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The differences between adjacent squares are the odd numbers:

0 1 4 9 16 25 …

1 3 5 7 9 …

The differences between adjacent odd numbers are 2 = 2! I found that this truth is more general: If the differences between adjacent powers of $n$ are written out, and the differences of those differences etc, the $(n+1)$th sequence is {n!, n!, …}

Is this a well-known theorem? Does it have an easy proof?

A.Γ.
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Toothrot
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  • Yeah, I think this is true; I can't recall a proof at the moment, though. – Akiva Weinberger Sep 18 '15 at 13:34
  • Yes. This is the field of "finite differences." – Thomas Andrews Sep 18 '15 at 13:35
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    By the way, some notation that you may or may not find useful: Define $\Delta f(x)$ to mean $f(x+1)-f(x)$. Define $\Delta^n f(x)$ to mean $\Delta(\dots(\Delta f(x))\dots)$, with $n\ $ $\Delta$s. Then you're asking if $\Delta^n x^n=n!$. – Akiva Weinberger Sep 18 '15 at 13:36
  • Related: $(x^n)'=nx^{n-1}$,...,$(x^n)^{(n)}=n!$. – A.Γ. Sep 18 '15 at 13:44
  • @user21820, can an older post be a duplicate of a newer? – Toothrot May 08 '19 at 08:25
  • @Toothrot: Yes. And in this case, since the linked question is better and has much more comprehensive answers, it is better to mark as duplicate in that direction, otherwise the better thread will have lower visibility. – user21820 May 08 '19 at 08:56

1 Answers1

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Good catch! In fact, a little more is true.

Take $f(x)$ a polynomial of degree $n$ (in your case, $f(x)=x^2$ or $f(x)=x^k$) and leading coefficient $a$ Then the polynomial $f_1(x)=f(x+1)-f(x)$ has degree $n-1$ and leading coefficient $an$. This is because $$(x+1)^n=x^n+nx^{n-1}+(\text{terms of degree $< n-1$}).$$ Iterating this procedure (called finite differences) you obtain $f_2(x)$ of degree $n-2$ and leading coefficient $an(n-1)$, $f_3(x)$ of degree $n-3$ and leading coefficient $an(n-1)(n-2)$, etc. After $n$ iterations you get to $f_n(x)$ of degree $n-n=0$ and leading coefficient $an(n-1)\dots 2\cdot 1$, that is $an!$.

Sonner
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