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YUFEI ZHAO: Last time we started
talking about Roth's theorem,

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and we showed a Fourier
analytic proof of Roth's theorem

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in the finite field model.

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So Roth's theorem
in F3 to the N.

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And I want to today show
you how to modify that proof

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to work in integers.

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And this will be basically
Roth's original proof

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of his theorem.

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OK.

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So what we'll prove
today is the statement

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that the size of the largest
3AP-free subset of 1 through N

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is, at most, N divided
by log log N. OK,

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so we'll prove a
bound of this form.

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The strategy of this proof
will be very similar to the one

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that we had from last time.

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So let me review for you
what is the strategy.

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So from last time, the
proof had three main steps.

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In the first step, we
observed that if you

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are in the 3AP-free set then
there exists a large Fourier

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coefficient.

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From this Fourier
coefficient, we

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were able to extract
a large subspace where

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there is a density increment.

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I want to modify that
strategy so that we

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can work in the integers.

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Unlike an F2 to
the N, where things

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were fairly nice and clean,
because you have subspaces,

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you can take a
Fourier coefficient,

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pass it down to a subspace.

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There is no subspaces, right?

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There are no subspaces
in the integers.

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So we have to do something
slightly different,

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but in the same spirit.

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So you'll find a large
Fourier coefficient.

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And we will find that there
is density increment when

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you restrict not to
subspaces, but what

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could play the role of subspaces
when it comes to the integers?

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So I want something which
looks like a smaller

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version of the original space.

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So instead of it
being integers, if we

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restrict to a subprogression,
so to a smaller arithmetic

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progression.

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I will show that you can
restrict to a subprogression

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where you can obtain
density increment.

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So we'll restrict integers
to something smaller.

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And then, same as last time,
we can iterate this increment

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to obtain the
conclusion that you

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have an upper bound on the
size of this 3AP-free set.

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OK, so that's the strategy.

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So you see the same strategy
as the one we did last time,

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and many of the ingredients
will have parallels,

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but the execution will
be slightly different,

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especially in the
second step where,

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because we no longer
have sub spaces, which

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are nice and clean,
so that's why

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we started with a
finite fuel model, just

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to show how things work in
a slightly easier setting.

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And today, we'll see how to
do a same kind of strategy

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here, where there is going
to be a bit more work.

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Not too much more,
but a bit more work.

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OK, so before we
start, any questions?

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All right.

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So last time I used the
proof Roth's theorem

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as an excuse to introduce
Fourier analysis.

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And we're going to see basically
the same kind of Fourier

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analysis, but it's going to take
on a slightly different form,

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because we're not working
at F3 to the N. We're

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working inside the integers.

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And there's a general theory of
Fourier analysis on the group,

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on a billion groups.

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I don't want to go into that
theory, because that's--

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I want to focus on
the specific case,

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but the point is that
given the billion groups,

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you always have a dual
group of characters.

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And they play the role
of Fourier transform.

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Specifically in
the case of Z, we

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have the following
Fourier transform.

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So the dual group of Z
turns out to be the torus.

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So real numbers mod one.

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And the Fourier
transform is defined

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as follows, starting with a
function on the integers, OK.

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If you'd like, let's say
it's finitely supported,

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just to make our
lives a bit easier.

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Don't have to deal
with technicalities.

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But in general, the
following formula holds.

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We have this Fourier
transform defined

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by setting f hat of theta
to be the following sum.

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OK, where this e is actually
somewhat standard notation,

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additive combinatorics.

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It's e to the 2
pi i t, all right?

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So it goes a fraction, t,
around the complex unit circle.

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OK.

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So that's the Fourier
transform on the integers.

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OK, so you might have seen this
before under a different name.

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This is usually
called Fourier series.

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All right.

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You know, the notation
may be slightly different.

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OK, so that's what
we'll see today.

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And this Fourier transform plays
the same role as the Fourier

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transform from last time, which
was on the group F3 to the N.

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And just us in--

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so last time, we had a number
of important identities,

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and we'll have the same
kinds of identities here.

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So let me remind
you what they are.

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And the proofs are all
basically the same,

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so I won't show you the proofs.

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f hat of 0 is simply the
sum of f over the domain.

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We have this Plancherel Parseval
identity, which tells us

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that if you look at
the inner product

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by linear form in
the physical space,

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it equals to the inner
product in the Fourier space.

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OK.

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So in the physical
space now, you sum.

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In the frequency space,
you take integral

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over the torus, or the
circle, in this case.

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It's a one-dimensional torus.

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There is also the Fourier
inversion formula,

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which now says that f of x
is equal to f hat of theta.

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E of x theta, you integrate
theta from 0 to 1.

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Again, on the torus,
on the circle.

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And third-- and finally, there
was this identity last time

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that related three-term
arithmetic progressions

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to the Fourier transform, OK?

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So this last one was
slightly not as--

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I mean, it's not as standard
as the first several, which are

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standard Fourier identities.

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But this one will
be useful to us.

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So the identity relating the
Fourier transform, the 3AP, now

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has the following form.

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OK, so if we define
lambda of f, g,

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and h to be the
following sum, which

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sums over all 3APs
in the integers,

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then one can write
this expression

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in terms of the Fourier
transform as follows.

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OK.

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All right.

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So comparing this formula to the
one that we saw from last time,

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it's the same formula,
where different domains,

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where you're summing
or integrating,

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but it's the same formula.

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And the proof is the same.

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So go look at the proof.

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It's the same proof.

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OK.

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So these are the key Fourier
things that we'll use.

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And then we'll try to
follow on those with--

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the same as the proof as last
time, and see where we can get.

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So let me introduce
one more notation.

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So I'll write lambda sub 3
of f to be lambda of f, f, f,

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three times.

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OK.

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So at this point, if you
understood the lecture

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from last time, none of
anything I've said so far

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should be surprising.

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We are working
integers, so we should

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look at the corresponding
Fourier transform in integers.

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And if you follow your
notes, this is all the things

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that we're going to use.

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OK, so what was one
of the first things

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we mentioned regarding
the Fourier transform

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from last time after this point?

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OK.

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AUDIENCE: The counting lemma.

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YUFEI ZHAO: OK.

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So let's do a counting lemma.

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So what should the
counting lemma say?

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Well, the spirit of the counting
lemma is that if you have two

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functions that are close to
each other-- and now, "close"

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means close in Fourier--

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then their corresponding number
of 3APs should be similar, OK?

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00:12:03,910 --> 00:12:06,150
So that's what we want to say.

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And indeed, right, so
the counting lemma for us

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will say that if f and g
are functions on Z, and--

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such that their L2 norms are
both bounded by this M. OK,

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the sum of the squared absolute
value entries are both bounded.

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Then the difference
of their 3AP counts

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should not be so different
from each other if f

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00:13:08,570 --> 00:13:10,790
and g are close in Fourier, OK?

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And that means that if all
the Fourier coefficients of f

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minus g are small,
then lambda 3ff,

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which considers 3AP counts
in f, is close to that of g.

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OK.

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00:13:38,110 --> 00:13:42,744
Same kind of 3AP counting
lemma from last time.

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00:13:42,744 --> 00:13:45,706
OK, so let's prove it, OK?

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00:13:51,520 --> 00:13:53,620
As with the counting
lemma proofs

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00:13:53,620 --> 00:13:56,350
you've seen several times
already in this course,

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we will prove it by first
writing this difference

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00:14:00,070 --> 00:14:01,450
as a telescoping sum.

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00:14:07,420 --> 00:14:12,920
The first term
being f minus g f f,

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00:14:12,920 --> 00:14:25,045
and then g of f minus g f and
lambda g, g, f, f minus g.

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00:14:25,045 --> 00:14:27,790
OK, and we will like to show
that each of these terms

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00:14:27,790 --> 00:14:34,273
is small if f minus g has
small Fourier coefficients.

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00:14:39,690 --> 00:14:40,190
OK.

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00:14:40,190 --> 00:14:41,900
So let's bound the first term.

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00:14:50,077 --> 00:14:55,340
OK, so let me bound this first
term using the 3AP identity,

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00:14:55,340 --> 00:14:57,950
relating 3AP to
Fourier coefficients,

200
00:14:57,950 --> 00:15:05,150
we can write this lambda
as the following integral

201
00:15:05,150 --> 00:15:07,118
over Fourier coefficients.

202
00:15:24,070 --> 00:15:25,730
And now, let me--

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00:15:25,730 --> 00:15:27,550
OK, so what was the
trick last time?

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00:15:27,550 --> 00:15:32,600
So we said let's pull
out one of these guys

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00:15:32,600 --> 00:15:37,245
and then use triangle inequality
on the remaining factors.

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00:15:37,245 --> 00:15:38,150
OK, so we'll do that.

207
00:15:56,410 --> 00:15:58,852
So far, so good.

208
00:15:58,852 --> 00:16:02,295
And now you see this integral.

209
00:16:02,295 --> 00:16:03,773
Apply Cauchy-Schwartz.

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00:16:18,090 --> 00:16:20,890
OK, so apply Cauchy-Schwartz
to the first factor,

211
00:16:20,890 --> 00:16:25,533
you got this l2 sum,
this l2 integral.

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00:16:25,533 --> 00:16:26,950
And then you apply
Cauchy-Schwartz

213
00:16:26,950 --> 00:16:28,295
to the second factor.

214
00:16:34,633 --> 00:16:35,550
You get that integral.

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00:16:38,826 --> 00:16:40,920
OK, now what do we do?

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00:16:44,912 --> 00:16:46,409
Yep?

217
00:16:46,409 --> 00:16:49,980
AUDIENCE: [INAUDIBLE]

218
00:16:49,980 --> 00:16:50,980
YUFEI ZHAO: OK, so yeah.

219
00:16:50,980 --> 00:16:54,700
So you see an l2 of Fourier,
the first automatic reaction

220
00:16:54,700 --> 00:16:56,890
should be to use a
Plancherel or Parseval, OK?

221
00:16:56,890 --> 00:17:04,589
So apply Plancherel identity
to each of these factors.

222
00:17:04,589 --> 00:17:13,520
We find that each
of those factors

223
00:17:13,520 --> 00:17:20,040
is equal to this l2 sum
in the physical space.

224
00:17:23,750 --> 00:17:25,609
OK, so this square
root, the same thing.

225
00:17:25,609 --> 00:17:27,980
Square root again.

226
00:17:27,980 --> 00:17:28,520
OK.

227
00:17:28,520 --> 00:17:34,750
And then we find
that because there

228
00:17:34,750 --> 00:17:36,720
was an hypothesis--
in the hypothesis,

229
00:17:36,720 --> 00:17:40,430
there was a bound M on
this sum of squares.

230
00:17:43,610 --> 00:17:46,090
You have that down there.

231
00:17:46,090 --> 00:17:48,080
And similarly, with
the other two terms.

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00:18:01,526 --> 00:18:02,050
OK.

233
00:18:02,050 --> 00:18:03,511
So that proves the
counting lemma.

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00:18:07,359 --> 00:18:08,802
Question?

235
00:18:08,802 --> 00:18:12,760
AUDIENCE: Last time, the
term on the right-hand side

236
00:18:12,760 --> 00:18:16,525
was the maximum over
non-zero frequency?

237
00:18:16,525 --> 00:18:17,730
YUFEI ZHAO: OK.

238
00:18:17,730 --> 00:18:18,230
OK.

239
00:18:18,230 --> 00:18:22,230
So the question is, last time
we had a counting lemma that

240
00:18:22,230 --> 00:18:24,120
looked slightly different.

241
00:18:24,120 --> 00:18:26,730
But I claimed they're all
really the same counting lemma.

242
00:18:26,730 --> 00:18:28,320
They're all the same proofs.

243
00:18:28,320 --> 00:18:30,810
If you run this
proof, it won't work.

244
00:18:30,810 --> 00:18:33,150
If you take what
we did last time,

245
00:18:33,150 --> 00:18:35,260
it's the same kind of proofs.

246
00:18:35,260 --> 00:18:36,760
So last time we had
a counting lemma

247
00:18:36,760 --> 00:18:44,120
where we had the same
f, f, f essentially.

248
00:18:44,120 --> 00:18:48,660
We know have-- I allow you
to essentially take three

249
00:18:48,660 --> 00:18:49,640
different things, and--

250
00:18:49,640 --> 00:18:52,980
OK, so both-- in
both cases, you're

251
00:18:52,980 --> 00:18:54,750
running through
this calculation,

252
00:18:54,750 --> 00:18:57,378
but they look
slightly different.

253
00:18:57,378 --> 00:19:02,555
AUDIENCE: [INAUDIBLE]

254
00:19:02,555 --> 00:19:03,430
YUFEI ZHAO: So, yeah.

255
00:19:03,430 --> 00:19:04,435
So I agree.

256
00:19:04,435 --> 00:19:05,810
It doesn't look
exactly the same,

257
00:19:05,810 --> 00:19:07,520
but if you think about
what's involved in the proof,

258
00:19:07,520 --> 00:19:08,520
they're the same proofs.

259
00:19:11,232 --> 00:19:12,630
OK.

260
00:19:12,630 --> 00:19:16,170
Any more questions?

261
00:19:16,170 --> 00:19:19,060
All right.

262
00:19:19,060 --> 00:19:20,670
So now, we have
this counting lemma,

263
00:19:20,670 --> 00:19:25,660
so let's start our proof of
Roth's theorem in the integers.

264
00:19:25,660 --> 00:19:27,820
As with last time, there
will be three steps,

265
00:19:27,820 --> 00:19:32,175
as mentioned up there, OK?

266
00:19:32,175 --> 00:19:37,590
In the first step, let us
show that if you are 3AP-free,

267
00:19:37,590 --> 00:19:40,130
then we can obtain a density--

268
00:19:40,130 --> 00:19:42,198
a large Fourier coefficient.

269
00:19:56,640 --> 00:19:59,200
Yeah, so in this course,
this counting lemma,

270
00:19:59,200 --> 00:20:01,450
we actually solved this--

271
00:20:01,450 --> 00:20:03,970
basically this kind of proof
for the first time when we

272
00:20:03,970 --> 00:20:07,000
discussed graph counting
lemma, back in the chapter

273
00:20:07,000 --> 00:20:09,810
on Szemerédi's regularity lemma.

274
00:20:09,810 --> 00:20:12,430
And sure, they all
look literally--

275
00:20:12,430 --> 00:20:14,740
not exactly the same,
but they're all really

276
00:20:14,740 --> 00:20:16,400
the same kind of proofs, right?

277
00:20:16,400 --> 00:20:17,050
So I want--

278
00:20:17,050 --> 00:20:19,540
I'm showing you the same thing
in many different guises.

279
00:20:19,540 --> 00:20:20,873
But they're all the same proofs.

280
00:20:23,690 --> 00:20:28,540
So if you are a set
that is 3AP-free--

281
00:20:37,630 --> 00:20:41,480
and as with last time,
I'm going to call

282
00:20:41,480 --> 00:20:46,550
alpha the density of A now
inside this progression,

283
00:20:46,550 --> 00:20:49,820
this length N progression.

284
00:20:49,820 --> 00:20:54,560
And suppose N is large enough.

285
00:20:59,460 --> 00:21:07,580
OK, so the conclusion now is
that there exists some theta

286
00:21:07,580 --> 00:21:19,170
such that if you look
at this sum over here

287
00:21:19,170 --> 00:21:23,930
as a sum over both integers--

288
00:21:23,930 --> 00:21:26,680
actually, let me do
the sum only from 1

289
00:21:26,680 --> 00:21:35,040
to uppercase N. Claim that--

290
00:21:35,040 --> 00:21:39,510
OK, so-- so it's saying
what this title says.

291
00:21:39,510 --> 00:21:43,205
If you are 3AP-free
and this N is large

292
00:21:43,205 --> 00:21:45,580
enough relative to the density,
you think of this density

293
00:21:45,580 --> 00:21:49,990
alpha is a constant, then
I can find a large Fourier

294
00:21:49,990 --> 00:21:50,860
coefficient.

295
00:21:50,860 --> 00:21:54,010
Now, there's a small
difference, and this

296
00:21:54,010 --> 00:21:56,140
is related to what you
were asking earlier,

297
00:21:56,140 --> 00:21:59,710
between how we set things up now
versus what happened last time.

298
00:21:59,710 --> 00:22:05,530
So last time, we just looked
for a Fourier coefficient

299
00:22:05,530 --> 00:22:10,080
corresponding to a non-zero r.

300
00:22:10,080 --> 00:22:13,340
Now, I'm not
restricting non-zero,

301
00:22:13,340 --> 00:22:15,700
but I don't start with
an indicator function.

302
00:22:15,700 --> 00:22:19,280
I start with the demeaned
indicator function.

303
00:22:19,280 --> 00:22:24,340
I take out the mean so that
the zeroth coefficient,

304
00:22:24,340 --> 00:22:28,450
so to speak, which corresponds
to the mean, is already 0.

305
00:22:28,450 --> 00:22:32,790
So you don't get to use
that for your coefficient.

306
00:22:32,790 --> 00:22:34,840
So if you didn't do
this, if you just

307
00:22:34,840 --> 00:22:36,420
tried to do this
last time, I mean,

308
00:22:36,420 --> 00:22:38,045
you can also do
exactly the same setup.

309
00:22:38,045 --> 00:22:40,180
But if you don't
demean it, then--

310
00:22:40,180 --> 00:22:42,400
if you don't have this
term, then this statement

311
00:22:42,400 --> 00:22:46,960
is trivially true, because I
can take theta equal to 0, OK?

312
00:22:46,960 --> 00:22:48,300
But I don't want.

313
00:22:48,300 --> 00:22:52,900
I want an actual significant
Fourier improvement.

314
00:22:52,900 --> 00:22:53,930
So I take--

315
00:22:53,930 --> 00:22:58,540
I do this demean, and then
I consider its Fourier

316
00:22:58,540 --> 00:22:59,583
coefficient.

317
00:23:02,260 --> 00:23:02,760
OK.

318
00:23:02,760 --> 00:23:06,220
Any questions about
the statement?

319
00:23:06,220 --> 00:23:09,810
Yeah, so this demeaning is
really important, right?

320
00:23:09,810 --> 00:23:12,530
So that's something that's a
very common technique whenever

321
00:23:12,530 --> 00:23:14,140
you do these kind of analysis.

322
00:23:14,140 --> 00:23:16,510
So make sure you're--

323
00:23:16,510 --> 00:23:19,940
so that you're-- yeah, so you're
looking at functions with mean

324
00:23:19,940 --> 00:23:22,777
0.

325
00:23:22,777 --> 00:23:23,610
Let's see the proof.

326
00:23:28,190 --> 00:23:33,290
We have the
following information

327
00:23:33,290 --> 00:23:39,670
about all 3AP counts in A.
Because A is 3AP-free, OK,

328
00:23:39,670 --> 00:23:43,320
so what is the value of lambda
sub 3 of the indicator of A?

329
00:23:46,100 --> 00:23:49,440
Lambda of 3, if you
look at the expression,

330
00:23:49,440 --> 00:23:55,770
it basically sums over all
3APs, but A has no 3APs,

331
00:23:55,770 --> 00:23:58,500
except for the trivial ones.

332
00:23:58,500 --> 00:24:01,020
So we'll only consider
the trivial 3APs,

333
00:24:01,020 --> 00:24:06,210
which has size exactly
the size of A, which

334
00:24:06,210 --> 00:24:10,820
is alpha N from trivial 3APs.

335
00:24:14,870 --> 00:24:20,030
On the other hand, what
do we know about lambda 3

336
00:24:20,030 --> 00:24:25,256
of this interval from 1 to N?

337
00:24:25,256 --> 00:24:28,410
OK, so how many 3APs are there?

338
00:24:28,410 --> 00:24:33,640
OK, so roughly, it's going
to be about N squared over 2.

339
00:24:33,640 --> 00:24:38,470
And in fact, it will be
at least N squared over 2,

340
00:24:38,470 --> 00:24:41,860
because to generate
a 3AP, I just

341
00:24:41,860 --> 00:24:49,070
have to pick a first
term and a third term,

342
00:24:49,070 --> 00:24:51,760
and I'm OK as long as
they're the same parity.

343
00:24:58,195 --> 00:25:00,410
And then you have a 3AP.

344
00:25:00,410 --> 00:25:04,220
So the same parity
cuts you down by half,

345
00:25:04,220 --> 00:25:11,190
so you have at least N squared
over 2 3APs from 1 through N.

346
00:25:11,190 --> 00:25:14,930
So now, let's look at
how to apply the counting

347
00:25:14,930 --> 00:25:16,480
lemma all to the setting.

348
00:25:16,480 --> 00:25:18,420
So we have the counting
lemma up there,

349
00:25:18,420 --> 00:25:19,930
where I now want to apply it--

350
00:25:22,600 --> 00:25:28,960
so apply counting
to, on one hand,

351
00:25:28,960 --> 00:25:34,780
the indicator function of A
so we get the count 3APs in A,

352
00:25:34,780 --> 00:25:43,770
but also compared to
the normalized indicator

353
00:25:43,770 --> 00:25:45,252
on the interval.

354
00:25:45,252 --> 00:25:46,940
OK, so maybe this is
a good point for me

355
00:25:46,940 --> 00:25:50,425
to pause and remind you that
the spirit of this whole proof

356
00:25:50,425 --> 00:25:54,950
is understanding structure
versus pseudorandomness, OK?

357
00:25:54,950 --> 00:25:57,680
So as was the case last time.

358
00:25:57,680 --> 00:26:03,620
So we want to understand, in
what ways is A pseudorandom?

359
00:26:03,620 --> 00:26:06,020
And here, "pseudorandom,"
just as with last time,

360
00:26:06,020 --> 00:26:08,540
means having small
Fourier coefficients,

361
00:26:08,540 --> 00:26:11,856
being Fourier uniform.

362
00:26:11,856 --> 00:26:16,850
If A is pseudorandom,
which here, means f and g

363
00:26:16,850 --> 00:26:18,830
are close to each other.

364
00:26:18,830 --> 00:26:20,420
That's what being
pseudorandom means,

365
00:26:20,420 --> 00:26:21,920
then the counting
lemma will tell us

366
00:26:21,920 --> 00:26:25,790
that f and g should
have similar AP counts.

367
00:26:25,790 --> 00:26:31,190
But A has basically no AP
count, so they should not

368
00:26:31,190 --> 00:26:35,970
be close to each other.

369
00:26:35,970 --> 00:26:38,610
So that's the strategy, to
show that A is not pseudorandom

370
00:26:38,610 --> 00:26:41,490
in this sense, and thereby
extracting a large Fourier

371
00:26:41,490 --> 00:26:43,240
coefficient.

372
00:26:43,240 --> 00:26:47,490
So we apply counting
to these two functions,

373
00:26:47,490 --> 00:26:49,350
and we obtain that.

374
00:26:54,860 --> 00:26:55,360
OK.

375
00:26:55,360 --> 00:27:03,000
So this quantity, which
corresponds to lambda 3 of g,

376
00:27:03,000 --> 00:27:17,300
minus alpha N. So these were
lambda 3 of g, lambda 3 of F.

377
00:27:17,300 --> 00:27:18,760
So it is upper-bounded.

378
00:27:18,760 --> 00:27:23,897
The difference is up
rebounded by the--

379
00:27:28,010 --> 00:27:29,760
using the counting
lemma, we find

380
00:27:29,760 --> 00:27:31,500
that their difference
is upper-bounded

381
00:27:31,500 --> 00:27:32,670
by the following quantity.

382
00:27:43,760 --> 00:27:46,850
Namely, you look at the
difference between f and g

383
00:27:46,850 --> 00:27:52,082
and evaluate its maximum
Fourier coefficient.

384
00:27:52,082 --> 00:27:53,010
OK.

385
00:27:53,010 --> 00:27:58,760
So if A is pseudorandom,
meaning that Fourier uniform--

386
00:27:58,760 --> 00:28:04,840
this l infinity
norm is small, then

387
00:28:04,840 --> 00:28:10,010
I should expect lots
and lots of 3APs in A,

388
00:28:10,010 --> 00:28:12,390
but because that
is not the case,

389
00:28:12,390 --> 00:28:14,510
we should be able to
conclude that there is

390
00:28:14,510 --> 00:28:16,745
some large Fourier coefficient.

391
00:28:20,000 --> 00:28:21,090
All right, so thus--

392
00:28:33,131 --> 00:28:36,720
so rearranging the equation
above, we have that--

393
00:28:44,030 --> 00:28:45,270
so this should be a square.

394
00:28:51,480 --> 00:28:52,100
OK.

395
00:28:52,100 --> 00:28:54,260
So we have this expression here.

396
00:28:54,260 --> 00:28:56,690
And now we are--

397
00:28:56,690 --> 00:29:03,320
OK, so let me simplify
this expression slightly.

398
00:29:03,320 --> 00:29:07,550
And now we're using that N
is sufficiently large, OK?

399
00:29:07,550 --> 00:29:12,290
So we're using N is
sufficiently large.

400
00:29:12,290 --> 00:29:23,202
So this quantity is at least
a tenth of alpha squared N.

401
00:29:23,202 --> 00:29:24,910
OK, and that's the
conclusion, all right?

402
00:29:24,910 --> 00:29:28,937
So that's the conclusion
of this step here.

403
00:29:28,937 --> 00:29:29,770
What does this mean?

404
00:29:29,770 --> 00:29:33,600
This means there
exists some theta

405
00:29:33,600 --> 00:29:37,050
so that the Fourier
coefficient at theta

406
00:29:37,050 --> 00:29:38,750
is at least the
claimed quantity.

407
00:29:42,530 --> 00:29:43,834
Any questions?

408
00:29:51,420 --> 00:29:52,050
All right.

409
00:29:52,050 --> 00:29:53,960
So that finishes step 1.

410
00:29:53,960 --> 00:29:57,720
So now let me go on step 2.

411
00:29:57,720 --> 00:30:02,120
In step 2, we wish to show that
if you have a large Fourier

412
00:30:02,120 --> 00:30:11,956
coefficient, then one can
obtain a density increment.

413
00:30:21,200 --> 00:30:26,720
So last time, we were working
in a finite field vector space.

414
00:30:26,720 --> 00:30:30,470
A Fourier coefficient, OK,
so which is a dual vector,

415
00:30:30,470 --> 00:30:34,020
corresponds to some hyperplane.

416
00:30:34,020 --> 00:30:37,610
And having a large
Fourier coefficient

417
00:30:37,610 --> 00:30:40,380
then implies that
the density of A

418
00:30:40,380 --> 00:30:43,140
on the co-sets of
those hyperplanes

419
00:30:43,140 --> 00:30:46,720
must be not all
close to each other.

420
00:30:46,720 --> 00:30:48,570
All right, so one
of the hyperplanes

421
00:30:48,570 --> 00:30:53,380
must have significantly
higher density than the rest.

422
00:30:53,380 --> 00:30:55,300
OK, so we want to do
something similar here,

423
00:30:55,300 --> 00:30:57,580
except we run into this
technical difficulty

424
00:30:57,580 --> 00:31:01,250
where there are no
subspaces anymore.

425
00:31:01,250 --> 00:31:05,243
So the Fourier character, namely
corresponding to this theta,

426
00:31:05,243 --> 00:31:06,160
is just a real number.

427
00:31:06,160 --> 00:31:07,510
It doesn't divide up your space.

428
00:31:07,510 --> 00:31:09,070
It doesn't divide
up your 1 through N

429
00:31:09,070 --> 00:31:11,050
very nicely into sub chunks.

430
00:31:11,050 --> 00:31:15,820
But we still want to use this
theta to chop up 1 through N

431
00:31:15,820 --> 00:31:20,860
into smaller spaces so
that we can iterate and do

432
00:31:20,860 --> 00:31:24,730
density increment.

433
00:31:24,730 --> 00:31:25,910
All right.

434
00:31:25,910 --> 00:31:28,920
So let's see what we can do.

435
00:31:28,920 --> 00:31:36,980
So given this theta,
what we would like to do

436
00:31:36,980 --> 00:31:50,260
is to partition this 1 through
N into subprogressions.

437
00:31:53,564 --> 00:32:07,030
OK, so chop up 1 through
N into sub APs such that

438
00:32:07,030 --> 00:32:12,230
if you evaluate for--
so this theta is fixed.

439
00:32:12,230 --> 00:32:17,990
So on each sub AP,
this function here

440
00:32:17,990 --> 00:32:28,000
is roughly constant
on each of your parts.

441
00:32:31,150 --> 00:32:34,000
Last time, we had this
Fourier character,

442
00:32:34,000 --> 00:32:37,120
and then we chopped it up
using these three hyperplanes.

443
00:32:37,120 --> 00:32:40,390
And each hyperplane,
the Fourier character

444
00:32:40,390 --> 00:32:43,078
is literally constant, OK?

445
00:32:43,078 --> 00:32:46,670
So you have-- and so
that's what we work with.

446
00:32:46,670 --> 00:32:49,450
And now, you cannot get
them to be exactly constant,

447
00:32:49,450 --> 00:32:53,020
but the next best thing we can
hope for is to get this Fourier

448
00:32:53,020 --> 00:32:56,097
character to be
roughly constant.

449
00:32:56,097 --> 00:32:58,430
OK, so we're going to do some
positioning that allows us

450
00:32:58,430 --> 00:33:00,860
to achieve this characteristic.

451
00:33:00,860 --> 00:33:03,740
And let me give
you some intuition

452
00:33:03,740 --> 00:33:04,910
about why this is true.

453
00:33:04,910 --> 00:33:07,570
And this is not exactly
a surprising fact.

454
00:33:07,570 --> 00:33:10,040
The intuition is
just that if you

455
00:33:10,040 --> 00:33:11,900
look at what this
function behaves like--

456
00:33:17,444 --> 00:33:19,950
all right, so what's
going on here?

457
00:33:19,950 --> 00:33:27,600
You are on the unit circle,
and you are jumping by theta.

458
00:33:31,288 --> 00:33:42,080
OK, so you just keep
jumping by theta and so on.

459
00:33:42,080 --> 00:33:49,370
And I want to show that I
can sharp up my progression

460
00:33:49,370 --> 00:33:57,690
into a bunch of almost periodic
pieces, where in each part,

461
00:33:57,690 --> 00:34:01,370
I'm staying inside a small arc.

462
00:34:01,370 --> 00:34:06,560
So in the extreme case of this
where it is very easy to see

463
00:34:06,560 --> 00:34:18,159
is if x is some rational number,
a over b, with b fairly small,

464
00:34:18,159 --> 00:34:22,010
then we can--

465
00:34:22,010 --> 00:34:34,190
so then, this character is
actually constant on APs

466
00:34:34,190 --> 00:34:35,659
with common difference b.

467
00:34:43,063 --> 00:34:43,563
Yep?

468
00:34:43,563 --> 00:34:46,708
AUDIENCE: Is theta
supposed to be [INAUDIBLE]??

469
00:34:46,708 --> 00:34:48,875
YUFEI ZHAO: Ah, so theta,
yes. so theta-- thank you.

470
00:34:48,875 --> 00:34:50,270
So theta 2 pi.

471
00:34:54,174 --> 00:34:57,370
AUDIENCE: Like, is x
equal to your theta?

472
00:34:57,370 --> 00:34:58,078
YUFEI ZHAO: Yeah.

473
00:34:58,078 --> 00:34:59,054
Thank you.

474
00:34:59,054 --> 00:35:00,520
So theta equals-- yeah.

475
00:35:00,520 --> 00:35:06,180
So if theta is some rational
with some small denominator--

476
00:35:06,180 --> 00:35:13,420
so then you are literally
jumping in periodic steps

477
00:35:13,420 --> 00:35:15,690
on the unit circle.

478
00:35:15,690 --> 00:35:21,230
So if you partition N according
to the exact same periods,

479
00:35:21,230 --> 00:35:26,300
you have that this character
is exactly constant in each

480
00:35:26,300 --> 00:35:28,190
of your progressions.

481
00:35:28,190 --> 00:35:32,210
Now, in general, the theta
you get out of that proof

482
00:35:32,210 --> 00:35:35,400
might not have this
very nice form,

483
00:35:35,400 --> 00:35:38,340
but we can at
least approximately

484
00:35:38,340 --> 00:35:39,840
achieve the desired effect.

485
00:35:42,630 --> 00:35:43,130
OK.

486
00:35:46,890 --> 00:35:47,700
Any questions?

487
00:36:15,560 --> 00:36:16,260
OK.

488
00:36:16,260 --> 00:36:20,100
So to achieve approximately the
desired effect, what we'll do

489
00:36:20,100 --> 00:36:25,370
is to find something
so that b times theta

490
00:36:25,370 --> 00:36:29,485
is not quite an integer, but
very close to an integer.

491
00:36:29,485 --> 00:36:31,610
OK, so this, probably many
of you have seen before.

492
00:36:31,610 --> 00:36:37,470
It's a classic
pigeonhole-type result.

493
00:36:37,470 --> 00:36:41,510
It's usually attributed
to Dirichlet.

494
00:36:45,020 --> 00:36:54,990
So if you have theta, a
real number, and a delta,

495
00:36:54,990 --> 00:37:00,300
kind of a tolerance,
then there exists

496
00:37:00,300 --> 00:37:09,350
a positive integer d at
most 1 over delta such

497
00:37:09,350 --> 00:37:19,030
that d times theta is
very close to an integer.

498
00:37:19,030 --> 00:37:23,720
OK, so this norm
here is distance

499
00:37:23,720 --> 00:37:25,850
to the closest integer.

500
00:37:33,215 --> 00:37:38,390
All right, so the proof is
by pigeonhole principle.

501
00:37:38,390 --> 00:37:44,560
So if we let N be 1
over delta rounded down

502
00:37:44,560 --> 00:37:51,970
and consider the numbers 0,
theta, 2 theta, 3 theta, and so

503
00:37:51,970 --> 00:37:55,630
on, to N theta--

504
00:37:55,630 --> 00:38:08,040
so by pigeonhole, there
exists i theta and j theta, so

505
00:38:08,040 --> 00:38:12,660
two different terms
of the sequence such

506
00:38:12,660 --> 00:38:20,700
that they differ by
less than-- at most

507
00:38:20,700 --> 00:38:22,590
delta in their fractional parts.

508
00:38:30,260 --> 00:38:36,920
OK, so now take d to be
difference between i and j.

509
00:38:36,920 --> 00:38:37,820
OK, and that works.

510
00:38:48,060 --> 00:38:48,560
OK.

511
00:38:48,560 --> 00:38:52,720
So even though you don't
have exactly rational,

512
00:38:52,720 --> 00:38:55,210
you have approximately rational.

513
00:38:55,210 --> 00:38:56,200
So this is a--

514
00:38:56,200 --> 00:38:58,970
it's a simple rational
approximation statement.

515
00:38:58,970 --> 00:39:00,890
And using this
rational approximation,

516
00:39:00,890 --> 00:39:05,660
we can now try to do
the intuition here,

517
00:39:05,660 --> 00:39:10,490
pretending that we're working
with rational numbers, indeed.

518
00:39:16,442 --> 00:39:20,670
OK, so if we take
eta between 0 and 1

519
00:39:20,670 --> 00:39:34,850
and theta irrational and
suppose N is large enough--

520
00:39:34,850 --> 00:39:37,090
OK, so here, C
means there exists

521
00:39:37,090 --> 00:39:40,490
some sufficiently large--

522
00:39:40,490 --> 00:39:45,420
some constant C such that
the statement is true, OK?

523
00:39:45,420 --> 00:39:49,420
So suppose you think
a million here.

524
00:39:49,420 --> 00:39:51,160
That should be fine.

525
00:39:51,160 --> 00:39:55,740
So then there exists--

526
00:39:55,740 --> 00:40:12,330
so then one can partition
1 through N into sub-APs,

527
00:40:12,330 --> 00:40:14,960
which we'll call P i.

528
00:40:14,960 --> 00:40:24,250
And each having length
between cube root of N

529
00:40:24,250 --> 00:40:33,110
and twice the cube root of
N such that this character

530
00:40:33,110 --> 00:40:35,700
that we want to stay
roughly constant indeed

531
00:40:35,700 --> 00:40:39,550
does not change very much.

532
00:40:39,550 --> 00:40:47,610
If you look at two terms in
the same AP, in the sub-AP,

533
00:40:47,610 --> 00:41:00,670
then the value of this
character on each P sub i

534
00:41:00,670 --> 00:41:03,080
is roughly the same.

535
00:41:03,080 --> 00:41:08,920
So they don't vary by
more than eta on each P i.

536
00:41:08,920 --> 00:41:16,350
So here, we're partitioning this
1 through N into a sub-A piece

537
00:41:16,350 --> 00:41:20,355
so that this guy here
stays roughly constant.

538
00:41:25,940 --> 00:41:26,440
OK.

539
00:41:26,440 --> 00:41:29,470
Any questions?

540
00:41:29,470 --> 00:41:29,970
All right.

541
00:41:29,970 --> 00:41:33,728
So think about how
you might prove this.

542
00:41:33,728 --> 00:41:34,770
Let's take a quick break.

543
00:41:37,580 --> 00:41:41,160
So you see, we are basically
following the same strategy

544
00:41:41,160 --> 00:41:44,770
as the proof from last
time, but this second step,

545
00:41:44,770 --> 00:41:47,370
which we're on right now,
needs to be somewhat modified

546
00:41:47,370 --> 00:41:52,230
because you cannot cut this
space up into pieces where

547
00:41:52,230 --> 00:41:54,330
your character is constant.

548
00:41:54,330 --> 00:41:58,140
Well, if they're roughly
constant then we're go to go,

549
00:41:58,140 --> 00:42:01,300
so that's what we're doing now.

550
00:42:01,300 --> 00:42:03,000
So let's prove the
statement up there.

551
00:42:15,350 --> 00:42:15,860
All right.

552
00:42:15,860 --> 00:42:18,950
So let's prove this
statement over here.

553
00:42:18,950 --> 00:42:37,600
So using Dirichlet's lemma, we
find that there exists some d.

554
00:42:37,600 --> 00:42:39,615
OK, so I'll write down
some number for now.

555
00:42:39,615 --> 00:42:40,490
Don't worry about it.

556
00:42:40,490 --> 00:42:45,090
It will come up shortly why I
write this specific quantity.

557
00:42:47,670 --> 00:43:06,330
So there exists some d which is
not too big, such that d theta

558
00:43:06,330 --> 00:43:09,830
is very close to an integer.

559
00:43:09,830 --> 00:43:12,540
So now, I'm literally
applying Dirichlet's lemma.

560
00:43:16,050 --> 00:43:17,680
OK.

561
00:43:17,680 --> 00:43:24,826
So given such d--

562
00:43:24,826 --> 00:43:27,640
so how big is this d?

563
00:43:27,640 --> 00:43:32,740
You see that because I assumed
that N is sufficiently large,

564
00:43:32,740 --> 00:43:48,800
if we choose that C large
enough, d is at most root N. So

565
00:43:48,800 --> 00:43:53,570
given such d, which
is at most root N,

566
00:43:53,570 --> 00:44:04,180
you can partition 1 through
N into subprogressions

567
00:44:04,180 --> 00:44:09,580
with common difference d.

568
00:44:09,580 --> 00:44:14,550
Essentially, look at--
let's do classes mod d.

569
00:44:14,550 --> 00:44:19,130
So they're all going to have
length by basically N over d.

570
00:44:19,130 --> 00:44:25,140
And I chop them up a
little bit further to get--

571
00:44:25,140 --> 00:44:33,390
so a piece of length
between cube root of N

572
00:44:33,390 --> 00:44:40,990
and twice cube root of N.

573
00:44:40,990 --> 00:44:41,490
OK.

574
00:44:41,490 --> 00:44:44,520
So I'm going to make
sure that all of my APs

575
00:44:44,520 --> 00:44:48,000
are roughly the same length.

576
00:44:48,000 --> 00:44:56,090
And now, inside each
subprogression--

577
00:45:00,790 --> 00:45:06,610
let me call this subprogression
P prime, subprogression P,

578
00:45:06,610 --> 00:45:09,430
let's look at how much
this character value can

579
00:45:09,430 --> 00:45:13,119
vary inside this progression.

580
00:45:19,346 --> 00:45:20,304
All right.

581
00:45:26,010 --> 00:45:29,420
OK, so how much can this vary?

582
00:45:29,420 --> 00:45:36,330
Well, because theta is such
that d times theta is very close

583
00:45:36,330 --> 00:45:41,778
to an integer and the length
of each progression is not too

584
00:45:41,778 --> 00:45:42,570
large-- so here's--

585
00:45:42,570 --> 00:45:44,710
I want some control
on the length.

586
00:45:44,710 --> 00:45:49,020
So we find that the
maximum variation

587
00:45:49,020 --> 00:45:56,695
is, at most, the size of
P, the length of P, times--

588
00:45:59,580 --> 00:46:04,530
so this-- that
difference over there.

589
00:46:04,530 --> 00:46:08,920
So all of these are exponential,
so I can shift them.

590
00:46:08,920 --> 00:46:19,995
Well, the length of P is at
most twice cube root of N. And--

591
00:46:19,995 --> 00:46:22,160
OK, so what is this quantity?

592
00:46:22,160 --> 00:46:25,470
So the point is that if
this fractional part here

593
00:46:25,470 --> 00:46:29,080
is very close to an
integer, then e to that,

594
00:46:29,080 --> 00:46:31,560
e to the 2 pi times that
i times some number should

595
00:46:31,560 --> 00:46:36,630
be very close to 1, because
what is happening here?

596
00:46:36,630 --> 00:46:42,050
This is the distance
between those two points

597
00:46:42,050 --> 00:46:45,230
on the circle, which
is at most bounded

598
00:46:45,230 --> 00:46:46,490
by the length of the arc.

599
00:46:57,460 --> 00:47:02,530
OK, so cord length,
at most of the arc.

600
00:47:02,530 --> 00:47:05,560
So now, you put
everything here together,

601
00:47:05,560 --> 00:47:09,230
and apply the bound
that we got on d theta.

602
00:47:09,230 --> 00:47:11,410
So this is the reason for
choosing that weird number

603
00:47:11,410 --> 00:47:13,960
up there.

604
00:47:13,960 --> 00:47:30,050
We find that the variation
within each progression

605
00:47:30,050 --> 00:47:31,903
is at most eta, right?

606
00:47:31,903 --> 00:47:33,320
So the variation
of this character

607
00:47:33,320 --> 00:47:36,175
within each progression
is not very large, OK?

608
00:47:36,175 --> 00:47:37,050
And that's the claim.

609
00:47:39,830 --> 00:47:41,780
Any questions?

610
00:47:41,780 --> 00:47:45,920
All right, so this is the
analogous claim to the one

611
00:47:45,920 --> 00:47:46,780
that we had--

612
00:47:46,780 --> 00:47:48,500
the one that we used
last time, where

613
00:47:48,500 --> 00:47:51,950
we said that the character
is constant on each coset

614
00:47:51,950 --> 00:47:52,915
of the hyperplane.

615
00:47:52,915 --> 00:47:55,430
They're not exactly constant,
but almost good enough.

616
00:48:05,610 --> 00:48:06,300
All right.

617
00:48:06,300 --> 00:48:10,020
So the goal of step 2
is to show an energy--

618
00:48:10,020 --> 00:48:13,020
show a density increment, that
if you have a large Fourier

619
00:48:13,020 --> 00:48:15,180
coefficient, then
we want to claim

620
00:48:15,180 --> 00:48:18,720
that the density
goes up significantly

621
00:48:18,720 --> 00:48:21,940
on some subprogression.

622
00:48:21,940 --> 00:48:24,910
And the next part,
the next lemma,

623
00:48:24,910 --> 00:48:27,280
will get us to that goal.

624
00:48:27,280 --> 00:48:29,920
And this part is very
similar to the one

625
00:48:29,920 --> 00:48:34,600
that we saw from last time,
but with this new partition

626
00:48:34,600 --> 00:48:35,980
in mind, like I said.

627
00:48:35,980 --> 00:48:47,860
If you have A that is
3AP-free with density alpha,

628
00:48:47,860 --> 00:48:57,990
and N is large
enough, then there

629
00:48:57,990 --> 00:49:08,980
exists some subprogression
P. So by subprogression,

630
00:49:08,980 --> 00:49:11,560
I just mean that I'm starting
with original progression

631
00:49:11,560 --> 00:49:15,340
1 through N, and I'm zooming
into some subprogression,

632
00:49:15,340 --> 00:49:22,160
with the length of P fairly
long, so the length of P

633
00:49:22,160 --> 00:49:29,430
is at least cube root
of N, and such that A,

634
00:49:29,430 --> 00:49:32,760
when restricted to
this subprogression,

635
00:49:32,760 --> 00:49:36,295
has a density increment.

636
00:49:41,995 --> 00:49:44,950
OK, so originally, the
density of A is alpha,

637
00:49:44,950 --> 00:49:47,300
so we're zooming into
some subprogression P,

638
00:49:47,300 --> 00:49:48,980
which is a pretty
long subprogression,

639
00:49:48,980 --> 00:49:50,870
where the density
goes up significantly

640
00:49:50,870 --> 00:49:53,360
from A to essentially A--

641
00:49:53,360 --> 00:49:56,330
from alpha to roughly
alpha plus alpha squared.

642
00:50:00,240 --> 00:50:00,740
OK.

643
00:50:07,120 --> 00:50:10,660
So we start with
A, a 3AP-free set.

644
00:50:10,660 --> 00:50:25,110
So from step 1, there exists
some theta with large--

645
00:50:25,110 --> 00:50:27,932
so that corresponds to a
large Fourier coefficient.

646
00:50:31,790 --> 00:50:46,070
So this sum here is large.

647
00:50:48,782 --> 00:50:54,710
OK, and now we use--

648
00:50:54,710 --> 00:51:00,510
OK, so-- so step 1 obtains us,
you know, this consequence.

649
00:51:00,510 --> 00:51:14,780
And from this theta, now we
apply the lemma up there to--

650
00:51:14,780 --> 00:51:16,760
so we apply lemma
with, let's say,

651
00:51:16,760 --> 00:51:21,350
eta being alpha squared over 30.

652
00:51:21,350 --> 00:51:23,690
OK, so the exact constants
are not so important.

653
00:51:23,690 --> 00:51:31,010
But when we apply the
lemma to partition,

654
00:51:31,010 --> 00:51:37,720
and into a bunch
of subprogressions,

655
00:51:37,720 --> 00:51:40,570
which we'll call P1 through Pk.

656
00:51:43,462 --> 00:51:48,080
And each of these
progressions have

657
00:51:48,080 --> 00:51:59,750
length between cube root of
N and twice cube root of N.

658
00:51:59,750 --> 00:52:04,070
And I want to understand what
happens to the density of A

659
00:52:04,070 --> 00:52:07,680
when restricted to
these progressions.

660
00:52:07,680 --> 00:52:14,030
So starting with this
inequality over here,

661
00:52:14,030 --> 00:52:17,150
which suggests to us that
there must be some deviation.

662
00:52:30,191 --> 00:52:34,700
OK, so starting
with what we saw.

663
00:52:34,700 --> 00:52:40,950
And now, inside each
progression this e x theta

664
00:52:40,950 --> 00:52:44,050
is roughly constant.

665
00:52:44,050 --> 00:52:46,760
So if you pretend them
as actually constant,

666
00:52:46,760 --> 00:52:54,230
I can break up the sum,
depending on where the x's lie.

667
00:52:54,230 --> 00:52:57,530
So i from 1 to k.

668
00:52:57,530 --> 00:53:01,492
And let me sum inside
each progression.

669
00:53:13,560 --> 00:53:17,610
So by triangle inequality, I
can upper bound the first sum

670
00:53:17,610 --> 00:53:23,540
by where I now cut the sum into
progression by progression.

671
00:53:23,540 --> 00:53:29,520
And on each progression, this
character is roughly constant.

672
00:53:29,520 --> 00:53:34,760
So let me take out the
maximum possible deviations

673
00:53:34,760 --> 00:53:36,440
from them being constant.

674
00:53:36,440 --> 00:53:48,525
So upper bound-- again, you'll
find that we can essentially

675
00:53:48,525 --> 00:53:49,025
pretend--

676
00:53:52,190 --> 00:53:55,410
all right, so if
each exponential

677
00:53:55,410 --> 00:53:57,450
is constant on each
subprogression,

678
00:53:57,450 --> 00:54:00,750
then I might as well
just have this sum here.

679
00:54:00,750 --> 00:54:03,510
But I lose a little bit, because
it's not exactly constant.

680
00:54:03,510 --> 00:54:04,680
It's almost constant.

681
00:54:04,680 --> 00:54:06,140
So I loose a little bit.

682
00:54:06,140 --> 00:54:11,350
And that little bit is this eta.

683
00:54:11,350 --> 00:54:13,440
So you lose that
little bit of eta.

684
00:54:13,440 --> 00:54:18,960
And so on each
progression, P i, you

685
00:54:18,960 --> 00:54:22,830
lose at most something that's
essentially of alpha squared

686
00:54:22,830 --> 00:54:24,265
times the length of P i.

687
00:54:27,240 --> 00:54:29,160
OK.

688
00:54:29,160 --> 00:54:32,880
Now, you see, I've chosen
the error parameter

689
00:54:32,880 --> 00:54:37,260
so that everything
I've lost is not

690
00:54:37,260 --> 00:54:41,250
so much more than the
initial bound I began with.

691
00:54:41,250 --> 00:54:47,100
So in particular,
we see that even

692
00:54:47,100 --> 00:54:49,800
if we had pretended that the
characters were constant,

693
00:54:49,800 --> 00:54:54,090
on each progression we would
have still obtained some lower

694
00:54:54,090 --> 00:54:56,880
bound of the total deviation.

695
00:55:08,680 --> 00:55:09,860
OK.

696
00:55:09,860 --> 00:55:15,130
And what is this
quantity over here?

697
00:55:15,130 --> 00:55:18,880
Oh, you see, I'm
restricting each sum

698
00:55:18,880 --> 00:55:23,590
to each subprogression,
but the sum

699
00:55:23,590 --> 00:55:26,440
here, even though it's
the sum, but it's really

700
00:55:26,440 --> 00:55:31,990
counting how many elements
of A are in that progression.

701
00:55:31,990 --> 00:55:35,710
So this sum over here
is the same thing.

702
00:55:35,710 --> 00:55:38,543
OK, so let me write
it in a new board.

703
00:55:43,480 --> 00:55:45,213
Oh, we don't need
step 1 anymore.

704
00:55:49,077 --> 00:55:51,270
All right.

705
00:55:51,270 --> 00:55:52,440
So what we have--

706
00:56:01,386 --> 00:56:05,431
OK, so left-hand side over
there is this quantity here,

707
00:56:05,431 --> 00:56:07,690
all right?

708
00:56:07,690 --> 00:56:15,890
We see that the right-hand side,
even though you have that sum,

709
00:56:15,890 --> 00:56:19,010
it is really just counting
how many elements of A

710
00:56:19,010 --> 00:56:23,360
are in each progression versus
how many you should expect

711
00:56:23,360 --> 00:56:28,750
based on the overall
density of A. OK,

712
00:56:28,750 --> 00:56:32,220
so that should look similar
to what we got last time.

713
00:56:32,220 --> 00:56:35,280
I know the intuition
should be that, well,

714
00:56:35,280 --> 00:56:42,210
if the average deviation
is large, then one of them,

715
00:56:42,210 --> 00:56:47,447
one of these terms, should
have the density increment.

716
00:56:51,190 --> 00:56:53,950
If you try to do the next
step somewhat naively,

717
00:56:53,950 --> 00:56:56,840
you run into an issue,
because it could be--

718
00:56:56,840 --> 00:56:59,290
now, here you have k terms.

719
00:56:59,290 --> 00:57:07,270
It could be that you have all
the densities except for one

720
00:57:07,270 --> 00:57:12,640
going up only slightly, and one
density dropping dramatically,

721
00:57:12,640 --> 00:57:15,580
in which case you may not
have a significant density

722
00:57:15,580 --> 00:57:18,210
increment, all right?

723
00:57:18,210 --> 00:57:20,840
So we want to show that
on some progression

724
00:57:20,840 --> 00:57:24,350
the density increases
significantly.

725
00:57:24,350 --> 00:57:26,870
So far, from this
inequality, we just

726
00:57:26,870 --> 00:57:29,720
know that there is some
subprogression where

727
00:57:29,720 --> 00:57:32,450
the density changes
significantly.

728
00:57:32,450 --> 00:57:34,850
But of course, the overall
density, the average density,

729
00:57:34,850 --> 00:57:36,120
should remain constant.

730
00:57:36,120 --> 00:57:38,950
So if some goes up,
others must go down.

731
00:57:38,950 --> 00:57:41,180
But if you just try to
do an averaging argument,

732
00:57:41,180 --> 00:57:42,890
you have to be careful, OK?

733
00:57:42,890 --> 00:57:45,290
So there was a trick last
time, which we didn't really

734
00:57:45,290 --> 00:57:49,910
need last time, but now,
it's much more useful, where

735
00:57:49,910 --> 00:57:52,220
I want to show
that if this holds,

736
00:57:52,220 --> 00:57:55,850
then some P i sees
a large energy--

737
00:57:55,850 --> 00:57:58,520
sees a large density increment.

738
00:57:58,520 --> 00:58:08,990
And to do that, let me rewrite
the sum as the following,

739
00:58:08,990 --> 00:58:11,660
so I keep the same expression.

740
00:58:14,880 --> 00:58:18,980
And I add a term, which
is the same thing,

741
00:58:18,980 --> 00:58:20,560
but without the absolute value.

742
00:58:29,434 --> 00:58:35,000
OK, so you see these
guys, they total to 0,

743
00:58:35,000 --> 00:58:40,350
so adding that term doesn't
change my expression.

744
00:58:40,350 --> 00:58:45,460
But now, the summand
is always non-negative.

745
00:58:45,460 --> 00:58:49,380
So it's either 0 or
twice this number,

746
00:58:49,380 --> 00:58:51,500
depending on the
sign of that number.

747
00:58:54,610 --> 00:58:55,110
OK.

748
00:58:55,110 --> 00:58:57,600
So comparing left-hand
side and right-hand side,

749
00:58:57,600 --> 00:59:00,790
we see that there
must be some i.

750
00:59:00,790 --> 00:59:08,730
So hence, there exists some eye
such that the left-hand side--

751
00:59:13,410 --> 00:59:15,270
the i-th term on
the left hand side

752
00:59:15,270 --> 00:59:18,272
is less than or equal
to the i-th term

753
00:59:18,272 --> 00:59:19,230
on the right-hand side.

754
00:59:24,940 --> 00:59:28,640
And in particular, that
term should be positive,

755
00:59:28,640 --> 00:59:31,460
so it implies--

756
00:59:31,460 --> 00:59:35,740
OK, so how can you
get this inequality?

757
00:59:35,740 --> 00:59:40,540
It implies simply that the
restriction of a to this P i

758
00:59:40,540 --> 00:59:50,810
is at least alpha plus alpha
squared over 40 times P i.

759
00:59:50,810 --> 00:59:54,320
So this claim here just says
that on the i-th progression,

760
00:59:54,320 --> 00:59:58,570
there's a significant
energy increment.

761
00:59:58,570 --> 01:00:02,342
If it's more decrement,
that term would have been 0.

762
01:00:02,342 --> 01:00:04,610
So remember that.

763
01:00:16,040 --> 01:00:17,280
OK.

764
01:00:17,280 --> 01:00:20,610
So this achieves what we
were looking for in step 2,

765
01:00:20,610 --> 01:00:24,720
namely to find that there
is a density increment

766
01:00:24,720 --> 01:00:28,041
on some long subprogression.

767
01:00:28,041 --> 01:00:30,630
OK, so now we can go to
step 3, which is basically

768
01:00:30,630 --> 01:00:35,460
the same as what we saw
last time, where now we

769
01:00:35,460 --> 01:00:38,929
want to iterate this
density increment.

770
01:00:47,911 --> 01:00:55,900
OK, so it's basically the
same argument as last time,

771
01:00:55,900 --> 01:01:00,300
but you start with
density alpha,

772
01:01:00,300 --> 01:01:04,140
and each step in the
iteration, the density

773
01:01:04,140 --> 01:01:05,600
goes up quite a bit.

774
01:01:09,740 --> 01:01:14,030
And we want to control
the total number of steps,

775
01:01:14,030 --> 01:01:20,704
knowing that the final
density is at most 1, always.

776
01:01:20,704 --> 01:01:23,370
OK.

777
01:01:23,370 --> 01:01:25,020
So how many steps can you take?

778
01:01:25,020 --> 01:01:28,080
Right, so this was the same
argument that we saw last time.

779
01:01:28,080 --> 01:01:38,450
We see that starting with
alpha, alpha 0 being alpha,

780
01:01:38,450 --> 01:01:41,810
it doubles after a certain
number of steps, right?

781
01:01:41,810 --> 01:01:45,590
So we double after--

782
01:01:45,590 --> 01:01:48,710
OK, so how many
steps do you need?

783
01:01:48,710 --> 01:01:53,430
Well, I want to get
from alpha to 2 alpha.

784
01:01:53,430 --> 01:02:00,960
So I need at most
alpha over 40 steps.

785
01:02:00,960 --> 01:02:03,770
OK, so last time I
was slightly sloppy.

786
01:02:03,770 --> 01:02:09,630
And so there's basically a
floor, upper floor down--

787
01:02:09,630 --> 01:02:11,060
rounding up or down situation.

788
01:02:11,060 --> 01:02:13,620
But I should add a plus 1.

789
01:02:13,620 --> 01:02:14,120
Yeah.

790
01:02:14,120 --> 01:02:16,120
AUDIENCE: Shouldn't
it be 40 over alpha?

791
01:02:16,120 --> 01:02:17,340
YUFEI ZHAO: 40-- thank you.

792
01:02:17,340 --> 01:02:19,930
40 over alpha, yeah.

793
01:02:19,930 --> 01:02:25,080
So you double after at
most that many steps.

794
01:02:25,080 --> 01:02:31,560
And then now, you add
density at least 2 alpha.

795
01:02:31,560 --> 01:02:39,880
So we double after at most 20
over alpha steps, and so on.

796
01:02:39,880 --> 01:02:44,690
And we double at most--

797
01:02:44,690 --> 01:02:51,690
well, basically, log sub
2 of 1 over alpha times.

798
01:02:55,658 --> 01:02:58,600
OK, so anyway, putting
everything together,

799
01:02:58,600 --> 01:03:04,350
we see that the
total number of steps

800
01:03:04,350 --> 01:03:08,790
is at most on the
order of 1 over alpha.

801
01:03:12,340 --> 01:03:15,490
When you stop, you can
only stop for one reason,

802
01:03:15,490 --> 01:03:18,210
because in the--

803
01:03:18,210 --> 01:03:18,710
yeah.

804
01:03:18,710 --> 01:03:19,880
So, yeah.

805
01:03:19,880 --> 01:03:24,790
So in step 1,
remember, the iteration

806
01:03:24,790 --> 01:03:28,762
said that the
process terminates.

807
01:03:28,762 --> 01:03:37,589
The process can always go
on, and then terminates

808
01:03:37,589 --> 01:03:43,170
if the length--

809
01:03:43,170 --> 01:03:45,590
so you're now at
step i, so let N i be

810
01:03:45,590 --> 01:03:54,580
the length of the progression,
that step i is at most C times

811
01:03:54,580 --> 01:03:56,990
alpha i to the minus 12th.

812
01:03:56,990 --> 01:04:01,180
So we have a--

813
01:04:01,180 --> 01:04:04,980
right, so provided
that N is large enough,

814
01:04:04,980 --> 01:04:07,020
you can always pass
to a subprogression.

815
01:04:07,020 --> 01:04:09,540
And here, when you pass to
subprogression, of course,

816
01:04:09,540 --> 01:04:12,030
you can re-label
that subprogression.

817
01:04:12,030 --> 01:04:15,140
And it's now, you
know, 1 through N i.

818
01:04:15,140 --> 01:04:16,553
Right, so I can-- it's--

819
01:04:16,553 --> 01:04:17,970
all the progressions
are basically

820
01:04:17,970 --> 01:04:21,373
the same as the first
set of positive integers.

821
01:04:21,373 --> 01:04:23,850
I'm sorry, prefix of
the positive integers.

822
01:04:23,850 --> 01:04:26,590
So when we stop a
step i, you must

823
01:04:26,590 --> 01:04:31,290
have N sub i being at most
this quantity over here, which

824
01:04:31,290 --> 01:04:36,510
is at most C times the initial
density raised to this minus

825
01:04:36,510 --> 01:04:37,450
12th.

826
01:04:37,450 --> 01:04:49,050
So therefore, the initial length
N of the space is bounded by--

827
01:04:49,050 --> 01:04:55,595
well, each time we went
down by a cube root at most.

828
01:04:55,595 --> 01:04:57,560
Right, so the fine--

829
01:04:57,560 --> 01:05:00,980
if you stop a step i,
then the initial length

830
01:05:00,980 --> 01:05:07,670
is at most N sub i to the
3 times 3 to the power of i

831
01:05:07,670 --> 01:05:10,922
each time you're
doing a cube root.

832
01:05:10,922 --> 01:05:13,135
OK, so you put
everything together.

833
01:05:20,280 --> 01:05:22,360
At most that many
iterations when you stop

834
01:05:22,360 --> 01:05:24,781
the length is at most this.

835
01:05:24,781 --> 01:05:26,680
So you put them
together, and then you

836
01:05:26,680 --> 01:05:37,320
find that the N must be at
most double exponential in 1

837
01:05:37,320 --> 01:05:40,140
over the density.

838
01:05:40,140 --> 01:05:47,670
In other words, the
density is at most 1

839
01:05:47,670 --> 01:05:55,665
over log log N, which is what
we claimed in Roth's theorem,

840
01:05:55,665 --> 01:05:57,060
so what we claimed up there.

841
01:06:01,460 --> 01:06:01,960
OK.

842
01:06:01,960 --> 01:06:03,228
So that finishes the proof.

843
01:06:12,210 --> 01:06:13,307
Any questions?

844
01:06:18,780 --> 01:06:22,230
So the message here is that it's
the same proof as last time,

845
01:06:22,230 --> 01:06:24,660
but we need to do
a bit more work.

846
01:06:24,660 --> 01:06:26,220
And none of this
work is difficult,

847
01:06:26,220 --> 01:06:27,780
but there are more technical.

848
01:06:27,780 --> 01:06:29,580
And that's often
the theme that you

849
01:06:29,580 --> 01:06:31,500
see in additive combinatorics.

850
01:06:31,500 --> 01:06:34,650
This is part of the reason
why the finite field model is

851
01:06:34,650 --> 01:06:38,130
a really nice playground,
because there, things

852
01:06:38,130 --> 01:06:41,850
tend to be often cleaner,
but the idea's often similar,

853
01:06:41,850 --> 01:06:43,140
or the same ideas.

854
01:06:43,140 --> 01:06:43,890
Not always.

855
01:06:43,890 --> 01:06:46,242
Next lecture, we'll
see one technique

856
01:06:46,242 --> 01:06:47,700
where there's a
dramatic difference

857
01:06:47,700 --> 01:06:50,940
between the finite field vector
space and over the integers.

858
01:06:50,940 --> 01:06:53,170
But for many things in
additive combinatorics,

859
01:06:53,170 --> 01:06:54,720
the finite field
vector space is just

860
01:06:54,720 --> 01:06:58,172
a nicer place to be in to try
all your ideas and techniques.

861
01:07:04,440 --> 01:07:09,090
Let me comment on some analogies
between these two approaches

862
01:07:09,090 --> 01:07:10,200
and compare the bounds.

863
01:07:13,090 --> 01:07:15,280
So on one hand--

864
01:07:15,280 --> 01:07:20,130
OK, so we saw last time
this proof in F3 to the N,

865
01:07:20,130 --> 01:07:26,770
and now in the integers
inside this interval of length

866
01:07:26,770 --> 01:07:33,280
N. So let me write
uppercase N in both cases

867
01:07:33,280 --> 01:07:38,205
to be the size of the
overall ambient space.

868
01:07:38,205 --> 01:07:43,310
OK, so what kind of bounds
do we get in both situations?

869
01:07:43,310 --> 01:07:46,050
So last time, for--

870
01:07:46,050 --> 01:07:51,720
in F3 to the N, we
got a bound which

871
01:07:51,720 --> 01:08:06,220
is of the order N over log N,
whereas today, the bound is

872
01:08:06,220 --> 01:08:08,560
somewhat worse.

873
01:08:08,560 --> 01:08:09,920
It's a little bit worse.

874
01:08:09,920 --> 01:08:13,690
Now we lose an extra log.

875
01:08:13,690 --> 01:08:16,810
So where do we lose an
extra log in this argument?

876
01:08:16,810 --> 01:08:19,060
So where does these
two argument--

877
01:08:19,060 --> 01:08:22,980
where do these two arguments
differ, quantitatively?

878
01:08:26,240 --> 01:08:26,740
Yep?

879
01:08:26,740 --> 01:08:29,569
AUDIENCE: When you're dividing
by 3 versus [INAUDIBLE]??

880
01:08:29,569 --> 01:08:31,560
YUFEI ZHAO: OK, so
you're dividing by--

881
01:08:31,560 --> 01:08:41,960
so here, in each
iteration, over here,

882
01:08:41,960 --> 01:08:46,399
your size of the iteration--

883
01:08:46,399 --> 01:08:48,300
I mean, each iteration
the size of the space

884
01:08:48,300 --> 01:08:53,279
goes down by a factor of
3, whereas over here, it

885
01:08:53,279 --> 01:08:56,979
could go down by a cube root.

886
01:08:56,979 --> 01:08:58,399
And that's precisely right.

887
01:08:58,399 --> 01:09:02,420
So that explains for this
extra log in the balance.

888
01:09:05,630 --> 01:09:08,490
So while this is
a great analogy,

889
01:09:08,490 --> 01:09:10,760
it's not a perfect analogy.

890
01:09:10,760 --> 01:09:13,279
You see there is this
divergence here between the two

891
01:09:13,279 --> 01:09:14,790
situations.

892
01:09:14,790 --> 01:09:16,430
And so then you
might ask, is there

893
01:09:16,430 --> 01:09:21,240
some way to avoid the
loss, this extra log factor

894
01:09:21,240 --> 01:09:22,950
loss over here?

895
01:09:22,950 --> 01:09:25,260
Is there some way to
carry out the strategy

896
01:09:25,260 --> 01:09:28,689
that we did last
time in a way that

897
01:09:28,689 --> 01:09:31,300
is much more faithful to
that strategy of passing down

898
01:09:31,300 --> 01:09:33,060
to subspaces.

899
01:09:33,060 --> 01:09:36,090
So here, we pass
to progressions.

900
01:09:36,090 --> 01:09:40,649
And because we have to do this
extra pigeonhole-type argument,

901
01:09:40,649 --> 01:09:41,970
it was somewhat lost--

902
01:09:41,970 --> 01:09:48,770
we lost a power, which
translated into this extra log.

903
01:09:48,770 --> 01:09:51,910
So it turns out there
is some way to do this.

904
01:09:51,910 --> 01:09:53,600
So let me just
briefly mention what's

905
01:09:53,600 --> 01:09:55,640
the idea that is
involved, all right?

906
01:09:55,640 --> 01:09:59,930
So last time, we
went down from--

907
01:09:59,930 --> 01:10:02,660
so the main objects
that we were passing

908
01:10:02,660 --> 01:10:05,180
would start with a vector
space and pass down

909
01:10:05,180 --> 01:10:08,310
to a subspace, which is
also a vector space, right?

910
01:10:08,310 --> 01:10:17,490
So you can define subspaces in
F3 to the N by the following.

911
01:10:17,490 --> 01:10:24,310
So I can start with some set
of characters U, and I define--

912
01:10:24,310 --> 01:10:26,630
some set of characters
S, and I define

913
01:10:26,630 --> 01:10:40,822
U sub S to be basically the
orthogonal complement of S.

914
01:10:40,822 --> 01:10:42,972
OK, so this is a subspace.

915
01:10:42,972 --> 01:10:44,430
And these were the
kind of subspace

916
01:10:44,430 --> 01:10:48,168
that we saw last time, because
the S's or the R's that

917
01:10:48,168 --> 01:10:50,460
came out of the proof last
time, every time we saw one,

918
01:10:50,460 --> 01:10:51,390
we threw it in.

919
01:10:51,390 --> 01:10:56,330
We cut down to a smaller
subspace, and we repeat.

920
01:10:56,330 --> 01:10:59,018
But the progressions, they
don't really look like this.

921
01:10:59,018 --> 01:11:00,560
So the question is,
is there some way

922
01:11:00,560 --> 01:11:02,935
to do this argument so that
you end up with progressions,

923
01:11:02,935 --> 01:11:04,010
and looked like that?

924
01:11:04,010 --> 01:11:06,300
And it turns out there is a way.

925
01:11:06,300 --> 01:11:08,150
And there are these
objects, which

926
01:11:08,150 --> 01:11:13,215
we'll see more later in this
course, called Bohr sets.

927
01:11:13,215 --> 01:11:18,030
OK, so they were
used by Bourgain

928
01:11:18,030 --> 01:11:23,220
to mimic this Machoulin argument
that we saw last time more

929
01:11:23,220 --> 01:11:26,010
faithfully into
the integers, where

930
01:11:26,010 --> 01:11:30,840
we're going to come up with some
set of integers that resemble--

931
01:11:30,840 --> 01:11:34,950
much more closely resemble
this notion of subspaces

932
01:11:34,950 --> 01:11:37,290
in the finite field setting.

933
01:11:37,290 --> 01:11:41,020
And for this, it's much
easier to work inside a group.

934
01:11:41,020 --> 01:11:42,900
So instead of working
in the integers,

935
01:11:42,900 --> 01:11:45,110
let's work inside and Z mod nZ.

936
01:11:45,110 --> 01:11:47,580
So we can do Fourier
transform in Z mod nZ, so

937
01:11:47,580 --> 01:11:50,040
the discrete Fourier
analysis here.

938
01:11:50,040 --> 01:11:52,335
So in Z mod nZ, we define--

939
01:11:55,510 --> 01:11:59,260
so given a S, let's
define this Bohr

940
01:11:59,260 --> 01:12:14,080
set to be the set of
elements of Z mod nZ such

941
01:12:14,080 --> 01:12:20,470
that if you look at what
really is supposed to resemble

942
01:12:20,470 --> 01:12:26,280
this thing over here, OK?

943
01:12:26,280 --> 01:12:34,280
If this quantity is
small for all S--

944
01:12:34,280 --> 01:12:36,930
OK, so we put that element
into this Bohr set.

945
01:12:39,670 --> 01:12:44,870
OK, so these sets, they function
much more like subspaces.

946
01:12:44,870 --> 01:12:48,890
So there are the
analog of subspaces

947
01:12:48,890 --> 01:12:53,420
inside Z mod nZ, which,
you know, n is prime,

948
01:12:53,420 --> 01:12:54,430
has no subgroups.

949
01:12:54,430 --> 01:12:56,990
It has no natural
subspace structure.

950
01:12:56,990 --> 01:12:58,880
But by looking at
these Bohr sets,

951
01:12:58,880 --> 01:13:03,020
they provide a
natural way to set up

952
01:13:03,020 --> 01:13:05,330
this argument so that you can--

953
01:13:05,330 --> 01:13:07,910
but with much more
technicalities,

954
01:13:07,910 --> 01:13:13,630
repeat these kind of arguments
more similar to last time,

955
01:13:13,630 --> 01:13:17,220
but passing not to
subspaces but to Bohr sets.

956
01:13:17,220 --> 01:13:22,270
And then with quite
a bit of extra work,

957
01:13:22,270 --> 01:13:32,370
one can obtain bounds of the
quantity N over a poly log N.

958
01:13:32,370 --> 01:13:36,510
So the current best bound
I mentioned last time

959
01:13:36,510 --> 01:13:40,080
is of this type, which is
through further refinements

960
01:13:40,080 --> 01:13:41,347
of this technique.

961
01:13:50,610 --> 01:13:52,230
The last thing I
want to mention today

962
01:13:52,230 --> 01:13:56,040
is, so far we've been
talking about 3APs.

963
01:13:56,040 --> 01:13:59,600
So what about four term
arithmetic progressions?

964
01:13:59,600 --> 01:14:04,998
OK, do any of the things that we
talk about here work for 4APs?

965
01:14:04,998 --> 01:14:07,290
And there's an analogy to be
made here compared to what

966
01:14:07,290 --> 01:14:09,085
we discussed with graphs.

967
01:14:09,085 --> 01:14:11,770
So in graphs, we had a triangle
counting lemma and a triangle

968
01:14:11,770 --> 01:14:12,930
removal lemma.

969
01:14:12,930 --> 01:14:15,210
And then we said
that to prove 4APs,

970
01:14:15,210 --> 01:14:18,390
we would need the hypergraph
version, the simplex removal

971
01:14:18,390 --> 01:14:20,580
lemma, hypergraph
regularity lemma.

972
01:14:20,580 --> 01:14:22,500
And that was much
more difficult.

973
01:14:22,500 --> 01:14:24,170
And that analogy
carries through,

974
01:14:24,170 --> 01:14:27,040
and the same kind of
difficulties that come up.

975
01:14:27,040 --> 01:14:30,700
So it can be done, but
you need something more.

976
01:14:30,700 --> 01:14:34,590
And the main message I
want you to take away

977
01:14:34,590 --> 01:14:42,940
is that 4APs, while we
had a counting lemma that

978
01:14:42,940 --> 01:14:45,980
says that the
Fourier coefficients,

979
01:14:45,980 --> 01:14:54,540
so the Fourier transform,
controls 3AP counts,

980
01:14:54,540 --> 01:14:59,100
it turns out the same
is not true for 4APs.

981
01:14:59,100 --> 01:15:05,200
So the Fourier does
not control 4AP counts.

982
01:15:10,410 --> 01:15:12,340
Let me give you some--

983
01:15:12,340 --> 01:15:12,840
OK.

984
01:15:12,840 --> 01:15:14,910
So in fact, in the
homework for this week,

985
01:15:14,910 --> 01:15:19,870
there's a specific example of
a set where it has uniformly

986
01:15:19,870 --> 01:15:21,490
small Fourier coefficients.

987
01:15:21,490 --> 01:15:22,990
But that's the wrong
number of 4APs.

988
01:15:27,130 --> 01:15:28,500
So the following-- it is true--

989
01:15:28,500 --> 01:15:32,610
OK, so it is true that you
have Szemerédi's term in--

990
01:15:32,610 --> 01:15:34,770
let's just talk about
the finite field setting,

991
01:15:34,770 --> 01:15:36,840
where things are a
bit easier to discuss.

992
01:15:36,840 --> 01:15:42,510
So it is true that the size
of the biggest subset of F5

993
01:15:42,510 --> 01:15:45,180
to the N is a tiny
fraction-- it's a little, one

994
01:15:45,180 --> 01:15:46,530
fraction of the entire space.

995
01:15:46,530 --> 01:15:50,460
OK, I use F5 here,
because if I set F3,

996
01:15:50,460 --> 01:15:53,690
it doesn't make sense
to talk about 4APs.

997
01:15:53,690 --> 01:15:58,990
So F5, but it doesn't really
matter which specific field.

998
01:15:58,990 --> 01:16:03,550
So you can prove this using
hypergraph removal, same proof,

999
01:16:03,550 --> 01:16:07,330
verbatim, that we saw earlier,
if you have hypergraph removal.

1000
01:16:07,330 --> 01:16:10,610
But if you want to try to prove
it using Fourier analysis,

1001
01:16:10,610 --> 01:16:15,640
well, it doesn't work quite
using the same strategy.

1002
01:16:15,640 --> 01:16:17,590
But in fact, there
is a modification

1003
01:16:17,590 --> 01:16:20,800
that would allow
you to make it work.

1004
01:16:20,800 --> 01:16:23,670
But you need an extension
of Fourier analysis.

1005
01:16:23,670 --> 01:16:30,790
And it is known as higher
order Fourier analysis, which

1006
01:16:30,790 --> 01:16:35,470
was an important development in
modern additive combinatorics

1007
01:16:35,470 --> 01:16:38,590
that initially arose
in Gowers' work

1008
01:16:38,590 --> 01:16:41,190
where he gave a new proof
of similarities theorem.

1009
01:16:41,190 --> 01:16:42,880
So Gowers didn't
work in this setting.

1010
01:16:42,880 --> 01:16:44,020
He worked in integers.

1011
01:16:44,020 --> 01:16:47,140
But many of the ideas
originated from his paper,

1012
01:16:47,140 --> 01:16:49,120
and then subsequently
developed by a lot

1013
01:16:49,120 --> 01:16:51,250
of people in various settings.

1014
01:16:51,250 --> 01:16:53,877
I just want to give you one
specific statement, what

1015
01:16:53,877 --> 01:16:55,710
this high-order Fourier
analysis looks like.

1016
01:16:55,710 --> 01:16:58,450
So it's a fancy term,
and the statements often

1017
01:16:58,450 --> 01:17:00,070
get very technical.

1018
01:17:00,070 --> 01:17:04,140
But I just want to give you one
concrete thing to take away.

1019
01:17:04,140 --> 01:17:05,790
All right, so for
a Fourier piece,

1020
01:17:05,790 --> 01:17:08,220
higher-order Fourier
analysis, roughly--

1021
01:17:08,220 --> 01:17:12,484
OK, so it also goes by the name
quadratic Fourier analysis.

1022
01:17:16,436 --> 01:17:22,370
OK, so let me give you
a very specific instance

1023
01:17:22,370 --> 01:17:23,570
of the theorem.

1024
01:17:27,090 --> 01:17:31,710
And this can be sometimes
called an inverse theorem

1025
01:17:31,710 --> 01:17:34,272
for quadratic Fourier analysis.

1026
01:17:40,790 --> 01:17:48,700
OK, so for every delta,
there exists some c such

1027
01:17:48,700 --> 01:17:51,490
that the following is true.

1028
01:17:51,490 --> 01:18:03,390
If A is a subset of a F5 to the
N with density alpha and such

1029
01:18:03,390 --> 01:18:05,132
that it's--

1030
01:18:05,132 --> 01:18:07,750
OK, so now lambda
sub 4, so this is

1031
01:18:07,750 --> 01:18:10,380
the 4AP density,
so similar to 3AP,

1032
01:18:10,380 --> 01:18:11,710
but now you write four terms.

1033
01:18:11,710 --> 01:18:18,630
The 4AP density of A differs
from alpha to the fourth

1034
01:18:18,630 --> 01:18:21,542
by a significant amount.

1035
01:18:21,542 --> 01:18:25,050
OK, so for 3APs, then we said
that now A has a large Fourier

1036
01:18:25,050 --> 01:18:26,750
coefficient, right?

1037
01:18:29,700 --> 01:18:33,640
So for-- OK.

1038
01:18:33,640 --> 01:18:37,530
For 4APs, that may not be
true, but the following

1039
01:18:37,530 --> 01:18:38,375
is true, right?

1040
01:18:38,375 --> 01:18:48,668
So then there exists a
non-zero quadratic polynomial.

1041
01:18:55,500 --> 01:19:08,620
F and N variables over F5
such that the indicator

1042
01:19:08,620 --> 01:19:13,280
function of A correlates with
this quadratic exponential

1043
01:19:13,280 --> 01:19:13,780
face.

1044
01:19:23,200 --> 01:19:24,833
So Fourier analysis,
the conclusion

1045
01:19:24,833 --> 01:19:26,250
that we got from
counting lemma is

1046
01:19:26,250 --> 01:19:28,440
that you have some
linear function

1047
01:19:28,440 --> 01:19:33,650
F, such that this
quantity is large,

1048
01:19:33,650 --> 01:19:35,630
this large Fourier coefficient.

1049
01:19:35,630 --> 01:19:37,980
OK, so that is not true 4APs.

1050
01:19:37,980 --> 01:19:39,870
But what is true
is that now you can

1051
01:19:39,870 --> 01:19:44,860
look at quadratic exponential
faces, and then it is true.

1052
01:19:44,860 --> 01:19:48,720
So that's the content
of higher order Fourier.

1053
01:19:48,720 --> 01:19:51,660
I mean, that's the example of
higher-order Fourier analysis.

1054
01:19:51,660 --> 01:19:56,010
And you can imagine with
this type of result,

1055
01:19:56,010 --> 01:19:58,200
and with quite a
bit more work, you

1056
01:19:58,200 --> 01:20:01,560
can try to follow a
similar density increment

1057
01:20:01,560 --> 01:20:07,110
strategy to prove
similarities term for 4APs.