634 lines
18 KiB
Plaintext
634 lines
18 KiB
Plaintext
---
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title: "Of Infinitesimals and Exponents"
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description: |
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Is there such a thing as an raising a number to an infinitesimal power?
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format:
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html:
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html-math-method: katex
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date: "2021-04-16"
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date-modified: "2025-07-31"
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categories:
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- algebra
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- power series
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---
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Infinitesimal quantities, useful as they were to the development of calculus,
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have been deprecated in favor of limits.
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However, it is not the case that their existence is completely unjustified.
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In fact, it is rather easy to devise matrices whose powers "die off" at different rates:
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$$
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\begin{gather*}
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\epsilon_2 = \begin{pmatrix}
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0 & 1 \\
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0 & 0
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\end{pmatrix}, ~~
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\epsilon_3 = \begin{pmatrix}
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0 & 1 & 0 \\
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0 & 0 & 1 \\
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0 & 0 & 0
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\end{pmatrix}, ~~
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\epsilon_3^2 = \begin{pmatrix}
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0 & 0 & 1 \\
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0 & 0 & 0 \\
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0 & 0 & 0
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\end{pmatrix}
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\\
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\epsilon_2^2 = \bf{0}, ~~
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\epsilon_3^3 = \bf{0}
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\end{gather*}
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$$
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For the purposes of this post, I wish to narrow the kind of infinitesimal under consideration strictly
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to $\varepsilon = \epsilon_2$.
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Their role in justifying calculus is at this point spent.
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However, it is also fun (and often interesting) to devise new "numbers"
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to see how they interact with preexisting structures.
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For example, what does it mean to take a number to an infinitesimal power
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(or more generally, a matrix power)?
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A Frank Evaluation
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------------------
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Since exponentials and logarithms are inverses, it might make sense to argue:
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$$
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x^\varepsilon
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= e^{\ln{x^\varepsilon}}
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= e^{\varepsilon \ln{x}}
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$$
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$e^x$ has a very nice power series, so we can try plugging this expression in:
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$$
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\begin{align*}
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e^x &= \sum_{n=0}^\infty {x^n \over n!}
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= 0 + x + {x^2 \over 2!} + {x^3 \over 3!} + ...
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\\
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e^{\varepsilon x} &= \sum_{n=0}^\infty {(\varepsilon x)^n \over n!}
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= 1 + {\varepsilon x} + {(\varepsilon x)^2 \over 2!} + {(\varepsilon x)^3 \over 3!} + ...
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\\
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&= 1 + {\varepsilon x}
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\\
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e^{\varepsilon \ln x} &= 1 + {\varepsilon \ln x}
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\end{align*}
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$$
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Befitting its nature "opposite" to the infinity, the infinitesimal $\varepsilon$ transforms
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an infinite sum into a finite one.
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$\varepsilon$ is realized as a square matrix (since otherwise powers could not exist),
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so evaluating the exponential more directly:
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$$
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\begin{align*}
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&\phantom{=\ }
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e^{\varepsilon \ln{x}}
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= \exp{\begin{pmatrix}
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0 & \ln{x} \\
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0 & 0
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\end{pmatrix}}
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\\
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\exp{
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\begin{pmatrix}
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0 & \ln{x} \\
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0 & 0
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\end{pmatrix} }
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&= \begin{pmatrix}
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0 & \ln{x} \\
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0 & 0
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\end{pmatrix}^0
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+ \begin{pmatrix}
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0 & \ln{x} \\
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0 & 0
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\end{pmatrix}^1
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+ {1 \over 2}\begin{pmatrix}
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0 & \ln{x} \\
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0 & 0
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\end{pmatrix}^2
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+ ...
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\\
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&= \begin{pmatrix}
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1 & 0 \\
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0 & 1
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\end{pmatrix} +
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\begin{pmatrix}
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0 & \ln{x} \\
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0 & 0
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\end{pmatrix} +
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{1 \over 2} \begin{pmatrix}
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0 & \ln{x} \\
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0 & 0
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\end{pmatrix}
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\begin{pmatrix}
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0 & \ln{x} \\
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0 & 0
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\end{pmatrix} + ...
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\\
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&= \begin{pmatrix}
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1 & \ln{x} \\
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0 & 1
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\end{pmatrix}
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= 1 + \varepsilon \ln x
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\end{align*}
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$$
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Of course, since $\varepsilon$ and its matrix form are equivalent, the answer is the same as before.
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There are multiple problems with this argument, which are signified by reexamining the earlier equation:
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$$
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x^\varepsilon
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\stackrel{?_1}{=} e^{\ln{x^\varepsilon}}
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\stackrel{?_2}{=} e^{\varepsilon \ln{x}}
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$$
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1. It is not obvious that $x^\varepsilon$ is reconcilable with $e^x$ and the natural logarithm,
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i.e., that composing them is still an identity with respect to $\varepsilon$
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2. The power identity for logarithms may not be obeyed by $\varepsilon$
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Additionally, there were only two terms used in the series for $e^x$;
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there are many more power series that begin with terms $1, 1...$, and the natural logarithm('s series)
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is simply the one that corresponds to the inverse of $e^x$.
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Are there other compositions of a series and its inverse that can be considered?
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Teeming with a lot of News
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--------------------------
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Fortunately, there is not precisely one way to identify the value of $x^\varepsilon$,
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and one in particular has much less handwaving.
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The binomial theorem is a very useful tool for writing the power of a sum of numbers:
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$$
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(x + 1)^n = \sum_{r=0}^n {n \choose r}x^r
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= \sum_{r=0}^n {n! \over {r!(n-r)!}}x^k
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$$
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If the binomial coefficient is asserted to be 0 for $r > n$, then the binomial theorem can also be written
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as an infinite sum.
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However, the denominator of ${n! \over (n-r)!}$ doesn't make sense, since it will be a negative factorial
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in this circumstance.
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On the other hand, multiplying *n* with the r numbers immediately below it can be assigned
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a new symbol $(n)_r$ (named the [
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Pochhammer symbol
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](https://en.wikipedia.org/wiki/Falling_and_rising_factorials)).
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This falling factorial satisfies the 0 rule since if *n* is an integer smaller than *r*,
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then the product will include 0, annihilating all other terms.
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$$
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\begin{gather*}
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(n)_0 = 1,~ (n)_r = (n - r + 1)(n)_{r-1}
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\\
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(x + 1)^n = \sum_{r=0}^\infty {(n)_r \over r!} x^r
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\end{gather*}
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$$
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However, if *n* is not an integer, then the series will miss 0, and continue indefinitely.
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This directly gives the series for square root:
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$$
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\begin{align*}
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\sqrt{x + 1} &= (x + 1)^{1/2} =
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\sum_{r=0}^\infty {(1/2)_r \over r!} x^r
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\\
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&= 1 + {1 \over 2}x
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+ \left({1 \over 2!}\right)
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\left({1 \over 2}\right)
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\left({1 \over 2} - 1\right)x^2
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+ ...
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\\
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&= 1 + {1 \over 2}x + \left({1 \over 2!}\right)
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\left(-{1 \over 4}\right)x^2
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+ \left({1 \over 3!}\right)
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\left(-{1 \over 4} \right)
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\left({1 \over 2} - 2 \right)x^3
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+ ...
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\\
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&= 1 + {1 \over 2}x - {1 \over 8}x^2
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+ \left({1 \over 3!}\right)
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\left({3 \over 8} \right)x^3
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+ \left({1 \over 4!}\right)
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\left({3 \over 8} \right)
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\left({1 \over 2} - 3 \right)x^4
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+ ...
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\\
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&= 1 + {1 \over 2}x - {1 \over 8}x^2
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+ {1 \over 16}x^3 - {5 \over 128}x^4
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+ ...
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\end{align*}
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$$
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Tumbling down Infinitesimals
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----------------------------
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Since this definition works for rational numbers as well as integers,
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is it possible to assign a value to $(\varepsilon)_r$?
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Indeed it is, since this symbol's definition only requires that integers can be subtracted from it
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and that it can multiply with other numbers.
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In other words, it works in any integer ring, a property which the matrices underlying $\varepsilon$
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very fortunately have.
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$$
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\begin{align*}
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(\varepsilon)_0 &= 1,~
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(\varepsilon)_r = (\varepsilon - r + 1)(\varepsilon)_{r-1}
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\\
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(\varepsilon)_1 &= (\varepsilon - 0)(1)
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= \varepsilon
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\\
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(\varepsilon)_2 &= (\varepsilon - 1)(\varepsilon)
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= \varepsilon^2 - \varepsilon = -\varepsilon
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\\
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(\varepsilon)_3 &= (\varepsilon - 2)(-\varepsilon)
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= -\varepsilon^2 + 2\varepsilon = 2\varepsilon
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\\
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(\varepsilon)_4 &= (\varepsilon - 3)(2\varepsilon)
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= 2\varepsilon^2 - 6\varepsilon = -6\varepsilon
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\\
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& ...
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\\
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(\varepsilon)_r &= (-1)^{r-1}(r-1)!\varepsilon,~ r > 0
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\end{align*}
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$$
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It is easy to check that this can also be directly computed from the matrix underlying $\varepsilon$.
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We can now plug this into the binomial formula:
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$$
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\begin{align*}
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(x + 1)^\varepsilon
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&= \sum_{r=0}^\infty {(\varepsilon)_r \over r!} x^r
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= 1 + \sum_{r=1}^\infty {(\varepsilon)_r \over r!} x^r
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\\
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&= 1 + \sum_{r=1}^\infty {(-1)^{r-1}(r-1)!\varepsilon \over r!} x^r
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\\
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&= 1 + \varepsilon \sum_{r=1}^\infty {(-1)^{r-1} \over r} x^r
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\end{align*}
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$$
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The final sum may be familiar, but let's hold off from hastily cross-referencing a table.
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The term inside the sum looks a lot like an integral, so simplifying:
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$$
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\begin{align*}
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\sum_{r=1}^\infty {(-1)^{r-1} \over r} x^r
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&= \sum_{r=1}^\infty \int {(-1)^{r-1}} x^{r-1} dx
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= \int \sum_{r=0}^\infty {(-1)^{r}} x^{r} dx
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\\
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&= \int \sum_{r=0}^\infty (-x)^{r} dx
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= \int {dx \over 1 + x}
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\end{align*}
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$$
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This is looking very promising!
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Substituting this expression for the previous sum, we can now conclude:
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$$
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\begin{align*}
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(x + 1)^\varepsilon &= 1 + \varepsilon \int {1 \over 1 + x}
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\\
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x^\varepsilon &= 1 + \varepsilon \int {1 \over x}
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\\
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&= 1 + \varepsilon \ln x
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\end{align*}
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$$
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Fortunately, this slightly more grounded approach agrees with the initial result.
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Other Elementary Algebraic Powers
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---------------------------------
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Analogously, the Pochhammer symbol can be extended to other algebraic objects.
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### Complex
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For example, the original approach with the imaginary unit *i* yields:
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$$
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x^i = e^{\ln{x^i}} = e^{i\ln x} = \cos(\ln x) + i\sin(\ln x)
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$$
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To be slightly more rigorous, we can calculate the real and imaginary components of
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$(i)_r$ and tabulate the results:
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| r | Real | Imaginary |
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|---|------|-----------|
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| 0 | 1 | 0 |
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| 1 | 0 | 1 |
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| 2 | -1 | -1 |
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| 3 | 3 | 1 |
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| 4 | -10 | 0 |
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| 5 | 40 | -10 |
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The real component corresponds to [OEIS A003703](http://oeis.org/A003703),
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and the imaginary to [OEIS A009454](http://oeis.org/A009454),
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which state that they have exponential generating functions
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$\cos(\ln x)$ and $\sin(\ln x)$ respectively, agreeing with the intuitive series above.
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In fact, reexamining the binomial formula, it obviously produces exponential generating functions
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(in $x - 1$) based on the series given by the Pochhammer symbol.
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$$
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\begin{gather*}
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(x + 1)^n = \sum_{r=0}^\infty {(n)_r \over r!} x^r
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\Longleftrightarrow
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x^n = \sum_{r=0}^\infty {(n)_r \over r!} (x-1)^r
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\\
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(x - 1)^n = \sum_{r=0}^\infty {(-1)^r (n)_r \over r!} x^r
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\Longleftrightarrow
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x^n = \sum_{r=0}^\infty {(-1)^r (n)_r \over r!} (x+1)^r
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\end{gather*}
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$$
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### Split-complex
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How about the hyperbolic analogue of the "circular" functions above?
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We want a series that equals $\cosh(\ln x) + j\sinh(\ln x)$, for some algebraic *j*.
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Unlike with *i*, the exponential series does not need to alternate, but merely be partitioned
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into even and odd components.
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Therefore, *j* has the property that $j^2 = 1,~ j \neq 1$.
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It has a very simple matrix presentation and Pochhammer sequence.
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$$
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\begin{align*}
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j &:= \begin{pmatrix}
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0 & 1 \\
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1 & 0
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\end{pmatrix}
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\\
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(j)_2 &= j(j - 1) = 1 - j
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\\
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(j)_3 &= (1 - j)(j - 2) = -3 + 3j = -3(1 - j)
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\end{align*}
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$$
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With a little guesswork, this can be turned into a recurrence relation.
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$$
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\begin{align*}
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\\
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(j)_r &\stackrel{?}{=} f(r)(1 - j)
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\\
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(j)_{r+1} &= f(r+1)(1 - j) = f(r)(1 - j)(j - r)
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\\
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&= f(r) \left( (1 - j)j - (1 - j)r^{\vphantom{1}} \right)
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= f(r)(j - 1 - r + jr)
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\\
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&= f(r) \left( -(r + 1) + j(r + 1)^{\vphantom{1}} \right)
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\\
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&= -f(r)(r + 1)(1 - j)
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\\
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\implies f(r) &= -rf(r - 1)
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\\
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&= {(-1)^r r! \over 2},~ r \ge 2
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\end{align*}
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$$
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Since this Pochhammer sequence contains $r!$, it (pleasingly) cancels out with the denominator
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of the binomial coefficient.
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$$
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\begin{align*}
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x^j &= \sum_{r=0}^\infty {(j)_r \over r!} (x-1)^r
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\\
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&= 1 + j(x-1) + (1-j)\sum_{r=2}^\infty {r!(-1)^r \over 2r!} (x - 1)^r
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\\
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&= \phantom{+ j} \left(1 + {1 \over 2}\sum_{r=2}^\infty (-1)^r (x - 1)^r \right)
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\\
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&\phantom{=} \vphantom{0} + j \left(
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x - 1 - {1 \over 2}\sum_{r=2}^\infty (-1)^r (x - 1)^r
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\right)
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\end{align*}
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$$
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The remaining infinite sum is the same in the real and *j* components.
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It is very close to a geometric sum, so we can solve for the complete expression:
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$$
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\begin{align*}
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\sum_{r=2}^\infty (-1)^r (x-1)^r
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&= \sum_{r=0}^\infty (-1)^r (x-1)^r ~-~ \sum_{r=0}^1 (-1)^r (x-1)^r
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\\
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&= {1 \over 1 - (x - 1)} - (1 - (x - 1)) = {1 \over x} + x - 2
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\\
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&= {x^2 - 2x + 1 \over x}
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\\
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x^j &= \left(1 + {x^2 - 2x + 1 \over 2x} \right)
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+ j\left(x - 1 - {x^2 - 2x + 1 \over 2x} \right)
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\\
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&= {x^2 + 1 \over 2x} + j{x^2 - 1 \over 2x}
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\end{align*}
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$$
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This expression is equivalent to the expression in cosh and sinh after writing them in terms of
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the exponential function and simplifying the natural logarithm.
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Notably, its components parametrize the unit hyperbola $x^2 - y^2 = 1$.
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The components of $t^i$ (switching variables to avoid confusion) also do so for the unit circle
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$x^2 + y^2 = 1$ (since the range of the natural logarithm is over all reals),
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but fail to do so rationally.
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### Golden
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The golden ratio also has a very simple relationship with its square, in particular $\phi^2 = \phi + 1$.
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This minimal polynomial has degree 2, so numbers can be represented by 2-tuples of the form $a + b\phi$.
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The Pochhammer sequence of $\phi$ is:
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| r | Real | Phi |
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|---|------|------|
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| 0 | 1 | 0 |
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| 1 | 0 | 1 |
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| 2 | 1 | 0 |
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| 3 | -2 | 1 |
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| 4 | 7 | -4 |
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| 5 | -32 | -19 |
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Removing the alternation, both sequences can be located in the OEIS; the real component is
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[OEIS A265165](http://oeis.org/A265165), and the imaginary component is
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[OEIS A306183](http://oeis.org/A306183) (alternating at [A323620](http://oeis.org/A323620)).
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### Opposite Infinitesimals
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If we're working with exponential-type series, wouldn't it be nice if, for some ω, $x^\omega$
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was an expression in $e^x$ ?
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This would match to match the relationship between $x^\varepsilon$ and the natural logarithm.
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The simplest "solution" is by allowing some algebraic element all of whose Pochhammer symbols
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(besides the 0th) are itself.
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In this case, we obtain:
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$$
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\sum_{r=0}^\infty {(\omega)_r \over r!} (x-1)^r
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= 1 + \omega \sum_{r=1}^\infty {(x-1)^r \over r!}
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= 1 + \omega (e^{x-1}-1)
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$$
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But the representation of $\omega$ is a problem.
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It must satisfy all of the all of the following equations:
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$$
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\begin{align*}
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(\omega)_2 &= (\omega - 1)\omega = \omega^2 - \omega = \omega
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\\
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(\omega)_3 &= (\omega - 2)\omega = \omega^2 - 2\omega = \omega
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\\
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(\omega)_4 &= (\omega - 3)\omega = \omega^2 - 3\omega = \omega
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\\
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...
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\\
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(\omega)_r &= (\omega - r)\omega = \omega^2 - r\omega = \omega
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\\[10pt]
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\implies \omega^2 &= (r + 1)\omega,~~ r \ge 1
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\end{align*}
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$$
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The simplest element that satisfies these relationships is obviously 0.
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Trying a little harder, it could make sense if $\omega$ is some sort of infinite fixed point,
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such that multiplying it by any integer (or itself) is just itself.
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Whatever $\omega$ actually represents, it cannot be captured in a matrix.
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Infinitesimal Duality
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---------------------
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Of course, there is another infinitesimal that we should not forget to consider:
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the transpose of $\epsilon_2$ (denoted hence $\varepsilon'$).
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However, ε and ε' do do not play nice with each other.
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Notice that $\varepsilon + \varepsilon' = j$.
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Is it the case that
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$$
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x^\varepsilon x^{\varepsilon'} \stackrel{?}{=} x^{\varepsilon + \varepsilon'} = x^j
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$$
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Surprisingly, no.
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Starting by converting the left hand side to matrices, we get two matrices which do not commute:
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$$
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\begin{align*}
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\begin{pmatrix}
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1 & \ln x \\
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0 & 1
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\end{pmatrix}
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\begin{pmatrix}
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1 & 0 \\
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\ln x & 1
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\end{pmatrix}
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&= \begin{pmatrix}
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(\ln x)^2 + 1 & \ln x \\
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\ln x & 1
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\end{pmatrix}
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\\
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= 1 + j\ln x + \begin{pmatrix}
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(\ln x)^2 & 0 \\
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0 & 0
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\end{pmatrix}
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&\neq 1 + j\ln x + \begin{pmatrix}
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0 & 0 \\
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0 & (\ln x)^2
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\end{pmatrix}
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\\
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= \begin{pmatrix}
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1 & \ln x \\
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\ln x & (\ln x)^2 + 1
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\end{pmatrix}
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&= \begin{pmatrix}
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1 & 0 \\
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\ln x & 1
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\end{pmatrix}
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\begin{pmatrix}
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1 & \ln x \\
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0 & 1
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\end{pmatrix}
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\end{align*}
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$$
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Thus, numbers of the form $a + b\varepsilon$ will not commute with numbers of the form $a + b\varepsilon'$.
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Neither of these expressions are $x^j$.
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Note also that $\varepsilon = {i + j \over 2}$, but *i* and *j* do not commute.
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The placement of the $(\ln x)^2$ term is the phantom of another matrix *k* that obeys
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$k^2 = 1,~ k \neq j \neq 1$.
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$$
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\begin{align*}
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k &:= \begin{pmatrix}
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1 & 0 \\
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0 & -1
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\end{pmatrix}
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\\
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1 + j\ln x + \begin{pmatrix}
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(\ln x)^2 & 0 \\
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0 & 0
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\end{pmatrix}
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&= 1 + j\ln x + (1 + k)\left( {(\ln x)^2 \over 2} \right)
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\\
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1 + j\ln x + \begin{pmatrix}
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0 & 0 \\
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0 & (\ln x)^2
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\end{pmatrix}
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&= 1 + j\ln x + (1 - k)\left( {(\ln x)^2 \over 2} \right)
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\end{align*}
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$$
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### Skew-Infinitesimals
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What about 1 + *j* (or equivalently 1 + *k*)?
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In either case, the result has 0 determinant, and in *k*'s, it is another kind of infinitesimal
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along the main diagonal.
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Its Pochhammer sequence also terminates.
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$$
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\begin{align*}
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(j + 1)_1 &= j + 1
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\\
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(j + 1)_2 &= (j + 1)j = j + 1
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\\
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(j + 1)_3 &= (j + 1)(j - 1) = 0
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\end{align*}
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$$
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This also makes sense by analyzing $x^{j + 1} = x(x^j)$, showing that the addition of exponents is preserved
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in some circumstances.
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$$
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\begin{align*}
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x(x^j) &= {x^2+1 \over 2} + j{x^2-1 \over 2}
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\\
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x^{1 + j} &= \sum_{n=0}^\infty {(j + 1)_n \over n!}(x - 1)^n
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\\
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&= 1 + (1 + j)(x - 1) + {1 \over 2}(1 + j)(x - 1)^2
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\\
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&= {(x - 1)^2 \over 2} + x + j\left({(x -1)^2 \over 2} + x - 1 \right) = x(x^j)
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\end{align*}
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$$
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Intriguingly, this clears the *x* from the denominator of $x^j$,
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so the result no longer necessitates power series in *x*.
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Since it has only even components, this expression also has a square root:
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|
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|
$$
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\sqrt{x^{j + 1}} = 1 + {j + 1 \over 2}(x - 1)
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|
= {x+1 \over 2} + j{x-1 \over 2}
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|
$$
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Closing
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|
-------
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It is good to remember that the natural logarithm of *x* grows slower than any positive power of *x*.
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Similarly, infinitesimals can be imagined as smaller than any positive rational number.
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In some way, taking a number to an infinitesimal power is related to a function that grows slower
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than all rational powers, albeit only within an infinitesimal "space".
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Generalizing slightly, it becomes obvious that algebraic powers can be assigned a series,
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which can have an interesting representation in transcendental functions.
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