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I am in a first semester numerical analysis course and we are going over numerical integration and more specifically quadrature forms. So far we have gone over standard quadrature as well as Gaussian quadrature. The problem is:

Show that the quadrature of the form

$\displaystyle \int_a^b f(x)dx \approx c_1f(x_1)+c_2f(x_2)+c_3f(x_3)+c_4f(x_4)$

cannot be exact for all polynomials of degree 4.

I seem to be having an issue with the fact that through Gaussian quadrature exact accuracy is achieved for polynomials of $2k-1$ where $k$ denotes the number of nodes. In this case we are using 4 nodes so we can accurately integrate a polynomial of degree 7. I don't know how to prove this because of this. As well I've tested it for standard polynomials when using Gaussian quadrature (i.e. $x^4$) and it held.

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so would we start by saying f(x)=x? This may be a dumb question but isn't that not a degree 4 polynomial? – user29921 Apr 25 '12 at 5:51
Would we have to show that the integral from a to b of 1 holds.. Then the integral from a to b of x holds.. down to x^4 ? – user29923 Apr 25 '12 at 6:04
Well I thought that as well but i don't know what I need to justify. since x_1 ... x_4 are arbitrary. – trmpt08 Apr 25 '12 at 6:52
If $a$ and $b$ are fixed then you are correct that Gaussian quadrature gives a formula of the stated form for polynomials up to degree 7. But allowing $a$ and $b$ to vary does not make much sense. – kiwi Jun 29 '12 at 23:20

Demand that: $$f(x_i)=a_i$$ There are infinitely many such polynomials, since there are infinite polynomials of degree $n$ that go through $n$ points. If the quadrature were exact, all of these polynomials would have the exact same integral, which is easily proved to be false.

For instance, if $a_i=0$, then any polynomial of the form $f(x)=\prod_{i=1}^{4}{A(x-x_i)}$ has the same "exact" quadrature rule.

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