Show a multivariable function is nonvanishing when it is subject to constraints
 
  
 
  
 
 
 
 
 
 
    
        
        asked 7 years ago  
 
 Andres Mejia gravatar image  
 
 
 
 
    
        
        updated 7 years ago  
 
 
 
 
 
say we have a function  f:R3→R given by
 
f(x,y,z)=sin(x)sin(y)sin(z)
 
suppose further that there constraints x,y,z∈(0,π/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 (
    
        7 years ago
    )
Maybe, but it is unclear how to extend this to the multivariable case.
Andres Mejia gravatar imageAndres Mejia (
    
        7 years ago
    )
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