I am reading MIT's numerical analysis notes. On page 6 they say it is a good exercise to show the explicit Euler method is of order 1. I thought this would be a good exercise because I am not familiar.
Given a numerical scheme $$y_{n+1} = \Psi(h,y_n,t_n)$$ they defined it to have order $p$ if the local error $e_{n+1}(h)=\Psi(h,y_n,t_n)-y(t_{n+1})$ is of order $O(h^{p+1})$. If we plug the forward Euler into this definition we get (note that $t_n = nh$ for some step $h$)
$$e_{n+1}(h)=y_n+hf(t_n,y_n)-y(t_{n+1})$$ They suggest we use 'elementary' Taylor expansion to show that $\Psi$ is of order 1. I dont know what they mean by elementary, other than a first order expansion. They also dont say what to take a Taylor expansion of, but I assume it is of $y(t_{n+1})$. Also not sure where to take the centre of the expansion, so I leave it as $t$ for now. Then $$ y(t_{n+1}) \approx y(t) + y^{\prime}(t) t_{n+1} $$ We can use the initial value problem $y'(t) = f(t,y(t))$ to re-write the prime term to get $$ y(t_{n+1}) \approx y(t) + f(t,y(t)) t_{n+1} $$I thought that I would probably need to put this into the local error formula and do some algebra to get a $h^2$ term, my attempt goes as follows
\begin{align*} e_{n+1}(h) &\approx y_n + hf(t_n,y_n) - y(t) -f(t,y(t)) t_{n+1}\\ &= y_n - y(t) + h(f(t_n,y_n) - (n+1) f(t,y(t))) \end{align*} from here I cant see how to argue that $e_{n+1}$ is of order $O(h^2)$.
Suppose your step size is $h$, so $t_{n+1}=t_n+h$. Then Taylor expanding $y(t_n+h)$ gives \begin{align*} y(t_{n+1})=y(t_n+h)&=y(t_n)+hy'(t_n)+O(h^2)\\ &=y_n+hf(t_n,y_n)+O(h^2)\\ &=y_{n+1}+O(h^2), \end{align*} which tells you that the local truncation error is $O(h^2)$. Remember that when computing the local truncation error, you implicitly assume that you got your previous steps right, i.e. $y_n=y(t_n)$.