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The Derivative of Bessel Function of the Second Kind

  1. Jul 24, 2009 #1
    Hello,

    What is [tex]\frac{d}{dx}K_v\left(f(x)\right)=?[/tex]

    Thanks in advance
     
  2. jcsd
  3. Jul 24, 2009 #2

    Mute

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    Homework Helper

    For [itex]\nu[/itex] not necessarily an integer, [itex]C_{\nu}(y) = e^{\nu \pi i}K_{\nu}(y)[/itex] satisfies the identity

    [tex]2\frac{dC_\nu}{dy} = C_{\nu-1}(y) + C_{\nu + 1}(y)[/tex]

    Then let [itex]y = f(x)[/itex] and use the chain rule.

    Identities like this can usually be found on wikipedia, and for something as studied and used as frequently as the Bessel Functions, are generally correct.

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

    For a more 'official' reference, see something like http://www.math.sfu.ca/~cbm/aands/page_437.htm (scans of a reference book).
     
  4. Jul 24, 2009 #3
    I know that
    [tex]z\frac{d}{dz}K_v(z)+vK_v(z)=-zK_{v-1}(z)[/tex]
    but I was confused when we have more complicated arguments such as
    [tex]z=\sqrt{x^2+x}[/tex].
    But after your posting, I have now a simple method to move from simple to more complicated arguments. So, I can say the following:

    [tex]\frac{f(z)}{f'(z)}\frac{d}{dz}K_v(f(z))+vK_v(f(z))=-f(z)K_{v-1}(f(z))[/tex]

    Am I right?

    Best regards
     
  5. Jul 27, 2010 #4
    Do you get the answer?
     
  6. Jul 27, 2010 #5
    let [tex]y=f(x)[/tex] and then use the chain rule.
     
  7. Jul 27, 2010 #6
    i think this is right
     
  8. Jul 27, 2010 #7
    thanks a lot
     
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