Roots of polynomial

  1. rock.freak667

    rock.freak667 6,231
    Homework Helper

    Considering the roots of a cubic polynomial([itex]ax^3+bx^2+cx+d[/itex]),[itex]\alpha,\beta,\gamma[/itex]

    [tex]\sum \alpha=\frac{-b}{a}[/tex]

    [tex]\sum \alpha\beta=\frac{c}{a}[/tex]

    [tex]\sum \alpha\beta\gamma=\frac{-d}{a}[/tex]

    If I have those sums of roots..and I am told to find [itex]\alpha^9+\beta^9+\gamma^9[/tex] is there any easy way to find this without having to expand?

    and also for a quartic polynomial
    when I expand [itex](x-\alpha)(x-\beta)(x-\gamma)(x-\delta)[/itex]
    I get:
    [tex]x^4-(\alpha+\beta+\alpha\gamma+\beta\gamma+\alpha\delta+\beta\gamma)x^3+(\alpha\beta+\gamma\delta+\alpha\gamma+\beta\gamma+\alpha\delta+\beta\delta)x^2 -(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\gamma\delta\beta)x+\alpha\beta\gamma\delta[/tex]
    for -x^3 I am supposed to get the sum of the roots...yet I expanded correctly, where did i go wrong?
     
  2. jcsd
  3. Ben Niehoff

    Ben Niehoff 1,727
    Science Advisor
    Gold Member

    On the last bit, you obviously didn't expand correctly, but with no intermediate steps, I don't see how one could say where you went wrong, exactly.
     
  4. rock.freak667

    rock.freak667 6,231
    Homework Helper

    Oh I thought I typed it out well this is it

    [tex](x-\alpha)(x-\beta)(x-\gamma)(x-\delta)

    =(x^2-(\alpha+\beta)+)(x^2-(\gamma\delta)+(\gamma\delta)[/tex]

    =[tex]x^4-(\alpha+\beta)x^3+\alpha\beta x^2
    -(\alpha+\beta)(\gamma+\delta)x^3+(\alpha+\beta)(\gamma+\delta)x^2-\alpha\beta(\gamma+\delta)x

    +\alpha\gamma x^2-\gamma\delta(\alpha+\beta)x+\alpha\beta\gamma\delta[/tex]

    =

    [tex]x^4-(\alpha+\beta+\alpha\gamma+\beta\gamma+\alpha\delt a+\beta\gamma)x^3+(\alpha\beta+\gamma\delta+\alpha \gamma+\beta\gamma+\alpha\delta+\beta\delta)x^2 -(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\ delta+\gamma\delta\beta)x+\alpha\beta\gamma\delta[/tex]
     
  5. Something is wrong in your expansion. Try:
    [tex](x-a)(x-b)(x-c)(x-d) = (x^2 - (a+b)x + ab)(x^2 - (c+d)x + cd) [/tex]
    [tex] = x^4 - (c+d)x^3 + cdx^2 - (a+b)x^3 + (a+b)(c+d)x^2 - cd(a+b)x + abx^2 - ab(c+d)x + abcd [/tex]
    [tex] = x^4 - (a+b+c+d)x^3 + (ab + cd + ac + ad + bc + bd)x^2 - (acd + bcd + abc + abd)x + abcd[/tex]
    which is what you'd expect.

    For the first problem, try using the Newton's sums trick.
     
  6. Ben Niehoff

    Ben Niehoff 1,727
    Science Advisor
    Gold Member

    The very first line is your mistake. This should be

    [tex](x-\alpha)(x-\beta)(x-\gamma)(x-\delta)

    =(x^2-(\alpha+\beta)x+\alpha\beta)(x^2-(\gamma+ \delta)x+ \gamma\delta)[/tex]
     
  7. rock.freak667

    rock.freak667 6,231
    Homework Helper

    No, I typed it wrongly, on paper I expanded it and found my error...so thanks...

    but is there any general formula that will give me the sums of the roots in a form that I need rather than having to expand it?
     
  8. Read the page I linked to about Newton sums. There's nothing faster than that, I think; that method allows you to calculate that kind of stuff pretty quickly though.
     
  9. Integral

    Integral 7,351
    Staff Emeritus
    Science Advisor
    Gold Member

    I have to admit that I do not understand your notation. If the 3 roots are [itex] \alpha, \beta, \gamma [/itex] Then what do your sums mean?
     
  10. rock.freak667

    rock.freak667 6,231
    Homework Helper

    Oh well...
    [itex]\sum \alpha[/itex] is simply the sum of the roots taking one at a time, i.e.[itex]\alpha+\beta+\gamma[/itex]

    and well [itex]\sum \alpha\beta[/itex] is the sum of the roots taking two at a time, i.e. [itex]\alpha\beta+\alpha\gamma+\beta\gamma[/itex]

    and for newton's sums

    I get up to the 3rd sum formula

    but I dont get how I would find an expression to find S[itex]_9[/itex] or for 4 and higher
     
  11. So, in the notation of that link, [tex]a_{n-k}[/tex] is the sum of the products of roots taking [tex]k[/tex] at a time. In your cubic equation, [tex]a_3 = a, a_2 = b, a_1 = c, a_0 = d[/tex]. Using the Newton sum equations, you can find [tex]S_1, S_2[/tex] and so on, up through [tex]S_9[/tex], which is what you asked for.

    [tex]aS_1 + b = 0[/tex]
    [tex]aS_2 + bS_1 + 2c = 0[/tex]
    [tex]aS_3 + bS_2 + cS_1 + 3d = 0[/tex]
    [tex]aS_4 + bS_3 + cS_2 + dS_1 = 0[/tex] (there's nothing after [tex]d[/tex])
    [tex]aS_5 + bS_4 + cS_3 + dS_2 = 0[/tex]
    ....
    So you should be able to get all the way to [tex]S_9[/tex] on your own this way.
     
  12. rock.freak667

    rock.freak667 6,231
    Homework Helper

    ah ok...but if there was something after d it would be
    [tex]aS_5 + bS_4 + cS_3 + dS_2 + eS_1 = 0[/tex] ?
     
  13. right
     
  14. rock.freak667

    rock.freak667 6,231
    Homework Helper

    oh thanks, then...this is a real help...Now i can do my roots of polynomials questions even faster now

    Edit: so in general the sums would be like this

    [tex]aS_n + bS_{n-1}+cS_{n-2}+...+ n[/tex]*(The term independent of x in polynomial)
     
    Last edited: Dec 1, 2007
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