Iterative Examples

8/12/2019 Iterative Examples

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MATLABExamples of Iterative operations

Ch E 111David A. Rockstraw, Ph.D., P.E.

New Mexico State UniversityChemical Engineering Department


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Nested Loop Statements

% Nested loopsfori=1:4

forj=1:3

disp([i j i*j])end

end


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clear Afori=1:n

forj=1:n

ifi < j

A(i,j)=-1;

elseifi > jA(i,j)=0;

else

A(i,j)=1;

end

endend

A


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clear Afori=1:n

forj=1:n

ifi < j

A(i,j)=-1;

elseifi > jA(i,j)=0;

else

A(i,j)=1;

end

endend

A AA=eye(n)-triu(ones(n),1)compare with the built-in function statement:


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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 Solutionsum1 = 0;

sum2 = 0;

fori = 1:100

sum1 = sum1 + i;sum2 = sum2 + i^2;

end

sum1sum2


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Example 2

Determine the largest value of nsuch that

6

n

1i

210i


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Example 2 Solution

sum = 0;i = 0;

while( sum < 1000000 )

i = i + 1;

sum = sum + i^2;

end

i


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Example 3

Find the integer value of nbetween 0 and 100

such that

minimumicosn

0i


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Example 3 SolutionSolve assuming that minimum occurs when n= 0. Then, for each

n= 1, 2, ..., 100, compare the absolute value of the running sumwith 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)

forn = 1:100running_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


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Example 4

Continue subtracting (to a max of 1000 times)efrom until a value less than -10 is obtained.

x = pi

fori=1:1000

x = x - exp(1)

ifx < -10

break;

end

end


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Example 4

Calculate the sum of those entries of the matrixM=[123;456;789]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;

fori = 1:3

forj = i:3sum = sum + M(i, j);

end

endsum


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average.mclear

% 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; % countdata = 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 resultdisp(['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)

functiony = intsin(N)

% INTSIN - integrate sin(x) from 0 to pi

y = (sin(0) + sin(pi))* pi/(2*N);

forn = 1 : N-1y = y + sin(n*pi/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 pi2over6function[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


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grcodi script

Write a program that, given two positive integers N1 and N2,calculates their greatest common divisor. The greatestcommon divisor can be calculated easily by iterativelyreplacing the largest of the two numbers by the difference ofthe two numbers until the smallest of the two numbers reaches

zero. When the smallest number becomes zero, the other givesthe greatest common divisor.

You will need to use a while loop and an if-else statement. Thelargest of two numbers can be obtained using the built-infunction 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

N1=input('first number = ');N2=input('second number = ');

while(min(N1,N2)>0)

if(N1 > N2)

N1=N1-N2;else

N2=N2-N1;

end

enddisp(sprintf('%s%g','GCD is ',max(N1,N2)))

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Iterative Examples