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If I have a holomorphic function $f(z)$, then I can write it as $$f(z)=\sum_{n=0}^\infty a_n z^n.$$ So these functions can be viewed as a generating function of the coefficients $\{a_n\}$ which have a certain structure. There are nice finite differential problems, solved exactly by some sequences $\{a_n\}$ encoded in a generating function. From what I see, they are like a set of numbers with incredibly low Kolmogorov complexity (see comments). The property of being holomorphic is tied to the derivation $f'(z)$ being defined/existent/computable for the function. And this derivation is also used in the "algorithm" for generating the $a_n$'s, namely the series expansion. Is this a valuable perspective?

In this same vain, I wonder if there is a class of functions, let's denote one by $F$, for which there are unique $A$, such that

$$F(z)=\int A(t)z^t\ \mathbb{d}t.$$

Thoughts: This reminds me of the Laplace transform, which is essentiall the same thing, only with $z\equiv\mathbb{e}^{-s}$. Hence, in the sense of the discussion abov, $F(z)$ would encode information $A(t)$, which solves some problem. One of my motivations really is just to understand the Laplace transform better. Some differential equations are solved via Laplace transform, but to see how their information relates to the differential equations is not so straight forward to me as in the case with the countable series expansion numbers $a_n$. I assume there must be something coming from the fact that the exponential function is tied to the derivative operation.

Trying to "grok it", I've come across some (category theoretical) approaches to these integral transforms, which reseble the Pontryagin duality construction. This approach seems sensible to me. (And after all, I find it kind of arbitrary that this theory is exclusively centered around the hom's to the circle group.)

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I don't understand the first paragraph. What is the "certain structure"? What "incredibly low Kolmogorov complexity"? –  Qiaochu Yuan Jan 21 '13 at 10:03
    
@Ilya: That's just mean. –  NiftyKitty95 Jan 21 '13 at 10:10
    
@NickKidman: I apologize then, but if it's not on purpose - it rather means that both the OP and the comments are written without being well-thought about, which is not something welcomed. –  Ilya Jan 21 '13 at 10:12
    
@QiaochuYuan: I mean the infinite number of coefficients $a_n$ can be encoded in one expresstion for the function, e.g. the string "$x/(1-x-x^2)$" as input for the "trylor expansion algorithm" generates the Fibinacci numbers $0,\;1,\;1,\;2,\;3,\;5,\;8,\;13,\;21,\;34,\;55,\;89,\;144$. All complex differentiable functions, i.e. all strings with which the algorithm could work with, generate some numbers $a_n$ this way. –  NiftyKitty95 Jan 21 '13 at 10:12
    
@Ilya: That's kind of mean too, but I try to take it as constructive criticism. –  NiftyKitty95 Jan 21 '13 at 10:17
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