taylor xe^{-3x}

Lösung

taylor

x e^{- 3 x}

x - 3 x^{2} + \frac{9}{2} x^{3} - \frac{9}{2} x^{4} + \frac{27}{8} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{- 6}{2 !} x^{2} + \frac{27}{3 !} x^{3} + \frac{- 108}{4 !} x^{4} + \frac{405}{5 !} x^{5} + \dots

Fasse zusammen

= x - 3 x^{2} + \frac{9}{2} x^{3} - \frac{9}{2} x^{4} + \frac{27}{8} x^{5} + \dots

y^{^{'}} - y = 3 t e^{2 t}

\frac{\partial}{\partial x} (3 x + 2 y)

\int \frac{1}{9 x ^{2} + 6 x - 8} d x

n = 0 \sum \infty n \frac{1}{5 ^{n}}

y ln (x) \frac{d y}{d x} = (\frac{( y + 1 )}{x})^{2}