Classifying Order of Convergence for ODE Approximation Basic Question Different approximation methods yield different orders of convergence/accuracy, e.g., Euler's method should have an order of 1, while the fourth order Runge-Kutta should have an order of 4. 
Using the linear shooting method for a boundary-value problem, say I get this:

and it is supposed to have an order of convergence of 4. How do I get that?
 A: In general, if $u_{1,i}$ and $v_{1,i}$ are $O(h^n)$ approximations to $y_1(x_i)$ and $y_2(x_i)$, respectively, for each $i = 0, 1, \ldots, N$, then it can be shown that $w_{1,i}$ will be an $O(h^n)$ approximation to $y(x_i)$.
In "Analysis of Numerical Methods" by Isaacson and Keller, this can be shown to be:
$$\tag 1 \displaystyle |w_{1,i} - y(x_i)| \le K h^n \left|1 + \frac{v_{1,i}}{v_{1,N}}\right|.$$
These methods behave like this generally, but bad things can happen if there is cancellation of significant figures due to rounding errors, but you can actually usually see that in the intermediate calculations and take measures to correct (more digits or another method, for example).
From your table, you are using Runge-Kutta, which is expected to be a fourth order accurate method, that is $O(h^4)$ to the solution. Using $(1)$, you should be able to choose a $K$ that satisfies:
$$\displaystyle |w_{1,i} - y(x_i)| \le K h^n \left|1 + \frac{v_{1,i}}{v_{1,N}}\right|.$$
Plug in your $h$, which appears to be $\displaystyle \frac{1}{10}$ and $n = 4$ and very that the inequality is satisfied for ALL iterates, that is:
$$\displaystyle |w_{1,i} - y(x_i)| \le K \left(\frac{1}{10}\right)^4 \left|1 + \frac{v_{1,i}}{v_{1,N}}\right|.$$
If this works for a particular $K$, you are fourth-order accurate.
Lastly, if this method has horrible performance due to the rounding errors, other techniques must be used.  
