taylor 1/(1-x),4

Lösung

taylor

\frac{1}{1 - x}, 4

- \frac{1}{3} + \frac{1}{9} (x - 4) - \frac{1}{27} (x - 4)^{2} + \frac{1}{81} (x - 4)^{3} - \frac{1}{243} (x - 4)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= - \frac{1}{3} + \frac{\frac{d}{d x} ( \frac{1}{1 - x} ) ( 4 )}{1 !} (x - 4) + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{1 - x} ) ( 4 )}{2 !} (x - 4)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{1 - x} ) ( 4 )}{3 !} (x - 4)^{3} + \dots

Evaluate Derivatives

= - \frac{1}{3} + \frac{\frac{1}{9}}{1 !} (x - 4) + \frac{- \frac{2}{27}}{2 !} (x - 4)^{2} + \frac{\frac{2}{27}}{3 !} (x - 4)^{3} + \frac{- \frac{8}{81}}{4 !} (x - 4)^{4} + \dots

Fasse zusammen

= - \frac{1}{3} + \frac{1}{9} (x - 4) - \frac{1}{27} (x - 4)^{2} + \frac{1}{81} (x - 4)^{3} - \frac{1}{243} (x - 4)^{4} + \dots

x \to 0 lim (6 x^{2})

(x^{2} + 1) y^{^{'}} + 5 x y = 5 x, y (0) = 6

\frac{d y}{d x} = 4 + y - 4 x + 6

\frac{d}{d x} (2 x^{2})

\int sec (x) \cdot tan^{2} (x) d x