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On nonnegative-order first-kind Bessel functions with large argument

  1. Jul 10, 2013 #1
    Hello.

    I'm not terribly proficient with Bessel functions, but I know that those of the first kind are given by
    [tex]
    \begin{eqnarray}
    J_n(x) & = & \left(\frac{x}{2}\right)^n\,\sum_{\ell=0}^\infty\frac{(-1)^\ell}{\ell!\,\Gamma(n+\ell+1)}\,\left(\frac{x}{2}\right)^{2\ell},
    \end{eqnarray}
    [/tex]where [itex]\Gamma[/itex] is the gamma function.

    I found in a book that for large [itex]x[/itex] the following is true:
    [tex]
    \begin{eqnarray}
    J_n(x) & = & \sqrt{\frac{2}{\pi x}}\,\cos\left(x-\frac{n\pi}{2}-\frac{\pi}{4}\right)\,\left[1+\mathcal{O}(x^{-1})\right]\\
    & \approx & \sqrt{\frac{2}{\pi x}}\,\cos\left[x-(2n+1)\,\frac{\pi}{4}\right].
    \end{eqnarray}
    [/tex]The second line of that last equation is obvious (they just removed every term except the first one from the first line because [itex]1/x^\ell[/itex] is negligible for large [itex]x[/itex] and [itex]\ell\ge1[/itex]), but where does the cosine come from? How would I begin to prove this?

    I'm working with [itex]n\in\mathbb{N}\cup\{0\}[/itex] only, so perhaps this expression is only true for Bessel functions of natural (or zero) order; I wouldn't know. All I know is that for such values of [itex]n[/itex] the gamma function becomes [itex]\Gamma(n+\ell+1)=(n+\ell)![/itex].

    Thanks in advance for any help!
     
  2. jcsd
  3. Jul 10, 2013 #2
    [tex]\cos(x) = \sum^{\infty}_{k=0} (-1)^{k} \frac{x^{2k}}{(2k)!}[/tex]

    Try using that.
     
  4. Jul 12, 2013 #3
    Thanks, Millennial. I will.
     
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