Exponential of two different derivatives
I know that
$$begin{align*}
expleft(alphafrac{d}{dx}right)f(x)=f(x+alpha),,
end{align*} tag{1}$$
but I am looking for a definition for
$$begin{align*}
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right)f(x,y),,
end{align*} tag{2}$$
so, the first question is: Is there a definition for the previous expression?.
Now, in the particular case where $f(x,y)$ is a product of two Gaussian functions centered in $y_{0}$ and $x_{0}$, that is $f(x,y)=expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]$ (with $k_{1}$ and $k_{2}$ being real or complex constants), so, the second question is: Is it valid to apply equation (1) in order to displace first a Gaussian function?, I mean
$$begin{align*}
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right)expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right],,
end{align*} tag{3}$$
taking $hat{c} = alphafrac{partial}{partial x}$, the above expresión is
$$begin{align*}
expleft(hat{c}frac{partial}{partial y}right)expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
,,
end{align*} tag{4}$$
Using equation (1) the $y$ variable is displaced
$$begin{align*}
expleft[-k_{2}(left[y+hat{c}right]-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
,,
end{align*} tag{5}$$
or
$$begin{align*}
=expleft[-k_{2}left(left[y+alphafrac{partial}{partial x}right]-y_{0}right)^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
end{align*} tag{6}$$
then, the first exponential of above equation will become an operator that will act on the second exponential. I think that this procedure is wrong because if the exponentials of equation (3) are swapping, I would get
$$begin{align*}
expleft[-k_{1}(x-x_{0})^{2}right]
expleft[-k_{2}left(left[y+alphafrac{partial}{partial x}right]-y_{0}right)^{2}right]
end{align*} tag{7}$$
which clearly is not the same as the equation (6).
calculus derivatives power-series taylor-expansion exponential-function
asked 2 days ago
Julio Abraham Mendoza Fierro
61
New contributor
Julio Abraham Mendoza Fierro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I know that
$$begin{align*}
expleft(alphafrac{d}{dx}right)f(x)=f(x+alpha),,
end{align*} tag{1}$$
but I am looking for a definition for
$$begin{align*}
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right)f(x,y),,
end{align*} tag{2}$$
so, the first question is: Is there a definition for the previous expression?.
Now, in the particular case where $f(x,y)$ is a product of two Gaussian functions centered in $y_{0}$ and $x_{0}$, that is $f(x,y)=expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]$ (with $k_{1}$ and $k_{2}$ being real or complex constants), so, the second question is: Is it valid to apply equation (1) in order to displace first a Gaussian function?, I mean
$$begin{align*}
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right)expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right],,
end{align*} tag{3}$$
taking $hat{c} = alphafrac{partial}{partial x}$, the above expresión is
$$begin{align*}
expleft(hat{c}frac{partial}{partial y}right)expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
,,
end{align*} tag{4}$$
Using equation (1) the $y$ variable is displaced
$$begin{align*}
expleft[-k_{2}(left[y+hat{c}right]-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
,,
end{align*} tag{5}$$
or
$$begin{align*}
=expleft[-k_{2}left(left[y+alphafrac{partial}{partial x}right]-y_{0}right)^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
end{align*} tag{6}$$
then, the first exponential of above equation will become an operator that will act on the second exponential. I think that this procedure is wrong because if the exponentials of equation (3) are swapping, I would get
$$begin{align*}
expleft[-k_{1}(x-x_{0})^{2}right]
expleft[-k_{2}left(left[y+alphafrac{partial}{partial x}right]-y_{0}right)^{2}right]
end{align*} tag{7}$$
which clearly is not the same as the equation (6).
calculus derivatives power-series taylor-expansion exponential-function
asked 2 days ago
Julio Abraham Mendoza Fierro
61
New contributor
Julio Abraham Mendoza Fierro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Basically what you want to do is to find the fundamental solution of the associated second order PDE which is an evolution equation defined by that operator $partial_{xy}$. Then your operator is just given by convolution with that fundamental solution.
– Shalop
2 days ago
You can solve the equation in Fourier space though I doubt the fundamental solutions will be function-valued, and I also doubt there is an explicit expression for inverting that Fourier transform. then again it could be possible.
– Shalop
2 days ago
add a comment |
I know that
$$begin{align*}
expleft(alphafrac{d}{dx}right)f(x)=f(x+alpha),,
end{align*} tag{1}$$
but I am looking for a definition for
$$begin{align*}
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right)f(x,y),,
end{align*} tag{2}$$
so, the first question is: Is there a definition for the previous expression?.
Now, in the particular case where $f(x,y)$ is a product of two Gaussian functions centered in $y_{0}$ and $x_{0}$, that is $f(x,y)=expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]$ (with $k_{1}$ and $k_{2}$ being real or complex constants), so, the second question is: Is it valid to apply equation (1) in order to displace first a Gaussian function?, I mean
$$begin{align*}
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right)expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right],,
end{align*} tag{3}$$
taking $hat{c} = alphafrac{partial}{partial x}$, the above expresión is
$$begin{align*}
expleft(hat{c}frac{partial}{partial y}right)expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
,,
end{align*} tag{4}$$
Using equation (1) the $y$ variable is displaced
$$begin{align*}
expleft[-k_{2}(left[y+hat{c}right]-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
,,
end{align*} tag{5}$$
or
$$begin{align*}
=expleft[-k_{2}left(left[y+alphafrac{partial}{partial x}right]-y_{0}right)^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
end{align*} tag{6}$$
then, the first exponential of above equation will become an operator that will act on the second exponential. I think that this procedure is wrong because if the exponentials of equation (3) are swapping, I would get
$$begin{align*}
expleft[-k_{1}(x-x_{0})^{2}right]
expleft[-k_{2}left(left[y+alphafrac{partial}{partial x}right]-y_{0}right)^{2}right]
end{align*} tag{7}$$
which clearly is not the same as the equation (6).
calculus derivatives power-series taylor-expansion exponential-function
asked 2 days ago
Julio Abraham Mendoza Fierro
61
New contributor
Julio Abraham Mendoza Fierro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I know that
$$begin{align*}
expleft(alphafrac{d}{dx}right)f(x)=f(x+alpha),,
end{align*} tag{1}$$
but I am looking for a definition for
$$begin{align*}
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right)f(x,y),,
end{align*} tag{2}$$
so, the first question is: Is there a definition for the previous expression?.
Now, in the particular case where $f(x,y)$ is a product of two Gaussian functions centered in $y_{0}$ and $x_{0}$, that is $f(x,y)=expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]$ (with $k_{1}$ and $k_{2}$ being real or complex constants), so, the second question is: Is it valid to apply equation (1) in order to displace first a Gaussian function?, I mean
$$begin{align*}
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right)expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right],,
end{align*} tag{3}$$
taking $hat{c} = alphafrac{partial}{partial x}$, the above expresión is
$$begin{align*}
expleft(hat{c}frac{partial}{partial y}right)expleft[-k_{2}(y-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
,,
end{align*} tag{4}$$
Using equation (1) the $y$ variable is displaced
$$begin{align*}
expleft[-k_{2}(left[y+hat{c}right]-y_{0})^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
,,
end{align*} tag{5}$$
or
$$begin{align*}
=expleft[-k_{2}left(left[y+alphafrac{partial}{partial x}right]-y_{0}right)^{2}right]expleft[-k_{1}(x-x_{0})^{2}right]
end{align*} tag{6}$$
then, the first exponential of above equation will become an operator that will act on the second exponential. I think that this procedure is wrong because if the exponentials of equation (3) are swapping, I would get
$$begin{align*}
expleft[-k_{1}(x-x_{0})^{2}right]
expleft[-k_{2}left(left[y+alphafrac{partial}{partial x}right]-y_{0}right)^{2}right]
end{align*} tag{7}$$
which clearly is not the same as the equation (6).
calculus derivatives power-series taylor-expansion exponential-function
calculus derivatives power-series taylor-expansion exponential-function
asked 2 days ago
Julio Abraham Mendoza Fierro
61
New contributor
Julio Abraham Mendoza Fierro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Julio Abraham Mendoza Fierro
61
New contributor
Julio Abraham Mendoza Fierro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Julio Abraham Mendoza Fierro
61
New contributor
Julio Abraham Mendoza Fierro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Julio Abraham Mendoza Fierro
61
asked 2 days ago
Julio Abraham Mendoza Fierro
61
61
New contributor
Julio Abraham Mendoza Fierro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Julio Abraham Mendoza Fierro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Julio Abraham Mendoza Fierro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Basically what you want to do is to find the fundamental solution of the associated second order PDE which is an evolution equation defined by that operator $partial_{xy}$. Then your operator is just given by convolution with that fundamental solution.
– Shalop
2 days ago
You can solve the equation in Fourier space though I doubt the fundamental solutions will be function-valued, and I also doubt there is an explicit expression for inverting that Fourier transform. then again it could be possible.
– Shalop
2 days ago
add a comment |
1
Basically what you want to do is to find the fundamental solution of the associated second order PDE which is an evolution equation defined by that operator $partial_{xy}$. Then your operator is just given by convolution with that fundamental solution.
– Shalop
2 days ago
You can solve the equation in Fourier space though I doubt the fundamental solutions will be function-valued, and I also doubt there is an explicit expression for inverting that Fourier transform. then again it could be possible.
– Shalop
2 days ago
1
1
Basically what you want to do is to find the fundamental solution of the associated second order PDE which is an evolution equation defined by that operator $partial_{xy}$. Then your operator is just given by convolution with that fundamental solution.
– Shalop
2 days ago
Basically what you want to do is to find the fundamental solution of the associated second order PDE which is an evolution equation defined by that operator $partial_{xy}$. Then your operator is just given by convolution with that fundamental solution.
– Shalop
2 days ago
You can solve the equation in Fourier space though I doubt the fundamental solutions will be function-valued, and I also doubt there is an explicit expression for inverting that Fourier transform. then again it could be possible.
– Shalop
2 days ago
You can solve the equation in Fourier space though I doubt the fundamental solutions will be function-valued, and I also doubt there is an explicit expression for inverting that Fourier transform. then again it could be possible.
– Shalop
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
Expressions like $exp(alpha T),$ where $T$ is some operator and $alpha$ is a constant, are defined by the Maclaurin expansion:
$$
exp(alpha T) = sum_{n=0}^{infty} frac{1}{n!} alpha^n T^n
$$
Therefore,
$$
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right) f(x,y)
= sum_{n=0}^{infty} frac{1}{n!} alpha^n left( frac{partial}{partial x} frac{partial}{partial y} right)^n f(x,y)
= sum_{n=0}^{infty} frac{1}{n!} alpha^n left( frac{partial}{partial x} right)^n left( frac{partial}{partial y} right)^n f(x,y),
$$
where the last equality is valid since partial derivatives commute if $f$ is smooth.
answered 2 days ago
md2perpe
7,53311027
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
Expressions like $exp(alpha T),$ where $T$ is some operator and $alpha$ is a constant, are defined by the Maclaurin expansion:
$$
exp(alpha T) = sum_{n=0}^{infty} frac{1}{n!} alpha^n T^n
$$
Therefore,
$$
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right) f(x,y)
= sum_{n=0}^{infty} frac{1}{n!} alpha^n left( frac{partial}{partial x} frac{partial}{partial y} right)^n f(x,y)
= sum_{n=0}^{infty} frac{1}{n!} alpha^n left( frac{partial}{partial x} right)^n left( frac{partial}{partial y} right)^n f(x,y),
$$
where the last equality is valid since partial derivatives commute if $f$ is smooth.
answered 2 days ago
md2perpe
7,53311027
add a comment |
Expressions like $exp(alpha T),$ where $T$ is some operator and $alpha$ is a constant, are defined by the Maclaurin expansion:
$$
exp(alpha T) = sum_{n=0}^{infty} frac{1}{n!} alpha^n T^n
$$
Therefore,
$$
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right) f(x,y)
= sum_{n=0}^{infty} frac{1}{n!} alpha^n left( frac{partial}{partial x} frac{partial}{partial y} right)^n f(x,y)
= sum_{n=0}^{infty} frac{1}{n!} alpha^n left( frac{partial}{partial x} right)^n left( frac{partial}{partial y} right)^n f(x,y),
$$
where the last equality is valid since partial derivatives commute if $f$ is smooth.
answered 2 days ago
md2perpe
7,53311027
add a comment |
Expressions like $exp(alpha T),$ where $T$ is some operator and $alpha$ is a constant, are defined by the Maclaurin expansion:
$$
exp(alpha T) = sum_{n=0}^{infty} frac{1}{n!} alpha^n T^n
$$
Therefore,
$$
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right) f(x,y)
= sum_{n=0}^{infty} frac{1}{n!} alpha^n left( frac{partial}{partial x} frac{partial}{partial y} right)^n f(x,y)
= sum_{n=0}^{infty} frac{1}{n!} alpha^n left( frac{partial}{partial x} right)^n left( frac{partial}{partial y} right)^n f(x,y),
$$
where the last equality is valid since partial derivatives commute if $f$ is smooth.
answered 2 days ago
md2perpe
7,53311027
Expressions like $exp(alpha T),$ where $T$ is some operator and $alpha$ is a constant, are defined by the Maclaurin expansion:
$$
exp(alpha T) = sum_{n=0}^{infty} frac{1}{n!} alpha^n T^n
$$
Therefore,
$$
expleft(alphafrac{partial}{partial x}frac{partial}{partial y}right) f(x,y)
= sum_{n=0}^{infty} frac{1}{n!} alpha^n left( frac{partial}{partial x} frac{partial}{partial y} right)^n f(x,y)
= sum_{n=0}^{infty} frac{1}{n!} alpha^n left( frac{partial}{partial x} right)^n left( frac{partial}{partial y} right)^n f(x,y),
$$
where the last equality is valid since partial derivatives commute if $f$ is smooth.
answered 2 days ago
md2perpe
7,53311027
answered 2 days ago
md2perpe
7,53311027
answered 2 days ago
md2perpe
7,53311027
answered 2 days ago
md2perpe
7,53311027
7,53311027
add a comment |
add a comment |
Julio Abraham Mendoza Fierro is a new contributor. Be nice, and check out our Code of Conduct.
Julio Abraham Mendoza Fierro is a new contributor. Be nice, and check out our Code of Conduct.
Julio Abraham Mendoza Fierro is a new contributor. Be nice, and check out our Code of Conduct.
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1
Basically what you want to do is to find the fundamental solution of the associated second order PDE which is an evolution equation defined by that operator $partial_{xy}$. Then your operator is just given by convolution with that fundamental solution.
– Shalop
2 days ago
You can solve the equation in Fourier space though I doubt the fundamental solutions will be function-valued, and I also doubt there is an explicit expression for inverting that Fourier transform. then again it could be possible.
– Shalop
2 days ago