So this question had two main parts that I got stuck on:
Suppose that (X,d) is a complete metric space and $f : X \rightarrow X$ is a map.
Parts a) & b) just asked for the definition of a contraction and to prove that $f$ has at most one fixed point without using Banach's fixed point theorem, which I was fine with.
(c) Prove that $f : \mathbb{R} \rightarrow \mathbb{R}, x\mapsto f(x)= $ $\frac{1}{20} \frac{1}{1+x^4}$ is a contraction.
(d) Use the Banach fixed point theorem to prove that the polynomial equation $x^5 + 3x − 1 = 0$ has exactly one real solution and compute this solution numerically to 3 decimal places.
So for part c) I have:
For $C^1$ functions $|f(x)-f(y)|\leqslant M|x-y|$ if $|f'(x)|\leqslant M$.
We compute
$$f'(x)= -\frac{x^3}{5(x^4+1)^2}$$
$$=-\frac{x^3}{(x^4+1)^2}\cdot\frac{1}{5}$$ $$\leqslant \frac{1}{5}$$
Therefore $$|f(x)-f(y)|\leqslant\frac{1}{5}|x-y|$$and hence $f$ is a contraction.
If somebody could tell me if this is correct I would appreciate it a lot!
Part d) I am completely stuck on and don't really know how to tackle it! All I managed to do was compute the root to be 0.332 by iterating.
Yes the first part is (mostly) fine, you have by the mean value theorem $$d(f(x),f(y)) = |f(x) - f(y)| \leq \sup_{z\in\mathbb{R}}|f'(z)||x-y|\leq \frac{1}{5}|x-y| = \frac{1}{5}d(x,y)$$ Therefore it is a contraction. For the next part, consider
$$g(x) := \frac{1}{3+x^4}$$ If $g$ has a unique fixed point $x_0$ (by the Banach fixed point theorem), then $$g(x_0) = x_0 = \frac{1}{3+x_0^4} \iff x_0^5 + 3x_0 - 1 =0$$ as in the proof of the Banach fixed point theorem, you may take any point, say $x=0$, then $$x_0 = \lim_{n\rightarrow\infty}g^n(x)$$ Where I am using the notation $g^{n}(x) = g^{n-1}(g(x))$.