Euler’s Method — Question 9

Question 9 – Solution

Strategy. Treat rounding as a new perturbation at each step; it can create fixed points absent from the exact-arithmetic recurrence.

Step 1: Compute both updates. Exact Euler gives $Z_{n+1}=0.9Z_n$, so $Z_n=0.9^n$. The rounded implementation uses $Y_{n+1}=\operatorname{round}_{0.1}(0.9Y_n)$ and yields $$\begin{array}{c|rrrrrrrrr} n&0&1&2&3&4&5&6&7&8\\\hline Y_n&1.0&0.9&0.8&0.7&0.6&0.5&0.5&0.5&0.5 \end{array}$$ Once it reaches $0.5$, its unrounded next value is $0.45$, which rounds back to $0.5$. Both $Z_n$ and the exact solution $e^{-0.1t}$ tend to zero, whereas this implementation stays at $0.5$.

Step 2: Identify all spurious fixed points. Write a nonnegative grid value as $Y=k/10$, with integer $k\ge 0$. In units of one tenth, the unrounded update is $0.9k$. For $k>0$, rounding back to $k$ occurs exactly when $$k-\tfrac 12\le 0.9k<k+\tfrac 12,$$ which reduces to $k\le 5$. Zero is also fixed. Thus the complete set is $$\boxed{0,\ 0.1,\ 0.2,\ 0.3,\ 0.4,\ 0.5}.$$

Step 3: Bound accumulated rounding. Writing $Y_{n+1}=0.9Y_n+\rho_n$ gives $|\rho_n|\le 0.05$. Therefore $$D_{n+1}=0.9D_n+\rho_n,\qquad D_0=0,$$ so $$\boxed{|D_n|\le 0.05\sum_{j=0}^{n-1}0.9^j =0.5(1-0.9^n)}.$$ This bound approaches a nonzero limit, consistent with the observed plateau.

Step 4: Interpret the stable but biased computation. A factor $0.9$ damps old perturbations, but fresh rounding occurs at every step and can balance the intended decrease. Stability does not remove continually injected errors. Rounding only the final reported answer would leave the preceding exact-arithmetic Euler trajectory unchanged and would introduce only one final reporting error, rather than altering subsequent slopes.