. . ..
2.(1)???1?n?1??n?1?x2n?1n?11?x?,;(2) ??,??lna??1???????,(?a,a];
22n?1?2n?1?!na??n?1?2nnxn?1?2x?(3)?,???,???;(4)???1?,???,???。
2(2n)!n?1n?1?2n?1?!2n?13.(1)????1?n?1n3??1????n?;(2)x?3x??x????????????1?n??2n?n2n?1;
n?03n?02(2n)!?3?n?02(2n?1)!?3? (3)????1?n2nn?02n?1?x?1?
B组
1.x???(?1)n(2n?1)!!(2n)!!(2n?1)x2n?1,x?[?1,1]
n?1
习题9.6 A组
1.2.9926
2.0.15643,误差不超过10?5 3.0.545 4.5e 5.略 B组
1.32
n2.(1)略;(2)??22cosn?xn?。 n?04n!
习题9.7 A组
1.(1)在x?n?(n?Z)处的展开式为12?2???1?1?2sin?2n?1?x,n?0?2nx?n?(n?Z)处收敛于
12 。 (2)
24???1?n?1????2?x???);
n?14n?1cosnx,(??.v .. ..
在 . . ..
(3)
?2?4??n?0?1?2n?1?n?12cos?2n?1?t, (???t???);
2.(1)
183????1?n?1??nsinnx; 29n?1n?1??1???1?n?(b?a)??1?(b?a)?a?b????(2)????cosnx?sinnx?,(x??2n?1??,24n?nn?1????n?0,?1,?2,???) 13.?sinnx
n?1n?4.(1)
111cos4?xcos6?x?2(cos2?x???),(???x???); 2212?23?n???4sinn?1?2(?1)?n?x2sin (2)??22??n??2n?1?n???2,(???x???,x??2,?6,)
5. 展开成正弦级数:f(x)?级数收敛于
sin??n?1?1?cosn?2sinn?x,x?(0,1)?(1,2),当x?0时,n2?1?1?0?f(0)?1;展开成余弦级数: 2f(x)?12??2?n?1?n?2cosn?x, x?[0,1)?(1,2)。 n2B组
4E?1?1?1.u?t??, ????t???? ?cosnt???2??2n?14n?1??cosnx?21?4?(?1)?2.,x?(??,?);? 223n8n?1n?1(2n?1)?2?n3.?1?sinn?x,0?x?2 ?n?1n2?4.证明略。
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. . ..
总复习题九 一、基础知识 1.(1)错;(2)对;(3)对;(4)对;(5)错 2.(1)发散;(2)收敛;(3)收敛;(4)收敛,发散 3.(1)发散;(2)发散;(3)收敛;(4)a?1时收敛,a?1时发散,a?1时,s?1收敛,s?1时发散。 4.(1)绝对收敛;(2)条件收敛
5.(1)????15,1?5??,(2)????1e,1?e??;(3)??2,0?;(4)??3,3?,
26.(1)
2?x?,2?x2?2,??2,2?;(2)arctanx,[?1,1];(3)
x?1?2?x?2(4)12(1?x2?1xln(1?x)?xln(1?x)),[?1,1];
7.(1)??(?1)nx2n?,x????,??;(2)?nn?1n?0n!n?1x,x???2,2? n?12二、技能拓展
1.0
2.(1)2e;(2)12?cos1?sin1? 3.正弦级数:
4l??x13?x1??2??sinl?32sinl?5?x52sinl???,?0?x?l?;?余弦级数:l212?2n?1??x4?l?2?n?1?2n?1?2cosl,?0?x?l?. 4.证明略
5.1?12ln(1?x2),x?1,极大值f(0)?1。
三、探究应用
1.4?(?1)m?1???1)2sin(2m?1)x,(???x???) m?1(2m2.m 13.(1)2n?1;(2)
43 4.f(x)展开成的傅里叶级数为??14?m?12m?1sin(2m?1)x。
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?0,2?;