4

I'm a student of computer science and was reading through my algorithms textbook about matrix chain multiplication. It brought up Catalan numbers and I was hoping to prove the lower bounds on it. This is not an assignment.

It's pretty well-known that the (tightest?) lower bound is $\Omega(4^n n^{-3/2})$. How can I prove this? I suspect that Stirling's approximation would be used, but this is seemingly a different beast altogether.

erip
  • 145
  • Isn't $C_n = \frac1{n+1}\binom{2n}n$? Then you can expand $\binom{2n}n = \frac{2n!}{n!n!}$ and use Stirling's approximation for the factorials. – MJD Dec 01 '14 at 14:06

1 Answers1

6

As MJD said, just use the fact that $C_n=\frac1{n+1}\binom{2n}n$ and Stirling’s approximation:

$$C_n=\frac{(2n)!}{(n+1)n!^2}\sim\frac{\sqrt{4\pi n}(2n/e)^{2n}}{2\pi n(n+1)(n/e)^{2n}}=\frac{4^n}{\sqrt{\pi n}(n+1)}\sim\frac1{\sqrt\pi}\cdot4^nn^{-3/2}\;.$$

Brian M. Scott
  • 631,399