The Taylors expansions of Butcher's B-series (for Runge-Kutta order conditions)

Question

In the Butchher's book numerical methods for ODE I do not understand in the proof how do we obtain the coefficient in the formula $311d$ at $F(t)(y(x_0))h^{|t|}$ from $$\frac{1}{t!}\xi^{|t|}$$ where $t=[t_1t_2...t_k]$

EDIT Lemma 310B

Answer 1

Term-wise integration is easier to justify than term-wise differentiation, but as the context is formal and abstract anyway it makes no difference if you develop the theory on the original (always sufficiently smooth and analytical) differential equation $y'(x)=f(y(x))$ or on its integral form $y(x)=y_0+\int_0^xf(y(s))\,ds$.

Form of the expansion

Exploration in the low order terms shows that the expansion of the solution $y(x+h)$ has terms that contain polynomial expressions $F(t)$ of the derivatives of the ODE function. Taking the reduced commutativity of the derivative terms for systems of ODE into account, one ends up with the parametrization by single-rooted trees $t$ (with ordered sub-trees). So in a first view one gets $$ y(x+h)=y(x)+\sum_{t:\rm tree}h^{d(t)}\alpha(t)F(t) $$ (see answer with links for Butchers B-series). Now insert into the differential equation, using $h$ as independent variable.

Taylor formula

For the expansion of the right side as first step consider $$ f(y(x+h))=f(y(x))+\sum \frac{h^m}{m!}f^{(m)}(y(x))[(y(x+h)-y(x))^m]. $$

Multinomial expansion of the power of the series

The powers are a simple multiplication in the scalar case and symmetric tensor products in the systems case. Algebraically there is no difference for this step. Applying multinomial expansion one has $$ \frac1{m!}(a_1+a_2+...)^m = \sum_{\substack{j_1<j_2<...<j_k\\ m_1+...+m_k=m}} \frac1{m_1!...m_k!}a_{j_1}^{m_1}...a_{j_k}^{m_k}. $$

Expansion of the full differential equation

Inserting into the differential equation this gives (powers in tuples indicate repetitions) \begin{multline} \sum_{t:\rm tree}d(t)h^{d(t)-1}\alpha(t)F(t) \\= \sum_{t=[t_1^{m_1},...,t_k^{m_k}]}h^{m_1d(t_1)+...m_kd(t_k)}\cdot \prod_{j=1}^k\frac{\alpha(t_j)^{m_j}}{m_j!}\cdot f^{(m_1+..+m_k)}(y(x))[F(t_1)^{m_1},...,F(t_k)^{m_k}] \end{multline} Comparing coefficients for the same tree gives the recursions \begin{align} d(t)&=1+m_1d(t_1)+...+m_kd(t_k)\\ \alpha(t)=\frac1{d(t)}\prod_{j=0}^k\frac{\alpha(t_j)^{m_j}}{m_j!}\\ \end{align} The factor $\alpha(t)$ can now be split up into a symmetry term $\sigma(t)$ and a differentiation term $\gamma(t)$, $$ \alpha(t)=\frac1{\sigma(t)\gamma(t)}\\ \sigma(t)=\prod_{j=1}^km_j!\sigma(t_j)^{m_j}\\ \gamma(t)=d(t)\prod_{j=1}^k\gamma(t_j)^{m_j} $$ The symmetry term is universal to all iterations that are related to differential equations. The differentiation term is, from a certain point of view, a generalization of the factorial, and gives the factorial for trees that are stems, have no multiple branches.

Example implicit 1-stage method

So for instance for a first-order method $y_{n+1}=y_n+hk_1$, $k_1=f(y_n+hak_1)$ one can as well set $$ hk_1=\sum_{t:\rm trees}h^{d(t)}\frac{\Phi(t)}{\sigma(t)}F(t) $$ to get the expansion $$ k_1=\sum_{t=[t_1^{m_1},...,t_k^{m_k}]}\frac{(ah)^{d(t)-1}}{\sigma(t)}\prod_{j=1}^k\Phi(t_j)F(t) $$ so that $\Phi(t)=a^{d(t)-1}\prod\Phi(t_j)^{m_j}$. For the root $\tau$ with $F(\tau)=f(y_n)$ the coefficient is obviously $\Phi(\tau)=1$