Question

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.

$\sin x,\pi ,\mathrm{ℝ}$

Short answer

The required Lagrange form of the remainder is$\left|R_n\left(x\right)\right|=\frac{|x-\pi {|}^{n+1}}{(n+1)!}$

Step-by-step solution

Given Information

Consider the function $f\left(x\right)=\sin x$

Lagrange's form

If $f\left(x\right)=\sin x$, we know that for every $n\ge 0$, $f^{(e+1)}\left(x\right)$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

role="math" localid="1650275703897" $\left|R_n\left(x\right)\right|=\left|\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{x}^{n+1}\right|$

Calculation

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

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

Therefore,

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