Why is $bar{p}$ a critical point of $nabla f(p)$?
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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 $f(p)$
I wanted to know if $(p-bar{p})$ is a critical point since $nabla f(p) = 2n(p-bar{p})$?
calculus multivariable-calculus
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up vote
1
down vote
favorite
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 $f(p)$
I wanted to know if $(p-bar{p})$ is a critical point since $nabla f(p) = 2n(p-bar{p})$?
calculus multivariable-calculus
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 $f(p)$
I wanted to know if $(p-bar{p})$ is a critical point since $nabla f(p) = 2n(p-bar{p})$?
calculus multivariable-calculus
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 $f(p)$
I wanted to know if $(p-bar{p})$ is a critical point since $nabla f(p) = 2n(p-bar{p})$?
calculus multivariable-calculus
calculus multivariable-calculus
asked Dec 5 at 2:46
K.M
651312
651312
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Writing the function as $$f(p) =sum_{i=1}^nlangle p-p_i,p-p_irangle,$$we have $$Df(p)(v) =sum_{i=1}^n 2langle p-p_i,vrangle.$$So, write this as $$Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$$to conclude that $$nabla f(p) = 2np-sum_{i=1}^n2p_i. $$ So $nabla f (p) $ is the zero vector if and only if $$p= frac{1}{n}sum_{i=1}^n p_i. $$To see that this gives the minimum, compute the Hessian: $$D^2f(overline{p})(v,w) = 2nlangle v,wrangle .$$Since $D^2f (overline{p})$ is positive-definite, we are done.
I was wondering how we get that $nabla f(p) = 2np-sum_{i=1}^n2p_i.$ from $Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$?
– K.M
Dec 6 at 4:07
1
$nabla f (p)$ is the unique vector for which $Df(p)(v)=langle nabla f (p),vrangle $ for all $v$. So if you found a vector with this property, this is the gradient.
– Ivo Terek
Dec 6 at 7:07
add a comment |
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1 Answer
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Writing the function as $$f(p) =sum_{i=1}^nlangle p-p_i,p-p_irangle,$$we have $$Df(p)(v) =sum_{i=1}^n 2langle p-p_i,vrangle.$$So, write this as $$Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$$to conclude that $$nabla f(p) = 2np-sum_{i=1}^n2p_i. $$ So $nabla f (p) $ is the zero vector if and only if $$p= frac{1}{n}sum_{i=1}^n p_i. $$To see that this gives the minimum, compute the Hessian: $$D^2f(overline{p})(v,w) = 2nlangle v,wrangle .$$Since $D^2f (overline{p})$ is positive-definite, we are done.
I was wondering how we get that $nabla f(p) = 2np-sum_{i=1}^n2p_i.$ from $Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$?
– K.M
Dec 6 at 4:07
1
$nabla f (p)$ is the unique vector for which $Df(p)(v)=langle nabla f (p),vrangle $ for all $v$. So if you found a vector with this property, this is the gradient.
– Ivo Terek
Dec 6 at 7:07
add a comment |
up vote
1
down vote
Writing the function as $$f(p) =sum_{i=1}^nlangle p-p_i,p-p_irangle,$$we have $$Df(p)(v) =sum_{i=1}^n 2langle p-p_i,vrangle.$$So, write this as $$Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$$to conclude that $$nabla f(p) = 2np-sum_{i=1}^n2p_i. $$ So $nabla f (p) $ is the zero vector if and only if $$p= frac{1}{n}sum_{i=1}^n p_i. $$To see that this gives the minimum, compute the Hessian: $$D^2f(overline{p})(v,w) = 2nlangle v,wrangle .$$Since $D^2f (overline{p})$ is positive-definite, we are done.
I was wondering how we get that $nabla f(p) = 2np-sum_{i=1}^n2p_i.$ from $Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$?
– K.M
Dec 6 at 4:07
1
$nabla f (p)$ is the unique vector for which $Df(p)(v)=langle nabla f (p),vrangle $ for all $v$. So if you found a vector with this property, this is the gradient.
– Ivo Terek
Dec 6 at 7:07
add a comment |
up vote
1
down vote
up vote
1
down vote
Writing the function as $$f(p) =sum_{i=1}^nlangle p-p_i,p-p_irangle,$$we have $$Df(p)(v) =sum_{i=1}^n 2langle p-p_i,vrangle.$$So, write this as $$Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$$to conclude that $$nabla f(p) = 2np-sum_{i=1}^n2p_i. $$ So $nabla f (p) $ is the zero vector if and only if $$p= frac{1}{n}sum_{i=1}^n p_i. $$To see that this gives the minimum, compute the Hessian: $$D^2f(overline{p})(v,w) = 2nlangle v,wrangle .$$Since $D^2f (overline{p})$ is positive-definite, we are done.
Writing the function as $$f(p) =sum_{i=1}^nlangle p-p_i,p-p_irangle,$$we have $$Df(p)(v) =sum_{i=1}^n 2langle p-p_i,vrangle.$$So, write this as $$Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$$to conclude that $$nabla f(p) = 2np-sum_{i=1}^n2p_i. $$ So $nabla f (p) $ is the zero vector if and only if $$p= frac{1}{n}sum_{i=1}^n p_i. $$To see that this gives the minimum, compute the Hessian: $$D^2f(overline{p})(v,w) = 2nlangle v,wrangle .$$Since $D^2f (overline{p})$ is positive-definite, we are done.
edited Dec 6 at 7:08
answered Dec 5 at 3:35
Ivo Terek
45.2k951139
45.2k951139
I was wondering how we get that $nabla f(p) = 2np-sum_{i=1}^n2p_i.$ from $Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$?
– K.M
Dec 6 at 4:07
1
$nabla f (p)$ is the unique vector for which $Df(p)(v)=langle nabla f (p),vrangle $ for all $v$. So if you found a vector with this property, this is the gradient.
– Ivo Terek
Dec 6 at 7:07
add a comment |
I was wondering how we get that $nabla f(p) = 2np-sum_{i=1}^n2p_i.$ from $Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$?
– K.M
Dec 6 at 4:07
1
$nabla f (p)$ is the unique vector for which $Df(p)(v)=langle nabla f (p),vrangle $ for all $v$. So if you found a vector with this property, this is the gradient.
– Ivo Terek
Dec 6 at 7:07
I was wondering how we get that $nabla f(p) = 2np-sum_{i=1}^n2p_i.$ from $Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$?
– K.M
Dec 6 at 4:07
I was wondering how we get that $nabla f(p) = 2np-sum_{i=1}^n2p_i.$ from $Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$?
– K.M
Dec 6 at 4:07
1
1
$nabla f (p)$ is the unique vector for which $Df(p)(v)=langle nabla f (p),vrangle $ for all $v$. So if you found a vector with this property, this is the gradient.
– Ivo Terek
Dec 6 at 7:07
$nabla f (p)$ is the unique vector for which $Df(p)(v)=langle nabla f (p),vrangle $ for all $v$. So if you found a vector with this property, this is the gradient.
– Ivo Terek
Dec 6 at 7:07
add a comment |
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