The relation between Frechet derivative and differentiation in $mathbbR^2$
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In the book of Complex Made Simple by Ullrich, at page 4, it is given that
However, if we are considering $mathbbC = mathbbR^2$, then $Df$ is a $2times 2$ matrix, and not a complex number, but when we consider use the "complex differentiation", $f'(x)$ is a complex numbers, so is the author claiming that
$$T(h) = a * h, quad a in mathbbC = mathbbR^2, quad T: mathbbC to mathbbC quad textand $T$ is $mathbbR-linear$ $$
for any unit $h in mathbbC$ ?
real-analysis complex-analysis analysis derivatives frechet-derivative
asked Aug 15 at 4:40
onurcanbektas
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In the book of Complex Made Simple by Ullrich, at page 4, it is given that
However, if we are considering $mathbbC = mathbbR^2$, then $Df$ is a $2times 2$ matrix, and not a complex number, but when we consider use the "complex differentiation", $f'(x)$ is a complex numbers, so is the author claiming that
$$T(h) = a * h, quad a in mathbbC = mathbbR^2, quad T: mathbbC to mathbbC quad textand $T$ is $mathbbR-linear$ $$
for any unit $h in mathbbC$ ?
real-analysis complex-analysis analysis derivatives frechet-derivative
asked Aug 15 at 4:40
onurcanbektas
3,1221834
Well, from the excerpt, all he is saying is that the notion of differentiation with respect to a complex scalar is different to differentiating with respect to $mathbbR^2$. Of course, you can express any complex differentiable function as the latter, in which case $Df(z)$ will be a $2 times 2$ matrix that corresponds to multiplication by a complex scalar, and hence has a very restrictive format (cf. the Cauchy Riemann equations). But the existence of $Df(z)$ in general does not mean that the function has a complex differentiable equivalent.
– copper.hat
Aug 15 at 4:50
@copper.hat what do you mean by "complex differentiable equivalent" ?
– onurcanbektas
Aug 15 at 4:56
Every complex differentiable $f$ can be written as an $mathbbR^2$ differentiable $f$, but not vice versa. The matrix $Df(x)$ will look like $beginbmatrix u_x & -v_x \ v_x & u_x endbmatrix$. The function $f(x,y) = x^2+y^2$ does not have a complex equivalent (otherwise the Cauchy Riemann equations would not hold).
– copper.hat
Aug 15 at 5:30
@copper.hat First of all, thanks for your responses; however, you comment does focus on different points than that I'm raising in my question. Plus, I have just started Complex Analysis, and no idea what Cauchy-Riemann equation is.
– onurcanbektas
Aug 15 at 6:51
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In the book of Complex Made Simple by Ullrich, at page 4, it is given that
However, if we are considering $mathbbC = mathbbR^2$, then $Df$ is a $2times 2$ matrix, and not a complex number, but when we consider use the "complex differentiation", $f'(x)$ is a complex numbers, so is the author claiming that
$$T(h) = a * h, quad a in mathbbC = mathbbR^2, quad T: mathbbC to mathbbC quad textand $T$ is $mathbbR-linear$ $$
for any unit $h in mathbbC$ ?
real-analysis complex-analysis analysis derivatives frechet-derivative
asked Aug 15 at 4:40
onurcanbektas
3,1221834
In the book of Complex Made Simple by Ullrich, at page 4, it is given that
However, if we are considering $mathbbC = mathbbR^2$, then $Df$ is a $2times 2$ matrix, and not a complex number, but when we consider use the "complex differentiation", $f'(x)$ is a complex numbers, so is the author claiming that
$$T(h) = a * h, quad a in mathbbC = mathbbR^2, quad T: mathbbC to mathbbC quad textand $T$ is $mathbbR-linear$ $$
for any unit $h in mathbbC$ ?
real-analysis complex-analysis analysis derivatives frechet-derivative
asked Aug 15 at 4:40
onurcanbektas
3,1221834
asked Aug 15 at 4:40
onurcanbektas
3,1221834
asked Aug 15 at 4:40
onurcanbektas
3,1221834
asked Aug 15 at 4:40
onurcanbektas
3,1221834
3,1221834
Well, from the excerpt, all he is saying is that the notion of differentiation with respect to a complex scalar is different to differentiating with respect to $mathbbR^2$. Of course, you can express any complex differentiable function as the latter, in which case $Df(z)$ will be a $2 times 2$ matrix that corresponds to multiplication by a complex scalar, and hence has a very restrictive format (cf. the Cauchy Riemann equations). But the existence of $Df(z)$ in general does not mean that the function has a complex differentiable equivalent.
– copper.hat
Aug 15 at 4:50
@copper.hat what do you mean by "complex differentiable equivalent" ?
– onurcanbektas
Aug 15 at 4:56
Every complex differentiable $f$ can be written as an $mathbbR^2$ differentiable $f$, but not vice versa. The matrix $Df(x)$ will look like $beginbmatrix u_x & -v_x \ v_x & u_x endbmatrix$. The function $f(x,y) = x^2+y^2$ does not have a complex equivalent (otherwise the Cauchy Riemann equations would not hold).
– copper.hat
Aug 15 at 5:30
@copper.hat First of all, thanks for your responses; however, you comment does focus on different points than that I'm raising in my question. Plus, I have just started Complex Analysis, and no idea what Cauchy-Riemann equation is.
– onurcanbektas
Aug 15 at 6:51
add a comment |Â
Well, from the excerpt, all he is saying is that the notion of differentiation with respect to a complex scalar is different to differentiating with respect to $mathbbR^2$. Of course, you can express any complex differentiable function as the latter, in which case $Df(z)$ will be a $2 times 2$ matrix that corresponds to multiplication by a complex scalar, and hence has a very restrictive format (cf. the Cauchy Riemann equations). But the existence of $Df(z)$ in general does not mean that the function has a complex differentiable equivalent.
– copper.hat
Aug 15 at 4:50
@copper.hat what do you mean by "complex differentiable equivalent" ?
– onurcanbektas
Aug 15 at 4:56
Every complex differentiable $f$ can be written as an $mathbbR^2$ differentiable $f$, but not vice versa. The matrix $Df(x)$ will look like $beginbmatrix u_x & -v_x \ v_x & u_x endbmatrix$. The function $f(x,y) = x^2+y^2$ does not have a complex equivalent (otherwise the Cauchy Riemann equations would not hold).
– copper.hat
Aug 15 at 5:30
@copper.hat First of all, thanks for your responses; however, you comment does focus on different points than that I'm raising in my question. Plus, I have just started Complex Analysis, and no idea what Cauchy-Riemann equation is.
– onurcanbektas
Aug 15 at 6:51
Well, from the excerpt, all he is saying is that the notion of differentiation with respect to a complex scalar is different to differentiating with respect to $mathbbR^2$. Of course, you can express any complex differentiable function as the latter, in which case $Df(z)$ will be a $2 times 2$ matrix that corresponds to multiplication by a complex scalar, and hence has a very restrictive format (cf. the Cauchy Riemann equations). But the existence of $Df(z)$ in general does not mean that the function has a complex differentiable equivalent.
– copper.hat
Aug 15 at 4:50
Well, from the excerpt, all he is saying is that the notion of differentiation with respect to a complex scalar is different to differentiating with respect to $mathbbR^2$. Of course, you can express any complex differentiable function as the latter, in which case $Df(z)$ will be a $2 times 2$ matrix that corresponds to multiplication by a complex scalar, and hence has a very restrictive format (cf. the Cauchy Riemann equations). But the existence of $Df(z)$ in general does not mean that the function has a complex differentiable equivalent.
– copper.hat
Aug 15 at 4:50
@copper.hat what do you mean by "complex differentiable equivalent" ?
– onurcanbektas
Aug 15 at 4:56
@copper.hat what do you mean by "complex differentiable equivalent" ?
– onurcanbektas
Aug 15 at 4:56
Every complex differentiable $f$ can be written as an $mathbbR^2$ differentiable $f$, but not vice versa. The matrix $Df(x)$ will look like $beginbmatrix u_x & -v_x \ v_x & u_x endbmatrix$. The function $f(x,y) = x^2+y^2$ does not have a complex equivalent (otherwise the Cauchy Riemann equations would not hold).
– copper.hat
Aug 15 at 5:30
Every complex differentiable $f$ can be written as an $mathbbR^2$ differentiable $f$, but not vice versa. The matrix $Df(x)$ will look like $beginbmatrix u_x & -v_x \ v_x & u_x endbmatrix$. The function $f(x,y) = x^2+y^2$ does not have a complex equivalent (otherwise the Cauchy Riemann equations would not hold).
– copper.hat
Aug 15 at 5:30
@copper.hat First of all, thanks for your responses; however, you comment does focus on different points than that I'm raising in my question. Plus, I have just started Complex Analysis, and no idea what Cauchy-Riemann equation is.
– onurcanbektas
Aug 15 at 6:51
@copper.hat First of all, thanks for your responses; however, you comment does focus on different points than that I'm raising in my question. Plus, I have just started Complex Analysis, and no idea what Cauchy-Riemann equation is.
– onurcanbektas
Aug 15 at 6:51
add a comment |Â
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Well, from the excerpt, all he is saying is that the notion of differentiation with respect to a complex scalar is different to differentiating with respect to $mathbbR^2$. Of course, you can express any complex differentiable function as the latter, in which case $Df(z)$ will be a $2 times 2$ matrix that corresponds to multiplication by a complex scalar, and hence has a very restrictive format (cf. the Cauchy Riemann equations). But the existence of $Df(z)$ in general does not mean that the function has a complex differentiable equivalent.
– copper.hat
Aug 15 at 4:50
@copper.hat what do you mean by "complex differentiable equivalent" ?
– onurcanbektas
Aug 15 at 4:56
Every complex differentiable $f$ can be written as an $mathbbR^2$ differentiable $f$, but not vice versa. The matrix $Df(x)$ will look like $beginbmatrix u_x & -v_x \ v_x & u_x endbmatrix$. The function $f(x,y) = x^2+y^2$ does not have a complex equivalent (otherwise the Cauchy Riemann equations would not hold).
– copper.hat
Aug 15 at 5:30
@copper.hat First of all, thanks for your responses; however, you comment does focus on different points than that I'm raising in my question. Plus, I have just started Complex Analysis, and no idea what Cauchy-Riemann equation is.
– onurcanbektas
Aug 15 at 6:51