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December 9th, 2017, 09:24 PM  #1 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 614 Thanks: 83 Math Focus: Electrical Engineering Applications  Method of finding polynomial series coefficients?
Recently, at work, an application came up where I had a series that I suspected was generated by successive terms of a polynomial and I needed to find the coefficients of that polynomial, similar to the problem in this thread (Reference 1). I give another sample calculation in Reference 1, with a little bit more explanation than presented below relating to the coefficients of the numbers raised to a power. I was aware of taking the differences until a constant is reached (due to previous posts on this forum), which gives the order of the polynomial. I did this and used the linear equations to get the result (there was a cubic term and a linear term). Then I started playing around and I think I found a way to get the coefficients from the difference triangle. The method relies on two properties of Pascal's Triangle (PT), which I ask about here (Reference 2). While I am not certain of these properties of PT for all rows, I have calculated them to be true up to row 100 (for the values that I am interested in) for the purposes of the method that I am about to describe. This is a lot higher than any practical polynomial degree that I would need to solve. Anyway, I would like to attempt to describe the method and show by a fourth order polynomial why it just might work.  So, suppose we have the following series given by row 0, and the differences: $\displaystyle \large {\begin{array}{c c} & n & 1 & & 2 & & 3 & & 4 & & 5 & & 6 \\ row \\ 0 & & 28 & & 101 & & 320 & & 823 & & 1796 & & 3473 \\ 1 & & & 73 & & 219 & & 503 & & 973 & & 1677 \\ 2 & & & & 146 & & 284 & & 470 & & 704 \\ 3 & & & & & 138 & & 186 & & 234 \\ 4 & & & & & & 48 & & 48 \\ \end{array}}$ We can see that we have a fourth order polynomial since 4 subtractions were needed to get a constant term (I would take more terms just to be sure in a series that I was not familiar with). So we are looking for: $\displaystyle \large a \cdot n^4 + b \cdot n^3 + c \cdot n^2 + d \cdot n + e$ I am going to go term by term to get to the first (leftmost) constant of 48, and honestly, you can probably skip on down to where you start seeing braces inside $\LaTeX$.  $\displaystyle \large {\begin{align*} 973 &= a \cdot 5^4 + b \cdot 5^3 + c \cdot 5^2 +d \cdot 5 +e  \left (a \cdot 4^4 + b \cdot 4^3 + c \cdot 4^2 +d \cdot 4 +e \right) \\ &= a (5^4  4^4) + b (5^34^3)+c(5^24^2)+d \end{align*}}$  $\displaystyle \large 503 = a (4^4  3^4) + b (4^33^3)+c(4^23^2)+d$  $\displaystyle \large 470 = a (5^4  2 \cdot 4^4 + 3^4) + b (5^3  2 \cdot 4^3 + 3^3)+c(5^2  2 \cdot 4^2 + 3^2)$  $\displaystyle \large 219 = a (3^4  2^4) + b (3^32^3)+c(3^22^2)+d$  $\displaystyle \large 284 = a (4^4  2 \cdot 3^4 + 2^4) + b (4^3  2 \cdot 3^3 + 2^3)+c(4^2  2 \cdot 3^2 + 2^2)$  $\displaystyle \large 186 = a (5^4  3 \cdot 4^4 + 3 \cdot 3^4  2^4) + b (5^3  3 \cdot 4^3 + 3 \cdot 3^3  2^3)+c(5^2  3 \cdot 4^2 + 3 \cdot 3^2  2^2)$  $\displaystyle \large 73 = a (2^4  1^4) + b (2^31^3)+c(2^21^2)+d$  $\displaystyle \large 146 = a (3^4  2 \cdot 2^4 + 1^4) + b (3^3  2 \cdot 2^3 + 1^3)+c(3^2  2 \cdot 2^2 + 1^2)$  $\displaystyle \large 138 = a (4^4  3 \cdot 3^4 + 3 \cdot 2^4  1^4) + b (4^3  3 \cdot 3^3 + 3 \cdot 2^3  1^3)+c(4^2  3 \cdot 3^2 + 3 \cdot 2^2  1^2)$  $\displaystyle \large { \begin{align*}48 &= a \ \underbrace{(5^4  4 \cdot 4^4 + 6 \cdot 3^4  4 \cdot 2^4 + 1^4)}_{4! \text{ per Reference 2}} \\ &+ b \ \underbrace{(5^3  4 \cdot 4^3 + 6 \cdot 3^3  4 \cdot 2^3 + 1^3)}_{0 \text{ per Reference 2}} \\ &+c \ \underbrace{(5^2  4 \cdot 4^2 + 6 \cdot 3^2  4 \cdot 2^2 + 1^2)}_{0 \text{ per Reference 2}} \end{align*}}$ So: $\displaystyle a=\frac{48}{4!}=2$  Back to 138 with $a=2$: $\displaystyle \large { \begin{align*} 138 &= \underbrace{2(4^4  3 \cdot 3^4 + 3 \cdot 2^4  1^4)}_{120} \\ &+ b \ \underbrace{(4^3  3 \cdot 3^3 + 3 \cdot 2^3  1^3)}_{3! \text{ per Reference 2}} \\ &+ c \ \underbrace{(4^2  3 \cdot 3^2 + 3 \cdot 2^2  1^2)}_{0 \text{ per Reference 2}} \end{align*}}$ So to get $b$: $\displaystyle \large b=\frac{138120}{3!}=3$  Back to 146 with $a=2$ and $b=3$: $\displaystyle \large { \begin{align*} 146 &= \underbrace{2(3^4  2 \cdot 2^4 + 1^4) + 3(3^3  2 \cdot 2^3 + 1^3)}_{136} \\ &+ c \ \underbrace{(3^2  2 \cdot 2^2 + 1^2)}_{2! \text{ per Reference 2}} \end{align*}}$ So to get $c$: $\displaystyle \large c=\frac{146136}{2!}=5$  $\displaystyle \large 73 = 2(2^4  1^4) + 3(2^31^3)+5(2^21^2)+d$ $\displaystyle \large d=7$  $\displaystyle \large 28 = 2+3+5+7+e$ $\displaystyle \large e=11$  So the generating polynomial is: $\displaystyle 2 \cdot n^4 + 3 \cdot n^3 + 5 \cdot n^2 + 7 \cdot n + 11$ which is easily verified. 

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coefficients, finding, method, polynomial, series 
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