# Explanation of ultimate factorial value in differences between x^n integer series

1. Sep 4, 2010

### ershi

In the situation where differences between consecutive squares, (or consecutive cubes, consecutive x^4, etc.) are calculated,
then the differences between those differences are calculated, and then the differences of those differences, and so on until you reach a constant number at a deep enough level,
which is equal to n! (n being the exponent that produced the initial numbers)

Is there some type of proof or explanation why it happen to be a factorial value?

Is it involved with calculus, since it is similar to transforming a function into a derivitive function, and continuing to find the derivitive?

Example:
F(x)=x^5
F'(x)=5x^4
F''(x)=5*4x^3
F'''(x)=5*4*3x^2
F''''(x)=5*4*3*2x
F'''''(x)=5*4*3*2*1=120=5!

2. Sep 4, 2010

### JCVD

My tex is not working properly, so I have separated what should be in tex but left out the delineating commands so that someone may hopefully be able to present this post with the proper tex.

Given any finite list X(n)_0 = (x_0, x_1, ..., x_n) of at least 2 integers, we define the first difference iteration of X(n)_0 to be the list X(n)_1 = (x_1-x_0, x_2-x_1, ..., x_n-x_{n-1}), and we inductively define the (j+1)st difference iteration of X(n)_0 to be the first difference iteration X(n)_{j+1} of X(n)_j. Note that by this construction, for j between 0 and n, X(n)_j will have length n-j+1.

Let x be any nonnegative integer. We show that for any positive integer n, the nth difference iteration of the list X(n)_0 = (x^n, (x+1)^n, ..., (x+n)^n) is the single number n!.

Let n = 1. Then X(1)_0 = (x, x+1) and X(1)_1 = 1 = 1!.

Assume inductively that for any j between 1 and n-1, the jth difference iteration of the list X(j)_0 = (x^j, (x+1)^j, ..., (x+j)^j) is the single number j!. By the binomial theorem, the first difference iteration of X(n)_0 = (x^n, (x+1)^n, ..., (x+n)^n) is

X(n)_1 = (\sum_{j=0}^{n-1}\begin{pmatrix}n \\ j\end{pmatrix}x^j, \sum_{j=0}^{n-1}\begin{pmatrix}n \\ j\end{pmatrix}(x+1)^j, ..., \sum_{j=0}^{n-1}\begin{pmatrix}n \\ j\end{pmatrix}(x+n-1)^j).

Applying the induction hypothesis, if we write an entry in the kth difference iteration X(n)_k as the summation of n terms a(k)_0, a(k)_1, ..., a(k)_{n-1}, with each a(k)_j consisting of the terms indexed by j in the summations of X(n)_1 appropriately added or subtracted (so that all entries in X(n)_1 have a(1)_0 = 1 and all entries in X(n)_k have a(k)_0 = 0 for k>1), all entries in X(n)_j will have

a(j)_{j-1} = \begin{pmatrix}n \\ j-1\end{pmatrix}(j-1)!.

so all entries in X(n)_k will have a(k)_j = 0 for k > j+1. Thus the nth difference iteration X(n)_n will be left with the single term

a(n)_{n-1} = \begin{pmatrix}n \\ n-1\end{pmatrix}(n-1)! = n!.

3. Sep 4, 2010

### JCVD

Now it seems to be kind of working.

Given any finite list X(n)_0 = (x_0, x_1, ..., x_n) of at least 2 integers, we define the first difference iteration of X(n)_0 to be the list X(n)_1 = (x_1-x_0, x_2-x_1, ..., x_n-x_{n-1}), and we inductively define the (j+1)st difference iteration of X(n)_0 to be the first difference iteration X(n)_{j+1} of X(n)_j. Note that by this construction, for j between 0 and n, X(n)_j will have length n-j+1.

Let x be any nonnegative integer. We show that for any positive integer n, the nth difference iteration of the list X(n)_0 = (x^n, (x+1)^n, ..., (x+n)^n) is the single number n!.

Let n = 1. Then X(1)_0 = (x, x+1) and X(1)_1 = 1 = 1!.

Assume inductively that for any j between 1 and n-1, the jth difference iteration of the list X(j)_0 = (x^j, (x+1)^j, ..., (x+j)^j) is the single number j!. By the binomial theorem, the first difference iteration of X(n)_0 = (x^n, (x+1)^n, ..., (x+n)^n) is

$$X(n)_1 = (\sum_{j=0}^{n-1}\begin{pmatrix}n \\ j\end{pmatrix}x^j, \sum_{j=0}^{n-1}\begin{pmatrix}n \\ j\end{pmatrix}(x+1)^j, ..., \sum_{j=0}^{n-1}\begin{pmatrix}n \\ j\end{pmatrix}(x+n-1)^j).$$

Applying the induction hypothesis, if we write an entry in the kth difference iteration X(n)_k as the summation of n terms a(k)_0, a(k)_1, ..., a(k)_{n-1}, with each a(k)_j consisting of the terms indexed by j in the summations of X(n)_1 appropriately added or subtracted (so that all entries in X(n)_1 have a(1)_0 = 1 and all entries in X(n)_k have a(k)_0 = 0 for k>1), all entries in X(n)_j will have

$$a(j)_{j-1} = \begin{pmatrix}n \\ j-1\end{pmatrix}(j-1)!.$$

so all entries in X(n)_k will have a(k)_j = 0 for k > j+1. Thus the nth difference iteration X(n)_n will be left with the single term

$$a(n)_{n-1} = \begin{pmatrix}n \\ n-1\end{pmatrix}(n-1)! = n!.$$

4. May 22, 2011

### jmh2o

Hello JCVD, I have been trying to figure out the reason to the why the nth difference is equal to n! for years. Thank you for resolving what I knew to be true. I wish I knew your name so that I could explain who revealed this great mystery to me. I would not like to pretend that I figured this out myself, nor am I do I completely understand but I am going to run it by a professor to get a better understanding. It has been very important to me. I am really thankful that ershi asked you that question.