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:

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

  • $\begingroup$ Use another character instead of $\large{\rm e}$. People will be confused. $\endgroup$ Jul 2 '14 at 4:35
  • $\begingroup$ 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). $\endgroup$
    – Silynn
    Jul 2 '14 at 4:54
  • $\begingroup$ @Silynn Where did you look? $\endgroup$
    – mvw
    Jul 2 '14 at 5:00
  • 1
    $\begingroup$ I am not sure why is this tagged (functional-analysis). $\endgroup$ Jul 2 '14 at 5:31
  • 1
    $\begingroup$ @Martin Sleziak, I was hoping there was some way to resolve this using techniques such as infinite dimensional matrices etc... $\endgroup$ 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 $$ f(n)=f_0(1)+\sum_{k=1}^{n-1}\frac1k,\qquad n=2,3,\dots\,. $$

In particular, the "general solution" in the OP is not general but only special.

  • $\begingroup$ How did you arrive at those results? $\endgroup$
    – mvw
    Jul 2 '14 at 15:43
  • $\begingroup$ 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 $f(x)=f(x-1)+(x-1)^{-1}=f(x-2)+(x-2)^{-1}+(x-1)^{-1}=f_0(x-2)+(x-2)^{-1}+(x-1)^{-1}$ $\endgroup$
    – Vladimir
    Jul 2 '14 at 15:49
  • $\begingroup$ Thank you for that information! $\endgroup$
    – mvw
    Jul 2 '14 at 16:01

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