## Exploring 0^0, x^x, and z^z.

My Youtube home page knows that I’m geeky enough to watch math videos.  Today it suggested Eddie Woo’s video about $$0^0$$.

Mr Woo, who has great enthusiasm, and must be an awesome teacher to have in person.  He reminds his class about the exponent laws, which allow for an interpretation that $$0^0$$ would be equal to 1.  He points out that $$0^n = 0$$ for any positive integer, which admits a second contradictory value for $$0^0$$, if this was true for $$n=0$$ too.

When reviewing the exponent laws Woo points out that the exponent law for subtraction $$a^{n-n}$$ requires $$a$$ to be non-zero.  Given that restriction, we really ought to have no expectation that $$0^{n-n} = 1$$.

To attempt to determine a reasonable value for this question, resolving the two contradictory possibilities, neither of which we actually have any reason to assume are valid possibilities, he asks the class to perform a proof by calculator, computing a limit table for $$x \rightarrow 0+$$. I stopped at that point and tried it by myself, constructing such a table in Mathematica. Here is what I used

griddisp[labelc1_, labelc2_, f_, values_] := Grid[({
({{labelc1}, values}) // Flatten,
({ {labelc2}, f[#] & /@ values} ) // Flatten
}) // Transpose,
Frame -> All]
decimalFractions[n_] := ((10^(-#)) & /@ Range[n])
With[{m = 10}, griddisp[x, x^x, #^# &, N[decimalFractions[m], 10]]]
With[{m = 10}, griddisp[x, x^x, #^# &, -N[decimalFractions[m], 10]]]


Observe that I calculated the limits from both above and below. The results are

and for the negative limit

Sure enough, from both below and above, we see numerically that $$\lim_{\epsilon\rightarrow 0} \epsilon^\epsilon = 1$$, as if the exponent law argument for $$0^0 = 1$$ was actually valid.  We see that this limit appears to be valid despite the fact that $$x^x$$ can be complex valued — that is ignoring the fact that a rigorous limit argument should be valid for any path neighbourhood of $$x = 0$$ and not just along two specific (real valued) paths.

Let’s get a better idea where the imaginary component of $$(-x)^{-x}$$ comes from.  To do so, consider $$f(z) = z^z$$ for complex values of $$z$$ where $$z = r e^{i \theta}$$. The logarithm of such a beast is

\label{eqn:xtox:20}
\begin{aligned}
\ln z^z
&= z \ln \lr{ r e^{i\theta} } \\
&= z \ln r + i \theta z \\
&= e^{i\theta} \ln r^r + i \theta z \\
&= \lr{ \cos\theta + i \sin\theta } \ln r^r + i r \theta \lr{ \cos\theta + i \sin\theta } \\
&= \cos\theta \ln r^r – r \theta \sin\theta
+ i r \lr{ \sin\theta \ln r + \theta \cos\theta },
\end{aligned}

so
\label{eqn:xtox:40}
z^z =
e^{ r \lr{ \cos\theta \ln r – \theta \sin\theta}} \times
e^{i r \lr{ \sin\theta \ln r + \theta \cos\theta }}.

In particular, picking the $$\theta = \pi$$ branch, we have, for any $$x > 0$$
\label{eqn:xtox:60}
(-x)^{-x} = e^{-x \ln x – i x \pi } = \frac{e^{ – i x \pi }}{x^x}.

Let’s get some visual appreciation for this interesting $$z^z$$ beastie, first plotting it for real values of $$z$$


Manipulate[
Plot[ {Re[x^x], Im[x^x]}, {x, -r, r}
, PlotRange -> {{-r, r}, {-r^r, r^r}}
, PlotLegends -> {Re[x^x], Im[x^x]}
], {{r, 2.25}, 0.0000001, 10}]


From this display, we see that the imaginary part of $$x^x$$ is zero for integer values of $$x$$.  That’s easy enough to verify explicitly: $$(-1)^{-1} = -1, (-2)^{-2} = 1/4, (-3)^{-3} = -1/27, \cdots$$.

The newest version of Mathematica has a few nice new complex number visualization options.  Here’s two that I found illuminating, an absolute value plot that highlights the poles and zeros, also showing some of the phase action:

Manipulate[
ComplexPlot[ x^x, {x, s (-1 – I), s (1 + I)},
PlotLegends -> Automatic, ColorFunction -> "GlobalAbs"], {{s, 4},
0.00001, 10}]

We see the branch cut nicely, the tendency to zero in the left half plane, as well as some of the phase periodicity in the regions that are in the intermediate regions between the zeros and the poles.  We can also plot just the phase, which shows its interesting periodic nature


Manipulate[
ComplexPlot[ x^x, {x, s (-1 – I), s (1 + I)},
PlotLegends -> Automatic, ColorFunction -> "CyclicArg"], {{s, 6},
0.00001, 10}]


I’d like to take the time to play with some of the other ComplexPlot ColorFunction options, which appears to be a powerful and flexible visualization tool.

## Motivation

I initially thought that I might submit a problem set solution for ece1228 using Geometric Algebra. In order to justify this, I needed to add an appendix to that problem set that outlined enough of the ideas that such a solution might make sense to the grader.

I ended up changing my mind and reworked the problem entirely, removing any use of GA. Here’s the tutorial I initially considered submitting with that problem.

## Geometric Algebra in a nutshell.

Geometric Algebra defines a non-commutative, associative vector product

\label{eqn:gaTutorial:20}
\begin{aligned}
\Ba \Bb \Bc
&=
(\Ba \Bb) \Bc \\
&=
\Ba (\Bb \Bc),
\end{aligned}

where the square of a vector equals the squared vector magnitude

\label{eqn:gaTutorial:40}
\Ba^2 = \Abs{\Ba}^2,

In Euclidean spaces such a squared vector is always positive, but that is not necessarily the case in the mixed signature spaces used in special relativity.

There are a number of consequences of these two simple vector multiplication rules.

• Squared unit vectors have a unit magnitude (up to a sign). In a Euclidean space such a product is always positive

\label{eqn:gaTutorial:60}
(\Be_1)^2 = 1.

• Products of perpendicular vectors anticommute.

\label{eqn:gaTutorial:80}
\begin{aligned}
2
&=
(\Be_1 + \Be_2)^2 \\
&= (\Be_1 + \Be_2)(\Be_1 + \Be_2) \\
&= \Be_1^2 + \Be_2 \Be_1 + \Be_1 \Be_2 + \Be_2^2 \\
&= 2 + \Be_2 \Be_1 + \Be_1 \Be_2.
\end{aligned}

A product of two perpendicular vectors is called a bivector, and can be used to represent an oriented plane. The last line above shows an example of a scalar and bivector sum, called a multivector. In general Geometric Algebra allows sums of scalars, vectors, bivectors, and higher degree analogues (grades) be summed.

Comparison of the RHS and LHS of \ref{eqn:gaTutorial:80} shows that we must have

\label{eqn:gaTutorial:100}
\Be_2 \Be_1 = -\Be_1 \Be_2.

It is true in general that the product of two perpendicular vectors anticommutes. When, as above, such a product is a product of
two orthonormal vectors, it behaves like a non-commutative imaginary quantity, as it has an imaginary square in Euclidean spaces

\label{eqn:gaTutorial:120}
\begin{aligned}
(\Be_1 \Be_2)^2
&=
(\Be_1 \Be_2)
(\Be_1 \Be_2) \\
&=
\Be_1 (\Be_2
\Be_1) \Be_2 \\
&=
-\Be_1 (\Be_1
\Be_2) \Be_2 \\
&=
-(\Be_1 \Be_1)
(\Be_2 \Be_2) \\
&=-1.
\end{aligned}

Such “imaginary” (unit bivectors) have important applications describing rotations in Euclidean spaces, and boosts in Minkowski spaces.

• The product of three perpendicular vectors, such as

\label{eqn:gaTutorial:140}
I = \Be_1 \Be_2 \Be_3,

is called a trivector. In \R{3}, the product of three orthonormal vectors is called a pseudoscalar for the space, and can represent an oriented volume element. The quantity $$I$$ above is the typical orientation picked for the \R{3} unit pseudoscalar. This quantity also has characteristics of an imaginary number

\label{eqn:gaTutorial:160}
\begin{aligned}
I^2
&=
(\Be_1 \Be_2 \Be_3)
(\Be_1 \Be_2 \Be_3) \\
&=
\Be_1 \Be_2 (\Be_3
\Be_1) \Be_2 \Be_3 \\
&=
-\Be_1 \Be_2 \Be_1
\Be_3 \Be_2 \Be_3 \\
&=
-\Be_1 (\Be_2 \Be_1)
(\Be_3 \Be_2) \Be_3 \\
&=
-\Be_1 (\Be_1 \Be_2)
(\Be_2 \Be_3) \Be_3 \\
&=

\Be_1^2
\Be_2^2
\Be_3^2 \\
&=
-1.
\end{aligned}

• The product of two vectors in \R{3} can be expressed as the sum of a symmetric scalar product and antisymmetric bivector product

\label{eqn:gaTutorial:480}
\begin{aligned}
\Ba \Bb
&=
\sum_{i,j = 1}^n \Be_i \Be_j a_i b_j \\
&=
\sum_{i = 1}^n \Be_i^2 a_i b_i
+
\sum_{0 < i \ne j \le n} \Be_i \Be_j a_i b_j \\ &= \sum_{i = 1}^n a_i b_i + \sum_{0 < i < j \le n} \Be_i \Be_j (a_i b_j - a_j b_i). \end{aligned} The first (symmetric) term is clearly the dot product. The antisymmetric term is designated the wedge product. In general these are written $$\label{eqn:gaTutorial:500} \Ba \Bb = \Ba \cdot \Bb + \Ba \wedge \Bb,$$ where \label{eqn:gaTutorial:520} \begin{aligned} \Ba \cdot \Bb &\equiv \inv{2} \lr{ \Ba \Bb + \Bb \Ba } \\ \Ba \wedge \Bb &\equiv \inv{2} \lr{ \Ba \Bb - \Bb \Ba }, \end{aligned} The coordinate expansion of both can be seen above, but in \R{3} the wedge can also be written $$\label{eqn:gaTutorial:540} \Ba \wedge \Bb = \Be_1 \Be_2 \Be_3 (\Ba \cross \Bb) = I (\Ba \cross \Bb).$$ This allows for an handy dot plus cross product expansion of the vector product $$\label{eqn:gaTutorial:180} \Ba \Bb = \Ba \cdot \Bb + I (\Ba \cross \Bb).$$ This result should be familiar to the student of quantum spin states where one writes $$\label{eqn:gaTutorial:200} (\Bsigma \cdot \Ba) (\Bsigma \cdot \Bb) = (\Ba \cdot \Bb) + i (\Ba \cross \Bb) \cdot \Bsigma.$$ This correspondence is because the Pauli spin basis is a specific matrix representation of a Geometric Algebra, satisfying the same commutator and anticommutator relationships. A number of other algebra structures, such as complex numbers, and quaterions can also be modelled as Geometric Algebra elements.

• It is often useful to utilize the grade selection operator
$$\gpgrade{M}{n}$$ and scalar grade selection operator $$\gpgradezero{M} = \gpgrade{M}{0}$$
to select the scalar, vector, bivector, trivector, or higher grade algebraic elements. For example, operating on vectors $$\Ba, \Bb, \Bc$$, we have

\label{eqn:gaTutorial:580}
\begin{aligned}
&= \Ba \cdot \Bb \\
&=
\Ba (\Bb \cdot \Bc)
+
\Ba \cdot (\Bb \wedge \Bc) \\
&=
\Ba (\Bb \cdot \Bc)
+
(\Ba \cdot \Bb) \Bc

(\Ba \cdot \Bc) \Bb \\
\Ba \wedge \Bb \\
\Ba \wedge \Bb \wedge \Bc.
\end{aligned}

Note that the wedge product of any number of vectors such as $$\Ba \wedge \Bb \wedge \Bc$$ is associative and can be expressed in terms of the complete antisymmetrization of the product of those vectors. A consequence of that is the fact a wedge product that includes any colinear vectors in the product is zero.

## Example: Helmholz equations.

As an example of the power of \ref{eqn:gaTutorial:180}, consider the following Helmholtz equation derivation (wave equations for the electric and magnetic fields in the frequency domain.)

Application of \ref{eqn:gaTutorial:180} to
Maxwell equations in the frequency domain for source free simple media gives

\label{eqn:emtProblemSet1Problem6:340}
\label{eqn:emtProblemSet1Problem6:360}
\spacegrad \BE = -j \omega I \BB

\label{eqn:emtProblemSet1Problem6:380}
\spacegrad I \BB = -j \omega \mu \epsilon \BE.

These equations use the engineering (not physics) sign convention for the phasors where the time domain fields are of the form $$\boldsymbol{\mathcal{E}}(\Br, t) = \textrm{Re}( \BE e^{j\omega t}$$.

Operation with the gradient from the left produces the Helmholtz equation for each of the fields using nothing more than multiplication and simple substitution

\label{eqn:emtProblemSet1Problem6:400}
\label{eqn:emtProblemSet1Problem6:420}
\spacegrad^2 \BE = – \mu \epsilon \omega^2 \BE

\label{eqn:emtProblemSet1Problem6:440}
\spacegrad^2 I \BB = – \mu \epsilon \omega^2 I \BB.

There was no reason to go through the headache of looking up or deriving the expansion of $$\spacegrad \cross (\spacegrad \cross \BA )$$ as is required with the traditional vector algebra demonstration of these identities.

Observe that the usual Helmholtz equation for $$\BB$$ doesn’t have a pseudoscalar factor. That result can be obtained by just cancelling the factors $$I$$ since the \R{3} Euclidean pseudoscalar commutes with all grades (this isn’t the case in \R{2} nor in Minkowski spaces.)

## Example: Factoring the Laplacian.

There are various ways to demonstrate the identity

\label{eqn:gaTutorial:660}

such as the use of (somewhat obscure) tensor contraction techniques. We can also do this with Geometric Algebra (using a different set of obscure techniques) by factoring the Laplacian action on a vector

\label{eqn:gaTutorial:700}
\begin{aligned}
&=
&=
&=
+
%+
&=
+
\end{aligned}

Should we wish to express the last term using cross products, a grade one selection operation can be used
\label{eqn:gaTutorial:680}
\begin{aligned}
&=
&=
&=
&=
&=
\end{aligned}

Here coordinate expansion was not required in any step.