Q. 36 - Calculus

Question

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 $\underset{n\to \mathrm{\infty }}{lim} R_n\left(x\right)=0$ on the specified interval.  

 

        $\ln (1+x),(-1/2,1/2)$

Step-by-Step Solution

Answer

Ans:  The value of the limit is zero, this is because the quotient $c^{n+1}\frac{|x{|}^{n+1}}{(n+1)!}\to 0\text{ when n→∞ }$

Step 1. Given information.

given,

        $\ln (1+x),(-1/2,1/2)$

Step 2. Consider the given function,

 For $n\ge 0\text{, if }\left|f^{(n+1)}\left(c\right)\right|\le 1$ for every value of x, so using the language's from the remainder,

We have

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

Step 3. So, let us first construct the table of the Maclaurin series for the function f ( x ) = ln ( 1 + x )

Step 4. Therefore, the Maclaurin series for the function f ( x ) = ln ( 1 + x ) is

     $0+1\cdot x+\frac{-1}{2!}{x}^2+\frac{2}{3!}{x}^3+\frac{(-6)}{4!}{x}^4+\dots$

Or, we can write it as 

    $f\left(x\right)=\ln 1+\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^{k-1}(k-1)!}{k!}{x}^k$

Since $\ln 1=0$, so the Maclaurin series for the function $f\left(x\right)=\ln (1+x)$ can also be written as

      $f\left(x\right)=\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^{k-1}(k-1)!}{k!}{x}^k$

 Hence, the Lagrange’s form for the remainder is 

       $\begin{array}{r}{R}_n\left(x\right)=\frac{(-1{)}^{n+1-1}\frac{(n+1-1)}{(n+1)!}{c}^{n+1}}{(n+1)!}{x}^{n+1}\\ =\frac{(-1{)}^{n}n!c^{n+1}}{\left[\right(n+1)!{]}^2}{x}^{n+1} \end{array}$

Step 5. Now we take the limit

 $\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim} \left|R_n\left(x\right)\right|\le \underset{n\to \mathrm{\infty }}{lim} c^{n+1}\frac{|x{|}^{n+1}}{(n+1)!}\\ =0 \end{array}$

Here, the value of the limit is zero, this is because the quotient $c^{n+1}\frac{|x{|}^{n+1}}{(n+1)!}\to 0\text{ when n→∞ }$