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elgersma
Posts:
3
From:
Minnesota, USA
Registered:
11/26/12


Re: Find N(eps) so that n^(1/n)  1 < eps for n > N(eps)
Posted:
Nov 26, 2012 5:04 PM


If we can restrict eps to lie in the range 0 < eps < 1, then N(eps) = 1.2*(1 + ln(1/eps)) / eps gives n^(1/n)  1 < eps for n > N(eps) Derivation: n^(1/n) = exp(ln(n)/n) = 1 + ln(n)/n + ... so epsilon = n^(1/n)  1 = ln(n)/n + ... so n = ln(n)/epsilon + ... iterate: n_2 = ln(ln(n_1)/epsilon)/epsilon + ... Iteration converges very rapdily. Set n_1 = exp(1) giving n_2 = (1 + ln(1/epsilon))/epsilon Numerical results show that if we multiply by 1.2 then two iterations is enough: N(ep) = 1.2*(1 + ln(1/ep)) / ep This gives epsilon = n^(1/n)  1 < ep for n > N(ep)



