the For loop

is i put

put a=(1:1:10)

k=a*b

end

but when i run it,work space say a min and max value is both 10. why? is there anyway other way to do it?

Answer 1

[Walter Roberson]

What shape is b ? Unless b is a scalar constant, the "*" would be interpreted as matrix multiplication; you would use .* for element-by-element multiplication. For more information please see the reference material for mtimes and times .

Also, "put" is not MATLAB command. If the actual word you have there is "for", as in

for a=(1:1:10)
  k=a*b
end

then you have a different situation. The for loop sets the loop variable ("a" in this case) to each value in turn, so each execution of the loop would involve a scalar value for "a", multiplied by "b". The loop would then overwrite the current value of k (as it appears alone on the left hand side), so at the end of the loop the value of k would be the same as if the loop had only been executed once with "a" being the maximum value (10) . Possibly the code you want is

A = 1:1:10;
for a = 1:length(A);
  k(a) = A(a) .* b;
end

At the end of a "for" loop, the loop variable will only have a single scalar value, the value that it had at the time the loop finished.

Possibly the code you want is

a = 1:1:10;
k = a .* b;

with no loop.

[kevin]

yea i mean

for a=(1:1:10)

k=a*b

end

but i dont think

a = 1:1:10;

k = a .* b;

will work as the actual equation is more like

for a from 1 to 10

b= cos(a)*12

c= sin(a)/2

for d from 2 to 5

e=a*b*5/d

so for minum of e , what is a and b? the actually equation is more complicated than that but how would u do it if it is like above?

[Image Analyst]

Did you ever look at my code I wrote for you below?

Answer 2

[Image Analyst]

kevin: Try this code. It will calculate e for all those possibilities of a, and d and plot e. It will then find the index where e is minimum and report the values of a and b there at the min (for just the specific, discrete values of a and b used, not for any continuous-values of a and b):

e_index = 1;
for a = 1 : 10
  for d = 2 : 5
    % Get and store a and b for this iteration.
    a1(e_index) = a;
    b(e_index) = cos(a)*12;
    % kevin never uses c at all!
    c = sin(a)/2;
    % Calculate e for these values of a, b, and d.
    e(e_index) = a * b(e_index) * 5 / d;
    % Increment to the next element in e.
    e_index = e_index + 1;
  end % of "d" loop
end % of "a" loop
plot(e);
grid on;
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1])

[min_e_value, min_e_index] = min(e);
bAtMinE = b(min_e_index);
aAtMinE = a1(min_e_index);
fprintf('e has a min value of %f at element %d,\nwhich occurs for a = %d and b = %f.\n',...
  min_e_value, min_e_index, aAtMinE, bAtMinE);

Results in command window:

e has a min value of -251.721459 at element 37,
which occurs for a = 10 and b = -10.068858.