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consider rewriting a taylor series f(z)/f(0) as a kind of taylor product :
f(z)/f(0) = 1 + a x + b x^2 + ... = (1 + a' x)(1 + b'x^2) ...
if we want (1 + a' x)(1 + b'x^2) ... to be valid at least where the taylor series is we have to " take care of the zero's "
(1 + c x^n)*... has as zero's (-1/c)^(1/n) so the taylor series must have these zero's too.
despite that simple condition , its not quite easy to me how to find
f(z)/f(0) = 1 + a x + b x^2 + ... = (1 + a' x)(1 + b'x^2) ...
with both expressions converging in the same domain ( or all of C ).
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05/31/2011, 02:02 PM
(This post was last modified: 05/31/2011, 02:02 PM by bo198214.)
(05/31/2011, 12:26 PM)tommy1729 Wrote: despite that simple condition , its not quite easy to me how to find
f(z)/f(0) = 1 + a x + b x^2 + ... = (1 + a' x)(1 + b'x^2) ...
with both expressions converging in the same domain ( or all of C ).
Ya basically this is a polynomial approximation different from the usual powerseries one, which is easiest to handle.
I was
aware of the different domain of convergence already when discussing the
Mittag-Leffler expansion/star. This is a polynomial approximatino that converges in the star-region, instead of a disk for the powerseries approximation.
I wonder what approximation implies what region of convergence.
For example is there a polynomial approximation that implies a quadratic region of convergence? But perhaps this more a question for mathoverflow.
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Be also aware that there is a corpus of theory about products, different from yours:
)
For example:
 = \pi z \prod_{n=0}^\infty (1+ z^2/n^2))
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yeah , i was somewhat aware of all that.
more specifically a theorem by hadamard or was it weierstrass :
any entire function can be written as
f(z) = exp(taylor(z)) * (z - a_1)(z - a_2)..(z - a_n)...
i guess this implies if f(z)/f(0) is entire then
f(z)/f(0) = 1 + a_1 z + a_2 z^2 + ... = (1 + b_1 z)(1 + b_2 z^2) ...
and the product form also converges for all z apart from the set of points (-1/b_n)^(1/n) (unless they truely give 0 ) IFF those zero's are nowhere dense ( so that they do not form a " natural boundary " if there are oo many zero's ).
to avoid bounded non-cauchy sequences , i do assume a simple summability method is used ( averaging ).
( or "productability method " -> exp ( summability ( sum log (c_n) ) ) )
tommy1729
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06/01/2011, 05:31 PM
(This post was last modified: 06/01/2011, 05:31 PM by tommy1729.)
i have to add some conditions and ideas :
if f(z) = f(g(z)) then we have product terms of the form 1 + b_n g(z)^n.
those product terms have different zero's despite f(z) = f(g(z)).
also it seems if f ' (z) = 0 our product form does not work.
so i propose the following condition.
the product form of f(z) converges to the correct value in Q if f(z) =/= 0 and f(z) is univalent in Q.
***
it would be nice to consider product forms of f^[n](z).
=z+\frac{1}{2} n z^2 + \frac{1}{12} (3n^2-n) z^3 + \frac{1}{48} (6n^3-5n^2+n) z^4 + \cdots)
is the expression for f(z) = e^z - 1 and it would be nice to have a similar looking product form ...
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perhaps a good question for gottfried :
for taylor series we use carleman matrices.
what matrices are we suppose to use for these product forms (and how) , if any ??
