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 Searching for an asymptotic to exp[0.5] tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 08/15/2015, 10:22 PM (This post was last modified: 08/15/2015, 10:38 PM by tommy1729.) I think Let s = exp^[1/2] For suffic Large positive real x Exp(X + DC) > S( [1 + C / ( x - AC )] s(x) ) > exp(x + BC). Were A , B , C and D are real constants. And A,B,D depend on C. Also suff Large depends on C too ; X >> (a^2 + b^2 + c^2)^4 At least i think. The inspiration and arguments came from fake function theory , hence no new thread. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 08/16/2015, 02:49 PM Update on previous post Im able to prove For sufficiently Large x >> C : S( [1 + C/x^2] s(x) ) = exp(x + o(1)) However this was without fake function theory and not so elegant imho. o is little-o notation here. Regards Tommy1729 The master tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 08/16/2015, 03:23 PM First i considered fake function theory. Then An elementary argument. Those 2 led to the last 2 Posts. However a simple argument is S(x [1 + 1/t(x)] ) > s(x) [1 + 1/t(x)]. Omg So simple. Probably wrote similar stuff before , so my apologies. However S( x [1 + 1/u(x)] ] < s(x) [1 + 1/t(x)] Remains An issue and the Ideas from the last few posts remain intresting. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 08/26/2015, 07:36 PM In the context of TPID 17 1) does this conjecture even hold ?? We have waay to little supporting examples. We have the sqrt(n) factor in semi-exp and exp but for instance double exponential is not confirmed. For instance for SUM x^n / n^(n^n). Im Hoping you guys can help. --- assuming its true --- I noticed Min ( f^2) = min ( f )^2 Yet (1 + f)^2 = 1 + 2f + f^2. therefore A_n = min ( f(x)/x^n ) Is improved by b_n = A_n + 2 sqrt A_n. That is only a minor improvement. I tried to use the same arguments and as you might have guessed Replacing ^2 and sqrt by say ^3 and cuberoot. However it seems to violate ... I even hesitated to post this because intuition can get you in trouble here. Clearly there is something missing here. Yes i know , we approximate the LHS in P(x) < min f With a polynomial degree n , so if we take ^(n/2) we end Up with the largest naively possible root. Because O(x^2)^(n/2) = O(x^n) We cant ALLOW growth smaller then O (x^2) since this violates the conditions f>0,f ' > 0 , f '' > 0. But that does not explain enough. It does show the upperbound factor < O (n/2) But that is very close from O ( ln(n) sqrt(n) ). All intuïtieve logic fails. However i believe we can repeat the b_n argument and thus arrive at the improved C_n = a_n + 2 sqrt a_n + 2 sqrt(a_n + 2 sqrt a_n) + ... ~~ a_n + 2 ln(n) sqrt(a_n). But that is still far from the desired C ( a_n ln(n+1) sqrt(n) ) Its getting weird , I know. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 09/03/2015, 10:31 PM Ok trying to clarity the previous post. G (x) ~ g(0) + f(x)^2 = g0 + g1 x + ... (Taylor expansion) F(x) = f1 x + f2 x^2 + ... From this we can say gn = f1 f#n-1# + f2 f#n-2# + ... + (f#n/2#)^2. However since (min [f / x^(n/2) ]) ^2 = min[f^2 / x^n] , gn is wrongly estimated as gn ~ (f#n/2#)^2. Assuming Ass 1 f#k# f#n-k# < ( f#n/2# )^2 ( notice this depends on the convergeance speed of fn ! ) we thus get the improved estimate g'n ~< (n/2) (f#n/2#)^2. Hence a correcting factor upper bound : n/2. --- Well that is the idea. Handwaving informal and sketchy ... Yes i admit. But still. Issues ?? 1) n is not even. 2) repeating the argument ... As in (q^2)^2. Leading to arbitrary correcting factors ?? 3) similar to 2); replacing ^2 with other functions such as ^5 , also leading to arbitrary correcting factors. Solutions to 1) 2) and 3) are considered but not formal. I hope 1) 2) and 3) make clear what i meant with intuition failure. Also the correcting factor n/2 is far from sqrt ( ln(n) n ). --- Here i considered the n th Taylor polynomials with sqrt. Clearly i Cannot meaningfully take anything beyond ^(2/n) since ( X^2 )^(n/2) = x^n. --- Hope this clarifies a bit. Sorry for the late reply , but this subject is tricky ! Since Ass 1 depends on converg speed im not even sure that TPID 17 is correct. On the other hand it holds for semi-exp and exp and sheldon believes in a very similar variant. I must be missing something. The importance of the decending chain condition for the derivatives perhaps ?? --- Maybe i know someone who could help us here. --- Pls inform me if its much simpler then i think but i guess it is not. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 09/05/2015, 08:16 AM About issues 2) and 3) and in general ... Suppose we want the correcting factors c_n for f(x). Since a_n depends on the truncated fake Taylor polynomial of degree n , At best we can take ^(2/n) < as explained before >. But there is another upperbound. F(x)^q needs to satisfy the fundamental conditions D^a [F(x)^q] > 0 for a E (0,1,2). In particular a = 2. This places An upper bound on q INDEP of n but DEP on the values and rate of descent of the a_n. And this is the balance we look for : the faster a_n descends , the larger q is. And vice versa. There we cannot " repeat the argument " as much as we want , nor choose any m-th root we want ( or other function ). This gives hope for proving results of type Correcting factors ~< O ( ( ln(n) n)^gamma ). gamma ~ 1/2 is then close to a proof of TPID 17. I call Q = 1/q the power level of f(x). Guess this clarifies alot. Im not sure how this relates to sheldon's integrals , Hadamard products and zeration [ min,- algebra ] yet. Although I have Some Ideas ... Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 09/09/2015, 12:17 PM (This post was last modified: 09/09/2015, 12:19 PM by tommy1729.) However Some functions have a power level of infinity. And this makes it nontrivial to even decide if this strategy is helpful. More investigation is neccessary. Unfortunately i lack time. For example exp(x) has a power level of infinity. Exp(x)^a = exp(a x). The analogue idea of using semi-logarithms instead of sqrt comes to mind. Another idea is to write F(x) = exp(a(x)) b(x) With b(x) growing slower than exp. Then repeat if necc with a(x) until we get functions a*(x),b(x) that grow slower then exp , and then use the power level tricks on them. [ this assumes F(x) grows slower then some power tower exp^[k](x). ] Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 09/12/2015, 01:14 AM (This post was last modified: 09/12/2015, 01:25 AM by tommy1729.) To give An example that should work. F(x) = Sum a_n x^n With a_n = exp(-n^2) To compute the fake a_n we consider F_n(x) = Sum^n a_i x^i Now we solve (Alpha x^2)^[b] = a_n x^n So alpha is close to a_n ^ (log_2(n-1))^(-1). Or alpha ~ exp( - n^2 / log_2(n-1) ) And b ~ ln(n-1)/ln(2). -- Max_x Integral_0^x exp(- t^2) exp(- (x-t)^2 ) dt = Exp(- x^2/2 ) C. Where C is a constant ( upper bound constant with respect to x ). Therefore the correcting factor for taking sqrt is C. And thus the correcting factor for taking r th root is C^log_2{r}. Or r^( ln{C}\ln(2) ). From the computation of b we get max{r} ~ 2^b. Therefore D_n = (n-1)^( ln{C}\ln(2) ) Where D_n is An upper bound on the (final) correcting factor for a_n. -- So we should have Exp(- n^2) =< D_n min( f(x) / x^n ). -- Something like that. It looks a bit like using convolution and fourrier analysis , but its different. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 09/14/2015, 01:30 AM (This post was last modified: 09/14/2015, 01:38 AM by tommy1729.) Let f(x) be a real-entire function with all derivatives > 0 and f(0) >= 0. Let C be the Cauchy constant ( 1/ 2 pi i ). The taylor coëfficiënts are given by the contour integral [1] C * $\oint x^{-n-1}f(x) \; \;$ The estimate from fake function theory, Min (f(x) x^{-n}) can also be given by a contour integral Let g(x,n) = f(x) / x^n. then [2] Min (f(x) x^{-n}) = C * $\oint g(x,n) g ' ' (x,n)/g ' (x,n)$ So the correcting factors are given by Cor(n) = [1]\[2] = $\frac{ \oint x^{-n-1}f(x) \; \; }{ \oint g(x,n) g ' ' (x,n)/g ' (x,n)}$ So the question becomes to estimate , bound or simplify [1]\[2]. Not sure how to proceed here. But now we have a reformulation in terms of more standard calculus ; in terms of (contour) integration. I call this the " ratio formulation " and TPID 17 can be expressed in it. Im aware I did not mention alot of related things such as the specification of the contours , numerical methods , Laplace etc etc. Certainly special cases can be solved but a general idea is missing. I was able to prove / disprove the expressibility in similar cases , but contour integration is a bit trickier then " ordinary " integrals. Ideal would be if we could express this ratio as a single contour. But im not sure if that is possible. While considering that, the idea of " contour derivative " [1]\[2] Comes to mind. For Some of you - or most - this was already clear I assume. But for completeness I make this post. Also Sheldon has similar ideas and I am not sure how exactly they relate ... Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 09/18/2015, 11:31 PM I think Mick did a good job posting the problem on MSE. Here is the link : http://math.stackexchange.com/questions/...r-dx-n-n-0 No votes or reactions yet. Regards Tommy1729 « Next Oldest | Next Newest »

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