# How should I do this? Euler's Method?

I am approximating a solution to a first order LODE using Euler's method. I made two tables, one using a step size of .01 and another using .05 ( I had to start at x=0 and end at x=1). I am not understanding the directions for the second part of my assignment:

It states that the order of numerical methods (like Euler's) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by $Ch^n$, where $C$ is a generally unknown constant and $n$ is the order. For Euler's method, plot the points: $$(0.1, \phi(1)-y_{10}),$$ $$(.05, \phi(1)-y_{20}),$$ $$(.025, \phi(1)-y_{40}),$$ $$(.0125, \phi(1)-y_{80}),$$ $$(.00625, \phi(1)-y_{160})$$ And then fit a line to the above data of the form $Ch$. I don't understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?

Any clarification is appreciated.

Thanks

Edit:

The LODE I am given is $y'=x+2y, y(0)=1$ and the exact solution I found was $\phi(x)=\frac{1}{4}(-2x+5e^{2x}-1)$.

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what is $\phi$, what is $y$? – user31280 Nov 26 '12 at 2:04
@F'OlaYinka I have edited my original post. – A A Nov 26 '12 at 2:12

From the way I see it, if $\phi$ is the exact solution, the teacher probably wants you to compare the errors at $x=1$ for different step sizes.
Take for example the one before the last $$(0.0125, \phi(1)-y_{80})$$ the step size to use here is $h$ such that $n=80$ and $0+nh=1\implies h=1/80=0.0125$ (I didn't see that coming).
The first number in the bracket is the step size to use for each point. For each point evaluate the approximated value of $\phi(1)\approx y_n$ Then ploy the graph of points $(h_n,\epsilon_n)$ where $\epsilon_n=\phi(1)-y_n$ and $h_n$ is the equivalent step size.
@AA: as you say, you expect the error to be of the form $Ch^n$, where $h$ is the stepsize. You are checking that assumption. The points you are to plot are (stepsize, error). Then see if you can fit them to a form $Ch^n$ – Ross Millikan Nov 26 '12 at 2:24