# Tetration Forum

Full Version: Improper integrals at MSE
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Last Friday while playing chess I talked to my MSE friend mick.

You might like this :

http://math.stackexchange.com/questions/...-integrals

regards

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 "
(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
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