Energy functions and convex combinations

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There are situations (in optimization problems) where problems are formulated as

$$
f(E_1,E_2 ; lambda) = E_1 + lambda E_2
$$

where $E_1,E_2$ are some energy functions, and $lambda$ is a real value that "weights" the contributions of such energies. I wonder if there's any relationship with a formula of the form

$$
g(E_1,E_2;lambda) = (1-lambda)E_1+lambda E_2
$$

is there maybe some reparametrization that would allow to switch from one form to the other?

Thank you

calculus optimization soft-question

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asked Jul 20 at 17:59

user8469759

1,4291513

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up vote
0
down vote

favorite

There are situations (in optimization problems) where problems are formulated as

$$
f(E_1,E_2 ; lambda) = E_1 + lambda E_2
$$

where $E_1,E_2$ are some energy functions, and $lambda$ is a real value that "weights" the contributions of such energies. I wonder if there's any relationship with a formula of the form

$$
g(E_1,E_2;lambda) = (1-lambda)E_1+lambda E_2
$$

is there maybe some reparametrization that would allow to switch from one form to the other?

Thank you

calculus optimization soft-question

share|cite|improve this question

asked Jul 20 at 17:59

user8469759

1,4291513

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

There are situations (in optimization problems) where problems are formulated as

$$
f(E_1,E_2 ; lambda) = E_1 + lambda E_2
$$

where $E_1,E_2$ are some energy functions, and $lambda$ is a real value that "weights" the contributions of such energies. I wonder if there's any relationship with a formula of the form

$$
g(E_1,E_2;lambda) = (1-lambda)E_1+lambda E_2
$$

is there maybe some reparametrization that would allow to switch from one form to the other?

Thank you

calculus optimization soft-question

share|cite|improve this question

asked Jul 20 at 17:59

user8469759

1,4291513

There are situations (in optimization problems) where problems are formulated as

$$
f(E_1,E_2 ; lambda) = E_1 + lambda E_2
$$

where $E_1,E_2$ are some energy functions, and $lambda$ is a real value that "weights" the contributions of such energies. I wonder if there's any relationship with a formula of the form

$$
g(E_1,E_2;lambda) = (1-lambda)E_1+lambda E_2
$$

is there maybe some reparametrization that would allow to switch from one form to the other?

Thank you

calculus optimization soft-question

share|cite|improve this question

asked Jul 20 at 17:59

user8469759

1,4291513

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 17:59

user8469759

1,4291513

asked Jul 20 at 17:59

user8469759

1,4291513

asked Jul 20 at 17:59

user8469759

1,4291513

1,4291513

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

In general, the optimal solution (but not the objective value) of $min_x f(x)$ and $min_x cf(x)$ are the same for $c > 0$. Therefore,
$$f(E_1,E_2 ; lambda_f) = E_1 + lambda_f E_2$$
and
$$h(E_1,E_2 ; lambda_f) = frac11+lambda_f E_1 + fraclambda_f1+lambda_f E_2$$
have the same solution (take $c=1/(1+lambda_f)>0$). You can see $h$ as $g$ where the $lambda_g$ in $g$ equals $lambda_f / (1+lambda_f)$.

share|cite|improve this answer

edited Jul 20 at 23:41

answered Jul 20 at 18:21

LinAlg

5,4061319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't get your notation, what's $lambda^f$?
  – user8469759
  Jul 20 at 22:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user8469759 it is the $lambda$ in your function $f$; I noticed it wasn't consistent, so I fixed that.
  – LinAlg
  Jul 20 at 23:40
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

In general, the optimal solution (but not the objective value) of $min_x f(x)$ and $min_x cf(x)$ are the same for $c > 0$. Therefore,
$$f(E_1,E_2 ; lambda_f) = E_1 + lambda_f E_2$$
and
$$h(E_1,E_2 ; lambda_f) = frac11+lambda_f E_1 + fraclambda_f1+lambda_f E_2$$
have the same solution (take $c=1/(1+lambda_f)>0$). You can see $h$ as $g$ where the $lambda_g$ in $g$ equals $lambda_f / (1+lambda_f)$.

share|cite|improve this answer

edited Jul 20 at 23:41

answered Jul 20 at 18:21

LinAlg

5,4061319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't get your notation, what's $lambda^f$?
  – user8469759
  Jul 20 at 22:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user8469759 it is the $lambda$ in your function $f$; I noticed it wasn't consistent, so I fixed that.
  – LinAlg
  Jul 20 at 23:40
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

In general, the optimal solution (but not the objective value) of $min_x f(x)$ and $min_x cf(x)$ are the same for $c > 0$. Therefore,
$$f(E_1,E_2 ; lambda_f) = E_1 + lambda_f E_2$$
and
$$h(E_1,E_2 ; lambda_f) = frac11+lambda_f E_1 + fraclambda_f1+lambda_f E_2$$
have the same solution (take $c=1/(1+lambda_f)>0$). You can see $h$ as $g$ where the $lambda_g$ in $g$ equals $lambda_f / (1+lambda_f)$.

share|cite|improve this answer

edited Jul 20 at 23:41

answered Jul 20 at 18:21

LinAlg

5,4061319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't get your notation, what's $lambda^f$?
  – user8469759
  Jul 20 at 22:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user8469759 it is the $lambda$ in your function $f$; I noticed it wasn't consistent, so I fixed that.
  – LinAlg
  Jul 20 at 23:40
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

In general, the optimal solution (but not the objective value) of $min_x f(x)$ and $min_x cf(x)$ are the same for $c > 0$. Therefore,
$$f(E_1,E_2 ; lambda_f) = E_1 + lambda_f E_2$$
and
$$h(E_1,E_2 ; lambda_f) = frac11+lambda_f E_1 + fraclambda_f1+lambda_f E_2$$
have the same solution (take $c=1/(1+lambda_f)>0$). You can see $h$ as $g$ where the $lambda_g$ in $g$ equals $lambda_f / (1+lambda_f)$.

share|cite|improve this answer

edited Jul 20 at 23:41

answered Jul 20 at 18:21

LinAlg

5,4061319

In general, the optimal solution (but not the objective value) of $min_x f(x)$ and $min_x cf(x)$ are the same for $c > 0$. Therefore,
$$f(E_1,E_2 ; lambda_f) = E_1 + lambda_f E_2$$
and
$$h(E_1,E_2 ; lambda_f) = frac11+lambda_f E_1 + fraclambda_f1+lambda_f E_2$$
have the same solution (take $c=1/(1+lambda_f)>0$). You can see $h$ as $g$ where the $lambda_g$ in $g$ equals $lambda_f / (1+lambda_f)$.

share|cite|improve this answer

edited Jul 20 at 23:41

answered Jul 20 at 18:21

LinAlg

5,4061319

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 23:41

edited Jul 20 at 23:41

edited Jul 20 at 23:41

answered Jul 20 at 18:21

LinAlg

5,4061319

answered Jul 20 at 18:21

LinAlg

5,4061319

answered Jul 20 at 18:21

LinAlg

5,4061319

5,4061319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't get your notation, what's $lambda^f$?
  – user8469759
  Jul 20 at 22:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user8469759 it is the $lambda$ in your function $f$; I noticed it wasn't consistent, so I fixed that.
  – LinAlg
  Jul 20 at 23:40
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't get your notation, what's $lambda^f$?
  – user8469759
  Jul 20 at 22:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user8469759 it is the $lambda$ in your function $f$; I noticed it wasn't consistent, so I fixed that.
  – LinAlg
  Jul 20 at 23:40
  
  

  
  
  
  

I don't get your notation, what's $lambda^f$?
– user8469759
Jul 20 at 22:46

I don't get your notation, what's $lambda^f$?
– user8469759
Jul 20 at 22:46

@user8469759 it is the $lambda$ in your function $f$; I noticed it wasn't consistent, so I fixed that.
– LinAlg
Jul 20 at 23:40

@user8469759 it is the $lambda$ in your function $f$; I noticed it wasn't consistent, so I fixed that.
– LinAlg
Jul 20 at 23:40

add a comment | 

 

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