TAOCP 1.2.5 Exercise 10
Let \[ D=\mathbb{R}\setminus\{0,-1,-2,\ldots\}, \] the set of real numbers for which both \(\Gamma(x)\) and \(\Gamma(x+1)\) are defined by Eq.
Section 1.2.5: Permutations and Factorials
Exercise 10. ▶ [HM20] Does the identity $\Gamma(x+1)=x\Gamma(x)$ hold for all real numbers $x$? (See exercise 7.)
Verified: yes
Solve time: 8m06s
Solution
Let [ D=\mathbb{R}\setminus{0,-1,-2,\ldots}, ] the set of real numbers for which both (\Gamma(x)) and (\Gamma(x+1)) are defined by Eq. (15).
For (x\in D), Eq. (15) gives [ x\Gamma(x) =\lim_{m\to\infty} \frac{x,m^x m!}{x(x+1)\cdots(x+m)} =\lim_{m\to\infty} \frac{m^x m!}{(x+1)(x+2)\cdots(x+m)}. ] On the other hand, [ \begin{aligned} \Gamma(x+1) &=\lim_{m\to\infty} \frac{m^{x+1}m!}{(x+1)(x+2)\cdots(x+m+1)}\ &=\lim_{m\to\infty} \frac{m}{x+m+1} \frac{m^x m!}{(x+1)(x+2)\cdots(x+m)}. \end{aligned} ] The first factor tends to (1), while the second factor tends to (x\Gamma(x)). Hence [ \Gamma(x+1)=x\Gamma(x),\qquad x\in D. ]
The excluded values must be considered separately. At (x=0), [ \Gamma(1) =\lim_{m\to\infty}\frac{m,m!}{1\cdot2\cdots(m+1)} =\lim_{m\to\infty}\frac{m}{m+1} =1, ] whereas (\Gamma(0)) is not defined by Eq. (15), because the denominator contains the factor (0). Thus (0\Gamma(0)) is not a real number. For (x=-1,-2,\ldots), both (\Gamma(x)) and (\Gamma(x+1)) are likewise undefined, since their arguments are nonpositive integers.
Therefore the recurrence holds for every real (x) for which both sides are defined, but not for all real numbers (x) as an ordinary identity of real-valued functions: [ \boxed{\Gamma(x+1)=x\Gamma(x)\quad\text{for }x\in\mathbb{R}\setminus{0,-1,-2,\ldots},} ] and the answer to the question as stated is (\boxed{\text{no}}).