What is taylor f(x)=sqrt(x) ?

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

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taylor f(x)=sqrt(x)

taylor

f (x) = x

1 + \frac{1}{2} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{16} (x - 1)^{3} - \frac{5}{128} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{\frac{1}{2}}{1 !} (x - 1) + \frac{- \frac{1}{4}}{2 !} (x - 1)^{2} + \frac{\frac{3}{8}}{3 !} (x - 1)^{3} + \frac{- \frac{15}{16}}{4 !} (x - 1)^{4} + \dots

Refine

= 1 + \frac{1}{2} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{16} (x - 1)^{3} - \frac{5}{128} (x - 1)^{4} + \dots

\frac{\partial}{\partial y} (y^{2} + 1)

y^{^{''}} + 2 y^{^{'}} + y = 2 e^{- t}, y (0) = - 1, y^{^{'}} (0) = 3

g (x) = x^{\frac{5}{2}}

\int_{0}^{\infty} \frac{1}{4 + x ^{2}} d x

\frac{d}{d x} (e^{x} + x)

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