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Suppose you have the following analytical information about a function $f(x)$ on $[-1,1]$:

\begin{align*} f(x) &= \frac{1}{x+3}\\ f'(x) & = -\frac{1}{(x+3)^2}\\ f^{''}(x) &= \frac{2}{(x+3)^3} \end{align*}

We are asked to find the linear polynomial that is the minimax (best) approximation to $f(x)$ on the interval $[-1,1]$

Thus we must find three points that have equal by alternating sign of the error. Let $q_1(x) = \alpha x + \beta$. Since $f(x)$ is monotonically increasing on $[-1,1]$ i.e. $f'(x) < 0$ we therefore I am told that we need to look for an interior point $c$ such that $$e'(c) = f'(c) - \alpha = 0 = -(c+3)^{-2}$$

I have been trying to understand why this is the case as in why do we need to find an interior point when $f'(x) < 0$? What would happen if $f(x)$ was monotoncally increasing instead?

Any suggestions or references are greatly appreciated. I am using the Numerical Analysis by Quarteroni, et al. but this particular section is short and does not go into much detail.

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  • $\begingroup$ I solved the problem by my own means, which might not be what your professor had in mind, but in fact solved the problem as stated. Optimal $\alpha = -0.12500, \beta = 0.35355$ . $\endgroup$ – Mark L. Stone Aug 17 '16 at 0:09

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