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三阶麦克劳林公式其余项为26R3xex4423所以当0x1时,按公式ex1xxx计算ex的误差262
2R3xex431400045001442
1
ee211112113164522262
1
9应用三阶泰勒公式求下列各数的近似值并估计误差13302si
18解1设fx3x则fx在x027点展开成三阶泰勒公式为
fx3x3271273x2712273x27232925
11
f110273x2731803x274介于27与x之间327481于是33032712733122733211027333329327258
8
11
311153107243633310其误差为
R330180481
111133418027334
481
801881054311
2已知
si
4si
xx1x3x介于0与x之间34所以si
18si
13030901010310
其误差为
si
4R310410si
64203104410
10利用泰勒公式求下列极限1lim3x33x24x42x3
x
2lim
x0
x2xl
1x
xcosxe2
2

11x21x223lim2x0cosxexsi
x2
解1lim3x33x24x42x3lim
x313
x
412xxlim313t412tt01tx
因为313t1tot412t11tot所以2
1tot11totot32lim3x33x24x42x3limlim3xt0t02tt2
12
f2lim
x0x2xl
1x
xcosxe2
2
11x21x4ox411x211x4ox424224lim
x0
x
3
11l
1xx
ox41x3x00lim121x01e11l
1xx
11x211x23x4ox411x21x222423limlim2x0cosxexsi
x2x0111x2x4ox41x21x4ox4x2242
3ox43x4ox4344x4limlim414x034116x0311ox312xxx2ox4x222242224x
习题341判定函数fxarcta
xx单调性解因为fx121120且仅当x0时等号成立所以fx在内1x1x单调减少2判定函数fxxcosx0x2的单调性解因为fx1si
x0所以fxxcr
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