Exercise 8.4.21

Question

\begin{enumerate}[label=(\alph*)] 
\item Use the convexity of $\log(f(x))$ and the three intervals $[n-1,n]$, $[n, n+x]$, and $[n, n+1]$ to show
$$x \log(n) \leq \log (f(n+x)) - \log (n!) \leq x \log (n+1)$$
\item Show $\log(f(n+x)) = \log (f(x)) + \log((x+1)(x+2) \cdots (x+n))$.
\item Now establish that
$$0 \leq \log(f(x)) - \log\left(\frac{n^x n!}{(x+1)(x+2)\cdots(x+n)}\right) \leq x \log (1 + \frac{1}{n})$$
\item Conclude that
$$f(x) = \lim_{n \to \infty} \frac{n^x n!}{(x+1)(x+2) \cdots (x+n)}, \text{\quad for all } x \in (0, 1].$$
\item Finally, show that the conclusion in $d$ holds for all $x \geq 0 $.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item Since the slopes are increasing between these three intervals,
$$\log(f(n)) - \log(f(n-1)) \leq \frac{\log(f(n+x)) - \log(f(n))}{x} \leq \log(f(n+1)) - \log(f(n)) $$
$$x \log\left(\frac{f(n)}{f(n-1)}\right) \leq \log(f(n+x)) - \log(n!) \leq x \log\left(\frac{f(n+1)}{f(n)}\right) $$
$$x \log(n) \leq \log(f(n+x)) - \log(n!) \leq x \log (n+1)$$
\item By property (ii),
$$f(n+x) = f(x) \prod^n_{i=1} (x+i) $$
so
$$\log(f(n+x)) = \log(f(x)) + \log\left(\prod^n_{i=1}(x+i)\right) $$
\item Plug (b) into (a):
$$x \log(n) \leq \log(f(x)) - \left(\log(n!) - \log\left(\prod^n_{i=1}(x+i)\right) \right) \leq x \log (n+1)$$
$$0 \leq \log(f(x)) - \left(\log(n!) - \log\left(\prod^n_{i=1}(x+i)\right)  + \log\left(n^x\right)\right) \leq x \log (n+1) - x \log(n)$$
$$0 \leq
\log(f(x)) - \log\left(\frac{n^x n!}{\prod^n_{i=1}(x+i)}\right)
\leq x \log \left(\frac{n+1}{n}\right) = x \log \left(n + \frac{1}{n}\right)$$
 \end{enumerate}

\item $\lim_{n \to \infty} \log \left(1 + \frac{1}{n}\right) = 0$ so by the Squeeze Theorem,
$$\log(f(x)) - \lim_{n \to \infty} \log\left(\frac{n^x n!}{\prod^n_{i=1}(x+i)}\right) =0
$$
and since $\log$ is continuous,
$$f(x) = \lim_{n \to \infty} \frac{n^x n!}{\prod^n_{i=1}(x+i)}$$

\item We only need to show this equation holds for $x = 0$. We have $f(0) = 1$ by property (i), and
$$\lim_{n \to \infty} \frac{n^0 n!}{\prod^{n}_{i=1} i} = \lim_{n \to \infty} \frac{n!}{n!} = 1 = f(0)$$

Exercise 8.4.21

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Exercise 8.4.21

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