Q. 46 - Calculus

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.

$e^x,1,\mathrm{ℝ}$

Step-by-Step Solution

Answer

The required Lagrange's form is $\left|R_n\left(x\right)\right|=\frac{e|x-1{|}^{n+1}}{(n+1)!}$

Step 1 Given Information

Consider the function $f\left(x\right)=e^x$. The Taylor series of the function $f$ around $x=1$ is $P_n\left(x\right)=\sum _{k=0}^{\infty }\frac{e}{k!}(x-1{)}^k$

The objective is to find the Lagrange's form for the remainder $R_n\left(x\right)$.

Step 2 Calculation

For $n\ge 0,f^{(x+1)}\left(x\right)$ is $e^x$.

The Lagrange's form for the remainder $R_n\left(x\right)$ is $R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{x}^{n+1}$. Thus,

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

Therefore, the Lagrange's form is $\left|R_n\left(x\right)\right|=\frac{e|x-1{|}^{n+1}}{(n+1)!}$