I'm working with a proof of local and global truncation errors for the euler method in numerical methods, and I can't seem to understand the equality and the less/equal seen below. At first I thought it had to do with geometrical sums, but I still don't understand it, any help would be appreciated:

$$1+e^{Lh}+e^{2Lh}+...+e^{(i-1)Lh} = \frac{e^{iLh}-1}{e^{Lh}-1} \le \frac{e^{L(t_i-a)}-1}{Lh}$$

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    $\begingroup$ What are $L,t_i$ and $a$ ? $\endgroup$ – Yves Daoust Mar 31 '17 at 16:42

The formula $$\sum^i_{n=0} x^n = \frac{x^{i+1}-1}{x-1}$$ holds for any $x\ne 1$ [this is easily seen for $i=1$ and then proven by induction]. Just plug in $x=e^{Lh}$. The inequality follows since $e^{Lh} -1 \ge Lh$ and presumably $ih = t_i -a$.


The equality is indeed the sum of the geometric series (although $i$ is somewhat confusing: it is the number of intervals, not $\sqrt{-1}$ :-) ).

The inequality arises from two things: $iLh = L(t_i - a)$, so the numerators are equal. For the denominators, $Lh$ is apparently assumed small, so you can approximate $e^{Lh}$ by the first two terms of its Taylor expansion. Since the rest of the expansion is positive, leaving it out reduces the total:

$$ e^{Lh} -1 \ge Lh$$

so inverting inverts the inequality.

EDIT: The inequality is true even if $Lh$ is not small, but when it is small, the inequality is "good" in the sense that it provides a tighter bound.


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