What is the significance of Bessel function quotients?

In summary, the conversation discusses Bessel function quotients and finding their values for x tending to zero. The speaker's supervisor suggested using a squeezing technique, but upper and lower bounds are needed. The conversation also explains the series expansion of Bessel functions and clarifies the use of the Landau's symbol.
  • #1
rj_brown
4
0
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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  • #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))
 
  • #3
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:
  • #4
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.
 
  • #5
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.
 
  • #6
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
 

FAQ: What is the significance of Bessel function quotients?

1. What are Bessel function quotients?

Bessel function quotients are mathematical functions that are used to solve various problems in physics and engineering. They are named after the mathematician Friedrich Bessel who first studied them.

2. How are Bessel function quotients calculated?

Bessel function quotients are calculated using a series of mathematical equations that involve the Bessel functions and other related functions. These equations can be solved using numerical methods or specialized software.

3. What are the applications of Bessel function quotients?

Bessel function quotients have many applications in the fields of physics and engineering. They are commonly used to solve problems involving wave propagation, heat transfer, and vibration analysis.

4. Are Bessel function quotients only used in theoretical calculations?

No, Bessel function quotients are also used in practical applications. They are often used to design and optimize various engineering systems, such as antennas, filters, and signal processing algorithms.

5. Are Bessel function quotients difficult to understand?

Bessel function quotients can be complex and challenging to understand, especially for those without a strong background in mathematics. However, with proper training and practice, they can be understood and utilized effectively in various applications.

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