---
title: "Generating Polynomials, Part 2: Ghostly Chains"
description: |
  What do polygons without distance still know about planar geometry?
format:
  html:
    html-math-method: katex
date: "2021-08-19"
date-modified: "2025-06-24"
categories:
  - algebra
  - linear algebra
  - generating functions
  - graph theory
  - python
---

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.figure-img {
  max-width: 512px;
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}

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}
</style>

In the [previous post](../1), I tied the geometry regular polygons to a sequence of polynomials
  though some clever algebraic manipulation.
But let's deign to ask a very basic question: what is a polygon?


Loops without Distance
----------------------

Fundamentally, a polygon is just a collection of vertices and edges.
For polygons in a Euclidean setting, the position of points matters,
  as well as the lines connecting them -- a rectangle is different from a trapezoid or a kite.
But at its simplest, this is just a tabulation of points and adjacencies.

![
  Topologically, all of these are indistinguishable since they all correspond to
    the description "4 points in a loop".
](./quadrilaterals.png)

Only examining these figures by their connectedness is precisely the kind of thing
  *graph theory* deals with.
"Graph" is a potentially confusing term, since it has nothing to do with "graphs of functions",
  but the name is supposed to evoke the fact that they are "drawings".
For the graphs we're interested in, there's some additional terminology:

- Vertices themselves are sometimes instead called *nodes*
- Edge in the graph have no direction in how they connect nodes, so the graph is called *undirected*.
- If the nodes in a graph can be arranged so that no edges appear to intersect,
  the graph is *planar*.
    - For example, lower-right figure in the above diagram appears to have intersecting edges,
      but the nodes can be rearranged to look like the other graphs, so it is planar.

It's easiest to study families of graphs, rather than isolated examples.
If the graph is a simple loop of nodes, it is called a
  [*cycle graph*](https://en.wikipedia.org/wiki/Cycle_graph).
They are denoted by $C_n$, where *n* is the number of nodes.
In a cycle graph, since all nodes are identical to each other (they all connect to two edges)
  and all edges are identical to each other (they connect identical vertices),
  the best geometric interpretation is a shape which is

- Regular, so that each edge and each angle (vertex) are of equal measure
- Convex, so that no edge meets another without creating a vertex (or node)

In other words, $C_3$ is analogous to an equilateral triangle, $C_4$ is analogous to a square, and so on.


### Encoding Graphs

There are two primary ways to store information about a graph.
The first is by labelling each node (for example, with integers), then recording the edges as
  a list of pairs of connected nodes.
In the case of an undirected graph, these are unordered pairs.
While such a list is convenient, it doesn't convey a lot of information about the graph
  besides the number of edges.

Alternatively, these pairs can also be interpreted as addresses in a square matrix,
  called an *adjacency matrix*.
Each column and row correspond to a specific node, and an entry is 1
  when the nodes of a row and column of are joined by an edge (and 0 otherwise).
For undirected graphs, these matrices are symmetric, since it is possible
  to traverse an edge in either direction.

$$
\begin{align*}
  C_3 := \begin{matrix}[
    (0, 1), \\
    (1, 2), \\
    (2, 0)
  ]\end{matrix} & \cong
    \begin{pmatrix}
      0 & 1 & 1 \\
      1 & 0 & 1 \\
      1 & 1 & 0
    \end{pmatrix}
    \\ \\
  C_4 := \begin{matrix} [
    (0, 1), \\
    (1, 2), \\
    (2, 3), \\
    (3, 0)
  ]\end{matrix} & \cong
    \begin{pmatrix}
      0 & 1 & 0 & 1 \\
      1 & 0 & 1 & 0 \\
      0 & 1 & 0 & 1 \\
      1 & 0 & 1 & 0
    \end{pmatrix}
    \\ \\
  C_5 := \begin{matrix}[
    (0, 1), \\
    (1, 2), \\
    (2, 3), \\
    (3, 4), \\
    (4, 0)
  ]\end{matrix} &\cong
    \begin{pmatrix}
      0 & 1 & 0 & 0 & 1 \\
      1 & 0 & 1 & 0 & 0 \\
      0 & 1 & 0 & 1 & 0 \\
      0 & 0 & 1 & 0 & 1 \\
      1 & 0 & 0 & 1 & 0
    \end{pmatrix}
\end{align*}
$$

Swapping the labels on two nodes is will exchange two rows and two columns
  of the adjacency matrix.
Just one of these swaps would flip the sign of its determinant, but since they occur in pairs,
  the determinant is invariant of the labelling (equally, a graph invariant).


Prismatic Recurrence
--------------------

The determinant of a matrix is also the product of its eigenvalues, which are another matrix invariant.
The set of eigenvalues is also called its *spectrum*, and the study of the spectra of graphs is called
  [*spectral graph theory*](https://en.wikipedia.org/wiki/Spectral_graph_theory)[^1],

[^1]: It is also among the most mystifying names in math to read without any context

Eigenvalues are the roots of the characteristic polynomial of a matrix.
The matrix $C_5$ is sufficiently large enough to generalize to $C_n$, and its characteristic polynomial by
  [Laplace expansion](https://en.wikipedia.org/wiki/Laplace_expansion) is:

$$
\begin{gather*}
  Ax = \lambda x \implies (\lambda I - A)x = 0
  \\ \\
  c_5(\lambda) = |\lambda I - C_5|
    = \left |
    \begin{matrix}
      \lambda &      -1 &       0 &       0 &      -1 \\
           -1 & \lambda &      -1 &       0 &       0 \\
            0 &      -1 & \lambda &      -1 &       0 \\
            0 &       0 &      -1 & \lambda &      -1 \\
           -1 &       0 &       0 &      -1 & \lambda
    \end{matrix}
    \right |
  \\
  = \lambda m_{1,1}
    + \overbrace{(-1)}^\text{entry}\overbrace{(-1)^{1 + 2 \ }}^\text{sign} m_{1, 2}
    + \overbrace{(-1)}^\text{entry}\overbrace{(-1)^{1 + 5 \ }}^\text{sign} m_{1, 5}
\end{gather*}
$$

Note that every occurrence of "5" generalizes to higher *n*.
The first [minor](https://en.wikipedia.org/wiki/Matrix_minor)
  is easily expressed in terms of *another* matrix's characteristic polynomial.

$$
m_{1, 1}[C_5]
  = \left |
  \begin{matrix}
    \lambda &      -1 &       0 &       0 \\
         -1 & \lambda &      -1 &       0 \\
          0 &      -1 & \lambda &      -1 \\
          0 &       0 &      -1 & \lambda
  \end{matrix}
  \right |
  = |\lambda I - P_4| = p_{5 - 1}(\lambda)
$$

We will come to the meaning of the $P_n$ in a moment.
The other minors require extra expansions, but ones that (thankfully) quickly terminate.

$$
\begin{matrix}
  m_{1, 2}[C_5]
    &= \left |
    \begin{matrix}
      -1 &      -1 &       0 &       0 \\
       0 & \lambda &      -1 &       0 \\
       0 &      -1 & \lambda &      -1 \\
      -1 &       0 &      -1 & \lambda
    \end{matrix}
    \right |
    &=& (-1)
    \left |
    \begin{matrix}
      \lambda &      -1 &       0 \\
           -1 & \lambda &      -1 \\
            0 &      -1 & \lambda
    \end{matrix}
    \right |
    &+& (-1)(-1)^{1 + 4}
    \left |
    \begin{matrix}
           -1 &       0 &  0 \\
      \lambda &      -1 &  0 \\
           -1 & \lambda & -1
    \end{matrix}
    \right |
  \\
  &&=& (-1)|\lambda I - P_3|
    &+& (-1)\overbrace{(-1)^{5}(-1)^{5 - 2}}^{\text{even power, even when $\scriptsize n \neq 5$}}
  \\
  &&=& (-1)p_{5 - 2}(\lambda) &+& (-1)
  \\
  &&=& -(p_{5 - 2}(\lambda) &+& 1)
\end{matrix}
$$

The "1 + 4" exponent when evaluating this minor comes from the address of the lower-left -1, (i.e., (1, 4)).
This entry exists for all $C_n$.
The determinant of the rightmost matrix is just the product of the -1's on the diagonal, so it will always
  have a power of the same parity as *n*, which cancels out with the sign of the minor.
Meanwhile, another $P$-type matrix appears in the other term, this time of two lower orders.

$$
\begin{matrix}
  \\ \\
  m_{1, 5}[C_5] &=
    \left |
    \begin{matrix}
      -1 & \lambda &      -1 &       0 \\
       0 &      -1 & \lambda &      -1 \\
       0 &       0 &      -1 & \lambda \\
      -1 &       0 &       0 &      -1
    \end{matrix}
    \right |
    &=& (-1)
    \left |
    \begin{matrix}
      -1 & \lambda &      -1 \\
       0 &      -1 & \lambda \\
       0 &       0 &      -1
    \end{matrix}
    \right |
    &+& (-1)(-1)^{5 - 2}
    \left |
    \begin{matrix}
      \lambda &      -1 &       0 \\
           -1 & \lambda &      -1 \\
            0 &      -1 & \lambda
    \end{matrix}
    \right |
  \\
  &&=& (-1)(-1)^{5-2} &+& (-1)(-1)^{5 - 2}|\lambda I - P_3|
  \\
  &&=& (-1)^{5-1}((-1)(-1) &+& (-1)(-1)p_{5 - 2}(\lambda))
  \\
  &&=& (-1)^{5-1}(1 &+& p_{5 - 2}(\lambda))
\end{matrix}
$$

A third $P$-type matrix appears, just like the other minor.
Unfortunately, this minor *does* depend on the parity of *n*.

All together, this produces a characteristic polynomial in terms of the polynomials $p_n(\lambda)$:

$$
\begin{align*}
  && c_5(\lambda) &= \lambda p_{5 - 1}
    + (-1)(p_{5 - 2} + 1)
    + (-1)\overbrace{(-1)^{5 - 1} (-1)^{5 - 1}}^{\text{even, even when $\scriptsize n \neq 5$}}(p_{5 - 2} + 1)
  \\
  &&&= \lambda p_{5 - 1}
    - (p_{5 - 2} + 1)
    - (p_{5 - 2} + 1)
  \\
  &&&= \lambda p_{5 - 1}
    - 2(p_{5 - 2} + 1)
  \\
  && \implies c_n(\lambda) &= \lambda p_{n - 1}(\lambda)
    - 2(p_{n - 2}(\lambda) + 1)
\end{align*}
$$

Fortunately, the minor whose determinant depended on the parity of *n* cancels with $(-1)^{1 + 5}$,
  and the resulting expression seems to generically apply across all *n*.
Further, this resembles a recurrence relation, which is great for building a rule.
But it is meaningless without knowing $p_n(\lambda)$.


Powerful Chains
---------------

The various $P_n$ are in fact the adjacency matrices of a path on *n* nodes.

![
  Example path graphs of orders 2, 3, and 4
](./path_graphs.png){.wide}

$$
\begin{align*}
  P_2 &:=
    \begin{matrix}[
      (0, 1)
    ]\end{matrix}
    \cong \begin{pmatrix}
      0 & 1 \\
      1 & 0
    \end{pmatrix}
  \\
  P_3 &:=
    \begin{matrix}[
      (0, 1), \\
      (1, 2)
    ]\end{matrix}
    \cong \begin{pmatrix}
      0 & 1 & 0 \\
      1 & 0 & 1 \\
      0 & 1 & 0
    \end{pmatrix}
  \\
  P_4 &:=
    \begin{matrix}[
      (0, 1), \\
      (1, 2), \\
      (2, 3)
    ]\end{matrix}
    \cong \begin{pmatrix}
      0 & 1 & 0 & 0 \\
      1 & 0 & 1 & 0 \\
      0 & 1 & 0 & 1 \\
      0 & 0 & 1 & 0
    \end{pmatrix}
\end{align*}
$$

These matrices are similar to the ones for cycle graphs, but lack the entries in bottom-left
  and upper-right corners.
Consequently, the characteristic polynomials of $P_n$ are much easier to solve for.

$$
\begin{gather*}
  p_4(\lambda) = |\lambda I - P_4|
    = \left |
    \begin{matrix}
      \lambda &      -1 &       0 &       0 \\
           -1 & \lambda &      -1 &       0 \\
            0 &      -1 & \lambda &      -1 \\
            0 &       0 &      -1 & \lambda
    \end{matrix}
    \right |
  \\ \\
  = \lambda \left |
    \begin{matrix}
      \lambda &      -1 &       0 \\
           -1 & \lambda &      -1 \\
            0 &      -1 & \lambda
    \end{matrix}
    \right | + (-1)(-1)^{1+2} \left |
    \begin{matrix}
      -1 &      -1 &       0 \\
       0 & \lambda &      -1 \\
       0 &      -1 & \lambda
    \end{matrix}
    \right |
  \\ \\
  = \lambda |\lambda I - P_3| + \left (
    (-1) \left |
    \begin{matrix}
      \lambda &      -1 \\
           -1 & \lambda
    \end{matrix}
    \right |
    + (-1)(-1) \left |
    \begin{matrix}
      0 &      -1 \\
      0 & \lambda
    \end{matrix}
    \right |
    \right)
  \\ \\
  = \lambda |\lambda I - P_3| - |\lambda I - P_2|
  \\
  = \lambda p_{4 - 1}(\lambda) - p_{4 - 2}(\lambda)
  \\
  \implies p_{n}(\lambda) = \lambda p_{n - 1}(\lambda) - p_{n - 2}(\lambda)
\end{gather*}
$$

While the earlier equation for $c_n$ in terms of $p_n$ reminded of a recurrence relation,
  *this* actually is one (and it should look familiar).

Since the recurrence has order 2, it requires two initial terms: $p_0$ and $p_1$.
The graph corresponding to $p_1$ is a single node, not connected to anything.
Therefore, its adjacency matrix is a 1x1 matrix with 0 as its only entry,
  and its characteristic polynomial is $\lambda$.
By the recurrence, $p_2 = \lambda p_1 -\ p_0 = \lambda^2 -\ p_0$.
Equating terms with the characteristic polynomial of $P_2$, it is obvious that

$$
|\lambda I - P_2|
= \begin{pmatrix}
  \lambda &      -1 \\
       -1 & \lambda
\end{pmatrix}
= \lambda^2 - 1 = \lambda p_1 - p_0 \\
\implies p_0 = 1
$$

which makes sense, since $p_0$ should have degree zero.
Therefore, the sequence of polynomials $p_n(\lambda)$ is:

$$
\begin{gather*}
  p_0(\lambda) &=&& && 1
  \\
  p_1(\lambda) &=&& && \lambda
  \\
  p_2(\lambda) &=&& \lambda \lambda - 1
    &=& \lambda^2 - 1
  \\
  p_3(\lambda) &=&& \lambda (\lambda^2 - 1) - \lambda
    &=& \lambda(\lambda^2 - 2)
  \\
  p_4(\lambda) &=&& \lambda (\lambda(\lambda^2 - 2)) - (\lambda^2 - 1)
    &=& \lambda^4 - 3\lambda^2 + 1
  \\
  \vdots & && \vdots && \vdots
\end{gather*}
$$

But wait, we've seen these before (if you read the previous post, that is).
These are just the Chebyshev polynomials of the second kind, evaluated at $\lambda / 2$.
Indeed, their recurrence relations are identical, so the characteristic polynomial of $P_n$ is $U_n(\lambda / 2)$.
Effectively, this connects an *n*-path to a regular *n+1*-gon.


### Back to Cycles

Since the generating function of $U_n$ is known, the generating function for the $c_n$
  (which prompted this) is also easily determined.
For ease of use, let

$$
P(x; \lambda) = {B(x; \lambda / 2) \over x} = {1 \over 1 - \lambda x +\ x^2}
$$

Discarding the initial $c_0$ and $c_1$ by setting them to zero[^2], the generating function is

[^2]: It's a good idea to ask why we can do this.
      Try examining $c_2$ and $c_3$.

$$
\begin{align*}
  c_{n+2}(\lambda) &= \lambda p_{n+1}(\lambda) - 2(p_n(\lambda) + 1)
  \\[14pt]
  {C(x; \lambda) - c_0(\lambda) - x c_1(\lambda) \over x^2}
    &= \lambda \left( {P(x; \lambda) - 1 \over x} \right)
    - 2\left( P(x; \lambda) + {1 \over 1 - x} \right)
  \\
  C &= x \lambda (P - 1) -\
    2x^2\left( P + {1 \over 1 - x} \right)
  \\
  C{(1 - x) \over P} &= x \lambda \left(1 - {1 \over P} \right)(1 - x) -\
    2x^2\left( (1 - x) + {1 \over P} \right)
  \\
  &= x^4 \lambda - 2 x^4 - x^3 \lambda^2 + x^3 \lambda
    + 2 x^3 + x^2 \lambda^2 - 4 x^2
  \\
  &= x^2 (\lambda - 2) (x^2 - \lambda x - x + \lambda + 2)
  \\[14pt]
  C(x; \lambda) &= x^2 (\lambda - 2)
    {(x^2 - (\lambda + 1) x + \lambda + 2)
    \over (1 - x)(1 - \lambda x + x^2)}
\end{align*}
$$

While the numerator is considerably more complicated than the one for P,
  the factor $\lambda - 2$ drops out of the entire series.
This pleasantly informs that 2 is an eigenvalue of all $C_n$.


Off the Beaten Path
-------------------

When we use Laplace expansion on the adjacency matrices, we were very fortunate that the minors
  *also* looked like adjacency matrices undergoing expansion.
This let us terminate early and recurse.
From the perspective of the graph, Laplace expansion almost looks like removing a node,
  but requires special treatment for the nodes connected to the one being removed.
For example, in cycle graphs, the first stage of expansion had three minors:

- The node itself, on the main diagonal
  - Being on the main diagonal, this immediately produced another adjacency matrix.
- Either neighbor connected to it, which are on opposite sides of
  a path after the node is removed
  - Both of these nodes required second expansion to get the *λ*s back on the main diagonal.

For "good enough" graphs that are nearly paths (including paths themselves),
  this gives a second-order recurrence relation.


### Trees

Another simple family of graphs are [*trees*](https://en.wikipedia.org/wiki/Tree_%28graph_theory%29).
In some sense, they are the opposite of cycle graphs, since by definition they contain no cycles.

Paths are degenerate trees, but we can make them slightly more interesting by instead adding
  exactly one node and edge to (the middle of) a path.

![
  Nondegenerate tree graphs based on 3-, 4-, 5-, and 6-paths
](./tree_graphs.png){.wide}

In this notation, the subscripts denote the consituent paths if the "added" node and the
  one it is connected to are both removed.
It's easy to see that $T_{a,b} \cong T_{a,b}$, since this just swaps the arms.
Also, $T_{a, 0} \cong T_{0, a} \cong P_{a + 2}$.

Let's try dissecting some of the larger trees.
The adjacency matrices for $T_{1,3}$ and $T_{2,2}$ are:

$$
\begin{align*}
  T_{1,3} := \begin{matrix}[
    (0, 1), \\
    (1, 2), \\
    (2, 3), \\
    (3, 4), \\
    (1, 5)
  ]\end{matrix} &\cong
    \begin{pmatrix}
      0 & 1 & 0 & 0 & 0 & 0 \\
      1 & 0 & 1 & 0 & 0 & 1 \\
      0 & 1 & 0 & 1 & 0 & 0 \\
      0 & 0 & 1 & 0 & 1 & 0 \\
      0 & 0 & 0 & 1 & 0 & 0 \\
      0 & 1 & 0 & 0 & 0 & 0
    \end{pmatrix}
    \\ \\
  T_{2,2} := \begin{matrix}[
    (0, 1), \\
    (1, 2), \\
    (2, 3), \\
    (3, 4), \\
    (2, 5)
  ]\end{matrix} &\cong
    \begin{pmatrix}
      0 & 1 & 0 & 0 & 0 & 0 \\
      1 & 0 & 1 & 0 & 0 & 0 \\
      0 & 1 & 0 & 1 & 0 & 1 \\
      0 & 0 & 1 & 0 & 1 & 0 \\
      0 & 0 & 0 & 1 & 0 & 0 \\
      0 & 0 & 1 & 0 & 0 & 0
    \end{pmatrix}
\end{align*}
$$

Starting with $T_{1,3}$, its characteristic polynomial is:

$$
\begin{align*}
  |I \lambda - T_{1,3}|
    &= \left |
    \begin{matrix}
      \lambda &      -1 &       0 &       0 &       0 &       0 \\
           -1 & \lambda &      -1 &       0 &       0 &      -1 \\
            0 &      -1 & \lambda &      -1 &       0 &       0 \\
            0 &       0 &      -1 & \lambda &      -1 &       0 \\
            0 &       0 &       0 &      -1 & \lambda &       0 \\
            0 &      -1 &       0 &       0 &       0 & \lambda
    \end{matrix}
    \right |
  \\
  &= \lambda (-1)^{6 + 6} m_{6,6} + (-1) (-1)^{2 + 6} m_{2,6}
\end{align*}
$$

It's easy to see that $m_{6,6}$ is just $p_5(\lambda)$, since the rest of the graph
  other than the additional node is a 5-path.
But the other minor is trickier.

$$
\begin{align*}
  m_{2,6}
    &= \left |
      \begin{matrix}
      \lambda &      -1 &       0 &       0 &       0 \\
            0 &      -1 & \lambda &      -1 &       0 \\
            0 &       0 &      -1 & \lambda &      -1 \\
            0 &       0 &       0 &      -1 & \lambda \\
            0 &      -1 &       0 &       0 &       0
    \end{matrix}
    \right |
  \\
  &= (-1) (-1)^{5 + 2}
    \left |
    \begin{matrix}
      \lambda &       0 &       0 &       0 \\
            0 & \lambda &      -1 &       0 \\
            0 &      -1 & \lambda &      -1 \\
            0 &       0 &      -1 & \lambda \\
    \end{matrix}
    \right |
\end{align*}
$$

Through one extra expansion, the determinant of this final matrix can be written as
  a product of $\lambda$ and $p_3(\lambda)$.

Before making any conjectures, let's do the same thing to $T_{2,2}$.

$$
\begin{align*}
  |I \lambda - T_{2,2}|
    &= \left |
    \begin{matrix}
      \lambda &      -1 &       0 &       0 &       0 &       0 \\
           -1 & \lambda &      -1 &       0 &       0 &       0 \\
            0 &      -1 & \lambda &      -1 &       0 &      -1 \\
            0 &       0 &      -1 & \lambda &      -1 &       0 \\
            0 &       0 &       0 &      -1 & \lambda &       0 \\
            0 &       0 &      -1 &       0 &       0 & \lambda
    \end{matrix}
    \right |
  \\
  &= (-1)^{6 + 6} \lambda m_{6,6} + (-1) (-1)^{3 + 6} m_{3,6}
  \\ \\
  m_{3,6}
    &= \left |
      \begin{matrix}
      \lambda &      -1 &       0 &       0 &       0 \\
           -1 & \lambda &      -1 &       0 &       0 \\
            0 &       0 &      -1 & \lambda &      -1 \\
            0 &       0 &       0 &      -1 & \lambda \\
            0 &       0 &      -1 &       0 &       0
    \end{matrix}
    \right |
  \\
  &= (-1) (-1)^{5 + 3}
    \left |
    \begin{matrix}
      \lambda &      -1 &       0 &       0 \\
           -1 & \lambda &       0 &       0 \\
            0 &       0 & \lambda &      -1 \\
            0 &       0 &      -1 & \lambda \\
    \end{matrix}
    \right |
\end{align*}
$$

Here we get something similar: a combination of $p_5(\lambda)$ and an extra term.
In this case, the final determinant can be written as $p_2(\lambda)^2$.

Now it can be observed that the extra terms are the polynomials corresponding
  to $P_a$ and $P_b$ (recall that $p_1(\lambda) = \lambda$, after all).
In both cases, the second expansion was necessary to get rid of the symmetric -1
  entries added to the matrix.
The sign of this extra term is always negative, since the -1 entries cancel
  and one of the signs of the minors along the two expansions must be negative.

Therefore, the expression for these characteristic polynomials should be:

$$
t_{a,b}(\lambda) = \lambda p_{a + b + 1}(\lambda) - p_a(z) p_b(\lambda)
$$

Note that if *b* is 0, this agrees with the recurrence for $p_n(\lambda)$.


### Examining Small Trees

Due to the subscript of the first term of the RHS, this recurrence is harder to turn into
  a generating function.
Instead, let's look at a few smaller trees to see what kind of polynomials they build.
We'll also change the variable of the polynomial to *z* for simplicity.

The first tree of note is $T_{1,1}$.
This has characteristic polynomial

$$
\begin{align*}
t_{1,1}(z) &= z p_{3}(z) - p_1(z) p_1(z)
  \\
  &= z (z^3 - z^2) - z^2
  \\
  &= z^2 (z^2 - 3)
\end{align*}
$$

Next, we have both $T_{2,1}$ and $T_{1,2}$.
By symmetry, these are the same graph, so we have characteristic polynomial

$$
\begin{align*}
t_{1,2}(z) &= z p_{4}(z) - p_1(z) p_2(z)
  \\
  &= z (z^4 - 3z^2 + 2) - z \cdot (z^2 - 1)
  \\
  &= z (z^4 - 4z^2 + 2)
\end{align*}
$$

Finally, let's look at $T_{1,3}$ and $T_{2,2}$, the trees we used to derive the rule.

$$
\begin{align*}
  t_{1,3}(z) &= z p_{5}(z) - p_1(z) p_3(z)
  \\
  &= z (z^5 - 4z^3 + 3z) - z \cdot (z^3 - 2z)
  \\
  &= z^2 (z^4 - 5z^2 + 5)
  \\[10pt]
  t_{2,2}(z) &= z p_{5}(z) - p_2(z) p_2(z)
  \\
  &= z (z^5 - 4z^3 + 3z) - ( z^2 - 1 )^2
  \\
  &= (z^2 - 1)(z^4 - 4z^2 + 1)
\end{align*}
$$

Many of these expressions factor surprisingly nicely.
Further, some of these might seem familiar.
From the last post, we saw that $z^4 - 5z^2 + 5$ is a factor of $p_9(z)$, from which we know
  it is the minimal polynomial of $2 \cos(\pi / 10)$.

This is also true for:

- In $t_{1,2}$, the factor $z^4 - 4z^2 + 2$, $p_7(z)$, and $2 \cos(\pi / 8)$, respectively
- In $t_{2,2}$, the factor $z^4 - 4z^2 + 1$, $p_11(z)$, and $2 \cos(\pi / 12)$, respectively

We established that the subscripts of the tree (*a* and *b*) indicate constituent *n*-paths,
  which we know to correspond to *n+1*-gons.
But these trees also seem to "know" about higher polygons.


### Some Extra Trees

$T_{2,3}$ is the first tree not to partition two equal paths or a path and a single node.
In this regard, the next such tree is $T_{2,4}$.

These graphs turn out to have characteristic polynomials whose factors we haven't seen before.

$$
\begin{align*}
  t_{2,3}(z)
    &= z (z^{6} - 6 z^{4} + 9 z^{2} - 3)
  \\
  t_{2,4}(z)
    &= z^{8} - 7 z^{6} + 14 z^{4} - 8 z^{2} + 1
\end{align*}
$$

Searching the OEIS for the coefficients of $t_{2,4}$ returns sequence
  [A228786](http://oeis.org/A228786), which informs that it is the minimal polynomial
  of $2\sin( \pi/15 )$.
This sequence also informs that the unknown factor in the other polynomial is
  the minimal polynomial of $2\sin( \pi / 9 )$:

In fact, both of these polynomials show up in factorizations of Chebyshev polynomials
  of the *first* kind (specifically, $2T_15(z / 2)$ and $2T_9(z / 2)$).
Perhaps this is not surprising since we were already working with those of the second kind.
However, it is interesting to see them appear from the addition of a single node.


Closing
-------

Regardless of whether chains or polygons are more fundamental, it is certainly interesting
  that they are just an algebraic stone's (a *calculus*'s?) toss away from one another.
Perhaps Euler skipped such stones from the bridges of Koenigsberg which inspired him
  to initiate graph theory.

Trees are certainly more complicated than either, and we only investigated those removed
  from a path by a single node.
Regardless, they still related to Chebyshev polynomials, albeit through their factors.

In fact, I was initially prompted to look into them due to a remarkable correspondence between
  certain trees and Platonic solids.
I have since reorganized these thoughts, as from the perspective of this article, the relationship
  is tangential at best.
