taylor ln(1+2x),0

Lösung

taylor

ln (1 + 2 x), 0

2 x - 2 x^{2} + \frac{8}{3} x^{3} - 4 x^{4} + \frac{32}{5} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{2}{1 !} x + \frac{- 4}{2 !} x^{2} + \frac{16}{3 !} x^{3} + \frac{- 96}{4 !} x^{4} + \frac{768}{5 !} x^{5} + \dots

Fasse zusammen

= 2 x - 2 x^{2} + \frac{8}{3} x^{3} - 4 x^{4} + \frac{32}{5} x^{5} + \dots

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