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

Is there a not identically zero, real-analytic function $f\colon\mathbb R\to\mathbb R$, which satisfies

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

and $a\in \mathbb R$

I saw a special case when $a=0$

I try to solve it by :

$$f(x)=e^{cx}$$ $$f(n)=e^{nc}$$ $$f^{(n)}(x)=c^ne^{cx}$$ $$f^{(n)}(a)=c^ne^{ca}$$ so $$e^{nc}=c^ne^{ca}$$ so $$c=\frac{nW(\frac{a-n}{n})}{a-n}$$

the problem is we always see n with c but the special case when a=0 give

$$c=\frac{nW(\frac{0-n}{n})}{0-n}$$ $$c=\frac{W(\frac{-1}{1})}{-1}=-W(-1)$$

I think there is no solution when $a\neq 0$

may be there is another function can solve it

Is there any solution in general?

thanks for all

Solution 1:

I can't explicitly find an example, so perhaps turning to an existence proof, that can also be used to construct an example. To do so, consider the operator $\mathcal L_a: C^\infty \times \mathbb N \to \mathbb R$, where $\mathcal L_a\{f,n\} = f(n) - f^{(n)}(a)$. Then inspect Banach fixed-point theorem, if you can choose a norm in $C^\infty \times \mathbb N$ where $\mathcal L_a$ is a contraction, then jack-pot. Or, if it is not a contraction, for many reasonable norms, then you can say there is probably no such function in Banach space.
Hope this helps.

Solution 2:

So, an easy answer to a seemingly related problem:

$f(x)=\sin(x)$ has the property that for each $n\in\mathbb{N}$, $f^{(n)}(a)=f\left(\frac{\pi}{2}n\right)$, where $a=2\pi k$ for any $k\in\mathbb{Z}$.

What one would like to do is just replace $f$ with $g(x)=\sin(\frac{\pi}{2}x)$, but of course this messes up the derivative. I do not immediately see how to fix this, but I also do not see why solving the equation $f^{(n)}(a)=f(\frac{\pi}{2}n)$ should be fundamentally different than solving $f^{(n)}(a)=f(n)$. Perhaps this can shed light on the question though.