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

  • $\begingroup$ Does the relation have to hold for all $u \in \mathbb{R}$ at once, or are you looking for it to hold at some specific value of $u$? $\endgroup$ – coffeemath May 20 '13 at 19:00
  • $\begingroup$ @coffeemath if we don't have solution for all $u\in R$ so I'm looking at some specific value of u $\endgroup$ – mhd.math May 20 '13 at 19:10
  • $\begingroup$ Yes, if it held on an interval $I$ for example, $f'(u)=f(1)$ would force $f$ to be constant on $I$. Nice question, maybe can use the case $u=0$ and some kind of shifted and integrated function related to original solution, but I haven't been able to make that work, say by shifting backwards or forwards by 1. +1 for question. $\endgroup$ – coffeemath May 20 '13 at 20:12
  • $\begingroup$ @coffeemath thanks alot but can you give me example but with full answer please ? $\endgroup$ – mhd.math May 20 '13 at 20:32
  • $\begingroup$ I don't have an example with fixed nonzero $u$. I'd like to see one... Given the complexity of the answer to the $u=0$ case you linked to in the question, there may be another way to do this using that method, however the procedure with $W$ function is a bit beyond my knowledge. Maybe someone else can do it with a nonzero $u$. $\endgroup$ – coffeemath May 21 '13 at 0:54

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.

  • $\begingroup$ Of course, OP wanted real-analytic, not just $C^\infty$ (which I think would make the problem fairly trivial). $\endgroup$ – user43208 Sep 8 '13 at 23:02

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.


I don't know about a general solution. But if we assume the function is invertible, then: $$ f(n) = f^n(a)$$ $$f^{-1}f(n) = n= f^{n-1}(a)$$

Then, $$f^0(a) = a = 1$$ $$f(1) = f(a)= 2$$ $$f(2) = f^2(1) = 3$$ and soon. In this case $f$ is an increment function.

  • 4
    $\begingroup$ The question concerns the $n$-th derivative $f^{(n)}$ of $f$, not the $n$-th power $f^n$ of $f$. $\endgroup$ – Servaes Sep 6 '13 at 1:57

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