Explained: "Find the Maclaurin series for this function: f(x)=x5cos⁡(πx)."

sf2or5eek8

Answered question

2023-01-07

Find the Maclaurin series for this function:

f (x) = x^{5} \cos (π x)

scernenefmet

Beginner2023-01-08Added 7 answers

We are looking for a Maclaurin series

f (x) = x^{5} \cos (π x)

We start by using the well studied Maclaurin series for

\cos x

\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!} + \frac{x^{8}}{8!} + \dots

So that:

\cos π x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{{(π x)}^{2 n}}{2 n!}

= 1 - \frac{{(π x)}^{2}}{2!} + \frac{{(π x)}^{4}}{4!} - \frac{{(π x)}^{6}}{6!} + \frac{{(π x)}^{8}}{8!} + \dots

And finally:

x^{5} \cos π x = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{5} \frac{{(π x)}^{2 n}}{2 n!}

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{5} π^{2 n} x^{2 n}}{2 n!}

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{π^{2 n} x^{2 n + 5}}{2 n!}

= 1 - \frac{π^{2} x^{7}}{2} + \frac{π^{4} x^{9}}{24} - \frac{π^{6} x^{11}}{720} + \dots

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