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Given the function: $$F=\frac{\exp(-iTz)}{z}\prod_{k=1}^N(1-\frac{z}{z_k})^{-1}$$ with $z\in C$ is it possible to give a closed expression of the $n^{th}$ derivative of $F$ with respect to $z$?: $$\frac{d^n}{dz^n}F$$ Thanks in advance.

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What do you mean by closed form? Will you settle for an expression with sum can products? Actually, the most difficult part is to find a closed expression of the $j$-th derivative of the product. It will involve combinatorial arguments. – Davide Giraudo Sep 18 '12 at 9:30
@Davide Giraudo: yes. I mean a fomula containing eventually products and sum. – Riccardo.Alestra Sep 18 '12 at 9:37
up vote 3 down vote accepted

First we note that the $j$-th derivative of $\frac 1{1-\frac z{z_k}} = \frac{z_k}{z_k - z}$ is \[ \frac{d^j}{dz^j} \frac{z_k}{z_k - z} = \frac{j!\cdot z_k}{(z_k - z)^{j+1}} \] Now, for the product, we use (a generalisation of) Leibniz' formula, we have \[ \frac{d^j}{dz^j} \prod_{k=1}^N \frac{z_k}{z_k - z} = \sum_{j_1 + \cdots + j_N = j} \frac{j!}{j_1! \cdots j_N!} \prod_{k=1}^N \frac{d^{j_k}}{dz^{j_k}} \frac{z_k}{z_k - z} = \sum_{j_1 + \cdots + j_N = j} \frac{j!}{j_1! \cdots j_N!} \prod_{k=1}^N \frac{{j_k}! z_k}{(z_k - z)^{j_k + 1}} \] Using Leibniz again, we have \[ \frac{d^n}{dz^n}F = \sum_{k_1 + k_2 + k_3 = n} \frac{n!}{k_1!k_2!k_3!} (-iT)^{k_1}\exp(-iTz) (-1)^{k_2}k_2! \frac 1{z^{k_2+1}} \sum_{j_1 + \cdots + j_N = k_3} \frac{k_3!}{j_1! \cdots j_N!} \prod_{k=1}^N \frac{{j_k}! z_k}{(z_k - z)^{j_k + 1}}. \]

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