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

The answer to taylor 1/(sqrt(x)),1 is 1-1/2 (x-1)+3/8 (x-1)^2-5/16 (x-1)^3+35/128 (x-1)^4+\ldots

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taylor 1/(sqrt(x)),1

taylor

\frac{1}{x}, 1

1 - \frac{1}{2} (x - 1) + \frac{3}{8} (x - 1)^{2} - \frac{5}{16} (x - 1)^{3} + \frac{35}{128} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{- \frac{1}{2}}{1 !} (x - 1) + \frac{\frac{3}{4}}{2 !} (x - 1)^{2} + \frac{- \frac{15}{8}}{3 !} (x - 1)^{3} + \frac{\frac{105}{16}}{4 !} (x - 1)^{4} + \dots

Refine

= 1 - \frac{1}{2} (x - 1) + \frac{3}{8} (x - 1)^{2} - \frac{5}{16} (x - 1)^{3} + \frac{35}{128} (x - 1)^{4} + \dots

f (x) = x^{2} - x, at x = 2

n = 1 \sum \infty \frac{2}{n ( n + 2 )}

\int \frac{sin ( x )}{cos ^{7} ( x )} d x

y^{^{'}} + 8 y - 6 = 0

\int r 1 + r^{2} d r

What is taylor 1/(sqrt(x)),1 ?
The answer to taylor 1/(sqrt(x)),1 is 1-1/2 (x-1)+3/8 (x-1)^2-5/16 (x-1)^3+35/128 (x-1)^4+\ldots