经济数学第五章级数与拉普拉斯变换 下载本文

因为u?t?是周期为2?的偶函数,所以有 bn?0 ?n?1,2,3?, ... an? =

2?E??02?tu??tcosnt=dt?Esincosntdt

?02???0??1?1???sinn?t?sinn????t?dt ??22????????1?1????cosn?tcosn?t??????E2?2???? =???11???n?n???22??0??E?11? =??? 11??n?n???22? =?4E?4n2?1?? ?n?0,1,2,3? ,...将求得的an代入余弦级数,得u?t?的傅里叶级数展开式为

4E?1?1? u?t?? ???? ?cosntt???????2??2n?14n?1?

习题5.4

1下列周期函数f?x?的周期为2?,试将f?x?展开成傅里叶级数,如果f?x?在???,??上的表达式为:

(1)f?x??3x2?1 ????x??? (2)f?x??e2x ????x???

?bx,???x?0(3)f?x??? ?a,b 为常数,且a?b?0??ax,0?x??2 将下列函数f?x?展开成傅里叶级数: (1)f?x??2sinx ????x??? 3?ex,???x?0 (2)f?x???

?1,0?x??x (3)f?x??cos ????x???

23 设f?x?是周期为2?的周期函数,它在???,??上的表达式为

????,???x???22???? f?x???x,??x?

22?????2,2?x???将f?x?展开成傅里叶级数.

4 将函数f?x??

??x2 ?0?x??? 展开成正弦级数.