Derivation of normalizing transform for GLMs $\newcommand{\E}{\mathbb{E}}$How is the $A(\cdot) = \displaystyle\int\frac{du}{V^{1/3}(\mu)}$ normalizing transform for the exponential family derived?  
More specifically: I tried to follow the Taylor expansion sketch on page 3, slide 1 here but have several questions. With $X$ from an exponential family, transformation $h(X)$, and $\kappa _i$ denoting the $i^{th}$ cumulant, the slides argue that:
$$
\kappa _3(h(\bar{X})) \approx h'(\mu)^3\frac{\kappa _3(\bar{X})}{N^2} + 3h'(\mu)^2h''(\mu)\frac{\sigma^4}{N} + O(N^{-3}),
$$
and it remains to simply find $h(X)$ such that the above evaluates to 0. 


*

*My first question is about arithmetic: my Taylor expansion has different coefficients, and I can't justify their having dropped many of the terms. 
\begin{align}
 \text{Since }h(x) &\approx h(\mu) + h'(\mu)(x - \mu) + \frac{h''(x)}{2}(x - \mu)^2\text{, we have:}  \\
h(\bar{X}) - h(u) &\approx h'(u))(\bar{X} - \mu) + \frac{h''(x)}{2}(\bar{X} - \mu)^2 \\
\E\left(h(\bar{X}) - h(u)\right)^3 &\approx h'(\mu)^3 \E(\bar{X}-\mu)^3 + \frac{3}{2}h'(\mu)^2h''(\mu) \E(\bar{X} - \mu)^4 +  \\
  &\quad \frac{3}{4}h'(\mu)h''(\mu)^2 \E(\bar{X}-\mu)^5 + \frac{1}{8}h''(\mu)^3 \E(\bar{X} - \mu)^6.
\end{align}
I can get to something similar by replacing the central moments by their cumulant equivalents, but it still doesn't add up. 

*The second question: why does the analysis start with $\bar{X}$ instead of $X$, the quantity we actually care about?
 A: $\blacksquare$ 1.Why can't I get the same result by approximating in terms of noncentral moments $\mathbb{E}\bar{X}^k$ and then calculate the central moments $\mathbb{E}(\bar{X}-\mathbb{E}\bar{X})^k$using the approximating noncentral moments?
Because you change the derivation arbitrarily and drop the residue term which is important. If you are not familiar with the big O notation and relevant results, a good reference is [Casella&Lehmann].
$$h(\bar{X}) - h(u) \approx h'(u)(\bar{X} - \mu) + \frac{h''(x)}{2}(\bar{X} - \mu)^2 +O[(\bar{X} - \mu)^3]$$
$$\mathbb{E}[h(\bar{X}) - h(u)] \approx h'(u)\mathbb{E}(\bar{X} - \mu) + \frac{h''(x)}{2}\mathbb{E}(\bar{X} - \mu)^2+(?) $$
But even if you do not drop the residue by arguing that you are always doing $N\rightarrow \infty$(which is not legal...), the following step:
$$
\E\left(h(\bar{X}) - h(u)\right)^3 \approx h'(\mu)^3 \E(\bar{X}-\mu)^3 + \frac{3}{2}h'(\mu)^2h''(\mu) \E(\bar{X} - \mu)^4 + \frac{3}{4}h'(\mu)h''(\mu)^2 \E(\bar{X}-\mu)^5 + \frac{1}{8}h''(\mu)^3 \E(\bar{X} - \mu)^6.
(1)$$ is saying that $$\int [h(x)-h(x_0)]^3dx=\int [h'(x_0)(x-x_0)+\frac{1}{2}h''(x_0)(x-x_0)^2+O((x-x_0)^3)]^3dx=(1)$$
if this is still not clear, we can see the algebra of expanding the integrand goes as
$[h'(x_0)(x-x_0)+\frac{1}{2}h''(x_0)(x-x_0)^2+O((x-x_0)^3)]^3(2)$
Letting $A=h'(x_0)(x-x_0)$,$B=\frac{1}{2}h''(x_0)(x-x_0)^2$,$C=O((x-x_0)^3)$
$(2)=[A+B+C]^3$
$\color{red}{\neq}[A^3+3A^2 B+3A B^2+B^3]=[A+B]^3=(1)$
Your mistake is to omit the residue before expansion, which is a "classical" mistake in big O notation and later became a criticism of the usage of big O notation.
$\blacksquare$ 2.Why does the analysis start with $\bar{X}$ instead of $X$, the quantity we actually care about?
Because we want to base our analysis on the sufficient statistics of the exponential model we are introducing. If you have a sample of size 1 then there is no difference whether you analyze with $\bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_i$ OR $X_1$. 
This is a good lesson in big O notation though it is not relevant to GLM...
Reference
[Casella&Lehmann]Lehmann, Erich Leo, and George Casella. Theory of point estimation. Springer Science & Business Media, 2006.
