2.C0=0
From the Basic Introduction, we thus have
and
as composition won't add a constant element. Applying composition, we have
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:
Taking \(k=1\) and the second equality, one can get
This is a way to compute them. A way that I found more useful was to take k=1 and The first equality.
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
Injecting it, we get
which is equivalent to
As this must be true for all \(C_1\), we must have
Assuming
we have the following method of computation:
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
We thus can model \(V\) as
The second equation then becomes
\(U\) 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 | \(2C_1C_2^2+C_1(1-C_1)C_3\) | \(2C_1C_2^2\) | \(2C_1C_2^2+C_1^2(C_1-1)C_3\) |