Why is $nabla f(p) = sum_{i=1}^n 2(p - p_i) = 0$?












0












$begingroup$


Given $n$ points in space, $p_1,p_2, cdots , p_n$, find the point $p$ for which $f(p) = sum_{i=1}^n |p-p_j|^2$.



I have the following in my notes:



(1) $f(p) = sum_{i=1}^n |p-p_i|^2 = sum_{i=1}^n (p-p_i)cdot(p-p_i)$



=$sum_{i=1}^n p cdot p - 2p cdot p_i + p_{i} cdot p_{i}$



.........................................



(2) Find $p$ to minimize $f(p)$.



$nabla f(p) = sum_{i=1}^n 2(p - p_i) = 0$



$np = sum_{i=1}^n p_i rightarrow frac{1}{n} sum_{i=1}^n p_i = bar{p}$



$bar{p}$ is a critical point for $f(p)$



.........................................



(3) $D^2 f = 2I_n$



.........................................



(4) $f(p)-f(bar{p}) = sum_{i=1}^n |p-bar{p_i}|^2 - |bar{p} - p_i|^2$



= $sum_{i=1}^n |p|^2 -2p cdot p_i + |p_i|^2 - |bar{p}|^2 +2bar{p} cdot p_i - |p_i|^2$



= $sum_{i=1}^n |p|^2 - |bar{p}|^2 + 2(bar{p}-p) cdot p_i$



=$2bar{p} cdot p_i = bar{p} cdot (sum p_i) cdot 2 = bar{p} cdot nbar{p} cdot 2 = 2n|bar{p}|^2$



= $n(|p|^2 + |bar{p}|^2 -2npcdot bar{p} = n(|p-bar{p}|^2) ge 0$



I find that part (1) is self-explanatory.



For part (2), why do we do $nabla f(p) = sum_{i=1}^n 2(p cdot p_i) = 0$?






calculus multivariable-calculus







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$endgroup$








  • 1




    $begingroup$
    the dot should be a minus sign; maybe you transcribed something incorrectly or wherever you got the notes has a typo
    $endgroup$
    – T_M
    Dec 5 '18 at 0:49
















0












$begingroup$


Given $n$ points in space, $p_1,p_2, cdots , p_n$, find the point $p$ for which $f(p) = sum_{i=1}^n |p-p_j|^2$.



I have the following in my notes:



(1) $f(p) = sum_{i=1}^n |p-p_i|^2 = sum_{i=1}^n (p-p_i)cdot(p-p_i)$



=$sum_{i=1}^n p cdot p - 2p cdot p_i + p_{i} cdot p_{i}$



.........................................



(2) Find $p$ to minimize $f(p)$.



$nabla f(p) = sum_{i=1}^n 2(p - p_i) = 0$



$np = sum_{i=1}^n p_i rightarrow frac{1}{n} sum_{i=1}^n p_i = bar{p}$



$bar{p}$ is a critical point for $f(p)$



.........................................



(3) $D^2 f = 2I_n$



.........................................



(4) $f(p)-f(bar{p}) = sum_{i=1}^n |p-bar{p_i}|^2 - |bar{p} - p_i|^2$



= $sum_{i=1}^n |p|^2 -2p cdot p_i + |p_i|^2 - |bar{p}|^2 +2bar{p} cdot p_i - |p_i|^2$



= $sum_{i=1}^n |p|^2 - |bar{p}|^2 + 2(bar{p}-p) cdot p_i$



=$2bar{p} cdot p_i = bar{p} cdot (sum p_i) cdot 2 = bar{p} cdot nbar{p} cdot 2 = 2n|bar{p}|^2$



= $n(|p|^2 + |bar{p}|^2 -2npcdot bar{p} = n(|p-bar{p}|^2) ge 0$



I find that part (1) is self-explanatory.



For part (2), why do we do $nabla f(p) = sum_{i=1}^n 2(p cdot p_i) = 0$?






calculus multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    the dot should be a minus sign; maybe you transcribed something incorrectly or wherever you got the notes has a typo
    $endgroup$
    – T_M
    Dec 5 '18 at 0:49














0












0








0





$begingroup$


Given $n$ points in space, $p_1,p_2, cdots , p_n$, find the point $p$ for which $f(p) = sum_{i=1}^n |p-p_j|^2$.



I have the following in my notes:



(1) $f(p) = sum_{i=1}^n |p-p_i|^2 = sum_{i=1}^n (p-p_i)cdot(p-p_i)$



=$sum_{i=1}^n p cdot p - 2p cdot p_i + p_{i} cdot p_{i}$



.........................................



(2) Find $p$ to minimize $f(p)$.



$nabla f(p) = sum_{i=1}^n 2(p - p_i) = 0$



$np = sum_{i=1}^n p_i rightarrow frac{1}{n} sum_{i=1}^n p_i = bar{p}$



$bar{p}$ is a critical point for $f(p)$



.........................................



(3) $D^2 f = 2I_n$



.........................................



(4) $f(p)-f(bar{p}) = sum_{i=1}^n |p-bar{p_i}|^2 - |bar{p} - p_i|^2$



= $sum_{i=1}^n |p|^2 -2p cdot p_i + |p_i|^2 - |bar{p}|^2 +2bar{p} cdot p_i - |p_i|^2$



= $sum_{i=1}^n |p|^2 - |bar{p}|^2 + 2(bar{p}-p) cdot p_i$



=$2bar{p} cdot p_i = bar{p} cdot (sum p_i) cdot 2 = bar{p} cdot nbar{p} cdot 2 = 2n|bar{p}|^2$



= $n(|p|^2 + |bar{p}|^2 -2npcdot bar{p} = n(|p-bar{p}|^2) ge 0$



I find that part (1) is self-explanatory.



For part (2), why do we do $nabla f(p) = sum_{i=1}^n 2(p cdot p_i) = 0$?






calculus multivariable-calculus







share|cite|improve this question











$endgroup$




Given $n$ points in space, $p_1,p_2, cdots , p_n$, find the point $p$ for which $f(p) = sum_{i=1}^n |p-p_j|^2$.



I have the following in my notes:



(1) $f(p) = sum_{i=1}^n |p-p_i|^2 = sum_{i=1}^n (p-p_i)cdot(p-p_i)$



=$sum_{i=1}^n p cdot p - 2p cdot p_i + p_{i} cdot p_{i}$



.........................................



(2) Find $p$ to minimize $f(p)$.



$nabla f(p) = sum_{i=1}^n 2(p - p_i) = 0$



$np = sum_{i=1}^n p_i rightarrow frac{1}{n} sum_{i=1}^n p_i = bar{p}$



$bar{p}$ is a critical point for $f(p)$



.........................................



(3) $D^2 f = 2I_n$



.........................................



(4) $f(p)-f(bar{p}) = sum_{i=1}^n |p-bar{p_i}|^2 - |bar{p} - p_i|^2$



= $sum_{i=1}^n |p|^2 -2p cdot p_i + |p_i|^2 - |bar{p}|^2 +2bar{p} cdot p_i - |p_i|^2$



= $sum_{i=1}^n |p|^2 - |bar{p}|^2 + 2(bar{p}-p) cdot p_i$



=$2bar{p} cdot p_i = bar{p} cdot (sum p_i) cdot 2 = bar{p} cdot nbar{p} cdot 2 = 2n|bar{p}|^2$



= $n(|p|^2 + |bar{p}|^2 -2npcdot bar{p} = n(|p-bar{p}|^2) ge 0$



I find that part (1) is self-explanatory.



For part (2), why do we do $nabla f(p) = sum_{i=1}^n 2(p cdot p_i) = 0$?






calculus multivariable-calculus




calculus multivariable-calculus






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share|cite|improve this question













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share|cite|improve this question








edited Dec 5 '18 at 2:00







K.M

















asked Dec 5 '18 at 0:43









K.MK.M

686412




686412








  • 1




    $begingroup$
    the dot should be a minus sign; maybe you transcribed something incorrectly or wherever you got the notes has a typo
    $endgroup$
    – T_M
    Dec 5 '18 at 0:49














  • 1




    $begingroup$
    the dot should be a minus sign; maybe you transcribed something incorrectly or wherever you got the notes has a typo
    $endgroup$
    – T_M
    Dec 5 '18 at 0:49








1




1




$begingroup$
the dot should be a minus sign; maybe you transcribed something incorrectly or wherever you got the notes has a typo
$endgroup$
– T_M
Dec 5 '18 at 0:49




$begingroup$
the dot should be a minus sign; maybe you transcribed something incorrectly or wherever you got the notes has a typo
$endgroup$
– T_M
Dec 5 '18 at 0:49










1 Answer
1






active

oldest

votes


















1












$begingroup$

Take



$$
f({bf p}) = sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_iright)
$$



The gradient is just



begin{eqnarray}
nabla f({bf p}) &=& frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) \
&=& sum_i left( 2{bf p} - 2{bf p}_i right) = 2sum_i {bf p} - 2sum_i{bf p}_i \
&=& 2n {bf p} - 2sum_i {bf p}_i = 2n left({bf p}-frac{1}{n}sum_i{bf p}_iright) \
&=& 2n ({bf p} - overline{{bf p}})
end{eqnarray}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is $frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) = sum_i left( 2{bf p} - 2{bf p}_i right)$?
    $endgroup$
    – K.M
    Dec 5 '18 at 1:01












  • $begingroup$
    @K.M Apply the product rule of the derivative
    $endgroup$
    – caverac
    Dec 5 '18 at 1:11











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Take



$$
f({bf p}) = sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_iright)
$$



The gradient is just



begin{eqnarray}
nabla f({bf p}) &=& frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) \
&=& sum_i left( 2{bf p} - 2{bf p}_i right) = 2sum_i {bf p} - 2sum_i{bf p}_i \
&=& 2n {bf p} - 2sum_i {bf p}_i = 2n left({bf p}-frac{1}{n}sum_i{bf p}_iright) \
&=& 2n ({bf p} - overline{{bf p}})
end{eqnarray}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is $frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) = sum_i left( 2{bf p} - 2{bf p}_i right)$?
    $endgroup$
    – K.M
    Dec 5 '18 at 1:01












  • $begingroup$
    @K.M Apply the product rule of the derivative
    $endgroup$
    – caverac
    Dec 5 '18 at 1:11
















1












$begingroup$

Take



$$
f({bf p}) = sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_iright)
$$



The gradient is just



begin{eqnarray}
nabla f({bf p}) &=& frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) \
&=& sum_i left( 2{bf p} - 2{bf p}_i right) = 2sum_i {bf p} - 2sum_i{bf p}_i \
&=& 2n {bf p} - 2sum_i {bf p}_i = 2n left({bf p}-frac{1}{n}sum_i{bf p}_iright) \
&=& 2n ({bf p} - overline{{bf p}})
end{eqnarray}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is $frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) = sum_i left( 2{bf p} - 2{bf p}_i right)$?
    $endgroup$
    – K.M
    Dec 5 '18 at 1:01












  • $begingroup$
    @K.M Apply the product rule of the derivative
    $endgroup$
    – caverac
    Dec 5 '18 at 1:11














1












1








1





$begingroup$

Take



$$
f({bf p}) = sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_iright)
$$



The gradient is just



begin{eqnarray}
nabla f({bf p}) &=& frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) \
&=& sum_i left( 2{bf p} - 2{bf p}_i right) = 2sum_i {bf p} - 2sum_i{bf p}_i \
&=& 2n {bf p} - 2sum_i {bf p}_i = 2n left({bf p}-frac{1}{n}sum_i{bf p}_iright) \
&=& 2n ({bf p} - overline{{bf p}})
end{eqnarray}






share|cite|improve this answer









$endgroup$



Take



$$
f({bf p}) = sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_iright)
$$



The gradient is just



begin{eqnarray}
nabla f({bf p}) &=& frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) \
&=& sum_i left( 2{bf p} - 2{bf p}_i right) = 2sum_i {bf p} - 2sum_i{bf p}_i \
&=& 2n {bf p} - 2sum_i {bf p}_i = 2n left({bf p}-frac{1}{n}sum_i{bf p}_iright) \
&=& 2n ({bf p} - overline{{bf p}})
end{eqnarray}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 0:56









caveraccaverac

14.5k31130




14.5k31130












  • $begingroup$
    Why is $frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) = sum_i left( 2{bf p} - 2{bf p}_i right)$?
    $endgroup$
    – K.M
    Dec 5 '18 at 1:01












  • $begingroup$
    @K.M Apply the product rule of the derivative
    $endgroup$
    – caverac
    Dec 5 '18 at 1:11


















  • $begingroup$
    Why is $frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) = sum_i left( 2{bf p} - 2{bf p}_i right)$?
    $endgroup$
    – K.M
    Dec 5 '18 at 1:01












  • $begingroup$
    @K.M Apply the product rule of the derivative
    $endgroup$
    – caverac
    Dec 5 '18 at 1:11
















$begingroup$
Why is $frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) = sum_i left( 2{bf p} - 2{bf p}_i right)$?
$endgroup$
– K.M
Dec 5 '18 at 1:01






$begingroup$
Why is $frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right) = sum_i left( 2{bf p} - 2{bf p}_i right)$?
$endgroup$
– K.M
Dec 5 '18 at 1:01














$begingroup$
@K.M Apply the product rule of the derivative
$endgroup$
– caverac
Dec 5 '18 at 1:11




$begingroup$
@K.M Apply the product rule of the derivative
$endgroup$
– caverac
Dec 5 '18 at 1:11


















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