What is taylor (1+x)^{1/2},0 ?

The answer to taylor (1+x)^{1/2},0 is 1+1/2 x-1/8 x^2+1/16 x^3-5/128 x^4+\ldots

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taylor (1+x)^{1/2},0

taylor

(1 + x)^{\frac{1}{2}}, 0

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 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{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^{^{'}} = (y + 5) (x + 2), y (0) = - 1

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What is taylor (1+x)^{1/2},0 ?
The answer to taylor (1+x)^{1/2},0 is 1+1/2 x-1/8 x^2+1/16 x^3-5/128 x^4+\ldots