My Math Forum Factorial expressed as the sum of a series - new ?
 User Name Remember Me? Password

 Number Theory Number Theory Math Forum

 May 13th, 2009, 05:33 AM #1 Newbie   Joined: May 2009 Posts: 25 Thanks: 0 Factorial expressed as the sum of a series - new ? I noted this result when I was a child If you take the series 1^n , 2^n, 3^n ........ Then repeatedly take the differences, after n operations you end up with n! I.e for n=2 then 1,4,9,16,25 3,5,7,9 2,2,2 n=3 1,8.27.64,125, 216 7,19,37,61,91 12,18,24,30 6,6,6 From this by considering the general series 1^n, 2^n, 3^n,........r^n...and taking differences I derived the general result obtaining the alternating series n! = sum (r = 0 to n) (-1)^r * nCr * (n+1-r)^n Where C is the combinatorial function nCr = n!/r!(n-r)! or sum(r= 0 to n) (-1)^r * (1/(r!(n-r)!) * (n+1-r)^n = 1 for any n i.e 5! = 6^5 - 5 * 5^5 + 10 * 4^5 - 10 * 3^5 + 5 * 2^5 -1 = 7776 - 15625 + 10240 - 2430 + 160 -1 = 120 Is this result known about elsewhere or am I just showing something trivial like n! = n! ?
 May 13th, 2009, 08:50 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 932 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Factorial expressed as the sum of a series - new ? You've discovered an important result: $x'=1!$ $(x^2)''=2!$ $(x^3)'''=3!$ . . . $(x^n)^{(n)}=n!$
 May 13th, 2009, 09:46 AM #3 Senior Member   Joined: Nov 2007 Posts: 258 Thanks: 0 Re: Factorial expressed as the sum of a series - new ? That last one is nice. Here it is in LaTeX (more readable) $\sum_{j=0}^n(-1)^j{n \choose j}(n+1-j)^n = n!$
May 14th, 2009, 01:36 AM   #4
Senior Member

Joined: Oct 2008

Posts: 215
Thanks: 0

Re: Factorial expressed as the sum of a series - new ?

Quote:
 Originally Posted by brunojoyal That last one is nice. Here it is in LaTeX (more readable) $\sum_{j=0}^n(-1)^j{n \choose j}(n+1-j)^n = n!$
We could have
$\sum_{j=0}^n(-1)^j{n \choose j}(n+x-j)^n = n!$
Given n+x boxes and n balls, how much different ways there're to put the n balls into the n+x boxes and none of the first n boxes is empty?
Using Inclusion-Exclusion Principle, the result is the leftside of the equation. And it is obvious the result is n! too.

 May 14th, 2009, 02:29 AM #5 Newbie   Joined: May 2009 Posts: 25 Thanks: 0 Re: Factorial expressed as the sum of a series - new ? How can I find out if this series result has been obtained before, as seems fairly likely ? And more importantly why would we expect an alternating set of powers of n+1 -> 1 raised to the n to produce n!
October 31st, 2009, 02:29 AM   #6
Newbie

Joined: Oct 2009

Posts: 1
Thanks: 0

Re: Factorial expressed as the sum of a series - new ?

Quote:
Originally Posted by duz
Quote:
 Originally Posted by brunojoyal That last one is nice. Here it is in LaTeX (more readable) $\sum_{j=0}^n(-1)^j{n \choose j}(n+1-j)^n = n!$
We could have
$\sum_{j=0}^n(-1)^j{n \choose j}(n+x-j)^n = n!$
Given n+x boxes and n balls, how much different ways there're to put the n balls into the n+x boxes and none of the first n boxes is empty?
Using Inclusion-Exclusion Principle, the result is the leftside of the equation. And it is obvious the result is n! too.
For $x \in \mathbb{N}$ this is intuitive and clear. But it works also for $x \in \mathbb{R}$.

This is, for me, less clear...

 Tags expressed, factorial, series, sum

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# sum is n factorial

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post nothingisimpossible Calculus 1 August 11th, 2013 03:21 PM Issler Real Analysis 1 March 14th, 2012 12:56 PM Red5 Calculus 1 February 27th, 2010 09:12 PM karel747 Number Theory 10 January 9th, 2008 10:09 AM nothingisimpossible Algebra 1 January 1st, 1970 12:00 AM

 Contact - Home - Forums - Top