將下列各周期函數(shù)展開(kāi)成傅里葉級(jí)數(shù)(下面給出函數(shù)在一個(gè)...

1、221/2L(1-x2)01/2dx4J2(1-x2)(-1)n+1n2冗2(n=1,2,),由于f(x)在(-8,+8)內(nèi)連續(xù)，所以f(x)=11+丄為co2n冗x,xw(-8,+8).12兀2n2n=1x-1x0f(x)10 x2；一1丄x1I2Af(x)dx=J。xdxJ2dx-J1dx=-21-1-1a=Pf(x)con冗xdxJ。xcn-1-1on兀xdxJ2con兀xd:con兀xdx21-(-1)n+2sin(n=1,2,),2b=Pf(x)sinn冗xdx=J。xsinn冗xdxJ2sinn冗xdx-n-1-1J:sinn冗xdx習(xí)題1181.將下列各周期函數(shù)展開(kāi)成傅里葉級(jí)數(shù)(

2、下面給出函數(shù)在一個(gè)周期內(nèi)的表達(dá)式)：(1)f(X)1x2(丄x丄)；22解因?yàn)閒(x)=1-x2為偶函數(shù)，所以bn=0(n=1,2,),而1/22(1-X2)dx4J2(1-X2)dx11622A(=1，2而在(也+8)上f(x)的間斷點(diǎn)為x2k,2k+1,k0,1,2,2f(x)-4+丫3n+4n1n冗2sincon冗1-2cosrn冗x+sim冗xn冗(Z,x豐2k+1,k0,1,2,).f(x)2x+11f(x)dx(2x+1)dx+dx一1,一3一3dx+dx13n冗x,f(x)cosdx3一331(1)n(n1,2,.)，n2冗2,11*3、.n冗x7lo小1、-n冗x73.n冗x,

3、.bJf(x)sindxJ(2x+1)sindx+Jsindxn333333032(1)n(n1,2,.),n冗而在(-8,+8)上,f(x)的間斷點(diǎn)為x3(2k+1),k0,1,2,.,故f(x)1+Z61(1)nco+(1)n+1Sin,TOC o 1-5 h z2n2兀23n兀3n1(x豐3(2k+1),k0,1,2,).2.將下列函數(shù)分別展開(kāi)成正弦級(jí)數(shù)和余弦級(jí)數(shù)：x0 x(1)f(x)l2“xxl2解正弦級(jí)數(shù)：對(duì)f(x)進(jìn)行奇延拓，則函數(shù)的傅氏系數(shù)為a00(n0,1,2,),4-sin(n1,2,)n2兀221.n冗x/n冗x,.J2xsindx+J(lx)sindxl0ll故f(x)

4、，址丄sinsinnxx0,l冗2n22ln，1余弦級(jí)數(shù)：對(duì)f(x)進(jìn)行偶延拓，則函數(shù)的傅氏系數(shù)為，J2xdx+l0(1-x)dx，右22vf1n冗x,門/?、n冗x,.TOC o 1-5 h za，J2xcosdx+J(l一x)cosdxnl0l1l22CO爐-1-(-1)n(n=l,2)n2冗22bn，0(n，12,)故f(x)，-+里2cO爐一1-(1)ncOxW0,4冗2n22ln，1(2f(x)，x2(0女W2)解正弦級(jí)數(shù)：對(duì)f(x)進(jìn)行奇延拓，則函數(shù)的傅氏系數(shù)為a0=0(n=0,1,2,)b，-J2x2sin迪dx，(-Dn+1-8+16(-1)n-1,n202n冗(n冗)3故f(x)，(-1)n+1丄+西(-1)n-1sinn冗(n冗)32n=1,8+2(-1)n-1sin,x0,2)冗nn3冗22n=1余弦級(jí)數(shù)：對(duì)f(x)進(jìn)行偶延拓，則函數(shù)的傅氏系數(shù)為a，x2cos兀xdx，(-1)n16(n=1,2n