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Yunfei duan Hui Pan.  A parabola through three points f(a) f(b) f(c)  for the derivation of this formula is from  denominator should not be zero.

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Presentation on theme: "Yunfei duan Hui Pan.  A parabola through three points f(a) f(b) f(c)  for the derivation of this formula is from  denominator should not be zero."— Presentation transcript:

1 Yunfei duan Hui Pan

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3  A parabola through three points f(a) f(b) f(c)  for the derivation of this formula is from  denominator should not be zero

4  six function points  a, b, u, v, w and x,  The minimum is bracketed between a and b;  x is the point with the very least function value found  w is the point with the second least function value; v u is the previous value of w;  Xm is the value of the f

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6  Brent's Method:  This method takes advantage of the fact that, near a minimum, the function is approximately quadratic. It conisists of the following steps  Find the minimum of the 2nd order polynomial defined by the three bracket points  Use this minimum to define the new point x  Re-define the bracket  Of course, there also need to be checks on this algorithm, when it is implemented in practice, to ensure that:

7  function f interpolating parabola and a bracketing triplet of abscissas (a b c)  a f(b) { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/12/3422668/slides/slide_7.jpg", "name": " function f interpolating parabola and a bracketing triplet of abscissas (a b c)  a f(b)

8  one dimensional search – what’s the problem?  determine the minimizer of a function f : R ->R  typically over a closed interval [a0,b0]  assumption: function is unimodal (at least over this interval) typical approaches: 1 bisection method 2 golden section search 3 Newton’s method 4 secant method

9 basic idea start with interval [a0,b0] (1st derivative not available)  split the interval into three subintervals:  pick two new values a1 and b1 and a0 < a1 < b1 < b0  there are two possible outcomes:  f(a0) >f(a1) > f(b1)< f(b0)  f(a0) > f(a1) < f(b1) < f(b0)

10  F ( x )=3 x^ 4−4 x^ 3−12 x^ 2+3  We cannot find regions of which f is increasing or decreasing, relative maxima or minima.  Graphing by hand is tedious and imprecise. Even the use of a graphing program will only give us an approximation for the locations and values of maxima and minima. We can use the first derivative of f, however, to find all these things quickly and easily.

11 F ( x )=3 x^ 4−4 x^ 3−12 x^ 2+3 has first derivative F’ ( x ) = 12 x^ 3−12 x^ 2−24 x = 12 x ( x^ 2− x −2) =12 x ( x +1)( x −2)  f ( x ) is increasing on (−1,0) ∪ (2 , ∞) and decreasing on(− ∞ , −1) ∪ (0 , 2).

12  use derivative information only as follows  The sign of the derivative at the central point of the bracketing triplet ( a b c )  the next test point should be taken in the interval [a b] or [a c]  The value of this derivative and of the derivative at the second-best-so-far point are extrapolated to zero by the secant method  We impose the same sort of restrictions on this new trial point as in Brent’s method.  If the trial point must be rejected, we bisect the interval under scrutiny.


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