taylor ln(1-x)

Lösung

taylor

ln (1 - x)

- x - \frac{1}{2} x^{2} - \frac{1}{3} x^{3} - \frac{1}{4} x^{4} - \frac{1}{5} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{- 1}{1 !} x + \frac{- 1}{2 !} x^{2} + \frac{- 2}{3 !} x^{3} + \frac{- 6}{4 !} x^{4} + \frac{- 24}{5 !} x^{5} + \dots

Fasse zusammen

= - x - \frac{1}{2} x^{2} - \frac{1}{3} x^{3} - \frac{1}{4} x^{4} - \frac{1}{5} x^{5} + \dots

\int \frac{y ^{5}}{1 - y ^{4}} d y

\frac{\partial}{\partial t} (\frac{s}{t})

x \to 4 lim (2 x^{4} - x^{2} - 8 x)

\int_{0}^{2} x 3 4 + x^{2} d x

\frac{\partial}{\partial x} (\frac{3}{3 x + 5 y})