5

Usually we define the function $a^b$ over the complexes using the exponential function, as $e^{b\log a}$. This function has some issues with multivalued-ness, but it still more or less satisfies $a^ba^c=a^{b+c}$ and $(a^b)^c=a^{bc}$ give or take a multiple of $2\pi i$. But we often treat this as an extension of the integer power function defined by $$a\uparrow 0:=1\qquad a\uparrow(n+1):=(a\uparrow n)\cdot a\qquad a\uparrow-n:=\frac1{a\uparrow n}.$$

This function is not defined for $0\uparrow -n$, but $0\uparrow 0=1$ and $0\uparrow n=0$ for $n\in\Bbb N$. Yet $0^a$ is not defined at all since it involves $\log 0$ which is undefined.

My question is about what the "right" definition of $0^z$ is that continues to satisfy most of the properties of the integer power function and still extends it. Obviously the extension requirement demands $0^0=1$ and $0^n=0$ for positive integer $n$, and if we use the root function as well to extend the integer power function to rationals when $a$ is a nonnegative real, then we must also demand $0^r=0$ for positive rationals $r$, so the reasonable extension by continuity places $0^x=0$ for all positive reals. For other complex numbers, the equation $0^z=0^{z-1}\cdot 0$ demands that $0^z$ be undefined at negative integers (since $0^0=1$ would violate that equation), but otherwise would seem to demand that $0^z=0$ for all complex $z\notin\Bbb Z$.

Does this match any literature definitions of the general complex power function? What are the literature definitions of the power function?

Edit: The expression $e^{b\log a}$ is most well-behaved in the region $a\in\Bbb R^+$, $b\in\Bbb C$. In this region there are no branch cut issues, and it is continuous and satisfies $(ab)^c=a^cb^c$ and $(a^b)^c=a^{bc}$ when all quantities stay in this domain. A reasonable approach to a proper definition of $0^b$ is to take the limit of $a^b$ as $a\to 0$ from the positive real direction.

Now for $a\in\Bbb R^+$, $|a^b|=e^{\Re[b]\log a}$, and for $\Re[b]>0$ we have $\Re[b]\log a\to-\infty$ and so $a^b\to0$. Thus the above convention of $0^b=0$ is justified for all complex $b$ with $\Re[b]>0$. However, when $\Re[b]<0$, this expression spirals off to infinity as $a\to 0$, and when $\Re[b]=0$ it oscillates with no limit unless $b=0$. Thus this would motivate the definition of $0^b$ undefined when $\Re[b]\le0$ unless $b=0$, in which case again we have $0^0=1$.

Mario Carneiro
  • 28,221
  • 1
    You have illustrated why it is not possible to define $0^z$ in a continuous maner. You require both $0^r=0$ for positive rationals, yet you also require $0^0=1$. – MPW Aug 02 '14 at 03:38
  • @MPW Of course I don't require continuity everywhere, because it is impossible - $e^{b\log a}$ has no limit as $a,b\to 0$, so no continuous function can agree with the usual definition on a dense set. But continuity on a set of full measure (excepting $0$ and the logarithm branch cut) is possible. – Mario Carneiro Aug 02 '14 at 04:41
  • @MPW For any positive real $x$, there is a neighborhood disjoint from zero; the only values which are defined in this neighborhood are $0^r=0$, for positive rational $r$; this set is also dense in the neighborhood, thus the only continuous function on this neighborhood sets $0^x=0$. This is what I mean by "continuous extension": as you can see it does not produce a continuous function, but it is continuous at the extension itself (the irrational numbers). – Mario Carneiro Aug 02 '14 at 15:41
  • @MPW often, Zeta function/Bernoulli numbers come to help when you need to interpolate a function which is otherwise zero but one at one point. This is because $B_n(1)-B_n(0)$ is zero except when $n=1$. – Anixx Jun 27 '22 at 10:46
  • @MPW An example of such kind of interpolation which works perfectly: https://mathoverflow.net/questions/396076/asymptotically-similar-functions-with-opposite-parity-were-they-considered-are – Anixx Jun 27 '22 at 10:48
  • @MPW so, the discontinuity is not of principal importance, because one still can interpolate the function in natural manner. – Anixx Jun 27 '22 at 10:50

2 Answers2

1

In the context of the Riemann sphere, \begin{align} 0^s = \begin{cases} 0 & \Re(s) > 0 \\ \infty & \Re(s) < 0 \\ 1 & s = 0 \end{cases} \end{align}

The case $\Re(s) = 0 \land s \neq 0$ is trickier. The best I can say is

\begin{align} |0^s| = 1 \end{align} for such $s$, since

\begin{align} \lim_{\varepsilon \to 0} |\varepsilon^s| = 1 \end{align}

But the limit of the argument no longer exists. Perhaps a generalized limit might give us an answer there. See the following plot for a visualization:

With[{\[Epsilon] = .0000001}, 
 ComplexPlot[\[Epsilon]^s, {s, -2 - 2 I, 2 + 2 I}, 
  ColorFunction -> "GlobalAbs", Epilog -> {
    Point[{0, 0}],
    Point[{0, 1}]
    }]]

enter image description here

$s = 0$ (the bottom dot) retains the same value no matter the value of $\varepsilon$, but this is not true for $s = i$ (the top dot).

user76284
  • 6,408
0

The function $0^z$ will be constant zero at $\Re (z)>0$, infinite or oscillating otherwise. But there is a way to represent this function in "natural" form (kinda) using divergent integrals. So, let's represent the function $0^x$ in integral form.

For that purpose, I armed myself with divergent integrals, Laplace transforms, etc, etc. Of course, negative powers of zero cannot be represented even as divergent integrals (they have opposite signs around zero), but we can speak about germs of $s^x$ as $s$ approaches $0$ from the above.

So, using the identities from divergent integrals theory,

$$\underset{s\to0^+}{\operatorname{germ}} s^{-n}=\frac{\omega_+^{n+1}-\omega_-^{n+1}}{(n+1)!}$$

where

$$\omega_-^n=B_n(0)+n\int_0^\infty B_{n-1}(x)dx$$

and

$$\omega_+^n=B_n(1)+n\int_0^\infty B_{n-1}(x+1)dx$$

Inserting the values of the divergent integrals, we have:

$$\underset{s\to0^+}{\operatorname{germ}} s^{x}=\frac{B_{1-x}(1)-B_{1-x}+\int_0^{\infty } (x-1) x t^{-x-1} \, dt}{\Gamma (2-x)}$$

We can further rewrite the Bernoulli numbers as Zeta functions:

$$\underset{s\to0^+}{\operatorname{germ}} s^{x}=\frac{(x-1) \zeta (x,1)-(x-1) \zeta (x,0)+\int_0^{\infty } (x-1) x t^{-x-1} \, dt}{\Gamma (2-x)}$$

The formula with Bernoulli polynomials works well in Mathematica at positive and negative integer $x$, as well as at zero, returning the expected divergent integrals or finite values (particularly, $\underset{s\to0^+}{\operatorname{germ}} s^{0}=1$), but the formula with Hurwitz Zeta fails at positive integers, returning expression with the symbol ComplexInfinity.

Anixx
  • 10,161