>> There is no restriction on
>> http://www.sciface.com/support/doc/40/en/combinat/countingFunctions.html
>> so let me take f = x + x^2 + x^3 + ... = 1/(1-x) -1 and g=1
>>
>> g(f) = 1 and f(g) = \infty
>
> In theory, there is no problem. Both f pleth g and g pleth f are \infty.
BTW, I write g(f) if I replace every x in g by f. So since there is no x
in g, I get g(f) = 1.
If (f pleth g) means the definition of Kerber then there is nothing to
replace in g so (f pleth g) = 1.
If (f pleth g) is defined as sum_{k=0}^\infty g_k*f(z^k) I also get 1
since the only non-zero term is for k=0.
But (g pleth f) is
sum_{k=0}^\infty f_k*g(z^k) = \infty + 0*x + 0*x^2 + 0*x^2 + ...
Do I do something wrong?
Another question. Is for (arbitrary power series) f and g always
f(g) = (f pleth g) = sum_{k=0}^\infty g_k*f(z^k)?
If "pleth" is commutative then that is clearly wrong. Take
f=x+x^2, g=x^2. Then
f(g) = x^2 + (x^2)^2
g(f) = (x+x^2)^2
If plethysm corresponds to substitution then there is something I don't
understand.
Aldor-combinat implements for two species F and G
(F\circ G)(x) = F(G(x))
for the exponential generating series as is given in BLL Theorem 1.4.2.
Where does the plethysm of _univariate_ series come into play in
Mupad-Combinat?
> Ok, in practice, the program cannot deal with g(0) <> 0 or f(0) <> 0
> (as those require an infinite sum which you cannot compute in
> general). There is a precondition in the code checking this, but it
> was missing in the doc. Fixed. Thanks for the report.
I reloaded the page
http://www.sciface.com/support/doc/40/en/combinat/countingFunctions.html
nothing has changed. Also not on
http://mupad-combinat.sourceforge.net/doc/en/combinat/countingFunctions.html
:-(
Ralf
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