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ee231_ass5_sol

# ee231_ass5_sol - EE 231_Bi Numerical Analysis for...

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EE 231_Bi Numerical Analysis for Electrical and Computer Engineers Course Instructor: Dr. Sergiy Vorobyov 7.4 (a) The function can be plotted C // -50 100 -2 -1.5 -1 -0.5 C 0.5 1 1.5 (b) The function can be differentiated twice to give f”z—45x 4 +24x 2 The second derivate is positive for small values of x, therefore it is not concave. (e) Differentiating the function and setting the result equal to zero results in the following roots problem to locate the maximum f’(x)=O=—9x 5 +8x 3 +12 A plot of this function can be developed 200 150 100 50 0 -, -50 -100 -150 -200 -25C -2 -1.5 -1 -0.5 0 0.5 1 1.5

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EE 231_Bi Numerical Analysis for Electrical and Computer Engineers Course Instructor: Dr. Sergiy Vorobyov A technique such as secant method can be employed to determine the root. Here are the first 2 iterations: f’(x) = —9x 5 + 8x 3 +12 0 f(x 0 )(x 1 x 0 ) f(1 .5)(0 —0.5) = 15 29.3438* (—0.5) 1-±363 f(x 1 ) - f(x 0 ) f(1) - f(1 .5) 11- (-29.343 8) 2 _f(x 1 )(x 0 =1.1363— f(1.1363)(1.5—1.1363) 11363 6.6867*0,3637 =1.2038 f(x 0 ) f(x 1 ) f(1 .5) f(1 .1363) 29.3438 6.6867 ea 1x,÷i —x = 11.2038—1.13631 0.0675 7.5 The problem of finding a maximum of the function f(x) = —1.5x 6 + 2x 4 +12x is equivalent to the problem of finding a minimum of the function —f(x)=l.5x 6 —2x 4 —12x. Using the golden ratio we find d, x 1 , x 2 , and evaluate the function at these points, that is, d=1(x_x/)=1(2_o)1.2361, x 1 =0+1.2361=1.2361, x2 =2—1.2361=0.7639 —f(x 1 ) =—14.1516, —f(x 2 )=—9.5503 Because f(x 1 ) <—f(x
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