# General Solution of Functional Equation

What is the general solution to:

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

Obviously the solution to this for the limiting case of $h\to 0$ is $f(x) = \ln(x) + c$

Attempting to solve the case of $h=1$ poses some difficulties.

But I imagine that once I can cover the case of $h=1$ the rest shall become simpler

A related problem is that the solution to:

$$\frac{f(x+h)-f(x)}{h} = f(x) \tag{2}$$

Can be assumed to have exponential form. Yielding:

$$a^{x+h} - a^{x} = ha^x \rightarrow a^h = 1+h \rightarrow a=(1+h)^{\frac{1}{h}}$$

Thus the general solution to this similar problem is:

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

• Use another character instead of $\large{\rm e}$. People will be confused. Jul 2 '14 at 4:35
• I looked up a few of the solutions and they all seem to be related to the digamma function (in fact, for $h=1$ that's just the recurrence relation for the digamma function). Jul 2 '14 at 4:54
• @Silynn Where did you look?
– mvw
Jul 2 '14 at 5:00
• I am not sure why is this tagged (functional-analysis). Jul 2 '14 at 5:31
• @Martin Sleziak, I was hoping there was some way to resolve this using techniques such as infinite dimensional matrices etc... Jul 2 '14 at 5:33

Obviously the general solution for $f(x+1)-f(x)=1/x$ (assuming $f(x)$ is required to be defined for $x>0$) is as follows. $f(x)=f_0(x)$ can be an arbitrary function for $x\in(0,1]$ and then, for noninteger $x$, $$f(x) =f_0(x-[x])+\sum_{n=1}^{[x]}\frac 1{x-n},$$ where $[x]$ is the integer part of $x$, while $$f(n)=f_0(1)+\sum_{k=1}^{n-1}\frac1k,\qquad n=2,3,\dots\,.$$
• Rewrite the equation in the form $f(x+1)=f(x)+1/x$ and use induction on $[x]$. E.g. if $x\in(2,3]$, then $f(x)=f(x-1)+(x-1)^{-1}=f(x-2)+(x-2)^{-1}+(x-1)^{-1}=f_0(x-2)+(x-2)^{-1}+(x-1)^{-1}$ Jul 2 '14 at 15:49