Lagrange polynomial $x^n$ coefficient












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How can we show using Lagrange interpolation polynomial that
$$sum_{i=0}^n y_i prod_{j=0, jne i}^n frac{1}{x_i-x_j}$$
is the coefficient of $x^{n}$?



I know that $f[x_0,x_1, .. x_n]$ is the coefficient of $x^{n}$ in Newton polynomial $$sum_{i=0}^n f[x_0,x_1, .. x_i] prod_{j=0, }^{i-1} (x-x_i)$$
So $$bigg[f[x_0,x_1,..x_n]bigg] * x^n $$should be equal (?) to
$$bigg[sum_{i=0}^n y_i prod_{j=0, jne i}^n frac{1}{x_i-x_j}bigg] * x^n $$ There is possible to show that using above facts?






numerical-methods interpolation lagrange-interpolation







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    1












    $begingroup$


    How can we show using Lagrange interpolation polynomial that
    $$sum_{i=0}^n y_i prod_{j=0, jne i}^n frac{1}{x_i-x_j}$$
    is the coefficient of $x^{n}$?



    I know that $f[x_0,x_1, .. x_n]$ is the coefficient of $x^{n}$ in Newton polynomial $$sum_{i=0}^n f[x_0,x_1, .. x_i] prod_{j=0, }^{i-1} (x-x_i)$$
    So $$bigg[f[x_0,x_1,..x_n]bigg] * x^n $$should be equal (?) to
    $$bigg[sum_{i=0}^n y_i prod_{j=0, jne i}^n frac{1}{x_i-x_j}bigg] * x^n $$ There is possible to show that using above facts?






    numerical-methods interpolation lagrange-interpolation







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      How can we show using Lagrange interpolation polynomial that
      $$sum_{i=0}^n y_i prod_{j=0, jne i}^n frac{1}{x_i-x_j}$$
      is the coefficient of $x^{n}$?



      I know that $f[x_0,x_1, .. x_n]$ is the coefficient of $x^{n}$ in Newton polynomial $$sum_{i=0}^n f[x_0,x_1, .. x_i] prod_{j=0, }^{i-1} (x-x_i)$$
      So $$bigg[f[x_0,x_1,..x_n]bigg] * x^n $$should be equal (?) to
      $$bigg[sum_{i=0}^n y_i prod_{j=0, jne i}^n frac{1}{x_i-x_j}bigg] * x^n $$ There is possible to show that using above facts?






      numerical-methods interpolation lagrange-interpolation







      share|cite|improve this question











      $endgroup$




      How can we show using Lagrange interpolation polynomial that
      $$sum_{i=0}^n y_i prod_{j=0, jne i}^n frac{1}{x_i-x_j}$$
      is the coefficient of $x^{n}$?



      I know that $f[x_0,x_1, .. x_n]$ is the coefficient of $x^{n}$ in Newton polynomial $$sum_{i=0}^n f[x_0,x_1, .. x_i] prod_{j=0, }^{i-1} (x-x_i)$$
      So $$bigg[f[x_0,x_1,..x_n]bigg] * x^n $$should be equal (?) to
      $$bigg[sum_{i=0}^n y_i prod_{j=0, jne i}^n frac{1}{x_i-x_j}bigg] * x^n $$ There is possible to show that using above facts?






      numerical-methods interpolation lagrange-interpolation




      numerical-methods interpolation lagrange-interpolation






      share|cite|improve this question















      share|cite|improve this question













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      edited Feb 5 at 9:18









      Max

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      9211318










      asked Jan 27 at 13:15









      MarkMark

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          $begingroup$

          The (interpolation)polynomial $P_n$ can be written as $sum_{i=0}^nC_ix^i$ (Monomial basis),



          and $C_n$ can be expressed as: $C_n = sum^n_{i=0}y_iprod^n_{j=0,jneq i}frac{1}{x_i-x_j}$



          Proof from Langrange-interpolation:



          $L_n = sum^n_{i=0}y_iprod^n_{j=0,jneq i}frac{x-x_j}{x_i-x_j}$



          If we expand the product $(x-x_0)...(x-x_{n_1})$ to $x^n + ...x^{n-1} + etc.$



          And we only look at the $x^n$ component, $C_n$ follows.



          The same holds up for Newton-interpolation:



          $N_n = sum_{i=0}^nf[x_0,...,x_i](x-x_0)...(x-x_{i-1})$



          Then we get: $C_n = f[x_0,...,x_n]$.



          Thus, written differently:



          $f[x_0,...,x_n] = sum^n_{i=0}frac{y_i}{prod^n_{j=0,jneq i}(x_i-x_j)}$ (https://en.wikipedia.org/wiki/Divided_differences)






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            1 Answer
            1






            active

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            active

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            active

            oldest

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            0












            $begingroup$

            The (interpolation)polynomial $P_n$ can be written as $sum_{i=0}^nC_ix^i$ (Monomial basis),



            and $C_n$ can be expressed as: $C_n = sum^n_{i=0}y_iprod^n_{j=0,jneq i}frac{1}{x_i-x_j}$



            Proof from Langrange-interpolation:



            $L_n = sum^n_{i=0}y_iprod^n_{j=0,jneq i}frac{x-x_j}{x_i-x_j}$



            If we expand the product $(x-x_0)...(x-x_{n_1})$ to $x^n + ...x^{n-1} + etc.$



            And we only look at the $x^n$ component, $C_n$ follows.



            The same holds up for Newton-interpolation:



            $N_n = sum_{i=0}^nf[x_0,...,x_i](x-x_0)...(x-x_{i-1})$



            Then we get: $C_n = f[x_0,...,x_n]$.



            Thus, written differently:



            $f[x_0,...,x_n] = sum^n_{i=0}frac{y_i}{prod^n_{j=0,jneq i}(x_i-x_j)}$ (https://en.wikipedia.org/wiki/Divided_differences)






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The (interpolation)polynomial $P_n$ can be written as $sum_{i=0}^nC_ix^i$ (Monomial basis),



              and $C_n$ can be expressed as: $C_n = sum^n_{i=0}y_iprod^n_{j=0,jneq i}frac{1}{x_i-x_j}$



              Proof from Langrange-interpolation:



              $L_n = sum^n_{i=0}y_iprod^n_{j=0,jneq i}frac{x-x_j}{x_i-x_j}$



              If we expand the product $(x-x_0)...(x-x_{n_1})$ to $x^n + ...x^{n-1} + etc.$



              And we only look at the $x^n$ component, $C_n$ follows.



              The same holds up for Newton-interpolation:



              $N_n = sum_{i=0}^nf[x_0,...,x_i](x-x_0)...(x-x_{i-1})$



              Then we get: $C_n = f[x_0,...,x_n]$.



              Thus, written differently:



              $f[x_0,...,x_n] = sum^n_{i=0}frac{y_i}{prod^n_{j=0,jneq i}(x_i-x_j)}$ (https://en.wikipedia.org/wiki/Divided_differences)






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The (interpolation)polynomial $P_n$ can be written as $sum_{i=0}^nC_ix^i$ (Monomial basis),



                and $C_n$ can be expressed as: $C_n = sum^n_{i=0}y_iprod^n_{j=0,jneq i}frac{1}{x_i-x_j}$



                Proof from Langrange-interpolation:



                $L_n = sum^n_{i=0}y_iprod^n_{j=0,jneq i}frac{x-x_j}{x_i-x_j}$



                If we expand the product $(x-x_0)...(x-x_{n_1})$ to $x^n + ...x^{n-1} + etc.$



                And we only look at the $x^n$ component, $C_n$ follows.



                The same holds up for Newton-interpolation:



                $N_n = sum_{i=0}^nf[x_0,...,x_i](x-x_0)...(x-x_{i-1})$



                Then we get: $C_n = f[x_0,...,x_n]$.



                Thus, written differently:



                $f[x_0,...,x_n] = sum^n_{i=0}frac{y_i}{prod^n_{j=0,jneq i}(x_i-x_j)}$ (https://en.wikipedia.org/wiki/Divided_differences)






                share|cite|improve this answer









                $endgroup$



                The (interpolation)polynomial $P_n$ can be written as $sum_{i=0}^nC_ix^i$ (Monomial basis),



                and $C_n$ can be expressed as: $C_n = sum^n_{i=0}y_iprod^n_{j=0,jneq i}frac{1}{x_i-x_j}$



                Proof from Langrange-interpolation:



                $L_n = sum^n_{i=0}y_iprod^n_{j=0,jneq i}frac{x-x_j}{x_i-x_j}$



                If we expand the product $(x-x_0)...(x-x_{n_1})$ to $x^n + ...x^{n-1} + etc.$



                And we only look at the $x^n$ component, $C_n$ follows.



                The same holds up for Newton-interpolation:



                $N_n = sum_{i=0}^nf[x_0,...,x_i](x-x_0)...(x-x_{i-1})$



                Then we get: $C_n = f[x_0,...,x_n]$.



                Thus, written differently:



                $f[x_0,...,x_n] = sum^n_{i=0}frac{y_i}{prod^n_{j=0,jneq i}(x_i-x_j)}$ (https://en.wikipedia.org/wiki/Divided_differences)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 4 at 15:31









                MaxMax

                9211318




                9211318






























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