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Breaking a polynomial in a sum of polynomials

Source: CIIM 2020 P4

October 26, 2020
algebrapolynomialderivative

Problem Statement

For each polynomial P(x)P(x) with real coefficients, define P0=P(0)P_0=P(0) and Pj(x)=xjP(j)(x)P_j(x)=x^j\cdot P^{(j)}(x) where P(j)P^{(j)} denotes the jj-th derivative of PP for j1j\geq 1. Prove that there exists one unique sequence of real numbers b0,b1,b2,b_0, b_1, b_2, \dots such that for each polynomial P(x)P(x) with real coefficients and for each xx real, we have P(x)=b0P0+k1bkPk(x)=b0P0+b1P1(x)+b2P2(x)+P(x)=b_0P_0+\sum_{k\geq 1}b_kP_k(x)=b_0P_0+b_1P_1(x)+b_2P_2(x)+\dots