taylor 3/(x^2-x-2),1

Lösung

taylor

\frac{3}{x ^{2} - x - 2}, 1

- \frac{3}{2} - \frac{3}{4} (x - 1) - \frac{9}{8} (x - 1)^{2} - \frac{15}{16} (x - 1)^{3} - \frac{33}{32} (x - 1)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= - \frac{3}{2} + \frac{- \frac{3}{4}}{1 !} (x - 1) + \frac{- \frac{9}{4}}{2 !} (x - 1)^{2} + \frac{- \frac{45}{8}}{3 !} (x - 1)^{3} + \frac{- \frac{99}{4}}{4 !} (x - 1)^{4} + \dots

Fasse zusammen

= - \frac{3}{2} - \frac{3}{4} (x - 1) - \frac{9}{8} (x - 1)^{2} - \frac{15}{16} (x - 1)^{3} - \frac{33}{32} (x - 1)^{4} + \dots

\frac{\partial}{\partial y} (4 - x^{2} - 6 y^{2})

(8 t + y + 2) d t - d y = 0

\int x^{2} 1 + 2 x d x

\int x^{2} e^{- a x} d x

s l o p e in t erce pt (3, - 29), (- 1, 7)