Solutions to Prelim 1 Review Questions


QUESTION 2 ------------------------------------------------------

"M Questions" from Insight.  You can check these on your own 
using Matlab. Ask a course staff member if you have questions.




QUESTION 3 ------------------------------------------------------

amt = input('Enter the Income: ');

if (amt > 60000)
	tax = (25/100)*30000+(30/100)*30000+(40/100)*(amt-60000);
elseif (amt > 30000)
	tax = (25/100)*30000+(30/100)*(amt-30000);
else
	tax = (25/100)*(amt);
end

fprintf('The tax to be paid for an income of %d is %d',amt,tax);




QUESTION 4 -----------------------------------------------------

% Sec 2.1 Problem 8
% The Euler constant

disp('    n            E_n');
disp('-------------------------');

kMax = 100;
step = 100;
% Initialization
total = 0;          % Keep track of the summation
for n=1:kMax*step
    % Calculate the sum
    total = total + 1/n;
    if rem(n,step) == 0
        % Calculate E_n
        E_n = total - log(n);
        fprintf('  %5d      %.8f\n', n, E_n)
    end
end

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% Sec 2.2 Problem 8
% Two sequences that approximate pi

n = 0;
a = 6/sqrt(3)*(-1)^n/3^n/(2*n+1);
fprintf('a%d = %.7f\n',0, a)

while abs(a - pi) > .000001 
    n = n + 1;
    a = a + 6/sqrt(3)*(-1)^n/3^n/(2*n+1);
    fprintf('a%d = %.7f\n',n, a)
end 

n = 0;
b = 16 * (-1)^n / 5^(2*n+1) / (2*n+1) - 4 * (-1)^n / 239^(2*n + 1);
fprintf('\nb%d = %.7f\n',0, b)

while abs(b - pi) > .000001 
    n = n + 1;
    b = b + 16 * (-1)^n / 5^(2*n+1) / (2*n+1) - 4 * (-1)^n / 239^(2*n + 1);
    fprintf('b%d = %.7f\n',n, b)
end 

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% Section 3.1 Problem 8
% Draw ASCII figure
%      *
%     * *
%    *   *
%   *     *
%  *       *
%   *     *
%    *   *
%     * *
%      *

n= input('Input n, n>3: ');

% Top half 
for r= 1:n
    % A general row has spaces, star, spaces, star, \n
    for c= 1:n-r
        fprintf(' ');
    end
    fprintf('*');
    if (r > 1)
        for c = 1:2*(r-1)-1
            fprintf(' ');
        end
        fprintf('*');
    end
    fprintf('\n')
end

% Bottom half
for r= n-1:-1:1
    for c= 1:n-r
        fprintf(' ');
    end
    fprintf('*');
    if (r > 1)
        for c = 1:2*(r-1)-1
            fprintf(' ');
        end
        fprintf('*');
    end
    fprintf('\n')
end

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%Section 3.1 Problem 10
%Print a sequence a(2), a(3), ..., a(n) where n is smallest integer 
%such that abs(a(n-1)-a(n))<=.01.

% Initialization
tol= 0.01;        % Stopping tolerance
n= 1;   
a_current= 0.5;  % a(n), i.e., a(1)
a_previous= 0;   % a(n-1) 

while abs(a_previous - a_current) > tol
    
    % Update
    n= n+1;
    a_previous= a_current;
   
    % Calculate current term a(n), which is a summation
    a_current = 0; 
    for j = 1:n^2
        a_current = a_current + (n/(n^2+j^2)); 
    end
    fprintf('%4d  %10.6f\n', n, a_current)
end

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% Section 3.2 Problem 7
% Print t_1 to t_26

disp('  n         t_n')
disp('--------------------')

n = 1;
while n<=26
    
    % Initialization
    current = 1;
    k = n;
    
    % Calculate from inside to outside.
    % current is the value of the kth squre root
    while k>1 || (n==1 && k>0)
        % Update...
        current = sqrt(1+k*current);
        k = k-1;
    end
    fprintf(' %2d      %10.8f \n',n,current)
    
    % Update...
    n = n+1;
end



QUESTION 5 ------------------------------------------------------

% Assume variables nBig and nSm are given.
% Outer loop to iterate from nBig to nSm (while-loop used here; 
% you can use for loop).

while nBig >= nSm 
	% Check nBig to see if it is prime
	d=2; % divisor
	% Iterate until first proper divisor is found
	while (mod(nBig,d) ~= 0);
	        d = d + 1;
	end
	if (d == nBig)
		fprintf('%d is a prime\n',nBig);
	else
		fprintf('%d\n',nBig);
	end
	nBig = nBig – 1;
end



QUESTION 6 -----------------------------------------------------

% Example program for computing the mode of a sequence of integers.

disp('Determine mode of a set of nonnegative integers.')
disp('Use a negative number to quit.')

previous = -1;          % Previous number seen
count = 1;              % Count of current number
mode = -1;              % Mode seen so far
modeCount = 0;          % Count for mode so far
number = input('Give me a number:  ');

while number >= 0   % Quit when negative number is encountered

    if number == previous     % same run, so increment count
        count = count + 1;
    else                      % new run, so reset count
        count = 1;
    end

    if count > modeCount
        mode = number;
        modeCount = count;
    end

    previous = number;
    number = input('Another number not smaller than the previous: ');    
end

if mode == -1
    disp('Mode is undefined')
else
    fprintf('Mode is %d which occurred %d times.\n', mode, modeCount)
end