What is the largest real number one can write within 200 characters?

[ComplexVar89]

I don't know about you, but I plan to live forever.

"Speak for yourself, sir. I plan to live forever."
-- William T. Riker, Star Trek: Generations

 

[DaTario]

I would post the number Y, such that

 1 + 1/2 + 1/3 + 1/4 + ... + 1/Y = G,
where G is the Graham's number

 

[micromass]

I would post the number Y, such that

 1 + 1/2 + 1/3 + 1/4 + ... + 1/Y = G,
where G is the Graham's number

So something like $e^G$?

 

[m k]

LLTTBTBA number

(Least Less Than Too Big To Be Accepted)

VMVSI number > 0

(Very Many Very Small Increments)

 

[DevilsAvocado]

It's time to put the nail in the coffin:

Alternative 1:

[itex]
y=\Sigma(G^{\text{Googolyottaplex}^{\text{TREE}(3)^{\text{SCG}(13)^{\text{Rayo}(\text{Googol})^{\text{FOOT}^{10}(\text{Googol})}}}}})\\x=y![y]\\Ref:googology.wikia.com/wiki/List\_of\_googologisms
[/itex]

(195 characters including LaTeX)

There's a reference to everything in the included link, but to make it easier, here's direct links:
http://googology.wikia.com/wiki/Googol
http://googology.wikia.com/wiki/Googolyottaplex
http://googology.wikia.com/wiki/BB(n)
http://googology.wikia.com/wiki/Graham's_number
http://googology.wikia.com/wiki/TREE(3)
http://googology.wikia.com/wiki/SCG(13)
http://googology.wikia.com/wiki/Rayo's_number
http://googology.wikia.com/wiki/BIG_FOOT
http://googology.wikia.com/wiki/Nested_Factorial_Notation

Alternative 2:

Call Max Tegmark and ask him to check with the Mathematical Universe, if there's a bigger number than Alternative 1.

(Sorry Drakkith, it seems like this universe isn't enough... there's a risk of a Big Rip... the only hope is that Stephen Hawking, Alan Guth, Andrei Linde, etc are right that Multiverse do exist... and that you could migrate to new fresh one... )

 

[DaTario]

So something like $e^G$?

Kind of Euler-Mascheroni apart of.

 

[nolxiii]

[itex]
y=\Sigma(G^{\text{Googolyottaplex}^{\text{TREE}(3)^{\text{SCG}(13)^{\text{Rayo}(\text{Googol})^{\text{FOOT}^{10}(\text{Googol})}}}}})\\x=y![y]
\\(13)_{x}\\Ref:
bit.ly/1Y6OXRR[/itex]

or if bit.ly not allowed, then x + 13

 

[DevilsAvocado]

or if bit.ly not allowed, then x + 13

I think bit.ly is allowed, but not stealing MY ideas!

()

 

[DevilsAvocado]

Seriously nolxiii, I called Max Tegmark the other day (April 1st) and asked him to comment on x, and after consulting the Mathematical Universe Supreme Court, he returned with the following remarks:

- Rayo's number and BIG FOOT are nice big numbers, however they can be risky due to the philosophical/self-referential (bigger than the smaller yada yada) nature. It's better to feed that eager beaver's mouth, with what you get ... and also utilize the layers of Graham in a more 'brutal' way.

So here's my final Alternative 3:

[itex]
z=\Sigma(G^{\text{Googolyottaplex}^{\text{TREE}(3)^{\text{SCG}(13)^{\text{SCG}(13)^{\text{SCG}(13)}}}}})\\y=\Sigma(g_{z})\\x=\Sigma(y![y])\\\text{Ref: googology.wikia.com/wiki/List_of_googologisms}
[/itex]

(197 characters including LaTeX)

Looks like a killer to me ...

 

[nolxiii]

On the shoulders of giants..

[itex]
z=\Sigma({\text{SGC}(13)^{\text{TREE}(3)^{\text{SGC}(13)^{\text{SCG}(13)^{\text{SCG}(13)^{\text{SCG}(13)}}}}}})\\y=\Sigma(g_{z})\\x=\Sigma(y![y])\\\text{Ref: googology.wikia.com/wiki/List_of_googologisms}
[/itex]

 

[nolxiii]

better off just stringing together a lot of busy beavers though? i lost track

 

[DevilsAvocado]

On the shoulders of giants..

Maybe ... and almost insanely* large ... now we're just awaiting for OP to tell us (at least ) how many positions/powers the top contesters have ... ??

*I get dizzy just trying to 'visualize' ...

 

[micromass]

Maybe ... and almost insanely* large ... now we're just awaiting for OP to tell us (at least ) how many positions/powers the top contesters have ... ??

*I get dizzy just trying to 'visualize' ...

The OP got scared of the numbers

 

[Kyx]

∞^∞!

 

[ProfuselyQuarky]

∞^∞!

Infinity is not a number

 

[ProfuselyQuarky]

Are you going to declare a winner, @micromass?

 

[micromass]

I think the biggest number posted is by DevilsAvocado. Any use of the bigfoot functions is superior to any other number posted. So:

[itex]
y=\Sigma(G^{\text{Googolyottaplex}^{\text{TREE}(3)^{\text{SCG}(13)^{\text{Rayo}(\text{Googol})^{\text{FOOT}^{10}(\text{Googol})}}}}})\\x=y![y]\\Ref:googology.wikia.com/wiki/List\_of\_googologisms
[/itex]

 

[DevilsAvocado]

Thanks man!

... where do I collect the Prize?

()

 

[DevilsAvocado]

The OP got scared of the numbers

No worries, there's nothing to fear but fear itself, this is not the end. It is not even the beginning of the end of numbers. But it is, perhaps, the end of the beginning of numbers.

Nonetheless, it's hard (impossible) to get any form of perspective on the extremely large numbers we have in this thread.

But, what almost drive me nuts, is that the 'small' and ridiculously general π (in physics & elsewhere), have an infinite number of decimals, that never repeats itself (only subsets), that is 'static random', i.e. we get the same sequence every time we look, but there is no way of predicting the next decimal (without calculations), and (if π is a normal number) — the entire works of William Shakespeare is encoded (A=01, B=02, etc) somewhere in this never ending story...

This is insane... who 'wrote' the algorithm for this craziness?? And where is this bugger 'stored'!?

So, if we start by trying to visualize one million. That's easy, right? If we print one million digits in one row, with a 8pt font — how long would that paper be; 10, 20, 40 or maybe 100 meters?

Wrong. Check spoiler ...

Spoiler: One Million Digits

I'm just waiting for Monty Python's Flying Circles ...

If you like this on, check out The Making of a Mile of Pi.
​

If we were to expand this technique a little bit — how many digits would we get in a turn around the Earth? The Earth radius is 6,378 km (equatorial), so let's check it out:

● $(2\pi\times6378/1.693)\times10^{6}=23,670,499,639\ digits$

23.67 Billion Digits on One Turn Around The Earth​

● $10^{23,670,499,639}$

The size of 23 Billion Digits (Bigger than Googol! But smaller than Googolplex ...)​

● $10^{12}/23,670,499,639=42.24\ turns$

The number of Turns to get One Trillion Digits​

● $10^{10^{12}} = 10^{1,000,000,000,000}$

The size of 42.24 Turns Around The Earth (Much bigger than Googol, but still smaller than Googolplex)​

● $10^{100}/23,670,499,639=4.224\times10^{89}\ turns$

Number of turns to get One Googol Digits​

● $10^{10^{100}} = 10^{10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000}$

The size of 4.224 x 10⁸⁹ Turns Around The Earth (That's a Googolplex!)​

10⁸⁹ turns ... that's a neat little number, but there's something missing here, right?

If we assume that the paper is 0.1 mm thick, how thick (radius) would this pile of π be? Well, 0.1 mm is 0.0000001 km, and:

$4.224\times10^{89}\times10^{-7} = 4.224\times10^{82}\ km$

That's a BIG roll of π! And it's a 'little' bit wrong as well ...

While the π roll expands, there will be room for more numbers on every turn, so the correct number of turns to get One Googol Digits is $\color{Blue}{1.645\times10^{50}}$:

$(2\pi\times\color{Blue}{1.645\times10^{50}}\times10^{-7} / 1.693)\times10^{6}\times\color{Blue}{1.645\times10^{50}} = 10^{100}$

(This is actually wrong as well, the calculated circumference is for the last biggest turn, but I don't have the time/energy to do the exact calculation, the truth is somewhere between 10⁸² and 10⁵⁰, so this is a 'generous' solution, but if there's a true magician mathematician out there, please be my guest and present the correct formula!)

The radius (for the 'generous' solution) is $\color{Blue}{1.645\times10^{50}}\times10^{-7} = 1.645\times10^{43}\ km$

Gosh, 10⁴³ km ... it's 150,000,000 km to the Sun ... and 7,000,000,000 km to Pluto ...

And one light-year is 9.4607 × 10¹⁵ m, or 9.461 trillion kilometres, or 9.461 × 10¹² km.

$1.645\times10^{43} / (9.461\times10^{12}) = 1.738\times10^{30}\ ly$

OMG, a π roll with a radius of 1.738 x 10³⁰ light-years!?

Wikipedia:
45.7 x 10⁹ ly = The comoving distance from the Earth to the edge of the visible universe is about 45.7 billion light-years in any direction; this is the comoving radius of the observable universe.


No comment.I think we got this covered, so let's move on to see what Ron Graham has to say.

​

So, if your head now feels like a Black Hole — that's perfectly normal!

​

 

[nolxiii]

Also like how once we started getting into ∑ with BIG FOOTs and TREE(3)s as inputs, graham's number started to seem so tiny in my head.

 

[collinsmark]

Could the biggest number be:

40



Seriously though, I found the video quite interesting. It's worth a watch.

Although most of the video describes transfinite (if not infinite) numbers, outside the scope of this thread, it's at least indirectly relevant in a few places.

 

[DevilsAvocado]

Could the biggest number be:

40

Some say the ultimate answer to everything is: