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Finding the roots of a high degree polynomial equation

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    y(6) - 3y(4) + 3y''-y = 0

    2. Relevant equations



    3. The attempt at a solution

    The characteristic equation of that differential equation is:

    r^6 - 3r^4 + 3r^2 - r = 0

    But how am I expected to solve such a high degree polynomial (and thus the DE?)
     
  2. jcsd
  3. Apr 16, 2013 #2

    Bacle2

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    First, notice you can rewrite you char. equation as:

    r(r^5-3r^3+3r-1)=0

    Then notice too, that, inside your parenthesis: 1-3+3-1=0. What does this tell you?
     
  4. Apr 16, 2013 #3

    tiny-tim

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    Hi 1MileCrash! :smile:

    (try using the X2 button just above the Reply box :wink:)
    nooo :redface:

    r6 - 3r4 + 3r2 - 1 = 0 :wink:
     
  5. Apr 16, 2013 #4

    SammyS

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    Following tiny-tim's corrected characteristic equation:
    r6 - 3r4 + 3r2 - 1 = 0​

    Expand (a - b)3 .
     
  6. Apr 16, 2013 #5

    epenguin

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    I don't know at what point certain things are supposed to be familiar but I would say pretty early there is a rather familiar pattern to be discerned in that last formula.
     
  7. Apr 16, 2013 #6

    HallsofIvy

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    [itex]r^6- 3r^4+ 3r^2- 1[/itex] has only even powers of r. Let [itex]x= r^2[/itex] and that becomes [itex]x^3- 3x^2+ 3x- 1[/itex]. And, as SammyS suggests, that is [itex](x- 1)^3[/itex].
     
  8. Apr 16, 2013 #7

    Bacle2

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    Worse comes to worse and you cannot see this pattern, it always makes sense to try for the simple roots,like 0,1 and -1 . Checking for 1 as a root comes down to adding the coefficients and seeing if the sum is zero; similar for -1.
     
  9. Apr 17, 2013 #8

    HallsofIvy

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    Let me point out that Bacle2 is not just choosing "simple roots" at random. Since the leading coefficient is 1 and the constant term is 1, by the "rational root theorem" the only possible rational number roots are 1 and -1. (I don't know why he mentions "0".)
     
  10. Apr 17, 2013 #9

    Bacle2

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    Right, my bad. I thought the characteristic equation had no constant term.
     
  11. Apr 17, 2013 #10

    LCKurtz

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    All this help and the OP is nowhere in sight.
     
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