Bruno Marchal wrote: > ... > 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

## Advertising

I would have thought that my previous result captures this: > > 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) > > If we express my "f(i,m,n)" as your "m [i] n", then this would be m [i] n = m [i-1] (m [i-1] (...(m [i-1] n)...) ( [i-1] taken n times ) Or if we just look at "m [i] m" to keep it simpler as you suggest, m [i] m = m [i-1] (m [i-1] (...(m [i-1] m)...) ( [i-1] taken n times ) In terms of a program, in a sort of pseudocode, to compute m [i] n, initialize result to (m [i-1] n) do the following n-1 times set result to (m [i-1] result) end do The input is (m [i-1] n), the end "result" is m [i] n. If we simply want m [i] m, then set the input to (m [i-1] m). Of course in a real computer language you would have to worry about numerical representation and storage. Tom X-Google-Language: ENGLISH,ASCII-7-bit Received: by 10.11.53.63 with SMTP id b63mr128186cwa; Mon, 22 May 2006 09:20:25 -0700 (PDT) X-Google-Token: _qvdtQwAAACtJ1G4Hlrpr8kdXHeLayAm Received: from 199.46.245.234 by j73g2000cwa.googlegroups.com with HTTP; Mon, 22 May 2006 16:20:25 +0000 (UTC) From: "Tom Caylor" <[EMAIL PROTECTED]> To: "Everything List" <everything-list@googlegroups.com> Subject: Re: Smullyan Shmullyan, give me a real example Date: Mon, 22 May 2006 09:20:25 -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]> <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> User-Agent: G2/0.2 X-HTTP-UserAgent: Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.0),gzip(gfe),gzip(gfe) X-HTTP-Via: 1.0 tus-gate5.raytheon.com:8080 (Squid/2.4.STABLE7) Mime-Version: 1.0 Content-Type: text/plain Bruno Marchal wrote: > ... > 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 I would have thought that my previous result captures this: > > 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) > > If we express my "f(i,m,n)" as your "m [i] n", then this would be m [i] n = m [i-1] (m [i-1] (...(m [i-1] n)...) ( [i-1] taken n times ) Or if we just look at "m [i] m" to keep it simpler as you suggest, m [i] m = m [i-1] (m [i-1] (...(m [i-1] m)...) ( [i-1] taken n times ) In terms of a program, in a sort of pseudocode, to compute m [i] n, initialize result to (m [i-1] n) do the following n-1 times set result to (m [i-1] result) end do The input is (m [i-1] n), the end "result" is m [i] n. If we simply want m [i] m, then set the input to (m [i-1] m). Of course in a real computer language you would have to worry about numerical representation and storage. Tom --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-list@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list -~----------~----~----~----~------~----~------~--~---