# Revision history [back]

### Show a function is nonvanishing in sage 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.

### Show a function is nonvanishing in sage 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

### Show a function is nonvanishing in sage 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

### Show a function is nonvanishing in sage 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

### Show a multivariable function is nonvanishing in sage 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)
f=sin(x)*sin(y)*sin(z)
solve(f==0,x,y,z)
solve(f=0,x,y,z)


but this does not work ( I don't think I understand the solve functionfunction)