Mixing
Infinite sum of 1/sin^2 and theta function
$begingroup$
In studying some physical propagator, I came across the following sum
$$
sum_{n = -infty}^{+infty} frac{ a^n }{ sin^2(z + n pi tau) } .
$$
Obviously, my question is how to evaluate this sum.
To some extent, I understand the result when $a = 1$. Loosely speaking, without properly regularizing, we have
$$
sum_{n in mathbb{Z}} frac{ 1 }{ sin^2(z + n pi tau) } = - sum_{n in mathbb{Z}} partial_z partial_z ln sin(z + npitau) = - partial_z^2 ln prod_{n in mathbb{Z}} sin(z+npitau) .
$$
The final infinite product can be identified with $theta_1(z/pi|tau)$, where $q = e^{2pi i tau}$, so up to regularization issue, we have
$$
sum_{nin mathbb{Z}} frac{ 1 }{ sin^2(z + npi tau) } = - partial_z partial_z ln theta_1(z/pi|tau)
$$
However in the presence of $a^n$, I can't pull off this trick again (as far as I can see).
Suggestions on literature/references and more tricks are welcome!
sequences-and-series complex-analysis theta-functions
asked Jan 26 at 9:30
LelouchLelouch
1958
$endgroup$
add a comment |
$begingroup$
In studying some physical propagator, I came across the following sum
$$
sum_{n = -infty}^{+infty} frac{ a^n }{ sin^2(z + n pi tau) } .
$$
Obviously, my question is how to evaluate this sum.
To some extent, I understand the result when $a = 1$. Loosely speaking, without properly regularizing, we have
$$
sum_{n in mathbb{Z}} frac{ 1 }{ sin^2(z + n pi tau) } = - sum_{n in mathbb{Z}} partial_z partial_z ln sin(z + npitau) = - partial_z^2 ln prod_{n in mathbb{Z}} sin(z+npitau) .
$$
The final infinite product can be identified with $theta_1(z/pi|tau)$, where $q = e^{2pi i tau}$, so up to regularization issue, we have
$$
sum_{nin mathbb{Z}} frac{ 1 }{ sin^2(z + npi tau) } = - partial_z partial_z ln theta_1(z/pi|tau)
$$
However in the presence of $a^n$, I can't pull off this trick again (as far as I can see).
Suggestions on literature/references and more tricks are welcome!
sequences-and-series complex-analysis theta-functions
asked Jan 26 at 9:30
LelouchLelouch
1958
$endgroup$
$begingroup$
The Fourier series in $tau$ will have coefficients of the form $sum_{k | m}a^{m/k} k e^{2kz} $, so it is close to the inverse Mellin transform (in $s$) of $Li_{s-1}(e^{2z})Li_s(a)$
$endgroup$
– reuns
Jan 26 at 9:43
add a comment |
$begingroup$
In studying some physical propagator, I came across the following sum
$$
sum_{n = -infty}^{+infty} frac{ a^n }{ sin^2(z + n pi tau) } .
$$
Obviously, my question is how to evaluate this sum.
To some extent, I understand the result when $a = 1$. Loosely speaking, without properly regularizing, we have
$$
sum_{n in mathbb{Z}} frac{ 1 }{ sin^2(z + n pi tau) } = - sum_{n in mathbb{Z}} partial_z partial_z ln sin(z + npitau) = - partial_z^2 ln prod_{n in mathbb{Z}} sin(z+npitau) .
$$
The final infinite product can be identified with $theta_1(z/pi|tau)$, where $q = e^{2pi i tau}$, so up to regularization issue, we have
$$
sum_{nin mathbb{Z}} frac{ 1 }{ sin^2(z + npi tau) } = - partial_z partial_z ln theta_1(z/pi|tau)
$$
However in the presence of $a^n$, I can't pull off this trick again (as far as I can see).
Suggestions on literature/references and more tricks are welcome!
sequences-and-series complex-analysis theta-functions
asked Jan 26 at 9:30
LelouchLelouch
1958
$endgroup$
In studying some physical propagator, I came across the following sum
$$
sum_{n = -infty}^{+infty} frac{ a^n }{ sin^2(z + n pi tau) } .
$$
Obviously, my question is how to evaluate this sum.
To some extent, I understand the result when $a = 1$. Loosely speaking, without properly regularizing, we have
$$
sum_{n in mathbb{Z}} frac{ 1 }{ sin^2(z + n pi tau) } = - sum_{n in mathbb{Z}} partial_z partial_z ln sin(z + npitau) = - partial_z^2 ln prod_{n in mathbb{Z}} sin(z+npitau) .
$$
The final infinite product can be identified with $theta_1(z/pi|tau)$, where $q = e^{2pi i tau}$, so up to regularization issue, we have
$$
sum_{nin mathbb{Z}} frac{ 1 }{ sin^2(z + npi tau) } = - partial_z partial_z ln theta_1(z/pi|tau)
$$
However in the presence of $a^n$, I can't pull off this trick again (as far as I can see).
Suggestions on literature/references and more tricks are welcome!
sequences-and-series complex-analysis theta-functions
sequences-and-series complex-analysis theta-functions
asked Jan 26 at 9:30
LelouchLelouch
1958
asked Jan 26 at 9:30
LelouchLelouch
1958
asked Jan 26 at 9:30
LelouchLelouch
1958
asked Jan 26 at 9:30
LelouchLelouch
1958
asked Jan 26 at 9:30
LelouchLelouch
1958
1958
$begingroup$
The Fourier series in $tau$ will have coefficients of the form $sum_{k | m}a^{m/k} k e^{2kz} $, so it is close to the inverse Mellin transform (in $s$) of $Li_{s-1}(e^{2z})Li_s(a)$
$endgroup$
– reuns
Jan 26 at 9:43
add a comment |
$begingroup$
The Fourier series in $tau$ will have coefficients of the form $sum_{k | m}a^{m/k} k e^{2kz} $, so it is close to the inverse Mellin transform (in $s$) of $Li_{s-1}(e^{2z})Li_s(a)$
$endgroup$
– reuns
Jan 26 at 9:43
$begingroup$
The Fourier series in $tau$ will have coefficients of the form $sum_{k | m}a^{m/k} k e^{2kz} $, so it is close to the inverse Mellin transform (in $s$) of $Li_{s-1}(e^{2z})Li_s(a)$
$endgroup$
– reuns
Jan 26 at 9:43
$begingroup$
The Fourier series in $tau$ will have coefficients of the form $sum_{k | m}a^{m/k} k e^{2kz} $, so it is close to the inverse Mellin transform (in $s$) of $Li_{s-1}(e^{2z})Li_s(a)$
$endgroup$
– reuns
Jan 26 at 9:43
add a comment |
1 Answer
1
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oldest
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$begingroup$
The following is probably not mathematically rigorous, and is loosely based on the Ramanujan's identity
begin{align}
sum_{n = -infty}^infty frac{ (A;q)_n }{ (B;q)_n }X^n
= frac{(q;q) (B/A;q) (AX;q) (q/(AX))}{(B;q)(q/A;q)(X;q)(B/(AX);q)} .
end{align}
The original problem can be rephrased (assuming $partial_z$ can be moved into the sum),
begin{align}
F(z) equiv & sum_{n in mathbb{Z}} frac{1}{sin^2(frac{z}{2} + n pi tau)} a^n = -2 partial_z sum_{n in mathbb{Z}}frac{cos(frac{z}{2} + npi tau)}{sin(frac{z}{2} + npi tau)} a^n equiv -2 partial_z G(z).
end{align}To compute $G(z)$, we reorganize it
begin{align}
G(z) = - i sum_{n in mathbb{Z}}
frac{e^{i(z + 2 npi tau)}}{1-e^{i(z + n2pi tau)}} a^n
- i sum_{n in mathbb{Z}}
frac{1}{1 - e^{i(z + 2 npi tau)}} a^n .
end{align}
Defining $x equiv e^{iz}$, $q = e^{2 pi i tau}$, we have
begin{align}
G(z) = -i frac{x}{1-x} sum_n frac{(x;q)_n}{(qx;q)_n} (aq)^n - i frac{1}{1-x} sum_n frac{(x;q)_n}{(qx;q)_n} a^n ,
end{align}
where we used
begin{align}
frac{1}{1-xq^n} = frac{1}{1-x} frac{(x;q)_n}{(qx;q)_n} .
end{align}Now Ramanujan's identity comes in, and using shift properties of the $q$-Pochhammer symbols, the two sums actually are equal and they add. The final result is
begin{align}
G(z) = 2 eta(tau)^3 frac{ vartheta_1(mathfrak{a} + frac{ z}{ 2pi }|tau) }{ vartheta_1(frac{ z}{ 2pi }|tau)vartheta_1(mathfrak{a}|tau )} .
end{align}The above computation is not rigorous because the Ramanujan's identities requires $|q| < 1$ ,$|B/A| < |X| < 1$ for convergence. However, these requirements applied to the two sums at the end of step 2 are not compatible: one sum requires $|a| < 1$, the other $|a| > 1$. Besides, we need to move the $partial_z$ into a sum.
However the final answer seems physically reasonable, since it does produce physical results that we expect, despite the above issue.
answered Feb 1 at 6:18
LelouchLelouch
1958
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The following is probably not mathematically rigorous, and is loosely based on the Ramanujan's identity
begin{align}
sum_{n = -infty}^infty frac{ (A;q)_n }{ (B;q)_n }X^n
= frac{(q;q) (B/A;q) (AX;q) (q/(AX))}{(B;q)(q/A;q)(X;q)(B/(AX);q)} .
end{align}
The original problem can be rephrased (assuming $partial_z$ can be moved into the sum),
begin{align}
F(z) equiv & sum_{n in mathbb{Z}} frac{1}{sin^2(frac{z}{2} + n pi tau)} a^n = -2 partial_z sum_{n in mathbb{Z}}frac{cos(frac{z}{2} + npi tau)}{sin(frac{z}{2} + npi tau)} a^n equiv -2 partial_z G(z).
end{align}To compute $G(z)$, we reorganize it
begin{align}
G(z) = - i sum_{n in mathbb{Z}}
frac{e^{i(z + 2 npi tau)}}{1-e^{i(z + n2pi tau)}} a^n
- i sum_{n in mathbb{Z}}
frac{1}{1 - e^{i(z + 2 npi tau)}} a^n .
end{align}
Defining $x equiv e^{iz}$, $q = e^{2 pi i tau}$, we have
begin{align}
G(z) = -i frac{x}{1-x} sum_n frac{(x;q)_n}{(qx;q)_n} (aq)^n - i frac{1}{1-x} sum_n frac{(x;q)_n}{(qx;q)_n} a^n ,
end{align}
where we used
begin{align}
frac{1}{1-xq^n} = frac{1}{1-x} frac{(x;q)_n}{(qx;q)_n} .
end{align}Now Ramanujan's identity comes in, and using shift properties of the $q$-Pochhammer symbols, the two sums actually are equal and they add. The final result is
begin{align}
G(z) = 2 eta(tau)^3 frac{ vartheta_1(mathfrak{a} + frac{ z}{ 2pi }|tau) }{ vartheta_1(frac{ z}{ 2pi }|tau)vartheta_1(mathfrak{a}|tau )} .
end{align}The above computation is not rigorous because the Ramanujan's identities requires $|q| < 1$ ,$|B/A| < |X| < 1$ for convergence. However, these requirements applied to the two sums at the end of step 2 are not compatible: one sum requires $|a| < 1$, the other $|a| > 1$. Besides, we need to move the $partial_z$ into a sum.
However the final answer seems physically reasonable, since it does produce physical results that we expect, despite the above issue.
answered Feb 1 at 6:18
LelouchLelouch
1958
$endgroup$
add a comment |
$begingroup$
The following is probably not mathematically rigorous, and is loosely based on the Ramanujan's identity
begin{align}
sum_{n = -infty}^infty frac{ (A;q)_n }{ (B;q)_n }X^n
= frac{(q;q) (B/A;q) (AX;q) (q/(AX))}{(B;q)(q/A;q)(X;q)(B/(AX);q)} .
end{align}
The original problem can be rephrased (assuming $partial_z$ can be moved into the sum),
begin{align}
F(z) equiv & sum_{n in mathbb{Z}} frac{1}{sin^2(frac{z}{2} + n pi tau)} a^n = -2 partial_z sum_{n in mathbb{Z}}frac{cos(frac{z}{2} + npi tau)}{sin(frac{z}{2} + npi tau)} a^n equiv -2 partial_z G(z).
end{align}To compute $G(z)$, we reorganize it
begin{align}
G(z) = - i sum_{n in mathbb{Z}}
frac{e^{i(z + 2 npi tau)}}{1-e^{i(z + n2pi tau)}} a^n
- i sum_{n in mathbb{Z}}
frac{1}{1 - e^{i(z + 2 npi tau)}} a^n .
end{align}
Defining $x equiv e^{iz}$, $q = e^{2 pi i tau}$, we have
begin{align}
G(z) = -i frac{x}{1-x} sum_n frac{(x;q)_n}{(qx;q)_n} (aq)^n - i frac{1}{1-x} sum_n frac{(x;q)_n}{(qx;q)_n} a^n ,
end{align}
where we used
begin{align}
frac{1}{1-xq^n} = frac{1}{1-x} frac{(x;q)_n}{(qx;q)_n} .
end{align}Now Ramanujan's identity comes in, and using shift properties of the $q$-Pochhammer symbols, the two sums actually are equal and they add. The final result is
begin{align}
G(z) = 2 eta(tau)^3 frac{ vartheta_1(mathfrak{a} + frac{ z}{ 2pi }|tau) }{ vartheta_1(frac{ z}{ 2pi }|tau)vartheta_1(mathfrak{a}|tau )} .
end{align}The above computation is not rigorous because the Ramanujan's identities requires $|q| < 1$ ,$|B/A| < |X| < 1$ for convergence. However, these requirements applied to the two sums at the end of step 2 are not compatible: one sum requires $|a| < 1$, the other $|a| > 1$. Besides, we need to move the $partial_z$ into a sum.
However the final answer seems physically reasonable, since it does produce physical results that we expect, despite the above issue.
answered Feb 1 at 6:18
LelouchLelouch
1958
$endgroup$
add a comment |
$begingroup$
The following is probably not mathematically rigorous, and is loosely based on the Ramanujan's identity
begin{align}
sum_{n = -infty}^infty frac{ (A;q)_n }{ (B;q)_n }X^n
= frac{(q;q) (B/A;q) (AX;q) (q/(AX))}{(B;q)(q/A;q)(X;q)(B/(AX);q)} .
end{align}
The original problem can be rephrased (assuming $partial_z$ can be moved into the sum),
begin{align}
F(z) equiv & sum_{n in mathbb{Z}} frac{1}{sin^2(frac{z}{2} + n pi tau)} a^n = -2 partial_z sum_{n in mathbb{Z}}frac{cos(frac{z}{2} + npi tau)}{sin(frac{z}{2} + npi tau)} a^n equiv -2 partial_z G(z).
end{align}To compute $G(z)$, we reorganize it
begin{align}
G(z) = - i sum_{n in mathbb{Z}}
frac{e^{i(z + 2 npi tau)}}{1-e^{i(z + n2pi tau)}} a^n
- i sum_{n in mathbb{Z}}
frac{1}{1 - e^{i(z + 2 npi tau)}} a^n .
end{align}
Defining $x equiv e^{iz}$, $q = e^{2 pi i tau}$, we have
begin{align}
G(z) = -i frac{x}{1-x} sum_n frac{(x;q)_n}{(qx;q)_n} (aq)^n - i frac{1}{1-x} sum_n frac{(x;q)_n}{(qx;q)_n} a^n ,
end{align}
where we used
begin{align}
frac{1}{1-xq^n} = frac{1}{1-x} frac{(x;q)_n}{(qx;q)_n} .
end{align}Now Ramanujan's identity comes in, and using shift properties of the $q$-Pochhammer symbols, the two sums actually are equal and they add. The final result is
begin{align}
G(z) = 2 eta(tau)^3 frac{ vartheta_1(mathfrak{a} + frac{ z}{ 2pi }|tau) }{ vartheta_1(frac{ z}{ 2pi }|tau)vartheta_1(mathfrak{a}|tau )} .
end{align}The above computation is not rigorous because the Ramanujan's identities requires $|q| < 1$ ,$|B/A| < |X| < 1$ for convergence. However, these requirements applied to the two sums at the end of step 2 are not compatible: one sum requires $|a| < 1$, the other $|a| > 1$. Besides, we need to move the $partial_z$ into a sum.
However the final answer seems physically reasonable, since it does produce physical results that we expect, despite the above issue.
answered Feb 1 at 6:18
LelouchLelouch
1958
$endgroup$
The following is probably not mathematically rigorous, and is loosely based on the Ramanujan's identity
begin{align}
sum_{n = -infty}^infty frac{ (A;q)_n }{ (B;q)_n }X^n
= frac{(q;q) (B/A;q) (AX;q) (q/(AX))}{(B;q)(q/A;q)(X;q)(B/(AX);q)} .
end{align}
The original problem can be rephrased (assuming $partial_z$ can be moved into the sum),
begin{align}
F(z) equiv & sum_{n in mathbb{Z}} frac{1}{sin^2(frac{z}{2} + n pi tau)} a^n = -2 partial_z sum_{n in mathbb{Z}}frac{cos(frac{z}{2} + npi tau)}{sin(frac{z}{2} + npi tau)} a^n equiv -2 partial_z G(z).
end{align}To compute $G(z)$, we reorganize it
begin{align}
G(z) = - i sum_{n in mathbb{Z}}
frac{e^{i(z + 2 npi tau)}}{1-e^{i(z + n2pi tau)}} a^n
- i sum_{n in mathbb{Z}}
frac{1}{1 - e^{i(z + 2 npi tau)}} a^n .
end{align}
Defining $x equiv e^{iz}$, $q = e^{2 pi i tau}$, we have
begin{align}
G(z) = -i frac{x}{1-x} sum_n frac{(x;q)_n}{(qx;q)_n} (aq)^n - i frac{1}{1-x} sum_n frac{(x;q)_n}{(qx;q)_n} a^n ,
end{align}
where we used
begin{align}
frac{1}{1-xq^n} = frac{1}{1-x} frac{(x;q)_n}{(qx;q)_n} .
end{align}Now Ramanujan's identity comes in, and using shift properties of the $q$-Pochhammer symbols, the two sums actually are equal and they add. The final result is
begin{align}
G(z) = 2 eta(tau)^3 frac{ vartheta_1(mathfrak{a} + frac{ z}{ 2pi }|tau) }{ vartheta_1(frac{ z}{ 2pi }|tau)vartheta_1(mathfrak{a}|tau )} .
end{align}The above computation is not rigorous because the Ramanujan's identities requires $|q| < 1$ ,$|B/A| < |X| < 1$ for convergence. However, these requirements applied to the two sums at the end of step 2 are not compatible: one sum requires $|a| < 1$, the other $|a| > 1$. Besides, we need to move the $partial_z$ into a sum.
However the final answer seems physically reasonable, since it does produce physical results that we expect, despite the above issue.
answered Feb 1 at 6:18
LelouchLelouch
1958
answered Feb 1 at 6:18
LelouchLelouch
1958
answered Feb 1 at 6:18
LelouchLelouch
1958
answered Feb 1 at 6:18
LelouchLelouch
1958
1958
add a comment |
add a comment |
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$begingroup$
The Fourier series in $tau$ will have coefficients of the form $sum_{k | m}a^{m/k} k e^{2kz} $, so it is close to the inverse Mellin transform (in $s$) of $Li_{s-1}(e^{2z})Li_s(a)$
$endgroup$
– reuns
Jan 26 at 9:43