高等数学上复旦第三版 课后习题答案(4)

故f?x??x2????1????1?n?2n?1?!!?n?1?2n?!!x2n??

?x2?????1?n?2n?1?!!2?n?1? (-1≤x≤1) n?1?2n?!!x(5)f?x??x?13x2

1?3?x?n?x2?n3????1?n?0??3???nx2n?1????1?n?1??n?03x?3(6)由ex???xn!，x∈(-∞,+∞)

n?0n得?x????1?n?xne?，x∈(-∞,+n?0n!∞)

所以f?x??12?ex?e?x?

?1??xn???1?nxn?2????n?0n!??n?0n!?1?xn?2??1???1n?n?0????n!??1??x2n???n?0?2n?1?!x????,(7)因为excosx为ex?cosx?isinx??e?1?i?x的实部，而?e?1?i?x??1!??1?i?x?n

n?0n??xn??1?i?nn?0n!?n??xn??ππ?n?0n!??2??cos4?isin?4???????xnn?22??cosnπ?nπ?n?0n!?4isin4?? 298

nnπ取上式的实部．得?22cosexcosx??4 (-∞

n?0n!?xn(8)由于

1??1?1?x?2??nxn |x|<1

n?0而f?x??14?1?，所以

?x?2?1?2??n?1??1f?x??1??x???n?xn4??n??n?1 (|x|<2) n?0?2?n?0214．将f?x??1x2?3x?2展开成(x+4)的幂级数．

解：1x2?3x?2?1x?1?1x?2

而

11x?1??3??x?4???13?11?x?43n??1??x?4??3???n?0?3??x?4?3?1??????x?4?n??3n?1??7?x?1?n?0

又

1x?2?1?2??x?4???1121?x?42??1??x?4n?2????n?0??x?42??2?1???????x?4?n?2n?1??6?x??2?n?0

299

所以f?x??1x2?3x?2??x?4?n??x?4?n???n?1??32n?1n?0n?01??1???n?1?n?1??x?4?n3?n?0?2?

??6?x??2?15．将函数f?x??x3展开成(x-1)的幂级数． 解

?1?x?m?1?mx?1：

?m?1?2因

???为

n??m??????x!nm1?????x??

2m1m!所

f?x??x3??1??x?1??32以

3?3?3?3??33??3???1???1???2????n?1?22n??2???1?2?x?1???2??x?1?2????2??2x?1???1!2!n!

(-1

33?13?1???3?1???1????3????2n?5?f?x??1??x?1??2?x?1?2?3?1?x?1?3???x?1?n??n22?2!2?3!2?n!?3?1???1???5?2n?n?1?????0?x?2?x?1n2?n!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）

1?x??x3x5x2n?1解：(1)ln?2?x????????，x∈(-1,1)

1?x352n?1??令

1?x1?3，可得x????1,1?， 1?x2 300

1?1故ln3?ln2?2?1?1?1?2?13?23?15?25???1?2n?1??22n?1????? 2又

r??11?n?2??2n?1??22n?1??2n?3??22n?3?????2??2n?1??22n?1?2n?1??22n?1??2n?1?22n?1??1??2n?3??22n?3??2n?5??22n?5?????2??2n?1?22n?1??1?122?124??????2?2n?1?22n?1?11?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??1111?2?3?23?5?25????11?211???1.0986 ?π?2?π?4?π?2n(2)cos20?cosπ??90??90?n??90??90?1??2!???4!?????1??2n?!??

?24∵?π??90??π???6?10?4；??90??2!4!?10?8 ?2故?π?90?cos20?1???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．

301

2n?1x3x5nx??，解：由于arctanx?x???????1?（-1≤x≤1） 352n?1故?00.50.5?arctanx?x2x4x2ndx???1???????1????dx 0x352n?1??3570.5??xxx??x??????92549??01111111???3??5??7??292252492111111而?3?0.0139，?5?0.0013，?7?0.0002． 922524920.5arctanx11111dx???3???0.487 因此?0x292252518．判别下列级数的敛散性：

?(1)?n?1n1n?nn1??n???n??；

nx????n?cos3??； (2)?n2n?12(3)

?n?1?ln?n?2?1??3???n??n．

解：(1)∵

nn?1nn1??n???n??2n??n? ??n2?1??1?n??n???n???n??1?n2?1?n2nn2nn而lim???????1??lim1?????n???1?n2?n????1?n2?n????1?0

?n2?故级数??发散，由比较审敛法知原级数发散． 2?n?1?1?n?nx?n?cos??3???n (2)∵0?2n2n2nx??ncos????n3??由比值审敛法知级数?n收敛，由比较审敛法知，原级数?n2n?1n?122 302

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