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I know that $a^*$ is the second largest root of the following equation:

$$32\left(\sqrt{50-10\sqrt{5}}\right)a^3-\frac{48}{5}(5+\sqrt{5}) \pi a + 8\sqrt{10(5+\sqrt{5})} = 0 $$

Note that $\pi$ is the constant, not a variable.

Q1: What would you call such an equation?

Perhaps a cubic equation with irrational coefficients?

I also want to express this as neatly as possible, maybe writing something like: $$a^* = \text{Root}\left(32\left(\sqrt{50-10\sqrt{5}}\right)a^3-\frac{48}{5}(5+\sqrt{5}) \pi a + 8\sqrt{10(5+\sqrt{5})} = 0,2\right).$$

Q2: Does an operator (i.e. symbol denoting a mathematical operation, not this) like that exist?

I'm sure some will think it more elegant to just define $a^*$ as I did to start this question, and others will think it better to give the solution in radicals. The problem with the former is that, while more aesthetically appealing, it's a little more difficult to use in tables and perhaps a bit annoying if used repetitively. The problem with the latter is that it is very messy (relative to the operator approach I try above), even after simplifying as much as possible. That's why I'd prefer to define $a^*$ with an operator if an appropriate one exists.

Note: An ideal answer would answer both Q1 and Q2. It may be that such an answer would simply be "Yes and no, respectively." But, if that is the case, I would appreciate any commentary on how you would recommend representing the root in a journal article.

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    $\begingroup$ Q2 : what do you mean exactly by "operator"? $\endgroup$
    – paf
    Aug 9, 2016 at 13:43
  • $\begingroup$ @paf I meant it as a symbol to denote a mathematical operation, not a mapping between vector spaces and modules. Thanks for pointing out that potential confusion. $\endgroup$
    – Shane
    Aug 9, 2016 at 17:57

2 Answers 2

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Before answering the questions, it is desirable to convert the equation to the foreseeable mind. In this case we use the golden section Mauger constant $$\varphi=\dfrac{\sqrt5+1}2\approx1.618,$$ then $$\dfrac{\sqrt5-1}2=\dfrac1\varphi,$$ and original equation takes the form $$\dfrac{64\cdot5^{3/4}}{\sqrt\varphi}a^3-\dfrac{96\pi\varphi}{\sqrt5}a+16\cdot5^{3/4}\sqrt\varphi = 0,$$ or $$4a^3-\dfrac{6\pi\varphi^{3/2}}{5^{5/4}}a+\varphi=0.$$ Number $\pi$ is irrational (transcendental), multiplication and division on rational and irrational numbers does not change the situation. Consequently,

$$\boxed{\text{Q1: yes, it's a cubical equation with irrational (transcendental) coefficients.}}$$

Both the original and changed forms give the same result.

For an elegant answer to the second question, we introduce the coefficient $$\mu=\dfrac{\sqrt{2\pi}\varphi^{3/4}}{5^{5/8}}\approx1.315,$$ then equation takes the form $$4a^3-3\mu^2a+\varphi=0.\qquad(1)$$

The precise mathematical formula for recording the second-highest root is unknown to me. At the same time, the Mathcad package required value is given a convenient and understandable expression of the form $$\boxed{\text{Q2a:}\qquad a^* = \mathrm{polyroots}\left( \begin{pmatrix} \varphi\\ -3\mu^2\\ 4 \end{pmatrix} \right)_1}$$ The highest degree coefficient places below others and the result of polyroots() function is ordered vector of roots wherein the root with the greatest real part has index $0.$

Also the equation $(1)$ can be presented in the form $$4\left(\dfrac{a}\mu\right)^3-3\dfrac{a}\mu=-\dfrac\varphi{\mu^3},$$ or $$\mathrm T_3\left(\dfrac{a}\mu\right) = -\dfrac\varphi{\mu^3},$$ where $$\mathrm T_3(x)=4x^3-3x$$ is a Chebyshev polynomial of the first kind. Note that $$\left|-\dfrac\varphi{\mu^3}\right|\approx 0.711 <1,$$ so we can use the Chebyshev polynomial in form $$\mathrm T_3(x) = \cos(3\arccos x)$$ and get the equation $$\cos\left(3\arccos \left(\dfrac{a}\mu\right)\right) = -\dfrac\varphi{\mu^3}$$ with the ordered roots as in Mathcad $$a_k=\mu\cos\left(\dfrac13\arccos\dfrac\varphi{\mu^3}+\dfrac{2k-1}3\pi\right),\quad k=0,1,2.$$ Thus, $$\boxed{\text{Q2b:}\quad a^*=\mu\cos\left(\dfrac13\arccos\dfrac\varphi{\mu^3}+\pi/3\right)}$$

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  • $\begingroup$ Where does 'no' for Q1 comes from? Yes, the coefficients are transcendental (one of them), but they are irrational as well. As for the rest of the answer, I like it, +1 $\endgroup$
    – Yuriy S
    Aug 14, 2016 at 23:20
  • $\begingroup$ @YuriyS Of course, you're right. I would like to emphasize that the roots of the equation are not algebraic numbers, but did it badly. $\endgroup$ Aug 15, 2016 at 6:10
  • $\begingroup$ Thanks for this -- this is a neat approach and I like that you've illuminated a connection to the Golden ratio constant! $\endgroup$
    – Shane
    Aug 16, 2016 at 12:24
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    $\begingroup$ @Shane I like it too) $\endgroup$ Aug 16, 2016 at 14:45
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Q1 : yes, it's a cubic equation with irrational coefficients.

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    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ Aug 9, 2016 at 14:06
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    $\begingroup$ This is precisely an answer to Q1 which asks "what would you call such an equation?" and to which I answer "this is a cubic equation with irrational coefficients". $\endgroup$
    – paf
    Aug 9, 2016 at 14:23

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