Talk:Formal power series

Untitled
I've slightly improved (in my opinion) the sentence on k!ak in the section on differentiation. However, I'd probably prefer k! to be regarded as a natural number, and to view this term as module multiplication of the additive group of R regarded as a module over the integers. I'm not sure how to state that concisely. How about


 * here nak = ak+...+ak (n summands)

? —Preceding unsigned comment added by 160.45.45.134 (talk • contribs)

Formal power series as functions
I think this article is quite well written, although I don't know if the introduction of the metric d is standard and/or necessary for the notation $$\sum a_kX^K$$, which I was taught to be a mere notation for the sequence $$(a_k)$$.

I like this approach, but it may lead to confusion. Indeed, this metric is the only one introduced on this page, while the section "Formal power series as functions", starting with

In mathematical analysis, every convergent power series defines a function with values in the real or complex numbers.

clearly does not refer to this metric (for "convergent"), but to the topology in C (or in some complex function space), for the series of partial sums (of numbers or functions?) associated to the power series in the "obvious" way (but a precise define is not so immediate at all...).

As it stands, this phrase is meaningless, because any power series is convergent according to what precedes, even if it has convergence radius of zero. So I think we should change it, so that this subsection becomes as "perfect" as the rest of the article.

If the author of this article can find a more precise introduction for this subsection in the same style as what precedes, I would appreciate. (I don't know what is the most "gentle" way to go from a power series to the associated analytic function.) &mdash; MFH: Talk 12:53, 26 Apr 2005 (UTC)

Multiplication of power series
Would someone add a section on the coefficients of two multiplied power series? I.e., if we have $$f(z) := \sum_{k=0}^{\infty}{a_kz^k}$$ and $$g(z) := \sum_{k=0}^{\infty}{b_kz^k}$$, then what is $$f(z)g(z)$$? --Rocketman768 14:49, 30 April2007 (UTC)


 * I find the answer to that question appearing in this article TWICE in its present form (although it does not have an entire section devoted to it). Michael Hardy 16:44, 30 April 2007 (UTC)

completion
There's something a little annoying on this page. In several places it refers to completions with respect to I-adic topologies. Unfortunately, the article completion discusses only metric completion, whereas the most natural way to express I-adic completion is not via a metric. (I'm not even sure if it can always be described in terms of a metric.) A more natural way instead is to use sequences that are coherent with respect to a nested sequence of subgroups (i.e. the ideals I, I^2, I^3 etc). Any opinions on what to do? Dmharvey Talk 01:09, 26 Jun 2005 (UTC)


 * It seems that completion in the sense of topology is most generally defined for uniform spaces using Cauchy nets. Ring-theoretic completion (and in particular l-adic completion) seems to be a separate concept. The link, therefore, may be inappropriate in this case. Maybe someone will get around to writing an article on this other kind of completion. - Gauge 07:26, 26 Jun 2005 (UTC)


 * Thanks for the info. Unfortunately my topology is a bit weak to write this up. Maybe someone else will. Dmharvey Talk 12:24, 26 Jun 2005 (UTC)

Elementary questions
I don't follow the advanced stuff in this article, so my question is elementary. Does this mean that 1+2+4+8+... = &minus;1 and 1&minus;1+1&minus;1+1&minus;1+... = 1/2 ? Most people object against this, but I don't understand why. The algebraic argument is that S = 1+2+4+8+... satisfies the first-order equation 2S = 2(1+2+4+8+...) = 2+4+8+... = S&minus;1 which has the unique solution S = &minus;1. T=1&minus;1+1&minus;1+1&minus;1+... satisfies T&minus;1 = &minus;T. The geometric series 1/(1&minus;x) = 1+x+x^2+... is true as an algebraic identity whenever x is not = 1. The binomial theorem provides more examples of divergent series having values. Bo Jacoby 14:23, 3 October 2005 (UTC)
 * The equation 2S=2(1+2+4+8+...) = 2+4+8+... = S&minus;1 is the same as
 * 2&infin; = &infin;&minus;1,
 * as S=&infin;. Then, the usual rules of algebra don't apply, and you can't subtract S from both sides. As such one cannot deduce that S= &minus;1. Does that help? Oleg Alexandrov 23:33, 3 October 2005 (UTC)

Legibility experiment:


 * 2S=2(1+2+4+8+...)=2+4+8+...=S-1
 * 2S = 2(1 + 2 + 4 + 8 + ...) = 2 + 4 + 8 + ... = S &minus; 1

(This seemed like a good opportunity to demonstrate this. But ignore this if you want to stay on topic.) Michael Hardy 00:02, 4 October 2005 (UTC)

Thank you very much, Oleg and Michael. I agree with Michael on the improved legibility. I also agree with Oleg that if S = &infin; then algebra doesn't work, but consider the alternative: that algebra does work. Then S = &minus;1. Does this lead to contradiction ? I think not, because formal power series constitute a ring, and computation in a ring does not lead to contradiction. The case x=1/2 gives the Achilles-and-tortoise-series: A = 1 + 1/2 + 1/4 + 1/8 + ... = 2, and everybody seems to accept that A = 1 + A/2 without bothering about convergence. So I think that the concept of convergence as a necessary and sufficient condition for a series to define a number is somewhat overrated. The article on divergent series states that 1 &minus; 1 + 1 &minus; 1 + ... = 1/2 is 'cesaro summable', but the series 1 + 2 + 4 + 8 + ... = -1 is not treated there. Computer people let the 8-bit byte 11111111 mean &minus;1, and computing with these 'signed integers' work. Bo Jacoby 07:39, 4 October 2005 (UTC)


 * Let us still clarify things a bit. The algebra never works with &infin;. In the definition of a formal power series, things are formal. One never actually adds 1+2+4+.... The concept of addition is totally different in a ring of formal series. And you never actually replace X with 2, it stays X, and it is a formal variable.


 * The formal series 1+X+X^2+X^3+... is nothing but the sequence (1, 1, 1, ..., 1, ...) in the ring of formal series. As such you are right that the algebra works and there is no contradiction, but it is because you operate in a totally different universe than the set of real numbers, you operate in an abstract ring of power series, where there is no &infin; and X is not a number, neither X is 2 nor is it 1/2. Oleg Alexandrov 17:09, 4 October 2005 (UTC)

Thanks again. The power series
 * f(X)=1+X+X^2+X^3+...=(1,1,1,1,...)

satisfies the formal equation
 * (1&minus;X)f(X)=1

or
 * (1&minus;X+0X^2+0X^3+...)(1+X+X^2+X^3+...)=1+0X+0X^2+0X^3+...

or
 * (1,&minus;1,0,0,...)(1,1,1,1,...)=(1,0,0,0,...).

For polynomials the substitution of a number for the formal variable X makes sense. For some other power series this substitution does not make sense. f(2)=1+2+4+8+... is undefined. Right?

Now the identity
 * (1&minus;X)f(X) = 1

provides the information needed to define the undefined. Formally substituting X=2 gives
 * (&minus;1)f(2) = 1

or
 * f(2) = &minus;1.

Why not accept this definition ? Some elementary particle physicists, bothering that the permutation series for the masses diverge, might welcome this kind of definition. Bo Jacoby 09:41, 5 October 2005 (UTC)


 * You are right when you say that f(2)=1+2+4+8+... is undefined. This because here the "+" sign here does not mean either the "+" for real numbers, nor the "+" for formal series. It just does not make sence.


 * So, substitutions valid for polynomials are not valid any longer for formal series. What you are trying to do, with


 * f(2) = &minus;1


 * has no mathematical justification. It is just an extrapolation from formal power series by using the example of polynomials, but it is an extrapolation which is not correct mathematically.


 * And by the way, for computers, 111....1 equals &minus;1 because the computers work modulo some power of two. Computers basically set any large enough power of two to equal zero. So, no connection with formal series. Oleg Alexandrov 18:46, 5 October 2005 (UTC)

Surely, a new definition does not have a mathematical justification. The question is: is there a mathematical objection ? Our discussion is similar to the renaissance discussion regarding imaginary numbers. It was argued that a square root of minus one is an extrapolation which is not correct mathematically. There are plenty of philosophical objections: it is impossible, imaginary, does not exist, is neither positive, negative nor zero, and so on. But the question is: are there mathematical objections against it ? The answer turned out to be 'no', and useful calculations with imaginary numbers are now performed. Will my definition 1+2+4+8+...=-1 lead to contradiction ? If not, let us define it and perform useful calculations. Bo Jacoby 14:48, 6 October 2005 (UTC)


 * There are no objections to the complex numbers because they were mathematically justified. You need to first define a mathematical model (maybe similar to the formal series) where your addition can be rigorously perfomed. Without that model, what you say is just a game in "let's pretend". Oleg Alexandrov 19:09, 6 October 2005 (UTC)

That's nice. The formal power series defined by f(X)=(1-X)^(-n) define the function f(x)=(1-x)^(-n), even for values x where the series is not convergent. (Here X is a formal variable and x is a complex number.) This definition can be further generalized, but I'll take one step at a time. Bo Jacoby 08:31, 7 October 2005 (UTC)


 * You are basically talking about analytic continuation. That works indeed for the function
 * $$f(x)=\frac{1}{1-x}.$$


 * However, if you look at


 * $$\log (1+x) = x - \frac{x^2}{2}+\frac{x^3}{3}-\dots $$


 * you run into trouble. There is no unique way of extending this function in a unique manner for x a complex number, or even for x<0. So, again, your tricks with diverging series have no mathematical justfication. Oleg Alexandrov (talk) 18:19, 7 October 2005 (UTC)

Yes sir ! That is exactly why I didn't try to extend the definition beyond the single valued algebraic functions $$f(x)=(1-x)^{-n}$$ where it works 'without trouble'. But of cause I want to. If a formal power series satisfy an irreducible algebraic equation, then it is reasonable to identify the series with the (set of) roots of the equation. This is pure algebra with no analysis. The series may converge to one of the roots, and still algebraically define all of the roots. The binomial series for $$\sqrt{1+x}$$is one example. My point is that algebra provides values to (divergent) series. This point of view seems to be new. Everybody else talks about convergence. (PS Your logarithmic series converges for |x|<1, even if x<0. ) Bo Jacoby 12:21, 10 October 2005 (UTC)
 * I don't really see your "contribution" and the "value" of your contribution. :) Oleg Alexandrov (talk) 14:38, 10 October 2005 (UTC)


 * Bo Jacoby, it is good that you are thinking about these things, although I'm not sure if this is the best place to ask questions about them. In any case, I think part of the problem is that you are treating formal power series as if they were functions when a priori they are not, precisely because they diverge at certain values. Given that your series may be defined somewhere (like in the interval (-1, 1)) as a function, you may try to extend the domain of the function by using various criteria to define what f(2), etc should be. Insofar as you have chosen a way to extend this domain for a particular function, I have no objection. The problem is that most people (including me) will only consider these extended definitions worthwhile if they satisfy nice properties (e.g. if the function remains analytic on the extended domain). This is what Oleg was referring to with analytic continuation. In summary: there's nothing wrong with trying to extend the domain of your function in this way, but there isn't a good reason for doing it either. Hope that helps. - Gauge 04:43, 11 October 2005 (UTC)

Thanks to Oleg and Gauge for answers. I don't know if I am contributing value to anybody else, but I wonder why the concept of convergence seems to be the only legitimite way of assigning values to series. I ask at this place because it is hard for me to understand the section Formal_power_series.

The substitution Sa of a complex number a, in a series f with rational coefficients, must map the field Q[X] of formal power series into the field C of complex numbers.
 * Sa(f)=f(a).

This substitution must be a homomorphism:
 * Sa(f+g)=Sa(f)+Sa(g)
 * Sa(fg)=Sa(f)Sa(g)

or
 * (f+g)(a)= f(a)+g(a)
 * (fg)(a)=f(a)g(a)

I the formal power series is the root of a polynomial P, so that the equation P(f(X))=0 is a formal identity, then the equation P(z)=0 defines the complex roots
 * z1,..,zn,

(where n is the degree of the polynomial P), then the homomorphism is established. The function f(x) is algebraic, and so analytic. The standard example is f(X)=1+X+X^2+X^3+.... It solves the equation z(1-X)-1=0. This equation is of degree 1 in z. The unique solution is z=f(X)=1/(1-X). Substituting 2 for X gives S2(f)=f(2)=-1. Bo Jacoby 09:43, 11 October 2005 (UTC)
 * You wrote
 * The substitution Sa of a complex number a, in a series f with rational coefficients, must map the field Q[X] of formal power series into the field C of complex numbers.
 * That is incorrect. First of all, the formal series form a ring, not a field. Second, given a complex number a, the morphism f&rarr; f(a) is not defined for all f. Oleg Alexandrov (talk) 09:40, 14 October 2005 (UTC)

You are right, it is not a field. Thank you. But the ring operations suffice to compute polynomials. For example P(f)=fg-1=0 where f(X)=1+X+X^2+... and g(X)=1-X. The substitution is defined for g (because g(X) is a polynomial), and the definition of the substitution can be extended by requiring that it still be a homomorphism. Bo Jacoby 11:48, 14 October 2005 (UTC)
 * You cannot create a morphism from the ring of formal power series to the complex numbers by using substitution. It will not be defined everywhere. I don't know how to prove that you cannot, but I have a very strong feeling that you cannot. If you wish that anything in here make sence, you need to prove that such a morphism is possible. Oleg Alexandrov (talk) 02:22, 15 October 2005 (UTC)

I agree. The substitution will not be defined everywhere. I didn't intend it to be defined everywhere. But the definition can be extended beyond convergent series. Some divergent series can be assigned values by their algebraic properties. Equations of degree one have unique solutions. So the extension of the substitution to formal power series satisfying equations of degree one is unique. The substitution of X=1 into the power series z=f(X)=1+X+X^2+...=1/g(X)=1/(1&minus;X) make no sense, because then the equation g(1)f(1)=0z=1 is of degree zero and has no solution. So 1+1+1+... is senseless. That is the proof you asked for. The details on how wide the extension can be made may require further research. Perhaps the answer is in the article, which alas I don't understand. Bo Jacoby 09:10, 17 October 2005 (UTC)
 * That's the thing. We are talking about extending the substitution, but it will not be defined all the time, and we don't know when it will be defined and when it will not. I doubt one can do anything with this. Oleg Alexandrov (talk) 10:01, 17 October 2005 (UTC)

The value of the power series of (1+X/n)^n, negative integer n, is a useful beginning. Not all power series are convergent, so the value is not defined for all power series. That was not an objection against the concept of convergence. Nor should it be an objection against the extension. Bo Jacoby 10:18, 17 October 2005 (UTC)
 * Hey guys: (a) this page is for discussion about the article on formal power series, not for discussions about formal power series. (b) Even if your discussions would lead to anything, it would have to be considered original research so it can't go in the article anyway.--345Kai (talk) 23:15, 24 March 2009 (UTC)
 * Sorry, you're wrong. Your distinction (a) is nonsensical; such discussions are necessary. (b) Would their discussions (that I haven't read carefully) lead to something meaningful, with all probability that would be already in the literature. So the next step would be just finding a reference to it. --pma (talk) 16:36, 18 July 2009 (UTC)

R xy Versus R x,y
The article mentions that the two are not the same topologically, because while associating a topology with R x, one takes discrete toplogy on R and then the product topology on R x. However, if R is itself a topological ring, one could avoid this problem by considering the topology of R x to be the product of the topology of the topological ring R. Also, if R is not a topological ring, then associate discrete topology with it to make it a topological ring? Any thoughts? Ustad NY (talk) 05:44, 12 December 2007 (UTC)

Composition of series
The article currently states the following in the section on "Composition of series":
 * The critical point here is that this operation is only valid when f(X) has no constant term, so that the series for g(f(X)) converges in the topology of R X  . In other words, each cn depends on only a finite number of coefficients of f(X) and g(X).

The word "only" here is inaccurate, as the operation is also valid when g is finite, i.e. a polynomial. I'm not sure right now if these are the only two cases in which it's valid; if I did, I'd change the article ... Tsoeto (talk) 03:37, 15 March 2009 (UTC)


 * Right, but if you interpret the term operation in that sentence as an operation that has to be defined on the whole R X  the restriction is necessary e.g if the constant term of f  is 1: then g(f(X)) can't be defined for all g&isin;R  X  . Certainly there are cases of g(f(X)) being well defined as a composition of formal power series  when f does have a nonzero constant term: for instance when g is a polynomial  as you remark. Also, if g is any formal series, and the constant term of f is nilpotent in the (commutative) ring R, we still have a well defined g(f(X)) with coefficients defined by means of finite sums. I'm not sure if these details may be of interest for the article though. --pm a  10:12, 19 January 2010 (UTC)

Functional Inverse
Some talk about the functional inverse would be an interesting addition to this article:

http://www.mathpages.com/home/kmath625/kmath625.htm

S243a (talk) 22:09, 4 August 2009 (UTC)John Creighton
 * What exactly from that talk may be of any interest for this article, in your opinion?--pm a 10:17, 19 January 2010 (UTC)

Formula for composition is wrong
The formula for integral powers of a power series should involve the binomial coefficients and hence so should the formula for composition.

If $$f(X)=\sum_{p=1}^{\infty} f_p X^p$$, then $$(f(X))^m = \sum_{q=1}^{\infty} a_{m,q} X^q$$, where

$$ a_{m,q} = \sum_{(i_1+\dots+i_r)=m} \sum_{(i_1j_1+\dots+i_rj_r)=q} \frac{m!}{i_1!\dots i_r!} f_{j_1}^{i_1}\dots f_{j_r}^{i_r} $$

It follows that if $$g(Y)=\sum_{s=0}^{\infty} g_s Y^s$$, then $$g(f(X))=\sum_{t=0}^{\infty} b_t X^t$$, where $$b_t = \sum_{m=0}^{t} g_m a_{m,t}$$. (The sum stops at $$t$$ since we note that $$a_{m,t}=0$$ for $$t<m$$.)

Formal Laurent series
I don't think it is very serious to generalize Laurent series with infinite negative powers, as such formal sums don't have a natural ring structure. So unless someone provides an official source where this terminology is used, I would suggest to only consider the standard definition of formal Laurent  power series (finitely many negative powers). Algebraically, the ring of such series is just the localization of $$RX$$ with respect to the positive powers of $$X$$. 77.233.114.244 (talk) 22:48, 28 December 2010 (UTC)


 * For instance, in here, here (bottom line in page 7) and in Parshin's ICM talk (top line in page 2), Laurent formal power series all mean finitely many negative powers. 77.233.114.244 (talk) 10:40, 29 December 2010 (UTC)


 * Actually, I would put the material on Laurent formal series in a subsection "Extensions" rather than "Generalizations". Possibly, within the existing section "Operations". It is true that the notion of Laurent formal series generalizes the notion of formal power series, but if we want to keep its origin, a Laurent series is, first of all, the result of an operation on formal power series, giving rise to an extension. What do you think? --pm a 20:02, 13 April 2011 (UTC)


 * First a word on negative powers: it is absolutely essential that formal Laurent series have only finitely many negative power terms. The absence of a ring structure without that restriction means that usual ways to designate the series would become ambiguous and therefore useless. For instance as a Laurent series $$\textstyle\frac1{1-X^{-1}}$$ might suggest a series $$\textstyle\sum_{i\geq0}X^{-i}$$, but it already designates the ordinary formal power series $$\textstyle\frac{-X}{1-X}=-\sum_{i\geq1}X^i$$. In fact most expressions have both an interpretation when negative powers are restricted and when positive powers are restricted (i.e., as formal Laurent series in X−1) and their coefficients (at monomials they both contain) do not coincide. I even seem to recall that R. Stanley used this observation to explain certain dualities in combinatorics.
 * Then on the terminology "Extensions" versus "Generalizations", I was at first surprised by the distinction made, though I would more naturally call formal Laurent series generalizations than extensions of formal power series. But I then realized (by imagining what I would call for instance rational functions relative to polynomials) that it depends on what one is considering: I would never call rings of formal Laurent series a generalization of rings of formal power series, since they aren't, but indeed they are extensions of them. So the main point would be, is one more interested in the structure (more rich than just a ring in fact) or in the individual series. Looking at the article as it is, there is no clear bias, both individual and structural properties are discussed. The article title does not refer to the structure of course. My personal guess would be that, in contrast to most algebraic topics, interest in formal power series (for instance in combinatorics) often focusses on individuals more than on structure (although the latter needs consideration to know what is allowed when manipulating individual series). In the end, I think I would vote for " Generalization", as it currently is stated. Marc van Leeuwen (talk) 06:59, 15 April 2011 (UTC)
 * Ok, you got the point very clearly. I'm personally more inclined to look at the structure, but it is true that the article focuses more on the individual, which is not bad, so I think you convinced me: let's keep "Generalization".  --pm a  20:01, 16 April 2011 (UTC)

Laurent polynomial
Is "formal Laurent series" the same as Laurent polynomial? If so should the article link to it? Thanks! ciphergoth (talk) 08:03, 15 October 2012 (UTC)
 * No, a Laurent polynomial has only finitely many terms (of positive or negative degree), whereas a formal Laurent series may have infinitely many terms of positive degree. (The difference is similar to the difference between polynomials and formal power series.)  However, it might still be nice to have a link to Laurent polynomial if there's an appropriate place for one.  --JBL (talk) 19:06, 15 October 2012 (UTC)

Towards a final form?
I've read quite carefully the whole article, and it seems to me that it has improved a lot in the past years. I found everything correct; the only section I did not check carefully, though it is well written, is Algebraic properties of the formal power series ring. It contains interesting material and maybe could be expanded, and it is the main place still wanting references. Up to this point, it seems to me that we may be close to remove the initial tag "Refimprove", dating back to 2009. pm a 00:06, 28 November 2012 (UTC)

Lagrange inversion formula
In the "very short computation" I don't understand the second equals sign. Maybe the derivation is too short, a hint might help. It somehow enters that f(g(X)) = X, and powers of X, f, g and f' get mixed up in a fascinating way. radical_in_all_things (talk) 08:42, 15 June 2013 (UTC)
 * Actually the second equal sign is a plain application of the above rule iv. Should you need more explanations, you may like to post further questions to the math reference desk --pm a 14:18, 3 May 2014 (UTC)
 * Actually, this talk page is exactly the place to discuss the article and in particular whether the article explains a point adequately. And I tend to agree that the second equality as currently written is not adequately explained.  Deltahedron (talk) 16:03, 3 May 2014 (UTC)

commutativity
The ring of formal power series is defined here over a commutative ring. Why do we assume the coefficient ring to be commutative? I suggest do take any coefficient ring (maybe unital). Spaetzle (talk) 13:43, 20 March 2015 (UTC)

Some confusion about characteristic whether 0 vs. finite
In the article I find under Formal power series:
 * “A more explicit description of these coefficients is provided by Faà di Bruno's formula, at least in the case where the coefficient ring is a field of characteristic 0.”

But when looking at Faà di Bruno's formula I find only the multinomial $${\frac{n!}{m_1!\,1!^{m_1}\,m_2!\,2!^{m_2}\,\cdots\,m_n!\,n!^{m_n}}}$$ which is always an integer. Thus, there seems to be no reason to request characteristic 0.

On the other hand, in the subsequent section Formal power series I find:
 * “Whenever a formal series $$\textstyle f(X)=\sum_k f_k X^k \in RX$$ has f0 = 0 and f1 being an invertible element of R, there exists a series $$\textstyle g(X)=\sum_k g_k X^k$$ that is the composition inverse of $$f$$.”

This, in my opinion, has kind of a counterexample in the following: Let $$p\in \mathbb P$$ and $$R:=\mathbb F_p$$ be the prime field of characteristic p. Then $$\textstyle f(X) := X-X^p$$ is certainly $$ \in \mathbb F_pX$$ and has f0 = 0 and f1 = 1. However, f can not have a functional inverse in $$ \mathbb F_pX$$, because $$\forall x\in \mathbb F_p : f(x) = 0$$. --Nomen4Omen (talk) 13:41, 8 May 2018 (UTC)


 * On the first point, the formula in question involves division by arbitrary integers, so it does not make sense as written over positive characteristic. On the other hand, one could ask to first compute that integer, then to map it to an element of the finite field; that might make sense and give the right answer (and that probably explains the hedged wording) but I haven't thought hard about it.  A definitive source would be nice.
 * On the second point, $$ x + x^2 + x^4 + x^8 + x^{16} + \ldots$$ is the compositional inverse of $$x - x^2$$ over the field with two elements, and similarly for other primes. Formal power series are not functions. --JBL (talk) 17:41, 8 May 2018 (UTC)


 * On the first point, the division by arbitrary integers in the formula in question can be completely ignored, because the result is PROVED to be integer against all optical appearance.
 * On the second point, it may indeed happen to be the case that the compositional inverse of every formal power series with coefficients having denominators ≠ 1 has only denominators dividing these denominators and vice versa. This would be a very interesting fact, but would have to be proved. Your example $$\scriptstyle x + x^2 + x^4 + x^8 + x^{16} + \ldots$$, however, does NOT! --Nomen4Omen (talk) 18:50, 8 May 2018 (UTC)


 * On the first point, you are repeating yourself, and I already understand the substance of your comment. What is needed to move forward is a reliable source that definitively states that Faà di Bruno's formula (interpreted in the way you are saying) is the correct thing to do over finite characteristic.
 * On the second point, I have no idea what your response means. The statement you quoted is true, and your proposed counter-example is not a counter-example; it is easy to check that the other series is a compositional inverse.  --JBL (talk) 19:18, 8 May 2018 (UTC)


 * On the first point, from your mention of "division by arbitrary integers" and "written over positive characteristic" I did not at all get the impression that you understand the substance of my comment. But now it's OK, although the reliable source seems to be obvious to me.


 * On the second point, I agree that my "counter-example" is not really one. But your example, although as you say is easily found valid (even by me!), is not a PROOF neither. Assume a function $$\scriptstyle f(x) = x + x^2/d + \ldots$$ with $$\scriptstyle d \ne 1$$ . How do YOU guarantee that the inverse $$\scriptstyle g(x) $$ does NOT have a coefficient with a denominator $$\scriptstyle c $$ which does not divide $$\scriptstyle d $$ ? Because if there is such a $$\scriptstyle c $$, take it as characteristic and try to invert $$\scriptstyle f $$. --Nomen4Omen (talk) 20:47, 8 May 2018 (UTC)


 * I do not understand the talk about denominators. The coefficients lie in some ring R; there is no division.  Given any formal power series f(x) without constant term and with leading coefficient a unit, one can extract coefficients in g(f(x)), set them successively equal to 1, 0, 0, 0, ..., and solve for the coefficients of g one by one.  The only division that ever arises is by the leading coefficient of f, which by hypothesis is a unit.  (This is exactly the procedure in e.g. Stanley's Enumerative Combinatorics, v2, prop 5.4.1, although he states the result only when the coefficients are a field.)  --JBL (talk) 00:47, 9 May 2018 (UTC)


 * Sorry, I do not have access to Stanley's Enumerative Combinatorics. Still it may be you are right. And Stanley's procedure could make that point clear, although − as you already admit − even "he states the result only when the coefficients are a field". But the current wiki texts around Lagrange inversion theorem, Bell polynomials etc involve a much more confusing set of divisions "by arbitrary integers" (= your words on my first point) than the relatively simple multinomial in Faà di Bruno's formula. (And at the same time allow as coefficients any rational fraction with any denominator as long as it can be an element of the original ring.)
 * So, if you are right and "the only division that ever arises is by the leading coefficient of f", I question why nobody dares to state all these theorems for simple unital rings without even mentioning any fields nor characteristics. --Nomen4Omen (talk) 09:33, 9 May 2018 (UTC)
 * Ok, I finally understand: you are absolutely correct that the Lagrange inversion theorem is not valid over arbitrary fields. Nevertheless, the statement you quoted is valid over arbitrary coefficient rings.  In fact this is exactly what the section "Composition inverse" of this article says.  --JBL (talk) 10:29, 9 May 2018 (UTC)

As far as I found out, there is NO EFFECTIVE division in the Lagrange inversion algorithm. In the following, the $$a_i$$ are the coefficients of $$\textstyle f(X) = 1+\sum_{n=2}^\infty a_n X^n $$ with $$a_0=0, a_1=1$$, $$B_{n,k}$$ the Bell polynomials in the $$a_i$$, and $$b_i$$ are the coefficients of the compositional inverse $$\textstyle g(X) = 1+\sum_{n=2}^\infty b_n X^n $$ expressed as polynomials in the $$a_i$$:

$$\begin{array}{ll} \scriptstyle b_1=1 \\ \scriptstyle B_{1,1}=a_2 \\ \scriptstyle b_2=-a_2 \\ \scriptstyle B_{2,1}=2\,a_3 \\ \scriptstyle B_{2,2}=a_2^{\,2} \\ \scriptstyle b_3=2\,a_2^{\,2}-a_3 \\ \scriptstyle B_{3,1}=6\,a_4 \\ \scriptstyle B_{3,2}=6\,a_2\,a_3 \\ \scriptstyle B_{3,3}=a_2^{\,3} \\ \scriptstyle b_4=5\,a_2\,a_3-5\,a_2^{\,3}-a_4 \\ \scriptstyle B_{4,1}=24\,a_5 \\ \scriptstyle B_{4,2}=12\,a_3^{\,2}+24\,a_2\,a_4 \\ \scriptstyle B_{4,3}=12\,a_2^{\,2}\,a_3 \\ \scriptstyle B_{4,4}=a_2^{\,4} \\ \scriptstyle b_5=3\,a_3^{\,2}-a_5+14\,a_2^{\,4}+6\,a_2\,a_4-21\,a_2^{\,2}\,a_3 \\ \scriptstyle B_{5,1}=120\,a_6 \\ \scriptstyle B_{5,2}=120\,a_2\,a_5+120\,a_3\,a_4 \\ \scriptstyle B_{5,3}=60\,a_2\,a_3^{\,2}+60\,a_2^{\,2}\,a_4 \\ \scriptstyle B_{5,4}=20\,a_2^{\,3}\,a_3 \\ \scriptstyle B_{5,5}=a_2^{\,5} \\ \scriptstyle b_6=7\,a_2\,a_5-42\,a_2^{\,5}-a_6+7\,a_3\,a_4-28\,a_2\,a_3^{\,2}-28\,a_2^{\,2}\,a_4+84\,a_2^{\,3}\,a_3 \\ \scriptstyle B_{6,1}=720\,a_7 \\ \scriptstyle B_{6,2}=360\,a_4^{\,2}+720\,a_2\,a_6+720\,a_3\,a_5 \\ \scriptstyle B_{6,3}=120\,a_3^{\,3}+360\,a_2^{\,2}\,a_5+720\,a_2\,a_3\,a_4 \\ \scriptstyle B_{6,4}=120\,a_2^{\,3}\,a_4+180\,a_2^{\,2}\,a_3^{\,2} \\ \scriptstyle B_{6,5}=30\,a_2^{\,4}\,a_3 \\ \scriptstyle B_{6,6}=a_2^{\,6} \\ \scriptstyle b_7=4\,a_4^{\,2}-12\,a_3^{\,3}-a_7+132\,a_2^{\,6}+8\,a_2\,a_6+8\,a_3\,a_5-36\,a_2^{\,2}\,a_5+120\,a_2^{\,3}\,a_4-330\,a_2^{\,4}\,a_3-72\,a_2\,a_3\,a_4+180\,a_2^{\,2}\,a_3^{\,2} \\ \scriptstyle B_{7,1}=5040\,a_8 \\ \scriptstyle B_{7,2}=5040\,a_2\,a_7+5040\,a_3\,a_6+5040\,a_4\,a_5 \\ \scriptstyle B_{7,3}=2520\,a_2\,a_4^{\,2}+2520\,a_3^{\,2}\,a_4+2520\,a_2^{\,2}\,a_6+5040\,a_2\,a_3\,a_5 \\ \scriptstyle B_{7,4}=840\,a_2\,a_3^{\,3}+840\,a_2^{\,3}\,a_5+2520\,a_2^{\,2}\,a_3\,a_4 \\ \scriptstyle B_{7,5}=210\,a_2^{\,4}\,a_4+420\,a_2^{\,3}\,a_3^{\,2} \\ \scriptstyle B_{7,6}=42\,a_2^{\,5}\,a_3 \\ \scriptstyle B_{7,7}=a_2^{\,7} \\ \scriptstyle b_8=9\,a_2\,a_7-429\,a_2^{\,7}-a_8+9\,a_3\,a_6+9\,a_4\,a_5+165\,a_2\,a_3^{\,3}-45\,a_2\,a_4^{\,2}-45\,a_3^{\,2}\,a_4-45\,a_2^{\,2}\,a_6+165\,a_2^{\,3}\,a_5-495\,a_2^{\,4}\,a_4+1287\,a_2^{\,5}\,a_3-90\,a_2\,a_3\,a_5-990\,a_2^{\,3}\,a_3^{\,2}+495\,a_2^{\,2}\,a_3\,a_4 \\ \scriptstyle B_{8,1}=40320\,a_9 \\ \scriptstyle B_{8,2}=20160\,a_5^{\,2}+40320\,a_2\,a_8+40320\,a_3\,a_7+40320\,a_4\,a_6 \\ \scriptstyle B_{8,3}=20160\,a_3\,a_4^{\,2}+20160\,a_3^{\,2}\,a_5+20160\,a_2^{\,2}\,a_7+40320\,a_2\,a_3\,a_6+40320\,a_2\,a_4\,a_5 \\ \scriptstyle B_{8,4}=1680\,a_3^{\,4}+6720\,a_2^{\,3}\,a_6+10080\,a_2^{\,2}\,a_4^{\,2}+20160\,a_2\,a_3^{\,2}\,a_4+20160\,a_2^{\,2}\,a_3\,a_5 \\ \scriptstyle B_{8,5}=1680\,a_2^{\,4}\,a_5+3360\,a_2^{\,2}\,a_3^{\,3}+6720\,a_2^{\,3}\,a_3\,a_4 \\ \scriptstyle B_{8,6}=336\,a_2^{\,5}\,a_4+840\,a_2^{\,4}\,a_3^{\,2} \\ \scriptstyle B_{8,7}=56\,a_2^{\,6}\,a_3 \\ \scriptstyle B_{8,8}=a_2^{\,8} \\ \scriptstyle b_9=55\,a_3^{\,4}-a_9+5\,a_5^{\,2}+1430\,a_2^{\,8}+10\,a_2\,a_8+10\,a_3\,a_7+10\,a_4\,a_6-55\,a_3\,a_4^{\,2}-55\,a_3^{\,2}\,a_5-55\,a_2^{\,2}\,a_7+220\,a_2^{\,3}\,a_6-715\,a_2^{\,4}\,a_5+2002\,a_2^{\,5}\,a_4-5005\,a_2^{\,6}\,a_3-110\,a_2\,a_3\,a_6-110\,a_2\,a_4\,a_5-1430\,a_2^{\,2}\,a_3^{\,3}+330\,a_2^{\,2}\,a_4^{\,2}+5005\,a_2^{\,4}\,a_3^{\,2}+660\,a_2\,a_3^{\,2}\,a_4+660\,a_2^{\,2}\,a_3\,a_5-2860\,a_2^{\,3}\,a_3\,a_4 \\ \end{array}$$

This means that the Lagrange inversion algorithm works for all unital commutative rings $$R$$ of all characteristics and the above formulae are kind of universal. The respective wiki-articles (especially the section in Lagrange inversion theorem starting with „If f is a formal power series, then the above formula does not give ...“) could be formulated without any effective division (except the one by $$a_1$$) (and as far as I see even the recurrence to the Bell polynomials could remain!) so that every reader is able to immediately see this conclusion – a modification which would be no big effort at all. The only remaining problem being the so far missing qualified source (although there are lots of articles around on the subject, and I will continue to look for a qualified one which contains the universal message).

Finally, I interpret one of your last sentences „the Lagrange inversion theorem is not(?) valid over arbitrary fields“ as kind of a misunderstanding: the Lagrange inversion theorem IS valid, and not only over arbitrary fields, but even over the mentioned arbitrary rings. --Nomen4Omen (talk) 06:58, 11 May 2018 (UTC)

Is { X+X²(...) in Z[[X]] } a group for composition?
It seems that if T(0) = 0, T'(0) = 1 for $$T\in\ZX$$, then Res(1/T^k) is always divisible by k and therefore the compositional inverse of T exists in $$\ZX$$ and has again these properties. Is this well-known? Is there a simple proof for this? &mdash; MFH:Talk 15:08, 16 December 2022 (UTC)
 * If you write explicitely the composition as
 * $$(X+a_2X^2+\cdots) + b_2(X+a_2X^2+\dots)^2 +\cdots,$$
 * you can see immediately that the coefficient of degree $n$ of the composition is $$a_n + \cdots +b_n,$$ where the dots represents a polynomial in the $$a_i$$ and $$b_i$$ for which $$i<n.$$ So, the compositional inverse can be computed recursively by equating to zero these polynomials and supposing that the $$a_i$$ are known and the $$b_i$$ are unknown (or vice versa). This does not only provides a computational procedure, but also proves the existence of inverses (as a series of this type has a left inverse and a right inverse, they are necessary equal). D.Lazard (talk) 19:09, 16 December 2022 (UTC)
 * you can see immediately that the coefficient of degree $n$ of the composition is $$a_n + \cdots +b_n,$$ where the dots represents a polynomial in the $$a_i$$ and $$b_i$$ for which $$i<n.$$ So, the compositional inverse can be computed recursively by equating to zero these polynomials and supposing that the $$a_i$$ are known and the $$b_i$$ are unknown (or vice versa). This does not only provides a computational procedure, but also proves the existence of inverses (as a series of this type has a left inverse and a right inverse, they are necessary equal). D.Lazard (talk) 19:09, 16 December 2022 (UTC)

QUESTION: division theorem
I have expected some words on division theorem: for a formal power series f and a polynomial P there exist unique power series q and polynomial R with deg(R)<deg(P) such that f=qP+R. Is this not true or of no use? I thought this could help defining f(A) for a square matrix A, in particular exp(A). And lots of nice functions after, including log and Baker-Champbell-Hausdorff formula... I hope this will trigger an add-on to this article... if so, PLEASE signal to me 0ctavte0 (talk) 07:06, 30 May 2023 (UTC)


 * This is just false. --JBL (talk) 20:41, 30 May 2023 (UTC)

Formal power series raised to rational powers
I would like to suggest that formal power series can also be raised to any rational number, and that this follows from the fact of being able to raise them to integral powers.

When I first considered adding this concept to the formal power series page, I thought it wouldn't be allowed as it appears that it has not previously been published. However, it is a matter of fairly simple algebra to verify it, so I am asking for a ruling on whether it can be included.

I have also discovered in the History section of the Binomial series page, that Newton and Wallis achieved similar results for more restricted series. So, this more generalized result is not entirely new. Pjpline (talk) 03:00, 19 October 2023 (UTC)


 * The problem is not new and not as simple as you suggest. For example, the ⇭⇭⇭th root of a formal power series is not a series if the constant term is zero. In fact, the subject is considered in Puiseux series. D.Lazard (talk) 07:53, 19 October 2023 (UTC)
 * Thank you, that is a very interesting page, which is worthy of study.
 * I see that I wasn't clear enough. I am suggesting that it is possible to derive $$c_m$$ as a solution to
 * $$ \left( \sum_{k=0}^\infty a_k X^k \right)^{\!\frac{1}{n}} =\, \sum_{m=0}^\infty c_m X^m,$$
 * where n, k and m are integers, with similar restrictions to those specified for power series raised to powers. For example:
 * $$ \left( 1 + X + X^2 \right)^{\!\frac{1}{3}} =\ 1 + \frac{X}{3} + \frac{2X^2}{9} - \frac{13X^3}{81} + \frac{8X^4}{243} + \frac{37X^5}{729} - \frac{379X^6}{6561} + \cdots$$ Pjpline (talk) 06:06, 20 October 2023 (UTC)
 * "Similar restrictions" is not sufficient, since there is no nth root in terms of power series if $$a_0=0.$$ D.Lazard (talk) 10:38, 20 October 2023 (UTC)
 * Restrictions:
 * $$c_0 = a_0^\frac{1}{n}$$ exists in the ring of coefficients.
 * $$c_0^{n-1}$$ exists and is invertible in the ring of coefficients.
 * n is invertible in the ring of coefficients. Pjpline (talk) 20:38, 20 October 2023 (UTC)
 * Since there isn't an immediate "No" here is an explanation of the method. It uses some concepts which appear complicated but which are actually quite simple.
 * The first is an extension to the partitions of an integer, making fixed length partitions by padding shorter partitions with zero(s). The partitions have an added restriction that no part is allowed to exceed a specified value. I use $$\overset{k}{\underset{n}{ \operatorname{pe}(m)}}$$ to represent the set of partitions of m with length n, where each part is limited to k.
 * $$\begin{align} \overset{k}{\underset{n}{ \operatorname{pe}(0)}} &= \{ 0^n \}, \ \ k \geq 0 , n \geq 1 && n \text{-partitions of zero} \\ \overset{k}{\underset{1}{ \operatorname{pe}(m)}} &= \{ m^1 \} , \ \  k \geq m , m \geq 0 && 1 \text{-partitions of } m \text{ with limit } k \\ \overset{k}{\underset{n}{ \operatorname{pe}(m)}} &= \bigcup_{i = \left \lceil \frac{m}{n} \right \rceil}^{\min(k,m)} \left ( \bigcup_{P \in \left ( \overset{i}{\underset{n - 1}{ \operatorname{pe}(m - i)}} \right ) } \{ i P \} \right ) , \ \ n > 1 , m \leq kn && n \text{-partitions of } m \text{ with limit } k \\ \overset{3}{\underset{4}{ \operatorname{pe}(5)}} &= \{ 3^12^10^2, 3^11^20^1, 2^21^10^1, 2^11^3 \} && \text{example} \end{align}$$
 * The second is a function that takes a partition set and an ordered list of coefficients and outputs a new coefficient:
 * $$\begin{align} tx \left ( \mathcal{P}, (a_k) \right ) &= \sum_{P \in \mathcal{P}} \left ( \left ( \sum_{p \in P} m(p) \right ) ! \prod_{p \in P} \frac{a_p^{m(p)}}{m(p)!} \right ) && m(p) \text{ is the count of part } p \\ tx \left ( \overset{3}{\underset{4}{ \operatorname{pe}(5)}} , (a_k) \right ) &= 4!\frac{a_3a_2a_0^2}{2!} + 4!\frac{a_3a_1^2a_0}{2!} + 4!\frac{a_2^2a_1a_0}{2!} + 4!\frac{a_2a_1^3}{3!} && \text{example} \\ \end{align}$$
 * Note that when $$\mathcal{P}$$ is a set of n-partitions, $$\sum_{p \in P} m(p)$$ is always n.
 * Using these, we have:
 * $$\begin{align} \left( \sum_{k=0}^\infty a_k X^k \right)^{\!n} &= \sum_{m=0}^\infty c_m X^m \\ c_m &= \operatorname{tx} \left ( \overset{m}{\underset{n}{ \operatorname{pe}(m)}}, (a_k) \right ) \\ \end{align}$$
 * Now, to find $$c_m$$ to satisfy
 * $$\begin{align} \left( \sum_{k=0}^\infty a_k X^k \right) ^\frac{1}{n} &= \sum_{m=0}^\infty c_m X^m, n>1 && \text{potentially multiple solutions} \\ \sum_{k=0}^\infty a_k X^k  &= \left( \sum_{m=0}^\infty c_m X^m \right) ^{\!n} && \text{as a possibility} \\ a_m &= \operatorname{tx} \left ( \overset{m}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right ) && (c_j) \text{ to be solved} \\ a_0 &= \operatorname{tx} \left ( \overset{0}{\underset{n}{ \operatorname{pe}(0)}} , (c_j) \right ) \\ &= n!\frac{c_0^n}{n!} \\ c_0 &= a_0^\frac{1}{n} && \text{possible multi-choice, obvious restrictions} \\ a_1 &= \operatorname{tx} \left ( \overset{1}{\underset{n}{ \operatorname{pe}(1)}} , (c_j) \right ) \\ &= n!\frac{c_1c_0^{n - 1}}{{(n - 1)}!} \\ c_1 &= \frac{a_1}{nc_0^{n-1}} && \text{obvious restrictions} \\ a_2 &= \operatorname{tx} \left ( \overset{2}{\underset{n}{ \operatorname{pe}(2)}} , (c_j) \right ) \\ &= n!\frac{c_2c_0^{n - 1}}{{(n - 1)}!} + n!\frac{c_1^2c_0^{n - 2}}{{2!(n - 2)}!} \\ c_2 &= \frac{a_2 - \frac{n(n - 1)c_1^2c_0^{n-2}}{2}}{nc_0^{n-1}} && \text{obvious restrictions} \\ \overset{m}{\underset{n}{ \operatorname{pe}(m)}} &= \{m0^{n-1}\} \cup \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}} , \ \ n \geq 1, m  \geq 1 && \text{verify from definition}\\ a_m &= \operatorname{tx} \left ( \{m0^{n-1}\} \cup \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}} , (c_j) \right )  , \ \ n \geq 1, m  \geq 1 \\ &= \operatorname{tx} \left ( \{m0^{n-1}\} , (c_j) \right ) + \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}} , (c_j) \right ) \\ &= nc_mc_0^{n - 1} + \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}} , (c_j) \right ) \\ nc_mc_0^{n - 1} &= a_m - \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}} , (c_j) \right ) \\ c_m &= \frac{a_m - \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}} , (c_j) \right )}{nc_0^{n - 1}} && \text{exact for chosen } c_0 \\ \end{align}$$
 * The question is, would it be allowable to add a less verbose version of this to the Formal power series page? Pjpline (talk) 04:43, 1 November 2023 (UTC)
 * These formulas seem to be WP:OR, as you do not provide any published soure for them. So, it is not allowable to add them to the article, even in a simplified form. D.Lazard (talk) 10:44, 1 November 2023 (UTC)
 * I can't say that was unexpected. I guess the world will have to wait for someone else to discover and publish. Thank you for considering. Pjpline (talk) 23:14, 4 November 2023 (UTC)
 * If it is new, why don't you publish it (outside Wikipedia)? - Jochen Burghardt (talk) 09:46, 5 November 2023 (UTC)
 * Since I don't have any formal education, I am unaware of any means of doing so. Pjpline (talk) 22:07, 5 November 2023 (UTC)