Login

KAR · 02/21/2008, 11:04 PM

quickfur Wrote:This isn't really related to zeration per se, but some time ago I searched for an operator ♦ over which addition is distributive (i.e., a+(b♦c) = (a+b)♦(a+c)). (I guess my motivation was that just as exponentiation distributes over multiplication, and multiplication distributes over addition, addition should also distribute over an operation that may be regarded to be, in some sense, "before" addition.)

Eventually, I came upon the definition: $a\diamond b = \ln(e^a + e^b)$ . As you can verify, $a+(b\diamond c) = \ln\ e^a + \ln(e^b + e^c) = \ln(e^a(e^b + e^c)) = \ln(e^{a+b}+e^{a+c})=(a+b)\diamond(a+c)$ .

Yes, you are right, that it is possible to construct this operator. It is a classical homomorphism of addition with using function exp(x). I researched in 90th years application of this operator and of some others for upgrade numerical solution systems of the differential equations. Good the outcome was with a homomorphism on the basis function ln(x).
However, this operator (a homomorphism of addition) is not Zeration. Moreover, application Zeration has allowed to achieve more simple algorithms of a homomorphism methods numerical solution of the differential equations and their systems. These researches are available in my monography (in Russian).

bo198214 Wrote:Can you btw show me, how you derived the commutativity of zeration?

Thanks for a problem on commutability Zeration.

Originally, I would like to note, that violation of logic is correctly noted:

$\;\; a\;[4]\;1\; = {}^1a = a$
$\;\;\;\; a^1 = a$
$\;\;\;\; a \cdot 1 = a$
$\;\;\;\; a + 1 \ne a$

For the last case we have $a + 0 = a$ . If to accept existence Zeration this fact means violation of a rule of construction of operation of a following rank $n = 1$ .

Zeration it has been created at learning Tetration in 1987.
I developed algorithm evaluation superradical of the radical similar to an evaluation:
$\sqrt x \approx \frac{{\frac{x}{a} + a}}{2}$ $\Rightarrow$ $ssqrt(x) \approx \sqrt {\log _a \left( x \right) \cdot a}$

After a successful evaluation of the superradical on this algorithm, I have decided to calculate x/2:

$\frac{x}{2} \approx \left( {x - a{\rm{?}}a} \right) - 2$ $\Rightarrow$ $\frac{x}{2} \approx \left( {x - a \circ a} \right) - 2$

Here it is necessary to apply operation by a level below addition.

At definition there were problems:

1. $a + 1 \ne a$ . It means, that for Zeartion we have violation in typical representation of number 1 as "units"
2. How to install or refute a commutability?
3. How to solve the equation: $a \circ x = a,\;\;\;\;x = ?$

The hint on a problem 2 and 3 has Tetration.
For this purpose we shall recollect a known limit:

$\lim\limits_{n \to \infty } \left( {{}^n\left( {\sqrt[a]{a}} \right)} \right) = a,\;\;\;a \le e,\;\;\sqrt[a]{a} > e^{ - e}$

From it follows, that the superradical from $a$ superpower aspiring infinity has solution as $\sqrt[a]{a}$ .
Similar association can be observed, reducing a rank of operation:

$\lim\limits_{n \to \infty } \left( {\sqrt[n]{a}} \right) = \frac{a}{a} = 1$

$\lim\limits_{n \to \infty } \left( {\frac{a}{n}} \right) = a - a = 0$

Here it is necessary to pay attention, that at a limit there is an operation of type "radical", in summary type "radical", but on a rank below. As there is "paradox" in operation "addition" on number 1, and on number of repetitions 2 and more - is not present, it is necessary to assume, that on infinite number of repetitions on operation "addition" there will be no paradox. We shall write:
$\lim\limits_{n \to \infty } \left( {a - n} \right) = a\Delta a = \left( { - \infty } \right)$ , where " $\Delta$ "-expected inverse Zeration operation of type "radical".

Further, from the given legitimacies it is possible to install, that Zeration it is commutative. For this purpose we shall consider two functions at various values $a \ge 2$ :

$\;\;\;\sqrt[a]{a} = {\rm{var}}\;\;\;\;$ and $\;\;\;\;\log _a \left( a \right) = 1 = {\rm{const}}$ .

For operations "division" and "subtraction":

$\;\;\;\frac{a}{a} = 1 = {\rm{const}}\;\;\;\;$ and $\;\;\;\;a - a = 0 = {\rm{const}}$ .

From the reduced example, follows, that inverses operations of type "log" at identical values of argument have constant value. Inverse operation of type "radical" has variable values at a modification of identical arguments and is equal to a constant in case of coincidence with operation of type "log", that is commutabilities.

From a limit $\lim \limits_{n \to \infty } \left( {a - n} \right) = a\Delta a = \left( { - \infty } \right)$ follows, that inverse operation of type "radical" for Zeration has unique solution for identical arguments, that is Zeration is commutative.

If Zeration it is commutative $a\Delta a = \left( { - \infty } \right)\;\; \Rightarrow \;\left( { - \infty } \right) \circ a = a$ .

bo198214 Wrote:No, thats not a strict consequence. The following is a strict consequence:
We start with the axiom
ao(ao(ao......))) = a+n for integer n>1 (and a being real or complex)
then we conclude
ao(a+n)=a+n+1

You write n> 1 and safely expand "n" up to complex? You can seriously make matching of any complex number with 1?

KAR

P.s. I bring an apology, for quality of computer translation on English.

Login
Username:
Password:	Lost Password?
	Remember me

Possibly Related Threads...
Thread		Author	Replies	Views	Last Post
	Zeration reconsidered using plusation.	tommy1729	1	3,123	10/23/2015, 03:39 PM Last Post: MphLee
	Is this close to zeration ?	tommy1729	0	2,151	03/30/2015, 11:34 PM Last Post: tommy1729
	[2015] 4th Zeration from base change pentation	tommy1729	5	6,190	03/29/2015, 05:47 PM Last Post: tommy1729
	[2015] New zeration and matrix log ?	tommy1729	1	3,357	03/24/2015, 07:07 AM Last Post: marraco
	Zeration = inconsistant ?	tommy1729	20	23,435	10/05/2014, 03:36 PM Last Post: MphLee