Implicit function theorem and taylor expansion

Hoi, I'm realy stuck on this and i hope someone can look a little into this. I'll explain the problem first.

We have a probability function $f:\mathbb{R}_{+}\to \mathbb{R}_+$ for the random variable $X$, and denote for convenience $g(y):=L(f)(y)$ its laplace transform. We have the following rule $$\mathbb{E}[X^n]=\mu_n = (-1)^n \frac{d^n g}{d^n y}(0)$$ and also $$(-1)^n\frac{d^n \log g}{d^n y}(0)= \kappa_n \ \ \text{cumulants of } X$$ We have$\kappa_1=\mu_1$ and $\kappa_2 = \mu_2-\mu_1^2$.

Suppose we have an implicit relation $$x\cdot g(y)= 1$$ which defines $y:=y(x)$ implicitly as function of $x$. Observe that the point $(x,y(x))=(1,0)$ satisfies the relation since $g(0)=1$. Using a suitable implicit function theorem, i want to show that $y=y(x)$ is $C^k$ for all $k$, in an open neighborhood of $x=1$.

So that is, we consider $g$ as $g(x,y(x))$ a function of 2 variables.

Then we have by totally differentiating to $x$ that $$\frac{dg}{dy}\cdot y'(x)+\frac{dg}{dx}= 0$$

(not sure if it makes much difference if we differentiate $g(y)=1/x$ instead). We can continue differentiating, but does that imply that high order derivatives of $y(x)$ exist in some open neighborhood of 1?

Also i want to show that $y(x)$ admits an expansion in $\log x$ as long as $\log x$ is close to 0 as follows $$y(x) = \frac{1}{\mu_1}\cdot \log x+ \frac{\mu_2 -\mu_1^2}{2\cdot \mu_1^3}\cdot (\log x)^2 + \mathcal{O}(\log x)^3$$

I dont realy see the point in expanding into $\log x$ at all, not that it matters...i just want to show that i can. But how do we make such expansion in $\log x$? I see how i can make an expansion around $x=1$, using the implicit relations and calculating $y'(1)$ and $y''(1)$, but here i dont know how to start..

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Well, your function $y$ is $C^k$ and $\log$ is a analitical function. Use the rule for Taylor expansion fo order $k$ for composit functions in terms of Taylor expancion of $y$ and $\log$.
Sorry im trying to grasp what u mean: So $y(x) = y'(1)(x-1)+\frac{y''(1)}{2}(x-1)^2 + O((x-1)^3)$. Then we have to make a similar expansion, but then in terms of $\log(x)$. To be quite honoust, i dont see how to do this... – DinkyDoe Jan 19 '13 at 13:56