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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:

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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).

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 ??