Math 170B HW #4 Partial Solutions
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1. Computational.

2. Computational.


3a. Just check p(p_i) = f(p_i) for i = 0,1,2.

3b. The roots are x = -0.5, 1.

3c. We choose p_3 = 1 since it is closest to the last guess, p_2.


4. The roots should be -4, -2, 0, 1, 2. (check these)


5. Define R(x) = P(x) - Q(x). Then R(x) is a polynomial and deg(R) <= n. 
   Also R(x_i) = 0 for i = 0, 1,..., n. Hence R(x) has n+1 roots. So by 
   the Fundamental Theorem of Algebra, deg(R) != 1, 2,..., n. But since 
   deg(R) <= n, this forces deg(R) = 0 which means R(x) = c, for some 
   constant c. But since R(x_i) = 0, we get c = 0; i.e., R(x) = 0 and thus 
   P(x) = Q(x) for all x.  QED


6a.  p(x) = x + 2.

6b.  p(x) = 2.3403x + 2.7083.


7a.  p(x) = x^2.

7b.  p(x) = 0.854136x^2 + 1.48626x + 1.


8. approx = 1.00000000000015,
        
          abs(p(n+1)-1)/(p(n)-1)^2         
 
   n = 1,     0.01906365280598
   n = 2,     0.10066343755305
   n = 3,     0.12449253625223
   n = 4,     0.12468093583555