What is taylor (1-x^2)^{-1/2},0 ?

The answer to taylor (1-x^2)^{-1/2},0 is 1+1/2 x^2+3/8 x^4+5/16 x^6+\ldots

Popular >

taylor (1-x^2)^{-1/2},0

taylor

(1 - x^{2})^{- \frac{1}{2}}, 0

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

Solution steps

Apply the Taylor 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

x \to 3 lim (- 5)

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

\frac{( t + 2 t ^{2} )}{1 + t ^{2}}

5 + 6 x

n = 0 \sum \infty \frac{( x ^{n} )}{n !}

What is taylor (1-x^2)^{-1/2},0 ?
The answer to taylor (1-x^2)^{-1/2},0 is 1+1/2 x^2+3/8 x^4+5/16 x^6+\ldots