Is there a formula for the taylor coefficients of
$$g(w) := \frac{1}{(w-z_1)^{a_1}\cdots (w-z_n)^{a_n} }$$
for general $z_i$ and $a_i \in (0,2)$?
My goal would be integrating this function numerically, so not a very high accuracy is required.
General formula for taylor coefficients
0
$\begingroup$
power-series
taylor-expansion
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0Around which value of $w$ do you wnt the expansion ? What would be the bounds for integration ? – 2017-02-07
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0In the end I'd like to visualize [Schwarz-Christoffel-transformations](https://en.wikipedia.org/wiki/Schwarz%E2%80%93Christoffel_mapping), so the expantion around 0 would be natural, but any other would work too I think. – 2017-02-07
1 Answers
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Just an idea to be explored (may be) .
$$g=\frac 1 {\prod_{i=1}^n (w-z_i)^{a_i}}\implies \log(g)=-\sum_{i=1}^n a_i \log(w-z_i)$$ Now, using Taylor expansion around $w=0$ $$\log(g)=-\sum_{i=1}^n a_i \log(z_i)+w \sum_{i=1}^n \frac{a_i}{z_i}+w^2\sum_{i=1}^n \frac{a_i}{2z_i^2}+w^3\sum_{i=1}^n \frac{a_i}{3z_i^3}+\cdots+w^k\sum_{i=1}^n \frac{a_i}{kz_i^k}+\cdots$$ that is to say $$\log(g)=\sum_{i=0}^k A_i w^{i}$$ Now, Taylor again (limiting to very first terms) $$g=e^{\log(g)}=e^{A_0}+e^{A_0} A_1 w+\frac{1}{2} e^{A_0} \left(A_1^2+2 A_2\right) w^2+\frac{1}{3} e^{A_0} \left(2 A_1 A_2+\frac{1}{2} A_1 \left(A_1^2+2 A_2\right)+3 A_3\right) w^3+O\left(w^4\right)$$