Euler’s Method — Question 7

Question 7 – Solution

Strategy. Exploit the contracting error recurrence to turn a local defect bound into a rigorous global guarantee.

Step 1: Solve both continuous and discrete problems. The exact solution is $y(t)=1-e^{-t}$. Euler gives $Y_{n+1}=(1-h)Y_n+h$, so $$\boxed{Y_n=1-(1-h)^n},\qquad E_n=e^{-nh}-(1-h)^n\ge 0.$$ For $0<h<1$, the sign follows from $1-h<e^{-h}$; at $h=1$, it follows directly for $n\ge 1$. At $n=0$, both solutions are zero.

Step 2: Isolate the local defect. Starting Euler at the exact value at $t_n$, its overshoot of the exact next value is $$d_n=y(t_n)+h[1-y(t_n)]-y(t_n+h) =e^{-t_n}(e^{-h}-1+h).$$ Therefore $0\le d_n\le h^2/2$, and subtracting the exact update from the numerical one gives $$\boxed{E_{n+1}=(1-h)E_n+d_n,\qquad E_0=0}.$$

Step 3: Sum the propagated defects. Since $0\le 1-h<1$, iteration gives $$0\le E_n\le\frac{h^2}{2}\sum_{j=0}^{n-1}(1-h)^j =\frac h2[1-(1-h)^n]\le\frac h2.$$ Thus $1/(2N)\le 0.005$ is sufficient. The smallest integer supplied by this simpler bound is $\boxed{N=100}$, with $h=0.01$. This is not a claim that no smaller $N$ could meet the tolerance using a sharper analysis.

Step 4: Evaluate the actual error. For that grid, $$\boxed{E_{100}=e^{-1}-(0.99)^{100}\approx 0.00184710<0.005}.$$ The proven bound controls the error without assuming a particular observed cancellation or leading asymptotic term. Its conservatism explains the smaller actual error. Both the recurrence and this guarantee assumed exact arithmetic; step-by-step rounding would add another error term and require a separate bound.