数学分析泰勒公式-4-2

2008／11／27§4.2带Peano余项的Taylor定理微分：)(xf用一次多项式逼近)())((')()(0000xxoxxxfxfxf一背景?))((,20xxo使误差是用二次多项式近似?))((,0nxxon使误差是次多项式近似用二导出，设])[()()()(20200xxoxxCxxBAxf))((.0存在设、、求xfCBA).(00xfAxx，令⑴⑵)(])[()()()()(020000xxxxoxxCBxxxfxf).('0xfB⑶202020000)(])[()())((')()(xxxxoCxxxxxfxfxf).(21)(2)(')('lim0000xfxxxfxfCxx⑷令二次多项式)'(.0)();,()(lim20020HospitalLxxxxfTxfxx))((')();,(00002xxxfxfxxfT200))((21xxxf⑸200000)(!2)())((')(xxxfxxxfxfknkknxxkxfxxfT)(!)();,(000)(0nnxxnxf)(!)(00)(.0多项式阶处的在称为Taylornxf,0阶的导数有直到在点设函数nxf三Taylor定理(Peano余项)定理],)[();,()(00nnxxoxxfTxf)(0xx证明：归纳法：.,1微分定义时n即：时成立设,kn.0)();,()(lim000kkxxxxxxfTxf则：阶的导数有直到在点设函数,0nxf：考虑1kn);,'();,('001xxfTxxfTkk1001)();,()(lim0kkxxxxxxfTxfkkxxxxkxxfTxf))(1();,'()('lim0000四Taylor公式的意义⑴200000)(!2)())((')()(xxxfxxxfxfxf].)[()(!)(000)(nnnxxoxxnxf)()(,0xfxTxn逼近用的小邻域内在以简代繁⑵])[()()()(0nnnxxoxTxfxR误差：余项Peano⑶)(,00麦克劳林展开称为时取Maclaurinx)(!)0(!2)0()0(')0()()(2nnnxoxnfxfxffxf)(!)0(0)(nnkkkxoxkf五常用展开式1.)(!!212nnxxonxxxe证：.,1)0(,)()()(可得由nxnfexf2.)()!12()1(!7!5!3sin2121753nnnxoxnxxxxx12,)1(2,0)20sin()0(1)(knknnfkn熟记证：3.)()!2()1(!6!4!21cos122642nnnxoxnxxxx4.)()1(32)1ln(132nnnxoxnxxxx)(]32[)1ln(32nnxonxxxxx证：)!1()1()0(,)1()!1()1()(1)(1)(kfxkxfkkkkk5.)1(,)1()(xxxf)(0nnkkkxoxC)(!)1()1(0nnkkxoxkk证：kkxkxf)1)(1()1()()()1()1()0()(kfk广义二项式特例：)(!)121()121(2110nnkkxoxkkxnkkkxkk01!)!2(!)!32()1()121()121(21!1kkak!)!2(!)!32()1(!2)32(31)1(11kkkkkkk)(nxo)()1(11132nnnxoxxxxx)()1(0nnkkkxox)(!)!2(!)!12()1(110nnkkkxoxkkx例1.arctan的麦克劳林展开式求xy解：:)0()(nf关键是求出⑴,11)('2xxf.1)0('f⑵.,1)(')1(2阶导数两边求nxfx0)(22)1()(2)()1()1()()1(2xfnnxxfnxfxnnn)0()1()0(,0)1()1(nnfnnfx取12,)!2()1(2,0)0()(knkknfkn)()12()1(2212nnnxoxn753arctan753xxxxx例2xxfarcsin)(类似地，应用1极值问题定理,0处）阶导数（在有设xkf,0)(,0)()()('0)(0)1(00xfxfxfxfkk12不是极值点为奇数时0,xk是极值点为偶数时0,xk极大极小0)(0)(0)(0)(xfxfkk六应用证：])[()(!)()()(000)(0kkkxxoxxkxfxfxfkkkkxxxxokxfxxxfxf)(])[(!)()()()(000)(00,00时当xx.)()()()(0)(00符号相同与xfxxxfxfkk两侧变号，在为奇数时00)()(,xxfxfk不是极值点；极大极小)()(,0)()()(,0)(00)(00)(xfxfxfxfxfxfkk,为偶数时k应用2求极限例24202coslimxexxx解：)(!4!21cos442xoxxx])2[()2(!21212222222xoxxex)(821442xoxx121)(12lim4440xxoxx原式＝例3解：xxxxxsintanarctansinlim0xxxxxsincossinsintanxxxxxcos1)cos1(sin)1cos1(sin,2~sintan3xxx),(!3sin33xoxxx)(3arctan33xoxxx312)(61lim3330xxoxx原式例4解：xxxxexx30sin)1(sinlim)](6)][(21[sin3322xoxxxoxxxex)(12266353423xoxxxxxx)(3332xoxxx可不写出来)(3)1(sin33xxxxxex31)(3lim3330xxoxx原式＝例5解：)]11ln([lim2xxxx01,xx.21)]11ln([lim2xxxx)1()211()11ln(222oxxxxxxx)1()1(211)11ln(22xoxxx)1(21o例6解：nnnannannancos2coscoslim232323cos2coscosnnananaPn,coslnln123nknnkaP)(21cos32xoxx)()(211cos29332223nkonaknka)]()(211ln[cosln29332223nkonaknka)()1ln(xoxx)]([)](21[29332293322nkonkonkonak)1(2322nonkanknknonkanka1322123)]1()2[(cosln)1(6)12)(1(232onnnna,6coslnlimlnlim2123ankaPnknnn.62ae原式作业(数学分析习题集)习题4.2Taylor公式A2、1),4),6),8);3;4、2),3);5、2),3),8),9).