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Let's suppose I'm working with some set of functions $f_k(n)$. $f_1(n)$ is essentially the root of my functions, and could be nearly anything, and then $f_k(n) = (f_1(n) * f_{k-1}(n))$ for some convolution $*$ that won't be described here. $f_k(n)$ takes non-negative integers and yields reals. $f_k(n)$ has as a given the following properties.

1) $f_0(n) = 1$

2) ($f_j(n) * f_k(n)) = f_{j+k}(n)$. This mirrors, of course, the usual rules about raising numbers to powers ($x^j \cdot x^k = x^{j+k}$). More broadly, $a f_j(n) * b f_k(n)) = (ab)f_{j+k}(n)$

3) $f_k(n)$ structurally must be 0 once $ k > \log n$, or thereabouts.

4) * is both associative and commutative.

Suppose I can't guarantee almost anything else useful about $f_k(n)$, because it's such an irritable function. Integration and derivation don't work. $f_k(n) \cdot f_k(m)$ and $f_k(n) + f_k(m)$ are manually computable but in wildly unpredictable ways, and $f_k(n+m)$ has no obvious or easy relationship to $f_k(m) + f_k(n)$. There are no guarantees about values being greater or less than other values in any direction.

(I realize that might be insufficiently precise, so let me know if there needs to be any clarification)

So here's my question. Only given the four properties above, which identities from the world of power series will still be valid, if we replace the usual powers of x with my convolution here? Or, more generally, is there a way to tell which identities will be valid?

I know by hand that many series identities seem to work regardless of the nature of $f_1(n)$ - so, for example, if we define $cosf(n) = 1 - \frac{f_2(n)}{2!}+\frac{f_4(n)}{4!}-\frac{f_6(n)}{6!}+...$, and do something similar for $sinf(n)$, then $cosf(n)*cosf(n) + sinf(n)*sinf(n) = 1$, if $cosf(n)*cosf(n)$ is taken to mean the term-wise convolution of its power series with itself.

An awful lot of other series relationships seem to work from this perspective.

On the other hand, $sinf(n+ \frac{\pi}{2})$ usually isn't $cosf(n)$ (for the various $f_1(n)$ I'm working with). Trig identities dealing with things like half angle formulas I've tried don't generally work, either.

Is there an easy way for me to tell which power series relationships will still be valid? Am I stumbling on an obvious and well-covered area of research here? It would be really handy for me if I were...

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I don't see the relevance of property 4) especially since you haven't described what set $f_k(n)$ takes values in. What seems to be much more relevant is the question of whether * is associative and/or commutative. – Qiaochu Yuan Nov 17 '11 at 22:59
* is both associative and commutative. I updated my question and changed property #4 to say as much... And $f_k(n)$ operates operates on non-negative integers and produces reals. I'll add that to my question too. – Nathan McKenzie Nov 17 '11 at 23:28
It sounds like you are looking for some kind of functional calculus, i.e. a way of calculating $P(f_1)$ for any polynomial $P$ (which is just trival algebra) or, more generally, for $P$ an analytic function on some domain. If your space of functions were, for example, a Banach algebra, then you could use the holomorphic functional calculus for Banach algebras. If your space is not a Banach algebra, then you need to specify further topological properties; e.g. in what sense does the series for $\cos f_1$ converge? You definitely need some kind of extra analysis. – Zen Harper Nov 18 '11 at 3:45

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