# Why is Resolve not reducing my set?

Posted 4 months ago
634 Views
|
6 Replies
|
1 Total Likes
|
 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!
6 Replies
Sort By:
Posted 4 months ago
 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 4 months ago
 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