Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) ) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:15:35Z http://mathoverflow.net/feeds/question/111066 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111066/is-there-any-techniques-for-solving-a-differential-equation-including-iterated-fu Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) ) frigen 2012-10-30T10:19:11Z 2012-11-02T16:40:13Z <p>I am trying to solve this differential equation but I have no idea how.</p> <p>$f ' (x) = f( f( x ) )$ </p> <p>Although I don't think this differential equation is solvable, I'd like to know if there is any interesting approach on solving a differential equation of this kind, or at least a non-trivial solution of the equation.</p> <p>P.S. I don't think chain rule is useful for this</p> http://mathoverflow.net/questions/111066/is-there-any-techniques-for-solving-a-differential-equation-including-iterated-fu/111073#111073 Answer by Brendan McKay for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) ) Brendan McKay 2012-10-30T11:16:39Z 2012-10-30T11:16:39Z <p>I don't know, but one answer is $f(x)=ax^c$ where $a=\frac12(\sqrt {3}+i){ e^{\frac16\pi\sqrt {3}}}$ and $c=\frac12+\frac12i\sqrt{3}$. Another is obtained by taking the complex conjugate of both $a$ and $b$.</p> http://mathoverflow.net/questions/111066/is-there-any-techniques-for-solving-a-differential-equation-including-iterated-fu/111078#111078 Answer by Alex Gavrilov for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) ) Alex Gavrilov 2012-10-30T12:37:45Z 2012-10-30T12:37:45Z <p>For what I know, the standard method is the Taylor series expansion at a fixed point, i.e. at a point $x=a$ such that $f(a)=a$. </p> http://mathoverflow.net/questions/111066/is-there-any-techniques-for-solving-a-differential-equation-including-iterated-fu/111092#111092 Answer by fedja for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) ) fedja 2012-11-01T02:36:50Z 2012-11-01T02:36:50Z <p>Nothing is new under the Moon... </p> <p><a href="http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&amp;t=321705" rel="nofollow">http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&amp;t=321705</a></p> http://mathoverflow.net/questions/111066/is-there-any-techniques-for-solving-a-differential-equation-including-iterated-fu/111096#111096 Answer by Anixx for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) ) Anixx 2012-11-01T03:06:48Z 2012-11-01T11:02:49Z <p>There are two closed form solutions:</p> <p>$$\displaystyle f_1(x) = e^{\frac{\pi}{3} (-1)^{1/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$ $$\displaystyle f_2(x) = e^{\frac{\pi}{3} (-1)^{11/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$</p> <p>The solution technique can be found in <a href="http://faculty.kfupm.edu.sa/math/akca/papers/cheng.pdf" rel="nofollow">this paper</a>.</p> <p>For a general case, solution of the equation</p> <p>$$f'(z)=f^{[m]}(z)$$</p> <p>has the form</p> <p>$$f(z)=\beta z^\gamma$$</p> <p>where $\beta$ and $\gamma$ should be obtained from the system</p> <p>$$\gamma^m=\gamma-1$$ $$\beta^{\gamma^{m-1}+...+\gamma}=\gamma$$</p> <p>In your case $m=2$.</p> http://mathoverflow.net/questions/111066/is-there-any-techniques-for-solving-a-differential-equation-including-iterated-fu/111179#111179 Answer by Anixx for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) ) Anixx 2012-11-01T17:19:43Z 2012-11-02T16:40:13Z <p>And regarding real solutions to the question, Alex Gavrilov is completely correct. A Taylor expansion at fixed point $p$ gives us the real solution. Existence of this solution is proven in <a href="http://faculty.kfupm.edu.sa/math/akca/papers/cheng.pdf" rel="nofollow">the paper</a> which I already referenced from my another answer.</p> <p>$$f(z)=\sum_{n=0}^\infty \frac{d_n (z-p)^n}{n!}$$</p> <p>where $d_n$ is defined as follows:</p> <p>$$d_0=p$$ $$d_{n+1}=\sum _{k=0}^n d_k \operatorname{B}_{n,k}(d_1,...,d_{n-k+1})$$</p> <p>where $B_{n,k}$ are the <a href="http://en.wikipedia.org/wiki/Bell_polynomials" rel="nofollow">Bell polynomials</a></p> <p>This gives the following starting coefficients:</p> <p>$$d_1=p^2$$ $$d_2=p^3+p^4$$ $$d_3=p^4 + 4 p^5 + p^6 + p^7$$ $$d_4=p^5 + 11 p^6 + 11 p^7 + 8 p^8 + 4 p^9 + p^{10} + p^{11}$$</p> <p>etc.</p> <p>The fixed point $p$ here serves as a parameter, which determines the family of solutions. According the linked theorem, the expansion should converge in the neighborhood of $p$ for $0 &lt; |p| &lt; 1$ or $p$ being a Siegel number.</p>