Explaining Ultimate Factorial Value in x^n Integer Series

[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!

 

[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!.

 

[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

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

 

[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.

 

[blue_raver22]

I am intrigued by this phenomenon and would like to provide a response to your questions. The concept of calculating differences between consecutive terms in a series and then taking the differences of those differences is known as finite differences. This method is often used in mathematics to find patterns and make predictions in sequences and series.

In the case of the x^n integer series, where n is the exponent that produces the initial numbers, we can see that the differences between consecutive terms will always result in a constant value of n. This is because each term in the series is multiplied by n, which is why we see the constant value of n at deeper levels.

Now, why does this constant value end up being equal to n!? This can be explained by the binomial theorem, which states that (x+y)^n = Σ(nCr)x^(n-r)y^r, where nCr represents the combination of n objects taken r at a time. When we expand this equation for x^n, we get x^n = Σ(nCr)x^(n-r). If we take the differences between consecutive terms, we end up with (nCr)x^(n-r-1). Continuing this process, we eventually reach a constant value of n!, which is the number of ways to arrange n objects.

So, there is a mathematical proof for this phenomenon, and it is not necessarily related to calculus. However, the concept of finite differences is similar to taking derivatives, as you mentioned. This is because both methods involve finding the rate of change or the slope of a function at a given point. However, in the case of finite differences, we are looking at the discrete changes between consecutive terms, whereas in calculus, we are looking at the continuous changes in a function.

In conclusion, the explanation for why the ultimate factorial value is found in the x^n integer series lies in the binomial theorem and the concept of finite differences. It is not necessarily related to calculus, but there are some similarities in the methods used. I hope this response has provided some clarity on this interesting phenomenon.