Find a real number $x$ such that $x^5 − x − 1 = 0$.

Question

Find a real number $x$ such that $x^5 − x − 1 = 0$.

I have already proven that such a number exists, I now try to find which number it is.

Answer 1

You can use numerical methods such as Newtons method .

Define $h(x) = x^5-x-1$

$h'(x) = 5x^4-1$

$x_{n+1} = x_n-\dfrac{h(x_n)}{h'(x_n)}$

$x_1 = x_0 - \dfrac{h(x_0)}{h'(x_0)}$

With $x_0 = 1$ we get the following;

$\begin{pmatrix}n&&&&x_n\\0&&&&1\\1&&&&1.25\\2&&&&1.178459\\3&&&&1.167537\\4&&&&1.167304\\\vdots\end{pmatrix}$

Answer 2

You can use an iteration to find its approximate value:

Rearrange for:$$x^5=x+1\to x=\sqrt[5]{x+1}$$ Then iterate: $$x_{n+1}=\sqrt[5]{x_n+1}, x_1=1$$

Calculator accuracy gives me $x\approx1.167303978$