Upper bound on $f(x)= \frac{\lfloor x\rfloor!}{\lfloor x+h\rfloor!} 2^{\lfloor x+h\rfloor!-\lfloor x\rfloor!}$

Question

I am trying to show that the following function is bounded: \begin{align} f(x)= \frac{\lfloor x\rfloor!}{\lfloor x+h\rfloor!} 2^{\lfloor x+h\rfloor!-\lfloor x\rfloor!} \end{align} for some $h>0$ and $x \ge 0$.

I did a few simulations and looks like it is bounded.

Answer 1

This function is never bounded. For instance, if $h=1$, then whenever $x$ is an integer we have $$f(x)=\frac{x!}{(x+1)!}2^{(x+1)!-x!}=\frac{2^{x\cdot x!}}{x+1}$$ which grows very very fast. In general, $f(x)$ will have the form $\frac{y!}{(y+n)!}2^{(y+n)!-y!}$ where $y=\lfloor x\rfloor$ and $n$ can be either $\lfloor h\rfloor$ or $\lceil h\rceil$ (depending on the fractional part of $x$). As long as $n>0$, this expression grows very fast: $\frac{y!}{(y+n)!}=\frac{1}{(y+1)(y+2)\dots(y+n)}$ is roughly $\frac{1}{y^n}$ when $y$ is very large while $2^{(y+n)!-y!}>2^{y!}$ grows much faster than $y^n$.

(I am rather baffled by your "simulations" that found $f(x)$ was bounded. Unless you were choosing values of $x$ such that $\lfloor x\rfloor=\lfloor x+h\rfloor$ and so $f(x)=1$, it should be really obvious from any examples that $f(x)$ is growing very fast.)

Answer 2

By contradiction:

Let $0<h<1$ be the searched constant. Then let $x:=n-\dfrac h2$ where $n$ is an integer. For this $x$,

$$f(x)=\frac{(n-1)!}{n!}2^{n!-(n-1)!}$$ and the RHS is clearly unbounded.

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Note that if $h\ge1$, $f(x)$ is unbounded for any $x$.