您好,欢迎访问三七文档
习题1171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)3x21(x)解因为)1(2)13(1)(1220dxxdxxfadxnxfancos)(12212)1(cos)13(1ndxnxn(n12)dxnxfbnsin)(10sin)13(12dxnx(n12)所以f(x)的傅里叶级数展开式为)(cos)1(121)(122xnxnxfnn(2)f(x)e2x(x)解因为21)(12220eedxedxxfaxdxnxfancos)(1)4()()1(2cos12222needxnenx(n12)dxnxfbnsin)(1)4()()1(sin12222neendxnenx(n12)所以f(x)的傅里叶级数展开式为1222)sincos2(4)1(41[)(nnnxnnxneexf(x(2n1)n012)(3)xaxxbxxf00)((ab为常数且ab0)解因为)(211000baaxdxbxdxa]cos1cos100nxdxaxnxdxbxannnab)1(1[2(n12)00sin1sin1nxdxaxnxdxbxbnnban1)1((n12)所以f(x)的傅里叶级数展开式为112}sin)()1(cos)]()1(1[{)(4)(nnnnxnbanxnabbaxf(x(2n1)n012)2将下列函数f(x)展开成傅里叶级数(1)3sin2)(xxf(x)解将f(x)拓广为周期函数F(x)则F(x)在()中连续在x间断且)()]()([21fFF)()]()([21fFF故F(x)的傅里叶级数在()中收敛于f(x)而在x处F(x)的傅里叶级数不收敛于f(x)计算傅氏系数如下因为3sin2x(x)是奇函数所以an0(n012)00])31cos()31[cos(2sin3sin22dxxnxnnxdxxbn19318)1(21nnn(n12)所以12119sin)1(318)(nnnnxnxf(x)(2)xxexfx010)(解将f(x)拓广为周期函数F(x)则F(x)在()中连续在x间断且)()]()([21fFF)()]()([21fFF故F(x)的傅里叶级数在()中收敛于f(x)而在x处F(x)的傅里叶级数不收敛于f(x)计算傅氏系数如下edxdxeax1][1000)1()1(1]coscos[1200nenxdxnxdxeanxn(n12)]sinsin[100nxdxnxdxebxn})1(11])1(1[{12nnennn(n12)所以21)(exf122}]sin)1(11)1([cos1)1(1{1nnnnnxnnnennxne(x)3设周期函数f(x)的周期为2证明f(x)的傅里叶系数为20cos)(1nxdxxfan(n012)20sin)(1nxdxxfbn(n12)证明我们知道若f(x)是以l为周期的连续函数则laadxxf)(的值与a无关且llaadxxfdxxf0)()(因为f(x)cosnxsinnx均为以2为周期的函数所以f(x)cosnxf(x)sinnx均为以2为周期的函数从而2cos)(1cos)(1nxdxxfnxdxxfan20cos)(1nxdxxf(n12)同理20sin)(1nxdxxfbn(n12)4将函数2cos)(xxf(x)展开成傅里叶级数解因为2cos)(xxf为偶函数故bn0(n12)而0cos2cos2cos2cos1nxdxxnxdxxan0])21cos()21[cos(1dxxnxn1414)1(21nn(n12)由于2cos)(xxf在[]上连续所以121cos141)1(422cosnnnxnx(x)5设f(x)的周期为2的周期函数它在[)上的表达式这xxxxxf222222)(将f(x)展开成傅里叶级数解因为f(x)为奇函数故an0(n012)而]sin2sin[2sin)(22200nxdxnxdxxnxdxxfbn2sin2)1(2nnnn(n12)又f(x)的间断点为x(2n1)n012所以nxnnnxfnnsin]2sin2)1([)(121(x(2n1)n012)6将函数2)(xxf(0x)展开成正弦级数解作奇延拓得F(x)0)(000)()(xxfxxxfxF再周期延拓F(x)到()则当x(0]时F(x)f(x))0(20)0(fF因为an0(n012)而nnxdxxbn1sin220(n12)故nxnxfnsin1)(1(0x)级数在x0处收敛于07将函数f(x)2x2(0x)分另别展开成正弦级数和余弦级数解对f(x)作奇延拓则an0(n012)而]2)2()1[(4sin2232302nnnnxdxxbnn(n12)故正弦级数为nxnnnxfnnsin]2)2()1[(4)(1323(0x)级数在x0处收敛于0对f(x)作偶延拓则bn0(n12)而20203422dxxa2028)1(cos22nnxdxxann(n12)故余弦级数为nxnxfnncos)1(832)(122(0x)8设周期函数f(x)的周期为2证明(1)如果f(x)f(x)则f(x)的傅里叶系数a00a2k0b2k0(k12)解因为020200)(1)(1)(1adttfdxtfdxxfaxt令所以a00因为dxtktfkxdxxfaxtk)(2cos)(12cos)(1202令kaktdttf2202cos)(1所以a2k0同理b2k0(k12)(2)如果f(x)f(x)则f(x)的傅里叶系数a2k10b2k10(k12)解因为)12cos()(112xdxkxfakdxtktfxt))(12cos()(120令1220)12cos()(1katdtktf所以a2k10(k12)同理b2k10(k12)
本文标题:117
链接地址:https://www.777doc.com/doc-6513251 .html