Question

b8-taylor and maclaurin series: proble
(2 points)
find the maclaurin series for ( f(x)=cos (5 x) ).
cos (5 x)=sum_{n = 0}^{infty}
on what interval is the expansion valid? give your answer using interval notation. if you need
is the only point in the interval of convergence, you would answer with 0.
the expansion is valid on
note: you can earn partial credit on this problem.
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Explanation:

Step1: Recall the Maclaurin series of $\cos t$

The Maclaurin series of $\cos t=\sum_{n = 0}^{\infty}\frac{(-1)^{n}t^{2n}}{(2n)!}$, where $t\in(-\infty,\infty)$

Step2: Substitute $t = 5x$

Let $t = 5x$. Then $\cos(5x)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}(5x)^{2n}}{(2n)!}$.
Since the Maclaurin series of $\cos t$ converges for all $t\in(-\infty,\infty)$, when $t = 5x$, for any real - valued $x$, the series $\sum_{n = 0}^{\infty}\frac{(-1)^{n}(5x)^{2n}}{(2n)!}$ converges. So the interval of convergence is $(-\infty,\infty)$

Answer:

$\cos(5x)=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}(5x)^{2n}}{(2n)!}$; The expansion is valid on $(-\infty,\infty)$