Asymptotic of rational function

  • #1
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given a rational function [tex] R(x,y,z) [/tex] that has no poles on the R^3 plane my question is if for big [tex] (x,y,z) \rightarrow \infty [/tex] it has the following asymptotic expansion

[tex] R(x,y,z) \sym \sum_{m,n,l= -\infty}^{N}a_{m,n,l}x^{m}y^{n}z^{l} [/tex]
 

Answers and Replies

  • #2
mathman
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It looks like you didn't finish your sentence. What are you asking?
 
  • #3
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i am asking if for any multi-vriable rational function the asymptotics can be obtained in the form of a multiple Laurent series as i have pointed

provided there are no singularities for real (x,y,z) on the function R(x,y,z)
 
  • #4
mathman
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Your statements are confusing - R^3 plane????? (R^3 is 3 dim. plane is 2 dim.). Poles are things defined in complex function theory. Your function seems to be a real(?) function of three real (?) variables.
 
  • #5
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[tex] R(x,y,z) [/tex] is a function of 3 variable x,y and z

the Rational function (quotient of polynomials) has NO poles on the R^3 only zeros (it may have poles inside the complex plane)

my question is can [tex] R(x,y,z) [/tex] be expanded into a convergent Laurent series ??
 
  • #6
mathman
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A rational function is a ratio of polynomials. The denominator has zeros which implies that you do need terms for the poles even though they may not be real. You then need inverse power terms, but I suspect they can be finite.

This whole question is a bit outside anything I have ever worked on. Is R a scalar function? What is the complex plane you keep referring to - you have 3 independent variables?
 
  • #7
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I mean a Laurent series on (x,y,z)

since Laurent series use Complex variables representation you would need to define a Laurent series on CxCXC

here [tex] w= R (x,y,z) [/tex] is an SCALAR function of three variables (x,y,z) for example [tex] \frac{x+y+xz}{x^{2}+y^{2}+1} [/tex]
 
  • #8
mathman
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Some thoughts to simplify the problem:

1) The numerator polynomial doesn't matter, so one can concentrate on the problem of 1 divided by a polynomial.
2) I am not familiar with the problem as such, but it looks to me if you can solve it for 1/(x2+a2), the general problem can easily be resolved.
 
Last edited:
  • #9
mathman
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Possible solution:
1) Set the numerator polynomial aside till the end and concentrate on the denominator polynomial, i.e. consider g(x,y,z)=1/P(x,y,z) where P is a polynomial.
2) Let X=1/x and Y=1/y and Z=1/z, so g(x,y,z)=h(X,Y,Z).
3) Get the 3 variable power series for h as sums of powers of X, Y, Z.
4) Replace X, Y, Z by 1/x, 1/y, 1/z to give you a series of inverse powers.
5) Multiply through by the original numerator to get the expression you want.

The major question regarding this procedure is whether or not step 3 can always be carried out.
 

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