The Aesthetics of Polynomials, by Vinyasi In loving thanks to Helen and Charlie Lutes Table of Contents Introduction Partial Solution Methods For: Polynomials In One Unknown Riddle One Continued Fractions Special Case Music Theory Geometry Riddle Two Algebraic Expressions Riddle Three: e Polynomials In Two Unknowns (Multiple?) Appendix: General Theory and Tables For: Riddle One Special Case Music Theory Geometry Riddle Two Riddle Three Speculation Glossary Bibliography Index Introduction Polynomials are pretty neat. If we wanted to factor a composite number as integer factors, we would simply have a * b = c. But if we set a and b equal to zero as, x = a and y = b >>>> x * y = c x - a = 0 and y - b = 0 >>>> (x - a) * (y - b) = 0 This last polynomial is nice, but it >>>> xy - bx - ay + ab = 0 doesn't give us a form that is easy to solve. Solutions are much easier if the polynomial is in one unknown. So let's change the setup. The above example takes two different polynomial systems, x and y, and combines them. X and y are dif- ferent variable names from the start because they were not intended to be com- bined into one system. So let's start them off on a better footing. Let's assume that they are one system making two alternate attempts at solving for either a or b seperately, x = a or x = b either, x * b = c or a * x = c x - a = 0 or x - b = 0 >>>> (x - a) * (x - b) = 0 x^2 - ax - bx + ab = 0 This last polynomial is very easy to >>>> x^2 + (-a - b)x + ab = 0 solve for either a or b seperately, but not both at the same time. It may be more ploddingly methodical, but makes solving it more assured. The first section of this text deals with polynomials of this type. ? The second section deals with polynomials in multiple unknowns. Riddle One Given the following criteria: A) A set of integers >>>> (a, b, c, ...) B) Can a polynomial be constructed in one unknown of integer exponents and coefficients? integer exponentÄÄÄ¿ ÚÄÄÄÄrational coefficient Set: (a, b, c) ³ ³ Polynomial: 0 = (x - a) * (x - b) * (x - c) = ³ ³ x^3 + (-a - b - c) * x^2 + ((-a * -b) + (-a * -c) + (-b * -c)) * x^1 + (-a * -b * -c) * x^0 ³ ÄÄÄ ÄÄÄÄÄ ÀÄÄ 1.1 factorization for polynomials in one or simply, or omit unknown x For a greater challenge, add one more criteria: C) The factors of the polynomial are not the elements of the set unless the set has only one member, but are derived from it using: a) the Euclidean algorithm b) infinite convergent series D) Derive the continued fraction for the polynomial's factor of greatest absolute value(to be used in the following chapter) Ready to takle it? Set: (a) Procedure: Using the Euclidean algorithm, a = a * 1 The partial quotient is: [1;] Using the partial quotient fed back into the Euclidean algorithm and replacing "a" with 1 to derive a finite series, x term: 0 1 x term refers to x raised to a power in a polynomial: series #1: 1 a for example, 0 = x^1 + ? * x^0 for a polynomial of as yet unknown coefficients: x^1 = a; and x^0 = 1; giving 0 = a + ? * 1 0 = x^1 + ? * x^0 0 = a + ? * 1 using 1.1 from above, 0 = a - a * 1 thus, 0 = x^1 - a * x^0 or simply, 0 = x - a this is the polynomial we are searching for in the first degree; its continued fraction is, x = a That was the easiest subtext to this riddle. It gradually becomes more difficult with the increase of the number of elements in the starting set of integers. Set: (a, b) and a > b; With the help of the Euclidean algorithm, set (a, b) is actually the terminating set for a finite series of sets that begin with (?, 0) as set zero, and terminating with (a, b) as set n. Each set is designed to approximate a polynomial with greater accuracy. A and b are always integers throughout all sets. The Euclidean algorithm is algebraically synonymous to a continued frac- tion. A continued fraction is a partial, or approximate, solution of a polyno- mial. Since solving for two unknowns, a and b, give an infinite n is the index to a and b; example, b(n) is the n-th term of b; the index of a and b is unknown, so consequently a(n) will be made equivalent to a and b(n) = b, or set (a, b) = set (a(n), b(n)); a(0) is unknown; b(0) = 0 Procedure: Using the Euclidean algorithm, we will create a finite convergent series beginning with (a, b) and receeding in reverse to zero: a(n) = p * a(n-1) + q * b(n-1) b(n) = a(n-1) a(n-1) = p * a(n-2) + q * b(n-2) b(n-1) = a(n-2) etc., to a(0) = p * a(n-?) + q * b(n-?), or a(0) = 0 + q * b(n-?) 0 = b(0) = a(n-?) To find p and q, we will be looking for integer solutions in two unknowns; there will be only one such case. Of rational solutions though, there will be an infinite amount, and an even greater number of irra- tional solutions, relatively speaking. The reason for focusing on an integer solution is that the Euclidean algorithm uses only integers as its multipliers. Also, there is only one such case. Procedure: Bear in mind: x^2 + (-p) * x^1 + (-q) * x^0 = 0 a/b is not only the n-th convergent of the Euclidean algorithm, but it is also the n-th convergent of a continued fraction. In general, a continued fraction is a partial solution method for find- ing numerical approximations to the factors of polynomials in one unknown. A convergent is an approximation; it is part of an infinite series of approxima- tions resulting in ever greater accuracy of representing a factor. The standard procedure for listing factors is to expect to list them from greatest absolute value to least, once they are found. Every even numbered factor in the polyno- mial's list will be negative. These rules apply equally to situations involving imaginary factors. So, a/b is the n-th convergent of the first factor of the above polyno- mial, while b/a * -q is the n-th convergent of the second factor. To search for an integer solution: x^2 + (-p) * x^1 + (-q) * x^0 = 0 x^2 = p * x + q; dividing both sides by x leaves, x = p + q/x; but approx., 0 = (x - a/b) * (x - (-q * b/a)) -p = (-a/b) + -(-q * b/a) -p = (-a^2 + q * b^2) / (a * b) p = (a^2 - q * b^2) / ab Initially set q equal to 0 and solve for p; Then set q equal to p and solve for the next version of p; Repeat until p is equivalent to q to some degree of accuracy; (1) Round p up to the nearest integer; Replace p in the equation with this best guess and solve for q; (2) Round q down to the nearest integer; Test by using a/b, or -b/a, as the first or second roots of the polynomial for relative correctness; if either one or both are not approximately right, then Go back to (1), but this time round p down to the nearest integer and at (2) round q up; We now have p and q for the polynomial that we have been looking for: x^2 + (-p) * x^1 + (-q) * x^0 = 0 (if p >= q, then p is rounded up and q is rounded down; otherwise, the reverse is the situation) Before continuing with this process to greator degree let's put it to some use. Definition: polynomial in multiple unknowns---a partially integrated system of relations/ratios/factors polynomial in one unknown---a fully integrated system of factors whose equiva- lence to a polynomial in multiple unknowns is as a special case aesthetics---is the rapport among factors of a polynomial, in one unknown, of rational coefficients golden---are the ratios, or factors, of an otherwise golden polynomial in one unknown, whose coefficients are rational, and all of the partial quo- tients(such as p and q of the previous example), resulting from Eucli- dean (algorithm) division performed on the polynomial's factor of greatest absolute value, are all ones; there will be the same number of partial quotients as are the number of factors of the polynomial consequence---there is only one polynomial that is golden for each degree semi-aesthetics/semi-golden---a polynomial with some zero coefficients and partial quotients hybrid aesthetics---ratio/factors, and their polynomials, produced between different geometric forms A parallel occurs in geometry which exemplifies the situation. Golden polynomials have their factors stored within regular odd-sided polygons. They are thus associated with the circle that circumscribes them. Semi-aesthetic polynomials will store their factors within regular even-sided polygons. Hybrid-aesthetic polynomials store their factors as relations between, rather than within, regular polygons. examples, Some golden polynomials and their associated geometric forms: x - 1 = 0 the triangle x^2 - x - 1 = 0 the pentagon x^3 - 2x^2 - x + 1 = 0 the heptagon x^4 - 2x^3 - 3x^2 + x + 1 = 0 the nonagon x^5 - 3x^4 - 3x^3 + 4x^2 + x - 1 = 0 the 11-gon x^6 - 3x^5 - 5x^4 + 4x^3 + 4x^2 - x - 1 = 0 the 13-gon x^7 - 4x^6 + 2x^5 +10x^4 - 11x^3 + 2x^2 - x + 1 = 0 the 15-gon etc. Their factors: x - 1 = 0 x = 1 x^2 - x - 1 = 0 x = (1 +-(5^1/2))/2 >>> Root1 = 1.6180339 Root2 = -0.6180339 x^3 - 2x^2 - x + 1 = 0 Some semi-golden polynomials and their associated geometric forms: 2x^2 - 1 = 0 square 3x^4 - 4x^2 + 1 = 0 hexagon 16x^10 - 72x^8 + 64x^6 - 72x^4 + 20x^2 - 2 octagon 6x^14 - 59x^12 + 106x^10 + 265x^8 - 286x^6 - 213x^4 + 238x^2 - 49 dodecagon ?: corollary consequence---the factors of each of these polynomials is equivalent to the ratios taken between sides of certain associated polynomials