Re: infinity , C , C * , C** , tetration and iterations ( by tommy1729 )

David wrote :

On Mon, 16 Jun 2008 18:14:53 EDT, amy666
<tommy1729@xxxxxxxxxxx>
wrote:

*** about single valued functions and equations ***

the equation f(x) = 0.

usually x is a finite complex number.

x E C.

now consider these 2 equations :

exp(x) = 0
gamma(x) = 0

now x = - oo. ( and no other solutions exist )

Yes, if we're talking about real values and
interpreting f(-oo) in the obvious way as a
limit then exp(-oo) = 0.

indeed.


No, even in that context gamma does not vanish
at -oo.

it does not vanish no.

but it approaches 0 better and better.



C U oo is called C*.

It's sometimes called that, yes. Let's call it that,
fine.

:)


if a function has parameters in C , then that
functions usually has a zero in C*.

You've given a few counterexamples to this. The
function exp
does not vanish at oo, in the sense the terminology
is used
in complex analysis.

you probably meant something different from what you wrote , you probably want to replace oo with -oo

or exp with gamma

not ?



if not , after analytic continuation of the function
considered it usually has.

And you don't get anything extra from analytic
continuation of exp,
since it's already an entire function.

exp(z) does not require analytic continuation.

It has no zero
in C*, period.
(It has an "essential singularity" at oo.)

once again i assume you mean - oo.

exp(-oo) = 0

no doubt about it.

exp(-oo + i) = 0 too btw.



HOWEVER the story does not end here.

That's a relief. What you've said so far isn't so.

consider the following equation :

2 ^ [x/(x-1)] = 0.

.. take your time ...

have you found a solution in C* ?

No, and neither have you.

exactly.



if you think you did , you probably aimed for the
simple x = 1.

since you know 2 ^ - oo = 0 => x/(x-1) = -oo

and indeed most math software would agree with you ,
x = 1.

it is clear that 1 is the only " candidate " since

2^ (+ oo + bi) =/= 0 , 2 ^ (c + d oo i) =/= 0
and 2^ ( finite complex ) =/= 0.

thus we reduce the equation to x / (x-1) = -oo + b i

(where b is any number in R U oo ( R*? ) )

now we know abs(x) = oo will not satisfy that
equation.

thus we need a finite number x.

so we need a zero of (x-1).

but x - 1 has only 1 zero , trivially -> 1.

and when take values approaching 1 starting at 0 ,
we keep getting closer to our solution.

if we plot x / (x - 1) is appears x = 1 gives - oo.

For heaven's sake, if you said this in a calculus
class you'd
flunk. If you plot x/(x-1) you see that it approaches
-oo on one side of x=1 and approaches oo on the
other side (which is exactly why x=1 is _not_ a zero
of the function 2^(x/(x-1)), by the way.)

possible, but consider my viewpoint ; the function is single valued and f(1) gives +oo , so cannot also give -oo , unless we add infinitesimals ...



thus 1 is our " solution ".

defining the solution as a limit " seems " to
approach to 1 ; f(0.99999999999) gives a number very
close to 0.

BUT !!!

if you fill in x = 1 , you get 1 / (1-1) which is
oo.

and not - oo !!!

same appears if we plot x/(x-1)

if we look with our eyes for - oo we might conclude
1 gives -oo ...

if we look closer we see 1 seems to give + oo.

it cannot give both , and if we fill in 1 in x/(x-1)
we get +oo.

some have called this ( and be considered idiots for
it ) : actual infinity and potential infinity.

so x/(x-1) contains
actual + infinity and potential - infinity.

conclusion : 2 ^ (x/(x-1)) = 0
does not have a solution in C*.

note that analytic continuation or other logical
continuations do not resolve the issues ;

if we define
F(x) = x / (x-1) for x =/= 1
F(x) = -oo for x = 1.

nice ...

If you say so.

and 2^ x/(x-1) is smooth and has a solution ...

Of course you meant 2^F(x). No, 2^F(x) is _not_
smooth.

no not smooth. sorry.



You should try actually drawing those plots.
The try again, this time drawing them correctly.

but then if F(1) = -oo , then were does it equal +
oo ???

conclusion : to garantee a zero for all welldefined
functions ( not sin(1/x) at 0 , see divergeance and
singularity theory ) we need to extend C*.

what will this extension look like ?

answer : infinitesimal extension.

solve x/(x-1) = -oo => x = 1 - h

(1 - h) / (1 - h - 1) = 1/-h + 1 = -1/h = -oo

( although one can argue that this -oo is potential
rather than actual because i wrote -1/h and not -1/0
, that is another discussion )

lets call this extension C ** which is basicly :

C** = C U oo U [ C U oo ]^-1

That's exactly the same as C*.

thus 1 - h is an element of C* ?

but then we do have a zero in C* for 2^(x/(x+1)) !!!

namely 1 - h.


----

this might clear up things for many people about
infinity , equations and that misunderstood and
underrated potential vs actual infinity ...

and these concepts relate to many branches of math
such as set theory and calculus.

( without weird axioms or cantor ! )

now how does this relate to iterations ? you might
wonder


(and of course since this relates to iterations , it
is something worthwhile considering while dealing
with tetration !!!! )

( btw " disproofs " of tetration in the trend of "
it has radius of convergeance 0 " is not a disproof
at all , since we have e.g. summability methods , the
well known illusionary taylor series for exp(x^2) at
0 also known as " too flat " , functions defined by
signomials instead of taylor series , meromorphic
functions , singularity points etc )

consider the following iteration :

f(x+1) = Q(f(x)) where Q(x) = 2^(x/(x-1))

let f(0) = 0.

then f(1) = 1.

f(2) = oo.

f(3) = 2.

f(4) = 4.

etc

knowing about C** , we now know that for the
negative iterations we dont have values in C*.

f(-1) = 1 - h !!!

but there is more !!

since 1 - h is infinitesimaly close to 1 AND

Q(1 - h) is very different from Q(1) , we now that

f(n) is discontinu at integer values of n !!!

***

in fact iterations of a certain well-defined
function
A(z) , in general , increases the number of poles
and discontinu points.

and quite often do we need extentions : C** to give
consistant meaning to an iteration.

regards

tommy1729

David C. Ullrich

regards

tommy1729
.

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