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Consider the differences between powers of successive natural numbers. Note that the difference between two consecutive squares is always odd. Here is my picture of this relationship:

0 - 0
       - 1
1 - 1      - 2
       - 3
2 - 4      - 2
       - 5
3 - 9      - 2
       - 7
4 - 16     - 2
       - 9
5 - 25

Etc. But this sort of pattern is not limited to the squares. Notice the "last common difference" in the third powers of the integers:

0 - 0
        - 1
1 - 1        - 6
        - 7       - 6
2 - 8        - 12
        - 19      - 6
3 - 27       - 18
        - 37      - 6
4 - 64       - 24
        - 61      - 6
5 - 125      - 30
        - 91
6 - 216

If one continues this process for the fourth, fifth, etc. powers of numbers, the "last common difference" is the factorial of the power in question. Thus any power of a natural number is expressible as a simple sum. This process leads to the question: What about the 1.5 power? For an answer, see the gamma function, which is a close relative to the factorial.

It is also possible to apply this process to generic polynomials. As applied to a given polynomial, the process has effects very similar to that of a derivative, and indeed, the results are exactly the secant line slopes, as opposed to the tangent line slopes from a true derivative. See the mean value theorem for how these relate. Here is one level of a "difference tree" for a generic fourth-degree polynomial, i.e., p(x) = ax^4 + bx^3 + cx^2 + dx + e:

p(x+0) = ax^4 + bx^3 + cx^2 + dx + e
p(x+1) = ax^4 + (4a+b)x^3 + (6a+3b+c)x^2 + (4a+3b+2c+d)x + a+b+c+d+e
and thus the difference between p(x+1) and p(x) is:
p(x+1)-p(x) = 4ax^3 + (6a+3b)x^2 + (4a+3b+2c)x + a+b+c+d
Using the same process, we can find each "successive difference" until we get a constant. Let p`(x) (read "p back-tick of x") represent this type of "pseudo-derivative" of p(x). Then we have the following:
p`(x+1)-p`(x) = p``(x) = 12ax^2 + (24a+6b)x + 14a+6b+2c
p``(x+1)-p``(x) = p```(x) = 24ax + 36a+6b
p```(x+1)-p```(x) = p````(x) = 24a
Just as with the numbers-only version, we can see that the final result is a*(deg(p)!), where deg(p) is the degree of polynomial p.

Returning to the idea of constructing a value of p(x) using only a sum of its pseudo-derivatives, it is clear that we have constructed p`(x)=p(x+1)-p(x), so p(x+1)=p(x)+p`(x). What about p(x+2)? We would use p(x+2)=p(x+1)+p`(x+1)=p(x)+p`(x)+p`(x)+p``(x) = p(x)+2p`(x)+p``(x). Thus it is clear that any value of p(x) can be constructed given only its values at x=0 for p(x) and all its pseudo-derivatives.


I have discovered that the methods shown here are more rigorously defined on wikipedia under the title Finite_Difference. I won't do much more in the way of updating this unless I find something interesting that expands on it.

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