taylor 1/(1+2x),0

Lösung

taylor

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

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

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{- 2}{1 !} x + \frac{8}{2 !} x^{2} + \frac{- 48}{3 !} x^{3} + \frac{384}{4 !} x^{4} + \dots

Fasse zusammen

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

\frac{d}{d x} ((ln (x)) (\frac{x ^{7} + 3}{x ^{3} - 10}))

\int (3 x - 7)^{9} d x

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d er i v a t i v e f (x) = x^{- 1} - \frac{1}{x}