This post is a continuation of the previous post on sieve theory, which is an ongoing part of the Polymath8 project. As the previous post was getting somewhat full, we are rolling the thread over to the current post.
In this post we will record a new truncation of the elementary Selberg sieve discussed in this previous post (and also analysed in the context of bounded prime gaps by Graham-Goldston-Pintz-Yildirim and Motohashi-Pintz) that was recently worked out by Janos Pintz, who has kindly given permission to share this new idea with the Polymath8 project. This new sieve decouples the parameter that was present in our previous analysis of Zhang’s argument into two parameters, a quantity
that used to measure smoothness in the modulus, but now measures a weaker notion of “dense divisibility” which is what is really needed in the Elliott-Halberstam type estimates, and a second quantity
which still measures smoothness but is allowed to be substantially larger than
. Through this decoupling, it appears that the
type losses in the sieve theoretic part of the argument can be almost completely eliminated (they basically decay exponential in
and have only mild dependence on
, whereas the Elliott-Halberstam analysis is sensitive only to
, allowing one to set
far smaller than previously by keeping
large). This should lead to noticeable gains in the
quantity in our analysis.
To describe this new truncation we need to review some notation. As in all previous posts (in particular, the first post in this series), we have an asymptotic parameter going off to infinity, and all quantities here are implicitly understood to be allowed to depend on
(or to range in a set that depends on
) unless they are explicitly declared to be fixed. We use the usual asymptotic notation
relative to this parameter
. To be able to ignore local factors (such as the singular series
), we also use the “
-trick” (as discussed in the first post in this series): we introduce a parameter
that grows very slowly with
, and set
.
For any fixed natural number , define an admissible
-tuple to be a fixed tuple
of
distinct integers which avoids at least one residue class modulo
for each prime
. Our objective is to obtain the following conjecture
for as small a value of the parameter
as possible:
Conjecture 1 (
) Let
be a fixed admissible
-tuple. Then there exist infinitely many translates
of
that contain at least two primes.
The twin prime conjecture asserts that holds for
as small as
, but currently we are only able to establish this result for
(see this comment). However, with the new truncated sieve of Pintz described in this post, we expect to be able to lower this threshold
somewhat.
In previous posts, we deduced from a technical variant
of the Elliot-Halberstam conjecture for certain choices of parameters
,
. We will use the following formulation of
:
Conjecture 2 (
) Let
be a fixed
-tuple (not necessarily admissible) for some fixed
, and let
be a primitive residue class. Then
for any fixed
, where
,
are the square-free integers whose prime factors lie in
, and
is the quantity
and
is the set of congruence classes
and
is the polynomial
The conjecture is currently known to hold whenever
(see this comment and this confirmation). Actually, we can prove a stronger result than
in this regime in a couple ways. Firstly, the congruence classes
can be replaced by a more general system of congruence classes obeying a certain controlled multiplicity axiom; see this post. Secondly, and more importantly for this post, the requirement that the modulus
lies in
can be relaxed; see below.
To connect the two conjectures, the previously best known implication was the folowing (see Theorem 2 from this post):
Theorem 3 Let
,
and
be such that we have the inequality
where
is the first positive zero of the Bessel function
, and
are the quantities
and
Then
implies
.
Actually there have been some slight improvements to the quantities ; see the comments to this previous post. However, the main error
remains roughly of the order
, which limits one from taking
too small.
To improve beyond this, the first basic observation is that the smoothness condition , which implies that all prime divisors of
are less than
, can be relaxed in the proof of
. Indeed, if one inspects the proof of this proposition (described in these three previous posts), one sees that the key property of
needed is not so much the smoothness, but a weaker condition which we will call (for lack of a better term) dense divisibility:
Definition 4 Let
. A positive integer
is said to be
-densely divisible if for every
, one can find a factor of
in the interval
. We let
denote the set of positive integers that are
-densely divisible.
Certainly every integer which is -smooth (i.e. has all prime factors at most
is also
-densely divisible, as can be seen from the greedy algorithm; but the property of being
-densely divisible is strictly weaker than
-smoothness, which is a fact we shall exploit shortly.
We now define to be the same statement as
, but with the condition
replaced by the weaker condition
. The arguments in previous posts then also establish
whenever
.
The main result of this post is then the following implication, essentially due to Pintz:
Theorem 5 Let
,
,
, and
be such that
where
and
and
Then
implies
.
This theorem has rather messy constants, but we can isolate some special cases which are a bit easier to compute with. Setting , we see that
vanishes (and the argument below will show that we only need
rather than
), and we obtain the following slight improvement of Theorem 3:
Theorem 6 Let
,
and
be such that we have the inequality
where
Then
implies
.
This is a little better than Theorem 3, because the error has size about
, which compares favorably with the error in Theorem 3 which is about
. This should already give a “cheap” improvement to our current threshold
, though it will fall short of what one would get if one fully optimised over all parameters in the above theorem.
Returning to the full strength of Theorem 5, let us obtain a crude upper bound for that is a little simpler to understand. Extending the
summation to infinity and using the Taylor series for the exponential, we have
We can crudely bound
and then optimise in to obtain
Because of the factor in the integrand for
and
, we expect the ratio
to be of the order of
, although one will need some theoretical or numerical estimates on Bessel functions to make this heuristic more precise. Setting
to be something like
, we get a good bound here as long as
, which at current values of
is a mild condition.
Pintz’s argument uses the elementary Selberg sieve, discussed in this previous post, but with a more efficient estimation of the quantity , in particular avoiding the truncation to moduli
between
and
which was the main source of inefficiency in that previous post. The basic idea is to “linearise” the effect of the truncation of the sieve, so that this contribution can be dealt with by the union bound (basically, bounding the contribution of each large prime one at a time). This mostly avoids the more complicated combinatorial analysis that arose in the analytic Selberg sieve, as seen in this previous post.
— 1. Review of previous material —
In this section we collect some results from previous posts which we will need.
We first record an asymptotic for multiplicative functions. For any natural number , define a
-dimensional multiplicative function to be a multiplicative function
which obeys the asymptotic
for all . The following result is Lemma 8 from this previous post:
Lemma 7 Let
. Let
be a fixed positive integer, and let
be a multiplicative function of dimension
. Then for any fixed compactly supported, Riemann-integrable function
, and any
for some fixed
, one has
Next, we record a criterion for , which is Lemma 7 from this previous post:
Lemma 8 (Criterion for DHL) Let
. Suppose that for each fixed admissible
-tuple
and each congruence class
such that
is coprime to
for all
, one can find a non-negative weight function
, fixed quantities
, a quantity
, and a fixed positive power
of
such that one has the upper bound
for all
, and the key inequality
holds. Then
holds. Here
is defined to equal
when
is prime and
otherwise.
— 2. Pintz’s argument —
We can now prove Theorem 5. Fix to obey the hypotheses of this theorem. Let
be a congruence class with
coprime to
for all
(this class exists by the admissibility of
). We apply Lemma 8 with
the elementary Selberg sieve defined by
are the multiplicative functions
and
the sieve level is given by the formula
is a fixed smooth function supported on
, and
is a certain subset of
to be chosen shortly. (The constraints
and
are redundant given that
, but we retain them for emphasis.) We will assume that
non-negative and non-increasing on
. In this previous post, we considered this sieve with
equal to either all of
, or the subset
consisting of smooth numbers. For now, we will discuss the estimates as far as we can without having to explicitly specify
.
We first consider the asymptotic (5). By arguing exactly as in Section 2 (or Section 3) of this previous post, we can write the left-hand side of (5), up to errors of , as
The summand here is non-negative, so we may crudely replace by all of
and apply Lemma 7 to obtain (5) with
Now we turn to the more difficult asymptotic (6). The left-hand side expands as
As observed in Section 2 of this previous post, we have
where
and
Now let be a small fixed constant to be chosen later, and suppose the following claim holds:
Claim 1 (Dense divisibility of moduli) Whenever
and
are non-zero, then either
or else
.
Then from the Bombieri-Vinogradov theorem (for the moduli) or the hypothesis
(for the larger moduli, noting that
) and standard arguments (cf. Proposition 5 of this post) we have
for any fixed and any multiplicative function
of a fixed dimension
, where
ranges only over those integers of the form
with
non-zero. From this we easily see (arguing as in Section 2 of this previous post) that the contribution of the
error term is
, and we are left with establishing the lower bound
up to errors of (henceforth referred to as negligible errors).
As in Section 2 of the previous post, we can write the left-hand side as
where is the
-dimensional multiplicative function
So we would like to select small enough that Claim (1) holds, but large enough that we can lower bound (9) by
up to negligible errors.
Pintz’s idea is to choose to be the set of all elements
of
with the property that
Let us first verify Claim 1 with this definition. Suppose that is non-zero, then from (8) and the support of
there are multiples
,
of
,
respectively with
(in particular
) and
. Since
, we have
and thus
Similarly for and
. Multiplying, we obtain
On the other hand,
and so on dividing we have
We conclude that either is less than
, or else
The latter conclusion implies that is
-densely divisible. Indeed, for any
we multiply together all the prime divisors of
between
and
one at a time until just before one reaches or exceeds
, or if one runs out of such prime divisors. In the former case one ends up at least as large as
, and in the latter case one has reached
. Next, one multiplies to this the prime divisors of
less than
until one reaches or exceeds
; this is possible thanks to (11) in the former case and is automatic in the latter case, and gives a divisor between
and
as required.
Now we need to obtain a lower bound for (9). If we write
and
(the minus sign being to compensate for the non-positive nature of ) then we have
and thus
Note that this inequality replaces the quadratic expression with a linear expression in the truncation error
, which will be more tractable for computing the effect of that error. We may thus lower bound (9) by the difference of
In Section 2 of this post it is shown that
which implies that (12) is equal to
up to negligible errors. Similar considerations show that (13) is equal to
up to negligible errors. To upper bound (14), we need to upper bound
For this we need to catalog the ways in which can fail to be in
. In order for this to occur, at least one of the following three statements must hold:
- (i)
could be divisible by a prime
with
.
- (ii)
could be divisible by a prime
with
.
- (iii)
lies in
, but
.
We consider the contributions of (i), (ii), (iii) to (14). We begin with the contribution of (i). This is bounded above by
Applying Lemma 7, one can simplify this modulo negligible errors as
which by another application of Lemma 7 is equal to
where we adopt the notation
Applying Mertens’ theorem and summation by parts, this expression is equal up to negligible errors to
where
Now we turn to the contribution of (ii) to (14). This is bounded above by
By Lemma 7 we have
and so we may bound this contribution up to negligible errors by
which by Lemma 7 again is equal (up to negligible errors) to
By definition, . By Mertens’ theorem, we can thus write the above expression up to negligible errors as
Finally, we turn to the contribution of case (iii) to (14). By Lemma 7 we have
so we may bound this contribution up to negligible errors by
where is as in case (iii).
We introduce the quantities
and
so that case (iii) consists of those in
such that
From the support of we may also take
. This implies that we may factor
for some primes
and some coprime to
, with
The contribution of this case may thus be bounded by
Evaluating the inner sum using Lemma 7, we obtain (up to negligible errors)
where is a truncation of
:
Again we have . By Mertens’ theorem we may write this (up to negligible errors) as
Putting all this together, we have obtained the lower bound (6) with
where
and
We now place upper bounds on . In this previous post, the bounds
and
for are proven. Thus we have
These are already quite small for , say, which would correspond to
.
For we will use the crude estimate
this may surely be improved, but we will not do so here to simplify the exposition. Then we may bound
The point here is that the first term is exponentially decaying in
, which can compensate for the second term if
which is currently the case in the regime of interest.
One can do a bit better than this. For any parameter , one has
since the left-hand side vanishes for . This gives the bound
If we insert these bounds into (7), send to zero, and optimise in
using Theorem 14 from this previous post, we obtain Theorem 5.
82 comments
Comments feed for this article
18 June, 2013 at 6:43 pm
Andrew Sutherland
Has anyone worked out the “cheap” gain k0 mentioned after Theorem 6? (or better, a fully optimized k0, but even a cheap gain will give those working on optimizing H(k0) something to play with — we’re not used to working with the same k0 for more than 2 days at time :)).
18 June, 2013 at 7:39 pm
v08ltu
If I typed it all correctly,
allows
.
Bz(k)=k+1.85575708149*k^(1/3)+1.033150/k^(1/3);
DELTA(om)=(1/4-87*om)/17;
KAPPA1(om,k0)=local(d);d=DELTA(om);intnum(t=4*d/(1+4*om),1,(1-t)^((k0-1)/2)/t);
KAPPA2(om,k0)=local(d);\
d=DELTA(om);(k0-1)*intnum(t=4*d/(1+4*om),1,(1-t)^((k0-1))/t);
INEQ(om,k0)=(1+4*om)*(1-2*KAPPA1(om,k0)-2*KAPPA2(om,k0))-Bz(k0-2)^2/k0/(k0-1);
INEQ(2/733,5937)
18 June, 2013 at 7:42 pm
Andrew Sutherland
Thanks!
18 June, 2013 at 9:01 pm
v08ltu
If I use the above
bound and assume the
-quotient really is
, I think that I get that
and
with
to satisfy the previous constraint, one can get
. I should write a better program to optimize this though.
18 June, 2013 at 9:27 pm
Terence Tao
From my other comment, I now think the G-quotient is going to be closer to
, where
. I can try to get more precise asymptotics for the
term, but perhaps it will be simpler just to compute the integral numerically (it will serve as a good sanity check to compare the numerics against the asymptotics though, and the asymptotics can give a rough idea of what region of parameter space to search in).
18 June, 2013 at 9:44 pm
v08ltu
In the range of interest, I seem to have
and the other
are also
and
respectively. So none of them matter. This assumes I am not making any errors or typos with formulas.
18 June, 2013 at 9:55 pm
v08ltu
I am really not sure where I came up with the above numbers, there is no obstacle to making
closer to 1/348. With the sech/tanh estimate (ignoring o(1)) I can take
and get
with
, and
is not of import again (make the delta-difference smaller if it is). I am not sure I trust all the numerics though (instability), but this is approaching the limit of previous analysis.
18 June, 2013 at 10:04 pm
v08ltu
Looks like
with
and
works. If you try to take
smaller, then
becomes not negligible, and you have the balance the delta-difference. Too hard to look at all parameters right now.
Ooops, looks like
and
and
works (with all
in play).
18 June, 2013 at 10:06 pm
v08ltu
You can also make
in this last one, and then
is irrelevant (
in sech/tanh), while
is then slightly bigger (5x), but not enough to vitiate key inequality.
18 June, 2013 at 10:10 pm
v08ltu
Managed to squeeze one more,
(larger than previous!) with a
-difference of (say)
allows
.
18 June, 2013 at 11:38 pm
v08ltu
Slightly nicer form is
with
and
. I never checked that making
smaller than forced by
would not gain, but I don’t see why it would (probably follows by monotonicity).
19 June, 2013 at 8:27 am
Terence Tao
Cross-checking these computations with Maple. With
and
we have
. Setting
(say) we have
We need
to not exceed
I get
which looks good so far. Now for
. Using the crude bound
the exponential factor here is
, which looks scary. On the other hand, the G-ratio here is predicted to be roughly
with
(I made a slight mistake earlier in not having the square root and factor of 2 here but it fortunately turns out not to matter), and this is is completely negligible. In order to get an independent confirmation of the smallness of the G-ratio is, the Bessel function
at
takes the value
, and at
takes the value
, so it looks abundantly clear that the G-ratio is indeed ridiculously small in this regime.
19 June, 2013 at 4:30 pm
v08ltu
You have an error in your
expression in the Maple check, it is
in the denominator of the sqrt inside the exp, not
. I think the sech-sqrt fix breaks
, I will check.
19 June, 2013 at 6:51 pm
Terence Tao
Oops, sorry about that. Well, one can hope to squeeze something out of the full version of Theorem 5 (or at least the first display after Theorem 5) in which there is also the A parameter to optimise over. One can probably pick up some other ways to make the estimate for
more efficient, but at the cost of simplicity. Pintz’s own notes have a slightly different way of estimating
, I can try to have a look at that instead.
There were two corrections I had to make; one (which worked against our favour) was the replacing of
with its square root, but the other (which works for us) is the doubling of the exponential from
to
. I had thought that the two effects roughly cancelled each other out, but I guess not quite…
19 June, 2013 at 4:40 pm
v08ltu
Currently, I can only get
with
and
. You can note the incredible impact that
suddenly has if you change “6” to “5” in the numerator of the subtrahend of
. So saving a few more in
could occur from its more benevolent estimation.
19 June, 2013 at 8:14 pm
Terence Tao
I just realised that the estimate
that I used below Theorem 6 can be improved to
(by replacing
with infinity and replacing
with
). This extra exponential factor makes it harder to optimise in A, but it should give a significant savings, hopefully enough to push back to 5453 and maybe even a bit further.
18 June, 2013 at 8:25 pm
Terence Tao
I’m recording here some asymptotics of Bessel functions to help understand the ratio
in the regime where
is fixed and
is large. Here we have made the change of variable
.
Write
. By 9.3.23 of Abramowitz and Stegun (see e.g. this link) we have the asymptotic
for bounded
, where
is the Airy function. This shows that the zero
has the asymptotic
where
is the first negative zero of the Airy function. It also suggests that
Meanwhile, 9.3.7 of Abramowitz and Stegun (Debye asymptotic) tells us that
for any
(this is also consistent with the exponential decay of the Airy function to the right), so if we set
(EDIT: this should be
), the ratio (*) should take the form
(EDIT: this should be
) which is exponentially decaying in
as claimed in the post. (Unfortunately, we cannot directly use these asymptotics to get explicit values of
because we need control on the error terms, which standard sources (e.g. Abramowitz-Stegun) do not supply.)
19 June, 2013 at 5:22 pm
Eytan Paldi
Please note that the upper bound for the integration in the numerator is defined in theorem 5 to be
(not its square), and in the presentation of Pintz argument it is defined (for a general pair
) to be
, so it is not clear which is the correct one.
[Corrected, thanks -T.]
19 June, 2013 at 6:25 pm
Eytan Paldi
In my last message I did not notice that the integrals are after change of variable, but I think that in the numerator
should be replaced by its square root. And more importantly, the integral (at least for even
) can be precisely evaluated as a finite linear combination of squares of Bessel functions by finding the indefinite integral using the recursive formula (11.3.33) and the formula (11.3.34) in Abramowitz and Stegun.
, then the recursive formula can be used to compute the integral for odd
as well.)
(If there is a corresponding formula for
19 June, 2013 at 5:49 am
coco
typo: divsibility -> divisibility, in second paragraph.
[Corrected, thanks – T.]
19 June, 2013 at 7:07 am
Gergely Harcos
The nonnegativity property I tried to prove in the previous thread (https://terrytao.wordpress.com/2013/06/11/further-analysis-of-the-truncated-gpy-sieve/) is equivalent to
Unfortunately, SAGE found the following counterexample for
and
:
More precisely, with these parameters we have
This result is verified by the SAGE commands
A={20/100, 21/100, 24/100, 25/100, 31/100}
print sum([((-1)^U.cardinality())*sum([min(max(0,1-sum(S.union(U))),max(0,1-sum(A.difference(S))))^2 for S in Subsets(A.difference(U))]) for U in Subsets(A)])
19 June, 2013 at 7:41 am
Terence Tao
A minor technical point: there is one tiny way in which
-densely divisible numbers are worse than
-smooth numbers: the behaviour is not quite hereditary with respect to factors. If
and
is
-smooth, then
is automatically
-smooth as well. On the other hand, if
is merely
-densely divisible, then
is only
-densely divisible, there is a loss of
(basically because
could suck up all the useful small primes that are giving the dense divisibility property). This makes the Type III analysis a little more complicated as there is a part in which one forces the frequency parameter
to be coprime to the modulus
by Mobius inversion, dividing both
and
by the common factor
. However even with the worsening of
by
, one can check that the net power of
remains favorable and the coprime case still dominates, so this is not actually an issue. (But something to look out for when we do the final writing up of the arguments.)
19 June, 2013 at 9:10 am
Terence Tao
For reference, I have uploaded (with permission) Pintz’s original notes on this sieve here:
Section 1
Section 2
Section 3
19 June, 2013 at 8:27 pm
v08ltu
OK, I think with
and
and
this now recovers
.
19 June, 2013 at 8:29 pm
v08ltu
And
with
seems to give
with the same
.
20 June, 2013 at 9:35 am
Terence Tao
I verified that this works with the Maple code below (using crude upper and lower bounds on the Bessel integral after verifying that the Bessel function
was decreasing from
to
, and was on the other side of the maximum at
). By using the
bound in the display after Theorem 6 rather than perform any further estimations, there is actually a little bit of margin:
is allowed to be as large as
but is only of size
, with
dominating,
, and
. It seems likely that we could squeeze
by one or two more by changing the free parameters
a bit, but it is very tight and I didn’t try too hard to do so.
varpi := 1/348 - 4/2000000;
k0 := 5452;
deltap := 1/820;
A := 9500;
delta := (1/4 - 87 * varpi) / 17;
theta := deltap / (1/4 + varpi);
thetat := (deltap - delta + varpi) / (1/4 + varpi);
deltat := delta / (1/4 + varpi);
j := BesselJZeros(k0-2,1);
eps := 1 - j^2 / (k0 * (k0-1) * (1+4*varpi));
kappa1 := int( (1-t)^((k0-1)/2)/t, t = theta..1, numeric);
kappa2 := (k0-1) * int( (1-t)^(k0-1)/t, t=theta..1, numeric);
e := exp( A + (k0-1) * int( exp(-A*t)/t, t=deltat..theta, numeric ) );
s1 := BesselJ(k0-2,0.999*j);
s2 := BesselJ(k0-2,0.998*j);
s3 := BesselJ(k0-2,0.997*j);
# check that s3 > s2 > s1 to ensure Bessel function decreases from 0.998 j to 0.999 j
# crude lower bound on the denominator integral
gd := s1^2 * 0.001 * j * 0.998 * j;
# crude upper bound on the numerator integral
tn := sqrt(thetat)*j;
s0 := BesselJ(k0-2,tn);
gn := s0^2 * tn * tn / 2;
# check that s0 < s3 to ensure that Bessel function decreases up to tn (but this is true with massive amounts of room to spare)
kappa3 := (gn/gd) * e;
eps2 := 2*(kappa1+kappa2+kappa3);
# we win if eps2 < eps
20 June, 2013 at 9:37 am
abqmathteacher
A quick nontechnical question. Any idea of the likelihood of pushing this line of reasoning down to H < 1476, the largest known (I think) "maximal prime gap" (see http://oeis.org/A005250, http://oeis.org/A005250/a005250.txt)? That would be an exciting landmark.
20 June, 2013 at 10:21 am
Terence Tao
Actually it is known that prime gaps can be arbitrarily large. For instance, if n is a natural number, then
are all composite, creating a prime gap of length at least
. I think 1476 may be the smallest H for which the location of the first prime gap of length H is known, though.
My feeling though is that we are going to run out of improvements before we reach that level;
looks like a more realistic target at this stage.
20 June, 2013 at 10:47 am
abqmathteacher
Thanks! Indeed the term “maximal prime gap” used in the OEIS entry is confusing, sorry to just quote it. It means “maximal gap for a given size integer”. So 1476 is (according to that table) the *largest* H for which the location of the first prime gap of length H is known.
Anyway, even H = 10^5 is quite exciting.
David Metzler
20 June, 2013 at 1:12 pm
Dave
If I’m reading the polymath wiki correctly, isn’t the best known H value already below $10^5$?
20 June, 2013 at 1:16 pm
Terence Tao
Er, sorry, I meant
would be a reasonable end target for this project. :)
20 June, 2013 at 1:29 pm
Gergely Harcos
1. If
denotes the first zero of
, then it is known that
see Section 15.3 in Watson: A treatise on the theory of Bessel functions (2nd ed., Cambridge University Press, 1944). Hence for
and
we can bound
On the other hand, by an orthogonality relation of Bessel functions (see Wikipedia)
It follows that
SAGE tells me that for
and
the right hand side is less than
.
2. Some misprints: "analyhsis", "systetm".
[Corrected, thanks. These calculations clean up the rather ad hoc Maple code I had previously! -T]
20 June, 2013 at 1:48 pm
Eytan Paldi
Please note that if
is even , than by formulas (11.3.33) and (11.3.34) in Abramowitz & Stegun both integrals can be expressed (and thereby precisely computed) as a simple finite linear combinations of squares of Bessel functions! (For odd k, one needs an analogous formula to (11.3.34) in which
replaces
.)
20 June, 2013 at 2:54 pm
Gergely Harcos
Thank you! Actually this works for
with any
, see my latest comment below.
20 June, 2013 at 2:44 pm
Gergely Harcos
Actually we can calculate the ratio
explicitly in terms of Bessel values. According to Lemma 5.4 in Folland: Fourier analysis and its applications,
where we can express
as
. Using this, SAGE now tells me (for
and
) that the numerator is
, and the denominator is
, so that
20 June, 2013 at 2:48 pm
Gergely Harcos
Small correction: “
as
” should be “
as
“.
20 June, 2013 at 3:00 pm
Eytan Paldi
Nice ! (such simple expression is quite surprising!)
20 June, 2013 at 6:24 pm
Eytan Paldi
It seems that there is a slightly simpler expression (for integer order n):
This is formula (8) in Auluck’s arxiv paper :
http://arxiv.org/abs/1006.4417
20 June, 2013 at 6:55 pm
Gergely Harcos
Very good. Actually this formula holds for all
(not only for integers): it can be derived from Folland’s lemma, using
and
.
20 June, 2013 at 7:48 pm
Eytan Paldi
I think that it even holds (by analytic continuation) for all complex
(since the integrand is entire as a function of
for each fixed nonzero
.)
21 June, 2013 at 8:39 am
Gergely Harcos
Eytan: in extending the formula to all complex
, one needs to be careful, e.g. one needs to check if the integral converges (hopefully locally uniformly, the difficulty coming from
). Already the extension of
is tricky to all complex
. Certainly everything is fine for
.
21 June, 2013 at 10:12 am
Eytan Paldi
I agree that the integral can’t be entire in
, since for any fixed
, (9.1.10) in A&S implies that the integral diverges as
tends (from the right) to
– which must therefore be a singular point for any analytic continuation.
21 June, 2013 at 10:28 am
Gergely Harcos
Eytan: Right. Probably there is no problem for
, as the integral converges absolutely and locally uniformly there. I sticked to
just to be safe, because the definition beyond that range is less straightforward (according to Watson’s book). BTW I am also quite pleased and surprised that so many transparent features of Bessel functions play a role in this project. I come from automorphic forms where this phenomenon is wide-spread, but (I think) it is new for sieve theory where this project belongs.
21 June, 2013 at 11:31 am
Eytan Paldi
There is another interesting “sub-multiplicative” property of
(shared also by many entire functions of order
) which is very useful for improving the current bounds on the
's. (I'm preparing the details.)
21 June, 2013 at 9:30 am
Exciting real-time developments about prime numbers | Mountain of Math
[…] I asked about this at Terry Tao’s blog, and he indicated that H = 10,000 would be a reasonable target for the polymath8 project. And in […]
22 June, 2013 at 7:28 am
Eytan Paldi
In the sum defining
, the sum variable J is upper bounded by
, but the original range of J (17 lines above) is
. So it is not clear which upper bound for J is the correct one.
[The correct range is
. I was not able to locate any instance where
was used instead, although sometimes the tilde is almost obscured by the dividing bar between the numerator and denominator. -T]
22 June, 2013 at 11:16 am
Gergely Harcos
1. I have difficulty in deriving the fourth display below (8), i.e. that the left hand side of (5) is up to negligible error
This formula should come from
where
. Writing
and
, the inner sum becomes
Without the condition
in the innermost sum, we would get in the end
which is admissible. However, I don’t see how to handle the condition
(it was automatic for the earlier choices of
).
2. I don’t see in (9) the condition
. Certainly
is the product of
and
, but why does it lie in
?
22 June, 2013 at 1:39 pm
Terence Tao
Oops, I think I mis-stated the constraint in (8) which is causing both of these problems: it should be
that lies in
rather than
. I’ll edit the post accordingly.
22 June, 2013 at 3:54 pm
Gergely Harcos
Thank you, it is all clear now. I just posted a small improvement, I hope it is ok.
22 June, 2013 at 3:50 pm
Gergely Harcos
1. I think the display below (11) can be improved to
By assumption,
such that
for some
.
It follows that
whence
Multiplying by
and using
we obtain
Therefore either
or
Hopefully this improvement lowers
somewhat.
2. "Section 2 of the previous section" should be "Section 2 of this previous blog".
22 June, 2013 at 4:38 pm
Terence Tao
Thanks for the correction and improved argument! I edited the main post accordingly. The net effect is to improve the quantity
in Theorem 5 from
to
which should hopefully squeeze a bit more out of
starting from the current constraint
since it delays the time when the
error starts exploding and so
may be taken a bit larger than previously. Presumably some perturbation of v08ltu's choice of parameters
should work here to nudge
below 1467…
22 June, 2013 at 6:59 pm
v08ltu
For instance,
.
22 June, 2013 at 7:49 pm
Terence Tao
Confirmed! H is now 12006.[Retracted, see below. -T]varpi := 1/148 - 12/4000000;
k0 := 1466;
deltap := 1/180;
A := 1200;
delta := (1 - 148 * varpi) / 33;
theta := deltap / (1/4 + varpi);
# Gergely's improved value for thetat
thetat := ((deltap - delta)/2 + varpi) / (1/4 + varpi);
deltat := delta / (1/4 + varpi);
j := BesselJZeros(k0-2,1);
eps := 1 - j^2 / (k0 * (k0-1) * (1+4*varpi));
kappa1 := int( (1-t)^((k0-1)/2)/t, t = theta..1, numeric);
kappa2 := (k0-1) * int( (1-t)^(k0-1)/t, t=theta..1, numeric);
e := exp( A + (k0-1) * int( exp(-A*t)/t, t=deltat..theta, numeric ) );
# using Gergely's exact expression for denominator
gd := (j^2/2) * BesselJ(k0-3,j)^2;
# using Eytan's exact expression for numerator
tn := sqrt(thetat)*j;
gn := (tn^2/2) * (BesselJ(k0-2,tn)^2 - BesselJ(k0-3,tn)*BesselJ(k0-1,tn));
kappa3 := (gn/gd) * e;
eps2 := 2*(kappa1+kappa2+kappa3);
# we win if eps2 < eps
22 June, 2013 at 8:14 pm
Hannes
With that script I get eps2-eps=1.268156732*10^251 (not good).
22 June, 2013 at 8:33 pm
Hannes
(I used Maple 14)
22 June, 2013 at 8:37 pm
Terence Tao
Hmm, you’re right! My version of maple seems to be a bit glitchy, but it is now reporting the same answer as you – once again
is blowing up. So I have to retract the confirmation for now…
23 June, 2013 at 12:53 am
v08ltu
I found my error, that I divided
by 2 also in the
numerator, not only the
difference.
23 June, 2013 at 1:02 am
v08ltu
The main issue that you need to take
or less to have a hope from Bessel zeroes with
, and this is not able to work, as the
part dominates the
numerator (at least currently).
22 June, 2013 at 4:49 pm
Gergely Harcos
Let
and
, then the paragraph below (11) seems to work only when
(e.g. for
the first step gives nothing). For
we can do the following. We have
, hence there is a divisor of
in the interval
. Multiplying this divisor by
yields a divisor of
in the interval
.
[Corrected, thanks – T.]
22 June, 2013 at 7:44 pm
Gergely Harcos
1. In the post and in the comments we used the bound (valid for any
)
Here is an alternate way to bound the left hand side, which probably performs better in practice without any auxiliary parameter:
2. Some typos:
2a. "
,
respectively" should be "
,
respectively".
2b. "Proposition 7" should be "Lemma 7".
2c. Exponent
should be
(at least four occurrences).
2d. Link is missing under "In this previous post".
2e. "exceeds
. or" should be "exceeds
or"
22 June, 2013 at 8:19 pm
Terence Tao
Thanks for the corrections!
The alternate estimate you propose, namely
seems to be a bit worse actually than the current estimate
for (say) v08ltu’s most recent values: the latter is about
times the G-ratio but the former seems to be something like
, peaking at
about 4200. Somehow the problem is that
is so small that the double factorial
in the denominator has a hard time counteracting the factors of
in the numerator. (The current estimate also blows up as
, but the blowup seems to be milder somehow.)
22 June, 2013 at 10:53 pm
Gergely Harcos
Thanks for the information! I haven’t experimented with the numerics lately. Another idea is to optimize
for each
individually, i.e. choose
such that
.
22 June, 2013 at 10:56 pm
Gergely Harcos
Here I meant to use your upper bound for the
-integral, then optimize in
.
23 June, 2013 at 6:22 am
Eytan Paldi
From my next message (on a sub multiplicative property of
), it follows that one can take
only in the integral (i.e. without the constant factor
)
23 June, 2013 at 1:50 pm
Gergely Harcos
Actually keeping the
-integral (i.e. not bounding it) gives much better results. I also experimented with replacing 
in the
-term. We can gain a factor of
this way, but for
we would need a gain of
.
by
23 June, 2013 at 6:52 pm
Eytan Paldi
[Edited, Jun 25, to incorporate corrections by the author – T.]
A sub-multiplicative property of f and new bounds for the
‘s:
1. Let
, be non-vanishing on
(with the possible vanishing at
.) Consider the following inequality for
:
(1)
Where for each fixed
is (obviously) the best (uniform) multiplier
.
Clearly, a sufficient condition for
to hold for each fixed
, is that the function
should be non-increasing (as a function of
) on
.
) should be non-increasing on
. Or equivalently, the logarithmic derivative of
should be non-positive on
.
That is, its logarithm (as a function of
It follows that the above sufficient condition is implied by
(2)
is concave on
.
To prove the implication, observe that for a fixed
, the derivative of
is
where
is the logarithmic derivative of
.
(as a non-increasing function on
) satisfies
whenever
, which proves the sufficiency of (2) for (1) with
.
Now (2) implies that
This motivates the following
[For some reason, LaTeX is not working on the line below, I was unable to fix it – T.]
Definition: A function g \in C^1[0,1] will be called nice ( on [0,1] ) if it is non-vanishing on [0, 1) and satisfy
(3)
and
is concave
on
Consequently, any nice function g on [0, 1] satisfies
(4)
, whenever ![t, \tau, t + \tau \in [0, 1]](https://s0.wp.com/latex.php?latex=t%2C+%5Ctau%2C+t+%2B+%5Ctau+%5Cin+%5B0%2C+1%5D&bg=ffffff&fg=545454&s=0&c=20201002)
(4')
on
.
Where (4') follows from the concavity of
on
.
It is easy to verify the following properties
(i) A finite product of nice functions is nice.
is a sequence of nice and analytic functions on
, converging locally uniformly on
to a function g, than g is (obviously) analytic and nice on
.
(ii) If
Examples: If
and
, then the functions
Theorem: Let
be an entire function of order
, having its zeros in
and satisfying
, than
is a nice function on
and, consequently, it satisfies properties (4)-(4').
Proof: denote by
the genus and order of g respectively. Since
we have two possibilities
: implying that the canonical product representation of g
where
denote
, since (by the first example), each canonical factor
) is nice and the canonical product converges locally uniformly to
on
, it follows (from the above properties) that g is nice on
.
: implying that the canonical product representation of g is given similarly by
with the same proof.
(i)
is given by
the zeros of
(i.e.
(ii)
2. Applications:
Let
for
, then the product representation (9.5.10) in A&S shows that f is an entire function of genus
having its zeros in
.
is nice on
. hence properties (4) and (4') imply
Hence by the above theorem
(5)
whenever the
's are non-negative and
.
is modified to
on
.
Remark: the last limitation on the sum is unnecessary if
(5')
for each ![t \in [0,1]](https://s0.wp.com/latex.php?latex=t+%5Cin+%5B0%2C1%5D&bg=ffffff&fg=545454&s=0&c=20201002)
for
, (5') holds obviously for each
.
(with the above modification of
Remark: it is easy to verify that![f'(0)/f(0) = - J_{k_0 -2}^2/[4 (k_0 - 1)]](https://s0.wp.com/latex.php?latex=f%27%280%29%2Ff%280%29+%3D+-+J_%7Bk_0+-2%7D%5E2%2F%5B4+%28k_0+-+1%29%5D&bg=ffffff&fg=545454&s=0&c=20201002)
is asymptotic to
.
which for large
Using (5), (5') it is easy to verify that for each
Where
Since
is asymptotic (for large
) to
, it seems that the last exponential bound is inferior to the current bound
which determines the upper bound on
, similar situation exists for the new bound
relative to the current bound
the situation seems better for the new bound on
:
Using properties (5)-(5') of
, it follows that
It is easy to verify that the last exponential bound implies the current bound with with the parameter
in the integrand, but without the constant multiplier
.
23 June, 2013 at 8:15 pm
Gergely Harcos
These are very good ideas. Some thoughts:
1. Your (4) and (4′) can be proved directly for
divided by
initially to make its value
at
. All we need is that
is decreasing on
, i.e. that
on
. We have that
and
, so we need
This in turn follows from Auluck’s formula that you quoted earlier:
2. So we have now a direct proof that
where
. This means that Terry’s
can be improved to
for any
.
23 June, 2013 at 8:36 pm
Gergely Harcos
Well, perhaps I should have restricted to
,
, and the interval
instead of
.
23 June, 2013 at 9:08 pm
Eytan Paldi
This is a nice observation! I’m quite surprised to see that the log-concavity holds also for
, though at least in a sufficiently small neighborhood of a zero, the dominating logarithmic singularity is concave, the problem is that the logarithmic derivative is positive sufficiently close to the right of each zero, this is exactly the reason why the interval of interest should not have zeros of
.
known to be entire, log-concave and monotonically decreasing to the right of its maximizer (which is in
) – in this case the zeros are to the left of the “interesting interval”!
Another well known example is
23 June, 2013 at 8:35 pm
Gergely Harcos
The new
that arises from Eytan’s ideas seems to allow
. Using
,
,
, I am getting
,
,
,
. These values might not be optimal, and they should be double checked.
23 June, 2013 at 10:02 pm
Terence Tao
Maple seems to confirm it. Nice work by both of you!
.
varpi := 1/148 - 1/318000;
k0 := 1466;
deltap := 1/200;
A := 1360;
delta := (1 - 148 * varpi) / 33;
theta := deltap / (1/4 + varpi);
# Gergely's improved value for thetat
thetat := ((deltap - delta)/2 + varpi) / (1/4 + varpi);
deltat := delta / (1/4 + varpi);
j := BesselJZeros(k0-2,1);
eps := 1 - j^2 / (k0 * (k0-1) * (1+4*varpi));
kappa1 := int( (1-t)^((k0-1)/2)/t, t = theta..1, numeric);
kappa2 := (k0-1) * int( (1-t)^(k0-1)/t, t=theta..1, numeric);
# using Gergely and Eytan's improved kappa_3
alpha := j^2 / (4 * (k0-1));
e := exp( A + (k0-1) * int( exp(-(A+2*alpha)*t)/t, t=deltat..theta, numeric ) );
# using Gergely's exact expression for denominator
gd := (j^2/2) * BesselJ(k0-3,j)^2;
# using Eytan's exact expression for numerator
tn := sqrt(thetat)*j;
gn := (tn^2/2) * (BesselJ(k0-2,tn)^2 - BesselJ(k0-3,tn)*BesselJ(k0-1,tn));
kappa3 := (gn/gd) * e;
eps2 := 2*(kappa1+kappa2+kappa3);
# we win if eps2 < eps
23 June, 2013 at 10:33 pm
v08ltu
In your Maple worksheet, you integrate, deltat to theta, but the formula listed by Harcos is
.
[Well spotted! But I think Gergely’s formula has a typo, the integral here (which is counting primes up to
) should go up to
rather than
. In any case the exponential decay of the integrand renders the upper limit of the integral more or less irrelevant. -T.]
24 June, 2013 at 4:53 am
Gergely Harcos
Indeed I had a typo, thanks for pointing it out! It was inherited by copy+paste from Terry’s earlier comment (https://terrytao.wordpress.com/2013/06/18/a-truncated-elementary-selberg-sieve-of-pintz/#comment-235766), but my SAGE worksheet contained the right endpoint
. Also, as Terry explained, the figures would still be fine with integration up to
.
24 June, 2013 at 4:58 am
Gergely Harcos
I mean that https://terrytao.wordpress.com/2013/06/18/a-truncated-elementary-selberg-sieve-of-pintz/#comment-235986 was a response to https://terrytao.wordpress.com/2013/06/18/a-truncated-elementary-selberg-sieve-of-pintz/#comment-235937
24 June, 2013 at 7:18 am
Terence Tao
Oops! Sorry about that; too late, now, but I edited the comment to fix it.
25 June, 2013 at 4:39 pm
Eytan Paldi
By (9.5.4) in A&S, it follows that
, so this ratio may be used for
to check the relative numerical error.
23 June, 2013 at 9:14 pm
The distribution of primes in densely divisible moduli | What's new
[…] The difference between this conjecture and the stronger conjecture is that the modulus is constrained to be -densely divisible rather than -smooth (note that is no longer constrained to lie in ). This relaxation of the smoothness condition improves the Goldston-Pintz-Yildirim type sieving needed to deduce from ; see this previous post. […]
25 June, 2013 at 11:08 pm
Terence Tao
I wrote a wiki page at
http://michaelnielsen.org/polymath1/index.php?title=Dickson-Hardy-Littlewood_theorems
recording most of the techniques we have for converting results of the form
or
to
. Nothing new on that page, but it at least collects all the different implications that we have found in a single location (though they are all basically obsolete now that we have Pintz’s sieve).
30 June, 2013 at 12:39 pm
Bounded gaps between primes (Polymath8) – a progress report | What's new
[…] “dense divisibility”, replacing the conjecture by a variant that we have called ); see this post for details. Finally, by optimising the remaining steps in Zhang’s argument (see below), we […]
5 July, 2013 at 4:41 pm
Eytan Paldi
The current definition of
ensures (under theorem 5 hypotheses) that
, but without the factor
in its definition, we need to add this constraint to theorem 5 hypotheses.
5 July, 2013 at 5:43 pm
Gergely Harcos
I don’t think we need this constraint. If we replace
by
, then the function
does not change, hence the condition of Theorem 5 remains the same, while its proof works verbatim, i.e. the original stated conclusion holds.
27 July, 2013 at 6:55 pm
An improved Type I estimate | What's new
[…] We use the Pintz sieve from this post, repeating the proof of Theorem 5 from that post (and using the explicit formulae for and from […]