There are many interesting things about polynomials whose coefficients are taken from slices of Pascal's triangle. (These are a form of what's called Chebyshev polynomials.) For example, the numbers 1,9,28,35,15,1 are taken from the 11th diagonal of Pascal's triangle as shown below: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1/ 1 6 15 20 15/ 1 7 21 35/ 1 8 28/ 1 9/ 1/ The polynomial with these coefficients is x^5 + 9x^4 + 28x^3 + 35x^2 + 15x + 1 = 0 which has five roots given by / k pi \ x = -4 cos^2 | ---- | k=1,2,3,4,5 \ 11 / These same roots can also be expressed in terms of radicals, i.e., if Q denotes any 11th root of 1, then (1 + Q)^2 x = - ----------- Q To find the roots corresponding to the nth diagonal, simply replace 11 by n in the above formulas. There's an interesting connection between these "diagonal polynomials" and a special class of functions called Mobius transformations. These functions (also known as linear fractional transformations) have the general form ax + b f(x) = -------- cx + d where a,b,c,d are constants. If you iterate functions of this form you find that some of them lead to periodic cycles. For example, one of the simplest Mobius transformations is f(x)=1/x, which we get by setting a=d=0 and b=c=1. Iterating this function obviously gives a repeating cycle with a period of 2. A slightly less obvious example is the function 1 + x f(x) = -------- 1 - x Beginning with, say, x=2, you can repeatedly iterate on this function to generate the sequence of values 2, -3, -1/2, 1/3, 2, -3, -1/2, 1/3, 2, ... Thus, iterating this function gives a repeating cycle with a period of 4. If you play around with these for a while you can come up with simple functions that have other periods. However, most Mobius functions do not give repeating cycles. For example, the function 2 + x f(x) = -------- 2 - x never repeates, the iterates just meander around in a seemingly aimless way. (It actually isn't aimless; the density of the iterates on the real line approaches a Cauchy distribution, and you can compute the value of pi from these iterates...but that's another story.) So the question is, which Mobius transformations give repeating cycles of iterates? Also, can we find functions whose iterates repeat with any desired period? It turns out that the iterates of a Mobius transformation (az+b)/(cz+d) repeat with period N if and only if the quantity (a+d)^2 ------- (ad-bc) equals one of the roots of the Nth "diagonal polynomial" from Pascal's triangle. Another interesting thing about "diagonal polynomials" is the way they can be factored. Let P_N[x] denote the Nth diagonal polynomial. In general, if N = mn, then P_N[x] is divisible by P_n[x] and P_m[x]. On the other hand, if N is a prime, then P_N[x] is irreducible, i.e., it can't be factored. However, if N is a prime, it IS possible to factor the polynomial P_N[-x^2], which is the polynomial formed by substituting -x^2 in place of x. For example, the quintic polynomial corresponding to the 11th diagonal becomes x^10 - 9x^8 + 28x^6 - 35x^4 + 15x^2 - 1 = (x^5 - x^4 - 4x^3 + 3x^2 + 3x - 1)(x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1) = (x^5 - 4x^3 + 3x)^2 - (x^4 - 3x^2 + 1)^2 It shouldn't be surprising that this factors, remembering that the roots of P_N[x] are -4cos^2(k pi/N), from which it follows that the roots of P_N[-x^2] are just 2cos(k pi/N). Well, enough about diagonals. How about horizontals? Of course, if we construct a polynomial whose coefficients are all the numbers on the Nth horizontal row of Pascal's triangle, then the roots are all -1's, because the polynomial is just (1+x)^N. However, if you take every OTHER number from a horizontal row, you get an interesting result. For example, the 7th row is 1 7 21 35 35 21 7 1 and if you take every other number from this row you have the cubic f(x) = x^3 + 21x^2 + 35x + 7 The three roots of this polynomial are x = -tan^2(k pi/7) k = 1,2,3 Taking advantage of the identity tan^2(u) = 1/tan^2(pi/2 - u) the roots of the two complementary polynomials from the 7th "horizontal" row can be summarized as / k=1,3,5 for 7x^3 + 35x^2 + 21x + 1 -tan^2(k pi/14) ( \ k=2,4,6 for x^3 + 21x^2 + 35x + 7 Notice that we have 2N instead of N in the denominator. This also works nicely for even numbered rows. For example, the 8th row of Pascal's triangle is 1 8 28 56 70 56 28 8 1 and the roots of the two complementary polynomials from this row are given by / k=1,3,5,7 for x^4 + 28x^3 + 70x^2 + 28x + 1 -tan^2(k pi/16) ( \ k=2,4,6 for 8x^3 + 56x^2 + 56x + 8 You can also define other arrays of integers, similar to Pascal's triangle, and determine the roots of the corresponding diagonal and horizontal polynomials. It generally turns out that these roots determine the values of some parameter for which a Mobius transformation is periodic.

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