how do I calculate the first three Hermite Polynomials ?
thank you
ive got to the point where i am integrating
1 and 2x and the weighting function e^x^-2 (sori i dont know how to use LaTex) between the intervals infinity and minus infinity but i dont understand how the integral equals to zero if the intervals do not have a value. i have integrated the polynomials and i got -e^x^-2, hope you understand thanks
Or just note that the integrand $\displaystyle 2xe^{-x^2}$ is an odd function, and hence :
$\displaystyle \int_0^\infty 2xe^{-x^2} ~ dx=-\int_{-\infty}^0 2xe^{-x^2} ~ dx$
Since $\displaystyle \int_{-\infty}^{\infty}=\int_0^\infty+\int_{-\infty}^0$, the integral is 0.
[LEFT]The Actual Question
Using the definition of orthogonality, verify that the polynomials found in 1.(i.e. $\displaystyle 1,2x, 4x^2 -2$[FONT=CMR12])are mutually
orthogonal. You may use the following integrals without proof
$\displaystyle \int_{-\infty}^{\infty}e^{-x^2}~dx = \sqrt\pi $
i think my actual problem is that i missed the class on this topic and i am totally confused, the book i am using is showing me a different way to do it, if you could please explain in a very simple way, thanks