Show that ∑ n = 0 ∞ [ ( − y ) n / n ! ] D n exp ⁡ ( − 2 π x 2 ) = exp ⁡ [ − 2 π ( x − y ) 2 ]

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Show that       ∑          n      =      0        ∞    [  (  −  y      )    n        /    n  !  ]      D    n    exp  ⁡  (  −  2  π      x    2    )  =  exp  ⁡  [  −  2  π  (  x  −  y      )    2    ]

recajossikpfmq · Accepted Answer

The differential operator against the gaussian can be expressed in terms of Hermite polynomials,            D        n        e          −      2      π              x        2              =                    d        n                    e                  −          2          π                      x            2                                      d              x        n              =  (  −  1      )    n    (  2  π      )    n        H    n    (  2  π  x  )      e          −      2      π              x        2            You can prove this by induction. The sum becomes,  S  =      ∑          n      =      0              ∞                  (      2      π      y              )        n                    H        n            (      2      π      x      )              e                  −          2          π                      x            2                                      n      !        =      e          −      2      π              x        2                  ∑          n      =      0              ∞                  (      2      π      y              )        n                    H        n            (      2      π      x      )              n      !      The exponential generating function can be expressed in terms of Hermite polynomials,      e          2      x      t      −              t        2              =      ∑          n      =      0              ∞                          H        n            (      x      )              t        n                    n      !      Setting  t  =  2  π  ythe desired result is obtained.

Show that <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> n =

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