"game" - to calculate 1.00...001 ^^ 0.5 = ? - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: "game" - to calculate 1.00...001 ^^ 0.5 = ? (/showthread.php?tid=272) Pages: 1 2 3 "game" - to calculate 1.00...001 ^^ 0.5 = ? - nuninho1980 - 04/17/2009 Hi! I already investigated to know the parts of cases decimais of number. to remember - 1.000000000000000000001^0.5 =~ 1.000000000000000000000499999999999999999999875\ 00000000000000000006249999999999999999996093750\ 00000000000000273437499999999999999794921875000\ 000000000161132812499999999999869079589843750000000109100342 1.00000000000000000001 ^^ 0.5 = ?? (in 100 cases decimal) 1.000000000000000000001 ^^ 0.5 = ?? (in 100 cases decimal) you do calcuate. good luck! but I already got results with almost exact. !!warning: x ^^ 0.5 =/= ssqrt(x). RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - bo198214 - 04/18/2009 nuninho1980 Wrote:1.00000000000000000001 ^^ 0.5 = ?? (in 100 cases decimal) 1.000000000000000000001 ^^ 0.5 = ?? (in 100 cases decimal) you do calcuate. good luck! but I already got results with almost exact. But this depends strongly on the method you use, doesnt it? RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - nuninho1980 - 04/18/2009 bo198214 Wrote:But this depends strongly on the method you use, doesnt it? you do calculate by new "regular slog". sorry but you are a bit failure because I help this solution and it's "game". lol RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - bo198214 - 04/18/2009 nuninho1980 Wrote:you do calculate by new "regular slog". There are several methods: 1. Regular iteration (which is actually the oldest, was already considered in 1870 by Schröder) 2. via natural slog. 3. matrix power method. RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - nuninho1980 - 04/18/2009 bo198214 Wrote:There are several methods: 1. Regular iteration (which is actually the oldest, was already considered in 1870 by Schröder) 2. via natural slog. 3. matrix power method. 1. you do not because it's old method is NOT equal to new method. 2. no! the base is any number. 3. no! ops! not "regular slog" but yes new "Regular iteration". sorry! http://en.wikipedia.org/wiki/Talk:Tetration/Summary#Evaluation_methods RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - bo198214 - 04/18/2009 Sorry, I dont know what you mean. I never read about a "new" and an "old" regular iteration method. I also dont know what do you want to say with "no". Do you want to say that this is not the method which should be used. Or do you want to say that this method doesnt work. Or that it shouldnt be used generally? *head scratch* RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - nuninho1980 - 04/18/2009 There are several methods: 1. Regular iteration (which is actually the oldest, was already considered in 1870 by Schröder) the "old" Regular iteration is by Schröder until now!?? no, now "new" Regular iteration is by Ansus. you know Ansus (he is the user of this site) 2. via natural slog. no! the base is any number. 3. matrix power method. no! you do calculate by "Regular iteration by Ansus". RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - bo198214 - 04/18/2009 nuninho1980 Wrote:1. Regular iteration (which is actually the oldest, was already considered in 1870 by Schröder) the "old" Regular iteration is by Schröder until now!?? no, now "new" Regular iteration is by Ansus. you know Ansus (he is the user of this site)Its not by Ansus, its by Schröder and Levy. Thatswhy there is the name Schröder function for $\chi(z)=\lim_{n\to\infty} \frac{f^{[n]}(z) - \lambda}{f'(\lambda)^n}$ if the fixed point is $\lambda$. I described the method here, its the same as what Ansus wrote down, except that the focus is on slog and not on sexp. Quote:2. via natural slog. no! the base is any number. Natural slog is a method that works for arbitrary bases $b>1$. It was first described by Walker and rediscovered by Andrew Robbins (andydude on this forum). You can find a description somewhere on the forum. Though I am inclined to rename the method to "intuitive slog/iteration", because in a context like "natural iteration" it can be easily mistaken to mean "iteration by a natural number". Quote:3. matrix power method. no! What means "no"? Does it mean "matrix power method is bad"? Haha, then Gottfried would get quite angry with you! PS: You can quote someone by sourounding the quoted text by Code:[quote] and Code:[/quote]. If you press "reply" it is already sourounded and you only need to edit whats inside the quote. RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - nuninho1980 - 04/18/2009 bo198214 Wrote:Its not by Ansus, its by Schröder and Levy. Thatswhy there is the name Schröder function for $\chi(z)=\lim_{n\to\infty} \frac{f^{[n]}(z) - \lambda}{f'(\lambda)^n}$ if the fixed point is $\lambda$. I described the method here, its the same as what Ansus wrote down, except that the focus is on slog and not on sexp.ok. bo198214 Wrote:Natural slog is a method that works for arbitrary bases $b>1$. It was first described by Walker and rediscovered by Andrew Robbins (andydude on this forum). You can find a description somewhere on the forum. Though I am inclined to rename the method to "intuitive slog/iteration", because in a context like "natural iteration" it can be easily mistaken to mean "iteration by a natural number".I know it. natural superlog is slog_e (x) or sln (x). lol bo198214 Wrote:What means "no"? Does it mean "matrix power method is bad"? Haha, then Gottfried would get quite angry with you!ok. but if you do caculate by Andrew' slog then you don't get or you get result with problem. bo198214 Wrote:PS: You can quote someone by sourounding the quoted text by Code:[quote] and Code:[/quote]. If you press "reply" it is already sourounded and you only need to edit whats inside the quote.I already know it!! lol if you have some difficulty for caculate then I will do 1 solution of result here. RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - bo198214 - 04/18/2009 nuninho1980 Wrote:bo198214 Wrote:Natural slog is a method that works for arbitrary bases $b>1$. It was first described by Walker and rediscovered by Andrew Robbins (andydude on this forum). You can find a description somewhere on the forum. Though I am inclined to rename the method to "intuitive slog/iteration", because in a context like "natural iteration" it can be easily mistaken to mean "iteration by a natural number".I know it. natural superlog is slog_e (x) or sln (x). lolNo, thatswhy I wrote "is a method that works for arbitrary bases $b>1$"!!! Quote:ok. but if you do caculate by Andrew' slog then you don't get or you get result with problem. I would get a result. But actually its not clear why these values are interesting. Quote:I already know it!! lol So why dont you use it then!?