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I know that by a theorem, $A(B(x))$ is a formal power series if $b_0=0$.
I wasn't sure if it works the other way around.

Can I also say that if $b_0\ne0$, $A(B(x))$ is not a formal power series?

If it doesn't work the other way around, what other techniques can I perform to determine if $A(B(x))$ is a formal power series when $b_0\ne0$?

Bill Dubuque
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Haxify
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  • What if $A$ is just a polynomial? – Akiva Weinberger Jan 22 '16 at 02:35
  • @AkivaWeinberger As far as I know, I don't think that case is considered but what happens if $A$ is just a polynomial? – Haxify Jan 22 '16 at 02:38
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    First off, $A$ would still be a formal series; just most of the coefficients would be zero. Right? In any case, I would think that $A(B(x))$ would be defined, because it would just be a finite sum. Like, if $A(x)=x^2+2x$, then $A(B(x))$ would be $B(x)^2+2B(x)$. We know that $B(x)^2$ exists (since we can take the product of two formal power series) and we know $2B(x)$ exists, so the sum should exist (since we can add together any two formal power series). – Akiva Weinberger Jan 22 '16 at 02:41
  • That is very interesting. Thank you very much! – Haxify Jan 22 '16 at 02:47
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    @BillDubuque: You marked this as a duplicate of a question about products of power series, but this concerns composition of power series, a much more subtle issue. Following the link to your Answer of the other Question does not enlighten about why you insist on using your Golden Hammer to force a round peg into a square hole. – hardmath May 21 '17 at 20:50
  • @hardmath Please read more closely. The answer I gave there apples here. Alas, there is much confusion on this and related matters (see the deleted answer there). – Bill Dubuque May 21 '17 at 22:31

2 Answers2

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The problem is that an infinite sum of nonzero elements of the coefficient ring (which you would need to do to compute the composition, unless $A$ had finitely many terms) is not defined without adding additional structure to the ring. Thus it is correct to say the composition is not necessarily a formal power series if $A$ has infinitely many terms (in particular plugging in a nonzero constant is invalid).

Matt Samuel
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  • Your remark about convergence is incorrect - see my answer. – Bill Dubuque May 02 '16 at 20:18
  • @Bill I've edited my answer. Convergence of any sort, topological or not, requires us to augment the structure of the ring and does not come from the basic axioms. – Matt Samuel May 02 '16 at 21:19
  • I believe we can define the convergence by topology using "basic neighborhoods" generated by powers of the variable/indeterminate $X$. – hardmath May 21 '17 at 20:52
  • @hardmath Regardless, that doesn't necessarily tell you what happens when you plug $1$ in to $\sum_{k=0}^\infty{x^k}$. Even with a topology there are very likely to be compositions you can't do. – Matt Samuel May 21 '17 at 21:47
  • We agree about the fact that there are compositions that "you can't do," and agree about the case you describe where $b_0 = 1$. I wanted to get the Question re-opened because (a) it was not a duplicate as Bill claimed and (b) I feel more can be said to sharpen the issue of existence by giving a formal definition. – hardmath May 21 '17 at 23:48
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There is a simple, natural notion of convergence that settles this and related matters, and it can be presented in a elementary way that requires no knowledge of topology. For example, see the discussion below (excerpted from Stanley's classic $ $ Enumerative Combinatorics I). $ $ In particular, the final paragraph shows how this applies to your problem of composition of formal power series.

Note $ $ Beware that there is widespread confusion between formal vs. analytic power series, and this often clouds discussion of this and related matters (as Rota often remarked, even some eminent mathematicians have published nonsense based on such confusion). For example, see this closely related question on products of formal power series (which was initially clouded by such confusion - see esp. the deleted answer and comments). It seems that - just as for polynomials - it proves difficult for some to shed analytic (function) bias and pass to purely algebraic (formal) abstraction.

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Bill Dubuque
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  • I've read the book you quote cover to cover and I still think you're missing the point. Even with this definition you can just plug an arbitrary element into a power series. – Matt Samuel May 21 '17 at 23:53
  • @Matt Did someone make such a claim? – Bill Dubuque May 21 '17 at 23:55
  • The question is whether the composition is always a power series. The answer is no because of that. – Matt Samuel May 21 '17 at 23:56
  • @Matt In my opinion the above viewpoint provides a much deeper understanding of the matter. – Bill Dubuque May 21 '17 at 23:58
  • Bill, you are missing the point of the Question (I am making that claim). The OP asks about the conditions under which a composition $A(B(x))$ of formal power series may be said to itself be a formal power series. Having a definition of "convergence" in a topological sense or otherwise, you should address the Question being asked. It's your obligation when posting an Answer. Matt has said $A(1)$ is not defined for formal power series when $A$ is not polynomial. Would it not be best to agree on this point? – hardmath May 22 '17 at 00:00
  • @hardmath I think perhaps we may disagree on the essence of the matter. – Bill Dubuque May 22 '17 at 00:04