Huge numbers

[PWrong]

Ok, these numbers are getting a bit ridiculous . I'm actually more interested in the use of these operators at low values and non-integer values. For instance, what number tetrated to 2 gives 2?

i.e. what is x if x^x=2?

It's about 1.5596104, but I don't think there's an exact value. We need a power tower version of a square root.

[Geosphere]

Just one question: How do I become a Tetraspace citizen?

Post a lot.

Note that posting for the sake of posting will lead to deleting for the sake of deleting.

[Euclid]

Just one question: How do I become a Tetraspace citizen?

You must submit a gaggol of posts.

[jinydu]

What's a gaggol?

[Euclid]

What's a gaggol?

Some enormously huge number posted on another thread. Don't be concerned with the rating system, you will see many here with high rankings, yet weak and relatively useless posts. Likewise, you will find some who post rarely, but post great wisdom. Learn from them.

[Keiji]

To be precise, it's 2048.

Anyway, I've looked at that site and found it to be extremely interesting. Still, I prefer the name "hash" for the hyper4 operator.

[jinydu]

Interestingly, it also introduces hyper5, which is just repeated "hyper4-ing". Then, it generalizes the whole thing into a function with three input variables (one variable for the "base" number, one for the "exponent [or analogue], and one to determine whether its addition [1], multiplication, [2], exponentiation [3], hyper-4 [4], etc.).

Something like (100,100,100) would make googolplex look like nothing.

[elpenmaster]

is it possible to be a tetraspace citizen? or does it only go up to realmspace?

it seems like you could go off to hyper-googleplex (#g#)

some enormous number like (googleplex#g#googleplex)#g#(googleplex#G#googleplex) might defeat even the best quantum computer.

that number makes any other number look like less than nohing, i do believe
:twisted:

[Geosphere]

is it possible to be a tetraspace citizen? or does it only go up to realmspace?

Alkaline (site owner) is Tetra.

(googleplex#g#googleplex)#g#(googleplex#G#googleplex) might defeat even the best quantum computer.

Two ways to go with this:
1) No, that's what scientific notation is for.
2) Well, 9*9 defeats the best quantum computer.

[PWrong]

(googleplex#g#googleplex)#g#(googleplex#G#googleplex) might defeat even the best quantum computer.

Two ways to go with this:
1) No, that's what scientific notation is for.
2) Well, 9*9 defeats the best quantum computer.

Scientific notation isn't enough to compress that number, but the other notations on that webpage can.

As far as I know, the record for the best quantum computer is 17 qubits.
17 qubits is equivalent to 2^17= 131072 bits.

It's not good enough to compete with normal computers yet, but it probably won't take long before they get up to hundreds of qubits.

but nothing could express that number in ordinary notation

[jinydu]

is it possible to be a tetraspace citizen? or does it only go up to realmspace?

If I remember correctly, there is a thread somewhere, buried deep in this forum, that says 256 posts are needed to become a tetraspace visitor and 1024 posts are needed to become a tetraspace citizen! I'm certain I saw it once, but never thought to record the URL at the time. I spent over an hour searching for it today, but couldn't find it!

[PWrong]

Ok, I've figured out some more stuff for tetration.

the inverse of tetration can only be done with the Newton-rhapson algorithm.

Example, to find the square "tetroot" of 2:
2=x^x
0=x^x-2

By Newton-Rhapson's algorithm, if x is a good approximation for f(x)=0, then x-f(x)/f'(x) is a better approximation.

x^x is equivalent to e^(x*lnx)
By the product rule, the derivative is
(d/dx(x)*lnx + x*d/dx(lnx)) e^(x*lnx)
= (lnx + x/x) * e^(x*lnx)
= (lnx + 1)x^x

so f'(x)=(lnx + 1)x^x

Iterate the following:
xi+1=xi-(xi^xi-2)/(d/dx(xi^xi)

x(i+1)=xi-(xi^xi-2)/((lnx+1)xi^xi

If you do this only a few times, starting with x1 = 2, you'll get to
1.5596... in only a few iterations. The only way to give an exact value is to say tetroot(2).

It's much more difficult for higher powers, because the derivatives are more complicated.

That's the fastest way to find the tetroot of a number. If
You can work out x tetrated to a rational number p/q in two different ways, implying that there are 2 possible answers.

You can find the tetroot of (x^p), or you can find the tetroot of x, then tetrate this number to p. Several examples show that these are not always equal.

Tell me if any of this doesn't make sense. I'm pretty sure it's right, since I've checked extensively.

I'm working on some general laws for tetration and tetroots, so I'll keep you informed.[/i]

[elpenmaster]

i found an interesting idea in a book ive been reading

n$=n!^n!^n!^n!. . . n! times

this is a very huge number if, say, n=100

i was wondering which was bigger: Googleplex$ or 100#G#100

or you could write: (Googleplex#G#Googleplex)$

how much bigger or smaller about might (Googleplex#G#Googleplex)$ be than Googleplex$ times 100#G#100?

or you could have the function n'G$, so that n'G$=n!#G#n!#G#n!. . .
n$ times
:?

[Keiji]

1$ = 1!^1! = 1
2$ = 2!^2!^2! = 4^2 = 16
3$ = 3!^3!^3!^3! = 6^6^6^6 = 1.2041208676482351082020900568573e+168 (according to Windows Calculator)

Yep, you're right

[pat]

For another interesting approach to writing really big numbers, take a look at Knuth's arrow notation.

3↑↑↑3 is more than seven trillion and 3↑↑↑↑3 is mind-bogglingly bigger than that it's (if I've got my notation on straight) 3^{3^{3^...}} with more than seven trillion 3's.