Question

Show that when the trapezoid scheme given in formula (8) is used to approximate the solution$\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}\mathbf{}\mathbf{of}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$ , at x = 1, then we get ${\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}\left(\frac{\mathbf{1}\mathbf{+}\frac{\mathbf{h}}{\mathbf{2}}}{\mathbf{1}\mathbf{-}\frac{\mathbf{h}}{\mathbf{2}}}\right){\mathbf{y}}_{\mathbf{n}}$,n = 0, 1, 2, . . . , which leads to the approximation ${\left(\frac{1+\frac{h}{2}}{1-\frac{h}{2}}\right)}^{\frac{\mathbf{1}}{\mathbf{h}}}$for the constant e.Compute this approximation for h= 1,${\mathbf{10}}^{\mathbf{-}\mathbf{1}}\mathbf{,}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{,}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{,}\mathbf{and}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$and compare your results with those in Tables 3.4 and 3.5.

Short answer

h	y(1)=e
1	3
0.1	2.72055
0.01	2.71830
0.001	2.71828
0.0001	2.71828

Step-by-step solution

Apply iteration rule for the trapezoidal scheme

Here $\mathrm{y}\text{'}=\mathrm{y},\mathrm{y}\left(0\right)=1$

Apply the iteration rule for the trapezoidal scheme is

${\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\frac{\mathrm{h}}{2}\left[\mathrm{f}\right({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}})+\mathrm{f}({\mathrm{x}}_{\mathrm{n}+1},{\mathrm{y}}_{\mathrm{n}+1}\left)\right]$

In our case of differential equation our equation is

${\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}\left(\frac{1+\frac{\mathrm{h}}{2}}{1-\frac{\mathrm{h}}{2}}\right)$

The interval is

$$\begin{gathered} \frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{1-0}{\mathrm{n}} \\ =\frac{1}{\mathrm{n}} \\ =\mathrm{h} \\ \mathrm{n}=\frac{1}{\mathrm{h}} \end{gathered}$$

Find the values of f(x) = ex

h	y(1)=e
1	3
0.1	2.72055
0.01	2.71830
0.001	2.71828
0.0001	2.71828

Substitute in the iteration rule for the trapezoidal scheme

Now,

$$\begin{gathered} \mathrm{y}\left(1\right)=\mathrm{y}(0+\mathrm{nh})={\mathrm{y}}_{\mathrm{n}} \\ {\mathrm{y}}_1=\mathrm{y}\left(0\right)\left(\frac{1+\frac{\mathrm{h}}{2}}{1-\frac{\mathrm{h}}{2}}\right)=\left(\frac{1+\frac{\mathrm{h}}{2}}{1-\frac{\mathrm{h}}{2}}\right) \\ {\mathrm{y}}_2={\left(\frac{1+\frac{\mathrm{h}}{2}}{1-\frac{\mathrm{h}}{2}}\right)}^2 \\ . \\ . \\ . \\ {\mathrm{y}}_{\mathrm{n}}={\left(\frac{1+\frac{\mathrm{h}}{2}}{1-\frac{\mathrm{h}}{2}}\right)}^{\mathrm{n}} \\ {\mathrm{y}}_{\mathrm{n}}={\left(\frac{1+\frac{\mathrm{h}}{2}}{1-\frac{\mathrm{h}}{2}}\right)}^{\frac{1}{\mathrm{h}}} \end{gathered}$$

Hence the solution is

h	e
1	3
0.1	2.72055
0.01	2.71830
0.001	2.71828
0.0001	2.71828