taylor 1/(2-x),0

Lösung

taylor

\frac{1}{2 - x}, 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}{2 - x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{2 - x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{2 - x} ) ( 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

\frac{d}{d x} (\frac{4 x ^{2}}{x ^{2} - 1})

t an g e n t x^{3} + 7 x - 6

d er i v a t i v e 3 x

\int \frac{1}{z ^{2} - 1} d z

d er i v a t i v e f (x) = (6 x^{2} - 10 x) e^{x}