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How do you solve an inequality with the factorial of a variable?

Example: Determine the interval of $n \in \Bbb N$ for which the following inequality holds:

$$n! \leq 157788 \cdot 10^{10} $$

Can this be solved using algebraic techniques?

If not, what calculus techniques can I use to solve this inequality?

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    $\begingroup$ Stirlings formula is helpful for this, I don't remember the exact details, but I think it is something like $n!\approx \left({n\over e}\right)^n$. I'll look it up. $\endgroup$
    – abiessu
    Sep 30, 2013 at 23:54
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    $\begingroup$ Using the Gamma function, its easy to see $n \leq \lfloor\Gamma^{-1}(157788 \cdot 10^{10})\rfloor - 1$. But computing the inverse is still a problem. $\endgroup$ Sep 30, 2013 at 23:56
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    $\begingroup$ It is actually $n!\approx \left({n\over e}\right)^n\sqrt{\pi n}$. $\endgroup$
    – abiessu
    Sep 30, 2013 at 23:57
  • $\begingroup$ @abiessu Computing the inverse of that function is hard too. $\endgroup$ Sep 30, 2013 at 23:58
  • $\begingroup$ @pratyushsarkar: true, but a rough estimation can be garnered from taking $\log_n k$ and using the above formula to check how close the guess is. $\endgroup$
    – abiessu
    Oct 1, 2013 at 0:00

1 Answer 1

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If you want to think about it algebraically, you would probably try to factor. So, notice the 10^10. 10s can only come from a combination of evens and multiples of 5. (namely, you will be limited by multiples of 5, as there are fewer of them). So, 10^10 implies that you have 10 multiples of 5, but not 11. Thinking about this, however, we note that 50! is way too big, so we have to be more rudimentary.

Thought of a different way, our answer is on the scale of $10^{16}$. This means that it is about 15 things the size of 10 multiplied together. namely 2*5, 3*4,6*7,8,9, 10, etc. to get 15 places this way, we'd likely have to go up to about 16ish. Being realistic, we could try some numbers in this area.

That gives us a good place to start, so we play around in that area, we find that it has to be 17!. (the comment below was absolutely right, I messed up and said 15!)

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    $\begingroup$ As a rule-of-thumb method, this seems pretty good, but I think $17! < 157788 \cdot 10^{10} < 18!$. Or at least so I gather from an online calculator. $\endgroup$
    – user43208
    Oct 1, 2013 at 0:12
  • $\begingroup$ @user43208 This estimation method was sufficient for me because $n \in \Bbb N$ allowed me to apply the squeeze theorem to show the inequality holds for $0 \leq n \leq 17$. Thanks Th_ $\endgroup$ Oct 1, 2013 at 0:30

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