Logical gremlins

Hello alternative tuners!

I have been doing a tad of experimenting, and I have found some strange
behaviour, and concepts.

A friend that I have just met, on the Internet, and I have discovered:

O
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O

What is uncanny is that your e-mail came just when I started thinking about
this stuff again. I just sat through a lecture last week where I spent the
whole time trying to calculate 2^65536 (2&5) on paper. Of course, by the end
of class I had only the lowest five digits (65536, interestingly enough) and
I determined that it would take me no less than 87 years to figure out the
rest. (The calculater on my computer can do it in 2.5 seconds. Of course, I
was originally interested in 2&65536, which would be ludicrous to find out
with any machine. At any rate, now that I think about it, the last five
digits to 2&65536 are 65536. In fact, the last five digits to any superpower
of 2 where the superexponent is greater than 3 are 65536. Strange.)

Sarn's Reply:

Yes, as J.B. Haldans said:"Not only is the Universe weirder than we think,
it's weirder than we CAN think".
I agree, but what may be even weirder is a thing called "The Polyverse".

At the Big Bang (TIME=0------>TIME=10^-43), there is evidence that the
Universe split into many (infinity...) Universe's, which is still splitting
up as we speak.
A theory fom quantum mechanics suggests, that whenever we make a decision,
e.g. toss a coin, the Universe we inhabit splits into three (HEADS OR TAILS
OR "ON THE SIDE" :o) ).

OK, then, I believe that at the moment of creation, all the possible
Universes were created, our Universe, and all the other "un-Universes", from
variations on a theme of ours (being very similar to ours), to Universes
with completely different laws of science.
Empty Universes, with different propertys, and laws of science, like
different games in a toystore.
And also, variations on ours, e.g. Todd with green hair, Todd with a striped
shirt, Sarn's with a purple computer and three eyes and also of "tri-sexual"
orientation-i.e. Sarn will "try anything for a laugh"....
:o) ).

Mabey this is where the Superpowers could be used in theoretical physics and
practical engineering one day.

Actually, I want to use curves created from ((((x^x)^x)^x)^x)...... and
......((((x^x)^x)^x)), for variokinetic weight training FVT curves.

But anyway, about the pattern of 65536 in the last digits where and
superexponent of 2 is greater than three.

I wounder if there are any such other interesting propertys of numbers a&b
where a=3, 4, 5, 6, 7...... and b=2, 3, 4, 5, 6, 7......?

I don't believe that there is a formula for prime numbers, no matter how
high we go into the super-super-super-super...........power stack.

I believe its been proven.

I checked out if (5&0.25)&0.25= 5&0.5, when we use a&b=.....(x^(x^x)) and it
wasen't to be the case.

However, what you may, also find really interesting, is that is is ALSO NOT
THE CASE when a&b=((x^x)^x)..... either!

Here the 4th QRoot reverse 5=1.545921430......., and its quiggle
square=2.12937248 (as normal quiggles), but and THIS fellows quiggle square
(or 5=(((x^x)^x)^x)= 1.960960668757......, and this did not = x, when x^x=5
(5's quiggle square), or 2.12937248.....

Mabey, ((5&0.25)&0.25)=
5&(0.25^0.25)=5&0.707106781=(5&10000000000)&-root(707106781), bigger
numbers=more accuracy.

O
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O

I'm guessing that left to right should be the order for hyperquiggles also. I
think this is because when you look at something like this x^(x^(x^x)) you
aren't really going from left to right anyway, you are starting with the second
to last, then going right, then going back again to the second x, then going
back to your original result of x^x, in a sort of zig-zag. If you order things
left to right, you are going the way which the exponent ascociates just by
itself.

Sarn's Reply:

Yes, this is true.

I have managed to base some microtonal musical temperaments on this point,
thanks to you!!!!!, also, which I will explain below.
Thank's for pointing this out to me.

I will be composing using David Cope's SARA in LISP, a VERY GOOD music
creation AI type programme, which has fooled music lovers.
You can hardly tell the difference from human brain composed music.

Actually, from this logical inconsistency, we can actually MAKE this have
some very interesting results!

(Much like a famous New Zealand drink, "Lemon And Paeroa" was a fluke").

Say we take 7(&)7=p.

I put the little & symbol in brackest, because it:

a)Looked lonely, and a little cold, :o)
b)Because its a general way of specifing and generalizeing what I'm about to
show you.

From this, we can have:

p=7th (QuiggleRoot type_n) 7, where (QuiggleRoot_n)=

1.(((x^x)^x)^x)^(x^(x^x))
2.((x^x)^x)^(x^(x^(x^x)))
3.((x^x)^(x^x))^(x^(x^x)) ect.......

Actually, there are n*(n-1)*(n-2)*.....(n-r+1) when they are selected r at a
time, thus ALL the possible ways of defineing a&n=

(n*(n-1)*(n-2)*......(n-r+1))
E
r

Where n=kth root, and r=1, 2, 3, ....n

so, for the 7th quiggle root,of 7, there are:

(7*6*5*4*3*2*1)+(6*5*4*3*2*1*2)+(5*4*3*2*1*3)+(4*3*2*1*4)+(1*2*3*3)+(1*2*2)+
(1*1)=

5040+1440+360+96+18+4+1=

6959 different ways of defineing the 7th quiggle root of any number.

Which one is correct?

Also, what if we line these up in Square, of 83.42 unit side lengths, and
apply them to the color spectrum of a photograph gridded off like so, to
translate them into another color spectrum?

The grid could be ordered by rank of size: 1 2 3 4 5 6
19 20--------- 7
18 |8
17 <-----|9
16 14 13 12 11 10

spiraling inwards.

I want to do this with sonograms, and to see what happens to the sound of a
sample.

Anyway, I woundered, what would the graph of:

y=xth(Quiggleroot_z)w as a 3D solid in hyperspace looklike?

Assumeing here, that the z's are ordered by rank, what would the solid looklike?

O
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O

Plus look at this pattern:

3*3 = 3+3+3 = 9
3^3 = 3*3*3 = 3+3+3+3+3+3+3+3+3 = 3*9 = 27
3(&.)3 = (3^3)^3 = 3*3*3*3*3*3*3*3*3 = 3^9 = it doesn't matter

Notice that the reverse quiggle fits in with the pattern, whereas, a regular
quiggle:
3&3 = 3^(3^3) = 3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3 = 3^27

On that same token, the hyperquiggle (the reverse hyperreversequiggle) should
be: 3@3 = (3(&.)3)(&.)3 = 3^3^3^3^3^3^3^3^3 = 3&9

So, from here on out, maybe the reverse quiggle should be changed to just
quiggle, and the quiggle should become the reverse quiggle.

Sarn's Reply:

Yes, perhaps that's correct.

I used to define what I now call reverse quiggle as quiggle, until someone
pointed out to me that it was just exponentional exponents.

However, using this as the rule, the only operator we'd ever need is addition!

I guess its just a matter of practicality!

O
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O
Hmmmm...x^0=1 because 1 is the starting point of all multiplication (anything
times 1 equals itself) so x&0 should be the starting point of all
exponentation. I think it would be 1 again, because you can raise things to the
power of one and never change them. But that only follows for this form x^1=x,
it's not true that 1^x=x, because 1^x=1. Maybe it should be different for each
number. 1.414...^2=2 so 2&0=1.414... x^3=3 so 3&0 = x. Negative
superexponents should be reverse exponentation on the starting point (roots).
For example: 3&-1 would be the third root of 3&0 (the third root of the third
root of 3) just like 3^-1 would be 3 divided by 3 divided by 3.

Sarn's reply:

Well, this is interesting!

So, 3&0=3^(1/3), because (3^3)^(1/3)=3, but (1/3)^(3^3)=1.311372652*10^-13.

And, it thus follows that x&-n=xth quiggle root(x&(n+1)) until n=0.

O
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Of course, something like (-2)&0 would be the negative second root of -2, which
would be the square root of one over -2, which would be imaginary. I suppose it
is possible that there would be some equations which couldn't be solved by
imaginary numbers. Perhaps we would have to create superimaginary
numbers...ludricrous numbers, or something like that.

What is a number raised to the power of an imaginary number anyway? 2^i=? You
multiply 2 by itself an imaginary number of times.

Sarn's Reply:

Yes, FASINATING!!!!!!!!

Fancy-An imaginary number multiplyed by itself an imaginary number of times.

Yes, using the x&-n=Q-root (x&(n+1))---->n----->0 definition, (-2)&0 should
be equal to
(-2)^2=2, or i(2^(1/2)) or i1.414213562.

I wounder what i2&0 would give?

((i2)^2)=-1.4141213562

* |+2 | -2 | -i2 | i2 |
--------------------------------------------
+2 | 4 | -4 | -i4 | i4 |
---------------------------------------------
-2 |-4 | 4 | 4i | -4i |
---------------------------------------------
-i2 |-i4 | 4i | -4 | 4 |
---------------------------------------------
i2 | i4 | -4i | 4 | -4 |
---------------------------------------------

Yes, there are some equations that can't be solved by reals, negative reals,
+/- imaginary.

I believe they use "Quarternians", "Octernians", or a combination of there
and complex numbers.

You can't have anything higher, like hyper-hyper-hyper-complex numbers either.

The proof exists, but is not nice, at all.

O
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O

Sarn's reply:

Yes, I wounder.

Using my table above, we could have, say:

1^i=i1, -i1, +1, -1??

Supposedly, we could have:

Z | i | -i | 1 | -1 | Where Z=Operator above
multiplication: ^, &, hyper&, hyper/hyper&
--------------------------------
i | | | | |
--------------------------------
-i | | | | |
--------------------------------
-1 | | | | |
--------------------------------
+1 | | | | |
---------------------------------

Any more comments?

SARN GOES ON FURTHER TO SAY:

This is all very peculiar.

Since there are:

---- -------- ^2
| (n*(n-1)*(n-2)*(n-3)*.......(n-r+1))|
| E |
| r |
| |
---- ------
r=1, 2, 3.......n

Different temperaments that can be gained from a 12 tower high exponent
series, I should have a lot to play with.

Any ideas?

Sarn Ursell.