How to solve this system of four non-linear equations: $a+b=1, a x+b y=\frac{1}{2}, a x^2+b y^2=\frac{1}{3}, a x^3+b y^3=\frac{1}{4}$

Question

I've tried a lot of substitutions and combinations of equations to manually solve this system for $a$, $b$, $x$, $y$, however nothing seems to bring me closer to a solution.

\begin{align*} a + b &= 1 \tag{1}\\ a x + b y &= \frac{1}{2} \tag{2}\\ a x^2 + b y^2 &= \frac{1}{3} \tag{3}\\ a x^3 + b y^3 &= \frac{1}{4} \tag{4}\\ \end{align*}

In case it matters, I'm trying to calculate the points and weights of a Gaussian quadrature with $n=2$. I'm aware the manual solving would become easier for integrating in $(-1, 1)$ instead of $(0, 1)$, however if possible I'd like to learn how to solve the exact system above. Expected solution is $a=\frac{1}{2}$, $b=\frac{1}{2}$, $x=\frac{1}{2} - \frac{1}{2\sqrt{3}}$, $y=\frac{1}{2} + \frac{1}{2\sqrt{3}}$. WolframAlpha link is here.

Answer 1

\begin{align*} (1)x-(2)\Rightarrow b(x-y)&=x-\frac{1}{2}\tag{5}\\ (2)x-(3)\Rightarrow by(x-y)&=\frac{x}{2}-\frac{1}{3}\tag{6}\\ (3)x-(4)\Rightarrow by^2(x-y)&=\frac{x}{3}-\frac{1}{4}\tag{7}\\ (5)\rightarrow(6)\Rightarrow y(x-\frac{1}{2})&=\frac{x}{2}-\frac{1}{3}\tag{8}\\ (6)\rightarrow(7)\Rightarrow y(\frac{x}{2}-\frac{1}{3})&=\frac{x}{3}-\frac{1}{4}\tag{9}\\ \frac{(8)}{(9)}\Rightarrow \frac{x-\frac{1}{2}}{\frac{x}{2}-\frac{1}{3}}&=\frac{\frac{x}{2}-\frac{1}{3}}{\frac{x}{3}-\frac{1}{4}}\\ 6x^2-6x+1&=0 \end{align*}

Answer 2

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\begin{align}a + b &= 1 \tag{1}\label{1},\\ a x + b y &= \frac{1}{2} \tag{2}\label{2},\\ a x^2 + b y^2 &= \frac{1}{3} \tag{3}\label{3},\\ a x^3 + b y^3 &=\frac{1}{4} \tag{4}\label{4}.\end{align}

The system of \eqref{1}-\eqref{4} is perfectly suitable for Prony's method.

\begin{align} a\, \e^{0\cdot\lambda_x} + b\,\e^{0\cdot\lambda_x} &= 1 \tag{5}\label{5} ,\\ a\, \e^{1\cdot\lambda_x} + b\, \e^{1\cdot\lambda_y} &= \frac{1}{2} \tag{6}\label{6} ,\\ a\, \e^{2\cdot\lambda_x} + b\, \e^{2\cdot\lambda_y} &= \frac{1}{3} \tag{7}\label{7} ,\\ a\, \e^{3\cdot\lambda_x} + b\, \e^{3\cdot\lambda_y} &= \frac{1}{4} \tag{8}\label{8} , \end{align}

where $\lambda_x=\ln x$, $\lambda_y=\ln y$.

The pair $x,y$ is found as a solution to quadratic equation

\begin{align} z^2-a_0\,z-a_1&=0 \tag{9}\label{9} , \end{align}

where

\begin{align} \begin{bmatrix} a_0 \\ a_1 \end{bmatrix} &= \begin{bmatrix} 1/2 & 1 \\ 1/3 & 1/2 \end{bmatrix}^{-1} \begin{bmatrix} 1/3 \\ 1/4 \end{bmatrix} = \begin{bmatrix} -6 & 12 \\ 4 & -6 \end{bmatrix} \begin{bmatrix} 1/3 \\ 1/4 \end{bmatrix} = \begin{bmatrix} 1 \\ -1/6 \end{bmatrix} \tag{10}\label{10} , \end{align}

so \eqref{9} becomes \begin{align} z^2-z+\frac16&=0 \tag{11}\label{11} ,\\ x,y&=\frac12 \pm \frac{\sqrt3}6 \tag{12}\label{12} . \end{align}

And from the $2\times2$ system \eqref{1}, \eqref{2},

\begin{align} a &= \tfrac12\cdot\frac{1-2y}{x-y} =\frac12 \tag{13}\label{13} ,\\ b &= \tfrac12\cdot\frac{2x-1}{x-y} =\frac12 \tag{14}\label{14} . \end{align}

$\endgroup$