755 lines
23 KiB
Plaintext
755 lines
23 KiB
Plaintext
---
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title: "Generating Polynomials, Part 1: Regular Constructibility"
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description: |
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What kinds of regular polygons are constructible with compass and straightedge?
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format:
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html:
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html-math-method: katex
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date: "2021-08-18"
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date-modified: "2025-06-17"
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categories:
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- geometry
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- generating functions
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- algebra
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- python
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---
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<style>
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.figure-img {
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max-width: 512px;
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object-fit: contain;
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height: 100%;
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}
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.figure-img.wide {
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max-width: 768px;
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}
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</style>
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```{python}
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#| echo: false
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from math import comb
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from IPython.display import Markdown
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from tabulate import tabulate
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import sympy
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from sympy.abc import z
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```
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[Recently](/posts/math/misc/platonic-volume), I used coordinate-free geometry to derive
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the volumes of the Platonic solids, a problem which was very accessible to the ancient Greeks.
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On the other hand, they found certain problems regarding which figures can be constructed via
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compass and straightedge to be very difficult. For example, they struggled with problems
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like [doubling the cube](https://en.wikipedia.org/wiki/Doubling_the_cube)
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or [squaring the circle](https://en.wikipedia.org/wiki/Squaring_the_circle),
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which are known (through circa 19th century mathematics) to be impossible.
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However, before even extending planar geometry by a third dimension or
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calculating the areas of circles, a simpler problem becomes apparent.
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Namely, what kinds of regular polygons are constructible?
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Regular Geometry and a Complex Series
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-------------------------------------
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When constructing a regular polygon, one wants a ratio between the length of a edge
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and the distance from a vertex to the center of the figure.
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{.wide}
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In a convex polygon, the total central angle is always one full turn, or 2π radians.
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The central angle of a regular *n*-gon is ${2\pi \over n}$ radians,
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and the green angle above (which we'll call *θ*) is half of that.
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This means that the ratio we're looking for is $\sin(\theta) = \sin(\pi / n)$.
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We can multiply by *n* inside the function on both sides to give
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$\sin(n\theta) = \sin(\pi) = 0$.
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Therefore, constructing a polygon is actually equivalent to solving this equation,
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and we can rephrase the question as how to express $\sin(n\theta)$ (and $\cos(n\theta)$).
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### Complex Recursion
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Thanks to [Euler's formula](https://en.wikipedia.org/wiki/Euler%27s_formula)
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and [de Moivre's formula](https://en.wikipedia.org/wiki/De_Moivre%27s_formula),
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the expressions we're looking for can be phrased in terms of the complex exponential.
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$$
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\begin{align*}
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e^{i\theta}
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&= \text{cis}(\theta) = \cos(\theta) + i\sin(\theta)
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& \text{ Euler's formula}
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\\
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\text{cis}(n \theta) = e^{i(n\theta)}
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&= e^{(i\theta)n} = {(e^{i\theta})}^n = \text{cis}(\theta)^n
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\\
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\cos(n \theta) + i\sin(n \theta)
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&= (\cos(\theta) + i\sin(\theta))^n
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& \text{ de Moivre's formula}
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\end{align*}
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$$
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De Moivre's formula for $n = 2$ gives
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$$
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\begin{align*}
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\text{cis}(\theta)^2
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&= (\text{c} + i\text{s})^2
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\\
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&= \text{c}^2 + 2i\text{cs} - \text{s}^2 + (0 = \text{c}^2 + \text{s}^2 - 1)
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\\
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&= 2\text{c}^2 + 2i\text{cs} - 1
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\\
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&= 2\text{c}(\text{c} + i\text{s}) - 1
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\\
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&= 2\cos(\theta)\text{cis}(\theta) - 1
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\end{align*}
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$$
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This can easily be massaged into a recurrence relation.
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$$
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\begin{align*}
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\text{cis}(\theta)^2
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&= 2\cos(\theta)\text{cis}(\theta) - 1
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\\
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\text{cis}(\theta)^{n+2}
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&= 2\cos(\theta)\text{cis}(\theta)^{n+1} - \text{cis}(\theta)^n
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\\
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\text{cis}((n+2)\theta)
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&= 2\cos(\theta)\text{cis}((n+1)\theta) - \text{cis}(n\theta)
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\end{align*}
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$$
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Recurrence relations like this one are powerful.
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Through some fairly straightforward summatory manipulations,
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the sequence can be interpreted as the coefficients in a Taylor series,
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giving a [generating function](https://en.wikipedia.org/wiki/Generating_function).
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Call this function *F*. Then,
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$$
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\begin{align*}
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\sum_{n=0}^\infty \text{cis}((n+2)\theta)x^n
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&= 2\cos(\theta) \sum_{n=0}^\infty \text{cis}((n+1)\theta) x^n
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- \sum_{n=0}^\infty \text{cis}(n\theta) x^n
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\\
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{F(x; \text{cis}(\theta)) - 1 - x\text{cis}(\theta) \over x^2}
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&= 2\cos(\theta) {F(x; \text{cis}(\theta)) - 1 \over x}
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- F(x; \text{cis}(\theta))
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\\[10pt]
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F - 1 - x\text{cis}(\theta)
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&= 2\cos(\theta) x (F - 1)
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- x^2 F
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\\
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F - 2\cos(\theta) x F + x^2 F
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&= 1 + x(\text{cis}(\theta) - 2\cos(\theta))
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\\[10pt]
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F(x; \text{cis}(\theta))
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&= {1 + x(\text{cis}(\theta) - 2\cos(\theta)) \over
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1 - 2\cos(\theta)x + x^2}
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\end{align*}
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$$
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Since $\text{cis}$ is a complex function, we can separate *F* into real and imaginary parts.
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Conveniently, these correspond to $\cos(n\theta)$ and $\sin(n\theta)$, respectively.
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$$
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\begin{align*}
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\Re[ F(x; \text{cis}(\theta)) ]
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&= {1 + x(\cos(\theta) - 2\cos(\theta)) \over 1 - 2\cos(\theta)x + x^2}
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\\
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&= {1 - x\cos(\theta) \over 1 - 2\cos(\theta)x + x^2} = A(x; \cos(\theta))
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\\
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\Im[ F(x; \text{cis}(\theta)) ]
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&= {x \sin(\theta) \over 1 - 2\cos(\theta)x + x^2} = B(x; \cos(\theta))\sin(\theta)
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\end{align*}
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$$
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In this form, it becomes obvious that the even though the generating function *F* was originally
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parametrized by $\text{cis}(\theta)$, *A* and *B* are parametrized only by $\cos(\theta)$.
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Extracting the coefficients of *x* yields an expression for $\cos(n\theta)$ and $\sin(n\theta)$
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in terms of $\cos(\theta)$ (and in the latter case, a common factor of $\sin(\theta)$).
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If $\cos(\theta)$ in *A* and *B* is replaced with the parameter *z*, then all trigonometric functions
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are removed from the equation, and we are left with only polynomials[^1].
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These polynomials are [*Chebyshev polynomials*](https://en.wikipedia.org/wiki/Chebyshev_polynomial)
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*of the first (A) and second (B) kind*.
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In actuality, the polynomials of the second kind are typically offset by 1
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(the x in the numerator of *B* is omitted).
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However, retaining this term makes indexing consistent between *A* and *B*
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(and will make things clearer later).
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[^1]:
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This can actually be observed as early as the recurrence relation.
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$$
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\begin{align*}
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\text{cis}(\theta)^{n+2}
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&= 2\cos(\theta)\text{cis}(\theta)^{n+1} - \text{cis}(\theta)^n
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\\
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a_{n+2}
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&= 2 z a_{n+1} - a_n
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\\
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\Re[ a_0 ]
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&= 1,~~ \Im[ a_0 ] = 0
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\\
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\Re[ a_1 ]
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&= z,~~ \Im[ a_1 ] = 1 \cdot \sin(\theta)
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\end{align*}
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$$
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We were primarily interested in $\sin(n\theta)$, so let's tabulate
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the first few polynomials of the second kind (at $z / 2$).
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```{python}
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#| echo: false
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#| label: tbl-chebyshevu
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#| tbl-cap: "[OEIS A049310](http://oeis.org/A049310)"
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#| classes: plain
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Markdown(tabulate(
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[
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[
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n,
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"$" + sympy.latex(poly) + "$",
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"$" + sympy.latex(sympy.factor(poly)) + "$",
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]
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for n in range(0, 11)
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for poly in [sympy.chebyshevu_poly(n - 1, z / 2) if n > 0 else sympy.sympify(0)]
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],
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headers=[
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"*n*",
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"$[x^n]B(x; z / 2) = U_{n - 1}(z / 2)$",
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"Factored",
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],
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numalign="left",
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stralign="left",
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))
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```
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Evaluating the polynomials at $z / 2$ cancels the 2 in the denominator (and recurrence),
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making these expressions much simpler.
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This evaluation has an interpretation in terms of the previous diagram --
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recall we used *half* the length of a side as a leg of the right triangle.
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For a unit circumradius, the side length itself is then $2\sin( {\pi / n} )$.
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To compensate for this doubling, the Chebyshev polynomial must be evaluated at half its normal argument.
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### Back on the Plane
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The constructibility criterion is deeply connected to the Chebyshev polynomials.
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In compass and straightedge constructions, one only has access to linear forms (lines)
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and quadratic forms (circles).
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This means that a figure is constructible if and only if the root can be expressed using
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normal arithmetic (which is linear) and square roots (which are quadratic).
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#### Pentagons
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Let's look at a regular pentagon.
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The relevant polynomial is
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$$
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[x^5]B ( x; z / 2 )
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= z^4 - 3z^2 + 1
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= (z^2 - z - 1) (z^2 + z - 1)
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$$
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According to how we derived this series, when $z = 2\cos(\theta)$, the roots of this polynomial
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correspond to when $\sin(5\theta) / \sin(\theta) = 0$.
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This relation itself is true when $\theta = \pi / 5$, since $\sin(5 \pi / 5) = 0$.
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One of the factors must therefore be the minimal polynomial of $2\cos(\pi / 5 )$.
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The former happens to be correct correct, since $2\cos( \pi / 5 ) = \varphi$, the golden ratio.
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Note that the second factor is the first evaluated at -*z*.
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#### Heptagons
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An example of where constructability fails is for $2\cos( \pi / 7 )$.
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$$
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\begin{align*}
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[x^7]B ( x; z / 2 )
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&= z^6 - 5 z^4 + 6 z^2 - 1
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\\
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&= ( z^3 - z^2 - 2 z + 1 ) ( z^3 + z^2 - 2 z - 1 )
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\end{align*}
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$$
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Whichever is the minimal polynomial (the former), it is a cubic, and constructing
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a regular heptagon is equivalent to solving it for *z*.
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But there are no (nondegenerate) cubics that one can produce via compass and straightedge,
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and all constructions necessarily fail.
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#### Decagons
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One might think the same of $2\cos(\pi /10 )$
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$$
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\begin{align*}
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[x^{10}]B ( x; z / 2 )
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&= z^9 - 8 z^7 + 21 z^5 - 20 z^3 + 5 z
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\\
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&= z ( z^2 - z - 1 )( z^2 + z - 1 )( z^4 - 5 z^2 + 5 )
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\end{align*}
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$$
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This expression also contains the polynomials for $2\cos( \pi / 5 )$.
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This is because a regular decagon would contain two disjoint regular pentagons,
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produced by connecting every other vertex.
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The polynomial which actually corresponds to $2\cos( \pi / 10 )$ is the quartic,
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which seems to suggest that it will require a fourth root and somehow decagons are not constructible.
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However, it can be solved by completing the square...
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$$
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\begin{align*}
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z^4 - 5z^2 &= -5
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\\
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z^4 - 5z^2 + (5/2)^2 &= -5 + (5/2)^2
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\\
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( z^2 - 5/2)^2 &= {25 - 20 \over 4}
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\\
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( z^2 - 5/2) &= {\sqrt 5 \over 2}
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\\
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z^2 &= {5 \over 2} + {\sqrt 5 \over 2}
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\\
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z &= \sqrt{ {5 + \sqrt 5 \over 2} }
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\end{align*}
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$$
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...and we can breathe a sigh of relief.
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The Triangle behind Regular Polygons
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------------------------------------
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Preferring *z* to be halved in $B(x; z/2)$ makes something else more evident.
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Observe these four rows of the Chebyshev polynomials
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```{python}
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#| echo: false
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#| classes: plain
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Markdown(tabulate(
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[
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[
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n,
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"$" + sympy.latex(poly) + "$",
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k,
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int(poly.coeff(z, k)), # type: ignore
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]
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for n, k in zip(range(4, 8), range(3, -1, -1))
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for poly in [sympy.chebyshevu_poly(n - 1, z / 2) if n > 0 else sympy.sympify(0)]
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],
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headers=[
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"*n*",
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"$[x^n]B(x; z / 2)$",
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"*k*",
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"$[z^{k}][x^n]B(x; z / 2)$",
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],
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numalign="left",
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stralign="left",
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))
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```
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The last column looks like an alternating row of Pascal's triangle
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(namely, ${n - \lfloor {k / 2} \rfloor - 1 \choose k}(-1)^k$).
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This resemblance can be made more apparent by listing the coefficients of the polynomials in a table.
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```{python}
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#| echo: false
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#| classes: plain
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rainbow_classes = [
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"",
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"red",
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"orange",
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"yellow",
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"green",
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"cyan",
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"aqua",
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"blue",
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"purple"
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"",
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"",
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]
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rainbow_class = lambda x, color: f"<span style=\"color: {rainbow_classes[color]}\">{x}</span>"
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Markdown(tabulate(
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[
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[
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n,
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*[" " for _ in range(1, 11 - n)], # offset for terms of the polynomial
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*[
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0 if k % 2 == 1
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else rainbow_class(
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comb(n - (k // 2) - 1, k // 2) * (-1)**(k // 2),
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n - (k // 2) - 1,
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)
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for k in range(n)
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]
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]
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for n in range(1, 11)
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],
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headers=[
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"n",
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*[f"$z^{nm}$" for nm in reversed(range(2, 10))],
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"$z$",
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"$1$",
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],
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numalign="right",
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stralign="right",
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))
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```
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Though they alternate in sign, the rows of Pascal's triangle appear along diagonals,
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which I have marked in rainbow.
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Meanwhile, alternating versions of the naturals (1, 2, 3, 4...),
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the triangular numbers (1, 3, 6, 10...),
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the tetrahedral numbers (1, 4, 10, 20...), etc.
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are present along the columns, albeit spaced out by 0's.
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The relationship of the Chebyshev polynomials to the triangle is easier to see if
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the coefficient extraction of $B(x; z / 2)$ is reversed.
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In other words, we extract *z* before extracting *x*.
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$$
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\begin{align*}
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B(x; z / 2) &= {x \over 1 - zx + x^2}
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= {x \over 1 + x^2 - zx}
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= {x \over 1 + x^2}
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\cdot {1 \over {1 + x^2 \over 1 + x^2} - z{x \over 1 + x^2}}
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\\[10pt]
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[z^n]B(x; z / 2) &= {x \over 1 + x^2} [z^n] {1 \over 1 - z{x \over 1 + x^2}}
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= {x \over 1 + x^2} \left( {x \over 1 + x^2} \right)^n
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\\
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&= \left( {x \over 1 + x^2} \right)^{n+1}
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= x^{n+1} (1 + x^2)^{-n - 1}
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\\
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&= x^{n+1} \sum_{k=0}^\infty {-n - 1 \choose k}(x^2)^k
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\quad \text{Binomial theorem}
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\end{align*}
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$$
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While the use of the binomial theorem is more than enough to justify
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the appearance of Pascal's triangle (along with explaining the 0's),
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I'll simplify further to explicitly show the alternating signs.
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$$
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\begin{align*}
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{(-n - 1)_k} &= (-n - 1)(-n - 2) \cdots (-n - k)
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\\
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&= (-1)^k (n + k)(n + k - 1) \cdots (n + 1)
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\\
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&= (-1)^k (n + k)_k
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\\
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\implies {-n - 1 \choose k}
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&= {n + k \choose k}(-1)^k
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\\[10pt]
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[z^n]B(x; z / 2)
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&= x^{n+1} \sum_{k=0}^\infty {n + k \choose k} (-1)^k x^{2k}
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\end{align*}
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$$
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Squinting hard enough, the binomial coefficient is similar to the earlier
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which gave the third row of Pascal's triangle.
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If k is fixed, then this expression actually generates the antidiagonal entries
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of the coefficient table, which are the columns with uniform sign.
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The alternation instead occurs between antidiagonals (one is all positive,
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the next is 0's, the next is all negative, etc.).
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The initial $x^{n+1}$ lags these sequences so that they reproduce the triangle.
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### Imagined Transmutation
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The generating function of the Chebyshev polynomials resembles other two term recurrences.
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For example, the Fibonacci numbers have generating function
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$$
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\sum_{n = 0}^\infty \text{Fib}_n x^n = {1 \over 1 - x - x^2}
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$$
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This resemblance can be made explicit with a simple algebraic manipulation.
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$$
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\begin{align*}
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B(ix; -iz / 2)
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&= {1 \over 1 -\ (-i z)(ix) + (ix)^2}
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= {1 \over 1 -\ (-i^2) z x + (i^2)(x^2)}
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\\
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&= {1 \over 1 -\ z x -\ x^2}
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\end{align*}
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$$
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If $z = 1$, these two generating functions are equal.
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The same can be said for $z = 2$ with the generating function of the Pell numbers,
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and so on for higher recurrences (corresponding to metallic means) for higher integral *z*.
|
|
|
|
In terms of the Chebyshev polynomials, this series manipulation removes the alternation in
|
|
the coefficients of $U_n$, restoring Pascal's triangle to its nonalternating form.
|
|
Related to the previous point, it is possible to find the Fibonacci numbers (Pell numbers, etc.)
|
|
in Pascal's triangle, which you can read more about
|
|
[here](http://users.dimi.uniud.it/~giacomo.dellariccia/Glossary/Pascal/Koshy2011.pdf).
|
|
|
|
|
|
Manipulating the Series
|
|
-----------------------
|
|
|
|
Look back to the table of $U_{n - 1}(z / 2)$ (@tbl-chebyshevu).
|
|
When I brought up $U_{10 - 1}(z / 2)$ and decagons, I pointed out their relationship to pentagons
|
|
as an explanation for why $U_{5 -\ 1}(z / 2)$ appears as a factor.
|
|
Conveniently, $U_{2 -\ 1}(z / 2) = z$ is also a factor, and 2 is likewise a factor of 10.
|
|
|
|
This pattern is present throughout the table; $n = 6$ contains factors for
|
|
$n = 2 \text{ and } 3$ and the prime numbers have no smaller factors.
|
|
If this observation is legitimate, call the newest term $f_n(z)$
|
|
and denote $p_n(z) = U_{n -\ 1}( z / 2 )$.
|
|
|
|
|
|
### Factorization Attempts
|
|
|
|
The relationship between $p_n$ and the intermediate $f_d$, where *d* is a divisor of *n*,
|
|
can be made explicit by a [Möbius inversion](https://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula).
|
|
|
|
$$
|
|
\begin{align*}
|
|
p_n(z) &= \prod_{d|n} f_n(z)
|
|
\\
|
|
\log( p_n(z) )
|
|
&= \log \left( \prod_{d|n} f_d(z) \right)
|
|
= \sum_{d|n} \log( f_d(z) )
|
|
\\
|
|
\log( f_n(z) ) &= \sum_{d|n} { \mu \left({n \over d} \right)}
|
|
\log( p_d(z) )
|
|
\\
|
|
f_n(z) &= \prod_{d|n} p_d(z)^{ \mu (n / d) }
|
|
\\[10pt]
|
|
f_6(z) = g_6(z)
|
|
&= p_6(z)^{\mu(1)}
|
|
p_3(z)^{\mu(2)}
|
|
p_2(z)^{\mu(3)}
|
|
\\
|
|
&= {p_6(z) \over p_3(z) p_2(z)}
|
|
\end{align*}
|
|
$$
|
|
|
|
Unfortunately, it's difficult to apply this technique across our whole series.
|
|
Möbius inversion over series typically uses more advanced generating functions such as
|
|
[Dirichlet series](https://en.wikipedia.org/wiki/Dirichlet_series#Formal_Dirichlet_series)
|
|
or [Lambert series](https://en.wikipedia.org/wiki/Lambert_series).
|
|
However, naively reaching for these fails for two reasons:
|
|
|
|
- We built our series of polynomials on a recurrence relation, and these series
|
|
are opaque to such manipulations.
|
|
- To do a proper Möbius inversion, we need these kinds of series over the *logarithm*
|
|
of each polynomial (*B* is a series over the polynomials themselves).
|
|
|
|
Ignoring these (and if you're in the mood for awful-looking math) you may note
|
|
the Lambert equivalence[^2]:
|
|
|
|
[^2]:
|
|
This equivalence applies to other polynomial series obeying the same factorization rule
|
|
such as the [cyclotomic polynomials](https://en.wikipedia.org/wiki/Cyclotomic_polynomial).
|
|
|
|
$$
|
|
\begin{align*}
|
|
\log( p_n(z) )
|
|
&= \sum_{d|n} \log( f_d(z) )
|
|
\\
|
|
\sum_{n = 1}^\infty \log( p_n ) x^n
|
|
&= \sum_{n = 1}^\infty \sum_{d|n} \log( f_d ) x^n
|
|
\\
|
|
&= \sum_{k = 1}^\infty \sum_{m = 1}^\infty \log( f_m ) x^{m k}
|
|
\\
|
|
&= \sum_{m = 1}^\infty \log( f_m ) \sum_{k = 1}^\infty (x^m)^k
|
|
\\
|
|
&= \sum_{m = 1}^\infty \log( f_m ) {x^m \over 1 - x^m}
|
|
\end{align*}
|
|
$$
|
|
|
|
Either way, the number-theoretic properties of this sequence are difficult to ascertain
|
|
without advanced techniques.
|
|
If research has been done, it is not easily available in the OEIS.
|
|
|
|
|
|
### Total Degrees
|
|
|
|
It can be also be observed that the new term is symmetric ($f(z) = f(-z)$), and is therefore
|
|
either irreducible or the product of polynomial and its reflection (potentially negated).
|
|
For example,
|
|
|
|
$$
|
|
p_9(z) = \left\{
|
|
\begin{matrix}
|
|
(z - 1)(z + 1)
|
|
& \cdot
|
|
& (z^3 - 3z - 1)(z^3 - 3z + 1)
|
|
\\
|
|
\shortparallel && \shortparallel
|
|
\\
|
|
f_3(z)
|
|
& \cdot
|
|
& f_9(z)
|
|
\\
|
|
\shortparallel && \shortparallel
|
|
\\
|
|
g_3(z) \cdot g_3(-z)
|
|
& \cdot
|
|
& g_9(z) \cdot -g_9(-z)
|
|
\end{matrix}
|
|
\right.
|
|
$$
|
|
|
|
These factor polynomials $g_n$ are the minimal polynomials of $2\cos( \pi / n )$.
|
|
|
|
Multiplying these minimal polynomials by their reflection can be observed in the Chebyshev polynomials
|
|
for $n = 3, 5, 7, 9$, strongly implying that it occurs on the odd terms.
|
|
Assuming this is true, we have
|
|
|
|
$$
|
|
f_n(z) = \begin{cases}
|
|
g_n(z) & \text{$n$ is even}
|
|
\\
|
|
g_n(z)g_n(-z)
|
|
& \text{$n$ is odd and ${\deg(f_n) \over 2}$ is even}
|
|
\\
|
|
-g_n(z)g_n(-z)
|
|
& \text{$n$ is odd and ${\deg(f_n) \over 2}$ is odd}
|
|
\end{cases}
|
|
$$
|
|
|
|
Without resorting to any advanced techniques, the degrees of $f_n$ are
|
|
not too difficult to work out.
|
|
The degree of $p_n(z)$ is $n -\ 1$, which is also the degree of $f_n(z)$ if *n* is prime.
|
|
If *n* is composite, then the degree of $f_n(z)$ is $n -\ 1$ minus the degrees
|
|
of the divisors of $n -\ 1$.
|
|
This leaves behind how many numbers less than *n* are coprime to *n*.
|
|
Therefore $\deg(f_n) = \phi(n)$, the
|
|
[Euler totient function](https://en.wikipedia.org/wiki/Euler_totient_function) of the index.
|
|
|
|
The totient function can be used to examine the parity of *n*.
|
|
If *n* is odd, it is coprime to 2 and all even numbers.
|
|
The introduced factor of 2 to 2*n* removes the evens from the totient, but this is compensated by
|
|
the addition of the odd multiples of old numbers coprime to *n* and new primes.
|
|
This means that $\phi(2n) = \phi(n)$ for odd *n* (other than 1).
|
|
|
|
The same argument can be used for even *n*: there are as many odd numbers from 0 to *n* as there are
|
|
from *n* to 2*n*, and there are an equal number of numbers coprime to 2*n* in either interval.
|
|
Therefore, $\phi(2n) = 2\phi(n)$ for even *n*.
|
|
|
|
This collapses all cases of the conditional factorization of $f_n$ into one,
|
|
and the degrees of $g_n$ are
|
|
|
|
$$
|
|
\begin{align*}
|
|
\deg( g_n(z) )
|
|
&= \begin{cases}
|
|
\deg( f_n(z) )
|
|
= \phi(n)
|
|
& n \text{ is even} & \implies \phi(n) = \phi(2n) / 2
|
|
\\
|
|
\deg( f_n(z) ) / 2
|
|
= \phi(n) / 2
|
|
& n \text{ is odd} & \implies \phi(n) / 2 = \phi(2n) / 2
|
|
\end{cases}
|
|
\\
|
|
&= \varphi(2n) / 2
|
|
\end{align*}
|
|
$$
|
|
|
|
Though they were present in the earlier Chebyshev table,
|
|
the $g_n$ themselves are presented again, along with the expression for their degree
|
|
|
|
```{python}
|
|
#| echo: false
|
|
#| classes: plain
|
|
|
|
def poly_to_rising_power_list(poly, var):
|
|
"""
|
|
Convert a polynomial to a list in rising powers.
|
|
E.g., x^2 + x - 1 will be converted to [-1, 1, 1].
|
|
"""
|
|
ret = []
|
|
for term in reversed(poly.as_ordered_terms()):
|
|
deg = sympy.degree(term, var)
|
|
if deg > len(ret):
|
|
ret.extend(0 for _ in range(int(deg) - len(ret)))
|
|
|
|
if deg == 0:
|
|
ret.append(term)
|
|
else:
|
|
ret.append(term.coeff(z**deg))
|
|
|
|
return ret
|
|
|
|
|
|
Markdown(tabulate(
|
|
[
|
|
[
|
|
n,
|
|
sympy.totient(2*n) / 2,
|
|
"$" + sympy.latex(g) + "$",
|
|
str(poly_to_rising_power_list(g, z)),
|
|
]
|
|
for n in range(2, 11)
|
|
for g in [
|
|
factor # the first factor polynomial with matching degree and negative second term
|
|
for factor in sympy.chebyshevu_poly(n - 1, z / 2).factor(z).as_ordered_factors() # type: ignore
|
|
for second_term in [
|
|
0 if len(factor.as_ordered_terms()) == 1 # if there's only one term, pass it through
|
|
else factor.as_ordered_terms()[1].as_ordered_factors()[0] # type: ignore
|
|
] if n == 2
|
|
or (
|
|
sympy.degree(factor) == sympy.totient(2*n) / 2
|
|
and isinstance(second_term, sympy.Integer)
|
|
and second_term < 0
|
|
)
|
|
]
|
|
] + [[
|
|
"-",
|
|
"[OEIS A055034](http://oeis.org/A055034)",
|
|
"-",
|
|
"[OEIS A187360](http://oeis.org/A187360)",
|
|
]],
|
|
headers=[
|
|
"n",
|
|
"$\\varphi(2n)/2$",
|
|
"$g_n(z)$",
|
|
"Coefficient list, rising powers",
|
|
],
|
|
numalign="right",
|
|
stralign="left",
|
|
))
|
|
```
|
|
|
|
|
|
Closing
|
|
-------
|
|
|
|
My initial jumping off point for writing this article was completely different.
|
|
However, in the process of writing, its share of the article shrank and shrank until its
|
|
introduction was only vaguely related to what preceded it.
|
|
But alas, the introduction via geometric constructions flows better coming off my
|
|
[post about the Platonic solids](/posts/math/misc/platonic-volume).
|
|
Also, it reads better if I rely less on "if you search for this sequence of numbers"
|
|
and more on how to interpret the definition.
|
|
|
|
Consider reading [the follow-up](../2) to this post if you're interested in another way
|
|
one can obtain the Chebyshev polynomials.
|
|
I have since rederived the Chebyshev polynomials without the complex exponential,
|
|
which you can read about in [this post](/posts/math/stereo/2).
|
|
|
|
Diagrams created with GeoGebra.
|