What is taylor sqrt(x+1) ?

The answer to taylor sqrt(x+1) is 1+1/2 x-1/8 x^2+1/16 x^3-5/128 x^4+\ldots

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taylor sqrt(x+1)

taylor

x + 1

1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} - \frac{5}{128} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{\frac{1}{2}}{1 !} x + \frac{- \frac{1}{4}}{2 !} x^{2} + \frac{\frac{3}{8}}{3 !} x^{3} + \frac{- \frac{15}{16}}{4 !} x^{4} + \dots

Refine

= 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} - \frac{5}{128} x^{4} + \dots

\int_{1}^{2} \frac{x ^{2} + 8}{3 x - x ^{2}} d x

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

\frac{7 + 2 x}{5 - 8 x}

\frac{\partial}{\partial x} (\frac{1}{tan ( x y )})

\frac{d y}{d x} = (x + y + 2)^{2}

What is taylor sqrt(x+1) ?
The answer to taylor sqrt(x+1) is 1+1/2 x-1/8 x^2+1/16 x^3-5/128 x^4+\ldots