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February 6

Bring radicals

Bring radicals are used to solve quintic equations. BR($a$) is defined as the unique real solution of the equation $x^5+x+a=0.$ The following suggestions for the article were made on the talk page in October of 2015, but no one has carried them out yet:

• Show a table and/or graph of the function. (And then of its derivative, integral, etc.)
• Show expressions for $\mathrm{BR}(a+b)$, $\mathrm{BR}(ab)$, $\mathrm{BR}(a^b)$, etc., or the impossibility thereof.

Does anyone have a reference for either of these? Loraof (talk) 21:21, 6 February 2018 (UTC)

A table

  7j2":(,.~0 _1 0 0 0 _1&p.)-0.2*i:10
_34.00   2.00
_20.70   1.80
_12.09   1.60
 _6.78   1.40
 _3.69   1.20
 _2.00   1.00
 _1.13   0.80
 _0.68   0.60
 _0.41   0.40
 _0.20   0.20
  0.00   0.00
  0.20  _0.20
  0.41  _0.40
  0.68  _0.60
  1.13  _0.80
  2.00  _1.00
  3.69  _1.20
  6.78  _1.40
 12.09  _1.60
 20.70  _1.80
 34.00  _2.00

Bo Jacoby (talk) 07:48, 7 February 2018 (UTC)

... and the graph of $y=\mathrm{BR}(x)$ is the graph of $y=x^5+x$ rotated anticlockwise by 90 degrees. Gandalf61 (talk) 09:34, 7 February 2018 (UTC)

Thanks, both of you! Is there anything about the second bulleted question? I’m guessing that at best it’s like the function $\sqrt{x}$ in the sense that $\sqrt{ab}$ can be written in terms of $\sqrt{a}$ and $\sqrt{b},$ but $\sqrt{a+b}$ cannot be. Loraof (talk) 16:13, 7 February 2018 (UTC)

• Sounds dubious to me: exponentiation (including taking the square root) is distributive over multiplication, but "Bring radicaling" is not (otherwise, you could reconstruct all Bring radicals from BR(2), and I am pretty sure this would be known and mentioned in the article).

Probably an expert of the Abel-Ruffini theorem could say more - I have a feeling but no proof that any low-degree polynomial relation between BR(a) and BR(b) (for any a,b) would violate it (general idea: craft a polynomial of degree 5 with a known rational root and which can be reduced to a "Bring quintic", that gives you BR(a) (= the rational root) for some a (which is going to depend on the polynomial reduction process but that involves only an algebraic solution), derive BR(b) for all/all but a countable number of b from the polynomial relation, and you solved the fifth degree by radicals, which should be impossible). Tigraan^{Click here to contact me} 13:15, 9 February 2018 (UTC)