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MATLAB Examples of Iterative operations

MATLAB Examples of Iterative operations. Ch E 111 David A. Rockstraw, Ph.D., P.E. New Mexico State University Chemical Engineering Department. % Nested loops for i=1:4 for j=1:3 disp([i j i*j]) end end. Nested Loop Statements. clear A for i=1:n for j=1:n if i < j

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MATLAB Examples of Iterative operations

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  1. MATLABExamples of Iterative operations Ch E 111 David A. Rockstraw, Ph.D., P.E. New Mexico State UniversityChemical Engineering Department

  2. % Nested loops for i=1:4 for j=1:3 disp([i j i*j]) end end Nested Loop Statements

  3. clear A for i=1:n for j=1:n if i < j A(i,j)=-1; elseif i > j A(i,j)=0; else A(i,j)=1; end end end A Nested Loop Statements

  4. clear A for i=1:n for j=1:n if i < j A(i,j)=-1; elseif i > j A(i,j)=0; else A(i,j)=1; end end end A Nested Loop Statements compare with the built-in function statement: AA=eye(n)-triu(ones(n),1)

  5. Write a for loop which calculates the sum the integers from 1 to 100 and the sum of the squares of the integers from 1 to 100. Print out only the results. Example 1

  6. sum1 = 0; sum2 = 0; for i = 1:100 sum1 = sum1 + i; sum2 = sum2 + i^2; end sum1 sum2 Example 1 Solution

  7. Determine the largest value of n such that Example 2

  8. sum = 0; i = 0; while ( sum < 1000000 ) i = i + 1; sum = sum + i^2; end i Example 2 Solution

  9. Find the integer value of n between 0 and 100 such that Example 3

  10. Solve assuming that minimum occurs when n = 0. Then, for each n = 1, 2, ..., 100, compare the absolute value of the running sum with that of the minimum absolute running sum currently found. If it is less, update the two variables. minimum_n = 0; % the sum when n = 0 minimum_abs_sum = 1; % initially, the absolute value of |cos(0)| running_sum = 1; % cos(0) + ... + cos(n) for n = 1:100 running_sum = running_sum + cos(n); if ( abs( running_sum ) < minimum_abs_sum ) minimum_n = n; minimum_abs_sum = abs( running_sum ); end end minimum_n minimum_abs_sum Example 3 Solution

  11. Continue subtracting (to a max of 1000 times) e from π until a value less than -10 is obtained. x = pi for i=1:1000 x = x - exp(1) if x < -10 break; end end Example 4

  12. Calculate the sum of those entries of the matrix M=[123;456;789] which lie in the upper-triangular portion (i.e., on or above diagonal). Example 4

  13. M = [1 2 3; 4 5 6; 7 8 9]; sum = 0; for i = 1:3 for j = i:3 sum = sum + M(i, j); end end sum Example 4 Solution

  14. clear % initialize - prepare to read 1st datum i = 1; % read and count data values data = input('Enter datum ("Enter" to stop): '); while ~isempty(data) %data? y(i) = data; % - yes: store i = i+1; % count data = input('Enter datum ("Enter" to stop): '); end % no more data - compute average sumY = sum(y); % compute sum [dummy, n] = size(y); % determine # values = # columns averageY = sumY/n; % print result disp(['the average of the 'num2str(n) ' values is ' num2str(averageY)]) average.m

  15. Write a program that approximates as N is the number of points used in the integration. You may need to use a for loop from 1 to N-1. Calculate the approximation using N = 5, 10, 20, and 50. Integrate the sine function

  16. function y = intsin(N) % INTSIN - integrate sin(x) from 0 to pi y = (sin(0) + sin(pi))* pi/(2*N); for n = 1 : N-1 y = y + sin(n*pi/N)*pi/N; end function intsin(N)

  17. Use a while loop to calculate the summation until the deviation from the exact answer is less than 0.1. Determine up to what n-value is needed to carry out the summation so that this deviation is achieved? How about a deviation of 0.01 and 0.001? Infinite sum

  18. function [y,n] = pi2over6(tol) % PI2OVER6 - Approximates (pi)^2/6 y=0; aim=pi*pi/6; n=0; while (abs(aim-y)>tol) n=n+1; y=y+1/(n*n); disp(sprintf('%g%s%g%s%g',n,' ',y,' ',aim-y)) end function pi2over6

  19. Write a program that, given two positive integers N1 and N2, calculates their greatest common divisor. The greatest common divisor can be calculated easily by iteratively replacing the largest of the two numbers by the difference of the two numbers until the smallest of the two numbers reaches zero. When the smallest number becomes zero, the other gives the greatest common divisor. You will need to use a while loop and an if-else statement. The largest of two numbers can be obtained using the built-in function MAX(A,B). The smallest of two numbers can be obtained using the built-in function MIN(A,B). grcodi script

  20. % GRCODI - determine greatest common divisor N1=input('first number = '); N2=input('second number = '); while (min(N1,N2)>0) if (N1 > N2) N1=N1-N2; else N2=N2-N1; end end disp(sprintf('%s%g','GCD is ',max(N1,N2))) grcodi script

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