Math 449 - Homework  2 - Solutions

1) 3 (b) and 3 (d), section 2.1, page 50: see graphs 1 and 2.

 
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2)  Algorithm 1 (b), section 2.1, page 51. 

After saving the function files fixpt.m and g.m, where g(x)=cos(sin(x)),
we can enter the following command (I've chosen the initial point p0=1
after inspecting the graph.)

>>[k, p, err, P]=fixpt('g',1,5*10^(-12),100)

The above gives the value p=0.76816915673757, with a relative error of

3.2098e-12. Therefore, the approximation for the fixed point holds to
12 significant digits. 

See graph 3.

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3)  9 of section 2.2, p. 61.

(a) The function f(x)=1/(x-2) is positive over the interval [3, 7]. The 
bisection method does not apply.

(b) The limits of a_n and b_n as n goes to infinity are both equal to 2,
even though the function is not defined at 2. 

 
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4)  11 of section 2.2, p. 61.

The bisection theorem (Theorem 2.4) gives |r-c_n| less than or equal
to (b-a)/2^(n+1). For this number to be less than 5*10^(-9) we need

            (7-2)/2^(n+1)< 5*10^(-9).
      
Solving for n gives: n+1 > 9/log 2 (where this log has base 10.) That is,
n > 28.89     Therefore, n=29.       


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5)  Problem 2 in "Algorithms and Programs" of section 2.3, p. 69.

We first define a function file with the function to be investigated.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function y=g(x)
y=1000000*x.^3 - 111000*x.^2 + 1110*x - 1;

%%%%The above is saved in the Matlab folder to a file named g.m%%%%%%%%%%%


The program for approximate location of roots (page 68) given below should
also be saved to a function file.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function R = approot(X,epsilon)

% Input  - g is the object function saved as an M-file named g.m
%        - X is the vector of abscissas
%        - epsilon is the tolerance
% Output - R is the vector of approximate roots

Y=g(X);
yrange = max(Y) - min(Y);
epsilon2 = yrange*epsilon;
n=length(X);
m=0;
X(n+1)=X(n);
Y(n+1)=Y(n);

for k=2:n
    if Y(k-1)*Y(k)<=0
        m=m+1;
        R(m)=(X(k-1)+X(k))/2;
    end
    s=(Y(k)-Y(k-1))*(Y(k+1)-Y(k));
    if (abs(Y(k))) & (s<=0)
        m=m+1;
        R(m)=X(k);
    end
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

To obtain candidates for the roots to four decimal places over the 
interval [-2, 2] we use the program approot as in example 2.10:

>> X=-2:.001:2;
>> approot(X,0.00001)

This gives the numbers

    0.0015    0.0050    0.0095    0.0690    0.1005    2.0000

A closer examination of the graph near these values shows that only 

               0.0015, 0.0095, and 0.1005 

lie near a root. (There are at most three roots of a third degree 
polynomial, so these are all the roots.)  Further examination gives
the intervals where the function changes sign:

[0.000, 0.002], [0.009, 0.011], [0.095, 0.105].

We can now apply, say, the bisection method to find the roots to
any desired degree of accuracy.



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6)  Exercise 2 of section 2.4, p. 85.

(a) p_(k+1) = p_k - (p_k^2 - p_k -3)/(2*p_k - 1).

(b) Starting with p_0=1.6 gives: 

          p_1=2.5273, p_2=2.3152, p_3=2.3028, p_4=2.3028

(c) Starting with p_0=0.0 gives:
 
p_1=-3.0, p_2=-1.7143, p_3=-1.3410, p_4=-1.3032  p_5=-1.3028, p_6=-1.3028. 

Conjecture that it converges to -1.3028.
    
    
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7) Exercise 11 of section 2.4, p. 86.

This follows from a direct application of the Newton-Raphson iteration
formula and a little algebra. 

  
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