mersenneforum.org [Curiosity] Binary logarithm of a Mersenne number
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 2013-11-29, 15:40 #1 jnml   Feb 2012 Prague, Czech Republ AB16 Posts [Curiosity] Binary logarithm of a Mersenne number The binary logarithm[0] $L$ of a Mersenne number $M_n,\ n \in N$, having enough precision to reconstruct $M_n$ exactly after rounding $2^L$ to an integer, ie. $\left| M_n - 2^L \right|$ < ${1} \over {2}$ is $\ \ \ \ \ L = n - 2^{1-n}$. ---- The integral part of $L$ is $n-1$. The fractional part of $L$ consists of $n-1$ binary ones. For example: Code: n L L (base 2) ------------------------------- 1 0 0 2 1.5 1.1 3 2.75 10.11 4 3.875 11.111 5 4.9375 100.1111 ... [0]: http://en.wikipedia.org/wiki/Binary_logarithm
 2013-11-29, 22:15 #2 ewmayer ∂2ω=0     Sep 2002 República de California 22·32·17·19 Posts So 3 = 2*sqrt(2), then? Interesting - had not realized that. Learn something new every day around here.
2013-11-30, 03:39   #3
jnml

Feb 2012
Prague, Czech Republ

17110 Posts

Quote:
 Originally Posted by ewmayer So 3 = 2*sqrt(2), then? Interesting - had not realized that. Learn something new every day around here.
Check again the "having enough precision" part of the definition of L.

 2013-11-30, 05:35 #4 LaurV Romulan Interpreter     Jun 2011 Thailand 25×5×59 Posts Puzzle: Does the error $e=M_n - 2^L$ (why would you need the absolute of it? Mn is always bigger) converges? And if so, to what? (Hint: $\int_1^2\ln x\mathrm{d}x=0.386294361.....$) (grrr, why \dif doesn't work here? also, I can't hide $\TeX$ stuff?) Last fiddled with by LaurV on 2013-11-30 at 06:12

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