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[complex functions] polynomial roots in a convex hull

  1. Mar 25, 2009 #1
    Hi everyone,

    I've got this problem to solve:

    My problem is that I don't fully understand the question.

    I have found such definition of convex hull:
    So I do have to prove, that all the roots of [tex]W'(z)[/tex] [let's denote them as [tex]z'_k[/tex]] must be able to be written in such form:
    [tex]z'_k = \sum_{k=1}^n \beta_k z_k[/tex], where [tex]\beta_k[/tex] are satysfying the conditionsof convex hull and [tex]z_k[/tex] are the roots of [tex]W(z)[/tex].
    Am i right?

    If so, I thought about this kind of sollution:
    we assume that what we have to prove is true, so we can write the roots of [tex]W'(z)[/tex] as:
    [tex]z'_j = \sum_{k=1}^n \beta_k^{(j)} z_k[/tex]
    Now we write down [tex]W'(z)[/tex] using viete's formulas:
    [tex]W'(z) = n a_n z^{n-1} - n a_n ( \sum_j \sum_k \beta_k^{(j)} z_k ) z^_{n-2} + n a_n \sum_{i<j} ( \sum_k \beta_k^{(i)} z_k ) ( \sum_k \beta_k^{(j)} z_k ) z^{n-3} - ... + n a_n \prod_j \sum_k \beta_k^{(j)} z_k[/tex] [dont know why the second part of the equation landed little higher that the first one, sorry for that]
    and compare it to the polynomial we get when multyplying [tex]W(z)[/tex] by [tex]\sum_k {1} / (z-z_k)[/tex]. What do you think of it? I've tried to do this, but the calculus grow pretty vast and I feel that there is a simplier method of proving this.

    I would appreciate if you could tell me wheter I understand the question right and if my idea of solving it looks fine.

    thanks for your time,
    rahl.
     
    Last edited: Mar 25, 2009
  2. jcsd
  3. Apr 19, 2009 #2
    I have found the solution. If no one deletes this thread, I will write how to solve it, maybe it will help someone some day.
     
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