Question

In Exercises 31–34 in Section 8.2 you were asked to find the Maclaurin series for the specified function. Now 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.

$e^x,\mathrm{ℝ}$

Short answer

We've proved that$\underset{n\to \infty }{\lim }{R}_n\left(x\right)=0$

Step-by-step solution

Given Information  

Given equation : $e^x,\mathrm{ℝ}$

Theory used : For $n\mathit{>}\mathit{0},\mathrm{if}\mathit{\left|}{f}^{\mathit{(}n\mathit{+}\mathit{1}\mathit{)}}\mathit{\left(}c\mathit{\right)}\mathit{\right|}\mathit{\le }\mathit{1}$for every value of $x$ then using the Lagrange's form for the remainder, we have

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

Finding the Lagrange’s form for the remainder and proving limn→∞Rn(x)=0

We get the Lagrange form of remainder by :

$R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{(x-x_0)}^{n+1}$

Where, $c$lies between $x\mathrm{and}{x}_0$

But,

$$\begin{gathered} f\left(x\right)=e^x \\ \Rightarrow f^{(n+1)}c=e^c\forall n\ge 0 \end{gathered}$$

Also, since the series is Maclaurin's. So, :

$$\begin{gathered} \left|R_n\left(x\right)\right|=\left|\frac{e^c}{(n+1)!}{x}^{n+1}\right| \\ \Rightarrow \left|R_n\left(x\right)\right|\le e^{\left|x\right|}\frac{{\left|x\right|}^{n+1}}{(n+1)!} \end{gathered}$$

Taking the limit, we have :

$$\begin{gathered} \underset{n\to \infty }{\lim }{R}_n\left(x\right)\le e^{\left|x\right|}\frac{{\left|x\right|}^{n+1}}{(n+1)!} \\ =0 \end{gathered}$$

as the quotient$\frac{{\left|x\right|}^{n+1}}{(n+1)!}\to 0\mathrm{when}n\to 0$