taylor 1/(x+2),0

Lösung

taylor

\frac{1}{x + 2}, 0

\frac{1}{2} - \frac{1}{4} x + \frac{1}{8} x^{2} - \frac{1}{16} x^{3} + \frac{1}{32} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

= \frac{1}{2} - \frac{1}{4} x + \frac{1}{8} x^{2} - \frac{1}{16} x^{3} + \frac{1}{32} x^{4} + \dots

k = 2 \sum \infty \frac{1}{k} - \frac{1}{4 ^{k}}

(1 + ln (x) + \frac{y}{x}) d x - (1 - ln (x)) d y = 0

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

y^{^{'}} + 2 x y = 0, y (0) = 3

\int sec^{6} (x) tan^{6} (x) d x