What is taylor sqrt(x-1),0 ?

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

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taylor sqrt(x-1),0

taylor

x - 1, 0

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

Solution steps

Apply the Taylor Formula

= i + \frac{\frac{d}{d x} ( x - 1 ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( x - 1 ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x - 1 ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

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

Refine

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

x \to 0 lim (\frac{2}{3 + 4 ^{\frac{1}{0}}})

y = \frac{16 x}{x ^{2} + 16}, (- 2, - \frac{8}{5})

y^{^{''}} - 9 y = 0, y (0) = 7

\int \frac{3}{( x + 3 ) ^{2}} d x

f (x) = 3^{x^{2} - x + 5}

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