taylor 1/(1+x^{-1)}

Lösung

taylor

\frac{1}{1 + x ^{- 1}}

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

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

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

y^{^{''}} - 5 y^{^{'}} = 0, y (0) = 6, y^{^{'}} (0) = - 9

\frac{\partial}{\partial x} (ln (x^{4} - y^{4}))

\frac{d}{d x} (I n^{4} x)

d er i v a t i v e \frac{5 x + 1}{5 x - 1}

a re a y = x, y = x^{6}