31
$\begingroup$

Let $A=1+\sum_{n=1}^\infty \alpha_nx^n\in\mathbb Z[[x]]$ and $B=\frac{1}{A}=1+\sum_{n=1}^\infty\beta_n x^n$ two mutually inverse power series having bounded integral coefficients (ie. $\vert \alpha_n\vert,\vert \beta_n\vert<C$ for some constant $C$ and for all $n$).

Examples are given by $A=\frac{P}{Q}$ where $P$ and $Q$ are both finite products of cyclotomic polynomials having only simple roots.

Are there other, exotic, examples?

More generally, consider again $A=1+\sum_{n=1}^\infty \alpha_nx^n\in\mathbb Z[[x]]$ with inverse $B=\frac{1}{A}=1+\sum_{n=1}^\infty\beta_n x^n$ and require that the integral coefficients $\alpha_n,\beta_n$ have at most polynomial growth (ie. there exists a constant $C$ such that $\vert\alpha_n\vert,\vert\beta_n\vert<Cn^C+C$ for all $n$).

Examples are now arbitrary rational fractions $A=\frac{P}{Q}$ involving only cyclotomic polynomials. Again, I know of no other examples. Do they exist?

Added: This question is closely linked to The sum of integers being a bijection, see Zaimi's comment after Venkataramana's anwser.

$\endgroup$
  • 1
    $\begingroup$ unfortunately my references aren't within reach at the moment, but I believe many modular forms for congruence subgroups have integral coefficients and they all satisfy the Ramanujan-Petersen bound, which should imply polynomial growth in the coefficients of their q-expansions... $\endgroup$ – Will Chen Feb 16 '15 at 21:21
  • 3
    $\begingroup$ Yes, but $1/f$ should also have polynomial growth and this is very stringent (and $1/f$ is generally not a modular from what I remember). $\endgroup$ – Roland Bacher Feb 16 '15 at 21:31
21
$\begingroup$

Consider the set $S$ of nonnegative integers whose $2$-adic expansion involves only square powers of two (e.g. $n=1+16$, and $n=2^{25}+2^{49}+2^{64}$ belong to $S$). Let $T$ be the set of integers whose $2$-adic expansion never involves any square power of two. Then, $(S\cup\{0\})+(T\cup\{0\})$ represents each positive integer only once. Write $f(x)=\sum _{k\in S\cup\{0\}}x^k$ and $g(x)=\sum _{k\in T\cup\{0\}}x^k$. Clearly, $f(x)g(x)=\frac{1}{1-x}$. This shows that $f(x)$ and $g(x)(1-x)$ have bounded integer coefficients, but are not ratios of products of cyclotomic polynomials.

$\endgroup$
  • 1
    $\begingroup$ You'd better include $0$ in $S$ and $T$: otherwise, $2^{n^2}$ gets no representation in $S+T$. Then $(fg)(x) = 1/(1-x)$. $\endgroup$ – Vesselin Dimitrov Feb 17 '15 at 2:51
  • $\begingroup$ @Vesselin Thanks. I have included this in the edited version $\endgroup$ – Venkataramana Feb 17 '15 at 2:56
  • 8
    $\begingroup$ In some sense these are also "products of cyclotomic polynomials", you just need to allow for infinite products. See my answer here mathoverflow.net/questions/50798/… $\endgroup$ – Gjergji Zaimi Feb 17 '15 at 3:43
  • $\begingroup$ Is a "square power of 2" a square number that's a power of 2, or is it the exponent that's a square? If the latter, then $n=4+16$ doesn't qualify, as $4=2^2$. If the former, the other example doesn't qualify, as $2^{25}$ is not a square number. $\endgroup$ – Gerry Myerson Feb 17 '15 at 4:37
  • 4
    $\begingroup$ This example can even be simplified: $S$ given by $0$ and the union of all integers involving only even powers of two, $T$ union of $0$ and integers involving only odd powers of $2$. Since both $S$ and $T$ have arbitrary large gaps, their characteristic functions are not algebraic and multplying one of them by $1-x$, one gets an example (one might even replace even, odd numbers by an arbitrary partition of $\mathbb N$ into two infinite subsets). $\endgroup$ – Roland Bacher Feb 17 '15 at 11:09
3
$\begingroup$

Not a complete answer, just to suggest that there should be many examples: assume $f(z)$ is a holomorphic function on the open unit disc $D$, with continuous non-vanishing extension up to $\overline D$ and $f(0)=1$. Then $g(z):=1/f(z)$ is also continuous and non-vanishing up to $\overline D$, and by the Cauchy path integral formula, both have bounded coefficients $\alpha_n$, respectively $\beta_n$. If moreover for some reason the $\alpha_n$ are integers, then so are the $\beta_n$, since $\alpha_0=1$ and $\sum_{j=0}^n\alpha_j\beta_{n-j}=0$ for $n\ge1$.

In other words, a source of examples (hopefully not too empty) are the coefficients $(\alpha_n)$ of the power series expansions of the invertible elements of the algebra $H(D)\cap C^0(\overline D)$, whenever they are integers with $\alpha_0=1$.

$\endgroup$
  • $\begingroup$ And the Hardy space $H^\infty$ should work as well. $\endgroup$ – Pietro Majer Feb 16 '15 at 23:02
  • 4
    $\begingroup$ This can only produce polynomials: The $\alpha_j$ can be viewed as Fourier coefficients of the boundary function, so $\alpha_j\in\ell^2$ if the boundary values are in $H^{\infty}$. $\endgroup$ – Christian Remling Feb 16 '15 at 23:40
0
$\begingroup$

More examples follow from the paper of Duffin and Schaeffer, 1945.enter link description here

$\endgroup$
  • 5
    $\begingroup$ The above paper seems to give a correct answer to a different question, if I am not mistaken. $\endgroup$ – Roland Bacher Feb 16 '15 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.