Thus these two pairs are of dimensionality one.
(1/10)! and (2/10)! are sufficient to express (N/10)! for all N.
(1/12)! and (2/12)! are sufficient to express (N/12)! for all N.
(1/3)! and (1/4)! are sufficient to express (N/12)! for all N.
Thus the three cases above are of dimensionality two.
PROBLEM: Find some order to this dimensionality business.
The reflection and multiplication formulas:
pi Z Z! (-Z)! = --------- sin(pi Z) (N-1)/2 -NZ-1/2 (2 pi) N (NZ)! = Z! (Z-1/N)! (Z-2/N)! ... (Z-(N-1)/N)!
On the line:
4 3 2 X + F X + G X + H X + I is (as the discriminant of 2 2 A X + B X + C is B - 4 A C): 4 3 3 3 3 2 2 2 2 - 27 H + 18 F G H - 4 F H - 4 G H + F G H 2 2 2 2 3 4 2 3 + I * [144 G H - 6 F H - 80 F G H + 18 F G H + 16 G - 4 F G ] 2 2 2 4 + I * [- 192 F H - 128 G + 144 F G - 27 F ] 3 - 256 I
/===\ ! ! 2 ! ! (ROOT - ROOT ) = square of determinant whose i,j element is ! ! i j i < j i-1 ROOT . j(The discriminant is the lowest degree symmetric function of the roots which is 0 when any two are equal.)
= X + Y + Z + ...and B is the second symmetric function of N variables
= X Y + X Z + ... + Y Z + ...(B = sum of pairs), then
2 2 2 2 X + Y + Z + ... = A - 2 B. 3 3 3 3 X + Y + Z + ... = A - 3 A B + 3 C. 4 4 4 4 2 2 X + Y + Z + ... = A - 4 A B + 2 B + 4 A C - 4 D.
/ 0 if I > N f(I;X,Y,...) = ! 1 if I = 0 \ X*f(I-1;Y,Z...) + f(I;Y,Z,...) (N-1 variables)The generating function is simply
N ==== \ I > F(I; X, Y, Z) S = (1 + S X) (1 + S Y) (1 + S Z) ... / ==== I=0
3 2 F(X) = X - 3 B X + C X + D = 0are
------------------------------ / ------------------ / F(B) / F(B) 2 F'(B) 3 B - K * 3/ ---- + / [----] + [-----] V 2 V 2 3 ------------------------------ / ------------------ 2 / F(B) / F(B) 2 F'(B) 3 - K * 3/ ---- - / [----] + [-----] V 2 V 2 3where K is one of the three cuberoots of 1:
1, (-1+sqrt(-3))/2, (-1-sqrt(-3))/2.
4 2 X + B X + C X + D = 0,then 2 X = sqrt(Z1) + sqrt(Z2) + sqrt(Z3), where Z1, Z2, Z3 are roots of
3 2 2 2 Z + 2 B Z + (B - 4 D) Z - C = 0.The choices of square roots must satisfy
sqrt(Z1) sqrt(Z2) sqrt(Z3) = -C.
3 -4 X + 3 X - a = 0is X = sin((arcsin a)/3).
In a similar manner, the general quintic can be solved exactly by use of the elliptic modular function and its inverse. See Davis: Intro. to Nonlinear Differential and Integral Equations (Dover), p. 172. Unfortunately, there exists >= 1 typo, since his eqs. (7) and (13) are inconsistent.
Show that all such conformal maps are generated by these operations for any N. If the one-to-one and onto conditions are removed, then for N = 2, conformal maps can be obtained by analytic functions. Show that for N > 2, no new conformal maps exist.