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 Mizugadro, pentation, Book MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 02/10/2015, 12:12 PM Ok, let's try it. LEFT SIDE Substituting it in $F(z+1)$ it gives $\displaystyle F(z) = L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N)$ $\displaystyle F(z+1) = L+\sum_{n=1}^{N} a_n {e^{k(z+1)n}} + o({e^{k(z+1)N}})$ $\displaystyle F(z+1)=L+\sum_{n=1}^{N}(a_n{e^{kn}})\varepsilon^n+ o(\varepsilon^N{e^{kN}})$ Let's pretend for a second that I know how the little-o notation works (even if I don't even know what it is), now we have a new series that instead of the coefficents $a_n$ has the coefficents $b_n=a_n e^{kn}$ that looks really similar to the one we started with $\displaystyle F(z+1)=L+\sum_{n=1}^{N}b_n\varepsilon^n+ o(\varepsilon^N{e^{kN}})$ LEFT --- RIGHT SIDE Now I substitute it in $T(F(z))$ $\displaystyle \exp_b(F(z)) = \exp_b(L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N))$ Im not really good with series but maybe I could use the definition of exponentiation (for T=exp_b) and obtain this $\displaystyle b^{ F(z)} =e^{\ln(b)\cdot F(z)}= \exp_b(L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N))$ $\displaystyle \sum_{n=0}^{\infty}{\ln(b)^i\over i!}F(z)^i= \sum_{n=0}^{\infty}{\ln(b)^i\over i!}\left (L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N) \right)^i$ At this point with my poor knowledge of the subject I can't continue, so looking at your book it says that series has the property that (6.4) $\displaystyle T(F(z)) = L+T^{[1]}(L)\sum_{n=1}^{\infty}a_n {\varepsilon^n} +{T^{[2]}(L)\over 2}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^2+...$ The summation in (6.4) seems an infinite series because I saw the dots “...”, so if I understand well the formulas, assuming that the sequence 1,2,6... is the factorial and with transfer function $T=\exp_b$ (with $T^{[n]}=T^{[n]}(L)$) the final form should be something like that $\displaystyle T(F(z)) =L+\sum_{i=1}^{\infty}{T^{[i]}(L)\over i!}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^i$ RIGHT --- So the question is how can be the left side the same as the right side? $\displaystyle F(z+1)=L+\sum_{n=1}^{N}b_n\varepsilon^n+ o(\varepsilon^N{e^{kN}})=L+\sum_{i=1}^{\infty}{T^{[i]}(L) \over i!}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^i=T(F(z))$ So what the book is probably trying to say is that is possible to manipulate algebraically the RIGHT side in order to obtain a series of the form $\displaystyle T(F(x))=L+\sum_{n=1}^{\infty}\tau_n\varepsilon^n$ with the coefficients $\tau_n$ defined in the formulas (6.5), (6.6) and (6.7) of your book $\tau_1=T'(L)$ $\tau_2=T'(L)a_2+T^{[2]}(L)/2$ $\tau_3=T'(L)a_3+T^{[2]}(L)a_2+T^{[3]}(L)/6$ and probably $\tau_n=T'(L)a_n+T^{[2]}(L)a_{n-1}+...+T^{[n-1]}(L)a_2+T^{[n]}(L)/n!$ I don't know the rules that makes you able to turn the series on the RIGHT in a series with coefficients $\tau_n$ but if is possible and we assume that $T(F(z))=F(z+1)$ then we should have that $F(z+1)=L+b_1\varepsilon+b_2\varepsilon^2+b_3\varepsilon^3+...=L+ {{\tau}}_1 \varepsilon + {\tau}_2 \varepsilon^2 + {\tau}_3 \varepsilon^3+...=T(F(z))$ and thus $b_n=a_n e^{kn}={\tau}_n$. This makes us able to find $k$ $b_1=a_1 e^{k}=T'(L)=\exp'_b(L)=\ln(b)b^L$ so $k=\ln({\ln(b)b^L\over a_1})$ Is this correct? And since we have $k$ and $a_1$ the other $a_n$ are given by your formulas (6.8 ), (6.9) and (6.10)... Thank you alot for the effort and your help. I'm really curious about this but I'm also sorry if is so hard, I'm a donkey in analysis and im not good with powerseries as well. MathStackExchange account:MphLee « Next Oldest | Next Newest »