Question

In Exercises $49-54$in Section $8.2$you were asked to find the Taylor series for the specified function at the given value of $x_0$. In Exercises $45-50$find the Lagrange's form for the remainder $R_n\left(x\right)$, and show that $\underset{n\to \infty }{\lim }{R}_n\left(x\right)=0$on the specified interval.

$\cos x,\frac{\pi }{2},\mathrm{ℝ}$

Short answer

That, which is$\left|R_n\left(x\right)\right|=\frac{{\left|x-\frac{\pi }{2}\right|}^{n+1}}{(n+1)!}$proved.

Step-by-step solution

Given Information

Let us consider the function$f\left(x\right)=\cos x$

Lagrange's Form

If $f\left(x\right)=\cos x$, we know that for every $n\ge 0,f^{(n+1)}\left(x\right)$is one of the four function $\pm \sin x$and $\pm \cos x$. So, for any of the four functions, $\left|f^{(n+1)}\left(c\right)\right|\le 1$, for every value of $x$, so using the Lagrange's form for the remainder, we have

$\left|R_n\left(x\right)\right|=\left|\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{x}^{n+1}\right|$

Proof

Since the Taylor series for the function $f\left(x\right)=\cos x$at $x=\frac{\pi }{2}$is

$P_n\left(x\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^{n+1}}{(2n+1)!}{\left(x-\frac{\pi }{2}\right)}^{2n+1}$

Therefore,

$\left|R_n\left(x\right)\right|=\frac{{\left|x-\frac{\pi }{2}\right|}^{n+1}}{(n+1)!}$