Why is $nabla f(p) = sum_{i=1}^n 2(p - p_i) = 0$?
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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}
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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votes
$begingroup$
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}
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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}
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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}
$endgroup$
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}
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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