Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

High order derivitative of exponential function

  1. Mar 12, 2009 #1

    KFC

    User Avatar

    I am looking a formula to compute the derivitative of [tex]e^{a x^2}[/tex] with respect to x n times, where a is a constant.

    [tex]\frac{d^n}{dx^n}e^{a x^2}[/tex]

    I am going to find the result of above derivitative when x -> 0.
     
  2. jcsd
  3. Mar 13, 2009 #2
    well, take a couple of first derivatives, and see if you notice a pattern. Then if you do notice a pattern, try to prove it by induction....and after that take the limit as x-->0

    P.S. I would have tried to take some derivatives myself, but it is bed time now...lol...
     
  4. Mar 15, 2009 #3
    The answer to this question would most probably be (2xa)^n*(e)^a(x)^2
     
  5. Mar 15, 2009 #4

    Mute

    User Avatar
    Homework Helper

    Most certainly not! You're going to have to use the chain rule every time, which is going to result in some polynomail times [itex]e^{ax^2}[/itex]. In fact, the result is going to be related to the Hermite polynomials, defined by

    [tex]H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}[/tex].

    Getting the form desired in the OP amounts to setting [itex]x \rightarrow iax[/itex].

    http://en.wikipedia.org/wiki/Hermite_polynomials

    (Note I used the "physicist definition" of the Hermite polynomials defined on the wikipedia page).
     
    Last edited: Mar 15, 2009
  6. Mar 15, 2009 #5
    We define a new variable u = i sqrt(a) x:

    [tex]\frac{d^n}{dx^n}e^{a x^2} |_{x = 0} = (i \sqrt(a))^n \frac{d^n}{du^n}e^{-u^2}|_{u = 0} =
    (i \sqrt(a))^n (-1)^n e^{-u^2}H_n (u) |_{u = 0}
    =(-2 i \sqrt(a))^n \frac{\sqrt{\pi }}{\Gamma \left(\frac{1-n}{2}\right)} [/tex]

    where the last step has used the form of the Hermite numbers:

    http://mathworld.wolfram.com/HermiteNumber.html

    Note also that when n is odd the Gamma function blows up and so the value we seek is zero, when n is even the imaginary unit disappears and we obtain a real result.
     
    Last edited: Mar 15, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: High order derivitative of exponential function
Loading...