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.

$\ln (1+x),(-1/2,1/2)$

Short answer

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

Step-by-step solution

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 localid="1650646884870" $c^{n+1}\frac{|x{|}^{n+1}}{(n+1)!}\to 0\text{whenn→∞}$