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

x ↓ 2 1½

A Generalisation of Ackermann’s Function f(a + 1, b + 1, c) = f (a, f(a + 1, b, c),c) f(a + 1, b + 1, c) = f (a, f(a + 1, b, c),c) Initial conditions Initial conditions f (0, b, c) = b + 1 f (0, b, c) = b + 1 f (1, 0, c) = c f (1, 0, c) = c f (2, 0, c) = 0 f (2, 0, c) = 0 f (a + 1, 0, c) = 1 for a>0 f (a + 1, 0, c) = 1 for a>0 Easy to verify Easy to verify f (0, b, c) = b + 1 Successor Function f (0, b, c) = b + 1 Successor Function f (1, b, c) = b + c Addition f (1, b, c) = b + c Addition f (2, b, c) = b x c Multiplication f (2, b, c) = b x c Multiplication f (3, b, c) = c b Exponentiation f (3, b, c) = c b Exponentiation f (4, b, c) = c c b times Tetration ( b c ) f (4, b, c) = c c b times Tetration ( b c ) We seek f (3/2, b, c) ? 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 ) = f ( 6, 3, 2 ) = ….

But: Inverse Ackerman Function - a function that grows very very slowly used in Computational Complexity Taken 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 ⓐ c i.e b ⓪ c = b + 1 b ① c = b + c b ② c = b x c b ③ c = c b c b ④ c = c c b times ( b c ) etc What 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 Function Fractional differentiation

An Aside F(a, b, c) not generally well defined for fractional b either. We could define p/q 2 = x such that q x = p 2 If we were to define p/q r, where (p, q) = 1 it would not be continuous in b. If ½ 2 is solution of 2 x = 2 → x = … If ½ 2 is solution of 2 x = 2 → x = … But 2/4 2 is solution of 4 x = 2 2 → x = … We can define 1/∞ 2. It is √2 = 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 a 1 = G(a,b), b 1 = A (a,b) a n+1 = G (a n, b n ), b n+1 = A(a n, b n ) M (a, b) = Lt a n = Lt b n n→∞ n→∞ n→∞ n→∞

Example G ( 2, 128 ) = 16, A ( 2, 128 ) = 65 G ( 16, 65 ) = 32.25, A ( 16, 65 ) = 40.5 G ( 32.25, 40.5 ) = 36.14, A ( 32.25, 40.5 ) = G ( 36.14, ) = 36.26, A ( 36.14, ) = Hence M ( 2, 128 ) = a + b = A ( a, b ) x 2 = A ( a, b ) ② 2 a x b = G ( a, b ) 2 = G ( a, b ) ③ 2 Let 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 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 Difficulty A Difficulty

A Functional Equation e (a+b) = e a x e b Let ff (x) = e x 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) = e x f(x) is a function ‘between’ x and e x Insist (for x≥0) (i) x < f(x) < e x (ii) f(x) monotonic strictly increasing (iii) f(x) continuous and infinitely differentiable (iv) derivatives are monotonically strictly increasing

set p = 0.49 (say) set p = 0.49 (say) f (0) = p = 0.49 f (0) = p = 0.49 f (p) = e 0 = 1 f (p) = e 0 = 1 f (1) = e p = 1.63 f (1) = e p = 1.63 f (1.63) = e 1 = 2.72 f (1.63) = e 1 = 2.72 f (2.72) = e e = 5.12 f (2.72) = e e = 5.12 etc etc

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