02/14/2012, 10:48 PM

@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: http://en.wikipedia.org/wiki/Tetration#Examples 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:

2^^65536=(2^^65535)^2=(2^^65534)^4=(2^^65533)^16=(2^^65532)^65536~=

(2^^65531)^2\\4\19728\20035299304068464649790723515603~=

(2^^65530)^3\\4\19727\60312260626302953715645484996575~=

65533\\4\19727\60312260626302953715645484996575 (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=

(3^^?)^2\\1\12\7625597484987~=

(3^^?)^3\\1\12\3638334640024\1258014290627491317860391~=

(3^^?)^4\\1\12\3638334640023\6002253567994547349966182~=

???\\1\12\3638334640023\6002253567994547349966182

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

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: http://en.wikipedia.org/wiki/Tetration#Examples 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:

2^^65536=(2^^65535)^2=(2^^65534)^4=(2^^65533)^16=(2^^65532)^65536~=

(2^^65531)^2\\4\19728\20035299304068464649790723515603~=

(2^^65530)^3\\4\19727\60312260626302953715645484996575~=

65533\\4\19727\60312260626302953715645484996575 (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=

(3^^?)^2\\1\12\7625597484987~=

(3^^?)^3\\1\12\3638334640024\1258014290627491317860391~=

(3^^?)^4\\1\12\3638334640023\6002253567994547349966182~=

???\\1\12\3638334640023\6002253567994547349966182

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