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 Woon's expansion bo198214 Administrator Posts: 1,539 Threads: 97 Joined: Aug 2007 08/10/2007, 11:53 PM (This post was last modified: 08/13/2007, 05:48 PM by bo198214.) I wonder about Woon's expansion as you gave it: $ \begin{array}{rl} f^{[t]}(x) & = (w + (f - w))^{[t]}(z) = w^t(1 + (f/w - 1))^{[t]}(x) \\ & = w^t\left(\sum_{n=0}^{\infty} \left({t \atop n}\right) \left[\frac{1}{w}f - 1\right]^{[n]}\right)(x) \\ & = w^t\left(\sum_{n=0}^{\infty} \left({t \atop n}\right) \sum_{m=0}^{n} \left({n \atop m}\right)(-1)^{n-m}\left(\frac{1}{w}f\right)^{[m]}\right)(x) \\ & = w^t \sum_{n=0}^{\infty} \left({t \atop n}\right) \sum_{m=0}^{n} \left({n \atop m}\right)(-1)^{n-m} w^{-m} f^{[m]}(x) \end{array}$ What is the parameter $w$??? I mean the fractional iteration of a formal power series f is unique as long as $f_0=0$ and $f_1>0$ and the above derivation would contradict this uniqueness. Let me show the uniqueness of fractional iteration for the example of a compositional square root. Take an arbitrary formal powerseries $F$ and look for the compositional square root $f$, i.e. a formal powerseries such that $f^{\circ 2}=f\circ f=F$. For this we need a formula for the composition of two formal powerseries. If we innocently start computing it: $f\circ g(x)=f(g(x))=\sum_{n=0}^\infty f_n \left(\sum_{k=0}^\infty g_k x^k\right)^n$ we realize that we need the $n$-th power of the powerseries g, i.e. at least a formula for multiplication: $fg(x)=f(x)g(x)=\left(\sum_{n=0}^\infty f_n x^n\right)\left(\sum_{k=0}^\infty g_k x^k\right) =\sum_{n=0}^\infty \sum_{k=0}^\infty f_n g_k x^{n+k} =\sum_{m=0}^\infty \left(\sum_{k+n=m} f_n g_k\right) x^m$ If we generalize this to the multiplication of an arbitrary number of series we get for the coefficients of the $n$-th power $(f^n)_m = \quad\quad\sum_{m_1+\dots+m_n=m} f_{m_1}\dots f_{m_n}$ Then we put this into our composition computation $f\circ g(x)=\sum_{n=0}^\infty f_n \quad \sum_{m=0}^\infty x^m\sum_{m_1+\dots+m_n=m} g_{m_1}\dots g_{m_n} =\sum_{m=0}^\infty x^m \sum_{n=0}^\infty f_n\sum_{\;m_1+\dots+m_n=m} g_{m_1}\dots g_{m_n}$ If we now assume that $g_0=0$ then the sum $\sum_{n=0}^\infty f_n \sum_{\;m_1+\dots+m_n=m} g_{m_1}\dots g_{m_n}$ is finite because for $n>m$ at least one $m_i=0$, which causes $g_{m_i}=0$ and hence the whole product $g_{m_1}\dots g_{m_n}=0$. This gives $f\circ g(x)=\sum_{m=0}^\infty x^m \sum_{n=0}^m f_n\sum_{\;m_1+\dots+m_n=m} g_{m_1}\dots g_{m_n}$ Now lets go back to the solution of $f\circ f=F, f_0=0$ . By the above formula the coefficients must satisfy the following equations $\sum_{n=1}^m f_n\sum_{\;m_1+\dots+m_n=m} f_{m_1}\dots f_{m_n}=F_m$ For $m=1$, the composition formula reduces to $f_1 f_1 = F_1$. If we assume $f_1>0$ then $f_1$ is uniquely determined as $f_1 = \sqrt{F_1}$. For $m\ge 2$ the $m_i$ are all smaller than m, except for n=1, because otherwise all other $m_j=0, j\neq i$. So the only terms containing $f_m$ that can occur on the left side is $f_1f_m$ and $f_mf_1\dots f_1$. And this gives us a mean to recursively define $f_m$, i.e. by $f_m = \frac{1}{f_1^m+f_1}\left(F_m - \sum_{n=2}^{m-1} f_n\sum_{\;m_1+\dots+m_n=m} f_{m_1}\dots f_{m_n}\right)$. This reasoning can be extended to arbitrary natural exponents, and shows us that in the domain of formal powerseries f with $f_0=0$ and $f_1>0$ the fractional iteration is unique. bo198214 Administrator Posts: 1,539 Threads: 97 Joined: Aug 2007 08/18/2007, 09:59 PM (This post was last modified: 08/18/2007, 10:00 PM by bo198214.) Another strange thing with the derivation: $f^{[t]}(x) = (w + (f - w))^{[t]}(z) = w^t(1 + (f/w - 1))^{[t]}(x)$ This is not true for example take $t=2$ $f(f(x)) \neq w f(\frac{f(x)}{w}) = w^2 \frac{f(\frac{f(x)}{w})}{w} = w^2 (1+(f/w-1))^{[2]}(x)$. Is this Woon's derivation? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 09/05/2007, 08:04 PM Yes, the main difference between the one above and Woon's original expansion is that his expansion uses two (-1) factors, whereas I use one (-1) with two exponents... basically the same. For his original paper, you can download it from arxiv.org: http://arxiv.org/abs/hep-th/9707206 (click on PS or PDF) and his formulas are at #18 (for the D operator), #71 (for any operator). Andrew Robbins bo198214 Administrator Posts: 1,539 Threads: 97 Joined: Aug 2007 09/07/2007, 03:13 PM (This post was last modified: 09/07/2007, 03:19 PM by bo198214.) andydude Wrote:Yes, the main difference between the one above and Woon's original expansion is that his expansion uses two (-1) factors, whereas I use one (-1) with two exponents... basically the same. For his original paper, you can download it from arxiv.org: http://arxiv.org/abs/hep-th/9707206 (click on PS or PDF) and his formulas are at #18 (for the D operator), #71 (for any operator). Yes I already had a look at it. His derivation was meant for operators not for functions. His paper mainly deals with fractional differentiation, there you have the differentiation operator D. An operator is usually a linear map in a function space, in this case D maps a function onto its derivative and it is linear as it has the property $D(f+g)=D(f)+D(g)$ and $D(\alpha f)=\alpha D(f)$. So the derivation of Woon is not directly applicable to functions instead of operators. Of course one can chose the power derivation matrix as operator. However the coefficients of the continuous iteration of power series with fixed point at 0 can be obtained in a finite manner. I.e. with no limits involved. Which is not the case in Woons expansion. There it merely works if $A$ is upper triangular with a diagonal of 1's and $w=1$. Because in this (parabolic) case a truncated $A-I$ is nilpotent $(A-I)^n=0$ and so the involved sum is finite. But already it can not be used already for the hyperbolic case. Where you can use the diagonalization instead to obtain a finite solution. Also with infinite sums there is always the question of convergence, which can sometimes not be established for matrices though for example $(1+x)^t$ can be defined also for real $x>1$ despite the series is converging for only $|x|<1$. « Next Oldest | Next Newest »

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