Need help showing Riemann's Functional equation for negative numbers and complex numbers
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Riemann's Functional equation: $zeta(-z)$=${-2*z!over(2pi)^{z+1}}$$sin({pi zover2})$$zeta(z+1)$
This formulas expresses $zeta(-z)$ in terms of $zeta(z+1)$
Note: I read that the author said, suppose $z=x+iy$ is a complex number in the right half plane, meaning $x ge 0$ then $-z$ is in the left half plane. Since $z+1$ is in the right half plane, Riemann's functional equation allows us to indirectly define $zeta(-z)$. For example, if we take $z=2+3i$, the point $z+1=3+3i$ is located in the half plane where $x ge 1$ but we already know how to compute $zeta(z+1)$ using the zeta series and a computer and it will show that $zeta(3+3i)approx 0.94+0.008i$
My first question is how do I compute $zeta(z+1)$ using the zeta series or a computer for fact checking?
My 2nd question is how do I compute Riemann's functional equation when $z$ is equal to a negative number because I can solve this when $z ge 0$
For example, when $z=2$ in the Riemann functional equation we have:
$zeta(-2)$=${-2*2!over(2pi)^3}$$sin({2pi over2})$$zeta(3)$
Here we have $sinpi=0$ which indicates that $zeta(-2)=0$
What if we had $z=-2$, how do I solve this?
Can someone show me in detail when $z=-2$ because the equation would now look like:
$zeta(2)$=${(-2)(-2!)over(2pi)}$$sin({-2pi over2})$$zeta(-1)$
trigonometry factorial riemann-zeta zeta-functions
asked Jan 22 at 22:31
DDPDDP
166
$endgroup$
add a comment |
$begingroup$
Riemann's Functional equation: $zeta(-z)$=${-2*z!over(2pi)^{z+1}}$$sin({pi zover2})$$zeta(z+1)$
This formulas expresses $zeta(-z)$ in terms of $zeta(z+1)$
Note: I read that the author said, suppose $z=x+iy$ is a complex number in the right half plane, meaning $x ge 0$ then $-z$ is in the left half plane. Since $z+1$ is in the right half plane, Riemann's functional equation allows us to indirectly define $zeta(-z)$. For example, if we take $z=2+3i$, the point $z+1=3+3i$ is located in the half plane where $x ge 1$ but we already know how to compute $zeta(z+1)$ using the zeta series and a computer and it will show that $zeta(3+3i)approx 0.94+0.008i$
My first question is how do I compute $zeta(z+1)$ using the zeta series or a computer for fact checking?
My 2nd question is how do I compute Riemann's functional equation when $z$ is equal to a negative number because I can solve this when $z ge 0$
For example, when $z=2$ in the Riemann functional equation we have:
$zeta(-2)$=${-2*2!over(2pi)^3}$$sin({2pi over2})$$zeta(3)$
Here we have $sinpi=0$ which indicates that $zeta(-2)=0$
What if we had $z=-2$, how do I solve this?
Can someone show me in detail when $z=-2$ because the equation would now look like:
$zeta(2)$=${(-2)(-2!)over(2pi)}$$sin({-2pi over2})$$zeta(-1)$
trigonometry factorial riemann-zeta zeta-functions
asked Jan 22 at 22:31
DDPDDP
166
$endgroup$
1
$begingroup$
The functional equation is $zeta(s) = chi(s) zeta(1-s)$ where $chi(s) = pi^{s-1} 2^s Gamma(1-s) sin(pi s/2)$. This is an equality of meromorphic functions. From the zeros-poles locations of $Gamma(1-s)$ you know those of $chi(s)$. It is always true that $zeta(s) = lim_{z to s}chi(z) zeta(1-z)$ and $zeta(1-s) = lim_{z to s}frac{1}{chi(z)} zeta(z)$ and depending if $chi$ has no pole or zero at $s$ then you can remove the $lim_{z to s}$ and use direct evaluation $zeta(s) = chi(s) zeta(1-s),zeta(1-s) = frac{1}{chi(s)} zeta(s)$
$endgroup$
– reuns
Jan 23 at 0:10
$begingroup$
$chi(s)$ has a zero at $s=-2$ so $zeta(-2) = chi(-2) zeta(3) = 0$. And $frac{1}{chi(s)}$ has a pole at $s=-2$ so $zeta(3) = lim_{s to -2}zeta(1-s)=lim_{s to -2} frac{zeta(s)}{chi(s)} $, as the pole is simple the latter $=frac{zeta'(-2)}{chi'(-2)} $ whose only the denominator has a closed-form
$endgroup$
– reuns
Jan 23 at 0:16
add a comment |
$begingroup$
Riemann's Functional equation: $zeta(-z)$=${-2*z!over(2pi)^{z+1}}$$sin({pi zover2})$$zeta(z+1)$
This formulas expresses $zeta(-z)$ in terms of $zeta(z+1)$
Note: I read that the author said, suppose $z=x+iy$ is a complex number in the right half plane, meaning $x ge 0$ then $-z$ is in the left half plane. Since $z+1$ is in the right half plane, Riemann's functional equation allows us to indirectly define $zeta(-z)$. For example, if we take $z=2+3i$, the point $z+1=3+3i$ is located in the half plane where $x ge 1$ but we already know how to compute $zeta(z+1)$ using the zeta series and a computer and it will show that $zeta(3+3i)approx 0.94+0.008i$
My first question is how do I compute $zeta(z+1)$ using the zeta series or a computer for fact checking?
My 2nd question is how do I compute Riemann's functional equation when $z$ is equal to a negative number because I can solve this when $z ge 0$
For example, when $z=2$ in the Riemann functional equation we have:
$zeta(-2)$=${-2*2!over(2pi)^3}$$sin({2pi over2})$$zeta(3)$
Here we have $sinpi=0$ which indicates that $zeta(-2)=0$
What if we had $z=-2$, how do I solve this?
Can someone show me in detail when $z=-2$ because the equation would now look like:
$zeta(2)$=${(-2)(-2!)over(2pi)}$$sin({-2pi over2})$$zeta(-1)$
trigonometry factorial riemann-zeta zeta-functions
asked Jan 22 at 22:31
DDPDDP
166
$endgroup$
Riemann's Functional equation: $zeta(-z)$=${-2*z!over(2pi)^{z+1}}$$sin({pi zover2})$$zeta(z+1)$
This formulas expresses $zeta(-z)$ in terms of $zeta(z+1)$
Note: I read that the author said, suppose $z=x+iy$ is a complex number in the right half plane, meaning $x ge 0$ then $-z$ is in the left half plane. Since $z+1$ is in the right half plane, Riemann's functional equation allows us to indirectly define $zeta(-z)$. For example, if we take $z=2+3i$, the point $z+1=3+3i$ is located in the half plane where $x ge 1$ but we already know how to compute $zeta(z+1)$ using the zeta series and a computer and it will show that $zeta(3+3i)approx 0.94+0.008i$
My first question is how do I compute $zeta(z+1)$ using the zeta series or a computer for fact checking?
My 2nd question is how do I compute Riemann's functional equation when $z$ is equal to a negative number because I can solve this when $z ge 0$
For example, when $z=2$ in the Riemann functional equation we have:
$zeta(-2)$=${-2*2!over(2pi)^3}$$sin({2pi over2})$$zeta(3)$
Here we have $sinpi=0$ which indicates that $zeta(-2)=0$
What if we had $z=-2$, how do I solve this?
Can someone show me in detail when $z=-2$ because the equation would now look like:
$zeta(2)$=${(-2)(-2!)over(2pi)}$$sin({-2pi over2})$$zeta(-1)$
trigonometry factorial riemann-zeta zeta-functions
trigonometry factorial riemann-zeta zeta-functions
asked Jan 22 at 22:31
DDPDDP
166
asked Jan 22 at 22:31
DDPDDP
166
edited Jan 23 at 2:53
DDP
asked Jan 22 at 22:31
DDPDDP
166
asked Jan 22 at 22:31
DDPDDP
166
asked Jan 22 at 22:31
DDPDDP
166
166
1
$begingroup$
The functional equation is $zeta(s) = chi(s) zeta(1-s)$ where $chi(s) = pi^{s-1} 2^s Gamma(1-s) sin(pi s/2)$. This is an equality of meromorphic functions. From the zeros-poles locations of $Gamma(1-s)$ you know those of $chi(s)$. It is always true that $zeta(s) = lim_{z to s}chi(z) zeta(1-z)$ and $zeta(1-s) = lim_{z to s}frac{1}{chi(z)} zeta(z)$ and depending if $chi$ has no pole or zero at $s$ then you can remove the $lim_{z to s}$ and use direct evaluation $zeta(s) = chi(s) zeta(1-s),zeta(1-s) = frac{1}{chi(s)} zeta(s)$
$endgroup$
– reuns
Jan 23 at 0:10
$begingroup$
$chi(s)$ has a zero at $s=-2$ so $zeta(-2) = chi(-2) zeta(3) = 0$. And $frac{1}{chi(s)}$ has a pole at $s=-2$ so $zeta(3) = lim_{s to -2}zeta(1-s)=lim_{s to -2} frac{zeta(s)}{chi(s)} $, as the pole is simple the latter $=frac{zeta'(-2)}{chi'(-2)} $ whose only the denominator has a closed-form
$endgroup$
– reuns
Jan 23 at 0:16
add a comment |
1
$begingroup$
The functional equation is $zeta(s) = chi(s) zeta(1-s)$ where $chi(s) = pi^{s-1} 2^s Gamma(1-s) sin(pi s/2)$. This is an equality of meromorphic functions. From the zeros-poles locations of $Gamma(1-s)$ you know those of $chi(s)$. It is always true that $zeta(s) = lim_{z to s}chi(z) zeta(1-z)$ and $zeta(1-s) = lim_{z to s}frac{1}{chi(z)} zeta(z)$ and depending if $chi$ has no pole or zero at $s$ then you can remove the $lim_{z to s}$ and use direct evaluation $zeta(s) = chi(s) zeta(1-s),zeta(1-s) = frac{1}{chi(s)} zeta(s)$
$endgroup$
– reuns
Jan 23 at 0:10
$begingroup$
$chi(s)$ has a zero at $s=-2$ so $zeta(-2) = chi(-2) zeta(3) = 0$. And $frac{1}{chi(s)}$ has a pole at $s=-2$ so $zeta(3) = lim_{s to -2}zeta(1-s)=lim_{s to -2} frac{zeta(s)}{chi(s)} $, as the pole is simple the latter $=frac{zeta'(-2)}{chi'(-2)} $ whose only the denominator has a closed-form
$endgroup$
– reuns
Jan 23 at 0:16
1
1
$begingroup$
The functional equation is $zeta(s) = chi(s) zeta(1-s)$ where $chi(s) = pi^{s-1} 2^s Gamma(1-s) sin(pi s/2)$. This is an equality of meromorphic functions. From the zeros-poles locations of $Gamma(1-s)$ you know those of $chi(s)$. It is always true that $zeta(s) = lim_{z to s}chi(z) zeta(1-z)$ and $zeta(1-s) = lim_{z to s}frac{1}{chi(z)} zeta(z)$ and depending if $chi$ has no pole or zero at $s$ then you can remove the $lim_{z to s}$ and use direct evaluation $zeta(s) = chi(s) zeta(1-s),zeta(1-s) = frac{1}{chi(s)} zeta(s)$
$endgroup$
– reuns
Jan 23 at 0:10
$begingroup$
The functional equation is $zeta(s) = chi(s) zeta(1-s)$ where $chi(s) = pi^{s-1} 2^s Gamma(1-s) sin(pi s/2)$. This is an equality of meromorphic functions. From the zeros-poles locations of $Gamma(1-s)$ you know those of $chi(s)$. It is always true that $zeta(s) = lim_{z to s}chi(z) zeta(1-z)$ and $zeta(1-s) = lim_{z to s}frac{1}{chi(z)} zeta(z)$ and depending if $chi$ has no pole or zero at $s$ then you can remove the $lim_{z to s}$ and use direct evaluation $zeta(s) = chi(s) zeta(1-s),zeta(1-s) = frac{1}{chi(s)} zeta(s)$
$endgroup$
– reuns
Jan 23 at 0:10
$begingroup$
$chi(s)$ has a zero at $s=-2$ so $zeta(-2) = chi(-2) zeta(3) = 0$. And $frac{1}{chi(s)}$ has a pole at $s=-2$ so $zeta(3) = lim_{s to -2}zeta(1-s)=lim_{s to -2} frac{zeta(s)}{chi(s)} $, as the pole is simple the latter $=frac{zeta'(-2)}{chi'(-2)} $ whose only the denominator has a closed-form
$endgroup$
– reuns
Jan 23 at 0:16
$begingroup$
$chi(s)$ has a zero at $s=-2$ so $zeta(-2) = chi(-2) zeta(3) = 0$. And $frac{1}{chi(s)}$ has a pole at $s=-2$ so $zeta(3) = lim_{s to -2}zeta(1-s)=lim_{s to -2} frac{zeta(s)}{chi(s)} $, as the pole is simple the latter $=frac{zeta'(-2)}{chi'(-2)} $ whose only the denominator has a closed-form
$endgroup$
– reuns
Jan 23 at 0:16
add a comment |
1 Answer
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$begingroup$
Your problem boils down to finding $zeta(2k)$. In fact, it can be shown that $$zeta(2k)=(-1)^{k+1}frac{B_{2k} 2^{2k-1}pi^{2k}}{(2k)!}$$
Here are some proofs. No known values of the zeta function exist except for the even positive integers and negative integers.
EDIT: $zeta(z)$ is equal to $0$ at the notorious non-trivial zeroes as well as negative even integers.
answered Jan 22 at 22:38
aledenaleden
2,409511
$endgroup$
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What about for odd numbers? I looked at the proof section also
$endgroup$
– DDP
Jan 22 at 22:42
$begingroup$
@DDP no known values have been found for odd positive integers.
$endgroup$
– aleden
Jan 22 at 22:43
$begingroup$
okay. I was just fact checking for all real numbers(negative numbers included) in the functional equation for $z$
$endgroup$
– DDP
Jan 22 at 22:45
add a comment |
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1 Answer
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$begingroup$
Your problem boils down to finding $zeta(2k)$. In fact, it can be shown that $$zeta(2k)=(-1)^{k+1}frac{B_{2k} 2^{2k-1}pi^{2k}}{(2k)!}$$
Here are some proofs. No known values of the zeta function exist except for the even positive integers and negative integers.
EDIT: $zeta(z)$ is equal to $0$ at the notorious non-trivial zeroes as well as negative even integers.
answered Jan 22 at 22:38
aledenaleden
2,409511
$endgroup$
$begingroup$
What about for odd numbers? I looked at the proof section also
$endgroup$
– DDP
Jan 22 at 22:42
$begingroup$
@DDP no known values have been found for odd positive integers.
$endgroup$
– aleden
Jan 22 at 22:43
$begingroup$
okay. I was just fact checking for all real numbers(negative numbers included) in the functional equation for $z$
$endgroup$
– DDP
Jan 22 at 22:45
add a comment |
$begingroup$
Your problem boils down to finding $zeta(2k)$. In fact, it can be shown that $$zeta(2k)=(-1)^{k+1}frac{B_{2k} 2^{2k-1}pi^{2k}}{(2k)!}$$
Here are some proofs. No known values of the zeta function exist except for the even positive integers and negative integers.
EDIT: $zeta(z)$ is equal to $0$ at the notorious non-trivial zeroes as well as negative even integers.
answered Jan 22 at 22:38
aledenaleden
2,409511
$endgroup$
$begingroup$
What about for odd numbers? I looked at the proof section also
$endgroup$
– DDP
Jan 22 at 22:42
$begingroup$
@DDP no known values have been found for odd positive integers.
$endgroup$
– aleden
Jan 22 at 22:43
$begingroup$
okay. I was just fact checking for all real numbers(negative numbers included) in the functional equation for $z$
$endgroup$
– DDP
Jan 22 at 22:45
add a comment |
$begingroup$
Your problem boils down to finding $zeta(2k)$. In fact, it can be shown that $$zeta(2k)=(-1)^{k+1}frac{B_{2k} 2^{2k-1}pi^{2k}}{(2k)!}$$
Here are some proofs. No known values of the zeta function exist except for the even positive integers and negative integers.
EDIT: $zeta(z)$ is equal to $0$ at the notorious non-trivial zeroes as well as negative even integers.
answered Jan 22 at 22:38
aledenaleden
2,409511
$endgroup$
Your problem boils down to finding $zeta(2k)$. In fact, it can be shown that $$zeta(2k)=(-1)^{k+1}frac{B_{2k} 2^{2k-1}pi^{2k}}{(2k)!}$$
Here are some proofs. No known values of the zeta function exist except for the even positive integers and negative integers.
EDIT: $zeta(z)$ is equal to $0$ at the notorious non-trivial zeroes as well as negative even integers.
answered Jan 22 at 22:38
aledenaleden
2,409511
edited Jan 22 at 22:46
answered Jan 22 at 22:38
aledenaleden
2,409511
answered Jan 22 at 22:38
aledenaleden
2,409511
answered Jan 22 at 22:38
aledenaleden
2,409511
2,409511
$begingroup$
What about for odd numbers? I looked at the proof section also
$endgroup$
– DDP
Jan 22 at 22:42
$begingroup$
@DDP no known values have been found for odd positive integers.
$endgroup$
– aleden
Jan 22 at 22:43
$begingroup$
okay. I was just fact checking for all real numbers(negative numbers included) in the functional equation for $z$
$endgroup$
– DDP
Jan 22 at 22:45
add a comment |
$begingroup$
What about for odd numbers? I looked at the proof section also
$endgroup$
– DDP
Jan 22 at 22:42
$begingroup$
@DDP no known values have been found for odd positive integers.
$endgroup$
– aleden
Jan 22 at 22:43
$begingroup$
okay. I was just fact checking for all real numbers(negative numbers included) in the functional equation for $z$
$endgroup$
– DDP
Jan 22 at 22:45
$begingroup$
What about for odd numbers? I looked at the proof section also
$endgroup$
– DDP
Jan 22 at 22:42
$begingroup$
What about for odd numbers? I looked at the proof section also
$endgroup$
– DDP
Jan 22 at 22:42
$begingroup$
@DDP no known values have been found for odd positive integers.
$endgroup$
– aleden
Jan 22 at 22:43
$begingroup$
@DDP no known values have been found for odd positive integers.
$endgroup$
– aleden
Jan 22 at 22:43
$begingroup$
okay. I was just fact checking for all real numbers(negative numbers included) in the functional equation for $z$
$endgroup$
– DDP
Jan 22 at 22:45
$begingroup$
okay. I was just fact checking for all real numbers(negative numbers included) in the functional equation for $z$
$endgroup$
– DDP
Jan 22 at 22:45
add a comment |
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$begingroup$
The functional equation is $zeta(s) = chi(s) zeta(1-s)$ where $chi(s) = pi^{s-1} 2^s Gamma(1-s) sin(pi s/2)$. This is an equality of meromorphic functions. From the zeros-poles locations of $Gamma(1-s)$ you know those of $chi(s)$. It is always true that $zeta(s) = lim_{z to s}chi(z) zeta(1-z)$ and $zeta(1-s) = lim_{z to s}frac{1}{chi(z)} zeta(z)$ and depending if $chi$ has no pole or zero at $s$ then you can remove the $lim_{z to s}$ and use direct evaluation $zeta(s) = chi(s) zeta(1-s),zeta(1-s) = frac{1}{chi(s)} zeta(s)$
$endgroup$
– reuns
Jan 23 at 0:10
$begingroup$
$chi(s)$ has a zero at $s=-2$ so $zeta(-2) = chi(-2) zeta(3) = 0$. And $frac{1}{chi(s)}$ has a pole at $s=-2$ so $zeta(3) = lim_{s to -2}zeta(1-s)=lim_{s to -2} frac{zeta(s)}{chi(s)} $, as the pole is simple the latter $=frac{zeta'(-2)}{chi'(-2)} $ whose only the denominator has a closed-form
$endgroup$
– reuns
Jan 23 at 0:16