Approximation of product of numbers close to 1

Question

In one of the papers I came across to the approximation below:

$$\prod_{k=1}^K (1-x_k) \approx 1 - \sum_{k=1}^K x_k,$$

where $x_k$s are very small positive numbers.

I am aware that by the binomial approximation, we can write that

$$(1-x)^K \approx 1-Kx$$

, where $|x|<1$. I assume that these two approximations are related to each other but I am not able to solid connection between them.

Can you explain how to calculate the first approximation and when it is valid?

Answer 1

Use it for $K=3$ $$\prod_{k=1}^3 (1-x_k)=1- (x_1+x_2+x_3)+(x_1x_2+x_1x_3+x_2x_3)-x_1x_2x_3$$

You could do it in another way

$$P_K=\prod_{k=1}^K (1-x_k)\implies \log(P_K)=\sum_{k=1}^K \log(1-x_k)$$ Now, by Taylor $$\log(1-x_k)=-x_k-\frac{x_k^2}{2}+O\left(x_k^3\right)$$ $$\log(P_K)=-\sum_{k=1}^K x_k-\frac 12\sum_{k=1}^K x^2_k$$ $$P_k=e^{\log(P_K)}=1-\sum_{k=1}^K x_k-\cdots$$

Answer 2

Let $0<x_k<<1$, then $(1-x^2_k)<1 \implies 1-x_k <\frac{1}{1+x_k}$ Use Weirstrass Inequality: $$\prod_{1}^{n} (1-x_k) \ge 1+\sum_{k=1}^{n} x_k=1+S_n$$ $$\implies \prod_{1}^{n} (1-x_k) \le \frac{1}{1+S_n} \approx 1-S_n$$ lastlt we have isede Binomial approximation provided $S_n<1.$