Chapter 12: Problem 16
Prove the least squares approximation property of Legendre polynomials [see (9.5) and (9.6)] as follows. Let \(f(x)\) be the given function to be approximated. Let the functions \(p_{l}(x)\) be the normalized Legendre polynomials, that is, $$p_{l}(x)=\sqrt{\frac{2 l+1}{2}} P_{l}(x) \quad \text { so that } \quad \int_{-1}^{1}\left[p_{l}(x)\right]^{2} d x=1$$ Show that the Legendre series for \(f(x)\) as far as the \(p_{2}(x)\) term is $$f(x)=c_{0} p_{0}(x)+c_{1} p_{1}(x)+c_{2} p_{2}(x) \quad \text { with } \quad c_{l}=\int_{-1}^{1} f(x) p_{l}(x) d x$$ Write the quadratic polynomial satisfying the least squares condition as \(b_{0} p_{0}(x)+\) \(b_{1} p_{1}(x)+b_{2} p_{2}(x)\) (by Problem 5.14 any quadratic polynomial can be written in this form). The problem is to find \(b_{0}, b_{1}, b_{2}\) so that $$I=\int_{-1}^{1}\left[f(x)-\left(b_{0} p_{0}(x)+b_{1} p_{1}(x)+b_{2} p_{2}(x)\right)\right]^{2} d x$$ is a minimum. Square the bracket and write \(I\) as a sum of integrals of the individual terms. Show that some of the integrals are zero by orthogonality, some are 1 because the \(p_{t}\) 's are normalized, and others are equal to the coefficients \(c_{l}\). Add and subtract \(c_{0}^{2}+c_{1}^{2}+c_{2}^{2}\) and show that $$I=\int_{-1}^{1}\left[f^{2}(x)+\left(b_{0}-c_{0}\right)^{2}+\left(b_{1}-c_{1}\right)^{2}+\left(b_{2}-c_{2}\right)^{2}-c_{0}^{2}-c_{1}^{2}-c_{2}^{2}\right] d x$$ Now determine the values of the \(b\) 's to make \(I\) as small as possible. (Hint: The smallest value the square of a real number can have is zero.) Generalize the proof to polynomials of degree \(n\).
Short Answer
Step by step solution
Key Concepts
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