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I've been messing around with generating functions, power series, and related series, and I've come across a simple method of using two "roots of unity filters" to calculate the Hadamard product. A professor at a local university has offered to comment on my ideas, but I'm still searching for more information on Hadamard products. I'd like to know the history of them if possible, and Wikipedia and searches haven't turned up much. I'm especially interested in direct calculations of the Hadamard product. I seem to have a method that works for all generating functions. So I'd like to know if this is public knowledge, or if there is a chance I've discovered something new.

Again, my method essentially uses two generating functions to generate a complete generating function that is the Hadamard product. Is this new?

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There's a pretty well-known integral formula for the Hadamard product; I vaguely recall seeing it in Companion to Concrete Mathematics, and Qiaochu Yuan notes it at math.stackexchange.com/questions/4744/… . As that thread points out, too, the Hadamard product of two elementary (or even algebraic) series isn't necessarily elementary/algebraic, so 'closed forms' simpler than an integral seem unlikely. –  Steven Stadnicki Aug 12 '11 at 17:01
    
@Steven Stadnicki:I'm familiar with this method. Perhaps you'd be interested in my ideas, too. I get a closed form, with one (perhaps unwanted) byproduct; each series must be in a different variable, and the result is a function or series in both variables. The closed form is not particularly elegant; it is at least six times the size of both the orignal closed forms. But it is a closed form that can be found without the problems of integration. –  Matt Groff Aug 12 '11 at 17:11
    
A very small piece of "history": According to Stanley's Enumerative Combinatorics, I, p. 597 of the July 15 2011 edition, the Hadamard product was introduced in Hadamard's Théorème sur les séries entières, Acta Math. 22 (1) (1899), 55-63. –  t.b. Aug 12 '11 at 20:37
    
@Matt Groff: I'm a little bit curious, at least, though with different variables I'm not sure that it qualifies as a 'Hadamard Product' per se. Maybe an examle would be in order: to use the classic example, what closed-form does your formula provide for the Hadamard product of $e^{x/2}$ and $e^{-y/2}$? –  Steven Stadnicki Aug 12 '11 at 21:51
    
@Steven Stadnicki:$\frac{1}{3}(\frac{1}{2}(e^{\frac{x}{2}-\frac{y}{2}}+e^{-\frac{x}{2}+\‌​frac{y}{2}})$+$1/2(e^{1/2e^{2i\pi /3}x-1/2e^{-2i\pi /3}y}$+$e^{-1/2e^{2i\pi /3}x+1/2e^{-2i\pi /3}y})$+$1/2(e^{1/2e^{-2i\pi /3}x-1/2e^{2i\pi /3}y}$+$e^{-1/2e^{-2i\pi /3}x+1/2e^{2i\pi /3}y}))$. Sorry for the slow reply. I think this formula is correct, if it displays ok... –  Matt Groff Aug 13 '11 at 1:28

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