Feb
13
comment Finding the area of a circle given the inscribed figure inside it
yes, I see this now.
Feb
13
comment Finding the area of a circle given the inscribed figure inside it
I think this is simpler. If we arrange the triangles in this alternating manner and call the vertexes of the hexagon $A,B,\ldots,F$ then $A,C,E$ is an equilateral triangle with $AC=\sqrt{3}r$. $M$ is the center of the circle and $\angle(A,M,C)=\frac{2 \pi}{3}$ and so $\angle(A,B,C)=\frac{2 \pi}{3}$. Using the law of cosine we get ${AC}^2=1^2+2^2-2 \cdot 1 \cdot 2 \cdot \cos{\frac{2 \pi}{3}}=7$. So $3r^2=7$
Feb
13
answered Finding the area of a circle given the inscribed figure inside it
Feb
12
revised Function F(x,y) which is high when (both x and y are high) and (x is close to y)
grammar
Feb
12
comment If a is an arbitrary integer, then 6|a(a^2+11)
using Maxima makelist(mod(a*(a^2+11),6),a,0,5); gives [0,0,0,0,0,0]
Feb
10
comment Show that $x^4 + 8$ is irreducible over Z
@neofoxmulder No, he picked $x=9$ for the polynomial $x^4+8$ because $9$ is greater than all coefficients of the polynom. Also the coefficients have to be not negative to apply Cohn's test. For the polynomial $x^3+1$ you have to select an $x \ge 2$ for Cohn's test.
Feb
9
answered How to quickly find $\sqrt{x^4+4x^3+2x^2-4x+1}$ or anything similar
Feb
3
revised Gateaux Derivative.
correct log function in latex
Feb
3
comment proving zeros of a polynomial are not real
see math.stackexchange.com/questions/494243/… there I use Sturm's theorem to show this. This method is straightforward and is implemented in some CAS (e.g. Maxima) to find the numbers of real roots in an interval
Feb
2
comment A dynamic dice game
I think this is the same question as math.stackexchange.com/questions/661166/…
Feb
2
comment What is a good strategy for this dice game?
I think this is the same question as math.stackexchange.com/questions/660523/a-dynamic-dice-game
Feb
2
comment What is a good strategy for this dice game?
@String I see no reason to assume that they don't count
Feb
2
comment Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it's an integer
-1 It is true that $$a^2+b^2={{a}\over{b}}\,\left(a\,b+1\right)+\left(b^2-{{a}\over{b}}\right)$$ but one cannot conclude that the remainder $b^2-a/b=0$ if one divides $a^2+b^2$ by an arbitrary number.
Feb
2
revised Finding the Robot
added 454 characters in body
Feb
2
revised Finding the Robot
extending the statement
Feb
2
revised Finding the Robot
generalize the reesult for n boxes
Feb
2
revised Finding the Robot
changed the text to adopt the new function
Feb
2
revised Finding the Robot
a better program, some output
Feb
2
answered Finding the Robot
Feb
2
revised Numerical puzzle
deleted 47 characters in body