Step by step solution to "Find the Maclaurin series for the function f(x)=2sin⁡x3..."

foass77W

Answered question

2021-02-25

Find the Maclaurin series for the function

f (x) = 2 {\sin x}^{3}

Use the table of power series for elementary functions

Aamina Herring

Skilled2021-02-26Added 85 answers

To find the Maclaurin series for the function

f (x) = 2 {\sin x}^{3}

Solution:
From the given table, we can find that Maclaurin series of the function

\sin x

is:

\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots

Now, if we replace x by

x^{3}

then:

{\sin x}^{3} = x^{3} - \frac{{(x^{3})}^{3}}{3!} + \frac{{(x^{3})}^{5}}{5!} - \frac{{(x^{3})}^{7}}{7!} + \dots

= x^{3} - \frac{x^{9}}{3!} + \frac{x^{15}}{5!} - \frac{x^{21}}{7!} + \dots

Therefore, Maclaurin series of the function

f (x) = 2 {\sin x}^{3}

will be:

2 {\sin x}^{3} = 2 [x^{3} - \frac{x^{9}}{3!} + \frac{x^{15}}{5!} - \frac{x^{21}}{7!} + \dots]

= 2 x^{3} - \frac{2 x^{9}}{6} + \frac{2 x^{15}}{120} - \frac{2 x^{21}}{5040} + \dots

= 2 x^{3} - \frac{x^{9}}{3} + \frac{x^{15}}{60} - \frac{x^{21}}{2520} + \dots

Hence, Maclaurin series of the given function is

f (x) = 2 x^{3} - \frac{x^{9}}{3} + \frac{x^{15}}{60} - \frac{x^{21}}{2520} + \dots

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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