Mixing
Is the infinite product of $-1 times -1 times -1 timesdots = -i$?
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So I woke up this morning and I was thinking about the infinite product $-1 times -1 times -1 timesdots$, and what it equals. I came to the conclusion that it equals $-i$. Alternatively stated,
$
{displaystyle prod_{i}^{infty} (-1)} = -i
$
Here's how I reached this:
$${prod_{i}^{infty} (-1)} = e^{ln({displaystyle prod_{i}^{infty} (-1)})} = e^{displaystyle sum_{i}^{infty}{ln(-1)}}= e^{displaystyle sum_{i}^{infty}{ipi}}=e^{ipidisplaystyle sum_{i}^{infty}{1}}$$
Now, here's where I'm a little hesitant. I want to say that, from $zeta(0)=-frac{1}{2}$, we can conclude that
$$e^{ipidisplaystyle sum_{i}^{infty}{1}} = e^{-frac{1}{2}ipi} = -i$$.
I have been told before that the sum $displaystyle sum_{i}^{infty}{1}$ is not actually $-frac{1}{2}$, but I'm not really sure why. It would seem that if this is the case, then my product would in fact not be $-i$. Though, I must say that $-i$ sort of makes sense, because multiplying complex numbers is essentially rotating them, and so rotating by $180$ every time will get you $180+180+180+...$ is the same as $180*(1+1+1+...)$ which is (if my premise is right) $180*(-frac{1}{2})=-90$. $-90$ degrees on the complex plane turns out to be $-i$.
So my question is, is there a hole in my logic? I know what not accounting for $zeta(0)=-frac{1}{2}$, the sum $1+1+1+...$ is divergent, but taking that into account, can I say with confidence that $-1 times -1 times -1 timesdots = -i$?
complex-analysis riemann-zeta infinite-product
asked Apr 25 '18 at 17:54
user3210986user3210986
974
$endgroup$
add a comment |
$begingroup$
So I woke up this morning and I was thinking about the infinite product $-1 times -1 times -1 timesdots$, and what it equals. I came to the conclusion that it equals $-i$. Alternatively stated,
$
{displaystyle prod_{i}^{infty} (-1)} = -i
$
Here's how I reached this:
$${prod_{i}^{infty} (-1)} = e^{ln({displaystyle prod_{i}^{infty} (-1)})} = e^{displaystyle sum_{i}^{infty}{ln(-1)}}= e^{displaystyle sum_{i}^{infty}{ipi}}=e^{ipidisplaystyle sum_{i}^{infty}{1}}$$
Now, here's where I'm a little hesitant. I want to say that, from $zeta(0)=-frac{1}{2}$, we can conclude that
$$e^{ipidisplaystyle sum_{i}^{infty}{1}} = e^{-frac{1}{2}ipi} = -i$$.
I have been told before that the sum $displaystyle sum_{i}^{infty}{1}$ is not actually $-frac{1}{2}$, but I'm not really sure why. It would seem that if this is the case, then my product would in fact not be $-i$. Though, I must say that $-i$ sort of makes sense, because multiplying complex numbers is essentially rotating them, and so rotating by $180$ every time will get you $180+180+180+...$ is the same as $180*(1+1+1+...)$ which is (if my premise is right) $180*(-frac{1}{2})=-90$. $-90$ degrees on the complex plane turns out to be $-i$.
So my question is, is there a hole in my logic? I know what not accounting for $zeta(0)=-frac{1}{2}$, the sum $1+1+1+...$ is divergent, but taking that into account, can I say with confidence that $-1 times -1 times -1 timesdots = -i$?
complex-analysis riemann-zeta infinite-product
asked Apr 25 '18 at 17:54
user3210986user3210986
974
$endgroup$
8
$begingroup$
What you've actually found is the value of $e^{i pi zeta(0)}$, where $zeta$ is the Riemann Zeta Function
$endgroup$
– Naweed G. Seldon
Apr 25 '18 at 17:59
5
$begingroup$
Infinite products are (usually) defined as the limit of the sequence of partial products. There this infinite product is not convergent. In fact it agrees with some of your observations since the parity of $sum_{k=1}^n1$ alternate between odd and even as n changes.
$endgroup$
– Bumblebee
Apr 25 '18 at 18:12
1
$begingroup$
In you interpret $prod_{n=1}^infty (-1)$ as the ratio of two zeta-regularized product, $prod_{n=1}^infty (e^{ipi} n)$ and $prod_{n=1}^infty n$, the value of your product do equal to $e^{ipizeta(0)} = -i$.
$endgroup$
– achille hui
Apr 25 '18 at 18:18
add a comment |
$begingroup$
So I woke up this morning and I was thinking about the infinite product $-1 times -1 times -1 timesdots$, and what it equals. I came to the conclusion that it equals $-i$. Alternatively stated,
$
{displaystyle prod_{i}^{infty} (-1)} = -i
$
Here's how I reached this:
$${prod_{i}^{infty} (-1)} = e^{ln({displaystyle prod_{i}^{infty} (-1)})} = e^{displaystyle sum_{i}^{infty}{ln(-1)}}= e^{displaystyle sum_{i}^{infty}{ipi}}=e^{ipidisplaystyle sum_{i}^{infty}{1}}$$
Now, here's where I'm a little hesitant. I want to say that, from $zeta(0)=-frac{1}{2}$, we can conclude that
$$e^{ipidisplaystyle sum_{i}^{infty}{1}} = e^{-frac{1}{2}ipi} = -i$$.
I have been told before that the sum $displaystyle sum_{i}^{infty}{1}$ is not actually $-frac{1}{2}$, but I'm not really sure why. It would seem that if this is the case, then my product would in fact not be $-i$. Though, I must say that $-i$ sort of makes sense, because multiplying complex numbers is essentially rotating them, and so rotating by $180$ every time will get you $180+180+180+...$ is the same as $180*(1+1+1+...)$ which is (if my premise is right) $180*(-frac{1}{2})=-90$. $-90$ degrees on the complex plane turns out to be $-i$.
So my question is, is there a hole in my logic? I know what not accounting for $zeta(0)=-frac{1}{2}$, the sum $1+1+1+...$ is divergent, but taking that into account, can I say with confidence that $-1 times -1 times -1 timesdots = -i$?
complex-analysis riemann-zeta infinite-product
asked Apr 25 '18 at 17:54
user3210986user3210986
974
$endgroup$
So I woke up this morning and I was thinking about the infinite product $-1 times -1 times -1 timesdots$, and what it equals. I came to the conclusion that it equals $-i$. Alternatively stated,
$
{displaystyle prod_{i}^{infty} (-1)} = -i
$
Here's how I reached this:
$${prod_{i}^{infty} (-1)} = e^{ln({displaystyle prod_{i}^{infty} (-1)})} = e^{displaystyle sum_{i}^{infty}{ln(-1)}}= e^{displaystyle sum_{i}^{infty}{ipi}}=e^{ipidisplaystyle sum_{i}^{infty}{1}}$$
Now, here's where I'm a little hesitant. I want to say that, from $zeta(0)=-frac{1}{2}$, we can conclude that
$$e^{ipidisplaystyle sum_{i}^{infty}{1}} = e^{-frac{1}{2}ipi} = -i$$.
I have been told before that the sum $displaystyle sum_{i}^{infty}{1}$ is not actually $-frac{1}{2}$, but I'm not really sure why. It would seem that if this is the case, then my product would in fact not be $-i$. Though, I must say that $-i$ sort of makes sense, because multiplying complex numbers is essentially rotating them, and so rotating by $180$ every time will get you $180+180+180+...$ is the same as $180*(1+1+1+...)$ which is (if my premise is right) $180*(-frac{1}{2})=-90$. $-90$ degrees on the complex plane turns out to be $-i$.
So my question is, is there a hole in my logic? I know what not accounting for $zeta(0)=-frac{1}{2}$, the sum $1+1+1+...$ is divergent, but taking that into account, can I say with confidence that $-1 times -1 times -1 timesdots = -i$?
complex-analysis riemann-zeta infinite-product
complex-analysis riemann-zeta infinite-product
asked Apr 25 '18 at 17:54
user3210986user3210986
974
asked Apr 25 '18 at 17:54
user3210986user3210986
974
asked Apr 25 '18 at 17:54
user3210986user3210986
974
asked Apr 25 '18 at 17:54
user3210986user3210986
974
asked Apr 25 '18 at 17:54
user3210986user3210986
974
974
8
$begingroup$
What you've actually found is the value of $e^{i pi zeta(0)}$, where $zeta$ is the Riemann Zeta Function
$endgroup$
– Naweed G. Seldon
Apr 25 '18 at 17:59
5
$begingroup$
Infinite products are (usually) defined as the limit of the sequence of partial products. There this infinite product is not convergent. In fact it agrees with some of your observations since the parity of $sum_{k=1}^n1$ alternate between odd and even as n changes.
$endgroup$
– Bumblebee
Apr 25 '18 at 18:12
1
$begingroup$
In you interpret $prod_{n=1}^infty (-1)$ as the ratio of two zeta-regularized product, $prod_{n=1}^infty (e^{ipi} n)$ and $prod_{n=1}^infty n$, the value of your product do equal to $e^{ipizeta(0)} = -i$.
$endgroup$
– achille hui
Apr 25 '18 at 18:18
add a comment |
8
$begingroup$
What you've actually found is the value of $e^{i pi zeta(0)}$, where $zeta$ is the Riemann Zeta Function
$endgroup$
– Naweed G. Seldon
Apr 25 '18 at 17:59
5
$begingroup$
Infinite products are (usually) defined as the limit of the sequence of partial products. There this infinite product is not convergent. In fact it agrees with some of your observations since the parity of $sum_{k=1}^n1$ alternate between odd and even as n changes.
$endgroup$
– Bumblebee
Apr 25 '18 at 18:12
1
$begingroup$
In you interpret $prod_{n=1}^infty (-1)$ as the ratio of two zeta-regularized product, $prod_{n=1}^infty (e^{ipi} n)$ and $prod_{n=1}^infty n$, the value of your product do equal to $e^{ipizeta(0)} = -i$.
$endgroup$
– achille hui
Apr 25 '18 at 18:18
8
8
$begingroup$
What you've actually found is the value of $e^{i pi zeta(0)}$, where $zeta$ is the Riemann Zeta Function
$endgroup$
– Naweed G. Seldon
Apr 25 '18 at 17:59
$begingroup$
What you've actually found is the value of $e^{i pi zeta(0)}$, where $zeta$ is the Riemann Zeta Function
$endgroup$
– Naweed G. Seldon
Apr 25 '18 at 17:59
5
5
$begingroup$
Infinite products are (usually) defined as the limit of the sequence of partial products. There this infinite product is not convergent. In fact it agrees with some of your observations since the parity of $sum_{k=1}^n1$ alternate between odd and even as n changes.
$endgroup$
– Bumblebee
Apr 25 '18 at 18:12
$begingroup$
Infinite products are (usually) defined as the limit of the sequence of partial products. There this infinite product is not convergent. In fact it agrees with some of your observations since the parity of $sum_{k=1}^n1$ alternate between odd and even as n changes.
$endgroup$
– Bumblebee
Apr 25 '18 at 18:12
1
1
$begingroup$
In you interpret $prod_{n=1}^infty (-1)$ as the ratio of two zeta-regularized product, $prod_{n=1}^infty (e^{ipi} n)$ and $prod_{n=1}^infty n$, the value of your product do equal to $e^{ipizeta(0)} = -i$.
$endgroup$
– achille hui
Apr 25 '18 at 18:18
$begingroup$
In you interpret $prod_{n=1}^infty (-1)$ as the ratio of two zeta-regularized product, $prod_{n=1}^infty (e^{ipi} n)$ and $prod_{n=1}^infty n$, the value of your product do equal to $e^{ipizeta(0)} = -i$.
$endgroup$
– achille hui
Apr 25 '18 at 18:18
add a comment |
1 Answer
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$$prod_{n=1}^{infty}(1+c)=sum_{n=0}^{infty}(2n)!/(n!)^2*(-c/4)^n=frac{1}{sqrt{(1+c)}}$$
The hole in your logic is found within the fact that $-1=e^{ipi}$ why wouldn't it be $-1=e^{3ipi}$?
answered Jan 14 at 13:19
GerbenGerben
13413
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$begingroup$
$$prod_{n=1}^{infty}(1+c)=sum_{n=0}^{infty}(2n)!/(n!)^2*(-c/4)^n=frac{1}{sqrt{(1+c)}}$$
The hole in your logic is found within the fact that $-1=e^{ipi}$ why wouldn't it be $-1=e^{3ipi}$?
answered Jan 14 at 13:19
GerbenGerben
13413
$endgroup$
add a comment |
$begingroup$
$$prod_{n=1}^{infty}(1+c)=sum_{n=0}^{infty}(2n)!/(n!)^2*(-c/4)^n=frac{1}{sqrt{(1+c)}}$$
The hole in your logic is found within the fact that $-1=e^{ipi}$ why wouldn't it be $-1=e^{3ipi}$?
answered Jan 14 at 13:19
GerbenGerben
13413
$endgroup$
add a comment |
$begingroup$
$$prod_{n=1}^{infty}(1+c)=sum_{n=0}^{infty}(2n)!/(n!)^2*(-c/4)^n=frac{1}{sqrt{(1+c)}}$$
The hole in your logic is found within the fact that $-1=e^{ipi}$ why wouldn't it be $-1=e^{3ipi}$?
answered Jan 14 at 13:19
GerbenGerben
13413
$endgroup$
$$prod_{n=1}^{infty}(1+c)=sum_{n=0}^{infty}(2n)!/(n!)^2*(-c/4)^n=frac{1}{sqrt{(1+c)}}$$
The hole in your logic is found within the fact that $-1=e^{ipi}$ why wouldn't it be $-1=e^{3ipi}$?
answered Jan 14 at 13:19
GerbenGerben
13413
answered Jan 14 at 13:19
GerbenGerben
13413
answered Jan 14 at 13:19
GerbenGerben
13413
answered Jan 14 at 13:19
GerbenGerben
13413
13413
add a comment |
add a comment |
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8
$begingroup$
What you've actually found is the value of $e^{i pi zeta(0)}$, where $zeta$ is the Riemann Zeta Function
$endgroup$
– Naweed G. Seldon
Apr 25 '18 at 17:59
5
$begingroup$
Infinite products are (usually) defined as the limit of the sequence of partial products. There this infinite product is not convergent. In fact it agrees with some of your observations since the parity of $sum_{k=1}^n1$ alternate between odd and even as n changes.
$endgroup$
– Bumblebee
Apr 25 '18 at 18:12
1
$begingroup$
In you interpret $prod_{n=1}^infty (-1)$ as the ratio of two zeta-regularized product, $prod_{n=1}^infty (e^{ipi} n)$ and $prod_{n=1}^infty n$, the value of your product do equal to $e^{ipizeta(0)} = -i$.
$endgroup$
– achille hui
Apr 25 '18 at 18:18