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From: John on 13 Mar 2010 09:03
Let f be analytic in C\{-1,3}, and -1, 3 are simple poles of f.
can
sum{n = - oo, n = -1} a_n (z-2)^n
be a Laurent series of f in 1<|z-2|<3 ?

(probably the answer should be "no")
From: José Carlos Santos on 13 Mar 2010 13:19
On 13-03-2010 14:03, John wrote:

> Let f be analytic in C\{-1,3}, and -1, 3 are simple poles of f.
> can
> sum{n = - oo, n = -1} a_n (z-2)^n
> be a Laurent series of f in 1<|z-2|<3 ?
>
> (probably the answer should be "no")

Yes, probably the answer is "no".

Let g(z) = (z + 1)(z - 3)f(z). Then _g_ can be extended to an entire
function (which I will also denote by _g_). But

(z + 1)(z - 3) = ((z - 2) + 3)((z - 2) - 1)

= (z - 2)^2 + 2(z - 2) - 3.

So g(z) is equal to ((z - 2)^2 + 2(z - 2) - 3) times

(a_{-1}(z - 2)^{-1} + a_{-2}(z - 2)^{-2} + ...

Therefore, since _g_ is an entire function and since the coefficient of
(z - 2)^{-n} in this product is -3a_{-n} + 2a_{-n - 1} + a_{-n - 2}, all
these numbers are 0. So, the sequence (a_{-n})_n is defined recursively;
each term is a linear combination of the two previous one. It seems to
me (but I did not check it), that then the series

sum_{n < 0}a_n(z - 2)^n

will converge only when |z - 2| > 3. But, I repeat, I did not check it
thoroughly.

Best regards,

Jose Carlos Santos
From: A N Niel on 13 Mar 2010 15:18
In article
<70dff403-e027-4c85-ad82-d34820417344(a)g7g2000yqe.googlegroups.com>,
John <to1mmy2(a)yahoo.com> wrote:

> Let f be analytic in C\{-1,3}, and -1, 3 are simple poles of f.
> can
> sum{n = - oo, n = -1} a_n (z-2)^n
> be a Laurent series of f in 1<|z-2|<3 ?
>
> (probably the answer should be "no")

The Laurent series valid in 1 < |z-3| < 3 is unique, and
does not converge beyond the poles. It must therefore have
terms with exponents going to infinity and terms with
exponents going to -infinity.
From: W^3 on 13 Mar 2010 16:22
In article
<70dff403-e027-4c85-ad82-d34820417344(a)g7g2000yqe.googlegroups.com>,
John <to1mmy2(a)yahoo.com> wrote:

> Let f be analytic in C\{-1,3}, and -1, 3 are simple poles of f.
> can
> sum{n = - oo, n = -1} a_n (z-2)^n
> be a Laurent series of f in 1<|z-2|<3 ?
>
> (probably the answer should be "no")

No, because if that series converges for 1 < |z-2| < 3, it converges
for 1 < |z-2| and defines a function analytic on the larger region.
That rules out a pole (or essential singularity) at -1.
From: John on 14 Mar 2010 16:11
On Mar 13, 11:22 pm, W^3 <aderamey.a...(a)comcast.net> wrote:
> In article
> <70dff403-e027-4c85-ad82-d34820417...(a)g7g2000yqe.googlegroups.com>,
>
>  John <to1m...(a)yahoo.com> wrote:
> > Let f be analytic in C\{-1,3}, and -1, 3 are simple poles of f.
> > can
> >   sum{n = - oo, n = -1} a_n (z-2)^n
> > be a Laurent series of f in  1<|z-2|<3 ?
>
> > (probably the answer should be "no")
>
> No, because if that series converges for 1 < |z-2| < 3, it converges
> for 1 < |z-2| and defines a function analytic on the larger region.
> That rules out a pole (or essential singularity) at -1.


Thank you !!!
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