Chapter 10, Problem 60RE

(a)

To determine

The first four terms and general term of taylor series f(x) at x=3.

Answer to Problem 60RE

The first four non-zero terms of the series are $T_0=1T_1=-\left(x-3\right)T_2={\left(x-3\right)}^2{T}_3=-{\left(x-3\right)}^3$

The general term of the series is given by $T_n=\frac{{\left(x-2\right)}^n}{n!}$

Explanation of Solution

Given information:

The given function is

  $f\left(x\right)=\frac{1}{\left(x-2\right)}$

Formula used:

  $f\left(x\right)=f\left(a\right)+\left[f\text{'}\left(a\right)\right]\left(x-a\right)+\frac{\left[f"\left(a\right)\right]{\left(x-a\right)}^2}{2!}+\frac{\left[f"\text{'}\left(a\right)\right]{\left(x-a\right)}^3}{3!}+\mathrm{...}$

Calculation:

  $f\left(x\right)=\frac{1}{\left(x-2\right)}$ series of a function $f\left(x\right)$ is given by at $x=a$

  $f\left(x\right)=f\left(a\right)+\left[f\text{'}\left(a\right)\right]\left(x-a\right)+\frac{\left[f"\left(a\right)\right]{\left(x-a\right)}^2}{2!}+\frac{\left[f"\text{'}\left(a\right)\right]{\left(x-a\right)}^3}{3!}+\mathrm{...}$

To determine the first four terms of the series we need to evaluate f(x) its first three derivatives

at x = 3 f (3)=1

  $\begin{array}{l}f\text{'}\left(x\right)=-{\left(x-2\right)}^{-2}\\ \Rightarrow f\text{'}\left(3\right)=-1\\ f"\left(x\right)=2{\left(x-2\right)}^{-3}\\ \Rightarrow \frac{f"\left(3\right)}{2!}=1\\ f"\text{'}\left(x\right)=-6x{\left(x-2\right)}^{-4}\\ \Rightarrow \frac{f"\text{'}\left(3\right)}{3!}=-1\\ f^{\left(n\right)}\left(x\right)=n!{\left(-1\right)}^n{\left(x-2\right)}^{-\left(n+1\right)}\\ \Rightarrow \frac{f^{\left(n\right)}\left(3\right)}{n!}={\left(-1\right)}^n\end{array}$

Therefore, series of a function f (x) is given by at x = 3

  $f\left(x\right)=1-\left(x-3\right)+{\left(x-3\right)}^2-{\left(x-3\right)}^3+\mathrm{......}+{\left(-1\right)}^n{\left(x-3\right)}^n+\mathrm{.....}$

Therefore, the first four non-zero terms of the series are

  $T_0=1T_1=-\left(x-3\right)T_2={\left(x-3\right)}^2{T}_3=-{\left(x-3\right)}^3$

The general term of the series is given by

  $T_n=\frac{{\left(x-2\right)}^n}{n!}$

Conclusion:

The first four non-zero terms of the series are $T_0=1T_1=-\left(x-3\right)T_2={\left(x-3\right)}^2{T}_3=-{\left(x-3\right)}^3$

The general term of the series is given by $T_n=\frac{{\left(x-2\right)}^n}{n!}$

(b)

To determine

The first four terms and general term of taylor series at x=3.

Answer to Problem 60RE

The general term of the series is given by $T_n={\left(-1\right)}^n\frac{{\left(x-3\right)}^{n+1}}{n+1}$

Explanation of Solution

Given information:

Integrating term by term $\ln \left|x-2\right|={\displaystyle \underset{3}{\overset{x}{\int }}\frac{1}{t-2}dt}$

Formula used:

Integration is used.

Calculation:

  $\ln \left|x-2\right|={\displaystyle \underset{3}{\overset{x}{\int }}\frac{1}{t-2}dt}$

  $\begin{array}{l}={\displaystyle \underset{3}{\overset{x}{\int }}\left(1-\left(t-3\right)+{\left(t-3\right)}^2-{\left(t-3\right)}^3+\mathrm{.....}+{\left(-1\right)}^n{\left(t-3\right)}^n+\mathrm{....}\right)dt}\\ ={\left[t-\frac{1}{2}{\left(t-3\right)}^2+\frac{1}{3}{\left(t-3\right)}^3-\frac{1}{4}{\left(t-3\right)}^4+\mathrm{.....}+\frac{1}{n}{\left(-1\right)}^n{\left(t-3\right)}^n+\mathrm{....}\right]}_3^x\\ =\left(x-3\right)-\frac{{\left(x-3\right)}^2}{2}+\frac{{\left(x-3\right)}^3}{3}-\frac{{\left(x-3\right)}^4}{4}+\mathrm{....}+{\left(-1\right)}^n\frac{{\left(x-3\right)}^{n+1}}{n+1}+\mathrm{...}\end{array}$

Therefore, the first four non-zero terms of the series are

  $\begin{array}{l}{T}_0=\left(x-3\right)\\ T_1=\frac{-{\left(x-3\right)}^2}{2}\\ T_2=\frac{{\left(x-3\right)}^3}{3}\\ T_3=\frac{-{\left(x-3\right)}^4}{4}\end{array}$

The general term of the series is given by $T_n={\left(-1\right)}^n\frac{{\left(x-3\right)}^{n+1}}{n+1}$

Conclusion:

The general term of the series is given by $T_n={\left(-1\right)}^n\frac{{\left(x-3\right)}^{n+1}}{n+1}$

(c)

To determine

To determine: The value of $\ln \left(\frac{2}{3}\right)$

Answer to Problem 60RE

The value is $\ln \left(\frac{2}{3}\right)\approx 0.375$

Explanation of Solution

Given information:

The function is $\ln \left|x-2\right|={\displaystyle \underset{3}{\overset{x}{\int }}\frac{1}{t-2}dt}$

Formula used: Alternating Series Estimation Theorem is used.

Calculation:

  $\ln \left|x-2\right|=\left(x-3\right)-\frac{{\left(x-3\right)}^2}{2}+\frac{{\left(x-3\right)}^3}{3}-\frac{{\left(x-3\right)}^4}{4}+\mathrm{....}+{\left(-1\right)}^n\frac{{\left(x-3\right)}^{n+1}}{n+1}+\mathrm{...}$

Calculating the values of $\ln \left|x-2\right|\text{at}x=3.5\ln \left(\frac{3}{2}\right)=\frac{1}{2}-\frac{1}{2^2.2}+\frac{1}{2^3.3}-\frac{1}{2^4.4}+\mathrm{....}$

By the Alternating Series Estimation Theorem,

Since the size of the third term is $\frac{1}{24}<0.05$

Therefore first two terms will be sufficient

Therefore,

  $\ln \left(\frac{2}{3}\right)\approx 0.375$

Conclusion:

The value is $\ln \left(\frac{2}{3}\right)\approx 0.375$