内容发布更新时间 : 2024/11/17 18:40:26星期一 下面是文章的全部内容请认真阅读。
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． 300 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