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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.

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Here's an AoPS thread with some fairly simple 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 $h$.

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