Derivative of function with 2 summations and powers












0














I'm trying to compute the gradient of a complex loss function but I'm stuck on calculating the derivative of part of the expression,namely:



with respect to $w_i$



$$sum_{n=1}^N(y^n -w_0 - sum_{j=1}^p A_{nj}w_j)^2$$



A - design matrix;
w - vector; y - vector; $w_0$ - first element of w



Edit
Where it said k it should have been n - now fixed






derivatives partial-derivative







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  • How can $y$ be a vector? Should it be $y_n$ rather than $y^n$?
    – smcc
    Nov 25 '18 at 11:14
















0














I'm trying to compute the gradient of a complex loss function but I'm stuck on calculating the derivative of part of the expression,namely:



with respect to $w_i$



$$sum_{n=1}^N(y^n -w_0 - sum_{j=1}^p A_{nj}w_j)^2$$



A - design matrix;
w - vector; y - vector; $w_0$ - first element of w



Edit
Where it said k it should have been n - now fixed






derivatives partial-derivative







share|cite|improve this question
























  • How can $y$ be a vector? Should it be $y_n$ rather than $y^n$?
    – smcc
    Nov 25 '18 at 11:14














0












0








0







I'm trying to compute the gradient of a complex loss function but I'm stuck on calculating the derivative of part of the expression,namely:



with respect to $w_i$



$$sum_{n=1}^N(y^n -w_0 - sum_{j=1}^p A_{nj}w_j)^2$$



A - design matrix;
w - vector; y - vector; $w_0$ - first element of w



Edit
Where it said k it should have been n - now fixed






derivatives partial-derivative







share|cite|improve this question















I'm trying to compute the gradient of a complex loss function but I'm stuck on calculating the derivative of part of the expression,namely:



with respect to $w_i$



$$sum_{n=1}^N(y^n -w_0 - sum_{j=1}^p A_{nj}w_j)^2$$



A - design matrix;
w - vector; y - vector; $w_0$ - first element of w



Edit
Where it said k it should have been n - now fixed






derivatives partial-derivative




derivatives partial-derivative






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share|cite|improve this question













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edited Nov 25 '18 at 11:19

























asked Nov 25 '18 at 10:16









Sreten Jocić

174




174












  • How can $y$ be a vector? Should it be $y_n$ rather than $y^n$?
    – smcc
    Nov 25 '18 at 11:14


















  • How can $y$ be a vector? Should it be $y_n$ rather than $y^n$?
    – smcc
    Nov 25 '18 at 11:14
















How can $y$ be a vector? Should it be $y_n$ rather than $y^n$?
– smcc
Nov 25 '18 at 11:14




How can $y$ be a vector? Should it be $y_n$ rather than $y^n$?
– smcc
Nov 25 '18 at 11:14










1 Answer
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For $iin{1,ldots,p}$:



$$begin{align*}&phantom{=} , frac{partial}{partial w_i}sum_{n=1}^Nleft[left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)^2right]\
&=sum_{n=1}^Nfrac{partial}{partial w_i}left[left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)^2right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)frac{partial}{partial w_i}left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)left( -sum_{j=1}^p A_{nj}frac{partial w_j}{partial w_i}right)right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)left( - A_{ni}right)right]\
&=-2sum_{n=1}^Nleft[A_{ni}left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)right]
end{align*}$$






share|cite|improve this answer























  • I'm so sorry I just noticed my mistake where it sais k it should be n, will substituting it everywhere be sufficient?
    – Sreten Jocić
    Nov 25 '18 at 11:20












  • Yes, it does not matter whether the upper limit of the sum is called $N$ or $K$.
    – smcc
    Nov 25 '18 at 11:22










  • I meant "k" as the index, sorry
    – Sreten Jocić
    Nov 25 '18 at 11:22










  • Changing $k$ to $j$ in $A_{kj}$ just changes the last lines. I have updated my answer.
    – smcc
    Nov 25 '18 at 11:26











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1 Answer
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For $iin{1,ldots,p}$:



$$begin{align*}&phantom{=} , frac{partial}{partial w_i}sum_{n=1}^Nleft[left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)^2right]\
&=sum_{n=1}^Nfrac{partial}{partial w_i}left[left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)^2right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)frac{partial}{partial w_i}left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)left( -sum_{j=1}^p A_{nj}frac{partial w_j}{partial w_i}right)right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)left( - A_{ni}right)right]\
&=-2sum_{n=1}^Nleft[A_{ni}left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)right]
end{align*}$$






share|cite|improve this answer























  • I'm so sorry I just noticed my mistake where it sais k it should be n, will substituting it everywhere be sufficient?
    – Sreten Jocić
    Nov 25 '18 at 11:20












  • Yes, it does not matter whether the upper limit of the sum is called $N$ or $K$.
    – smcc
    Nov 25 '18 at 11:22










  • I meant "k" as the index, sorry
    – Sreten Jocić
    Nov 25 '18 at 11:22










  • Changing $k$ to $j$ in $A_{kj}$ just changes the last lines. I have updated my answer.
    – smcc
    Nov 25 '18 at 11:26
















0














For $iin{1,ldots,p}$:



$$begin{align*}&phantom{=} , frac{partial}{partial w_i}sum_{n=1}^Nleft[left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)^2right]\
&=sum_{n=1}^Nfrac{partial}{partial w_i}left[left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)^2right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)frac{partial}{partial w_i}left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)left( -sum_{j=1}^p A_{nj}frac{partial w_j}{partial w_i}right)right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)left( - A_{ni}right)right]\
&=-2sum_{n=1}^Nleft[A_{ni}left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)right]
end{align*}$$






share|cite|improve this answer























  • I'm so sorry I just noticed my mistake where it sais k it should be n, will substituting it everywhere be sufficient?
    – Sreten Jocić
    Nov 25 '18 at 11:20












  • Yes, it does not matter whether the upper limit of the sum is called $N$ or $K$.
    – smcc
    Nov 25 '18 at 11:22










  • I meant "k" as the index, sorry
    – Sreten Jocić
    Nov 25 '18 at 11:22










  • Changing $k$ to $j$ in $A_{kj}$ just changes the last lines. I have updated my answer.
    – smcc
    Nov 25 '18 at 11:26














0












0








0






For $iin{1,ldots,p}$:



$$begin{align*}&phantom{=} , frac{partial}{partial w_i}sum_{n=1}^Nleft[left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)^2right]\
&=sum_{n=1}^Nfrac{partial}{partial w_i}left[left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)^2right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)frac{partial}{partial w_i}left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)left( -sum_{j=1}^p A_{nj}frac{partial w_j}{partial w_i}right)right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)left( - A_{ni}right)right]\
&=-2sum_{n=1}^Nleft[A_{ni}left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)right]
end{align*}$$






share|cite|improve this answer














For $iin{1,ldots,p}$:



$$begin{align*}&phantom{=} , frac{partial}{partial w_i}sum_{n=1}^Nleft[left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)^2right]\
&=sum_{n=1}^Nfrac{partial}{partial w_i}left[left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)^2right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)frac{partial}{partial w_i}left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)left( -sum_{j=1}^p A_{nj}frac{partial w_j}{partial w_i}right)right]\
&=sum_{n=1}^Nleft[2left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)left( - A_{ni}right)right]\
&=-2sum_{n=1}^Nleft[A_{ni}left(y^n -w_0 - sum_{j=1}^p A_{nj}w_jright)right]
end{align*}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 '18 at 11:25

























answered Nov 25 '18 at 11:13









smcc

4,297517




4,297517












  • I'm so sorry I just noticed my mistake where it sais k it should be n, will substituting it everywhere be sufficient?
    – Sreten Jocić
    Nov 25 '18 at 11:20












  • Yes, it does not matter whether the upper limit of the sum is called $N$ or $K$.
    – smcc
    Nov 25 '18 at 11:22










  • I meant "k" as the index, sorry
    – Sreten Jocić
    Nov 25 '18 at 11:22










  • Changing $k$ to $j$ in $A_{kj}$ just changes the last lines. I have updated my answer.
    – smcc
    Nov 25 '18 at 11:26


















  • I'm so sorry I just noticed my mistake where it sais k it should be n, will substituting it everywhere be sufficient?
    – Sreten Jocić
    Nov 25 '18 at 11:20












  • Yes, it does not matter whether the upper limit of the sum is called $N$ or $K$.
    – smcc
    Nov 25 '18 at 11:22










  • I meant "k" as the index, sorry
    – Sreten Jocić
    Nov 25 '18 at 11:22










  • Changing $k$ to $j$ in $A_{kj}$ just changes the last lines. I have updated my answer.
    – smcc
    Nov 25 '18 at 11:26
















I'm so sorry I just noticed my mistake where it sais k it should be n, will substituting it everywhere be sufficient?
– Sreten Jocić
Nov 25 '18 at 11:20






I'm so sorry I just noticed my mistake where it sais k it should be n, will substituting it everywhere be sufficient?
– Sreten Jocić
Nov 25 '18 at 11:20














Yes, it does not matter whether the upper limit of the sum is called $N$ or $K$.
– smcc
Nov 25 '18 at 11:22




Yes, it does not matter whether the upper limit of the sum is called $N$ or $K$.
– smcc
Nov 25 '18 at 11:22












I meant "k" as the index, sorry
– Sreten Jocić
Nov 25 '18 at 11:22




I meant "k" as the index, sorry
– Sreten Jocić
Nov 25 '18 at 11:22












Changing $k$ to $j$ in $A_{kj}$ just changes the last lines. I have updated my answer.
– smcc
Nov 25 '18 at 11:26




Changing $k$ to $j$ in $A_{kj}$ just changes the last lines. I have updated my answer.
– smcc
Nov 25 '18 at 11:26


















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