Exercise 7

We have seen that the error in approximating the solution to an initial value problem is proportional to $\Delta t$. That is, if $E_{\Delta t}$ is the Euler's method approximation to the solution to an initial value problem at $\overline{t}$, then $$\begin{gathered} y(\overline{t})-E_{\Delta t} \approx K\Delta t \end{gathered}$$ for some constant of proportionality $K$.

In this problem, we will see how to use this fact to improve our estimates, using an idea called accelerated convergence.

1. We will create a new approximation by assuming the error is exactly proportional to $\Delta t$, according to the formula $$\begin{gathered} y(\overline{t})-E_{\Delta t} =K\Delta t \end{gathered}$$. Using our earlier results from the initial value problem $dy/dt = y$ and $y(0)=1$ with $\Delta t = 0.2$ and $\Delta t = 0.1$, we have $$\begin{aligned} y(1) - 2.4883 =\mathstrut & 0.2K \\ y(1) - 2.5937 =\mathstrut & 0.1K \end{aligned}$$. This is a system of two linear equations in the unknowns $y(1)$ and $K$. Solve this system to find a new approximation for $y(1)$. (You may remember that the exact value is $y(1) = e = 2.71828\ldots$.)

2. Use the other data, $E_{0.05} = 2.6533$ and $E_{0.025} = 2.6851$ to do similar work as in (a) to obtain another approximation. Which gives the better approximation? Why do you think this is?

3. Let's now study the initial value problem $$\begin{gathered} \frac{dy}{dt} = t-y, \ y(0) = 0 \end{gathered}$$. Approximate $y(0.3)$ by applying Euler's method to find approximations $E_{0.1}$ and $E_{0.05}$. Now use the idea of accelerated convergence to obtain a better approximation. (For the sake of comparison, you want to note that the actual value is $y(0.3) = 0.0408$.)