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==Global truncation error==
The [[global truncation error]] is the error at a fixed time <math>t_i</math>, after however many steps the method needs to take to reach that time from the initial time. The global truncation error is the cumulative effect of the local truncation errors committed in each step.<ref>{{harvnb|Atkinson|1989|p=344}}</ref> The number of steps is easily determined to be <math display="inline">\frac{t_i-t_0}{h}</math>, which is proportional to <math display="inline">\frac{1}{h}</math>, and the error committed in each step is proportional to <math>h^2</math> (see the previous section). Thus, it is to be expected that the global truncation error will be proportional to <math>h</math>.<ref>{{harvnb|Butcher|2003|p=49}}</ref>

This intuitive reasoning can be made precise. If the solution <math>y</math> has a bounded second derivative and <math>f</math> is [[Lipschitz continuity|Lipschitz continuous]] in its second argument, then the global truncation error (denoted as <math> |y(t_i)-y_i| </math>) is bounded by

:<math> |y(t_i)-y_i| \le \frac{hM}{2L}\left(e^{L(t_i-t_0)}-1\right) </math>

where <math>M</math> is an upper bound on the second derivative of <math>y</math> on the given interval and <math>L</math> is the Lipschitz constant of <math>f</math>.<ref>{{harvnb|Atkinson|1989|p=346}}; {{harvnb|Lakoba|2012|loc=equation (1.16)}}</ref> Or more simply, when <math>y'(t) = f(t,y) </math>, the value <math display="inline">L = \text{max}\bigl(|\frac{d}{dy}\bigl[f(t,y)\bigr] | \bigr) </math> (such that <math>t </math> is treated as a constant). In contrast, <math display="inline">M = \text{max}\bigl( | \frac{d^2}{dt^2}\bigl[y(t)\bigr] |  \bigr) </math> where function <math>y(t) </math> is the exact solution which only contains the <math>t </math> variable.

The precise form of this bound is of little practical importance, as in most cases the bound vastly overestimates the actual error committed by the Euler method.<ref>{{harvnb|Iserles|1996|p=7}}</ref> What is important is that it shows that the global truncation error is (approximately) proportional to <math>h</math>. For this reason, the Euler method is said to be first order.<ref>{{harvnb|Butcher|2003|p=63}}</ref>

=== Example ===


If we have the differential equation <math>y'=1+ ( t-y)^2</math>, and the exact solution <math>y=t+ \frac{1}{t-1}</math>, and we want to find <math>M</math> and <math>L</math> for when <math>2 \le t \le 3</math>. <math display="block">L = \text{max}\bigl( | \frac{d}{dy}\bigl[f(t,y)\bigr] | \bigr) = \max_{2 \le t \le 3}\bigl( | \frac{d}{dy}\bigl[1+ ( t-y)^2 \bigr] | \bigr) = \max_{2 \le t \le 3}\bigl( | 2( t-y) | \bigr) = \max_{2 \le t \le 3}\bigl( | 2( t-[t+ \frac{1}{t-1}]) | \bigr) = \max_{2 \le t \le 3}\bigl(  | - \frac{2}{t-1}  |  \bigr) = 2 </math><math display="block">M = \text{max}\bigl(|\frac{d^2}{dt^2}\bigl[y(t)\bigr]|\bigr) = \max_{2 \le t \le 3} \left( | \frac{d^2}{dt^2}  \bigl[ t  +  \frac{1}{1-t} \bigr] |\right) = \max_{2 \le t \le 3}\left( | \frac{2}{(-t+1)^3} |  \right) = 2 </math>Thus we can find the error bound at t=2.5 and h=0.5:

<math display="block"> \text{error bound} = \frac{hM}{2L}\left(e^{L(t_i-t_0)}-1\right) = \frac{0.5 \cdot 2}{2 \cdot 2}\left(e^{2(2.5-2)}-1\right) = 0.42957 </math>Notice that t<sub>0</sub> is equal to 2 because it is the lower bound for t in <math>2 \le t \le 3</math>.