What lies between + and and beyond?

What lies between + and and beyond?

H.P.Williams

London School of Economics

56789

1011121314151617

2 + 3

2 x 3

23

22

↓ 2 1½ 3

22

1 2 3 4

A Generalisation of Ackermann’s Function

A Generalisation of Ackermann’s Function

f(a + 1, b + 1, c) = f (a, f(a + 1, b, f(a + 1, b + 1, c) = f (a, f(a + 1, b, c),c)c),c)

Initial conditionsInitial conditions

f (0, b, c) = b + 1f (0, b, c) = b + 1

f (1, 0, c) = cf (1, 0, c) = c

f (2, 0, c) = 0f (2, 0, c) = 0

f (a + 1, 0, c) = 1 for a>0f (a + 1, 0, c) = 1 for a>0

Easy to verifyEasy to verify

f (0, b, c) = b + 1 Successor Functionf (0, b, c) = b + 1 Successor Function

f (1, b, c) = b + c Additionf (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) = cf (3, b, c) = cbb Exponentiation Exponentiation

f (4, b, c) = cf (4, b, c) = ccc b times Tetration ( b times Tetration ( bbc )c )

We seek f (3/2, 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 ) = 65536 f ( 6, 3, 2 ) = ….

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 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.

But: Inverse Ackerman Function

- a function that grows very very slowlyused 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 c = b + 1⓪ b ① c = b + c b ② c = b x c b ③ c = cb

c

b ④ c = cc b times ( bc )

etc

What is f (3/2, b, c ) = b 1½ c ?

Attention will be confined to non-negative reals.

Will denote

f ( a, b, c ) by b ⓐ c

i.e b c = b + 1⓪ b ① c = b + c b ② c = b x c b ③ c = cb

c

b ④ c = cc b times ( bc )

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

Other Integer Functions definable

for Real values.

e.g n! by Gamma Function

Fractional differentiation

An AsideAn Aside

F(a, b, c) not generally well defined for fractional b either.F(a, b, c) not generally well defined for fractional b either.

We could define We could define p/qp/q2 = x such that 2 = x such that qqx = x = pp22

If we were to define If we were to define p/q p/q r, where (p, q) = 1 it would not be r, where (p, q) = 1 it would not be continuous in b.continuous in b.

If If ½½2 is solution of 2 is solution of 22xx = 2 = 2 → x = 1.5596…→ x = 1.5596…

But But 2/42/42 is solution of 2 is solution of 44xx = 2= 222 → x = 1.6204… → x = 1.6204…

We can define We can define 1/∞1/∞2. It is √22. It is √2

= 2= 2

⋰⋰

√√22√√22

sincesince

Gauss’ Arithmetic – Geometric MeanGauss’ Arithmetic – Geometric Mean

A(a,b) = Arithmetic Mean A(a,b) = Arithmetic Mean a + ba + b 22G (a,b) = Geometric Mean G (a,b) = Geometric Mean √(a x b)√(a x b)

M (a,b) = Gauss’ Mean ‘halfway between’ M (a,b) = Gauss’ Mean ‘halfway between’ A(a,b) & G(a,b)A(a,b) & G(a,b)

Let aLet a1 1 = G(a,b), b= G(a,b), b1 1 = A (a,b)= A (a,b)

aan+1 n+1 = G (a= G (ann, b, bnn), b), bn+1 n+1 = A(a= A(ann, b, bnn))

M (a, b) = Lt aM (a, b) = Lt an n = Lt b= Lt bnn

n→∞ n→∞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 ) = 36.26

Hence M ( 2, 128 ) = 36.26

a + b = A ( a, b ) x 2 = A ( a, b ) 2②a x b = G ( a, b ) 2 = G ( a, b ) 2③

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.

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 ) = 36.26

Hence M ( 2, 128 ) = 36.26

a + b = A ( a, b ) x 2 = A ( a, b ) 2②a x b = G ( a, b ) 2 = G ( a, b ) 2③

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, a M(a, M(a, b)) M(a, b) M(M(a, b),b) bb),b) b

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

A DifficultyA Difficulty

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

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)set p = 0.49 (say)

f (0) = p = 0.49f (0) = p = 0.49

f (p) = ef (p) = e0 0 = 1= 1

f (1) = ef (1) = ep p = 1.63= 1.63

f (1.63) = ef (1.63) = e1 1 = 2.72= 2.72

f (2.72) = ef (2.72) = eee = 5.12 = 5.12

etcetc

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Let f(0) = p 0Let f(0) = p 0≤p ≤1≤p ≤1

f(p) = ff (0) = 1f(p) = ff (0) = 1

f(1) = ff (p) = ef(1) = ff (p) = epp

f(e f(epp)) = ff (1) = e= ff (1) = e

f(ef(e ) = ff (e) = ff (epp) = e) = eee

etc.etc.

Gradient >1 → Gradient >1 → 1-p1-p > 1 → p < 0.5 > 1 → p < 0.5 p p

Gradient increasing → Gradient increasing → eepp – 1 – 1 > > 1 – p1 – p → p > → p > 0.46954499310.4695449931 1 – p p 1 – p p

Hence 0.469 < p < 0.5Hence 0.469 < p < 0.5

An infinite numbers of functions f(x) possible.An infinite numbers of functions f(x) possible.

pp