answerkey4_4

# With the condition x 2 we have x 1 x f x f 00 x 2 f 2

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With the condition x 0 = 2, we have x 1 = x 0 - f 0 ( x 0 ) f 00 ( x 0 ) = 2 - f 0 (2) f 00 (2) = 0 . Next we plug in x 1 = 0 to the formula to get x 2 = 0 - f 0 (0) f 00 (0) = 4 / 5 . (b) (Ans) This time we chose a different starting value. Similarly we can get x 1 and x 2 from the iteration, now with x 0 = 3. x 1 = x 0 - f 0 ( x 0 ) f 00 ( x 0 ) = 3 - f 0 (3) f 00 (3) = 5 x 2 = x 1 - f 0 ( x 1 ) f 00 ( x 1 ) = 5 - f 0 (5) f 00 (5) = 21 / 5 . (c) (Ans) Under some regularity conditions, we expect that x n will approach to the point x * at which the local minimum (or maximum) is obtained. To be precise, x n converges to x * where f 0 ( x * ) = 0. To explain this, let us take the limit on each side of the Gauss-Newton formula. Then we have the equation, lim x n +1 = lim x n - lim f 0 ( x n ) f 00 ( x n ) . Assuming the limit of the sequence { x n } exists (which we will denote as x * ), and f 0 ( x ) and f 00 ( x ) are continuous. Then the above equation can be written as x * = x * - f 0 ( x * ) f 00 ( x * ) . 2

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Here we have used the fact that lim x n +1 = lim x n = x * , and lim g ( x n ) = g (lim x n ) for any continuous function g ( x ). Therefore, we conclude that f 0 ( x * ) is zero. In this example, the Gauss-Newton optimization process pushes the sequence towards two different limits depending on starting value. The limit of x n is 1 with the initial value x 0 = 2 however, x n converges to 4 with the initial value 3. The limit of { x n } can be sensitive to initial values especially when there are multiple solutions of x in the equation f 0 ( x ) = 0 . We can tell the direction which { x n } converges to by looking at the signs of f 0 ( x n ) and f 00 ( x n ). Let’s go back to our example. From the form of derivatives f 0 ( x ) and f 00 ( x ), you will see that f 0 ( x n ) > 0 and f 00 ( x n ) < 0 when x n < 1 . Therefore, f 0 ( x n ) f 00 ( x n ) < 0 and thus we know from the Gauss-Newton formula that the next value in the sequence is larger than x n , i.e., x n +1 > x n . This implies that as we repeat the iteration, we will get larger value as long as the sequence moves along the range in x n < 1. Likewise, we can find where the sequence moves to for all x n :
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