
X^2 is the sum of three periodic functions - ColinWright
http://gotmath.com/?p=760
======
ColinWright
I hesitate to say this, but ...

If you think this belongs on the front page of HN then you need to up-vote it.
Any moment now it will trigger the flame-war detector, get a penalty in the
ranking score, and disappear without trace. Anything with 40 comments and
fewer points than comments gets such a penalty, and while it's a good proxy
for flame-war detection, is does get quite a lot of false positives.

If you want this to avoid that fate, you need to up-vote it. If you don't
really care, then that's fine. I think it's interesting, but not everyone
does.

 _Added in edit - something has giving this item a penalty[0] - maybe there
are people who 've flagged it as inappropriate. Certainly it doesn't seem to
have tripped the "flame-war" detector, but who can tell._

[0] [http://hnrankings.info/7056295/](http://hnrankings.info/7056295/)

~~~
thaumasiotes
> Anything with 40 comments and fewer points than comments gets such a
> penalty, and while it's a good proxy for flame-war detection

Is this really true? There's a strong implication there that the typical HN
account is more likely to upvote an article than to comment on one. I upvote
comments all the time, and comment frequently, but I almost never upvote
articles (though I did find this one to be a great mix of interestingness with
accessibility).

~~~
ColinWright
See the section "Controversy" in this article:

[http://www.righto.com/2013/11/how-hacker-news-ranking-
really...](http://www.righto.com/2013/11/how-hacker-news-ranking-really-
works.html)

Discussed at length here:

[https://news.ycombinator.com/item?id=6799854](https://news.ycombinator.com/item?id=6799854)

One side-effect is that it's starting to stifle genuine discussion, as people
decline to comment/reply for fear of an interesting item getting penalized off
the front page.

------
losvedir
> It is clear that a non-constant polynomial cannot be expressed as a finite
> sum of continuous periodic functions,

But it _can_ be expressed as an _infinite_ sum of continuous periodic
functions, right? I seem to remember that you could use all the sine functions
as bases for the vector space of functions and (almost?) any function could be
expressed as an infinite sum of sines.

It's been a while since I've thought about these things, is that recollection
correct?

~~~
ot
Yes, but only on a finite open interval, that is (a, b) with a, b real
numbers.

In the post the sum of periodic functions is equal to x^2 on the whole real
line. This is impossible with sums of continuous periodic functions, such as
Fourier series, because continuous periodic functions are bounded, while x^2
is unbounded.

~~~
vbuterin
> This is impossible with sums of continuous periodic functions, such as
> Fourier series, because continuous periodic functions are bounded, while x^2
> is unbounded.

But infinite sums of continuous periodic functions are not necessarily
bounded: sin(x) + sin(x/2) + sin(x/4) + ... is an example. So there has to be
something else to the proof.

~~~
ot
You are right, but I was referring to either finite sums or Fourier series,
your example is not one because sum of the squares of the coefficients is
infinite. I don't know what can be said about arbitrary countable/uncountable
sums of continuous periodic functions.

------
jbert
I think the thing which is defeating my intuition here is the continuity
issue.

"If a periodic function is continuous and nonconstant, then it has a least
period, and all other periods are positive integer multiples of the least
period."

I think my naive intuition is modelling "periodic function" as "continuous and
periodic". Basically, I'm not exercising the the full freedom of the "it's
periodic" concept.

There's a good Feynman anecdote on this (from Surely You're Joking...):

\------

"I had a scheme, which I still use today when somebody is explaining something
that I'm trying to understand: I keep making up examples.

For instance, the mathematicians would come in with a terrific theorem, and
they're all excited. As they're telling me the conditions of the theorem, I
construct something which fits all the conditions. You know, you have a set
(one ball)-- disjoint (two balls). Then the balls turn colors, grow hairs, or
whatever, in my head as they put more conditions on.

Finally they state the theorem, which is some dumb thing about the ball which
isn't true for my hairy green ball thing, so I say "False!" [and] point out my
counterexample."

\------

So I think it's useful to consider the _most extreme_ thing which meets your
criteria, instead of a "representative" example.

To come back to tech I think a similar mindset is also useful for things like
system failure mode analysis. "Yes, but what if _that_ switch dies at the same
time...?"

~~~
Tloewald
The Axiom of Choice defeats all intuition.

------
simias
It's very interesting but why isn't there any practical example? What are the
equations of three periodic functions that would add up to y = x^2 for
instance?

How would one even construct those three equations? If I'm not mistaken TFA
demonstrates that these functions exist, not how to find them.

I wish I hadn't stopped with maths in high school...

~~~
impendia
Brief answer: the author invokes the Axiom of Choice, so the three equations
will be _really_ screwball and there will be no way to write them down
explicitly.

This is, in some sense, regarded as cheating. (The Axiom of Choice is
generally accepted as a legitimate mathematical axiom, but not universally.)

[http://en.wikipedia.org/wiki/Axiom_of_choice](http://en.wikipedia.org/wiki/Axiom_of_choice)

------
gpvos
It was when the article introduced "Lebesgue measurable periodic functions"
without any explanation whatsoever that I realized that I was not going to
understand this article at all (at least not without spending a day on
Wikipedia).

------
baking
Can someone who understands this please make a graph? Pics or it didn't
happen.

~~~
username42
Disclosure: the full article is a joke.

~~~
ColinWright
Why do you say that?

Disclosure - the article is not a joke, but exploits the fact that when you
think you have a definition that captures some intuitive concept, sometimes it
allows things you didn't expect.

------
adsche
Heads up: In section "A polynomial of degree n is the sum of n+1 periodic
functions" it says 'cdots' in two formulas. (Probably a backslash missing in
the TeX(?) source.)

------
nmc
Fun fact about considering R as a vector space over Q: the Hamel basis A
mentioned in the post is not only infinite, but even uncountable.

~~~
scotty79
I can see now why some mathematicians would like to do without axiom of
choice.

~~~
ColinWright
Consider the product of two sets, A and B:

    
    
      A x B = { (a,b): a in A, b in B }
    

Pretty obviously if A and B are both non-empty, the product is non-empty.

The axiom of choice says that this is still true even when you take the
product of infinitely many sets. Without the axiom of choice, the product of
infinitely many non-empty sets could be empty.

Is that a better choice?

~~~
scotty79
Infinty is unreal. Any formal intuition we build around it is free to choose.
If you can imagine base of R then surely you can imagine infinite number of
nonempty sets that have empty cartesian product. They might all be not empty
but be sort of asymptotically empty, like probably not empty but with that
probabiliy going to zero as number of sets goes to infinity.

The question is, does any good physics come from the axiom of choice?

~~~
ColinWright

      > If you can imagine base of R then surely you can
      > imagine infinite number of nonempty sets that have
      > empty cartesian product.
    

So you think it's OK to have infinitely may sets and not be able to choose one
from each of them, even though they're all non-empty? That's a choice you're
free to make, of course, but to me it seems more perverse than the
alternatives.

    
    
      > The question is, does any good physics come from
      > the axiom of choice?
    

Does it matter? Once you're dealing with uncountable infinities you're already
outside physics and pursuing things because they're interesting, not because
they have immediate practical implications. Of course, bits of pure math has a
tendency to crop up and be useful decades later, and you never know which bits
that will be.

~~~
Tloewald
How do you feel about well ordering the reals? Because the Axiom of Choice
gives you that too and so much more. I'm not a zealot one way or another, but
don't argue for it based on intuition, it plays both ways.

~~~
ColinWright
As the old joke goes: We all know that the Axiom of Choice is true, Well-
Ordering of the Reals is false, and who knows about Zorn's Lemma.

I use AC when I need it, avoid it when I can do without it, and sometimes look
at what it allows versus what it implies.

------
pcvarmint
"By repeating this, we find that e^x is the sum of 1 periodic function. But
this is absurd."

No, it's not absurd.

e^x is periodic with period 2 Pi I.

Maybe the article was limiting itself to real functions and the existence
thereof, but this example raised eyebrows.

~~~
DerekL
The first sentence: "A real function is said to be periodic if there exists a
real number P > 0 so that f(x+P) = f(x) for all x." Complex numbers aren't
mentioned at all.

------
impendia
Another fun (and seemingly nonsensical) thing you can prove if you use the
Axiom of Choice:

[http://en.wikipedia.org/wiki/Banach-
Tarski_paradox](http://en.wikipedia.org/wiki/Banach-Tarski_paradox)

------
bitL
What is a neat characteristics of N^2 where N is integer is that its value is
the sum of the first N odd integers.

1^2 = 1 2^2 = 1 + 3 = 4 3^2 = 1 + 3 + 5 = 9 4^2 = 1 + 3 + 5 + 7 = 16 ...

I am wondering if there is a relation of this to the article.

~~~
ufo
Not really. First of all, the article is about real-valued functions while
your fact is only interesting for integers. Secondly, the article is about
polynomials in general, meaning that the proof still works if you multiply
everything by a constant factor and add terms of a lower degree. The reason
this matters is that sum(i^k for i from 1 to N) will always be a polynomial of
degree (k+1) and the only special thing about your sum in this case (not
having a constant factor or lower order tems) is not something tha the proof
in the article cares about.

