an??1???f(x)cosnxdx?????1?0xcosnxdx1n2?cosnx?11xsinnx?0n?n?1??0sinnxdx??0
?2,当n为奇数时，???2(cosn??1)??n2?n??当n为偶数时；?0, 1?1?bn??f(x)sinnxdx??xsinnxdx????0?11?(?1)n?11?(?1)n?1??xcosnx?cosnxdx??2?cosnxdx?.?000n?n?nnn?所以在开区间(??,?)上
??(2cosx?sinx)?1sin2x?(2cos3x?1sin3x)?. f(x)?
4?29?3在x???时,上式右边收敛于
f(??0)?f(???0)??0???.
222于是,在[??,?]上f的傅里叶级数的图象如图15-4所示(注意它与图15-3的差别). □
例2 把下列函数展开成傅里叶级数:
?x2,0?x??,? f(x)??0,x??,??x2,??x?2?.? 解 f及其周期延拓的图形如图15-5所示．显然f是按段光滑的，因此它可以展开
成傅里叶级数．
在(10)中令c=0来计算傅里系数如下：
'a0????0?27?23?312?f(x)dx?1???0x2dx?1???2?(?x2)dx
??2?2.an?11??2?0f(x)cosnxdx?1???0x2cosnxdx?1???2?(?x2)cosnxdx?1x222? x222x2x?[(?3)sinnx?2cosnx]?[(?3)sinnx?2cosnx]0?nn??nnnn4[(?1)n?1],2n12?1bn??f(x)sinnxdx???0???0xsinnxdx?2???12?(?x2)sinnxdx?12?x222xx222x?[(??3)cosnx?2sinnx]?[(??3)cosnx?2sinnx]
0???nnnnnn122?2?2?{?(?3)[1?(?1)n]}.?nnn所以当x?(0,?)?(?,2?)时,
242?2?2nf(x)?????{2[(?1)?1]cosnx?[?(?3)(1?(?1)n)]sinnx}?nnnn?1n112 ???2?8(cosx?2cos3x?2cos5x??)?{(3?2?4)sinx 35?22?3?24??sin2x?(?3)sin3x?sin4x??}.23342?当x??时,由于
所以
f(??0)?f(??0)?0
22 0????8(当x?0或2?是，由于
因此
?2?由(14)或(15)都可推得
2111????). (14) 22213511(f(0?0)?f(0?0))?(?4?2?0)??2?2, 22111???2?8(2?2?2??). (15)
1352111?. □ 2?2?2???1358
作业布置：Ｐ７０ １（１）；３（１）；４；７（１）．