emlightened wrote:Unfortunately, for virtually any two notations in googology, we can only ever use approximations, and never exact values.

That's true. But in most notations, you can get as close as you like to any number you wish. So it really isn't that different from the way that we are forced to round numbers when we use scientific notation (say, by writing 7^7^7 ~ 3.7598x10

^{695 974}).

This, by the way, is why I think it is vital to agree on some kind of standard notation (similar to scientific notation) for our numbers on this thread. We've been using my "F-notation" for this up until now and it worked well... But we are getting close to the limit of this system already. So we'll need something else, soon.

Sabrar wrote:I generally like my math to be precise so it irks me a little to see the approximations described in

this post but I understand why they are necessary.

They're not really necessary. You can calculate any googological number to arbitrary precision, even though it is impractical to write its decimal expansion in full.

The rough-and-quick approximations just make things easier to follow. Think of how easy it was for you to submit <9,3,2> and be sure that it is in the ballpark of F729. But you

could have also done it the hard way, and gotten a nice accurate figure: 2.8423735885 F 728.

But there's a deeper reason for learning to use these "rough-and-quick" methods: Once the argument gets large enough, these rough-and-quick approximations actually give you very precise answers.

For example, let's say that I use WolframAlpha to calculate 7^^7:

7^^7 = 7^7^7^7^7^7^7 ~ 10^10^10^10^10^10^5.842593328962333 = 5.842593328962333 F 6

In this case WA gives the answer with a precision of 16 digits. Not bad, right?

Now, I'll use my rough rule of "a^x ~ 10^x when a is small and x is larger" to deduce 7^^8:

7^^8 = 7^(7^^7) ~ 7^(5.842593328962333 F 6) ~ 10^(5.842593328962333 F 6) = 5.842593328962333 F 7

This may seem like a "rough approximation", but the truth is that it is far more accurate than the original result given by WolframAlpha. In fact:

7^(5.84259332896233 F 6) > 5.84259332896232999999...9999 F 7 (with more 9's then there are atoms in the entire observable universe).

So if you feel okay with your calculator rounding things to 16 (or 100 or a million) digits, you should have no qualms about using these methods (as long as you're dealing with numbers larger than about F4).

Anyway:

<5,13,2> ~ 8.9310685950807299484102 F 1 220 703 125The calculation (all the given digits are correct):