megafugagoogolplex

We're going to bound the megafugagoogolplex within Hyper-E Notation. We can begin by trying to imagine a power tower of googolplexes a googolplex terms high:

googolplex^googolplex^ ... ... ^googolplex^googolplex

Now replace every term but the topmost one with 10:

10^10^ ... ... ^10^googolplex

This must be less than a megafugagoogolplex. Note that it will contain exactly googolplex-1 10s. Now write this in Hyper-E:

E(googolplex)#(googolplex-1)

Now we use the properties of Hyper-E to simplify:

E(googolplex)#(googolplex-1) = E(10^10^100)#(googolplex-1)

= E100#(googolplex-1+2) = E100#(googolplex+1) > E100#(googolplex) = E100#2#2 = googolplexidex.

So we have proven that:

googolplexidex < megafugagoogolplex

In fact we've shown that:

10^googolplexidex < megafugagoogolplex

Next we can show that megafugagoogolplex < E100#(googolplex+2). Proving the upper-bound is more difficult because we can't simply ignore all of the googolplexes in the power tower, but must take account of them.

This can be done most easily in stages through an accumulator method.

Begin with a googolplex. Next simplify googolplex^googolplex:

(10^10^100)^(10^10^100) = 10^(10^100*10^10^100) = 10^10^(100+10^100)

Next we simplify googolplex^^3:

(10^10^100)^(10^10^(100+10^100)) = 10^(10^100 * 10^10^(100+10^100)) =

10^10^(100+10^(100+10^100))

At this point we set up an accumulator. Let X=100+10^(100+10^100).

Then we simplify googolplex^^4:

(10^10^100)^(10^10^X) = 10^(10^100+10^10^X)

Note that as long as X>2, then adding in 10^100 has almost no effect on 10^10^X. Observe that:

10^100+10^10^X < 10^10^X+10^10^X = 2*10^10^X < 10*10^10^X = 10^(1+10^X) < 10^10^(X+1)

Put altogether we have that:

googolplex^^4 < 10^10^10^(X+1)

Thus we have increased the accumulator, X, by 1. Subsequently we can show that if:

googolplex^^(n) < E(X+n-3)#(n-1)

Then:

googolplex^^(n+1) = googolplex^googolplex^^(n) <

10^(10^100 * E(X+n-3)#(n-1)) <

10^(E(X+n-2)#(n-1)) =

E(X+n-2)#n

Increasing the accumulator is accomplished through the technique of the ascending one. We can think of the effect of 10^100 as traveling up the power tower as a +1 to eventually get added to the accumulator. This accounts for all the effects of the larger base. In fact it's always a gross overestimate, but it's more than sufficient for our purposes. So what happens when we reach a megafugagoogolplex? The accumulator gains googolplex-3. Let's work out the simplification:

googolplex^^googolplex < E(X+googolplex-3)#(googolplex-1)

Recall that X = 100+10^(100+10^100)

E(100+10^(100+10^100)+10^10^100-3)#(googolplex-1)

= E(97+10^10^100+10^(100+10^100))#(googolplex-1)

< E(10^(1+10^100)+10^(100+10^100))#(googolplex-1)

< E(10^(101+10^100))#(googolplex-1)

< E(10^10^101)#(googolplex-1)

< E(10^10^10^100)#(googolplex-1)

= E100#(googolplex-1+3)

= E100#(googolplex+2)

Thus it's been demonstrated that:

E100#(googolplex+1) < megafugagoogolplex < E100#(googolplex+2)

Not bad for a number this big. Note that this also proves that:

googolplexidex < megafugagoogolplex < googolduplexidex

What about (E100#2#2)^(E100#2#2)? This turns out to be an extremely close call. Luckily it's not so close as to be indeterminate. It turns out that a megafugagoogolplex is ever so slightly larger.

We can begin by placing a very fine upperbound on (E100#2#2)^(E100#2#2).

Observe that:

[E100#(googolplex)]^[E100#(googolplex)]

= 10^(E100#(googolplex-1) * E100#(googolplex))

< 10^(E100#(googolplex))^2

= 10^10^(2*E100#(googolplex-1))

< 10^10^(10*E100#(googolplex-1))

= 10^10^10^(1+E100#(googolplex-2))

< 10^10^10^10^(1+E100#(googolplex-3))

< ... <

< E(1+E100#1)#(googolplex)

= E(1+10¹⁰⁰)#(googolplex)

Remember that this is an upperbound on (E100#2#2)^(E100#2#2), so it's actually much smaller than this. On the other hand we can produce a lower-bound on a megafugagoogolplex which is only slightly larger than this. Simply turn every googolplex in googolplex^^googolplex into a 10 except the topmost 2:

10^10^10^ ... ... ^10^10^googolplex^googolplex =

E(googolplex^googolplex)#(googolplex-2)

= E(10^100 * 10^10^100)#(googolplex-1)

= E(100+10^100)#(googolplex)

This is a lowerbound on a megafugagoogolplex. So we discover that a megafugagoogolplex just barely surpasses a googolplexidex tetrated to the 2nd. This bound is even finer than the previous bound since a googolplexidex^^2 is obviously larger than 10^googolplexidex = E100#(googolplex+1).

I have now justified my placement of the megafugagoogolplex on my Ultimate Large Number List.