A few months ago I posted a question about analytic functions that I received from a bright high school student, which turned out to be studied and resolved by de Bruijn. Based on this positive resolution, I thought I might try my luck again and list three further questions that this student asked which do not seem to be trivially resolvable.
- Does there exist a smooth function
which is nowhere analytic, but is such that the Taylor series
converges for every
? (Of course, this series would not then converge to
, but instead to some analytic function
for each
.) I have a vague feeling that perhaps the Baire category theorem should be able to resolve this question, but it seems to require a bit of effort. (Update: answered by Alexander Shaposhnikov in comments.)
- Is there a function
which meets every polynomial
to infinite order in the following sense: for every polynomial
, there exists
such that
for all
? Such a function would be rather pathological, perhaps resembling a space-filling curve. (Update: solved for smooth
by Aleksei Kulikov in comments. The situation currently remains unclear in the general case.)
- Is there a power series
that diverges everywhere (except at
), but which becomes pointwise convergent after dividing each of the monomials
into pieces
for some
summing absolutely to
, and then rearranging, i.e., there is some rearrangement
of
that is pointwise convergent for every
? (Update: solved by Jacob Manaker in comments.)
Feel free to post answers or other thoughts on these questions in the comments.
31 comments
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14 September, 2021 at 5:04 pm
MyMathYourMath
Dr Tao, small question if the map was from complex to complex then this wouldn’t even be a question correct?
14 September, 2021 at 5:35 pm
Rex
For question 2, do we know how to construct such a function meeting, not all real polynomials, but just the linear ones
for
?
15 September, 2021 at 12:29 am
Anonymous
Indeed, if
and
, then
forms a discrete subset of R (since in some small neighborhood of $x_a$ the function
has no zeroes other than
and hence this neighborhood contains no
for
), and so only countably many polynomials of the form
can be accommodated. This argument also does not require any global continuity or smoothness of f.
15 September, 2021 at 12:38 am
Aditya Guha Roy
This argument looks nice.
15 September, 2021 at 9:22 pm
Anonymous
Why are we looking at the zeroes of $ latex f$?
14 September, 2021 at 5:48 pm
Guilherme Rocha de Rezende
Maybe we can answer second one using interpolation polynomials….
14 September, 2021 at 5:49 pm
Guilherme Rocha de Rezende
x_{0} selects the polynomials.
Em ter., 14 de set. de 2021 às 22:48, Guilherme Rocha de Rezende escreveu:
> Maybe we can answer second one using interpolation polynomials…. > > Em ter., 14 de set. de 2021 às 21:34, What’s new <
14 September, 2021 at 6:23 pm
Guilherme Rocha de Rezende
I am a physicist and have good intuition but your technique is much better than mine.
Em ter., 14 de set. de 2021 às 23:18, Guilherme Rocha de Rezende escreveu:
> We must pass to continuous coefficients…..perhaps discretizing the > coefficients and take the limit to approximate the continuous case. > > Em ter., 14 de set. de 2021 às 22:49, Guilherme Rocha de Rezende 1434903@etfbsb.edu.br> escreveu: > >> x_{0} selects the polynomials. >> >> Em ter., 14 de set. de 2021 às 22:48, Guilherme Rocha de Rezende > 1434903@etfbsb.edu.br> escreveu: >> >>> Maybe we can answer second one using interpolation polynomials…. >>> >>> Em ter., 14 de set. de 2021 às 21:34, What’s new <
14 September, 2021 at 7:25 pm
Adam Denchfield
For 3), physicists encounter this situation of a power series diverging everywhere except x=0 fairly often. The concept of resummation has been refined tremendously over the decades, since every physical problem of interest involves power series whose coefficients grow as fast or faster than n!. One of the simplest models exhibiting this is doing perturbation theory for the lowest eigenvalue of the anharmonic oscillator, talked about extensively by folks like Carl Bender. Borel resummation can be used here, involving taking the Taylor series coefficients, dividing them by n!, evaluating the modified power series then, and performing an integration against a Borel kernel to get the ‘resummed’ form of the function. For more: https://en.wikipedia.org/wiki/Borel_summation
Perhaps this process could be reinterpreted as the desired above reorganization of the Taylor series terms. On the other hand, I think the equivalence of summation methods that preserve order of summation is proved in Bender & Orszag’s text, and I don’t recall if Borel resummation falls under this roof.
14 September, 2021 at 7:43 pm
Aleksei Kulikov
Answer to the second question is negative, below are 1,5 proofs of it:
1)Cheap argument via Sard’s theorem — you can not even serve all the constant polynomials. Indeed, if
,
, then
is a critical value of
, so the measure of possible
‘s is
and it can’t be all of
. Excluding constants also wouldn’t work as
for all
can’t work either (consider
, then the same argument works for
, and
gives us at most one more polynomial).
1.5)Here’s an argument which is more appealing to my analysis soul, but which I’m too lazy to make rigorous since we already have an easy proof.
Consider the curve
— it’s a smooth
-dimensional curve in
(since
, so no space-filling curves possible). Each polynomial
also gives us a smooth curve in this space. By the assumption these curves intersect. Now it remains to pick a
-dimensional family of polynomials which is generic enough, then there’s no way our smooth curve intersect every curve in this family just by dimension counting (rigorous argument here would probably repeat many steps from the proof of Sard’s theorem, though).
15 September, 2021 at 12:22 am
Aditya Guha Roy
This was also the argument I thought in mind; but for applying Sard’s theorem we need
to be smooth everywhere, but I don’t see any easy way to infer that about
from the given conditions.
15 September, 2021 at 11:05 am
Terence Tao
Nice argument! Here is a variant that works for any two-parameter set of polynomials, such as the linear polynomials
for sake of argument, and avoids Sard’s theorem: if a smooth function
was tangent to every linear polynomial
, then the smooth map
would be a surjective smooth map from
onto
. But this is impossible (Lipschitz maps cannot increase Hausdorff dimension).
14 September, 2021 at 10:06 pm
Harry West
A negative answer to 1 is a theorem of Boas: https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-41/issue-4/A-theorem-on-analytic-functions-of-a-real-variable/bams/1183498133.full
14 September, 2021 at 10:37 pm
Jacques
No it does not answer the question. The result you cite consider the case where the radii of convergence are uniformly lower bounded by a positive constant.
This tells at least that if such a function exists then there is no possible lower bound of these radii of convergence.
14 September, 2021 at 10:22 pm
Alexander Shaposhnikov
Let
be the set of
such that for all
,
. It is clear that
is closed and
equals the union of all
. Then by the Baire category theorem there exists
such that
contains an interval
. Without loss of generality one may assume that
. Now one can see that
is analytic on
, e.g. using the the Lagrange form for the remainder in Taylor's theorem.
15 September, 2021 at 11:05 am
Terence Tao
Nice! A good textbook use of the Baire category theorem.
14 September, 2021 at 11:48 pm
Aditya Guha Roy
Negative answer to (2.)
exists. Then for each constant polynomial
one must have a solution
to the system of equations
and thus each real number must be a critical value of
; but according to Sard’s theorem the set of critical values of
must have measure 0, a contradiction!
Assume to the contrary that such a function
15 September, 2021 at 12:21 am
Aditya Guha Roy
Now, I find that my solution above is same as one posted much earlier in the comments.
Unfortunately, this argument won’t hold if our function
is not smooth.
15 September, 2021 at 4:23 pm
Nkv
Does Sard theorem work if restrict the domain to smooth points?
15 September, 2021 at 7:51 pm
Aditya Guha Roy
No, we need the domain to be a manifold.
15 September, 2021 at 4:15 am
Aditya Guha Roy
Reblogged this on Aditya Guha Roy's weblog.
15 September, 2021 at 11:31 am
David Fry
Terry, I think its great that you are open to the younger generation presenting mathematical concepts to you and the above-commenters who are all experts within their discipline. With schools at a virtual standstill because of COVID, I know that you can think of additional innovative ways to get younger students involved in math/science on a much larger scale. Thank you,
David
15 September, 2021 at 8:01 pm
Aditya Guha Roy
I really like the kind of mathematics questions which are asked by this high school student. Back during my high school days, I used to try solving analysis problems from contests such as the Putnam or the notoriously difficult Miklos Schweitzer competition, and literally that led to several interesting encounters with challenging problems and sometimes slight modifications of the problems (often done by me) seemed very difficult to solve; I did not know about places to ask these questions until quite long, and so they remained unattended by audience (until much later when I learned the right place to ask these questions). I am happy that this high school student has been identified and that he or she is receiving support with his or her questions.
15 September, 2021 at 11:32 am
Anonymous
A well-known example for a smooth nowhere analytic function on
is the Fabius function (which has many interesting properties), see
https://en.wikipedia.org/wiki/Fabius_function
Perhaps a modification of this function may answer the first question.
16 September, 2021 at 3:02 pm
Terry Tao teaches – The nth Root
[…] A few months ago I posted a question about analytic functions that I received from a bright high school student, which turned out to be studied and resolved by de Bruijn. Based on this positive resolution, I thought I might try my luck again and list three further questions that this student asked which do not seem to be trivially resolvable. … (What’s New) […]
21 September, 2021 at 3:23 pm
Jacob Manaker
The answer to (3) is yes. Let
be any divergent power series, and fix a conditionally convergent power series
(i.e.,
). For each
, there is a rearrangement
of
such that
OTOH, there is also a rearrangement
summing to 0. Now we have the following rearrangement heuristic:
which converges everywhere.
Of course, this does not quite address the problem, because we need to linearize the order of summation from ordinal
to ordinal
. But that is easy enough to solve, as shown below.
Fix any power series
that converges everywhere and has positive coefficients (i.e.
). Define a “
-graded” summation order
as follows: at level k, we sum enough terms of the form
with
so that
(a) the partial sum is a polynomial
of degree at most k with
and
(b) we've used all terms of the form
where
.
Since
is a conditionally convergent series, there are only finitely many terms necessary to achieve both effects.
Now suppose we apply
. Then, at the completion of each level, our partial sum is (for any x) at most
Suppose we now sum more terms. Any coming from
are at most
, so the proof of the Riemann rearrangement theorem guarantees that (1) will continue to hold. Thus
holds pointwise, as desired.
21 September, 2021 at 3:30 pm
Jacob Manaker
Also worth mentioning: this shows that in fact, we can take _any_ power series and split+rearrange it to get any _other_ power series, with pointwise convergence if the “destination” series is convergent. (Also probably if it isn’t, but I’d need to think about it more to be sure.)
27 September, 2021 at 9:34 am
Terence Tao
Nice solution!
27 September, 2021 at 11:58 pm
Jacob Manaker
Aww, thank you!
11 November, 2021 at 1:03 am
Christian Weiland
Hello everyone!
even for polynomials of the form
. Suppose there is such an
and let
. Then
has the same cardinality as
. The number of isolated points in
is at most countably infinite. So most of the points in
are limit points of
. But
for each such limit point
rendering it impossible for $f$ to exist.
I’d like to propose the following solution for problem 2: There can’t be such a function
16 November, 2021 at 2:03 am
Christian Weiland
Hi again, I’d like to retract my “solution” because it contains a mistake.