Answers to Exam 1(b) Wed 23 Feb 2005

Suppose the basis of our floating-point numbers is ten and all numbers have a fraction (mantissa) of 4 decimal places long, with exponents ranging between -22 and +22.

What is the smallest positive floating-point number?
Answer:
```
        +.1000E-22 = 10^(-23)
```

What is the result of 228.5 + 1.704 in this number system?
Answer:

      228.5 = +.2285E+3
      1.704 = +.1704E+1 = +.001704E+1

       +.2285E+3
       +.001704E+1
      -------------
       +.230204E+3 = +.2302E+3 = 230.2

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Below is the plot of g(x) = exp(0.25*x). Starting at x(0) = 8, illustrate on the plot below how to produce three more points defined by x(k+1) = g(x(k)), k=0,1,...

Answer: See the handout on paper.

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Show the convergence rate of Newton's method is (m-1)/m when applied to f(x) = (x-1)^m.
Answer:

       x(k+1) = x(k) - f(x(k))/f'(x(k))
      => g(x) = x - f(x)/f'(x)

                    f'(x)f'(x) - f(x)f"(x)
       g'(x) = 1 - ------------------------
                          [ f'(x) ]^2

                    m^2(x-1)^(m-1)(x-1)^(m-1) - (x-1)^m m(m-1)(x-1)^(m-2)
             = 1 - -------------------------------------------------------
                               m^2 (x-1)^(m-1) (x-1)^(m-1)

                    m^2 - m^2 + m
             = 1 - ---------------
                         m^2

                    1      m-1
             = 1 - --- =  -----
                    m       m

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Consider f(x) = x^4 + exp(-x) over the interval [0,1].

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.

Answer:
```
        [a1,b1] = [3.82E-1,1.00E+0]
        [a2,b2] = [3.82E-1,7.64E-1]
        [a3,b3] = [3.82E-1,6.18E-1]
```

Write the values for x1, x2, f(x1), and f(x2) (4 decimal places, scientific notation):
Answer:

   +--------+-----------+-----------+-----------+------------+
   |  step  |    x1     |     x2    |   f(x1)   |   f(x2)    |
   +========+===========+===========+===========+============+
   |    0   |  3.820E-1 |  6.180E-1 |  7.038E-1 |  6.849E-1  |
   |    1   |  6.180E-1 |  7.639E-1 |  6.849E-1 |  8.064E-1  |
   |    2   |  5.279E-1 |  6.180E-1 |  6.675E-1 |  6.849E-1  |
   +--------+-----------+-----------+-----------+------------+

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Consider

             [  5.041E-01   2.143E-01   7.506E-01  ]
         A = [  6.483E-01   5.604E-01   8.562E-01  ]
             [  9.167E-01   2.797E-01   9.629E-01  ]

Compute the LU decomposition of A with partial pivoting. Use 4 decimal places with rounding, and write all floating-point numbers in scientific format.
Answer:

                           [  9.167E-1    2.797E-1    9.629E-1  ] 3
  A  ------------------->  [  6.483E-1    5.604E-1    8.562E-1  ] 2
                           [  5.041E-1    2.143E-1    7.506E-1  ] 1
             6.483
 R2 := R2 - ------- R1
             9.167         [  9.167E-1    2.797E-1    9.629E-1  ] 3
------------------------>  [  7.072E-1    3.626E-1    1.752E-1  ] 2
             5.041         [  5.499E-1    6.050E-2    2.211E-1  ] 1
 R3 := R3 - ------- R1     
             9.167         

            0.6050
 R3 := R3 - ------- R2
             3.626         [  9.167E-1    2.797E-1    9.629E-1  ] 3
------------------------>  [  7.072E-1    3.626E-1    1.752E-1  ] 2
                           [  5.499E-1    1.669E-1    1.919E-1  ] 1


        [ 0 0 1 ]          [  1.000E+0    0.000E+0    0.000E+0  ]
    P = [ 0 1 0 ]      L = [  7.072E-1    1.000E+0    0.000E+0  ]
        [ 1 0 0 ]          [  5.499E-1    1.669E-1    1.000E+0  ]

                           [  9.167E-1    2.797E-1    9.629E-1  ]
                       U = [  0.000E+0    3.626E-1    1.752E-1  ]
                           [  0.000E+0    0.000e+0    1.919E-1  ]

What is the determinant of A?
Answer:

         det(A) = det(P L U)
                = det(P)*det(L)*det(U)
                = (-1)*1*(9.167E-1)*(3.626E-1)*(1.919E-1)
                = -6.379E-2

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