Proof of Radau IA method being of order 3 I am trying to proof directly that the Radau IA method
\begin{align}
k_1 &= f\left(t_0, x_o+h\left(\frac14 k_1 - \frac14 k_2\right)\right) \\
k_2 &= f\left(t_0+\frac23 h, x_o+h\left(\frac14 k_1 + \frac5{12} k_2\right)\right) \\
x_1 &= x_0 + h\left(\frac14 k_1 + \frac34 k_2\right)
\end{align}
is of order 3 (so the local order of convergence is of order 4).
I think the idea is to compute the derivatives (this is where I am struggeling) of
$$\varphi(h,t,x)=x_0 + h\left(\frac14 k_1 + \frac34 k_2\right)$$
w.r.t $h$ up to order 4 and then using Taylor expansion at $h=0$ and compare it with the expansion of the flow. Is that more or less correct? How do I actually do this? Or is there a better way for a direct proof?
 A: The order conditions for
$$
\begin{array}{c|cc}
c_1 & a_{11}& a_{12}\\
c_2 & a_{21}& a_{22}\\
\hline
& b_1& b_2
\end{array}
~~=~~
\begin{array}{c|cc}
0 & \frac14& -\frac14\\
\frac23 & \frac14& \frac5{12}\\
\hline
& \frac14& \frac34
\end{array}
$$
are
\begin{align}
\text{order 1}\\
1&=b_1+b_2&&=\frac14+ \frac34\\
\hline
\text{order 2}\\
\frac12&=b_1c_1+b_2c_2&&=\frac14·0+ \frac34·\frac23\\
\hline
\text{order 3}\\
\frac13&=b_1c_1^2+b_2c_2^2&&=\frac34·\frac49\\
\frac16&=\sum b_ia_{ij}c_j&&=(b_1a_{12}+b_2a_{22})c_2&&=\left(\frac14·-\frac14+ \frac34·\frac5{12}\right)·\frac23\\
\hline
\text{order 4?}\\
\frac14&\overset{?}{=}\sum b_ic_i^3=\frac34·\frac{8}{27}&&\text{No}\\
\frac18&=\sum b_ic_ia_{ij}c_j\\
\frac1{12}&=\sum b_ia_{ij}c_j^2\\
\frac1{24}&=\sum b_ia_{ij}a_{jk}c_k
\end{align}
So indeed, order 3 holds and in order 4 the interpolation condition can not be satisfied.

So the goal is to get a match for as many terms as possible in $y'=F(y)$,
\begin{multline}
y(x+h)=y(x)+Fh+\tfrac12F'Fh^2+\tfrac16(F''[F,F]+F'F'F)h^3
\\+\tfrac1{24}(F'''[F,F,F]+3F''[F'F,F]+F'F''[F,F]+F'F'F'F)h^4+...
\end{multline}
It seems obvious that some kind of shorthand for the expressions in $F$ is needed. The order of the derivative corresponds to the number of arguments and there is only one function in play, so $F'[F''[F,F]]$ can be conveniently replaced by $[[*,*]]$ or $[[,]]$ or more graphical representations of rooted trees. Order 3 is still low enough that this is not needed, if one were to compute the order 4 error coefficients, this shorthand would substantially reduce space.
The order 1 consistency conditions should be trivial to motivate, so $c_i=\sum a_{ij}$ and $1=\sum b_i$. For the stages one gets implicit equations
\begin{align}
k_i&=F+F'[a_{ij}k_j]h+\tfrac12F''[a_{ij}k_j,a_{il}k_l]h^2+...\\
&=F+a_{ij}F'[k_j]h+\tfrac12a_{ij}a_{il}F''[k_j,k_l]h^2+...
\end{align}
This means that $k_i=F+O(h)$, inserting this gives immediately $k_i=F+c_iF'[F]h+O(h^2)$, and then inserting this
$$
k_i=F+c_iF'[F]h+a_{ij}c_jF'[F'[F]]h^2+\tfrac12c_i^2F''[F,F]h^2+O(h^3)
$$
Now put this into the assembled formula
$$
y_{+1}=y+b_ik_ih=y+Fh+b_ic_iF'Fh^2+b_ia_{ij}c_jF'[F'[F]]h^3+\tfrac12b_ic_i^2F''[F,F]h^3+O(h^4)
$$
Comparing both step-$h$ expressions results in the cited order conditions.
