# How to calculate the fraction factorial without doing it manually?

We were asked to list the first $$3$$ terms of $$\frac1{\sqrt{4+3x}} = \frac12\left(1+\frac{3x}4\right)^{-\frac12}$$ which is straightforward and easy.

Using the following with $$n=-1/2$$:

$$(1+a)^n = {n \choose 0} + {n \choose 1}a + {n \choose 2}a^n + ...$$

and the coefficients are... $${n \choose 0} = 1$$ $${n \choose 1} = n$$ $${n \choose 2} = \frac{n(n-1)}{2!}$$ $${n \choose 3} = \frac{n(n-1)(n-2)}{3!}$$

and you end up with

$$\frac12 - \frac3{16}x+\frac{27}{256}x^2+\cdots$$

However, how are you able to determine the coefficient of the 60th term without doing it manually? My scientific calculator is not able to compute a fraction factorial, it returns a math error.

$${n \choose 60} = \frac{n(n-1)(n-2)\cdots(n-58)(n-59)}{60!}$$

• I think if you organize the calculation as $(n/60)*((n-1)/59)*\cdots$ your computer could handle it. We avoid overflow this way. Commented Aug 21, 2019 at 5:34
• Of course $$\binom{-\frac12}n=\frac{(-1)^n}{4^n}\binom{2n}n$$ but I don't think that helps any.
– bof
Commented Aug 21, 2019 at 7:47

\begin{align}\binom{-\frac12}{60}&=\frac{-\frac12\left(-\frac12-1\right)\left(-\frac12-2\right)\cdots\left(-\frac12-58\right)\left(-\frac12-59\right)}{1\cdot2\cdot3\cdot\cdots\cdot59\cdot60}\\&=\frac{\left(1-\frac12\right)\left(2-\frac12\right)\left(3-\frac12\right)\cdots\left(59-\frac12\right)\left(60-\frac12\right)}{1\cdot2\cdot3\cdot\cdots\cdot59\cdot60}\\&=\prod_{i=1}^{60}\left(1-\frac1{2i}\right)\\&=\exp\left(\sum_{i=1}^{60}\ln\left(1-\frac1{2i}\right)\right)\end{align}