Question about Taylor Series While loop.

I'm completely stuck on this While Loop using Taylor Series.

x = input('Input the angle in radians: ');
Cos_Estimate = 0;
k = 0;
Errrr = 1
while Errrr > .000001
if mod(k,2)
      Sign = 1;
  else
      Sign = -1;
  end
  k = k + 2;
  Cos_Estimate = Cos_Estimate + (x^k/(factorial(k)*Sign));
  Errrr = abs(Cos_Estimate - cos(x));
end
err = abs(Cos_Estimate - cos(x));
fprintf('The estimated cosine value based on the Taylor Series is: %0.6f \n',Cos_Estimate)
fprintf('The actual cosine value is : %0.6f \n',cos(x))
fprintf('The estimation error is: %0.6f \n',err)
fprintf('The number of terms required was: \n',term)

Now I am getting NaN for my variable. I'm stuck.

[Azzi Abdelmalek]

Now your variable is Nan

[Derryn]

How do I fix this?

Answer 1

[Azzi Abdelmalek]

Try this

x = input('Input the angle in radians: ');
Cos_Estimate = 1;
Errrr=1
Sign = 1;
k = 0;
while Errrr > .000001 & k<60
  Sign =-Sign;
   k = k + 2;
   Cos_Estimate = Cos_Estimate + (x^k/(factorial(k)*Sign))
   Errrr = abs(Cos_Estimate - cos(x))
end
display(Cos_Estimate)

[Derryn]

THANKS

Answer 2

[Babak]

add this line to the beginning of your code:

Errrr =1;

[Derryn]

Added it, but now I get this as my output:

EDU>> Lab7_Prob3Script

Input the angle in radians: pi

The estimated cosine value based on the Taylor Series is: NaN

The actual cosine value is : -1.000000

The estimation error is: NaN

The number of terms required was:

Answer 3

[Azzi Abdelmalek]

Your code never enter in the loop because Errrr is not defined

[Derryn]

Added it, but now I get this as my output:

EDU>> Lab7_Prob3Script

Input the angle in radians: pi

The estimated cosine value based on the Taylor Series is: NaN

The actual cosine value is : -1.000000

The estimation error is: NaN

The number of terms required was:

[Derryn]

How do I fix my variable from being NaN?

This is the first time this problem has happened to me!

[Matt Kindig]

The issue is because the factorial() function, at very high values of k, exceeds the limit for double-precision numbers (specifically, you get a value of Inf). To demonstrate, run this at the command line:

>> factorial(160:180)

Notice that after the 11th column (i.e. factorial(170)) the factorial no longer is defined. When dividing by Inf, you get NaN in Cos_Estimate.

To fix this, the easiest way might be to add a second condition to your while loop, to terminate when k exceeds 170 or so.

Also, you don't define 'term' anywhere in your code, so the last line throws an error.

[Matt Kindig]

Also, what exactly is the point of the "estimation error" calculation (calculation of 'err')? By definition, won't this be equal to Errrr, because you have defined it that way? Also, won't 'err' always be less than 0.000001, because of the way you have defined your loop?

[Derryn]

Ah ok thanks. Does this look better?

x = input('Input the angle in radians: ');

Cos_Estimate = 0; k = 0;

count = 0;

Errrr = 1;

while Errrr > .000001

    count = count + 1;

    if mod(k,2)

        Sign = 1;

    else

        Sign = -1;

    end

    k = k + 2;

    Cos_Estimate = Cos_Estimate + (x^k/(factorial(k)*Sign));

    Errrr = abs(Cos_Estimate - cos(x));

    if k > 170

        break

    end

end

err = abs(Cos_Estimate - cos(x));

fprintf('The estimated cosine value based on the Taylor Series is: %0.6f \n',Cos_Estimate)

fprintf('The actual cosine value is : %0.6f \n',cos(x))

fprintf('The estimation error is: %0.6f \n',err)

fprintf('The number of terms required was: %0.0f \n',count)

I'm still getting the wrong value for Cos_Estimate