Taylor expansion on Images which are warped












1












$begingroup$


I'm currently reading following paper: https://www.ri.cmu.edu/pub_files/pub3/baker_simon_2001_2/baker_simon_2001_2.pdf



Here I struggle to follow the Taylor expansion from equation 2 to equation 3. I mean I know in general how Taylor expansion works, but in this case it is not clear to me.



Starting equation:



$$
sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} + Deltamathbf{p})) - mathit{T}(mathbf{x})]^2
$$



Taylor expansion



$$
sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} )) + nablamathit{I}frac{partialmathbf{W}}{partialmathbf{p}}Deltamathbf{p} - mathit{T}(mathbf{x})]^2
$$



Maybe somebody of you has more experience and can elaborate little more on this?






taylor-expansion







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I'm currently reading following paper: https://www.ri.cmu.edu/pub_files/pub3/baker_simon_2001_2/baker_simon_2001_2.pdf



    Here I struggle to follow the Taylor expansion from equation 2 to equation 3. I mean I know in general how Taylor expansion works, but in this case it is not clear to me.



    Starting equation:



    $$
    sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} + Deltamathbf{p})) - mathit{T}(mathbf{x})]^2
    $$



    Taylor expansion



    $$
    sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} )) + nablamathit{I}frac{partialmathbf{W}}{partialmathbf{p}}Deltamathbf{p} - mathit{T}(mathbf{x})]^2
    $$



    Maybe somebody of you has more experience and can elaborate little more on this?






    taylor-expansion







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I'm currently reading following paper: https://www.ri.cmu.edu/pub_files/pub3/baker_simon_2001_2/baker_simon_2001_2.pdf



      Here I struggle to follow the Taylor expansion from equation 2 to equation 3. I mean I know in general how Taylor expansion works, but in this case it is not clear to me.



      Starting equation:



      $$
      sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} + Deltamathbf{p})) - mathit{T}(mathbf{x})]^2
      $$



      Taylor expansion



      $$
      sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} )) + nablamathit{I}frac{partialmathbf{W}}{partialmathbf{p}}Deltamathbf{p} - mathit{T}(mathbf{x})]^2
      $$



      Maybe somebody of you has more experience and can elaborate little more on this?






      taylor-expansion







      share|cite|improve this question









      $endgroup$




      I'm currently reading following paper: https://www.ri.cmu.edu/pub_files/pub3/baker_simon_2001_2/baker_simon_2001_2.pdf



      Here I struggle to follow the Taylor expansion from equation 2 to equation 3. I mean I know in general how Taylor expansion works, but in this case it is not clear to me.



      Starting equation:



      $$
      sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} + Deltamathbf{p})) - mathit{T}(mathbf{x})]^2
      $$



      Taylor expansion



      $$
      sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} )) + nablamathit{I}frac{partialmathbf{W}}{partialmathbf{p}}Deltamathbf{p} - mathit{T}(mathbf{x})]^2
      $$



      Maybe somebody of you has more experience and can elaborate little more on this?






      taylor-expansion




      taylor-expansion






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 1 '18 at 17:42









      antibusantibus

      1083




      1083






















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          $begingroup$

          You just need to apply the expansion twice. First for ${bf W}$ up to first order in $Delta {bf p}$, we get



          $$
          {bf W}({bf x}; {bf p} + Delta {bf p}) approx {bf W}({bf x}; {bf p}) + frac{partial {bf W}}{partial {bf p}}Delta {bf p} equiv {bf W} + Delta{bf W}tag{1}
          $$



          Now for $I$ up to first order in $Delta {bf W}$,



          $$
          I({bf W} + Delta{bf W}) approx I({bf W}) + nabla I Delta{bf W} = I({bf W}) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p} tag{2}
          $$



          Putting everything together



          $$
          I({bf W}({bf x}; {bf p} + Delta {bf p})) approx I({bf W}({bf x}; {bf p})) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p}
          $$






          share|cite|improve this answer









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            $begingroup$

            You just need to apply the expansion twice. First for ${bf W}$ up to first order in $Delta {bf p}$, we get



            $$
            {bf W}({bf x}; {bf p} + Delta {bf p}) approx {bf W}({bf x}; {bf p}) + frac{partial {bf W}}{partial {bf p}}Delta {bf p} equiv {bf W} + Delta{bf W}tag{1}
            $$



            Now for $I$ up to first order in $Delta {bf W}$,



            $$
            I({bf W} + Delta{bf W}) approx I({bf W}) + nabla I Delta{bf W} = I({bf W}) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p} tag{2}
            $$



            Putting everything together



            $$
            I({bf W}({bf x}; {bf p} + Delta {bf p})) approx I({bf W}({bf x}; {bf p})) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p}
            $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              You just need to apply the expansion twice. First for ${bf W}$ up to first order in $Delta {bf p}$, we get



              $$
              {bf W}({bf x}; {bf p} + Delta {bf p}) approx {bf W}({bf x}; {bf p}) + frac{partial {bf W}}{partial {bf p}}Delta {bf p} equiv {bf W} + Delta{bf W}tag{1}
              $$



              Now for $I$ up to first order in $Delta {bf W}$,



              $$
              I({bf W} + Delta{bf W}) approx I({bf W}) + nabla I Delta{bf W} = I({bf W}) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p} tag{2}
              $$



              Putting everything together



              $$
              I({bf W}({bf x}; {bf p} + Delta {bf p})) approx I({bf W}({bf x}; {bf p})) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p}
              $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                You just need to apply the expansion twice. First for ${bf W}$ up to first order in $Delta {bf p}$, we get



                $$
                {bf W}({bf x}; {bf p} + Delta {bf p}) approx {bf W}({bf x}; {bf p}) + frac{partial {bf W}}{partial {bf p}}Delta {bf p} equiv {bf W} + Delta{bf W}tag{1}
                $$



                Now for $I$ up to first order in $Delta {bf W}$,



                $$
                I({bf W} + Delta{bf W}) approx I({bf W}) + nabla I Delta{bf W} = I({bf W}) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p} tag{2}
                $$



                Putting everything together



                $$
                I({bf W}({bf x}; {bf p} + Delta {bf p})) approx I({bf W}({bf x}; {bf p})) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p}
                $$






                share|cite|improve this answer









                $endgroup$



                You just need to apply the expansion twice. First for ${bf W}$ up to first order in $Delta {bf p}$, we get



                $$
                {bf W}({bf x}; {bf p} + Delta {bf p}) approx {bf W}({bf x}; {bf p}) + frac{partial {bf W}}{partial {bf p}}Delta {bf p} equiv {bf W} + Delta{bf W}tag{1}
                $$



                Now for $I$ up to first order in $Delta {bf W}$,



                $$
                I({bf W} + Delta{bf W}) approx I({bf W}) + nabla I Delta{bf W} = I({bf W}) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p} tag{2}
                $$



                Putting everything together



                $$
                I({bf W}({bf x}; {bf p} + Delta {bf p})) approx I({bf W}({bf x}; {bf p})) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p}
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 '18 at 21:17









                caveraccaverac

                14.2k21130




                14.2k21130






























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