In Banach algebea $A$ if $ab=ba$ prove that $e^{a+b}=e^ae^b$
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Let $A$ be aBanach algebra
if $ab=ba$ then prove that $e^{a+b}=e^ae^b$
I've started by $e^a=sum _{n=0}^{infty }frac{a^n}{n!}$, Iwant to khow this is correct way?
Any help will be greatly appreciated.
banach-algebras
asked Nov 19 at 17:51
user62498
1,888613
add a comment |
up vote
0
down vote
favorite
Let $A$ be aBanach algebra
if $ab=ba$ then prove that $e^{a+b}=e^ae^b$
I've started by $e^a=sum _{n=0}^{infty }frac{a^n}{n!}$, Iwant to khow this is correct way?
Any help will be greatly appreciated.
banach-algebras
asked Nov 19 at 17:51
user62498
1,888613
yes, and then multiply it with $e^{b}$. Do you know how to multiply two convergent series? There’s a we’ll known formula
– Fakemistake
Nov 19 at 17:58
@Dear Fakemistake, I think I have to youse Cachy formula
– user62498
Nov 19 at 18:09
Correct it‘s Cauchy formula
– Fakemistake
Nov 19 at 18:10
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A$ be aBanach algebra
if $ab=ba$ then prove that $e^{a+b}=e^ae^b$
I've started by $e^a=sum _{n=0}^{infty }frac{a^n}{n!}$, Iwant to khow this is correct way?
Any help will be greatly appreciated.
banach-algebras
asked Nov 19 at 17:51
user62498
1,888613
Let $A$ be aBanach algebra
if $ab=ba$ then prove that $e^{a+b}=e^ae^b$
I've started by $e^a=sum _{n=0}^{infty }frac{a^n}{n!}$, Iwant to khow this is correct way?
Any help will be greatly appreciated.
banach-algebras
banach-algebras
asked Nov 19 at 17:51
user62498
1,888613
asked Nov 19 at 17:51
user62498
1,888613
asked Nov 19 at 17:51
user62498
1,888613
asked Nov 19 at 17:51
user62498
1,888613
asked Nov 19 at 17:51
user62498
1,888613
1,888613
yes, and then multiply it with $e^{b}$. Do you know how to multiply two convergent series? There’s a we’ll known formula
– Fakemistake
Nov 19 at 17:58
@Dear Fakemistake, I think I have to youse Cachy formula
– user62498
Nov 19 at 18:09
Correct it‘s Cauchy formula
– Fakemistake
Nov 19 at 18:10
add a comment |
yes, and then multiply it with $e^{b}$. Do you know how to multiply two convergent series? There’s a we’ll known formula
– Fakemistake
Nov 19 at 17:58
@Dear Fakemistake, I think I have to youse Cachy formula
– user62498
Nov 19 at 18:09
Correct it‘s Cauchy formula
– Fakemistake
Nov 19 at 18:10
yes, and then multiply it with $e^{b}$. Do you know how to multiply two convergent series? There’s a we’ll known formula
– Fakemistake
Nov 19 at 17:58
yes, and then multiply it with $e^{b}$. Do you know how to multiply two convergent series? There’s a we’ll known formula
– Fakemistake
Nov 19 at 17:58
@Dear Fakemistake, I think I have to youse Cachy formula
– user62498
Nov 19 at 18:09
@Dear Fakemistake, I think I have to youse Cachy formula
– user62498
Nov 19 at 18:09
Correct it‘s Cauchy formula
– Fakemistake
Nov 19 at 18:10
Correct it‘s Cauchy formula
– Fakemistake
Nov 19 at 18:10
add a comment |
1 Answer
1
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1
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Yes. Expand as power series, and you can rearrange the $a$'s and $b$'s because they commute (so $(a+b)^2=a^2+2ab+b^2$ for example), so you can pretend you are doing high-school maths with numbers. You might want to read about the BCH (Baker Campbell Hausdorff) expansion. A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it gets harder ...
answered Nov 19 at 17:56
Richard Martin
1,63918
@DRichard Martinear , please explain a little more about, A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it
– user62498
Nov 19 at 18:13
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
1
down vote
Yes. Expand as power series, and you can rearrange the $a$'s and $b$'s because they commute (so $(a+b)^2=a^2+2ab+b^2$ for example), so you can pretend you are doing high-school maths with numbers. You might want to read about the BCH (Baker Campbell Hausdorff) expansion. A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it gets harder ...
answered Nov 19 at 17:56
Richard Martin
1,63918
@DRichard Martinear , please explain a little more about, A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it
– user62498
Nov 19 at 18:13
add a comment |
up vote
1
down vote
Yes. Expand as power series, and you can rearrange the $a$'s and $b$'s because they commute (so $(a+b)^2=a^2+2ab+b^2$ for example), so you can pretend you are doing high-school maths with numbers. You might want to read about the BCH (Baker Campbell Hausdorff) expansion. A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it gets harder ...
answered Nov 19 at 17:56
Richard Martin
1,63918
@DRichard Martinear , please explain a little more about, A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it
– user62498
Nov 19 at 18:13
add a comment |
up vote
1
down vote
up vote
1
down vote
Yes. Expand as power series, and you can rearrange the $a$'s and $b$'s because they commute (so $(a+b)^2=a^2+2ab+b^2$ for example), so you can pretend you are doing high-school maths with numbers. You might want to read about the BCH (Baker Campbell Hausdorff) expansion. A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it gets harder ...
answered Nov 19 at 17:56
Richard Martin
1,63918
Yes. Expand as power series, and you can rearrange the $a$'s and $b$'s because they commute (so $(a+b)^2=a^2+2ab+b^2$ for example), so you can pretend you are doing high-school maths with numbers. You might want to read about the BCH (Baker Campbell Hausdorff) expansion. A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it gets harder ...
answered Nov 19 at 17:56
Richard Martin
1,63918
answered Nov 19 at 17:56
Richard Martin
1,63918
answered Nov 19 at 17:56
Richard Martin
1,63918
answered Nov 19 at 17:56
Richard Martin
1,63918
1,63918
@DRichard Martinear , please explain a little more about, A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it
– user62498
Nov 19 at 18:13
add a comment |
@DRichard Martinear , please explain a little more about, A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it
– user62498
Nov 19 at 18:13
@DRichard Martinear , please explain a little more about, A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it
– user62498
Nov 19 at 18:13
@DRichard Martinear , please explain a little more about, A slightly harder result holds when $a$ and $b$ both commute with their commutator $[a,b]$, and if that fails then it
– user62498
Nov 19 at 18:13
add a comment |
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yes, and then multiply it with $e^{b}$. Do you know how to multiply two convergent series? There’s a we’ll known formula
– Fakemistake
Nov 19 at 17:58
@Dear Fakemistake, I think I have to youse Cachy formula
– user62498
Nov 19 at 18:09
Correct it‘s Cauchy formula
– Fakemistake
Nov 19 at 18:10