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Bessel functions of the first kind

  1. Jul 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Can anyone tell me if:
    [tex]
    \frac{d}{dx}J_k(ax)=aJ'_k(x)
    [/tex]
    where a is a real positive constant and
    [tex]
    J_k(x)
    [/tex]
    is the Bessel function of the first kind.
    Regards
    John


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 5, 2010 #2

    Char. Limit

    User Avatar
    Gold Member

    I believe that the answer to that is no. I believe that...

    [tex]\frac{d}{dx}J_k(ax) = a J_k^{'}(ax)[/tex]
     
  4. Jul 5, 2010 #3
    But doesn't
    [tex]
    J_k(ax)=aJ_k(x)?
    [/tex]
     
  5. Jul 5, 2010 #4

    Char. Limit

    User Avatar
    Gold Member

    Sorry, but that is not the case.
     
  6. Jul 5, 2010 #5
    Then that is my error.
    To find the derivative of
    [tex]
    J_k(ax)
    [/tex]
    presumably one has to differentiate the series expansion?
    John
     
  7. Jul 5, 2010 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can express the derivative of a bessel function in terms of other bessel functions. See the "Selected identities" list at the end of http://en.wikipedia.org/wiki/Bessel_function
     
  8. Jul 5, 2010 #7
    Hi Dick
    The question I am asking derives from proving the integral property of Bessel functions of the first kind. This amounts to showing the orthogonal properties of Bessel functions.
    Part way through the proof I need to show that:
    [tex]
    u'_1(x)=r_ix^(\frac{1}{2})J'_k(r_ix)+\frac{1}{2}x^{\frac{-1}{2}}J_(r_ix)
    [/tex]
    where
    [tex]
    r_i
    [/tex]
    is a distinct positive zero of
    [tex]
    J_k(x)
    [/tex]
     
  9. Jul 5, 2010 #8
    Hi Dick
    I need to show that if:
    [tex]
    u(x)=x^{\frac{1}{2}}J_k(r_ix)
    [/tex]
    Then:
    [tex]
    u'=r_ix^{\frac{1}{2}}J'_k(r_ix)+\frac{1}{2}x^{\frac{-1}{2}}J_k(r_ix)
    [/tex]
    This, of course, uses product rule but the derivative of:
    [tex]
    J_k(r_ix)
    [/tex]
    I am unclear about?
    Incidentally
    [tex]
    r_i
    [/tex]
    is the ith distinct positive zero of
    [tex]
    J_k(x)
    [/tex]
    This is part of proving the orthogonality of Bessel functions
    [tex]
    J_k(r_ix)
    [/tex]
    with respect to the weight function x.
    John
     
  10. Jul 5, 2010 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's just the chain rule. If h(x)=f(a*x) then h'(x)=f'(a*x)*a. It's a special case of the general chain rule, if h(x)=f(g(x)) then h'(x)=f'(g(x))*g'(x). It's just like saying the derivative of sin(2*x) is cos(2*x)*2.
     
  11. Jul 5, 2010 #10
    Thanks Dick
    John
     
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