91Ó°ÊÓ

Let \(p_n(x)\) be the nth orthogonal polynomial. The highest power of x in \(p_n(x)\) is \(x^n\), and the lead coefficient is \(a_n\). So, we can write \(p_n(x)\) as: \[p_n(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\]

Calculate the square of the norm of the polynomial \(p_n(x)\) as follows: \[\|p_n\|^2 = \langle p_n, p_n \rangle\] Where \(\langle p_n, p_n \rangle\) is the inner product of \(p_n(x)\) with itself.

Calculate the inner product of \(x^n\) and \(p_n(x)\) as follows: \[\langle x^n, p_n \rangle\]

To show that the square of the norm is equal to the product of the lead coefficient and the inner product of \(x^n\) and \(p_n(x)\), we must prove: \[\|p_n\|^2 = a_n \langle x^n, p_n \rangle\] Since the sequence of orthogonal polynomials is orthogonal, their inner products are zero except when they are multiplied by themselves. Therefore, when calculating the inner product of \(x^n\) and \(p_n(x)\), only the term with the highest power of x in \(p_n(x)\) will contribute to the inner product. That is: \[\langle x^n, p_n \rangle = a_n \langle x^n, x^n \rangle\] Now, let's calculate the square of the norm of \(p_n(x)\): \[\|p_n\|^2 = \langle p_n, p_n \rangle = \langle a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0, a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \rangle\] Since the polynomials are orthogonal, all the terms in the above inner product will be zero, except for the term of the highest power of x: \[\|p_n\|^2 = a_n^2 \langle x^n, x^n \rangle\] Now, let's compare the expressions for the square of the norm and the inner product: \[\|p_n\|^2 = a_n^2 \langle x^n, x^n \rangle = a_n (a_n \langle x^n, x^n \rangle)\] Thus, we've proven that: \[\|p_n\|^2 = a_n \langle x^n, p_n \rangle\] The proof is complete.