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 Laws and Orders GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 04/27/2008, 10:18 PM By the way, I have no sponsor at all! My wife says that I am no more able to bring money at home. Unfortunately ..., she is completely right! GFR bo198214 Administrator Posts: 1,394 Threads: 91 Joined: Aug 2007 04/28/2008, 08:09 AM GFR Wrote:However, I have always doubts, about ... everything. Sorry about that. "Ich bin der Geist der stehts verneint". Oh and I thought that was me! Apparently it looks more like that, doesnt it *ggg* !? Quote:By the way, I have no sponsor at all! My wife says that I am no more able to bring money at home. Unfortunately ..., she is completely right! So thats luck! I have not even a wife, not to mention a sponsor (*whispering* But perhaps you can tell me, what you do with all the money before you reach your home!) Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 04/28/2008, 01:39 PM Ivars Wrote:I think I know what is "imagination" . It is , even if I know you will say it can not be , related to ordering/enumeration in time. As opposed and complementing to ordering in space, or dimension(s) represented by real hyperoperations. Ivars Just to make a note here that this thinking has lead me to causal sets, adn that seems nice as someone else has already thought to bring time into mathematics, so it is reading again..... Causal Sets bo198214 Administrator Posts: 1,394 Threads: 91 Joined: Aug 2007 04/28/2008, 07:00 PM Ivars Wrote:I think I know what is "imagination" . It is , even if I know you will say it can not be , related to ordering/enumeration in time. As opposed and complementing to ordering in space, or dimension(s) represented by real hyperoperations. .... Just to make a note here that this thinking has lead me to causal sets, adn that seems nice as someone else has already thought to bring time into mathematics, so it is reading again..... Causal Sets However we exploring new grounds in mathematics and not in physics. The only relation to physics I can see so far is that you have an axis $ict$ in special relativity. However mostly physics uses mathematics and not vice versa. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 04/29/2008, 07:46 AM bo198214 Wrote:However we exploring new grounds in mathematics and not in physics. The only relation to physics I can see so far is that you have an axis $ict$ in special relativity. However mostly physics uses mathematics and not vice versa. You can always choose $ct=1$ and drop dimensionality. For me, it helps to understand math by "imagining" $I$ as continuos imaginary time that has certain relationships with other numbers which also might have physical interpretations. It might not work always, but also its not forbidden to use such analogy for certain dynamic understanding of mathematics, I guess. Ivars GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 04/30/2008, 10:49 AM bo198214 Wrote:(... perhaps you can tell me, what you do with all the money before you reach your home!)It (the money) just collapses and I implode!! It is like the my hypothetical "sublime mathematical nonsense" operator, such as: Smn(x) := a[0]<0>(x) >< x (neq x). The Smn() operator would be such that it could modifiy any variable x, only when (IFF) it IS NOT applied to it (or applied with ZERO iterations). Actually, this could give an idea of what "imagination" might be. Unfortunately, it was "imagined" at the rank s=0 level. Smn(x) would not even be the contrary of the identity operator, because: Id(x) = x, i.e.: all x's are fixpoints of Id(x) ... Nid(x) >< x, i. e.: Nid(x) has no fixpoint ! ... In fact, both Id() and Nid() are applied at least once, while Smn() is supposed to refer to a fhypothetical a[0]() operator, applied only (so to say) ZERO times. Sublime nonsense indeed!!! Sorry about that. Having said that, I remain, Confusingly yours GFR Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 05/01/2008, 12:21 PM (This post was last modified: 05/01/2008, 12:28 PM by Ivars.) GFR Wrote:Smn(x) := a[0]<0>(x) >< x (neq x). The Smn() operator would be such that it could modifiy any variable x, only when (IFF) it IS NOT applied to it (or applied with ZERO iterations). Actually, this could give an idea of what "imagination" might be. Unfortunately, it was "imagined" at the rank s=0 level. Smn(x) would not even be the contrary of the identity operator, because: Id(x) = x, i.e.: all x's are fixpoints of Id(x) ... Nid(x) >< x, i. e.: Nid(x) has no fixpoint ! ... In fact, both Id() and Nid() are applied at least once, while Smn() is supposed to refer to a fhypothetical a[0]() operator, applied only (so to say) ZERO times. Sublime nonsense indeed!!! Hi GFR, I think "imagination" will reveal more about imaginary unit than the fact that it is +- sqrt(-1) although already this is telling more than we appreceate. When I think about I as imaginary continuos time it is like time which is applied to events which You imagine in Your head- so its nor really applied to real events, but still the imaginary sequency of e.g memories or future vision has its own time. In this sense, the copies or partial reflections of real or potentially real or even unreal events in Your mind will have Imaginary timing which will partly correspond to real time , partly have imaginary component which has no real counterpart. In that sense (which is unclear) its a little similar to Your operator? Another way to find out what "imagination""could be is to take real continuous (!) operation (we do not have them yet) , add small imaginary part to it, then slowly reduce real part to 0 and try to understand what it means. Or , start with interpretation of real continuous iterations t, ( that is possible) , add small imaginary part, grow it to I and than reduce real part to 0, so that t=I in the end. With this mathematically and generally confusing statement, I'll be offline till May 7th. Ivars GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 05/02/2008, 05:55 PM bo198214 Wrote:GFR Wrote:Concerning the DL, I prefer my "implication", i.e.: IF a[s]x = x, THEN x = a[s+1]oo Gianfranco, that is not true! Take $a=\sqrt{2}$ and $x=4$ then surely $a[3]x=x$. But $4\neq 2 = a[4]\infty$. You can not bend mathematics to suit your wishes.I am not bending anything and I don't have any particular wish to satisfy, in this context. Could you please explain to me why, from: a[3]x = x, we should not have, by simple substitutions: x = a[3](a[3](a[3](a[3]x))), and therefore: x = a[4]oo ??? Then, why I cannot say that: a[3]x = x ----> x = a[4]oo ?? Concerning your classical example, by putting: a = sqrt(2), we indeed have: (sqrt(2))^x = x ----> x = (sqrt(2))#oo = h = plog(-ln x) / (-ln x). We know that, for 1 < a < e^(1/e), h has two real solutions, one for plog(z) = W/-1 (z) and one for plog(z) = W/0 (z). These two solutions can be easily calculated by Mathematica and they are: h/inf = 2 and h/sup = 4. In my non-orthodox notation, I would put: h = {2,4}. Value h/inf = 2 is an asymptotical value of x = a#n, "reached" when n->oo. Value h/sup = 4 is an old mystery that has nothing to do with my mathematical weaknesses. It is not "reachable", but you must accept that the two values satisfy the selfroot condition: selfrt(4) = selfrt(2) = sqrt(2). You see, Henryk, you are certainly right to agitate the flag of mathematical orthodoxy. But,alway remaining calm and cool !!! GFR bo198214 Administrator Posts: 1,394 Threads: 91 Joined: Aug 2007 05/02/2008, 09:02 PM (This post was last modified: 05/02/2008, 09:05 PM by bo198214.) GFR Wrote:Could you please explain to me why, from: a[3]x = x, we should not have, by simple substitutions: x = a[3](a[3](a[3](a[3]x))), and therefore: x = a[4]oo ??? Because $a[4]\infty = \lim_{n\to\infty} \exp_a^{\circ n}(1)$ (note that $a[4]n = \exp_a^{\circ n}(1)$) and usually $a[4]\infty \neq \lim_{n\to\infty} \exp_a^{\circ n}(x)$. Quote:Concerning your classical example, by putting: a = sqrt(2), we indeed have: (sqrt(2))^x = x ----> x = (sqrt(2))#oo. Gianfranco, I exactly explained before that if you put $x=4$ this is not true. I really dont know how to explain this even more simple. The left side is an equation with multiple solutions for $x$ on the right side of the implication there is a limit it can take only one value (out of the multiple solutions of the left side). In the case of a[4] the right side is the lower real fixed point of $a^x$, if existing, otherwise it is $\infty$. So if you chose the upper real fixed point or a complex fixed point, then the right side is no more true (though the left is). This really has nothing to do with orthodox or non-orthodox mathematics but with clear thinking. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 05/02/2008, 09:47 PM OK! Let me stop here. Thank you very much. GFR « Next Oldest | Next Newest »

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