For dual numbers, what does $\epsilon^\epsilon$ equal?

Question

$$\epsilon^\epsilon=?$$ Where $\epsilon^2=0$, $\epsilon\notin\mathbb R$. There is a formula for exponentiation of dual numbers, namely: $$(a+b\epsilon)^{c+d\epsilon}=a^c+\epsilon(bca^{c-1}+da^c\ln a)$$ However, this formula breaks down in multiple places for $\epsilon^\epsilon$, yielding many undefined expressions like $0^0$ and $\ln 0$. So, here's my question: what is $\epsilon^\epsilon$ equal to for dual numbers?

Answer 1

The problem is that you can't isolate $\epsilon^\epsilon$ as a standalone expression like what you're doing.

You need to consider the whole expression so it keeps its sense. So if we go back to your expression: \begin{align} (a+b\epsilon)^{c+d\epsilon}&=\left(a+b\epsilon\right)^c\left(a+b\epsilon\right)^{d\epsilon}\\ &=\left(a^c+ca^{c-1}b\epsilon\right)\left(1+d\log(a)\epsilon\right)\\ &=a^c+a^{c-1}\left(ad\log(a)+bc\right)\epsilon \end{align}

We obtain the relations in the second line by looking at the Taylor expansions for small $\epsilon$ and using the fact that $\epsilon^2=0$.

Answer 2

The function $x^x$ is undefined at $x = \epsilon$ in the dual numbers.

To see this, we note that it's equal to $\exp(x \log x) = 1 + x \log x + \frac{x^2 \log^2 x}{2!} + ...$.

If we put $x = \epsilon$ into that series, the second-order and higher terms vanish. So treating this as a purely formal expression, we would get $\exp(x \log x) = 1 + \epsilon \log \epsilon$.

The problem is that $\log \epsilon$ is not defined for the dual numbers, much for the same reason that $1/\epsilon$ isn't defined, or $\sqrt \epsilon$, etc. There are plenty of other non-Archimedian rings (or fields, usually) where quantities like $\epsilon \log \epsilon$ can exist, but the dual numbers are not one of them. Basically, $\epsilon \log \epsilon$ should be an infinitesimal that's in a greater Archimedean class than $\epsilon$ itself, but in the dual numbers there are no such classes. The problem is pretty similar to asking what $\sqrt x$ evaluates to when $x = \epsilon$; the answer is that you need to go beyond the dual numbers to extend the function $\sqrt x$ to $x = \epsilon$.

Answer 3

To prove that $\epsilon^{\epsilon}$ does not exist

If $\epsilon^{\epsilon}$ has a value in the set of dual numbers and if $$\epsilon^{\epsilon}\not=\epsilon$$ then $$\epsilon^{\epsilon}-\epsilon=a+b\epsilon$$ and either $a$ or $b$ is different from zero. From here $$\epsilon\left[\epsilon^{\epsilon}-\epsilon\right]=\epsilon^2\left[\epsilon^{\epsilon-1}-1\right]=a\epsilon$$ so $a=0$ and $b\not=0$, then $$\epsilon^{\epsilon-1}-1=b\epsilon.$$

Multiplying both sides by $\epsilon$ we get that $$\epsilon^{\epsilon}=\epsilon.$$ This is a contradiction if $\epsilon^{\epsilon}$ exists at all.

---

On the other hand, if we assume again that $\epsilon^{\epsilon}$ exists in the set of dual numbers and if $$\epsilon^{\epsilon}=\epsilon$$ then multiplying both sides with $\epsilon$ we get that $$\epsilon\epsilon^{\epsilon}=0.$$ Since $\epsilon^{\epsilon}\not=0$ according to the indirect hypotheses, $\epsilon$ has to be $0$. That is, $$\epsilon^{\epsilon}\not= \epsilon.$$ This is another contradiction showing that $\epsilon^{\epsilon}$ cannot exist.

Answer 4

Since $f(\varepsilon)=f(0)+f'(0)\varepsilon$,

formally we have

$\varepsilon^\varepsilon=1+\varepsilon+\varepsilon\ln 0$

The quantity $\ln 0$ is infinite with negative sign. We could stop here, but if you want you can write it as a divergent integral:

$\varepsilon^\varepsilon=1+\varepsilon-\varepsilon\int_0^1\frac{dx}x=1+\varepsilon-\varepsilon\gamma -\varepsilon\int_1^\infty\frac{dx}x$

I have no ideas where we can advance further.