---
format:
  html:
    html-math-method: katex
---


Of Infinitesimals and Exponents
===============================

Infinitesimal quantities, useful as they were to the development of calculus, have been deprecated in favor of limits. However, it is not the case that their existence is completely unjustified. In fact, it is rather easy to devise matrices whose powers "die off" at different rates:

$$
\epsilon_2 = \begin{pmatrix}
0 & 1 \\ 0 & 0
\end{pmatrix}, ~~
\epsilon_3 = \begin{pmatrix}
0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0
\end{pmatrix}, ~~
\epsilon_3^2 = \begin{pmatrix}
0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0
\end{pmatrix} \\
\epsilon_2^2 = \bf{0},~~ \epsilon_3^3 = \bf{0}
$$

For the purposes of this post, I wish to narrow the kind of infinitesimal under consideration strictly to $\varepsilon = \epsilon_2$.

Their role in justifying calculus is at this point spent, but attempting to devise new "numbers" to see how they interact with preexisting structures is often interesting. For example, what does it mean to take a number to an infinitesimal power (or more generally, a matrix power)?


A Frank Evaluation
------------------

It is understood that exponentials and logarithms are inverses, therefore it might make sense to argue:

$$
x^\varepsilon =
e^{\ln{x^\varepsilon}} =
e^{\varepsilon \ln{x}}
$$

Since $e^x$ has a very nice power series, we can try plugging this expression in:

$$
\begin{align*}
e^x &= \sum_{n=0}^\infty {x^n \over n!}
= 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + ... \\
e^{\varepsilon x} &= \sum_{n=0}^\infty {(\varepsilon x)^n \over n!}
= 1 + {\varepsilon x} + {(\varepsilon x)^2 \over 2!} + {(\varepsilon x)^3 \over 3!} + ... \\
&= 1 + {\varepsilon x} \\
e^{\varepsilon \ln x} &= 1 + {\varepsilon \ln x}
\end{align*}
$$

As an infinitesimal, $\varepsilon$ very naturally transforms an infinite sum into a finite one. $\varepsilon$ is realized as a square matrix (since otherwise powers could not exist), so evaluating the exponential more directly:

$$
\begin{align*}
&\phantom{=\ }
e^{\varepsilon \ln{x}} = \exp{\begin{pmatrix}0 & \ln{x} \\ 0 & 0\end{pmatrix}} \\
\exp{\begin{pmatrix}0 & \ln{x} \\ 0 & 0 \end{pmatrix} } &=
\begin{pmatrix}0 & \ln{x} \\ 0 & 0 \end{pmatrix}^0 +
\begin{pmatrix}0 & \ln{x} \\ 0 & 0 \end{pmatrix}^1 +
{1 \over 2}\begin{pmatrix}0 & \ln{x} \\ 0 & 0\end{pmatrix}^2 + ... \\ &=
\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} +
\begin{pmatrix}0 & \ln{x} \\ 0 & 0\end{pmatrix} +
{1 \over 2}
\begin{pmatrix}0 & \ln{x} \\ 0 & 0\end{pmatrix}
\begin{pmatrix}0 & \ln{x} \\ 0 & 0\end{pmatrix} + ... \\ &=
\begin{pmatrix}1 & \ln{x} \\ 0 & 1\end{pmatrix} = 1 + \varepsilon \ln x
\end{align*}
$$

Of course, since $\varepsilon$ and its matrix form are equivalent, the answer is the same as before.

There are multiple problems with this argument, which are signified by reexamining the earlier equation:

$$
x^\varepsilon \stackrel{?_1}{=}
e^{\ln{x^\varepsilon}} \stackrel{?_2}{=}
e^{\varepsilon \ln{x}}
$$

1. It is not obvious that $x^\varepsilon$ is reconcilable with $e^x$ and the natural logarithm, i.e., that composing them is still an identity with respect to $\varepsilon$
2. The power identity for logarithms may not be obeyed by $\varepsilon$

Additionally, there were only two terms used in the series for $e^x$; there are many more power series that begin with terms $1, 1...$, and the natural logarithm('s series) is simply the one that corresponds to the inverse of $e^x$. Are there other compositions of a series and its inverse that can be considered?


Teeming with a lot of News
--------------------------

Fortunately, there is not precisely one way to identify the value of $x^\varepsilon$, and one in particular has much less handwaving. The binomial theorem is a very useful tool for writing the power of a sum of numbers:

$$
(x + 1)^n = \sum_{r=0}^n {n \choose r}x^r = \sum_{r=0}^n {n! \over {r!(n-r)!}}x^k
$$

If the binomial coefficient is asserted to be 0 for $r > n$, then the binomial theorem can also be written as an infinite sum. However, the denominator of ${n! \over (n-r)!}$ doesn't make sense, since it will be a negative factorial in this circumstance. On the other hand, multiplying *n* with the r numbers immediately below it can be assigned a new symbol $(n)_r$ (named the [Pochhammer symbol](https://en.wikipedia.org/wiki/Falling_and_rising_factorials)). This falling factorial satisfies the 0 rule since if *n* is an integer smaller than *r*, then the product will include 0, annihilating all other terms.

$$
(n)_0 = 1,~ (n)_r = (n -\ r + 1)(n)_{r-1} \\
(x + 1)^n = \sum_{r=0}^\infty {(n)_r \over r!} x^r
$$

However, if *n* is not an integer, then the series will miss 0, and continue indefinitely. This directly gives the series for square root:

$$
\begin{align*}
\sqrt{x + 1} &= (x + 1)^{1/2} =
\sum_{r=0}^\infty {(1/2)_r \over r!} x^r \\ &=
1 + {1 \over 2}x
+ \left({1 \over 2!}\right)
\left({1 \over 2}\right)
\left({1 \over 2} -\ 1\right)x^2 + ... \\ &=
1 + {1 \over 2}x + \left({1 \over 2!}\right)
\left(-{1 \over 4}\right)x^2 +
\left({1 \over 3!}\right)
\left(-{1 \over 4} \right)
\left({1 \over 2} -\ 2 \right)x^3 + ...\\ &=
1 + {1 \over 2}x -\ {1 \over 8}x^2 +
\left({1 \over 3!}\right)
\left({3 \over 8} \right)x^3 +
\left({1 \over 4!}\right)
\left({3 \over 8} \right)
\left({1 \over 2} -\ 3 \right)x^4 + ...\\ &=
1 + {1 \over 2}x -\ {1 \over 8}x^2 +
{1 \over 16}x^3 -\ {5 \over 128}x^4 + ...
\end{align*}
$$


Tumbling down Infinitesimals
----------------------------

Since this definition works for rational numbers as well as integers, is it possible to assign a value to $(\varepsilon)_r$? Indeed it is, since this symbol's definition only requires that integers can be subtracted from it and that it can multiply with other numbers. In other words, it works in any integer ring, a property which the matrices underlying $\varepsilon$ very fortunately have.

$$
\begin{align*}
(\varepsilon)_0 &= 1,~ (\varepsilon)_r = (\varepsilon -\ r + 1)(\varepsilon)_{r-1} \\
(\varepsilon)_1 &= (\varepsilon -\ 0)(1) = \varepsilon\\
(\varepsilon)_2 &= (\varepsilon -\ 1)(\varepsilon)
= \varepsilon^2 -\ \varepsilon = -\varepsilon \\
(\varepsilon)_3 &= (\varepsilon -\ 2)(-\varepsilon)
= -\varepsilon^2 + 2\varepsilon = 2\varepsilon \\
(\varepsilon)_4 &= (\varepsilon -\ 3)(2\varepsilon)
= 2\varepsilon^2 -\ 6\varepsilon = -6\varepsilon \\
& ... \\
(\varepsilon)_r &= (-1)^{r-1}(r-1)!\varepsilon,~ r > 0
\end{align*}
$$

It is easy to check that this can also be directly computed from the matrix underlying $\varepsilon$. With this expression in mind, it can now be plugged into the binomial formula:

$$
\begin{align*}
(x + 1)^\varepsilon &= \sum_{r=0}^\infty {(\varepsilon)_r \over r!} x^r
= 1 + \sum_{r=1}^\infty {(\varepsilon)_r \over r!} x^r \\ &=
1 + \sum_{r=1}^\infty {(-1)^{r-1}(r-1)!\varepsilon \over r!} x^r \\ &=
1 + \varepsilon \sum_{r=1}^\infty {(-1)^{r-1} \over r} x^r
\end{align*}
$$

The final sum may be familiar, but let's hold off from hastily cross-referencing a table. The term inside the sum looks a lot like an integral, so simplifying:

$$
\begin{align*}
\sum_{r=1}^\infty {(-1)^{r-1} \over r} x^r &=
\sum_{r=1}^\infty \int {(-1)^{r-1}} x^{r-1} =
\int \sum_{r=0}^\infty {(-1)^{r}} x^{r} \\ &=
\int \sum_{r=0}^\infty (-x)^{r} =
\int {1 \over 1 + x}
\end{align*}
$$

This is looking very promising! Substituting this expression for the previous sum, we can now conclude:

$$
\begin{align*}
(x + 1)^\varepsilon &= 1 + \varepsilon \int {1 \over 1 + x} \\
x^\varepsilon &= 1 + \varepsilon \int {1 \over x} \\
&= 1 + \varepsilon \ln x
\end{align*}
$$

Fortunately, this slightly more grounded approach agrees with the initial result.


Other Elementary Algebraic Powers
---------------------------------


### Complex

Extending the Pochhammer symbol to other algebraic objects works in an analogous way. For example, the original approach with the imaginary unit *i* yields:

$$
x^i = e^{\ln{x^i}} = e^{i\ln x} = \cos(\ln x) + i\sin(\ln x)
$$

Since sine is odd and cosine is even, taking the reciprocal of $x^i$ (conjugating the exponent) just conjugates the result of the exponentiation. To be slightly more rigorous, calculating the real and imaginary components of $(i)_r$ and tabulating the results:

| r | Real | Imaginary |
|---|------|-----------|
| 0 |    1 |         0 |
| 1 |    0 |         1 |
| 2 |   -1 |        -1 |
| 3 |    3 |         1 |
| 4 |  -10 |         0 |
| 5 |   40 |       -10 |

The real component corresponds to [OEIS A003703](http://oeis.org/A003703), and the imaginary to [OEIS A009454](http://oeis.org/A009454), which state that they have exponential generating functions $\cos(\ln x)$ and $\sin(\ln x)$ respectively, agreeing with the intuitive series above. In fact, reexamining the binomial formula, it obviously produces exponential generating functions (in $x -\ 1$) based on the series given by the Pochhammer symbol.

$$
(x + 1)^n = \sum_{r=0}^\infty {(n)_r \over r!} x^r \Leftrightarrow
x^n = \sum_{r=0}^\infty {(n)_r \over r!} (x-1)^r \\
(x -\ 1)^n = \sum_{r=0}^\infty {(-1)^r (n)_r \over r!} x^r \Leftrightarrow
x^n = \sum_{r=0}^\infty {(-1)^r (n)_r \over r!} (x+1)^r
$$


### Split-complex

How about the hyperbolic analogue of the above circular approach? We want a series that equates to $\cosh(\ln x) + j\sinh(\ln x)$, for some algebraic element *j*. Unlike with *i*, the exponential series does not need to alternate, but merely be partitioned into even and odd components. Therefore, *j* has the property that $j^2 = 1,~ j \neq 1$. It has a very simple matrix presentation and Pochhammer sequence.

$$
\begin{align*}
j &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\
(j)_2 &= j(j -\ 1) = 1 -\ j \\ (j)_3 &= (1 -\ j)(j -\ 2) = -3 + 3j \\
(j)_r &= f(r)(1 -\ j) \\
(j)_{r+1} &= f(r+1)(1 -\ j) = f(r)(1 -\ j)(j -\ r) \\
&= f(r) \left( (r+1)j - (r+1)^{\vphantom{2}} \right)
= -f(r)(r+1)(1 -\ j) \\
f(r) &= -rf(r -\ 1)(1 -\ j) \implies \text{$f$ alternates and contains $r!$} \\
&= {r!(-1)^r \over 2}, r \ge 2
\end{align*}
$$

Since this Pochhammer sequence contains $r!$, it pleasingly cancels out with the denominator of the binomial coefficient.

$$
\begin{align*}
x^j &= \sum_{r=0}^\infty {(j)_r \over r!} (x-1)^r \\
&= 1 + j(x-1) + (1-j)\sum_{r=2}^\infty {r!(-1)^r \over 2r!} (x -\ 1)^r \\
&= \phantom{+ j} \left(1 + {1 \over 2}\sum_{r=2}^\infty (-1)^r (x -\ 1)^r \right) \\
&\phantom{=} + j \left(x -\ 1 -\ {1 \over 2}\sum_{r=2}^\infty (-1)^r (x -\ 1)^r \right) \\
\sum_{r=2}^\infty (-1)^r (x-1)^r &=
\sum_{r=0}^\infty (-1)^r (x-1)^r ~-~ \sum_{r=0}^1 (-1)^r (x-1)^r \\
&= {1 \over 1 -\ (x -\ 1)} -\ (1 -\ (x -\ 1)) = {1 \over x} + x -\ 2 \\
&= {x^2 -\ 2x + 1 \over x} \\
x^j &= \left(1 + {x^2 -\ 2x + 1 \over 2x} \right)
+ j\left(x -\ 1 -\ {x^2 -\ 2x + 1 \over 2x} \right) \\
&= {x^2 + 1 \over 2x} + j{x^2 -\ 1 \over 2x}
\end{align*}
$$

The final expression is equivalent to the expression in cosh and sinh after writing them in terms of the exponential function and simplifying the natural logarithm. Notably, the components of $t^j$ (switching variables to avoid confusion) parametrize the hyperbola $x^2 -\ y^2 = 1$. The components of $t^i$ also do so for the circle $x^2 + y^2 = 1$ (since the range of the natural logarithm is over all reals), but fail to do so rationally.


### Golden

The golden ratio also has a very simple relationship with its square, in particular $\phi^2 = \phi + 1$. This minimal polynomial has degree 2, so numbers can be represented by 2-tuples of the form $a + b\phi$. The Pochhammer sequence of $\phi$ is:

| r | Real |  Phi |
|---|------|------|
| 0 |    1 |    0 |
| 1 |    0 |    1 |
| 2 |    1 |    0 |
| 3 |   -2 |    1 |
| 4 |    7 |   -4 |
| 5 |  -32 |  -19 |

Removing the alternation, both sequences can be located in the OEIS; the real component is [OEIS A265165](http://oeis.org/A265165), and the imaginary component is [OEIS A306183](http://oeis.org/A306183) (alternating at [A323620](http://oeis.org/A323620)).


Infinitive
----------

If we're working with exponential-type series, wouldn't it be nice if $x^\omega$ was an expression in $e^x$ to match $x^\varepsilon$ for some $\omega$? The simplest "solution" is by allowing some algebraic element all of whose Pochhammer symbols (besides the 0th) are itself. In this case, we obtain:

$$
\sum_{r=0}^\infty {(\omega)_r \over r!} (x-1)^r =
1 + \omega \sum_{r=1}^\infty {(x-1)^r \over r!} =
1 + \omega (e^{x-1}-1)
$$

But the representation of $\omega$ is a problem. If all but one of its Pochhammer symbols are 1, then it satisfies all of the following:

$$
\begin{align*}
(\omega)_2 &= (\omega -\ 1)\omega = \omega^2 -\ \omega = \omega \\
(\omega)_3 &= (\omega -\ 2)\omega = \omega^2 -\ 2\omega = \omega \\
(\omega)_4 &= (\omega -\ 3)\omega = \omega^2 -\ 3\omega = \omega \\
... \\
(\omega)_r &= (\omega -\ r)\omega = \omega^2 -\ r\omega = \omega \\
&= \omega^2 -\ (r+1)\omega = 0,~ r \ge 1
\end{align*}
$$

The simplest element that satisfies these relationships is obviously 0. Trying a little harder, maybe it is more the case that $\omega$ is some sort of infinite fixed point, such that $\omega^2$ is "bigger" than $\omega$ such that subtraction by a multiple of the latter is still positive. In fact, it is so much bigger that is unchanged, similarly to $\omega$'s relationship to the natural numbers, and it makes sense to normalize it down to its square root. Whatever $\omega$ actually represents, it cannot be captured in a matrix.


Infinitesimal Duality
---------------------

Of course, there is another infinitesimal that we should not forget to consider: the transpose of $\epsilon_2$, denoted hence $\varepsilon'$. Notice that $\varepsilon + \varepsilon' = j$. However, these two quantities do not  play nice with each other. For example, is it the case that $x^\varepsilon x^{\varepsilon'} \stackrel{?}{=} x^{\varepsilon + \varepsilon'} = x^j$? Surprisingly, no. Starting with converting the left hand side to a matrix product:

$$
\begin{align*}
\begin{pmatrix} 1 & \ln x \\ 0 & 1\end{pmatrix}
\begin{pmatrix} 1 & 0 \\ \ln x & 1\end{pmatrix} &=
\begin{pmatrix} (\ln x)^2 + 1 & \ln x \\ \ln x & 1 \end{pmatrix} = \\
1 + j\ln x + \begin{pmatrix} (\ln x)^2 & 0 \\ 0 & 0 \end{pmatrix} &\neq
1 + j\ln x + \begin{pmatrix} 0 & 0 \\ 0 & (\ln x)^2 \end{pmatrix} \\ =
\begin{pmatrix} 1 & \ln x \\ \ln x & (\ln x)^2 + 1 \end{pmatrix} &=
\begin{pmatrix} 1 & 0 \\ \ln x & 1 \end{pmatrix}
\begin{pmatrix} 1 & \ln x \\ 0 & 1 \end{pmatrix}
\end{align*}
$$

Numbers of the form $a + b\varepsilon$ do not commute with numbers of the form $a + b\varepsilon'$; moreover, neither of these expressions are $x^j$. Since $\varepsilon = {i + j \over 2}$, but *i* and *j* do not commute, the main diagonal (where the identity occupies) becomes polluted. The placement of the $(\ln x)^2$ term is the phantom of another matrix *k* that obeys $k^2 = 1,~ k \neq j \neq 1$.

$$
\begin{align*}
k &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\
1 + j\ln x + \begin{pmatrix} (\ln x)^2 & 0 \\ 0 & 0 \end{pmatrix} &=
1 + j\ln x + {(\ln x)^2 \over 2} + k{(\ln x)^2 \over 2} \\
1 + j\ln x + \begin{pmatrix} 0 & 0 \\ 0 & (\ln x)^2 \end{pmatrix} &=
1 + j\ln x + {(\ln x)^2 \over 2} -\ k{(\ln x)^2 \over 2}
\end{align*}
$$

What about 1 + *j* (or equivalently 1 + *k*)? In either case, the result has 0 determinant, and in *k*'s, it is a skew-infinitesimal along the main diagonal. The Pochhammer sequence terminates, which makes sense:  the *x* term is cleared from the denominator, eliminating the need for a series.

$$
\begin{align*}
(j + 1)_1 &= (j + 1), ~ (j + 1)_2 = (j + 1), ~ (j + 1)_3 = 0 \\
x(x^j) &= {x^2+1 \over 2} + j{x^2-1 \over 2} \\
x^{1 + j} &= 1 + (1 + j)(x -\ 1) + {1 \over 2}(1 + j)(x -\ 1)^2 \\
&= {(x -\ 1)^2 \over 2} + x + j\left({(x -1)^2 \over 2} + x -\ 1 \right) = x(x^j)
\end{align*}
$$

Fortunately, this implies that addition of exponents is preserved in some circumstances. This expression also has a square root (since 1 + *j* has all even components):

$$
\sqrt{x^{j + 1}} = 1 + {j + 1 \over 2}(x -\ 1) =
{x+1 \over 2} + j{x-1 \over 2}
$$

It is good to remember that the natural logarithm of *x* grows slower than any positive power of *x*. Similarly, infinitesimals can be imagined as smaller than any positive rational number. In some way, taking a number to an infinitesimal power is related to a function that grows slower than all rational powers, albeit only within an infinitesimal "space".

Generalizing slightly, it becomes obvious that algebraic powers can be assigned a series, which can have an interesting representation in transcendental functions.
