解:将f?x?在???,??外作周期延拓,使其满意定理条件.可先考虑取g?x??x2,
an?a0?2?2???04??1?xcosnxdx?,n?1,2,? n2n2??02x2dx??2, bn?0,n?1,2,?
3故g?x???23?4?n?122???1?ncosnx,x????,??
n2n???1?2?2所以f?x????x??4?2cosnx,x????,??
3n?1n例2:设在区间???,??上f?x?为可积的偶函数,且f?的展开式中系数a2n?0.
???????x???f??x?,证明在f?x??2??2?证明:由f?x?为可积的偶函数,故展开式为余弦级数.因此
a2n?2????0???22?2??????fxcos2nxdx???fxcos2nxdx???fxcos2nxdx???I1?I2?
??02???令t?????x,I1??2f??t?cos?n??2nt?dt
02?2?令t?x?????,I2??2f??t?cos?nx?2nt?dt,又,cos?n??2nt??cos?n??2nt?
02?2???故a2n?2??20?????f?t??2?????2????f??t??cos?n??2nt?dt?0. ??2???1???1?例3:将f?x??x在区间?0,2??上展开成傅里叶级数,并求?2,?n2n?1nn?1解: f?x?在?0,2??外作周期延拓,使其满意定理条件.
n?1.
an?1?1?2?041xcosnxdx?2,n?1,2,? a0?n?2?2?08?2xdx?,
32bn???2?0x2sinnxdx??4?,n?1,2,? n42?4cosnx4?sinnx?x2故 ??????223nnn?1?2?x??0,2??
x?0,2?4?2?4令x?0, 由2????2得
3n?1n21?2, ??26nn?1?4?2?4??1?令x??, 由??得 ??23nn?1n2?n?1???1?n?1??2.
n212例4:怎样才能将在?0,
???
?内可积的函数f?x?延拓到???,??,使其傅里叶展开式为?2?
??Ansin?2n?1?x.
n?1解: 展开式中只有正弦项,故可取f?x???f??x?,又sin2nx项不b2n?0,n?1,2,?因此
b???2n?2??0fx?sin2nxdx?2????2f?x?sin2nxdx???2nxdx??0?f?x?sin?
2????? ?2????20f?xsin2nxdx??20f???t?sin2n???t?dt?? ? ?2???20?f?x??f???x??sin2nxdx f?x??f???x?则b2n?0,n?1,2,?,故当f?x?满足f?x???f??x?,f?x?,就使展开式满足要求.即
???f?x???,???x???,?2??F?x????f??x?,??x?0,?2 ??f?x?,0?x???2,??f???x?,?2?x??.5:在区间?0,2?上将f?x????x,0?x?1,??2?x,1?x?2展开成余弦级数,并求?12,
n?1n:先在??2,0?内作偶延拓, ,然后在??2,2?外作周期延拓,使其满足收敛定理要求.
a2n?22?0f?x?cosn?1n?22xdx??0xcos2xdx??1?2?x?cosn?2xdx ,故
x?出现取????时例解0,n?2m?1,?4?n??n? ?22?2cos?1???1????2 m???1?1,n?2m.2n????m2?2???n?1,2,?
又,
2m2?2?0,m?2k,????1?n?1???4,m?2k?1.
22???2k?1???1222a0??f?x?dx??xdx???2?x?dx?1,
012014所以,f?x???22???2k?1?k?1?12cos2?2k?1??x, x??0,2?
2?14又,由f?0???22?1, ?2n?1n????2k?1?k?1?2?12?0得?k?11?2k?1?2??28,
??2k?1???2n?,
n?1n?1?112收敛,故
???111??11?21?1???????????2?22?222?2n??n?1?2n?1?n?1?2n?84n?1nn?1nn?1??2n?1?1?2因此,?2?.
6n?1n?例6: 已知f?x?是以2?为周期的函数, an,bn为其傅里叶系数, 试将
F?x??1???f?t?f?x?t?dt展开成傅里叶级数.
??解:由F?x?2???????1?f?t?f?x?2??t?dt?f?t?f?x?t?dt?F?x?知:F?x?为以????1?2?为周期的函数. 1?1 由F??x???f?t?f??x?t?dt?(令u??x?t)
????????x??xf?u?x?f?u?du
??f?u?f?u?x?du?F?x?知F?x?为偶函数,故
???A0?F?x?的傅里叶级数为余弦级数:??Ancosnx,其中
2n?1An? ? ?2?1?F?x?cosnxdx???02?2?0?1??????f?t?f?x?t?dt?cosnxdx ?????????1f?t??????f?x?t?cosnxdx??dt?2???????f?t?(???t???tf?v?cosn?v?t?dv)dt
1?2????f?t?[?f?v?(cosnvcosnt?sinnvsinnt)dv]dt
??? ?f?t??a????1?n22cosnt?bnsinnt?dt?an?bn, n?0,1,2,?
a2?22因此, F?x?的傅里叶级数为??an?bncosnx。
2n?1??