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This is a generalization of this question :: Which is larger? $20!$ or $2^{40}$?.

No explicit general solution was presented there and I'm just curious :D


Edit :: I want a most-general solution lfor arbitrary $x$ and $y$; not some specfic cases which can then be solved by direct computation. Below, Ahaan-Rungta shows that the case $x < y$ is the one to be considered.

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Since $y!$ does not involve $x$, this question is not well posed. – lhf Dec 16 '13 at 15:42
The solution may require some relation between $x$ and $y$ ---as I have no solution, I am unable to `pose' the question in a more suitable fashion. – Musa Al-hassy Dec 18 '13 at 11:42

4 Answers 4

up vote 4 down vote accepted

"Just compute them and compare" is the only fully failsafe method.

In most cases, however, estimating the logarithms of both of $x^y$ and $y!$ using Stirling's formula will yield a conclusive result without needing to compute the two values in high detail.

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Makkolm : As usually stated, Stirling's formula contains "big-O" notation that seems to make it useless to answer a specific problem with specific numbers such as the OP's. How can you remedy this? This is not a nitpick, I really want to know how to use Stirling's formula in problems like this. Where/how can one find guaranteed error bounds for the formula? – Stefan Smith Dec 17 '13 at 15:56
@Stefan: The Wikipedia article gives the error bound: $$1 \leq \frac{n!}{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n} \leq \frac{e}{\sqrt{2\pi}} = 1.0844375514...$$ In other words, an absolute error bound for the logarithm of $n!$ is $1-\log \sqrt{2\pi} \approx 0.081$. – Henning Makholm Dec 17 '13 at 16:03
thanks, I spent a minute or so scanning the article and I did not see such a bound. – Stefan Smith Dec 17 '13 at 19:10

For a fixed $x,$ $y!$ will eventually be larger, as you can tell by using Stirling's formula.

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On the other hand, for a fixed $y$, as $x$ increases, $x^y$ will eventually be larger. It's really not clear from the question whether to keep $x$ fixed and let $y$ increase (as you do), or keep $y$ fixed and let $x$ increase (as I just did), or have them both increase in some way (in which case it matters exactly how they increase). – ShreevatsaR Dec 17 '13 at 7:12
SreevatsaR :: yes; I am not choosing either constraint. I would a most-general solution. Thank-you. – Musa Al-hassy Dec 18 '13 at 11:44


In general, for positive integer $x,y$ we have:

$y!=y(y-1)(y-2) \cdot\cdot\cdot2\cdot1$

$x^y=xxx\cdot\cdot\cdot xx$

  1. If $x\geq y$; then $ x^y\ge y! $
  2. If $x<y $, ?! (Now this is the question!)


If $ \left\lceil{\frac{y}{2}}\right\rceil+1< x<y$; then $x^y\ge y!.$ (easy proof)

Now, we have only the case $ x\le\lceil{\frac{y}{2}}\rceil +1 $

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$$\frac{20!}{2^{40}}= \frac{20!}{4^{20}} = \left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right) \left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)\cdots \left(\frac{20}{4}\right)>\cdots$$

$$ \left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right) \left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)=\frac{9!}{4^9}=\frac{2835}{2048}>1,$$

so $20!>2^{40}$.

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Cool; now what about arbitrary $x$ and $y$? – Musa Al-hassy Dec 18 '13 at 11:45
That answer probably belongs to…. – lhf Dec 18 '13 at 11:48
@Moses : the inequality Henning Makholm cites in his comment under his own answer to this question will work for many $x$ and $y$, but not all. – Stefan Smith Dec 19 '13 at 1:59

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