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I need to write the following (which is the result of a Solve command I ran in Mathematica) in an academic journal paper:

Root[-4 q2^2 - 4 q2^3 + (7 q2 + 19 q2^2) #1 - 17 q2 #1^2 + 2 #1^3 &, 1]

I still don't have a good idea of what this Root function means. Can someone write this in an academically acceptable (journal paper-"plain math") format? Thanks for your help. I have quite a few of these expressions that I need to interpret and write up.

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  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$
    – Michael E2
    Commented Mar 9, 2016 at 19:37
  • $\begingroup$ You can format inline code and code blocks by selecting the code and clicking the {} button above the edit window. The edit window help button ? is also useful for learning how to format your questions and answers. You may also find this this meta Q&A helpful $\endgroup$
    – Michael E2
    Commented Mar 9, 2016 at 19:37
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    $\begingroup$ It's the "first" root of the equation -4 q2^2 - 4 q2^3 + (7 q2 + 19 q2^2) X - 17 q2 X^2 + 2 X^3 == 0. Which is first depends on the parameter q2. $\endgroup$
    – Michael E2
    Commented Mar 9, 2016 at 19:38
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    $\begingroup$ Have a look at ToRadicals which is also linked to in the docs for Root $\endgroup$
    – Lukas
    Commented Mar 9, 2016 at 19:46

3 Answers 3

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TeXForm@ToRadicals@Root[-4 q2^2 - 4 q2^3 + (7 q2 + 19 q2^2) #1 - 17 q2 #1^2 + 2 ^3 &, 1]

$$\frac{1}{6} \sqrt[3]{2222 \text{q2}^3-855 \text{q2}^2+3 \sqrt{3} \sqrt{-15633 \text{q2}^6+2190 \text{q2}^5-7225 \text{q2}^4+2744 \text{q2}^3}}-\frac{42 \text{q2}-175 \text{q2}^2}{6 \sqrt[3]{2222 \text{q2}^3-855 \text{q2}^2+3 \sqrt{3} \sqrt{-15633 \text{q2}^6+2190 \text{q2}^5-7225 \text{q2}^4+2744 \text{q2}^3}}}+\frac{17 \text{q2}}{6}$$

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    $\begingroup$ Or you can use TraditionalForm@ToRadicals@ $\endgroup$
    – BlacKow
    Commented Mar 9, 2016 at 20:06
  • $\begingroup$ Thanks! That's precisely what I was looking for - ToRadicals $\endgroup$
    – user917983
    Commented Mar 9, 2016 at 20:07
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You are fortunate here that the polynomial you were considering is a cubic, and thus you have recourse to an alternative radical representation via ToRadicals[]. But recall why Root[] is needed in the first place: there are algebraic numbers (roots of polynomials of degree 5 or higher) that do not admit radical representations. What to do in those cases?

The Wolfram Functions site suggests a possible notation for Root[]:

$$\texttt{Root[a0 + a1 # + … + an #^n &, k]}=\left(z; a_0 + a_1 z + \cdots a_n z^n\right)_{k}^{-1}$$

This is of course nonstandard notation, so you will have to explain it if you do use it in a paper with something like

…where $\left(z; z^3+z+1\right)_1^{-1}\approx -0.68232$ is the adopted notation for the first root of the polynomial $z^3+z+1$ in Mathematica's ordering…

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This seems "mathematical" to me:

$${\frak R}(f,k) \equiv {\frak R}(f(z),k) = z_k\,,$$ where $z_k$ is the $k^{th}$ root of $f(z) = 0$, sorted according to Mathematica's ordering. Real roots precede complex roots increasing order. Complex roots are sorted with increasing real part; in the case of ties, they are sorted according to the magnitude of the imaginary part; thus conjugates will be adjacent, the one with a negative imaginary part preceding the other.

Alternatively,

$${\frak R}_k(f)$$

or

$$Z_k(f) \quad \text{($k^{th}$ Zero),}$$

or something like

$$\{f^{-1}(0)\}_k\,$$ where $f^{-1}(0)$ is understood to denote the multiset inverse image.

The Vieta map, $v \colon {\Bbb C}^n \rightarrow P_n(1)$ from the zeros $(z_1,\dots,z_n)$ to a monic polynomial with exactly the roots $z_k$, might be used to frame the definition, depending on the context. The inverse of $v$ can be made a function $P_n(1)\rightarrow {\Bbb C}^n/\Sigma_n$ onto the symmetric quotient of ${\Bbb C}^n$ by the symmetric group $\Sigma_n$ of degree $n$. Then ${\frak R}_k$ is the projection $\pi_k$ onto the $k^{th}$ sorted factor of quotient space. More explicitly, for $f(z) = a_nz^n+\cdots+a_0$, $${\frak R}_k(f) = \pi_k\left(v^{-1}\bigl(\,f(z)/a_n\bigr)\right)$$

On the other hand, the k-th root according to some ordering might be a sufficiently clear definition in some cases, even without specifying the ordering. For instance a numeric approximation of sufficient precision to isolate the root would identify it and not clutter the page with complicated formulas in cases where the formulas were unimportant.

My advice would be to be as simple as possible while still conveying all the information you need to convey to make yourself clear.

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  • $\begingroup$ Now, now. It is for the Esquimo Journal of Qualitative Cetacean Studies. :) $\endgroup$ Commented Mar 11, 2016 at 2:32
  • $\begingroup$ @Dr. bel, they prefer to be called "Inuit" these days, just so you know. :) $\endgroup$ Commented Mar 11, 2016 at 3:22
  • $\begingroup$ @J.M. I've read this one when I was seven. Nice people. Deserved better than they got $\endgroup$ Commented Mar 11, 2016 at 3:33
  • $\begingroup$ @Dr.belisarius They still call it the World Eskimo-Indian Olympics $\endgroup$
    – Michael E2
    Commented Mar 11, 2016 at 3:58
  • $\begingroup$ @MichaelE2 Hey! They are quite tough $\endgroup$ Commented Mar 11, 2016 at 4:20

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