b8-taylor and maclaurin series: proble\n(2 points)\nfind the maclaurin series for ( f(x)=cos (5 x) ).\n cos…

b8-taylor and maclaurin series: proble\n(2 points)\nfind the maclaurin series for ( f(x)=cos (5 x) ).\n cos (5 x)=sum_{n = 0}^{infty} \non what interval is the expansion valid? give your answer using interval notation. if you need\nis the only point in the interval of convergence, you would answer with 0.\nthe expansion is valid on\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 15 attempts remaining.

b8-taylor and maclaurin series: proble\n(2 points)\nfind the maclaurin series for ( f(x)=cos (5 x) ).\n cos (5 x)=sum_{n = 0}^{infty} \non what interval is the expansion valid? give your answer using interval notation. if you need\nis the only point in the interval of convergence, you would answer with 0.\nthe expansion is valid on\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 15 attempts remaining.

Answer

Answer:

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

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 $(-\infty,\infty)$