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In this question, one asks whether tropical polynomials are differentiable, and the answer is "yes" in the classical sense.

However, I'm wondering whether tropical polynomials are "tropically differentiable". This means that is there any tropical differentiation operator $\partial^t$ which acts like derivation?

In my opinion, if there exists such $\partial^t:\mathbb{T}[x]\to\mathbb{T}[x]$ where $\mathbb{T}$ is tropical semiring (here superscript $t$ means t of tropical.), it should satisfy:

$$\partial^t\circ\operatorname{trop}=\operatorname{trop}\circ\frac{d}{dx}$$

where $\operatorname{trop}$ is tropicalization operator. This intuition is validated from diagram between ring of polynomials and semiring of tropical polynomials.

With my calculation, I had found:

$$\partial^t(x^{\odot n}) = \operatorname{val}(n)\cdot x^{\odot(n-1)}$$

should be satisfied.

Furthermore, if there is such operator $\partial^t$, it should satisfy:

  1. $\partial^t$ is classically subadditive. That is, $\partial^t$ is tropically submultiplicative (I doubt whether there is such word, submultiplicative.) - in the min-plus algebraic sense.

  2. $\partial^t$ is not tropically linear. Therefore, it does not satisfy Leibniz Law in the tropical sense.

My question is

  • Are there any papers or theories that deals with such operation? If so, can you introduce some basic materials?

  • What can be said about this operator?

Thanks.

Ryoungwoo Jang
  • 199

1 Answers1

1

(First Edit)

I concluded that one can establish a tropical derivation, but its properties are not well-behaving as well as reasonable. My logic is as follows.

First, consider following diagram:

$$\require{AMScd}\begin{CD}p(x) = a_0 + a_1x+\cdots+a_nx^n@>{\frac{d}{dx}}>>p'(x)=a_1+2a_2x+\cdots+na_nx^{n-1}\\ @V{\text{trop}}VV @V{\text{trop}}VV\\ \text{trop}(p)=\text{val}(a_0)\oplus\text{val}(a_1)\odot x^{\odot}\oplus\cdots\oplus\text{val}(a_n)\odot x^{\odot n}@>{\partial^t}>>? \end{CD}$$

To define tropical derivation, $?$ should be tropicalization of $p'$. Therefore,

$$?=\text{trop}(p')=\text{val}(a_1)\oplus\text{val}(2)\odot\text{val}(a_2)\odot x^{\odot1}\oplus\cdots\oplus\text{val}(n)\odot\text{val}(a_n)\odot x^{\odot(n-1)}$$

If we define in this way, the following holds:

  • $\partial^t$ is classically subadditive. That is, $\partial^t$ is tropically submultiplicative. $$\longrightarrow\partial^t(x^{\odot n}+x^{\odot m})=\partial^t(x^{\odot n}\odot x^{\odot m})\leq \partial^t(x^{\odot n})\odot \partial(x^{\odot m})=\partial^t(x^{\odot n})+\partial(x^{\odot m})$$

Unfortunately, linearlity breaks. If we observe $\partial^t(x^{\odot n}\oplus x^{\odot m})$:

  1. Inside-first:

$$\partial^t(x^{\odot n}\oplus x^{\odot m})=\partial^t(\max(n\cdot x, m\cdot x))=\begin{cases}\text{val}(n)\odot x^{n-1}&x\geq0\\\text{val}(m)\odot x^{m-1}&x<0\end{cases}$$

  1. Outside-first:

$$\partial^t(x^{\odot n}\oplus x^{\odot m})=\partial^t(x^{\odot n})\oplus\partial^t(x^{\odot m})=\text{val}(n)\odot x^{n-1}\oplus \text{val }(m)\odot x^{m-1}$$

These two does not yield same result. Therefore this tropical derivation lacks common sense.

Ryoungwoo Jang
  • 199