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 real slog developed at a fixed point bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 12/30/2008, 12:28 PM Kouznetsov Wrote:bo198214 Wrote:Didnt you say that the series does not converge? Or not to the values of slog?Not to the values of slog. The radius of convergence of the sub-series $S(z)$ of terms with integer poers of $z$ seems to be $|L|$. Oh I just see, the construction I made above is the regular slog. I.e. the function Knesers starts with. Thatswhy it doesnt yield the values of the kslog, especially it is not real at the real axis. It is the powerseries development of the regular Abel function, while Kneser works with the limit formula of the regular Abel/Schroeder function. So the continuation of this series to the upper half plane has a singularity at $\exp^n(0)$ for each $n\ge 0$. Here the singularity at 0 limits the convergence radius to $|L|$. Lets call this super logarithm rslog while we call your super logarithm kslog. Then we know already that $\text{kslog}(\text{rslog}^{-1}(z))=z + \phi(z)$ or $\text{kslog}(z)=\text{rslog}(z)+\phi(\text{rslog}(z))$ where $\phi$ is 1-periodic. By the above consideration we can develop $\text{rslog}(z)=\log_L(z-L) + p(z-L)$ for a powerseries $p(z)=p_1 z + p_2 z^2 + \dots$. And every 1-periodic holomorphic function can be developed as: $\phi(z) = \sum_{n=-\infty}^\infty \varrho_n e^{2\pi i n z}$ Then $\phi(\text{rslog}(z))=\sum_{n=-\infty}^\infty \varrho_n e^{2\pi i n \left(\log_L(z-L)+p(z-L)\right)}=\sum_{n=-\infty}^\infty \varrho_n (z-L)^{2\pi i n/L}e^{2\pi i n p(z-L)}$ Now the exponential of a powerseries is again a powerseries: $e^{2\pi i n p(z-L)} = \sum_{m=0}^\infty d_{m,n} (z-L)^m$ inserted: $\phi(\text{rslog}(z)) =\sum_{n=-\infty}^{\infty} \varrho_n (z-L)^{2\pi i n/L}\sum_{m=0}^\infty d_{m,n} (z-L)^m$ thatswhy we can finally write: $\text{kslog}(z)=\text{rslog}(z)+\phi(\text{rslog}(z))=\log_L(z-L) + \sum_{m=0,n=-\infty}^\infty c_{m,n} (z-L)^{\frac{2\pi i n}{L} + m}$ which is slightly more unspecific than Knesers's formula. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 04/06/2009, 09:06 AM (This post was last modified: 04/06/2009, 09:11 AM by andydude.) @Henryk Did you forget a binomial coefficient somewhere? More specifically: Quote:Now we know that $\eta^{\cdot j}(z)=\el^j (e^z-1)^j=\el^j\sum_{k=0}^j C_{jk} e^{kz}(-1)^{j-k}$ $\left(\eta^{\cdot j}\right)_n=\el^j \sum_{k=0}^j C_{jk}\frac{k^n}{n!}(-1)^{j-k}=\el^j \sum_{k=1}^j C_{jk} \frac{k^n}{n!}(-1)^{j-k}$ Andrew Robbins bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/06/2009, 10:10 PM andydude Wrote:@Henryk Did you forget a binomial coefficient somewhere? More specifically: Quote:Now we know that $\eta^{\cdot j}(z)=\el^j (e^z-1)^j=\el^j\sum_{k=0}^j C_{jk} e^{kz}(-1)^{j-k}$ $\left(\eta^{\cdot j}\right)_n=\el^j \sum_{k=0}^j C_{jk}\frac{k^n}{n!}(-1)^{j-k}=\el^j \sum_{k=1}^j C_{jk} \frac{k^n}{n!}(-1)^{j-k}$ Andrew Robbins Yes, you are right. I will correct it in my post. However I think the coefficients $r_n$ are still ok (because I computed them by other means.) « Next Oldest | Next Newest »

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