Trouble with little o notation.
$begingroup$
Problem:
Compute the differential of $phi(t) = e^{A+tB}$ at $0$ where $A$ and $B$ are commuting square matrices of size $ntimes n.$
We proceed in the usual manner:
$$phi(0+h) = e^{A+hB} = e^{A}e^{hB}$$ since $A$ and $hB$ commute. Thus
$$phi(h) = e^{A}left(I+hB+o(hB)right)$$
$$ = e^A+h e^AB+e^{A}o(hB) = phi(0)+he^{A}B+o(h).$$
I am not sure why $e^{A}o(hB)$ is $o(h).$ Perhaps someone can explain?
real-analysis asymptotics
asked Jan 8 at 17:12
model_checkermodel_checker
4,44821931
$endgroup$
add a comment |
$begingroup$
Problem:
Compute the differential of $phi(t) = e^{A+tB}$ at $0$ where $A$ and $B$ are commuting square matrices of size $ntimes n.$
We proceed in the usual manner:
$$phi(0+h) = e^{A+hB} = e^{A}e^{hB}$$ since $A$ and $hB$ commute. Thus
$$phi(h) = e^{A}left(I+hB+o(hB)right)$$
$$ = e^A+h e^AB+e^{A}o(hB) = phi(0)+he^{A}B+o(h).$$
I am not sure why $e^{A}o(hB)$ is $o(h).$ Perhaps someone can explain?
real-analysis asymptotics
asked Jan 8 at 17:12
model_checkermodel_checker
4,44821931
$endgroup$
add a comment |
$begingroup$
Problem:
Compute the differential of $phi(t) = e^{A+tB}$ at $0$ where $A$ and $B$ are commuting square matrices of size $ntimes n.$
We proceed in the usual manner:
$$phi(0+h) = e^{A+hB} = e^{A}e^{hB}$$ since $A$ and $hB$ commute. Thus
$$phi(h) = e^{A}left(I+hB+o(hB)right)$$
$$ = e^A+h e^AB+e^{A}o(hB) = phi(0)+he^{A}B+o(h).$$
I am not sure why $e^{A}o(hB)$ is $o(h).$ Perhaps someone can explain?
real-analysis asymptotics
asked Jan 8 at 17:12
model_checkermodel_checker
4,44821931
$endgroup$
Problem:
Compute the differential of $phi(t) = e^{A+tB}$ at $0$ where $A$ and $B$ are commuting square matrices of size $ntimes n.$
We proceed in the usual manner:
$$phi(0+h) = e^{A+hB} = e^{A}e^{hB}$$ since $A$ and $hB$ commute. Thus
$$phi(h) = e^{A}left(I+hB+o(hB)right)$$
$$ = e^A+h e^AB+e^{A}o(hB) = phi(0)+he^{A}B+o(h).$$
I am not sure why $e^{A}o(hB)$ is $o(h).$ Perhaps someone can explain?
real-analysis asymptotics
real-analysis asymptotics
asked Jan 8 at 17:12
model_checkermodel_checker
4,44821931
asked Jan 8 at 17:12
model_checkermodel_checker
4,44821931
asked Jan 8 at 17:12
model_checkermodel_checker
4,44821931
asked Jan 8 at 17:12
model_checkermodel_checker
4,44821931
asked Jan 8 at 17:12
model_checkermodel_checker
4,44821931
4,44821931
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One definition of $o(h)$ is as follows:
A matrix valued function $f(h)$ is $o(h)$ if $lim_{h to 0} frac{|f(h)|}{h} = 0$ (for some matrix norm $|cdot|$).
With this definition, I think you should find the result easy to prove.
answered Jan 8 at 17:17
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
$begingroup$
But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
$endgroup$
– model_checker
Jan 8 at 17:19
$begingroup$
The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
$endgroup$
– Omnomnomnom
Jan 8 at 17:22
$begingroup$
Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
$endgroup$
– Omnomnomnom
Jan 8 at 17:24
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One definition of $o(h)$ is as follows:
A matrix valued function $f(h)$ is $o(h)$ if $lim_{h to 0} frac{|f(h)|}{h} = 0$ (for some matrix norm $|cdot|$).
With this definition, I think you should find the result easy to prove.
answered Jan 8 at 17:17
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
$begingroup$
But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
$endgroup$
– model_checker
Jan 8 at 17:19
$begingroup$
The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
$endgroup$
– Omnomnomnom
Jan 8 at 17:22
$begingroup$
Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
$endgroup$
– Omnomnomnom
Jan 8 at 17:24
add a comment |
$begingroup$
One definition of $o(h)$ is as follows:
A matrix valued function $f(h)$ is $o(h)$ if $lim_{h to 0} frac{|f(h)|}{h} = 0$ (for some matrix norm $|cdot|$).
With this definition, I think you should find the result easy to prove.
answered Jan 8 at 17:17
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
$begingroup$
But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
$endgroup$
– model_checker
Jan 8 at 17:19
$begingroup$
The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
$endgroup$
– Omnomnomnom
Jan 8 at 17:22
$begingroup$
Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
$endgroup$
– Omnomnomnom
Jan 8 at 17:24
add a comment |
$begingroup$
One definition of $o(h)$ is as follows:
A matrix valued function $f(h)$ is $o(h)$ if $lim_{h to 0} frac{|f(h)|}{h} = 0$ (for some matrix norm $|cdot|$).
With this definition, I think you should find the result easy to prove.
answered Jan 8 at 17:17
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
One definition of $o(h)$ is as follows:
A matrix valued function $f(h)$ is $o(h)$ if $lim_{h to 0} frac{|f(h)|}{h} = 0$ (for some matrix norm $|cdot|$).
With this definition, I think you should find the result easy to prove.
answered Jan 8 at 17:17
OmnomnomnomOmnomnomnom
129k793187
answered Jan 8 at 17:17
OmnomnomnomOmnomnomnom
129k793187
answered Jan 8 at 17:17
OmnomnomnomOmnomnomnom
129k793187
answered Jan 8 at 17:17
OmnomnomnomOmnomnomnom
129k793187
129k793187
$begingroup$
But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
$endgroup$
– model_checker
Jan 8 at 17:19
$begingroup$
The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
$endgroup$
– Omnomnomnom
Jan 8 at 17:22
$begingroup$
Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
$endgroup$
– Omnomnomnom
Jan 8 at 17:24
add a comment |
$begingroup$
But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
$endgroup$
– model_checker
Jan 8 at 17:19
$begingroup$
The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
$endgroup$
– Omnomnomnom
Jan 8 at 17:22
$begingroup$
Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
$endgroup$
– Omnomnomnom
Jan 8 at 17:24
$begingroup$
But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
$endgroup$
– model_checker
Jan 8 at 17:19
$begingroup$
But the function $f(h)= e^{A}o(hB).$ How do I deal with the $o(hB)$ term?
$endgroup$
– model_checker
Jan 8 at 17:19
$begingroup$
The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
$endgroup$
– Omnomnomnom
Jan 8 at 17:22
$begingroup$
The $o(hB)$ term is a stand-in for a function $g(h)$ satisfying $$ lim_{h to 0} frac{|g(h)|}{|hB|} = 0 $$ However, this is equivalent to saying that $g$ is $o(h)$. So, we can simply replace $o(hB)$ with $o(h)$.
$endgroup$
– Omnomnomnom
Jan 8 at 17:22
$begingroup$
Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
$endgroup$
– Omnomnomnom
Jan 8 at 17:24
$begingroup$
Typically, however, I wouldn't use the notation $o(hB)$. Instead, I would say that $$ e^{hB} = I + hB + o(h) $$
$endgroup$
– Omnomnomnom
Jan 8 at 17:24
add a comment |
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