Suppose that we have the Taylor expansions : $$f(x)=\sum_{n=1}^{\infty}\frac{a_{n}}{n!}x^{n}$$ $$g(x)=\sum_{n=1}^{\infty}\frac{b_{n}}{n!}x^{n}$$ Then we have the standard result : $$g(f(x))=\sum_{n=1}^{\infty}\left(\sum_{k=1}^{n}b_{k}B_{n,k}(a_{1},...,a_{n-k+1})\right)\frac{x^{n}}{n!}\ \ \ \ \ (1)$$ Where $B_{n,k}(\cdot)$ are the partial Bell polynomials. Now suppose we have the expansion : $$h(x)=\sum_{n=1}^{\infty}B_{n}(a_{1},...,a_{n})\frac{x^{n}}{(n!)^{2}}\ \ \ \ \ (2)$$ Where $B_{n}(\cdot)$ are the complete Bell polynomials. We want to express $h(x)$ in closed form as a composite function. Thus, we use the relation above, and we set: $$\sum_{k=1}^{n}b_{k}B_{n,k}(a_{1},...,a_{n-k+1})=\frac{B_{n}(a_{1},...,a_{n})}{n!}\ \ \ \ \ (3)$$ How can we solve for $b_{k}$ !?

To be more specific, I'm interested in the relation between the two expansions: $$\sum_{n=0}^{\infty}\frac{B_{n}\left(a_{1},a_{2},...,a_{n}\right)}{n!}x^{n}\ \ \ \ \ (4)$$ And: $$\sum_{n=0}^{\infty}\frac{B_{n}\left(a_{1},2!a_{2},...,n!a_{n}\right)}{(n!)^{2}}x^{n}\ \ \ \ \ (5)$$ Using the reasoning above - function composition - or any other method for that matter.


Let us use the notion

$$a_i=\frac{d^i}{dx^i}f(x)\bigg|_0=f_i,\ \ \ \ \ b_k=\frac{d^k}{dx^k}g(x)\bigg|_0=g_k,\ \ \ \ \ \frac{d^n}{dx^n}h(x)\bigg|_0=h_n.$$

Faà di Bruno's formula is then:


Equation 2 in connection with equation 1 means $g_k=\frac{1}{n!}$. That means $g_{k_1}\neq g_{k_2}$ for $k_1=k_2$ for different $n$. $g_k$ cannot be a function value of the $k$-th derivative of $g$ therefore. Faà di Bruno's formula is therefore not applicable here.

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If e.g. $f(x)=cx$ where $c$ is a constant, $h(x)=I_0(2\sqrt{cx})$, wherein $I$ is the Modified Bessel function of the first kind.


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