Proof of Successive Differences and the Degree of a Polynomial
Date: 06/26/2007 at 20:13:59 From: Craig Subject: Why do the sequence differences define the polynomial Given a sequence of numbers, and finding the differences between them, one can get a recurring number, for instance in this basic example: Sequence: 1 2 10 25 47 76 1st differences: 1 8 15 22 29 2nd differences: 7 7 7 7 As you can see, at the second set of differences one gets a recurring 7. I have always been taught that this means that the equation to find the nth term will be a quadratic. And thus the number of differences one takes to get to the constantly recurring number is the same as the degree of the polynomial that defines the nth term. What is the proof for this?
Date: 06/26/2007 at 23:07:02 From: Doctor Peterson Subject: Re: Why do the sequence differences define the polynomial Hi, Craig. Suppose that the initial sequence is a polynomial of degree n: u[x] = ax^n (We'll talk in a minute about polynomials with more than one term.) The sequence of first differences will be u'[x] = a(x+1)^n - ax^n = a[(x+1)^n - x^n] Notice the difference of nth powers; this can always be factored: a^n - b^n = [a - b][a^(n-1) + a^(n-2)b + a^(n-3)b^2 + ... + b^(n-1)] Applying this, we find u'[x] = a[(x+1) - x][a polynomial in x with degree n-1] = a*1*[a polynomial in x with degree n-1] So taking differences once reduces the degree of the polynomial by 1. If the polynomial had other terms, they too would be reduced in degree, so the degree of the difference would still be n-1. After taking differences n times, the degree will be reduced to 0, making it a constant! As an example (I'll try to keep it simple), suppose the original sequence is u[x] = x^2 - 3x + 5 This is the sequence (starting x at 1): 3, 3, 5, 9, 15, ... The first difference is u'[x] = [(x+1)^2 - 3(x+1) + 5] - [x^2 - 3x + 5] = [x^2 + 2x + 1 - 3x - 3 + 5] - [x^2 - 3x + 5] = [x^2 - x + 3] - [x^2 - 3x + 5] = 2x - 2 which has degree 1. This is the sequence 0, 2, 4, 6, ... The second difference is u''[x] = [2(x+1) - 2] - [2x - 2] = [2x] - [2x - 2] = 2 which has degree 0. This is the sequence 2, 2, 2, ... which is constant, as we knew. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum