Polynomial representation: Marsden's Identity.












2












$begingroup$


Marsden's Identity states that:



For every $tau$ in $mathbb{R }$:



$(cdot -tau)^{k-1}=sum_jPsi_{j,k}B_{j,k,t}$



with: $Psi_{j,k}=(t_j-tau)times...times(t_{j+k-1}-tau)$



Following de Boor's notation $B_{j,k,t}$ stands for the $j-th$ B-spline of order $k$ defined over the knot vector $t$ i.e. $(t_{j+k}-t_j)cdot [t_j,...,t_{j+k}](cdot - x)_+^{k-1}$



Also define the space spanned by the B-splines as:



$$_{k,t}:={sumalpha_jB_{j,k,t} : alpha_jinmathbb{R}}$



Technically using Marsden's Identity I should be able to show that $mathbb{P}_k$ that stands for the space of all polynomials of degree $<k-1$ is contained in $$_{k,t}$ by putting $alpha_j=Psi_{j,k}$.
But by doing so, I don't really see how this expression is describing all possible polynomials of degree $k-1$.




Doesn't $(cdot -tau)^{k-1}$ represents all polynomials of degree
$k-1$ where $tau$ is a root with multiplicity $k-1$ which is a subset
of $mathbb{P}_k$? Am I missing something here?







linear-algebra polynomials recurrence-relations recursion spline







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Marsden's Identity states that:



    For every $tau$ in $mathbb{R }$:



    $(cdot -tau)^{k-1}=sum_jPsi_{j,k}B_{j,k,t}$



    with: $Psi_{j,k}=(t_j-tau)times...times(t_{j+k-1}-tau)$



    Following de Boor's notation $B_{j,k,t}$ stands for the $j-th$ B-spline of order $k$ defined over the knot vector $t$ i.e. $(t_{j+k}-t_j)cdot [t_j,...,t_{j+k}](cdot - x)_+^{k-1}$



    Also define the space spanned by the B-splines as:



    $$_{k,t}:={sumalpha_jB_{j,k,t} : alpha_jinmathbb{R}}$



    Technically using Marsden's Identity I should be able to show that $mathbb{P}_k$ that stands for the space of all polynomials of degree $<k-1$ is contained in $$_{k,t}$ by putting $alpha_j=Psi_{j,k}$.
    But by doing so, I don't really see how this expression is describing all possible polynomials of degree $k-1$.




    Doesn't $(cdot -tau)^{k-1}$ represents all polynomials of degree
    $k-1$ where $tau$ is a root with multiplicity $k-1$ which is a subset
    of $mathbb{P}_k$? Am I missing something here?







    linear-algebra polynomials recurrence-relations recursion spline







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Marsden's Identity states that:



      For every $tau$ in $mathbb{R }$:



      $(cdot -tau)^{k-1}=sum_jPsi_{j,k}B_{j,k,t}$



      with: $Psi_{j,k}=(t_j-tau)times...times(t_{j+k-1}-tau)$



      Following de Boor's notation $B_{j,k,t}$ stands for the $j-th$ B-spline of order $k$ defined over the knot vector $t$ i.e. $(t_{j+k}-t_j)cdot [t_j,...,t_{j+k}](cdot - x)_+^{k-1}$



      Also define the space spanned by the B-splines as:



      $$_{k,t}:={sumalpha_jB_{j,k,t} : alpha_jinmathbb{R}}$



      Technically using Marsden's Identity I should be able to show that $mathbb{P}_k$ that stands for the space of all polynomials of degree $<k-1$ is contained in $$_{k,t}$ by putting $alpha_j=Psi_{j,k}$.
      But by doing so, I don't really see how this expression is describing all possible polynomials of degree $k-1$.




      Doesn't $(cdot -tau)^{k-1}$ represents all polynomials of degree
      $k-1$ where $tau$ is a root with multiplicity $k-1$ which is a subset
      of $mathbb{P}_k$? Am I missing something here?







      linear-algebra polynomials recurrence-relations recursion spline







      share|cite|improve this question











      $endgroup$




      Marsden's Identity states that:



      For every $tau$ in $mathbb{R }$:



      $(cdot -tau)^{k-1}=sum_jPsi_{j,k}B_{j,k,t}$



      with: $Psi_{j,k}=(t_j-tau)times...times(t_{j+k-1}-tau)$



      Following de Boor's notation $B_{j,k,t}$ stands for the $j-th$ B-spline of order $k$ defined over the knot vector $t$ i.e. $(t_{j+k}-t_j)cdot [t_j,...,t_{j+k}](cdot - x)_+^{k-1}$



      Also define the space spanned by the B-splines as:



      $$_{k,t}:={sumalpha_jB_{j,k,t} : alpha_jinmathbb{R}}$



      Technically using Marsden's Identity I should be able to show that $mathbb{P}_k$ that stands for the space of all polynomials of degree $<k-1$ is contained in $$_{k,t}$ by putting $alpha_j=Psi_{j,k}$.
      But by doing so, I don't really see how this expression is describing all possible polynomials of degree $k-1$.




      Doesn't $(cdot -tau)^{k-1}$ represents all polynomials of degree
      $k-1$ where $tau$ is a root with multiplicity $k-1$ which is a subset
      of $mathbb{P}_k$? Am I missing something here?







      linear-algebra polynomials recurrence-relations recursion spline




      linear-algebra polynomials recurrence-relations recursion spline






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 16:31







      RScrlli

















      asked Dec 7 '18 at 16:04









      RScrlliRScrlli

      649114




      649114






















          1 Answer
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          active

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          0












          $begingroup$

          I found the solution after some research, hence I'll post it here in case anyone have curiosity:



          Marsden's Identity states that for all $tau$ in $mathbb{R}$ it holds that:
          $$(cdot -tau)^{k-1}=sum_jPsi_{j,k}B_{j,k} , ,$$



          It's straightforward to show that $((cdot-tau_j)^{k-1}:j=1,...,k)$ , $tau_1<...<tau_k$, is a basis for the space $Pi_{<k}$ (the space of polynomials of degree smaller than $k$).



          Hence any polynomials of degree $<k$ can be written as a linear combination of the elements in the basis, i.e.:



          $$beta_1(cdot-tau_1)^{k-1}+...+beta_k(cdot-tau_k)^{k-1},$$



          By Marsden's Identity we have that the latter equals:
          $$sum_jbeta_1Psi_{j,k}(tau_1)B_{j,k} +...+sum_jbeta_kPsi_{j,k}(tau_k)B_{j,k}$$



          Reordering it yelds to:
          $$sum_j(beta_1Psi_{j,k}(tau_1)+...+beta_kPsi_{j,k}(tau_k))B_{j,k}$$



          Letting $alpha_j=(beta_1Psi_{j,k}(tau_1)+...+beta_kPsi_{j,k}(tau_k))$ we have shown that any polynomial in $Pi_{<k}$ is contained in $$_{j,k}$.






          share|cite|improve this answer









          $endgroup$













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            0












            $begingroup$

            I found the solution after some research, hence I'll post it here in case anyone have curiosity:



            Marsden's Identity states that for all $tau$ in $mathbb{R}$ it holds that:
            $$(cdot -tau)^{k-1}=sum_jPsi_{j,k}B_{j,k} , ,$$



            It's straightforward to show that $((cdot-tau_j)^{k-1}:j=1,...,k)$ , $tau_1<...<tau_k$, is a basis for the space $Pi_{<k}$ (the space of polynomials of degree smaller than $k$).



            Hence any polynomials of degree $<k$ can be written as a linear combination of the elements in the basis, i.e.:



            $$beta_1(cdot-tau_1)^{k-1}+...+beta_k(cdot-tau_k)^{k-1},$$



            By Marsden's Identity we have that the latter equals:
            $$sum_jbeta_1Psi_{j,k}(tau_1)B_{j,k} +...+sum_jbeta_kPsi_{j,k}(tau_k)B_{j,k}$$



            Reordering it yelds to:
            $$sum_j(beta_1Psi_{j,k}(tau_1)+...+beta_kPsi_{j,k}(tau_k))B_{j,k}$$



            Letting $alpha_j=(beta_1Psi_{j,k}(tau_1)+...+beta_kPsi_{j,k}(tau_k))$ we have shown that any polynomial in $Pi_{<k}$ is contained in $$_{j,k}$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              I found the solution after some research, hence I'll post it here in case anyone have curiosity:



              Marsden's Identity states that for all $tau$ in $mathbb{R}$ it holds that:
              $$(cdot -tau)^{k-1}=sum_jPsi_{j,k}B_{j,k} , ,$$



              It's straightforward to show that $((cdot-tau_j)^{k-1}:j=1,...,k)$ , $tau_1<...<tau_k$, is a basis for the space $Pi_{<k}$ (the space of polynomials of degree smaller than $k$).



              Hence any polynomials of degree $<k$ can be written as a linear combination of the elements in the basis, i.e.:



              $$beta_1(cdot-tau_1)^{k-1}+...+beta_k(cdot-tau_k)^{k-1},$$



              By Marsden's Identity we have that the latter equals:
              $$sum_jbeta_1Psi_{j,k}(tau_1)B_{j,k} +...+sum_jbeta_kPsi_{j,k}(tau_k)B_{j,k}$$



              Reordering it yelds to:
              $$sum_j(beta_1Psi_{j,k}(tau_1)+...+beta_kPsi_{j,k}(tau_k))B_{j,k}$$



              Letting $alpha_j=(beta_1Psi_{j,k}(tau_1)+...+beta_kPsi_{j,k}(tau_k))$ we have shown that any polynomial in $Pi_{<k}$ is contained in $$_{j,k}$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I found the solution after some research, hence I'll post it here in case anyone have curiosity:



                Marsden's Identity states that for all $tau$ in $mathbb{R}$ it holds that:
                $$(cdot -tau)^{k-1}=sum_jPsi_{j,k}B_{j,k} , ,$$



                It's straightforward to show that $((cdot-tau_j)^{k-1}:j=1,...,k)$ , $tau_1<...<tau_k$, is a basis for the space $Pi_{<k}$ (the space of polynomials of degree smaller than $k$).



                Hence any polynomials of degree $<k$ can be written as a linear combination of the elements in the basis, i.e.:



                $$beta_1(cdot-tau_1)^{k-1}+...+beta_k(cdot-tau_k)^{k-1},$$



                By Marsden's Identity we have that the latter equals:
                $$sum_jbeta_1Psi_{j,k}(tau_1)B_{j,k} +...+sum_jbeta_kPsi_{j,k}(tau_k)B_{j,k}$$



                Reordering it yelds to:
                $$sum_j(beta_1Psi_{j,k}(tau_1)+...+beta_kPsi_{j,k}(tau_k))B_{j,k}$$



                Letting $alpha_j=(beta_1Psi_{j,k}(tau_1)+...+beta_kPsi_{j,k}(tau_k))$ we have shown that any polynomial in $Pi_{<k}$ is contained in $$_{j,k}$.






                share|cite|improve this answer









                $endgroup$



                I found the solution after some research, hence I'll post it here in case anyone have curiosity:



                Marsden's Identity states that for all $tau$ in $mathbb{R}$ it holds that:
                $$(cdot -tau)^{k-1}=sum_jPsi_{j,k}B_{j,k} , ,$$



                It's straightforward to show that $((cdot-tau_j)^{k-1}:j=1,...,k)$ , $tau_1<...<tau_k$, is a basis for the space $Pi_{<k}$ (the space of polynomials of degree smaller than $k$).



                Hence any polynomials of degree $<k$ can be written as a linear combination of the elements in the basis, i.e.:



                $$beta_1(cdot-tau_1)^{k-1}+...+beta_k(cdot-tau_k)^{k-1},$$



                By Marsden's Identity we have that the latter equals:
                $$sum_jbeta_1Psi_{j,k}(tau_1)B_{j,k} +...+sum_jbeta_kPsi_{j,k}(tau_k)B_{j,k}$$



                Reordering it yelds to:
                $$sum_j(beta_1Psi_{j,k}(tau_1)+...+beta_kPsi_{j,k}(tau_k))B_{j,k}$$



                Letting $alpha_j=(beta_1Psi_{j,k}(tau_1)+...+beta_kPsi_{j,k}(tau_k))$ we have shown that any polynomial in $Pi_{<k}$ is contained in $$_{j,k}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 8 '18 at 12:25









                RScrlliRScrlli

                649114




                649114






























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