泰勒 ln(1-2y)

解答

泰勒

ln (1 - 2 y)

- 2 y - 2 y^{2} - \frac{8}{3} y^{3} - 4 y^{4} - \frac{32}{5} y^{5} + \dots

求解步骤

应用泰勒公式

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

求导数

= 0 + \frac{- 2}{1 !} y + \frac{- 4}{2 !} y^{2} + \frac{- 16}{3 !} y^{3} + \frac{- 96}{4 !} y^{4} + \frac{- 768}{5 !} y^{5} + \dots

整理后得

= - 2 y - 2 y^{2} - \frac{8}{3} y^{3} - 4 y^{4} - \frac{32}{5} y^{5} + \dots

\frac{\partial}{\partial x} (2 x + 2 x y)

(x + 2 y) d x + (2 x + y) d y = 0

\int 42 sec^{4} (x) d x

a re a x^{2}, 0, 2

\int x \frac{1}{1 - x ^{2}} d x