Why is $nabla f({bf p}) =sum_i left( 2{bf p} - 2{bf p}_i right) $?












0












$begingroup$


If



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



Why is the gradient of $f(p)$ equal to



$
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) $






calculus multivariable-calculus







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$endgroup$

















    0












    $begingroup$


    If



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



    Why is the gradient of $f(p)$ equal to



    $
    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) $






    calculus multivariable-calculus







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      If



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



      Why is the gradient of $f(p)$ equal to



      $
      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) $






      calculus multivariable-calculus







      share|cite|improve this question









      $endgroup$




      If



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



      Why is the gradient of $f(p)$ equal to



      $
      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) $






      calculus multivariable-calculus




      calculus multivariable-calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 5 '18 at 1:08









      K.MK.M

      693412




      693412






















          1 Answer
          1






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          1












          $begingroup$

          You can do it in components, consider the first term for example



          $$
          {bf p}cdot {bf p} = p_1p_1 + p_2p_2 + cdots
          $$



          So the gradient of this term is



          begin{eqnarray}
          frac{partial {bf p}cdot {bf p}}{partial p_1} &=& 2p_1 \
          frac{partial {bf p}cdot {bf p}}{partial p_2} &=& 2p_2 \
          &vdots &
          end{eqnarray}



          or in general



          $$
          frac{partial {bf p}cdot {bf p}}{partial {bf p}} = 2{bf p}
          $$



          Similarly you can prove that



          $$
          frac{partial {bf p}cdot {bf p}_i}{partial {bf p}} = {bf p}_i
          $$






          share|cite|improve this answer











          $endgroup$













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            1 Answer
            1






            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            You can do it in components, consider the first term for example



            $$
            {bf p}cdot {bf p} = p_1p_1 + p_2p_2 + cdots
            $$



            So the gradient of this term is



            begin{eqnarray}
            frac{partial {bf p}cdot {bf p}}{partial p_1} &=& 2p_1 \
            frac{partial {bf p}cdot {bf p}}{partial p_2} &=& 2p_2 \
            &vdots &
            end{eqnarray}



            or in general



            $$
            frac{partial {bf p}cdot {bf p}}{partial {bf p}} = 2{bf p}
            $$



            Similarly you can prove that



            $$
            frac{partial {bf p}cdot {bf p}_i}{partial {bf p}} = {bf p}_i
            $$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              You can do it in components, consider the first term for example



              $$
              {bf p}cdot {bf p} = p_1p_1 + p_2p_2 + cdots
              $$



              So the gradient of this term is



              begin{eqnarray}
              frac{partial {bf p}cdot {bf p}}{partial p_1} &=& 2p_1 \
              frac{partial {bf p}cdot {bf p}}{partial p_2} &=& 2p_2 \
              &vdots &
              end{eqnarray}



              or in general



              $$
              frac{partial {bf p}cdot {bf p}}{partial {bf p}} = 2{bf p}
              $$



              Similarly you can prove that



              $$
              frac{partial {bf p}cdot {bf p}_i}{partial {bf p}} = {bf p}_i
              $$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                You can do it in components, consider the first term for example



                $$
                {bf p}cdot {bf p} = p_1p_1 + p_2p_2 + cdots
                $$



                So the gradient of this term is



                begin{eqnarray}
                frac{partial {bf p}cdot {bf p}}{partial p_1} &=& 2p_1 \
                frac{partial {bf p}cdot {bf p}}{partial p_2} &=& 2p_2 \
                &vdots &
                end{eqnarray}



                or in general



                $$
                frac{partial {bf p}cdot {bf p}}{partial {bf p}} = 2{bf p}
                $$



                Similarly you can prove that



                $$
                frac{partial {bf p}cdot {bf p}_i}{partial {bf p}} = {bf p}_i
                $$






                share|cite|improve this answer











                $endgroup$



                You can do it in components, consider the first term for example



                $$
                {bf p}cdot {bf p} = p_1p_1 + p_2p_2 + cdots
                $$



                So the gradient of this term is



                begin{eqnarray}
                frac{partial {bf p}cdot {bf p}}{partial p_1} &=& 2p_1 \
                frac{partial {bf p}cdot {bf p}}{partial p_2} &=& 2p_2 \
                &vdots &
                end{eqnarray}



                or in general



                $$
                frac{partial {bf p}cdot {bf p}}{partial {bf p}} = 2{bf p}
                $$



                Similarly you can prove that



                $$
                frac{partial {bf p}cdot {bf p}_i}{partial {bf p}} = {bf p}_i
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 5 '18 at 1:51

























                answered Dec 5 '18 at 1:45









                caveraccaverac

                14.5k31130




                14.5k31130






























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