taylor f(x)=e^{-2x}

Lösung

taylor

f (x) = e^{- 2 x}

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

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{- 2}{1 !} x + \frac{4}{2 !} x^{2} + \frac{- 8}{3 !} x^{3} + \frac{16}{4 !} x^{4} + \dots

Fasse zusammen

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

\frac{1}{x ^{2}} \frac{d y}{d x} = 25 + 14 y

t an g e n t f (x) = x x, at x = 16

a re a sec^{2} (x), 8 cos (x)

x \to 3 lim (c x^{2} - 4 x)

x \to - 8 lim (10 x + 1)