The previous answer does a great job of explaining the derivation up until the even and odd recursion formulae. The iterated coefficients are from 3 to 8 are correct, but if you check, the recursive formulae do not match the iterated examples. The correct recursion formulae are:
$$a_{2j}=\frac{a_2}{(j-1)!}$$
and
$$a_{2j+1}=\frac{2^{2j-1} (j-1)! a_1}{ (2j-1)!}$$
To derive the approximation for $$a_j$$, you can use either formula, but the even formula is easier. In case you would like to derive the other as well, the general process is the same. Remembering that the approximation is in the regime of very large $$j$$, you can invoke <PERSON>'s approximation for very large factorials. 915
This is:
$$n!\approx \sqrt[]{2\pi}\, n^{n+\frac{1}{2}}e^{-n}$$
So, for the even coefficients:
$$ a_{2j}\approx\ \ \frac{a_2}{\sqrt[]{2\pi}\, {(j-1)}^{(j-\frac{1}{2})} e^{(-j+1)}}\approx $$
$$\frac{a_2}{{\sqrt[]{2\pi}\,e^1{j}^{j}e^{-j}}} \approx$$
$$\frac{a_2}{{\sqrt[]{2\pi}\,e^1{j}^{j+\frac{1}{2}}e^{-j}}}\approx$$
$$\frac{a_2}{{e\,(j)!}} = \frac{C}{{(j)!}}$$
Note that some of the fractions in the exponents were dropped or added since, in the approximation for very large numbers, these numbers become negligible; it has no effect on the approximation. Lastly, this approximation was carried out for the coefficient $$a_{2j}$$. To find it for any arbitrary coefficient, you can perform a change of variables:
$$n\equiv 2j$$
so
$$a_n = a_{2j} \approx \frac{C}{{(j)!}} = \frac{C}{{(\frac{n}{2})!}}$$
One could derive the same result for the odd formula, but it is a little bit of a lengthy derivation; it has the same procedure, but one extra step. Use <PERSON>'s approximation (it will lead to the same form as the even approximation) and then perform a change of variables. Once the variables have been changed, it will require one more <PERSON> approximation.
As far as your third question:
$$h(\xi) = \sum\, a_n \xi^{n} \approx C\, \sum\, \frac{\xi^{n}}{{(\frac{n}{2})!}}$$
Performing a change of variables:
$$m\equiv \frac{n}{2}\,\,\, \Rightarrow\,\,\, 2m=n$$
So:
$$h(\xi) = \sum\, a_n \xi^{n} \approx C\, \sum\, \frac{\xi^{2m}}{{(m)!}} $$
$$=\,C\, \sum\, \frac{({\xi^{2}})^{m}}{{(m)!}}\, =\, C\,e^{\xi^2}$$
Hope this helped! 374