= Solution
{c}
Rearrange the <polynomial> equation as a <fixed point> equation:
$$
\boxed{g(x)=\frac{1-x^5}{7},\qquad [a,b]=[0,1/7].}
$$
First establish existence and uniqueness of the target <root of a polynomial>. Since $f$ is a <continuous function>, $f(0)=-1<0$ and $f(1/7)=1/7^5>0$, the <intermediate value theorem> gives a <root of a polynomial> $\xi\in(0,1/7)$. Moreover $f'(x)=5x^4+7>0$ for all real $x$, so $f$ is <strictly increasing> and this is its only real <root of a polynomial>.
For $x\in[0,1/7]$,
$$
0<\frac{1-7^{-5}}7\leq g(x)\leq\frac17,
$$
so $g$ maps the <closed interval> into itself. Also
$$
|g'(x)|=\frac{5x^4}{7}\leq\frac5{7^5}=q<1.
$$
The <mean value theorem> therefore gives $|g(x)-g(y)|\leq q|x-y|$ on the <closed interval>: $g$ is a <contraction mapping>. For any starting value $x_0\in[0,1/7]$, all iterates remain in that <closed interval>, and since $g(\xi)=\xi$,
$$
|x_k-\xi|=|g(x_{k-1})-g(\xi)|\leq q|x_{k-1}-\xi|\leq q^k|x_0-\xi|\longrightarrow0.
$$
Thus \b[the <fixed-point iteration> converges to the unique real <root of a polynomial> from every $x_0\in[0,1/7]$]. This explicit error estimate proves the needed convergence directly.
Back to article page
