Multiple values for a variable in a for loop

Hi, I'm just trying to do something really simple but it doesn't work and I don't know why despite the error message. I'm new to matlab so be gentle.

x is a column vector (10,1) with values ranging from 0.2 to 2 with 0.2 incraments. I would like for the eq to go through each of the x values and then possibly, if someone is willing to help, to store these 10 calculated values in a new (10,1) column vector.

for x2 = x(1:10)

y1 = ((fg*x2^2)/(24*E*I))*(x2^2 - 4*L*x2 + 6*L^2)

end

Thanks in advance!

Regards Christoffer

Answer 1

[Guillaume]

Assuming that fg, E, I, and L are scalar, the reason your code does not work (and you should have explained what does not work mean) is because you're always assigning the output to the same element of the same variable: y1(1) (if you don't put an index, it's assumed to be one). To solve, one possible way

for idx = 1:numel(x)
   y(idx) = ((fg*x(idx)^2)/(24*E*I))*(x(idx)^2 - 4*L*x(idx) + 6*L^2);
end

However, the loop is not needed. Simply use the vectorising ability of matlab and just one line is needed:

y = ((fg*x.^2)/(24*E*I))*(x.^2 - 4*L*x + 6*L^2)

Note that I've replaced the ^ with .^ to tell matlab to apply the function elementwise.

[Christoffer Thornvall]

oh that makes perfect sense, can't believe I missed that with the(y1). Thank you so much for explaining this. I will make sure to describe the error in detail too next time

Answer 2

[Purushottama Rao]

You dont need a for loop for doing it. you can try somthing like

x(1:10)=0.2:0.2:2
y1 = ((fg*x2^2)/(24*E*I))*(x2^2 - 4*L*x2 + 6*L^2)
..................................
which should result in a column vector of same size as x

[Guillaume]

For that matter, you don't need to index the result of an assignment. So it's just

x = 0.2:0.2:2;

which also has the advantage of not having to worry between size mismatch between destination and source.

And of course, the code will not result in a column vector but an error because

• you forgot to rename x2
• you're using ^ with a vector

[Christoffer Thornvall]

I will try to see if this works too, thank you!!