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Using Reduce function to solve inequality q_2 < c ?

Posted 8 years ago

Hello! These are two screenshots from the same Notebook page as you can see:

enter image description here

enter image description here

As you might see, I use the Reduce command to solve the inequality $q_2 < c$ in three different cases:

  • Case 1: $b>0$ and $b<r_1$
  • Case 2: $b>r_1$ and $b<a^2$
  • Case 3: $b>0$ and $b<a^2$

Case (3) should in fact summarize case (1) and case (2), which I think is also the case when looking at the outputs. Now, I would love to use the output from case (3) since this avoid the annoying root object that bounds $a$. I'm not sure if I can do that since $q_2$ can be different depending on the boundaries for $b$.Any help?

Thanks in advance! Cheers

POSTED BY: Cédric Cavents
5 Replies
Posted 8 years ago

I am worried that you have created q2=Root[ ,2] without a detailed understanding of the numbering of Root. I would guess that by your choosing to number your Root[ ,2] you may have forced Mathematica to assume things about your root that are incompatible with you then imposing additional constraints on b. If I were going to try this I would tell Mathematica what the constraints were, look at the results and try to decide which of those was the one I wanted.

If I ignore all that and just try

q2 = Root[(48 a^2 + 16 b) #^4 - (40 a^3 + 168 a b) #^3 + (-45 a^4 + 
        225 a^2 b + 72 b^2) #^2 + (27 a^3 b - 162 a b^2) # + 27 b^3 &, 2];
Simplify[Reduce[q2<c<0 && 0<b<a^2 && a<0, b, Reals], q2<c<0 && 0<b<a^2 && a<0]

then I get

3 a<4 c && b > Root[-45 a^4 c^2 - 40 a^3 c^3 + 48 a^2 c^4 + (27 a^3 c +
225 a^2 c^2 - 168 a c^3 + 16 c^4) #1 + (-162 a c + 72 c^2) #1^2 + 27 #1^3 &, 1]

so, because of the b > Root, I believe it knows that b is real and there is a constraint on the value of b.

I really doubt this resolves your uncertainties about the result.

For your second question, about whether r1<a^2, I would try to give Simplify that question and provide all the domain information you have that is applicable as the second argument to Simplify. Unfortunately I can't tell exactly what that would be from your question.

POSTED BY: Bill Simpson

I was not expecting an easy answer and I'm really glad that someone takes the time to answer. This is the code that I ran in Mathematica:

  q1 = Root[(48 a^2 + 16 b) #^4 - (40 a^3 + 168 a b)#^3 + (-45 a^4 + 225 a^2 b + 72 b^2) #^2 + (27a^3 b - 162 a b^2) #+ 27 b^3 &,1]
q2 = Root[(48 a^2 + 16 b) #^4 - (40 a^3 + 168 a b)#^3 + (-45 a^4 + 225 a^2 b + 72 b^2) #^2 + (27a^3 b - 162 a b^2) #+ 27 b^3 &,2]
r1 = Root[375 a^8 - 3600 a^6 # + 8088 a^4 #^2 - 6912 a^2 #^3 + 2048 #^4 &,1]
r2 = Root[375 a^8 - 3600 a^6 # + 8088 a^4 #^2 - 6912 a^2 #^3 + 2048 #^4 &,2]

Reduce[q2 < c && b > 0 && b < a^2 && a <0 && c <0,b,Reals]
Reduce[q2 < c && b > 0 && b < r1 && a < 0 && c < 0,b,Reals]
Reduce[q2 < c && b > r1 && b < a^2 && a < 0 && c < 0,b,Reals]

In fact, I ask Mathematica to solve an inequality with a Root Object ( $q_2$). I'm not sure if it's able to do it (although I get an answer). Above that, I simply implemented $q_2$ as the 'second' root of a certain quartic polynomial. The problem is that since $a$ and $b$ are symbolic, this quartic polynomial can either have $4$ complex roots, $4$ reals or $2$ reals and $2$ complex roots. More precisely, for $a<0$

  • The quartic has four real roots if 0 < b < r_1
  • The quartic has 2 reals and 2 complex roots if r_1 < b < a^2

(The above can easily be verified with Mathematica as well). This is also the reason that I impose those constraints on $b$ when solving the inequality, because I want to let Mathematica know about which roots I'm speaking. This still leads to many unanswered questions: can Mathematica correctly handle $q_2$? Does Mathematica understand the constraints that are additionally imposed on $b$? I know that $r_1 < a^2$ but does Mathematica know this? etc...

POSTED BY: Cédric Cavents
Posted 8 years ago

I understand you are annoyed with Root objects.

If you could paste neatly formatted code into your post instead of pictures then others would be able to paste that code back into Mathematica and see if they could help find a way to make any progress with your problem.

I suspect it may be very difficult to make progress when your constructed Root objects are built on multivariate polynomials of other variables. If someone gave you a quartic equation that contained not just x and exact coefficients, but other variables and demanded to know exactly what the roots were, not using Mathematica, I'm not sure that you could provide any answer to this unless the quartic happened to be a very special form. You might try such an exercise and see if you can make progress. Anything you learn from that exercise might be helpful.

POSTED BY: Bill Simpson

Thanks for the comment but the annoying root object that bounds $a$ still remains. Unfortunately, Mathematica can't reduce it further. However, it does not answer my question why there is such a difference in output? Is there perhaps someone who can enlighten me more?

POSTED BY: Cédric Cavents
Posted 8 years ago

Sometimes Simplify can help with the output from Reduce, particularly if you give your assumptions as the second argument to Simplify. Thus

Simplify[Reduce[stuff], 0<b<a^2]

or even

Simplify[Reduce[stuff], 0<b<a^2 && c<0 && a<4*c/3 && b<a^2]

or similar techniques might really help.

Use this cautiously.

POSTED BY: Bill Simpson
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