將下列各周期函數(shù)展開成傅里葉級數(shù)下面給出函數(shù)在一個

個人收集整理僅供參照學習習題1181將以下各周期函數(shù)張開成傅里葉級數(shù)(下面給出函數(shù)在一個周期內的表達式)(1)f(x)1x2(1x1)22解由于f(x)1x2為偶函數(shù)所以bn0(n12)而11a022(1x2)dx42(1x2)dx111/2006212nxan2(1dx1/2x)cos01/212(1)n142(1x)co2snxdx(n12)n220由于f(x)在()內連續(xù)所以f(x)111(1)n1xx()122n1n2co2snx1x0(2)f(x)10x1211x1210111解anxdx2dxf(x)dx1dx11022an1xdx0xdx11f(x)consxcons2consxdx1consxdx11021n2nn22[1(1)]nsin2(n12)1011bxdxxdx2sinnxdxf(x)sinnxsinn1sinnxdxn11022n1(n12)ncosn2個人收集整理僅供參照學習而在()上f(x)的中止點為x2k2k1k0122nn1n故11(1)2sin2cosf(x){[2]consx2sin}4n22nnxn1(x2kx2k1k02(3)f(x)2x13x010x3解a131[01)dx31f(x)dx(2xdx]033330an13nxdx1[0nxdx3nxdx]3f(x)cos3(2x1)coscos33330361)n](n12)22[1(nbn13nxdx10nxdx3nxdx]3f(x)sin3[(2x1)sinsin3333306(1)n(n12)n而在()上f(x)的中止點為x3(2k1)k012故f(x)1{6[1(1)nnx(1)n16nx}2]cosnsinn1n2233(x3(2k1)k0將以下函數(shù)分別張開成正弦級數(shù)和余弦級數(shù)x0xl(1)f(x)2llxxl2解正弦級數(shù)對f(x)進行奇延拓則函數(shù)的傅氏系數(shù)為a00(n012)21nxlnx4lnbn[dx1(ldx]l2xsinx)sinn22sin(n120l2l2

12)12))個人收集整理僅供參照學習故4l1nnxx[0l]f(x)2n1n2sin2sinl余弦級數(shù)對f(x)進行偶延拓則函數(shù)的傅氏系數(shù)為2[1lla02xdx1(lx)dx]l022an2[l

120

xcosnxdxlx)cosnxdx]1(ll2l2ln1(1)n](n12)n22[2cos2bn0(n12)故f(x)l2l1n1(1)nnxx[0l]42n1n2[2cos]cosl2f(x)x2(0x2)解正弦級數(shù)對f(x)進行奇延拓則函數(shù)的傅氏系數(shù)為a00(n012)b222sinnxdx(1)n1816[(1)n1]xn202n(n)3故f(x){(1)n1816[(1)nnxn(n)31]}sin2n18{(1)n12[(1)n1]nxx[02)nn32}sin2n1余弦級數(shù)對f(x)進行偶延拓則函數(shù)的傅氏系數(shù)為a2228xdx0203an22x2cosnxdx(1)n16(n12)202(n)2bn0