What is maclaurin f(x)=cos(2x) ?

The answer to maclaurin f(x)=cos(2x) is 1-2x^2+2/3 x^4-4/45 x^6+2/315 x^8+\ldots

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maclaurin f(x)=cos(2x)

maclaurin

f (x) = cos (2 x)

1 - 2 x^{2} + \frac{2}{3} x^{4} - \frac{4}{45} x^{6} + \frac{2}{315} x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 4}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{16}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 64}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{256}{8 !} x^{8} + \dots

Refine

= 1 - 2 x^{2} + \frac{2}{3} x^{4} - \frac{4}{45} x^{6} + \frac{2}{315} x^{8} + \dots

f (x) = (x^{3} - 4)^{2} (2 x + 5)^{3}

\frac{d}{d x} (ln (5 x^{4} + 4 x))

\int \frac{6 x ^{2} + 5 x + 6}{( x ^{2} + 1 ) ^{2}} d x

y^{^{'}} = \frac{3 ^{2} y ^{2} + 1}{3}, y (0) = - 0.5

\int_{0}^{5} π (25 - y^{2}) d y

What is maclaurin f(x)=cos(2x) ?
The answer to maclaurin f(x)=cos(2x) is 1-2x^2+2/3 x^4-4/45 x^6+2/315 x^8+\ldots