Can I take a derivative of any complex function so long as I treat the complex numbers as matrices?
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Complex numbers can be represented as matrices, for example
$$
a+bi leftrightarrow pmatrix{a &b\-b&a}
$$
Only some functions of a complex variable have a derivative that is a complex number (those which are holomorphic). However, I can take derivatives of all smooth maps from matrices to matrices, even those that correspond to functions that are not holomorphic. What gives?
matrices complex-analysis
asked Jan 15 at 16:31
Display NameDisplay Name
34010
$endgroup$
|
$begingroup$
Complex numbers can be represented as matrices, for example
$$
a+bi leftrightarrow pmatrix{a &b\-b&a}
$$
Only some functions of a complex variable have a derivative that is a complex number (those which are holomorphic). However, I can take derivatives of all smooth maps from matrices to matrices, even those that correspond to functions that are not holomorphic. What gives?
matrices complex-analysis
asked Jan 15 at 16:31
Display NameDisplay Name
34010
$endgroup$
5
$begingroup$
Only some functions of a complex variable have a derivative that is a complex number, yes, and only some matrices represent complex numbers. When you take the derivative of a smooth map from matrices to matrices that does not correspond to a holomorphic function, you get a matrix that does not represent a complex number.
$endgroup$
– Rahul
Jan 15 at 16:37
|
$begingroup$
Complex numbers can be represented as matrices, for example
$$
a+bi leftrightarrow pmatrix{a &b\-b&a}
$$
Only some functions of a complex variable have a derivative that is a complex number (those which are holomorphic). However, I can take derivatives of all smooth maps from matrices to matrices, even those that correspond to functions that are not holomorphic. What gives?
matrices complex-analysis
asked Jan 15 at 16:31
Display NameDisplay Name
34010
$endgroup$
Complex numbers can be represented as matrices, for example
$$
a+bi leftrightarrow pmatrix{a &b\-b&a}
$$
Only some functions of a complex variable have a derivative that is a complex number (those which are holomorphic). However, I can take derivatives of all smooth maps from matrices to matrices, even those that correspond to functions that are not holomorphic. What gives?
matrices complex-analysis
matrices complex-analysis
asked Jan 15 at 16:31
Display NameDisplay Name
34010
asked Jan 15 at 16:31
Display NameDisplay Name
34010
asked Jan 15 at 16:31
Display NameDisplay Name
34010
asked Jan 15 at 16:31
Display NameDisplay Name
34010
asked Jan 15 at 16:31
Display NameDisplay Name
34010
34010
5
$begingroup$
Only some functions of a complex variable have a derivative that is a complex number, yes, and only some matrices represent complex numbers. When you take the derivative of a smooth map from matrices to matrices that does not correspond to a holomorphic function, you get a matrix that does not represent a complex number.
$endgroup$
– Rahul
Jan 15 at 16:37
|
5
$begingroup$
Only some functions of a complex variable have a derivative that is a complex number, yes, and only some matrices represent complex numbers. When you take the derivative of a smooth map from matrices to matrices that does not correspond to a holomorphic function, you get a matrix that does not represent a complex number.
$endgroup$
– Rahul
Jan 15 at 16:37
5
5
$begingroup$
Only some functions of a complex variable have a derivative that is a complex number, yes, and only some matrices represent complex numbers. When you take the derivative of a smooth map from matrices to matrices that does not correspond to a holomorphic function, you get a matrix that does not represent a complex number.
$endgroup$
– Rahul
Jan 15 at 16:37
$begingroup$
Only some functions of a complex variable have a derivative that is a complex number, yes, and only some matrices represent complex numbers. When you take the derivative of a smooth map from matrices to matrices that does not correspond to a holomorphic function, you get a matrix that does not represent a complex number.
$endgroup$
– Rahul
Jan 15 at 16:37
|
1 Answer
1
active
oldest
votes
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That's because of what the derivative is: A linear approximation to the function at a point.
Let's think about the complex plane as a 2-dimensional real plane, and take a function $f:Bbb R^2toBbb R^2$. Its derivative at, say, the origin is an $Bbb R$-linear transformation $f'_{(0,0)}:Bbb R^2toBbb R^2$ such that $f(x,y)approx f(0,0)+f_{(0,0)}'(x,y)$ (with some formal requirements on $approx$). Any linear transformation of the plane can appear as the derivative; any $2times2$ real matrix.
Now think about the same function as $f:Bbb CtoBbb C$. This time the derivative at $0$ is a $Bbb C$-linear transformation $f_0':Bbb CtoBbb C$ such that $f(z)approx f(0)+f'_0(z)$. The available linear transformations here are only scalings (equally much in all directions) and rotations (i.e. multiplication by a complex number). No mirroring, no skew transformations, and so on. It's just more restrictive.
answered Jan 15 at 16:41
ArthurArthur
123k7122211
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Isn't there a factor missing next to $f'$ in both Taylor approximations?
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– Klaas van Aarsen
Jan 15 at 16:43
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@ILikeSerena First off, those are not Taylor approximations, they are the definition of the derivative (along with rigourously constraining $approx$, of course). Second, no. That's what it's supposed to be.
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– Arthur
Jan 15 at 16:47
|
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That's because of what the derivative is: A linear approximation to the function at a point.
Let's think about the complex plane as a 2-dimensional real plane, and take a function $f:Bbb R^2toBbb R^2$. Its derivative at, say, the origin is an $Bbb R$-linear transformation $f'_{(0,0)}:Bbb R^2toBbb R^2$ such that $f(x,y)approx f(0,0)+f_{(0,0)}'(x,y)$ (with some formal requirements on $approx$). Any linear transformation of the plane can appear as the derivative; any $2times2$ real matrix.
Now think about the same function as $f:Bbb CtoBbb C$. This time the derivative at $0$ is a $Bbb C$-linear transformation $f_0':Bbb CtoBbb C$ such that $f(z)approx f(0)+f'_0(z)$. The available linear transformations here are only scalings (equally much in all directions) and rotations (i.e. multiplication by a complex number). No mirroring, no skew transformations, and so on. It's just more restrictive.
answered Jan 15 at 16:41
ArthurArthur
123k7122211
$endgroup$
$begingroup$
Isn't there a factor missing next to $f'$ in both Taylor approximations?
$endgroup$
– Klaas van Aarsen
Jan 15 at 16:43
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@ILikeSerena First off, those are not Taylor approximations, they are the definition of the derivative (along with rigourously constraining $approx$, of course). Second, no. That's what it's supposed to be.
$endgroup$
– Arthur
Jan 15 at 16:47
|
$begingroup$
That's because of what the derivative is: A linear approximation to the function at a point.
Let's think about the complex plane as a 2-dimensional real plane, and take a function $f:Bbb R^2toBbb R^2$. Its derivative at, say, the origin is an $Bbb R$-linear transformation $f'_{(0,0)}:Bbb R^2toBbb R^2$ such that $f(x,y)approx f(0,0)+f_{(0,0)}'(x,y)$ (with some formal requirements on $approx$). Any linear transformation of the plane can appear as the derivative; any $2times2$ real matrix.
Now think about the same function as $f:Bbb CtoBbb C$. This time the derivative at $0$ is a $Bbb C$-linear transformation $f_0':Bbb CtoBbb C$ such that $f(z)approx f(0)+f'_0(z)$. The available linear transformations here are only scalings (equally much in all directions) and rotations (i.e. multiplication by a complex number). No mirroring, no skew transformations, and so on. It's just more restrictive.
answered Jan 15 at 16:41
ArthurArthur
123k7122211
$endgroup$
$begingroup$
Isn't there a factor missing next to $f'$ in both Taylor approximations?
$endgroup$
– Klaas van Aarsen
Jan 15 at 16:43
$begingroup$
@ILikeSerena First off, those are not Taylor approximations, they are the definition of the derivative (along with rigourously constraining $approx$, of course). Second, no. That's what it's supposed to be.
$endgroup$
– Arthur
Jan 15 at 16:47
|
$begingroup$
That's because of what the derivative is: A linear approximation to the function at a point.
Let's think about the complex plane as a 2-dimensional real plane, and take a function $f:Bbb R^2toBbb R^2$. Its derivative at, say, the origin is an $Bbb R$-linear transformation $f'_{(0,0)}:Bbb R^2toBbb R^2$ such that $f(x,y)approx f(0,0)+f_{(0,0)}'(x,y)$ (with some formal requirements on $approx$). Any linear transformation of the plane can appear as the derivative; any $2times2$ real matrix.
Now think about the same function as $f:Bbb CtoBbb C$. This time the derivative at $0$ is a $Bbb C$-linear transformation $f_0':Bbb CtoBbb C$ such that $f(z)approx f(0)+f'_0(z)$. The available linear transformations here are only scalings (equally much in all directions) and rotations (i.e. multiplication by a complex number). No mirroring, no skew transformations, and so on. It's just more restrictive.
answered Jan 15 at 16:41
ArthurArthur
123k7122211
$endgroup$
That's because of what the derivative is: A linear approximation to the function at a point.
Let's think about the complex plane as a 2-dimensional real plane, and take a function $f:Bbb R^2toBbb R^2$. Its derivative at, say, the origin is an $Bbb R$-linear transformation $f'_{(0,0)}:Bbb R^2toBbb R^2$ such that $f(x,y)approx f(0,0)+f_{(0,0)}'(x,y)$ (with some formal requirements on $approx$). Any linear transformation of the plane can appear as the derivative; any $2times2$ real matrix.
Now think about the same function as $f:Bbb CtoBbb C$. This time the derivative at $0$ is a $Bbb C$-linear transformation $f_0':Bbb CtoBbb C$ such that $f(z)approx f(0)+f'_0(z)$. The available linear transformations here are only scalings (equally much in all directions) and rotations (i.e. multiplication by a complex number). No mirroring, no skew transformations, and so on. It's just more restrictive.
answered Jan 15 at 16:41
ArthurArthur
123k7122211
edited Jan 15 at 16:44
answered Jan 15 at 16:41
ArthurArthur
123k7122211
answered Jan 15 at 16:41
ArthurArthur
123k7122211
answered Jan 15 at 16:41
ArthurArthur
123k7122211
123k7122211
$begingroup$
Isn't there a factor missing next to $f'$ in both Taylor approximations?
$endgroup$
– Klaas van Aarsen
Jan 15 at 16:43
$begingroup$
@ILikeSerena First off, those are not Taylor approximations, they are the definition of the derivative (along with rigourously constraining $approx$, of course). Second, no. That's what it's supposed to be.
$endgroup$
– Arthur
Jan 15 at 16:47
|
$begingroup$
Isn't there a factor missing next to $f'$ in both Taylor approximations?
$endgroup$
– Klaas van Aarsen
Jan 15 at 16:43
$begingroup$
@ILikeSerena First off, those are not Taylor approximations, they are the definition of the derivative (along with rigourously constraining $approx$, of course). Second, no. That's what it's supposed to be.
$endgroup$
– Arthur
Jan 15 at 16:47
$begingroup$
Isn't there a factor missing next to $f'$ in both Taylor approximations?
$endgroup$
– Klaas van Aarsen
Jan 15 at 16:43
$begingroup$
Isn't there a factor missing next to $f'$ in both Taylor approximations?
$endgroup$
– Klaas van Aarsen
Jan 15 at 16:43
$begingroup$
@ILikeSerena First off, those are not Taylor approximations, they are the definition of the derivative (along with rigourously constraining $approx$, of course). Second, no. That's what it's supposed to be.
$endgroup$
– Arthur
Jan 15 at 16:47
$begingroup$
@ILikeSerena First off, those are not Taylor approximations, they are the definition of the derivative (along with rigourously constraining $approx$, of course). Second, no. That's what it's supposed to be.
$endgroup$
– Arthur
Jan 15 at 16:47
|
5
$begingroup$
Only some functions of a complex variable have a derivative that is a complex number, yes, and only some matrices represent complex numbers. When you take the derivative of a smooth map from matrices to matrices that does not correspond to a holomorphic function, you get a matrix that does not represent a complex number.
$endgroup$
– Rahul
Jan 15 at 16:37