Is there a function with the property $f(n)=f^{(n)}(0)$?

Is there a not identically zero, real-analytic function $f:\mathbb{R}\rightarrow\mathbb{R}$, which satisfies

$$f(n)=f^{(n)}(0),\quad n\in\mathbb{N} \text{ or } \mathbb N^+?$$

What I got so far:



then for $n=0$ this works anyway and else we have


Now $a_1=\sum_{k=0}^\infty\frac{a_k}{k!}1^k=a_0+a_1+\sum_{k=2}^\infty\frac{a_k}{k!},$ so


For $n=2$ we find


The first case was somehow special since $a_1$ cancelled out, but now I have to juggle around with more and more stuff.

I could express $a_1$ in terms of the higher $a’s$, and then for $n=3$ search for $a_2$ and so on. I didn’t get far, however. Is there a closed expression? My plan was to argue somehow, that if I find such an algorythm to express $a$’s in terms of higher $a$’s, that then, in the limit, the series of remaining sum or sums would go to $0$ and I’d eventually find my function.

Or maybe there is a better approach to such a problem.


Let complex number $c$ be a solution of $e^c=c$. For example $c = -W(-1)$, where $W$ is the Lambert W function. Then since function $f$ defined by

f(x) = \sum_{n=0}^\infty \frac{e^{cn}x^n}{n!}

evaluates to $e^{e^c x} = e^{cx}$, we have $f(n) = e^{cn} = f^{(n)}(0)$. For a real solution, let $c = a+bi$ be real and imaginary parts and let $g(x)$ be the real part of $f(x)$. More explicitly:

g(x) = \sum_{n=0}^\infty \frac{e^{an}\cos(bn)\;x^n}{n!}

evaluates to $e^{ax}\cos(bx)$. With the principal branch of Lambert W, this is approximately:

g(x) = e ^{0.3181315052 x} \operatorname{cos} (1.337235701 x)

Source : Link , Question Author : Nikolaj-K , Answer Author : GEdgar

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