Use a Maclaurin series in this table (attached) toobtain the Maclaurin series for the given function. f(x)=5cos(frac{pi x}{8})

Question
Series
asked 2021-02-04
Use a Maclaurin series in this table (attached) toobtain the Maclaurin series for the given function.
\(\displaystyle{f{{\left({x}\right)}}}={5}{\cos{{\left({\frac{{\pi{x}}}{{{8}}}}\right)}}}\)

Answers (1)

2021-02-05
According to the given information, it is required to calculate the Maclaurin series of the given function.
\(\displaystyle{f{{\left({x}\right)}}}={5}{\cos{{\left({\frac{{\pi{x}}}{{{8}}}}\right)}}}\)
By the given table maclaurin series of \(\displaystyle{\cos{{\left({x}\right)}}}\) is:
\(\displaystyle{\cos{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{\left(-{1}\right)}^{{n}}{\frac{{{x}^{{{2}{n}}}}}{{{\left({2}{n}\right)}!}}}={1}-{\frac{{{x}^{{2}}}}{{{2}!}}}+{\frac{{{x}^{{4}}}}{{{4}!}}}-{\frac{{{x}^{{6}}}}{{{6}!}}}+\ldots\)
In the given series replace x by \(\displaystyle{\left({\frac{{\pi{x}}}{{{8}}}}\right)}\):
\(\displaystyle{\cos{{\left({\frac{{\pi{x}}}{{{8}}}}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{\left(-{1}\right)}^{{n}}{\frac{{{\left({\frac{{\pi{x}}}{{{8}}}}\right)}^{{{2}{n}}}}}{{{\left({2}{n}\right)}!}}}={1}-{\frac{{{\left({\frac{{\pi{x}}}{{{8}}}}\right)}^{{2}}}}{{{2}!}}}+{\frac{{{\left({\frac{{\pi{x}}}{{{8}}}}\right)}^{{4}}}}{{{4}!}}}-{\frac{{{\left({\frac{{\pi{x}}}{{{8}}}}\right)}^{{6}}}}{{{6}!}}}+\ldots\)
\(\displaystyle={\sum_{{{n}={0}}}^{\infty}}{\left(-{1}\right)}^{{n}}{\frac{{\pi^{{{2}{n}}}{x}^{{{2}{n}}}}}{{{8}^{{{2}{n}}}{\left({2}{n}\right)}!}}}={1}-{\frac{{\pi^{{2}}{x}^{{2}}}}{{{128}}}}+{\frac{{\pi^{{4}}{x}^{{4}}}}{{{98304}}}}-{\frac{{\pi^{{6}}{x}^{{6}}}}{{{8}^{{6}}{\left({6}!\right)}}}}+\ldots\)
\(\displaystyle{5}{\cos{{\left({\frac{{\pi{x}}}{{{8}}}}\right)}}}={5}{\sum_{{{n}={0}}}^{\infty}}{\left(-{1}\right)}^{{n}}{\frac{{\pi^{{{2}{n}}}{x}^{{{2}{n}}}}}{{{8}^{{{2}{n}}}{\left({2}{n}\right)}!}}}={5}{\left({1}-{\frac{{\pi^{{2}}{x}^{{2}}}}{{{128}}}}+{\frac{{\pi^{{4}}{x}^{{4}}}}{{{98304}}}}-{\frac{{\pi^{{6}}{x}^{{6}}}}{{{8}^{{6}}{\left({6}!\right)}}}}+\ldots\right)}\)
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