What is taylor f(x)=sqrt(1+x),x=0 ?

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

taylor f(x)=sqrt(1+x),x=0

taylor

f (x) = 1 + x, x = 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 ) ( 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

\frac{d}{d x} (2 1 - x^{2})

x \to \infty lim (x^{2} + x)

f (x) = 3 x^{4}

x \to + (- 1) lim (- x^{2} + 4 x - 5)

\frac{d y}{d x} = 4 x 1 - y^{2}

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