superexponentiation is not elementary

superexponentiation is not elementary

In this entry, we will show that the superexponetial

function

f

:

ℕ

2

→

ℕ

, given by

f

⁢

(

m

,

0

)

=

m

,

f

⁢

(

m

,

n

+

1

)

=

m

f

⁢

(

m

,

n

)

is not

elementary recursive

(we set

f

⁢

(

0

,

n

)

:=

0

for all

n

). We will use the

properties

of

f

(listed here (

http://planetmath.org/PropertiesOfSuperexponentiation

)) to

complete

this task.

The idea behind the proof is to find a property satisfied by all elementary recursive functions but not by

f

. The particular property we have in mind is the “growth rate” of a function. We want to demonstrate that

f

, in some way, grows faster than any elementary function

g

. This idea is

similar

to showing that

2

x

is larger than, say,

x

100

for large enough

x

. Formally,

Definition

. A function

h

:

ℕ

2

→

ℕ

is said to

majorize

g

:

ℕ

k

→

ℕ

if there is a

b

∈

ℕ

, such that for any

a

1

,

…

,

a

k

∈

ℕ

:

g

⁢

(

a

1

,

…

,

a

k

)

<

h

⁢

(

a

,

b

)

,

where

⁢

a

=

max

⁡

{

a

1

,

…

,

a

k

}

>

1

.

It is easy to see that no binary function majorizes itself:

Proposition 1

.

h

:

ℕ

2

→

ℕ

does not majorize

h

.

Proof.

Otherwise, there is a

b

such that for any

x

,

y

,

h

⁢

(

x

,

y

)

<

h

⁢

(

a

,

b

)

where

a

=

max

⁡

{

x

,

y

}

>

1

. Let

c

=

max

⁡

{

a

,

b

}

>

1

. Then

h

⁢

(

c

,

b

)

<

h

⁢

(

max

⁡

{

b

,

c

}

,

b

)

=

h

⁢

(

c

,

b

)

, a

contradiction

.
∎

Let

ℰ

⁢

ℛ

be the set of all elementary recursive functions.

Proposition 2

.

Let

A

be the set of all functions majorized by

f

. Then

E

⁢

R

⊆

A

.

Proof.

We simply show that

𝒜

contains the

addition

,

multiplication

,

difference

,

quotient

, and the projection functions, and that

𝒜

is

closed under

composition

,

bounded sum

, and bounded product. And since

ℰ

⁢

ℛ

is the smallest such set, the proof completes.

•

For addition, multiplication, and difference: suppose

t

=

max

⁡

{

x

,

y

}

>

1

. Then

x

+

y

≤

2

⁢

t

=

2

⁢

f

⁢

(

t

,

0

)

≤

f

⁢

(

t

,

1

)

<

f

⁢

(

t

,

2

)

, and

x

⁢

y

≤

t

2

=

f

⁢

(

t

,

0

)

2

≤

f

⁢

(

t

,

1

)

<

f

⁢

(

t

,

2

)

. Moreover,

|

x

-

y

|

≤

t

=

f

⁢

(

t

,

0

)

<

f

⁢

(

t

,

1

)

, and

quo

⁡

(

x

,

y

)

≤

t

=

f

⁢

(

t

,

0

)

<

f

⁢

(

t

,

1

)

.

•

For projection functions

p

m

k

, suppose

t

=

max

⁡

{

x

1

,

…

,

x

k

}

>

1

. Then

p

m

k

⁢

(

𝒙

)

=

x

m

≤

t

=

f

⁢

(

t

,

0

)

<

f

⁢

(

t

,

1

)

.

•

Suppose

g

1

,

…

,

g

m

∈

A

are

n

-ary, and

h

∈

A

is

m

-ary. Let

u

=

h

⁢

(

g

1

,

…

,

g

m

)

and suppose

x

=

{

x

1

,

…

,

x

n

}

>

1

. Given

u

⁢

(

𝒙

)

=

h

⁢

(

g

1

⁢

(

𝒙

)

,

…

,

g

m

⁢

(

𝒙

)

)

, let

z

=

max

⁡

{

g

1

⁢

(

𝒙

)

,

…

,

g

m

⁢

(

𝒙

)

}

. We have two cases:

(a)

z

≤

1

. Let

y

=

max

⁡

{

h

⁢

(

y

1

,

…

,

y

m

)

∣

y

i

∈

{

0

,

1

}

}

. Then

u

⁢

(

𝒙

)

≤

y

<

f

⁢

(

x

,

y

)

.

(b)

z

>

1

. Then, for some

i

,

z

=

g

i

⁢

(

𝒙

)

<

f

⁢

(

x

,

s

)

for some

s

, since

g

i

∈

A

. Then

u

⁢

(

𝒙

)

=

h

⁢

(

g

1

⁢

(

𝒙

)

,

…

,

g

m

⁢

(

𝒙

)

)

≤

f

⁢

(

z

,

t

)

for some

t

since

h

∈

𝒜

. Now,

f

⁢

(

z

,

t

)

=

f

⁢

(

g

i

⁢

(

𝒙

)

,

t

)

<

f

⁢

(

f

⁢

(

x

,

s

)

,

t

)

≤

f

⁢

(

x

,

s

+

2

⁢

t

)

. As a result,

u

⁢

(

𝒙

)

<

f

⁢

(

x

,

s

+

2

⁢

t

)

.

In either case, let

r

=

max

⁡

{

y

,

s

+

2

⁢

t

}

. We see that

u

⁢

(

𝒙

)

<

f

⁢

(

x

,

r

)

.

•

For the next two parts, suppose

g

∈

A

is

(

n

+

1

)

-ary. For any

𝒙

=

(

x

1

,

…

,

x

n

)

, let

x

=

max

⁡

{

x

1

,

…

,

x

n

}

, and for any

y

, assume

z

=

max

⁡

{

x

,

y

}

>

1

. Since

g

∈

A

, let

t

∈

ℕ

be such that

g

⁢

(

𝒙

,

y

)

≤

f

⁢

(

z

,

t

)

, where

z

is as described above.

Let

g

s

⁢

(

𝒙

,

y

)

:=

∑

i

=

0

y

g

⁢

(

𝒙

,

i

)

. We break this down into cases:

(a)

x

>

1

. Then

g

⁢

(

𝒙

,

i

)

<

f

⁢

(

z

i

,

t

)

where

z

i

=

max

⁡

{

x

,

i

}

>

1

for each

i

. Let

f

⁢

(

z

j

,

t

)

be the maximum among the

f

⁢

(

z

i

,

t

)

. Then

g

s

⁢

(

𝒙

,

y

)

≤

(

y

+

1

)

⁢

f

⁢

(

z

j

,

t

)

≤

(

y

+

1

)

⁢

f

⁢

(

z

,

t

)

, as

j

≤

y

. Since

y

+

1

≤

z

+

1

<

2

⁢

z

=

2

⁢

f

⁢

(

z

,

0

)

≤

f

⁢

(

z

,

1

)

, we see that

g

s

⁢

(

𝒙

,

y

)

<

f

⁢

(

z

,

1

)

⁢

f

⁢

(

z

,

t

)

≤

f

⁢

(

z

,

t

1

)

, where

t

1

=

1

+

max

⁡

{

1

,

t

}

.

(b)

x

=

1

. Then

y

>

1

. So

g

s

⁢

(

𝒙

,

y

)

=

h

⁢

(

𝒙

)

+

∑

i

=

2

y

g

⁢

(

𝒙

,

i

)

, where

h

⁢

(

𝒙

)

=

g

⁢

(

𝒙

,

0

)

+

g

⁢

(

𝒙

,

1

)

. Let

v

=

max

⁡

{

h

⁢

(

v

1

,

…

,

v

n

)

∣

v

i

∈

{

0

,

1

}

}

. Then

g

s

⁢

(

𝒙

,

y

)

≤

v

+

∑

i

=

2

y

g

⁢

(

𝒙

,

i

)

. As before,

g

⁢

(

𝒙

,

i

)

≤

f

⁢

(

z

i

,

t

)

for each

i

≤

2

, so pick the largest

f

⁢

(

z

j

,

t

)

among the

f

⁢

(

z

i

,

t

)

. Then

∑

i

=

2

y

g

⁢

(

𝒙

,

i

)

≤

(

y

-

1

)

⁢

f

⁢

(

z

j

,

t

)

≤

(

y

-

1

)

⁢

f

⁢

(

z

,

t

)

<

z

⁢

f

⁢

(

z

,

t

)

=

f

⁢

(

z

,

0

)

⁢

f

⁢

(

z

,

t

)

≤

f

⁢

(

z

,

t

+

1

)

. Therefore,

g

s

⁢

(

𝒙

,

y

)

<

v

+

f

⁢

(

z

,

t

+

1

)

<

f

⁢

(

z

,

v

)

+

f

⁢

(

z

,

t

+

1

)

≤

f

⁢

(

z

,

t

2

)

, where

t

2

=

1

+

max

⁡

{

v

,

t

+

1

}

.

In each case, pick

t

3

=

max

⁡

{

t

1

,

t

2

}

, so that

g

s

⁢

(

𝒙

,

y

)

<

f

⁢

(

z

,

t

3

)

.

•

Let

g

p

⁢

(

𝒙

,

y

)

:=

∏

i

=

0

y

g

⁢

(

𝒙

,

i

)

. We again break down the proof into cases:

(a)

x

>

1

. Then each

g

⁢

(

𝒙

,

i

)

<

f

⁢

(

z

i

,

t

)

where

z

i

=

max

⁡

{

x

,

i

}

>

1

. Let

f

⁢

(

z

j

,

t

)

be the maximum among the

f

⁢

(

z

i

,

t

)

. Then

g

s

⁢

(

𝒙

,

y

)

≤

f

⁢

(

z

j

,

t

)

(

y

+

1

)

≤

f

⁢

(

z

,

t

)

(

y

+

1

)

. Since

y

+

1

≤

z

+

1

<

2

⁢

z

=

2

⁢

f

⁢

(

z

,

0

)

≤

f

⁢

(

z

,

1

)

, we see that

g

s

⁢

(

𝒙

,

y

)

<

f

⁢

(

z

,

t

)

f

⁢

(

z

,

1

)

≤

f

⁢

(

z

,

t

1

)

, where

t

1

=

2

+

max

⁡

{

1

,

t

}

.

(b)

x

=

1

. Then

y

>

1

. So

g

p

⁢

(

𝒙

,

y

)

=

h

⁢

(

𝒙

)

⁢

∏

i

=

2

y

g

⁢

(

𝒙

,

i

)

, where

h

⁢

(

𝒙

)

=

g

⁢

(

𝒙

,

0

)

⁢

g

⁢

(

𝒙

,

1

)

. Let

v

=

max

⁡

{

h

⁢

(

v

1

,

…

,

v

n

)

∣

v

i

∈

{

0

,

1

}

}

. Then

g

p

⁢

(

𝒙

,

y

)

≤

v

⁢

∏

i

=

2

y

g

⁢

(

𝒙

,

i

)

. As before, each

g

⁢

(

𝒙

,

i

)

≤

f

⁢

(

z

i

,

t

)

, so pick the largest

f

⁢

(

z

j

,

t

)

among the

f

⁢

(

x

i

,

t

)

. Then

∏

i

=

2

y

g

⁢

(

𝒙

,

i

)

≤

f

⁢

(

z

j

,

t

)

(

y

-

1

)

≤

f

⁢

(

z

,

t

)

(

y

-

1

)

<

f

⁢

(

z

,

t

)

z

=

f

⁢

(

z

,

t

)

f

⁢

(

z

,

0

)

≤

f

⁢

(

z

,

t

+

2

)

. Therefore,

g

p

⁢

(

𝒙

,

y

)

<

v

⁢

f

⁢

(

z

,

t

+

2

)

<

f

⁢

(

z

,

v

)

⁢

f

⁢

(

z

,

t

+

2

)

≤

f

⁢

(

z

,

t

2

)

, where

t

2

=

1

+

max

⁡

{

v

,

t

+

2

}

.

In each case, pick

t

3

=

max

⁡

{

t

1

,

t

2

}

, so that

g

p

⁢

(

𝒙

,

y

)

<

f

⁢

(

z

,

t

3

)

.

As a result,

ℰ

⁢

ℛ

⊆

𝒜

. In other words, every elementary function is majorized by

f

.
∎

In

conclusion

, we have

Corollary 1

.

f

is not elementary.

Proof.

If it were, it would majorize itself, which is impossible. ∎

Remark

. Although

f

is not elementary recursive, it is easy to see that, for any

n

, the function

f

n

:

ℕ

→

ℕ

defined by

f

n

⁢

(

m

)

:=

f

⁢

(

m

,

n

)

is elementary. This can be done by

induction

on

n

:

f

0

⁢

(

m

)

=

f

⁢

(

m

,

0

)

=

m

=

p

1

1

⁢

(

m

)

is elementary, and if

f

n

⁢

(

m

)

is elementary, so is

f

n

+

1

⁢

(

m

)

=

f

⁢

(

m

,

n

+

1

)

=

exp

⁡

(

m

,

f

⁢

(

m

,

n

)

)

=

exp

⁡

(

p

1

1

⁢

(

m

)

,

f

n

⁢

(

m

)

)

, since

exp

is elementary, and elementary recursiveness

preserves

composition.

Using this fact, one may in fact show that

ℰ

⁢

ℛ

=

𝒜

∩

𝒫

⁢

ℛ

, where

𝒫

⁢

ℛ

is the set of all

primitive recursive functions

.

Title

superexponentiation is not elementary

Canonical

name

SuperexponentiationIsNotElementary

Date of creation

2013-03-22 19:07:14

Last modified on

2013-03-22 19:07:14

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

29

Author

CWoo (3771)

Entry type

Result

Classification

msc 03D20

Related topic

PropertiesOfSuperexponentiation

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