Newton's Method problem－香港介紹ppt

標題:

Newton's Method problem

發問:

1. Find all values of x such that x^4 = x + 16 to 5 decimal places. 2. Use Newton's Method with x0 = 2 to solve 2x^3 = 9x^2+5. Make some useful observations on the properties of this value of x0.

最佳解答:

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(1) Let f(x) be x4 - x - 16 and now, we are going to find out all x's satisfyng f(x) = 0. From the derivative f'(x) = 4x3 - 1, it can be found that there's only one turning point [satisfying f'(x) = 0] of f(x) and therefore there are only 2 real roots for f(x) = 0. With the curfe sketch of y = f(x) below: 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/May08/Crazynewton1-1.jpg We can make initial guess for the 2 roots x0 = 2 and -2. For x0 = 2: x1 = x0 - f(x0)/f'(x0) = 2.064516 x2 = x1 - f(x1)/f'(x1) = 2.061532 x3 = x2 - f(x2)/f'(x2) = 2.061525 x4 = x3 - f(x3)/f'(x3) = 2.061525 So the first root value is 2.06153 corr. to 5 d.p. For x0 = -2: x1 = x0 - f(x0)/f'(x0) = -1.939394 x2 = x1 - f(x1)/f'(x1) = -1.936531 x3 = x2 - f(x2)/f'(x2) = -1.936525 x4 = x3 - f(x3)/f'(x3) = -1.936525 So the second root value is -1.93653 corr. to 5 d.p. (2) Let f(x) be 2x3 - 9x2 - 5 and find out all x's satisfyng f(x) = 0. From the sketch of y = f(x) below: 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/May08/Crazynewton2.jpg x = 2 (the initial guess) lies between two extreme points which are both below the x-axis, i.e. the neighbourhood of x = 2 must not be the roots of f(x) = 0. Also with x = 2 being the initial guess, the values of first few iteration steps will fluctuate much as follows: x0 = 2: x1 = x0 - f(x0)/f'(x0) = -0.083333 x2 = x1 - f(x1)/f'(x1) = 3.201201 x3 = x2 - f(x2)/f'(x2) = 11.383182 x4 = x3 - f(x3)/f'(x3) = 8.276450 x5 = x4 - f(x4)/f'(x4) = 6.320998 x6 = x5 - f(x5)/f'(x5) = 5.205368 x7 = x6 - f(x6)/f'(x6) = 4.722995 x8 = x7 - f(x7)/f'(x7) = 4.621644 x9 = x8 - f(x8)/f'(x8) = 4.617273 x10 = x9 - f(x9)/f'(x9) = 4.617266 x11 = x10 - f(x10)/f'(x10) = 4.617266 So the root value is 4.61727 corr. to 5 d.p.

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