The Orthogonality of Hermite Polynomials

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In this video I prove that Hermite polynomials are orthogonal with respect to a weighted inner product.

For more videos in this series, visit:
  1. Physics and Math Lectures
  2. Hermite
  3. Hermite Polynomial
  4. Orthogonal
  5. Orthogonality
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Комментарии
Автор

When I figured this out before dinner yesterday, I thought I was wrong because what if m < m? This video pointed out to me that m and n can switched so I had figured the solution correctly all along.

MisterTutor
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I screamed at my computer at how easy that was. I tried to do it unsuccessfully on my own for a Thank you so much.

FunnCubes
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Great Video and thank you for the clear explanation!

sanderscheel
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At 5:29, why is it just exp(- x^2)? I know the result will still be 0, but not quite following that jump. Thanks

ventrue
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One question, why the evaluation of the first term is 0 because of the gaussian? We don't have only to take into account the infinity but the central part, where the density is concentred

MrMagraden
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you are assuming H_n is poly in the limit, but was that shown?

captainfartolini
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why can you rewrite the hermitian pol as a derivative
?

domenicagarzon