Exercise 8.4.8

Question

Inspired by the fact that $0! = 1$ and $1! = 1$, let $h(x)$ satisfy
\begin{enumerate}[label=(\roman*)] 
\item $h(x) = 1$ \quad for all $0 \leq x \leq 1$, and
\item $h(x) = x h(x-1) $ \quad for all $x \in \mathbf{R}$.
 \end{enumerate}
\begin{enumerate}[label=(\alph*)] 
    \item Find a formula for $h(x)$ on $[1,2]$, $[2,3]$, and $[n, n+1]$ for arbitrary $n \in \mathbf{N}$.
    \item Now do the same for $[-1, 0]$, $[-2, -1]$, and $[-n, -n+1]$.
    \item Sketch $h$ over the domain $[-4, 4]$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item Over $[1,2]$, $h(x) = x h(x-1) = x$. Over $[2,3]$, $h(x) = x h(x-1) = x(x-1)$. Over $[n, n+1]$, $h(x) = \prod^{n-1}_{i=0} x-i$, which can readily be proven with induction.
\item Over $[-1, 0]$, $(x+1) h(x) = h(x+1) = 1 \implies h(x) = \frac{1}{x + 1}$. Over $[-2, 1]$, $(x+1) h(x) = h(x+1) = \frac{1}{(x+1)} \implies h(x) = \frac{1}{(x+1)^2}$. Over $[-n, -n+1]$, $h(x) = (x+1)^{-n}$.
\item This is an exercise best left for your favourite graphing utility.
 \end{enumerate}

Exercise 8.4.8

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Exercise 8.4.8

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