1,126 reputation
816
bio website embeddedrelated.com/blogs-1/…
location Arizona
age
visits member for 4 years, 5 months
seen Dec 9 at 1:41

Apr
5
asked constructing “pseudonoise” sequences other than (2^n)-1? (low cyclical autocorrelation)
Mar
7
comment How can adding an infinite number of rationals yield an irrational number?
Infinite sums have meaning, sure. But I have a finite lifetime, and each addition step takes a finite time -- therefore I cannot add infinitely many numbers. Neither can a computer, by the same reasoning. However I can use reasoning in some cases to compute the limit of an infinite sum (as could a computer in some other cases, with sufficient programming). The difference may be a bit pedantic, but this is mathematics, after all.
Feb
24
awarded  Taxonomist
Dec
26
comment how to distribute n red and m blue balls in some containers to maximize probability of random picking a red one from them?
off-topic. Best fit is probably math.stackexchange (NOT mathoverflow.net)
Oct
6
awarded  Nice Answer
Jul
21
awarded  Yearling
Jun
22
accepted order-independent accumulator operations?
May
2
comment order-independent accumulator operations?
That's reasonable. I was going to put only + and XOR in my original post, but the context for this is an "accumulator" where you start with a = 0 and keep applying a = f(a,x) for some function/operation f. The functions f(a,x) = a+x, f(a,x) = a^x, and f(a,x) = a-x were the only ones I could think of where the order didn't matter.
May
2
comment order-independent accumulator operations?
cool! your constructive approach in the first paragraph is interesting, although in a practical sense, you'd have to be mapping back and forth between regular integers and permuted sets of integers. (which I suppose explains your "tongue-in-cheek" comment)
May
2
comment order-independent accumulator operations?
0 - a - b - c - d = 0 - b - c - a - d
May
2
asked order-independent accumulator operations?
Apr
2
comment There are infinitely many primes such that $p^n \equiv 1 \pmod{b^m}$ implies $b^{m-1} \mid n$
@Arturo: OK, I follow. (though your restatement is slightly off: the claim is that for fixed m and n, for any odd b, there are infinitely many primes p such that (if p^n congruent to 1 (mod b^m), then b^(m-1) | n). (my parentheses as per the original problem statement))
Apr
2
comment There are infinitely many primes such that $p^n \equiv 1 \pmod{b^m}$ implies $b^{m-1} \mid n$
:confused: b is any given odd positive integer, that's a condition on b... I'm just not following why the contrapositive is false.
Apr
2
comment There are infinitely many primes such that $p^n \equiv 1 \pmod{b^m}$ implies $b^{m-1} \mid n$
that may be, but why would that imply the OP's theorem isn't true?
Feb
14
revised LFSR with limited numbers of runs?
added 216 characters in body
Feb
14
asked LFSR with limited numbers of runs?
Dec
11
awarded  Quorum
Nov
3
asked characterizing noise PDF
Nov
3
revised What are the three cube roots of -1?
edited tags
Oct
30
comment Does Stirling's formula give the correct number of digits for $n!\phantom{}$?
Please edit the title to be useful.