# k-th Difference of product of multiple arithmetic progression powers

I found a "surprising" result involving the $k$-th Difference of product of multiple arithmetic progression powers but am not sure how to prove it. Hope somebody here can help me to prove.

Given $n$ arithmetic progressions with initial terms $a_{i}$ and the common differences of successive members $d_i$ and powers $p_i$ $(i=1, \dots, n)$, we can form a product as following:

$a_{1}^{p_1}\times \dots \times a_{n}^{p_n}, (a_{1}+d_1)^{p_1}\times \dots \times (a_{n}+d_n)^{p_n}, (a_{1}+2d_1)^{p_1}\times \dots \times (a_{n}+2d_n)^{p_n},\dots, (a_{1}+j\cdot d_1)^{p_1}\times \dots \times (a_{n}+j\cdot d_n)^{p_n},\dots$

Now one can form $1$-difference(the difference between this term and the previous term), $2$-difference (the $1$-difference of $1$-difference), ..., $k$-difference on such product. What I found it that: after $p=\sum_{i=1}^np_{i}$-difference, we will get back a series with constant terms which are all $p!\prod_{i=1}^nd_{i}^{p_i}$.

Directly writing out every term and do the difference up to $p$-th is very cumbersome. There must be some trick to do this, but I am not seeing how.

Thank you for your help.

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Given any polynomial $q(x)$ of degree $p$, the sequence $q(0),q(1),...,q(j),...$ has this property.

You have a particular $q(x)=\prod_{i=1}^n (a_i+xd_i)^{p_i}$, which is a polynomial of degree $p=\sum_{i=1}^n p_i$ in $x$.

You can prove the more general result by noting that if $q(x)$ is a polynomial of degree $p$ then $(\Delta q)(x)=q(x+1)-q(x)$ is a polynomial of degree $p-1$. So by induction, $(\Delta^p q)(x)$ is a constant.

Using the finite Taylor series: $$q(x+1)=\sum_{i=0}^p \frac{q^{(i)}(x)}{i!}$$

where $q^{(i)}(x)$ is the $i$th derivative of $q$. So we see that $$(\Delta q)(x) = q(x+1)-q(x) = \sum_{i=1}^p \frac{q^{(i)}(x)}{i!}$$

But if $a(x), b(x)$ are polynomials, then $\Delta(a+b)=\Delta a + \Delta b$.

And, since $q^{(i)}(x)$ is of degree $p-i$, then $\Delta^{p-1} p^{(i)}$ is zero if $i>1$. So we get that:

$$\Delta^p q(x) = \Delta^{p-1} q'(x)$$

And therefore $\Delta^p q(x) = q^{(p)}(x)$.

But $q^{(p)}(x)$ is just $p!$ times the initial coefficient of $q$.

In your instance, that would be $p!\prod d_i^{p_i}$.

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How to get the constant even if you can prove it is one. :) –  Qiang Li Mar 5 '12 at 20:37
Okay, completed the proof. –  Thomas Andrews Mar 5 '12 at 20:47
(Unfortunately, the derivative and Taylor series are not really "pre-calculus" ideas.) You can prove it more directly by noting that if $q$ is a polynomial of degree $p$ with initial coefficient $A$ then $\Delta q$ is a polynomial of degree $p-1$ with initial coefficient $pA$. –  Thomas Andrews Mar 5 '12 at 21:16