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Bessel Function and cosine

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that
    [itex]\cos x[/itex]=[itex]J_{0}[/itex]+[itex]2[/itex][itex]\sum(-1)^{n}[/itex][itex]J_{2n}[/itex]
    where the summation range from n=1 to +inf

    2. Relevant equations
    Taylor series for cosine?
    series expression for bessel function?

    3. The attempt at a solution
    My approach is to start from R.H.S.
    I would like to express all bessel functions in the series form, then compare it to the taylor series of cosine.

    I notice that the summation can be written as
    -[itex]J_{2}[/itex]+[itex]J_{4}[/itex]-[itex]J_{6}[/itex]+[itex]J_{8}[/itex]+...
    Using the recurrence relation, we have - [itex]2J'_{3}[/itex]-[itex]2J_{7}[/itex]-[itex]2J_{11}[/itex]-...

    Therefore, R.H.S can be written as[itex]J_{0}+[/itex] [itex]4J'_{3}[/itex]-[itex]4J_{7}[/itex]-[itex]4J_{11}[/itex]-...

    But it seems it will be extremely difficult to deal with it. Since each term itself is a series. We are now summing up infinity many series.

    I wonder if we have a better way to finish this question
     
  2. jcsd
  3. Nov 9, 2012 #2
    How about just a few? Can we even get that far with it? I don't know. Say the first three terms:

    [tex]1-x^2/2+x^4/24[/tex]

    Ok, just that much. Can you start writing out say 3 or 4 or five of those series and see what the first term of them looks like and then notice a trend and then use induction to equate coefficients on each side of the expression and conclude the general expression for each constant? Each will be a sum of course so that we'd have:

    [tex]1=\sum_1^{\infty} a_n[/tex]

    [tex] -1/2=\sum_1^{\infty}b_n[/tex]

    [tex] 1/24=\sum_1^{\infty}c_n[/tex]

    and so forth.
     
  4. Nov 9, 2012 #3
    I manage to match the first two terms.
    But when doing the n=k step in MI, I have no idea now to express it.
    There are so many terms on the right contain x^(2k)
     
  5. Nov 9, 2012 #4
    After some works, I discovered the coefficient of x^2n on RHS is like this (shown in the picture below).

    But it seems it is impossible for them to be the same.
     

    Attached Files:

  6. Nov 9, 2012 #5
    Ok. But that's not easy to read. Also, I'm not sure that's the best approach ok but it's essential in order to become good at math: to try things and if they don't work, well, you try something else so I admire you're willingness to at least try it and even if someone comes with a better way, you're on your way to becommng better at math. :)
     
  7. Nov 9, 2012 #6
    I discovered a formula very similar to my question.
    I don't know if I can work on it to get the answer. I am still trying.
    Comments are very welcome!
    attachment.php?attachmentid=52815&d=1352510432.gif

    L.H.S. is [itex]e^{i x sinθ}[/itex]
     

    Attached Files:

  8. Nov 9, 2012 #7
    By putting θ=Pi/2, then done!
     
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