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 Searching for an asymptotic to exp[0.5] tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 11/03/2014, 01:26 PM Fake function theory has been used mainly for non-entire functions. But it can also be used for entire functions with some negative derivatives. (as mentioned in the previous post) And it can even be used to estimate entire functions that have all derivatives positive. For those reasons , it seems likely that fake function theory will also have applications in number theory. Lets take the simplest case : exp(x). Using sheldon's post 9 (S9) a_n x^n = exp(x) ln(a_n) + n ln(x) = x ln(a_n) = min ( x - n ln(x) ) d/dx [ x - n ln(x) ] = 1 - n/x 1 - n/x = 0 x = n min ( x - n ln(x) ) = n - n ln(n) ln(a_n) = n - n ln(n) a_n = e^n/n^n This approximates 1/n! Although not so good. My Q9 gives a different result. Further investigation is needed. Error term theorems are wanted. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 11/03/2014, 10:49 PM I noticed Q9 must give about the same result as S9 for the exp(x) in the sense that There exists fixed positive real constants A,B such that A < S9(n)/Q9(n) < B for all positive integer n. [ S9(n) is the sheldon post 9 estimate of a_n and Q9(n) is my Q9 estimate of a_n ] Therefore it seems intuitive to conjecture The SQ conjecture : A < S9(n)/Q9(n) < B where the fixed positive real values A,B only depend on the function considered. *** I wonder what S9 and Q9 give for the J(x) function. ( the J(x) function that estimates the binary partition function discussed before ... J for Jay D fox ) If there exists a counterexample to the SQ conjecture that is entire , then it probably has to do alot with convergeance speed. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 11/03/2014, 11:34 PM (This post was last modified: 11/03/2014, 11:35 PM by tommy1729.) For entire functions with positive derivatives such that 0 < a_(n+1) =< a_n conjecture P : There exists a fixed polynomial P(n) such that (S9(n) / D^n f(x))^2 < P(n). If I recall correctly for f(x) = exp(x) we have P(n) = e n. Analogue for Q9(n) ... which follows from the SQ conjecture posted before. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 11/03/2014, 11:39 PM Crazy idea : deg P(n) = growth f(x) ? regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 11/04/2014, 09:41 AM One wonders how well fake function theory will approximate the bell numbers ... regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 11/04/2014, 10:38 PM a_n x^n = exp(exp(x)-1) ln(a_n) + n ln(x) = exp(x)-1 ln(a_n) = Min[exp(x) - 1 - n ln(x)] d/dx exp(x) - 1 - n ln(x) = exp(x) - n/x exp(x) = n/x x = W(n) ln(a_n) = exp(W(n)) - 1 - n ln(W(n)) ln(a_n) = exp(W(n)) - 1 - n ln(n) - n W(n) *** Need to compare this [ ln(a_n) ] to ln(B_n) - lngamma(n+1) for which many estimates exist. However its not immediately clear which one to use. Also a simplification of exp(W(n)) or exp(exp(W(n))) would be useful. Does that exist ? --- Many years ago I considered many forms of generalized bell numbers. Together with " galathaea " and Han de Bruin this was partially discussed on sci.math. One of those generalizations was B_n(m) = D^n dexp^[m](x) where n and m are positive integers , dexp is the decremented exponential [exp(x)-1] , D^n denotes the n th derivative and ^[m] is the m th iterate. This might intrest you. Its probably the most logical generalization from the viewpoint of tetration. I notice that NOT MUCH IS EVER WRITTEN ABOUT TRIPLE EXPONENTIALS IN MATH ! --- More investigation is needed. It seems hard to automate these ideas into a program or such. On the other hand there is a clear direction to these ideas and inspiration is just around the corner. regards tommy1729 " Truth is what does not go away when you stop believing in it " tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 11/05/2014, 11:58 PM POST 102 correction ! ln(a_n x^n) = ln(exp(x)sqrt(x)) ln(a_n) + n ln(x) = x + 0.5 ln(x) ln(a_n) = min ( x - (n-1/2) ln(x) ) d/dx ( x - (n-1/2) ln(x) ) = 1 - (n-1/2) / x 1 = (n-1/2) / x x = n-1/2 ln(a_n) = n - 1/2 - (n - 1/2) ln(n - 1/2) NOTICE this is equivalent to the solution for exp , but n replaced by n - 1/2 ... which is the analogue of the mittag-leffler function solution !! BTW notice Im aware of sheldon's scaling from post 9 , however working with that is not so easy yet. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 11/07/2014, 12:27 AM I have considered many more things then I could ever post ... I probably said it before but Im considering " fake analytic number theory ". That might take some time to develop. Here are some other ideas that inspire me and for which Im currently not sure how to continue. --- Jay's function that approximates the binary partition function has become popular here. I wonder about variants of type $\Sigma (2^{n*(n-1)*(n+1)/6} n!)^{-1} x^n$ DO these variants of " cubic type " solve anything ? --- *** Initially fake function methods start with an OVERESTIMATE. It is possible to use a method that is both GOOD and starts with an UNDERESTIMATE ? *** ### Im wondering about fake fourier series and fake integrals. A logical way , but possibly not the best , is to consider the fake function methods as fake n'th derivatives. Then the fake integral is computed as the fake first derivative of the true second integral. Or something like that ... ### ||| Lets try Jay's function J(x). a_n x^n = J(x) ln(a_n) + n ln(x) = ln(J(x)) ln(a_n) = min[ ln(J(x)) - n ln(x) ] d/dx [ ln(J(x)) - n ln(x) ] = J(x/2)/J(x) - n/x J(x/2)/J(x) = n/x I really like the shape of this equation. And ofcourse I wonder how good a_n will be compared to $(2^{n*(n-1)/2} n!)^{-1}$ ALthough Im not completely stuck here , Im also not completely sure how to proceed. Use asymptotics , invent new special function , use contour integrals , numerical methods , ... ? @@@ Under some trivial conditions I consider some ideas to improve finding a fake function without actually changing the method ... example Find fake exp(x). Note : We already discussed the idea that we already have an entire function with positive derivatives , yet this is intresting. And just an example , it applies to non-entire functions too. Instead of a_n x^n = exp(x) We solve for a_n x^n = 2 exp(x) / (1+x) Then we multiply our result (taylor series with a_n , possibly scaled ) with (1+x)/2. Notice multiplication by (1+x)/2 does not change the signs of the derivatives and is easy to compute ! The q-variant of this is also possible of course. I call these methods sx9(n) and qx9(n). So for instance $qx9 := (1+x)/2 * Qfake[ 2 f(x)/(1+x) ]$ Much more work needs to be done ! @@@ Comments and help is appreciated. regards tommy1729 " Formally define useful and useless , but beware : take into account we are plain mortals and your an atheist who claims to be not obsessed by money nor by ego " tommy1729 @sci.math tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 03/28/2015, 11:11 PM Im not sure but have we considered showing that the zero's of the fake exp^[1/2] all have multiplicity 1 ? regards tommy1729 marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 03/29/2015, 12:59 PM Anybody estimated the fourier series of $\vspace{15}{^xa \,\,\, for \,\,\, -1?? « Next Oldest | Next Newest »

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