1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook