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Solving an equation with Root Objects?

Cédric Cavents, Student
Posted 9 years ago

Hello!

Given $a = ((k-3) \sqrt{v})/s$ where $k$ and $v$ are known. I need to solve the following equation for $s$: $$q(a,b) = s\sqrt{v}, \qquad (*),$$ where $b = 1.08148a^2+\epsilon$ and $\epsilon>0$ a very small number (for instance $\epsilon = 10^{-6}$). Furthermore, $q(a,b)$ is the greatest real root of the polynomial $$(48a^2+16b)x^4 - (40a^3+168ab)x^3 + (-45a^4+225a^2 b+72b^2)x^2 + (27a^3 b - 162 ab^2)x+27b^3.$$ How can I solve $(*)$ with Mathematica? I know it should be possible theoretically speaking, since I can determine the roots of the quartic in function of $a$ and I know the expression for $a$. However, I think the expression $q(a,b)$ might be quite ugly. Perhaps, I should use a numerical algorithm? Is this possible with Wolfram Mathematica?

Thanks in advance! Cheers
POSTED BY: Cédric Cavents
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POSTED BY: Cédric Cavents

Indeed it seems that my condition for real roots can be simplified to

b < 1/32 (27 - 3 Sqrt[6] - Sqrt[111 + 46 Sqrt[6]]) a^2

The coefficient is 0.146741, which is different from what you found. Perhaps we are not working on the same equation.
POSTED BY: Gianluca Gorni

With version 10.4 I get this expression for q[a,b]:

ConditionalExpression[
 Max[Root[27 b^3 + (27 a^3 b - 162 a b^2) #1 + (-45 a^4 + 225 a^2 b + 
        72 b^2) #1^2 + (-40 a^3 - 168 a b) #1^3 + (48 a^2 + 
        16 b) #1^4 &, 1], 
  Root[27 b^3 + (27 a^3 b - 162 a b^2) #1 + (-45 a^4 + 225 a^2 b + 
        72 b^2) #1^2 + (-40 a^3 - 168 a b) #1^3 + (48 a^2 + 
        16 b) #1^4 &, 2], 
  Root[27 b^3 + (27 a^3 b - 162 a b^2) #1 + (-45 a^4 + 225 a^2 b + 
        72 b^2) #1^2 + (-40 a^3 - 168 a b) #1^3 + (48 a^2 + 
        16 b) #1^4 &, 3], 
  Root[27 b^3 + (27 a^3 b - 162 a b^2) #1 + (-45 a^4 + 225 a^2 b + 
        72 b^2) #1^2 + (-40 a^3 - 168 a b) #1^3 + (48 a^2 + 
        16 b) #1^4 &, 4]], 
 b - Root[375 a^8 - 3600 a^6 #1 + 8088 a^4 #1^2 - 6912 a^2 #1^3 + 
      2048 #1^4 &, 1] < 0]
POSTED BY: Gianluca Gorni

Thanks for the answer Gianluca! I copied your code (q[a,b] = ...) in Mathematica but Mathematica returns ('no global maxima found')? Could you perhaps give the complete output?

The error I made is that it should be $b = 1.0848a^2 - \epsilon$ (instead of $+\epsilon$). I verified that the quartic polynomial has real roots $q$ if $0<b<1.08148a^2$ (probably other regions for $b$ are possible as wel but I only need the positive values for $b$). The $1.08148a^2$ comes from the fact that this is the greatest real root of $$2048x^4 - 6912 a^2 x^3+8088 a^4 x^2-3600 a^6 x + 375 a^8$$ (the root object occurring in the condition for $b$ you obtained). The smallest root of the above polynomial is $0.14674a^2$. So, it seems that you obtained $b<0.14764a^2$ as a condition.

If I have an explicit expression for the root $q$ which is not too messy than I can perhaps use it.
POSTED BY: Cédric Cavents

The expression for q is not terribly ugly:

q[a_, b_] = 
 Max[x /. Solve[((48*a^2 + 16*b)*x^4 - (40*a^3 + 168*a*b)*x^3) + 
              (-45*a^4 + 225*a^2*b + 72*b^2)*
       x^2 + (27*a^3*b - 162*a*b^2)*x + 27*b^3 == 0, x, 
          Reals]]

but it has a condition b - Root[375 a^8 - 3600 a^6 #1 + 8088 a^4 #1^2 - 6912 a^2 #1^3 + 2048 #1^4 &, 1] < 0, which does not seem to be compatible with your b=1.08148a^2+?:

Show[RegionPlot[
  b - Root[375 a^8 - 3600 a^6 #1 + 8088 a^4 #1^2 - 6912 a^2 #1^3 + 
       2048 #1^4 &, 1] < 0, {a, -2, 2}, {b, -2, 2}], 
 Plot[1.08148 a^2, {a, -2, 2}]]

unless I misunderstood or made a mistake.
POSTED BY: Gianluca Gorni
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