pan?1?n?1?ln?n?1?解: 由??lim?lim?1知R?1, pn??an??nlnnn??n?p?1?1???,故由?当p?0时,limun??1??lim,?发散知这时该级数的
nn??n??lnnn?2nplnnn?2nplnn收敛域为??1,1?.
?当p?0时,由???1?n收敛??1(11?1n?2lnn,
n?2lnnlnn?n?n?2?,?发散)发散知这时该级数的收n?2n敛域为??1,1?.
n?1?????当0?p?1时, 由于???1?1nplnnn?2nplnn收敛,?n?2nplnn发散??lim???,这时该级数的收?n??1??n??敛域为??1,1?.
当p?1时,由于
????1?n收敛,
??1(n?2nlnnn?2nlnn???12xlnxdx?lnlnx??2???)发散,数的收敛域为??1,1?.
?n?当p?1时,由于u?1n??1nplnn?1?n,?1?, 故???1?,1pn?2np收敛n?2nplnn?n?2nplnn均收敛时该级数的收敛域为??1,1?. 例2: :求下列幂级数的和函数:
?(1)?xnn!; n?01解:??liman?1n??a?lim?n?1?!n??1?lim1n??n?1?0,所以R???.当x????,???时,nn!?nS?x???x,S??x?????xn????xn???xn?1?xnn?1n!????n?0n!?????n?0??n!?????n?1?n?1?!??n?0n!?S?x? 故由?S??x??S?x?dx??dx知lnS?x??x?c,S?0??1,故S?x??ex,因此?xn?ex. n?0n!,这这时该级记 xn(2)?;
nn?1?解: ??liman?1n??an1?n?1??limn?1,所以R?1.当x???1,1?时,记 ?limn??n??n?11n?xx?x1xnn?1S?x??????xdx???xn?1dx??dx??ln?1?x?,x???1,1?.
0001?xn?1nn?1n?1??(3)
?nxn?1n;
解??liman?1n?1?lim?1,所以R?1.当x???1,1?时,记
n??an??nn???S?x???nxn?x?nxn?1?x?xnn?1n?1n?1?????x??n??x??x??x??x?,x???1,1?. ??21?x?1?x????n?1?(4)
?n2n?1?n2x3n?1;
n?12un?1?x??n?1?2x3n?23解: 该级数为缺项级数,直接用比值法判别:lim, ?lim?2xnn??u?x?n??nn22x3n?1当2x?1,即x?3162时,级数绝对收敛,当x?n162时,limun?x??0 ,级数发散,
n???1?21??11?3n?123n?1当x???,6?时,记S?x???n2x??23nx??263n?13n?122?n?1?n?????x??
3nn?n3??1?12x2x23?? ?(?2x)???3??3n?13?1?2x?1?2x3????2.
(5)
nnx; ?n?1n?1?解:
??liman?1n??ann?1?limn?2?1,所以R?1.当x???1,1?时,记 n??nn?1?nn?1??????nnnxxx1x?????x?0? S?x???x?x??x??x??x??????n?1n?1n?1n?1n?1n?1?n?1n?1??xn?1n?1???n?1???n????1?xn1xx1?1x?n??????x(??xdx)??x???xdx??x??dx??x???x?ln?1?x???
xn?10?x0n?1??x01?x??x?x????1?ln?1?x???x???11?x?1xln?1?x?,x???1,0???0,1? S?0??0.
?n?1(6):计算
???1?n2n?12n?1; ?n?1解:由于
???1?n2?2n?1?2n?1??nx,考虑n2xn?1; n?1n?1x??1?2n?1 ??liman??a?limn?1?2n?1??1,所以R?1.当x??nn??n2?1,1?时,记 ????S?x???n2xn?1??n?xn?????x???x?xn?????????n????n?1?n?1?nxn?1????n?1???n?1????x??x???n?1????? ???x???1?x???1?x?2????1?x?3, 1?1?n?12因此,
???1?n?S????1?2???2??4. n?12n?1?1327?1??2??例3:将下列函数展开成关于x的幂级数(即泰勒级数). (1)ax
解: ax?exlna????lna?nn!xn,x????,???
n?0(2)
1x2?3x?2;
解: 1111111x2?3x?2??x?2??x?1??x?2?x?1?1?x?2? 1?x2?x??x?1?x??????????????x?11?x?1??? ??xn???????1?n?1?xn,由?x得x?1.
?12n?0?2?2?n?0n?0??2???n?(3):ln?1?x?;
n?1??x?1nnnxnnxln?1?x???dx?????1?xdx????1??xdx????1?01?x00n?1n?0n?0n?0x解:,
x???1,1?
(4):arctanx; 解:arctanx?22?x204n?2?2?x2?1n2nnx2nnx, dt?????1?tdt????1??tdt????1?002n?11?t2n?0n?0n?0x???1,1?
(5):f?x???x?1??ln?1?x??1?;
xn?1解:f??x??ln?1?x???????1?
n?1n?0?n f?x??f?0???x0n?1?xn?1nxxf??x?dx??1??[???1?]dx??1????1??dx
00n?1n?1n?0n?0x?nxn?2 ??1????1?, x???1,1?.
????n?1n?2n?0?n4?x2(6):f?x??arctan;
4?x2解:f??x??1?4?x1???4?x2?2????2?2x?4?x2?4?x2???2x???4?x?n???22?8xx??16?x421?x?1????2?4
x?n?x? ????1???2n?0?2? f?x??f?0??4nx4n?1????1?4n?1
2n?0??x04n?1x4n?1??nxxf??x?dx??????1?4n?1dx?????1??4n?1dx
0240n?04n?02?x?nx4n?2 ?????1?4n?1,x???2,2?
4n?02?4n?2???n三、傅里叶级数
a0?n?n?1、??ancosx?bnsinx称为三角级数,
2n?1ll若an?0?n?0,1,2,??,则称
?bnsinn?1?n?x为正弦级数, la0?n?若bn?0?n?1,2,??,则称??ancosx为余弦级数
2n?1l:设f?x?是周期为2l的周期函数,若取
1l1ln?1ln?a0??f?x?dx, an??f?x?cosxdx,bn??f?x?sinxdx,n?1,2,?
l?ll?lll?ll上述三角级数称为傅里叶级数,其中系数a0,an,bn称为傅里叶系数. 2.函数展开成傅里叶级数的条件(收敛定理)
设f?x?是周期为2l的周期函数,如果它在??l,l?上连续,或只有有限个第一类间断点,并且至多有有限个极值点,则f?x?的傅里叶级数收敛,并且当x为连续点时,级数收敛于
1fx??fx?. 22ln?xdx?n?1,2,??,此时展3.当f?x?为奇函数时,an?0?n?0,1,2,??,bn??f?x?sin0llf?x?本身, 当x为间断点时,级数收敛于
??????开的傅里叶级数为正弦级数.
2ln?xdx?n?0,1,2,??,bn?0?n?1,2,??此时展当f?x?为偶函数时, an??f?x?cosl0l开的傅里叶级数为余弦级数.
4.周期延拓:若f?x?仅在??l,l?有定义,且满足定理条件,此时可将f?x?在??l,l?外作周期延拓(即作以2l为周期的周期函数F?x?,使x???l,l?,F?x??f?x?),然后将F?x?展开成傅里叶级数,限制x???l,l?,即得到f?x?的傅里叶级数.
5.奇(偶)延拓:若f?x?仅在?0,l?有定义,且满足定理条件,此时可将f?x?先在??l,0?内作奇(偶)延拓, ,然后在??l,l?外作周期延拓(即作以2l为周期的奇(偶)函数F?x?,使
x??0,l?,F?x??f?x?),然后将F?x?展开成傅里叶级数,限制x??0,l?,即得到f?x?的正(余)
弦级数.
例1:将f?x????x????x???展开成傅里叶级数.
22