product functions - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: product functions (/showthread.php?tid=641) product functions - tommy1729 - 05/31/2011 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 ). RE: product functions - bo198214 - 05/31/2011 (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. RE: product functions - bo198214 - 05/31/2011 Be also aware that there is a corpus of theory about products, different from yours: $\prod_{n=0}^\infty (1+a_n z)$ For example: $\sin(\pi z) = \pi z \prod_{n=0}^\infty (1+ z^2/n^2)$ RE: product functions - tommy1729 - 05/31/2011 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 RE: product functions - tommy1729 - 06/01/2011 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). $f^n(z)=z+\frac{1}{2} n z^2 + \frac{1}{12} (3n^2-n) z^3 + \frac{1}{48} (6n^3-5n^2+n) z^4 + \cdots$ is the expression for f(z) = e^z - 1 and it would be nice to have a similar looking product form ... RE: product functions - tommy1729 - 06/01/2011 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 ??