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I looked on Wikipedia for a formula for roots of a 5th degree polynomial, but it said that by Abel's theorem it isn't possible. The Abel's theorem states that you can't solve specific polynomials of the 5th degree using basic operations and root extractions.

Can you find the roots of a specific quintic with only real irrational roots (e.g. $f(x)=x^5+x+2$) using other methods (such as logarithms, trigonometry, or convergent sums of infinite series, etc.)?

Basically, how can the exact values of the roots of such functions be expressed other than a radical (since we know that for some functions it is not a radical)?

If no, is numerical solving/graphing the only way to solve such polynomials?

Edit: I found a link here that explains all the ways that the above mentioned functions could be solved.

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Stepanich02
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    I don't think Abel's theorem states that you can't solve specific polynomials (consider the specific polynomial $(x-1)(x-2)(x-3)(x-4)(x-5)$ for example). Abel's theorem states that there is no general formula (i.e. no analogue of the quadratic formula) that will work for all quintic equations. – Sam Weatherhog Dec 02 '15 at 01:00
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    You can solve a quintic equation in terms of roots only when it's Galois group is solvable. – Sam Weatherhog Dec 02 '15 at 01:01
  • @SamWeatherhog there are specific polynomials that cannot be solved in the described way. Of course not every polynomial is such a polynomial, only specific ones. – quid Dec 02 '15 at 01:02
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    Did you see the section "Beyond radicals" on the WIkipedia page? – quid Dec 02 '15 at 01:04
  • @quid I think I'm missing your point. Is there an error in what I said? – Sam Weatherhog Dec 02 '15 at 02:47
  • Wikipedia has articles on how to solve quintic, sextic, and septic equations if you allow the use of functions that cannot be expressed in radicals. – Matt Samuel Dec 02 '15 at 04:47
  • Perhaps this paper contains the answer to your question: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwjC2eKyr7zJAhXLWogKHZ-3DnYQFggcMAA&url=http%3A%2F%2Fwww.cem.uvm.edu%2F~dummit%2Fquintics%2Fsolvable.pdf&usg=AFQjCNG5tqHqlaRSoHEhfLV9EJFuwnmrrA&sig2=nz7aqgYGfH7xSrbp8VoLYg&bvm=bv.108538919,d.cGU – jbuddenh Dec 02 '15 at 04:47
  • @SamWeatherhog the result does give there exist specific polynomials that one cannot solve with radicals, contrary to what you say in you first comment. This is stronger than there not being a general formula (that works for all polynomials). – quid Dec 02 '15 at 09:07
  • @quid The "Beyond Radicals" section defines one of the solutions to a set of quintics with no radical solutions(aka "unsolvable" polynomials) as a set of constants, which are then used to solve other "unsolvable" quintic polynomials. But is there a way to find the exact value of the roots of such functions using elementary algebra( i.e. can the roots be represented using something like a logarithm or another algebraic function)? – Stepanich02 Dec 03 '15 at 01:23
  • Did you read the full section? As explained there they can be expressed using theta functions. – quid Dec 03 '15 at 01:29
  • A useful starting point for approximate solutions is that if $n>0$ and $0=p(z)=z^n+\sum_{j=0}^{j=n-1}a_jz^j$ then $|z|<1+\max (|a_0|,...,|a_{n-1}|)$. Applying this to $z^np(1/z)$ when $a_0\ne 0$ also gives us a lower bound on $|z|$. – DanielWainfleet Dec 03 '15 at 03:30

3 Answers3

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Here is a summary of posts that should address related questions:

  1. Reducing the general quintic to Bring-Jerrard form.
  2. Solving (1) using elliptic functions.
  3. Reducing the general quintic to Brioschi form.
  4. Solving (3) using trigonometric and special functions.
  5. Solving the Brioschi form using $R(q)$ (or the Rogers-Ramanujan continued fraction).
Community
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Tito Piezas III
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As mentioned above, no general formula to find all the roots of any 5th degree equation exists, but various special solution techniques do exist. My own favourite: - By inspection, see if the polynomial has any simple real solutions such as x = 0 or x = 1 or -1 or 2 or -2. If so, divide the poly by (x-a), where a is the found root, and then solve the resultant 4th degree equation by Ferrari's rule. - If no obvious real root exists, one will have to be found. This can be done by noting that if f(p) and f(-p) have different signs, then a root must lie between x=p and x= -p. We now try the halfway point between p and -p, say q. We then repeat the above procedure, continually decreasing the interval in which the root can be found. When the interval is small enough, we have found a root. - This is the bisection method; when such a root has been isolated we divide the polynomial by that root, producing a 4th degree equation which can again be solved by Ferrari or any another method.

Peter Bokhout
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The explicit elementary numbers are generated from the rational numbers by applying finite numbers of $\exp$, $\ln$ and/or radicals.
Clearly, the elementary numbers contain all algebraic numbers and the explicit elementary numbers contain all explicit algebraic numbers (the numbers representable by radicals).

Chow proves in [Chow 1999] his
Corollary 1:
"If Schanuel's conjecture is true, then the algebraic numbers in" the explicit elementary numbers "are precisely the roots of polynomial equations with integer coefficients that are solvable in radicals."

That means, the quintics that are not solvable in radicals cannot be solved by elementary numbers (means by applying elementary functions).

IV_
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