Zeta like function summing over Gaussian integers in the first quadrant
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Let $x$ be a real number and let
$$f(x)=sum_{ z = re^{theta i} inmathbb{Z}[i] \ r le x \ 0le theta le pi/2}frac{1}{z^s}$$
Is it possible to compute in (terms of the $zeta$ function perhaps) $$xi
(s)=lim_{xtoinfty} f_s(x)$$
It looks just by toying around that $xi(2)$ is divergent. Does $xi$ converge for larger $s$?
Here's some empirical evidence that $xi(2)$ diverges.
Let $tau(n) : mathbb{N} to mathbb{N}[i]$ be a pairing function and let $tau(n)=x_n+y_ni$.
We are looking for $frac{1}{tau(n)^2}=frac{1}{(x_n+y_ni)^2}= frac{(x_n-y_ni)^2}{(x_n^2+y_n^2)^2}= frac{x_n^2-2x_ny_ni-y_n^2}{(x_n^2+y_n^2)^2}=frac{x_n^2-y_n^2}{(x_n^2+y_n^2)^2}+ifrac{2x_ny_n}{{(x_n^2+y_n^2)^2}}$
We have then $$sum_{ }frac{1}{z^2} =sum_{n=1 \ x+yi=tau(n)}^inftyfrac{x^2-y^2}{(x^2+y^2)^2}+2isum_{n=1 \ x+yi=tau(n)}^inftyfrac{xy}{(x^2+y^2)^2}$$
By numerical methods it seems to me that the sum diverges.
zeta-functions
asked Dec 4 at 0:35
Mason
1,8411527
add a comment |
up vote
2
down vote
favorite
Let $x$ be a real number and let
$$f(x)=sum_{ z = re^{theta i} inmathbb{Z}[i] \ r le x \ 0le theta le pi/2}frac{1}{z^s}$$
Is it possible to compute in (terms of the $zeta$ function perhaps) $$xi
(s)=lim_{xtoinfty} f_s(x)$$
It looks just by toying around that $xi(2)$ is divergent. Does $xi$ converge for larger $s$?
Here's some empirical evidence that $xi(2)$ diverges.
Let $tau(n) : mathbb{N} to mathbb{N}[i]$ be a pairing function and let $tau(n)=x_n+y_ni$.
We are looking for $frac{1}{tau(n)^2}=frac{1}{(x_n+y_ni)^2}= frac{(x_n-y_ni)^2}{(x_n^2+y_n^2)^2}= frac{x_n^2-2x_ny_ni-y_n^2}{(x_n^2+y_n^2)^2}=frac{x_n^2-y_n^2}{(x_n^2+y_n^2)^2}+ifrac{2x_ny_n}{{(x_n^2+y_n^2)^2}}$
We have then $$sum_{ }frac{1}{z^2} =sum_{n=1 \ x+yi=tau(n)}^inftyfrac{x^2-y^2}{(x^2+y^2)^2}+2isum_{n=1 \ x+yi=tau(n)}^inftyfrac{xy}{(x^2+y^2)^2}$$
By numerical methods it seems to me that the sum diverges.
zeta-functions
asked Dec 4 at 0:35
Mason
1,8411527
I like to use $mathbb{N}[i]$ for the Gaussian integers with real and imaginary parts non-negative.
– Mason
Dec 4 at 0:35
1
No to your first question. Do you know Hecke L-functions ? For example $L(s,psi^n)$ with $psi(a+ib) = frac{a+ib}{a-ib}$, for $n$ even it is a Hecke character depending only on the embeddings $mathbb{Z}[i] to mathbb{C}$ and not on the quotient rings $mathbb{Z}[i]/(n)$. We need to sum over all the $L(s,psi^n),n in mathbb{Z}$ to obtain the completely multiplicative function $f(a+ib) = i^m$ if $arg(i^{-m} (a+ib)) in [0,pi/2)$
– reuns
Dec 4 at 0:58
I don't know about Hecke L-functions. But I guess I'll do some reading
– Mason
Dec 4 at 1:02
Oops of course $f$ isn't multiplicative. But summing over all the $L(s,psi^n)$ make it appear.
– reuns
Dec 4 at 1:07
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $x$ be a real number and let
$$f(x)=sum_{ z = re^{theta i} inmathbb{Z}[i] \ r le x \ 0le theta le pi/2}frac{1}{z^s}$$
Is it possible to compute in (terms of the $zeta$ function perhaps) $$xi
(s)=lim_{xtoinfty} f_s(x)$$
It looks just by toying around that $xi(2)$ is divergent. Does $xi$ converge for larger $s$?
Here's some empirical evidence that $xi(2)$ diverges.
Let $tau(n) : mathbb{N} to mathbb{N}[i]$ be a pairing function and let $tau(n)=x_n+y_ni$.
We are looking for $frac{1}{tau(n)^2}=frac{1}{(x_n+y_ni)^2}= frac{(x_n-y_ni)^2}{(x_n^2+y_n^2)^2}= frac{x_n^2-2x_ny_ni-y_n^2}{(x_n^2+y_n^2)^2}=frac{x_n^2-y_n^2}{(x_n^2+y_n^2)^2}+ifrac{2x_ny_n}{{(x_n^2+y_n^2)^2}}$
We have then $$sum_{ }frac{1}{z^2} =sum_{n=1 \ x+yi=tau(n)}^inftyfrac{x^2-y^2}{(x^2+y^2)^2}+2isum_{n=1 \ x+yi=tau(n)}^inftyfrac{xy}{(x^2+y^2)^2}$$
By numerical methods it seems to me that the sum diverges.
zeta-functions
asked Dec 4 at 0:35
Mason
1,8411527
Let $x$ be a real number and let
$$f(x)=sum_{ z = re^{theta i} inmathbb{Z}[i] \ r le x \ 0le theta le pi/2}frac{1}{z^s}$$
Is it possible to compute in (terms of the $zeta$ function perhaps) $$xi
(s)=lim_{xtoinfty} f_s(x)$$
It looks just by toying around that $xi(2)$ is divergent. Does $xi$ converge for larger $s$?
Here's some empirical evidence that $xi(2)$ diverges.
Let $tau(n) : mathbb{N} to mathbb{N}[i]$ be a pairing function and let $tau(n)=x_n+y_ni$.
We are looking for $frac{1}{tau(n)^2}=frac{1}{(x_n+y_ni)^2}= frac{(x_n-y_ni)^2}{(x_n^2+y_n^2)^2}= frac{x_n^2-2x_ny_ni-y_n^2}{(x_n^2+y_n^2)^2}=frac{x_n^2-y_n^2}{(x_n^2+y_n^2)^2}+ifrac{2x_ny_n}{{(x_n^2+y_n^2)^2}}$
We have then $$sum_{ }frac{1}{z^2} =sum_{n=1 \ x+yi=tau(n)}^inftyfrac{x^2-y^2}{(x^2+y^2)^2}+2isum_{n=1 \ x+yi=tau(n)}^inftyfrac{xy}{(x^2+y^2)^2}$$
By numerical methods it seems to me that the sum diverges.
zeta-functions
zeta-functions
asked Dec 4 at 0:35
Mason
1,8411527
asked Dec 4 at 0:35
Mason
1,8411527
asked Dec 4 at 0:35
Mason
1,8411527
asked Dec 4 at 0:35
Mason
1,8411527
asked Dec 4 at 0:35
Mason
1,8411527
1,8411527
I like to use $mathbb{N}[i]$ for the Gaussian integers with real and imaginary parts non-negative.
– Mason
Dec 4 at 0:35
1
No to your first question. Do you know Hecke L-functions ? For example $L(s,psi^n)$ with $psi(a+ib) = frac{a+ib}{a-ib}$, for $n$ even it is a Hecke character depending only on the embeddings $mathbb{Z}[i] to mathbb{C}$ and not on the quotient rings $mathbb{Z}[i]/(n)$. We need to sum over all the $L(s,psi^n),n in mathbb{Z}$ to obtain the completely multiplicative function $f(a+ib) = i^m$ if $arg(i^{-m} (a+ib)) in [0,pi/2)$
– reuns
Dec 4 at 0:58
I don't know about Hecke L-functions. But I guess I'll do some reading
– Mason
Dec 4 at 1:02
Oops of course $f$ isn't multiplicative. But summing over all the $L(s,psi^n)$ make it appear.
– reuns
Dec 4 at 1:07
add a comment |
I like to use $mathbb{N}[i]$ for the Gaussian integers with real and imaginary parts non-negative.
– Mason
Dec 4 at 0:35
1
No to your first question. Do you know Hecke L-functions ? For example $L(s,psi^n)$ with $psi(a+ib) = frac{a+ib}{a-ib}$, for $n$ even it is a Hecke character depending only on the embeddings $mathbb{Z}[i] to mathbb{C}$ and not on the quotient rings $mathbb{Z}[i]/(n)$. We need to sum over all the $L(s,psi^n),n in mathbb{Z}$ to obtain the completely multiplicative function $f(a+ib) = i^m$ if $arg(i^{-m} (a+ib)) in [0,pi/2)$
– reuns
Dec 4 at 0:58
I don't know about Hecke L-functions. But I guess I'll do some reading
– Mason
Dec 4 at 1:02
Oops of course $f$ isn't multiplicative. But summing over all the $L(s,psi^n)$ make it appear.
– reuns
Dec 4 at 1:07
I like to use $mathbb{N}[i]$ for the Gaussian integers with real and imaginary parts non-negative.
– Mason
Dec 4 at 0:35
I like to use $mathbb{N}[i]$ for the Gaussian integers with real and imaginary parts non-negative.
– Mason
Dec 4 at 0:35
1
1
No to your first question. Do you know Hecke L-functions ? For example $L(s,psi^n)$ with $psi(a+ib) = frac{a+ib}{a-ib}$, for $n$ even it is a Hecke character depending only on the embeddings $mathbb{Z}[i] to mathbb{C}$ and not on the quotient rings $mathbb{Z}[i]/(n)$. We need to sum over all the $L(s,psi^n),n in mathbb{Z}$ to obtain the completely multiplicative function $f(a+ib) = i^m$ if $arg(i^{-m} (a+ib)) in [0,pi/2)$
– reuns
Dec 4 at 0:58
No to your first question. Do you know Hecke L-functions ? For example $L(s,psi^n)$ with $psi(a+ib) = frac{a+ib}{a-ib}$, for $n$ even it is a Hecke character depending only on the embeddings $mathbb{Z}[i] to mathbb{C}$ and not on the quotient rings $mathbb{Z}[i]/(n)$. We need to sum over all the $L(s,psi^n),n in mathbb{Z}$ to obtain the completely multiplicative function $f(a+ib) = i^m$ if $arg(i^{-m} (a+ib)) in [0,pi/2)$
– reuns
Dec 4 at 0:58
I don't know about Hecke L-functions. But I guess I'll do some reading
– Mason
Dec 4 at 1:02
I don't know about Hecke L-functions. But I guess I'll do some reading
– Mason
Dec 4 at 1:02
Oops of course $f$ isn't multiplicative. But summing over all the $L(s,psi^n)$ make it appear.
– reuns
Dec 4 at 1:07
Oops of course $f$ isn't multiplicative. But summing over all the $L(s,psi^n)$ make it appear.
– reuns
Dec 4 at 1:07
add a comment |
1 Answer
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Yes, $xi(2)$ diverges. Note that for $z$ in this quadrant, $-text{Im}(1/z^2) ge 0$. So $-text{Im} sum_{z in mathbb N[i]: 0 < |z| < r} 1/z^2$ can be approximated by
$$int_1^r drho int_0^{pi/2} dtheta; r frac{cos(theta)}{r^2}
sim {text {const}} cdot log(r)$$ (essentially the sum is a two-dimensional Riemann sum for the integral).
answered Dec 4 at 4:13
Robert Israel
316k23206457
I, of course, appreciate this answer and will see whether I can use it to answer the question that I proposed but I should just comment that I am not accepting as the answer because it doesn't address my question which was: Does $xi$ converge for $s>2$
– Mason
Dec 8 at 16:17
add a comment |
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1 Answer
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Yes, $xi(2)$ diverges. Note that for $z$ in this quadrant, $-text{Im}(1/z^2) ge 0$. So $-text{Im} sum_{z in mathbb N[i]: 0 < |z| < r} 1/z^2$ can be approximated by
$$int_1^r drho int_0^{pi/2} dtheta; r frac{cos(theta)}{r^2}
sim {text {const}} cdot log(r)$$ (essentially the sum is a two-dimensional Riemann sum for the integral).
answered Dec 4 at 4:13
Robert Israel
316k23206457
I, of course, appreciate this answer and will see whether I can use it to answer the question that I proposed but I should just comment that I am not accepting as the answer because it doesn't address my question which was: Does $xi$ converge for $s>2$
– Mason
Dec 8 at 16:17
add a comment |
up vote
0
down vote
Yes, $xi(2)$ diverges. Note that for $z$ in this quadrant, $-text{Im}(1/z^2) ge 0$. So $-text{Im} sum_{z in mathbb N[i]: 0 < |z| < r} 1/z^2$ can be approximated by
$$int_1^r drho int_0^{pi/2} dtheta; r frac{cos(theta)}{r^2}
sim {text {const}} cdot log(r)$$ (essentially the sum is a two-dimensional Riemann sum for the integral).
answered Dec 4 at 4:13
Robert Israel
316k23206457
I, of course, appreciate this answer and will see whether I can use it to answer the question that I proposed but I should just comment that I am not accepting as the answer because it doesn't address my question which was: Does $xi$ converge for $s>2$
– Mason
Dec 8 at 16:17
add a comment |
up vote
0
down vote
up vote
0
down vote
Yes, $xi(2)$ diverges. Note that for $z$ in this quadrant, $-text{Im}(1/z^2) ge 0$. So $-text{Im} sum_{z in mathbb N[i]: 0 < |z| < r} 1/z^2$ can be approximated by
$$int_1^r drho int_0^{pi/2} dtheta; r frac{cos(theta)}{r^2}
sim {text {const}} cdot log(r)$$ (essentially the sum is a two-dimensional Riemann sum for the integral).
answered Dec 4 at 4:13
Robert Israel
316k23206457
Yes, $xi(2)$ diverges. Note that for $z$ in this quadrant, $-text{Im}(1/z^2) ge 0$. So $-text{Im} sum_{z in mathbb N[i]: 0 < |z| < r} 1/z^2$ can be approximated by
$$int_1^r drho int_0^{pi/2} dtheta; r frac{cos(theta)}{r^2}
sim {text {const}} cdot log(r)$$ (essentially the sum is a two-dimensional Riemann sum for the integral).
answered Dec 4 at 4:13
Robert Israel
316k23206457
answered Dec 4 at 4:13
Robert Israel
316k23206457
answered Dec 4 at 4:13
Robert Israel
316k23206457
answered Dec 4 at 4:13
Robert Israel
316k23206457
316k23206457
I, of course, appreciate this answer and will see whether I can use it to answer the question that I proposed but I should just comment that I am not accepting as the answer because it doesn't address my question which was: Does $xi$ converge for $s>2$
– Mason
Dec 8 at 16:17
add a comment |
I, of course, appreciate this answer and will see whether I can use it to answer the question that I proposed but I should just comment that I am not accepting as the answer because it doesn't address my question which was: Does $xi$ converge for $s>2$
– Mason
Dec 8 at 16:17
I, of course, appreciate this answer and will see whether I can use it to answer the question that I proposed but I should just comment that I am not accepting as the answer because it doesn't address my question which was: Does $xi$ converge for $s>2$
– Mason
Dec 8 at 16:17
I, of course, appreciate this answer and will see whether I can use it to answer the question that I proposed but I should just comment that I am not accepting as the answer because it doesn't address my question which was: Does $xi$ converge for $s>2$
– Mason
Dec 8 at 16:17
add a comment |
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I like to use $mathbb{N}[i]$ for the Gaussian integers with real and imaginary parts non-negative.
– Mason
Dec 4 at 0:35
1
No to your first question. Do you know Hecke L-functions ? For example $L(s,psi^n)$ with $psi(a+ib) = frac{a+ib}{a-ib}$, for $n$ even it is a Hecke character depending only on the embeddings $mathbb{Z}[i] to mathbb{C}$ and not on the quotient rings $mathbb{Z}[i]/(n)$. We need to sum over all the $L(s,psi^n),n in mathbb{Z}$ to obtain the completely multiplicative function $f(a+ib) = i^m$ if $arg(i^{-m} (a+ib)) in [0,pi/2)$
– reuns
Dec 4 at 0:58
I don't know about Hecke L-functions. But I guess I'll do some reading
– Mason
Dec 4 at 1:02
Oops of course $f$ isn't multiplicative. But summing over all the $L(s,psi^n)$ make it appear.
– reuns
Dec 4 at 1:07