Construction of a Regular HeptadecagonDate: 12/27/2009 at 04:58:01 From: David Subject: Construction of a regular Heptadecagon Gauss derived a finite algebraic expression for sin(pi/17) which lead to an algorithm for the construction of the regular 17-gon. The expression was: 1/8 * sqrt(34-2sqrt(17)-2sqrt(2)sqrt(17-sqrt(17))- 2sqrt(68+12sqrt(17)+2sqrt(2)(sqrt(17)-1)sqrt(17-sqrt(17))- 16sqrt(2)sqrt(17+sqrt(17))))) Daunting I know. But that's basically my point. I see that such an expression must exist since 17 is a Fermat number, but I don't see how to derive it. If I wanted to try to figure out how to construct a 257- or 65537-gon, I'd be at a loss since without Gauss' formula, I wouldn't even be able to construct a heptadecagon. Going from a trigonometric function to a rational expression in itself seems impossible. I've tried using power series expansions of trigonometric functions evaluated at specific points to try to derive rational expressions to known-to-be-rational ordinates, but haven't been able to generate anything finite. Date: 12/29/2009 at 05:19:27 From: Doctor Jacques Subject: Re: Construction of a regular Heptadecagon Hi David, I'm afraid I cannot give you the full derivation here (there is a lot of algebra involved), but I will try to give you an idea of the concepts involved. The best way to solve this problem would be to use Galois theory; if you are familiar with it, you will find a link at the end of this answer. Actually, Galois theory was not yet discovered when Gauss produced his result; however, I believe that he used implicitly some of the symmetry concepts of the theory. The key ingredient is a result of Lagrange. Assume that f(x) is a polynomial with rational coefficients of degree n, with distinct (possibly complex) roots x_1, ..., x_n. Let y = g(x_1,...,x_n) be a polynomial expression (with rational coefficients) in the roots of f (not all roots need appear explicitly). There may be up to n! ways of permuting the x_i in g, and each of these permutations may produce a value of y. Lagrange's result says that, if y can take k distinct values for all those permutations, then y is a root of a polynomial of degree k with rational coefficients (more precisely, whose coefficients are rational functions of the coefficients of the original polynomial). Actually, there is a little more to it... For example, consider the quadratic equation f(x) = x^2 - 2x - 1, whose roots are x_1 = 1 + sqrt(2) and x_2 = 1 - sqrt(2). The expression y = x_1 + x_2 takes the same value when we interchange x_1 and x_2; it is therefore a root of a rational equation of degree 1, which means that it is a rational number (2 in this case). On the other hand, the expression z = x_1 - x_2 can take two distinct values when we permute x_1 and x_2; it is therefore a root of a polynomial of degree 2 (x^2 - 8). Another example of such an expression is simply x_1: this expression can take the values x_1 and x_2 when we permute the roots, and it is a root of an equation of degree 2 (the original equation). Lagrange used this method to give another derivation of the formulas for the solution of the cubic and quartic equations (the formulas were known before, but Lagrange's method does not involve the kind of "black magic" used in the previous methods). How can we use those ideas for solving the problem at hand? From an algebraic point of view, constructing the regular 17-sided polygon in a unit circle amounts to solving the equation: z^17 - 1 = 0 [1] in the complex plane, by solving a set of quadratic equations (since these correspond to the ruler and compass constructions). This equation has an obvious root z = 1; if we divide it by (z - 1), we obtain an equation of degree 16: z^16 + z^15 + ... + z + 1 = 0 [2] The roots of this equation (called "roots of unity") are exp(2*m*pi*i/17), where m is any integer not divisible by 17. If z is any of the roots, all the roots are given by z^k, where k is any integer not divisible by 17; as z^17 = 1, the exponent is defined modulo 17. Note also that, as |z| = 1, 1/z^k = z^(17-k) is the complex conjugate of z^k; this implies that z^k + 1/z^k is real; actually, if z^k = exp(2*m*pi*i/17), we have: z^k + 1/z^k = 2*cos(2*m*pi/17) [3] and this is sufficient to construct the polygon. Consider now the expressions: u_0 = z + z^9 + z^13 + z^15 + z^16 + z^8 + z^4 + z^2 u_4 = z^3 + z^10 + z^5 + z^11 + z^14 + z^7 + z^12 + z^6 (The reason for the subscripts in u_0 and u_4 will become apparent later). These are polynomial expressions in one of the roots z. If we permute the roots of [2], only the image of z matters, since it is the only root that actually appears in u_0 and u_4. Now, replacing z by another root means replacing z by z^k (and reducing the exponents modulo 17). If you try a few cases, you will see that such a substitution either leaves u_0 and u_4 invariant (for example z -> z^9), or exchanges them (for example, z -> z^3). By the remark above, this implies that u_0 and u_4 are roots of a quadratic equation. We can easily find that equation. Note first that u_0 + u_4 is the sum of all the powers of z; by [2], this is equal to -1. With some patience, you can compute the 64 terms of the expansion of u_0 * u_4 (still using the fact that z^17 = 1), and you will find that this product equals -4. The fact that these expressions are rational is to be expected: because a permutation of the roots of [2] at most interchanges u_0 and u_4, both expressions remain the same under such a permutation, and, as we have seen, this implies that they are rational. To summarize, we have: u_0 + u_4 = 1 u_0 * u_4 = -4 and, as we know the sum and product of the u_i, we can find them by solving the quadratic equation: u^2 - u - 4 = 0 [4] (the fact that the discriminant of that equation is equal to 17 is no accident). We can now repeat the process. We separate each of u_0 and u_4 into two sets: v_0 = z + z^13 + z^16 + z^4 v_2 = z^9 + z^15 + z^8 + z^2 v_4 = z^3 + z^5 + z^14 + z^12 v_6 = z^10 + z^11 + z^7 + z^6 such that u_0 = v_0 + v_2 and u_4 = v_4 + v_6. We now look at permutations of the roots that leave each of u_0 and u_4 invariant, i.e., substitutions of the form z -> z^(9k). If we look at v_0 and v_2, we can verify that such a permutation either leaves both of them invariant, or exchanges them (the same is true for v_4 and v_6). Using the same kind of argument (and much work), we obtain: v_0 + v_ 2 = u_0 v_0 * v_2 = -1 and this shows that v_0 and v_2 are the roots of the quadratic: v^2 - u_0 * v - 1 = 0 Now, there is no algebraic way to distinguish u_0 from u_4: they are defined by the same equation (this is really the basic idea of Galois theory). Therefore, if we replace u_0 by u_4 in the above equation, the resulting equation will produce v_4 and v_6. We can write the single equation: v^2 - uv - 1 = 0 [5] where u stands for either root of [4]. For each root of [4], [5] will give two values of v, for a total of 4. As you might suspect, the next step is to separate each of the v_i in two parts: w_0 = z + z^16 w_1 = z^13 + z^4 w_2 = z^9 + z^8 w_3 = z^15 + z^2 w_4 = z^3 + z^14 w_5 = z^5 + z^12 w_6 = z^10 + z^7 w_7 = z^11 + z^6 Note that each of the w_i is the sum of two complex conjugate numbers, and is therefore real. As the v_i and u_i are sums of these values, they are also real, which is really helpful when you attempt the actual geometric construction. In this case, however, we are faced with a problem. If we attempt the construction using w_0 and w_1, for example, we obtain: w_0 + w_1 = v_0 w_0 * w_1 = v_4 The reason why this is a problem is that the resulting equation would contain v_0 and v_4; as these correspond to distinct values of u, we cannot relate them together. We can describe the procedure as a tree: | --------------------------- u0 u4 | | ----------- ----------- v0 v2 v4 v6 | | | | ------- ------- ------- ------- w0 w1 w2 w3 w4 w5 w6 w7 In the construction of any root, we are allowed to use only "ancestors" of that root; for example, to compute w_0, we may use v_0 and u_0. We can also use "uncles", for example, to compute w0, we could use v_2, because we have v_2 = u_0 - v_0, and both u_0 and v_0 are ancestors of w_0. However, we cannot use "cousins": we cannot use v_4 in the computation of w_0, because v_4 corresponds to a different choice of u in [4], i.e., a different equation, and we cannot isolate one specific root of that equation. However, with (heavy) algebraic manipulations (which did not deter Gauss), it is possible to express v_4 in terms of v_0, giving: v_4 = -(v_0^3 -6v_0 + 3)/2 and this allows us to solve the final equation: w^2 - vw - (v^3 -6v + 3)/2 = 0 [6] where v is either root of [5]. To summarize, we solve three quadratic equations ([4], [5], and [6]) with real roots, which can be done using ruler and compass. Each equation uses one root of a previous equation. As each equation has two roots, we get a total of 8 solutions for w, corresponding to the values of 2*cos(2k*pi/17). In practice, we can transform the u, v, and w (for example, by adding constants) to make the geometric construction look "nicer". As I said before, the proper way to formulate this would be in the context of Galois theory. If you are familiar with that, you may want to read a paper I write on the subject some time ago: Construction of Regular Polygons http://home.scarlet.be/j-willekens/store/p17.pdf The paper also contains the coefficients of the equations to be solved for a 257-sided polygon; as you will see, it is hopeless to try to find those coefficients by hand, or to attempt an actual geometric construction (although it is theoretically possible). Please feel free to write back if you want to discuss this further. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
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