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1d
comment Expression for negating every other odd number index
PairedAlt is very nice !
1d
revised Expression for negating every other odd number index
update with new knowledge about a function that has support size 1.
1d
comment Expression for negating every other odd number index
The $(-1)^{2^{n(n-1)}}$ idea is very nice !
1d
comment Expression for negating every other odd number index
Yes and no. You can get rid of the floor functions if you accept the other definition of a quasi-polynomial: A polynomial with periodic coefficients. Then "periodic" can be implemented with modular arithmetic, so the conditionals are hidden there.
1d
comment Prove that an equation has solution in R
It is always good to give some background on your question. What have you tried, where did you get stuck? In a few minutes somebody will just tell you the answer, but you missed an opportunity to learn problem solving strategy.
1d
answered Expression for negating every other odd number index
Jul
28
revised Why can mathematical induction only be used with natural numbers?
Remove rare acronym.
Jul
28
suggested approved edit on Why can mathematical induction only be used with natural numbers?
Jul
28
comment Why do remainders show cyclic pattern?
Start at Fermat's little theorem, then Euler's theorem.
Jul
28
comment Given $N$, is there a formula for $card( \{(m,n)\, s.t.\, m\cdot n \leq N \} )$?
@GerryMyerson Granted. I removed the adjective "unsolved".
Jul
28
revised Given $N$, is there a formula for $card( \{(m,n)\, s.t.\, m\cdot n \leq N \} )$?
remove unsolved
Jul
28
answered Given $N$, is there a formula for $card( \{(m,n)\, s.t.\, m\cdot n \leq N \} )$?
Jul
28
comment Given $N$, is there a formula for $card( \{(m,n)\, s.t.\, m\cdot n \leq N \} )$?
You probably also want to make some restrictions on $m,n$. Are they positive integers? Then the formula may be off. For $N=3$, I get $\lfloor 3 \rfloor + \lfloor 3/2 \rfloor + \lfloor 1 \rfloor = 5$ while the only solutions are $(1,1), (1,2), (2,1)$.
Jul
28
comment Given $N$, is there a formula for $card( \{(m,n)\, s.t.\, m\cdot n \leq N \} )$?
The sum of floor functions is a formula! It is called a quasi-polynomial. See here for more on this fascinating piece of mathematics: oberlin.edu/faculty/kwoods/research/ubiquitous_qps_full.pdf
Jul
25
answered How prove $\deg (f^{m} - g^{n})\ge \frac{mn - m - n}{n}\deg f + 1$?
Jul
21
revised How prove $\deg (f^{m} - g^{n})\ge \frac{mn - m - n}{n}\deg f + 1$?
fix typographical errors
Jul
21
suggested approved edit on How prove $\deg (f^{m} - g^{n})\ge \frac{mn - m - n}{n}\deg f + 1$?
Jul
21
comment How prove $\deg (f^{m} - g^{n})\ge \frac{mn - m - n}{n}\deg f + 1$?
Probably you need to assume that $f,g$ are not constant. Otherwise you can choose any $m,n$ and get $0\ge 1$.
Jun
28
answered Minimization on compact region
Jun
28
comment Minimization on compact region
Your critical equations are not all. You need to include $x^2+y^2+z^2=1$ and $x+y+z=0$ too. In the method of Lagrange multipliers they arise from differentiating with respect to $\lambda_i$. If you do that, you get a polynomial system with isolated solutions. I would compute them using Gröbner bases, but there are probably other methods too.