Finding Expression for terms in Lagrange Polynomial Interpolation












1












$begingroup$


So for a paper I am writing I am using Lagrange polynomial interpolation:



$$P(x)=sum_{i=0}^{N}f(x_i) cdot prod_{j=0;jneq i}^{N}frac{x-x_j}{x_i-x_j}$$



And I need to find an expression that describes each term. Like I put $k = 1$ and get the constant term, then $k = 2$ for the $x$ - term, etc. I don't mind if it's really complicated but I just need an expression for the coefficient for an arbitrary term. Just wondering if this is even possible, or if someone could point me in the right direction..



Any help is greatly appreciated, Thank you!






analysis functions polynomials interpolation lagrange-interpolation







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    So for a paper I am writing I am using Lagrange polynomial interpolation:



    $$P(x)=sum_{i=0}^{N}f(x_i) cdot prod_{j=0;jneq i}^{N}frac{x-x_j}{x_i-x_j}$$



    And I need to find an expression that describes each term. Like I put $k = 1$ and get the constant term, then $k = 2$ for the $x$ - term, etc. I don't mind if it's really complicated but I just need an expression for the coefficient for an arbitrary term. Just wondering if this is even possible, or if someone could point me in the right direction..



    Any help is greatly appreciated, Thank you!






    analysis functions polynomials interpolation lagrange-interpolation







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      So for a paper I am writing I am using Lagrange polynomial interpolation:



      $$P(x)=sum_{i=0}^{N}f(x_i) cdot prod_{j=0;jneq i}^{N}frac{x-x_j}{x_i-x_j}$$



      And I need to find an expression that describes each term. Like I put $k = 1$ and get the constant term, then $k = 2$ for the $x$ - term, etc. I don't mind if it's really complicated but I just need an expression for the coefficient for an arbitrary term. Just wondering if this is even possible, or if someone could point me in the right direction..



      Any help is greatly appreciated, Thank you!






      analysis functions polynomials interpolation lagrange-interpolation







      share|cite|improve this question











      $endgroup$




      So for a paper I am writing I am using Lagrange polynomial interpolation:



      $$P(x)=sum_{i=0}^{N}f(x_i) cdot prod_{j=0;jneq i}^{N}frac{x-x_j}{x_i-x_j}$$



      And I need to find an expression that describes each term. Like I put $k = 1$ and get the constant term, then $k = 2$ for the $x$ - term, etc. I don't mind if it's really complicated but I just need an expression for the coefficient for an arbitrary term. Just wondering if this is even possible, or if someone could point me in the right direction..



      Any help is greatly appreciated, Thank you!






      analysis functions polynomials interpolation lagrange-interpolation




      analysis functions polynomials interpolation lagrange-interpolation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 5 at 8:20









      mrtaurho

      6,08771641




      6,08771641










      asked Jan 1 at 7:55









      ToadfutureToadfuture

      505




      505






















          2 Answers
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          0












          $begingroup$

          To go about finding the coefficients for the various powers of $x$, we first look at the product portion. Looking at
          $$prod_{j=0; j neq i}^{N}frac{x-x_i}{x_j-x_i}$$
          we find that the full product of the denominator will be in the coefficient of any power of $x$. Pulling that out and looking at the numerator, we find
          $$prod_{j=0; j neq i}^{N} x-x_i = (x-x_i)^N$$
          Here we can pick out the various powers of $x$. The coefficient for each term in this polynomial is
          $$binom{N}{k}(-x_i)^{N-k}$$ where $k$ is the power of the corresponding $x$.



          Now taking into account the product that we initially ignored and the sum out front we find
          $$C(k)=sum_{i=0}^N binom{N}{k}(-x_i)^{N-k}f(x_i)prod_{j=0; j neq i}^{N}frac{1}{x_j-x_i}.$$



          I hope this helps!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
            $endgroup$
            – Max
            Feb 5 at 8:20



















          0












          $begingroup$

          To find the coefficients in monomial basis.



          $C_k = (-1)^{n-i}sum^n_{j=0}y_jfrac{e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)}{prod^n_{k=0,kneq j}(x_j-x_k)}$



          Where $e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)$ is the elementary symmetric polynomial (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).



          P.S. This formula follows directly if we use the following property: $prod _{{j=1}}^{n}(lambda -X_{j})=lambda ^{n}-e_{1}(X_{1},ldots ,X_{n})lambda ^{{n-1}}+e_{2}(X_{1},ldots ,X_{n})lambda ^{{n-2}}+cdots +(-1)^{n}e_{n}(X_{1},ldots ,X_{n}).$






          share|cite|improve this answer











          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            To go about finding the coefficients for the various powers of $x$, we first look at the product portion. Looking at
            $$prod_{j=0; j neq i}^{N}frac{x-x_i}{x_j-x_i}$$
            we find that the full product of the denominator will be in the coefficient of any power of $x$. Pulling that out and looking at the numerator, we find
            $$prod_{j=0; j neq i}^{N} x-x_i = (x-x_i)^N$$
            Here we can pick out the various powers of $x$. The coefficient for each term in this polynomial is
            $$binom{N}{k}(-x_i)^{N-k}$$ where $k$ is the power of the corresponding $x$.



            Now taking into account the product that we initially ignored and the sum out front we find
            $$C(k)=sum_{i=0}^N binom{N}{k}(-x_i)^{N-k}f(x_i)prod_{j=0; j neq i}^{N}frac{1}{x_j-x_i}.$$



            I hope this helps!






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
              $endgroup$
              – Max
              Feb 5 at 8:20
















            0












            $begingroup$

            To go about finding the coefficients for the various powers of $x$, we first look at the product portion. Looking at
            $$prod_{j=0; j neq i}^{N}frac{x-x_i}{x_j-x_i}$$
            we find that the full product of the denominator will be in the coefficient of any power of $x$. Pulling that out and looking at the numerator, we find
            $$prod_{j=0; j neq i}^{N} x-x_i = (x-x_i)^N$$
            Here we can pick out the various powers of $x$. The coefficient for each term in this polynomial is
            $$binom{N}{k}(-x_i)^{N-k}$$ where $k$ is the power of the corresponding $x$.



            Now taking into account the product that we initially ignored and the sum out front we find
            $$C(k)=sum_{i=0}^N binom{N}{k}(-x_i)^{N-k}f(x_i)prod_{j=0; j neq i}^{N}frac{1}{x_j-x_i}.$$



            I hope this helps!






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
              $endgroup$
              – Max
              Feb 5 at 8:20














            0












            0








            0





            $begingroup$

            To go about finding the coefficients for the various powers of $x$, we first look at the product portion. Looking at
            $$prod_{j=0; j neq i}^{N}frac{x-x_i}{x_j-x_i}$$
            we find that the full product of the denominator will be in the coefficient of any power of $x$. Pulling that out and looking at the numerator, we find
            $$prod_{j=0; j neq i}^{N} x-x_i = (x-x_i)^N$$
            Here we can pick out the various powers of $x$. The coefficient for each term in this polynomial is
            $$binom{N}{k}(-x_i)^{N-k}$$ where $k$ is the power of the corresponding $x$.



            Now taking into account the product that we initially ignored and the sum out front we find
            $$C(k)=sum_{i=0}^N binom{N}{k}(-x_i)^{N-k}f(x_i)prod_{j=0; j neq i}^{N}frac{1}{x_j-x_i}.$$



            I hope this helps!






            share|cite|improve this answer









            $endgroup$



            To go about finding the coefficients for the various powers of $x$, we first look at the product portion. Looking at
            $$prod_{j=0; j neq i}^{N}frac{x-x_i}{x_j-x_i}$$
            we find that the full product of the denominator will be in the coefficient of any power of $x$. Pulling that out and looking at the numerator, we find
            $$prod_{j=0; j neq i}^{N} x-x_i = (x-x_i)^N$$
            Here we can pick out the various powers of $x$. The coefficient for each term in this polynomial is
            $$binom{N}{k}(-x_i)^{N-k}$$ where $k$ is the power of the corresponding $x$.



            Now taking into account the product that we initially ignored and the sum out front we find
            $$C(k)=sum_{i=0}^N binom{N}{k}(-x_i)^{N-k}f(x_i)prod_{j=0; j neq i}^{N}frac{1}{x_j-x_i}.$$



            I hope this helps!







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 1 at 9:02









            Cameron KuchtaCameron Kuchta

            12




            12












            • $begingroup$
              The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
              $endgroup$
              – Max
              Feb 5 at 8:20


















            • $begingroup$
              The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
              $endgroup$
              – Max
              Feb 5 at 8:20
















            $begingroup$
            The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
            $endgroup$
            – Max
            Feb 5 at 8:20




            $begingroup$
            The indexes of the product (in the Lagrange-formula) were 'switched', thus your elaboration won't hold.
            $endgroup$
            – Max
            Feb 5 at 8:20











            0












            $begingroup$

            To find the coefficients in monomial basis.



            $C_k = (-1)^{n-i}sum^n_{j=0}y_jfrac{e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)}{prod^n_{k=0,kneq j}(x_j-x_k)}$



            Where $e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)$ is the elementary symmetric polynomial (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).



            P.S. This formula follows directly if we use the following property: $prod _{{j=1}}^{n}(lambda -X_{j})=lambda ^{n}-e_{1}(X_{1},ldots ,X_{n})lambda ^{{n-1}}+e_{2}(X_{1},ldots ,X_{n})lambda ^{{n-2}}+cdots +(-1)^{n}e_{n}(X_{1},ldots ,X_{n}).$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              To find the coefficients in monomial basis.



              $C_k = (-1)^{n-i}sum^n_{j=0}y_jfrac{e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)}{prod^n_{k=0,kneq j}(x_j-x_k)}$



              Where $e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)$ is the elementary symmetric polynomial (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).



              P.S. This formula follows directly if we use the following property: $prod _{{j=1}}^{n}(lambda -X_{j})=lambda ^{n}-e_{1}(X_{1},ldots ,X_{n})lambda ^{{n-1}}+e_{2}(X_{1},ldots ,X_{n})lambda ^{{n-2}}+cdots +(-1)^{n}e_{n}(X_{1},ldots ,X_{n}).$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                To find the coefficients in monomial basis.



                $C_k = (-1)^{n-i}sum^n_{j=0}y_jfrac{e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)}{prod^n_{k=0,kneq j}(x_j-x_k)}$



                Where $e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)$ is the elementary symmetric polynomial (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).



                P.S. This formula follows directly if we use the following property: $prod _{{j=1}}^{n}(lambda -X_{j})=lambda ^{n}-e_{1}(X_{1},ldots ,X_{n})lambda ^{{n-1}}+e_{2}(X_{1},ldots ,X_{n})lambda ^{{n-2}}+cdots +(-1)^{n}e_{n}(X_{1},ldots ,X_{n}).$






                share|cite|improve this answer











                $endgroup$



                To find the coefficients in monomial basis.



                $C_k = (-1)^{n-i}sum^n_{j=0}y_jfrac{e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)}{prod^n_{k=0,kneq j}(x_j-x_k)}$



                Where $e_{n-i}(x_0,...,x_{j-1},x_{j+1},...,x_n)$ is the elementary symmetric polynomial (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).



                P.S. This formula follows directly if we use the following property: $prod _{{j=1}}^{n}(lambda -X_{j})=lambda ^{n}-e_{1}(X_{1},ldots ,X_{n})lambda ^{{n-1}}+e_{2}(X_{1},ldots ,X_{n})lambda ^{{n-2}}+cdots +(-1)^{n}e_{n}(X_{1},ldots ,X_{n}).$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 6 at 14:44

























                answered Feb 4 at 15:50









                MaxMax

                9211319




                9211319






























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