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This is a question I encountered in my own research on Generalized Hyperbolic Secant (GHS) distributions. It is known that the Laplace transform of the basis measure for this family is $$L\left( \theta\right) =\left( \cos \theta\right) ^{-\sigma}$$ where $\sigma >0$ is the scale parameter and $\vert \theta \vert < \pi/2$.

Now I need to compute $L^{\left( n\right) }\left( 0\right) =\left. \frac{d^{n}}{d\theta^{n}}L\left( \theta\right) \right\vert _{\theta=0}$, i.e., the $n$th derivative of $L\left( \theta\right) $ at $0$ for each even $n$. Note that $L\left( \theta\right) $ is the Laplace transform of a density $f$ that is symmetric around $0$ and $$ L^{\left( n\right) }\left( 0\right) =\int_{-\infty}^{\infty}x^{n}f\left( x\right) dx. $$ We know $L^{\left( n\right) }\left( 0\right) =0$ if $n$ is odd.

Since $\cos^{\left( j\right) }\left( 0\right) =0$ whenever $j$ is odd, in Faa di Bruno's formula for $L^{\left( n\right) }\left( 0\right) $ as long as $\cos^{\left( j\right) }\left( \theta\right) $ with $j$ odd appears, the corresponding summand is $0$. But I was not able to single out the nonzero terms from this formula.

I also tried induction on $L^{\left( n\right) }\left( 0\right) $ with $n$ even but did not find a pattern yet. (Also tried integration by parts, but computing the primitive $G$ of $f$ and the primitive of $G$ is much involved; also tried to directly compute $\int_{-\infty}^{\infty}x^{n+2}f\left( x\right) dx$ with $n$ even using contour integral (involving Gamma functions) but failed)

Any suggestions? Thank you.

Updates: found this paper "The Curious History of Faa di Bruno's Formula" by Warren P. Johnson. Its page 222 provides Faa di Bruno's determinant formula, a beautiful formua that at least provides a rather complicated recursion for $n$ even.

For the composite function considered here, the diagonals in the upper left triangle of this matrix alternate between being zero and none-zero, which simplies the computation a lot but still is quite messy.

Update 12/11/2017: In compliance with a reply in the comment (whose user name I am able to decipher), Theorem 2 and Theorem 3 of the paper "Natural Exponential Families with Quadratic Variance Functions" by Carl N. Morris in 1982 gives the same type of formula for computing this. (QED)

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    $\begingroup$ Some information about the triangle of numbers$$\begin{array}{rl}L(0)&=1\\L^{(2)}(0)/\sigma&=2+3\sigma,\\L^{(4)}(0)/\sigma&=16+30\sigma+15\sigma^2,\\L^{(6)}(0)/\sigma&=272+588\sigma+420\sigma^2+105\sigma^3,\\ \cdots \end{array}$$can be found at the OEIS A085734 page $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 11, 2017 at 6:40
  • $\begingroup$ Thank you. If you know an analytic strategy instead of computing case by case, could you please point me to it? I do not get how triangular numbers come into play yet. $\endgroup$
    – Chee
    Commented Dec 11, 2017 at 7:01
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    $\begingroup$ Just look at the link - there seems to be a (more or less) explicit formula:$$L^{(2n)}(0)=\sum_{m=1}^nT_{n,m}\sigma^m$$with$$T_{n,m}=(-1)^{n+m}\sum_{k=1}^n(-1)^k\frac{S1(k, m)}{2^{k-1}k!}\sum_{j=1}^k(-1)^j\binom{2k}{k-j}j^{2n}$$where $S1(k,m)$ are the Stirling numbers of the first kind. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 11, 2017 at 8:22
  • $\begingroup$ en.wikipedia.org/wiki/Faà_di_Bruno%27s_formula $\endgroup$
    – Max Alekseyev
    Commented Dec 11, 2017 at 13:02
  • $\begingroup$ Now I need to compute $L^{(n)}(0)$. What is the exact meaning of the verb "to compute" in this sentence? $\endgroup$
    – fedja
    Commented Dec 14, 2017 at 2:41

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