Classifying All Transducer Degrees Below $N^{3}$

Let $f$, $g$, $\bm{\alpha}$ be as above, and suppose $\bm{\alpha}=\langle\alpha_{0},\alpha_{1},\ldots,\alpha_{m-1}\rangle$ has more than one weight. By shifting $g$ as necessary, we can assume WLOG that $\alpha_{0}=\langle a_{0},a_{1},\ldots,a_{p},b\rangle$ has the shortest length of all of the weights in $\bm{\alpha}$. Let $L=\sum_{i=0}^{m-1}(|\alpha_{i}|-1)$.

We have for all $n\equiv 0$ mod $m$,

Now by assumption, $\alpha_{0}$ had the shortest length of the weights in $\bm{\alpha}$, which means that if we take this length (which is $p+1$) times the number of weights $m$, this must be less than or equal to $L$. Therefore we can construct the following weight $\beta$:

$\beta=\frac{1}{m^{k}}(a_{0},0,0,\cdots,a_{1},0,0,\cdots,0,0,a_{p},0,0,\cdots,0,bm^{k})$

where each $a_{i}$ is in position $im+1$, and after $a_{p}$ we have $L-(pm+1)$ zeroes, so that this weight has length L. Then for all $n\equiv 0$ mod $m$, $(\langle\beta\rangle\otimes f)(n)=(\langle\alpha\rangle\otimes f)(n)=g(n)$. However, all three expressions in this equation are polynomials. Since they are equal for infinitely many $n$, they must be equal for all $n$, and thus $\beta$ is indeed the desired weight. ∎

The determinant of this matrix is $9m^{6}(a-b)(b-c)(a-c)$, which is indeed nonzero if $a,b,c$ are distinct and $m\neq 0$ as assumed. ∎

Since $M$ has rank 3, we can assume WLOG that the first three columns are linearly independent. Then we can write $M$ in block form as $[B\mid C]$, where $B$ is an invertible 3 by 3 matrix, and $C$ is everything else. We can also split $U$ into blocks, and writing these blocks in terms of $B,C,P$ gives us $U=[B^{-1}(P-CW)\mid W]^{T}$.

Now if $M_{\epsilon}=[B_{\epsilon}\mid C_{\epsilon}]$ is close enough to $M$, $B_{\epsilon}$ is still invertible, and we can define a vector $U_{\epsilon}=[B_{\epsilon}^{-1}(P-C_{\epsilon}W)\mid W]^{T}$, and by construction $M_{\epsilon}U_{\epsilon}=P$. Further, since $U_{\epsilon}$ converges to $U$, the first three entries can be made strictly positive with a sufficiently small $\epsilon$, and its other entries (the ”W” block) are equal to the entries of $U$ and thus nonnegative.

∎

The basic idea of this proof is that we want to narrow our focus to only three columns of our matrix (and three nonzero rows of our vector) in order to replicate the proof method from Theorem 32 in [4].

Let $f(n)=\sum_{i=0}^{m-1}a_{i}(mn+t+i)^{3}$, where each $a_{i}$ is nonnegative, and at least 3 are strictly positive. By definition we have $M_{m}((n+t)^{3})U(\alpha)=V(f(n))$, where $\alpha$ is the weight $\langle a_{0},a_{1},...,a_{m-1},0\rangle$. We will assume for now that it is the first three $a_{i}$’s which are nonnegative, and justify this assumption later.

We want to prove that any cubic polynomial can be transduced into $f(n)$. Let $q(n)$ be an arbitrary cubic polynomial, and let $q_{\epsilon}(n)$ be as defined in Lemma 4.9. Consider the matrix $M_{m}(q_{\epsilon}(n+t))$. We have the right conditions to apply Lemma 4.11 and obtain a vector $U_{\epsilon}$ as stated in the lemma. In particular, we have that $M_{m}(q_{\epsilon}(n+t))U_{\epsilon}=V(f(n))$ and $U_{\epsilon}$ has all nonnegative entries. But this means that $U_{\epsilon}$ is not just a vector but in fact a weight vector, since we can form a weight out of the entries of $U_{\epsilon}$. Then the equation $M_{m}(q_{\epsilon}(n+t))U_{\epsilon}=V(f(n))$ corresponds to a weight product, and in fact what it says is that $q_{\epsilon}(n+t)$ can be transduced into $f(n)$ via the weight corresponding to the vector $U_{\epsilon}$. Combining this with the way $q_{\epsilon}$ was defined, we have for an arbitrary cubic polynomial $q(n)$, $\langle q(n)\rangle\equiv\langle q(n+t)\rangle\geq\langle q_{\epsilon}(n+t)\rangle\geq\langle f(n)\rangle$. Therefore $f(n)$ is in the bottom degree for cubics.

However, we do still have to justify why we can assume WLOG that our weight had the first three entries strictly positive. In general, the nonzero entries of the weight could be in any location. But the fundamental argument behind Lemma 4.11 works the same regardless, as long as an invertible matrix can be formed by the columns which correspond to the nonzero entries of the vector $U$ (i.e. if $U$’s strictly positive entries were in the $ith,jth,kth$ positions, the $ith,jth,kth$ columns of $M$ need to be linearly independent). If the nonzero entries are in different places we simply have a messier block matrix to draw and more details to write out. And because Lemma 4.9 guarantees that any 3 columns of $M_{m}((n+t)^{3})$ can form an invertible 3 by 3 matrix, we are justified in using our assumption to simplify the proof.

∎

We first observe that all 2-transforms of $n^{3}$ are of the form $a(pn+r)^{3}+b(pn+s)^{3}$, with $a,b$ both positive rational numbers, $0\leq r,s<p,r\neq s$, and $p,r,s\in\mathbb{N}$.

We need the following equality:

where $a_{1}=\frac{dj(-1+k+jk)}{k^{2}(jk-1)(jk+k-2)},a_{2}=\frac{dj}{(k+1)k^{2}(jk+k-2)},a_{3}=\frac{1}{k^{3}}(1-\frac{d(jk+k-1)}{(k-1)(jk-1)})$

This can be easily verified, and holds for all $d,j,k$. We also observe that if $d$ and $j$ are fixed positive numbers, for a large enough value of $k$ each $a_{i}$ is strictly positive.

Now we proceed to the proof, which essentially consists of applying algebraic operations to this equation to end up with a 2-transform on the left hand side, and a 3-transform (or higher) on the right hand side. Then applying Thm 5.1 lets us conclude both sides of the equation are in the bottom degree for cubics.

Let $a(pn+r)^{3}+b(pn+s)^{3}$ be a 2-transform of $n^{3}$. Assume WLOG $s>r$.

We have two cases.

Case 1: $r>0$

If r $>$ 0, in the previous equation let $d=\frac{b}{a}$, let $j=s-r$, and choose an integer k large enough so each $a_{i}$ is positive.Then on both sides of the equation replace $n$ with $(n+r)$ to obtain:

$(n+r)^{3}+d(n+j+r)^{3}=a_{1}(kn+rk+jk-1)^{3}+a_{2}(kn+rk-k+1)^{3}+a_{3}(kn+rk)^{3}$

Now we use that $r+j=s$, and also replace $n$ with $pn$:

$(pn+r)^{3}+d(pn+s)^{3}=a_{1}(pkn+sk-1)^{3}+a_{2}(pkn+(r-1)k+1)^{3}+a_{3}(pkn+rk)^{3}$

Multiplying through by $a$ gives us what we want on the left hand side:

$a(pn+r)^{3}+b(pn+s)^{3}=aa_{1}(pkn+sk-1)^{3}+aa_{2}(pkn+(r-1)k+1)^{3}+aa_{3}(pkn+rk)^{3}$

Now let’s look at the right hand side. Since $p>s$ and $s>1$, $pk>sk-1>0$. Since $r>0$ and $r$ is an integer, $r\geq 1$, so $pk>(r-1)k+1>0$. Finally, $p>r$ so $pk>rk>0$.

Then we can rewrite the right hand side as the following: $\sum_{i=0}^{pk-1}c_{i}(pkn+i)^{3}$, where $c_{i}=0$ for all i except for $c_{sk-1}=aa_{1},c_{(r-1)k+1}=aa_{2},c_{rk}=aa_{3}$. Since $a$ and the $a_{i}$’s are all positive, this is 3-transform of $n^{3}$. Therefore it is in the bottom degree, and so is the left hand side.

Case 2: $r=0$

This will proceed similarly to the first case, except at the end we get a transformation of $(n-k)^{3}$. Again let $d=\frac{b}{a}$, $j=s$, and choose an integer k large enough so each $a_{i}$ is positive. We have:

$n^{3}+d(n+s)^{3}=a_{1}(kn+sk-1)^{3}+a_{2}(kn-k+1)^{3}+a_{3}(kn)^{3}$

Now rewrite the right hand side as follows:

$n^{3}+d(n+s)^{3}=a_{1}(kn+(sk+k-1)-k)^{3}+a_{2}(kn+1-k)^{3}+a_{3}(kn+k-k)^{3}$

Multiply both sides by $a$ and replace $n$ with $pn$ to obtain the desired left hand side:

$a(pn)^{3}+b(pn+s)^{3}=aa_{1}(pkn+(s+1)k-1-k)^{3}+aa_{2}(pkn+1-k)^{3}+aa_{3}(pkn+k-k)^{3}$

Again we examine the right hand side. Since $p>s$ and both are integers, $p\geq s+1$ and thus $pk>(s+1)k-1>0$, and we also have $pk>1>0$, $pk>k>0$. So we can rewrite the right hand side into the following sum: $\sum_{i=0}^{pk-1}c_{i}(pkn+i-k)^{3}$, where $c_{i}=0$ for all i except for $c_{(s+1)k-1}=aa_{1},c_{1}=aa_{2},c_{k}=aa_{3}$. Since $a$ and the $a_{i}$’s are all positive, this is a 3-transform of $(n-k)^{3}$. Therefore it is in the bottom degree by Theorem 5.2, and so is the left hand side. ∎

We will proceed by contradiction. Assume that there is some 1-transform of $n^{3}$ which is equal to the bottom degree. From the definition of the weight product, all 1-transforms of $n^{3}$ are of the form $(an+b)^{3}$, with $a,b\in\mathbb{N},b<a,a\neq 0$. To say that such a function is in the bottom degree is to say that all other cubic polynomials $f$ satisfy $\langle f\rangle\geq\langle(an+b)^{3}\rangle$. In particular, we can choose f to be a 3-transform of $n^{3}$, say $f=\alpha\otimes n^{3}$ for some weight $\alpha$ with three nonzero entries.

This means that we have $S^{n_{0}}((an+b)^{3})=\beta\otimes S^{m_{0}}(\alpha\otimes n^{3})$.

A bit of algebra gives us that $\beta\otimes S^{m_{0}}(\alpha\otimes n^{3})=S^{m}(\beta\otimes\alpha\otimes n^{3})=S^{m}(\gamma\otimes n^{3})$, where $m=m_{0}*|\beta|$ and $\gamma=\beta\otimes\alpha$ is the composition of weights (Lemma 12, [3]). The key thing to note here is that because $\alpha$ had three nonzero entries, $\gamma$ also must have at least three nonzero entries (The composition of a $k$-transform with an $l$-transform is a $kl$-transform). Let $|\gamma|=j$.

Therefore we have:

$S^{n_{0}}((an+b)^{3})=S^{m}(\gamma\otimes n^{3})$

and thus

$(an+b+an_{0})^{3}=\sum_{i=0}^{j-1}c_{i}(jn+i+jm)^{3}$

where at least 3 of the $c_{i}^{\prime}s$ are nonzero. We can simplify this equation with the goal of getting just $n^{3}$ on the left hand side. First we multiply both sides by $j^{3}$, as well as multiplying the right hand side by $\frac{a^{3}}{a^{3}}$ and distributing the top $a^{3}$ to the inside of the parentheses. This gives:

$(ajn+jb+ajn_{0})^{3}=\sum_{i=0}^{j-1}\frac{j^{3}}{a^{3}}c_{i}(ajn+ai+ajm)^{3}$

We can replace $n$ with $\frac{n}{aj}$ to obtain:

$(n+jb+ajn_{0})^{3}=\sum_{i=0}^{j-1}\frac{j^{3}}{a^{3}}c_{i}(n+ai+ajm)^{3}$

Now replace $n$ with $n-jb-ajn_{0}$:

$n^{3}=\sum_{i=0}^{j-1}\frac{j^{3}}{a^{3}}c_{i}(n+ai+ajm-jb-ajn_{0})^{3}$

Replace $n$ with $ajn$, and then divide both sides by $a^{3}j^{3}$ (this leaves the left hand side unchanged):

$n^{3}=\sum_{i=0}^{j-1}\frac{1}{a^{6}}c_{i}(ajn+ai+ajm-jb-ajn_{0})^{3}$

Finally, define $c^{\prime}_{ak}=\frac{c_{k}}{a^{6}}$ for all $k<j$ and 0 otherwise, and let $ajm-jb-ajn_{0}=t$.

Then we have:

$n^{3}=\sum_{i=0}^{aj-1}c^{\prime}_{i}(ajn+i+t)^{3}$

Now the right hand side is clearly equal to a weight transformation of $(n+t)^{3}$, and since at least three of the $c_{i}$’s were nonzero, at least three of the $c^{\prime}_{i}$’s will be nonzero. This would mean that $n^{3}$ is equal to a function which is in the bottom degree for cubics, by Thm 5.1. However, it has been shown that this is not the case in (Theorem 23, [4]), therefore this contradiction means that $\langle(an+b)^{3}\rangle$ is not equal to the bottom degree. ∎

We prove this via double inequality. One direction is trivial:

$\langle(an+b)^{3}\rangle\geq\langle(abn+b)^{3}\rangle=\langle b^{3}(an+1)^{3}\rangle\equiv\langle(an+1)^{3}\rangle$

The other direction requires a bit more work. First, we prove the following: For all $i>1\in{N}$, $\langle(an+b^{i})^{3}\rangle\geq\langle(an+b)^{3}\rangle$

Proof: $\langle(an+b^{i})^{3}\rangle\geq\langle(ab^{i-1}n+b^{i})^{3}\rangle=\langle b^{3(i-1)}(an+b)^{3}\rangle\equiv\langle(an+b)^{3}\rangle$

Now we want to show that for some $i$, $\langle(an+1)^{3}\rangle\equiv\langle(an+b^{i})^{3}\rangle$. Since $b$ is relatively prime to $a$, it is invertible in the multiplicative group $\mathbb{Z}^{*}_{a}$, which means for some $i$, $b^{i}\equiv 1$ mod $a$. So there exists some number $m$ with $am+1=b^{i}$. Then we have:

$\langle(an+1)^{3}\rangle\equiv\langle(a(n+m)+1)^{3}\rangle=\langle(an+am+1)^{3}\rangle=\langle(an+b^{i})^{3}\rangle$

Since we proved that $\langle(an+b^{i})^{3}\rangle\geq\langle(an+b)^{3}\rangle$ earlier, this completes the proof. ∎

The reverse implication is relatively straightforward: if $a|b$, then let $b=ac$ and we have $\langle(an+1)^{3}\rangle\geq\langle(acn+1)^{3}\rangle\equiv\langle(bn+1)^{3}\rangle$.

Now for the forward implication. The key observation is that if $\langle(an+1)^{3}\rangle\geq\langle(bn+1)^{3}\rangle$, this means that there exists $n_{0},m_{0}$, and a weight $\alpha$ such that $S^{n_{0}}(bn+1)^{3}=S^{m_{0}}(\alpha\otimes(an+1)^{3})$, and this weight $\alpha$ must have only one nonzero entry. Otherwise, the right hand side of the previous equation would be an $m$-transform of $n^{3}$ for $m\geq 2$, which (by Thms 5.1 and 5.2) would put it in the bottom degree. ($(an+1)^{3}$ is a 1-transform of $n^{3}$, and an $m$-transform of a 1-transform is also an $m$-transform of the original function, via weight product composition.) To say that $\alpha$ must have only one nonzero entry means that $\alpha\otimes(an+1)^{3}$ has the form $(a(cn+d)+1)^{3}=(acn+ad+1)^{3}$. Plugging this into the above equation means that