Solving a differential equation of infinite order

Question

I would like to solve the following differential equation

$$\frac{1}{x+1} = \sum_{n=1}^{\infty} \frac{f^{(n)}(x)}{n!}.$$

Does anyone know how to solve this, or if a solution even exists?

My guess is that the solution to this differential equation is the partial sum of the harmonic series $H_n$, or using the variable shown in the equation above, $H_x$$\ldots$

P.S.: I just finished Calculus IV — differential equations & Fourier series/Laplace transforms — so I’m still not as skilled as I’m sure some of y’all are, so please be patient.

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Andrei · Answer 1 · 2025-12-18T20:20:08.510

17

The right hand side looks like a Taylor series: $$f(x+y)=f(x)+\sum_{n=1}^\infty\frac{f^{(n)}(x)}{n!}y^n$$ Now just use $y=1$, and your equation becomes $$\frac{1}{x+1}=f(x+1)-f(x)$$

edited Dec 18 '25 at 20:20

answered Dec 18 '25 at 14:57

Andrei

40,527

11

The solution to this functional equation is the digamma function $\psi$: $$f(x)=\psi(x+1)+\gamma+C,$$ where $\gamma$ is the Euler-Mascheroni constant and $C$ is a constant set by the intitial conditions – Caesar.tcl Dec 18 '25 at 15:01
6

This is not quite correct to say 'THE SOLUTION, since you can add to it a function arbitrarily with period one and obtain an infinity of solutions.. – Letac Gérard Dec 19 '25 at 15:43

Rodrigo de Azevedo · Answer 2 · 2025-12-20T14:49:30.633

$$ \frac{1}{x+1} = \sum_{n=1}^{\infty} \frac{1}{n!} f^{(n)}(x) = \left( \sum_{n=0}^{\infty} \frac{1}{n!} f^{(n)}(x) \right) - f(x) $$

Taking the (bilateral) Laplace transform,

$$ {\cal L} \left\{ \sum_{n=0}^{\infty} \frac{1}{n!} f^{(n)} \right\} = \underbrace{\left( \sum_{n=0}^{\infty} \frac{s^n}{n!} \right)}_{=: e^s} {\hat f} = e^s {\hat f} $$

and, taking the inverse (bilateral) Laplace transform (or consulting a table of (bilateral) Laplace transform pairs),

$$ \sum_{n=0}^{\infty} \frac{1}{n!} f^{(n)} (x) = {\cal L}^{-1} \left\{ e^s {\hat f} (s) \right\} = f (x+1) $$

as, in the Laplace domain, multiplication by exponentials of $s$ corresponds to shifts in the dual, non-Laplace domain. Hence, the original equation can be rewritten as the functional equation

$$ \frac{1}{x+1} = f (x+1) - f (x) $$

Related

Exponential of the differential operator and DDE

Doesn't the Laplace transform of a derivative involve extra terms that you're not accounting for here? That is, $\mathcal{L}\left{f'(t)\right} = s\hat{f}(s)-f(0)$. — mathperson314, Dec 20 '25 at 14:17
@mathperson314 Think of the bilateral Laplace transform instead. Here, $s$ is almost synonymous with $\frac{{\rm d}}{{\rm d}x}$ but avoids scary-looking operator-theoretic notation — Rodrigo de Azevedo, Dec 20 '25 at 14:53

Solving a differential equation of infinite order

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