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Periodicity of exponential function in a skew field
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If an exponential function can be defined in a field $mathbb F$ as:
$$
exp(z)=sum_n=0^infty fracz^nn! qquad z in mathbbF qquad (1)
$$
we can prove (using the binomial formula and the Cauchy rule for the product of series) that
$$
exp(x+y)=exp(x)exp(y)qquad (2)
$$
and, if in the field there is an element $tau ne 0$ such that $exp(tau)=1$ than we see that the exp function is periodic with period $tau$ because:
$$
exp(z+tau)=exp(z)exp(tau)=exp(z)
$$
and we can have also an element $sigma= tau/2$ such that $exp(sigma)=-1$ and an element $iota=sigma/2$ such that $[exp(iota)]^2=-1$
The classical example is $mathbbF=mathbbC$ where $tau=2ipi$ and $exp(iota)=i$.
Now, if we use $(1)$ to define the exp function in a non commutative ring, the property $(2)$ is not valid in general, and we can have different elements whose square is $-1$, as in the quaternion ring $mathbbH$ where $mathbf i^2=mathbf j^2 =mathbf k^2=-1$.
What can we say in this case about the periodicity of the exp function?
This question is connected to : Periodicity of the exponential function in a field , that has no answer.
ring-theory field-theory exponential-function lie-groups
asked Jul 16 at 20:13
Emilio Novati
50.2k43170
 |Â
show 2 more comments
up vote
0
down vote
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If an exponential function can be defined in a field $mathbb F$ as:
$$
exp(z)=sum_n=0^infty fracz^nn! qquad z in mathbbF qquad (1)
$$
we can prove (using the binomial formula and the Cauchy rule for the product of series) that
$$
exp(x+y)=exp(x)exp(y)qquad (2)
$$
and, if in the field there is an element $tau ne 0$ such that $exp(tau)=1$ than we see that the exp function is periodic with period $tau$ because:
$$
exp(z+tau)=exp(z)exp(tau)=exp(z)
$$
and we can have also an element $sigma= tau/2$ such that $exp(sigma)=-1$ and an element $iota=sigma/2$ such that $[exp(iota)]^2=-1$
The classical example is $mathbbF=mathbbC$ where $tau=2ipi$ and $exp(iota)=i$.
Now, if we use $(1)$ to define the exp function in a non commutative ring, the property $(2)$ is not valid in general, and we can have different elements whose square is $-1$, as in the quaternion ring $mathbbH$ where $mathbf i^2=mathbf j^2 =mathbf k^2=-1$.
What can we say in this case about the periodicity of the exp function?
This question is connected to : Periodicity of the exponential function in a field , that has no answer.
ring-theory field-theory exponential-function lie-groups
asked Jul 16 at 20:13
Emilio Novati
50.2k43170
3
In most fields $F$, infinite sums don't make sense. In many fields $F$ neither does $z^n/n!$.
– Lord Shark the Unknown
Jul 16 at 20:14
Yes, I see. But the question obviously refers tu situations where the serie is defined and absolutely convergent.
– Emilio Novati
Jul 16 at 20:18
Perhaps a more natural context for this question is Lie groups rather than fields?
– Lord Shark the Unknown
Jul 16 at 20:21
I add the tag. Maybe that we have to add a new tag as ''exponential-rings'' ?
– Emilio Novati
Jul 16 at 20:25
1
The problem is you talked about this series being "absolutely convergent". This term is completely meaningless in the context of fields. Convergence requires at least a topology. So the least you need to give any sense to your question is a limitation to topological fields of characteristic zero. Even that restriction might not be enough, not sure.
– Hamed
Jul 16 at 21:26
 |Â
show 2 more comments
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0
down vote
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up vote
0
down vote
favorite
If an exponential function can be defined in a field $mathbb F$ as:
$$
exp(z)=sum_n=0^infty fracz^nn! qquad z in mathbbF qquad (1)
$$
we can prove (using the binomial formula and the Cauchy rule for the product of series) that
$$
exp(x+y)=exp(x)exp(y)qquad (2)
$$
and, if in the field there is an element $tau ne 0$ such that $exp(tau)=1$ than we see that the exp function is periodic with period $tau$ because:
$$
exp(z+tau)=exp(z)exp(tau)=exp(z)
$$
and we can have also an element $sigma= tau/2$ such that $exp(sigma)=-1$ and an element $iota=sigma/2$ such that $[exp(iota)]^2=-1$
The classical example is $mathbbF=mathbbC$ where $tau=2ipi$ and $exp(iota)=i$.
Now, if we use $(1)$ to define the exp function in a non commutative ring, the property $(2)$ is not valid in general, and we can have different elements whose square is $-1$, as in the quaternion ring $mathbbH$ where $mathbf i^2=mathbf j^2 =mathbf k^2=-1$.
What can we say in this case about the periodicity of the exp function?
This question is connected to : Periodicity of the exponential function in a field , that has no answer.
ring-theory field-theory exponential-function lie-groups
asked Jul 16 at 20:13
Emilio Novati
50.2k43170
If an exponential function can be defined in a field $mathbb F$ as:
$$
exp(z)=sum_n=0^infty fracz^nn! qquad z in mathbbF qquad (1)
$$
we can prove (using the binomial formula and the Cauchy rule for the product of series) that
$$
exp(x+y)=exp(x)exp(y)qquad (2)
$$
and, if in the field there is an element $tau ne 0$ such that $exp(tau)=1$ than we see that the exp function is periodic with period $tau$ because:
$$
exp(z+tau)=exp(z)exp(tau)=exp(z)
$$
and we can have also an element $sigma= tau/2$ such that $exp(sigma)=-1$ and an element $iota=sigma/2$ such that $[exp(iota)]^2=-1$
The classical example is $mathbbF=mathbbC$ where $tau=2ipi$ and $exp(iota)=i$.
Now, if we use $(1)$ to define the exp function in a non commutative ring, the property $(2)$ is not valid in general, and we can have different elements whose square is $-1$, as in the quaternion ring $mathbbH$ where $mathbf i^2=mathbf j^2 =mathbf k^2=-1$.
What can we say in this case about the periodicity of the exp function?
This question is connected to : Periodicity of the exponential function in a field , that has no answer.
ring-theory field-theory exponential-function lie-groups
asked Jul 16 at 20:13
Emilio Novati
50.2k43170
edited Jul 18 at 12:07
asked Jul 16 at 20:13
Emilio Novati
50.2k43170
asked Jul 16 at 20:13
Emilio Novati
50.2k43170
asked Jul 16 at 20:13
Emilio Novati
50.2k43170
50.2k43170
3
In most fields $F$, infinite sums don't make sense. In many fields $F$ neither does $z^n/n!$.
– Lord Shark the Unknown
Jul 16 at 20:14
Yes, I see. But the question obviously refers tu situations where the serie is defined and absolutely convergent.
– Emilio Novati
Jul 16 at 20:18
Perhaps a more natural context for this question is Lie groups rather than fields?
– Lord Shark the Unknown
Jul 16 at 20:21
I add the tag. Maybe that we have to add a new tag as ''exponential-rings'' ?
– Emilio Novati
Jul 16 at 20:25
1
The problem is you talked about this series being "absolutely convergent". This term is completely meaningless in the context of fields. Convergence requires at least a topology. So the least you need to give any sense to your question is a limitation to topological fields of characteristic zero. Even that restriction might not be enough, not sure.
– Hamed
Jul 16 at 21:26
 |Â
show 2 more comments
3
In most fields $F$, infinite sums don't make sense. In many fields $F$ neither does $z^n/n!$.
– Lord Shark the Unknown
Jul 16 at 20:14
Yes, I see. But the question obviously refers tu situations where the serie is defined and absolutely convergent.
– Emilio Novati
Jul 16 at 20:18
Perhaps a more natural context for this question is Lie groups rather than fields?
– Lord Shark the Unknown
Jul 16 at 20:21
I add the tag. Maybe that we have to add a new tag as ''exponential-rings'' ?
– Emilio Novati
Jul 16 at 20:25
1
The problem is you talked about this series being "absolutely convergent". This term is completely meaningless in the context of fields. Convergence requires at least a topology. So the least you need to give any sense to your question is a limitation to topological fields of characteristic zero. Even that restriction might not be enough, not sure.
– Hamed
Jul 16 at 21:26
3
3
In most fields $F$, infinite sums don't make sense. In many fields $F$ neither does $z^n/n!$.
– Lord Shark the Unknown
Jul 16 at 20:14
In most fields $F$, infinite sums don't make sense. In many fields $F$ neither does $z^n/n!$.
– Lord Shark the Unknown
Jul 16 at 20:14
Yes, I see. But the question obviously refers tu situations where the serie is defined and absolutely convergent.
– Emilio Novati
Jul 16 at 20:18
Yes, I see. But the question obviously refers tu situations where the serie is defined and absolutely convergent.
– Emilio Novati
Jul 16 at 20:18
Perhaps a more natural context for this question is Lie groups rather than fields?
– Lord Shark the Unknown
Jul 16 at 20:21
Perhaps a more natural context for this question is Lie groups rather than fields?
– Lord Shark the Unknown
Jul 16 at 20:21
I add the tag. Maybe that we have to add a new tag as ''exponential-rings'' ?
– Emilio Novati
Jul 16 at 20:25
I add the tag. Maybe that we have to add a new tag as ''exponential-rings'' ?
– Emilio Novati
Jul 16 at 20:25
1
1
The problem is you talked about this series being "absolutely convergent". This term is completely meaningless in the context of fields. Convergence requires at least a topology. So the least you need to give any sense to your question is a limitation to topological fields of characteristic zero. Even that restriction might not be enough, not sure.
– Hamed
Jul 16 at 21:26
The problem is you talked about this series being "absolutely convergent". This term is completely meaningless in the context of fields. Convergence requires at least a topology. So the least you need to give any sense to your question is a limitation to topological fields of characteristic zero. Even that restriction might not be enough, not sure.
– Hamed
Jul 16 at 21:26
 |Â
show 2 more comments
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3
In most fields $F$, infinite sums don't make sense. In many fields $F$ neither does $z^n/n!$.
– Lord Shark the Unknown
Jul 16 at 20:14
Yes, I see. But the question obviously refers tu situations where the serie is defined and absolutely convergent.
– Emilio Novati
Jul 16 at 20:18
Perhaps a more natural context for this question is Lie groups rather than fields?
– Lord Shark the Unknown
Jul 16 at 20:21
I add the tag. Maybe that we have to add a new tag as ''exponential-rings'' ?
– Emilio Novati
Jul 16 at 20:25
1
The problem is you talked about this series being "absolutely convergent". This term is completely meaningless in the context of fields. Convergence requires at least a topology. So the least you need to give any sense to your question is a limitation to topological fields of characteristic zero. Even that restriction might not be enough, not sure.
– Hamed
Jul 16 at 21:26