804 lines
24 KiB
Plaintext
804 lines
24 KiB
Plaintext
---
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title: "Generating Polynomials, Part 2: Ghostly Chains"
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description: |
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"What do polygons without distance still know about planar geometry?"
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format:
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html:
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html-math-method: katex
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date: "2021-08-19"
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date-modified: "2025-06-24"
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categories:
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- geometry
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- algebra
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- graph theory
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---
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In the [previous post](../1), I tied the geometry regular polygons to a sequence of polynomials
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though some clever algebraic manipulation.
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But let's deign to ask a very basic question: what is a polygon?
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Loops without Distance
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----------------------
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Fundamentally, a polygon is just a collection of vertices and edges.
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For polygons in a Euclidean setting, the position of points matters,
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as well as the lines connecting them -- a rectangle is different from a trapezoid or a kite.
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But at its simplest, this is just a tabulation of points and adjacencies.
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Only examining these figures by their connectedness is precisely the kind of thing
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*graph theory* deals with.
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"Graph" is a potentially confusing term, since it has nothing to do with "graphs of functions",
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but the name is supposed to evoke the fact that they are "drawings".
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For the graphs we're interested, there's some additional terminology:
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- Vertices themselves are sometimes instead called *nodes*
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- Edge in the graph have no direction associated to them, so the graph is called *undirected*.
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- Additionally, these graphs are *planar* since the nodes can be arranged
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so that no edges appear to intersect.
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- The lower-right figure in the above diagram has intersecting edges,
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but the nodes can be rearranged to look like the other graphs, so it is planar.
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Graphs themselves typically come in families.
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If the graph is a simple loop, it is called a [*cycle graph*](https://en.wikipedia.org/wiki/Cycle_graph).
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They are denoted by $C_n$, where *n* is the number of nodes.
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In a cycle graph, since all nodes are identical to each other (they all connect to two edges)
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and all edges are identical to each other (they connect identical vertices),
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the best geometric interpretation is a shape which is
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- Regular, so that each edge and each angle (vertex) are of equal measure
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- Convex, so that no edge meets another without creating a vertex (or node)
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In other words, $C_3$ is analogous to an equilateral triangle, $C_4$ is analogous to a square, and so on.
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### Encoding Graphs
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There are two primary ways to store information about a graph.
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The first is by labelling each node (for example, with integers), then recording the edges as
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a list of pairs of connected nodes.
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In the case of an undirected graph, these are unordered pairs.
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While such a list is convenient, it doesn't convey a lot of information about the graph
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besides the number of edges.
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Alternatively, these pairs can also be interpreted as addresses in a matrix,
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called an *adjacency matrix*.
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$$
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\begin{align*}
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C_3 := \begin{matrix}[
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(0, 1), \\
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(1, 2), \\
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(2, 0)
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]\end{matrix} & \cong
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\begin{pmatrix}
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0 & 1 & 1 \\
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1 & 0 & 1 \\
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1 & 1 & 0
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\end{pmatrix}
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\\ \\
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C_4 := \begin{matrix} [
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(0, 1), \\
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(1, 2), \\
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(2, 3), \\
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(3, 0)
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]\end{matrix} & \cong
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\begin{pmatrix}
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0 & 1 & 0 & 1 \\
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1 & 0 & 1 & 0 \\
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0 & 1 & 0 & 1 \\
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1 & 0 & 1 & 0
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\end{pmatrix}
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\\ \\
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C_5 := \begin{matrix}[
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(0, 1), \\
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(1, 2), \\
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(2, 3), \\
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(3, 4), \\
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(4, 0)
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]\end{matrix} &\cong
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\begin{pmatrix}
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0 & 1 & 0 & 0 & 1 \\
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1 & 0 & 1 & 0 & 0 \\
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0 & 1 & 0 & 1 & 0 \\
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0 & 0 & 1 & 0 & 1 \\
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1 & 0 & 0 & 1 & 0
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\end{pmatrix}
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\end{align*}
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$$
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Each adjacency matrix is square, where each column and row refer to a specific node.
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An entry is 1 when the nodes corresponding to the row and column of its address are joined by an edge (and zero otherwise).
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For undirected graphs, these matrices are symmetric, since it is possible to traverse an edge in either direction.
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Swapping the labels on two nodes is as simple as exchanging two rows and two columns.
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Just one of these swaps would flip the sign of the determinant of the adjacency matrix.
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However, since they occur in pairs, the determinant is invariant of the labelling (equally, a graph invariant).
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Prismatic Recurrence
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--------------------
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The determinant of a matrix is also the product of its eigenvalues, which are another matrix invariant.
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The set of eigenvalues is also called its *spectrum*, and the study of the spectra of graphs is called
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[*spectral graph theory*](https://en.wikipedia.org/wiki/Spectral_graph_theory)[^1],
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[^1]: It is also among the most mystifying names in math to read without any context
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Eigenvalues are the roots of the characteristic polynomial of a matrix.
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The matrix $C_5$ is sufficiently large enough to generalize to $C_n$, and its characteristic polynomial by
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[Laplace expansion](https://en.wikipedia.org/wiki/Laplace_expansion) is:
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$$
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\begin{gather*}
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Ax = \lambda x \implies (\lambda I - A)x = 0
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\\ \\
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c_5(\lambda) = |\lambda I - C_5|
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= \left |
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\begin{matrix}
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\lambda & -1 & 0 & 0 & -1 \\
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-1 & \lambda & -1 & 0 & 0 \\
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0 & -1 & \lambda & -1 & 0 \\
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0 & 0 & -1 & \lambda & -1 \\
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-1 & 0 & 0 & -1 & \lambda
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\end{matrix}
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\right |
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\\
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= \lambda m_{1,1}
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+ \overbrace{(-1)}^\text{entry}\overbrace{(-1)^{1 + 2 \ }}^\text{sign} m_{1, 2}
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+ \overbrace{(-1)}^\text{entry}\overbrace{(-1)^{1 + 5 \ }}^\text{sign} m_{1, 5}
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\end{gather*}
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$$
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Note that every occurrence of "5" generalizes to higher *n*.
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The first [minor](https://en.wikipedia.org/wiki/Matrix_minor)
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is easily expressed in terms of *another* matrix's characteristic polynomial.
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$$
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m_{1, 1}[C_5]
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= \left |
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\begin{matrix}
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\lambda & -1 & 0 & 0 \\
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-1 & \lambda & -1 & 0 \\
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0 & -1 & \lambda & -1 \\
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0 & 0 & -1 & \lambda
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\end{matrix}
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\right |
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= |\lambda I - P_4| = p_{5 - 1}(\lambda)
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$$
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We will come to the meaning of the $P_n$ in a moment.
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The other minors require extra expansions, but ones that (thankfully) quickly terminate.
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$$
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\begin{matrix}
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m_{1, 2}[C_5]
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&= \left |
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\begin{matrix}
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-1 & -1 & 0 & 0 \\
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0 & \lambda & -1 & 0 \\
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0 & -1 & \lambda & -1 \\
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-1 & 0 & -1 & \lambda
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\end{matrix}
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\right |
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&=& (-1)
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\left |
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\begin{matrix}
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\lambda & -1 & 0 \\
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-1 & \lambda & -1 \\
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0 & -1 & \lambda
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\end{matrix}
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\right |
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&+& (-1)(-1)^{1 + 4}
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\left |
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\begin{matrix}
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-1 & 0 & 0 \\
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\lambda & -1 & 0 \\
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-1 & \lambda & -1
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\end{matrix}
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\right |
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\\
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&&=& (-1)|\lambda I - P_3|
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&+& (-1)\overbrace{(-1)^{5}(-1)^{5 - 2}}^{\text{even power, even when $\scriptsize n \neq 5$}}
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\\
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&&=& (-1)p_{5 - 2}(\lambda) &+& (-1)
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\\
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&&=& -(p_{5 - 2}(\lambda) &+& 1)
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\end{matrix}
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$$
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The "1 + 4" exponent when evaluating this minor comes from the address of the lower-left -1, (i.e., (1, 4)).
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This entry exists for all $C_n$.
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The determinant of the rightmost matrix is just the product of the -1's on the diagonal, so it will always
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have a power of the same parity as *n*, which cancels out with the sign of the minor.
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Meanwhile, another $P$-type matrix appears in the other term, this time of two lower orders.
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$$
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\begin{matrix}
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\\ \\
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m_{1, 5}[C_5] &=
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\left |
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\begin{matrix}
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-1 & \lambda & -1 & 0 \\
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0 & -1 & \lambda & -1 \\
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0 & 0 & -1 & \lambda \\
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-1 & 0 & 0 & -1
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\end{matrix}
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\right |
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&=& (-1)
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\left |
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\begin{matrix}
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-1 & \lambda & -1 \\
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0 & -1 & \lambda \\
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0 & 0 & -1
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\end{matrix}
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\right |
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&+& (-1)(-1)^{5 - 2}
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\left |
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\begin{matrix}
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\lambda & -1 & 0 \\
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-1 & \lambda & -1 \\
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0 & -1 & \lambda
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\end{matrix}
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\right |
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\\
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&&=& (-1)(-1)^{5-2} &+& (-1)(-1)^{5 - 2}|\lambda I - P_3|
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\\
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&&=& (-1)^{5-1}((-1)(-1) &+& (-1)(-1)p_{5 - 2}(\lambda))
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\\
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&&=& (-1)^{5-1}(1 &+& p_{5 - 2}(\lambda))
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\end{matrix}
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$$
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A third $P$-type matrix appears, just like the other minor.
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Unfortunately, this minor *does* depend on the parity of *n*.
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All together, this produces a characteristic polynomial in terms of the polynomials $p_n(\lambda)$:
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$$
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\begin{align*}
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&& c_5(\lambda) &= \lambda p_{5 - 1}
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+ (-1)(p_{5 - 2} + 1)
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+ (-1)\overbrace{(-1)^{5 - 1} (-1)^{5 - 1}}^{\text{even, even when $\scriptsize n \neq 5$}}(p_{5 - 2} + 1)
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\\
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&&&= \lambda p_{5 - 1}
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- (p_{5 - 2} + 1)
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- (p_{5 - 2} + 1)
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\\
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&&&= \lambda p_{5 - 1}
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- 2(p_{5 - 2} + 1)
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\\
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&& \implies c_n(\lambda) &= \lambda p_{n - 1}(\lambda)
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- 2(p_{n - 2}(\lambda) + 1)
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\end{align*}
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$$
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Fortunately, the minor whose determinant depended on the parity of *n* cancels with $(-1)^{1 + 5}$,
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and the resulting expression seems to generically apply across all *n*.
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Further, this resembles a recurrence relation, which is great for building a rule.
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But it is meaningless without knowing $p_n(\lambda)$.
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Powerful Chains
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---------------
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The various $P_n$ are in fact the adjacency matrices of a path on *n* nodes.
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$$
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\begin{align*}
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P_2 &:=
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\begin{matrix}[
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(0, 1)
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]\end{matrix}
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\cong \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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P_3 &:=
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\begin{matrix}[
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(0, 1), \\
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(1, 2)
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]\end{matrix}
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\cong \begin{pmatrix}
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0 & 1 & 0 \\
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1 & 0 & 1 \\
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0 & 1 & 0
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\end{pmatrix}
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\\
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P_4 &:=
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\begin{matrix}[
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(0, 1), \\
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(1, 2), \\
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(2, 3)
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]\end{matrix}
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\cong \begin{pmatrix}
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0 & 1 & 0 & 0 \\
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1 & 0 & 1 & 0 \\
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0 & 1 & 0 & 1 \\
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0 & 0 & 1 & 0
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\end{pmatrix}
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\end{align*}
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$$
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These matrices are similar to the ones for cycle graphs, but lack the entries in bottom-left
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and upper-right corners.
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Consequently, the characteristic polynomials of $P_n$ are much easier to solve for.
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$$
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\begin{gather*}
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p_4(\lambda) = |\lambda I - P_4|
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= \left |
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\begin{matrix}
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\lambda & -1 & 0 & 0 \\
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-1 & \lambda & -1 & 0 \\
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0 & -1 & \lambda & -1 \\
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0 & 0 & -1 & \lambda
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\end{matrix}
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\right |
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\\ \\
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= \lambda \left |
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\begin{matrix}
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\lambda & -1 & 0 \\
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-1 & \lambda & -1 \\
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0 & -1 & \lambda
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\end{matrix}
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\right | + (-1)(-1)^{1+2} \left |
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\begin{matrix}
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-1 & -1 & 0 \\
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0 & \lambda & -1 \\
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0 & -1 & \lambda
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\end{matrix}
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\right |
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\\ \\
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= \lambda |\lambda I - P_3| + \left (
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(-1) \left |
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\begin{matrix}
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\lambda & -1 \\
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-1 & \lambda
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\end{matrix}
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\right |
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+ (-1)(-1) \left |
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\begin{matrix}
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0 & -1 \\
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0 & \lambda
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\end{matrix}
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\right |
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\right)
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\\ \\
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= \lambda |\lambda I - P_3| - |\lambda I - P_2|
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\\
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= \lambda p_{4 - 1}(\lambda) - p_{4 - 2}(\lambda)
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\\
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\implies p_{n}(\lambda) = \lambda p_{n - 1}(\lambda) - p_{n - 2}(\lambda)
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\end{gather*}
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$$
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While the earlier equation for $c_n$ in terms of $p_n$ reminded of a recurrence relation,
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*this* actually is one (and it should look familiar).
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Since the recurrence has order 2, it requires two initial terms: $p_0$ and $p_1$.
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The graph corresponding to $p_1$ is a single node, not connected to anything.
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Therefore, its adjacency matrix is a 1x1 matrix with 0 as its only entry,
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and its characteristic polynomial is $\lambda$.
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By the recurrence, $p_2 = \lambda p_1 -\ p_0 = \lambda^2 -\ p_0$.
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Equating terms with the characteristic polynomial of $P_2$, it is obvious that
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$$
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|\lambda I - P_2|
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= \begin{pmatrix}
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\lambda & -1 \\
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-1 & \lambda
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\end{pmatrix}
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= \lambda^2 - 1 = \lambda p_1 - p_0 \\
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\implies p_0 = 1
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$$
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which makes sense, since $p_0$ should have degree zero.
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Therefore, the sequence of polynomials $p_n(\lambda)$ is:
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$$
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\begin{gather*}
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p_0(\lambda) &= && 1
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\\
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p_1(\lambda) &= && \lambda
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\\
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p_2(\lambda) &= \lambda \lambda - 1
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&=& \lambda^2 - 1
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\\
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p_3(\lambda) &= \lambda (\lambda^2 - 1) - \lambda
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&=& \lambda(\lambda^2 - 2)
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\\
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p_4(\lambda) &= \lambda (\lambda(\lambda^2 - 2)) - (\lambda^2 - 1)
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&=& \lambda^4 - 3\lambda^2 + 1
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\\
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\vdots & \vdots && \vdots
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\end{gather*}
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$$
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But wait, we've seen these before (if you read the previous post, that is).
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These are just the Chebyshev polynomials of the second kind, evaluated at $\lambda / 2$.
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Indeed, their recurrence relations are identical, so the characteristic polynomial of $P_n$ is $U_n(\lambda / 2)$.
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Effectively, this connects an *n*-path to a regular *n+1*-gon.
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### Back to Cycles
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Since the generating function of $U_n$ is known, the generating function for the $c_n$
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(which prompted this) is also easily determined.
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For ease of use, let
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$$
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P(x; \lambda) = {B(x; \lambda / 2) \over x} = {1 \over 1 - \lambda x +\ x^2}
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$$
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Discarding the initial $c_0$ and $c_1$ by setting them to 0[^2], the generating function is
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[^2]: It's a good idea to ask why we can do this.
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Try examining $c_2$ and $c_3$.
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$$
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\begin{align*}
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c_{n+2}(\lambda) &= \lambda p_{n+1}(\lambda) - 2(p_n(\lambda) + 1)
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\\ \\
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{C(x; \lambda) - c_0(\lambda) - x c_1(\lambda) \over x^2}
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&= \lambda \left( {P(x; \lambda) - 1 \over x} \right)
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- 2\left( P(x; \lambda) + {1 \over 1 - x} \right)
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\\
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C &= x \lambda (P - 1) -\
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2x^2\left( P + {1 \over 1 - x} \right)
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\\
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C{(1 - x) \over P} &= x \lambda \left(1 - {1 \over P} \right)(1 - x) -\
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2x^2\left( (1 - x) + {1 \over P} \right)
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\\
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&= x^4 \lambda - 2 x^4 - x^3 \lambda^2 + x^3 \lambda
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+ 2 x^3 + x^2 \lambda^2 - 4 x^2
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\\
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&= x^2 (\lambda - 2) (x^2 - \lambda x - x + \lambda + 2)
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\\ \\
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C(x; \lambda) &= x^2 (\lambda - 2)
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{(x^2 - (\lambda + 1) x + \lambda + 2)
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\over (1 - x)(1 - \lambda x + x^2)}
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\end{align*}
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$$
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While the numerator is considerably more complicated than the one for P,
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the factor $\lambda - 2$ drops out of the entire series.
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This pleasantly informs that 2 is an eigenvalue of all $C_n$.
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Off the Beaten Path
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-------------------
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When we use Laplace expansion on the adjacency matrices, we were very fortunate that the minors
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*also* looked like adjacency matrices undergoing expansion.
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This let us terminate early and recurse.
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From the perspective of the graph, Laplace expansion almost looks like removing a node,
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but requires special treatment for the nodes connected to the one being removed.
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For example, in cycle graphs, the first stage of expansion had three minors:
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- The node itself, on the main diagonal
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- Being on the main diagonal, this immediately produced another adjacency matrix.
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- Either neighbor connected to it, which are on opposite sides of
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a path after the node is removed
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- Both of these nodes required second expansion to get the *λ*s back on the main diagonal.
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For "good enough" graphs that are nearly paths (including paths themselves),
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this gives a second-order recurrence relation.
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### Trees
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Another simple family of graphs are [*trees*](https://en.wikipedia.org/wiki/Tree_%28graph_theory%29).
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In some sense, they are the opposite of cycle graphs, since by definition they contain no cycles.
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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.
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|
|
|
|
|
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|
$$
|
|
\begin{align*}
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|
T_{1,1} := \begin{matrix}[
|
|
(0, 1), \\
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|
(1, 2), \\
|
|
(1, 3)
|
|
]\end{matrix} & \cong
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|
\begin{pmatrix}
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|
0 & 1 & 0 & 0 \\
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|
1 & 0 & 1 & 1 \\
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|
0 & 1 & 0 & 0 \\
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|
0 & 1 & 0 & 0
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|
\end{pmatrix}
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|
\\ \\
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|
T_{1,2} := \begin{matrix} [
|
|
(0, 1), \\
|
|
(1, 2), \\
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|
(2, 3), \\
|
|
(1, 4)
|
|
]\end{matrix} & \cong
|
|
\begin{pmatrix}
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|
0 & 1 & 0 & 0 & 0 \\
|
|
1 & 0 & 1 & 0 & 1 \\
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|
0 & 1 & 0 & 1 & 0 \\
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|
0 & 0 & 1 & 0 & 0 \\
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|
0 & 1 & 0 & 0 & 0
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|
\end{pmatrix}
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|
\\ \\
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|
T_{1,3} := \begin{matrix}[
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|
(0, 1), \\
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|
(1, 2), \\
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|
(2, 3), \\
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|
(3, 4), \\
|
|
(1, 5)
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|
]\end{matrix} &\cong
|
|
\begin{pmatrix}
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|
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}
|
|
\\ \\
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|
T_{2,2} := \begin{matrix}[
|
|
(0, 1), \\
|
|
(1, 2), \\
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|
(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 \\
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|
0 & 0 & 0 & 1 & 0 & 0 \\
|
|
0 & 0 & 1 & 0 & 0 & 0
|
|
\end{pmatrix}
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|
\end{align*}
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|
$$
|
|
|
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The subscripts denote the consituent paths if the "added" node and the
|
|
one it is connected to are both removed.
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It's easy to see that $T_{a,b} \cong T_{a,b}$, since this just swaps the arms.
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Also, $T_{a, 0} \cong T_{0, a} \cong P_{a + 2}$.
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|
|
|
Let's try dissecting one of the larger trees.
|
|
|
|
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|
$$
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|
\begin{align*}
|
|
|I \lambda - T_{1,3}|
|
|
&= \left |
|
|
\begin{matrix}
|
|
\lambda & -1 & 0 & 0 & 0 & 0 \\
|
|
-1 & \lambda & -1 & 0 & 0 & -1 \\
|
|
0 & -1 & \lambda & -1 & 0 & 0 \\
|
|
0 & 0 & -1 & \lambda & -1 & 0 \\
|
|
0 & 0 & 0 & -1 & \lambda & 0 \\
|
|
0 & -1 & 0 & 0 & 0 & \lambda
|
|
\end{matrix}
|
|
\right |
|
|
\\
|
|
&= (-1)^{6 + 6} \lambda m_{6,6} + (-1)^{2 + 6} (-1) m_{2,6}
|
|
\end{align*}
|
|
$$
|
|
|
|
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.
|
|
|
|
$$
|
|
\begin{align*}
|
|
m_{2,6}
|
|
&= \left |
|
|
\begin{matrix}
|
|
\lambda & -1 & 0 & 0 & 0 \\
|
|
0 & -1 & \lambda & -1 & 0 \\
|
|
0 & 0 & -1 & \lambda & -1 \\
|
|
0 & 0 & 0 & -1 & \lambda \\
|
|
0 & -1 & 0 & 0 & 0
|
|
\end{matrix}
|
|
\right |
|
|
\\
|
|
&= (-1)^{5 + 2} (-1)
|
|
\left |
|
|
\begin{matrix}
|
|
\lambda & 0 & 0 & 0 \\
|
|
0 & \lambda & -1 & 0 \\
|
|
0 & -1 & \lambda & -1 \\
|
|
0 & 0 & -1 & \lambda \\
|
|
\end{matrix}
|
|
\right |
|
|
\end{align*}
|
|
$$
|
|
|
|
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 $T_{2,2}$.
|
|
|
|
$$
|
|
\begin{align*}
|
|
|I \lambda - T_{2,2}|
|
|
&= \left |
|
|
\begin{matrix}
|
|
\lambda & -1 & 0 & 0 & 0 & 0 \\
|
|
-1 & \lambda & -1 & 0 & 0 & 0 \\
|
|
0 & -1 & \lambda & -1 & 0 & -1 \\
|
|
0 & 0 & -1 & \lambda & -1 & 0 \\
|
|
0 & 0 & 0 & -1 & \lambda & 0 \\
|
|
0 & 0 & -1 & 0 & 0 & \lambda
|
|
\end{matrix}
|
|
\right |
|
|
\\
|
|
&= (-1)^{6 + 6} \lambda m_{6,6} + (-1)^{3 + 6} (-1) m_{3,6}
|
|
\\ \\
|
|
m_{3,6}
|
|
&= \left |
|
|
\begin{matrix}
|
|
\lambda & -1 & 0 & 0 & 0 \\
|
|
-1 & \lambda & -1 & 0 & 0 \\
|
|
0 & 0 & -1 & \lambda & -1 \\
|
|
0 & 0 & 0 & -1 & \lambda \\
|
|
0 & 0 & -1 & 0 & 0
|
|
\end{matrix}
|
|
\right |
|
|
\\
|
|
&= (-1)^{5 + 3} (-1)
|
|
\left |
|
|
\begin{matrix}
|
|
\lambda & -1 & 0 & 0 \\
|
|
-1 & \lambda & 0 & 0 \\
|
|
0 & 0 & \lambda & -1 \\
|
|
0 & 0 & -1 & \lambda \\
|
|
\end{matrix}
|
|
\right |
|
|
\end{align*}
|
|
$$
|
|
|
|
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 $P_b$ ($p_1(\lambda) = \lambda$, 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:
|
|
|
|
$$
|
|
t_{a,b}(z) = z \cdot p_{a + b + 1}(z) - p_a(z) p_b(z)
|
|
$$
|
|
|
|
Note that if *b* is 0, this coincides with the recurrence for $p_n(z)$.
|
|
|
|
|
|
### 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.
|
|
|
|
The first tree of note is $T_{1,1}$.
|
|
This has characteristic polynomial
|
|
|
|
$$
|
|
\begin{align*}
|
|
t_{1,1}(z) &= z \cdot p_{3} - p_1(z) p_1(z)
|
|
\\
|
|
&= z (z^3 - z^2) - z^2
|
|
\\
|
|
&= z^2 (z^2 - 3)
|
|
\end{align*}
|
|
$$
|
|
|
|
Next, we have both $T_{2,1}$ and $T_{1,2}$.
|
|
By symmetry, these are the same graph, so we have characteristic polynomial
|
|
|
|
$$
|
|
\begin{align*}
|
|
t_{1,2}(z) &= z \cdot p_{4} - p_1(z) p_2(z)
|
|
\\
|
|
&= z (z^4 - 3z^2 + 2) - z (z^2 - 1)
|
|
\\
|
|
&= z (z^4 - 4z^2 + 2)
|
|
\end{align*}
|
|
$$
|
|
|
|
Finally, let's look at $T_{1,3}$ and $T_{2,2}$, the trees we used to derive the rule.
|
|
|
|
$$
|
|
\begin{align*}
|
|
t_{1,3}(z) &= z \cdot p_{5} - p_1(z) p_3(z)
|
|
\\
|
|
&= z (z^5 - 4z^3 + 3z) - z \cdot (z^3 - 2z)
|
|
\\
|
|
&= z^2 (z^4 - 5z^2 + 5)
|
|
\\[10pt]
|
|
t_{2,2}(z) &= z \cdot p_{5} - p_2(z) p_2(z)
|
|
\\
|
|
&= z (z^5 - 4z^3 + 3z) - ( z^2 - 1 )^2
|
|
\\
|
|
&= (z^2 - 1)(z^4 - 4z^2 + 1)
|
|
\end{align*}
|
|
$$
|
|
|
|
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 $t_{1,2}$,
|
|
- $z^4 - 4z^2 + 2$, $p_7(z)$, and $2 \cos(\pi / 8)$, respectively
|
|
- In $t_{2,2}$,
|
|
- $z^4 - 4z^2 + 1$, $p_11(z)$, 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.
|