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4 Improve case coverage
• Each $b_n$ is a linear combination of $a_0, \dots, a_n$; i.e. we have an upper-triangular infinite matrix $F$, not depending on $a_n$, such that (writing $\vec{a} = (a_n), \vec{b} = (b_n)$) $\vec{b} = F \vec{a}$. $F$ is injective, since if Suppose for now that $F p(0), r'(0) \vec{a} = neq 0$; then $p(x) A(r(x)) = 0$F$has a nonzero diagonal, and so$p(x) = r(x) = 0$. Since its rows are finite, this means that it is invertible. • Now we remove the assumptions that$p(0), r'(0) \neq 0$. If$x^k$divides$p(x)$, then replacing$p(x)$by$p(x)/x^k$does not change positivity of the coefficients. Now suppose the bottom exponent of$r(x)$is$x^m$with positive coefficient; then in$\mathbb{R}[[x]]$we can write$r(x) = s(x)^m$, and if we denote$A_m(x) = A(x^m)$, we have$A(r(x)) = A_m(s(x))$. Clearly,$A_m$has nonnegative coefficients if and only if$A$does, and$s'(0) \neq 0$, so the previous proof applies and$s(x)$is a multiple of$x$, i.e.$r(x)$is a multiple of$x^m$, and$p(x)$is a multiple of$x^k$. • 3 Rewrite and change conclusion I sort of worked It is impossible, and not just for rational functions. To see thisout as I wrote it, so the answer is at let's consider the bottom. Consider it a whodunnit coefficients$b_n$of sorts. Let us do some explicit computations. Suppose$p(x) = \sum_{n \geq 0} p_n x^n$and A(r(x))$ as functions of the $r(x) = \sum_{n \geq 0} r_n x^n$, with a_n$, the coefficients of$r_0 A(x)$. Since$r(0) = 0$. Now, (as it must be) we just grindsee that: • Each$p(x) A(r(x)) = (\sum_k p_k x^k) \sum _n a_n (\sum_l r_l x^l)^n = b_n$is a linear combination of$a_0, \sum_n a_n (\sum_k p_k x^k) dots, a_n$; i.e. we have an upper-triangular infinite matrix$F$, not depending on$a_n$, such that (\sum_l r_l x^l)^n$writing $\vec{a} = \sum_n a_n \sum_m (\sum_{k + l_1, \dots, l_n = m} p_k r_{l_1} a_n), \dots r_{l_n}) x^m vec{b} = \sum_m (\sum_n f(m,n) a_n) x^m$,

where I have written b_n)$)$f(m,n) = \sum_{k + l_1, \dots, l_n \vec{b} = m} p_k r_{l_1} F \dots r_{l_n}$. I require only one fact:$f(m,n) = 0$if vec{a}$. $n > m$F$is injective, since$r_0 = 0$. Thus, if we write$F \vec{a} = [f(m,n)]$as an infinite upper-triangular matrix and 0$ then $\bar{a} p(x) A(r(x)) = [a_n]$ as an infinite column vector0$, we have so$p(x) A(r(x)= r(x) = \sum_m (F \bar{a})_m x^m$0$. You seek the condition:Since its rows are finite, this means that it is invertible.

• If $F \bar{a} = \vec{b} \bar{b}$ (equivalently, geq 0$for all$\bar{a} = F^{-1} \vec{a} \bar{b}$), geq 0$, then the entries in particular this is true of $\bar{a}$ are nonnegative if and only if those the columns of $\bar{b}$ are.

Choosing $\bar{a}$ and F$, taking$\bar{b}$\vec{a}$ to be the infinite "basis" vectors. Conversely, we see have equivalently that $\vec{a} = F^{-1} \vec{b}$, so if $\vec{a} \geq 0$ for all $\vec{b} \geq 0$ this means that must be true of the entries columns of $F^{-1}$. We conclude that both $F$ and $F^{-1}$ are necessarily all have nonnegative entries.It is easy (for me

• Lemma in linear algebra: ) ) to prove by induction that if $F$ is therefore diagonal (with positive entries there).

How can this happen? We must upper-triangular and both it and $F^{-1}$ have nonnegative real entries, then $\sum_{k + l_1 + \dots + l_n = m} p_k r_{l_1} \dots r_{l_n} = 0$ if F$is diagonal. Proof by induction: true for$m 1 \neq n$times 1$ matrices vacuously. In general, by induction we may assume that the upper-left and $> 0$ if lower-right $(n - 1) \times (n = m$.

For - 1)$blocks of$m = n$F$ are diagonal, we get $p_0 r_1^n > 0$ for all $n$, so only the $p_0, r_1 > 0$.

For (1,n)$entry of$m = n + 1$, F$ is nonzero off the diagonal. Then we get $p_1 r_1^n + p_0 r_1^{n - 1} r_2 = 0$, and therefore have $r_2 (F^{-1})_{1n} = -(p_1/p_0) r_1$F_{1n} F_{nn}/F_{11}$. For$m = n + 2$, we get Since$p_2 r_1^n + p_1 r_1^{n - 1} r_2 + p_0(r_1^{n - 1} r_3 + r_1^{n - 2} r_2^2) = 0$, F_{11}$ and we observe the occurrence of $(p_1 r_1 + p_0 r_2)r_1^{n - 2} r_2 = f(n + 1F_{nn}$ are both positive, n)r_1^{-1} r_2 = $F_{1n}, (F^{-1})_{1n} \geq 0$ , which we eliminate to obtain implies $p_2 r_1^n + p_0 r_1^{n - 1} r_3 F_{1n} = 0$.Solving

• This is also true of infinite matrices, since we get $r_3 = -(p_2/p_0) r_1$can compute the finite upper-left blocks independently of the rest of the matrix.

More generally

• However, if $F$ is diagonal then we (or I, at any rate) havesee that $f(n + k, n) p(x)A(r(x)) = p_k r_1^n + \sum_{i = 1}^{k - 1} f(n + i, nsum a_n p(x) r_1^{-1} r_{k - i + 1} + p_0 r_1^{n - 1r(x)^n = \sum F_{nn} r_{k + 1}$,a_n x^n$for all choices of$a_n$, so by induction we have (for example, taking$r_{k + 1} a_n = -(p_k/p_0) r_1$. It follows that: The pairs of functions t^n$ for a new variable $p(x), r(x)$ answering your question are exactly those of the form t$and rewriting both sides as power series in$t$) we have$p(x) r(x)^n = p_0(1 + xq(x))$F_{nn} x^n$ for all $n$ (and some $F_{nn} > 0$). That is, $r(x) (r(x)/x)^n = r_1(x - x^2q(x))$ with F_{nn} p(x)^{-1}$for all$p_0, r_1 > 0$n$, so in fact $r(x)/x = F_{11}/F_{00}$ is constant, and finally, $q(x) \in \mathbb{Q}[[x]]$.p(x)$is constant as well. • 2 Fix an index I sort of worked this out as I wrote it, so the answer is at the bottom. Consider it a whodunnit of sorts. Let us do some explicit computations. Suppose$p(x) = \sum_{n \geq 0} p_n x^n$and$r(x) = \sum_{n \geq 0} r_n x^n$, with$r_0 = 0$. Now, we just grind:$p(x) A(r(x)) = (\sum_k p_k x^k) \sum _n a_n (\sum_l r_l x^l)^n = \sum_n a_n (\sum_k p_k x^k) (\sum_l r_l x^l)^n= \sum_n a_n \sum_m (\sum_{k + l_1, \dots, l_n = m} p_k r_{l_1} \dots r_{l_n}) x^m = \sum_m (\sum_n f(m,n) a_n) x^m$, where I have written$f(m,n) = \sum_{k + l_1, \dots, l_n = m} p_k r_{l_1} \dots r_{l_n}$. I require only one fact:$f(m,n) = 0$if$n > m$, since$r_0 = 0$. Thus, if we write$F = [f(m,n)]$as an infinite upper-triangular matrix and$\bar{a} = [a_n]$as an infinite column vector, we have$p(x) A(r(x)) = \sum_m (F \bar{a})_m x^m$. You seek the condition: If$F \bar{a} = \bar{b}$(equivalently,$\bar{a} = F^{-1} \bar{b}$), then the entries of$\bar{a}$are nonnegative if and only if those of$\bar{b}$are. Choosing$\bar{a}$and$\bar{b}$to be the infinite "basis" vectors, we see that this means that the entries of$F$and$F^{-1}$are necessarily all nonnegative. It is easy (for me :) ) to prove by induction that$F$is therefore diagonal (with positive entries there). How can this happen? We must have$\sum_{k + l_1 + \dots + l_n = m} p_k r_{l_1} \dots r_{l_n} = 0$if$m \neq n$, and$> 0$if$n = m$. For$m = n$, we get$p_0 r_1^n > 0$for all$n$, so$p_0, r_1 > 0$. For$m = n + 1$, we get$p_1 r_1^n + p_0 r_1^{n - 1} r_2 = 0$, and therefore$r_2 = -(p_1/p_0) r_1$. For$m = n + 2$, we get$p_2 r_1^n + p_1 r_1^{n - 1} r_2 + p_0(r_1^{n - 1} r_3 + r_1^{n - 2} r_2^2) = 0$, and we observe the occurrence of$(p_1 r_1 + p_0 r_2)r_1^{n - 2} r_2 = f(n + 1, n)r_1^{-1} r_2 = 0$, which we eliminate to obtain$p_2 r_1^n + p_0 r_1^{n - 1} r_3 = 0$. Solving, we get$r_3 = -(p_2/p_0) r_1$. More generally, we (or I, at any rate) have$f(n + k, n) = p_k r_1^n + \sum_{i = 1}^{k - 1} f(n + i, n) r_1^{-1} r_{i r_{k - i + 1} + p_0 r_1^{n - 1} r_{k + 1}$, so by induction we have$r_{k + 1} = -(p_k/p_0) r_1$. It follows that: The pairs of functions$p(x), r(x)$answering your question are exactly those of the form$p(x) = p_0(1 + xq(x))$,$r(x) = r_1(x - x^2q(x))$with$p_0, r_1 > 0$, and$q(x) \in \mathbb{Q}[[x]]\$.

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