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Midterm

The document discusses using various numerical methods to find the maximum of the function f(x)=4x-1.8x^2+1.2x^3-0.3x^4, including golden section search, parabolic interpolation, the secant method, and Newton's method. It provides the inputs and outputs of each method over multiple iterations to find the maximum, such as applying golden section search over the interval [-2,4] and Newton's method starting at x0=3.

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NOV DAVANN
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89 views1 page

Midterm

The document discusses using various numerical methods to find the maximum of the function f(x)=4x-1.8x^2+1.2x^3-0.3x^4, including golden section search, parabolic interpolation, the secant method, and Newton's method. It provides the inputs and outputs of each method over multiple iterations to find the maximum, such as applying golden section search over the interval [-2,4] and Newton's method starting at x0=3.

Uploaded by

NOV DAVANN
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Name: NOV DAVANN

ID: e20190712 Midterm


Numerical Method and Optimization

Employ the following methods to find the maximum of f ( x )=4 x−1.8 x 2+ 1.2 x 3−0.3 x 4

a) Golden section search ( x l=−2 , xu =4 , ε s =1% )

i xl f (x l) x2 f (x 2) x1 f (x 1) xu f (x u) d
1 -2 -29.6 0.3 1.067 1.700 4.987 4 -12.8 3.708
2 0.3 1.067 1.700 4.987 1.165 3.561 1.70 4.987 0.865
3 0.3 1.067 0.386 1.338 1.700 4.987 1.165 3.561 0.779
4 0.386 1.338 1.70 4.987 1.646 4.856 1.165 3.561 0.481

b) Parabolic interpolation ( x 0=1.75 , x 1=2 , x 2=2.5 )

i x0 f (x 0) x1 f ( x 1) x2 f ( x 2) x3 f ( x 3)
1 1.75 5.105 2 5.6 2.5 5.78 2.333 5.885
2 2 5.6 2.33 5.885 2.5 5.78 2.310 5.884
3 2.310 5.884 2.33 5.885 2.5 5.78 2.327 5.885
4 2.310 5.884 2.327 5.885 2.33 5.885 2.318 5.885

 Secant method for interval a=1 , b=2

i Xn X n+1 X n+2 error

1 1 2 1.1680 1.7090
2 2 1.1680 1.5381 1.1670
3 1.168 1.2854 1.5382 1.0171
4 1.2854 1.5382 1.4205 1.9629

c) Newton’s Method ( x 0=3 , ε =1 %)

i x f ( x) f ' (x) f ' ' (x)


0 3 3.9 -6.8 -14.4
1 2.5277 5.7434 -1.4366 -8.3470
2 2.3517 5.8832 -0.1636 -6.5775
3 2.3268 5.8853 -0.00283 -6.3374
4 2.3263 5.8853 0.00033 -6.3226

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