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Re: borderline of finite with infinite Chapt13.40085 Maxwell
Equations placing demands on mathematics #631 New Physics #751 ATOM TOTALITY
5th ed
Posted:
Jun 16, 2012 6:00 PM
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On Jun 15, 11:18 am, Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
> And for short, let me call the number simply as 3x10^603.
A better name for the number is pi*10^603.
An even better name for the number is floor(pi*10^603).
> The same sort of question I am asking about pi, if perchance the first
> time that pi is evenly cube rooted is [floor(pi*10^603)].
It isn't a perfect cube. For trial division reveals this number is a
multiple of three, yet not nine. Therefore floor(pi*10^603) can't be a
perfect power at all.
As for perfect powers among the numbers of the form floor(pi*10^n),
let
us begin with squares. Here's a simple heuristic: as d(sqrt(x))/dx is
1/(2sqrt(x)), the probability that a natural number of approximate
size
x is a square would be 1/(2sqrt(x)). Thus the probability that a
number
of the form floor(pi*10^n) is a square is 1/(2sqrt(pi*10^n)), which we
then sum as n ranges from zero to infinity.
This is a geometric series with initial term 1/(2sqrt(pi)) and a
common
ratio of 1/sqrt(10). So the expected number of squares is:
.412556664...
As this is less than one, we shouldn't be surprised if there are _no_
squares among 3, 31, 314, 3141, 31415, 314159, ..., and very surprised
if there were two or more squares.
Replacing pi with e in the formula above doesn't give much
improvement:
.443518052...
As cubes and higher powers are much rarer than squares, we don't even
need to calculate to determine that the expected number of cubes among
3, 31, 314, 3141, 31415, 314159, ..., is close to zero. That 27 is a
cube is a fluke -- we'd be surprised to find any more cubes in the
sequence of 271, 2718, 27182, 271828, ..., and so on.
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