Mixing
Derivative of Euclidean norm (L2 norm)
$begingroup$
Let:$x=[x_1, x_2]$ and $y = [y_1, y_2]$
What is the derivative of the square of the Euclidean norm of $y-x $?
I'm not sure if I've worded the question correctly, but this is what I'm trying to solve:
$$
frac{d}{dx}(||y-x||^2)
$$
It has been a long time since I've taken a math class, but this is what I've done so far:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}(||[y_1,y_2]-[x_1,x_2]||^2)
$$
Subtracting $x $ from $y$:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}(||[y_1-x_1,y_2-x_2]||^2)
$$
Taking the norm:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}((y_1-x_1)^2+(y_2-x_2)^2)
$$
Then at this point do I take the derivative independently for $x_1$ and $x_2$?
This is where I am guessing:
$$
frac{d}{dx}(||y-x||^2)=[frac{d}{dx_1}((y_1-x_1)^2+(y_2-x_2)^2),frac{d}{dx_2}((y_1-x_1)^2+(y_2-x_2)^2)]
$$
Which would result in:
$$
frac{d}{dx}(||y-x||^2)=[2x_1-2y_1,2x_2-2y_2]
$$
Is this correct? Thank you for your time.
derivatives norm
asked May 23 '18 at 4:12
Trevor Danniel CowanTrevor Danniel Cowan
113
$endgroup$
add a comment |
$begingroup$
Let:$x=[x_1, x_2]$ and $y = [y_1, y_2]$
What is the derivative of the square of the Euclidean norm of $y-x $?
I'm not sure if I've worded the question correctly, but this is what I'm trying to solve:
$$
frac{d}{dx}(||y-x||^2)
$$
It has been a long time since I've taken a math class, but this is what I've done so far:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}(||[y_1,y_2]-[x_1,x_2]||^2)
$$
Subtracting $x $ from $y$:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}(||[y_1-x_1,y_2-x_2]||^2)
$$
Taking the norm:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}((y_1-x_1)^2+(y_2-x_2)^2)
$$
Then at this point do I take the derivative independently for $x_1$ and $x_2$?
This is where I am guessing:
$$
frac{d}{dx}(||y-x||^2)=[frac{d}{dx_1}((y_1-x_1)^2+(y_2-x_2)^2),frac{d}{dx_2}((y_1-x_1)^2+(y_2-x_2)^2)]
$$
Which would result in:
$$
frac{d}{dx}(||y-x||^2)=[2x_1-2y_1,2x_2-2y_2]
$$
Is this correct? Thank you for your time.
derivatives norm
asked May 23 '18 at 4:12
Trevor Danniel CowanTrevor Danniel Cowan
113
$endgroup$
add a comment |
$begingroup$
Let:$x=[x_1, x_2]$ and $y = [y_1, y_2]$
What is the derivative of the square of the Euclidean norm of $y-x $?
I'm not sure if I've worded the question correctly, but this is what I'm trying to solve:
$$
frac{d}{dx}(||y-x||^2)
$$
It has been a long time since I've taken a math class, but this is what I've done so far:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}(||[y_1,y_2]-[x_1,x_2]||^2)
$$
Subtracting $x $ from $y$:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}(||[y_1-x_1,y_2-x_2]||^2)
$$
Taking the norm:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}((y_1-x_1)^2+(y_2-x_2)^2)
$$
Then at this point do I take the derivative independently for $x_1$ and $x_2$?
This is where I am guessing:
$$
frac{d}{dx}(||y-x||^2)=[frac{d}{dx_1}((y_1-x_1)^2+(y_2-x_2)^2),frac{d}{dx_2}((y_1-x_1)^2+(y_2-x_2)^2)]
$$
Which would result in:
$$
frac{d}{dx}(||y-x||^2)=[2x_1-2y_1,2x_2-2y_2]
$$
Is this correct? Thank you for your time.
derivatives norm
asked May 23 '18 at 4:12
Trevor Danniel CowanTrevor Danniel Cowan
113
$endgroup$
Let:$x=[x_1, x_2]$ and $y = [y_1, y_2]$
What is the derivative of the square of the Euclidean norm of $y-x $?
I'm not sure if I've worded the question correctly, but this is what I'm trying to solve:
$$
frac{d}{dx}(||y-x||^2)
$$
It has been a long time since I've taken a math class, but this is what I've done so far:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}(||[y_1,y_2]-[x_1,x_2]||^2)
$$
Subtracting $x $ from $y$:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}(||[y_1-x_1,y_2-x_2]||^2)
$$
Taking the norm:
$$
frac{d}{dx}(||y-x||^2)=frac{d}{dx}((y_1-x_1)^2+(y_2-x_2)^2)
$$
Then at this point do I take the derivative independently for $x_1$ and $x_2$?
This is where I am guessing:
$$
frac{d}{dx}(||y-x||^2)=[frac{d}{dx_1}((y_1-x_1)^2+(y_2-x_2)^2),frac{d}{dx_2}((y_1-x_1)^2+(y_2-x_2)^2)]
$$
Which would result in:
$$
frac{d}{dx}(||y-x||^2)=[2x_1-2y_1,2x_2-2y_2]
$$
Is this correct? Thank you for your time.
derivatives norm
derivatives norm
asked May 23 '18 at 4:12
Trevor Danniel CowanTrevor Danniel Cowan
113
asked May 23 '18 at 4:12
Trevor Danniel CowanTrevor Danniel Cowan
113
asked May 23 '18 at 4:12
Trevor Danniel CowanTrevor Danniel Cowan
113
asked May 23 '18 at 4:12
Trevor Danniel CowanTrevor Danniel Cowan
113
asked May 23 '18 at 4:12
Trevor Danniel CowanTrevor Danniel Cowan
113
113
add a comment |
add a comment |
1 Answer
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$begingroup$
Sure, that's right.
Some sanity checks: the derivative is zero at the local minimum $x=y$, and when $xneq y$,
$$frac{d}{dx}|y-x|^2 = 2(x-y)$$
points in the direction of the vector away from $y$ towards $x$: this makes sense, as the gradient of $|y-x|^2$ is the direction of steepest increase of $|y-x|^2$, which is to move $x$ in the direction directly away from $y$.
answered May 23 '18 at 4:27
user7530user7530
35.1k761114
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Sure, that's right.
Some sanity checks: the derivative is zero at the local minimum $x=y$, and when $xneq y$,
$$frac{d}{dx}|y-x|^2 = 2(x-y)$$
points in the direction of the vector away from $y$ towards $x$: this makes sense, as the gradient of $|y-x|^2$ is the direction of steepest increase of $|y-x|^2$, which is to move $x$ in the direction directly away from $y$.
answered May 23 '18 at 4:27
user7530user7530
35.1k761114
$endgroup$
add a comment |
$begingroup$
Sure, that's right.
Some sanity checks: the derivative is zero at the local minimum $x=y$, and when $xneq y$,
$$frac{d}{dx}|y-x|^2 = 2(x-y)$$
points in the direction of the vector away from $y$ towards $x$: this makes sense, as the gradient of $|y-x|^2$ is the direction of steepest increase of $|y-x|^2$, which is to move $x$ in the direction directly away from $y$.
answered May 23 '18 at 4:27
user7530user7530
35.1k761114
$endgroup$
add a comment |
$begingroup$
Sure, that's right.
Some sanity checks: the derivative is zero at the local minimum $x=y$, and when $xneq y$,
$$frac{d}{dx}|y-x|^2 = 2(x-y)$$
points in the direction of the vector away from $y$ towards $x$: this makes sense, as the gradient of $|y-x|^2$ is the direction of steepest increase of $|y-x|^2$, which is to move $x$ in the direction directly away from $y$.
answered May 23 '18 at 4:27
user7530user7530
35.1k761114
$endgroup$
Sure, that's right.
Some sanity checks: the derivative is zero at the local minimum $x=y$, and when $xneq y$,
$$frac{d}{dx}|y-x|^2 = 2(x-y)$$
points in the direction of the vector away from $y$ towards $x$: this makes sense, as the gradient of $|y-x|^2$ is the direction of steepest increase of $|y-x|^2$, which is to move $x$ in the direction directly away from $y$.
answered May 23 '18 at 4:27
user7530user7530
35.1k761114
answered May 23 '18 at 4:27
user7530user7530
35.1k761114
answered May 23 '18 at 4:27
user7530user7530
35.1k761114
answered May 23 '18 at 4:27
user7530user7530
35.1k761114
35.1k761114
add a comment |
add a comment |
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