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

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

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taylor 1/(sqrt(1+x))

taylor

\frac{1}{1 + x}

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

x \to 3 + lim (x^{2})

\int \frac{1}{x ^{3} + 3 x ^{2}} d x

y^{^{''}} - y^{^{'}} - 2 y = - 6 t + 10 t^{2}

\frac{1 + e ^{- s π}}{s ^{2} + 1}

y = 3 e^{3 e^{x} + x}

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