TAOCP 1.2.5 Exercise 5

Exercise 5. [**] [15] Estimate $8!$ using the following more exact version of Stirling's approximation:

$$ n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+\frac{1}{12n}\right). $$

x 6. [17] Using Eq. (8), write $20!$ as a product of prime factors.

Verified: yes
Solve time: 37s

Applying the refined form of Stirling’s approximation with $n=8$ gives

$$ 8! \approx \sqrt{2\pi\cdot 8}\left(\frac{8}{e}\right)^8\left(1+\frac{1}{12\cdot 8}\right). $$

Since

$$ \sqrt{16\pi}=4\sqrt{\pi},\qquad \left(\frac{8}{e}\right)^8=2^{24}e^{-8}, \qquad 1+\frac{1}{96}=\frac{97}{96}, $$

we obtain

$$ 8!\approx 4\sqrt{\pi}\left(\frac{8}{e}\right)^8\frac{97}{96} =2^{26}\sqrt{\pi},e^{-8}\frac{97}{96}. $$

Using $\sqrt{\pi}\approx1.77245$ and $e^{-8}\approx0.00033546$,

$$ 8!\approx 67108864\cdot1.77245\cdot0.00033546\cdot\frac{97}{96} \approx 40318. $$

Since $8!=40320$ exactly, the estimate differs by only about $2$. Therefore

$$ \boxed{8!\approx 40318}. $$