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

The answer to maclaurin f(x)=sin(2x) is 2x-4/3 x^3+4/15 x^5-8/315 x^7+4/2835 x^9+\ldots

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maclaurin f(x)=sin(2x)

maclaurin

f (x) = sin (2 x)

2 x - \frac{4}{3} x^{3} + \frac{4}{15} x^{5} - \frac{8}{315} x^{7} + \frac{4}{2835} x^{9} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{2}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 8}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{32}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 128}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{512}{9 !} x^{9} + \dots

Refine

= 2 x - \frac{4}{3} x^{3} + \frac{4}{15} x^{5} - \frac{8}{315} x^{7} + \frac{4}{2835} x^{9} + \dots

\int t^{3} sin (n t) d t

\int - \frac{2}{x ^{2} - 1} d x

x \to 5 lim (\frac{( x - 5 )}{x - 5})

\int 18 x^{2} (2 x^{3} + 3)^{5} d x

x \to 7 lim (7 - 7)

What is maclaurin f(x)=sin(2x) ?
The answer to maclaurin f(x)=sin(2x) is 2x-4/3 x^3+4/15 x^5-8/315 x^7+4/2835 x^9+\ldots