Is there a closed form expression for the Taylor series of exp((f(z))?

Question

Given a holomorphic function $f(z) = \sum_{k=0}^\infty f_k z^k/k!$, is there a readable formula for the Taylor series of $\exp(f(z))$? Using the chain and product rules, one can obtain $$\partial_z e^f = f' e^f, \partial_z^2 e^f = (f''+f'^2)e^f, \partial_z^3 e^f = (f'''+2f''f'+(f''+f'^2)f')e^f,...$$ to obtain $$\exp(f(z)-f_0) = 1 + f_1z + (f_2+f_1^2)z^2/ + (f_3+3f_2f_2+f_1^3)z^3/3! + \mathcal O(z^4)$$ but that's tedious and only yields the first view terms. Is the a more general formula to directly obtain all coefficients of this series?

Answer 1

The answer is as special case of Faà di Bruno's formula for power series, called the Exponential Formula:

$$\exp\left(\sum_{k=0}^\infty f_k\frac{z^k}{k!}\right) = e^{f_0}\cdot\sum_{n=0}^\infty \frac{B_n(f_1,f_2,...)}{n!}z^n$$

where $B_n$ is the $n$-the complete Bell polynomial

$$\begin{align} B_n(f_1,f_2,...) &= \sum_{k=1}^n B_{n,k}(f_1,f_2,...), \\ B_{n,k}(f_1,f_2,...) &= \sum_{(*)}{n! \over j_1!j_2!\cdots j_{n-k+1}!} \left({f_1\over 1!}\right)^{j_1}\left({f_2\over 2!}\right)^{j_2}\cdots\left({f_{n-k+1} \over (n-k+1)!}\right)^{j_{n-k+1}}, \end{align}$$

where the sum $(*)$ is taken over all sequences $j_1,...,j_{n-k+1}$ with $\sum\limits_{i=1}^{n-k+1}j_i = k$ and $\sum\limits_{i=1}^{n-k+1}ij_i=n$.

Answer 2

We can express it with the Carlemanmatrix-notation for functional composition.

Let $F_1$ be the Carleman-matrix for $f_1(x) = f(x)-f_0 $ (as shown below at the end), $E_1$ that for the decremented exponentiation $\exp(x)-1=\exp(x)-e_0$ and $P$ that for the addition of 1
(and let them all be transposed relative to the version in the wikipedia article such that with a vector of powers of x $V(x) = [1,x,x^2,x^3,...]$ we have for instance $V(x) \cdot E_1 = V(\exp(x)-1) $) then you ask for $$ V(x) \cdot F_1 \cdot E_1 \cdot P =V(\exp(f_1(x)) \tag 1$$

Full explanation of the matrixproduct:

In the columns of $F_1$ we have the coefficients of the powers of the formal power series of $f_1(x)$ beginning with that of $f_1(x)^0 = columnvector([1,0,0,0,...]$ ,that of $f_1(x)^1 = columnvector([0,f_1/1!,f_2/2!,f_3/3!,...])$ and in the next column that of $f_1(x)^2$ and so on.

The matrix $E_1$ contains the Stirling numbers 2'nd kind, (similarity) scaled by the factorials $E_1 = C^{-1} \cdot S_2 \cdot C \tag 2$ where $C=\operatorname{diagonalmatrix}[0!,1!,2!,3!,...]$ (see for instance Abramowitz&Stegun).

The matrix $P$ is the upper triangular pascal-matrix, which performs $V(x) \cdot P = V(x+1) $ such that $$V(x) \cdot (E_1 \cdot P) = V(\exp(x)-1) \cdot P = V(\exp(x)-1+1)= V(\exp(x)) \tag 3 $$.

Short form of the matrix-product: For the final result we need only the entries of the second column of $F_1 \cdot E_1 \cdot P$ or $F_1 \cdot B$ where $B$ is the columnvector containing $[1,1,1/2!,1/3!,1/4!,...]$ only. This makes things much easier: then $$V(x) \cdot (F_1 \cdot B) = \exp(f_1(x)) \tag 4$$

The complete Faá di Bruno-formula is hidden in the mechanism of the dot-product of the matrix $F_1$ and $B$ and "can be forgotten" if one becomes familiar with that Carleman-matrix-notation.

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Here are the top-left excerpts of the used Carleman-matrices. (note, that products of carleman-matrixes are again Carleman). $$ F_1= \small \begin{bmatrix} 1 & . & . & . & . & \cdots\\ 0 & f_1 & . & . & . & \cdots\\ 0 & 1/2! f_2 & f_1^2 & . & .& \cdots \\ 0 & 1/3! f_3 & f_2 f_1 & f_1^3 & . & \cdots\\ 0 & 1/4! f_4 & 1/3 f_3 f_1+1/4 f_2^2 & 3/2 f_2 f_1^2 & f_1^4 & \cdots\\ \vdots & \vdots&&&& \ddots \end{bmatrix} $$ Here the rational fractions at the coefficients should be expanded such that the column $c$ has a cofactor $c!$ and the rows $r$ has a cofactor $1/r!$ (indexes beginning at zero). The second columsn has the coefficients of the formal power series of $f_1(x)$, the third column that of $f_1(x)^2$, and so on.

The matrix $S_2$ has the top-left entries: $$ S_2 = \small \begin{bmatrix} 1 & . & . & . & . &\cdots \\ 0 & 1 & . & . & . &\cdots \\ 0 & 1 & 1 & . & . &\cdots \\ 0 & 1 & 3 & 1 & . &\cdots \\ 0 & 1 & 7 & 6 & 1&\cdots \\ \vdots &&&&& \ddots \end{bmatrix}$$ and that of the matrix $E_1$ are $$ E_1=\small \begin{bmatrix} 1 & . & . & . & . &\cdots \\ 0 & 1 & . & . & . &\cdots \\ 0 & 1/2 & 1 & . & . &\cdots \\ 0 & 1/6 & 1 & 1 & . &\cdots \\ 0 & 1/24 & 7/12 & 3/2 & 1 &\cdots \\ \vdots & & & & & \ddots \end{bmatrix}$$
Finally, the required column-vector $B$ reduces to the second column of $E_1$ plus the $1$ in the first row: $$B = \operatorname{col}([1,1,1/2!,1/3!,1/4!,\ldots]$$ Then $$F_1 \cdot B =\small \left[ \begin{array} {lll} 1 \\ f_1 \\ 1/2 f_1^2+1/2 f_2 \\ 1/6 f_1^3+1/2 f_2 f_1+1/6 f_3 \\ 1/24 f_1^4+1/4 f_2 f_1^2+1/6 f_3 f_1+(1/8 f_2^2+1/24 f_4) \\ \vdots \end{array} \right] $$ and the power series occurs, when we prefix each row wit the according power of the parameter $z$ which gives the result, which I've documented above.