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# PolyDiv-PartFrac-LagrangeIntPoly

Published on: Mar 4, 2016

#### Transcripts - PolyDiv-PartFrac-LagrangeIntPoly

• 1. Let p(x) be a polynomial and p(x) = q(x)d(x) + r(x) where d(x) is the divisor, q(x), the quotient, and r(x) the remainder in polynomial division. Further assume that d(x) = a; d(a)=0(x − a). It may seem that we lose generality with the assumption that our divisor factors linearly, but in practice, our divisor would be constructed rather than factored in this way. Naturally,∀a p(a) = r(a). Then r(x) d(x) = a; d(a)=0 n i=1 ai (x − a)i where n satises ∀k n d(k) (a) = 0, and d(n) (a) = 0, i.e. a is a root of multiplicity n. This double summation is the partial fraction expansion of the left-hand-side with ai = 1 (n − i)! limx→a dn−i dxn−i r(x)(x − a)n d(x) where p(x) may be substituted for r(x) since they agree at all zeroes of the divisor up to the required derivatives. Now we may multiply both sides of the partial fraction expansion by the divisor to obtain r(x) = a; d(a)=0 n i=1 aid(x) (x − a)i precisely the Lagrange interpolating polynomial of degree deg(d)−1 for p(x) if d(x) has all roots of multiplicity one, i.e., ∀a n = 1, and may be considered a generalisation of said interpolating polynomial that agrees with p(k) (a)∀k n, i.e., it agrees up to the (n − 1)th derivative at the a's. This formula could then be applied to any function, not just a polynomial, that has the required derivatives at the desired points. Of course, in general, if the function is not a polynomial, then division becomes meaningless, and r(x) is not a remainder but only our generalised Lagrange interpolating polyno- mial.