What is taylor (1+0.5x)/(1-0.5x) ?

The answer to taylor (1+0.5x)/(1-0.5x) is 1+x+1/2 x^2+0.25x^3+1/8 x^4+\ldots

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taylor (1+0.5x)/(1-0.5x)

taylor

\frac{1 + 0.5 x}{1 - 0.5 x}

1 + x + \frac{1}{2} x^{2} + 0.25 x^{3} + \frac{1}{8} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( \frac{1 + 0.5 x}{1 - 0.5 x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1 + 0.5 x}{1 - 0.5 x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1 + 0.5 x}{1 - 0.5 x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{1}{1 !} x + \frac{1}{2 !} x^{2} + \frac{1.5}{3 !} x^{3} + \frac{3}{4 !} x^{4} + \dots

Refine

= 1 + x + \frac{1}{2} x^{2} + 0.25 x^{3} + \frac{1}{8} x^{4} + \dots

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

\frac{\partial}{\partial x} (\frac{1}{2} (x^{2} + 6 y^{2}))

f (x) = x (x^{2} - 4 x + 5)^{4}, at x = 2

\frac{1}{s ^{2} + 2 s - 3}

\int \frac{e ^{- s t}}{s} d t

What is taylor (1+0.5x)/(1-0.5x) ?
The answer to taylor (1+0.5x)/(1-0.5x) is 1+x+1/2 x^2+0.25x^3+1/8 x^4+\ldots