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

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

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

taylor

(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 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

\frac{d}{d t} (e^{- t^{2}})

\int x ln^{2} (5 x) d x

\int 2 x + 3 y d x

(2 y + 2 t - 3) d t + (8 y + 2 t - 1) d y = 0, y (- 1) = 2

\frac{d}{d x} (lo g_{8} (\frac{x ^{3} 8 ^{x}}{3 x + 1}))

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