Mixing
Why isn't $zeta(3) = 2^3pi ^{3-1}sinleft( frac{pi 3}{2}right)Gamma(1-3)zeta(1-3) = 0$?
$begingroup$
The zeta function (for $Re(s)>1$) is given by the definition:
- $zeta(s) = sum _{n=1}^{infty} left [frac{1}{n^s} right]$
The zeta function also can be given by the following functional equation:
- $zeta(s) = 2^spi ^{s-1}sinleft( frac{pi s}{2}right)Gamma(1-s)zeta(1-s)$
Using the above functional equation, we also know than for all negative even integers (i.e. $s=-2, -4, -6...$) the zeta function has trivial zeroes i.e.
$zeta(-2) = 0$, $zeta(-4) = 0$, $zeta(-6) = 0$ and so on.
Substituting $s=3$ in the functional equation we have:
a. $zeta(3) = sum _{n=1}^{infty} left [frac{1}{n^3} right]$
b. $zeta(3) = 2^3pi ^{3-1}sinleft( frac{pi 3}{2}right)Gamma(1-3)zeta(1-3)$
Since, $zeta(1-3)=zeta(-2)=0$, using this in the R.H.S of the functional equation (b.) seems to state that $zeta(3) = 0$. But, using (a.) we know that the $zeta(3)$ is convergent but not equal to zero.
So, these two values are different. What is it that I am missing in the functional definition and where is the discrepancy?
P.S. I am from a C.S. background and a beginner in analysis still trying to learn.
real-analysis complex-analysis riemann-zeta
edited Feb 2 at 14:28
user21820
40.2k544162
asked Feb 2 at 9:19
TheoryQuest1TheoryQuest1
1498
$endgroup$
add a comment |
$begingroup$
The zeta function (for $Re(s)>1$) is given by the definition:
- $zeta(s) = sum _{n=1}^{infty} left [frac{1}{n^s} right]$
The zeta function also can be given by the following functional equation:
- $zeta(s) = 2^spi ^{s-1}sinleft( frac{pi s}{2}right)Gamma(1-s)zeta(1-s)$
Using the above functional equation, we also know than for all negative even integers (i.e. $s=-2, -4, -6...$) the zeta function has trivial zeroes i.e.
$zeta(-2) = 0$, $zeta(-4) = 0$, $zeta(-6) = 0$ and so on.
Substituting $s=3$ in the functional equation we have:
a. $zeta(3) = sum _{n=1}^{infty} left [frac{1}{n^3} right]$
b. $zeta(3) = 2^3pi ^{3-1}sinleft( frac{pi 3}{2}right)Gamma(1-3)zeta(1-3)$
Since, $zeta(1-3)=zeta(-2)=0$, using this in the R.H.S of the functional equation (b.) seems to state that $zeta(3) = 0$. But, using (a.) we know that the $zeta(3)$ is convergent but not equal to zero.
So, these two values are different. What is it that I am missing in the functional definition and where is the discrepancy?
P.S. I am from a C.S. background and a beginner in analysis still trying to learn.
real-analysis complex-analysis riemann-zeta
edited Feb 2 at 14:28
user21820
40.2k544162
asked Feb 2 at 9:19
TheoryQuest1TheoryQuest1
1498
$endgroup$
$begingroup$
"The zeta function is given by the definition: 1. $zeta(s) = sum _{n=1}^{infty} left [frac{1}{n^s} right]$" For the record, this is the definition of $zeta(s)$ only when $Re s>1$.
$endgroup$
– Did
Feb 2 at 9:43
$begingroup$
Yes. I was a bit loose with the definition.
$endgroup$
– TheoryQuest1
Feb 2 at 9:50
add a comment |
$begingroup$
The zeta function (for $Re(s)>1$) is given by the definition:
- $zeta(s) = sum _{n=1}^{infty} left [frac{1}{n^s} right]$
The zeta function also can be given by the following functional equation:
- $zeta(s) = 2^spi ^{s-1}sinleft( frac{pi s}{2}right)Gamma(1-s)zeta(1-s)$
Using the above functional equation, we also know than for all negative even integers (i.e. $s=-2, -4, -6...$) the zeta function has trivial zeroes i.e.
$zeta(-2) = 0$, $zeta(-4) = 0$, $zeta(-6) = 0$ and so on.
Substituting $s=3$ in the functional equation we have:
a. $zeta(3) = sum _{n=1}^{infty} left [frac{1}{n^3} right]$
b. $zeta(3) = 2^3pi ^{3-1}sinleft( frac{pi 3}{2}right)Gamma(1-3)zeta(1-3)$
Since, $zeta(1-3)=zeta(-2)=0$, using this in the R.H.S of the functional equation (b.) seems to state that $zeta(3) = 0$. But, using (a.) we know that the $zeta(3)$ is convergent but not equal to zero.
So, these two values are different. What is it that I am missing in the functional definition and where is the discrepancy?
P.S. I am from a C.S. background and a beginner in analysis still trying to learn.
real-analysis complex-analysis riemann-zeta
edited Feb 2 at 14:28
user21820
40.2k544162
asked Feb 2 at 9:19
TheoryQuest1TheoryQuest1
1498
$endgroup$
The zeta function (for $Re(s)>1$) is given by the definition:
- $zeta(s) = sum _{n=1}^{infty} left [frac{1}{n^s} right]$
The zeta function also can be given by the following functional equation:
- $zeta(s) = 2^spi ^{s-1}sinleft( frac{pi s}{2}right)Gamma(1-s)zeta(1-s)$
Using the above functional equation, we also know than for all negative even integers (i.e. $s=-2, -4, -6...$) the zeta function has trivial zeroes i.e.
$zeta(-2) = 0$, $zeta(-4) = 0$, $zeta(-6) = 0$ and so on.
Substituting $s=3$ in the functional equation we have:
a. $zeta(3) = sum _{n=1}^{infty} left [frac{1}{n^3} right]$
b. $zeta(3) = 2^3pi ^{3-1}sinleft( frac{pi 3}{2}right)Gamma(1-3)zeta(1-3)$
Since, $zeta(1-3)=zeta(-2)=0$, using this in the R.H.S of the functional equation (b.) seems to state that $zeta(3) = 0$. But, using (a.) we know that the $zeta(3)$ is convergent but not equal to zero.
So, these two values are different. What is it that I am missing in the functional definition and where is the discrepancy?
P.S. I am from a C.S. background and a beginner in analysis still trying to learn.
real-analysis complex-analysis riemann-zeta
real-analysis complex-analysis riemann-zeta
edited Feb 2 at 14:28
user21820
40.2k544162
asked Feb 2 at 9:19
TheoryQuest1TheoryQuest1
1498
edited Feb 2 at 14:28
user21820
40.2k544162
asked Feb 2 at 9:19
TheoryQuest1TheoryQuest1
1498
edited Feb 2 at 14:28
user21820
40.2k544162
edited Feb 2 at 14:28
user21820
40.2k544162
edited Feb 2 at 14:28
user21820
40.2k544162
40.2k544162
asked Feb 2 at 9:19
TheoryQuest1TheoryQuest1
1498
asked Feb 2 at 9:19
TheoryQuest1TheoryQuest1
1498
asked Feb 2 at 9:19
TheoryQuest1TheoryQuest1
1498
1498
$begingroup$
"The zeta function is given by the definition: 1. $zeta(s) = sum _{n=1}^{infty} left [frac{1}{n^s} right]$" For the record, this is the definition of $zeta(s)$ only when $Re s>1$.
$endgroup$
– Did
Feb 2 at 9:43
$begingroup$
Yes. I was a bit loose with the definition.
$endgroup$
– TheoryQuest1
Feb 2 at 9:50
add a comment |
$begingroup$
"The zeta function is given by the definition: 1. $zeta(s) = sum _{n=1}^{infty} left [frac{1}{n^s} right]$" For the record, this is the definition of $zeta(s)$ only when $Re s>1$.
$endgroup$
– Did
Feb 2 at 9:43
$begingroup$
Yes. I was a bit loose with the definition.
$endgroup$
– TheoryQuest1
Feb 2 at 9:50
$begingroup$
"The zeta function is given by the definition: 1. $zeta(s) = sum _{n=1}^{infty} left [frac{1}{n^s} right]$" For the record, this is the definition of $zeta(s)$ only when $Re s>1$.
$endgroup$
– Did
Feb 2 at 9:43
$begingroup$
"The zeta function is given by the definition: 1. $zeta(s) = sum _{n=1}^{infty} left [frac{1}{n^s} right]$" For the record, this is the definition of $zeta(s)$ only when $Re s>1$.
$endgroup$
– Did
Feb 2 at 9:43
$begingroup$
Yes. I was a bit loose with the definition.
$endgroup$
– TheoryQuest1
Feb 2 at 9:50
$begingroup$
Yes. I was a bit loose with the definition.
$endgroup$
– TheoryQuest1
Feb 2 at 9:50
add a comment |
1 Answer
1
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oldest
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$begingroup$
The function $Gamma$ has a pole at every non-positive integer. The pole of $Gamma$ at $-2$ "cancels out" the zero of $zeta$ at $-2$ since they are both simple.
edited Feb 2 at 9:42
Did
249k23228466
answered Feb 2 at 9:27
asdasd
30729
$endgroup$
add a comment |
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$begingroup$
The function $Gamma$ has a pole at every non-positive integer. The pole of $Gamma$ at $-2$ "cancels out" the zero of $zeta$ at $-2$ since they are both simple.
edited Feb 2 at 9:42
Did
249k23228466
answered Feb 2 at 9:27
asdasd
30729
$endgroup$
add a comment |
$begingroup$
The function $Gamma$ has a pole at every non-positive integer. The pole of $Gamma$ at $-2$ "cancels out" the zero of $zeta$ at $-2$ since they are both simple.
edited Feb 2 at 9:42
Did
249k23228466
answered Feb 2 at 9:27
asdasd
30729
$endgroup$
add a comment |
$begingroup$
The function $Gamma$ has a pole at every non-positive integer. The pole of $Gamma$ at $-2$ "cancels out" the zero of $zeta$ at $-2$ since they are both simple.
edited Feb 2 at 9:42
Did
249k23228466
answered Feb 2 at 9:27
asdasd
30729
$endgroup$
The function $Gamma$ has a pole at every non-positive integer. The pole of $Gamma$ at $-2$ "cancels out" the zero of $zeta$ at $-2$ since they are both simple.
edited Feb 2 at 9:42
Did
249k23228466
answered Feb 2 at 9:27
asdasd
30729
edited Feb 2 at 9:42
Did
249k23228466
edited Feb 2 at 9:42
Did
249k23228466
edited Feb 2 at 9:42
Did
249k23228466
249k23228466
answered Feb 2 at 9:27
asdasd
30729
answered Feb 2 at 9:27
asdasd
30729
answered Feb 2 at 9:27
asdasd
30729
30729
add a comment |
add a comment |
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$begingroup$
"The zeta function is given by the definition: 1. $zeta(s) = sum _{n=1}^{infty} left [frac{1}{n^s} right]$" For the record, this is the definition of $zeta(s)$ only when $Re s>1$.
$endgroup$
– Did
Feb 2 at 9:43
$begingroup$
Yes. I was a bit loose with the definition.
$endgroup$
– TheoryQuest1
Feb 2 at 9:50