# Re: Smullyan Shmullyan, give me a real example

```I've been working on this off and on when I get a chance, even before
my "first guess".  My version of this defines an operation as a
recursive function f(N,m,n), where N is the degree of the operation.
m is one of the operands.   n is the other operand, which is the
counting operand.  n is the number iterations that the recursive
function is evaluated.```
```
I'll call addition the operation of degree 1. So addition can be
defined as follows.

Initial value:
f(1,m,0) = m

Recursion rule:
f(1,m,k) = f(1,m,k-1) + 1

So for a given n,
f(1,m,n) = f(1,...f(1,m,0) + 1,... + 1)   (f(1) taken n times)
= f(1,m,0) + 1 + ... + 1        (1 added n times)
= f(1,m,0) + n
= m + n

Note that counting can be separately defined as the operation of degree
0, but this didn't add to the current argument.  Counting is also

Multiplication is defined in a similar manner, as the operation of
degree 2.

Initial value:
f(2,m,0) = 0

Recursion rule:
f(2,m,k) = f(2,m,k-1) + m

So for a given n,
f(2,m,n) = f(2,...f(2,m,0) + m,... + m)   (f(2) taken n times)
= f(2,m,0) + m + ... + m        (m added n times)
= f(2,m,0) + m * n
= m * n

Exponentiation is the operation of degree 3.

f(3,m,0) = 1
f(3,m,k) = f(3,m,k-1) * m
f(3,m,n) = f(3,...f(3,m,0) * m,...* m)   (f(3) taken n times)
= f(3,m,0) * m * ... * m        (m multiplied n times)
= f(3,m,0) * m ^ n
= m ^ n

Just for kicks, I tried to define "hyper-nentiation" as the operation
of degree 4, with operator symbol "@".  I was interested in what
the initial value of this would be: the mth root of m.  Any further
than this gets too weird for me.

f(4,m,0) = m^(1/m)
f(4,m,1) = m
f(4,m,k) = f(4,m,k-1) ^ m
f(4,m,n) = f(4,...f(4,m,0) ^ m,...^ m)   (f(4) taken n times)
= m ^ m ^ ... ^ m                (m exponentiated n times)
= m @ n

Looking closer specifically at multiplication, we see that it is
defined in terms of addition, which is the preceding function in the
sequence of functions.

f(2,m,n) = f(2,m,n-1) + m
= f(1,m,f(2,m,n-1))
= f(1,m,f(2,m,n-2)+m)
= f(1,m,f(1,m,f(2,m,n-2)))
= f(1,m,f(1,m,f(2,n-3,m)+m))
= f(1,m,f(1,m,f(1,m,f(2,m,n-3))))
= f(1,m,...f(2,m,n-n)+m)...)     (f(1) taken n-1 times,
f(2) taken 1 time)
= f(1,m,...f(2,m,n-n))              (f(1) taken n times,
f(2) taken 1 time)
= f(1,m,...f(2,m,0))                 (f(1) taken n times,
f(2) taken 1 time)
= f(1,m,...f(1,m,0))                 (f(1) taken n times)
= m * n

The above is a formal way of saying that multiplication of m and n is
adding n m's together.  We knew that.

Generalizing this, given the function in the sequence corresponding to
the operation of degree N.

f(N,m,n) = f(N-1,m,...f(N-1,m,n))         (f(N-1) taken n times)

The above is a formal way of saying that f(N)ing of m and n together is
the same as f(N-1)ing n m's together.

The sequence of ever growing functions is defined as f(N,m,n) for N=
1,2,3,...
Given a function of degree N, I take the growth of f(N,m,n) as defined
as its magnitude as n approaches infinity.

So here's a thought toward finding a function that's bigger than
any function in this sequence.  Define the following function.

d(m,n) = f(1,m,...f(n,m,n))

Note that n has been placed not only as the counting operation, but
also as the degree!  So now as n approaches infinity, the degree
approaches infinity. (!)  So here is a single function that has a
degree of operation that is higher than any function of a given degree
of operation.

Am I on the right track?

Tom

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Sun, 21 May 2006 07:08:25 +0000 (UTC)
From: "Tom Caylor" <[EMAIL PROTECTED]>
Subject: Re: Smullyan Shmullyan, give me a real example
Date: Sun, 21 May 2006 00:08:25 -0700
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I've been working on this off and on when I get a chance, even before
my "first guess".  My version of this defines an operation as a
recursive function f(N,m,n), where N is the degree of the operation.
m is one of the operands.   n is the other operand, which is the
counting operand.  n is the number iterations that the recursive
function is evaluated.

I'll call addition the operation of degree 1. So addition can be
defined as follows.

Initial value:
f(1,m,0) = m

Recursion rule:
f(1,m,k) = f(1,m,k-1) + 1

So for a given n,
f(1,m,n) = f(1,...f(1,m,0) + 1,... + 1)   (f(1) taken n times)
= f(1,m,0) + 1 + ... + 1        (1 added n times)
= f(1,m,0) + n
= m + n

Note that counting can be separately defined as the operation of degree
0, but this didn't add to the current argument.  Counting is also

Multiplication is defined in a similar manner, as the operation of
degree 2.

Initial value:
f(2,m,0) = 0

Recursion rule:
f(2,m,k) = f(2,m,k-1) + m

So for a given n,
f(2,m,n) = f(2,...f(2,m,0) + m,... + m)   (f(2) taken n times)
= f(2,m,0) + m + ... + m        (m added n times)
= f(2,m,0) + m * n
= m * n

Exponentiation is the operation of degree 3.

f(3,m,0) = 1
f(3,m,k) = f(3,m,k-1) * m
f(3,m,n) = f(3,...f(3,m,0) * m,...* m)   (f(3) taken n times)
= f(3,m,0) * m * ... * m        (m multiplied n times)
= f(3,m,0) * m ^ n
= m ^ n

Just for kicks, I tried to define "hyper-nentiation" as the operation
of degree 4, with operator symbol "@".  I was interested in what
the initial value of this would be: the mth root of m.  Any further
than this gets too weird for me.

f(4,m,0) = m^(1/m)
f(4,m,1) = m
f(4,m,k) = f(4,m,k-1) ^ m
f(4,m,n) = f(4,...f(4,m,0) ^ m,...^ m)   (f(4) taken n times)
= m ^ m ^ ... ^ m                (m exponentiated n times)
= m @ n

Looking closer specifically at multiplication, we see that it is
defined in terms of addition, which is the preceding function in the
sequence of functions.

f(2,m,n) = f(2,m,n-1) + m
= f(1,m,f(2,m,n-1))
= f(1,m,f(2,m,n-2)+m)
= f(1,m,f(1,m,f(2,m,n-2)))
= f(1,m,f(1,m,f(2,n-3,m)+m))
= f(1,m,f(1,m,f(1,m,f(2,m,n-3))))
= f(1,m,...f(2,m,n-n)+m)...)     (f(1) taken n-1 times,
f(2) taken 1 time)
= f(1,m,...f(2,m,n-n))              (f(1) taken n times,
f(2) taken 1 time)
= f(1,m,...f(2,m,0))                 (f(1) taken n times,
f(2) taken 1 time)
= f(1,m,...f(1,m,0))                 (f(1) taken n times)
= m * n

The above is a formal way of saying that multiplication of m and n is
adding n m's together.  We knew that.

Generalizing this, given the function in the sequence corresponding to
the operation of degree N.

f(N,m,n) = f(N-1,m,...f(N-1,m,n))         (f(N-1) taken n times)

The above is a formal way of saying that f(N)ing of m and n together is
the same as f(N-1)ing n m's together.

The sequence of ever growing functions is defined as f(N,m,n) for N=
1,2,3,...
Given a function of degree N, I take the growth of f(N,m,n) as defined
as its magnitude as n approaches infinity.

So here's a thought toward finding a function that's bigger than
any function in this sequence.  Define the following function.

d(m,n) = f(1,m,...f(n,m,n))

Note that n has been placed not only as the counting operation, but
also as the degree!  So now as n approaches infinity, the degree
approaches infinity. (!)  So here is a single function that has a
degree of operation that is higher than any function of a given degree
of operation.

Am I on the right track?

Tom

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