第五节函数展开成幂级数第十一章二、典型例题一、主要内容三、同步练习四、同步练习解答一、主要内容(一)函数的幂级数展开式——泰勒(Taylor)展开式定义,为区间若)(,)()(00IIxxxaxfnnn∈−=∑∞=.)()(0的幂级数上可以展开成在则称xxIxf−问题:(1)如果能展开,an是什么?(2)展开式是否唯一?(3)在什么条件下才能展开成幂级数?1.函数展开成幂级数2.定理(展开式的唯一性)若在邻域U(x0,R)内任意阶可导的函数f(x)能展成幂级数:),(,)()(000RxUxxxaxfnnn∈−=∑∞=,),2,1,0(!)(0)(L==nnxfann则其系数且展开式是唯一的.麦克劳林级数(x0=0):+)0(f+′xf)0(2!2)0(xf′′LL+++nnxnf!)0()(称处具有任意阶导数,则在设0)(xxfnnnxxnxf)(!)(000)(−∑∞=+=)(0xf+−′))((00xxxf200)(!2)(xxxf−′′处的在为0)(xxfLL+−++nxx)0nnxf(!)(0)(泰勒系数.泰勒级数3.定义(泰勒级数)~)(xf(2)收敛域?(3)在收敛域I内,级数是否一定收敛到f(x)?4.泰勒级数基本问题;)(!)()1(000)(nnnxxnxf−∑∞=构造即Ixxxnxfxfnnn∈−=∑∞=,)(!)()(000)(?答:不一定.设f(x)在区间I上具有各阶导数,(二)函数展开成幂级数的充分必要条件则f(x)在I上能展开成泰勒级数,即)(0)(limIxxRnn∈∀=∞→定理11.14Ixxxnxfxfnnn∈−=∑∞=,)(!)()(000)(⇔10)1()()!1()()(++−+=nnnxxnfxRξ其中的泰勒公式中的余项)(xf之间),在0(xxξ(三)函数展开成幂级数的方法(1)直接展开法的幂级数的展开成xxf)(1º求f(n)(x),f(n)(0),n=0,1,2,···;2º写出幂级数3º判断?0)(lim=∞→xRnn展开方法直接展开法—用泰勒公式间接展开法—用已有展开式∑∞=0)(,!)0(nnnxnf并求收敛半径R;),(RRx−∈步骤:(2)间接展开法根据展开式的唯一性,利用常见展开式,通过变量代换,四则运算,恒等变形,逐项求导,逐项积分等方法,求展开式.二、典型例题例1将xexf=)(展开成x的幂级数.,)()(xnexf=),,1,0(1)0()(L==nfn解1收敛半径+∞=∞→=nRlim!1n!)1(1+nx+2!21x+3!31x+LL+++nxn!1的麦克劳林级数写出xeLL+++′′+′+nnxxnfxfxffe!)0(!2)0()0()0()(2~即o1o2),(+∞−∞收敛区间:xe余项满足)(=xRnξe!)1(+n1+nx),,(+∞−∞∈∀xQ3ºLL+++++=nxnxxxS!1!211)(2=?),(+∞−∞∈x)0(之间与介于xξxe!)1(1++nxn∞→n0)(0时当通项收敛级数的∞→→nun,!1!31!21132LL++++++=nxxnxxxe),(+∞−∞∈x例2将xxfsin)(=展开成x的幂级数.解),2sin()()(πnxxfn+=,2sin)0()(πnfn=,0)0()2(=∴kf,)1()0()12(kkf−=+),2,1,0(L=k==!)0()(nfann⎪⎩⎪⎨⎧+=+−=12,)!12()1(2,0knkknk),2,1,0(L=k~sinx,)!12()1(012∑∞=++−kkkkx收敛半径.+∞=Ro1o2!)1(1++nxn∞→n0余项满足),,(+∞−∞∈∀x3°),(∞+−∞∈xxsinLL++−+−+−=+1253)!12(1)1(!51!31nnxnxxx1)1()!1()()(+++=nnnxnfxRξ=]2)1(sin[πξ++n!)1(+n1+nx例3将mxxf)1()(+=展开成x的幂级数(m:任意常数).o1,)0(m解f=′,)1()0(−=′′mmf)2)(1()0()(−−=mmmfn2°麦克劳林级数++mx1L+−2!2)1(xmm1lim+∞→=nnnaaRnmnn−+=∞→1lim1=LL++−−+nxnnmmm!)1()1()1,1(−∈xLL,)1(+−nm,1)0(=fL+−2!2)1(xmmLL++−−+nxnnmmm!)1()1([xmmxF111)('−+=++=xmxF1)(]LLL+−+−−++−1!)1()1()1(nxnnmm11,)(−xxF3°设和函数为.)1()(:mxxF+=下证]LL+−+−−+nxnnmm!)1()1()1([L+−+=′211)(xmxmxFx!)()1(nnmm−−L!)1()1()1(−+−−+nnmmL!)1()1(nnmmm+−−=L)()1(xFx′+[1m=xm+L+−+2!2)1(xmm]LL++−−+nxnnmmm!)1()1()(xFm=)()1(xFx′+),(xmF=,∫∫+=′xxxxmxxFxF00d1d)()(),1ln()0(ln)(lnxmFxF+=−1)0(=FL+−2!2)1(xmmLL++−−+nxnnmmm!)1()1(++=xmxF1)()1,1()1()(−∈+=∴xxxFm,)1,1(−∈xL+−+2!2)1(xmmmx)1(+)11(−xLL++−−+nxnnmmm!)1()1(二项展开式:xm+=1注1°.1的取值有关处收敛性与在mx±=2°m为正整数时,得二项式定理:++=+mxxm1)1(L+−2!2)1(xmmmx+2121,−=m时二项展开式分别为3°11=+x2421x⋅−364231x⋅⋅⋅+)11(≤≤−xL+⋅⋅⋅⋅⋅−48642531x111=+x24231x⋅⋅+3642531x⋅⋅⋅⋅−)11(≤−xL−⋅⋅⋅⋅⋅⋅+486427531xx21−x21+mx)1(+nnxnnmmm!)1()1(1+−−∑∞=L+=1)11(−x将cosx展开成x的幂级数.解xsin120)!12(1)1(+∞=+−=∑nnnxn),(∞+−∞∈x逐项求导:)(sin′=xxcosnnnxn20)!2(1)1(−=∑∞=LL+−+−+−=)!2()1(!41!211242nxxxnn),(∞+−∞∈x例4例5将)1ln()(xxf+=展开成x的幂级数.解xx+=+11)]'1[ln()11()1(0−−=∑∞=xxnnn=+xx0)1n(l,d)1(d11000xxxxxnnnx∫∑∫∞=−=+1x=+)1ln(x,∑∞=++−=011)1(nnnxn连续,.11≤−x因右端幂级数在x=1收敛,1)1ln(=+xx在而故展开式对x=1也成立,收敛域为.11≤−x注取x=1得,LL++−+11)1(nn−+−=312112ln11≤−x,∑∞=++−=011)1(nnnxn)1ln(x+例6将L++−−++)!12(4112kπxkk)()(xsin展成4πx−解xsin)cos(sin44ππx−=[])sin()cos(4421ππxx−+−=⎢⎣⎡=21⎜⎝⎛⎟⎠⎞⎜⎝⎛+−−−−−+=L32)4(!31)4(!21)4(121πxπxπx)(+∞∞−x的幂级数.2)4(!21πx−−L−−+4)4(!41πx⎟⎠⎞1⎜⎝⎛+⎥⎦⎤)4(πx−3)4(!31πx−−L+⎟⎠⎞)sin(cos44ππx−+[])(sin44ππx−+=L+−−+)!2()4(12kπxkk)(.42312的幂级数展开成将+++xxx例7解2312++xx()()211++=xx2111+−+=xxnnx∑∞=⎟⎠⎞⎜⎝⎛+−=03431∑∞=⎟⎠⎞⎜⎝⎛++02421nnx34+x2312++∴xx()nnnnx4)3121(011+−=∑∞=++)26(−−x)1(21)1(31−+−−=34+x24+x24+x24+x的幂级数展开成将xxarcsin()例8解211arcsinxx−=′得取二项展开式中,21−=m22321!221xx⋅+−=32!!6!!5!!4!!321xxx−+−=[]212)(1−−+=xL+⋅⋅−3252321!3x()()()LL+−−++nnxnn1!!2!!12()()()()1,1!!2!!12111−−−+=∑∞=nnnxnn()211−+x得代替xx2−422!!4!!32111xxx++=−()()nnxnn21!!2!!121∑∞=−+=xxxxd11arcsin02∫−==xarcsin)1,1(−∈x()()LL+−++nxnn2!!2!!12()()(),12!!2!!12121+−++∞=∑nxnnxnn()()()()1,1!!2!!1211−−−∑∞=nnnxnn()+=+−1121x时,当1=x∑∞=+−+1121!)!2(!)!12(1nnnn121!)!2(!)!12(+⋅−=nnnun则令,!)!2(!)!12(nnvn−=nnvn2124321−⋅=L11,0++bababa则若1211+⋅=nvn,1212+nvn121+nvn1225432+⋅nnL23)12(1+n又因121+⋅=nvunnQ23)12(1+n,212323n收敛而∑∞=1232321nn收敛,∑∞=∴1nnu收敛故∑∞=+−+1121!)!2(!)!12(1nnnn时,当1−=x收敛]121!)!2(!)!12(1[1∑∞=+−+−nnnn],1,1[−∈∴x展开式成立的范围是即=xarcsin]1,1[−∈x()()(),12!!2!!12121+−++∞=∑nxnnxnn例9解()xxxf23ln−=将()()xxxf23lnln−−=()]11ln[−+=x()()∑∞=−−−=1111nnnnx()()nnnnxn12111−+−=∑∞=−时21=x.1的幂级数展成−x.23时发散=x拆配化一展范围],[2321∈x()]121ln[−−−x()[]∑∞=−−−112nnnx⎟⎠⎞⎜⎝⎛−211x(),212111收敛nnnnn⎟⎠⎞⎜⎝⎛−+−∑∞=−三、同步练习1.将在x=0处展为幂级数.)32ln()(2xxxf−+=2.将3412++xx展成x-1的幂级数.3.将函数展开成x的幂级数:xxxf−+=11arctan)(()21lnarctanxxxxf+−=将4..的幂级数展成x四、同步练习解答1.将在x=0处展为幂级数.)32ln()(2xxxf−+=解)1ln(2ln)1ln()(23xxxf+++−==−)1ln(x∑∞=−1nnnx)11(≤−x)1ln(23x+nnnxn)(23)1(11−−∞=∑=)(3232≤−x2ln)(故=xf∑∞=−1nnnxnnnxn)()1(2311∑∞=−−+nnnxnxf])(1[12ln)(231−+−=∑∞=)(3232≤−x)32)(1(322xxxx+−=−+2.将3412++xx展成x-1的幂级数.解)3)(1(13412++=++xxxx)3(21)1(21xx+−+=()141+=21−x⎥⎦⎤⎢⎣⎡=4121−−xL+−+222)1(xL+−−+nnnx2)1()1(⎥⎦⎤⎢⎣⎡−81141−−xL+−+224)1(xL+−−+nnnx4)1()1(()nnnnnx)1(2121)1(3220−−−=++∞=∑)31(−x)21(−x()181+−41−x13.将函数展开成x的幂级数:xxxf−+=11arctan)(解=′)(xf211x+,)1(02∑∞=−=nnnx)1,1(−∈x=−)0()(fxf∑∫∞=−002d)1(nxnnxx∑∞=++−=01212)1(nnnxnx=±1时,级数条件收敛,,4)0(πf=,12)1(4)(012∑∞=++−+=nnnxnπxf]1,1[−∈x故问题:f(x)的导数易展?积分()21lnarctanxxxxf+−=将4