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

The answer to maclaurin (1-x^2)^{-1/2} is 1+1/2 x^2+3/8 x^4+5/16 x^6+\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{3}{8} x^{4} + \frac{5}{16} x^{6} + \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{9}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{225}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \dots

Refine

= 1 + \frac{1}{2} x^{2} + \frac{3}{8} x^{4} + \frac{5}{16} x^{6} + \dots

\int 2 x (x^{2} + 3)^{- 4} d x

\int \frac{x - 17}{x ^{2} + x - 12} d x

\frac{d}{d y} (\frac{1 + sin ( y )}{1 - sin ( y )})

\int 20 e^{- 0.6 t} d t

\int_{0}^{4} (2 x + 3) d x

What is maclaurin (1-x^2)^{-1/2} ?
The answer to maclaurin (1-x^2)^{-1/2} is 1+1/2 x^2+3/8 x^4+5/16 x^6+\ldots