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Is there a way to solve the following equation without using Newton's method: $$x^5-x-1=0$$

Carl Christian
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Andrea
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  • There are many other numerical methods out there for finding roots. – Ross Millikan Apr 29 '22 at 13:36
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    $x^5-x-1=0 \Rightarrow x=(x+1)^{\frac15}$. Let $x_1=1$, $x_{i+1}=(x_i+1)^\frac15$. Then $x_n$ tends to root of equation. Several iterations give numerical value $r_1=1.167303978261418$. Then one can divide polynomial $x^5-x-1$ by $(x-r_1)$ to obtain $x^4+1.167303978261418 x^3 + 1.362598577664933 x^2+ 1.590566740481627 x + 0.8566748838545004=0$ which I suppose has no real roots. – Ivan Kaznacheyeu Apr 29 '22 at 14:08
  • @Andrea: See this for a list of different methods: https://en.wikipedia.org/wiki/Root-finding_algorithms#:~:text=Most%20numerical%20root%2Dfinding%20methods,accurate%20approximation%20to%20the%20root. – Moo Apr 29 '22 at 16:43

2 Answers2

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Consider that you look for one zero of function $$f(x)=x^5-x-1$$

We know that there is one root between $x=1$ (because $f(1)=-1$) and $x=1.2$ (because $(1+0.2)^5 >1+0.2\times 5=2$)

So rewrite (this is exact) $$f(x)=x^5-x-1$$ as $$g(t)=t^5+6 t^4+\frac{72 }{5}t^3+\frac{432 }{25}t^2+\frac{1171 }{125}t+\frac{901}{3125}$$ where $t=x-\frac 65$.

Now, consider that this is a series expansion truncated to $O(t^k)$ with $k>5$ and use series reversion. For $k=6$ (the simplest case), this will give as an approximation

$$x=\frac{6}{5}+y-\frac{2160 y^2}{1171}+\frac{7223400 y^3}{1371241}-\frac{28652670750 y^4}{1605723211}+\frac{124688551398625 y^5}{1880301880081}+O\left(y^6\right)$$ where $$y=\frac{125 }{1171}\left(g(t)-\frac{901}{3125}\right)$$ and we want $g(t)=0$. So, the simplest approximation is $$x_{(6)}=\frac{377560146782854729128701445836786}{323446224424728422628706478984375}$$ Now, varying $k$ and writing the decimal representation $$\left( \begin{array}{cc} n & x_{(n)} \\ 6 & \color{red}{1.16730}423257953 \\ 7 & \color{red}{1.16730}401051128 \\ 8 & \color{red}{1.1673039}8247031 \\ 9 & \color{red}{1.167303978}82297 \\ 10 & \color{red}{1.167303978}33766 \\ 11 & \color{red}{1.1673039782}7192 \\ 12 & \color{red}{1.16730397826}288 \\ 13 & \color{red}{1.167303978261}62 \\ 14 & \color{red}{1.1673039782614}5 \\ 15 & \color{red}{1.16730397826142} \end{array} \right)$$

Claude Leibovici
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Of course, any method to solve algebraic equations is a combination of algebraic and numerical methods. You can solve some equations by radicals, but you can take the square root of 2 only numerically. So radicals are overrated. Already for cubic equations, using trigonometric or hyperbolic functions would be advantageous. Quintic equations can be solved with elliptic functions, especially in this case (it's already in Bring–Jerrard normal form, without any Tschirnhaus transformation). But that's rather unwieldy (enjoy How to solve fifth-degree equations by elliptic functions?), and you need those elliptic integrals and functions (Mathematica, or Wolfram Alpha). I don't know (and quite possibly, I don't want to know) where the "without using Newton's method" comes from. Be that as it may, you can find zeroes of analytical functions without any iterative methods, even with complex integrals, though that's not in the list of methods mentioned in comments.

If you know residues, you know that for a closed contour $C$ in the complex plane and an analytical function $f$ in the interior of $C$, $$\frac1{2\pi i}\oint_C\frac{f'(z)}{f(z)}\,dz\tag1$$ is the number of zeroes of $f$ inside $C$. If that number happens to be $1$, $$\frac1{2\pi i}\oint_Cz\,\frac{f'(z)}{f(z)}\,dz\tag2$$ would be exactly that zero.

Integrals like (1) or (2), at least if $C$ is a smooth curve, like a circle, can be easily evaluated by the trapezoidal rule. (1) will show if there's really exactly one root within $C$, and how precise the trapezoidal rule is (as the result must be an integer). In your case of $f(z)=z^5-z-1$, one may try circles around the zeroes of $z^5-1$ with radius $1/2$, and a trapezoidal rule with 128 nodes, and (1) will give $1$, and (2) will give

$1.167303978261419+1.3968281853471986E-17*I$
$0.1812324444698753+1.0839541013177103*I$
$-0.7648844336005847+0.35247154603172665*I$
$-0.764884433600585-0.3524715460317263*I$
$0.18123244446987535-1.0839541013177107*I$

So you have all 5 solutions of your equation, with reasonable accuracy.