Let $P_n$ be the orthogonal Legendre polynomial with a degree of $n$, meaning it satisfies the following recursive formula: $$(n+1)P_{n+1}(x)-(2n+1)xP_n(x)+nP_{n-1}(x)=0$$ where $P_0(x) = 1$ and $P_1(x) = x$. Let $w(x)=1$ be the weight function, and the interval $[-1,1]$. $I_n(f)=\sum_{k=1}^nw_k f(x_k)$ is the integration formula. For $n\geq1$ we define $$q_n(y) =\frac{P_{2n+1}(\sqrt{y})}{\sqrt{y}}.$$

Problem: prove that $q_n$ is a polynomial of degree $n$, determine the orthogonality relations, weight function (for the dot product $\int_{-1}^1w(x)f(x)g(x)dx$) and the interval where they form a set of orthogonal polynomials. Express the weights and nodes of the formula $I_n^q =\sum_{k=1}^n v_k f(y_k)$ using $w_k$ and $x_k$ from the integration formula $I_{2n+1}$.

My attempt:

I proved that $q_n$ is a polynomial of degree $n$ using induction and the fact that Legendre polynomial of degree $n$ is even/odd if $n$ is even/odd.

Next, I determined that they are orthogonal with the weight function being $x \mapsto \sqrt{x}$ and the interval $[0,1]$, by using the fact that Legendre polynomials are orthogonal and some substitutions.

Now, given that the nodes in the formula $I_n$ are roots of the n-th orthogonal polynomial we're using, when we look at the formula $I_{2n+1}$ we're looking for roots of $P_{2n+1}$. Seeing as it's a polynomial of degree $2n+1$ that is odd, it must have $n$ positive roots (let's denote them by $x_1,\ldots x_n$, $n$ negative roots and $0$. (there's also a theorem that states that the roots of these polynomials are all different)

Now, I think I should take $y_k = x_k^2$, $k = 1,\ldots,n$.

But I'm not sure how to express $v_k$ using $w_k$.

Any hints would be appreciated!


I don't see an elegant way to do this so we will apply brute force starting from this formula $$w_k=\frac{a_n}{a_{n-1}}\frac{\int_a^bw(x)p_{n-1 }(x)^2dx}{p_n^{\prime}(x_k)p_{n-1}(x_k)}$$ In our case we write it as $$v_k=\frac{b_m}{b_{m-1}}\frac{\int_0^1\sqrt y\,q_{m-1}(y)^2dy}{q_m^{\prime}(y_k)q_{m-1}(y_k)}$$ Now $b_m$ is the leading coefficient of $q_m(y)=\frac{P_{2m+1}(\sqrt y)}{\sqrt y}$ so $b_m=a_{2m+1}$, the leading coefficient of $P_{2m+1}(x)$. This is a good start! Now, $b_{m-1}=a_{2m-1}$: not quite what we want but hopefully it will work itself out. We have $$q_m^{\prime}(y_k)=\frac{P_{2m+1}^{\prime}(\sqrt{y_k})}{\sqrt{y_k}}\frac1{2\sqrt{y_k}}-\frac12\frac{P_{2m+1}(y_k)}{y_k^{3/2}}=\frac{P_{2m+1}^{\prime}(x_k)}{2x_k^2}$$ $$q_{m-1}(y_k)=\frac{P_{2m-1}(\sqrt{y_k})}{\sqrt{y_k}}=\frac{P_{2m-1}(x_k)}{x_k}$$ We can adjust this equation by taking $n=2m$ and $x=x_k$ the the three term recurrence relation to get $$(2m+1)P_{2m+1}(x_k)-(4m+1)x_kP_{2m}(x_k)+2mP_{2m-1}(x_k)=0$$ Solve for $P_{2m-1}(x_k)$ and apply to the previous equation to get $$q_{m-1}(y_k)=\frac{4m+1}{2m}P_{2m}(x_k)$$ Which is the form we want. Now, $$\begin{align}\int_0^1\sqrt y\,q_{m-1}(y)^2dy&=\int_0^1\sqrt y\,\frac{P_{2m-1}(\sqrt y)^2}{y}dy=\int_0^1x\frac{P_{2m-1}(x)^2}{x^2}\cdot2x\,dx\\ &=2\int_0^1P_{2m-1}(x)^2dx=2\int_0^{-1}P_{2m-1}(-u)^2(-du)\\ &=2\int_{-1}^0\left(-P_{2m-1}(u)\right)^2du=\int_{-1}^1P_{2m-1}(x)^2dx\end{align}$$ We go bck to the three term recurrence relation with $n=2m$, multiply by $P_{2m-1}(x)$ and integerate from $-1$ to $1$ with respect to $x$ to get $$(2m+1)\int_{-1}^1P_{2m+1}(x)P_{2m-1}(x)dx-(4m+1)\int_{-1}^1P_{2m}(x)\left[xP_{2m-1}(x)\right]dx+2m\int_{-1}^1P_{2m-1}(x)^2dx=0$$ The first integral above is $0$ by the orthogonality of the Legendre polynomials and $$xP_{2m-1}(x)=\frac{a_{2m-1}}{a_{2m}}P_{2m}(x)+R_{2m-1}(x)$$ By the division algorithm where $R_{2m-1}(x)$ is a polynomial in $x$ of degree at most $2m-1$. Thus by the orthogonality of the Legendre polynomials, $$\begin{align}\int_{-1}^1P_{2m}(x)\left[xP_{2m-1}(x)\right]dx&=\int_{-1}^1P_{2m}(x)\left[\frac{a_{2m-1}}{a_{2m}}P_{2m}(x)+R_{2m-1}(x)\right]dx\\ &=\frac{a_{2m-1}}{a_{2m}}\int_{-1}^1P_{2m}(x)^2dx\end{align}$$ So we have arrived at $$-(4m+1)\frac{a_{2m-1}}{a_{2m}}\int_{-1}^1P_{2m}(x)^2dx+2m\int_{-1}^1P_{2m-1}(x)^2dx=0$$ And we can use this result to achieve $$\int_0^1\sqrt y\,q_{m-1}(y)^2dy=\frac{(4m+1)a_{2m-1}}{2ma_{2m}}\int_{-1}^1P_{2m}(x)^2dx$$ So now we have $$\begin{align}v_k&=\frac{a_{2m+1}}{a_{2m-1}}\frac{\frac{(4m+1)a_{2m-1}}{2ma_{2m}}\int_{-1}^1P_{2m}(x)^2dx}{\frac{P_{2m+1}^{\prime}(x_k)}{2x_k^2}\frac{(4m+1)P_{2m}(x_k)}{2m}}\\ &=2x_k^2\frac{a_{2m+1}}{a_{2m}}\frac{\int_{-1}^1P_{2m+1}(x)^2dx}{P_{2m+1}^{\prime}(x_k)P_{2m}(x_k)}\\ &=2x_k^2w_k\end{align}$$ Mmm... now that we have an answer, it looks like an elegant proof should be possible, but I don't see it just now. It's getting late so I leave the reader with the above brute force approach.

EDIT: It's late but not that late. We start with the most horrible possible way of computing the weights $v_k$: $$I_j=\int_0^1\sqrt y\,y^jdy=\sum_{k=1}^mv_ky_k^j=\sum_{k=1}^mv_kx_k^{2j}$$ For $0\le j\le m-1$, where we have applied our Gauss-Jacobi quadrature formula, exact for polynomials of degree at most $2m-1$, to find $m$ linear equations for its $m$ weights. Since the matrix of coefficients is a Vandermonde matrix, the weights are uniquely determined by this system.

Now we transform variables first to get $$\begin{align}I_j&=\int_0^1\sqrt y\,y^jdy=\int_0^1x\cdot x^{2j}\cdot 2x\,dx=\int_{-1}^1x^{2j+2}dx\\ &=2\sum_{k=1}^mw_kx_k^{2j+2}=\sum_{k=1}^m(2x_k^2w_k)x_k^{2j}\end{align}$$ For $0\le j\le m-1$. We showed in our complicated proof how to fold the integral over the origin to change the interval to $[-1,1]$ and the Gauss-Legendre quadrature formula was again exact for polynomials of degree at most $4m+1$. We know already that the zeros of the Legendre polynomial of degree $2m+1$ are symmetrically disposed about the origin and that the weight corresponding to $-x_k$ is the same as that corresponding to $x_k$, due to the symmetry of the weight function and interval in Gauss-Legendre quadrature about the origin. Also the weight for the zero node doesn't enter in because it always gets multiplied by zero in every equation.

Thus we see that $v_k$ and $2x_k^2w_k$ satisfy the same nonsingular system of $m$ equations in $m$ unknowns, so they are the same.

  • $\begingroup$ That's pretty awesome. Thanks a lot! $\endgroup$ – Collapse Jun 10 '19 at 14:58

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