# If $f(x) = \sum \limits_{n=0}^{\infty} \frac{x^n}{2^{n(n-1)/2} n!}$ then $f^{-1}(f(x)-f(x-1))-\frac{x}{2}$ is bounded

For every $x>0$, let $$f(x) = \sum \limits_{n=0}^{\infty} \dfrac{x^n}{2^{n(n-1)/2} n!}.$$ Let $f^{-1}$ be the functional inverse of $f$.

How to show there exists a positive real constant $C$ such that, for all $x$, $$\left(f^{-1}\left(f(x)-f(x-1)\right)-\frac{x}{2}\right)^2 < C$$

Edit : I believe this is true because $f'(x) = f(x/2)$.

• Any suggestions for improving the title ?
– mick
Oct 21, 2014 at 22:15
• Why do you think this is true? Oct 21, 2014 at 23:17
• Because $f ′ (x)=f(x/2)$ . That is an analogue where the difference operator is replaced by a differential operator. @martycohen
– mick
Oct 22, 2014 at 22:47
• Do you mean "there exists a fixed positive real constant $C$ such that, for all $x$, ..."?
– Did
Oct 23, 2014 at 22:03
• @Did thats what I said in reverse order right ?
– mick
Oct 23, 2014 at 22:04

Assume that $x\gt1$ then $f'(t)=f\left(\frac{t}2\right)$ for every $t\gt0$ and $f$ is increasing on the interval $\left(\frac{x-1}2,\frac{x}2\right)$ hence $$f(x)-f(x-1)=\int_{x-1}^xf'(t)\mathrm dt=\int_{x-1}^xf\left(\tfrac{t}2\right)\mathrm dt,$$ yields $$f\left(\tfrac{x-1}2\right)\leqslant f(x)-f(x-1)\leqslant f\left(\tfrac{x}2\right),$$ that is, $$\tfrac{x-1}2\leqslant f^{-1}\left(f(x)-f(x-1)\right)\leqslant\tfrac{x}2,$$ in particular, $$\left(f^{-1}\left(f(x)-f(x-1)\right)-\tfrac{x}2\right)^2\leqslant\tfrac14.$$
• I think that as x grows to + infinity we get the limit :$\left(f^{-1}\left(f(x)-f(x-1)\right)-\frac{x}{2}\right)^2 = 0$