0

Let $f=(f_0,f_1,f_2...)$ and $g=(g_0,g_1,g_2,...)$ be sequences in $F^{\infty}$. We define multiplication $fg$ by expressing the $n$-th component $(fg)_n=\sum_{i=0}^ng_if_{n-i}$. If $h=(h_0,h_1,h_2,...)$ is also in $F^{\infty}$, we want to show multiplication is associative. Hoffman and Kunze give the following calculation:

\begin{align} [(fg)h]_n&=\sum_{i=0}^n(fg)_ih_{n-i}\\ &=\sum_{i=0}^n(\sum_{j=0}^if_jg_{i-j})h_{n-i}\\ &=\sum_{i=0}^n\sum_{j=0}^if_ig_{i-j}h_{n-i}\\ &=\sum_{j=0}^nf_j\sum_{i=0}^{n-j}g_ih_{n-i-j}\\ &=\sum_{j=0}^nf_j(gh)_{n-j}=[f(gh)]_n. \end{align} My question is regarding the second to last equality. I'm getting $\sum_{j=0}^n\sum_{i=0}^{n-j}f_{i+j}g_ih_{n-i-j}$. Is my calculation wrong, or is the one in the book wrong? If the book is wrong, it doesn't look like we have associativity, so I'm a bit confused.

Eugene Zhang
  • 16,805
user124910
  • 3,007

2 Answers2

2

You are wrong. It should be as \begin{align} [(fg)h]_n&=\sum_{i=0}^n(\sum_{j=0}^if_jg_{i-j})h_{n-i} \\ &=\sum_{i=0}^n\sum_{j=0}^if_jg_{i-j}h_{n-i} \\ &=\sum_{j=0}^nf_j\sum_{i=j+1}^{n}g_{i-j}h_{n-i} \\ &=\sum_{j=0}^nf_j\sum_{k=0}^{n-j}g_kh_{n-k-j}\tag{$i-j=k$} \\ &=\sum_{j=0}^nf_j(gh)_{n-j}=[f(gh)]_n \end{align}

Eugene Zhang
  • 16,805
2

\begin{align} \sum_{i=0}^n \biggl( \sum_{j=0}^if_j g_{i-j} \biggr) h_{n-i} &= \sum_{i=0}^n \sum_{j=0}^i f_j g_{i-j} h_{n-i} \\ &= \sum_{j=0}^n \sum_{i=j}^n f_j g_{i-j} h_{n-i} && \text{(swap order of summation)} \\ &= \sum_{j=0}^n \sum_{k=0}^{n-j} f_j g_k h_{n-j-k} && \text{(reindex $i = j+k$)} \\ &= \sum_{j=0}^n f_j \biggl( \sum_{k=0}^{n-j} g_k h_{n-j-k} \biggr) \\ \end{align}

BTW the multiplication corresponds to multiplication of power series $$ \biggl( \sum_{n=0}^\infty f_n t^n \biggr) \biggl( \sum_{n=0}^\infty g_n t^n \biggr) = \sum_{n=0}^\infty \biggl( \sum_{i=0}^n f_i g_{n-i} \biggr) t^n $$ which may help your intuition.

arkeet
  • 6,695