Chapter 5, Problem 1E

Linear systems are so easy to work with that engineers often construct linear models of real (nonlinear) systems to assist in analysis and design. Such models are often surprisingly accurate over a limited range. For example, consider the simple exponential function e^x. The Taylor series representation of this function is

$e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{2}+\mathrm{.....}$

(a) Construct a linear model for this function by truncating the Taylor series expansion alter the linear (first-order) term. (b) Evaluate your model function at x = 0.000005,.0.0005,0.05,0.5, and 5.0. (c) For which values of x does your model yield a “reasonable” approximation to e^x? Explain your reasoning.

To determine

Construct a linear model for the given function by truncating the Taylor series expansion after the linear (first-order) term.

Answer to Problem 1E

Linear function for the given function is $f_{linear}=1+x$.

Explanation of Solution

Given Data:

The Taylor series expansion of the given function is

$e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\mathrm{..........}$

Calculation:

A linear function is a simple function composed of constant terms and simple variable without exponent.

To make given function linear, remove all terms for which power of $x$ is greater than $1$,

So,

$f_{linear}=1+x$        (1)

Conclusion:

Thus, linear function for the given function is $f_{linear}=1+x$.

To determine

Evaluate linear function for given values of $x$.

Answer to Problem 1E

The percentage change in given function for $x=0.000005$ is $0\%$, percentage change in given function for $x=0.0005$ is ${10}^{-5}\%$, percentage change in given function for $x=0.05$ is $0.12\%$, percentage change in given function for $x=0.5$ is $9\%$ and percentage change in given function for $x=5$ is $96\%$.

Explanation of Solution

Given Data:

Values of $x$ are $0.000005$, $0.0005$, $0.05$, $0.5$ and $5$.

Calculation:

The expression for percentage change in given function is as follows.

$\Delta f=\frac{e^x-f_{linear}}{e^x}\times 100$        (2)

Here,

$\Delta f$ is percentage change in given function,

$x$ is a variable and

$f_{linear}$ is modified linear function.

For $x=0.000005$,

Substitute $0.000005$ for $x$ in equation (1),

$\begin{array}{c}{f}_{linear}=1+0.000005\\ =1.000005\end{array}$

Substitute $0.000005$ for $x$ and $1.000005$ for $f_{linear}$ in equation (2),

$\begin{array}{c}\Delta f=\frac{e^{0.000005}-1.000005}{e^{0.000005}}\times 100\\ =\frac{1.000005-1.000005}{1.000005}\times 100\\ =0\%\end{array}$

So percentage change in given function for $x=0.000005$ is $0\%$.

For $x=0.0005$,

Substitute $0.0005$ for $x$ in equation (1),

$\begin{array}{c}{f}_{linear}=1+0.0005\\ =1.0005\end{array}$

Substitute $0.0005$ for $x$ and $1.0005$ for $f_{linear}$ in equation (2),

$\begin{array}{c}\Delta f=\frac{e^{0.0005}-1.0005}{e^{0.0005}}\times 100\\ =\frac{1.000500125-1.0005}{1.000500125}\times 100\\ ={10}^{-5}\%\end{array}$

So percentage change in given function for $x=0.0005$ is ${10}^{-5}\%$.

For $x=0.05$,

Substitute $0.05$ for $x$ in equation (1),

$\begin{array}{c}{f}_{linear}=1+0.05\\ =1.05\end{array}$

Substitute $0.05$ for $x$ and $1.05$ for $f_{linear}$ in equation (2),

$\begin{array}{c}\Delta f=\frac{e^{0.05}-1.05}{e^{0.05}}\times 100\\ =\frac{1.05127-1.05}{1.05127}\times 100\\ =0.12\%\end{array}$

So percentage change in given function for $x=0.05$ is $0.12\%$.

For $x=0.5$,

Substitute $0.5$ for $x$ in equation (1),

$\begin{array}{c}{f}_{linear}=1+0.5\\ =1.5\end{array}$

Substitute $0.5$ for $x$ and $1.5$ for $f_{linear}$ in equation (2),

$\begin{array}{c}\Delta f=\frac{e^{0.5}-1.5}{e^{0.5}}\times 100\\ =\frac{1.6487-1.5}{1.6487}\times 100\\ =9\%\end{array}$

So percentage change in given function for $x=0.5$ is $9\%$.

For $x=5$,

Substitute $5$ for $x$ in equation (1),

$\begin{array}{c}{f}_{linear}=1+5\\ =6\end{array}$

Substitute $5$ for $x$ and $6$ for $f_{linear}$ in equation (2),

$\begin{array}{c}\Delta f=\frac{e^5-6}{e^5}\times 100\\ =\frac{148.413-6}{148.413}\times 100\\ =96\%\end{array}$

So percentage change in given function for $x=5$ is $96\%$.

Conclusion:

Thus, the percentage change in given function for $x=0.000005$ is $0\%$, percentage change in given function for $x=0.0005$ is ${10}^{-5}\%$, percentage change in given function for $x=0.05$ is $0.12\%$, percentage change in given function for $x=0.5$ is $9\%$ and percentage change in given function for $x=5$ is $96\%$.

To determine

For which values of $x$ value of linear function is approximately equal to $e^x$.

Answer to Problem 1E

The values for which the approximation yields a reasonable result is $x=0.000005$, $x=0.0005$ and $x=0.05$.

Explanation of Solution

Calculation:

For $x=0.000005$, percentage change in given function is $0\%$, for $x=0.0005$, percentage change in given function is ${10}^{-5}\%$ and for $x=0.05$, percentage change in given function

is $0.12\%$.

As for $x=0.000005$, $x=0.0005$ and $x=0.05$ percentage change in values is very less means for these values of $x$ linear function is approximately equal to $e^x$.

Conclusion:

Thus, the values for which the approximation yields a reasonable result is $x=0.000005$, $x=0.0005$ and $x=0.05$.