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 equivalent to adding with m=0. 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 X-Google-Language: ENGLISH,ASCII-7-bit Received: by 10.11.53.63 with SMTP id b63mr96488cwa; Sun, 21 May 2006 00:08:25 -0700 (PDT) X-Google-Token: MV9CDAwAAABW1UuTDcJqFpeal26hqLve Received: from 207.200.116.67 by y43g2000cwc.googlegroups.com with HTTP; Sun, 21 May 2006 07:08:25 +0000 (UTC) From: "Tom Caylor" <[EMAIL PROTECTED]> To: "Everything List" <everything-list@googlegroups.com> Subject: Re: Smullyan Shmullyan, give me a real example Date: Sun, 21 May 2006 00:08:25 -0700 Message-ID: <[EMAIL PROTECTED]> References: <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> User-Agent: G2/0.2 X-HTTP-UserAgent: Mozilla/4.0 (compatible; MSIE 6.0; AOL 9.0; Windows NT 5.1; Q312461; SV1; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) X-HTTP-Via: HTTP/1.1 (Velocity/1.3.32 [uScMs f p eN:t cCMp s ]), HTTP/1.1 Turboweb [ntc-tc091 8.4.0], HTTP/1.0 cache-ntc-ab03.proxy.aol.com[CFC87443] (Traffic-Server/6.1.0 [uScM]) Mime-Version: 1.0 Content-Type: text/plain 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 equivalent to adding with m=0. 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? 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