[Mathematica] How to calculate residues if poles not simple?

[earth2]

Hey guys,

i have the following situation:

I have a function which looks like

$$\frac{(a+bx)^3}{(x-y)^6(x-z)^6}$$

As one can easily see this function has poles at y and z of order 6. Now, I know how to calculate the residue of this function for instance at y, but how do I implement this into Mathematica? If i use "Residue" on this fella, Mathematica gives me zero, but I've done the calculation by hand and have shown that the result is not zero. So, my question boils down to:

How can i use mathematica to (easily) evaluate residues of poles of order n.

Thanks!
earth2

Edit: Sorry, forget to write: x, y, z are complex variables.

 

[Bill Simpson]

Unless I misunderstand,
Residue[(a + b x)^3/((x - y)^6(x - z)^6), {x, y}]
doesn't seem to give a zero residue.

http://www.wolframalpha.com/input/?i=Residue[(a+++b+x)^3/((x+-+y)^6(x+-+z)^6),+{x,+y}]
agrees

Perhaps you can show a fresh calculation of the expression you used that gave zero and that will help track down the discrepancy

 

[earth2]

Thanks for the reply!
Ok, if I enter a function (a similar one, to be more specific but the core question remains) in its original form
$$\frac{(Q*X+q*x)}{((x-B/A)^2-(B^2-AC)/A^2+i\frac{\epsilon}{A})^3}$$
Mathematica gives me as the residue zero. But I know (from calculating it myself and a paper) that it is not. Is this form of the expression maybe too complicated?

x is the variable in question. Q,X,A,B are just numbers and epsilon tends to zero.
cheers!

 

[Bill Simpson]

Please show the exact Residue expression you typed into Mathematica and make certain you include any assignments you made to variables that are used in that expression.

"I typed something and the answer is wrong" just doesn't provide enough information

 

[earth2]

Hi, thanks for the answer :)

The poles of the denominator are located at

$$x^\pm= \frac{B \pm \sqrt{d} \mp i \epsilon }{A}$$

and i want to evaluate the residues at $$x^-$$.
So i type into Mathematica

Residue[$$<br /> \frac{(Q*X+q*x)}{((x-B/A)^2-(B^2-AC)/A^2+i\frac{\epsilon}{A})^3}<br /> <br /> ,\{x,\frac{B-\sqrt{d}+ i \epsilon}{A}\}]$$.
with d=B^2-AC.
So, where do I go wrong? :)
Thanks for the answer!

Cheers,
earth2

 

[Bill Simpson]

I apologize if I'm making some silly mistake here.

If I try

Residue[(Q*X+q*x)/((x-B/A)^2-(B^2-A*C)/A^2+i*e/A)^3, {x, (B-Sqrt[B^2-A*C]+i*e)/A}]

I get zero, which I presume is what you are doing and getting.

If I look for zeros of your denominator

Solve[((x - B/A)^2 - (B^2 - A*C)/A^2 + i*e/A)^3 == 0, x]

seems to tell me that one pole will be at

(B-Sqrt[B^2-A*C-A*e*i])/A rather than (B-sqrt[B^2-A*C]+i*e)/A

Maybe you meant the i*e to be under the radical and this is just a failure of my trying to understand the typesetting.

If I try

Residue[(Q*X+q*x)/((x-B/A)^2-(B^2-A*C)/A^2+i*e/A)^3, {x, (B-Sqrt[B^2-A*C-A*e*i])/A}]

I get

(-3*A^4*(B*q+A*Q*X))/(16*(B^2-A*C-A*e*i)^(5/2))

Is that the correct residue or are there still errors?
Thanks

 

[earth2]

Hi again!

Hm, you are right.
What I forgot to write, is that I (or precisely the guys in the paper) am/are doing a Taylor expansion of the solution the zeros of the denominator in epsilon and evaluate the residues at the Taylor expanded expression. In that sense "my" epsilon (after Taylor expansion) is not the original one anymore. And that's where I go wrong :) Thanks for the illumination!

Btw, the paper is http://prd.aps.org/abstract/PRD/v29/i8/p1699_1 and the calculation I am talking about (although yet again with another but similar expression) is in appendix A.

Thanks!