bobxp wrote:MY GOD. That's a BIG number. But what about hfac(3.5)?
RQ wrote:PWrong wrote:i.e. what is x if x^x=2?
X=logx2
We're looking for an operation halfway between addition and multiplication. Any ideas?
bobxp wrote:Edit: I think this might not be very good at all. Look at the graph I produced, and you'll see why.
The main problem showing up after one has defined 1/nm to be the n-th order hyperroot of the real number x, is that there are two "natural" ways to define [sup]m/n[/sup]x, either as [sup]m[/sup]([sup]1/n[/sup]x) or [sup]1/n[/sup]([sup]m[/sup]x).
jinydu wrote:However, it looks like exponentiation isn't associative either.
PWrong wrote:Presumably there is also a "pentaroot" and "hexaroot" and so on. If we redefined elementary functions to include all of these roots, down to infinity, I wonder if we could possibly express pi or e using these functions.
bobxp wrote:I remember posting a new notation on this forum capable of storing huge numbers with erratic lengths required for each. Maybe that could be used to store numbers in a computer? And also, to display numbers that are too big to be calculated. Also, I made a program in C++ a while ago that can do addition, subtraction and multiplication (that's all so far) on HUGE whole numbers (any number of bytes, only limited by the width of the screen) but only in hexadecimal.
To PWrong: When I asked if anyone could solve:
x^x + x - 3 = 0
using elementary functions, I meant to include tetration and the square tetraroot (I don't think you would need cube tetraroots) as elementary functions (not in general, just for the purposes of solving that equation).
In order to help simplify expressions containing tetration, it would help if we could find some property that is true for tetration. But so far, I haven't found any that are worth mentioning.
bobxp wrote:There are the triangle numbers, 1, 3, 6, 10, 15, etc., where they are formed by 1+2+3+4+...+n. Then there are the factorial numbers formed by 1*2*3*4*...*n. What about the third series - 1^(2^(3^(4^(...))))?
Jinydu wrote:Its possible that an exact answer is not possible in terms of exponentiation, but perhaps tetration can do the job!
Jinydu wrote:You would have to expand that (x tetra 1000) into exponential form, and differentiate that ! Unless, of course, you're smart enough to differentiate that function in tetra form.
PWrong wrote:
Jinydu, I can't work out how you got your answers
PWrong wrote:
I don't know if you noticed equation 9 on the page you linked to, about Stirling's approximation:
n! = n^n * e^-n * sqrt(2 pi n)
Looks like tetration has already done the job!
PWrong wrote:
And have a look at this limit for e. It's equation 5 on
http://mathworld.wolfram.com/e.html
Not only does it use tetration 4 times, but the limit is apparently connected to prime numbers somehow!
PWrong wrote:
You should be able to do that for x^^1000, provided you have a supercomputer handy.
bobxp wrote:What about 5.5! for example? :?
Galidakis was presented as the leading researcher in tetration.
PWrong wrote:
Interestingly, the simplest top-function for continuity is linear, the top-function for the 1st derivative is a quadratic, and the top-function for the 2nd derivative is a cubic.
jinydu wrote:I would also like to know, how do you calculate (x tetra -1) and (x tetra -2)?
bobxp wrote:x tetra 0 = x
x tetra -1 = sqrt(x) (NOT SQTRT!!!!)
x tetra -2 = rt[sub]3[/sub](x)
x tetra -3 = rt[sub]4[/sub](x)
x tetra -n = rt[sub]n[/sub](x)
I don't know if that is right though.
jinydu wrote:I guess this means that if you want the first n derivatives to be continuous, you would need a polynomial of degree n+1.
jinydu wrote:Here's an important question: Do you always get exactly the same answer for (x tetra y), regardless of whether you use the linear, quadratic or cubic "top-function"? This is important because we (x tetra y) should have a unique value.
jinydu wrote:Also, does your extension of tetration satisfy:
(x tetra 1/2) = sqtrt(x)?
(x tetra 1/3) = cbtrt(x)?
etc.
jinydu wrote:I would also like to know, how do you calculate (x tetra -1) and (x tetra -2)?
PWrong wrote:Besides, even if I do that, and the next one, I'd then have to solve a quintic, which Godel proved is impossible using elementary functions. But I have no intention of going that far for a few years anyway. Maybe I'll be able to do it with "non-elementary" functions one day.
PWrong wrote:That derivative technique is very interesting. Do you think that would work for any operation?
Well, first of all, it was Abel, not Godel, who proved that the general quintic is unsolvable by radicals.
Keep in mind that in order for (n tetra x) to be differentiable at all, it must be a continuous function (at least in a certain interval)! In other words, if this derivative is correct, then it is possible to define (a tetra b) for all real values of b within a certain (possibly infinite!) interval.
PWrong wrote:Differentiation implies that the function is already defined for reals. I don't think you could find the derivative first, and use that to define it for reals.
Specifically:
(n^^x)' = the limit of [n^^(x+h) - n^^x] / h
as h approaches 0
Maybe that would be a better starting point.
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