4 ms·
Pick any `b` you want, even infinity, and evaluate: sum_n^b 2^n (floor(0/2^n) mod 2) (floor(-1/2^n) mod 2) sum_n^b 2^n * 0 * 1 sum_n^b 0 0
by Strilanc 10y ago
Pick any `b` you want, even infinity, and evaluate:
sum_n^b 2^n (floor(0/2^n) mod 2) (floor(-1/2^n) mod 2)
sum_n^b 2^n * 0 * 1
sum_n^b 0
0
- thedufer 10y agoExcellent. I'm glad we agree. You'll notice that the stated identity f(x,x) = x fails to hold over x < 0. I hope this is sufficient to make it clear that the equation does not consider the negatives?
- Strilanc 10y agoSeems to work fine for -1: f(-1, -1) = sum_{n=0}^b 2^n (floor(-1/2^n) mod 2) (floor(-1/2^n) mod 2) = sum_{n=0}^b 2^n * 1 * 1 = 1 + 10 + 100 + 1000 + ... [in binary] = ...11111 [in the 2-adics] = -1