Mixing
How to calculate a Fréchet derivative?
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What is the standard algorithm for calculating a Fréchet derivative? i.e.
$f(x,y)=x^2y$
for $(x_0,y_0)inmathbb{R}^2$
derivatives partial-derivative
asked Jun 22 '15 at 23:32
CommanderPirxCommanderPirx
11
$endgroup$
add a comment |
$begingroup$
What is the standard algorithm for calculating a Fréchet derivative? i.e.
$f(x,y)=x^2y$
for $(x_0,y_0)inmathbb{R}^2$
derivatives partial-derivative
asked Jun 22 '15 at 23:32
CommanderPirxCommanderPirx
11
$endgroup$
2
$begingroup$
For smooth functions $mathbb{R}^n to mathbb{R}$, the Fréchet derivative is just the appropriate matrix of partial derivatives. For example, the above has $Df((x,y)) = (2xy, x^2)$.
$endgroup$
– copper.hat
Jun 22 '15 at 23:38
$begingroup$
What do you mean by algorithm? To the best of my knowledge, there is no real algorithm for this, any more than there is a standard algorithm for taking the "standard" derivative.
$endgroup$
– qaphla
Jun 22 '15 at 23:39
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@qaphla well, I mean something like a step-by-step solution which can be applied on a particular type of function to calculate the Fréchet derivative, hence my example
$endgroup$
– CommanderPirx
Jun 22 '15 at 23:48
add a comment |
$begingroup$
What is the standard algorithm for calculating a Fréchet derivative? i.e.
$f(x,y)=x^2y$
for $(x_0,y_0)inmathbb{R}^2$
derivatives partial-derivative
asked Jun 22 '15 at 23:32
CommanderPirxCommanderPirx
11
$endgroup$
What is the standard algorithm for calculating a Fréchet derivative? i.e.
$f(x,y)=x^2y$
for $(x_0,y_0)inmathbb{R}^2$
derivatives partial-derivative
derivatives partial-derivative
asked Jun 22 '15 at 23:32
CommanderPirxCommanderPirx
11
asked Jun 22 '15 at 23:32
CommanderPirxCommanderPirx
11
asked Jun 22 '15 at 23:32
CommanderPirxCommanderPirx
11
asked Jun 22 '15 at 23:32
CommanderPirxCommanderPirx
11
asked Jun 22 '15 at 23:32
CommanderPirxCommanderPirx
11
11
2
$begingroup$
For smooth functions $mathbb{R}^n to mathbb{R}$, the Fréchet derivative is just the appropriate matrix of partial derivatives. For example, the above has $Df((x,y)) = (2xy, x^2)$.
$endgroup$
– copper.hat
Jun 22 '15 at 23:38
$begingroup$
What do you mean by algorithm? To the best of my knowledge, there is no real algorithm for this, any more than there is a standard algorithm for taking the "standard" derivative.
$endgroup$
– qaphla
Jun 22 '15 at 23:39
$begingroup$
@qaphla well, I mean something like a step-by-step solution which can be applied on a particular type of function to calculate the Fréchet derivative, hence my example
$endgroup$
– CommanderPirx
Jun 22 '15 at 23:48
add a comment |
2
$begingroup$
For smooth functions $mathbb{R}^n to mathbb{R}$, the Fréchet derivative is just the appropriate matrix of partial derivatives. For example, the above has $Df((x,y)) = (2xy, x^2)$.
$endgroup$
– copper.hat
Jun 22 '15 at 23:38
$begingroup$
What do you mean by algorithm? To the best of my knowledge, there is no real algorithm for this, any more than there is a standard algorithm for taking the "standard" derivative.
$endgroup$
– qaphla
Jun 22 '15 at 23:39
$begingroup$
@qaphla well, I mean something like a step-by-step solution which can be applied on a particular type of function to calculate the Fréchet derivative, hence my example
$endgroup$
– CommanderPirx
Jun 22 '15 at 23:48
2
2
$begingroup$
For smooth functions $mathbb{R}^n to mathbb{R}$, the Fréchet derivative is just the appropriate matrix of partial derivatives. For example, the above has $Df((x,y)) = (2xy, x^2)$.
$endgroup$
– copper.hat
Jun 22 '15 at 23:38
$begingroup$
For smooth functions $mathbb{R}^n to mathbb{R}$, the Fréchet derivative is just the appropriate matrix of partial derivatives. For example, the above has $Df((x,y)) = (2xy, x^2)$.
$endgroup$
– copper.hat
Jun 22 '15 at 23:38
$begingroup$
What do you mean by algorithm? To the best of my knowledge, there is no real algorithm for this, any more than there is a standard algorithm for taking the "standard" derivative.
$endgroup$
– qaphla
Jun 22 '15 at 23:39
$begingroup$
What do you mean by algorithm? To the best of my knowledge, there is no real algorithm for this, any more than there is a standard algorithm for taking the "standard" derivative.
$endgroup$
– qaphla
Jun 22 '15 at 23:39
$begingroup$
@qaphla well, I mean something like a step-by-step solution which can be applied on a particular type of function to calculate the Fréchet derivative, hence my example
$endgroup$
– CommanderPirx
Jun 22 '15 at 23:48
$begingroup$
@qaphla well, I mean something like a step-by-step solution which can be applied on a particular type of function to calculate the Fréchet derivative, hence my example
$endgroup$
– CommanderPirx
Jun 22 '15 at 23:48
add a comment |
1 Answer
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oldest
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$begingroup$
Let $f:mathbb R^n to mathbb R^m$ be (Frechet) differentiable at $x in mathbb R^n$. The Frechet derivative of $f$ at $x$, which is a linear transformation from $mathbb R^n$ to $mathbb R^m$, is represented (with respect to the standard bases of $mathbb R^n$ and $mathbb R^m$) by the matrix
begin{equation*}
f'(x) = begin{bmatrix}
frac{partial f_1(x)}{partial x_1} & cdots & frac{partial f_1(x)}{partial x_n} \
vdots & ddots & vdots \
frac{partial f_m(x)}{partial x_1} & cdots & frac{partial f_m(x)}{partial x_n}
end{bmatrix}.
end{equation*}
Here $f_i$ is the $i$th component function of $f$. In the example you gave, you can compute these partial derivatives explicitly.
answered Jun 23 '15 at 0:28
littleOlittleO
29.9k646109
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1 Answer
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active
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votes
1 Answer
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active
oldest
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$begingroup$
Let $f:mathbb R^n to mathbb R^m$ be (Frechet) differentiable at $x in mathbb R^n$. The Frechet derivative of $f$ at $x$, which is a linear transformation from $mathbb R^n$ to $mathbb R^m$, is represented (with respect to the standard bases of $mathbb R^n$ and $mathbb R^m$) by the matrix
begin{equation*}
f'(x) = begin{bmatrix}
frac{partial f_1(x)}{partial x_1} & cdots & frac{partial f_1(x)}{partial x_n} \
vdots & ddots & vdots \
frac{partial f_m(x)}{partial x_1} & cdots & frac{partial f_m(x)}{partial x_n}
end{bmatrix}.
end{equation*}
Here $f_i$ is the $i$th component function of $f$. In the example you gave, you can compute these partial derivatives explicitly.
answered Jun 23 '15 at 0:28
littleOlittleO
29.9k646109
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb R^n to mathbb R^m$ be (Frechet) differentiable at $x in mathbb R^n$. The Frechet derivative of $f$ at $x$, which is a linear transformation from $mathbb R^n$ to $mathbb R^m$, is represented (with respect to the standard bases of $mathbb R^n$ and $mathbb R^m$) by the matrix
begin{equation*}
f'(x) = begin{bmatrix}
frac{partial f_1(x)}{partial x_1} & cdots & frac{partial f_1(x)}{partial x_n} \
vdots & ddots & vdots \
frac{partial f_m(x)}{partial x_1} & cdots & frac{partial f_m(x)}{partial x_n}
end{bmatrix}.
end{equation*}
Here $f_i$ is the $i$th component function of $f$. In the example you gave, you can compute these partial derivatives explicitly.
answered Jun 23 '15 at 0:28
littleOlittleO
29.9k646109
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb R^n to mathbb R^m$ be (Frechet) differentiable at $x in mathbb R^n$. The Frechet derivative of $f$ at $x$, which is a linear transformation from $mathbb R^n$ to $mathbb R^m$, is represented (with respect to the standard bases of $mathbb R^n$ and $mathbb R^m$) by the matrix
begin{equation*}
f'(x) = begin{bmatrix}
frac{partial f_1(x)}{partial x_1} & cdots & frac{partial f_1(x)}{partial x_n} \
vdots & ddots & vdots \
frac{partial f_m(x)}{partial x_1} & cdots & frac{partial f_m(x)}{partial x_n}
end{bmatrix}.
end{equation*}
Here $f_i$ is the $i$th component function of $f$. In the example you gave, you can compute these partial derivatives explicitly.
answered Jun 23 '15 at 0:28
littleOlittleO
29.9k646109
$endgroup$
Let $f:mathbb R^n to mathbb R^m$ be (Frechet) differentiable at $x in mathbb R^n$. The Frechet derivative of $f$ at $x$, which is a linear transformation from $mathbb R^n$ to $mathbb R^m$, is represented (with respect to the standard bases of $mathbb R^n$ and $mathbb R^m$) by the matrix
begin{equation*}
f'(x) = begin{bmatrix}
frac{partial f_1(x)}{partial x_1} & cdots & frac{partial f_1(x)}{partial x_n} \
vdots & ddots & vdots \
frac{partial f_m(x)}{partial x_1} & cdots & frac{partial f_m(x)}{partial x_n}
end{bmatrix}.
end{equation*}
Here $f_i$ is the $i$th component function of $f$. In the example you gave, you can compute these partial derivatives explicitly.
answered Jun 23 '15 at 0:28
littleOlittleO
29.9k646109
edited Jun 23 '15 at 0:48
answered Jun 23 '15 at 0:28
littleOlittleO
29.9k646109
answered Jun 23 '15 at 0:28
littleOlittleO
29.9k646109
answered Jun 23 '15 at 0:28
littleOlittleO
29.9k646109
29.9k646109
add a comment |
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$begingroup$
For smooth functions $mathbb{R}^n to mathbb{R}$, the Fréchet derivative is just the appropriate matrix of partial derivatives. For example, the above has $Df((x,y)) = (2xy, x^2)$.
$endgroup$
– copper.hat
Jun 22 '15 at 23:38
$begingroup$
What do you mean by algorithm? To the best of my knowledge, there is no real algorithm for this, any more than there is a standard algorithm for taking the "standard" derivative.
$endgroup$
– qaphla
Jun 22 '15 at 23:39
$begingroup$
@qaphla well, I mean something like a step-by-step solution which can be applied on a particular type of function to calculate the Fréchet derivative, hence my example
$endgroup$
– CommanderPirx
Jun 22 '15 at 23:48