Info

This question is closed. Reopen it to edit or answer.

What is wrong with this code?

1 view (last 30 days)
matlabnoob
matlabnoob on 18 Feb 2019
Closed: MATLAB Answer Bot on 20 Aug 2021
I am new to Matlab, which is probably very apparent from the above picture. I am trying to display that the left side of the equation on line 4 is equal to the function on the right after the n and x have been defined by the user. Any help would be much appreciated. Thanks!

Answers (2)

madhan ravi
madhan ravi on 18 Feb 2019
Edited: madhan ravi on 18 Feb 2019
Remove the single quotes from the last line, upload the code instead of a picture.
a/rw^2 == (cos(x)+(n^2*cos(2*x)+sin(x)^4))/((n^2-sin(x)^2)^(3/2))
% ^-^---------^-^------^-^^^------------^-^-- see the changes
You can simply use isequal() to compare if they are the same.
Note: You should define rw as a number like x and n.

KSSV
KSSV on 18 Feb 2019
Edited: KSSV on 18 Feb 2019
'a/rw^2'==
The above is completely wrong........variable shoould be simply:
a = your expression
There are lot of typos...you need to use something like below:
a = (cos(x)+(n^2*cos(2*x)+sin(x)^4))/((n^2-sin(x)^2)^(3/2))

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!