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







    share|cite|improve this question
























      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







      share|cite|improve this question













      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






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      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.






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          • 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











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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.






          share|cite|improve this answer























          • 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















          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.






          share|cite|improve this answer























          • 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













          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          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


















          • 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


















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