Proper Way To Compute An Upper Bound
I regard to the proof of Lemma 10 in "A remark on a conjecture of Chowla" by M. R. Murty, A. Vatwani, J. Ramanujan Math. Soc., 33, No. 2, 2018, 111-123,
the authors used the average value $(log x)^c$, $c$ constant, of the number of divisors function $tau(d)=sum_{d|n}1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
$$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)right | ll x (log x)^{2c},$$
where $2 delta <1/2$.
The questions are these:
Is the main result invalid? The upper bound should be
$$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)right | ll x ^{1+2delta}.$$
This is the best unconditional upper bound, under any known result, including Proposition 3.It is true that the proper upper bound $tau(d)^2 ll x^{2epsilon}$, $epsilon >0$, is not required here?
Can we use this as a precedent to prove other upper bounds in mathematics?
nt.number-theory analytic-number-theory
edited 6 mins ago
GH from MO
57.9k5142219
asked 6 hours ago
r. t.
161
New contributor
r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I regard to the proof of Lemma 10 in "A remark on a conjecture of Chowla" by M. R. Murty, A. Vatwani, J. Ramanujan Math. Soc., 33, No. 2, 2018, 111-123,
the authors used the average value $(log x)^c$, $c$ constant, of the number of divisors function $tau(d)=sum_{d|n}1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
$$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)right | ll x (log x)^{2c},$$
where $2 delta <1/2$.
The questions are these:
Is the main result invalid? The upper bound should be
$$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)right | ll x ^{1+2delta}.$$
This is the best unconditional upper bound, under any known result, including Proposition 3.It is true that the proper upper bound $tau(d)^2 ll x^{2epsilon}$, $epsilon >0$, is not required here?
Can we use this as a precedent to prove other upper bounds in mathematics?
nt.number-theory analytic-number-theory
edited 6 mins ago
GH from MO
57.9k5142219
asked 6 hours ago
r. t.
161
New contributor
r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Advices for the new year: (1) Please be very careful in mathematics, especially in analytic number theory. (2) Mathematical proofs are not based on precedents, but on axioms and the rules of first order logic. I wish you a Happy New Year!
– GH from MO
7 mins ago
add a comment |
I regard to the proof of Lemma 10 in "A remark on a conjecture of Chowla" by M. R. Murty, A. Vatwani, J. Ramanujan Math. Soc., 33, No. 2, 2018, 111-123,
the authors used the average value $(log x)^c$, $c$ constant, of the number of divisors function $tau(d)=sum_{d|n}1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
$$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)right | ll x (log x)^{2c},$$
where $2 delta <1/2$.
The questions are these:
Is the main result invalid? The upper bound should be
$$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)right | ll x ^{1+2delta}.$$
This is the best unconditional upper bound, under any known result, including Proposition 3.It is true that the proper upper bound $tau(d)^2 ll x^{2epsilon}$, $epsilon >0$, is not required here?
Can we use this as a precedent to prove other upper bounds in mathematics?
nt.number-theory analytic-number-theory
edited 6 mins ago
GH from MO
57.9k5142219
asked 6 hours ago
r. t.
161
New contributor
r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I regard to the proof of Lemma 10 in "A remark on a conjecture of Chowla" by M. R. Murty, A. Vatwani, J. Ramanujan Math. Soc., 33, No. 2, 2018, 111-123,
the authors used the average value $(log x)^c$, $c$ constant, of the number of divisors function $tau(d)=sum_{d|n}1$ as an upper bound for $tau(d)^2$, where $d leq x$. To be specific, they claim that
$$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)right | ll x (log x)^{2c},$$
where $2 delta <1/2$.
The questions are these:
Is the main result invalid? The upper bound should be
$$sum_{q leq x^{2delta}}tau(q)^2 left | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)right | ll x ^{1+2delta}.$$
This is the best unconditional upper bound, under any known result, including Proposition 3.It is true that the proper upper bound $tau(d)^2 ll x^{2epsilon}$, $epsilon >0$, is not required here?
Can we use this as a precedent to prove other upper bounds in mathematics?
nt.number-theory analytic-number-theory
nt.number-theory analytic-number-theory
edited 6 mins ago
GH from MO
57.9k5142219
asked 6 hours ago
r. t.
161
New contributor
r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 mins ago
GH from MO
57.9k5142219
asked 6 hours ago
r. t.
161
New contributor
r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 mins ago
GH from MO
57.9k5142219
edited 6 mins ago
GH from MO
57.9k5142219
edited 6 mins ago
GH from MO
57.9k5142219
57.9k5142219
asked 6 hours ago
r. t.
161
New contributor
r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
r. t.
161
asked 6 hours ago
r. t.
161
161
New contributor
r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
r. t. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Advices for the new year: (1) Please be very careful in mathematics, especially in analytic number theory. (2) Mathematical proofs are not based on precedents, but on axioms and the rules of first order logic. I wish you a Happy New Year!
– GH from MO
7 mins ago
add a comment |
Advices for the new year: (1) Please be very careful in mathematics, especially in analytic number theory. (2) Mathematical proofs are not based on precedents, but on axioms and the rules of first order logic. I wish you a Happy New Year!
– GH from MO
7 mins ago
Advices for the new year: (1) Please be very careful in mathematics, especially in analytic number theory. (2) Mathematical proofs are not based on precedents, but on axioms and the rules of first order logic. I wish you a Happy New Year!
– GH from MO
7 mins ago
Advices for the new year: (1) Please be very careful in mathematics, especially in analytic number theory. (2) Mathematical proofs are not based on precedents, but on axioms and the rules of first order logic. I wish you a Happy New Year!
– GH from MO
7 mins ago
add a comment |
1 Answer
1
active
oldest
votes
Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
begin{align*}
sum_{q leq x^{2delta}}tau(q)^2 bigg | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)bigg | &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} |mu(m)| \
&le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} 1 \
&ll sum_{q leq x^{2delta}}tau(q)^2 frac xq = x sum_{q leq x^{2delta}} frac{tau(q)^2}q.
end{align*}
And this remaining sum is indeed $ll_delta (log x)^{2c}$ for some constant $c$; indeed, it's not hard to show that
$$
sum_{q leq y} frac{tau(q)^2}q sim frac{(log y)^4}{4pi^2}.
$$
answered 4 hours ago
Greg Martin
8,09313458
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
r. t. is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f319773%2fproper-way-to-compute-an-upper-bound%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
begin{align*}
sum_{q leq x^{2delta}}tau(q)^2 bigg | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)bigg | &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} |mu(m)| \
&le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} 1 \
&ll sum_{q leq x^{2delta}}tau(q)^2 frac xq = x sum_{q leq x^{2delta}} frac{tau(q)^2}q.
end{align*}
And this remaining sum is indeed $ll_delta (log x)^{2c}$ for some constant $c$; indeed, it's not hard to show that
$$
sum_{q leq y} frac{tau(q)^2}q sim frac{(log y)^4}{4pi^2}.
$$
answered 4 hours ago
Greg Martin
8,09313458
add a comment |
Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
begin{align*}
sum_{q leq x^{2delta}}tau(q)^2 bigg | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)bigg | &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} |mu(m)| \
&le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} 1 \
&ll sum_{q leq x^{2delta}}tau(q)^2 frac xq = x sum_{q leq x^{2delta}} frac{tau(q)^2}q.
end{align*}
And this remaining sum is indeed $ll_delta (log x)^{2c}$ for some constant $c$; indeed, it's not hard to show that
$$
sum_{q leq y} frac{tau(q)^2}q sim frac{(log y)^4}{4pi^2}.
$$
answered 4 hours ago
Greg Martin
8,09313458
add a comment |
Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
begin{align*}
sum_{q leq x^{2delta}}tau(q)^2 bigg | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)bigg | &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} |mu(m)| \
&le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} 1 \
&ll sum_{q leq x^{2delta}}tau(q)^2 frac xq = x sum_{q leq x^{2delta}} frac{tau(q)^2}q.
end{align*}
And this remaining sum is indeed $ll_delta (log x)^{2c}$ for some constant $c$; indeed, it's not hard to show that
$$
sum_{q leq y} frac{tau(q)^2}q sim frac{(log y)^4}{4pi^2}.
$$
answered 4 hours ago
Greg Martin
8,09313458
Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that
begin{align*}
sum_{q leq x^{2delta}}tau(q)^2 bigg | sum_{substack{m leq x+2\
m equiv a bmod q}} mu(m)bigg | &le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} |mu(m)| \
&le sum_{q leq x^{2delta}}tau(q)^2 sum_{substack{m leq x+2\ m equiv a bmod q}} 1 \
&ll sum_{q leq x^{2delta}}tau(q)^2 frac xq = x sum_{q leq x^{2delta}} frac{tau(q)^2}q.
end{align*}
And this remaining sum is indeed $ll_delta (log x)^{2c}$ for some constant $c$; indeed, it's not hard to show that
$$
sum_{q leq y} frac{tau(q)^2}q sim frac{(log y)^4}{4pi^2}.
$$
answered 4 hours ago
Greg Martin
8,09313458
answered 4 hours ago
Greg Martin
8,09313458
answered 4 hours ago
Greg Martin
8,09313458
answered 4 hours ago
Greg Martin
8,09313458
8,09313458
add a comment |
add a comment |
r. t. is a new contributor. Be nice, and check out our Code of Conduct.
r. t. is a new contributor. Be nice, and check out our Code of Conduct.
r. t. is a new contributor. Be nice, and check out our Code of Conduct.
r. t. is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f319773%2fproper-way-to-compute-an-upper-bound%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Advices for the new year: (1) Please be very careful in mathematics, especially in analytic number theory. (2) Mathematical proofs are not based on precedents, but on axioms and the rules of first order logic. I wish you a Happy New Year!
– GH from MO
7 mins ago