Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to prove bessel function J1/2(x) = sqrt(2/πx)sinx;

  1. May 1, 2009 #1
    I really have no idea.

    I started with the frobenius method. Until the recurrence formula.

    I got that already. But I just don't know where to plug in the 1/2 into the equation. Can anyone help? I just need to know where to put in the 1/2?

    Or can i use the normal bessel function which in general.

    [tex]
    J_m(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{m+2n}n!(n+m)!}x^{m+2n} = x^m \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{m+2n}n!(n+m)!}(x^2)^n
    [/tex]

    to prove that function?

    need advice thanks..
     
  2. jcsd
  3. May 1, 2009 #2
    using series from the general case actually works.

    to plug the 1/2 into this equation, you will need to study a special case of the gamma function for the factorial. Once you do this, the whole 2^(m+2n)∙n!∙(n+m)! should simplify nicely for m=1/2.

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

    to prove that J1/2(x) = sqrt(2/πx)sinx, you would better factorize x^(m-1) out of the sum and remain with x^(2n+1), this should simplify more easily.
     
  4. May 1, 2009 #3
    the whole thing will become this . but how does it simplify to sinx?


    [tex]

    J_m(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{1/2+2n}n!(n+1/2)!}x^{1/2+2n} = x^1/2 \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{1/2+2n}n!(n+1/2)!}(x^2)^n

    [/tex]

    i mean how to simplyfy the (n+1/2)!. thanks
     
  5. May 2, 2009 #4
    just use (n+1/2)! = Γ(n+1/2 + 1 )
     
  6. May 2, 2009 #5
    with that Γ(n+1/2 + 1 )

    am I able to simplify it to sinx ?
     
  7. May 3, 2009 #6
    thanks for the advice..

    i have got the solutions.. thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to prove bessel function J1/2(x) = sqrt(2/πx)sinx;
  1. Integrate Sqrt[x^2-a]? (Replies: 3)

  2. Integrate Sqrt[x^2-a] (Replies: 4)

Loading...