taylor e^x5

Lösung

taylor

e^{x} 5

5 + 5 x + \frac{5}{2} x^{2} + \frac{5}{6} x^{3} + \frac{5}{24} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 5 + \frac{\frac{d}{d x} ( e ^{x} \cdot 5 ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{x} \cdot 5 ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{x} \cdot 5 ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 5 + \frac{5}{1 !} x + \frac{5}{2 !} x^{2} + \frac{5}{3 !} x^{3} + \frac{5}{4 !} x^{4} + \dots

Fasse zusammen

= 5 + 5 x + \frac{5}{2} x^{2} + \frac{5}{6} x^{3} + \frac{5}{24} x^{4} + \dots

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

\int_{- 1}^{1} \frac{1}{x ^{2} - x - 2} d x

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

n \to \infty lim (- 1)

\int_{1}^{2} - \frac{1}{16 z} + \frac{1}{4 z} d z