What is maclaurin (1-x)^{-1/2},4 ?

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

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maclaurin (1-x)^{-1/2},4

maclaurin

(1 - x)^{- \frac{1}{2}}, 4

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 Maclaurin 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

y = x^{2} - 3

f (x) = x sin (\frac{4}{x})

\frac{d}{d x} (3 x^{2} - e^{- x})

x = y^{2}, x = y^{2} + 2

arcsin (cos (x^{3}))

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