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How do I take derivative that involves double summation?

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1

$begingroup$

Suppose I have a function like this:

$$E(U,V) =sum_(u,i)in M (M_u,i -U^mathrm T_u V_i)^2 = sum_(u,i)in M left( M_u,i -sum_k=1^r U_u,k V_i,k right)^2.$$

How do I take the partial derivative of $E$ with respect to $U_u, i$?

Could anyone point me to some reference or give me some suggestion on how to solve this?

calculus linear-algebra summation partial-derivative

share|cite|improve this question

edited Mar 10 at 19:23

Rócherz

2,9262821

asked Nov 13 '15 at 4:58

JennyJenny

213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For future reference, there is an option to embed images (little picture icon of a mountain). All you need to provide is the URL.
  $endgroup$
  – zahbaz
  Nov 13 '15 at 5:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You need to clarify I think. What's $M_u,i$ etc? Functions or constants? Matrix entries?
  $endgroup$
  – Benjamin Lindqvist
  Nov 13 '15 at 8:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't matter. It's just a way of specifying variables. But for the sake of the argument, in this context, it's a matrix entry.
  $endgroup$
  – Jenny
  Nov 13 '15 at 12:55
  

  
  
  
  

add a comment | 

1

$begingroup$

Suppose I have a function like this:

$$E(U,V) =sum_(u,i)in M (M_u,i -U^mathrm T_u V_i)^2 = sum_(u,i)in M left( M_u,i -sum_k=1^r U_u,k V_i,k right)^2.$$

How do I take the partial derivative of $E$ with respect to $U_u, i$?

Could anyone point me to some reference or give me some suggestion on how to solve this?

calculus linear-algebra summation partial-derivative

share|cite|improve this question

edited Mar 10 at 19:23

Rócherz

2,9262821

asked Nov 13 '15 at 4:58

JennyJenny

213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For future reference, there is an option to embed images (little picture icon of a mountain). All you need to provide is the URL.
  $endgroup$
  – zahbaz
  Nov 13 '15 at 5:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You need to clarify I think. What's $M_u,i$ etc? Functions or constants? Matrix entries?
  $endgroup$
  – Benjamin Lindqvist
  Nov 13 '15 at 8:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't matter. It's just a way of specifying variables. But for the sake of the argument, in this context, it's a matrix entry.
  $endgroup$
  – Jenny
  Nov 13 '15 at 12:55
  

  
  
  
  

add a comment | 

$begingroup$

Suppose I have a function like this:

$$E(U,V) =sum_(u,i)in M (M_u,i -U^mathrm T_u V_i)^2 = sum_(u,i)in M left( M_u,i -sum_k=1^r U_u,k V_i,k right)^2.$$

How do I take the partial derivative of $E$ with respect to $U_u, i$?

Could anyone point me to some reference or give me some suggestion on how to solve this?

calculus linear-algebra summation partial-derivative

share|cite|improve this question

edited Mar 10 at 19:23

Rócherz

2,9262821

asked Nov 13 '15 at 4:58

JennyJenny

213

$endgroup$

share|cite|improve this question

edited Mar 10 at 19:23

Rócherz

2,9262821

asked Nov 13 '15 at 4:58

JennyJenny

213

share|cite|improve this question

edited Mar 10 at 19:23

Rócherz

2,9262821

asked Nov 13 '15 at 4:58

JennyJenny

213

edited Mar 10 at 19:23

Rócherz

2,9262821

edited Mar 10 at 19:23

Rócherz

2,9262821

asked Nov 13 '15 at 4:58

JennyJenny

213

asked Nov 13 '15 at 4:58

JennyJenny

213

1 Answer
1

active

oldest

votes

1

$begingroup$

Lets say you want the partial derivative with respect to $U_mn$. Most of the summands are treated as constants so derivation yields:

$$fracpartial Epartial U_mn(U, V) = fracpartialpartial U_mn sum_i (M_mi-U_mnV_in)^2$$

Using the linearity of the derivation operator and the chain rule we have:

$$=sum_ifracpartialpartial U_mn (M_mi-U_mnV_in)^2 = sum_i 2(M_mi-U_mnV_in)(-V_in).$$
Which can be written as
$$=-2(MV)_mn + 2U_mn ||V_n||^2$$
where $(MV)_mn$ is the $mn$-th entry of the matrix product $MV$ and $||V_n||$ is the euclidean norm of the $n$-th column of $V$.

share|cite|improve this answer

edited Nov 13 '15 at 14:59

answered Nov 13 '15 at 14:45

testmantestman

52729

$endgroup$

add a comment | 

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1

$begingroup$

Lets say you want the partial derivative with respect to $U_mn$. Most of the summands are treated as constants so derivation yields:

$$fracpartial Epartial U_mn(U, V) = fracpartialpartial U_mn sum_i (M_mi-U_mnV_in)^2$$

Using the linearity of the derivation operator and the chain rule we have:

$$=sum_ifracpartialpartial U_mn (M_mi-U_mnV_in)^2 = sum_i 2(M_mi-U_mnV_in)(-V_in).$$
Which can be written as
$$=-2(MV)_mn + 2U_mn ||V_n||^2$$
where $(MV)_mn$ is the $mn$-th entry of the matrix product $MV$ and $||V_n||$ is the euclidean norm of the $n$-th column of $V$.

share|cite|improve this answer

edited Nov 13 '15 at 14:59

answered Nov 13 '15 at 14:45

testmantestman

52729

$endgroup$

add a comment | 

1

$begingroup$

Lets say you want the partial derivative with respect to $U_mn$. Most of the summands are treated as constants so derivation yields:

$$fracpartial Epartial U_mn(U, V) = fracpartialpartial U_mn sum_i (M_mi-U_mnV_in)^2$$

Using the linearity of the derivation operator and the chain rule we have:

$$=sum_ifracpartialpartial U_mn (M_mi-U_mnV_in)^2 = sum_i 2(M_mi-U_mnV_in)(-V_in).$$
Which can be written as
$$=-2(MV)_mn + 2U_mn ||V_n||^2$$
where $(MV)_mn$ is the $mn$-th entry of the matrix product $MV$ and $||V_n||$ is the euclidean norm of the $n$-th column of $V$.

share|cite|improve this answer

edited Nov 13 '15 at 14:59

answered Nov 13 '15 at 14:45

testmantestman

52729

$endgroup$

add a comment | 

$begingroup$

Lets say you want the partial derivative with respect to $U_mn$. Most of the summands are treated as constants so derivation yields:

$$fracpartial Epartial U_mn(U, V) = fracpartialpartial U_mn sum_i (M_mi-U_mnV_in)^2$$

Using the linearity of the derivation operator and the chain rule we have:

$$=sum_ifracpartialpartial U_mn (M_mi-U_mnV_in)^2 = sum_i 2(M_mi-U_mnV_in)(-V_in).$$
Which can be written as
$$=-2(MV)_mn + 2U_mn ||V_n||^2$$
where $(MV)_mn$ is the $mn$-th entry of the matrix product $MV$ and $||V_n||$ is the euclidean norm of the $n$-th column of $V$.

share|cite|improve this answer

edited Nov 13 '15 at 14:59

answered Nov 13 '15 at 14:45

testmantestman

52729

$endgroup$

share|cite|improve this answer

edited Nov 13 '15 at 14:59

answered Nov 13 '15 at 14:45

testmantestman

52729

answered Nov 13 '15 at 14:45

testmantestman

52729

answered Nov 13 '15 at 14:45

testmantestman

52729