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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 $f'(x) = f(x/2)$.

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  • $\begingroup$ Any suggestions for improving the title ? $\endgroup$
    – mick
    Oct 21 '14 at 22:15
  • $\begingroup$ Why do you think this is true? $\endgroup$ Oct 21 '14 at 23:17
  • $\begingroup$ Because $f ′ (x)=f(x/2)$ . That is an analogue where the difference operator is replaced by a differential operator. @martycohen $\endgroup$
    – mick
    Oct 22 '14 at 22:47
  • $\begingroup$ Do you mean "there exists a fixed positive real constant $C$ such that, for all $x$, ..."? $\endgroup$
    – Did
    Oct 23 '14 at 22:03
  • $\begingroup$ @Did thats what I said in reverse order right ? $\endgroup$
    – mick
    Oct 23 '14 at 22:04
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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, $$\left(f^{-1}\left(f(x)-f(x-1)\right)-\tfrac{x}2\right)^2\leqslant\tfrac14.$$

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  • $\begingroup$ I think that as x grows to + infinity we get the limit :$\left(f^{-1}\left(f(x)-f(x-1)\right)-\frac{x}{2}\right)^2 = 0$ $\endgroup$
    – mick
    Oct 23 '14 at 22:34
  • $\begingroup$ Im sorry , I wanted to give you the bounty but I had no time for MSE :( $\endgroup$
    – mick
    Nov 7 '14 at 23:10

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