What is maclaurin (3+x^2)^{-1} ?

The answer to maclaurin (3+x^2)^{-1} is 1/3-1/9 x^2+1/27 x^4-1/81 x^6+1/243 x^8+\ldots

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maclaurin (3+x^2)^{-1}

maclaurin

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

\frac{1}{3} - \frac{1}{9} x^{2} + \frac{1}{27} x^{4} - \frac{1}{81} x^{6} + \frac{1}{243} x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= \frac{1}{3} + \frac{0}{1 !} x + \frac{- \frac{2}{9}}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{\frac{8}{9}}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- \frac{80}{9}}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{\frac{4480}{27}}{8 !} x^{8} + \dots

Refine

= \frac{1}{3} - \frac{1}{9} x^{2} + \frac{1}{27} x^{4} - \frac{1}{81} x^{6} + \frac{1}{243} x^{8} + \dots

n = 0 \sum \infty e^{(- 2 n)}

x \to 0 lim (2^{- \frac{1}{x ^{2}}})

\frac{\partial}{\partial x} (x y + yz + z x)

\frac{\partial}{\partial y} (x^{2} y - x y^{2})

4 y^{^{''}} - 12 y^{^{'}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = 5

What is maclaurin (3+x^2)^{-1} ?
The answer to maclaurin (3+x^2)^{-1} is 1/3-1/9 x^2+1/27 x^4-1/81 x^6+1/243 x^8+\ldots