Find the Maclaurin series for the function f(x)=2sin x^3 Use the table of power series for elementary functions

Question
Series
asked 2021-02-25
Find the Maclaurin series for the function \(\displaystyle{f{{\left({x}\right)}}}={2}{{\sin{{x}}}^{{3}}}\)
Use the table of power series for elementary functions

Answers (1)

2021-02-26
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
\(\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{3}}-{\frac{{{x}^{{9}}}}{{{3}}}}+{\frac{{{x}^{{{15}}}}}{{{60}}}}-{\frac{{{x}^{{{21}}}}}{{{2520}}}}+\ldots\)
0

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