Legendre - orthogonality related proof












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How can I proceed to prove that there are constants $α_0, α_1, ..., α_n$ such that
$x^n = α_0P_0(x) + α_1P_1(x) + ... + α_nP_n(x)$ where $P_n$ is legendre polynomial.



I guess that this has to do with orthogonality property and the fact that $P_n$'s are linearly independent, but I have no clue where to start. Can anybody help?






legendre-polynomials







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  • $begingroup$
    for every $x$ ? that I don't know but if $|x| leq 1$ then the Legendre polynomials is an orthogonal basis of the space $L^2([-1,1])$ which is a Hilbert space and it just happens that $x^n in L^2([-1,1]) $ therefore it can be written as a Legendre polynomials series
    $endgroup$
    – rapidracim
    Jan 18 at 13:59
















0












$begingroup$


How can I proceed to prove that there are constants $α_0, α_1, ..., α_n$ such that
$x^n = α_0P_0(x) + α_1P_1(x) + ... + α_nP_n(x)$ where $P_n$ is legendre polynomial.



I guess that this has to do with orthogonality property and the fact that $P_n$'s are linearly independent, but I have no clue where to start. Can anybody help?






legendre-polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    for every $x$ ? that I don't know but if $|x| leq 1$ then the Legendre polynomials is an orthogonal basis of the space $L^2([-1,1])$ which is a Hilbert space and it just happens that $x^n in L^2([-1,1]) $ therefore it can be written as a Legendre polynomials series
    $endgroup$
    – rapidracim
    Jan 18 at 13:59














0












0








0





$begingroup$


How can I proceed to prove that there are constants $α_0, α_1, ..., α_n$ such that
$x^n = α_0P_0(x) + α_1P_1(x) + ... + α_nP_n(x)$ where $P_n$ is legendre polynomial.



I guess that this has to do with orthogonality property and the fact that $P_n$'s are linearly independent, but I have no clue where to start. Can anybody help?






legendre-polynomials







share|cite|improve this question











$endgroup$




How can I proceed to prove that there are constants $α_0, α_1, ..., α_n$ such that
$x^n = α_0P_0(x) + α_1P_1(x) + ... + α_nP_n(x)$ where $P_n$ is legendre polynomial.



I guess that this has to do with orthogonality property and the fact that $P_n$'s are linearly independent, but I have no clue where to start. Can anybody help?






legendre-polynomials




legendre-polynomials






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share|cite|improve this question













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edited Jan 18 at 13:51









idriskameni

749321




749321










asked Jan 18 at 12:48









user635977user635977

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63












  • $begingroup$
    for every $x$ ? that I don't know but if $|x| leq 1$ then the Legendre polynomials is an orthogonal basis of the space $L^2([-1,1])$ which is a Hilbert space and it just happens that $x^n in L^2([-1,1]) $ therefore it can be written as a Legendre polynomials series
    $endgroup$
    – rapidracim
    Jan 18 at 13:59


















  • $begingroup$
    for every $x$ ? that I don't know but if $|x| leq 1$ then the Legendre polynomials is an orthogonal basis of the space $L^2([-1,1])$ which is a Hilbert space and it just happens that $x^n in L^2([-1,1]) $ therefore it can be written as a Legendre polynomials series
    $endgroup$
    – rapidracim
    Jan 18 at 13:59
















$begingroup$
for every $x$ ? that I don't know but if $|x| leq 1$ then the Legendre polynomials is an orthogonal basis of the space $L^2([-1,1])$ which is a Hilbert space and it just happens that $x^n in L^2([-1,1]) $ therefore it can be written as a Legendre polynomials series
$endgroup$
– rapidracim
Jan 18 at 13:59




$begingroup$
for every $x$ ? that I don't know but if $|x| leq 1$ then the Legendre polynomials is an orthogonal basis of the space $L^2([-1,1])$ which is a Hilbert space and it just happens that $x^n in L^2([-1,1]) $ therefore it can be written as a Legendre polynomials series
$endgroup$
– rapidracim
Jan 18 at 13:59










2 Answers
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Any function on $[-1,1]$ whether continuous or not can be expanded in a series of Legendre polynomials. That is the same as saying that ${P_n}$ form a basis for this kind of expansions. Now ${P_n}$ is also an orthogonal basis, meaning that $int_{-1}^{1} P_n(x)P_m(x)dx = 0$ if $n neq m$.
This helps us determining the values $alpha_j$: suppose $f(x) = sum_{j=0}^{infty} alpha_j P_j(x)$, then



$$int_{-1}^{1} P_n(x)f(x)dx = int_{-1}^{1} P_n(x)sum_{j=0}^{infty} alpha_j P_j(x) dx = sum_{j=0}^{infty} alpha_j int_{-1}^{1} P_n(x)P_j(x)dx = \alpha_n int_{-1}^{1} P_n^2(x)dx = frac{2alpha_n}{2n+1}$$



Hence



$$alpha_j = frac{2j+1}{2} int_{-1}^{1} P_j(x)f(x)dx$$



Now plug in $x^n$ for $f(x)$ and evaluate the integrals - I leave that part to you. You also need to show that $alpha_j = 0$ for $j > n$.






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    While Maestro’s answer works, he doesn’t show that you produce $x^n$, only that you approximate arbitrarily closely it by orthogonal polynomials. Rather, notice that the first n legendre polynomials are linearly independent and contained in the sub space of n degree polynomials. Since the dimensions are equal, the legendre polynomials span the subpace of up-to-n dimensional polynomials, of which $x^n$ is. Orthogonality doesn’t actually matter, only linear independence.






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      2 Answers
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      2 Answers
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      0












      $begingroup$

      Any function on $[-1,1]$ whether continuous or not can be expanded in a series of Legendre polynomials. That is the same as saying that ${P_n}$ form a basis for this kind of expansions. Now ${P_n}$ is also an orthogonal basis, meaning that $int_{-1}^{1} P_n(x)P_m(x)dx = 0$ if $n neq m$.
      This helps us determining the values $alpha_j$: suppose $f(x) = sum_{j=0}^{infty} alpha_j P_j(x)$, then



      $$int_{-1}^{1} P_n(x)f(x)dx = int_{-1}^{1} P_n(x)sum_{j=0}^{infty} alpha_j P_j(x) dx = sum_{j=0}^{infty} alpha_j int_{-1}^{1} P_n(x)P_j(x)dx = \alpha_n int_{-1}^{1} P_n^2(x)dx = frac{2alpha_n}{2n+1}$$



      Hence



      $$alpha_j = frac{2j+1}{2} int_{-1}^{1} P_j(x)f(x)dx$$



      Now plug in $x^n$ for $f(x)$ and evaluate the integrals - I leave that part to you. You also need to show that $alpha_j = 0$ for $j > n$.






      share|cite|improve this answer











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        0












        $begingroup$

        Any function on $[-1,1]$ whether continuous or not can be expanded in a series of Legendre polynomials. That is the same as saying that ${P_n}$ form a basis for this kind of expansions. Now ${P_n}$ is also an orthogonal basis, meaning that $int_{-1}^{1} P_n(x)P_m(x)dx = 0$ if $n neq m$.
        This helps us determining the values $alpha_j$: suppose $f(x) = sum_{j=0}^{infty} alpha_j P_j(x)$, then



        $$int_{-1}^{1} P_n(x)f(x)dx = int_{-1}^{1} P_n(x)sum_{j=0}^{infty} alpha_j P_j(x) dx = sum_{j=0}^{infty} alpha_j int_{-1}^{1} P_n(x)P_j(x)dx = \alpha_n int_{-1}^{1} P_n^2(x)dx = frac{2alpha_n}{2n+1}$$



        Hence



        $$alpha_j = frac{2j+1}{2} int_{-1}^{1} P_j(x)f(x)dx$$



        Now plug in $x^n$ for $f(x)$ and evaluate the integrals - I leave that part to you. You also need to show that $alpha_j = 0$ for $j > n$.






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          Any function on $[-1,1]$ whether continuous or not can be expanded in a series of Legendre polynomials. That is the same as saying that ${P_n}$ form a basis for this kind of expansions. Now ${P_n}$ is also an orthogonal basis, meaning that $int_{-1}^{1} P_n(x)P_m(x)dx = 0$ if $n neq m$.
          This helps us determining the values $alpha_j$: suppose $f(x) = sum_{j=0}^{infty} alpha_j P_j(x)$, then



          $$int_{-1}^{1} P_n(x)f(x)dx = int_{-1}^{1} P_n(x)sum_{j=0}^{infty} alpha_j P_j(x) dx = sum_{j=0}^{infty} alpha_j int_{-1}^{1} P_n(x)P_j(x)dx = \alpha_n int_{-1}^{1} P_n^2(x)dx = frac{2alpha_n}{2n+1}$$



          Hence



          $$alpha_j = frac{2j+1}{2} int_{-1}^{1} P_j(x)f(x)dx$$



          Now plug in $x^n$ for $f(x)$ and evaluate the integrals - I leave that part to you. You also need to show that $alpha_j = 0$ for $j > n$.






          share|cite|improve this answer











          $endgroup$



          Any function on $[-1,1]$ whether continuous or not can be expanded in a series of Legendre polynomials. That is the same as saying that ${P_n}$ form a basis for this kind of expansions. Now ${P_n}$ is also an orthogonal basis, meaning that $int_{-1}^{1} P_n(x)P_m(x)dx = 0$ if $n neq m$.
          This helps us determining the values $alpha_j$: suppose $f(x) = sum_{j=0}^{infty} alpha_j P_j(x)$, then



          $$int_{-1}^{1} P_n(x)f(x)dx = int_{-1}^{1} P_n(x)sum_{j=0}^{infty} alpha_j P_j(x) dx = sum_{j=0}^{infty} alpha_j int_{-1}^{1} P_n(x)P_j(x)dx = \alpha_n int_{-1}^{1} P_n^2(x)dx = frac{2alpha_n}{2n+1}$$



          Hence



          $$alpha_j = frac{2j+1}{2} int_{-1}^{1} P_j(x)f(x)dx$$



          Now plug in $x^n$ for $f(x)$ and evaluate the integrals - I leave that part to you. You also need to show that $alpha_j = 0$ for $j > n$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 18 at 20:26

























          answered Jan 18 at 20:15









          Maestro13Maestro13

          1,106724




          1,106724























              0












              $begingroup$

              While Maestro’s answer works, he doesn’t show that you produce $x^n$, only that you approximate arbitrarily closely it by orthogonal polynomials. Rather, notice that the first n legendre polynomials are linearly independent and contained in the sub space of n degree polynomials. Since the dimensions are equal, the legendre polynomials span the subpace of up-to-n dimensional polynomials, of which $x^n$ is. Orthogonality doesn’t actually matter, only linear independence.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                While Maestro’s answer works, he doesn’t show that you produce $x^n$, only that you approximate arbitrarily closely it by orthogonal polynomials. Rather, notice that the first n legendre polynomials are linearly independent and contained in the sub space of n degree polynomials. Since the dimensions are equal, the legendre polynomials span the subpace of up-to-n dimensional polynomials, of which $x^n$ is. Orthogonality doesn’t actually matter, only linear independence.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  While Maestro’s answer works, he doesn’t show that you produce $x^n$, only that you approximate arbitrarily closely it by orthogonal polynomials. Rather, notice that the first n legendre polynomials are linearly independent and contained in the sub space of n degree polynomials. Since the dimensions are equal, the legendre polynomials span the subpace of up-to-n dimensional polynomials, of which $x^n$ is. Orthogonality doesn’t actually matter, only linear independence.






                  share|cite|improve this answer









                  $endgroup$



                  While Maestro’s answer works, he doesn’t show that you produce $x^n$, only that you approximate arbitrarily closely it by orthogonal polynomials. Rather, notice that the first n legendre polynomials are linearly independent and contained in the sub space of n degree polynomials. Since the dimensions are equal, the legendre polynomials span the subpace of up-to-n dimensional polynomials, of which $x^n$ is. Orthogonality doesn’t actually matter, only linear independence.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 at 22:40









                  AlexAlex

                  17713




                  17713






























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