The Unapologetic Mathematician

Mathematics for the interested outsider

Power Series

Prodded by some comments, I think I’ll go even further afield from linear algebra. It’s a slightly different order than I’d originally thought of, but it will lead to some more explicit examples when we’re back in the realm of linear algebra, so it’s got its own benefits.

I’ll note here in passing that mathematics actually doesn’t proceed in a straight line, despite the impression most people get. The lower-level classes are pretty standard, yes — natural-number arithmetic, fractions, algebra, geometry, calculus, and so on. But at about this point where most people peter out, the subject behaves more like an alluvial fan — many parallel rivulets carry off in different directions, but they’re all ultimately part of the same river. So in that metaphor, I’m pulling a bit of an avulsion.

Anyhow, power series are sort of like polynomials, except that the coefficients don’t have to die out at infinity. That is, when we consider the algebra of polynomials \mathbb{F}[X] as a vector space over \mathbb{F} it’s isomorphic to the infinite direct sum

\displaystyle\mathbb{F}[X]\cong\bigoplus\limits_{k=0}^\infty\mathbb{F}X^k

but the algebra of power series — written \mathbb{F}[[X]] — is isomorphic to the infinite direct product

\displaystyle\mathbb{F}[[X]]\cong\prod\limits_{k=0}^\infty\mathbb{F}X^k

It’s important to note here that the X^i do not form a basis here, since we can’t write an arbitrary power series as a finite linear combination of them. But really they should behave like a basis, because they capture the behavior of every power series. In particular, if we specify that \mu(X^m,X^n)=X^{m+n} then we have a well-defined multiplication extending that of power series.

I don’t want to do all the fine details right now, but I can at least sketch how this all works out, and how we can adjust our semantics to talk about power series as if the X^i were an honest basis. The core idea is that we’re going to introduce a topology on the space of polynomials.

So what polynomials should be considered “close” to each other? It turns out to make sense to consider those which agree in their lower-degree terms to be close. That is, we should have the space of tails

\displaystyle\bigoplus\limits_{k=n+1}^\infty\mathbb{F}X^k

as an open set. More concretely, for every polynomial p with degree n there is an open set U_p consisting of those polynomials q so that X^{n+1} divides the difference q-p.

Notice here that any power series defines, by cutting it off after successively higher degree terms, a descending sequence of these open sets. More to the point, it defines a sequence of polynomials. If the power series’ coefficients are zero after some point — if it’s a polynomial itself — then this sequence stops and stays at that polynomial. But if not it never quite settles down to any one point in the space. Doesn’t this look familiar?

Exactly. Earlier we had sequences of rational numbers which didn’t converge to a rational number. Then we completed the topology to give us the real numbers. Well here we’re just doing the same thing! It turns out that the topology above gives a uniform structure to the space of polynomials, and we can complete that uniform structure to give the vector space underlying the algebra of power series.

So here’s the punch line: once we do this, it becomes natural to consider not just linear maps, but continuous linear maps. Now the images of the X^k can’t be used to uniquely specify a linear map, but they will specify at most one value for a continuous linear map! That is, any power series comes with a sequence converging to it — its polynomial truncations — and if we know the values f(X^k) then we have uniquely defined images of each of these polynomial truncations since each one is a finite linear combination. Then continuity tells us that the image of the power series must be the limit of this sequence of images, if the limit exists.

About these ads

August 18, 2008 - Posted by | Algebra, Power Series, Ring theory

9 Comments »

  1. While the introduction of topology here is easy to motivate (just as you explained), it’s also an instance of a general phenomenon, where topology and topological completions are used to extend dualities between structures ‘of finite type’ to more general dualities between “pro-objects” and “ind-objects”. A very famous example (which I am blogging about, very gradually) is Stone duality: there is a duality of finite type between finite sets and finite Boolean algebras, and this is extended to a duality between topological projective limits of finite sets (called Stone spaces) and general inductive limits of finite Boolean algebras, which are general Boolean algebras.

    In the present case, we have a perfect duality between finite-dimensional algebras and finite-dimensional coalgebras. Then, on the ind-object side, the space of polynomials P carries a coalgebra structure (with comultiplication x^n \mapsto \sum_{i + j = n} x^i \otimes x^j), which can be construed as an inductive limit of finite-dimensional coalgebras.

    On the pro-object side, the vector space dual P^* of the space of polynomials is, as you say in the post, the space of formal power series. The algebra structure on P^* can be gotten either purely algebraically, defining the multiplication as a composite

    P^* \otimes P^* \to (P \otimes P)^* \to P^*

    where the first arrow is a canonical map and the second is the transpose of the comultiplication on P. Or, in line with the general phenomenon, the multiplication can be defined topologically (just as you have done), and this corresponds to taking a (dual) projective limit of (dual) finite-dimensional algebras, equipped with suitable uniform structures.

    Comment by Todd Trimble | August 18, 2008 | Reply

  2. This is all true, except I haven’t defined the space of power series as the dual of the space of polynomials, though an astute reader could figure that much out from the general yoga of direct products and sums. Even so, it wouldn’t really help, since I haven’t remotely talked about comultiplications.

    Still, good points.

    Comment by John Armstrong | August 18, 2008 | Reply

  3. [...] a little while we’re going to want to talk about “evaluating” a power series like we did when we considered polynomials as functions. But when we try to map into our base [...]

    Pingback by Some Topological Fields « The Unapologetic Mathematician | August 26, 2008 | Reply

  4. [...] that a power series is like an infinite polynomial. In fact, we introduced a topology so we could see in any power [...]

    Pingback by Evaluating Power Series « The Unapologetic Mathematician | August 27, 2008 | Reply

  5. [...] Series Expansions Up to this point we’ve been talking about power series like , where “power” refers to powers of . This led to us to show that when we evaluate [...]

    Pingback by Power Series Expansions « The Unapologetic Mathematician | September 15, 2008 | Reply

  6. [...] of Power Series Formally, we defined the product of two power series to be the series you get when you multiply out all the terms and [...]

    Pingback by Products of Power Series « The Unapologetic Mathematician | September 22, 2008 | Reply

  7. [...] and there I decided to stop what I was working on about linear algebra. Instead, I set off on power series and how power series expansions can be used to express analytic functions. Then I showed how power [...]

    Pingback by Pi: A Wrap-Up « The Unapologetic Mathematician | October 16, 2008 | Reply

  8. Hello,

    I’ve posted in sci.math.research the follwoing question: Consider a power series sum a_n x^n that is convergent for all real x, thus defining a function f: R \to R.
    Are there criteria for the a_n to decide whether f is bounded?

    Can the topology you defined above give answers to the question how “sparse” the subspace of all bounded power series is within the set of al convergent power series (in tems of cardinality or measure or Baire category or dimensionality)?

    Thank you!
    Andreas

    Comment by Andreas Rüdinger | April 2, 2009 | Reply

  9. I can’t say as I know, Andreas. But maybe another commenter around here might?

    Comment by John Armstrong | April 2, 2009 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 393 other followers

%d bloggers like this: