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The book Numerical Analysis by Richard L. Burden has the following exercise:

Suppose $ p^{*} $ must approximate $ p $ with relative error at most $ 10^{-3} $. Find the largest interval in which $ p^{*} $ must lie for each value of $ p $.

a) 150 $\quad$ b) 900 $\quad$ c) 1500 $\quad$ d) 90

My answer for c):

$ \dfrac{\vert p-p^{*} \vert}{\vert p \vert} \leq 10^{-3}$.

As $ p =1500 $, $ \dfrac{\vert 1500-p^{*} \vert}{1500} \leq 10^{-3}$. (I put $ \leq $ by the phrase at most).

$ \Rightarrow \vert 1500-p^{*} \vert \leq 1.5$

So my answer is $ [1498.5, 1501.5] $.

Analogously for a), b) and d) my answers are

$ [149.85,150.15] $, $ [899.1, 900.9] $ y $ [89.91, 90.09] $, respectively.

The answers at the end of the book are

$ (149.85, 150.15) $, $ (899.1,900.9) $, $ (1498.5, 1501.5) $ y $ (89.91, 90.09) $, respectively.

Could someone explain to me why p * can not take the extreme values of the intervals?

miguel
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  • How comes the intervals are different for b) and d)? the fact that the intervals are open is probably because the book assumes $|p^-p|/|p|<10^{-3}$ instead of $|p^-p|/|p|\leq 10^{-3}$. – Surb Oct 31 '18 at 22:44
  • Thanks, I have corrected. – miguel Oct 31 '18 at 23:03
  • The notation a ≤ b means that a is less than or equal to b (or, equivalently, not greater than b, or at most b) https://en.wikipedia.org/wiki/Inequality_(mathematics) – miguel Nov 02 '18 at 10:58
  • I understand your point, but I think that you may be wasting your time on an, in this context, insignificant detail. – Surb Nov 02 '18 at 11:32

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