Learning Lab

Example 1

Find the solution of \(e^{-x}=x\) for \(0<x<1\) to three decimal places.

Solution:

First define the function:22 Note that you can also use \(f\left(x\right)=x-e^{-x}\) as it is equal to zero at the same value of \(x\) as \(f\left(x\right)=e^{-x}-x.\) \[\begin{align*} f\left(x\right) & =e^{-x}-x. \end{align*}\] We want to find \(x\) where \(f\left(x\right)=0\). To do this we need an initial guess \(x_{0}\) of the root. This is easily done by graphing both \(y=x\) and \(y=e^{-x}.\) The point where the graphs intersect is the point where \(f\left(x\right)=0.\) The graphs are shown below.

A suitable value for \(x_{0}\) would be \(0.5\) and

\[\begin{align*} f'\left(x\right) & =-e^{-x}-1 \end{align*}\]

Using eqn \(\left(1\right)\), we can start the method \[\begin{align*} x_{1} & =x_{0}-\frac{f\left(x_{0}\right)}{f'\left(x_{0}\right)}\\ & =0.5-\frac{f\left(0.5\right)}{f'\left(0.5\right)}\\ & =0.5-\frac{e^{-0.5}-0.5}{-e^{-0.5}-1}\\ & =0.5663\\ x_{2} & =0.5663-\frac{f\left(0.5663\right)}{f'\left(0.5663\right)}\\ & =0.5663-\frac{e^{-0.5663}-0.5663}{-e^{-0.5663}-1}\\ & =0.5671\\ x_{3} & =x_{2}-\frac{f\left(x_{2}\right)}{f'\left(x_{2}\right)}\\ & =0.5671-\frac{e^{-0.5671}-0.5671}{-e^{-0.5671}-1}\\ & =0.5671. \end{align*}\] There has been no change to the first four places so we stop and the answer is \(x=0.567\,.\)

Example 2

Find the solution of \(-\ln\left(x\right)=x^{3}\) correct to five decimal places with an initial guess of \(x_{0}=1.\)

Solution:

Define the function \[\begin{align*} f\left(x\right) & =x^{3}+\ln\left(x\right) \end{align*}\] then \[\begin{align*} f'\left(x\right) & =3x^{2}+\frac{1}{x}. \end{align*}\] In this case we have a value for \(x_{0}\) so there is no need to graph the functions. But we do it anyway.33 It is a good idea to graph the functions because it allows us to get an idea of the answer. If Newton’s method gives a result that is different to what we see in the graph, we have to consider that a mistake has been made or Newton’s method has found another root that we did not see on our graph.

From the graph we see the answer should be about \(0.7\).

Using Newton’s method: \[\begin{align*} x_{1} & =x_{0}-\frac{f\left(x_{0}\right)}{f'\left(x_{0}\right)}\\ & =1-\frac{1^{3}+\ln\left(1\right)}{3\cdot1^{2}+\left(\frac{1}{1}\right)}\\ & =1-\frac{1+0}{3+1}\\ & =0.75\\ x_{2} & =0.75-\frac{\left(0.75\right)^{3}+\ln\left(0.75\right)}{3\cdot\left(0.75\right)^{2}+\left(\frac{1}{0.75}\right)}\\ & =0.7055775\\ x_{3} & =0.7055775-\frac{\left(0.7055775\right)^{3}+\ln\left(0.7055775\right)}{3\cdot\left(0.7055775\right)^{2}+\left(\frac{1}{0.7055775}\right)}\\ & =0.7047098\\ x_{4} & =0.7047098-\frac{\left(0.7047098\right)^{3}+\ln\left(0.7047098\right)}{3\cdot\left(0.7047098\right)^{2}+\left(\frac{1}{0.7047098}\right)}\\ & =0.7047095. \end{align*}\] As we have no change in the first five places after the decimal point we stop and take the answer as \(x=0.70471\,.\) Note that we round up the fifth digit.