megiston
The Megiston
The Megiston
With that sobering fact in mind, lets now look at the Megiston. It turns out that the Mega doesn't even tap into the full power of the circle operator, Megiston on the other hand:
Megiston = circle(10) =
square(square(square(square(square(square(square(square(square(square(10))))))))))
As you can see we are in some real trouble now. Using Subscript notation we can make it a little easier to proceed:
square10(10) = square9(square(10)) = square9(triangle10(10)) =
square9(triangle9(10^10)) =
square9(triangle8(10^10^11)) =
square9(triangle7((10^10^11)^(10^10^11)))
At this point we begin to have trouble writing out the exact value, and an estimate becomes desirable. We know from our previous work that:
(10^10^11)^(10^10^11) ~ 10^10^10^11
We can surmise from this that triangle10(10) should be approximately 10^10^10^10^10^10^10^10^10^10^11, or E11#10 for short. But this presents a problem at the next stage because next we must find:
square9(E11#10) = square8(triangleE11#10(E11#10))
How do we apply E11#10 triangles to E11#10?! We can approximate that each additional triangle adds another 10 to the power tower. So roughly speaking:
triangleE11#10(E11#10) ~ E11#(10+E11#10) ~ E11#(E11#10) = E11#10#2
With these concepts in mind the Megiston can be estimated:
Megiston ~ E11#10#10
This number is much much bigger than a Mega. In fact it's much much bigger even than our massive over estimate of 256^^(2^256), for while that is a power tower with only about E77 terms, the Megiston would require a power tower with about E11#10#9 terms ... inconceivably bigger.
But how do we know whether our estimate is an over-estimate or under-estimate? How can we be sure it's any good at all? To bound the Megiston just as we did the Mega requires some more careful analysis, but it's made fairly routine by applying the Left Associative Polyates Lemma (LAPL). This is an important Lemma for working with Ackermann class numbers. I will develop a proof of it later in chapter 3.3. The details of this Lemma are beyond our scope here, so we simply use the result here. Basically for our purposes we can say that:
b^^(p+q-1) < (b^^p)^^q < b^^(p+q)
"p" and "q", which I refer to as the tetrates, simply can be added to provide an upper-bound. Subtract 1 from their total to obtain a lower-bound. With this in mind we can now come up with a definite lower bound for the Megiston. Let's begin with 10^10^11, as this is the exact value of triangle2(10). We need 8 additional applications of the triangle to apply the first square:
triangle3(10) = (10^10^11)^(10^10^11) > 10^10^10^11
triangle4(10) > (10^10^10^11)^(10^10^10^11) > 10^10^10^10^11
...
triangle10(10) > 10^10^10^10^10^10^10^10^10^10^11 = E11#10
Next we proceed to the 2nd square:
square2(10) > square(E11#10) = triangleE11#10(E11#10) > E11#(10+E11#10) > E11#(E11#10) = E11#10#2
As you can see, our original estimate was actually an underestimate. So we can say:
E11#10#10 < Megiston
The upper-bound is again a little more work because we can't simply sweep the complicated bits under the rug. However we can carefully account for them without letting them balloon out of control. Again we begin with 10^10^11:
triangle3(10) = (10^10^11)^(10^10^11) = 10^(10^11 * 10^10^11) = 10^10^(11+10^11)
Let N= 11+10^11
triangle4(10) = (10^10^N)^(10^10^N) = 10^10^(N+10^N) =
10^10^(11+10^11+10^(11+10^11)) <
10^10^(10^12 + 10^(11+10^11)) <
10^10^10^(12+10^11)
Let N = 12+10^11
triangle5(10) < (10^10^10^N)^(10^10^10^N) = 10^10^(10^N + 10^10^N) <
10^10^10^(1+10^N) <
10^10^10^10^(13+10^11)
...
triangle10(10) < 10^10^10^10^10^10^10^10^10^(18+10^11) = E(18+10^11)#9
< E(10^12)#9 = E12#10
Now proceeding with the second square:
square2(10) < square(E12#10) = triangleE12#10(E12#10)
This is a little trickier to deal with than the lower-bound, but let's look at what happens when we apply just one of those E12#10 triangles:
triangle(E12#10) = (E12#10)^(E12#10) = 10^(E12#9 * E12#10) =
10^10^(E12#8 + E12#9) =
10^10^(E12#8 + 10^(E12#8))
< 10^10^10^(1+E12#8)
We can continue onto the next level by assigning N = 1+E12#8 ...
triangle2(E12#10) < (10^10^10^N)^(10^10^10^N)
= 10^10^(10^N + 10^10^N)
< 10^10^10^(1+10^N)
Now let N2 = 1+10^N
We can continue in this manner, and what we will find is that at each level a +1 will occur one level below
the +1 from before. A zipper technique can then be used, moving the +1s up, collecting together until we can account for all of these increases by changing the leading exponent to 13. Thus we can say that:
triangleE12#10(E12#10) < E13#(10+E12#10) < E13#(E13#10) = E13#10#2
Using the same basic technique for 8 additional applications of the square we obtain the upper-bound E21#10#10. So we can say that:
E11#10#10 < Megiston < E21#10#10
We can actually improve on this slightly, by using square(10) < E(18+10^11)#9 instead. Then using our zipper technique we could show that:
square2(10) < E(19+10^11)#(9+E(18+10^11)#9)
square3(10) < E(20+10^11)#(9+E(18+10^11)#9 + E(19+10^11)#(9+E(18+10^11)#9))
...
square10(10) < E(27+10^11)#(9+E(18+10^11)#9 + E(19+10^11)#(9+E(18+10^11)#9 + ... )
< E12#(10+E(18+10^11)#9 + E(19+10^11)#(9+E(18+10^11)#9 + ... )
< E12#(E(19+10^11)#9 + E(19+10^11)#(9+E(18+10^11)#9 + ... )
< E12#(E12#10 + E(19+10^11)#(9+E(18+10^11)#9 + ... )
< E12#(E12#(10+E(18+10^11)#9 + ... )
< ...
< E12#10#10
Despite the technical difficulties involved I'm fairly certain that:
E11#10#10 < Megiston < E12#10#10
Are both correct and relatively precise bounds. I had also developed these bounds previously:
E1#11#10 < Megiston < E1#21#10
But these are actually far less precise.
The Megiston is so large, and the distance between it's bounds so vast that my previous illustrations using powers are no longer quite adequate. We can however make very rough estimates.
Mega^x = Megiston
What power , x, do we need to raise a Mega to get a megiston? Roughly speaking the logarithm of a Megiston. ie.
x ~ E11#(E11#10#9 - 1)
At which point your probably wondering ... what the heck does that mean?! So this doesn't really help us much to understand how much bigger it is. A Megiston simply transcends comparison with the Mega. How far apart are our bounds from each other:
(E1#11#10)^y = E1#21#10
To find y we observe that:
E1#11#10 = 10^^10^^10^^10^^10^^10^^10^^10^^10^^10^^11
Now we know from LAPL that
(b^^p)^^q ~< b^^(p+q)
so we can say that:
(b^^p)^^q ~ (b^^q)^^p
In this case...
(10^^(E1#11#9))^^(E1#21#9) ~ (10^^(E1#21#9))^^(E1#11#9) ~ 10^^(E1#21#9) = E1#21#10
Furthermore:
(10^^(E1#11#9))^^(E1#21#9)
= (10^^(E1#11#9))^[ (10^^(E1#11#9))^^(E1#21#9 -1)]
Therefore:
(E1#11#10)^[ (E1#11#10)^^(E1#21#9 -1) ] ~ E1#21#10
or more crudely
(E1#11#10)^[ 10^^(E1#21#9 - 1) ] ~ E1#21#10
Thus we conclude:
y ~ E1#(E1#21#9 - 1)
As you can see we've transcended the kinds of numbers where this sort of illustration might make any kind of sense. Basically both x and y become unimaginably vast power towers.
Leo Moser & Polygon Notation
As amazing as all that was, that's not the end of the story. Eventually a mathematician by the name of Leo Moser, decided to take Steinhaus' game even further. To begin Leo Moser replaced Steinhaus' circle-operator and replaced it with a pentagon, ... the next polygon in the sequence triangle, square, etc. Next he declared that any polygon could be used as an operator, and each was defined as repetitions of the operation below.
Thus Mega and Megiston can now be written alternatively as:
Mega = pentagon(2)
Megiston = pentagon(10)
But now we can easily define numbers much larger. How big would these be:
hexagon(2)
heptagon(2)
octagon(2)
...
This is already mind-boggling, but this is really just to set the stage for Moser's reveal. Moser now defines a new number as "two in a Mega-gon". That is, place two inside a Mega-sided polygon!!!
This massive number has come to be known as the Moser.
It's clearly impractical to try and draw out a "Mega-gon" around a 2 to represent this number. Thankfully a much more convenient notation has been devised. Back in 1998 Susan Stepney developed an elegantly simple bracket notation for handling these sorts of expressions. She begins by defining:
n[3] = triangle(n)
n[4] = square(n)
n[5] = pentagon(n)
n[6] = hexagon(n)