高等数学下(修订版)高纯一(复旦出版社) 习题十二答案详解 下载本文

11?x?1?3??x?4?11???31?x?431??x?4??????3n?0?3?n?x?4??1???3?

?x?4?n???3n?1n?0???7?x?1?又

11?x?2?2??x?4?11??21?x?42n1??x?4??????2n?0?2??x?4??1???2?

?x?4?n???2n?1n?0???6?x??2?所以f?x??1x2?3x?2??x?4?n??x?4?n?????n?132n?1n?0n?01??1???n?1?n?1??x?4?n3?n?0?2???6?x??2?15.将函数f?x??解:因为

x3展开成(x-1)的幂级数.

?1?x?m?1?所以

mm?m?1?2m?m?1???m?n?1?nx?x???x????1?x?1? 1!2!n!f?x??x3??1??x?1??323?3?3?3??33???3??1?1?2?n?1????????2?2?2?2??22n??2??2?1??x?1???x?1????x?1???1!2!n!

(-1

298

33?13?1???3?1???1????3????2n?5?f???1??????2??1??3????n??x2x?122?2!x?123?3!x?12n?n!??1??3?1???1???5?2n??x?n?0?x?2?n?12n?n!?116.利用函数的幂级数展开式,求下列各数的近似值:

(1)ln3(误差不超过0.0001); (2)cos20(误差不超过0.0001)

解:(1)ln1?x?x3x5x2n?1?1?x?2??x?3?5???2n?1????,x∈(-1,1)

令1?x1?x?3,可得x?12???1,1?, 1?1故ln3?ln2?11?1?2??2?111?3?23?5?25????2n?1??22n?1???? 2又

r2?11n????2n?1??22n?1??2n?3??22n?3???????2n?1??22n?1?2?2n?1?22n?1??1??2n?1??22n?1??2n?3??22n?3??2n?5??22n?5?????2??2n?1?22n?1??1?122?124??????21?2n?1?22n?1?1?14

?13?2n?1?22n?2故r15?3?11?28?0.00012 r16?3?13?210?0.00003.

因而取n=6则

ln3?2??1?2?111?3?23?5?25???11?211???1.0986 ?242n(2)cos20?cosπ?π??90??π??π????90??n??90??90?1?2!?4!?????1??2n?!?? x?1299

?24?π??π?∵?90??2!?6?10?4??;?90?4!?10?8

?2?π?故cos20?1??90??2!?1?0.0006?0.9994 17.利用被积函数的幂级数展开式,求定积分

?0.5arctanx0xdx(误差不超过0.001)的近似值. 2解:由于arctanx?x?x3x5n?1nx3?5?????1?2n?1??,(-1≤x≤1)故

?0.5arctanx0.50x??0??1?x23?x45?????1?x2ndx??2n?1????dx ?x3x5x70.5???x?9?25?49?????0?1111112?9?123?25?25?49?27??而19?1111123?0.0139,25?25?0.0013,49?27?0.0002. 因此?0.5arctanx111110xdx?2?9?23?25?25?0.487

18.判别下列级数的敛散性:

2??n??nx?(1)

?nn?1n;

(2)n?1???cos3??;

?n?1?nn?12n?n???(3)

?ln?n?2?n?1?n.

??3?1?n??解:(1)∵nn?1n?nnn????n2? ?1?n1?n?1?n2???n?n?????n?n??n??n?1?n2?而lim??1?n2n???n2??1?n2???lim??n??????1??1?1?n2????1?0

??n故级数??n2?n?1??1?n2??发散,由比较审敛法知原级数发散.

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2n??cosnx?(2)∵0??3???n2n2n ?nx?2??n?cos?由比值审敛法知级数?n?3?n?12n收敛,由比较审敛法知,原级数?收敛.

n?12n(3)∵0?ln?n?2?ln?n?2??n??3n ?3?1?n??由limUn?1n??U?limln?n?3?n?1?3n nn??3ln?n?2??1ln?n?3?3limn??ln?n?2??13?1??知级数?ln?n?2?ln?n?2n?13n收敛,由比较审敛法知,原级数???n收敛. n?1?1??3?n??19.若lim2?n??nUn存在,证明:级数

?Un收敛.

n?1证:∵limn??n2Un存在,∴?M>0,使|n2Un|≤M,

即n2|Un|≤M,|Un|≤

Mn2 ???而Mn2收敛,故?Un绝对收敛. n?1n?1??20.证明,若

?U2Unn收敛,则

绝对收敛. n?1?n?1nU2n?1证:∵

Unn2n?1n?Un??122U2n?12?1n2

?2?而由

?Un收敛,

n?1?1收敛,知

n?1n2 301

?Un?1211?收敛, ??Un??2?收敛,故?2n?n?1?2n?1n?因而

Un绝对收敛. ?nn?1?21.若级数?a?n与

?b?n都绝对收敛,则函数项级数

??a?ncosnx?bnsinnx?在R上一致

n?1n?1n?1收敛.

证:Un(x)=ancosnx+bnsinnx,?x∈R有

Un?x??ancosnx?bnsinnx?ancosnx?bnsinnx?an?bn

?由于

?an与

n?1???bn都绝对收敛,故级数

n?bn?收敛.

n?1??an?1??由魏尔斯特拉斯判别法知,函数项级数

?ancosnx?bnsinnx?在R上一致收敛.n?122.计算下列级数的收敛半径及收敛域:

??3nn?(1) ???1??xn;

(2)

πnn?1?n?1??sinn2n?x?1?; ?1??x?1?n(3) ??1n2?2n

n解:(1)??liman?1n??a

n??n?1n?lim?3n?3?1n??????n?2??n?1??3n?1??3n?3?1n?3?n?1??1?n?limn??n?2?lim?n???n?1??n?2???limn????3n?1???3?e?1?e?3∴R?1??33, 又当x??3?3时,级数变为??n?1?3n?1?n?n?1??????3?n?n3??????1?n?n?1?3n?3??3n?3?, ??3n?n3?3因为lim3n???3??3n?3???e?0

302