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Roots of polynomial

  1. Nov 29, 2007 #1

    rock.freak667

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    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. Nov 29, 2007 #2

    Ben Niehoff

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    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. Nov 29, 2007 #3

    rock.freak667

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    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. Nov 29, 2007 #4
    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 http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums" [Broken] trick.
     
    Last edited by a moderator: May 3, 2017
  6. Nov 30, 2007 #5

    Ben Niehoff

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    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. Nov 30, 2007 #6

    rock.freak667

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    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. Nov 30, 2007 #7
    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. Nov 30, 2007 #8

    Integral

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    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. Nov 30, 2007 #9

    rock.freak667

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    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. Nov 30, 2007 #10
    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. Dec 1, 2007 #11

    rock.freak667

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    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. Dec 1, 2007 #12
    right
     
  14. Dec 1, 2007 #13

    rock.freak667

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    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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