Lower bound for sum of Hecke eigenvalues
up vote
2
down vote
favorite
Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.
How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?
Here is the background:
The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.
In Xu, section 3.1 something similar happens. Here the hints $lambda(p)^2=lambda(p^2)+1$ (which leads to $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.
number-theory eigenvalues-eigenvectors estimation modular-forms upper-lower-bounds
asked Nov 12 at 13:17
Nodt Greenish
30013
add a comment |
up vote
2
down vote
favorite
Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.
How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?
Here is the background:
The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.
In Xu, section 3.1 something similar happens. Here the hints $lambda(p)^2=lambda(p^2)+1$ (which leads to $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.
number-theory eigenvalues-eigenvectors estimation modular-forms upper-lower-bounds
asked Nov 12 at 13:17
Nodt Greenish
30013
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.
How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?
Here is the background:
The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.
In Xu, section 3.1 something similar happens. Here the hints $lambda(p)^2=lambda(p^2)+1$ (which leads to $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.
number-theory eigenvalues-eigenvectors estimation modular-forms upper-lower-bounds
asked Nov 12 at 13:17
Nodt Greenish
30013
Let $lambda$ be weakly multiplicative, $lambda(n)geq0$, $p$ prime and $S(x)=sum_{nleq x}lambda(n)log(frac{x}{n})$ for real $x$.
How can I show $S(x)gg left(sum_{pleq sqrt{x/3}}lambda(p)right)^2-left(sum_{pleq sqrt{x/3}}1right)$?
Here is the background:
The question is coming from IKS, section 3. $lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $pnotmid N$, thus giving multiplicativity.
In Xu, section 3.1 something similar happens. Here the hints $lambda(p)^2=lambda(p^2)+1$ (which leads to $lambda(p)geq1$ and $|lambda(n)|leqsigma_0(n)$ (divisor function) are given.
number-theory eigenvalues-eigenvectors estimation modular-forms upper-lower-bounds
number-theory eigenvalues-eigenvectors estimation modular-forms upper-lower-bounds
asked Nov 12 at 13:17
Nodt Greenish
30013
asked Nov 12 at 13:17
Nodt Greenish
30013
edited Nov 20 at 14:14
asked Nov 12 at 13:17
Nodt Greenish
30013
asked Nov 12 at 13:17
Nodt Greenish
30013
asked Nov 12 at 13:17
Nodt Greenish
30013
30013
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
I found a way which should do the trick. I think the division by 3 is just a technical detail and 2 would be already sufficient to pull $log(frac{x}{x/2})$ out of the sum. You can take $S(x)=sum_{nleq x}lambda(n)log(x/n)geq sum_{pqleq x/2}lambda(pq)$. Furthermore we use the formula $lambda(p)^2=lambda(p^2)+1$, which is (1.2) in IKS., to estimate the case $p=q$. That's where the subtracted sum of 1's comes from.
answered Nov 20 at 14:12
Nodt Greenish
30013
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I found a way which should do the trick. I think the division by 3 is just a technical detail and 2 would be already sufficient to pull $log(frac{x}{x/2})$ out of the sum. You can take $S(x)=sum_{nleq x}lambda(n)log(x/n)geq sum_{pqleq x/2}lambda(pq)$. Furthermore we use the formula $lambda(p)^2=lambda(p^2)+1$, which is (1.2) in IKS., to estimate the case $p=q$. That's where the subtracted sum of 1's comes from.
answered Nov 20 at 14:12
Nodt Greenish
30013
add a comment |
up vote
0
down vote
accepted
I found a way which should do the trick. I think the division by 3 is just a technical detail and 2 would be already sufficient to pull $log(frac{x}{x/2})$ out of the sum. You can take $S(x)=sum_{nleq x}lambda(n)log(x/n)geq sum_{pqleq x/2}lambda(pq)$. Furthermore we use the formula $lambda(p)^2=lambda(p^2)+1$, which is (1.2) in IKS., to estimate the case $p=q$. That's where the subtracted sum of 1's comes from.
answered Nov 20 at 14:12
Nodt Greenish
30013
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I found a way which should do the trick. I think the division by 3 is just a technical detail and 2 would be already sufficient to pull $log(frac{x}{x/2})$ out of the sum. You can take $S(x)=sum_{nleq x}lambda(n)log(x/n)geq sum_{pqleq x/2}lambda(pq)$. Furthermore we use the formula $lambda(p)^2=lambda(p^2)+1$, which is (1.2) in IKS., to estimate the case $p=q$. That's where the subtracted sum of 1's comes from.
answered Nov 20 at 14:12
Nodt Greenish
30013
I found a way which should do the trick. I think the division by 3 is just a technical detail and 2 would be already sufficient to pull $log(frac{x}{x/2})$ out of the sum. You can take $S(x)=sum_{nleq x}lambda(n)log(x/n)geq sum_{pqleq x/2}lambda(pq)$. Furthermore we use the formula $lambda(p)^2=lambda(p^2)+1$, which is (1.2) in IKS., to estimate the case $p=q$. That's where the subtracted sum of 1's comes from.
answered Nov 20 at 14:12
Nodt Greenish
30013
edited Nov 22 at 8:40
answered Nov 20 at 14:12
Nodt Greenish
30013
answered Nov 20 at 14:12
Nodt Greenish
30013
answered Nov 20 at 14:12
Nodt Greenish
30013
30013
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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%2fmath.stackexchange.com%2fquestions%2f2995298%2flower-bound-for-sum-of-hecke-eigenvalues%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
