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If we know that there exist a positive integer $n$ in an interval $[a,b]$. Here $a,b$ are real numbers. How we can know the bounds for the factorial $n!$.

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  • $\begingroup$ What's wrong with $b!$? $\endgroup$
    – Arthur
    Dec 5 '19 at 13:29
  • $\begingroup$ @Arthur: Here $a,b$ are real numbers. $\endgroup$
    – Safwane
    Dec 5 '19 at 13:30
  • $\begingroup$ And with $\lfloor a\rfloor!$ and $\lceil b\rceil!\,$? $\endgroup$
    – Bernard
    Dec 5 '19 at 13:33
  • $\begingroup$ @Gabe Yes you can: the other natural idea $a\cdot (a-1)\cdot \ldots$ is sometimes better (say $a=5.8$, then $\Gamma(6.8)<497$ whereas $5.8\times 4.8\times\ldots>533$). $\endgroup$ Dec 5 '19 at 13:37
  • $\begingroup$ Ah yes, I didn’t notice that n was an integer. Then the bound would simply be ceil(a)! And floor(b)! then, no? $\endgroup$
    – Gabe
    Dec 5 '19 at 13:54
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There are two natural things to say:

  • It's between $\Gamma(a+1)$ and $\Gamma(b+1)$ where $\Gamma$ is defined here.
  • It's between $a\cdot (a-1)\cdot\ldots\, $ (stop when you reach a factor that is less than $2$) and $b\cdot (b-1)\cdot\ldots\, $ (same).

In fact it seems$^{\color{blue}{\left[\underline{\text{reference needed}}\right]}^{\star}}$ that $\Gamma(x+1)$ is always less than or equal to $x\times (x-1)\times\ldots$, meaning that the tightest couple of bounds would be $$a\cdot(a-1)\cdot\ldots\cdot (a-\lfloor a\rfloor+1)<n!<\Gamma(b+1)$$


$^\star$Edit: Peter Foreman proved the inequality $\Gamma(x+1)\leq \prod_{k=0}^{\lfloor x\rfloor-1}(x-k)$ in the comments below.

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    $\begingroup$ Using the recursive defintion of $\Gamma(x+1)$ we have$$\Gamma(x+1)=\Gamma(x-\lfloor x\rfloor+1)\prod_{k=0}^{\lfloor x\rfloor-1}(x-k)$$but, as $1\le x-\lfloor x\rfloor+1\lt2$, we have $0.885603\lt\Gamma(x-\lfloor x\rfloor+1)\le1$ which implies that$$\Gamma(x+1)\le\prod_{k=0}^{\lfloor x\rfloor-1}(x-k)$$ $\endgroup$ Dec 5 '19 at 13:55
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    $\begingroup$ @PeterForeman Nice. I'll include this in the body of the answer. $\endgroup$ Dec 5 '19 at 13:58
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Similar to what Aurnaud Mortier said , I think the answer is that $\lceil a \rceil !\le n! \le \lfloor b \rfloor ! $ , where $\lceil a \rceil $ is the ceiling function of $a$ and $\lfloor b \rfloor$ is the floor function of b. If $a$ and $b$ are integers, it's obvious why that is. If one of them is a real number, you can only have a factorial of a natural number, and you check the one imediatly above or down.

This is slighty more precise than between $a⋅(a−1)\cdot{...}$ and $b⋅(b−1)\cdot{...}$ , at least if you are looking for factorial for naturals. If you want to "extend" it to the reals, the gama function solution is the right one, this is for n as a positive integer.

*Edit

This is tighter than other solutions because for all $x$:

$\lfloor x \rfloor ! \le Γ(x+1)$

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