2019-12-16 12:59:25 +0200 received badge ● Popular Question (source) 2017-09-26 22:14:01 +0200 received badge ● Student (source) 2017-09-23 20:18:01 +0200 commented question Show a multivariable function is nonvanishing when it is subject to constraints Maybe, but it is unclear how to extend this to the multivariable case. 2017-09-23 17:01:25 +0200 commented answer Call error for integers (when I haven't declared any.) sigh I can't believe that I did not see this. Thank you 2017-09-23 17:01:09 +0200 received badge ● Scholar (source) 2017-09-23 16:58:36 +0200 received badge ● Editor (source) 2017-09-23 16:56:19 +0200 asked a question Show a multivariable function is nonvanishing when it is subject to constraints say we have a function $f:\mathbb R^3 \to \mathbb R$ given by $f(x,y,z)=\sin(x)\sin(y)\sin(z)$ suppose further that there constraints $x,y,z \in (0, \pi/2)$ and $z>x+y$. Clearly this function is nonvanishing with these constraints. Is there a way to get sage to show this? I've tried fiddling around, but I'm not sure how to do it. I've tried  var('x,y,z') assume(pi/2>x>0) assume(pi/2>y>0) assume(pi/2>z>x+y) f=sin(x)*sin(y)*sin(z) solve(f=0,x,y,z)  but this does not work ( I don't think I understand the solve function) 2017-09-23 03:20:04 +0200 asked a question Callback Error for integers (when I didn't declare any) So, I declared some variables assume(k,'real') assume(pi/2>k>0) assume(j,'real') assume(pi/2>j>0) assume(N,'real') assume(N>0)  and I wanted to define a function (that I eventually want to solve for zeroes of.) Basically, the function is f=sin(2*k*2*pi/N)*sin((k+1)*2*pi/N)*sin((k-1)*2*pi/N)*sin((2*j+2)*2*pi/N)*sin^2(j*2*pi/N)*sin(6*pi/N)*sin(4*pi/N)*sin(2*pi/N)+sin(2*j*2*pi/N)*sin((j+1)*2*pi/N)*sin((j-1)*2*pi/N)*sin((2*k+2)*2*pi/N)*sin(k*2*pi/N)*sin(k*2*pi/N)*sin(6*pi/N)*sin(4*pi/N)*sin(2*pi/N)+sin((k-1)*2*pi/N)*sin((2*k+2)*2*pi/N)*sin(k*2*pi/N)*sin((2*j+2)*2*pi/N)*sin(j*2*pi/N)*sin(j*2*pi/N)*sin(12*pi/N)*sin(4*pi/N)*sin(2*pi/N)+sin((j-1)*2*pi/N)*sin((2*j+2)*2*pi/N)*sin(j*2*pi/N)*sin((2*k+2)*2*pi/N)*sin(k*2*pi/N)*sin(k*2*pi/N)*sin(12*pi/N)*sin(4*pi/N)*sin(2*pi/N)-sin((k-1)*2*pi/N)*sin((j-1)*2*pi/N)*sin((2*k+2)*2*pi/N)*sin((2*j+2)*2*pi/N)*sin(k*2*pi/N)*sin(j*2*pi/N)*sin(6*pi/N)*sin(2*pi/N)* sin(8*pi/N)-sin(12*pi/N)*sin(4*(pi/N))*sin(4*(pi/N))*sin((2*k+2)*2*pi/N)*sin((2*j+2)*2*pi/N)*sin(j*2*pi/N)*sin(j*2*pi/N)*sin(k*2*pi/N)*sin(k*2*pi/N)  But I'm getting 'sage.rings.integer.Integer' object is not callable which just do not understand. 2017-09-23 03:17:11 +0200 asked a question Call error for integers (when I haven't declared any.) I'm trying to define a pretty atrocious function (that I eventually want to solve for zeroes) in three variables, k,j,N. I tried to just use f=sin(2*k*2*pi/N)*sin((k+1)*2*pi/N)*sin((k-1)*2*pi/N)*sin((2*j+2)*2*pi/N)*sin^2(j*2*pi/N)*sin(6*pi/N)*sin(4*pi/N)*sin(2*pi/N)+sin(2*j*2*pi/N)*sin((j+1)*2*pi/N)*sin((j-1)*2*pi/N)*sin((2*k+2)*2*pi/N)*sin(k*2*pi/N)*sin(k*2*pi/N)*sin(6*pi/N)*sin(4*pi/N)*sin(2*pi/N)+sin((k-1)*2*pi/N)*sin((2*k+2)*2*pi/N)*sin(k*2*pi/N)*sin((2*j+2)*2*pi/N)*sin(j*2*pi/N)*sin(j*2*pi/N)*sin(12*pi/N)*sin(4*pi/N)*sin(2*pi/N)+sin((j-1)*2*pi/N)*sin((2*j+2)*2*pi/N)*sin(j*2*pi/N)*sin((2*k+2)*2*pi/N)*sin(k*2*pi/N)*sin(k*2*pi/N)*sin(12*pi/N)*sin(4*pi/N)*sin(2*pi/N)-sin((k-1)*2*pi/N)*sin((j-1)*2*pi/N)*sin((2*k+2)*2*pi/N)*sin((2*j+2)*2*pi/N)*sin(k*2*pi/N)*sin(j*2*pi/N)*sin(6*pi/N)*sin(2*pi/N)* sin(8*pi/N)-sin(12*pi/N)*sin(4*(pi/N))*sin(4*(pi/N))*sin((2*k+2)*2*pi/N)*sin((2*j+2)*2*pi/N)*sin(j*2*pi/N)*sin(j*2*pi/N)*sin(k*2*pi/N)*sin(k*2*pi/N)  but I'm getting: TypeError: 'sage.rings.integer.Integer' object is not callable why is this, I already had that assume(k,'real') assume(pi/2>k>0) assume(j,'real') assume(pi/2>j>0) assume(N,'real') assume(N>0)  so what can I do to fix this error? 2017-08-07 23:10:42 +0200 asked a question Is there a way to check whether or not this is a floating point error? I have the following functions defined: var('k,j,N') s(k)=sin(4*pi*k/N)*sin(2*pi/N*(k+1))*sin(2*pi/N*(k-1)) w(k)=(sin(2*pi/N*(k-1))/sin(2*pi/N*k)*sin(4*pi/N))*sin(2*pi/N) AA(N,j,k)=s(k)+s(j)+sin(2*pi*6/N)*sin(2*pi*2/N)/(sin(2*pi*3/N)*sin(2*pi/N))*(w(k)+w(j))-sin(4*pi*2/N)*sin(2*2*pi/N)/sin(2*pi/N)*(w(k)*w(j))-sin(2*pi*6/N)*sin(2*pi*2/N)/(sin(2*pi*3/N)*sin(2*pi/N)) assume(j>1,k>1,N>8*(j+k+1),N>10)  Now, if I use the solve function: sage: solve(AA(N,j,k)==0,N) I get the output [sin(4*pi*k/N) == (sin(8*pi/N)*sin(6*pi/N)*sin(4*pi/N)^3*sin(2*pi/N)^2*sin(-2*(pi - pi*j)/N)*sin(-2*(pi - pi*k)/N) - sin(2*pi*j/N)*sin(12*pi/N)*sin(4*pi/N)^2*sin(2*pi/N)*sin(-2*(pi - pi*k)/N) - (sin(4*pi*j/N)*sin(2*pi*j/N)*sin(6*pi/N)*sin(2*pi/N)*sin(2*(pi + pi*j)/N)*sin(-2*(pi - pi*j)/N) + sin(12*pi/N)*sin(4*pi/N)^2*sin(2*pi/N)*sin(-2*(pi - pi*j)/N) - sin(2*pi*j/N)*sin(12*pi/N)*sin(4*pi/N))*sin(2*pi*k/N))/(sin(2*pi*j/N)*sin(2*pi*k/N)*sin(6*pi/N)*sin(2*pi/N)*sin(2*(pi + pi*k)/N)*sin(-2*(pi - pi*k)/N))]  However, it is my hope that this equation has no solutions. Indeed, if I add to the assumption that AA(N,j,k)>0, I obtain a contradiction (inconsistent assumptions), but if I add AA(N,j,k)==0, I don't get inconsistent assumptions. Is there a way to check if this is a floating point error, or if there really is a solution with my assumptions? 2017-07-29 05:02:34 +0200 asked a question Sage not returning roots of polynomimal I have a polynomial w=q^32 - q^30 + 3q^28 - 3q^26 + 6q^24 - 6q^22 + 9q^20 - 9q^18 + 12q^16 - 9q^14 + 9q^12 - 6q^10 + 6q^8 - 3q^6 + 3*q^4 - q^2 + 1 and I tried using solve(w==0,q). But sage only returns [0 == q^32 - q^30 + 3q^28 - 3q^26 + 6q^24 - 6q^22 + 9q^20 - 9q^18 + 12q^16 - 9q^14 + 9q^12 - 6q^10 + 6q^8 - 3q^6 + 3*q^4 - q^2 + 1] why is this? I'm looking for complex roots.