Exercise 8.4.20

Question

\begin{enumerate}[label=(\alph*)] 
\item Show that $x!$ is an infinitely differentiable function on $(0, \infty)$ and produce a formula for the $n^{th}$ derivative. In particular show that $(x!)'' > 0$.
\item Use the integration-by-parts formula employed earlier to show that $x!$ satisfies the functional equation
$$(x+1)! = (x+1)x!$$
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Before continuing, note that $t^x$ currently isn't defined at $t=0$; this can easily be solved by defining $0^x = 0$. We show that $x!$ is infinitely differentiable at $x_0 > 0$ by repeatedly applying Theorem 8.4.9, over the domain $D = \{(x,t) : x_0 / 2 \leq x \leq x_0 + 1, 0 \leq t\}$. Denote the $n$'th derivative of $t^x e^{-t}$ with respect to $x$ as $f_n(x,t) = t^x e^{-t} \log^n t$ (except at $t = 0$, where $f(x, t) = 0$). The work in Exercise 8.4.19 (a) means that to show $f_n$ is continuous over $D$, we only need to show that $t^x$ is continuous. % We can use the same technique as was used to show $g(\alpha, t) = e^{-\alpha t}$ is continuous in Exercise 8.4.19 (a). Note that for any fixed $(x_1, t_1) \in D$, $$ \left|t_1^{x_1}- t^x \right| \leq \left|t_1^{x_1} - t^{x_1}\right| + \left|t^x\right|\left|t^{x_1 - x} - 1\right|$$ Since $g(t) = t^{x_1}$ is continuous, we can require $\left|t_1 - t\right| < \delta_t$ to ensure the first term is less than $\epsilon/2$. Since $t^x$ is bounded for finite $t$ and $x$, let $M_1$ be an upper bound for $|t^x|$ over $V_1(x_1, t_1) \cap D$. Let $M_2$ be an upper bound for $|\log t|$ over $V_{t_1 / 2}(t_1)$. We can find $\delta_1$ so that when $\left|(x_1 - x) \log t\right| < \delta_1$, $\left|e^{(x_1 - x) \log t} - 1\right| < \frac{\epsilon}{2 M_1}$. Then requriring $|x_1 - x| < \frac{\delta_1}{M_2} = \delta_x$ guarentees that $$\left|t^x\right|\left|t^{x_1 - x} - 1\right| \leq M_1 \left|e^{(x_1 - x) \log t} - 1\right| < \frac{\epsilon}{2}$$ Putting it all together, $$\|(x, t) - (x_0, t_0)\| < \min\left\{\delta_t, \delta_x, 1, \frac{t_1}{2}\right\} \implies \left|t^x - t_0^{x_0}\right| < \epsilon $$ proving $t^x$ is continuous. We also need to show that $\int_0^\infty f_n(x,t) dt$ converges uniformly. Now, since $x > \log x \geq 0$ for $x \geq 1$, we can show $\int_1^\infty f_n(x,t) dt$ converges uniformly by comparison against $\int_1^\infty t^{x + n} e^{-t} dt$. We also have $f_n(x,t)$ continuous in $t$ over $[0, 1]$ so $\int_0^1 f_n(x,t) dt$ is well defined. Thus, repeatedly applying Theorem 8.4.9 lets us conclude that $$(x!)^{(n)} = \int_0^\infty t^x e^{-t} \left(\log t\right)^n dt$$ Note that for $(x!)''$, the term under the integral is always positive, so $(x!)'' > 0$. \item $$x! = \int_0^\infty t^x e^{-t} dt = 0^x e^0 - \lim_{t \to \infty} t^x e^{-t} + \int_0^\infty x t^{x-1} e^t dt = x (x-1)! $$ which is equivalent to saying $(x+1)! = (x+1) x!$. \end{enumerate}

Exercise 8.4.20

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Exercise 8.4.20

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