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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.

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  • $\begingroup$ What is known about $f$? $\endgroup$ – Daniel Fischer Jul 9 '13 at 21:24
  • $\begingroup$ Sorry, forgot to write it. Fixed now. $\endgroup$ – Oria Gruber Jul 9 '13 at 21:25
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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.

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  • $\begingroup$ Could you please elaborate on how you showed that the function phi is a contraction? We didn't do it in class with a derivative. we showed that |f(x)-f(y)| < k|x-y|, that is a derivative only when x and y are very close $\endgroup$ – Oria Gruber Jul 9 '13 at 21:48
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    $\begingroup$ By the mean value theorem, $\Phi(x)-\Phi(y)=\Phi'(\xi)\cdot(x-y)$ with $\xi$ between $x$ and $y$. Thus a bound for $|\Phi'|$ can be taken as $k$. $\endgroup$ – Hagen von Eitzen Jul 9 '13 at 21:55
  • $\begingroup$ Hagen, thank you so so much. $\endgroup$ – Oria Gruber Jul 9 '13 at 22:01

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