Find the minimum value of $$\displaystyle \frac{2\cos^{-1}(x)}{\pi(1 - x)} , x \in [-1,+1) $$
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Extremely sorry, corrected now. – sherlock Mar 27 '14 at 14:06
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1(+1) What have you tried so far? – imranfat Mar 27 '14 at 14:08
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What do you deduce? – evil999man Mar 27 '14 at 14:08
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Substituting $x = cos^{-1}(y)$ doesn't take me anywhere. I tried to simplify the function to differentiate it to apply the standard theory of Maxima-minima. – sherlock Mar 27 '14 at 14:10
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I have some doubts that an exact value can be found. Is this a problem from some calc book? – imranfat Mar 27 '14 at 14:14
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No idea, this was a question from my college assignment in Algorithm course. – sherlock Mar 27 '14 at 14:16
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@Holmes.Sherlock if this just for an algorithm you probably don't need the exact minima. A weak lower bound would work. – Guy Mar 27 '14 at 14:26
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@Sabyasachi But the question seems to be asking for an exact value, not a lower bound. – sherlock Mar 27 '14 at 14:27
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@Holmes.Sherlock I doubt it. Wolfram Alpha is unable to find an exact value. – Guy Mar 27 '14 at 14:33
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@Sabyasachi Oh yes! I forgot about WolframAlpha – sherlock Mar 27 '14 at 15:11
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WolframAlpha plots a graph for ((2 cos^(-1)(x))/(pi*(1-x))) in [-1,1) – sherlock Mar 27 '14 at 15:27
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I have almost solved the problem. Can anyone tell me how WolframAlpha arrives at a Numerical Solution of -0.689 for this equation – sherlock Mar 27 '14 at 17:49
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May be : http://en.wikipedia.org/wiki/Newton's_method – evil999man Mar 28 '14 at 07:48