Math 170B HW #6 Partial Solutions
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1.  Consider p_n, the Lagrange interpolatory polynomial passing through 
    x_0, x_1, ..., x_n. Also, consider p_{n+1}, the Newton interpolatory 
    divided difference form of the Lagrange interpolatory polynomial with 
    the additional data point at z. So we have that 
    
     (p_{n+1})(t) = (p_n)(t) + f[x_0, x_1, ... , x_n, z](t-x_0)(t-x_1)...(t-x_n).

    Since p_{n+1} = f at the n+2 points x_0, x_1, ..., x_n, z we have that 
 
     f(z) = p_{n+1}(z) = (p_n)(z) + f[x_0, x_1, ... , x_n, z](z-x_0)(z-x_1)...(z-x_n)

    which implies that 

   (1)  f(z) - (p_n)(z) = f[x_0, x_1, ... , x_n, z](z-x_0)(z-x_1)...(z-x_n).

    But since p_n is the Lagrange interpolatory polynomial passing through x_0,..., x_n, 
    we can use the remainder theorem to get 

   (2)  f(z) - (p_n)(z) = [f^{n+1}(s(z))/(n+1)!](z-x_0)...(z-x_n).

    Equating the right hand sides of (1) and (2) gives the desired result.  QED


2.  Some of these are in the book - they mostly follow from properties of the L_n,k.


3a. H(x) = 0 + (x+1) + 0*(x+1)^2 - (1/2)(x-1)(x+1)^2.

3b. H(x) = (x+1) - (1/2)(x-1)(x+1)^2 - (1/2)(x+1)^2(x-1)^2 - (3/2)x(x+1)^2(x-1)^2.

3c. f(1/2) ~ H(1/2) = 1.35938 and f'(1/2) ~ H'(1/2) = 1.65625.


4.  H(x) = x^2 - x^2(x-1) + (1/2)x^2(x-1)^2 and f(3/2) ~ H(3/2) = 1.40625.


5a. Verify that H(x_i) = f(x_i) and H'(x_i) = f'(x_i) for i = 0, 1.

5b. You should get H(x) - 4x^2(x-1)^2 + (25/4)x^2(x-1)^2(x-2), where H(x) is the 
    polynomial given in 5a. Check this, I may have made a mistake. Again, just like 
    in HW #5 8b, when adding one more data point, we add onto the old polynomial 
    rather than computing everything over again. This is the power of divided 
    differences!


6a. Note that x_1 = 2pi/10 < 1 < 4pi/10 = x_2. Hence we use the formula for error 
    in the book to get (take n = 1 since H is cubic) 

                       
     |f(1) - H(1)| <= [    max sin(y)      ]*(1/4!)*(1 - pi/5)^2(1 - 2pi/5)^2
                      [ pi/5 <= y <= 2pi/5 ]

    On [pi/5, 2pi/5], sin(y) takes its maximum value at 2pi/5. Hence we have that

     |f(1) - H(1)| <= 0.00036056. 

6b. We need n >= 449.