Skip to main content
MathOverflow

Return to Answer

replaced http://mathoverflow.net/ with https://mathoverflow.net/
Source Link
Community Bot
Community Bot
  • 1
  • 2
  • 3

Following on a little from Andy's commentAndy's comment, Hermite polynomials (multiplied by a Gaussian factor) give a basis of eigenvectors for the FT as an operator on $L^2({\mathbb R})$

Following on a little from Andy's comment, Hermite polynomials (multiplied by a Gaussian factor) give a basis of eigenvectors for the FT as an operator on $L^2({\mathbb R})$

Following on a little from Andy's comment, Hermite polynomials (multiplied by a Gaussian factor) give a basis of eigenvectors for the FT as an operator on $L^2({\mathbb R})$

Source Link
Yemon Choi
Yemon Choi
  • 25.8k
  • 9
  • 69
  • 156

Following on a little from Andy's comment, Hermite polynomials (multiplied by a Gaussian factor) give a basis of eigenvectors for the FT as an operator on $L^2({\mathbb R})$