What is taylor x/(1+x),1 ?

The answer to taylor x/(1+x),1 is 1/2+1/4 (x-1)-1/8 (x-1)^2+1/16 (x-1)^3-1/32 (x-1)^4+\ldots

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taylor x/(1+x),1

taylor

\frac{x}{1 + x}, 1

\frac{1}{2} + \frac{1}{4} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{16} (x - 1)^{3} - \frac{1}{32} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{2} + \frac{\frac{1}{4}}{1 !} (x - 1) + \frac{- \frac{1}{4}}{2 !} (x - 1)^{2} + \frac{\frac{3}{8}}{3 !} (x - 1)^{3} + \frac{- \frac{3}{4}}{4 !} (x - 1)^{4} + \dots

Refine

= \frac{1}{2} + \frac{1}{4} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{16} (x - 1)^{3} - \frac{1}{32} (x - 1)^{4} + \dots

\int_{0}^{\frac{9}{2}} 81 - x^{2} d x

\frac{d x}{d t} = 0.6 - \frac{3 x}{500}

\frac{d y}{d x} = 8 x (y - 3)

\int \frac{3}{4} x^{2} - 5 x^{2} d x

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

What is taylor x/(1+x),1 ?
The answer to taylor x/(1+x),1 is 1/2+1/4 (x-1)-1/8 (x-1)^2+1/16 (x-1)^3-1/32 (x-1)^4+\ldots