What is maclaurin (1-x)^{-1/2} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

maclaurin (1-x)^{-1/2}

maclaurin

(1 - x)^{- \frac{1}{2}}

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

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

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

f (x) = x^{2} e^{2 x}

(cos (2 x))^{^{'}}

\frac{d}{d x} (2 \cdot x)

\frac{d}{d x} (x + 2 x)

\frac{d}{d x} (6 e^{3 x})

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