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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 $C(x)$:

$$y(x)=y_1(x)+C(x)$$

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:

$$y(x)=C(x)y_1(x)$$

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.

UPDATE

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

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 $C(x)$:

$$y(x)=y_1(x)+C(x)$$

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:

$$y(x)=C(x)y_1(x)$$

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.

UPDATE

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

ADDENDUM

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.

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 $C(x)$:

$$y(x)=y_1(x)+C(x)$$

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:

$$y(x)=C(x)y_1(x)$$

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.

UPDATE

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.

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 $C(x)$:

$$y(x)=y_1(x)+C(x)$$

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:

$$y(x)=C(x)y_1(x)$$

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.

UPDATE

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