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

The answer to maclaurin (1-x^2)^{1/2} is 1-1/2 x^2-1/8 x^4-1/16 x^6-5/128 x^8+\ldots

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

maclaurin

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

1 - \frac{1}{2} x^{2} - \frac{1}{8} x^{4} - \frac{1}{16} x^{6} - \frac{5}{128} x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 1}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- 3}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 45}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{- 1575}{8 !} x^{8} + \dots

Refine

= 1 - \frac{1}{2} x^{2} - \frac{1}{8} x^{4} - \frac{1}{16} x^{6} - \frac{5}{128} x^{8} + \dots

2 x y \frac{d y}{d x} = 3

\int \frac{ln ( x ) - 1}{x} d x

x \to 1 lim (3 x - 6)

\frac{2 s e ^{- 2 s}}{5 s ^{2} + 45}

What is maclaurin (1-x^2)^{1/2} ?
The answer to maclaurin (1-x^2)^{1/2} is 1-1/2 x^2-1/8 x^4-1/16 x^6-5/128 x^8+\ldots