What is taylor sqrt(x),x=1 ?

The answer to taylor sqrt(x),x=1 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 sqrt(x),x=1

taylor

x, x = 1

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

y = 3 x^{3} + x^{2} - 8 x + 4, (- 2, 0)

\int \frac{3 x}{7 - x ^{2}} d x

\frac{d}{d x} (- \frac{4 x}{( x ^{2} + 4 ) ^{2}})

\int sin (θ) + cos (θ) d θ

x \to 5 lim (4)

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