Hello. Very good questions. 1- the keyword 'return' exits the execution of the current script and returns to the MATLAB command prompt. The 'break' statement is used to quit the current loop (while or for loops). Note that the statements 'exit' and 'quit' terminate the whole MATLAB program itself. 2- Nested if-conditions are usually used when there are more than one level of conditions. For example { if x >= 0 ... if sqrt(x)

thank you. Could you make a video about nested if? And explain it with details.

Good question Mouad. The root of an equation is defined as the point of intersection between the curve of the equation and the x-axis, and at that point y(x) = 0. According to this definition, a root is approached if y(x1) is zero or very close to zero. In numerical methods, it is almost impossible to get the exact zero, so it is assumed that the root is found if y(x1) is inside the range of -1.0E-6 to +1.0E-6. This range can be easily be tested by using the absolute value function and the positive part of the range in the condition statement: if abs(y(x1)) < 1.0E-6. I hope I could answer your question.

There are two alternative ways to define a function in MATLAB, as I know. First, you can use inline() definition method, which is an obsolete method and not recommended to be used anymore. The second method, is to use the function definition method which needs to be coded a new .m file. This method is recommended if you do not define the function in body of your bisection code, or if you have a function has multiple lines of code.

@@mechtutorcom Thank you sir and really i want to know more about programming loops and different tips about it. It will be wonderful if you make videos about it.

karim electronics When while loop is used, the whole if condition {if abs(y(x1)) < 1.0E-6 ... break ... end} should removed and the condition of the while loop is written as following while abs(y(x1)) >= 1.0E-6 ... end The other statements remain unchanged. If the number of bisections at the solution is required, a counter {i=i+1} can be added inside the loop.

Is it correct instead of while abs(y(x1)) >= 1.0E-6 ... end we can put while abs(y(x2)) >= 1.0E-6 ... end ?

Yes, it is correct. Because during bisections both x1 and x2 approach to the root from both sides and y(x1) and y(x2) accordingly approach to zero.

Thank you sir

The total number of bisections means how many times the intervals need to be divided until the root is approached. I hope I could answer your question. Thanks for your comment.

@@mechtutorcom ouh i see, thank you very much, sir

Yes, equation can be changed.

Thank you for the comment. I couldn't understand what you mean by processes, but the steps can be written as following; 1- Input two values of x that embrace the interval where the root is expected 2- Calculate corresponding values for y 3- Check for the sign difference between y-values 4- In case of same signs, stop 5- Calculate the value of x in the half of the interval 6- Check for the sign difference between the y-values first half interval 7- In case of opposite signs, calculate the value of x in the half of the interval 8- In case of similar signs, take the x-values bounding the second half 9- If the values of y approaches zero, print the x-value and stop 10- Else repeat steps 6 to 10

Actually, in MATLAB/Octave videos, I used Octave. It has very similar syntax to MATLAB. You should take into consideration that these videos were recorded back in 2017 and 2018, so there must be many changes in the languages ever since.

It's IDLE, downloaded for free from python.org

As iterations proceed, the values of x1 and x2 get closer and closer to the root, and consequently y(x1) and y(2) get closer and closer to zero. It is difficult to predict which one will be within the given tolerance (1.0E-6) first: abs(y(x1)) or abs(y(x2)). For that reason, either one will work well -- no difference. Also, you can modify the code to use xh, so each new value can be tested before assigning it to x1 or x2. You need to move the test condition up just under the bisection line: . . xh = (x1+x2)/2; %bisection if abs(y(xh)) < 1.0E-6 break if y(x1) *y(xh) < 0 . . This may be better.

Please reply if anyone knows

or you just copy it by typing it out, its not hard...