taylor f(x)=(x-ln(1+x))/(x^2)

Lösung

taylor

f (x) = \frac{x - ln ( 1 + x )}{x ^{2}}

1 - ln (2) + \frac{4 ln ( 2 ) - 3}{2} (x - 1) + \frac{- 24 ln ( 2 ) + 17}{8} (x - 1)^{2} + \frac{192 ln ( 2 ) - 134}{48} (x - 1)^{3} + \frac{- 1920 ln ( 2 ) + 1334}{384} (x - 1)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 - ln (2) + \frac{\frac{4 l n ( 2 ) - 3}{2}}{1 !} (x - 1) + \frac{\frac{- 24 l n ( 2 ) + 17}{4}}{2 !} (x - 1)^{2} + \frac{\frac{192 l n ( 2 ) - 134}{8}}{3 !} (x - 1)^{3} + \frac{\frac{- 1920 l n ( 2 ) + 1334}{16}}{4 !} (x - 1)^{4} + \dots

Fasse zusammen

= 1 - ln (2) + \frac{4 ln ( 2 ) - 3}{2} (x - 1) + \frac{- 24 ln ( 2 ) + 17}{8} (x - 1)^{2} + \frac{192 ln ( 2 ) - 134}{48} (x - 1)^{3} + \frac{- 1920 ln ( 2 ) + 1334}{384} (x - 1)^{4} + \dots

ma c l a u r in 2 e^{- \frac{x}{2}} + e^{- x}

x \to + \frac{π}{2} + lim (e^{s e c (x)})

\frac{d}{d x} (\frac{4 x ^{2} + 2 x + 6}{x})

\int \frac{( ln ( x ) ) ^{40}}{x} d x

\frac{d}{d x} ((\frac{x}{1 + 1 - x ^{2}})^{n})