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Problem while playing with Bessel functions

  1. Nov 30, 2015 #1
    1. The problem statement, all variables and given/known data
    I have run in to a number of problems while working through problems regarding Bessel and Modified Bessel Functions. At one point I run in to [itex]i^{m}e^{\frac{im\pi}{2}}[/itex] and it needs to equal [itex](-1)^m[/itex] but I'm not sure how it does. This came up while trying to solve an identity for a modified Bessel function in problem 14.5.2 in Arfken and Weber 'Mathematical Methods for Physicists'. Any help would be greatly appreciated, thank you.

    2. Relevant equations
    [itex]i^{m}e^{\frac{im\pi}{2}}[/itex]=[itex](-1)^m[/itex] but how??

    3. The attempt at a solution
    I solved for this by expanding the exponent in to cos and sin but I wind up with [itex]i^{m} [cos\frac{m\pi}{2}+(-1)^{m}\sin\frac{m\pi}{2}][/itex] and don't know how that can equal [itex](-1)^m[/itex]
     
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  3. Nov 30, 2015 #2

    Samy_A

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    There is an error in [itex]i^{m} [cos\frac{m\pi}{2}+(-1)^{m}\sin\frac{m\pi}{2}][/itex], that should be [itex]i^{m} [cos\frac{m\pi}{2}+i\sin\frac{m\pi}{2}][/itex]

    An easy way to solve your exercise it to consider four case separately: ##m=0, m=1,m=2, m=3##.
     
  4. Nov 30, 2015 #3
    You're right, I just typed it out incorrectly. Okay, I'll try and consider those cases; Thank you.
     
  5. Nov 30, 2015 #4

    SteamKing

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    If you express i in polar form, you have i = e(iπ/2). From there, you should be able to find im in polar form quite easily and then to find the product of im and e(imπ/2) in polar form.

    Soon after that, a vist from your Uncle Bob can be expected.
     
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