Exercise 8.4.22

Question

\begin{enumerate}[label=(\alph*)] 
\item Where does $g(x) = \frac{x}{x! (-x)!}$ equal zero? What other familiar function has the same set of roots?
\item The function $e^{-x^2}$ provides the raw material for the all-important Gaussian bell curve from probability, where it is known that $\int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}$. Use this fact (and some standard integration techniques) to evaluate $(1/2)!$.
\item Now use (a) and (b) to conjecture a striking relationship between the factorial function and a well-known function from trigonometry.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\item $g(x)$ has roots at all integers, as does $\sin \pi x$.
\item Apply integration by parts:
$$\left(\frac{1}{2}\right)! = \int_0^\infty \sqrt(t) e^{-t} dt = \left(-\lim_{t \to \infty} \sqrt{t} e^{-t}\right) - (-\sqrt{0} e^{0}) + \int_0^\infty e^{-t} \frac{1}{2\sqrt{t}} dt = \int_0^\infty e^{-t} \frac{1}{2 \sqrt{t}} dt $$
Apply change-of-variable formula (Theorem 8.1.10), with $u = \sqrt{t}$ taking advantage of $t \geq 0$:
$$\int_0^\infty e^{-t} \frac{1}{2\sqrt{t}}dt = \int_0^\infty e^{-u^2} du = \frac{1}{2}\int_{-\infty}^\infty e^{-u^2} du = \frac{\sqrt{\pi}}{2}$$
\item We have $(-1/2)! = 2 (1/2)! = \sqrt{\pi}$, so from part (a) $g(1/2) = \frac{1}{\pi}$. The conjecture is
$$\frac{x}{x! (-x)!} = \frac{\sin(\pi x)}{\pi}$$

Exercise 8.4.22

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Exercise 8.4.22

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