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Graham's Number

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Cougar Draven

PostPosted: Sat Jan 20, 2007 6:25 am


All right, this is one of the main reasons I came here, to get some help with this.

I've been trying to calculate Graham's Number for about a week and a half now. Essentially, what I know is that this first image is true.

User Image - Blocked by "Display Image" Settings. Click to show.

Also, that 3^27 = 7,625,597,484,987.

Now, g_n = 3(^g_n-1)3, which is insanely impossible, considering that g_1 is impossibly large in itself. Eventually, I would like to chain together the g_1 chain, and attempt g_2, because Graham's number itself is g_64.

Any help would be awesome.
PostPosted: Mon Jan 22, 2007 9:18 am


The only help I can offer you is to stop now while you still can.

jestingrabbit


The_Bartner

PostPosted: Tue Jan 23, 2007 2:11 pm


Cougar Draven
I've been trying to calculate Graham's Number


I don't get it... You are trying to calculate this:
Wikipedia
It is too large to be written in scientific notation because even the digits in the exponent would exceed the number of particles in the visible universe, so it needs special notation to write down.
PostPosted: Sun Jan 28, 2007 12:30 am


The_Bartner
Cougar Draven
I've been trying to calculate Graham's Number


I don't get it... You are trying to calculate this:
Wikipedia
It is too large to be written in scientific notation because even the digits in the exponent would exceed the number of particles in the visible universe, so it needs special notation to write down.


Yes, I am. I challenge anything and everything I read, especially if it says I can't do something.

Cougar Draven


jestingrabbit

PostPosted: Sun Jan 28, 2007 12:46 pm


Cougar Draven
Yes, I am. I challenge anything and everything I read, especially if it says I can't do something.


I think you've gotta pick your battles a little more carefully than that. When mathematicians say you can't do something, they usually mean it. For instance, you can't write down a formula to solve every quintic polynomial (ie if x^5+ax^4+bx^3+cx^2+dx+e=0, in general you can't write a formula for x in terms of a,b,c,d and e using only plus, times, divide, subtract and nth roots). You can try as much as you like, but it just won't happen.

Some people speak a lot of nonsense about what can and can't be done, mathematicians tend to speak a lot of sense about the limits of knowledge.
PostPosted: Sat Feb 10, 2007 9:04 pm


As much as I want your life to fall apart as obsession and lack of resources eat at you from both sides, as the drive to succeed battles lack of computational power and time constraints, as you lie on your death bed still attempting to write down g_1, I really have to agree with jestingrabbit.
You are doomed.

Let's take the mass of the Earth: we get 6.0 x 10^24 kg from here. The calculations here are reasonable; you shouldn't have any reason to doubt them.
So let's suppose that the entire Earth can be converted into electrons for the storage of numbers. Let's even assume that these electrons can hold decimal digits, which they obviously can't in real life.

We have the mass of the electron to be 9.109 382616 × 10–31 kg, according to wikipedia. No calculations here, but hopefully you'll accept this number without too much hassle; if you want, you can set up one of many devices and measure the value yourself.

Anyway, suppose that the entire Earth is electrons. Then we get that there are 6.6*10^54 electrons, all capable of holding a single decimal digit. Let's round that up to 10^55 just for good measure.
So we can write a number of 10^55 digits, but not much more.

Note that 3 is slightly less than the square root of 10; therefore 10^55 should be somewhat more than 3^110, but not by all that much. Let's write down 3^200 just to be safe.
Note that 3^^4 > 3^200 > 10^55, and that 3^^6 > 10^(3^^4). Note that these bounds are horrible estimates; 3^^4 exceeds 3^200, and thus 10^55, by several billion orders of magnitude. Thus you can't write down 3^^6 even if you stick a decimal digit onto every hadronic particle on Earth (note that I'm not mentioning non-hadronic particles; you have no chance of writing on them, so I'm not concerned).

Now note that 3^^^3 > 3^^6, as 3^^^3 = 3^^(3^^3), and 3^^3 > 6 by a lot. Thus 3^^^3 absolutely cannot be written down. And since g_1 = 3^^^(3^^^3) > 3^^^3, well, let's just say that I wish you good luck and that I feel sorry for anyone who requires you to do absolutely anything else ever.

Note that the visible universe does not have particles on the order of several billions of magnitudes, as light does not travel that fast and the universe is not that old. So even if we perfected space travel, and could write decimal on, say, neutrinos, you're still doomed.

Layra-chan
Crew


Aetherius Lamia

PostPosted: Fri Mar 02, 2007 7:38 pm


Wow. This forum makes me feel stupid.

And here I was, thinking I was something special for being in Calculus III. They were right when they told me I hadn't had any real math yet. crying
PostPosted: Mon Mar 05, 2007 8:45 am


Dontcha worry, Calculus is the gateway to all higher maths! Before you know it, you'll be up to your neck in ODE's and linear transforms and vector spaces...

Forcedtojoin


33ars

PostPosted: Tue Mar 20, 2007 2:13 pm


This reminds me of the xkcd number (from the webcomic xkcd.) Calling the Ackermann function with Graham's number as the argument.
PostPosted: Tue Mar 20, 2007 4:41 pm


Aetherius Lamia
Wow. This forum makes me feel stupid.

And here I was, thinking I was something special for being in Calculus III. They were right when they told me I hadn't had any real math yet. crying


yeah that happens a lot to me too.

nonameladyofsins


altik_0

PostPosted: Thu May 24, 2007 7:41 pm


Cougar Draven
The_Bartner
Cougar Draven
I've been trying to calculate Graham's Number


I don't get it... You are trying to calculate this:
Wikipedia
It is too large to be written in scientific notation because even the digits in the exponent would exceed the number of particles in the visible universe, so it needs special notation to write down.


Yes, I am. I challenge anything and everything I read, especially if it says I can't do something.
In that case, I want to help! xd
PostPosted: Thu May 24, 2007 8:25 pm


Altik_0
Cougar Draven
The_Bartner
Cougar Draven
I've been trying to calculate Graham's Number


I don't get it... You are trying to calculate this:
Wikipedia
It is too large to be written in scientific notation because even the digits in the exponent would exceed the number of particles in the visible universe, so it needs special notation to write down.


Yes, I am. I challenge anything and everything I read, especially if it says I can't do something.
In that case, I want to help! xd


The only way to help him is to get him to stop. Unless the universe is several billion orders of magnitude larger than we think it is, and unless he lives several trillion orders of magnitude longer than the average human, he is going to fail terribly, and I'm not going to be able to laugh because we'll all be dead long before then.

Layra-chan
Crew

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Mathematics

 
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