taylor (1+x)^{-1}

Lösung

taylor

(1 + x)^{- 1}

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

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

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

t an g e n t y = - 3 x^{3} - 2 x^{2} - x + 3, (- 1, 5)

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

\frac{\partial}{\partial x} (6 x^{4 y})

\int x^{3} 1 - x d x

\frac{d}{d x} (4 - \frac{25}{6 x ^{\frac{1}{6}}})