f(n+) (5)R,(x)f(x)-P,(x)(x-x)n+1(x-xo)")(n + 1)!p(x)证 由于f(x)在区间内有n+1阶导9(x)=(x-x)"+1令R,(x) = f(x) - P,(x)p'(x)=(n+1)(x-x,)"R( F'(x) - P(),R"(x)= F"(x) - P"(x)g"(xd =(n+1)n(xX-xo)"-1R(n(x) = f(n)(x) - P("(x)p(")(x) = (n+1)n...2(xo- xo)p(n+1)(x) = (n+ 1)!R(n+1)(x) = f(n+1)(x),由要求 P(k)(x) = f(k)(x)k = 0,1,2,..,n
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 L R (x) f (x) P (x), n n = − n (x) (n 1)( x x ) = + − 0 1 0 ( ) ( 1) ( ) − = + − n x n n x x L ( ) ( ) ( ), ( ) ( ) ( ) R x f x P x n n n n n = − ( ) ( 1) 2( ) 0 ( ) x n n x x n = + L − ( ) ( ), ( 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 ( ) = = L 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
R,(x)f(n+1)()(x-x)"+)(n + 1)!p(x)设x > Xo, R,(x),p(x)在[xo,x]上连续,在(xo,x)内可导,且β(x)±0. 用1次柯西定理R,(x)R'(5)R,(x)- R,(xo)(在x,与x之间(x)-(xo) ()p(x)R,(x)及β(x)在[xo,多]上连续,在(xo,)内可导,且β"(x)± 0.用2次柯西定理R'(5) R(S)- R'(xo)R'(525,在x,与引之间β'()p'()-(xo)Φ"(52)R(x) = R,(xo) = R"(x) = ... = R(n(xo) = 0P(x) = (x) = β"(x) = ...= β(n)(x) = 0
, 设x x0 ( , ) , ( ) 0. 在 x0 x 内可导 且 x 柯西定理 ( ) ( ) 1 1 Rn = ( ) ( ) x Rn x ( ), ( ) [ , ] , Rn x x 在 x0 x 上连续 ( ) = 1 在x0 与x之间 − − ( ) ( ) x Rn x = − +1 0 ( ) ( ) n n x x R x ( 1)! ( ) ( 1) + + n f n ( ) ( ) [ , ] , Rn x 及 x 在 x0 1 上连续 = ( ) ( ) 1 1 Rn ( ) ( ) 2 2 Rn ( ) 2 在x0 与 1 之间 在(x0 , 1 )内 柯西定理 用1次 用2次 可导,且(x) 0. LL 1 0 ( ) + − n x x (x) ( ) ( ) ( ) ( ) 0 0 ( ) R x0 = R x0 = R x0 = = R x = n n n n L n ( ) Rn x0 ( ) 0 x ( ) ( ) ( ) ( 0 ) 0 ( ) x0 = x0 = x0 = = x = n L = − − ( ) ( ) 1 1 Rn ( ) Rn x0 ( ) 0 x
f(n+1)R,(x)()(x-x,)n+1(n + 1)!如此下去用n+1次西定理R(n+1)()R,(x)R("(5n) _ R("(n)-R("(xo)得β(n+l)()β(n)(,)(n)(,)-(n)(x)p(x)(在x,与,之间也在x,与x之间)R(n+1) (3)R,(x)即(x-x,)n+i(n + 1)!+(S(x-xo)n+1 (5在x,与x之间)R,(x) =可得(n +1)!R(n+1)(x) = f(n+1)(x),p(n+1)(x) =(n+1)
= − +1 0 ( ) ( ) n n x x R x ( ) 在x0 与 n 之间也在x0 与x之间 如此下去, 得 1 0 ( 1) ( ) ( 1)! ( ) ( ) + + − + = n n n x x n f R x 可得 = − +1 0 ( ) ( ) n n x x R x ( 1)! ( ) ( 1) + + n f n ( ) 在x0 与x之间 = ( ) ( ) ( ) ( ) n n n n Rn ( ) ( ) ( 1) ( 1) + + n n Rn = − − ( ) ( ) ( ) ( ) 0 ( ) ( ) 0 ( ) ( ) x R R x n n n n n n n n 即 用n+1次柯西定理, = ( ) ( ) x Rn x ( 1)! ( ) ( 1) + + n R n n ( ) ( ), ( ) ( 1)! ( 1) ( 1) ( 1) = = + + + + R x f x x n n n n n