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So, I need to find the second-degree polynomial that is the best approximation for $f(x) = cos(x)$ in $L^2_w[a, b]$, where $w(x) = e^{-x}$, $a=0$, $b=\infty$.

"Best approximation" for f is a function $\hat{\varphi} \in \Phi$ such that:

$||f - \hat{\varphi}|| \le ||f - \varphi||,\; \forall \varphi \in \Phi$

I have several methods available:

  • Lagrange interpolation
  • Hermite interpolation

Which would be the most appropriate?

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    @AlexBecker I updated the answer with a definition for what is meant b$y$ "best approximation".2012-05-24

2 Answers 2

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In your $L^2$ space the Laguerre polynomials form an orthonormal family, so if you use the polynomial $ P(x)=\sum_{i=0}^n a_i L_i(x), $ you will get the approximation error $ ||P(x)-\cos x||^2=\sum_{i=0}^n(a_i-b_i)^2+\sum_{i>n}b_i^2, $ (Possibly you need to add a constant to account for the squared norm of the component of cosine, if any, that is orthogonal to all the polynomials. If the Laguerre polynomials form a complete orthonormal family, then this extra term is not needed. Anyway, having that extra term will not affect the solution of this problem.) where $ b_k=\langle L_k(x)|\cos x\rangle=\int_0^{\infty}L_k(x)\cos x e^{-x}\,dx. $ I recommend that you calculate $b_0$, $b_1$ and $b_2$, and then try and figure out how you should select the numbers $a_i$ to minimize the error and meet your degree constraint.

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    You're right. There was a typo in my course. Laguerre polynomials are orthogonal for all \alpha > -1, not +1. Thanks.2012-05-25
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You can directly compute $\| f - g\|^2 = \int_0^\infty (\cos(x) - (a + b x + c x^2))^2 e^{-x}\ dx$, giving you a quadratic function of $a$, $b$, $c$, and then minimize that.

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    I did get the correct results, but it turns out that I was supposed to make use of the orthogonality of the Laguerre polynomials.2012-05-26