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Can anyone prove that $\lfloor\log n\rfloor !$ is an exponential function?

I've tried a lot but i didn't find anything relating to the solution except e number that i guess it can help.

jimjim
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sajjad
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    $\log{n}$ is never an integer except for $n=1$ so what do you mean by $(\log{n})!$. Exponential function means $\exists a,(\log{n})!=a^n$ or is it a matter of limit at infinity? You need to give more context. – marwalix Jan 08 '17 at 08:26
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    @marwalix I believe the brackets are supposed to denote the greatest integer function... – aras Jan 08 '17 at 08:28
  • @marwalix I guess the "[ ]" stands for integer part function (some books use this notation) – AlienRem Jan 08 '17 at 08:28
  • I had guessed so this is why in my question I replaced with brackets. I will edit using \lfloor and \rfloor. We still need more context about exponential. – marwalix Jan 08 '17 at 08:33
  • Another way out is using the $\Gamma$ function. – zoli Jan 08 '17 at 08:35
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    "Please someone help me quickly"... What makes this question so urgent? Is the factorial of the natural logarithm going to save the world? – pseudoeuclidean Jan 08 '17 at 08:51
  • What do you call "an exponential function" ? Mine are of the form $a^n$ for some real $a$ and follow the recurrence $f(n+1)=af(n)$. This is clearly not matched by your expression. Improve the question. –  Jan 08 '17 at 11:34

3 Answers3

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$$\log(\lfloor \log n\rfloor!)=\sum_{k=2}^{\lfloor\log n\rfloor}\log k<\int_2^{\lfloor\log n\rfloor}\log(x+1)\,dx\\ =(\lfloor\log n\rfloor+1)(\log(\lfloor\log n\rfloor+1)-1)-3\log3+3.$$

Taking the antilogarithm, you get an $O(n^{\log\log n})$ expression.

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An elementary disproof

The family of exponential functions is characterized by the following functional equation

$$f(x+y)=f(x)f(y).$$

For our case, this means that we should be able to show that

$$\lfloor\log(n+m)\rfloor !=\lfloor\log(n)\rfloor !\lfloor\log(m)\rfloor !.$$

Assume for the sake simplicity that the base of our $\log$ is $10$.

For the integers $N<M$ let $n=10^N, m=10^M$. Then

$$\lfloor\log(10^{N}+10^M)\rfloor!<\lfloor\log(2\times 10^{M})\rfloor!=\lfloor\log(2)+ M\rfloor!= M!$$ and

$$\lfloor\log(10^{N})\rfloor!\lfloor\log(10^{M})\rfloor!=(N!)(M!)>M!.$$

So, our function is not exponential.

zoli
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  • neat, can this method be used for logarithmic and polynomial functions? any idea what is analogous to f(x+y)=f(x)f(y) for super exponentials or teteration functions? – jimjim Jan 08 '17 at 21:42
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Let $f(n)= [\log n] !$ for $n\geq 1.$ We have $f(n) \leq (1+\log n)!=(\log ne)! .$

From Stirling's Formula $x!=(x/e)^x\sqrt {2\pi x}\;(1+d(x))$ where $\lim_{x\to \infty}d(x)=0$ we have $$f(n)\leq ((\log ne)/e)^{\log ne}\sqrt {2 \pi \log ne}\;(1+d(\log ne)).$$ Taking logs we have $$\log f(n)\leq (\log ne)^2 -\log ne +\log \sqrt {2 \pi}+ (1/2)\log \log ne+\log (1+d(\log ne)).$$ For any $k>0$ we have therefore, $$\lim_{n\to \infty} (\log f(n))/\log e^{kn}=(\log f(n))/k\log n=0$$ because $(\log ne)^2/n\to 0$ as $n\to \infty.$

So as $n\to \infty,$ the function $f(n)$ grows more slowly than any increasing exponential function in $n $.

Footnotes: (i). FYI, for $n\geq 1$ we have $0<d(n)<1+1/6n.$ (ii). It is not hard to show there exists $L>0$ such that $L=\lim_{n\to \infty} (\log n!)-( (n+1/2) \log n -n),$ which suffices for this Q . Proving that $L=\log \sqrt {2\pi}$ is more difficult.

DanielWainfleet
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