Re: X -> LOG|X|

Bill Owen wrote:
NILS BÖRJESSON wrote:

START WITH A NUMBER, FOR EXAMPLE: 2!!!!!!!!!!!!!!!!!!!!!!!!!!!

CALCULATE LOG|2| = B_1

CALCULATE LOG|B_1| = B_2

CALCULATE LOG|B_2| = B_3

ETC. ETC. ETC.!!!!!!!!!!!!!!!!!!!!!!!!!

THE BASE OF THE LOGARITHM IS D!!!!!!!!!!!!!!!!!!!

THE NUMBER I START WITH CAN'T
BEE 0!!!!!!!!!!!!!!!!!!!!!!!!!
OR 1 OR -1!!!!!!!!!!!!!!!!!!!!!!!!!!!!
OR E OR -E OR 1/E OR -1/E!!!!!!!!!!!!!!!!!!!!!!!
ETC. ETC. ETC.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I.E. B CAN'T BEE 0!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

THE NUMBER I START WITH CAN'T
BEE SUCH THAT B IS PERIODIC

CALCULATE THE ARITIMETIC MEAN OF THE NUMBERS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
B_1 B_2 B_3 B_4.. ..B_N !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

=C_N

Hypothesis :

THAT THE LIMIT OF C_N/N
EXIST AND IS INDEPENDENT OF THE NUMBER I START WITH!!!!!!!!!!!!!!!
BUT DEPENDENT ON THE BASE OF THE LOGARITHM D!!!!!!!!!!!!!!!!!!!!!!!!!

NOTE: THE:!!!!!!!!!!!!!!!!! ARE ARE!!!!!!!!!!!!!!!!!!!!!!!!!


Um, no.

The domain of the log function (I don't care which base) y = log x is
x > 0. Furthermore, log x <= 0 when x <= 1. Finally, log x < x for any
base > 1.

So whatever positive number B_1 you pick, B_2 < B_1, B_3 < B_2, etc.,
until you come to some B_i < 1. Then B_(i+1) < 0 and B_(i+2) does not
exist.

The ratio B_i/i is not independent of B_1. A small change in B_1 will
produce a small change in all the subsequent B's without changing the value of i. Therefore there will be a small change in C_i.

There is no "limit of C_N/N" because C_N is not defined for N > i.

-- Bill Owen


The original poster pointed out to me privately that I missed the absolute value signs. Actually B_(i+1) = log |B_i|, not log B_i.
This makes for a much more interesting problem.

-- Bill Owen

.

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