Show a multivariable function is nonvanishing when it is subject to constraints

asked 2017-09-23 16:56:19 +0100

Andres Mejia gravatar image

updated 2017-09-23 20:18:49 +0100

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)

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Comments

Is the following enough?

sage: f = sin(x)
sage: f.find_root( 0, pi/2 )
0.0
sage: try:
....:     f.find_root( 0.00001, pi/2 )
....: except RuntimeError, e:
....:     print "no root...", e
....:     
no root... f appears to have no zero on the interval
dan_fulea gravatar imagedan_fulea ( 2017-09-23 19:56:27 +0100 )edit

Maybe, but it is unclear how to extend this to the multivariable case.

Andres Mejia gravatar imageAndres Mejia ( 2017-09-23 20:18:01 +0100 )edit
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