MATLAB Examples of Iterative operations

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MATLAB Examples of Iterative operations

• Ch E 111
• David A. Rockstraw, Ph.D., P.E.
• New Mexico State UniversityChemical Engineering
  Department

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Nested Loop Statements

• Nested loops
• for i14
• for j13
• disp(i j ij)
• end
• end

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Nested Loop Statements

• clear A
• for i1n
• for j1n
• if i lt j
• A(i,j)-1
• elseif i gt j
• A(i,j)0
• else
• A(i,j)1
• end
• end
• end
• A

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Nested Loop Statements

• clear A
• for i1n
• for j1n
• if i lt j
• A(i,j)-1
• elseif i gt j
• A(i,j)0
• else
• A(i,j)1
• end
• end
• end
• A

compare with the built-in function statement
AAeye(n)-triu(ones(n),1)
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Example 1

• 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.

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Example 1 Solution

• sum1 0
• sum2 0
• for i 1100
• sum1 sum1 i
• sum2 sum2 i2
• end
• sum1
• sum2

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Example 2

• Determine the largest value of n such that

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Example 2 Solution

• sum 0
• i 0
• while ( sum lt 1000000 )
• i i 1
• sum sum i2
• end
• i

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Example 3

• Find the integer value of n between 0 and 100
  such that

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Example 3 Solution

• 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 1100
• running_sum running_sum cos(n)
• if ( abs( running_sum ) lt minimum_abs_sum )
• minimum_n n
• minimum_abs_sum abs( running_sum )
• end
• end
• minimum_n
• minimum_abs_sum

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Example 4

• Continue subtracting (to a max of 1000 times) e
  from p until a value less than -10 is obtained.
• x pi
• for i11000
• x x - exp(1)
• if x lt -10
• break
• end
• end

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Example 4

• Calculate the sum of those entries of the matrix
  M 1 2 3 4 5 6 7 8 9 which lie in the
  upper-triangular portion (i.e., on or above
  diagonal).

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Example 4 Solution

• M 1 2 3 4 5 6 7 8 9
• sum 0
• for i 13
• for j i3
• sum sum M(i, j)
• end
• end
• sum

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average.m

• 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 i1 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))

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Integrate the sine function

• 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.

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function intsin(N)

• function y intsin(N)
• INTSIN - integrate sin(x) from 0 to pi
• y (sin(0) sin(pi)) pi/(2N)
• for n 1 N-1
• y y sin(npi/N)pi/N
• end

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Infinite sum

• 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?

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function pi2over6

• function y,n pi2over6(tol)
• PI2OVER6 - Approximates (pi)2/6
• y0
• aimpipi/6
• n0
• while (abs(aim-y)gttol)
• nn1
• yy1/(nn)
• disp(sprintf('gsgsg',n,' ',y,' ',aim-y))
• end

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grcodi script

• 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).

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grcodi script

• GRCODI - determine greatest common divisor
• N1input('first number ')
• N2input('second number ')
• while (min(N1,N2)gt0)
• if (N1 gt N2)
• N1N1-N2
• else
• N2N2-N1
• end
• end
• disp(sprintf('sg','GCD is ',max(N1,N2)))