讲高数17堂专题课公式及其应用

(一)

专题 定理1（型余项f(x)f(x)f(x)(xx)1f(x)(xx)2L1f(n)(x)(xx)no[(xx)n 常称R(x)o(xx)n为型余项 若x00，则得林f(x)f(0)f(0)x

f(0)x2L

f(n)(0)xno(xnf(xx0的开区间(a,bn1x(a,bf(x)f(x)f(x)(xx)1f(x)(xx L1f(n)(x)(xx)nR

其中Rn(x)n1(xx0 ex1x

Lx

o(xn 3sinxx3

xL(1)

(2n

2n1xcosx1

L

o(x2n

ln(1x)xx2

L

n1n

o(xn(1x)1x(1)x2L(1)L(n1)xn(xn (二)本质及两个的异同 f(xx0的开区间(a,bn1型余项：R(x)o(xx)n；n(三)的应

n1(xx0

【例1】若f(x)f(x)Lf(n1)(x)0,f(n)(x) f(xxf(nx0f(nx0n f(xx0,3在[0,1]上的最大值不超过2

f(x1f1】计算e的近似值，使误差不超过【解】ex1x

Lx

(nRn((n

xe(nxe(n

e111L

R (n (nn10时，误差不超过106.得e

x2 (14 42】f(xx0的某邻域内二阶可导，且limsin3xf(x

x2（1）求f(0),f(0),f [f(0)3,f(0)0,f(0) ]（2）求lim3f(x) ]x0

x2 【例1】(2015年2)函数f(x)x22x在x0处的n阶导数f(n)(0) 1】f(4x)

f

1，试证：f(x) (x0) |f(x|a|f(x|babc是(0,1)写出f(x)在点c处带日型余项的一阶证明|f(c|2ab2【证】（1）f(x)f(cf(c)(xcf((x1(01(0f(0)f(c)f(c)(0c)f(1)f(c)f(c)(1c)f(2)(1

f(1)f(0)f(c)1[f

)(1c)2f(1)c2|f(c)|f(0)

f(1)1[|f

)|(1c)2|f()|c212ab[(1c)2c22又因为当c(0,1(1c)2c21,故|f(c|2ab2【证】由林f(x)f(0)f(0)x

f(0)x2

其中介于0与x之间，x[1,1].分别令x1和x1，并结合已知条件，得0f(1)f(0)1f(0)1f(), 10 1f(1)f(0)1f(0)1f( 0

m1[f()f()]M f()1[f()f()] 4（20012）f(x的区间[aa](a0)f(0)0写出f(x)的带日余项的一阶林证明在[aa]上至少存在一点a3f()3

f(x)d【解】（1）x[af(x)f(0)f(0)xf()x2

f(0)xf(x2，其中在0x之间2a

f(x)dxa

f(0)xdx

xf()dx1ax2f()d a 2f(x在[aa]x[aa]，有mf(xMMmf(x在[aamax2dxaf(x)dx1ax2f()dxMax2dx 2 2 ma3af(x)dxM

2f(x的连续性知，至少存在一点[aaf() af(x)da3 a3f()3

f(x)d【例5】设f(x)在[0,1]上二阶可导,f(0)0,1)f(16

f(1)0,maxf(x)2.试证存在点0试证11x1x2 (x 设f(x)[a,b]上连续，在(a,b)内二阶可导，试证存在a,b)，使f(b)2f(ab)f(a)(b

f设f(x)三阶可