Exercise 8.2

Question

ewcommand\seqfin[2]{\left(#1_1, \ldots, #1_{#2}ight)}

ewcommand\seqinf[1]{\left(#1_1, #1_2, \ldotsight)}
\def{ho}

Let $\seqinf{b}$ be an infinite sequence of real numbers.
We define the product $\prod_{k=1}^{n} b_k$ by the equations
\begin{align*}
\prod_{k=1}^{1} b_k &= b_1 \,, \
\prod_{k=1}^{n} b_k &= \left(\prod_{k=1}^{n-1} b_kight) \cdot b_n \hspace{1cm} 	ext{for $n > 1$.}
\end{align*}
Use Theorem~8.4 to define the product rigorously.
We sometimes denote the product $\prod_{k=1}^{n} b_k$ by the symbol $b_1 b_2 \cdots b_n$.

Dan Whitman · Accepted Answer

ewcommand\seqfin[2]{\left(#1_1, \ldots, #1_{#2}ight)}

ewcommand\seqinf[1]{\left(#1_1, #1_2, \ldotsight)}
\def{ho}
\defest{estriction}

First, for any function $f : \left\{1, \ldots, might\} 	o {\mathbb{R}}$, define $$ by $(f) = f(m) \cdot b_{m+1}$.
Then, by the recursion theorem (Theorem~8.4), there is a unique function $p: {{\mathbb{Z}}_+} 	o {\mathbb{R}}$ such that
\begin{align*}
p(1) &= b_1 \,, \
p(n) &= (p est \left\{1, \ldots, n-1ight\}) \hspace{1cm} 	ext{for $n > 1$.}
\end{align*}
Then we define $\prod_{k=1}^{n} b_k = p(n)$ so that we have $\prod_{k=1}^{1} b_k = p(1) = b_1$ and
\begin{gather*}
\prod_{k=1}^{n} b_k = p(n) = (p est \left\{1, \ldots, n-1ight\}) = p(n-1) \cdot b_{(n-1)+1} = \left(\prod_{k=1}^{n-1} b_kight) \cdot b_n
\end{gather*}
for $n>1$ as desired.

Exercise 8.2

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Exercise 8.2

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