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Bessel function quotients

  1. Nov 22, 2011 #1
    Hey guys!

    I'm having to complete a piece of work for which I have to consider Bessel function quotients. By that I mean:

    Kn'(x)/Kn(x) and In'(x)/In(x)

    By Kn(x) I mean a modified Bessel function of the second kind of order n and by Kn'(x) I mean the derivative of Kn(x) with respect to the arguement x.

    Simurlaly, In(x) is a modified Bessel function of the first kind of order n and In'(x) is its derivative.

    Basically what I need to find is Kn'(x)/Kn(x) and In'(x)/In(x) for x tending to zero, this obviously gives rise to alot of "infinity/infinity" and "0/0" situations, so I need to perform some analysis on these.

    My supervisor has suggested using a squeezing technique, which I think would work but would require upper and lower bounds of the quotients.

    I know it's a bit of an involved question, but does anyone have any advice (I'm killing myself with this one!).

    Thanks guys!
     
  2. jcsd
  3. Nov 22, 2011 #2
    Case n>0 :
    I[n,x] = ((x/2)^n)(1+O(x))
    I'[n,x] = (n/2)((x/2)^(n-1))(1+O(x))
    I'/I = (n/(2x))(1+O(x))
    x -> 0 then I'/I tends to infinity with sign of x
    Case n=0 :
    I[0,x] = 1+(x/2)²+O(x^4)
    I'[0,x] = (x/2)+O(x^3)
    I'/I = (x/2)+O(x^3)
    x -> 0 then I'/I tends to 0

    Case n>0 :
    K[n,x) = (1/2)((n-1)!)((2/x)^n)(1+O(x))
    k'[n,x] = (1/2)((n-1)!)((2/x)^n)(-1/x)(1+O(x))
    K'/K = (-n/x)(1+O(x))
    x -> 0 tnen K'/K tends to infinity with sign of (-x)
    Case n=0 :
    K[0,x] = ln(2/x) -g +O(x²ln(x))
    g = Euler's constant
    K'[0,x] = -1/x +O(x ln(x))
    K'/K = (-1/x +O(x ln(x)))/(ln(x)+O(1))/
    K'/K = -1/(x ln(x)) +O(x ln(x))
    x -> 0 then K'/K tends to infinity with sign of (-x)

    In all the above, ln(x) means ln(abs(x))
     
  4. Nov 22, 2011 #3
    Hey thanks for that I really appreciate it! Could you just explain your initial form of K[x,n] and I[x,n] as its not something I'm familiar with. Thanks.
     
    Last edited: Nov 22, 2011
  5. Nov 22, 2011 #4
    What I wrote is the first terms of the series expansion of the Bessel functions.
    In order to solve a problem related to Bessel functions, of course, some properties of these functions are supposed to be known.
    If you don't known these formulas, you have to use another method related to what you are supposed to know about the basic properties of those functions. This wasn't stated in your first post.
     
  6. Nov 22, 2011 #5
    Sorry I'm an applied mathematician not a Bessel expert. My main sticking point is just what O(x) means, the only place I've seen this before is in discussing orders of magnitude.
     
  7. Nov 23, 2011 #6
    Hello!
    O(x) is the Landau's symbol. Yes, it is a matter of order of magnituide. Roughly, f(x)=1+O(x) means that f(x) tends to 1 and O(x) tends to 0 while x tends to 0.
    http://mathworld.wolfram.com/LandauSymbols.html
    http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html
    http://mathworld.wolfram.com/BesselFunctionoftheSecondKind.html
     
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