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

Lösung

taylor

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

2 x + \frac{2}{3} x^{3} + \frac{2}{5} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{2}{1 !} x + \frac{0}{2 !} x^{2} + \frac{4}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{48}{5 !} x^{5} + \dots

Fasse zusammen

= 2 x + \frac{2}{3} x^{3} + \frac{2}{5} x^{5} + \dots

8 (ln (y)) y^{^{'}} - x^{7} y = 0

\frac{d}{d t} (cos (5 t))

\int \frac{( 1 - x ^{2} ) ^{\frac{1}{2}}}{x ^{4}} d x

- 6 x^{2} y + y^{^{'}} = - 10 x^{2}

in v erse l a pl a ce \frac{5}{s ^{2} - 12} - \frac{4 s}{s ^{2} + 8}