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 Improper integrals at MSE tommy1729 Ultimate Fellow Posts: 1,424 Threads: 346 Joined: Feb 2009 12/21/2014, 12:48 AM Last Friday while playing chess I talked to my MSE friend mick. You might like this : http://math.stackexchange.com/questions/...-integrals regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,424 Threads: 346 Joined: Feb 2009 12/23/2014, 11:31 PM (This post was last modified: 12/23/2014, 11:41 PM by tommy1729.) Sheldon's answer on MSE is nice. Thank you Sheldon. I made an intresting observation relating things to complex dynamics. The main thing is the mysterious looking change $x$ -> $x - 2^t / x.$ My observation can be considered positive or negative , intresting or dissapointing , it depends on taste I guess and the hope for nontrivial analogues. But the idea of having some function $g(x)$ for which every real iterate $g^{[t]}(x)$ " works " is found , though it might not be as nontrivial as mick hoped. ( not saying a nontrivial case cannot exist ). - Maybe variants of this exist in calculus textbooks / papers but its very " dynamical " in nature - Anyway here it is : $f_x(t) = x - 2^t / x$ $f_x(f_x^{[-1]}(t) + 1)$ => $x - 2^T / x$ with $T = f_x^{[-1]}(t) + 1$ => $x - 2/x * 2^{f_x^{-1} (t)}$ => $x - 2/x * Solve(q,x - q/x = t)$ Solve .. => $q = x(x-t)$ Thus : $x - 2/x *x(x-t) = x - 2(x-t) = x - 2x + t = -x + t$ Which is trivial. Reminds me of this quote : " Young man, in mathematics you don't understand things. You just get used to them. " John von Neumann. Btw I considered doing the things (steps above) in reverse : showing $x$ -> $x - t/x$ is valid from the validity of $x -> -x + t.$ regards tommy1729 " the master " sheldonison Long Time Fellow Posts: 664 Threads: 23 Joined: Oct 2008 12/24/2014, 03:08 PM (This post was last modified: 12/25/2014, 12:01 AM by sheldonison.) (12/23/2014, 11:31 PM)tommy1729 Wrote: ....function $g(x)$ for which every real iterate $g^{[t]}(x)$ " works " is found , though it might not be as nontrivial as mick hoped. ( not saying a nontrivial case cannot exist ). ... $f_x(t) = x - 2^t / x$ $f_x(f_x^{[-1]}(t) + 1)$ => $x - 2^T / x$ with $T = f_x^{[-1]}(t) + 1$ => $x - 2/x * 2^{f_x^{-1} (t)}$ => $x - 2/x * Solve(q,x - q/x = t)$ Solve .. => $q = x(x-t)$ Thus : $x - 2/x *x(x-t) = x - 2(x-t) = x - 2x + t = -x + t$ .... Hey Tommy, Not sure I understood all of that ... But it inspired me to consider the following sequence of functions $f(x)=x-\frac{1}{2x}\;\;\; g(x)=f^{o2}(x)$ $f(x)=x-\frac{1}{4x}\;\;\; g(x)=f^{o4}(x)\;\;\;$ $f(x)=x-\frac{1}{8x}\;\;\; g(x)=f^{o8}(x)\;\;\;$ $f(x)=x-\frac{1}{16x}\;\;\; g(x)=f^{o16}(x)\;\;\;$ ... $\lim_{n \to \infty} f(x)=x-\frac{1}{2^nx}\;\;\; g(x)=f^{o2^n}(x)$ Does g(x) converge, and is it a solution of interest to Mick? If g(x) converges, and it is analytic, then it has a Taylor/Laurent series.... Update:, by brute force, using a lot of computer cycles to estimate the limit, and then turn the coefficents it back into a fraction with power's of 2's... I get the following Laurent series, as the function that Mick might be looking for. $g(x)= x - \frac{1}{x} - \frac{1}{2x^{3}} - \frac{1}{2x^{5}} - \frac{5}{8x^{7}} - \frac{7}{8x^{9}} - \frac{21}{16x^{11}} - \frac{33}{16x^{13}} - \frac{429}{128x^{15}} - \frac{715}{128x^{17}} - \frac{2431}{256x^{19}} - \frac{4199}{256x^{21}} - \frac{29393}{1024x^{23}} - \frac{52003}{1024x^{25}} - \frac{185725}{2048x^{27}} - \frac{334305}{2048x^{29}} - ...$ It would probably be normally expressed as $f(x)=\frac{1}{g(1/x)$ $f(x) = x + x^3 + \frac{3 x^5 }{2 } + \frac{5 x^7 }{2 } + \frac{35 x^9 }{8 } + \frac{63 x^{11} }{8 } + \frac{231 x^{13} }{16 } + \frac{429 x^{15} }{16 } + \frac{6435 x^{17} }{128 } + \frac{12155 x^{19} }{128 } + \frac{46189 x^{21} }{256 } + \frac{88179 x^{23} }{256 } + \frac{676039 x^{25} }{1024 } + \frac{1300075 x^{27} }{1024 } + ...$ update2: This would be compactly expressed via the Abel function as: $\alpha(z)=\frac{-1}{2z^2}\;\;\;\alpha(f(z))=\alpha(z)+1\;\;\;\alpha^{-1}(z)=-\sqrt{\frac{-1}{2z}}$ And then we get: $f(z)=\alpha^{-1}(\alpha(z)+1) \;=\; \frac{z}{\sqrt{1-2z^2}$ Finally, Mick's desired function in closed form would be as follows. With a little algebra, we generate all of the fractional iterates of g(z) as well. Then, using Mick's notation we have the desired g(z,t) function, which has all fractional iterates defined as: $g(z,t)=\; \frac{1}{f(1/z)}\; = \; sqrt{z^2-2t}\; = \; z - \frac{t}{z} - \frac{t^2}{2z^{3}} - \frac{t^3}{2z^{5}} - \frac{5t^4}{8z^{7}} - \frac{7t^5}{8z^{9}} - \frac{21t^6}{16z^{11}} - ... \;\;\;$ for t=1, this is the same as the Laurent series above - Sheldon tommy1729 Ultimate Fellow Posts: 1,424 Threads: 346 Joined: Feb 2009 12/28/2014, 04:40 PM I once had a few threads here were I discussed the need for limits of the form ( a + f(n)/(bn)) ^[n] = C or similar. This seems very much like your limit, maybe you got inspired from me too. Anyways I must say that thread was looking for tetration type functions / limits so in that sense your limit is more " classical ". I was not able to find the threads again but this one is somewhat similar : http://math.eretrandre.org/tetrationforu...ight=limit regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,424 Threads: 346 Joined: Feb 2009 01/11/2015, 02:23 AM This leads to the question Is there a solution F(x) = super of ( x + a - 2^t/(x+b) ) for some real a,b (super with respect to t) such that integral f(F(x)) dx = integral f(x) ? regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,424 Threads: 346 Joined: Feb 2009 01/16/2015, 10:10 PM I think this is an underrated thread. If it turns out true , this gives an intresting connection between dynamical systems and calculus ! regards tommy1729 « Next Oldest | Next Newest »

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