Mixing
Lower bound for sum of Hecke eigenvalues
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 '18 at 13:17
Nodt Greenish
29813
add a comment |
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 '18 at 13:17
Nodt Greenish
29813
add a comment |
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 '18 at 13:17
Nodt Greenish
29813
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 '18 at 13:17
Nodt Greenish
29813
asked Nov 12 '18 at 13:17
Nodt Greenish
29813
edited Nov 20 '18 at 14:14
asked Nov 12 '18 at 13:17
Nodt Greenish
29813
asked Nov 12 '18 at 13:17
Nodt Greenish
29813
asked Nov 12 '18 at 13:17
Nodt Greenish
29813
29813
add a comment |
add a comment |
1 Answer
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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 '18 at 14:12
Nodt Greenish
29813
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
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 '18 at 14:12
Nodt Greenish
29813
add a comment |
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 '18 at 14:12
Nodt Greenish
29813
add a comment |
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 '18 at 14:12
Nodt Greenish
29813
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 '18 at 14:12
Nodt Greenish
29813
edited Nov 22 '18 at 8:40
answered Nov 20 '18 at 14:12
Nodt Greenish
29813
answered Nov 20 '18 at 14:12
Nodt Greenish
29813
answered Nov 20 '18 at 14:12
Nodt Greenish
29813
29813
add a comment |
add a comment |
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