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When reducing and simplifying a complex algebraic inequality I get an expression containing

Root[-a + b #1^3 &, 1] < β < (a/b)

How can I interpret the first term in the inequality? The 1st root of some function? Why not putting it explicitly?

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1 Answer 1

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Why not putting it explicitly?

From help,

Root "Represents the exact k^th root of the polynomial equation f[x]==0"

This can be rewritten as follows

 expr = Root[-a+b#1^3&,1]< \[Beta] < (a/b)
 expr // ToRadicals

enter image description here

From help on ToRadicals it says

attempts to express all Root objects in expr in terms of radicals.

To try to answer the comment:

how the expression Root[-a+b#1^3&,1] results in the fraction shown above?

The above answer comes from, when rewriting Root[-a + b #1^3 &, 1] as

 Solve[-a + b*x^3 == 0, x]

Where #1^3 becomes x^3. Now Root[....,1] says the first root. i.e. the first root of -a + b*x^3 == 0. Which is

 Solve[-a + b*x^3 == 0, x]

enter image description here

Now the question might be, since there are 3 roots to the cubic, why the first root was choosing x -> ((-1)^(2/3) a^(1/3))/b^(1/3) and not x -> a^(1/3)/b^(1/3) ? It looks like some internal ordering is used to decide which is the first root. As Root[-a + b #1^3 &, 3] // ToRadicals gives a^(1/3)/b^(1/3)

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  • $\begingroup$ Dear @Nasser, you actually answered my question, but may I ask you to add a couple of words on how the expression Root[-a+b#1^3&,1] results in the fraction shown above? I'm a novice in Mathematica and I have hard time trying to decode this... $\endgroup$
    – Dmitry
    Jul 16, 2020 at 10:13
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    $\begingroup$ @Dmitry I tried to answer you comment in the post. $\endgroup$
    – Nasser
    Jul 16, 2020 at 10:33
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    $\begingroup$ Ah, now I understand where this root comes from. That was the missing piece in the puzzle :) $\endgroup$
    – Dmitry
    Jul 16, 2020 at 10:44

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