# global error in Euler's method

I learned how to find local error in Euler's method and it is proportional to $$\frac {h^2} {2}$$ . I can understand this. But when we calculate the global error, why do we just multiply by the number of steps and say global error is proportional to $$h$$? I mean I've been taught that global error is proportional to $$\frac {h^2} 2 \frac {t_f} h$$ where $$\frac {t_f} h$$. I can see $$\frac {t_f} h$$ is the number of steps. But I think the global error should be $$\frac {h^2} 2 l_1 +\frac {h^2} 2l_2 + ... +\frac {h^2} 2l_n$$ where $$n$$ is the number of steps. So, I think the global error is just proportional to $$\frac {h^2} 2$$ not $$h$$. Could you explain why the global error is proportional to $$h$$? Thank you!

If you posit that for the exact solution you get the formula $$y(t_k+h)=y(t_k)+hf(t_k,y(t_k))+\frac{h^2}{2}l_k$$ where $$l_k=y''(t_k+\theta_kh)$$, $$θ_k\in(0,1)$$, then the error $$e_k=y_k-y(t_k)$$ propagates as $$|e_{k+1}|=\left|e_k+h[f(t_k,y_k)-f(t_k,y(t_k))]-\frac{h^2}{2}l_k\right| \le |e_k|+hL|e_k|+\frac{h^2}{2}|l_k|$$ so that $$|e_k|\le\sum_{j=0}^{k-1}(1+Lh)^{k-j-1}\frac{h^2}{2}|l_j|$$ You can now interpret this sum after further relaxing $$(1+Lh)\le e^{Lh}$$ as a Riemann sum for $$|e_k|\lessapprox\frac{h}2\int_{t_0}^{t_k} e^{L(t_k-s)}|y''(s)|\,ds$$ or use a bound $$M_2$$ on the second derivative $$y''(t)=f_t(t,y(t))+f_x(t,y(t))f(t,y(t))$$ and the geometric sum formula $$|e_k|\le\frac{(1+Lh)^k-1}{(1+Lh)-1}\frac{h^2}2M_2=\frac{M_2}{2L}[(1+Lh)^k-1]h \le\frac{M_2}{2L}[e^{L(t_k-t_0)}-1]h.$$