What is taylor sqrt(x) ?

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

taylor sqrt(x)

taylor

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{d}{d x} (\frac{1}{x ^{7}} + \frac{1}{x ^{8}})

n = 1 \sum \infty n e^{7 n}

d er i v a t i v e \frac{x}{( 1 + x ^{2} )}

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

\frac{d y}{d x} = x y^{2}

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