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

Rational functions

  1. Aug 10, 2007 #1
    In this article, how do I rewrite (2) to get the third equation R(z)=... ?

    thank you
     
  2. jcsd
  3. Aug 10, 2007 #2
    remember the rational function will produce a quotient, which when multiplied by the divisor will yeild the original R(z). So the second expression is in the form R(z) = quotient x divisor.

    More specifically (z-a_j)^uj X S_j(z) .. Where as they said S_j(z) is the rational fuction ( which was the quotient ). It is not that much about "deriving" the third form but more showing that the complex function R(z) is a product of the number of roots ( z-a_j) and the quotient S_j(z).

    My 2 cents - correct if neccesary.
     
    Last edited: Aug 10, 2007
  4. Aug 11, 2007 #3
    thanks for answering!

    But what do you mean by qoutient produced by a rational function? The rational is a quotient of two polynomials, so what quotient is it producing?
     
  5. Aug 11, 2007 #4

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Take
    [tex]
    S_j(z) = \frac{a_0(z-\alpha_1)^{\mu_1}(z-\alpha_2)^{\mu_2}\cdots(z-\alpha_{j - 1})^{\mu_{j - 1}}(z-\alpha_{j + 1})^{\mu_{j + 1}}\cdots(z-\alpha_r)^{\mu_r}} {b_0(z-\beta_1)^{\nu_1}(z-\beta_2)^{\nu_2}\ldots(z-\beta_s)^{\nu_s}},
    [/tex]
    :smile:
     
  6. Aug 11, 2007 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The equation labled (3) is not derived directly from equation (2). What they have done is write the product of all terms in the numerator of (2) as P(z) and the product of all terms in the denominator as Q(z):
    R(z)= P(z)/Q(z).

    Then they look at R(z)- c= P(z)/Q(z)- c. Getting the common denominator (Q(z)) you have P(z)/Q(z)- cQ(z)/Q(z)= (P(z)- cQ(z))/Q(z)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?