Given the Legendre Polynomial $L_n(x)$ of degree $n$, and a quadracture rule for approximating $\int_{0}^{1} f(x) dx$ using $n$ points,

Prove that the absolute error for applying the quadrature rule to $f(x)=\sin x$ is not larger than $\frac{1}{(2n+1)!}$

I totally have no idea how to deal with absolute error of quadrature rule. I'm not allowed to use the error estimate term to prove this.

Please help, thanks.

  • $\begingroup$ What quadrature rule are you using? Without that, the problem is impossible. $\endgroup$ – user14717 May 2 '17 at 17:52
  • $\begingroup$ @user14717 imgur.com/a/TXUk4 heres the exact question $\endgroup$ – Little Rookie May 2 '17 at 18:07
  • $\begingroup$ Context is clearly missing, as the Legendre polynomial has nothing to do with the question as stated. $\endgroup$ – user14717 May 2 '17 at 18:20

We assumed the fact that the Gauss-Legendre rule having a degree of precision of $2n+1$ for $n+1$ points is known.

Applying Taylor expansion to $\sin{x}$, we obtain

$\int_{0}^1 \sin{x}\,dx = \int_{0}^1 x-\frac{x^3}{3!}+\dots+ (-1)^n \frac{x^{2n-1}}{(2n-1)!} + f^{(2n+1)}(\xi)\frac{x^{2n+1}}{(2n+1)!} \,dx, \text{ for some } \xi \in(0,1)$

Since the quadrature is exact for $x^k, k\leq2n+1$, and we know that $f^{(2n+1)}(\xi) \leq 1$, the error term is just $\int_{0}^1f^{(2n+1)}(\xi)\frac{x^{2n+1}}{(2n+1)!} \,dx =\frac{f^{(2n+1)}(\xi)}{(2n+2)!}\leq \frac{1}{(2n+2)!}<\frac{1}{(2n+1)!} $, as required.


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