Are there any solutions to this functional/differential equation involving an infinite series?

Question

I'm am interested in how to find solutions related the equation

$$\sum_{i=1}^n \frac {\mathrm d^i}{\mathrm dx^i}(f) = f.$$

If $n$ goes to infinity it seems like I can use taylor series to help find solutions. What I am really interested in however is, is there a method for finding exact solutions to this equation for arbitrary $n$? Also, is there a maximum or minimum $n$ for which solutions exist? Are there any conditions I am forgetting? I have tried looking at different questions or researching on the internet but "differential equation as a series" in searches returns something else.

Answer 1

This is an $n$th order linear ODE, and is solvable by considering the characteristic equation. Note that the equation can be simplified to $$\sum_{i=1}^n f^{(i)} = 0,$$ a homogeneous equation which has a characteristic polynomial $$\sum_{i=1}^n r^i = r\sum_{i=1}^n r^{i-1} = 0.$$ Clearly $r = 0$ is a root, as $r$ is a factor. The remaining polynomial, $\sum_{i=1}^n r^{i-1}$, is a cyclotomic polynomial. Note that $$(r - 1)\sum_{i=1}^n r^{i-1} = r^n - 1,$$ so the roots of $\sum_{i=1}^n r^{i-1}$ are all the strictly complex $n$th roots of $1$, which we denote $\omega_1, \ldots, \omega_{n-1}$. The most general solution to the ODE is therefore, $$f(x) = C_0 + C_1e^{\omega_1 x} + \ldots + C_{n-1}e^{\omega_{n-1} x},$$ for arbitrary constants $C_0, C_1, \ldots, C_{n-1}$.

I didn't use Taylor series here, but I honestly think this is the most straightforward approach.

Answer 2

Consider $$\sum_{i=1}^{n}f^{(i)}(x)=f(x).$$ We let $f(x)=e^{rx}$ so that $$\sum_{i=1}^{n}r^{i}=1\implies \frac{r(1-r^{n})}{1-r}-1=0\implies r^{n+1}-2r+1=0.$$ In the case $n\rightarrow\infty$ $$\sum_{i=1}^{\infty}r^{i}=1\implies \frac{r}{1-r}-1=0\implies r=\frac{1}{2},$$ so it looks like there is only one solution of the form $e^{rx}$ with $r=1/2$. Note that for finite $n$ there is always one solution with $x=1$.