### Optimization on 1D Ch10.3-4

```Yunfei duan Hui Pan
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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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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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Brent's Method:
of the fact that, near a
minimum, the function is
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
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Of course, there also need to
be checks on this algorithm,
when it is implemented in
practice, to ensure that:
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function f
interpolating
parabola and a
bracketing triplet
of abscissas (a b c)
a<b<c & f(a) >f(b)
<f(c)
(1 2 3) (1 2 4) (1 2 5)
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
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basic idea
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)
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F(x)=3x^4−4x^3−12x^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.
F(x)=3x^4−4x^3−12x^2+3 has first derivative
F’(x) = 12x^3−12x^2−24x = 12x(x^2−x−2)
=12x(x+1)(x−2)
 f(x) is increasing on (−1,0) ∪(2， ∞) and
decreasing on(− ∞ ，−1) ∪(0，2).
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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.
```