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

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    Orthogonal Polynomials
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Discussion Overview

The discussion revolves around the orthogonality of the Hermite polynomials Ho(x) and H1(x) with respect to H2(x) under the weight function e^(-x^2) over the interval from negative to positive infinity. Participants explore the integration techniques required to demonstrate this property.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states the need to prove that Ho(x) and H1(x) are orthogonal to H2(x) using the weight function e^(-x^2).
  • Another participant clarifies that Hn(x) refers to the Hermite polynomial for n = 0, 1, 2, 3, etc.
  • A suggestion is made to evaluate the integral 2∫₀^∞ (4x² - 2)e^(-x²) dx, proposing the use of integration by parts.
  • One participant expresses uncertainty about the integration process, particularly when integrating by parts with e^(x²).
  • Another participant questions the validity of using e^(x²) in the integral, arguing that it would not converge.
  • A participant insists that the integral must be divided by the weight function e^(-x²).
  • Contradictory claims arise regarding whether the integral should involve e^(-x²) or e^(x²), with concerns about convergence being raised.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correct formulation of the integral and whether it should involve e^(-x²) or e^(x²). The discussion remains unresolved with multiple competing views on the integration process and convergence.

Contextual Notes

There are unresolved mathematical steps related to the integration process, particularly concerning the use of integration by parts and the implications of the weight function on convergence.

nadeemo88
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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)...
 
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in this question Hn(x) is the herite polynomial...where n = 0, 1, 2 ,3 etc
 
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.
 
LCKurtz said:
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)...
 
Why do you say that? The integral won't even converge with a positive exponential in there.
 
you are supposed to divide by the weight function...which is e^(-x^2)
 
No you aren't. And like I said, the integral wouldn't converge.
 

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