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Other than $x=1$, is it possible to find definite solutions to $x^n - 2x^{(n-1)} + 1 = 0$ for any integers $n>3$, or must the zeros always be approximated?

Damon
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    By "definite solution" you mean a closed-form algebraic solution in terms of the coefficients, I presume? – H. sapiens rex Aug 21 '23 at 06:28
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    I have very good approximations of the largest root when $n$ starts to be large. But, to get any answer or help, please show us some work (even if, at a poin, you are stuck. By the way, Welcome to the site ! – Claude Leibovici Aug 21 '23 at 06:47
  • Thanks, everyone. Peter, does this line in the answer to the linked question mean that there's no solution by radicals for any n > 5? "So contains a five cycle and an element with a cycle decomposition equal to the product of two 2-cycles. It follows contains 5 and is not solvable. Consequently, will not be solvable by radicals." – Damon Aug 21 '23 at 07:07
  • Thank you, Claude; it's great to be here. I must confess I don't understand most of what I encounter here, but I still manage to learn a thing or two. – Damon Aug 21 '23 at 07:11
  • Yes it does but only for the example the answerer was using. Thus there is no general solution. – Peter Phipps Aug 21 '23 at 07:16
  • Is it for the lagest root ? If yes, is it "close" to some number ? Please, write your thoughts. – Claude Leibovici Aug 21 '23 at 07:36
  • @PeterPhipps. But we can be so close to the solution by simple approximations. – Claude Leibovici Aug 21 '23 at 07:38
  • @Claude Leibovici, the solution I'm most interested in is the one that approaches 2 as n grows larger, but I thought it'd be best to pose it as a more general question. The problem came about as I was playing with Fibonacci type sequences, increasing the number of prior terms that are summed. As you're aware, for the Fibonacci sequence, the ratio of a(k+1) to a(k) as k goes to Inf is the golden ratio. I was curious what the ratio would be if a(k) = a(k-1) + a(k-2) + ... + a(k-n). The ratio approaches 2 as n grows larger, and I wondered if determining the exact values is possible. – Damon Aug 21 '23 at 08:28
  • Exact not but very good approximation. For example, for $n=5$, with a minimum effort, I have $x=1.927562124$ instead of $1.927561975$ and we can do much better – Claude Leibovici Aug 21 '23 at 09:59
  • Have a look AT thé other post where I wrote an answer – Claude Leibovici Aug 21 '23 at 15:10

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