assumptions about smooth functions ...
amy666
. because there are many more smooth functions then analytic functions ...
or does such a general series expansion exist ?
what about subclasses of smooth functions R -> R, like not including bump functions ?
and what about smooth continuation ?
for clarity with smooth i mean C^oo.
i hope my question is clear.
i feel a bit silly asking this :/
subsets of smooth functions seem confusing :s
thanks in advance
regards
tommy1729
ps : this might relate to tetration ... or not ...
Dave L. Renfro
> i assume that there does not exist a series expansion
> for smooth functions R -> R ?
>
> . because there are many more smooth functions then
> analytic functions ...
>
> or does such a general series expansion exist ?
[snip]
> for clarity with smooth i mean C^oo.
In May 2002 I posted a 2-part essay on this topic. I posted
from the Math Forum sci.math web pages, and for some reason
these two posts never got archived at google's sci.math
web pages. Also, some accented letters (e.g. in some French
words) have gotten garbled somehow in one of the Math
Forum's site migrations since then. (The accents were
fine a year or two after the posts appeared.)
ESSAY ON NOWHERE ANALYTIC C-INFINITY FUNCTIONS (9 May 2002)
http://mathforum.org/kb/thread.jspa?messageID=387148
ESSAY ON NOWHERE ANALYTIC C-INFINITY FUNCTIONS (19 May 2002)
http://mathforum.org/kb/message.jspa?messageID=387149
Dave L. Renfro
_ Prof. Jonez _
werds
"amy666" <tomm...@hotmail.com> wrote in message
news:15219605.1218985908...@nitrogen.mathforum.org...
>i assume that there does not exist a blah blah blah blah blah blah blah
amy666
> Eat balls, you spamming mathforum imbecile. and
> lern two rite ingilsh
> werds
>
> "amy666" <tomm...@hotmail.com> wrote in message
> news:15219605.1218985908033.JavaMail.jakarta@nitrogen.
> mathforum.org...
> >i assume that there does not exist a blah blah blah
> blah blah blah blah
>
>
yeah sure , your a real prof !
as if.
amy666
> amy666 wrote:
>
> > i assume that there does not exist a series
> expansion
> > for smooth functions R -> R ?
> >
> > . because there are many more smooth functions then
> > analytic functions ...
> >
> > or does such a general series expansion exist ?
>
> [snip]
>
> > for clarity with smooth i mean C^oo.
>
> In May 2002 I posted a 2-part essay on this topic.
thanks for that Dave.
i cant believe that didnt get a reply on a mathforum !
typical sci.math , when you post some REAL math you dont get replies.
only insults or misunderstanding cantors diagonal arguments gets over 500 replies.
but in Dave's case , he was ' loonely ' in his thread.
above sci.math level i assume.
oh well.
I
> posted
> from the Math Forum sci.math web pages, and for some
> reason
> these two posts never got archived at google's
> sci.math
> web pages.
tss typical !
Also, some accented letters (e.g. in some
> French
> words) have gotten garbled somehow in one of the Math
> Forum's site migrations since then. (The accents were
> fine a year or two after the posts appeared.)
>
> ESSAY ON NOWHERE ANALYTIC C-INFINITY FUNCTIONS (9 May
> 2002)
> http://mathforum.org/kb/thread.jspa?messageID=387148
>
> ESSAY ON NOWHERE ANALYTIC C-INFINITY FUNCTIONS (19
> May 2002)
> http://mathforum.org/kb/message.jspa?messageID=387149
>
> Dave L. Renfro
>
thank you Dave.
although my questions are not really answered.
after a very brief reading , is it correct to conclude that the set of smooth functions is not much bigger then that of the analytic functions ?
didnt i read there are never an uncountable amount of intersections between a smooth function and an analytic one ?
thus in theory such a series expansion could exist ?
for smooth functions were the left lim derivatives = the right lim derivatives, could that series be of the form
triple sum n = 1 -> oo , m = 1 -> oo , i = 1 -> oo
a_n [x + a_i] * b_m [x + b_i]
where a_n is a signomial, b_m is the mth natural logaritm and a_i and b_i are constants ?
and thus smooth functions where for every real x the left lim derivatives = the right lim derivatives, has cardinality R ^ ( N ^ 3 ) = R ?
high regards
tommy1729
Robert Israel
> after a very brief reading , is it correct to conclude that the set of
> smooth functions is not much bigger then that of the analytic functions ?
No it isn't. Almost all smooth functions (in the Baire category sense) are
nowhere analytic.
> didnt i read there are never an uncountable amount of intersections between
> a smooth function and an analytic one ?
I don't know if you read it, but it's not true. For example, see the thread
"C^infinity function c-to-1" from 2005
<http://groups.google.ca/group/sci.math.research/browse_thread/thread/8095a6884b78f7b2>
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
amy666
> amy666 <tomm...@hotmail.com> writes:
>
>
> > after a very brief reading , is it correct to
> conclude that the set of
> > smooth functions is not much bigger then that of
> the analytic functions ?
>
> No it isn't. Almost all smooth functions (in the
> Baire category sense) are
> nowhere analytic.
does ' nowhere ' mean measure 0 or aleph_0 ?
what if we dont allow bump functions ?
( may i call all smooth functions that dont have identical left and right limits for all their derivatives bump functions or is that the wrong term ?
its what i meant anyways, i assume both are equivalent ? )
>
> > didnt i read there are never an uncountable amount
> of intersections between
> > a smooth function and an analytic one ?
>
> I don't know if you read it, but it's not true. For
> example, see the thread
> "C^infinity function c-to-1" from 2005
> <http://groups.google.ca/group/sci.math.research/brows
> e_thread/thread/8095a6884b78f7b2>
thanks.
again what if we restrict to non-bump functions ?
> --
> Robert Israel
> isr...@math.MyUniversitysInitials.ca
> Department of Mathematics
> http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver,
> BC, Canada
regards
tommy1729
The World Wide Wade
<26303724.1219008933...@nitrogen.mathforum.org>,
amy666 <tomm...@hotmail.com> wrote:
> Robert Israel wrote :
>
> > amy666 <tomm...@hotmail.com> writes:
> >
> >
> > > after a very brief reading , is it correct to
> > conclude that the set of
> > > smooth functions is not much bigger then that of
> > the analytic functions ?
> >
> > No it isn't. Almost all smooth functions (in the
> > Baire category sense) are
> > nowhere analytic.
>
> does ' nowhere ' mean measure 0 or aleph_0 ?
Neither. It means "nowhere", as in "at no point".
> what if we dont allow bump functions ?
>
> ( may i call all smooth functions that dont have identical left and right
> limits for all their derivatives bump functions or is that the wrong term ?
> its what i meant anyways, i assume both are equivalent ? )
Not sure what you're trying to say here. If f is C^oo on R, bump
function or not, then each derivative of f is continuous on R, hence
each derivative of f has identical left and right limits at each point
of R.
David C. Ullrich
wrote:
>Robert Israel wrote :
>
>> amy666 <tomm...@hotmail.com> writes:
>>
>>
>> > after a very brief reading , is it correct to
>> conclude that the set of
>> > smooth functions is not much bigger then that of
>> the analytic functions ?
>>
>> No it isn't. Almost all smooth functions (in the
>> Baire category sense) are
>> nowhere analytic.
>
>does ' nowhere ' mean measure 0 or aleph_0 ?
It means nowhere.
>what if we dont allow bump functions ?
>
>( may i call all smooth functions that dont have identical left and
>right limits for all their derivatives bump functions or is that the wrong term ?
>its what i meant anyways, i assume both are equivalent ? )
The word "smooth" is used in various different ways.
We've been assuming you meant "infinitely differentiable",
since that's probably the most common meaning. But
of course an infinitely differentiable function and all
its derivatives do have the same right and left limits
at every point.
What _do_ you mean by "smooth" here?
>
>>
>> > didnt i read there are never an uncountable amount
>> of intersections between
>> > a smooth function and an analytic one ?
>>
>> I don't know if you read it, but it's not true. For
>> example, see the thread
>> "C^infinity function c-to-1" from 2005
>> <http://groups.google.ca/group/sci.math.research/brows
>> e_thread/thread/8095a6884b78f7b2>
>
>thanks.
>
>again what if we restrict to non-bump functions ?
>
>
>> --
>> Robert Israel
>> isr...@math.MyUniversitysInitials.ca
>> Department of Mathematics
>> http://www.math.ubc.ca/~israel
>> University of British Columbia Vancouver,
>> BC, Canada
>
>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.)
David C. Ullrich
<18005268.1219004899...@nitrogen.mathforum.org>,
amy666 <tomm...@hotmail.com> wrote:
> [...]
>
> didnt i read there are never an uncountable amount of intersections between a
> smooth function and an analytic one ?
A much simpler (and less interesting) "no" than the
one Israel gave: The 0 function is analytic, and
any closed subset of R is the zero set of some
smooth function.
(At least for every definition of "smooth" I've ever seen;
as of this morning we don't know what _you_ mean by
"smooth" here.)
> thus in theory such a series expansion could exist ?
>
> for smooth functions were the left lim derivatives = the right lim
> derivatives, could that series be of the form
>
> triple sum n = 1 -> oo , m = 1 -> oo , i = 1 -> oo
> a_n [x + a_i] * b_m [x + b_i]
>
> where a_n is a signomial, b_m is the mth natural logaritm and a_i and b_i are
> constants ?
>
> and thus smooth functions where for every real x the left lim derivatives =
> the right lim derivatives, has cardinality R ^ ( N ^ 3 ) = R ?
>
>
> high regards
>
> tommy1729
--
David C. Ullrich
amy666
ill have to make a restatement ...
regards
tommy1729
amy666
first a conjecture :
a function f(z) is entire apart from potential removable singularities if
1) it converges for all of z
2) it has no poles
3) it has no essential singularities
4) it is Coo
5) its first derivate is nowhere oo
6) it is Goo , analogue to Coo where G is gaussian curvature
7) the equation f(z) = A has a solution for all A apart from perhaps a single exception
8) f'(z) has the same 7 properties above
9) G_1 f(z) has the same 8 properties as above
regards
tommy1729
David C. Ullrich
wrote:
>> i will have to be more clear and specific about my
>> questions ...
>>
>> ill have to make a restatement ...
>>
>> regards
>>
>> tommy1729
>
>first a conjecture :
Not really, just nonsense.
>a function f(z) is entire apart from potential removable singularities if
What is a "potential" removable singularity?
>1) it converges for all of z
What does it mean for a function to converge at z?
>2) it has no poles
>3) it has no essential singularities
>4) it is Coo
I can't imagine a class of functions for which it really
makes sense to "assume" all of the previous three assumptions.
>5) its first derivate is nowhere oo
>6) it is Goo , analogue to Coo where G is gaussian curvature
>7) the equation f(z) = A has a solution for all A apart from perhaps a single exception
>8) f'(z) has the same 7 properties above
>9) G_1 f(z) has the same 8 properties as above
>
>
>
>regards
>
>tommy1729
David C. Ullrich
amy666
> On Wed, 01 Oct 2008 17:05:43 EDT, amy666
> <tomm...@hotmail.com>
> wrote:
>
> >> i will have to be more clear and specific about my
> >> questions ...
> >>
> >> ill have to make a restatement ...
> >>
> >> regards
> >>
> >> tommy1729
> >
> >first a conjecture :
>
> Not really, just nonsense.
>
> >a function f(z) is entire apart from potential
> removable singularities if
>
> What is a "potential" removable singularity?
>
> >1) it converges for all of z
>
> What does it mean for a function to converge at z?
>
> >2) it has no poles
> >3) it has no essential singularities
> >4) it is Coo
>
> I can't imagine a class of functions for which it
> really
> makes sense to "assume" all of the previous three
> assumptions.
why not ?????
David C. Ullrich
wrote:
>David wrote :
>
>> On Wed, 01 Oct 2008 17:05:43 EDT, amy666
>> <tomm...@hotmail.com>
>> wrote:
>>
>> >> i will have to be more clear and specific about my
>> >> questions ...
>> >>
>> >> ill have to make a restatement ...
>> >>
>> >> regards
>> >>
>> >> tommy1729
>> >
>> >first a conjecture :
>>
>> Not really, just nonsense.
>>
>> >a function f(z) is entire apart from potential
>> removable singularities if
>>
>> What is a "potential" removable singularity?
>>
>> >1) it converges for all of z
>>
>> What does it mean for a function to converge at z?
>>
>> >2) it has no poles
>> >3) it has no essential singularities
>> >4) it is Coo
>>
>> I can't imagine a class of functions for which it
>> really
>> makes sense to "assume" all of the previous three
>> assumptions.
>
>why not ?????
You answer my question first.
amy666
'potential' as in 'might exist'
David C. Ullrich
wrote:
Because we talk about poles and essential singularities in
the context of holomorphic functions, which _are_ Coo;
if we have a holomophic function then "assuming" that
it's Coo makes no sense.
>> You answer my question first.
>>
>> >
>> >>
>> >> >5) its first derivate is nowhere oo
Same comment here, given that we've already
assumed there are no isolated singularities.
amy666
no we dont assume holomorphic from the start !
we conjecture holomorphic follows !
Denis Feldmann
> sigh !
>
> no we dont assume holomorphic from the start !
>
> we conjecture holomorphic follows !
Then it would be good 1) to learn what it means 2) to express yourself
more clearly 3) to learn a few theorems (like Cauchy's) 4) to sigh a
little cless...
amy666
i know what it means !!
you guys are confused by it , not me !
Achava Nakhash, the Loving Snake
Temper, temper.
The problem with what you write, Tommy, is that use often write
phrases that make no sense to those who actually know the mathematical
technical meanings of what you write, and so it becomes really
difficult to tell what you are really on about. We have to guess, and
only you can tell if they are accurate guesses. With that in mind,
lets go through your list, which I will paste in here since it wasn't
in what you replied to:
>a function f(z) is entire apart from potential removable singularities if
>1) it converges for all of z
This one, for example, makes no sense as stated. Sequences converge,
series converge, you can talk about f(z) converging to a limit as z
goes to a. This last is the only one that has any hope of being what
you mean, when put into the form, f(z) coverges to some limit
(depending on z, of course) as z goes to a for every complex a.
Is that what you meant? Or did you mean something to the effect that
f(z) is defined as a power series, and that series converges for every
complex z. A complex-valued function of a complex variable is, after
all, just a set of ordered pairs of complex numbers and can be
provided in a very large variety of ways. If you meant that the power
series converges everywhere, then there is nothing to discuss, as then
the function is already known to be entire.
>2) it has no poles
I have not seen the concept of pole discussed outside of the context
of a function that holomorphic at a lot of places. The pole is then
an isolated singularity, the center of some circle of radius say, r,
inside of which, except for the center, it is holomorphic. It is
unclear to me what a pole is for the most general complex-valued
function of a complex variable. What do you mean by it? Without that
information, no one can proceed.
>3) it has no essential singularities
A comment similar to the one for condition 2). What do YOU mean by an
essential singularity?
>4) it is Coo
This one is clear enough. Obviously that is a necessary condition,
and you understand that it is not a sufficient condition .
>5) its first derivate is nowhere oo
I assume you mean derivative. If a function is Coo, then 5) is
already excluded, as it would not even be differentiable at such a
point. I have to assume that you mean that it is Coo outside of some
isolated "potential removable singularities." Since there is no such
thing as a potential removable singularity, as a previous poster
noted, I assume that is just a bit of tangled English and you really
meant to allow "some removable singularities." There is still a
problem, as it is unclear what a removable singularity might be in
the context of a general complex-valued function of a complex
variable. How does this sound as a definition of removable
singularity? A removable singularity is a point a in the complex
plane where f(a) is defined, and the limit as z approaches a of f(z)
is b, where b is not equal to f(a). You should then specify what you
mean by "some" removable singularities. A finite set? A countable
set? What?
>6) it is Goo , analogue to Coo where G is gaussian curvature
How do you define the Gaussian curvature of your function? I have
only seen Gaussian curvature in the context of surfaces in the plane
which really doesn't work here. It is always a real number in the
definitions I have seen. However, I am a babe in the woods as far as
differential geometry is concerned so I am willing to concede that my
ignorance of such things is do to my ignorance and not to yours. But
I won't concede it unless you present a definition or a reference to a
place where it can be found.
That being said, your version of Gaussian curvature would have to give
you a complex number associated with every complex number which is to
say a function g(z) which is also a complex-valued function of a
complex variable.
>7) the equation f(z) = A has a solution for all A apart from perhaps a single exception
So you have seen Picard's little theorem and make its conclusion about
entire functions a part of your precondition for their existence.
Nothing wrong with that.
>8) f'(z) has the same 7 properties above
Nothing wrong with this one.
>9) G_1 f(z) has the same 8 properties as above
I assume you mean that G_1 f(z) is your version of the Gaussian
curvature of f at z. In other words, it is what I called g(z).
I hope that, after having read my detailed comments about the
conditions of your conjecture you are beginning to realize the
inadequacy of your presentation. It is impossible to discuss your
conjecture because, as stated, it doesn't make sense. There is
nothing to work with because nobody can figure out what you mean. You
have given some evidence of actually being interested in mathematics
which is why I am giving you this help. I don't think that you are a
troll. I hope you can appreciate how frustrating and annoying it can
be to deal with language such as yours which is so loose as to be
impossible to turn into anything meaningful without further
clarification and which looks like it might be impossible to fix up
anyway even with greater clarificationl. In other words, if you try
to fix up the problems I have discussed, you find that your thinking
was flawed and you will have to go back to the drawing board.
Regards,
Achava
David C. Ullrich
<27683514.1223479567...@nitrogen.mathforum.org>,
amy666 <tomm...@hotmail.com> wrote:
> sigh !
>
> no we dont assume holomorphic from the start !
If not then saying "no poles" and "no essential singularities"
simple makes no sense.
> we conjecture holomorphic follows !
--
David C. Ullrich
amy666
there is no guessing.
the goal and meaning of my post is unique and quite well defined.
the details might not been given , but they can be achieved by some work without the need of guessing.
i am very well aware of the words i use , it might not sound simple , but thats tough , i wont change the words , in fact i cant ; without changing the meaning and nontriviality of the question.
if you want easier questions , im not your guy ...
if you want traditional questions , get a math book.
my questions are uncommon and original and thus might sound weird and unfamiliar but they are correct !
my style is implicit rather than explicit...
>
> >a function f(z) is entire apart from potential
> removable singularities if
>
> >1) it converges for all of z
>
> This one, for example, makes no sense as stated.
> Sequences converge,
> series converge, you can talk about f(z) converging
> to a limit as z
> goes to a. This last is the only one that has any
> hope of being what
> you mean, when put into the form, f(z) coverges to
> some limit
> (depending on z, of course) as z goes to a for every
> complex a.
> Is that what you meant?
when i said convergeance , i indeed meant convergeance
too a ? to whatever f(z) converges !
when i said no poles ; no poles -> no convergeance to oo.
( since hey pole of 1/x is oo right ? )
Or did you mean something to
> the effect that
> f(z) is defined as a power series, and that series
> converges for every
> complex z. A complex-valued function of a complex
> variable is, after
> all, just a set of ordered pairs of complex numbers
> and can be
> provided in a very large variety of ways. If you
> meant that the power
> series converges everywhere, then there is nothing to
> discuss, as then
> the function is already known to be entire.
>
> >2) it has no poles
>
> I have not seen the concept of pole discussed outside
> of the context
> of a function that holomorphic at a lot of places.
do you know any ' finite poles ' ?
again pole = oo
see previous reply.
note that no other poles exist !
apart from the north and south pole and perhaps the citizens of poland ...
> The pole is then
> an isolated singularity, the center of some circle of
> radius say, r,
> inside of which, except for the center, it is
> holomorphic. It is
> unclear to me what a pole is for the most general
> complex-valued
> function of a complex variable. What do you mean by
> it? Without that
> information, no one can proceed.
>
> >3) it has no essential singularities
>
> A comment similar to the one for condition 2). What
> do YOU mean by an
> essential singularity?
the same it has always been " extreme behaviour " around a point such as taking almost all values of z in its infinitesimal neighbourhood
>
> >4) it is Coo
>
> This one is clear enough. Obviously that is a
> necessary condition,
> and you understand that it is not a sufficient
> condition .
of course i do.
>
> >5) its first derivate is nowhere oo
>
> I assume you mean derivative.
as far as i recall derivate = derivative ?
maybe its like the " us word " and the " british word " for the same thing ...
im not a native speaker srr ...
If a function is Coo,
> then 5) is
> already excluded,
keep in mind the other conditions and the recursiveness !
5) its like f'(x) has no poles ( oo ) either
i admit this is implicit.
this is the hardest part and core.
i have not given a method to compute it.
but the analogue has to exist !
im not sure but i guess
real part curvature = curvature ( real f(z) )
and analogue for the imaginary part.
and the curvature has to be continu (and Coo of course , which already follows from the other conditions btw )
,
>
> That being said, your version of Gaussian curvature
> would have to give
> you a complex number associated with every complex
> number which is to
> say a function g(z) which is also a complex-valued
> function of a
> complex variable.
>
> >7) the equation f(z) = A has a solution for all A
> apart from perhaps a single exception
>
> So you have seen Picard's little theorem and make its
> conclusion about
> entire functions a part of your precondition for
> their existence.
> Nothing wrong with that.
nope , nothing wrong with that.
>
> >8) f'(z) has the same 7 properties above
>
> Nothing wrong with this one.
nope 8) is perfect :)
>
> >9) G_1 f(z) has the same 8 properties as above
>
> I assume you mean that G_1 f(z) is your version of
> the Gaussian
> curvature of f at z. In other words, it is what I
> called g(z).
yes , in analogue of D_1 f(z) which is the first derivate , G_1 f(z) is the first curvature.
>
>
> I hope that, after having read my detailed comments
> about the
> conditions of your conjecture you are beginning to
> realize the
> inadequacy of your presentation.
i have realized nothing new.
i knew i was implicit rather than explicit , and it might be desirable to explicitly define the ' complex curvature ' but the question is nevertheless well stated and adequate.
It is impossible to
> discuss your
> conjecture because, as stated, it doesn't make sense.
> There is
> nothing to work with because nobody can figure out
> what you mean.
their loss if they dont.
most of my posts contain intresting questions or answers.
You
> have given some evidence of actually being interested
> in mathematics
> which is why I am giving you this help. I don't
> think that you are a
> troll. I hope you can appreciate how frustrating and
> annoying it can
> be to deal with language such as yours which is so
> loose as to be
> impossible to turn into anything meaningful without
> further
> clarification and which looks like it might be
> impossible to fix up
> anyway even with greater clarificationl.
well , hard questions can be frustration.
and my implicit way of stating things might be annoying , but that is not intentional ; keep in mind that asking " out of book " questions and introducing unfamiliar abstract concepts naturally leads to a nessarily need to do effort just to comprehend , let alone solve the matter.
In other
> words, if you try
> to fix up the problems I have discussed, you find
> that your thinking
> was flawed and you will have to go back to the
> drawing board.
i have answered all of your questions , my thinking was not flawed and i dont need to go to the drawing board.
if you understand me now , and are really into math , you are the one running hasty to a drawing board to investigate the matter.
>
>
> Regards,
> Achava
>
>
regards
the loving amy
aka
tommy1729
David C. Ullrich
wrote:
Ok, answer me this: georfial quiddle not blig?
I guess that question's too hard for you, eh?
>if you want traditional questions , get a math book.
>
>my questions are uncommon and original and thus might sound weird and unfamiliar but they are correct !
>
>
>my style is implicit rather than explicit...
>
>
>>
>> >a function f(z) is entire apart from potential
>> removable singularities if
>>
>> >1) it converges for all of z
>>
>> This one, for example, makes no sense as stated.
>> Sequences converge,
>> series converge, you can talk about f(z) converging
>> to a limit as z
>> goes to a. This last is the only one that has any
>> hope of being what
>> you mean, when put into the form, f(z) coverges to
>> some limit
>> (depending on z, of course) as z goes to a for every
>> complex a.
>> Is that what you meant?
>
>when i said convergeance , i indeed meant convergeance
>
>too a ? to whatever f(z) converges !
Saying "f(z) converges" is meaningless. Adding an exclamation
point doesn't help.
It may well be that there is something that you meant when you
wrote that. But there's not way to tell what you meant. What
is your _definition_ of "f(z) converges"?
>when i said no poles ; no poles -> no convergeance to oo.
>
>( since hey pole of 1/x is oo right ? )
>
>
>
> Or did you mean something to
>> the effect that
>> f(z) is defined as a power series, and that series
>> converges for every
>> complex z. A complex-valued function of a complex
>> variable is, after
>> all, just a set of ordered pairs of complex numbers
>> and can be
>> provided in a very large variety of ways. If you
>> meant that the power
>> series converges everywhere, then there is nothing to
>> discuss, as then
>> the function is already known to be entire.
>>
>> >2) it has no poles
>>
>> I have not seen the concept of pole discussed outside
>> of the context
>> of a function that holomorphic at a lot of places.
>
>do you know any ' finite poles ' ?
>
>again pole = oo
>
>see previous reply.
>
>note that no other poles exist !
In standard terminology it's only holomorphic functions
that have poles. So when you say "f has no poles" but
you insist that you're not assuming that f is holomorphic
you're simply not making sense. (The problem is
much worse for "essential singularity".)
[...]
[...]
>
>i have answered all of your questions ,
You've answered none of the questions. When he asks what you
mean by "gaussian curvature" here you really think "it has to exist!"
is an acceptable _definition_?
> my thinking was not flawed and i dont need to go to the drawing board.
>
>if you understand me now , and are really into math , you are the one running hasty to a drawing board to investigate the matter.
>>
>>
>> Regards,
>> Achava
>>
>>
>
>regards
>
>the loving amy
>
>aka
>
>tommy1729
David C. Ullrich