What is the correct method for composing Taylor series at a non-zero point?

[mxbob468]

in general I'm trying to figure out a way to work with taylor series more efficiently. this means i want to be able to write down the taylor series of a complicated function just by knowing the taylor series(es?) of the component functions. I've figured out how to do products and quotients pretty well but I'm having trouble with compositions, specifically at points other than 0. without further ado:

i'd like to find the taylor series at 1 of [itex]x^{1/(1-x)}[/itex]

according to my understanding what i'll need (all series expanded at 1):

[tex]Log(x)=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3+O((x-1)^4)[/tex]
[tex]Exp(x)=e(1+(x-1)+\frac{1}{2}(x-1)^2+\frac{1}{6}(x-1)^3 +O((x-1)^4)[/tex]

hence

[tex]x^{1/(1-x)}=Exp(\frac{Log(x)}{1-x})[/tex]

[tex]=Exp(\frac{(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3+O((x-1)^4)}{1-x})[/tex]

[tex]=Exp(-1(1-\frac{1}{2}(x-1)+\frac{1}{3}(x-1)^2+O((x-1)^3)))[/tex]

[tex]=Exp(-1+\frac{1}{2}(x-1)-\frac{1}{3}(x-1)^2+O((x-1)^3)))[/tex]

[tex]=e^{-1}Exp(\frac{1}{2}(x-1)-\frac{1}{3}(x-1)^2+O((x-1)^3)))[/tex]

then i make the replacement [itex]x \rightarrow (\frac{1}{2}(x-1)-\frac{1}{3}(x-1)^2+O((x-1)^3))[/itex] in the expansion for Exp(x) at 1 and i get the wrong answer. the part i don't understand is that if i make the replacement [itex](x-1)\rightarrow (\frac{1}{2}(x-1)-\frac{1}{3}(x-1)^2+O((x-1)^3))[/itex] I'm an [itex] e^{-1}[/itex] off from the right answer.

also if i try do this same exact thing, composing taylor series for [itex](x+1)^{-1/x}[/itex] but taylor expanding at zero (which should also give me the taylor series for [itex]x^{1/(1-x)}[/itex] at 1 right?) i do get the right answer.

so what gives? my feeling is I'm missing something very blatantly obvious.

 

[mxbob468]

mathematica saves the day:

Two series can only meaningfully be composed when the point about which the first series is expanded corresponds to the limiting value of the second series at its expansion point.