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Topic: Find N(eps) so that n^(1/n) - 1 < eps for n > N(eps)
Replies: 1   Last Post: Nov 26, 2012 5:04 PM

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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
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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)



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