Goliath & Leviathan Numbers - Numberphile 

tree goliath

Hey we love you America Numberphile and I want to say the UK is in my thoughts as you deal with the loss of the 3 aid workers accidently hit. Love you guys

@@WhataMensch I sat on the joystick by accident. sorry.

The log base 10 of that but point stands.

I thought that was the nickname of the notorious hacker known as "4chan"? /jk

That’s a cold response ngl

Well, you would not expect a demon to be called Nigel.

I wonder what Ivan Panin found on this? Hummm

Factorials only have roughly the same strength as exponentials. So a power tower (with some factorials thrown in) of factorial height is only about a triple up arrow in strength. I’d image G1 is bigger than Tony’s leviathan number.

No matter how big N is, if hit with with *0, it falls.

Ivn Panin! Have you heard of him?

@@17.11Acts No. Should I?

Yes another commenter says it refers to Mark 5:9.

Glad you enjoyed it

You can't check that many powers. That exponent is big, as in "bigger than the number of particles in the universe" big. There's no way to check them one by one - there has to be some way to generate possible candidates.

I think it's ᚠ (Fehu) from Futhark (Norse runes).

Especially since ᚺ and ᛞ are also Nordic runes.

Cirth does certainly resemble fuþorc.

@@wbfaulk makes sense then.... _sad Gandalf noises_

It’s a statistical argument. Think of rolling five dice. There’s only a finite number of possible outcomes, right? Well if you roll the dice enough times, you would expect to get every possible combination eventually. And if you kept rolling, everything would be a repeat of something that came before since every combination has already shown up. If you just keep rolling over and over again, it would be weird if certain combinations repeated but others magically never showed up again. In the case of five dice, there aren’t that many combinations, so it wouldn’t really take that long in the grand scheme of things to start seeing repeats again. If instead of rolling 5 dice you start talking about ways to arrange atoms in a finite region of space, note that there is still a finite number of possible combinations. If you have a big region of space, there could be a truly vast number of ways to arrange atoms in that space, but it is still finite. So, if you randomly arrange atoms in that space over and over again an even more extremely vast number of times, eventually you will hit every combination. After that point everything will have to be a repeat of something before, and if you run through that cycle multiple times, it would be weird if certain things repeated but others seemed to magically get skipped. That wouldn’t be very random. Granted, the universe itself isn’t completely random. There are big clumps of matter in stars and then vast empty spaces between them. However, there is some randomness in the universe, and that provides a bit of wiggle room in these arguments about duplicates in sufficiently large universes.

@@friiq0 Wow! Many thanks for taking the time to give a detailed answer, I appreciate it. I do understand what you mean. I think that when people talk about this they often forget to stress that it's a statistical argument, entirely predicated on a random process to generate the combinations of stuff. In the case of physics I suppose that's reasonable, although when it comes to things like every finite digit sequence appearing in the decimal expansion of some transcendental number, it gets much tougher because mathematical assertions require rigour.

@@macronencer You’re welcome! I love this kind of stuff. I see what you mean about the mathematical arguments. I think that mathematicians have proved that pi and other transcendental numbers will never settle into any kind of regular pattern. Even though the digits are fixed, that’s why you can think of the digits of pi as “random”. The argument is also different in the case of pi, because the digits go on literally forever. Suppose we consider pi in binary. We expect that pi will eventually write out the entire play Hamlet by Shakespeare at some point. Why are we so sure? Well, suppose that Hamlet never ever shows up in pi even once. That would mean that pi goes on for an actual eternity and somehow magically dodges Hamlet for ever and ever and ever-even though Hamlet is just a finite string of digits like any other. This is why mathematicians are quite confident that pi will eventually contain Hamlet. They are confident there is no special ordered pattern to the digits, and so there is no special Hamlet-avoiding pattern in the digits either. The same argument works for any play, book, poem, etc.

@@friiq0 I know what you mean. Is "confident" the same as "certain", though? We're getting into the gritty areas of metamathematics if we start down that road, perhaps...

@@macronencer I don’t know the math well enough to say that we know for absolute certain. Maybe someone has proved it, but I couldn’t tell you specifically.

Nice.

Exactly what they are, and I don't really care about them

It's a funny exercise, but also serve as a sandbox for the development of mathematical methods. For example, how do you find a sequence 666 in a power of two? Then you go on trying to break this problem into smaller steps, find shortcuts in calculations, use approximations for lower and upper bounds, and so on. The funny exercise may not have a purpose by itself, but give a well defined problem to work on.

He mentions that in the first video of this trilogy. This is all just for fun. It can be generalized to other sequences of digits, but he doesn't care about any potential future application (or lack thereof).

His life really is in perpetual decline

Don’t compare a mathematician to that loser

?

You know, the jimquisition guy. He dresses like a clown and "is" pregnant.

Don't know about the video game guy, but the mathematician is Stirling, not Sterling.

Hmm, on average a 6660 digit long number should have about 666 6s. Then you are close to 2^22k. You could make a better estimate about how much sooner it could occur since there are lots of lower powers that could by accident have an overrepresentation of 2s. But I bet that it isn't far off the 22k-th power of two. A computer can do that brute force for you...

I checked it. 2^20674 has 666 6s in it. A bit smaller than I expected.

@@landsgevaer, read my comment again. I didn't say 666 6's. I said 666 666's, non-consecutively.

@@pacman52280 Ah, but then you can surely extrapolated my estimate to this problem too? I'll leave it as an exercise for the reader that that occurs around 2^2M. 😉 Takes a bit more effort for a computer to do, but still possible...

Since the next two also look like Norse runes, I'd expect that was the intention.

It's the letter G in the Cirth alphabet. Mathematicians be getting bored :)

I had that thought through the video as well, and I'm surprised it wasn't mentioned.

Long sequences of repeated digits in a given base are interesting, however you choose to find them can be extended to other bases. Plus, you didn't seem to think about the choice of powers of 2 also being arbitrary.

@@Nebukanezzer You're right, I didn't question the exponent's base being 2 because it was stated in the definition of the number. Yes, it's arbitrary, but it's clearly defined. Also, in the video they mentioned that you could consider other bases for the exponent and do the same thing. On the other hand, the "in base 10" part was never explicitly stated and was simply assumed throughout the video; that's the part I have a problem with.

​@@JGMeador444The choice of 666 was already arbitrary, and the base 10 came from this. The number and base are just a motivation for the problem, making them funny to work with, but the mathematical tools applied are not limited by them. You could ask "what is the smallest power of 3 where a sequence 3457 (base 8) appears", and the methods would still be the same. But nobody cares about 3457 in base 8, so there would be no fun doing this xD

Yeah that one threw me for loop as well. It's definitely larger than a googol (10^100), but definitely not a googolplex (10^[10^100])

But it's bigger than log gplex, which was the point

@@pauldubois0 Dang you're right. I get it now, thanks!

Sorry still had sleep in my eyes from just coming on shift.

Don't be sorry, it's the 3*10⁹$ you just gave them and the NATO arms stockpiles in the region that allowed it to happen. Biden could stop making angry remarks tomorrow if he only started heading back the military support

@@IDFpartyboi972The angry mob is at your gates, Dr. Frankenstein.

Sequels always start unlisted and will be de-unlisted after a while.

@@unvergebeneid fair enough

@@deadmanrang Maybe for statistics? See how many people click the links?

My assumption is that since this is a series, it makes sense for public recommendations to not include later parts.

​@@Tahgtahvit's called a 'bug' in the program

And it's ticking away, ticking away with their sanity...