Improving convergence of Andrew's slog
#14
quickfur Wrote:
bo198214 Wrote:[...]
what makes exponentiation and multiplication unique are the "distributive laws":
\( a^{x+y}=a^xa^y \) and \( a(x+y)=ax+ay \) which are an extension of
\( a^{x+1}=aa^x \) and \( a(x+1)=ax+a \), however we know it doesnt work for tetration.
Hmm, I wasn't aware of this property being referred to as the "distributive law"... I am more familiar with distributive laws being of the form \( a\diamond(b\circ c) = (a\diamond b)\circ(a\diamond c) \) (and/or vice versa for non-commutative operators).
right, thatswhy I put it into quotation marks. These laws are generally not referred to as distribution laws. However I also dont know any other term for them.

Quote:
Quote:I'm not entirely convinced, though, that it's not possible to find some kind of "natural" uniqueness property that must hold for tetration, which would uniquely determine it on the reals. It may require operations (or rather, binary functions) other than addition, multiplication, or exponentiation to be expressed, but surely there must be some such property that would allow us to have uniqueness.
Nobody said, that it is not *possible*. The situation is simply that there *is* no yet.

It directly results from multiplication being associative.
...
Non-associativity also causes the negation of the so-called "distributive laws", which are essentially a consequence of the associativity of multiplication in the case of exponentiation.

But there are also non-associative operations, that satisfy the "distributive laws", however not on the natural numbers. Because the addition on the natural numbers is associative and hence
a#(x+y)=a#x * a#y (* and # arbitrary operations) would cause
a#(x+(y+z))=a#x * (a#y * a#z) the operation * also to be associative.
However if we start on a domain with an addition that is not associative, for example the domain of real functions f together with exponentiation as addition then we can easily define higher operations
f[1]g=f^g
f[2](g^h)=f[2]g ^ f[2]h, f[2]x=f
f[3](g^h)=(f[3]g) [2] (f[3]h), f[3]x=f
etc.

as long as you can decompose every function into exponentiations of x (identity function) which is the 1-Element of that domain, for example:

x^x [1] x^x = (x^x)^(x^x)=x^(xx^x)
x [2] x^x = (x [2] x) ^ (x[2]x) = x^x
x^x [2] x^x = (x^x[2]x) ^ (x^x[2]x)= x^x ^ (x^x)
x^x [2] x^(x^x) = (x^x[2]x) ^ (x^x[2]x ^ x^x[2]x) = (x^x)^((x^x)^(x^x))

It is easily to prove that the operation [2], though not commutative, is associative, and \( f[2]g=g\circ f \).

In that way you can start a hierarchy of operations with "distributive law" on the set of real functions, that are made up of exponentiations of the identity function.


Messages In This Thread
RE: Improving convergence of Andrew's slog - by bo198214 - 02/25/2008, 02:00 PM

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