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160    INTERPOLATION AND CURVE FITTING

                       k         0        1      2      3       4

                       x k      −π      −π/2     0    +π/2     +π
                       f(x k )   −1       0      1      0       −1

                (b) Find the Lagrange/Newton polynomial of degree 4 matching the fol-
                   lowing five points and plot the resulting polynomial on the same graph
                   that has the result of (a).


                    k          0            1       2       3           4
                          π cos(9π/10)π cos(7π/10) 0 π cos(3π/10)π cos(π/10)
                    x k
                    f(x k )  −0.9882     −0.2723    1    −0.2723     −0.9882

                (c) Find the Chebyshev polynomial of degree 4 for cos x over [−π, +π]
                   and plot the resulting polynomial on the same graph that has the result
                   of (a) and (b).
            3.4 Chebyshev Nodes
                The current speed/pressure of the liquid flowing in the pipe, which has irreg-
                ular radius, will be different from place to place. If you are to install seven
                speed/pressure gauges through the pipe of length 4 m as depicted in Fig.
                P3.4, how would you determine the positions of the gauges so that the max-
                imum error of estimating the speed/pressure over the interval [0, 4] can
                be minimized?




                                                                  x
                           0         1       2         3        4
                                    Figure P3.4 Chebyshev nodes.
            3.5 Pade Approximation
                For the Laplace transform
                                          F(s) = e −sT                  (P3.5.1)

                representing the delay of T [seconds], we can write its Maclaurin series
                expansionuptofifth order as

                                      (sT ) 2  (sT ) 3  (sT ) 4  (sT ) 5
                           ∼
                     Mc(s) = 1 − sT +       −      +       −            (P3.5.2)
                                        2!     3!      4!      5!
                (a) Show that we can solve Eq. (3.4.4) and use Eq. (3.4.1) to get the Pade
                   approximation as
                                        q 0 + q 1 s  1 − (T /2)s
                             ∼                              ∼   −Ts
                         F(s) = p 1,1 (s) =     =           = e         (P3.5.3)
                                        1 + d 1 s  1 + (T /2)s
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