Q. 35 - Calculus

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.

$\sin x, \mathrm{ℝ}$

Step-by-Step Solution

Answer

The Lagrange’s form for the remainder is $\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{c}}^{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{3}}}{\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{)}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{!}}{\mathit{x}}^{\mathbf{n}\mathbf{+}\mathbf{1}}$and we've shown that the limit tends to zero.

Step 1 : Given Information

Given equation : $\sin x, \mathrm{ℝ}$

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  

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

Step 2 : Finding the Lagrange’s form for the remainder

The Maclaurin series for $f\left(x\right)=\sin x$ is :

$$\begin{gathered} 0+1·x+\frac{0}{2!}{x}^2+\frac{(-1)}{3!}{x}^3+... \\ \Rightarrow f\left(x\right)=\sum _{k=0}^{\infty }{(-1)}^k\frac{1}{(2k+1)!}{x}^{2k+1} \end{gathered}$$

Therefore, the Lagrange’s form for the remainder is :

$$\begin{gathered} R_n\left(x\right)=\frac{{(-1)}^{n+1}\frac{1}{\left(2\right(n+1)+1)!}{c}^{2(n+1)+1}}{(n+1)!}{x}^{n+1} \\ =\boxed{\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{c}}^{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{3}}}{\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{)}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{!}}{\mathit{x}}^{\mathbf{n}\mathbf{+}\mathbf{1}}} \end{gathered}$$

Step 3 : Proving lim n → ∞ R n ( x ) = 0

As $f\left(x\right)=\sin x$ and the function's derivative goes through the cycle :

$\sin x, \cos x, -\sin x \mathrm{and} -\cos x$

So, the required limit :

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

Hence proved.