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How would you take the limit of this?

  1. May 21, 2012 #1
    I was just thinking about the golden ratio, and how it is a solution to the equation x^2 - x - 1 = 0.

    So I wondered what it would be like to make a similar higher degree equation.

    x^3 - x^2 - x - 1 = 0

    To which the positive solution is 1.84.. is there any significance to this ratio?

    I kept adding more terms.

    x^4 - x^3 - x^2 - x - 1 = 0
    x^5 - x^4 - x^3 - x^2 - x - 1 = 0
    Etc.

    These solutions, as I add more terms, grow more and more slowly and approach 2. I haven't found it to go past 2.

    How can I show that the solution to the equation approaches 2 as the number of terms in the format I've shown grows without bound?

    Another weird question from me, but I feel compelled to know the answers to these questions I have. :)
     
  2. jcsd
  3. May 21, 2012 #2
    rewriting the equation as x^5 - 1 = x^4 + x^3 + x^2 + x is always a difference of 1 when x = 2, because X^5 - 1 = x^4 + x^3 + x^2 + x + 1. this is true for binary.
     
  4. May 21, 2012 #3

    micromass

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    One can make the problem easier by remembering the partial sum of a geometric series:

    [tex]1+x+x^2+...+x^{n-1}=\frac{1-x^{n+1}}{1-x}[/tex]

    Can you do it now??
     
  5. May 22, 2012 #4

    haruspex

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  6. May 22, 2012 #5

    D H

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    Ahem.
    [tex]1+x+x^2+...+x^{n-1}=\frac{1-x^n}{1-x}[/tex]
     
  7. May 22, 2012 #6

    micromass

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    Oops :redface:
     
  8. May 22, 2012 #7
    To formalize what you're getting at:

    {x∈R^+ : lim_(n→∞)⁡ (x^n-∑_(k=0)^(n-1) (x^k) )=0}={2} forgive the lack of "real" notation
     
  9. May 22, 2012 #8
    Taking the infinite sum on the left, x= inf sum, you get phi. Did this the other week actually
     
  10. May 23, 2012 #9
    Something interseting useing the sum of powers rules mentioned earlier:

    We began with:
    xn-xn-1-xn-2-...-x-1 = 0

    Rearrange to get:

    xn = xn-1+xn-2+...+x+1

    Then using sum of consecutive powers:

    xn = (1-xn) / (1-x)

    xn(x-1) = xn-1

    xn+1-xn = xn-1

    ∴ xn+1-2xn+1 = 0

    However this sort of question (to me) sounds a little before it's time as it calls for the ability to solve very high degree polynomials. But what do I know? Answer: "All I know is I know nothing."
     
    Last edited: May 23, 2012
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