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# 12.01 - Last time we started on the problem of...

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Last time we started on the problem of approximating the minimum value of the function x 2 – sin x . We showed that there is a unique value x * for which this function achieves its global minimum on (– , ), and that this value is the unique real number satisfying the equation 2 x – cos x = 0. Let’s estimate this number with Newton’s method. Trial and error gives x * to be about 1/2, so take x 1 = 0.5. With g ( x ) = 2 x – cos x , we have g ( x ) = 2 + sin x . x n +1 = x n – (2 x n – cos x n ) / (2 + sin x n ). With x 1 = 0.5 , we get x 2 = 0.450626693077243046568, x 3 = 0.450183647577774742501 , x 4 = 0.45018361129487381641, x 5 = 0.45018361129487357304, x 6 = 0.45018361129487357304, ... So after just a few iterations, the process stabilizes, and we’ve found x * to twenty significant figures. Hence the minimum value taken on by x 2 – sin x is about ( x 4 ) 2 – sin x 4 = –0.232465575. .. In general, when Newton’s method works, the error at the n +1st stage is approximately equal to the square of the error at the n th stage; hence each iteration gives you about

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twice as many correct digits as the iteration before: x 1 = 0 .5 , we get x 2 = 0.450 626693077243046568, x 3 = 0.4501836 47577774742501, x 4 = 0.450183611294873 81641, etc. Questions on Section 4.6? Section 4.7: Antiderivatives
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12.01 - Last time we started on the problem of...

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