Newton’s Method for One Dimensional Optimization

Major: All Engineering Majors Authors: Autar Kaw, Ali Yalcin 4/24/2020 http://nm.mathforcollege.com

Transforming Numerical Methods Education for STEM Undergraduates http://nm.mathforcollege.com

1

Newton’s Method

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Newton’s Method-Overview

    Open search method A good initial estimate of the solution is required The objective function must be twice differentiable Unlike Golden Section Search method • • Lower and upper search boundaries are not required (open vs. bracketing) May not converge to the optimal solution http://nm.mathforcollege.com

3

4  

Newton’s Method-How it works

f

at the function’s maximum and minimum.

 

f

'  0 The minima and the maxima can be found by applying the Newton-Raphson method to the derivative, essentially obtaining

x i

 1 

x i



f f

' (

x i

) '' (

x i

) http://nm.mathforcollege.com

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Newton’s Method-Algorithm

Initialization: Step 1 Step 2

f

. Determine and .

'

f

Determine a reasonably good estimate for ''  

x i x

0

f

  . Substitute (initial estimate for the first iteration) '

x i

 1 

x i



f f

'' ( (

x i x i

) ) to determine

Step 3.

x i

 1 and the function value in iteration i .

If the value of the first derivative of the function is zero then you have reached the optimum (maxima or minima). Otherwise, repeat Step 2 with the new value of

i

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6

Example

.

2 2   2 The cross-sectional area A of a gutter with equal base and edge length of 2 is given by

A

 4 sin  ( 1  cos  ) Find the angle  which maximizes the cross-sectional area of the gutter. http://nm.mathforcollege.com

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Solution

The function to be maximized is

f

(  )  4 sin  ( 1  cos  )

f

 (  )  4 (cos   cos 2   sin 2  )

f

 (  )   4 sin  ( 1  4 cos  ) Iteration 1: Use as the initial estimate of the solution 0    1   4  4 (cos   4 4 sin   cos 2 4 ( 1   4  sin 4 cos 2   4 ) 4 )  1 .

0466

f

( 1 .

0466 )  5 .

196151 http://nm.mathforcollege.com

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Solution Cont.

Iteration 2:

 2  1 .

0466  4 (cos 1 .

0466  cos 2 1 .

0466  sin 2 1 .

0466 )  4 sin 1 .

0466 ( 1  4 cos 1 .

0466 )  1 .

0472

Summary of iterations

Iteration 1 2 3 4 5  0.7854

1.0466

1.0472

1.0472

1.0472

f

' (  ) 2.8284

0.0062

1.06E-06 3.06E-14 1.3322E-15

f

'' (  ) -10.8284

-10.3959

-10.3923

-10.3923

-10.3923



estimate

1.0466

1.0472

1.0472

1.0472

1.0472

f

(  ) 5.1962

5.1962

5.1962

5.1962

5.1962

Remember that the actual solution to the problem is at 60 degrees or 1.0472 radians. http://nm.mathforcollege.com

Additional Resources

For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit http://nm.mathforcollege.com/topics/opt_newtons_method.html

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