The simplification of $(1+a_1\cdot h) \cdot \ldots \cdot (1+a_n\cdot h)$

Question

I would like to simplify the expression $(1+a_1\cdot h) \cdot \ldots \cdot (1+a_n\cdot h)$. One will later anyway truncate all terms with $h^2$ and greater by considering $h << 1$, but I would like to write down a formula before doing $h<<1$.

So far I have: \begin{align} (1+a_1\cdot h) \cdot \ldots \cdot (1+a_n\cdot h) = 1 + h \sum_{j=1}^n a_j + h^2 \sum_{j\ne k}^n a_ja_k + \ldots \end{align}

Answer 1

Denoting with $[n]=\{1,2,\ldots,n\}$ we obtain \begin{align*} \color{blue}{\prod_{j=1}^n}\color{blue}{\left(1+a_jh\right)} &=\sum_{k=0}^n\sum_{{S\subseteq [n]}\atop{|S|=k}}\left(\prod_{j\in S}a_j\right)h^k\\ &=\sum_{{S\subseteq [n]}\atop{|S|=0}}\left(\prod_{j\in S}a_j\right)h^0+\sum_{{S\subseteq [n]}\atop{|S|=1}}\left(\prod_{j\in S}a_j\right)h^1+\sum_{k=2}^n\sum_{{S\subseteq [n]}\atop{|S|=k}}\left(\prod_{j\in S}a_j\right)h^k\\ &\,\,\color{blue}{=1+\sum_{j=1}^na_jh +\sum_{k=2}^n\sum_{{S\subseteq [n]}\atop{|S|=k}}\left(\prod_{j\in S}a_j\right)h^k}\\ \end{align*}

Answer 2

Hint:

$$\prod_{i=1}^n\left[1+a_i h \right]=\prod_{i=1}^na_i\left[h + \dfrac{1}{a_i} \right]=\prod_{i=1}^na_i\prod_{i=1}^n\left[h + \dfrac{1}{a_i} \right]$$

The last product can be interpreted as a polynomial in $h$ with degree $n$ in which $-1/a_i$ are the roots of the polynomial. The coefficients can be written down by Vieta's formulas.