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Math Help - Roots and linear combinations

  1. #1
    Senior Member I-Think's Avatar
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    Roots and linear combinations

    Let f(x)=x^3+x-1 and \theta be its real root

    Express \frac{\theta^7+4}{\theta+1} as a linear combination of 1,\theta and \theta^2
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Roots and linear combinations

    Quote Originally Posted by I-Think View Post
    Let f(x)=x^3+x-1 and \theta be its real root

    Express \frac{\theta^7+4}{\theta+1} as a linear combination of 1,\theta and \theta^2
    Hi I-Think!

    Since \theta is a root, it follows that \theta^3=1-\theta.
    Substitute and divide?
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  3. #3
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    Re: Roots and linear combinations

    note that:

    \theta^3 = 1 - \theta.

    hence:

    (\theta + 1)^3 = \theta^3 + 3\theta^2 + 3\theta + 1 = 1 - \theta + 3\theta^2 + 3\theta + 2 = 3\theta^2 + 2\theta + 2

    = 3\theta^2 + 6\theta + 3 - 4\theta - 1 = 3(\theta + 1)^2 - 4(\theta + 1) + 3

    that is:

    1 = \frac{1}{3}(4(\theta + 1) - 3(\theta + 1)^2 + (\theta + 1)^3)

    so that evidently:

    \frac{1}{\theta + 1} = \frac{1}{3}(4 - 3(\theta + 1) + (\theta + 1)^2) = \frac{1}{3}(2 - \theta + \theta^2)

    this is probably the hard part.

    now:

    \theta^7 + 4 = (\theta^3)^2\theta + 4 = (1 - \theta)^2\theta + 4 = \theta - 2\theta^2 + \theta^3 + 4

    =\theta - 2\theta^2 + 1 - \theta + 4 = 5 - 2\theta^2

    thus:

    \frac{\theta^7 +4}{\theta + 1} = \frac{1}{3}(5 - 2\theta^2)(2 - \theta + \theta^2)

    = \frac{1}{3}(10 - 5\theta + \theta^2 + 2\theta^3 - 2\theta^4) = \frac{1}{3}(10 - 5\theta + \theta^2 + 2 - 2\theta - 2(1 - \theta)\theta)

     = \frac{1}{3}(12 - 7\theta + \theta^2 - 2\theta + 2\theta^2) = \frac{1}{3}(12 - 9\theta + 3\theta^2) = 4 - 3\theta + \theta^2

    ...if...my arithmetic is ok....

    Quote Originally Posted by ILikeSerena View Post
    Hi I-Think!

    Since \theta is a root, it follows that \theta^3=1-\theta.
    Substitute and divide?
    the trouble with this is you still have some remainder \frac{r}{\theta + 1} so finding the inverse of \frac{1}{\theta + 1} appears to be unavoidable.
    Last edited by Deveno; January 28th 2013 at 12:33 PM.
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Roots and linear combinations

    Yep. Your arithmetic is ok.

    Alternatively:
    \theta^7 +4 = {(\theta^3)^2\theta +4} = {(1 - \theta)^2\theta +4} = {\theta^3-2\theta^2+\theta +4}

    Now do a long division by \theta + 1 to find (remainder turns out to be zero):
    {\theta^7 +4 \over \theta + 1} = {\theta^3-2\theta^2+\theta +4 \over \theta + 1} = \theta^2-3\theta+4. \quad \Box
    Last edited by ILikeSerena; January 28th 2013 at 12:39 PM.
    Thanks from Deveno
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  5. #5
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    Re: Roots and linear combinations

    i'm curious, how did you know to stop at a cubic?
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  6. #6
    Super Member ILikeSerena's Avatar
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    Re: Roots and linear combinations

    Quote Originally Posted by Deveno View Post
    i'm curious, how did you know to stop at a cubic?
    The power of the numerator had to go down.
    I could easily get it down to a cubic.
    Division by a first order polynomial would yield a second order as requested... I just tried - it seemed to be the type of problem that would have a neat solution.
    Last edited by ILikeSerena; January 28th 2013 at 12:54 PM.
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