Answers to Exam 1(b) Mon 7 Oct 2005

  1. Consider a floating-point number system with base 10. There are five digits in the fraction (mantissa) and the exponents range between -8 and +9.
    1. What is the smallest positive floating-point number in this number system? 
          +.10000E-8 = 10^(-9)
    2. What is the result of 42.873 + 0.079872 in this number system? 
          42.873 = +.42873E+2
    0.079872 = +.79872E-1 = +.00079872E+2 (denormalize)

    +.42873E+2
    +.00079872E+2
   ---------------
    +.42952872E+2 = +.42953E+2 = 42.953

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  2. Derive the formula for the secant method for one equation f(x) = 0.
    Illustrate the working of the secant method with a plot. 
       The line through (x(k),f(x(k))) and (x(k+1),f(x(k+1))) has equation

                    f(x(k+1)) - f(x(k))
      y - f(x+1) = --------------------- ( x - x(k+1) ).
                       x(k+1) - x(k)

   To obtain x(k+2) we solve for y = 0:

                       x(k+1) - x(k)
      x - x(k+1) = --------------------- ( 0 - f(x(k+1) ).
                    f(x(k+1)) - f(x(k))

   Setting x = x(k+2) we obtain

                           x(k+1) - x(k)
     x(k+2) = x(k+1) - --------------------- f(x(k+1)).
                        f(x(k+1)) - f(x(k))

   For an illustration of the method, see the solution handout.

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  3. Below is the plot of g(x) = -0.3x^2 + 0.8 x + 2.5. Starting at x(0) = -3, illustrate on the plot below how to produce three more points defined by x(k+1) = g(x(k)), k=0,1,...

    click here to see the plot
    See the solution handout for the answer. 

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  4. Consider f(x) = x^2 - 0.5 x - 0.24 over the interval [0,1].
    1. Starting with [0,1], apply two steps of the golden section search method, and indicate on the graph below where you do the function evaluations.
      In addition, mark the new intervals as [a1,b1], [a2,b2], and [a3,b3] on the x-axis.

      click here to see the graph
      See the solution handout for the answer.

    2. Write the values for x1, x2, f(x1), and f(x2) (4 decimal places, scientific notation): 
      +-------+------------+------------+------------+------------+
| step  |     x1     |     x2     |    f(x1)   |    f(x2)   |
+-------+------------+------------+------------+------------+
|   0   |  3.820E-1  |  6.180E-1  | -2.851E-1  | -1.671E-1  |
|   1   |  2.361E-1  |  3.820E-1  | -3.023E-1  | -2.851E-1  |
|   2   |  1.459E-1  |  2.361E-1  | -2.917E-1  | -3.023E-1  |
+-------+------------+------------+------------+------------+

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  5. Consider 
                 [  9.229E-02  -1.324E+00   1.976E+00  ]
         A = [ -6.501E-01   1.201E+00  -3.308E-01  ]. 
             [  2.245E+00  -1.265E+00  -1.277E+00  ]
    1. Compute the LU decomposition of A with partial pivoting. Use 4 decimal places with rounding, and write all floating-point numbers in scientific format. 
                [  2.245E+00  -1.265E+00  -1.277E+00  ] 3
  A ----> [ -6.501E-01   1.201E+00  -3.308E-01  ] 2
          [  9.229E-02  -1.324E+00   1.976E+00  ] 1

            -.6501      [                                     ]
 R2 := R2 - ------ R1   [  2.245E+00  -1.265E+00  -1.277E+00  ] 3
             2.245      [                                     ]
 ---------------------> [ -2.896E-01   8.347E-01  -7.006E-01  ] 2
            0.09229     [                                     ]
 R3 := R3 - ------- R1  [  4.111E-02  -1.272E+00   2.028E+00  ] 1
             2.245      [                                     ]

                        [  2.245E+00  -1.265E+00  -1.277E+00  ] 3
 ---------------------> [  4.111E-02  -1.272E+00   2.028E+00  ] 1
                        [ -2.896E-01   8.347E-01  -7.006E-01  ] 2

             .8347      [                                     ]
 R3 := R3 - ------ R2   [  2.245E+00  -1.265E+00  -1.277E+00  ] 3
            -1.272      [                                     ]
 ---------------------> [  4.111E-02  -1.272E+00   2.028E+00  ] 1
                        [                                     ]
                        [ -2.896E-01  -6.562E-01   6.302E-01  ] 2
                        [                                     ]

     [ 0 0 1 ]       [     1           0            0      ]
 P = [ 1 0 0 ]  L =  [  4.111E-02      1            0      ]
     [ 0 1 0 ]       [ -2.896E-01  -6.562E-01       1      ]

                     [  2.245E+00  -1.265E+00  -1.277E+00  ]
                U =  [     0       -1.272E+00   2.028E+00  ]
                     [     0           0        6.302E-01  ]
    2. What is the determinant of A? 
       det(A) = det(P*L*U)
          = det(P)*det(L)*det(U)
          = (+1)*(+1)*(2.245E+00)*(-1.272E+00)*(6.302E-01)
          = -1.800E+00

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