泰勒公式f"(xof(x)= f(x)+ f(x)(x-xo)+2!c-xo)" +R,(x)nl(n+1)()其中R,(x)一x)n+1 (在x,与x之间)X(n + 1)!分析即证R,(x)= f(x)-P(x).n+l((在x,与x之间)(x-x)n+1R,(x)(n + 1)!也即证f(n+1)()R,(x)fClf-(P)(x)(x-x,)n+1((n+)n+1(n + 1)!16
16 分析 R (x) f (x) P (x). 即证 n n 1 0 ( 1) ( ) ( 1)! ( ) ( ) n n n x x n f R x ( ). 在x0与x之间 1 0 ( ) ( ) n n x x R x ( 1)! ( ) ( 1) n f n 也即证 ( 1)! ( ) ( 1) n f n 1 0 ( ) ( ) ( ) n n x x f x P x 1 0 ( 1) ( ) ( 1)! ( ) ( ) n n n x x n f R x 其中 ( ). 在x0与x之间 0 2 0 0 0 0 ( ) ( ) ( ) ( )( ) ( ) 2! f x f x f x f x x x x x ( ) 0 0 ( ) ( ) ( ) ! n n n f x x x R x n 泰勒公式
豪勃公式R,(x)f(n+l ()f(x)-P,(x)(x-x,)n+1(x-x)"t(n + 1)!p(x)证 由于f(x)在区间内有n+1阶导9()=(x-x)"+1令R,(x) = f(x) - P,(x)p'(x) =(n+1)(x-x)"R(= F'(x) - P(x),R"(x)= F"(x) - P(x),p"(x) =(n +1)n(x-x)n-1R("(x)= f(")(x) - P("(x),p("(x) = (n + 1)n ..2(x-xo)p(n+I)(x) =(n+ 1)!R(n+1)(x) = f(n+1)(x),由要求 P()(x)= f(k)(x)k = 0,1,2,,n17
17 证 由于f (x)在区间内有n 1阶导. R (x) f (x) P (x) n n R (x) f (x) P (x), n n 1 0 ( ) ( ) n x x x R (x) f (x) P (x), n n n (x) (n 1)(x x ) 0 1 0 ( ) ( 1) ( ) n x n n x x ( ) ( ) ( ), ( ) ( ) ( ) R x f x P x n n n n n ( ) ( 1) 2( ) 0 ( ) x n n x x n ( ) ( ), ( 1) ( 1) R x f x n n n ( ) ( 1)! ( 1) x n n 令 1 0 ( ) ( ) n n x x R x 1 0 ( ) ( ) ( ) n n x x f x P x ( 1)! ( ) ( 1) n f n 1 0 ( ) n x x (x) P x f x k n k k n ( ) ( ) 0,1,2, , 0 ( ) 0 由要求 ( ) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 泰勒公式