Exercise 7.1.6

Answers

For $a = b = 1$ , we have $2^{N} = \sum_{k = 0}^{N} (\binom{N}{k}) = r (N, N)$ . The coefficients $(\binom{N}{k})$ sum up to $2^{N}$

If we fold $R^{N}$ space $N$ times, we have $2^{N}$ pieces. For $r (N, m)$ with $m \geq N$ , we always get $2^{N}$ pieces.

niuers

2020-03-20 00:00