taylor e^{4x}

Lösung

taylor

e^{4 x}

1 + 4 x + 8 x^{2} + \frac{32}{3} x^{3} + \frac{32}{3} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{4}{1 !} x + \frac{16}{2 !} x^{2} + \frac{64}{3 !} x^{3} + \frac{256}{4 !} x^{4} + \dots

Fasse zusammen

= 1 + 4 x + 8 x^{2} + \frac{32}{3} x^{3} + \frac{32}{3} x^{4} + \dots

\int \frac{1}{x ( 1 + 4 ln ^{2} ( x ) )} d x

\frac{d y}{d x} = 2 x + 2 x y

x \to \infty lim (3 x^{5})

y^{2} \frac{d y}{d x} + y^{3} = 1

\frac{d}{d x} (\frac{x ^{2} + 3}{x ^{2} - 3})