How do we solve for $x$ in $x^5-x-1=0$?

Question

What is the procedure to finding the simplest exact (or atleast a verifiable approximation to desired level of precision) correct answer to this quintic equation, and more generally to other polynomial equations of degree- 4 and higher yielding non-rational zeroes? And in this particular case, what are all the (exact or approximate) solutions? On desmos there appears to be one real x-intercept at ~1.167; I would imagine that there are up to four other non-real solutions (or fewer, should some have multiplicity greater than 1---is this possible for non-real complex roots?)

Answer 1

Just to add to the comment, the theory of finding roots of a polynomial equation of any degree $d$ in a exact way can be summarized in the following points:

• If the corresponding Galois group is solvable, then a formula using radical expressions exists. This is true for any Galois group obtained from equations of degree $1$, $2$, $3$, $4$, and sometimes for the groups of equations of higher degree. The sometimes can be checked depending on the nature of the specific group (see this).

• If the corresponding Galois group is not solvable, then the general form of the equations can be solved using non-radical expressions that involves theta functions, elliptic functions, hyperelliptic functions and more generally modular functions. In the case of the quintic equation, this can be solved using Jacobi theta functions, as proved by Hermite. In addition, one can always try to reduce the number of terms in the equation using a Tschirnhausen transformation, and indeed this approach is successful in the quintic. Thus, a general quintic can be transformed into a Bring quintic, that can be solved exactly in a complicated way, but is simpler than other approaches (see this).

For the example in the question, which is a Bring quintic, the associated Galois group is not solvable (in this case, it gives the symmetric group $S_{5}$, for our bad luck), so a solution involving theta functions can be found.

Of course, that means that for higher degrees the computational burden is big (see this), and therefore, approximate methods are suitable, like the Aberth method mentioned in the comments.

Finally, a interesting result by Jordan said that any polynomial equation of any degree $d$ can be solved using modular functions.

A good survey of the theory can be found in the book of Bruce King "Beyond the Quartic Equation"

Answer 2

Similar equations can be solved by suitable power series, or by elliptic functions. The equation you propose can be recast in the form:

$x^5+x+a=0$

i.e. : the Bring normal form. Such equation defines a function $x(a)$, the so-called Bring radical or Ultraradical (not an elementary function):

https://en.wikipedia.org/wiki/Bring_radical

General equations of 5th degree can be solved by means of such function.

Answer 3

Using Newton - Raphson

$$f(x) = x^5 - x - 1$$

$$\text{At}\quad x= 1.00,\space f(x) = -1.00 $$

$$\text{At}\quad x= 2.00 ,\space f(x) = 29.00 $$

$$f′(x)=5x^4-1$$

$$x_0=2$$

$$x_1=x_0-\frac{f\left(x_0\right)}{f'\left(x_0\right)}$$

$$x_1≈1.6329$$

$$x_2≈1.3731$$

$$x_3≈1.2236$$

$$x_4≈1.1727$$

$$x_5≈1.1674$$

$$x_6≈1.167$$

Answer 4

If you have a calculator that'll take fifth roots, the iteration

$$x_{n+1}=(1+x_n)^{1/5}$$

with $x_0=1$ gives

$$\begin{align} x_1&\approx1.148698355\\ x_2&\approx1.1652928729\\ x_3&\approx1.1670872626\\ x_4&\approx1.1672806328\\ x_5&\approx1.1673014635\\ x_6&\approx1.1673037074\\ x_7&\approx1.1673039491\\ x_8&\approx1.1673039751\\ x_9&\approx1.1673039779\\ x_{10}&\approx1.1673039782\\ x_{11}&\approx1.1673039783 \end{align}$$