taylor ln(1-e^{-x})

Lösung

taylor

ln (1 - e^{- x})

ln (1 - \frac{1}{e}) + \frac{1}{e - 1} (x - 1) - \frac{e}{2 ( e - 1 ) ^{2}} (x - 1)^{2} + \frac{e ^{2} + e}{6 ( - 1 + e ) ^{3}} (x - 1)^{3} - \frac{e ^{3} + 4 e ^{2} + e}{24 ( - 1 + e ) ^{4}} (x - 1)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

= ln (1 - \frac{1}{e}) + \frac{1}{e - 1} (x - 1) - \frac{e}{2 ( e - 1 ) ^{2}} (x - 1)^{2} + \frac{e ^{2} + e}{6 ( - 1 + e ) ^{3}} (x - 1)^{3} - \frac{e ^{3} + 4 e ^{2} + e}{24 ( - 1 + e ) ^{4}} (x - 1)^{4} + \dots

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