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I need help on solving ∫e^(z^2)* fn(z) dz, where fn(z) = . . .

  1. Jul 26, 2007 #1
    I need to work on solving ∫e^(z^2)* fn(z) dz.

    where fn(z) = d^n/dz^n * e^(-z^2).

    Thanks alot!
     
  2. jcsd
  3. Jul 26, 2007 #2
    Try working out (inductively) the form of fn(z).
     
  4. Jul 26, 2007 #3
    you mean to find the derivative for the
    fn(z) = d^n/dz^n * e^(-z^2).

    and then substitue it?
     
  5. Jul 26, 2007 #4
    Indeed, yes. Try f1, then f2 and so on to see what's going on.
     
  6. Jul 26, 2007 #5
    the derivative for e^-(Z^2) is
    -2 Z e^-(z^2)

    is not that true?
     
  7. Jul 26, 2007 #6
    Yes it is. Do you notice anything special about the integral?
     
  8. Jul 27, 2007 #7
    Now if I will substitute the derivative of fn(z)
    Which is:
    -2 z * e^(-z^2)
    So, the integral equation will be like this:
    -2 z * e^(z^2) * e^(-z^2) dz
    here I get confused and I couldn’t find the integral?
     
  9. Jul 27, 2007 #8
    What is [itex]e^{a}\cdot e^{b}[/itex] equal to?
     
  10. Jul 27, 2007 #9
    you mean it will be somthing like this
    e^[(z^2)*(-z^2)]

    or it maybe e^(a+b)
     
    Last edited: Jul 27, 2007
  11. Jul 27, 2007 #10
    There's no guessing involved there! It's [itex]e^a*e^b=e^{a+b}[/itex].
     
  12. Jul 27, 2007 #11
    So, the integral it will be like this:
    -2 Z dz

    is not that right?
     
  13. Jul 27, 2007 #12
    That's correct.
     
  14. Jul 27, 2007 #13
    Thanks alot!!
     
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