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Why is Resolve not reducing my set?

Posted 4 months ago
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I tried to run the following get the region where (v,b,r) satisfies some complicated set of inequalities.

Resolve[Exists[{d}, (v + b (1 - r) (d)^0.5 - 
    d - ((v + b (1 - r) (d)^0.5 - d)^2 - v^2)^0.5) (v + 
    b (1 + r) (d)^0.5 - 
    d + ((v + b (1 - r) (d)^0.5 - d)^2 - v^2)^0.5) - 
 4 (v/2 + r^2 b^2/8)^2 == 0 && d > 0 && d < v&& ((3/4 b^2 r^2 < v < 3/4 b^2 && 
   r < (8 v*b^2 + b^4)/(4 v + b^4)) || (v > 3/4 b^2 && 
   r < ((1 + b^2) v + b^4/4)/(4 v + b^4))) && 0 < b < 1 && v > 0 &&
r > 0],Reals]

However, Mathematica always return the same set of existence conditions without numerical approximation. There is no sign of any evaluation.

Does anyone know why Resolve fails to work in this situation? What other commands can I try? Thanks!

POSTED BY: Yue Li
Answer
6 Replies

I think you need to specify the variables if you restrict to Reals

Posted 4 months ago

Thanks for your reply. I changed the code to

Resolve[%, {v, b, r}, Reals]

But the output is still the same. Do you know how I can get a numerical approximation of this set? Thanks!

POSTED BY: Yue Li
Answer

I tried Reduce, which is more powerful than Resolve and got a message that it can't be done with Reduce.

In[3]:= Reduce[ Exists[{d}, (v + b (1 - r) (d)^0.5 - d - ((v + b (1 - r) (d)^0.5 - d)^2 - v^2)^0.5) (v + b (1 + r) (d)^0.5 - d + ((v + b (1 - r) (d)^0.5 - d)^2 - v^2)^0.5) - 4 (v/2 + r^2 b^2/8)^2 == 0 && d > 0 && d < v && ((3/4 b^2 r^2 < v < 3/4 b^2 && r < (8 v*b^2 + b^4)/(4 v + b^4)) || (v > 3/4 b^2 && r < ((1 + b^2) v + b^4/4)/(4 v + b^4))) && 0 < b < 1 && v > 0 && r > 0], {d, v, b, r}, Reals]

During evaluation of In[3]:= Reduce::nsmet: This system cannot be solved with the methods available to Reduce.

Out[3]= Reduce[\!\(
\*SubscriptBox[\(\[Exists]\), \({d}\)]\((\(-4\)\ 
\*SuperscriptBox[\((
\*FractionBox[\(
\*SuperscriptBox[\(b\), \(2\)]\ 
\*SuperscriptBox[\(r\), \(2\)]\), \(8\)] + 
\*FractionBox[\(v\), \(2\)])\), \(2\)] + \((\(-d\) + b\ 
\*SuperscriptBox[\(d\), \(0.5`\)]\ \((1 - r)\) + v - 
\*SuperscriptBox[\((\(-
\*SuperscriptBox[\(v\), \(2\)]\) + 
\*SuperscriptBox[\((\(-d\) + b\ 
\*SuperscriptBox[\(d\), \(0.5`\)]\ \((1 - r)\) + 
               v)\), \(2\)])\), \(0.5`\)])\)\ \((\(-d\) + b\ 
\*SuperscriptBox[\(d\), \(0.5`\)]\ \((1 + r)\) + v + 
\*SuperscriptBox[\((\(-
\*SuperscriptBox[\(v\), \(2\)]\) + 
\*SuperscriptBox[\((\(-d\) + b\ 
\*SuperscriptBox[\(d\), \(0.5`\)]\ \((1 - r)\) + 
               v)\), \(2\)])\), \(0.5`\)])\) == 0 && d > 0 && 
    d < v && \((\((
\*FractionBox[\(3\ 
\*SuperscriptBox[\(b\), \(2\)]\ 
\*SuperscriptBox[\(r\), \(2\)]\), \(4\)] < v < 
\*FractionBox[\(3\ 
\*SuperscriptBox[\(b\), \(2\)]\), \(4\)] && r < 
\*FractionBox[\(
\*SuperscriptBox[\(b\), \(4\)] + 8\ 
\*SuperscriptBox[\(b\), \(2\)]\ v\), \(
\*SuperscriptBox[\(b\), \(4\)] + 4\ v\)])\) || \((v > 
\*FractionBox[\(3\ 
\*SuperscriptBox[\(b\), \(2\)]\), \(4\)] && r < 
\*FractionBox[\(
\*FractionBox[
SuperscriptBox[\(b\), \(4\)], \(4\)] + \((1 + 
\*SuperscriptBox[\(b\), \(2\)])\)\ v\), \(
\*SuperscriptBox[\(b\), \(4\)] + 4\ v\)])\))\) && 0 < b < 1 && v > 0 && 
    r > 0)\)\), {d, v, b, r}, Reals]
Posted 3 months ago

Hi, i have the same problem . Did you find a solution plz

Posted 3 months ago

No. I can only plot the regions using RegionPlot3D. Resolve will not reduce the set no matter what I tried.

POSTED BY: Yue Li
Answer
Posted 3 months ago

VRTb = Resolve[Exists[{x, y, b, n}, FullSimplify[V && IR && Not[T]]], Reals] // FullSimplify;

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