AREA
amy666
how to express the area of D ?
David C. Ullrich
wrote:
>a function f(z) is holomorhic in (and converges in) domain D.
That's getting closer to making sense. Saying f "converges in D"
is meaningless.
This one is close enough to making sense that I can guess what
you meant. My conjecture is that what you meant is simply
"a function f is holomorphic in domain D"
Or maybe there's something you're not saying, what you have
in mind is some _representation of_ f, as a sum or a product,
that converges in D.
>how to express the area of D ?
But this is still an utterly silly question.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
amy666
> On Wed, 24 Sep 2008 18:46:27 EDT, amy666
> <tomm...@hotmail.com>
> wrote:
>
> >a function f(z) is holomorhic in (and converges in)
> domain D.
>
> That's getting closer to making sense. Saying f
> "converges in D"
> is meaningless.
yes ...
>
> This one is close enough to making sense that I can
> guess what
> you meant. My conjecture is that what you meant is
> simply
>
> "a function f is holomorphic in domain D"
agreed.
>
> Or maybe there's something you're not saying, what
> you have
> in mind is some _representation of_ f, as a sum or a
> product,
> that converges in D.
>
> >how to express the area of D ?
>
> But this is still an utterly silly question.
why is it a silly question to ask for the area of D ???
may i conclude and translate that as : david doesnt know and doesnt care ?
or do you have a valid reason ?
i said holomorhic so its not to general , you can use the parameters of a taylor expansion of any point in D.
unless perhaps the domain D is not connected.
Lets assume D is connected.
perhaps that is your objection ...
> David C. Ullrich
>
> "Understanding Godel isn't about following his formal
> proof.
> That would make a mockery of everything Godel was up
> to."
> (John Jones, "My talk about Godel to the post-grads."
> in sci.logic.)
regards
tommy1729
Robert Israel
amy666 <tomm...@hotmail.com> writes:
No, the objection is that it's a silly question, because the existence
of a function holomorphic in D does not tell us anything useful about D.
It could be any domain at all.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
amy666
not quite , i mean the domain depends on f(z).
> --
> Robert Israel
> isr...@math.MyUniversitysInitials.ca
> Department of Mathematics
> http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver,
> BC, Canada
regards
tommy1729
Robert Israel
Yes, maybe...
OK, the way I can make some sense of your question is this.
Perhaps we are given a holomorphic function element (f,D) and
ask for a maximal connected region of the plane in which
there is a holomorphic function that agrees with f on D. Sometimes
such a thing does exist and has finite area (say if there is a
natural boundary enclosing D).
David C. Ullrich
Which is of course not what he said - _could_ be what he
meant, (or rather it may well be a precise formulation of
what he had in mind).
And of course we should note that it remains a silly question,
asking us to "express" the area of D here.
amy666
good.
that is similar to what i had in mind.
now that the question is ( almost ) understood...
maybe we can really start this thread and find the answer ??
regards
tommy1729
Robert Israel
Before finding the answer, we still need to have a well-formulated
question.
If the question is:
Given a function element, perhaps specified by a Taylor series, find
any natural boundaries for its analytic continuation
I'm pretty sure there is no good general answer. There are various
ways to produce a holomorphic function with a given natural boundary,
and there are certain classes of Taylor series which are known to have
the circle at the radius of convergence as a natural boundary (look up
"Hadamard gaps"). I think it can be shown that a "generic" (in the Baire
category sense) function holomorphic in a given region D will have the
boundary of D as a natural boundary.
amy666
"express" might be understood as " compute " or " give a formula for "
formula for what ? -> area of D.
formula of what variables ? -> i guess the taylor coefficients at a certain point in D.
>
> David C. Ullrich
>
> "Understanding Godel isn't about following his formal
> proof.
> That would make a mockery of everything Godel was up
> to."
> (John Jones, "My talk about Godel to the post-grads."
> in sci.logic.)
regards
tommy1729
David C. Ullrich
wrote:
First, you seemed to have missed the subtleties: If we're given, say,
a Taylor series centered at the origin in general that does _not_
determine a largest region D in which the function is holomotphic.
For example, consider the power series for log(1-z) centered
at the origin:
f(z) = log(1-z) = sum z^n/n
(or maybe there are some minus signs). That has radius of
convergence 1; the disk about the origin with radius 1
is the largest disk centered at the origin where the function
is holomorphic. But in this example there is simply no such
thing as the largest region D such that f extends holmorphically
to D; there are regions D_1 and D_2 such that f extends to
D_1 and f also extends to D_2 but f does not extend to the
union of D_1 and D_2.
It's impossible to say for certain that the sort of thing you're
asking for does not exist, but I would be very surprised if
there was a reasonable way to determine from the coefficients
whether there is such a thing as that largest region D, much
less a formula for the area of D when it does exist.
David C. Ullrich
Seems plausible. We should parhaps add for the benefit of
anyone who might misunderstand that it's not going to be
possible to determine from, say, the coefficients whether
a function element _is_ in that class of "generic" functions.
Where "impossible" means "surely impossible, in any
reasonable way" - of course the answer is determined by
those coefficients, but not in a way that allows a human
being to actually find the answer given the coefficients.
amy666
sigh.
if the origin is part of D , then no , the taylor series does not 'directly' determine the largest region D.
trivial because of concepts like monodromy theory and continuation of a line.
BUT the taylor series does so 'indirectly' ; the taylor series does describe the function f(z) , and that all it needs to do !!
since with f(z) there is a region D defined !!
talk about subtle !!!
>
> For example, consider the power series for log(1-z)
> centered
> at the origin:
no need , the above already shows my point.
>
> f(z) = log(1-z) = sum z^n/n
sum from where to where ?
who always speaks of being more clear and formal ?
besides its wrong , what you had in mind is probably this :
- log ( 1 - z ) = sum n = 1 -> oo [ z^n / n ]
for abs(z) =< 1 APART FROM x = 1
>
> (or maybe there are some minus signs).
didnt you call yourself a professor in math ??
how come you dont know that then ?
sure , you could say , i didnt think about it or didnt compute it , but it aint so hard and a professor in math should know that by heart, simply because of often use !
That has
> radius of
> convergence 1; the disk about the origin with radius
> 1
> is the largest disk centered at the origin where the
> function
> is holomorphic.
not holomorhic for z = 1.
( which has annoyed me many times when considering tetration subjects ...)
But in this example there is simply
> no such
> thing as the largest region D such that f extends
> holmorphically
> to D; there are regions D_1 and D_2 such that f
> extends to
> D_1 and f also extends to D_2 but f does not extend
> to the
> union of D_1 and D_2.
if f(z) is onto :
strategy =
continuation along a line -> all lines ! ( including lines from other lines ) -> surface D.
( if started at a point in D and D is connected )
>
> It's impossible to say for certain that the sort of
> thing you're
> asking for does not exist, but I would be very
> surprised if
> there was a reasonable way to determine from the
> coefficients
> whether there is such a thing as that largest region
> D, much
> less a formula for the area of D when it does exist.
therefore it is intresting.
many of my questions are not so trivial or meaningless as you always assumed.
and they are not just designed to be hard , they are fundamental and might lead to new branches when answered.
>
> >
> >>
> >> David C. Ullrich
> >>
> >> "Understanding Godel isn't about following his
> formal
> >> proof.
> >> That would make a mockery of everything Godel was
> up
> >> to."
> >> (John Jones, "My talk about Godel to the
> post-grads."
> >> in sci.logic.)
> >
> >regards
> >
> >tommy1729
>
> David C. Ullrich
>
> "Understanding Godel isn't about following his formal
> proof.
> That would make a mockery of everything Godel was up
> to."
> (John Jones, "My talk about Godel to the post-grads."
> in sci.logic.)
regards
tommy1729
" statisticly i dont exist " tommy1729
David C. Ullrich
wrote:
That's simply not so. Adding two exclamation points doesn't change
anything.
>talk about subtle !!!
>
>
>>
>> For example, consider the power series for log(1-z)
>> centered
>> at the origin:
>
>no need , the above already shows my point.
>
>
>>
>> f(z) = log(1-z) = sum z^n/n
>
>sum from where to where ?
>
>who always speaks of being more clear and formal ?
>
>besides its wrong , what you had in mind is probably this :
>
>- log ( 1 - z ) = sum n = 1 -> oo [ z^n / n ]
>
>for abs(z) =< 1 APART FROM x = 1
>
>
>>
>> (or maybe there are some minus signs).
>
>didnt you call yourself a professor in math ??
>
>how come you dont know that then ?
I assumed we all knew what that power series was.
Your supposed "corrections" indicate that you _did_
understand exactly what I meant, so it's hard to see
what you're complaining about.
Now answer the question: For that power series,
what is the region D?
Robert Israel
Yes. Mind you, this may also depend on how the original coefficients are
"given": in general there is no hope of computing the limits required for the
determination of a radius of convergence. But maybe we can try using
numerical methods to make a good (non-rigourous) approximation of the radius
of convergence. Then if the answer is "no", i.e. the function element does
have an analytic continuation outside D, there will be some point p of D such
that the radius of convergence of the Taylor series at p is greater than
the distance from p to the boundary of D, and we may expect that our
numerical method will indicate that (and a search should eventually find a
suitable p).
Robert Israel
For your amusement, you might consider the following class of examples.
Let sum_n a_n z^n be your favourite example of a Taylor series which has
the unit circle as its natural boundary. Given any Turing machine M,
let b(M,n) = 1 if M goes at least n steps without halting, 0 otherwise.
Then the series sum_n a_n b(M,n) z^n is either a polynomial (and thus entire)
or has the unit circle as its natural boundary, and there is no algorithm
to determine which it is.
quasi
<isr...@math.MyUniversitysInitials.ca> wrote:
Out of curiosity, how about this question ...
What connected open sets D are possible as maximal domains?
More precisely, which connected open subsets D of C are such that
there is a holomorphic function f: D --> C which cannot be extended to
a holomorphic function g: E --> C where E is a connected open subset
of C and D is a proper subset of E.
quasi
amy666
log is a multivalued function.
im talking about single-valued functions.
regards
tommy1729
Robert Israel
All of them. Given any D, you can construct a sequence {z_n} of points
of D such that the set of limit points of {z_n} is the boundary of D.
Then take a non-constant analytic function f on D such that all f(z_n) = 0.
> quasi
Denis Feldmann
All of them, because of the conformal mapping theorem
>
> quasi
quasi
<isr...@math.MyUniversitysInitials.ca> wrote:
That's wild.
So for example, consider a simple closed curve S in R^2, which is
nowhere smooth and such that the origin is interior to S. Let D be the
region interior to S. Then based on what you say above, there is a
holomorphic function f: D --> C (now regarding D as a subset of C)
which cannot be extended to a holomorphic function in a larger
connected open set. Thus essentially, f "determines" S. But isn't f
completely determined by its power series centered at 0? If so, then
the countably many coefficients of that power series determine f, and
hence indirectly also determine S (a wild curve).
But intuitively, some such sets S should require uncountable many data
values for their specification -- which suggests that I probably
misunderstand the situation.
One of these days I'm going to have to learn some complex analysis.
quasi
quasi
<denis.feldm...@neuf.fr> wrote:
>
>quasi a écrit :
>>
>> Out of curiosity, how about this question ...
>>
>> What connected open sets D are possible as maximal domains?
>>
>> More precisely, which connected open subsets D of C are such that
>> there is a holomorphic function f: D --> C which cannot be extended to
>> a holomorphic function g: E --> C where E is a connected open subset
>> of C and D is a proper subset of E.
>
>All of them, because of the conformal mapping theorem
Thanks.
Ok, to explore the issue, let's take a specific region ...
Let D be the square open region in R^2 defined by
-1 < x < 1
-1 < y < 1
As claimed above, there exists a holomorphic function f : D --> C such
f cannot be extended to a holomorphic function whose domain is an open
connected set E containing D as a proper subset.
2 questions ...
(1) Is there a simple expression for such an f?
(2) Is it necessarily true that, for each z0 in D, the power series
for f, centered at z0, has a disk of convergence which shares a
boundary point with the boundary of D?
quasi
David C. Ullrich
wrote:
The function defined by that power series is an
ordinary single-valued function in the unit disk.
You _seem_ to be agreeing that for that function
there is no largest region D that it extends to -
hard to see why you get so hysterical when I
point this out.
David C. Ullrich
<isr...@math.MyUniversitysInitials.ca> wrote:
Heh-heh. Yes, that's extremely amusing.
For a second it seemed like this might be cheating - have we really
"specified" a power series here? After all, if we say f = 0 if ZF is
consistent and something bad otherwise we've defined f in some
sense, but...
But come to think of it yes, we've specified a power series.
Given M, we have sum_n a_n b(M,n) z^n = sum c_n z^n,
and there _is_ a simple algorithm to calculate c_n for any
given n.
Heh-heh.
David C. Ullrich
Yes, the situation seems strange at first. It may seem less
counterintuitive if you note that it's easy to see that the plane
has exactly c open subsets. (Say O is the collection of open
subsets of the plane and let V be the set of disks with rational
center and rational endpoints. There's an obvious one-to-one
mapping from O into the power set of V.)
> -- which suggests that I probably
>misunderstand the situation.
>
>One of these days I'm going to have to learn some complex analysis.
>
>quasi
David C. Ullrich
Denis Feldmann
> On Mon, 29 Sep 2008 08:26:25 +0200, Denis Feldmann
> <denis.feldm...@neuf.fr> wrote:
>> quasi a écrit :
>>> Out of curiosity, how about this question ...
>>>
>>> What connected open sets D are possible as maximal domains?
>>>
>>> More precisely, which connected open subsets D of C are such that
>>> there is a holomorphic function f: D --> C which cannot be extended to
>>> a holomorphic function g: E --> C where E is a connected open subset
>>> of C and D is a proper subset of E.
>> All of them, because of the conformal mapping theorem
>
> Thanks.
>
> Ok, to explore the issue, let's take a specific region ...
>
> Let D be the square open region in R^2 defined by
>
> -1 < x < 1
> -1 < y < 1
>
> As claimed above, there exists a holomorphic function f : D --> C such
> f cannot be extended to a holomorphic function whose domain is an open
> connected set E containing D as a proper subset.
>
> 2 questions ...
>
> (1) Is there a simple expression for such an f?
Not in elementary terms. See "Riemann mapping theorem", at
http://en.wikipedia.org/wiki/Riemann_mapping_theorem ; especially
http://en.wikipedia.org/wiki/Riemann_mapping_theorem#Why_is_this_theorem_impressive.3F
>
> (2) Is it necessarily true that, for each z0 in D, the power series
> for f, centered at z0, has a disk of convergence which shares a
> boundary point with the boundary of D?
Yes (this, otoh, is very elementary, using only Abel's lemma (or
plausible arguments of convergence))
>
> quasi
Robert Israel
> On Mon, 29 Sep 2008 08:26:25 +0200, Denis Feldmann
> <denis.feldm...@neuf.fr> wrote:
> >
> >quasi a écrit :
> >>
> >> Out of curiosity, how about this question ...
> >>
> >> What connected open sets D are possible as maximal domains?
> >>
> >> More precisely, which connected open subsets D of C are such that
> >> there is a holomorphic function f: D --> C which cannot be extended to
> >> a holomorphic function g: E --> C where E is a connected open subset
> >> of C and D is a proper subset of E.
> >
> >All of them, because of the conformal mapping theorem
>
> Thanks.
>
> Ok, to explore the issue, let's take a specific region ...
>
> Let D be the square open region in R^2 defined by
>
> -1 < x < 1
> -1 < y < 1
>
> As claimed above, there exists a holomorphic function f : D --> C such
> f cannot be extended to a holomorphic function whose domain is an open
> connected set E containing D as a proper subset.
>
> 2 questions ...
>
> (1) Is there a simple expression for such an f?
Let g(z) = sum_{n=0}^infty z^{2^n}, converging in the open unit disk and with
a natural boundary on the unit circle (or if you want a "named" function, use
one of the Jacobi theta functions, which also has the unit circle as natural
boundary). Let
f(z) = g(z/(z+2i)) + g(z/(z-2i)) + g(z/(z+2)) + g(z/(z-2)).
> (2) Is it necessarily true that, for each z0 in D, the power series
> for f, centered at z0, has a disk of convergence which shares a
> boundary point with the boundary of D?
Yes.
Robert Israel
> quasi a écrit :
> > Out of curiosity, how about this question ...
> >
> > What connected open sets D are possible as maximal domains?
>
>
>
> >
> > More precisely, which connected open subsets D of C are such that
> > there is a holomorphic function f: D --> C which cannot be extended to
> > a holomorphic function g: E --> C where E is a connected open subset
> > of C and D is a proper subset of E.
>
>
>
> All of them, because of the conformal mapping theorem
> >
> > quasi
I can see how that takes care of simply-connected D, but how do you deal
with the general case?
Denis Feldmann
> Denis Feldmann <denis.feldm...@neuf.fr> writes:
>
>> quasi a écrit :
>
>>> Out of curiosity, how about this question ...
>>>
>>> What connected open sets D are possible as maximal domains?
>>
>>
>>> More precisely, which connected open subsets D of C are such that
>>> there is a holomorphic function f: D --> C which cannot be extended to
>>> a holomorphic function g: E --> C where E is a connected open subset
>>> of C and D is a proper subset of E.
>>
>>
>> All of them, because of the conformal mapping theorem
>>> quasi
>
> I can see how that takes care of simply-connected D, but how do you deal
> with the general case?
amy666
the function log is multivalued , but ok lets consider the taylor series which converges to a single value.
since that taylor series converges in the unit disk , it has AREA pi , at least so it seems ...
but we might expand the function elsewhere ...
and perhaps get ' another ' D.
i guess this is your issue ?
the AREA is oo if we do not have a natural boundary.
hmm
intresting ;
seems i could say:
-------------
f(z) is only meromorphic in a single connected domain D without any holes.
given such a f(z) by a taylor expansion at a non-pole point in D , what is the area of domain D ?
-------------
regards
tommy1729
David C. Ullrich
wrote:
The issue is that the only way we've been able to make sense of
your question is to assume you're talking about extending the
function to the _largest_ domain D, and in this case (which
is actually very common) there is no such largest set.
>the AREA is oo if we do not have a natural boundary.
>
>hmm
>
>intresting ;
>
>seems i could say:
>
>-------------
>
>f(z) is only meromorphic in a single connected domain D without any holes.
Meaning that there's only one such domain D? That never happens.
amy666
i see your still confused about the " largest D ".
ok , ill give an example :
suppose f(z) expanded at x = 0 has a radius of convergeance r = 1.
lets call that set of points A.
and that same f(z) expanded at x = 1 has a radius of convergeance r = sqrt(2).
lets call that set of points B.
now if f(z) is expanded at any other point then those in set A or B then f(z) does not converge.
( = condition of the example , not a conclusion that follows of course )
thus the Mittag-Leffler star is the domain D and has an area of A U B ; the area of the two circles minus its overlapping area.
regards
tommy1729
David C. Ullrich
<24702684.1222801834...@nitrogen.mathforum.org>,
amy666 <tomm...@hotmail.com> wrote:
You have no idea how funny it is, for you to say that I'm
confused about that.
> ok , ill give an example :
>
> suppose f(z) expanded at x = 0 has a radius of convergeance r = 1.
>
> lets call that set of points A.
>
> and that same f(z) expanded at x = 1 has a radius of convergeance r =
> sqrt(2).
>
> lets call that set of points B.
>
> now if f(z) is expanded at any other point then those in set A or B then f(z)
> does not converge.
> ( = condition of the example , not a conclusion that follows of course )
You're still not quite making sense, talking about "expanding f" at other
points.
Never mind that. What you meant to say was that f cannot be extended
to a domain larger than A union B. Fine, if we _assume_ that this happens
then D = A union B is that largest domain that we've been trying to
talk about.
I understand that. What you may or may not yet understand is that
often it _doesn't work this way_.
> thus the Mittag-Leffler star is the domain D and has an area of A U B ; the
> area of the two circles minus its overlapping area.
If you want to talk about the M-L star then say so. The M-L star does
always exist (it's _not_ the largest domain in which the function
is holomorphic).
> regards
>
> tommy1729
--
David C. Ullrich
amy666
> In article
> <24702684.1222801834...@nitrogen.math
if you understand that , you have to admit my question is not meaningless.
What you may or may not yet
> understand is that
> often it _doesn't work this way_.
i may or may not understand ...
depends also on what your talking about with " doesnt work ".
it doesnt work that way if there are several unconnected areas , but we assume f(z) is meromorphic in a single connected domain D (thus nowhere else).
also , 2 unions might be insufficient , i am well aware of that and also circles are 'naive'...
but it was a very simple example , just to show what is meant by ' domain D ' , ' f(z) ' and AREA.
>
> > thus the Mittag-Leffler star is the domain D and
> has an area of A U B ; the
> > area of the two circles minus its overlapping area.
>
> If you want to talk about the M-L star then say so.
> The M-L star does
> always exist (it's _not_ the largest domain in which
> the function
> is holomorphic).
how do you mean its not the largest domain in which the function is holomorphic ?
if f(z) is holomorphic ( has no poles anywhere ) and converges only in a single and connected domain D , which subset E of D is not an element of the M-L star ???
>
> > regards
> >
> > tommy1729
>
> --
> David C. Ullrich
regards
tommy1729
Denis Feldmann
> ok , ill give an example :
>
> suppose f(z) expanded at x = 0 has a radius of convergeance r = 1.
>
> lets call that set of points A.
>
> and that same f(z) expanded at x = 1 has a radius of convergeance r = sqrt(2).
>
> lets call that set of points B.
>
> now if f(z) is expanded at any other point then those in set A or B then f(z) does not converge.
> ( = condition of the example , not a conclusion that follows of course )
Especially as there is no way to show f is like that by inspection of
its taylor series (actually, I bet you could never *construct* such an f)
Denis Feldmann
Just that. Perhpas you should try learning some of that stuff...
>
> if f(z) is holomorphic ( has no poles anywhere ) and converges only in a single and connected domain D
Exacttly. But this doesntwok for all holomorphic functions. Hint :use
the function defined by f(z)=sum z^n/n (n>0), start around 0, and try to
determinate the M-L star. Then , find D ...
David C. Ullrich
wrote:
>David wrote :
>
>> In article
>> <24702684.1222801834...@nitrogen.math
>> forum.org>,
>> amy666 <tomm...@hotmail.com> wrote:
>>[...]
Nope. The statement that I understood had various _hypotheses_ that
were not in the original question. In particular, in the example you
gave there _is_ a largest domain D in which the function is
holomorphic; that simply doesn't happen in general, so when
you start with that power series and ask for the area of the
largest domain where the function is holomorphic the
question is meaningless.
>
> What you may or may not yet
>> understand is that
>> often it _doesn't work this way_.
>
>i may or may not understand ...
>
>depends also on what your talking about with " doesnt work ".
>
>it doesnt work that way if there are several unconnected areas , but we
>assume f(z) is meromorphic in a single connected domain D (thus nowhere else).
You continually change the assumptions.
>also , 2 unions might be insufficient , i am well aware of that and also circles are 'naive'...
>
>but it was a very simple example , just to show what is meant by ' domain D ' , ' f(z) ' and AREA.
>
>
>
>
>>
>> > thus the Mittag-Leffler star is the domain D and
>> has an area of A U B ; the
>> > area of the two circles minus its overlapping area.
>>
>> If you want to talk about the M-L star then say so.
>> The M-L star does
>> always exist (it's _not_ the largest domain in which
>> the function
>> is holomorphic).
>
>how do you mean its not the largest domain in which the function is holomorphic ?
I mean the M-L star is not always, in fact not typically, the largest
domain in which the function is holomorphic. Exactly what I
said. If S is the M-L star it will typically happen that there exists
D such that S is a proper subset of D and the function is holomorphic
in D.
>if f(z) is holomorphic ( has no poles anywhere ) and converges only in a single
>and connected domain D , which subset E of D is not an element of the M-L star ???
You never seem to tire of talking about things you don't understand -
the fact that you don't understand something never stops you from
issuing patronizing corrections to people who do understand.
That's not the answer to your question, just a comment inspired
by the question: the question _shows_ that you don't understand
the definition of the M-:L star, (or that you seriously misunderstand
some other aspect of all this, hard to be certain). To answer your
question:
Say D is the union of the two disks defined by |z| < 1 and
|z - 9/5| < 1. Say we're talking about the M-L star with
respect to the origin. Say f has the boundary of D for a
natural boundary. Any point z in D with the property that
the line segment from 0 to z does not lie entirely in D is
not in the M-L star. (That's exactly the _definition_!
You have to learn not to use words in mathematics if
you don't know the definition, if you want to stop making
such a fool of yourself.)
>
>>
>> > regards
>> >
>> > tommy1729
>>
>> --
>> David C. Ullrich
>
>regards
>
>tommy1729
David C. Ullrich
amy666
functions like 1/x or log(x) are defined for all z,
thus their area = oo.
they DONT even have a boundary !
get it now ?
>
>
> , which subset E of D is not an element of the M-L
> -L star ???
> >
> >
> >>> regards
> >>>
> >>> tommy1729
> >> --
> >> David C. Ullrich
> >
> > regards
> >
> > tommy1729
regards
tommy1729
amy666
no that is your misunderstanding.
In particular, in
> the example you
> gave there _is_ a largest domain D in which the
> function is
> holomorphic;
thus showing my question is not meaningless afterall !
and my example shows that i do know what im talking about.
> that simply doesn't happen in general.
given the conditions i gave , that does happen.
> but if it does happen
> so when
> you start with that power series and ask for the area
> of the
> largest domain where the function is holomorphic the
> question is meaningless.
so even if it happens , and you know it happens , you still consider the question meaningless ??
you are so stubborn.
even when i give a valid example that even you can understand , you still wont admit !
>
> >
> > What you may or may not yet
> >> understand is that
> >> often it _doesn't work this way_.
> >
> >i may or may not understand ...
> >
> >depends also on what your talking about with "
> doesnt work ".
> >
> >it doesnt work that way if there are several
> unconnected areas , but we
> >assume f(z) is meromorphic in a single connected
> domain D (thus nowhere else).
>
> You continually change the assumptions.
no , i use different words in attempts of making it clear to you.
im still with the original idea of the OP.
thrust me , you are the one who doesnt understand.
wrong again david.
sure you understand the M-L star.
that is clear.
but so do i.
so in your answer you talked about the M-L star constructed from point 0.
you talk about the line segments from 0 to z.
indeed that is the way to construct the M-L star from the point 0.
and indeed that is not the domain D.
so what is the difference between your construction and mine ?
well , for start i construct M-L star from every point.
and the unions of those M-L stars give the domain D.
so for instance , lets start at 0.
we make a line segment 0 till z1 by continuation along a line.
now i also make L-F star starting from z1.
thus i will also have the line segment z1 too z2.
when repeating oo i will have a L-F star from the points
0 , z1 , z2 , ...
and together they form domain D.
and then we have AREA of domain D.
its that simple.
>
> >
> >>
> >> > regards
> >> >
> >> > tommy1729
> >>
> >> --
> >> David C. Ullrich
> >
> >regards
> >
> >tommy1729
>
> David C. Ullrich
>
> "Understanding Godel isn't about following his formal
> proof.
> That would make a mockery of everything Godel was up
> to."
> (John Jones, "My talk about Godel to the post-grads."
> in sci.logic.)
regards
tommy1729
amy666
> amy666 a écrit :
> > ok , ill give an example :
> >
> > suppose f(z) expanded at x = 0 has a radius of
> convergeance r = 1.
> >
> > lets call that set of points A.
> >
> > and that same f(z) expanded at x = 1 has a radius
> of convergeance r = sqrt(2).
> >
> > lets call that set of points B.
> >
> > now if f(z) is expanded at any other point then
> those in set A or B then f(z) does not converge.
> > ( = condition of the example , not a conclusion
> that follows of course )
>
> Especially as there is no way to show f is like that
> by inspection of
> its taylor series (actually, I bet you could never
> *construct* such an f)
( even stronger , i bet you can construct such an f yourself , though it might look ' ugly ' )
>
> >
> > thus the Mittag-Leffler star is the domain D and
> has an area of A U B ; the area of the two circles
> minus its overlapping area.
> >
> > regards
> >
> > tommy1729
regards
tommy1729
David C. Ullrich
wrote:
Over and over you say things that are false or meaningless,
then you say that what you meant was something else.
It's like if I told you that hgu frtoble seg. You correctly
point out that that's nonsense. I "explain" that when I
said that I meant ghrtlble semiflop. You point out that
that's nonsense. I say no, you just don't understand -
what I meant was that if n is odd and m is even then
n + m is even. You point out that that makes sense
but it's false. I explain again that you just don't
understand - by "odd" I mean 0, and by "even"
I mean 3.
So now somehow that fact that 0 + 3 = 3 is
supposed to mean that I was right all along when
I aaid hgu frtoble seg.
amy666
its pathetic , you still dont understand.
i assumed you understood M-L star , apparently you dont , or dont want too ...
David C. Ullrich
wrote:
Sorry, guy. For most of this thread the things you said made
no sense, because you were talking about things that simply
don't exist. When you mentioned the M-L star it seemed possible
that that's what you really meant to be talking about all the time.
Now when you start talking about "the M-L star with respect
to all points" you're really back to talking about an arbitrary
open set again. And one that doesn't exist.