Global minimum of $f(x)=sum_v=1^k m_v|x-x_v|_2^2, x,x_v in mathbbR^n, m_v in mathbbR>0$
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$$f(x)=sum_v=1^k m_v|x-x_v|_2^2, x,x_v in mathbbR^n, m_v in mathbbR>0$$
($|.|_2^2$ is the euclidean norm but squared)
I need to determine the global minimum of $f(x)$ and reason why it truly is a global minimum.
What I tried:
As $m_v>0$ and because $|x-x_v|_2^2$ is a norm it clear that $f(x)=sum_v=1^k m_v|x-x_v|_2^2 ge 0 forall x in mathbbR^n$ and therefore the global minimum of $f$ is$x=x_v$.
Because I don't know if that's sufficient I also tried this:
$f(x)=sum_v=1^k m_v|x-x_v|_2^2=sum_v=1^k m_v sum_i=1^n(x_i-x_v_i)^2$ and therefore
$f'(x)=2sum_v=1^k m_v sum_i=1^n(x_i-x_v_i)$.
So $f'(x)=0 Leftrightarrow x_i=x_v_i forall i=1...n forall v=1...k$
$f''(x)=2sum_v=1^k m_v sum_i=1^n1>0$
And therefore $x=x_v$ truly is the global minimum.
Is that correct? Thanks in advance!
analysis optimization norm
asked Aug 28 at 12:21
user586087
635
add a comment |Â
up vote
0
down vote
favorite
$$f(x)=sum_v=1^k m_v|x-x_v|_2^2, x,x_v in mathbbR^n, m_v in mathbbR>0$$
($|.|_2^2$ is the euclidean norm but squared)
I need to determine the global minimum of $f(x)$ and reason why it truly is a global minimum.
What I tried:
As $m_v>0$ and because $|x-x_v|_2^2$ is a norm it clear that $f(x)=sum_v=1^k m_v|x-x_v|_2^2 ge 0 forall x in mathbbR^n$ and therefore the global minimum of $f$ is$x=x_v$.
Because I don't know if that's sufficient I also tried this:
$f(x)=sum_v=1^k m_v|x-x_v|_2^2=sum_v=1^k m_v sum_i=1^n(x_i-x_v_i)^2$ and therefore
$f'(x)=2sum_v=1^k m_v sum_i=1^n(x_i-x_v_i)$.
So $f'(x)=0 Leftrightarrow x_i=x_v_i forall i=1...n forall v=1...k$
$f''(x)=2sum_v=1^k m_v sum_i=1^n1>0$
And therefore $x=x_v$ truly is the global minimum.
Is that correct? Thanks in advance!
analysis optimization norm
asked Aug 28 at 12:21
user586087
635
Careful however that $xin mathbb R^n$ so that you can't write "$f'(x)$", you should use a partial derivative. Also, is $v$ your index? Don't you have $x_vneq x_v+1$?
– Mefitico
Aug 28 at 12:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$f(x)=sum_v=1^k m_v|x-x_v|_2^2, x,x_v in mathbbR^n, m_v in mathbbR>0$$
($|.|_2^2$ is the euclidean norm but squared)
I need to determine the global minimum of $f(x)$ and reason why it truly is a global minimum.
What I tried:
As $m_v>0$ and because $|x-x_v|_2^2$ is a norm it clear that $f(x)=sum_v=1^k m_v|x-x_v|_2^2 ge 0 forall x in mathbbR^n$ and therefore the global minimum of $f$ is$x=x_v$.
Because I don't know if that's sufficient I also tried this:
$f(x)=sum_v=1^k m_v|x-x_v|_2^2=sum_v=1^k m_v sum_i=1^n(x_i-x_v_i)^2$ and therefore
$f'(x)=2sum_v=1^k m_v sum_i=1^n(x_i-x_v_i)$.
So $f'(x)=0 Leftrightarrow x_i=x_v_i forall i=1...n forall v=1...k$
$f''(x)=2sum_v=1^k m_v sum_i=1^n1>0$
And therefore $x=x_v$ truly is the global minimum.
Is that correct? Thanks in advance!
analysis optimization norm
asked Aug 28 at 12:21
user586087
635
$$f(x)=sum_v=1^k m_v|x-x_v|_2^2, x,x_v in mathbbR^n, m_v in mathbbR>0$$
($|.|_2^2$ is the euclidean norm but squared)
I need to determine the global minimum of $f(x)$ and reason why it truly is a global minimum.
What I tried:
As $m_v>0$ and because $|x-x_v|_2^2$ is a norm it clear that $f(x)=sum_v=1^k m_v|x-x_v|_2^2 ge 0 forall x in mathbbR^n$ and therefore the global minimum of $f$ is$x=x_v$.
Because I don't know if that's sufficient I also tried this:
$f(x)=sum_v=1^k m_v|x-x_v|_2^2=sum_v=1^k m_v sum_i=1^n(x_i-x_v_i)^2$ and therefore
$f'(x)=2sum_v=1^k m_v sum_i=1^n(x_i-x_v_i)$.
So $f'(x)=0 Leftrightarrow x_i=x_v_i forall i=1...n forall v=1...k$
$f''(x)=2sum_v=1^k m_v sum_i=1^n1>0$
And therefore $x=x_v$ truly is the global minimum.
Is that correct? Thanks in advance!
analysis optimization norm
asked Aug 28 at 12:21
user586087
635
asked Aug 28 at 12:21
user586087
635
asked Aug 28 at 12:21
user586087
635
asked Aug 28 at 12:21
user586087
635
635
Careful however that $xin mathbb R^n$ so that you can't write "$f'(x)$", you should use a partial derivative. Also, is $v$ your index? Don't you have $x_vneq x_v+1$?
– Mefitico
Aug 28 at 12:28
add a comment |Â
Careful however that $xin mathbb R^n$ so that you can't write "$f'(x)$", you should use a partial derivative. Also, is $v$ your index? Don't you have $x_vneq x_v+1$?
– Mefitico
Aug 28 at 12:28
Careful however that $xin mathbb R^n$ so that you can't write "$f'(x)$", you should use a partial derivative. Also, is $v$ your index? Don't you have $x_vneq x_v+1$?
– Mefitico
Aug 28 at 12:28
Careful however that $xin mathbb R^n$ so that you can't write "$f'(x)$", you should use a partial derivative. Also, is $v$ your index? Don't you have $x_vneq x_v+1$?
– Mefitico
Aug 28 at 12:28
add a comment |Â
1 Answer
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1
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It can't be $x_v$ because $v$ is an index and there are multiple of them.
$$f(x) = sum_v=1^k m_v|x-x_v|^2$$
Let's differentiate it,
$$nabla f(x) = 2 sum_v=1^k m_v(x-x_v) = 0$$
$$xsum_v=1^k m_v = sum_v=1^k m_vx_v$$
$$x = fracsum_v=1^k m_vx_vsum_v=1^k m_v$$
The optimal solution is the weighted average.
answered Aug 28 at 12:28
Siong Thye Goh
81k1453103
Thanks. Stricly speaking this only gives us a candidate for an extremum, right? Do I now need to differentiate $nabla f(x)$ and show that is greater than $0$ so we can be sure that it is a minimum? Or can I also argue that $f(x) ge 0$ and because for the x you found $f(x)=0$ and therefore it is the global minimum?
– user586087
Aug 28 at 12:49
It's a good exercise to check that it is indeed positive definite and after you check it, you can conclude that it is indeed the global minimum. The optimal solution can't attain $f(x)=0$ if that $x_v$ values are distinct.
– Siong Thye Goh
Aug 28 at 12:56
And is it really $nabla f(x)=2 sum_v=1^k m_v(x-x_v)$ instead of $nabla f(x)=2 sum_v=1^k m_v(x-x_v)^2$?
– user586087
Aug 28 at 14:46
It might help to write out special cases, say when $k=2$, and all $m_v=n=1$, that is $f(x) = (x-x_1)^2 + (x-x_2)^2$, would you like to try and see which case do you get?
– Siong Thye Goh
Aug 28 at 14:54
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It can't be $x_v$ because $v$ is an index and there are multiple of them.
$$f(x) = sum_v=1^k m_v|x-x_v|^2$$
Let's differentiate it,
$$nabla f(x) = 2 sum_v=1^k m_v(x-x_v) = 0$$
$$xsum_v=1^k m_v = sum_v=1^k m_vx_v$$
$$x = fracsum_v=1^k m_vx_vsum_v=1^k m_v$$
The optimal solution is the weighted average.
answered Aug 28 at 12:28
Siong Thye Goh
81k1453103
Thanks. Stricly speaking this only gives us a candidate for an extremum, right? Do I now need to differentiate $nabla f(x)$ and show that is greater than $0$ so we can be sure that it is a minimum? Or can I also argue that $f(x) ge 0$ and because for the x you found $f(x)=0$ and therefore it is the global minimum?
– user586087
Aug 28 at 12:49
It's a good exercise to check that it is indeed positive definite and after you check it, you can conclude that it is indeed the global minimum. The optimal solution can't attain $f(x)=0$ if that $x_v$ values are distinct.
– Siong Thye Goh
Aug 28 at 12:56
And is it really $nabla f(x)=2 sum_v=1^k m_v(x-x_v)$ instead of $nabla f(x)=2 sum_v=1^k m_v(x-x_v)^2$?
– user586087
Aug 28 at 14:46
It might help to write out special cases, say when $k=2$, and all $m_v=n=1$, that is $f(x) = (x-x_1)^2 + (x-x_2)^2$, would you like to try and see which case do you get?
– Siong Thye Goh
Aug 28 at 14:54
add a comment |Â
up vote
1
down vote
accepted
It can't be $x_v$ because $v$ is an index and there are multiple of them.
$$f(x) = sum_v=1^k m_v|x-x_v|^2$$
Let's differentiate it,
$$nabla f(x) = 2 sum_v=1^k m_v(x-x_v) = 0$$
$$xsum_v=1^k m_v = sum_v=1^k m_vx_v$$
$$x = fracsum_v=1^k m_vx_vsum_v=1^k m_v$$
The optimal solution is the weighted average.
answered Aug 28 at 12:28
Siong Thye Goh
81k1453103
Thanks. Stricly speaking this only gives us a candidate for an extremum, right? Do I now need to differentiate $nabla f(x)$ and show that is greater than $0$ so we can be sure that it is a minimum? Or can I also argue that $f(x) ge 0$ and because for the x you found $f(x)=0$ and therefore it is the global minimum?
– user586087
Aug 28 at 12:49
It's a good exercise to check that it is indeed positive definite and after you check it, you can conclude that it is indeed the global minimum. The optimal solution can't attain $f(x)=0$ if that $x_v$ values are distinct.
– Siong Thye Goh
Aug 28 at 12:56
And is it really $nabla f(x)=2 sum_v=1^k m_v(x-x_v)$ instead of $nabla f(x)=2 sum_v=1^k m_v(x-x_v)^2$?
– user586087
Aug 28 at 14:46
It might help to write out special cases, say when $k=2$, and all $m_v=n=1$, that is $f(x) = (x-x_1)^2 + (x-x_2)^2$, would you like to try and see which case do you get?
– Siong Thye Goh
Aug 28 at 14:54
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It can't be $x_v$ because $v$ is an index and there are multiple of them.
$$f(x) = sum_v=1^k m_v|x-x_v|^2$$
Let's differentiate it,
$$nabla f(x) = 2 sum_v=1^k m_v(x-x_v) = 0$$
$$xsum_v=1^k m_v = sum_v=1^k m_vx_v$$
$$x = fracsum_v=1^k m_vx_vsum_v=1^k m_v$$
The optimal solution is the weighted average.
answered Aug 28 at 12:28
Siong Thye Goh
81k1453103
It can't be $x_v$ because $v$ is an index and there are multiple of them.
$$f(x) = sum_v=1^k m_v|x-x_v|^2$$
Let's differentiate it,
$$nabla f(x) = 2 sum_v=1^k m_v(x-x_v) = 0$$
$$xsum_v=1^k m_v = sum_v=1^k m_vx_v$$
$$x = fracsum_v=1^k m_vx_vsum_v=1^k m_v$$
The optimal solution is the weighted average.
answered Aug 28 at 12:28
Siong Thye Goh
81k1453103
answered Aug 28 at 12:28
Siong Thye Goh
81k1453103
answered Aug 28 at 12:28
Siong Thye Goh
81k1453103
answered Aug 28 at 12:28
Siong Thye Goh
81k1453103
81k1453103
Thanks. Stricly speaking this only gives us a candidate for an extremum, right? Do I now need to differentiate $nabla f(x)$ and show that is greater than $0$ so we can be sure that it is a minimum? Or can I also argue that $f(x) ge 0$ and because for the x you found $f(x)=0$ and therefore it is the global minimum?
– user586087
Aug 28 at 12:49
It's a good exercise to check that it is indeed positive definite and after you check it, you can conclude that it is indeed the global minimum. The optimal solution can't attain $f(x)=0$ if that $x_v$ values are distinct.
– Siong Thye Goh
Aug 28 at 12:56
And is it really $nabla f(x)=2 sum_v=1^k m_v(x-x_v)$ instead of $nabla f(x)=2 sum_v=1^k m_v(x-x_v)^2$?
– user586087
Aug 28 at 14:46
It might help to write out special cases, say when $k=2$, and all $m_v=n=1$, that is $f(x) = (x-x_1)^2 + (x-x_2)^2$, would you like to try and see which case do you get?
– Siong Thye Goh
Aug 28 at 14:54
add a comment |Â
Thanks. Stricly speaking this only gives us a candidate for an extremum, right? Do I now need to differentiate $nabla f(x)$ and show that is greater than $0$ so we can be sure that it is a minimum? Or can I also argue that $f(x) ge 0$ and because for the x you found $f(x)=0$ and therefore it is the global minimum?
– user586087
Aug 28 at 12:49
It's a good exercise to check that it is indeed positive definite and after you check it, you can conclude that it is indeed the global minimum. The optimal solution can't attain $f(x)=0$ if that $x_v$ values are distinct.
– Siong Thye Goh
Aug 28 at 12:56
And is it really $nabla f(x)=2 sum_v=1^k m_v(x-x_v)$ instead of $nabla f(x)=2 sum_v=1^k m_v(x-x_v)^2$?
– user586087
Aug 28 at 14:46
It might help to write out special cases, say when $k=2$, and all $m_v=n=1$, that is $f(x) = (x-x_1)^2 + (x-x_2)^2$, would you like to try and see which case do you get?
– Siong Thye Goh
Aug 28 at 14:54
Thanks. Stricly speaking this only gives us a candidate for an extremum, right? Do I now need to differentiate $nabla f(x)$ and show that is greater than $0$ so we can be sure that it is a minimum? Or can I also argue that $f(x) ge 0$ and because for the x you found $f(x)=0$ and therefore it is the global minimum?
– user586087
Aug 28 at 12:49
Thanks. Stricly speaking this only gives us a candidate for an extremum, right? Do I now need to differentiate $nabla f(x)$ and show that is greater than $0$ so we can be sure that it is a minimum? Or can I also argue that $f(x) ge 0$ and because for the x you found $f(x)=0$ and therefore it is the global minimum?
– user586087
Aug 28 at 12:49
It's a good exercise to check that it is indeed positive definite and after you check it, you can conclude that it is indeed the global minimum. The optimal solution can't attain $f(x)=0$ if that $x_v$ values are distinct.
– Siong Thye Goh
Aug 28 at 12:56
It's a good exercise to check that it is indeed positive definite and after you check it, you can conclude that it is indeed the global minimum. The optimal solution can't attain $f(x)=0$ if that $x_v$ values are distinct.
– Siong Thye Goh
Aug 28 at 12:56
And is it really $nabla f(x)=2 sum_v=1^k m_v(x-x_v)$ instead of $nabla f(x)=2 sum_v=1^k m_v(x-x_v)^2$?
– user586087
Aug 28 at 14:46
And is it really $nabla f(x)=2 sum_v=1^k m_v(x-x_v)$ instead of $nabla f(x)=2 sum_v=1^k m_v(x-x_v)^2$?
– user586087
Aug 28 at 14:46
It might help to write out special cases, say when $k=2$, and all $m_v=n=1$, that is $f(x) = (x-x_1)^2 + (x-x_2)^2$, would you like to try and see which case do you get?
– Siong Thye Goh
Aug 28 at 14:54
It might help to write out special cases, say when $k=2$, and all $m_v=n=1$, that is $f(x) = (x-x_1)^2 + (x-x_2)^2$, would you like to try and see which case do you get?
– Siong Thye Goh
Aug 28 at 14:54
add a comment |Â
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Careful however that $xin mathbb R^n$ so that you can't write "$f'(x)$", you should use a partial derivative. Also, is $v$ your index? Don't you have $x_vneq x_v+1$?
– Mefitico
Aug 28 at 12:28