Re: Table #48 in Beiler's book on "least exponents" i.e. ORDER of p,
to any base, e.g. ORD4 (7) = 3 or 4^3 == is congruent to 1 MOD 7.
Thus, 3 is the "least exponent" for base 4, N=7.
I've been working on a system of polynomials which come up with the
same cycle lengths, but the nth polynomial in each set matches the
n^2 result in Beiler's chart. I'm trying to figure out why n^2 and
not n, if anybody knows, I would appreciate it (better yet, a proof).
We'll go through several of the polys, so get out your pocket
calulator. By way of example, given seed is Cos 2 Pi/N, N=7,
Chebychev Poly #2 is 2x^2 - 1. Plug in the seed, take iterates, and
the cycle length = 3. Similarly (you can take my word for this); the
nth poly in each set matches cycle length for n^2, but not n.
The Chebychev polys are: T1 = x, T2 = 2x^2 - 1, T3 = 4x^3 - 3x, T4 =
8x^4 - 8x^2 + 1, T5 = 16x^5 -20x^3 + 5x ...That's enough.
Case II, the Lucas Polys. L1 = x, L2 = x^2 - 2, L3 = x^3 - 3x, L4 =
x^4 -4x^2 +2, that's enough for now. So let's take L2 which ix x^2 -
2. Seeds for Lucas are 2 Cos 2 Pi/N. Taking N=7, we get the 3 cycle.
Case III. Seeds Sin^2 2 Pi/N, 2nd poly is 4x(1 - x). Again, 3 cycle.
Case IV. 2x/(1 - x^2), seed is Tan 2 Pi/N. Again, 3 cycle.
There are infinite sets of infinite sets, but the above examples are
sufficient to establish the principles and the missing element.: Why
the nth poly in each set applies to n^2 entry in the least exponents
chart, but not the nth row.
Procedure for derivation of the formulas: We can use an algebraic
trick which bypasses trig derivations, as follows.
For angle A, the pythagorean relationships are
A Cos Sin Hyp.
x y 1
2A x^2-y^2 2xy 1
We take our seed, and find the algebraic equivalent for 2A.
Therefore, the formulas given above, #2 in each set, apply to the
double angle, while 3rd poly = the triple angle, etc.
In essence, the double angle formulas are taking a complex term and
successively squaring it.
If anybody knows why nth poly corresponds to n^2 row in the least
exponents chart, let me know. Hopefully a proof. Thanks a lot.
Sincerely, Gary W. Adamson
A previous contributor was kind enough to provide a proof for the
following. If ORDb (N) is odd, then ORDb^2 (N) is the same. But if
ORDb (N) is even, then ORDb^2 is half the amount.
Example: ORD2 (7) = 3, ORD4 (7) = 3; but ORD2 (11) = 10, with ORD4
(11) = 5.