Taylor expansion of $\prod_{i=1}^n(1-x_i)^{-1/2}$ around 0

Question

What is the Taylor expansion of $$\prod_{i=1}^n\frac{1}{\sqrt{1-x_i}}$$ around $(0,0,...,0)$？I know that we can write it as $$\prod_{i=1}^n\left(1+\sum_{k=1}^{\infty}\frac{(2k-1)!!x_i^k}{(2k)!!}\right)$$ But how to simplify it?

Answer 1

Maybe a bit silly, but what about:

$$\prod_{i=1}^n\frac{1}{\sqrt{1-x_i}}=\dfrac{1}{\sqrt{1-\left(\displaystyle\prod_{i=1}^n(1-x_i)-1\right)}}$$

Which gives you then:

$$1+\sum_{k=1}^{\infty}\frac{(2k-1)!!y^k}{(2k)!!}$$

with

$$ y = -1+\prod_{i=1}^n(1-x_i) = -\sum_{i=1}^nx_i + \sum_{i<j=1}^n x_i x_j - \sum_{i<j<k=1}^n x_i x_j x_k + ... + (-1)^n \prod_{i=1}^n x_i $$

Answer 2

Note that by the distributivity of multipliaction over addition $$ \prod_{i=1}^n \sum_{k_i=0}^\infty f(i,k_i) = \sum_{k_1,\dots, k_n=0}^\infty \prod_{i=1}^n f(i,k_i) $$ so in your case $$ \prod_{i=1}^n \frac{1}{\sqrt{1-x_i}} = \sum_{k_1,\dots, k_n=0}^\infty \prod_{i=1}^n (a_{k_i} x_i^k ) = \sum_{k_1,\dots, k_n=0}^\infty c_{k_1\dots k_n} \left(\prod_{i=1}^nx_i^k \right)$$ where $$c_{k_1\dots k_n} = \prod_{i=1}^n a_{k_i}$$ $$ a_{k} = \left\{\begin{array}{ll} 1 & \text{for }k=0 \\ \frac{(2k-1)!!}{(2k)!!} & \text{for }k\ge 1\end{array} \right.$$ You can't really get anything much better than this, coefficients $c_{k_1\dots k_n}$ cannot be further simlified.

Another method would be to note that $$ \prod_{i=1}^n \frac{1}{\sqrt{1-x_i}} = \exp\Big(-\frac12\sum_{i=1}^n \ln(1-x_i)\Big) = \exp\Big(\sum_{i=1}^n\sum_{k=1}^\infty \frac{x_i^k}{2k}\Big) $$ and use the Taylor series of $\exp(y)$, but I don't think it gives a simpler result.