taylor x*e^{-x+1},1

Lösung

taylor

x \cdot e^{- x + 1}, 1

1 - \frac{1}{2} (x - 1)^{2} + \frac{1}{3} (x - 1)^{3} - \frac{1}{8} (x - 1)^{4} + \frac{1}{30} (x - 1)^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

= 1 - \frac{1}{2} (x - 1)^{2} + \frac{1}{3} (x - 1)^{3} - \frac{1}{8} (x - 1)^{4} + \frac{1}{30} (x - 1)^{5} + \dots

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

\int (5 e^{x}) d x

t an g e n t f (x) = \frac{1}{x - 2}, at x = 7

\int \frac{x}{36 x ^{2} - 1} d x

d er i v a t i v e f (x) = 6 e^{x} + \frac{2}{3 x}