Generalizing...
$begingroup$
I am trying to generalize the fact that, for $k>frac12$,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
To reach this I start off with the Fourier series
$$
t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt),qquad |t|leqpi\
sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt)=frac{t^2}4-frac{pi^2}{12}
$$
integrate both sides from $0$ to $x$:
$$
sum_{ngeq1}frac{(-1)^n}{n^2}int_0^xcos(nt)dt=frac{x^3}{12}-frac{pi^2x}{12}\
sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}
$$
plugging in $x=m^{-2k}$ for $mgeq1$, and $k>1/2$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12m^{6k}}-frac{pi^2}{12m^{2k}}$$
Then applying $sumlimits_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
With the same process, we have
$$begin{align}
sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^5}sin(n/m^{2k})&=frac{pi^2}{72}zeta(6k)-frac1{240}zeta(10k)-frac{7pi^4}{720}zeta(2k)\
end{align}$$
I am trying to find a general form in terms of $zeta$ values of
$$begin{align}
S_j(k)&=sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^j}sin(n/m^{2k}),qquad text{j is odd},quad j>0\
end{align}$$
And as you've seen, I've found up to $j=5$, but I would like to know if a general form exists. Any help is appreciated.
real-analysis sequences-and-series fourier-series riemann-zeta
asked Jan 1 at 10:30
clathratusclathratus
4,9641438
$endgroup$
|
show 6 more comments
$begingroup$
I am trying to generalize the fact that, for $k>frac12$,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
To reach this I start off with the Fourier series
$$
t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt),qquad |t|leqpi\
sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt)=frac{t^2}4-frac{pi^2}{12}
$$
integrate both sides from $0$ to $x$:
$$
sum_{ngeq1}frac{(-1)^n}{n^2}int_0^xcos(nt)dt=frac{x^3}{12}-frac{pi^2x}{12}\
sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}
$$
plugging in $x=m^{-2k}$ for $mgeq1$, and $k>1/2$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12m^{6k}}-frac{pi^2}{12m^{2k}}$$
Then applying $sumlimits_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
With the same process, we have
$$begin{align}
sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^5}sin(n/m^{2k})&=frac{pi^2}{72}zeta(6k)-frac1{240}zeta(10k)-frac{7pi^4}{720}zeta(2k)\
end{align}$$
I am trying to find a general form in terms of $zeta$ values of
$$begin{align}
S_j(k)&=sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^j}sin(n/m^{2k}),qquad text{j is odd},quad j>0\
end{align}$$
And as you've seen, I've found up to $j=5$, but I would like to know if a general form exists. Any help is appreciated.
real-analysis sequences-and-series fourier-series riemann-zeta
asked Jan 1 at 10:30
clathratusclathratus
4,9641438
$endgroup$
$begingroup$
Sorry, what does $zeta$ stand for?
$endgroup$
– Omojola Micheal
Jan 1 at 11:08
$begingroup$
@Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
$endgroup$
– mrtaurho
Jan 1 at 11:56
1
$begingroup$
@Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
$endgroup$
– mrtaurho
Jan 1 at 12:25
1
$begingroup$
@clathratus: Thanks for that!
$endgroup$
– Omojola Micheal
Jan 1 at 19:48
1
$begingroup$
How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
$endgroup$
– Shashi
Jan 1 at 22:43
|
show 6 more comments
$begingroup$
I am trying to generalize the fact that, for $k>frac12$,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
To reach this I start off with the Fourier series
$$
t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt),qquad |t|leqpi\
sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt)=frac{t^2}4-frac{pi^2}{12}
$$
integrate both sides from $0$ to $x$:
$$
sum_{ngeq1}frac{(-1)^n}{n^2}int_0^xcos(nt)dt=frac{x^3}{12}-frac{pi^2x}{12}\
sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}
$$
plugging in $x=m^{-2k}$ for $mgeq1$, and $k>1/2$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12m^{6k}}-frac{pi^2}{12m^{2k}}$$
Then applying $sumlimits_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
With the same process, we have
$$begin{align}
sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^5}sin(n/m^{2k})&=frac{pi^2}{72}zeta(6k)-frac1{240}zeta(10k)-frac{7pi^4}{720}zeta(2k)\
end{align}$$
I am trying to find a general form in terms of $zeta$ values of
$$begin{align}
S_j(k)&=sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^j}sin(n/m^{2k}),qquad text{j is odd},quad j>0\
end{align}$$
And as you've seen, I've found up to $j=5$, but I would like to know if a general form exists. Any help is appreciated.
real-analysis sequences-and-series fourier-series riemann-zeta
asked Jan 1 at 10:30
clathratusclathratus
4,9641438
$endgroup$
I am trying to generalize the fact that, for $k>frac12$,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
To reach this I start off with the Fourier series
$$
t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt),qquad |t|leqpi\
sum_{ngeq1}frac{(-1)^n}{n^2}cos(nt)=frac{t^2}4-frac{pi^2}{12}
$$
integrate both sides from $0$ to $x$:
$$
sum_{ngeq1}frac{(-1)^n}{n^2}int_0^xcos(nt)dt=frac{x^3}{12}-frac{pi^2x}{12}\
sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}
$$
plugging in $x=m^{-2k}$ for $mgeq1$, and $k>1/2$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12m^{6k}}-frac{pi^2}{12m^{2k}}$$
Then applying $sumlimits_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sin(n/m^{2k})=frac1{12}zeta(6k)-frac{pi^2}{12}zeta(2k)$$
With the same process, we have
$$begin{align}
sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^5}sin(n/m^{2k})&=frac{pi^2}{72}zeta(6k)-frac1{240}zeta(10k)-frac{7pi^4}{720}zeta(2k)\
end{align}$$
I am trying to find a general form in terms of $zeta$ values of
$$begin{align}
S_j(k)&=sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^j}sin(n/m^{2k}),qquad text{j is odd},quad j>0\
end{align}$$
And as you've seen, I've found up to $j=5$, but I would like to know if a general form exists. Any help is appreciated.
real-analysis sequences-and-series fourier-series riemann-zeta
real-analysis sequences-and-series fourier-series riemann-zeta
asked Jan 1 at 10:30
clathratusclathratus
4,9641438
asked Jan 1 at 10:30
clathratusclathratus
4,9641438
edited Jan 2 at 2:06
clathratus
asked Jan 1 at 10:30
clathratusclathratus
4,9641438
asked Jan 1 at 10:30
clathratusclathratus
4,9641438
asked Jan 1 at 10:30
clathratusclathratus
4,9641438
4,9641438
$begingroup$
Sorry, what does $zeta$ stand for?
$endgroup$
– Omojola Micheal
Jan 1 at 11:08
$begingroup$
@Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
$endgroup$
– mrtaurho
Jan 1 at 11:56
1
$begingroup$
@Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
$endgroup$
– mrtaurho
Jan 1 at 12:25
1
$begingroup$
@clathratus: Thanks for that!
$endgroup$
– Omojola Micheal
Jan 1 at 19:48
1
$begingroup$
How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
$endgroup$
– Shashi
Jan 1 at 22:43
|
show 6 more comments
$begingroup$
Sorry, what does $zeta$ stand for?
$endgroup$
– Omojola Micheal
Jan 1 at 11:08
$begingroup$
@Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
$endgroup$
– mrtaurho
Jan 1 at 11:56
1
$begingroup$
@Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
$endgroup$
– mrtaurho
Jan 1 at 12:25
1
$begingroup$
@clathratus: Thanks for that!
$endgroup$
– Omojola Micheal
Jan 1 at 19:48
1
$begingroup$
How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
$endgroup$
– Shashi
Jan 1 at 22:43
$begingroup$
Sorry, what does $zeta$ stand for?
$endgroup$
– Omojola Micheal
Jan 1 at 11:08
$begingroup$
Sorry, what does $zeta$ stand for?
$endgroup$
– Omojola Micheal
Jan 1 at 11:08
$begingroup$
@Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
$endgroup$
– mrtaurho
Jan 1 at 11:56
$begingroup$
@Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
$endgroup$
– mrtaurho
Jan 1 at 11:56
1
1
$begingroup$
@Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
$endgroup$
– mrtaurho
Jan 1 at 12:25
$begingroup$
@Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
$endgroup$
– mrtaurho
Jan 1 at 12:25
1
1
$begingroup$
@clathratus: Thanks for that!
$endgroup$
– Omojola Micheal
Jan 1 at 19:48
$begingroup$
@clathratus: Thanks for that!
$endgroup$
– Omojola Micheal
Jan 1 at 19:48
1
1
$begingroup$
How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
$endgroup$
– Shashi
Jan 1 at 22:43
$begingroup$
How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
$endgroup$
– Shashi
Jan 1 at 22:43
|
show 6 more comments
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$begingroup$
Sorry, what does $zeta$ stand for?
$endgroup$
– Omojola Micheal
Jan 1 at 11:08
$begingroup$
@Mike I assume that this should be the Riemann Zeta Function $zeta(s)$ hence you can see how the $frac1{4m^{4k}}$ term of the RHS within the line $$sum_{ngeq1}frac{(-1)^n}{n^2}cos(n/m^{2k})=frac1{4m^{4k}}-frac{pi^2}{12}$$ becomes $frac14zeta(4k)$ after summing over all integer $mgeq 1$ which equals the defintion of the Riemann Zeta Function for $operatorname{Re}(4k)>1$.
$endgroup$
– mrtaurho
Jan 1 at 11:56
1
$begingroup$
@Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times?
$endgroup$
– mrtaurho
Jan 1 at 12:25
1
$begingroup$
@clathratus: Thanks for that!
$endgroup$
– Omojola Micheal
Jan 1 at 19:48
1
$begingroup$
How is $$sum_{mgeq 1}left( frac{1}{4m^{2k}} -frac{pi^2}{12}right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-infty$ if I'm not mistaken.
$endgroup$
– Shashi
Jan 1 at 22:43