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泰勒 |x|

解答

泰勒

∣ x ∣

解答

1 + (x - 1) + \dots

求解步骤

应用泰勒公式

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

求导数

= 1 + \frac{1}{1 !} (x - 1) + \frac{0}{2 !} (x - 1)^{2} + \frac{0}{3 !} (x - 1)^{3} + \frac{0}{4 !} (x - 1)^{4} + \frac{0}{5 !} (x - 1)^{5} + \frac{0}{6 !} (x - 1)^{6} + \frac{0}{7 !} (x - 1)^{7} + \frac{0}{8 !} (x - 1)^{8} + \frac{0}{9 !} (x - 1)^{9} + \frac{0}{10 !} (x - 1)^{10} + \dots

整理后得

= 1 + (x - 1) + \dots

流行的例子

derivative of 1/6 (e^{3x}+e^{-3x})

\frac{d}{d x} (\frac{1}{6} (e^{3 x} + e^{- 3 x}))

(dx)/(dt)=3x(5-x)

\frac{d x}{d t} = 3 x (5 - x)

normal f(x)= 1/(x^2),(-1,1)

n or ma l f (x) = \frac{1}{x ^{2}}, (- 1, 1)

y^'=0.3y(1-y/(2500))-120

y^{^{'}} = 0.3 y (1 - \frac{y}{2500}) - 120

inverselaplace (6s+3)/(s^4+5s^2+4)

in v erse l a pl a ce \frac{6 s + 3}{s ^{4} + 5 s ^{2} + 4}