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I am somewhat new to summations and products, but I know that the sum of the first positive n integers is given by:

$$\sum_{k=1}^n k = \frac{n(n+1)}{2} = \frac{n^2+n}{2}$$

However, I know that no such formula is known for the most simple product (in my opinion) - the factorial:

$$n!=\prod_{k=1}^n k = \prod_{k=2}^n k$$

I don't know if that is how products work, but I would really like to know!

So my question is why is there no explicit formula (in terms of n, other than n(n-1)...2*1) for the product of the first n integers? Is there a proof that says that one cannot exist or is it that one has not been discovered?

By explicit formula I mean a non-functional equation that does not require n amount of steps to calculate - just like the summation formula does not require n additions.

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  • $\begingroup$ Are you familiar with Stirling approximation? $\endgroup$
    – Kunnysan
    Commented Jul 29, 2013 at 8:31
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    $\begingroup$ This really depends on what one would accept as explicit formula. $\endgroup$ Commented Jul 29, 2013 at 8:36
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    $\begingroup$ Such a formula exists, but it involves integration :) $$n!=\int_0^{\infty}x^{n}e^{-x}dx.$$ $\endgroup$ Commented Jul 29, 2013 at 8:36
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    $\begingroup$ @O.L. That's not an improvement—instead of finding it in $n$ steps, you're finding it after an infinite number of steps!! $\endgroup$ Commented Nov 14, 2014 at 16:37
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    $\begingroup$ @columbus8myhw I never saw it that way until you pointed it out. Nice observation! $\endgroup$ Commented Nov 15, 2014 at 2:38

5 Answers 5

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(The following is a joke.)

Put $a_n:={1\over 6}n(n+1)(n+2)$ and $b_n:={1\over2}n(n+1)$, and define the magic constant $$\xi:=\sum_{n=1}^\infty {n!\over 2^{a_n}}\ =\ 0.630882266676063396815526621896\ldots\quad .$$ Then $$n!=\bigl\lfloor\> 2^{a_n}\xi\>\bigr\rfloor\quad \bigl({\rm mod}\ \ 2^{b_n}\bigr)\qquad(n\geq1)\ .$$ Try it out!

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  • $\begingroup$ Is this an encoding of an infinite sequence of integers into a real number? Hahaha $\endgroup$
    – Memming
    Commented Jul 29, 2013 at 11:52
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    $\begingroup$ I wonder if this magic constant has any other uses. $\endgroup$ Commented Jul 29, 2013 at 13:55
  • $\begingroup$ How did you arrive at the formula? $\endgroup$
    – BigbearZzz
    Commented Nov 14, 2015 at 2:31
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Since nobody seems to be willing to answer, let me try to summarize the hints from the comments.

First, we can write factorial as $$n!=\int_0^{\infty}x^n e^{-x}dx.$$ One of the consequences of this formula is Stirling's approximation: as $n\rightarrow\infty$, $$n!\sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n}.\tag{1}$$ At first sight, if there was an explicit algebraic formula for $n!$ in terms of $n$, in such expression there would be no place for $e$ or $\pi$; since they are both present in the asymptotics, one is tempted to conclude that no such algebraic formula can exist. Actually this is not quite true: as was pointed to me by Michalis, one can e.g. obtain $e$ as the asymptotics of $\left(1+\frac{1}{n}\right)^n$.

To show non-existence rigorously, you should first rigorously define what do you mean by "explicit formula" $n!=f(n)$, i.e. to define the class of "acceptable" functions $f$. For example, rational $f$ are immediately ruled out by the above asymptotics.

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    $\begingroup$ Hm. The existence of $e$ in asymtotics doesn't exclude an algebraic formula. For example $(n+1)^n\sim n^n e$. $\endgroup$
    – MichalisN
    Commented Jul 29, 2013 at 9:58
  • $\begingroup$ @Michalis You are right. Let me think how to say this more correctly. $\endgroup$ Commented Jul 29, 2013 at 10:06
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    $\begingroup$ The sum of the first $n$ integers takes us from addition up to the next operator, multiplication. So it seems reasonable to look for a formula for $n!$ involving exponentiation. $\endgroup$
    – Jack M
    Commented May 8, 2015 at 0:40
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I thought about Fermat theory and how there must be an algebraic version of

$n!=\frac{{\mathrm d}^n}{{\mathrm d}^nx}\,x^n$

and indeed it let me to a formula of the form

$n!=\sum_{k=0}^nR_{nk}\,k^n$

namely with $R_{nk}=(-1)^{n+k}{n\choose k}$. Of course, you would use factorials to compute the binomials, so there is no computational speedup, but it still makes for some nice looking equations:

0! = + 0^0 1! = - 0^1 + 1*1^1, 2! = + 0^2 - 2*1^2 + 2^2, 3! = - 0^3 + 3*1^3 - 3*2^3 + 3^3, 4! = + 0^4 - 4*1^4 + 6*2^4 - 4*3^4 + 4^4, 5! = - 0^5 + 5*1^5 - 10*2^5 + 10*3^5 - 5*4^5 + 5^5, 6! = + 0^6 - 6*1^6 + 15*2^6 - 20*3^6 + 15*4^6 - 6*5^6 + 6^6, 7! = - 0^7 + 7*1^7 - 21*2^7 + 35*3^7 - 35*4^7 + 21*5^7 - 7*6^7 + 7^7, 8! = + 0^8 - 8*1^8 + 28*2^8 - 56*3^8 + 70*4^8 - 56*5^8 + 28*6^8 - 8*7^8 + 8^8

The theorem underlying here is that, for all $n$

$\sum_{k=0}^n\dfrac{(-1)^k (-k)^n}{k!\,(n - k)!}=1$

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These two formulas give n! This was discovered by Euler. Reference this link for further reading. http://eulerarchive.maa.org/hedi/HEDI-2007-09.pdf

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If I am not mistaken, Manjul Bhargava generalized the factorial function to include numbers such as (pi)! and (2+√3)!

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    $\begingroup$ This generalization is tightly related to so called gamma function, which has been discovered long before Bhargava was born, and this is in no way related to the question. en.m.wikipedia.org/wiki/Gamma_function $\endgroup$
    – Wojowu
    Commented Nov 14, 2015 at 7:34
  • $\begingroup$ Oh interesting. I didn't see that the original poster (OP) asked for an explicit formula. $\endgroup$
    – Yunus Syed
    Commented Nov 14, 2015 at 14:46
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    $\begingroup$ But this is the only thing which OP asks for. This is even said in the question title. $\endgroup$
    – Wojowu
    Commented Nov 14, 2015 at 15:25
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    $\begingroup$ I misread it, sorry. $\endgroup$
    – Yunus Syed
    Commented Nov 14, 2015 at 16:01

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