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Summation of the polynomial and division

  1. May 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Let [tex]p(z) = \sum_{j=0}^{n} a_{n-j}z^j[/tex] be a polynomial of at least degree 1 thus
    [tex]n \geq 1[/tex].
    Show that if [tex]z\neq 0[/tex] then 1/z is a root of the polynomial p.

    2. Relevant equations

    Fundamental theorem of Algebra

    3. The attempt at a solution

    If a expand the polynomial above then

    [tex]p(z) = \sum_{j=0}^{n} a_{n-j}z^j = a_n + a_{n-1}z + a_{n-2}z^2 + \cdots + a_0 z^n[/tex]

    to test if 1/z is a root of p then this must be equal too

    [tex]p(1/z) = \sum_{j=0}^{n} a_{n-j}z^j = a_n + a_{n-1}(1/z) + a_{n-2}(1/z)^2 + \cdots + a_0 (1/z)^n = 0[/tex]

    Then as I see it its my task to show that the new sum

    [tex]\sum_{n=0}^{\infty} a_n (\frac{1}{z})^n[/tex] converges to zero. Then by the ratio-test

    [tex]\frac{a_{n+1}(\frac{1}{z})^{n+1}}{a_n (\frac{1}{z})^{n}} \rightarrow 0[/tex] as [tex]z \rightarrow \infty[/tex] therefore 1/z is a root of p.

    Am I on the right track here?
     
    Last edited: May 9, 2010
  2. jcsd
  3. May 9, 2010 #2
    I think your problem should be formulated as follows. If

    [tex]q(z) = \sum_{j=0}^{n}a_{j}z^{j}[/tex]

    and we define p(z) in the way you have written down, then if q(z)=0 for z not equal to zero, then we have that p(1/z) = 0.
     
  4. May 9, 2010 #3
    Hi Count Iblis and thank you for your answer,

    So what you are saying (please correct if I missunderstand) that if I show that if q(z) doesn't converge to zero then p(1/z) = 0 is a root of p?

    thats easy cause if n is of degree at least 1 then then in my understanding
    [tex]\lim_{n \rightarrow 1} q(z) \rightarrow 1[/tex]
     
    Last edited: May 9, 2010
  5. May 9, 2010 #4
    q(z) is assumed to be zero for some z. And then you can see that
    p(1/z) must be zero as well. This, of course, for z itself not equal to zero, otherwise 1/z would not exist.

    I don't see how convergence plays any role here, as we're dealing with polynomials....
     
  6. May 9, 2010 #5
    I see that now :) But is this really that easily? No need to show anything to show that p(1/z) is a root of p?
     
  7. May 9, 2010 #6
    You do need to show that p(1/z) is zero. You can take the expression for p(1/z) you wrote down:

    a_n + a_{n-1} 1/z + ...+a_0 1/z^n

    If you multiply this with z^n, you get...?
     
  8. May 9, 2010 #7
    I get 1, but if I can't use convergense how else am I suppose to show that its zero?
     
  9. May 9, 2010 #8
    What I mean is:

    (a_n + a_{n-1} 1/z + ...+a_0 1/z^n) z^n =

    a_n z^n+ a_{n-1} z^(n-1) + ...+a_0
     
  10. May 9, 2010 #9
    I am such a dumb person Count...

    If you multiply the above by z^n then you get the original p(z)........

    where q(z) = z^n for a set not equal to zero then p(1/z) = 0 and thus 1/z is a root of p :) Right ?
     
    Last edited: May 9, 2010
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