Q. 48 - 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.

$\sqrt{x}, 1, (1/2, 3/2)$

Step-by-Step Solution

Answer

The required Lagrange's form for the given remainder is,$\left|R_n\left(x\right)\right|={\frac{\left|\mathrm{x}-\mathrm{\pi }\right|}{(n+1)!}}^{n+1}$

Step 1. Given Information

The function is$f\left(x\right)=\sqrt{x}$

Step 2 . Calculation

If $f\left(x\right)=\sqrt{x}$, we know that for every $\sqrt{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

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

Step 3. Simplification

Since the Taylor series for the function $f\left(x\right)=\sqrt{x}$ at $x=1$ is

$P_n\left(x\right)=1+\frac{1}{2}(x-1)+\sum _{k=2}^{\infty }{(-1)}^{k+1}\frac{k.1.3.5...(2k-3)}{2kk!}{(x-1)}^k$

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