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So the problem is to find $f(x)$ such that: $$f(x+1)-f(x)=1/(x+1)$$

I have found that $\ln x$ is a good approximation for large values of $x$.

$f(x)$ not differentiable at $x=-1$.

George Law
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NickThyname
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  • This is called a functional equation. –  Aug 24 '17 at 16:13
  • The first thing that comes to mind is that there might be more than one function satisfying this. For instance, both the functions $x$ and $\lfloor x\rfloor$ satisfy $f(x+1)-f(x)=1$. –  Aug 24 '17 at 16:13
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    This equation is undefined for $x=-1$ so you should say exactly for what values of $x$ the equation needs to hold. Note that adding a constant to a solution gives us another solution. – hardmath Aug 24 '17 at 16:15
  • $f(x)=\psi(x+1)+C$ would be a solution, where $\psi$ is the digamma function. –  Aug 24 '17 at 16:17
  • You shouldn't write $F(x)$ if you mean $f(x). \qquad$ – Michael Hardy Aug 24 '17 at 19:32

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This is a simple recurrence, that can be written

$$f(x)=f(x-1)+\frac1x.$$

By induction,

$$f(x)=f(x-n)+\sum_{k=0}^{n-1}\frac1{x-k}.$$

If we assume the initial condition that $f$ is known in $[0,1)$, we have

$$f(x)=f(\{x\})+\sum_{k=0}^{\left\lfloor x\right\rfloor-1}\frac1{x-k}.$$

For large $x$, it indeeds tends to an Harmonic series, with a perturbation term.