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
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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
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]
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... $\endgroup$– DmitryJul 16, 2020 at 10:13
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1$\begingroup$ @Dmitry I tried to answer you comment in the post. $\endgroup$– NasserJul 16, 2020 at 10:33
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1$\begingroup$ Ah, now I understand where this root comes from. That was the missing piece in the puzzle :) $\endgroup$– DmitryJul 16, 2020 at 10:44