What is taylor 1/(x^{0.5)} ?

The answer to taylor 1/(x^{0.5)} is 1-1/2 (x-1)+3/8 (x-1)^2-5/16 (x-1)^3+35/128 (x-1)^4+\ldots

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

taylor 1/(x^{0.5)}

taylor

\frac{1}{x ^{0.5}}

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

\frac{\partial}{\partial y} ((5 x^{3} y + 4 y^{5})^{3})

x^{2} + 3 x - 2

\int_{- π}^{π} cos (3 x) sin (3 x) d x

x \to 0 lim (\frac{k}{x ( x + 1 )})

x \to 3 + lim (\frac{x + 1}{3 - x})

What is taylor 1/(x^{0.5)} ?
The answer to taylor 1/(x^{0.5)} is 1-1/2 (x-1)+3/8 (x-1)^2-5/16 (x-1)^3+35/128 (x-1)^4+\ldots