Euler one-step defect (${\delta }_h=y(x+h)-y(x)-hf(x,y(x))=\frac{1}{2}{h}^2{y}^{\prime \prime }(x)+O(h^3)$)

For a smooth solution of $y^{\prime }=f(x,y)$, Taylor's theorem gives the displayed local truncation error when an Euler method step starts at an exact solution value. Thus the one-step defect is $O(h^2)$. Dividing it by $h$ gives a normalized local truncation error of $O(h)$; these are different conventions for reporting the same residual. Also $y^{\prime \prime }=f_x+f_{y}f$ along the solution by the chain rule. A global $O(h)$ error conclusion additionally requires the usual regularity and stability control over successive steps.