$$\prod_x f(x)= e^{\sum_x \ln f(x)}$$

This means that we can obtain the solution to the equation $\frac{y(x+1)}{y(x)}=F(x)$ in the following form:

$$y(x)=e^{\sum_x \ln F(x)}$$

This allows us to select the distinguished solution by another method, that is taking the natural solution to the sum and taking exponent of it. The result will have a constant factor, but it is unevident whether it will be discrete-analytic or not, and as such, whether the both distinguished solutions coincide.


Due to the answer by David Speyer it is evident now that counter-examples exist among complex-valued functions and also there are instances when function $f(x)$ is discrete-analytic while the Newton series of its exponent does not converge.

So the question should be formulated more precisely: we assume that $f(x)$ is real-valued and Newton series for its exponent converges.

I started a bounty for this question


It would be even more great if somebody could prove a more general theorem about a composition of monotonous discrete-analytic functions. Whether the composition is also discrete-analytic and under what conditions.

share|cite|improve this question

No. Take $f(x) = 2 \pi i x$.

There is also a more subtle way I can cheaply answer the question: It is easy to give functions $f(x)$ such that the Newton series of $f$ converges to $f$, but the Newton series of $e^f$ diverges. Take $f(x)=x^2$ or, more subtly, $f(x) = \cos (2 \pi x/8)$. I assume that the right formulation of the question is "If $f(x)$ is real, the Newton series of $f$ converges to $f$, and the Newton series of $e^f$ converges, does the Newton series of $e^f$ converge to $e^f$?

share|cite|improve this answer
Thanks but what if we limit ourselves to only real-valued functions? – Anixx Mar 9 '12 at 18:39
Sorry, that was a little obnoxious of me. It is a nice question, which I've been thinking about for the last hour without success. – David Speyer Mar 9 '12 at 20:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.