How to find Frechet derivative of $f(x)=|Ax-b|^2$ at any $x^*$?
$begingroup$
Given a real $m times n$ matrix $A$ and $b in mathbb{R}^m$, let $f(x)=|Ax-b|^2$ for any $x in mathbb{R}^n$.
Find Frechet derivative of $f(x)=|Ax-b|^2$ at any $x^*$?
Actually I am wondering how to use the following to find the Frechet derivative, i.e., $J$:
$$lim_{h rightarrow 0} frac{|f(x+h)-f(x)-Jh|}{|h|} =0$$
Another question is what the difference between what we would get from Frechet derivative and the gradient $nabla f(x)=2A^T(Ax-b)$?
Please explain your reasons in detail, especially, what is the difference between gradient of $f$ and Frechet derivative. Also, explain when they might be identical.
derivatives frechet-derivative
edited Dec 9 '18 at 21:02
Saeed
969310
asked Dec 9 '18 at 19:37
SepideSepide
2788
$endgroup$
add a comment |
$begingroup$
Given a real $m times n$ matrix $A$ and $b in mathbb{R}^m$, let $f(x)=|Ax-b|^2$ for any $x in mathbb{R}^n$.
Find Frechet derivative of $f(x)=|Ax-b|^2$ at any $x^*$?
Actually I am wondering how to use the following to find the Frechet derivative, i.e., $J$:
$$lim_{h rightarrow 0} frac{|f(x+h)-f(x)-Jh|}{|h|} =0$$
Another question is what the difference between what we would get from Frechet derivative and the gradient $nabla f(x)=2A^T(Ax-b)$?
Please explain your reasons in detail, especially, what is the difference between gradient of $f$ and Frechet derivative. Also, explain when they might be identical.
derivatives frechet-derivative
edited Dec 9 '18 at 21:02
Saeed
969310
asked Dec 9 '18 at 19:37
SepideSepide
2788
$endgroup$
$begingroup$
Write $f$ as a composition of an affine function and a bilinear function.
$endgroup$
– Will M.
Dec 10 '18 at 4:05
add a comment |
$begingroup$
Given a real $m times n$ matrix $A$ and $b in mathbb{R}^m$, let $f(x)=|Ax-b|^2$ for any $x in mathbb{R}^n$.
Find Frechet derivative of $f(x)=|Ax-b|^2$ at any $x^*$?
Actually I am wondering how to use the following to find the Frechet derivative, i.e., $J$:
$$lim_{h rightarrow 0} frac{|f(x+h)-f(x)-Jh|}{|h|} =0$$
Another question is what the difference between what we would get from Frechet derivative and the gradient $nabla f(x)=2A^T(Ax-b)$?
Please explain your reasons in detail, especially, what is the difference between gradient of $f$ and Frechet derivative. Also, explain when they might be identical.
derivatives frechet-derivative
edited Dec 9 '18 at 21:02
Saeed
969310
asked Dec 9 '18 at 19:37
SepideSepide
2788
$endgroup$
Given a real $m times n$ matrix $A$ and $b in mathbb{R}^m$, let $f(x)=|Ax-b|^2$ for any $x in mathbb{R}^n$.
Find Frechet derivative of $f(x)=|Ax-b|^2$ at any $x^*$?
Actually I am wondering how to use the following to find the Frechet derivative, i.e., $J$:
$$lim_{h rightarrow 0} frac{|f(x+h)-f(x)-Jh|}{|h|} =0$$
Another question is what the difference between what we would get from Frechet derivative and the gradient $nabla f(x)=2A^T(Ax-b)$?
Please explain your reasons in detail, especially, what is the difference between gradient of $f$ and Frechet derivative. Also, explain when they might be identical.
derivatives frechet-derivative
derivatives frechet-derivative
edited Dec 9 '18 at 21:02
Saeed
969310
asked Dec 9 '18 at 19:37
SepideSepide
2788
edited Dec 9 '18 at 21:02
Saeed
969310
asked Dec 9 '18 at 19:37
SepideSepide
2788
edited Dec 9 '18 at 21:02
Saeed
969310
edited Dec 9 '18 at 21:02
Saeed
969310
edited Dec 9 '18 at 21:02
Saeed
969310
969310
asked Dec 9 '18 at 19:37
SepideSepide
2788
asked Dec 9 '18 at 19:37
SepideSepide
2788
asked Dec 9 '18 at 19:37
SepideSepide
2788
2788
$begingroup$
Write $f$ as a composition of an affine function and a bilinear function.
$endgroup$
– Will M.
Dec 10 '18 at 4:05
add a comment |
$begingroup$
Write $f$ as a composition of an affine function and a bilinear function.
$endgroup$
– Will M.
Dec 10 '18 at 4:05
$begingroup$
Write $f$ as a composition of an affine function and a bilinear function.
$endgroup$
– Will M.
Dec 10 '18 at 4:05
$begingroup$
Write $f$ as a composition of an affine function and a bilinear function.
$endgroup$
– Will M.
Dec 10 '18 at 4:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $mathrm{H}$ is a Hilbert space, then every continuous linear function $u:mathrm{H} to mathbf{R}$ can be represented by means of scalar product with respect to a unique vector, here denoted as $x_u:$
$$u(y) = (y mid x_u).$$
Hence, if $f:mathrm{H} to mathbf{R}$ is a differentiable function, then its derivative $u = f'(a)$ at $a$ is a continuous linear function. The vector $x_u$ is denoted $nabla f(a)$ in this case. And we have the fundamental relation:
$$f'(a) cdot h = (nabla f(a) mid h).$$
In regards to your particular $f,$ we can write $f(x) = (Ax -b mid Ax - b)$ and by the products and chain rules,
$$f'(x) cdot h = (Ax - b mid Ah) + (Ah mid Ax - b) = 2(Ah mid Ax - b).$$
If $mathrm{H} = mathbf{R}^d,$ and we are dealing with the standard Euclidean inner product, we can write further $f'(x) cdot h = (2A^intercal (Ax - b) mid h),$ this signifies $nabla f(x) = 2A^intercal (Ax - b).$ Q.E.D.
answered Dec 11 '18 at 5:13
Will M.Will M.
2,615315
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $mathrm{H}$ is a Hilbert space, then every continuous linear function $u:mathrm{H} to mathbf{R}$ can be represented by means of scalar product with respect to a unique vector, here denoted as $x_u:$
$$u(y) = (y mid x_u).$$
Hence, if $f:mathrm{H} to mathbf{R}$ is a differentiable function, then its derivative $u = f'(a)$ at $a$ is a continuous linear function. The vector $x_u$ is denoted $nabla f(a)$ in this case. And we have the fundamental relation:
$$f'(a) cdot h = (nabla f(a) mid h).$$
In regards to your particular $f,$ we can write $f(x) = (Ax -b mid Ax - b)$ and by the products and chain rules,
$$f'(x) cdot h = (Ax - b mid Ah) + (Ah mid Ax - b) = 2(Ah mid Ax - b).$$
If $mathrm{H} = mathbf{R}^d,$ and we are dealing with the standard Euclidean inner product, we can write further $f'(x) cdot h = (2A^intercal (Ax - b) mid h),$ this signifies $nabla f(x) = 2A^intercal (Ax - b).$ Q.E.D.
answered Dec 11 '18 at 5:13
Will M.Will M.
2,615315
$endgroup$
add a comment |
$begingroup$
If $mathrm{H}$ is a Hilbert space, then every continuous linear function $u:mathrm{H} to mathbf{R}$ can be represented by means of scalar product with respect to a unique vector, here denoted as $x_u:$
$$u(y) = (y mid x_u).$$
Hence, if $f:mathrm{H} to mathbf{R}$ is a differentiable function, then its derivative $u = f'(a)$ at $a$ is a continuous linear function. The vector $x_u$ is denoted $nabla f(a)$ in this case. And we have the fundamental relation:
$$f'(a) cdot h = (nabla f(a) mid h).$$
In regards to your particular $f,$ we can write $f(x) = (Ax -b mid Ax - b)$ and by the products and chain rules,
$$f'(x) cdot h = (Ax - b mid Ah) + (Ah mid Ax - b) = 2(Ah mid Ax - b).$$
If $mathrm{H} = mathbf{R}^d,$ and we are dealing with the standard Euclidean inner product, we can write further $f'(x) cdot h = (2A^intercal (Ax - b) mid h),$ this signifies $nabla f(x) = 2A^intercal (Ax - b).$ Q.E.D.
answered Dec 11 '18 at 5:13
Will M.Will M.
2,615315
$endgroup$
add a comment |
$begingroup$
If $mathrm{H}$ is a Hilbert space, then every continuous linear function $u:mathrm{H} to mathbf{R}$ can be represented by means of scalar product with respect to a unique vector, here denoted as $x_u:$
$$u(y) = (y mid x_u).$$
Hence, if $f:mathrm{H} to mathbf{R}$ is a differentiable function, then its derivative $u = f'(a)$ at $a$ is a continuous linear function. The vector $x_u$ is denoted $nabla f(a)$ in this case. And we have the fundamental relation:
$$f'(a) cdot h = (nabla f(a) mid h).$$
In regards to your particular $f,$ we can write $f(x) = (Ax -b mid Ax - b)$ and by the products and chain rules,
$$f'(x) cdot h = (Ax - b mid Ah) + (Ah mid Ax - b) = 2(Ah mid Ax - b).$$
If $mathrm{H} = mathbf{R}^d,$ and we are dealing with the standard Euclidean inner product, we can write further $f'(x) cdot h = (2A^intercal (Ax - b) mid h),$ this signifies $nabla f(x) = 2A^intercal (Ax - b).$ Q.E.D.
answered Dec 11 '18 at 5:13
Will M.Will M.
2,615315
$endgroup$
If $mathrm{H}$ is a Hilbert space, then every continuous linear function $u:mathrm{H} to mathbf{R}$ can be represented by means of scalar product with respect to a unique vector, here denoted as $x_u:$
$$u(y) = (y mid x_u).$$
Hence, if $f:mathrm{H} to mathbf{R}$ is a differentiable function, then its derivative $u = f'(a)$ at $a$ is a continuous linear function. The vector $x_u$ is denoted $nabla f(a)$ in this case. And we have the fundamental relation:
$$f'(a) cdot h = (nabla f(a) mid h).$$
In regards to your particular $f,$ we can write $f(x) = (Ax -b mid Ax - b)$ and by the products and chain rules,
$$f'(x) cdot h = (Ax - b mid Ah) + (Ah mid Ax - b) = 2(Ah mid Ax - b).$$
If $mathrm{H} = mathbf{R}^d,$ and we are dealing with the standard Euclidean inner product, we can write further $f'(x) cdot h = (2A^intercal (Ax - b) mid h),$ this signifies $nabla f(x) = 2A^intercal (Ax - b).$ Q.E.D.
answered Dec 11 '18 at 5:13
Will M.Will M.
2,615315
answered Dec 11 '18 at 5:13
Will M.Will M.
2,615315
answered Dec 11 '18 at 5:13
Will M.Will M.
2,615315
answered Dec 11 '18 at 5:13
Will M.Will M.
2,615315
2,615315
add a comment |
add a comment |
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$begingroup$
Write $f$ as a composition of an affine function and a bilinear function.
$endgroup$
– Will M.
Dec 10 '18 at 4:05