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I'm doing a exam problem in the subject numerical methods with the book "Numerical Mathematics and Computing" 7th edition by Ward Cheney and David Kincaid

The problem is in the picture below enter image description here

the relevant part of the solution which I have question about is below here.

enter image description here

I do not understand why symmetric nodes around the midpoint of the integral implies that the weights have to be symmetric. Could you guys explain why symmetric nodes implies symmetric weights?

My second question is a bit more general about quadrature rule.

In the proposed solutions (which I have not included) they proceeds working out what $ \int_{-1}^{1}1dx=2=2(c_{0}f(-1)+c_{1}f(\frac{-2}{3})+\dots+c_{4}f(1))$ (here f=1) $ \int_{-1}^{1}xdx=0=2(c_{0}f(-1)+c_{1}f(\frac{-2}{3})+\dots+c_{4}f(1))$ (here f=x)

$\vdots$

$ \int_{-1}^{1}x^{4}dx=\frac{2}{5}=2(c_{0}f(-1)+c_{1}f(\frac{-2}{3})+\dots+c_{4}f(1))$

And you end up getting a $Ax=b$ problem with 5 unknowns.

What I'm wondering about is why it is suffices to integrate these 5 functions. I am aware of these polynomials are a linear combination of all polynomials with degree 4. And this is also how they justify it on page 227 in the book.

So why does integrating a linear combination of your n degree polynomial imply getting a polynomial that integrates exactly then P(x) is of degree n?

Postskjerm
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1 Answers1

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For the second part the values of the integrals you quote is not correct. For example, $\int_{-1}^1x^4=\frac 25,$ not $2$. The integrals of $x,x^3$ are zero. That will just change the weights that come out of your simultaneous equations. The reason solving the equations for $1,x,x^2,x^3,x^4$ is sufficient is that the polynomials up to degree four form a vector space. You can express the integral of any polynomial up to degree four as the polynomial of the integrals. As $1,x,x^2,x^3,x^4$ forms a basis for the vector space if the weights are exact for these they are exact for all polynomials of degree up to four. You could use any other set of five polynomials that form a basis and get the same weights. These are just easy to integrate and easy to see they are linearly independent.

The symmetric nodes implies symmetric weights seems "obvious" but I don't see an easy proof. You can derive it just by considering the $x$ and $x^3$ integrals in the second part. They both integrate to zero because the functions are odd.

Ross Millikan
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