What is taylor (1+x)^{-1/2} ?

The answer to taylor (1+x)^{-1/2} is 1-1/2 x+3/8 x^2-5/16 x^3+35/128 x^4+\ldots

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taylor (1+x)^{-1/2}

taylor

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

1 - \frac{1}{2} x + \frac{3}{8} x^{2} - \frac{5}{16} x^{3} + \frac{35}{128} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{- \frac{1}{2}}{1 !} x + \frac{\frac{3}{4}}{2 !} x^{2} + \frac{- \frac{15}{8}}{3 !} x^{3} + \frac{\frac{105}{16}}{4 !} x^{4} + \dots

Refine

= 1 - \frac{1}{2} x + \frac{3}{8} x^{2} - \frac{5}{16} x^{3} + \frac{35}{128} x^{4} + \dots

\frac{\partial}{\partial x} (x^{6} y^{7} - x^{5} y^{6})

\frac{d}{d x} (\frac{7 + 2 ln ( x )}{x})

\int (csc (x) cot (x)) d x

\frac{1}{( s ^{2} + s + 1 ) s}

y = e^{x}

What is taylor (1+x)^{-1/2} ?
The answer to taylor (1+x)^{-1/2} is 1-1/2 x+3/8 x^2-5/16 x^3+35/128 x^4+\ldots