close
標題:
F.6 Polynomials
發問:
For n=0,1,2...,P (X) is a polynomial in X of degree n. _____________n Proved by induction that every polynomial P( X ) of degree n can be expressed as n Σ__α_P (X) j=0__j__j where α € R _______j
For n = 0, P_0 (x) is a polynomial in x of degree 0. i.e. P_0 (x) = r ,where r is a non-zero real constant. Then every polynomial P(x) of degree 0 (i.e. P(x) = s, where k is a real number) can be expressed as P(x) = (s / r) P_0 (x), where s / r is real. Thus it is true for n = 0. Assume that every polynomial P(x) of degree p ≤ k can be expressed as P(x) = Σ[from j = 0 to p] α_j P_j(x), where a_j is real. For n = k + 1, let Q(x) be a polynomial of degree k + 1. Let c be the leading coefficient of Q(x). Note that P_(k + 1) (x) is also a polynomial of degree k + 1. Let d be the leading coefficient of P_(k + 1) (x). Then Q(x) - (c / d)P_(k + 1) (x) is a polynomial of degree at most k. By the induction hypothesis, Q(x) - (c / d)P_(k + 1) (x) = Σ[from j = 0 to k] α_j P_j(x). Thus Q(x) = (c / d)P_(k + 1) (x) + Σ[from j = 0 to k] α_j P_j(x) = Σ[from j = 0 to k + 1] α_j P_j(x), where α_(k + 1) = c / d, which is real. Thus it is true for n = k + 1. By the SECOND principle of induction, it is true for all non-negative integers n.
其他解答:
F.6 Polynomials
發問:
For n=0,1,2...,P (X) is a polynomial in X of degree n. _____________n Proved by induction that every polynomial P( X ) of degree n can be expressed as n Σ__α_P (X) j=0__j__j where α € R _______j
此文章來自奇摩知識+如有不便請留言告知
最佳解答:For n = 0, P_0 (x) is a polynomial in x of degree 0. i.e. P_0 (x) = r ,where r is a non-zero real constant. Then every polynomial P(x) of degree 0 (i.e. P(x) = s, where k is a real number) can be expressed as P(x) = (s / r) P_0 (x), where s / r is real. Thus it is true for n = 0. Assume that every polynomial P(x) of degree p ≤ k can be expressed as P(x) = Σ[from j = 0 to p] α_j P_j(x), where a_j is real. For n = k + 1, let Q(x) be a polynomial of degree k + 1. Let c be the leading coefficient of Q(x). Note that P_(k + 1) (x) is also a polynomial of degree k + 1. Let d be the leading coefficient of P_(k + 1) (x). Then Q(x) - (c / d)P_(k + 1) (x) is a polynomial of degree at most k. By the induction hypothesis, Q(x) - (c / d)P_(k + 1) (x) = Σ[from j = 0 to k] α_j P_j(x). Thus Q(x) = (c / d)P_(k + 1) (x) + Σ[from j = 0 to k] α_j P_j(x) = Σ[from j = 0 to k + 1] α_j P_j(x), where α_(k + 1) = c / d, which is real. Thus it is true for n = k + 1. By the SECOND principle of induction, it is true for all non-negative integers n.
其他解答:
文章標籤
全站熱搜
留言列表
