(1/4)! and (3/4)! are interexpressible.

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:

- they do not converge
- there is a dense set of points which involve division by zero
- there is a dense set of points which loop, but roundoff error propagates so all loops are unstable
- being on the line is also unstable

(if the roots are imaginary and you are on the real axis, you may be doing exact computation of the imaginary part (0), hence stay on the line. Example: X^2 + 1 = 0, X0 = random real floating point number.)

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

= X + Y + Z + ...

= X Y + X Z + ... + Y Z + ...

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)

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.

- Translate N-space.
- Expand N-space about one of its points.
- Stereographically project N-space onto an N-sphere, rotate the sphere, then project back onto N-space.

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.