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

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

Popular >

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

taylor

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

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

\frac{d}{d x} ((x^{\frac{1}{2}} + y^{\frac{1}{2}})^{2})

(2 x^{3} - y) d x - (x + 2 y) d y = 0

y^{^{'}} = \frac{y}{t} + 2 t + 1

\int \frac{6 x}{x ^{2} + 4 x + 4} d x

\frac{e ^{- 3 s}}{s ^{3} + 4 s}

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