Gus Wiseman's suggestion UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 08/15/2007, 09:30 PM I've been looking at Gus' suggestion, here for some time, and I think I am finally starting to understand it. It doesn't appear to work, at least as he states it. He says (verbatim): By taking derivatives of both sides of the equation $T(e^x)=e^{T(x)}$ and evaluating at a fixed point (such that $x=e^x$), we can construct a Taylor series with a parameter a satisfying $T^a(x)={^x}e$. Well, such a fixed point exists. It is indeed equal to $a=-W(-1)=0.3181315049-1.337235701i$, where W is Lambert's function. So, in effect, if I understand him correctly, we are looking for the Taylor series: ${^x}e=T^a(x)=T(a)+D^{(1)}(T)(a)*(x-a)+D^{(2)}(T)(a)*(x-a)^2/2!+D^{(3)}(T)(a)*(x-a)^3/3!+\cdots$. So we embark on a journey to determine $D^{(n)}(T)(a)$. This already presents severe problems. Let's see: $ T(e^x)=e^{T(x)}\Rightarrow\\ T'(e^x)*e^x=T'(x)*e^{T(x)}$ Indeed, if we stuff x=a in the last equation above, (remember $e^a=a$), we get immediately: $ T'(a)*a=T'(a)*e^{T(a)}\Rightarrow\\ a=e^{T(a)}\Rightarrow\\ T(a)=log(a)=log(-W(-1))$ We got the first term of the series. What about $T'(a)$? Herein lies the problem. The first derivatives of T cancel each other as above (after evaluation at x=a), so there's no way to get T' from the functional relation $T(e^x)=e^{T(x)}$. Not being able to get T', automatically kills the entire process for the rest of the derivatives, because upon differentiating this functional equation n times, it is easy to see that all derivatives up to the n-th might show up and thus the calculation of $D^{(n)}(T)(a)$ implicitly depends on the existence of $D^{(n-1)}(T)(a)$. Unless I am not seeing something else obvious, this means that his suggestion doesn't work and the Taylor series cannot be constructed. (And all this besides the fact that even if this series could be constructed, we'd need to check its radius, etc. etc). bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/15/2007, 09:47 PM Hm, not sure. It is a strange mixture from several techniques. 1. technique: search for a fixed point and take the unqiue $t$-th iterate there. This approach is due to Kneser, however yields complex values for real arguments, because all the fixed points are complex. I dont know whether it even yields the same result for each fixed point of exp. 2. technique: piecewise construction such that all the derivates are continuous at the joining points, this was performed by Andrew Robbins, however on the slog. So this approach seems to be a mixture from 1. and 2. for sexp. Makes this sense? I mean it was for good reason that Andrew worked with the slog and not with sexp. And why Gus then needs a fixed point to develop a series? « Next Oldest | Next Newest »