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What is $f(2s+1)$ when $f(s)=sum_n=0^infty frac(-1)^n(2n+1)^s=1-frac13^s+frac15^s-frac17^s+dots$?
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Is there an exact form of
$$f(s)=sum_n=0^infty frac(-1)^n(2n+1)^s=1-frac13^s+frac15^s-frac17^s+dots$$
when $s$ is odd?
Discussion
I have been exploring infinite series and will be spending my evening looking for patterns in this particular class. I invite the interested reader to join me and the not so interested to just move along. I will be updating this question with relevant facts as the evening unfolds.
There must (there must!) be some closed form in terms of $pi$ and $s$ when $s$ is odd and there will definitely be something we can say about how this relates to the generalized zeta function.
$f(1)=fracpi4$
$f(2)=$Catalan. [I will leave a remark about this below.]
$f(3)= frac164 (ζ(3, 1/4) - ζ(3, 3/4))=fracpi^332$
$f(4)= frac1256 (ζ(4, 1/4) - ζ(4, 3/4))$
$f(5)= frac11024(ζ(5, 1/4) - ζ(5, 3/4)) =frac5pi^51536 $
$f(6)= frac14096(ζ(6, 1/4) - ζ(6, 3/4)$
$f(7)=frac116384(ζ(7, 1/4) - ζ(7, 3/4))= frac61 À^7184320$
I thought about posting in Meta asking about this type of question. It's a "call to adventure" question: Come look at this with me if you so please. If you're not into it... downvote the question/let me know in the comments/move on to some other question that you do enjoy.
Update 1: It looks like $$f(s)= frac12^2s Bigg(zeta(s, frac14)-zeta(s, frac34) Bigg)$$
Update 2: A remark on Catalan's number and on $s$ even in general.
The wiki page claims it to be unknown whether this Catalan's constant is irrational or transcendental. Come on guys? What do we pay you for? Let me just state for the conjectural record that $sum_n=1^inftyfraca_nn^s$ for a periodic sequence of integers $a_n$ has just must be transcendental (it must!). I am very confident this is the case when $a_n$ has period of prime $p$ and for $s=1$. That appears in this. It's surprising to me that I would need these conditions. Note that for $f(2)$ the numerators of the series would be $1,0,-1,0 dots$ and that's not a prime period and also $s neq 1$ so we cannot use any of those tools to make any statements about Catalan's number but also... one cannot deny the conjecture isn't really too bold. Most numbers should be transcendental and this periodic numerators of these series must be a push in the transcendental direction.
sequences-and-series power-series
asked Aug 4 at 2:27
Mason
1,1271223
 |Â
show 2 more comments
up vote
3
down vote
favorite
Is there an exact form of
$$f(s)=sum_n=0^infty frac(-1)^n(2n+1)^s=1-frac13^s+frac15^s-frac17^s+dots$$
when $s$ is odd?
Discussion
I have been exploring infinite series and will be spending my evening looking for patterns in this particular class. I invite the interested reader to join me and the not so interested to just move along. I will be updating this question with relevant facts as the evening unfolds.
There must (there must!) be some closed form in terms of $pi$ and $s$ when $s$ is odd and there will definitely be something we can say about how this relates to the generalized zeta function.
$f(1)=fracpi4$
$f(2)=$Catalan. [I will leave a remark about this below.]
$f(3)= frac164 (ζ(3, 1/4) - ζ(3, 3/4))=fracpi^332$
$f(4)= frac1256 (ζ(4, 1/4) - ζ(4, 3/4))$
$f(5)= frac11024(ζ(5, 1/4) - ζ(5, 3/4)) =frac5pi^51536 $
$f(6)= frac14096(ζ(6, 1/4) - ζ(6, 3/4)$
$f(7)=frac116384(ζ(7, 1/4) - ζ(7, 3/4))= frac61 À^7184320$
I thought about posting in Meta asking about this type of question. It's a "call to adventure" question: Come look at this with me if you so please. If you're not into it... downvote the question/let me know in the comments/move on to some other question that you do enjoy.
Update 1: It looks like $$f(s)= frac12^2s Bigg(zeta(s, frac14)-zeta(s, frac34) Bigg)$$
Update 2: A remark on Catalan's number and on $s$ even in general.
The wiki page claims it to be unknown whether this Catalan's constant is irrational or transcendental. Come on guys? What do we pay you for? Let me just state for the conjectural record that $sum_n=1^inftyfraca_nn^s$ for a periodic sequence of integers $a_n$ has just must be transcendental (it must!). I am very confident this is the case when $a_n$ has period of prime $p$ and for $s=1$. That appears in this. It's surprising to me that I would need these conditions. Note that for $f(2)$ the numerators of the series would be $1,0,-1,0 dots$ and that's not a prime period and also $s neq 1$ so we cannot use any of those tools to make any statements about Catalan's number but also... one cannot deny the conjecture isn't really too bold. Most numbers should be transcendental and this periodic numerators of these series must be a push in the transcendental direction.
sequences-and-series power-series
asked Aug 4 at 2:27
Mason
1,1271223
chat.stackexchange.com/rooms/81142/mason-korb
– Mason
Aug 4 at 2:38
Does the sum start at $n=0$ or $n=1$?
– Simply Beautiful Art
Aug 4 at 2:46
Should be $n=0$. But this said $n=1$ originally. I edited. But apparently not the title. Thanks. I changed it.
– Mason
Aug 4 at 2:47
1
Note this is the Dirichlet beta function.
– Simply Beautiful Art
Aug 4 at 4:04
1
Almost-duplicate of math.stackexchange.com/questions/850442/…
– Jack D'Aurizio♦
2 days ago
 |Â
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Is there an exact form of
$$f(s)=sum_n=0^infty frac(-1)^n(2n+1)^s=1-frac13^s+frac15^s-frac17^s+dots$$
when $s$ is odd?
Discussion
I have been exploring infinite series and will be spending my evening looking for patterns in this particular class. I invite the interested reader to join me and the not so interested to just move along. I will be updating this question with relevant facts as the evening unfolds.
There must (there must!) be some closed form in terms of $pi$ and $s$ when $s$ is odd and there will definitely be something we can say about how this relates to the generalized zeta function.
$f(1)=fracpi4$
$f(2)=$Catalan. [I will leave a remark about this below.]
$f(3)= frac164 (ζ(3, 1/4) - ζ(3, 3/4))=fracpi^332$
$f(4)= frac1256 (ζ(4, 1/4) - ζ(4, 3/4))$
$f(5)= frac11024(ζ(5, 1/4) - ζ(5, 3/4)) =frac5pi^51536 $
$f(6)= frac14096(ζ(6, 1/4) - ζ(6, 3/4)$
$f(7)=frac116384(ζ(7, 1/4) - ζ(7, 3/4))= frac61 À^7184320$
I thought about posting in Meta asking about this type of question. It's a "call to adventure" question: Come look at this with me if you so please. If you're not into it... downvote the question/let me know in the comments/move on to some other question that you do enjoy.
Update 1: It looks like $$f(s)= frac12^2s Bigg(zeta(s, frac14)-zeta(s, frac34) Bigg)$$
Update 2: A remark on Catalan's number and on $s$ even in general.
The wiki page claims it to be unknown whether this Catalan's constant is irrational or transcendental. Come on guys? What do we pay you for? Let me just state for the conjectural record that $sum_n=1^inftyfraca_nn^s$ for a periodic sequence of integers $a_n$ has just must be transcendental (it must!). I am very confident this is the case when $a_n$ has period of prime $p$ and for $s=1$. That appears in this. It's surprising to me that I would need these conditions. Note that for $f(2)$ the numerators of the series would be $1,0,-1,0 dots$ and that's not a prime period and also $s neq 1$ so we cannot use any of those tools to make any statements about Catalan's number but also... one cannot deny the conjecture isn't really too bold. Most numbers should be transcendental and this periodic numerators of these series must be a push in the transcendental direction.
sequences-and-series power-series
asked Aug 4 at 2:27
Mason
1,1271223
Is there an exact form of
$$f(s)=sum_n=0^infty frac(-1)^n(2n+1)^s=1-frac13^s+frac15^s-frac17^s+dots$$
when $s$ is odd?
Discussion
I have been exploring infinite series and will be spending my evening looking for patterns in this particular class. I invite the interested reader to join me and the not so interested to just move along. I will be updating this question with relevant facts as the evening unfolds.
There must (there must!) be some closed form in terms of $pi$ and $s$ when $s$ is odd and there will definitely be something we can say about how this relates to the generalized zeta function.
$f(1)=fracpi4$
$f(2)=$Catalan. [I will leave a remark about this below.]
$f(3)= frac164 (ζ(3, 1/4) - ζ(3, 3/4))=fracpi^332$
$f(4)= frac1256 (ζ(4, 1/4) - ζ(4, 3/4))$
$f(5)= frac11024(ζ(5, 1/4) - ζ(5, 3/4)) =frac5pi^51536 $
$f(6)= frac14096(ζ(6, 1/4) - ζ(6, 3/4)$
$f(7)=frac116384(ζ(7, 1/4) - ζ(7, 3/4))= frac61 À^7184320$
I thought about posting in Meta asking about this type of question. It's a "call to adventure" question: Come look at this with me if you so please. If you're not into it... downvote the question/let me know in the comments/move on to some other question that you do enjoy.
Update 1: It looks like $$f(s)= frac12^2s Bigg(zeta(s, frac14)-zeta(s, frac34) Bigg)$$
Update 2: A remark on Catalan's number and on $s$ even in general.
The wiki page claims it to be unknown whether this Catalan's constant is irrational or transcendental. Come on guys? What do we pay you for? Let me just state for the conjectural record that $sum_n=1^inftyfraca_nn^s$ for a periodic sequence of integers $a_n$ has just must be transcendental (it must!). I am very confident this is the case when $a_n$ has period of prime $p$ and for $s=1$. That appears in this. It's surprising to me that I would need these conditions. Note that for $f(2)$ the numerators of the series would be $1,0,-1,0 dots$ and that's not a prime period and also $s neq 1$ so we cannot use any of those tools to make any statements about Catalan's number but also... one cannot deny the conjecture isn't really too bold. Most numbers should be transcendental and this periodic numerators of these series must be a push in the transcendental direction.
sequences-and-series power-series
asked Aug 4 at 2:27
Mason
1,1271223
edited Aug 4 at 3:20
asked Aug 4 at 2:27
Mason
1,1271223
asked Aug 4 at 2:27
Mason
1,1271223
asked Aug 4 at 2:27
Mason
1,1271223
1,1271223
chat.stackexchange.com/rooms/81142/mason-korb
– Mason
Aug 4 at 2:38
Does the sum start at $n=0$ or $n=1$?
– Simply Beautiful Art
Aug 4 at 2:46
Should be $n=0$. But this said $n=1$ originally. I edited. But apparently not the title. Thanks. I changed it.
– Mason
Aug 4 at 2:47
1
Note this is the Dirichlet beta function.
– Simply Beautiful Art
Aug 4 at 4:04
1
Almost-duplicate of math.stackexchange.com/questions/850442/…
– Jack D'Aurizio♦
2 days ago
 |Â
show 2 more comments
chat.stackexchange.com/rooms/81142/mason-korb
– Mason
Aug 4 at 2:38
Does the sum start at $n=0$ or $n=1$?
– Simply Beautiful Art
Aug 4 at 2:46
Should be $n=0$. But this said $n=1$ originally. I edited. But apparently not the title. Thanks. I changed it.
– Mason
Aug 4 at 2:47
1
Note this is the Dirichlet beta function.
– Simply Beautiful Art
Aug 4 at 4:04
1
Almost-duplicate of math.stackexchange.com/questions/850442/…
– Jack D'Aurizio♦
2 days ago
chat.stackexchange.com/rooms/81142/mason-korb
– Mason
Aug 4 at 2:38
chat.stackexchange.com/rooms/81142/mason-korb
– Mason
Aug 4 at 2:38
Does the sum start at $n=0$ or $n=1$?
– Simply Beautiful Art
Aug 4 at 2:46
Does the sum start at $n=0$ or $n=1$?
– Simply Beautiful Art
Aug 4 at 2:46
Should be $n=0$. But this said $n=1$ originally. I edited. But apparently not the title. Thanks. I changed it.
– Mason
Aug 4 at 2:47
Should be $n=0$. But this said $n=1$ originally. I edited. But apparently not the title. Thanks. I changed it.
– Mason
Aug 4 at 2:47
1
1
Note this is the Dirichlet beta function.
– Simply Beautiful Art
Aug 4 at 4:04
Note this is the Dirichlet beta function.
– Simply Beautiful Art
Aug 4 at 4:04
1
1
Almost-duplicate of math.stackexchange.com/questions/850442/…
– Jack D'Aurizio♦
2 days ago
Almost-duplicate of math.stackexchange.com/questions/850442/…
– Jack D'Aurizio♦
2 days ago
 |Â
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
As you've noticed, we have
$$4^sf(s)=sum_n=0^inftyleft(frac1(n+1/4)^s-frac1(n+3/4)^sright)$$
which, for odd $s$, can be written as
$$beginalign
&(s-1)!4^sf(s)\
&=lim_xto1/4fracd^s-1dx^s-1sum_n=0^inftyleft(frac1n+x+frac1x+1-nright)\
&=lim_xto1/4fracd^s-1dx^s-1sum_n=-infty^inftyfrac1n+x
\
&=lim_xto1/4fracd^s-1dx^s-1picot(pi x)
endalign$$
which gives exact values for $f(s)$ when $s$ is odd.
It also turns out this is the Dirichlet beta function with the special values of
$$f(s)=beta(s)=frac(-1)^kE_2k4^k+1(2k)!pi^2k+1$$
where $s=2k+1$ and $E_k$ are the Euler numbers.
answered Aug 4 at 3:02
Simply Beautiful Art
49k571169
What happens when $s$ is even? Are those values not exact also?
– Mark Viola
Aug 4 at 3:10
The operations $lim_xrightarrow 1/4$, $d^s-1/dx^s-1$ and $sum_n=0^infty$ have been interchanged. These interchanges must be done carefully at each step.
– i707107
Aug 4 at 3:15
Why does your development fail for even values of $s$?
– Mark Viola
Aug 4 at 3:26
@MarkViola one can write it as $$frac4^sf(s)(s-1)!=psi^(s-1)(1/4)-psi^(s-1)(3/4)$$using polygamma functions, which can be reduced to evaluating either polygamma function due to the reflection formula (or by following the process in my answer), reducing it to finding $sum_n=0^inftyfrac1(4n+1)^s$, which doesn't reduce any further AFAIK.
– Simply Beautiful Art
Aug 4 at 3:32
As per why we cannot just apply the result from my answer, note the derivatives of $frac1n+x$ alternates signs, while the derivatives of $frac1n-x$ do not. So the signs only line up nicely for odd $s$.
– Simply Beautiful Art
Aug 4 at 3:34
add a comment |Â
up vote
1
down vote
You can write this as
$$ -fraci2 left( it polylog left( s,i right) -it polylog left( s,-
i right) right)
$$
EDIT: It does seem that for odd $s$, $f(s)$ is a rational multiple of $pi^s$.
See OEIS sequence A053005 and A046976 and references there.
answered Aug 4 at 2:45
Robert Israel
303k22199438
The answer has been improved a bit, but there is still no explanation, and the only parts of the answer that provides some explanation are links. I shouldn't have to mention the help center, not to a user such as you. Until then, I'll keep my downvote.
– Simply Beautiful Art
Aug 4 at 4:18
en.wikipedia.org/wiki/Polylogarithm
– Mason
2 days ago
add a comment |Â
up vote
0
down vote
This is not an answer. Honestly these are just notes for me which you can ignore. Or not. If this is culturally bad form: down vote it+ leave me a comment. That's why this exists. I put together a table below that suggests we should see similar patterns for other series. This will hopefully be a follow up question once I have shopped it up to the standards of this site. There is quite a lot to study here before I can competently ask a question. The claim that is made above should have a nice generalization but I don't quite see through exactly what that is...
Let $a_1, a_2, dots $ be a sequence of integers with period $k$. We will write the first $k$ symbols as a vector: $veca =(a_1 ,a_2, dots, a_k)$ and write $$f(s,veca)= sum_n=1^inftyfraca_nn^s $$
Questions:
1) What is going on here?!?!
1) Is there somewhere I can read more about $f$?
2) Do we get neat results only when $a_n$ is a dirichlet character or should this happen in general?
3) For which $veca$ should I expect this series to converge to $alphapi^s$ for some algebraic $alpha$? Is there a way to find $alpha$ in terms of $veca$?
Discussion:
$$
beginarrayc
f(s,veca) & veca=(1,-1)
& veca=(1,0,-1,0)
& veca=(1,1,0,-1,-1,0)
& veca=(1,0,1,0,-1,0,-1,0) \
hline
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=1
& ln(2)
& fracpi4
& frac2 pi3sqrt3
& fracpi2sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=3
& frac34zeta(3)
& fracpi^332
& frac5 pi^381sqrt3
& frac3pi^364sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=5
& frac1516zeta(5)
& frac5 pi^51536
& frac17 pi^52916sqrt3
& frac19 pi^54096 sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=7
& frac6364zeta(7)
& frac61pi^7184320
& frac91 pi^7157464sqrt3
& frac307 pi^7655360sqrt2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
endarray$$
Column(1)
Column(2)
Column(3)
Column(4)
Projects:
1) Write $f(s,veca)$ as the sum of polylogs.
2) Learn more about Euler's Numbers and Dirichlet. They seem... relevant.
3) Take the advice from the comments above and apply the functional equation found here and here.
answered 2 days ago
Mason
1,1271223
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
As you've noticed, we have
$$4^sf(s)=sum_n=0^inftyleft(frac1(n+1/4)^s-frac1(n+3/4)^sright)$$
which, for odd $s$, can be written as
$$beginalign
&(s-1)!4^sf(s)\
&=lim_xto1/4fracd^s-1dx^s-1sum_n=0^inftyleft(frac1n+x+frac1x+1-nright)\
&=lim_xto1/4fracd^s-1dx^s-1sum_n=-infty^inftyfrac1n+x
\
&=lim_xto1/4fracd^s-1dx^s-1picot(pi x)
endalign$$
which gives exact values for $f(s)$ when $s$ is odd.
It also turns out this is the Dirichlet beta function with the special values of
$$f(s)=beta(s)=frac(-1)^kE_2k4^k+1(2k)!pi^2k+1$$
where $s=2k+1$ and $E_k$ are the Euler numbers.
answered Aug 4 at 3:02
Simply Beautiful Art
49k571169
What happens when $s$ is even? Are those values not exact also?
– Mark Viola
Aug 4 at 3:10
The operations $lim_xrightarrow 1/4$, $d^s-1/dx^s-1$ and $sum_n=0^infty$ have been interchanged. These interchanges must be done carefully at each step.
– i707107
Aug 4 at 3:15
Why does your development fail for even values of $s$?
– Mark Viola
Aug 4 at 3:26
@MarkViola one can write it as $$frac4^sf(s)(s-1)!=psi^(s-1)(1/4)-psi^(s-1)(3/4)$$using polygamma functions, which can be reduced to evaluating either polygamma function due to the reflection formula (or by following the process in my answer), reducing it to finding $sum_n=0^inftyfrac1(4n+1)^s$, which doesn't reduce any further AFAIK.
– Simply Beautiful Art
Aug 4 at 3:32
As per why we cannot just apply the result from my answer, note the derivatives of $frac1n+x$ alternates signs, while the derivatives of $frac1n-x$ do not. So the signs only line up nicely for odd $s$.
– Simply Beautiful Art
Aug 4 at 3:34
add a comment |Â
up vote
3
down vote
accepted
As you've noticed, we have
$$4^sf(s)=sum_n=0^inftyleft(frac1(n+1/4)^s-frac1(n+3/4)^sright)$$
which, for odd $s$, can be written as
$$beginalign
&(s-1)!4^sf(s)\
&=lim_xto1/4fracd^s-1dx^s-1sum_n=0^inftyleft(frac1n+x+frac1x+1-nright)\
&=lim_xto1/4fracd^s-1dx^s-1sum_n=-infty^inftyfrac1n+x
\
&=lim_xto1/4fracd^s-1dx^s-1picot(pi x)
endalign$$
which gives exact values for $f(s)$ when $s$ is odd.
It also turns out this is the Dirichlet beta function with the special values of
$$f(s)=beta(s)=frac(-1)^kE_2k4^k+1(2k)!pi^2k+1$$
where $s=2k+1$ and $E_k$ are the Euler numbers.
answered Aug 4 at 3:02
Simply Beautiful Art
49k571169
What happens when $s$ is even? Are those values not exact also?
– Mark Viola
Aug 4 at 3:10
The operations $lim_xrightarrow 1/4$, $d^s-1/dx^s-1$ and $sum_n=0^infty$ have been interchanged. These interchanges must be done carefully at each step.
– i707107
Aug 4 at 3:15
Why does your development fail for even values of $s$?
– Mark Viola
Aug 4 at 3:26
@MarkViola one can write it as $$frac4^sf(s)(s-1)!=psi^(s-1)(1/4)-psi^(s-1)(3/4)$$using polygamma functions, which can be reduced to evaluating either polygamma function due to the reflection formula (or by following the process in my answer), reducing it to finding $sum_n=0^inftyfrac1(4n+1)^s$, which doesn't reduce any further AFAIK.
– Simply Beautiful Art
Aug 4 at 3:32
As per why we cannot just apply the result from my answer, note the derivatives of $frac1n+x$ alternates signs, while the derivatives of $frac1n-x$ do not. So the signs only line up nicely for odd $s$.
– Simply Beautiful Art
Aug 4 at 3:34
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
As you've noticed, we have
$$4^sf(s)=sum_n=0^inftyleft(frac1(n+1/4)^s-frac1(n+3/4)^sright)$$
which, for odd $s$, can be written as
$$beginalign
&(s-1)!4^sf(s)\
&=lim_xto1/4fracd^s-1dx^s-1sum_n=0^inftyleft(frac1n+x+frac1x+1-nright)\
&=lim_xto1/4fracd^s-1dx^s-1sum_n=-infty^inftyfrac1n+x
\
&=lim_xto1/4fracd^s-1dx^s-1picot(pi x)
endalign$$
which gives exact values for $f(s)$ when $s$ is odd.
It also turns out this is the Dirichlet beta function with the special values of
$$f(s)=beta(s)=frac(-1)^kE_2k4^k+1(2k)!pi^2k+1$$
where $s=2k+1$ and $E_k$ are the Euler numbers.
answered Aug 4 at 3:02
Simply Beautiful Art
49k571169
As you've noticed, we have
$$4^sf(s)=sum_n=0^inftyleft(frac1(n+1/4)^s-frac1(n+3/4)^sright)$$
which, for odd $s$, can be written as
$$beginalign
&(s-1)!4^sf(s)\
&=lim_xto1/4fracd^s-1dx^s-1sum_n=0^inftyleft(frac1n+x+frac1x+1-nright)\
&=lim_xto1/4fracd^s-1dx^s-1sum_n=-infty^inftyfrac1n+x
\
&=lim_xto1/4fracd^s-1dx^s-1picot(pi x)
endalign$$
which gives exact values for $f(s)$ when $s$ is odd.
It also turns out this is the Dirichlet beta function with the special values of
$$f(s)=beta(s)=frac(-1)^kE_2k4^k+1(2k)!pi^2k+1$$
where $s=2k+1$ and $E_k$ are the Euler numbers.
answered Aug 4 at 3:02
Simply Beautiful Art
49k571169
edited 2 days ago
answered Aug 4 at 3:02
Simply Beautiful Art
49k571169
answered Aug 4 at 3:02
Simply Beautiful Art
49k571169
answered Aug 4 at 3:02
Simply Beautiful Art
49k571169
49k571169
What happens when $s$ is even? Are those values not exact also?
– Mark Viola
Aug 4 at 3:10
The operations $lim_xrightarrow 1/4$, $d^s-1/dx^s-1$ and $sum_n=0^infty$ have been interchanged. These interchanges must be done carefully at each step.
– i707107
Aug 4 at 3:15
Why does your development fail for even values of $s$?
– Mark Viola
Aug 4 at 3:26
@MarkViola one can write it as $$frac4^sf(s)(s-1)!=psi^(s-1)(1/4)-psi^(s-1)(3/4)$$using polygamma functions, which can be reduced to evaluating either polygamma function due to the reflection formula (or by following the process in my answer), reducing it to finding $sum_n=0^inftyfrac1(4n+1)^s$, which doesn't reduce any further AFAIK.
– Simply Beautiful Art
Aug 4 at 3:32
As per why we cannot just apply the result from my answer, note the derivatives of $frac1n+x$ alternates signs, while the derivatives of $frac1n-x$ do not. So the signs only line up nicely for odd $s$.
– Simply Beautiful Art
Aug 4 at 3:34
add a comment |Â
What happens when $s$ is even? Are those values not exact also?
– Mark Viola
Aug 4 at 3:10
The operations $lim_xrightarrow 1/4$, $d^s-1/dx^s-1$ and $sum_n=0^infty$ have been interchanged. These interchanges must be done carefully at each step.
– i707107
Aug 4 at 3:15
Why does your development fail for even values of $s$?
– Mark Viola
Aug 4 at 3:26
@MarkViola one can write it as $$frac4^sf(s)(s-1)!=psi^(s-1)(1/4)-psi^(s-1)(3/4)$$using polygamma functions, which can be reduced to evaluating either polygamma function due to the reflection formula (or by following the process in my answer), reducing it to finding $sum_n=0^inftyfrac1(4n+1)^s$, which doesn't reduce any further AFAIK.
– Simply Beautiful Art
Aug 4 at 3:32
As per why we cannot just apply the result from my answer, note the derivatives of $frac1n+x$ alternates signs, while the derivatives of $frac1n-x$ do not. So the signs only line up nicely for odd $s$.
– Simply Beautiful Art
Aug 4 at 3:34
What happens when $s$ is even? Are those values not exact also?
– Mark Viola
Aug 4 at 3:10
What happens when $s$ is even? Are those values not exact also?
– Mark Viola
Aug 4 at 3:10
The operations $lim_xrightarrow 1/4$, $d^s-1/dx^s-1$ and $sum_n=0^infty$ have been interchanged. These interchanges must be done carefully at each step.
– i707107
Aug 4 at 3:15
The operations $lim_xrightarrow 1/4$, $d^s-1/dx^s-1$ and $sum_n=0^infty$ have been interchanged. These interchanges must be done carefully at each step.
– i707107
Aug 4 at 3:15
Why does your development fail for even values of $s$?
– Mark Viola
Aug 4 at 3:26
Why does your development fail for even values of $s$?
– Mark Viola
Aug 4 at 3:26
@MarkViola one can write it as $$frac4^sf(s)(s-1)!=psi^(s-1)(1/4)-psi^(s-1)(3/4)$$using polygamma functions, which can be reduced to evaluating either polygamma function due to the reflection formula (or by following the process in my answer), reducing it to finding $sum_n=0^inftyfrac1(4n+1)^s$, which doesn't reduce any further AFAIK.
– Simply Beautiful Art
Aug 4 at 3:32
@MarkViola one can write it as $$frac4^sf(s)(s-1)!=psi^(s-1)(1/4)-psi^(s-1)(3/4)$$using polygamma functions, which can be reduced to evaluating either polygamma function due to the reflection formula (or by following the process in my answer), reducing it to finding $sum_n=0^inftyfrac1(4n+1)^s$, which doesn't reduce any further AFAIK.
– Simply Beautiful Art
Aug 4 at 3:32
As per why we cannot just apply the result from my answer, note the derivatives of $frac1n+x$ alternates signs, while the derivatives of $frac1n-x$ do not. So the signs only line up nicely for odd $s$.
– Simply Beautiful Art
Aug 4 at 3:34
As per why we cannot just apply the result from my answer, note the derivatives of $frac1n+x$ alternates signs, while the derivatives of $frac1n-x$ do not. So the signs only line up nicely for odd $s$.
– Simply Beautiful Art
Aug 4 at 3:34
add a comment |Â
up vote
1
down vote
You can write this as
$$ -fraci2 left( it polylog left( s,i right) -it polylog left( s,-
i right) right)
$$
EDIT: It does seem that for odd $s$, $f(s)$ is a rational multiple of $pi^s$.
See OEIS sequence A053005 and A046976 and references there.
answered Aug 4 at 2:45
Robert Israel
303k22199438
The answer has been improved a bit, but there is still no explanation, and the only parts of the answer that provides some explanation are links. I shouldn't have to mention the help center, not to a user such as you. Until then, I'll keep my downvote.
– Simply Beautiful Art
Aug 4 at 4:18
en.wikipedia.org/wiki/Polylogarithm
– Mason
2 days ago
add a comment |Â
up vote
1
down vote
You can write this as
$$ -fraci2 left( it polylog left( s,i right) -it polylog left( s,-
i right) right)
$$
EDIT: It does seem that for odd $s$, $f(s)$ is a rational multiple of $pi^s$.
See OEIS sequence A053005 and A046976 and references there.
answered Aug 4 at 2:45
Robert Israel
303k22199438
The answer has been improved a bit, but there is still no explanation, and the only parts of the answer that provides some explanation are links. I shouldn't have to mention the help center, not to a user such as you. Until then, I'll keep my downvote.
– Simply Beautiful Art
Aug 4 at 4:18
en.wikipedia.org/wiki/Polylogarithm
– Mason
2 days ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can write this as
$$ -fraci2 left( it polylog left( s,i right) -it polylog left( s,-
i right) right)
$$
EDIT: It does seem that for odd $s$, $f(s)$ is a rational multiple of $pi^s$.
See OEIS sequence A053005 and A046976 and references there.
answered Aug 4 at 2:45
Robert Israel
303k22199438
You can write this as
$$ -fraci2 left( it polylog left( s,i right) -it polylog left( s,-
i right) right)
$$
EDIT: It does seem that for odd $s$, $f(s)$ is a rational multiple of $pi^s$.
See OEIS sequence A053005 and A046976 and references there.
answered Aug 4 at 2:45
Robert Israel
303k22199438
edited Aug 4 at 2:57
answered Aug 4 at 2:45
Robert Israel
303k22199438
answered Aug 4 at 2:45
Robert Israel
303k22199438
answered Aug 4 at 2:45
Robert Israel
303k22199438
303k22199438
The answer has been improved a bit, but there is still no explanation, and the only parts of the answer that provides some explanation are links. I shouldn't have to mention the help center, not to a user such as you. Until then, I'll keep my downvote.
– Simply Beautiful Art
Aug 4 at 4:18
en.wikipedia.org/wiki/Polylogarithm
– Mason
2 days ago
add a comment |Â
The answer has been improved a bit, but there is still no explanation, and the only parts of the answer that provides some explanation are links. I shouldn't have to mention the help center, not to a user such as you. Until then, I'll keep my downvote.
– Simply Beautiful Art
Aug 4 at 4:18
en.wikipedia.org/wiki/Polylogarithm
– Mason
2 days ago
The answer has been improved a bit, but there is still no explanation, and the only parts of the answer that provides some explanation are links. I shouldn't have to mention the help center, not to a user such as you. Until then, I'll keep my downvote.
– Simply Beautiful Art
Aug 4 at 4:18
The answer has been improved a bit, but there is still no explanation, and the only parts of the answer that provides some explanation are links. I shouldn't have to mention the help center, not to a user such as you. Until then, I'll keep my downvote.
– Simply Beautiful Art
Aug 4 at 4:18
en.wikipedia.org/wiki/Polylogarithm
– Mason
2 days ago
en.wikipedia.org/wiki/Polylogarithm
– Mason
2 days ago
add a comment |Â
up vote
0
down vote
This is not an answer. Honestly these are just notes for me which you can ignore. Or not. If this is culturally bad form: down vote it+ leave me a comment. That's why this exists. I put together a table below that suggests we should see similar patterns for other series. This will hopefully be a follow up question once I have shopped it up to the standards of this site. There is quite a lot to study here before I can competently ask a question. The claim that is made above should have a nice generalization but I don't quite see through exactly what that is...
Let $a_1, a_2, dots $ be a sequence of integers with period $k$. We will write the first $k$ symbols as a vector: $veca =(a_1 ,a_2, dots, a_k)$ and write $$f(s,veca)= sum_n=1^inftyfraca_nn^s $$
Questions:
1) What is going on here?!?!
1) Is there somewhere I can read more about $f$?
2) Do we get neat results only when $a_n$ is a dirichlet character or should this happen in general?
3) For which $veca$ should I expect this series to converge to $alphapi^s$ for some algebraic $alpha$? Is there a way to find $alpha$ in terms of $veca$?
Discussion:
$$
beginarrayc
f(s,veca) & veca=(1,-1)
& veca=(1,0,-1,0)
& veca=(1,1,0,-1,-1,0)
& veca=(1,0,1,0,-1,0,-1,0) \
hline
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=1
& ln(2)
& fracpi4
& frac2 pi3sqrt3
& fracpi2sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=3
& frac34zeta(3)
& fracpi^332
& frac5 pi^381sqrt3
& frac3pi^364sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=5
& frac1516zeta(5)
& frac5 pi^51536
& frac17 pi^52916sqrt3
& frac19 pi^54096 sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=7
& frac6364zeta(7)
& frac61pi^7184320
& frac91 pi^7157464sqrt3
& frac307 pi^7655360sqrt2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
endarray$$
Column(1)
Column(2)
Column(3)
Column(4)
Projects:
1) Write $f(s,veca)$ as the sum of polylogs.
2) Learn more about Euler's Numbers and Dirichlet. They seem... relevant.
3) Take the advice from the comments above and apply the functional equation found here and here.
answered 2 days ago
Mason
1,1271223
add a comment |Â
up vote
0
down vote
This is not an answer. Honestly these are just notes for me which you can ignore. Or not. If this is culturally bad form: down vote it+ leave me a comment. That's why this exists. I put together a table below that suggests we should see similar patterns for other series. This will hopefully be a follow up question once I have shopped it up to the standards of this site. There is quite a lot to study here before I can competently ask a question. The claim that is made above should have a nice generalization but I don't quite see through exactly what that is...
Let $a_1, a_2, dots $ be a sequence of integers with period $k$. We will write the first $k$ symbols as a vector: $veca =(a_1 ,a_2, dots, a_k)$ and write $$f(s,veca)= sum_n=1^inftyfraca_nn^s $$
Questions:
1) What is going on here?!?!
1) Is there somewhere I can read more about $f$?
2) Do we get neat results only when $a_n$ is a dirichlet character or should this happen in general?
3) For which $veca$ should I expect this series to converge to $alphapi^s$ for some algebraic $alpha$? Is there a way to find $alpha$ in terms of $veca$?
Discussion:
$$
beginarrayc
f(s,veca) & veca=(1,-1)
& veca=(1,0,-1,0)
& veca=(1,1,0,-1,-1,0)
& veca=(1,0,1,0,-1,0,-1,0) \
hline
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=1
& ln(2)
& fracpi4
& frac2 pi3sqrt3
& fracpi2sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=3
& frac34zeta(3)
& fracpi^332
& frac5 pi^381sqrt3
& frac3pi^364sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=5
& frac1516zeta(5)
& frac5 pi^51536
& frac17 pi^52916sqrt3
& frac19 pi^54096 sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=7
& frac6364zeta(7)
& frac61pi^7184320
& frac91 pi^7157464sqrt3
& frac307 pi^7655360sqrt2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
endarray$$
Column(1)
Column(2)
Column(3)
Column(4)
Projects:
1) Write $f(s,veca)$ as the sum of polylogs.
2) Learn more about Euler's Numbers and Dirichlet. They seem... relevant.
3) Take the advice from the comments above and apply the functional equation found here and here.
answered 2 days ago
Mason
1,1271223
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is not an answer. Honestly these are just notes for me which you can ignore. Or not. If this is culturally bad form: down vote it+ leave me a comment. That's why this exists. I put together a table below that suggests we should see similar patterns for other series. This will hopefully be a follow up question once I have shopped it up to the standards of this site. There is quite a lot to study here before I can competently ask a question. The claim that is made above should have a nice generalization but I don't quite see through exactly what that is...
Let $a_1, a_2, dots $ be a sequence of integers with period $k$. We will write the first $k$ symbols as a vector: $veca =(a_1 ,a_2, dots, a_k)$ and write $$f(s,veca)= sum_n=1^inftyfraca_nn^s $$
Questions:
1) What is going on here?!?!
1) Is there somewhere I can read more about $f$?
2) Do we get neat results only when $a_n$ is a dirichlet character or should this happen in general?
3) For which $veca$ should I expect this series to converge to $alphapi^s$ for some algebraic $alpha$? Is there a way to find $alpha$ in terms of $veca$?
Discussion:
$$
beginarrayc
f(s,veca) & veca=(1,-1)
& veca=(1,0,-1,0)
& veca=(1,1,0,-1,-1,0)
& veca=(1,0,1,0,-1,0,-1,0) \
hline
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=1
& ln(2)
& fracpi4
& frac2 pi3sqrt3
& fracpi2sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=3
& frac34zeta(3)
& fracpi^332
& frac5 pi^381sqrt3
& frac3pi^364sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=5
& frac1516zeta(5)
& frac5 pi^51536
& frac17 pi^52916sqrt3
& frac19 pi^54096 sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=7
& frac6364zeta(7)
& frac61pi^7184320
& frac91 pi^7157464sqrt3
& frac307 pi^7655360sqrt2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
endarray$$
Column(1)
Column(2)
Column(3)
Column(4)
Projects:
1) Write $f(s,veca)$ as the sum of polylogs.
2) Learn more about Euler's Numbers and Dirichlet. They seem... relevant.
3) Take the advice from the comments above and apply the functional equation found here and here.
answered 2 days ago
Mason
1,1271223
This is not an answer. Honestly these are just notes for me which you can ignore. Or not. If this is culturally bad form: down vote it+ leave me a comment. That's why this exists. I put together a table below that suggests we should see similar patterns for other series. This will hopefully be a follow up question once I have shopped it up to the standards of this site. There is quite a lot to study here before I can competently ask a question. The claim that is made above should have a nice generalization but I don't quite see through exactly what that is...
Let $a_1, a_2, dots $ be a sequence of integers with period $k$. We will write the first $k$ symbols as a vector: $veca =(a_1 ,a_2, dots, a_k)$ and write $$f(s,veca)= sum_n=1^inftyfraca_nn^s $$
Questions:
1) What is going on here?!?!
1) Is there somewhere I can read more about $f$?
2) Do we get neat results only when $a_n$ is a dirichlet character or should this happen in general?
3) For which $veca$ should I expect this series to converge to $alphapi^s$ for some algebraic $alpha$? Is there a way to find $alpha$ in terms of $veca$?
Discussion:
$$
beginarrayc
f(s,veca) & veca=(1,-1)
& veca=(1,0,-1,0)
& veca=(1,1,0,-1,-1,0)
& veca=(1,0,1,0,-1,0,-1,0) \
hline
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=1
& ln(2)
& fracpi4
& frac2 pi3sqrt3
& fracpi2sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=3
& frac34zeta(3)
& fracpi^332
& frac5 pi^381sqrt3
& frac3pi^364sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=5
& frac1516zeta(5)
& frac5 pi^51536
& frac17 pi^52916sqrt3
& frac19 pi^54096 sqrt2 \
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s=7
& frac6364zeta(7)
& frac61pi^7184320
& frac91 pi^7157464sqrt3
& frac307 pi^7655360sqrt2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
endarray$$
Column(1)
Column(2)
Column(3)
Column(4)
Projects:
1) Write $f(s,veca)$ as the sum of polylogs.
2) Learn more about Euler's Numbers and Dirichlet. They seem... relevant.
3) Take the advice from the comments above and apply the functional equation found here and here.
answered 2 days ago
Mason
1,1271223
edited 23 hours ago
answered 2 days ago
Mason
1,1271223
answered 2 days ago
Mason
1,1271223
answered 2 days ago
Mason
1,1271223
1,1271223
add a comment |Â
add a comment |Â
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chat.stackexchange.com/rooms/81142/mason-korb
– Mason
Aug 4 at 2:38
Does the sum start at $n=0$ or $n=1$?
– Simply Beautiful Art
Aug 4 at 2:46
Should be $n=0$. But this said $n=1$ originally. I edited. But apparently not the title. Thanks. I changed it.
– Mason
Aug 4 at 2:47
1
Note this is the Dirichlet beta function.
– Simply Beautiful Art
Aug 4 at 4:04
1
Almost-duplicate of math.stackexchange.com/questions/850442/…
– Jack D'Aurizio♦
2 days ago