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Given dual numbers, what would be the value of $0^\varepsilon$ and $\varepsilon^\varepsilon$?

Micah
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Anixx
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2 Answers2

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I suspect that neither one of your expressions has a canonical value. Exponentials can be tricky because the operation of exponentiation conflates two (and a half) loosely-related concepts: the family of power functions $f_n(x):x\to x^n$ that raise $x$ to the $n$th power (and more abstractly, the notion of 'exponential object' in a category that considers $A^B$ as the collection of all the functions from $A$ to $B$) and the (canonical) exponential map $\operatorname{exp}(x):x\to e^x$ that (in a very abstract sense) maps 'local to global'.

For any ring $\mathcal{R}$, the family of functions $f_*$ has a straightforward, canonical definition as a set of functions from $\mathcal{R}$ to $\mathcal{R}$: $f_n$ is just the $n$-times multiplication of $x$ with itself. (On the other hand, these functions don't necessarily have inverses: $f_2$ - that is, the squaring function - is a map from $\mathbb{Z}\to\mathbb{Z}$, but there's no $n\in\mathbb{Z}$ with $f_2(n)=2$.) More broadly, this idea leads to the aforementioned notion of an exponential object in a category as a means of organizing functions from $A$ to $B$ - in this sense, $n^2$ just counts the way you can pick two (ordered, not necessarily distinct) numbers between $1$ and $n$, and likewise $2^n$ counts the number of ways you can pick $n$ 'bits'.

On the other hand, the exponential map is a fundamentally distinct concept; whereas the power maps have range a subset of their domains (since multiplication is a map $\mathcal{R}\times\mathcal{R}\to\mathcal{R}$), the range of an exponential map can be entirely distinct from its domain. For instance, the exponentiation map on a Lie algebra takes elements of that algebra to their 'exponentiations' in a Lie group; see http://en.wikipedia.org/wiki/Exponential_map#Lie_theory for some details on this case. While the two structures are related, they're fundamentally distinct; in a sense the Lie algebra studies local structure, while the Lie group looks at global structure, and exponentiation serves as a sort of 'integration' to show how the local structure extends out into a global structure.

A classic example of this is rotation (for simplicity, about the origin): on a 'local' (or 'instantaneous') level a rotation in three dimensions picks out an axis (the axis being rotated around) and a velocity; in two dimensions it just picks out a velocity (the speed of rotation). The exponential map then takes this to a transformation of the underlying two-dimensional space; it serves as a map (e.g.) $\operatorname{rot}:\mathbb{R}\to SL_2(\mathbb{R})$ from real numbers to $2\times 2$ matrices of real numbers with determinant $1$ that integrates out the 'instantaneous rotation' to the resulting transformation after one (abstract) unit of time. Note that even though this can be written as $e^X$ (and that it happily can be computed using the power series for exponential in this case), it's not a function that takes the two arguments $e$ and $X$; and raises one to the power of the other; it's fundamentally a function that takes a single argument $X$ and gives you its exponential.

The fact that these two distinct concepts of exponentiation (essentially) happen to agree over the real numbers (and that there's a real number $e$ such that $\operatorname{exp}(x) = e^x$) is, in some sense, a deep coincidence; it's not something that you can expect to happen anywhere else. In particular, this means that you shouldn't expect expressions like $\varepsilon^\varepsilon$ to be defined on the dual numbers, because the definition of exponentiation that shows up in the duals is essentially the exponential map rather than an exponential object; its exponentiation is a one-argument function, not a two-argument one, so the expression itself doesn't make sense.

Steven Stadnicki
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  • I see your point but I believe that the dual numbers are defined enough well so that exponentiation $x^y$ (in non-boundary cases) is defined. Am I wrong? Does square function of constant e differ in dual numbers from exp(2)? – Anixx Nov 01 '12 at 19:03
  • After all if such things would not be defined there, the set possibly would not be called a number set. – Anixx Nov 01 '12 at 19:08
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    People call a lot of things 'numbers' that aren't necessarily closed under exponential operations; consider the ring of rational numbers $\mathbb{Q}$, for instance. $\operatorname{exp}(2)$ is the same as $e^2$ over the dual numbers (since both are just real numbers and the exponential map on duals restricted to reals corresponds with the exponential map on the reals), but that doesn't mean that one can define $x^y$ for arbitrary dual numbers $x$ and $y$, any more than one can define $x^y$ on $\mathbb{Q}$ just because it can be defined on $\mathbb{Z}$. – Steven Stadnicki Nov 01 '12 at 19:56
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    well at least any smooth function of one argument can be extended to dual numbers as well following the rule: $f(a+b\varepsilon)=f(a)+f'(a)b\varepsilon$. So $p^{a+b\varepsilon}=p^a+p^a \ln p \varepsilon$ and $(a+b\varepsilon)^p=a^p+p a^{p-1} b \varepsilon$ so the both notions of exponentiation are well defined for the most of the dual numbers. – Anixx Nov 01 '12 at 20:48
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    @Anixx: You can actually extend the two-variable function $f(x,y)=x^y$ to the dual numbers in a similar fashion, generalizing both of your functions. Unfortunately, this only works when $x>0$, and attempts to continuously extend it through the line $x=0$ will fail violently. So this isn't useful. – Micah Nov 01 '12 at 20:50
  • @Micah that is I was asking about. Whether the limits exist and whether they agree in all directions. – Anixx Nov 01 '12 at 20:51
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    @Anixx: Notice that your expression for $p^{a+b\epsilon}$ already contains a $\ln p$. This is sufficient to tell you that the limits you need won't exist. The general expression you get when $a>0$ is $(a+b\varepsilon)^{c+d\varepsilon}=a^c+\varepsilon(b \frac{c}{a} a^c+d a^c \ln a)$; if you play around with this, it's not hard to see that not only is it discontinuous at $a=0$, it's pretty hard to even find a path ending at $a=0$ along which it's continuous. – Micah Nov 01 '12 at 20:55
  • @Micah this is interesting information, it would be great if you could summarize this as an answer. – Anixx Nov 01 '12 at 20:57
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I suspect that $0^\epsilon=0$.

If we define the exponent $\exp(x)$ in the usual way $\exp(x):= 1 + x +\frac{x^2}{2!} + \ldots$ then in the duals for $ k \cdot \epsilon$ is $\exp(k\, \epsilon) = 1 + k\,\epsilon $. http://en.wikipedia.org/wiki/Dual_number

Then if we use $a^b= \exp (\log(a) b) $ we get $\epsilon^\epsilon= 1+ \log (\epsilon) \epsilon$. However, I am not sure that $\log( \epsilon)$ is well defined.

Anixx
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user48672
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