What is maclaurin sqrt(1-x) ?

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

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maclaurin sqrt(1-x)

maclaurin

1 - x

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 Maclaurin Formula

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

y^{^{''}} + 8 y^{^{'}} + 16 y = 1521 e^{2 t} cos (9 t)

cos^{2} (1 t)

f (x) = \frac{1}{x - 1}, at x = 5

6 e^{- 5 x}

(8 x)^{l n (8 x)}

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