Re: Extending the reals


cbrown@xxxxxxxxxxxxxxxxx schrieb:

hagman wrote:
One needs to distinguish R[[X]] from R(X), i.e.
one need a power series that is not a rational function,
for example the exponential (contrary to my former post, I view X as
the infinitesimal unit here, not the infinite unit; but that's just to
simplify notation).

Yes; I understand: we are simply considering R[[X]] and R(X) as
polynomials with some total ordering.

I should note that the only exposure I've gotten in any depth regarding
formal power series was from a quick introduction in Stanley's
"Enumerative Combinatorics", where he just needs to describe enough of
their properties to make them useful for describing generating
functions; and to establish a calculus such that F(X) + G(X),
F(X)*G(X), and F(G(X)) all have meaning.

So just to make sure I remember this correctly (since you refer to it
below):

In Stanley's description, the topology is one where a sequence of
elements of R[[X]] {Fi(X)} = (F1(X), F2(X), ..., Fn(X), ...) converges
to the element F(X) according to the following rules:

Let a_(i, n) be the coeffcient of X^n in Fi(X); and a_n be the similar
coefficient in F(X).

Then for every n, there is a natural j such that for all k > j, a_(k,
j) = a_j.

This should probably read:
Then for every n, there is a natural j such that for all k > j, a_(k,n)
= a_n.


This is different than requiring that the limit k->oo {a_(k, j)} = a_j;

.... at least if one takes the standard topology on R.
If you take discrete topology on R, both notions coincide.
In other words, for power series one uses pointwise convergence of
coefficients with respect
to discrete topology on the base ring.

and I presume this is equivalent to your statement below "the fact that
powers of X are small"?

Exactly.
For all n there is a j_n such that for all k>j_n, a_(k,n) = a_n
<=>
For all n there is a m (namely m=max{j_0, j_1, ..., j_n}) such that
for all k>m, Fk(X)-F(X) is a multiple of X^n.

Usually one would want the latter to read
For all eps>0 there is a m such that for all k>m
|Fk(X)-F(X)| < eps.

Therefore one may say "something is mindbogglingly small"
instead of "something is divisible by a mindbogglingly high power of
X".

Or do you use a different notion of convergence
than Stanley defines?

Fortunately not.
When I started reading your post I got shocked that I might have
erroneously ignored the topology of R in R[[X]].
Well, partly I have as we do use the topology of R in R(X).
Interestingly, this poses no problem - see below.


Let S be the subset of R(X) consisting of all f(X)/(1+X*g(X)).
Since 1+X*g(X) is invertible in R[[X]], ...

... because an element of R[[X]] is invertible iff its "constant
coefficient" is non-zero...

exactly. And of course we can hide any non-zero constant in f if it
should be
different from 1.


...we can view S as (part of) the
intersection of R(X) and R[[X]]...

... because surely (1 + X*g(X) designates a particular element in R(X).

("Fortunately", the topology on R[[X]]
is also defined by the fact that powers of X are small).

See above. In any case, I assume you imply that the open sets of S =
R[[X]] intersect R(X) are the open sets of R(X) intersect S, which are
the same as the open sets of R[[X]] intersect S; and so if {Fi(X)}
converges to F in R[[X]], it also converges to the equivalent member of
R(X) in R(X).

Well, at least I thought so.
But since R(X) respects the standard topology of R and R[[X]]
implicitly uses discrete topology on R, a sequence converging in R(X)
need not necessarily converge in R[[X]], or does it?
Well, a zero sequence in R(X) will have to have almost all members
smaller than eps for any given positive eps.
But every eps>0 can be written as
r*X^m*(1+X*f(X))/(1+X*g(X)) with m in Z, r>0, f(X), g(X) polynomials
(check that!).
But being <eps in absolute value
- implies being divisible by X^m and
- is implied by being divisible by X^(m+1)
Thus respecting or ignoring the standard topology of R made no
difference.
Lucky me.

Note that this implies that 1/n does not converge to 0.
Indeed, 1/n will never be less than X.
But that does not show incompleteness, as 1/n is no Cauchy sequence...


One can show that the series (a_n) with a_n = sum_{k=0,...,n} X^n/n!
is a Cauchy sequence.

.... because differences between "late" elements of the sequence are
divisible
by high powers of X.

Assume it converges against an element f(X)/g(X) of R(X) where f(X) and
g(X) have no irreducible factors in common.
Obviously, the limit has to be finite (namely, very close to a_0 = 1
but obviuously different from a_0).
Hence we can assume that g(X) = 1 + X*h(X).
But then f(X)/g(X) is in S and we must obtain the same limit in R[[X]].
But there we have exp(X) as limit.
Hence f(X) = g(X)*exp(X) as formal power series.

With you so far...

The equality must still hold when we insert a complex number for X as
long as we have convergence.

I am definitely missing a theorem in your toolbox. Can you amplify on
this assertion?

Maybe a less invasive tool would have sufficed if it had come to my
mind in time:.
The (or at least almost all) coefficients of the formal power series of
f(X)/g(X) fulfill a linear recursion (in a similar fashion as the
decimal expansion of fractions is eventually periodic).
This is clear from writing down the condition that the coefficient of
X^n in g(X) * (f(X)/g(X)) must be 0 if n exceeds deg(f).
Thus automatically f(X)/g(X) cannot equal exp(X), where the
coefficients do /not/ follow a linear recursion (needs to be checked).
This much simpler trick works with any formal power series instead of
exp(X),
as long as there is no linear recursion for the coefficients.
It works even for those series that do not converge when evaluated at
any nonzero number.
Now that I've thought of it I'm much happier with this proof method
than my previous one...

A much simpler power series that /obviously/ doesn't allow a linear
recursion for its coefficients would be sum_(n>=0) X^(n^2), which has
arbitrary long blocks of 0's.
If I had that used that series in the proof, the proof would end a
couple of lines earlier
and without using properties of R or C.


Are you saying that f(X) and g(X) /must/ belong to the subset of R[[X]]
where every series converges in C when X is replaced by any complex
number?

I admit that I magically (but legally) went from R[[X]] to C[[X]].
What we do have is:
If z is in C, then in C[[X]] we have the subring of series converging
at z.
Then setting X=z is a homomorphism of that subring to C.

Fortunately, f(X) and g(X) are polynomials and exp is a very happy
converger.

But according to my remarks above you may ignore this part anyway.


Since exp converges everywhere and has no
zeroes, we see that f(X) and g(X) have the same zeroes (in same
multiplicity) in C, hence have the same irreducible factors.

Since they have no irreducible factors in common, it follows that they
f(X) and g(X) are constant, i.e. the limit is indeed real.
But as the limit is infinitesimal close to but different from 1, we
have a contradiction.


Cheers - Chas

.

Relevant Pages

  • Re: Extending the reals
    ... We want to show that Rin not complete under the order topology. ... formal power series was from a quick introduction in Stanley's ... I see now what you meant by: Divisibility by high powers of X in R ... The coefficients of the formal power series of ...
    (sci.math)
  • Re: Extending the reals
    ... We want to show that Rin not complete under the order topology. ... formal power series was from a quick introduction in Stanley's ... I see now what you meant by: Divisibility by high powers of X in R ... The coefficients of the formal power series of ...
    (sci.math)
  • inverse of power series with prime power coefficients
    ... The power series with prime coefficients ... Consider instead the power series with prime power coefficients: ... absolute value of the ratio of consecutive coefficients of QQ ...
    (sci.math)
  • Re: need some help with limits
    ... David C. Ullrich wrote: ... so you know that it's given by a power series. ... limit of the coefficients in the power series for f_k; ...
    (sci.math)
  • Re: Infinite Series in Infinite Series
    ... infinite series. ... coefficients a_n, ... If we sere to write has a power series, ... subscript n) be the coefficients of h. ...
    (sci.math)
Quantcast