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I am reading "Introduction to Combinatorial Mathematics" by C. L. Liu.

The author wrote about generating functions as follow:

However, because $x$ is just a formal variable, there is no need to question whether the series converges.

The author also wrote as follows:

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I think the author used the fact $\sum_{i=0}^{\infty}\binom {2i}{i}x^i$ converges for some $x\neq 0$ and $(1-4x)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\binom {2i}{i}x^i$ holds.
I think the author also used this fact.
So, I don't think "there is no need to question whether the series converges".
Am I wrong?

I think if $\sum_{i=0}^{\infty}\binom {2i}{i}x^i$ didn't converge for any $x\neq 0$, then we could not derive any useful results from this power series.

Is the power series like $\sum_{n=0}^{\infty} n!x^n$ usuful in combinatorics?

tchappy ha
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    On what step do you think the author used the fact that $\sum_{i=0}^\infty {2i\choose i}x^i$ converges for some $x\ne 0$? Again, these are not functions in any sense; these are formal power series. In this context, $f=(1-4x)^{-1/2}$ doesn't refer to a function, merely the unique (up to sign) formal power series $f$ such that $\frac1{f^2}=1-4x$. – Christian E. Ramirez May 13 '23 at 09:02
  • @C-RAM Thank you very much for your answer. How to prove that there exists a formal power series such that $\frac{1}{f^2}=1-4x$? – tchappy ha May 13 '23 at 09:06
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    Now that I'm re-reading the proof, I see where the confusion lies. The way you would usually prove that $\sum_{i=0}^\infty {2i\choose i}x^i=(1-4x)^{-1/2}$ is to either use the binomial sum identity you're trying to prove (which would be circular here), or appealing to the fact that $\sum_{i=0}^\infty {2i\choose i}x^i=(1-4x)^{-1/2}$ does indeed make sense as an equality between functions analytic around $0$, and therefore the equality is also true as an equality between generating functions; in this second case, you do need something like the theorem you linked. – Christian E. Ramirez May 13 '23 at 09:30
  • @C-RAM Thank you very much. – tchappy ha May 13 '23 at 09:32
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    No problem. Hopefully someone can provide a better and more detailed explanation than I did as a full answer; my explanation is definitely lacking, and there is definitely more to be said about why we often use formal power series instead of just using non-formal power series. – Christian E. Ramirez May 13 '23 at 09:52
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    I explained in detail why formal power series work (including inverses, rational powers, exponentials and logarithms) in my Algebraic Combinatorics notes from Spring 2021 (Sections 3.1--3.8). They do require "convergence" to be checked, but it's a very formal and simple form of convergence (essentially "you're not adding one and the same power of $x$ infinitely often"). This form of convergence is clearly satisfied in your example. – darij grinberg May 13 '23 at 23:54
  • @darijgrinberg Thank you very much for your very nice book. I was not able to find a book which explains why formal power series work before. Thank you very much. – tchappy ha May 14 '23 at 01:39

1 Answers1

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The answer to the question

Are power series like $\sum_n n! x^n$ useful in combinatorics?

is "yes". A simple example is that if you want to count connected graphs on $n$ labeled vertices, you do it using the exponential formula: if $c_n$ is the number of connected graphs, then $$ \exp\left(\sum_{n \geq 1} c_n \frac{x^n}{n!}\right) = \sum_{n \geq 0} 2^{\binom{n}{2}} \frac{x^n}{n!} $$ (because the RHS is the exponential GF for all graphs by number of vertices) and we can extract $c_n$ by taking logarithms $$ c_n = n! \cdot [x^n] \log\left(\sum_{n \geq 0} 2^{\binom{n}{2}} \frac{x^n}{n!}\right).$$ Relationships of this kind are extremely useful, even where (as here) the series diverges badly if you try to evaluate. An example involving the specific series you asked about is the enumeration of indecomposable permutations.

Here are a couple other places where some version of this question has been asked:

JBL
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