I want to prove that this formula ||f-r||=max(|a/b-1|,|a/(b+c)|) does not have a minimum? for a/b=1
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Since a/b = 1, the maximum is |a/(b+c)| and the minimum is 0.
William Elliot
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thank you for the answer, but I mean the minimum of ||f-r||, and I want to show that no minimum exists. – chayma Dec 10 '19 at 09:39
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As ||f-r|| = max(|a/b-1|,|a/(b+c)|) the minimum of |f-r|| is still 0. – William Elliot Dec 10 '19 at 14:33
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me too I find it 0, but the professor says: If you choose a/b=1, then you can do something with parameter c to show that no minimum exists. – chayma Dec 10 '19 at 18:42
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Of course |a/(b+c)| has no minimum, which is superflous as it is always larger than |a/b-1|. – William Elliot Dec 11 '19 at 01:46
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how you find that |a/(b+c)| has no minimum or it is evident ?? or what we should do for c to prove that there is no minimum? thank you so much – chayma Dec 11 '19 at 09:46
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Sent c off to infinity. – William Elliot Dec 11 '19 at 11:34
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when c goes to infinity, |a/(b+c)| goes to 0 , right ? how it does not have a minimum ? sorry for a lot of question, and thank you for the help – chayma Dec 11 '19 at 11:41
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It never gets there. – William Elliot Dec 11 '19 at 15:28
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can you explain more? I didn't understand how – chayma Dec 11 '19 at 17:45