Mixing
Exponential of the product between $x$ the derivative operator of $x$ acting in a $f(x)$
$begingroup$
The question
I'm stuck here trying to figure out how to compute and prove, the following operator action in a function:
$exp(varepsilon x partial_x) f(x) = f(x exp(varepsilon) )$
where $varepsilon$ is a constant.
I saw this result and I failed in my attempt to reproduced it. What I did was to expand $exp(varepsilon x partial_x)$ in Taylor's series as:
$begin{align*}
exp(varepsilon x partial_x)f(x)& = sum_{m=0}^{infty}frac{1}{m!}(varepsilon x partial_x)^m f(x)\
&= sum_{m=0}^{infty}frac{1}{m!}(varepsilon x)^mfrac{partial^m}{partial x^m}f(x) \
end{align*}$
I took this procedure because I already know how to compute $e^{partial_x}f(x)$. Let me show you what I did in this case:
The translation operator
The Taylor series of a function f is
begin{equation}
f(x)=sum_{n=0}^inftyfrac{(partial_x^nf)(a)}{n!}(x-a)^n
end{equation}
Expanding about $x+b$ and letting $a=x$:
begin{equation}
f(x+b)=sum_{n=0}^inftyfrac{(partial_x^nf)(x)}{n!}b^n=sum_{n=0}^inftyfrac{((bpartial_x)^nf)(x)}{n!}
end{equation}
By definition:
begin{equation}
e^{bpartial_x}=sum_{n=0}^inftyfrac{(bpartial_x)^n}{n!}
end{equation}
Hence
begin{equation}
f(x+b)=(e^{bpartial_x}f)(x)
end{equation}
Returning to my question
I tried to generalize or make something similar for the previous case I discussed but I didn't get anywhere. Anyone can give me a tip or recommend a book or paper?
Thanks!!
taylor-expansion lie-algebras operator-algebras differential-operators
asked Feb 1 at 13:43
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
$endgroup$
add a comment |
$begingroup$
The question
I'm stuck here trying to figure out how to compute and prove, the following operator action in a function:
$exp(varepsilon x partial_x) f(x) = f(x exp(varepsilon) )$
where $varepsilon$ is a constant.
I saw this result and I failed in my attempt to reproduced it. What I did was to expand $exp(varepsilon x partial_x)$ in Taylor's series as:
$begin{align*}
exp(varepsilon x partial_x)f(x)& = sum_{m=0}^{infty}frac{1}{m!}(varepsilon x partial_x)^m f(x)\
&= sum_{m=0}^{infty}frac{1}{m!}(varepsilon x)^mfrac{partial^m}{partial x^m}f(x) \
end{align*}$
I took this procedure because I already know how to compute $e^{partial_x}f(x)$. Let me show you what I did in this case:
The translation operator
The Taylor series of a function f is
begin{equation}
f(x)=sum_{n=0}^inftyfrac{(partial_x^nf)(a)}{n!}(x-a)^n
end{equation}
Expanding about $x+b$ and letting $a=x$:
begin{equation}
f(x+b)=sum_{n=0}^inftyfrac{(partial_x^nf)(x)}{n!}b^n=sum_{n=0}^inftyfrac{((bpartial_x)^nf)(x)}{n!}
end{equation}
By definition:
begin{equation}
e^{bpartial_x}=sum_{n=0}^inftyfrac{(bpartial_x)^n}{n!}
end{equation}
Hence
begin{equation}
f(x+b)=(e^{bpartial_x}f)(x)
end{equation}
Returning to my question
I tried to generalize or make something similar for the previous case I discussed but I didn't get anywhere. Anyone can give me a tip or recommend a book or paper?
Thanks!!
taylor-expansion lie-algebras operator-algebras differential-operators
asked Feb 1 at 13:43
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
$endgroup$
add a comment |
$begingroup$
The question
I'm stuck here trying to figure out how to compute and prove, the following operator action in a function:
$exp(varepsilon x partial_x) f(x) = f(x exp(varepsilon) )$
where $varepsilon$ is a constant.
I saw this result and I failed in my attempt to reproduced it. What I did was to expand $exp(varepsilon x partial_x)$ in Taylor's series as:
$begin{align*}
exp(varepsilon x partial_x)f(x)& = sum_{m=0}^{infty}frac{1}{m!}(varepsilon x partial_x)^m f(x)\
&= sum_{m=0}^{infty}frac{1}{m!}(varepsilon x)^mfrac{partial^m}{partial x^m}f(x) \
end{align*}$
I took this procedure because I already know how to compute $e^{partial_x}f(x)$. Let me show you what I did in this case:
The translation operator
The Taylor series of a function f is
begin{equation}
f(x)=sum_{n=0}^inftyfrac{(partial_x^nf)(a)}{n!}(x-a)^n
end{equation}
Expanding about $x+b$ and letting $a=x$:
begin{equation}
f(x+b)=sum_{n=0}^inftyfrac{(partial_x^nf)(x)}{n!}b^n=sum_{n=0}^inftyfrac{((bpartial_x)^nf)(x)}{n!}
end{equation}
By definition:
begin{equation}
e^{bpartial_x}=sum_{n=0}^inftyfrac{(bpartial_x)^n}{n!}
end{equation}
Hence
begin{equation}
f(x+b)=(e^{bpartial_x}f)(x)
end{equation}
Returning to my question
I tried to generalize or make something similar for the previous case I discussed but I didn't get anywhere. Anyone can give me a tip or recommend a book or paper?
Thanks!!
taylor-expansion lie-algebras operator-algebras differential-operators
asked Feb 1 at 13:43
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
$endgroup$
The question
I'm stuck here trying to figure out how to compute and prove, the following operator action in a function:
$exp(varepsilon x partial_x) f(x) = f(x exp(varepsilon) )$
where $varepsilon$ is a constant.
I saw this result and I failed in my attempt to reproduced it. What I did was to expand $exp(varepsilon x partial_x)$ in Taylor's series as:
$begin{align*}
exp(varepsilon x partial_x)f(x)& = sum_{m=0}^{infty}frac{1}{m!}(varepsilon x partial_x)^m f(x)\
&= sum_{m=0}^{infty}frac{1}{m!}(varepsilon x)^mfrac{partial^m}{partial x^m}f(x) \
end{align*}$
I took this procedure because I already know how to compute $e^{partial_x}f(x)$. Let me show you what I did in this case:
The translation operator
The Taylor series of a function f is
begin{equation}
f(x)=sum_{n=0}^inftyfrac{(partial_x^nf)(a)}{n!}(x-a)^n
end{equation}
Expanding about $x+b$ and letting $a=x$:
begin{equation}
f(x+b)=sum_{n=0}^inftyfrac{(partial_x^nf)(x)}{n!}b^n=sum_{n=0}^inftyfrac{((bpartial_x)^nf)(x)}{n!}
end{equation}
By definition:
begin{equation}
e^{bpartial_x}=sum_{n=0}^inftyfrac{(bpartial_x)^n}{n!}
end{equation}
Hence
begin{equation}
f(x+b)=(e^{bpartial_x}f)(x)
end{equation}
Returning to my question
I tried to generalize or make something similar for the previous case I discussed but I didn't get anywhere. Anyone can give me a tip or recommend a book or paper?
Thanks!!
taylor-expansion lie-algebras operator-algebras differential-operators
taylor-expansion lie-algebras operator-algebras differential-operators
asked Feb 1 at 13:43
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
asked Feb 1 at 13:43
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
asked Feb 1 at 13:43
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
asked Feb 1 at 13:43
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
asked Feb 1 at 13:43
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
163
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are assuming that multiplication by $x$ and the derivative commute, and that's not the case. For clarity, let me write $M_x$ for the operator of multiplication by $x$. If $f(x)=x^k$, then
$$
[M_xpartial_x f](x)=kx^k, [(M_xpartial_x)^2f](x)=k^2x^k, cdots , [(M_xpartial_x)^mf](x)=k^mx^k.
$$
Then
begin{align*}
[exp(varepsilon x partial_x)f](x)& = sum_{m=0}^{infty}frac{1}{m!}(varepsilon x partial_x)^m f(x)\
&= sum_{m=0}^{infty}frac{1}{m!}(varepsilon )^mk^mx^k \
&=x^k,exp(varepsilon k)=x^k (exp(varepsilon)^k\
&=f(xexp(varepsilon)).
end{align*}
Thus we get by linearity that, for any polynomial $p$,
$$tag1
[exp(varepsilon x partial_x)p](x)=p(xexp(varepsilon)).
$$
Analytic functions are differentiable term by term, so the differential operator gets inside the series. This allows us to extend $(1)$ to $f$ analytic.
answered Feb 1 at 18:24
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
When I tried to perform the solution, I didn't notice the commutation relation between $x$ and $partial_x$. Very thank you for the solution, Martin.
$endgroup$
– Alexssandre de Oliveira J
Feb 1 at 19:30
add a comment |
$begingroup$
Before Martin Argerami came up with a solution I friend of mine thought in something but he's not so sure if it's right... I like to think that it is a physicist solution. I'll explain why, but first, let me show you what he thought.
If we assume that $f(x)$ has a Fourier Transform, we can write:
begin{align*}
e^{varepsilon x partial_x}f(x) &= int d^3k , f(k)e^{varepsilon x partial_x}e^{ikx}\
&= int d^3k, f(k) sum_{n,m} frac{(varepsilon x partial_x)}{m!}frac{ikx}{n!}
end{align*}
Notice that
begin{align*}
(x partial_x)^m x^n = n^m x^n , , ,
end{align*}
where commutation relation is respect! And then we can resum the series in $m$ and $n$ to obtain:
begin{align*}
e^{varepsilon x partial_x}f(x) &= int d^3k, f(k) sum_{n} e^{(varepsilon)^n}frac{(ikx)^n}{n!}\
&= int d^3k, f(k) sum_{n} frac{(e^{varepsilon}ikx)^n}{n!} \
&= int d^3k , f(k) e^{ike^varepsilon x} = f(e^epsilon x)
end{align*}
It's a physicist solution because we're assuming that every operation is well defined.
answered Feb 1 at 20:01
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
$endgroup$
add a comment |
Your Answer
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
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oldest
votes
$begingroup$
You are assuming that multiplication by $x$ and the derivative commute, and that's not the case. For clarity, let me write $M_x$ for the operator of multiplication by $x$. If $f(x)=x^k$, then
$$
[M_xpartial_x f](x)=kx^k, [(M_xpartial_x)^2f](x)=k^2x^k, cdots , [(M_xpartial_x)^mf](x)=k^mx^k.
$$
Then
begin{align*}
[exp(varepsilon x partial_x)f](x)& = sum_{m=0}^{infty}frac{1}{m!}(varepsilon x partial_x)^m f(x)\
&= sum_{m=0}^{infty}frac{1}{m!}(varepsilon )^mk^mx^k \
&=x^k,exp(varepsilon k)=x^k (exp(varepsilon)^k\
&=f(xexp(varepsilon)).
end{align*}
Thus we get by linearity that, for any polynomial $p$,
$$tag1
[exp(varepsilon x partial_x)p](x)=p(xexp(varepsilon)).
$$
Analytic functions are differentiable term by term, so the differential operator gets inside the series. This allows us to extend $(1)$ to $f$ analytic.
answered Feb 1 at 18:24
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
When I tried to perform the solution, I didn't notice the commutation relation between $x$ and $partial_x$. Very thank you for the solution, Martin.
$endgroup$
– Alexssandre de Oliveira J
Feb 1 at 19:30
add a comment |
$begingroup$
You are assuming that multiplication by $x$ and the derivative commute, and that's not the case. For clarity, let me write $M_x$ for the operator of multiplication by $x$. If $f(x)=x^k$, then
$$
[M_xpartial_x f](x)=kx^k, [(M_xpartial_x)^2f](x)=k^2x^k, cdots , [(M_xpartial_x)^mf](x)=k^mx^k.
$$
Then
begin{align*}
[exp(varepsilon x partial_x)f](x)& = sum_{m=0}^{infty}frac{1}{m!}(varepsilon x partial_x)^m f(x)\
&= sum_{m=0}^{infty}frac{1}{m!}(varepsilon )^mk^mx^k \
&=x^k,exp(varepsilon k)=x^k (exp(varepsilon)^k\
&=f(xexp(varepsilon)).
end{align*}
Thus we get by linearity that, for any polynomial $p$,
$$tag1
[exp(varepsilon x partial_x)p](x)=p(xexp(varepsilon)).
$$
Analytic functions are differentiable term by term, so the differential operator gets inside the series. This allows us to extend $(1)$ to $f$ analytic.
answered Feb 1 at 18:24
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
When I tried to perform the solution, I didn't notice the commutation relation between $x$ and $partial_x$. Very thank you for the solution, Martin.
$endgroup$
– Alexssandre de Oliveira J
Feb 1 at 19:30
add a comment |
$begingroup$
You are assuming that multiplication by $x$ and the derivative commute, and that's not the case. For clarity, let me write $M_x$ for the operator of multiplication by $x$. If $f(x)=x^k$, then
$$
[M_xpartial_x f](x)=kx^k, [(M_xpartial_x)^2f](x)=k^2x^k, cdots , [(M_xpartial_x)^mf](x)=k^mx^k.
$$
Then
begin{align*}
[exp(varepsilon x partial_x)f](x)& = sum_{m=0}^{infty}frac{1}{m!}(varepsilon x partial_x)^m f(x)\
&= sum_{m=0}^{infty}frac{1}{m!}(varepsilon )^mk^mx^k \
&=x^k,exp(varepsilon k)=x^k (exp(varepsilon)^k\
&=f(xexp(varepsilon)).
end{align*}
Thus we get by linearity that, for any polynomial $p$,
$$tag1
[exp(varepsilon x partial_x)p](x)=p(xexp(varepsilon)).
$$
Analytic functions are differentiable term by term, so the differential operator gets inside the series. This allows us to extend $(1)$ to $f$ analytic.
answered Feb 1 at 18:24
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
You are assuming that multiplication by $x$ and the derivative commute, and that's not the case. For clarity, let me write $M_x$ for the operator of multiplication by $x$. If $f(x)=x^k$, then
$$
[M_xpartial_x f](x)=kx^k, [(M_xpartial_x)^2f](x)=k^2x^k, cdots , [(M_xpartial_x)^mf](x)=k^mx^k.
$$
Then
begin{align*}
[exp(varepsilon x partial_x)f](x)& = sum_{m=0}^{infty}frac{1}{m!}(varepsilon x partial_x)^m f(x)\
&= sum_{m=0}^{infty}frac{1}{m!}(varepsilon )^mk^mx^k \
&=x^k,exp(varepsilon k)=x^k (exp(varepsilon)^k\
&=f(xexp(varepsilon)).
end{align*}
Thus we get by linearity that, for any polynomial $p$,
$$tag1
[exp(varepsilon x partial_x)p](x)=p(xexp(varepsilon)).
$$
Analytic functions are differentiable term by term, so the differential operator gets inside the series. This allows us to extend $(1)$ to $f$ analytic.
answered Feb 1 at 18:24
Martin ArgeramiMartin Argerami
129k1184185
answered Feb 1 at 18:24
Martin ArgeramiMartin Argerami
129k1184185
answered Feb 1 at 18:24
Martin ArgeramiMartin Argerami
129k1184185
answered Feb 1 at 18:24
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
$begingroup$
When I tried to perform the solution, I didn't notice the commutation relation between $x$ and $partial_x$. Very thank you for the solution, Martin.
$endgroup$
– Alexssandre de Oliveira J
Feb 1 at 19:30
add a comment |
$begingroup$
When I tried to perform the solution, I didn't notice the commutation relation between $x$ and $partial_x$. Very thank you for the solution, Martin.
$endgroup$
– Alexssandre de Oliveira J
Feb 1 at 19:30
$begingroup$
When I tried to perform the solution, I didn't notice the commutation relation between $x$ and $partial_x$. Very thank you for the solution, Martin.
$endgroup$
– Alexssandre de Oliveira J
Feb 1 at 19:30
$begingroup$
When I tried to perform the solution, I didn't notice the commutation relation between $x$ and $partial_x$. Very thank you for the solution, Martin.
$endgroup$
– Alexssandre de Oliveira J
Feb 1 at 19:30
add a comment |
$begingroup$
Before Martin Argerami came up with a solution I friend of mine thought in something but he's not so sure if it's right... I like to think that it is a physicist solution. I'll explain why, but first, let me show you what he thought.
If we assume that $f(x)$ has a Fourier Transform, we can write:
begin{align*}
e^{varepsilon x partial_x}f(x) &= int d^3k , f(k)e^{varepsilon x partial_x}e^{ikx}\
&= int d^3k, f(k) sum_{n,m} frac{(varepsilon x partial_x)}{m!}frac{ikx}{n!}
end{align*}
Notice that
begin{align*}
(x partial_x)^m x^n = n^m x^n , , ,
end{align*}
where commutation relation is respect! And then we can resum the series in $m$ and $n$ to obtain:
begin{align*}
e^{varepsilon x partial_x}f(x) &= int d^3k, f(k) sum_{n} e^{(varepsilon)^n}frac{(ikx)^n}{n!}\
&= int d^3k, f(k) sum_{n} frac{(e^{varepsilon}ikx)^n}{n!} \
&= int d^3k , f(k) e^{ike^varepsilon x} = f(e^epsilon x)
end{align*}
It's a physicist solution because we're assuming that every operation is well defined.
answered Feb 1 at 20:01
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
$endgroup$
add a comment |
$begingroup$
Before Martin Argerami came up with a solution I friend of mine thought in something but he's not so sure if it's right... I like to think that it is a physicist solution. I'll explain why, but first, let me show you what he thought.
If we assume that $f(x)$ has a Fourier Transform, we can write:
begin{align*}
e^{varepsilon x partial_x}f(x) &= int d^3k , f(k)e^{varepsilon x partial_x}e^{ikx}\
&= int d^3k, f(k) sum_{n,m} frac{(varepsilon x partial_x)}{m!}frac{ikx}{n!}
end{align*}
Notice that
begin{align*}
(x partial_x)^m x^n = n^m x^n , , ,
end{align*}
where commutation relation is respect! And then we can resum the series in $m$ and $n$ to obtain:
begin{align*}
e^{varepsilon x partial_x}f(x) &= int d^3k, f(k) sum_{n} e^{(varepsilon)^n}frac{(ikx)^n}{n!}\
&= int d^3k, f(k) sum_{n} frac{(e^{varepsilon}ikx)^n}{n!} \
&= int d^3k , f(k) e^{ike^varepsilon x} = f(e^epsilon x)
end{align*}
It's a physicist solution because we're assuming that every operation is well defined.
answered Feb 1 at 20:01
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
$endgroup$
add a comment |
$begingroup$
Before Martin Argerami came up with a solution I friend of mine thought in something but he's not so sure if it's right... I like to think that it is a physicist solution. I'll explain why, but first, let me show you what he thought.
If we assume that $f(x)$ has a Fourier Transform, we can write:
begin{align*}
e^{varepsilon x partial_x}f(x) &= int d^3k , f(k)e^{varepsilon x partial_x}e^{ikx}\
&= int d^3k, f(k) sum_{n,m} frac{(varepsilon x partial_x)}{m!}frac{ikx}{n!}
end{align*}
Notice that
begin{align*}
(x partial_x)^m x^n = n^m x^n , , ,
end{align*}
where commutation relation is respect! And then we can resum the series in $m$ and $n$ to obtain:
begin{align*}
e^{varepsilon x partial_x}f(x) &= int d^3k, f(k) sum_{n} e^{(varepsilon)^n}frac{(ikx)^n}{n!}\
&= int d^3k, f(k) sum_{n} frac{(e^{varepsilon}ikx)^n}{n!} \
&= int d^3k , f(k) e^{ike^varepsilon x} = f(e^epsilon x)
end{align*}
It's a physicist solution because we're assuming that every operation is well defined.
answered Feb 1 at 20:01
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
$endgroup$
Before Martin Argerami came up with a solution I friend of mine thought in something but he's not so sure if it's right... I like to think that it is a physicist solution. I'll explain why, but first, let me show you what he thought.
If we assume that $f(x)$ has a Fourier Transform, we can write:
begin{align*}
e^{varepsilon x partial_x}f(x) &= int d^3k , f(k)e^{varepsilon x partial_x}e^{ikx}\
&= int d^3k, f(k) sum_{n,m} frac{(varepsilon x partial_x)}{m!}frac{ikx}{n!}
end{align*}
Notice that
begin{align*}
(x partial_x)^m x^n = n^m x^n , , ,
end{align*}
where commutation relation is respect! And then we can resum the series in $m$ and $n$ to obtain:
begin{align*}
e^{varepsilon x partial_x}f(x) &= int d^3k, f(k) sum_{n} e^{(varepsilon)^n}frac{(ikx)^n}{n!}\
&= int d^3k, f(k) sum_{n} frac{(e^{varepsilon}ikx)^n}{n!} \
&= int d^3k , f(k) e^{ike^varepsilon x} = f(e^epsilon x)
end{align*}
It's a physicist solution because we're assuming that every operation is well defined.
answered Feb 1 at 20:01
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
answered Feb 1 at 20:01
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
answered Feb 1 at 20:01
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
answered Feb 1 at 20:01
Alexssandre de Oliveira JAlexssandre de Oliveira J
163
163
add a comment |
add a comment |
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