Euler’s Method — Question 2

Question 2 – Solution

Strategy. Solve the discrete recurrence exactly, then compare its logarithm with the continuous exponential growth.

Step 1: Find and bound the endpoint. Euler gives $Y_{n+1}=(1+h)Y_n$, so $$\boxed{Y_N=(1+1/N)^N}.$$ The exact solution is $e^t$. Since $\ln(1+h)<h$ for $h>0$, $N\ln(1+1/N)<1$ and hence $Y_N<e$. This is an underestimate for every grid, not merely for the three grids below.

Step 2: Define the one-step defect. Starting with the exact value $e^{t_n}$, the difference between the exact next value and one Euler step is $$d_n=e^{t_n+h}-e^{t_n}(1+h) =e^{t_n}(e^h-1-h)=\tfrac 12e^{t_n}h^2+O(h^3).$$ On $0\le t_n\le 1$, this is uniformly $O(h^2)$. Here defect means an error in solution value; dividing it by $h$ would use a different convention.

Step 3: Derive the accumulated error. Because $Nh=1$, $$\ln Y_N=\frac{\ln(1+h)}h=1-\frac h2+O(h^2).$$ Exponentiating gives $Y_N=e(1-h/2+O(h^2))$, so $$\boxed{e-Y_N=\frac e2h+O(h^2)}.$$ There are $N=1/h$ steps over the fixed interval, and their defects propagate through later updates. An $O(h^2)$ one-step defect therefore need not yield an $O(h^2)$ endpoint error; here the accumulated error is first order.

Step 4: Compare the grids. $$\begin{array}{c|c|c} N&Y_N&e-Y_N\\\hline 2&2.250000&0.468282\\ 4&2.441406&0.276876\\ 8&2.565785&0.152497 \end{array}$$ The errors approach a factor-of-two reduction under each doubling of $N$, because their leading term is proportional to $h$. Exact halving is not promised on coarse grids. The number of slope evaluations doubles, exposing the first-order accuracy versus work tradeoff.