Chapter 4, Problem 1P

The following series can be used to approximate $e^x$:

$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdot \cdot \cdot +\frac{x^n}{n!}$

(a) Prove that this Maclaurin series expansion is a special case of the Taylor series expansion $\left[\text{Eq}\text{.}\left(4.7\right)\right]\text{ with }{x}_i=0\text{ and }h=x$.

(b) Use the Taylor series to estimate $f\left(x\right)=e^{-x}\text{ at }{x}_{i+1}=1\text{ for }{x}_i=0.2$. Employ the zero-, first-, second-, and third-order versions and compute the $\left|{\epsilon }_t\right|$ for each case.

To determine

To prove: The Maclaurin series $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdots +\frac{x^n}{n!}$ is the special case of the Taylor series expansion $f\left(x_{i+1}\right)=f\left(x_i\right)+f\text{'}\left(x_i\right)h+\frac{f\text{'}\text{'}\left(x_i\right)}{2!}{h}^2+\frac{f^{\left(3\right)}\left(x_i\right)}{3!}{h}^3+\cdots +\frac{f^{\left(n\right)}\left(x_i\right)}{n!}{h}^n+R_n$ where the remainder term $R_n$ equals $\frac{f^{\left(n+1\right)}\left(\xi \right)}{\left(n+1\right)!}{h}^{n+1}$ with $x_i=0$ and $h=x$.

Explanation of Solution

Given Information:

The Maclaurin series of exponential function $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdots +\frac{x^n}{n!}$.

Formula Used:

The Taylor series expansion

$f\left(x_{i+1}\right)=f\left(x_i\right)+f\text{'}\left(x_i\right)h+\frac{f\text{'}\text{'}\left(x_i\right)}{2!}{h}^2+\frac{f^{\left(3\right)}\left(x_i\right)}{3!}{h}^3+\cdots +\frac{f^{\left(n\right)}\left(x_i\right)}{n!}{h}^n+R_n$.

Proof:

Let the function $f\left(x\right)$ equal $e^x$.

For this function,

$\begin{array}{c}{f}^{\left(n\right)}\left(x\right)=f\left(x\right)\\ =e^x\end{array}$

For all integer n.

Now apply this in the Taylor series expansion to obtain:

$\begin{array}{c}f\left(x_{i+1}\right)=f\left(x_i\right)+f\text{'}\left(x_i\right)h+\frac{f\text{'}\text{'}\left(x_i\right)}{2!}{h}^2+\frac{f^{\left(3\right)}\left(x_i\right)}{3!}{h}^3+\cdots +\frac{f^{\left(n\right)}\left(x_i\right)}{n!}{h}^n+R_n\\ e^{x_{i+1}}=e^{x_i}+h{e}^{x_i}+\frac{h^2}{2}{e}^{x_i}+\frac{h^3}{3!}{e}^{x_i}+\cdots \frac{h^n}{n!}{e}^{x_i}+\frac{h^{n+1}}{\left(n+1\right)!}{e}^{\xi }\end{array}$

Now replace $x_i$ with 0 and h with x to get,

$\begin{array}{c}{e}^x=e^0+x{e}^0+\frac{x^2}{2}{e}^0+\frac{x^3}{3!}{e}^0+\cdots \frac{x^n}{n!}{e}^0+\frac{x^{n+1}}{\left(n+1\right)!}{e}^{\xi }\\ =1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdots +\frac{x^n}{n!}+\frac{x^{n+1}}{\left(n+1\right)!}{e}^{\xi }\end{array}$

Note that the remainder term $\frac{x^{n+1}}{\left(n+1\right)!}{e}^{\xi }$ would tend to 0 as n tends to infinity.

This gives,

$e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdots +\frac{x^n}{n!}$

Hence, proved that the provided Maclaurin series is an approximation of $e^x$ and this is a generalisation of the Taylor series expansion.

To determine

To calculate: The approximate value of $f\left(x\right)=e^{-x}$ at the point $x_{i+1}=1$ when $x_i=0.2$ using the Taylor series with the simultaneous computation of $\left|\xi \right|$ at the zero-, first-, second- and third order approximates.

Answer to Problem 1P

Solution:

The zero first, second and third order Taylor series approximations for the function $f\left(x\right)=e^{-x}$ at the point $x_{i+1}=1$ when $x_i=0.2$ are $0.81873$, $0.16375$, $0.42574$ and $0.35587$ with the relative percentage error at the respective stages as $122.55\%$, $55.49\%$, $15.73\%$ and $3.26\%$ respectively.

Explanation of Solution

Given Information:

The function $f\left(x\right)=e^{-x}$ with points $x_{i+1}=1$ and $x_i=0.2$.

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 1 for $x_{i+1}$ and 0.2 for $x_i$ to obtain,

$\begin{array}{c}f\left(1\right)=f\left(0.2\right)\\ =e^{-0.2}\\ =0.81873\end{array}$

The zero order approximation gives 0.81873.

The exact value of the function at 1 would be:

$\begin{array}{c}f\left(1\right)=e^{-1}\\ =0.36788\end{array}$

Thus,

$\begin{array}{c}\left|{\xi }_t\right|=\left|\frac{0.36788-0.81873}{0.36788}\right|\times 100\%\\ =122.55\%\end{array}$

The relative percentage error at this stage is 122.55%.

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 1 for $x_{i+1}$ and 0.2 for $x_i$ to obtain,

$\begin{array}{c}f\left(1\right)=f\left(0.2\right)+f\text{'}\left(0.2\right)\left(1-0.2\right)\\ =e^{-0.2}-e^{-0.2}\left(0.8\right)\\ =0.16375\end{array}$

The first order approximation gives 0.16375.

Thus,

$\begin{array}{c}\left|{\xi }_t\right|=\left|\frac{0.36788-0.16375}{0.36788}\right|\times 100\%\\ =55.49\%\end{array}$

The relative percentage error at this stage is 55.49%.

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 1 for $x_{i+1}$ and 0.2 for $x_i$ to obtain,

$\begin{array}{c}f\left(1\right)=f\left(0.2\right)+f\text{'}\left(0.2\right)\left(1-0.2\right)+\frac{f\text{'}\text{'}\left(0.2\right)}{2}{\left(1-0.2\right)}^2\\ =e^{-0.2}-e^{-0.2}\left(0.8\right)+e^{-0.2}\left(0.32\right)\\ =0.42574\end{array}$

The second order approximation gives 0.42574.

Thus,

$\begin{array}{c}\left|{\xi }_t\right|=\left|\frac{0.36788-0.42574}{0.36788}\right|\times 100\%\\ =15.73\%\end{array}$

The relative percentage error at this stage is 15.73%.

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 1 for $x_{i+1}$ and 0.2 for $x_i$ to obtain,

$\begin{array}{c}f\left(1\right)=f\left(0.2\right)+f\text{'}\left(0.2\right)\left(1-0.2\right)+\frac{f\text{'}\text{'}\left(0.2\right)}{2}{\left(1-0.2\right)}^2+\frac{f^{\left(3\right)}\left(0.2\right)}{3!}{\left(1-0.2\right)}^3\\ =e^{-0.2}-e^{-0.2}\left(0.8\right)+e^{-0.2}\left(0.32\right)-e^{-0.2}\left(0.08533\right)\\ =0.35587\end{array}$

The third order approximation gives 0.35587.

Thus,

$\begin{array}{c}\left|{\xi }_t\right|=\left|\frac{0.36788-0.35587}{0.36788}\right|\times 100\%\\ =3.26\%\end{array}$

The relative percentage error at this stage is $3.26\%$.