Modified Hermite interpolation












7












$begingroup$


I asked similar questions here and here, but I tried to formulate this
one in a sharper way.



Is anyone aware of some literature on polynomial interpolation where we sample the function and its derivatives, but perhaps we sample the derivative at a point where we have not sampled the function?



For instance, given a function $f$, suppose we choose a polynomial $p$ of degree at most 3 where
$$p(a_1)=f(a_1)\p(a_2)=f(a_2)\p'(a_3)=f'(a_3)\p'(a_4)=f'(a_4).$$



Finding such a polynomial $alpha_3 x^3 + alpha_2 x^2 + alpha_1 x + alpha_0$ is equivalent to solving
$$left[
begin{matrix}
1&a_1 & a_1^2 & a_1^3\
1&a_2 & a_2^2 & a_2^3\
0&1 & 2a_3 & 3a_3^2\
0&1 & 2a_4 & 3a_4^2
end{matrix}
right]
left[
begin{matrix}
alpha_0 \ alpha_1 \ alpha_2 \ alpha_3
end{matrix}
right] =
left[
begin{matrix}
f(a_1) \ f(a_2) \ f'(a_3) \ f'(a_4)
end{matrix}
right].
$$



I claim that the determinant of that matrix is $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}$. Therefore, $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}neq 0$ is a necessary and sufficient condition for such a polynomial to exist uniquely.



In particular, I am interested in estimating the error of such a polynomial interpolation, since I don't see how to modify the normal Hermite estimation.










share|cite|improve this question











$endgroup$

















    7












    $begingroup$


    I asked similar questions here and here, but I tried to formulate this
    one in a sharper way.



    Is anyone aware of some literature on polynomial interpolation where we sample the function and its derivatives, but perhaps we sample the derivative at a point where we have not sampled the function?



    For instance, given a function $f$, suppose we choose a polynomial $p$ of degree at most 3 where
    $$p(a_1)=f(a_1)\p(a_2)=f(a_2)\p'(a_3)=f'(a_3)\p'(a_4)=f'(a_4).$$



    Finding such a polynomial $alpha_3 x^3 + alpha_2 x^2 + alpha_1 x + alpha_0$ is equivalent to solving
    $$left[
    begin{matrix}
    1&a_1 & a_1^2 & a_1^3\
    1&a_2 & a_2^2 & a_2^3\
    0&1 & 2a_3 & 3a_3^2\
    0&1 & 2a_4 & 3a_4^2
    end{matrix}
    right]
    left[
    begin{matrix}
    alpha_0 \ alpha_1 \ alpha_2 \ alpha_3
    end{matrix}
    right] =
    left[
    begin{matrix}
    f(a_1) \ f(a_2) \ f'(a_3) \ f'(a_4)
    end{matrix}
    right].
    $$



    I claim that the determinant of that matrix is $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}$. Therefore, $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}neq 0$ is a necessary and sufficient condition for such a polynomial to exist uniquely.



    In particular, I am interested in estimating the error of such a polynomial interpolation, since I don't see how to modify the normal Hermite estimation.










    share|cite|improve this question











    $endgroup$















      7












      7








      7


      1



      $begingroup$


      I asked similar questions here and here, but I tried to formulate this
      one in a sharper way.



      Is anyone aware of some literature on polynomial interpolation where we sample the function and its derivatives, but perhaps we sample the derivative at a point where we have not sampled the function?



      For instance, given a function $f$, suppose we choose a polynomial $p$ of degree at most 3 where
      $$p(a_1)=f(a_1)\p(a_2)=f(a_2)\p'(a_3)=f'(a_3)\p'(a_4)=f'(a_4).$$



      Finding such a polynomial $alpha_3 x^3 + alpha_2 x^2 + alpha_1 x + alpha_0$ is equivalent to solving
      $$left[
      begin{matrix}
      1&a_1 & a_1^2 & a_1^3\
      1&a_2 & a_2^2 & a_2^3\
      0&1 & 2a_3 & 3a_3^2\
      0&1 & 2a_4 & 3a_4^2
      end{matrix}
      right]
      left[
      begin{matrix}
      alpha_0 \ alpha_1 \ alpha_2 \ alpha_3
      end{matrix}
      right] =
      left[
      begin{matrix}
      f(a_1) \ f(a_2) \ f'(a_3) \ f'(a_4)
      end{matrix}
      right].
      $$



      I claim that the determinant of that matrix is $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}$. Therefore, $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}neq 0$ is a necessary and sufficient condition for such a polynomial to exist uniquely.



      In particular, I am interested in estimating the error of such a polynomial interpolation, since I don't see how to modify the normal Hermite estimation.










      share|cite|improve this question











      $endgroup$




      I asked similar questions here and here, but I tried to formulate this
      one in a sharper way.



      Is anyone aware of some literature on polynomial interpolation where we sample the function and its derivatives, but perhaps we sample the derivative at a point where we have not sampled the function?



      For instance, given a function $f$, suppose we choose a polynomial $p$ of degree at most 3 where
      $$p(a_1)=f(a_1)\p(a_2)=f(a_2)\p'(a_3)=f'(a_3)\p'(a_4)=f'(a_4).$$



      Finding such a polynomial $alpha_3 x^3 + alpha_2 x^2 + alpha_1 x + alpha_0$ is equivalent to solving
      $$left[
      begin{matrix}
      1&a_1 & a_1^2 & a_1^3\
      1&a_2 & a_2^2 & a_2^3\
      0&1 & 2a_3 & 3a_3^2\
      0&1 & 2a_4 & 3a_4^2
      end{matrix}
      right]
      left[
      begin{matrix}
      alpha_0 \ alpha_1 \ alpha_2 \ alpha_3
      end{matrix}
      right] =
      left[
      begin{matrix}
      f(a_1) \ f(a_2) \ f'(a_3) \ f'(a_4)
      end{matrix}
      right].
      $$



      I claim that the determinant of that matrix is $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}$. Therefore, $frac{partial^2}{partial x_3partial x_4}left( prod_{i<j}(x_i-x_j) right) left. right|_{x_i=a_i,,forall i}neq 0$ is a necessary and sufficient condition for such a polynomial to exist uniquely.



      In particular, I am interested in estimating the error of such a polynomial interpolation, since I don't see how to modify the normal Hermite estimation.







      calculus reference-request polynomials derivatives interpolation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 13 '17 at 12:20









      Community

      1




      1










      asked Sep 8 '14 at 21:25









      Eric AuldEric Auld

      13.3k433112




      13.3k433112






















          2 Answers
          2






          active

          oldest

          votes


















          1





          +50







          $begingroup$

          See my answer at your previous question.



          If $V(x_1, ldots, x_n)$ is the Vandermonde matrix whose determinant is
          $$det V(x_1, ldots, x_n),$$ the fact that $det V(x_1,ldots,x_n)$ is
          linear as a function of the $i$'th row tells you that the partial derivative
          of the determinant w.r.t. $x_i$ is the determinant of the matrix with row $i$ replaced by its derivative w.r.t. $x_i$. If you prefer, you could also get this
          from the Laplace expansion (aka minor expansion) along the $i$'th row:



          $$ det V = sum_k (-1)^{i+k} v_{ik} M_{ik} $$
          where the $M_{ik}$ don't depend on $x_i$, so
          $$ dfrac{partial}{partial x_i} det V = sum_{k} (-1)^{i+k} dfrac{partial v_{ik}}{partial x_i} M_{ik} $$
          which is the determinant of the matrix obtained from $V$ by replacing
          each entry of row $i$ by its partial derivative wrt $x_i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
            $endgroup$
            – Eric Auld
            Sep 16 '14 at 21:05












          • $begingroup$
            Well, now you've taken care of the first question in the other post. :)
            $endgroup$
            – Eric Auld
            Sep 17 '14 at 1:22










          • $begingroup$
            I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
            $endgroup$
            – Eric Auld
            Sep 18 '14 at 17:33



















          0












          $begingroup$

          Robert Israel has sagely pointed out that we should not expect a convenient error term, since sampling the derivative where you have not sampled the function does not improve the approximation:



          Estimate the difference between $f$ and $p$ interpolating $f$



          I have shown that a sufficient condition for the polynomial to exist uniquely is that we are sampling in Hermite fashion. I did it without resorting to divided differences, which is the usual way.



          There is a unique polynomial interpolating $f$ and its derivatives






          share|cite|improve this answer











          $endgroup$














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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1





            +50







            $begingroup$

            See my answer at your previous question.



            If $V(x_1, ldots, x_n)$ is the Vandermonde matrix whose determinant is
            $$det V(x_1, ldots, x_n),$$ the fact that $det V(x_1,ldots,x_n)$ is
            linear as a function of the $i$'th row tells you that the partial derivative
            of the determinant w.r.t. $x_i$ is the determinant of the matrix with row $i$ replaced by its derivative w.r.t. $x_i$. If you prefer, you could also get this
            from the Laplace expansion (aka minor expansion) along the $i$'th row:



            $$ det V = sum_k (-1)^{i+k} v_{ik} M_{ik} $$
            where the $M_{ik}$ don't depend on $x_i$, so
            $$ dfrac{partial}{partial x_i} det V = sum_{k} (-1)^{i+k} dfrac{partial v_{ik}}{partial x_i} M_{ik} $$
            which is the determinant of the matrix obtained from $V$ by replacing
            each entry of row $i$ by its partial derivative wrt $x_i$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
              $endgroup$
              – Eric Auld
              Sep 16 '14 at 21:05












            • $begingroup$
              Well, now you've taken care of the first question in the other post. :)
              $endgroup$
              – Eric Auld
              Sep 17 '14 at 1:22










            • $begingroup$
              I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
              $endgroup$
              – Eric Auld
              Sep 18 '14 at 17:33
















            1





            +50







            $begingroup$

            See my answer at your previous question.



            If $V(x_1, ldots, x_n)$ is the Vandermonde matrix whose determinant is
            $$det V(x_1, ldots, x_n),$$ the fact that $det V(x_1,ldots,x_n)$ is
            linear as a function of the $i$'th row tells you that the partial derivative
            of the determinant w.r.t. $x_i$ is the determinant of the matrix with row $i$ replaced by its derivative w.r.t. $x_i$. If you prefer, you could also get this
            from the Laplace expansion (aka minor expansion) along the $i$'th row:



            $$ det V = sum_k (-1)^{i+k} v_{ik} M_{ik} $$
            where the $M_{ik}$ don't depend on $x_i$, so
            $$ dfrac{partial}{partial x_i} det V = sum_{k} (-1)^{i+k} dfrac{partial v_{ik}}{partial x_i} M_{ik} $$
            which is the determinant of the matrix obtained from $V$ by replacing
            each entry of row $i$ by its partial derivative wrt $x_i$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
              $endgroup$
              – Eric Auld
              Sep 16 '14 at 21:05












            • $begingroup$
              Well, now you've taken care of the first question in the other post. :)
              $endgroup$
              – Eric Auld
              Sep 17 '14 at 1:22










            • $begingroup$
              I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
              $endgroup$
              – Eric Auld
              Sep 18 '14 at 17:33














            1





            +50







            1





            +50



            1




            +50



            $begingroup$

            See my answer at your previous question.



            If $V(x_1, ldots, x_n)$ is the Vandermonde matrix whose determinant is
            $$det V(x_1, ldots, x_n),$$ the fact that $det V(x_1,ldots,x_n)$ is
            linear as a function of the $i$'th row tells you that the partial derivative
            of the determinant w.r.t. $x_i$ is the determinant of the matrix with row $i$ replaced by its derivative w.r.t. $x_i$. If you prefer, you could also get this
            from the Laplace expansion (aka minor expansion) along the $i$'th row:



            $$ det V = sum_k (-1)^{i+k} v_{ik} M_{ik} $$
            where the $M_{ik}$ don't depend on $x_i$, so
            $$ dfrac{partial}{partial x_i} det V = sum_{k} (-1)^{i+k} dfrac{partial v_{ik}}{partial x_i} M_{ik} $$
            which is the determinant of the matrix obtained from $V$ by replacing
            each entry of row $i$ by its partial derivative wrt $x_i$.






            share|cite|improve this answer











            $endgroup$



            See my answer at your previous question.



            If $V(x_1, ldots, x_n)$ is the Vandermonde matrix whose determinant is
            $$det V(x_1, ldots, x_n),$$ the fact that $det V(x_1,ldots,x_n)$ is
            linear as a function of the $i$'th row tells you that the partial derivative
            of the determinant w.r.t. $x_i$ is the determinant of the matrix with row $i$ replaced by its derivative w.r.t. $x_i$. If you prefer, you could also get this
            from the Laplace expansion (aka minor expansion) along the $i$'th row:



            $$ det V = sum_k (-1)^{i+k} v_{ik} M_{ik} $$
            where the $M_{ik}$ don't depend on $x_i$, so
            $$ dfrac{partial}{partial x_i} det V = sum_{k} (-1)^{i+k} dfrac{partial v_{ik}}{partial x_i} M_{ik} $$
            which is the determinant of the matrix obtained from $V$ by replacing
            each entry of row $i$ by its partial derivative wrt $x_i$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 11 '18 at 4:12









            Aloizio Macedo

            23.8k24088




            23.8k24088










            answered Sep 16 '14 at 20:55









            Robert IsraelRobert Israel

            329k23218472




            329k23218472












            • $begingroup$
              Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
              $endgroup$
              – Eric Auld
              Sep 16 '14 at 21:05












            • $begingroup$
              Well, now you've taken care of the first question in the other post. :)
              $endgroup$
              – Eric Auld
              Sep 17 '14 at 1:22










            • $begingroup$
              I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
              $endgroup$
              – Eric Auld
              Sep 18 '14 at 17:33


















            • $begingroup$
              Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
              $endgroup$
              – Eric Auld
              Sep 16 '14 at 21:05












            • $begingroup$
              Well, now you've taken care of the first question in the other post. :)
              $endgroup$
              – Eric Auld
              Sep 17 '14 at 1:22










            • $begingroup$
              I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
              $endgroup$
              – Eric Auld
              Sep 18 '14 at 17:33
















            $begingroup$
            Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
            $endgroup$
            – Eric Auld
            Sep 16 '14 at 21:05






            $begingroup$
            Thanks for teaching me how to find the determinant. In this question, I suppose I am interested in seeing if I can estimate the error as in the normal Hermite interpolation problem. Also, in showing sufficient conditions for $frac{partial^k}{partial x_{i_1}dotsb partial x_{i_k}}det(V)$ to be nonzero (e.g. the condition for standard Hermite interpolation: that if we sample $f^{(n)}$ at $x$, we are also sampling $f^{(n-1)}, f^{(n-2)}, dotsc, f^{(0)}$ there.)
            $endgroup$
            – Eric Auld
            Sep 16 '14 at 21:05














            $begingroup$
            Well, now you've taken care of the first question in the other post. :)
            $endgroup$
            – Eric Auld
            Sep 17 '14 at 1:22




            $begingroup$
            Well, now you've taken care of the first question in the other post. :)
            $endgroup$
            – Eric Auld
            Sep 17 '14 at 1:22












            $begingroup$
            I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
            $endgroup$
            – Eric Auld
            Sep 18 '14 at 17:33




            $begingroup$
            I believe I solved the problem of why the matrix must be invertible here: math.stackexchange.com/questions/933124/…
            $endgroup$
            – Eric Auld
            Sep 18 '14 at 17:33











            0












            $begingroup$

            Robert Israel has sagely pointed out that we should not expect a convenient error term, since sampling the derivative where you have not sampled the function does not improve the approximation:



            Estimate the difference between $f$ and $p$ interpolating $f$



            I have shown that a sufficient condition for the polynomial to exist uniquely is that we are sampling in Hermite fashion. I did it without resorting to divided differences, which is the usual way.



            There is a unique polynomial interpolating $f$ and its derivatives






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Robert Israel has sagely pointed out that we should not expect a convenient error term, since sampling the derivative where you have not sampled the function does not improve the approximation:



              Estimate the difference between $f$ and $p$ interpolating $f$



              I have shown that a sufficient condition for the polynomial to exist uniquely is that we are sampling in Hermite fashion. I did it without resorting to divided differences, which is the usual way.



              There is a unique polynomial interpolating $f$ and its derivatives






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Robert Israel has sagely pointed out that we should not expect a convenient error term, since sampling the derivative where you have not sampled the function does not improve the approximation:



                Estimate the difference between $f$ and $p$ interpolating $f$



                I have shown that a sufficient condition for the polynomial to exist uniquely is that we are sampling in Hermite fashion. I did it without resorting to divided differences, which is the usual way.



                There is a unique polynomial interpolating $f$ and its derivatives






                share|cite|improve this answer











                $endgroup$



                Robert Israel has sagely pointed out that we should not expect a convenient error term, since sampling the derivative where you have not sampled the function does not improve the approximation:



                Estimate the difference between $f$ and $p$ interpolating $f$



                I have shown that a sufficient condition for the polynomial to exist uniquely is that we are sampling in Hermite fashion. I did it without resorting to divided differences, which is the usual way.



                There is a unique polynomial interpolating $f$ and its derivatives







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Apr 13 '17 at 12:21









                Community

                1




                1










                answered Sep 18 '14 at 17:36









                Eric AuldEric Auld

                13.3k433112




                13.3k433112






























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