taylor e^xln(1+x)

Lösung

taylor

e^{x} ln (1 + x)

x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \frac{3}{40} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{1}{2 !} x^{2} + \frac{2}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{9}{5 !} x^{5} + \dots

Fasse zusammen

= x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \frac{3}{40} x^{5} + \dots

x \to 4 lim (\frac{3 - x}{x - 4})

x \to 0 lim (2 x + c)

\int cos^{3} (π x - 1) d x

t an g e n t f (x) = \frac{1}{2 - 3 x}

\frac{d}{d x} ((\frac{x ^{2} - 9}{4 x + 5})^{4})