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12 lines
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"markdown": "---\ntitle: \"Generating Polynomials, Part 1: Regular Constructibility\"\ndescription: |\n What kinds of regular polygons are constructible with compass and straightedge?\nformat:\n html:\n html-math-method: katex\ndate: \"2021-08-18\"\ndate-modified: \"2025-06-17\"\ncategories:\n - geometry\n - generating functions\n - algebra\n - python\n---\n\n<style>\n.figure-img {\n max-width: 512px;\n object-fit: contain;\n height: 100%;\n}\n\n.figure-img.wide {\n max-width: 768px;\n}\n</style>\n\n\n\n[Recently](/posts/misc/platonic-volume), I used coordinate-free geometry to derive\n the volumes of the Platonic solids, a problem which was very accessible to the ancient Greeks.\nOn the other hand, they found certain problems regarding which figures can be constructed via\n compass and straightedge to be very difficult. For example, they struggled with problems\n like [doubling the cube](https://en.wikipedia.org/wiki/Doubling_the_cube)\n or [squaring the circle](https://en.wikipedia.org/wiki/Squaring_the_circle),\n which are known (through circa 19th century mathematics) to be impossible.\nHowever, before even extending planar geometry by a third dimension or\n calculating the areas of circles, a simpler problem becomes apparent.\nNamely, what kinds of regular polygons are constructible?\n\n\nRegular Geometry and a Complex Series\n-------------------------------------\n\nWhen constructing a regular polygon, one wants a ratio between the length of a edge\n and the distance from a vertex to the center of the figure.\n\n{.wide}\n\nIn a convex polygon, the total central angle is always one full turn, or 2π radians.\nThe central angle of a regular *n*-gon is ${2\\pi \\over n}$ radians,\n and the green angle above (which we'll call *θ*) is half of that.\nThis means that the ratio we're looking for is $\\sin(\\theta) = \\sin(\\pi / n)$.\nWe can multiply by *n* inside the function on both sides to give\n $\\sin(n\\theta) = \\sin(\\pi) = 0$.\nTherefore, constructing a polygon is actually equivalent to solving this equation,\n and we can rephrase the question as how to express $\\sin(n\\theta)$ (and $\\cos(n\\theta)$).\n\n\n### Complex Recursion\n\nThanks to [Euler's formula](https://en.wikipedia.org/wiki/Euler%27s_formula)\n and [de Moivre's formula](https://en.wikipedia.org/wiki/De_Moivre%27s_formula),\n the expressions we're looking for can be phrased in terms of the complex exponential.\n\n$$\n\\begin{align*}\n e^{i\\theta}\n &= \\text{cis}(\\theta) = \\cos(\\theta) + i\\sin(\\theta)\n & \\text{ Euler's formula}\n \\\\\n \\text{cis}(n \\theta) = e^{i(n\\theta)}\n &= e^{(i\\theta)n} = {(e^{i\\theta})}^n = \\text{cis}(\\theta)^n\n \\\\\n \\cos(n \\theta) + i\\sin(n \\theta)\n &= (\\cos(\\theta) + i\\sin(\\theta))^n\n & \\text{ de Moivre's formula}\n\\end{align*}\n$$\n\nDe Moivre's formula for $n = 2$ gives\n\n$$\n\\begin{align*}\n \\text{cis}(\\theta)^2\n &= (\\text{c} + i\\text{s})^2\n \\\\\n &= \\text{c}^2 + 2i\\text{cs} - \\text{s}^2 + (0 = \\text{c}^2 + \\text{s}^2 - 1)\n \\\\\n &= 2\\text{c}^2 + 2i\\text{cs} - 1\n \\\\\n &= 2\\text{c}(\\text{c} + i\\text{s}) - 1\n \\\\\n &= 2\\cos(\\theta)\\text{cis}(\\theta) - 1\n\\end{align*}\n$$\n\nThis can easily be massaged into a recurrence relation.\n\n$$\n\\begin{align*}\n \\text{cis}(\\theta)^2\n &= 2\\cos(\\theta)\\text{cis}(\\theta) - 1\n \\\\\n \\text{cis}(\\theta)^{n+2}\n &= 2\\cos(\\theta)\\text{cis}(\\theta)^{n+1} - \\text{cis}(\\theta)^n\n \\\\\n \\text{cis}((n+2)\\theta)\n &= 2\\cos(\\theta)\\text{cis}((n+1)\\theta) - \\text{cis}(n\\theta)\n\\end{align*}\n$$\n\nRecurrence relations like this one are powerful.\nThrough some fairly straightforward summatory manipulations,\n the sequence can be interpreted as the coefficients in a Taylor series,\n giving a [generating function](https://en.wikipedia.org/wiki/Generating_function).\nCall this function *F*. Then,\n\n$$\n\\begin{align*}\n \\sum_{n=0}^\\infty \\text{cis}((n+2)\\theta)x^n\n &= 2\\cos(\\theta) \\sum_{n=0}^\\infty \\text{cis}((n+1)\\theta) x^n\n - \\sum_{n=0}^\\infty \\text{cis}(n\\theta) x^n\n \\\\\n {F(x; \\text{cis}(\\theta)) - 1 - x\\text{cis}(\\theta) \\over x^2}\n &= 2\\cos(\\theta) {F(x; \\text{cis}(\\theta)) - 1 \\over x}\n - F(x; \\text{cis}(\\theta))\n \\\\[10pt]\n F - 1 - x\\text{cis}(\\theta)\n &= 2\\cos(\\theta) x (F - 1)\n - x^2 F\n \\\\\n F - 2\\cos(\\theta) x F + x^2 F\n &= 1 + x(\\text{cis}(\\theta) - 2\\cos(\\theta))\n \\\\[10pt]\n F(x; \\text{cis}(\\theta))\n &= {1 + x(\\text{cis}(\\theta) - 2\\cos(\\theta)) \\over\n 1 - 2\\cos(\\theta)x + x^2}\n\\end{align*}\n$$\n\nSince $\\text{cis}$ is a complex function, we can separate *F* into real and imaginary parts.\nConveniently, these correspond to $\\cos(n\\theta)$ and $\\sin(n\\theta)$, respectively.\n\n$$\n\\begin{align*}\n \\Re[ F(x; \\text{cis}(\\theta)) ]\n &= {1 + x(\\cos(\\theta) - 2\\cos(\\theta)) \\over 1 - 2\\cos(\\theta)x + x^2}\n \\\\\n &= {1 - x\\cos(\\theta) \\over 1 - 2\\cos(\\theta)x + x^2} = A(x; \\cos(\\theta))\n \\\\\n \\Im[ F(x; \\text{cis}(\\theta)) ]\n &= {x \\sin(\\theta) \\over 1 - 2\\cos(\\theta)x + x^2} = B(x; \\cos(\\theta))\\sin(\\theta)\n\\end{align*}\n$$\n\nIn this form, it becomes obvious that the even though the generating function *F* was originally\n parametrized by $\\text{cis}(\\theta)$, *A* and *B* are parametrized only by $\\cos(\\theta)$.\nExtracting the coefficients of *x* yields an expression for $\\cos(n\\theta)$ and $\\sin(n\\theta)$\n in terms of $\\cos(\\theta)$ (and in the latter case, a common factor of $\\sin(\\theta)$).\n\nIf $\\cos(\\theta)$ in *A* and *B* is replaced with the parameter *z*, then all trigonometric functions\n are removed from the equation, and we are left with only polynomials[^1].\nThese polynomials are [*Chebyshev polynomials*](https://en.wikipedia.org/wiki/Chebyshev_polynomial)\n *of the first (A) and second (B) kind*.\nIn actuality, the polynomials of the second kind are typically offset by 1\n (the x in the numerator of *B* is omitted).\nHowever, retaining this term makes indexing consistent between *A* and *B*\n (and will make things clearer later).\n\n[^1]:\n This can actually be observed as early as the recurrence relation.\n\n $$\n \\begin{align*}\n \\text{cis}(\\theta)^{n+2}\n &= 2\\cos(\\theta)\\text{cis}(\\theta)^{n+1} - \\text{cis}(\\theta)^n\n \\\\\n a_{n+2}\n &= 2 z a_{n+1} - a_n\n \\\\\n \\Re[ a_0 ]\n &= 1,~~ \\Im[ a_0 ] = 0\n \\\\\n \\Re[ a_1 ]\n &= z,~~ \\Im[ a_1 ] = 1 \\cdot \\sin(\\theta)\n \\end{align*}\n $$\n\n\nWe were primarily interested in $\\sin(n\\theta)$, so let's tabulate\n the first few polynomials of the second kind (at $z / 2$).\n\n::: {#tbl-chebyshevu .cell .plain tbl-cap='[OEIS A049310](http://oeis.org/A049310)' execution_count=3}\n\n::: {.cell-output .cell-output-display .cell-output-markdown execution_count=2}\n*n* $[x^n]B(x; z / 2) = U_{n - 1}(z / 2)$ Factored\n----- --------------------------------------------- -------------------------------------------------------------------------------------------------\n0 $0$ $0$\n1 $1$ $1$\n2 $z$ $z$\n3 $z^{2} - 1$ $\\left(z - 1\\right) \\left(z + 1\\right)$\n4 $z^{3} - 2 z$ $z \\left(z^{2} - 2\\right)$\n5 $z^{4} - 3 z^{2} + 1$ $\\left(z^{2} - z - 1\\right) \\left(z^{2} + z - 1\\right)$\n6 $z^{5} - 4 z^{3} + 3 z$ $z \\left(z - 1\\right) \\left(z + 1\\right) \\left(z^{2} - 3\\right)$\n7 $z^{6} - 5 z^{4} + 6 z^{2} - 1$ $\\left(z^{3} - z^{2} - 2 z + 1\\right) \\left(z^{3} + z^{2} - 2 z - 1\\right)$\n8 $z^{7} - 6 z^{5} + 10 z^{3} - 4 z$ $z \\left(z^{2} - 2\\right) \\left(z^{4} - 4 z^{2} + 2\\right)$\n9 $z^{8} - 7 z^{6} + 15 z^{4} - 10 z^{2} + 1$ $\\left(z - 1\\right) \\left(z + 1\\right) \\left(z^{3} - 3 z - 1\\right) \\left(z^{3} - 3 z + 1\\right)$\n10 $z^{9} - 8 z^{7} + 21 z^{5} - 20 z^{3} + 5 z$ $z \\left(z^{2} - z - 1\\right) \\left(z^{2} + z - 1\\right) \\left(z^{4} - 5 z^{2} + 5\\right)$\n:::\n:::\n\n\nEvaluating the polynomials at $z / 2$ cancels the 2 in the denominator (and recurrence),\n making these expressions much simpler.\nThis evaluation has an interpretation in terms of the previous diagram --\n recall we used *half* the length of a side as a leg of the right triangle.\nFor a unit circumradius, the side length itself is then $2\\sin( {\\pi / n} )$.\nTo compensate for this doubling, the Chebyshev polynomial must be evaluated at half its normal argument.\n\n\n### Back on the Plane\n\nThe constructibility criterion is deeply connected to the Chebyshev polynomials.\nIn compass and straightedge constructions, one only has access to linear forms (lines)\n and quadratic forms (circles).\nThis means that a figure is constructible if and only if the root can be expressed using\n normal arithmetic (which is linear) and square roots (which are quadratic).\n\n\n#### Pentagons\n\nLet's look at a regular pentagon.\nThe relevant polynomial is\n\n$$\n[x^5]B ( x; z / 2 )\n = z^4 - 3z^2 + 1\n = (z^2 - z - 1) (z^2 + z - 1)\n$$\n\nAccording to how we derived this series, when $z = 2\\cos(\\theta)$, the roots of this polynomial\n correspond to when $\\sin(5\\theta) / \\sin(\\theta) = 0$.\nThis relation itself is true when $\\theta = \\pi / 5$, since $\\sin(5 \\pi / 5) = 0$.\n\nOne of the factors must therefore be the minimal polynomial of $2\\cos(\\pi / 5 )$.\nThe former happens to be correct correct, since $2\\cos( \\pi / 5 ) = \\varphi$, the golden ratio.\nNote that the second factor is the first evaluated at -*z*.\n\n\n#### Heptagons\n\nAn example of where constructability fails is for $2\\cos( \\pi / 7 )$.\n\n$$\n\\begin{align*}\n [x^7]B ( x; z / 2 )\n &= z^6 - 5 z^4 + 6 z^2 - 1\n \\\\\n &= ( z^3 - z^2 - 2 z + 1 ) ( z^3 + z^2 - 2 z - 1 )\n\\end{align*}\n$$\n\nWhichever is the minimal polynomial (the former), it is a cubic, and constructing\n a regular heptagon is equivalent to solving it for *z*.\nBut there are no (nondegenerate) cubics that one can produce via compass and straightedge,\n and all constructions necessarily fail.\n\n\n#### Decagons\n\nOne might think the same of $2\\cos(\\pi /10 )$\n\n$$\n\\begin{align*}\n [x^{10}]B ( x; z / 2 )\n &= z^9 - 8 z^7 + 21 z^5 - 20 z^3 + 5 z\n \\\\\n &= z ( z^2 - z - 1 )( z^2 + z - 1 )( z^4 - 5 z^2 + 5 )\n\\end{align*}\n$$\n\nThis expression also contains the polynomials for $2\\cos( \\pi / 5 )$.\nThis is because a regular decagon would contain two disjoint regular pentagons,\n produced by connecting every other vertex.\n\n\n\nThe polynomial which actually corresponds to $2\\cos( \\pi / 10 )$ is the quartic,\n which seems to suggest that it will require a fourth root and somehow decagons are not constructible.\nHowever, it can be solved by completing the square...\n\n$$\n\\begin{align*}\n z^4 - 5z^2 &= -5\n \\\\\n z^4 - 5z^2 + (5/2)^2 &= -5 + (5/2)^2\n \\\\\n ( z^2 - 5/2)^2 &= {25 - 20 \\over 4}\n \\\\\n ( z^2 - 5/2) &= {\\sqrt 5 \\over 2}\n \\\\\n z^2 &= {5 \\over 2} + {\\sqrt 5 \\over 2}\n \\\\\n z &= \\sqrt{ {5 + \\sqrt 5 \\over 2} }\n\\end{align*}\n$$\n\n...and we can breathe a sigh of relief.\n\n\nThe Triangle behind Regular Polygons\n------------------------------------\n\nPreferring *z* to be halved in $B(x; z/2)$ makes something else more evident.\nObserve these four rows of the Chebyshev polynomials\n\n::: {#273cfec0 .cell .plain execution_count=4}\n\n::: {.cell-output .cell-output-display .cell-output-markdown execution_count=3}\n*n* $[x^n]B(x; z / 2)$ *k* $[z^{k}][x^n]B(x; z / 2)$\n----- ------------------------------- ----- ---------------------------\n4 $z^{3} - 2 z$ 3 1\n5 $z^{4} - 3 z^{2} + 1$ 2 -3\n6 $z^{5} - 4 z^{3} + 3 z$ 1 3\n7 $z^{6} - 5 z^{4} + 6 z^{2} - 1$ 0 -1\n:::\n:::\n\n\nThe last column looks like an alternating row of Pascal's triangle\n (namely, ${n - \\lfloor {k / 2} \\rfloor - 1 \\choose k}(-1)^k$).\nThis resemblance can be made more apparent by listing the coefficients of the polynomials in a table.\n\n::: {#92d420a6 .cell .plain execution_count=5}\n\n::: {.cell-output .cell-output-display .cell-output-markdown execution_count=4}\n n $z^9$ $z^8$ $z^7$ $z^6$ $z^5$ $z^4$ $z^3$ $z^2$ $z$ $1$\n--- ------------------------------ ------------------------------------ ------------------------------------- ----------------------------------- ----------------------------------- ----------------------------------- ------------------------------------ ------------------------------------- ------------------------------------- -------------------------------------\n 1 <span style=\"color: \">1</span>\n 2 <span style=\"color: red\">1</span> 0\n 3 <span style=\"color: orange\">1</span> 0 <span style=\"color: red\">-1</span>\n 4 <span style=\"color: yellow\">1</span> 0 <span style=\"color: orange\">-2</span> 0\n 5 <span style=\"color: green\">1</span> 0 <span style=\"color: yellow\">-3</span> 0 <span style=\"color: orange\">1</span>\n 6 <span style=\"color: cyan\">1</span> 0 <span style=\"color: green\">-4</span> 0 <span style=\"color: yellow\">3</span> 0\n 7 <span style=\"color: aqua\">1</span> 0 <span style=\"color: cyan\">-5</span> 0 <span style=\"color: green\">6</span> 0 <span style=\"color: yellow\">-1</span>\n 8 <span style=\"color: blue\">1</span> 0 <span style=\"color: aqua\">-6</span> 0 <span style=\"color: cyan\">10</span> 0 <span style=\"color: green\">-4</span> 0\n 9 <span style=\"color: purple\">1</span> 0 <span style=\"color: blue\">-7</span> 0 <span style=\"color: aqua\">15</span> 0 <span style=\"color: cyan\">-10</span> 0 <span style=\"color: green\">1</span>\n 10 <span style=\"color: \">1</span> 0 <span style=\"color: purple\">-8</span> 0 <span style=\"color: blue\">21</span> 0 <span style=\"color: aqua\">-20</span> 0 <span style=\"color: cyan\">5</span> 0\n:::\n:::\n\n\nThough they alternate in sign, the rows of Pascal's triangle appear along diagonals,\n which I have marked in rainbow.\nMeanwhile, alternating versions of the naturals (1, 2, 3, 4...),\n the triangular numbers (1, 3, 6, 10...),\n the tetrahedral numbers (1, 4, 10, 20...), etc.\n are present along the columns, albeit spaced out by 0's.\n\nThe relationship of the Chebyshev polynomials to the triangle is easier to see if\n the coefficient extraction of $B(x; z / 2)$ is reversed.\nIn other words, we extract *z* before extracting *x*.\n\n$$\n\\begin{align*}\n B(x; z / 2) &= {x \\over 1 - zx + x^2}\n = {x \\over 1 + x^2 - zx}\n = {x \\over 1 + x^2}\n \\cdot {1 \\over {1 + x^2 \\over 1 + x^2} - z{x \\over 1 + x^2}}\n \\\\[10pt]\n [z^n]B(x; z / 2) &= {x \\over 1 + x^2} [z^n] {1 \\over 1 - z{x \\over 1 + x^2}}\n = {x \\over 1 + x^2} \\left( {x \\over 1 + x^2} \\right)^n\n \\\\\n &= \\left( {x \\over 1 + x^2} \\right)^{n+1}\n = x^{n+1} (1 + x^2)^{-n - 1}\n \\\\\n &= x^{n+1} \\sum_{k=0}^\\infty {-n - 1 \\choose k}(x^2)^k\n \\quad \\text{Binomial theorem}\n\\end{align*}\n$$\n\nWhile the use of the binomial theorem is more than enough to justify\n the appearance of Pascal's triangle (along with explaining the 0's),\n I'll simplify further to explicitly show the alternating signs.\n\n$$\n\\begin{align*}\n {(-n - 1)_k} &= (-n - 1)(-n - 2) \\cdots (-n - k)\n \\\\\n &= (-1)^k (n + k)(n + k - 1) \\cdots (n + 1)\n \\\\\n &= (-1)^k (n + k)_k\n \\\\\n \\implies {-n - 1 \\choose k}\n &= {n + k \\choose k}(-1)^k\n \\\\[10pt]\n [z^n]B(x; z / 2)\n &= x^{n+1} \\sum_{k=0}^\\infty {n + k \\choose k} (-1)^k x^{2k}\n\\end{align*}\n$$\n\nSquinting hard enough, the binomial coefficient is similar to the earlier\n which gave the third row of Pascal's triangle.\nIf k is fixed, then this expression actually generates the antidiagonal entries\n of the coefficient table, which are the columns with uniform sign.\nThe alternation instead occurs between antidiagonals (one is all positive,\n the next is 0's, the next is all negative, etc.).\nThe initial $x^{n+1}$ lags these sequences so that they reproduce the triangle.\n\n\n### Imagined Transmutation\n\nThe generating function of the Chebyshev polynomials resembles other two term recurrences.\nFor example, the Fibonacci numbers have generating function\n\n$$\n\\sum_{n = 0}^\\infty \\text{Fib}_n x^n = {1 \\over 1 - x - x^2}\n$$\n\nThis resemblance can be made explicit with a simple algebraic manipulation.\n\n$$\n\\begin{align*}\n B(ix; -iz / 2)\n &= {1 \\over 1 -\\ (-i z)(ix) + (ix)^2}\n = {1 \\over 1 -\\ (-i^2) z x + (i^2)(x^2)}\n \\\\\n &= {1 \\over 1 -\\ z x -\\ x^2}\n\\end{align*}\n$$\n\nIf $z = 1$, these two generating functions are equal.\nThe same can be said for $z = 2$ with the generating function of the Pell numbers,\n and so on for higher recurrences (corresponding to metallic means) for higher integral *z*.\n\nIn terms of the Chebyshev polynomials, this series manipulation removes the alternation in\n the coefficients of $U_n$, restoring Pascal's triangle to its nonalternating form.\nRelated to the previous point, it is possible to find the Fibonacci numbers (Pell numbers, etc.)\n in Pascal's triangle, which you can read more about\n [here](http://users.dimi.uniud.it/~giacomo.dellariccia/Glossary/Pascal/Koshy2011.pdf).\n\n\nManipulating the Series\n-----------------------\n\nLook back to the table of $U_{n - 1}(z / 2)$ (@tbl-chebyshevu).\nWhen I brought up $U_{10 - 1}(z / 2)$ and decagons, I pointed out their relationship to pentagons\n as an explanation for why $U_{5 -\\ 1}(z / 2)$ appears as a factor.\nConveniently, $U_{2 -\\ 1}(z / 2) = z$ is also a factor, and 2 is likewise a factor of 10.\n\nThis pattern is present throughout the table; $n = 6$ contains factors for\n $n = 2 \\text{ and } 3$ and the prime numbers have no smaller factors.\nIf this observation is legitimate, call the newest term $f_n(z)$\n and denote $p_n(z) = U_{n -\\ 1}( z / 2 )$.\n\n\n### Factorization Attempts\n\nThe relationship between $p_n$ and the intermediate $f_d$, where *d* is a divisor of *n*,\n can be made explicit by a [Möbius inversion](https://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula).\n\n$$\n\\begin{align*}\n p_n(z) &= \\prod_{d|n} f_n(z)\n \\\\\n \\log( p_n(z) )\n &= \\log \\left( \\prod_{d|n} f_d(z) \\right)\n = \\sum_{d|n} \\log( f_d(z) )\n \\\\\n \\log( f_n(z) ) &= \\sum_{d|n} { \\mu \\left({n \\over d} \\right)}\n \\log( p_d(z) )\n \\\\\n f_n(z) &= \\prod_{d|n} p_d(z)^{ \\mu (n / d) }\n \\\\[10pt]\n f_6(z) = g_6(z)\n &= p_6(z)^{\\mu(1)}\n p_3(z)^{\\mu(2)}\n p_2(z)^{\\mu(3)}\n \\\\\n &= {p_6(z) \\over p_3(z) p_2(z)}\n\\end{align*}\n$$\n\nUnfortunately, it's difficult to apply this technique across our whole series.\nMöbius inversion over series typically uses more advanced generating functions such as\n [Dirichlet series](https://en.wikipedia.org/wiki/Dirichlet_series#Formal_Dirichlet_series)\n or [Lambert series](https://en.wikipedia.org/wiki/Lambert_series).\nHowever, naively reaching for these fails for two reasons:\n\n- We built our series of polynomials on a recurrence relation, and these series\n are opaque to such manipulations.\n- To do a proper Möbius inversion, we need these kinds of series over the *logarithm*\n of each polynomial (*B* is a series over the polynomials themselves).\n\nIgnoring these (and if you're in the mood for awful-looking math) you may note\n the Lambert equivalence[^2]:\n\n[^2]:\n This equivalence applies to other polynomial series obeying the same factorization rule\n such as the [cyclotomic polynomials](https://en.wikipedia.org/wiki/Cyclotomic_polynomial).\n\n$$\n\\begin{align*}\n \\log( p_n(z) )\n &= \\sum_{d|n} \\log( f_d(z) )\n \\\\\n \\sum_{n = 1}^\\infty \\log( p_n ) x^n\n &= \\sum_{n = 1}^\\infty \\sum_{d|n} \\log( f_d ) x^n\n \\\\\n &= \\sum_{k = 1}^\\infty \\sum_{m = 1}^\\infty \\log( f_m ) x^{m k}\n \\\\\n &= \\sum_{m = 1}^\\infty \\log( f_m ) \\sum_{k = 1}^\\infty (x^m)^k\n \\\\\n &= \\sum_{m = 1}^\\infty \\log( f_m ) {x^m \\over 1 - x^m}\n\\end{align*}\n$$\n\nEither way, the number-theoretic properties of this sequence are difficult to ascertain\n without advanced techniques.\nIf research has been done, it is not easily available in the OEIS.\n\n\n### Total Degrees\n\nIt can be also be observed that the new term is symmetric ($f(z) = f(-z)$), and is therefore\n either irreducible or the product of polynomial and its reflection (potentially negated).\nFor example,\n\n$$\np_9(z) = \\left\\{\n\\begin{matrix}\n (z - 1)(z + 1)\n & \\cdot\n & (z^3 - 3z - 1)(z^3 - 3z + 1)\n \\\\\n \\shortparallel && \\shortparallel\n \\\\\n f_3(z)\n & \\cdot\n & f_9(z)\n \\\\\n \\shortparallel && \\shortparallel\n \\\\\n g_3(z) \\cdot g_3(-z)\n & \\cdot\n & g_9(z) \\cdot -g_9(-z)\n\\end{matrix}\n\\right.\n$$\n\nThese factor polynomials $g_n$ are the minimal polynomials of $2\\cos( \\pi / n )$.\n\nMultiplying these minimal polynomials by their reflection can be observed in the Chebyshev polynomials\n for $n = 3, 5, 7, 9$, strongly implying that it occurs on the odd terms.\nAssuming this is true, we have\n\n$$\nf_n(z) = \\begin{cases}\n g_n(z) & \\text{$n$ is even}\n \\\\\n g_n(z)g_n(-z)\n & \\text{$n$ is odd and ${\\deg(f_n) \\over 2}$ is even}\n \\\\\n -g_n(z)g_n(-z)\n & \\text{$n$ is odd and ${\\deg(f_n) \\over 2}$ is odd}\n\\end{cases}\n$$\n\nWithout resorting to any advanced techniques, the degrees of $f_n$ are\n not too difficult to work out.\nThe degree of $p_n(z)$ is $n -\\ 1$, which is also the degree of $f_n(z)$ if *n* is prime.\nIf *n* is composite, then the degree of $f_n(z)$ is $n -\\ 1$ minus the degrees\n of the divisors of $n -\\ 1$.\nThis leaves behind how many numbers less than *n* are coprime to *n*.\nTherefore $\\deg(f_n) = \\phi(n)$, the\n [Euler totient function](https://en.wikipedia.org/wiki/Euler_totient_function) of the index.\n\nThe totient function can be used to examine the parity of *n*.\nIf *n* is odd, it is coprime to 2 and all even numbers.\nThe introduced factor of 2 to 2*n* removes the evens from the totient, but this is compensated by\n the addition of the odd multiples of old numbers coprime to *n* and new primes.\nThis means that $\\phi(2n) = \\phi(n)$ for odd *n* (other than 1).\n\nThe same argument can be used for even *n*: there are as many odd numbers from 0 to *n* as there are\n from *n* to 2*n*, and there are an equal number of numbers coprime to 2*n* in either interval.\nTherefore, $\\phi(2n) = 2\\phi(n)$ for even *n*.\n\nThis collapses all cases of the conditional factorization of $f_n$ into one,\n and the degrees of $g_n$ are\n\n$$\n\\begin{align*}\n \\deg( g_n(z) )\n &= \\begin{cases}\n \\deg( f_n(z) )\n = \\phi(n)\n & n \\text{ is even} & \\implies \\phi(n) = \\phi(2n) / 2\n \\\\\n \\deg( f_n(z) ) / 2\n = \\phi(n) / 2\n & n \\text{ is odd} & \\implies \\phi(n) / 2 = \\phi(2n) / 2\n \\end{cases}\n \\\\\n &= \\varphi(2n) / 2\n\\end{align*}\n$$\n\nThough they were present in the earlier Chebyshev table,\n the $g_n$ themselves are presented again, along with the expression for their degree\n\n::: {#835a5369 .cell .plain execution_count=6}\n\n::: {.cell-output .cell-output-display .cell-output-markdown execution_count=5}\nn $\\varphi(2n)/2$ $g_n(z)$ Coefficient list, rising powers\n--- --------------------------------------- ------------------------- ---------------------------------------\n2 1 $z$ [0, 1]\n3 1 $z - 1$ [-1, 1]\n4 2 $z^{2} - 2$ [-2, 0, 1]\n5 2 $z^{2} - z - 1$ [-1, -1, 1]\n6 2 $z^{2} - 3$ [-3, 0, 1]\n7 3 $z^{3} - z^{2} - 2 z + 1$ [1, -2, -1, 1]\n8 4 $z^{4} - 4 z^{2} + 2$ [2, 0, -4, 0, 1]\n9 3 $z^{3} - 3 z - 1$ [-1, -3, 0, 1]\n9 3 $z^{3} - 3 z + 1$ [1, -3, 0, 1]\n10 4 $z^{4} - 5 z^{2} + 5$ [5, 0, -5, 0, 1]\n- [OEIS A055034](http://oeis.org/A055034) - [OEIS A187360](http://oeis.org/A187360)\n:::\n:::\n\n\nClosing\n-------\n\nMy initial jumping off point for writing this article was completely different.\nHowever, in the process of writing, its share of the article shrank and shrank until its\n introduction was only vaguely related to what preceded it.\nBut alas, the introduction via geometric constructions flows better coming off my\n [post about the Platonic solids](/posts/misc/platonic-volume).\nAlso, it reads better if I rely less on \"if you search for this sequence of numbers\"\n and more on how to interpret the definition.\n\nConsider reading [the follow-up](../2) to this post if you're interested in another way\n one can obtain the Chebyshev polynomials.\n\nDiagrams created with GeoGebra.\n\n<!--\nUpdate: I have since rederived the Chebyshev polynomials without the complex exponential,\n which you can read about in [this post]().\n-->\n\n",
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