泰勒 ln(1-3y)

解答

泰勒

ln (1 - 3 y)

- 3 y - \frac{9}{2} y^{2} - 9 y^{3} - \frac{81}{4} y^{4} - \frac{243}{5} y^{5} + \dots

求解步骤

应用泰勒公式

= 0 + \frac{\frac{d}{d y} ( ln ( 1 - 3 y ) ) ( 0 )}{1 !} y + \frac{\frac{d ^{2}}{d y ^{2}} ( ln ( 1 - 3 y ) ) ( 0 )}{2 !} y^{2} + \frac{\frac{d ^{3}}{d y ^{3}} ( ln ( 1 - 3 y ) ) ( 0 )}{3 !} y^{3} + \dots

求导数

= 0 + \frac{- 3}{1 !} y + \frac{- 9}{2 !} y^{2} + \frac{- 54}{3 !} y^{3} + \frac{- 486}{4 !} y^{4} + \frac{- 5832}{5 !} y^{5} + \dots

整理后得

= - 3 y - \frac{9}{2} y^{2} - 9 y^{3} - \frac{81}{4} y^{4} - \frac{243}{5} y^{5} + \dots

\int \frac{( x ^{2} )}{x ^{2} + 9} d x

\int \frac{tan ^{2} ( θ )}{sec ( θ )} d θ

- y + \frac{d y}{d x} ln (x) = 0

\frac{d}{d x} (arcsin (x^{2} + 1))

t an g e n t y = \frac{x}{36 + x ^{2}}, (0, 0)