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Can't find the roots of a seemingly simple characteristic polynomial

$\lambda^8 -2\lambda^7+1=0$

I failed factoring.

Is it possible to find the roots? Can you show me how?

David Lee
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1 Answers1

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The factorisation into irreducible factors in $\Bbb Q[x]$ is given by $$x^8-2x^7+1=(x-1)(x^7-x^6-x^5-x^4-x^3-x^2-x-1).$$ The real roots are $x=1$ and $x=1.99196419661$. For details see this post:

Irreducibility of an infinite sequence of polynomials

Indeed, $f(x) = x^n-x^{n-1}-x^{n-2}-\cdots-x^2-x-1=0$ has exactly one real root, which is between $2$ and $2-\frac{2}{n+1}$, so the root approaches $2$ for large $n$.

Dietrich Burde
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  • Thank you for answering. The polynomial relates to flipping a coin and 7 heads in a row. ($y_n=2y_{n-1} - y_{n-8}$) So it's not a typo. – David Lee May 19 '23 at 06:11
  • Because $y_n$ means an integer coefficient of some polynomial, I thought the roots of $x^8-2x^7+1 = 0$ must be in Q in the first place. How can I get the complex roots? – David Lee May 19 '23 at 06:26
  • My goal is to set up $y_n = a_1\lambda_1^n + ... +a_8\lambda_8^n$. If I get the other complex roots, is it possible with converting complex to trigonometric(Wikipedia says)? – David Lee May 19 '23 at 06:34
  • There is a lot known about the roots of $x^n-x^{n-1}-\cdots -x-1$, see for example here, but to compute them you will need a numerical method, I suppose. – Dietrich Burde May 19 '23 at 07:46
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    And for coins with $7$ in a row see this post: "The asymptotic growth of $a_n$ is controlled by the largest positive root of the characteristic polynomial $x^7 = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$, which is a little less than $2$." This answer should be helpful. You don't need the complex roots. You should add this context about coins in your post! – Dietrich Burde May 19 '23 at 09:26
  • Thank you very much! The post you mentioned is like a perfect answer. – David Lee May 22 '23 at 08:24