Math 449 - Homework  1 - Solutions

1) 14 (a)  of section 1.1, page 12.

We look for the coefficients of the polynomial

Q(x)=b_4x^3 +b_3 x^2 +b_2 x + b_1, and b_0=P(c),  

such that

P(x)=(x-3)Q(x) + b_0.

The coefficients of P(x) are a_4=1, a_3=1, a_2=-13, a_1=-1, a_0=-12.

The algorithm for Horner's method is:
 
b_4=a_4
   =1 
   
b_3=a_3+c*b_4
   =1+3*1
   =4 
    
b_2=a_2+c*b_3
   =-13+3*4
   =-1 
    
b_1=a_1+3*b_2
   =-1+3*(-1)
   =-4
   
b_0=a_0+3*b_1
   =-12+3*(-4)
   =-24
    
So P(3)=-24. 
 
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2)  2 (d)  of section 1.2, page 23.
 
1000000111_(two)=2^9+2^2+2^1+2^0=512+4+2+1=519.
 
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3)  3 (d)  of section 1.2, page 23.
 
0.110110110_(two)=2^(-1)+2^(-2)+2^(-4)+2^(-5)+2^(-7)+2^(-8)
                 =219/256
                 =0.85546875.
 
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4)  1 (c) of section 1.3, pages 37-38.

Absolute error:
E_z=|z-z^|=0.00008

Relative error:
R_z=|z-z^|/|z|=0.000008/0.000068=0.117

Since 0.5*10^(-1) < R_z < 0.5*10^(-0), z^ approximates z to 1 significant digit.


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5)  5 (c)  of section 1.3, pages 37-38.

To avoid dealing with differences, we can use the identity 
cos^2(x)-sin^2(x)=cos(2x).

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6)  10  of section 1.3, pages 37-38.

exp(h) = 1 + h + h^2/2! + h^3/3! + h^4/4! + O(h^5)
sin(h) = h - h^3/3! + O(h^5)

Then 

exp(h) + sin(h) = 1 + 2*h + h^2/2! + h^4/4! + O(h^5)
exp(h)*sin(h) = h + h^2 + h^3/3 + O(h^5).


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7) Algorithms and Programs: 2 (b) of section 1.3, page 39.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% We want to generate the sequence {1/2^n : n = 1, 2, ... , m} using the 
% recursive scheme
%
%                  p(n) = (3/2)*p(n-1) - (1/2)*p(n-2)
%
% with two initial conditions: First, p0 = 1 and p1 = 1/2. This gives the 
% correct sequence 1/2^n. Second, with p0 = 1 and p1 = 0.497, which 
% introduces a small error. One possible MATLAB script for this is 
% as follows.
 
m = 10; % This is the number of terms in the sequence. 

p0 = 1;        % These are the conditions for the first case. For the second,
p1 = 0.5;      % we take p1 = 0.497. 

p=[p0 ; p1];   % This creates a column vector of dimension two containing p0 and p1. 
               % Notice that p(1)=p_0 and p(2)=p_1. 
               
A=[3/2 -1/2 ; 1 0]; % The matrix A multiplied by the vector [p(n-1) ; p(n-2)] gives
                    % the vector [p(n) ; p(n-1)].
                    
for n=1:m-2
  q = [p(n+1) ; p(n)];
  q = A*q;
  p = [p ; q(1)]; % Adds to the vector p a new entry equal to the top entry of q.
end                   

% We compare this sequence with what is obtained by direct calculation of x(n+1)=1/2^n

i = 0:1:m-1;
x = 2.^(-i);

% The sequence of errors is now 
e = x' - p; % The apostrophe indicates transpose.


% In the first case (p1 = 0.5) the resulting table is:

%  iteration     x         p         e
  
    1.0000    1.0000    1.0000         0
    2.0000    0.5000    0.5000         0
    3.0000    0.2500    0.2500         0
    4.0000    0.1250    0.1250         0
    5.0000    0.0625    0.0625         0
    6.0000    0.0312    0.0312         0
    7.0000    0.0156    0.0156         0
    8.0000    0.0078    0.0078         0
    9.0000    0.0039    0.0039         0
   10.0000    0.0020    0.0020         0


% In the second case (p1 = 0.497) the resulting table is:

%  iteration     x         p         e
    
    0.0000    1.0000    1.0000         0
    1.0000    0.5000    0.4970    0.0030
    2.0000    0.2500    0.2455    0.0045
    3.0000    0.1250    0.1198    0.0052
    4.0000    0.0625    0.0569    0.0056
    5.0000    0.0312    0.0254    0.0058
    6.0000    0.0156    0.0097    0.0059
    7.0000    0.0078    0.0019    0.0060
    8.0000    0.0039   -0.0021    0.0060
    9.0000    0.0020   -0.0040    0.0060


% The graph for the error e as a function of the iteration number is
% obtained by

plot(i, e' , '--rs') 
grid
xlabel('iteration number')
ylabel('error = x - p')
title('Error for iteration scheme of HW 01, problem 7 - Your name here')
  
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