2.C0=0

From the Basic Introduction, we thus have

\[ p(x)=\sum_{i=1}^{\infty}C_ix^i \]

and

\[ p_j(x)=\sum_{i=1}^{\infty}C_{i,j}x^i \]

as composition won't add a constant element. Applying composition, we have

\[ p_{j+k}(x)=p_k(p_j(x))=\sum_n C_{n,k}\left(\sum_i C_{i,j} x^i\right)^n=\sum_n\sum_i C_{n,k}\hat B_{i,n}(C_{r,j})x^i \]

With \(\hat B(n,k)\), the Ordinary Bell polynomial. \(C_{r,j}= C_{r,j}|_{r=1}^\infty\), an ordered list as input to the polynomial. This leads us to the following defining equation:

\[ C_{n,j+k}=\sum_i C_{i,j} \hat B_{n,i}(C_{r,k})=\sum_i C_{i,k} \hat B_{n,i}(C_{r,j}) \]

Taking \(k=1\) and the second equality, one can get

\[ C_{N,j}=\frac{1}{C_2(N-2)}\sum_{i=1}^{N-2}\left(C_2C_{i+1,j}C_{N-i,j}+C_{i+2}\hat B_{N+1,i+2}(C_{r,j})-C_{i+1,j}\hat B_{N+1,i+1}(C_r)\right) \]

This is a way to compute them. A way that I found more useful was to take k=1 and The first equality.

\[ C_{n,j+1}=\sum_i C_{i,j} \hat B_{n,i}(C) \]

The case \(C_1=1\) is especially nice given \(C\) 's become polynomial in \(j\). This case will thus be studied in its own page that you can go to now if this is the case that interests you.

Staying in the case \(C_1\ne1\), here is some examples of \(C\)'s:

\(i\)	\(C_{i,j}\)
1	\(C_1^j\)
2	\(C_2\frac{C_1^{2j}-C_1^j}{C_1^2-C_1}\)
3	\(\frac{(C_1-1)C_1C_3+2C_2^2}{C_1^2(C_1-1)^2(C_1+1)}C_1^{3j}-\frac{2C_2^2}{(C_1^2-C_1)^2}C_1^{2j}+\frac{(1-C_1)C_3+2C_2^2}{(C_1-1)^2C_1(C_1+1)}C_1^j\)

From those, we can guess the model

\[ C_{i,j}=\sum_{k=1}^i C_1^{kj}V_{i,k}(C_r|_{r=1}^i) \]

Injecting it, we get

\[ \sum_k^iC_{1}^{kj+k}V_{i,k} = \sum_{k=1}^i\sum_{l=k}^i \hat B_{i,l}C_1^{kj}V_{l,k} \]

which is equivalent to

\[ \sum_{k=1}^iC_1^{kj}\left(C_1^kV_{i,k}-\sum_{l=k}^i \hat B_{i,l} V_{l,k}\right)=0 \]

As this must be true for all \(C_1\), we must have

\[ C_1^kV_{i,k}=\sum_{l=k}^i \hat B_{i,l}V_{l,k} \]

Assuming

\[ \sum_{k=1}^{i}V_{i,k}=0, \]

we have the following method of computation:

\[ \left\{ \begin{array}{ll} V_{1,1}=1\\ (C_1^k-C_1^i)V_{i,k}=\sum_{l=k}^{i-1}\hat B_{i,l}V_{l,k}\\ V_{i,i} = -\sum_{k=1}^{i-1}V_{i,k} \end{array} \right. \]

The third equations gives us the initial condition of the second one. The first gives the initial condition of the system. One can see that each \(V_i,k\) will introduce a new factor \((C_1^k-C_1^i)^{-1}\) so we have

\[ V_{i,k}\xrightarrow{\prod_{l=k+1}^i(C_1^k-C_1^l)}V_{k,k}\xrightarrow{\prod_{l=1}^{k-1}(C_1^l-C_1^k)}V_{1,1}=1 \]

We thus can model \(V\) as

\[ V_{i,k}=\frac{U_{i,k}}{\prod_{l=1\ne k}^i(C_1^k-C_1^l)} \]

The second equation then becomes

\[ U_{i,k}=\sum_{l=k}^{i-1}\hat B_{i,l} U_{l,k}\prod_{h=l+1}^{i-1}(C_1^k-C_1^h) \]

\(U\) has the advantage of being a "simple" polynomial and should thus be easier to then model and solve.

\(U_{a,b}\)	b=1	2	3
a=1	\(1\)
2	\(C_2\)	\(C_2\)
3	\(2C_1C_2^2+C_1(1-C_1)C_3\)	\(2C_1C_2^2\)	\(2C_1C_2^2+C_1^2(C_1-1)C_3\)