I have a question regarding Numerical Analysis. I've never been asked these sorts of questions before and don't even know where to begin.
The goal of this exercise is to find a value alpha such that: $$f(\alpha)=0$$
When $f$ is:
$$f(x) = x-0.2\sin(x)-0.5$$
Do the following steps to achieve this goal:
1) Build a contractive function $\Phi$ that satisifies: $$\Phi(\alpha)=\alpha$$ 2) Calculate $\alpha$ using the fixed-point method.
I don't know how to do step 1 and would like some help.
The $\Phi$ almost suggests itself: $$f(\alpha)=0\iff \alpha-0.2\sin\alpha-0.5=0\iff \alpha=0.2\sin\alpha+0.5 $$ so try $\Phi(x)=0.2\sin x+0.5$. Fortunately, $|\Phi'(x)|=|0.2\cos x|\le 0.2<1$, so this is a contraction.