习题11?4 1? 求函数f(x)?cos x的泰勒级数? 并验证它在整个数轴上收敛于这函数? 解 f(n)(x)?cos(x?n??) (n?1? 2? ? ? ?)? 2? f(n)(x0)?cosx(0?n?) (n?1? 2? ? ? ?)? 2从而得f(x)在x0处的泰勒公式 f(x)?cosx0?cos(x0??)(x?x0)?2cosx(0??)2!(x?x0)2? ? ? ? ?cosx(0?n!n?)2(x?x)n? R(x)? 0n 因为|Rn(x)|?|而级数??cos[x0??(x?x0)?(n?1)!n?1?]|x?x0|n?1n?12(x?x0)|?(n?1)!(0???1)? |x?x0|n?1(n?1)!总是收敛的? 故lim|x?x0|n?1(n?1)!n??n???0? 从而lim|Rn(x)|?0? n?? 因此f(x)?cosx0?cos(x0??)(x?x0)?2cos(x0??)2!(x?x0)2? ? ? ? ?cosx(0?n!n?)2(x?x)n? ? ? ?? x?(??? ??)? 0 2? 将下列函数展开成x的幂级数? 并求展开式成立的区间? ex?e?x(1)shx?2? 解 因为 xne??n?0n!x??? x?(??? ??)? xnn!所以 e?x?n?0?(?1)n? x?(??? ??)? 故
?nn1?xn?1?x2n?1nxnxshx?[???(?1)]??[1?(?1)]??2n?0n!n?0n!2n?0n!n?0(2n?1)!? x?(??? ??)?
(2)ln(a?x)(a?0)? 解 因为ln(a?x)?lna(1?x)?lna?ln(1?x)? aa xn?1ln1(?x)??(?1)n?1n?0n? (?1?x?1)? 所以 ?(?1)nxn?11xn?1ln(a?x)?lna??(?1)()?lna??n?1n?1an?0n?0(n?1)a?n(?a?x?a)? (3)ax? 解 xn因为e??n?0n!x?? x?(??? ??)? ?所以 a?exxlna?e?xn?0?(xlna)nn!?n?0??(lna)nn!xn? x?(??? ??)? (4)sin2x? 解 因为sin2x?1?1cos2x? 22?2n cosx??(?1)nx? x?(??? ??)? n?0(2n)!所以 sin22n?2n?111??x2nn(2x)n2x???(?1)??(?1)22n?0(2n)!n?1(2n)! x?(??? ??)? (5)(1?x)ln(1?x)? ?n?1 解 因为ln(1?x)??(?1)nx (?1?x?1)? n?0n?1所以 xn?1(1?x)ln1(?x)?(1?x)?(?1)n?1n?0n?? ?n?0?(?1)n??xn?1?xn?2xn?1?xn?1??(?1)n?x??(?1)n??(?1)n?1n?1n?0n?1n?1n?1nn?1 ?x??[n?1(?1)nn?1?(?1)n?1n]xn?1?x???(?1)n?1n?1n(n?1)xn?1 (?1?x?1)?
(6) 解 x1?x2? (?1?x?1)? ??1n(2n?1)!!2n因为?1?(?1)x?(2n)!!(1?x2)1/2n?1所以 x1?x2?x??(?1)nn?1?(2n?1)!!(2n)!!x2n?1?x??(?1)nn?12?(2n)!x2n?1(?1?x?1)? ()22(n!) 3? 将下列函数展开成(x?1)的幂级数? 并求展开式成立的区间? (1)x3? 解 因为 (1?x)m?1?mx?所以 x33?[1?(x?1)]2m(m?1)2!x2? ? ? ? ?m(m?1) ? ? ? (m?n?1)n!xn? ? ? ? (?1?x?1)? 33333(?1)(?1) ? ? ? (?n?1)32?1?(x?1)?22(x?1)2? ? ? ? ?22(x?1)n? ? ? ? 22!n! (?1?x?1?1)? 即 x3?1?3?1?(?1)?(?3) ? ? ? (5?2n)33?1(x?1)?2(x?1)2? ? ? ? ?(x?1)n? ? ? ? n22?2!2?n! (0?x?2)? 上术级数当x?0和x?2时都是收敛的? 所以展开式成立的区间是[0? 2]? (2)lg x? 解 lnx11lgx??ln[1?(x?1)]?ln10ln10ln101lgx?ln10?n?1?(?1)?n?1(x?1)nn (?1?x?1?1)? 即 n?1?(?1)n?1(x?1)nn (0?x?2)? 4? 将函数f(x)?cos x展开成(x??)的幂级数? 3 解 cosx?cos[(x??)??]?cos(x??)cos??sin(x??)sin? 333333
?3? ?1cosx(?)?sinx(?) 2323 1?(?1)?3??(x?)2n?2n?0(2n)!32nn?0(2n?1)!??(?1)n(x??3)2n?1 1?1?3???(?1)n[(x?)2n?(x?)2n?1] (???x???)? 2n?0(2n)!3(2n?1)!3 5? 将函数f(x)?1展开成(x?3)的幂级数? x 解 11111nx?3nx?3????(?1)n() (?1??1)? x3?x?331?x?33n?0333n即 1?1?(?1)n(x?3)n (0?x?6)? x3n?03 6? 将函数f(x)? 解 f(x)?21x2?3x?2展开成(x?4)的幂级数? ? 111??x?3x?2x?1x?2而 1?x?11111?x?4nx?4?????() (||?1)? ?3?(x?4)31?x?43n?0333n即 ?(x?4)1??? (?7?x??1)? n?1x?13n?0 11111?x?4nx?4??????() (||?1)? x?2?2?(x?4)21?x?42n?0222?(x?4)1??? (?6?x??2)? n?1x?22n?0n即 n因此 ?(x?4)?(x?4)1f(x)?2?????n?1n?1x?3x?23n?0n?02n 11n?)(x?4) (?6?x??2)? ??(n?1n?1n?0?23