Q. 49 - Calculus

Question

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.

$\ln \left(x\right), 3, (2,4)$

Step-by-Step Solution

Answer

The Lagrange's form for the remainder $R_n\left(x\right)$ on the specified interval is $\left|R_n\left(x\right)\right|=\frac{1}{(n+1) c^{n+1}}{(x-3)}^{n+1}$

Step 1. Given Information

The function is, $f\left(x\right)=\ln \left(x\right)$

Phow that $\underset{n\to \infty }{\lim }{R}_n\left(x\right)=0$ on the specified interval.

Step 2. Calculation

If $f\left(x\right)=\ln \left(x\right)$, we know that for every $n\ge 0, \left|f^{n+1}\left(c\right)\right|\le 1$, for every value of $x$, so using the Lagrange's form for the remainder, we have

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

Step 3. Simplification

Since the Taylor series for the function $f\left(x\right)=\ln \left(x\right)$, at $x=3$ is

$P_n\left(x\right)=1+\frac{1}{2}(x-1)+\sum _{k=2}^{\infty }{(-1)}^{k+1}\frac{1.2.3...(k-1)}{3kk!}{(x-3)}^k$

Therefore, $\left|R_n\left(x\right)\right|=\frac{1}{(n+1) c^{n+1}}{(x-3)}^{n+1}$