What is taylor sqrt((1-x^2)) ?

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

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taylor sqrt((1-x^2))

taylor

(1 - x^{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 Taylor Formula

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

\int 8 \frac{tan ^{3} ( 2 x )}{x} d x

n = 2 \sum \infty n (n - 1) x^{n}

- x + 5, 0, 2

\int - x e^{x} sin (x) d x

x \to - 2 lim (\frac{x ^{2} - 3}{x + 1})

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