Need help making a script!!! any help?please

This is the problem to make the scrip: http://i.imgur.com/EkSTC0x.png?1

That is the scrip that someone recommended to me but my professor say that was wrong, and that I have to use the student version of matlab to do it, with the basic thing!!

x_deg = input('Input angle in degrees: ');
x = degtorad(x_deg);
sumTerms = 0;
n = 0;
E = Inf;
while E > 1e-6
    a = ((-1)^n * x^(2 * n + 1)) / factorial(2 * n + 1);
    prevSum = sumTerms;
    sumTerms = sumTerms + a;
    if n ~= 0
        E = abs((sumTerms - prevSum) / prevSum);
    end
    n = n + 1;
end
fprintf('Value of sin(%0.1f) is: %0.5f\n', x_deg, sumTerms

I have the student version, so something in this program is not letting me run the script, But also my professor say that was wrong and that I have to make the program stop when E<=0.000001?

[sixwwwwww]

What prof say?. what is wrong with this code?

[Walter Roberson]

Your professor might want you to structure the loop differently:

while TEST
  ...
end

can be rewritten as

while true
  ...
  if ~TEST; break; end
end

[Leonardo Concepcion]

sixwwwwww your code was good, but my prof say that was missing a part, that the program has to stop when E<=0.000001 like it says in the problem.

Answer 1

[Image Analyst]

degtorad() is not part of base MATLAB. Use x_deg = x * pi/180 instead. Also, take a look at your duplicate question, which has been answered. Here is the corrected code, along with the improved fprintf() that I gave in your original duplicate question (which you ignored for some reason):

clc;
clear all;
x_deg = input('Input angle in degrees: ');
x = x_deg*pi/180
sumTerms = 0;
n = 0;
E = Inf;
while E > 1e-6
    a = ((-1)^n * x^(2 * n + 1)) / factorial(2 * n + 1);
    prevSum = sumTerms;
    sumTerms = sumTerms + a;
    if n ~= 0
        E = abs((sumTerms - prevSum) / prevSum);
    end
    n = n + 1;
end
fprintf('The value of the Taylor series after %d terms is %.7f.\nThe true value of sind(%0.1f) is: %0.7f\n', ...
  n, sumTerms, x_deg, sind(x_deg))

[Leonardo Concepcion]

This is excellent, but the part that says that has to stop when E<=0.000001?

[Image Analyst]

It does! Look at the condition. The while loop executes only when E is > 1e-6. This means it stops when E is < 1e-6, just like you wanted. Do you not agree? If not, explain your confusion more explicitly.

[Leonardo Concepcion]

ok,excuse me if I don't understand to much, is because I am new in this, but when I put a number lower than that the program does not stop, is another way to put like a display or something or an elseif that says stop or something? Also I want to know how to make the program to keep running? for example if I put 45 the program give me the result and then I can put another number without running the program again?

[Image Analyst]

Lower than what? Are you inputting a degree (angle) less than the error 1e-6? That would not make sense. So what number did you put in? I entered 45 for 45 degrees. Are you saying that the loop never exits for you and it hangs because you have an infinite loop? That simply cannot be the case. Did you copy and paste my code? Did you step through it with the debugger? For 45 degrees, the error is less than 1e-6 after 5 terms, and the loop exits. How can it not exit on your system?????????

Maybe you're totally thrown off by not understanding how "continue while E is more than 1e-6" is equivalent to "stop looping when E is less than 1e-6". If so, you can put these totally unnecessary lines at the bottom of the while loop, just before the end statement:

if E < 1e-6
  break; % break out of the while loop.
end

but like I said, it's totally unnecessary since the loop will exit anyway when it fails the "while E > 1e-6" test at the top of the loop. But if it makes you feel more secure, then go for it.

If you want to make it keep asking you for angles until you say to quit, then use this code:

n = 0;
while n < 20 % Max number of questions you want to ask
        % This is a failsafe so you don't get into an infinite loop.
  x_deg = input('Input angle in degrees: ');
  x = x_deg*pi/180
  sumTerms = 0;
  E = Inf;
  while E > 1e-6
    a = ((-1)^n * x^(2 * n + 1)) / factorial(2 * n + 1);
    prevSum = sumTerms;
    sumTerms = sumTerms + a;
    if n ~= 0
      E = abs((sumTerms - prevSum) / prevSum);
    end
    n = n + 1;
  end
  fprintf('The value of the Taylor series after %d terms is %.7f.\nThe true value of sind(%0.1f) is: %0.7f\n', ...
    n, sumTerms, x_deg, sind(x_deg))
  promptMessage = sprintf('Do you want to Continue processing,\nor Cancel to abort processing?');
  titleBarCaption = 'Continue?';
  button = questdlg(promptMessage, titleBarCaption, 'Continue', 'Cancel', 'Continue');
  if strcmpi(button, 'Cancel')
    break;
  end
end

[Leonardo Concepcion]

Can't you give me an output photo, because I put .00000001 that is less than 1e-6 and give me this:

[Image Analyst]

Here it is:

The loop executes twice, which it has to if we're going to check the difference between one iteration and the prior one. It only looks like pure 0 because I only put %.7f and the difference is too small to be seen at that precision. Try %.13f or something to see the actual difference.

I'm really baffled as to why you're confused. Are you somehow getting confused between the value of the error (which must be less than 1e-6 before we can quit), and the value of the angle ? Why do you happen to be entering an angle that is exactly the same as the max allowable error? Is that just a coincidence, or are you confusing the two. Remember, one is error and one is angle - two different entities and concepts.

[Leonardo Concepcion]

I am confused with the value of the error I don't know why?

[Leonardo Concepcion]

there is no other way to do this, more basic?

[Leonardo Concepcion]

And what I don't understand is that the problem says stop adding terms when E<=0.000001.???

[Image Analyst]

You could get rid of the temporary variable "a", but that's about the only way to simplify it.

Let me ask you, when do you think the while loop should quit ? When E < 0.000001? Well that's what it does. Let me ask in a different way. How long do you think the loop should keep going ? Should it keep going while the error is large ? Larger than 0.000001? Well that's what it does. Can you answer those two questions for me?

I hope someone else contributes here to explain it in a different way because I just don't know what else to say to him to make it clear.

[Walter Roberson]

Change

fprintf('The value of the Taylor series after %d terms is %.7f.\nThe true value of sind(%0.1f) is: %0.7f\n',

to

fprintf('The value of the Taylor series after %d terms is %012g.\nThe true value of sind(%0.12g) is: %0.12g\n',