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Connection between roots of polynomials of degree n

  1. May 26, 2012 #1
    1. The problem statement, all variables and given/known data
    The two polynomial eqns have the same coefficients, if switched order:
    a_0 x_n+ a_1 x_n-1 + a_2 x_n-2 + … + a_n-2 x_2 + a_n-1 x + a_n = 0 …….(1)
    a_n x_n+ a_n-1 x_n-1 + a_n-2 x_n-2 + … + a_2 x_2 + a_1 x + a_0 = 0 …….(2)
    what is the connection between the roots of the eqns?

    2. The attempt at a solution

    i don't know how to do this at all!
    i thought maybe divide through by the first terms coeff... but that doesn't seem to help. maybe there is some kowledge i'm missing. something about how to work out the roots?
    please help!
     
  2. jcsd
  3. May 26, 2012 #2
    what if [itex]n=1[/itex], what's the relation between the roots?

    Then try [itex]n=2[/itex].
     
  4. May 26, 2012 #3

    Dick

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    Don't divide by the coefficient. Divide by x^n.
     
  5. May 27, 2012 #4
    algebrat:
    at n=1 it is just the reciprocal
    at n =2 it's: x_1 = (a_1 +- sqrt(a_1^2 -4a_2a_0))/2a_0
    and x_2 = (a_1 +- sqrt(a_1^2 -4a_2a_0))/2a_2
    so everything above the line is he same, just the a_0 and a_2 change
    i'll find a formula for n =3 and try that...

    dick: i tried this, but don't see where it leads,
    a_0 + a_1/x + a_2/x^2 + a_3/x^3 +...+ a_n/x^n
    and
    a_n + a_n-1/x + a_n-2/x^2 + a_n-3/x^3 + ...+a_0/x^n
     
  6. May 27, 2012 #5
    So here you have

    [itex]f(x) = a_n + \frac{a_{n-1}}{x} + \frac{a_{n-2}}{x^2} + .....+\frac{a_0}{x^n}[/itex]

    Make this a function of a new variable [itex]t=1/x[/itex]

    Can you relate this function to the first one? then find how roots depend on each other?
     
  7. May 27, 2012 #6
    :) you rock my world! are you really still in high school?
     
  8. May 27, 2012 #7
    Yep, I'm still learning the ropes. Two more years to go further! :biggrin:

    Glad you got it. :smile:
     
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