Euler’s Method: Question 2-9.8 & Solution

Question 8

For $y'=-y$ , $y(0)=1$ , let $Y_h$ denote the Euler approximation at $t=1$ with equal step $h$ . Compare $h=1/2$ and $h=1/4$ . Assume exact arithmetic. For the extrapolation argument, suppose $Y_h=y(1)+Ch+O(h^2)$ with a coefficient $C$ independent of $h$ as $h\to 0$ .

Tasks

Compute $Y_{1/2}$ , $Y_{1/4}$ , and $R=2Y_{1/4}-Y_{1/2}$ .
Use the stated error expansion to explain why $R$ cancels the leading first-order error. Relate the difference between the two Euler values to an estimate of the finer-grid error.
Compare the actual errors of the two Euler values and $R$ with the exact endpoint. Is it a contradiction if $R$ lies outside the interval between the Euler values?
Test whether grid agreement certifies accuracy using a different IVP, $z'=\sin^2(4\pi t)$ , $z(0)=0$ . Compute both Euler endpoints on the same grids and the exact value $z(1)$ , and explain the failure of the grid-difference indicator.

Original worksheet page 1: question and worked solution for 2-9-008

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Question 8 – Solution

Strategy. Separate asymptotic error cancellation from a rigorous certificate: two grids can share the same sampling blind spot.

Step 1: Compute the three approximations. For decay, $Y_h=(1-h)^{1/h}$ . Hence $\boxed{Y_{1/2}=1/4,\quad Y_{1/4}=81/256,\quad R=49/128=0.3828125}.$

Step 2: Cancel the leading error term. The assumed expansions give $2Y_{h/2}-Y_h=y(1)+O(h^2),\qquad Y_{h/2}-Y_h=-Ch/2+O(h^2).$ Thus $|Y_{h/2}-Y_h|$ estimates the leading magnitude of the finer-grid error, while the extrapolated value removes that term. This argument is asymptotic; it supplies no explicit bound for the omitted terms on these particular grids.

Step 3: Compare actual errors. The exact endpoint is $e^{-1}$ . Absolute errors are $\begin{array}{c|ccc} \text{value}&Y_{1/2}&Y_{1/4}&R\\\hline \text{absolute error}&0.117879&0.051473&0.014933 \end{array}$ The finer-grid estimate is $17/256=0.06640625$ , not its exact error. Both Euler values are below the truth, while $R$ is above it. Extrapolation is not interpolation, so lying outside their interval is consistent with its construction and does not by itself invalidate it.

Step 4: Exhibit shared undersampling. At all left nodes of both specified grids, $\sin^2(4\pi t)=0$ . Both Euler approximations for $z(1)$ are therefore zero, and their difference is zero. But $\boxed{z(1)=\int_0^1\sin^2(4\pi t)\,dt=\frac 12}.$

See the diagram in the original worksheet below.

The nonnegative forcing is missed between the sample nodes. Agreement of two grids does not certify accuracy; an explicit error bound or further analysis of the resolved time scales is needed.

Original worksheet page 2: question and worked solution for 2-9-008

Euler’s Method — Question 8

Question 8

Question 8 – Solution