Talk:Formal power series

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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)

?

Contents

[edit] 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 (ak).

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.)

— MFH: Talk 12:53, 26 Apr 2005 (UTC)

[edit] 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 April 2007 (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)

[edit] 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 Image:User_dmharvey_sig.png 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 Image:User_dmharvey_sig.png Talk 12:24, 26 Jun 2005 (UTC)

[edit] 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+... = −1 and 1−1+1−1+1−1+... = 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−1 which has the unique solution S = −1. T=1−1+1−1+1−1+... satisfies T−1 = −T. The geometric series 1/(1−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−1 is the same as
2∞ = ∞−1,
as S=∞. 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= −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 − 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 = ∞ then algebra doesn't work, but consider the alternative: that algebra does work. Then S = −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 − 1 + 1 − 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 −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 ∞. 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 ∞ 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−X)f(X)=1

or

(1−X+0X^2+0X^3+...)(1+X+X^2+X^3+...)=1+0X+0X^2+0X^3+...

or

(1,−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−X)f(X) = 1

provides the information needed to define the undefined. Formally substituting X=2 gives

(−1)f(2) = 1

or

f(2) = −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) = −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 −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#Interpreting_formal_power_series_as_functions .

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→ 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−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)


[edit] R[[x]][[y]] 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)