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Answered question

2021-02-04

Use a Maclaurin series in this table (attached) toobtain the Maclaurin series for the given function.

f (x) = 5 \cos (\frac{π x}{8})

Faiza Fuller

Skilled2021-02-05Added 108 answers

According to the given information, it is required to calculate the Maclaurin series of the given function.

f (x) = 5 \cos (\frac{π x}{8})

By the given table maclaurin series of

\cos (x)

is:

\cos (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{2 n}}{(2 n)!} = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \dots

In the given series replace x by

(\frac{π x}{8})

\cos (\frac{π x}{8}) = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{{(\frac{π x}{8})}^{2 n}}{(2 n)!} = 1 - \frac{{(\frac{π x}{8})}^{2}}{2!} + \frac{{(\frac{π x}{8})}^{4}}{4!} - \frac{{(\frac{π x}{8})}^{6}}{6!} + \dots

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{π^{2 n} x^{2 n}}{8^{2 n} (2 n)!} = 1 - \frac{π^{2} x^{2}}{128} + \frac{π^{4} x^{4}}{98304} - \frac{π^{6} x^{6}}{8^{6} (6!)} + \dots

5 \cos (\frac{π x}{8}) = 5 \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{π^{2 n} x^{2 n}}{8^{2 n} (2 n)!} = 5 (1 - \frac{π^{2} x^{2}}{128} + \frac{π^{4} x^{4}}{98304} - \frac{π^{6} x^{6}}{8^{6} (6!)} + \dots)

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