Does a generating function for $zeta(2k+1)$ exist?
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I know that a generating function for the Zeta function at the even integers already exists, but how about the Zeta function at the odd integers?
I've done some research, and found some alternative formulas for the harmonic numbers that allowed me to create a generating function for $zeta(2k+1)$, but I'd like to know if it'd be new.
That would be nice if the answer is no, after all, it gets me really frustrated when I find out that all my discoveries in Math are actually just rediscoveries. And that's been the case in like 99% of the times I found something.
Just found out the answer is yes, but my formula is different anyway, less bad.
riemann-zeta
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up vote
2
down vote
favorite
I know that a generating function for the Zeta function at the even integers already exists, but how about the Zeta function at the odd integers?
I've done some research, and found some alternative formulas for the harmonic numbers that allowed me to create a generating function for $zeta(2k+1)$, but I'd like to know if it'd be new.
That would be nice if the answer is no, after all, it gets me really frustrated when I find out that all my discoveries in Math are actually just rediscoveries. And that's been the case in like 99% of the times I found something.
Just found out the answer is yes, but my formula is different anyway, less bad.
riemann-zeta
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I know that a generating function for the Zeta function at the even integers already exists, but how about the Zeta function at the odd integers?
I've done some research, and found some alternative formulas for the harmonic numbers that allowed me to create a generating function for $zeta(2k+1)$, but I'd like to know if it'd be new.
That would be nice if the answer is no, after all, it gets me really frustrated when I find out that all my discoveries in Math are actually just rediscoveries. And that's been the case in like 99% of the times I found something.
Just found out the answer is yes, but my formula is different anyway, less bad.
riemann-zeta
I know that a generating function for the Zeta function at the even integers already exists, but how about the Zeta function at the odd integers?
I've done some research, and found some alternative formulas for the harmonic numbers that allowed me to create a generating function for $zeta(2k+1)$, but I'd like to know if it'd be new.
That would be nice if the answer is no, after all, it gets me really frustrated when I find out that all my discoveries in Math are actually just rediscoveries. And that's been the case in like 99% of the times I found something.
Just found out the answer is yes, but my formula is different anyway, less bad.
riemann-zeta
edited Aug 19 at 5:34
asked Aug 19 at 4:14
JR S.
177
177
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add a comment |Â
3 Answers
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active
oldest
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up vote
2
down vote
Yes, it exists and it is now new.
$$sum_ngeq 1zeta(2n+1)z^2n+1 = sum_ngeq 1sum_mgeq 1left(fraczmright)^2n+1=sum_mgeq 1fracfracz^3m^31-fracz^2m^2=sum_mgeq 1fracz^3m(m-z)(m+z)$$
equals
$$ -gamma z+psi(1+z)+psi(1-z)=-fracz2left(H_z+H_-zright)$$
(where $gamma$ is the Euler-Mascheroni constant and $psi(z)=fracGamma'(z)Gamma(z)$) due to
$$ sum_ngeq 0frac1(n+a)(n+b)=fracpsi(a)-psi(b)a-b. $$
Are you interested in the exponential generating function?
@HansMusgrave: this is for sure an answer to Does a generating function for ö(2k+1) exist?
â Jack D'Aurizioâ¦
Aug 19 at 15:19
@HansMusgrave: I am not getting what the issue is. Is there a generating function for $zeta(2n+1)$? Yes, the shown one. Is it well-known? Yes. Are there other OGFs? No, the OGF is unique.
â Jack D'Aurizioâ¦
Aug 19 at 15:50
I have it in a different form, more elegant than the existing one in a way. It has to with the fact that my formula for $H_k(n)$ also bypasses $psi$.
â JR S.
Aug 19 at 17:47
$H_n$ and $psi(n+1)$ are essentially the same thing so I do not see what it is to bypass there.
â Jack D'Aurizioâ¦
Aug 19 at 17:49
If you don't you're near sighted. Perhaps study a little more to broaden your horizons. Or have a look at my findings when I release them. $psi$ is by no means the only way to express $H_k(n)$, to think otherwise is ignorant.
â JR S.
Aug 19 at 17:53
 |Â
show 4 more comments
up vote
1
down vote
Based on this answer, it wouldn't be much work to construct your generating function from what has been known at least as far back as 2013 (and probably much further).
My experience with research in classical mathematics is that not only are the interesting results known, they've been known for so long that the results aren't digitized. This can introduce somewhat of a challenge in proving novelty for a new idea.
I'm usually extremely open to the DIY approach, but starting out in mathematical research is one area where having a professional opinion would be valuable. Many (definitely not all -- be respectful) professors are willing to sit down and talk about research for hours, especially if it has anything whatsoever to do with their interests. Such conversations can be a good way to help gauge whether an idea is novel or not and hopefully to spark interest in other ideas as well.
That said, novelty isn't everything. If your research doesn't tie in to the current body of literature well, it will likely need to be more exciting than just another formula (not to disparage the result. I don't know the field well enough) to be publishable. Reading through current papers and using those as starting points is likely to be a fruitful area of exploration. Assuming recent papers represent the cutting edge, anything beyond that which you discover is almost certainly going to be both new and relevant to what other researchers (i.e., your target audience) are interested in.
1
I totally agree with you. This is the only really great website for asking Math questions online, so if the people here doesn't know, nobody knows.
â JR S.
Aug 19 at 4:38
This is not an answer. The ordinary generating function for $zeta(2n+1)$ exists and it is pretty straightforward to find.
â Jack D'Aurizioâ¦
Aug 19 at 13:37
@JackD'Aurizio I beg to differ, I liked his answer. One answer is better than no answer, unless it's a rude answer.
â JR S.
Aug 19 at 17:55
@JRS.: then accept it. In my opinion it does not bring any actual content, but you are the asker, not me.
â Jack D'Aurizioâ¦
Aug 19 at 18:04
@JackD'Aurizio Suggestion accepted
â JR S.
Aug 19 at 18:07
add a comment |Â
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$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
With $dsvertsz < 1$:
beginalign
sum_n = 1^inftyzetapars2n + 1z^n & =
sum_n = 1^infty
brackszetapars2n + 1 - 1parspm z^1/2^2n +
sum_n = 1^inftyz^n
\[5mm] & =
sum_n = 1^infty
brackszetapars2n + 1 - 1parspm z^1/2^2n +
z over 1 - z
endalign
The first sum can be evaluated with the
A & S $dsmathbfcolorblack6.3.15$ identity. Namely,
beginalign
&sum_n = 1^inftyzetapars2n + 1z^n
\[2mm] = &
bracks!!1 over 2parspmrootz - 1 over 2,picotpars!pibrackspmrootz!! -
1 over 1 - z + 1 - gamma - Psipars! 1 pm !rootz!!!
\[2mm] & phantombracksA+ z over 1 - z
\[5mm] = &
pm,1 over 2rootz mp 1 over 2,picotparspirootz - gamma - Psipars1 pm rootz
endalign
where $dsgamma$ is the Euler-Mascheroni Constant and $dsPsi$ is the Digamma Function. By adding the expressions for both signs $ds~pm~$and dividing by two:
$$
bbxsum_n = 1^inftyzetapars2n + 1z^n =
-gamma -
Psipars1 + rootz + Psipars1 - rootz over 2
$$
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Yes, it exists and it is now new.
$$sum_ngeq 1zeta(2n+1)z^2n+1 = sum_ngeq 1sum_mgeq 1left(fraczmright)^2n+1=sum_mgeq 1fracfracz^3m^31-fracz^2m^2=sum_mgeq 1fracz^3m(m-z)(m+z)$$
equals
$$ -gamma z+psi(1+z)+psi(1-z)=-fracz2left(H_z+H_-zright)$$
(where $gamma$ is the Euler-Mascheroni constant and $psi(z)=fracGamma'(z)Gamma(z)$) due to
$$ sum_ngeq 0frac1(n+a)(n+b)=fracpsi(a)-psi(b)a-b. $$
Are you interested in the exponential generating function?
@HansMusgrave: this is for sure an answer to Does a generating function for ö(2k+1) exist?
â Jack D'Aurizioâ¦
Aug 19 at 15:19
@HansMusgrave: I am not getting what the issue is. Is there a generating function for $zeta(2n+1)$? Yes, the shown one. Is it well-known? Yes. Are there other OGFs? No, the OGF is unique.
â Jack D'Aurizioâ¦
Aug 19 at 15:50
I have it in a different form, more elegant than the existing one in a way. It has to with the fact that my formula for $H_k(n)$ also bypasses $psi$.
â JR S.
Aug 19 at 17:47
$H_n$ and $psi(n+1)$ are essentially the same thing so I do not see what it is to bypass there.
â Jack D'Aurizioâ¦
Aug 19 at 17:49
If you don't you're near sighted. Perhaps study a little more to broaden your horizons. Or have a look at my findings when I release them. $psi$ is by no means the only way to express $H_k(n)$, to think otherwise is ignorant.
â JR S.
Aug 19 at 17:53
 |Â
show 4 more comments
up vote
2
down vote
Yes, it exists and it is now new.
$$sum_ngeq 1zeta(2n+1)z^2n+1 = sum_ngeq 1sum_mgeq 1left(fraczmright)^2n+1=sum_mgeq 1fracfracz^3m^31-fracz^2m^2=sum_mgeq 1fracz^3m(m-z)(m+z)$$
equals
$$ -gamma z+psi(1+z)+psi(1-z)=-fracz2left(H_z+H_-zright)$$
(where $gamma$ is the Euler-Mascheroni constant and $psi(z)=fracGamma'(z)Gamma(z)$) due to
$$ sum_ngeq 0frac1(n+a)(n+b)=fracpsi(a)-psi(b)a-b. $$
Are you interested in the exponential generating function?
@HansMusgrave: this is for sure an answer to Does a generating function for ö(2k+1) exist?
â Jack D'Aurizioâ¦
Aug 19 at 15:19
@HansMusgrave: I am not getting what the issue is. Is there a generating function for $zeta(2n+1)$? Yes, the shown one. Is it well-known? Yes. Are there other OGFs? No, the OGF is unique.
â Jack D'Aurizioâ¦
Aug 19 at 15:50
I have it in a different form, more elegant than the existing one in a way. It has to with the fact that my formula for $H_k(n)$ also bypasses $psi$.
â JR S.
Aug 19 at 17:47
$H_n$ and $psi(n+1)$ are essentially the same thing so I do not see what it is to bypass there.
â Jack D'Aurizioâ¦
Aug 19 at 17:49
If you don't you're near sighted. Perhaps study a little more to broaden your horizons. Or have a look at my findings when I release them. $psi$ is by no means the only way to express $H_k(n)$, to think otherwise is ignorant.
â JR S.
Aug 19 at 17:53
 |Â
show 4 more comments
up vote
2
down vote
up vote
2
down vote
Yes, it exists and it is now new.
$$sum_ngeq 1zeta(2n+1)z^2n+1 = sum_ngeq 1sum_mgeq 1left(fraczmright)^2n+1=sum_mgeq 1fracfracz^3m^31-fracz^2m^2=sum_mgeq 1fracz^3m(m-z)(m+z)$$
equals
$$ -gamma z+psi(1+z)+psi(1-z)=-fracz2left(H_z+H_-zright)$$
(where $gamma$ is the Euler-Mascheroni constant and $psi(z)=fracGamma'(z)Gamma(z)$) due to
$$ sum_ngeq 0frac1(n+a)(n+b)=fracpsi(a)-psi(b)a-b. $$
Are you interested in the exponential generating function?
Yes, it exists and it is now new.
$$sum_ngeq 1zeta(2n+1)z^2n+1 = sum_ngeq 1sum_mgeq 1left(fraczmright)^2n+1=sum_mgeq 1fracfracz^3m^31-fracz^2m^2=sum_mgeq 1fracz^3m(m-z)(m+z)$$
equals
$$ -gamma z+psi(1+z)+psi(1-z)=-fracz2left(H_z+H_-zright)$$
(where $gamma$ is the Euler-Mascheroni constant and $psi(z)=fracGamma'(z)Gamma(z)$) due to
$$ sum_ngeq 0frac1(n+a)(n+b)=fracpsi(a)-psi(b)a-b. $$
Are you interested in the exponential generating function?
edited Aug 19 at 15:20
answered Aug 19 at 13:36
Jack D'Aurizioâ¦
272k32267632
272k32267632
@HansMusgrave: this is for sure an answer to Does a generating function for ö(2k+1) exist?
â Jack D'Aurizioâ¦
Aug 19 at 15:19
@HansMusgrave: I am not getting what the issue is. Is there a generating function for $zeta(2n+1)$? Yes, the shown one. Is it well-known? Yes. Are there other OGFs? No, the OGF is unique.
â Jack D'Aurizioâ¦
Aug 19 at 15:50
I have it in a different form, more elegant than the existing one in a way. It has to with the fact that my formula for $H_k(n)$ also bypasses $psi$.
â JR S.
Aug 19 at 17:47
$H_n$ and $psi(n+1)$ are essentially the same thing so I do not see what it is to bypass there.
â Jack D'Aurizioâ¦
Aug 19 at 17:49
If you don't you're near sighted. Perhaps study a little more to broaden your horizons. Or have a look at my findings when I release them. $psi$ is by no means the only way to express $H_k(n)$, to think otherwise is ignorant.
â JR S.
Aug 19 at 17:53
 |Â
show 4 more comments
@HansMusgrave: this is for sure an answer to Does a generating function for ö(2k+1) exist?
â Jack D'Aurizioâ¦
Aug 19 at 15:19
@HansMusgrave: I am not getting what the issue is. Is there a generating function for $zeta(2n+1)$? Yes, the shown one. Is it well-known? Yes. Are there other OGFs? No, the OGF is unique.
â Jack D'Aurizioâ¦
Aug 19 at 15:50
I have it in a different form, more elegant than the existing one in a way. It has to with the fact that my formula for $H_k(n)$ also bypasses $psi$.
â JR S.
Aug 19 at 17:47
$H_n$ and $psi(n+1)$ are essentially the same thing so I do not see what it is to bypass there.
â Jack D'Aurizioâ¦
Aug 19 at 17:49
If you don't you're near sighted. Perhaps study a little more to broaden your horizons. Or have a look at my findings when I release them. $psi$ is by no means the only way to express $H_k(n)$, to think otherwise is ignorant.
â JR S.
Aug 19 at 17:53
@HansMusgrave: this is for sure an answer to Does a generating function for ö(2k+1) exist?
â Jack D'Aurizioâ¦
Aug 19 at 15:19
@HansMusgrave: this is for sure an answer to Does a generating function for ö(2k+1) exist?
â Jack D'Aurizioâ¦
Aug 19 at 15:19
@HansMusgrave: I am not getting what the issue is. Is there a generating function for $zeta(2n+1)$? Yes, the shown one. Is it well-known? Yes. Are there other OGFs? No, the OGF is unique.
â Jack D'Aurizioâ¦
Aug 19 at 15:50
@HansMusgrave: I am not getting what the issue is. Is there a generating function for $zeta(2n+1)$? Yes, the shown one. Is it well-known? Yes. Are there other OGFs? No, the OGF is unique.
â Jack D'Aurizioâ¦
Aug 19 at 15:50
I have it in a different form, more elegant than the existing one in a way. It has to with the fact that my formula for $H_k(n)$ also bypasses $psi$.
â JR S.
Aug 19 at 17:47
I have it in a different form, more elegant than the existing one in a way. It has to with the fact that my formula for $H_k(n)$ also bypasses $psi$.
â JR S.
Aug 19 at 17:47
$H_n$ and $psi(n+1)$ are essentially the same thing so I do not see what it is to bypass there.
â Jack D'Aurizioâ¦
Aug 19 at 17:49
$H_n$ and $psi(n+1)$ are essentially the same thing so I do not see what it is to bypass there.
â Jack D'Aurizioâ¦
Aug 19 at 17:49
If you don't you're near sighted. Perhaps study a little more to broaden your horizons. Or have a look at my findings when I release them. $psi$ is by no means the only way to express $H_k(n)$, to think otherwise is ignorant.
â JR S.
Aug 19 at 17:53
If you don't you're near sighted. Perhaps study a little more to broaden your horizons. Or have a look at my findings when I release them. $psi$ is by no means the only way to express $H_k(n)$, to think otherwise is ignorant.
â JR S.
Aug 19 at 17:53
 |Â
show 4 more comments
up vote
1
down vote
Based on this answer, it wouldn't be much work to construct your generating function from what has been known at least as far back as 2013 (and probably much further).
My experience with research in classical mathematics is that not only are the interesting results known, they've been known for so long that the results aren't digitized. This can introduce somewhat of a challenge in proving novelty for a new idea.
I'm usually extremely open to the DIY approach, but starting out in mathematical research is one area where having a professional opinion would be valuable. Many (definitely not all -- be respectful) professors are willing to sit down and talk about research for hours, especially if it has anything whatsoever to do with their interests. Such conversations can be a good way to help gauge whether an idea is novel or not and hopefully to spark interest in other ideas as well.
That said, novelty isn't everything. If your research doesn't tie in to the current body of literature well, it will likely need to be more exciting than just another formula (not to disparage the result. I don't know the field well enough) to be publishable. Reading through current papers and using those as starting points is likely to be a fruitful area of exploration. Assuming recent papers represent the cutting edge, anything beyond that which you discover is almost certainly going to be both new and relevant to what other researchers (i.e., your target audience) are interested in.
1
I totally agree with you. This is the only really great website for asking Math questions online, so if the people here doesn't know, nobody knows.
â JR S.
Aug 19 at 4:38
This is not an answer. The ordinary generating function for $zeta(2n+1)$ exists and it is pretty straightforward to find.
â Jack D'Aurizioâ¦
Aug 19 at 13:37
@JackD'Aurizio I beg to differ, I liked his answer. One answer is better than no answer, unless it's a rude answer.
â JR S.
Aug 19 at 17:55
@JRS.: then accept it. In my opinion it does not bring any actual content, but you are the asker, not me.
â Jack D'Aurizioâ¦
Aug 19 at 18:04
@JackD'Aurizio Suggestion accepted
â JR S.
Aug 19 at 18:07
add a comment |Â
up vote
1
down vote
Based on this answer, it wouldn't be much work to construct your generating function from what has been known at least as far back as 2013 (and probably much further).
My experience with research in classical mathematics is that not only are the interesting results known, they've been known for so long that the results aren't digitized. This can introduce somewhat of a challenge in proving novelty for a new idea.
I'm usually extremely open to the DIY approach, but starting out in mathematical research is one area where having a professional opinion would be valuable. Many (definitely not all -- be respectful) professors are willing to sit down and talk about research for hours, especially if it has anything whatsoever to do with their interests. Such conversations can be a good way to help gauge whether an idea is novel or not and hopefully to spark interest in other ideas as well.
That said, novelty isn't everything. If your research doesn't tie in to the current body of literature well, it will likely need to be more exciting than just another formula (not to disparage the result. I don't know the field well enough) to be publishable. Reading through current papers and using those as starting points is likely to be a fruitful area of exploration. Assuming recent papers represent the cutting edge, anything beyond that which you discover is almost certainly going to be both new and relevant to what other researchers (i.e., your target audience) are interested in.
1
I totally agree with you. This is the only really great website for asking Math questions online, so if the people here doesn't know, nobody knows.
â JR S.
Aug 19 at 4:38
This is not an answer. The ordinary generating function for $zeta(2n+1)$ exists and it is pretty straightforward to find.
â Jack D'Aurizioâ¦
Aug 19 at 13:37
@JackD'Aurizio I beg to differ, I liked his answer. One answer is better than no answer, unless it's a rude answer.
â JR S.
Aug 19 at 17:55
@JRS.: then accept it. In my opinion it does not bring any actual content, but you are the asker, not me.
â Jack D'Aurizioâ¦
Aug 19 at 18:04
@JackD'Aurizio Suggestion accepted
â JR S.
Aug 19 at 18:07
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Based on this answer, it wouldn't be much work to construct your generating function from what has been known at least as far back as 2013 (and probably much further).
My experience with research in classical mathematics is that not only are the interesting results known, they've been known for so long that the results aren't digitized. This can introduce somewhat of a challenge in proving novelty for a new idea.
I'm usually extremely open to the DIY approach, but starting out in mathematical research is one area where having a professional opinion would be valuable. Many (definitely not all -- be respectful) professors are willing to sit down and talk about research for hours, especially if it has anything whatsoever to do with their interests. Such conversations can be a good way to help gauge whether an idea is novel or not and hopefully to spark interest in other ideas as well.
That said, novelty isn't everything. If your research doesn't tie in to the current body of literature well, it will likely need to be more exciting than just another formula (not to disparage the result. I don't know the field well enough) to be publishable. Reading through current papers and using those as starting points is likely to be a fruitful area of exploration. Assuming recent papers represent the cutting edge, anything beyond that which you discover is almost certainly going to be both new and relevant to what other researchers (i.e., your target audience) are interested in.
Based on this answer, it wouldn't be much work to construct your generating function from what has been known at least as far back as 2013 (and probably much further).
My experience with research in classical mathematics is that not only are the interesting results known, they've been known for so long that the results aren't digitized. This can introduce somewhat of a challenge in proving novelty for a new idea.
I'm usually extremely open to the DIY approach, but starting out in mathematical research is one area where having a professional opinion would be valuable. Many (definitely not all -- be respectful) professors are willing to sit down and talk about research for hours, especially if it has anything whatsoever to do with their interests. Such conversations can be a good way to help gauge whether an idea is novel or not and hopefully to spark interest in other ideas as well.
That said, novelty isn't everything. If your research doesn't tie in to the current body of literature well, it will likely need to be more exciting than just another formula (not to disparage the result. I don't know the field well enough) to be publishable. Reading through current papers and using those as starting points is likely to be a fruitful area of exploration. Assuming recent papers represent the cutting edge, anything beyond that which you discover is almost certainly going to be both new and relevant to what other researchers (i.e., your target audience) are interested in.
answered Aug 19 at 4:35
Hans Musgrave
1,484111
1,484111
1
I totally agree with you. This is the only really great website for asking Math questions online, so if the people here doesn't know, nobody knows.
â JR S.
Aug 19 at 4:38
This is not an answer. The ordinary generating function for $zeta(2n+1)$ exists and it is pretty straightforward to find.
â Jack D'Aurizioâ¦
Aug 19 at 13:37
@JackD'Aurizio I beg to differ, I liked his answer. One answer is better than no answer, unless it's a rude answer.
â JR S.
Aug 19 at 17:55
@JRS.: then accept it. In my opinion it does not bring any actual content, but you are the asker, not me.
â Jack D'Aurizioâ¦
Aug 19 at 18:04
@JackD'Aurizio Suggestion accepted
â JR S.
Aug 19 at 18:07
add a comment |Â
1
I totally agree with you. This is the only really great website for asking Math questions online, so if the people here doesn't know, nobody knows.
â JR S.
Aug 19 at 4:38
This is not an answer. The ordinary generating function for $zeta(2n+1)$ exists and it is pretty straightforward to find.
â Jack D'Aurizioâ¦
Aug 19 at 13:37
@JackD'Aurizio I beg to differ, I liked his answer. One answer is better than no answer, unless it's a rude answer.
â JR S.
Aug 19 at 17:55
@JRS.: then accept it. In my opinion it does not bring any actual content, but you are the asker, not me.
â Jack D'Aurizioâ¦
Aug 19 at 18:04
@JackD'Aurizio Suggestion accepted
â JR S.
Aug 19 at 18:07
1
1
I totally agree with you. This is the only really great website for asking Math questions online, so if the people here doesn't know, nobody knows.
â JR S.
Aug 19 at 4:38
I totally agree with you. This is the only really great website for asking Math questions online, so if the people here doesn't know, nobody knows.
â JR S.
Aug 19 at 4:38
This is not an answer. The ordinary generating function for $zeta(2n+1)$ exists and it is pretty straightforward to find.
â Jack D'Aurizioâ¦
Aug 19 at 13:37
This is not an answer. The ordinary generating function for $zeta(2n+1)$ exists and it is pretty straightforward to find.
â Jack D'Aurizioâ¦
Aug 19 at 13:37
@JackD'Aurizio I beg to differ, I liked his answer. One answer is better than no answer, unless it's a rude answer.
â JR S.
Aug 19 at 17:55
@JackD'Aurizio I beg to differ, I liked his answer. One answer is better than no answer, unless it's a rude answer.
â JR S.
Aug 19 at 17:55
@JRS.: then accept it. In my opinion it does not bring any actual content, but you are the asker, not me.
â Jack D'Aurizioâ¦
Aug 19 at 18:04
@JRS.: then accept it. In my opinion it does not bring any actual content, but you are the asker, not me.
â Jack D'Aurizioâ¦
Aug 19 at 18:04
@JackD'Aurizio Suggestion accepted
â JR S.
Aug 19 at 18:07
@JackD'Aurizio Suggestion accepted
â JR S.
Aug 19 at 18:07
add a comment |Â
up vote
0
down vote
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
With $dsvertsz < 1$:
beginalign
sum_n = 1^inftyzetapars2n + 1z^n & =
sum_n = 1^infty
brackszetapars2n + 1 - 1parspm z^1/2^2n +
sum_n = 1^inftyz^n
\[5mm] & =
sum_n = 1^infty
brackszetapars2n + 1 - 1parspm z^1/2^2n +
z over 1 - z
endalign
The first sum can be evaluated with the
A & S $dsmathbfcolorblack6.3.15$ identity. Namely,
beginalign
&sum_n = 1^inftyzetapars2n + 1z^n
\[2mm] = &
bracks!!1 over 2parspmrootz - 1 over 2,picotpars!pibrackspmrootz!! -
1 over 1 - z + 1 - gamma - Psipars! 1 pm !rootz!!!
\[2mm] & phantombracksA+ z over 1 - z
\[5mm] = &
pm,1 over 2rootz mp 1 over 2,picotparspirootz - gamma - Psipars1 pm rootz
endalign
where $dsgamma$ is the Euler-Mascheroni Constant and $dsPsi$ is the Digamma Function. By adding the expressions for both signs $ds~pm~$and dividing by two:
$$
bbxsum_n = 1^inftyzetapars2n + 1z^n =
-gamma -
Psipars1 + rootz + Psipars1 - rootz over 2
$$
add a comment |Â
up vote
0
down vote
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
With $dsvertsz < 1$:
beginalign
sum_n = 1^inftyzetapars2n + 1z^n & =
sum_n = 1^infty
brackszetapars2n + 1 - 1parspm z^1/2^2n +
sum_n = 1^inftyz^n
\[5mm] & =
sum_n = 1^infty
brackszetapars2n + 1 - 1parspm z^1/2^2n +
z over 1 - z
endalign
The first sum can be evaluated with the
A & S $dsmathbfcolorblack6.3.15$ identity. Namely,
beginalign
&sum_n = 1^inftyzetapars2n + 1z^n
\[2mm] = &
bracks!!1 over 2parspmrootz - 1 over 2,picotpars!pibrackspmrootz!! -
1 over 1 - z + 1 - gamma - Psipars! 1 pm !rootz!!!
\[2mm] & phantombracksA+ z over 1 - z
\[5mm] = &
pm,1 over 2rootz mp 1 over 2,picotparspirootz - gamma - Psipars1 pm rootz
endalign
where $dsgamma$ is the Euler-Mascheroni Constant and $dsPsi$ is the Digamma Function. By adding the expressions for both signs $ds~pm~$and dividing by two:
$$
bbxsum_n = 1^inftyzetapars2n + 1z^n =
-gamma -
Psipars1 + rootz + Psipars1 - rootz over 2
$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
With $dsvertsz < 1$:
beginalign
sum_n = 1^inftyzetapars2n + 1z^n & =
sum_n = 1^infty
brackszetapars2n + 1 - 1parspm z^1/2^2n +
sum_n = 1^inftyz^n
\[5mm] & =
sum_n = 1^infty
brackszetapars2n + 1 - 1parspm z^1/2^2n +
z over 1 - z
endalign
The first sum can be evaluated with the
A & S $dsmathbfcolorblack6.3.15$ identity. Namely,
beginalign
&sum_n = 1^inftyzetapars2n + 1z^n
\[2mm] = &
bracks!!1 over 2parspmrootz - 1 over 2,picotpars!pibrackspmrootz!! -
1 over 1 - z + 1 - gamma - Psipars! 1 pm !rootz!!!
\[2mm] & phantombracksA+ z over 1 - z
\[5mm] = &
pm,1 over 2rootz mp 1 over 2,picotparspirootz - gamma - Psipars1 pm rootz
endalign
where $dsgamma$ is the Euler-Mascheroni Constant and $dsPsi$ is the Digamma Function. By adding the expressions for both signs $ds~pm~$and dividing by two:
$$
bbxsum_n = 1^inftyzetapars2n + 1z^n =
-gamma -
Psipars1 + rootz + Psipars1 - rootz over 2
$$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandroot[2],sqrt[#1],#2,,
newcommandtotald[3]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
With $dsvertsz < 1$:
beginalign
sum_n = 1^inftyzetapars2n + 1z^n & =
sum_n = 1^infty
brackszetapars2n + 1 - 1parspm z^1/2^2n +
sum_n = 1^inftyz^n
\[5mm] & =
sum_n = 1^infty
brackszetapars2n + 1 - 1parspm z^1/2^2n +
z over 1 - z
endalign
The first sum can be evaluated with the
A & S $dsmathbfcolorblack6.3.15$ identity. Namely,
beginalign
&sum_n = 1^inftyzetapars2n + 1z^n
\[2mm] = &
bracks!!1 over 2parspmrootz - 1 over 2,picotpars!pibrackspmrootz!! -
1 over 1 - z + 1 - gamma - Psipars! 1 pm !rootz!!!
\[2mm] & phantombracksA+ z over 1 - z
\[5mm] = &
pm,1 over 2rootz mp 1 over 2,picotparspirootz - gamma - Psipars1 pm rootz
endalign
where $dsgamma$ is the Euler-Mascheroni Constant and $dsPsi$ is the Digamma Function. By adding the expressions for both signs $ds~pm~$and dividing by two:
$$
bbxsum_n = 1^inftyzetapars2n + 1z^n =
-gamma -
Psipars1 + rootz + Psipars1 - rootz over 2
$$
answered Aug 24 at 21:08
Felix Marin
65.7k7105136
65.7k7105136
add a comment |Â
add a comment |Â
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