What is taylor sqrt((1+x)/(1-x))-1 ?

The answer to taylor sqrt((1+x)/(1-x))-1 is x+1/2 x^2+1/2 x^3+3/8 x^4+3/8 x^5+\ldots

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taylor sqrt((1+x)/(1-x))-1

taylor

\frac{1 + x}{1 - x} - 1

x + \frac{1}{2} x^{2} + \frac{1}{2} x^{3} + \frac{3}{8} x^{4} + \frac{3}{8} x^{5} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{1}{2 !} x^{2} + \frac{3}{3 !} x^{3} + \frac{9}{4 !} x^{4} + \frac{45}{5 !} x^{5} + \dots

Refine

= x + \frac{1}{2} x^{2} + \frac{1}{2} x^{3} + \frac{3}{8} x^{4} + \frac{3}{8} x^{5} + \dots

f (x) = sec (x^{2})

\frac{d}{d x} (9 + 3 x + sin (3 x))

f (x) = 2 x^{- 1}, at x = 5

\int_{\frac{1}{2}}^{\frac{3}{2}} (- 2 x + 4) d x

\int (2 x - 5) e^{x^{2} - 5 x} d x

What is taylor sqrt((1+x)/(1-x))-1 ?
The answer to taylor sqrt((1+x)/(1-x))-1 is x+1/2 x^2+1/2 x^3+3/8 x^4+3/8 x^5+\ldots