Hilbert space for integral












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Find numbers $alpha,beta, gamma in mathbb{C}$ so that the integral $$int_{-1}^{1}|x^3-alpha-beta x-gamma x^2|^2dx$$ is minimal.
Here I want to apply Hilbert space and projection lemma by considering the integral as the distance square between a point and a closed linear subspace. So how can I start my work?






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    Find numbers $alpha,beta, gamma in mathbb{C}$ so that the integral $$int_{-1}^{1}|x^3-alpha-beta x-gamma x^2|^2dx$$ is minimal.
    Here I want to apply Hilbert space and projection lemma by considering the integral as the distance square between a point and a closed linear subspace. So how can I start my work?






    functional-analysis







    share|cite|improve this question

























      1












      1








      1







      Find numbers $alpha,beta, gamma in mathbb{C}$ so that the integral $$int_{-1}^{1}|x^3-alpha-beta x-gamma x^2|^2dx$$ is minimal.
      Here I want to apply Hilbert space and projection lemma by considering the integral as the distance square between a point and a closed linear subspace. So how can I start my work?






      functional-analysis







      share|cite|improve this question













      Find numbers $alpha,beta, gamma in mathbb{C}$ so that the integral $$int_{-1}^{1}|x^3-alpha-beta x-gamma x^2|^2dx$$ is minimal.
      Here I want to apply Hilbert space and projection lemma by considering the integral as the distance square between a point and a closed linear subspace. So how can I start my work?






      functional-analysis




      functional-analysis






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      asked Mar 2 '16 at 2:22









      UserAbUserAb

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          If $mathcal{M}$ is the subspace of polynomials $p(x)=alpha + beta x + gamma x^2$, then you are looking for $pin mathcal{M}$ closest to $x^3$. Equivalently, $(x^3-p)perpmathcal{M}$, which gives equations
          $$
          int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)dx =0 \
          int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)xdx = 0 \
          int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)x^2dx = 0.
          $$
          That's 3 equations in the 3 unknowns $alpha,beta,gamma$.






          share|cite|improve this answer





















          • Another way consists of constructing an ONB ${u,v,w}$ via Gram-Schmidt from ${1,x,x^2}$ for $M$ (which is easy since $1$ and $x$ are already orthogonal) and then simply projecting $x^3$ onto $M$: $P(x^3) = langle x^3,urangle u + langle x^3,vrangle v + langle x^3,wrangle w$.
            – Friedrich Philipp
            Mar 2 '16 at 5:06












          • @FriedrichPhilipp : Not much is gained in this case. The second equation gives $beta$ directly. The other two equations involve only $alpha$, $gamma$. If you want the Gram-Schmidt version, then can also look up the normalized Legendre polynomials and use them.
            – DisintegratingByParts
            Mar 2 '16 at 5:18










          • Yes. Also, when you have the projection, you have to go back to the coefficients which means to solve a 3x3 linear system anyway.
            – Friedrich Philipp
            Mar 2 '16 at 5:25



















          4














          As the Hilbert space, choose the span of the functions $x^0,x^1,x^2,x^3$ (on $(-1,1)$), equipped with the scalar product
          $$
          langle f,grangle = int_{-1}^1 f(x)overline{g(x)},dx.
          $$
          The subspace $M$ that you want to project onto should be the span of $x^0,x^1,x^2$. In this setting you are looking for that function $f_0$ in $M$ which has the smallest distance to $x^3$.






          share|cite|improve this answer





















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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

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            2














            If $mathcal{M}$ is the subspace of polynomials $p(x)=alpha + beta x + gamma x^2$, then you are looking for $pin mathcal{M}$ closest to $x^3$. Equivalently, $(x^3-p)perpmathcal{M}$, which gives equations
            $$
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)dx =0 \
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)xdx = 0 \
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)x^2dx = 0.
            $$
            That's 3 equations in the 3 unknowns $alpha,beta,gamma$.






            share|cite|improve this answer





















            • Another way consists of constructing an ONB ${u,v,w}$ via Gram-Schmidt from ${1,x,x^2}$ for $M$ (which is easy since $1$ and $x$ are already orthogonal) and then simply projecting $x^3$ onto $M$: $P(x^3) = langle x^3,urangle u + langle x^3,vrangle v + langle x^3,wrangle w$.
              – Friedrich Philipp
              Mar 2 '16 at 5:06












            • @FriedrichPhilipp : Not much is gained in this case. The second equation gives $beta$ directly. The other two equations involve only $alpha$, $gamma$. If you want the Gram-Schmidt version, then can also look up the normalized Legendre polynomials and use them.
              – DisintegratingByParts
              Mar 2 '16 at 5:18










            • Yes. Also, when you have the projection, you have to go back to the coefficients which means to solve a 3x3 linear system anyway.
              – Friedrich Philipp
              Mar 2 '16 at 5:25
















            2














            If $mathcal{M}$ is the subspace of polynomials $p(x)=alpha + beta x + gamma x^2$, then you are looking for $pin mathcal{M}$ closest to $x^3$. Equivalently, $(x^3-p)perpmathcal{M}$, which gives equations
            $$
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)dx =0 \
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)xdx = 0 \
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)x^2dx = 0.
            $$
            That's 3 equations in the 3 unknowns $alpha,beta,gamma$.






            share|cite|improve this answer





















            • Another way consists of constructing an ONB ${u,v,w}$ via Gram-Schmidt from ${1,x,x^2}$ for $M$ (which is easy since $1$ and $x$ are already orthogonal) and then simply projecting $x^3$ onto $M$: $P(x^3) = langle x^3,urangle u + langle x^3,vrangle v + langle x^3,wrangle w$.
              – Friedrich Philipp
              Mar 2 '16 at 5:06












            • @FriedrichPhilipp : Not much is gained in this case. The second equation gives $beta$ directly. The other two equations involve only $alpha$, $gamma$. If you want the Gram-Schmidt version, then can also look up the normalized Legendre polynomials and use them.
              – DisintegratingByParts
              Mar 2 '16 at 5:18










            • Yes. Also, when you have the projection, you have to go back to the coefficients which means to solve a 3x3 linear system anyway.
              – Friedrich Philipp
              Mar 2 '16 at 5:25














            2












            2








            2






            If $mathcal{M}$ is the subspace of polynomials $p(x)=alpha + beta x + gamma x^2$, then you are looking for $pin mathcal{M}$ closest to $x^3$. Equivalently, $(x^3-p)perpmathcal{M}$, which gives equations
            $$
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)dx =0 \
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)xdx = 0 \
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)x^2dx = 0.
            $$
            That's 3 equations in the 3 unknowns $alpha,beta,gamma$.






            share|cite|improve this answer












            If $mathcal{M}$ is the subspace of polynomials $p(x)=alpha + beta x + gamma x^2$, then you are looking for $pin mathcal{M}$ closest to $x^3$. Equivalently, $(x^3-p)perpmathcal{M}$, which gives equations
            $$
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)dx =0 \
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)xdx = 0 \
            int_{-1}^{1}(x^3-alpha-beta x-gamma x^2)x^2dx = 0.
            $$
            That's 3 equations in the 3 unknowns $alpha,beta,gamma$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 2 '16 at 4:47









            DisintegratingByPartsDisintegratingByParts

            58.7k42579




            58.7k42579












            • Another way consists of constructing an ONB ${u,v,w}$ via Gram-Schmidt from ${1,x,x^2}$ for $M$ (which is easy since $1$ and $x$ are already orthogonal) and then simply projecting $x^3$ onto $M$: $P(x^3) = langle x^3,urangle u + langle x^3,vrangle v + langle x^3,wrangle w$.
              – Friedrich Philipp
              Mar 2 '16 at 5:06












            • @FriedrichPhilipp : Not much is gained in this case. The second equation gives $beta$ directly. The other two equations involve only $alpha$, $gamma$. If you want the Gram-Schmidt version, then can also look up the normalized Legendre polynomials and use them.
              – DisintegratingByParts
              Mar 2 '16 at 5:18










            • Yes. Also, when you have the projection, you have to go back to the coefficients which means to solve a 3x3 linear system anyway.
              – Friedrich Philipp
              Mar 2 '16 at 5:25


















            • Another way consists of constructing an ONB ${u,v,w}$ via Gram-Schmidt from ${1,x,x^2}$ for $M$ (which is easy since $1$ and $x$ are already orthogonal) and then simply projecting $x^3$ onto $M$: $P(x^3) = langle x^3,urangle u + langle x^3,vrangle v + langle x^3,wrangle w$.
              – Friedrich Philipp
              Mar 2 '16 at 5:06












            • @FriedrichPhilipp : Not much is gained in this case. The second equation gives $beta$ directly. The other two equations involve only $alpha$, $gamma$. If you want the Gram-Schmidt version, then can also look up the normalized Legendre polynomials and use them.
              – DisintegratingByParts
              Mar 2 '16 at 5:18










            • Yes. Also, when you have the projection, you have to go back to the coefficients which means to solve a 3x3 linear system anyway.
              – Friedrich Philipp
              Mar 2 '16 at 5:25
















            Another way consists of constructing an ONB ${u,v,w}$ via Gram-Schmidt from ${1,x,x^2}$ for $M$ (which is easy since $1$ and $x$ are already orthogonal) and then simply projecting $x^3$ onto $M$: $P(x^3) = langle x^3,urangle u + langle x^3,vrangle v + langle x^3,wrangle w$.
            – Friedrich Philipp
            Mar 2 '16 at 5:06






            Another way consists of constructing an ONB ${u,v,w}$ via Gram-Schmidt from ${1,x,x^2}$ for $M$ (which is easy since $1$ and $x$ are already orthogonal) and then simply projecting $x^3$ onto $M$: $P(x^3) = langle x^3,urangle u + langle x^3,vrangle v + langle x^3,wrangle w$.
            – Friedrich Philipp
            Mar 2 '16 at 5:06














            @FriedrichPhilipp : Not much is gained in this case. The second equation gives $beta$ directly. The other two equations involve only $alpha$, $gamma$. If you want the Gram-Schmidt version, then can also look up the normalized Legendre polynomials and use them.
            – DisintegratingByParts
            Mar 2 '16 at 5:18




            @FriedrichPhilipp : Not much is gained in this case. The second equation gives $beta$ directly. The other two equations involve only $alpha$, $gamma$. If you want the Gram-Schmidt version, then can also look up the normalized Legendre polynomials and use them.
            – DisintegratingByParts
            Mar 2 '16 at 5:18












            Yes. Also, when you have the projection, you have to go back to the coefficients which means to solve a 3x3 linear system anyway.
            – Friedrich Philipp
            Mar 2 '16 at 5:25




            Yes. Also, when you have the projection, you have to go back to the coefficients which means to solve a 3x3 linear system anyway.
            – Friedrich Philipp
            Mar 2 '16 at 5:25











            4














            As the Hilbert space, choose the span of the functions $x^0,x^1,x^2,x^3$ (on $(-1,1)$), equipped with the scalar product
            $$
            langle f,grangle = int_{-1}^1 f(x)overline{g(x)},dx.
            $$
            The subspace $M$ that you want to project onto should be the span of $x^0,x^1,x^2$. In this setting you are looking for that function $f_0$ in $M$ which has the smallest distance to $x^3$.






            share|cite|improve this answer


























              4














              As the Hilbert space, choose the span of the functions $x^0,x^1,x^2,x^3$ (on $(-1,1)$), equipped with the scalar product
              $$
              langle f,grangle = int_{-1}^1 f(x)overline{g(x)},dx.
              $$
              The subspace $M$ that you want to project onto should be the span of $x^0,x^1,x^2$. In this setting you are looking for that function $f_0$ in $M$ which has the smallest distance to $x^3$.






              share|cite|improve this answer
























                4












                4








                4






                As the Hilbert space, choose the span of the functions $x^0,x^1,x^2,x^3$ (on $(-1,1)$), equipped with the scalar product
                $$
                langle f,grangle = int_{-1}^1 f(x)overline{g(x)},dx.
                $$
                The subspace $M$ that you want to project onto should be the span of $x^0,x^1,x^2$. In this setting you are looking for that function $f_0$ in $M$ which has the smallest distance to $x^3$.






                share|cite|improve this answer












                As the Hilbert space, choose the span of the functions $x^0,x^1,x^2,x^3$ (on $(-1,1)$), equipped with the scalar product
                $$
                langle f,grangle = int_{-1}^1 f(x)overline{g(x)},dx.
                $$
                The subspace $M$ that you want to project onto should be the span of $x^0,x^1,x^2$. In this setting you are looking for that function $f_0$ in $M$ which has the smallest distance to $x^3$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 2 '16 at 2:27









                Friedrich PhilippFriedrich Philipp

                3,323314




                3,323314






























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