Exists a closed-form formula for $n!$? Otherwise, why not?

I recognize that the Fibonacci series can be defined using the Binet's formula.

Nonetheless, I was asking yourself if there was a comparable formula for $n!$.

Is this feasible? Otherwise, why not?

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2019-05-06 22:12:08
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Answers: 4

This is a riff on several of the comments concerning what could comprise an "solution" and also what "closed form" could suggest : although it is rather sarcastic, it is planned to motivate ideas concerning these concerns.

Our base - 10 number system analyzes a string ${a_n}{a_{n-1}} \cdots {a_0}$ as the amount $\sum_{i=0}^{n} a_i 10^i $ (which can be, and also is, computed recursively as $a_0 + 10 \left( a_1 + 10 \left( \cdots + 10 a_n \right) \cdots \right)$). If you take the previous to be an appropriate "closed form" depiction, after that why not make use of a mild alteration of this number system? Especially, analyze the very same string as equivalent to $a_0 + 2 \left( a_1 + 3 \left( \cdots + (n+1) a_n \right) \cdots \right)$ and also call for that $0 \le a_0 \le 1, 0 \le a_1 \le 2, \ldots, 0 \le a_n \le n$. In this "factorial" number system, $n! = 10 \cdots 0$ is stood for as a straightforward $n$ - figure string: it is "shut"!

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2019-05-17 11:55:55
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Contrast these 2 approaches for computing n!:

  • Counting permutations of a n - component set one - by - one (this is what n! matters), vs.
  • Multiplying with each other the numbers 1,2, ..., n.

Therefore n! stands for an extremely reliable method for counting permutations! So I assume (and also I do not assume I'm the just one) that n! need to possibly be taken into consideration a closed form (unless you have a few other clear definition of what comprises a "shut" kind).

For more analysis, I advise : H. S. Wilf, What is a solution?, Amer. Mathematics. Month-to-month, 89 (1982 ), pp. 289-- 292, DOI: 10.2307/2321713, JSTOR.

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2019-05-17 09:13:58
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If you're eager to approve an indispensable as a solution, after that $n! = \int_0^\infty t^n e^{-t} \: dt$.

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2019-05-08 20:42:22
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The loved one mistake of Stirling's approximation obtains randomly tiny as n obtains bigger.

$$n!\sim\sqrt{2\pi n} \left(\frac{n}{e}\right)^n$$

However, it is just an estimate, not a shut - kind of $n!$

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2019-05-08 18:37:52
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