Consistency Trapezoidal rule I want to prove that the consistency order of the trapezoidal rule is actually second order, that means that the error of the actual solution $x(t_{k+1})$ where we can restrict ourselves to the equation $x'(t)=\lambda x(t)$ and the approximation in each step $x_{k+1}=x(t_k)+\frac{\lambda \tau}{2}(x(t_k)+x_{k+1})$ is $O(\tau^3)$. I guess one has to use taylor approximation somewhere, but this does not work. 
 A: This is a sketch which shows that solving an ODE with the trapezoidal rule has global error $O(h^2)$.
Consider the differential equation $\dfrac{dx}{dt}=\lambda x$. Integrating this from $t_n$ to $t_{n+1}=t_n+h$ gives
$$
\int_{t_n}^{t_{n+1}}\dfrac{dx}{dt}dt=\int_{t_n}^{t_{n+1}}\lambda x\,dt \Longrightarrow x(t_{n+1})-x(t_n)=\lambda\int_{t_n}^{t_{n+1}} x\,dt.\tag{1}
$$
Evaluating the integral on the RHS:
$$
\int_{t_n}^{t_{n+1}} x\,dt=\frac{h}{2}(x(t_{n+1})+x(t_n))-\frac{h^3}{12} x''(\xi),
$$
where $t_n\leq\xi\leq t_{n+1}$, provided that $x(t)\in \mathcal{C}^2$.
This gives the update formula $(1)$ as
 $$
x(t_{n+1})=x(t_n)+\dfrac{h\lambda}{2}\left(x(t_{n+1})+x(t_n)\right)-\dfrac{h^3}{12} x''(\xi).
$$
Solving for $x(t_{n+1})$ gives
$$
x(t_{n+1})=\dfrac{2+h\lambda}{2-h\lambda}x(t_n)-\dfrac{h^3}{6(2-h\lambda)}x''(\xi)=\dfrac{2+h\lambda}{2-h\lambda}x(t_n)+O(h^3).
$$
Thus, the local error is $O(h^3)$.
Considering the global error, integrating the DE from $0$ to $t_n$:
$$
\int_{t_0}^{t_{n}}\dfrac{dx}{dt}dt=\int_{t_0}^{t_{n}}\lambda x\,dt \Longrightarrow x(t_{n})-x(t_0)=\lambda\int_{t_0}^{t_{n}} x\,dt.
$$
The integral can be computed using the trapezoidal rule
$$
\int_{t_0}^{t_n}x\,dt=\frac{h}{2}\left[x(t_0)+2\sum\limits_{i=1}^{n-1}x(t_i)+x(t_n)\right]-\frac{(t_n-t_0)h^2}{12}x''(\xi),
$$
where $t_0\leq\xi\leq t_n$. Following similarly as above, the global error can be shown to be $O(h^2)$.
This is one example of a Runge-Kutta method, this one from the second order family.
