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Given a quadrature formula for functions $f: \Bbb R \rightarrow \Bbb R$ of the form

$$Q[f] := 2/3f(x) + 1/3f(y) \approx \int_0^1 f(z) dz,$$

determine every $(x,y)^T \in \Bbb R^2$ such that the formula integrates polynomials of maximal degree exactly, so

$$Q[p] = \int_0^1 p(z) dz.$$

I know that a quadrature formula can integrate polynomials of maximal degree $2(n-1)$, so in this case, we would have to determine the formula for a base of the form $1, x, x^2, x^3$. Now I know how I would have to determine the formula when I have to find the weights $w$, but I don't know how I would go about this when I have to find the $x, y$. Can anybody explain to me how I would have to proceed in such a case?

Julian
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1 Answers1

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Assuming this integration formula exact for polynomials $1,X,X^2$ we have $$Q[1]=1,\quad Q[X]=\frac23 x+\frac13 y,\quad Q[X^2]=\frac23 x^2+\frac13 y^2$$ While $$\int_0^11dt=1,\quad\int_0^1tdt=\frac12,\quad\int_0^1t^2dt=\frac13,$$ So, saying that tge formula is exact for $1,X,X^2$ is equivalent to following system of two equations: $$\eqalign{ 2x+y&=\frac32\cr 2x^2+y^2&=1 }$$ Solving for $x$ and $y$, we get two solutions: $$ (x,y)=\left(\frac12-\frac{\epsilon}{2\sqrt6},\frac12+\frac{\epsilon}{\sqrt6}\right),\quad\epsilon\in\{-1,+1\}.$$ and in both cases the obtained formula is not exact for $X^3$ as one can check easily.$\square$

Felix Klein
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