Question by Student 201612150 Professor, I think I found something. We can recall "Formula V" used in the last lecture: $\phi_{n+1} = \phi_{n} + \Delta tf(t_{n+1/2}, \phi_{n}+\frac{\Delta t}{2}f(t_{n},\phi_{n})+O(\Delta t^{2})) + O(\Delta t^{3})$ Looking closely, We multiply $\Delta t$ by $\phi_{n}+\frac{\Delta t}{2}f(t_{n},\phi_{n})+O(\Delta t^{2})$. Therefore, $\Delta t$ times $O(\Delta t^{2})$ is $O(\Delta t^{3})$. * Note: I figured out this from the progress to analyze global error. So due to this, the result of global error analysis is unaffected - since we multiply $\Delta t$ by not only the term $\phi_{n}+\frac{\Delta t}{2}f(t_{n},\phi_{n})$, but also error term $O(\Delta t^{2})$! Therefore, we now can sure that the modified Euler's method is of order two. Although I'm not sure if my deduction is correct, but I think this may be an answer.
It's not so simple because you need to show that $\Delta t f(t_{n+1/2}, \phi_{n}+\frac{\Delta t}{2}f(t_{n},\phi_{n}+O(\Delta t^2))$ scales with $O(\Delta t^3)$. Note that you can not simply take $O(\Delta t^2)$ out of $f$ as you did. This needs to be done more carefully.
 $\pi$