# Multiplicative inverses of formal series with non-negative coefficients

What are the formal series $f$ with non-negative integer coefficients and constant term equal to $1$ whose multiplicative inverse $1/f$ has all coefficients, apart from a finite subset, all non-positive?

In fact, assume the series converges in some disk if you want...

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I had to solve an annoying captcha to ask this question... – Mariano Suárez-Alvarez Dec 20 '10 at 6:48

For example, $f(x) = (1 - a x)/(1 - b x)$ with $b > a > 0$ has all coefficients nonnegative while $1/f(x)$ has all coefficients nonpositive after the constant term.

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More generally, suppose $f(z)$ is meromorphic in a disk $\{z: |z| < r\}$, with $f(0) = 1$, $f(\overline{z}) = \overline{f(z)}$, and $f$'s closest pole and zero to the origin are a simple pole at $q$ and a simple zero at $p$ with $0 < p < q < r$. Then all but finitely many coefficients of the Maclaurin series of $f(z)$ are positive, and all but finitely many coefficients of the Maclaurin series of $1/f(z)$ are negative. – Robert Israel May 5 '11 at 15:42

Here is a non-rigorous attempt at an answer, but perhaps someone can make this into a formal argument. I will assume your series is centered at $0$. I will also denote your function by $g$ for reasons you will see as you read on.

Suppose $g(x) = \sum_{n=0}^{\infty} a_n x^n$, where $a_n \geq0$ for every $n$.

We look for a series for $h(x) = \frac{1}{g(x)} = \sum_{n=0}^{\infty} b_n$.

By taking formal derivatives $b_n = D^n(h(0))/n!$ (as in the Taylor expansion).

And then do a Google search... Behold the Faá di Bruno's Formula for derivatives of composite functions:

Let $f(x) = 1/x$ so that $h(x) = f(g(x))$. Comparing with the terms in this formula, we see that the signs of the $b_n$ must alternate (or be zero). This is because the derivatives of $1/x$ alternate in sign and the derivatives of $g(x)$ are non-negative by your hypothesis.

Now, I don't trust any of this, however, since I've completely neglected matters of convergence; but when dealing with formal power series, maybe this point is moot?

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This can't quite be right. For example, try quadratic polynomials with complex roots $g(x) = (x - \omega)(x - \overline{\omega})$. The series for $1/g(x)$ involves the real parts of powers of $\omega$, and these don't necessarily alternate in sign. – Robert Israel May 4 '11 at 18:45

Assume the powerseries of f(x) as $f(x)=1+a x+bx^2+cx^3+dx^4...$ where we deal with the explication of the problem just up to the coefficient $d$. Then I have a description of the $p$'th power of $f$ with $p$ as a variable, which may be best displayed with the help of a table.

Remark: I adopted a binomial notation, where in my original derivation is a pochammer-symbol. So for $p*(p-1)*(p-2)/3!$ in the original I write $\binom p3$ for shortness here. Thus for negative $p$ the binomial-expressions must respectively be interpreted!

Here is the beginning of the table for $f(x)^p$ with positive $p$ :
$$\small \begin{array} {rllll} f(x)^p=x^0* & \binom{p}{0}*(1)\\ +x^1*[& \binom{p}{1}*(1a) ] \\ +x^2*[& \binom{p}{1}*(1b)&+ \binom{p}{2}* (1a^2) ] \\ +x^3*[& \binom{p}{1}*(1c)&+ \binom{p}{2}* (2ab)&+ \binom{p}{3}* (1a^3)] \\ +x^4*[& \binom{p}{1}*(1d)&+ \binom{p}{2}* (2ac+1b^2)&+ \binom{p}{3}* (3a^2b)&+ \binom{p}{4}* &(a^4)] \\ +\ldots&\ldots \\ \end{array}$$

If we insert $p=1$ for $f(x)^1$ we get the obvious correct result
$\qquad \begin{array} {rllll} f(x)^1=x^0* & 1*(1)\\ +x^1*[& 1*(1a) ] \\ +x^2*[& 1*(1b) ] \\ +x^3*[& 1*(1c) ] \\ +x^4*[& 1*(1d) ] \\ +\ldots&\ldots \\ \end{array}$

If the above table is indeed a complete description for all $p$, then using $p=-1$ for the multiplicative inverse $\frac{1}{f(x)}$ , this would give:
$\qquad \begin{array} {rllll} f(x)^{-1}=x^0* & 1*(1)\\ +x^1*[& -1*(1a) ] \\ +x^2*[& -1*(1b)&+1*(1a^2) ] \\ +x^3*[& -1*(1c)&+1*(2ab)&-1*(1a^3)] \\ +x^4*[& -1*(1d)&+1*(2ac+1b^2)&-1*(3a^2b)&+ 1*(a^4)] \\ +\ldots&\ldots \\ \end{array}$

Then to have nonnegative coefficients in $f(x)$ as well as in $f(x)^{-1}$ means to have
1) $a=0$ because of $x^1$
2) then $b=0$ because of $x^2$ where also $a^2=0$
3) and so on.

So if the above description (which was derived from positive powers $f(x)^p$ only) is sufficient also for the negative powers, then this should be acceptable as a proof, that only the constant function $f(x)=1$ has nonnegative coefficients in the formal powerseries of $f(x)$ and $f(x)^{-1}$ simultanously.

[Update]: I apply the correction according to Robert Israel, and get the following result.

Because in &f(x)& all coefficients are nonnegative, I can, without loss of generality rewrite
$\qquad f(x)=1+ax+b^2x^2+c^3x^3+d^4x^4+...$

Then the table for $f(x)^{-1}$ looks like

$\qquad \begin{array} {rllll} f(x)^{-1}=x^0* & 1*(1)\\ +x^1*[& -1*(1a) ] \\ +x^2*[& -1*(1b^2)&+1*(1a^2) ] \\ +x^3*[& -1*(1c^3)&+1*(2ab^2)&-1*(1a^3)] \\ +x^4*[& -1*(1d^4)&+1*(2ac^3+1b^4)&-1*(3a^2b^2)&+ 1*(a^4)] \\ +\ldots&\ldots \\ \end{array}$
and from
1) $x^2$ follows that the minimal possible b equals a, so $b\ge a$
2) $x^3$ follows that the minimal possible c equals a, so $c\ge b\ge a$
3) and so on

So we have $0 < a \le b \le c \le d \le \ldots$

If all coefficients after a take their minimum, then $0<a=b=c=d=e= \ldots$ then the coefficients at all $x^k, k>1$ simplify to binomial coefficients whose sum is zero at the same power $a^k$, so the complete coefficient is zero as well. Thus $f(x)^{-1} = 1-ax$ , then $f(x)={1 \over 1-ax}$ and we have nonnegative coefficients in $f(x)$ and nonpositive in $f(x)^{-1}$ after the constant.

If we take $b>a$ at $x^2$ then at $x^3$ follows, that $c^3=b^3$ suffices to give a negative value at $x^3$ in $f(x)^{-1}$ .

Analoguos reasoning inherits to all following coefficients $0<a<b=c\le d \le e \le \ldots$ leading to $f(x)$ has positive coefficients only and $f(x)^{-1}$ has negative coefficients after the constant, but I didn't look yet at the coefficients $d,e, \ldots$ to arrive at Robert Israel's solution.

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Indeed, if $f(x) = 1 + b_k x^k + O(x^{k+1})$ with $b_k > 0$, then $1/f(x) = 1 - b_k x^k + O(x^{k+1})$, so if all coefficients of both $f(x)$ and $1/f(x)$ are nonnegative, $f(x) = 1$. But that's not what the original problem asked about. – Robert Israel May 4 '11 at 22:15
Aaarrrgh.... Misreading of the question. Even it was not too late, so why? Anyway it has been a nice exercise. – Gottfried Helms May 5 '11 at 1:50
Just updated the answer and included consideration of the nonpositive case for $f(x)^{-1}$ – Gottfried Helms May 5 '11 at 3:02