Unformatted text preview: Problem 5.10 Instructions and changes in problems Write short Matlab programs for bi‐section and for false position. Generate a table as given in the answers below to compare the progress of bisection and false position. In plot have your axis to (xmin xmax ymin ymax) so that you can see that there is actually a root at 1.5 and not asymptotic at 0. ‐ Also make sure that when you calculate error is ea and not et. ‐ Make sure you have an initial guess in the fzero function 5.10 mod 5.13 6.2 6.2 mod 6.16 7.2 7.4 Compare with the convergence of Matlab fzero with the same interval. Use or modify the bi‐section program you wrote for 5.10. Do one iteration by hand to prepare for exam. For the fixed point iteration, there are two ways of creating the iteration (one involving a square root). Write both, and calculate the derivative of the right hand side in order to select which one to use. Do one iteration by hand to prepare for exam. Compare the convergence in 6.2 (with a table showing the iteration history) for Matlab functions fzero and roots starting with the same initial guess. Do not use an interval for fzero. Use fzero. When doing part (a) make sure you take a good guess, do not put a range in the fzero call. Part (a) by hand, part (b) with Matlab. For part (c), use any rootfinding method you
want.
Note: Use the equation
f (x) = −1.5x − 2x +12x instead of the equation
given in the book for problems 7.4 through 7.6. 6 4 ‐ When you take the second derivative of the function make sure you analyze the expression and you will see that for any value of x the expression will always be negative which means.... ‐ For part a make sure you have a small interval so you can see the maximum more clearly 7.5 Put your intermediate results in a table like in the answers below. ‐ Remember that you are looking for a maximum and the example in the book is for a minimum 7.6 Put your intermediate results in a table like in the answers below. Will need to cancel a common factor in denominator and numerator to avoid division by zero at optimum (s=1). Express one of the dimensions in terms of the other two by using the cross‐sectional area. The textbook solution uses d and s=tan(theta) as the two remaining variables. Find the minimum first graphically, by using contours. Then use fminsrch to home in on the exact solution. Solve the same problem with fmincon using all three variables and the area constraint. It is easier to solve the problem numerically if the area requirement is posed as an inequality constraint rather than an equality constraint. 7.28 7.32 7.32mod 5.10. Result for bisection i
1 x f(x ) l l 0.5 2 1.5 3 1.5 4 1.5 5 1.5 6 1.5 7 1.5 8 1.5 5.13
x = 268.328
ymax = 0.51519 x u f(x )
u x r −32.258
2.5 0.81303
2
2
−0.030
2.5 0.8130
2
95
32
−0.030
2 0.6018
1.75
95
09
−0.030
1.75 0.3789
1.625
95
09
−0.030
1.625 0.2069 1.5625
95
27
−0.030 1.5625 0.0979 1.5312
95
56
5
−0.030 1.5312 0.0362 1.5156
95
5
61
25
−0.030 1.5156 0.0033 1.5078
95
25
83
125 f(x )
r 1.5 ε a −0.0309
46
0.6018 25.00%
09
0.3789 14.29%
09
0.2069
7.69%
27
0.0979
4.00%
56
0.0362
2.04%
61
0.0033
1.03%
83
0.52%
−0.013
595 6.2
For Newton’s method you should get the following convergence: i
0
1 x i f(x)
5 3.353
659 2 2.801
332 3 2.721
108 4 2.719
341 5 2.719
341 2.710
44
0.305
06
0.006
44
3.1E06
7.4E13 f'(x)
13.5
4.907
32
3.802
66
3.642
22
3.638
68
3.638
68 ε a 8.2
49.09
%
19.72
%
2.95%
0.06%
0.00% 6.16 Ta = 1684.3
7.2
X=4.
7.4
(a) (b)
Show that the second derivative will always be negative and hence the function will be concave for all
values of x. (c) x = 0.91692
7.5 7.6 7.28 T=3, s=1
7.32 Theta=60o ...
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 Spring '09
 RAPHAELHAFTKA
 Derivative, initial guess, fixed point iteration, short Matlab programs, xmin xmax ymin, false position. Generate

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