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help me please with this question: definition of Chebyshev's polynomials it's given by $T_n(x)=\cos(n\arccos(x)) $

  1. Find by Gauss Quadrature method $\displaystyle\int_{-1}^1 \sqrt{1-x^2} \; dx$ with null error

  2. Estimate $\displaystyle\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}} \; dx$ accuracy of 5 decimal places

Thanks!

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    Ah: Chebyshev-Gauss quadrature. With $n=1$, the second case at http://en.wikipedia.org/wiki/Chebyshev%E2%80%93Gauss_quadrature gives the exact result for $\int_{-1}^1 \sqrt{1-x^2} f(x)\ dx$ where $f$ is a polynomial of degree $\le 1$.2012-04-02

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As Robert mentions, you do not want to use the Chebyshev polynomials of the first kind for your first integral; what you need are the Chebyshev polynomials of the second kind,

$U_n(x)=\frac{\sin((n+1)\arccos\,x)}{\sqrt{1-x^2}}$

which are orthogonal with respect to the inner product

$\langle f,g\rangle=\int_{-1}^1 \sqrt{1-u^2}f(u)g(u)\mathrm du$

Knowing that $\sin\,k\pi=0$ if $k$ is an integer ought to be a big hint on how to generate the nodes for Gauss-Chebyshev quadrature. From the theory, this quadrature rule is designed to give exact results for integrands of the form $\sqrt{1-x^2}p(x)$, where $p(x)$ is a polynomial, and since constant functions are effectively polynomials...


For the second one: just keep increasing the number of nodes until your error estimate (that I presume was mentioned in your textbook) gives something less than $10^{-5}$; one convenient thing about Gauss-Chebyshev is that the weights stay the same, and all one does is to change the nodes. (P.S. for what values of $u$ is $\cos\,u$ equal to zero?)

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    @Antonio: that, unfortunately, is a typo and not another permutation of old man Pafnuty's surname. I'll fi$x$, a$n$d tha$n$ks for checking!2012-04-10