Does This Condition Characterize $e^z$? The following is a question from a Complex Analysis qualifying exam I was studying from:
Does there exist an entire function $f$, distinct from $e^z$, such that $f(0)=1$ and $f'(n)=f(n)$ for all $n\geq 1$?
My instinct is that such a function should exist, although I have tried cooking one up and I have been unable to do so. I then decided my instinct may have been incorrect, so I tried proving $e^z$ is the only such function, but I have been unable to do so.
The roadblock to one potential method: A common way to show two analytic functions are the same is showing they agree on a set in their domains with a limit point. If I could show any function $f$ and $e^z$ agreed on the positive integers, then they would agree with $\infty$ as the limit point (under the transformation $z\mapsto 1/z$, this is the same as saying the limit point is zero) but $e^z$ is not a function on $\mathbb{C}_{\infty}$ (i.e., $e^{1/z}$ has an essential singularity at $z=0$).
Another method I haven't been able to make work: Showing the coefficients of their respective power series are equal. 
 A: When you want to find out whether a condition uniquely determines an entire function, it is often fruitful to assume one has two (not necessarily different) functions satisfying the condition and find out what one can determine about their difference or quotient. In particular, if one function satisfying the condition is explicitly known.
Here, we have a condition that is satisfied by the exponential function, a function that is rather well known and has several very convenient properties. One of the convenient properties is that it has no zeros, hence the quotient $\dfrac{f(z)}{e^z}$ is entire with $f$. The fact that the exponential function satisfies the differential equation $y' = y$ is also often very useful.
So suppose $f$ is an entire function with
$$f(0) = 1\quad \text{and }\quad f'(n) = f(n),\; n \in \mathbb{Z}^+\tag{1}$$
To find out whether necessarily $f(z) = e^z$, let us consider $h(z) = f(z)e^{-z}$ and see what we can say about that. First, since $f(0) = e^0 = 1$, we have $h(0) = 1$. Differentiating, we find
$$h'(z) = f'(z)e^{-z} + f(z)\bigl(-e^{-z}\bigr) = \bigl(f'(z) - f(z)\bigr)e^{-z},$$
so $h'(n) = 0$ for all positive integers $n$.
On the other hand, if $g(0) = 1$ and $g'(n) = 0$ for all positive integers $n$, then we find for $f(z) = g(z)e^z$ that $f(0) = 1$ and since $f'(z) = g'(z)e^z + g(z)e^z = \bigl(g'(z) + g(z)\bigr)e^z$, that $f'(n) = \bigl(g'(n) + g(n)\bigr)e^n = g(n)e^n = f(n)$ for positive integers $n$. So the solutions to $(1)$ are in correspondence to the entire functions $h$ with $h(0) = 1$ and $h'(n) = 0,\; n \in \mathbb{Z}^+$.
Given only $h'(n) = 0,\; n \in \mathbb{Z}^+$, we can always achieve $h(0) = 1$ by adding a constant, so the question is:
Are there entire functions $g$ with $g(n) = 0,\, n \in \mathbb{Z}^+$ other than $g \equiv 0$?
Any primitive of such a function (since $\mathbb{C}$ is simply connected, all entire functions have a primitive) gives rise to a solution of $(1)$.
A non-constant function that vanishes in all $n \in \mathbb{Z}$ (not only in the positive integers, but that doesn't hurt of course) is $\sin (\pi z)$ - of course all (nonzero) multiples of that have the same property. Choosing the factor $-\pi$, we see that $h(z) = \cos (\pi z)$ gives rise to the solution $f(z) = \cos (\pi z)e^z$ of $(1)$ which evidently is different from $e^z$.
There are many more solutions to $(1)$, but few can be expressed as simply as the above.
