How do we conclude that $nabla f(p) = 2np-sum_{i=1}^n2p_i$?












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I'm trying to show that $bar{p}$ is a minimum




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: $$D^2f(overline{p})(v,w) = Clangle v,wrangle $$for some $C>0$. Since $D^2f (overline{p})$ is positive-definite, we are done.




For the part $$Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$$How do we conclude that $$nabla f(p) = 2np-sum_{i=1}^n2p_i$$since we can't divide by a vector?






calculus multivariable-calculus derivatives







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    I'm trying to show that $bar{p}$ is a minimum




    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: $$D^2f(overline{p})(v,w) = Clangle v,wrangle $$for some $C>0$. Since $D^2f (overline{p})$ is positive-definite, we are done.




    For the part $$Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$$How do we conclude that $$nabla f(p) = 2np-sum_{i=1}^n2p_i$$since we can't divide by a vector?






    calculus multivariable-calculus derivatives







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      $begingroup$


      I'm trying to show that $bar{p}$ is a minimum




      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: $$D^2f(overline{p})(v,w) = Clangle v,wrangle $$for some $C>0$. Since $D^2f (overline{p})$ is positive-definite, we are done.




      For the part $$Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$$How do we conclude that $$nabla f(p) = 2np-sum_{i=1}^n2p_i$$since we can't divide by a vector?






      calculus multivariable-calculus derivatives







      share|cite|improve this question









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      I'm trying to show that $bar{p}$ is a minimum




      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: $$D^2f(overline{p})(v,w) = Clangle v,wrangle $$for some $C>0$. Since $D^2f (overline{p})$ is positive-definite, we are done.




      For the part $$Df(p)(v) =leftlangle 2np-sum_{i=1}^n2p_i,v rightrangle$$How do we conclude that $$nabla f(p) = 2np-sum_{i=1}^n2p_i$$since we can't divide by a vector?






      calculus multivariable-calculus derivatives




      calculus multivariable-calculus derivatives






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      asked Dec 6 '18 at 3:26









      K.MK.M

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          There is an integral Theorem in Analysis linking the directional derivative to the derivative. To be more precise:



          Let $f : mathbb{R}^n rightarrow mathbb{R}$ be continous differentiable in $x_0$. The directional derivative $Df(x_0)(v)$ exists for every $v in mathbb{R}^n setminus {0}$ and one has $Df(x_0)(v) = langle nabla f(x_0),v rangle$.



          So for your problem you immediately get that $nabla f(p) = 2np-sum_{i=1}^n2p_i$






          share|cite|improve this answer









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            $begingroup$

            There is an integral Theorem in Analysis linking the directional derivative to the derivative. To be more precise:



            Let $f : mathbb{R}^n rightarrow mathbb{R}$ be continous differentiable in $x_0$. The directional derivative $Df(x_0)(v)$ exists for every $v in mathbb{R}^n setminus {0}$ and one has $Df(x_0)(v) = langle nabla f(x_0),v rangle$.



            So for your problem you immediately get that $nabla f(p) = 2np-sum_{i=1}^n2p_i$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              There is an integral Theorem in Analysis linking the directional derivative to the derivative. To be more precise:



              Let $f : mathbb{R}^n rightarrow mathbb{R}$ be continous differentiable in $x_0$. The directional derivative $Df(x_0)(v)$ exists for every $v in mathbb{R}^n setminus {0}$ and one has $Df(x_0)(v) = langle nabla f(x_0),v rangle$.



              So for your problem you immediately get that $nabla f(p) = 2np-sum_{i=1}^n2p_i$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                There is an integral Theorem in Analysis linking the directional derivative to the derivative. To be more precise:



                Let $f : mathbb{R}^n rightarrow mathbb{R}$ be continous differentiable in $x_0$. The directional derivative $Df(x_0)(v)$ exists for every $v in mathbb{R}^n setminus {0}$ and one has $Df(x_0)(v) = langle nabla f(x_0),v rangle$.



                So for your problem you immediately get that $nabla f(p) = 2np-sum_{i=1}^n2p_i$






                share|cite|improve this answer









                $endgroup$



                There is an integral Theorem in Analysis linking the directional derivative to the derivative. To be more precise:



                Let $f : mathbb{R}^n rightarrow mathbb{R}$ be continous differentiable in $x_0$. The directional derivative $Df(x_0)(v)$ exists for every $v in mathbb{R}^n setminus {0}$ and one has $Df(x_0)(v) = langle nabla f(x_0),v rangle$.



                So for your problem you immediately get that $nabla f(p) = 2np-sum_{i=1}^n2p_i$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 '18 at 4:46









                RiquelmeRiquelme

                9316




                9316






























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