Chapter 4, Problem 6P

Use zero- through fourth-order Taylor series expansions to predict $f\left(2.5\right)\text{ for }f\left(x\right)\text{=In}$ x using a base point at $x=1$. Compute the true percent relative error ${\epsilon }_t$ for each approximation. Discuss the meaning of the results.

To determine

To calculate: The approximate value of $f\left(x\right)=\ln x$ at the point $x=2.5$ with stating iteration at $x=1$ using the Taylor series with the simultaneous computation of the true percentage relative error at the zero-, first-, second-, third- and fourth order approximates.

Answer to Problem 6P

Solution:

The zero first, second, third and fourth order Taylor series approximations for the function $f\left(x\right)=\ln x$ at the point $x=2.5$ are 0, 1.5, 0.375, 1.5 and 0.2344 with the true relative percentage error at the respective stages as 100 %, 63.7 %, 59.07 %, 63.7%, and 74.42 % respectively.

The result shows that theTaylor series approximation could be generalised as,

$f\left(3\right)=0+\frac{1.5}{1}-\frac{{\left(1.5\right)}^2}{2}+\frac{{\left(1.5\right)}^3}{3}-\frac{{\left(1.5\right)}^4}{4}+\cdots$

Explanation of Solution

Given Information:

The function $f\left(x\right)=\ln x$ with points $x=2.5$ and the starting iteration at $x=1$.

Formula Used:

The Taylor series approximation of $n^{th}$ order is:

$f\left(x_{i+1}\right)=f\left(x_i\right)+f\text{'}\left(x_i\right)\left(x_{i+1}-x_i\right)+\frac{f\text{'}\text{'}\left(x_i\right)}{2!}{\left(x_{i+1}-x_i\right)}^2+\cdots +\frac{f^{\left(n\right)}\left(x_i\right)}{n!}{\left(x_{i+1}-x_i\right)}^n$.

Calculation:

Consider the zero-order approximation for the provided function,

$f\left(x_{i+1}\right)=f\left(x_i\right)$

Now replace 2.5 for $x_{i+1}$ and 1 for $x_i$ to obtain,

$\begin{array}{c}f\left(2.5\right)=f\left(1\right)\\ =\ln \left(1\right)\\ =0\end{array}$

The zero order approximation gives 0.

The exact value of the function at 2.5 would be:

$\begin{array}{c}f\left(2.5\right)=\ln \left(2.5\right)\\ =0.9163\end{array}$

Thus, the true relative percentage error would be:

$\begin{array}{c}{\epsilon }_t=\left(\frac{\left|\text{true}-\text{approximate}\right|}{\text{true}}\times 100\right)\%\\ =\left(\frac{\left|0.9163-0\right|}{\text{0}\text{.9163}}\times 100\right)\%\\ =100\%\end{array}$

The relative percentage error at this stage is 100%.

The first-order Taylor series approximation would be:

$f\left(x_{i+1}\right)=f\left(x_i\right)+f\text{'}\left(x_i\right)\left(x_{i+1}-x_i\right)$

Now replace 2.5 for $x_{i+1}$ and 1 for $x_i$ to obtain,

$\begin{array}{c}f\left(3\right)=f\left(1\right)+f\text{'}\left(1\right)\left(2.5-1\right)\\ =0+\frac{1}{\left(1\right)}\left(1.5\right)\\ =1.5\end{array}$

The first order approximation gives 1.5.

Thus,

$\begin{array}{c}{\epsilon }_t=\left(\frac{\left|\text{true}-\text{approximate}\right|}{\text{true}}\times 100\right)\%\\ =\left(\frac{\left|0.9163-1.5\right|}{\text{0}\text{.9163}}\times 100\right)\%\\ =63.7\%\end{array}$

The relative percentage error at this stage is 38.91%.

The second-order Taylor series approximation would be:

$f\left(x_{i+1}\right)=f\left(x_i\right)+f\text{'}\left(x_i\right)\left(x_{i+1}-x_i\right)+\frac{f\text{'}\text{'}\left(x_i\right)}{2}{\left(x_{i+1}-x_i\right)}^2$

Now replace 2.5 for $x_{i+1}$ and 1 for $x_i$ to obtain,

$\begin{array}{c}f\left(3\right)=f\left(1\right)+f\text{'}\left(1\right)\left(2.5-1\right)+\frac{f\text{'}\text{'}\left(1\right)}{2}{\left(2.5-1\right)}^2\\ =0+\frac{1}{\left(1\right)}\left(1.5\right)-\frac{1}{2{\left(1\right)}^2}{\left(1.5\right)}^2\\ =0.375\end{array}$

The second order approximation gives 0.375.

Thus,

$\begin{array}{c}{\epsilon }_t=\left(\frac{\left|\text{true}-\text{approximate}\right|}{\text{true}}\times 100\right)\%\\ =\left(\frac{\left|0.9163-0.375\right|}{\text{0}\text{.9163}}\times 100\right)\%\\ =59.07\%\end{array}$

The relative percentage error at this stage is 59.07%.

The third-order Taylor series approximation would be:

$f\left(x_{i+1}\right)=f\left(x_i\right)+f\text{'}\left(x_i\right)\left(x_{i+1}-x_i\right)+\frac{f\text{'}\text{'}\left(x_i\right)}{2}{\left(x_{i+1}-x_i\right)}^2+\frac{f^{\left(3\right)}\left(x_i\right)}{3!}{\left(x_{i+1}-x_i\right)}^3$

Now replace 2.5 for $x_{i+1}$ and 1 for $x_i$ to obtain,

$\begin{array}{c}f\left(3\right)=f\left(1\right)+f\text{'}\left(1\right)\left(2.5-1\right)+\frac{f\text{'}\text{'}\left(1\right)}{2}{\left(2.5-1\right)}^2+\frac{f^{\left(3\right)}\left(1\right)}{3!}{\left(2.5-1\right)}^3\\ =0+\frac{1}{\left(1\right)}\left(1.5\right)-\frac{1}{2{\left(1\right)}^2}{\left(1.5\right)}^2+\frac{2}{6{\left(1\right)}^3}{\left(1.5\right)}^3\\ =1.5\end{array}$

The third order approximation gives 1.5.

Thus,

$\begin{array}{c}{\epsilon }_t=\left(\frac{\left|\text{true}-\text{approximate}\right|}{\text{true}}\times 100\right)\%\\ =\left(\frac{\left|0.9163-1.5\right|}{\text{0}\text{.9163}}\times 100\right)\%\\ =63.7\%\end{array}$

The relative percentage error at this stage is 63.7%

The fourth-order Taylor series approximation would be:

$f\left(x_{i+1}\right)=f\left(x_i\right)+f\text{'}\left(x_i\right)\left(x_{i+1}-x_i\right)+\frac{f\text{'}\text{'}\left(x_i\right)}{2}{\left(x_{i+1}-x_i\right)}^2+\frac{f^{\left(3\right)}\left(x_i\right)}{3!}{\left(x_{i+1}-x_i\right)}^3+\frac{f^{\left(4\right)}\left(x_i\right)}{4!}{\left(x_{i+1}-x_i\right)}^4$

Now replace 2.5 for $x_{i+1}$ and 1 for $x_i$ to obtain,

$\begin{array}{c}f\left(3\right)=f\left(1\right)+f\text{'}\left(1\right)\left(2.5-1\right)+\frac{f\text{'}\text{'}\left(1\right)}{2}{\left(2.5-1\right)}^2+\frac{f^{\left(3\right)}\left(1\right)}{3!}{\left(2.5-1\right)}^3+\frac{f^{\left(4\right)}\left(1\right)}{3!}{\left(2.5-1\right)}^4\\ =0+\frac{1}{\left(1\right)}\left(1.5\right)-\frac{1}{2{\left(1\right)}^2}{\left(1.5\right)}^2+\frac{2}{6{\left(1\right)}^3}{\left(1.5\right)}^3-\frac{6}{24{\left(1\right)}^4}{\left(1.5\right)}^4\\ =0.2344\end{array}$

The third order approximation gives 0.2344.

Thus,

$\begin{array}{c}{\epsilon }_t=\left(\frac{\left|\text{true}-\text{approximate}\right|}{\text{true}}\times 100\right)\%\\ =\left(\frac{\left|0.9163-0.2344\right|}{\text{0}\text{.9163}}\times 100\right)\%\\ =74.42\%\end{array}$

The relative percentage error at this stage is 74.42%.

This implies that the Taylor series approximation could be generalised as,

$f\left(3\right)=0+\frac{1.5}{1}-\frac{{\left(1.5\right)}^2}{2}+\frac{{\left(1.5\right)}^3}{3}-\frac{{\left(1.5\right)}^4}{4}+\cdots$