Let $k$ be a large even integer. The following polynomial has exactly one real positive root.
$$x^{k/2} - x - 1$$
How can one determine what it is asymptotic to, as a function of $k$?
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There is in general no closed form of that root. – DINEDINE Apr 06 '19 at 19:05
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A possible approach: (i) For sufficiently large $k$ show the root $x^(k)$ satisfies $1<x^(k)<2$; (ii) Find $\lim_{k\rightarrow\infty} x^*(k)$. – Michael Apr 06 '19 at 19:12
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Three related questions: one, two, three. – Antonio Vargas Apr 07 '19 at 08:34
1 Answers
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Let $f_k(x) =x^k-x-1 $.
$f_k(0) = -1$, $f_k(1) = -1$, and $f_k(2) =2^k-3 \gt 0 $ for $k \gt \log_2(3) \approx 1.58 $.
If $x = 1+y/k$, then, for large $k$,
$\begin{array}\\ f_k(x) &=(1+y/k)^k-(1+y/k)-1\\ &\approx e^y-2-y/k\\ &=e^y-2-y/k\\ &\approx e^y-2\\ &=0 \qquad\text{for } y = \ln(2)\\ \end{array} $
Therefore $f_k(1+\ln(2)/k) \approx 0$.
As a check,
$\begin{array}\\ f_k(1+\ln(2)/k) &=(1+\ln(2)/k)^k-(1+\ln(2)/k)-1\\ &\approx 2-2-\ln(2)/k\\ &=O(1/k)\\ \end{array} $
marty cohen
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