设 a = 2 ln ( 1.01 ) , b = ln ( 1.02 ) , c = 1.04 − 1 a=2\ln(1.01),b=\ln(1.02),c=\sqrt{1.04}-1 a=2ln(1.01),b=ln(1.02),c=1.04 −1,则( \quad )
A . a < b < c B . b < c < a C . b < a < c D . c < a < b A. aA.a<b<cB.b<c<aC.b<a<cD.c<a<b
解:
ln ( 1 + x ) \qquad \ln(1+x) ln(1+x)带皮亚诺余项的 3 3 3阶麦克劳林展开为
ln ( 1 + x ) = x − x 2 2 + x 3 3 + o ( x 3 ) \ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+o(x^3) ln(1+x)=x−2x2+3x3+o(x3)
则
ln ( 1.01 ) ≈ 0.01 − 0.0 1 2 2 + 0.0 1 3 3 \ln(1.01)\approx 0.01-\dfrac{0.01^2}{2}+\dfrac{0.01^3}{3} ln(1.01)≈0.01−20.012+30.013
ln ( 1.02 ) ≈ 0.02 − 0.0 2 2 2 + 0.0 2 3 3 \ln(1.02)\approx0.02-\dfrac{0.02^2}{2}+\dfrac{0.02^3}{3} ln(1.02)≈0.02−20.022+30.023
a = 2 ln ( 1.01 ) ≈ 0.02 − 0.0 1 2 + 2 3 × 0.0 1 3 \qquad a=2\ln(1.01)\approx 0.02-0.01^2+\dfrac 23\times 0.01^3 a=2ln(1.01)≈0.02−0.012+32×0.013
b = ln ( 1.02 ) ≈ 0.02 − 2 × 0.0 1 2 + 3 4 × 0.0 1 3 \qquad b=\ln(1.02)\approx 0.02-2\times 0.01^2+\dfrac 34\times 0.01^3 b=ln(1.02)≈0.02−2×0.012+43×0.013
1 + x \qquad \sqrt{1+x} 1+x 带皮亚诺余项的 3 3 3阶麦克劳林展开为
1 + x = 1 + 1 2 x − 1 8 x 2 + 1 16 x 3 + o ( x 3 ) \sqrt{1+x}=1+\dfrac 12x-\dfrac 18x^2+\dfrac{1}{16}x^3+o(x^3) 1+x =1+21x−81x2+161x3+o(x3)
c = 1.04 − 1 = 1 + 1 2 × 0.4 − 1 8 × 0. 4 2 + 1 16 × 0.0 4 3 − 1 = 0.2 − 2 × 0.0 1 2 + 4 × 0.0 1 3 \qquad c=\sqrt{1.04}-1=1+\dfrac 12\times 0.4-\dfrac 18\times 0.4^2+\dfrac{1}{16}\times 0.04^3-1=0.2-2\times 0.01^2+4\times 0.01^3 c=1.04 −1=1+21×0.4−81×0.42+161×0.043−1=0.2−2×0.012+4×0.013
\qquad 所以 a > c > b a>c>b a>c>b,选 B B B。