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\section{Euler's Method}
The object of \emph{Euler's method} is to obtain approximations to the well-posed initial-value problem
\[
\frac{\D y}{\D t}=f(t,y),\quad a\leq t\leq b,\quad y(a)=\alpha
\]
We will use Taylor's Theorem to derive Euler's method. Suppose that $y(t)$, the unique solution has two continuous derivations on $[a,b]$, so that for each $i=0,1,\cdots,N-1$
\[
y(t_{i+1})=y(t_i)+(t_{i+1}-t_i)y'(t_i)+\frac{(t_{i+1}-t_i)^2}{2}y''(\xi_i)
\]
for some number $\xi_i$ in $t_i,t_{i+1}$. Because $h=t_{i+1}-t_i$, we have
\[
y(t_{i+1})=y(t_i)+hy'(t_i)+\frac{h)^2}{2}y''(\xi_i)
\]
Euler's method constructs $\omega_i\approx y(t_i)$, for each $i=1,2,\cdots,N$, by deleting the remainder term, then Euler's method is
\begin{align*}
  \begin{cases}
    \omega_0=\alpha & \\
    \omega_{i+1}=\omega_i+hf(t_i,\omega_i) & for i=0,1,\cdots,N-1
  \end{cases}
\end{align*}
\subsection{Errors Bounds for Euler's Method}
\begin{theo}
Suppose $f$ is continuous and satisfies a Lipschitz condition with constant $L$ on
\[
D=\left\{(t,y)|a\leq t\leq b,y\in\mathbb{R}\right\}
\]
and that a constant $M$ exists with
\[
\abs{y''(t)}\leq M,\quad \text{for all $t\in [a,b]$}
\]
where $y(t)$ denotes the unique solution to the initial-value problem
\[
\frac{\D y}{\D t}=f(t,y),\quad a\leq t\leq b,\quad y(a)=\alpha
\]
Let $\omega_0,\cdots,\omega_N$ be the approximations generated by Euler's method for some positive integer $N$, then for each $i=0,1,\cdots,N$
\[
\abs{y(t_i)-\omega_i}\leq \frac{hM}{2L}\left[e^{L(t_i-a)}-1\right].
\]
\end{theo}
\begin{proof}
\begin{align*}
  \abs{y_{i+1}-\omega_{i+1}} &\leq \abs{y_i-\omega_i}+h\abs{f(t_i,y_i)-f(t_i,\omega_i)}+\frac{h^2}{2}\abs{y''(\xi_i)} \\
  &\leq (1+hL)\abs{y_i-\omega_i}+\frac{h^2M}{2} \\
  &\leq e^{(i+1)hL}(\abs{y_0-\omega_0}+\frac{h^2M}{2hL})-\frac{h^2M}{2hL} \\
  &= \frac{hM}{2L}\left(e^{(t_{i+1}-a)L}-1\right).
\end{align*}
\end{proof} 
\begin{theo}
If $u_0,u_1,\cdots,u_N$ be the approximations and $\abs{\delta_i}<\delta$, then
\[
\abs{y(t_i)-u_i}\leq\frac{1}{L}\left(\frac{hM}{2}+\frac{\delta}{h}\right)\left[e^{L(t_i-a)}-1\right]+\delta e^{L(t_i-a)}
\]
The minimal value of $E(f)$ occurs when $h=\sqrt{\frac{2\delta}{M}}$
\end{theo}