FINDING THE DIGITS OF BLASPHEMORGULMINEXIPLEX
To find the converging digits we will use the function:
[10^10^(-x)-1]*10^x
and we let x approach infinity.
Recall that ln(1+x) --> x as x-->0
This can be seen because d/dx [ln(1+x)] = 1/(1+x)
As x-->0 the slope of ln(1+x) approaches 1/(1+0) = 1. As a result the tangent line y=x becomes a better and better approximation of ln(1+x) as x-->0.
Given this, we now adjust it to our purposes:
ln(1+x) --> x
implies
x --> ln(1+x) As x-->0
e^x --> e^ln(1+x) As x-->0
e^x --> 1+x As x-->0
10^x = e^(xln10) --> 1+xln10 As x-->0
:: 10^x --> 1+xln10 As x-->0
From this we can gather the following:
10^10^(-x) = 10^(1/10^x)
1/10^x will be very small when x is very large, therefore:
10^(1/10^x) --> 1+ln10/10^x
So we can conclude:
10^10^(-x) ~ 1+ln10/10^x For large x
Going back to our original expression we obtain:
[10^10^(-x)-1]*10^x
~ [1+ln10/10^x-1]*10^x
= [ln10/10^x]*10^x
= ln10
= 2.30258509299...
So we know that the digits converge to ln10. This means a blasphemorgulminexiplex will be roughly 1+ln10/blasphemorgulplex. When I say roughly I really mean almost exactly, but not exactly. It's the sort of difference only a mathematician would care about. Not only do we have this number accurate to a blasphemorgulus place values, but we can also be sure that the digits quickly converge to ln10, so that something on the order of a blasphemorgulus digits will match with ln10. That's more digits than we could ever hope to see anyway. Sure it's incorrect in an infinite number of decimal places but, eh, we got this level of accuracy with fairly little work so not much to complain about. In any case we can be sure this number is really really REALLY CLOSE TO 1!!!