# Given a generating function with "zeros", can one derive the function for ONLY the "zeros"?

If I have an generating function (GF) --- ordinary or exponential --- defining a series with at least one coefficient equal to zero, is there a general method to find the "inverse GF", i.e., the GF defining the series which only includes those zeros?

For example [a very simplified one!], if I have the "odd number exponential GF"

$EG(2n+1;x) = x^1 + x^3 + x^5 + x^7 + \dots,$

how can I use it to derive the "even number exponential GF"

$EG'(2n;x) = 1 + x^2 + x^4 + \dots$

?

Thanks! Kieren.

Addendum: For my trivial example, above, the inversion is (d'oh!) simply to subtract that series from the "unit series" (where all coefficients are 1), i.e.,

$(1 + x^1 + x^2 + x^3 + x^4 + x^5 + \dots) - (x^1 + x^3 + x^5 + \dots) = 1 + x^2 + x^4 + \dots.$

Ergo, this "inversion" is trivial to implement --- and the problem is reduced to the open question referenced below (i.e., to determine whether either the original GF or its "inverse" has any zeros).

Now my question is more specific: For arbitrary coefficients, is there a reasonable algorithm to derive the "inverted GF"?

• Why would you want to do this? Aug 27 '13 at 2:01
• @Qiaochu Yuan I have the opposite objection: it seems that such an operation (if efficient to implement) would be too useful to actually exist Aug 27 '13 at 2:11
• What do you want as coefficients for the zero terms? Do you always want 1? Aug 27 '13 at 2:13
• Benjamin: Ultimately, it would be useful --- perhaps, as John points out, too useful to actually exist --- if the coefficients of the original sequence could be arbitrary (as opposed to all 1, as in my simple example), and the inversion would be all 1s. However, any inversion would be quite useful, in my opinion. Do you have an idea for an inversion algorithm? Aug 27 '13 at 3:15