• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 fascinating average tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/29/2011, 12:30 PM im currently fascinated by these sequences of " tetration averages ". for integer a > 2 exp^[n]((ln^[n](a) + ln^[n](a+1))/2) for n = 0 its AM for n = 1 its GM the average , real part and im part , inf real , sup im , infinite sum over n etc all seem intresting to me. in particular for a = 17. maybe there are simple answers. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/02/2011, 03:05 PM (03/29/2011, 12:30 PM)tommy1729 Wrote: im currently fascinated by these sequences of " tetration averages ". for integer a > 2 exp^[n]((ln^[n](a) + ln^[n](a+1))/2) for n = 0 its AM for n = 1 its GM the average , real part and im part , inf real , sup im , infinite sum over n etc all seem intresting to me. in particular for a = 17. maybe there are simple answers. using logarithmic semi-operators, for 0 <= q <= 1, your "fascinating average" is given as: (n {q} n+1) {1+q} 1/2 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/06/2011, 03:55 PM with respect , but i prefer my notation , also because it is similar to the notation others have used here and is hence imho somewhat standard notation on this forum. have you found anything intresting about it ? regards tommy1729 JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/06/2011, 08:51 PM (This post was last modified: 04/06/2011, 08:57 PM by JmsNxn.) yes, actually I have. I've proved, using logarithmic semi operator notation (0<= q < 1): if {-q} is "lowered addition", {1-q} is "lowered multiplication", {2-q} is lowered "exponentiation"; All follow the law of recursion. Also }-q{ is "lowered subtraction", }1-q{ is "lowered division", }2-q{ is "lowered roots". and if S(p) is the identity function, such that, p E R x {p} S(p) = x we find, just as 0 (the additive identity) multiplied by x is 0, S(-q) (the "lowered" additive identity) {1-q} multiplied by x is S(-q), so it behaves just like multiplication by zero. This means, the inverse function }1-q{ division has a pole at S(-q), just like division by zero. Therefore we can create a new calculus as such q:d/dx f(x) = lim h -> S(-q) [f(x {-q} h) }-q{ f(x)] }1-q{ h which is basically just the normal difference quotient with lowered operators and h approaching a lowered additive identity. using this and a lot of further digging, I can prove that q:d/dx e^x = e^x If you care I can explain how (it involves a generalization of taylor series). I didn't mean to tell you to use my notation or anything, I just saw what you posted and that's what popped into my head. some of your ideas seem to gravitate around logarithmic semi operators tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/06/2011, 11:10 PM (04/06/2011, 08:51 PM)JmsNxn Wrote: yes, actually I have. I've proved, using logarithmic semi operator notation (0<= q < 1): if {-q} is "lowered addition", {1-q} is "lowered multiplication", {2-q} is lowered "exponentiation"; All follow the law of recursion. Also }-q{ is "lowered subtraction", }1-q{ is "lowered division", }2-q{ is "lowered roots". and if S(p) is the identity function, such that, p E R x {p} S(p) = x we find, just as 0 (the additive identity) multiplied by x is 0, S(-q) (the "lowered" additive identity) {1-q} multiplied by x is S(-q), so it behaves just like multiplication by zero. This means, the inverse function }1-q{ division has a pole at S(-q), just like division by zero. again with respect , but that is trivial and already known to me and i believe all regular posters. then again it might not have been explictly posted and some newbies might learn from it. ( i dont wanna sound unthankfull for a well-intented and true fact posted ) (04/06/2011, 08:51 PM)JmsNxn Wrote: Therefore we can create a new calculus as such q:d/dx f(x) = lim h -> S(-q) [f(x {-q} h) }-q{ f(x)] }1-q{ h which is basically just the normal difference quotient with lowered operators and h approaching a lowered additive identity. using this and a lot of further digging, I can prove that q:d/dx e^x = e^x If you care I can explain how (it involves a generalization of taylor series). I didn't mean to tell you to use my notation or anything, I just saw what you posted and that's what popped into my head. some of your ideas seem to gravitate around logarithmic semi operators if i got the notation right , i think i know this as well , but im not sure because of the notation. as for the generalization of the taylor series i had some ideas , possibly similar possibly not , of generalizing taylor series too in the hyperoperator context. so im all ear for your taylor series - i intended to post about it but canceled it for some reasons ( e.g. doubt of use ) - but i cannot promise i will like it or not. its easier for me if you work with more commen notation. regards tommy1729 JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/07/2011, 01:02 AM (This post was last modified: 04/07/2011, 01:05 AM by JmsNxn.) (04/06/2011, 11:10 PM)tommy1729 Wrote: (04/06/2011, 08:51 PM)JmsNxn Wrote: yes, actually I have. I've proved, using logarithmic semi operator notation (0<= q < 1): if {-q} is "lowered addition", {1-q} is "lowered multiplication", {2-q} is lowered "exponentiation"; All follow the law of recursion. Also }-q{ is "lowered subtraction", }1-q{ is "lowered division", }2-q{ is "lowered roots". and if S(p) is the identity function, such that, p E R x {p} S(p) = x we find, just as 0 (the additive identity) multiplied by x is 0, S(-q) (the "lowered" additive identity) {1-q} multiplied by x is S(-q), so it behaves just like multiplication by zero. This means, the inverse function }1-q{ division has a pole at S(-q), just like division by zero. again with respect , but that is trivial and already known to me and i believe all regular posters. then again it might not have been explictly posted and some newbies might learn from it. ( i dont wanna sound unthankfull for a well-intented and true fact posted ) (04/06/2011, 08:51 PM)JmsNxn Wrote: Therefore we can create a new calculus as such q:d/dx f(x) = lim h -> S(-q) [f(x {-q} h) }-q{ f(x)] }1-q{ h which is basically just the normal difference quotient with lowered operators and h approaching a lowered additive identity. using this and a lot of further digging, I can prove that q:d/dx e^x = e^x If you care I can explain how (it involves a generalization of taylor series). I didn't mean to tell you to use my notation or anything, I just saw what you posted and that's what popped into my head. some of your ideas seem to gravitate around logarithmic semi operators if i got the notation right , i think i know this as well , but im not sure because of the notation. as for the generalization of the taylor series i had some ideas , possibly similar possibly not , of generalizing taylor series too in the hyperoperator context. so im all ear for your taylor series - i intended to post about it but canceled it for some reasons ( e.g. doubt of use ) - but i cannot promise i will like it or not. its easier for me if you work with more commen notation. regards tommy1729 The first part was meant to be very simple. It was just an introduction. Alright, well, i'll try to use standard notation, but you'll see, right from the start how clunky it is and how inappropriate it is but I'll see what I can come up with q:d/dx f(x) = lim h -> exp^[-q](0) exp^[-q]((exp^[q](f(exp^[-q](exp^[q](x) + exp^[q](h)))) - exp^[q](f(x)))/exp^[q](h))) I cannot adopt standard notation to explain the extended taylor series however: therefore q, p E R; 0 <= q < 1 p:ln(x) = exp^[-p](x) x {-q} y = exp^[-q](exp^[q](x) + exp^[q](y)) x {1-q} y = exp^[q](exp^[-q](x) + exp^[-q](y)) x {2-q} y = exp^[q](exp^[-q](x)*y) if $\sum_{n=0}^{\f}\${p} a_n = a_0 {p} a_1 {p} .... a_f Let us first give a few lemmas since {-q} is defined as x {-q} y = q:ln(-q:lnx + -q:lny) x {-q} x = q:ln(-q:ln(x) * 2) = x {1-q} q:ln(2) and by induction x {1-q} q:ln(y) = $\sum_{n=1}^{\y}\${-q} x therefore {-q} is not "proper recursive" with this we can now approach the polynomial knowing the definition of q:d/dx f(x) it is simple to see that it is distributive across {-q} (or lowered addition), and that there is a product rule for {1-q}. therefore, we can create polynomials defined as f(x) = $\sum_{n=0}^{\f}\${-q} a_n {1-q} [x {2-q} n] = a_0 {-q} (a_1 {1-q} x) {-q} (a_2 {1-q} (x {2-q} 2)) ... and therefore to find the q:d/dx of f(x) we only need the q:derivative of x {2-q} n first q:d/dx x {2-q} 2 = lim h -> S(-q) [(x{-q}h){1-q}(x{-q}h) }-q{ x {2-q} 2] }1-q{ h and since {1-q} is distributable across {-q} like multiplication across division it becomes q:d/dx x {2-q} 2 = lim h -> S(-q) [(x {2-q} 2) {-q} (x {1-q} h) {-q} (x {1-q} h) {-q} (h {2-q} 2) }-q{ (x {2-q} 2]) }1-q{ h and if you can siv through that mess you'll see 2 terms (x {1-q} h), but since they're spread across {-q} and not addition, they add up to (x {1-q} h) {-q} (x {1-q} h) = q:ln(2) {1-q} (x {1-q} h) therefore, everything cancels out normally and we're left with q:d/dx x {2-q} 2 = q:ln(2) {1-q} x this can be generalized to q:d/dx x {2-q} n = q:ln(n) {1-q} (x {2-q} (n-1)) therefore, if q:n! = q:ln(1) {1-q} q:ln(2) {1-q} q:ln(3) {1-q} ... q:ln(n) = q:ln(n!) we find that an infinite termed polynomial f(x) = $\sum_{n=0}^{\infty}\${-q} (S(1-q) }1-q{ q:n!) {1-q} (x {2-q} n) would be it's own q:derivative. It is of me to prove that e^x is this infinite termed polynomial. e^x = $\sum_{n=0}^{\infty}\$1/n! * x^n q:ln(e^x) = q:ln($\sum_{n=0}^{\infty}\$1/n! * x^n) = $\sum_{n=0}^{\infty}\$ {-q} (S(1-q) }1-q{ q:n!) {1-q} (q:ln(x) {2-q} n) therefore q:ln(e^(-q:ln(x))) = $\sum_{n=0}^{\infty}\${-q} (S(1-q) }1-q{ q:n!) {1-q} (x {2-q} n) = f(x) therefore f(x) = e^x « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post the " average restart " of a number. tommy1729 0 1,566 03/26/2014, 01:02 AM Last Post: tommy1729

Users browsing this thread: 1 Guest(s)