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I want to show that for $D>0, Dequiv 0,1 pmod4$, $z in mathbbC$ with $ operatornameIm(z)>0$ and $k>1$ the sum
$$sum_a,b,c in mathbbZ \ b^2-4ac=D(az^2+bz+c)^-k$$ converges absolutely. I tried to take the absolute value and wrote $z=x+iy$. However, after calculating everything I just got $$|(az^2+bz+c)|^-k=sqrta^2l_1+al_2+acl_3+a^2cl_4+ac^2l_5+a^2c^2l_6+ac^3l_7+c^2l_8+c^4+l_9^-k,$$ where $l_i$ only depends on $x,y,D$ which can be seen as fixed. However, I have no idea how to go on.
sequences-and-series complex-analysis convergence
asked Aug 2 at 17:57
Deavor
568412
add a comment |Â
up vote
2
down vote
favorite
I want to show that for $D>0, Dequiv 0,1 pmod4$, $z in mathbbC$ with $ operatornameIm(z)>0$ and $k>1$ the sum
$$sum_a,b,c in mathbbZ \ b^2-4ac=D(az^2+bz+c)^-k$$ converges absolutely. I tried to take the absolute value and wrote $z=x+iy$. However, after calculating everything I just got $$|(az^2+bz+c)|^-k=sqrta^2l_1+al_2+acl_3+a^2cl_4+ac^2l_5+a^2c^2l_6+ac^3l_7+c^2l_8+c^4+l_9^-k,$$ where $l_i$ only depends on $x,y,D$ which can be seen as fixed. However, I have no idea how to go on.
sequences-and-series complex-analysis convergence
asked Aug 2 at 17:57
Deavor
568412
What does $Dequiv 0,1 pmod4$ mean?
– zhw.
Aug 2 at 20:40
$D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
– Deavor
Aug 2 at 20:50
You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
– i707107
Aug 4 at 19:38
Oh yes thanks for noticing the typo
– Deavor
Aug 4 at 20:18
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I want to show that for $D>0, Dequiv 0,1 pmod4$, $z in mathbbC$ with $ operatornameIm(z)>0$ and $k>1$ the sum
$$sum_a,b,c in mathbbZ \ b^2-4ac=D(az^2+bz+c)^-k$$ converges absolutely. I tried to take the absolute value and wrote $z=x+iy$. However, after calculating everything I just got $$|(az^2+bz+c)|^-k=sqrta^2l_1+al_2+acl_3+a^2cl_4+ac^2l_5+a^2c^2l_6+ac^3l_7+c^2l_8+c^4+l_9^-k,$$ where $l_i$ only depends on $x,y,D$ which can be seen as fixed. However, I have no idea how to go on.
sequences-and-series complex-analysis convergence
asked Aug 2 at 17:57
Deavor
568412
I want to show that for $D>0, Dequiv 0,1 pmod4$, $z in mathbbC$ with $ operatornameIm(z)>0$ and $k>1$ the sum
$$sum_a,b,c in mathbbZ \ b^2-4ac=D(az^2+bz+c)^-k$$ converges absolutely. I tried to take the absolute value and wrote $z=x+iy$. However, after calculating everything I just got $$|(az^2+bz+c)|^-k=sqrta^2l_1+al_2+acl_3+a^2cl_4+ac^2l_5+a^2c^2l_6+ac^3l_7+c^2l_8+c^4+l_9^-k,$$ where $l_i$ only depends on $x,y,D$ which can be seen as fixed. However, I have no idea how to go on.
sequences-and-series complex-analysis convergence
asked Aug 2 at 17:57
Deavor
568412
edited 2 days ago
asked Aug 2 at 17:57
Deavor
568412
asked Aug 2 at 17:57
Deavor
568412
asked Aug 2 at 17:57
Deavor
568412
568412
What does $Dequiv 0,1 pmod4$ mean?
– zhw.
Aug 2 at 20:40
$D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
– Deavor
Aug 2 at 20:50
You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
– i707107
Aug 4 at 19:38
Oh yes thanks for noticing the typo
– Deavor
Aug 4 at 20:18
add a comment |Â
What does $Dequiv 0,1 pmod4$ mean?
– zhw.
Aug 2 at 20:40
$D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
– Deavor
Aug 2 at 20:50
You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
– i707107
Aug 4 at 19:38
Oh yes thanks for noticing the typo
– Deavor
Aug 4 at 20:18
What does $Dequiv 0,1 pmod4$ mean?
– zhw.
Aug 2 at 20:40
What does $Dequiv 0,1 pmod4$ mean?
– zhw.
Aug 2 at 20:40
$D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
– Deavor
Aug 2 at 20:50
$D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
– Deavor
Aug 2 at 20:50
You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
– i707107
Aug 4 at 19:38
You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
– i707107
Aug 4 at 19:38
Oh yes thanks for noticing the typo
– Deavor
Aug 4 at 20:18
Oh yes thanks for noticing the typo
– Deavor
Aug 4 at 20:18
add a comment |Â
1 Answer
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1
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This can be done with Theory of Quadratic Forms. The constants $C$ appearing in this solution may not be the same. We begin with
Definition
Let $ax^2+bxy+cy^2$ be a quadratic form with $a, b, cin mathbbZ$ and $b^2-4ac=D$. We say that a quadratic form $a_1x_1^2+b_1x_1y_1+c_1y_1^2$ with $a_1,b_1,c_1inmathbbZ$ and $b_1^2-4a_1c_1=D$ is equivalent to $ax^2+bxy+cy^2$ if there is $beginbmatrixp & q \ r & s endbmatrixinmathrmSL_2(mathbbZ)$ such that
$$
a_1x_1^2+b_1x_1y_1+c_1y_1^2=a(px_1+qy_1)^2+b(px_1+qy_1)(rx_1+sy_1)+c(rx_1+sy_1)^2.
$$
This can be written again with matrices
$$
beginbmatrixa_1 & fracb_12\ fracb_12 & c_1 endbmatrix = beginbmatrixp & r \ q & s endbmatrix beginbmatrixa & fracb2\ fracb2 & c endbmatrix beginbmatrixp & q \ r & s endbmatrix
$$
and
$$
beginbmatrixx\yendbmatrix=beginbmatrixp & q \ r & s endbmatrixbeginbmatrixx_1\y_1endbmatrix.
$$
The next one is the reducing step.
Lemma 1
Any integer quadratic form with discriminant $D$ is equivalent to $ax^2+bxy+cy^2$ with $|b|leq |a|leq |c|$.
The proof is a repeated application of the matrices in $mathrmSL_2(mathbbZ)$.
$$
beginbmatrix 1 & 1 \ 0 & 1 endbmatrix, beginbmatrix 0 & -1 \ 1 & 0endbmatrix.
$$
Corollary 1
The number of equivalence classes of integer quadratic forms with discriminant $D$ is finite.
If $b^2-4ac=D$ and $|b|leq |a|leq |c|$, then $|D|geq 4|ac|-|b^2| geq 3|ac|$ gives only finitely many choices for $a$ and $c$, subsequently for $b$.
Then it suffices to prove the following for $A=beginbmatrixa& b/2\ b/2&cendbmatrix$ with $b^2-4ac=D$ and $|b|leq |a|leq |c|$, as the sum can be written as a finite sum of the following kind.
The sum is absolutely convergent for $k>1$:
$$
sum_gammain mathrmSL_2(mathbbZ) frac1left( beginbmatrixz&1endbmatrixgamma^T A gamma beginbmatrixz\1endbmatrixright)^k.
$$
It suffices to show the following sum is convergent instead, as it differs only by a positive constant.
$$
sum_gammain mathrmSL_2(mathbbZ) frac1left( beginbmatrixz&1endbmatrixgamma^T beginbmatrixalpha \ betaendbmatrix beginbmatrixalpha' &beta'endbmatrix gamma beginbmatrixz\1endbmatrixright)^k
$$
where $alpha=alpha'=2a$, $beta=b+sqrtD$, and $beta'=b-sqrtD$.
We establish the following uniform inequality:
Lemma 2
We have $$
left|beginbmatrixalpha' &beta'endbmatrix gamma beginbmatrixz\1endbmatrixright|geq C (|pz+q||rz+s|)^1/2
$$
where $C>0$ is an absolute constant depending at most on $a,b,c$, and $gamma=beginbmatrix p&q\r&sendbmatrixinmathrmSL_2(mathbbZ)$.
The proof is by AM-GM inequality.
Now, it suffices to prove the convergence of
$$
sum_gammain mathrmSL_2(mathbbZ)frac1 (*)
$$
Consider $Gamma=mathrmSL_2(mathbbZ)$ and $Gamma_infty=leftpmbeginbmatrix 1&m\0&1endbmatrix, minmathbbZright$.
Fix coprime integers $r$, $s$. Let $p_0,q_0inmathbbZ$ satisfy $p_0s-q_0r=1$. Then it is well-known that all integers $p,q$ satisfying $ps-qr=1$ are $p=p_0+rt$, $q=q_0+st$ with $tinmathbbZ$. Varying $t$ if necessary, we may choose $p_0, q_0$ such that a coset representative $beginbmatrixp_0&q_0\r&sendbmatrix$ in $Gamma_inftybackslash Gamma$ with $-frac12leqfracp_0z+q_0rz+s<frac12$. We have the above sums over $r,s$ with $|t|geq 1$, is convergent since
$$
sum_(r,s)=1 frac1rz+sleq Csum_sfrac1
$$
is convergent where $C>0$ is an absolute constant that depends only on $z$.
We also have that the area of a parallelogram with vertices $0$, $pz+q$, $rz+s$ is the same as that of vertices $0$, $1$, $z$, this is because of $ps-qr=1$. Call this area $C$. Then
$$
|pz+q||rz+s| |sin theta_gamma| = C.
$$
Here $theta_gamma$ is the angle of the corner formed by the three points $pz+q$, $0$, $rz+s$. This angle is also written as $theta_gamma=arg fracpz+qrz+s=arg gamma z$.
Then the convergence of above sum (*) over $r,s$ with $t=0$ follows from that of
$$
sum_gammainGamma_inftybackslashGamma|sin theta_gamma|^k (**)
$$
Lemma 3
Let $Y>0$. We have a constant $C>0$ depending on $z$ such that
$$
# gammainGamma_inftybackslashGamma biggvert leq 4+frac4CY.
$$
This is proved by applying a classical lemma
$$
# gammainGamma_inftybackslashGamma biggvert Im(gamma z)geq Yleq 1+frac CY.
$$
Now, the convergence of (**) follows by partial summation with application of Lemma 3 for $Y=frac1n$, $n=1,2,3,ldots$.
answered 17 hours ago
i707107
11.2k21241
Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $beginbmatrixalpha \ betaendbmatrix * beginbmatrixalpha' &beta'endbmatrix$, since by multiplying them I got $4a*beginbmatrixa & fracb-sqrtD2 \ fracb+sqrtD2 & cendbmatrix$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
– Deavor
11 hours ago
Since we can interpret quadratic forms as matrices:$$beginbmatrixx¥dbmatrixbeginbmatrixa&b\c&dendbmatrixbeginbmatrixx\yendbmatrix=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
– i707107
9 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This can be done with Theory of Quadratic Forms. The constants $C$ appearing in this solution may not be the same. We begin with
Definition
Let $ax^2+bxy+cy^2$ be a quadratic form with $a, b, cin mathbbZ$ and $b^2-4ac=D$. We say that a quadratic form $a_1x_1^2+b_1x_1y_1+c_1y_1^2$ with $a_1,b_1,c_1inmathbbZ$ and $b_1^2-4a_1c_1=D$ is equivalent to $ax^2+bxy+cy^2$ if there is $beginbmatrixp & q \ r & s endbmatrixinmathrmSL_2(mathbbZ)$ such that
$$
a_1x_1^2+b_1x_1y_1+c_1y_1^2=a(px_1+qy_1)^2+b(px_1+qy_1)(rx_1+sy_1)+c(rx_1+sy_1)^2.
$$
This can be written again with matrices
$$
beginbmatrixa_1 & fracb_12\ fracb_12 & c_1 endbmatrix = beginbmatrixp & r \ q & s endbmatrix beginbmatrixa & fracb2\ fracb2 & c endbmatrix beginbmatrixp & q \ r & s endbmatrix
$$
and
$$
beginbmatrixx\yendbmatrix=beginbmatrixp & q \ r & s endbmatrixbeginbmatrixx_1\y_1endbmatrix.
$$
The next one is the reducing step.
Lemma 1
Any integer quadratic form with discriminant $D$ is equivalent to $ax^2+bxy+cy^2$ with $|b|leq |a|leq |c|$.
The proof is a repeated application of the matrices in $mathrmSL_2(mathbbZ)$.
$$
beginbmatrix 1 & 1 \ 0 & 1 endbmatrix, beginbmatrix 0 & -1 \ 1 & 0endbmatrix.
$$
Corollary 1
The number of equivalence classes of integer quadratic forms with discriminant $D$ is finite.
If $b^2-4ac=D$ and $|b|leq |a|leq |c|$, then $|D|geq 4|ac|-|b^2| geq 3|ac|$ gives only finitely many choices for $a$ and $c$, subsequently for $b$.
Then it suffices to prove the following for $A=beginbmatrixa& b/2\ b/2&cendbmatrix$ with $b^2-4ac=D$ and $|b|leq |a|leq |c|$, as the sum can be written as a finite sum of the following kind.
The sum is absolutely convergent for $k>1$:
$$
sum_gammain mathrmSL_2(mathbbZ) frac1left( beginbmatrixz&1endbmatrixgamma^T A gamma beginbmatrixz\1endbmatrixright)^k.
$$
It suffices to show the following sum is convergent instead, as it differs only by a positive constant.
$$
sum_gammain mathrmSL_2(mathbbZ) frac1left( beginbmatrixz&1endbmatrixgamma^T beginbmatrixalpha \ betaendbmatrix beginbmatrixalpha' &beta'endbmatrix gamma beginbmatrixz\1endbmatrixright)^k
$$
where $alpha=alpha'=2a$, $beta=b+sqrtD$, and $beta'=b-sqrtD$.
We establish the following uniform inequality:
Lemma 2
We have $$
left|beginbmatrixalpha' &beta'endbmatrix gamma beginbmatrixz\1endbmatrixright|geq C (|pz+q||rz+s|)^1/2
$$
where $C>0$ is an absolute constant depending at most on $a,b,c$, and $gamma=beginbmatrix p&q\r&sendbmatrixinmathrmSL_2(mathbbZ)$.
The proof is by AM-GM inequality.
Now, it suffices to prove the convergence of
$$
sum_gammain mathrmSL_2(mathbbZ)frac1 (*)
$$
Consider $Gamma=mathrmSL_2(mathbbZ)$ and $Gamma_infty=leftpmbeginbmatrix 1&m\0&1endbmatrix, minmathbbZright$.
Fix coprime integers $r$, $s$. Let $p_0,q_0inmathbbZ$ satisfy $p_0s-q_0r=1$. Then it is well-known that all integers $p,q$ satisfying $ps-qr=1$ are $p=p_0+rt$, $q=q_0+st$ with $tinmathbbZ$. Varying $t$ if necessary, we may choose $p_0, q_0$ such that a coset representative $beginbmatrixp_0&q_0\r&sendbmatrix$ in $Gamma_inftybackslash Gamma$ with $-frac12leqfracp_0z+q_0rz+s<frac12$. We have the above sums over $r,s$ with $|t|geq 1$, is convergent since
$$
sum_(r,s)=1 frac1rz+sleq Csum_sfrac1
$$
is convergent where $C>0$ is an absolute constant that depends only on $z$.
We also have that the area of a parallelogram with vertices $0$, $pz+q$, $rz+s$ is the same as that of vertices $0$, $1$, $z$, this is because of $ps-qr=1$. Call this area $C$. Then
$$
|pz+q||rz+s| |sin theta_gamma| = C.
$$
Here $theta_gamma$ is the angle of the corner formed by the three points $pz+q$, $0$, $rz+s$. This angle is also written as $theta_gamma=arg fracpz+qrz+s=arg gamma z$.
Then the convergence of above sum (*) over $r,s$ with $t=0$ follows from that of
$$
sum_gammainGamma_inftybackslashGamma|sin theta_gamma|^k (**)
$$
Lemma 3
Let $Y>0$. We have a constant $C>0$ depending on $z$ such that
$$
# gammainGamma_inftybackslashGamma biggvert leq 4+frac4CY.
$$
This is proved by applying a classical lemma
$$
# gammainGamma_inftybackslashGamma biggvert Im(gamma z)geq Yleq 1+frac CY.
$$
Now, the convergence of (**) follows by partial summation with application of Lemma 3 for $Y=frac1n$, $n=1,2,3,ldots$.
answered 17 hours ago
i707107
11.2k21241
Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $beginbmatrixalpha \ betaendbmatrix * beginbmatrixalpha' &beta'endbmatrix$, since by multiplying them I got $4a*beginbmatrixa & fracb-sqrtD2 \ fracb+sqrtD2 & cendbmatrix$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
– Deavor
11 hours ago
Since we can interpret quadratic forms as matrices:$$beginbmatrixx¥dbmatrixbeginbmatrixa&b\c&dendbmatrixbeginbmatrixx\yendbmatrix=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
– i707107
9 hours ago
add a comment |Â
up vote
1
down vote
This can be done with Theory of Quadratic Forms. The constants $C$ appearing in this solution may not be the same. We begin with
Definition
Let $ax^2+bxy+cy^2$ be a quadratic form with $a, b, cin mathbbZ$ and $b^2-4ac=D$. We say that a quadratic form $a_1x_1^2+b_1x_1y_1+c_1y_1^2$ with $a_1,b_1,c_1inmathbbZ$ and $b_1^2-4a_1c_1=D$ is equivalent to $ax^2+bxy+cy^2$ if there is $beginbmatrixp & q \ r & s endbmatrixinmathrmSL_2(mathbbZ)$ such that
$$
a_1x_1^2+b_1x_1y_1+c_1y_1^2=a(px_1+qy_1)^2+b(px_1+qy_1)(rx_1+sy_1)+c(rx_1+sy_1)^2.
$$
This can be written again with matrices
$$
beginbmatrixa_1 & fracb_12\ fracb_12 & c_1 endbmatrix = beginbmatrixp & r \ q & s endbmatrix beginbmatrixa & fracb2\ fracb2 & c endbmatrix beginbmatrixp & q \ r & s endbmatrix
$$
and
$$
beginbmatrixx\yendbmatrix=beginbmatrixp & q \ r & s endbmatrixbeginbmatrixx_1\y_1endbmatrix.
$$
The next one is the reducing step.
Lemma 1
Any integer quadratic form with discriminant $D$ is equivalent to $ax^2+bxy+cy^2$ with $|b|leq |a|leq |c|$.
The proof is a repeated application of the matrices in $mathrmSL_2(mathbbZ)$.
$$
beginbmatrix 1 & 1 \ 0 & 1 endbmatrix, beginbmatrix 0 & -1 \ 1 & 0endbmatrix.
$$
Corollary 1
The number of equivalence classes of integer quadratic forms with discriminant $D$ is finite.
If $b^2-4ac=D$ and $|b|leq |a|leq |c|$, then $|D|geq 4|ac|-|b^2| geq 3|ac|$ gives only finitely many choices for $a$ and $c$, subsequently for $b$.
Then it suffices to prove the following for $A=beginbmatrixa& b/2\ b/2&cendbmatrix$ with $b^2-4ac=D$ and $|b|leq |a|leq |c|$, as the sum can be written as a finite sum of the following kind.
The sum is absolutely convergent for $k>1$:
$$
sum_gammain mathrmSL_2(mathbbZ) frac1left( beginbmatrixz&1endbmatrixgamma^T A gamma beginbmatrixz\1endbmatrixright)^k.
$$
It suffices to show the following sum is convergent instead, as it differs only by a positive constant.
$$
sum_gammain mathrmSL_2(mathbbZ) frac1left( beginbmatrixz&1endbmatrixgamma^T beginbmatrixalpha \ betaendbmatrix beginbmatrixalpha' &beta'endbmatrix gamma beginbmatrixz\1endbmatrixright)^k
$$
where $alpha=alpha'=2a$, $beta=b+sqrtD$, and $beta'=b-sqrtD$.
We establish the following uniform inequality:
Lemma 2
We have $$
left|beginbmatrixalpha' &beta'endbmatrix gamma beginbmatrixz\1endbmatrixright|geq C (|pz+q||rz+s|)^1/2
$$
where $C>0$ is an absolute constant depending at most on $a,b,c$, and $gamma=beginbmatrix p&q\r&sendbmatrixinmathrmSL_2(mathbbZ)$.
The proof is by AM-GM inequality.
Now, it suffices to prove the convergence of
$$
sum_gammain mathrmSL_2(mathbbZ)frac1 (*)
$$
Consider $Gamma=mathrmSL_2(mathbbZ)$ and $Gamma_infty=leftpmbeginbmatrix 1&m\0&1endbmatrix, minmathbbZright$.
Fix coprime integers $r$, $s$. Let $p_0,q_0inmathbbZ$ satisfy $p_0s-q_0r=1$. Then it is well-known that all integers $p,q$ satisfying $ps-qr=1$ are $p=p_0+rt$, $q=q_0+st$ with $tinmathbbZ$. Varying $t$ if necessary, we may choose $p_0, q_0$ such that a coset representative $beginbmatrixp_0&q_0\r&sendbmatrix$ in $Gamma_inftybackslash Gamma$ with $-frac12leqfracp_0z+q_0rz+s<frac12$. We have the above sums over $r,s$ with $|t|geq 1$, is convergent since
$$
sum_(r,s)=1 frac1rz+sleq Csum_sfrac1
$$
is convergent where $C>0$ is an absolute constant that depends only on $z$.
We also have that the area of a parallelogram with vertices $0$, $pz+q$, $rz+s$ is the same as that of vertices $0$, $1$, $z$, this is because of $ps-qr=1$. Call this area $C$. Then
$$
|pz+q||rz+s| |sin theta_gamma| = C.
$$
Here $theta_gamma$ is the angle of the corner formed by the three points $pz+q$, $0$, $rz+s$. This angle is also written as $theta_gamma=arg fracpz+qrz+s=arg gamma z$.
Then the convergence of above sum (*) over $r,s$ with $t=0$ follows from that of
$$
sum_gammainGamma_inftybackslashGamma|sin theta_gamma|^k (**)
$$
Lemma 3
Let $Y>0$. We have a constant $C>0$ depending on $z$ such that
$$
# gammainGamma_inftybackslashGamma biggvert leq 4+frac4CY.
$$
This is proved by applying a classical lemma
$$
# gammainGamma_inftybackslashGamma biggvert Im(gamma z)geq Yleq 1+frac CY.
$$
Now, the convergence of (**) follows by partial summation with application of Lemma 3 for $Y=frac1n$, $n=1,2,3,ldots$.
answered 17 hours ago
i707107
11.2k21241
Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $beginbmatrixalpha \ betaendbmatrix * beginbmatrixalpha' &beta'endbmatrix$, since by multiplying them I got $4a*beginbmatrixa & fracb-sqrtD2 \ fracb+sqrtD2 & cendbmatrix$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
– Deavor
11 hours ago
Since we can interpret quadratic forms as matrices:$$beginbmatrixx¥dbmatrixbeginbmatrixa&b\c&dendbmatrixbeginbmatrixx\yendbmatrix=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
– i707107
9 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This can be done with Theory of Quadratic Forms. The constants $C$ appearing in this solution may not be the same. We begin with
Definition
Let $ax^2+bxy+cy^2$ be a quadratic form with $a, b, cin mathbbZ$ and $b^2-4ac=D$. We say that a quadratic form $a_1x_1^2+b_1x_1y_1+c_1y_1^2$ with $a_1,b_1,c_1inmathbbZ$ and $b_1^2-4a_1c_1=D$ is equivalent to $ax^2+bxy+cy^2$ if there is $beginbmatrixp & q \ r & s endbmatrixinmathrmSL_2(mathbbZ)$ such that
$$
a_1x_1^2+b_1x_1y_1+c_1y_1^2=a(px_1+qy_1)^2+b(px_1+qy_1)(rx_1+sy_1)+c(rx_1+sy_1)^2.
$$
This can be written again with matrices
$$
beginbmatrixa_1 & fracb_12\ fracb_12 & c_1 endbmatrix = beginbmatrixp & r \ q & s endbmatrix beginbmatrixa & fracb2\ fracb2 & c endbmatrix beginbmatrixp & q \ r & s endbmatrix
$$
and
$$
beginbmatrixx\yendbmatrix=beginbmatrixp & q \ r & s endbmatrixbeginbmatrixx_1\y_1endbmatrix.
$$
The next one is the reducing step.
Lemma 1
Any integer quadratic form with discriminant $D$ is equivalent to $ax^2+bxy+cy^2$ with $|b|leq |a|leq |c|$.
The proof is a repeated application of the matrices in $mathrmSL_2(mathbbZ)$.
$$
beginbmatrix 1 & 1 \ 0 & 1 endbmatrix, beginbmatrix 0 & -1 \ 1 & 0endbmatrix.
$$
Corollary 1
The number of equivalence classes of integer quadratic forms with discriminant $D$ is finite.
If $b^2-4ac=D$ and $|b|leq |a|leq |c|$, then $|D|geq 4|ac|-|b^2| geq 3|ac|$ gives only finitely many choices for $a$ and $c$, subsequently for $b$.
Then it suffices to prove the following for $A=beginbmatrixa& b/2\ b/2&cendbmatrix$ with $b^2-4ac=D$ and $|b|leq |a|leq |c|$, as the sum can be written as a finite sum of the following kind.
The sum is absolutely convergent for $k>1$:
$$
sum_gammain mathrmSL_2(mathbbZ) frac1left( beginbmatrixz&1endbmatrixgamma^T A gamma beginbmatrixz\1endbmatrixright)^k.
$$
It suffices to show the following sum is convergent instead, as it differs only by a positive constant.
$$
sum_gammain mathrmSL_2(mathbbZ) frac1left( beginbmatrixz&1endbmatrixgamma^T beginbmatrixalpha \ betaendbmatrix beginbmatrixalpha' &beta'endbmatrix gamma beginbmatrixz\1endbmatrixright)^k
$$
where $alpha=alpha'=2a$, $beta=b+sqrtD$, and $beta'=b-sqrtD$.
We establish the following uniform inequality:
Lemma 2
We have $$
left|beginbmatrixalpha' &beta'endbmatrix gamma beginbmatrixz\1endbmatrixright|geq C (|pz+q||rz+s|)^1/2
$$
where $C>0$ is an absolute constant depending at most on $a,b,c$, and $gamma=beginbmatrix p&q\r&sendbmatrixinmathrmSL_2(mathbbZ)$.
The proof is by AM-GM inequality.
Now, it suffices to prove the convergence of
$$
sum_gammain mathrmSL_2(mathbbZ)frac1 (*)
$$
Consider $Gamma=mathrmSL_2(mathbbZ)$ and $Gamma_infty=leftpmbeginbmatrix 1&m\0&1endbmatrix, minmathbbZright$.
Fix coprime integers $r$, $s$. Let $p_0,q_0inmathbbZ$ satisfy $p_0s-q_0r=1$. Then it is well-known that all integers $p,q$ satisfying $ps-qr=1$ are $p=p_0+rt$, $q=q_0+st$ with $tinmathbbZ$. Varying $t$ if necessary, we may choose $p_0, q_0$ such that a coset representative $beginbmatrixp_0&q_0\r&sendbmatrix$ in $Gamma_inftybackslash Gamma$ with $-frac12leqfracp_0z+q_0rz+s<frac12$. We have the above sums over $r,s$ with $|t|geq 1$, is convergent since
$$
sum_(r,s)=1 frac1rz+sleq Csum_sfrac1
$$
is convergent where $C>0$ is an absolute constant that depends only on $z$.
We also have that the area of a parallelogram with vertices $0$, $pz+q$, $rz+s$ is the same as that of vertices $0$, $1$, $z$, this is because of $ps-qr=1$. Call this area $C$. Then
$$
|pz+q||rz+s| |sin theta_gamma| = C.
$$
Here $theta_gamma$ is the angle of the corner formed by the three points $pz+q$, $0$, $rz+s$. This angle is also written as $theta_gamma=arg fracpz+qrz+s=arg gamma z$.
Then the convergence of above sum (*) over $r,s$ with $t=0$ follows from that of
$$
sum_gammainGamma_inftybackslashGamma|sin theta_gamma|^k (**)
$$
Lemma 3
Let $Y>0$. We have a constant $C>0$ depending on $z$ such that
$$
# gammainGamma_inftybackslashGamma biggvert leq 4+frac4CY.
$$
This is proved by applying a classical lemma
$$
# gammainGamma_inftybackslashGamma biggvert Im(gamma z)geq Yleq 1+frac CY.
$$
Now, the convergence of (**) follows by partial summation with application of Lemma 3 for $Y=frac1n$, $n=1,2,3,ldots$.
answered 17 hours ago
i707107
11.2k21241
This can be done with Theory of Quadratic Forms. The constants $C$ appearing in this solution may not be the same. We begin with
Definition
Let $ax^2+bxy+cy^2$ be a quadratic form with $a, b, cin mathbbZ$ and $b^2-4ac=D$. We say that a quadratic form $a_1x_1^2+b_1x_1y_1+c_1y_1^2$ with $a_1,b_1,c_1inmathbbZ$ and $b_1^2-4a_1c_1=D$ is equivalent to $ax^2+bxy+cy^2$ if there is $beginbmatrixp & q \ r & s endbmatrixinmathrmSL_2(mathbbZ)$ such that
$$
a_1x_1^2+b_1x_1y_1+c_1y_1^2=a(px_1+qy_1)^2+b(px_1+qy_1)(rx_1+sy_1)+c(rx_1+sy_1)^2.
$$
This can be written again with matrices
$$
beginbmatrixa_1 & fracb_12\ fracb_12 & c_1 endbmatrix = beginbmatrixp & r \ q & s endbmatrix beginbmatrixa & fracb2\ fracb2 & c endbmatrix beginbmatrixp & q \ r & s endbmatrix
$$
and
$$
beginbmatrixx\yendbmatrix=beginbmatrixp & q \ r & s endbmatrixbeginbmatrixx_1\y_1endbmatrix.
$$
The next one is the reducing step.
Lemma 1
Any integer quadratic form with discriminant $D$ is equivalent to $ax^2+bxy+cy^2$ with $|b|leq |a|leq |c|$.
The proof is a repeated application of the matrices in $mathrmSL_2(mathbbZ)$.
$$
beginbmatrix 1 & 1 \ 0 & 1 endbmatrix, beginbmatrix 0 & -1 \ 1 & 0endbmatrix.
$$
Corollary 1
The number of equivalence classes of integer quadratic forms with discriminant $D$ is finite.
If $b^2-4ac=D$ and $|b|leq |a|leq |c|$, then $|D|geq 4|ac|-|b^2| geq 3|ac|$ gives only finitely many choices for $a$ and $c$, subsequently for $b$.
Then it suffices to prove the following for $A=beginbmatrixa& b/2\ b/2&cendbmatrix$ with $b^2-4ac=D$ and $|b|leq |a|leq |c|$, as the sum can be written as a finite sum of the following kind.
The sum is absolutely convergent for $k>1$:
$$
sum_gammain mathrmSL_2(mathbbZ) frac1left( beginbmatrixz&1endbmatrixgamma^T A gamma beginbmatrixz\1endbmatrixright)^k.
$$
It suffices to show the following sum is convergent instead, as it differs only by a positive constant.
$$
sum_gammain mathrmSL_2(mathbbZ) frac1left( beginbmatrixz&1endbmatrixgamma^T beginbmatrixalpha \ betaendbmatrix beginbmatrixalpha' &beta'endbmatrix gamma beginbmatrixz\1endbmatrixright)^k
$$
where $alpha=alpha'=2a$, $beta=b+sqrtD$, and $beta'=b-sqrtD$.
We establish the following uniform inequality:
Lemma 2
We have $$
left|beginbmatrixalpha' &beta'endbmatrix gamma beginbmatrixz\1endbmatrixright|geq C (|pz+q||rz+s|)^1/2
$$
where $C>0$ is an absolute constant depending at most on $a,b,c$, and $gamma=beginbmatrix p&q\r&sendbmatrixinmathrmSL_2(mathbbZ)$.
The proof is by AM-GM inequality.
Now, it suffices to prove the convergence of
$$
sum_gammain mathrmSL_2(mathbbZ)frac1 (*)
$$
Consider $Gamma=mathrmSL_2(mathbbZ)$ and $Gamma_infty=leftpmbeginbmatrix 1&m\0&1endbmatrix, minmathbbZright$.
Fix coprime integers $r$, $s$. Let $p_0,q_0inmathbbZ$ satisfy $p_0s-q_0r=1$. Then it is well-known that all integers $p,q$ satisfying $ps-qr=1$ are $p=p_0+rt$, $q=q_0+st$ with $tinmathbbZ$. Varying $t$ if necessary, we may choose $p_0, q_0$ such that a coset representative $beginbmatrixp_0&q_0\r&sendbmatrix$ in $Gamma_inftybackslash Gamma$ with $-frac12leqfracp_0z+q_0rz+s<frac12$. We have the above sums over $r,s$ with $|t|geq 1$, is convergent since
$$
sum_(r,s)=1 frac1rz+sleq Csum_sfrac1
$$
is convergent where $C>0$ is an absolute constant that depends only on $z$.
We also have that the area of a parallelogram with vertices $0$, $pz+q$, $rz+s$ is the same as that of vertices $0$, $1$, $z$, this is because of $ps-qr=1$. Call this area $C$. Then
$$
|pz+q||rz+s| |sin theta_gamma| = C.
$$
Here $theta_gamma$ is the angle of the corner formed by the three points $pz+q$, $0$, $rz+s$. This angle is also written as $theta_gamma=arg fracpz+qrz+s=arg gamma z$.
Then the convergence of above sum (*) over $r,s$ with $t=0$ follows from that of
$$
sum_gammainGamma_inftybackslashGamma|sin theta_gamma|^k (**)
$$
Lemma 3
Let $Y>0$. We have a constant $C>0$ depending on $z$ such that
$$
# gammainGamma_inftybackslashGamma biggvert leq 4+frac4CY.
$$
This is proved by applying a classical lemma
$$
# gammainGamma_inftybackslashGamma biggvert Im(gamma z)geq Yleq 1+frac CY.
$$
Now, the convergence of (**) follows by partial summation with application of Lemma 3 for $Y=frac1n$, $n=1,2,3,ldots$.
answered 17 hours ago
i707107
11.2k21241
answered 17 hours ago
i707107
11.2k21241
answered 17 hours ago
i707107
11.2k21241
answered 17 hours ago
i707107
11.2k21241
11.2k21241
Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $beginbmatrixalpha \ betaendbmatrix * beginbmatrixalpha' &beta'endbmatrix$, since by multiplying them I got $4a*beginbmatrixa & fracb-sqrtD2 \ fracb+sqrtD2 & cendbmatrix$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
– Deavor
11 hours ago
Since we can interpret quadratic forms as matrices:$$beginbmatrixx¥dbmatrixbeginbmatrixa&b\c&dendbmatrixbeginbmatrixx\yendbmatrix=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
– i707107
9 hours ago
add a comment |Â
Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $beginbmatrixalpha \ betaendbmatrix * beginbmatrixalpha' &beta'endbmatrix$, since by multiplying them I got $4a*beginbmatrixa & fracb-sqrtD2 \ fracb+sqrtD2 & cendbmatrix$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
– Deavor
11 hours ago
Since we can interpret quadratic forms as matrices:$$beginbmatrixx¥dbmatrixbeginbmatrixa&b\c&dendbmatrixbeginbmatrixx\yendbmatrix=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
– i707107
9 hours ago
Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $beginbmatrixalpha \ betaendbmatrix * beginbmatrixalpha' &beta'endbmatrix$, since by multiplying them I got $4a*beginbmatrixa & fracb-sqrtD2 \ fracb+sqrtD2 & cendbmatrix$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
– Deavor
11 hours ago
Thank you for your detailed answer. I hope you can answer two more questions for my comprehension. The first is, what is the motivation behind substituting $A$ by $beginbmatrixalpha \ betaendbmatrix * beginbmatrixalpha' &beta'endbmatrix$, since by multiplying them I got $4a*beginbmatrixa & fracb-sqrtD2 \ fracb+sqrtD2 & cendbmatrix$. And $D$ has not to be a square number. Did I make a mistake? The second question is why do we have to inspect the case $t=0$ seperately.
– Deavor
11 hours ago
Since we can interpret quadratic forms as matrices:$$beginbmatrixx¥dbmatrixbeginbmatrixa&b\c&dendbmatrixbeginbmatrixx\yendbmatrix=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
– i707107
9 hours ago
Since we can interpret quadratic forms as matrices:$$beginbmatrixx¥dbmatrixbeginbmatrixa&b\c&dendbmatrixbeginbmatrixx\yendbmatrix=ax^2+(b+c)xy+dy^2$$, we may have that decomposition. Please postpone accepting, because I found an error in Lemma 2 and need to fix it.
– i707107
9 hours ago
add a comment |Â
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What does $Dequiv 0,1 pmod4$ mean?
– zhw.
Aug 2 at 20:40
$D equiv 0 pmod 4$ or $D equiv 1 pmod 4$ sorry if that was cloudy
– Deavor
Aug 2 at 20:50
You might wanted to put $b^2-4ac$ in two places, one under the sum, the other after your attempted proof.
– i707107
Aug 4 at 19:38
Oh yes thanks for noticing the typo
– Deavor
Aug 4 at 20:18