第五节 函数展开成幂级数VIP免费

一、泰勒级数二、函数展开成幂级数三、欧拉公式四、小结第五节函数的泰勒级数一、泰勒级数上节例题)11()1ln()1(11xxnxnnnnnnxxaxf)()(00存在幂级数在其收敛域内以f(x)为和函数问题:1.如果能展开,是什么?na2.展开式是否唯一?3.在什么条件下才能展开成幂级数?证明即内收敛于在),()()(000xfxuxxannnnnxxaxxaaxf)()()(0010定理1如果函数)(xf在),(0xU内具有任意阶导数,且在),(0xU内能展开成)(0xx的幂级数,即nnnxxaxf)()(00则其系数),2,1,0()(!10)(nxfnann且展开式是唯一的.)(23)1(!)(01)(xxannanxfnnn即得令,0xx),2,1,0()(!10)(nxfnann泰勒系数是唯一的,.)(的展开式是唯一的xf10021)()(2)(nnxxnaxxaaxf逐项求导任意次,得泰勒系数如果)(xf在点0x处任意阶可导,则幂级数nnnxxnxf)(!)(000)(称为)(xf在0xx处的泰勒级数.nnnxnf0)(!)0(称为)(xf的麦克劳林级数.问题nnnxxnxfxf)(!)()(000)(?定义泰勒级数在收敛区间是否收敛于f(x)?不一定.0,00,)(21xxexfx例如),2,1,0(0)0()(nfn且00)(nnxxf的麦氏级数为.0)(),(xs内和函数该级数在可见).()(,0xfxfs于的麦氏级数处处不收敛外除在x=0点任意可导,定理2)(xf在点0x的泰勒级数,在),(0xU内收敛于)(xf在),(0xU内0)(limxRnn.证明必要性)()(!)()(000)(xRxxixfxfninii),()()(xsxfxRnn,)(能展开为泰勒级数设xf)()(limxfxsnn)(limxRnn)]()([limxsxfnn;0充分性),()()(xRxsxfnn)]()([limxsxfnn)(limxRnn,0),()(limxfxsnn即).()(xfxf的泰勒级数收敛于定理3设)(xf在)(0xU上有定义,0M,对),(00RxRxx,恒有Mxfn)()(),2,1,0(n,则)(xf在),(00RxRx内可展开成点0x的泰勒级数.证明10)1()()!1()()(nnnxxnfxR,)!1(10nxxMn),(00RxRxx,),()!1(010收敛在nnnxx,0)!1(lim10nxxnn,0)(limxRnn故.0的泰勒级数可展成点x),(00RxRxx二、函数展开成幂级数1.直接法(泰勒级数法)步骤:;!)()1(0)(nxfann求,)(0lim)2()(MxfRnnn或讨论).(xf敛于则级数在收敛区间内收例1解.)(展开成幂级数将xexf,)()(xnexf),2,1,0(.1)0()(nfn!1!)0()(nnfann,0M上在],[MMxnexf)()(Me),2,1,0(nnxxnxxe!1!2112由于M的任意性,即得),(!1!2112xxnxxenx例2.sin)(的幂级数展开成将xxxf解),2sin()()(nxxfn,2sin)0()(nfn,0)0()2(nf,)1()0()12(nnf),2,1,0(n)()(xfn且)2sin(nx1),(x)!12()1(!51!31sin1253nxxxxxnn),(x例3.)()1()(的幂级数展开成将xRxxf解,)1)(1()1()()(nnxnxf),1()1()0()(nfn),2,1,0(nnxnnxx!)1()1(!2)1(12nnnaa1lim1nn,1,1R若内在,)1,1(nxnnxxs!)1()1(1)(1)!1()1()1()1()(nxnnxxsnxnnxxxsx)!1()1()1()1()(2!)1()1(!)()1()!1()1()1(nnmmmnnmmnnmm利用)()1(xsx1222!)1()1(!2)1(nxnnxx)(xs,1)()(xxsxs.1)0(s且两边积分,1)()(00dxxdxxsxsxx)1,1(x得),1ln()0(ln)(lnxsxs即,)1ln()(lnxxs,)1()(xxs)1,1(xnxnnxxx!)1()1(!2)1(1)1(2)1,1(x牛顿二项式展开式注意:.1的取值有关处收敛性与在x有时当,21,1)1,1()1(11132nnxxxxx]1,1[!)!2(!)!32()1(642314...