ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sat, 23 Sep 2017 20:18:01 +0200Show a multivariable function is nonvanishing when it is subject to constraintshttps://ask.sagemath.org/question/38950/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)Sat, 23 Sep 2017 16:56:19 +0200https://ask.sagemath.org/question/38950/show-a-multivariable-function-is-nonvanishing-when-it-is-subject-to-constraints/Comment by dan_fulea for <p>say we have a function $f:\mathbb R^3 \to \mathbb R$ given by</p>
<p>$f(x,y,z)=\sin(x)\sin(y)\sin(z)$</p>
<p>suppose further that there constraints $x,y,z \in (0, \pi/2)$ and $z>x+y$.</p>
<p>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. </p>
<p>I've tried </p>
<pre><code> 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)
</code></pre>
<p>but this does not work ( I don't think I understand the solve function)</p>
https://ask.sagemath.org/question/38950/show-a-multivariable-function-is-nonvanishing-when-it-is-subject-to-constraints/?comment=38955#post-id-38955Is 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 intervalSat, 23 Sep 2017 19:56:27 +0200https://ask.sagemath.org/question/38950/show-a-multivariable-function-is-nonvanishing-when-it-is-subject-to-constraints/?comment=38955#post-id-38955Comment by Andres Mejia for <p>say we have a function $f:\mathbb R^3 \to \mathbb R$ given by</p>
<p>$f(x,y,z)=\sin(x)\sin(y)\sin(z)$</p>
<p>suppose further that there constraints $x,y,z \in (0, \pi/2)$ and $z>x+y$.</p>
<p>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. </p>
<p>I've tried </p>
<pre><code> 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)
</code></pre>
<p>but this does not work ( I don't think I understand the solve function)</p>
https://ask.sagemath.org/question/38950/show-a-multivariable-function-is-nonvanishing-when-it-is-subject-to-constraints/?comment=38956#post-id-38956Maybe, but it is unclear how to extend this to the multivariable case.Sat, 23 Sep 2017 20:18:01 +0200https://ask.sagemath.org/question/38950/show-a-multivariable-function-is-nonvanishing-when-it-is-subject-to-constraints/?comment=38956#post-id-38956