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 Second derivitive of a dynamical system Daniel Fellow Posts: 91 Threads: 33 Joined: Aug 2007 08/23/2007, 07:43 PM Without lose of generality, set the fixed point of $f(x)$ at zero. It is well known that $D f^t(0)=f'(0)^t$. Consider that $D^2 f(g(x)) = f''(g(x)) g'(x)^2 + f'(g(x)) g''(x)$, then $D^2 f(f^{t-1}(x)) = f''(f^{t-1}(x)) (D f^{t-1}(x))^2 + f'(f^{t-1}(x)) D^2 f^{t-1}(x)$. But this gives a functional equation for a geometrical progression solved at $x=0$ by $D^2 f^t(0) = f''(0) \sum_{k=0}^{t-1} f'(0)^{2t-k-2}$. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 08/23/2007, 08:16 PM Daniel Wrote:It is well known that $D f^t(0)=f'(0)^t$. This is not well-known but one requirement for uniqueness. Daniel Fellow Posts: 91 Threads: 33 Joined: Aug 2007 08/24/2007, 06:44 PM The idea here is that the Taylor series for an arbitrary dynamical system $f^t(x)$ can be derived. The second derivitive's dependancy on a geometrical progression results in the different types of fixed points. The higher derivitives also are built up from nested geometrical progressions. The regular formula for geometrical progressions breaks down when the roots of unity are involved which gives an explaination for why therre are different types of fixed points. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 08/24/2007, 07:00 PM Daniel Wrote:The regular formula for geometrical progressionsWhat do you mean by this? Daniel Fellow Posts: 91 Threads: 33 Joined: Aug 2007 08/24/2007, 10:34 PM bo198214 Wrote:Daniel Wrote:The regular formula for geometrical progressionsWhat do you mean by this? That $\sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x}$ unless $x^m=1$ for some integer $m$. For $x=1$ we have $\sum_{k=0}^n x^k = n+1$. Here $x$ is the Lyponouv characteristic - the first derivitive at a fixed point. Note that the regular formula for geometrical progressions blows up for $x=1$ which happens at $a=e^{\frac{1}{e}}$. « Next Oldest | Next Newest »

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