Cubic Interpolation Report

The Cubic Interpolation Method

Erika Paula L. MoisesJoe Mari D. Reyes

Math 288 (Numerical Optimization)

May 4, 2015

Erika Paula L. Moises, Joe Mari D. Reyes The Cubic Interpolation Method

Presentation Outline



The cubic interpolation method approximates the objectivefunction

g() = f (x + d)

by a cubic polynomial, where f is continuously differentiable. Bythe chain rule, we have

g () =dg()

d= f (x + d)Td

We assume that g (0) = f (x)Td < 0, that is, d is the descentdirection at x .



To construct the cubic polynomial p(t), four interpolationconditions are required.

For example, we may use:

function values at four points, or

function values at three points and a derivative value at onepoint; or

function values and derivative values at two points



In this report, we will focus on interpolation using function valuesand derivative values at two points a and b.

To guarantee that the appropriate minimum occurs in the interval[a, b], the following condition should hold:

g (a) < 0 and g (b) > 0



The central idea is to use the derivative information in the endpoints of the interval. Together with the function values, theminimum of the given function is approximated by the minimum ofa cubic polynomial.



We construct a cubic polynomial of the form

p(t) = c1(t a)3 + c2(t a)2 + c3(t a) + c4 (1)

where ci are the coefficients of the polynomial which are chosensuch that

p(a) = c4 = g(a),

p(a) = c3 = g (a),

p(b) = c1(b a)3 + c2(b a)2 + c3(b a) + c4 = g(b),p(b) = 3c1(b a)2 + 2c2(b a) + c3 = g (b).

(2)



From the sufficient condition of the minimizer, we have

p(t) = 3c1(t a)2 + 2c2(t a) + c3 = 0 (3)

andp(t) = 6c1(t a) + 2c2 > 0 (4)

Solving Equation (3) yields

t = a +c2

c22 3c1c3

3c1, if c1 6= 0 (5)

t = a c32c2

, if c1 = 0. (6)



In order to guarantee that condition (4) holds, we only take thepositive in Equation (5).

By equation (5), we get:

t a =c2 +

c22 3c1c3

3c1=

c3c2 +

c22 3c1c3

(7)



When c1 = 0, Equation (7) is just Equation (6). Then theminimizer of p(t) is

t = a c3c2 +

c22 3c1c3

(8)

The minimizer in Equation (8) is represented by c1,c2 and c3.

We hope to represent t by g(a),g(b), g (a) and g (b) directly.



We now let

s = 3g(b) g(a)

b az = s g (a) g (b),w2 = z2 g (a)g (b) (9)



By use of Equation (2), we get

s = 3g(b) g(a)

b a = 3[c1(b a)2 + c2(b a) + c3],

z = s g (a) g (b) = c2(b a) + c3,w2 = z2 g (a)g (b) = (b a)2(c22 3c1c3).



Then

(b a)c2 = z c3c22 3c1c3 =

w

b a

And so

c2 +

c22 3c1c3 =z + w c3

b a (10)



Using c3 = g(a) and substituting Equation (10) into Equation (8),

we get:

t a = (b a)g(a)

z + w g (a) , (11)

which is

t a = (b a)g(a)g (b)

(z + w g (a))g (b) =(b a)(z2 w2)

g (b)(z + w) (z2 w2)

=(b a)(w z)g (b) z + w (12)



Note that the denominator in Equation (12) may be equal to zeroor sufficiently small. We can overcome this by the use of Equation(11) on (12). We get

t a = (b a)g(a)

z + w g (a) =(b a)(w z)g (b) z + w

=(b a)(g (a) + w z)

g (b) g (a) + 2w

= (b a)(1 g(b) + z + w

g (b) g (a) + 2w ) (13)



This gives us the iterates of the Cubic Interpolation Method:

t = a + (b a) w g(a) z

g (b) g (a) + 2w (14)

where

s = 3g(b) g(a)

b a , z = s g(a) g (b)

w2 = z2 g (a)g (b)



In Equation (13) and (14), the denominator

g (b) g (a) + 2w 6= 0

.In fact, since g (a) < 0 and g (b) > 0, then

w2 = z2 g (a)g (b) > 0

.Taking w > 0, it follows that the denominator

g (b) g (a) + 2w > 0

.


Remark

Cubic interpolation has the danger of lack of convergencenumerically, but the theoretical convergence of the iterate is veryfast.

It can be shown that the cubic interpolation method convergessuperlinearly.


Cubic Interpolation Steps

We now lay down the steps for the Cubic Interpolation Method

Step 1 (Determination of the Initial Interval)

Let s > 0 be some scalar.

Evaluate g(t) and g (t) at the points t = 0, s, 2s, 4s, 8s, ... untiltwo successive points a and b are found such that either g (b) 0or g(b) g(a).

Then, it can be seen that a local minimum of g exists within theinterval (a, b].



Step 2 (Updating of the Current Interval)

Given the current interval [a, b], a cubic polynomial is fitted to thefour values g(a), g (a), g(b) and g (b).

The cubic can be shown to have a unique minimum t in theinterval (a, b] given by

t = a + (b a) w g(a) z

g (b) g (a) + 2w ,

where z = 3g(b)g(a)ba g (a) g (b), w =

z2 g (a)g (b)



Step 2 (Updating of the Current Interval)

If g (t) 0 or g (t) g(a) replace b by t.

If g (t) < 0 and g (t) < g(a) replace a by t.

Stopping Criterion

In practice, the computation is terminated once the length of thecurrent interval becomes smaller than a specified tolerance or elsewe obtain t = b.


Example 1

Cubic interpolation is applied to find the minimum f (x) = x + 16x+1with starting interval [2, 4.5] and accuracy tol = 0.01.


Example 1

One iteration after reaching the stopping criterion has been givenin Table 1. For this case, the interval changes very fast around theminimum point.

k ak bk xk f(xk) f(xk)

1 2.000 4.500 3.024 7.0001 0.0122 2.000 3.024 2.997 7.0000 -0.0013 2.997 3.024 3.000 7.0000 0.0004 2.997 3.000 3.000 7.0000 0.000

Table: Cubic interpolation for f (x) = x + 16x+1 ,[a0, b0] = [2, 4.5],tol = 0.01


References:

Bertsekas, Dimitri P. 1999. Nonlinear Programming. 2ndEdition. pp 723-724

Sun, Wenyu. Yuan, Ya-Xiang. 2006. Optimization Theoryand Methods (Nonlinear Programming). Volume 1. pp 98-101


Thank you


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