- #1

- 391

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter zetafunction
- Start date

- #1

- 391

- 0

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

- #2

mathman

Science Advisor

- 7,969

- 508

It looks like you didn't finish your sentence. What are you asking?

- #3

- 391

- 0

provided there are no singularities for real (x,y,z) on the function R(x,y,z)

- #4

mathman

Science Advisor

- 7,969

- 508

- #5

- 391

- 0

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

Science Advisor

- 7,969

- 508

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

- 391

- 0

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

Science Advisor

- 7,969

- 508

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/(x^{2}+a^{2}), the general problem can easily be resolved.

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/(x

Last edited:

- #9

mathman

Science Advisor

- 7,969

- 508

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.

Share: