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{d y}{d t} + y = \frac{1}{1 + e ^{2 t}}

\int \frac{4 x ^{2} + 26 x + 48}{( x + 5 ) ( x + 2 ) ^{2}} d x

\frac{d}{d x} (sin (\frac{x}{3}))

\frac{d ^{2}}{d x ^{2}} (x^{2} ln (x))

\frac{3 t}{t ^{2} + 4}

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