taylor 1/(1-x+\frac{x^2){2}},0

Lösung

taylor

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

1 + x + \frac{1}{2} x^{2} - \frac{1}{4} x^{4} - \frac{1}{4} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{1}{1 !} x + \frac{1}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- 6}{4 !} x^{4} + \frac{- 30}{5 !} x^{5} + \dots

Fasse zusammen

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

\frac{d}{d x} (x^{8} (x - 1)^{7})

\int (2 x - 4) d x

\int (- \frac{3}{7} x^{5} + \frac{1}{5} x^{3} - \frac{7}{4} x^{2}) d x

\frac{d}{d x} (6 x + 3)

x \to 3 lim (\frac{x ^{2} + 1}{x + 3})