When suitably normalized, the polynomial solutions to Equation (11.3.13) are called the Hermite polynomials, and are denoted by HN (x). (a) Use Equation (11.3.13) to show that HN (x) satisfies (ex2 H N ) + 2N ex2 HN = 0. (11.3.14) [Hint: Replace with N in Equation (11.3.13) and multiply the resulting equation by ex2 .] (b) Use Equation (11.3.14) to prove that the Hermite polynomials satisfy ex2 HN (x)HM (x)dx = 0, M = N. (11.3.15) [Hint: Follow the steps taken in proving orthogonality of the Legendre polynomials. You will need to recall that lim x ex2 p(x) = 0, for any polynomial p.] (c) Let p(x) be a polynomial of degree N. Then we can write p(x) = N k=1 ak Hk (x). (11.3.16) Given that ex2 H2 N (x)dx = 2N N! , use (11.3.15) to prove that the constants in (11.3.16) are given by a j = 1 2 j j! ex2 Hj(x)p(x)dx.

A l l e n | 1 Quanda Allen Michelle Stafford LUL-SHB18 September 19, 2016 Academic: Balance is important in academics because the course load in college varies from high school and in order to adapt to the changes students must alter how they approach their studies. Personally I work hard to organize my courses and the different deadlines. I also...