Testwiki:Reference desk/Archives/Mathematics/2021 May 9

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May 9

Deriving polynomial coefficients "synthetically"

Suppose you wanted to find the unknown coefficients the quadratic polynomial p(x) = Ax^2 + Bx + C, which happens to be 5x^2 + 13x^1 + 17. First, observe that the constant term C can be extracted by simply evaluating the polynomial at zero, ie: C = p(0) = 17. Furthermore, the sum of all coefficients is p(1) = 35 and the difference between even and odd terms of the polynomial given by p(-1) = 9. Remove the constant term from both to obtain the sum and difference of A and B, as in K = p(1) - p(0) = 35 - 17 = 5 + 13 = 18 and D = p(-1) - p(0) = 9 - 17 = 5 - 13 = -8, then solve with A = ((K + D)/2) = ((18 + (-8)) / 2) = 5 and B = ((K - D)/2) = ((18 - (-8)) / 2) = 13. Could this be extended to higher polynomials? Earl of Arundel (talk) 18:04, 9 May 2021 (UTC)

Yes. For any set of pairs $(x_1,y_1),(x_2,y_2),...,(x_n,y_n)$ in which all $x_i$ values are distinct, there is a unique polynomial $P(x)$ of degree at most $n-1$ such that $P(x_i)=y_i,i=1,2,...,n.$ See Template:Section link. Taking the $n$ coefficients of the polynomial $P(x)=\sum _j{c}_j{x}^j$ as unknowns, the $n$ equations

$\sum _{j=0}^{n-1}{c}_j{x}_i^j=y_i,i=1,2,...,n$

form a linear system of $n$ equations with $n$ unknowns. See Template:Section link.  --Lambiam 19:03, 9 May 2021 (UTC)

Awesome, thank you! Earl of Arundel (talk) 21:11, 9 May 2021 (UTC)

FYI, there's a systematic way to solve this system of linear equations, which is the Lagrange interpolation formula. It can come in handy a lot of times. Duckmather (talk) 20:18, 10 May 2021 (UTC)

Thank you! Yes, it looks like matrices are the way to go with that. Earl of Arundel (talk) 00:49, 12 May 2021 (UTC)