How to make it formally correct?












2














Can someone help me formalizing this statement:



$$
z= x^0 +ix^1
$$
And therefore
$$
frac{partial}{partial z} = frac{partial}{partial (x^0 +ix^1)} = frac{partial}{partial x^0} + frac{1}{i} frac{partial}{partial x^1}
$$



My problem is with the last equality, I see it's right, but I'm not sure I'm allowed to do it in that way. Is there a way to procede more formally?






calculus analysis multivariable-calculus partial-derivative formal-proofs







share|cite|improve this question


















  • 1




    It is customary to use $x$ and $y$ rather than $x^0$ and $x^1$ when talking about complex numbers. I've mostly seen $x^k$ notation when making arguments about unknown / arbitrary bases, not with concrete bases like this one. That being said, I don't really see anything wrong with it.
    – Arthur
    Nov 27 at 12:15












  • @Arthur Should I edit it using $x$ and $y$ to put it in a standard notation or it's fine?
    – Run like hell
    Nov 27 at 12:25












  • Nah. It's fine. I just found it a small curiousity. Some times unusual notation is bad. This time it's not.
    – Arthur
    Nov 27 at 12:29












  • @Runlikehell I have removed my comment, as I feel that the newly posted answer explains the situation more correctly. My apologies.
    – MisterRiemann
    Nov 27 at 12:39


















2














Can someone help me formalizing this statement:



$$
z= x^0 +ix^1
$$
And therefore
$$
frac{partial}{partial z} = frac{partial}{partial (x^0 +ix^1)} = frac{partial}{partial x^0} + frac{1}{i} frac{partial}{partial x^1}
$$



My problem is with the last equality, I see it's right, but I'm not sure I'm allowed to do it in that way. Is there a way to procede more formally?






calculus analysis multivariable-calculus partial-derivative formal-proofs







share|cite|improve this question


















  • 1




    It is customary to use $x$ and $y$ rather than $x^0$ and $x^1$ when talking about complex numbers. I've mostly seen $x^k$ notation when making arguments about unknown / arbitrary bases, not with concrete bases like this one. That being said, I don't really see anything wrong with it.
    – Arthur
    Nov 27 at 12:15












  • @Arthur Should I edit it using $x$ and $y$ to put it in a standard notation or it's fine?
    – Run like hell
    Nov 27 at 12:25












  • Nah. It's fine. I just found it a small curiousity. Some times unusual notation is bad. This time it's not.
    – Arthur
    Nov 27 at 12:29












  • @Runlikehell I have removed my comment, as I feel that the newly posted answer explains the situation more correctly. My apologies.
    – MisterRiemann
    Nov 27 at 12:39
















2












2








2







Can someone help me formalizing this statement:



$$
z= x^0 +ix^1
$$
And therefore
$$
frac{partial}{partial z} = frac{partial}{partial (x^0 +ix^1)} = frac{partial}{partial x^0} + frac{1}{i} frac{partial}{partial x^1}
$$



My problem is with the last equality, I see it's right, but I'm not sure I'm allowed to do it in that way. Is there a way to procede more formally?






calculus analysis multivariable-calculus partial-derivative formal-proofs







share|cite|improve this question













Can someone help me formalizing this statement:



$$
z= x^0 +ix^1
$$
And therefore
$$
frac{partial}{partial z} = frac{partial}{partial (x^0 +ix^1)} = frac{partial}{partial x^0} + frac{1}{i} frac{partial}{partial x^1}
$$



My problem is with the last equality, I see it's right, but I'm not sure I'm allowed to do it in that way. Is there a way to procede more formally?






calculus analysis multivariable-calculus partial-derivative formal-proofs




calculus analysis multivariable-calculus partial-derivative formal-proofs






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 27 at 12:10









Run like hell

1405




1405








  • 1




    It is customary to use $x$ and $y$ rather than $x^0$ and $x^1$ when talking about complex numbers. I've mostly seen $x^k$ notation when making arguments about unknown / arbitrary bases, not with concrete bases like this one. That being said, I don't really see anything wrong with it.
    – Arthur
    Nov 27 at 12:15












  • @Arthur Should I edit it using $x$ and $y$ to put it in a standard notation or it's fine?
    – Run like hell
    Nov 27 at 12:25












  • Nah. It's fine. I just found it a small curiousity. Some times unusual notation is bad. This time it's not.
    – Arthur
    Nov 27 at 12:29












  • @Runlikehell I have removed my comment, as I feel that the newly posted answer explains the situation more correctly. My apologies.
    – MisterRiemann
    Nov 27 at 12:39
















  • 1




    It is customary to use $x$ and $y$ rather than $x^0$ and $x^1$ when talking about complex numbers. I've mostly seen $x^k$ notation when making arguments about unknown / arbitrary bases, not with concrete bases like this one. That being said, I don't really see anything wrong with it.
    – Arthur
    Nov 27 at 12:15












  • @Arthur Should I edit it using $x$ and $y$ to put it in a standard notation or it's fine?
    – Run like hell
    Nov 27 at 12:25












  • Nah. It's fine. I just found it a small curiousity. Some times unusual notation is bad. This time it's not.
    – Arthur
    Nov 27 at 12:29












  • @Runlikehell I have removed my comment, as I feel that the newly posted answer explains the situation more correctly. My apologies.
    – MisterRiemann
    Nov 27 at 12:39










1




1




It is customary to use $x$ and $y$ rather than $x^0$ and $x^1$ when talking about complex numbers. I've mostly seen $x^k$ notation when making arguments about unknown / arbitrary bases, not with concrete bases like this one. That being said, I don't really see anything wrong with it.
– Arthur
Nov 27 at 12:15






It is customary to use $x$ and $y$ rather than $x^0$ and $x^1$ when talking about complex numbers. I've mostly seen $x^k$ notation when making arguments about unknown / arbitrary bases, not with concrete bases like this one. That being said, I don't really see anything wrong with it.
– Arthur
Nov 27 at 12:15














@Arthur Should I edit it using $x$ and $y$ to put it in a standard notation or it's fine?
– Run like hell
Nov 27 at 12:25






@Arthur Should I edit it using $x$ and $y$ to put it in a standard notation or it's fine?
– Run like hell
Nov 27 at 12:25














Nah. It's fine. I just found it a small curiousity. Some times unusual notation is bad. This time it's not.
– Arthur
Nov 27 at 12:29






Nah. It's fine. I just found it a small curiousity. Some times unusual notation is bad. This time it's not.
– Arthur
Nov 27 at 12:29














@Runlikehell I have removed my comment, as I feel that the newly posted answer explains the situation more correctly. My apologies.
– MisterRiemann
Nov 27 at 12:39






@Runlikehell I have removed my comment, as I feel that the newly posted answer explains the situation more correctly. My apologies.
– MisterRiemann
Nov 27 at 12:39












1 Answer
1






active

oldest

votes


















2














Not sure it's correct, as usually one defines:
$$
df(z)=frac{partial f}{partial z}dz+frac{partial f}{partial z^*}dz^*
$$
where $frac{partial}{partial z}$ and $frac{partial}{partial z^*}$ are called Wirtinger derivatives and are defined by:
$$
frac{partial}{partial z}:=frac{1}{2}left(frac{partial}{partial x^0}-ifrac{partial}{partial x^1}right)
$$
$$
frac{partial}{partial z^*}:=frac{1}{2}left(frac{partial}{partial x^0}+ifrac{partial}{partial x^1}right)
$$



Compared to your proposal, there is a 1/2 missing factor.




  • this is ok for holomorphic functions (thanks to Cauchy-Riemann)

  • but not in the general case


I can copy derivation from An Introduction to Complex Differentials and
Complex Differentiability p8-9, however this tutorial is really well written and I think it is better to directly read it.




From $F(x,y)=U(x,y)+iV(x,y)$ the bivariate function associated to $f(z)$, the total differential is:
$$
dF=frac{partial}{partial x}F(x,y)dx+frac{partial}{partial y}F(x,y)dy
$$
which can be rewritten as follows (have a look at the cited reference):
$$
dF=frac{1}{2}left[frac{partial U}{partial x}+frac{partial V}{partial y}+ileft(frac{partial V}{partial x}-frac{partial U}{partial y}right)right]dz+frac{1}{2}left[frac{partial U}{partial x}-frac{partial V}{partial y}+ileft(frac{partial V}{partial x}+frac{partial U}{partial y}right)right]dz^*
$$
Assuming $f$ holomorphic and using Cauchy-Riemann conditions ($frac{partial U}{partial x}=frac{partial V}{partial y}$ and $frac{partial U}{partial y}=-frac{partial V}{partial x}$)
we get your suggestion:
$frac{partial}{partial z}=frac{partial}{partial x^0}-ifrac{partial}{partial x^1}$ and $frac{partial}{partial z^*}=0$




One of the big advantage of Wirtinger calculus is:




For the Wirtinger derivatives, the common rules for differentiation
known from real-valued analysis concerning the sum, product, and
composition of two functions hold as well...




in brief you can do differential calculus like in $mathbb{R}^n$. This is very handy for automatic differentiation, gradient computation... etc.






share|cite|improve this answer



















  • 1




    Thank you, so I guess my intuition was wrong since I didn't have the factor $1/2$ which, as the link provided shows, comes from taking the differential of $dx^0$ and $dx^1$ as a function of $dz$ and $dz^*$
    – Run like hell
    Nov 27 at 12:44













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1 Answer
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active

oldest

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2














Not sure it's correct, as usually one defines:
$$
df(z)=frac{partial f}{partial z}dz+frac{partial f}{partial z^*}dz^*
$$
where $frac{partial}{partial z}$ and $frac{partial}{partial z^*}$ are called Wirtinger derivatives and are defined by:
$$
frac{partial}{partial z}:=frac{1}{2}left(frac{partial}{partial x^0}-ifrac{partial}{partial x^1}right)
$$
$$
frac{partial}{partial z^*}:=frac{1}{2}left(frac{partial}{partial x^0}+ifrac{partial}{partial x^1}right)
$$



Compared to your proposal, there is a 1/2 missing factor.




  • this is ok for holomorphic functions (thanks to Cauchy-Riemann)

  • but not in the general case


I can copy derivation from An Introduction to Complex Differentials and
Complex Differentiability p8-9, however this tutorial is really well written and I think it is better to directly read it.




From $F(x,y)=U(x,y)+iV(x,y)$ the bivariate function associated to $f(z)$, the total differential is:
$$
dF=frac{partial}{partial x}F(x,y)dx+frac{partial}{partial y}F(x,y)dy
$$
which can be rewritten as follows (have a look at the cited reference):
$$
dF=frac{1}{2}left[frac{partial U}{partial x}+frac{partial V}{partial y}+ileft(frac{partial V}{partial x}-frac{partial U}{partial y}right)right]dz+frac{1}{2}left[frac{partial U}{partial x}-frac{partial V}{partial y}+ileft(frac{partial V}{partial x}+frac{partial U}{partial y}right)right]dz^*
$$
Assuming $f$ holomorphic and using Cauchy-Riemann conditions ($frac{partial U}{partial x}=frac{partial V}{partial y}$ and $frac{partial U}{partial y}=-frac{partial V}{partial x}$)
we get your suggestion:
$frac{partial}{partial z}=frac{partial}{partial x^0}-ifrac{partial}{partial x^1}$ and $frac{partial}{partial z^*}=0$




One of the big advantage of Wirtinger calculus is:




For the Wirtinger derivatives, the common rules for differentiation
known from real-valued analysis concerning the sum, product, and
composition of two functions hold as well...




in brief you can do differential calculus like in $mathbb{R}^n$. This is very handy for automatic differentiation, gradient computation... etc.






share|cite|improve this answer



















  • 1




    Thank you, so I guess my intuition was wrong since I didn't have the factor $1/2$ which, as the link provided shows, comes from taking the differential of $dx^0$ and $dx^1$ as a function of $dz$ and $dz^*$
    – Run like hell
    Nov 27 at 12:44


















2














Not sure it's correct, as usually one defines:
$$
df(z)=frac{partial f}{partial z}dz+frac{partial f}{partial z^*}dz^*
$$
where $frac{partial}{partial z}$ and $frac{partial}{partial z^*}$ are called Wirtinger derivatives and are defined by:
$$
frac{partial}{partial z}:=frac{1}{2}left(frac{partial}{partial x^0}-ifrac{partial}{partial x^1}right)
$$
$$
frac{partial}{partial z^*}:=frac{1}{2}left(frac{partial}{partial x^0}+ifrac{partial}{partial x^1}right)
$$



Compared to your proposal, there is a 1/2 missing factor.




  • this is ok for holomorphic functions (thanks to Cauchy-Riemann)

  • but not in the general case


I can copy derivation from An Introduction to Complex Differentials and
Complex Differentiability p8-9, however this tutorial is really well written and I think it is better to directly read it.




From $F(x,y)=U(x,y)+iV(x,y)$ the bivariate function associated to $f(z)$, the total differential is:
$$
dF=frac{partial}{partial x}F(x,y)dx+frac{partial}{partial y}F(x,y)dy
$$
which can be rewritten as follows (have a look at the cited reference):
$$
dF=frac{1}{2}left[frac{partial U}{partial x}+frac{partial V}{partial y}+ileft(frac{partial V}{partial x}-frac{partial U}{partial y}right)right]dz+frac{1}{2}left[frac{partial U}{partial x}-frac{partial V}{partial y}+ileft(frac{partial V}{partial x}+frac{partial U}{partial y}right)right]dz^*
$$
Assuming $f$ holomorphic and using Cauchy-Riemann conditions ($frac{partial U}{partial x}=frac{partial V}{partial y}$ and $frac{partial U}{partial y}=-frac{partial V}{partial x}$)
we get your suggestion:
$frac{partial}{partial z}=frac{partial}{partial x^0}-ifrac{partial}{partial x^1}$ and $frac{partial}{partial z^*}=0$




One of the big advantage of Wirtinger calculus is:




For the Wirtinger derivatives, the common rules for differentiation
known from real-valued analysis concerning the sum, product, and
composition of two functions hold as well...




in brief you can do differential calculus like in $mathbb{R}^n$. This is very handy for automatic differentiation, gradient computation... etc.






share|cite|improve this answer



















  • 1




    Thank you, so I guess my intuition was wrong since I didn't have the factor $1/2$ which, as the link provided shows, comes from taking the differential of $dx^0$ and $dx^1$ as a function of $dz$ and $dz^*$
    – Run like hell
    Nov 27 at 12:44
















2












2








2






Not sure it's correct, as usually one defines:
$$
df(z)=frac{partial f}{partial z}dz+frac{partial f}{partial z^*}dz^*
$$
where $frac{partial}{partial z}$ and $frac{partial}{partial z^*}$ are called Wirtinger derivatives and are defined by:
$$
frac{partial}{partial z}:=frac{1}{2}left(frac{partial}{partial x^0}-ifrac{partial}{partial x^1}right)
$$
$$
frac{partial}{partial z^*}:=frac{1}{2}left(frac{partial}{partial x^0}+ifrac{partial}{partial x^1}right)
$$



Compared to your proposal, there is a 1/2 missing factor.




  • this is ok for holomorphic functions (thanks to Cauchy-Riemann)

  • but not in the general case


I can copy derivation from An Introduction to Complex Differentials and
Complex Differentiability p8-9, however this tutorial is really well written and I think it is better to directly read it.




From $F(x,y)=U(x,y)+iV(x,y)$ the bivariate function associated to $f(z)$, the total differential is:
$$
dF=frac{partial}{partial x}F(x,y)dx+frac{partial}{partial y}F(x,y)dy
$$
which can be rewritten as follows (have a look at the cited reference):
$$
dF=frac{1}{2}left[frac{partial U}{partial x}+frac{partial V}{partial y}+ileft(frac{partial V}{partial x}-frac{partial U}{partial y}right)right]dz+frac{1}{2}left[frac{partial U}{partial x}-frac{partial V}{partial y}+ileft(frac{partial V}{partial x}+frac{partial U}{partial y}right)right]dz^*
$$
Assuming $f$ holomorphic and using Cauchy-Riemann conditions ($frac{partial U}{partial x}=frac{partial V}{partial y}$ and $frac{partial U}{partial y}=-frac{partial V}{partial x}$)
we get your suggestion:
$frac{partial}{partial z}=frac{partial}{partial x^0}-ifrac{partial}{partial x^1}$ and $frac{partial}{partial z^*}=0$




One of the big advantage of Wirtinger calculus is:




For the Wirtinger derivatives, the common rules for differentiation
known from real-valued analysis concerning the sum, product, and
composition of two functions hold as well...




in brief you can do differential calculus like in $mathbb{R}^n$. This is very handy for automatic differentiation, gradient computation... etc.






share|cite|improve this answer














Not sure it's correct, as usually one defines:
$$
df(z)=frac{partial f}{partial z}dz+frac{partial f}{partial z^*}dz^*
$$
where $frac{partial}{partial z}$ and $frac{partial}{partial z^*}$ are called Wirtinger derivatives and are defined by:
$$
frac{partial}{partial z}:=frac{1}{2}left(frac{partial}{partial x^0}-ifrac{partial}{partial x^1}right)
$$
$$
frac{partial}{partial z^*}:=frac{1}{2}left(frac{partial}{partial x^0}+ifrac{partial}{partial x^1}right)
$$



Compared to your proposal, there is a 1/2 missing factor.




  • this is ok for holomorphic functions (thanks to Cauchy-Riemann)

  • but not in the general case


I can copy derivation from An Introduction to Complex Differentials and
Complex Differentiability p8-9, however this tutorial is really well written and I think it is better to directly read it.




From $F(x,y)=U(x,y)+iV(x,y)$ the bivariate function associated to $f(z)$, the total differential is:
$$
dF=frac{partial}{partial x}F(x,y)dx+frac{partial}{partial y}F(x,y)dy
$$
which can be rewritten as follows (have a look at the cited reference):
$$
dF=frac{1}{2}left[frac{partial U}{partial x}+frac{partial V}{partial y}+ileft(frac{partial V}{partial x}-frac{partial U}{partial y}right)right]dz+frac{1}{2}left[frac{partial U}{partial x}-frac{partial V}{partial y}+ileft(frac{partial V}{partial x}+frac{partial U}{partial y}right)right]dz^*
$$
Assuming $f$ holomorphic and using Cauchy-Riemann conditions ($frac{partial U}{partial x}=frac{partial V}{partial y}$ and $frac{partial U}{partial y}=-frac{partial V}{partial x}$)
we get your suggestion:
$frac{partial}{partial z}=frac{partial}{partial x^0}-ifrac{partial}{partial x^1}$ and $frac{partial}{partial z^*}=0$




One of the big advantage of Wirtinger calculus is:




For the Wirtinger derivatives, the common rules for differentiation
known from real-valued analysis concerning the sum, product, and
composition of two functions hold as well...




in brief you can do differential calculus like in $mathbb{R}^n$. This is very handy for automatic differentiation, gradient computation... etc.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 at 13:01

























answered Nov 27 at 12:35









Picaud Vincent

1,20838




1,20838








  • 1




    Thank you, so I guess my intuition was wrong since I didn't have the factor $1/2$ which, as the link provided shows, comes from taking the differential of $dx^0$ and $dx^1$ as a function of $dz$ and $dz^*$
    – Run like hell
    Nov 27 at 12:44
















  • 1




    Thank you, so I guess my intuition was wrong since I didn't have the factor $1/2$ which, as the link provided shows, comes from taking the differential of $dx^0$ and $dx^1$ as a function of $dz$ and $dz^*$
    – Run like hell
    Nov 27 at 12:44










1




1




Thank you, so I guess my intuition was wrong since I didn't have the factor $1/2$ which, as the link provided shows, comes from taking the differential of $dx^0$ and $dx^1$ as a function of $dz$ and $dz^*$
– Run like hell
Nov 27 at 12:44






Thank you, so I guess my intuition was wrong since I didn't have the factor $1/2$ which, as the link provided shows, comes from taking the differential of $dx^0$ and $dx^1$ as a function of $dz$ and $dz^*$
– Run like hell
Nov 27 at 12:44




















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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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