Sequence problem, find root the equation $x^3-5x+1=0$ has a root in $(0,1)$. Using a proper sequence for which $$|a(n+1)-a(n)|\le c|(a(n)-a(n-1)|$$ with $0<c<1$ , find the root with an approximation of $10^{-4}$.
 A: You can attempt this: define a sequence $x_n$ by
$$
x_{n+1} = \frac 15\left(1 + x_n^3\right).
$$
We know that if $x_1 \in (0, 1)$, then $x_n \in (0, 1)$ for all $n$. Next, we show that it satisfies the said condition:
\begin{align*}
x_{n+1} - x_n & = \frac 15(1 + x_n^3 - 5x_n)\\
& = \frac 15\left(1 + x_n^3 - (1 + x_{n-1}^3)\right)\\
& = \frac 15\left(x_n^3 - x_{n-1}^3\right) \\
& = \frac 15\left(x_n - x_{n-1}\right)\left(x_n^2 + x_nx_{n-1} + x_{n-1}^2\right) \\
\therefore
\left|x_{n+1} - x_n\right|
& = \frac 15\left|\left(x_n - x_{n-1}\right)\left(x_n^2 + x_nx_{n-1} + x_{n-1}^2\right)\right| \\
& \le \frac 15\left|x_n - x_{n-1}\right| \cdot 3 \\
& \le \frac 35\left|x_n - x_{n-1}\right|.
\end{align*}
From this, we see that the sequence is Cauchy, hence convergent. Let $x$ be the limit of the sequence. Take the limit $n \to \infty$ in the first equation (which defines $x_{n+1}$ from $x_n$) to get
$$
x = \frac 15(1 + x^3).
$$
This means $x$ is a root of $f(x) = x^3 - 5x + 1$.
To compute $x$ numerically within a given error tolerance $\epsilon$, we need to find $x_N$ such that $|x_N - x| \le \epsilon$. However, we do not know $x$, so instead, we can use the following criterion to find $N$: for all $n > N$, $|x_{n} - x_N| < \epsilon$.
This criterion is sufficient because if the inequality holds for all $n > N$,
it will follow that $\lim_{n \to \infty} |x_n - x_N| = |x - x_N| \le \epsilon$.
One convenient way to guarantee $|x_n - x_N| < \epsilon$ for all $n > N$ is via the triangle inequality:
$$
|x_n - x_N| \le \sum_{k=N}^{n-1} \left|x_{k+1} - x_k\right|
\le \sum_{k=N}^\infty \left|x_{k+1} - x_k\right|
\le \frac{\left(\frac 35\right)^N}{1 - \frac 35} = \frac 52\left(\frac 35\right)^N.
$$
So, if we can find $N$ such that $\frac 52\left(\frac 35\right)^N < \epsilon$, we will get $|x_N - x| \le \epsilon$.
A: Wouldn't a bisect algorithm do something like this: 
Let $f(x)=x^3-5x+1$. Then we find the root by:
f(0)>0, f(1)<0
f(0.5)<0 (hence, there is a root between 0 and 0.5)
f(0.25)<0
f(0.125)>0
f(0.1875)
and so on...
The sequence would be (0.5, 0.25,0.125, 0.1875...) and c=1/2
Edit: and the algorithm works as f is continuous :)
