What is the significance of Bessel function quotients?

AI Thread Summary
The discussion focuses on the significance of Bessel function quotients, specifically Kn'(x)/Kn(x) and In'(x)/In(x), as x approaches zero. The analysis reveals that for n>0, both quotients tend to infinity, while for n=0, In'(x)/In(x) approaches zero. Participants discuss using a squeezing technique to establish upper and lower bounds for these quotients. Clarifications on the series expansions of Bessel functions and the meaning of the Landau symbol O(x) are also provided. Understanding these properties is essential for solving related mathematical problems effectively.
rj_brown
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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 argument 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 a lot 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!
 
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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))
 
JJacquelin said:
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))

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:
rj_brown said:
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.

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.
 
JJacquelin said:
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.

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.
 
rj_brown said:
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.

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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