Math 449 - Homework 7 - Solutions

1) Exercise 2 (b), section 5.1, page 259.

The data points are:

x =[-6    -2     0     2     6]

y =[-5.3000   -3.5000   -1.7000    0.2000    4.0000]

I will use the notation e = [1 1 1 1 1] and u.v for the dot product of two

vectors u and v. The constants A and B for the least squares line 

y = Ax + B satisfy the linear system

x.x A  + x.e B  = x.y

x.e A  +   5 B  = y.e

From the above data we obtain the numerical values for the coefficients:

x.x = 80, x.e = 0, x.y = 63.2, y.e = -6.3

This immediately yields 80 A = 63.2 and 5 B = -6.3, or A = 0.79, B = -1.26

Therefore, the least squares line is y = 0.79 x - 1.26

The error E_2(f), using Matlab notation, is given by

E_2(f) = sqrt((1/5)*sum((A*x+B - y).^2)) = 0.5299

************************************************************************** 
 
2) Exercise 8 (a), section 5.1, page 261.

We want to find the power fits y = A x^2 and Y = B x^3 for the data:

x = [2.0000    2.3000    2.6000    2.9000    3.2000]

y = [5.1000    7.5000   10.6000   14.4000   19.0000]

and find the best fit based on the error E_2(f).

A and B are given by

A = sum(x.^2.*y)/sum(x.^4) = 1.6866     

B = sum(x.^3.*y)/sum(x.^6) = 0.5902

The respective errors are 

E_2(Ax^2) = sqrt((1/5)*sum((A*x.^2 - y).^2)) = 1.2971

E_2(Bx^3) = sqrt((1/5)*sum((B*x.^3 - y).^2)) = 0.2870

Therefore, the cubic polynomial is a better fit for the data.

************************************************************************** 
 
3)  Algorithms and programs 2 (a), section 5.1, page 262.

We wish to estimate the gravitational constant g from the following data

time,       t = [0.2000    0.4000    0.6000    0.8000    1.0000] (in seconds)

distance,   d = [0.1960    0.7835    1.7630    3.1345    4.8975] (in meters)

These variables are related by  the formula : d = (1/2)*g*t^2

We set (1/2)*g = A and use the expression for A that obtains best quadratic fit:

A = sum(t.^2.*d)/sum(t.^4) = 4.8975. This gives g = 2*A = 9.7950 m/sec^2.
     
************************************************************************** 

4) Exercise 2 (b), section 5.2, page 275.

We wish to find the least squares parabola y = A x^2 + B x + C for the data set:

x = [-2    -1     0     1     2]

y = [2.8000    2.1000    3.2500    6.0000   11.5000]

The values of A, B, C are obtained by solving the linear system L*[A B C]'= Y,

where Y = [sum(y.*x.^2) sum(y.*x) sum(y)]' = [65.3000   21.3000   25.6500]'

and L is the symmetric matrix whose entries are:

                       L(1,1)                   = sum(x.^4) = 34 

                       L(1,2) = L(2,1)          = sum(x.^3) =  0
                       
                       L(1,3) = L(3,1) = L(2,2) = sum(x.^2) = 10
                       
                       L(2,3) = L(3,2)          = sum(x)    =  0
                       
                       L(3,3)                               =  5 

This gives the system

   | 34     0    10||A|   |65.30|
   |  0    10     0||B| = |21.30|
   | 10     0     5||C|   |25.65|


We obtain the solution: A = 1.0000, B = 2.1300, C = 3.1300.

Therefore, the least squares parabola for the given data is 

y = x^2 + 2.13 x + 3.13

************************************************************************** 

5)  Exercise 6 (a), section 5.2, page 276.

We assume that a population size as function of time, is modeled by the expression

P(t) = L/(1 + Ce^(At)). The value of L is assumed known, L = 1000.

We wish to find A and C that give best fit for the data

t = [0     1     2     3     4]

P = [200   400   650   850   950]

Set new coordinates and constants: y = log(1000/P - 1), x = t, B = log(C)

x = [0     1     2     3     4]

y = [1.3863    0.4055   -0.6190   -1.7346   -2.9444]

Then A and B are obtained by standard linear least squares. Using the notations

of the first problem above, the constants A and B for the least squares line 

y = Ax + B satisfy the linear system

x.x A  + x.e B  = x.y

x.e A  +   5 B  = y.e

Evaluating the coefficients:

30 A + 10 B = -17.8142

10 B +  5 B = -3.5063

Solving the linear system yields: A = -1.0802, B = 1.4591. Thus C = exp(B) = 4.3019.

Therefore, P(t) = 1000/(1 + 4.3019 exp(-1.0802 t)) 

************************************************************************** 

6) Exercise 7 (b), section 5.3, page 294.

This is essentially the same as the previous problem, with the data set:

t = [0     2     4     6     8]

P = [76.1000  106.5000  132.6000  180.7000  226.5000]

and L= 800 (million)

We set x = t, y = log(800./P - 1), B=log(C). 

x = [0     2     4     6     8]

y = [2.2526    1.8736    1.6161    1.2318    0.9290]

The system of equations for A, B is:

120 A + 20 B = 25.0340

 20 A +  5 B = 7.9030
 
This yields A = -0.1645, B = 2.2384, C = exp(B) = 9.3784.
 
Therefore P(t) = 800/(1 + 9.3784*exp(-0.1645*t))

The year 2000 corresponds to t = 10. Then P(10) = 284.6824 million.

(The US Census Bureau web site gives the number 281,421,906.)

************************************************************************** 

7) Exercise 14, section 5.2, page 277.

The coordinate change: X = x, Y = y^(-1/2) turns the equation

y = (A x + B)^(-2) into the linear equation Y = A X + B.

************************************************************************** 

8) Exercise 2 (a, b), section 5.3, page 294.

We wish to find the cubic polynomial S(x) = a_0 + a_1x + a_2x^2 +a_3x^3

satisfying the conditions: S(1) = 3, S'(1) = -4, S(2) = 1, S'(2) = 2.

These conditions immediately give the equations, respectively,

a_0 +  a_1 +  a_2 +   a_3 =  3

       a_1 + 2a_2 +  3a_3 = -4
      
a_0 + 2a_1 + 4a_2 +  8a_3 =  1   

       a_1 + 4a_2 + 12a_3 =  2   
       
The solution to this system is easily obtained (by hand or computer) to

be [a_0 a_1 a_2 a_3] = [5 2 -6 2]. Therefore, S(x) = 5 + 2x -6x^2 + 2x^3.

See graph on separate page.       

************************************************************************** 

9) Exercise 2, section 5.4, page 307.

Find the Fourier series representation of the function f(x) defined by:

f(x) = pi/2 + x on interval -pi <= x < 0 and 

f(x) = pi/2 - x on interval   0 <= x < pi

and graph on the same coordinate system f(x) and the partial sums S_2(x), 

S_3(x), S_4(x).


I will denote the integral of a function g(x) from a to b by the symbol Int_[a,b] 

g(x) dx. Since f(x) is an even function of period 2pi, we can use a cosine 

Fourier series of the form f(x) = a_0/2 + a_1cos(x) + a_2cos(2x) + ...

The coefficients a_j are given by a_j = (2/pi)Int_[0 pi] f(x)*cos(jx) dx for 

j= 0, 1, 2, ...

This is an elementary integral, whose value is a_j = 4/(j^2pi) if j is odd and 

a_j = 0 if j is even (including 0). Therefore, the Fourier series of f(x) is

f(x) =  (4/pi^2)(cos(x) + cos(3x)/3^2 + cos(5x)/5^2 + ...)

See graphs on the separate page. (Note that S_2(x) = S_1(x) and S_4(3) = S_3(x).)

************************************************************************** 

10) Exercise 7, section 5.4, page 307.

It immediately follows from the previous problem, by substituting

0 for x, that  pi/2 = (4/pi)(1 + 1/3^2 + 1/5^2 + 1/7^2 ...)

Therefore,

pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + ...

************************************************************************** 

11) Exercise 1 (b), section 5.4, page 308.

Enter on the Matlab command line:

k=1:12;

X=-pi+2*pi*k/12;

Y=(X<0).*(X + pi/2) + (X>=0).*(-X +pi/2); %function of exe. 2 sect. 5.4.

M=5; %Degree of the trigonometric polynomial

%Now, obtain A, B using the function tpcoeff (see below)

[A, B]=tpcoeff(X,Y,M)

%This produces the vectors of coefficients:

%A = [0    1.4212    0.0000    0.1904    0.0000    0.1020]

%The vector B is nearly zero.

Therefore, the trigonometric polynomial has the form:

T_5(x) = 1.4212*cos(x) + 0.1904*cos(3*x) + 0.1020*cos(5*x)

%To plot the graphs, we do the following:

x=-pi:.01:pi;

y=tp(A,B,x,M); %See the function tp below.

plot(x,y,X,Y,'--o')



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [A,B]=tpcoeff(X,Y,M)
%Input  - X is a vector of equally spaced abscissas in [-pi, pi]
%       - Y is a vector of ordinates
%       - M is the degree of the trigonometric polynomial
%Output - A is a vector containing the coefficients of cos(jx)
%       - B is a vector containing the coefficients of sin(jx)
%

N=length(X)-1;
max1=fix((N-1)/2);

if M>max1
    M=max1;
end

A=zeros(1,M+1);
B=zeros(1,M+1);
A(1)=sum(Y);

for j=1:M
    A(j+1)=cos(j*X)*Y';
    B(j+1)=sin(j*X)*Y';
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function z=tp(A,B,x,M)
z=A(1);
for j=1:M
    z=z+A(j+1)*cos(j*x)+B(j+1)*sin(j*x);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%