# Iterative sequence converging

Consider the sequence

$$1^4,2^4,3^4,\ldots,k^4,\ldots$$

Form a new sequence, whose terms consist of the difference of the above sequence.

$$2^4-1^4,3^4-2^4,4^4-3^4,\ldots$$

Repeat the process with the terms of this new sequence. When this is done sufficiently many times, you will eventually get the sequence

$$24, 24, 24, 24, \ldots$$

Why is this the case fundamentally, from a mathematical perspective? Why 24 exactly? I suspect it might have something to do with the fact that $24 = 4!$, though this could be completely off. Will it still work if we used the same algorithm with any arbitrary exponent $n$ instead of $4$? I can not answer this question because I can't seem to formalize this process in a way that allows me to see obvious convergence to a constant sequence.

• if $f(x)$ is a polynomial of degree $n$ with leading coefficient $1$, then $f(x+1)-f(x)$ is a polynomial of degree $n-1$ with leading coefficient $n$. If you start from a polynomial of degree $n$ with leading coefficient $1$ and taking such a finite difference $n$ times. you get $n!$. Feb 4, 2018 at 6:59

This is the calculus of finite differences. You are starting with a function $f(x)$ (here $f(x)=x^4$) and defining a new one by $g(x)=f(x+1)-f(x)$, then iterating the procedure.
In general if $f(x)=a_nx^n+\cdots+a_0$ is a polynomial of degree $n$, then $g(x)=na_nx^{n-1}+\cdots$ is also a polynomial. Iterating then gives $n(n-1)a_n x^{n-2}+\cdots$ and after $n$ stages we get the constant $n(n-1)(n-2)\cdots 1a_n$.