taylor 1+e^x

Lösung

taylor

1 + e^{x}

2 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \frac{1}{24} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

= 2 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \frac{1}{24} x^{4} + \dots

t an g e n t f (x) = 5 x^{2}, at x = 2

\frac{d}{d x} (\frac{1 + sin ( x )}{x cos ( x )})

x \to 1 lim (3^{(\frac{1}{x})})

l a pl a ce t r an s f or m 3 sin (3 t + 2)

\int 4 x^{3} e^{2 x^{4}} d x