Today we discuss some generating functions.
Problem 1: Give a finite set of positive integers T. Let Tn be the collection of sequences (t1,t2,...,tm), such that ∑mi=1tm=n, and each ti∈T. Let an=|Tn| for n≥1 and a0=1, and f(x)=∑∞n=0anxn. Find out what is f(x) explicitly.
Solution: It is not hard to note the recursive relation an=∑t∈Tan−t for n≥1, if we set ai=0 for negative i. So that f(x)=1+∑t∈Txtf(x) and f(x)=1/(1−∑t∈Txt), which is a rational function.
Variantion 1: If T is infinite, what would happen? Would f(x) still be rational?
We first analyze simple cases. If T=N+, it is expected that f(x)=1+∑∞t=1xtf(x)=1+f(x)x/(1−x), so that f(x)=(1−x)/(1−2x)=1+∑∞n=12n−1xn. Indeed, in this case, counting the sequences amounts to divide a sequence of n objects arbitrarily. You can choose to divide between the ith and i+1th for 1≥i≥n−1, so in all 2n−1 choices, justifying the expansion.
I think it is equivalent to f being rational.
Theorem: Qp(t)∩Q[[t]]=Q(t)
It is very interesting that this theorem is used for the rationality of ζ-functions for algebraic varieties, which is part of the Weil conjectures.
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