3
$\begingroup$

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:

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

$\endgroup$
11
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.