Mark, your sqwiirol gives me an idea. It's certainly not a number
since it doesn't have a precise and constant value but it can be a
good unit of measure. (I keep on thinking about how big 3^^^3 is).
Imagine the sqwiirol as a spaceship that accelerates according to
his iterated definition:
Every second the sqwiirol travels 1 with a sqwiirol zeros meters.
- At second 1 the sqwiirol has travelled 1 meter
- At second 2 the sqwiirol has travelled 10 meters
- At second 3 the sqwiirol has travelled 10^10 meters, that's nearly
the distance to the sun. In between, the sqwiirol has surpassed the
speed of light.
- At 3.0000001 seconds or so the squiirol has already reached the
other side of the universe ! Beyond that point, talking about
distance is useless. It surpasses also the number of particules in
the universe then the googol (10^100) in a blink of an eye. This
gives small idea about how big the acceleration of the sqwiirol is.
It's not just constant acceleration but a gigantic exponential
acceleration.
- Just after 4 seconds, the sqwiirol surpasses the googolplex, which
is 10^googol.
So, with that fantastically growing acceleration, how long will it
take for the sqwiirol to surpass 3^^^3 ?
Just consider the sqwiirol reaching 3^^^3 and let's play the film
backwards:
- 1 second before, the sqwiirol equaled log10(3^^^3) = 0,477 * (3^(3^
(3^�&3))), where 3 is repeated 7,625,597,484,987-1 times
- 2 seconds before, the sqwiirol equalled :
log10(0,477 * (3^(3^(3^�&3))), where 3 is repeated
7,625,597,484,987-1 times)
= log10(0,477) + log10((3^(3^(3^�&3))), where 3 is repeated
7,625,597,484,987-1 times)
the first term, log10(0,477), is so small, considering the second
one, it can be neglected.
So, 2 seconds before, the sqwiirol equalled 0,477 * (3^(3^(3^�&3))),
where 3 is repeated 7,625,597,484,987-2 times.
The iteration then becomes simple, 7,625,597,484,987 seconds are
needed.
=> The sqwiirol surpasses 3^^^3 after 241,640 years !
Alexis (spending his precious time in frivolous calculations).
since it doesn't have a precise and constant value but it can be a
good unit of measure. (I keep on thinking about how big 3^^^3 is).
Imagine the sqwiirol as a spaceship that accelerates according to
his iterated definition:
Every second the sqwiirol travels 1 with a sqwiirol zeros meters.
- At second 1 the sqwiirol has travelled 1 meter
- At second 2 the sqwiirol has travelled 10 meters
- At second 3 the sqwiirol has travelled 10^10 meters, that's nearly
the distance to the sun. In between, the sqwiirol has surpassed the
speed of light.
- At 3.0000001 seconds or so the squiirol has already reached the
other side of the universe ! Beyond that point, talking about
distance is useless. It surpasses also the number of particules in
the universe then the googol (10^100) in a blink of an eye. This
gives small idea about how big the acceleration of the sqwiirol is.
It's not just constant acceleration but a gigantic exponential
acceleration.
- Just after 4 seconds, the sqwiirol surpasses the googolplex, which
is 10^googol.
So, with that fantastically growing acceleration, how long will it
take for the sqwiirol to surpass 3^^^3 ?
Just consider the sqwiirol reaching 3^^^3 and let's play the film
backwards:
- 1 second before, the sqwiirol equaled log10(3^^^3) = 0,477 * (3^(3^
(3^�&3))), where 3 is repeated 7,625,597,484,987-1 times
- 2 seconds before, the sqwiirol equalled :
log10(0,477 * (3^(3^(3^�&3))), where 3 is repeated
7,625,597,484,987-1 times)
= log10(0,477) + log10((3^(3^(3^�&3))), where 3 is repeated
7,625,597,484,987-1 times)
the first term, log10(0,477), is so small, considering the second
one, it can be neglected.
So, 2 seconds before, the sqwiirol equalled 0,477 * (3^(3^(3^�&3))),
where 3 is repeated 7,625,597,484,987-2 times.
The iteration then becomes simple, 7,625,597,484,987 seconds are
needed.
=> The sqwiirol surpasses 3^^^3 after 241,640 years !
Alexis (spending his precious time in frivolous calculations).
--- In [email protected], "Mark" <secret.squirrel@...> wrote:
>
> There you are. Thought you were dead!
>
> In the spirit of "googol" I've made up my own number. I've called
it
> a "skwiirol" which is a really good name.
>
> It's defined as "1 followed by a skwiirol of zeroes". The best
thing
> about it is that it is recursive - the more you try to work out
how
> many zeroes it actually has, the bigger it gets!
>
> So its reality (or state) is defined by the observer and the kind
of
> observation they do - very much like as it is with quantum
physics.
> It also obeys Scheissenberg's Uncertainty Principle: "Um, I'm not
> really sure how big it is". However, it is known to be large
enough
> to have a black hole in the middle of the central "0".
>
> Mark