(2m)x2x4mx???(?1)?(x2m) ③cosx?1?2!4!2m!nx2x3n?1x??(xn) ④ln(x?1)?x?????(?1)23n
⑤(1?x)??1??x?⑥
?(??1)2!x2????(??1)(??2)?(??n?1)n!xn??(xn)
1?1?x?x2???xn??(xn) 1?x验证，② ④
k??x) k?0,1,2,?n 证明②?(sinx)(k)?sin(2?f(2k)(0)?0 f(2k?1)(0)?sin(k??)?(?1)k?1
2?(sinx)(2k)|x?0?0?a2k?(2k)!? k?0,1,2,?n ?k?1(?1)?a?2k?1?(2k?1)!?代入即证例②
例2 写出f(x)?e例3 求
lnx?x22(98)(99)f(0)f(0). 的麦克劳林公式，并求与
在x=2处的泰勒公式.
?x22cosx?e. x?0x4二 、带Lagrange型余项的Taylor公式
例4 求极限lim 定理6.9 若函数f在?a,b?上存在直至n阶连续导数，在?a,h?内有n?1阶导数，则对任意给定的x,x0??a,b?至少存在一点???a,b?，使得
f???x0?f?n??x?x0???+?x?x0?n f?x? =f?x0??f??x0??x?x0??2!n!f?n?1??x??x?x0?n?1?0(x?x0)n?1 +
?n?1?!??证： 作辅助函数
??f???t?f?n??t?2 F?t??f?x???f?t??f??x??x?t???x?t?????x?t?n?
2!n!??
G?t???x?t?n?1，于是 要证
??f???t?f?n??t?2?x?t?????x?t?n? F?t??f?x???f?t??f??x??x?t??2!n!??f?n?1????f?n?1????n?1 =?x?x0??G?x0?
?n?1?!?n?1?!F?x0?fn?1???或 ，不防设x0?x ?G?x0?(n?1)!?F,G在?x0,x?连续，在?x0,x?可导，而
???f???t?? F??t???f?t???f?t??f??t??x?t??+ ?x?t?2?f?t??x?t?3??2!3!???f?n??t??x?t?n ??n!? =?0??f??t????f??t??f???t??x?t?????f???t??x?t??
?f????t?f????t?f4?t?223?? + ?x?t????????x?t?x?t??2!2!3!????f?n?f?n?1??t?n?1n??????????x?t?x?t??n?1?!?n!??
f?n?1??t??x?t?n??n!nG??t????n?1??x?t??0,t??x0,x?
??
而F?x??G?x??0，由Cauchy中值定理
?f?n?1????n?x???F?x0?F?x0??F?x?F????n!??? nG?x0?G?x0??G?x?G??????n?1??x???
f?n?1????,???x0,x???a,b? =
?n?1?!f?n?1?????x?x0??n?1? 这里 Rn?x??f?x??Tn?x???n?1?!
f?n?1??x0???x?x0?? =?x?x0?n?1,0???1
?n?1?!注①
当n?0时 Taylor公式即为Lagrange中值公式
②当x0?0时，Taylor公式为Maclaurin公式
f???0?2f?n??0?nf?n?1???x?n?1 f?x??f?0??f??0?x?x???x?x
?n?1?!2!n! ?0???1?
③六个须记忆的基本初等函数带Lagrange余项的Maclaurin公
式，其中?均满足0???1
x2xne?x1)e?1?x????xn?1,ex2!n!?n?1?!x???n?1?ex
?2m?1x3x5x2m?1m?1mcos?x2)sinx?x??????1????1?x
?2m?1?!?2m?1?!3!5!?sinx??2m?1??m???sin??2m?1??x????1?cosx2??2mx2x4mxm?1cos?x3)cosx?1???????1????1?x2m?22!4!?2m?! ?2m?2?!?cosx??2m?2??cos??m?1???x????1??n?1?m?1cosx
4)?ln?1?x?????1?n!?1?x?n??n?1?
??n?1?nx2x3n?1xn?1??x??ln?1?x??x???????1????1?23n?1 nxn?1
5)?1?x??????n?1??????1????2?????n??1?x????n?1???1?x??1??x?????1?2!x?2????1????2?3!x???3????1?????n?1?n!xn?????1?????n??n?1?!?1??x???n?1xn?1
?n?1?6)??1?!?1?x????n?1??1?x?n?21xn?1?1?x?1?x?x2???xn??1??x?n?2 、在近似计算中的应用
例6 （1）计算e的值，使其误差不超过10?6
（2）证明e为无理数
（1）ex?1?x?x2xne?x解 2!???n!??n?1?!xn?1
当x?1时
e?1?1?111ea2!?3!???n!??n?1?! （*）?Rea3n?1???n?1?!??n?1?! 当n?9时 ?9?1?!?10!?3628800
而
33628800?1100000?10?6 约去R9?1?
e?1?1?12!?13!???19!?2.718285
（2）由（*）得
n!e??n!?n!?3.4?n???4.5?n?1??ean?1
ea即 n!e??n!?n!?3.4?n???4.5?n?1??n?1若e为有理数 即e?pq p,q?N? 则当n?q时
n!e 为正整数，从而（**）左边为整数
ea而右边n?1?en?1?3n?1为非整数n?2时 例7 证明x?0 则
**）三 （
①x?1?x?111 其中?Q?x??
422x?Q?x?11 ②limQ?x??,limQ?x??
x?04x???2证：（1）考察函数f?t??t 在?x,x?1?上使用Lagrange中值定理 有
x?1?x?11??x?1?x??
2x?Q?x?2x?Q?x?1?x?1?x
2x?Q?x?而 2x?Q?x??设 4?x?Q?x????x?1??x?2x?x?1?
?Q?x??11?2x?x?1??2x 4?? ?11?x42?x?1?x
??x?x?1?x皆大于0
?Q?x??1 4又 x??x?1?x?1 2?x?x?1?xx1?
x?x2?Q?x??综合以上得
11?Q?x?? 42（2）由（1）
?11limQ?x??lim??xx?0x?042??1limQ?x??lim??x???x????42???1x?1x???4?????????1x1??lim??= x????x?1?x??11??2???1???2????xx??????例8 设f?C?a,b? f在?a,b?可导，且ab?0，证明存在???a,b? 使得