Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
A notation for really big numbers
When I first tried to write down the tetrations of 2, I noticed how fast the numbers grew. It's quite easy to write down 2^^5 in scientific notation: ~2.0035*10^19728
Or like calculators do it: 2.0035e+19728
But the next number (2^^6) would be like: 10^(6.0312*10^19727)
And then you get more and more 10^(10^(10^... which gets quite tedious.

I know now there are other methods for writing down numbers beyond the scope of scientific notation but I made up my own notation method. I basically wanted to have something that extends upon scientific notation.

There are computer systems that use the "e+" or a (back)slash as a shorthand of the usual "*10^" and since the exponent is more significant than the mantissa I decided to write the aforementioned 2^^5 as follows with the exponent first:

Notice that I omit the dot. The exponent already implies the "2" in this number has a value of 2*10^19728. Unnormalized form is not allowed in my notation.

When the number gets bigger you can add a secondary exponent. For example:
3^^4 ~= 12\3638334640024\12580

And now to show you the real power of my backslash notation we can write the complete exponent stack, of which the top should be a single digit number, and add the super exponent:

The super exponent (3 in this case) tells us the height of the exponent stack. This number becomes necessary when the exponent is so big that you can't write it down fully, like with 2^^6:
Note that the mantissa isn't displayed here because the primary exponent can't even be fully displayed, unless you have lots of space for it but even then calculating such a number to that high precision takes a long time.

If the super exponent ever gets too large you can extend it with its own exponents and super exponents. Make sure you also add a higher level hyper exponent with 3 backslashes. I even managed to write down G1 (3^^^^3, the first step in reaching Graham's number) in backslash notation:
3^^^^3 ~= 2\\\\2\\\2\\1\12\7625597484986\\1\12\3638334640023\60023
Graham's number is still too big for my backslash notation, though.
If you are interested in the subject, you can take a look at this (unfortunately it's in Italian, but the meaning would be clear):
Using my notation, 2[k(k(1))]2>>Graham's number.
So, IMO, the best notation is the "slowest one" for the size of the numbers/operators we are talking about.

Obviously there are more "powerful" techniques to obtain super-large numbers in only a few steps (for example the 0(n) based hierarchy, etc).

Let G(n) be a generic reverse-concatenated sequence. If G(1)≠{2, 3, 7}, [G(n)^^G(n)](mod 10^d)≡[G(n+1)^^G(n+1)](mod 10^d), ∀n∈N\{0} (La strana coda della serie n^n^...^n, 60).
(02/12/2012, 10:16 PM)Tai Ferret Wrote: When I first tried to write down the tetrations of 2, I noticed how fast the numbers grew. It's quite easy to write down 2^^5 in scientific notation: ~2.0035*10^19728
Perhaps you're also interested in Robert Munafo's page on big numbers and his hypercalc-calculator.

Gottfried Helms, Kassel
(02/13/2012, 11:24 AM)nuninho1980 Wrote: @marcokrt: Attention!! Due to spam. But please you should edit to remove message in one selected of your 2 posts.
Thank! Wink

@Marcokrt, I know there are many methods such as Conway's chained arrow notation and Bower's array notation to reach extremely high numbers. I was aiming more to the tetration level of numbers and I wanted a method that allowed easy comparison and any desired precision like with scientific notation.

After I invented my notation method I found out about another system that serves the same purpose, namely iterated exponential notation, which can also be used in base 10, like on this Wikipedia page: What I like about mine is that you can easily compare numbers by looking at the different exponent and hyper exponent levels.

@Gottfried, I think I've actually read those before. I like the Hypercalc program. I read about the power tower paradox, which is something I noticed too when I experimented with tetration.

When I first calculated the tetrations of 2 I was using my old TI calculator which doesn't go beyond 9.9999*10^99 so I had to use logarithms a lot to represent the numbers. To determine 2^^5 I simply had to multiply log 2 (base 10) by 65536 (which is 2^^4). Then I could split up the integer part, which is 19728, and use exponential (base 10) on the fractional part. 2^^6 was a bit trickier since I had to use the log of 2^^5 that fit on my calculator and I figured I had to add log(log 2) to obtain the log of the log of the answer. Then I wanted to calculate 2^^7 and that was the fun part. I had to imagine it to figure it out. I thought about the log of 2^^6 which had to have 19728 digits on the integer part and then adding log(log 2) which is a really small number so I figured it didn't matter, because it was far beyond the precision the calculator could handle.

With my backslash notation you can easily see what happens. When you raise x to the power of a number in this notation you literally raise the stack up a level. The mantissa becomes the primary exponent and is multiplied by log x. The primary exponent becomes the secondary exponent and will adjust depending on the change of number of digits through the previous operation, or if the last number was beyond our precision level, just add log (log x) if that's even within the precision level. The secondary exponent becomes the tertiary exponent and may adjust depending on the last operation if within precision level and so on. (And of course the super exponent increases.) Higher numbers will just raise levels and not change exponents because (the bottom of) the secondary exponent will soon fall out of the precision level.

I think it's pretty neat that the most significant part of the exponent stack converges. This makes tetration with a high super exponent very easy. Once the exponent stack converges you can add the rest to the super exponent. Thus for example:
(The rest happens way outside of our chosen precision.)

This even allows us to determine the top of the exponent stack of Graham's number even though we don't know the super exponents or any higher hyper exponents of that number:
G=3^^(something extremely large)=(3^^?)^3=(3^^?)^27=

So it turns out we can determine more of this number than just the final digits. Tongue (Yeah, I know. I guess I'm a bit obsessed with Graham's number.)

Possibly Related Threads...
Thread Author Replies Views Last Post
  Composition, bullet notation and the general role of categories MphLee 7 917 02/06/2021, 12:01 AM
Last Post: JmsNxn
  A Notation Question (raising the highest value in pow-tower to a different power) Micah 8 8,804 02/18/2019, 10:34 PM
Last Post: Micah
  Spiral Numbers tommy1729 9 13,155 03/01/2016, 10:15 PM
Last Post: tommy1729
  Fractionally dimensioned numbers marraco 3 6,115 03/01/2016, 09:45 PM
Last Post: tommy1729
  A new set of numbers is necessary to extend tetration to real exponents. marraco 7 14,701 03/19/2015, 10:45 PM
Last Post: marraco
  Tommy's conjecture : every positive integer is the sum of at most 8 pentatope numbers tommy1729 0 3,154 08/17/2014, 09:01 PM
Last Post: tommy1729
  Number theoretic formula for hyper operators (-oo, 2] at prime numbers JmsNxn 2 5,967 07/17/2012, 02:12 AM
Last Post: JmsNxn
  Tetration and imaginary numbers. robo37 2 7,799 07/13/2011, 03:25 PM
Last Post: robo37
  Iteration-exercises: article on Bell-numbers Gottfried 0 3,207 05/31/2008, 10:32 AM
Last Post: Gottfried

Users browsing this thread: 1 Guest(s)