泰勒 x^{1/3},8

解答

泰勒

x^{\frac{1}{3}}, 8

2 + \frac{1}{12} (x - 8) - \frac{1}{288} (x - 8)^{2} + \frac{5}{20736} (x - 8)^{3} - \frac{5}{248832} (x - 8)^{4} + \dots

求解步骤

应用泰勒公式

= 2 + \frac{\frac{d}{d x} ( x ^{\frac{1}{3}} ) ( 8 )}{1 !} (x - 8) + \frac{\frac{d ^{2}}{d x ^{2}} ( x ^{\frac{1}{3}} ) ( 8 )}{2 !} (x - 8)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ^{\frac{1}{3}} ) ( 8 )}{3 !} (x - 8)^{3} + \dots

求导数

= 2 + \frac{\frac{1}{12}}{1 !} (x - 8) + \frac{- \frac{1}{144}}{2 !} (x - 8)^{2} + \frac{\frac{5}{3456}}{3 !} (x - 8)^{3} + \frac{- \frac{5}{10368}}{4 !} (x - 8)^{4} + \dots

整理后得

= 2 + \frac{1}{12} (x - 8) - \frac{1}{288} (x - 8)^{2} + \frac{5}{20736} (x - 8)^{3} - \frac{5}{248832} (x - 8)^{4} + \dots

\frac{d}{d x} (\frac{1}{x ^{2} - 2})

\frac{d}{d x} (x^{2} + \frac{128000}{x})

a re a y = 4 - x^{2}, y = x^{2} - 4 x - 2

n = 2 \sum \infty \frac{3 ^{n}}{1 0 ^{n}}

t a y l or \frac{1}{1 + sin ( π \cdot x )}