Find local truncation error We have the method $$
y^{n+1}=y^n+h \cdot [\rho \cdot f(t^n,y^n)+(1-\rho) \cdot f(t^{n+1},y^{n+1})] ,\quad n=0,\ldots,N-1 \\ y^0=y_0
$$ 
I have to calculate the local truncation error for the method for rho=0.5 and rho=0 and to compare the "order" of the local error with the "order" of the total error.
To do that do we have to find the local truncation error for each $t_i$ ? 
Also how can we find the "order" of the local error using Matlab? Or is it only possible to find it by hand?
 A: For the purposes of order determination, it is more comfortable to use the autonomous formulation of ODE, $y'=f(y)$.
The method is $y_1=y_0+h·Φ(y_0,h)$ with
$$
Φ(y_0,h)=ρ⋅f(y_0)+(1−ρ)⋅f(y_0+h·Φ(y_0,h))
$$
Using the Taylor expansion and inserting it into itself recursively, one obtains up to terms quadratic in $h$
\begin{align}
Φ&=ρ⋅f_0+(1−ρ)⋅\Bigl(f_0+h·f'_0[Φ]+\frac{h^2}2f''_0[Φ,Φ]+O(h^3)\Bigr)\\
&=f_0+(1−ρ)⋅h·\Bigl(f'_0\bigl[\,f_0+(1−ρ)⋅h·f'_0[Φ]\,\bigr]+\frac{h}2f''_0[f_0,f_0]\Bigr)+O(h^3)\\\
&=f_0+(1−ρ)⋅h·f'_0[f_0]+(1−ρ)⋅h^2·\Bigl((1−ρ)·f'_0\bigl[\,f'_0[f_0]\,\bigr]+\frac12f''_0[f_0,f_0]\Bigr)+O(h^3)
\end{align}
On the other hand, for the IVP with $y'(0)=y_0$ one gets
\begin{align}
\frac{y(h)-y(0)}{h}
&=y'(0)+\frac{h}2·y''(0)+\frac{h^2}{6}y'''(0)+O(h^3)\\
&=f_0+\frac{h}2·f'_0[f_0]+\frac{h^2}{6}·\Bigl(f'_0\bigl[\,f'_0[f_0]\,\bigr]+f_0''[f_0,f_0]\Bigr)+O(h^3)
\end{align}
The difference is dominated by
$$
(\tfrac12-ρ)⋅h·f'_0[f_0]
$$
for $\rho\ne \frac12$ and
$$
\frac{h^2}{12}·\Bigl(f'_0\bigl[\,f'_0[f_0]\,\bigr]+f''_0[f_0,f_0]\Bigr)
$$
for $\rho= \frac12$.
