How is additive error handled in this simple algorithm? 'Product of all elements'

Question

Say we have two unit vectors $\hat{u}, \hat{v} \in \mathbb{R}^n$ where $\hat{u} = (u_1,...,u_n)$ and $\hat{v}$ approximates $\hat{u}$. $~\hat{v} = (u_1+\epsilon, ...,u_n+\epsilon)$ where $\epsilon = \frac{1}{poly(n)}$.

We have an algorithm which takes the product of all elements and outputs the product. For the exact case (with $\hat{u}$), $x = \prod_i^nu_i$. For the approximate case (with $\hat{v}$), $x^* = \prod_i^n(u_i + \epsilon)$.

What is the error of $x^*$ with respect to $x$? How do we bound the error of $x^*$ in terms of epsilon? How is the error analysis typically handled in this situation?

Answer 1

Use Taylor series expansion for the function $\prod_{i=1}^{n} (u_i + y)$ around $y=0$ to obtain the error as $\epsilon (\prod_{j=1}^{n} u_j) \sum_{i=1}^{n} u_i^{-1} + O(\epsilon^2)$. Note that Taylor series works for $\epsilon = O(1/poly(n))$, because $\mathbf{u}$ is a unit normed vector and the product $\prod_{j=1}^{n} u_j < 1$.