To find the Maclaurin series for the function \(\displaystyle{f{{\left({x}\right)}}}={2}{{\sin{{x}}}^{{3}}}\)

Solution:

From the given table, we can find that Maclaurin series of the function \(\displaystyle{\sin{{x}}}\) is:

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

Now, if we replace x by \(\displaystyle{x}^{{3}}\) then:

\(\displaystyle{{\sin{{x}}}^{{3}}=}{x}^{{3}}-{\frac{{{\left({x}^{{3}}\right)}^{{3}}}}{{{3}!}}}+{\frac{{{\left({x}^{{3}}\right)}^{{5}}}}{{{5}!}}}-{\frac{{{\left({x}^{{3}}\right)}^{{7}}}}{{{7}!}}}+\ldots\)

\(\displaystyle={x}^{{3}}-{\frac{{{x}^{{9}}}}{{{3}!}}}+{\frac{{{x}^{{{15}}}}}{{{5}!}}}-{\frac{{{x}^{{{21}}}}}{{{7}!}}}+\ldots\)

Therefore, Maclaurin series of the function \(\displaystyle{f{{\left({x}\right)}}}={2}{{\sin{{x}}}^{{3}}}\) will be:

\(\displaystyle{2}{{\sin{{x}}}^{{3}}=}{2}{\left[{x}^{{3}}-{\frac{{{x}^{{9}}}}{{{3}!}}}+{\frac{{{x}^{{{15}}}}}{{{5}!}}}-{\frac{{{x}^{{{21}}}}}{{{7}!}}}+\ldots\right]}\)

\(\displaystyle={2}{x}^{{3}}-{\frac{{{2}{x}^{{9}}}}{{{6}}}}+{\frac{{{2}{x}^{{{15}}}}}{{{120}}}}-{\frac{{{2}{x}^{{{21}}}}}{{{5040}}}}+\ldots\)

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

Hence, Maclaurin series of the given function is

\(\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{3}}-{\frac{{{x}^{{9}}}}{{{3}}}}+{\frac{{{x}^{{{15}}}}}{{{60}}}}-{\frac{{{x}^{{{21}}}}}{{{2520}}}}+\ldots\)

Solution:

From the given table, we can find that Maclaurin series of the function \(\displaystyle{\sin{{x}}}\) is:

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

Now, if we replace x by \(\displaystyle{x}^{{3}}\) then:

\(\displaystyle{{\sin{{x}}}^{{3}}=}{x}^{{3}}-{\frac{{{\left({x}^{{3}}\right)}^{{3}}}}{{{3}!}}}+{\frac{{{\left({x}^{{3}}\right)}^{{5}}}}{{{5}!}}}-{\frac{{{\left({x}^{{3}}\right)}^{{7}}}}{{{7}!}}}+\ldots\)

\(\displaystyle={x}^{{3}}-{\frac{{{x}^{{9}}}}{{{3}!}}}+{\frac{{{x}^{{{15}}}}}{{{5}!}}}-{\frac{{{x}^{{{21}}}}}{{{7}!}}}+\ldots\)

Therefore, Maclaurin series of the function \(\displaystyle{f{{\left({x}\right)}}}={2}{{\sin{{x}}}^{{3}}}\) will be:

\(\displaystyle{2}{{\sin{{x}}}^{{3}}=}{2}{\left[{x}^{{3}}-{\frac{{{x}^{{9}}}}{{{3}!}}}+{\frac{{{x}^{{{15}}}}}{{{5}!}}}-{\frac{{{x}^{{{21}}}}}{{{7}!}}}+\ldots\right]}\)

\(\displaystyle={2}{x}^{{3}}-{\frac{{{2}{x}^{{9}}}}{{{6}}}}+{\frac{{{2}{x}^{{{15}}}}}{{{120}}}}-{\frac{{{2}{x}^{{{21}}}}}{{{5040}}}}+\ldots\)

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

Hence, Maclaurin series of the given function is

\(\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{3}}-{\frac{{{x}^{{9}}}}{{{3}}}}+{\frac{{{x}^{{{15}}}}}{{{60}}}}-{\frac{{{x}^{{{21}}}}}{{{2520}}}}+\ldots\)