What is taylor f(x)=ln(((1+x))/((1-x))) ?

The answer to taylor f(x)=ln(((1+x))/((1-x))) is 2x+2/3 x^3+2/5 x^5+\ldots

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

taylor

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

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

Solution steps

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

Refine

= 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_{- 2}^{2} x^{3} - 4 x d x

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

\int 4 r - r^{2} sin (θ) d r

What is taylor f(x)=ln(((1+x))/((1-x))) ?
The answer to taylor f(x)=ln(((1+x))/((1-x))) is 2x+2/3 x^3+2/5 x^5+\ldots