A quadrature formula, proving weights are equal to 1?

Question

I am given a quadrature formula on the interval [-1,1] of the form:

$$\int_{-1}^{1} f(x) dx \approx w_{0}f(- \alpha) + w_{1}f(\alpha)$$

which uses the quadrature points $x_{0}= - \alpha$ and $x_{1}= \alpha$ where $0 <\alpha \leq 1$.

I would like to show that the weights, $$w_{0}=w_{1}=1$$ (which are independant of the value $\alpha$).

I understand that the formula is required to be exact whenever f is a polynomial of degree 1, but I am unsure how to begin with this problem.

I would also like to show that there is one particular value of α, such that the formula is exact also for all polynomials of degree 2. After this, I would then like to find this α, and show that, for this value, the formula is also exact for all polynomials of degree 3.

Answer 1

All polynomials of degree 1 can be represented as $f(x) = ax+b$. Since \begin{align*} \int_{-1}^{1} ax + b \, \mathrm{d}x = 2b \end{align*} then \begin{align*} w_{0}(-a\alpha + b) + w_{1}(a\alpha + b) = (w_{1} - w_{0})a\alpha + (w_{0}+w_{0})b = 2b \end{align*} From this we see that $w_{1} = w_{0} = 1$.

That should be enough to help you get the next part of your question. (Hint: $\alpha = 1/\sqrt{3}$.)

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Check with $\ds{\,\mrm{f}\pars{x} = x^{0},x^{1},x^{2}}$:

$$ \left.\begin{array}{rcl} \ds{2} & \ds{=} & \ds{w_{0} + w_{1}} \\[2mm] \ds{0} & \ds{=} & \ds{w_{0}\pars{-\alpha} + w_{1}\alpha} \\[2mm] \ds{2 \over 3} & \ds{=} & \ds{w_{0}\pars{-\alpha}^{2} + w_{1}\alpha^{2}} \end{array}\right\} \implies w_{0} = w_{1} = 1\,,\quad \alpha = {\root{3} \over 3} $$

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$$ \int_{-1}^{1}\mrm{f}\pars{x}\,\dd x \approx \mrm{f}\pars{-\,{\root{3} \over 3}} + \mrm{f}\pars{\root{3} \over 3} $$