Are Ho(x) and H1(x) orthogonal to H2(x) with respect to e^(-x^2)?

[nadeemo88]

In my third year math class we were asked a question to prove that Ho(X) and H1(x) are orthogonal to H2(x), with respect to the weight function e^(-x^2) over the interval negative to positive infinity

where Ho(x) = 1
H1(x) = 2x
H2(x) = (4x^2) - 2

i know that i have to multiply Ho(x) by H2(x) and divide by the weight function and integrate..but i get lost when it comes to integrating by parts with e^(x^2)...

 

[nadeemo88]

in this question Hn(x) is the herite polynomial...where n = 0, 1, 2 ,3 etc

 

[LCKurtz]

Since you have symmetry in x you can do this integral:

2\int_0^\infty (4x^2-2)e^{-x^2}\,dx

Try breaking it into two parts and on the first part use integration by parts with:

u = 2x\ dv = 2xe^{-x^2}dx

and if I'm not mistaken, nice things will happen.

 

[nadeemo88]

Since you have symmetry in x you can do this integral:

2\int_0^\infty (4x^2-2)e^{-x^2}\,dx

.

well that looks right..but it should be e^(x^2)...

 

[LCKurtz]

Why do you say that? The integral won't even converge with a positive exponential in there.

 

[nadeemo88]

you are supposed to divide by the weight function...which is e^(-x^2)

 

[LCKurtz]

No you aren't. And like I said, the integral wouldn't converge.