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

<style>
.red {
  color: red;
}

.orange {
  color: orange;
}

.yellow {
  color: yellow;
}

.aqua {
  color: aqua;
}

.green {
  color: green;
}

.cyan {
  color: cyan;
}

.blue {
  color: blue;
}

.purple {
  color: purple;
}
</style>


Generating Polynomials, Part 1: Regular Constructibility
========================================================

[Recently](), I used coordinate-free geometry to derive the volumes of the Platonic solids, a problem which was very accessible to the ancient Greeks. On the other hand, they found certain problems regarding which figures can be constructed via compass and straightedge to be very difficult. For example, they struggled with problems like [doubling the cube](https://en.wikipedia.org/wiki/Doubling_the_cube) or [squaring the circle](https://en.wikipedia.org/wiki/Squaring_the_circle), which are known (through circa 19th century mathematics) to be impossible . However, before extending geometry with a third dimension or including the areas of circles in geometry, a simpler problem becomes apparent; namely, what kinds of regular polygons are constructible?


Regular Geometry and a Complex Series
-------------------------------------

When constructing a regular polygon, one wants ratios relating a side, an inradius, and the circumradius. If one of these measurements is declared to be a unit, then only a single ratio is needed, since the other is available through the Pythagorean theorem. This is best exemplified by the following figure

![]()

In a convex polygon, the total central angle is always one full turn, or $2\pi$ radians. The central angle of a regular *n*-gon is ${2\pi \over n}$ radians, and the green angle above is half of it. This means that the ratio of half of the side and circumradius is $\sin(\pi / n)$ radians. But how can this quantity be derived?

Let's turn this question on its head. Let $\theta = \pi / n$, meaning $n\theta = \pi$. Applying sine over this gives $\sin(n\theta) = \sin(\pi) = 0$. Therefore, constructing a polygon is equivalent to solving this equation. It then becomes a question of how to express $\sin(n\theta)$ (and ￼$\cos(n\theta)$).

Thanks to [Euler's formula](https://en.wikipedia.org/wiki/Euler%27s_formula) and [de Moivre's formula](https://en.wikipedia.org/wiki/De_Moivre%27s_formula), this can be phrased in terms of the complex exponential.

$$
\begin{align*}
e^{i\theta} &= \text{cis}(\theta) = \cos(\theta) + i\sin(\theta)
& \text{ Euler's formula} \\
\text{cis}(n \theta) = e^{i(n\theta)} &= e^{(i\theta)n} = {(e^{i\theta})}^n = \text{cis}(\theta)^n  \\
\cos(n \theta) + i\sin(n \theta) &= (\cos(\theta) + i\sin(\theta))^n
& \text{ de Moivre's formula}
\end{align*}
$$

De Moivre's formula for $n = 2$ gives

$$
\begin{align*}
\text{cis}(\theta)^2 &= (\text{c} + i\text{s})^2 \\
&= \text{c}^2 + 2i\text{cs} - \text{s}^2 + (0 = \text{c}^2 + \text{s}^2 - 1) \\
&= 2\text{c}^2 + 2i\text{cs} - 1 \\
&= 2\text{c}(\text{c} + i\text{s}) - 1 \\
&= 2\cos(\theta)\text{cis}(\theta) - 1 \\
\end{align*}
$$

This can easily be massaged into a recurrence relation.

$$
\begin{align*}
\text{cis}(\theta)^2 &= 2\cos(\theta)\text{cis}(\theta) - 1 \\
\text{cis}(\theta)^{n+2} &= 2\cos(\theta)\text{cis}(\theta)^{n+1} - \text{cis}(\theta)^n \\
\text{cis}((n+2)\theta) &= 2\cos(\theta)\text{cis}((n+1)\theta) - \text{cis}(n\theta) \\
\end{align*}
$$

Through some fairly straightforward summatory manipulations, the sequence can be interpreted as the coefficients in a Taylor series, giving a [generating function](https://en.wikipedia.org/wiki/Generating_function).

$$
\begin{align*}
\sum_{n=0}^\infty \text{cis}((n+2)\theta)x^n
&= 2\cos(\theta) \sum_{n=0}^\infty \text{cis}((n+1)\theta) x^n
 - \sum_{n=0}^\infty \text{cis}(n\theta) x^n \\
{F(x; \text{cis}(\theta)) - 1 - x\text{cis}(\theta) \over x^2}
&= 2\cos(\theta) {F(x; \text{cis}(\theta)) - 1 \over x}
 - F(x; \text{cis}(\theta)) \\ \\
F - 1 - x\text{cis}(\theta)
&= 2\cos(\theta) x (F - 1)
 - x^2 F \\
F - 2\cos(\theta) x F + x^2 F
&= 1 + x(\text{cis}(\theta) - 2\cos(\theta)) \\ \\
F(x; \text{cis}(\theta))
&= {1 + x(\text{cis}(\theta) - 2\cos(\theta)) \over
1 - 2\cos(\theta)x + x^2}
\end{align*}
$$

Even though the generating function *F* is parametrized by the complex exponential of $\theta$, separating the real and imaginary components reveals an expression in terms of $\cos(\theta)$. This is useful because the real and imaginary parts of *F* correspond to $\cos(n\theta)$ and $\sin(n\theta)$, respectively.

$$
\begin{align*}
\Re[ F(x; \text{cis}(\theta)) ] &= {1 + x(\cos(\theta) - 2\cos(\theta)) \over
1 - 2\cos(\theta)x + x^2} \\
&= {1 - x\cos(\theta) \over 1 - 2\cos(\theta)x + x^2} = A(x; \cos(\theta)) \\
\Im[ F(x; \text{cis}(\theta)) ]
&= {x \sin(\theta) \over 1 - 2\cos(\theta)x + x^2} = B(x; \cos(\theta))\sin(\theta)
\end{align*}
$$

Extracting the coefficients of *x* in *A* and *B* yields an expression for $\cos(n\theta)$ and $\sin(n\theta)$ in terms of $\cos(\theta)$ (and in the latter case, a common factor of $\sin(\theta)$). If $\cos(\theta)$ in *A* and *B* is replaced with the parameter *z*, then all of the trigonometric functions are removed from the equation, and it becomes evident that this relationship is polynomial. This can actually be observed as early as the recurrence relation.

$$
\begin{align*}
\text{cis}(\theta)^{n+2} &= 2\cos(\theta)\text{cis}(\theta)^{n+1} - \text{cis}(\theta)^n \\
a_{n+2} &= 2 z a_{n+1} - a_n \\
\Re[ a_0 ] &= 1,~~ \Im[ a_0 ] = 0 \\
\Re[ a_1 ] &= z,~~ \Im[ a_1 ] = 1 \cdot \sin(\theta)
\end{align*}
$$

In other words, *A* starts off with $1, z, ...$, and *B* starts off with $0, 1, ...$. These polynomials are [*Chebyshev polynomials*](https://en.wikipedia.org/wiki/Chebyshev_polynomial) *of the first (A) and second (B) kind*. Actually, the polynomials of the second kind are typically not offset (the x in the numerator of *B* is omitted). However, this form makes explicit which polygon (remember? we were working with those) we are considering, and will be important later. For future reference, the first few polynomials of the second kind (at $z / 2$) are:

|  n | $[x^n]B(x; z / 2) = U_{n -\ 1}(z / 2)$  | Factored                                          |
|----|-----------------------------------------|---------------------------------------------------|
|  0 | 0                                       | 0                                                 |
|  1 | 1                                       | 1                                                 |
|  2 | $z$                                     | $z$                                               |
|  3 | $z^2 -\ 1$                              | $(z -\ 1)(z + 1)$                                 |
|  4 | $z^3 -\ 2z$                             | $z(z^2 -\ 2)$                                     |
|  5 | $z^4 -\ 3z^2 + 1$                       | $(z^2 -\ z -\ 1)(z^2 + z -\ 1)$                   |
|  6 | $z^5 -\ 4 z^3 + 3z$                     | $z(z^2 -\ 1)(z^2 -\ 3)$                           |
|  7 | $z^6 -\ 5 z^4 + 6 z^2 -\ 1$             | $(z^3 -\ z^2 -\ 2 z + 1)(z^3 + z^2 -\ 2 z -\ 1)$  |
|  8 | $z^7 -\ 6 z^5 + 10 z^3 -\ 4 z$          | $z(z^2 -\ 2)(z^4 -\ 4 z^2 + 2)$                   |
|  9 | $z^8 -\ 7 z^6 + 15 z^4 -\ 10 z^2 + 1$   | $(z^2 -\ 1)(z^3 -\ 3 z -\ 1)(z^3 -\ 3 z + 1)$     |
| 10 | $z^9 -\ 8 z^7 + 21 z^5 -\ 20 z^3 + 5 z$ | $z(z^2 -\ z -\ 1)(z^2 + z -\ 1)(z^4 -\ 5z^2 + 5)$ |

[OEIS A049310](http://oeis.org/A049310)

Evaluating the polynomials at $z / 2$ cancels the 2 in the denominator (and recurrence), making these expressions much simpler. This evaluation can also be interpreted intuitively by studying the the previous diagram. Since the side length was bisected by the inradius, the side length in terms of a unit circumradius is $2\sin ({\pi / n} )$. To compensate for this doubling, the Chebyshev polynomial must be evaluated at half its normal argument.


### Back on the Plane

The constructibility criterion is deeply connected to the Chebyshev polynomials. In compass and straightedge constructions, one only has access to linear forms (lines) and quadratic forms (circles). This means that a figure is constructible if and only if the root can be achieved through normal arithmetic (which is linear) and square roots (which are quadratic).

Let's look at a regular pentagon. The relevant polynomial is

$$
[x^5]B \left( x; {z \over 2} \right)
= z^4 - 3z^2 + 1
= (z^2 - z - 1) (z^2 + z - 1)
$$

As a reminder, when $z = 2\cos( \pi / 5 )$, the polynomial evaluates to 0. Also, either factor is the other evaluated at -*z*. Thus, one of the factors in terms is the minimal polynomial of $2\cos(\pi / 5 )$. The former is correct, since $2\cos( \pi / 5 ) = \phi$, the golden ratio.

An example of where constructability fails is for $2\cos( \pi / 7 )$.

$$
\begin{align*}
[x^7]B \left( x; {z \over 2} \right)
&= z^6 - 5 z^4 + 6 z^2 - 1 \\
&= ( z^3 - z^2 - 2 z + 1 ) (z^3 + z^2 - 2 z - 1 )
\end{align*}
$$

Whichever is the minimal polynomial (the former), it is a cubic, and constructing a regular heptagon is equivalent to solving it for *z*. But there are no (nondegenerate) cubics that one can produce via compass and straightedge, and the construction fails.

One might think the same of $2\cos(\pi /10 )$

$$
\begin{align*}
[x^{10}]B \left( x; {z \over 2} \right)
&= z^9 - 8 z^7 + 21 z^5 - 20 z^3 + 5 z \\
&= z ( z^2 - z - 1 )( z^2 + z - 1 )( z^4 - 5 z^2 + 5 )
\end{align*}
$$

This expression also contains the polynomials for $2\cos( \pi / 5 )$. This is because a regular decagon would contain two disjoint regular pentagons, produced by connecting every other vertex.

![]()

The polynomial which actually corresponds to $2\cos( \pi / 10 )$ is the quartic, which seems to suggest that it will require a fourth root and somehow decagons will not be constructible. However, it can be solved by completing the square, and we can breathe a sigh of relief.

$$
\begin{align*}
z^4 - 5z^2 &= -5 \\
z^4 - 5z^2 + (5/2)^2 &= -5 + (5/2)^2 \\
( z^2 - 5/2)^2 &= {25 - 20 \over 4} \\
( z^2 - 5/2) &= {\sqrt 5 \over 2} \\
z^2 &= {5 \over 2} + {\sqrt 5 \over 2} \\
z &= \sqrt{ {5 + \sqrt 5 \over 2} }
\end{align*}
$$


The Triangle behind Regular Polygons
------------------------------------

Preferring *z* to be halved in $B(x; z/2)$ makes something else more evident. Observe these four rows of the Chebyshev polynomials

|  n | $[x^n]B(x; z / 2)$          | k | *m* = 2*k* + 1 | *n* - *m* | $[z^{n -\ m}][x^n]B(x; z / 2)$ |
|----|-----------------------------|---|----------------|-----------|--------------------------------|
|  4 | $z^3 -\ 2z$                 | 0 | 1              | 3         | 1                              |
|  5 | $z^4 -\ 3z^2 + 1$           | 1 | 3              | 2         | -3                             |
|  6 | $z^5 -\ 4 z^3 + 3z$         | 2 | 5              | 1         | 3                              |
|  7 | $z^6 -\ 5 z^4 + 6 z^2 -\ 1$ | 3 | 7              | 0         | -1                             |

The last column looks like an alternating row of Pascal's triangle, and can be expressed as ${n -\ k -\ 1 \choose k}(-1)^k$. This resemblance can be made more apparent by listing the coefficients of the polynomials in a table.


```{python}
#| echo: false

from IPython.display import Markdown
from tabulate import tabulate
from math import comb

rainbow = [
  "",
  "red",
  "orange",
  "yellow",
  "aqua",
  "green",
  "cyan",
  "blue",
  "purple"
  "",
  "",
]

entry = lambda x, color: f"<span class=\"{rainbow[color]}\">{x}</span>"

Markdown(tabulate(
  [
    [
      n,
      *[" " for nm in range(1, 11 - n)],
      *[
        0 if k % 2 == 1
          else entry(
            comb(n - (k // 2) - 1, nm - (k // 2)) * (-1)**(k // 2),
            n - (k // 2) - 1,
          )
        for nm, k in zip(range(n), range(10))
      ]
    ]
    for n in range(1, 11)
  ],
  headers=[
    "n",
    *[f"$z^{nm}$" for nm in reversed(range(2, 10))],
    "$z$",
    "$1$",
  ],
  numalign="right",
  stralign="right",
))
```

Though they alternate in sign, the rows of Pascal's triangle appear along diagonals, which I have marked in rainbow. Meanwhile, alternating versions of the naturals (1, 2, 3, 4...), the triangular numbers (1, 3, 6, 10...), the tetrahedral numbers (1, 4, 10, 20...), etc. are present along the columns, albeit spaced out by 0's.

The relationship of the Chebyshev polynomials to the triangle is easier to see if the coefficient extraction of $B(x; z / 2)$ is reversed.

$$
\begin{align*}
B(x; z / 2) &= {x \over 1 - zx + x^2}
= {x \over 1 + x^2 - zx}
= {x \over 1 + x^2} \cdot
{1 \over {1 + x^2 \over 1 + x^2} - z{x \over 1 + x^2}} \\
[z^n]B(x; z / 2) &= {x \over 1 + x^2} [z^n] {1 \over 1 - z{x \over 1 + x^2}}
= {x \over 1 + x^2} \left( {x \over 1 + x^2} \right)^n \\
&= \left( {x \over 1 + x^2} \right)^{n+1}
= x^{n+1} (1 + x^2)^{-n - 1} \\
&= x^{n+1} \sum_{k=0}^\infty {-n - 1 \choose k}(x^2)^k
\end{align*}
$$

While the use of the binomial theorem in $1 + x^2$ is more than enough to justify the appearance of Pascal's triangle (along with explaining the 0's), I will press onward until it becomes excruciatingly obvious.

$$
\begin{align*}
{(-n - 1)_k} &= (-n - 1)(-n - 2) \cdots (-n - k) \\
&= (-1)^k (n + k)(n + k - 1) \cdots (n + 1) \\
&= (-1)^k (n + k)_k \\
\implies {-n - 1 \choose k}
&= {n + k \choose k}(-1)^k \\ \\
[z^n]B(x; z / 2) &= x^{n+1} \sum_{k=0}^\infty {n + k \choose k} (-1)^k x^{2k} \\
\end{align*}
$$

Squinting hard enough, the binomial coefficient is similar to the earlier which gave the third row of Pascal's triangle. If k is fixed, then this expression actually generates the antidiagonal entries of the coefficient table, which are the columns with uniform sign. The alternation instead occurs between antidiagonals (one is all positive, the next is 0's, the next is all negative, etc.). The initial $x^{n+1}$ lags these sequences so that they reproduce the triangle.


### Imagined Transmutation

The generating function of the Chebyshev polynomials resembles other two term recurrences. This resemblance can be made explicit with a simple algebraic manipulation.

$$
\begin{align*}
B(ix; -iz / 2) &= {1 \over 1 -\ (-i z)(ix) + (ix)^2}
= {1 \over 1 -\ (-i^2) z x + (i^2)(x^2)} \\
&= {1 \over 1 -\ z x -\ x^2}
\end{align*}
$$

The poles of this expression are the reciprocals and additive inverses of the metallic means. For example, if $z = 1$, then the denominator is $1 -\ x -\ x^2$, which has $1 / \phi$ as a root. Equally well, it is also a generating function of the Fibonacci numbers. The same can be said for the silver ratio and Pell numbers, and so on for higher $z$.

In terms of the Chebyshev polynomials, this removes the alternation in the coefficients of $U_n$, and restores 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
-----------------------

### Total Degrees

Look back to the table of $U_{n -\ 1}(z / 2)$. When I brought up decagons, I pointed out their relationship to pentagons as an explanation for why $U_{5 -\ 1}(z / 2)$ appears as a factor. Conveniently, $U_{2 -\ 1}(z / 2)$ is also a factor, and 10 is an even number. 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 factor polynomials $f_n(z)$ and denote $p_n(z) = U_{n -\ 1}( z / 2 )$.

It can also be observed that a factor is either symmetric ($g(z) = g(-z)$), or is the product of another polynomial and its reflection, potentially negated. For example, $p_9(z) = f_3(z) \cdot g_9(z) \cdot -g_9(-z)$, where $g_9(z) = z^3 -\ 3z -\ 1$. These reflections are necessary for $n = 3, 5, 7, 9$, strongly implying that it occurs on the odd terms.

In other words, if $f_n$ is the new polynomial introduced by $p_n$ of, then denote its conditional factorization $g_n$.

$$
f_n(z) = \begin{cases}
g_n(z) & n \text{ is even} \\
g_n(z)g_n(-z) & \text{$n$ is odd and ${\deg(f_n) \over 2}$ is even } \\
-g_n(z)g_n(-z) & \text{otherwise}
\end{cases}
$$

The final case seems to be tied to [OEIS A004614](http://oeis.org/A004614), which are primes of the form $4k+3$ and products thereof.

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 $\deg(f_n) = \varphi(n)$, the Euler totient function.

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 $\varphi(2n) = \varphi(n)$ for odd *n* ($n \neq 1$).

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, $\varphi(2n) = 2\varphi(n)$ for even *n*.

This collapses the cases of the conditional factorization of $f_n$ into one, and the degrees of $g_n$ are

$$
\begin{align*} \deg( g_n(z) ) &= \begin{cases}
\deg( f_n(z) ) = \varphi(n) & n \text{ is even} & \implies \varphi(2n) / 2 \\
\deg( f_n(z) ) / 2= \varphi(n) / 2 & n \text{ is odd} & \implies \varphi(2n) / 2 \\
\end{cases} \\
&= \varphi(2n) / 2
\end{align*}
$$

Though they were present in the earlier Chebyshev table, the $g_n$ themselves are presented again, along with the expression for their degree


| n  |                         $\varphi(2n)/2$ |               $g_n(z)$ | Coefficient List, Rising Powers         |
|----|-----------------------------------------|------------------------|-----------------------------------------|
|  2 | 1                                       |                    $z$ | [0, 1]                                  |
|  3 | 1                                       |               $z -\ 1$ | [-1, 1]                                 |
|  4 | 2                                       |             $z^2 -\ 2$ | [-2, 0, 1]                              |
|  5 | 2                                       |        $z^2 -\ z -\ 1$ | [-1, -1, 1]                             |
|  6 | 2                                       |             $z^2 -\ 3$ | [-2, 0, 1]                              |
|  7 | 3                                       | $z^3 -\ z^2 -\ 2z + 1$ | [1, -2, -1, 1]                          |
|  8 | 4                                       |      $z^4 -\ 4z^2 + 2$ | [2, 0, -4, 0, 1]                        |
|  9 | 3                                       |        $z^3 -\ 3z + 1$ | [1, -3, 0, 1]                           |
| 10 | 4                                       |      $z^4 -\ 5z^2 + 5$ | [5, 0, -5, 0, 1]                        |
|    | [OEIS A055034](http://oeis.org/A055034) |                        | [OEIS A187360](http://oeis.org/A187360) |


### Attempts to Factor

The relationship between $p_n$ and the intermediate $f_d$, where *d* is a divisor of *n*, can be made explicit by a [Moebius inversion](https://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula).

$$
\begin{align*}
p_n(z) &= \prod_{d|n} f_n(z) \\
\log( p_n(z) ) &= \log \left( \prod_{d|n} f_d(z) \right)
= \sum_{d|n} \log( f_d(z) ) \\
\log( f_n(z) ) &= \sum_{d|n} {\mu \left({n \over d}\right)}
\log( p_d(z) ) \\
f_n(z) &= \prod_{d|n} p_d(z)^{\mu \left({n \over d}\right)} \\ \\
f_6(z) = g_6(z) &= p_6(z)^{\mu(1)}
p_3(z)^{\mu(2)}
p_2(z)^{\mu(3)} \\
&= {p_6(z) \over p_3(z) p_2(z)} \\
\end{align*}
$$

Mobius inversion by means of [Dirichlet generating functions](https://en.wikipedia.org/wiki/Dirichlet_series#Formal_Dirichlet_series) is algebraically clear. But moving from $f_n$ to $g_n$ is not easy. $\varphi(n)$ can also be easily expressed by in the Dirichlet world (it is $\zeta(s -\ 1) / \zeta(s)$, where $\zeta$ is the Riemann zeta function), which is useful for knowing when to negate a term. However, the move which doubles the argument of $\varphi$ is difficult to perform in terms of a Dirichlet series, and leaves me stuck.

Even reformatting as a [Lambert series](https://en.wikipedia.org/wiki/Lambert_series) (another kind amenable to Moebius inversion) doesn't provide any intuition, since the product relationship between the polynomials necessitates a logarithm. If you're in the mood for awful-looking math, it is

$$
\begin{align*}
\log( p_n(z) ) &= \sum_{d|n} \log( f_d(z) ) \\
\sum_{n = 1}^\infty \log( p_n ) x^n
&= \sum_{n = 1}^\infty \sum_{d|n} \log( f_d ) x^n \\
&= \sum_{k = 1}^\infty \sum_{m = 1}^\infty \log( f_m ) x^{m k} \\
&= \sum_{m = 1}^\infty \log( f_m ) \sum_{k = 1}^\infty (x^m)^k \\
&= \sum_{m = 1}^\infty \log( f_m ) {x^m \over 1 - x^m} \\
\prod_{n = 1}^\infty p_n(z)^{(x^n)} &= \prod_{n = 1}^\infty f_n(z)^{\left({x^n \over 1 - x^n}\right)}
\end{align*}
$$

Either way, the number-theoretic properties of this sequence are very difficult to ascertain without advanced techniques. If research has been done, it is not easily available in the OEIS.


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](). As well, 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]() to this post if you're interested in another way one can obtain the Chebyshev polynomials.

Diagrams created with GeoGebra.

Update: I have since rederived the Chebyshev polynomials without the complex exponential, which you can read about in [this post]().
