Exponential of two different derivatives












1














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







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Julio Abraham Mendoza Fierro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
















1














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







share|cite|improve this question







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














1












1








1







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







share|cite|improve this question







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






share|cite|improve this question







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.











share|cite|improve this question







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.









share|cite|improve this question




share|cite|improve this question






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




61




New contributor




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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.






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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














  • 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










1 Answer
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oldest

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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.






share|cite|improve this answer





















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    1 Answer
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    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.






    share|cite|improve this answer


























      1














      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.






      share|cite|improve this answer
























        1












        1








        1






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        md2perpe

        7,53311027




        7,53311027






















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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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