I dont know how to solve the below question. I'm completely lost on where to even begin. I understand the fprintf, just the sin and cos always throw me off. Any direction would even be greatly appreciated

For 0 ≤ x ≤ 1 with an increment of 0.1 in x, compute f( x )=sin*x/1+x. For this computation, set x=0:0.1:1 and then perform element-by-element multiplication (.*) operation. Please avoid plugging each x to the function. Using fprintf, print out x and f(x) in two columns with each variable having 8 decimal places.

[Adam]

f( x )=sin*x/1+x

is a highly suspicious definition of a function. For a start sin has no argument. I assume it should be sin( x ). Also I assume the

1 + x

should be

( 1 + x )

otherwise you are just dividing by 1 (pointless) then adding x to the result.

Apart from that I don't really understand what your confusion is though. If you don't know what sin is then that is a very basic maths problem that you need to look up in a book or online. Assuming you do know what sin is then

doc sin

will take you to the Matlab help page on the function that implements sine.

Most of the rest of it is explicitly stated as to how to do it.

[James Niemiec]

So what I have completed done so far is like such

• X = ( 0 : 0.1 : 1) ' ;
• X1= sin(X);
• F_x=((X1)/(1+x));
• fprintf('Variable X: %1.8 f | Variable F(x): %1.8f', [X,F(x)].');
• This causes my code to appear in two different columns like i want but with a lot of other random data in it.
• Variable X = 0.00000000 | Variable F(x)= 0.00000000
• Variable X = 0.00000000 | Variable F(x)= 0.00000000
• Variable X = 0.00000000 | Variable F(x)= 0.00000000
• Variable X = 0.00000000 | Variable F(x)= 0.00000000
• Variable X = 0.00000000 | Variable F(x)= 0.00000000
• Variable X = 0.00000000 | Variable F(x)= 0.00000000
• Variable X = 0.10000000 | Variable F(x)= 0.00000000
• Variable X = 0.00000000 | Variable F(x)= 0.04991671
• It then repeats this cycle again 7 more times till Variable X=0.20000 and so on over and over again.

Answer 1

[monika shivhare]

x=0:0.1:1;
f=sin(x./(1+x));
for i=1:length(x)
fprintf('%0.8f %0.8f\n',x(i),f(i));
end

[James Niemiec]

Hello Monika, I've seen videos of people using this i=1 code but cant really pinpoint its use exactly. This ended up working exactly how i wanted it too. Thanks very much. Could you possibly explain what the function i=1: length(x) means in layman's terms. I found the help section of Matlab/videos to be lacking in detail.

[Rik]

You need to have a look at the for function. For-loops are pretty standard in many coding languages, any tutorial should cover this.

[Stephen23]

Note that you can easily avoid the loop entirely:

fprintf('%0.8f %0.8f\n',[x;f])

[James Niemiec]

I understand the for loops and if else statements. The statement that I cant seem to get a lot of clarity on by itself is the i=1:length(x) function. Many of the for statements ive dealt with in the past were used in C# or C++ coding. There for statements were used to repeat actions until the specific argument became false. Only thing that just doesn't make sense to me is the i itself since there is no variable labeled i so it seems like it would repeat constantly forever.

[Rik]

A for-loop in Matlab executes the block of code until the matching end, with i taking on the values of the columns of the supplied array. You don't have to actually use the loop index for the for-loop to work.

In this case, i will take on the values 1 through 11.

There is a big difference between for and while.

[Stephen23]

"Many of the for statements ive dealt with in the past were used in C# or C++ coding"

Sure, but MATLAB is not C or C++. If you want to learn about MATLAB, then you will need to read the MATLAB documentation: for is described as "for loop to repeat specified number of times" (it does not say "to repeat forever"). The documentation then shows how to specify the values to loop over.