# Explanation of Error in Euler's Method (first order differential equations)

Can anyone provide a simple explanation why halving the step size tends to decrease the numerical error in Euler's method by one-half?

I've looked at some online sources but they do provide very complex explanations.

In short: the error added in one step of length $$h$$ is $$h^2\frac{y''(c)}{2}$$, where $$c$$ is some point in the interval we stepped over. To cover a certain fixed distance $$A$$, we need about $$\frac{A}{h}$$ steps, so the total error looks like a constant times $$h$$.
There's also compounded error - in later steps, our $$y$$ values will be in error, and that will affect our estimated $$y'$$ values. With some calculation, we can show that this basically only makes the proportionality constant bigger; the error will still be proportional to $$h$$.