Question

【求助高等数学】求解第一小问，无穷级数相关，求详解，比较急，谢谢

Answer 1

f(x) = a0/2 + ∑<n=1,∞> [ancos(nπx/2) + bnsin(nπx/2)]
a0 = (1/2)∫<-2, 2> (x^2-x)dx (奇函数在对称区间积分为0)
= ∫<0, 2> x^2dx = 8/3.
an = (1/2)∫<-2, 2> (x^2-x)cos(nπx/2)dx
= ∫<0, 2> x^2cos(nπx/2)dx
= [2/(nπ)]∫<0, 2> x^2dsin(nπx/2)
= [2/(nπ)]{[x^2dsin(nπx/2)]<0, 2> - ∫<0, 2> 2xsin(nπx/2)}
= [8/(nπ)^2]∫<0, 2> xdcos(nπx/2)}
= [8/(nπ)^2]{[xcos(nπx/2)]<0, 2> - ∫<0, 2>cos(nπx/2)dx}
= [8/(nπ)^2]{2cosnπ - [2/(nπ)][sin(nπx/2)]<0, 2> }
= [16(-1)^n/(nπ)^2] , n = 1, 2, 3, ...;
bn = (1/2)∫<-2, 2> (x^2-x)sin(nπx/2)dx
= -∫<0, 2> xsin(nπx/2)dx = [2/(nπ)]∫<0, 2> xdcos(nπx/2)
= [2/(nπ)]{[xcos(nπx/2)]<0, 2> - ∫<0, 2> cos(nπx/2)dx}
= [2/(nπ)]{2cosnπ - [2/(nπ)][sin(nπx/2)]<0, 2>}
= 4(-1)^n/(nπ), n = 1, 2, 3, ...
代入第一行展开式即得。