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I have to approximate the root of the equation $x^3 -4x - 9 = 0$ by using the Bisection method from my numerical methods class. The question also states that I have to iterate the process until my relative approximation error gets below a specified error ($\epsilon_s$) of 0.1%. I have no problem of solving the question, but what I wonder is the step of finding the relative approximate error.

The formula for calculating relative approximate error is: $$\epsilon_a = \frac{current\,approximate-previous\,approximation}{current\,approximation} $$

I wonder why relative approximate error isn't calculated by dividing by the previous approximation instead? I think it is a way to see how improved our current approximation is compared to the previous one.

Makeda
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    Even if the approximation was improved significantly by a step, the resulting approximation could still be awfully bad. Why should we consider the relative approximation error to be small in this case ? Because it is relatively small (compared to the much worse approximation before) ? That would not really make sense. – Peter Nov 29 '21 at 10:29
  • Perhaps this is a substitute for the usual relative error because the exact solution is not known? If you knew the exact solution, you would not to have to compute approximations of it? Except of course when testing a method, then one would like to construct the test problems so that the exact solution is known, see for instance the "Method of manufactured solutions". – Lutz Lehmann Nov 29 '21 at 11:57
  • How do you define your approximate error. Is it the difference between the current and the previous approximation? – Carl Christian Nov 30 '21 at 08:45
  • I define it as the current approximation minus the previous one. I use this formula when wanting to find the root of a certain function by doing the Bisection method. The process of approximating the root iterates until my relative approximate error gets below a specified error. – Makeda Dec 01 '21 at 14:35
  • @Peter I see! My stated reason for comparing the error to the previous approximation would not make sense now. – Makeda Dec 01 '21 at 14:57

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Let's say you have some quantity $x\ne 0$ and an approximation $\tilde x$. The relative error is given by $$ |\varepsilon_x| = \frac{|x-\tilde x|}{|x|}. $$

Normally, we do not have access to $x$, but we are able to get an upper bound for $|x -\tilde x|$. For this reason, one usually computes $\varepsilon_{\tilde x}$ instead. This is justified by the fact that

$$ |\varepsilon_{\tilde x}| = |\varepsilon_x|-|\varepsilon_x|^2+|\varepsilon_x|^3-\cdots $$

PierreCarre
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