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Definition: My motivation for this question stems from the following definition: Define the deterministic finite automata generated by the nonconstant* polynomial $f(x_0 , \dots , x_n) \in \mathbb{Z} [x_0 , x_1 , \dots , x_n]$, denoted $\text{DFA}(f)$, to be the $5$-tuple $$\text{DFA}(f) = (\partial f, \Sigma, f, \delta, \equiv0),$$ where

  • $\partial f = \{ \text{all partial derivatives of } f \}$ is the set of states,
  • $\text{alphabet } \Sigma = \Bigg \{ \frac{\partial}{\partial x_0}, \frac{\partial}{\partial x_1}, ... , \frac{\partial}{\partial x_n} \Bigg \},$
  • $f = f(x_0 , x_1 , \dots , x_n )$ is the unique start state,
  • transition (partial, as in not defined for all possible inputs) function $\delta : \Sigma \times \partial f \longrightarrow \partial f, \bigg( \frac{ \partial}{\partial x_i} , p \bigg) \mapsto \frac{\partial p}{\partial x_i}$, where we only keep those elements $\bigg( \frac{\partial}{\partial x_i} , p \bigg)$ that satisfy either $\frac{\partial p}{\partial x_i} \neq 0$ or $\frac{\partial p }{\partial x_i} = 0 \text{ for all } i$ and $p \not\equiv 0$,
  • $\text{and } \equiv0$, the zero polynomial, is the unique accept state.

The cardinality of $\partial f$ is the title of the question.

gxcc95
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    I don't expect there is a general formula. Even for a general monomial this seems like a hard problem. Can you restrict the problem? – blue May 22 '16 at 00:18
  • @blue Hmmmm okay thanks... I initially considered the following as a first step toward a solution to the original question, not wanting to ask because it should be easy; is there a nice formula for the number of derivatives of the $k^{th}$ elementary function in $\mathbb{Z} [x_1 , ... , x^n ]$ depending only on $k$ and $n$? This really is too restrictive though... – gxcc95 May 22 '16 at 00:44
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    @blue I have privovided an answer for a general monomial – ASKASK May 22 '16 at 02:06
  • @ASKASK Thanks; to be sure I understand, your proof is stating a bijection between the set of $n$-tuples with $i^{th}$ entry $a_i$ restricted by $0 \leq a_i \leq n_i$ and the set of partial derivatives of the original monomial, and then you counted the former. Correct? – gxcc95 May 22 '16 at 02:31
  • @gxcc95 that is correct. – ASKASK May 22 '16 at 02:32
  • @gxcc95 by construction is is a subjection and it is an injection because any two elements in the image of the map have different degrees (where you take degree as the ordered tuple of the degrees of the individual variables) – ASKASK May 22 '16 at 02:33
  • @gxcc95 just to be clear though, as far as I know, the technique only applies to monomials. I'm not quite sure how to generalize it to polynomials – ASKASK May 22 '16 at 02:34
  • @ASKASK Oh right. Thanks for correcting me. I came up with the same construction but for some reason (mostly staying awake for too long) I thought that this would not be injective. – blue May 22 '16 at 10:18
  • It looks like the answer for the elementary symmetric function of degree $k$ in $n$ variables is ${n \choose n} + \sum_{i=1}^{k} {n \choose i-1} + 1$. Can anyone verify that this is correct? – gxcc95 May 22 '16 at 20:02
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    @gxcc95 It looks correct. There are three special values: $f$, $0$, $1$ (the result of differentiating against any subset of $k$ variables). All other non-zero derivatives come from a mixture of $0<i<k$ distinct partials. Given any such derivative, the partials are uniquely determined as the variables which don't appear in the derivative (it's easy to see that every other variable survives the differentiation). Thus these are in correspondence to the subsets of size $i$ in $n$ variables. This gives $3 + \sum_{i=1}^{k-1} {n\choose i}$, which is the same as your expression. – Erick Wong May 30 '16 at 20:17
  • @ErickWong Great, thanks! – gxcc95 May 30 '16 at 21:32

1 Answers1

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For monomials, this is not to difficult to solve:

In particular, consider $x_1^{n_1}x_2^{n_2}\cdots x_k^{n_k}$

Then any sequence $(a_1,a_2,\ldots,a_k) \in \{0,1,...,n_1\}\times\{0,1,...,n_2\}\times \cdots \{0,1,\ldots,n_k\}$

Map it to the monomial $\dfrac{d^{a_1+a_2+\cdots+a_k}}{dx_1^{a_1} dx_2^{a_2}\cdots dx_k^{a_k}}x_1^{n_1}x_2^{n_2}\cdots x_k^{n_k}$

Each monomial is unique and covers every possible monomial, except $0$, is of that form. Hence the total is: $$ (n_1+1)(n_2+1)\cdots(n_k+1)+1$$

In your example, $n_1=n_2=1$ so the total number is $(1+1)(1+1)+1=5$.

ASKASK
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