Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability / 6 / d

Write the derivative equation as $f^{\prime }(x)=\Phi (x,f(x))$ with initial approximation $y_0$. The forward Euler method uses the slope at the start of the step:If the derivative is a known function of $x$ alone, this is $y_{i+1}=y_i+h{f}^{\prime }(x_i)$.

For $f^{\prime }=\Phi (x,f)$, the Backward Euler method uses the slope at the end of the step:This generally requires solving an implicit equation for $y_{i+1}$. For a derivative depending only on $x$, it reduces to $y_{i+1}=y_i+h{f}^{\prime }(x_{i+1})$.

For $f^{\prime }=\Phi (x,f)$, the standard explicit midpoint method isThe implicit midpoint rule instead uses $y_{i+1}=y_i+h\Phi (x_i+h/2,(y_i+y_{i+1})/2)$. When the derivative depends only on $x$, both give $y_{i+1}=y_i+h{f}^{\prime }(x_i+h/2)$.

The endpoint-only request in (e) suggests another possible convention: averaging the endpoint slopes, $y_{i+1}=y_i+\frac{h}{2}[\Phi (x_i,y_i)+\Phi (x_{i+1},y_{i+1})]$. This is the trapezoidal rule, and generally differs from the implicit midpoint rule. A further use of “midpoint” is the two-step explicit midpoint linear multistep method, $y_{i+1}=y_{i-1}+2h\Phi (x_i,y_i)$, which needs a second starting value. The printed question supplies no convention to distinguish these meanings; (e) spells out the resulting alternatives.