Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rationnal functions

  1. Feb 26, 2006 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Consider [itex]p_n(z)[/itex] and [itex]q_d(z)[/itex] two polynomials over [itex]\mathbb{C}[/tex], which can be factorized like so:

    [tex]p_n(z) = a_n (z-z_1)^{n_1}...(z-z_{k})^{n_k}[/tex]
    [tex]q_d(z) = b_d(z-\zeta_1)^{d_1}...(z-\zeta_{m})^{d_m}[/tex]

    ([itex]\sum^k n_i =n \ \ \ \sum^m d_i =d[/itex])

    and the rationnal function [itex]R: \mathbb{C}\cup \{\infty\} \rightarrow \mathbb{C}\cup \{\infty\}[/itex] defined by

    [tex]R(z) = \frac{p_n(z)}{q_d(z)}[/tex] if [tex]z \neq \zeta_i, \infty[/tex]

    [tex] R(\zeta_i) = \infty[/tex]

    [tex]R(\infty) = \left\{ \begin{array}{rcl}
    \infty & \mbox{if}
    & n>d \\ \frac{a_n}{b_n} & \mbox{if} & n=d \\
    0 & \mbox{if} & n<d
    \end{array}\right[/tex]

    I fail to see why R(z) has exactly [itex]max\{n,d\}[/itex] roots and poles. It seems to me the number of roots is equal to k or k+1 in the case of n<d and the number of poles is m or m+1 in the case of n>d.
     
    Last edited: Feb 26, 2006
  2. jcsd
  3. Feb 26, 2006 #2

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    They are counting with multiplicity, e.g. z^2 has 2 zeros at z=0.
     
  4. Feb 26, 2006 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Don't forget that you are assuming that [itex]z_i \neq \zeta_j[/itex]. (But that has nothing to do with your confusion)
     
    Last edited: Feb 26, 2006
  5. Feb 26, 2006 #4

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I also investigated that possibility. But even so, counting with multiplicity, R has n or n+1 roots and d or d+1 poles.
     
  6. Feb 26, 2006 #5

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You have to count multiplicity at infinity too.
     
  7. Feb 26, 2006 #6

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What does that mean?
     
  8. Feb 26, 2006 #7

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your function may have a multiple root/pole at infinity, just like it may have a multiple root/pole at any other number. You have to count the multiplicity of the root/pole at infinity, just like you have to count the multiplicity of the roots/poles at all the other numbers.
     
  9. Feb 26, 2006 #8

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    But for the roots/poles in [itex]\mathbb{C}[/itex], I know what their order of multiplicity are by looking at the number [itex]n_i/d_i[/itex] respectively. How do I know what the multiplicity is at infinity?!
     
  10. Feb 26, 2006 #9

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't know how your book defines the multiplicity of a root/pole at infinity. What does its definition say?
     
  11. Feb 26, 2006 #10

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It is not defined. I am using the definition from a linear algebra book my Lay, which says that the order of multiplicity of an eigenvalue [itex]a[/itex] is the power of [itex](\lambda-a)[/itex] in the caracteristic polynomial.
     
  12. Feb 26, 2006 #11

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, as you could guess from the answer, a function that looks asymptotically like x^k has a pole of order k at infinity, and similarly for one that looks like x^-k.
     
  13. Feb 26, 2006 #12

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Is that a formal definition?
     
  14. Feb 26, 2006 #13

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't remember what the formal definition is. I just remember that that's what you want to get out of it.
     
  15. Feb 26, 2006 #14

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Thanks Hurky, but it seems unlike our teacher to just throw stuff at us that we can't prove for ourself very easily. I'll ask him for more details.
     
  16. Feb 27, 2006 #15

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    The order of the pole/zero at infinity of [tex]f(z)[/tex] is usually defined to be the order of the pole/zero of [tex]f\left(\frac{1}{z}\right)[/tex] at zero.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Rationnal functions
Loading...