Math 170B HW #3 Partial Solutions
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1a. p_n converges to 0 linearly with asymptotic error constant 1.

1b. p_n converges to 1 linearly with asymptotic error constant 1/10.

1c. p_n converges to 5 quadratically with asymptotic error constant 1. 
    (this one was a little tricky!)


2.  When a = 1, the ratio |p_{n+1} - p|/|p_n - p|^a = 2^{-2n-1} which converges 
    to 0.

    When a > 1, this gets more complicated! We will define that ratio to be q_n. 
    After simplification, we get q_n = 2^{(a-1)n^2 - 2n - 1}. Let C > 0 be arbitrary.
    Consider the parabola w(n) := (a-1)n^2 - 2n - 1 - C in the variable n. Remember 
    a > 1 is fixed and C > 0 is fixed so the variable is n. Using the quadratic 
    formula on w(n) we can get the two real distinct roots; call them R1 and R2 
    respectively with R1 < R2. Since a > 1, a - 1 > 0 so the parabola w(n) is convex 
    and hence for all n > R2, w(n) > 0. Equivalently, (a-1)n^2 - 2n - 1 - C > 0 for 
    all n > R2; or even more, 

                    (a-1)n^2 - 2n - 1 > C    for all   n > R2.

    Thus, raising 2 to these powers we get, for all n > R2, 

                    (a-1)n^2 - 2n - 1        C
                   2                    >   2   .

    Since C was arbitrary, we have the statement: For any C > 0, there exists N such 
    that for all n > N, q_n > 2^C.  This is precisely what it means for q_n --> infinity; 
    ie, for q_n to not converge.  QED


3.  All multiplicities should be 2. (Use the derivative checking theorem)


4a. p_n = [(1 - 1/k)^n] p_0.

4b. You must show that it converges linearly but also does not converge for any a > 1. 
    To do this, set up the standard ratio, |p_(n+1) - p| / |p_n - p|^a and use the 
    fact that (1 - 1/k) < 1. Examine the two cases - when a = 1 and a > 1. Note that 
    p = 0 since we are looking for the root of the polynomial x^k.

4c. Use the fact that f(x) = x^k and f'(x) = k x^(k-1) so h(x) = x/k.

4d. This is easy if you plug in f(x) = x^k and f'(x) = k x^(k-1).


5, 6, 7, 8 are computational.