What is maclaurin (1+x^2)^{-1} ?

The answer to maclaurin (1+x^2)^{-1} is 1-x^2+x^4-x^6+x^8+\ldots

Popular >

maclaurin (1+x^2)^{-1}

maclaurin

(1 + x^{2})^{- 1}

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

Solution steps

Apply the Maclaurin Formula

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

(- 1.1) (1.6)

f (x) = x^{6}

\int \frac{x + 1}{x ^{2} + 4 x + 7} d x

y = ln (\frac{e ^{x} + 1}{e ^{x} - 1})

h (v) = 4 v^{2} - 2

What is maclaurin (1+x^2)^{-1} ?
The answer to maclaurin (1+x^2)^{-1} is 1-x^2+x^4-x^6+x^8+\ldots