What is maclaurin 1/(1+4x^2) ?

The answer to maclaurin 1/(1+4x^2) is 1-4x^2+16x^4-64x^6+256x^8+\ldots

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maclaurin 1/(1+4x^2)

maclaurin

\frac{1}{1 + 4 x ^{2}}

1 - 4 x^{2} + 16 x^{4} - 64 x^{6} + 256 x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 8}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{384}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 46080}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{10321920}{8 !} x^{8} + \dots

Refine

= 1 - 4 x^{2} + 16 x^{4} - 64 x^{6} + 256 x^{8} + \dots

f (x) = e^{x} - cot (x)

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

\int_{- 1}^{5} (y^{2} - 5) - 4 y d y

\int_{0}^{2} 20000 e^{0.03 x} d x

k = 1 \sum \infty k \cdot 2^{- k}

What is maclaurin 1/(1+4x^2) ?
The answer to maclaurin 1/(1+4x^2) is 1-4x^2+16x^4-64x^6+256x^8+\ldots