Let $x_{0} = 2$ and $x_{n+1} = x_{n} + \ln(x_{n})$, how can I find an asymptotic equivalent of this sequence say, to the third term? (This is not homework, it was a problem in the Oral Examination 2010 of Ecole Centrale).
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My approach would be:
- Integrate the "continuous version", to get $x(t) = t \log t - t.$ So your $x_n = n \log n - n + \epsilon(n),$ then plug into the recurrence to see what it tells you about $\epsilon(n).$
Igor Rivin
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but why we should have $x_n = n \log n - n + \epsilon(n)$ ? – user22323 Dec 13 '13 at 18:52
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Because the recurrence is a "discrete version" of the ODE $$\frac{dy}{dx} = \log x.$$ – Igor Rivin Dec 13 '13 at 19:38
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I'm interested in the completion of your method. My approach indicates $x_n = n\log n + n\log\log n - n + o(n)$, though I'm having trouble justifying it rigorously. – Antonio Vargas Dec 13 '13 at 19:44
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@AntonioVargas OK, I will try to work it out... – Igor Rivin Dec 13 '13 at 19:45
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Ah, I think I see what the issue is. The system $x_{n+1} - x_n = \log x_n$ isn't a discrete version of $\frac{dx}{dt} = \log t$, it's a discrete version of $\frac{dx}{dt} = \log x$. – Antonio Vargas Dec 13 '13 at 20:11
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@AntonioVargas Oops, you are completely right (not thinking straight today). Of course, that justifies your asymptotics idea. – Igor Rivin Dec 13 '13 at 20:18