For $f(x) = tan(pi cdot x)$, find the Interpolation $Q(x) = b_0 + b_1 x + b_2 frac{1}{x-frac{1}{2}}$












0












$begingroup$


Progress so far:



In a previous task, I determined a polynomial interpolation using a system of linear equations.



The data points to be used were $(0, f(0)), (frac{1}{6}, f(frac{1}{6})), (frac{1}{4}, f(frac{1}{4}))$



The linear equation used was of the form:



begin{pmatrix}0^2&0^1&0^0\ :frac{1}{6}^2&frac{1}{6}^1&frac{1}{6}^0\ :frac{1}{4}^2&frac{1}{4}^1&frac{1}{4}^0end{pmatrix}begin{pmatrix}a_2\ :a_1\ :a_0end{pmatrix}=begin{pmatrix}0\ frac{1}{sqrt{3}}\ 1end{pmatrix}



and the polynomial of the form



$$ p_2(x) = a_0 + a_1 x + a_2 x^2$$ was determined to have the coefficients



$a_0 = 0, a_1 = -8 + frac{18}{sqrt{3}}, a_2 = 48 - frac{72}{sqrt{3}}$



I am not aware of how I would use a system of linear equations to determine $Q(x)$. Is it perhaps possible to derive the coefficients from $p_2$? Or is there some other method of interpolation I should pursue?






interpolation interpolation-theory







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  • $begingroup$
    In my answer, I assumed that you would also use $x=frac 12$. If this is not the case, for $Q(x)$ you still have three linear equations.
    $endgroup$
    – Claude Leibovici
    Dec 3 '18 at 11:26
















0












$begingroup$


Progress so far:



In a previous task, I determined a polynomial interpolation using a system of linear equations.



The data points to be used were $(0, f(0)), (frac{1}{6}, f(frac{1}{6})), (frac{1}{4}, f(frac{1}{4}))$



The linear equation used was of the form:



begin{pmatrix}0^2&0^1&0^0\ :frac{1}{6}^2&frac{1}{6}^1&frac{1}{6}^0\ :frac{1}{4}^2&frac{1}{4}^1&frac{1}{4}^0end{pmatrix}begin{pmatrix}a_2\ :a_1\ :a_0end{pmatrix}=begin{pmatrix}0\ frac{1}{sqrt{3}}\ 1end{pmatrix}



and the polynomial of the form



$$ p_2(x) = a_0 + a_1 x + a_2 x^2$$ was determined to have the coefficients



$a_0 = 0, a_1 = -8 + frac{18}{sqrt{3}}, a_2 = 48 - frac{72}{sqrt{3}}$



I am not aware of how I would use a system of linear equations to determine $Q(x)$. Is it perhaps possible to derive the coefficients from $p_2$? Or is there some other method of interpolation I should pursue?






interpolation interpolation-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    In my answer, I assumed that you would also use $x=frac 12$. If this is not the case, for $Q(x)$ you still have three linear equations.
    $endgroup$
    – Claude Leibovici
    Dec 3 '18 at 11:26














0












0








0





$begingroup$


Progress so far:



In a previous task, I determined a polynomial interpolation using a system of linear equations.



The data points to be used were $(0, f(0)), (frac{1}{6}, f(frac{1}{6})), (frac{1}{4}, f(frac{1}{4}))$



The linear equation used was of the form:



begin{pmatrix}0^2&0^1&0^0\ :frac{1}{6}^2&frac{1}{6}^1&frac{1}{6}^0\ :frac{1}{4}^2&frac{1}{4}^1&frac{1}{4}^0end{pmatrix}begin{pmatrix}a_2\ :a_1\ :a_0end{pmatrix}=begin{pmatrix}0\ frac{1}{sqrt{3}}\ 1end{pmatrix}



and the polynomial of the form



$$ p_2(x) = a_0 + a_1 x + a_2 x^2$$ was determined to have the coefficients



$a_0 = 0, a_1 = -8 + frac{18}{sqrt{3}}, a_2 = 48 - frac{72}{sqrt{3}}$



I am not aware of how I would use a system of linear equations to determine $Q(x)$. Is it perhaps possible to derive the coefficients from $p_2$? Or is there some other method of interpolation I should pursue?






interpolation interpolation-theory







share|cite|improve this question









$endgroup$




Progress so far:



In a previous task, I determined a polynomial interpolation using a system of linear equations.



The data points to be used were $(0, f(0)), (frac{1}{6}, f(frac{1}{6})), (frac{1}{4}, f(frac{1}{4}))$



The linear equation used was of the form:



begin{pmatrix}0^2&0^1&0^0\ :frac{1}{6}^2&frac{1}{6}^1&frac{1}{6}^0\ :frac{1}{4}^2&frac{1}{4}^1&frac{1}{4}^0end{pmatrix}begin{pmatrix}a_2\ :a_1\ :a_0end{pmatrix}=begin{pmatrix}0\ frac{1}{sqrt{3}}\ 1end{pmatrix}



and the polynomial of the form



$$ p_2(x) = a_0 + a_1 x + a_2 x^2$$ was determined to have the coefficients



$a_0 = 0, a_1 = -8 + frac{18}{sqrt{3}}, a_2 = 48 - frac{72}{sqrt{3}}$



I am not aware of how I would use a system of linear equations to determine $Q(x)$. Is it perhaps possible to derive the coefficients from $p_2$? Or is there some other method of interpolation I should pursue?






interpolation interpolation-theory




interpolation interpolation-theory






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asked Dec 3 '18 at 10:38









OscarOscar

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  • $begingroup$
    In my answer, I assumed that you would also use $x=frac 12$. If this is not the case, for $Q(x)$ you still have three linear equations.
    $endgroup$
    – Claude Leibovici
    Dec 3 '18 at 11:26


















  • $begingroup$
    In my answer, I assumed that you would also use $x=frac 12$. If this is not the case, for $Q(x)$ you still have three linear equations.
    $endgroup$
    – Claude Leibovici
    Dec 3 '18 at 11:26
















$begingroup$
In my answer, I assumed that you would also use $x=frac 12$. If this is not the case, for $Q(x)$ you still have three linear equations.
$endgroup$
– Claude Leibovici
Dec 3 '18 at 11:26




$begingroup$
In my answer, I assumed that you would also use $x=frac 12$. If this is not the case, for $Q(x)$ you still have three linear equations.
$endgroup$
– Claude Leibovici
Dec 3 '18 at 11:26










1 Answer
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$begingroup$

Instead of working with $Q(x)$
$$Q(x) = b_0 + b_1 x + frac{b_2}{x-frac{1}{2}}$$
work with $Phi(x)$ an $Psi(x)$
$$Phi(x)=left(x-frac 12right) Q(x)=left(b_2-frac{b_0}{2}right)+x left(b_0-frac{b_1}{2}right)+b_1 x^2=alpha+beta x+gamma x^2$$



$$Psi(x)=left(x-frac 12right) F(x)=left(x-frac 12right) tan(pi x)$$ $Psi(x)$ is a very nice function for the range of interest.



Compute $alpha,beta ,gamma$ and go back to $b_0,b_1,b_2$.






share|cite|improve this answer









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    $begingroup$

    Instead of working with $Q(x)$
    $$Q(x) = b_0 + b_1 x + frac{b_2}{x-frac{1}{2}}$$
    work with $Phi(x)$ an $Psi(x)$
    $$Phi(x)=left(x-frac 12right) Q(x)=left(b_2-frac{b_0}{2}right)+x left(b_0-frac{b_1}{2}right)+b_1 x^2=alpha+beta x+gamma x^2$$



    $$Psi(x)=left(x-frac 12right) F(x)=left(x-frac 12right) tan(pi x)$$ $Psi(x)$ is a very nice function for the range of interest.



    Compute $alpha,beta ,gamma$ and go back to $b_0,b_1,b_2$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Instead of working with $Q(x)$
      $$Q(x) = b_0 + b_1 x + frac{b_2}{x-frac{1}{2}}$$
      work with $Phi(x)$ an $Psi(x)$
      $$Phi(x)=left(x-frac 12right) Q(x)=left(b_2-frac{b_0}{2}right)+x left(b_0-frac{b_1}{2}right)+b_1 x^2=alpha+beta x+gamma x^2$$



      $$Psi(x)=left(x-frac 12right) F(x)=left(x-frac 12right) tan(pi x)$$ $Psi(x)$ is a very nice function for the range of interest.



      Compute $alpha,beta ,gamma$ and go back to $b_0,b_1,b_2$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Instead of working with $Q(x)$
        $$Q(x) = b_0 + b_1 x + frac{b_2}{x-frac{1}{2}}$$
        work with $Phi(x)$ an $Psi(x)$
        $$Phi(x)=left(x-frac 12right) Q(x)=left(b_2-frac{b_0}{2}right)+x left(b_0-frac{b_1}{2}right)+b_1 x^2=alpha+beta x+gamma x^2$$



        $$Psi(x)=left(x-frac 12right) F(x)=left(x-frac 12right) tan(pi x)$$ $Psi(x)$ is a very nice function for the range of interest.



        Compute $alpha,beta ,gamma$ and go back to $b_0,b_1,b_2$.






        share|cite|improve this answer









        $endgroup$



        Instead of working with $Q(x)$
        $$Q(x) = b_0 + b_1 x + frac{b_2}{x-frac{1}{2}}$$
        work with $Phi(x)$ an $Psi(x)$
        $$Phi(x)=left(x-frac 12right) Q(x)=left(b_2-frac{b_0}{2}right)+x left(b_0-frac{b_1}{2}right)+b_1 x^2=alpha+beta x+gamma x^2$$



        $$Psi(x)=left(x-frac 12right) F(x)=left(x-frac 12right) tan(pi x)$$ $Psi(x)$ is a very nice function for the range of interest.



        Compute $alpha,beta ,gamma$ and go back to $b_0,b_1,b_2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 11:12









        Claude LeiboviciClaude Leibovici

        120k1157132




        120k1157132






























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