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

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

Popular >

taylor 1/(x^2+1)

taylor

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

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

Solution steps

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( \frac{1}{x ^{2} + 1} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{x ^{2} + 1} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{x ^{2} + 1} ) ( 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 cos (x) (8 + 4 sin^{2} (x)) d x

y = cos (x^{3})

x \to 5 lim (\frac{x ^{2} - 3 x - 10}{x - 5})

\frac{d}{d x} ((7 x - 8)^{5})

\frac{\partial}{\partial x} (ln (2 y e^{x y}))

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