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On Concrete Mathematics section 5.5, which is teaching the hypergeometric functions, generalized factorials is defined as:

\[ \frac 1 {z!} = \lim_{n \to \infty} \binom{n+z}{n}n^{-z} \]

where

\[ \binom r k = \left\{ \begin{array}{ll} r^{\underline k} / k! = r(r-1) \cdots (r-k+1) / k! & k > 0 \\ 1 & k = 0 \\ 0 & k < 0 \end{array} \right. \]

follows the ordinary definition.

So $z! = z(z-1)!$ for all complex $z$ (except negative integers), then we can check $0! = 1$ and $n! = n(n-1) \cdots 1$ for $n > 0$.

Then, a binomial coefficient can be written

\[ \binom z w = \lim_{\zeta \to z} \lim_{\omega \to w} \frac{\zeta!}{\omega!(\zeta-\omega)!} \]

Let $t_k = \dbinom r k \dbinom s {n-k}$.

However, the succeeding paragraph says that

\[ t_k = \frac{r!}{(r-k)!k!} \frac{s!}{(s-n+k)!(n-k)!}\] and we are no longer too shy to use generalized factorials in these expressions.

without limits (it is said that we must use appropriate limiting values when these formulas give $\infty / \infty$) and considers the ratio $t_{k+1} / t_k$ for all $t_k \neq 0$ and cancels some factorials using the property $z! = z(z-1)!$

I'm "too shy" and my question remains: why can we do such canceling?

To observe closely, we take a variety of an example from section 5.7:

Considering indefinite summation \[ \sum \binom n {-k} \delta k, \qquad n < 0 \]

Let $t(k) = \dbinom n {-k} = \dfrac{n!}{(-k)!(n+k)!}$, we have \[ \frac{t(k+1)}{t(k)} = \frac{n!}{(-k-1)!(n+k+1)!} \frac{(-k)!(n+k)!}{n!} = -\frac{k}{n+k+1} \]

Let $n = -1$, we have $t(k+1) / t(k) = -1$ for $t(k) \neq 0$. But it's wrong for $k = 0$, where $t(1) = 0$ and $t(0) = 1$.

To see how the error happens, we resume the $\lim$ notation:

\begin{align*} t(k+1) &= \binom n {-k-1} \\ &= \lim_{z_2 \to 0} \lim_{z_1 \to 0} \frac{(n+z_2)!}{(-k-1+z_1)!(n+k+1-z_1+z_2)!} \\ &= \lim_{z_2 \to 0} \lim_{z_1 \to 0} \frac{-k+z_1}{n+k+1-z_1+z_2} \frac{(n+z_2)!}{(-k+z_1)!(n+k-z_1+z_2)!} \\ &= \binom n {-k} \lim_{z_2 \to 0} \lim_{z_1 \to 0} \frac{-k+z_1}{n+k+1-z_1+z_2} \end{align*} So when $n = -1$ and $k = 0$, we have $\lim_{z_2 \to 0} \lim_{z_1 \to 0} (-k+z_1)/(n+k+1-z_1+z_2) = 0$ not $-k/(n+k+1)$.

Another example (also from section 5.5) is:

\[ \lim_{x \to -1} \frac{x!}{(x-1)!} = \lim_{x \to -1} x = -1 \] but \[ \lim_{x \to -1} \frac{x!}{(2x)!} = -2 \] because of $(-z)! \Gamma(z) = \pi / \sin(z\pi)$, so expression $(-2)! / (-1)!$ is illegal.

My question is: in the frame of Concrete Mathematics, how to prevent such errors?

Thanks a lot.

  • 2
    How is $\binom{n+z}{n}$ defined during the definition of $1/z!$ for complex $z$?2012-05-03
  • 0
    @anon Related to the normal definition for binomial coefficents.2012-05-03
  • 0
    @anon I've added something. Fixed.2012-05-03

1 Answers 1