What is taylor 1/(1+x^2),0 ?

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

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taylor 1/(1+x^2),0

taylor

\frac{1}{1 + x ^{2}}, 0

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_{0}^{1} (1 - x^{7})^{2} d x

x^{3} - 2 x^{2} - 4 x + 6

y^{^{''}} - 2 y^{^{'}} = x + 2 e^{x}, y (0) = 18, y^{^{'}} (0) = \frac{31}{4}

\frac{d}{d x} (2 cos (x) + 3 sin (x))

\frac{s + 3}{s ^{2} + 2 s + 2}

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