Hi Tom,

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`Apparently you have (re)discover Ackermann function which indeed`

`provide formally a sequence of more and more growing functions, similar`

`to the sequence I was pointing too.`

`I will present it in a easier way for the benefit of the others, but`

`also for using a presentation which will facilitate the future`

`successive diagonalizations. In your second recent post you got the`

`first diagonalization right (or almost right) but the diag is quite`

`hidden and it would be hard to continue the process.`

(Note that you could have define f(0, m, n) as the successor function)

`So your sequence of functions are, in your "Ackerman like parametrized`

`presentation", the functions`

`f(1, m, n) is m + n (or m [1] n, writting "+"`

`as [1] ; for first function)`

f(2, m, n) is m * n (or m [2] n) f(3, m, n) is m ^ n (or m [3] n) f(4, m, n) is m [4] n f(5, m, n) is m [5] n etc

`It will be easier to diagonalize functions of one argument, that is x +`

`x, x * x, x ^ x, x [4] x, x[5]x, etc.`

Let us see their values on 10: 1) 10 + 10 = 20 2) 10 * 10 = 10+10+10+10+10+10+10+10+10 = 100 (ten sums) 3) 10 ^ 10 = 10*10*10*10*10*10*10*10*10 = 10000000000 (ten products)

`4) 10 [4] 10 = 10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^`

`10))))))))) (ten exponentiations) Note the parenthesis:`

`schoolboys/girls know that a ^ (b ^ c) is in general different and`

`bigger than (a ^ b) ^ c .`

`Note also that 10 [4] 10 is already so big that the known observed part`

`of the physical universe is not enough big to write its digits. [4] is`

`called tetration, and 10 [4] 10 can be read 10 tetrated up to 10 (like`

`10 ^ 10 can be read as 10 up to the power of 10). Each function`

`(operation) is the iteration of the preceding one. So:`

`5) 10 [5] 10 = 10 [4] (10 [4] (10 [4] (10 [4] (10 [4] (10 [4] (10 [4]`

`(10 [4] (10 [4] (10 [4] 10))))))))))`

`6) 10 [6] 10 = 10 [5] (10 [5] (10 [5] (10 [5] (10 [5] (10 [5] (10 [5]`

`(10 [5] (10 [5] (10 [5] 10))))))))))`

etc.

`Those are unimaginably BIG numbers. Archimedes biggest number 10^63 is`

`far greater than 10 [3] 10 (that is 10 ^ 10), but "infinitesimal"`

`compared to 10 [4] 10, which is itself "infinitesimal" compared to 10`

`[5] 10, not to speak of 10 [1000] 10 ....`

`So if you give something like (10 [10] 10) [100] (10 [10] 10),`

`having defined the [i] for i = 1 to 100, then you are specifying a`

`*very very very* huge number.`

`I give, for all, one last exercise before introducing diagonalization:`

`define recursively in an explicit way the operation [i+1] from the`

`preceding operation [i]. If you know a "computer language" (Fortran,`

`Lisp, Prolog, c++, Java, whatever ...) write the program. If you don't`

`know any such language, read my "combinators posts" and program those`

`function with S and K, (if you have the time). Well, just be sure you`

`follow the idea.`

Must leave now. Bruno Le 21-mai-06, à 09:08, Tom Caylor a écrit :

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

http://iridia.ulb.ac.be/~marchal/ X-Google-Language: ENGLISH,ASCII Received: by 10.54.160.18 with SMTP id i18mr191684wre; Mon, 22 May 2006 05:29:35 -0700 (PDT) Return-Path: <[EMAIL PROTECTED]> Received: from bonito.ulb.ac.be (bonito.ulb.ac.be [164.15.59.220]) by mx.googlegroups.com with ESMTP id v23si963617cwb.2006.05.22.05.29.34; Mon, 22 May 2006 05:29:35 -0700 (PDT) Received-SPF: pass (googlegroups.com: best guess record for domain of [EMAIL PROTECTED] designates 164.15.59.220 as permitted sender) Received: from mach.vub.ac.be (maxi.ulb.ac.be [164.15.128.8]) by bonito.ulb.ac.be (Postfix) with ESMTP id 9EDD72F for <everything-list@googlegroups.com>; Mon, 22 May 2006 14:29:33 +0200 (CEST) Received: by mach.vub.ac.be (Postfix, from userid 21099) id 7E9C78D27; Mon, 22 May 2006 14:29:33 +0200 (CEST) Received: from [164.15.10.83] (post.ulb.ac.be [164.15.10.83]) by mach.vub.ac.be (Postfix) with ESMTP id 1AE668D02 for <everything-list@googlegroups.com>; Mon, 22 May 2006 14:29:32 +0200 (CEST) Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable In-Reply-To: <[EMAIL PROTECTED]> References: <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> Message-Id: <[EMAIL PROTECTED]> From: Bruno Marchal <[EMAIL PROTECTED]> Subject: Re: Smullyan Shmullyan, give me a real example Date: Mon, 22 May 2006 14:29:26 +0200 To: everything-list@googlegroups.com X-Mailer: Apple Mail (2.623) X-Spam-Checker-Version: maxi.ulb.ac.be SA 2.63 (2004-01-11) X-Spam-Status: No, hits=0.0, level Hi Tom,

`Apparently you have (re)discover Ackermann function which indeed`

`provide formally a sequence of more and more growing functions, similar`

`to the sequence I was pointing too.`

`I will present it in a easier way for the benefit of the others, but`

`also for using a presentation which will facilitate the future`

`successive diagonalizations. In your second recent post you got the`

`first diagonalization right (or almost right) but the diag is quite`

`hidden and it would be hard to continue the process.`

(Note that you could have define f(0, m, n) as the successor function)

`So your sequence of functions are, in your "Ackerman like parametrized`

`presentation", the functions`

`f(1, m, n) is m + n (or m [1] n, writting "+"`

`as [1] ; for first function)`

f(2, m, n) is m * n (or m [2] n) f(3, m, n) is m ^ n (or m [3] n) f(4, m, n) is m [4] n f(5, m, n) is m [5] n etc

`It will be easier to diagonalize functions of one argument, that is x +`

`x, x * x, x ^ x, x [4] x, x[5]x, etc.`

Let us see their values on 10: 1) 10 + 10 = 20 2) 10 * 10 = 10+10+10+10+10+10+10+10+10 = 100 (ten sums) 3) 10 ^ 10 = 10*10*10*10*10*10*10*10*10 = 10000000000 (ten products)

`4) 10 [4] 10 = 10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^ (10 ^`

`10))))))))) (ten exponentiations) Note the parenthesis:`

`schoolboys/girls know that a ^ (b ^ c) is in general different and`

`bigger than (a ^ b) ^ c .`

`Note also that 10 [4] 10 is already so big that the known observed part`

`of the physical universe is not enough big to write its digits. [4] is`

`called tetration, and 10 [4] 10 can be read 10 tetrated up to 10 (like`

`10 ^ 10 can be read as 10 up to the power of 10). Each function`

`(operation) is the iteration of the preceding one. So:`

`5) 10 [5] 10 = 10 [4] (10 [4] (10 [4] (10 [4] (10 [4] (10 [4] (10 [4]`

`(10 [4] (10 [4] (10 [4] 10))))))))))`

`6) 10 [6] 10 = 10 [5] (10 [5] (10 [5] (10 [5] (10 [5] (10 [5] (10 [5]`

`(10 [5] (10 [5] (10 [5] 10))))))))))`

etc.

`Those are unimaginably BIG numbers. Archimedes biggest number 10^63 is`

`far greater than 10 [3] 10 (that is 10 ^ 10), but "infinitesimal"`

`compared to 10 [4] 10, which is itself "infinitesimal" compared to 10`

`[5] 10, not to speak of 10 [1000] 10 ....`

`So if you give something like (10 [10] 10) [100] (10 [10] 10),`

`having defined the [i] for i = 1 to 100, then you are specifying a`

`*very very very* huge number.`

`I give, for all, one last exercise before introducing diagonalization:`

`define recursively in an explicit way the operation [i+1] from the`

`preceding operation [i]. If you know a "computer language" (Fortran,`

`Lisp, Prolog, c++, Java, whatever ...) write the program. If you don't`

`know any such language, read my "combinators posts" and program those`

`function with S and K, (if you have the time). Well, just be sure you`

`follow the idea.`

Must leave now. Bruno Le 21-mai-06, à 09:08, Tom Caylor a écrit :

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

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