infinity , C , C * , C** , tetration and iterations ( by tommy1729 )
amy666
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 )
C U oo is called C*.
if a function has parameters in C , then that functions usually has a zero in C*.
if not , after analytic continuation of the function considered it usually has.
HOWEVER the story does not end here.
consider the following equation :
2 ^ [x/(x-1)] = 0.
.. take your time ...
have you found a solution in C* ?
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.
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 ...
and 2^ x/(x-1) is smooth and has a solution ...
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
----
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
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.
No, even in that context gamma does not vanish
at -oo.
>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.
>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. It has no zero in C*, period.
(It has an "essential singularity" at oo.)
>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.
>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.)
>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.
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*.
David C. Ullrich
amy666
> On Mon, 16 Jun 2008 18:14:53 EDT, amy666
> <tomm...@hotmail.com>
> 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 ...
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.
regards
tommy1729
mike3
> *** about single valued functions and equations ***
>
> the equation f(x) = 0.
>
> usually x is a finite complex number.
>
> x E C.
>
OK...
> now consider these 2 equations :
>
> exp(x) = 0
> gamma(x) = 0
>
exp(x) = 0 has no solution in C, nor goes gamma(x).
> now x = - oo. ( and no other solutions exist )
>
-oo is not a member of C.
> C U oo is called C*.
>
But -oo is not a member of C*. oo in C* is unsigned,
or projective, infinity, not affine infinity, which
means one can reach it by going in any direction.
exp(oo) in C* is undefined -- exp has an essential
singularity at oo. So does gamma(oo). As a singularity,
the behavior of exp as we go off to oo depends on the
path. There's no limit of exp at oo, or of gamma at oo.
> if a function has parameters in C , then that functions usually has a zero in C*.
>
??? That doesn't make any sense. How does that
follow? f(z) = 1 has parameters "in C", yet has
no zeros in C OR C*.
> if not , after analytic continuation of the function considered it usually has.
>
Again, how's this? See above for my objection.
> HOWEVER the story does not end here.
>
> consider the following equation :
>
> 2 ^ [x/(x-1)] = 0.
>
> .. take your time ...
>
> have you found a solution in C* ?
>
> 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.
>
No, because in C* if x = 1 we get 2^oo, which is undefined
(see above about the singularity at oo and the non-affine
nature of oo.). It has no solutions in C*.
> 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.
>
Again you are using an AFFINE infinity but that is
NOT what is used in C*.
<SNIP, see above.>
mike3
> David wrote :
<snip>
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
>
Because the usage of "oo" in complex analysis and especially
in relation to the Riemann sphere C* means something different
from the way you are using it. Hint hint: it's unsigned,
that is, zoo = oo for all nonzero z. 0oo is undefined.
> 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.
>
?
So you're agreeing with him there, then.
> It has no zero
>
> > in C*, period.
> > (It has an "essential singularity" at oo.)
>
> once again i assume you mean - oo.
>
<snip>
No he means oo, since -oo = oo just as -0 = 0, in
C*. C* has an unsigned (projective) infinity, not
a signed (affine) infinity.
Note that if one approaches oo by going along the
positive axis, exp(z) converges toward oo. If one
goes along the negative axis, it converges toward
0. If one goes along the imaginary axis, it
wobbles forever around the unit circle and does not
converge on anything. All these differing behaviors
mean there is no way to reasonably define a value at
z = oo (that's unsigned, projective infinity, mind
you.).
David C. Ullrich
wrote:
>David wrote :
>
>> On Mon, 16 Jun 2008 18:14:53 EDT, amy666
>> <tomm...@hotmail.com>
>> 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.
No it doesn't.
>
>>
>> >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 ?
Not. I meant exactly what I wrote.
Note the words "in the sense the terminology
is used in complex analysis". When you say
C* is C union infinity there is no oo and -oo
any more, as there are when you're talking
about reals - there's just the one "point at
infinity". And exp(z) simply does not tend to 0
as z tends to infinity.
>
>>
>> >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.
Assume what you want. If we're talking
about complex analysis there simply _is_
no "-oo".
You should consider the facts instead of telling me to consider
your viewpoint. It's simply not true that f(1) gives +oo.
(As x -> 1 from one direction f(x) -> +infinity. As x -> 1
from the other direction f(x) -> -oo.)
No. _If_ C** = C U oo U [ C U oo ]^-1 as
you claim then in fact C** = C* and it does
not contain this mysterious h.
David C. Ullrich
amy666
>
> Again you are using an AFFINE infinity but that is
> NOT what is used in C*.
>
> <SNIP, see above.>
ok.
so i am using AFFINE infinity.
so perhaps i should write C' then instead of C* ?
and this is rather algebra then calculus or analysis.
so C = finite complex
C' is C U oo which accepts AFFINE infinities like - oo and (1+2i)oo
C'' is then both AFFINE and infinitesimal
C'' = C' U 1/C'
C'' has as elements h + i , oo + hi , (2 + h) - (1 -h)i
then exp(x) = 0 has a solution in C' : -oo.
and i also accept gamma(-oo) = 0 although that is discussible ,
i see it as for " some -oo " gamma(-oo) = 0.
no if you reread the OP and keep in mind " some -oo " and ( affine ) C' and C'' as discussed above ;
you will probably find the OP to contain no ( further ) mistakes.
and 1-h is indeed a solution towards 2^(x/(x-1)) = 0.
and the function that is the iteration of that function must be discontinuous.
and my viewpoint that tetration and iterational concepts have to be aware of C''.
regards
tommy1729
David C. Ullrich
<4105666.12137921357...@nitrogen.mathforum.org>,
amy666 <tomm...@hotmail.com> wrote:
> > >
> >
> > Again you are using an AFFINE infinity but that is
> > NOT what is used in C*.
> >
> > <SNIP, see above.>
>
> ok.
>
> so i am using AFFINE infinity.
>
> so perhaps i should write C' then instead of C* ?
>
> and this is rather algebra then calculus or analysis.
>
> so C = finite complex
>
> C' is C U oo which accepts AFFINE infinities like - oo and (1+2i)oo
>
> C'' is then both AFFINE and infinitesimal
>
> C'' = C' U 1/C'
>
> C'' has as elements h + i , oo + hi , (2 + h) - (1 -h)i
>
> then exp(x) = 0 has a solution in C' : -oo.
If you could actually give coherent definitions of what
you mean that might be true.
> and i also accept gamma(-oo) = 0 although that is discussible ,
But _that_ is _still_ simply _false_.
It's funny how _long_ you can hold onto ideas that are
simply wrong.
Look. If "gamma(-oo) = 0" then gamma(x) should be very small
if x is a large negative number, right? Tell me, what is
gamma(-1000000000)? Or just give me a clue approximately
how small it is.
Easier question, which you should probably work out first:
What is gamma(0)? Next: what is gamma(-1)?
> i see it as for " some -oo " gamma(-oo) = 0.
>
> no if you reread the OP and keep in mind " some -oo " and ( affine ) C' and
> C'' as discussed above ;
>
> you will probably find the OP to contain no ( further ) mistakes.
>
> and 1-h is indeed a solution towards 2^(x/(x-1)) = 0.
>
> and the function that is the iteration of that function must be
> discontinuous.
>
> and my viewpoint that tetration and iterational concepts have to be aware of
> C''.
>
> regards
>
> tommy1729
--
David C. Ullrich