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Solving an equation with Root Objects?

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
5 Replies

If you want I will explain a little bit more background. In fact, the original problem was to solve the system of equations $$\begin{cases} 16\mu^4 -40 a \mu^3+(15a^2+24b)\mu^2-18ab\mu + 3b^2 = 0 \\ 5a\mu-4\mu^2-b >0 \\ 15a \mu-20 \mu^2 - 3b < 0 \\ 4\mu^3 - 8 c \mu^2 + 5 a c \mu - c b < 0, \end{cases}$$ where $a<0, c<0$ and $b$ all real numbers. In fact, I'm not really interested in a solution of the system but I'm interested in the constraints on $a,b$ and $c$ such that the system yields a real solution. Solving with Mathematica and summarizing leads to the following constraints: $$a<0, \begin{cases} \qquad 0<b \leq a^2, \quad q_2<c<0 \\ a^2 < b < 1.08148a^2, \quad q_1<c<0 \end{cases}$$ where $q_1$ is the smallest real root and $q_2$ is the greatest real root of the quartic polynomial from my original post. Now, the parameters $a,b$ and $c$ are given by $$a = \frac{(k-3) \sqrt{v}}{s}, \qquad b =\frac{v}{s}(w-10s), \qquad c = s \sqrt{v}.$$ In my calculations I know the values of $s<0, k>3, v>0$ and $w<0$ and hence I know $a,b,$ and $c$. However, this does not necessarily mean that $a,b$ and $c$ satisfy the constraints. Now, most of the time the value for $b$ or $c$ will not satisfy the constraints ( $a$ always does). Therefore, if the constraints are not satisfied, how can I modify $s$ and $w$ such that they are satisfied?

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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