Mixing
Sum of Series Zeros Riemann function
$begingroup$
It is posible to prove the equality of the following series
$$-sum _{k=1}^{infty } left(frac{2^{2 k+1} pi ^{2 k}}{4 k^2 (k+2) (-1)^k (2 k)! zeta (2 k+1)}+frac{4}{2 left(rho _k-4right) left(rho _kright){}^2 zeta 'left(rho _kright)}right)+frac{1}{4}+frac{45}{4 pi ^4}=log (2 pi )$$ Numerically for a few Zeros it seem to work
calculus
asked Jan 18 at 20:57
capea perezcapea perez
335
$endgroup$
|
show 1 more comment
$begingroup$
It is posible to prove the equality of the following series
$$-sum _{k=1}^{infty } left(frac{2^{2 k+1} pi ^{2 k}}{4 k^2 (k+2) (-1)^k (2 k)! zeta (2 k+1)}+frac{4}{2 left(rho _k-4right) left(rho _kright){}^2 zeta 'left(rho _kright)}right)+frac{1}{4}+frac{45}{4 pi ^4}=log (2 pi )$$ Numerically for a few Zeros it seem to work
calculus
asked Jan 18 at 20:57
capea perezcapea perez
335
$endgroup$
1
$begingroup$
Where does it come from?
$endgroup$
– Seewoo Lee
Jan 18 at 22:33
2
$begingroup$
Is that you, Ramanujan?
$endgroup$
– lzralbu
Jan 18 at 23:09
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@Capea If you don't explain with which CAS and formulas you are playing we can't help you. In this form your questions do not fit the requirements of a math website.
$endgroup$
– reuns
Jan 20 at 2:19
$begingroup$
@reuns what mean CAS
$endgroup$
– capea perez
Jan 20 at 9:37
$begingroup$
mathematica, mapple, sage, pari...
$endgroup$
– reuns
Jan 22 at 13:50
|
show 1 more comment
$begingroup$
It is posible to prove the equality of the following series
$$-sum _{k=1}^{infty } left(frac{2^{2 k+1} pi ^{2 k}}{4 k^2 (k+2) (-1)^k (2 k)! zeta (2 k+1)}+frac{4}{2 left(rho _k-4right) left(rho _kright){}^2 zeta 'left(rho _kright)}right)+frac{1}{4}+frac{45}{4 pi ^4}=log (2 pi )$$ Numerically for a few Zeros it seem to work
calculus
asked Jan 18 at 20:57
capea perezcapea perez
335
$endgroup$
It is posible to prove the equality of the following series
$$-sum _{k=1}^{infty } left(frac{2^{2 k+1} pi ^{2 k}}{4 k^2 (k+2) (-1)^k (2 k)! zeta (2 k+1)}+frac{4}{2 left(rho _k-4right) left(rho _kright){}^2 zeta 'left(rho _kright)}right)+frac{1}{4}+frac{45}{4 pi ^4}=log (2 pi )$$ Numerically for a few Zeros it seem to work
calculus
calculus
asked Jan 18 at 20:57
capea perezcapea perez
335
asked Jan 18 at 20:57
capea perezcapea perez
335
asked Jan 18 at 20:57
capea perezcapea perez
335
asked Jan 18 at 20:57
capea perezcapea perez
335
asked Jan 18 at 20:57
capea perezcapea perez
335
335
1
$begingroup$
Where does it come from?
$endgroup$
– Seewoo Lee
Jan 18 at 22:33
2
$begingroup$
Is that you, Ramanujan?
$endgroup$
– lzralbu
Jan 18 at 23:09
$begingroup$
@Capea If you don't explain with which CAS and formulas you are playing we can't help you. In this form your questions do not fit the requirements of a math website.
$endgroup$
– reuns
Jan 20 at 2:19
$begingroup$
@reuns what mean CAS
$endgroup$
– capea perez
Jan 20 at 9:37
$begingroup$
mathematica, mapple, sage, pari...
$endgroup$
– reuns
Jan 22 at 13:50
|
show 1 more comment
1
$begingroup$
Where does it come from?
$endgroup$
– Seewoo Lee
Jan 18 at 22:33
2
$begingroup$
Is that you, Ramanujan?
$endgroup$
– lzralbu
Jan 18 at 23:09
$begingroup$
@Capea If you don't explain with which CAS and formulas you are playing we can't help you. In this form your questions do not fit the requirements of a math website.
$endgroup$
– reuns
Jan 20 at 2:19
$begingroup$
@reuns what mean CAS
$endgroup$
– capea perez
Jan 20 at 9:37
$begingroup$
mathematica, mapple, sage, pari...
$endgroup$
– reuns
Jan 22 at 13:50
1
1
$begingroup$
Where does it come from?
$endgroup$
– Seewoo Lee
Jan 18 at 22:33
$begingroup$
Where does it come from?
$endgroup$
– Seewoo Lee
Jan 18 at 22:33
2
2
$begingroup$
Is that you, Ramanujan?
$endgroup$
– lzralbu
Jan 18 at 23:09
$begingroup$
Is that you, Ramanujan?
$endgroup$
– lzralbu
Jan 18 at 23:09
$begingroup$
@Capea If you don't explain with which CAS and formulas you are playing we can't help you. In this form your questions do not fit the requirements of a math website.
$endgroup$
– reuns
Jan 20 at 2:19
$begingroup$
@Capea If you don't explain with which CAS and formulas you are playing we can't help you. In this form your questions do not fit the requirements of a math website.
$endgroup$
– reuns
Jan 20 at 2:19
$begingroup$
@reuns what mean CAS
$endgroup$
– capea perez
Jan 20 at 9:37
$begingroup$
@reuns what mean CAS
$endgroup$
– capea perez
Jan 20 at 9:37
$begingroup$
mathematica, mapple, sage, pari...
$endgroup$
– reuns
Jan 22 at 13:50
$begingroup$
mathematica, mapple, sage, pari...
$endgroup$
– reuns
Jan 22 at 13:50
|
show 1 more comment
1 Answer
1
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oldest
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$begingroup$
Iff all the zeros are simple then the residue theorem gives
$$0=lim_{sigmato infty} frac{1}{2ipi}int_{sigma-iinfty}^{sigma+iinfty} frac{1}{zeta(s)} frac1{(s-4)s^2}ds=frac{1}{2ipi}int_{5-iinfty}^{5+iinfty} frac{1}{zeta(s)} frac1{(s-4)s^2}ds$$ $$ =
Res_{s=0}(frac{1}{zeta(s)} frac1{(s-4)s^2})+Res_{s=4}(frac{1}{zeta(s)} frac1{(s-4)s^2})$$
$$+sum_rho Res_{s=rho}(frac{1}{zeta(s)} frac1{(s-4)s^2})+sum_{k=1}^infty Res_{s=-2k}(frac{1}{zeta(s)} frac1{(s-4)s^2})$$
$$ = frac{1}{zeta(0)(-4)}-frac{1}{zeta(0)(-4)^2}-frac{zeta'(0)}{zeta(0)^2(-4)}+frac{1}{zeta(4) 4^2}$$
$$ + sum_rho frac{1}{zeta'(rho)} frac1{(rho-4)rho^2}+sum_{k=1}^infty frac{1}{zeta'(-2k)} frac1{(-2k-4)(-2k)^2}$$
Proving the convergence of the series over the non-trivial zeros needs a lot of estimates (density of zeros, Phragmén–Lindelöf, Hadamard 3 circles...)
The functional equation is $zeta(s) = chi(s) zeta(1-s)$ with the closed-form for $chi'(s)$ we have $$zeta'(-2k) = chi'(-2k)zeta(2k+1)= (-1)^k frac {(2k)!} {2 (2pi)^{2k}} zeta (2k+1)$$
$$frac{zeta'(0)}{zeta(0)} = log (2pi)$$
answered Jan 19 at 2:09
reunsreuns
20.7k21148
$endgroup$
$begingroup$
See also en.wikipedia.org/wiki/Riesz_function for how to see the Riemann hypothesis in those kind of series over the trivial zeros
$endgroup$
– reuns
Jan 19 at 2:17
add a comment |
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$begingroup$
Iff all the zeros are simple then the residue theorem gives
$$0=lim_{sigmato infty} frac{1}{2ipi}int_{sigma-iinfty}^{sigma+iinfty} frac{1}{zeta(s)} frac1{(s-4)s^2}ds=frac{1}{2ipi}int_{5-iinfty}^{5+iinfty} frac{1}{zeta(s)} frac1{(s-4)s^2}ds$$ $$ =
Res_{s=0}(frac{1}{zeta(s)} frac1{(s-4)s^2})+Res_{s=4}(frac{1}{zeta(s)} frac1{(s-4)s^2})$$
$$+sum_rho Res_{s=rho}(frac{1}{zeta(s)} frac1{(s-4)s^2})+sum_{k=1}^infty Res_{s=-2k}(frac{1}{zeta(s)} frac1{(s-4)s^2})$$
$$ = frac{1}{zeta(0)(-4)}-frac{1}{zeta(0)(-4)^2}-frac{zeta'(0)}{zeta(0)^2(-4)}+frac{1}{zeta(4) 4^2}$$
$$ + sum_rho frac{1}{zeta'(rho)} frac1{(rho-4)rho^2}+sum_{k=1}^infty frac{1}{zeta'(-2k)} frac1{(-2k-4)(-2k)^2}$$
Proving the convergence of the series over the non-trivial zeros needs a lot of estimates (density of zeros, Phragmén–Lindelöf, Hadamard 3 circles...)
The functional equation is $zeta(s) = chi(s) zeta(1-s)$ with the closed-form for $chi'(s)$ we have $$zeta'(-2k) = chi'(-2k)zeta(2k+1)= (-1)^k frac {(2k)!} {2 (2pi)^{2k}} zeta (2k+1)$$
$$frac{zeta'(0)}{zeta(0)} = log (2pi)$$
answered Jan 19 at 2:09
reunsreuns
20.7k21148
$endgroup$
$begingroup$
See also en.wikipedia.org/wiki/Riesz_function for how to see the Riemann hypothesis in those kind of series over the trivial zeros
$endgroup$
– reuns
Jan 19 at 2:17
add a comment |
$begingroup$
Iff all the zeros are simple then the residue theorem gives
$$0=lim_{sigmato infty} frac{1}{2ipi}int_{sigma-iinfty}^{sigma+iinfty} frac{1}{zeta(s)} frac1{(s-4)s^2}ds=frac{1}{2ipi}int_{5-iinfty}^{5+iinfty} frac{1}{zeta(s)} frac1{(s-4)s^2}ds$$ $$ =
Res_{s=0}(frac{1}{zeta(s)} frac1{(s-4)s^2})+Res_{s=4}(frac{1}{zeta(s)} frac1{(s-4)s^2})$$
$$+sum_rho Res_{s=rho}(frac{1}{zeta(s)} frac1{(s-4)s^2})+sum_{k=1}^infty Res_{s=-2k}(frac{1}{zeta(s)} frac1{(s-4)s^2})$$
$$ = frac{1}{zeta(0)(-4)}-frac{1}{zeta(0)(-4)^2}-frac{zeta'(0)}{zeta(0)^2(-4)}+frac{1}{zeta(4) 4^2}$$
$$ + sum_rho frac{1}{zeta'(rho)} frac1{(rho-4)rho^2}+sum_{k=1}^infty frac{1}{zeta'(-2k)} frac1{(-2k-4)(-2k)^2}$$
Proving the convergence of the series over the non-trivial zeros needs a lot of estimates (density of zeros, Phragmén–Lindelöf, Hadamard 3 circles...)
The functional equation is $zeta(s) = chi(s) zeta(1-s)$ with the closed-form for $chi'(s)$ we have $$zeta'(-2k) = chi'(-2k)zeta(2k+1)= (-1)^k frac {(2k)!} {2 (2pi)^{2k}} zeta (2k+1)$$
$$frac{zeta'(0)}{zeta(0)} = log (2pi)$$
answered Jan 19 at 2:09
reunsreuns
20.7k21148
$endgroup$
$begingroup$
See also en.wikipedia.org/wiki/Riesz_function for how to see the Riemann hypothesis in those kind of series over the trivial zeros
$endgroup$
– reuns
Jan 19 at 2:17
add a comment |
$begingroup$
Iff all the zeros are simple then the residue theorem gives
$$0=lim_{sigmato infty} frac{1}{2ipi}int_{sigma-iinfty}^{sigma+iinfty} frac{1}{zeta(s)} frac1{(s-4)s^2}ds=frac{1}{2ipi}int_{5-iinfty}^{5+iinfty} frac{1}{zeta(s)} frac1{(s-4)s^2}ds$$ $$ =
Res_{s=0}(frac{1}{zeta(s)} frac1{(s-4)s^2})+Res_{s=4}(frac{1}{zeta(s)} frac1{(s-4)s^2})$$
$$+sum_rho Res_{s=rho}(frac{1}{zeta(s)} frac1{(s-4)s^2})+sum_{k=1}^infty Res_{s=-2k}(frac{1}{zeta(s)} frac1{(s-4)s^2})$$
$$ = frac{1}{zeta(0)(-4)}-frac{1}{zeta(0)(-4)^2}-frac{zeta'(0)}{zeta(0)^2(-4)}+frac{1}{zeta(4) 4^2}$$
$$ + sum_rho frac{1}{zeta'(rho)} frac1{(rho-4)rho^2}+sum_{k=1}^infty frac{1}{zeta'(-2k)} frac1{(-2k-4)(-2k)^2}$$
Proving the convergence of the series over the non-trivial zeros needs a lot of estimates (density of zeros, Phragmén–Lindelöf, Hadamard 3 circles...)
The functional equation is $zeta(s) = chi(s) zeta(1-s)$ with the closed-form for $chi'(s)$ we have $$zeta'(-2k) = chi'(-2k)zeta(2k+1)= (-1)^k frac {(2k)!} {2 (2pi)^{2k}} zeta (2k+1)$$
$$frac{zeta'(0)}{zeta(0)} = log (2pi)$$
answered Jan 19 at 2:09
reunsreuns
20.7k21148
$endgroup$
Iff all the zeros are simple then the residue theorem gives
$$0=lim_{sigmato infty} frac{1}{2ipi}int_{sigma-iinfty}^{sigma+iinfty} frac{1}{zeta(s)} frac1{(s-4)s^2}ds=frac{1}{2ipi}int_{5-iinfty}^{5+iinfty} frac{1}{zeta(s)} frac1{(s-4)s^2}ds$$ $$ =
Res_{s=0}(frac{1}{zeta(s)} frac1{(s-4)s^2})+Res_{s=4}(frac{1}{zeta(s)} frac1{(s-4)s^2})$$
$$+sum_rho Res_{s=rho}(frac{1}{zeta(s)} frac1{(s-4)s^2})+sum_{k=1}^infty Res_{s=-2k}(frac{1}{zeta(s)} frac1{(s-4)s^2})$$
$$ = frac{1}{zeta(0)(-4)}-frac{1}{zeta(0)(-4)^2}-frac{zeta'(0)}{zeta(0)^2(-4)}+frac{1}{zeta(4) 4^2}$$
$$ + sum_rho frac{1}{zeta'(rho)} frac1{(rho-4)rho^2}+sum_{k=1}^infty frac{1}{zeta'(-2k)} frac1{(-2k-4)(-2k)^2}$$
Proving the convergence of the series over the non-trivial zeros needs a lot of estimates (density of zeros, Phragmén–Lindelöf, Hadamard 3 circles...)
The functional equation is $zeta(s) = chi(s) zeta(1-s)$ with the closed-form for $chi'(s)$ we have $$zeta'(-2k) = chi'(-2k)zeta(2k+1)= (-1)^k frac {(2k)!} {2 (2pi)^{2k}} zeta (2k+1)$$
$$frac{zeta'(0)}{zeta(0)} = log (2pi)$$
answered Jan 19 at 2:09
reunsreuns
20.7k21148
edited Jan 19 at 3:43
answered Jan 19 at 2:09
reunsreuns
20.7k21148
answered Jan 19 at 2:09
reunsreuns
20.7k21148
answered Jan 19 at 2:09
reunsreuns
20.7k21148
20.7k21148
$begingroup$
See also en.wikipedia.org/wiki/Riesz_function for how to see the Riemann hypothesis in those kind of series over the trivial zeros
$endgroup$
– reuns
Jan 19 at 2:17
add a comment |
$begingroup$
See also en.wikipedia.org/wiki/Riesz_function for how to see the Riemann hypothesis in those kind of series over the trivial zeros
$endgroup$
– reuns
Jan 19 at 2:17
$begingroup$
See also en.wikipedia.org/wiki/Riesz_function for how to see the Riemann hypothesis in those kind of series over the trivial zeros
$endgroup$
– reuns
Jan 19 at 2:17
$begingroup$
See also en.wikipedia.org/wiki/Riesz_function for how to see the Riemann hypothesis in those kind of series over the trivial zeros
$endgroup$
– reuns
Jan 19 at 2:17
add a comment |
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1
$begingroup$
Where does it come from?
$endgroup$
– Seewoo Lee
Jan 18 at 22:33
2
$begingroup$
Is that you, Ramanujan?
$endgroup$
– lzralbu
Jan 18 at 23:09
$begingroup$
@Capea If you don't explain with which CAS and formulas you are playing we can't help you. In this form your questions do not fit the requirements of a math website.
$endgroup$
– reuns
Jan 20 at 2:19
$begingroup$
@reuns what mean CAS
$endgroup$
– capea perez
Jan 20 at 9:37
$begingroup$
mathematica, mapple, sage, pari...
$endgroup$
– reuns
Jan 22 at 13:50