# Trying to revert a series with problematic log term

I'm stuck on a problem which I'm not sure has a solution. I have the first few terms of a series I want to invert,

$y(x)=\ln(x)+a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+\cdots$

I know the inverse exists but I don't know how to invert this series to get at it, that is write $x$ as a series in $y$. Of course if the $\ln(x)$ term was not present this would be a straight forward task. I was wondering if anyone knows of a way to get around this, perhaps a clever change of variable?

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Do you know anything about coefficients $a_i$? I don't have much faith on finding the inverse if you have no idea what the coefficients are... unless magic occurs somewhere. –  Patrick Da Silva Nov 19 '11 at 18:34
In the limit $x\to+\infty$? –  Did Nov 19 '11 at 18:43
@Silva: The coefficients are non zero. –  mark Nov 19 '11 at 19:54
@Piau: Yes the series is asymptotic –  mark Nov 19 '11 at 19:54

No series in $y$ can yield $x$. To see this, first note that, replacing $y(x)$ by $y(x)-a_0$, one can assume without loss of generality that $a_0=0$. Introduce the series $A(u)=\sum\limits_{n=0}^{+\infty}a_nu^n$. Thus, one looks for an expression of $x$ as a function of $y(x)=\log(x)+A(\frac1x)$.

A solution is to note that $\mathrm e^{y(x)}=x\mathrm e^{A(1/x)}$, hence $$x=\mathrm e^{y(x)}B(\mathrm e^{-y(x)}),$$ where the series $B(v)=\sum\limits_{n=0}^{+\infty}b_nv^n$ solves the relation $$\mathrm e^{-A(u)}=\sum\limits_{n=0}^{+\infty}b_nu^n\mathrm e^{-nA(u)}.$$ Finally, $b_0=1$ hence $$\color{red}{x=\mathrm e^{y(x)}+\sum\limits_{n=0}^{+\infty}b_{n+1}\mathrm e^{-ny(x)}}.$$ Note 1: The series $B$ solves the relation $C(u)=uB(C(u))$ with $C(u)=u\mathrm e^{-A(u)}$.

Note 2: Using the inverse $A^{-1}$ of $A$, one can characterize the series $B$ through its inverse $B^{-1}$ by the relation $$B^{-1}(w)=wA^{-1}(-\log w).$$ Note 3: To compute the polynomial $B_N(v)=\sum\limits_{n=0}^{N}b_nv^n$, one can use the polynomial $A_N(u)=\sum\limits_{n=0}^{N}a_nu^n$ and solve the finite relation $$\mathrm e^{-A_N(u)}=\sum\limits_{n=0}^{N}b_nu^n\mathrm e^{-nA_N(u)}+o(u^N),$$ that is, $$\sum\limits_{k=0}^N\frac{(-1)^k}{k!}\left(\sum\limits_{i=0}^{N}a_iu^i\right)^k=\sum\limits_{k=0}^N\frac{(-1)^k}{k!}\left(\sum\limits_{i=0}^{N}a_iu^i\right)^k\sum\limits_{n=0}^{N}b_nn^ku^n+o(u^N).$$ For example, $b_1=-a_1$, $b_2=-a_2-\frac12a_1^2$.

Note 4: Assume that $a_n=\frac1n$ for every $n\geqslant1$, then $A(u)=-\log(1-u)$ and $B(v)=\frac12+\sqrt{\frac14-v}$, hence $x=\frac12\mathrm e^{y(x)}\left(1+\sqrt{1-4\mathrm e^{-y(x)}}\right)$.

Note 5: This is to answer a question asked by the OP in a comment. Fix $x$ and let $z=\mathrm e^{-y(x)}$. Then $x^{-1}\mathrm e^{-A(1/x)}=z$ and $x=z^{-1}B(z)$, hence $\mathrm e^{-A(1/x)}=xz=B(z)=B(x^{-1}\mathrm e^{-A(1/x)})$. Thus, for every $u$, $\mathrm e^{-A(u)}=B(u\mathrm e^{-A(u)})$.

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@Piau: Thank you very much, this to me at least, seems very clever. Can I ask you however, how did you deduce B should satisfy the relationship $e^{−A(u)}=B(u e^{−A(u)})$? –  mark Nov 21 '11 at 18:58
See Note 5.  –  Did Nov 21 '11 at 20:28
Thanks, I seen it eventually. –  mark Nov 22 '11 at 0:02
@Piau: I have a related problem. I was wondering if you could have a look at this for me please. I couldn't find a way to alter this method if there is an additional $x$ term. math.stackexchange.com/questions/84457/… –  mark Nov 22 '11 at 4:25

You might try to use $x = z-1$ and $\log_e(1+z) = z - z^2/2 +z^3/3 - z^4/4 +\cdots$

to give something like

$y(z-1) = (a_0-a_1+a_2-a_3+a_4-\cdots) + (1+a_1-2a_2+3a_3-4a_4+\cdots)z$

$\qquad\qquad\qquad\quad +(-1/2 + a_2 - 3a_3 +6a_4 - \cdots)z^2 + \cdots$

but I am not sure that will make things much easier.

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Seems like I am clutching at straws! –  mark Nov 19 '11 at 18:29