What lies between + and  and beyond?

What lies between + and  and beyond? H.P.Williams London School of Economics

2 22 23 2 1½ 3 ↓ 2 x 3 2 + 3 12 3 4

f(a + 1, b + 1, c) = f (a, f(a + 1, b, c),c) Initial conditions f (0, b, c) = b + 1 f (1, 0, c) = c f (2, 0, c) = 0 f (a + 1, 0, c) = 1 for a>0 Easy to verify f (0, b, c) = b + 1 Successor Function f (1, b, c) = b + c Addition f (2, b, c) = b x c Multiplication f (3, b, c) = cb Exponentiation f (4, b, c) = ccb times Tetration ( bc ) We seek f (3/2, b, c) ? A Generalisation of Ackermann’s Function c ⋰

Ackermann’s Function is usually expressed as a function of 2 arguments by fixing c at (say) 2. It is a doubly recursive function which grows faster than any primitive recursive function.I.e In order to evaluate f(a + 1, …,..) we need to evaluate f (a + 1,…,… )for smaller arguments and f (a,…,…) for much larger arguments.Example f ( 0, 3, 2 ) = 4 f ( 1, 3, 2 ) = 5 f ( 2, 3, 2 ) = 6 f ( 3, 3, 2 ) = 8 f ( 4, 3, 2 ) = 16 f ( 5, 3, 2 ) = 65536 f ( 6, 3, 2 ) = ….

But:Inverse Ackerman Function- a function that grows very very slowlyused in Computational ComplexityTaken further we get Goodstein Numbers which converge to zero more slowly then any finite recursive proof can show. I.e convergence not provable by Peano’s Axioms.

Will denote f ( a, b, c ) by b ⓐ ci.e b ⓪ c = b + 1 b ① c = b + c b ② c = b x c b ③ c = cb c b ④ c = cc b times (bc ) etcWhat is f (3/2, b, c ) = b 1½ c ?Attention will be confined to non-negative reals. ⋰

Other Integer Functions definable for Real values.e.g n! by Gamma FunctionFractional differentiation

An Aside F(a, b, c) not generally well defined for fractional b either. We could define p/q2 = x such that qx = p2 If we were to define p/q r, where (p, q) = 1 it would not be continuous in b. If ½2 is solution of 2x= 2 → x = 1.5596… But 2/42 is solution of 4x= 22 → x = 1.6204… We can define 1/∞2. It is √2 = 2 √2 √2 ⋰ since

Gauss’ Arithmetic – Geometric Mean A(a,b) = Arithmetic Mean a + b 2 G (a,b) = Geometric Mean √(a x b) M (a,b) = Gauss’ Mean ‘halfway between’ A(a,b) & G(a,b) Let a1 = G(a,b), b1 = A (a,b) an+1 = G (an, bn), bn+1 = A(an, bn) M (a, b) = Lt an = Lt bn n→∞ n→∞

Example G ( 2, 128 ) = 16, A ( 2, 128 ) = 65G ( 16, 65 ) = 32.25, A ( 16, 65 ) = 40.5G ( 32.25, 40.5 ) = 36.14, A ( 32.25, 40.5 ) = 36.75G ( 36.14, 36.37 ) = 36.26, A ( 36.14, 36.37 ) = 36.26Hence M ( 2, 128 ) = 36.26a + b = A ( a, b ) x 2 = A ( a, b ) ② 2a x b = G ( a, b ) 2 = G ( a, b ) ③ 2Let a 1½ b = M ( a, b ) 2½ 2 = M ( a, b) 1½ M( a, b )M ( a, b ) has an analytic solution in terms of Elliptic integrals.

A Difficulty a M(a, M(a, b)) M(a, b) M(M(a, b),b) b But M(a, b) ≠ M(M(a,M(a,b)), M(M(a,b),b))

A Functional Equation e(a+b) = ea x eb Let ff (x) = ex Let f (a 1½ b) = f(a) x f(b) Hence a 1½ b = f –1(f(a) x f(b)) = f (f –1 (a) + f -1 (b)) Hence we seek solutions of ff(x) = ex f(x) is a function ‘between’ x and ex Insist (for x≥0) (i) x < f(x) < ex(ii) f(x) monotonic strictly increasing (iii) f(x) continuous and infinitely differentiable (iv) derivatives are monotonically strictly increasing

set p = 0.49 (say) f (0) = p = 0.49 f (p) = e0 = 1 f (1) = ep = 1.63 f (1.63) = e1 = 2.72 f (2.72) = ee = 5.12 etc • • • • •

Let f(0) = p 0≤p ≤1 f(p) = ff (0) = 1 f(1) = ff (p) = ep f(ep)= ff (1) = e f(e) = ff (ep) = ee etc. Gradient >1 → 1-p > 1 → p < 0.5 p Gradient increasing → ep – 1 > 1 – p → p > 0.4695449931 1 – p p Hence 0.469 < p < 0.5 An infinite numbers of functions f(x) possible. p

What lies between + and  and beyond?