What is taylor 1/((1+x^2)) ?

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

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taylor 1/((1+x^2))

taylor

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

1 - x^{2} + x^{4} - x^{6} + x^{8} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 2}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{24}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 720}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{40320}{8 !} x^{8} + \dots

Refine

= 1 - x^{2} + x^{4} - x^{6} + x^{8} + \dots

\int \frac{x ^{2}}{9 - x ^{2}} d x

\frac{d}{d x} (ln (sin (3^{x^{2}})))

(x^{2} + y^{2}) d x - 2 x y d y = 0

1 - ln (3 x)

f (x) = 3 x^{2}, at x = 1

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